This pattern breaks, but for a good reason | Moser's circle problem

Подписаться 6 млн

Просмотров 1,9 млн

50% 1

An apparent pattern that breaks, and the reason behind it.
Summer of math exposition: 3blue1brown.substack.com/p/so...
Learn more at some.3b1b.co/
Help fund future projects: / 3blue1brown
An equally valuable form of support is to simply share the videos.
For the long-time viewers among you, if this sounds familiar, it's because it's a remake of one of the earliest videos on the channel. It's such a wonderful problem, and the audio/pacing in earlier videos was really suboptimal, so I wanted to freshen it up a little here.
Timestamps
0:00 - The pattern
2:20 - Counting chords
4:03 - Counting intersection points
6:20 - Euler's characteristic formula
11:30 - Connection with Pascal's triangle
15:10 - Reflections
Correction at 8:56 - The number of the regions should of course be (1, 2, 3, 4, 5), instead of (0,1,2,3,5)
Thanks to these viewers for their contributions to translations
Arabic: mouhsiiin
French: Jonhfing
Hebrew: Omer Tuchfeld
Hungarian: Fabó Bence
Russian: fedor
Spanish: Paweł Cesar Sanjuan Szklarz
Thai: @korakot
------------------
These animations are largely made using a custom python library, manim. See the FAQ comments here:
www.3blue1brown.com/faq#manim
github.com/3b1b/manim
github.com/ManimCommunity/manim/
You can find code for specific videos and projects here:
github.com/3b1b/videos/
Music by Vincent Rubinetti.
www.vincentrubinetti.com/
Download the music on Bandcamp:
vincerubinetti.bandcamp.com/a...
Stream the music on Spotify:
open.spotify.com/album/1dVyjw...
------------------
3blue1brown is a channel about animating math, in all senses of the word animate. And you know the drill with RU-vid, if you want to stay posted on new videos, subscribe: 3b1b.co/subscribe
Various social media stuffs:
Website: www.3blue1brown.com
Twitter: / 3blue1brown
Reddit: / 3blue1brown
Instagram: / 3blue1brown
Patreon: / 3blue1brown
Facebook: / 3blue1brown

Опубликовано:

18 май 2024

Ссылка:

Скачать:

Готовим ссылку...

Добавить в:

Мой плейлист

Посмотреть позже

Комментарии : 1,9 тыс.

@someknave 10 месяцев назад

There's another great sequence that goes, 1,2,4,8,16,30,60,96... The number of divisors of n!

@faastex 10 месяцев назад

Is that supposed to be an exclamation mark or a factorial? 😅

@physiologic187 10 месяцев назад

@@faastexfactorial.

@lilasarkany3381 10 месяцев назад

@@faastex God, that reminds me of a time, when I got an exercise with an exclamation mark right at the end of the equation and I interpreted it as a factorial. Making it 10x more complicated and unsolvable.

@jogadorjnc 10 месяцев назад

@@lilasarkany3381I once had an exam with a differential equation that we had to solve, but the differential equation was in quotation marks I didn't notice the initial quotation marks so I just assumed the last term was supposed to be the second derivative of what it actually was The problem got harder, but was still solvable, and I did solve it. But the prof still gave me 0 points for it even after I contested the grade

@divisix024 10 месяцев назад

That’s so dumb, why would anyone put quotation marks in a differential equation?

@iteerrex8166 10 месяцев назад

“Simple yet difficult” problems are always lots of fun. You go through pages and pages of papers, and suddenly you’re like.. wait a minute.. OH COME ON.. that shoulda been easier 😂

@piranha031091 10 месяцев назад

Have you heard of the Collatz conjecture? ^^ One of the simplest problems there is. You could easily explain it to a 6-year old. Nearly a century since it was first formulated, and still unsolved. Despite the efforts of many brilliant mathematicians. Erdös went so far as to say "Mathematics may not yet be ready for such problems".

@KiLLJoYYouTube 10 месяцев назад

@@piranha031091I do wish amateur Mathematicians would give Collatz a rest smh

@hoebare 10 месяцев назад

@@KiLLJoYRU-vid It's not just amateurs. Erdos famously took a stab at it too and said "Mathematics is not yet ready for such problems." Some day math _will_ be ready, so it's fine if everyone keeps poking at it to see if that time has come.

@vigilantcosmicpenguin8721 10 месяцев назад

There's no surer way to occupy a mathematician than by giving them a problem that seems simple.

@iteerrex8166 10 месяцев назад

@@vigilantcosmicpenguin8721 I’m actually working on a simple engineering problem right now, which I thought of it myself, and got myself down the rabbit hole of prime numbers. Seemed so promising a few days ago 😂

@shmoola 10 месяцев назад

13:18 Love how in the merging animation size of a number ends up proportional to its value. I.e. when 1 and 4 merge to 5, then 1 takes 20% and 4 takes 80%

@donsanderson 10 месяцев назад

Thanks for pointing that out. I had totally missed it, but it's a fun little flourish

@Un-TedxTalks 4 месяца назад

yeah, i noticed it after your comment, i made me happy 🤧🤧

@mohammadmoin-ud-din8387 4 месяца назад

It showed how much effort was put into this. Love when people put so much thought into helping others learn. What a great and intuitive learning experience!

@NachitenRemix 3 месяца назад

WOW thats super impressive, and I thing its more impressive hoy you noticed it, superb attention to detail!!

@pendragon7600 Месяц назад

How on earth did you notice that

@danishazhar86 4 месяца назад

I'm just impressed that you were able to make a poem AND a FREAKING SONG about this problem as a CHILD.

@matthewkendrick8280 2 месяца назад

I can’t believe no one else is talking about that. The poem was incredibly impressive, and now I really want to know the melody of that song. I’ve never heard of poetry about math but hey I guess there’s something for everyone.

@apotatoman4862 2 месяца назад

@@matthewkendrick8280 hallelujah

@Gozieaaa 2 месяца назад

@@apotatoman4862 😐

@iitn8437 2 месяца назад

@@matthewkendrick8280 poems have been always made by mathematicians but we didn't got taught. I learned about one rishi named pingala who wrote even verses describing pascals triangle. But at that time as decimal number notations were not in ecistance he used letters to denote that idea.💡

@alvarito8496 2 месяца назад

Actually Easy for me...

@andrew_ray 10 месяцев назад

13:10 I love that in this animation, when two numbers are combining, each turns into some portion of the sum proportional to the value it contributes to that sum (so that when 1 + 1 turns into 2, each 1 turns into half of the two, but when 3 + 1 turns into 4, the 3 turns into three quarters of the four and the one into only one quarter). You absolutely didn't have to do that, but I find it strangely pleasing that you did.

@OnDragi 10 месяцев назад

Ah, so cool! Didn't notice it myself, but totally agree-it's so needless, but sooo satisfying, haha.

@521Undertaker 10 месяцев назад

Good eye, my friend!

@bigguyherealt 10 месяцев назад

AA someone else noticed too!! i thought that was so neat!

@ericfingerman 10 месяцев назад

How did you see that!? 🤯

@mspeir 9 месяцев назад

His animations, as much as, if not more than, his explanations, are what make his videos so easy to follow and understand.

@InactiveAccount325 10 месяцев назад

A more intuitive way to understand the formula for number of areas from n chords, is that for each additional chord, it creates a new area, and another new area for each chord it intersects. So the total number of areas is the original circle + the number of chords + the number of chord intersections.

@JimBalter 10 месяцев назад

Nice. That also covers the cases where more than two chords intersect at the same point.

@sszone-yt6vb 10 месяцев назад

Was just wondering about a more direct way to prove the formula and here it is!

@bachirblackers7299 10 месяцев назад

@@JimBalter this is even more beautiful Maths problem and it will lead to a very complicated formula no one dare to come out with a video covering it . Much beautiful Maths hidden in there . ❤❤❤❤

@3blue1brown 10 месяцев назад

Nice! I hadn't thought about that.

@xelaxander 10 месяцев назад

Of course it’s induction!

@HunterJE 10 месяцев назад

The resulting integer series are also known as the "Parker powers of 2"

@slimsh8dy 9 месяцев назад

It was a Parker square of a pattern

@peorakef 5 месяцев назад

i just noticed that parker powers are equivalent to the botez gambit.

@locrianphantom3547 4 месяца назад

@@peorakef-9 point score 😢

@Nathan-wp1ir 7 месяцев назад

Being able to understand the math behind this problem is a true human blessing, let alone the fact that a stranger on the internet shared it with me

@j_sum1 10 месяцев назад

Well done Grant. My 11 year old son was able to follow this and his curiosity was satisfied. For context, his formal maths education has not included functions, nor algebraic manipulation, nor factorials, nor the multiplicative principle for counting permutations. He had met Pascal's triangle but knew very few of its patterns. And a huge amount of the terminology was brand new: graph, edge, permutation, function etc. In other words, you have done a superlative job of communicating. The fact that he was able to follow and understand the reasoning when so much was new to him is testament to a job well done.

@theiigotriangularround4880 10 месяцев назад

oh come on hes 11 not a kid wht else do u expect if he doesnt get the all the parts other than eulers formulae then he lacks stuff

@francescoalexgiacalone878 10 месяцев назад

@@theiigotriangularround4880how can someone be so arrogant? Do you not know kids or just want to show off?

@Eagle3302PL 10 месяцев назад

@@francescoalexgiacalone878That dude was being so sheltered, many 11yo are still struggling with division and exponents from what I've seen.

@hhoopplaa 10 месяцев назад

guys I'm not sure if the first reply is ironic but I liked it bc it's funny

@delete7316 10 месяцев назад

Props to your son for taking an interest in maths at such a young age.

@wolfelkan8183 10 месяцев назад

I don't know much about diophantine equations, but if I've learned anything from Matt Parker, the first step is to write some dodgy Python code to brute force check if there's another power of 2. Obviously, we can't go to infinity, but this will at least identify if the conjecture is obviously false. I can confirm that other than the cases already discussed, the number of regions will never be a power of 2 if the number of points is less than 15 billion.

@oliviapg 10 месяцев назад

The second step is to make a video about your code so that someone else can rewrite it so it takes milliseconds instead of months

@Howtheheckarehandleswit 10 месяцев назад

@@oliviapgAlready on it

@Pystro 10 месяцев назад

@@oliviapg And the third step seems to be to wait until you asking the question has spawned an international collaboration of mathematicians all working on the problem. I'm not 100% sure on that one. I haven't watched the video on the "Great Collatz Collab" yet. And I don't know if having the last name "Collatz" is a strict requirement for this to work.

@lukeabby8927 10 месяцев назад

So far I'm at n = 2 billion and haven't found any greater than n = 10.

@Howtheheckarehandleswit 10 месяцев назад

@@lukeabby8927 I've just passed n=2^32 with no hits Edit: typed the wrong power of 2

@tonitete 3 месяца назад

well, there is not another power of 2 in at least the first 10,000,000 numbers, maybe later on there is another, but my CPU is crying so i better stop here.

@marcusscience23 10 месяцев назад

The reason Euler’s formula works for both polyhedra and planar graphs is because the former can be transformed into the latter by stereographic projection, preserving the relations of vertices, edges, and faces. Additionally, the reason Pascal’s triangle shows combinations can be understood this way. With n items, there are a certain number of ways to choose k of them. With (n+1), you can either include the new item and end up with (k+1), or not include it and have k. Both of these are carried onto the next row.

@canyoupoop 3 месяца назад

or cuz it's Euler's formula

@pohnprayin 2 месяца назад

Thank you! I got stumped by this briefly, as well as when he converted the segmented circle into V-E+R, since he included the arcs. But then I worked it out and realized that the formula continues to work for curved segments and could also work in 3-D space

@warasilawombat 10 месяцев назад

Euler really was such a gift to humanity

@davidjennings2179 10 месяцев назад

It is a great word the Germans gave us.

@JasperWilliams42 10 месяцев назад

Euler was gay and I have proof. In 1987 he freaked a man. You can tell because his name starts with the letter E.

@shahram6869 10 месяцев назад

Eh, if it wasn't Euler it would have been someone else.

@markjenkins9424 9 месяцев назад

@@shahram6869Very true

@thanewolf740 7 месяцев назад

@@shahram6869yet it was he who was known for it, like a race if you remove the one who got first someone else is bound to finish but we still held the person in first place in high regard. In short its because euler did it first is what is important. Also you are scum for trying to remove value to a person achievement. And in your logic some other sperm could've fertilized your mother's egg before the sperm that created you thus everything you are in your logic is unimportant.

@pakkiufung883 10 месяцев назад

The fact that 3blue1brown wrote a poem about this problem when he was a kid is mind-blowing Edit: Ok, I know there are some angry replies stating that ‘when he was younger’ doesn’t mean kid, but to an 11-year old, it’s mind-blowing all the same.

@FG-Supercharged 10 месяцев назад

... And certifiably a nerd 🤣 Seriously, just what sent him down that path, right? 🤔

@feliksporeba5851 10 месяцев назад

I'd give you a like, but that would break a beautiful 69

@arnet95 10 месяцев назад

What? That's probably the least surprising thing I've heard today.

@Osskyw 10 месяцев назад

built different

@phenixcrow 10 месяцев назад

@@OsskywSteered different, more than likely.

@RichardHolmesSyr 10 месяцев назад

For additional powers of 2: You can consider a generalization, where you look at sums of the first m numbers in the nth row of Pascal's triangle. As in the m=5 case you get m powers of 2, followed by 2^m-1, and then the (2m-1)st entry is 2^{2m-2}. For the m=1 case the numbers you get are 1, 1, 1, 1, 1... (all powers of 2!) and for m=2 you get the natural numbers, so trivially you get all the powers of 2. Beyond that, for m=3 the 90th entry is 4096, for m=4 the 23rd entry is 2048, and otherwise for the first 800 entries for each m through 800 I find no other powers of 2. You can take these numbers and color the even and odd numbers differently, and you find they form a Sierpinski triangle, just they do with Pascal's triangle itself: There are increasingly large triangles of even numbers outlined by odd numbers. So as n gets large the density of even numbers increases. However the density of powers of 2 within the counting numbers decreases much faster. So my otherwise uninformed guess is that for any m an infinite number of powers of 2 occur, but (for m>4) even the first one probably appears for n much larger than 800.

@thebighg 7 месяцев назад

Found this channel an hour ago from the short presenting this problem. Immediately hooked to see why. I finally found a place that can explain math so well and visually beautiful that people are drawn to it and can love math again. And this is the power of right education. Thank you.

@jucom756 10 месяцев назад

This is a fun problem to give to high schoolers right before they get binomial, and then revisit after, because it uses a lot of the concepts you learn with binomials and polynomial approximations.

@Secret_Moon 4 месяца назад

I fear you'd make them quit school...

@lwade225 3 месяца назад

@@Secret_Moonyeah most of my classmates want to quit school and they haven’t even gotten close to this

@larsprins3200 10 месяцев назад

To solve Moser's circle problem you don't need to use Euler's theorem for graphs (but the detours to Euler and the Pascal triangle certainly make the video a lot more educational). Once you know that there are (n choose 2) chords and (n choose 4) intersections, you continue as follows: The original circle without chords is 1 region. Now start drawing chords, one after the other. When you start drawing a chord from a point on the circle, the chord starts cutting an existing region up into two new regions right away, thus adding one region. Whenever a chord crosses a chord drawn earlier at an intersection, it finishes cutting up a region and starts cutting up a further region into two new regions, thus adding one further region. The total number of regions is thus 1 + "the number of chords" + "the number of intersections". The answer to Moser's circle problem is thus 1 + (n choose 2) + (n choose 4). Yet another great video by 3Blue1Brown, thanks!

@TimothyZhou0 10 месяцев назад

This is the "morally correct" combinatorial proof, makes it much more intuitive to interpret the formula. Thanks for posting!

@NeilGirdhar 10 месяцев назад

This is just the proof of Euler's theorem for graphs.

@klemo2529 10 месяцев назад

would have been cool if he added this at the end like he normally does for us to understand something intuitively great and clever way to see this problem, thanks for posting

@mystic3549 10 месяцев назад

awesome!! this is the most elegant intuition ive come across so far for this :) tysm !!

@adityaazad1939 Месяц назад

I checked 10,000 iteration of Pascal's triangle and found that 10 was the last iteration where the sum of first 5 values was a power of 2. I have checked 100,000,000 iterations using the Euler's Characteristic Formula and have found .............. that 10 is still the last iteration that gives a power of 2😞

@Aditya_196 22 дня назад

We have infinity more to go

@prestonfisher2632 17 дней назад

Can find any algorithmicaly?

@justarandomdood 10 месяцев назад

3:55 music and the chords lighting up line up so perfectly I'm glad you redid this video. Original was great but this one has just a little something on top that makes it fantastic

@emperor-pavoom3929 10 месяцев назад

Was NOT expecting that joke at 16:07 very glad I stayed to the end of the video

@widmur 10 месяцев назад

Pretty sure that’s a classic algorithmic problem.

@Vaaaaadim 10 месяцев назад

@@widmur if you are not just trolling and you actually think it is a reference to 3SAT, you are mistaken.

@andrewfleenor7459 10 месяцев назад

I'm sure there are some interesting combinatorics problems to be solved.

@jayvaghela9888 10 месяцев назад

3 some 😂

@widmur 10 месяцев назад

@@Vaaaaadim I was making a bad joke about 3SOME being a homophone for 3SUM.

@billyjoethethird8436 10 месяцев назад

The explanation of Euler's characteristic is simply mindblowing. It's much easier to understand than the inductive proof!

@newbie4789 10 месяцев назад

It actually IS the inductive proof you know . Just shows with diagrams and really intuitive language

@Lait_au_Miel 10 месяцев назад

Grant provides an inductive proof, though

@christianbraune807 10 месяцев назад

@@Lait_au_Mieljust what I was thinking... Such a clever way to hide an inductive proof :)

@norude 10 месяцев назад

That IS the inductive proof

@Smitology 10 месяцев назад

It IS an inductive proof: starts with a base case (trivial graph) and shows that it holds true when the edge count is incremented for an arbitrary graph

@IhabFahmy 10 месяцев назад

_The thought and care that you put into the logical sequencing of the concepts, the animations, the selection of how to illustrate a spoken statement, are beyond fantastic. The improved intuitive understanding that results is thus extremely solid and greatly satisfying. That's exactly how I would wish to perfectly explain anything to anyone. Thank for doing this! (that's an exclamation point, not a factorial 😂)_

@111games. 4 месяца назад

-thanks for the additional information, I was wondering what is a factorial of "this".-

@thegurw1994 4 месяца назад

The italicized emoji makes me uncomfortable

@thunderboltpoplays1177 Месяц назад

_m_

@apollo6409 10 месяцев назад

I love videos like this because I did further maths at GCSE and maths at A level but there was NEVER any explanation for stuff like how the n choose 2 stuff could be implemented in problems. The entire point of the curriculum is to turn us into the fastest calculators we could be- Not to actually critically think about puzzles, or show us how the calculations we were doing applied to challenges like this. I wouldn't have hated them so much if we got to see cool stuff like this!

@sebgamingkid 10 месяцев назад

I love the bit at 8:30 being 3 blue faces and 1 brown face. attention to detail & i love it

@themathhatter5290 10 месяцев назад

Also interesting to notes he labels the faces 0,1,2,3... 5?

@nathanbekin4660 10 месяцев назад

I didn’t even notice that, thanks for poynting that out!

@SimonBuchanNz 10 месяцев назад

@@themathhatter5290it's so you spend the next hour looking for region 4 😋

@EinChris75 10 месяцев назад

What a nice problem. Somehow I got goosebumps when he explained the reason for the missing 1 in 31. It suddenly all made sense.

@nipurnshakya9312 10 месяцев назад

What are your qualifications in mathematics ?

@EinChris75 10 месяцев назад

@@nipurnshakya9312 nothing fancy Just ordinary engineering mathematics of a masters degree in CS.

@dmlppspringtraplover 10 месяцев назад

I still don't fully understand why 1 is missing in 31 is it because not a math is a factor 2. Sometimes it's different.

@bencreagh5074 3 месяца назад

I love how you wrote a poem over this problem it really shows how invested you are in it and I read the poem, I have tried many times in school to write a poem, and my brain just can’t do it

@romanceano7519 10 месяцев назад

It is so nice to feel again being a kid in math class, marveling on the the magic of how everything connects and wondering what's next. Thank you Grant for all this hapiness.

@Luke-me9qe 10 месяцев назад

I love that he does not just get satisfied to answer the issue but explains also why it is the way it is. Plus in a nice and calm way.❤

@trillionalyssa 10 месяцев назад

Another way to see that the sum of a row in Pascal’s triangle is always a power of two uses the fact that each entry is of the form n choose k. If you look at a given row, it’s going to be the sum of all n choose k where n is a constant and k goes from 0 to n. But if you’re looking at, say, the number of ways to choose 0, 1, 2, 3, or 4 elements from a set of size 4, that’s just the number of subsets, which is 2^n (because for each element in the larger set, it can either be in the subset or not, and all of these choices are independent). I never actually noticed this before!

@krngl421 10 месяцев назад

Nice ❤👌

@klemo2529 10 месяцев назад

clever

@CharlieDavis-ww4zf 10 месяцев назад

Such lucidity of explanation! Such a reassuring voice! 3B1B is a bulwark against chaos.

@MaxCareyPlus 10 месяцев назад

I actually did the thing in a 3b1b video where I "paused and pondered" and got to the right answer! ...but through different reasononing. So then I paused and pondered some more and realised that our reasonings were equivalent! The satisfaction this gave me can't be expressed in words, but warrants an appreciative comment. 🙏❤️

@_AvaGlass 10 месяцев назад

I just watched the original video yesterday! I'm so excited for the remake!

@limepop340 10 месяцев назад

Fantastic work with the rewrite! This version flows very cleanly, and it’s easier to follow compared to the original

@markuskeller4281 10 месяцев назад

Thank you Grant, your videos are always worth a watch even if the topics are over my head for the moment.

@TQuantP 10 месяцев назад

It's nice that the first thing that dawned on me when you started introducing Europe's Characteristic was a differential topology argument, but it never occurred to me that it can be used in a simple combinatorial problem. Kudos for using math that I thought during my masters have no tangible applications to issues like this one. Nice video !

@PowerhouseCell 10 месяцев назад

*Amazing as always! Thanks for inspiring a generation of learners to love math and push education forward :)*

@umbrellajack 10 месяцев назад

I'm so happy you are posting on RU-vid more!!! You are the best!

@user-qy7mu1yo1e 10 месяцев назад

I'm a kid in Korea, I love math, and I'm really impressed. I once wrote a poem about Ptolemy's theorem, so I'm loving this stuff.

@rahimeozsoy4244 3 месяца назад

A Poem? What about finding a proof

@jo555444 8 месяцев назад

This is a fantastic video: You explain every single detail and give reasons for every open question which raise from the original problem. Even explaining Euler's formula and not just referencing to some other video. Thank you! It was a pleasure to watch and understand. Every math teacher should watch this and learn how to present proofs.

@Dziaji 10 месяцев назад

Holy shit, Grant. Just holy shit. You have somehow outdone yourself. I can’t even on this one. And you made the Hallelujah parody to tease it?!?!?! This was the most masterful rollout and execution of a thing that I have ever come across. I’m the opposite of speechless; I could talk about the amazingness of this video infinitely! WELL DONE SIR

@harrymoschops 10 месяцев назад

This might be my favourite 3blue1brown video yet, I should watch more

@ineffabletryx6528 7 месяцев назад

This video was even more satisfying to me because my calculus class JUST learned about how series work, and so it was gratifying to be able to test out the equations as they came up in my own series of the problem

@poisongame2625 10 месяцев назад

This solution was so much more satisfying than I thought it'd be and I'm so glad I chose to watch both the explanation and the song

@Nethershaw 10 месяцев назад

You are a rock star for posting this gem at this time of day while I am bored and looking for something to learn in the middle of the night.

@torinor6703 10 месяцев назад

This whole video is beautiful man. Keep up the good work

@scikick 5 месяцев назад

I watch all of 3b1b videos, but this is one my brain was able to follow fully. Such elegance!

@clubsandwich559 10 месяцев назад

31… it better not be computing pi

@Dayanto 10 месяцев назад

7:03 An important caveat to this is that it only applies when you have exactly one connected "island". If you remove the last vertex or add new shapes that aren't connected to everything else, it breaks down. In a sense, the formula counts the number of islands (+1 for for the background region).

@matiasnovabaza8208 10 месяцев назад

at 7:33 "I should add this is all with respect to connected graphs", so yes, all is one 'island', but is a nice observation, it let you count how many regions are in a not connected graph (if you know the number of conexed components)

@MsAora 10 месяцев назад

This was a very interesting problem. I wish we had a mathematics space where such problems were dished out perhaps on a weekly or daily basis and members would try solving them on their own before the answer was revealed.

@alexkonopatski429 10 месяцев назад

I love, just love your channel ❤ I want to thank you for what you are doing. Really. Each and every time you inspire me again and again, and my love for math just grows and grows. Seeing the beauty of math is something truly wonderful and you have most definitely helped me achieve that several times. Thank you

@danbromberg 7 месяцев назад

Fascinating problem and, as usual, your graphics are superb!

@diamondblade3490 10 месяцев назад

I remember once in my middle school math class, we were seeing how many pieces we could cut a cake into with a certain number of straight lines. The first two were obvious. One line gave two, two lines gave four. We almost thought that three would be six, but then I was like "hey wait you can offset the diagonal to make a seventh slice" because we explicitly did not care about how big the slices were, just how many there were. Needless to say, it got kinda crazy after that.

@josephward4918 10 месяцев назад

If you make the third cut horizontal instead of vertical, you can get 8 slices!

@matheusmotadegodoy1572 8 месяцев назад

@@josephward4918how come?

@josephward4918 8 месяцев назад

@@matheusmotadegodoy1572 Make two normal vertical cuts to get four slices. Then a horizontal cut to split each one of the four slices, making eight. If only the top of the cake has frosting, then the top four pieces will have frosting and the bottom four pieces won't.

@matheusmotadegodoy1572 8 месяцев назад

@@josephward4918 the third cut can only go through 3 slices at most, so that would be 7 instead of 8

@josephward4918 8 месяцев назад

@@matheusmotadegodoy1572 It might help to imagine taking the cake out of the pan first. The last cut doesn't touch the top or bottom of the cake at all, it just goes from the right side to the left side of the cake. The cake after the cuts ends up looking like a Rubik's cube with only 2 squares on each side; it has a top layer with 4 pieces, a bottom layer with 4 pieces, and a horizontal cut straight through the middle.

@khrysarorosindustries 10 месяцев назад

To answer your question about if the pattern ever shows another power of 2, it doesn't. For any given row of Pascal's Triangle, The first value is always 1. The second value is always n (n choose 1). After doing this up until n choose 4, which would be the fifth value, becomes n(n-1)(n-2)(n-3)/4!. In the end, after combining the 2nd-5th terms and expanding out the polynomial, you get (n^4-2n^3+11n^2+14n)/24. Now you need to find an integer value for n that equals one less than a power of 2. This works well for the first rows, you end up with the correct value. Plugging in 0 for n gives you 0, representing the first point. Plugging in 1 gives you 1-2+11+14, which is 24, divided by 24, making the one you need. This explains why it only works for the first group of numbers. The equation repeatedly inflects for the first 5 values, and then goes through once more at n=9, giving that 256 value. afterward, the equation slowly diverges, and at n=18, the equation is at 4048. and appears to never hit the 2^x graph again.

@deltalima6703 10 месяцев назад

"Appears" 👋

@khrysarorosindustries 10 месяцев назад

@@deltalima6703 someone more experienced in multivariable analysis may be able to prove me wrong.

@Soul-Burn 10 месяцев назад

I watched the previous version of this video and uncharacteristically for these videos, I almost fell asleep. This version is *much* better, easier to understand and follow. Goes to show how much you grew as a science educator, and how hard it is to actually convey ideas in an engaging manner.

@charliepalmer484 10 месяцев назад

It always amazes me how much time and effort goes into every animation. Did anyone else notice that when Grant talks about the layers of Pascal’s triangle donating two of itself, the number occupies the correct ratio in e new number. So 1 and 3 making 4: 1 fills in only a fourth of the 4. It’s like that for every one. How many other things did I miss!

@secondthread-uc9bd 10 месяцев назад

Ur doing gods work , showing the true beauty of math to people , i wish i had this channel in my school days , im sure your videos alone have inspired many people to pursue math or related fields ,and then those people do great in their life , your impact only grows exponentially its amazing , honestly that is the biggest achievement ,bigger than any medal or any paycheck , any prize , you really are changing the world video at a time .

@emanuele1750 10 месяцев назад

I came across this video by chance. Not only I enjoyed the content, but I also appreciated the format on how everything is described, words and animations. Also, the pace of exposition is perfectly timed with my capacity to process the information that was just provided, so I was able to follow without pausing the video. Thank you for this great content, you got a new Subscriber.

@dadush4 9 месяцев назад

The amazing display of the problem of induction... thanks for the vid!

@boonjabby 5 месяцев назад

Really well explained. I thought I was getting lost until the end and it all came together for me. Thanks 3b1b

@PurityXD 10 месяцев назад

I really just can't get enough of 3b1b videos. Thank you so much for your work, I'm happy to see all of your success as a long time viewer :D

@byronwatkins2565 10 месяцев назад

At 13:20, Pascal's triangle is used to expand binomials to the nth power. So consider 2^n = (1+1)^n and expand it. The coefficients in each row must add to 2^n.

@erikvanderheide1440 9 месяцев назад

This should have been shown at my first calculus course in university - directly captivating, plus explaining the concept of n choose 2 in an intuitive way.

@whiskeyfur 6 месяцев назад

usually with these math based videos.. they're either too simple and I get bored.. or you need a calculus course just to understand them. This video hit the sweet spot for me and it made it UNDERSTANDABLE, from start to finish. Thank you!

@emperor-pavoom3929 10 месяцев назад

I remember the original video! Looking forward to seeing how it's remade here!

@JasperWilliams42 10 месяцев назад

Delete this comment, because nobody asked.

@siddharthnemani5301 10 месяцев назад

Hey Grant. Eagerly waiting for your statistics series which you promised you would do this year. I know I shouldn't ask here, but I don't know where else to ask. Please do one like your Linear algebra one, it was really helpful for a foundation for QM for me.

@PackerFanGamer 10 месяцев назад

Not a series, but his last several videos have all been statistics-y.

@siddharthnemani5301 10 месяцев назад

@@PackerFanGamer I was talking about a course, like the linear algebra one.

@wavingbuddy3535 9 месяцев назад

As usual a fantastic video. This is why we love you man. Your videos are always a pleasure truly.

@paeng1935 7 месяцев назад

Bro I have been the most atupid student at maths for my whole life. But I was able to follow through this all and even recall almost all of the little I learned back in secondary school. You've even managed to spark a little interest into it. Kudos to you, man!🎉

@donaldasayers 10 месяцев назад

When I was 8 or 9 my teacher showed us this. He showed N=2 through 5 and showed that it doubled each time. Being an awkward sod I drew the case for N=6, counted 31 and wasted half an hour looking for the one I missed. I tried to ask for help, but the class moved on to other things and I never got my answer til now. My teacher was in all other matters quite brilliant, except this one time.

@Radu 10 месяцев назад

Very nice video, combining many elementary things in an interesting way. Thanks!

@richardfredlund8846 10 месяцев назад

the explanation at the end is very satisfying.

@hectorosbornerodriguez9683 3 месяца назад

This was a beautiful way to explain a problem I wasn’t aware of. Love this way of explaining.

@anirudhmenon1961 5 месяцев назад

Such an interesting concept that can be viewed in several ways. Love your videos.

@alejandrosaldarriaga6061 10 месяцев назад

Great video as always! For algebra fans, there is an underrated tool for combinatorics problems like those that can generalize this type of results. These tools are the family of polynomials Hk(X) = X choose k = X(X-1)...(X-k+1)/k! and the discrete derivative D(P)(X) = P(X+1)-P(X). The Hk are a basis of the space of polynomials, and it is easy to verify that D(H_(k+1))(X)=Hk(X), noting how related it is to Pascal's triangle formula. Knowing those simple facts, one can decompose any polynomial P of degree d into p0H0(X) +p1H1(X) +... +pdHd(X). Since H0(0)=1 and Hk(0)=0 for k > 0, we immediately have p0=P(0). Using the shifting property of the discrete derivative over the Hk and iterating until reaching the degree d, we have pk=D^k(P)(0) where D^k is the successive application of the discrete derivative k times in a row. Going a little bit deeper into the expression of pk, it is possible to recursively show that D^k(P)(0)= sum_(l=0)^k { kCl * (-1)^(k-l) * P(l) }, which is very similar to Newton's binomial formula. This might seem at first very heavy and not very useful in practice, but look at what happens when P is of degree d and has the property that P(k) = a^k for k = 0 to d : D^k(P)(0)=sum_(l=0)^k { kCl * (-1)^(k-l) * a^l } = (a-1)^k since this is exactly Newton's formula! This shows that when P follows a power sequence for the d first terms, it's simpler representation is using those special polynomials Hk. In this particular problem, the sequence follows a shifted version of power sequence, which using a similar technique by expanding the nCk term using Pascal's formula shows that D^k(P)(0) is 0 for k odd and 1 for k even.

@thisismycoolnickname 10 месяцев назад

Mathologer also made a video starting from the sequence 1,2,3,4,8,16,31 but he had a completely different approach to solve it, basically discovering "sequence calculus".

@technofeeling2462 10 месяцев назад

Can you tell me the title of the video? Would like to watch it

@Blaisem Месяц назад

13:27 "circling" back to our original question. I see what you did there. But my favorite part of the video is how you hunt for the underlying "why." That's exactly how I like to learn, and I appreciate that you delve into that for just about every component of the explanation.

@stephenweigel 10 месяцев назад

This video is so elegant I nearly wept on the spot

@evancrow2191 10 месяцев назад

Any one who has played 2048 knows a lot of power of 2's

@rainbobow_8125 10 месяцев назад

Wow I was very surprised to see this sequence as I know it to be the sequence obtained from the 4th degree Lagrange polynomial for 2^n. If anyone is interested it's 1 + 7n/12 + 11n^2/24 - n^3/12 + n^4/24 It coincides with the terms 0-4 but gets one unit below at the term n = 5 and this pattern goes the same for every degree of said Lagrange polynomial ! Don't hesitate to ask more if you're interested PS : LaTex formula (for desmos by example) : \frac{x^{4}}{24}-\frac{x^{3}}{12}+\frac{11x^{2}}{24}+\frac{7x}{12}+1

@DrZip 5 месяцев назад

I found the same equation! But... I used a different method, one I invented to solve any sequences that is generated by a polynomial (which... probably already exists, is obvious or equivalent to something else out there already known, circa probably 1600-1800s, and has a fancy name...) >.

@mus3equal 8 месяцев назад

I have such a back log of awesome vids to catch up on, this helped big time with something I'm working on related to harmonic cords and coordinates, thanks for another invaluable lesson!

@OrigamiGuyII 10 месяцев назад

Thanks for the great Math Lesson. I'm always amazed that youtube Math gods like you can make me enjoy lessons from the subject I hated the most in highschool.

@Maths_3.1415 10 месяцев назад

6:05 wow that's beautiful

@newbie4789 10 месяцев назад

I actually remember learning about this exact problem in a math summer programme as a kid. It always fascinated me. And I do remember it having a rather convoluted equation but a nice looking pattern that comes from Adding natural numbers. It basically starts with the normal 1,2, 3... Pattern. You do one special process on the sequence and you get a new one. You repeat this process and at some point, you will reach this pattern

@siddanthvenkatesh2744 3 месяца назад

you can start with 1,1,1,1,1 to, because if you go one layer up you get the 1,2,3 pattern.

@ashisinteresting4806 10 месяцев назад

This is the first time I’ve really intuitively understood the nCr formulas thank you

@footlover9416 3 месяца назад

this is so fire bro i have always loved these videos like i cannot express my appreciation and you always go far enoigh that it scratches the itch fr but in a way that everyone can understand and its pretty and satisfying too and also your voice is mad calming

@toricon8070 10 месяцев назад

So the ultimate solution is (n C 0) + (n C 2) + (n C 4). Is there a pattern whose solution is (n C 0) + (n C 2) + (n C 4) + (n C 6)? What about [...] + (n C 8)? What about just (n C 0) + (n C 2)? I'll have to look into this.

@KiLLJoYYouTube 10 месяцев назад

I’m more curious that intersections of 3 or more are excluded, perhaps it’s something to do with that. Like a sum across a line of (n k) that picks out different iterations and sums those instead.

@coaster1235 10 месяцев назад

It seems feasible to me that the pattern might continue with slicing a sphere by planes in general position, and so on.

@minerscale 10 месяцев назад

Careful about extracting a linear progression from the pattern!!!

@ShankarSivarajan 10 месяцев назад

8:43 Demonstrating the famous German sense of humor.

@AlilAlilA 2 месяца назад

Best math video I've ever watched. Thank you for such a wonderful video!❤

@BrotherInTears Месяц назад

It's refreshing to hear mathematics being taught in such a calm and measured manner, with a good balance of creativity and humor. This video has the potential to heal generations of trauma. 🙏😂

@benjamingross3384 10 месяцев назад

I solved this problem many years ago in undergrad. Still one of my all time favorite problems. I used another path which was probably not so generalizable, but it was fun.

@brunolevilevi5054 10 месяцев назад

13:20 I feel like a more intuitive explanation for why the rows add up to powers of 2, is that if you have n objects, and you want to know how many ways there are of picking 0, or 1 or 2 or ... n-1 or n objects out the n, thats the same as having a binary choice for each n, so its 2^n. Its a simple concept but it seems crazy that the sum of n choose i from i going from 0 to n equals 2^n if you just write it down and look at the equation, I never got how to algebraicly manipulate one to get the other, only by reframing it does it make sense they're both the same

@prismaphotonic2710 10 месяцев назад

To get the result algebraically, simply use binomial expansion on (1+1)^n

@MohammedFarhan-wd6ye 4 месяца назад

I have never liked maths that much, but his voice is so soothing and calm, that I watched through the entire video without even realising.

@nocil134 10 месяцев назад

I ended up using convolutions to calculate the increase in areas per point added. It worked but not as elegant as the one in the video. So cool to see this other way

@Dezdichado1000 10 месяцев назад

One reason why middle school math competitions can be really useful. Those competitions are full of these combinatorics/counting flavor problems that require you to really engage fully and do small cases, formulate + conjecture and prove etc etc. This was a contest problem for me in 6th grade once and I was so certain to have proved the general term to be 2^n via induction.

@Nabbo1501 10 месяцев назад

It's 3AM my brain is not comprehending

@breeze3126 9 месяцев назад

This is so beautiful 😭

@marcioviannajunior9684 10 месяцев назад

That was beautiful, thank you so much for this video

@angeldude101 10 месяцев назад

When we got the final formula of 1 + (n choose 2) + (n choose 4), I recognized that immediately from Clifford Algebra as the size of the even subalgebra... _but only for dimensions less than 6._ Why? Because the actual formula would include (n choose 6), (n choose 8), and so on, which also happens to be 2^(n - 1). On the 6th iteration, you're missing that (n choose 6) part needed to bring it back to a power of 2. Yet again, Pascal's Triangle appears, because it is absolutely everywhere.

@sethhansen1825 10 месяцев назад

you mentioned a couple years ago that a video about the nonexistence of a general quintic solution was in the works or at least being talked about. is that still something we could expect in the future? love your videos so much 😊

@cocowitt24 9 месяцев назад

I had seen n choose k before in several videos, but I never knew how to calculate it. Thanks for the clear explanation! I was even able to find the general form by playing around with it a bit.