Sketch each function and determine its Fourier series. $f(x)=x,\, 1<x<2$

Question

Sketch each function and determine its Fourier series . $f(x)=x$, $1<x<2$

Could you help me with this problem. My doubt is the interval given to me that is $1 <x <2$, it is not a symmetric interval as they are in many exercises ranging from $[-\pi ,\pi ]$ or $[-1 ,1 ]$ etc, if I have the Fourier coefficients:

\begin{align} a_0 &=\frac{1}{L}\int_{-L}^{L}f(x)dx \\ a_n &=\frac{1}{L}\int_{-L}^{L}f(x) \cos\left ( \frac{n\pi x}{L} \right )dx \\ b_{n} &=\frac{1}{L}\int_{-L}^{L}f(x) \sin\left ( \frac{n\pi x}{L} \right )dx \end{align}

I see that the limits of integration are for a symmetric interval, how can I solve this problem. Would you have to modify the equations of Fourier coefficients?

Answer 1

Please notice that in Fourier coefficient, you just need to calculate integration in one period, it's not necessary to be symmetric around origin. i.e.

\begin{align} a_0 &=\frac{1}{L}\int_{<T>}f(x)dx \\ a_n &=\frac{1}{L}\int_{<T>}f(x) \cos\left ( \frac{n\pi x}{L} \right )dx \\ b_{n} &=\frac{1}{L}\int_{<T>}f(x) \sin\left ( \frac{n\pi x}{L} \right )dx \end{align}

here $T = 2-1=1$ and $L = \frac{T}{2}=\frac{1}{2}$. The function is depicted in the figure below

Therefore the coefficients are ($\cos (n\pi) = (-1)^n$)

\begin{align} a_0 &=\frac{1}{\frac{1}{2}}\int_{1}^{2}xdx =3 \\ a_{n} &=\frac{1}{\frac{1}{2}}\int_{1}^{2} x\cos\left ( n\pi x\right )dx = \frac{x\sin(n\pi x)}{(n\pi)}+\frac{\cos(n\pi x)}{(n\pi)^2} \Bigl|_{1}^{2}=\frac{2(1-(-1)^n)}{(n\pi)^2}\\ b_{n} &=\frac{1}{\frac{1}{2}}\int_{1}^{2}x \sin\left ( n\pi x\right )dx = \frac{-x\cos(n\pi x)}{(n\pi)}+\frac{\sin(n\pi x )}{(n\pi)^2} \Bigl|_{1}^{2}=\frac{-2(2-(-1)^n)}{n\pi} \end{align}

If you want to find the coefficients in another interval, you can simply find the equation of the function in that interval and calculate the integrals. Here is an exercise. Show that the below integrals yields the same results

\begin{align} a_0 &=\frac{1}{\frac{1}{2}}\int_{0}^{1}(x+1)dx \\ a_{n} &=\frac{1}{\frac{1}{2}}\int_{0}^{1} (x+1)\cos\left ( n\pi x\right )dx \\ b_{n} &=\frac{1}{\frac{1}{2}}\int_{0}^{1} (x+1)\sin\left ( n\pi x\right )dx \end{align}

Since

$$f(x) = \begin{cases}\vdots & \\ x+2 &-1\le x< 0 \\ x+1 &0\le x< 1 \\ x &1\le x< 2 \\ x-1 &2\le x< 3 \\ \vdots & \end{cases}$$

Answer 2

If you assume the period is $T=1$ then

$$f(x) \sim \frac{a_0}{2} + \sum_{k=0}^\infty a_k \cos 2k\pi x + \sum_{k=0}^\infty b_k \sin 2k\pi x.$$

$$a_0 = 2 \int_1^2 x \, dx =3,$$

$$a_k = 2 \int_1^2 x \cos 2 k \pi x \, dx =0,$$

$$b_k = 2 \int_1^2 x \sin 2 k \pi x \, dx= -\frac{1}{k\pi}.$$

The function $f(x)$ is piecewise continuous and the series converges to $f(x)$ at points of continuity.

At the points of discontinuity (integers $n$), the series converges to the average of the left and right limits, namely $\frac{1}{2}(\lim_{x\uparrow n} f(x) + \lim_{x\downarrow n} f(x))= 3/2$.

K. K. McDonald's plot is a nice sketch, assuming $T=1.$

Please also see another example.