Simplifying Fourier series coefficients with cos(nπ) terms

[teme92]

Homework Statement

I'm calculating the coefficients for the Fourier series and I got to part where I can't simplify a_n any further but I know I have to.

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

Homework Equations

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.

 

[Orodruin]

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

 

[teme92]

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

 

[Orodruin]

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

 

[teme92]

$$(-1)^n$$

 

[Orodruin]

So how can you apply this to your problem?

 

[teme92]

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

 

[Ray Vickson]

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

Yes it does; start again.

 

[teme92]

When I sub in and simplify I'm getting:

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

 

[Orodruin]

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

 

[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?

 

[Orodruin]

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

Is this correct?

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

 

[teme92]

What would I be checking for?

 

[Orodruin]

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

 

[teme92]

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

 

[Orodruin]

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

 

[teme92]

$$n=0: a_n = 0$$

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

$$n=2: a_n = \frac{1}{12π}$$

 

[Orodruin]

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

 

[teme92]

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

 

[Orodruin]

Yes. But you want to check that you got the right result first. You asked:

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

Is this correct?

If there is any n for which your new expression does not equal the old, then it is not correct.