Why Is the Equality in This Spectral Analysis Proof Correct?

[Ma Xie Er]

I'm reading "Time Series Analysis and Its Applications with R examples", 3rd edition, by Shumway and Stoffer, and I don't really understand a proof. This is not for homework, just my own edification.

It goes like this:
Σ_t=1ⁿ cos²(2πtj/n) = ¼ ∑_t=1ⁿ (e^2πitj/n - e^2πitj/n)² = ¼∑_t=1ⁿe^4πtj/n + 1 + 1 + e^-4πtj/n = n/2.

I'm don't see how the last equality follows. I think, somehow, that the e^4πtj/n and e^-4πtj/n terms cancel out, but how?

Any ideas?

 

[Dale]

I think, somehow, that the e4πtj/n and e-4πtj/n terms cancel out, but how?

They don't need to cancel out if they are each 0 on their own.

 

[Ma Xie Er]

They don't need to cancel out if they are each 0 on their own.

How are they zero on their own? If this is by De Moivre's theorem, then that doesn't apply to non-integers powers, i.e. (cos(x)+isin(x))ⁿ ≠(cos(nx) + i sin(nx)) for n ∉ℤ.

 

[Dale]

How are they zero on their own?

Why don't you try it by hand for small n. Try writing out the sum for n=4.

 

[Ma Xie Er]

Why don't you try it by hand for small n. Try writing out the sum for n=4.

Yes, I can see that, for n=4, j=1, but it doesn't work for j=2, n=4.
∑_t=1⁴ e^{4π i t 2/4} =∑_t=1⁴ e^{π i (2t)} = (-1)² + (-1)⁴ + (-1)⁶ + (-1)⁸ ≠ 0.

 

[Dale]

Umm. Doesn't $j=\sqrt{-1}$ always?

 

[Ma Xie Er]

Umm. Doesn't $j=\sqrt{-1}$ always?

No. n is a positive integer, and j= 1, ..., [[n/2]], where [[n/2]] is the floor or greatest integer function of n/2.

 

[Ma Xie Er]

Umm. Doesn't $j=\sqrt{-1}$ always?

This text denotes i as √(-1)

 

[Ma Xie Er]

Here's a link to the text http://www.stat.pitt.edu/stoffer/tsa3/tsa3.pdf. I was trying to solve Problem 2.10 on pg 77 (pg 87 of pdf). I don't quite understand footnote⁹, which is why I posted. I'm completely new to Fourier decomposition, so I'm having a hard time with this.

 

[Dale]

This text denotes i as √(-1)

Oh, then your summand is written wrong. You wrote.
$$ \sum _{t=1}^n e^{4\pi t j/n} + 1 + 1 + e^{-4\pi t j/n} $$
but it should be
$$ \sum _{t=1}^n e^{4\pi i t j/n} + 1 + 1 + e^{-4\pi i t j/n} $$

I'm not sure that fixes the proof, but it is important to write the problem clearly.

 

[Dale]

OK, so I don't think that they individually sum to 0, but you can recombine them to get
$$ \sum _{t=1}^n \cos(4\pi t j/n) + 2 $$

 

[Ma Xie Er]

OK, so I don't think that they individually sum to 0, but you can recombine them to get
$$ \sum _{t=1}^n \cos(4\pi t j/n) + 2 $$

I think $e^{ix}-e^{-ix}= 2 cos(x)$. In this case, $e^{4 \pi t j/n}+ e^{- 4 \pi t j/n} = 2 cos(4 \pi t j/n)$, so shouldn't it be $\sum_{t=1}^n 2 (1 + cos(4 \pi t j/n)$ ?

And after this I'm still not sure how the series sums to 0.

 

[Dale]

Right now I agree with you on that. It doesn't appear to work for j = n/2

 

[Dale]

Just looked at the textbook. It specifically excludes the cases j = 0 and j = n/2. I think it works for all other j.

 

[Ma Xie Er]

Oops. You I forgot that case.

For $j=1,.,,[[n/2]]-1$, I still don't see why it's true.