ull Asiacell4:26 PM81% 4a docs.google.comمسائي*Find the Fourier Series for the following periodic functions1.f(x) =1–x²2.1, Add fileSubmitNever submit passwords through Google Forms.This form was created inside of University of Baghdad. ReportAbuseGooale Carme->4

Answered: Find the Fourier Series for the…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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ull Asiacell
4:26 PM
81% 4
a docs.google.com
مسائي
*
Find the Fourier Series for the following periodic functions
1.
f(x) =1–x²
2.
1, Add file
Submit
Never submit passwords through Google Forms.
This form was created inside of University of Baghdad. Report
Abuse
Gooale Carme
->
4

Expert Solution

Step 1

Given, $f (x) = 1 - x^{2}; 0 \leq x \leq π$

we know that function $f (x) = 1 - x^{2} i s a n e v e n f u n c t i o n a s f (- x) = 1 - x^{2}$

If a function f $(x)$ is defined and periodic in the interval $(a, b)$ then it's fourier series is given by,

$f (x) = \frac{a_{0}}{2} + \sum_{n = 1}^{\infty} [a_{n} \cos (\frac{2 n πx}{b - a}) + b_{n} \sin (\frac{2 n πx}{b - a})] w h e r e, a_{0} = \frac{2}{b - a} \int_{a}^{b} f (x) d x a_{n} = \frac{2}{b - a} \int_{a}^{b} f (x) \cos (\frac{2 n πx}{b - a}) d x a n d b_{n} = \frac{2}{b - a} \int_{a}^{b} f (x) \sin (\frac{2 n πx}{b - a}) d x$

for even function $b_{n} = 0$

Step 2

To find the Fourier series of $f (x)$ we have to find $a_{0}, a_{n} a n d b_{n}$

$\begin{array}{rcl} N o w, a_{0} & = & \frac{2}{π - 0} \int_{0}^{π} 1 - x^{2} d x \\ = & \frac{2}{π} {[x - \frac{x^{3}}{3}]}_{0}^{π} \\ = & \frac{2}{π} [π - \frac{π^{3}}{3}] \\ a_{0} & = & 2 [1 - \frac{π^{2}}{3}] \end{array}$

$\begin{array}{rcl} a_{n} & = & \frac{2}{b - a} \int_{a}^{b} f (x) \cos (\frac{2 n πx}{b - a}) d x \\ = & \frac{2}{π} \int_{0}^{π} (1 - x^{2}) \cos (\frac{2 n πx}{π}) d x \\ = & \frac{2}{π} \int_{0}^{π} (1 - x^{2}) \cos (2 nx) d x \\ B y i n t e g r a t i o n b y p a r t s \\ = & \frac{2}{π} [{(1 - x^{2})}_{0}^{π} {(\frac{\sin 2 n x}{2 n})}_{0}^{π} - \int_{0}^{π} (- 2 x) \frac{\sin (2 n x)}{2 n}] \\ = & \frac{2}{π} [0 + \frac{1}{n} \int_{0}^{π} x \sin (2 n x) d x] d o a g a i n b y p a r t s \\ = & \frac{2}{nπ} [{(\frac{- x \cos (n x)}{2})}_{0}^{π} + \int_{0}^{π} \frac{\cos (2 n x)}{2} d x] \\ = & \frac{2}{n π} [\frac{- πcos (nπ)}{2} + {(\frac{\sin (2 n x)}{4 n})}_{0}^{π}] \\ = & \frac{2}{n π} [\frac{- π {(- 1)}^{n}}{2} + 0] \\ a_{n} & = & \frac{{(- 1)}^{n + 1}}{n} \end{array}$

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