Help - Derivation of formula for Pi (1 Viewer)

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
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:



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:



Where:



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:



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





:(

What did I do wrong?

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



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?
 
Last edited:

seanieg89

Well-Known Member
Joined
Aug 8, 2006
Messages
2,662
Gender
Male
HSC
2007
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:



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:



Where:



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:



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





:(

What did I do wrong?

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



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

Well-Known Member
Joined
Aug 8, 2006
Messages
2,662
Gender
Male
HSC
2007
Okay, so your initial expression for



is incorrect. To check this chuck in

.

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

.
 

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
Okay, so your initial expression for



is incorrect. To check this chuck in

.

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

.
Yep there was a mistake, in the denominator it is supposed to be: not cos :s

The integral then falls out as:



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



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



Sub in theta = pi /2




And it falls out perfectly :)

Thanks for the help sean and anyone else who contributed
 

RealiseNothing

what is that?It is Cowpea
Joined
Jul 10, 2011
Messages
4,591
Location
Sydney
Gender
Male
HSC
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

what is that?It is Cowpea
Joined
Jul 10, 2011
Messages
4,591
Location
Sydney
Gender
Male
HSC
2013
This is what I would do though, leaving out the limits to make the latexing easier:











Let 'n' approach infinity in the fraction:





 

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
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.
Nevermind lol:

Cant I just say that

And hence they cancel out?
 

Carrotsticks

Retired
Joined
Jun 29, 2009
Messages
9,494
Gender
Undisclosed
HSC
N/A
This is what I would do though, leaving out the limits to make the latexing easier:











Let 'n' approach infinity in the fraction:







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

Counterexample:



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

RealiseNothing

what is that?It is Cowpea
Joined
Jul 10, 2011
Messages
4,591
Location
Sydney
Gender
Male
HSC
2013
Nevermind lol:

Cant I just say that

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

Retired
Joined
Jun 29, 2009
Messages
9,494
Gender
Undisclosed
HSC
N/A
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 , then .

But similarly, would you not say that ?

So then, by your argument, we should have , which is most certainly not the case.

BONUS: Try to find the limit I posed above.
 
Last edited:

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
Your assumption was that because , then .

But similarly, would you not say that ?

So then, by your argument, we should have , 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:



Yes?
 

Carrotsticks

Retired
Joined
Jun 29, 2009
Messages
9,494
Gender
Undisclosed
HSC
N/A
Yep, that is correct, though it is best to actually prove, rather than assume, result such as

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

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
Yep, that is correct, though it is best to actually prove, rather than assume, result such as

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



Using the same conjugate technique I arrive at:



Not sure why I didn't see that earlier lol
 

RealiseNothing

what is that?It is Cowpea
Joined
Jul 10, 2011
Messages
4,591
Location
Sydney
Gender
Male
HSC
2013


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

Counterexample:



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

Premium Member
Joined
Apr 5, 2010
Messages
2,246
Gender
Male
HSC
2011
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

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
===========
Another thing:

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



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: , 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:



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

Carrotsticks

Retired
Joined
Jun 29, 2009
Messages
9,494
Gender
Undisclosed
HSC
N/A
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.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top