# Fourier Series Of $a^2-x^2$

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 '18 at 13:31
• I think $x\in (-a,a)$. – Nosrati Sep 14 '18 at 13:32
• @Nosrati $x\in (-\pi,\pi)$ but the function vanishes at $|x|\geq 1$ – newhere Sep 14 '18 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 '18 at 13:45
• I agree with this. – Nosrati Sep 14 '18 at 13:48