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Determine the function to which the Fourier series converges for $f(x)$ given the following

$$f(x)=x,~~~~~-\pi<x<\pi$$

Solution: \begin{align*} a_0&=\frac{1}{\pi}\int_{-\pi}^{\pi} x dx\\ &=\frac{1}{\pi}\left[\frac{x^2}{2}\right]_{-\pi}^{\pi}\\ &=0. \end{align*} Since $f(x)$ is odd so $a_n=0$.

\begin{align*} b_n&=\frac{1}{\pi}\int_{-\pi}^\pi f(x)\sin(nx)dx\\ &=\frac{1}{\pi}\int_{-\pi}^\pi x\sin(nx)dx\\ &=\frac{1}{\pi}\left[\frac{x\cos (nx)}{n}+\dfrac{\sin(nx)}{n^{2}}\right]_{-\pi}^{\pi}\\ &=-\frac{1}{n\pi}[\pi\cos(nx)-(-\pi)\cos(-n\pi)]\\ &= -\frac{1}{n\pi}\cdot 2\pi\cos(n\pi)\\ & = \frac{2\cdot(-1)^{n+1}}{n} \end{align*}

Therefore, the required fourier series is $$f(x)=\sum_{n=1}^\infty \frac{2\cdot(-1)^{n+1}}{n} \sin nx.$$

Help with the second part! Thanks in advance!

Unknown
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    At $\pm \pi$ the series has all terms $0$ and so the sum is $0$. At other points it converges to $x$ by basic results on convergence of Fourier series. – geetha290krm Oct 04 '22 at 06:05
  • Yeah right! can you please solve it for me, I am a bit confused – Unknown Oct 04 '22 at 06:07
  • To be precise the series converges to the periodic extension of $x$ outside $(-\pi,\pi)$ and is zero at all integer multiples of $\pi$. The graph looks like a sawtooth. – lcv Oct 04 '22 at 07:43
  • $$ \sum\limits_{k = 1}^\infty {\frac{{2( - 1)^{n + 1} }}{n}\sin (nx)} = 2\operatorname{Im} \sum\limits_{k = 1}^\infty {\frac{{( - 1)^{n + 1} }}{n}e^{ixn} } = 2\operatorname{Im} \log (1 + e^{ix} ) = x + 2\pi \left\lfloor \frac{\pi-x}{2\pi} \right\rfloor $$ – Gary Oct 04 '22 at 08:40
  • Shouldn't one be more careful with the series $\sum_{n \ge 1}\frac{(-1)^{n + 1}z}{n}$? Because radius of convergence of this series is $1$ and there is a singularity at $z = -1$. – Pavel Gubkin Oct 05 '22 at 06:17

1 Answers1

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Your question follows from the following general theorem:

If the function $F$ is differentiable at the point $x_0$ then its Fourier series at the point $x_0$ converges to $F(x_0)$.

Gary
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Pavel Gubkin
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    Probably you want to show the convergence by hands? For example, one way to show the convergence (not necessarily to $f(x)$) of the series is the Dirichet test for the monotone sequence $a_n = \frac{1}{n}$ and the sequence $b_n = (-1)^{n+1}\sin(nx)$ satisfying $\sup_N|\sum_{n = 1}^Nb_n| < \infty$. – Pavel Gubkin Oct 04 '22 at 07:09
  • Thanks Pavel! Please look the title, which convergent I need to find. Can I just show that Foueries series of $f(x)$ converges or converges to $f(x)$? Actually my English is weak. – Unknown Oct 05 '22 at 05:11
  • It does converge to $f(x)$ on $[-\pi, \pi)$ and to the periodic continuation of $f$ everywhere else. – Pavel Gubkin Oct 05 '22 at 06:18
  • The function is piecewise smooth and, in the interval $(-\pi,\pi)$ its periodic extension is same as original function. Hence the Fourier series converges to $f(x)$ for all $x\in(-\pi,\pi)$. Is this OK? – Unknown Oct 05 '22 at 14:50
  • Yes, it is absolutely correct – Pavel Gubkin Oct 05 '22 at 18:42