# Exercise from basic Fourier Analysis

1. Aug 30, 2004

### broegger

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

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

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

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

[Tex] cot(x/2) = 2\sum_{j=1}^{\infty}\sin(jx) [/Tex] 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 [Tex] s_0,s_1,s_2,\ldots [/Tex] converges against a given limit L (the sequence itself doesn't nescessarily) - that is, the sequence [Tex] s_0, (s_0 + s_1)/2, (s_0 + s_1 + s_2)/3,\ldots \rightarrow L [/Tex].

2. Aug 30, 2004

### arildno

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

3. Aug 30, 2004

### 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.

4. Aug 31, 2004

### 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)...

5. Aug 31, 2004

### Wong

Did you mean (ii)?

6. Aug 31, 2004

### broegger

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

7. Aug 31, 2004

### Wong

hi broegger.

I did not do part (ii), but I think all they want you to realise 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.

Last edited: Aug 31, 2004
8. Aug 31, 2004

### 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).

9. Aug 31, 2004

### Wong

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

10. Aug 31, 2004

### 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 :\

11. Aug 31, 2004

### 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.

12. May 11, 2007

### 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.

Last edited: May 11, 2007