2
$\begingroup$

Can you prove or disprove the following?

The periodic function $f:[0,2\pi]\to \mathbb{R}$ \begin{align*} f(x):=\cos(x)\sum_{k=0}^\infty \bigg(\prod_{l=1}^{k}\frac{2l}{2l+1}\bigg)\sin^{2k+1}(x) \end{align*} is continuous and attains therefore its maximum on the interval $[0,2\pi]$. Just to make notation clear: The empty product $\prod_{l=1}^0$ is set to 1.

For $x\ne\frac{\pi}{2}$, $f$ is continuous because the sum converges. This follows since the geometric series $\sum_{k=0}^\infty \sin^k(x)$ converges. We have to show continuity for $x=\frac{\pi}{2}$.

EDIT 2:

We claim that it is possible to show that \begin{align*} \cos(x)\sum_{k=0}^\infty \bigg(\prod_{l=1}^{k}\frac{2l}{2l+1}\bigg) \sin^{2k+1}(x)\leq \cos(x)\sum_{k=0}^\infty \frac{\sin^{2k+1}(x)}{\sqrt{2k+1}}=:g(x). \end{align*} Hence, it is left to prove that $g(x)$ is contiuous in $\frac{\pi}{2}$.

It is known that the function \begin{align*} \cos(x)\sum_{k=0}^\infty \sin^k(x)=\begin{cases} \frac{\cos(x)}{1-\sin(x)} & \text{for }\quad x\ne \frac{\pi}{2}\\ 0 & \text{for }\quad x= \frac{\pi}{2} \end{cases} \end{align*} is not continous.

Notice, in EDIT 1, there was a mistake. It should be but the square root as it is now, not the k-root.

$\endgroup$
  • 1
    $\begingroup$ EDIT: It is sufficient to show the function is continuous in $\frac{\pi}{2}$ $\endgroup$ – user3154270 Jun 9 at 21:50
2
$\begingroup$

Maybe the Cauchy product helps. Define $$ h(x)=\sum_{k=0}^\infty a_k = \sum_{k=0}^\infty \bigg(\prod_{l=1}^{k}\frac{2l}{2l+1}\bigg)\sin^{2k+1}(x) $$ and $$ \cos(x)=\sum_{k=0}^\infty b_k=\sum_{k=0}^\infty (-1)^k\frac{x^{2k}}{(2k)!} $$ Then, $$ \cos(x)h(x) = \sum_{k=0}^\infty a_k \sum_{k=0}^\infty b_k = \sum_{k=0}^\infty c_k $$ where $ c_k=\sum_{i=0}^k a_i b_{k-i}$.

We calculate $c_k$ in this case $$ c_k=\sin^{2k+1}(x)\sum_{i=0}^k (-1)^i\frac{x^{2i}}{\sin^{2i}(x)}\frac{1}{(2i)!}\bigg(\prod_{l=1}^{k-i}\frac{2l}{2l+1}\bigg). $$ The hardest problem is now to show the convergence of $\sum_{k=0}^\infty c_k$.

Since is it known that $\cos(x)h(x)$ is continous for $x<\frac{\pi}{2}$, it is sufficient to show that the series $\sum_{k=0}^\infty c_k$ converges for $x=\frac{\pi}{2}$.

$\endgroup$
  • 1
    $\begingroup$ This seems to be a nasty calculation $\endgroup$ – J G Jun 9 at 21:48
  • 1
    $\begingroup$ Maybe $\lim_{x\to o}\frac{sin(x)}{x}=1$ helps $\endgroup$ – user3154270 Jun 9 at 21:58

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.