Wait, is this really true?
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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
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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.
I don't know what to say about the square root of twos thing, but I'm sure there's a reason for it.
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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
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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.
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 phlip
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Re: Wait, is this really true?
Yeah, that's infinity * nonzero, which is infinity.
However, the limit with just the square roots and without the 2^{n} 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 phi1.)
However, the limit with just the square roots and without the 2^{n} 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 phi1.)
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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.
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.
Re: Wait, is this really true?
Definitely false.
Just the part with the square root evaluates to the root of a quartic ([imath]x^44x^2x+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”.
Just the part with the square root evaluates to the root of a quartic ([imath]x^44x^2x+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”.
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 jestingrabbit
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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.
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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:
[math]\lim_{n\to \infty } \text{$\, $ }2^{n1}\sqrt{2\sqrt{2+\sqrt{2+\text{...} +\sqrt{2}}}}= \pi[/math]
... with n1 square roots nested in there.
http://functions.wolfram.com/Constants/Pi/09/
(The second from the bottom.)
EDIT
That is to say, this:
[math]\lim_{n\to \infty } \text{$\, $ }2^{n1}\sqrt{2\sqrt{2+\sqrt{2+\text{...} +\sqrt{2}}}}= \pi[/math]
... with n1 square roots nested in there.
Last edited by svensen on Mon Aug 30, 2010 3:46 am UTC, edited 1 time in total.
Re: Wait, is this really true?
I expect MathFacts was going for Vieta's formula:
[math]{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[/math]
[math]{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[/math]
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:
[math]\lim_{n\to \infty } \text{$\, $ }2^{n1}\sqrt{2\sqrt{2+\sqrt{2+\text{...} +\sqrt{2}}}}= \pi[/math]
... with n1 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?
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^(n1) 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?
Re: Wait, is this really true?
Holy shit, it actually works.svensen wrote:EDIT
That is to say, this:
[math]\lim_{n\to \infty } \text{$\, $ }2^{n1}\sqrt{2\sqrt{2+\sqrt{2+\text{...} +\sqrt{2}}}}= \pi[/math]
... with n1 square roots nested in there.
nash: The plus signs make a huge difference; now the value under the radical goes to 0 as n goes to infinity.
Re: Wait, is this really true?
++$_ wrote:nash: The plus signs make a huge difference; now the value under the radical goes to 0 as n goes to infinity.
Didn't notice that; pretty cool.
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Re: Wait, is this really true?
++$_ wrote:Proof
Neat.
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 jestingrabbit
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Re: Wait, is this really true?
Very nice. I've put the right thing in the WP article too.
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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:
[math]\lim_{n\to \infty } \text{$\, $ }2^{n1}\sqrt{2\sqrt{2+\sqrt{2+\text{...} +\sqrt{2}}}}= \pi[/math]
... with n1 square roots nested in there.
Very nice.
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Re: Wait, is this really true?
++$_ wrote:Proof
Cool.
That process equaling pi is actually one of the coolest things I have found out.
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Re: Wait, is this really true?
OK, so if I'm reading that right, the linked proof can be manipulated into saying:[math]2^{n+1}\sin^{1}(\frac 12 \underbrace{\sqrt{2\sqrt{2+\sqrt{2+\cdots+\sqrt{2}}}}}_n) = \pi[/math]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.
Is that right? 'Cause that's pretty awesome.
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Re: Wait, is this really true?
phlip wrote:OK, so if I'm reading that right, the linked proof can be manipulated into saying:[math]2^{n+1}\sin^{1}(\frac 12 \underbrace{\sqrt{2\sqrt{2+\sqrt{2+\cdots+\sqrt{2}}}}}_n) = \pi[/math]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.
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?
I have it right?
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.
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Re: Wait, is this really true?
I find this one the most intriguing because of how simple it:
[math]\pi = \lim_{n\rightarrow \infty }\frac{2(2n)!!^{2}}{(2n+1)(2n1)!!^{2}}[/math]
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.
[math]\pi = \lim_{n\rightarrow \infty }\frac{2(2n)!!^{2}}{(2n+1)(2n1)!!^{2}}[/math]
Last edited by Eastwinn on Mon Aug 30, 2010 4:39 pm UTC, edited 3 times in total.
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Re: Wait, is this really true?
mikel 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.
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.
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Re: Wait, is this really true?
Ive always used x!! as (x!)!
I know its used that way in statistical mechanics.
I know its used that way in statistical mechanics.
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.
EDIT I typed it into Wolfram Alpha using !! as double factorial, and yes, the equation works.
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.
[math]\pi =\lim_{n \to \infty}\frac{n!^{4}2^{4n+1}}{(2n)!(2n+1)!}[/math]
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.
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.
[math]\pi =\lim_{n \to \infty}\frac{n!^{4}2^{4n+1}}{(2n)!(2n+1)!}[/math]
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.
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Re: Wait, is this really true?
Very nice, above poster!
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](2k1)!!=\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.
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 Talith
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Re: Wait, is this really true?
How about [imath](2n)!(2n+1)!=(2n+1)(2n)!^2[/imath].

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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.
[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.
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Re: Wait, is this really true?
Use [math], not [imath].
Re: Wait, is this really true?
$$\pi = \lim_{n\rightarrow\infty} \frac{2^{4n+1}}{{2n \choose n}^2 (2n+1)}$$
or $$
or $$
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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.
http://mathworld.wolfram.com/PiFormulas.html
The continued fraction version is also nice. But continued fractions are total phi groupies.
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