Help - Derivation of formula for Pi (3 Viewers)

Sy123 · Dec 29, 2012

seanieg89 said:
You want to be taking definite integrals for these sorts of questions. The output of an indefinite integral is an equivalence class of functions which differ by a constant, the output of a definite integral is a number. We want to show that our integral is small, this notion is well defined for the latter but not for the former. (A notion of size could possible be given for the set of such equivalence classes of functions but it would be beyond mx2 and less straightforward).

Well I tried to make my integral definite, the limits being the same as yours -> theta and pi
However I ended up on the RHS with one term being:

$\tan \frac{\pi}{2}$

Which I can't do. So I decided to put in the only other one that leave P on the LHS by itself and that was theta and 0.

So I ended up with:

$\sum_{k=1}^{n} \frac{\sin k\theta}{k} = \theta /2 - \tan \theta /2 + 0.25I_n - 0.25I_{n+1}$

Where:

$I_n = \int _{0}^{\theta} \frac{\cos n\phi}{\cos^2 \phi /2} \ d\ phi$

I then integrated I_n by parts, instead of letting u = cos n phi like I usually do (because it was easier to integrate dv = sec^2 phi/2). Instead I switched them around in order to get 1/n so I can then tend the terms to zero as n is very large.

I got:

$I_n= \frac{\sin n \phi}{n \cos^2 \phi /2}|_{0}^{\theta} - \frac{1}{n} \int_{0}^{\theta} \frac{\sin n \phi}{\cos^3 \phi /2} \ d\phi$

Now we have n in the denominator for the first term, therefore it tends to zero (as numerator is bounded by -1 and 1)
And the second term is a definite integral where at the front it is 1/n. Hence I can say that the integral tends to zero as well? (not sure if my reasoning is correct here).

And since I have proved that I_n and I_(n+1) tend to zero as n approaches infinity, I simply plug in theta = pi /2

$\sum_{k=1}^{\infty} \frac{\sin k \pi /2}{k} = \frac{\pi}{2} - \tan \frac{\pi}{4}$

$1 - 1/3 + 1/5 - 1/7 + ... = \frac{\pi}{4} - 1 \\ \\ \pi/4 = 2 -1/3+1/5-1/7+ ....$

What did I do wrong?

EDIT: Hang on hang on hang on,
If I need to evaluate:

$\lim_{n \to \infty} I_n - I_{n+1}$

At infinity I_n = I_{n+1} right?

Then that limit should be zero anyway yeah? And if so, why do I still get the wrong answer?

seanieg89 · Dec 29, 2012

Sy123 said:
Well I tried to make my integral definite, the limits being the same as yours -> theta and pi
However I ended up on the RHS with one term being:

$\tan \frac{\pi}{2}$

Which I can't do. So I decided to put in the only other one that leave P on the LHS by itself and that was theta and 0.

So I ended up with:

$\sum_{k=1}^{n} \frac{\sin k\theta}{k} = \theta /2 - \tan \theta /2 + 0.25I_n - 0.25I_{n+1}$

Where:

$I_n = \int _{0}^{\theta} \frac{\cos n\phi}{\cos^2 \phi /2} \ d\ phi$

I then integrated I_n by parts, instead of letting u = cos n phi like I usually do (because it was easier to integrate dv = sec^2 phi/2). Instead I switched them around in order to get 1/n so I can then tend the terms to zero as n is very large.

I got:

$I_n= \frac{\sin n \phi}{n \cos^2 \phi /2}|_{0}^{\theta} - \frac{1}{n} \int_{0}^{\theta} \frac{\sin n \phi}{\cos^3 \phi /2} \ d\phi$

Now we have n in the denominator for the first term, therefore it tends to zero (as numerator is bounded by -1 and 1)
And the second term is a definite integral where at the front it is 1/n. Hence I can say that the integral tends to zero as well? (not sure if my reasoning is correct here).

And since I have proved that I_n and I_(n+1) tend to zero as n approaches infinity, I simply plug in theta = pi /2

$\sum_{k=1}^{\infty} \frac{\sin k \pi /2}{k} = \frac{\pi}{2} - \tan \frac{\pi}{4}$

$1 - 1/3 + 1/5 - 1/7 + ... = \frac{\pi}{4} - 1 \\ \\ \pi/4 = 2 -1/3+1/5-1/7+ ....$

What did I do wrong?

EDIT: Hang on hang on hang on,
If I need to evaluate:

$\lim_{n \to \infty} I_n - I_{n+1}$

At infinity I_n = I_{n+1} right?

Then that limit should be zero anyway yeah? And if so, why do I still get the wrong answer?

Will respond to this post in a second but first a response to your edit: your manipulation with I_n and I_{n+1} is only valid if I_n tends to some finite limit.

I will go through your other working carefully soon.

seanieg89 · Dec 29, 2012

Okay, so your initial expression for

$\sum_{k=1}^n \cos(k\theta)$

is incorrect. To check this chuck in

$\theta=\pi, n=1$ .

Re-do this part and simplify and you will end up with my expression for

$S_n(\phi)$ .

Sy123 · Dec 29, 2012

seanieg89 said:
Okay, so your initial expression for

$\sum_{k=1}^n \cos(k\theta)$

is incorrect. To check this chuck in

$\theta=\pi, n=1$ .

Re-do this part and simplify and you will end up with my expression for

$S_n(\phi)$ .

Yep there was a mistake, in the denominator it is supposed to be: $4\sin^2 \phi /2$ not cos :s

The integral then falls out as:

$\sum_{k=1}^{n}k^{-1} \sin k\theta = \int_{\pi}^{\phi} \frac{\cos \phi -1}{4\sin^2 \phi /2} + I_n - I_{n+1}$

Where the I_n is same as before but sin at the bottom instead of cos.

$I_n= \frac{\sin n \phi}{n \sin^2 \phi/2}|_{\pi}^{\theta} + \frac{1}{n} \int_{\pi}^{\theta} \frac{\sin n \phi \cos \phi /2}{\sin^3 \phi/2} \ d\phi$

So as usual first term converges to zero, and I am allowed to say this about the integral too right?
If so then when we integrate the first term of the sum of sin thing at the start

$\lim_{n \to \infty}\sum_{k=1}^{n} \frac{\sin k \theta}{k} = \frac{\pi}{2} - \frac{\theta}{2} + 0 - 0$

Sub in theta = pi /2

$\frac{\pi}{4} = 1- 1/3 +1/5-1/7+ ...$

And it falls out perfectly

Thanks for the help sean and anyone else who contributed

RealiseNothing · Jan 1, 2013

This is outside of the syllabus, but I would try using the Taylor Series expansions of sin(x+h) and sin(x), I think that would be a lot easier.

Unless of course you are trying to derive it with HSC methods.

RealiseNothing · Jan 1, 2013

This is what I would do though, leaving out the limits to make the latexing easier:

$\sqrt{n^2-1} - n$

$\sqrt{n^2-1} - \sqrt{n^2}$

$-\sqrt{n^2}(1 - \frac{\sqrt{n^2-1}}{\sqrt{n^2}})$

$-\sqrt{n^2}(1-\sqrt{\frac{n^2-1}{n^2}})$

$-\sqrt{n^2}(1-\sqrt{1-\frac{1}{n^2}})$

Let 'n' approach infinity in the fraction:

$-\sqrt{n^2}(1-\sqrt{1-0})$

$-\sqrt{n^2}(1-1)$

$=0$

Sy123 · Jan 1, 2013

RealiseNothing said:
This is outside of the syllabus, but I would try using the Taylor Series expansions of sin(x+h) and sin(x), I think that would be a lot easier.

Unless of course you are trying to derive it with HSC methods.

Nevermind lol:

Cant I just say that $\lim_{n \to \infty} \sqrt{n^2-1} = \lim_{n \to \infty}n$

And hence they cancel out?

Sy123 · Jan 1, 2013

RealiseNothing said:
This is what I would do though, leaving out the limits to make the latexing easier:

$\sqrt{n^2-1} - n$

$\sqrt{n^2-1} - \sqrt{n^2}$

$-\sqrt{n^2}(1 - \frac{\sqrt{n^2-1}}{\sqrt{n^2}})$

$-\sqrt{n^2}(1-\sqrt{\frac{n^2-1}{n^2}})$

$-\sqrt{n^2}(1-\sqrt{1-\frac{1}{n^2}})$

Let 'n' approach infinity in the fraction:

$-\sqrt{n^2}(1-\sqrt{1-0})$

$-\sqrt{n^2}(1-1)$

$=0$

Nice proof there

Carrotsticks · Jan 1, 2013

RealiseNothing said:
This is what I would do though, leaving out the limits to make the latexing easier:

$\sqrt{n^2-1} - n$

$\sqrt{n^2-1} - \sqrt{n^2}$

$-\sqrt{n^2}(1 - \frac{\sqrt{n^2-1}}{\sqrt{n^2}})$

$-\sqrt{n^2}(1-\sqrt{\frac{n^2-1}{n^2}})$

$-\sqrt{n^2}(1-\sqrt{1-\frac{1}{n^2}})$

Let 'n' approach infinity in the fraction:

$-\sqrt{n^2}(1-\sqrt{1-0})$

$-\sqrt{n^2}(1-1)$

$=0$

$\\ \sqrt{n^2-1} - n \\\\ = \frac{n^2-1-n^2}{\sqrt{n^2-1}+n} \\\\ = \frac{-1}{\sqrt{n^2-1}+n} \rightarrow 0 $ as $ n \rightarrow \infty$

Also, you can't just take the limit as n -> infinity for individual things.

Counterexample:

$\\ \lim_{n \rightarrow \infty} \left ( 1 + \frac{1}{n}\right )^n \\\\ $By that logic, the limit becomes $ \lim_{n \rightarrow \infty} \left ( 1 + 0 \right )^n = 1 $ when in fact the limit is $ e$

Taking a limit requires 'teamwork' in the sense that ALL terms must approach infinity together, not just individual components.

RealiseNothing · Jan 1, 2013

Sy123 said:
Nevermind lol:

Cant I just say that $\lim_{n \to \infty} \sqrt{n^2-1} = \lim_{n \to \infty}n$

And hence they cancel out?

I would think that is enough, but it depends on how rigorous you want to be. At a higher level of maths (ie a few years into uni) I doubt that would be enough justification. But I can't really speak considering I'm still in high school, so I'll wait until seanie or carrot answer.

Carrotsticks · Jan 1, 2013

Sy123 said:
Nevermind lol:

Cant I just say that $\lim_{n \to \infty} \sqrt{n^2-1} = \lim_{n \to \infty}n$

And hence they cancel out?

$\\ $Consider this limit instead $ \lim_{n \rightarrow \infty} \sqrt{n^2-n} - n$

Sy123 · Jan 1, 2013

Carrotsticks said:
$\\ $Consider this limit instead $ \lim_{n \rightarrow \infty} \sqrt{n^2-n} - n$

I this is a contradiction to my assumption or something to consider in my proof? Because I can't see how to use that.

Carrotsticks · Jan 1, 2013

Sy123 said:
I this is a contradiction to my assumption or something to consider in my proof? Because I can't see how to use that.

Your assumption was that because $\lim_{n \rightarrow \infty} \sqrt{n^2-1} = \lim_{n \rightarrow \infty} n$ , then $\lim_{n \rightarrow \infty} \sqrt{n^2-1} - n = 0$ .

But similarly, would you not say that $\lim_{n \rightarrow \infty} \sqrt{n^2-n} = \lim_{n \rightarrow \infty} n$ ?

So then, by your argument, we should have $\lim_{n \rightarrow \infty} \sqrt{n^2-n} - n = 0$ , which is most certainly not the case.

BONUS: Try to find the limit I posed above.

Sy123 · Jan 1, 2013

Carrotsticks said:
Your assumption was that because $\lim_{n \rightarrow \infty} \sqrt{n^2-1} = \lim_{n \rightarrow \infty} n$ , then $\lim_{n \rightarrow \infty} \sqrt{n^2-1} - n = 0$ .

But similarly, would you not say that $\lim_{n \rightarrow \infty} \sqrt{n^2-n} = \lim_{n \rightarrow \infty} n$ ?

So then, by your argument, we should have $\lim_{n \rightarrow \infty} \sqrt{n^2-n} - n = 0$ , which is most certainly not the case.

BONUS: Try to find the limit I posed above.

By multiplying by 1, where we pick the 1 to be the conjugate of the expression, we arrive at:

$\lim_{n \to \infty} \frac{-n}{\sqrt{n^2+n}+n} \\ \\ =\lim_{n \to \infty} \frac{-\sqrt{n}}{\sqrt{n-1}+\sqrt{n}} = -\frac{1}{2}$

Yes?

Carrotsticks · Jan 1, 2013

Yep, that is correct, though it is best to actually prove, rather than assume, result such as $\lim_{n \rightarrow \infty} \sqrt{n^2-k} - n = 0, \qquad \forall k \in \mathbb{R}$

So for example, in your last step, I would have actually rationalised the denominator to arrive at -1/2.

Sy123 · Jan 1, 2013

Carrotsticks said:
Yep, that is correct, though it is best to actually prove, rather than assume, result such as $\lim_{n \rightarrow \infty} \sqrt{n^2-k} - n = 0, \qquad \forall k \in \mathbb{R}$

So for example, in your last step, I would have actually rationalised the denominator to arrive at -1/2.

Ah I see, alright.

$d/dx \sin x = \sin x (\lim_{n \to \infty} \sqrt{n^2-1}-n) + \cos x$

Using the same conjugate technique I arrive at:

$d/dx \sin x = \sin x (\lim_{n \to \infty} \frac{-1}{\sqrt{n^2-1}+n}) + \cos x = \cos x$

Not sure why I didn't see that earlier lol

RealiseNothing · Jan 1, 2013

Carrotsticks said:
$\\ \sqrt{n^2-1} - n \\\\ = \frac{n^2-1-n^2}{\sqrt{n^2-1}+n} \\\\ = \frac{-1}{\sqrt{n^2-1}+n} \rightarrow 0 $ as $ n \rightarrow \infty$

Also, you can't just take the limit as n -> infinity for individual things.

Counterexample:

$\\ \lim_{n \rightarrow \infty} \left ( 1 + \frac{1}{n}\right )^n \\\\ $By that logic, the limit becomes $ \lim_{n \rightarrow \infty} \left ( 1 + 0 \right )^n = 1 $ when in fact the limit is $ e$

Taking a limit requires 'teamwork' in the sense that ALL terms must approach infinity together, not just individual components.

Well you can take the limit of everything and still arrive at 0, it was just easier to show/explain it by taking the limit of one thing (also just made latexing easier).

AAEldar · Jan 1, 2013

RealiseNothing said:
Well you can take the limit of everything and still arrive at 0, it was just easier to show/explain it by taking the limit of one thing (also just made latexing easier).

Wait, what.

You may arrive at the same answer by taking the limit of what was just in the fraction and say that it's easier to show, but it is by no means correct!

Sy123 · Jan 4, 2013

===========
Another thing:

Ok, I want to find the probability of making n heads from 2n tosses of a coin.
Using binomial probability:

$\binom{2n}{n} (1/2)^{2n}$

Now I can show via algebra, that this is largest value, hence it is most likely to have n heads n tails (this was meant to be the very basis of a marathon question)

I wanted to find the limit as n approaches infinity, now I couldn't find it manually, so I plugged it in Wolfram Alpha, and it says it approaches zero

The problem is, I put the limit in for a general term: $\binom{2n}{k} (1/2)^{2n}$ , put the limit in Wolfram Alpha and it outputs zero...

But surely all the probabilities should add up to 1 right? Is this something to do with how we treat infinity again?

EDIT: I had a hunch that it was zero from this:

$\binom{2n}{n} (1/2)^{2n} = \frac{(2n)(2n-1)(2n-2)...(n+1)}{(4n)(4n-4)(4n-8)...(4)}$

And it looks like that approaches zero, I can't really prove that rigourously but eh.

Carrotsticks · Jan 4, 2013

But you didn't sum up the probabilities. What you just did was find the probability of getting k heads out of 2n tosses, then made the number of tosses approach infinity.

Of course, if I take a million tosses, then the probability of getting say 20 heads/tails would approach zero.

Help - Derivation of formula for Pi (3 Viewers)

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