I don't get why this troll physics is wrong.

  • Thread starter Thread starter LogicX
  • Start date Start date
  • Tags Tags
    Physics
Click For Summary
The discussion centers on the flaws in a troll physics scenario that claims to approximate the perimeter of a circle by removing corners from a square. Participants argue that this method fails to converge to the actual circumference because the perimeter remains constant at 4, regardless of the number of corners removed. They emphasize that true convergence requires the path to approach a smooth curve, which this method does not achieve. Additionally, the conversation touches on the definitions of perimeter and the implications of using different norms, concluding that the example misrepresents the relationship between perimeter and pi. Ultimately, the flawed logic leads to an incorrect assertion that pi equals 4, which is fundamentally incorrect.
  • #61
Limits don't always commute with other operations.

In other words, if s_i is a sequence, it is it not necessarily true that f(lim s_i) = lim f(s_i). In this case in particular, the limit of the perimeters isn't necessarily the perimeter of the limit.

Why is this? It just is.

Let's put it this way. You expect there should be a law that says the perimeter of the limit is the limit of the perimeters. But you can't just assume it's true. You have to PROVE it's true.

The problem is that you can't prove it to be true, because it's false. Why is it false? Because we can find a counter example. What counter example is that? The circle-square limits.

Another trivial example is to take the sign function: sgn(x) = 1 if x is positive, -1 if x is negative, or 0 if x is 0.

Now, take the sequence s_i = (-1)^i (1/i). So the sequence starts off -1, 1/2, -1/3, 1/4.

The limit of s_i is 0, because the numbers get closer and closer to zero, as close as we want if we go far enough. Thus, sgn(lim s_i) = 0.

BUT, the sign of each element in the sequence alternates. sgn(s_1) = -1, sgn(s_2) = 1, sgn(s_3) = -1. In fact, the sequence sgn(s_i) doesn't even converge! It has no limit. Thus, sgn(lim s_i) cannot equal lim sgn(s_i).
 
Mathematics news on Phys.org
  • #62
1MileCrash said:
Maybe I'm not seeing it, but I picture repeating this process "infinitely" just yields a diamond or sideways square, never a circle.

It does if you cut out squares, then you can get another trick about the lengths of sides of a square or right-tiangle, like alphachapmtl said.

But here the diagram shows you have to use rectangles so the new internal corners touch the circle.
 
  • #63
EDIT: THIS IS WRONG. Working on the correct one now.

There is always going to be area between where the "ridges" are and where the outside of the circle lies. That area is given by:

A = 1 - \pi\ - \frac{4n+4}{n(1 - 2\sqrt{2})}

Where n is like this (and A is the blue area):
QH6Uk.png


The limit of the above function as n -> infinity is .0461.

P.S. My constant of 1 - \pi\ may be off. If someone wants to check/correct me on that equation that would be great, I'm a little tired at the moment. If not, I can go over this tomorrow.
 
Last edited:
  • #64
it is not true that lim(lengths) = length(limits).

the (euclidean) length of a curve is not well-defined on the set of all possible curves. the notion of length, can be counter-intuitive, it depends on 2 things: what you are measuring, and how you measure it. the various fractal curves give examples of how length can be "worse than it looks" (the koch snowflake, for example. its length doesn't appear to be infinite).

if you were to produce a specific parametrization (piece-wise) for the "boxed" approximation, you would find that in the limit, it is not differentiable. the differentiability of the parametrizations are a key hypothesis in proving this independence.
 
  • #65
Deveno said:
if you were to produce a specific parametrization (piece-wise) for the "boxed" approximation, you would find that in the limit, it is not differentiable. the differentiability of the parametrizations are a key hypothesis in proving this independence.

Not only that, the derivatives of the parametrizations must themselves converge uniformly to a continuous function. At least this is a sufficient condition, and counter-examples to "limit(length) = length(limit)" where this is not satisfied (while your condition is) exists. E.g: One could easily make the jagged lines around the circle smooth by substituting the tip with a quarter of a circle or something similar, but the lengths would still not converge to the correct value.

And it is not true, as suggested by someone eariler, that the jagged curves around the circle does not converge to the circle. The curves does converge uniformly to the circle, the problem is just that the lengths does not.
 
  • #66
Ugh, finally got it. Maybe. Could someone try to do this out themselves?

n is the number of points in each quadrant where there is a point touching the circle. A is the area of the blue. See below image for sample. Basically, there is always going to be area surrounding the circle that is blue.
VgJkD.png


Equation:
[PLAIN]http://www4b.wolframalpha.com/Calculate/MSP/MSP67019f5i3id7f23f46800006a6bfeede1b8g0ih?MSPStoreType=image/gif&s=30&w=215&h=61

lim x->infinity = 1-(pi/4) = original area of blue when n=0. But I don't understand how that could be the case. I think the above equation must be wrong. I'll have another look tomorrow. I suppose it could make sense when there are thousands of points touching the circle, but I don't think so.

Graph:
[PLAIN]http://www4b.wolframalpha.com/Calculate/MSP/MSP96719f5hfca6ccfc80700004fie2ha6eb8fca2b?MSPStoreType=image/gif&s=10&w=300&h=166&cdf=Coordinates&cdf=Tooltips
http://www.wolframalpha.com/input/?...sqrt(x^2+1)-x)^2)/(2+(x^2+1))+from+x=0+to+x=5
 
Last edited by a moderator:
  • #67
I'm quite interested in seeing a derivation of that formula if it's correct. From my knowledge this is quite a difficult problem. In each removal, the areas of the squares removed are non-uniform. For example, for the third "cycle" (n=7 according to your notation), the area of the squares removed at the edges of each corner are larger than the squares removed at the interiors of the corner and this effect builds up as n increases. Did you manage to account for that?

edit: Upon reading your post again, it may seem like I've misunderstood. That formula seems to suggest that the area of the blue surrounding does not decrease which is certainly not the case. I believe that the area of blue surrounding does converge towards the circle and its just the perimeter that does not.
 
Last edited:
  • #68
Alright, I'll post the derivation I did. I'm almost 100% sure there's a mistake in here, so please try to help.

Let's say that n is the number of times a white "corner" touches the black circle in one of it's quadrants. See previous image above for a guide. For simplicity's sake, let's only focus on one quadrant for now. We can multiply it by 4 to get the total area lost later.

Now, to get the area lost (in each quadrant) by having n number of touches (in each quadrant).

I THINK we can break the whiteness up into even squares. The amount of squares along the top row is simply n, and the amount of squares along the outside column is also n. This means (I THINK) we can get the total number of squares for touches n by using the triangular number formula (but I could be wrong because the circle being a curved surface might change this).

\frac{n^2+n}{2}

So that is the total number of squares lost in each quadrant. Now, let's get the area of each of those squares. Let's focus now on the corner-to-circle touch on the edge of the quadrant. (in quadrant 1, it would be the farthest right touch).

First, we should simply get the angle on the circle where this touch is. cos(theta) gives us the distance from that touch to where the top edge of the original square was (if n=0 and there were no touches). sin(theta) gives us the distance from that touch to the right edge of the original square was (in quadrant 1, but that's all we have to worry about for now). That means, to get theta, all we have to do is:

n(sin(\theta)) = cos(\theta)

Now, let's get the distance from where theta lies on the outside of the circle to the edge of where the original square would be. This will give us the length of each side of each square. This equation is:

L = \frac{1}{2} - \frac{cos(\theta)}{2}

After some simplifying and combining the previous two equations, we get

L = \frac{1}{2} - \frac{n}{2\sqrt{n^2-1}}

The original area with just the original square and circle is just 1- pi/4. Let's square the L in the equation above to get the area of each square, then multiply it by the triangular number formula (giving us the total area subtracted from each quadrant), then multiply that by 4 to get the resulting area subtracted from the entire figure. After all this, we get this:
WolframAlpha equation

I'm almost certain my math is correct, it's just the concept. I'm not sure if we can count every square as having the same area, and I'm not sure if we can simply use the triangular number equation due to the nature of the circle being curved.

If someone could take a look at this that would be great, because I'm stuck, and not to mention kind of sick of working on this problem myself.

Thanks.
 
  • #69
As I've mentioned, for each iteration of removing squares (which I will call cycles), the area will differ. (i.e. the second and third squares removed will be smaller than the first square removed). In fact it gets more complicated than that. For each removal cycle you will remove 2 squares near the edge of the quadrant and you will remove squares interior to the quadrant. These squares will also have different areas in fact. This effect will build up and for later cycles there may be several different sizes in a single cycle (the two most interior squares share an area, then pairwise as you progress outwards of the quadrant). You cannot simply take every square to be the same size.
 
  • #70
Yuqing said:
As I've mentioned, for each iteration of removing squares (which I will call cycles), the area will differ. (i.e. the second and third squares removed will be smaller than the first square removed). In fact it gets more complicated than that. For each removal cycle you will remove 2 squares near the edge of the quadrant and you will remove squares interior to the quadrant. These squares will also have different areas in fact. This effect will build up and for later cycles there may be several different sizes in a single cycle (the two most interior squares share an area, then pairwise as you progress outwards of the quadrant). You cannot simply take every square to be the same size.

My equation takes all of that into account, except for the fact that the squares change size. If someone could derive the right equation I'd love to see it.
 
  • #71
guss said:
My equation takes all of that into account, except for the fact that the squares change size. If someone could derive the right equation I'd love to see it.

But the fact that the squares change size is precisely what makes it difficult.
 
  • #72
Yuqing said:
But the fact that the squares change size is precisely what makes it difficult.

I know.

You can think of the squares being removed as being in sort of rings of size. The smaller ones are closer to the edge, the next smallest are one in, and so on. The amount of rings in a certain layer can be given by 2(n-2k)-1 where n the n value discussed before and k is a constant that decreases incrementally.

Thinking about it like that was easier for me, but I still can't get it.
 
  • #73
It seems as if a proof of the "nonconvergence" of the jagged lengths to the circumference could be constructed by looking at the ratio of "Jaggy Lengths" to the circumference C. The jaggies between two touches on the circle are always two legs of a right triangle, while the hypotenuse is the chord length (which should converge to the arc length). I realize this "triangle" fact was pointed out a few times earlier in the thread, but my point is mainly that you don't need to explicitly evaluate the Jaggy Lengths to prove the inequality.
 
  • #74
Uh, not to oversimplify a discussion that is certainly interesting, it would be enough to say that the boundaries of the blue-area will never converge to be tangential to the circle except on the four cardinal points.

Without solving for the area of the blue-shaded region, you can at least conclude that the blue shaded region will never converge with the surface of the circle (no matter how many times the process is repeated).
 
  • #75
The problem with this proof for pi = 4 is that no matter how small the little squares get, the ratio of the length of the arc and the sum of the square edges that encompases it is always the same pi/4 no matter how small the square becomes. And from the triangle inequality we get that pi/4 < 1 therefore pi can not be equal to 4!

If the ratio was to converge to 1 then the proof could have been correct.
 
Last edited:
  • #76
To put the above posts into "time to teach a 6th grader why pi = 3.14 and not 4" speak:

It doesn't equal 4 because, no matter how many times you cut the squares smaller and smaller, there will always be area of the squares that is not touching the circle.

Here's a little picture I drew in paint to help visualize:

Squarevscircle.jpg


In the above picture, there is a section of it that is red. You'll notice that this area exists on the other pictures as well (but it's blue, I believe). This area, no matter how many times you cut away at the square, still exists to an extent. At some point, it's a very small extent, but the upper-right corner of the square will never touch the circle, there will always exist 2 more upper-right corners for every single upper-right corner that you cut away.
 
  • #77
Ryumast3r said:
To put the above posts into "time to teach a 6th grader why pi = 3.14 and not 4" speak:

It doesn't equal 4 because, no matter how many times you cut the squares smaller and smaller, there will always be area of the squares that is not touching the circle.

Here's a little picture I drew in paint to help visualize:

Squarevscircle.jpg


In the above picture, there is a section of it that is red. You'll notice that this area exists on the other pictures as well (but it's blue, I believe). This area, no matter how many times you cut away at the square, still exists to an extent. At some point, it's a very small extent, but the upper-right corner of the square will never touch the circle, there will always exist 2 more upper-right corners for every single upper-right corner that you cut away.

I agree with you completely, but the diameter is specified as being 1 and not 4, the perimeter is 4 :P I'm sure you realized this but just had a quick lapse and mistyped.
 
  • #78
Ryumast3r said:
It doesn't equal 4 because, no matter how many times you cut the squares smaller and smaller, there will always be area of the squares that is not touching the circle.

I don't think that counts as a reason. The area keeps getting smaller, and it approaches something, presumably it approaches the area of the circle.

The pi=4 comes from the perimeter, not the area. The perimeter seems to remain 4 no matter how many squares (or rectangles, which is what the original picture shows) you cut away.

But how's this for a simple proof:

pi < 4 (already been http://en.wikipedia.org/wiki/Proof_that_22/7_exceeds_%CF%80" )
therefore pi doesn't equal 4
 
Last edited by a moderator:
  • #79
Unrest said:
I don't think that counts as a reason. The area keeps getting smaller, and it approaches something, presumably it approaches the area of the circle.

The pi=4 comes from the perimeter, not the area. The perimeter seems to remain 4 no matter how many squares (or rectangles, which is what the original picture shows) you cut away.

But how's this for a simple proof:

pi < 4 (already been http://en.wikipedia.org/wiki/Proof_that_22/7_exceeds_%CF%80" )
therefore pi doesn't equal 4

See below as well, What I meant is that there is area unaccounted for (though it becomes infinitesimally small), which also means that there is a section of the perimeter that does not touch the circle.

Put more simply what I was trying to say: Every time you cut a corner, 2 more appear that do not touch the circle, ergo the perimeters of both objects will never touch at every point.

Eldar said:
I agree with you completely, but the diameter is specified as being 1 and not 4, the perimeter is 4 :P I'm sure you realized this but just had a quick lapse and mistyped.

Yeah, that's what I meant. I had a test and... well... you know what happens after tests. :P
 
Last edited by a moderator:
  • #80
Grep said:
haha That's a good one, I'll have to remember that next time I want to mess with someone.

I think of it this way. Since "removing" the corners like that doesn't change the perimeter at all, it will fail to converge on the perimeter of a circle. So it's rather unlike, say, increasing the number of sides of a polygon inside the circle. That one converges on the real perimeter. His example does not.

The fact that the perimeter never changes as he removes the corners is pretty much proof that the technique will never work. It needs to converge to a smooth curve (to it's limit) that equals the perimeter of the circle, which this will never do (being jagged).

Not sure if that's 100% clear, someone else can probably put it better.

Great explanation, actually. *thumbs up*

However, I will definitely use it to mess with ppl! :)


Edit: Wait, I'm confused. If this fails to prove that an area is really the sum of infinitely small rectangles, why should I trust taking the integral to give me a correct approximation?
 
  • #81
Femme_physics said:
Edit: Wait, I'm confused. If this fails to prove that an area is really the sum of infinitely small rectangles, why should I trust taking the integral to give me a correct approximation?
The jagged curve *does* converge to the circle in almost every1[/color] sense of the word "converge". The area *does* converge to that of a circle. The distance between any point on the jagged curve and the circle *does* converge to zero.

Think of the upper (or lower) half of the jagged curve / circle as a function of x. While the jagged semi-curve converges uniformly to the semi-circle, the derivatives of those curves are miles apart. In fact, the jagged semi-curve is nowhere differentiable in the limit N → ∞. Just because two functions converge to one another does not mean that their derivatives, or their lengths, do so.1[/color]Addendum: Well not quite every sense. The jagged curve does not converge smoothly to the circle. The jagged curve is nowhere differentiable in the limit N → ∞ while the circle (upper semicircle) is infinitely differentiable almost everywhere.
 
Last edited:
  • #82
It's the idea that with every step, the difference between areas of the jagged figure and the circle is constantly shrinking, until the difference is zero, and also that the jagged figure and the circle are the exact same shape. Because the jagged shape always has a perimeter of 4, the circumference of the circle is also 4, and not 3.14.

Without getting into differentials and tangents and other topics more advanced than the thought that went into the trolling, I prefer the simpler answer that the jagged shape never does become a circle. In math it's called an infinitesimal, where 1/\infty (or any other mind-bogglingly small number) is not zero. The difference between the jagged shape and the circle keeps shrinking, and you can choose to disagree with me and believe whether or not the difference becomes zero. I'm no math professor, so I won't claim to be more than pretty sure about this.
 
  • #83
hillzagold said:
Without getting into differentials and tangents and other topics more advanced than the thought that went into the trolling, I prefer the simpler answer that the jagged shape never does become a circle.

Approximation by polygons will also never become the circle, but still the perimeter of the polygons will approach the circle. The main problem here is that one will not be certain that length is preserved under the limit operation unless the (almost everywhere) derivatives of the sequence curves (such as the jagged curves in our example) has a (almost everywhere) continuous (maybe simply integrable, not sure) limit. If this condition is satisfied, and it is for polygons, we will have length preservation. And this is why we can trust the approximation by polygons, as someone commented on earlier.
 
  • #84
hillzagold said:
The difference between the jagged shape and the circle keeps shrinking

I think you're mixing up area with perimeter by using the vague word "shape". You can't say "the difference between these two shapes is less than the difference between those other two shapes". Difference implies subtraction, and subtraction isn't defined on "shapes". This confusion is what the original picture plays on. The area of the jagged shape does approach the area of the circle, but the length of the perimeter doesn't even begin to.

Here's an even more extreme example of how our intuition about the relationship between area and perimeter length doesn't work: http://en.wikipedia.org/wiki/Koch_snowflake"
 
Last edited by a moderator:
  • #85
How do you know it is true that the shape never becomes a circle?
 
  • #86
Well, any iteration of the jaggy shape is going to have lots of sharp corners on it, so it pretty clearly does not satisfy the definition of a circle.
 
  • #87
guss said:
How do you know it is true that the shape never becomes a circle?

The distance from corner to opposite corner will always be x+y, regardless of how short you make x and y. Even teeny tiny jags too small to see will still never make x+y equal to root(x2 +y2).

Try it. Make x=y = .0000000000000000000000000000000000000001
Now make x=y=.0000000000000000000000000000000000000000000000000000000000000000000000000000000000001

Not only will they will never reach it, they never even start towards it.


Another way of looking at it:

If the jags are the size of an atom, and the circle is 10 trillion atoms high, then each jag still contributes 1 atom's-worth of y, times 10 trillion equals the side of a square. Not a circle. This is true no matter how small you choose to go.

Thus, no matter how small you make the jags, they're still jags, and they still add to the perimeter of a square, not a circle.
 
Last edited:
  • #88
DaveC426913 said:
The distance from corner to opposite corner will always be x+y, regardless of how short you make x and y. Even teeny tiny jags too small to see will still never make x+y equal to root(x2 +y2).

Try it. Make x=y = .0000000000000000000000000000000000000001
Now make x=y=.0000000000000000000000000000000000000000000000000000000000000000000000000000000000001

Not only will they will never reach it, they never even start towards it.


Another way of looking at it:

If the jags are the size of an atom, and the circle is 10 trillion atoms high, then each jag still contributes 1 atom's-worth of y, times 10 trillion equals the side of a square. Not a circle. This is true no matter how small you choose to go.

Thus, no matter how small you make the jags, they're still jags, and they still add to the perimeter of a square, not a circle.

Sorry, I'm not following. What corner to what corner? What are you saying x and y are?
 
  • #89
guss said:
Sorry, I'm not following. What corner to what corner? What are you saying x and y are?

Simply put, when the circle is enclosed in a square, the square's perimeter is going to be 2x+2y, where x and y are both diameters of the circle. As you add more jags, the jags get smaller, but the perimeter does not decrease - it is still 2x+2y (A straight vertical line from top of square/circle to bottom of square/circle, no matter how much you subdivide it, will always add up to y).

No matter how small the jags get, the perimeter of the shape remains 2x+2y. Seriously, the jags could be microscopic, and yet the perimeter never wavers from 2x+2y.
 

Attachments

  • jaggies.png
    jaggies.png
    6.1 KB · Views: 541
Last edited:

Similar threads

  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 17 ·
Replies
17
Views
2K
Replies
15
Views
1K
  • · Replies 16 ·
Replies
16
Views
5K
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
10K
  • · Replies 55 ·
2
Replies
55
Views
6K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K