Question:solving Sqrt(y)=tanh(x) For Y - MaplePrimes
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Question: solving sqrt(y)=tanh(x) for yTags are words are used to describe and categorize your content. Combine multiple words with dashes(-), and seperate tags with spaces.
Question:solving sqrt(y)=tanh(x) for y
Posted: nm 11783 Product: Maple 2020
solve hyperbolic trig + Manage Tags October 31 2020 1 When solving this by hand sqrt(y)=tanh(x), I would square both sides to obtain y=tanh(x)^2 as solution.
Maple does this when RHS is tan,cos,sin, but not when RHS is hyperbolic.
Why is that, and how to best tell Maple to give the simple solution? Here is an example
restart; eq:=sqrt(y)=tanh(x); sol:=solve(eq,y);Compare to Mathematica:
I can post-process Maple solution like this
restart; eq:=sqrt(y)=tanh(x); sol:=solve(eq,y); sol:=convert(sol,trigh); sol:=simplify(sol)Which is tanh(x)^2.
But this is all too much work, since this is done in a program, and I do not know what specific convert to use at the time to try.
My question is, why does not maple simply solves sqrt(y) = f(x) giving y=f(x)^2? it works for generic function
restart; eq:=sqrt(y)=f(x); sol:=solve(eq,y);it works for non-hyper trig functions also
restart; eq:=sqrt(y)=tan(x); sol:=solve(eq,y);Is there a trick to make it work for hyperbolic functions also automatically?
ps. for now, to avoid all these issues, I am doing this
restart; the_rhs:=tanh(x); eq:=sqrt(y)=f; sol:=solve(eq,y); sol:=subs(f=the_rhs,sol);Since I know that the RHS contains no y before hand. The above bypasses any issues Maple has with hyper trig functions.
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