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Math Help - Pi = 4

  1. #1
    MHF Contributor alexmahone's Avatar
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    Pi = 4

    I found this on a website. Where's the fallacy?
    Attached Thumbnails Attached Thumbnails Pi = 4-pi-4.jpg  
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: Pi = 4

    I don't understand the last step: \pi=4!=24
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  3. #3
    MHF Contributor alexmahone's Avatar
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    Re: Pi = 4

    Quote Originally Posted by FernandoRevilla View Post
    I don't understand the the last step \pi=4!=24
    That's an exclamation mark, not a factorial sign. But where's the flaw in this "proof"?
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  4. #4
    Behold, the power of SARDINES!
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    Re: Pi = 4

    When Archimedes did his calculuations he used a lower bound as well. Then he used the squeeze theorem to say that the value of Pi must be inbetween the two values. The above diagram only gives you an upper bound on the value of Pi.

    @ Fernando haha that is an awsome observation!
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  5. #5
    MHF Contributor FernandoRevilla's Avatar
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    Re: Pi = 4

    Quote Originally Posted by alexmahone View Post
    That's an exclamation mark, not a factorial sign.
    Irony and Internet: bad marriage.

    But where's the flaw in this "proof"?
    Hint: Try 4 as a lower bound of \pi .
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  6. #6
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    Re: Pi = 4

    Quote Originally Posted by TheEmptySet View Post
    When Archimedes did his calculuations he used a lower bound as well. Then he used the squeeze theorem to say that the value of Pi must be inbetween the two values. The above diagram only gives you an upper bound on the value of Pi.
    Can't you have a similar polygonal path whose length is arbitrarily close to 4 inside the circle?
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  7. #7
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    Re: Pi = 4

    the polygonal path (at the limit) is not differentiable...this shows one must be careful as to how you define "length" of a curve. put another way:

    length(limit) ≠ limit(length).
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  8. #8
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    Re: Pi = 4

    Quote Originally Posted by emakarov View Post
    Can't you have a similar polygonal path whose length is arbitrarily close to 4 inside the circle?
    I do not think so. With a little bit of calculus you should be able to convice yourself that the side length of an inscribed square with maximum periemter would have length \dfrac{\sqrt{2}}{2} So the starting Perimeter would be 2\sqrt{2} \approx 2.83

    Again this just shows that pi is less than for but bigger than 2.83
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  9. #9
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    Re: Pi = 4

    For a similar fallacy, let the broken line from (0, 0) to (1/2, 0) to (1/2, 1/2) to (1, 1/2) to (1, 1) be an "approximation" to the straight line from (0, 0) to (1, 1)- it has length 1/2+ 1/2+ 1/2= 2. A better approximation would be the broken line from (0, 0) to (1/4, 0) to (1/4, 1/4) to (1/2, 1/4) to (1/2, 1/2) to (3/4, 1/2) to (3/4, 3/4) to (1, 3/4) to (1, 1) for a total length of 8(1/4)= 2 again. In general, you have 2n line segments each of length 1/n. "In the limit" this appears to converge to the straight line, which has length \sqrt{2}, but the total length is always 2.
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  10. #10
    Super Member TheChaz's Avatar
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    Re: Pi = 4

    Quote Originally Posted by FernandoRevilla View Post
    I don't understand the last step: \pi=4!=24
    All of this (including the wit) has been thoroughly discussed...
    Is value of $\pi$ = 4? - Mathematics - Stack Exchange
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  11. #11
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    Re: Pi = 4

    this is true. also on physicsfoums.com, and other places.

    this is the nicest "non-technical" explanation i have seen:

    when we approximate things in calculus, we always have a way of showing that the uncertainty, or error of our calculations goes to 0, in the limit.

    in this case, the length of the approximating curve, the zig-zagging polygon, always stays the same, so the error is always 4-pi.

    so, in the limit, all we know is that pi ≤ pi + 4-pi, which is true.
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