(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

For 0</= x </= pi, let f(x) be defined by

f(x) = { A for [x - pi/2] < alpha

{ 0 for alpha </= [x-pi/2]</= pi/2

Where A and alpha are positive constants with alpha <pi/2. I have used </ to represent less than or equal to. The square brackets are to represent single straight vertical lines.

Let f(x) be extended into an odd, periodic function of period 2pi. Sketch the graph of f(x) in the interval -3pi</= x </= 3pi, and show that the fourier series expansion of f(x) takes the form

f(x) = 4A/pi{Infinity}Sigma(m=1)((-1)^m+1)/(2m-1) sin(2m-1)alpha sin(2m-1)x

The part in bold should be sigma with infinity above and m=1 below

2. Relevant equations

3. The attempt at a solution

I see how to draw the function, i get y = A from 0 to pi at which point it drops to zero. Because it is odd and period 2pi it will be a square wave with three horizontal maxima (effectively) from x = 2npi - (2n+1)pi.

Is this correct?

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# Homework Help: Fourier series and draw function

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