What is wrong with this 'limit' proof?

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July 2nd 2011, 10:02 AM
MacstersUndead

What is wrong with this 'limit' proof?

I posted this thread here rather than the university thread because of its apparent simplicity. I just can't seem to figure it out what's wrong. too much summer slacking I suppose.

Start with a circle of diameter 1 inscribed in a square of side length 1
http://imageftw.com/uploads/20110702/circles6.gif

Remove corner squares such that the perimeter of the outside polygon doesn't change and doesn't include the area inside the circle.
http://imageftw.com/uploads/20110702...20-%20Copy.gif

Repeat the above process infinitely for each new corner generated by this process.
--

The perimeter of the polygon is always 4, but the limit of the perimeter as the number of steps approaches infinity is $\pi$ . As 4 = $\pi$ is absurd, what is wrong with this reasoning?
July 2nd 2011, 11:50 AM
Archie Meade

Re: What is wrong with this 'limit' proof?

Quote:

Originally Posted by MacstersUndead

I posted this thread here rather than the university thread because of its apparent simplicity. I just can't seem to figure it out what's wrong. too much summer slacking I suppose.

Start with a circle of diameter 1 inscribed in a square of side length 1
http://imageftw.com/uploads/20110702/circles6.gif

Remove corner squares such that the perimeter of the outside polygon doesn't change and doesn't include the area inside the circle.
http://imageftw.com/uploads/20110702...20-%20Copy.gif

Repeat the above process infinitely for each new corner generated by this process.
--

The perimeter of the polygon is always 4, but the limit of the perimeter as the number of steps approaches infinity is $\pi$ . As 4 = $\pi$ is absurd, what is wrong with this reasoning?

The area within the shape is changing, but the perimeter is a constant.
It may look like a circle at a certain stage,
but if we zoom in, we will see a series of straight lines whose lengths sum to 4.
July 2nd 2011, 08:09 PM
Deveno

Re: What is wrong with this 'limit' proof?

it is, in general, false to assume that because a sequence of curves $\{f_n\}$ converges to another curve g, that they will have equal lengths. if you think about the koch snowflake curve, it is seen to be possible that a convergent sequence of curves can even fail to HAVE a length.

in fact, the polygons actually converge uniformly to the circle, and are continuous. so even these assumptions aren't strong enough.

the limit of the polygons is rather like a fractal, and it is nowhere differentiable. the derivatives of the polygonal segments do not uniformly approach the tangent lines of the circles.

put another way: the limit of the perimeter ≠ the perimiter of the limit. limits can have properties that the terms of the limit do not.

for example, the following sequence: {1/n}. every term is positive, but the limit is not, it is 0. we can divide by any term of the sequence, we cannot divide by 0.

normally, when we find the perimeter of a curve (such as the circle), we compute derivatives (based on tiny slopes). that doesn't work with the polygon, the slopes are always at 0 or 90 degrees.
July 3rd 2011, 07:34 AM
LoblawsLawBlog

Re: What is wrong with this 'limit' proof?

The only thing wrong with the argument is the very last conclusion that pi=4, as opposed to pi<=4. This is just one half of a doomed proof by exhaustion (is it still called that if it doesn't deal with area?). Now if you can do something similar with polygons lying within the circle whose perimeters are less than the circumference and also approach 4, then there would be a serious problem.
July 3rd 2011, 09:20 AM
Archie Meade

1 Attachment(s)

Re: What is wrong with this 'limit' proof?

If the internal edges are all horizontal and vertical as shown
in the attachment,
then no matter how many edges there are,
and whether or not they have unequal lengths,
if the square perimeter is 4

$4\left[\sum_{i=1}^nv_i+\sum_{i=1}^nh_i\right]=4$

It's no more complex than that.

However, if you tilt the edges to form a polygon,
that's quite a different matter,
as the perimeter length is then undergoing reduction.
July 3rd 2011, 02:33 PM
MacstersUndead

Re: What is wrong with this 'limit' proof?

Thank you all for your detailed responses on this matter; it has been quite revealing on the nature of what's wrong with this proof. especially LoblawsLawBlog's post. of all that I learned, I should have at least recalled the Squeeze Theorem.
so the problem really is that giving an upper bound for the limit of the perimeter is insufficient information to conclude the lengths to be equal.

what a colossal mistake...
July 3rd 2011, 09:50 PM
Prove It

Re: What is wrong with this 'limit' proof?

Quote:

Originally Posted by MacstersUndead

I posted this thread here rather than the university thread because of its apparent simplicity. I just can't seem to figure it out what's wrong. too much summer slacking I suppose.

Start with a circle of diameter 1 inscribed in a square of side length 1
http://imageftw.com/uploads/20110702/circles6.gif

Remove corner squares such that the perimeter of the outside polygon doesn't change and doesn't include the area inside the circle.
http://imageftw.com/uploads/20110702...20-%20Copy.gif

Repeat the above process infinitely for each new corner generated by this process.
--

The perimeter of the polygon is always 4, but the limit of the perimeter as the number of steps approaches infinity is $\pi$ . As 4 = $\pi$ is absurd, what is wrong with this reasoning?

I expect you could use this process to evaluate $\displaystyle \pi$ if you concentrate on the areas rather than the perimeters.

For starters, I'll define the circle to have a radius of $\displaystyle 1$ instead of a diameter of $\displaystyle 1$ , so that its area is $\displaystyle \pi$ instead of $\displaystyle \frac{\pi}{4}$ . The area of the largest square is $\displaystyle 4$ .

Since the radius of the circle is $\displaystyle 1$ , the distance from the centre of the circle to the corner of the square is $\displaystyle \sqrt{2}$ by Pythagoras' Theorem. So when you cut off the smaller square in the corner, its diagonal is $\displaystyle \sqrt{2} - 1$ .

Using Pythagoras again to find the side lengths of the square

$\displaystyle \begin{align*} l^2 + l^2 &= (\sqrt{2} - 1)^2 \\ 2l^2 &= 3 - 2\sqrt{2} \\ l^2 &= \frac{3 - 2\sqrt{2}}{2} \\ l &= \sqrt{\frac{3 - 2\sqrt{2}}{2}}\end{align*}$

Therefore the area of the square you have cut off is $\displaystyle \left(\sqrt{\frac{3 - 2\sqrt{2}}{2}}\right)^2 = \frac{3 - 2\sqrt{2}}{2}$ .

Four of these squares would have an area of $\displaystyle 2(3 - 2\sqrt{2}) = 6 - 4 \sqrt{2}$ .

So subtracting these areas from the larger square gives

$\displaystyle 4 - (6 - 4\sqrt{2}) = 4\sqrt{2} - 4 = 2^2(\sqrt{2} - 1)$ .

I am unsure how to continue...