- #1
mathmari
Gold Member
MHB
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Hey! 
I want to find the Fourier series of the following function :
$$g: [-\pi, \pi]\rightarrow \mathbb{R} \\ g(x)=\left\{\begin{matrix}
-\frac{\pi+x}{2} & , -\pi \leq x \leq 0\\
\frac{\pi-x}{2} & , 0<x\leq \pi
\end{matrix}\right.$$
I have done the following:
$$g \sim \frac{a_0}{2}+\sum_{k=1}^{\infty}(a_k \cos (kx)+b_k \sin (kx))$$
$$a_k=\frac{1}{\pi}\int_{\pi}^{\pi} g(x) \cos (kx)dx=\frac{1}{\pi}\int_{-\pi}^0-\frac{\pi+x}{2}\cos (kx)dx+\frac{1}{\pi}\int_0^{\pi}\frac{\pi-x}{2}\cos (kx)dx=\dots=0, k=0, 1, 2, 3, 4, 5, \dots $$
$$b_k=\frac{1}{\pi}\int_{-\pi}^{\pi}g(x)\sin (kx)dx=\frac{1}{\pi}\int_{-\pi}^0-\frac{\pi+x}{2}\sin (kx)dx+\frac{1}{\pi}\int_0^{\pi}\frac{\pi-x}{2}\sin (kx)dx =\dots=\frac{1}{k}, k=1, 2, 3, 4, 5, \dots $$
So, $$g \sim \sum_{k=1}^{\infty}\frac{1}{k}\sin (kx)$$
Is this correct??
How can I find the function at which the Fourier series of $g$ converges?? Do we have to use the following theorem:
Let $f: [0, 2 \pi] \rightarrow \mathbb{R}$ piecewise $C^1[0, 2\pi]$, and $$f \sim \frac{a_0}{2}+\sum_{k=1}^{\infty}(a_k \cos (kx)+b_k \sin (kx))$$
Then $$\frac{a_0}{2}+\sum_{k=1}^{\infty}(a_k\cos (kx)+b_k \sin (kx))=\left\{\begin{matrix}
\frac{f(x^+)+f(x^-)}{2} & , \forall x \in (0,2\pi)\\
\frac{f(0^+)+f(2\pi^-)}{2} & , \forall x \in \{0, 2\pi\}
\end{matrix}\right.$$
?? (Wondering) Is it then as followed??
$$\sum_{k=1}^{\infty}\frac{1}{k}\sin (kx)=\left\{\begin{matrix}
\frac{g(x^+)+g(x^-)}{2}=g(x) & , \forall x \in (-\pi, 0)\\
\frac{g(x^+)+g(x^-)}{2}=g(x) & , \forall x \in (0, \pi)\\
\frac{g(0^+)+g(0^-)}{2}=\frac{\frac{\pi}{2}-\frac{\pi}{2}}{2}=0 & , x=0
\end{matrix}\right.$$
I want to find the Fourier series of the following function :
$$g: [-\pi, \pi]\rightarrow \mathbb{R} \\ g(x)=\left\{\begin{matrix}
-\frac{\pi+x}{2} & , -\pi \leq x \leq 0\\
\frac{\pi-x}{2} & , 0<x\leq \pi
\end{matrix}\right.$$
I have done the following:
$$g \sim \frac{a_0}{2}+\sum_{k=1}^{\infty}(a_k \cos (kx)+b_k \sin (kx))$$
$$a_k=\frac{1}{\pi}\int_{\pi}^{\pi} g(x) \cos (kx)dx=\frac{1}{\pi}\int_{-\pi}^0-\frac{\pi+x}{2}\cos (kx)dx+\frac{1}{\pi}\int_0^{\pi}\frac{\pi-x}{2}\cos (kx)dx=\dots=0, k=0, 1, 2, 3, 4, 5, \dots $$
$$b_k=\frac{1}{\pi}\int_{-\pi}^{\pi}g(x)\sin (kx)dx=\frac{1}{\pi}\int_{-\pi}^0-\frac{\pi+x}{2}\sin (kx)dx+\frac{1}{\pi}\int_0^{\pi}\frac{\pi-x}{2}\sin (kx)dx =\dots=\frac{1}{k}, k=1, 2, 3, 4, 5, \dots $$
So, $$g \sim \sum_{k=1}^{\infty}\frac{1}{k}\sin (kx)$$
Is this correct??
How can I find the function at which the Fourier series of $g$ converges?? Do we have to use the following theorem:
Let $f: [0, 2 \pi] \rightarrow \mathbb{R}$ piecewise $C^1[0, 2\pi]$, and $$f \sim \frac{a_0}{2}+\sum_{k=1}^{\infty}(a_k \cos (kx)+b_k \sin (kx))$$
Then $$\frac{a_0}{2}+\sum_{k=1}^{\infty}(a_k\cos (kx)+b_k \sin (kx))=\left\{\begin{matrix}
\frac{f(x^+)+f(x^-)}{2} & , \forall x \in (0,2\pi)\\
\frac{f(0^+)+f(2\pi^-)}{2} & , \forall x \in \{0, 2\pi\}
\end{matrix}\right.$$
?? (Wondering) Is it then as followed??
$$\sum_{k=1}^{\infty}\frac{1}{k}\sin (kx)=\left\{\begin{matrix}
\frac{g(x^+)+g(x^-)}{2}=g(x) & , \forall x \in (-\pi, 0)\\
\frac{g(x^+)+g(x^-)}{2}=g(x) & , \forall x \in (0, \pi)\\
\frac{g(0^+)+g(0^-)}{2}=\frac{\frac{\pi}{2}-\frac{\pi}{2}}{2}=0 & , x=0
\end{matrix}\right.$$