2
$\begingroup$

When I type(a^2)^s Mathematica does not give me $a^{2s}$ instead it gives ${(a^2)}^s$. Is there a way to make it print $a^{2s}$.

It made some real difference where I wanted to compute

Sum[((n \[Pi]/T)^2)^-s, {n, 1, Infinity}]

vs

Sum[((n \[Pi]/T))^(-2 s), {n, 1, Infinity}]

In the first case it gave me $\sum _{n=1}^{\infty } \pi ^{-2 s} \left(\frac{n^2}{T^2}\right)^{-s}$ but in the second case it gave me $\pi ^{-2 s} \left(\frac{1}{T}\right)^{-2 s} \zeta (2 s)$

I had to take the s-derivative of the result so I would like to get the result like the second one.

So, my question:

Is there any Mathematica function which helps me make Mathematica evaluate the first command and give back me the result like that of the second one?

$\endgroup$
3
  • 3
    $\begingroup$ Compare (a^2)^s /. {a -> -1, s -> 1/2} against a^(2 s) /. {a -> -1, s -> 1/2}. Also consider Simplify[(a^2)^s, Assumptions -> a > 0]. $\endgroup$ – JimB Feb 6 at 4:35
  • $\begingroup$ (a^2)^s // PowerExpand and Sum[((n \[Pi]/T)^2)^-s // PowerExpand, {n, 1, Infinity}] works. $\endgroup$ – Bill Watts Feb 6 at 6:39
  • 1
    $\begingroup$ To check if you are making some implied assumptions, use FindInstance to check for counterexamples, e.g., FindInstance[(a^2)^s != a^(2s), {a, s}, 5] and FindInstance[(a^2)^s != a^(2s), {a, s}, Reals, 5] $\endgroup$ – Bob Hanlon Feb 6 at 13:42
3
$\begingroup$
FullSimplify[Sum[((n \[Pi]/T)^2)^-s, {n, 1, Infinity}],Assumptions->{T>0}]

Take your First input and assume $T>0$, that gives you the same result as the second one.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.