Fourier series (I here)

[FrogPad]

Here is the question:
At $x= \frac{\pi}{2}$ the square wave equals 1. From the Fourier series at this point find the alternating sum that equals $\pi$.
$$\pi = 4(1 - \frac{1}{3}+\frac{1}{5}-\frac{1}{7} + \ldots$$

I do not understand what the question is asking. I'm not knowledgeable enough with Fourier series to understand it I believe. This is my best guess:

I need to find the Fourier series of the square wave starting at $\frac{\pi}{2}$ fand this will "magically" yield the alternating series for $\pi$.

But I don't understand what "the square wave equals 1" part means. Do I define the square wave to jump to 1 at $\frac{pi}{2}$ instead of the typical wave (typical for me) where the wave is -1 from -pi to 0, and jumps to 1 at 0 to pi ?

Any help clarifying would be swell. Thanks :)

What's up with the LaTeX by the way?

 

[quasar987]

Here is the question:
At $x= \frac{\pi}{2}$ the square wave equals 1. From the Fourier series at this point find the alternating sum that equals $\pi$.
$$\pi = 4(1 - \frac{1}{3}+\frac{1}{5}-\frac{1}{7} + \ldots$$

It says consider a square wave that is 1 at pi/2. The most casual one would be the one you mention later: the wave is -1 from -pi to 0, and jumps to 1 at 0 to pi. The Fourier series of a function at a point, if it converges towards the function, equals the function evaluated at that point. See what you get when you evaluate the Fourier series of the square wave at x=pi/2.

 

[FrogPad]

For a problem before this one, I just got done evaluating the Fourier series for a square wave with properties:

-1 , -pi to 0
1 , 0 to pi

$$\hat f(x) = \frac{4}{\pi}\left(\frac{\sin x}{1}+\frac{\sin 3x}{3}+\frac{\sin 5x}{5}+\ldots\right)$$

Evaluating the Fourier series at $\frac{\pi}{2}$ yields:
(It took me so long to realize what this means. :)

$$\hat f(\frac{\pi}{2}) = \frac{4}{\pi}\left(\frac{\sin \frac{\pi}{2}}{1}+\frac{\sin \frac{\pi}{2}}{3}+\frac{\sin \frac{\pi}{2}}{5}+\ldots \right)$$

$$\hat f(\frac{\pi}{2}) = \frac{4}{\pi}\left(1 + \frac{1}{3} + \frac{1}{5} + \ldots \right)$$

We know that the square wave at pi/2 is equal to 1 so:
$$1 = \frac{4}{\pi}\left(1 + \frac{1}{3} + \frac{1}{5} + \ldots \right)$$

This can then be solved for $\pi$. But! This is not correct. So is my Fourier series for the square wave wrong?

 

[shmoe]

$$\hat f(x) = \frac{4}{\pi}\left(\frac{\sin x}{1}+\frac{\sin 3x}{3}+\frac{\sin 5x}{5}+\ldots\right)$$

Evaluating the Fourier series at $\frac{\pi}{2}$ yields:
(It took me so long to realize what this means. :)

$$\hat f(\frac{\pi}{2}) = \frac{4}{\pi}\left(\frac{\sin \frac{\pi}{2}}{1}+\frac{\sin \frac{\pi}{2}}{3}+\frac{\sin \frac{\pi}{2}}{5}+\ldots \right)$$

sin(3*pi/2) is not sin(pi/2)!

 

[FrogPad]

hahahahah...
yeah, you are right !

man, I took this stupid cold medicine. I'm just going to go to bed and work on this in the morning, because obviously I'm making some really dumb mistakes!


thanks you guys :)
that was actually an interesting problem.