Fourier coefficients and series

Fourier series calculation example

Due to numerous requests on the web, we will make an example of calculation of the Fourier series of a piecewise defined function from an exercise submitted by one of our readers. The calculations are more laborious than difficult, but let's get on with it ...

It is asked to calculate the Fourier series of following picewise function

\( f(x)=\left\{\begin{matrix} x & si\; x \in [-1,0)\\ 1 & si\; x \in [0, 1] \\ 0 & si\; |x|>1 \end{matrix}\right.\)

The first thing of all is to remember that our formulas here seen refer to interval \([-\pi, \pi] \), so we perform a change of variable

\(t = \pi x\) to obtain

\( f(t)=\left\{\begin{matrix} \frac{t}{\pi} & si\; t \in [-\pi,0)\\ 1 & si\; t \in [0, \pi] \\ 0 & si\; |t|>\pi \end{matrix}\right.\)

We can already apply our formulas for the Fourier series that we remember was

$$ f(x)=\frac{A_{0}}{2} + \sum_{n=1}^{\infty}(A_n cos nx + B_nsin nx ) $$ with \( A_i, B_i \in \mathbb{R} \) and

(1)       \( A_0= \frac{1}{\pi} \int_{-\pi}^{\pi}f(x) dx\)

(2)      \( A_n= \frac{1}{\pi} \int_{-\pi}^{\pi}f(x) cos nx dx\)

(3)      \( B_n= \frac{1}{\pi} \int_{-\pi}^{\pi}f(x) sin nx dx\)

We start by calculating (1)

\( A_0=\frac{1}{\pi}\int_{-\pi}^{0} \frac{t}{\pi}dt + \frac{1}{\pi}\int_{0}^{\pi} dt = \frac{1}{\pi}( \pi-\frac{\pi}{2}) = \frac{1}{2} \)

The next, we calculate (2), for this our integral is separated into two pieces

\( A_n=\frac{1}{\pi}\int_{-\pi}^{0} \frac{t}{\pi} cos(nt)dt + \frac{1}{\pi}\int_{0}^{\pi} cos(nt)dt = I_1 + I_2\)

We start by calculating the first of the integrals I1, we do it by parts

\(\left\{\begin{matrix} u= \frac{t}{\pi} \Rightarrow du=\frac{dt}{\pi} \\ dv= cos(nt) dt \Rightarrow v=\frac{sin(nt)}{n} \end{matrix}\right.\)

Applying the method of integration by parts, whose formula we remember was

\(\int udv = uv-\int vdu\)


\( I_1 = 0 - \frac{1}{\pi^2} \int_{-\pi}^{0} \frac{sin(nt)}{n}dt = ... = \frac{1}{(\pi n)^2} (1-(-1)^n) \)

Now we calculate the second integral, we see that

\( I_2 = 0 \)

\( A_n= \frac{1}{(\pi n)^2} (1-(-1)^n) \)

Patience, it only remains to calculate the Bn, it is very similar to the previous calculation

\( b_n=\frac{1}{\pi}\int_{-\pi}^{0} \frac{t}{\pi} sin(nt)dt + \frac{1}{\pi}\int_{0}^{\pi} sin(nt)dt = J_1 + J_2\)

We start by calculating J1, we proceed as before by parts to obtain.

\( J_1 = -\frac{(-1)^n}{n\pi} \)

J2 is calculated directly

\( J_2 = \frac{1}{n \pi}(1- (-1)^n) \)


\( B_n = \frac{2-(-1)^n}{n\pi}\)

We already have all the ingredients for the Fourier series formula, it is as follows

\(f(t)=\frac{1}{4}+\sum_{n=1}^{\infty}\frac{1}{(\pi n)^2} (1-(-1)^n)cos(nt) + (\frac{2-(-1)^n}{n\pi}) sin(nt) \)

Almost everything is done, the only thing now is to undo the initial variable change to finally get:

\( f(x)=\frac{1}{4}+\sum_{n=1}^{\infty}\frac{1}{(\pi n)^2} (1-(-1)^n)cos(n \pi x) + (\frac{2-(-1)^n}{n\pi}) sin(n \pi x) \)

To test our result we can go to 快乐飞艇怎么玩:Fourier Series calculator on this same site, if we introduce our function it remains

if we click on calculate, with 8 coefficients we obtain

we see that numerical results coincide (except truncation error) with the obtained values, the graph of Fourier series remains

Was useful? want add anything?

快乐飞艇怎么玩:Post here

Post from other users

快乐飞艇怎么玩:Post here
快乐飞艇开奖 快乐飞艇做任务靠谱吗 熊猫乐园快乐飞艇 华创投资快乐飞艇靠谱吗 快乐飞艇综合走势图 快乐飞艇app首页 快乐飞艇官网 快乐飞艇是不是官方的 快乐飞艇计划 快乐飞艇彩票 快乐飞艇34567玩法 快乐飞艇app下载