# Homework Help: Fourier Series?

1. Sep 26, 2006

### pivoxa15

When a question asks Construct sine and cosine series for the function:
f(t)=t, 0<t<pi.

Should I assume the period of f(t) is pi? I think it must because the domain is discontinous at 0 and pi.

2. Sep 26, 2006

### StatusX

I think that's a safe assumption.

3. Sep 26, 2006

### pivoxa15

I got a fourier series
$$\frac{\pi}{2} - \sum_{n=1}^\infty \frac{1}{n}sin(2nt)$$

correct?

Last edited: Sep 26, 2006
4. Sep 26, 2006

### StatusX

That's hard to say...

5. Sep 26, 2006

### pivoxa15

Do you say this because the question was vague in that it could have been asking for an even and/or odd extension of the function, in which case my answer would be wrong because I assumed a period of $$\pi$$.

Last edited: Sep 26, 2006
6. Sep 26, 2006

### StatusX

Sorry, my mistake (I didn't see you defined f(t) above). That looks right.

7. Sep 26, 2006

### pivoxa15

What do you think of the point I made in post 5?

Last edited: Sep 26, 2006
8. Sep 26, 2006

### StatusX

I think if they didn't say anything about an even or odd extension of the function, you can assume the period is pi. Do you have any reason to think they might have implied otherwise?

9. Sep 26, 2006

### pivoxa15

In the answers section they quoted functions which were even and odd extentions. Sometimes the functions with period pi are the odd extenstion functions (although not this case) but the even functions never are. They should have mentioned about even and odd extensions in the question.

10. Sep 27, 2006

### HallsofIvy

No, it was not necessary to say that, it was already implied. Cosine is an even function, sine is an odd function. Any "cosine series" is necessarily an even function, any "sine series" must be an odd function. The period is necessarily $2\pi$ with one interval from $[\pi, - \pi]$.

11. Sep 27, 2006

### pivoxa15

The series I derived in post 3 has is a sin function but has only period $\pi$ and fits the domain in the question perfectly well but is not an odd extension of the function.

Last edited: Sep 27, 2006
12. Sep 27, 2006

### StatusX

If they meant a sine series and a cosine series, ie, two different series, then they were looking for even/odd extensions of f(t). If they were looking for a single series involving both sine and cosine terms, then what you got is correct (remember pi/2=pi/2*cos(0) is a cosine term).

Last edited: Sep 27, 2006
13. Sep 27, 2006

### pivoxa15

Good point. That was also probably what HallsofIvy meant as well. I guess a single consine term does not count as a series so it had to be a cosine series which corresponds to the even extension and one sine series corresponding to the even extension.