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Please take a look at below Fourier series :

$ f(x)=\begin{cases} x+1 &-1<x<0\\ 1-x & 0<x<1 \end{cases} $

I tried to solve it as follows :

$ a_n=\displaystyle \dfrac{1}{1}\int_{-1}^1f(x)\cos (n\pi x) dx=2\int_{0}^{1} (1-x)\cos( n\pi x) dx $

solving $a_n$ comes to :

$ a_n=2\left[\dfrac{(-1)^{n}}{n^2\pi^2}+\dfrac{1}{n^2\pi^2}\right] $

also solving $b_n=0$ then we have Fourier series as below :

$ \displaystyle f(x)=\sum_{n=1}^{\infty}\left(2\left[\dfrac{(-1)^{n}}{n^2\pi^2}+\dfrac{1}{n^2\pi^2}\right]\cos\frac{n\pi}{1}x\right) $

I have doubts about my recognition about function if it is odd or even. (Is my solution right?)

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  • $\begingroup$ So what is your question? Are you asking if your answer is correct? $\endgroup$
    – user296602
    Dec 28, 2015 at 4:27
  • $\begingroup$ Excuse me I typed question in a hurry to go somewhere...That's why I couldn't edit context too fast.please now take it out of "on hold". $\endgroup$
    – sajjad
    Dec 28, 2015 at 14:34
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    $\begingroup$ I found $a_0=1$ and $a_n=\frac{2}{n^2\pi^2}\left(1-(-1)^n\right)$ for all $n\geq 1$. $\endgroup$
    – Nicolas
    Dec 28, 2015 at 16:44
  • $\begingroup$ @Nicolas thank you.It's almost same except '+' sign. Can you explain it or leave an answer $\endgroup$
    – sajjad
    Dec 28, 2015 at 17:31
  • $\begingroup$ @sajjad I wrote my computations, please check if you agree with them. $\endgroup$
    – Nicolas
    Dec 28, 2015 at 21:32

1 Answer 1

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As asked in a comment, I compute the Fourier coefficients :

\begin{align} a_0(f) & = \int_{-1}^{1}f(x)\mathrm{d}x \\ & = \int_{-1}^{0}\left(x+1\right)\mathrm{d}x + \int_{0}^{1}\left(1-x\right)\mathrm{d}x \\ & = \left[\frac{x^2}{2}+x\right]_{-1}^{0} + \left[x-\frac{x^2}{2}\right]_{0}^{1} \\ & = \left[-\frac{1}{2}+1\right] + \left[1-\frac{1}{2}\right] \\ & = \frac{1}{2} + \frac{1}{2} \\ & = 1. \end{align}

For all integer $n\geq 1$, \begin{align} a_n(f) & = \int_{-1}^{1}f(x)\cos\left(n\pi x\right)\mathrm{d}x \\ & = \int_{-1}^{0}\left(x+1\right)\cos\left(n\pi x\right)\mathrm{d}x + \int_{0}^{1}\left(1-x\right)\cos\left(n\pi x\right)\mathrm{d}x \\ & = \left[\left(x+1\right)\frac{\sin\left(n\pi x\right)}{n\pi}\right]_{-1}^{0} - \frac{1}{n\pi}\int_{-1}^{0}\sin\left(n\pi x\right)\mathrm{d}x \\ & + \left[\left(1-x\right)\frac{\sin\left(n\pi x\right)}{n\pi}\right]_{0}^{1} + \frac{1}{n\pi}\int_{0}^{1}\sin\left(n\pi x\right)\mathrm{d}x \\ & = - \frac{1}{n\pi}\int_{-1}^{0}\sin\left(n\pi x\right)\mathrm{d}x + \frac{1}{n\pi}\int_{0}^{1}\sin\left(n\pi x\right)\mathrm{d}x \\ & = + \frac{1}{n\pi}\left[\frac{\cos\left(n\pi x\right)}{n\pi}\right]_{-1}^{0} - \frac{1}{n\pi}\left[\frac{\cos\left(n\pi x\right)}{n\pi}\right]_{0}^{1} \\ & = \frac{1}{n^2\pi^2}\left[1-(-1)^n\right] - \frac{1}{n^2\pi^2}\left[(-1)^n-1\right] \\ & = \frac{2}{n^2\pi^2}\left(1-(-1)^n\right). \end{align}

Since $f$ is odd (i.e. $b_n(f)=0$ for all integer $n\geq 1$), the Fourier's series is then formally given by \begin{align} S_n(f)(x) & = \frac{a_0(f)}{2} + \sum_{n=1}^{+\infty}a_n(f)\cos\left(n\pi x\right) \\ & = \frac{1}{2} + \sum_{n=1}^{+\infty}\frac{2}{n^2\pi^2}\left(1-(-1)^n\right)\cos\left(n\pi x\right) \\ & = \frac{1}{2} + 4\sum_{p=0}^{+\infty}\frac{\cos\left((2p+1)\pi x\right)}{(2p+1)^2\pi^2} \end{align} where the last inequality is due to the fact that $1-(-1)^n=0$ if $n$ is even.

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  • $\begingroup$ two thumps up.. $\endgroup$
    – sajjad
    Dec 29, 2015 at 15:04

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