- #1

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Check it out: http://imgur.com/EpYQv

Where's the trick?

Where's the trick?

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- Thread starter Johnny B.
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- #1

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Check it out: http://imgur.com/EpYQv

Where's the trick?

Where's the trick?

- #2

mathman

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The 10 line is never quite flat. In the limit it consists of a lot of infinitesimal wiggles.

- #3

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The 10 line is never quite flat. In the limit it consists of a lot of infinitesimal wiggles.

That is certainly not my intuition of the topic. The limit IS flat, and the functions mentioned in the example WILL converge uniformly to the flat line.

The only thing is that even uniform convergence does not imply convergence of the lengths. Indeed, calculating the length involves taking the derivative. And a uniform convergent sequence might not have a converging sequence of derivatives. That is the thing that's going on here!

- #4

olivermsun

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Isn't this kind of the same as the rectangle-circle problem that was discussed a little while back?

- #5

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I think the only trick is the bare assertion that 6 = 10.Check it out: http://imgur.com/EpYQv

Where's the trick?

The limit of the geometric progression is indeed the line segment. However, there's no reason to believe that the length of the limit of the geometric progression is equal to the limit of the length of the geometric progression...

- #6

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This kind of post would be greatly improved by a link...Isn't this kind of the same as the rectangle-circle problem that was discussed a little while back?

- #7

olivermsun

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This kind of post would be greatly improved by a link...

https://www.physicsforums.com/showthread.php?t=450364

- #8

- #9

HallsofIvy

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- #10

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Indeed, calculating the length involves taking the derivative. And a uniform convergent sequence might not have a converging sequence of derivatives.

That definitely makes sense. Thanks to all of you for replying!

- #11

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That is certainly not my intuition of the topic. The limit IS flat, and the functions mentioned in the example WILL converge uniformly to the flat line.

The only thing is that even uniform convergence does not imply convergence of the lengths. Indeed, calculating the length involves taking the derivative. And a uniform convergent sequence might not have a converging sequence of derivatives. That is the thing that's going on here!

is this the same explanation for the pi = 4 paradox? The cutting corners method will converge uniformly to the circle, but there may not exist a converging sequence of derivatives?

- #12

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is this the same explanation for the pi = 4 paradox? The cutting corners method will converge uniformly to the circle, but there may not exist a converging sequence of derivatives?

Yes, it's the same thing actually!

- #13

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I believe you can prove that the cutting corners thing *does* converge uniformly to the circle; you can define on a quadrant-by-quadrant basis functions f_n to represent the nth cut-corner spiky thing, and C to be the original circle ( or partial circle on each quadrant). Then, you can come up with a sequence of numbers M_n, which represent the distances between C and "bigger circles" ( and also engulfing, being bigger than the spiky thing ). You can make M_n converge, and so by the weierstrass M-test, the sequence {f_n} converges uniformly to the circle.

If the convergence of the spiky things is not even uniform, then there is no hope at all right? All that says, is that for some point on your spiky thing, after some n, the point will come arbitrarily close to the smooth curve. But, the ability to draw a picture like the one linked above, or the pi = 4 picture, with ALL points looking arbitrarily closer and closer to the smooth-curve, it seems like uniform convergence is guaranteed

- #14

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I believe you can prove that the cutting corners thing *does* converge uniformly to the circle; you can define on a quadrant-by-quadrant basis functions f_n to represent the nth cut-corner spiky thing, and C to be the original circle ( or partial circle on each quadrant). Then, you can come up with a sequence of numbers M_n, which represent the distances between C and "bigger circles" ( and also engulfing, being bigger than the spiky thing ). You can make M_n converge, and so by the weierstrass M-test, the sequence {f_n} converges uniformly to the circle.

If the convergence of the spiky things is not even uniform, then there is no hope at all right? All that says, is that for some point on your spiky thing, after some n, the point will come arbitrarily close to the smooth curve. But, the ability to draw a picture like the one linked above, or the pi = 4 picture, with ALL points looking arbitrarily closer and closer to the smooth-curve, it seems like uniform convergence is guaranteed

I think you're right here. I don't so any reason why there shouldn't be uniform convergence...

- #15

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thanks :)

- #16

olivermsun

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You can distill the argument to this: suppose you have one right triangle with sides 3, 4, 5 (or whatever). Now start shrinking the triangle. Obviously the hypotenuse is going to get "closer and closer" to the base, but does the ratio of the sides ever change? Basic geometry tells you no -- similar triangles are similar triangles, and the hypotenuse is always longer by the exact same ratio no matter how much you shrink the triangle. You can add or multiply the respective sides of as many little triangles as you want, but that won't change the ratios either.

- #17

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For 1 triangle there are 2 hypotenuses each of length 5.00

For 2 triangles there are 4 hypotenuses each of length 2.50

For 4 triangles there are 8 hypotenuses each of length 1.25 and so on.

There is a simple mathematical relationship between number of hypotenuses and length of one hypotenuse and the total length always comes out to be 5*2 no matter how many triangles are used.

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