Fourier series integration help required.

In summary, the conversation discusses finding the value of a0 for a periodic function with a period of 2. The function is defined as 1.7(x+1) for -1≤x<0 and 1.7(x-1) for 0<x≤1 with f(x+2) = f(x) for all x. The person initially struggles with integrating the function, but eventually realizes the mistake in removing parentheses. The expert suggests correcting the expressions to 1.7x+1.7 and 1.7x-1.7 and ultimately concludes that a0=0.
  • #1

Homework Statement



I have a periodic function which has a period 2, which is defined as

f(x) = ( 1.7(x + 1) , -1 ≤ x < 0 With f(x+2) = f(x) for all x
( 1.7(x - 1), 0 < x ≤ 1

I have to determine a0


The Attempt at a Solution



I have started this as

a0 = 1/T ∫ top limit = T/2 Bottom limit = -T/2 f(x) dx
= 1/2 ∫ top limit = 0 Bottom limit = -1. 1.7(x+1) dx + ∫Top limit 1, Bottom limit 0 1.7(x-1) dx.

Is this right what I am doing so far, the 1.7(x+1) and 1.7(x-1) part is confusing me should I make it 1.7x+1 and 1.7x-1 ? And how would I integrate them?
 
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  • #2
Have you drawn a picture of that function? Is it even, odd, or neither? Does that matter when calculating a0?
 
  • #3
I haven't drawn it yet I have an example from my lecturer which is similar up until that point, when I have to integrate using 1.7(x+1) on the example there is no mention of odd or even numbers. I couldn't tell you if it mattered or not.
 
  • #4
andycampbell1 said:
I haven't drawn it yet I have an example from my lecturer which is similar up until that point, when I have to integrate using 1.7(x+1) on the example there is no mention of odd or even numbers. I couldn't tell you if it mattered or not.

I wasn't talking about odd or even numbers. It was about odd or even functions. Have you studied half-range Fourier expansions yet? Do you know what odd or even functions are and what that has to do with the coefficients?
 
  • #5
No, we covered Fourier series over an hour we didn't really get into it much we got a couple of pages of notes and some examples nothing like that was mentioned.
 
  • #6
andycampbell1 said:

Homework Statement



I have a periodic function which has a period 2, which is defined as

f(x) = ( 1.7(x + 1) , -1 ≤ x < 0 With f(x+2) = f(x) for all x
( 1.7(x - 1), 0 < x ≤ 1

I have to determine a0


The Attempt at a Solution



I have started this as

a0 = 1/T ∫ top limit = T/2 Bottom limit = -T/2 f(x) dx
= 1/2 ∫ top limit = 0 Bottom limit = -1. 1.7(x+1) dx + ∫Top limit 1, Bottom limit 0 1.7(x-1) dx.

Is this right what I am doing so far, the 1.7(x+1) and 1.7(x-1) part is confusing me should I make it 1.7x+1 and 1.7x-1 ? And how would I integrate them?

andycampbell1 said:
No, we covered Fourier series over an hour we didn't really get into it much we got a couple of pages of notes and some examples nothing like that was mentioned.

OK. Well I was going to suggest a shortcut but you haven't covered it yet. So to answer your original question, first you need to correct the colored expressions above. You aren't removing parentheses correctly. a(x-1) = ax - a. Once you fix that your integrals should work. You should come up with a0= 0.
 
  • #7
Hi thanks for the help things are starting to make sense now so just so I'm clear my expressions should be 1.7x + 1.7 and 1.7x - 1.7
 

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