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Suppose $\displaystyle f(x)=\sum_{n=0}^{\infty}2^{-n}\sin(2^nx)$, evaluate $$ \lim_{x\to0+}\frac{f(x)-f(0)}{x}. $$

$f(x)$ converges uniformly on $\mathbb R$. But I don't know how to evaluate $$ \lim_{x\to0+}\sum_{n=0}^{\infty}2^{-n}\frac{\sin (2^nx)}{x}. $$

Knt
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1 Answers1

1

After defining

$$g_n(x) = \frac{\sin\left(2^nx\right)}{2^nx},$$

and $$g(x) = \sum_{n=0}^{+\infty}g_n(x),$$

observe that in the interval $$0<x<\frac{\pi}{2^N},$$

  1. For $0\leq j\leq m\leq N-1$, $g_j(x)\geq g_m(x)$.
  2. $g_n(x)$ is a monotonically decreasing function.

Consider a sequence $\{a_k\}$, with

$$\{a_k\}\to 0^+.$$

For large enough $k$, $$a_k< \frac{\pi}{2^N}$$ so that \begin{eqnarray} g\left(a_k\right) &=& \sum_{n=0}^{N-1}g_n\left(a_k\right) + \sum_{N}^{+\infty} g_n\left(a_k\right)\geq\\ &\geq& N\cdot g_{N-1}\left(a_k\right) - \sum_{n=N}^{+\infty}\frac1{2^n\pi}>\\ &>& N\cdot g_{N-1}\left(\frac{\pi}{2^N}\right) -\frac1{2^{N-1}\pi}=\\ &=&\frac{2N}{\pi} -\frac1{2^{N-1}\pi}, \end{eqnarray} where we took advantage of fact 1 in the first inequality, and of fact 2. in the second inequality. From the above chain we conclude that $g\left(a_k\right)$ can be made arbitrarily large by a suitable choice of $k$, thus

$$\left\{g\left(a_k\right)\right\} \to +\infty,$$ and therefore $$\lim_{x\to 0^+} \sum_{n=0}^{+\infty} \frac{\sin\left(2^nx\right)}{2^nx} = +\infty.$$

Edit

Prompted by Jake's comment I want to make facts 1. and 2. above more explicit.

  1. This fact is just a consequence of $$\left|\frac{\sin\left(2^m\alpha\right)}{2^m\alpha}\right| = \left|\frac{\sin\left(2^{m-1}\alpha\right)\cos\left(2^{m-1}\alpha\right)}{2^{m-1}\alpha}\right| \leq \left|\frac{\sin\left(2^{m-1}\alpha\right)}{2^{m-1}\alpha}\right|$$

  2. This fact can be derived by differentiation, since for $0 <\alpha<\frac{\pi}2$, $$\alpha < \tan \alpha$$ and therefore, in such interval we get $$\frac{d\left(\frac{\sin\alpha}{\alpha}\right)}{d\alpha}=\frac{\alpha\cos\alpha-\sin\alpha}{\alpha^2}<0$$

I do hope someone can give a simpler proof of the limit in OP's question, since I could not find one, so far.

dfnu
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  • Where did you use properties of the sine function near zero? This answer is very hard to read. This is a very simple problem once you replace the sine function with its linear approximation---and bound the error. – Jake Mirra Oct 21 '19 at 11:59
  • @JakeMirra if you mean the sine MacLaurin expansion, I did not use it explicitely. What do you find hard to follow? Do you have time to sketch the proof you are referring to? I tried other paths, before posting this one, but could not find one. I'd really be glad to learn it! Thanks. – dfnu Oct 21 '19 at 12:05
  • $ \sin(y) \geq y/2 $ for small $ y $ will suffice. – Jake Mirra Oct 21 '19 at 15:41
  • @JakeMirra thanks! Got it! Very simple indeed. – dfnu Oct 21 '19 at 17:06
  • When estimating the second part of the series you seem to be using the inequality $g_n(a_k) \geq - \frac{1}{2^n \pi}$ for $n \geq N$, which is wrong in general (counterexample: $g_{N+1} \left(\frac{3}{4} \frac{\pi}{2^N}\right) = - \frac{2}{3 \pi} < - \frac{1}{2^{N+1} \pi}$). You might want to assume $\frac{\pi}{2^{N+1}} \leq a_k < \frac{\pi}{2^N}$ to get $g_n(a_k) > - \frac{2^{N+1}}{2^n \pi}$ instead. This eventually leads to $g(a_k) > \frac{2}{\pi} (N - 2)$, which is still sufficient to conclude that $g(a_k) \to \infty$ as $k \to \infty$. – ComplexYetTrivial Oct 23 '19 at 16:29