Fractions and Iterations (extra) - Numberphile

From the main video with Holly Krieger at • A Fascinating Thing ab... --- Catch Holly on the Numberphile Podcast: • Champaign Mathematicia...
The Dynamical Uniform Boundedness Conjecture.
More videos with Holly: bit.ly/HollyKri... Holly's website: www.dpmms.cam....
She is the Corfield Lecturer at the University of Cambridge as well as a Fellow at Murray Edwards College.
Go deeper with this technical paper: doi.org/10.115...
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[@AlInGaP_Diode]

Why wasn't this included in the main video? I appreciate that you posted it here but this seems like it would have fit right in and explains the statement about larger sets at the end.

[@FightAgainstOurFears]

I've just watched this with my friend who has very little math exposure (and this was their first numberphile video!) - they commented how this video feels more advanced and behind the scenes but is glad it wasn't included since it was harder to grasp. I think depending on the level of mathematical understand it could be included, but I agree with it being bonus footage.

[@mihailmilev9909]

@@FightAgainstOurFears oh wow nice! I know it's a bit late but congratulations lol. How r u guys now? What do they think of math now? What was ur background and circumstance for showing this video?

[@mihailmilev9909]

"Guy" lol

[@STAR0SS]

I'm not sure why you guys cut that from the main video, it's only 3 minutes and kind of important.

[@heyandy889]

there's always a constant tension between including more interesting content and keeping it short and concise. all things equal short and concise is preferred; as the video gets longer and longer more people will abandon. so in the end it's just a creative decision. that's why the second channel exists, to include more interesting stuff that is not critical to the story told in the first channel.

[@Banan_Gutten]

@@heyandy889 More interested to know why this video is unlisted, Isn't the point of a second channel to not hide theese videos, just make them available another place?

[@Cosine_Wave]

@@Banan_Gutten it's unlisted temporarily to ensure people see the primary video first

[@dittbub]

youtube algorithm, i take it

[@scottmuck]

heyandy x that’s one KZbin theory. There’s an equally valid theory that long videos are the way to go because people skip ahead or close the video when they get board anyway, and with longer videos you tend to get longer average watch times. Some channels are very successful with 1 - 2 hour videos as the norm, even understanding that on average maybe only 15 mins is watched.

[@cachomanify]

i wish that you were my math teacher at the university, you explain math so well and clear.

[@mc101]

Terence Tao Collatz Conjecture Update PLEASE!!!

[@Maharani1991]

+

[@KaliFissure]

Hi Holly. Thank you for the videos. Unless I'm mistaken one of your interests is the pressure densities and structures within numbers. The most universal example would be the incredible pressure outward towards infinity, and inward towards 1 and 0. Then within the sphere of 1 there is an internal pressure in both directions. Outward towards 1 (the periphery) and 0 (centre). Almost any iterated function will head for this conformation. Any more convoluted but circular functions like toroids, ovoids, electron orbital shapes etc could be considered functions collapsing to sphere of 1 and then having a set of scaling vectors applied. If there is a membrane to the function then it has collapsed to sphere 1 + scaling vectors. I am an educator and wanted to introduce these ideas to my kids but wanted to make sure with someone who is knowledgeable before i pass on bad info. health and peace :)

[@Galakyllz]

Great video! I always enjoy these kinds of videos that involve sequences.

[@mihailmilev9909]

wym?

[@mihailmilev9909]

Also how's the 3 years been lol

[@Galakyllz]

@@mihailmilev9909 lol, my life has greatly improved over the last three years. Thanks for asking. To your previous question, I like videos about sequences because I like iterative processes. It motivates me to write some code and play around with some of the hyper variables of the process.

[@madwilliamflint]

"Oh, I couldn't tell you that." So....yes.

[@scottmuck]

2nd the other comments that this is the best bit of the video!

[@samharper5881]

Thanks for sharing this second part. Makes sense.

[@iboremytherapist]

on Unbreakable Kimmy Schmidt, one of the fictional characters is named Holly Krieger! also a redhead. coincidence?

[@stardust6773]

Maths are beautiful, and even more if Dr Krieger teach us something interesting

[@Gunstick]

Nobody expects the Fermat theorem!

[@trb6676]

but this these considerartions only apply if we have an iteration of length 1 because only then we can say z = z^2 + c, right?

[@rwanamo6307]

If we have a cycle of length 2 we can say that z = (z^2 + c)^2 + c, which gives a much more complicated equation to solve. You can continue the same way to extend it to longer cycles, getting a way more complicated equation on each step.

[@reset9668]

@@rwanamo6307 that makes no sense, because if you plug in the numbers we had previously, the thing cycles back to itself. say, we first had c = 1/4 and z = 1/2, (0.5^2 + 0.25)^2 + 0.25 = 0.5^2 + 0.25. so 1 periodic cycle = the same as 2 periodic cycles. The true polynomials the video is talking about, I don't know.

[@shokan7178]

I like this

[@kam1470]

Can this be connected with 196-no palindrome problem?

[@ericsmith1801]

The Holy Grail: finding an equation for the distribution of primes by studying the Mandelbrot Set ?

[@kevolegend]

Holly Grail

[@WylliamJudd]

I still want to know what the criteria is to get 3-cycles.

[@RJSRdg]

m/n = (((m/n)2 + (k/n))2 + (k/n))2 + (k/n)

[@Jajo372]

Excuse my bad english ;) i only do "german math" If i took a complex z, maybe the kth unit root, and define c=0 ... I think i could construct a lot of larger periods, can't i?

[@joeybf]

Yes, that's why we only take z and c to be rational

[@Jajo372]

@@joeybf Ahh okay, thanks, i missed that restriction

[@smg0003]

From a practical point of view if you have a sine wave or similar (where begins on line somewhere, gets taller and lower in middle then sine wave gets gets smaller again back to size of how started), then u can lift paper off table and curve this sheet of paper into 3D/4D? matrix and join lines up so end point equals start point. Way af cheating i mean Or is this just old or new thought in math? Top view is circle, side on view is more complex i mean

[@VorpalGun]

What about solutions for arbitrary real numbers instead of fractions? Can we find solutions then?

[@pythagorasaurusrex9853]

As the problem can be converted into polynomials, not every cycle can be satisfied with real numbers, as polynomials in the real number set do not need to have a solution. Instead as every polynomial has solutions with complex numbers, I am sure any arbitrary cycle can be satisfied with the right complex numbers.

[@willemvandebeek]

So zooming out of the Mandelbrot set will end up into a field of nothing?

[@kwinvdv]

But there should be solutions (not necessarily rational though) for any period, since it essentially comes down to solving for roots of polynomials? Also technically since there are rational solutions for two and three there are also rational solutions for 4, 8, 16, ect. and 9, 27, 81, etc. by looping multiple times through the same cycle. For c=0 one can also easily get (not necessarily rational) solutions by solving z=z^(2n) with n the period of the orbit, which gives z=0 and z=exp(2kπi/(n-1)). For rational c these solutions for z should at least be algebraic and those solutions might in general not have an analytical expression, since in general there are only analytical solutions to polynomials up to order four. This does makes me wonder what is so special about rationals numbers in this context.

[@yovliporat8608]

Polynomials don't necesserily have roots over the rational numbers.

[@pythagorasaurusrex9853]

This video is about fractions and polynomials only containing integers as the solution. I bet, in the set of the complex numbers any cycle can be satisfied, but not in this particular problem Holly is talking about.

[@pyglik2296]

I wonder, it's hard because we take rational numbers and we have to solve polynomials in integers, but if we took real numbers would it have infinitely many solutions for loop any size?

[@phiefer3]

I don't think so. I may be incorrect, but I think that the Reals aren't enough for this problem, as you can't always get Real solutions for polynomial equations either. Expanding to Reals may provide 'some' solutions for 'some' larger loop sizes, but I don't know if there would be infinite solutions for any size. As others have pointed out, that would likely require expanding to Complex numbers.

[@Sam-yf4kt]

Is that Fermat? Or Thermals Theorem? 😅

[@postbodzapism]

What about algebraic numbers?

[@levipoon5684]

Pick an arbitrary rational number c. Finding a cycle is equivalent to solving a polynomial equation in z, namely f^n(z) - z = 0 with f(z) = z^2+c. All non-constant polynomials have at least one complex root and these roots are by definition algebraic (since the coefficients will be some polynomial in c with integer coefficients).

[@DeclanMBrennan]

How about irrational numbers ? (Not with the sane approach obviously as there won't be any integers forming ratios ).

[@tothm129]

@@DeclanMBrennanthe algebraic numbers contain all rational numbers, some complex numbers and some irrational numbers. Since we are trying to solve polynomials, there are by definition, no irrational solutions that are not algebraic.

[@anastasiaklyuch2746]

what if you use numbers like a+b*i?

[@jonp3674]

I guess the fundamental theorem of algebra applies. If you compute the polynomial like Holly does at the end of the video then you know that in the complex numbers there will be a solutions. For example if z = z^2 + c then that is z^2 - z + c = 0 which, for any c, is a single variable polynomial in z and therefore must have two roots. Those roots are (1 +- sqrt(1 - 4c))/2 so if c = 1/2, for example, then those roots are 1/2 + 1/2i and 1/2 -1/2i. For longer cycles the polynomial will be higher degree and will therefore have more solutions, and they will be harder to compute, but they will exist.

[@anastasiaklyuch2746]

@@jonp3674 Thank you my science friend! My brain can rest in peace now, thank you :3

[@jpkotta]

z0 = 0, c = i

[@ScandGeek]

@@jonp3674 you have to remember that we are looking for fractional solutiobs (so I guess complex numbers with rational real and imaginary parts) which might not exist

[@joeybf]

@@ScandGeek Not even, we are restricting to just rationals, not complex numbers with rational coordinates. We can't apply the fundamental theorem of algebra at all

[@TheGovernancePage]

Sylvia Tilly

[@amirhoque655]

28

[@fedechiossone9932]

Niiiiice

[@anuppokhrel2357]

I wish you were my maths teacher in school.

[@robertpage4991]

Amy Adams would be the perfect actor to play you in (movie)^n ...

[@douro20]

Not allowed to tell you what she's working on...

[@PhilipSmolen]

"zed"

[@mikedoe1737]

Zee.

[@Jasper_4444]

Zeww...

[@SatisfyingWhirlpools]

Who else got here by clicking the video?

[@tommyzDad]

Holly, you are gorgeous.

[@scriptendo44]

Marry me Holly :D

[@coloradoron2296]

No offense intended, but you are very cute!

[@1996Pinocchio]

holly: _offended_