## Troll pi explained

Here's the "Troll Pi" or "Pi equals 4" image.

Here's the breakdown, as simple as I can make it.

All of the following facts are true:

1. Panels one to four describe a sequence of curves. (Here, "curve" is a generic term referring to any continuous line, be it straight or crooked or curved.)
2. Each curve in the sequence has a well-defined length of exactly 4.

These facts are also true:

1. The sequence of curves converges uniformly on a limit.
2. As panel five correctly states, the limit of the sequence is a circle.
• It is not a sawtoothed curve.
• It is not an "infinitely jagged" sawtoothed curve.
• It is not a "polygon with an infinite number of sides" or "infinigon".
• It is not a fractal.
• The limit is an ordinary, perfectly smooth perfect circle.
3. Thus, the length of the limit is exactly π (3.1 or so). (Because it is a perfect circle with diameter 1.)
4. It's not 4!

And so is this final fact:

1. None of these facts contradict each other.

#### Why?

The limit of a sequence isn't necessarily a member of that sequence.

Because of this, the limit of a sequence doesn't necessarily share any properties with the members of that sequence.

Here, you've seen a sequence of curves of length 4, whose limit does not have length 4.

You've also seen a sequence of jagged, right-angled curves whose limit is not jagged or right-angled at all, but smooth.

Here's another example of this:

1, 1/2, 1/3, 1/4, 1/5, ...

All of the numbers in this sequence are positive, but the limit, 0, is not.

This is not a problem. It's not a contradiction. It's just the way it is.

Breathe in. Breathe out. Carry on with whatever you were doing.

### Discussion (69)

#### 2010-12-09 01:43:11 by atomicthumbs:

You've disproven the Roundamajig Theorem!

#### 2010-12-09 01:55:59 by Randall:

Not that I disagree with the conclusion, but do you have a more rigorous proof that the limit curve is simply f_inf(x) = 0 for all x? It's pretty clear that that's the key step in the proof, and you hand-wave past it pretty quickly.

#### 2010-12-09 02:13:50 by Joseph:

Um, shouldn't the case for floor(nx) congruent to 1 mod 2 be 1/n + (1/n)floor(nx) - x? It doesn't match up otherwise. However, the rest of the math checks out. I think. Now, when I read "pointwise limit", I immediately thought, "Hey, is this a uniform convergence question?" After some reading, I believe I now understand what's going on here. f_n is actually uniformly convergent to its pointwise limit function f_inf. We see this because the distance between f_n(x) and f_inf(x) is, if I've done it correctly, the {sup(|f_n(x) - f_inf(x)|) over x in [0,1]} = 1/n -> 0 as n -> infinity. So f_n converges uniformly to f_inf = 0. I believe we **get** the pointwise limit by observing that the maximum of the function is given by 1/n, and as n -> infinity, since f_n is non-negative for all x in [0,1], the result follows by Squeeze Theorem. Because of this, we get a couple neat properties. Ie, the limit function is continuous (which in this case is trivial), and the integral of f_n over [0,1] does indeed converge to the integral of the limit function, ie the area under the curve is 0. **However!** We're looking for the **arc length** of the curve, which involves the derivative of f_n. Since f_n is NOT continuously differentiable (in fact, f_n isn't even C1 on [0,1]), we don't have that the derivative of f_n converges uniformly to the derivative of the limit. This is clear. Thus, since the **derivative** of f_n does not converge uniformly to the derivative of the limit, we have that the integral of the arc lengths of f_n does not converge to the integral of the arc lengths of the limit function, assuming that we can apply the same idea to a function of the limit function (... which is probably not a safe bet, but it still makes sense). Cool. First time poster, btw. I couldn't help myself.

#### 2010-12-09 02:39:24 by Ross:

Easy way to add TeX formatted equations to your blog: use a personal user page on Wikipedia as a scratchpad, with Wikitext. Save (or just preview), then grab the PNG. Example: http://en.wikipedia.org/wiki/User:Rpresser/fnord

#### 2010-12-09 09:22:15 by qntm:

Fixed the 1/n error in the definition of f_n(x).

#### 2010-12-09 09:36:54 by qntm:

Made the uniform convergence proof much more explicit.

#### 2010-12-09 09:39:11 by qntm:

Last time I tried to sketch out mathematical equations on my Wikipedia user page somebody just came along and blanked it because you're not supposed to do that.

#### 2010-12-09 11:56:48 by Sean:

A quick note on the convergence: I'm being a bit lazy on typing out the details of the proof (which is similar to one for the squeeze theorem), but it seems to me that any time you have a function sandwiched between two functions that uniformly converge to "f", the sandwiched function converges to f. That is, if for all (relevant) x and n, f_n(x)<=g_n(x)<=h_n(x), and (f_n) and (h_n) converge uniformly to f, then (g_n) converges uniformly to f. (Similarly, if they both converge pointwise, g converges pointwise.) Taking f_n and h_n to be horizontal lines at the infimum and supremum of the range of g_n (assuming such exist), you then have that any time the sequence of infima and the sequence of suprema of the range approach a constant value, the function uniformly converges to that value. Or, to put it in a less jargon-y way, when a function is sandwiched between horizonal lines that squeeze into a single limit, they squeeze the function into the same limit (since the convergence of the lines is obviously uniform, so is the convergence of the squeezed function). Sadly, this was the most obvious proof that sprang to mind when I was thinking of the uniform convergence of these sawtooth functions.

#### 2010-12-09 13:15:46 by Chris:

http://www.math.union.edu/~dpvc/jsMath/ jsMath is what my uni uses on its maths student forum. It's usually pretty reliable in my experience. No idea about setting it up though.

#### 2010-12-10 17:30:49 by pascal:

And then there's always MathML, but it still only seems to work in Firefox (although Webkit support is underway)

#### 2010-12-11 00:33:44 by Hans:

If you want TeX on your webpage, use MathJax. http://www.mathjax.org/ This what both mathoverflow.net and math.stackexchange.com run on.

#### 2010-12-11 04:17:40 by David:

>Last time I tried to sketch out mathematical equations on my Wikipedia user page somebody just came along and blanked it because you're not supposed to do that. Sam, you *could* just try having a blank LaTeX document around to recompile and take PNG snapshots of... not all that much more work than the Wikipedia solution. Of course this is a terrible kludge and I'm sure that, long-term, you'd be much happier with mathjax or something along those lines (and you've shown before that you're willing to invest time in getting this website to run). This also presumes that you've already installed LaTeX and have an easy way to compile documents. But that isn't too much of a stretch, I hope...? On the off chance you've not already installed LaTeX, and for the benefit of other people: If you're running Windows, try TeXworks (which comes bundled with MiKTeX); if on Linux, Kile works fairly well but I don't know how easy it is to get working. Both editors support compiling with one keystroke (ctrl+t, alt+6 respectively; at least I think it's alt-6).

#### 2010-12-12 17:35:14 by TeX:

For TeX you can use Google Charts API. Enter any LaTeX expression after this: http://chart.apis.google.com/chart?cht=tx&chl=

#### 2010-12-14 09:33:55 by Ross:

Sam, I explicitly stated "GRAB THE IMAGES". After you've got your formulas right you download the images they render to and server them from your own server. Then you can blank your own page because you don't need it anymore. But the Google Charts option is probably less work.

#### 2010-12-19 11:23:34 by FinDude:

Would WolframAlpha be of any use here?

#### 2010-12-24 23:31:07 by danny:

I can't believe how intellectual this website is. All the comments are properly formatted with good grammar and good form - no caps lock, no one-liners, no nothing. My Father offered a simpler explanation; I was wondering if it is correct: if you take the process of corner-concaving and repeat it infinitely on the circle, you will never end up with the curve itself, no matter how infinitesimal you get - you will end up with a series of right angles with legs of infinitesimal size with an arc (the curve of the circle) scrawled across. While the individual difference from arc to right-angle may be tiny, the number of these structures is, as the adjective implies, nearly infinite; and thus you still have a perimeter which properly adds up to four. Sorry for the bad English. I'm terrible at outputting my mental simulations.

#### 2010-12-25 09:16:24 by qntm:

Danny, I'm sorry but your father's argument doesn't make any mathematical sense. If you repeat the corner-concaving for a finite number of steps you NEVER get a circle and the perimeter is ALWAYS 4. But if you "repeat for an infinite number of steps" i.e. take the limit at infinity, you DO have a circle of perimeter pi.

#### 2011-01-03 03:54:24 by Mirdan:

Doesn't f_n have no derivative at the points where x-1/n[nx] = 1/n +1/n[nx]-x ? If so, doesn't that mean that you cannot use an integral to find the arc length of f_n?

#### 2011-01-03 12:32:54 by qntm:

There is a discontinuity, but because the discontinuity is isolated (i.e. there is a positive finite distance between it and the next discontinuity), the function is still integrable. The value of the function at this isolated point has no effect on the integral so the fact that it is undefined doesn't present a problem.

#### 2011-01-13 15:12:58 by Aether:

Pi is actually 4, all you need to do is rotate the surrounding square 1/4, within it a perfect circle is inscribed. Also, when you annotated the square you seem to have put '1' on each of it's sides -- this is incorrect. The left and right, and top and bottom, exist as two sets of 0 value pairs. (Left + Right, is the complete package, both cancel each other out to 0, neither holds significance and neither is individually '1', without 'one' the other cannot exist). With this interpretation you can see that the square is 3 dimensional. It has an inside, and an outside. A top and a bottom; and a left and a right = 3 sets of 0 value pairs that are reliant on each other to exist. It's purely down to perception, and this takes into account the symmetrical nature of the universe as oppose to the human created asymmetrical nature-matrix of innovation that we subside in. Have a good day.

#### 2011-01-21 22:26:23 by anon:

The point of troll science pictures are to get you to waste your time. You fell for it lol.

#### 2011-01-23 18:40:36 by quintopia:

another option for equation rendering for the web that doesn't involve slow-loading javascript interpreter is tex4ht http://www.tug.org/applications/tex4ht/mn.html Feed it a valid tex file and it will give you back an html file with all the equations converted to pngs.

#### 2012-02-09 03:31:42 by Prafull:

To each of your right angled triangles surrounding the circle, draw the hypotenuse as well (chord for circle). As you go on making the them smaller and infinite, the sum of all such chords equals perimeter of the circle, and from your arguments sum of other two sides for all such small triangles also equals the perimeter of circle. So in effect "the sum of two smaller sides now become equal to the larger side". This can only be true when your right angle goes on increasing to become obtuse, eventually ending up at angle pi (180 degree). So that these triangles are effectively straight lines tangent to the circle.

#### 2012-03-25 19:41:36 by anon:

a heuristic (not proof) to see where the problem is, is that every time the corners are folded in, the absolute distance from the square to the circle is decreased. This, I assume, is using the euclidean norm, not the 'taxicab norm.' For this reason, a contradiction arises. Using taxicab geometry, the 'circle' is actually a square rotated counterclockwise 45 degrees (which is stil a square, regardless). Now, is this were to use the Minkowski Norm, then the original square would be a ciricle in that defined space. However, seeing as this is trying to prove taht a square is a circle, we can assume that euclidean norms are used, while the implicit starting point is in the Taxicab Vector Space, using the Minkowski Norm. Therefore, three contradictions of space, metrics, and norms are made using this argument and pi does not equal 4

#### 2012-03-25 19:55:43 by qntm:

Nobody mentioned taxicab geometry or the Minkowski Norm.

#### 2012-05-14 15:52:01 by LawrenceColes:

By inspection the shape made from removing squares from the corners would be a diamond (the same square rotated 45') not a circle.

#### 2012-05-17 21:34:29 by SmexyFish:

People take Troll circle so seriously....

#### 2012-05-23 00:13:35 by Luke:

Actually 4!=24 so even if your 'proof' was right you managed to cock up at the end. Problem?

#### 2012-05-30 00:13:31 by Anonymous:

http://www.scribd.com/doc/95216567/Solution-to-Circle-Con-Nun-Drum A bit more rigorous... Well the proof is very hand-wavey, but I wanted it to be accessable.

#### 2012-05-30 17:21:22 by Ramki:

see http://infrastruct.blogspot.com/2012/05/why-pi-is-not-4-and-other-mysteries.html for the easy derivation of the actual length of the approximate figure. The approximate figure will never be the same as the circle.

#### 2012-07-03 12:25:18 by daleofcourse:

My first thought is that repeating that pattern would create a diamond, which would simply be the square at 45 degrees which would obviously have a perimeter of 4. Just my first thought, no calculations done or owt!

#### 2012-07-04 22:03:51 by Daniel:

I propose a rigorous proof against TrollPi be constructed as follows: 1) Show that the square begins with a perimeter larger than the circle. 2) Show that each succeeding shape maintains the perimeter measurement of the square, after any number (infinite) iterations of inverting the corners, as described by Troll. Since the shape begins with a larger measurement and maintains its measurement, it is larger than pi even after infinite iterations.

#### 2012-07-05 15:30:58 by JanWillem:

Vi Hart proved you wrong. In a very nice video: http://www.youtube.com/watch?v=D2xYjiL8yyE

#### 2012-07-22 20:53:05 by qntm:

Vi Hart doesn't know what she's talking about. That video seems informative but is actually highly misleading.

#### 2012-10-31 04:33:46 by Jakim:

Let use Cauchy's limit definition for a(n) sequence and g limit: For all e>0 exists such natural N that for all natural n>N satisfies a condition: |a(n) - g| < e Thus, pi cannot be the limit of sequence of constant 4 values, there is no natural N that for every n>N: 4 - pi < e for every e>0.

#### 2012-11-01 13:48:26 by Tim:

The author of this blog is wrong. The approximating figure is only ever tangent to the circle in four places. The perimeter never converges to that of a circle, only the area does.

#### 2012-11-01 14:03:59 by qntm:

The perimeter converges to the circle. But the length of the perimeter doesn't converge to the length of the circle's perimeter.

#### 2012-11-02 16:35:52 by Tim:

I suggest you look up the meaning of perimeter. It is defined as the length of an enclosing curve.

#### 2012-11-05 13:37:45 by qntm:

"Perimeter" can be used to mean the path or the length of the path. In this case, the path converges while the length of the path does not.

#### 2012-11-20 06:33:25 by Carl:

Isn't this a hell of a lot easier to disprove by just noting that this is a proof of induction, yet for induction to work, you need a finite n, and this "proof" relies on an infinite n

#### 2013-01-15 06:22:28 by Steven:

"The point of troll science pictures are to get you to waste your time. You fell for it lol. " -anon The problem with trying to 'troll' scientists and mathematicians with 'troll science pictures' is that we actually enjoy wasting our time on them

#### 2013-03-05 03:42:43 by pelrun:

This used to bug me when I was younger - not quite this formulation, but close enough - and how I eventually resolved the paradox was by realising how a limit is supposed to work. Each term of the sequence can be written as the limit plus an error term. Each successive term then has to reduce the error term by a percentage. But in the troll pi example, the sequence is chosen that the error term is evenly distributed across each of the subdivided units, which is invalid.

#### 2013-03-14 15:07:08 by PiThagaurus:

TrollPi also doubles to demonstrated the principle of "pi are squared"

#### 2013-07-05 19:44:09 by ANon:

For including nicely formatted LaTeX equations, you can use the Online LaTeX Equation Editor at: http://www.codecogs.com/latex/eqneditor.php It has a nice GUI (or you can just type your code in directly) and it dynamically produces a .png file which you can use or a snippet of HTML code you can include in your webpage without the need to save the image.

#### 2013-07-05 22:28:37 by benevolentwanderer:

I think the problem here is that the 'troll circle' (TC) created after an infinite number of iterations is in fact... not a circle. It's more closely related to the Koch snowflake than anything else, as far as I can tell. If you cut off the corners rather than subtracting a square, the estimation would work; but the troll way isn't actually approaching the slope of the surface of the circle so it's not approaching the perimeter of the circle either!

#### 2013-07-05 22:34:40 by qntm:

No, you're wrong. The limit curve is a perfectly smooth perfect circle. There is uniform convergence. Just as the limit of the lengths of each curve is different from the length of the limit curve, the limit of the slopes of each curve is different from the slope of the limit curve. (In fact, the limit of the slopes is only defined at four points: the top, bottom, left and right. For the rest of the points on the curve, the slopes do not converge. They just oscillate between vertical and horizontal forever.)

#### 2013-12-30 09:37:19 by JohnGabriel:

The perimeter of the square is never the same as the circle perimeter. I would expect the perimeter of 4 to be the same no matter how many times the process is repeated. All that is happening, is that the path is changing shape with each iteration (but the path length NEVER changes). Since infinity is unattainable, one can't say anything about what happens "at infinity", because there is no such thing. And since the path length never changes, it has NOTHING to do with limits. Nothing is changing in terms of length. I have seen other examples along the same lines, where the path is changed: Here is a similar question I was asked by a mathematics educator: Consider the following curve. It starts at the origin and goes straight up to the point (0,6). Then it makes a sharp right turn and goes straight to the point (6,6). We’ll call this C1. Its length is 12. Now, C2 follows a different path. It starts at the origin and goes straight up to (0,3). Then it makes a sharp right turn and goes straight to (3,3). Then up to (3,6), and then over to (6,6). Same two endpoints, and again the length is 12. C3 goes up 1.5, over 1.5, up 1.5, over 1.5, up 1.5, over 1.5, up 1.5, over 1.5. Same two endpoints, and again the length is 12. ...and so on, forever. If you draw these, you can see that they are getting closer and closer to a diagonal line. But the length of that line is not 12; it is 6sqrt(2). I can make this more “rigorous” but rotating the whole drawing by 45 degrees. Now each curve in the sequence is, in fact, a function. (Passes the vertical line test.) And this sequence of functions is approaching a straight horizontal line: in fact, converging uniformly toward it. But the arclengths are not approaching. What gives? My response: Hi K....., Consider that between any two points, there are an infinite number of paths. However, only one of these paths is the shortest distance, that is, a straight line. So this is a geometric explanation. The observation is found many times in the study of combinations (Combinatorial theory). In your example, you have found a special set containing the number of ways to make up a sum of 12 using distances that are a factor of 6. And this is an algebraic explanation. :) You could also define a general combination as 2k(6/2k) where k is a natural number. John Gabriel P.S. To interpret the diagonal as a "limit" would be completely wrong. http://thenewcalculus.weebly.com

#### 2014-07-05 07:22:48 by Coda:

I think I have to agree with the conclusion given by benevolentwanderer: The result of the iteration is not, in fact, a circle, at least not according to a parametric definition. Certainly the resulting figure contains every point on the circumference of the circle. By that definition, you might call it a circle. However, this figure has a derivative that is discontinuous everywhere, whereas a circle's derivative is continuous. That is to say, it is everywhere nondifferentiable, despite being everywhere continuous. It's like the Weierstrass function in this sense. You can't meaningfully define the tangent of the figure, whereas a true circle has a well-defined tangent.

#### 2014-07-05 11:05:15 by qntm:

If the figure contains every point on the circumference of the circle, and no points not on the circumference of the circle, then in what sense is it not a circle?

#### 2014-08-26 04:42:43 by Rodney:

Nah. Yer all have it wrong. Pi really equals approximately 2.828427125.... . The problem with the calculation which arrives at a value of 4 is that it is effectively measuring two equal sides of a very large number of tiny right angled triangles. Instead, the length of the hypotenuses of all those little triangles is what we want. The hypotenuses of the triangles are shorter than the sum of the two equal sides by the factor: root 2 divided by 2, which = 0.7071o6781. So, fixing the Pi = 4 calculation for this error, gives a value of Pi of 4 x o.707106781, which is 2.828.... . Quite easily done. And I don't see my error! Rodney.

#### 2014-08-26 04:58:56 by Rodney:

Strictly speaking there is a second adjustment that needs to be made: The hypotenuses of all the tiny triangles are slightly further from the centre of the circle than the circle radius, so the lengths of these hypotenuses are slightly greater than they would be at the exact distance of the radius. However, as the number of tiny triangles tends to infinity, the extra distance tends to zero, so this adjustment factor turns out to be zero x infinity. Rodney.

#### 2014-08-27 22:14:11 by Rodney:

Ah. But only four of the little triangles are isosceles. Hmmm. Maybe Pi isn't 2.828.... after all? So the difference between 3.142 and 2.828 is made up from the gradually diminishing differences in length between the hypotenuses and the sum of the other two sides, as the triangles get closer to 0°, 90°, 180°, and 270°. Rodney.

#### 2015-04-13 19:23:22 by Coda:

Revisiting this because I'm bored. :P I realized an intuitive point that I haven't seen expressed anywhere else. When you're performing your subdivisions, after each step you've increased the number of distinct vertices in the figure and decreased the distance between them. However, in the limit case, the distance has become zero, meaning that vertices that were once distinct now coincide -- meaning that the "final" step of the infinite process has intuitively speaking destroyed/merged/collapsed some of the points that you started with instead of preserving all of them. Obviously this is just an intuitive approximation of the problem, as there is no "final" step and the cardinality of the points in each figure is the same. But I believe that you might be able to formalize this with some effort by using the separability axioms of Hausdorff spaces (and a two-dimensional Euclidean plane is a Hausdorff space) but I'm not well enough versed in using these axioms to do it myself.

#### 2015-12-08 15:41:46 by Alexander Iurovetski:

Coda, you are very-very close, but in my opinion it is simpler, than you suspect. Yes, it is about cardinality. Let me put everything together: The real problem (sorry for the pun) is that the length of the polygon side tends to 0, and the final figure should consist of vertices only (otherwise, it is still a polygon, not a cricle). However, the number of vertices of each polygon is finite, and the 'number' of vertices of the final figure is countable (i.e. can be enumerated). On the other hand, the 'number' of points of a circle is continuum (incountable). Therefore, we simply cannot 'cover' circle with that final figure. It is like rational numbers on real line - dense everywhere, but significantly less. You can read more in Wikipedia about so called cardinal numbers and the life of Georg Cantor, who discovered how to compare infinities and proved that real numbers between 0 and 1 are incountable. To make ends meet I'd like to add that a circle is equivalent to (can be mapped 1-to-1 onto) a segment of real numbers between 0 and 1. I do understand that my reasoning is quite complex, but the case is seriously getting outside simple school maths.﻿

#### 2015-12-08 19:24:41 by qntm:

> in the limit case, the distance has become zero, meaning that vertices that were once distinct now coincide > the length of the polygon side tends to 0, and the final figure should consist of vertices only No, in the limit case there are no vertices, any more than there are any straight line segments.

#### 2015-12-09 00:34:49 by Alexander Iurovetski:

I don't think so, because every line should contain countable set of vertices, dense everywhere in that line. On the other hand, the sequence of perimeters has all members equal to 4, and therefore, the limit is 4. Yes, the limit does not have to belong to the sequence, but for any small neighbourhood of the limit only finite number of sequence elements lie outside that neighbourhood, so 4s cannot converge to 3.14..., but only to 4. And a simple practical approximate measurement will show us that the length of circumference is conspicuously less than 4.

#### 2015-12-09 01:30:04 by qntm:

> every line should contain countable set of vertices, dense everywhere in that line No. A vertex is a point where two straight line segments meet. In the limit case, there are no vertices and there are no straight lines. It's a circle.

#### 2015-12-09 12:44:37 by Alexander Iurovetski:

At every iteration only finite number of vertices 'land' on circumference. Any sequence is countable by definition. Hence, the set of dots at the limit will still be only countable. This is substantially less than continuum, although both are infinities. We can approximate closed curves with polygons and closed curves lengths by perimeters of polygons, but we cannot transform polygon into a closed curve sequentially. Secondly, every polygon considered here, has a perimeter of 4, and therefore the sequence of perimeters has 4 as its limit as well. If not, then, by definition of a limit, only finite number of polygons should have perimeter greater than, let's say, 3.5 (because eventually, the sequence should converge to 3.14...) Finally, the circumference of a circle is defined as a limit of perimeters of inscribed regular polygons, and therefore, if diameter is 1, then the circumference is pi (3.14...).

#### 2015-12-09 13:26:24 by qntm:

> Any sequence is countable by definition. Hence, the set of dots at the limit will still be only countable. The set of points on the circle's circumference in the limit case is uncountable. We're not talking about the limit of a series of sets of points-on-the-circumference. We're talking about the limit of a series of *curves*. The former is completely irrelevant here, and it's a very confusing thing to bring up.

#### 2015-12-09 20:49:09 by Alexander Iurovetski:

>We're not talking about the limit of a series of sets of points-on-the-circumference. We're talking about the limit of a series of *curves*. The former is completely irrelevant here, and it's a very confusing thing to bring up. If the limit of the sequence of these polygons is a circle, then the limit of those perimeters should be the circumference of a circle, which by definition is a limit of perimeters of inscribed regular polygons,l. Therefore, if diameter is 1, then the circumference is 3.14..., which was marked as pi. BUT every polygon, considered here, has a perimeter of 4, and therefore the sequence of perimeters has 4 as its limit as well. If not, then, by definition of a limit, only finite number of polygons should have perimeter greater than, let's say, 3.5 (because eventually, the sequence should converge from 4 to 3.14...). I am eager to see that number, starting from which, all perimeters are less than or equal to 3.5

#### 2015-12-09 21:23:06 by qntm:

> If the limit of the sequence of these polygons is a circle, then the limit of those perimeters should be the circumference of a circle No. This is a false statement.

#### 2016-06-14 19:26:51 by edward789121:

they both make sense, but https://qntm.org/files/trollpi/piequals4.png and https://qntm.org/files/trollpi/piequals3root3.jpg would create the pi to be 2 numbers, and make it 3.14........ and the universe would explode (;

#### 2016-09-26 15:08:19 by katesisco:

http://milesmathis.com/pi7.pdf Miles Mathis says Pi equal 4 when motion is involved.

#### 2016-09-26 15:34:10 by qntm:

Miles Mathis is clearly a crackpot.

#### 2018-03-03 19:18:43 by John Gabriel:

Miles Mathis generally knows very little about what he is talking when it comes to mathematics. But in spite of this, he knows much more than mainstream academics do. Chuckle.

#### 2019-04-14 23:42:27 by thingymajigabob:

In vihart's video, the description seems to make the point you made.

#### 2019-04-15 00:32:43 by qntm:

I hadn't seen that description before, but yeah, Vi Hart seems to agree that the video makes many false assertions and doesn't go any distance at all towards explaining the paradox. It would be cool if there were a rhapsody on the actual truth of the matter!

#### 2020-05-11 03:16:02 by Jonnyboyo:

He does hand-wave past the actual proof of the limit, but the drawn of the triangle is helpful, because you see that as you continue the process infinitely, the sizes of the triangular perimeter (while staying the same) doesn't conform as closely to the curve as it could. That is, the slope of a circle at theta=0 is infinity, and theta=π/2, 0, which is reflected by calculus, and not by this process.

#### 2021-03-28 00:41:14 by caydenja:

There is nothing wrong with this, it is true that pi=4... in taxicab geometry, which this actually is, instead of Euclidean geometry!

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