Comments by "EebstertheGreat" (@EebstertheGreat) on "Mathologer" channel.

  1. In high school, I was surprised by how by far the hardest parts of Calc I and II simply involved a lot of steps of algebra. Things like partial fraction decomposition are a major pain, but actually integrating the resulting rational functions was very straightforward--once you did the necessary algebra (completing the square, etc.). Then in Calc III, I found it was much the same. Vector algebra is obnoxious, but the calc part really isn't so bad. I will say though that it gets much worse. Nonlinear differential equations are way harder than anything you have to deal with in a high school algebra class. I'd sooner factor ten solvable quintics than stare at a system of nonlinear PDEs until my brain melts.
    1700
  2.  @topilinkala1594  Blessed be double precision
    45
  3. +Mathieu Bautista The problem is that the sum of a convergent series already has an accepted meaning, and using the same word in two mutually inconsistent ways is pretty unhelpful. If you told me "3+4=8," I would feel justified in telling you that you were wrong, even if you told me you had "redefined" addition such that it was true. The actual methods for finding "sums" of divergent series are numerous, used in a variety of ways, and do not always give the same results. So simply saying "the sum is -1/12" without giving the name of a specific summation technique is just wrong; you are using the wrong term according to mathematical convention.
    25
  4.  @WarmongerGandhi  Another problem with "New Math" was that it focused very strongly on the axiomatic method even in primary schools, where that isn't really appropriate. It tended to put theoretical foundations before practical examples, which is the opposite of how people normally learn math (and how it was historically developed). However, Common Core actually corrects these mistakes in a lot of way, focusing much more on comprehension and on solving problems in multiple ways. That still makes parents livid though, because now they complain "my student knows how to get the right answer, why does he have to do it a particular way? Isn't getting the right answer good enough?" As a tutor, I see these complaints all the time, and it is very frustrating. Because no, getting the right answer is definitely not the point. Nobody cares if you can, say, long-divide two decimals. Your calculator will always do it faster and better. People only care if you understand how the algorithm works, which most kids don't, and just following a list of instructions doesn't show you understand.
    12
  5.  @AlexBesogonov  Well, but you can sometimes. Much like solvable quintics.
    11
  6.  @CousinoMacul  For sure, it's not even close. Check out the problem of finding the distribution of the maximum of some iid discrete random variables. Then compare it to the absolutely continuous version.
    10
  7. He points out that the series diverges. He does not prove it by taking the limit, because the video was already long enough, but it should be obvious that the limit of that sequence is +∞. The zeta function does not "stop making sense" for Re[z] < 1; that's just where the Dirichlet series Σ(1/n^s) diverges. Instead, the zeta function is defined as the analytic continuation of that Dirichlet series over the complex plane (except at z=1). This fact is made explicit in the video several times and is kind of the whole point.
    8
  8. Hero's formula for the area of a triangle is one of those things introduced as a curiosity in a math textbook way back in middle school that I never really got a handle on. To a sixth grader, that is an impressively complex formula for an ancient to have discovered, and no proof was forthcoming. Through high school I saw it a couple more times, always in passing, a sort of neat oddity that seems compact but rarely gets used in practice. It was really neat to see an intuitive proof and motivation after all these years. That said, it doesn't exactly seem useful. Even if you somehow do know the lengths of a triangle but not its angles, this formula is still not the fastest way to find the area. Typically, if you're doing this by hand, you will either have a table of square roots (for Hero's method) or of logs and logs of sines (for the law of sines method). That method is still faster, because you skip all the multiplication steps. If you want to compute the area of the triangle with a computer, you can use Newton's method to get the square root, and I assume Heron's formula really is faster. But the thing is, you basically never know all the side lengths of a triangle (and nothing else) before trying to find its area. Rather, you probably have coordinates, in which case the shoelace formula is by far the fastest. So like, what is this formula actually good for? Is it just a novelty like the quartic formula? If it's never used, then no, I don't think it should be taught as part of a standard curriculum. The brief mentions in books for interested students are probably enough. There is so much I want to add to the math curriculum, and the curriculum is already packed as it is. It's hard to justify cramming in more random formulas to teach, prove, and memorize. (BTW, although the phrase "Heron's formula" is seen pretty often in mathematical texts, in pretty much all other contexts in English, "Hero" is far more common than "Heron." Similarly, we say "Plato" rather than "Platon." The practice of Latinizing ancient Greek names is pretty standard in English. In classical Latin, the nominative singular would be "HERO," and the genitive singular would be "HERONIS." Since the Latin stem is still Heron-, the English adjective would be "Heronic" rather than "Heroic." Again, that's like the adjective "Platonic" rather than "Platoic. Other examples include "Pluto/Plutonic" and "Apollo/Apollonic." Admittedly, there are some exceptions, like the word "gnomon.")
    8
  9.  @nishadthakur  If you already know there are 49 houses, this problem becomes much easier. You can just write and solve a single equation for x, though you do need to know or find a way to sum 1 + 2 + ... + k for any whole number k.
    7
  10. Casinos could choose to simply not offer a game which they know is beatable. That is, they could modify the rules of their blackjack tables such that even card counters could no longer win. They don't do this because most solutions require either slowing down hands or changing the rules so significantly that a significant number of people would stop betting at the tables. Casinos understand that their rules have mathematically predictable consequences, and their algorithm spits out all the details they actually work with in the end. If everybody was a card-counting master, blackjack either would not exist, or all the avenues for advantage play would be cut off by various house rules. But since card counting is so rare, casinos are actually better off pretending they don't exist and advertising to everyone else. They make more money by pushing the problem to the side than by doing literally anything about it. So blackjack remains a game where it is technically possible to make some money from the casino, even though the casino knows this can happen.
    6
  11. You do need a notion of "inside" of the shape, which is uncontroversial but does rely on some other theorems. Every simple closed (hyper)surface embedded in R^n partitions the space into three connected components: the surface itself, a bounded component called the interior, and an unbounded component called the exterior. This is a consequence of the Jordan–Brouwer separation theorem. So then we can say that a polytope is convex if it is simple and every line segment connecting two endpoints in its interior lies entirely in the interior (i.e. every point in the line segment is in the interior of the polytope).
    6
  12. +Karma Peny I haven't read all of your comments (as they are very long), but one mistake you make right away is to claim that "the whole point of algebra is that we can perform manipulations without knowing what value is going to be plugged into any particular variable." That is not only not "the whole point" of algebra, it isn't even true! For instance, consider the function f(x) = (x²+x)/x. You might be tempted to cancel out a factor of x and conclude that f(x) = x+1, but in fact that is true only for x≠0. If x=0, then that rational function has no defined value. The domain of f just doesn't include 0. We can, however, define a new piecewise function g(x) = f(x) when x≠0 and g(x) = 1 when x=0. This new function is indeed g(x) = x+1, and it is the only function that analytically continues f over all real x. The same thing is happening here. The function s(z) = Σ n^(-z) is defined only for Re[z] > 1, since the partial sums diverge elsewhere. However, we can analytically continue this function to all complex z≠1, and one method of computing values of this unique analytical continuation is given in the video. This function is ζ(z), and it is here that we finally get ζ(-1) = -1/12. These are provable, basic, indisputable mathematical facts. If you don't agree with the conclusions, there are three possibilities: 1. Mathematicians are mistaken on this point. 2. Mathematicians are lying on this point. 3. Mathematicians know something about math that you don't. Please take a minute to seriously consider possibility 3.
    5
  13. The main reason people don't teach these rules in high school is that they are never actually used for anything. We also don't teach how to compute products with a slide rule. (That said, I did learn how to find products using log tables in high school, which I thought was bizarre.) Divisibility rules are useful in primary school when learning to reduce fractions, and that's where I learned the rules for divisibility by 2, 3, and 5. Of course, if you think about it, these rules also give you the remainder if you divide by that number. Like, the divisibility rule for 5 says that if a number ends in a 0 or 5, then it is divisible by 5. I think most students realize pretty quickly that if a number ends in 1 or 6, it is one more than a multiple of 5, and if it ends in 2 or 7, it is two more, etc. This works for 3 in the same way. Although the divisibility rule itself is less obvious (that a number is divisible by 3 iff its digital root is 3, 6, or 9), the corollary still is. If 51 is a multiple of 3 (cause it's digital root is 6), then 52 must be one more than a multiple of 3, and 53 must be two more. That's basically all that's going on here. If you take a class on number theory, this basic idea comes up in more general form pretty quickly. But there are like fifty things I wish I had learned in high school math, and this barely even makes the list. There's just too much math to teach.
    5
  14.  @steffahn  Yeah, that checks out.
    4
  15. In general, for the polygon {p/q} with p≥3, p,q coprime, and 1≤q<p/2, the sum of interior angles defined in this way should be σ({p/q}) = (p-2q)π. So for the pentagram {5/2}, we get σ({5/2}) = (5-2*2)π = π, while for the pentagon {5/1} = {5}, we get σ({5/1}) = (5-2*1)π = 3π. Some more small examples: σ({3}) = π σ({4}) = 2π σ({5}) = 3π ­ ­ ­ ­ ­ σ({5/2}) = π σ({6}) = 4π σ({7}) = 5π ­ ­ ­ ­ ­ σ({7/2}) = 3π ­ ­ ­ ­ σ({7/3}) = π σ({8}) = 6π ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ σ({8/3}) = 2π σ({9}) = 7π ­ ­ ­ ­ ­ σ({9/2}) = 5π ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ σ({9/4}) = π σ({10}) = 8π ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ σ({10/3}) = 4π
    4
  16. The formula actually works fine for other values of p and q too, but the interpretation is slightly more involved.
    4
  17.  @Dziaji  Every high school has its own curriculum, and every state has its own standards, but there are federal standards on top of that. There are also national testing services like the College Board and ACT that strongly influence high school curricula, as well as highly influential textbooks like Pearson/Prentice Hall or MacMillan/McGraw-Hill that shape math curricula nationwide. The Euclidean algorithm is hardly ever taught. I'm not convinced that really matters too much, but any teacher would be aware that this was a national phenomenon.
    3
  18. The series diverges and therefore has no sum. It does however have a zeta regularized sum, which is what he is describing in this video. Sums and zeta regularized sums are not the same thing, a fact which greatly troubled physicists for decades. You can't just call them equal because it seems close enough; math is precise.
    3
  19.  @__8120  It's BC Calc. Same curriculum.
    3
  20.  @DTux5249  Maybe they should just be explicit that the policy is no smart people at the blackjack table.
    3
  21.  @drv255  Yeah, it's not an easy question, just much easier than the one in this video.
    3
  22.  @GianniCampanale  Well if circumference = 2π, then surely perimeter = 2p makes sense.
    3
  23. It's interesting that the symbol ∫ was long used for discrete sums as well. For instance, in Summae Potestatum, Jakob Bernoulli wrote ∫nn = ⅓n³ + ½nn + ⅙n. The interpretation is that the left side is the sum of the first n squares and the right side is a polynomial in n. It kind of annoys me that n is playing a dual role here (both as the index and the upper bound), but I guess it wasn't considered confusing at the time.
    3
  24. Interestingly (and perhaps not too surprisingly), while there is an infinite solution to Conway's Soldiers in a certain sense, there is no well-founded solution. That is, every solution involves sequences of steps which are forwards-infinite (no final move) and also ones which are backwards-infinite (no initial move). You can't do it in the intuitive way of having some function f(α) on the ordinals α≤κ (where κ is some large countable ordinal), where f returns a legal position for each ordinal in its domain, each proceeding from the last by a legal move, where f(0) is the starting position and f(κ) is the winning position. If we do allow arbitrary forwards- and backwards-infinite sequences in a solution, we can actually do better. We can make a soldier appear anywhere on an empty board! That's clearly not acceptable, so an additional requirement is necessary. Specifically, there is some number N such that no space on the board flips from empty to occupied more than N times. With that extra condition, the best we can do is reach row 5, using up every soldier in the process.
    3
  25. You need the square root for computing accelerations from position or distances from a plan or much much more. And when surveying land, you need square roots all over the place. But you will never once use Hero's formula. In particular, accurately measuring angles is easier than accurately measuring distances.
    3
  26.  @justinnanu4338  It specifically assigns them the value that analytically continues the zeta function. It arises in situations in which the analytic continuation of the zeta function arises. It's not really magic or just a bookkeeping technique, it's simply a function. In the case of renormalization in physics, I don't really understand it, but I know it is required to make quantum field theory self-consistent. I don't think it is expected to represent a real physical process. I guess calculations of charges and other properties in "dressed" particles are equal to each other at all scales even though measurement of the properties should be scale-dependent (or something like that). So in a sense, the theory predicts the wrong value. And renormalization gives a mathematically precise method to replace these predictions with observed values to very high degree of precision. The fact that the zeta function can be used to do this is not totally surprising, but again, you would have to ask a physicist. The main point is that this isn't some sort of physical proof that zeta summation is "correct."
    2
  27.  @MK-13337  You don't change your basic strategy when you count cards, you change your bet. But yes, they have algorithms to detect card counting.
    2
  28.  @normanstevens4924  The generalization of the Jordan Curve Theorem is precisely how we know Mathologer's definition will always work. Because there will always be a connected "interior" component to which we can apply his definition.
    2
  29. Even some articles written in English use p (but usually by authors who don't speak English natively). In English-language textbooks, it's usually a lowercase s. Rarely, you might see it with no s at all, in the fully expanded form in terms of a, b, and c.
    2
  30. So one magician got to see the five cards and the one picked, then had to arrange the remaining four cards while the other magician was off stage (or back turned or whatever)? How did you communicate the suit of the fifth card with just one bit of information? Wouldn't it require two bits for four suits?
    2
  31.  @blackmephistopheles2273  You can also work it out by feel as the OP suggests. There are only two possibilities, and they should depend only on the parity of the number of disks, because they depend only on the parity of some permutation in the computation. It all comes down to one choice, and it's deterministic, so either left first means left at the end or left first means right at the end. So you actually only need to check a single case. When there is n=1 disk, you move it onto the target peg. So if there are n=2k+1 disks, make your first move toward the target peg, and if n=2k, toward the other peg.
    2
  32. What exactly is your definition of an "algorithm" for a Rubik's Cube? You seem to be claiming that reversibility is a property of all algorithms, which is obviously not the case. For instance, there is an algorithm to simplify fractions, and it will simplify both 2/4 and 3/6 to 1/2. This resembles real Rubik's Cube algorithms. A typical set of algorithms will take any unsolved cube to the solved state, so they are definitely not reversible. Is the "algorithm" here just a fixed sequence of moves that does not depend on the current state of the cube?
    2
  33. This is great. Unfortunately, since I learned Pick's theorem, I know longer have a need for this one when teaching my ACT students. Clearly the shoelace formula is far more broadly applicable though.
    1
  34.  @NoNameAtAll2  I haven't found a real use yet, but I'd like to find out if there is a good use out there.
    1
  35. David Treeby's paper has the word "undoubtably" on the first page. Undoubtedly, the editor should have caught that.
    1
  36. Does the paper include a way to calculate the minimum Lipschitz constant for rectangular tables with legs of arbitrary (equal) length? Many tables aren't as tall as that, but I imagine they still can be balanced this way on most real-world surfaces.
    1
  37. This idea at the start of the video is referred to in Tetris as "parity," in this case referring to the single- or double-evenness of the light squares. Just like in this puzzle, you overlay a checkerboard pattern on the Tetris board and count the number of light squares covered. The parity (or double parity, rather) only changes when you clear a line or drop a "T" piece (the piece literally shaped like a T). Knowing this helps players of modern Tetris games set up boards they know can be perfect-cleared, or to assist them in resolving parts of boards. For instance, if you are trying to clear garbage on top of a clean well, considering the parity of that garbage can help in deciding the best way to clear it, including in deciding whether to burn a line. Since you usually know if a T is coming up with the multiple next boxes in these games, you can make a decision a few moves in advance.
    1
  38. I'm surprised nobody makes digital calculators that can survive autoclaving, but maybe the market is just too small.
    1
  39. This is a good video, but I really hope you go into the generalizations of this like the one published by Dexter Kozen and Alexandra Silva in 2013. There is a lot to unpack here.
    1
  40. Well, if you know the x- and y-coordinates of each point, there is a much easier method. You have A = ½ |det([[x₁,y₁,1],[x₂,y₂,1],[x₃,y₃,1]])| = ½|x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁-y₂)| by the shoelace formula. No messing around with square roots or anything. You do have to take one absolute value in case the triangle is negatively oriented, but that's still much faster than taking a square root.
    1
  41.  @Mathologer  Ah right, you can't use it unless you could project the triangle onto a plane (which takes a lot of time). Is there a better way to find the area of a triangle given three sets of 3D-coordinates?
    1
  42. This might be the ultimate algorithm for the Tower of Hanoi. But "the" ultimate algorithm is surely Marcus Hutter's algorithm that solves every well-defined problem. If the fastest possible algorithm to solve an arbitrary problem P(n) takes f(n) steps, then Hutter's search solves it in 5*f(n) + d time(n) + c steps, where c and d are constants depending on P but not n and time(n) is the time required to calculate f(n). Since time(n) << f(n) for almost all interesting problems, this is about 5*f(n) + c. (However, c is usually so large, this is considered a galactic algorithm for almost all interesting problems.)
    1
  43. It would never have occurred to me to write π²/6 as 2 𝖀/ɪ. 2 π² .
    1
  44.  @matejlieskovsky9625  There are, like, fifty different definitions of polyhedra. The subspace definition is used in linear programming for instance. They can also be defined as realizations of abstract polyhedra in R^3. But they can also be defined purely in geometric terms, like Cromwell's definition in terms of incident polygons. All of these definitions can be extended to higher dimensions, and most of them are compatible for nondegenerate convex polytopes (though not necessarily for other polytopes). However they are defined, the usual geometric characterisation of convexity that Mathologer gave should always work.
    1
  45. ​ @matejlieskovsky9625  I'm not sure I agree. It depends on where you come from. I would bet that the large majority of analysts couldn't provide an algebraic definition of "polyhedron" if asked but could provide a proof sketch of the Jordan-Brouwer theorem given enough time. And from a geometric perspective, it is immediately obvious (visually) that any polyhedron defined with an R^3 representation in mind must be closed, and that any convex polyhedron must be simple. But that's not obvious with more abstract definitions. If you take the perspective of Hilbert, building up from geometry, then the easiest way to define polyhedra is in terms of polygonal faces, and that is also historically how they have been understood. The Jordan-Brouwer theorem is unnecessarily heavy machinery. In R^3, you can prove everything you need geometrically if necessary. But it's a well-known theorem that is certainly sufficient to show that this definition of "convex" will always work for polyhedra.
    1
  46. That's definitely not just an Australian thing. In Ohio, USA, I read about Hero's formula in my textbooks a few times in middle school when first learning about geometry, but it wasn't even in any of the chapters of the book we covered in high school geometry. It's one of those "by the way" moments, or a "well how about that?" kind of thing. It's funny, because the various triangle center properties were considered essential parts of the curriculum, but those feel pretty much the same way. Especially the coincidence of angle bisectors at the incenter. Like seriously, so what?
    1
  47. The line segment and point cases are still cyclic. For the point, it's clearly cyclic, since everything coincides. And indeed, 0*0 + 0*0 = 0*0. For the line segment, the points lie on a "circle" of infinite radius. So the converse still holds true when properly interpreted.
    1
  48. I like the idea that your son, with mathematical training, is standing there telling you the whole 2 times tables.
    1
  49. I guess he decided against a follow-up on 9.99... at the end of the day. I think that's reasonable. The people who currently think it's wrong won't be convinced by more of the same sort of argument. They would only be convinced when confronted with geometric series in other context and the usefulness of defining infinite sums in terms of limits of partial sums.
    1
  50. I never knew Germans called trigonometry "Goniometrie." I guess it measures all gons, not just trigons.
    1