Finger multiplication on steroids @Mathologer

Uploaded October 2015 | Updated September 2026, 2 weeks ago

The Mathologer presents his very special take on two nifty ways to multiply using your fingers. After an introduction to the basic method buckle your seat belts for a wild ride involving fingers of the negative kind. Good news for everybody struggling to remember or teach the times tables and crazy mathematicians.

Enjoy :)

$How did Ramanujan solve the STRAND puzzle? Todays video is about making sense of an infinite fraction that pops up in an anecdote about the mathematical genius Srinivasa Ramanujan. 00:00 Intro 04:31 Chapter 1: Getting a feel for the puzzle 08:27 Chapter 2: Algebra autopilot 12:37 Chapter 3: Infinite fraction 17:51 Chapter 4: Root 2 21:19 Chapter 5: Euclidean algorithm 30:15 Chapter 6: The best of the best: 17/12 36:34 Chapter 7: Outramanujing Ramanujan This was supposed to be a short video but in the end turned out to be quite a tricky to sort out. Anyway, as it sometimes happens, I got carried away and now the video really covers a lot of ground : Pell equations, visualising continued fractions by dissecting rectangles into squares, the relationship between continued fractions and the Euclidean algorithm, the irrationality of root 2. Overall quite a few things that you wont find anywhere else :) The way I tell the anecdote in this video is based on the following account by Ramanujans friend Prasanta Mahalanobis: Current Science, Vol. 9 (3), pp. 74-75. On another occasion, I went to his room to have lunch with him. The First World War had started some time ago. I had in my hand a copy of the monthly Strand Magazine which at that time used to publish a number of puzzles to be solved by the readers. Ramanujan was stirring something in a pan over the fire for our lunch. I was sitting near the table, turning over the pages of the Strand Magazine. I got interested in a problem involving a relation between two numbers. I have forgotten the details but I remember the type of the problem. Two British officers had been billeted in Paris in two different houses in a long street; the two numbers of these houses were related in a special way; the problem was to find out the two numbers. It was not at all difficult; I got the solution in a few minutes by trial and error. In a joking way, I told Ramanujan, Now here is a problem for you. He said, What problem, tell me, and went on stirring the pan. I read out the question from the Strand Magazine. He promptly answered Please take down the solution and dictated a continued fraction. The first term was the solution which I had obtained. Each successive term represented successive solutions for the same type of relation between two numbers, as the number of houses in the street would increase indefinitely. I was amazed and I asked him how he got the solution in a flash. He said, Immediately I heard the problem it was clear that the solution should obviously be a continued fraction; I then thought, which continued fraction? And the answer came to my mind. It was just as simple as this. There is a complete digital archive of The Strand magazine. You can find the page with the puzzle here: https://tinyurl.com/y2lnb8xf (page 790) If you read the puzzle in the Strand youll find that the problem is actually phrased somewhat differently to what Mahalanobis remembers and Mahalanobis also does not spell out the infinite fraction that Ramanujan came up with. And if you do the math(s) some of the other things he says also dont quite sound right. What I am presenting in this video is my best guess for what really happened. In particular, the continued fraction that I am talking about in video is probably the most natural candidate for Ramanujans infinite fraction, but others have argued that it could have been a different continued fraction (which I dont buy :) You can find these other infinite fractions here: Ramanujans Continued Fraction for a Puzzle by Poo-Sung Park https://tinyurl.com/yyfdscgr and here On Ramanujan, continued fractions and an interesting street number by John Butcher https://tinyurl.com/yy6nv2yg Solution to the red cross puzzle from Dudeneys book Amusements in Mathematics p. 168 :) https://imgur.com/a/bBuLOZN Another interesting way to systematically search for solutions to the Strand puzzle is this: The equation we want to solve is 2 x^2=y^2+y. You can rewrite this as x^2 = y(y+1)/2. The formula on the right is just the formula for 1+2+3+...+y. So just keep adding 1+2+3+... and at every step check whether the number you get is a square ... :) Other short formulas: 1) Expanding (1+√ 2)^n gives a number a+b√2. Then a/b is the nth partial fraction. 2) Play with powers of the matrix {{2, 1}, {1, 0}} Some number Easter eggs are hiding on this slide https://youtu.be/V2BybLCmUzs?t=1402 :) Link to the unlisted Marching Squares video: https://youtu.be/f1yDExNAEMg Here is a version of the t-shirt I am wearing: https://tinyurl.com/y5vgo7zb This one is about that other famous Ramanujan anecdote: https://tinyurl.com/y626c86x actually features prominently in another one of my videos. The music in this video is by Chris Haugen, Fresh Fallen Snow (playing in the video) and Morning Mandolin (for the credits) and Nate Blaze Tis the season, all from the free YouTube music library Enjoy! Burkard 14.9.2021: Thank you very much Michael Didenko for your Russian subtitles.$

Toroflux paradox: making things (dis)appear with math

Explaining the bizarre pattern in making change for a googol dollars (infinite generating functions)

$The best A – A ≠ 0 paradox This video is about a new stunning visual resolution of a very pretty and important paradox that I stumbled across while I was preparing the last video on logarithms. 00:00 Intro 00:56 Paradox 03:52 Visual sum = ln(2) 07:58 Pi 11:00 Gelfonds number 14:22 Pi exactly 17:35 Riemanns rearrangement theorem 22:40 Thanks! Riemann rearrangement theorem. https://en.wikipedia.org/wiki/Riemann_series_theorem This page features a different way to derive the sums of those nice m positive/n negative term arrangements of the alternating harmonic series by expressing H(n) the sum of the first n harmonic numbers by ln(n) and the Euler–Mascheroni constant. That could also be made into a very nice visual proof along the lines that I follow in this video https://youtu.be/vQE6-PLcGwU?si=iWTasKqo_JFn4etG&t=1321. Gelfonds number e^π being approximate equal to 20 + π may not be a complete coincidence after all: @mathfromalphatoomega Theres actually a sort-of-explanation for why e^π is roughly π+20. If you take the sum of (8πk^2-2)e^(-πk^2), it ends up being exactly 1 (using some Jacobi theta function identities). The first term is by far the largest, so that gives (8π-2)e^(-π)≈1, or e^π≈8π-2. Then using the estimate π≈22/7, we get e^π≈π+(7π-2)≈π+20. I wouldnt be surprised if it was already published somewhere, but I havent been able to find it anywhere. I was working on some problems involving modular forms and I tried differentiating the theta function identity θ(-1/τ)=√(τ/i)*θ(τ). That gave a similar identity for the power series Σk^2 e^(πik^2τ). It turned out that setting τ=i allowed one to find the exact value of that sum. (@kasugaryuichi9767) I dont know if its new, but its certainly not well known. To quote the Wolfram MathWorld article Almost Integer: This curious near-identity was apparently noticed almost simultaneously around 1988 by N. J. A. Sloane, J. H. Conway, and S. Plouffe, but no satisfying explanation as to why e^π-π≈20 is true has yet been discovered. Ratio of the number of positive and negative terms Its interesting to look at the patterns of positive & negative terms when rearranging to Pi. We always only use one negative term before we switch. The first ten terms on the positive side are: 13, 35, 58, 81, 104, 127, 151, 174, 197, 220,... If you look at the differences between terms, you get: 22, 23, 23, 23, 23, 24, 23, 23, 23, 23, 23, 23, 23, 24,... The reason for this is that Gelfonds number is approximately equal to 23. It turns out that if an arrangement of our series has the sum pi, then the ratio of the numbers of positive to negative terms in the finite partial sums of the series converges to Gelfonds number. This is just one step up from what I said about us being able to get arbitrarily close to pi by turning truncations of the decimal expansion of Gelfonds number into fractions. Similarly for other target numbers. For example, to predict what the repeating pattern for e is, you just have to calculate e^e :) @penguincute3564 thus ln(0) = negative infinity (referring to +0/1-) Bug report: At the 1:18 mark, I say minus one sixth when I should have just said one sixth. Music: Silhouettes only-piano by Muted T-shirt: Pi Day Left Vs Right Brain Pie Math Geek T-Shirt https://tinyurl.com/3e3p5yeb Enjoy! Burkard$

PETRS MIRACLE: Why was it lost for 100 years? (Mathologer Masterclass)

$Eulers infinite pi formula generator Today we derive them all, the most famous infinite pi formulas: The Leibniz-Madhava formula for pi, John Walliss infinite product formula, Lord Brounckers infinite fraction formula, Eulers Basel formula and its infinitely many cousins. And we do this starting with one of Eulers crazy strokes of genius, his infinite product formula for the sine function. This video was inspired by Paul Levries one-page article Eulers wonderful insight which appeared in the Mathematical Intelligencer in 2012. Stop the video at the right spot and zoom in and have a close look at this article or download it from here https://link.springer.com/journal/283/34/4 Very pretty. If you are a regular and some of what I talk about in this video looks familiar thats not surprising since weve visited this territory before in Eulers real identity NOT e to the i pi = -1: https://youtu.be/yPl64xi_ZZA 0:00 Intro 1:49 A sine of madness. Eulers ingenious derivation of the product formula for sin x 7:43 Wallis product formula for pi: pi/2 = 2*2*4*4*6*6*.../1*3*3*5*5*... 9:16 Leibniz-Madhava formula for pi: pi/4=1-1/3+1/5-1/7+... 11:50 Brounckers infinite fraction formula for pi: 4/pi = ... 18:31 Eulers solution to the Basel problem: pi^2/6=1/1^2+1/2^2+1/3^2+... 21:51 More Basel formulas for pi involving pi^4/90=1/1^4+1/2^4+1/3^4+... , etc. Music (all from the free audio library that YouTube provides to creators): https://www.youtube.com/audiolibrary/music?nv=1 Take me to the Depth (chapter transitions) Fresh fallen snow Morning mandolin English country garden Enjoy! Burkard$ $The PROOF: e and pi are transcendental Today’s video is dedicated to introducing you to two of the holy grails of mathematics, proofs that e and pi are transcendental numbers. For the longest time I was convinced that these proofs were simply out of reach of a self-contained episode of Mathologer, and I even said so in a video on transcendental numbers last year. Well, I am not teaching any classes at uni this semester and therefore got a bit more time to spend on YouTube. And so I thought why not sink some serious time into trying to make this “impossible” video anyway. I hope you enjoy the outcome and please let me know in the comments which of the seven levels of enlightenment that make up this video you manage to conquer. Even if you just make it to the end of level one it will be an achievement and definitely worth it :) 0:00 Intro 1:48 Enter Transcendence 2:12 Level 1 - e - irrational 6:40 Level 2 - e - quadratic irrational 10:03 Level 3 - e - master proof 12:47 Level 4 - e - gamma 26:14 Level 5 - pi - Lindemann’s trick 29:59 Level 6 - pi - transcendence 35:35 Level 7 Today’s video would have been impossible without the help from my analyst colleague and friend Marty Ross who did most of the heavy lifting in terms of identifying and adapting the least crazy transcendence proofs for e and pi in existence for this video. Here is Marty’s writeup of the infinite series based proof that e is not a quadratic irrational http://www.qedcat.com/notes/e%20not%20quadratic.pdf and here is his formal write-ups of the proofs for the transcendence of pi and e that this video is based on http://www.qedcat.com/notes/e%20+%20pi%20transcendental.pdf. Marty’s version of the proofs is based on this paper by Steinberg and Redheffer https://projecteuclid.org/download/pdf_1/euclid.pjm/1103051870 which in turn has its origins in a proof by David Hilbert http://www.cs.toronto.edu/~yuvalf/Hilbert%20Ueber%20die%20Transzendenz%20der%20Zahlen%20e%20und%20pi.pdf. Apart from Marty I’d also like to thank MIchael Fraklin for his help with recording this episode of Mathologer, as well as Danil Diitriev who as usual will take care of the preparing amazing Russian subtitles for this video. The mysterious Buddhabrot fractal was discovered by Melinda Green. For an intro to this strange mathematical creature check out my Mandelbrot video. http://youtu.be/9gk_8mQuerg Finally, here is a playlist of all my videos on irrational and transcendent numbers. http://www.youtube.com/playlist?list=PLmNp3NTX4KXK6-xgFJdMAbavzgQaKjRzY Enjoy :) Two ways to support Mathologer Mathologer Patreon: https://www.patreon.com/mathologer Mathologer PayPal: paypal.me/mathologer (see the Patreon page for details)$

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Whats hiding beneath? Animating a mathemagical gem