Phi and the TRIBONACCI monster 

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

James Bond will stop him.

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

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.

He's a mad mathematician!

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? ;)

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.

It's like Neuron gym

Indeed

your feelings are irrational

satisfying is not a feeling

@@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

That's nice :)

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!

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

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

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.

Good idea :)

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

Switch #1 and #2

TacoDude314 why? He's right

2 is faster than 1

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

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

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

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

I have no soul!

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

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

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)

well, i starting to feel it.

@ 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

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. :)

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

expchrist souls don’t exist

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

That's absolutely right :)

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

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

we need an n-part series. B)

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.

That's also an interesting fact :)

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.

K van der Veen ((&,~>}\§\«|§>||℉°§¤¤¤℉°¤,¢\|>¤= c cgy%+

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.

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?

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.

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).

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. 👍

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.

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 :)

Yeah, it's just my own pet peeve.

Sure, absolutely no problem :)

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.

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

Very nice :)

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.

*"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.

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.

@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.

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

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)

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)

That works :)

1 plus phi*

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 :)

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 :)

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 :)

Santiago Arizti I found this, but I thought it was only a property of primes until I mistyped a 7 for an 8 and was lazy. Btw, look up phi infinite series, it clears up why this is.

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

:)

best giggle. B)

Sqrt(x+1) = x

Phodd Antomity What's even more interesting is if you start with 1,1,2,3 and then form new terms by adding last four terms to get quabonacci sequence. KABOOM😂

1,1,2,3,6,9,18,27,54,81,162,243,486,729 1,1,2,(3x1),(3x2),(3x3),(3x6),(3x9),(3x18),(3x27),(3x54),(3x81),(3x162),(3x243) 1,1,2,(3x1),(3x2),(3x3),(3x(3x2)),(3x(3x3)),(3x(3x(3x2))),(3x(3x(3x3))),(3x(3x(3x(3x2)))),(3x(3x(3x(3x3)))),(3x(3x(3x(3x(3x2))))),(3x(3x(3x(3x(3x3)))))

Phodd Antomity But the ratio between sequential numbers will always keep oscillating between 2 and 1.5, which is a bit more boring then the convergence to the golden ratio.

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 :)

Wunderbar :)

Failing to justify arbitrary assumptions is just about the best definition of mathematics.

rightwraith | ...within the extant axioms of arithmetic and algebra. Down on the ground!

Time to switch to 1/2 speed, RU-vid makes it possible :) Added attraction, I really sound quite funny at this speed.

Nice idea but not much help to someone like me with hardly any knowledge of algebra. When the experts say something like "Let's just play with that equation for a moment", I think well, you have fun and I'll trust you to get it right. But never mind. I always enjoy your videos and always take something away.

Yes, there's a characteristic polynomial for each n-bonacci sequence, which is a polynomial of degree n. Namely: p_n(x) = xⁿ - xⁿ⁻¹ - xⁿ⁻² - ... - 1 = xⁿ - (xⁿ⁻¹ + xⁿ⁻² + ... + 1) The limiting consecutive-term-ratio, r, is then the positive real root of p_n(x) = 0; i.e., p_n(r) = 0. Now if you take: q_n(r) = (r-1)p_n(r) = rⁿ⁺¹ - rⁿ - (rⁿ - 1) = rⁿ⁺¹ - 2rⁿ + 1 = rⁿ(r-2) + 1 Then q(r) = (r-1)p(r) = 0 So: rⁿ(r-2) + 1 = 0 rⁿ = -1/(r-2) = 1/(2-r) n ln r = ln[1/(2-r)] n = ln[1/(2-r)] / ln r and as n→∞, r→2⁻⁻

Very nice :)

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

What else have you watched ?

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...

Great, keep watching :)

Maurice Merry Exactly 😐

Because 0+1+1=2

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

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 :)

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

Actually, in some ways the best way to start the Fibonacci sequence is with 0, 1 (1,2,3,5...) and the best way to start the Tribonacci sequence is 0,0,1(1,2,4,...). For a number of reasons (historic, common usage, how things align, etc.) it's better to start a little bit into these sequences in this video :) You can indeed start with whatever non-negative seed numbers you want, you'll always end up with the same limiting ratios. That's definitely true for Fibonacci like sequences, but may very well also be true for tribonacci like sequences.

Duh... yes, you have a point there. Leading 0's makes even more sense than leading 1's. OK, point taken :). And the fact that the limit of these sequences seems to be independent from the starting numbers makes (at least intuitive) sense. If the numbers get big (and they all get big, don't they? :-) then the absolute values don't matter all that much -- I mean, the ratio of, say, 149:99 is almost the same as 151:101 for example, so the exact values are irrelevant and all that is left in the long run are their ratio's. And yes, I know this is not a scientific argument, but 90% of all maths starts with intuition anyway... Ha!

Absolutely, depends which formula you use to solve the cubic equation :)

If it had three eyes, it'd probably be a lizard... en.wikipedia.org/wiki/Parietal_eye

Definitely the generating functions approach is also supernice, and a video on generating functions is actually on my to-do list :)

It never left !!! :)

I've actually tried the multi-part approach before with these two videos ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-rAHcZGjKVvg.html and ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-cEhLNS5AHss.html Somehow did not work that well :)

Yes, it's really very easy to show this :)