Fourier series function

Question

$f(x) = x$ , $f(x+2\pi) = f(x) $ on $ [-\pi , \pi] $

How do I know that this function is even or odd? My book says odd, but I don't understand how to work this out?

also why does $a_0 = 0$ and $a_n = 0$?

since its an odd function I thought we use the even extension?

i.e $$ a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(nx)dx $$

but the answer is

$$ b_{n} = \frac{-2}{n}\cos(n\pi) = \frac{2(-1)^{n+1}}{n} $$

Answer 1

The integral of an odd function over symmetric limits is zero. To see this, we observe

$$\begin{align} \int_{-a}^af(x)dx&=\int_{-a}^0f(x)dx+\int_{0}^af(x)dx\tag1\\\\ &=-\int_{0}^af(x)dx+\int_{0}^af(x)dx\tag2\\\\ &=0 \end{align}$$

where we used the substitution $x \to -x$ in going from $(1)$ to $(2)$.

We also have that the product of an odd function and an even function is an odd function. We can show this as follows. Let $f$ be even and let $g$ be odd and let $h=fg$. Then

$$h(-x)=f(-x)g(-x)=f(x)(-g(x))=-h(x)$$

as was to be shown.

For the problem at hand, note that $\cos nx$ is even about $x=0$ and $\sin nx$ is odd about $x=0$. We also note that $x$ is odd about $x=0$. Thus, we have

$$a_n=\frac{1}{\pi}\int_{-\pi}^{\pi}x \cos (nx)\,dx=0$$

Answer 2

If $f(-x)=-f(x)$, then we say $f$ is odd. On the other hand if $f(-x)=f(x)$ we say $f$ is even.

The general Fourier representation of $f$ is

$$ f(x)=\frac{a_0}{2}+\sum_{n=1}^\infty \biggl[ a_{n}\cos\biggl(\frac{n\pi x}{L}\biggl)+b_{n} \sin\biggl(\frac{n\pi x}{L}\biggl) \biggl]\qquad for~-L\leq x\leq L $$ where \begin{align*} a_0&=\frac{1}{L}\int_{-L}^{L} f(x)dx \\ a_n&=\frac{1}{L}\int_{-L}^{L} f(x)\cos\biggl(\frac{n\pi x}{L}\biggl)dx \\ b_n&=\frac{1}{L}\int_{-L}^{L} f(x)\sin\biggl(\frac{n\pi x}{L}\biggl)dx \end{align*} And we represent even functions using Fourier cosine since cosine is an even function. For the same reason we use sine function to represent odd function.

In your case $f$ is an odd function that is why the coefficient of cosine becomes zero.

To calculate $b_n$, just plug in the values $L=\pi$ and $f(x)=x$ $$ b_n =\frac{1}{\pi}\int_{-\pi}^{\pi} x\sin(nx)dx $$ Hint: use integration by parts to evaluate the integral.

Hence the answer!