Exercise from basic Fourier Analysis

[broegger]

I really need help with this exercise (it's from a course in basic Fourier analysis). It consists of two parts:

(i) Let s_0 = 1/2 and s_n = 1/2 + \sum_{j=1}^{n}\cos(jx) for n \geq 1. By writing s_n = \left(\sum_{j=-n}^{n}e^{ijx}\right)/2 and summing geometric series show that (n+1)^{-1}\sum_{j=0}^{n}s_j \rightarrow 0 as n \rightarrow \infty for all x \neq 0~mod~2\pi, and so

0 = 1/2 + \sum_{j=1}^{\infty}\cos(jx) in the Cesáro sense.

(ii) Show similarly that, if x \neq 0~mod~2\pi, then

cot(x/2) = 2\sum_{j=1}^{\infty}\sin(jx) in the Cesáro sense.

In (i) I have tried to write out two geometric series and summing them, but I can't get the desired result. I have no idea on (ii).

"in the Cesáro sense" means (i think) that the average of a given sequence s_0,s_1,s_2,\ldots converges against a given limit L (the sequence itself doesn't nescessarily) - that is, the sequence s_0, (s_0 + s_1)/2, (s_0 + s_1 + s_2)/3,\ldots \rightarrow L.

 

[arildno]

I really need help with this exercise (it's from a course in basic Fourier analysis). It consists of two parts:

(i) Let s_0 = 1/2 and s_n = 1/2 + \sum_{j=1}^{n}\cos(jx) for n \geq 1. By writing s_n = \left(\sum_{j=-n}^{n}e^{ijx}\right)/2 and summing geometric series show that (n+1)^{-1}\sum_{j=0}^{n}s_j \rightarrow 0 as n \rightarrow \infty for all x \neq 0~mod~2\pi, and so

0 = 1/2 + \sum_{j=1}^{\infty}\cos(jx) in the Cesáro sense.

(ii) Show similarly that, if x \neq 0~mod~2\pi, then

cot(x/2) = 2\sum_{j=1}^{\infty}\sin(jx) in the Cesáro sense.

In (i) I have tried to write out two geometric series and summing them, but I can't get the desired result. I have no idea on (ii).

"in the Cesáro sense" means (i think) that the average of a given sequence s_0,s_1,s_2,\ldots converges against a given limit L (the sequence itself doesn't nescessarily) - that is, the sequence s_0, (s_0 + s_1)/2, (s_0 + s_1 + s_2)/3,\ldots \rightarrow L

Is this how you meant it?
(Use "tex" not "Tex")

 

[Wong]

For part i, I would say one should first convert s_n = \left(\sum_{j=-n}^{n}e^{ijx}\right)/2 into simpler form using the sum of geometric series formula. Then one should have no difficulty in finding \sum_{n} s_{n}. Moreover one should note that for all integer m, |e^{imx}|=1 for all real x.

 

[broegger]

Thanks for answering.. I have shown that (n+1)^{-1} \sum_{j=0}^{n} s_n \rightarrow 0 as n \rightarrow \infty, but I can't see exactly how that relates to solving (i)...

 

[Wong]

...but I can't see exactly how that relates to solving (i)...

Did you mean (ii)?

 

[broegger]

yes, I'm sorry, part (ii).. I have no idea on that one (i assume cot(x) = cos(x)/sin(x))

 

[Wong]

hi broegger.

I did not do part (ii), but I think all they want you to realize that sinx = \frac {1}{2i} (e^{ix}-e^{-ix}). Using this formula, find \sum_{j=0}^{n} sin(jx). Then all the steps are similar to what you did in part i), I think. Of course one should always remember cosx = \frac{1}{2} (e^{ix}+e^{-ix}) and sinx = \frac {1}{2i} (e^{ix}-e^{-ix}). Using these, one may find an expression for cot(x/2) in "complex" exponential.

I did not do it, but I think that may be the way to do it.

 

[broegger]

Hi. Thank you very much for taking the time to help me.

I'll try the method you advised for part (ii).. about part (i) I'm not sure my reasoning are correct; how can you apply the geometric series formula when the series in question are not infinite - ex. \sum_{j=-4}^{4}e^{ijx} = 1/2 + \sum_{j=1}^{4}\cos(jx). I don't know if you see what I mean (maybe I'm getting this all wrong).

 

[Wong]

broegger, do you know that 1+r+r^2+...r^n = (1-r^(n+1))/(1-r)?

 

[broegger]

Erm.. How can that be true?? In the limit n = \infty the sum you are mentioning is equal to 1/(1-r).

I'm really troubled by this :\

 

[Wong]

Let S(r) = 1+r..+r^n, then
r*S(r) = r+r^2+...+r^(n+1)
(1-r)*S(r) = 1-r^(n+1)
S(r) = (1-r^(n+1))/(1-r)

Yes, the limit of the expression as n tends to infinity tends to 1/(1-r), *if*|r|<1.

 

[Sonden]

Three years old, but now I'm trying to solve the problem ((i)).

I suppose

2s_n = \sum_{j=-n}^{n}e^{ijx} = \frac{1-e^{(i x)(n+1)}}{1-e^{i x}}+\frac{1-e^{(-i x)(n+1)}}{1-e^{-i x}}-1

so I have to show (?) that

\lim_{p\rightarrow\infty}\frac{1}{p+1}\sum_{n=0}^{p}s_n=\lim_{p\rightarrow\infty} \frac{1}{p+1}\sum_{n=0}^{p}\left(\frac{1-e^{(i x)(n+1)}}{1-e^{i x}}+\frac{1-e^{(-i x)(n+1)}}{1-e^{-i x}}-1\right)/2=0.

Apparently one should have no difficulty in showing that, but I do.