Proving a function is continuous and periodic

Question

Suppose we are given a function $$g\left ( x \right )= \sum_{n=1}^{\infty}\frac{\sin \left ( nx \right )}{10^{n}\sin \left ( x \right )},x\neq k\pi , k\in\mathbb{Z}$$ and $$g\left ( k\pi \right )=\lim _{x\rightarrow k\pi}g\left ( x \right )$$ I found that $\lim _{x\rightarrow k\pi}g\left ( x \right )= \sum_{n=1}^{\infty}\frac{1}{10^{n}}$ for both odd and even $k$. However, I am still unsure how to proceed from here. What does continuity even mean for functions like this? If I wanted to take the derivative of this function, surely I must prove first that it converges and then I would apply differentiation term by term. Proving periodicity for one term would be easy enough if it weren't for the $10^{n}$ term in the denominator. How to deal with it? EDIT: How to find the Fourier series of this function( by first proving that Dirichlet's conditions are met)

Answer 1

It is a basic precalculus fact that for arbitrary complex $z$ with $|z|<1$ one has $$\sum_{k=0}^\infty z^k={1\over 1-z}\ .$$ Furthermore, by Euler's equation, for arbitrary real $x$ we may write $\sin x={\rm Im}\>e^{ix}$ . Putting these two together we obtain $$\eqalign{\sum_{n=1}^\infty{\sin (nx)\over 10^n}&={\rm Im}\sum_{n=1}^\infty\left({e^{ix}\over10}\right)^n={\rm Im}{e^{ix}/10\over 1-e^{ix}/10}\cr &={\rm Im}{e^{ix}(10-e^{-ix})\over (10-e^{ix})(10-e^{-ix})}={10\sin x\over101-20\cos x}\ .\cr}$$ It follows that $$g(x)={10\over 101-20\cos x}\qquad(x\in{\mathbb R}\setminus \pi{\mathbb Z})\ .\tag{1}$$ It follows that the apparent singularities at integer $x$ are removable, so that $g$ has a natural $C^\infty$ extension to all of ${\mathbb R}$, and this extension is $2\pi$-periodic. Note that the initially given expression for $g$ is not its Fourier series. In order to obtain the latter one would have to compute the integrals $$a_k:={1\over\pi}\int_0^{2\pi}g(x)\>\cos(kx)\>dx\ ,$$ using the expression $(1)$ for $g$. The resulting expressions for the $a_k$ will not be simple.

Answer 2

I am not sure that I arrive to the same result $$g(k\pi)=\sum_{n=1}^{\infty}\frac{1}{10^{n}}=\frac{1}{9}$$

Using complex representation of $\sin(x)$, what I got is that $$g\left ( x \right )= \sum_{n=1}^{\infty}\frac{\sin \left ( nx \right )}{10^{n}\sin \left ( x \right )}=-\frac{10 e^{i x}}{\left(-10+e^{i x}\right) \left(-1+10 e^{i x}\right)}$$ So, $g\big(2k\pi\big)=\frac{10}{81}$, $g\big((2k+1)\pi\big)=\frac{10}{121}$.

Answer 3

You should first examine the series

$\sum\limits_{n = 1}^\infty {\frac{{\sin (nx)}}{{{{10}^n}}}} $ and check that this is a continuous function.

Looking at the form of the series, think of geometric series

$f(z) = \sum\limits_{n = 1}^\infty {\frac{{{z^n}}}{{{{10}^n}}}} $ . Then by the theory of power series, $f(z)$ converges in a disk of radius 10. Therefore, it converges uniformly to a differentiable function in the closed unit disk.

Now $\sum\limits_{n = 1}^\infty {\frac{{\sin (nx)}}{{{{10}^n}}}} $ is just the imaginary part of $f(z)$, i.e., $Im(f(z))$ on the unit circle. Now

$f(z) = \sum\limits_{n = 1}^\infty {\frac{{{z^n}}}{{{{10}^n}}}} = \frac{1}{{1 - {\textstyle{z \over {10}}}}} - 1 = \frac{z}{{10 - z}}$ and so

$f({e^{ix}}) = \frac{{{e^{ix}}}}{{10 - {e^{ix}}}} = \frac{{10\cos (x) - 1}}{{101 - 20\cos (x)}} + i\frac{{10\sin (x)}}{{101 - 20\cos (x)}}$.

Thus $\sum\limits_{n = 1}^\infty {\frac{{\sin (nx)}}{{{{10}^n}}}} $ converges uniformly to the differentiable function

$\frac{{10\sin (x)}}{{101 - 20\cos (x)}}$ on the whole real line.

Therefore, for $x$ not a multiple of $\pi $,

$g(x) = \frac{{10}}{{101 - 20\cos (x)}}$ .

Since $\mathop {\lim }\limits_{x \to k\pi } g(x) = \mathop {\lim }\limits_{x \to k\pi } \frac{{10}}{{101 - 20\cos (x)}} = \frac{{10}}{{101 - 20\cos (k\pi )}}$ , we may consider $g(x)$ as given by $g(x) = \frac{{10}}{{101 - 20\cos (x)}}$ by giving it the value of its limit at all multiple of $\pi $. Thus, taking the limit or not, $g(2k\pi ) = \frac{{10}}{{101 - 20\cos (2k\pi )}} = \frac{{10}}{{81}}$ and $g((2k + 1)\pi ) = \frac{{10}}{{101 - 20\cos ((2k + 1)\pi )}} = \frac{{10}}{{121}}$ .

You can now use $g(x)$ in this form to find its Fourier series.

g is an even function of period $ 2 \pi $. So it has a cosine Fourier series with coefficients $ {a_n}$ given by

${a_n} = \frac{1}{\pi }\int_0^{2\pi } {\frac{{10\cos (nx)}}{{101 - 20\cos (x)}}dx} = \frac{2}{{99 \times {{10}^{n - 1}}}}$ , for $ n = 0, 1 ,2, ... $

You can use complex contour integration to evaluate the integrals.

Answer 4

First check that function converge uniformly. It sure does, on this union: $x \in I = \bigcup_{k=-\infty}^{\infty} (\pi k,\, \pi (k+1))$ because: $$ \sum_{n=1}^{\infty} \frac{1}{10^n} \frac{\sin(nx)}{\sin(x)} \le \sum_{n=1}^{\infty} \frac{1}{10^n} \frac{|\sin(nx)|}{|\sin(x)|} \le \sum_{n=1}^{\infty} \frac{n}{10^n} $$ Now in singular points $\lim_{x \to k \pi} \; g(x) = \sum_{n \in \mathbb{N}} \frac{1}{10^n} \frac{\sin(nx)}{\sin(x)}$ . The interesting part here is just $\frac{\sin(n x)}{\sin(x)}$. Here is pretty handy formula that $\sin(x + y) = \sin(x)\cos(y) + \cos(x)\sin(y)$. You can now rewrite the limit: $$ \lim_{x \to k \pi} 10^{-n} \frac{\sin(xn)}{\sin(x)} = \lim_{\varepsilon \to 0} 10^{-n}\frac{\sin(n\varepsilon + n k \pi)}{\sin(\varepsilon + k \pi)} = \lim_{\varepsilon \to 0} 10^{-n}\frac{\sin(n\varepsilon)(-1)^{nk} + 0}{\sin(\varepsilon) (-1)^k + 0} = 10^{-n} (-1)^{-n} n $$ Now you can see that it is continous on whole $\mathbb{R}$.
Continuity: $\lim_{x \to x'} g(x) = g(x')$. 10^{-n} does not affect periodicity of this function. To prove periodicity, you just have to prove that there exists such $\exists T \in \mathbb{R}\forall t \in \mathbb{R}: g(t + T) = g(t)$. There is such $T$ and it is equal to $2 \pi$. Just use that trick that is above in section where I tried to estimate limit in $k \pi$. If you look at the function you can see that the function is even. So you dont have to calculate odd coeficients in fourier series. $$ a_m = \frac{2}{2 \pi} \int_{-\pi}^{\pi} \sum_{n \in \mathbb{N}} 10^{-n} \frac{\sin(nx)}{\sin(x)} \cos\bigg(m \frac{2 \pi x}{2\pi}\bigg) \mathrm{d}x = \frac{2}{\pi} \sum_{n \in \mathbb{N}} \int_{0}^{\pi} 10^{-n} \frac{\sin(nx)}{\sin(x)} \cos(m x) \mathrm{d}x $$