Phi and the TRIBONACCI monster

[@nitsanbenhanoch8691]

"you're allowed to do this?" "who's gonna stop me"

[@MagicGonads]

I burst out laughing, he said it so menacingly for someone so cheerful

[@skoockum]

James Bond will stop him.

[@a9238-w4x]

'Wait, you're allowed to do that?' 'Who's gonna stop me?' I love this.

[@SHA256HASH]

brudermüll /// I’m going to use this forever.

[@bryanroland9402]

I thought that was a great moment too. It reminds me that, historically, some of the best work was done by mathematicians who didn't take the rulebook too seriously.

[@kamoroso94]

He's a mad mathematician!

[@EightchBee]

He could easily apply for a role as a Bond villain. "Do you expect me to talk?" "No, Mr. Bond, I expect you to die." By the way, mighty Mathologer, do you own a white cat? ;)

[@robinbreslin1626]

Interestingly, in other videos he shows the dangers of guessing its going to be ok with infinite series, how often this blows up. cf 1+2+3 != -1/12. But in this case he is careful to explain "out of control" nature of the rounding approximation is at the start (small number) ends of the series and how it quickly damps down as a result of ref phi, and the coloured t's. He knows where to be gung-ho, and where to be careful.

[@AttilaAsztalos]

Please, never stop making these vids! Each one feels like it's Christmas... :)

[@SevenLovedFrench]

It's like Neuron gym

[@quantumgaming9180]

Indeed

[@Fire_Axus]

your feelings are irrational

[@The-cyber-imbiber]

As a graphics artist, whenever I am trying to make components in a scene fit together in an aesthetically pleasing way, I will often adjust their scales and proportions by phi. It feels extremely satisfying to actually know a bit of the maths behind this beautiful constant!!! Thanks for the awesome video.

[@Fire_Axus]

satisfying is not a feeling

[@The-cyber-imbiber]

@@Fire_Axus you're baiting me, I say "this feels satisfying" and you correct me say "satisfying is not a feeling" as if that's what I was claiming. Either you're baiting me into an argument or you genuinely need help

[@Mathologer]

Got distracted again and just had to make this video about my tribonacci friends. Will be interesting to see whether anybody comes up with a nice answer to the puzzle about the mutant tribonacci rabbit population at the end. Anyway, enjoy :) By popular demand here is a link a place that sells today's t-shirt: www.zazzle.com/fibonacci_parrots_t_shirt-235568206086965961

[@ChertineP]

Mathologer Love this! And where can we get the awesome t-shirt?

[@rohansharma1250]

Mathologer when are we getting the Riemann hypothesis video !?!?

[@fefeisbored1958]

I have no soul!

[@guest_informant]

2:27 Technical point. Is [x] being used as Floor(x)? In which case isn't it "rounded" to the integer _below_ rather than the nearest integer eg [1.9] = 1, and [1.9] != 2 (Related: Floor(x) or [x] is often described as "the integer part of". This definition works for positive numbers and zero, but does not work for negative numbers, so "the integer part of" -3.14 would (presumably) be (thought of as) -3 but [-3.14] is -4. Edit: Googled and [x] is being used as nint(x) so the statement is correct, but I thought I'd leave the comment anyway. mathworld.wolfram.com/NearestIntegerFunction.html

[@Mathologer]

+ChertineP For the t-shirt have a look here: www.zazzle.com/fibonacci_parrots_t_shirt-235568206086965961

[@Antediluvian137]

This was brilliant - love how it was laid out. A lot of information at first and then carefully breaking down how an equation like this could be discovered. Marvelous work put into these videos, much thanks!

[@robertdibenedetto6445]

At around 5:40 you mention "cheating a little bit" with the rounding for the first number. Instead of rounding to the nearest whole number, you can "round down, round up" successively, and you'll reach the correct integer every time.

[@Mathologer]

Good idea :)

[@fandibataineh4586]

this is because the 'tiny' number we omited (phi-red)^n is alternating its sign because phi-red itself is negative

[@eshel56765]

3 fastest things in the universe: 1. The speed of light 2. The expansion of the universe 3. Me clicking on a new mathologer video

[@TacoDude314]

Switch #1 and #2

[@cubicardi8011]

TacoDude314 why? He's right

[@alexwang982]

2 is faster than 1

[@alexwang982]

Universe expansion is faster than light, that’s why we can’t get to the other groups in the laniakea supercluster

[@yuvalpaz3752]

no he is not, expansion is not information so the it can be faster

[@daicon2k6]

The relationship between the icosahedron and three golden rectangles is my favorite detail from any of your videos, and that's saying a lot. What a beautiful geometric connection. Thank you.

[@yxlxfxf]

15:15 it's actually quite easy to see why the formula always spits out integers. Just by expanding both terms, we can notice all even powers of sqrt(5) from the 1st term will cancel with the evens from the second one. The remaining odd powers will all be some power of 5 multiplied by a factor of sqrt(5) which cancels out and the remaining numerators will be even (because adding two equal numerators), being divisible by 2.

[@Mathologer]

That's absolutely right :)

[@whygreen44]

I think I have a solution... You have three stages in the life cycle of the Tribonacci bunnies: 1. The adolescent bunnies have no babys 2. The mature bunnies have 1 baby 3. The extra-mature bunnies have twins!

[@Mathologer]

That's nice :)

[@andymcl92]

Technically we're always talking about pairs of bunnies, so the mature bunnies have twins and the extra-mature bunnies have quadruplets. Imagine all the birthday cards!

[@stylis666]

The first bunny clones itself twice, or clones itself once and has a baby with that clone... I don't think this can end well :p

[@michaels4340]

Perhaps they reproduce asexually? Maybe these are single-celled bunnies?

[@chrisg3030]

I suggest in my solution (I hope you can find it) that if you transpose the question into bugs (single dividing cells) rather than bunnies, some divide into 2 in each generation, some into 3, and some not at all. So instead of your three stages we have three levels of fertility. The numbers come out nicely if you do the family tree diagram with a blob for a cell and lines coming down from it to other blobs to signify splitting into further cells.

[@alexhancu969]

See? This is why I love maths. People might think it's all just adding and subtracting and all messy stuff, but if you get deep enough, you start finding beautiful patterns that amazingly link with each other in many ways, the Fibonacci numbers and the Luka numbers being one of many examples

[@sofia.eris.bauhaus]

you don't have to go "deep" into math to find amazing things. you just have to be willing to mess around by yourself instead of becoming a calculator for your math teacher. Jacob Turbaugh: math is real in the sense that it is reproducible at any place and time given the same procedure. just like physics. B)

[@ChefXbb]

well, i starting to feel it.

[@ChefXbb]

@ Jacob Turnbaugh based on laws of nature, even mentality must make perfect sense, right? btw, realitys like math, principles and sort of are designated as entities. give logic a chance ; p

[@sofia.eris.bauhaus]

math is built on computation. computation is a physical process that our brains happen to be able to do deliberately (to some extent). there is probably a degree of bias (human and cultural) in what computations we deem to be "proper mathematics", while others would seem completely alien to us. but then again, researching mathematics already led to some rather alien systems and counter-intuitive results. the major scale being built on the golden ratio does indeed sounds like nonsense, not sure where you got this from. phi is neither found on in 12 tone equal temperament nor pythagorean tuning. math and physics didn't change, it's our understanding of it that improved (and perhaps very rarely worsened). and it probably still can improve drastically. also how you count dimensions doesn't matter. the point of seeing them as dimensions is to make them interchangeable. :)

[@sofia.eris.bauhaus]

Luke Müller: wat. sorry but that sounds like complete word salad. XD

[@expchrist]

Anybody who does not agree that this mathologer video is wonderful does not have a soul!!!

[@alexwang982]

expchrist souls don’t exist

[@sofia.eris.bauhaus]

Pi: that's not true. gingers have souls..

[@Iridiumalchemist]

Of all the mathematics popularizers I follow on the internet, I think you do the best job at combining rigorous mathematics and sound explanations. Keep up the great work! (I'm doing my PhD in number theory at UF right now- so it's enjoyable to learn things far afield from what I'm doing.)

[@Mathologer]

Glad you think so, and thank you very much for saying so :)

[@voltrevo]

16:17 In general, you can derive formulas for any sequence of the form s_n = A*s_(n-1) + B*s_(n-2) using a similar technique. Step 1: suppose that s_n = X^n. Step 2: Derive a quadratic for X to obtain Xa and Xb. Step 3: Observe that s_n = C*Xa^n + D*Xb^n satisfies the definition of the sequence for any values of C and D. Step 5: Substitute two known values (probably the first two values) of the sequence to obtain two equations in the two unknowns (C and D). Step 6: Solve for C and D. Step 7: Now that Xa, Xb, C, and D are all known, s_n = C*Xa^n + D*Xb^n.

[@Fassislau]

Omg this was SO GOOD !!!! Hands down best mathematics video in the internet. Thank you so much for this. I absolutely loved it.

[@stevenvanhulle7242]

Granted, this is a great channel, but don't forget to have a look at 3Blue1Brown's channel as well. Even though the math he discusses is sometimes a bit more complex (no pun intended), it's always equally clearly explained.

[@kellsierliosan4404]

Hey, I've got to say that your channel is the closest thing to mathematical disneyland in youtube, if not in the internet as a whole. Amazing work, seriously! :)

[@JCOpUntukIndonesia]

I have ever thought about this variant of Fibonacci Sequence, also, how the ratio would be, but I didn't know that it already existed. Your explanation is on spot prof. It is easy to understand, and thank you for it.

[@buckleysangel7019]

You’re not cheating by rounding. You’re adding or subtracting one over phi to the same power as phi! So beautiful

[@blue_tetris]

Is there a general case for the n-onacci sequence? That is, a sequence of integers where you start with 0 ... 01, then add up the most recent n digits to get the next in the sequence?

[@blue_tetris]

I suppose the first n+1 digits of an n-onacci sequence just resemble powers of 2, before they start approaching the golden ratio by sums. But is there a generalized formula, like the one presented for finding the nth tribonacci, which could measure my tetranacci or icosanacci (etc) sequences?

[@TheDoubleEagle1]

I had the exact same question, so I did some math and turns out that the sequences converge to a series of powers of 2 (1, 2^0, 2^1, ..., 2^(n-2) ) and so the ratio of the n-onacci sequence will also converge to 2.

[@JordanMetroidManiac]

frogboy7000 I have a treat for you. Look at the coefficients on the right side of the equation in x^2=x+1. They’re 1 and 1 in order of descending powers of x. Let’s call them A and B. You use 1 * A + 1 * B to calculate the next term in the Fibonacci sequence. So you have 1,1,2 up to this point. Then do 1 * A + 2 * B. You get the new sequence 1,1,2,3. Continue it forever and you find that the ratio of consecutive terms approaches the golden ratio, which is no coincidence that it is the root of f(x)=x^2-x-1. This can be applied to almost any polynomial of absolutely any degree. For demonstration, I will show how to approximate the root of f(x)=x^7-2x^6+3x^2-6x-12 with the following steps: 1. Find the coefficients of each term, excluding x^7. They are -2, 0, 0, 0, 3, -6, -12. 2. Negate each number in the sequence. You get 2, 0, 0, 0, -3, 6, 12. Call this sequence C (for “coefficients”) 3. Start your new “nacci” sequence with as many ones as the degree of the polynomial function. You should have 1,1,1,1,1,1,1. Call this sequence S. 4. Generate S with S(n) = C(1)*S(n-1) + C(5)*S(n-5) + C(6)*S(n-6) + C(7)*S(n-7). Sequence S should now be 1,1,1,1,1,1,1,17,49,113,241,497,961,1889,3937,8417,18145,39281,84625,180305,381649,806609,1705249,3610049,7658305,16268993,34571713,73447121,155974897,331098161,702659761,... 5. Compute the ratio of successive terms. You’ll find 702659761/331098161=2.12220979687... which is a fairly close approximation of the only real root of the given polynomial. The coefficient of the highest power term must be 1. Try it yourself. If it doesn’t work for you, it probably means the real root of the polynomial is negative or between 1 and -1. If it’s negative, substitute x = -y, then approximate the root with y, and then solve for x. If the real root is between -1 and 1, then substitute x=1/y and then multiply the whole polynomial expression by y raised to the power of one less than the degree of the polynomial. Then approximate the root with y and solve for x. If the ratio of consecutive terms still diverges, then it is possible that there are no real roots (you could predict this by using Descartes' rule of signs).

[@JordanMetroidManiac]

Murdock Grewar That’s probably because I discovered it. I mean, I may not be the first one to have found it, considering that the coefficient sequence is very closely related to characteristic polynomials of matrices. However, I brought it to my mathematics professors, and they said they haven’t seen it before. So, I’m pretty proud of this. 👍

[@JordanMetroidManiac]

Murdock Grewar They probably just use Newton’s method. At least the ones that require you to give a close guess of the root do.

[@joaoalcantara6676]

Anxiously waiting for the video on the QUADRIBONACCI constant/sequence.

[@Mathologer]

You may be in for a long wait (unless you do it yourself). So many nice topics to talk about and so little time :)

[@sofia.eris.bauhaus]

we need an n-part series. B)

[@powerdriller4124]

The infinite family of equations: x=1, x^2 + x = 1, x^3 + x^2 + x = 1, x^4 + x^3 + x^2 + x = 1, ..... ; has for each equation one and only one positive real solution for x ; those solutions together form a remarkable family: 1, 1/phi, 1/t, ... *the inverse of the N-nacci Constants.* For n infinite, the solution for x is =0.5, which corresponds to 1/0.5 = 2, *2 is the Infinity-Nacci Constant.* Also noticeable, when n=infinity, the infinite complex solutions fill up the complex unit circle,... well not exactly, ... the infinite solutions occupy the positions in the unit circle for all 2*Pi * (j/m) where m goes to infinity, for all natural numbers j , from 1 to infinity. That circle is porous in the "irrational" angles.

[@modusvivendi2]

Always great to get more videos on phi, my favorite number. A few more related weird properties that are probably obvious but weren't explicitly mentioned in the video: what you call phi-red is just negative one over phi; one over phi (the reciprocal of phi) is phi minus one or phi-squared minus two. And a neat physical property (and this is something only an American would probably know): by a weird coincidence the number of kilometers in a mile is... not quite exactly phi, but really, really close to it, which means that (kilometers per mile) ~= ((miles per kilometer) - 1). In fact, we could even define a "mathematically non-shitty mile" (much as we have defined the "mathematically north pole," which is not at the north pole) to be EXACTLY phi kilometers, which would really make conversions a snap. To give you a sense of just how close the non-shitty phi-based mile is to an actual mile, it's approximately 5308 feet compared to 5280, or less than one percent longer than an actual mile.

[@kwinvdv]

By defining higher Fibonacci numbers as the sequence such that the next number in the sequence is equal to the sum of the previous k numbers. For k values from 1 to 5 the ratio between consecutive numbers of these sequences converges to the following numbers: 1, 1.6180, 1.8393, 1.9276, 1.9659. For larger and larger values for k this ratio converges to 2. This limit as k goes to infinity can be shown to be related to the geometric series with ratio 1/2.

[@Mathologer]

That's also an interesting fact :)

[@chrisg3030]

The recursion for a sequence with this ratio constant of 2 is a(n)=a(n-1)+a(n-1), compared to Fibonacci with a(n)=a(n-1)+a(n-2). Let's have another recursion a(n)=a(n-1)+a(n-3). The ratio constants for each respectively are 2, 1.618..., 1.4656..., which I'll designate by Delta, Phi, Moo. There's a nice common bracket shift identity: Delta^(0+1) = (Delta^0)+1, Phi^(1+1) = (Phi^1)+1, Moo^(2+1)=(Moo^2)+1.

[@bahkimi]

You are unbelievably good. Thanks for making the beauty of the beautiful math more clear and understandable for all.

[@luckyluckydog123]

Wonderful video, as usual! I still remember when, on my first year at university, I saw the explanation of the formula for Fibonacci numbers as a exercise in matrix diagonalization, it struck me as very neat. Almost 20 years have passed but I still remember visually the text of the exercise (there was a 2x2 matrix at the end of the page etc)... Some things are so beautiful that they are difficult to forget

[@Mathologer]

That's actually the way I also teach it one of the first year units :)

[@topilinkala1594]

About irrationality producing nice integers here's a sequence of polynomial functions: There's one each order starting from 1, which is y = x, the order two one is y = x^2 - 2, the third order one is y = x^3 - 3x, the fourth order one is y = x^4 - 4x^2 + 2, the fifth order one is y = x^5 - 5x^3 + 5x, the sixth one is y = x^6 - 6x^4 + 9x^2 - 2, etc. These are polynomial functions that have the following properties: 1. All local maxima and minima values for y are 2 or -2 and their corresponding x-values are between -2 and 2. 2. Each odd function goes through points (-2, -2) and (2, 2) and even function through (-2, 2) and (2, 2). These two properties mean that the polynomial oscilates between -2 and 2 when x goes from -2 and 2. Outside the range the function goes monotoniously to (minus) infinity. 3. The coefficients of the polynoms are integers and the leading coefficient is 1. This is something I find beautiful. That there are integer coefficient polynomials with leading coefficient 1 that have all zeroes inside small range (-2, 2) and that at the same region the polynomial function don't exceed -2 or 2. They can be generated by the parametric pair x = 2cos(t), y = 2cos(nt), where n is the order of the polynomial. The pair x = cos(t), y = cos(nt) gives the same first two properties scaled down by 2 but the polynomials leading coefficient is not 1 which is IMHO ugly. So that's why I like the scaled up versions.

[@RadicalCaveman]

What a marvelous video! I've seen many of your videos, and I can't believe I've only just discovered this one.

[@aformalevent]

Amazing video as always! It's very easy to see that you love the mathematics you're talking about. I will never forget the first time I saw the square root of five in the formula for fibonacci numbers and how crazy it was to get nice integer patterns from what seemed like an arbitrary formula. This has been on the back burner for things that I hope one day to understand and you've made very, very happy with this video. :) Thank you so much for your work! Just amazing.

[@aformalevent]

1100 likes and 13000 views shows you just how many people are in the same boat as me. I think one like in thirteen views is one of the highest ratio's I've ever noticed :D

[@away5534]

I had studied fibonacci before, but everytime I see the F(n) equation my mind blows, it's so beautiful. It's a true mistery why fibonacci is everywhere

[@cphVlwYa]

If you're like me and don't like rounding formulas, you can always take the (phi^x - (1-phi)^x)/sqrt(5) formula and solve for the real part and you get: (phi^x - cos(pi*x)(phi-1)^x)/sqrt(5) Plus, it's always nice to pull a few more constants into the mix :)

[@aymankhan3146]

one of the best of best educators on the youtube platform. Don't you guys agree? (If u don't, u don't have a soul, haha.)

[@nathanwhitten8950]

In addition to your great videos, I really like your T-shirts. The ones I've seen recently, at least, illustrate another bit of math profundity--which is a unique kind since no words are needed.

[@pythagorasaurusrex9853]

Thanks for deriving the Binet-Formula. I have seen (well... similar) other proofs, but yours has been the simpliest of them all. Yeah.. the Fiboncci sequence gives one such a deep insight into math. At university, I had most fun at a seminar about the Fibonacci sequence and still today after 25 years, I learn something new about it every time I see some videos here. (Grüße aus Deutschland)

[@Mathologer]

Wunderbar :)

[@hamidkazemiroodbali9506]

The most fantastic video of you which I've watched till now ;)

[@Mathologer]

What else have you watched ?

[@hamidkazemiroodbali9506]

Mathologer honestly few of them. Of course the I should revise that Conture's infinity is the most fantastic one. I will watch all them but this one was more exciting than these I watched before: 1. Win a small fortune with math 2. A simple trick to design your own solution for Rubik's cube 3. Why do mirrors flip left to right.... 4. Ramanuj infinite root... 5. Death by identity puzzles.... For example the infinity concept is purely interesting and also dangerous. But the general picture of this one brought the existence idea of comprehendible general math system (?!) that some corespondent mysteries have been showed. Maybe it is like a living system :) It was also fun and your presentation was the best in my watched list. Spiral out, keep doing...

[@Mathologer]

Great, keep watching :)

[@minoc2]

I love the interaction with the person behind the camera.. especially the laughs.

[@CristiNeagu]

I never liked the fact that the Fibonacci sequence starts with 1 1. It seems too contrived. I like to think that it starts with 0 1.

[@Mathologer]

Yes, I know, in many ways it makes more sense to start the Fibonacci sequence with 0, 1 and the tribonacci sequence with 0,0,1. Having said that doing so in this video would have made the results that I focus on in this video a bit more complicated to state and also conflicted with what most people are familiar with (starting the Fibonacci sequence with 1, 1 :)

[@CristiNeagu]

Yeah, it's just my own pet peeve.

[@Mathologer]

Sure, absolutely no problem :)

[@martinepstein9826]

I actually like starting it with 1 2 so that f(n) = n for the first three terms. But starting at 0 1 and indexing from 0 means f(0) = 0 and f(1) = 1 which is also nice.

[@c.ladimore1237]

in a quantum sense, there is no zero in nature, no true vacuum, so the closest (non-zero) integer is 1

[@beamathematician2487]

Please make one lecture on partition number, and love your series.

[@johnfedoruk4414]

What a video! Mathologer is the best KZbinr in town!!!

[@morris1818]

Why doesn't the tribonacci sequence start with 1 1 1?

[@AkshayAradhya]

Maurice Merry Exactly 😐

[@hypnovia]

Because 0+1+1=2

[@yakov9ify]

Because you can think of it having a an infinite tails of zeros before the first 1. So the first seed is 1, then the second seed is 1+0+0 = 1, then the third one is, 1+1+0=2

[@Mathologer]

Actually in some ways it makes more sense to start the Fibonacci sequence with 0, 1 and the tribonacci sequence with 0,0,1. Having said that doing so in this video would have made the results that I focus on in this video a bit more complicated to state and also conflicted with what most people are familiar with (starting the Fibonacci sequence with 1, 1 :)

[@blue_tetris]

It really just starts by adding a single 1. The Fibonacci sequence starts the same way. The sequences are preceded by all 0s.

[@simplyyummy92]

I actually figured this out myself not long ago and I was shouting at the screen 'I KNOW!'! :). But didn't know there was ALOT more to this... I love it!

[@reinholdkemper3411]

This Aussie makes Math fun. I'd wish I could go to school again.... I really really like this guy. He really can teach and he comes across so nice n easy...

[@jakeehrlich8113]

Out of curiosity what video editing software do you (Mathologer) use? Do you write up the latex separately and import images into the software or does it have some kind of built-in way of doing this?

[@Mathologer]

I use Adobe Premier to combine my slide show with the video of me dancing around in from of a blank screen. The animations for this video were done in Apple keynote and Mathematica. In particular, the animations of the formulas are based on importing the individual (in this case LaTeX) components into keynote and then using a feature called Magic Moves to have them move from place to place. Pretty easy conceptually, just very, very time consuming :)

[@traso56]

my mission this day is to explains this formula... dis gonna be gud

[@vitalnutrients744]

Sqrt(x+1) = x

[@chrisg3030]

Here's an answer to Mathologer's tribonacci question, but applied to cells, not rabbits (bugs not bunnies): Some cells take an hour to divide, some two. The first single cell is one of the latter, so the population growth goes 1, 1 Then it splits into 2 One of the daughter cells splits into 3, but the other doesn't split and stays 1. So a total of 4. The cell which didn't split at all last time now splits into 3. Of the daughters of the cell which split into 3 last time, one splits into 2 and the other two don't split. So a total of 3+2+1+1=7. I hope this is enough for others to construct a diagrammatic tree to carry on. When you get to 24 you should see 4 single cells, (4x1), 4 cell twin pairs (4x2), and 4 lots of triplets (4x3). Maybe this 1:1:1 ratio stays roughly constant throughout subsequent population growth. For example when there are 504 cells, are there 84 singles, 84 twin pairs, and 84 sets of triplets?

[@hadireg]

I feel so lucky I"m still understanding this and enjoying the mystery-magic numbers still hold 👍👍 Greatly explained but I'll still watch it some more times later

[@steffen5121]

9:05 - Worry not. Mitochondria is the powerhouse of the cell.

[@peterflom6878]

"Who’s going to stop me?" is so wonderfully mathematical!

[@whatthefunction9140]

that giggle though

[@Mathologer]

:)

[@sofia.eris.bauhaus]

best giggle. B)

[@kindlin]

For the bit at 14:00-14:17, you don't explain your very important conclusion very much. I didn't catch what you were trying to solve for, or why, I didn't understand what identify holds between the two equations, I generally had no idea what was happening or why it was important. I'm going to watch the rest of the video, so maybe I'll be able to figure it out, but for this being such a "really important conclusion", you didn't spend a lot of time on it. EDIT: Seeing the F(n-1) cancel out was nice, but it had no great impact, due to my OP. I was still trying to fish out how what you said explained why that canceling out worked so well. How did the -.618 solution come into play?

[@quantumpotato]

+Mathologer you don't need to cheat at 5:30, Phi ^ 0 = 1, right? Which gives 1 +2 = 3

[@strengthman600]

quantumpotato 2+3 still isn't 4 though

[@quantumpotato]

Oh, right :)

[@JorgetePanete]

quantumpotato 2+2=4 4-1=3 quick mafs

[@dillenpantua2449]

Where do you teach?

[@Mathologer]

Monash University in Melbourne, Australia :)

[@ninjakille316]

I'm sure I would be perusing at least a math minor if I had you teaching math classes at my University.

[@TheDoh007]

i found a way of doing tribonacci numbers (i've labeled them with numbers though) the rules are: there are three labels (1,2,3) you start with 1 a 1 becomes a 2 (1 -> 2) a 2 becomes a 2 and a 3 (2 -> 2 & 3) a 3 becomes a 1 and a 2 (3 -> 1 & 2) it goes like this: (n)(labels) (1) 1 (1) 2 (2) 2 3 (4) 2 3 1 2 (7) 2 3 1 2 2 2 3 (13)2 3 1 2 2 2 3 2 3 2 3 1 2 incidentally you can split this into 3 parts in the from when there are 4 labels/bunnies like this: 2 3|1|2 where the first part is a copy of the previous whole, the third part is equal to the whole previous to that, and the second part a copy of the whole prior to that: (n)(labels) (with letters for explanation) (1) 1c (1) 2d (2) 2a 3b (4) 2a 3b | 1c| 2d (7) 2 3 1 2 | 2 | 2 3 (13)2 3 1 2 2 2 3|2 3|2 3 1 2 this then continues to be true (although i don't know why)

[@Mathologer]

Very nice :)

[@stevethecatcouch6532]

I like this better than the baby, teen, adult models. It's more alien. I see the symbols in TheDoh007's chart as 1 earred, 2 earred and 3 earred alien rabbits.

[@stevethecatcouch6532]

*"this then continues to be true (although i don't know why)"* Because the 3 parts act independently and deterministically. The left part of the current line is a copy of the line above. It will produce a copy of the current line as the left part of the next line. So the left part of the new line will be a copy of the line above, i.e., the current line. The middle part of the current line is a copy of the line 3 above it. It will produce a copy of the line 2 above the current line as the middle of the new line.So the middle of the new line will be a copy of the line 3 above it. I'll leave the rest of the argument to the reader.

[@stevethecatcouch6532]

If you start with a 2, you get the tribonacci, but starting 1, 2, 4, 7 ... instead of 1, 1, 2, 4, 7 ... If you start with a 3, you get the tribonacci sequence, 1, 2, 3, 6, 11, 20 ... The lines are still split but ... differently.

[@TheDoh007]

@Mathologer thanks :) @Steve's Mathy Stuff, i don't quite understand your explanation, although i have found it to maybe be more understandable (at least for me) when i lay it out like this: for the first part: (1) 2 23 2312 2312223 2312223232312 231222323231223122312223 23122232323122312231222323122232312223232312 (made by copy-pasting the correct ones) here, it simply always creates the same initial conditions for the third part: (1) 2 23 2312 2312223 2312223232312 231222323231223122312223 ...2312223232312 here, it always creates the previous sets ending conditions for the second part: (least intuitive for me) 1 2 23 2312 2312223 2312223232312 231222323231223122312223 and the thing in-between is always a copy of the one 2 times ago but yeah, the 3 parts act independently after the split at step 4.

[@Peaserist]

this blew my mind duder

[@MyAce8]

7:43 since I couldn't find one in the comments I tried my self. First lets agree that both the luca numbers and the fib numbers follow the rule f(n) = f(n-1) + f(n-2). Then we can see that n*f(n) = n*(f(n-1) + f(n-2)). 5*f(3) = l(2) + l(4) and 5*f(4) = l(3) + l(5) both true so lets write this cool equation 5*f(3) + 5*f(4) = l(2) + l(3) + l(4) + l(5) this can be re written using the rule we stated at the beginning. 5*f(5) = l(4) + l(6). woah holy cow! now all we have to do is realize that we can repeat this process indefinitely by doing 5*f(n) = l(n-1) + l(n+1) which is the same as f(n) = (l(n-1) + l(n+1))/2. Pretty sure my math is right but I'm just a dumb high school student so what do I know. If any of you guys know how to write this more formally that would be cool

[@harrypanagiotidis7370]

so if φ is for golden rectangles, is t (tribonacci constant) for some sort of golden cuboid? Does it generalize to n dimensions?

[@dragoncurveenthusiast]

Interesting thought. I'm curious what others have to say, that's why I comment. But I don't think there is a cuboid that is analog to a golden rectangle, at least not in the way I think about it right now: Such a cuboid would not change the ratios of its edge lengths when you stick a cube to one of its faces (in the same way a golden rectangle does not change the ratios of its edge lengths when you stick a square to one of its edges). If you take any cuboid, two of its edges would need to be the same length, so that there's a square face to which you can glue a cube to (let's say a=b). original cuboid: a*a*c Then, edge c would get longer by adding the cube. So, c' = c+a = c+b. Second cuboid: a*a*c' = a*a*(c+a) However, now you can't add another cube to get the same edge length ratios, no matter how you do it and no matter which edge lengths you chose to start with. You need a square face to stick a cube to it. however, your faces are a*a and c'*a = (c+a)*a. So, the only direction you can stick the cube to is the same direction you did before, which would make the cuboid even longer. Third cuboid: a*a*c'' = a*a*(c+2a). This cuboid definitely has different side ratios than the second one. This shows that you can't have a golden cuboid, if you want to generate each iteration by sticking a cube to one of its faces while keeping the edge length ratios.

[@dragoncurveenthusiast]

I think it would work by adding a 3D L-shape instead of a cube with each iteration. A special L-shape that would turn (if a

[@Mathologer]

Something sort of in this direction: en.wikipedia.org/wiki/Padovan_sequence en.wikipedia.org/wiki/Padovan_cuboid_spiral Also check out Padovan Compare this the spiral of squares that is associated with golden rectangles :)

[@dragoncurveenthusiast]

Cool! I did not know that spiral! I'll have to have a closer look at it, but it seems that the side length ratios are not preserved. Still, a cool thing that I didn't know before! Vielen lieben Dank and greetings from Austria to Australia :-)

[@chrisg3030]

+ Harry Panagiotidis I asked a similar question in the comments column of a recent Numberphile vid about the orchard problem. The orchard is represented by a 2D lattice, in which a line with a gradient of Phi dodges the most number of trees, and is flanked by two "elastic" lines that catch on the first two Fibonacci terms at the lattice coordinates 1,1. kzbin.info?search_query=orchard+numberphile I'm wondering if such a gradient in a 3D lattice would be represented by the tribonacci sequence, with the flanking lines catchng on 1,1,1. Any ideas?

[@shoyuramenoff]

As of 5:40 pm EST on 12-11-17, 16 people are confirmed to have no soul. This video was quite captivating.

[@jacobwerner425]

phi to the n plus phi to the n+1= phi to the n plus 2 because it can be manipulated to phi to the n plus phi to the n times phi = phi to the n times phi squared and then the phis can be divided on both sides to simplify to 1 plus x = phi squared which is the quadratic equation to solve for phi which was seen with the golden rectangles

[@jacobwerner425]

1 plus phi*

[@sakhumzinelson6204]

Dope content 🔥.. How do you edit these videos?

[@DrJG9]

Why isn't a tau used to denote Tribonacci, to go with the phi for Fibonacci? Alternately one could do phi sub 2, which would set up the whole further denoting of series - phi sub m would be where nth number is sum of previous m numbers, if n>m.

[@SylwerFox]

always a pleasure to watch this videos........simple and elegant

[@ekadria-bo4962]

NthNacci number?

[@stevefrandsen]

Excellent video! Best of the season and 2018 to you.

[@yakov9ify]

For the tribonachi sequence you can think of the rabbits as having three stages of growth, first they can't make babies, then they make 1 pair, and then they make 2 pairs.

[@dragoncurveenthusiast]

My thoughts exactly. If every newborn rabbit pair is a 1, the middle stage a 2 and an adult pair a 3 you get: 1 -> 2 (growing up) 2 -> 3 1 (growing up and having a baby pair) 3 -> 3 1 1 (having two baby pairs)

[@dragoncurveenthusiast]

Here's an attempt of making a readable tree in a youtube comment: 1 = 1 pair 2 = 1 pair 3 1 = 2 pairs 3 1 1 2 = 4 pairs 3 1 1 2 2 3 1 = 7 pairs 3 1 1 2 2 3 1 3 1 3 1 1 2 = 13 pairs (each row is a generation, each pair stays within their column. The youngest children always appear directly right to the parent pair)

[@Mathologer]

That works :)

[@martinepstein9826]

I know in general you can solve the n-fibonacci problem by expressing the recurrence relation as a linear transformation in R^n and diagonalizing. But since we're basically given the characteristic formula right off the bat it looks you can skip a lot of that work.

[@Mathologer]

Absolutely there are a lot of ways to skin a cat (not that I would ever skin a cat :) There is also the generating function approach which is very nice too :)

[@martinepstein9826]

Thanks for the reply! Something that just occurred to me; an important step in this process is expressing integers as linear combinations of roots of a polynomial. This seems closely related to the theory of field extensions and treating them as vector spaces over Q, and that's the next topic in the abstract algebra text I'm working through :)

[@MrRyanroberson1]

15:00 i guess phi red also makes this equation work for negatives? because if f(1) and f(2) are 1, then f(0) must be 0, and f(-1) must be 1, f(-2) must be -1, so on, alternating polarity.

[@santhoshwagle9857]

Your videos are awesome... Great content... Nearly explained... I have only one request... Please change the back ground color from bright white to something soothing to eyes...

[@vkilgore11]

You top yourself every video; this was great.

[@nikolaalfredi3025]

Good explanation, I didn't knew about tribonacci numbers until now.

[@DarthCalculus]

This is great! Beautiful mathematics, you're clearly having a great time... It's like watching a jazz show

[@HarryFortyTwo]

Outstanding! Brings back the joy into mathematics!

[@Mathologer]

It never left !!! :)

[@__-cx6lg]

@0:50 personally, I prefer to think of it as starting with 0 and 1, but that's just me also, the tribonacci sequence would then start 0,0,1, which to me also seems less arbitrary. n-onacci would start with (n-1) zeroes, then a 1, instead of just copying off of the previous sequence.

[@avi12]

This shirt full of golden spirals is amazing!

[@amitliber75]

great video! very enlightning. i enjoyed the riddle in the end. maybe finish every video with a riddle?

[@Mathologer]

Definitely a great way to get people more engaged :)

[@amitliber75]

the answer i find is that the bunnies have 3 stages of growths instead of 2 and in the third stage they make two babes instead of 1. is this the only answer?

[@danielinfinito6304]

Amazing video. Thank you very much.

[@biometrix_]

Closed Capitions Error: At 14:01, it says "solutio,n" instead of "solution" Another at 17:19, the last word is "irrational" instead of "imaginary"

[@robertbright7581]

Love your videos! Thank you! Where does one find a T-shirt like yours?

[@Mathologer]

Here you go: www.zazzle.com.au/fibonacci_parrots_t_shirt-235568206086965961

[@ThatLoganGuyYT]

It would be crazy if you see this comment but Im just a sophomore in highschool and I had a debate on the probability of landing on nth space in a game after any amount of rolls, using either a 1, 2, or 3 jump where any of the three could be picked at random. My friend argued it would be 1/3, but then I brought into account that landing on the 2nd sqaure has 2 possible ways, either just the 1/3 pick of a 2, or by picking a 1, followed by another 1. He complied with my reasoning but said past the first sqaures it would fall to 1/3. I then brought out my notebook and wrote down all the possible outcomes, and saw a pattern emerge. The first sqaure had 1 possible way, 2nd had 2, 3rd had 4, 4th had 7, 5th had 13, ect. The amount of possible outcomes is equal to the sum of the previous 3 outcomes. I then tried to figure out the formula, but was stumped. I knew it would be in correlation to the Fibonacci sequence, but for weeks I tried and couldn't. Those weeks have led to now, and by chance it is exactly what I needed to know. Thank you for this great video.

[@dekay5555555]

Take the Fibonacci sequence that hasn't a beginning or end. . . . 13,-8,5,-3,2,-1,1,0,1,1,2,3,5,8,13 . . . Invert this to a column. Make a total of three columns of this shifting the first up one set of digits and the last down one set of digits. This will give you rows (13,-8,5),(-8,5,-3),(5,-3.2),(-3,2,-1),(2,-1,1),(-1,1,0),(1,0,1),(0,1,1),(1,1,2),(1,2,3),(2,3,5),(3,5,8),(5,8,13). Take the row (0,1,1), the middle number will always represent the amount of √5s We sum the number and divide by 2; 0+√5+1 = 3.236068, divided by 2 equals 1.618034 or phi or Ø^1. Row (-1,1,0) -1+√5+0 = 1.236068, divide by 2 equals .618034 or Ø^-1 I find this next row to be most important for it answers why Ø^0 =1 (1,0,1) 1+0√5+1 = 2 divided by 2 = 1.000000 You may see the Lucas numbers here, but like all great mathematicians, they already know what I am presenting here :)

[@mk-allard3788]

Hooray! New video!

[@stevekiley6121]

That is definitely the nicest T-shirt I've seen you wear.

[@Mathologer]

Definitely my nicest Fibonacci themed t-shirt, but I would not say my overall nicest one :)

[@Giacche]

I really like your videos, but I have to ask that... what comes first? The topic of a video or your t-shirts?

[@Mathologer]

Well I've got a collection of about 200 math themed t-shirts. So, usually when I decide on a topic I already have a t-shirt to go with it :)

[@asnierkishcowboy]

Makes one think about the p-bonacci sequence.

[@kennytran4]

Nice to see the QEDCat team back together :-)

[@gauravmanwani9148]

Great video man! Just wanted to suggest that you should have made this one in 3-4 parts of 10 min each. It would have produced a fantastic journey, and we would have gotten even more insight. :)

[@Mathologer]

I've actually tried the multi-part approach before with these two videos kzbin.info/www/bejne/qHKrlI19n7CJrMk and kzbin.info/www/bejne/mXbLfYGJaqZ7qdU Somehow did not work that well :)

[@MrRyanroberson1]

expanding the Fibonacci numbers into the negative Ns: -8 5 -3 2 -1 1 0 || 1 1 2 3 5 8, and that alternation is from the phi red. expanding the tribonacci: 7 5 -8 4 1 -3 2 0 -1 1 0 0 || 1 1 2 4 7 13 quite the monster in the bedroom, as it were. i'm stunned that such a normally passive sequence can become so sporadic.

[@anthonyfrantz884]

I am taking a bow too , I had solve so many times x^2=x+1 and do the analysis for it but I never thought it had that meaning ,my mind just blow ,I am taking a bow to you ! Thanks

[@MegaPhester]

I was just thinking about a fibonacci-like sequence the other day where you add up all the numbers in the sequence instead of just the last two. I was too dense to get it then but this video made me realize that would just make the powers of two, if you start with 1, 1. Then the n-bonacci sequences will actually approach the powers of two as n grows and their respective "golden ratios" will approach two. Pretty neat.

[@binary6666]

2:16 WHOA!!!!

[@in2infinitygeometry]

Do you have any explanation as t why Phi can be formed from a square, triangle and pentagon? Those just happen to be the faces of the Platonic Solids. Is there a relationship? I would be interested in knowing more about the snub cube relationship to the Tribboancci numbers. Lastly, what about the Plastic numbers. How do they relate?

[@Kram1032]

I love that Khan Academy dig! Also, great video all around :)

[@vivaldirules]

Higher order "onacci's"?

[@streak1burntrubber]

I was thinking about this too. Quadonacci, Quintonacci, Sextonacci, Heptonacci, etc. And you'd think that all these would have ratios converging to various numbers. My question is, if you take all those ratios, what number do those converge to, if they do converge?

[@AntiChangeling]

The SS?

[@Schindlabua]

Thinking about Quadronaccis makes me hungry.

[@livedandletdie]

AntiChangeling, haha, please Gestapo it. I'm dying.

[@robertdeaton4720]

They converge to 2 because the sum of 1 + all powers of 2 less than a given one sum to that power of two. The Infinonacci: 1, 1, 2, 4, 8, 16, 32, 64, 128, 256, etc.

[@yagalamaga]

at 6:11 about the Luca numbers - if you do a square root of 1.618... you get 1.272.. with the closest integer as 1. meaning the series would be: 1, 2, 3, 4, 7.... which works better than just changing arbitrarily a(i=1)=2 into a(i=1) =1. What do you say? @Mathologer

[@quentindimeoo]

My mind just got blown watching this video, this is why we f***ing love mathematics !

[@MrPictor]

Great video again and excellent prononciation of François Édouard Anatole Lucas. I wonder if you could make a video about the concept of limits. I think there's good stuff to be told, maybe how it evolved from its invention to today.

[@JimmyLundberg]

Hi Burkard! New subscriber here (few weeks), and I'm just curious about whether you play chess and (if so) what strength you're at. I think most would agree that both maths and chess involve logical thinking, but from my experience most chess players aren't that into maths and vice versa, which surprises me. You might have mentioned chess in an older video, I just haven't had time to watch them all yet. I imagine that a geometrician who picked up chess would become good very quickly.

[@KipIngram]

3:00 - That formula isn't really that bad. You only have to calculate that quantity [1+(19-3*sqrt(33))^(1/3)+(19+3*sqrt(33))^(1/3)]/3 once. Call it u. Then it's just (u/3)^n / [4*u-u^2-1]. If you stored that first value in your calculator somewhere, then it becomes quite easy.

[@miloszforman6270]

True. The formula is stated in this easier form once again at 17:51. This "obscure" quantity [1 + (19 - 3*sqrt(33))^(1/3) + (19 + 3*sqrt(33))^(1/3) ] / 3 simply being the one real root of x³ - x² - x - 1 = 0.

[@JohnDlugosz]

In my undergrad discrete math class, I recall a different derivation of the fib formula that was quite impressive. Starting with a recurrence relation, it finds a solution involving complex numbers. Then, removing the complex numbers produces a form that has trig functions and e. And that whole mess produces an integer, and doesn't need rounding to an integer.

[@Mathologer]

Yes, there are a couple of different and all very beautiful ways to "skin this cat". Eventually I'll also talk about some of these other methods :)

[@MichaelSartore]

TRIB-TASTIC!! Love your work!

[@gius8068]

Really love your videos! Keep it up!

[@xnick_uy]

Yet another cool fact is that the n-th term of any other Fibonacci or Tribonacci sequences (with different starting values) can be found by using the same formulas with 'minor' changes (we might have to give up on rounding and cope with the complete expression in some cases).