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(Dis)Proof pi = 2 (!)

  1. Aug 11, 2010 #1
    The proof is intuitively wrong, but I just can't figure out where.

    (Circumference = 2pi*r)

    1. Consider a rod of length L = 2. Draw a semicircle around it, which has radius R=1 and arclength C= pi
    2. Now draw two small semi circles, one going from the midpoint of the rod to the top and to the bottom. Each of these smaller circles has R=1/2 and C=pi/2, making the total length of the arcs = pi
    3. Then a 4-partitioned rod with 4 arcs would each have R=1/4 and C=pi/4, with a total arclength of pi.
    4. Repeat this process with a limit to infinity; so with infinitely many semicircles you can approximate the sum of the arclengths to be the actual length of the rod. This makes it seem pi (sum of infinite C) = 2 (original length of rod)

    !!!
     
  2. jcsd
  3. Aug 11, 2010 #2
    Step 4 is wrong. You have just proven that the arc lengths of a convergent sequence of paths does not necessarily converge to the arc length of the limit path.
     
  4. Aug 11, 2010 #3

    HallsofIvy

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    And you don't have to use arcs. Here is a much simpler version of the same thing:

    Go from (0, 0) to (1, 1) by going from (0, 0) to (1, 0) then to (1, 0) along the horizontal and vertical lines. The total distance is 2.

    Now go from (0, 0) to (1/2, 0) then to (1/2, 1/2) then to (1, 1/2) then to (1, 1), again by the staright lines connecting them. Now there are 4 intervals, each of length 1/2: total distance 2.

    For any positive integer n, go from (0, 0) to (1/n, 0), then to (1/n, 1/n), then to (1/n, 2/n), then to (2/n, 2/n), etc. always by horizontal and vertical straight lines. As n gets larger and larger, the path gets close and closer to the straight line from (0, 0) to (1, 1) but for any n, there are 2n intervals, each of length 1/n- total distance 2.
     
  5. Aug 11, 2010 #4
    adriank, I'm not sure when you say "necessarily" if that means that the statement is sometimes true? And what a limit path is exactly

    HallsOfIvy, I'm not sure but I think you're saying that the approximation is right. In your example going from (0,0) to (1,1) is a displacement of sqrt(2) while the distance moving 1 up on the x then y axes is 2. Then when you just keep breaking down that distance path into smaller intervals (the total of which should always still be 2), it then "approximates" to the displacement of sqrt(2) !
     
  6. Aug 11, 2010 #5
    The broken arc line that you get in the limiting process is not a piecewise continuously differentiable curve by the very definition because it has infinitely many cusp points.
     
  7. Aug 11, 2010 #6

    Mentallic

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    Not quite right. From a macroscopic point of view it looks like it's a straight line with distance [itex]sqrt{2)}[/itex] But with the idea of limits, which is the way it should be approached, the distance is still 2.
     
  8. Aug 11, 2010 #7
    It is sometimes true. A nontrivial example is if you are using piecewise linear approximations to a curve, with the vertices all on the curve. But in general, it's not true, as you've just given a counterexample.

    By limit path, I mean this: You have a sequence of (piecewise differentiable) paths [itex]\gamma_n(t)[/itex] (with say [itex]0 \le t \le 1[/itex]), and they converge to a limit [itex]\gamma(t)[/itex]; that is,
    [tex]\lim_{n \to \infty} \gamma_n(t) = \gamma(t),[/tex]
    And each path involved has a well-defined length
    [tex]L_n = \int_0^1 \lVert \gamma_n'(t) \rVert \;dt[/tex]
    [tex]L = \int_0^1 \lVert \gamma'(t) \rVert \;dt[/tex]
    because [itex]\gamma_n'(t)[/itex] exists except on a set of measure zero, and likewise for [itex]\gamma'(t)[/itex]. So even if
    [tex]\lim_{n \to \infty} \gamma_n(t) = \gamma(t),[/tex]
    it is not necessarily true that
    [tex]\lim_{n \to \infty} L_n = L.[/tex]
     
    Last edited: Aug 11, 2010
  9. Aug 11, 2010 #8
    Mentallic, I'm still confused. What it's saying is that the long way between the two points (ie. going 1 along x and 1 along y) just gets split into infinitesimal pieces that you can place along the straight line between the points. But these infinitesimal pieces are tiny triangles, so you're basically approximating the adjacent sides equal to the hypotenuse.

    This is a horrific pic, but if the semicircles alternated on each side, and didn't just "hop" along on one side, it wouldn't be piecewise because its continuous over the length.
    (|
    |)
    (|
    |)

    Do fractals fit in this anywhere?
     
  10. Aug 11, 2010 #9
    Regarding the semicircles attached to each side, now the functions are (continuously) differentiable everywhere, but the length of the limit curve is still different from the limit of the lengths. This is possible because the sequence of derivatives doesn't actually converge.

    In terms of my previous post, in general, the limit
    [tex]\lim_{n \to \infty} \gamma'_n(t).[/tex]
    doesn't exist. However: If the limit exists (and the convergence is uniform, which is certainly the case when the curves are continuously differentiable), then it is equal to [itex]\gamma'(t)[/itex].
     
    Last edited: Aug 11, 2010
  11. Aug 12, 2010 #10

    Mentallic

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    No matter how small the triangles are, the hypotenuse is still a ratio to the lengths of the other two sides. The length of travelling along the sides is 2 while travelling straight across the hypotenuse is [itex]\sqrt{2}[/itex]. Given a similar triangle with half each length, we have a triangle side [tex]\frac{2}{2}[/tex] and hypotenuse [tex]\frac{\sqrt{2}}{2}[/tex]. Keep making these triangles smaller and we still have the same rate.

    Now, limits aren't meant to be used in the way that if we say let the lengths be infinitesimal, then the hypotenuse is also infinitesimal and thus the hypotenuse equals the lengths. This is all wrong.

    The ratio of the hypotenuse to the lengths of the other two sides of each successively smaller triangle can be described by the limit: [tex]\lim_{n \to \infty}\frac{\sqrt{2}/n}{2/n}[/tex]

    Which gives us the indeterminate form [tex]\frac{0}{0}[/tex] but you're saying that this is equal to 1. This is not true. Looking at the problem, obviously the answer is [tex]1/\sqrt{2}[/tex] after some simplification.

    This is also the same mistake that the OP made with the semicircles problem.
     
  12. Aug 12, 2010 #11
    So, if the ratio of the hypotenuse to the side is [itex]\sqrt{2}[/itex], what is the length of the hypotenuse when the length of the side is 0?
     
  13. Aug 12, 2010 #12

    Char. Limit

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    [tex]0\sqrt{2}=0[/tex]
     
  14. Aug 12, 2010 #13

    Mentallic

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    Of course it's 0. With a side length of 0 the degenerate triangle becomes a point. This isn't how you should be thinking about limits though.
     
  15. Aug 12, 2010 #14
    But, the length of the broken line is not the limit you evaluated.
     
  16. Aug 12, 2010 #15

    Mentallic

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    Come again? I'm guessing the broken line is the sides other than the hypotenuse.

    I evaluated the limit of the length of the hypotenuse to the other two sides as they approach zero. I'm simply disproving SpY's claim that

     
  17. Aug 13, 2010 #16
    But, they don't approach zero.
     
  18. Aug 13, 2010 #17

    Mentallic

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    Read post #3 again. If they don't approach zero what do they do then?
     
  19. Aug 13, 2010 #18
    So, how exactly do [itex]2 n[/itex] intervals each of length [itex]1/n[/itex] approach a length of zero again?
     
  20. Aug 13, 2010 #19

    Mentallic

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    This is what all the fuss was about? I was talking about the ratio of the hypotenuse to the other two sides in a single triangle. Anyway, I believe we're both happy now, yes?
     
  21. Aug 13, 2010 #20
    He says that the broken line, consisting of n right angled triangles approximates the whole straight line (big hypothenuse) with length [itex]\sqrt{2}[/itex] when [itex]n \rightarrow \infty[/itex]. I don't see how your logic applies here.
     
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