:biggrin:
Hi all
This version is the result of what i learned
till now. Some procedures are now lighter.
Some strategies was corrected.
Now, we have not "division by 0" but infinity.
In this way, 1/(log(10)-1)* 0 gives "Indeterminate form"
and not "division by 0".
Now, it is time for corrections and improvements.
Meanwhile, the calculator should solve any expression. ;)
**** Here is calcula48 v1.08 ****
Calculation rules
1. terms inside parentheses or brackets
2. functions
3. exponents or powers
4. multiplication and division
5. addition and subtraction
note 1: if we end a pair of brackets with !
we must have a positive integer inside: (12-3)!
note 2: the exponents are treated as
a) real: it is the general case
b) integer: enclosed by brackets (1-log(3))^(-2)
c) rational: enclosed by brackets (1-log(3))^(-2/3)
note 3: roots are calculated following this rule:
1. put the expression inside brackets: (log(3)-log(2)*2)
2. put the rational exponent inside brackets: (n/k)
3. use "^"
(log(3)-log(2)*2)^(n/k)
square root: sqr(2*log(2)-log(3))=(2*log(2)-log(3))^(1/2)
= 0.353466740455591 = 0.353466740455591
note 4: exponents or powers are calculated starting at the end
4^3^2: first: 3^2 = 9 second: 4^9 4^3^2 = 262144.0
(4^3)^2 = 64.0 ^2 = 4096.0
We can define:
1. a set of 10 constants: type t=10; s=2.25;u=-2.5e-3;
2. a function: type f(x)=x^2/(1-x)
3. a derivative: type df(x)=2*x-1
4. up to 20 matrices, up to a 20*20 matrix:
a=[1,2,2]; vector column 3 columns
b=[1;2;3]; " line 3 lines
c=[1,2;2,3]; squared matrix
We can solve:
1. Any expression
2. Mean(x1,...,xn): mean(12,13,12,13,14)
3. Factorial and k-combination of n: 12!, comb(12,3)
4. Logarithms of any base: logb(15.2, 5)
5. Quadratic equations: aX^2+bX+c=c0
6. Systems of 2 equations: type aX+bY=c;fX+gY=h;
7. Systems of 3 equations: type aX+bY+cZ=d;fX+gY+hZ=i; jX+kY+mZ=l;
8. Systems of 4 equations: type aX+bY+cZ+dT=f;
gX+hY+iZ+jT=k;
lX+mY+nZ+oT=p;qX+rY+sZ+tT=u;
(note: a,b,c, ... are real numbers)
9. Conversion decimal-binary-hexadecimal - any 64-bit number
conv(11000) = 00002AF8H
conv(11000b)= 24
conv(11000h)= 69632
10. Linear interpolation: type: point(1,2; 3,5) and g(x=2)
or g(y=2)
11. The equation f(x)=0
a) root(x=-20, x=20; n=200)
b) root(x=-20, x=20; d=0.01)
c) root(x=-20, x=20; x=1.2)
note: in root(x=a, x=b; x=X0 ) a, b, X0 can be
-9pi,...,-1pi, -pi, pi, 2pi, 3pi,...,9pi,
-9e,...,-1e, -e, e, 2e, 3e,...,9e.
Matrix operations
. Copy
a=[1, 2; 3, 4]; [press ENTER/COMPUTE]
b=a; [press ENTER/COMPUTE]
Now we have a, b and a equal b
. Addiction
a1=[1, 2; 3, 4]; [press ENTER/COMPUTE]
a2=[5, 6; 7, 8]; [press ENTER/COMPUTE]
a=a1+a2; [press ENTER/COMPUTE]
. Subtraction
a1=[1, 2; 3, 4]; [press ENTER/COMPUTE]
a2=[5, 6; 7, 8]; [press ENTER/COMPUTE]
b=a1-a2; [press ENTER/COMPUTE]
. Multiplication
a1=[1, 2; 3, 4]; [press ENTER/COMPUTE]
a2=[5, 6; 7, 8]; [press ENTER/COMPUTE]
c=a1*a2; [press ENTER/COMPUTE]
. Scalar multiplication
a1=[1, 2; 3, 4]; [press ENTER/COMPUTE]
d=10 * a1; [press ENTER/COMPUTE]
f=+ a1; [press ENTER/COMPUTE]
g=- a1; [press ENTER/COMPUTE]
. Transpose matrix
a1=[1, 2; 3, 4]; [press ENTER/COMPUTE]
b=a1^t; [press ENTER/COMPUTE]
or
b=a1^T; [press ENTER/COMPUTE]
we get b=[ 1.0, 3.0; 2.0, 4.0];
. Inverse matrix
a1=[1, 2; 3, 4]; [press ENTER/COMPUTE]
b=a1^-1; [press ENTER/COMPUTE]
b=inv(a1)
we get b=[-2.0, 1.0; 1.5,-0.5];
. Determinant of a square matrix
. type
delta(a) [press ENTER/COMPUTE]
det(a)
where a is a matrix name
INFINITY and Indeterminate forms
Use 1/0 or (1/0) for generate the infinity
arcsec(1/0)= 1.570796326794897 arctan(1/0)= 1.570796326794897
arctan(1/0)= 1.570796326794897 arccot(1/0)= 0
1.570796326794897 = pi/2
0/0 = Indeterminate form
0^0 = Indeterminate form
(1/0)/(1/0) = Indeterminate form
(1/0)* 0 = Indeterminate form
Detailed questions about pi
Now, when we use pi we dont need to use *
In this way, we can type any expression with pi like
n pi OR +n pi OR -n pi ( n integer )
r pi OR +r pi OR -r pi ( r real, scientific notation)
It is the same as
n*pi, +n*pi, -n*pi ( we dont need to type * )
r*pi, +r*pi, -r*pi ( we dont need to type * )
Symbols
Type list and press ENTER/COMPUTE
Length of an expression or matrix
The calculator supports any expression or
matrix up to 3800 characters.
Memories
The calculator has 6 memories to save any
expression or matrix in the input box.
error function
Some functions use the system error = 1e-3
To set it to 1e-5, type error(1e-5)
and press ENTER/COMPUTE
printing on the paper
. print() prints the last results
. print(a) prints the matrix a
Constant name or Matrix name
. It must start with a letter and it can have
up to 8 characters;
. Type the constant name or the matrix name
and press ENTER/COMPUTE to see its definition or value.
Constants
Any integer or real number, scientific notation
e, +e, -e, pi, +pi, -pi
Keys
. ENTER to compute
. DELETE to clean the input box
. Use Home, End, Ctrl+Home, Ctrl+End etc.
when we have the focus in the input box.
Big expressions... or not
We can use a text editor to type it.
Then we copy and paste it into the edit box.
Some numbers
1700! = 2.99835320555842E+4755
e^10000 = 8.806818225662921E+4342
Try it and say something.
Good luck !
Thanks
Rui Loureiro
EDIT: i decided to replace calcula50 by the new calcula51
See the last post
I made some corrections in Calcula50
calcula51 has a new look
Good luck
Rui Loureiro
EDIT: now we have Calcula55 the version 3.00.1 with new functions and the file of rules RulesV2_20.txt
Please, see the last post
EDIT: I replaced Calcula58 by the new powerful Calcula59
It works with complex numbers. Please see the last post
EDIT: I replaced Calcula59 by the new powerful Calcula60
It works with complex numbers and complex functions.
Please see the last post
note: RulesV3_10_2I - rules in English
RulesV3_10_2P - regras em Português
EDIT: I replaced Calcula62 by the new powerful Calcula63
Please see the last post to get the examples
EDIT: I replaced calcula66 by calcula67
to correct some problems
DERIVATIVES : compute a derivative of any function
See the link in the last post
Hi all
News
The previous calcula48 doesnt work with
-INFINITY and +INFINITY.
Now i never test for overflow or division by 0
or other case than "invalid operation".
After getting "invalid operation" we examine
the case (we have 25 for powers).
To have no problems, we should use fclex before.
**** Here is calcula49 v1.09 ****
Now the calculator gives results from
-INFINITY to +INFINITY or "indeterminate form"
or ERROR.
It seems also that i solved all cases of
"indeterminate forms":
0/0 00/00 00-00 0*00
0^0 1^+00 1^-00 +00^0 -00^0
It gave me a lot of work to put it to work correctly
and to test each case.
Some cases:
1/0^3 = +INFINITY -(ln(e)-1)^-(1/0) = -INFINITY
-(-2/0)^(1/(2+1))= -(-INFINITY)^0.333333333333333 = ERROR
but
-(-2/0)^(1/3) = +INFINITY
(4*3-11)^-(3/(log(10)-1)) = Indeterminate form
Try it and say something.
Good luck !
Thanks
Rui Loureiro
raymond,
i added hyperbolic functions to calcula49
and now i want to test it.
I used your Ztest but i get this values
(it seems that something is wrong)
Ztest calcula50
arcsinh(1)= 0.881373587019543 0.881373587019543
arccosh(1)=0.000000000931323 0
arctanh(1)=21.48756259735830 +INFINITY
arccoth(1)=21.48756259735830 +INFINITY
arccsch(1)= 0.881373587019543 0.881373587019543
arcsech(1)=0-i0.000000000931323 0
arcsinh(-1234566677889e2345) = 0 -5428.096931407931
arcsinh(-1234566677889e3345) = 0 -INFINITY
arcsinh(+1234566677889e2345) = 0 5428.096931407931
arcsinh(+1234566677889e3345) = 0 +INFINITY
Hi all
Hyperbolic functions:
Can anyone confirm this values ?
Thanks ;)
sinh(x), tanh(x), coth(x), csch(x) are odd functions: sinh(-x)=-sinh(x)...
cosh(x), sech(x) are even functions: cosh(-x)=cosh(x)
sinh(-1/0) = -INFINITY
sinh(-1.234566677889e2375) = -INFINITY
sinh(-7200) = 4.161402346759365E+3126
sinh(-5) = -74.20321057778876
sinh(-1.0E-6) = -0.000001
sinh(0) = 0
sinh(+1.0E-6) = 0.000001
sinh(5) = 74.20321057778876
sinh(7200) =-4.161402346759365E+3126
sinh(1.234566677889e2375) = +INFINITY
sinh(1/0)= +INFINITY
cosh(-1/0) = +INFINITY
cosh(-1.234566677889e2375)= +INFINITY
cosh(-7200) = 4.161402346759365E+3126
cosh(-5) = 74.20994852478784
cosh(-1.0E-6) = 1.0000000000005
cosh(0) = 1.0
cosh(+1.0E-6) = 1.0000000000005
cosh(5) = 74.20994852478784
cosh(7200) = 4.161402346759365E+3126
cosh(1.234566677889e2375) = +INFINITY
cosh(1/0)= +INFINITY
tanh(-1/0) = -1.0
tanh(-1.234566677889e2375)= -1.0
tanh(-7200) =-1.0
tanh(-5) = -0.999909204262595
tanh(-1.0E-6) = -9.999999999996327E-0007
tanh(0) = 0
tanh(+1.0E-6) = 9.999999999996327E-0007
tanh(5) = 0.999909204262595
tanh(7200) = 1.0
tanh(1.234566677889e2375) = 1.0
tanh(1/0)= 1.0
coth(-1/0) = -1.0
coth(-1.234566677889e2375)= -1.0
coth(7200) = -1.0
coth(-5) = -1.000090803982019
coth(-1.0E-6) = -1000000.000000367
coth(0) = Undefined
coth(1.0E-6) = 1000000.000000367
coth(5) = 1.000090803982019
coth(7200) = 1.0
coth(1.234566677889e2375) =1.0
coth(1/0)= 1.0
sech(-1/0) = 0
sech(-1.234566677889e2375)= 0
sech(-7200) = 0
sech(-5) = 0.013475282221305
sech(-1.0E-6) = 0.9999999999995
sech(0) = 1.0
sech(+1.0E-6) = 0.9999999999995
sech(5) = 0.013475282221305
sech(7200) = 0
sech(1.234566677889e2375)= 0
sech(1/0)= 0
csch(-1/0) = 0
csch(-1.234566677889e2375)= 0
csch(-7200) = 0
csch(-5) = -0.013476505830589
csch(-1.0E-6) = -999999.9999998673
csch(0) = Undefined
csch( 1.0E-6) = 999999.9999998673
csch(5) = 0.013476505830589
csch(7200) = 0
csch(1.234566677889e2375)= 0
csch(1/0)= 0
Using the Windows calculator:
sinh(-1/0) = -INFINITY will not paste
sinh(-1.234566677889e2375) = -INFINITY too slow
sinh(-7200) = 4.161402346759365E+3126 -4.1614023467593632561951834091827e+3126
sinh(-5) = -74.20321057778876 -74.203210577788758977009471996065
-1.0e-6 :e
sinh(-1.0E-6) = -0.000001 -0.000001000000000000166666666666675
sinh(0) = 0 0
+1.0e-6 :e
sinh(+1.0E-6) = 0.000001 0.000001000000000000166666666666675
sinh(5) = 74.20321057778876 74.203210577788758977009471996065
sinh(7200) =-4.161402346759365E+3126 4.1614023467593632561951834091827e+3126
sinh(1.234566677889e2375) = +INFINITY invalid input for function
sinh(1/0)= +INFINITY will not paste
cosh(-1/0) = +INFINITY will not paste
cosh(-1.234566677889e2375)= +INFINITY invalid input for function
cosh(-7200) = 4.161402346759365E+3126 4.1614023467593632561951834091827e+3126
cosh(-5) = 74.20994852478784 74.209948524787844444106108044488
-1.0e-6 :e
cosh(-1.0E-6) = 1.0000000000005 1.0000000000005000000000000416667
cosh(0) = 1.0 1
+1.0e-6 :e
cosh(+1.0E-6) = 1.0000000000005 1.0000000000005000000000000416667
cosh(5) = 74.20994852478784 74.209948524787844444106108044488
cosh(7200) = 4.161402346759365E+3126 4.1614023467593632561951834091827e+3126
1.234566677889e2375 :e
cosh(1.234566677889e2375) = +INFINITY invalid input for function
cosh(1/0)= +INFINITY will not paste
tanh(-1/0) = -1.0 will not paste
tanh(-1.234566677889e2375)= -1.0 too slow
tanh(-7200) =-1.0 -1
tanh(-5) = -0.999909204262595 -0.99990920426259513121099044753447
-1.0e-6 :e
tanh(-1.0E-6) = -9.999999999996327E-0007 -0.0000009999999999996666666666668
tanh(0) = 0 0
+1.0e-6 :e
tanh(+1.0E-6) = 9.999999999996327E-0007 0.0000009999999999996666666666668
tanh(5) = 0.999909204262595 0.99990920426259513121099044753447
tanh(7200) = 1.0 1
tanh(1.234566677889e2375) = 1.0 invalid input for function
tanh(1/0)= 1.0 will not paste
do not know how to calc coth, trying cosh/sinh
coth(-1/0) = -1.0 will not paste
coth(-1.234566677889e2375)= -1.0 invalid input for function
coth(7200) = -1.0 1
coth(-5) = -1.000090803982019 -1.0000908039820193755366579205217
-1.0e-6 :e
coth(-1.0E-6) = -1000000.000000367 -1000000.0000003333333333333111111
coth(0) = +INFINITY 0
1.0e-6 :e
coth(1.0E-6) = 1000000.000000367 1000000.0000003333333333333111111
coth(5) = 1.000090803982019 1.0000908039820193755366579205217
coth(7200) = 1.0 1
coth(1.234566677889e2375) =1.0 invalid input for function
coth(1/0)= 1.0 will not paste
do not know how to calc sech, trying 1/cosh
sech(-1/0) = 0 will not paste
sech(-1.234566677889e2375)= 0 invalid input for functon
sech(-7200) = 0 2.4030360841670037099489877684017e-3127
sech(-5) = 0.013475282221305 0.013476505830589086655381881284338
-1.0e-6 :e
sech(-1.0E-6) = 0.9999999999995 0.99999999999950000000000020833333
sech(0) = 1.0 1
+1.0e-6 :e
sech(+1.0E-6) = 0.9999999999995 0.99999999999950000000000020833333
sech(5) = 0.013475282221305 0.013475282221304557305519138244882
sech(7200) = 0 2.4030360841670037099489877684017e-3127
sech(1.234566677889e2375)= 0 invalid input for function
sech(1/0)= 0 will not paste
do not know how to calc csch, trying 1/sinh
csch(-1/0) = 0 will not paste
csch(-1.234566677889e2375)= 0 too slow
csch(-7200) = 0 -2.4030360841670037099489877684017e-3127
csch(-5) = -0.013476505830589 -0.013476505830589086655381881284338
-1.0e-6 :e
csch(-1.0E-6) = -999999.9999998673 -999999.99999983333333333335277778
csch(0) = +INFINITY cannot divide by zero
1.0e-6 :e
csch( 1.0E-6) = 999999.9999998673 999999.99999983333333333335277778
csch(5) = 0.013476505830589 0.013476505830589086655381881284338
csch(7200) = 0 2.4030360841670037099489877684017e-3127
csch(1.234566677889e2375)= 0 invalid input for function
csch(1/0)= 0 will not paste
Hi Michael,
Thanks for reply :t
I corrected this:
sinh(-7200) = -4.161402346759365E+3126 «« copy-paste problem
sinh( 7200) = 4.161402346759365E+3126 «« copy-paste problem
sech(-7200) = 2.403036084167004E-3127
sech( 7200) = 2.403036084167004E-3127
coth(0) = Undefined This is undefined because it can be
csch(0) = Undefined -INFINITY or +INFINITY
It seems all results are correct
meanwhile you gave this result
coth(0) = +INFINITY 0
but it seems you want to write
coth(0) = +INFINITY cannot divide by zero
One question: the Windows calculator uses Real10 or another code?
Do you know, Michael ?
Many thanks
EDIT:
sinh(-1.0E-6) = -0.000001 Windows calculator=-0.000001000000000000166666666666675It seems i need to add a randon number generator to give 18 digits
+15 generated, no one knows if it is correct or not ! :greensml:
Hi all
Hyperbolic inverse functions:
Anyone can confirm this values ?
arcsinh(x)=ln(x+sqrt(x^2+1))
arcsinh(-1/0) = -INFINITY
arcsinh(-1.234566677889e2375) = -5469.543463081824
arcsinh(-7200) = -9.574983490386623
arcsinh(-5) = -2.312438341272753
arcsinh(-1.0e-6)= -9.999999999998556E-0007
arcsinh(0) = 0
arcsinh(1.0e-6) = 9.999999999998556E-0007
arcsinh(5) = 2.312438341272753
arcsinh(7200) = 9.574983490386623
arcsinh(1.234566677889e2375) = 5469.543463081824
arcsinh(1/0) = +INFINITY
arccosh(x)=ln(x+sqrt(x^2-1)) ; it is defined for x>=1
arccosh(-1/0) = Undefined
arccosh(-5) = Undefined
arccosh(0) = Undefined
arccosh(0.99) = Undefined
arccosh(1) = 0
arccosh(5) = 2.292431669561178
arccosh(7200) = 9.574983480741561
arccosh(1.234566677889e2375) = 5469.543463081824
arccosh(1/0) = +INFINITY
note1: if x is very large, arcsinh(x)=~arccosh(x)
arctanh(x)=2^-1* ln((1+x)/(1-x)) ; it is defined for |x|<1
arctanh(-1/0) = Undefined
arctanh(-1.234566677889e2375) = Undefined
arctanh(-5) = Undefined
arctanh(-1) = -INFINITY
arctanh(-0.1) = -0.100335347731076
arctanh(-0.01)= -0.010000333353335
arctanh(-0.001) = -0.001000000333334
arctanh(0) = 0
arctanh(0.001) = 0.001000000333334
arctanh(0.01)= 0.010000333353335
arctanh(0.1) = 0.100335347731076
arctanh(1) = +INFINITY
arctanh(1/0) = Undefined
arccoth(x)=2^-1* ln((x+1)/(x-1)) ; it is defined for |x|>1
arccoth(-1/0) = 0
arccoth(-5) = -0.202732554054082
arccoth(-1.99)= -0.552661861628048
arccoth(-1) = -INFINITY
arccoth(-0.99)= Undefined
arccoth(0) = Undefined
arccoth(0.99) = Undefined
arccoth(1) = +INFINITY
arccoth(1.99) = 0.552661861628048
arccoth(5) = 0.202732554054082
arccoth(1/0) = 0
arcsech(-1/0) = Undefined ; it is defined for 0 < x <= 1
arcsech(-5) = Undefined
arcsech(0) = Undefined +INFINITY / -INFINITY
arcsech(1.0E-8)= 19.11382792451231
arcsech(0.001)= 7.600902209541989
arcsech(0.01) = 5.298292365610485
arcsech(0.1) = 2.993222846126381
arcsech(0.99999) = 0.00447215458902
arcsech(0.999999999999) = 0.000001414213445
arcsech(1) = 0
arcsech(5) = Undefined
arcsech(1/0) = Undefined
arccsch(-1/0) = 0
arccsch(-1.234566677889e2375)= 0
arccsch(-7200)= -0.000138888888442
arccsch(-5) = -0.198690110349241
arccsch(-1) = -0.881373587019543
arccsch(-1.0E-8)= -19.11382792451231
arccsch(0) = Undefined +INFINITY / -INFINITY
arccsch(1.0E-8)= 19.11382792451231
arccsch(1) = 0.881373587019543
arccsch(5) = 0.198690110349241
arccsch(7200) = 0.000138888888442
arccsch(1.234566677889e2375)= 0
arccsch(1/0) = 0
Hi Rui,
The Windows calculator uses some sort of extended precision that can be very slow for some operations.
QuoteExtended Precision, a feature of Calculator, means that all operations are accurate to at least 32 digits. Calculator also stores rational numbers as fractions to retain accuracy. For example, 1/3 is stored as 1/3, rather than .333. However, errors accumulate during repeated operations on irrational numbers. For example, Calculator will truncate pi to 32 digits, so repeated operations on pi will lose accuracy as the number of operations increases.
Ok, Michael,
thank you :t
Hi all
**** Here is calcula50 v1.10 ****
Now, we have 57 functions
Now, the calculator gives results from
-INFINITY to +INFINITY
or
undefined e.g. ln(-1), arccos(-2), SIN(1/0)
or
"indeterminate form" e.g. 0/0
or
Operand too long e.g. 4444444444444444445
or
ERROR.
Now, the calculator solves all trigonometric functions,
all inverse trigonometric functions
all hyperbolic functions
and all inverse hyperbolic functions.
functions: atan2(y,x) and atan2d(y,x)
e.g. atan2d(24,sqr(3+1)) = 85.23635830927382
1+atan2d(24,sqr(-3+1)) = Undefined because sqr(-3+1) is Undefined
If you want to give any suggestion
please, answer this post
Try it and say something.
Good luck !
Thanks
Rui Loureiro
**** Here is calcula51 v1.11 ****
I made some corrections in Calcula50
calcula51 has a new look
Good luck
Rui Loureiro
Whoa, cool man, can I use it on my project? :t
Quote from: Farabi on June 30, 2012, 02:05:29 PM
Whoa, cool man, can I use it on my project? :t
Yes, Farabi, where you want no problems :t
If you know some erros, please tell me
i correct it
Hi all
***** Here is The calculator v2.20 *****
In this version we have:
1. conversions decimal-hexadecimal-binary
note: - => two's complement
type: -12d (decimal end with "d" or "D")
we get: unsigned: 18446744073709551604 / signed: -12
FFFFFFFFFFFFFFF4H
11111111111111111111111111111111
11111111111111111111111111110100B
type: 18446744073709551604d
we get: 18446744073709551604d
FFFFFFFFFFFFFFF4H
11111111111111111111111111111111
11111111111111111111111111110100B
type: FFFFFFFFFFFFFFF4H
we get: unsigned: 18446744073709551604 / signed: -12
FFFFFFFFFFFFFFF4H
11111111111111111111111111111111
11111111111111111111111111110100B
type: FFFFFFFFFFFFFFFFH
we get: unsigned: 18446744073709551615 / signed: -1
FFFFFFFFFFFFFFFFH
11111111111111111111111111111111
11111111111111111111111111111111B
type: 7FFF FFFF FFFF FFFFh
we get: 9223372036854775807
7FFFFFFFFFFFFFFFh
01111111111111111111111111111111
11111111111111111111111111111111B
type: 9223372036854775807d
we get: 9223372036854775807d
7FFFFFFFFFFFFFFFH
01111111111111111111111111111111
11111111111111111111111111111111B
2. logic expressions
first: shl and shr
next: or, and, xor (or = +)
- : invert
type: 21h shl 2d + 34h and -56h shl 3d and 99h xor 12h
we get: 18
00000012H
00000000000000000000000000010010B
type: -(21h shl 2d) + 34h and -56h shl 3d and 99h xor 12h
we get: 26
0000001AH
00000000000000000000000000011010B
type: -(21h shl 2d + (34h and -56h) shl 3d and 99h xor 12h)
we get: unsigned: 18446744073709551469 / signed: -147
FFFFFFFFFFFFFF6DH
11111111111111111111111111111111
11111111111111111111111101101101B
Now i need to do some more tests
but it seems it works correctly
Rui Loureiro
nice, Rui :t
hello,
what must be inputted to get the derivate of (e.g.) f(x)=x2?
For: df(x)=x^2 [ENTER] -> df(x=1) [ENTER] we get the output: X=1 and df(X)= 1.0, which is wrong.
regards, qWord
Rui,
I tested one of your expressions in test apps compiled with FreeBASIC and C (MSVC and GCC), and your expression evaluator produces a different result. Sorry, I didn't have time to test the other expressions.
dim as longint i
i = -(&h21 shl 2 + (&h34 and -&h56) shl 3 and &h99 xor &h12)
print i
print hex(i)
sleep
/'
Relevant parts of operator precedence table:
Operator Description Associativity
^ Exponentiate Left-to-Right
- Negate Right-to-Left
* Multiply Left-to-Right
/ Divide Left-to-Right
MOD Modulus Left-to-Right
SHL Shift left Left-to-Right
SHR Shift right Left-to-Right
+ Add Left-to-Right
- Subtract Left-to-Right
= Equal Left-to-Right
<> Not equal Left-to-Right
< Less than Left-to-Right
<= Less than or equal Left-to-Right
>= Greater than or equal Left-to-Right
> Greater than Left-to-Right
NOT Complement Right-to-Left
AND Conjunction Left-to-Right
OR Inclusive Disjunction Left-to-Right
EQV Equivalence Left-to-Right
IMP Implication Left-to-Right
XOR Exclusive Disjunction Left-to-Right
'/
-146
FFFFFFFFFFFFFF6E
#include <conio.h>
#include <stdio.h>
void main( void )
{
long long i;
i = -((0x21 << 2) + ((0x34 & -0x56) << 3) & 0x99 ^ 0x12);
printf("%I64d\n%I64X\n",i,i);
getch();
}
/*
Relevant parts of operator precedence table:
Operator Description Associativity
+ Unary plus Right-to-left
- Unary minus Right-to-left
~ Bitwise NOT Right-to-left
* Multiplication Left-to-right
/ Division Left-to-right
% Modulo (remainder) Left-to-right
+ Addition Left-to-right
- Subtraction Left-to-right
<< Bitwise left shift Left-to-right
>> Bitwise right shift Left-to-right
< Less than Left-to-right
<= Less than or equal to Left-to-right
> Greater than Left-to-right
>= Greater than or equal to Left-to-right
== Equal to Left-to-right
!= Not equal to Left-to-right
& Bitwise AND Left-to-right
^ Bitwise XOR (exclusive or) Left-to-right
| Bitwise OR (inclusive or) Left-to-right
= Direct assignment Right-to-left
*/
-146
FFFFFFFFFFFFFF6E
Dave,
:t
hello qWord,
We have not any way to get a derivative from f(x).
We should write f(x) and we should write
df(x) also.
In this way for f(x)=x^2 we should write
df(x)=2*x. For instance, for
f(x)=x^2+sin(x) we have df(x)=2*x+cos(x).
df(x) is used in one case in the root function.
Rules: If we define f(x)=x^2 it is wrong to
define the derivative of f(x) as
the function df(x)=x^2.
When we are typing df(x)= ...
we are defining it.
note: if f(x)=x^3/3 then df(x)= x^2.
best regards :t
Rui
MichaelW,
Thanks for your tests
When i thought to include logic expressions
the question was this: what operators to use.
May be someone should help me
In any way, i defined the next set of operators
operators what to do
------------- -----------------------
- (minus) invert the operand => -1100b => 0011b
(the same as NOT)
+ (plus) the same as OR
OR the same as "or" => 11 0101b or 00 1101b => 11 1101b
instruction
AND the same as "and" => 11 0101b and 00 1101b => 00 0101b
instruction
XOR the same as "xor" => 11 0101b xor 00 1101b => 11 1000b
instruction
SHL the same as "shl" => 11 0101b shl 2d => 1101 0100b
instruction
SHR the same as "shr" => 11 0101b shr 2d => 0000 1101b
instruction
We may use or, and, xor, shl, shr also.
The calculator doesnt use "not" or "NOT".
-------------------------------
How the calculator works
--------------------------------
First: any number with minus sign is inverted
before starting the operation.
In this way writing -1100b is the same
as writing 0011b (it should be).
Next: first operate SHL / SHR from left to right
and next all others. We may change this
rule using brackets.
If we want to change this rules we need to use
brackets. We may use as many as we want.
example:
This:
-(&h21 shl 2 + (&h34 and -&h56) shl 3 and &h99 xor &h12)
is this:
type: -(21h shl 2d + (34h and -56h) shl 3d and 99h xor 12h)
we get: unsigned: 18446744073709551469 / signed: -147
FFFFFFFFFFFFFF6DH
11111111111111111111111111111111
11111111111111111111111101101101B
So
1. We solve this: (34h and -56h)
32
00000020H
00000000000000000000000000100000B
Now the expression is:
-(21h shl 2d + 20h shl 3d and 99h xor 12h)
2. From left to right we need to solve 21h shl 2d
21h shl 2d = 132
00000084H
00000000000000000000000010000100B
Now the expression is:
-(84h + 20h shl 3d and 99h xor 12h)
3. From left to right we need to solve 20h shl 3d
20h shl 3d = 256
00000100H
00000000000000000000000100000000B
Now the expression is:
-(84h + 100h and 99h xor 12h) = -(84h or 100h and 99h xor 12h) =
Now we solve it from left to right:
84h or 100h = 388
00000184H
00000000000000000000000110000100B
= -(184h and 99h xor 12h)
184h and 99h = 128
00000080H
00000000000000000000000010000000B
= -(80h xor 12h)
80h xor 12h = 146
00000092H
00000000000000000000000010010010B
= -92h
Now we should invert it
= -00000000000000000000000010010010B
= 111111111111111111111111 1111 1111
111111111111111111111111 0110 1101b
To use the calculator we should use this: -92h or 0h
(because -92h gives the two's complement not the inverse)
The result is this:
unsigned: 18446744073709551469 / signed: -147
FFFFFFFFFFFFFF6DH
111111111111111111111111 1111 1111
111111111111111111111111 0110 1101B
I think that the problem is (or seems to be) the calculation rules
MichaelW,
From your data we have
Relevant parts of operator precedence table I:
Operator Description Associativity
^ Exponentiate Left-to-Right
- Negate Right-to-Left
* Multiply Left-to-Right
/ Divide Left-to-Right
MOD Modulus Left-to-Right
SHL Shift left Left-to-Right
SHR Shift right Left-to-Right
+ Add Left-to-Right
- Subtract Left-to-Right
= Equal Left-to-Right
<> Not equal Left-to-Right
< Less than Left-to-Right
<= Less than or equal Left-to-Right
>= Greater than or equal Left-to-Right
> Greater than Left-to-Right
NOT Complement Right-to-Left
AND Conjunction Left-to-Right
OR Inclusive Disjunction Left-to-Right
EQV Equivalence Left-to-Right
IMP Implication Left-to-Right
XOR Exclusive Disjunction Left-to-Right
Relevant parts of operator precedence table II:
Operator Description Associativity
+ Unary plus Right-to-left
- Unary minus Right-to-left
~ Bitwise NOT Right-to-left
* Multiplication Left-to-right
/ Division Left-to-right
% Modulo (remainder) Left-to-right
+ Addition Left-to-right
- Subtraction Left-to-right
<< Bitwise left shift Left-to-right
>> Bitwise right shift Left-to-right
< Less than Left-to-right
<= Less than or equal to Left-to-right
> Greater than Left-to-right
>= Greater than or equal to Left-to-right
== Equal to Left-to-right
!= Not equal to Left-to-right
& Bitwise AND Left-to-right
^ Bitwise XOR (exclusive or) Left-to-right
| Bitwise OR (inclusive or) Left-to-right
= Direct assignment Right-to-left
--------------------------------------------------------
The calculator operator precedence table III
--------------------------------------------------------
Operator Description Associativity Priority
SHL Shift left Left-to-Right 1
shl
SHR Shift right Left-to-Right 1
shr
- Complement Left-to-Right 0
AND Conjunction Left-to-Right 2
and
OR Inclusive Disjunction Left-to-Right 2
or
+ Inclusive Disjunction Left-to-Right 2
XOR Exclusive Disjunction Left-to-Right 2
xor
1. First are executed expressions inside brackets
2. All expressions are executed from left to right
3. Priority 0 means that it is executed before any other
4. All expressions with priority 1 are executed before
that with priority 2
NOTE: for decimal values we must end with "d" or "D"
For instance, 12345 should be 12345d or 12345D with spaces
or not
you could also add a few logic operations that CPU's don't generally support
NAND = NOT AND
NOR = NOT OR
XNOR = NOT XOR
when i design circuits, i find that NAND and NOR gates are more versatile than AND and OR gates :P
Quote from: dedndave on June 11, 2013, 09:01:01 PM
you could also add a few logic operations that CPU's don't generally support
NAND = NOT AND
NOR = NOT OR
XNOR = NOT XOR
when i design circuits, i find that NAND and NOR gates are more versatile than AND and OR gates :P
Yes it is true,
Ok, i will do it Dave :t
if f(x) = x2, then f'(x) = 2x, and f''(x) = 2 :biggrin:
Quote from: dedndave on June 11, 2013, 09:05:24 PM
if f(x) = x2, then f'(x) = 2x, and f''(x) = 2 :biggrin:
:biggrin:
Hey, Dave i wrote this:
Quote
In this way for f(x)=x^2 we should write
df(x)=2*x.
and you should know that 2X is equal to 2*X.
Dave about this i know what i am doing !
Quote from: dedndave on June 11, 2013, 09:01:01 PM
you could also add a few logic operations that CPU's don't generally support
NAND = NOT AND
NOR = NOT OR
XNOR = NOT XOR
when i design circuits, i find that NAND and NOR gates are more versatile than AND and OR gates :P
whats wrong with PANDN instruction?
Quote from: daydreamer2 on June 12, 2013, 10:20:03 PM
Quote from: dedndave on June 11, 2013, 09:01:01 PM
you could also add a few logic operations that CPU's don't generally support
NAND = NOT AND
NOR = NOT OR
XNOR = NOT XOR
when i design circuits, i find that NAND and NOR gates are more versatile than AND and OR gates :P
whats wrong with PANDN instruction?
NAND is: C = NOT (A AND B);
PANDN does: C = (NOT A) AND B
the PORN operation would be nice, also :badgrin:
i can't buy those gates in TTL or CMOS DIP packages
but, i can use them in ASIC designs :P
*** The Calculator v2.20 ***
The previous Calculator v2.10 doesnt solve
logic expressions correctly because i didnt write
the basic algorithm correctly - problems with brackets.
Now we have also the text file RulesV2_20.txt where i
say something about the calculator and not only.
You are there,
Read it and then ... say something
Sorry if i am not good in writing words, words, words ...
Good Luck
Thanks :t
RuiLoureiro
Files: Calcula54.zip & RulesV2_20.zip
Logic Expressions
--------------------------
The calculator operator precedence table
----------------------------------------------------
Operator Description Associativity Priority
SHL Shift left Left-to-Right 1
shl
SHR Shift right Left-to-Right 1
shr
- Complement (=NOT) Left-to-Right 0
AND Conjunction Left-to-Right 2
and
OR Inclusive Disjunction Left-to-Right 2
or
+ Inclusive Disjunction Left-to-Right 2
XOR Exclusive Disjunction Left-to-Right 2
xor
NAND not AND =-(x.y) Left-to-Right 2
nand
NOR not OR =-(x+y) Left-to-Right 2
nor
NXOR not XOR Left-to-Right 2
nxor
1. Expressions inside brackets are executed first
2. All expressions are executed from left to right
3. Priority 0 means that it is executed before any other
4. All expressions with priority 1 are executed before
that with priority 2
NOTE: for decimal values we must end with "d" or "D"
if it is the last value.
For instance, 12345 should be 12345d or 12345D with spaces
or not
------------
Examples
------------
type: 21h shl 2d + 34h and -56h shl 3d and 99h xor 12h
we get: 18
00000012H
00000000000000000000000000010010B
type: -(21h shl 2d) + 34h and -56h shl 3d and 99h xor 12h
we get: 26
0000001AH
00000000000000000000000000011010B
type: -(21h shl 2d + (34h and -56h) shl 3d and 99h xor 12h)
we get: unsigned: 18446744073709551469 / signed: -147
FFFFFFFFFFFFFF6DH
11111111111111111111111111111111
11111111111111111111111101101101B
type: 34h and 56h shl 3d + 24h nand 56 shr 2d nor 33 nxor 20h xor 11111b
we get: unsigned: 18446744073709551556 / signed: -60
FFFFFFFFFFFFFFC4H
11111111111111111111111111111111
11111111111111111111111111000100B
type: -(34h and (56h shl 3d) or (24h nand 56) shr 2d nor 33+22 nxor 20h) xor 11111b
we get: unsigned: 13835058055282163745 / signed: -4611686018427
C000000000000021H
11000000000000000000000000000000
00000000000000000000000000100001B
type: -(34h and (56h shl 3d) or (24h nand 56) shr 2d nor 33 or 22 nxor 20h) xor 11111b
unsigned: 13835058055282163745 / signed: -4611686018427
C000000000000021H
11000000000000000000000000000000
00000000000000000000000000100001B
type: -(34h and -(56h shl 3d) or -(24h nand 56) shr 2d nor 33+22 nxor 20h) xor 11111b
we get: unsigned: 18446744073709551593 / signed: -23
FFFFFFFFFFFFFFE9H
11111111111111111111111111111111
11111111111111111111111111101001B
type: -(1101b AND -(56h SHL 11b) OR -(24h nand 56) shr 2d nor 33+01100111b nxor 20h) xor 11111b
we get: unsigned: 18446744073709551560 / signed: -56
FFFFFFFFFFFFFFC8H
11111111111111111111111111111111
11111111111111111111111111001000B
type: -(000000000000000000000000000001101h AND -(56h SHL 11b) OR -(24h nand 56) shr 2d nor 33+01100111b nxor 20h) xor 11111b
we get: unsigned: 18446744073709547208 / signed: -4408
FFFFFFFFFFFFEEC8H
11111111111111111111111111111111
11111111111111111110111011001000B
type: (-(1101b AND -(56h SHL 11b) OR -(24h nand 56) shr 2d nor 33+01100111b nxor 20h) xor 11111b)
or
(34h and 56h shl 3d + 24h nand 56 shr 2d nor 33 nxor 20h xor 11111b)
and
-(21h shl 2d + 34h and -56h shl 3d and 99h xor 12h)
we get: unsigned: 18446744073709551564 / signed: -52
FFFFFFFFFFFFFFCCH
11111111111111111111111111111111
11111111111111111111111111001100B
-------------------------------------------------------
conversions decimal-hexadecimal-binary (64 bits)
-------------------------------------------------------
type: -12d (decimal end with "d" or "D")
we get: unsigned: 18446744073709551604 / signed: -12
FFFFFFFFFFFFFFF4H
11111111111111111111111111111111
11111111111111111111111111110100B
type: 18446744073709551604d
we get: 18446744073709551604d
FFFFFFFFFFFFFFF4H
11111111111111111111111111111111
11111111111111111111111111110100B
type: FFFFFFFFFFFFFFF4H
we get: unsigned: 18446744073709551604 / signed: -12
18446744073709551604
FFFFFFFFFFFFFFF4H
11111111111111111111111111111111
11111111111111111111111111110100B
type: FFFFFFFFFFFFFFFFH
we get: unsigned: 18446744073709551615 / signed: -1
FFFFFFFFFFFFFFFFH
11111111111111111111111111111111
11111111111111111111111111111111B
type: 7FFF FFFF FFFF FFFFh
we get: 9223372036854775807
7FFFFFFFFFFFFFFFh
01111111111111111111111111111111
11111111111111111111111111111111B
type: 9223372036854775807d
we get: 9223372036854775807d
7FFFFFFFFFFFFFFFH
01111111111111111111111111111111
11111111111111111111111111111111B
nice, as always, Rui :t
did you add the PORN operator ? :biggrin:
you could leave the "P" out
i think that indicates bitwise, which - all our operations are bitwise
so, it would be
NOT C = NOT(A)
AND C = A AND B
OR C = A OR B
XOR C = A XOR B
NAND C = NOT (A AND B)
NOR C = NOT (A OR B)
XNOR C = NOT (A XOR B) = A XOR (NOT B) = (NOT A) XOR B
ANDN C = A AND (NOT B)
ORN C = A OR (NOT B)
XORN is not needed - logically the same as XNOR
Quote from: dedndave on June 13, 2013, 03:46:02 AM
nice, as always, Rui :t
did you add the PORN operator ? :biggrin:
you could leave the "P" out
i think that indicates bitwise, which - all our operations are bitwise
so, it would be
NOT C = NOT(A)
AND C = A AND B
OR C = A OR B
XOR C = A XOR B
NAND C = NOT (A AND B)
NOR C = NOT (A OR B)
XNOR C = NOT (A XOR B) = A XOR (NOT B) = (NOT A) XOR B
ANDN C = A AND (NOT B)
ORN C = A OR (NOT B)
XORN is not needed - logically the same as XNOR
Hi Dave,
i will reply soon, i hope!
i want to post the new powerful calculator with
new interesting functions. But we need to wait
because i need to test it.
Yes it does ORN and ANDN also.
Replacing constants is now a very simple
and fast procedure.
We compute expressions to save as a new or old constant/variable
The same for matrices.
We have up to 20 real/logic/matrix names for constants/variables.
The Inverse matrix is all inside the edit box in any case 2x2 to 20x20.
Error messages are now better
Ola, RuiLoureiro!
ORN = (Not A) OR (Not B) = Not (A AND B) ?
NAND = NOT (A AND B)
NOR = NOT (A OR B)
ORN = A OR (NOT B)
Olá Mikl,
It's not right.
I use A ORN B = A or (not B) = A or -B as Dave wrote
Hi
In the next Calcula55, The powerfull Calculator v2.30,
we may solve any quadratic or any system of 2,3,4 equations
where any coeficient may be a real expression as large as
a train (11 300 characters)!
Well, we may solve any system of 5,6, ..,19,20 equations
using matrices;
We may define or redefine any one of 20 constants: real, logic
or matrix.
Each matrix element is a real constant or any real expression
as large as a train!
When we want to compute a real/logic expression, we may define or
redefine a constant:
(log(2)+56.38)/12.5 (=4.5344823996531185)
or
this = (log(2)+56.38)/12.5 (=4.5344823996531185)
(the expression may contain real constants)
Using 'list r' we may edit the real constants and we may redefine them.
Using 'list l' we may edit the logic constants and we may redefine them.
Any 20x20 matrix fits in th edit box, so we may copy it from or to
the edit box.
What do you think about this version ?
Thanks
Rui
-------------------------------------------------------------------------
TYPE: this = (log(2)+56.38)/12.5 ENTER/COMPUTE
TYPE: a=12.3;b=-1.5;c=15;d=3.24; ENTER/COMPUTE
Quadratic equation:
TYPE: ax^2+bx+c=d ENTER/COMPUTE
Root X0= 0.0609756097560975+ i 0.9758993472640925
Root X1= 0.0609756097560975- i 0.9758993472640925
--------------------------------------------------------------------------
Quadratic equation:
TYPE: this x^2 + bx + c=d ENTER/COMPUTE
Root X0= 0.1653992526373845+ i 1.6019061672211593
Root X1= 0.1653992526373845- i 1.6019061672211593
--------------------------------------------------------------------------
System of 2 equations:
TYPE: this x + by= c ; x - d y= 12;
X= 2.3196363400200491
Y= -2.9877665617222071
Determinant: -13.1917229748761039
****-- Here it is The Calculator v3.00.1 --****
This is the new calculator with new functions
and new better and faster procedures.
A. What it does ?
---------------------
1. Conversions
2. Real constants definitions
3. Logic constants definitions
4. Function definition
5. Derivative definition
6. Matrices definitions
7. Logic operations
8. Matrices operations
9. Systems of 2 linear equations x,y
10. Systems of 3 linear equations x,y,z
11. Systems of 4 linear equations x,y,z,t
12. Quadratic equation ax^2+bx+c=d
13. Solves any real expression
B. Special functions
------------------------
B.1. list shows the symbols defined
list r edit the r eal constans defined
list l edit the l ogic constants defined
list f edit the f unction defined
list d edit the d erivative defined
B.2. scan shows the values of a defined function f(x)
from x=x0 to x=x1 where x0,x1 are integers.
It shows also the zeros
B.3. root try to find the value x=x0 where the function f(x)=0
C. General rule about constants and matices
--------------------------------------------------------
1. Any constant or matrix is redefinable
2. The tables have space for 20 names
C. General rule about real constants
----------------------------------------------
1. The name cannot be any symbol already defined
like 'e' 'pi', 'log', 'sin', etc.
2. The name cannot be x, y, z, t
3. The name cannot be a matrix name
4. The matrix name cannot be any symbol already defined
and cannot be any real constant name
5. Any real constant is redefinable
6. We may define a real constant like this
name = real expression
D. General rule about logic constants
----------------------------------------------
1. The name cannot be any symbol already defined
2. The name cannot be 'd', 'h', 'b'
3. We may use real constant names or matrix names
because logic expressions doesnt use real values
or matrix names and real expressions doesnt use
logic constants
E. General rule about expressions
-------------------------------------------
We may use any constant already defined
F. Defining a matrix
-------------------------
F.1
matA =[1,2,3; 3,4,5; 2,5,9] (without semicolon)
F.2
a=2; b=3;
matB =[3*2, a, b; 1.2, 4.5,-5.23]; (with semicolon)
F.3
matC =[2, a, b; 1, 5,-5; 0,1,-1];
matC =matC^-1;
G. Defining a real constant
----------------------------------
G.1
a=2;b=-5;c=3;
G.2
a=2;b=3;
f=round(e^-(-3+sin(3*a+12)),6)+34*5-sin(pi/b)
H. Defining a logic constant
----------------------------------
H.1
l1=2d; l2=53h; l3=11100011b;
H.2
l1=2d; l2=53h; l3=11100011b;
l4= l1 + l2 shl 1d + 45h and l3
I. System of equations
-----------------------------
1. a=2;b=-3;c=4;d=-1;f=5;g=-2;h=-6;i=0;j=4;k=1;l=-3;
2. aX-bY=c; dY+fX=-a;
aX-bY+cZ=d; Z-dY+fX=-a; X+Y+Z=h;
kT+aX-bY+cZ=d; Z-dY+fX=-a; X+Y+Z-T=h; Y+lZ+fT=l;
J. Quadratic equation
---------------------------
a=2;b=-3;c=4;d=-1;
aX^2 -bX +c=d
K. root function
--------------------
1. f(x)=2*x^5+3*x^4-x^3+5*x^2+x+120
2. root(x=-2,x=2)
root(x=-2,x=2; n=1000)
root(x=-2,x=2; d=0.01)
df(x)=10*x^4+12*x^3-3*x^2+10*x+1
root(x=-2,x=2; x=-2)
L. scan function
--------------------
1. f(x)=x^5-5*x^3+4*x
2. scan(x=-10,x=10)
Now type 'list' and we can see all constants defined till now.
M. conversions
-------------------
Type 112233d or 112233h or 111000111b and press COMPUTE
note 1: type/copy and paste the examples and press compute
note 2: to get the polynomials zeros use scan() function
Try it and say something.
Good luck !
Thanks
Rui Loureiro
****-- The Calculator v3.00.2 --****
****-- The new version --****
--------------------------------
Three things A.1 A.2 A.3
--------------------------------
--------------------------------------------------------------------
A.1. Define a matrix as a direct operation of 2 matrices
---------------------------------------------------------------------
matA= -1.2*[1,2,3; 4,5,6; 7,8,9]
matB= [1,2,3; 4,5,6; 7,8,9]*[3; 6; 9]
matC= [1,2,3; 4,5,6; 7,8,9]-[-9,8,7; -4,0,6; 1,3,2]
Example 1
Solving a system of 5 linear equations:
1 X1 -2 X2 +3 X3 -1 X4 + 0 X5 = 123
0 X1 +3 X2 -2 X3 +5 X4 + 7 X5 = 15
3 X1 -4 X2 +0 X3 -2 X4 + 3 X5 = 9
-2 X1 -2 X2 +5 X3 +0 X4 + 0 X5 = -12
1 X1 +0 X2 -1 X3 +3 X4 + 9 X5 = 35
matC=[123; 15; 9; -12; 35]
matA=[1,-2,3,-1,0; 0,3,-2,5,7; 3,-4,0,-2,3; -2,-2,5,0,0; 1,0,-1,3,9]
matA1=matA^-1;
=[ 0.5230125523012552, 0.2761506276150627, 0.0711297071129707,
-0.2510460251046025,-0.2384937238493723; 0.6903765690376569,
-0.7154811715481171,-0.8661087866108786,-0.5313807531380753,
0.8451882845188284; 0.4853556485355648,-0.1757322175732217,
-0.3179916317991631,-0.11297071129707113, 0.2426778242677824;
-0.401673640167364, 1.1799163179916318, 0.8493723849372384,
0.4728033472803347,-1.200836820083682; 0.1297071129707113,
-0.4435146443514644,-0.3263598326359832,-0.1422594142259414,
0.5648535564853556];
multiplying by [123; 15; 9; -12; 35]
= [ 63.778242677824266; 102.347280334728032; 64.050209205020915;
-71.765690376569036; 27.841004184100418];
X1= 63.778242677824266
X2= 102.347280334728032
X3= 64.050209205020915
X4= -71.765690376569036
X5= 27.841004184100418
first equation:
63.778242677824266 -2* 102.347280334728032 + 3* 64.050209205020915
+71.765690376569036 = 122.999999999999983
--------------------
A.2 find function
--------------------
Now we have the scan function and the find function
to study one function.
The scan function uses integer points and is useful to
study any polynomial.
The find function try to get an interval (x0,x1) where
the function changes the sign. So it may be a discontinuity
or it may be a zero (it shows the f(x) values)
Example 2
f(x)=(x+1)/(x^2-x-1)
find(x=-10, x=10)
One zero was found
x= -1.0
f(x)= 0
Example 3
f(x)=(log(x-1)-x)/(x-5)
find(x=1, x=20)
The function change the sign in this interval
(x0,x1)=[ 4.9999 , 5.0 ]
f(x0)= 43978.50866169812
f(x1)= -INFINITY
Example 4
f(x)=x-log(1/(x-1))
find(x=1, x=5)
The function change the sign in this interval
(x0,x1)=[ 1.0826 , 1.0827 ]
f(x0)= -0.0004199526796177
f(x1)= 0.0002055095525466
Example 5
f(x)=(x-log(1/(x-1)))/(x^2-1)
find(x=1, x=10)
The function change the sign in this interval
(x0,x1)=[ 1.0826 , 1.0827 ]
f(x0)= -0.0004199526796177
f(x1)= 0.0002055095525466
Example 6
f(x)=x^3-log(x^2+x)+x-1
find(x=0,x=1)
The function change the sign in this interval
(x0,x1)=[ 0.1177 , 0.1178 ]
f(x0)= 0.0002288184583431
f(x1)= -0.000074703978969
scan(x=-2,x=2)
x= -2 , f(x)=-11.3010299956639812 ; x= -1 , f(x)=+INFINITY ;
x= 0 , f(x)=+INFINITY ; x= 1 , f(x)= 0.6989700043360188 ;
x= 2 , f(x)= 8.2218487496163564 ;
---------------------------------------------------------------------------
A.3 The limits of the root/find function may be real expressions
---------------------------------------------------------------------------
f(x)=x^3-x
root(x=log(2), x=log(3))
find(x=-log(3), x=log(3))
Good luck !
Rui Loureiro
****-- The Calculator v3.01.0 --****
---------------------------
COMPLEX NUMBERS
---------------------------
Now, we may solve any complex expression in the same way we do for
real expressions.
Any complex number is a number with a real part
and an imaginary part. For instance, z1=-2+i3 or z1=i3-2.
We may define it typing z1=(-2,3); or z1=-2+i3 for example.
To build a complex expression
we may define a set of real/complex constants z1, z2, z3, z4, ...
and then we write the expression
For instance,
(1-i2)*(2-i3) or (a-ib)*(c-id) where a=12;b=-4;c=1;d=-3;
are real constants
z1+z2*z3-z4 or z5=z1+z2*z3-z4 or z1=z1+z2*(z3/z4)+3-i2
Type: d=7;c=1;b=-3;a=2; press ENTER/COMPUTE
and (a-ib)*(c-id) press ENTER/COMPUTE
and Z1=(a-ib)*(c-id) press ENTER/COMPUTE
We get: 23.0-i11.0
| Z |= 25.495097567963924
Angle= -25.559965171823808 degrees
Complex Constant defined
If you have any suggestion, please tell me
Jul. 2013
Good luck !
Thanks
Rui Loureiro
EDIT: In the next version the calculator will do all
complex functions (e.g.: ln(2-i3), etc. )
Hi all
****-- The Calculator v3.10.1 --****
This is the last version and this is the powerful calculator
v3.10.1.
--------------------------------------------------------------
COMPLEX NUMBERS and COMPLEX FUNCTIONS
--------------------------------------------------------------
Now, we may solve any complex expression in the same way we do for
real expressions.
We may use any function:
conj(a+ib) = a-ib
inv(a+ib) = 1/(a+ib)
abs(a+ib) = sqr(a^2+b^2)+i0
sqr(a+ib) = (a+ib)^(1/2)
rnd(a+ib) = rnd(a)+i rnd(b)
rndi(a+ib) = rndi(a)+i rndi(b)
e^(a+ib) = e^a * e^ib = e^a * ( cos(b)+i sin(b))
(a+ib)^(c+id) = e^( (c+id)*ln(a+ib) )
ln(a+ib) = ln (|a+ib|* e^i angleR )
log(a+ib) = (ln (|a+ib|* e^i angleR )) / ln(10)
sin(a+ib)
cos(a+ib)
tan(a+ib)
sec(a+ib)
csc(a+ib)
cot(a+ib)
arcsin(a+ib)
arccos(a+ib)
arctan(a+ib)
arcsec(a+ib)
arccsc(a+ib)
arccot(a+ib)
sinh(a+ib)
cosh(a+ib)
tanh(a+ib)
sech(a+ib)
csch(a+ib)
coth(a+ib)
arcsinh(a+ib)
arccosh(a+ib)
arctanh(a+ib)
arcsech(a+ib)
arccsch(a+ib)
arccoth(a+ib)
round(a+ib, n) - round 'a' and 'b' to n decimal places
The operation rules are the same for real numbers
The result may be a complex number, INFINITY or indeterminate form.
We may use the division by 0 to generate the infinity.
We need to use brackets when we have powers of powers
(1-i2)^(1-i)^(i2) gives "Complex power too complex- use brackets"
We need to do ((1-i2)^(1-i))^(i2) or (1-i2)^((1-i)^(i2)).
Any complex number is a number with a real part 'a'
and an imaginary part 'b'. For instance, z1=-2+i3 or z1=i3-2.
We may define it typing z1=(-2,3); or z1=-2+i3 for example.
To build a complex expression
we may define a set of real/complex constants z1, z2, z3, z4, ...
and then we write the expression
For instance,
(1-i2)*(2-i3) or (a-ib)*(c-id) where a=12;b=-4;c=1;d=-3;
are real constants
z1+z2*z3-z4 or z5=z1+z2*z3-z4 or z1=z1+z2*(z3/z4)+3-i2
After defining any constant/matrix we may type the constant name
to see its value. We may use also list c or list r or list l or
list.
Example 1
Type: d=7;c=1;b=-3;a=2; press ENTER/COMPUTE
and (a-ib)*(c-id) press ENTER/COMPUTE
and Z1=(a-ib)*(c-id) press ENTER/COMPUTE
We get: 23.0-i11.0
| Z |= 25.495097567963924
Angle= -25.559965171823808 degrees
Angle= -0.4461055489434036 radians
Complex Constant defined
Example 2
conj(-1-i2)+inv(1-i3)
-0.9+i2.3
| Z |= 2.4698178070456938
Angle= 111.370622269343183 degrees
Angle= 1.9437840485949576 radians
Example 3
zx=arctan(-1-i3.3)+sin(1.2+i3e-2)*(i*cos(1-i))^(2-i)
-1.2156119789977935-i18.15086513723519
| Z |= 18.191525986392202
Angle= -93.831529845398134 degrees
Angle= -1.6376691379855234 radians
Complex Constant defined
Example 4
zy=(ln(zx)+1-i)* (ln(zx-1)+1+i)+e^(9.1-i2.3e-1) ; edit: ')' was not here
8732.8791809611277-i2054.6089708474515
| Z |= 8971.3207953094131
Angle= -13.2393627831433264 degrees
Angle= -0.231070471431851 radians
Complex Constant defined
Example 5
zz=zx*zy-zy/zx
-47989.335119512577-i156498.23124802209
| Z |= 163690.78370199185
Angle= -107.0478745161774 degrees
Angle= -1.8683378675690276 radians
Complex Constant defined
Example 6
zw=(e+i pi)*(pi-ie)/((ipi)* ie)
-2.0-i0.2904713703586566
| Z |= 2.0209833292231868
Angle= -171.73638751817216 degrees
Angle= -2.9973654076729973 radians
Complex Constant defined
Example 7
zr=round( (e+i pi)*(pi-ie)/((ipi)* ie), 3)
-2.0-i0.29
| Z |= 2.0209156340629363
Angle= -171.7496127710945 degrees
Angle= -2.9975962318809012 radians
Complex Constant defined
If you know any bug or something else, please post it
If you have any suggestion, please tell me
Jul. 2013
Good luck ! ;)
Thanks
Rui Loureiro
Trabalho excelente, rui.
Me ajudou muito há pouco quando estava testando uma função de atan2.
Olá guga,
obrigado !
Hi all,
(1) The previous The Calculator v3.10.1 (calcula59.exe)
dosn't solve trigonometric functions,
when we are working with angles in DEGREES:
sind(a+ib)
cosd(a+ib)
tand(a+ib)
secd(a+ib)
cscd(a+ib)
cotd(a+ib)
arcsind(a+ib)
arccosd(a+ib)
arctand(a+ib)
arcsecd(a+ib)
arccscd(a+ib)
arccotd(a+ib)
If we try, we get "Complex expression error"
(2) The results of some complex trigonometric functions
are not correct
because there is a bug in one internal procedure
(i wanted «fstp OperandZ» but i did «fstp OperandI» !!!)
(3) there is a problem when we try to define variables x1, x2, x3, etc
All these problems are solved in
the NEXT version v3.10.2 (calcula60.exe)
that i will post soon.
--------------------------------------------------------------------
These are some values that we get, when we use the
next version v3.10.2 (calcula60.exe)
--------------------------------------------------------------------
--------------------------------------
about trigonometric functions
inverse trigonometric functions
RADIANS
--------------------------------------
TYPE: z1=pi/4+i3
WE GET:
0.7853981633974483+i3.0
| Z |= 3.101104686247803
Angle= 75.329256947468645 degrees
Angle= 1.314743556814141 radians
Complex Constant defined
TYPE: z2=sin(z1)
WE GET:
7.1189120679085489+i7.0837072942502342
| Z |= 10.0427993139974647
Angle= 44.857978252688267 degrees
Angle= 0.7829194162974231 radians
Complex Constant defined
TYPE: w1=arcsin(z2)
WE GET:
0.7853981633974483+i3.0 ->>>> is = z1 (correct)
| Z |= 3.101104686247803
Angle= 75.329256947468645 degrees
Angle= 1.314743556814141 radians
Complex Constant defined
TYPE: z2=cos(z1)
WE GET:
7.1189120679085491-i7.0837072942502341
| Z |= 10.0427993139974647
Angle= -44.857978252688266 degrees
Angle= -0.7829194162974231 radians
Complex Constant defined
TYPE: w1=arccos(z2)
WE GET:
0.7853981633974483+i3.0 ->>>> is = z1 (correct)
| Z |= 3.101104686247803
Angle= 75.329256947468645 degrees
Angle= 1.314743556814141 radians
Complex Constant defined
TYPE: z2=tan(z1)
WE GET:
0.0049574738935603+i0.9999877116507955
| Z |= 1.0
Angle= 89.715956505376533 degrees
Angle= 1.5658388325948463 radians
Complex Constant defined
TYPE: w1=arctan(z2)
WE GET:
0.7853981633974411+i3.0000000000000079 ->>>> is = z1 (correct)
| Z |= 3.1011046862478089
Angle= 75.32925694746881 degrees
Angle= 1.3147435568141439 radians
Complex Constant defined
TYPE: z2=csc(z1)
WE GET:
0.0705836414544171-i0.0702345879617375
| Z |= 0.099573830834817
Angle= -44.857978252688267 degrees
Angle= -0.7829194162974231 radians
Complex Constant defined
TYPE: w1=arccsc(z2)
WE GET:
0.7853981633974483+i3.0000000000000004 ->>>> is = z1 (correct)
| Z |= 3.1011046862478034
Angle= 75.329256947468646 degrees
Angle= 1.3147435568141411 radians
Complex Constant defined
TYPE: z2=sec(z1)
WE GET:
0.0705836414544171+i0.0702345879617375
| Z |= 0.099573830834817
Angle= 44.857978252688266 degrees
Angle= 0.7829194162974231 radians
Complex Constant defined
TYPE: w1=arcsec(z2)
WE GET:
0.7853981633974482+i3.0000000000000004 ->>>> is = z1 (correct)
| Z |= 3.1011046862478034
Angle= 75.329256947468647 degrees
Angle= 1.3147435568141411 radians
Complex Constant defined
TYPE: z2=cot(z1)
WE GET:
0.0049574738935603-i0.9999877116507955
| Z |= 1.0
Angle= -89.715956505376533 degrees
Angle= -1.5658388325948463 radians
Complex Constant defined
TYPE: w1=arccot(z2)
WE GET:
0.7853981633974554+i3.0000000000000079 ->>>> is = z1 (correct)
| Z |= 3.1011046862478125
Angle= 75.329256947468554 degrees
Angle= 1.3147435568141395 radians
Complex Constant defined
--------------------------------------
about trigonometric functions
inverse trigonometric functions
RADIANS
--------------------------------------
--------------
First quadrant
--------------
z1=pi/4+i3 = 0.7853981633974483+i3.0
y1=sin(z1) = 7.1189120679085489+i7.0837072942502342
w1=arcsin(y1) = 0.7853981633974483+i3.0 = z1
y1=cos(z1) = 7.1189120679085491-i7.0837072942502341
w1=arccos(y1) = 0.7853981633974483+i3.0 = z1
y1=tan(z1) = 0.0049574738935603+i0.9999877116507955
w1=arctan(y1) = 0.78539816339744 11+i3.00000000000000 79 = z1
---------------
Second quadrant
---------------
z2=-pi/4+i3 = -0.7853981633974483+i3.0
y2=sin(z2) = -7.1189120679085489+i7.0837072942502342
w2=arcsin(y2) = -0.7853981633974483+i3.0 = z2
y2=cos(z2) = 7.1189120679085491+i7.0837072942502341
w2=arccos(y2) = -0.7853981633974483+i3.0 = z2
y2=tan(z2) = -0.0049574738935603+i0.9999877116507955
w2=arctan(y2) = -0.78539816339744 11+i3.00000000000000 79 = z2
--------------
Third quadrant
--------------
z3=-pi/4-i3 = -z1
y3=sin(z3) = -7.1189120679085489-i7.0837072942502342
w3=arcsin(y3) = -0.7853981633974483-i3.0 = z3
y3=cos(z3) = 7.1189120679085491-i7.0837072942502341
w3=arccos(y3) = 0.7853981633974483+i3.0 = z1
y3=tan(z3) = -0.0049574738935603-i0.9999877116507955
w3=arctan(y3) = -0.78539816339744 11-i3.00000000000000 79 = z3
---------------
Fourth quadrant
---------------
z4=pi/4-i3 = -z2
y4=sin(z4) = 7.1189120679085489-i7.0837072942502342
w4=arcsin(y4) = 0.7853981633974483-i3.0 = z4
y4=cos(z4) = 7.1189120679085491+i7.0837072942502341
w4=arccos(y4) = -0.7853981633974483+i3.0 = z2
y4=tan(z4) = 0.0049574738935603-i0.9999877116507955
w4=arctan(y4) = 0.78539816339744 11-i3.00000000000000 79 = z4
--------------------------------------
about trigonometric functions
inverse trigonometric functions
DEGREES
--------------------------------------
TYPE: z1=45+i3
WE GET:
| Z |= 45.099889135118725
Angle= 3.8140748342903542 degrees
Angle= 0.0665681637758238 radians
Complex Constant defined
TYPE: z2=sind(z1)
WE GET:
7.118912067908549+i7.0837072942502342
| Z |= 10.0427993139974647
Angle= 44.857978252688266 degrees
Angle= 0.7829194162974231 radians
Complex Constant defined
TYPE: w1=arcsind(z2)
WE GET:
45.0+i3.0 ->>>> is = z1 (correct)
| Z |= 45.099889135118724
Angle= 3.8140748342903542 degrees
Angle= 0.0665681637758238 radians
Complex Constant defined
TYPE: z2=cosd(z1)
WE GET:
7.118912067908549-i7.0837072942502342
| Z |= 10.0427993139974647
Angle= -44.857978252688266 degrees
Angle= -0.7829194162974231 radians
Complex Constant defined
TYPE: w1=arccosd(z2)
WE GET:
45.0+i3.0 ->>>> is = z1 (correct)
| Z |= 45.099889135118725
Angle= 3.8140748342903542 degrees
Angle= 0.0665681637758238 radians
Complex Constant defined
TYPE: z2=tand(z1)
WE GET:
0.0049574738935603+i0.9999877116507955
| Z |= 1.0
Angle= 89.715956505376533 degrees
Angle= 1.5658388325948463 radians
Complex Constant defined
TYPE: w1=arctand(z2)
WE GET:
44.99999999999959+i3.0000000000000079 ->>>> is = z1 (correct)
| Z |= 45.099889135118316
Angle= 3.8140748342903989 degrees
Angle= 0.0665681637758245 radians
Complex Constant defined
TYPE: z2=cscd(z1)
WE GET:
0.0705836414544171-i0.0702345879617375
| Z |= 0.099573830834817
Angle= -44.857978252688266 degrees
Angle= -0.7829194162974231 radians
Complex Constant defined
TYPE: w1=arccscd(z2)
WE GET:
45.000000000000002+i3.0000000000000004 ->>>> is = z1 (correct)
| Z |= 45.099889135118727
Angle= 3.8140748342903545 degrees
Angle= 0.0665681637758238 radians
Complex Constant defined
TYPE: z2=secd(z1)
WE GET:
0.0705836414544171+i0.0702345879617375
| Z |= 0.099573830834817
Angle= 44.857978252688266 degrees
Angle= 0.7829194162974231 radians
Complex Constant defined
TYPE: w1=arcsecd(z2)
WE GET:
44.999999999999998+i3.0000000000000004 ->>>> is = z1 (correct)
| Z |= 45.099889135118723
Angle= 3.8140748342903549 degrees
Angle= 0.0665681637758238 radians
Complex Constant defined
TYPE: z2=cotd(z1)
WE GET:
0.0049574738935603-i0.9999877116507955
| Z |= 1.0
Angle= -89.715956505376533 degrees
Angle= -1.5658388325948463 radians
Complex Constant defined
TYPE: w1=arccotd(z2)
WE GET:
45.00000000000041+i3.0000000000000079 ->>>> is = z1 (correct)
| Z |= 45.099889135119134
Angle= 3.8140748342903296 degrees
Angle= 0.0665681637758233 radians
Complex Constant defined
---------------------------------
about hyperbolic functions
inverse hyperbolic functions
---------------------------------
TYPE: z1=2.3+i0.25
WE GET:
2.3+i0.25
| Z |= 2.3135470602518549
Angle= 6.2034479016918352 degrees
Angle= 0.10827059086045604 radians
Complex Constant defined
TYPE: z2=sinh(z1)
WE GET:
4.783483618904492+i1.2462283322677304
| Z |= 4.9431569455636773
Angle= 14.6025145321957282 degrees
Angle= 0.2548619576571349 radians
Complex Constant defined
TYPE: w1=arcsinh(z2)
WE GET:
2.3+i0.25 ->>>> is = z1 (correct)
| Z |= 2.3135470602518549
Angle= 6.2034479016918353 degrees
Angle= 0.10827059086045604 radians
Complex Constant defined
TYPE: z2=cosh(z1)
WE GET:
4.8806256579738629+i1.2214238973804283
| Z |= 5.0311413367510157
Angle= 14.0502468762936514 degrees
Angle= 0.2452230687093726 radians
Complex Constant defined
TYPE: w1=arccosh(z2)
WE GET:
2.3+i0.25 ->>>> is = z1 (correct)
| Z |= 2.3135470602518549
Angle= 6.2034479016918353 degrees
Angle= 0.10827059086045604 radians
Complex Constant defined
TYPE: z2=tanh(z1)
WE GET:
0.9824664000943306+i0.0094701778128957
| Z |= 0.9825120414438294
Angle= 0.5522676559020767 degrees
Angle= 0.0096388889477623 radians
Complex Constant defined
TYPE: w1=arctanh(z2)
WE GET:
2.3000000000000002+i0.249999999999998 ->>>> is = z1 (correct)
| Z |= 2.3135470602518549
Angle= 6.2034479016917859 degrees
Angle= 0.10827059086045518 radians
Complex Constant defined
TYPE: z2=csch(z1)
WE GET:
0.1957651997848019-i0.0510021895924765
| Z |= 0.2022998684873777
Angle= -14.6025145321957282 degrees
Angle= -0.2548619576571349 radians
Complex Constant defined
TYPE: w1=arccsch(z2)
WE GET:
2.3000000000000003+i0.2499999999999997 ->>>> is = z1 (correct)
| Z |= 2.3135470602518552
Angle= 6.2034479016918292 degrees
Angle= 0.10827059086045593 radians
Complex Constant defined
TYPE: z2=sech(z1)
WE GET:
0.1928157309006095-i0.0482540063543098
| Z |= 0.1987620567713517
Angle= -14.0502468762936514 degrees
Angle= -0.2452230687093726 radians
Complex Constant defined
TYPE: w1=arcsech(z2)
WE GET:
2.3000000000000005+i0.2499999999999998 ->>>> is = z1 (correct)
| Z |= 2.3135470602518554
Angle= 6.2034479016918313 degrees
Angle= 0.10827059086045597 radians
Complex Constant defined
TYPE: z2=coth(z1)
WE GET:
1.017751950195062-i0.0098103018452776
| Z |= 1.0177992307661405
Angle= -0.5522676559020767 degrees
Angle= -0.0096388889477623 radians
Complex Constant defined
TYPE: w1=arccoth(z2)
WE GET:
2.3000000000000003+i0.2499999999999993 ->>>> is = z1 (correct)
| Z |= 2.3135470602518551
Angle= 6.2034479016918193 degrees
Angle= 0.10827059086045576 radians
Complex Constant defined
---------------------------------
hyperbolic functions
inverse hyperbolic functions
---------------------------------
TYPE: z1=2.3+i1.58 = Z + iI ( I > pi/2)
WE GET:
2.3+i1.58
| Z |= 2.790412155936825
Angle= 34.487372830078654 degrees
Angle= 0.6019182062477074 radians
Complex Constant defined
TYPE: y1=sinh(z1)
WE GET:
-0.045437541593796+i5.0370073053376468
| Z |= 5.0372122413306062
Angle= 90.516836405985953 degrees
Angle= 1.5798168237735256 radians
Complex Constant defined
TYPE: w1=arcsinh(y1)
WE GET:
-2.3+i1.5615926535897932 -> is not = z1
| Z |= 2.7800308659699467
Angle= 145.82532569791053 degrees
Angle= 2.5451320662216368 radians
Complex Constant defined
---------------------------------
about hyperbolic functions
inverse hyperbolic functions
---------------------------------
--------------
First quadrant
--------------
z1=2.3+i0.25
y1=sinh(z1) = 4.783483618904492+i1.2462283322677304
w1=arcsinh(y1) = 2.3+i0.25 = z1
y1=cosh(z1) = 4.8806256579738629+i1.2214238973804283
w1=arccosh(y1) = 2.3+i0.25 = z1
y1=tanh(z1) = 0.9824664000943306+i0.0094701778128957
w1=arctanh(y1) = 2.3000000000000002+i0.249999999999998 = z1
---------------
Second quadrant
---------------
z2=-2.3+i0.25 = -z4
y2=sinh(z2) = -4.783483618904492+i1.2462283322677304
w2=arcsinh(y2) = -2.3+i0.25 = z2
y2=cosh(z2) = 4.8806256579738629-i1.2214238973804283
w2=arccosh(y2) = 2.3-i0.25 = z4
y2=tanh(z2) = -0.9824664000943306+i0.0094701778128957
w2=arctanh(y2) = -2.3000000000000002+i0.249999999999998 = z2
--------------
Third quadrant
--------------
z3=-2.3-i0.25 = -z1
y3=sinh(z3) = -4.783483618904492-i1.2462283322677304
w3=arcsinh(y3) = -2.3-i0.25 = z3
y3=cosh(z3) = 4.8806256579738629+i1.2214238973804283
w3=arccosh(y3) = 2.3+i0.25 = z1
y3=tanh(z3) = -0.9824664000943306-i0.0094701778128957
w3=arctanh(y3) = -2.3000000000000002-i0.249999999999998 = z3
---------------
Fourth quadrant
---------------
z4=2.3-i0.25 = -z2
y4=sinh(z4) = 4.783483618904492-i1.2462283322677304
w4=arcsinh(y4) = 2.3-i0.25 = z4
y4=cosh(z4) = 4.8806256579738629-i1.2214238973804283
w4=arccosh(y4) = 2.3-i0.25 = z4
y4=tanh(z4) = 0.9824664000943306-i0.0094701778128957
w4=arctanh(y4) = 2.3000000000000002-i0.249999999999998 = z4
-------------------------------------------------------------------
about cos(w)=z and the inverse function w=arccos(z)
-------------------------------------------------------------------
By definition
cos(w)= (e^iw+e^-iw)/2 = z => w=arccos(z)
From
e^iw+e^-iw
z=-------------- doing t=e^iw
2
we get
1
2z= t + --- <=> 2z t = t^2 + 1 ( t<>0 )
t
<=> t^2 -2z t + 1 = 0
<=> t= z + SQRT(z^2-1) - positive solution I
or
t= z - SQRT(z^2-1) - negative solution I
From
SQRT(z^2-1)= SQRT[(-1).(1-z^2)]= i SQRT(1-z^2)
we have
t= z + i SQRT(1-z^2) - positive solution II
or
t= z - i SQRT(1-z^2) - negative solution II
From
t=e^iw we get w=-i ln(t)
Solutions:
w=arccos(z)= -i ln(z + SQRT(z^2-1)) - positive solution I
w=arccos(z)= -i ln(z - SQRT(z^2-1)) - negative solution I
w=arccos(z)= -i ln(z +i SQRT(1-z^2)) - positive solution II
w=arccos(z)= -i ln(z -i SQRT(1-z^2)) - negative solution II
Because cos(w)=z and cos(-w)=z
arccos(z)= w or -w
Here are some results
-----------------------------------
The calculator negative solution I
arccos(z) = -i * ln(z - sqrt(z^2-1)
-----------------------------------
z1=pi/4+i3 = 0.7853981633974483+i3 = -z3
y1=cos(z1)= 7.1189120679085491-i7.0837072942502341
w1=arccos(y1)= pi/4+i3 = z1
z2=-pi/4+i3 = -z4
y2=cos(z2)= 7.1189120679085491+i7.0837072942502341
w2=arccos(y2)= -pi/4+i3 = z2
z3=-pi/4-i3 = -z1
y3=cos(z3)= 7.1189120679085491-i7.0837072942502341
w3=arccos(y3)= pi/4+i3 = z1
z4=pi/4-i3 = -z2
y4=cos(z4)= 7.1189120679085491+i7.0837072942502341
w4=arccos(y4)= -pi/4+i3 = z2
------------------------------------
The calculator positive solution I
arccos(z) = -i * ln(z + sqrt(z^2-1))
------------------------------------
z1=pi/4+i3 = 0.7853981633974483+i3 = -z3
y1=cos(z1)= 7.1189120679085491-i7.0837072942502341
w1=arccos(y1)= -pi/4-i3 = z3
z2=-pi/4+i3 = -z4
y2=cos(z2)= 7.1189120679085491+i7.0837072942502341
w2=arccos(y2)= pi/4-i3 = z4
z3=-pi/4-i3 = -z1
y3=cos(z3)= 7.1189120679085491-i7.0837072942502341
w3=arccos(y3)= -pi/4-i3 = z3
z4=pi/4-i3 = -z2
y4=cos(z4)= 7.1189120679085491+i7.0837072942502341
w4=arccos(y4)= pi/4-i3 = z4
-------------------------------------
The calculator positive solution II
arccos(z) = -i * ln(z +i sqrt(1-z^2))
-------------------------------------
z1=pi/4+i3 = 0.7853981633974483+i3 = -z3
y1=cos(z1)= 7.1189120679085491-i7.0837072942502341
w1=arccos(y1)= pi/4+i3 = z1
z2=-pi/4+i3 = -z4
y2=cos(z2)= 7.1189120679085491+i7.0837072942502341
w2=arccos(y2)= pi/4-i3 = z4
z3=-pi/4-i3 = -z1
y3=cos(z3)= 7.1189120679085491-i7.0837072942502341
w3=arccos(y3)= pi/4+i3 = z1
z4=pi/4-i3 = -z2
y4=cos(z4)= 7.1189120679085491+i7.0837072942502341
w4=arccos(y4)= pi/4-i3 = z4
-------------------------------------
The calculator negative solution II
arccos(z) = -i * ln(z - i sqrt(1-z^2)
-------------------------------------
z1=pi/4+i3 = 0.7853981633974483+i3 = -z3
y1=cos(z1)= 7.1189120679085491-i7.0837072942502341
w1=arccos(y1)=-pi/4-i3 = z3
z2=-pi/4+i3 = -z4
y2=cos(z2)= 7.1189120679085491+i7.0837072942502341
w2=arccos(y2)= -pi/4+i3 = z2
z3=-pi/4-i3 = -z1
y3=cos(z3)= 7.1189120679085491-i7.0837072942502341
w3=arccos(y3)=-pi/4-i3 = z3
z4=pi/4-i3 = -z2
y4=cos(z4)= 7.1189120679085491+i7.0837072942502341
w4=arccos(y4)= -pi/4+i3 = z2
arccos(z) = +w or -w because cos(w)=z and cos(-w)=z
From these results,
the calculator uses the positive solution II.
w=arccos(z)= -i ln(z +i SQRT(1-z^2)) - positive solution II
Here some more results
--------------------------------------
The calculator negative solution
arccos(z) = -i * ln(z - sqrt(z^2-1))
--------------------------------------
x1=pi/4
x2=3 pi/4
z1=x1+ix1 = 0.7853981633974483+i0.7853981633974483
y1=cos(z1) = 0.9366400694314301-i0.6142431274865956
w1=arccos(y1)= 0.7853981633974482+i0.7853981633974482 = z1
z2=-x1+ix1 = -0.7853981633974483+i0.7853981633974483
y2=cos(z2) = 0.9366400694314301+i0.6142431274865956
w2=arccos(y2)= -0.7853981633974482+i0.7853981633974482 = z2
z3=-x1-ix1 = -0.7853981633974483-i0.7853981633974483
y3=cos(z3) = 0.9366400694314301-i0.6142431274865956
w3=arccos(y3)= 0.7853981633974482+i0.7853981633974482 = z1
z4=x1-ix1 = 0.7853981633974483-i0.7853981633974483
y4=cos(z4) = 0.9366400694314301+i0.6142431274865956
w4=arccos(y4)=-0.7853981633974482+i0.7853981633974482 = z2
------------------------------------------------------------
z1=x2+ix1 = 2.3561944901923449+i0.7853981633974483
y1=cos(z1) = -0.93664006943143-i0.6142431274865956
w1=arccos(y1)= -2.3561944901923449-i0.7853981633974482 = z3
z2=-x2+ix1 = -2.3561944901923449+i0.7853981633974483
y2=cos(z2) = -0.93664006943143+i0.6142431274865956
w2=arccos(y2)= 2.3561944901923449-i0.7853981633974482 = z4
z3=-x2-ix1 = -2.3561944901923449-i0.7853981633974483
y3=cos(z3) = -0.93664006943143-i0.6142431274865956
w3=arccos(y3)= -2.3561944901923449-i0.7853981633974482 = z3
z4=x2-ix1 = 2.3561944901923449-i0.7853981633974483
y4=cos(z4) = -0.93664006943143+i0.6142431274865956
w4=arccos(y4)= 2.3561944901923449-i0.7853981633974482 = z4
--------------------------------------
The calculator positive solution II
arccos(z) = -i * ln(z +i sqrt(1-z^2))
--------------------------------------
x1= pi/4 = 0.7853981633974483
x2=3 pi/4 = 2.3561944901923449
x3=5 pi/4 = 3.9269908169872415
x4=7 pi/4 = 5.4977871437821381
z1=x1+ix1 = 0.7853981633974483+i0.7853981633974483
y1=cos(z1) = 0.9366400694314301-i0.6142431274865956
w1=arccos(y1)= 0.7853981633974482+i0.7853981633974482 = z1
z2=-x1+ix1 = -0.7853981633974483+i0.7853981633974483
y2=cos(z2) = 0.9366400694314301+i0.6142431274865956
w2=arccos(y2)= 0.7853981633974482-i0.7853981633974482 = z4
z3=-x1-ix1 = -0.7853981633974483-i0.7853981633974483
y3=cos(z3) = 0.9366400694314301-i0.6142431274865956
w3=arccos(y3)= 0.7853981633974482+i0.7853981633974482 = z1
z4=x1-ix1 = 0.7853981633974483-i0.7853981633974483
y4=cos(z4) = 0.9366400694314301+i0.6142431274865956
w4=arccos(y4)= 0.7853981633974482-i0.7853981633974482 = z4
------------------------------------------------------------
z1=x2+ix1 = 2.3561944901923449+i0.7853981633974483
y1=cos(z1) = -0.93664006943143-i0.6142431274865956
w1=arccos(y1)= 2.3561944901923449+i0.7853981633974482 =z1
z2=-x2+ix1 = -2.3561944901923449+i0.7853981633974483
y2=cos(z2) = -0.93664006943143+i0.6142431274865956
w2=arccos(y2)= 2.3561944901923449-i0.7853981633974482 =z4
z3=-x2-ix1 = -2.3561944901923449-i0.7853981633974483
y3=cos(z3) = -0.93664006943143-i0.6142431274865956
w3=arccos(y3)= 2.3561944901923449+i0.7853981633974482 = z1
z4=x2-ix1 = 2.3561944901923449-i0.7853981633974483
y4=cos(z4) = -0.93664006943143+i0.6142431274865956
w4=arccos(y4)= 2.3561944901923449-i0.7853981633974482 = z4
------------------------------------------------------------
z1=x1+ix2 = 0.7853981633974483+i2.3561944901923449
y1=cos(z1) = 3.7637541395008347-i3.6967343997925613
w1=arccos(y1)= 0.7853981633974483+i2.3561944901923449 = z1
z2=-x1+ix2 = -0.7853981633974483+i2.3561944901923449
y2=cos(z2) = 3.7637541395008347+i3.6967343997925613
w2=arccos(y2)= 0.7853981633974483-i2.3561944901923449 = z4
z3=-x1-ix2 = -0.7853981633974483-i2.3561944901923449
y3=cos(z3) = 3.7637541395008347-i3.6967343997925613
w3=arccos(y3)= 0.7853981633974483+i2.3561944901923449 = z1
z4=x1-ix2 = 0.7853981633974483-i2.3561944901923449
y4=cos(z4) = 3.7637541395008347+i3.6967343997925613
w4=arccos(y4)= 0.7853981633974483-i2.3561944901923449 = z4
------------------------------------------------------------
z1=x2+ix2 = 2.3561944901923449+i2.3561944901923449
y1=cos(z1) = -3.7637541395008346-i3.6967343997925614
w1=arccos(y1)= 2.3561944901923449+i2.3561944901923449 = z1
z2=-x2+ix2 = -2.3561944901923449+i2.3561944901923449
y2=cos(z2) = -3.7637541395008346+i3.6967343997925614
w2=arccos(y2)= 2.3561944901923449-i2.3561944901923449 = z4
z3=-x2-ix2 = -2.3561944901923449-i2.3561944901923449
y3=cos(z3) = -3.7637541395008346-i3.6967343997925614
w3=arccos(y3)= 2.3561944901923449+i2.3561944901923449 = z1
z4=x2-ix2 = 2.3561944901923449-i2.3561944901923449
y4=cos(z4) = -3.7637541395008346+i3.6967343997925614
w4=arccos(y4)= 2.3561944901923449-i2.3561944901923449 = z4
-------------------------------------------------------------
x1= pi/4 = 0.7853981633974483
x2=3 pi/4 = 2.3561944901923449
x3=5 pi/4 = 3.9269908169872415
x4=7 pi/4 = 5.4977871437821381
z1=x3+ix1 = 3.9269908169872415+i0.7853981633974483
y1=cos(z1) = -0.9366400694314301+i0.6142431274865956
w1=arccos(y1)= 2.356194490192345-i0.7853981633974482 = x2-ix1
z2=-x3+ix1 = -3.9269908169872415+i0.7853981633974483
y2=cos(z2) = -0.9366400694314301-i0.6142431274865956
w2=arccos(y2)= 2.356194490192345+i0.7853981633974482 = x2+ix1
z3=-x3-ix1 = -3.9269908169872415-i0.7853981633974483
y3=cos(z3) = -0.9366400694314301+i0.6142431274865956
w3=arccos(y3)= 2.356194490192345-i0.7853981633974482 = x2-ix1
z4=x3-ix1 = 3.9269908169872415-i0.7853981633974483
y4=cos(z4) = -0.9366400694314301-i0.6142431274865956
w4=arccos(y4)= 2.356194490192345+i0.7853981633974482 = x2+ix1
------------------------------------------------------------------
z1=x3+ix2 = 3.9269908169872415+i2.3561944901923449
y1=cos(z1) = -3.7637541395008349+i3.6967343997925611
w1=arccos(y1)= 2.356194490192345-i2.3561944901923449 = x2-ix2
z2=-x3+ix2 = -3.9269908169872415+i2.3561944901923449
y2=cos(z2) = -3.7637541395008349-i3.6967343997925611
w2=arccos(y2)= 2.356194490192345+i2.3561944901923449 = x2+ix2
z3=-x3-ix2 = -3.9269908169872415-i2.3561944901923449
y3=cos(z3) = -3.7637541395008349+i3.6967343997925611
w3=arccos(y3)= 2.356194490192345-i2.3561944901923449 = x2-ix2
z4=x3-ix2 = 3.9269908169872415-i2.3561944901923449
y4=cos(z4) = -3.7637541395008349-i3.6967343997925611
w4=arccos(y4)= 2.356194490192345+i2.3561944901923449 = x2+ix2
; -------------------------------------------------------------------
x1= pi/4 = 0.7853981633974483
x2=3 pi/4 = 2.3561944901923449
x3=5 pi/4 = 3.9269908169872415
x4=7 pi/4 = 5.4977871437821381
z1=x4+ix1 = 5.4977871437821381+i0.7853981633974483 = -z3
y1=cos(z1) = 0.93664006943143+i0.6142431274865956
w1=arccos(y1)= 0.7853981633974483-i0.7853981633974482 = x1-ix1
z2=-x4+ix1 = -5.4977871437821381+i0.7853981633974483 = -z4
y2=cos(z2) = 0.93664006943143-i0.6142431274865956
w2=arccos(y2)= 0.7853981633974483+i0.7853981633974482 = x1+ix1
z3=-x4-ix1 = -5.4977871437821381-i0.7853981633974483 = -z1
y3=cos(z3) = 0.93664006943143+i0.6142431274865956
w3=arccos(y3)= 0.7853981633974483-i0.7853981633974482 = x1-ix1
z4=x4-ix1 = 5.4977871437821381-i0.7853981633974483 = -z2
y4=cos(z4) = 0.93664006943143-i0.6142431274865956
w4=arccos(y4)= 0.7853981633974483+i0.7853981633974482 = x1+ix1
----------------------------------------------------------------
z1=x4+ix2 = 5.4977871437821381+i2.3561944901923449
y1=cos(z1) = 3.7637541395008344+i3.6967343997925616
w1=arccos(y1)= 0.7853981633974483-i2.3561944901923449 =x1-ix2
z2=-x4+ix2 = -5.4977871437821381+i2.3561944901923449
y2=cos(z2) = 3.7637541395008344-i3.6967343997925616
w2=arccos(y2)= 0.7853981633974483+i2.3561944901923449 =x1+ix2
z3=-x4-ix2 = -5.4977871437821381-i2.3561944901923449
y3=cos(z3) = 3.7637541395008344+i3.6967343997925616
w3=arccos(y3)= 0.7853981633974483-i2.3561944901923449 =x1-ix2
z4=x4-ix2 = 5.4977871437821381-i2.3561944901923449
y4=cos(z4) = 3.7637541395008344-i3.6967343997925616
w4=arccos(y4)= 0.7853981633974483+i2.3561944901923449 =x1+ix2
------------------------------
z1=x4+ix3 = 5.4977871437821381+i3.9269908169872415
y1=cos(z1) = 17.951221702159904+i17.937289667062213
w1=arccos(y1)= 0.7853981633974484-i3.9269908169872415 =x1-ix3
z2=-x4+ix3 = -5.4977871437821381+i3.9269908169872415
y2=cos(z2) = 17.951221702159904-i17.937289667062213
w2=arccos(y2)= 0.7853981633974484+i3.9269908169872415 =x1+ix3
z3=-x4-ix3 = -5.4977871437821381-i3.9269908169872415
y3=cos(z3) = 17.951221702159904+i17.937289667062213
w3=arccos(y3)= 0.7853981633974484-i3.9269908169872415 =x1-ix3
z4=x4-ix3 = 5.4977871437821381-i3.9269908169872415
y4=cos(z4) = 17.951221702159904-i17.937289667062213
w4=arccos(y4)= 0.7853981633974484+i3.9269908169872415 =x1+ix3
----------------------------------------------------------------
x1= pi/4 = 0.7853981633974483
x2=3 pi/4 = 2.3561944901923449
x3=5 pi/4 = 3.9269908169872415
x4=7 pi/4 = 5.4977871437821381
z1=x4+ix4 = 5.4977871437821381+i5.4977871437821381
y1=cos(z1) = 86.321884181857327+i86.318987996303526
w1=arccos(y1)= 0.7853981633974483-i5.4977871437821381 =x1-ix4
z2=-x4+ix4 = -5.4977871437821381+i5.4977871437821381
y2=cos(z2) = 86.321884181857327-i86.318987996303526
w2=arccos(y2)= 0.7853981633974483+i5.4977871437821383 =x1+ix4
z3=-x4-ix4 = -5.4977871437821381-i5.4977871437821381
y3=cos(z3) = 86.321884181857327+i86.318987996303526
w3=arccos(y3)= 0.7853981633974483-i5.4977871437821381 =x1-ix4
z4=x4-ix4 = 5.4977871437821381-i5.4977871437821381
y4=cos(z4) = 86.321884181857327-i86.318987996303526
w4=arccos(y4)= 0.7853981633974483+i5.4977871437821383 =x1+ix4
Hi Rui,
only one question: What algorithm do you use by solving a linear equation system?
Gunther
Quote from: Gunther on November 10, 2013, 05:38:29 AM
Hi Rui,
only one question: What algorithm do you use by solving a linear equation system?
Gunther
Hi Gunther
Where is the linear equation system ?
Do you want to say, for example:
a X+ bY=c and dX+eY=f ? etc.
I use matrices.
Hi Rui,
the standard form for a linear equation is called the "slope-intercept" form:
Y=mX+b
m is the slope (=rise/run=(Y2-Y1)/(X2-X1))
b is the y-intercept (the value of Y when X=0)
many calculators let you enter 2 (X,Y) points (some allow m and b)
then, you can enter an X or Y value, and it will spit out the opposite
nearly any straight line can be described using the slope-intercept form
however, it gets a little tricky as the slope approaches infinity :P
those are lines that graph stright up and down, like the line: X=0
Quote from: dedndave on November 10, 2013, 06:20:54 AM
Hi Rui,
the standard form for a linear equation is called the "slope-intercept" form:
Y=mX+b
m is the slope (=rise/run=(Y2-Y1)/(X2-X1))
b is the y-intercept (the value of Y when X=0)
many calculators let you enter 2 (X,Y) points (some allow m and b)
then, you can enter an X or Y value, and it will spit out the opposite
nearly any straight line can be described using the slope-intercept form
however, it gets a little tricky as the slope approaches infinity :P
those are lines that graph stright up and down, like the line: X=0
Hi Dave, :t
i know all about that
Gunther asked for
linear equation system
Rui,
Quote from: RuiLoureiro on November 10, 2013, 05:46:50 AM
Do you want to say, for example:
a X+ bY=c and dX+eY=f ? etc.
I use matrices.
that's clear. The direction of my question was: Do you use the Gaussian elimination (https://en.wikipedia.org/wiki/Gaussian_elimination)?
Gunther
Gunther,
yes, the calculator uses that method.
sorry Rui :t
Rui,
Quote from: RuiLoureiro on November 10, 2013, 08:15:50 AM
yes, the calculator uses that method.
okay. Did you test some ill conditioned equation systems?
Gunther
Quote from: Gunther on November 10, 2013, 09:15:37 PM
Rui,
Quote from: RuiLoureiro on November 10, 2013, 08:15:50 AM
yes, the calculator uses that method.
okay. Did you test some ill conditioned equation systems?
Gunther
Hi Gunther,
use the calculator and see yourself the answers :t
Rui,
Quote from: RuiLoureiro on November 10, 2013, 11:01:57 PM
use the calculator and see yourself the answers :t
okay, I'll give you a report.
Gunther
Quote from: Gunther on November 10, 2013, 11:06:33 PM
Rui,
Quote from: RuiLoureiro on November 10, 2013, 11:01:57 PM
use the calculator and see yourself the answers :t
okay, I'll give you a report.
Gunther
Gunther,
ok, thank you ! :t
Quote from: dedndave on November 10, 2013, 11:45:47 AM
sorry Rui :t
i like you give me answers,
no problems with you, Dave :t :t
Hi all,
In Reply #39 on: November 10, 2013, 05:34:30 AM
i showed how to get the function arccos(z).
Now i show all other because i rarely use formulas
without knowing from where they come from.
The calculator uses them.
---------------------------------------------------------------------------------
about
trigonometric sin(w)=z and the inverse function w=arcsin(z)
trigonometric tan(w)=z and the inverse function w=arctan(z)
hyperbolic sinh(w)=z and the inverse function w=arcsinh(z)
hyperbolic cosh(w)=z and the inverse function w=arccosh(z)
hyperbolic tanh(w)=z and the inverse function w=arctanh(z)
---------------------------------------------------------------------------------
-----------------------------
Trigonometric functions
-----------------------------
By definition
sin(w)= (e^iw-e^-iw)/2i = z => w=arcsin(z)
From
e^iw-e^-iw
z=---------- doing t=e^iw
2i
we get
1
2iz= t - --- <=> 2iz t = t^2 - 1 ( t<>0 )
t
<=> t^2 -2iz t - 1 = 0
<=> t= iz + SQRT(1-z^2) - positive solution I
or
t= iz - SQRT(1-z^2) - negative solution I
From
SQRT(1-z^2)= SQRT[(-1).(z^2-1)]= i SQRT(z^2-1)
we have
t= iz + i SQRT(z^2-1) - positive solution II
or
t= iz - i SQRT(z^2-1) - negative solution II
From
t=e^iw we get w=-i ln(t)
Solutions:
w=arcsin(z)= -i ln(iz + SQRT(1-z^2)) - positive solution I
w=arcsin(z)= -i ln(iz - SQRT(1-z^2)) - negative solution I
w=arcsin(z)= -i ln(iz +i SQRT(z^2-1)) - positive solution II
w=arcsin(z)= -i ln(iz -i SQRT(z^2-1)) - negative solution II
note1: sin(w)=z and sin(-w)=-z
note2: the calculator uses positive solution I
----------------------------------------------------------------------------------
By definition
sin(w) e^iw - e^-iw e^2iw - 1
tan(w)=------- = --------------- = --------------= z
cos(w) i(e^iw + e^-iw) i(e^2iw + 1)
w= arctan(z)
Doing
(1) t=e^2iw we have
2iw= ln(t) => w = -i/2 ln(t) = i/2 ln(t^-1)
t - 1
(2) iz = ------- <=> iz (t+1) = t - 1 ( t<>-1 )
t + 1
izt + iz -t = -1 <=> (iz-1)t = - (iz + 1)
<=> (1-iz) t = (1 + iz)
(1 + iz) i (1 + iz) i - z
t= -----------= -----------=---------
(1 - iz) i (1 - iz) i + z
t^-1 = (i+z)/ (i-z)
=> w = arctan(z) = i/2 ln( (i+z)/(i-z) )
------------------------------------------------------------------------
-------------------------
Hyperbolic functions
-------------------------
By definition
sinh(w)= (e^w-e^-w)/2 = z => w=arcsinh(z)
From
e^w-e^-w
z=---------- doing t=e^w
2
we get
1
2z= t - --- <=> 2z t = t^2 - 1 ( t<>0 )
t
<=> t^2 -2z t - 1 = 0
<=> t= z + SQRT(z^2+1) - positive solution I
or
t= z - SQRT(z^2+1) - negative solution I
From
t=e^w we get w=ln(t)
Solutions:
w=arcsinh(z)= ln(z + SQRT(z^2+1)) - positive solution I
w=arcsinh(z)= ln(z - SQRT(z^2+1)) - negative solution I
note1: sinh(w)=z and sinh(-w)=-z
note2: the calculator uses positive solution I
-------------------------------------------------------------------------
By definition
cosh(w)= (e^w+e^-w)/2 = z => w=arccosh(z)
From
e^w+e^-w
z=---------- doing t=e^w
2
we get
1
2z= t + --- <=> 2z t = t^2 + 1 ( t<>0 )
t
<=> t^2 -2z t + 1 = 0
<=> t= z + SQRT(z^2-1) - positive solution I
or
t= z - SQRT(z^2-1) - negative solution I
From
SQRT(z^2-1)= SQRT[(-1).(1-z^2)]= i SQRT(1-z^2)
we have
t= z + i SQRT(1-z^2) - positive solution II
or
t= z - i SQRT(1-z^2) - negative solution II
From
t=e^w we get w=ln(t)
Solutions:
w=arccosh(z)= ln(z + SQRT(z^2-1)) - positive solution I
w=arccosh(z)= ln(z - SQRT(z^2-1)) - negative solution I
w=arccosh(z)= ln(z +i SQRT(1-z^2)) - positive solution II
w=arccosh(z)= ln(z -i SQRT(1-z^2)) - negative solution II
note1: Because cosh(w)=z and cosh(-w)=z
arccosh(z)= w or -w
note2: The calculator uses positive solution I
-----------------------------------------------------------------------------
By definition
sinh(w) e^w - e^-w e^2w - 1
tanh(w)=------- = --------------- = ---------- = z
cosh(w) (e^w + e^-w) (e^2w + 1)
w= arctanh(z)
Doing
(1) t=e^2w we have
2w= ln(t) => w = 1/2 ln(t)
t - 1
(2) z = ------- <=> z (t+1) = t - 1 ( t<>-1 )
t + 1
zt + z -t = -1 <=> (z-1)t = - (z + 1)
<=> (1 - z) t = (1 + z)
t = (1+z)/ (1-z)
=> w = arctanh(z) = 1/2 ln( (1+z)/(1-z) )
Quote from: Gunther on November 10, 2013, 11:06:33 PM
Rui,
Quote from: RuiLoureiro on November 10, 2013, 11:01:57 PM
use the calculator and see yourself the answers :t
okay, I'll give you a report.
Gunther
Hi Gunther,
I think you are talking about some cases like
0x+0y+0z=1; x+y+z=2;x+y+z=3;
or
0x+0y+0z=0; x+y+z=2;x+y+z=3;
or
0x+0y+0z=0; x+y+z=3;x+y+z=3;
or
x+y+z=1; x+y+z=2;x+y+z=3;
The calculator doesnt give a clear message because
i didnt test that cases.
I will do in the next version.
Thanks Gunther :t
Hi,
In the NEXT version v3.10.2 (calcula60.exe)
that i will post soon,
When the calculator gives the message
«The System of linear equations has no solution»
it means that «there isn't any solution»
When the calculator gives the message
«The System of linear equations has no unique solution»
it means that «there is infinite solutions»
The problem is completely solved now.
Gunther,
I don't know if you have any other question.
If you have, please give me the report.
Thanks :t
Rui
Here are some results
----------------------------
x+y+z=1; x+y+z=2; x+y+z=3;
The System of linear equations has no solution
Determinant: 0
x+y+z=1; x+y+z=3; x+y+z=2;
The System of linear equations has no solution
Determinant: 0
x+y+z=3; x+y+z=2; x+y+z=1;
The System of linear equations has no solution
Determinant: 0
0x+0y+0z=1; x+y+z=2; x+y+z=3;
The System of linear equations has no solution
Determinant: 0
0x+0y+0z=0; x+y+z=2; x+y+z=3;
The System of linear equations has no solution
Determinant: 0
0x+0y+0z=0; x+y+z=3; x+y+z=3;
The System of linear equations has no unique solution
Determinant: 0
0x+0y+0z=0; 0x+0y+0z=0; x+y+z=3;
The System of linear equations has no unique solution
Determinant: 0
0x+0y+0z=0; 0x+0y+0z=0; 0x+0y+0z=3;
The System of linear equations has no solution
Determinant: 0
0x+0y+0z=0; 0x+0y+0z=0; 0x+0y+0z=0;
The System of linear equations has no unique solution
Determinant: 0
x+y=1; x+y+z=2; y+x=3;
The System of linear equations has no solution
Determinant: 0
x+y=1; x+y+z=2; y+x=1;
The System of linear equations has no unique solution
Determinant: 0
; ----------------------------------------------------
x+y+z+t=1; x+y+z+t=2; x+y+z+t=1; x+y+z+t=2;
The System of linear equations has no solution
Determinant: 0
x+y+z+t=1; x+y+z+t=2; x+y+z+t=3; x+y+z+t=4;
The System of linear equations has no solution
Determinant: 0
x+y+z+t=4; x+y+z+t=4; x+y+z+t=4; x+y+z+t=4;
The System of linear equations has no unique solution
Determinant: 0
Rui,
sorry for the delay, but I'm sick and I'm confined to bed.
My question has that background:
Given is the following linear equation system A:
x - 2y = 1
x + y = 4
The solution is: x = 3 and y = 1
If we change the coefficients a bit, we've the system A':
x - 2y = 1
x + 1.2y = 4
The solution is now: x = 2.875 and y = 0.9375
But the relationship between the two solutions is still recognizable.
Lets have a look at the following system B:
x + 2y = 3
2x + 4.1y = 4
The solution is: x = 43 and y = -20
Again, we're changing only 1 coefficient by 0.2. That leads to system B':
x + 2y = 3
2x + 3.9y = 4
The solution is now: x =-37 and y = 20
There's no relationship between the two solutions; even the sign has been reversed. That's a simple example of an ill conditioned linear equation system.
Gunther
Hi Gunther,
Quote
sorry for the delay, but I'm sick and I'm confined to bed.
Oh, i hope you get better soon as possible :t
Quote
Lets have a look at the following system B:
x + 2y = 3
2x + 4.1y = 4
The solution is: x = 43 and y = -20
Again, we're changing only 1 coefficient by 0.2. That leads to system B':
x + 2y = 3
2x + 3.9y = 4
The solution is now: x =-37 and y = 20
(1) Ok, i think i understood the question
(2) If we multiply both systems by 10 we have this
Quote
10 x + 20 y= 30
20 x + 41 y= 40 solution B => vector (x,y)=(43,-20)
Quote
10 x + 20 y= 30
20 x + 39 y= 40 solution B' => vector (x,y)=(-37,20)
It seems that there are something "wrong" when
we change from 41 to 39
But from 41 to 39 we have "40"
Quote
this:
10 x + 20 y= 30
20 x + 40 y= 40
is equivalent to
20 x + 40 y = 60
20 x + 40 y = 40 solution C => has no solution
It behaves like a function that passes through
a discontinuity point: (its only an image)
Quote
10 x + 20 y= 30
20 x + 39 y= 40 solution B' => vector (x,y)=(-37,20)
10 x + 20 y= 30
20 x + 40 y= 40 solution C => has no solution
10 x + 20 y= 30
20 x + 41 y= 40 solution B => vector (x,y)=(43,-20)
Many years ago, i wrote hundreds of sheets of
exercises about this type of problems as examples
to my son when he was at the university studying
mathematics.
For instance: study the system
x+y+z=3
x-y+z=1
2x-2y+az=2
In this case, we have:
1. If a =2 the system has no unique solution
2. If a<>2 the system has a unique solution
I took this example from a book that i have here
about linear algebra and not from what i did because
i have no room to all my library in my house.
I think this is what you call
an ill conditioned linear equation system:
a linear equation system that depends on one or more
parameters.
I hope you recover soon as possible.
By the way, i have had some headaches
these days. Not too fine.
hope you're feeling better, Gunther :t
for those simple equations, i usually solve them by substitution :P
but, i'm sure Gunther is just using it as an example of a possibly more complex system
:biggrin:
Quote
those simple equations, i usually solve them by substitution :P
it is when we are on the first class, Dave :greensml:
I am playing with you Dave :t
playing ? kidding.
i sometimes "add" equations, which is what you are essentially doing with the matricies
Quote from: dedndave on November 13, 2013, 06:23:03 AM
i sometimes "add" equations, which is what you are essentially doing with the matricies
or multiply by c and add.
But the calculator cannot solve 2 systems
only 1 system at a time. All parameters in one system
are real constants not variables.
Hi Rui,
Quote from: RuiLoureiro on November 13, 2013, 06:08:44 AM
I think this is what you call
an ill conditioned linear equation system:
a linear equation system that depends on one or more
parameters.
no, not really. If you would draw both equations of system A into a coordinate system, you would see a clear intersection of both lines. That's our solution. The situation is total different with system B. You would see that both lines are smeared. That's the bad condition. Both equations are linear independent, but both are almost parallel.
Gunther
Gunther,
I don't know what you want the calculator should do.
When we type the following system B:
(a) x + 2y = 3
(b) 2x + 4.1y = 4
the calculator should give a pair of values for which
(a) and (b) are true. And this is exactly what i want.
The calculator doesn't study each coefficient or any relationship
between each equation. The calculator will never do this.
If we want to study the relationship between (a) and (b)
its another question you should solve.
The calculator doesn't do any geometric interpretation also
and will never do it.
If the angle between (a) and (b) are too small or too too small or not, the calculator does not compute it. It computes only the solution of the system.
So, the calculator will give only the solution: x = 43 and y = -20.
and the answer to your question
Quote
Did you test some ill conditioned equation systems?
seems to be this: no and never (if i understood whats behind it).
Rui,
Quote from: RuiLoureiro on November 13, 2013, 07:14:45 AM
So, the calculator will give only the solution: x = 43 and y = -20.
no offense, my posts wasn't meant as criticism. The Calculator brings the right result and that shows that it is numerically stable. :t The point with the coordinate system was only included to illustrate what's behind the term "ill conditioned". It's clear that the Calculator won't draw the system; that wouldn't be necessary and is in the most cases impossible. I apologize for the misunderstanding.
Gunther
Gunther,
«no offense, my posts wasn't meant as criticism.»
I didnt get it in that way. No problems :t
About the calculator, i wrote what i wrote
and i know what i want to do or if i want to
do something more ! In this particular case
of solving linear equation systems of 2,3,4
unknowns i don't want to know if the angle
between (a) and (b), (b) and (c) etc.
are so small or so so small or not. To me
it is irrelevant when the calculator
wants to solve linear equation systems of 2,3,4
unknowns. Only this.
Gunther,
I was working about complex functions
to be used in my "the calculator",
when you asked me for something that has
something to do with
"an ill conditioned linear equation system".
I used the calculator to see if there was
any problem when we try to solve a system.
And there was and i corrected the problem.
Meanwhile, my problem was to know
what you call a "an ill conditioned...",
thing that i never heard. But it is also
true that i don't follow this kind of
things there are many years. It could be
seen as a concept, but i don't give
importance to be included in the calculator.
Now, it seems we know what is the question
and, for me, it is solved. You give importance
and i think that it is out of context for
what i want the calculator should do.
I want to say that it seems that you have
a good background about this things...
The way you set the question, the way you
wrote about it, ... Nothing more to say.
I hope you give the best interpretation
to my replies.
I hope you recover soon as possible. :t
Hi Rui,
Quote from: RuiLoureiro on November 14, 2013, 04:02:27 AM
I hope you recover soon as possible. :t
thank you for the good wishes. The cold is very persistent.
Quote from: RuiLoureiro on November 14, 2013, 04:02:27 AM
I hope you give the best interpretation
to my replies.
Yes of course. I think you've made a good job with the Calculator. :t
Gunther
Hi Gunther,
take the correct tablets and wait ! :t
Dave,
What do you call "a possibly more complex system" ?
Don't forget that we are talking about "linear equation systems".
And to be solved by a computer procedure.
It seems that there are "linear equation systems" and
"linear complex equation systems"
or "complex linear equation systems"
or "linear equation complex systems" (don't know)
Where did you see it, Dave? :bgrin:
you can always make a line formula more complex
for example, one of the constants may be the sin(angle) - something like that
at a specific angle, the function is linear, but changes as the angle changes
you could also introduce time into the equation
at any given moment, the function is linear - and may be evaluated as such
but, the constants change with time
the reason i bring it up is....
let's say i evaluate a linear equation in a loop
at each pass - constants
but - the constants are different for different passes
the evaluation process should be able to handle any set of input values
Dave,
Rui's question is justified.
Quote from: dedndave on November 14, 2013, 05:34:00 AM
you can always make a line formula more complex
for example, one of the constants may be the sin(angle) - something like that
No, that wouldn't be a linear equation system, because the sine is bent. You're talking about a non-linear equation system. That's not so easy to solve and most of these systems can only be solved approximately. That's another point.
Gunther
Quote from: dedndave on November 14, 2013, 05:34:00 AM
you can always make a line formula more complex
for example, one of the constants may be the sin(angle) - something like that
at a specific angle, the function is linear, but changes as the angle changes
you could also introduce time into the equation
at any given moment, the function is linear - and may be evaluated as such
but, the constants change with time
Ok Dave,
now, i understood the idea.
Yes, if the constants are real
the system is a linear equation system.
The system is non-linear in x if we use sin(x) instead of x
or x^2, e^x, ...
For instance: 2 sin(3t) x+ y=5
2 x- 3y=10 for t={...}
--------------------------------------------------------------------
For each t, we get a real number A=2*sin(3t),
we have a linear equation system,
we may solve it, we may have or not a solution,
etc. etc.
---------------------------------------------------------------------
But this: 2 sin(3x)+ y=5
2 x- 3y=10
is a non-linear equation system.
Gunther....
sin(45) is a constant :P
Dave,
Quote from: dedndave on November 14, 2013, 07:44:55 AM
Gunther....
sin(45) is a constant :P
yes, that's true. But the value is sqrt(2)/2 which is irrational. :greensml:
Gunther
constants can be irrational - take my wife, for example :lol:
Dave,
Quote from: dedndave on November 14, 2013, 11:46:36 AM
constants can be irrational - take my wife, for example :lol:
I hope she didn't read your posts, otherwise ... :lol:
Gunther
Quote from: Gunther on November 14, 2013, 06:13:44 AM
No, that wouldn't be a linear equation system, because the sine is bent. You're talking about a non-linear equation system. That's not so easy to solve and most of these systems can only be solved approximately. That's another point.
Gunther
Gunther,
If the constants are real (like, 1, 1.2, pi, sqrt(5),...)
the system is a linear equation system.
The system is non-linear in x if we use sin(x) instead of x
or x^2, e^x, ...
For instance: 2 sin(3t) x+ y=5
2 x - 3y=10 for instance: t={1,2,...}
--------------------------------------------------------------------
For each t, we get a real number A=2*sin(3t),
we have a linear equation system,
we may solve it, we may have or not a solution,
etc. etc.
---------------------------------------------------------------------
But this: 2 sin(3x)+ y =5
2 x - 3y=10
is a non-linear equation system
and what you said is correct.
Quote from: dedndave on November 14, 2013, 11:46:36 AM
constants can be irrational - take my wife, for example :lol:
(...) :greensml:
:biggrin:
Gunther,
you tried to show us an example of
a linear equation
system A that you say "it has one property"
and another linear equation
system B that you say "it has another property".
Now, i show that they have the same:
the question is the starting point.
This is what you wrote
-----------------------------
Quote
My question has that background:
Given is the following linear equation system A:
x - 2y = 1
x + 1y = 4
The solution is: x = 3 and y = 1
If we change the coefficients a bit, we've the system A':
x - 2y = 1
x + 1.2y = 4
The solution is now: x = 2.875 and y = 0.9375
But the relationship between the two solutions is still recognizable.
Well,
Do this:
(1) replace +1y by +ay where "a" is a real number
we get:
x - 2y = 1
x + ay = 4
(2) study this system
(if you know how to do that.
(
explanations: $150 per hour. very cheap!).
we get:
if a = -2 the system has no solution
if a<>-2 the system has a unique solution
Using "The calculator", we get:
Quote
x-2y=1; x- 2y=4; The System of linear equations has no solution
x-2y=1; x-1.9y=4; X= 61.0 Y= 30.0 Determinant: 0.1
x-2y=1; x-2.1y=4; X= -59.0 Y= -30.0 Determinant: -0.1
Well, if we want to be far from a=-2, we may choose a=1 or a=200
x-2y=1; x+200y=4; X=1.0297029702970297 Y=0.01485148514851485
x-2y=1; x+201y=4; X=1.0295566502463054 Y=0.0147783251231527
...
Following what you call "an ill conditioned linear equation system",
i would say: that's a simple example of "an ill conditioned linear equation system",and this is your linear equation system A.
Well, you may say: the starting point is the first equation system.
And my question is: why to change "1" and no "-2" or "4" or other ?
-----------------------------------------------------------------------------------
This is what you wrote
----------------------------
Quote
Lets have a look at the following system B:
x + 2y = 3
2x + 4.1y = 4
Doing the same
we get:
x + 2y = 3
2x + ay = 4
studying the system:
if a = 4 the system has no solution
if a<>4 the system has a unique solution
Using "The calculator", we get:
Quote
x+2y=3; 2x+ 4y=4; The System of linear equations has no solution
x+2y=3; 2x+3.9y=4; X= -37.0 Y= 20.0 Determinant: -0.1
x+2y=3; 2x+4.1y=4; X= 43.0 Y= -20.0 Determinant: 0.1
---------------------------------------------------------------------------
my conclusion is this:
The calculator should solve the linear equation system with
real constant coefficients;
We may study the system, if and when we want;
You can call the name you want, for me, no problems.
---------------------------------------------------------------------------
I want to remember that many of us are here
doing this things to do something useful or
to pass the time, no to get more problems
for our lives
I will post the NEXT version v3.10.2 (calcula60.exe)
soon as possible
Hi Rui,
Quote from: RuiLoureiro on November 14, 2013, 10:57:34 PM
But this: 2 sin(3x)+ y =5
2 x - 3y=10
is a non-linear equation system
and what you said is correct.
no doubt about that.
Quote from: RuiLoureiro on November 15, 2013, 02:50:26 AM
I want to remember that many of us are here
doing this things to do something useful or
to pass the time, no to get more problems
for our lives
That's clear. It wasn't my matter of concern to make you problems.
Quote from: RuiLoureiro on November 15, 2013, 02:50:26 AM
Now, i show that they have the same:
That's not my point of view. I want you cause no further problems; therefore I won't answer in detail. Only that: The condition of a linear equation system is a measure for the behavior of the system. If we call M the matrix of the coefficients, it's defined as:
Quote
cond (M) = ||M|| ||M-1||
or formulated in a sentence: The condition of a linear equation system is the norm of the matrix M multiplied by the norm of the inverse matrix of M. The system A has a much better condition than system B. That's the difference.
But again, I don't want that you've further problems in your spare time. You may ignore this point. Anyway, your Calculator is a solid work. Thank you for that.
Gunther
Hi Gunther,
Quote
That's clear. It wasn't my matter of concern to make you problems.
I wanted to say the same to you
Quote
That's not my point of view.
...
The system A has a much better condition than system B.
That's the difference.
It's clear: now, i understand your point of view.
Well, nothing to say: i showed my point of view.
Thank you for that, also. :icon14:
RuiLoureiro
Quote
Anyway, your Calculator is a solid work.
:t
Hi all,
-----------------------------------------------------------------------
LASTEST VERSION
-----------------------------------------------------------------------
****-- The (powerful) Calculator v3.10.2 --****
(calcula60.exe)
This is the lastest version and this is the powerful calculator
v3.10.2
---------------------------------------------------------------
COMPLEX NUMBERS and COMPLEX FUNCTIONS
---------------------------------------------------------------
Now, we may solve any complex expression in the same way we do for
real expressions.
We may use any function:
z= a + i b (a,b are reals, i^2=-1)
= |z| e^(i.angleR)
conj(a+ib) = a-ib
inv(z) = 1/z
abs(a+ib) = sqr(a^2+b^2)+i0
sqr(a+ib) = (a+ib)^(1/2)
rnd(a+ib) = rnd(a)+i rnd(b)
rndi(a+ib) = rndi(a)+i rndi(b)
e^(a+ib) = e^a * e^ib = e^a * ( cos(b)+i sin(b))
(a+ib)^(c+id) = e^( (c+id)*ln(a+ib) )
ln(a+ib) = ln (|a+ib|* e^i angleR )
log(a+ib) = (ln (|a+ib|* e^i angleR )) / ln(10)
sin(z) = (e^iz - e^-iz)/2i
cos(z) = (e^iz + e^-iz)/2
tan(z) = (e^2iz - 1) / i (e^2iz + 1)
sec(z) = 1 / cos(z)
csc(z) = 1 / sin(z)
cot(z) = 1 / tan(z)
arcsin(z) = -i * ln(i*z + sqrt(1-z^2))
arccos(z) = -i * ln(z +i sqrt(1-z^2))
arctan(z) = i/2 * ln((i+z) / (i-z))
arcsec(z) = arccos(1/z)
arccsc(z) = arcsin(1/z)
arccot(z) = arctan(1/z)
sind(z)
cosd(z)
tand(z)
secd(z)
cscd(z)
cotd(z)
arcsind(z)
arccosd(z)
arctand(z)
arcsecd(z)
arccscd(z)
arccotd(z)
sinh(z) = (e^z - e^-z) / 2
cosh(z) = (e^z + e^-z) / 2
tanh(z) = (e^2z - 1) / (e^2z + 1)
sech(z) = 1 / cosh(z)
csch(z) = 1 / sinh(z)
coth(z) = 1 / tanh(z)
arcsinh(z) = ln(z + sqrt(z^2+1))
arccosh(z) = ln(z + sqrt(z^2-1))
arctanh(z) = 1/2 * ln((1+z) / (1-z))
arcsech(z) = arccosh(1/z)
arccsch(z) = arcsinh(1/z)
arccoth(z) = arctanh(1/z)
round(a+ib, n) - round 'a' and 'b' to n decimal places
The operation rules are the same for real numbers
The result may be a complex number, INFINITY or indeterminate form.
We may use the division by 0 to generate th infinity.
We need to use brackets when we have powers of powers
(1-i2)^(1-i)^(i2) gives "Complex power too complex- use brackets"
We need to do ((1-i2)^(1-i))^(i2) or (1-i2)^((1-i)^(i2)).
Any complex number is a number with a real part 'a'
and an imaginary part 'b'. For instance, z1=-2+i3 or z1=i3-2.
We may define it typing z1=(-2,3); or z1=-2+i3 for example.
To build a complex expression
we may define a set of real/complex constants z1, z2, z3, z4, ...
and then we write the expression
For instance,
(1-i2)*(2-i3) or (a-ib)*(c-id) where a=12;b=-4;c=1;d=-3;
are real constants
z1+z2*z3-z4 or z5=z1+z2*z3-z4 or z1=z1+z2*(z3/z4)+3-i2
If the expression uses '^', we may use brackets or constants
previously defined. See the following example:
z1=(e^i*(2-i3)+ e^-i*(2-i3) ) /2 = 1.0806046117362794-i1.6209069176044192
is not equal to
z2=(e^(i*(2-i3))+ e^(-i*(2-i3)) ) /2 = -4.1896256909688072+i9.1092278937553366
cos(2-i3)=-4.1896256909688072+i9.1092278937553366
but is equal to
x1=i*(2-i3)
x2=-x1
z3=(e^x1+e^x2)/2 = -4.1896256909688072+i9.1092278937553366
After defining any constant/matrix we may type the constant name
to see its value. We may use also list c or list r or list l or
list.
To delete all defined variables, type: del a (delete all)
If you know any bug or something else, please post it
If you have any suggestion, please send me a personal message or ...
I hope i have not did many mistakes!
What will coming next ?
Maybe the function norm(a), where a=matrix
and something about complex numbers.
---------------------------------------------------------
Nov. 2013
Good luck !
Rui Loureiro
note: RulesV3_10_2I - rules in English
RulesV3_10_2P - regras em Português
Rui,
impressive. You are a hard working man. :t Thank you for providing.
Gunther
Quote from: Gunther on November 20, 2013, 06:22:03 AM
Rui,
impressive. You are a hard working man. :t Thank you for providing.
Gunther
Thank you, Gunther :t
Hi all,
-----------------------------------------------------------------------
LASTEST VERSION-----------------------------------------------------------------------
****--
The (powerful) Calculator v3.11.1 --****
(
calcula61.exe)
This is the lastest version and this is the powerful calculator
v3.11.1
---------------------------------------------------------------------
LINEAR EQUATION SYSTEMS WITH COMPLEX VARIABLES
NORM of a MATRIX
---------------------------------------------------------------------
Now, we may solve linear equation systems with complex variables,
and we may compute the norm of a matrix.
++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
Systems of 2 complex variables x,y++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
-------------
example 1-------------
type the following lines (for each one, press COMPUTE)
or copy and paste:
z1=1-i2
z2=2+i3
x1=3;x2=5;x3=-7;
x1 X+ x2 Y=z1; x2 X- x3 Y=z2;
X= 0.75+i7.25
Y= -0.25-i4.75
Determinant: -4.0+i0
----------------------------------------
PROOF: type the following lines ( press COMPUTE):
X1= 0.75+i7.25
Y1= -0.25-i4.75
w1= x1*X1+ x2*Y1
= 1.0-i2.0 = z1
w2= x2*X1- x3*Y1
= 2.0+i3.0 = z2
---------------------------------------
------------
example 2------------
type the following lines ( press COMPUTE):
z1 X+ x2 Y=x1; x2 X- z2 Y=x2;
X= 0.9302752293577981+i0.3009174311926605
Y= 0.2935779816513761+i0.3119266055045871
Determinant: -33.0+i1.0
++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
Systems of 3 complex variables x,y,z++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
------------
example 3------------
type the following lines (for each one, press COMPUTE):
z1=1-i2
z2=2+i3
x1=3;x2=5;x3=-7;
x+y-x3 z=z1; x2 x-iy+2z=20; (x1+ix2)x+2y-z=8;
result:
X= 5.4763271162123386-i2.6248206599713056
Y= -10.3565279770444763-i9.0616929698708752
Z= 0.8400286944045911+i1.3837876614060258
Determinant: 42.0+i32.0
-------------------------------------------------
------------
example 4------------
type the following line ( press COMPUTE):
x+y-x3 z=z1; x2 x-iy+2z=20; z2 x+2y-z=x1+ix2;
result:
X= 4.5773011617515639-i1.0393208221626452
Y= -4.5656836461126005-i3.1689008042895442
Z= 0.1411974977658623+i0.3154602323503127
Determinant: 54.0+i21.0
PROOF: type the following lines ( press COMPUTE):
X1= 4.5773011617515639-i1.0393208221626452
Y1= -4.5656836461126005-i3.1689008042895442
Z1= 0.1411974977658623+i0.3154602323503127
w1= X1+ Y1-x3* Z1
=0.9999999999999995-i2.0000000000000005 = z1=1-i2
w2=x2*X1-i* Y1+2* Z1
=20.0-i9.9746599868666408e-17 = 20
w3=z2*X1+2* Y1- Z1
=3.0000000000000001+i5.0000000000000002 = x1+ix2 = 3+i5
++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
Systems of 4 complex variables x,y,z,t++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
--------------------------------
example 5: real variables--------------------------------
type the following line ( press COMPUTE):
x+y+z+t=10; -x-y+2z+2t=20; 3x+2y-z-3t=8; -2x+y+z-5t=18;
X= 7.6666666666666667
Y= -7.6666666666666667
Z= 15.1666666666666667
T= -5.1666666666666667
Determinant: -36.0
type the following line ( press COMPUTE):
matA=[1,1,1,1; -1,-1,2,2; 3,2,-1,-3; -2,1,1,-5];
det(matA)=-36.0
The triangular matrix is correct
--------------------------------------------------------
------------
example 6- same as example 5------------
type the following line ( press COMPUTE):
x+y+z+t=10; -x-y+2z+2t=20; 3x+2y-z-3t=8; -2x+y+z-5t=18
+i0;
X= 7.6666666666666667+i0
Y= -7.6666666666666667+i0
Z= 15.1666666666666667+i0
T= -5.1666666666666667+i0
Determinant: -36.0+i0
----------------------------------------------------------
------------
example 7------------
type the following line ( press COMPUTE):
x+iy+z+t=10; -x-y+2z+2t=20; 3x+2y-z-3t=8; -2x+y+z-5t=18+i0;
X= 5.1111111111111111-i2.5555555555555556
Y= 0+i7.6666666666666667
Z= 15.1666666666666667+i0
T= -2.6111111111111111+i2.5555555555555556
Determinant: 0-i36.0
----------------------------------------------
PROOF: type the following lines ( press COMPUTE):
X1=5.1111111111111111-i2.5555555555555556
Y1=0+i7.6666666666666667
Z1=15.1666666666666667+i0
T1=-2.6111111111111111+i2.5555555555555556
W1=X1+i*Y1+Z1+T1
=10.0+i0
W2=-X1-Y1+2*Z1+2*T1
=20.0+i0.0000000000000001
W3=3*X1+2*Y1-Z1-3*T1
=7.9999999999999999-i1.9949319973733282e-16
W4=-2*X1+Y1+Z1-5*T1
=18.0-i0.0000000000000001
----------------------------------------------
------------
example 8------------
type the following lines ( press COMPUTE):
z1=1-i2
z2=2+i3
x1=3;x2=5;x3=-7;
z1 x+iy+z+t=x1-ix2; -x-z2 y+2z+2t=20; 3x+2y-z1 z-3t=8; -2x+y+z-5t=18+i2;
X= 5.2599406528189911-i7.4712166172106825
Y= 5.7750741839762611+i2.2890207715133531
Z= 16.267062314540059+i3.4747774480712166
T= -1.2955489614243323+i3.7412462908011869
Determinant: -46.0-i68.0
----------------------------------------------
PROOF: type the following lines ( press COMPUTE):
X1= 5.2599406528189911-i7.4712166172106825
Y1= 5.7750741839762611+i2.2890207715133531
Z1= 16.267062314540059+i3.4747774480712166
T1= -1.2955489614243323+i3.7412462908011869
w1= z1*X1+ i* Y1+ Z1+ T1
= 2.9999999999999997-i5.0000000000000001 = x1-ix2 =3-i5
w2= -X1-z2 *Y1+2* Z1+2*T1
= 19.999999999999999+i0 = 20
w3= 3* X1+2* Y1-z1*Z1-3*T1
= 8.0000000000000002-i6.0108168442596366e-16 = 8
w4=-2* X1+ Y1+ Z1-5*T1
= 17.999999999999999+i2.0000000000000002 = 18+i2
;«««««««««««««««««««««««««««««««««««««««««««««««««««
NORM of a MATRIX;«««««««««««««««««««««««««««««««««««««««««««««««««««
type the following lines ( press COMPUTE):
matA=[1,-2; -3, 4];
norm(matA)
we get:
norm -max. of the sum of each line (matA)= 7.0
norm -max. of the sum of each column (matA)= 6.0
Hi Rui,
only one thing: Wow! :t
Gunther
:biggrin: :biggrin:
Hi Gunther
:t
Hi all,
This is only to say that "The powerful calculator" (calcula62)
is near to use matrices with COMPLEX NUMBERS.
I need only to do some tests.
RuiLoureiro
Hi qWord,
When i post calcula62, could you test it with matlab ?
Ok, thank you :t
See this:
Do:
matX=[1+i2,9-i3; 3-i5.4, 4+i1.2];
We get:
The complex matrix is defined
Do:
matY=matX^-1;
We get:
matY=[0.0095362094447181-i0.061193609684003, 0.0622844675909634+i0.1261524385952565;
0.0854036174255753+i0.0331480047856992,0.027359420085861 -i0.0187381237244];
The complex matrix is defined
Errors in the identity matrix are below 1e-18
The inverted matrix is correct
-------------------------------------------------
Do:
matA=[1, 2, 3, 2, 0, 2,1,-2, 3, 2,0,1 ; 1, 4,5, 6, 0, 3,-5, 0, 1, 2, 7,0 ; -3, 0, 7,-2,-3,5,1 ,-1, 2,9,-2,0;
9, 1, 1, 3, 4, 6,2, 5, 6, 8,2,3 ; 1,-1,1,-1, 8, 0, 4, 1,-1, 0, 5,1 ; 2,-1,-5, 2, 0,5,7 ,-9, 3,5,-3,1;
2, 3, 1, 3, 5, 6,4, 5,-6,-8,7,0 ;-1,-3,1,-7,-8, 0, 0, 2,-5, 3, 4,1 ; 12,-4, 5, 7, 1,0,2 , 0,-3,0,-4,-1;
1,-1, 1,-1,-9,-3,5,-7, 2, 2,1,-1 ; 0,-5,3,-2,-1,-7, 6,-1,-2,-5,-2,8 ; 0, 0, 1,-3, 0,1,-5,-2, 0,1, 5,1+i0];
We get:
The complex matrix is defined
Do:
matB=matA^-1;
We get:
matB=[0.3489475139105589+i0,-0.077717764135831+i0,-0.1305663289999746+i0,
-0.01189321415923582+i0, 0.030658417448943+i0,-0.01544063546736273+i0,
-0.032359956048502+i0, 0.12026503888873099+i0, 0.0563587763184989+i0,
-0.0469873206919989+i0,-0.0552819588513436+i0, 0.0028764341340867+i0;
1.2038253746424853+i0,-0.0737862490216782+i0,-0.3707567189349224+i0,
-0.1747472582414871+i0, 0.1591538921209745+i0,-0.0028790830112823+i0,
-0.1494282660592322+i0, 0.4412261789225444+i0, 0.01548276728050134+i0,
-0.2686901806756077+i0,-0.1567441075089298+i0,-0.2763391412739283+i0;
0.0187343364849522+i0,-0.0207830852282367+i0, 0.0772439172493358+i0,
-0.01210186072460702+i0, 0.0069412698217746+i0,-0.055170392364791+i0,
0.0258755788715113+i0,-0.0499767559328416+i0, 0.0257864718665081+i0,
0.0349311858466001+i0, 0.0159887049951585+i0, 0.0485851419165665+i0;
-0.3314381496330013+i0, 0.149112029548901+i0, 0.0633701813335249+i0,
0.0279434023893145+i0,-0.043087392918502+i0, 0.0414049652003719+i0,
0.01215734823711377+i0,-0.0944356492076842+i0,-0.0062092805019976+i0,
0.036984525440662+i0, 0.0541658475877536+i0,-0.058825516372493+i0;
0.1306324231186913+i0,-0.0212040294890572+i0,-0.0315844618695761+i0,
-0.0359397656450297+i0, 0.0930859455477016+i0, 0.006451085834335+i0,
-0.0357521157353953+i0, 0.0202766124971702+i0, 0.01438058210203112+i0,
-0.0674981469058792+i0,-0.0234276869891709+i0,-0.0083228389532894+i0;
-0.3692504255738974+i0,-0.021892117367119+i0, 0.1657167584891193+i0,
0.0705876315716985+i0,-0.1292871530040815+i0, 0.026429924473778+i0,
0.1266082036440829+i0,-0.164592953418614+i0,-0.01335398125911003+i0,
0.0718698527221532+i0, 0.0405050165976217+i0, 0.1594134514897885+i0;
0.0490849707981281+i0,-0.0171928398021129+i0,-0.00158264398711094+i0,
0.0022457425854745+i0, 0.0498990230369092+i0, 0.0029350115469339+i0,
0.0180160649764132+i0, 0.0216114527675041+i0,-0.01354256886325167+i0,
0.0354693568163014+i0,-0.0026549908694263+i0,-0.087100970997439+i0;
-0.2044568349995381+i0, 0.01297154108852978+i0, 0.0532421387390865+i0,
0.0737142372186105+i0,-0.0384820735367924+i0,-0.047213045317213+i0,
0.0295741565155512+i0,-0.0456359070616521+i0,-0.026393327801645+i0,
0.0326274528589774+i0, 0.0201915799013743+i0,-0.0406533648942984+i0;
-0.5932972720613709+i0,-0.0209918266600847+i0, 0.2026367180091205+i0,
0.1572055516251538+i0,-0.1259187657287346+i0,-0.0612563733893335+i0,
0.10269198782892028+i0,-0.3301642954853802+i0,-0.0588064472272272+i0,
0.2259098962133411+i0, 0.0763407165237221+i0, 0.1953977685856945+i0;
0.3182471547714092+i0, 0.0517233171449673+i0,-0.10861579851959029+i0,
-0.055216064571107+i0, 0.11886820932720037+i0, 0.0414594815056652+i0,
-0.1230670565368351+i0, 0.2047330641547552+i0, 0.01459943016238838+i0,
-0.12498644339395018+i0,-0.0577935633223318+i0,-0.1656982226986158+i0;
-0.2799018148394809+i0, 0.068480479218752+i0, 0.0509548448741404+i0,
0.0509176474488098+i0, 0.01172613482006036+i0,-0.0045568810249207+i0,
0.0402069822428226+i0,-0.0780196840570518+i0,-0.0297156907710767+i0,
0.10156931114386624+i0, 0.0259680263379561+i0, 0.062108712424104+i0;
0.2739747436075013+i0, 0.0358644611698261+i0,-0.10043750004637351+i0,
-0.01537054898749617+i0,-0.01188421088973993+i0, 0.0545292542113109+i0,
-0.0447509545683625+i0, 0.1334135675731232+i0,-0.0175808083399441+i0,
-0.1361058847726861+i0, 0.0610654582586491+i0,-0.0461320667215304+i0];
The complex matrix is defined
Errors in the identity matrix are below 1e-18
The inverted matrix is correct
-----------------------------------------------------------------
Do:
matR=[1, 2, 3, 2, 0, 2,1,-2, 3, 2,0,1 ; 1, 4,5, 6, 0, 3,-5, 0, 1, 2, 7,0 ; -3, 0, 7,-2,-3,5,1 ,-1, 2,9,-2,0;
9, 1, 1, 3, 4, 6,2, 5, 6, 8,2,3 ; 1,-1,1,-1, 8, 0, 4, 1,-1, 0, 5,1 ; 2,-1,-5, 2, 0,5,7 ,-9, 3,5,-3,1;
2, 3, 1, 3, 5, 6,4, 5,-6,-8,7,0 ;-1,-3,1,-7,-8, 0, 0, 2,-5, 3, 4,1 ; 12,-4, 5, 7, 1,0,2 , 0,-3,0,-4,-1;
1,-1, 1,-1,-9,-3,5,-7, 2, 2,1,-1 ; 0,-5,3,-2,-1,-7, 6,-1,-2,-5,-2,8 ; 0, 0, 1,-3, 0,1,-5,-2, 0,1, 5,1];
The real matrix is defined
NOTE: this matR is equal matA
***********************
Do:
matS=matR^-1;
The real matrix is defined
The identity matrix is correct
matS=[0.3489475139105589,-0.077717764135831,-0.1305663289999746,-0.01189321415923582,
0.030658417448943,-0.01544063546736273,-0.032359956048502, 0.12026503888873098,
0.0563587763184989,-0.0469873206919989,-0.0552819588513436, 0.0028764341340867;
1.2038253746424853,-0.0737862490216782,-0.3707567189349224,-0.1747472582414871,
0.1591538921209745,-0.0028790830112823,-0.1494282660592322, 0.4412261789225444,
0.01548276728050134,-0.2686901806756077,-0.1567441075089298,-0.2763391412739283;
0.0187343364849522,-0.0207830852282367, 0.0772439172493358,-0.01210186072460701,
0.0069412698217746,-0.055170392364791, 0.0258755788715113,-0.0499767559328416,
0.0257864718665081, 0.0349311858466001, 0.0159887049951585, 0.0485851419165665;
-0.3314381496330014, 0.149112029548901, 0.0633701813335249, 0.0279434023893145,
-0.043087392918502, 0.0414049652003719, 0.01215734823711377,-0.0944356492076842,
-0.0062092805019976, 0.036984525440662, 0.0541658475877536,-0.058825516372493;
0.1306324231186913,-0.0212040294890573,-0.0315844618695761,-0.0359397656450297,
0.0930859455477016, 0.006451085834335,-0.0357521157353953, 0.0202766124971702,
0.01438058210203112,-0.0674981469058792,-0.0234276869891709,-0.0083228389532894;
-0.3692504255738975,-0.021892117367119, 0.1657167584891193, 0.0705876315716985,
-0.1292871530040815, 0.026429924473778, 0.1266082036440829,-0.164592953418614,
-0.01335398125911003, 0.0718698527221532, 0.0405050165976217, 0.1594134514897885;
0.0490849707981281,-0.0171928398021129,-0.00158264398711094, 0.0022457425854745,
0.0498990230369092, 0.0029350115469339, 0.0180160649764132, 0.0216114527675041,
-0.01354256886325167, 0.0354693568163014,-0.0026549908694263,-0.087100970997439;
-0.2044568349995381, 0.01297154108852978, 0.0532421387390865, 0.0737142372186105,
-0.0384820735367924,-0.047213045317213, 0.0295741565155512,-0.0456359070616521,
-0.026393327801645, 0.0326274528589774, 0.0201915799013743,-0.0406533648942984;
-0.5932972720613709,-0.0209918266600847, 0.2026367180091205, 0.1572055516251538,
-0.1259187657287346,-0.0612563733893335, 0.10269198782892028,-0.3301642954853802,
-0.0588064472272272, 0.2259098962133411, 0.0763407165237221, 0.1953977685856945;
0.3182471547714092, 0.0517233171449673,-0.1086157985195903,-0.055216064571107,
0.11886820932720038, 0.0414594815056652,-0.1230670565368351, 0.2047330641547552,
0.01459943016238838,-0.12498644339395018,-0.0577935633223318,-0.1656982226986158;
-0.2799018148394809, 0.068480479218752, 0.0509548448741404, 0.0509176474488098,
0.01172613482006036,-0.0045568810249207, 0.0402069822428226,-0.0780196840570518,
-0.0297156907710767, 0.10156931114386624, 0.0259680263379561, 0.062108712424104;
0.2739747436075013, 0.0358644611698261,-0.10043750004637351,-0.01537054898749618,
-0.01188421088973993, 0.0545292542113109,-0.0447509545683625, 0.1334135675731232,
-0.0175808083399441,-0.1361058847726861, 0.0610654582586491,-0.0461320667215304];
:biggrin: :biggrin: :biggrin:
Hi
--**** The calculator v3.12.1 ****--
calcula62
This is the latest version.
Now, we may do almost all operations with complex numbers,
the same way we do for real numbers.
In the next version, the calculator will do more
matrix operations with real and complex numbers,
i hope. And, i hope to do a general revision too.
Dez. 2013
Good luck !
RuiLoureiro
1. matrix definition:
a11=2.453 ;a12=-5.236 ;a21=9.123 ;a22=-3.154;
z11=(2,3) ; z12=(3,-2); z21=(8,-3); z22=(0,-5);
matA=[a11+ia12, a12-ia21; 0, 2+ia22];
matB=[z11, z12; z21, z22];
2. Operations:
2.1 - ROUND
round(matA, 1) or
matC=round(matB, 1)
2.2 - CONJUGATE
conj(matA) or
matC=conj(matB)
2.3 - CHANGE
- lines: change l(matA;1,2) or
matX=change l(matA;1,2)
- columns: change c(matA;1,2) or
matY=change c(matA;1,2)
2.4 - ADD,SUB,MUL (x=+,-,*)
matX=matA x matB;
2.5 - SCALAR:
matX=-matB;
matX=-2.3*matB;
matX=(-2.3+i3.4)*matB;
2.6 - TRANSPOSE:
matX=matA^t;
2.7 - INVERSE:
matX=matA^-1;
2.8 - DETERMINANT:
det(matA) - Gauss
delta(matA) - Laplace (note: more than 12x12 use Gauss)
---------------------
Inverse example
20x20 matrix
---------------------
g=[1.23+i3,2,3,-4.45,5,6,7,-1.24,0,10,11,1.12,-13,14.11,15,-1.16,7,8,19,-4;
1,2.24,3,-4,5.23,6,7,8,9,-1.08,11,12,-13.23,14,15,16,-17,18.45,1,20;
1,0,3,4,5.23,-6,7,8,9,10,-2.11,0,-13.34,14,0,-16,17,18.58,19,-20;
1,2,1,0,0,6,7.25,8,9,10.12,0,12,13.45,14,15,16,17.21,18.29,-19.23,2;
-1,2.45,0,4,5,6,7.45,8,9,0,11,12,13,4,15,0,17.22,18.21,19,1;
1,2,3,4,5.24,6,0.27,8,9.89,10,11,2,-13,14,15,16,17,3,19,20;
20,9,18,17,16,5,14,13,12,11.38,10,9,8,7,6.34,5,4,3.25,2,-1;
1,2.23,-3,4,5,6,7,-1,9,10,11.45,12,13,14.89,15,-16,17,18,19,20;
1,0,3,-4,5,6,7.34,8.34,9,1,1,12.29,-13.59,14,5,16,17,18.34,19,20;
1,2,0,4,5,0,1,8,9,10,-11,12.24,13,14.22,15,16,17.45,18,19,20;
20,19.34,1.84,17,16,15,14.35,13,12.68,11,10,9,8.34,7,6.34,5,4,3,2,1;
1,2,3,4,5.24,6,7,8.24,9,10,11.23,12,13.34,14,15,-1,-17.23,1,19,1;
1,-2.81,1,4.23,5,6.21,7,8.89,9,10,11,0,13,14,15,16,17,18,19.23,20;
-1,2,3,4,5,-6.23,-7,8,9,10,1,12,13,-14,15.33,6,17,18,19,20;
1,0,0,4,5,6.34,7,8,9,10,-11.89,1.23,13,14.34,15.21,16.34,0,18,19,20;
1.34,2,3,4,5,6.23,7,0,9,0,11,0,1.35,14,15,16.62,17,18,1.92,0;
1,-2,3,4,5.45,6,1,-8,9,10,1.12,12,13,14,15,-16,17.43,2,19,0;
1,2,1,0,-1,6,7,8,9,10,11,-12.23,13,14,15,16,17,18,0,20.23;
-1.45,1,3,4.56,0,6,7,0,9,10,11,0,13.34,14,-15,16,7,18,9.34,20;
1,2,3,-4,5,6,7,8,9.45,10,0,12.48,13,14,-1.54,16.21,-17,0,19,-2];
det(g)=-7.4473766747107554e+24+i4.3514436660822637e+23
f=g^-1;
f=[-0.0194101390259696-i0.00113411916989698,-0.1370614197121183-i0.0080083910445378,
0.0066192461046662+i0.000386757348184,-0.0465182165804565-i0.0027180228386169,
-0.386511062558009-i0.0225835376468868,-0.368461737789722-i0.0215289297846751,
-0.1614524182309398-i0.009433537920411,-0.2385493532116904-i0.01393825124497746,
0.4595938754918398+i0.0268537089747356, 0.0388561956287034+i0.0022703369755786,
0.1837886974705918+i0.01073862916349639, 0.5117449650708355+i0.0299008561560835,
0.1620417483438316+i0.0094679720157843, 0.1277470356882338+i0.0074641589056987,
-0.1869878416007705-i0.01092555264098641, 0.1301116930261207+i0.0076023239756936,
0.1197250236072046+i0.0069954390438745, 0.0993442813080926+i0.005804608287473,
-0.0020717516763048-i0.00012105082236762,-0.2827179734166223-i0.0165189890137946;
0.0337604928916169+i0.0019725990690909, 0.0299769462062127+i0.01392917529187581,
-0.01306560097164575-i0.0006726957847483,-0.0192320231024803+i0.0047275184685492,
0.12136210891085013+i0.0392800567360801, 0.0866611744166858+i0.0374457534790036,
0.0386556888522107+i0.0164079654184201, 0.055693308951244+i0.0242431149745011,
-0.11648218123635306-i0.0467072621036945, 0.0604679439001707-i0.0039488483427679,
-0.0280615129773847-i0.01867794007322,-0.11348444865514418-i0.0520072339698397,
-0.1499236106659666-i0.0164678574176749,-0.0331475772238254-i0.01298258004956008,
0.0202305291773326+i0.0190030602964516,-0.0270031861935396-i0.01322289367416595,
-0.0440565862158477-i0.01216732501495661, 0.0358409776223077-i0.0100960862034869,
0.0154972153526858+i0.0002105464274417, 0.0667522074753097+i0.0287318504226455;
0.0406117941812262+i0.0023729152193714, 0.0818252740755585+i0.0167559402015824,
-0.01085170799121976-i0.0008092116085059,-0.01108316472790187+i0.0056869136256106,
0.1517248804988679+i0.0472514896246874, 0.0585950532450137+i0.0450449357517479,
0.0715022541055779+i0.0197377720948801,-0.0520828615153939+i0.0291629745696253,
-0.10913404205833358-i0.0561859603594483, 0.0430512462722827-i0.0047502214101024,
-0.0543945245926987-i0.0224684118332657,-0.12046089994367322-i0.0625615001741418,
-0.0724493563580234-i0.0198098184821957,-0.0432071231790731-i0.01561724441069838,
0.0189264569889868+i0.0228595114428722,-0.10750152552027906-i0.0159063269040366,
0.0513258689097457-i0.01463654280255493, 0.0355983503043648-i0.0121449700467418,
0.043286961062462+i0.0002532743880342, 0.0170873264137525+i0.0345626469245062;
-0.0159470029981016-i0.000931770853282,-0.0579278258913802-i0.0065795425692905,
-0.0002637164077883+i0.0003177525201018, 0.01317218340486517-i0.0022330761412033,
-0.0542926443389835-i0.0185542072666661,-0.0126062849965109-i0.0176877613994824,
-0.017181424227656-i0.0077504162797713,-0.0209795548031127-i0.01145140352135348,
0.0591087304758857+i0.022062499241108,-0.0074086132114467+i0.00186526590602018,
0.0163702885251149+i0.0088226545537184, 0.1342346535807686+i0.0245659777155073,
0.0000232256387358+i0.0077787066810621, 0.01368776224649405+i0.0061324117404943,
0.0219876624186343-i0.0089762273463687, 0.0539338060356428+i0.0062459255480201,
-0.0231183686896271+i0.0057473203698566,-0.0451954066975247+i0.0047689563500441,
0.029589433178144-i0.0000994530654642,-0.0979559737147994-i0.01357168884654252;
0.0270098173828801+i0.00157816240411039, 0.2822294053188393+i0.01114392737497949,
0.0135642886383116-i0.0005381849832173, 0.0039312280452027+i0.0037822140488184,
0.4288829609106601+i0.0314257011186627, 0.3581887821357926+i0.0299581811935978,
0.1830606920666091+i0.01312706396197744, 0.2244567837334123+i0.0193955138734373,
-0.6048031004824733-i0.0373677784837187, 0.0180793287694184-i0.0031592451257527,
-0.1548464757746993-i0.0149431393696068,-0.7404162256078907-i0.0416079793806236,
-0.00007960509715-i0.01317498008594397,-0.1638551599103601-i0.01038661128032024,
0.128850098868514+i0.01520324925263357,-0.2125982673139815-i0.01057887230968555,
-0.0449222722876399-i0.0097343728880726,-0.11590363813509191-i0.0080772945322047,
-0.0367118167199483+i0.0001684460168895, 0.3807303570510562+i0.0229866914407692;
0.0405289318993353+i0.0023680736413576, 0.11148214611394603+i0.0167217522158427,
-0.00129221501739713-i0.0008075605334483, 0.0019705997187974+i0.0056753103303257,
0.1254488014001815+i0.0471550800389505,-0.0034092780624095+i0.0449530283086203,
-0.0303372563008461+i0.0196975001278765,-0.1648226353587958+i0.0291034718889808,
-0.0618045619113692-i0.0560713213246684, 0.0192129713643234-i0.0047405292949555,
0.0484247009571903-i0.0224225684049586,-0.0763036052803377-i0.0624338528055026,
0.0500677141133544-i0.0197693995153322,-0.037371848055371-i0.01558537976312966,
0.0000199038379996+i0.0228128700344031,-0.1589038574370042-i0.0158738724269483,
0.1411181108148658-i0.01460667912969671, 0.0204365488505158-i0.01212019005482421,
0.0610045485222371+i0.000252757619589,-0.0517062399548468+i0.0344921270213575;
-0.0352221441445672-i0.0020580022031672,-0.2033129989221616-i0.01453223510451822,
-0.0291679975757334+i0.0007018199637047, 0.0474400439994978-i0.0049321950802055,
-0.12018171605391551-i0.0409806759873619, 0.0370009471101926-i0.0390669570753481,
0.0516429278153983-i0.0171183437677289, 0.2467030664092351-i0.0252927139685577,
0.1295205241152076+i0.0487294401683013,-0.1591280629174403+i0.004119812645524,
-0.0710703227173497+i0.0194865963507867, 0.0969597598894756+i0.0542588728584819,
-0.11046223479751792+i0.0171808287746164, 0.0406517006168604+i0.01354465728157462,
0.12341720850775279-i0.0198257925646494, 0.199315428173636+i0.01379537521845265,
-0.2055876840688113+i0.01269410600450729,-0.0599417938584268+i0.01053319347844866,
-0.0711746548999218-i0.000219661977101, 0.0624404120414577-i0.0299757879831741;
0.0081821082410326+i0.0004780741546421, 0.102675244302531+i0.0033758399295984,
0.0198194052144595-i0.000163032860384, 0.0004582798404677+i0.00114574949913589,
0.1515148995684158+i0.0095198158676266, 0.060127170119415+i0.0090752587385449,
0.0237902158205569+i0.0039765932772258,-0.0535693260866477+i0.0058755004394625,
-0.1269786180193445-i0.01131985470122105, 0.0560906730305373-i0.0009570329636974,
-0.01004175635606827-i0.0045267386317264,-0.1324901331847655-i0.01260434256762883,
0.0351354779174615-i0.003991108551699,-0.0604964952618838-i0.0031464254847934,
-0.0164770597133613+i0.0046055339522322,-0.1517478501646902-i0.0032046672911151,
0.0337269698867066-i0.0029488423227647, 0.0181214240660289-i0.0024468620879714,
0.01020547141042502+i0.0000510275032009, 0.0289199897365273+i0.0069633790856655;
-0.0855434614452994-i0.0049982372282164,-0.0876699811315491-i0.0352942083331155,
0.0171838885388476+i0.00170449898678236,-0.0013575635030036-i0.01197874376848046,
-0.1443737898739694-i0.0995291161701729,-0.0379226738097229-i0.0948812974770497,
-0.039423574272766-i0.0415750493238491, 0.0343289525511508-i0.0614280122566028,
0.1519604273907546+i0.11834841652963564,-0.10261154979135434+i0.01000572346640078,
0.0232597906075389+i0.0473267867166671, 0.1381659923690335+i0.1317776617852742,
-0.0403896939358725+i0.0417268056665535, 0.0722280711352362+i0.0328956937772025,
0.012981757502854-i0.0481505871679929, 0.134654076778117+i0.0335046084440366,
-0.004615171744153+i0.0308299734145118, 0.0301539335505019+i0.0255817995213824,
-0.0361964298334951-i0.0005334895511192,-0.00177590366985481-i0.0728017196541579;
0.0218453568013083+i0.00127640703080258,-0.0360687764178845+i0.0090131327518192,
0.0045033886416937-i0.0004352803581316, 0.036121924954747+i0.0030590290272651,
-0.0600614082205919+i0.0254168935663961, 0.0039853559555188+i0.0242299734209682,
-0.0094350279881629+i0.01061708015044768, 0.020246079162043+i0.0156869598525506,
0.01463167809985502-i0.0302228053702613,-0.043572986696341-i0.0025551759945851,
0.0060957258463732-i0.01208590960217478, 0.0295053933264537-i0.0336522510488103,
0.009592203958364-i0.01065583438598161, 0.0340152894450295-i0.0084006206394755,
0.0214516524167138+i0.01229628470844514, 0.0087630613671468-i0.0085561200538527,
-0.0201178115133616-i0.0078730946589707,-0.024229047535509-i0.006532860942522,
0.0061163698112344+i0.00013623799408,-0.00122346407094997+i0.0185914798714436;
0.0023717548784452+i0.0001385797737122, 0.00127000820674749+i0.0009785576756027,
-0.0004651536338222-i0.0000472584779584,-0.0059787913672044+i0.0003321194103035,
-0.0199869109976612+i0.0027595173591952,-0.017417003321501+i0.002630653195023,
-0.01114788616837892+i0.00115269857437899,-0.0076362675002467+i0.00170313645579978,
0.01147052620860822-i0.0032812961916431, 0.0035884156375167-i0.0002774159829735,
0.01018550301142924-i0.00131216968988529, 0.0327141889714081-i0.0036536318139193,
0.0338533248498052-i0.00115690612969722, 0.0087544684830588-i0.0009120571096577,
-0.0443652618458861+i0.00133501015841773,0.01190391112150391-i0.0009289397130411,
-0.00116041352008804-i0.0008547835055162, 0.0030220362544637-i0.0007092740554232,
0.0051412022124669+i0.00001479138702232,-0.01187840518949871+i0.0020184807913035;
-0.0021773302246684-i0.0001272196939799,-0.0413996878030306-i0.0008983403904993,
-0.01022822662108148+i0.000043384463279, 0.0200731016564679-i0.0003048939149761,
-0.059829128139176-i0.002533305868274,-0.0427014479813116-i0.002415005346547,
-0.0194138840209737-i0.00105820608560225,-0.0080547729396944-i0.00156352181064276,
0.0855139254474379+i0.0030123118704566, 0.0021761912802275+i0.0002546748021998,
0.01537435822730602+i0.00120460455321332, 0.10034410912545845+i0.0033541252726216,
0.0061856561978348+i0.0010620687264888, 0.0240180114421728+i0.0008372912097827,
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-0.0241403756810982+i0.0025208037290581,...];
det(f)=-1.3381861640951734e-25-i7.8189165959126229e-27
Hi
--**** The calculator v2014.01 ****--
(calcula63)
This is the latest powerful version.
In this new version, we may define up to
30 constants or matrices and matrices
20x21. Each matrix may be real or complex.
We may compute the determinant of any
defined matrix N by (N+1) - the same
as N by N.
We may do linear transformations also:
add lines, sub lines, etc.
See the examples in the files:
. Examples_real_matrix.txt
. Examples_complex_matrix.txt
. Examples_equations.txt
Finally, we may transfer all work done
to the file TheCalculator.txt. To do
this, press ToFile.
To start a new page, press Del File.
To transfer to the file ThisWork.txt
use
file(ThisWork)
Last note: now the calculator has a
unique table of erros with 190 messages.
Jan. 2014
Good luck !
RuiLoureiro
Rui,
awesome. :t
Gunther
Quote from: Gunther on January 08, 2014, 07:05:02 AM
Rui,
awesome. :t
Gunther
Hi Gunther
Thanks for your words :t
Hi
--****
The calculator v2014.02 ****--
(
calcula64)
This is the latest powerful version.
---------------------------------------------
Polynomials with integer coefficients
---------------------------------------------
General:
p(x)=I7 x^7 + I6 x^6 + ...+ I1 x + I0
where I7, I6, ..., I0 are integers.
The calculator tries to find integer roots only.
------------------------------------------------
Type:
p(x)=x^2+x-2
We get:
p(x)= (x+2 ) * (x-1 )
------------------------------------------------
Type:
p(x)=x^5 +2x^4 -x^3 -2x^2
We get:
p(x)= (x+2 ) * (x+1 ) * (x-0 ) * (x-0 ) * (x-1 )
------------------------------------------------
Type:
p(x)=x^2-2x+2
We get:
There is no integer roots in the table from -30 to 30
x^2-2 x+2= (x-1+i ) * (x-1-i )
------------------------------------------------
Type:
p(x)=x^6 -1
We get:
p(x)= (x+1 ) * (x-1 ) * ( x^4+ x^2+1 )
------------------------------------------------
Type:
p(x)=-5x^4 +5
We get:
p(x)= -5 * (x+1 ) * (x-1 ) * ( x^2+1 )
x^2+1= (x+i ) * (x-i )
------------------------------------------------
Type:
p(x)=-6x^6
We get:
p(x)= -6 * (x-0 ) * (x-0 ) * (x-0 ) * (x-0 ) * (x-0 ) * (x-0 )
Jan. 2014
Good luck !
RuiLoureiro
Hi all,
****--The Calculator v2014.03 by RuiLoureiro--****
(calcula65)
------------------------------------
DERIVATIVES
first version
------------------------------------
In this version, we may compute the derivative
of one given function.
1. Write an expression: x^2-x*sin(x-x^2)
2. Put the expression between brackets
(x^2-x*sin(x-x^2)) or [x^2-x*sin(x-x^2)]
3. end it
(x^2-x*sin(x-x^2))' or [x^-x*sin(x-x^2)]'
4. compute (we get)
2*x-[sin(x-x^2)+(1-2*x)*cos(x-x^2)*x]
In the same way, we have:
[-(x)*sin(x-x^2)]' = -[sin(x-x^2)+(1-2*x)*cos(x-x^2)*x]
------------
basic rule
------------
the calculator doesn't know how to solve
expressions with brackets inside brackets
--------------------
some messages
--------------------
1. «The expression is too complex»
means that the calculator doesnt solve that case
(it will solve later but not now)
2. «Expression error»
In many cases, it means that the calculator
doesnt solve that case
(it will solve later but not now)
[x^2 + 2 * sin(x) ]' = Expression error
but
[x^2 + (2) * sin(x) ]' = 2*x+cos(x)*(2)
-----------------------------------
how to read the expression
-----------------------------------
We must read the expression from left to right.
Example:
[arctan(x^2-1)^(-1/2)]' = (-1/2)*2*x/(1+(x^2-1)^2)*arctan(x^2-1)^(-3/2)
means this:
(-1/2)*2*x
[arctan(x^2-1)^(-1/2)]' = --------------- * arctan(x^2-1)^(-3/2)
(1+(x^2-1)^2)
-x
= --------------- * arctan(x^2-1)^(-3/2)
(1+(x^2-1)^2)
-x
= -----------------------------------------------
(1+(x^2-1)^2) * arctan(x^2-1)^(3/2)
------------
examples
------------
[sin(x^2-x)^(1/3) / (x^2-1)]'=[(1/3)*(2*x-1)*cos(x^2-x)*sin(x^2-x)^(-2/3)*(x^2-1) -
2*x*sin(x^2-x)^(1/3)]/(x^2-1)^2
[(x^2-1) * sin(x^2-x)^(1/3)]' = 2*x*sin(x^2-x)^(1/3) +
(1/3)*(2*x-1)*cos(x^2-x)*sin(x^2-x)^(-2/3)*(x^2-1)
[(x^2-1) / sin(x^2-x)^(1/3)]' = [2*x*sin(x^2-x)^(1/3) -
(1/3)*(2*x-1)*cos(x^2-x)*sin(x^2-x)^(-2/3)*(x^2-1)] /
sin(x^2-x)^(2/3)
This is the first version and this may contain some
bugs. I am sure you will call my attention to that
if you want to help me.
Thank you.
Good luck :t
Mar. 2014
RuiLoureiro
Hi Rui,
seems to be a great step forward. I'll test the new version next weekend. Thank you.
Gunther
Hi Gunther,
Thank you. :t
I also think it will be a great step forward:
we write a function, we have the derivative!
Only to us: i hope that Hutch will do some
differential calculus, now and quickly. ;)
:biggrin:
Hi all,
****--The Calculator v2014.03.01 by RuiLoureiro--****
(calcula65)
------------------------------------
DIFFERENTIAL CALCULUS
------------------------------------
I found a bug in the previous version v2014.03.
First, i want to say that i did a lot of tests before posting
the last version. For instance, i got
[cos(x^2-1)^(1/2) * sin(x-x^2)^(1/3)]' =
-(1/2)*2*x*sin(x^2-1)*cos(x^2-1)^(-1/2)*sin(x-x^2)^(1/3) +
(1/3)*(1-2*x)*cos(x-x^2)*sin(x-x^2)^(-2/3)*cos(x^2-1)^(1/2)
and it is correct. But probably i did some modifications in the model
and using the version i posted, we get
-(1/2)*2*x*sin(x^2-1)*cos(x^2-1)^(-1/2)*sin(x-x^2) +
(1/3)*(1-2*x)*cos(x-x^2)*sin(x-x^2)^(-2/3)*cos(x^2-1)
Here, we have sin(x-x^2) NOT sin(x-x^2)^(1/3)
and cos(x^2-1) NOT cos(x^2-1)^(1/2)
This is the bug.
In this new version we get
-x*sin(x^2-1)*cos(x^2-1)^(-1/2)*sin(x-x^2)^(1/3) +
(1/3)*(1-2*x)*cos(x-x^2)*sin(x-x^2)^(-2/3)*cos(x^2-1)^(1/2)
See this: -(1/2)*2*x ... in the previous
IS: -x ... now
This is the new model.
Another example I:
[sin(3.1*x^2-x)^(1/3) / cos(2.5*x^2-1)^(1/2)]'=
[(1/3)*(6.2*x-1)*cos(3.1*x^2-x)*sin(3.1*x^2-x)^(-2/3)*cos(2.5*x^2-1)^(1/2) -
(-5/2)*x*sin(2.5*x^2-1)*cos(2.5*x^2-1)^(-1/2)*sin(3.1*x^2-x)^(1/3)] / cos(2.5*x^2-1)
Another example II:
[sin(3.1*x^2-x)^(1/3) * cos(2.5*x^2-1)^(1/2)
+(x^2-x)^(3) * (x^2-1)
+exp(x^2-x) * ln(x^2-1)]'=
(1/3)*(6.2*x-1)*cos(3.1*x^2-x)*sin(3.1*x^2-x)^(-2/3)*cos(2.5*x^2-1)^(1/2) +
(-5/2)*x*sin(2.5*x^2-1)*cos(2.5*x^2-1)^(-1/2)*sin(3.1*x^2-x)^(1/3) +
3*(2*x-1)*(x^2-x)^2*(x^2-1) +
2*x*(x^2-x)^3 +
(2*x-1)*exp(x^2-x)*ln(x^2-1)+(2*x/(x^2-1))*exp(x^2-x)
Good luck !
Mar. 2014
RuiLoureiro
EDIT: we may use list or show to see the defined constants.
Rui,
thank you for the new archives. :t
Gunther
****--The Calculator v2014.04 by RuiLoureiro--****
( calcula66 )
--------------------------------------------
SOME DETECTED PROBLEMS
in the previous versions
--------------------------------------------
----------------------------
1. REAL EXPRESSIONS
----------------------------
If we try:
jr = 10.6;
tj = jr/1200
tj1 =(1200+jr)/1200
tj72 =tj1^72
we get:
pm=(10500.0 * ( jr / 1200) * (1+ jr / 1200)^72 ) / ( (1+tj)^72-1 )
=197.71325597595037
pm=(10500.0 * ( jr / 1200) * (1+ jr / 1200)^72 ) / ( tj72-1 )
=197.71325597595036
In the previous we GET: ERROR
HERE: +tj)^ <--- brackets not tested after variable name
Now it is solved.
----------------------------
2. COMPLEX NUMBERS
----------------------------
Try:
10^1+i0
It should be: 10+i0
(first power 10^1 and second ADD i0)
But we get: 1.0+i0
| Z |= 1.0
Angle= 0 degrees
Angle= 0 radians
Now we get: 10.0+i0
| Z |= 10.0
Angle= 0 degrees
Angle= 0 radians
Now it is solved.
------------------------------
3. LOGIC EXPRESSIONS
------------------------------
Try: xz=92h and 20h
we get: ERROR (doesnt detect 'and' operator)
I redid all procedures to compute
logic expressions and now
it is, better and faster.
We get: 0
00000000H
00000000000000000000000000000000B
Now it is solved.
--------------------------
5. CONVERSIONS
--------------------------
Now, '-' means 'negative number'
and NOT means 'complement'
Example 11
--------------
type: not -42d
we get: 41
00000029H
00000000000000000000000000101001B
--------------------------
4. DERIVATIVES
--------------------------
I did a lot of work about this, but it is
not completed yet.
NOTE: in many cases, when the calculator gives ERROR, it
means that it cannot solve the problem (it fails).
Good luck ! :t
Jun. 2014
RuiLoureiro
Rui,
I see, it's an error update. So the calculator becomes more and more stable. That's good news.
Gunther
Hi all,
****--The Calculator v2014.05 by RuiLoureiro--****
( calcula67 )
1. SOME DETECTED PROBLEMS
in the previous versions
were corrected;
-------------------------------------------------
2. Now, the calculator use SSE
instructions in some cases.
It is faster;
-------------------------------------------------
3. See the following files where
we can find a lot of examples
about matrices.
. Examples_real0_matrix.txt
. Examples_complex0_matrix.txt
If matCMPX is complex,
matREAL= cnvreal(matCMPX) is real.
-------------------------------------------------
I did these examples to show that
The calculator works correctly.
Good luck ! :t
Set. 2014
RuiLoureiro
Rui,
thank you for the new version. :t SSE accelerates much.
Gunther
Excelent work rui
You could do a library version. (dll and lib) as well. It would be very handfull
Quote from: guga on September 23, 2014, 05:51:45 AM
You could do a library version. (dll and lib) as well. It would be very handfull
This idea isn't so bad. I think it's a question of time.
Gunther
Hi all
:biggrin:
Ola, Rui!
(http://www.cyberforum.ru/images/smilies/yes.gif)
Quote from: Mikl__ on March 29, 2016, 05:25:53 PM
Ola, Rui!
(http://www.cyberforum.ru/images/smilies/yes.gif)
Olá Mikl__, very nice to see you here ! :t :icon14:
Hi Rui!
Have you thinked in symbolic solution of partial derivatives?
For example:
input box=> [a+b*x^c]'
solution box1=> der.a =1
solution box2=> der.b =x^c
solution box3=> der.c = b*c*x^(c-1) <- this, at least partially, explain Rui smile
solution box3=> der.c = b*x^c* ln(x) <- I think!
(more easy remember Rui calculator
that this things)
Regards, HSE
:biggrin:
Hi HSE,
Thanks for your post.
The answer is NO. I have no time now to do it
and my main computer is not working.
It needs to be repaired.
Meanwhile, we can see that the calculator
solves [3+5.34^x]' =ln(5.34)*5.34^x
but doesn't solve [3+2*5.34^x]'= Expression error.
So, i need to do some more work about derivatives because
it doesn't solve some cases yet.
Note: The calculator uses x or X as the variable (not a,b or c)
Regards :t
Rui
Hi HSE,
First of all, thanks for your question about partial derivatives.
Now, you need to correct your example because you made an error
in the third solution.
In your example, it is given a function of 3 variables a, b and c (x is a constant)
f(a,b,c)=[a+b*x^c]. And we want the partial derivative of f(a,b,c)
1. with respect to the variable a ( which means that b*x^c is a constant );
2. with respect to the variable b ( which means that a and x^c are constants );
3. with respect to the variable c ( which means that a, b and x are constants ).
This means that your two first solutions
solution box1=> der.a =1
solution box2=> der.b =x^c
are correct. But the third - solution box3=> der.c =b*c*x^(c-1) -
is not correct because x^c is not a power function
but an exponential function g(c)=x^c (c is a variable and x is a constant).
In this way, g'(c)=[e^(c*ln(x))]'= ln(x) * e^(c*ln(x)) = ln(x) * x^c
Particular cases:
1. If x= e then g'(c)= e^c ;
2. If x=10 then g'(c)=ln(10) * 10^c
So, the third solution is, in general:
solution box3=> der.c = b * ln(x) * x^c
EDIT:
or solution box3=> der.c = ln(x) * x^c * b etc
I will post the new version calcula68 to solve functions like f'(x)=[C1+...+C2*K^x]'
as soon as possible.
Good luck :t
Regards,
Rui Loureiro
EDIT: you say that «...this, at least partially, explain Rui smile».
No, it has nothing to do with it but with the question itself.
Thanks HSE :t
Thanks Rui!!
I change my post when I think in your smile! (just in time)
Quote from: HSE on June 02, 2016, 08:43:37 AM
Thanks Rui!!
I change my post when I think in your smile! (just in time)
Thanks HSE.
As i said before i need to do some more work about derivatives. My last post was in 2014 -
September 05, 2014.
Hi HSE,
If we try to get [x-cos(x)]' the calculator gives 1-[-sin(x)].
Do you like the solution ? I dont like ! In the next version
the solution will be 1+sin(x) and NOT -[- ... ].
:t
Hi Rui!
I have tried [2.5+3*sin(2*3.14*x/5+4)]', of course very complex.
But also is too complex [3+2*sin(x)]' . You have a lot of work to do! :biggrin:
Regards. HSE
Quote from: HSE on June 07, 2016, 10:21:41 PM
Hi Rui!
I have tried [2.5+3*sin(2*3.14*x/5+4)]', of course very complex.
But also is too complex [3+2*sin(x)]' . You have a lot of work to do! :biggrin:
Regards. HSE
Hi HSE
Thanks for your reply :t
I am working in a new version: calcula68 v2016.01
If we use calcula67 we cannot get the solution of
[3+2*sin(x)]'. But using the new version we get 2*cos(x) and this (for example):
[-x-x^2-1/x-e^x -x^x -ln(x)-log(x)-sqr(x)-sin(x)-cos(x)- arcsin(x) -arccos(x) -arctan(x) -arccsc(x) -arcsec(x) -arccot(x) ]'
=-1-2*x+1/(x^2)-e^x-(1+ln(x))*exp(x*ln(x))-1/x-1/(ln(10)*x)-1/(2*sqr(x))-cos(x)+sin(x)-1/sqr(1-x^2)-1/sqr(1-x^2)-1/(1+x^2)
-1/(abs(x)*sqr(x^2-1))-1/(abs(x)*sqr(x^2-1))-1/(1+x^2)Yes, i have a lot of work to do. I need to test a lot of expressions. But i dont want to solve 2*3.14*x/5 but only C * x where C= 2*3.14/5
Thanks :t
Hi Rui!
Perhaps the equation parser that qWord developed for SmplMath is usefull for parenthesis problem.
In practice, most of the equations are short but with parenthesis. Now I trying to write derivatives that are used in Levenberg-Marquardt method to solve non linear models:
Equation:
Quote
Y = a+b*sin(c+2*3.14159*(X/d));
First partial derivatives:
Quote
DER.a = 1;
DER.b = sin(c+(2*3.14159*X/d));
DER.c = b*cos(c+(2*3.14159*X/d));
DER.d = b*cos(c+(2*3.14159*X/d))*(2*3.14159*X*(-1)/d**2);
Second partial derivatives:
Quote
DER.a.a=0;
DER.a.b=0;
DER.a.c=0;
DER.a.d=0;
DER.b.a=0;
DER.c.a=0;
DER.d.a=0;
DER.b.b=0;
DER.b.c=cos(c+ (2*3.14159*X/d))*((2*3.14159*X/d)+1);
DER.b.d=cos(c+(2*3.14159*X/d))*2*3.14159*-1/d**2;
DER.c.b=cos(c+ (2*3.14159*X/d))*((2*3.14159*X/d)+1);
DER.d.b=cos(c+(2*3.14159*X/d))*2*3.14159*-1/d**2;
DER.c.c=b*(-1)*sin(c+(2*3.14159*X/d));
DER.c.d= b*-sin(c+(2*3.14159*X/d))*(2*3.14159*X*(-1)/d**2);
DER.d.c= b*-sin(c+(2*3.14159*X/d))*(2*3.14159*X*(-1)/d**2);
DER.d.d=b*2*3.14159*X*(-1)*( -sin(c+2*3.14159*X/d**2)*(2*3.14159*X*(-1)/d**2)+cos(c+2*3.14159*X/d**2)*-2*d**-3);
I can find the solution, but the convergence is very slow, perhaps some error. At least I'm learning a litlle of derivation (because I have my old notes, but I think another person write them :biggrin:)
Quote from: HSE on June 09, 2016, 12:32:59 AM
Perhaps the equation parser that qWord developed for SmplMath is usefull for parenthesis problem.
Please DON'T - he should use something well known, e.g. the Shunting-yard algorithm.
Whatever the case may be, seems like that his current problem is to bring the equation into a canonical form before applying derivation rules.
Quote from: qWord on June 09, 2016, 02:32:26 AM
Please DON'T - ... the Shunting-yard algorithm.
Just an idea. Very nice, that idea lead to other idea.
Next week I will try to find what is a canonical form. 8)
Quote from: HSE on June 09, 2016, 05:13:37 AM
Quote from: qWord on June 09, 2016, 02:32:26 AM
Please DON'T - ... the Shunting-yard algorithm.
Just an idea. Very nice, that idea lead to other idea.
Next week I will try to find what is a canonical form. 8)
Hi qWord, how are you ?
Hi HSE
I think that if we have a procedure that solves [c+K*x/d]' for any variable c OR x OR d
we have the problem solved for f(c+K*x/d). In this way, if the argument is in this "canonical form"
it is solved: sin(c+K*x/d)'= [c+K*x/d]' cos(c+K*x/d).
Hi Rui it's Good work :t
I have the concept!! (I think)
Thanks.
Quote from: Force on June 10, 2016, 08:37:07 AM
Hi Rui it's Good work :t
Thanks :t
My problem now is not only to solve the derivative of but to show it.
For example i dont want to show solutions like [...]'= 2*1/(...) but 2/(...)
Or 2*[-1/(...)] but -2/(...).
And i have no problems to solve expressions like C1*C2*C3/C4/C5*X/C6*C7 etc
where Cn are constants and X is the variable. But may be The calculator will solve it soon.
Don't Know now.
Thanks for your reply :t
Hi HSE,
Could you post the first partial derivative of sin(a-2.0*9/2*x/b) with respect to the variable b and a ? It is just to see the solution.
Hi Rui!
df/da = cos(a-9*x/b)
df/db = 9*x*b^(-2)*cos(a-9*x/b)
It`s my solution, perhaps not the solution. :biggrin:
Quote from: HSE on June 14, 2016, 01:01:27 AM
Hi Rui!
df/da = cos(a-9*x/b)
df/db = 9*x*b^(-2)*cos(a-9*x/b)
It`s my solution, perhaps not the solution. :biggrin:
:biggrin:
Hi HSE,
Yes it is the solution. Is it the solution shown by your
calculator ? It should be:
DER.a = cos(a-9*x/b)
DER.b = 9*x*b^(-2)*cos(a-9*x/b)
In any way why
(-2
) ? Why not 9*x*b^-2
? After '^' the sign must belong
to the next number (or literal constant) or it is an error. No
?
Hi Rui!
DER.a, etc is the way to charge derivatives in the package that solve Levenberg-Marquardt. I'm my calculator(OK, the table of rules is the calculator 8)).
After to read about Shunting-Yard the Saturday, on Sunday I tried to write a solver :biggrin:. I give up at this point:
Quotepsal -> [2+4*2/(1-5)^2^3]' = 0
I put the parenthesis because is more easy to read. 1/b^2 also is nice. Not very important.
Yesterday I downloaded a couple of solvers in Java, don't work yet because they need extra libraries, perhaps take some time to see what they do.
RuiLoureiro,
just for personal interest: how do you store the parsed expressions? As tree?
regards
Quote from: qWord on June 15, 2016, 08:57:42 AM
RuiLoureiro,
just for personal interest: how do you store the parsed expressions? As tree?
regards
Hi qWord,
i store parsed expressions in tables one after another. The length of
each line is a constante (2^N). It is saved from top to bottom
and removed from bottom to top (it is like push, push, push and pop, pop, pop).
In some cases i store the function itself to be solved later. So
i have one table for derivatives and another for functions.
I have one table of basic functions (Fnc1, Fnc2,...,FncN) and tables
of basic derivatives (Drv1,Drv2,...,DrvN): for argument X, for -X, 1/X, ...
The first step is to solve functions. The last is to solve something
like 1+x-x^2+x^-5+e^x, etc. (expressions without brackets).
Could you post an expression to be solved ?
Thanks
regards :t
Hi,
not so easy to find some expression, because nested functions (e.g. sin(cos(x))) are currently not implemented.
One thing I found is that
(-2^x)'
fails with
(-2^x)' = -2 ↯
Alos to note that editing expression is very unintuitive, because the DEL key clears the control rather than removing just one character (definitively breaks usability).
regards
Quote from: qWord on June 19, 2016, 02:57:16 AM
Hi,
not so easy to find some expression, because nested functions (e.g. sin(cos(x))) are currently not implemented.
One thing I found is that
(-2^x)'
fails with
(-2^x)' = -2 ↯
Alos to note that editing expression is very unintuitive, because the DEL key clears the control rather than removing just one character (definitively breaks usability).
regards
Hi qWord,
many thanks for your reply
Tomorrow i will see what about that expression.
I replaced my copy procedure but the last has a bug !
Thank you so much :t
Hi qWord,
First of all, many thanks for your reply.
Five things:
1. I got calcula67 from the first post in this forum.
It is «The Calculator v2014.05 by RuiLoureiro»
Using it i get this:
(a) (2^x)' = ln(2)*2^x
AND THIS
(b) (-2^x)' = ln(2)*2^x
NOTE: the correct solution for (b) would be -ln(2)*2^x
because (-2^x)'= -1*(2^x)'
and [2^x]'=[e^(x*ln(2))]' = [x*ln(2)]' * 2^x = ln(2) * 2^x
It means that the case (b) is wrong but it gives a solution and does't fail.
I don't know why you don't get a solution!
2. I know that the copy procedure used by calcula67 has a bug. I'm working now
in calcula68, «The Calculator v2016.01» and some bugs was corrected.
3. About nested functions, like sin( cos(x) ) or sin( cos(x)+x^2-1 ),
it is not implemented in the calcula67. I will try to implement it
in the next calcula68 if i have time to do it. At least some cases.
4. About the DEL key, i will correct it, thank you for the suggestion.
5. About the new version
In the previous calcula67, we get things like this:
(-ln(x))' = -[1/x] OR (-arcsin(x))'=-[1/sqr(1-x^2)]
I don't like these solutions. In the new version, it will be:
(-ln(x))' = -1/x and (-arcsin(x))'=-1/sqr(1-x^2) and ...
Thank you so much. :t
EDIT:
qWord,
Your suggestions are always good.
I don't forget that i developed some things with your suggestions - see the old forum.
qWord,
I want to write a general procedure to identify a general power function
(or exponential).
The PowerFuncX must be: _BaseArgumentX ^ _ExpoArgumentY
where
_BaseArgumentX and _ExpoArgumentY are strings;
K, N are integers; X,Y are real10
and
_BaseArgumentX - must be INTEGER K constant
REAL X "
10 - particular case K=10
x
e
FncX(x)
expressionX
_ExpoArgumentY - must be INTEGER N constant
RATIONAL (K/N) "
REAL Y "
x
e
FncY(x)
expressionY
Note: In general, the left side is X, the right side is Y, so we have
X operation Y, where operation is a code for: +, -, *, /, ^.
Example 1: we may type [-x ^ (x-1)]' => [ - x ^ expressionY ]'
Example 2: we may type [10.54 ^ sin(x-1)]' => [ REAL X ^ FncY ]'
Do you want to give me any suggestion ?
Thanks :t
you might see if this code works for you...
http://masm32.com/board/index.php?topic=222.0 (http://masm32.com/board/index.php?topic=222.0)
the IntExp_1 download
Quote from: RuiLoureiro on June 20, 2016, 08:42:08 AM
I want to write a general procedure to identify a general power function
[...]
Do you want to give me any suggestion ?
I would, but I'm not sure about your question. And without knowing the exact data structure it hard to give any help.
You have parsed the expression in some table(s), respecting the operator precedence and associativity for the purpose of evaluation (not necessarily for derivation). For common evaluation (means no variables) you traverse the table from bottom to top, substituting the entries in row
n-1 with results from row
n.
The problem is that you must apply the rules of derivation from top to bottom in table. Furthermore the expression grows due to the rules (e.g.
(f ⋅g)' = f'⋅g + f⋅g'), so you may need to create a new table when applying derivation rules (or the table "grows into the third dimension"). If
f' is some basic function like
sin,
cos,
exp,
... you use tables to look up
f' for particular arguments (does not sound practical because the set of possible arguments is infinite).
Did I get it right so far? You might show a graphic of your data structure.
regards
Quote from: dedndave on June 20, 2016, 10:54:50 AM
you might see if this code works for you...
http://masm32.com/board/index.php?topic=222.0 (http://masm32.com/board/index.php?topic=222.0)
the IntExp_1 download
Hi
Dave How are you , my old friend ?
About the code, i am writing one.
I have a lot of work done.
See this:
[sin(x^2-x)^(2/3) * cos(x-x^2)^(1/2) ]'=
(2/3)
*(2*x-1)
*cos(x^2-x)
*sin(x^2-x)^(-1/3)
*cos(x-x^2)^(1/2)
-(1/2)
*(1-2*x)
*sin(x-x^2)
*cos(x-x^2)^(-1/2)
*sin(x^2-x)^(2/3)
I am using The calculator - calcula67.
Thank you so much, Dave :t
Quote from: qWord on June 21, 2016, 09:40:27 AM
Quote from: RuiLoureiro on June 20, 2016, 08:42:08 AM
I want to write a general procedure to identify a general power function
[...]
Do you want to give me any suggestion ?
I would, but I'm not sure about your question. And without knowing the exact data structure it hard to give any help.
You have parsed the expression in some table(s), respecting the operator precedence and associativity for the purpose of evaluation (not necessarily for derivation). For common evaluation (means no variables) you traverse the table from bottom to top, substituting the entries in row n-1 with results from row n.
The problem is that you must apply the rules of derivation from top to bottom in table. Furthermore the expression grows due to the rules (e.g. (f ⋅g)' = f'⋅g + f⋅g'), so you may need to create a new table when applying derivation rules (or the table "grows into the third dimension"). If f' is some basic function like sin, cos, exp, ... you use tables to look up f' for particular arguments (does not sound practical because the set of possible arguments is infinite).
Did I get it right so far? You might show a graphic of your data structure.
regards
Hi qWord
Thanks for your reply
No, i dont do what you are saying.
To solve this: [1-x^2+e^-x]' i dont use any table. The general procedure do this.
To solve this: [
x-x^2 - sin(x^2-x)]' i solve [sin(x^2-x)]' and the result is saved in one table for 20
functions (i defined 20 but it can be 50 or 100 or ...). The last step is to solve [
x-x^2]'
The result is added with the result that i get from the table.
To solve [sin(x^2-x)]' i get cos(Y) from a derivative table and i replace Y by x^2-x to get cos(x^2-x).
I call the general procedure that solves [x^2-x]' and then i multiply (2*x-1) by cos(x^2-x). This result
is saved in the derivative table.
The derivative table for simple functions is something like this:
Function Number Function Name Derivative1 ln(x) 1/x
...
5 cos(x) -sin(x)
...
N arctan(x) ...
When we find '
cos(???)' we get the function number
5. If the function number is
5 the derivative is
-sin(???)
Thanks :t
regards
OK, so you do parsing on the fly and not as separate step (that was my thought).
regards
Quote from: qWord on June 23, 2016, 06:32:31 AM
OK, so you do parsing on the fly and not as separate step (that was my thought).
regards
Ok :t
Thanks
Hi all
Square Brackets or Not Brackets ?In the last calcula67 that i post in this forum we get things like this
input box=> [-x*cos(2.6*x^2-1)
-2*sin(x)
+5*cos(x^2-x)
-sin(x)*cos(x)]'
solution box1=>
-[cos(2.6*x^2-1)-5.2*x*sin(2.6*x^2-1)*x
] -2*cos(x)
-5*(2*x-1)*sin(x^2-x)
-[cos(x)*cos(x)-sin(x)*sin(x)
]In the new version calcula68 v2016.01 we get
input box=> [-x*cos(2.6*x^2-1)
-2*sin(x)
+5*cos(x^2-x)
-sin(x)*cos(x)]'
solution box1=> -cos(2.6*x^2-1)+5.2*
x*sin(2.6*x^2-1)
*x -2*cos(x)
-5*(2*x-1)*sin(x^2-x)
-cos(x)*cos(x)+sin(x)*sin(x)
OR
input box=> [-x*tan(2.6*x^2-1)]'
solution box1=> -tan(2.6*x^2-1)-5.2*
x*sec(2.6*x^2-1)^2
*xNow, we have not square brackets. What do you say ?
What is better ?
Thanks :t
Quote from: qWord on June 19, 2016, 02:57:16 AM
Hi,
not so easy to find some expression, because nested functions (e.g. sin(cos(x))) are currently not implemented.
One thing I found is that
(-2^x)'
fails with
(-2^x)' = -2 ↯
Alos to note that editing expression is very unintuitive, because the DEL key clears the control rather than removing just one character (definitively breaks usability).
regards
Hi
qWord,
In the new version v2016.01 -
calcula68 - the calculator do
derivatives of nested functions.
Today i got the first result:
input box=> [
sin(
x-
cos( ln(x-1)-
x )+
x^2 )]'
solution box1=>
cos(x-cos
( ln(x-1)-x
)+x^2 )
*[1-(-
sin(ln(x-1)-x)
*(1/(x-1)-
1)
)+
2*x ]Is there any problem ?
I will post the new versiom as soon as possible.
regards :t
Fantastic Rui!!
Perhaps you can replace in the final string "-(-sin" by "+(sin"
Quote from: HSE on July 05, 2016, 09:12:50 AM
Fantastic Rui!!
Perhaps you can replace in the final string "-(-sin" by "+(sin"
Many thanks
HSE :t
Here it is:
input box=> [
sin(
x-cos(
ln(x-1)
-x)+
x^2)]'
solution box1=>
cos(x-cos(ln(x-1)-x)+x^2)
*[1+sin(ln(x-1)-x)
*(
1/(x-1)-1)
+2*x]But it gave me a lot of work but it is better, we have not brackets !
Thanks
Are you working in your calculator ? Is it running ?
:icon14:
Perfect now :t
My derivator sleep. Last week I open the file but close without changes, and begin to make little modifications in HJWasm to see how many macro levels I'm using, and that take me some time. The connection is: I think to obtain derivatives through object recursion, and that perhaps increase macro levels very fast.
Quote from: HSE on July 06, 2016, 10:02:34 AM
Perfect now :t
My derivator sleep. Last week I open the file but close without changes, and begin to make little modifications in HJWasm to see how many macro levels I'm using, and that take me some time. The connection is: I think to obtain derivatives through object recursion, and that perhaps increase macro levels very fast.
Ok, good work
Now i have the solutions for 2 nested functions: one inside onother
input box=> [
sin(x^2-
cos(x^2-x))]'
solution box1=>
cos(x^2-cos(x^2-x))*[2*x+
sin(x^2-x)*(2*x-1)]
It starts to work !
Thanks :t
Hi all
Here we have the solutions for the sum of 6 nested functions:
Six functions and six derivatives in the format type I
input box=> [arcsin(cos(x^2-x)+x^2-x+1)
+arcsin(x^2-cos(x^2-x))
+arcsin(x^2-cos(x^2-x)+x^2-x+1)
+arccos(cos(x^2-x)+x^2-x+1)
+arccos(x^2-cos(x^2-x))
+arccos(x^2-cos(x^2-x)+x^2-x+1)]'
solution box1=> 1/sqr(1-(cos(x^2-x)+x^2-x+1)^2)*[-sin(x^2-x)*(2*x-1)+2*x-1]
+1/sqr(1-(x^2-cos(x^2-x))^2)*[2*x+sin(x^2-x)*(2*x-1)]
+1/sqr(1-(x^2-cos(x^2-x)+x^2-x+1)^2)*[2*x+sin(x^2-x)*(2*x-1)+2*x-1]
-1/sqr(1-(cos(x^2-x)+x^2-x+1)^2)*[-sin(x^2-x)*(2*x-1)+2*x-1]
-1/sqr(1-(x^2-cos(x^2-x))^2)*[2*x+sin(x^2-x)*(2*x-1)]
-1/sqr(1-(x^2-cos(x^2-x)+x^2-x+1)^2)*[2*x+sin(x^2-x)*(2*x-1)+2*x-1]
The same Six functions and six derivatives in the format type II
input box=> [arcsin(cos(x^2-x)+x^2-x+1)
+arcsin(x^2-cos(x^2-x))
+arcsin(x^2-cos(x^2-x)+x^2-x+1)
+arccos(cos(x^2-x)+x^2-x+1)
+arccos(x^2-cos(x^2-x))
+arccos(x^2-cos(x^2-x)+x^2-x+1)]'
solution box1=> [-sin(x^2-x)*(2*x-1)+2*x-1]/sqr(1-(cos(x^2-x)+x^2-x+1)^2)
+[2*x+sin(x^2-x)*(2*x-1)]/sqr(1-(x^2-cos(x^2-x))^2)
+[2*x+sin(x^2-x)*(2*x-1)+2*x-1]/sqr(1-(x^2-cos(x^2-x)+x^2-x+1)^2)
+[sin(x^2-x)*(2*x-1)-2*x+1]/sqr(1-(cos(x^2-x)+x^2-x+1)^2)
+[-2*x-sin(x^2-x)*(2*x-1)]/sqr(1-(x^2-cos(x^2-x))^2)
+[-2*x-sin(x^2-x)*(2*x-1)-2*x+1]/sqr(1-(x^2-cos(x^2-x)+x^2-x+1)^2)
Do you like the solutions ?
Good luck ;)
JUL 2016
More results:
input box=> [arctan(cos(x^2-x)+x^2-x+1)
+arctan(x^2-cos(x^2-x))
+arctan(x^2-cos(x^2-x)+x^2-x+1)
+arccot(cos(x^2-x)+x^2-x+1)
+arccot(x^2-cos(x^2-x))
+arccot(x^2-cos(x^2-x)+x^2-x+1)]'
solution box1=> [-sin(x^2-x)*(2*x-1)+2*x-1]/(1+(cos(x^2-x)+x^2-x+1)^2)
+[2*x+sin(x^2-x)*(2*x-1)]/(1+(x^2-cos(x^2-x))^2)
+[2*x+sin(x^2-x)*(2*x-1)+2*x-1]/(1+(x^2-cos(x^2-x)+x^2-x+1)^2)
+[sin(x^2-x)*(2*x-1)-2*x+1]/(1+(cos(x^2-x)+x^2-x+1)^2)
+[-2*x-sin(x^2-x)*(2*x-1)]/(1+(x^2-cos(x^2-x))^2)
+[-2*x-sin(x^2-x)*(2*x-1)-2*x+1]/(1+(x^2-cos(x^2-x)+x^2-x+1)^2)
input box=> [arcsec(cos(x^2-x)+x^2-x+1)
+arcsec(x^2-cos(x^2-x))
+arcsec(x^2-cos(x^2-x)+x^2-x+1)
+arccsc(cos(x^2-x)+x^2-x+1)
+arccsc(x^2-cos(x^2-x))
+arccsc(x^2-cos(x^2-x)+x^2-x+1)]'
solution box1=> [-sin(x^2-x)*(2*x-1)+2*x-1]/(abs(os(x^2-x)+x^2-x+1)*sqr((cos(x^2-x)+x^2-x+1)^2-1))
+[2*x+sin(x^2-x)*(2*x-1)]/(abs(x^2-cos(x^2-x))*sqr((x^2-cos(x^2-x))^2-1))
+[2*x+sin(x^2-x)*(2*x-1)+2*x-1]/(abs(x^2-cos(x^2-x)+x^2-x+1)*sqr((x^2-cos(x^2-x)+x^2-x+1)^2-1))
+[sin(x^2-x)*(2*x-1)-2*x+1]/(abs(os(x^2-x)+x^2-x+1)*sqr((cos(x^2-x)+x^2-x+1)^2-1))
+[-2*x-sin(x^2-x)*(2*x-1)]/(abs(x^2-cos(x^2-x))*sqr((x^2-cos(x^2-x))^2-1))
+[-2*x-sin(x^2-x)*(2*x-1)-2*x+1]/(abs(x^2-cos(x^2-x)+x^2-x+1)*sqr((x^2-cos(x^2-x)+x^2-x+1)^2-1))
input box=> [arcsinh(cos(x^2-x)+x^2-x+1)
+arcsinh(x^2-cos(x^2-x))
+arcsinh(x^2-cos(x^2-x)+x^2-x+1)
+arccosh(cos(x^2-x)+x^2-x+1)
+arccosh(x^2-cos(x^2-x))
+arccosh(x^2-cos(x^2-x)+x^2-x+1)]'
solution box1=> [-sin(x^2-x)*(2*x-1)+2*x-1]/sqr((cos(x^2-x)+x^2-x+1)^2+1)
+[2*x+sin(x^2-x)*(2*x-1)]/sqr((x^2-cos(x^2-x))^2+1)
+[2*x+sin(x^2-x)*(2*x-1)+2*x-1]/sqr((x^2-cos(x^2-x)+x^2-x+1)^2+1)
+[sin(x^2-x)*(2*x-1)-2*x+1]/sqr((cos(x^2-x)+x^2-x+1)^2-1)
+[-2*x-sin(x^2-x)*(2*x-1)]/sqr((x^2-cos(x^2-x))^2-1)
+[-2*x-sin(x^2-x)*(2*x-1)-2*x+1]/sqr((x^2-cos(x^2-x)+x^2-x+1)^2-1)
input box=> [arcsech(cos(x^2-x)+x^2-x+1)
+arcsech(x^2-cos(x^2-x))
+arcsech(x^2-cos(x^2-x)+x^2-x+1)
+arccsch(cos(x^2-x)+x^2-x+1)
+arccsch(x^2-cos(x^2-x))
+arccsch(x^2-cos(x^2-x)+x^2-x+1)]'
solution box1=> [sin(x^2-x)*(2*x-1)-2*x+1]/(cos(x^2-x)+x^2-x+sqr(1-(cos(x^2-x)+x^2-x+1)^2))
+[-2*x-sin(x^2-x)*(2*x-1)]/((x^2-cos(x^2-x))*sqr(1-(x^2-cos(x^2-x))^2))
+[-2*x-sin(x^2-x)*(2*x-1)-2*x+1]/((x^2-cos(x^2-x)+x^2-x+1)*sqr(1-(x^2-cos(x^2-x)+x^2-x+1)^2))
+[sin(x^2-x)*(2*x-1)-2*x+1]/(abs(os(x^2-x)+x^2-x+1)*sqr(1+(cos(x^2-x)+x^2-x+1)^2))
+[-2*x-sin(x^2-x)*(2*x-1)]/(abs(x^2-cos(x^2-x))*sqr(1+(x^2-cos(x^2-x))^2))
+[-2*x-sin(x^2-x)*(2*x-1)-2*x+1]/(abs(x^2-cos(x^2-x)+x^2-x+1)*sqr(1+(x^2-cos(x^2-x)+x^2-x+1)^2))
input box=> [arctanh(cos(x^2-x)+x^2-x+1)
+arctanh(x^2-cos(x^2-x))
+arctanh(x^2-cos(x^2-x)+x^2-x+1)
+arccoth(cos(x^2-x)+x^2-x+1)
+arccoth(x^2-cos(x^2-x))
+arccoth(x^2-cos(x^2-x)+x^2-x+1)]'
solution box1=> [-sin(x^2-x)*(2*x-1)+2*x-1]/(1-(cos(x^2-x)+x^2-x+1)^2)
+[2*x+sin(x^2-x)*(2*x-1)]/(1-(x^2-cos(x^2-x))^2)
+[2*x+sin(x^2-x)*(2*x-1)+2*x-1]/(1-(x^2-cos(x^2-x)+x^2-x+1)^2)
+[sin(x^2-x)*(2*x-1)-2*x+1]/(1+(cos(x^2-x)+x^2-x+1)^2)
+[-2*x-sin(x^2-x)*(2*x-1)]/(1+(x^2-cos(x^2-x))^2)
+[-2*x-sin(x^2-x)*(2*x-1)-2*x+1]/(1+(x^2-cos(x^2-x)+x^2-x+1)^2)
input box=> [arcsech(cos(x^2-x)+x^2-x+1)
+arcsech(x^2-cos(x^2-x))
+arcsech(x^2-cos(x^2-x)+x^2-x+1)
+arccsch(cos(x^2-x)+x^2-x+1)
+arccsch(x^2-cos(x^2-x))
+arccsch(x^2-cos(x^2-x)+x^2-x+1)]'
solution box1=> [sin(x^2-x)*(2*x-1)-2*x+1]/(cos(x^2-x)+x^2-x+sqr(1-(cos(x^2-x)+x^2-x+1)^2))
+[-2*x-sin(x^2-x)*(2*x-1)]/((x^2-cos(x^2-x))*sqr(1-(x^2-cos(x^2-x))^2))
+[-2*x-sin(x^2-x)*(2*x-1)-2*x+1]/((x^2-cos(x^2-x)+x^2-x+1)*sqr(1-(x^2-cos(x^2-x)+x^2-x+1)^2))
+[sin(x^2-x)*(2*x-1)-2*x+1]/(abs(os(x^2-x)+x^2-x+1)*sqr(1+(cos(x^2-x)+x^2-x+1)^2))
+[-2*x-sin(x^2-x)*(2*x-1)]/(abs(x^2-cos(x^2-x))*sqr(1+(x^2-cos(x^2-x))^2))
+[-2*x-sin(x^2-x)*(2*x-1)-2*x+1]/(abs(x^2-cos(x^2-x)+x^2-x+1)*sqr(1+(x^2-cos(x^2-x)+x^2-x+1)^2))
Olá, Rui!
Quero parabenizá-lo pela vitória de Portugal no Campeonato Europeu de Futebol!(http://arcanumclub.ru/smiles/smile136.gif)(http://www.gifki.org/data/media/764/flag-frantsii-animatsionnaya-kartinka-0024.gif)
(http://arcanumclub.ru/smiles/smile210.gif)
Congrats from Italy :t
Quote from: Mikl__ on July 11, 2016, 10:48:34 AM
Olá, Rui!
Quero parabenizá-lo pela vitória de Portugal no Campeonato Europeu de Futebol!(http://arcanumclub.ru/smiles/smile136.gif)(http://www.gifki.org/data/media/764/flag-frantsii-animatsionnaya-kartinka-0024.gif)
(http://arcanumclub.ru/smiles/smile210.gif)
:biggrin:
Olá meu amigo Mikl__ !
Muito obrigado
Por acaso gosto de futebol e também eu joguei futebol
Obrigado :t
:icon14:
Quote from: jj2007 on July 11, 2016, 11:15:48 AM
Congrats from Italy :t
Thank you so much, my old friend Jochen !
I like your Italy team too :t
:icon14:
Hi
Here more interesting results:
input box=> [exp(ln(sin(x)))
+sin(cos(-tan(x)))
+sin(-cos(tan(x)))
+sin(-cos(-tan(x)))
-sin(-cos(-tan(x)))
+exp(-ln(sin(x)))]'
solution box1=> exp(ln(sin(x)))/(sin(x))*cos(x)
-cos(cos(-tan(x)))*sin(-tan(x))*(-sec(x)^2)
+cos(-cos(tan(x)))*sin(tan(x))*sec(x)^2
+cos(-cos(-tan(x)))*sin(-tan(x))*(-sec(x)^2)
-cos(-cos(-tan(x)))*sin(-tan(x))*(-sec(x)^2)
-exp(-ln(sin(x)))/(sin(x))*cos(x)
Good luck :t
Hi
Functions and derivatives solutions TYPE I
input box=> [exp(ln(sin(x)))
+sin(cos(-tan(x)))
+sin(-cos(tan(x)))
+sin(-cos(-tan(x)))
-sin(-cos(-tan(x)))
+exp(-ln(sin(x)))]'
solution box1=> exp(ln(sin(x)))*cos(x)/sin(x)
-cos(cos(-tan(x)))*sin(-tan(x))*(-sec(x)^2)
+cos(-cos(tan(x)))*sin(tan(x))*sec(x)^2
+cos(-cos(-tan(x)))*sin(-tan(x))*(-sec(x)^2)
-cos(-cos(-tan(x)))*sin(-tan(x))*(-sec(x)^2)
-exp(-ln(sin(x)))*cos(x)/sin(x)
The same Functions and derivatives better solutions TYPE II
input box=> [exp(ln(sin(x)))
+sin(cos(-tan(x)))
+sin(-cos(tan(x)))
+sin(-cos(-tan(x)))
-sin(-cos(-tan(x)))
+exp(-ln(sin(x)))]'
solution box1=> exp(ln(sin(x)))*cos(x)/sin(x)
+cos(cos(-tan(x)))*sin(-tan(x))*sec(x)^2
+cos(-cos(tan(x)))*sin(tan(x))*sec(x)^2
-cos(-cos(-tan(x)))*sin(-tan(x))*sec(x)^2
+cos(-cos(-tan(x)))*sin(-tan(x))*sec(x)^2
-exp(-ln(sin(x)))*cos(x)/sin(x)
Good luck ! :t
Impressive Rui :t
When TYPE III ? :biggrin:
solution box1=> cos(x)/sin(x)* ( exp(ln(sin(x)))-exp(-ln(sin(x))) )+
sec(x)^2*(
+cos(-cos(tan(x)))*sin(tan(x))
+sin(-tan(x))*( cos(cos(-tan(x)))-cos(-cos(-tan(x)))+cos(-cos(-tan(x))) )
)
Quote from: HSE on July 13, 2016, 05:51:13 AM
Impressive Rui :t
When TYPE III ? :biggrin:
solution box1=> cos(x)/sin(x)* ( exp(ln(sin(x)))-exp(-ln(sin(x))) )+
sec(x)^2*(
+cos(-cos(tan(x)))*sin(tan(x))
+sin(-tan(x))*( cos(cos(-tan(x)))-cos(-cos(-tan(x)))+cos(-cos(-tan(x))) )
)
Hi,
Many thanks
HSE ! :t
Yes i could give the TYPE III solutions without any problems
It will be [FncX(ArgX]' = [ArgX]' * FncX' = ...
But i do
[FncX(ArgX)]' = FncX' * [ArgX]' = ...
Example: [exp(ln(sin(x)))]' = [ln(sin(x))]' * exp(ln(sin(x)))
= [sin(x)]'/ sin(x) * exp(ln(sin(x)))
= cos(x)/sin(x) * exp(ln(sin(x)))
In the following examples, when we get the cases FncZ
/ FncY
/ FncX at the end
they are modified to FncZ
/ (FncY
* FncX ).
The example is this:
-exp(ln(sin(1/x)))*cos(1/x)
/ (x^2
*sin(1/x))
Here we have functions of
X,
-X,
1/X and
-1/X Quote
input box=> [exp(ln(sin(x)))
+sin(cos(-tan(x)))
+sin(-cos(tan(x)))
+sin(-cos(-tan(x)))
-sin(-cos(-tan(x)))
+exp(-ln(sin(x)))]'
solution box1=> exp(ln(sin(x)))*cos(x)/sin(x)
+cos(cos(-tan(x)))*sin(-tan(x))*sec(x)^2
+cos(-cos(tan(x)))*sin(tan(x))*sec(x)^2
-cos(-cos(-tan(x)))*sin(-tan(x))*sec(x)^2
+cos(-cos(-tan(x)))*sin(-tan(x))*sec(x)^2
-exp(-ln(sin(x)))*cos(x)/sin(x)
input box=> [exp(ln(sin(-x)))
+sin(cos(-tan(-x)))
+sin(-cos(tan(-x)))
+sin(-cos(-tan(-x)))
-sin(-cos(-tan(-x)))
+exp(-ln(sin(-x)))]'
solution box1=> -exp(ln(sin(-x)))*cos(-x)/sin(-x)
-cos(cos(-tan(-x)))*sin(-tan(-x))*sec(-x)^2
-cos(-cos(tan(-x)))*sin(tan(-x))*sec(-x)^2
+cos(-cos(-tan(-x)))*sin(-tan(-x))*sec(-x)^2
-cos(-cos(-tan(-x)))*sin(-tan(-x))*sec(-x)^2
+exp(-ln(sin(-x)))*cos(-x)/sin(-x)
input box=> [exp(ln(sin(1/x)))
+sin(cos(-tan(1/x)))
+sin(-cos(tan(1/x)))
+sin(-cos(-tan(1/x)))
-sin(-cos(-tan(1/x)))
+exp(-ln(sin(1/x)))]'
solution box1=> -exp(ln(sin(1/x)))*cos(1/x)/(x^2*sin(1/x))
-cos(cos(-tan(1/x)))*sin(-tan(1/x))*sec(1/x)^2/x^2
-cos(-cos(tan(1/x)))*sin(tan(1/x))*sec(1/x)^2/x^2
+cos(-cos(-tan(1/x)))*sin(-tan(1/x))*sec(1/x)^2/x^2
-cos(-cos(-tan(1/x)))*sin(-tan(1/x))*sec(1/x)^2/x^2
+exp(-ln(sin(1/x)))*cos(1/x)/(x^2*sin(1/x))
input box=> [exp(ln(sin(-1/x)))
+sin(cos(-tan(-1/x)))
+sin(-cos(tan(-1/x)))
+sin(-cos(-tan(-1/x)))
-sin(-cos(-tan(-1/x)))
+exp(-ln(sin(-1/x)))]'
solution box1=> exp(ln(sin(-1/x)))*cos(-1/x)/(x^2*sin(-1/x))
+cos(cos(-tan(-1/x)))*sin(-tan(-1/x))*sec(-1/x)^2/x^2
+cos(-cos(tan(-1/x)))*sin(tan(-1/x))*sec(-1/x)^2/x^2
-cos(-cos(-tan(-1/x)))*sin(-tan(-1/x))*sec(-1/x)^2/x^2
+cos(-cos(-tan(-1/x)))*sin(-tan(-1/x))*sec(-1/x)^2/x^2
-exp(-ln(sin(-1/x)))*cos(-1/x)/(x^2*sin(-1/x))
Good luck ! :t
:icon14:
NOTE: In reply #146 we have other solutions because
Z/Y*X is equivalent to
Z*X/Y and
Z / Y / X is equivalent to
Z/ ( Y * X )
Hi
Here we have more solutions: all positive and all negative
Quote
input box=> [exp(ln(sin(x)))+exp(ln(sin(-x)))+exp(ln(sin(1/x)))+exp(ln(sin(-1/x)))
+exp(ln(-sin(x)))+exp(ln(-sin(-x)))+exp(ln(-sin(1/x)))+exp(ln(-sin(-1/x)))
+exp(-ln(sin(x)))+exp(-ln(sin(-x)))+exp(-ln(sin(1/x)))+exp(-ln(sin(-1/x)))
+exp(-ln(-sin(x)))+exp(-ln(-sin(-x)))+exp(-ln(-sin(1/x)))+exp(-ln(-sin(-1/x)))]'
solution box1=> exp(ln(sin(x)))*cos(x)/sin(x)
-exp(ln(sin(-x)))*cos(-x)/sin(-x)
-exp(ln(sin(1/x)))*cos(1/x)/(x^2*sin(1/x))
+exp(ln(sin(-1/x)))*cos(-1/x)/(x^2*sin(-1/x))
+exp(ln(-sin(x)))*cos(x)/sin(x)
-exp(ln(-sin(-x)))*cos(-x)/sin(-x)
-exp(ln(-sin(1/x)))*cos(1/x)/(x^2*sin(1/x))
+exp(ln(-sin(-1/x)))*cos(-1/x)/(x^2*sin(-1/x))
-exp(-ln(sin(x)))*cos(x)/sin(x)
+exp(-ln(sin(-x)))*cos(-x)/sin(-x)
+exp(-ln(sin(1/x)))*cos(1/x)/(x^2*sin(1/x))
-exp(-ln(sin(-1/x)))*cos(-1/x)/(x^2*sin(-1/x))
-exp(-ln(-sin(x)))*cos(x)/sin(x)
+exp(-ln(-sin(-x)))*cos(-x)/sin(-x)
+exp(-ln(-sin(1/x)))*cos(1/x)/(x^2*sin(1/x))
-exp(-ln(-sin(-1/x)))*cos(-1/x)/(x^2*sin(-1/x))
Quote
input box=> [-exp(ln(sin(x)))-exp(ln(sin(-x)))-exp(ln(sin(1/x)))-exp(ln(sin(-1/x)))
-exp(ln(-sin(x)))-exp(ln(-sin(-x)))-exp(ln(-sin(1/x)))-exp(ln(-sin(-1/x)))
-exp(-ln(sin(x)))-exp(-ln(sin(-x)))-exp(-ln(sin(1/x)))-exp(-ln(sin(-1/x)))
-exp(-ln(-sin(x)))-exp(-ln(-sin(-x)))-exp(-ln(-sin(1/x)))-exp(-ln(-sin(-1/x)))]'
solution box1=> -exp(ln(sin(x)))*cos(x)/sin(x)
+exp(ln(sin(-x)))*cos(-x)/sin(-x)
+exp(ln(sin(1/x)))*cos(1/x)/(x^2*sin(1/x))
-exp(ln(sin(-1/x)))*cos(-1/x)/(x^2*sin(-1/x))
-exp(ln(-sin(x)))*cos(x)/sin(x)
+exp(ln(-sin(-x)))*cos(-x)/sin(-x)
+exp(ln(-sin(1/x)))*cos(1/x)/(x^2*sin(1/x))
-exp(ln(-sin(-1/x)))*cos(-1/x)/(x^2*sin(-1/x))
+exp(-ln(sin(x)))*cos(x)/sin(x)
-exp(-ln(sin(-x)))*cos(-x)/sin(-x)
-exp(-ln(sin(1/x)))*cos(1/x)/(x^2*sin(1/x))
+exp(-ln(sin(-1/x)))*cos(-1/x)/(x^2*sin(-1/x))
+exp(-ln(-sin(x)))*cos(x)/sin(x)
-exp(-ln(-sin(-x)))*cos(-x)/sin(-x)
-exp(-ln(-sin(1/x)))*cos(1/x)/(x^2*sin(1/x))
+exp(-ln(-sin(-1/x)))*cos(-1/x)/(x^2*sin(-1/x))
Good luck ! :t
EDIT:
Here we have the results of the last test with 12 functions
[-arcsin(cos(x^2-x)+x^2-x+1)
-arcsin(x^2-cos(x^2-x))
-arcsin(x^2-cos(x^2-x)+x^2-x+1)
-arccos(cos(x^2-x)+x^2-x+1)
-arccos(x^2-cos(x^2-x))
-arccos(x^2-cos(x^2-x)+x^2-x+1)
-arctan(cos(x^2-x)+x^2-x+1)
-arctan(x^2-cos(x^2-x))
-arctan(x^2-cos(x^2-x)+x^2-x+1)
-arccot(cos(x^2-x)+x^2-x+1)
-arccot(x^2-cos(x^2-x))
-arccot(x^2-cos(x^2-x)+x^2-x+1)
-arcsec(cos(x^2-x)+x^2-x+1)
-arcsec(x^2-cos(x^2-x))
-arcsec(x^2-cos(x^2-x)+x^2-x+1)
-arccsc(cos(x^2-x)+x^2-x+1)
-arccsc(x^2-cos(x^2-x))
-arccsc(x^2-cos(x^2-x)+x^2-x+1)
]'
Sunday, 17-07-2016 16:10:05
input box=> [-arcsin(cos(x^2-x)+x^2-x+1)-arcsin(x^2-cos(x^2-x))-arcsin(x^2-cos(x^2-x)+x^2-x+1)-arccos(cos(x^2-x)+x
^2-x+1)-arccos(x^2-cos(x^2-x))-arccos(x^2-cos(x^2-x)+x^2-x+1)-arctan(cos(x^2-x)+x^2-x+1)-arctan(x^2-cos(x^2-x))-ar
ctan(x^2-cos(x^2-x)+x^2-x+1)-arccot(cos(x^2-x)+x^2-x+1)-arccot(x^2-cos(x^2-x))-arccot(x^2-cos(x^2-x)+x^2-x+1)-arcs
ec(cos(x^2-x)+x^2-x+1)-arcsec(x^2-cos(x^2-x))-arcsec(x^2-cos(x^2-x)+x^2-x+1)-arccsc(cos(x^2-x)+x^2-x+1)-arccsc(x^2
-cos(x^2-x))-arccsc(x^2-cos(x^2-x)+x^2-x+1)]'
solution box1=> -[-sin(x^2-x)*(2*x-1)+2*x-1]/sqr(1-(cos(x^2-x)+x^2-x+1)^2)-[2*x-sin(x^2-x)*(2*x-1)]/sqr(1-(x^2-cos
(x^2-x))^2)-[2*x-sin(x^2-x)*(2*x-1)-2*x+1]/sqr(1-(x^2-cos(x^2-x)+x^2-x+1)^2)+[-sin(x^2-x)*(2*x-1)+2*x-1]/sqr(1-(co
s(x^2-x)+x^2-x+1)^2)+[2*x+sin(x^2-x)*(2*x-1)]/sqr(1-(x^2-cos(x^2-x))^2)+[2*x+sin(x^2-x)*(2*x-1)+2*x-1]/sqr(1-(x^2-
cos(x^2-x)+x^2-x+1)^2)-[-sin(x^2-x)*(2*x-1)+2*x-1]/(1+(cos(x^2-x)+x^2-x+1)^2)-[2*x-sin(x^2-x)*(2*x-1)]/(1+(x^2-cos
(x^2-x))^2)-[2*x-sin(x^2-x)*(2*x-1)-2*x+1]/(1+(x^2-cos(x^2-x)+x^2-x+1)^2)+[-sin(x^2-x)*(2*x-1)+2*x-1]/(1+(cos(x^2-
x)+x^2-x+1)^2)+[2*x+sin(x^2-x)*(2*x-1)]/(1+(x^2-cos(x^2-x))^2)+[2*x+sin(x^2-x)*(2*x-1)+2*x-1]/(1+(x^2-cos(x^2-x)+x
^2-x+1)^2)-[-sin(x^2-x)*(2*x-1)+2*x-1]/(abs(os(x^2-x)+x^2-x+1)*sqr((cos(x^2-x)+x^2-x+1)^2-1))-[2*x-sin(x^2-x)*(2*x
-1)]/(abs(x^2-cos(
solution box2=> x^2-x))*sqr((x^2-cos(x^2-x))^2-1))-[2*x-sin(x^2-x)*(2*x-1)-2*x+1]/(abs(x^2-cos(x^2-x)+x^2-x+1)*sqr
((x^2-cos(x^2-x)+x^2-x+1)^2-1))+[-sin(x^2-x)*(2*x-1)+2*x-1]/(abs(os(x^2-x)+x^2-x+1)*sqr((cos(x^2-x)+x^2-x+1)^2-1))
+[2*x+sin(x^2-x)*(2*x-1)]/(abs(x^2-cos(x^2-x))*sqr((x^2-cos(x^2-x))^2-1))+[2*x+sin(x^2-x)*(2*x-1)+2*x-1]/(abs(x^2-
cos(x^2-x)+x^2-x+1)*sqr((x^2-cos(x^2-x)+x^2-x+1)^2-1))
solution box1=> -[-sin(x^2-x)*(2*x-1)+2*x-1]/sqr(1-(cos(x^2-x)+x^2-x+1)^2)
-[2*x-sin(x^2-x)*(2*x-1)]/sqr(1-(x^2-cos(x^2-x))^2)
-[2*x-sin(x^2-x)*(2*x-1)-2*x+1]/sqr(1-(x^2-cos(x^2-x)+x^2-x+1)^2)
+[-sin(x^2-x)*(2*x-1)+2*x-1]/sqr(1-(cos(x^2-x)+x^2-x+1)^2)
+[2*x+sin(x^2-x)*(2*x-1)]/sqr(1-(x^2-cos(x^2-x))^2)
+[2*x+sin(x^2-x)*(2*x-1)+2*x-1]/sqr(1-(x^2-cos(x^2-x)+x^2-x+1)^2)
-[-sin(x^2-x)*(2*x-1)+2*x-1]/(1+(cos(x^2-x)+x^2-x+1)^2)
-[2*x-sin(x^2-x)*(2*x-1)]/(1+(x^2-cos(x^2-x))^2)
-[2*x-sin(x^2-x)*(2*x-1)-2*x+1]/(1+(x^2-cos(x^2-x)+x^2-x+1)^2)
+[-sin(x^2-x)*(2*x-1)+2*x-1]/(1+(cos(x^2-x)+x^2-x+1)^2)
+[2*x+sin(x^2-x)*(2*x-1)]/(1+(x^2-cos(x^2-x))^2)
+[2*x+sin(x^2-x)*(2*x-1)+2*x-1]/(1+(x^2-cos(x^2-x)+x^2-x+1)^2)
-[-sin(x^2-x)*(2*x-1)+2*x-1]/(abs(os(x^2-x)+x^2-x+1)*sqr((cos(x^2-x)+x^2-x+1)^2-1))
-[2*x-sin(x^2-x)*(2*x-1)]/(abs(x^2-cos(x^2-x))*sqr((x^2-cos(x^2-x))^2-1))
-[2*x-sin(x^2-x)*(2*x-1)-2*x+1]/(abs(x^2-cos(x^2-x)+x^2-x+1)*sqr((x^2-cos(x^2-x)+x^2-x+1)^2-1))
+[-sin(x^2-x)*(2*x-1)+2*x-1]/(abs(os(x^2-x)+x^2-x+1)*sqr((cos(x^2-x)+x^2-x+1)^2-1))
+[2*x+sin(x^2-x)*(2*x-1)]/(abs(x^2-cos(x^2-x))*sqr((x^2-cos(x^2-x))^2-1))
+[2*x+sin(x^2-x)*(2*x-1)+2*x-1]/(abs(x^2-cos(x^2-x)+x^2-x+1)*sqr((x^2-cos(x^2-x)+x^2-x+1)^2-1))
Hi
I am working on a procedure to replace the variable x by one expression.
I am improving it.
And i wrote a procedure to test the expression syntax and it is working
correctly till now.
These are 2 results to
replace x by
-sin(x).
Note that the result is simplified:
we have not (-sin(x))^2 but sin(x)^2
we have not abs(-sin(x)) but abs(sin(x))
arcsin(x^2-x+1) is replaced by arcsin(sin(x)^2+sin(x)+1)
Quote
input box=> [x +cos(x) -x^2 -(x)^3 +1/x -2*x -x +abs(x)]'
solution box1=> -sin(x)+cos(-sin(x))-sin(x)^2+sin(x)^3-1/sin(x)+2*sin(x)+sin(x)+abs(sin(x))
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
input box=> [-x +cos(x) +x^2 +(x)^3 -1/x +2*x +x +abs(x) -arcsin(x^2-x+1)-455.23]'
solution box1=> sin(x)+cos(-sin(x))+sin(x)^2-sin(x)^3+1/sin(x)-2*sin(x)-sin(x)+abs(sin(x))-arcsin(sin(x)^2+sin(x)+
1)-455.23
Good luck :t
Another example:
Quote
input box=> [(x^-25) *(x)^(-15) +x +cos(x) -x^2 -(x)^3 +1/x -2*x -x +abs(x)]'
solution box1=> (-sin(x)^-25)*(-sin(x))^(-15)-sin(x)+cos(-sin(x))-sin(x)^2+sin(x)^3-1/sin(x)+2*sin(x)+sin(x)+abs(s
in(x))
Better:
Quote
input box=> [(x^-25) *(x)^(-15)+x +cos(x) -x^2 -(x)^3 +1/x -2*x -x +abs(x)]'
solution box1=> sin(x)^-25*sin(x)^-15-sin(x)+cos(-sin(x))-sin(x)^2+sin(x)^3-1/sin(x)+2*sin(x)+sin(x)+abs(sin(x))
Hi
Here more results replacing x by -sin(x).
The expressions are simplified.
Quote
input box=> [(2+x) -(2-x) +(2*x) -(2/x) -(-10^x) +(-10^+x) -(-10^-x)]'
solution box1=> (2-sin(x)) -(2+sin(x)) -2*sin(x) +2/sin(x) +10^-sin(x) -10^-sin(x) +10^sin(x)
input box=> [+x -x +x^2 +x^3 -x^2 -x^3]'
solution box1=> -sin(x) +sin(x) +sin(x)^2 +sin(x)^3 -sin(x)^2 -sin(x)^3
input box=> [-2/x +cos(x)/x -2/x^2 -2/x^3 +2/x^2 +2/x^3]'
solution box1=> 2/sin(x) -cos(-sin(x))/sin(x) -2/sin(x)^2 +2/sin(x)^3 +2/sin(x)^2 -2/sin(x)^3
input box=> [-2*x +cos(x)*x -2*x^2 -2*x^3 +2*x^2 +2*x^3]'
solution box1=> 2*sin(x) -cos(-sin(x))*sin(x) -2*sin(x)^2 +2*sin(x)^3 +2*sin(x)^2 -2*sin(x)^3
input box=> [-2^x +cos(x)^x -2^x^2 -2^x^3 +2^x^2 +2^x^3]'
solution box1=> -2^-sin(x) +cos(-sin(x))^-sin(x) -2^-sin(x)^2 -2^-sin(x)^3 +2^-sin(x)^2 +2^-sin(x)^3
Good luck :t
Hi all
Here we have some tests on a procedure to replace x by an expression.
It doenst give any syntax error.
Replace x by x*e^x/x^2
input box=> [arcsinh(cos(x^2-x)+x^2-x+1)+arcsinh(x^2-cos(x^2-x))+arcsinh(x^2-cos(x^2-x)+x^2-x+1)+arccosh(cos(x^2-x
)+x^2-x+1)+arccosh(x^2-cos(x^2-x))+arccosh(x^2-cos(x^2-x)+x^2-x+1)+arcsech(cos(x^2-x)+x^2-x+1)+arcsech(x^2-cos(x^2
-x))+arcsech(x^2-cos(x^2-x)+x^2-x+1)+arccsch(cos(x^2-x)+x^2-x+1)+arccsch(x^2-cos(x^2-x))+arccsch(x^2-cos(x^2-x)+x^
2-x+1)+arctanh(cos(x^2-x)+x^2-x+1)+arctanh(x^2-cos(x^2-x))+arctanh(x^2-cos(x^2-x)+x^2-x+1)+arccoth(cos(x^2-x)+x^2-
x+1)+arccoth(x^2-cos(x^2-x))+arccoth(x^2-cos(x^2-x)+x^2-x+1)+arcsech(cos(x^2-x)+x^2-x+1)+arcsech(x^2-cos(x^2-x))+a
rcsech(x^2-cos(x^2-x)+x^2-x+1)+arccsch(cos(x^2-x)+x^2-x+1)+arccsch(x^2-cos(x^2-x))+arccsch(x^2-cos(x^2-x)+x^2-x+1)
]'
solution box1=> arcsinh(cos((x*e^x/x^2)^2-x*e^x/x^2)+(x*e^x/x^2)^2-x*e^x/x^2+1)
+arcsinh(x*e^x/x^2^2-cos(x*e^x/x^2^2-x*e^x/x^2))
+arcsinh(x*e^x/x^2^2-cos(x*e^x/x^2^2-x*e^x/x^2)+(x*e^x/x^2)^2-x*e^x/x^2+1)
+arccosh(cos((x*e^x/x^2)^2-x*e^x/x^2)+(x*e^x/x^2)^2-x*e^x/x^2+1)
+arccosh(x*e^x/x^2^2-cos(x*e^x/x^2^2-x*e^x/x^2))
+arccosh(x*e^x/x^2^2-cos(x*e^x/x^2^2-x*e^x/x^2)+(x*e^x/x^2)^2-x*e^x/x^2+1)
+arcsech(cos((x*e^x/x^2)^2-x*e^x/x^2)+(x*e^x/x^2)^2-x*e^x/x^2+1)
+arcsech(x*e^x/x^2^2-cos(x*e^x/x^2^2-x*e^x/x^2))
+arcsech(x*e^x/x^2^2-cos(x*e^x/x^2^2-x*e^x/x^2)+(x*e^x/x^2)^2-x*e^x/x^2+1)
+arccsch(cos((x*e^x/x^2)^2-x*e^x/x^2)+(x*e^x/x^2)^2-x*e^x/x^2+1)
+arccsch(x*e^x/x^2^2-cos(x*e^x/x^2^2-x*e^x/x^2))
+arccsch(x*e^x/x^2^2-cos(x*e^x/x^2^2-x*e^x/x^2)+(x*e^x/x^2)^2-x*e^x/x^2+1)
+arctanh(cos((x*e^x/x^2)^2-x*e^x/x^2)+(x*e^x/x^2)^2-x*e^x/x^2+1)
+arctanh(x*e^x/x^2^2-cos(x*e^x/x^2^2-x*e^x/x^2))
+arctanh(x*e^x/x^2^2-cos(x*e^x/x^2^2-x*e^x/x^2)+(x*e^x/x^2)^2-x*e^x/x^2+1)
+arccoth(cos((x*e^x/x^2)^2-x*e^x/x^2)+(x*e^x/x^2)^2-x*e^x/x^2+1)
+arccoth(x*e^x/x^2^2-cos(x*e^x/x^2^2-x*e^x/x^2))
+arccoth(x*e^x/x^2^2-cos(x*e^x/x^2^2-x*e^x/x^2)+(x*e^x/x^2)^2-x*e^x/x^2+1)
+arcsech(cos((x*e^x/x^2)^2-x*e^x/x^2)+(x*e^x/x^2)^2-x*e^x/x^2+1)
+arcsech(x*e^x/x^2^2-cos(x*e^x/x^2^2-x*e^x/x^2))
+arcsech(x*e^x/x^2^2-cos(x*e^x/x^2^2-x*e^x/x^2)+(x*e^x/x^2)^2-x*e^x/x^2+1)
+arccsch(cos((x*e^x/x^2)^2-x*e^x/x^2)+(x*e^x/x^2)^2-x*e^x/x^2+1)
+arccsch(x*e^x/x^2^2-cos(x*e^x/x^2^2-x*e^x/x^2))
+arccsch(x*e^x/x^2^2-cos(x*e^x/x^2^2-x*e^x/x^2)+(x*e^x/x^2)^2-x*e^x/x^2+1)
Replace x by -x
Quote
input box=> [arcsinh(cos(x^2-x)+x^2-x+1)+arcsinh(x^2-cos(x^2-x))+arcsinh(x^2-cos(x^2-x)+x^2-x+1)+arccosh(cos(x^2-x
)+x^2-x+1)+arccosh(x^2-cos(x^2-x))+arccosh(x^2-cos(x^2-x)+x^2-x+1)+arcsech(cos(x^2-x)+x^2-x+1)+arcsech(x^2-cos(x^2
-x))+arcsech(x^2-cos(x^2-x)+x^2-x+1)+arccsch(cos(x^2-x)+x^2-x+1)+arccsch(x^2-cos(x^2-x))+arccsch(x^2-cos(x^2-x)+x^
2-x+1)+arctanh(cos(x^2-x)+x^2-x+1)+arctanh(x^2-cos(x^2-x))+arctanh(x^2-cos(x^2-x)+x^2-x+1)+arccoth(cos(x^2-x)+x^2-
x+1)+arccoth(x^2-cos(x^2-x))+arccoth(x^2-cos(x^2-x)+x^2-x+1)+arcsech(cos(x^2-x)+x^2-x+1)+arcsech(x^2-cos(x^2-x))+a
rcsech(x^2-cos(x^2-x)+x^2-x+1)+arccsch(cos(x^2-x)+x^2-x+1)+arccsch(x^2-cos(x^2-x))+arccsch(x^2-cos(x^2-x)+x^2-x+1)
]'
solution box1=> arcsinh(cos(x^2+x)+x^2+x+1)
+arcsinh(x^2-cos(x^2+x))
+arcsinh(x^2-cos(x^2+x)+x^2+x+1)
+arccosh(cos(x^2+x)+x^2+x+1)
+arccosh(x^2-cos(x^2+x))
+arccosh(x^2-cos(x^2+x)+x^2+x+1)
+arcsech(cos(x^2+x)+x^2+x+1)
+arcsech(x^2-cos(x^2+x))
+arcsech(x^2-cos(x^2+x)+x^2+x+1)
+arccsch(cos(x^2+x)+x^2+x+1)
+arccsch(x^2-cos(x^2+x))
+arccsch(x^2-cos(x^2+x)+x^2+x+1)
+arctanh(cos(x^2+x)+x^2+x+1)
+arctanh(x^2-cos(x^2+x))
+arctanh(x^2-cos(x^2+x)+x^2+x+1)
+arccoth(cos(x^2+x)+x^2+x+1)
+arccoth(x^2-cos(x^2+x))
+arccoth(x^2-cos(x^2+x)+x^2+x+1)
+arcsech(cos(x^2+x)+x^2+x+1)
+arcsech(x^2-cos(x^2+x))
+arcsech(x^2-cos(x^2+x)+x^2+x+1)
+arccsch(cos(x^2+x)+x^2+x+1)
+arccsch(x^2-cos(x^2+x))
+arccsch(x^2-cos(x^2+x)+x^2+x+1)
Good luck :t
Hi
This is a test.
The proc replace x by sin(x), next by x/e^x, next by x-2
next by -sin(x), by -x/e^x, by -x+2
Quote
input box=> [ -x^3 -x^2 +x^3 +x^2]'
solution box1=> -sin(x)^3 -sin(x)^2 +sin(x)^3 +sin(x)^2
-(x/e^x)^3 -(x/e^x)^2 +(x/e^x)^3 +(x/e^x)^2
-(x-2)^3 -(x-2)^2 +(x-2)^3 +(x-2)^2
+sin(x)^3 -sin(x)^2 -sin(x)^3 +sin(x)^2
+(x/e^x)^3 -(x/e^x)^2 -(x/e^x)^3 +(x/e^x)^2
-(-x+2)^3 -(-x+2)^2 +(-x+2)^3 +(-x+2)^2
This is another test
Quote
input box=> [arcsin(cos(x^2-x)+x^2-x+1)
+arcsin(x^2-cos(x^2-x))
+arcsin(x^2-cos(x^2-x)+x^2-x+1)
+arccos(cos(x^2-x)+x^2-x+1)
+arccos(x^2-cos(x^2-x))
+arccos(x^2-cos(x^2-x)+x^2-x+1)]'
Replace x by -sin(x)
solution box1=> arcsin(cos(sin(x)^2+sin(x))+sin(x)^2+sin(x)+1)
+arcsin(sin(x)^2-cos(sin(x)^2+sin(x)))
+arcsin(sin(x)^2-cos(sin(x)^2+sin(x))+sin(x)^2+sin(x)+1)
+arccos(cos(sin(x)^2+sin(x))+sin(x)^2+sin(x)+1)
+arccos(sin(x)^2-cos(sin(x)^2+sin(x)))
+arccos(sin(x)^2-cos(sin(x)^2+sin(x))+sin(x)^2+sin(x)+1)
Replace x by x*e^x/x^2
+arcsin(cos((x*e^x/x^2)^2-x*e^x/x^2)+(x*e^x/x^2)^2-x*e^x/x^2+1)
+arcsin((x*e^x/x^2)^2-cos((x*e^x/x^2)^2-x*e^x/x^2))
+arcsin((x*e^x/x^2)^2-cos((x*e^x/x^2)^2-x*e^x/x^2)+(x*e^x/x^2)^2-x*e^x/x^2+1)
+arccos(cos((x*e^x/x^2)^2-x*e^x/x^2)+(x*e^x/x^2)^2-x*e^x/x^2+1)
+arccos((x*e^x/x^2)^2-cos((x*e^x/x^2)^2-x*e^x/x^2))
+arccos((x*e^x/x^2)^2-cos((x*e^x/x^2)^2-x*e^x/x^2)+(x*e^x/x^2)^2-x*e^x/x^2+1)
Replace x by -x^2+1/x
+arcsin(cos((-x^2+1/x)^2+x^2-1/x)+(-x^2+1/x)^2+x^2-1/x+1)
+arcsin((-x^2+1/x)^2-cos((-x^2+1/x)^2+x^2-1/x))
+arcsin((-x^2+1/x)^2-cos((-x^2+1/x)^2+x^2-1/x)+(-x^2+1/x)^2+x^2-1/x+1)
+arccos(cos((-x^2+1/x)^2+x^2-1/x)+(-x^2+1/x)^2+x^2-1/x+1)
+arccos((-x^2+1/x)^2-cos((-x^2+1/x)^2+x^2-1/x))
+arccos((-x^2+1/x)^2-cos((-x^2+1/x)^2+x^2-1/x)+(-x^2+1/x)^2+x^2-1/x+1)
Good luck
Hi Rui!
A lot of work :t, but try to upload some exe file. With that we can test our equations.
Quote from: HSE on August 30, 2016, 10:04:23 AM
Hi Rui!
A lot of work :t , but try to upload some exe file. With that we can test our equations.
Hi HSE
Thank you.
But you cannot change the expressions to replace x because the way i am using to test it.
The expressions are pre defined and i only write the expression to be replaced in the
input box. More: it opens a lot of windows to show each step because i included
the files that have procedures to debug it.
Sorry. I will post the calculator as soon as possible.
Example about x^C, x^function, x^x
Quote
Replace by sin(x)
Replace by x/e^x
Replace by x-2
input box=> [-x^-2.34 +x^cos(x) -x^x]'
solution box1=> -sin(x)^-2.34 +sin(x)^cos(sin(x)) -sin(x)^sin(x)
-(x/e^x)^-2.34 +(x/e^x)^cos(x/e^x) -(x/e^x)^(x/e^x)
-(x-2)^-2.34 +(x-2)^cos(x-2) -(x-2)^(x-2)
Replace by -sin(x)
Replace by -x/e^x
Replace by -x+2
-(-sin(x))^-2.34 +(-sin(x))^cos(-sin(x)) -(-sin(x))^-sin(x)
-(-x/e^x)^-2.34 +(-x/e^x)^cos(-x/e^x) -(-x/e^x)^-(x/e^x)
-(-x+2)^-2.34 +(-x+2)^cos(-x+2) -(-x+2)^(-x+2)
Good luck :t
More
Quote
[-2*x^-2.34 +2*x^cos(x) -3*x^x -3/x^x]'
solution box1=> -2*sin(x)^-2.34 +2*sin(x)^cos(sin(x)) -3*sin(x)^sin(x) -3/(sin(x))^sin(x)
-2*(x/e^x)^-2.34 +2*(x/e^x)^cos(x/e^x) -3*(x/e^x)^(x/e^x) -3/(x/e^x)^(x/e^x)
-2*(x-2)^-2.34 +2*(x-2)^cos(x-2) -3*(x-2)^(x-2) -3/(x-2)^(x-2)
-2*(-sin(x))^-2.34 +2*(-sin(x))^cos(-sin(x)) -3*(-sin(x))^-sin(x) -3/(-sin(x))^-sin(x)
-2*(-x/e^x)^-2.34 +2*(-x/e^x)^cos(-x/e^x) -3*(-x/e^x)^-(x/e^x) -3/(-x/e^x)^-(x/e^x)
-2*(-x+2)^-2.34 +2*(-x+2)^cos(-x+2) -3*(-x+2)^(-x+2) -3/(-x+2)^(-x+2)
Hi Rui!!
I have some doubt:
[x^n]' = n*x^(n-1) Perfect
but [x^u]' where u is a function u(x)
have a short solution than [u^v]' ??? where v is a function v(x)
Thanks
Quote from: HSE on September 06, 2016, 08:57:16 AM
Hi Rui!!
I have some doubt:
[x^n]' = n*x^(n-1) Perfect
but [x^u]' where u is a function u(x)
have a short solution than [u^v]' ??? where v is a function v(x)
Thanks
Hi HSE
Yes it has. Why ? First, when we have
v(x)^u(x) we transform it in
e^
ln[
v(x)^u(x)]
because
e^
ln(
A) =
A. Is there any other way ? Do you know any other way ?
So, we do
x^u = e^ln(
x^u)
= e^[
u * ln(
x)]. Now, starting from here, we have
[x^u(x)]'=[e^ (u * ln(x) )]'= [u(x)* ln(x)]' * e^ (u * ln(x) ) = [u(x)* ln(x)]' * x^u.
Of course, x must be a positive real to get a real solution.
Is there any other problem ?
Good luck :t
Fantastic!
I have [u^v]'= v*u^(v-1)*[v]'+ln(u)*u^v*[_u]' [_u] (to no activate the underline)
but in one table say something like "don't use this, instead take logaritms"
Thanks a lot!
Quote from: HSE on September 06, 2016, 10:12:32 AM
Fantastic!
I have [u^v]'= v*u^(v-1)*[v]'+ln(u)*u^v*[_u]' [_u] (to no activate the underline)
but in one table say something like "don't use this, instead take logaritms"
Thanks a lot!
Hi
HSEOk, thank you. :t
We get the general solution following these steps:
(1) First we do v(x)^u(x) = e^ln[v(x)^u(x)]
(2) so [v(x)^u(x)]' = [e^ln[v(x)^u(x)]]' = [e^[u(x)* ln(v(x))]' ( from 3.1)
(3.1) but ln[v(x)^u(x)] = u(x) * ln(v(x))
(3.2) and [e^y(x)]' = [y(x)]' * e^y(x) - particular case: [e^x]' = e^x
(3.3) and [ln[y(x)]' = [y(x)]'/y(x) - particular case: [ln(x)]' = 1/x
(3.4) If u and v are functions of x, we have [u*v]'= u' * v + u * v'
(4) From 3.2 we get [ e^[u(x)* ln(v(x))] ]' = [u(x)* ln(v(x))]' * e^[u(x)* ln(v(x))]
From 3.4 we get [u(x)* ln(v(x))]'= u'(x) * ln(v(x)) + u(x) * v'(x)/v(x)
(5) The solution is:
[ e^[u(x)* ln(v(x))] ]' = [u'(x) * ln(v(x)) + u(x) * v'(x)/v(x)] * e^[u(x)* ln(v(x))]
or
[ e^(u * ln(v)) ]' = [
u' * ln(
v)
+ u *
v'/v] * e^(
u* ln(
v)) = [
v^u]
'Note: in the solution we use e^(u* ln(v)) and not v^u because the first
is used to calculate the last v^u.
Note that we may use the particular cases to compute the general cases.
For e^u, we get e^x and the replace x by u and multiply by u'
For ln(u), we do the same thing: we get 1/x and replace x by u and multiply by u'.
As we are seeing,
this is the general rule:
1. get the
solution for the particular case;
2. replace x by the
Argument ( any function of x)
3. multiply by the
derivative of the Argument.
Now you are understanding why i need a replace x function
(the name of my procedure is ReplaceXbyArgument).
Another example: [-sin(-x^2+x-1)]'
The particular case is: [sin(x)]' = cos(x).
So, the starting solution is: -cos(x). Now we use ReplaceXbyArgument
( where Argument= -x^2+x-1 )
and multiply by the derivative of the Argument.
We have: -(-x^2+x-1)' * cos(-x^2+x-1)
See you !
Good luck
:icon14:
I see Rui!! :t
I have a lot to code. :icon_rolleyes:
Also I begin the interface, wich have some challenges. :eusa_boohoo:
Regards. HSE
Quote from: HSE on September 07, 2016, 11:17:09 PM
I see Rui!! :t
I have a lot to code. :icon_rolleyes:
Also I begin the interface, wich have some challenges. :eusa_boohoo:
Regards. HSE
Hi HSE,
good news :t
I don't know but you may solve the derivative of only one X expression.
If we have g(x,Y)= x*y+x-y, if we replace y by something taken as a constant we
need to solve g(x)= x* & + x -& where & is a constant and we have the parcial derivative with respect to X.
When we have the solution we replace & by y.
Then we do g(y)= &* y + & -y and replacing y by x we have g(x)= &*x+&-x and we use the same procedure
that solves for x. In the end we replace x by y again and & by x. In this way we have only one procedure to solve derivatives. If the procedure works correctly for x it works for y and the solution is correct.
What do you think about ?
Good luck :t
Quote from: RuiLoureiro on September 08, 2016, 09:36:44 AM
...we replace y by something taken as a constant...
If the procedure works correctly for x it works for y and the solution is correct.
What do you think about ?
Hi Rui!!
That it's the general idea.
I don't replace in the equation, but in the procedure. The process begin storing the name of the variable, at present only one character variables are posible. Except "e". I think that in equations the number need to be <e> to prevent issues with scientific notation: 1.3e-4. Perhaps, I'm thinking now, the option is replace for exp(1), exp(x), etc 8)
Quote from: HSE on September 08, 2016, 11:22:31 PM
Quote from: RuiLoureiro on September 08, 2016, 09:36:44 AM
...we replace y by something taken as a constant...
If the procedure works correctly for x it works for y and the solution is correct.
What do you think about ?
Hi Rui!!
That it's the general idea.
I don't replace in the equation, but in the procedure. The process begin storing the name of the variable, at present only one character variables are posible. Except "e". I think that in equations the number need to be <e> to prevent issues with scientific notation: 1.3e-4. Perhaps, I'm thinking now, the option is replace for exp(1), exp(x), etc 8)
Hi !
Your variables are a,b,c,d, and so on ? Is it ? Did you write a general procedure to solve any expression with one well known variable ? Each variable is only one letter ?
My procedure now clean some cases like this: (x)^... or 2*(x)^ or (( ... ((...)) ...)). And cases where we put +x and we dont need the sign like x^+x ou x^(-x). It gives x^x and x^-x. So we have only 2 cases, no more. So (x)^(-x) is x^-x. Or ((x))^((-x)) gives x^-x.
Hi Rui!!
Remember the original problem:
[a+b*x^c]' http://masm32.com/board/index.php?topic=175.msg57776#msg57776
The idea is: calling object.solution(x, a+b*x^c) solve der.x, calling object.solution(c, a+b*x^c) solve der.c , etc.
Regards. HSE
Quote from: HSE on September 13, 2016, 10:03:12 AM
Hi Rui!!
Remember the original problem:
[a+b*x^c]' http://masm32.com/board/index.php?topic=175.msg57776#msg57776 (http://masm32.com/board/index.php?topic=175.msg57776#msg57776)
The idea is: calling object.solution(x, a+b*x^c) solve der.x, calling object.solution(c, a+b*x^c) solve der.c , etc.
Regards. HSE
Hi HSE,
Ok i hope you have something done as soon as possible. :t
I want to try it .
Now iam working in some previous procedures to clean some cases.
If we write something like this:
[(-x)*2 + (-x)/2 +(-x+2) -x-2 +2*(-x) *(+x)^2 ]'
the expression to solve is this (not the previous)
[ -x*2 -x/2 -x+2 -x-2 -2*x *x^2 ]'
Now i am testing it.
:icon14:
Hi
These are the results of the last test.
The expression used has a lot of brackets.
They are cleaned before used by the procedure
that replace x by expression and all procedures
used after that operation.
Here one sample
Quote
input box=> [x^+x-x^(-x)+x^(x)-x^(+X)+((sin(x)+2))]'
expression cleaned: [x^x -x^-x +x^x -x^X +sin(x) +2 ]'
Replace by sin(x)
Replace by x/e^x
Replace by x-2
solution box1=> sin(x)^sin(x) -sin(x)^-sin(x) +sin(x)^sin(x) -sin(x)^sin(x) +sin(sin(x))+2
+(x/e^x)^(x/e^x) -(x/e^x)^-(x/e^x) +(x/e^x)^(x/e^x) -(x/e^x)^(x/e^x) +sin(x/e^x)+2
+(x-2)^(x-2) -(x-2)^-(x-2) +(x-2)^(x-2) -(x-2)^(x-2) +sin(x-2)+2
Replace by -sin(x)
Replace by -x/e^x
Replace by -x+2
+(-sin(x))^-sin(x) -(-sin(x))^sin(x) +(-sin(x))^-sin(x) -(-sin(x))^-sin(x) +sin(-sin(x))+2
+(-x/e^x)^-(x/e^x) -(-x/e^x)^(x/e^x) +(-x/e^x)^-(x/e^x) -(-x/e^x)^-(x/e^x) +sin(-x/e^x)+2
+(-x+2)^(-x+2) -(-x+2)^-(-x+2) +(-x+2)^(-x+2) -(-x+2)^(-x+2) +sin(-x+2)+2
Good luck
Hi Rui!!
In a couple of weeks this thing perhaps work.
input:
2*x^35e-1*arctg(sin(x+2))
output:
2*(35e-1*x^(35e-1-1)*arctg(sin(x+2))+x^35e-1*1/(1+(sin(x+2))^2)*((cos(x+2)*(+(1)+0))))
Quote from: HSE on September 28, 2016, 08:41:13 AM
Hi Rui!!
In a couple of weeks this thing perhaps work.
input:
2*x^35e-1*arctg(sin(x+2))
output:
2*(35e-1*x^(35e-1-1)*arctg(sin(x+2))+x^35e-1*1/(1+(sin(x+2))^2)*((cos(x+2)*(+(1)+0))))
Hi HSE
Good work
! Go on !
About your output result, you may:
(1) Simplify brackets (there are too much brackets)
(2) Remove things like (+(1)+0) ( is =1 )
(3) This: x^35e-1*1 / could be x^35e-1/
Quote
input:
2*x^35e-1*arctg(sin(x+2))
output:
2*( 35e-1*x^( 35e-1-1) * arctg(sin(x+2))
+x^35e-1*1 / ( 1+(sin(x+2) )^2 ) * ( (cos(x+2)* (+(1)+0) ) ) )
Good luck ! :t
Thanks Rui!!
Right now I'm adding parenthesis :biggrin:. It's almost the last task to complete derivator.
The idea is to make an indepent process that "clean" the result, even obtain common factor or some other operation (that I not imagine now) for a nicer solution.
Regards. HSE
Quote from: HSE on October 04, 2016, 10:11:06 AM
Thanks Rui!!
Right now I'm adding parenthesis :biggrin: . It's almost the last task to complete derivator.
The idea is to make an indepent process that "clean" the result, even obtain common factor or some other operation (that I not imagine now) for a nicer solution.
Regards. HSE
Hi
HSE,
Nothing at all, HSE.
About your last answer (to make an independent process that "clean" the result )
«
The calculator»
do it also. But it starts to clenning same cases too as i
said before.
Quote
About the procedure to solve an expression of simple functions of X
- without function names or polynomials-
like « C +x +C*x - x^C + x^x - x/C + x/C - C +...»
you may start the output string result as a null string and add each
particular result until we get the end. But before adding, test the result
and don't add if it is 0 - if it is 0 it is done and go to the next.
Each result must have sign (+ or -) unless it is the first
or any equivalent procedure ( this is only one hint! ). In
this way we dont need to clean zeros.
See you !
Good luck, (i have a lot of work to do!). :t
Hi Rui!
You can see in http://masm32.com/board/index.php?topic=5725.0 the present state of my project. Opinions are welcome (mostly nice and encouraging ones :biggrin: )
Regards. HSE
Hi HSE
Correct this:
Quote
EXPRESSION:
2*sin(3*x+5)+x^-0.0
RESULT:
+2*(cos(3*x+5)*(+3*(1)+0))+((-0.0)*x^(-0.0-1))
der.x = 2*(3(cos(5+3*x)))+-0.000000*(x^-1.000000)
Good work ! :t
Thanks Rui!!
I'm working in the interface, but next week I have another round with the cleaner :t
Quote from: HSE on October 20, 2016, 11:04:19 AM
Thanks Rui!!
I'm working in the interface, but next week I have another round with the cleaner :t
Hi
HSE The cleaner is working well but needs
some more work
Quote
EXPRESSION: x^1+(5/1)*x^2
RESULT:
+((1)*x^(1-1))+(5/1)*((2)*x^(2-1))
der.x = (x^0)+5*(2*(x^1)) <<<<<------ x^0= 1 and x^1= x
Good luck :t
:icon14:
Thanks Rui!
The power is solved making a second pass in the cleanner. In first pass there is parenthesis ^( ).
+((1)*a^(1-1))+(5/1)*((2)*a^(2-1))
(a^0)+5*(2*(a^1)) < result clean 1
1+5*(2*(a)) < result clean 2
1+5*(2*a) < result clean 3
der.a = 1+5*(2*a) < result clean4 ( no improvement)
Note that parenthesis isn't removed, solution is 1+10*a. I also have some very absurd problems: (-1-1) is solved like (0) :biggrin:, also (--) (+-), etc. But now I'm a little in the dark side (J2ME) improving a Horse' data base I maked for my cell phone.
Hi
HSE,
correct this:
[
cos(cos(x))]'
This is your result:
Quote
(cos(cos(x))*((-sin(x)))) <<<<- why a lot of brackets ?
+0 <--- why this ?
Do this:
cos(cos(x)) = cos( argX) where argX= cos(x) and [cos(x)]'= -sin(x)
So,
Quote
[cos(cos(x))]' = [cos(argX)]' * [argX]'
= -sin(argX) * (-sin(x))
= sin(argX) * sin(x)
= sin(cos(x)) * sin(x) <<<<--- give this result (remove brackets)
Good luck :t
Perfect :t
I will see next week!
Theoretically the cleanning process remove parenthesis, but perhaps not. Sometimes work so bad that program crash! Because that, last version can disable cleanning from "options". When active solution say "der.x = ".
Thanks Rui!
Solved (I think) and updated!
:biggrin:
Hi all
Hi
HSE,
This is to show the first set of tests that
i did today about derivatives following my own (and
new) algorithm that uses
local stack variables (more than
130 000 bytes in 13 structures +
some particular variables).
It is not ready yet because the calculator solves
each expression case by case.
As soon as possible i will post it.
See you
Good luck :t
:icon14:
some basic tests in 3 Fev 2017 [sqr(x)+log(x)+ln(x)+exp(x)+sin(x)+cos(x)+tan(x)+sec(x)+csc(x)+cot(x)]',
[sinh(x)+cosh(x)+tanh(x)+sech(x)+csch(x)+coth(x)]',
[sind(x)+cosd(x)+tand(x)+secd(x)+cscd(x)+cotd(x)]',
[arcsin(x)+arccos(x)+arctan(x)+arcsec(x)+arccsc(x)+arccot(x)]',
[arcsind(x)+arccosd(x)+arctand(x)+arcsecd(x)+arccscd(x)+arccotd(x)]'
[arcsinh(x)+arccosh(x)+arctanh(x)+arcsech(x)+arccsch(x)+arccoth(x)]'
[sqr(-x)+log(-x)+ln(-x)+exp(-x)+sin(-x)+cos(-x)+tan(-x)+sec(-x)+csc(-x)+cot(-x)]',
Quote
Friday, 03-02-2017 14:45:11
input box=> [sqr(x) +log(x) +ln(x) +exp(x) +sin(x) +cos(x) +tan(x) +sec(x) +csc(x) +cot(x)]'
solution box1=> 1/(2*sqr(x)) +1/(ln(10)*x) +1/x +exp(x) +cos(x) -sin(x) +sec(x)^2 +sec(x)*tan(x) -csc(x)*cot(x) -csc(x)^2
; ----------------------------------------------------------------------------------------------
Friday, 03-02-2017 16:20:26
input box=> [sqr(-x) +log(-x) +ln(-x) +exp(-x) +sin(-x) +cos(-x) +tan(-x) +sec(-x) +csc(-x) +cot(-x)]'
solution box1=> -1/(2*sqr(-x)) -1/(-ln(10)*x) +1/x -exp(-x) -cos(-x) +sin(-x) -sec(-x)^2 -sec(-x)*tan(-x) +csc(-x)*cot(-x) +csc(-x)^2
++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
Friday, 03-02-2017 14:51:53
input box=> [sinh(x)+cosh(x)+tanh(x)+sech(x)+csch(x)+coth(x)]'
solution box1=> cosh(x)+sinh(x)+sech(x)^2+sech(x)*tanh(x)-csch(x)*coth(x)-csch(x)^2
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++
Friday, 03-02-2017 14:53:53
input box=> [sind(x)+cosd(x)+tand(x)+secd(x)+cscd(x)+cotd(x)]'
solution box1=> cosd(x)-sind(x)+secd(x)^2+secd(x)*tand(x)-cscd(x)*cotd(x)-cscd(x)^2
; ##########################################################
Friday, 03-02-2017 15:04:05
input box=> [arcsin(x)+arccos(x)+arctan(x)+arcsec(x)+arccsc(x)+arccot(x)]'
solution box1=> 1/sqr(1-x^2)-1/sqr(1-x^2)+1/(1+x^2)+1/(abs(x)*sqr(x^2-1))-1/(abs(x)*sqr(x^2-1))-1/(1+x^2)
++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
Friday, 03-02-2017 15:05:51
input box=> [arcsind(x)+arccosd(x)+arctand(x)+arcsecd(x)+arccscd(x)+arccotd(x)]'
solution box1=> 1/sqr(1-x^2)-1/sqr(1-x^2)+1/(1+x^2)+1/(abs(x)*sqr(x^2-1))-1/(abs(x)*sqr(x^2-1))-1/(1+x^2)
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
Friday, 03-02-2017 15:08:02
input box=> [arcsinh(x)+arccosh(x)+arctanh(x)+arcsech(x)+arccsch(x)+arccoth(x)]'
solution box1=> 1/sqr(x^2+1)-1/sqr(x^2-1)+1/(1-x^2)
-1/(x*sqr(1-x^2))-1/(abs(x)*sqr(1+x^2))-1/(1+x^2)
; «««««««««««««««««««««««««««««««««««««««««««««««««««««««««««««««««««««««
Friday, 03-02-2017 22:31:57
input box=> [sin(x^2+x)]'
solution box1=> (2*x+1)*cos(x^2+x)
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
Friday, 03-02-2017 22:37:01
input box=> [e^x+sin(x^2+x)-x^-2+arccos(x^2+x)]'
solution box1=> e^x+(2*x+1)*cos(x^2+x)+2*x^-3-(2*x+1)*1/sqr(1-(x^2+x)^2)
++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
Friday, 03-02-2017 22:43:28
input box=> [3*x^4-2*x^3+x^2+x-1]'
solution box1=> 12*x^3-6*x^2+2*x+1
++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
Friday, 03-02-2017 22:46:04
input box=> [sin(3*x^4-2*x^3+x^2+x-1)]'
solution box1=> (12*x^3-6*x^2+2*x+1)*cos(3*x^4-2*x^3+x^2+x-1)
Fantastic Rui :t
Quote from: HSE on February 05, 2017, 02:37:54 AM
Fantastic Rui :t
Thanks HSE :t
Hi
3 examples of one nested function Good luck
Quote
Saturday, 04-02-2017 15:36:23
input box=> [sin(cos(x^2)+x)]'
solution box1=> (-2*x*sin(x^2)+1) * cos(cos(x^2)+x)
++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
Saturday, 04-02-2017 15:38:11
input box=> [sin(cos(x))]'
solution box1=> -sin(x)*cos(cos(x))
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
Saturday, 04-02-2017 21:54:50
input box=> [cos(x^-2-4*x+arctan(x^2+x-1)+x)]'
solution box1=> -(-2*x^-3-4+(2*x+1)*1/(1+(-x^2-x+1)^2)+1)* sin(x^-2-4*x+arctan(x^2+x-1)+x)
:biggrin: :biggrin:
Hi all
Hi
HSE Here are
examples of 3 functions:
2 nested functions and
3 nested functions As soon as possible,
i will post it ( wait some more weeks ).
And i will say something about the algorithm (it is
very simple ;) )
. To
HSE: then i will test your derivator (and you test mine).
To
Hutch: using
a table of pointers does the work more simple :t
See you
Good luck :t
Quote
Sunday, 05-02-2017 14:06:05
input box=> [sin(x^2-cos(tan(x^2-x)+x))]'
solution box1=> (2*x+((2*x-1)*sec(x^2-x)^2+1)*sin(tan(x^2-x)+x)) * cos(x^2+cos(tan(x^2-x)+x))
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
Sunday, 05-02-2017 14:12:11
input box=> [x^2-cos(tan(x^2-x)+x)]' <------ Derivative of the argument ABOVE
solution box1=> 2*x+((2*x-1)*sec(x^2-x)^2+1) * sin(tan(x^2-x)+x)
++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
Sunday, 05-02-2017 14:16:46
input box=> [cos(tan(x^2-x)+x)]' <------ Derivative of the function ABOVE
solution box1=> -((2*x-1)*sec(x^2-x)^2+1)*sin(tan(x^2-x)+x)
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
Sunday, 05-02-2017 14:19:50
input box=> [tan(x^2-x)+x]' <------ Derivative of the argument ABOVE
solution box1=> (2*x-1)*sec(x^2-x)^2+1
example of 4 functions: 3 nested functionsQuote
++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
Sunday, 05-02-2017 14:35:17
input box=> [sin(x^2-cos(tan(ln(x^2-x)-x^3)+x))]'
solution box1=> (2*x+(((2*x-1)*1/(x^2-x)-3*x^2)*sec(ln(x^2-x)-x^3)^2+1)*sin(tan(ln(x^2-x)- x^3)+x)) * cos(x^2+cos(tan(ln(x^2-x)-x^3)+x))
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
Monday, 06-02-2017 00:05:22
input box=> [sin(x^2-cos(x^2-x)-ln(x^3))+x]'
solution box1=> (2*x+(2*x-1)*sin(x^2-x)-3*x^2*1/x^3)*cos(x^2-cos(x^2-x)+ln(x^3))+1
:biggrin: :biggrin:
Hi
all examples of 4, 5 and 6 nested functions (max. number of nested functions
=6 -last argument x -
or 5 nested functions with last argument g(x) ).
This is because we have only tables ResA,ResB,ResC,ResD,ResE,ResF
to solve each case: first function-> Table ResA, and so on.
I think we dont need more than this and this is too much.
See you
Good luck :t
:icon14:
Quote
Monday, 06-02-2017 12:38:09
input box=> [cos(sin(tan(sec(csc(x)))))]'
solution box1=> csc(x)*cot(x)*sec(csc(x))*tan(csc(x))*sec(sec(csc(x)))^2*cos(tan(sec(csc(x)))) * sin(sin(tan(sec(csc(x)))))
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
Monday, 06-02-2017 14:55:54
input box=> [cos(sin(tan(sec(cot(csc(x))))))]'
solution box1=> -csc(x)*cot(x)*csc(csc(x))^2*sec(cot(csc(x)))*tan(cot(csc(x)))*sec(sec(cot(csc(x))))^2
* cos(tan(sec(cot(csc(x)))))*sin(sin(tan(sec(cot(csc(x))))))
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
Monday, 06-02-2017 15:10:01
input box=> [cos(sin(tan(sec(cot(arcsin(x^2-x))))))]'
solution box1=> (2*x-1)/sqr(1-(x^2-x)^2)*csc(arcsin(x^2-x))^2*sec(cot(arcsin(x^2-x)))*tan(cot(arcsin(x^2-x)))*sec(
sec(cot(arcsin(x^2-x))))^2*cos(tan(sec(cot(arcsin(x^2-x)))))*sin(sin(tan(sec(cot(arcsin(x^2-x))))))
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
Monday, 06-02-2017 15:13:05
input box=> [cos(sin(tan(sec(cot(arcsin(csc(x)))))))]'
solution box1=> -csc(x)*cot(x)/sqr(1-csc(x)^2)*csc(arcsin(csc(x)))^2*sec(cot(arcsin(csc(x))))*tan(cot(arcsin(csc(x
))))*sec(sec(cot(arcsin(csc(x)))))^2*cos(tan(sec(cot(arcsin(csc(x))))))*sin(sin(tan(sec(cot(arcsin(csc(x)))))))
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
Monday, 06-02-2017 15:27:54
input box=> [cos(sin(tan(sec(cot(arcsin(x^2-x)+x^2)-2*x)+5)-3*x^2))]'
solution box1=> -((-((2*x-1)/sqr(1-(x^2-x)^2)+2*x)*csc(arcsin(x^2-x)+x^2)^2-2)*sec(cot(arcsin(x^2-x)+x^2)-2*x)*tan
(cot(arcsin(x^2-x)+x^2)-2*x)*sec(sec(cot(arcsin(x^2-x)+x^2)-2*x)+5)^2-6*x)*cos(tan(sec(cot(arcsin(x^2-x)+x^2)-2*x)
+5)-3*x^2)*sin(sin(tan(sec(cot(arcsin(x^2-x)+x^2)-2*x)+5)-3*x^2))
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
Monday, 06-02-2017 15:36:15
input box=> [cos(sin(tan(sec(cot(arcsin(x^2-x))))))-cos(sin(tan(sec(cot(arcsin(x^2-x))))))]'
solution box1=> (2*x-1)/sqr(1-(x^2-x)^2)*csc(arcsin(x^2-x))^2*sec(cot(arcsin(x^2-x)))*tan(cot(arcsin(x^2-x)))*sec(
sec(cot(arcsin(x^2-x))))^2*cos(tan(sec(cot(arcsin(x^2-x))))) * sin(sin(tan(sec(cot(arcsin(x^2-x))))))
-(2*x-1)/sqr(1-(x^2-x)^2)*csc(arcsin(x^2-x))^2*sec(cot(arcsin(x^2-x)))*tan(cot(arcsin(x^2-x)))*sec(
sec(cot(arcsin(x^2-x))))^2*cos(tan(sec(cot(arcsin(x^2-x))))) * sin(sin(tan(sec(cot(arcsin(x^2-x))))))
++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
Monday, 06-02-2017 17:45:08
input box=> [sqr(x)+ten(x)+log(x)+ln(x)+exp(x)+sin(x)+cos(x)+tan(x)+sec(x)+csc(x)+cot(x)+sinh(x)+cosh(x)+
tanh(x)+sech(x)+csch(x)+coth(x)+sind(x)+cosd(x)+tand(x)+secd(x)+cscd(x)+cotd(x)+arcsin(x)
+arccos(x)+arctan(x)+arcsec(x)+arccsc(x)+arccot(x)+arcsind(x)+arccosd(x)+arctand(x)+arcsecd(x)
+arccscd(x)+arccotd(x)+arcsinh(x)+arccosh(x)+arctanh(x)+arcsech(x)+arccsch(x)+arccoth(x)]'
solution box1=> 1/(2*sqr(x))+ln(10)*ten(x)+1/(ln(10)*x)+1/x+exp(x)+cos(x)-sin(x)+sec(x)^2+sec(x)*tan(x)-csc(x)*cot
(x)-csc(x)^2+cosh(x)+sinh(x)+sech(x)^2+sech(x)*tanh(x)-csch(x)*coth(x)-csch(x)^2+cosd(x)-sind(x)+secd(x)^2+secd(x)
*tand(x)-cscd(x)*cotd(x)-cscd(x)^2+1/sqr(1-x^2)-1/sqr(1-x^2)+1/(1+x^2)+1/(abs(x)*sqr(x^2-1))-1/(abs(x)*sqr(x^2-1))
-1/(1+x^2)+1/sqr(1-x^2)-1/sqr(1-x^2)+1/(1+x^2)+1/(abs(x)*sqr(x^2-1))-1/(abs(x)*sqr(x^2-1))-1/(1+x^2)+1/sqr(x^2+1)-
1/sqr(x^2-1)+1/(1-x^2)-1/(x*sqr(1-x^2))-1/(abs(x)*sqr(1+x^2))-1/(1+x^2)
Hi
HSE,
see this given by the Derivator:
(the cleaner is too much good: cleans all things ;)
See you :t
:icon14:
Quote
expression:
cos(sin(arctan(x^2-x)))
result:
(-sin(sin(arctan(x^2-x)))*((cos(arctan(x^2-x))*(((((+0))))))))
+0
++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
TheCalculator
++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
Monday, 06-02-2017 23:45:48
input box=> [cos(sin(arctan(x^2-x)))]'
solution box1=> -(2*x-1)/(1+(x^2-x)^2)*cos(arctan(x^2-x))*sin(sin(arctan(x^2-x)))
:biggrin: because it's "arctg". I will make a list of sinonimus :t .
Is posible to treat sind as sin, or that kind of functions have different properties?
Quote from: HSE on February 07, 2017, 09:17:14 PM
:biggrin: because it's "arctg". I will make a list of sinonimus :t .
Is posible to treat sind as sin, or that kind of functions have different properties?
Hi HSE,
Oh sorry if it is not a function name (arctan) it should give a syntax error.
But ok, i replaced by arctg and
the result seems to be correct after i cleaned the unnecessary
brackets and make some calculations and simplifications. For common people the result
is unreadable, so i think that you should do some more work to clean brackets and to replace
things like 2-1 by 1, x^1 by x, (2) by only 2, ... Think about this.
Another thing is when we use "save as" the result, it should save the
expression and the resultbecause that result is for that particular expression and not for another.
About sin
d, cos
d, arcsin
d, and so on, they are the same functions as sin, cos, etc but the argument is in
degrees or the result is in degrees (arcsin
d, ...)- it is written in TheClaculator messages, run and read.
note: i use also the function ten(x) = 10^x ( exp(x)=e^x ).
See you :t
Hi Rui!
I think you have activated option "show only crude derivatives" (see in "options" menu). It's a development option.
Thanks for the ideas. :t
Hi This is to show the derivatives of a product:
integer *
function(x)
The calculator simplify some expressions but not
integer*integer ( 2*2 in the first example). But
soon it will do.
See you
Good luck :t
Quote
Wednesday, 08-02-2017 20:35:18
input box=> [2*sin(x^2)]'
solution box1=> 2*2*x*cos(x^2) <<---- 4*x*cos(x^2)
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
Wednesday, 08-02-2017 20:36:10
input box=> [2*sin(cos(x))]'
solution box1=> -2*sin(x)*cos(cos(x))
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
Wednesday, 08-02-2017 22:17:26
input box=> [2*sin(cos(tan(x^2-x+1)))]'
solution box1=> -2*(2*x-1)*sec(x^2-x+1)^2*sin(tan(x^2-x+1))*cos(cos(tan(x^2-x+1)))
++++++++++ OLD solution +++++++++++++++++++++++++++++++++++++++++++++++++
Monday, 06-02-2017 10:15:55
input box=> [x^2-cos(tan(ln(x^2-x)-ln(x^3))+x)]' <<<<<<<<<<<- SIMPLIFY SOLUTION
solution box1=> 2*x+(((2*x-1)/(x^2-x)-3*x^2/x^3)*sec(ln(x^2-x)+ln(x^3))^2+1)*sin(tan(ln(x^2-x)+ln(x^3))+x)
++++++++++ NEW solution +++++++++++++++++++++++++++++++++++++++++++++
Wednesday, 08-02-2017 23:08:38
input box=> [x^2-cos(tan(ln(x^2-x)-ln(x^3))+x)]'
solution box1=> 2*x+(((2*x-1)-3*x^-1)*sec(ln(x^2-x)-ln(x^3))^2+1)*sin(tan(ln(x^2-x)-ln(x^3))+x)
I have moved this topic because its a long term project, not a Campus style question.
Hi all,
This is the link to the new version 2017:
>>>> http://masm32.com/board/index.php?topic=6197.0 (http://masm32.com/board/index.php?topic=6197.0) <<<<<<
Good luck