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Obtaining pi through use of trigonometry and limits

  1. Jul 2, 2012 #1
    Whilst messing around with some geometry pertaining to the n-sided regular polygon, I stumbled upon this equation which I could not find anywhere on the internet.
    [itex]\pi = \lim_{n \to \infty} n \sin \frac{\pi}{n}[/itex]
    But if we take this to be true then, by substitution, this is also true:
    [itex]\pi = \lim_{n \to \infty} n \sin (\sin \frac{\pi}{n})[/itex]
    And ad infinitum:
    [itex]\pi = \lim_{n \to \infty} n \sin(\sin(\sin(\sin(...))))[/itex]
    However this makes no sense to me... at the heart of this seemingly infinite sea of sine functions is there a pi/n core? Or is there no centre and is each sine function ultimately a function of nothing?
    Or have I been misusing the maths?

    EDIT:
    Just blew my own mind with
    [itex]p = \lim_{n \to \infty} n \sin \frac{p}{n}[/itex]
     
    Last edited: Jul 2, 2012
  2. jcsd
  3. Jul 2, 2012 #2

    Curious3141

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    This is actually a very well-known limit. It's found in lots of places on the internet if you use the right search phrases. In fact, I rediscovered it myself as a kid, just like you. :biggrin:

    It's not all that useful as a starting point for calculating ∏, though.

    You're misusing math notation. This is akin to saying:

    [tex]2 = \frac{1}{\frac{1}{2}} = \frac{1}{\frac{1}{\frac{1}{\frac{1}{2}}}} = \frac{1}{\frac{1}{\frac{1}{...}}}[/tex]

    So what's at the "bottom" of that infinitely nested fraction? 2? 1? Nothing? This sort of confusion is just because we're using shoddy notation.
     
  4. Jul 2, 2012 #3
    Is there a notation that exists which deals with this "nesting"?
     
  5. Jul 2, 2012 #4

    Curious3141

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    And that's simply a consequence of the fact that [itex]\lim_{x \rightarrow 0} \sin x = x[/itex].
     
  6. Jul 2, 2012 #5

    Curious3141

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  7. Jul 2, 2012 #6
    Isn't it [itex]\lim_{x \rightarrow 0} \sin x = 0[/itex] ?
     
  8. Jul 2, 2012 #7

    Curious3141

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    Yes, it is. I should've been clearer.

    What I meant was [itex]\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1[/itex].

    Meaning that as x gets smaller, sin x is better approximated by x. This can be seen from the Taylor series for sin x. It can also be proved by L' Hopital's Rule.

    In relation to your question,

    [tex]\lim_{n \rightarrow \infty} n \sin(\frac{p}{n}) = p\lim_{n \rightarrow \infty} \frac{n}{p} \sin(\frac{p}{n}) = p\lim_{x \rightarrow 0} \frac{\sin x}{x} = p[/tex]

    after making the substitution [itex]x = \frac{p}{n}[/itex].
     
    Last edited: Jul 2, 2012
  9. Jul 2, 2012 #8
    Ah that makes more sense. I'm stuck in the mindset that a limit implies that the variable is equal to its limit. Thanks
     
  10. Jul 2, 2012 #9

    Curious3141

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    I made an edit which you might want to see.
     
  11. Jul 2, 2012 #10

    micromass

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    This makes no sense. It's not because there is a limit for every finite nesting of sine functions, that you can write an infinite nesting. In fact, there are subtle convergence issues most of the time.

    Here, it is easy to see that such an infinite nesting of since functions must equal 0. Let

    [tex]x=\sin(\sin(\sin(\sin(\sin(...)))))[/tex]

    then

    [tex]\sin(x)=\sin(\sin(\sin(\sin(...))))=x[/tex]

    Solving [itex]\sin(x)=x[/itex], gives us x=0.

    So the nesting of infinite sines is 0.

    Now, why can't you go from a finite nesting to an infinite nesting. Well, let

    [tex]f_n(x)=\sin(...\sin(x))[/tex]

    a nesting of n sine functions. Let [itex]f(x)=\lim_n f_n(x)[/itex] the infinite nesting.

    You have proven that

    [tex]\lim_m mf_n(\pi/m)=\pi[/tex]

    and thus

    [tex]\lim_n \lim_m mf_n(\pi/m)=\pi[/tex]

    This does not imply

    [tex]\pi=\lim_m\lim_n mf_n(\pi/m)=\lim_m mf(\pi/m)[/tex]

    since you cannot exchange two limits in general.
     
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