0
$\begingroup$

I have two infinite sums that forms an equality:

$$\sum_{n=1}^{\infty} \left(\zeta(2n)\frac{x^{2n}}{\pi^{2n}}\right) = \sum_{n=1}^\infty \left(\frac{B_{2n}}{(2n)!}\left(-\frac{1}{2}\right)(2ix)^{2n}\right)$$

I know that the entire sum of the left side is equals to the entire sum of the right side. But how do I know that if I expand the left side, the coefficient of index 1 is gonna be equals to the coefficient of index 1 in the right side? In other words, proof that:

$$\zeta(2n)\frac{x^{2n}}{\pi^{2n}} = \frac{B_{2n}}{(2n)!}\left(-\frac{1}{2}\right)(2ix)^{2n}$$

$\endgroup$
3
$\begingroup$

If a smooth function $f$ can be expressed as a power series around $0$: $$ f(x) = \sum_{n=0}^\infty a_n x^n, $$ then $a_n = \frac{1}{n!}f^{(n)}(0)$, where $f^{(n)}$ is the $n$-th derivative of $f$.

$\endgroup$
  • $\begingroup$ I am saying by taking the $n$-th derivative, dividing by $n!$, and substituting $x = 0$, you can "extract" the $n$-th coefficient of a power series. That is how you justify that coefficients of two equal power series will have to be equal. $\endgroup$ – Tunococ Sep 4 '13 at 23:32
  • $\begingroup$ Yep but I don't know how to expand each side as a power series $\endgroup$ – Lucas Zanella Sep 4 '13 at 23:33
  • $\begingroup$ I am not sure what you mean by "expand". $\endgroup$ – Tunococ Sep 4 '13 at 23:42
  • $\begingroup$ You're saying that both my left and my right sides of the equation are an infinite expansion of some function, right? $\endgroup$ – Lucas Zanella Sep 4 '13 at 23:54
  • 1
    $\begingroup$ If you assume the power series on both sides are convergent, then both sides are analytic functions near $0$. You can find, for each $n$, the $n$-th derivatives of both functions, which are also functions analytic near $0$. If you substitute $x = 0$ in the derivatives, you will get the $n$-th coefficients of the original power series (multiplied by $n!$) on both sides. In short, if the original power series are equal, their derivatives of all orders must be equal, and so are their coefficients. $\endgroup$ – Tunococ Sep 4 '13 at 23:58

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.