Fourier Series

1. Aug 3, 2016

teme92

1. The problem statement, all variables and given/known data
I'm calculating the coefficients for the Fourier series and I got to part where I can't simplify an any further but I know I have to.

$$a_n = \frac{1}{2π}\Big[\frac{cos(n-1)π}{n-1}-\frac{cos(n+1)π}{n+1}-\frac{1}{n-1}+\frac{1}{n+1}\Big]$$

2. Relevant equations

3. The attempt at a solution
I have found an example where the above line goes straight to:

$$a_n=-\frac{1}{2π}(1+(-1)^n)\frac{2}{(n-1)(n+1)}$$

I don't know how you get from one line to the next. If anyone could explain this to me, it would be greatly appreciated.

2. Aug 3, 2016

Orodruin

Staff Emeritus
What is $\cos(n\pi)$?

3. Aug 3, 2016

teme92

-1 for n = odd and 1 for n = even.

4. Aug 3, 2016

Orodruin

Staff Emeritus
So how can you write this on a more concise form?

5. Aug 3, 2016

teme92

$$(-1)^n$$

6. Aug 3, 2016

Orodruin

Staff Emeritus
So how can you apply this to your problem?

7. Aug 3, 2016

teme92

I've tried subbing (-1)n in for both of the cos parts but it doesn't simply to what is given.

8. Aug 3, 2016

Ray Vickson

Yes it does; start again.

9. Aug 3, 2016

teme92

When I sub in and simplify I'm getting:

$$a_n = \frac{1}{2π}\Big[\frac{(-1)^n + (-1)^n}{(n-1)(n+1)}\Big]$$

Last edited: Aug 3, 2016
10. Aug 3, 2016

Orodruin

Staff Emeritus
Observe that you have $\cos[(n-1)\pi]$ and that you have some other terms too.

11. Aug 3, 2016

teme92

$$a_n = \frac{1}{2π}\Big[\frac{(-1)^n}{n-1}-\frac{(-1)^n}{n+1}-\frac{1}{n-1}+\frac{1}{n+1}\Big]$$

Is this correct?

12. Aug 3, 2016

Orodruin

Staff Emeritus
Why don't you try inserting a few values to check?

13. Aug 3, 2016

teme92

What would I be checking for?

14. Aug 3, 2016

Orodruin

Staff Emeritus
That the values that come out of your formula agree with the original expression.

15. Aug 3, 2016

teme92

If I put integers in for n will the result not be a single number?

16. Aug 3, 2016

Orodruin

Staff Emeritus
Yes, but it will be different for different n of course. I said "inserting a few values".

17. Aug 3, 2016

teme92

$$n=0: a_n = 0$$

$$n=1: a_n = \frac{1}{2π}$$(has zero dividers though)

$$n=2: a_n = \frac{1}{12π}$$

18. Aug 3, 2016

Orodruin

Staff Emeritus
And if you use the same values in the original expression?

19. Aug 3, 2016

teme92

But do I not want there to be n's in my result?

20. Aug 4, 2016

Orodruin

Staff Emeritus
Yes. But you want to check that you got the right result first. You asked:
If there is any n for which your new expression does not equal the old, then it is not correct.