Power series expansion and $\pi$

[gugo82]

Exercise:

1. Let [tex]$f(z) =\text{Ln }\frac{1+z}{1-z}$[/tex], where [tex]$\text{Ln}$[/tex] is the principal branch of the complex logarithm.
Find the power series expansion of [tex]$f(z)$[/tex] centered in [tex]$z_0=0$[/tex].

2. Use the answer to 1. to prove that the following summation identity:

(a) [tex]$\sum_{n=1}^{+\infty} \frac{\sin (2n+1)\theta}{2n+1} =\frac{\pi}{4}$[/tex]

holds true for all [tex]$\theta \in ]0,\pi[$[/tex].

3. Can you prove (a) using Fourier series methods?

[gugo82]

"gugo82":2. Use the answer to 1. to prove that the following summation identity:

(a) [tex]$\sum_{n=1}^{+\infty} \frac{\sin (2n+1)\theta}{2n+1} =\frac{\pi}{4}$[/tex]

holds true for all [tex]$\theta \in ]0,\pi[$[/tex].

As Camillo wrote, we have:

[tex]$\text{Ln} \frac{1+z}{1-z} =2\sum_{n=0}^{+\infty} \frac{z^{2n+1}}{2n+1}$[/tex],

and (by Picard's theorem on power series) the RHS converges in the closed unit disc with the only exception of the points [tex]$\pm 1$[/tex], i.e. the series converges in [tex]$\overline{D}(0;1) \setminus \{ \pm 1\}$[/tex].

Now, if we set [tex]$z=e^{\imath \theta}$[/tex] with [tex]$\theta \in ]0,\pi[$[/tex] (so that [tex]$z$[/tex] belongs to the unit semicircumference lying in the upper halfplane), the power series expansion gives:

(*) [tex]$\frac{1}{2}\ \text{Ln} \frac{1+e^{\imath \theta}}{1-e^{\imath \theta}} =\sum_{n=0}^{+\infty} \frac{e^{\imath (2n+1)\theta}}{2n+1}$[/tex]
[tex]$=\sum_{n=0}^{+\infty} \frac{\cos (2n+1)\theta }{2n+1} +\imath \sum_{n=0}^{+\infty} \frac{\sin (2n+1)\theta }{2n+1}$[/tex],

and we can see that in the last line of (*) there is the numerical series which appears at the LHS of (a).
So one way to solve our problem is to explicitly find the value of [tex]$\text{Ln} \tfrac{1+e^{\imath \theta}}{1-e^{\imath \theta}}$[/tex].

Let us draw some pictures.
From:
[asvg]xmin=-1.5;xmax=1.5;ymin=-1.5;ymax=1.5;
axes("","");
stroke="lightgray"; arc([1,0],[-1,0],1);
marker="arrow";
stroke="red"; line([0,0],[0.5,0.87]); text([0.5,0.87],"z",aboveright);
stroke="blue"; line([0,0],[1.5,0.87]); text([1.5,0.87],"1+z",above);
marker="none";
stroke="orange"; line([0,0],[1,0]); line([1,0],[1.5,0.87]); line([1.5,0.87],[0.5,0.87]);[/asvg]
we infer:

[tex]$\text{Arg} (1+z) =\tfrac{1}{2}\ \text{Arg} (z) =\tfrac{1}{2}\ \theta$[/tex],

because the vector [tex]$1+z$[/tex] is the diagonal of the parallelogram having [tex]$0,1,1+z,z$[/tex] as vertices; in analogous manner, from the picture:
[asvg]xmin=-1.5;xmax=1.5;ymin=-1.5;ymax=1.5;
axes("","");
stroke="lightgray"; arc([1,0],[-1,0],1);
marker="arrow";
stroke="red"; line([0,0],[0.5,0.87]); text([0.5,0.87],"z",aboveright);
stroke="purple"; line([0,0],[-0.5,-0.87]); text([-0.5,-0.87],"-z",belowleft);
stroke="cyan"; line([0,0],[0.5,-0.87]); text([0.5,-0.87],"1-z",belowright);
marker="none";
stroke="pink"; line([0,0],[1,0]); line([1,0],[0.5,-0.87]); line([0.5,-0.87],[-0.5,-0.87]);[/asvg]
we infer:

[tex]$\text{Arg} (1-z) =\tfrac{1}{2}\ \text{arg} (-z) = \tfrac{1}{2}\ (\theta -\pi)$[/tex],

because the vector [tex]$1-z$[/tex] is the diagonal of the parallelogram having [tex]$0,1,1-z,-z$[/tex] as vertices; moreover a simple computation shows that:

[tex]$|1\pm z|^2 =2(1\pm \cos \theta) \quad \Rightarrow \quad |1\pm z|=\sqrt{2(1\pm \cos \theta)}$[/tex].

Hence:

[tex]$\text{Ln} \frac{1+z}{1-z} = \text{Ln} \left\{ \sqrt{\frac{1+\cos \theta}{1-\cos \theta}}\ \frac{e^{\imath \frac{\theta}{2}}}{e^{\imath (\frac{\theta}{2} -\frac{\pi}{2})}} \right\}$[/tex]
[tex]$=\text{Ln} \left\{ \sqrt{\frac{1+\cos \theta}{1-\cos \theta}}\ e^{\imath \frac{\pi}{2}} \right\}$[/tex]
[tex]$=\text{Ln} \left\{ \cot \left( \frac{\theta}{2}\right)\ e^{\imath \frac{\pi}{2}} \right\}$[/tex],

and, recalling that [tex]$\text{Ln} \zeta =\ln |\zeta| + \imath \text{Arg} (\zeta)$[/tex] (where [tex]$\ln$[/tex] is the real natural logarithm of the positive real number [tex]$|\zeta|$[/tex]), from the last computations equality:

(**) [tex]$\text{Ln} \frac{1+z}{1-z} = \ln \cot \left( \frac{\theta}{2}\right) \ +\imath \frac{\pi}{2}$[/tex]

readily follows.
Therefore a comparison of (*) and (**) yields (a). [tex]$\square$[/tex]

***

We explicitly remark that a comparison of (*) and (**) yields also the following summation identity:

(b) [tex]$\ln \cot \left( \frac{\theta}{2}\right) = 2 \sum_{n=0}^{+\infty} \frac{\cos (2n+1)\theta}{2n+1}$[/tex],

which gives the Fourier series expansion of the [tex]$2\pi$[/tex]-periodic even function obtained from the extension of [tex]$\ln \cot \left( \tfrac{\theta}{2} \right)$[/tex].
Since the sum is of class [tex]$C^\infty(]0,\pi[)$[/tex], the convergence of the series is uniform in every compact interval [tex]$[a,b] \subset ]0,\pi[$[/tex], hence we can take the exponential of both sides in (b) and write the following infinite product expansion:

(c) [tex]$\cot \left( \frac{\theta}{2} \right) =\prod_{n=0}^{+\infty} \exp \left( \frac{2}{2n+1}\ \cos (2n+1)\theta} \right)$[/tex]

for [tex]$\theta \in ]0,\pi[$[/tex], which for example gives:

[tex]$\prod_{n=0}^{+\infty} \exp \left( \frac{2}{2n+1}\ \cos \frac{(2n+1)\pi}{6}} \right) =2+\sqrt{3}$[/tex]

for [tex]$\theta =\tfrac{\pi}{6}$[/tex]...

[gugo82]

Hint for 2: What happens in [tex]\text{Ln} \frac{1+z}{1-z} =2\sum_{n=0}^{+\infty} \frac{z^{2n+1}}{2n+1}[/tex] if one takes [tex]$z$[/tex] s.t. [tex]$|z|=1$[/tex] and [tex]$\text{Arg} z\in ]0,2\pi[$[/tex]?

[gugo82]

Again, correct!

[Camillo]

Point 1

$ln((1+z)/(1-z)) =ln(1+z)-ln(1-z) $.
It is
$ln(1+z)= z-z^2/2+z^3/3-z^4/4 +... $
and $ ln(1-z )=-z-z^2/2-z^3/3-z^4/4-...$
Consequently $ ln((1+z)/(1-z)) = 2[ z+z^3/3+z^5/5+...]= 2 sum_(n=0 )^(+oo)z^(2n+1)/(2n+1) $.

[Camillo]

Ok inserted the missing $x $ , thanks

[gugo82]

Correct.
OK Camillo!

Just a little remark: you missed an [tex]$x$[/tex] in the summation formulae written in your last lines.


Now... Does someone want to try to answer questions 1 and 2?

[Camillo]

My solution for point 3 .

I consider the square wave function $f(x)= A $ for $0
The Fourier series of $ f(x)= A sgn(x) $ extended periodically is given by :

$ a_0/2+sum_(n=1)^(+oo) (a_k cos kx +b_k sin kx) $ , being by definition :

$a_k = 1/(pi) int_(-pi)^(pi) Asgn(x) coskx dx =0 $ since we have an integral of an odd function over a symmetric interval.

$b_k= 1/pi int_(-pi)^(pi) A sgn(x) sinkxdx= 2/pi int_0^pi A sin kx dx =(2A)/pi [ -coskx/k]_0^pi=(2A)/(kpi) [1-(-1)^k]= 0 $ if $ k=2n $
$ ; = (4A)/(pi(2n+1))$ for $k=2n+1$.

If $A=pi/4 $ we get finally for $0 If $ -pi

Cita 0 0

Segnala

[gugo82]

"Camillo":I explain reason of my statement :

in the initial issue it was indicated $ theta in ( 0, 2pi) $ and I tried for $theta = pi $, of course without success, since I got $ 0=pi/4 $

Yes, my fault... I just made a mistake in my computations!

To be honest, I noticed my mistake today morning (during a break, I was reading again the computations I made to solve the exercise) but I had to work and didn't turn on my PC to correct the text.
When I finally turned on my PC (at home, after dinner), the first thing I did was to correct part 2 of my exercise; and I read your post only after having edited mine.

[Camillo]

I explain reason of my statement :

in the initial issue it was indicated $ theta in ( 0, 2pi) $ and I tried for $theta = pi $, of course without success, since I got $ 0=pi/4 $

[gugo82]

"Camillo":I don't understand identity 2a).

I'm not sure of what to say...

About the meaning of formula (a), I can say it states that the sines series:

[tex]$\sum \frac{1}{2n+1}\ \sin (2n+1)\theta$[/tex]

converges to the constant function [tex]$u(\theta):=\tfrac{\pi}{4}$[/tex] in the whole interval [tex]$]0,\pi[$[/tex].

Otherwise, if you're worried about the meaning of [tex]$\sin (2n+1)\theta$[/tex]... Well, it simply means [tex]$\sin [(2n+1)\theta]$[/tex].
I thought the square brackets were unnecessary and ugly, so I omitted them. Don't you think the same?

[Camillo]

I don't understand identity 2a).