Confused computing sum of Fourier series

Question

I am having some issues understanding Fourier series and I am stuck trying to solve a problem.

So the function $u$ has period $2\pi$ and is defined as

$$u(x) = \begin{cases} 1 & 0 \leq x \lt \pi \\ 0 & \pi \leq x \lt 2\pi \\ \end{cases}, $$ And I should determine the Fourier series and compute the sum of the series at $x = 0$ and $x = \pi$.

So I managed to compute the Fourier series so that

$$u(x) = \frac{1}{2} + \sum_{-\infty,\space n \space \mathrm{odd}}^{\infty} \frac{e^{inx}}{in\pi}.$$

But now I am wondering, first of all, the sum at $x = 0$ should, according to the book, be $\frac{1}{2},$ but in the beginning it is stated that $u(0) = 1$. Is that just the Fourier representation doing a "bad job"?

Secondly, I keep failing when I am trying to compute the sum. Beginning with $x=0$, I tried to compute

$$u(0) = \frac{1}{2} + \frac{1}{i\pi} \sum_{-\infty}^{\infty} \frac{1}{2n-1}.$$ I wrote it like that because $n$ is odd. But doesn't $\sum_{-\infty}^{\infty} \frac{1}{2n-1}$ diverge?

Answer 1

Concerning the "bad job", please refer to the Dirichlet Fourier Series Conditions (Dirichlet's Theorem). As Bungo mentioned, to address the issue of the divergence of $\sum {1\over 2n-1}$, it is best to write the series as a limit: $$u(x)=\lim_{N\rightarrow \infty}u_N(x)$$ where $$u_N(x)=\frac 12 +\frac1{i\pi}\sum_{-N,\ n \text{ odd}}^N\frac {e^{inx}}{n}$$ $$u_N(0)=\frac 12+\frac 1 {i\pi} \sum_{-N, \ n \text{ odd}}^N \frac 1n$$

$$u_N(0)=\frac 12+\frac 1 {i\pi} \left(\sum_{0}^N \frac 1{2n+1}-\sum_{0}^N \frac 1{2n+1}\right)$$ $$u_N(0)=\frac 12$$ $$u(0)=\lim_{N\rightarrow \infty}u_N(0)=\frac12$$

Answer 2

$$a_0=\frac{1}{\pi}\int_{0}^{2\pi}f(x)dx=\frac{1}{\pi}\int_{0}^{\pi}1dx=1$$ for $n=1,2\cdots$ $$a_n=\frac{1}{\pi}\int_{0}^{2\pi}f(x)\cos nx\, dx=\frac{1}{\pi}\int_{0}^{\pi}\cos nx\,dx=0$$ $$b_n=\frac{1}{\pi}\int_{0}^{2\pi}f(x)\cos nx\, dx=\frac{1}{\pi}\int_{0}^{\pi}\sin nx\,dx=\frac{1+(-1)^{n+1}}{n\pi}$$ therefore $$f(x)=\frac{1}{2}+\sum_{n=1}^{\infty}\left(\frac{1+(-1)^{n+1}}{n\pi}\right)\sin nx$$

Answer 3

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \,\mathrm{u}\pars{x} & = \sum_{n = -\infty}^{\infty} \,\hat{\mathrm{u}}_{n}\expo{\ic n x}\ \imp\ \int_{0}^{2\pi}\,\mathrm{u}\pars{x}\expo{-\ic nx}\,{\dd x \over 2\pi} = \sum_{m = -\infty}^{\infty} \,\hat{\mathrm{u}}_{m}\int_{0}^{2\pi}\expo{\ic\pars{m - n}x}\,{\dd x \over 2\pi} = \,\hat{\mathrm{u}}_{n} \\[4mm] \,\hat{\mathrm{u}}_{n} & = \half\,\delta_{n0} + \bracks{n \not= 0}\int_{0}^{\pi}\expo{-\ic nx} \,{\dd x \over 2\pi} = \half + \bracks{n \not= 0}{\expo{-\ic n\pi} - 1 \over -2\pi\ic n} \\[4mm] & = \half\,\delta_{n0} - \bracks{n \not= 0}{\ic \over 2\pi}\,{1 - \pars{-1}^{n} \over n} \end{align}

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$$ \color{#f00}{\,\mathrm{u}\pars{x}} = \color{#f00}{\half -{\ic \over \pi} \sum_{n = -\infty}^{\infty}\,{\expo{\ic\pars{2n + 1}x} \over 2n + 1}} $$