# Fourier series integration help required.

• andycampbell1
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.

## 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?

Have you drawn a picture of that function? Is it even, odd, or neither? Does that matter when calculating a0?

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.

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?

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.

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.

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