@@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.
@@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
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.
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.
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.
@@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.💡
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%
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!
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 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 . ❤❤❤❤
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 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.
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.
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
“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 😂
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".
@@KiLLJoYTH-cam 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 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 😂
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!
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
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.
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.
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.
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 Asian kid whose main language isn’t English*, it’s mind-blowing all the same.
@@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.
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
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!
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
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. 🙏❤️
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.
@@arianvc8239 Well, because the n = 10 is the only simple one that you can see would be a power of two. If each line is a power of two than half of a line is a power of two due to symmetry, but you can clearly see that a quarter, eighth, etc. of the line wouldn't also necessarily be a power of two. Otherwise n=20,40, etc. would also be powers of two.
Are you sure you actually checked? The numbers get very large and you have to implement some tricks to handle those big numbers without integer overflow. That happens within the first couple hundred numbers. I'm to lazy too implement that right now.
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!
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.
_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 😂)_
I just came back to the 3Blue1Brown youtube channel because I remember they make nice videos of proofs with beautiful and understandable visualizations. It just blew my mind twice 😍Once, when they explained how this problem relates to the Euler formula and a second time, when they showed how the power of 2 pattern breaks by going through Pascal's triangle. Just awesome!
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.
I loved math in school, but it’s been many many years since I practiced. I expected this video to go over my head, but it caught my attention and I didn’t get lost! Great video, super satisfying explanation.
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
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.
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).
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)
So very interesting! I’m not sure what level of math this is, as I am still young, (not going to share my age) but I find this so fascinating how math is so important in our lives, or in different ways of understanding life! I love watching types of videos where people like you, take your time, and explain every little detail of something important. It helps many other succeed and make differences between people. Thank you so much for teaching and sharing me. ❤
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. 🙏😂
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.
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😞
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!
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.
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.
@@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 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.
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
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.
You can tell that something must be up, because if you place _zero_ points on the boundary of the circle, then the question still makes sense, and the number of regions is again 1, while the 2ⁿ formula would (nonsensically) suggest that the number of regions should be ½. If you realize that the sequence begins 1, 1, 2, 4, 8, 16, rather than thinking that it begins only 1, 2, 4, 8, 16, then it becomes less surprising that the next term isn't really 32. This approach to math problems, thinking about what comes _before_ the sequence that you're trying to extend, is sometimes humorously called ‘negative thinking’. It can be tricky to do, but sometimes it helps.
For the last question u get powers of 2 when ever in the line above 5 is 1/2^n (being n the number of the line with the power of 2 after line 4. For example for n=5 we have the (n+4 = 5+4) 9th line corresponding to a power of 2), cuz u have a power of 2 by another power of 2( 2^k (sum of the 5 numbers)/2^n (the ratio of this with the corresponding line)= 2^k-n, k€ IN which is a power of two). For example in the 9th u have 10 elements 5/10 is 1/2^1, so the power of 2 is half of the sum of all the elements. 1/2^n = 5/5(2^n) so the line that correspondes to a power of 2 is given by this formula 5(2^n) and cuz the line with the power of two is the 1 after the ratio 1/2^n, the line with that ratio is 5(2^n)-1. We can replace n with a number for example 5, we get tht we have a power of 2 in the line 5(2^5)=5×32=160. But we can go further and discover the correspoding power, which is 1/32 the sum of the 5 first numbers of the line above (5/5(2^5) = 1/2^5 = 1/32). (0C159 + 1C150 + 2C 159 + 3C159 + 4C159 + 5C159)/32 and that is the power of 2 corresponding of the line 160.
@@Sleepingcatgaming191 The most I remember: 3.14159265358979323846264338327950288419716939937510582097494459230781640628620899862803482534211706798214808651328230664709384460955058223172535940812848111745028410270193852110555964462294895493038196
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.
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
wow ... things I just did not know, may be never need in my all day life, but shows me how much of an own world (or of a deep deep rabbit hole) math is and how fascinating it can be. sometimes I feel a bit of a bummer for all the people not interested that much in math to watch those video.
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.
I've watched this 2 days ago and today I used the Euler's characteristic formula and that "n choose 3" thingy in an outside test in class The timing was just perfect!!
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.
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.
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
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.
I had an intuitive answer at 4:20 but I wanted to watch to the end to see if I was on the right track (not a 420 joke 😅). I think I was. My intuition was that while it looks like we're simply adding up verts and edges to create more regions we're actually dividing an already established space. If we were adding up to multiply it might have ended up as 1, 2, 4, 8, 16, 32, etc, but since we're actually dividing by adding it will inevitably fall out of that pattern when the two fall out of sync because the number of verts/lines added are no longer growing exponentially against one another. When you got to the triangle pattern and explained it, I think it confirmed this as the case. It's adding to divide which is visuallized by moving to the next row of the pattern and adding up the first 5 numbers, which intuitively makes sense as well since you do multiplication and division before addition and subtraction. You divide the space first with the lines which moves to the next row of the triangle, then you add up the areas by adding the first 5 numbers. How does this answer "why 5", though? As I watched and got to the "n choose 4" part, I felt like it held the next clue, but it doesn't say how. As soon as you explained it was the first 5 numbers of the triangle pattern I thought back to the "n choose 4", because in any "n choose 4" equation there are 5 points. The 4 points you choose from the circle, and the 1 point where all lines cross. Like you said you can choose a symmetrical pattern, but this is random numbers, so the 5 points holds up. I felt like the lines crossing was the answer because it's where the division of the shape becomes more than just %(number of dots). Meaning at the start you choose 2 points and draw a line, dividing the space by the number of dots. The next dot only divides one of those spaces by the new number of dots, 3, making 4. The next dot breaks the pattern. It begins crossing lines to divide as many areas as the number of dots each by 2, the definition of the line. And then, without any warning at dot number 5, we divide more areas by 2 than there are dots. Suddenly some new factor comes into play. From then on the pattern is more complex, and it starts at 5. I know this makes it feel like a story where an establishment was formed, then 4 decided to return to the way it was when there was only 2, but then 5 betrayed all of their solemn oath that each took as they entered the circle, opening their system up to a new established system of growing numbers, but it's not about that. It's the strange logical choke point that changes everything, and I don't think it's a coincidence that it starts at 5 and then only draws from the first 5 numbers of the triangle pattern. This is not an answer that shows any figures to prove anything like your proper answer does. I didn't "figure" anything out. It also feels like it doesn't follow any logic, but it's actually a singular point where logic remains intact while it twists and inverts and can lose you if you don't hold on for dear life. I think it's a solid answer. A Wisdom stat based answer rather than an Intelligence based one. But I feel I deserve an A on the assignment. 😂
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".
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!🎉
well, at 15:15 you ask will there ever be another power of 2... Come to think of it, when the sum is 1/4th of a row, 1/8th of a row and so on, we'll get powers of 2. So I guess, infinite more powers of 2
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.
thats such an intuitive way of seeing n choose k, i was looking for a way of understanding it intuitively, or how could i invent it from scratch. thats a perfect way.
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 😊
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.
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.
I actually tried to solve this myself before watching the full video, and now that I watched it it's interesting how different my approach was than the one used in the video! Instead of thinking how I could calculate the number of regions directly, the first question I tried to answer was how adding a new line changes the number of regions. If you draw a single line between two points and you count the regions before and after, the difference is equal to the number of lines that your new line intersects +1 (its easier to see on a drawing, the line basically splits in two each region between the intersections). How many lines each line intersects is related to how many points are on each side of the line: let's say that the line has 2 points on one side and 3 on the other, then the line intersecting the new line will be 3 lines for each of the two points, so 2⋅3 = 6, meaning that the line created 2⋅3+1 = 7 new regions. Then you can calculate the sequence iteratively, by summing all the new regions created adding one point at a time, I tried it with a python script and the output is the correct sequence.
I've done exactly the same! I came up with an algorithm to solve: def F(p): result = 1 for n in range(1, p+1): for i in range (1, n): result += (i - 1) * (n -i -1) + 1 # How much adding a new line adds new regions? it's equals the number of intersections + 1 -> (Num of points in the right of the line) * (num of points in the left of the line) + 1. return result then after it I was able to come up with the O(1) final formula, it took a while.
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.
I like your approach better than brute force "let's code it and see what happens". Because analysis, when feasible, is so much more convincing than raking through mountains of data ... Of course, smart search methods in numerical analysis do have a venerable history, as shown by the fact we still use some pioneered by Isaac Newton. And that's where we turn when logic and theory don't provide a clear answer.
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.
11:18 correct me if I am wrong but you cannot calculate n choose k if k>n. 2 choose 4 would for example be calculated like this: (2!)/(4!*((2-4)!))=(2!)/(4!*(-2)!) Factorial cannot be calculated for negative whole numbers. ("Lines that connect" made a video on how to calculate the factorial for other numbers than just natural numbers.)
By convention, (n choose k) for k > n would be 0. Which makes sense, how many ways can you choose a subset of size 5 from 3 elements? Zero! It's impossible. Incidentally, when you extend the definition of factorial to negative values, and just plug in the formula, it's consistent with this convention.
It seems to me like the only reason we should expect to never see another power of 2 from this sequence is if for some reason there *can't* be another power of 2 in this sequence. Otherwise I'd expect to see them pop up from time to time just out of pure coincidence Update: Another commenter provided some extra mathematical context in a reply and after considering that I actually don't expect to see another power of 2
the sequence grows like O(n^4) which means the probability that k is in the sequence is O(k^-3). For k = 2^r this is O(2^-3r) = O(8^-r). Summing this over all r, i. e. all powers of two, yields a finite number- in fact a very low number. Thus, we would not expect them to keep appearing ‘by chance’.
@@debblezthe set of powers of 2 greater than 256 that appear in the sequence could have measure zero and still be non empty so neither is a good argument.
@@debblez Not really. Very low is not of measure zero, so if it were indeed actually "very low" than "heuristically" EMAngel would be correct, were it correct to "heuristically" determine something in the first place.
@@debblez An interesting insight, though I wonder if summation is actually the correct approach here. I'd want to do a multiplication of 1-p and see if it approaches 0 instead; as to my understanding that would represent the probability of never seeing another power of 2. Interestingly, the result that wolfram alpha gives to me for that is just shy of 0.86, actually giving a fairly high likelihood. I suppose I'll have to update my expectation then.
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!
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
There's another great sequence that goes, 1,2,4,8,16,30,60,96... The number of divisors of n!
Is that supposed to be an exclamation mark or a factorial? 😅
@@faastexfactorial.
@@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.
@@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
That’s so dumb, why would anyone put quotation marks in a differential equation?
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.
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
@@theiigotriangularround4880how can someone be so arrogant? Do you not know kids or just want to show off?
@@francescoalexgiacalone878That dude was being so sheltered, many 11yo are still struggling with division and exponents from what I've seen.
guys I'm not sure if the first reply is ironic but I liked it bc it's funny
Props to your son for taking an interest in maths at such a young age.
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.
Ah, so cool! Didn't notice it myself, but totally agree-it's so needless, but sooo satisfying, haha.
Good eye, my friend!
AA someone else noticed too!! i thought that was so neat!
How did you see that!? 🤯
His animations, as much as, if not more than, his explanations, are what make his videos so easy to follow and understand.
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
I'm just impressed that you were able to make a poem AND a FREAKING SONG about this problem as a CHILD.
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.
@@matthewkendrick8280 hallelujah
@@apotatoman4862 😐
@@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.💡
Actually Easy for me...
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%
Thanks for pointing that out. I had totally missed it, but it's a fun little flourish
yeah, i noticed it after your comment, i made me happy 🤧🤧
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!
WOW thats super impressive, and I thing its more impressive hoy you noticed it, superb attention to detail!!
How on earth did you notice that
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.
Nice. That also covers the cases where more than two chords intersect at the same point.
Was just wondering about a more direct way to prove the formula and here it is!
@@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 . ❤❤❤❤
Nice! I hadn't thought about that.
Of course it’s induction!
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.
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
@@oliviapgAlready on it
@@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.
So far I'm at n = 2 billion and haven't found any greater than n = 10.
@@lukeabby8927 I've just passed n=2^32 with no hits
Edit: typed the wrong power of 2
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.
or cuz it's Euler's formula
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
Is that why we can't have them being symmetrical since it would hide an existing vertices?
The resulting integer series are also known as the "Parker powers of 2"
It was a Parker square of a pattern
i just noticed that parker powers are equivalent to the botez gambit.
@@peorakef-9 point score 😢
That’s my name🙂😁
“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 😂
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".
@@piranha031091I do wish amateur Mathematicians would give Collatz a rest smh
@@KiLLJoYTH-cam 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.
There's no surer way to occupy a mathematician than by giving them a problem that seems simple.
@@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 😂
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!
This is the "morally correct" combinatorial proof, makes it much more intuitive to interpret the formula. Thanks for posting!
This is just the proof of Euler's theorem for graphs.
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
awesome!! this is the most elegant intuition ive come across so far for this :) tysm !!
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.
I fear you'd make them quit school...
@@Secret_Moonyeah most of my classmates want to quit school and they haven’t even gotten close to this
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.
Schools are silent since this has dropped 🗣
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.
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 Asian kid whose main language isn’t English*, it’s mind-blowing all the same.
... And certifiably a nerd 🤣 Seriously, just what sent him down that path, right? 🤔
I'd give you a like, but that would break a beautiful 69
What? That's probably the least surprising thing I've heard today.
built different
@@OsskywSteered different, more than likely.
Euler really was such a gift to humanity
It is a great word the Germans gave us.
Euler was gay and I have proof. In 1987 he freaked a man. You can tell because his name starts with the letter E.
Eh, if it wasn't Euler it would have been someone else.
@@shahram6869Very true
@@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.
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.❤
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
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!
The explanation of Euler's characteristic is simply mindblowing. It's much easier to understand than the inductive proof!
It actually IS the inductive proof you know . Just shows with diagrams and really intuitive language
Grant provides an inductive proof, though
@@Lait_au_Mieljust what I was thinking... Such a clever way to hide an inductive proof :)
That IS the inductive proof
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
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. 🙏❤️
I love the bit at 8:30 being 3 blue faces and 1 brown face. attention to detail & i love it
Also interesting to notes he labels the faces 0,1,2,3... 5?
I didn’t even notice that, thanks for poynting that out!
@@themathhatter5290it's so you spend the next hour looking for region 4 😋
Grant wrote his own version of Hallelujah and it was mathematics inspired! Amazing 1:45
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.
But why?
I checked all the way to n = 91,756,093,502 and still no power of 2.
Wish I’d seen this before I did the first 100 by hand
@@arianvc8239 Well, because the n = 10 is the only simple one that you can see would be a power of two. If each line is a power of two than half of a line is a power of two due to symmetry, but you can clearly see that a quarter, eighth, etc. of the line wouldn't also necessarily be a power of two. Otherwise n=20,40, etc. would also be powers of two.
Are you sure you actually checked? The numbers get very large and you have to implement some tricks to handle those big numbers without integer overflow. That happens within the first couple hundred numbers. I'm to lazy too implement that right now.
What a nice problem. Somehow I got goosebumps when he explained the reason for the missing 1 in 31. It suddenly all made sense.
What are your qualifications in mathematics ?
@@nipurnshakya9312 nothing fancy
Just ordinary engineering mathematics of a masters degree in CS.
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.
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!
Nice ❤👌
clever
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.
_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 😂)_
-thanks for the additional information, I was wondering what is a factorial of "this".-
The italicized emoji makes me uncomfortable
_m_
*_-That one was an exclamation point, but this one actually is a factorial!-_*
@@thegurw1994 Well now I’m curious what an italicized emoji looks like
😁🧍♂️🌛⭕️ = before
_😁🧍♂️🌛⭕️_ = after
Edit: dude there’s no difference 🤦♀️
I just came back to the 3Blue1Brown youtube channel because I remember they make nice videos of proofs with beautiful and understandable visualizations. It just blew my mind twice 😍Once, when they explained how this problem relates to the Euler formula and a second time, when they showed how the power of 2 pattern breaks by going through Pascal's triangle. Just awesome!
Was NOT expecting that joke at 16:07 very glad I stayed to the end of the video
Pretty sure that’s a classic algorithmic problem.
@@widmur if you are not just trolling and you actually think it is a reference to 3SAT, you are mistaken.
I'm sure there are some interesting combinatorics problems to be solved.
3 some 😂
@@Vaaaaadim I was making a bad joke about 3SOME being a homophone for 3SUM.
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.
the m=3,n=90 and m=4,n=23 cases are interesting because they are the only ones found that aren't in a row where n2.
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.
A Poem? What about finding a proof
I loved math in school, but it’s been many many years since I practiced. I expected this video to go over my head, but it caught my attention and I didn’t get lost! Great video, super satisfying explanation.
I watch all of 3b1b videos, but this is one my brain was able to follow fully. Such elegance!
*Amazing as always! Thanks for inspiring a generation of learners to love math and push education forward :)*
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
I just watched the original video yesterday! I'm so excited for the remake!
6:00 after 5 months i finally counted them and he is not lying. There is indeed 3,921,225 quadruplets in a circle with 100 points.
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.
Such lucidity of explanation! Such a reassuring voice!
3B1B is a bulwark against chaos.
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).
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)
Fun fact: Euler is responsible for discovering every single formula in math history, all of which carry his name
There's an xkcd for this, I think.
So very interesting! I’m not sure what level of math this is, as I am still young, (not going to share my age) but I find this so fascinating how math is so important in our lives, or in different ways of understanding life! I love watching types of videos where people like you, take your time, and explain every little detail of something important. It helps many other succeed and make differences between people. Thank you so much for teaching and sharing me. ❤
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. 🙏😂
Fantastic work with the rewrite! This version flows very cleanly, and it’s easier to follow compared to the original
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.
There used to be a maths club on Quora that worked rather like that, maybe it's still there?
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😞
We have infinity more to go
Can find any algorithmicaly?
2^n=E-V+1?
n=log b2 (E-V+1)
Using N choose 2 and 4 appropriately as substitutes.
Whoops. I forgot to relate it to Pascal’s triangle which might make the computation easier honestly, since Pascal’s triangle is so simple to use.
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!
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.
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.
If you make the third cut horizontal instead of vertical, you can get 8 slices!
@@josephward4918how come?
@@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.
@@josephward4918 the third cut can only go through 3 slices at most, so that would be 7 instead of 8
@@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.
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
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.
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.
You can tell that something must be up, because if you place _zero_ points on the boundary of the circle, then the question still makes sense, and the number of regions is again 1, while the 2ⁿ formula would (nonsensically) suggest that the number of regions should be ½. If you realize that the sequence begins 1, 1, 2, 4, 8, 16, rather than thinking that it begins only 1, 2, 4, 8, 16, then it becomes less surprising that the next term isn't really 32.
This approach to math problems, thinking about what comes _before_ the sequence that you're trying to extend, is sometimes humorously called ‘negative thinking’. It can be tricky to do, but sometimes it helps.
For the last question u get powers of 2 when ever in the line above 5 is 1/2^n (being n the number of the line with the power of 2 after line 4. For example for n=5 we have the (n+4 = 5+4) 9th line corresponding to a power of 2), cuz u have a power of 2 by another power of 2( 2^k (sum of the 5 numbers)/2^n (the ratio of this with the corresponding line)= 2^k-n, k€ IN which is a power of two). For example in the 9th u have 10 elements 5/10 is 1/2^1, so the power of 2 is half of the sum of all the elements. 1/2^n = 5/5(2^n) so the line that correspondes to a power of 2 is given by this formula 5(2^n) and cuz the line with the power of two is the 1 after the ratio 1/2^n, the line with that ratio is 5(2^n)-1. We can replace n with a number for example 5, we get tht we have a power of 2 in the line 5(2^5)=5×32=160. But we can go further and discover the correspoding power, which is 1/32 the sum of the 5 first numbers of the line above (5/5(2^5) = 1/2^5 = 1/32). (0C159 + 1C150 + 2C 159 + 3C159 + 4C159 + 5C159)/32 and that is the power of 2 corresponding of the line 160.
so we get a power of 2 when n is 160 right?
I'm so happy you are posting on TH-cam more!!! You are the best!
31… it better not be computing pi
pi is everywhere pi is everywhere pi is everywhere pi is everywhere pi is everywhere
Rule 3.14: "If it exists, it's related to pi"
The most amount of Π that I remember: 3.1415926535897932384626433832795028841
@@Sleepingcatgaming191crazy that of a years span of this comment and only 3 comments i found you from only 8 hours ago
@@Sleepingcatgaming191
The most I remember: 3.14159265358979323846264338327950288419716939937510582097494459230781640628620899862803482534211706798214808651328230664709384460955058223172535940812848111745028410270193852110555964462294895493038196
This was a beautiful way to explain a problem I wasn’t aware of. Love this way of explaining.
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
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.
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.
8:43 Demonstrating the famous German sense of humor.
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
wow ... things I just did not know, may be never need in my all day life, but shows me how much of an own world (or of a deep deep rabbit hole) math is and how fascinating it can be. sometimes I feel a bit of a bummer for all the people not interested that much in math to watch those video.
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
Very nice video, combining many elementary things in an interesting way. Thanks!
the explanation at the end is very satisfying.
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.
Not a series, but his last several videos have all been statistics-y.
@@JackTheGamingGuy4REALZ I was talking about a course, like the linear algebra one.
Bro u explain this more wisely than my math teacher❤❤ great co ntent
I've watched this 2 days ago and today I used the Euler's characteristic formula and that "n choose 3" thingy in an outside test in class
The timing was just perfect!!
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.
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.
It seems feasible to me that the pattern might continue with slicing a sphere by planes in general position, and so on.
Careful about extracting a linear progression from the pattern!!!
8:35 Jokes? Argh please no, we don't do that here!
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
you can start with 1,1,1,1,1 to, because if you go one layer up you get the 1,2,3 pattern.
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.
I had an intuitive answer at 4:20 but I wanted to watch to the end to see if I was on the right track (not a 420 joke 😅). I think I was.
My intuition was that while it looks like we're simply adding up verts and edges to create more regions we're actually dividing an already established space. If we were adding up to multiply it might have ended up as 1, 2, 4, 8, 16, 32, etc, but since we're actually dividing by adding it will inevitably fall out of that pattern when the two fall out of sync because the number of verts/lines added are no longer growing exponentially against one another. When you got to the triangle pattern and explained it, I think it confirmed this as the case. It's adding to divide which is visuallized by moving to the next row of the pattern and adding up the first 5 numbers, which intuitively makes sense as well since you do multiplication and division before addition and subtraction. You divide the space first with the lines which moves to the next row of the triangle, then you add up the areas by adding the first 5 numbers.
How does this answer "why 5", though? As I watched and got to the "n choose 4" part, I felt like it held the next clue, but it doesn't say how. As soon as you explained it was the first 5 numbers of the triangle pattern I thought back to the "n choose 4", because in any "n choose 4" equation there are 5 points. The 4 points you choose from the circle, and the 1 point where all lines cross. Like you said you can choose a symmetrical pattern, but this is random numbers, so the 5 points holds up. I felt like the lines crossing was the answer because it's where the division of the shape becomes more than just %(number of dots). Meaning at the start you choose 2 points and draw a line, dividing the space by the number of dots. The next dot only divides one of those spaces by the new number of dots, 3, making 4. The next dot breaks the pattern. It begins crossing lines to divide as many areas as the number of dots each by 2, the definition of the line. And then, without any warning at dot number 5, we divide more areas by 2 than there are dots. Suddenly some new factor comes into play. From then on the pattern is more complex, and it starts at 5. I know this makes it feel like a story where an establishment was formed, then 4 decided to return to the way it was when there was only 2, but then 5 betrayed all of their solemn oath that each took as they entered the circle, opening their system up to a new established system of growing numbers, but it's not about that. It's the strange logical choke point that changes everything, and I don't think it's a coincidence that it starts at 5 and then only draws from the first 5 numbers of the triangle pattern.
This is not an answer that shows any figures to prove anything like your proper answer does. I didn't "figure" anything out. It also feels like it doesn't follow any logic, but it's actually a singular point where logic remains intact while it twists and inverts and can lose you if you don't hold on for dear life. I think it's a solid answer. A Wisdom stat based answer rather than an Intelligence based one. But I feel I deserve an A on the assignment. 😂
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".
Can you tell me the title of the video? Would like to watch it
I remember the original video! Looking forward to seeing how it's remade here!
Delete this comment, because nobody asked.
6:13 wasn't for the math, it was for the visual
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!🎉
The amazing display of the problem of induction... thanks for the vid!
It's 3AM my brain is not comprehending
well, at 15:15 you ask will there ever be another power of 2... Come to think of it, when the sum is 1/4th of a row, 1/8th of a row and so on, we'll get powers of 2. So I guess, infinite more powers of 2
Not true, the numbers get larger closer to the centre so you don’t actually get 1/4 or 1/8 the value
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.
thats such an intuitive way of seeing n choose k,
i was looking for a way of understanding it intuitively, or how could i invent it from scratch.
thats a perfect way.
I’ve been watching this video for 5 minutes and I already learned more than I did in a year of high school math classes
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 😊
6:05 wow that's beautiful
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.
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.
Really well explained. I thought I was getting lost until the end and it all came together for me.
Thanks 3b1b
I actually tried to solve this myself before watching the full video, and now that I watched it it's interesting how different my approach was than the one used in the video!
Instead of thinking how I could calculate the number of regions directly, the first question I tried to answer was how adding a new line changes the number of regions. If you draw a single line between two points and you count the regions before and after, the difference is equal to the number of lines that your new line intersects +1 (its easier to see on a drawing, the line basically splits in two each region between the intersections).
How many lines each line intersects is related to how many points are on each side of the line: let's say that the line has 2 points on one side and 3 on the other, then the line intersecting the new line will be 3 lines for each of the two points, so 2⋅3 = 6, meaning that the line created 2⋅3+1 = 7 new regions.
Then you can calculate the sequence iteratively, by summing all the new regions created adding one point at a time, I tried it with a python script and the output is the correct sequence.
I've done exactly the same! I came up with an algorithm to solve:
def F(p):
result = 1
for n in range(1, p+1):
for i in range (1, n):
result += (i - 1) * (n -i -1) + 1 # How much adding a new line adds new regions? it's equals the number of intersections + 1 -> (Num of points in the right of the line) * (num of points in the left of the line) + 1.
return result
then after it I was able to come up with the O(1) final formula, it took a while.
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.
"Appears"
👋
@@deltalima6703 someone more experienced in multivariable analysis may be able to prove me wrong.
I like your approach better than brute force "let's code it and see what happens". Because analysis, when feasible, is so much more convincing than raking through mountains of data ...
Of course, smart search methods in numerical analysis do have a venerable history, as shown by the fact we still use some pioneered by Isaac Newton. And that's where we turn when logic and theory don't provide a clear answer.
Conjecture is true: 1+C(n,2)+C(n,4) is never a multiple of 512=2^9 (just check 512 cases, all mod 512).
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.
11:18 correct me if I am wrong but you cannot calculate n choose k if k>n. 2 choose 4 would for example be calculated like this:
(2!)/(4!*((2-4)!))=(2!)/(4!*(-2)!)
Factorial cannot be calculated for negative whole numbers. ("Lines that connect" made a video on how to calculate the factorial for other numbers than just natural numbers.)
By convention, (n choose k) for k > n would be 0. Which makes sense, how many ways can you choose a subset of size 5 from 3 elements? Zero! It's impossible.
Incidentally, when you extend the definition of factorial to negative values, and just plug in the formula, it's consistent with this convention.
I actually despise maths in school, but this was really cool and fun to watch!
Because most often its taught badly.
It seems to me like the only reason we should expect to never see another power of 2 from this sequence is if for some reason there *can't* be another power of 2 in this sequence. Otherwise I'd expect to see them pop up from time to time just out of pure coincidence
Update: Another commenter provided some extra mathematical context in a reply and after considering that I actually don't expect to see another power of 2
the sequence grows like O(n^4) which means the probability that k is in the sequence is O(k^-3). For k = 2^r this is O(2^-3r) = O(8^-r). Summing this over all r, i. e. all powers of two, yields a finite number- in fact a very low number. Thus, we would not expect them to keep appearing ‘by chance’.
@@debblezthe set of powers of 2 greater than 256 that appear in the sequence could have measure zero and still be non empty so neither is a good argument.
@@overwrite_oversweet yes of course, but im saying that from a heuristic standpoint, we shouldn’t expect there to be any more.
@@debblez Not really. Very low is not of measure zero, so if it were indeed actually "very low" than "heuristically" EMAngel would be correct, were it correct to "heuristically" determine something in the first place.
@@debblez An interesting insight, though I wonder if summation is actually the correct approach here. I'd want to do a multiplication of 1-p and see if it approaches 0 instead; as to my understanding that would represent the probability of never seeing another power of 2.
Interestingly, the result that wolfram alpha gives to me for that is just shy of 0.86, actually giving a fairly high likelihood. I suppose I'll have to update my expectation then.
Thank you Grant, your videos are always worth a watch even if the topics are over my head for the moment.
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!
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
To get the result algebraically, simply use binomial expansion on (1+1)^n