Lucas Numbers and Root 5 - Numberphile

[@numberphile2]

Part 1 is at: kzbin.info/www/bejne/mZvbY2aXgsp1obM Check out some Numberphile T-Shirts and other stuff: teespring.com/stores/numberphile

[@tjreynolds685]

what if you didn't use root 5 for this? what if you did root 7 instead? root 6? what if instead of the square root, you used the cubic root? would you still be able to come up with some "fibonacci-esque" addition?

[@losthor1zon]

I ran into the root 5 relationship with the golden ratio back in high school, when I tried to figure out a very simple quadratic equation: x^2 - x - 1 = 0. For a geometrical representation, check out the ratios of the lines in a pentagram. The larger lines to the smaller are in golden ratio proportions.

[@PC_Simo]

@losthor1zon I figured that equation out. Very nice solutions, indeed 🙂.

[@wyattstevens8574]

5:20 Is the formula for the generalized sequence related to how many lots of the Fibonacci and Lucas numbers make it up? The Fibonacci and Lucas numbers are complementary Lucas sequences.

[@julian_ossuna]

A trilogy made of four parts ...a Parker trilogy.

[@nanigopalsaha2408]

Tell that to Douglas Adams

[@Write_with_me_gknotes]

a 4 sided triangle :)

[@olanmills64]

😂

[@ronniemouser9752]

Nearly spit out my drink

[@elij.514]

@@Write_with_me_gknotes Someone get Cliff on this immediately

[@MisterAppleEsq]

I'm really impressed that Matt managed to work out the formula for the arbitrary numbers completely by himself without using any other kind of tool.

[@General12th]

Brain like a supercomputer!

[@MrAthoOome]

Not complicated really, it's just a matter of solving a linear difference equation and that's pretty basic.

[@MisterAppleEsq]

+@@MrAthoOome True.

[@giuzeppeedreimeimban1019]

Hes my favourite math person of the parker squarish kind 😁

[@Surfinite]

It doesn't work though. It's wrong.

[@unvergebeneid]

"The golden ratio is the marketable version of root five." Matt does not get enough credit for this quote!

[@shruggzdastr8-facedclown]

It'll feature prominently in his next book, Humble Phi 😏

[@blacxthornE]

Parker Square flash at 7:28

[@audiocancer]

True marketing! 👌

[@montano0222]

Then the Parker Square T-Shirts link on top

[@41-Haiku]

I caught that and I was so happy.

[@nicholasleclerc1583]

“I [didn’t] get that reference”

[@00bean00]

WOW!

[@hariskayani4703]

"The golden ratio is the marketable version of the root 5" should be on a t-shirt

[@_1derscore]

marketale plushie

[@quahntasy]

It's so amazing to see Matt come up with formula from those numbers without using any computer or something.

[@TheGreatPurpleFerret]

I love that Matt pointed out where the rounding was hidden in the last video. I never would have caught that and honestly it helps explain this a whole let better just with that tidbit.

[@loreleihillard5078]

Is that a calculator still in the box in the background? New calculator unboxing video confirmed

[@ffggddss]

The *real* reason that √5 keeps popping up in "Fibonacci-esque" sequences is the iteration rule coefficients, A₊ = 1·A₀ + 1·A₋₋ ; i.e., 1·A₊ - 1·A₀ - 1·A₋₋ = 0 and the corresponding quadratic that comes from that: x² - x - 1 = 0, whose discriminant is b² - 4ac = 5, the radical of which appears in the quadratic solution. If you generate a sequence with different iteration coefficients, you will get a different limiting ratio, which is a zero of a different quadratic, with a different discriminant. Which might in itself, make an interesting further addendum to these two videos. How to make a "designer" pyrite* ratio . . . * pyrite = "Fool's Gold" Fred

[@gwahli9620]

And because it's the solutions to a quadratic equation, there are actually TWO golden ratios (sqrt(5)-1)/2 is too. Which caused some confusion for some people but it just means that as a ratio 2:1 is the same as 1:2 or 1:1/2 ... simply that the bigger value is twice as big as the smaller one.

[@ffggddss]

Well, actually, the other solution is negative. x² - x - 1 = 0 x = φ = ½(1 + √5) = 1.61803..., and x = ½(1 - √5) = -1/φ = 1-φ = -0.61803... In the Fibonacci sequence, the ratio of consecutive terms, Fᵢ₊₁/Fᵢ → φ as i → ∞, and Fᵢ₊₁/Fᵢ → -1/φ as i → -∞ Fᵢ = ... , -21, 13, -8, 5, -3, 2, -1, 1, 0, 1, 1, 2, 3, 5, 8, 13, 21, ... Oh, and the bigger solution is -φ² = -(φ+1) = -2.61803... times the smaller one, not twice. Fred

[@Maharani1991]

Sweet. :)

[@jessstuart7495]

How about A₊ = 3·A₀ - 1·A₋ That gives you a discriminant of 5 also. I vote for calling (3 + sqrt(5))/2 = 2.618 the "Pyrite ratio". Interestingly if you calculate this sequence starting with 1,1 You get every other Fibonacci number. 1,1,2,5,13,34,...

[@TimMaddux]

Back in the day it was originally called the quadratic fivemula, until the root 5 haters ruined it for everyone.

[@dalitas]

You being wrong isn't a theme, it's a meme!

[@anononomous]

Themes and memes are both closely related, they just happen to be expressed by a different sequence of letters.

[@silkwesir1444]

Also, actually the term "theme" fits much better (though "meme" isn't exactly wrong either).

[@pronounjow]

It's a running gag!

[@silkwesir1444]

Jo Reven yes, that term fits even better than the other two.

[@thechemuns74]

Theme VS Meme is the new Lucas vs Fibonacci.

[@MahraiZiller]

Should we then call the golden ratio “root Phive”?

[@Maharani1991]

:D

[@QuasarRiceMints]

*slow claps*

[@st.paulmn9159]

I’m here cuz I think one brake light is fine

[@PC_Simo]

@Mahrai Ziller Indeed we should 😁. I was actually thinking the exact same thing.

[@PC_Simo]

@St. Paul MN Don’t you mean: ”phine”?

[@Dankey_King]

3:34 Love how Matt's password is "PrkrSqr"

[@chaschtestark7973]

go watch the video "the parker square" from numberphile, and you will get the joke ;)

[@ObjectsInMotion]

@@chaschtestark7973 you're the one who isn't getting the joke.

[@BastienHell]

"six! who knew?" Laughing so much it's painful

[@WAMTAT]

We can dream.

[@recklessroges]

*knew

[@BastienHell]

Thanks, missed the typo

[@Zarrykotter]

Parker being brilliant as always! - still chuckling

[@rosepinkskyblue]

I’ve watched this too many times and laughed every time

[@pigeonfog]

I love the nod to the parker square when matt messes up the 5.

[@robinsparrow1618]

i actually memorized the first 9 digits of root 5, because of my interest in phi interestingly the first 9 digits of root 5 + 1 contains three palindromes in a row 3.23 606 797 which is one of the things that made it fairly easy to memorize

[@moneym0ney]

That is actually quite awesome but only holds true if you explicitly round down.

[@steffahn]

since they're saying "the first 9 digits are," and not "the number, rounded to 9 digits, is," I don't think there's anything one needs to be more "explicit" about missing, except for (maybe) use base 10 ^^

[@PC_Simo]

@@steffahn Exactly.

[@johnchancey3941]

5:06 - "That's my birthday!" 😂😂😂😂😂😂

[@MushroomManToad]

I always wanted to be first for a Matt Parker video. But here I am, second... Guess that makes this a Parker Comment

[@vittoriosellingerstiatti4619]

Cameraman: "Oh, that's my birthday :D" Matt: "there you go :)" Cameraman: "No, that's not" What a savage.

[@LeoStaley]

He never bothered memorizing the fibonaci or lucas numbers, but he clearly memorized the exact decimal values of the ratio between them. A true legend.

[@Ramiprops]

The expression he gave further simplifies if you substitute √5=2φ-1 (golden ratio definition with the √5 isolated) in all the numbers which have a √5 factor (including the 5s and one from the 10 in the denominator, since 10=2*√5^2) and becomes G_n= [φ^n*((2-φ)A+(φ-1)B)/(√5)], which is much much more visually appealing than the expression Matt gave.

[@fcturner]

Very nice 👌🏽

[@NikhilSingh42]

"If i say 'this is interesting' enough times, it will be" :D

[@aspiringcloudexpert5127]

Whoever came up with the name "Golden Trilogy" deserves a cookie.

[@julian_ossuna]

Interestingly, this trilogy is made of 4 parts. A Parker trilogy.

[@TheDWZemke]

Sure we would have PI and the cookie number...

[@Nossairito]

Root 5 is the Iniesta to Golden Ratio's Messi.

[@ig2d]

Or the Buzz Aldrin to the golden ratio's Neil Armstrong. (Buzz Aldrin, the only person famous for not being famous)

[@ericvilas]

The big reason why many mathematicians like Fibonacci numbers and phi is because of its continued fraction: 1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(...))))))). I'd say that reason alone makes it a big deal. Another thing: when you talk about the golden ratio, you should talk about 1-phi as well! I'd definitely consider it important enough that you should give it another name. Let's call it chi. Chi is also phi's conjugate, its negative reciprocal, AND, what do you get when you subtract chi from phi? The square root of 5! So, phi and chi combined encapsulate all the beauty that people give to the golden ratio, all the beauty of the Lucas AND Fibonacci numbers, AS WELL AS all the beauty of the square root of 5. I love those 2, phi and its little brother, being all mathy and stuff. EDIT: Another reason why you should've talked about chi: you don't need to make approximations and your precious Lucas numbers always look a lot neater! L(n) = phi^n + chi^n, exactly, always. F(n) = (phi^n - chi^n) / (phi - chi), also exactly. Ooh, you should make a video about ways to make non-radical numbers by using conjugates!

[@chrisg3030]

Another beautiful thing: (Phi^1)+1 = Phi^(1+1) and Phi+(1^1) = (Phi+1)^1. The golden bracket shift.

[@ericvilas]

@@chrisg3030 ooh! Well, the first one, anyway, the second one seems like it would always be true?

[@chrisg3030]

Dammit you're right. Ok, so just the first one.

[@matthewbertrand4139]

Well, since φ - χ is equal to √5, I would argue it should be expressed in that form in the denominator of your rule for F(n), but that's otherwise amazing, man. I do get that you were just emphasizing their prevalence in these rules, but for practical purposes I'd use the radical.

[@chrisg3030]

Eric Vilas Popular name: Geometric, Fibonacci, Narayana First few terms: 1 2 4 8 16 32, 1 1 2 3 5 8 13, 1 1 1 2 3 4 6 9 Recurrence relation: An = An-1 + An-1, An = An-1 + An-2, An = An-1 + An-3 Ratio value: 2, !.618. . ., 1.4656. . . Designation: 2, Phi, Mu Bracket shift equation: (2^0)+1=2^(0+1), (Phi^1)+1=Phi^(1+1), (Mu^2)+1=Mu^(2+1)

[@yaj126]

4:02 Wow that's some advanced math you just did there off the top of your head

[@jaipod4561]

Did anyone else notice the "blink and you miss it" Parker Square at around 7:25?

[@seedschi]

"Root five is more the Ghostwriter for the Golden Ratio...." absolutely fantastic :-D Very nice video, I enjoy your number-juggling a lot!

[@maartendas1358]

Probably no news to anyone who studies this but I just found this out, and I didn’t find any comment yet describing this so I decided to just put this out there. When you write down the Lucas numbers and start adding the first and third number, the second and fourth, fifth and seventh, sixth and eighth and so on, the sequence you get consists of multiples of fives, following the same pattern as the original sequence (each new number is generated by adding the previous two together). Like this: 5, 5, 10, 15, 25, 40… Kinda neat when you consider how root 5 is so essential to the Lucas numbers and to similar sequences.

[@JordanMetroidManiac]

Another interesting note is that a right triangle with an angle of π/5 yields sides with lengths defined by the golden ratio. The golden ratio can be obtained with 2cos(π/5). How might the power series expansion of cosine help determine why π and Φ are related in this way? And by a hypothetical syllogism, Φ must also be related to e. We could say e^(5icos^-1(Φ/2))+1=0, which relates e to Φ.

[@josephbentham3901]

Sneaking in a Parker square. Loved it!

[@jannieschluter9670]

I love the Golden Ratio and all and everything about it!

[@bunshichi93f]

"The Golden Ratio is the marketable version of sqrt(5)" Don Draper approves!

[@chrisg3030]

bunschichi93f I guess it would be sold in a spray can (sqrt)

[@defuncttobedeleted3675]

Beautiful maths in the previous video! Very pleasing arguments Matt :D

[@adnanchaudhary5905]

"If I say interesting enough times it will be" goes for everything in maths mate

[@fluffywhiteducky]

"Its like the ghost writer for the golden ratio" Hahahahahaha

[@TheScoobs]

What's do the numbers flashed at 7:27/7:28 mean? 29^2 1^2 47^2 41^2 37^2 1^2 23^2 41^2 29^2

[@sander_bouwhuis]

I also downloaded the video to have a look at the flashing subliminal message. It's a Parker message.

[@dubsed]

Favorite thing about root 5 is that it is the hypotenuse of the 1, 2 triangle, which itself forms the dihedral angle of a dodecahedron.

[@aok76_]

The general form is certainly pleasing. You're an inspiration Matt!

[@n.l.4025]

The Fibonacci Sequence is the base sequence for the Lucas Numbers and the other sequences featured here. The Fibonacci Sequence is the number 1 sequence with a basis of root 5.

[@frankharr9466]

I think it's neat that with this formula, you can find an A and B for the Fib and Lucas numbers.

[@legendgames128]

2:31 a few seconds before you revealed it, I guessed it was root five since it was kinda square root of some number like, and it was closer to 4.

[@honkynel]

Really enjoyed.

[@johnnye87]

The fact that 1/phi + 1 = phi makes it seem like the series starting with those three values should be significant somehow...

[@sander_bouwhuis]

As the video already stated... it doesn't matter with which (positive) numbers you begin. They all tend to the golden ratio, and they all result in splitting the components in two Fibonacci series. The johnye87 numbers : 1/φ+1 φ φ + 1(1/φ + 1) 2φ + 1(1/φ + 1) 3φ + 2(1/φ + 1) 5φ + 3(1/φ + 1) 8φ + 5(1/φ + 1) 13φ + 8(1/φ + 1)

[@davidgillies620]

A generalised Lucas sequence is x_n = A x_(n-1) - B x_(n-2). This gives you a characteristic equation of x^2 - A x + B = 0. Let D be A^2 - 4B and positive. Then the roots of the equation are a = (A + sqrt(D))/2 and b = (A - sqrt(D))/2. You can form two sequences of integers, U_n = (a^n - b^n)/sqrt(D) and V_n = a^n + b^n. For A = 1 and B = -1, U is the Fibonacci numbers and V is the Lucas numbers. But all choices of A and B for which D is positive converge on a ratio for successive values of a, which for the Fibonacci and Lucas numbers is (1 + sqrt(5)/2, the Golden Ratio.

[@n.l.4025]

Finally, I now get to understand this and get to see the “Parker #5”! For everything, there is a Parker Something! (Eccl. 3:1-4)

[@atrumluminarium]

I love how everytime a Parker Square flashes on the screen we get a PS T-shirt card lol

[@PC_Simo]

Root(5) is the core or the heart of the Golden Ratio.

[@KaroxNightshade]

They put in the Parker Square for one frame. Anger.

[@rebmcr]

The most amazing thing I saw in this video is the divisor in the generalised formula. That thing is a PURE ten: not an artefact of the base-10 counting system, but the real true value 10 in its own right. You don't see that very often!

[@sander_bouwhuis]

I agree. That threw me for a loop. Because 10 has the factors 2 and 5 it almost always gets factored out. Most numbers (60%) are divisible by either 2 or 5 or both.

[@usernamenotfound80]

I must disagree with the statement that φ is the marketable version of √5. Instead, I'd say that out of the two, φ is the fundamental one, whereas √5 is merely some fallout that you happen to get additionally. The main reason for this is that if you look at the field ℚ(φ) (or ℚ(√5), they're identical), its ring of integers is ℤ[φ] and not ℤ[√5]. For the same reason, I think that ζ₆ = (1+√-3)/2 (a root of unity!) is more relevant than √-3 and more generally (1+√d)/2 is more relevant than √d for square-free integers d that are congruent to 1 mod 4.

[@cubicardi8011]

How did you make those integer or rational number number range characters?

[@usernamenotfound80]

ℚ and ℤ are in Unicode. I just copy-pasted the symbols from Wikipedia: en.wikipedia.org/wiki/Mathematical_Alphanumeric_Symbols

[@cubicardi8011]

@@usernamenotfound80 thx

[@matthewbertrand4139]

I've got a math keyboard called dxMath which has those symbols on it as well as many others, if you don't want to have to do that every time.

[@cubicardi8011]

@@matthewbertrand4139 yes! That's what i always searched for. Thanks alot

[@EllieSleightholm]

Amazing. I'm a university maths youtube vlogger and I can't tell you how much numberphile has helped and inspired me over the years! :)

[@RadeticDaniel]

Matt: - Do you recognize it? - No one will, you know, no idea - and this, i think this is a little bit upsetting, no ever recognizes this number Don't worry, Matt, we never forget the first square root calculated by bisection on a calculator with no square root button. 2.2360679 will never be forgotten

[@captapraelium1591]

This series was apparently edited by a nerdy version of Tyler Durden.

[@recklessroges]

I think it was his brother Parker Durden.

[@Nateಠ_ಠ]

I am Brady’s complete lack of surprise.

[@JonathonV]

Of course there's a game of SET on the shelf behind Matt! I tried to explain that game to my math prof who did his doctoral thesis on Latin Squares, thinking he'd be cash at it, but actually he found it pretty difficult to follow. It's one of the few games where kids, en masse, tend to do BETTER than adults because their pattern recognition skills are at the forefront of their minds.

[@parreiraleonardo4189]

One more interesting fact about Lucas numbers (Ln) and Fibonacci numbers (Fn): Fn•Ln=F2n

[@Yupppi]

Congrats Matt Parker on doing Arbitrary numbers formula on the go!

[@andrewjetter7351]

The subliminal Parker Square when he messed up writing a 5 was perfect.

[@shruggzdastr8-facedclown]

Also, if you sequentially subtract the Fib #s from their corresponding Lucas #s and divide those differences by 2, you get another set of Fib #s (offset by one step).

[@DjVortex-w]

Root 2 gets all the glory. Root 5 doesn't get even nearly as much love as it deserves.

[@Petch85]

1:35 Can someone explain this notation to me. Way is this smart? What benefits dos it haver over the normal notation? What do one uses for multiplication? and so on.

[@stevethecatcouch6532]

Matt is just careless about his placement of decimal points.

[@captapraelium1591]

Φst

[@MisterAppleEsq]

You're closer to πst, really.

[@WhattheHectogon]

πth* c'mon, duh

[@captapraelium1591]

πrd* XD

[@98danielray]

fist :v

[@konstantinbachem9800]

eth

[@alexpotts6520]

A small caveat - for these "arbitrary numbers", the formula doesn't actually work for small n. I imagine what's going on here is that the round error gets smaller and smaller for larger n, but for small numbers it's significantly larger than 1 and therefore just rounding doesn't get the job done.

[@DaTux91]

There should be a Parker Square somewhere in London.

[@juustgowithit]

I love this video so much

[@meisam9592]

This is better than the original video.

[@shruggzdastr8-facedclown]

(@ 7:28 ): You didn't "fudge" your sqrt5, Matt -- you gave it a go!

[@scimatar5]

Root 5 is the drummer, the golden ratio is the front man. Everyone knows the front man of a band, but smart people know that he wouldn't be there without the drummer. In fact, most of the band wouldn't be there without the drummer.

[@WildEngineering]

Okay so idk if this means anything but if you try to mirror the numbers IE) after the 0th digit, make the -1st -2nd etc. They both become mirrored across the 0th digit but their signs alternate eachother.

[@slightlokii3191]

That Parker square sublim at 7:30 hahahaha

[@dfs-comedy]

Now we know who is secretly pulling all the strings.

[@Godisdeadamen]

How about instead of writing ((3√5 - 5)A + (5 - √5)B)/10 instead write (3A - B)/2√5 + (B - A)/2 ? I guess the former gets to the amount of A and B quicker, but the latter looks into the relationship between the two.

[@fuseteam]

so (3A-B)/2√5=(A-B)/2? which would mean 3A-B=(A-B)√5..........oh my

[@aka5]

@@fuseteam wtf are you talking about... That's not at all what OP implied

[@fuseteam]

i simply did some maths!

[@aka5]

@@fuseteam ok, I'm sorry for being rude... I thought you meant that OP had made a rearranging mistake. You can prove that ur last line is correct by multiplying both sides by zero :p

[@fuseteam]

lol and thanks for being sincere and verifying :3 -also multiplying both sides by zero even proves 1=2 :P- what i did was assume that the golden ratio-esque sequence number that we are looking for is 0 aka (3A-B)/2√5+(B-A)/2=0 in which case we get 3A-B=(A-B)√5 which if we work it out further brings us 3A-B=A√5-B√5 so that (3-√5)A=(1-√5)B which is interesting :3

[@ImAllInNow]

Matt, I can't believe you didn't mention that the Grafting Constant is also based on root five: 3 - sqrt(5). Also, not sure if you know this, but in base 5, there's another grafting constant (The only two perfect grafting constants for square roots in any base) and it's value is: (3 - sqrt(5)) / 2.

[@xCorvus7x]

Well, in a context where we're more concerned with the average between 1 and any other number than with that other number (and so much so that this average could replace the variable input that defines it), the Golden Ratio is the essence of sqrt(5).

[@westtech001]

Just imagine how the Pythagorians would have loved discovering this relationship!

[@FlapMeister]

I love the Parker square at 8:28! Numberphile easter eggs are the best.

[@jackscrivens9520]

3:36 hidden stand up maths reference :)

[@chaschtestark7973]

the Parker square-flash killed me :D

[@willgrannis2652]

When is Matt's next book coming out?

[@Xnoob545]

√5 - the real Doug φ - Carl

[@paulcleary9107]

Talking of the general Fibonacci series, did you know that the sum of the first 10 terms is divisible by the 7th term and the result is always 11. Where a and b can be any positive or negative number.

[@chrisg3030]

No I didn't know that. Cool. I started looking for similar examples. In the Pell sequence, which crops up in a recent Numberphile vid on the Silver ratio, the sum of the first 8 terms is divisible by the 5th and the result is always 24 for any pair of arbitrary starting terms. What's the rule I wonder.

[@paulcleary9107]

Chris G \there are an infinite number of terms where this works, there is one before the example I gave, at 6 terms and the divisor term is the 5th, after the example it happens again at the 14th term and divisor is the 9th term. for the example if we take the terms {a, b, a + b, a + 2 b, 2 a + 3 b, 3 a + 5 b, 5 a + 8 b, 8 a + 13 b, 13 a + 21 b, 21 a + 34 b}. and sum them we get (55 a + 88 b), the 7th term is (5a + 8b) dividing now cancels out the a and b leaving just the number 11.

[@chrisg3030]

Number of terms, position of divisor term, result. 2, 3, 1. 6, 5, 4. 10, 7, 11. 14, 9, 29. 18, 11, 76. 22, 13, 199. And so on. The integer results sequence here is a bisection of the Lucas, OEIS A002878. I'll look into the Pell and other sequences.

[@MattMcIrvin]

I think you can get rid of the rounding in all of these formulae by bringing in the powers of the other root of the characteristic equation, which here is -1/phi.

[@MrRyanroberson1]

All the rounding is unnecessary if you use both values of phi. (phi^n - ihp^n)/sqrt(5) = fib

[@andrewolesen8773]

Having read the first few chapters of things to make and do in the fourth dimension, I must ask have you generalized this to all bases?

[@fernandobarrera4569]

The real reason that \sqrt{5} appears in Fibonacci or the generalized form with A and B, is due to the fact that the generating matrix of the sequence has entries: Matrix[[1,1],[1,0]] and its eigenvalues are the golden ratio and its conjugate (conjugate in the sense that are roots of the same irreducible polynomial). A little bit more precise, called G_n the general term of the sequence, then [G_(n+2) G_(n+1)]=Matrix[[1,1],[1,0]][G_(n+1) G_n]. The initial condition is [A B].

[@MaximeJean94]

or there is an exact formula for the fibonacci numbers : Fn = [ ((1+sqrt(5))/2)^n - ((1-sqrt(5))/2)^n ] / sqrt(5) this is link to the recurring sequence Fn+2 = Fn+1 + Fn which is linked to the quadratic equation x² = x+1, for which the two solution are the golden ratio phi=(1+sqrt(5))/2 and the other one psi=(1-sqrt(5))/2, and the formula rewrites itsef Fn = [ phi^n - psi^n ] / sqrt(5)

[@stuartofblyth]

Doesn't the Fibonacci sequence start at 0 (0:28)?

[@thomassynths]

Where can I buy the T-shirt Matt is wearing?

[@GeodesicBruh]

I can’t believe that he worked out the arbitrary numbers in his head.

[@JoshuaHillerup]

I will be really impressed when Matt can make pi and/or e pop out.

[@PC_Simo]

Commenting this at 1:51. The ratio is tending towards 2φ - 1 = sqrt(5).

[@USBDriveNo2]

Why is there a nautical chart of the South Atlantic in the background at the start of the video? A retrospective of the Falklands war?

[@SnowTerebi]

Did I see something flashed at around 7:28?

[@nomekop777]

*praises Lucas numbers* "I've never bothered memorizing them"

[@WaltRBuck]

The magic of root 5. The question is, "How is it related, mathematically, to 42?"

[@theDevintage]

Edit: I'm wrong, thanks to Vojtěch Strnad for the correction. But for the sake of learning, this is what I wrote: There are infinite amounts of 42's in the digits of root 5, though I suppose that's not hard for an irrational number.

[@edwardpaddock2528]

Root 42 x Root 5 = 14.4913767. The digits of 14.4913767 sum to 42. Ab ovo usque ad mala

[@vojtechstrnad1]

@@theDevintage Is there some proof of this? Because you might have fallen for a known misconception. There is no number that we know is normal, although a lot of irrational constants are conjectured to be and a proof can be given that "almost all" numbers are normal.

[@theDevintage]

Thank you for pointing this out, you're right. You can have irrational numbers where certain numbers don't show up, my bad.

[@vojtechstrnad1]

It has been conjectured that every irrational algebraic number is normal, that includes root 5. So you might actually be right. It's just that I hear people assume this is proven too often, mostly in saying "the decimal expansion of pi contains every possible sequence of numbers".

[@ThePoxun]

so what happens if you switch out those sqrt(5) with something else.. say the square roots of other primes such as sqrt(3) or sqrt(7)? anything interesting?

and the Parker square flashed when he committed a mistake writing root 5 :P

[@rosiefay7283]

"the worst 5..." says Matt, trying to cover up the fact that he wrote a 2.

[@kinyutaka]

The generalized formula might not work with A being greatly larger than B. Tried 325 and 5, and ended up with ~631 for the fifth number, but straight addition gets me 665. Could there be a limit on how far apart the numbers can be?

[@AnonimityAssured]

Gerald Bieniek, perhaps you would like to try this exact formula (excuse the messy notation). If G[1] = A and G[2] = B, then G[n] = ( [ φ^{n‒3} + φ^{n‒1} ‒ (-1)^n · (φ^{3‒n} + φ^{1‒n}) ]A + [ φ^{n‒2} + φ^n + (-1)^n · (φ^{2‒n} + φ^-n) ]B ) / 5.

[@kinyutaka]

@@AnonimityAssured I think the problem comes about with the rounding, when there is a large gap between the starting numbers, there is a wider margin of error in the generalized formula. It was off by more than 30 because it was still fluctuating wildly. We have to play around with where we round it to in order to get it to work. Basically, remove the rounding entirely and say that it should be *about* the answer to the equation, I think would be the best answer. Saying that you're rounding the figure makes it sound more precise than it is.

[@AnonimityAssured]

Yes, the rounding is the problem for small values of n. That's what the exact formula avoids. Here it is again with improved notation: G[n] = ( ( φⁿ⁻³ + φⁿ⁻¹ ‒ (-1)ⁿ · (φ³⁻ⁿ + φ¹⁻ⁿ) )A + ( φⁿ⁻² + φⁿ + (-1)ⁿ · (φ²⁻ⁿ + φ⁻ⁿ) )B ) / 5. And here is a modified version that relies on L[n] = φⁿ + (1-φ)ⁿ (found by Jonathan50 using Maxima the computer algebra system): G[n] = ( ( φⁿ⁻³ + (1 ‒ φ)ⁿ⁻³ + φⁿ⁻¹ + (1 ‒ φ)ⁿ⁻¹ )A + ( φⁿ⁻² + (1 ‒ φ)ⁿ⁻² + φⁿ + (1 ‒ φ)ⁿ )B ) / 5. Finally, here is a formula for the sum of the first n terms of G[k]: Ʃ{k = 1 to n} G[k] = ( ( φⁿ⁻¹ + (1 ‒ φ)ⁿ⁻¹ + φⁿ⁺¹ + (1 ‒ φ)ⁿ⁺¹ )A + ( φⁿ + (1 ‒ φ)ⁿ + φⁿ⁺² + (1 ‒ φ)ⁿ⁺² ‒ 5 )B ) / 5.

[@35571113]

Matt, you'll need to change your password now.

[@SillieWous]

woohoo Matt Parker!!!!