Inverse trigonometric functions

[Loïc]

Hello,

Does anyone know if it's possible with Maxima to compute expressions
such as Machin's formula:
pi/4=4arctan(1/5)-arctan(1/239)

or formula such as:

cos(1/2*arccos(1/3))=sqrt(6)/3

If it's impossible, does anyone know if someone plan to implement it in
Maxima?

Best

Loïc

[Andrej Vodopivec]

Here is one possibility (note that logcontract is not always correct
for complex arguments).

(%i1) 4*atan(1/5)-atan(1/239);
(%o1) 4*atan(1/5)-atan(1/239)
(%i2) rectform(logcontract(ratsimp(logarc(%))));
(%o2) %pi/4

(%i3) cos(1/2*acos(1/3));
(%o3) cos(acos(1/3)/2)
(%i4) radcan(exponentialize(logarc(%)));
(%o4) (sqrt(2)*%i+2)/(sqrt(3)*sqrt(2*sqrt(2)*%i+1))
(%i5) rectform(%);
(%o5) sqrt(2)/sqrt(3)

Andrej

Post by Loïc
Hello,
Does anyone know if it's possible with Maxima to compute expressions
pi/4=4arctan(1/5)-arctan(1/239)
cos(1/2*arccos(1/3))=sqrt(6)/3
If it's impossible, does anyone know if someone plan to implement it in
Maxima?
Best
Loïc
_______________________________________________
Maxima mailing list
http://www.math.utexas.edu/mailman/listinfo/maxima

[Richard Fateman]

Post by Loïc
Hello,
Does anyone know if it's possible with Maxima to compute expressions

Please define what you mean by "compute expressions".

Do you mean "prove that identities are true"
or do you mean "invent new formulas"
or do you mean confirm them numerically?

Post by Loïc
pi/4=4arctan(1/5)-arctan(1/239)
cos(1/2*arccos(1/3))=sqrt(6)/3
If it's impossible,

In general, proving that f(x)=0 is recursively undecidable, as shown by
Daniel Richardson in 1968.
So yes, some interpretation of your question leads to an impossibility.

Post by Loïc
does anyone know if someone plan to implement it in
Maxima?
Best
Loïc
_______________________________________________
Maxima mailing list
http://www.math.utexas.edu/mailman/listinfo/maxima

[Loïc]

Post by Richard Fateman

Post by Loïc
Hello,
Does anyone know if it's possible with Maxima to compute expressions

Please define what you mean by "compute expressions".
Do you mean "prove that identities are true"
or do you mean "invent new formulas"
or do you mean confirm them numerically?

Why I write for example on some other computer algebra system:
arctan(2)+arctan(5)+arctan(8)

it returns directly, or after a "simplify" command, the value 5*pi/4



That's what I would like Maxima to do.
Is it possible?

[Richard Fateman]

Post by Loïc

Post by Richard Fateman

Post by Loïc
Hello,
Does anyone know if it's possible with Maxima to compute expressions

Please define what you mean by "compute expressions".
Do you mean "prove that identities are true"
or do you mean "invent new formulas"
or do you mean confirm them numerically?

arctan(2)+arctan(5)+arctan(8)
it returns directly, or after a "simplify" command, the value 5*pi/4
That's what I would like Maxima to do.
Is it possible?

Depending on what you mean by "example" some built-in simplification
programs will do some of the work.
Otherwise you are free to write your own new simplify command. You can
also use something like the
Inverse Symbolic Calculator.

http://oldweb.cecm.sfu.ca/projects/ISC/ISCmain.html

But, basically, writing an algorithm that will ALWAYS find the smallest
or "simplest" expression is
not possible unless you have a severely restricted class of expressions.