# Fourier series of a function

• Jul 27th 2010, 09:33 AM
arbolis
Fourier series of a function
The exercise states "Let $\displaystyle f(x)$ be a periodic function with period $\displaystyle 2 \pi$ defined in the interval $\displaystyle [- \pi , \pi]$ by $\displaystyle f(x)=1$ if $\displaystyle 0 \leq x \leq \pi$ and $\displaystyle f(x)=0$ if $\displaystyle -\pi \leq x <0$.
Calculate the Fourier coefficients of f and obtain its Fourier series and analyze the convergence of the series.
My attempt: First of all I think they made an error and meant to say the interval $\displaystyle [- \pi , \pi)$. Otherwise it's not making sense to me.

Ok so I read the definition of the nth coefficient of the Fourier series in my class notes and I read $\displaystyle \hat f(n)= \frac{1}{2 \pi} \int _{- \pi}^{\pi} f(\theta ) e^{-in\theta } d\theta$.
So I calculated $\displaystyle \hat f(0)$ to be worth $\displaystyle \frac{1}{2}$.
For all $\displaystyle n$ even and $\displaystyle n \neq 0$, I found out that $\displaystyle \hat f(n)=0$. While for all $\displaystyle n$ uneven, I found out that $\displaystyle \hat f(n)=-\frac{1}{\pi n}$.
Since the Fourier series is defined as $\displaystyle \sum _{- \infty}^{\infty} \hat f(n) e^{in\theta}$, I reach that the Fourier series is worth $\displaystyle \frac{1}{2}+ \sum _{n=-\infty}^{\infty} \frac{- \cos (n \theta)+ i \sin (n \theta)}{n \pi}$ where $\displaystyle n=2k+1$ with $\displaystyle k \in \mathbb{Z}$, $\displaystyle n\neq 0$.
To check out the convergence of the series, I guess I should split the series in 2, starting from $\displaystyle -\infty$ to $\displaystyle 0$ and from $\displaystyle 0$ to $\displaystyle +\infty$, and then apply the quotient test?
Is what I've done right? I'm not confident but can't find any error.

Edit: I forgot to multiply by "i" for the series of the uneven terms I think. I don't think the result change that much, except with an "i" factor.