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Given $a>1, b>0$, and $N \in \mathbb{N}$, such that $b^N < a$, I'd love your help with proving that the series $$\sum_{n=1}^{\infty} \frac{\sin(b^nx)}{a^n}$$ converges to a function which is $N$ times differentiable for $x\in (0,1)$.

I tried to bound it and use Dirichlet's M-test so there's a uniform convergence

$$\sum_{n=1}^{N}\frac{\sin(b^nx)}{a^n}=\sum_{n=1}^{N}\frac{\sin(b^nx)}{a^n}+\sum_{N}^{\infty}\frac{\sin(b^nx)}{a^n} ,\\\ \sum_{N}^{\infty}\left|\frac{\sin(b^nx)}{a^n}\right|<\sum_{N}^{\infty}\frac{b^nx}{a^n}<\sum_{N}^{\infty}\frac{ax}{a^n}=\sum_{N}^{\infty}\frac{x}{a^{n-1}}.$$

How should I prove the aforementioned claim?

Thanks a lot!

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You don't want to truncate the series at $N$. You want to differentiate $k$ times, for $0\leq k\leq N$. – bgins Feb 12 '12 at 10:55
I need to show convergence to a function which I'll be able to differentiate it N times,I don't understand your answer. what is the function? Did you use Fourier series? – Jozef Feb 12 '12 at 11:00
up vote 1 down vote accepted

You're interested in the (real coefficient of the) imaginary part of $$ D^{(k)} \sum_{n=1}^{\infty} \frac{e^{ib^nx}}{a^n} = \sum_{n=1}^{\infty} D^{(k)} \frac{e^{ib^nx}}{a^n} = \sum_{n=1}^{\infty} i^k b^{kn} \frac{e^{ib^nx}}{a^n} $$ which converges for $\left|\frac{b^k}{a}\right|<1$ by the root test, since: $$ \left|i^k b^{kn} \frac{e^{ib^nx}}{a^n}\right| = \left|\frac{b^{kn}}{a^n}\right| = \left|\frac{b^{k}}{a}\right|^n. $$ The convenience of looking at the augmented complex series $$ \sum_{n=1}^{\infty} \frac{e^{ib^nx}}{a^n} = \sum_{n=1}^{\infty} a^{-n}\left(\cos{b^nx}+i\sin{b^nx}\right) $$ is that we don't need to bother with the fact that the sine and cosine functions, and the sign, keep swapping/interchanging (rotating quadrants) with each higher derivative.

Without this "convenience", we would proceed as follows. The original function $$ f(x)=\sum_{n=1}^{\infty} a^{-n} \sin{b^nx} $$ converges because $|\frac{\sin{b^nx}}{a^n}|^\frac1n\leq\frac1a<1$ by the root test. Likewise, the first derivative $$ f'(x)=\sum_{n=1}^{\infty} \left(\frac{b}a\right)^n \cos{b^nx} $$ converges because $|\frac{b^n\sin{b^nx}}{a^n}|^\frac1n\leq\frac{b}a<\frac{a^{1/N}}a<1$ by the root test. Continuing, we find that $$ f^{(k)}(x)= \left\{ \begin{matrix} \sum_{n=1}^{\infty} (-1)^\frac{k}{2} \left(\frac{b^k}a\right)^n \sin{b^nx} &\quad& 0\leq k~\text{even}\\ \sum_{n=1}^{\infty} (-1)^\frac{k-1}{2} \left(\frac{b^k}a\right)^n \cos{b^nx} &\quad& 1\leq k~\text{odd} \end{matrix} \right. $$ converge for $0\leq k\leq N$ because $$ \left|\frac{b^{kn}}{a^n}\right|^\frac1n\leq\frac{b^k}a<\frac{a^{k/N}}a<1 $$ (since $0 < b^N < a$ and $a>1$, and the sine and cosine functions are bounded in magnitude by one).

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Can you please explain a little bit more? I don't really understand. – Jozef Feb 12 '12 at 10:34
Thank you bgins! – Jozef Feb 12 '12 at 11:39

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