## Wait, is this really true?

For the discussion of math. Duh.

Moderators: gmalivuk, Moderators General, Prelates

Mike_Bson
Posts: 252
Joined: Mon Jul 12, 2010 12:00 pm UTC

### Wait, is this really true?

Is that really true? Is that process really equal to pi? If so, I find it pretty amazing. Are there any proofs for it that I would be interested in, or articles, or anything about it? The idea itself seems pretty ridiculous and unbelievable.
Last edited by Mike_Bson on Mon Aug 30, 2010 2:13 am UTC, edited 1 time in total.

The Scyphozoa
Posts: 2871
Joined: Sun Jun 01, 2008 6:33 pm UTC
Location: Sector 5

### Re: Wait, is this really true?

Well, pi can get pretty tricky at times. I remember reading about something that said that if you have a rectangular board with parallel lines evenly spaced across the whole board, and you toss toothpicks whose length is the same as the distance between the lines, the probability of a toothpick landing to cross one of the lines was a number involving pi. It didn't seem to make sense at all until I realized that it had to do with the area of a circle with a diameter the length of the toothpick.

I don't know what to say about the square root of twos thing, but I'm sure there's a reason for it.
3rdtry wrote:If there ever is another World War, I hope they at least have the decency to call it "World War 2: Episode One"

doogly wrote:murder is a subset of being mean

nash1429
Posts: 190
Joined: Tue Nov 17, 2009 3:06 am UTC
Location: Flatland
Contact:

### Re: Wait, is this really true?

The Scyphozoa wrote:I remember reading about something that said that if you have a rectangular board with parallel lines evenly spaced across the whole board, and you toss toothpicks whose length is the same as the distance between the lines, the probability of a toothpick landing to cross one of the lines was a number involving pi. It didn't seem to make sense at all until I realized that it had to do with the area of a circle with a diameter the length of the toothpick.

This is not a particularly unusual appearance. It simply involves using an integral to find a probability.

And I doubt that the limit actually equals pi because the limit of 2^n as n approaches infinity is infinity, and infinity times anything should still be infinity.

Edit: infinity times anything positive
Last edited by nash1429 on Mon Aug 30, 2010 4:04 am UTC, edited 1 time in total.

jestingrabbit
Factoids are just Datas that haven't grown up yet
Posts: 5967
Joined: Tue Nov 28, 2006 9:50 pm UTC
Location: Sydney

### Re: Wait, is this really true?

nash1429 wrote:And I doubt that the limit actually equals pi because the limit of 2^n as n approaches infinity is infinity, and infinity times anything should still be infinity.

No it shouldn't. What if you multiply 2^n by 1/4^n ? Then you have the limit being 0. If you have something that you might lazily write as [imath]\infty\cdot 0[/imath] in a limit you have an indeterminate form.

As for the OP's question, no idea. I'd start by trying to work out what the function f(x) = \sqrt{ 2- \sqrt{2 + x}} is doing when you apply it several times.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.

phlip
Restorer of Worlds
Posts: 7573
Joined: Sat Sep 23, 2006 3:56 am UTC
Location: Australia
Contact:

### Re: Wait, is this really true?

Yeah, that's infinity * non-zero, which is infinity.

However, the limit with just the square roots and without the 2n part is the golden ratio... that's pretty cool. (Well, if you do the addition first, it's the golden ratio... if you do the subtraction first, as is written in the OP, it's phi-1.)

Code: Select all

enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};void ┻━┻︵​╰(ಠ_ಠ ⚠) {exit((int)⚠);}
[he/him/his]

Mike_Bson
Posts: 252
Joined: Mon Jul 12, 2010 12:00 pm UTC

### Re: Wait, is this really true?

http://en.wikipedia.org/wiki/List_of_li ... ial_limits

The limit above the one that I posted also seems to yield infinity. Is there something wrong with whoever tried to type out the notation?

EDIT- Never mind, disregard.
Last edited by Mike_Bson on Mon Aug 30, 2010 3:01 am UTC, edited 1 time in total.

Qaanol
The Cheshirest Catamount
Posts: 3069
Joined: Sat May 09, 2009 11:55 pm UTC

### Re: Wait, is this really true?

Definitely false.

Just the part with the square root evaluates to the root of a quartic ([imath]x^4-4x^2-x+2=0[/imath]), so it is finite and algebraic. Stick on the exponential multiplier and we go to infinity.

I don’t know what that person was going for, but I can tell you when the incorrect limit was added: 01:50, 1 December 2009, by a user named “MathFacts”.
wee free kings

jestingrabbit
Factoids are just Datas that haven't grown up yet
Posts: 5967
Joined: Tue Nov 28, 2006 9:50 pm UTC
Location: Sydney

### Re: Wait, is this really true?

Mike_Bson wrote:http://en.wikipedia.org/wiki/List_of_limits#Notable_special_limits

The limit above the one that I posted also seems to yield infinity. Is there something wrong with whoever tried to type out the notation?

EDIT- Never mind, disregard.

Yeah, that's a consequence of Stirling's approximation.

http://en.wikipedia.org/wiki/Stirling%27s_approximation

Very much worth remembering. Stirling's approximation, not the limit they showed.

Edit: I've remvoed the offending identity. Someone called MathFacts seems to have added it sometime last december.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.

svensen
Posts: 14
Joined: Wed Aug 11, 2010 12:38 am UTC

### Re: Wait, is this really true?

I think he may have been trying to get at this:

http://functions.wolfram.com/Constants/Pi/09/
(The second from the bottom.)

--EDIT--

That is to say, this:

$\lim_{n\to \infty } \text{\,  }2^{n-1}\sqrt{2-\sqrt{2+\sqrt{2+\text{...} +\sqrt{2}}}}= \pi$

... with n-1 square roots nested in there.
Last edited by svensen on Mon Aug 30, 2010 3:46 am UTC, edited 1 time in total.

++$_ Mo' Money Posts: 2370 Joined: Thu Nov 01, 2007 4:06 am UTC ### Re: Wait, is this really true? I expect MathFacts was going for Vieta's formula: ${2 \over \pi} = \sqrt{1 \over 2} \cdot \sqrt{{1 \over 2} + {1 \over 2}\sqrt{1 \over 2}} \cdot \sqrt{{1 \over 2} + {1 \over 2}\sqrt{{1 \over 2} + {1 \over 2}\sqrt{1 \over 2}}}\cdots$ Mike_Bson Posts: 252 Joined: Mon Jul 12, 2010 12:00 pm UTC ### Re: Wait, is this really true? svensen wrote:I think he may have been trying to get at this: http://functions.wolfram.com/Constants/Pi/09/ (The second from the bottom.) --EDIT-- That is to say, this: $\lim_{n\to \infty } \text{\, }2^{n-1}\sqrt{2-\sqrt{2+\sqrt{2+\text{...} +\sqrt{2}}}}= \pi$ ... with n-1 square roots nested in there. And that is 100% equal to pi? Is there any intuitive reason for this, or any proof I can see? nash1429 Posts: 190 Joined: Tue Nov 17, 2009 3:06 am UTC Location: Flatland Contact: ### Re: Wait, is this really true? While I agree that this is what MathFacts was trying to do, I don't see it as any more plausible than what he/she actually posted. the limit of 2^(n-1) as n approaches infinity should be infinity just like the limit of 2^n. Does someone have an explanation for something like that coming from wolfram? ++$_
Mo' Money
Posts: 2370
Joined: Thu Nov 01, 2007 4:06 am UTC

### Re: Wait, is this really true?

svensen wrote:--EDIT--

That is to say, this:

$\lim_{n\to \infty } \text{\,  }2^{n-1}\sqrt{2-\sqrt{2+\sqrt{2+\text{...} +\sqrt{2}}}}= \pi$

... with n-1 square roots nested in there.
Holy shit, it actually works.

nash: The plus signs make a huge difference; now the value under the radical goes to 0 as n goes to infinity.

nash1429
Posts: 190
Joined: Tue Nov 17, 2009 3:06 am UTC
Location: Flatland
Contact:

### Re: Wait, is this really true?

++$_ wrote:nash: The plus signs make a huge difference; now the value under the radical goes to 0 as n goes to infinity. Didn't notice that; pretty cool. ++$_
Mo' Money
Posts: 2370
Joined: Thu Nov 01, 2007 4:06 am UTC

### Re: Wait, is this really true?

phlip
Restorer of Worlds
Posts: 7573
Joined: Sat Sep 23, 2006 3:56 am UTC
Location: Australia
Contact:

++$_ wrote:Proof Neat. Code: Select all enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};void ┻━┻︵​╰(ಠ_ಠ ⚠) {exit((int)⚠);} [he/him/his] jestingrabbit Factoids are just Datas that haven't grown up yet Posts: 5967 Joined: Tue Nov 28, 2006 9:50 pm UTC Location: Sydney ### Re: Wait, is this really true? Very nice. I've put the right thing in the WP article too. ameretrifle wrote:Magic space feudalism is therefore a viable idea. Qaanol The Cheshirest Catamount Posts: 3069 Joined: Sat May 09, 2009 11:55 pm UTC ### Re: Wait, is this really true? svensen wrote:I think he may have been trying to get at this: http://functions.wolfram.com/Constants/Pi/09/ (The second from the bottom.) --EDIT-- That is to say, this: $\lim_{n\to \infty } \text{\, }2^{n-1}\sqrt{2-\sqrt{2+\sqrt{2+\text{...} +\sqrt{2}}}}= \pi$ ... with n-1 square roots nested in there. Very nice. wee free kings Mike_Bson Posts: 252 Joined: Mon Jul 12, 2010 12:00 pm UTC ### Re: Wait, is this really true? ++$_ wrote:Proof

Cool.

That process equaling pi is actually one of the coolest things I have found out.

phlip
Restorer of Worlds
Posts: 7573
Joined: Sat Sep 23, 2006 3:56 am UTC
Location: Australia
Contact:

### Re: Wait, is this really true?

OK, so if I'm reading that right, the linked proof can be manipulated into saying:$2^{n+1}\sin^{-1}(\frac 12 \underbrace{\sqrt{2-\sqrt{2+\sqrt{2+\cdots+\sqrt{2}}}}}_n) = \pi$Then for sufficiently large n, the sin-1 can be dropped, since sin(x)=x for sufficiently small x.

Is that right? 'Cause that's pretty awesome.

Code: Select all

enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};void ┻━┻︵​╰(ಠ_ಠ ⚠) {exit((int)⚠);}
[he/him/his]

Mike_Bson
Posts: 252
Joined: Mon Jul 12, 2010 12:00 pm UTC

### Re: Wait, is this really true?

phlip wrote:OK, so if I'm reading that right, the linked proof can be manipulated into saying:$2^{n+1}\sin^{-1}(\frac 12 \underbrace{\sqrt{2-\sqrt{2+\sqrt{2+\cdots+\sqrt{2}}}}}_n) = \pi$Then for sufficiently large n, the sin-1 can be dropped, since sin(x)=x for sufficiently small x.

Is that right? 'Cause that's pretty awesome.

Yeah, looks like I could drop the sin^-1(x) once you go through the process far enough.

But I tried to get a decent result:

http://www.wolframalpha.com/input/?i=2^6*%28sin^-1%28%28sqrt%282-%28sqrt%282%2B%28sqrt%282%2B%28sqrt%282%2B%28sqrt%282%29%29%29%29%29%29%29%29%29%29%2F2%29%29

Looks like it works. You have to copy and paste the link.

svensen
Posts: 14
Joined: Wed Aug 11, 2010 12:38 am UTC

++$_ wrote:Proof Nice! Mike_Bson Posts: 252 Joined: Mon Jul 12, 2010 12:00 pm UTC ### Re: Wait, is this really true? So, just let me get this straight: Philip's equation yields pi for any value for n, on account of the arcsine being present. And Svensen's equation yields pi ONLY as n goes to infinity, i.e. the limit is pi, because the arcsine does not need to be there because when n approaches infinity, the whole sqyare root sequence approaches zero, making the arcsine redundant as it becomes zero? I have it right? mike-l Posts: 2758 Joined: Tue Sep 04, 2007 2:16 am UTC ### Re: Wait, is this really true? Mike_Bson wrote:So, just let me get this straight: Philip's equation yields pi for any value for n, on account of the arcsine being present. And Svensen's equation yields pi ONLY as n goes to infinity, i.e. the limit is pi, because the arcsine does not need to be there because when n approaches infinity, the whole sqyare root sequence approaches zero, making the arcsine redundant as it becomes zero? I have it right? More or less. As the squareroot approaches zero, arcsine changes it less and less, so in the limit it doesn't change it at all. addams wrote:This forum has some very well educated people typing away in loops with Sourmilk. He is a lucky Sourmilk. Eastwinn Posts: 303 Joined: Thu Jun 19, 2008 12:36 am UTC Location: Maryland ### Re: Wait, is this really true? I find this one the most intriguing because of how simple it: $\pi = \lim_{n\rightarrow \infty }\frac{2(2n)!!^{2}}{(2n+1)(2n-1)!!^{2}}$ Plus where else would you find something as ridiculous as [imath]!!^{2}[/imath]? Look at how fast it grows.. Nvm, !! did not mean what I thought it did. Last edited by Eastwinn on Mon Aug 30, 2010 4:39 pm UTC, edited 3 times in total. http://aselliedraws.tumblr.com/ - surreal sketches and characters. Mike_Bson Posts: 252 Joined: Mon Jul 12, 2010 12:00 pm UTC ### Re: Wait, is this really true? mike-l wrote: Mike_Bson wrote:So, just let me get this straight: Philip's equation yields pi for any value for n, on account of the arcsine being present. And Svensen's equation yields pi ONLY as n goes to infinity, i.e. the limit is pi, because the arcsine does not need to be there because when n approaches infinity, the whole sqyare root sequence approaches zero, making the arcsine redundant as it becomes zero? I have it right? More or less. As the squareroot approaches zero, arcsine changes it less and less, so in the limit it doesn't change it at all. Alright, this makes more sense, then. still awesome. Above poster: also awesome. jaap Posts: 2094 Joined: Fri Jul 06, 2007 7:06 am UTC Contact: ### Re: Wait, is this really true? Eastwinn wrote:Plus where else would you find something as ridiculous as [imath]!!^{2}[/imath]? Look at how fast it grows.. I think you misunderstand what the "double" factorial notation means. BlackSails Posts: 5315 Joined: Thu Dec 20, 2007 5:48 am UTC ### Re: Wait, is this really true? Ive always used x!! as (x!)! I know its used that way in statistical mechanics. Mike_Bson Posts: 252 Joined: Mon Jul 12, 2010 12:00 pm UTC ### Re: Wait, is this really true? Wait, Eastwin, what did you use !! as? Is your equation still true, you just misunderstood what !! means? EDIT- I typed it into Wolfram Alpha using !! as double factorial, and yes, the equation works. Eastwinn Posts: 303 Joined: Thu Jun 19, 2008 12:36 am UTC Location: Maryland ### Re: Wait, is this really true? I interpreted n!! as (n!)!. That was my intuition because I'd never seen it used before. If you limit yourself to whole values of n then the limit is still really simple. Edit: n!! can be expressed in terms of !, so fractional n wouldn't make the limit crazy either. Edit: Using a little mathamagic applied to the limit I posted originally, I have a new limit for pi which doesn't require !!. This is the simplest one I've seen. $\pi =\lim_{n \to \infty}\frac{n!^{4}2^{4n+1}}{(2n)!(2n+1)!}$ This limit is incredibly fragile. I made a slight alteration in attempts to create a lower bound and an upper bound but the former converged at 0 and the latter went off to infinity. http://aselliedraws.tumblr.com/ - surreal sketches and characters. Mike_Bson Posts: 252 Joined: Mon Jul 12, 2010 12:00 pm UTC ### Re: Wait, is this really true? Very nice, above poster! Eastwinn Posts: 303 Joined: Thu Jun 19, 2008 12:36 am UTC Location: Maryland ### Re: Wait, is this really true? It's not hard to derive when you consider this identity I lifted off the Wiki page that was posted for my once ignorant self ( ): [imath](2k-1)!!=\frac{(2k)!}{k!2^k}[/imath]. I suppose it looked more elegant in the first place but I consider my version to be more simple. I'm going to talk to my Calculus teacher about simplifying it further. He may have some ideas. http://aselliedraws.tumblr.com/ - surreal sketches and characters. Talith Proved the Goldbach Conjecture Posts: 848 Joined: Sat Nov 29, 2008 1:28 am UTC Location: Manchester - UK ### Re: Wait, is this really true? How about [imath](2n)!(2n+1)!=(2n+1)(2n)!^2[/imath]. squareroot Posts: 548 Joined: Tue Jan 12, 2010 1:04 am UTC Contact: ### Re: Wait, is this really true? And n!^4/(2n!)^2 = (2n choose n)^-2? [imath]\pi = lim_{n->\infty} \frac{2^{4n+1}}{{2n \choose n}^2 (2n+1)}[/imath] Which for some reason is being rendered very small. <signature content="" style="tag:html;" overused meta /> Good fucking job Will Yu, you found me - __ - Blatm Posts: 638 Joined: Mon Jun 04, 2007 1:43 am UTC ### Re: Wait, is this really true? Use [math], not [imath]. Qaanol The Cheshirest Catamount Posts: 3069 Joined: Sat May 09, 2009 11:55 pm UTC ### Re: Wait, is this really true? $$\pi = \lim_{n\rightarrow\infty} \frac{2^{4n+1}}{{2n \choose n}^2 (2n+1)}$$ or$\$
wee free kings

dissonant
Posts: 63
Joined: Sat Jan 24, 2009 10:33 am UTC

### Re: Wait, is this really true?

My favourite mysteries are Gregory's series and the "particular case" of Wallis' formula given here:

http://mathworld.wolfram.com/PiFormulas.html

The continued fraction version is also nice. But continued fractions are total phi groupies.