Fibonacci = Pythagoras: Help save a beautiful discovery from oblivion

  Рет қаралды 236,932

Mathologer

Mathologer

Күн бұрын

In 2007 a simple beautiful connection Pythagorean triples and the Fibonacci sequence was discovered. This video is about popularising this connection which previously went largely unnoticed.
00:00 Intro
07:07 Pythagorean triple tree
13:44 Pythagoras's other tree
16:02 Feuerbach miracle
24:28 Life lesson
26:10 The families of Plato, Fermat and Pythagoras
30:45 Euclid's Elements and some proofs
37:57 Fibonacci numbers are special
40:38 Eugen Jost's spiral
41:20 Thank you!!!
42:27 Solution to my pearl necklace puzzle
The two preprints by H. Lee Price and and Frank R. Bernhart and another related paper by the same authors:
arxiv.org/abs/0809.4324
arxiv.org/abs/math/0701554
tinyurl.com/y6k4eyx5
The wiki page on Pythagorean triples is very good and very comprehensive
en.wikipedia.org/wiki/Pythago...
Wiki page on Pythagorean triple trees
en.wikipedia.org/wiki/Tree_of...
Mathematica code for the Pythagorean Christmas tree by chyanog tinyurl.com/2z66rfkb
Geoalgebra app for the Pythagoras tree fractal by Juan Carlos Ponce Campuzano
www.geogebra.org/m/VU4SUVUp
Connection to the Farey tree/Stern-Brocot tree in a paper by Shin-ichi Katayama
tinyurl.com/vmvcs729
David Pagni (on the extra special feature of the Fibonacci number)
www.jstor.org/stable/30215477
Eugen Jost's Fibonacci meets Pythagoras spiral (in German)
mathothek.de/katalog/fibonacc...
Bug report:
06:06 - right circle doesn't touch line (I mucked up :(
Puzzle time codes:
11:41 Puzzle 1: a) Fibonacci box of 153, 104, 185 b) path from from 3, 4, 5, to this triple in the tree
16:02 Puzzle 2: Area of gen 5 Pythagorean tree
25:55 Puzzle 3: Necklace puzzle
Some interesting tidbits:
Jakob Lenke put together an app that finds the route from 3,4,5 to your primitive Pythagorean triple of choice inside the tree. Thanks Jacob pastebin.com/T71NP8Z9
theoriginalstoney and Michael Morad observed that at 39:28 (last section, extra special Fibonacci) the difference between the two righthand numbers (4 and 5, 12 and 13, 30 and 34, 80 and 89) are also squares of the Fibonacci numbers: F_(2n+3) - 2 F_(n+1) F_(n+2)=(F_n)^2
Éric Bischoff comments that the trick to get a right angle at 25:40 is popularized in French under the name "corde d'arpenteur". This term refers to a circular rope with 12 equally spaced nodes. If you pull 3, 4 and 5-node sides so the rope is tense, you get a right angle. See article "Corde à nœuds" on Wikipedia
Various viewers told me what F.J.M. stands for: Fredericus Johannes Maria Barning, Freek, b. Amsterdam 03.10.1924, master's degree in mathematics Amsterdam GU 1954|a|, employee Mathematical Center (1954-), deputy director Mathematical Center, later Center for Mathematics and Informatics (1972-1988) Deceased. Amstelveen 27.06.2012, begr. Amsterdam (RK Bpl. Buitenveldert) 04.07.2012.
John Klinger remarks that if the four numbers in the box are viewed as fractions, the two fractions are equal to the tangents of half of each of the two acute angles of the triangle.
Colin Pountney: Here is another piece in the jigsaw. The link to Pascals triangle. It only works for the Fermat series of triples (ie the set of "middle children"). Choose any row in Pascals triangle. Multiply the odd entries by 1, 2, 4, 8, ..... and add to get the top left entry in a Fibonacci box. Do the same with the even entries to get the top right entry. For example taking the 1 5 10 10 5 1 row, we have top left number = 1*1 + 2*10 + 4*5 = 21. Top right number = 1*5 + 2*10 + 4*1 =29. For example taking the 1 6 15 20 15 6 1 row we have top left = 1*1 + 2*15 + 4*15 + 8*1 = 99. Top right = 1*6 + 2*20 + 4*6 =70. Not obviously useful, but it seems to make things more complete.
Ricardo Guzman: Another cool property of Fibonacci numbers: Take any 3 consecutive Fibonacci numbers: 55,89,144. The difference of squares of the larger two, divided by the smallest, is the next Fibonacci. .... Thus, in interesting ways the Fibonacci numbers are intertwined with the squares.
CM63: This suggested the attached figure to me.
drive.google.com/file/d/1yjp6...
In reply I suggested to extend this picture a spiral using these identities: phi^2=phi+1, phi^3=phi^2+phi, phi^4=phi^3+phi^2 :)
According to this note on the relevant wiki page tinyurl.com/yv3fnac2 if you take overlaps of the Pythagorean tree into consideration the area of the tree is finite.
Today's music: Antionetta by Boreís and Dark tranquility by Anno Domini Beats
Today's t-shirt: google "Fibonacci cat t-shirt" for a couple of different versions. I just bought this t-shirt from somewhere but I think the cat is supposed to be superimposed onto this type of Fibonacci spiral tinyurl.com/2s3p7e3v
Enjoy!
Burkard

Пікірлер: 1 100
@johnklinger2868
@johnklinger2868 Жыл бұрын
One neat fact that was left out is that when the four numbers in the box are viewed as fractions, the two fractions are equal to the tangents of half of each of the two acute angles of the triangle.
@Mathologer
@Mathologer Жыл бұрын
Glad you mentioned that one :)
@TillerSeeker
@TillerSeeker Жыл бұрын
Yes, John, that is a rather nifty find: u/v = Tan(Atan(Y/X)/2) and (v-u)/(v+u) = Tan(Atan(X/Y)/2)
@hydra147147
@hydra147147 Жыл бұрын
24:22 For the 3,4,5 triangle the line connecting the incenter and the Feuerbach point is parallel to the shortest side. Thus, the parent triangle of the 3,4,5 triangle is the degenerate 0,1,1 triangle (and its clear why this construction cannot be taken further).
@danielhmorgan
@danielhmorgan Жыл бұрын
degenerate as it seems, does it help to think of 2,1,1,0 as 1.618,1, 0.618,0.382 ?
@Mathologer
@Mathologer Жыл бұрын
That's it :)
@haniamritdas4725
@haniamritdas4725 Жыл бұрын
@@danielhmorgan There's the rub. Answering it from this perspective leads to analysis which apparently leads to centuries of excited confusion lol
@landsgevaer
@landsgevaer Жыл бұрын
And the parent of the (0,1,1) triangle is the (0,0,0) triangle, which remains its own parent, ad infinitum. Addendum: except it doesn't... 🙁 See below.
@misterlau5246
@misterlau5246 Жыл бұрын
@@landsgevaer lol man, empty set? With the trivial solution, 0? It's better to put a generator with epsilon instead of 0 0 0 Xd:dx/dt
@royalninja2823
@royalninja2823 Жыл бұрын
Regarding the tree puzzle at 15:50: The bottom square's area is 1. The Pythagorean theorem states the two squares attached it share the same total area, so they are each 1/2. The total area so far is then 2, with 1 contributed from the big square and 1/2 + 1/2 = 1 contributed from the small squares. The next level down, the tinier squares attached to one of the small squares has to add up to 1/2, so they are each 1/4. There are 4 of them, so the total area of these squares is 1, and the total area is now 3. You continue down the line, adding 1 to the total area for each iteration of the tree. There are 5 iterations, so the total area is 5.
@Mathologer
@Mathologer Жыл бұрын
That's it :)
@yanniking7350
@yanniking7350 Жыл бұрын
Well thought man
@vevericac3294
@vevericac3294 Жыл бұрын
@@yanniking7350 its not that hard
@justinfuller5912
@justinfuller5912 Жыл бұрын
Didn’t think to iron out such a simple problem myself, but for those in tow you did good work. 🤟🏻
@PC_Simo
@PC_Simo Жыл бұрын
I got the same answer, through the same reasoning 😃👍🏻.
@aeschynanthus_sp
@aeschynanthus_sp Жыл бұрын
My father was a carpenter. He built various buildings, and the last one was a cottage where I helped. We checked that the corner was at right angle using the 3-4-5 measurement.
@Mathologer
@Mathologer Жыл бұрын
My father was a civil engineer :)
@anteshell
@anteshell Жыл бұрын
Just remember to use multiplied triangle instead of measuring just up to five what ever units you're using. For example measure to 30, 40, and 50. The larger the triangle, the less chance there is for the inevitable measurement error when doing it haphazardly by hand.
@FLScrabbler
@FLScrabbler Жыл бұрын
In most cases it would be quicker & good enough - and in some cases better - to use a set square (wie ein Geodreieck) instead. If the walls don't meet at an exact right angle, then perhaps they are not completely straight as they go up either. In this case the measurement ought to be repeated at various heights...
@frankharr9466
@frankharr9466 Жыл бұрын
@@Mathologer Hey! So was my uncle!
@frankharr9466
@frankharr9466 Жыл бұрын
@@FLScrabbler And then you discover that it's square at three feet off the floor and at no other height.
@thephilosophyofhorror
@thephilosophyofhorror Жыл бұрын
Despite serious competition, Mathologer remains the greatest math channel imo ^^ Thanks for another awesome video!
@Mathologer
@Mathologer Жыл бұрын
Glad that you think so :)
@Filipnalepa
@Filipnalepa Жыл бұрын
IMO Mathologer perfectly hits the balance between presenting topic in depth and in interesting way. There are other channels more attractive in form, there are channels discussing math deeper, but here I can follow up what's going on, and I want to know where it's going on.
@PhilBagels
@PhilBagels Жыл бұрын
I'm not sure I agree, but I'd definitely put it in the top 3.
@thedominion6643
@thedominion6643 Жыл бұрын
Top three of NBA all-time best players are completely interchangeable. Same with math KZbin.
@DendrocnideMoroides
@DendrocnideMoroides Жыл бұрын
@@PhilBagels what are the other 2? (mine are 3Blue1Brown and Flammable Maths (I would say Flammable Maths produces absolute shit now but used to make the best content ever so I am also ranking him) and 3Blue1Brown's quality has dropped a bit, but it still good but his upload frequency is also low) while Mathologer quality is improving and better than ever
@johnmeyers1344
@johnmeyers1344 Жыл бұрын
I've been fascinated with Fibonacci numbers and Pythagorean triples since I discovered them when I was about 8. 45 years later you taught me some new things and helped me understand the "why" behind some of what I already knew. Thank you.
@Mathologer
@Mathologer Жыл бұрын
That's great :)
@leif1075
@leif1075 Жыл бұрын
@@Mathologer Why or how would anyone think of making these connections or manipulations though? Hope you can respond when you can.
@johndoe-rq1pu
@johndoe-rq1pu Жыл бұрын
YOU discovered them?? Thank you for your contributions to mathematics!
@markdoldon8852
@markdoldon8852 Жыл бұрын
7 y
@PC_Simo
@PC_Simo 9 ай бұрын
@@johndoe-rq1pu I think he means he came across them. Of course, I realize your comment is sarcasm, and you knew perfectly well, what he meant. Just thought to point that out.
@TigruArdavi
@TigruArdavi Жыл бұрын
Awesome, Mathologer! Another ab-so-lute-ly delightful journey. Apart from its important and most beneficial applications, Mathologer never ceases to amaze with another revelation on how maths has just this amazing beauty and harmony in itsself. This channel is such a gem on YT. ✨Thank you very much, indeed, Sir. 🙏
@Mathologer
@Mathologer Жыл бұрын
Glad you like the videos so much :)
@contrawise
@contrawise Жыл бұрын
I am not a studied mathematician by anyone's measure. Yet, I carry away so much from your videos! Thank you so much for your well-constructed presentations. This one was wonderfully startling!
@axisjayy7625
@axisjayy7625 Жыл бұрын
The beauty of the interconnectedness of mathematics
@jakoblenke3012
@jakoblenke3012 Жыл бұрын
16:00 I found the area of the tree to be 5, since we have a depth of 5 and for every iteration, the new squares sum up to 1, thus a a tree with infinite iterations has an infinite surface area (still counting overlapping surfaces)
@JaceMillerr
@JaceMillerr Жыл бұрын
In general, area of tree of depth x is x.
@kevinmartin7760
@kevinmartin7760 Жыл бұрын
What is interesting though is that, as the number of levels goes to infinity, so does the area of the tree, but it never leaves a finite bounding rectangle so it ends up overlapping itself infinitely.
@margue27
@margue27 Жыл бұрын
Is there a formula, which takes the overlapping regions into account, and calculates the real visible area?
@jakoblenke3012
@jakoblenke3012 Жыл бұрын
@@margue27 I imagine it to be complicated, but I’ll work on it. For n=6 the area is 6-1/16
@sixhundredandfive7123
@sixhundredandfive7123 Жыл бұрын
Whenever I'm finding myself lost or at a dead end with my own mathematical work you seemingly post a video that helps me along my path. Thank you.
@DeclanMBrennan
@DeclanMBrennan Жыл бұрын
This is amazing. Everything truly is connected to everything. I could hardly have been more surprised if Pascal's triangle had also made an appearance. 🙂
@jonadabtheunsightly
@jonadabtheunsightly Жыл бұрын
To see how Pascal's triangle relates to all of this, you have to also introduce Euler's constant (the base of the natural logarithm), tau (the ratio between a circle's circumference and its radius in the Euclidean plane), the golden ratio, the zeta function, and the distribution of prime numbers. At that point the only things left to connect are the planck length, the speed of light in a vacuum, quantum chromodynamics, and gravity; and those connections remain undiscovered, last I checked.
@DeclanMBrennan
@DeclanMBrennan Жыл бұрын
@@jonadabtheunsightly 🤣
@alexanderstohr4198
@alexanderstohr4198 Жыл бұрын
@@jonadabtheunsightly - dig out the totally volatile fine-structure-number and you will advance a good bit.
@jesuschristislord4538
@jesuschristislord4538 Жыл бұрын
I'm pretty sure 345 is connected to π. How sure? 110%
@PC_Simo
@PC_Simo 9 ай бұрын
@@jonadabtheunsightly Also, those undiscovered connections are physicists’ territory. Mathematicians can’t be arsed to discover them (which is, why they’re still undiscovered, I presume). 😅
@danielhmorgan
@danielhmorgan Жыл бұрын
I was moved by this, almost to tears. what a great treatment of a rich subject. Thank you, thnkyu, thku, thx...
@denelson83
@denelson83 Жыл бұрын
My favourite way to calculate primitive Pythagorean triples is to just use Euclid's formula that 3blue1brown showed us, using two coprime integers that are not both odd, and in fact, the two integers you need to run through Euclid's formula to get a specific Pythagorean triple can be found in the right column of the Fibonacci box corresponding to said primitive Pythagorean triple, and this always works for any such box you choose.
@richardfredlund8846
@richardfredlund8846 Жыл бұрын
11:56 challenge question: The Pythagorean triple (153, 104, 185) corresponding to the box [ 9, 4, 17, 13 ]. If you call the children A,B,C, the 153, 104, 185 is the A'th Child of 'CCC'
@richardfredlund8846
@richardfredlund8846 Жыл бұрын
@Mathologer another interesting maths fact, ... which i saw originally in a Norman Wildberger video, is that the conic tangents of a cubic or odd degree polynomial don't over lap. Actually he's asked the question if every point in the plane lies on one of these tangential conics and appears not to know. (but actually it fairly obviously follows from what he'd already written on the board. ) I'm a bit disorganised about keeping records and it was a long time ago I saw it, so it would take me quite a long time to find it exactly. But I did write some python code at the time to generate a picture of the tangent conics which I was able to find and will cut and paste in my next reply. I find it rather amazing that these tangent conics map the whole plane and so in some sense any cubic equation represents a mapping from one plane to another which is bijective on the whole plane. If it's interesting let me know and I will try and hunt down the original source video.
@wyattstevens8574
@wyattstevens8574 5 ай бұрын
​@@richardfredlund8846Maybe Numberphile's "Journey to 3264" would help- that (not the conjectured 6^5) is the maximum amount of tangents for 5 conics.
@richardfredlund8846
@richardfredlund8846 5 ай бұрын
@@wyattstevens8574the tangent conics of a cubic are quadratics... if you can get the python code to work, then you will see (although that's not a proof ) that they fill the plane. The irony is Wilderberger is pretty close to proving the result (i.e. the information he puts on the board, in the video) but when on of his students actually poses the question, he apparently doesn't know. And it is a rather remarkable result.
@richardfredlund8846
@richardfredlund8846 5 ай бұрын
@@wyattstevens8574 it's approx min 23 of Tangent conics and tangent quadrics | Differential Geometry 5 | NJ Wildberger (vidoe on youtube)
@richardfredlund8846
@richardfredlund8846 5 ай бұрын
@@wyattstevens8574 the argument for the cubic : 23:26 every point does have a tangent conic going through it because for any arbitrary error k at x, and non zero coefficient d, there exists an r s.t. d*(x-r)^3 = k . so there is a maping from coordinates x,y to the cubic reference frame, x,r to prove the 5th and higher odd power polynomials, probably follow from the fact you can get any y value by inputting the correct x value to an odd order polynomial ( unlike even powers ).
@iveharzing
@iveharzing Жыл бұрын
Every single video of yours has so many interesting mathematical connections! I always get excited while watching them!
@BrandonWillWin
@BrandonWillWin Жыл бұрын
I’m stunned. What a sublime concept, especially the animations that produce the cool little fractal trees
@jhrhew
@jhrhew Жыл бұрын
Truly amazing. Finding wonderful hidden connections among the known things.
@nilofido411
@nilofido411 7 ай бұрын
thoroughly enjoyed .... and I will never stop of being amazed and surprised of how a theorem that it's the exception to another has so many dimensions of its own.
@misterlau5246
@misterlau5246 Жыл бұрын
Very well explained, with those graphics, it's a question of watching this carefully. Bravo signore!
@jollyroger9286
@jollyroger9286 Жыл бұрын
Man I've been waiting FOREVER for someone to bring this up!! This pops everywhere in quantum mechanics, Apollonian Gaskets, Ford Circles, Fractals..You Name it! Always had the feeling that the Theory of Everything would somehow be related to this! We need to keep it alive at all costs! THANK YOU!
@brahmanandamperuri5721
@brahmanandamperuri5721 Жыл бұрын
Hello sir , Namaste, I am big fan of your teaching.
@crigsbe
@crigsbe Жыл бұрын
This is a MUST VIDEO for all professors in mathematics ! ❤ Mathologer makes my retirement very colorful. Thank you very much.
@robhillen8007
@robhillen8007 Жыл бұрын
After having spent years of my life watching math videos like this one, I've concluded that math only has 3 original ideas and the entire field is just their remixes.
@bentationfunkiloglio
@bentationfunkiloglio Жыл бұрын
Best content on The Tube (and elsewhere)!
@Mathologer
@Mathologer Жыл бұрын
Glad that you think so :)
@S.G.Wallner
@S.G.Wallner Жыл бұрын
At first, the fact that the tree contained every irreducible fraction broke my mind. Then all of a sudden it seemed obvious. I can't explain why though.
@robertunderwood1011
@robertunderwood1011 8 ай бұрын
What is the Dedekind cut? How does it relate to this sequence of irreducible fractions
@PC_Simo
@PC_Simo 6 ай бұрын
I know the feeling 😅.
@jusinocasino11
@jusinocasino11 8 ай бұрын
This is the coolest most beautiful math video I've ever seen. Thank you
@MrJepreis
@MrJepreis Жыл бұрын
Wow Mathologer and Standupmath videos on the same day! great work as always! many thanks and respects Mathologer!
@alexdemoura9972
@alexdemoura9972 Жыл бұрын
Fiboghoras and Pythanacci, my favorite duo
@wgoode97
@wgoode97 Жыл бұрын
Others have answered already, but the "153² + 104² = 185²" triple has a matrix [9 4; 17 13]. I enjoyed programming this one using Julia. The path to get there from (3,4,5) is right, right, right, left. Triples along this path: "3² + 4² = 5²" "5² + 12² = 13²" "7² + 24² = 25²" "9² + 40² = 41²" "153² + 104² = 185²" Thanks for the challenge and the very interesting video!
@charliearmour1628
@charliearmour1628 Жыл бұрын
As always, an excellent video. After watching one of your videos, I'm always left with an unformed thought, like an itch you cannot scratch. I feel we are seeing glimpses of some universal truth that we still cannot see completely or understand. It is a frustratingly delicious feeling. Thank you.
@bens4446
@bens4446 5 ай бұрын
You have done a great service in raising this from obscurity.
@sinecurve9999
@sinecurve9999 Жыл бұрын
24:13 The location of the center of Feuerbach circle for the 3-4-5 triangle is on the same horizontal line as the incircle, therefore the outlined procedure will produce a degenerate triangle (a line).
@bscutajar
@bscutajar Жыл бұрын
So a 1^2+0^2=1^2 triple, which corresponds to the first two terms of the Fibonacci sequence 0,1
@calvincrady
@calvincrady Жыл бұрын
@@bscutajar I think that's just a coincidence. If you follow the box construction for the numbers 0, 1, 1, 2 you get a Pythagorean triple 0²+2²=2², and if you try to follow the tree structure none of the three children are our 1, 1, 2, 3 box.
@AnonimityAssured
@AnonimityAssured Жыл бұрын
Mathologer videos are always such an inspiration to me. I'm no mathematician, but I enjoy a bit of mathematical dabbling. Most of my exploration is what might be called empirical mathematics. In short, I look for patterns without bothering too much about proofs. To test my pattern-finding ability, I paused the video at 29:01, to see whether I could identify the next few members of the family. I got the following: 9² + 40² = 41²; 11² + 60² = 61²; 13² = 84² = 85²; 15² + 112² = 113²; 17² + 144² = 145². The general pattern could be expressed as (2n + 1)² + (2(n² + n))² = (2(n² + n) + 1)², where n is a natural number. The pattern for the family shown at 30:12 was even easier to identify. The next few members were: 63² + 16² = 65²; 99² + 20² = 101²; 143² + 24² = 145²; 195² + 28² = 197². The general pattern for this could be expressed as (4n² − 1)² + (4n)² = (4n² + 1)², where n is a natural number.
@Mathologer
@Mathologer Жыл бұрын
Glad you are having fun :)
@sidds2004
@sidds2004 Жыл бұрын
This is the best channel on Mathematics
@karlslavinjak5904
@karlslavinjak5904 Жыл бұрын
Congretulations! One of your best videos!
@lennyvoget8725
@lennyvoget8725 Жыл бұрын
The 153, 104, 185 triple at 12:00 is the box 9, 4, 13, 17 and you get there by navigating right right right and left :)
@Mathologer
@Mathologer Жыл бұрын
Correct :)
@cstiger4
@cstiger4 Жыл бұрын
the factors of 153 and 104 made it rather easy!
@insertcreativenamehere492
@insertcreativenamehere492 Жыл бұрын
15:58 Since each new pair of squares corresponds to a right triangle with the hypotenuse of the previous square's side length, and due to Pythagoras, the sum of the areas of these squares is equal to the area of the square in the previous generation. Therefore each new generation of squares has a total area of 1. Since there are 5 total generations in this tree, the area of the tree is 5. It's always cool when fractals turn out to have infinite area.
@Mathologer
@Mathologer Жыл бұрын
Yes in terms of the 5. However from some point on (beyond 5) the leaves of the tree start overlapping and then the question is whether there is enough overlap to make the total are covered finite after all. Have to think about this/look it up at some point :)
@insertcreativenamehere492
@insertcreativenamehere492 Жыл бұрын
@@Mathologer This is actually a pretty interesting question. It's beyond me at the moment, but I'm sure that there's a very elegant solution somewhere out there.
@mrwess1927
@mrwess1927 Жыл бұрын
@@insertcreativenamehere492 would there be more or less overlap if it was translated into 3 dimensions?
@Pedritox0953
@Pedritox0953 Жыл бұрын
The videos where calculus, geometry combines are awesome
@valentinmihairusu6643
@valentinmihairusu6643 Жыл бұрын
Wow, this is so fascinating. To answer your challenge 9,4,13,17 and from the root you gotta go right 3 times then left once. Previous group is 1,4,5,9. Had so much fun!
@Mathologer
@Mathologer Жыл бұрын
That's great. My mission is accomplished as far as you are concerned then :)
@jimmy685
@jimmy685 Жыл бұрын
Wow, so this sequence stands as another proof of infinite pythagorean triples! (As the Fibonacci sequence is, itself, infinite) Neat!
@bot24032
@bot24032 Жыл бұрын
4. The necklace has 12 parts of equal length between the big pearls (? idk), which can be streched into a 3-4-5 triangle to check if an angle is right. Ropes like these were used in ancient Egypt to make right angles, though they weren't so cool-looking, just ropes with knots
@twitter.comelomhycy
@twitter.comelomhycy Жыл бұрын
Wow
@Geenimetsuri
@Geenimetsuri Жыл бұрын
Elegant!
@HatterTobias
@HatterTobias Жыл бұрын
Interesting stuff, this video and the one about Moessner miracle are some of the most insightful videos I've encountered in KZbin
@zaprod
@zaprod Жыл бұрын
This is the most compelling proof I've ever seen. It's truly miraculous as you say.
@reekrev
@reekrev Жыл бұрын
I am Dutch and a huge fan of the Mathologer way of explaining mathematics. F.J.M. Barning is Fredericus Johannes Maria Barning, in daily life he was called Freek Barning. He died in 2012.
@Mathologer
@Mathologer Жыл бұрын
Thank you very much for sharing this with me. Anything else you know about him?
@willemvandebeek
@willemvandebeek Жыл бұрын
My condolences and thank you for sharing this, I had been looking for this information myself and failed to find it.
@R4ndomNMBRS
@R4ndomNMBRS Жыл бұрын
From a reaction on another comment I found your update in the description with a translation of his info on this website, love your dedication! Want to add also: Thanks for bringing wonderful mathematical facts into my life :)
@willemvandebeek
@willemvandebeek Жыл бұрын
@@Mathologer you probably already know this with your German accent, but just in case: "Freek" is pronounced as "Frake" and rhymes with "Lake". Please don't mispronounce the Dutch name "Freek" as "Freak", this would be very wrong.
@y.herstein2199
@y.herstein2199 10 ай бұрын
@@Mathologer Very quickly: he was an employee of the Mathematics faculty of the University of Amsterdam and later onwards was the associate director of the same institute.
@KyleDB150
@KyleDB150 Жыл бұрын
Mind blowing BTW at 39:25 it looks like these fibonacci-generated pythagorean triples have a pattern you missed: A alternates between 1 over and 1 under B/2, in a similar way to those other families in the tree. Makes me wonder what other families of triples can be generated from other sequences compatible with Euclids formula
@maxweinstein1537
@maxweinstein1537 Жыл бұрын
This is actually one of a whole family of Fibonacci identities, related to the Cassini identity! In this case, what you’ve noticed is the identity (which does hold, so good eye): F(n+2)F(n+1) - F(n+3)F(n) = (-1)^n This is best seen as a special case of d’Ocagne’s identity, which has a simple proof (read: short with little algebra autopilot) using determinants. This general identity is: F(m)F(n+1) - F(m+1)F(n) = (-1)^n F(m-n)
@didierblasco8116
@didierblasco8116 Жыл бұрын
What an amazing contribution !
@beautifulsmall
@beautifulsmall Жыл бұрын
dimensionless pythagoran triples. Its connections like this that feel like foundations rather than amusements. Stunning.
@Ken-no5ip
@Ken-no5ip Жыл бұрын
Im letting this play when I fall asleep so I can have mathematical revelations in my dreams
@matthewschellenberg8969
@matthewschellenberg8969 Жыл бұрын
This was one of the best videos yet!
@Mathologer
@Mathologer Жыл бұрын
Glad that you think so :)
@jkid1134
@jkid1134 Жыл бұрын
Very, very cool stuff. I suspected that old ppt generator would be the mechanism behind this dense and elegant tree. The equivalence between adding two numbers and drawing a triangle remains fascinating thousands of years later. And the circle stuff is pure magic.
@nejcvranic3421
@nejcvranic3421 Жыл бұрын
11:50 ab = 153 2cd = 104 ad + bc = 185 For my fourth equation I decided to use the radious of the in circle: ac = r I have one small issue that being i completely forgot the formula for the in circle so i made my own: I remember how to graphically find the center of the in circle so knowing that I will create 2 function which each draw one of the lines that halve one of the angles of the traingle. Where my functions overlap is the center of the in circle. I imagined the triangle orientated as show in the video intro and set the origin of my coordinate system to the most left point of the triangle. My functions draw the lines which halve the alpha and beta angles. 1. equation: f1(x) = x*tan(alfa/2), alfa = arctan(153/104) 2. equation: f2(x) = -(x-104)*tan(pi/4) = -x + 104 Combining f1(x) = f2(x), solution x = 68 The radious r = f2(68) = 36 ac = r = 36 Using the 4 equations I found the 4 variables to be: 9, 4 17, 13 ------------------------------------------------------------------------------- Would it be much easier if I also remembered: a+c=d c+d = b ? Yes. Yes, it would have been much easier.
@thorntontarr2894
@thorntontarr2894 Жыл бұрын
I did recall the last two equations but the closed form looked to ugly to solve so I opted for the easy way out using Excel and arrived at your solution.
@vladimir_woland
@vladimir_woland Жыл бұрын
The solution is easier: d = a+c, b= c+d= 2c+a. The second equation from here: 2ac+a^2=153. The second equation from here: 2c(a+c)=104 => c(a+c)=52. Note: 52 = 2*2*13. Consequently c = 2 or 4. Then a = 24 or 9. The first equation is true if c = 4 and a = 9.
@Nusret15220
@Nusret15220 Жыл бұрын
Here it is... Another interesting and deep connection between two (seems) "basic" things in Math. Really, this is all about: It's about deep connections, not just like "Did you know that a cup and a torus are equal, technically?". Math is beautiful with all its concepts and etc. what really amazes me are these connections. Man, It is such an *art* . It IS worth to spend the entirety of life with Math, really... No matter how it can be challenging sometimes.
@Mathologer
@Mathologer Жыл бұрын
I am living the dream :)
@Nusret15220
@Nusret15220 Жыл бұрын
@@Mathologer Yeah, and I wanna live it too. *Wait for me sir, I'm on the way.*
@phiarchitect
@phiarchitect Жыл бұрын
The true gift in the gift to your wife would be the ability to create a right angle any time she pleased. Very thoughtful of you.
@Mathologer
@Mathologer Жыл бұрын
I am just such a thoughtful guy ... :)
@phiarchitect
@phiarchitect Жыл бұрын
@@Mathologer You are definitely a thoughful guy. This video is prime evidence. I have been peering into the golden ratio / fibonacci for a while and never heard any of these relationships.
@orchestrain88keys
@orchestrain88keys Жыл бұрын
Great stuff!! Thanks for such insightful videos.
@ruferd
@ruferd Жыл бұрын
I'm only halfway through the video but so far I've learned a new geometric fact every 30 seconds. This is absolutely insane.
@misterlau5246
@misterlau5246 Жыл бұрын
Master! You have to work a lot to make these awesome animations and it's a demoivre set
@leofranklin84
@leofranklin84 Жыл бұрын
As always, you continue to amaze
@johnrickert5572
@johnrickert5572 Жыл бұрын
Absolutely stunning. Thank you!
@seanm7445
@seanm7445 Жыл бұрын
@24:20 is a little tougher I think if you try to take it’s parent you get a square of 1,0,1,0 So I suspect the Feuerbach centre is directly on the incircle. In other words you get a 0,0,0 pythagorean triplet. Which... works? For some definition of work!
@bscutajar
@bscutajar Жыл бұрын
You get a 1^2+0^2=1^2 triple.
@cauchym9883
@cauchym9883 Жыл бұрын
I was thinking about that Pythagorean triple tree. If one associates going straight with a 1, going left with a 2 and going right with a 3, then one can encode every primitive Pythagorean triple (PPT) with a rational number between 0 and 1 in base 4 as follows: 0 corresponds to (3,4,5), 0.1 to (21,20,29), 0.2 to (5,8,17), 0.3 to (5,12,13), 0.12 to (77,36,85) and so forth. That way one gets a 1-1 correspondence with the finitely representable base 4 numbers between 0 (included) and 1 that do not contain the digit 0. The three families of PPTs correspond to the numbers 0.11111..., 0.22222 and 0.3333... I wonder if one can get anything geometrically meaningful with that correspondence. Can one interpret the periodic or irrational numbers as interesting infinite paths/families of PTTs in the tree? What about interpreting numerical manipulations like multiplication of 0.1111 with to to get 0.2222 in terms of the associated PTTs?
@KanaalJo
@KanaalJo Жыл бұрын
Nice idea! Similar to the base 3 representation of the Cantor set? If you present it like this, why not put them in Sierpinski's gasket?
@SaturnCanuck
@SaturnCanuck Жыл бұрын
Thanks for this. Always love your videos on a lazy Sunday afternoon
@dl9ar245
@dl9ar245 Жыл бұрын
Toll erklärtes KZbin Video. Ein echtes Highlight. Danke.
@gcewing
@gcewing Жыл бұрын
The necklace made me think of the jewel-division problem from an earlier video. I'm now imagining a band of burglar-mathematicians trying to divide a necklace into equal-value pythagorean triples.
@jacksonstarky8288
@jacksonstarky8288 Жыл бұрын
Question for Mathologer: Why does the Fibonacci sequence start with 1 1 and not with 0 1 ? I'm guessing it has something to do with the reputation had by zero at the time Fibonacci discovered his sequence.
@Mathologer
@Mathologer Жыл бұрын
Actually, some people start the sequence: 1, 2, 3, 5, 8, ... . The 1, 1, start is very natural in a number of different ways, but the 0,1 start is also perfectly fine in my books. Of course, you can also extend the sequence beyond 0 as far as you wish ..., -3, 2, -1 , 1, 0, 1, 1, 2, 3, 5, ... and so infinitely many more starting pairs are possible :)
@misterlau5246
@misterlau5246 Жыл бұрын
As long as you have the intervals, you also can go not only zero but negative numbers. It's not the "number" but the interval. If you can do a linear transformation, then you can get out of negative zone just by summing a number, like if you start at - 1, just add 2 and you will be in positive area, while keeping intact the intervals
@jkid1134
@jkid1134 Жыл бұрын
Where did the first baby rabbit come from? 🤔
@JohnDoe-ti2np
@JohnDoe-ti2np Жыл бұрын
If we let F_1 = 1 and F_2 = 1 then we have the wonderful fact that gcd(F_m, F_n) = F_gcd(m,n) where "gcd" stands for "greatest common divisor." If we let F_1 = 0 and F_2 = 1 then this formula doesn't look as pretty any more. But if you like, you can let F_0 = 0 and then you can start the sequence with 0 if you want.
@jacksonstarky8288
@jacksonstarky8288 Жыл бұрын
@@jkid1134 That's an excellent analogy. At some point its ancestors weren't rabbits anymore, and that's where we leave the natural numbers: at 0 and the negative integer extension of the sequence. 🙂
@johnrichardson7629
@johnrichardson7629 Жыл бұрын
This is the sort of math I love. Thanks for saving this result from obscurity!
@Mathologer
@Mathologer Жыл бұрын
It's great to be in a position where I can do something like this :)
@johnrichardson7629
@johnrichardson7629 Жыл бұрын
@@MathologerVery glad that you are in that position and use your position to enlighten us all while promoting the work of people most of us would never know about.
@rogerdudra178
@rogerdudra178 10 ай бұрын
Greetings from the BIG SKY. Even when I get old I find I enjoy a puzzle.
@seanm7445
@seanm7445 Жыл бұрын
The are of the tree @16:00 I’m assuming the angle is 45 degrees. The side length of the trunk is 1. (1x1=1) Focusing on just the first branch we know that the side length is sqrt(0.5). (sqrt0.5 ^2 + sqrt0.5 ^2 =1) So the area of one first branch is 1/2 (sqrt0.5 x sqrt0.5). But there are 2 of them so the total area of first branches is 1! By similar logic the area of all second branches is 1, and so on. Total area of tree =5.
@briourbi1058
@briourbi1058 Жыл бұрын
The angle actually doesn't matter! As long as it's a right triangle, the sum of the areas of the 2 small squares is equal to the area of the big square, so the area of each layer is equal to the area of the "root" square: 1 and the total area is the number of layers
@seanm7445
@seanm7445 Жыл бұрын
@@briourbi1058 Thanks!
@e11e7en
@e11e7en Жыл бұрын
Madness. I’m absolutely blown away
@sankarsana
@sankarsana Жыл бұрын
Excellent! Math and math videos are a treat. Thanks Mathologer.
@Mathologer
@Mathologer Жыл бұрын
You're very welcome!
@martinmontero3517
@martinmontero3517 Жыл бұрын
GENERAL FORMULA for puzzle 11:54: Using the information from the video and Poncelet's Theorem, I managed to create some formulas to calculate the elements of the box from any Pythagorean theorem. If the box is distributed like this: A B D C And Pythagoras like this: C1^2 + C2^2 = H^2 We can find the elements of the box as follows: X = H +_ sqrt{ H^2 - 2*C1*C2 } Y = C1 + C2 - H A = sqrt{ (C1*Y)/X } B = sqrt{ (X*Y)/(4*C1) } C = sqrt{ (C1*C2^2)/(X*Y) } D = sqrt{ (C1*X)/Y } So, for the problem 153^2 + 104^2 = 185^2: X = 136 or 234 Y = 72 But only x = 136 gives us integers so: A = sqrt{ (153*72)/136 } = 9 B = sqrt{ (136*72)/(4*153) } = 4 C = sqrt{ (153*104^2)/(136*72) } = 13 D = sqrt{ (153*136)/72 } = 17 Clearly, the solution could be reached with the tree going 3 times to the left and then to the right, but this method is much longer especially with large sides of the right triangle. In the same way C and D are not necessary to calculate because for the Fibonacci series it is only necessary to add the two previous ones, but it also seemed useful to me to have a general formula. Added to this, something curious is that since I use a quadratic equation to solve for X, we have 2 solutions, only one of which is integer, but the decimal one still works.
@M.Ghilas
@M.Ghilas Жыл бұрын
what does sqrt mean????
@martinmontero3517
@martinmontero3517 8 ай бұрын
@@M.Ghilas it means square root
@seanm7445
@seanm7445 Жыл бұрын
Here’s my answer for @12:00 9,4,13,17 And you go Right, right, right, left
@Mathologer
@Mathologer Жыл бұрын
Correct :)
@miguelangelsimonfernandez5498
@miguelangelsimonfernandez5498 Жыл бұрын
Excellent, thanks. You put so much work and I will be ver so grateful for your explanations
@antosandras
@antosandras Жыл бұрын
Observation: At 27:17 you ask "Are there any isosceles triangles with integer sides?", and then prove to the negative. But you already touched this earlier mentioning that the incircle does not touch the excircles. For a right-angled isosceles triangle, the incircle would touch an excircle in the midpoint of the hypotenuse.
@beardymonger
@beardymonger Жыл бұрын
Love your work! Thank you!
@ridefast0
@ridefast0 Жыл бұрын
Just WOW. Again. Last time it was odd numbers summing to square numbers. Thanks.
@sapegrillansa9531
@sapegrillansa9531 Жыл бұрын
Glad i found this, i'm going to explore everything on this channel! Looks fantastic! thanks :-)
@ericlester3056
@ericlester3056 Жыл бұрын
This is an incredible topology of harmonic relations!!
@juanluisclaure6485
@juanluisclaure6485 Жыл бұрын
Amazing released of videos this year, gracias por tanto saludos from Bo
@alexandrebaz4765
@alexandrebaz4765 Жыл бұрын
Amazing!!! Thank you so much!
@mlmimichaellucasmontereyin6765
@mlmimichaellucasmontereyin6765 8 ай бұрын
Wow!!! Bravo! Once again, you guys rock the "world" of maths to the core. For exampkle, I finally see the way to use geometry + graphic programmiong to find the exact location of each primal positive number ('prime') n (=p) in the sequence n + 1 of N => positive infinity. For example, since all p = 6n +/-1 and there are only primals, coprimes, and pseudo-primal composites at 6n +/-1 then, in any decan of N at magnitude/cardinality M/C, we can check for primality by using the pythagorean-fibonacci geometry (PTG) rule. In other words, by progressing along the number line of N+ (or R+), we can eliminate multiples of n & p, yet also check for primality at 6n +/-1 by using the PTG Rule. Voila! We find no mystery of primal numeric logic or locations of noncomopsites p, and no mysterious patterns of p (determined by the symmetries and regularities of the preceding composites n). Clearly, this verifies my 2017 insight (& mapping). The noncomposites p are gaps in the sequences of composites n, due to the result of dyadic arithmetic continuation of n + 1. This also confirms the intrinsic interdependence of geometry and "numbers" as expressions of geometric-numeric logic, enabled by the natural metalogical principles of being (the cosmos, or life). QED. For more extensive consideration andf/or discussion, see my preprints (at ResearchGate .net). Thanks & best of luck etc. ~ M
@geon79
@geon79 Жыл бұрын
11:47 The box is 9 - 4 -13 -17. To reach it from the bottom (1-1-2-3), go right (1-2-3-5), then right (1-3-4-7), then right again (1-4-5-9) then left (9-4-13-17). In summary, R-R-R-L. I'm in awe! Thank you for sharing all this wonderful math.
@bennyloodts5497
@bennyloodts5497 Жыл бұрын
So many alien tricks: keeps me wondering! Fascinating and mindblowing stuff.
@adamnealis
@adamnealis Жыл бұрын
Wow. Simple maths, but mind completely blown. Easily one of my favourite episodes. So, so beautiful.
@Muhahahahaz
@Muhahahahaz 4 ай бұрын
11:38 Challenge Question The triple (153, 104, 185) comes from the following box: 9 4 17 13 Bonus Question From the starting square, you would go right, right, right, then left. The first 2 destination squares are shown in your video, the 3rd right would look like this: 1 4 9 5 And the final “left” is already shown in my answer to the previous question (obtained by using the 9-4 diagonal)
@mandaparajosue
@mandaparajosue Жыл бұрын
Thank you. Absolutely beautiful.
@05degrees
@05degrees Жыл бұрын
This all indeed is very neat! 🎄🧡 This triple tree reminds me of a tree of Markov triples a bit, though their constructions aren’t at all analogous.
@jakobthomsen1595
@jakobthomsen1595 Жыл бұрын
Stunningly beautiful!
@hylens5111
@hylens5111 9 ай бұрын
Some of this stuff is just incredible.
@whydontiknowthat
@whydontiknowthat Жыл бұрын
The total area for the figure at 15:58 is 5. At first I didn’t know how much the areas got smaller with each new color, but it’s clear that each new color diminishes by the same ratio, and at the same time the number of squares doubles. For the total area, you would then have an expression like 1+2r+4r^2+8r^3+16r^4 for the total area. To figure out r, I saw that the angle at which the squares were branching out had to be 45 degrees, since two branches made a right angle, which meant that the side lengths were decreasing by a ratio of 1/sqrt(2). This means that the area decreases by a ratio of 1/2, and plugging in r=1/2 into the expression above gives 5. Basically, each color contributes the same amount of area (1), and since there are five levels you get a total area of 5.
@BartvandenDonk
@BartvandenDonk Жыл бұрын
The nice thing about this is that it doesn't change if you're using decimal, hexadecimal or which other number system you choose. It is always true. The magic with circles and triangles is very intriguing.
@Mathologer
@Mathologer Жыл бұрын
In general, mathematical equations and identities hold true irrespective of your choice of representing numbers :)
@martw3240
@martw3240 Жыл бұрын
the run towards a straight line of pythagorus triples is easy to see in either direction, 345 is the clue. A balanced rise over run would suggest an integer relationship of 1/root2 ,which is clearly untrue
@andersstrksonberge2874
@andersstrksonberge2874 Жыл бұрын
Wow Amazing 😀 Mind blowing visuals!
@ArtSeiders
@ArtSeiders Жыл бұрын
Super interesting and well explained
@binathiessen4920
@binathiessen4920 5 ай бұрын
the area of the tree at 16:00 is infinite assuming that overlaps are excluded. The first square has sides of length 1 and area 1. The next square has sides of length 1/sqrt(2) and area 1/2, and there are 2 of them, so the area of the next step is also 1. Each step the area of each square is divided by 2, and the quantity of triangles is multiplied by 2, these effects cancel out, and so the area of the tree is equal to the number of steps in the sequence, and so the area grows without a finite limit.
@ricardolichtler3195
@ricardolichtler3195 Жыл бұрын
Mathologer is the very best channel ever
@tahamuhammad1814
@tahamuhammad1814 Жыл бұрын
I would really like to thank you as your previous videos on fibonacci sequence, fibonacci-like sequences and the strand puzzle made me follow a deep rabbit hole and led me to discover a super generalisation of the fibonacci numbers with similar properties, I generalised the Binet formula and many fibonacci identities and used them to find general integer solutions to infinitely many pell-like equations all super fun Oh and I'm particularly waiting for the video you promised at the of the strand puzzle video
@Mathologer
@Mathologer Жыл бұрын
That's great. Glad you are enjoying the videos so much :)
@benji-o1204
@benji-o1204 Жыл бұрын
the area is 5- each new layer has an area of 1, since the area of each individual square goes down by 1/2 after using the 45-45-90 triangle and squaring, and the number of new squares goes up by 2 (1/2^n * 2^n where n=0 is the first square)
@howdy832
@howdy832 Жыл бұрын
The question from 11:44 is rather easy once the relationships are written out, I'm convinced the question is just warm-up or to drive engagement. Getting started, note that C=A+B and D=A+ 2B. This turns the square's relationships into: 1) 153 = A²+2AB 2) 104 = 2AB+2B² 3) 185 = A²+2AB+2B² From here it's a short step to the square [9, 4, 13, 17]
@Mathologer
@Mathologer Жыл бұрын
Yes, fairly easy, and, yes, mainly there to drive engagement :)
@mitchwyatt9230
@mitchwyatt9230 Жыл бұрын
Fantastic video!!!
BALLOON CHALLENGE
00:26
Natan por Aí
Рет қаралды 19 МЛН
Woman is hanging out with Silverback Gorillas at her sanctuary!
00:14
Что будет с моей рукой?😨
01:00
Аришнев
Рет қаралды 3 МЛН
مقلب الكأس الضخم للغاية 😅
00:15
One More Arabic
Рет қаралды 78 МЛН
Alien Enigma: How Close Are We To Discovering Extraterrestrial Life?"
10:15
Total Solar Eclipse 2024 - WHEN, WHERE, WHAT, HOW, & WHY
13:24
A New Way to Look at Fibonacci Numbers
15:51
Jacob Yatsko
Рет қаралды 579 М.
Powell’s Pi Paradox:  the genius 14th century Indian solution
27:29
How did Ramanujan solve the STRAND puzzle?
45:10
Mathologer
Рет қаралды 853 М.
Why did we forget this simple visual solution? (Lill's method)
26:14
Researchers thought this was a bug (Borwein integrals)
17:26
3Blue1Brown
Рет қаралды 3,1 МЛН
All possible pythagorean triples, visualized
16:58
3Blue1Brown
Рет қаралды 3,8 МЛН
BALLOON CHALLENGE
00:26
Natan por Aí
Рет қаралды 19 МЛН