{Edit-Solved} Confirmation requested on deriving functions from graphs

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  • Thread starter warhammer
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  • #1
warhammer
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TL;DR Summary
So I basically saw this graph specified as a particular waveform in my book while reading Fourier Series. I decided to try and derive its function since once I do that I can easily find the FS. Please find the photo and my attempt below, just need a small confirmation if I'm right/wrong.
So I thought that the graph tries to tell us that the function is periodic after 2π interval. So I tried to derive its function from the graph as follows using the point slope equation form for the points (0,0) & (a,π): ##y= ({a}/{π})*x##

I hope this function is alright and I just need to find its Fourier Series
 

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  • #2
anuttarasammyak
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Actually the graph shows
##y=\frac{a}{\pi}(x-2n\pi)## for ##0<(x-2n\pi)<\pi## for integer n
y=0 for others.
 
  • #3
warhammer
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28
Actually the graph shows
##y=\frac{a}{\pi}x## for ##2n\pi<x<(2n+1)\pi## for integer n
y=0 for others.
Oh I see. Now I realize how you have represented it an even general manner. Although I am a bit confused now, how should I find FS for the same, I mean I'm confused about the intervalbin which integration for FS will be carried out..
 
  • #4
anuttarasammyak
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Because the function, say f(x), has period of 2##\pi##, it is expressed as
[tex]f(x)=\frac{a_0}{2}+\sum_{k=1}^\infty (b_k \sin kx + a_k \cos kx)[/tex]
You may calculate ##a_k## and ##b_k## in a usual way.
 
  • #5
warhammer
138
28
Because the function, say f(x), has period of 2##\pi##, it is expressed as
[tex]f(x)=\frac{a_0}{2}+\sum_{k=1}^\infty (b_k \sin kx + a_k \cos kx)[/tex]
You may calculate ##a_k## and ##b_k## in a usual way.
Thank you so much! Now I get it🙏🏻
 
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