10 ms·
π does not equal 4
- nazgulnarsil 16y agotroll math/physics are the only 4chan releated things that consistently make me crack up.
- RiderOfGiraffes 16y agoNice and fun, and brings into focus the definition of "distance" and how it interacts with the concept of a metric. Using the L<sub>1</sub> metric gives a different value of pi than using the L<sub>2</sub> metric. Obviously. It's easy enough with arguments like this to show the "length" to be anything you choose. Such demonstrations are instructive. EDIT: Changed lower-case ell to upper-case for clarity.
- jacobolus 16y agoYour comment should be at the top. Wikipedia has a nice short article about the ℓ₁ norm (also called “taxicab distance” or “Manhattan distance”): http://en.wikipedia.org/wiki/Taxicab_geometry http://en.wikipedia.org/wiki/Taxicab_geometry It explains, > “The use of Manhattan distance leads to a strange concept: when the resolution of the Taxicab geometry is made larger, approaching infinity (the size of division of the axis approaches 0), it seems intuitive that the Manhattan distance would approach the Euclidean metric [...] but it does not. This is essentially a consequence of being forced to adhere to single-axis movement: when following the Manhattan metric, one cannot move diagonally (in more than one axis simultaneously).”
- gfodor 16y agoYes, you've provided the most succinct and accurate explanation here. The root observation of this thought experiment revolves around properties of the l1 norm, and is an important set of concepts to understand property in implementing distance metric-oriented ML algorithms like k-means clustering.
- csomar 16y agoBecause in Infinity you can't predict things with sight. Actually, it seems like the rectangles have become a lined curve, but in reality they aren't. They are just too small to be noticeable. 4 - π will be that small, tiny difference.
- pixcavator 16y agoJust because the conclusion is wrong it does not mean that there is no point here. What’s described is exactly what one has to deal with in digital image analysis. Here is a write-up and a link to a paper: http://inperc.com/wiki/index.php?title=Lengths_of_curves http://inperc.com/wiki/index.php?title=Lengths_of_curves. The paper proves essentially that there is no good way to approximate lengths of curves with any grid, even if the size of the mesh goes to 0.
- martincmartin 16y agoFor any given n, all the points of the "right angled fractal beast" lie within some distance epsilon(n) of the circle. As n increases, epsilon(n) approaches zero, so in the limit, you have a circle, and at every step along the way, the length is 4. It's just that the length of the resulting shape isn't the limit of the lengths of the sequence of shapes.
- kwantam 16y agoIn the limit you do not have a circle. If we write down area and perimeter error e_A and e_P as a function of n, in the limit of infinite n e_A is proportional to 1/n. However, e_P is independent of n, since the perimeter does not change at any step. Thus, if the perimeter was unequal to begin (clear by inspection), it does not get closer through this approximation. The mistake is in accepting that anything that looks like a circle must be a circle. The fractal beast has area arbitrarily close to that of a circle, but it's clear once we look at e_A and e_P that it is not an approximation of its perimeter.
- RiderOfGiraffes 16y agoI believe you are mistaken. Pointwise, the limit is a circle. Every point on the enclosing shape gets mapped to a sequence of points. Each of these sequences has a limit, that limit is on the circle. The resulting implied mapping of the original square to the circle is a continuous bijection. By every sense that we usually talk of the convergence of lines, this enclosing shape does approach the circle in the limit. The point is that the mapping involved is not a length-preserving mapping.
- defen 16y agoWouldn't you only get a countably infinite number of points on the circle? N.B. I'm not a mathematician
- RiderOfGiraffes 16y agoOnly countably many points "arrive" at their destination in a finite number of steps. There are uncountably many points that never "arrive," but whose limit is on the circle. The usual picture given to non-mathematicians about sequences and limits can end up being strongly misleading in cases like this. Just because a sequence never "gets there," the limit is still the limit. This is the same kind of murky area that talks about 0.999... recurring never "getting to" 1. It doesn't have to "get to" 1 because it's never travelling. It's also the kind of problems that arise when talking about proof by induction. Talking about dominoes falling down is, in the longer term, very misleading. We prove P(1), and we prove that P(n) => P(n+1), then they are all true. They don't become true one by one, they are simply all true - it is what it is. I hope that helps. I'm thinking of starting a blog to talk about things like this - it falls between the levels of the non-mathematician and true researcher.
- a-priori 16y agoYou can "disprove" the Pythagorean equation the same way, by taking the limit of the Manhattan distance between two points as the grid size approaches zero. It approaches d_x + d_y, not sqrt(d_x^2 + d_y^2).
- ADRIANFR 16y agoThe easiest way to informally prove that the demonstration is false is to imagine starting with a circle in a triangle instead of a circle in a square. Or with any other weird shape around the circle and follow the same "cutting" algorithm to infinity. This way you can prove that pi is equal to anything greater than pi.
- patio11 16y agoYou can also demonstrate this is incorrect by calculating the area of the parts of the square cut off, which is the sum of an infinite geometric series. You will find it doesn't equal the area of the unit square minus the area of the inscribed circle. Here, let's try: the square is 1x1. Consider one quarter of the circle: radius of the circle is 1/2, half a diagonal of the square is sqrt(2) / 2, diagonal of the removed square is (rt(2) - 1) / 2. Area of the square removed is ((rt(2) - 1) / 2) ^ 2. (This is trivial via the pythagorean theorem, saving some math.) Alright, that's the first square we accumulate. Now the magic happens: every step, we cut the square's side in half, but make two of them. Agree with me so far? Good. If we cut the side of a square in half, we cut its area to a quarter, but since we have two squares now the total area is 1/2 of the last square. Agree with me so far? Good. We can trivially sum infinite geometric series: t1 / (1 - r), where t1 is the first term and r is the fraction each term gets multiplied by. In this case, it turns out that in any one quadrant the sum of the series of squares removed is 2 * ((rt(2) - 1) / 2) ^ 2, or just (rt(2) - 1)^2 / 2. Multiply by 4 to get the picture over all four quadrants, and we get 2 * (rt(2) - 1) ^ 2. A little simplification and we get 2 * (2 - 2 rt(2) + 1) = 6 - 4 rt(2) So, we've got a unit square, so the area of the square is 1. If we subtract the area of the infinite series of squares, we get 4 rt(2) - 5 =~ .657. We expect the area of the circle to be pi / 4 =~ 0.7853975. Thus, the square minus and infinite series of squares doesn't approximate the known area of a circle at all.
- spicyj 16y agoThis isn't quite right; when you divide up your problem, your "quarter circle" doesn't look the same after each iteration, and so the sizes of the squares cut off don't follow a geometric sequence. In fact, I'm sure that the area of the figure does approximate the area of a circle (because no point inside the circle is ever removed, yet every point outside of it will be removed eventually). The problem lies not with the area, but the perimeter; the figure's area turns out to be unrelated for this problem.
- patio11 16y agoHmh. Its 1 AM here. Give me some time to sleep and think over that. You might be right: I can't remember why I thought the squares would go geometric. I thought I had a better reason than "It looks that way on the picture, kinda...", but it eludes me.
- raymondhome 16y agoi don't get the "remove corners. Perimeter is still 4!" part. shouldn't it be 4-(some perimeter)?
- augustl 16y agoThis is similar to Zeno's paradox of Achilles and the tortoise: http://en.wikipedia.org/wiki/Zenos_paradoxes#Achilles_and_the_tortoise http://en.wikipedia.org/wiki/Zenos_paradoxes#Achilles_and_th... The paradox lies in "infinity" and "never". Achilles will overtake the tortoise when he's one atom away from the tortoise, and similarly at one point your corner removal will reach the atom level, where you can no longer reduce it and maintain a square shape. At atom level, when your squares consist of three atoms in a L pattern, you can't reduce it further without distorting the squares Assuming, of course, that atoms are the smallest particles.
- kd0amg 16y agoThere is no smallest nonzero distance in Euclidean geometry.
- augustl 16y agoMy comment was regarding physics in general, not a subset of math/geometry. Are you saying my comparison to the Zeno paradox is wrong?
- JeremyBanks 16y agoI think the analogy is flawed. Using limits we can resolve Zeno's arrow paradox. For the situation described here limits cannot solve the problem because it turns out that it is not an accurate model of a circle.
- sleight42 16y agoThere... are... FOUR... lights!
- alanh 16y agoIs this a Star Trek TNG reference? Did you just leave that here because, hey, there’s the number four?
- vilhelm_s 16y agoOver at Everything2.com we discussed this 10 years ago: http://everything2.com/title/2%255E.5+%253D+2 http://everything2.com/title/2%255E.5+%253D+2
- user24 16y agomblase's explanation was a very clear way of explaining the problem, with the two parallel lines never quite touching because you can't get a line to zero length by cutting it in half repeatedly, no matter how many times you do it. Thanks for sharing.
- qntm 16y agoI actually found mblase's explanation to be just as inaccurate as the one in the original story. "Infinity is not a number and you can never get there" is never a good defence in well-formed arguments about limits. It just sidesteps the question and says that it is meaningless, incidentally doing the same to all of mathematical analysis. "Just because a series of curves tends to a limit curve doesn't mean that the series of lengths of those curves tends to the length of the limit curve" is the whole of the explanation.
- jcampbell1 16y agoWhy would anyone assume that because a shape converges, the arc length must also converge? A troll can also prove that 2 == 1 by continuously folding the peaks of an equilateral triangle down to the baseline. Another fun one: $1 = 100¢ $.1 = 10¢ $.1^2 = 10¢ ^ 2 $.01 = 100¢ thus $1 = 1¢ Trollface
- theli0nheart 16y agoGood one, took me a minute to realize that the mistake occurs in the step with the squares--you have to square the units as well. So what you really get is: 0.01 (dollars)^2 = 100 (cents)^2 0.01 (dollars)^2 = 100 (1/100 dollars)^2 Taking out the 1/100 makes both sides equal.
- senki 16y agoHonoured Sir, Understanding you to be a distinguished algebraist (that is, distinguished from other algebraists by different face, different height, etc.), I beg to submit to you a difficulty which distresses me much. If x and y are each equal to 1, it is plain that 2 * (x^2 - y^2) = 0, and also that 5 * (x - y) = 0. Hence 2 * (x^2 - y^2) = 5 * (x - y). Now divide each side of this equation by (x - y). Then 2 * (x + y) = 5. But (x + y) = (1 + 1), i.e. = 2. So that 2 * 2 = 5. Ever since this painful fact has been forced upon me, I have not slept more than 8 hours a night, and have not been able to eat more than 3 meals a day. I trust you will pity me and will kindly explain the difficulty to Your obliged, Lewis Carroll.
- xenophanes 16y ago> Now divide each side of this equation by (x - y) You can't divide by 0 you just get nonsense.
- kdeberk 16y agothat's not really the point because we are dealing with algebra and not the numberical values. The error is the assumption that (x-y)^2 equals (x^2-y^2), which is not the case.
- extension 16y agoIn a similar way, you can "prove" that sqrt(2) = 2 by approximating the diagonal of a unit square (length = sqrt(2)) by iteratively dividing two adjacent edges of the square (length = 2). One interesting and useless law that this reveals is that the perimiter of any convex blob of pixels is equal to the perimiter of its bounding rectangle.
- evandijk70 16y agoIn fact, just a few days ago, I discovered this law while working on my master's thesis. I'm modelling cells in a Monte Carlo model on a square grid, and I needed the contact area with other cells. The contact area seemed to be to large on diagonal interfaces. After thinking about it a bit, I found the problem, and I started using a different measure for contact area.
- cies 16y agotry to find the perimeter of australia.. depending on the zoom level you find a different perimeter. :) if the 'troll' from the article adds smoothing to his 'ad infinitum' squarish circle then he finds pi.
- cscheid 16y agoGeometric convergence is tricky. Specifically, the issue here is that uniform convergence does not imply convergence in length or area or any other such measures. (see figure 6 here: http://www.sci.utah.edu/~etiene/publications/verifiable-vis.pdf http://www.sci.utah.edu/~etiene/publications/verifiable-vis.... . disclaimer - I'm a co-author) Imagine a circle and its diameter. Now imagine two circles with half of the diameter, lined up so that the diameter lines align. Now split those two in four, etc. The circle becomes a snaking line whose total length doesn't change, and the snaking line converges uniformly to the line. Clearly, however, pi is not 1. What you need is convergence in position _and_ angle. A curve that converges in position and angle _does_ converge in length: the reference I know which shows this is reference [9] on the above-mentioned paper. (edit: in case you don't want to download the gigantic file --- yay for publishing in graphicsy places --- the reference is: K. Hildebrandt, K. Polthier, and M. Wardetzky. On the convergence of metric and geometric properties of polyhedral surfaces. Geometriae Dediacata, (123):89–112, 2006.)
- pixcavator 16y ago>>...uniform convergence does not imply convergence in length or area... Under uniform convergence, the limit of the integrals is equal to the integral of the limit. So, this works fine for areas, or are you talking about something else?
- cscheid 16y agoSorry, I should have been clearer. I was trying to say that convergence in area for two-dimensional surfaces in R^3 requires convergence in normals in the same way that convergence in length for one-dimensional curves in R^2 requires convergence in normals. For area in R^2, volume in R^3, and so on, you're definitely right.
- huhtenberg 16y agoA simpler and a bit more elegant version - http://i52.tinypic.com/2daesf4.png http://i52.tinypic.com/2daesf4.png
- jbapple 16y agoWhen I first saw a "proof" like this, the explanation of its incorrectness was something like "limits don't preserve curve length". I wasn't satisfied with that answer until I took a first course in real analysis, which explained (some of) the reasons behind calculus. Real analysis was very satisfying, in somewhat the same way that building low-level software or libraries is satisfying -- I got to understand the guts. It was also fun to learn about erroneous historical assumptions made due to insufficient rigor. IIRC, until at least the 1870s, it was believed that any continuous function must be differentiable almost everywhere, and that in fact this should be obvious. It turns out that one can construct continuous functions that are nowhere differentiable! This was one of my favorite topics in the class: http://en.wikipedia.org/wiki/Construction_of_the_real_numbers http://en.wikipedia.org/wiki/Construction_of_the_real_number...
- xyzzyz 16y agoWhat is even more tricky is that the set of functions differentiable in at least one point is first category, and thus nowhere differentiable continuous functions are dense in continuous functions space.
- drbaskin 16y agoNot only are they dense, but they are "most" continuous functions! (For an appropriate interpretation of "most", of course)
- lmkg 16y ago> Real analysis was very satisfying, in somewhat the same way that building low-level software or libraries is satisfying -- I got to understand the guts. That is exactly the reason why taking a course in real analysis made me switch majors. I was trying to get that feeling from physics, but only found it to my satisfaction in math.
- cousin_it 16y agoWhat... ouch. This is not math, this is... I don't even. Read the post closely. The author introduces a set S of squarey curves approximating a circle. This set has an obvious correspondence with the natural numbers: there is curve #1, curve #2, etc. Then the author defines a function f: S->S that takes curve #n to curve #n+1. Wait, that doesn't sound right! The function f has no interesting structure whatsoever, it's exactly equivalent to defining f(n)=n+1 on the naturals. Of course, taking the "limit" of a function f: S->S makes no sense at all. What would make sense is taking the limit of a certain function N->C, where N is the naturals and C is the set of all curves on the plane. That is, the limit of a sequence of curves (not of a function from curves to curves as the OP tried to say). To talk about such limits, you need to define what it means for a sequence of curves to converge - a "topology" on C. There are many ways to do that, some more outlandish than others. One way is pointwise convergence: assume a parameterization t->C_i(t) on each curves in the sequence, and require that C_i(t) converges co C_lim(t) for each t separately. Now, pretty much any reasonable notion of "convergence" on the space of all curves has to imply pointwise convergence. That is, pointwise convergence is a very "weak" notion of convergence: if we have a sequence of curves that has a limit in some reasonable sense, then it had better coincide with the pointwise limit, dontcha think? And here we come to the second facepalm moment in the post. Under pointwise convergence, the sequence of squarey curves under discussion does not converge to some "right angled fractal beast". It converges to the circle. As n grows, every point on S_n comes closer and closer to some point on the circle. Ain't nothing more to it. Now the correct explanation for the original puzzle. Pointwise convergence of curves doesn't imply that their lengths converge to the limit's length. Hell, we don't even need 2-dimensional space to show that! A simple straight line will do. Imagine a human traveling a straight road of 1km length in this fashion: he takes two steps forward, then one step back, then repeats. In the end he will have traveled about 3km instead of 1km. As we make the human and his steps tinier and tinier, his movement looks smoother and smoother to an external observer, but he still travels 3km in total instead of 1. Or maybe (going back to the 2D space) the human could take a step left, then forward, then right; this would make his path look like a fine comb that approximates the straight line more and more closely, but it's always 3x longer. Something like this is happening in the original puzzle. Finishing touch: there are notions of convergence where it's true that the length of the curves in the sequence always converges to the length of the limit. One such notion says that the direction of travel (velocity vector) must also pointwise converge to the velocity vector of the limit curve. Under this definition of convergence, the original sequence of curves does not converge, because it makes too many sharp turns.
- finin 16y agoObvously, the science is not settled™.
- gyom 16y agoSame issue as with fractals. What's the length of Britain's coast ? Infinity ?