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$$f(x)=\pi-x \qquad x \in [0,2 \pi[$$
$$a_0=\frac{1}{\pi} \ \int_0^{2 \pi}f(x) \ dx=\frac{1}{\pi} \ \int_0^{2 \pi} (\pi-x) \ dx=0$$


$$a_n=\frac{1}{\pi} \ \int_0^{2 \pi} \cos(nx) \ dx=\frac{1}{\pi} \ \int_0^{2 \pi}(\pi \ \cos(nx)-x \ \cos(nx)) \ dx=$$ $$\frac{1}{\pi} \ \Big( \ \Big[\frac{\pi}{n} \ \sin(nx) \Big]_0^{2 \pi}-\Big[\frac{x}{n} \ \sin(nx)+\frac{1}{n^2} \ \cos(nx) \Big]_0^{2 \pi} \Big)=0 $$


$$b_n=\frac{1}{\pi} \ \Big( \ \Big[\ -\frac{\pi}{n} \ \cos(nx) \Big]_0^{2 \pi}-\Big[-\frac{x}{n} \ \cos(nx)+\frac{1}{n^2} \ \sin(nx) \Big]_0^{2 \pi} \Big)=\frac{2}{n}$$

$$f(x)=2 \sum_{n=1}^{+\infty} \frac{1}{n} \ \sin(nx)$$

Is it correct?

Elsa
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    That is the correct approach. A plot of your answer also shows that the approximation is tending to the desired function http://www.wolframalpha.com/input/?i=plot+sum_%7Bn%3D1%7D%5E%7B20%7D+2sin(nx)%2Fn+for+x+in+(0,2pi) – Hugh Sep 30 '16 at 15:14

2 Answers2

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Slight errors: $$ \begin{align} % a_{0} &= \frac{1}{\pi} \int_{-\pi }^{\pi } f(x) \, dx = 2\pi \\[5pt] % b_{k} &= \frac{1}{\pi } \int_{-\pi }^{\pi } f(x) \sin (k x) \, dx = (-1)^k \frac{2}{k \pi} % \end{align} $$ The decay of the amplitudes is linear:

amps

The series expansion looks like $$ \pi - x = \pi - 2 \sin (x) + \sin (2x) - \frac{2}{3} \sin (3x) + \frac{1}{2} \sin \left( 4x \right) - \dots $$

Convergence sequence:

1 2 10 100

dantopa
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Any function $F(x)$ defined on $[0,2\pi]$ expands as the Fourier series $$ F(x) = \dfrac{A_0}2 + \sum^\infty_{n=1}A_n\cos(nx) + \sum^\infty_{n=1} B_n \sin(nx),$$ $$\text{where} \qquad A_0 = \frac1\pi\!\int_0^{2\pi}\!\!\!\!\!F(x)dx, \quad A_n = \frac1\pi\!\int_0^{2\pi}\!\!\!\!\!F(x)\cos(nx)dx, \quad B_n = \frac1\pi\!\int_0^{2\pi}\!\!\!\!\!F(x)\sin(nx)dx. $$ Thus, for $f(x) = \pi-x$, these coefficients are computed as follows: $$\begin{split} a_0 &= \frac1\pi\!\int_0^{2\pi}\!\!\!(\pi-x)dx = 0, \\ a_n &= \frac1\pi\!\int_0^{2\pi}\!\!\!(\pi-x)\cos(nx)dx = \frac{1-\cos(2n\pi)-n\pi\sin(2n\pi)}{n^2\pi}, \\ b_n &= \frac1\pi\!\int_0^{2\pi}\!\!\!(\pi-x)\sin(nx)dx = \frac{2\cos(n\pi)\big[n\pi\cos(n\pi)-\sin(n\pi)\big]}{n^2\pi}. \end{split}$$ Therefore, its Fourier series is given by $$\sum_{n=1}^\infty \frac1{n^2\pi} \Big\{ \big[ 1 -\cos(2n\pi) -n\pi\sin(2n\pi) \big] \cos(nx) + 2\cos(n\pi) \big[ n\pi\cos(n\pi) -\sin(n\pi) \big] \sin(nx) \Big\}.$$

phydev
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