Basic Spectral Analysis proof help

  • #1
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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:
Σt=1n cos2(2πtj/n) = ¼ ∑t=1n (e2πitj/n - e2πitj/n)2 = ¼∑t=1ne4πtj/n + 1 + 1 + e-4πtj/n = n/2.

I'm don't see how the last equality follows. I think, somehow, that the e4πtj/n and e-4πtj/n terms cancel out, but how?

Any ideas?
 

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  • #2
Dale
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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.
 
  • #3
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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.
 
  • #4
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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.
 
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  • #5
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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 e4π i t 2/4 =∑t=14 eπ i (2t) = (-1)2 + (-1)4 + (-1)6 + (-1)8 ≠ 0.
 
  • #7
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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.
 
  • #9
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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.
 
  • #10
Dale
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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.
 
  • #11
Dale
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OK, so I don't think that they individually sum to 0, but you can recombine them to get
$$ \sum _{t=1}^n \cos(4\pi t j/n) + 2 $$
 
  • #12
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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.
 
  • #13
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Right now I agree with you on that. It doesn't appear to work for j = n/2
 
  • #14
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Just looked at the textbook. It specifically excludes the cases j = 0 and j = n/2. I think it works for all other j.
 
  • #15
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Oops. You I forgot that case.

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

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