The calculator

RuiLoureiro · July 16, 2016, 09:03:30 AM

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

Code Select


[-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))

RuiLoureiro · August 01, 2016, 08:16:48 AM

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))

RuiLoureiro · August 05, 2016, 08:50:38 AM

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

RuiLoureiro · August 19, 2016, 08:28:14 AM

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

Code Select


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

RuiLoureiro · August 30, 2016, 08:44:17 AM

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

HSE · 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.

RuiLoureiro · September 04, 2016, 08:19:53 AM

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)

HSE · 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

RuiLoureiro · September 06, 2016, 09:29:06 AM

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

HSE · 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!

RuiLoureiro · September 07, 2016, 08:12:12 AM

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 HSE
Ok, 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:

HSE · 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.

Regards. HSE

RuiLoureiro · September 08, 2016, 09:36:44 AM

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.

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

HSE · 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)

RuiLoureiro · September 10, 2016, 08:58:27 AM

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.

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