Proof That Pi = 2: Intuitively Wrong?

[SpY]]

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)

!

 

[adriank]

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.

 

[HallsofIvy]

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.

 

[SpY]]

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) !

 

[Dickfore]

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.

 

[Mentallic]

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) !

Not quite right. From a macroscopic point of view it looks like it's a straight line with distance $sqrt{2)}$ But with the idea of limits, which is the way it should be approached, the distance is still 2.

 

[adriank]

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

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 $\gamma_n(t)$ (with say $0 \le t \le 1$), and they converge to a limit $\gamma(t)$; that is,
$$\lim_{n \to \infty} \gamma_n(t) = \gamma(t),$$
And each path involved has a well-defined length
$$L_n = \int_0^1 \lVert \gamma_n'(t) \rVert \;dt$$
$$L = \int_0^1 \lVert \gamma'(t) \rVert \;dt$$
because $\gamma_n'(t)$ exists except on a set of measure zero, and likewise for $\gamma'(t)$. So even if
$$\lim_{n \to \infty} \gamma_n(t) = \gamma(t),$$
it is not necessarily true that
$$\lim_{n \to \infty} L_n = L.$$

 

[SpY]]

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.
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Do fractals fit in this anywhere?

 

[adriank]

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
$$\lim_{n \to \infty} \gamma'_n(t).$$
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 $\gamma'(t)$.

 

[Mentallic]

But these infinitesimal pieces are tiny triangles, so you're basically approximating the adjacent sides equal to the hypotenuse.

No matter how small the triangles are, the hypotenuse is still a ratio to the lengths of the other two sides. The length of traveling along the sides is 2 while traveling straight across the hypotenuse is $\sqrt{2}$. Given a similar triangle with half each length, we have a triangle side $$\frac{2}{2}$$ and hypotenuse $$\frac{\sqrt{2}}{2}$$. 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: $$\lim_{n \to \infty}\frac{\sqrt{2}/n}{2/n}$$

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

This is also the same mistake that the OP made with the semicircles problem.

 

[Dickfore]

No matter how small the triangles are, the hypotenuse is still a ratio to the lengths of the other two sides. The length of traveling along the sides is 2 while traveling straight across the hypotenuse is $\sqrt{2}$. Given a similar triangle with half each length, we have a triangle side $$\frac{2}{2}$$ and hypotenuse $$\frac{\sqrt{2}}{2}$$. 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: $$\lim_{n \to \infty}\frac{\sqrt{2}/n}{2/n}$$

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

This is also the same mistake that the OP made with the semicircles problem.

So, if the ratio of the hypotenuse to the side is $\sqrt{2}$, what is the length of the hypotenuse when the length of the side is 0?

 

[Char. Limit]

So, if the ratio of the hypotenuse to the side is $\sqrt{2}$, what is the length of the hypotenuse when the length of the side is 0?

$$0\sqrt{2}=0$$

 

[Mentallic]

So, if the ratio of the hypotenuse to the side is $\sqrt{2}$, what is the length of the hypotenuse when the length of the side is 0?

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.

 

[Dickfore]

But, the length of the broken line is not the limit you evaluated.

 

[Mentallic]

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

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) !

 

[Dickfore]

I evaluated the limit of the length of the hypotenuse to the other two sides as they approach zero.

But, they don't approach zero.

 

[Mentallic]

Read post #3 again. If they don't approach zero what do they do then?

 

[Dickfore]

So, how exactly do $2 n$ intervals each of length $1/n$ approach a length of zero again?

 

[Mentallic]

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?

 

[Dickfore]

He says that the broken line, consisting of n right angled triangles approximates the whole straight line (big hypothenuse) with length $\sqrt{2}$ when $n \rightarrow \infty$. I don't see how your logic applies here.

 

[Mentallic]

The broken line is the sides of each triangle, which is discluding the hypotenuse yes? So each triangle will have 1/n length sides with 2 sides per triangle, and there are n triangles which even as n approaches infinite, is still equal to a total length of 2. This isn't approximating the length of the hypotenuse which is $\sqrt{2}$ and my logic comes from the basic foundation of limits.

 

[Dickfore]

The broken line is the sides of each triangle, which is discluding the hypotenuse yes? So each triangle will have 1/n length sides with 2 sides per triangle, and there are n triangles which even as n approaches infinite, is still equal to a total length of 2. This isn't approximating the length of the hypotenuse which is $\sqrt{2}$ and my logic comes from the basic foundation of limits.

Lol, of course it is 2 at every step. The point was that in the limiting process, one cannot distinguish between the broken line and the hypotenuse, and, therefore we would conclude that these two lines, looking identical, ought to have equal lengths. Thus, the "proof" that $\sqrt{2} = 2$.

I am arguing that the broken line in the infinite process does not have defined length because it is not a piecewise smooth curve, but I don't know if I am right.

 

[SpY]]

Okay this triangle thing sort of makes sense, but can we apply it to the semi circles problem? If I'm translating correctly...

The ratio of the arc (semi-circle) to the radius will always be $$\pi$$.
So for infinitesimally small radii, the semi-circle arc $$da$$, will still be $$\pi dr$$ (where dr is a small piece of the radius, da is a small piece of the sum of the small arcs (which always total pi))
An infinitesimally small radius will not equal that infinitesimally small arc; but when you make the approximation that $$dr \approx da$$ then $$0.00000001 \approx 0.000000314$$ for one element of the line of length 2 (r=1), it eventually totals an approximation of $$\pi \approx 2$$ (If that makes sense)

 

[Mentallic]

The point was that in the limiting process, one cannot distinguish between the broken line and the hypotenuse, and, therefore we would conclude that these two lines, looking identical, ought to have equal lengths. Thus, the "proof" that $\sqrt{2} = 2$.

Well yes we both know that failing to distinguish between them doesn't necessarily mean they're equal when we consider them on a microscopic level (that is, we can see the limiting process) which is why we would result in an invalid proof.

I am arguing that the broken line in the infinite process does not have defined length because it is not a piecewise smooth curve, but I don't know if I am right.

It seems logical

Okay this triangle thing sort of makes sense, but can we apply it to the semi circles problem? If I'm translating correctly...

The ratio of the arc (semi-circle) to the radius will always be $$\pi$$.
So for infinitesimally small radii, the semi-circle arc $$da$$, will still be $$\pi dr$$ (where dr is a small piece of the radius, da is a small piece of the sum of the small arcs (which always total pi))
An infinitesimally small radius will not equal that infinitesimally small arc; but when you make the approximation that $$dr \approx da$$ then $$0.00000001 \approx 0.000000314$$ for one element of the line of length 2 (r=1), it eventually totals an approximation of $$\pi \approx 2$$ (If that makes sense)

Which is why we wouldn't assume $$dr\approx da$$ since relatively they're much different ($\pi$ times different to be exact).

 

[adriank]

I am arguing that the broken line in the infinite process does not have defined length because it is not a piecewise smooth curve, but I don't know if I am right.

That's not it, though. Each broken line in the process has a well-defined length, since it is composed of finitely many line segments. The limit also has a well-defined length, since it's just a line segment. (In general, if the derivative of the path exists and is continuous except at finitely many points, then the length of the path is well-defined, by the usual formula like $L_\gamma = \int \lVert \gamma'(t) \rVert \;dt$.) However, there's no reason to believe that the lengths of the broken lines converge to the length of the limit. As I've already noted above, this is true if the derivatives of the paths in the sequence converge to the derivative of the limit path, but in the case of the broken lines (or the semicircles), the derivatives simply don't converge at all.

 

[Dickfore]

In general, if the derivative of the path exists and is continuous except at finitely many points, then the length of the path is well-defined, by the usual formula like $L_\gamma = \int \lVert \gamma'(t) \rVert \;dt$.

But, that's exactly it. The limit curve has infinitely many points where the derivative is not defined.

 

[adriank]

No, the limit curve is a single line segment.

 

[Dickfore]

No, the limit curve is a single line segment.

No, it is not.

 

[adriank]

Let's make things explicit. For each natural number $n$, let
$$\gamma_n(t) = \begin{cases} (t, t - m/2n) & m/2n \le t \le (m+1)/2n, \quad m \text{ even} \\ (t, (m+1)/2n - t_ & m/2n \le t \le (m+1)/2n, \quad m \text{ odd}, \end{cases}$$
where $0 \le t \le 1$. Verify that this formula gives you a path consisting of $2n$ line segments, arranged in a triangle pattern. In particular, this path is piecewise continuously differentiable, and the derivative is given at all but finitely many points by
$$\gamma_n'(t) = \begin{cases} (1, 1) & m/2n < t < (m+1)/2n, \quad m \text{ even} \\ (1, -1) & m/2n < t < (m+1)/2n, \quad m \text{ odd}. \end{cases}$$
Thus, at all but finitely many points, $\lVert \gamma_n'(t) \rVert = \sqrt{2}$, so the length of $\gamma_n$ is
$$L_n = \int_0^1 \lVert \gamma_n'(t) \rVert \;dt = \int_0^1 \sqrt{2} \;dt = \sqrt{2}.$$Let $\gamma(t) = (t, 0)$. I claim that $\lim_{n \to \infty} \gamma_n(t) = \gamma(t)$. Proof: observe that for each $n$,
$$\lVert \gamma_n(t) - \gamma(t) \rVert \le 1/2n,$$
which converges to 0. Thus, the sequence of paths $\gamma_n$ converges to $\gamma$, which is just a line segment.Now it's pretty clear that $\gamma'(t) = (1, 0)$, so the length of $\gamma$ is
$$L = \int_0^1 \lVert \gamma'(t) \rVert \;dt = \int_0^1 1 \;dt = 1.$$Note that the sequence $\gamma_n'(t)$ does not converge.

 

[Dickfore]

What I'm arguing is that

$$\gamma'_{n}(t) \rightarrow (1, 0), \; n \rightarrow \infty$$

is NOT valid.

EDIT:

I fixed the components of the "limit vector".

 

[adriank]

Of course that's not true, and I never said that.

 

[Dickfore]

Of course that's not true, and I never said that.

I edited my post.

 

[adriank]

Right, and it's precisely because $\gamma_n'(t)$ doesn't converge that the lengths don't converge to the length of the limit. But every length involved exists.

This is a classic example where some behaviour that is constant during some limiting process is not preserved once you take the actual limit. (In other words, the "length of a path function" is not "continuous".)

 

[Dickfore]

I was simply arguing that the lengths do not converge at all, not just to that particular value.

 

[adriank]

The derivatives don't converge, but the lengths certainly do, since they're the constant $\sqrt2$! They don't converge to 1, that's all.