Fourier Series 2: Solving Homework Problems

[robertjford80]

Homework Statement

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.

 

[clamtrox]

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

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.

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

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

I'm also having a difficult time converting sin(npi/2) into -1^(n-1)/2. where does the 2 come from in the denominator?

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.

 

[robertjford80]

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.

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.

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

Thanks, I understand now.

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.

I was able to get this before you answered.

 

[clamtrox]

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.

No it's not.

 

[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

 

[Infinitum]

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.

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

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

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

 

[robertjford80]

<deleted>

 

[robertjford80]

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

 

[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??

 

[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

 

[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}

 

[robertjford80]

i understand now