Fibonacci Factorial & the Most Unknown Mathematical Constant [ PyMath #5 ]

[@NPCooking69]

*Hot damn, I love that Constant! Check out the new STEMweek deals btw! 10% off the WHOLE Brand New Nonlinear Dynamics & Chaos Collection! :0* stemerch.com/collections/nonlinear-dynamics-chaos Also, if you're still looking for the perfect Christmas present make sure to try out Brilliant! brilliant.org/FlammableMaths

[@neurophilosophers994]

There is a list of fractal’s of increasing Haussdorff dimension on Wikipedia. But why is there no procedural way to generate fractal shapes of every dimension across the continuum ? In the same way Conway generates all 0 to 1 dimensional numbers with Surreal Numbers ? Is there a possible connection ?

[@wansichen3743]

you should try compute this in log space like a normal person does

[@wansichen3743]

you can also define long double instead of just double in python to avoid the issue you have there

[@KillianDefaoite]

16:48 "We need math.sqrt" **(1/2): am I a joke to you?

[@frozenmoon998]

Andrew from the basement: *your floor function is too loud, I can't sleep*

[@daveconerly1450]

Now this is the old flammy that is so entertaining. Even though ive watched you for years , and I've never received a response except for the infamous WTF I am one of your biggest fans. You motivate me to see life in a whole different light. " A stroke of genius " put me on my death bed.

[@PapaFlammy69]

[@tomkerruish2982]

The infinite product (1-t)(1-t^2)(1-t^3)••• shows up when dealing with the partition function. Mathologer recently made a video about this. Euler showed that the product resolves to 1-t-t^2+t^5+t^7-••• with the nonvanishing teen's3 having exponents that are pentagonal numbers. This should aid in the calculation of the Fibonorial constant.

[@hauntedmasc]

your python skills are improving papa! a nice trick you can abuse in python for "swap"-type functionality is the following: -- code -- a, b = 1, 1 for iterate in range(3, n): a, b = b, a + b -- /code -- simultaneous assignment abuse!

[@PapaFlammy69]

great stuff, I'll use that next time! =)

[@PapaFlammy69]

great stuff, I'll use that next time! =)

[@michaeljones1686]

python ez

[@Ricocossa1]

Here's another cosmetic change you can do: instead of having your first and second fibonacci floating around in the global scope, you could make your fib function a generator function. Simply replace "return" by "yield" and your function will remember the values of its variables between calls. :) Edit: I wrote this before realizing you were going for a for loop and not a function. But it's still useful stuff to know.

[@martinepstein9826]

"57isprime" Nice.

[@PapaFlammy69]

Finally someone noticed XD Good catch Martin! :D

[@chonpincher]

Yes, it's the famous Grothendieck prime.

[@steve2817]

Python blows up when number is greater than 1.79e+308, greater numbers are defined as 'inf'

[@calculusphysicschemaccountingt]

BTW, you need to sleep more and maintain your health because I saw your black circle under your eyes.

[@PapaFlammy69]

I'm still ill^^

[@ComBOT]

@@PapaFlammy69 don't lie, you're trying to prove twin prime We know

[@erencan1453]

@@PapaFlammy69 Good Besserung!

[@longphung4874]

@@PapaFlammy69 Papa I know you’re trying to prove Navier Stokes

[@Kdd160]

@@PapaFlammy69 get well soon 😀

[@youtuberdisguiser6075]

Guys i just want to remind you to upvote and comment on his videos, so that the algorithm treats his videos better. ^^

[@PapaFlammy69]

awwwww

[@josephmichail4428]

phi^1*phi^2*phi^3*...phi^n = phi^(1 + 2 + 3 + ... n) = phi^(-1/12)

[@smrtfasizmu6161]

Python is kind of similar to C++ C and Pascal programming language, I was able to follow this even though I never studied python

[@TobyAsE120]

python is pseudo code that actually runs

[@Nothingtonnobodson]

@@TobyAsE120 I use to use java, trying to shift to python

[@HAL-oj4jb]

Some guy probably: Asks you for your QnA on Sunday what your favourite constant is Papa Flammy: Makes a whole video about it

[@ivanoliveira9784]

0:32 I was expecting him to say "Hello Kitty Cats", but he didn't ;-;.

[@debblez]

“57 is prime” STOP 😭

[@maciejsiedlecki8473]

Great video as always! You may actually want to look for some "arbitrary precision calculations". For example - python "decimal" library. It will give you way better result (literally - waaay more digits after the comma). Looking forward for more PyMath videos :)

[@MrRyanroberson1]

fun fact: it converges so fast that prod(38) already gives you the highest precision result that default floats can compute.

[@MrRyanroberson1]

actually: if you plot it in desmos, it's even more incredible! instead of -1/phi^2, try the product from 1 to 100 of (1-x^n), it produces such an interesting curve. the fibonorial constant is the special output for x=-1/phi^2; the function converges to finite values for |x|

[@smrtfasizmu6161]

You could have also initialized fib = 1, s = 1 and then in the loop put p*= fib, f =s, s = fib before fib = s + f Then the loop runs from 2 instead of 3, because the product p that you get in first iteration is 1*1 which is the first Fibonacci number multiplied by the second one

[@luchisevera1808]

An infinite product of immortal rabbits getting it on

[@beardedboulderer2609]

I didn't know drinking water while watching Papa Flammy could be so dangerous until I saw the intro meme... Love it!

[@sicko_the_ew]

You could try out this (f,s) = (s, fib). If it doesn't work, that's too bad, but you might as well try it. It might be necessary to start out with a tuple, (f, s) instead of individual variables. (You could also try f,s = s,fib , without the parentheses, but that might be confusing to begin with). The aim is this: (var0, var1, var2 .., varN) = (eval0, eval1, .. evalN) - which you've probably encountered before. Nice for swapping values. Just write a, b = b, a

[@The1RandomFool]

SageMath can handle very large numbers, and custom functions for it are programmed in Python as well. With the built-in functions the Fibonacci factorial can be expressed as prod(fibonacci_sequence(1,50)), which multiplies the Fibonacci numbers in a certain range. The C constant can be computed to 100 digits with product(1-(-1)^k/golden_ratio^(2*k),k,1,300).n(digits=100). Interestingly, the Fibonacci factorial constant was already put on the Online Encyclopedia of Integer Sequences in 2001.

[@hbowman108]

In conventional factorials, the product of the first zero integers isn't zero, it's one.

[@rogergarridovilallave9341]

The fibonorial constant is the value of the Q-pochhamer symbol with z=1, q=-phi^{-2}, i.e. it is (1; -phi^{-2})_infty

[@zorrozalai]

Fun fact: if you draw a Pascal's triangle, pick up a point on it's side and you start to jump like a chess knight, then the sum of the visited numbers will always be a fibonacci number. Going downward: 1+3+1 = 5, 1+6+5+1=13, 1+10+15+7+1=34. Going upward: 1+2=3, 1+4+3 =8 ...

[@Demki]

Up next: gnuctorial - where you take the product of the number of groups of order k (or as Conway called it, gnu(k)) for k between 1 and n and the trivial gnuctorial - where you take the product gnu(k) for k between 0 and n the second one is trivial because there are 0 groups of order 0...

[@roberthrusecky9439]

I thought it was when you compile successive versions of the GNU compiler toolchain with the previous one

[@hotlinkster123]

It would have been a nice exercise to plot the convergence of both methods in python.

[@duncanw9901]

I think non-computable numbers might be less well known :D I did like phi=2 engineer++ method there at 8:48 and a quick python tip: in 3.x you can use the "double asterisk" operator for exponentiation, so you don't have to import math.sqrt() or math.pow() but just use 5 ** 0.5 (edit: you did this for the other part I look dumb now)

[@unknownpalooza8475]

Sometimes, mathematics can be very mysterious also... Oh boi, this gets pretty intense...

[@georgecantu856]

bruh I just started learning python, you are a psychic dude

[@dominikstepien2000]

0:31 I was expecting papa saying 'including those very neat, cute kitties'

[@a_llama]

liking for the algorithm!

[@BlackHermit]

Fibonacci was not insane! He was an Italian mathematician from the Republic of Pisa, which is the mark of sanity! Fibonacci was a guest of Emperor Frederick II, who enjoyed mathematics and science. I love him so much... Dear Fibonacci Your series is great Dear Fibonacci Please open the gate!

[@VaradMahashabde]

13:45 Nah your brain was thinking ten steps ahead and took the log already

[@garvett6660]

*wipes tears off *it’s beautiful...*

[@arkitson]

The Fibonacci factorial constant can be expressed in terms of the derivative of the Jocobi theta function of the first kind. See: arXiv:math/0608187

[@everettknag7243]

Great video!

[@jarogniewborkowski5284]

Hmm... Double chaotic pendulum. What do You think about 3 body problem? Is there also chaos or not? I have heard about Sundman, who probably has already solved this "chaos". Maybe You could make some video about it. Best regards

[@Kdd160]

Papa you are soo creative 😄😄

[@Ryan-gq2ji]

would there be a way to extend the Fibonorial to all real numbers greater than zero?

[@nullinf]

How’d you get the slidey-chalkboard in your room :0

[@Tomaplen]

Factoriacci

[@johnsalkeld1088]

Dominated convergence

[@ta1949]

flammy u should do a video trying out oxford or cambridge's mathematical admission test

[@sharpfang]

I wonder if there's a connection to (x+1)^n polynomial factors. The factors are nth line of the Pascal's Triangle, and there's a strong connection between the Pascal's Triangle and the Fibonacci Sequence.

[@vivovivi5703]

yay! flammy goes coding!!

[@nicholasthesilly]

Relevant: en.wikipedia.org/wiki/Euler_function C would be this function at -1/phi^2.

[@PapaFlammy69]

yup, C is simply a q-pochhammer symbol^^

[@angelmendez-rivera351]

Yes. This would mean that C is equal to the sum over every integer n in Z of (-1)^[(3n^2 + n)/2]/φ^(3n^2 - n). Also, you can manipulate the product definition to obtain different expressions as well. For instance, Φ(-1/φ^2) = Π{n >= 1; 1 - (-1)^n/φ^(2n)}, but (-1)^n = i^(2n), so C = Π{n >= 1; [1 - (i/φ)^n]·[1 + (i/φ)^n]}, and now it may be worth investigating whether the products Π{n >= 1; 1 - (i/φ)^n} and Π{n >= 1 + (i/φ)^n} converge individually. The first product is just Φ(i/φ) if it converges, and since the factors of the second product are the conjugates of the factors of the first product, I would expect that the convergence of the first product imply the convergence of the second.

[@mind-h4i]

Wow early to a flammy math video

[@pimmm]

you showed up on my recommended, in case you are curious.

[@djvalentedochp]

square root ov faaaav

[@Vaaaaadim]

I wouldn't necessarily trust Method 1 you show for computing the constant, given the floating point calculations have to round. Hm, something that might be neat would be to have some sort of library that does your computations, but also tallies up the accumulation of rounding errors.

[@jordanhansen6649]

How is this guy so cool ?

[@technoguyx]

Damn, someday I'll have that huge sexy dynamics rag at my office. First I'd need to get an office, though

[@samuelemorreale7510]

What IDE are you using in this video?

[@ichdu2757]

Almandeutsch: Flammable Maths

[@jakemoll]

Couldn’t you also find a Stirling approximation equivalent by saying ln(n!F) = integral(1, n){Fn dn} and use a continuous formula for Fn? Can’t think exactly how it would look or whether it would work

[@lieflundmark3564]

Good evening (where I live), fellow mathematicians!

[@prakharanand7012]

Where I live too

[@toniokettner4821]

@@gigagrzybiarz yes

[@Leo-rh6rq]

Sehr interessant und gut erklärt. Gibt es sonst noch mysteriöse Zahlenreihen oder ähnliches wie π, e bzw hier Phi? Ich verstehe nicht alles auf Universitätsniveau, aber interessant ist es trotzdem.

[@shusanniblack7513]

Can you help me solve a/b=b.a????

[@TheRemixDenuo]

How can phi* = -1/phi? Wouldn't that imply 0 < |phi|^2 = phi * -1/phi = -1?

[@ethansolly96]

It's not the complex conjugate, it's just notation for the conjugate root of x^2 - x - 1 = 0

[@angelmendez-rivera351]

If φ = [1 + sqrt(5)]/2, then φ = 1 + 1/φ, hence φ - 1 = 1/φ, hence 1 - φ = -1/φ, and 1 - φ = 1 - [1 + sqrt(5)]/2 = [1 - sqrt(5)]/2. This is the other solution to the equation x = 1 + 1/x, and so it is often denoted φ*.

[@mightloki8181]

we need to get the likes to

[@DarkMonolth]

Papa you should make a GitHub to post all your code in. So you look like a real software boi

[@eulermed]

It's not the limit when it goes to infinity, you didn't read Ababou's proof, it's the limit when it goes to the universe's age :(

[@chunkynorris100]

hmm yes dedekind eta or euler function and the golden ratio

[@moonlightcocktail]

I think this will approximate phi^(nth triangular number)

[@longphung4874]

IM GETTING THAT EULEROID SHIRT RIgHT NOw

[@PapaFlammy69]

Nice! :D

[@emredogan2102]

Ahh papa I see you're a man of culture as well😉

[@АлександрКовалев-о7я]

step by step think 'bout it

[@weebrahim]

Poggers got that twitter notification

[@wernerheisenberg7192]

3 Views, 3 Likes. Hell ya!

[@Trekistar]

25

[@hamsterdam1942]

ya know, 4 is also the mathematical constant

[@arielfuxman8868]

Let us wait and see if this constant will have any real world applications

[@guill3978]

Is that constant a transcendental number?

[@PapaFlammy69]

We don't know sadly :'(

[@guill3978]

Try to deduce it.

[@guill3978]

Find a proof or a disproof.

[@JB-ym4up]

Says Fn but writes Fu. Seems legit.

[@twrk139]

But how can something be most unknown?

[@cornelhanekom5689]

Proof by calculator🤣

[@atharvasharma3492]

Hello

[@mr.unknowngamer2109]

Hi!! Do u remember me . Last time u gave me a heart😁😁.

[@user-me7hx8zf9y]

anime deskpad >:(

[@toniokettner4821]

your numerik vorlesung was a bit too far in the past ;)

[@pearlsek]

The breaking up of the limit at 13:41 is not legal because you didn't prove that one of the limits exists... Makes for ok math entertainment but not sure you should give the impression this kind of invalid reasoning is acceptable

[@angelmendez-rivera351]

It's a heuristic. It's not difficult for the viewer to see that the limit exists.