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## The calculator

Started by RuiLoureiro, May 31, 2012, 10:59:09 PM

#### RuiLoureiro

#150
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  16:10:05input 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))-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)]'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(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

#151
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

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

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))+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=> 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

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

Hi Rui!
A lot of work :t, but try to upload some exe file. With that we can test our equations.
Equations in Assembly: SmplMath

#### RuiLoureiro

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.

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

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
Equations in Assembly: SmplMath

#### RuiLoureiro

#158
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

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!
Equations in Assembly: SmplMath

#### RuiLoureiro

#160
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

I see Rui!!  :t

I have a lot to code.  :icon_rolleyes:

Also I begin the interface, wich have some challenges.

Regards. HSE
Equations in Assembly: SmplMath

#### RuiLoureiro

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

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)
Equations in Assembly: SmplMath

#### RuiLoureiro

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.