3
$\begingroup$

Consider:

Cosh[a ArcSinh[z]] // FunctionExpand

Cosh[a ArcSinh[z]]

Which seems to suggest that there is no simplification that can be done here. However, there is an identity on the wolfram functions site that states

Cosh[a ArcSinh[z]] == ((1/2) (1 + (z + Sqrt[1 + z^2])^(2 a)))/(z + Sqrt[1 + z^2])^a

which is clearly a simplification. Why did FunctionExpand not recognize this and what should I do in future to make Mathematica see this sort of simplification directly without looking it up on a webpage?

$\endgroup$
  • $\begingroup$ Hi; it is a funny thing about simplifications...If I substitute random reals for a and z I get that the RHS is almost 3 times slower than the LHS to compute. $\endgroup$ – bobbym Mar 29 '17 at 21:03
  • $\begingroup$ @bobbym That makes sense. But analytically trigonometric functions are definitely of higher complexity than rational functions of powers of variables. Additionally, if I understand correctly, the idea behind FunctionExpand is to simplify special functions into less special ones. And again, arguably trigonometric functions are more special than rational ones with certain powers. $\endgroup$ – Kagaratsch Mar 29 '17 at 21:21
  • $\begingroup$ no argument with the math. I was just musing that if we were to get that simplification by some command and blindly put it inside some loop we would be shooting ourselves in the foot as far as speed is concerned. But good question +1 $\endgroup$ – bobbym Mar 29 '17 at 21:38
  • 4
    $\begingroup$ @Kagaratsch Try using TrigToExp. As @bobbym explains, I think it's clear why FunctionExpand doesn't use it. $\endgroup$ – Carl Woll Mar 29 '17 at 21:42
  • $\begingroup$ @Carl Woll That is pretty darn close to what the question asks for +1 $\endgroup$ – bobbym Mar 29 '17 at 22:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.