1. Find Fourier coefficients of $ f(x)= \begin{cases} a^2-x^2, \mid x\mid<1 \\ 0, \mid x\mid\geq1\\ \end{cases} $ when $x\in (-\pi,\pi)$

  2. What is the value at $x=1$

  3. for which values of $a$ the series absloute convergent

1) The function is even so we have to evaluate $a_0,a_n$

$$a_0=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)dx$$

But the function is zero for $\mid x\mid\geq1$ and so is it integral so we get:

$$a_0=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)dx=\frac{1}{\pi}\int_{-1}^{1}(a^2-x^2)dx=\frac{1}{\pi} \left(2a^2-\frac{2}{3}\right)$$

$$a_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x) \cos(nx) dx=\frac{1}{\pi}\int_{-1}^{1}(a^2-x^2) \cos(nx) dx=\frac{1}{\pi}[2a^2\frac{\sin x}{n}+\frac{4}{n^2}\cos x-2(\frac{2}{n^3}-\frac{1}{n})\sin x]=\\=\frac{2a^2n^2+2n^2-4}{n^3\pi}\sin x+\frac{4}{n^2\pi}\cos x$$

Is the limit of integration are correct or should I use:

$$a_0=\frac{2}{2}{\pi}\int_{-1}^{1}f(x)dx$$

and

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

  1. using dirichlet we can see the $lim_{x\to 1^{+/-}}f'(x)$ does not exist is it correct?

  2. How can I valuate it?

  • $x\in (-\pi,\pi)$ or $x\in (-1,1)$.? – Nosrati Sep 14 at 13:31
  • I think $x\in (-a,a)$. – Nosrati Sep 14 at 13:32
  • @Nosrati $x\in (-\pi,\pi)$ but the function vanishes at $|x|\geq 1$ – newhere Sep 14 at 13:33
  • @Nosrati it is more correct to say that it is $\int_{-\pi}^{\pi}=\int_{-\pi}^{-1}+\int_{-1}^{1}+\int_{1}^{\pi}=0+\int_{-1}^{1}+0$ – newhere Sep 14 at 13:45
  • 1
    I agree with this. – Nosrati Sep 14 at 13:48

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