Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Sketch several periods of the corresponding periodic function of period . Expand the periodic2π function in a sine-cosine Fourier series.

f(x)={0,-π<x<π21,π2<x<π,

Short Answer

Expert verified

The expansion is n=1+4cosnxn-n=3+4cosnxn-n=1+2msinnxn+2n=2+4msinnxnwherem=0,1,2,--.

Step by step solution

01

Given

The given function is f(x)={0,-π<x<01,0<x<π2,0,π2<x<π..

02

The concept of the Fourier series for the function f(x)

The Fourier series for the function :

f(x)=a02+n=1(ancosnx+bnsinnx)a0=1π-ππf(x)dxan=1π-ππf(x)cosnxdxbn=1π-ππf(x)sinnxdx

If is an even function:

bn=0f(x)=a02+n=1ancosnx

If is an odd function:

a0=aa=0f(x)=n=1bnsinnx

03

From the given information

The sketch of the given function is shown below.

Coefficients of :

a0=1π-ππf(x)dx=1π-ππ2dx=1πx0π2=12

Coefficients of :

an=1π-ππf(x)sinnxdx=1π0π2f(x)sindx=1πcosdx0π2=1π1-cosnπ2

04

Calculate Coefficients of bn

Coefficients of bnare 1π,22π,13π,0,15π,26π,17π.

Hence the expansion is

f(x)=14+1πn=1+4cosnxn-n=3+4cosnxn-n=1+2msinnxn+2n=2+4msinnxnwherem=0,1,2,--.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Study anywhere. Anytime. Across all devices.

Sign-up for free