A Fascinating Thing about Fractions - Numberphile

The Dynamical Uniform Boundedness Conjecture with Dr Holly Krieger.
Extra from this interview: • Fractions and Iteratio...
Dr Krieger on the Numberphile Podcast: • Champaign Mathematicia...
More links & stuff in full description below ↓↓↓
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...
Numberphile is supported by the Mathematical Sciences Research Institute (MSRI): bit.ly/MSRINumb...
We are also supported by Science Sandbox, a Simons Foundation initiative dedicated to engaging everyone with the process of science. www.simonsfoun...
And support from Math For America - www.mathforame...
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Animation by Pete McPartlan
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Brady's latest videos across all channels: www.bradyharanb...
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[@MozartJunior22]

It's amazing how after Brady did so many of these videos, for so many years, he has developed a mathematician's mind, and is asking EXACTLY the questions a mathematician would ask.

[@akhileshmittal3396]

cause he is one

[@pedroaugusto656]

Does not he have a PHD in math ?

[@fernandomargueirat6454]

I kind of agree, but I'm pretty sure there are discussions about the topics before they record the interviews and these things are probably already discussed. It doesn't take away any value, he still has to be able to understand the concepts and throw the question at the right time, which is a skill that is much harder than what people expect.

[@RodelIturalde]

@@pedroaugusto656 he is a journalist.

[@huh968]

i mean that's pretty much expected tho lol, one would assume he learned quite a bit over those years

[@ARVash]

I never knew I could be excited about fractions but here we are. Great job Holly, your enthusiasm is infectious.

[@dliessmgg]

What fascinates me in the example that she gave with three numbers in a loop is, the specific rationals can be related to music in just intonation tuning. The 5:4 ratio is a major third, and the 7:4 ratio is a harmonic seventh (there is meaning behind those musical terms, but there's also a lot of historical baggage, so don't worry about the details). If you combine those two with a root note (1:1 ratio) and a perfect fifth (3:2 ratio), you get a harmonic seventh chord, that occurs frequently in for example barbershop quartet singing. The perfect fifth can often be left out (because our ears/our culture often hears them as implied) but the root needs to be included. Now the third rational in the three number loop can be related to the ratio 1:4, which in musical terms is two octaves below the root. But due to octave equivalence is in the same pitch class as the root, and can be used as such. tl;dr: the three rationals in the loop she showed, when interpreted in musical terms, form a neat harmonic seventh chord.

[@doublespoonco]

This is intriguing

[@harrympharrison]

I would love a Numberphile video explaining some aspects of just intonation.

[@jonnyphenomenon]

Math in music is fascinating! So, is that just a coincidence? Or could it reveal other patterns in this equation exactly?

[@Tapecutter59]

Interesting observation.

[@doublespoonco]

@@harrympharrison maths in music is always really interesing

[@johnathancorgan3994]

Between Holly and James Grime it's hard to choose who has more infectious enthusiasm for math!

[@jacobshirley3457]

What about that klein bottle guy?

[@AdStellae-]

@@jacobshirley3457 Cliff is great!

[@RadicalCaveman]

@@jacobshirley3457 You mean the guy that's in one?

[@HilbertXVI]

@@jacobshirley3457 I think cliff is just on another level

[@bane2201]

@@HilbertXVI Yeah Cliff wins hands-down. Whenever I watch him, I feel like his enthusiasm about math is so visceral (like his dancing) that it's _only_ restrained by the limits of the human body. If he was able to _fully_ unleash his true power in some explosive burst, his house would be a crater, and his city probably would be too.

[@73honda350]

She's one of the better interviewees on this channel, and as an interviewer, Brady is good at asking insightful, relevant questions.

[@hi_pd]

What about Dr Hannah Fry?

[@TheMrvidfreak]

What a way to end the week, she's one of my favorites on this channel!

[@R_V_]

Holly Krieger > Hannah Fry.

[@BubbaJ18]

@@R_V_ why not both?

[@Ken.-]

Today's Sunday.

[@sharpnova2]

Are you capable of being honest about why?

[@antilogis6204]

@@sharpnova2 Extend that question to the 500+ likers.

[@LeonardNguyenSchyen]

Brady is a crazy good math interviewer wth...

[@Bronzescorpion]

Agreed, he asks great questions. Either to clear something up, that was so obvious to the mathematician, that they briefly forgot to explain it or to give them a new point to explain from and thus progress the interview. I really think much of the joy these videos give, are in fact due to Brady's skill as an interviewer.

[@CalvinHikes]

He's clearly a smart guy.

[@mfc4655]

he educates himself on the topic so he can ask in depth question

[@davidgillies620]

I'm often struck by how quickly he homes in on a subtlety or a generalisation of a problem. Obviously a very highly intelligent individual. I think that must be very rewarding for the people he's interviewing (there really isn't much that scientists/mathematicians/engineers like more than explaining something cool to an intelligent layman)..

[@rayu__641]

i agree!!!!

[@waynewelshans1172]

I'm 36 and I feel like a 7th grader crushing on his math teacher ever time I watch Holly.

[@evanw7878]

Creep

[@mannyheffley9551]

You are a deviant

[@iamtheiconoclast3]

Well that got out of hand quickly. Some people are really offended at some fairly ordinary things. :|

[@Palimbacchius]

@@iamtheiconoclast3 or pretend to be ...

[@fattestallenalive7148]

7:58

[@TC-bz9dz]

this video is perfect timing....this is my lesson for my students after Christmas break...im going to tell them all to watch this to get them ready...Thank you for the video !!

[@mohammadfahrurrozy8082]

You wouldn't steal a comment

[@Idunnowhoiam102]

Well that's a bad idea

[@asparkdeity8717]

I can't believe Holly will be my Complex Analysis lecturer next term in my second year of undergraduate maths degree!

[@rollcats1290]

Damnnn

[@yvessioui2716]

Brings me to think about modulo arithmetic, fractions behavior in this context, and use of fraction to leaves placement by plants which is also kind of modulo to go from height to the next while growing. We obviously find a return to the starting point (orientation wise) at some time in life development of some plants. I love this.

[@gentlemandude1]

I love that she laughs so much. It's fun to hear Dr. Krieger giggle with delight as reveals a surprising truth about fractions that on the surface seems quite mundane. It must be so much fun to be in one of her classes.

[@Codricmon]

1:52 - Matt Parker: “Finally, a worthy opponent! Our battle will be legendary!“

[@thejiminator8816]

haha

[@shambosaha9727]

@@thejiminator8816 I know you

[@djkm9558]

😂😂😂🤣🤣🤣🤣

[@lazypops3117]

explain this to me someone

[@Codricmon]

@@lazypops3117 It's a quote from _Kung Fu Panda._

[@zubmit700]

I love the 'easy to understand' math problems that ends up with complicated solutions that I will never understand. Makes math so much more interesting.

[@FFXIDragonli]

Non Non 1-(-1)= 2, 1+(-1)=0, (-1)-1=-2, and (-1)+1=0.

[@ScottStratton]

Non Non Non Non um ... no. That is some personally-invented symbol manipulation by you. Is math the ultimate, super-truth of the universe? Probably not - seems unlikely to me. But regardless, when one applies a cognitive system to the world, it has to make sense at least internally ... and what you are saying is just arbitrary and nonsensical.

[@whoeveriam0iam14222]

maths does that so often 1? easy! 2? still easy! 3? soo many but not as obvious 4? haha no

[@riccardoorlando2262]

Along with the other thing math does: "We found this cool, easy, simple process and we want to run it backwards. Lemme get back to you with 100 years of research, 27 books and 492 papers"

[@Patalenski]

No, no, they said: 1 is easy, 2 is natural, 3 is complicated and 4+ is impossible! My wife's on the same opinion... :-D

[@danielmalo1753]

TREE(n) 1? simple 2? of course 3? no way jose

[@someoneonyoutube8622]

Yeah no it checks out I even tried using imaginary numbers too and it seems to cycle in 3’s or 2’s or 1’s but never 4’s or anything more’s

[@Jacob-ye7gu]

No matter how many iterations you're looking for, it's just finding solutions to a polynomial

[@smoorej]

OK, I’ve been a patron for a while but I have to double my contribution immediately. Holly Krieger is the reason. She has an incredible ability to make difficult topics understandable. Please, please have her on more frequently. Plus between the Mandelbrot set and this periodic fraction stuff her topics are so incredibly interesting. I see Holly Krieger, I press like, then I watch.

[@vylbird8014]

This looks a lot like the equation for the Mandlebrot set, but sticking to the real number line so you don't get pretty pictures.

[@MagruderSpoots]

With the right numbers you get the logistic equation which is the mandelbrot set on the real number line.

[@simoncopar2512]

It is, and the periodic sequences are related to the centers of out-growing bulbs on the real axis of the Mandelbrot set. Numberphile is slipping, normally, they would mention such interesting connections.

[@Bollibompa]

@@simoncopar2512 Yeah, for sure. One small example means Numberphile is slipping.

[@Gruuvin1]

And, what does Holly know about the Mandelbrot set? Right?

[@Narokkurai]

As I understand it, it IS the Mandelbrot set. C and Z are are the axes of the plane, and the beautiful colors of the Mandelbrot Set correspond to whether any combination of [Z,C] is periodic, and if so how many iterations it takes to converge. This theorem seems to be saying that for any non-integer rational Z, there is some value C which becomes periodic in three or fewer steps, but we cannot say for sure if there are combinations which will converge in 5 or more.

[@theflaggeddragon9472]

I remember stumbling on this EXACT problem about a year ago and trying over and over to show that for a polynomial of degree d, there are no points of period larger than d+1 (any period less than or equal to d+1 can be solved directly by a system of d+1 equations with d+1 unknowns). Then hours later I googled to see that even in the quadratic case we're almost completely in the dark! What a wonderful problem in arithmetic dynamics.

[@jlunde35]

I surely want to leave a comment that your family would be happy to read. Great content and thank you.

[@jgcornell]

I understood that reference :)

[@Apocalymon]

@@jgcornell what's the reference!

[@fatilumkins8730]

I love the “whoosh” sound Holly makes at 6:23. I’m glad I am not the only one that does that when drawing long arrows 😅

[@zray2937]

Good questions by brady.

[@JMTavares7]

His questions have gotten better then they used to be. He's learned from previous guests of course, a lot of the sames themes repeat, such as proving that we can/can't as opposed to conjecturing and infinite # of ways to accomplish something, etc.

[@omikronweapon]

I've always been pleasantly surprised by Brady's active role in the video. He often asks the same questions I have. He doesn't just let things slide.

[@nicolasjacquinot4202]

No there's only 7.

[@shrirammaiya9867]

@Captain_Morgan they will be divisible by 7

[@bipcuds]

@Captain_Morgan Only 7 will be prime, because 77....777 = 7 * 11....111

[@numberphile]

See extra footage and math detail from this interview with Dr Holly Krieger about The Uniform Boundedness Conjecture: kzbin.info/www/bejne/rGWviHmwid6bprc And Holly on our latest podcast episode: kzbin.info/www/bejne/h57JgoSwn9WXpq8

[@jadenpeterson4881]

Really close to 3.14M Subscribers. I’m expecting a special episode.

[@Brooke-rw8rc]

lmk when it gets to 6.28. Then they can celebrate with whole pies instead of half-pies.

[@jeromeorji1057]

@@Brooke-rw8rc The Tau-ist vs the Pi-ous debate, circa 2019. Colorized

[@astroboy3002]

Imagine how much of an unknown genius the guy sorting the paper at the recycling plant is.

[@narutosaga12]

I’m glad she’s my inspiration to get through these finals right now

[@alger8181]

The maths are far above my head, and yet these presenters explain the complicated subjects with such joy. This channel has me digging out my old electronics books and equipment, re-learning the maths i'd let slip for many years, and applying them to building again. Thank you.

[@alephnull4044]

For 2- and 3-cycles, the solution is 'easy' because you just need to solve a quadratic and a quartic polynomial equation respectively. But for 4-cycles you need to solve an octic (eighth-degree) equation, and higher degrees for larger cycles. It is a well-known result that you cannot solve a degree-five or higher polynomial equation in radicals. So it is not a surprise that no one knows how to find 'nice' larger cycles (but of course purely numerical solutions can be easily computed).

[@riccardoorlando2262]

Hold on. A 1-cycle requires a degree 2 polynomial solution. A 2-cycle, degree 4; a 3-cycle already requires a degree 8 polynomial solution...

[@alephnull4044]

Riccardo Orlando Yeah sorry you’re right. Kinda surprising that they’ve got examples of 3-cycles then. But point still stands about higher cycles.

[@patrickhodson8715]

Aleph Null maybe it was just a guess-and-check situation. -7/4 isn’t that weird of a reaction if you’re just trying stuff to see what works

[@patrickt.4121]

What about irrationals? (Not mentioned in video and so not directly related to your comment) We know how to solve those polynomials, no need for radicals ...

[@redpepper74]

I’ve heard that you can’t express all degree 5 polynomial solutions with radicals, so I wonder what other ways you could express them. Is there another kind of operation/ system/object that mathematicians use there?

[@the_beemer]

This channel is about to hit 3.14 Million subscribers...Thats THE real milestone

[@ninosawbrzostowiecki1892]

this lady was a sub for a number theory class I took years ago.

[@seededsoul]

In what country?

[@ninosawbrzostowiecki1892]

@@seededsoul @ UIC (Chicago)

[@hereb4theend]

I'm so glad channels like these exists.

[@avikdas4055]

I want some math with James Grime. He is one of the first guys who boosted this channel. I really miss him. I have not seen him in this channel for a long time. Also it would be great if you got some initial members in this channel, like Hannah Fry or Simon Pampena.

[@echo5delta286]

He joined Matt Parker for a video on his channel, Standupmaths, 7 months ago. That was a fun one called Difference of Two Squares.

[@ChrisLuigiTails]

Shouldn't forget about Matt! Him and Cliff Stoll are my favourites!

[@L0j1k]

MORE SIMON!

[@Crissix100]

@@ChrisLuigiTails Oh I love Cliff, his enthusiasm is just amazing!

[@brianpoi5117]

@@frankwc0o We need a Statisticsphile to go along with Numberphile and Computerphile.

[@mediawatchstream8340]

Thanks!

[@garret1930]

Z=2 , c=-2 Z always equals 2 This is actually the start to a family of solutions where you just set c=-(Z^2-Z) For all positive integer values of Z. This seems like it has a close relationship with the first example Holly showed for the fractions where Z=1/2 and c=1/4 If c=-(Z^2-Z) then c=1/4 I think it applies to all values of 0

[@ABaumstumpf]

Was a really strange way they phrased it. It is still an infinite number of integer-combinations you can use - not very interesting but still they exist.

[@evanbelcher]

It doesn't just work for positive integers, it works for literally every value of Z because it's the definition of a 1-value cycle. Z' = Z^2 + C (original equation) Z' = Z (defines a 1-value cycle) Z = Z^2 + C (rephrased equation) C = Z - Z^2 (solve for C) which is just the simplified version of your formula.

[@ABaumstumpf]

@@evanbelcher "which is just the simplified version of your formula. " That one step of expanding the sign :P

[@riccardoorlando2262]

@Non Non I don't understand. Can you explain what you mean, or at least provide references where I may read?

[@garret1930]

@@ABaumstumpf lol yes I should've seen that.

[@homieknowme223]

6:23 fascinating how that fraction made woosh sound travelling towards bottom left. Math is always fascinating.

[@nicolageorgiev4350]

Yes

[@themri]

Classic Numberphile video. What a treat!

[@missrobinhoodie]

My math teachers back from my gymnasium days would be so proud of me passionately watching numberphile! Also they would be very surprised...

[@tscoffey1]

“Zed squared” When an American professor has gone over to the dark side.

[@jshariff786]

Ah, the dark side of pronouncing it the way that the people who invented the language (and the vast majority of those who speak it) do. Cheers from Canada.

[@mattbarnes3467]

@@jshariff786 but we kicked their asses twice and bailed them out twice. And as we all know, to the victor.goes the spoils. Z it is.

[@jshariff786]

Within your borders, sure. Everywhere else it is Zed. So really, majority rules (based on what the entire English-speaking world is doing). Also, you didn't write "Zee it is", you just wrote "Z it is".

[@shyambuddh5546]

Can't believe you all are literally having an argument about America Vs England based on how to pronounce the letter "Z" in the comment section of a math video.

[@justincronkright5025]

@@mattbarnes3467 Go watch, This Hour Has 22 Minutes - Apology to Americans. 'I mean when you're going up against a crazed dictator, you want to have your friends by your side. I realise it took more than two years before you guys pitched in against Hitler. But that was different, everyone knew he had weapons'!

[@MathAdam]

Amy Adams teaches maths!

[@Nihil975]

That was my first thought too

[@gabor6259]

Ginny Weasley teaches maths!

[@paulreader1777]

@@sockington1 'maths' is the more common terminology amongst English speaking countries outside the north American continent.

[@ScottKentEdu]

Even the laugh.

[@steffen5121]

Holly Krieger and Hannah Fry are my favorites. I love them.

[@jacobschiller4486]

Gee, I wonder why...

[@steffen5121]

@@jacobschiller4486 Me too. It's a mystery... 🤔

[@nathanokun8801]

Many catalysts in physics and chemistry are the loop kind where they change one or more times during their function, but the last step reverses this and puts them back where they started (carbon as the first step in a chain of elements inside of stars as an alternative method for changing hydrogen to helium, for example -- the "Solar Phoenix" process discovered by Hans Bethe). So this is not just an academic exercise.

[@gordonchan4801]

2:05 'horseshoe mathematics'

[@alexandersweeney6182]

Gordon Chan I love this reference

[@kyrlics6515]

@@alexandersweeney6182 ??

[@comradesero]

Nothing more endearing than seeing someone nerd out over math. Love it.

[@RobberStormfire]

I love the Holly videos. They are always interesting topics.

[@vikraal6974]

Na you love Holly

[@none_of_your_business]

@@vikraal6974 i am certainly guilty of this crime

[@jacobschiller4486]

😏

[@MATHguide]

Dr. Krieger has a fantastic ability to explain things very well.

[@wntu4]

Dr. Holly = Autolike. My favorite equation.

[@wafikiri_]

Start with 2. Use C=-2. Next number is 2x2 - 2 = 2. Periodic! Start with 3. Use C=-6. Next number is 3x3 - 6 = 3. Periodic! Start with 0. Use C=-1. Series is 0, -1, 0, -1, ... Periodic! Start with 1. Use C=-3. Series is 1, -2, 1, -2, .... Periodic! Start with S. Next number is S^2 + C, then next is (S^2 + C)^2 + C =S^4+2CS^2 +C + C^2, we want it to be equal to S. So, S^4 +2CS^2 -S +C^2 + C=0, or C^2+ (2S^2+1)C + (S^4 - S) = 0. Two roots. This is how I got lines 3 and 4 above, others are possible. Periodic! Maybe a period longer than 2 can be obtained. Too high a grade for my little time available.

[@alanvo4037]

"You're not going to do the next one?" "I think it's 677" BOOOOOOOOOOM!

[@LB31415]

3:32: can also have periodic starting from integers 2 and 3, when c=-7

[@dhampson545]

So weird to hear “zed” with an American accent.

[@FBDSG]

🇨🇦

[@UnabashedOops]

Dustin Boyd everyone I know says Zee FS

[@jsloan16]

Canadians say 'zed'.

[@CowmanCowman]

Massive respect for saying zed

[@t7612-v6h]

Big up to all the Canadians in the house.

[@vishalmishra3046]

With complex numbers, cycle of any length is possible. It is just a matter of solving a high order polynomial equation and getting c = f(z). Hence a cycle of 4 is easily computable by solving a 4th order polynomial equation. Higher order polynomial equations often require numeric analysis to solve them, so z and c become approximate, instead of exact transcendental expressions.

[@1ToTheInfinity]

4:47 She just hit the woah

[@chaossspy6723]

?

[@blackcat5771]

???

[@CalvinHikes]

Saw that.

[@xybersurfer]

oh. "Hit The Woah" seems to be a dance move. i was expecting something more interesting

[@sfbs]

1 to the Infinity lol with out even knowing

[@StreuB1]

The thing about Holly is that she is wickedly smart, more wickedly smart that most anyone would think if you just saw her walking down the street. People have this preconceived notion as to what a "mathematician" looks like or acts like and Holly breaks the mold with her ultra-bubbly personality and overall personna. The podcast that Brady just did with her is a testament to this and frankly, it was at LEAST an hour TOO SHORT!!! I could have listened to that (as well as Professor Frenkel's podcast 2 weeks ago!!!) for far longer than the podcast actually was. We need more Dr. Krieger!!! Oh and she's from Illinois so I might be biased there. ;-)

[@omikronweapon]

there's a PODCAST with Holly? I don't know if I can take that :/

[@matthewphilpott1702]

This math makes me feel uncomfortable but after its done i feel chill

[@crackedemerald4930]

"what happens is that the numbers get really *big* or they get really _smol_" I loved the way she said that.

[@TSBilliards86]

Videos like this make me want to go back for my PhD in mathematics

[@jdlech]

To me, the most fascinating thing about fractions is that you can easily and rationally describe numbers with perfect precision when decimals fail. 2/3, for instance. There are still numbers, such as Pi, which both methods of expression fail. But there are many times which expressing a fraction is clearly better than a decimal. Even so, I often see people expressing a decimal when a fraction would do better.

[@pierreabbat6157]

Holly Krieger: Here's a 3-cycle of z=z²+c James Yorke: Period three implies chaos!

[@Keldor314]

Yup! A 3-cycle also implies all other length cycles exist, but it looks like you need to move past rational numbers to the set of real numbers for this to work. Finding their exact locations may be impossible, though, since it involves solving polynomials of order greater than 5.

[@leapdrive]

Pierre Abbat, did you mean: f(z)= z^2+c?

[@Axacqk]

@@Keldor314 You don't find the "exact locations" of irrational quadratic roots either. When we say "square root of c", what we really mean is "the number that is the unique positive root of x^2 - c"; the former is just a shorthand notation for the latter, and using shorthand notation does not increase the "exactness" of the value's description; it's still the _same_ description. There is another single parameter, polynomial-root-giving function, "the unique real root of x^5 + x + c", that can be used to write solutions to quintic polynomials in closed form. This function is called the Bring radical, and the shorthand notation is BR(c). It is as easy to compute with Newton's method as the square root.

[@Keldor314]

@@Axacqk Hrmm, true. Perhaps the algebraic numbers are too narrow to cover the concept of "exact locations". Or too broad. Although the order of the polynomials you have to solve increases exponentially with the length of the cycle since we're finding the solutions to f(f(f(...f(z))))=z. Assuming that there isn't some shortcut produced by by the fact that f(z)=z^2+c, we need to solve huge polynomials. 4-cycle gives a 16th order polynomial, 5-cycle gives a 32nd order polynomial, and so forth. Does a relatively simple root finding function exist for arbitrarily high order polynomials?

[@williamtilt8991]

I just stumbled onto this channel and noticed pi million subscribers. I won’t be subscribing, don’t want to mess that up. Cheers.

[@jesusthroughmary]

11 minutes after posting is the longest I have ever taken to watch a Dr Krieger video.

[@hexane360]

It's interesting how the three numbers are equally spaced on the number line (-7/4, -1/4, and 5/4).

[@lucasnoritomi-hartwig3928]

Is there a place to see the proofs showing it is impossible for 4 and 5?

[@steveyankou4144]

Just watched this twice first thing in the morning. Thanks for turning my brain on.

[@xway2]

@Max Chatterji America, probably? Certainly on a sunday anyways.

[@uniqueusername_]

You can't just end there! You gotta give us the proofs!

[@numberphile]

No proofs but there is more detail in the second video on Numberphile2: kzbin.info/www/bejne/rGWviHmwid6bprc

[@sasha6454]

These proofs are left as exercises for the viewer.

[@scowell]

Unfortunately, there is not enough space in the margin... or this comment.

[@randomdude9135]

@Nighthawk814 wtf

[@ccgarciab]

@Nighthawk814 I'd imagine you ad a ^2 + c to the left side each time you try to prove a higher number of iterations?

[@IslandCave]

If we extend to complex numbers, then we can get any n steps where n is a positive whole number, by starting with: cos(2pi / n) + sin(2pi / n) i, and multiplying by cos(2pi / n) + sin(2pi / n)i each time, and in n steps we would get back to 1. That is it would take n steps to go around the unit circle centered at 0 on the complex plane, 1 / nth rotations at a time.

[@N0xium]

Christmas came early for all of us :]

[@danhalfhill9169]

I have no idea what this was about but I watched the whole thing! Fantastic!

[@t.d.mich.7064]

I got drunk and iterated all over the place, and the next day I was back to myself☺

[@unoewho]

I think Dr. Krieger is probably the best speaker that's ever been on Numberphile. Like, the most clear and concise vocally.

[@asconblake]

There is a point for me watching these kind of numberphile videos where I can't listen anymore, because the video made my math(s) mind going hyperactive and I understand a WHOLE lot more at once *math(s) giggles*

[@dalelange5406]

What a great video. It taught me something fascinating about fractions I never knew. More please, and thanks Numberphile.

[@oolveea]

can you please tell me what it is you find fascinating?

[@n3v3rg01ngback]

I like hitting the equal sign over and over on my calculator.

[@mtmath1123]

Some people think there’s nothing anymore to solve in maths, but even this kind of naive question is way difficult and actually enables us to go further like chaos and fractal. Sounds great.

[@OlafKnolaf]

It can be done with 4 using complex numbers. I've found a remarkable proof of this fact, but there is not enough space in the comment section to write it.

[@stefandries0]

Google docs + the link maybe?

[@yashthakre8106]

I would love to hear that

[@tarkus44]

hommage a Fermat?

[@RalphDratman]

It will be proved in about 400 years

[@x714n0____]

🤣🤣🤣

[@LesCish]

Early on, what is 26^2 +1? Since (X+1)^2 = X^2 + X + (X+1), 26^2= 25^2 + 25 + 25 + 1= 625 + 25 + 25 + 1 = 676. As Dr Krieger recalls, 26^2 + 1= 677. Fast easy sequential squares.

[@quantumhealing341]

Holly is a great teacher!

[@steve--smith]

I wish I presented as well as Holly. Clear, concise, memorable.... and talented.

[@JNCressey]

All of a sudden "fourths" started to be "quarters" for the rest of the video.

[@rosiefay7283]

No, quarters started out being quarters, then became "fourths". Fortunately, normality was restored later.

[@idjles]

And zeds

[@VndNvwYvvSvv]

@@idjles Zombies should never have been involved.

[@jshariff786]

Umm yeah? There are frequently redundant, interchangeable ways of saying things. I'm sure you'll get over it eventually...

[@greensteve9307]

Who cares?

[@chloe-un9cn]

you can also get loops for any integer z’s if c = (z-z^2) ie: z=2,c=-2 (2^2-2=2) z=3,c=-6 (3^2-6=3) i guess this would work for any values of c and z that satisfy the equation, but i just noticed it first for integers

[@leehutchinson7005]

There were no people as interesting as this when I did maths at uni

[@metalhos]

you mean, no grrls?

[@okie9025]

@@metalhos :DDDDDDDD

[@finlayhutchinson7370]

Dad

[@An.Individual]

"interesting". I know what you mean, nudge nudge wink wink.

[@TVIDS123]

It's Only Me did you see the size of the iterations on that?!

[@chillsahoy2640]

The number of letters in the word for a number tends towards 4. For example: Fifteeen, 7 letters; seven, 5 letters; five, 4 letters; four, 4 letters.

[@KurtRichterCISSP]

The Amy Adams of maths 😁

[@Fizzfaldt]

Obviously we restricted to the reals at some point (cause n=4 or higher is pretty trivial to do with complex numbers.. plus you've done videos about this that loop (or close to loop) with complex numbers). However I don't recall during this video that you ever *said* we're restricting to the reals. For anyone curious, this is relatively simple to do with complex numbers to have loops of any positive integer length N. Starting with N=4 (the first impossible one one if you restrict to reals) Let C=0 I'm using polar coordinates here just cause it makes it easier to read/explain. Start with z= (1, 2Pi/15) (1, 2Pi/15) => (1, 4Pi/15) => (1, 8Pi/15) => (1, 16 Pi/15) => (1, 32 Pi/15) 2Pi = 30Pi/15 so 32Pi/15 == 2Pi + (2Pi/15) so we're back to the beginning. If you want N numbers in the loop, choose as starting point (polar coordinates) (1, 2Pi / (2^N-1)) Since that's on the unit circle, it just doubles the angle each time you square it. So after N doublings it becomes (1, 2^N * 2Pi / (2^N-1)) or equiv (1, (1 + 1/(2^N-1)) * 2Pi) and you can throw away the 2Pi so you're back to the beginning. I *expect* that there are an infinite number of solutions (ignoring duplicates caused by redundancy of polar coordinates.. e.g. infinite number even if you specify in cartesian coordinates) but I don't have any intuition as to where to look next.

[@TheBaggyT]

I am very appreciative of Numberphile videos, but something I keep asking myself is: why do they always write on paper instead of using a whiteboard? Isn't this just a waste of paper? This is a genuine question - if someone can answer this, I'd appreciate it! Thanks.

[@skipfred]

I don't know but I think it's just tradition. They're writing on what appears to be recycled paper anyway (from the color and texture). The amount of paper they're going through is nothing compared to even a small company.

[@TheBaggyT]

@@skipfred I get that. But comparing the amount of paper isn't really the point... the vast majority of similar videos use other technology (whiteboards, tablets, etc.) and have zero paper usage.

[@MarcusCactus]

My thesis director told me: "Never be afraid of wasting paper." One among the reasons being that chalk- or whiteboards are not permanent. Another being that you should not clutter your next equations in the blank spaces between previous ones. A third would be that limitations are bad for free science.

[@TheBaggyT]

@@MarcusCactus I get that when you're doing research, or working towards something big like a thesis. But if I'm making a video about something I already know about (which I assume these people are, that they haven't just thought of the idea on the spot), I would use a whiteboard because there's no need to keep a more permanent record. If you organise a whiteboard properly, there's no need to write new equations in the space between others. And how is a whiteboard a limitation?!?

[@legendhero-eu1lc]

Thank you for the video! You friends are all super awesome!

[@RyokaSteamboat]

I cant be the only person who sees Amy Adams

[@petersellers9219]

I'm still trying to calculate the number of freckles. A teaser!

[@AndreyPutilov]

They did not stress out the important thing is that we are talking about fractions. They said it at the beginning but in in the problem. The higher-period equations should definitely have solutions in real numbers. But this might be a problem solving in rational set. For example, for period of two, we have: z1^2 + c = z2, z2^2 + c = z1; (z^2+c)^2 + c = z; z^4 + 2c.z^2 - z + c^2 + c = 0. For higher periods we will have equations for z^8, z^16, z^32, ... Probably it will have a solution but not in rational numbers, I think. That's the key of ythe problem.

[@joshuagoodman2620]

Has chalkboard in background Uses paper towel from school bathroom

[@JWentu]

Her videos are always very interesting! thank you!

[@OlbaidFractalium]

Z->Z^2 +C is really mysterious formula.

[@vikraal6974]

Laughs in Mandelbrot photo

[@aidarosullivan5269]

Hooray, Holly is back!!

[@timohuber536]

I‘d really love to see a 3b1b-Video on this!

[@hodymcsackschweid5012]

I would be interested in hearing about the proofs, showing which iterations work and which don’t. For example showing why four iterations isn’t possible.

[@ditrixgenesis781]

Numberphile comments were the last place I thought I'd see thirst comments

[@jshariff786]

Why?? There is significant overlap between the set of math nerds and the set of desperate men.

[@amaarquadri]

If we expand the set of considered values to any complex number instead of just real fractions an interesting thing happens. The problem of finding a cycle of length L can be reduced to a problem in number theory by considering complex numbers on the unit circle. Let c = 0, so the function we are iterating is z = z^2. Let the starting point be a complex number on the unit circle e^[(2*pi*i*)/k] for some integer k. The nth term in the sequence (counting from 0) is then e^[(2*pi*i*)*((2^n)/k)]. Thus, there is a cycle of length L where L is the smallest value for n that satisfies 2^n = 1 (mod k) if such a value exists (I'm not sure if there exist values of k for which there is no such L). If we let k = 2^L - 1, then when n = L, 2^n=2^L=1 mod (2^L - 1). Also, this is definitely the lowest such value that is congruent to 1 mod k. Thus, we can get a cycle of any length L by iterating z=z^2, starting with e^[(2*pi*i*)/(2^L - 1)].

[@chrisdaley2852]

It's the restriction of the function to the rational domain that makes this question interesting as Sharkovskii's Theorem solves the problem over the real domain too. We need a discontinuous domain to rule it out.

[@Mario-dr8gm]

Try C = -3 and start with 2 It creates a loop

[@andrewcheng1948]

2,1,-2,-3,6,33,906...

[@thiantromp6607]

Andrew Cheng 2, 1, -2, 1, -2, 1...

[@eddiep5676]

FINALLY!!!!! She’s back!!!! More please!!!!!!!!!!!!!!!!!!!!

[@Sillu129]

Kinda reminds me the circle of fifths..... ;)

[@Ryuuuuuk]

Yes, my mind immediately made the same link, interesting.

[@CGoldthorpe]

Try this IF a number is divisible by two, then divide by two, but if not, multiply by 3 and add 1. Also try divide by 3 (when evenly divisible) or multiply by 4, then subtract 1. 19 is "reentrant"

[@yuyiya]

The Goldthorpe Conjecture?

[@CGoldthorpe]

@@yuyiya No I cannot take credit. I think it is from Douglas Hofstadter

[@pradhyumnchoudhary7383]

Amy Adams at it again!

[@mateuszbez516]

Numberphile is getting very close to 3,14 milion subscribers.