Help - Derivation of formula for Pi (4 Viewers)

Trebla · Jan 17, 2013

With the first bit there should be a |z|⁴ in one of the terms. By the way the method you used to derive the series is quite unusual. What I do is rewrite the fraction as

$\dfrac{z^{2n} - 1}{z - z^{-1}}$

then sub z in which then leads to the result

Otherwise I think the working is fine other than the justification of the integral converging to zero.

seanieg89 · Jan 17, 2013

A brief lapse of my bos hiatus.

Sy, your ideas are quite good although I do have one significant issue with your third page. You attempt to show that the alpha-integral tends to zero as n->inf by showing that the integrand tends to zero for each theta in (0,pi) as n->inf. This is in general not valid! Whilst this integral obviously needs to tend to zero by our knowledge of the value of zeta(2), you will need to show it another way.

A simple demonstration of why "bringing the limit inside" is generally illegal:

$Consider the sequence of functions $f_n(x)=n^2x^n(1-x).$ A simple computation yields:\\ $\int_0^1 f_n(x)\, dx=\frac{n^2}{(n+1)(n+2)} $, but as $n\rightarrow\infty$, we have $f_n(x)\rightarrow 0$ for each $x\in [0,1]$.\\ So\\ $\int_0^1 \lim_{n\rightarrow\infty}f_n(x)\, dx=0\neq 1= \lim_{n\rightarrow\infty}\int_0^1f_n(x)\,dx.$$$

Sy123 · Jan 17, 2013

Trebla said:
With the first bit there should be a |z|⁴ in one of the terms. By the way the method you used to derive the series is quite unusual. What I do is rewrite the fraction as

$\dfrac{z^{2n} - 1}{z - z^{-1}}$

then sub z in which then leads to the result

Otherwise I think the working is fine other than the justification of the integral converging to zero.

Ah yes, sorry about that my 3rd line on the 1st page should be:

$= \frac{z-|z|^2z-z^{2n+1}+z^{2n-1}|z^2|^2}{|1-z^2|^2}$

The reason I didn't sub in cis theta earlier is due to it just being messier and longer to write for me. Its easier this way because I know |cis theta| = 1 so I have less terms to compute.

seanieg89 said:
A brief lapse of my bos hiatus.

Sy, your ideas are quite good although I do have one significant issue with your third page. You attempt to show that the alpha-integral tends to zero as n->inf by showing that the integrand tends to zero for each theta in (0,pi) as n->inf. This is in general not valid! Whilst this integral obviously needs to tend to zero by our knowledge of the value of zeta(2), you will need to show it another way.

A simple demonstration of why "bringing the limit inside" is generally illegal:

$Consider the sequence of functions $f_n(x)=n^2x^n(1-x).$ A simple computation yields:\\ $\int_0^1 f_n(x)\, dx=\frac{n^2}{(n+1)(n+2)} $, but as $n\rightarrow\infty$, we have $f_n(x)\rightarrow 0$ for each $x\in [0,1]$.\\ So\\ $\int_0^1 \lim_{n\rightarrow\infty}f_n(x)\, dx=0\neq 1= \lim_{n\rightarrow\infty}\int_0^1f_n(x)\,dx.$$$

Awww man :/
So right from the first step on page 3, it is an illegal step? What if we just pretend not to evaluate the limit until the end and do this:

$\lim_{n \to \infty} \int_{0}^{\pi} \left(\frac{1}{2n} \frac{\cos 2n\theta}{2\sin \theta}|^{\alpha}_{\pi/2} - \frac{1}{2n}\int_{\pi/2}^{\alpha} \frac{\cos 2n\theta \cos \theta}{2\sin^2 \theta} \ d\theta \right ) \ d\alpha$

$=\lim_{n \to \infty} \frac{1}{2n}\int_{0}^{\pi} \varepsilon (\alpha) \ d\alpha = 0$

???

Is this invalid as well?
So close, yet so far

EDIT: Or what about doing that thing I did above but by applying squeeze theorem ->

$\frac{-1}{2n}\int_{0}^{\pi} \left(\frac{1}{2\sin \alpha} - (-1)^n/2 - \int_{\pi/2}^{\alpha} \frac{\cos \theta}{2\sin^2 \theta} \ d\theta \right) \ d\alpha \leq \frac{1}{2n} \int_{0}^{\pi} \left (\frac{\cos 2n\alpha}{2\sin \alpha} - \frac{(-1)^n}{2} - \int_{\pi/2}^{\alpha} \frac{\cos \theta \cos 2n\theta}{2\sin^2 \theta} \ d\theta \right ) \ d\alpha \leq \frac{1}{2n}\int_{0}^{\pi} \left(\frac{1}{2\sin \alpha} - (-1)^n/2 - \int_{\pi/2}^{\alpha} \frac{\cos \theta}{2\sin^2 \theta} \ d\theta \right) \ d\alpha$

Now, if we compute the upper bound:

$\frac{1}{2n} \int_{0}^{\pi} \frac{1}{\sin \alpha} + \frac{(-1)^{n}}{2} \ d\alpha$

And the problem is

$\int_{0}^{\pi} \frac{1}{\sin x} \ dx$ does not converge....

How would I be able to prove that:

$\lim_{n \to \infty}\frac{1}{2n} \int_{0}^{\pi} \frac{1}{\sin x} \ dx = 0 \ \ \ \ ??$

seanieg89 · Jan 17, 2013

Sy123 said:
Ah yes, sorry about that my 3rd line on the 1st page should be:

$= \frac{z-|z|^2z-z^{2n+1}+z^{2n-1}|z^2|^2}{|1-z^2|^2}$

The reason I didn't sub in cis theta earlier is due to it just being messier and longer to write for me. Its easier this way because I know |cis theta| = 1 so I have less terms to compute.

Awww man :/
So right from the first step on page 3, it is an illegal step? What if we just pretend not to evaluate the limit until the end and do this:

$\lim_{n \to \infty} \int_{0}^{\pi} \left(\frac{1}{2n} \frac{\cos 2n\theta}{2\sin \theta}|^{\alpha}_{\pi/2} - \frac{1}{2n}\int_{\pi/2}^{\alpha} \frac{\cos 2n\theta \cos \theta}{2\sin^2 \theta} \ d\theta \right ) \ d\alpha$

$=\lim_{n \to \infty} \frac{1}{2n}\int_{0}^{\pi} \varepsilon (\alpha) \ d\alpha = 0$

???

Is this invalid as well?
So close, yet so far

Yep, that is the same thing just written differently unfortunately.

Trebla · Jan 17, 2013

Sy123 said:
Ah yes, sorry about that my 3rd line on the 1st page should be:

$= \frac{z-|z|^2z-z^{2n+1}+z^{2n-1}|z^2|^2}{|1-z^2|^2}$

The reason I didn't sub in cis theta earlier is due to it just being messier and longer to write for me. Its easier this way because I know |cis theta| = 1 so I have less terms to compute.

Actually, the way I proposed it is much quicker because you don't even need to expand anything since you immediately get the $\sin 2n\theta$ in the numerator and $2\sin\theta$ denominator. It's just a matter of removing the i's to realise the denominator.

Sy123 · Jan 17, 2013

seanieg89 said:
Yep, that is the same thing just written differently unfortunately.

How about this then:

$\frac{-1}{2n}\int_{0}^{\pi} \left(\frac{1}{2\sin \alpha} - (-1)^n/2 - \int_{\pi/2}^{\alpha} \frac{\cos \theta}{2\sin^2 \theta} \ d\theta \right) \ d\alpha \leq \frac{1}{2n} \int_{0}^{\pi} \left (\frac{\cos 2n\alpha}{2\sin \alpha} - \frac{(-1)^n}{2} - \int_{\pi/2}^{\alpha} \frac{\cos \theta \cos 2n\theta}{2\sin^2 \theta} \ d\theta \right ) \ d\alpha \leq \frac{1}{2n}\int_{0}^{\pi} \left(\frac{1}{2\sin \alpha} - (-1)^n/2 - \int_{\pi/2}^{\alpha} \frac{\cos \theta}{2\sin^2 \theta} \ d\theta \right) \ d\alpha$

Now, if we compute the upper bound:

$\frac{1}{2n} \int_{0}^{\pi} \frac{1}{\sin \alpha} + \frac{(-1)^{n}}{2} \ d\alpha$

And the problem is

$\int_{0}^{\pi} \frac{1}{\sin x} \ dx$ does not converge....

How would I be able to prove that:

$\lim_{n \to \infty}\frac{1}{2n} \int_{0}^{\pi} \frac{1}{\sin x} \ dx = 0 \ \ \ \ ??$

===============

If I am able to prove the last fact that I needed, I end up with:

$\frac{1}{2n} \int_{0}^{\pi} \frac{1}{\sin \alpha} \ d\alpha + \frac{1}{2n}\int_{0}^{\pi} \frac{(-1)^{n}}{2} \ d\alpha$

The second term converges to zero, and so does the first one according to Wolfram Alpha.... if I can prove that the first term converges to zero, then I can apply squeeze law right? Because I haven't carried in the limit which I assume is the cause of the rigour issues.

seanieg89 · Jan 17, 2013

The first term is not defined. As you noted earlier the integral does not converge, so how can we divide something that doesn't exist by 2n?

seanieg89 · Jan 17, 2013

Trebla said:
Actually, the way I proposed it is much quicker because you don't even need to expand anything since you immediately get the $\sin 2n\theta$ in the numerator and $2\sin\theta$ denominator. It's just a matter of removing the i's to realise the denominator.

You edited your post so I assume you saw it but the things you were sandwiching your integral between weren't finite.

Sy123 · Jan 17, 2013

seanieg89 said:
The first term is not defined. As you noted earlier the integral does not converge, so how can we divide something that doesn't exist by 2n?

Well I was thinking of the situation of:

$\lim_{x \to \infty} \frac{x^n}{e^{x}} = 0 \ \ \ \ $for all n$$
And I was thinking if we can do somethng similar here for the non-convergent integral divided by something that could have a faster convergence? (I'm not very familiar with the rigour of convergence so please excuse any garbage in this post).

EDIT: Now I know I can't justify it straight away, but coming back all the way to the beginning after I integrate once, how about I take the limit to infinity at that time:

$\sum_{k=1}^{\infty} \frac{\sin(2k-1)\theta}{2k-1} - \frac{\pi}{4} = \lim_{n \to \infty} \int_{\pi/2}^{\alpha} \frac{\sin 2n\theta}{2\sin \theta} \ d\theta$

I can already prove that the RHS side is zero, I just integrate again and its still zero (with same limits of 0 and pi). The only problem here is, I can't justify integrating infinite number of terms right? So how would I be able to justify it?

EDIT2: Ok, sorry for the big blob of text, but I after some reading I found that I am allowed to justify integrating term by term if and only if the series converges, right?

seanieg89 · Jan 17, 2013

No, convergence itself is not enough to justify the interchange of summation and integration. You need something stronger, like uniform convergence for example.

Typically the sort of interchanges you are talking about are done via Fubini's theorem (wiki it). Note that summation is just a special case of integration when we introduce the measure-theoretic definition of the latter.

The interchange you want to do cannot be justified by Fubini (unless I made a mistake, I only looked at it quite briefly), and I cannot think of any easy way to make the method work.

Sy123 · Jan 25, 2013

This is related to the current problem at hand, but I won't explain my question in case it turns out to be stupidly wrong:

Conjecture:

$f(\theta) = \frac{1}{2}\sum_{k=0}^{n} \binom{2n}{2k-1}(-1)^{k-1} \cos^{2n-2k} \theta \sin^{2k-2} \theta$

Is this true?

$f(\theta) = J(n) \cdot f''(\theta)$

Where J(n) is a constant in this case, however it is in terms of n (this is important).

If I can prove that this is true, a more rigorous proof of the Basel problem arises. (I will explain how so when I get an answer or I may give background later)

EDIT: It would increase the rigour in my statement if I can show that J(n) is a polynomial, or at least diverges to infinity as n approaches infinity.

Sy123 · Jan 27, 2013

Sy123 said:
This is related to the current problem at hand, but I won't explain my question in case it turns out to be stupidly wrong:

Conjecture:

$f(\theta) = \frac{1}{2}\sum_{k=0}^{n} \binom{2n}{2k-1}(-1)^{k-1} \cos^{2n-2k} \theta \sin^{2k-2} \theta$

Is this true?

$f(\theta) = J(n) \cdot f''(\theta)$

Where J(n) is a constant in this case, however it is in terms of n (this is important).

If I can prove that this is true, a more rigorous proof of the Basel problem arises. (I will explain how so when I get an answer or I may give background later)

EDIT: It would increase the rigour in my statement if I can show that J(n) is a polynomial, or at least diverges to infinity as n approaches infinity.

bump, is it possible to prove the result above if its even true?

Sy123 · Jan 30, 2013

Sy123 said:
How about this then:

$\frac{-1}{2n}\int_{0}^{\pi} \left(\frac{1}{2\sin \alpha} - (-1)^n/2 - \int_{\pi/2}^{\alpha} \frac{\cos \theta}{2\sin^2 \theta} \ d\theta \right) \ d\alpha \leq \frac{1}{2n} \int_{0}^{\pi} \left (\frac{\cos 2n\alpha}{2\sin \alpha} - \frac{(-1)^n}{2} - \int_{\pi/2}^{\alpha} \frac{\cos \theta \cos 2n\theta}{2\sin^2 \theta} \ d\theta \right ) \ d\alpha \leq \frac{1}{2n}\int_{0}^{\pi} \left(\frac{1}{2\sin \alpha} - (-1)^n/2 - \int_{\pi/2}^{\alpha} \frac{\cos \theta}{2\sin^2 \theta} \ d\theta \right) \ d\alpha$

Now, if we compute the upper bound:

$\frac{1}{2n} \int_{0}^{\pi} \frac{1}{\sin \alpha} + \frac{(-1)^{n}}{2} \ d\alpha$

And the problem is

$\int_{0}^{\pi} \frac{1}{\sin x} \ dx$ does not converge....

How would I be able to prove that:

$\lim_{n \to \infty}\frac{1}{2n} \int_{0}^{\pi} \frac{1}{\sin x} \ dx = 0 \ \ \ \ ??$

===============

If I am able to prove the last fact that I needed, I end up with:

$\frac{1}{2n} \int_{0}^{\pi} \frac{1}{\sin \alpha} \ d\alpha + \frac{1}{2n}\int_{0}^{\pi} \frac{(-1)^{n}}{2} \ d\alpha$

The second term converges to zero, and so does the first one according to Wolfram Alpha.... if I can prove that the first term converges to zero, then I can apply squeeze law right? Because I haven't carried in the limit which I assume is the cause of the rigour issues.

Reviving because there was a mistake in my working that eventually crashed everything, now that I have found my mistake, everything actually dissolves quite nicely =)

$I=\int_{0}^{\pi} \int_{\pi/2}^{\alpha} \frac{\sin 2n \theta}{2\sin \theta} \ d\theta \ d\alpha$

Integrating by parts, and I will do this step out in full because this is where I made the mistake:

$u=0.5 \csc \theta$

$du=\frac{-\cos^2 \theta}{2\sin \theta}$

$dv=\sin 2n\theta$

$v=\frac{-\cos 2n\theta}{2n}$

$I= \frac{1}{2n}\int_{0}^{\pi} \frac{-\cos 2n\alpha}{2\sin \alpha} + \frac{(-1)^n}{2} - \int_{\pi/2}^{\alpha} \frac{\cos 2n \theta \cos \theta}{2\sin^2 \theta} \ d\theta \ \ d\alpha$

Once more I will apply squeeze theorem, but notice this time I have a - where there was a plus before in the FIRST TERM, which was the mistake the caused the convergence issue.

$\frac{-1}{2n} \int_{0}^{\pi} \frac{-1}{2\sin \alpha} + C - \int_{\pi/2}^{\alpha} \frac{\cos \theta}{2\sin^2 \theta} \ d\theta \ d\alpha \leq I \leq \frac{1}{2n}\int_{0}^{\pi} \frac{-1}{2\sin \alpha} + C - \int_{\pi/2}^{\alpha} \frac{\cos \theta}{2\sin^2 \theta} \ d\theta \ d\alpha$

Now the cool thing is, when I evaluate the integrals of the upper and lower bound (the theta integral), the result is -1/2sin theta, and when we pair this with the alpha integral limit AND the minus on the outside of the integral, in the end it cancels out with the first term! (evaluate it for yourself), so in the end I get:

$\frac{-1}{2n} \int_{0}^{\pi} C \ d\alpha \leq I \leq \frac{1}{2n} \int_{0}^{\pi} C \ d\alpha$

$\therefore \lim_{n \to \infty} I = 0$

$\therefore \sum_{k=1}^{\infty} \frac{1}{(2k-1)^2} = \frac{\pi^2}{8}$

$\therefore \sum_{k=1}^{\infty} \frac{1}{k^2} = \frac{\pi^2}{6}$

$\square$

Now, I repeat, the only thing I did different is that in my stage 3 (the evalution of the integral), I used squeeze theorem, but I did it without any algebraic mistakes this time, resulting in a neat cancellation and when I take it to infinity due to the constant 1/2n at the front everything converges to zero including the integral.

Is everything completely valid now? Is the Basel problem proof complete?

Sy123 · Jan 30, 2013

Full proof attached

A check would be appreciated =)

View attachment SyBaselProblem.pdf

Trebla · Jan 31, 2013

There are a couple of typos like the sum for pi should be (-1)^k-1 rather than (-1)^2k-1.

So far the general idea seems to be fine to me.

(In b4 purists seanie or carrot point out any errors lol

)

Sy123 · Jan 31, 2013

Trebla said:
There are a couple of typos like the sum for pi should be (-1)^k-1 rather than (-1)^2k-1.

So far the general idea seems to be fine to me.

(In b4 purists seanie or carrot point out any errors lol )

Oops, haha hopefully there are no other typos or rigour issues. If there are no rigor issues now, I will retype it in full latex with better explanation of justification of the inequality

seanieg89 · Feb 1, 2013

Another quick visit to BoS. If you can convince me of page three you can convince me that your idea works, (after fixing up less important typos that may exist). From a glance right now I think the inequality part is dodgy. If you can explain each step of that part to me as if I knew no more than the average high school student I could better assess your work (ie agree that it works or pinpoint a mistake). Also brackets would help when it comes to the integrations so I know exactly what you are integrating in each, eg in your current inequality line you have thetas floating around OUTSIDE of the theta integral...

Sy123 · Feb 1, 2013

seanieg89 said:
Another quick visit to BoS. If you can convince me of page three you can convince me that your idea works, (after fixing up less important typos that may exist). From a glance right now I think the inequality part is dodgy. If you can explain each step of that part to me as if I knew no more than the average high school student I could better assess your work (ie agree that it works or pinpoint a mistake). Also brackets would help when it comes to the integrations so I know exactly what you are integrating in each, eg in your current inequality line you have thetas floating around OUTSIDE of the theta integral...

Ok, so we can already establish that:

$I=\int_0^\pi \left (\left ( \frac{-\cos(2n\theta)}{2n \cdot 2\sin \theta} \right )_{\pi/2}^\alpha - \frac{1}{2n} \int_{\pi/2}^{\alpha} \frac{\cos \theta \cos(2n\theta)}{2\sin^2 \theta} \ d\theta \right ) \ d\alpha$

Now we can also establish:

$-1 \leq \cos 2n\theta \leq 1$

This means, that if we take away the cos2ntheta the magnitude of the expression becomes greater than it was before, thus we can create the upper bound in this way (the fact that we take it from the integral as well doesn't matter anyway)
The point is the bounds of the function we are taking away, the lower bound is obvious, we just put the minus there to make the lower bound of opposite magnitude to the first so I must lie in those bounds.

$\frac{1}{2n} \int \left ( \left(\frac{-1}{2\sin \theta} \right)_{\pi/2}^{\alpha} - \int_{\pi/2}^{\alpha} \frac{\cos \theta}{2\sin^2 \theta} \ d\theta \right ) \ d\alpha = J$

It must follow that:"

$-J \leq I \leq J$

However on the chance that J is negative, then it must follow that:

$-J \geq I \geq J$

This doesn't change the result of the squeeze theorem and I suspect this is what I was missing?

Anyway, we just evaluate the integral, variables cancel out, and we have a constant to integrate, which again becomes a constant since its a definite alpha integral, when n goes to infinity the 1/2n dominates the constant hence making J approach zero

Is this succinct enough?

And thanks very much again

seanieg89 · Feb 1, 2013

But if alpha is greater than pi/2, then cos theta is negative on the interval of integration of the inside integral. So making the magnitude of that integrand bigger is making your inner integral smaller (more negative). Whereas if alpha is smaller than pi/2 then the opposite is true.

Eg the integral of x*cos(x) is less than the integral of x when integrated between 0 and 1, but GREATER when integrated between -1 and 0.

I just don't buy your justification for that inequality yet if it is even true. Also, I would prefer you to be the opposite of succinct here. Because of succinctness, it is difficult to know exactly what logic you are using to get from each line to the next. Do one thing at a time if possible for the key progression: from your expression for I to your claim that -J = < I = < J.

Sy123 · Feb 1, 2013

seanieg89 said:
But if alpha is greater than pi/2, then cos theta is negative on the interval of integration of the inside integral. So making the magnitude of that integrand bigger is making your inner integrand smaller (more negative). Whereas if alpha is smaller than pi/2 then the opposite is true.

Eg the integral of x*cos(x) is less than the integral of x when integrated between 0 and 1, but GREATER when integrated between -1 and 0.

Ok so what I need to do to fix this, is:

1) Consider domain of Alpha, I am allowed to restrict it to 0 to pi right? because that is the only alpha that I will ever deal with?

2) Consider all the cases of cos2ntheta in this domain.

3) Find inequalities for all these cases?

Would that fix it?

Help - Derivation of formula for Pi (4 Viewers)

Administrator

Well-Known Member

This too shall pass

Well-Known Member

Administrator

This too shall pass

Well-Known Member

Well-Known Member

This too shall pass

Well-Known Member

This too shall pass

This too shall pass

This too shall pass

This too shall pass

Administrator

This too shall pass

Well-Known Member

This too shall pass

Well-Known Member

This too shall pass

Users Who Are Viewing This Thread (Users: 0, Guests: 4)