Question

Calculate the Fourier Series of f(x) = |x|
Assume the period to be -π to π.


(Necessary Formulae and Steps to be included along with explanation).

Answer 1

Question-

Calculate the Fourier Series of f(x) = |x|

Assume the period to be -π to π.

Solution-

General Formula for Fourier Series -

$\boxed{\tt f(x)=\frac{a_{0}}{2}+\sum_{n=1}^{\infty}a_{n}cos\;nx+\sum_{n=1}^{\infty}b_{n}sin\;nx}$

$\tt {Where }$

$\tt \: a_o = \frac{2}{\pi} \int_{0}^{\pi} f(x) dx$

$\tt \: a_n = \frac{2}{\pi} \int_{0}^{\pi} f(x)cos\;nx\;dx$

$\tt \: b_n = \frac{2}{\pi} \int_{0}^{\pi}f(x)sin\;nx\;dx$

$\tt \: n = 1, 2, 3…..$

Checking if the function is even or odd-

f(x) = |x| = x

f(-x) = |-x| = x

f(x) = f(-x) = x

$\implies$Since f(x) gives same result in negetive and positive value, therefore it is even.

Since function is even

$\boxed{b_n = 0}$

Now we will find $\sf\red{a_o}$

$\boxed{a_o= \frac{2}{ \pi} \int_{0}^{ \pi} f(x) dx}$

$\implies {a_o= \frac{2}{ \pi} \int_{0}^{ \pi} x \: dx}$

$\implies \frac{2}{ \pi} [\frac{x^2}{2}]_{0}^{\pi}$

$\implies \frac{2}{ \pi} [\frac{{\pi}^2}{2}]_{0}^{\pi}$

$\implies{\pi}$

$\boxed{a_o = \pi}$

Now we will find $\sf\pink{a_n}$

$\boxed{ a_n = \frac{2}{\pi} \int_{0}^{\pi} f(x)cos\;nx\;dx}$

$\implies a_n = \frac{2}{\pi} \int_{0}^{\pi} x \:cos\;nx\;dx$

Using Integration product rule -

$\implies \large a_n = \frac{2}{\pi} [ x\frac{ sin\;nx}{n}- \frac{-cos \:nx}{n^2}]_{0}^{\pi}$

$\implies \large \frac{2}{\pi} [ \frac{ cos\;n \pi}{n^2}-\frac{1}{n^2}]$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \tt (sin \:n \pi = 0 )$

$\implies \large \frac{2}{\pi}[ \frac{(-1)^n}{n^2}-\frac{1}{n^2}]$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \tt (cos \:n \pi = (-1)^n)$

$\implies \large \frac{2}{\pi n^2} [(-1)^n-1]$

$a_n = \large \{_{\frac{-4}{{\pi}n^2} \: \sf \: if \: n \: is \: odd}^{0 \: \sf \: \: \: \: \: if \: n \: is \: even}$

Now putting values of $\sf{a_o,a_n\: and\: b_n}$ in fourier series formula-

$\tt f(x)=\frac{\pi}{2} +\frac{-4}{{\pi}}\sum_{n=1}^{\infty} \frac{1}{n^2}cos\;nx+\sum_{n=1}^{\infty}(0)sin\;nx$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \tt (considering\: n\: to\: be\; odd)$

$\boxed {\tt f(x)=\frac{\pi}{2}+ \frac{-4}{{\pi}}(\frac{cos x}{1^2} + \frac{cos 3x}{3^2} + \frac{cos 5x}{5^2}+ . . . . . .. )}$

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Note-

Originally,

$\tt \: a_o = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) dx$

$\tt \: a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x)cos\;nx\;dx$

$\tt \: b_n = \frac{1}{\pi} \int_{-\pi}^{\pi}f(x)sin\;nx\;dx$

These were modified by Integration technique.

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Answer 2

Step-by-step explanation:

Calculate the Fourier Series of f(x) = |x|

Assume the period to be -π to π.