I have worked through the following question

with a period of 4

and got

4/n^2*pi^2 sin (n*pi/2) - 2/n*pi cos (n*pi)

Can someone tell me why the cos part of the answer is n*pi and not (-1)^n?

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- Feb 9th 2011, 12:27 AMhunterage2000question about answer to fourier series question
I have worked through the following question

with a period of 4

and got

4/n^2*pi^2 sin (n*pi/2) - 2/n*pi cos (n*pi)

Can someone tell me why the cos part of the answer is n*pi and not (-1)^n? - Feb 9th 2011, 06:53 AMAckbeet
Well,

You can convince yourself of this by looking at the graph of the cosine function.

Does that answer your question? - Feb 9th 2011, 09:02 AMhunterage2000
so why would this be (-1)^2 and not the another?

- Feb 9th 2011, 09:08 AMAckbeet
I don't understand your question. Can you please rephrase?

- Feb 9th 2011, 10:15 AMhunterage2000
The step before I got the answer of (-1)^2 was cos(n*pi) for post 3. The step before post 1 was cos(n*pi) and stayed cos(n*pi) and I dont know why.

- Feb 9th 2011, 10:17 AMAckbeet
Where did the (-1)^2 come from? I thought we were talking about (-1)^n. Did you mean (-1)^n in all your posts?

- Feb 9th 2011, 10:46 AMhunterage2000
yeah sorry I do

- Feb 9th 2011, 10:54 AMAckbeet
Ah. So the answer to your question is post # 5 is that I have no clue. Post # 2 still stands. It could be a preference of the author, though I find the exponential expression to be simpler than the trig expression.

- Feb 9th 2011, 12:16 PMhunterage2000
so the answer can be both in post #2

- Feb 9th 2011, 12:18 PMAckbeet
They're entirely equivalent. It's a preference thing.

- Feb 9th 2011, 12:26 PMhunterage2000
or right, we were never told that in class. Thats helped me, thanks for that.

- Feb 9th 2011, 12:29 PMAckbeet
You're very welcome. Have a good one!