# Homework Help: Fourier series 2

1. Jun 14, 2012

### robertjford80

1. The problem statement, all variables and given/known data

3. The attempt at a solution

I don't see how if n = 0 then the answer is 1/2. By my reckoning

1/pi * sin nx/n =

1/pi * 0 = 0

I also don't see where the 2 comes from when sin(npi/2) first shows up.

Last edited: Jun 14, 2012
2. Jun 14, 2012

### clamtrox

If n=0, you should calculate the integral differently. Also lim x->0 sin(x)/x is not 0, so you get the correct result even if you just do take the limit.

The upper limit substitution is always zero. You get pi/2 from the lower limit.

It comes from the fact that then they are equal... Work it out. Plug in values n=1,3,5,... and see how sine behaves, then see what the 2 does in the denominator.

3. Jun 14, 2012

### robertjford80

I still don't get this.

Also lim x->0 sin(x)/x is not 0 - That's correct, but 1/pi * sin 0/0 = 1/pi * 1/0 = undefined.

Thanks, I understand now.

I was able to get this before you answered.

4. Jun 14, 2012

### clamtrox

No it's not.

5. Jun 14, 2012

### robertjford80

Do the n's cancel? If not, then 1/pi * sin nx/n must be 1/pi * 1/0

If yes, then 1/pi * sin pi/2 = 1/pi

6. Jun 14, 2012

### Infinitum

Look at the graph of sinx/x. It is a continuous and differentiable function at x=0.

http://www.wolframalpha.com/input/?i=graph++sinx/x

n's don't cancel. In integrals, the points that you are evaluating is the limit of the function at those values.

7. Jun 14, 2012

### robertjford80

<deleted>

8. Jun 14, 2012

### robertjford80

even if sin x/x is continuous and differentiable I still don't see how they got 1/pi * pi/2

9. Jun 14, 2012

### Infinitum

You are trying to find,

$$\frac{1}{\pi }\int_{\pi/2}^{\pi} cos(nx) dx$$

At n=0,

$$\frac{1}{\pi }\int_{\pi/2}^{\pi} cos(0)dx = \frac{1}{\pi }\int_{\pi/2}^{\pi}1\cdot dx$$

What do you get if you integrate this??

10. Jun 14, 2012

### Karamata

For $n=0$ you have

$\dfrac{1}{\pi}\displaystyle\int\limits_{\frac{\pi}{2}}^{\pi} \cos (0 \cdot x) dx = \dfrac{1}{\pi}\displaystyle\int\limits_{\frac{\pi}{2}}^{\pi} 1 dx = \dfrac{1}{\pi} \cdot \dfrac{\pi}{2}$

Edit: Oh, Infinitum was faster :) Sorry

11. Jun 14, 2012

### clamtrox

$$\frac{1}{\pi} \int_{\pi/2}^\pi \cos(0x) dx = \frac{1}{\pi} \int_{\pi/2}^\pi dx = \frac{1}{\pi}(\pi - \frac{\pi}{2})$$

Alternatively, you can take the limit:
$$\lim_{n\rightarrow 0} \frac{\sin(n \pi/2)}{n \pi} = \frac{1}{2}\lim_{n\rightarrow 0} \frac{\sin(n \pi/2)}{(n \pi/ 2)} = \frac{1}{2}$$

12. Jun 14, 2012

### robertjford80

i understand now