How many terms are needed for Fourier Isometry to be under 5%?

kbooras
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Hey guys. I just started a class on Fourier Analysis and I'm having a difficult time understanding this question. Any help would be much appreciated!

Homework Statement


Verify that the Fourier Isometry holds on [−π, π] for f(t) = t. To do this, a) calculate
the coefficients of the orthogonal Fourier series from the orthogonal series representation, b)
calculate the sum of the squared coefficients, and c) Calculate the norm of the function as
∫ |f(t)|2 dt. They must be equal. How many terms in the Fourier series are necessary to have the isometry be under 5%? How many until you are under 3%, or 1%.

Homework Equations


Since f(t) = t is an odd function, you only need to calculate the sine coefficients.
bk = 1/\sqrt{}π∫f(t)dt


The Attempt at a Solution


I solved for bk and got (-2\sqrt{}π / k)*cos(k π)
Then I tried solving for the sum of bk ^2 but the series diverges, so now I'm stuck. Also, I'm not even sure what I'm trying to show for this problem.
 
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kbooras said:

The Attempt at a Solution


I solved for bk and got (-2\sqrt{}π / k)*cos(k π)
Then I tried solving for the sum of bk ^2 but the series diverges, so now I'm stuck. Also, I'm not even sure what I'm trying to show for this problem.
No, the series doesn't diverge. Note that ##\cos(k\pi) = (-1)^k##. Then (assuming your formula is correct; I haven't checked),
$$b_k^2 = \frac{4\pi}{k^2}$$
and so
$$\sum b_k^2 = 4\pi \sum \frac{1}{k^2}$$
which converges. (To what is another question.)
 
I got it! Thank you very much for pointing that out jbunniii. I can't believe I forgot to square the k.

Now that I've shown the Fourier Series is an isometry, I need to determine how many terms in the Fourier series are necessary to have the isometry be under 5%. I think I have to use the mean squared error formula for this but I'm not quite sure. If anyone could point me in the right direction I would really appreciate it.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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