Want another pascal's triangle pattern? Take the top right diagonal (1, 1, 1, ...) and treat it as if it was a decimal number with the decimal point between the first two numbers (1.11111...). Now square that number. You get 1.234567... which equals the second diagonal. Cube it and you get 1.3717421... Which is the third diagonal (you have to smush the numbers together the same was as 1,5,10,10,5,1 = 161,051). Now this works in bases other than 10. If you treat 1.11111 as a binary number it equals 2 (same way 1.9999... = 2 in decimal). Now 2^2 = 4. So we know that 1.234567... in binary equals 4. Written as an infinite sequence, 1.23456 in binary is 1 + 2/2 + 3/4 + 4/8 + 5/16 + 6/32 + ... which indeed equals 4! This pattern even has a connection to another Numberphile video, Grafting Numbers, and it's truly remarkable, although this comment section is too small to contain it.
For anyone interested in why these patterns emerge: *Pascal's Triangle encodes n choose k:* This arises from the fact that _C(n, k) = C(n-1, k-1) + C(n-1, k)._ This can be solved algebraically (note that _C(n, k) = n!/(k!(n-k)!)),_ but there's a combinatorial argument to it as well. Say you have _n_ different ice cream flavors and you want to choose _k_ of them for your super tall ice cream cone. Then you can consider two distinct cases: the combinations with chocolate, and the combinations without chocolate. If you include chocolate, then you have _n-1_ flavors left and you need to choose the remaining _k-1_ flavors. If you don't include chocolate, then you still have _n-1_ flavors, but you still have _k_ flavors to choose from as well. Hence, when you add them together, you should get the total _C(n, k)_ combinations. Because of this identity, you can inductively show that Pascal's Triangle encodes _n_ choose _k._ *The rows are consecutive powers of 2:* Remember that if you want to choose _k_ objects from _n_ items, you go down to the _n_ -th row in the triangle and you go over by _k._ This means that the _n_ -th row will be the numbers _C(n, 0), C(n, 1), C(n, 2), ..., C(n, n)._ Now, consider the total number of ways to choose _n_ objects, regardless of the number of items you choose. This will be the sum of all of the cases where you choose _i_ objects for _0
Arbitrary Renaissance Thanks! I'm going to try to come up with an explanation myself for the ones I did not know before actually reading your comment though.
When I say it takes a bit of complex induction to work out the Fibonacci pattern, I just mean that it's tough to rigorously describe. The intuition is really easy like you said: the first diagonal grabs all the left sums and the second diagonal grabs all the right sums. Regarding Fermat, that's really interesting! Thanks for taking the time to share your discoveries.
I found a simpler way to describe it: Consider a string of ones and zeros as both a Pascals triangle row (in mod 2) and as a binary number. Given that take the following string 1000…0001 (between the 1’s on either side are all zeros) Each 1 creates its own mini triangle due to the zeros on either side (this partly explains Siepinski’s triangle) at a certain point the two sub triangles will meet, they will meet when the both rows are all the 1’s (i.e. like 1111….111) Proof: If you consider a row of all 1’s, 1+1=0 (mod 2) so all will be zeros in the next row apart from extra 1’s placed at either side I.e it creates 1000…0001 so the previous row to 1000…0001 will be all 1’s with 1 less of them. Now the row with the 1’s can be called row X (which means it has X terms) now if you take each sub triangle to that row (the 1000…0001 is row 1 of the mini triangle) so at this point you will at row 2X so 2X terms so when you put the two row X’s together there are no 0’s in between) Now that you’ve done 2X rows you have all 1’s again so it creates another 1000…0001 Now as binary numbers each row can be factorised as the sub triangle row (as a binary number) multiplied by the 1000…0001 and that represents multiplying that number by all previous rows Finally to prove that the numbers you multiply each past term are Fermat numbers Proof: First the number is in the form 1000…0001 (in binary) which means it’s 1 more than a power of 2 (which in binary is 1000….000) Second you take the “111…111” number and duplicate it so the number of 1’s is doubled and if you add 1 it becomes a power of 2 with the same number as zeros as the number that had 1 added had 1’s, add 1 again and it has the same number of digits, and adding the first 1 increased it digits by 1 so if the 111…111 was row X then 1000…0001 (the number obtained by adding 2) it the next row of the Triangle and the number in the middle(not the triangle the value since there is difference is 2) is a power of 2 with a number of 0’s as explained above Since you started with 1 on its own (at the top of the triangle) and each time you doubled it (1,11,1111…etc) it means that the number of 0’s in the power of two (and therefore the power) is a power of 2 so it is in the form 2^(2^N) since this number has one added to create the number you multiply by, its (2^(2^N))+1 which is the definition of a Fermat number
Another way to prove the Hockey Stick Theorem: The corresponding identity is: C(n,n) + C(n+1,n) + ... + C(n+k,n) = C(n+k+1,n+1) (n,k∈N) (The other direction is just symmetry of Pascal's triangle C(n,r)=C(n,n-r)) The identity can be proved using a combinatorics argument: Number of ways to choose n + 1 items from a set of n + k + 1 elements: C(n+k+1,n+1) But this can be computed in another way by considering cases: including element 1 - C(n+k, n) excluding element 1, including element 2 - C(n+k-1, n) excluding elements 1, 2 including element 3 - C(n+k-2, n) ... Keep going with this pattern until: excluding elements 1, 2, ..., k, including element k + 1 - C(n, n) If keep going excluding more than k elements then it is impossible to choose n + 1 from n + k +1 elements. So we have covered all cases here. Equating the 2 methods of computing number of ways to choose n + 1 items from a set of n + k + 1 elements gives the identity.
The first diagonal is all 1's. The second diagonal is the natural numbers. The third diagonal is the triangular numbers. The fourth diagonal is the tetrahedral numbers. The fifth diagonal is the pentachoronal numbers. etc. The nth diagonal is the n-1-dimensional simplex numbers.
IceMetalPunk Sure it can. Since Pascal's triangle is of order Aleph(0), the set of things it encodes is of the size of the set of the real numbers, Aleph(1). So at the very least, it must encode almost everything.
IceMetalPunk's conjecture: "Pascal's triangle encodes literally everything that can be encoded." would be very funny to see in a paper. You should try to proove.
One of my favourite Numberphile videos in a long time. I knew Pascal's triangle, but I had no idea it showcased this many math phenomenons. Sierpiński's triangle is what blew my mind the most.
The nth row in the Pascal triangle discribes the geometrical properties of an n dimensional simplex, for instance, the 3D simplex is a tetrahedron, a tetrahedron has 4 vertices, 6 edges, and 4 faces, which is the third row. (**to get the full row count 1 zero dimensional shape, or an 'empty' shape, and 1 3D cell, that gets you to 1,4,6,4,1)
I think it's pretty cool how, with Pascal mod 2, you can see how adding odds and evens work. 0 + 0 = 0 1 + 0 = 1 1 + 1 = 0 Which maps to the classic rules of adding evens and odds. What's more remarkable is that it also follows adding in single digits in binary. 1 + 1 = 10 in binary but if you make the results only the ones place then it's 0
I love Casandra Monroe's zeal for mathematics. This video was inspiring and enriching. I love Pascal's Triangle and all the mathematical intricacies it reveals!
I absolutely love this channel. Everytime the poorly thought out school curriculum kills my interest in math, one of these videos fires it up again. Thanks guys!
Reminds a childhood discovery - To calculate 5th row for example, without adding up previous ones, just multiply 1 * 5/1 * 4/2 * 3/3 * 2/4 * 1/5. Note the numerator decreasing while denominator increases. So generally, k-th number in the n-th row is n!/(k!(n-k)!)
We know that the 1st number in the 5th row is 1. So let's try and see if that is what your formula produces. 5!/(1!(4!)) = 5. Oops. 5 ≠ 1. Are you considering the 1 to be in the k=0 position and the 5 to be the k=1 position? It also appears that you are considering the top of the triangle to be of row n=0?
Yosef MacGruber Yes. That is actually the definition of the Pascal triangle. The topmost row is the 0th row and the leftmost diagonal is the 0th diagonal. The adding recursion is not part of the definition, that's just an easy construction that works due to the properties that choose(n, k) as a function satisfies.
It is really quite a remarkable thing. So many patterns within it! Pascal's triangle really exemplifies mathematical beauty to me, it has so much going on within a simple rule.
Around the 8th grade in US we learn Pascal's triangle to solve n choose k problems. In high school we learn the short cut formula for each entry: n choose k equals n! / (k!(n-k!)). In university we learned binomial formula containing the n choose k term. This is by far the most practical use for Pascal's triangle.
i did not learn pascal's triangle in 8th grade in california, or ever in school for that matter. i also never learned the choose function in school, definitely not high school. i did learn binomial theorem in calculus 2 later on tho
Nick Kravitz I learned Pascal's triangle and binomial theorem as a junior in HS to expand expressions. I always chose to use pascals in my work. Never understood binomial theorem.
One more interesting thing for Pascal's triangle of (x-1)^n, change sign from 3rd term on at every row turn out is trivial zero of zeta function at s=1, -2 = 1 -2 - 1, -4 = 1 - 3 -3 +1, -6 = 1 - 4 - 6 + 4 - 1, -8 = 1 - 5 - 10 + 10 - 5 + 1,etc...-2n, .nontrivial zero which are extension of trivial zero obey same rule have 2^n series, (2*n)!/(n!)^2 of moment of nontrivial zero 1,2,6,20.. right at middle line of triangle, let bottom of triangle from 0 to 1 at x-axis, 1,2,6,20.. right at x = 1/2 line as Riemann hypothesis predicted.(Euler product of (p-1)/p is 0.04875 from 2 to 99991, take 2^9632 - 1 of mod(10^10,po)/po get 34490000, 0.04875*10^10/34490000 =14.13 po is all possible combination from 2 to 99991, 2nd 0 ,21.02 = 487500000/23190000 get without 2., 487500000/19500000 = 25 3rd zero of zeta function without 2 and 3, so on...25*(1/2)(2/3)(4/5) + 1/2 - 1/6 - 5/10 + 25/30 + 1/3 - 10/15 +0/5.+3 -1 = 9 prime number counting until 25.)
Great stuff! I knew most of the properties from the beginning of the video. But I NEVER heard about the fact that Fibonacci's row is encoded in the Pascal triangle. Mind blowing!!
We've been learning about it's application in chemistry as part of my A-level course - In a high resolution Nuclear Magnetic Resonance imaging spectrum, the number of peaks and their sizes have relative areas described exactly by Pascal's triangle - a peak with 4 splits in it suggests 3 Hydrogen atoms are adjacent to the unique environment of hydrogen observed, and each peak has relative area 1:3:3:1
If you go down the diagonals, you get n dimensional triangle numbers. For example, 0 dimensional triangle numbers are 1, 1, 1, 1, ... etc 1 dimensional triangle numbers would be 1, 2, 3, 4, 5, 6, 7, etc 2 dimensional triangle numbers would be 1, 3, 6, 10, 15, 21, 28, 36, etc Then the 3 dimensional triangle (pyramidal) numbers would be 1, 4, 10, 20, 35, 56, etc and so on. You can keep going in higher dimensions if you just keep going along the next diagonal.
pascal newton and Einstein were playing hide and seek. Einstein said to newton "found you!" but Newton went and stood in a square of length one meter and said "Hey I am newton per meter square..you found pascal" HAHAHAHAHAH...ha..ha
i once made up a "simplex operator": i generalized triangle numbers for any dimension: (triangle (2-simplex), tetrahedron (3-simplex), pentachoron (4-simplex)). the first operand was the edge length and the second operand was the number of dimension. x △ 2 was a triangle number for edge length x and so on. as i wrote down the Cayley table (like a multiplication table) i noticed quite a bit of a pattern. suddenly it struck me that i was writing down Pascal's triangle sideways. this was one of the coolest things that ocurred to me. :)
Here's how I heard about the relation between Pascal's triangle and Fibonacci numbers. Problem: You're climbing a ladder n rungs high and you always have a choice of climbing 1 or 2 rungs at a time (let's call these actions single steps and double steps). How many ways are there to climb the ladder? One approach to the problem is to use combinatorics: first find the number of ways to climb if you never take a double step, then the number of ways if you only take 1 double step, then 2, etc and then add it all up. This sum corresponds to adding terms along a shallow diagonal of Pascal's triangle. A second approach is to use recurrence: to climb n rungs you must first climb n-1 rungs and then take a single step OR climb n-2 rungs and then take a double step. So if f(n) is the number of ways to climb n rungs then we have f(n) = f(n-1) + f(n-2) which gives us the Fibonacci sequence.
I remember reading that this problem was first solved by composers of Indian classical music who wanted to know how many ways there are to, say, fill a bar of music with 8th and 16th notes. Not in those western terms of course.
Alright, i have a question if anyone is still reading the comments on this video. I was doing a problem for my Intro Quantum Mechanics class about spin-1/2 particles, and after doing a bit of math, ended up getting what is essentially a 3d square pyramid of numbers (idk if that's right, its 4 triangles, each making a side), except there is no bottom, it just goes on indefinitely as far as you want to extend it (like pascal's triangle). In fact, my professor noticed that the outer triangle of each side of the pyramid IS Pascal's triangle. Which made me curious if there was some overarching recursion relation (or other relation) to predict future rows/squares of the pyramid. If you look at one of the outside triangles, then remove it and look at the triangle beneath it, and continue doing this, this is what you find: Triangle 1: 1, 1 1, 1 2 1, 1 3 3 1, 1 4 6 4 1, 1 5 10 10 5 1, etc (Pascals Triangle) Triangle 2: 0, 1 1, 4 0 4, 9 2 2 9, 16 10 0 10 16, 25 27 5 5 27 25, 36 56 28 0 28 56 36, etc (what pattern?) Triangle 3: 4, 4 4, 1 12 1, 1 15 15 1, 16 8 40 8 16, 64 0 56 56 0 64, etc (my computer cant compute any more) Triangle 4: 0, 9 9, 36 0 36, 64 24 24 64, etc (computer cant do anything layer 10 or below) Triangle 5: 36, 36 36, etc That is all my computer can do, but as soon as you get to layer 10 of the pyramid, i go beyond the Integer limit in C# and i haven't fixed that problem yet, so my computer just gives me either 64 or null. weird bug, but yeah. if you can figure out some pattern, that would be awesome. And for anyone wondering what these numbers are here is the slightly longer story: Each layer, or square, of the pyramid corresponds to the spin of a particle, starting with zero. row zero is a spin-0 particle, row 1 is a spin-1/2 particle, row 2 is a spin-1 particle, row 3 a spin-3/2 particle etc. and the numbers each column/row correspond to the probability that the spin will be measured at that magnitude in a direction orthogonal to the currently known spin (the closes thing Quantum Mechanically to "random"). for example, row 3, column 5, in layer 7, of this pyramid corresponds to the probability that a spin-7/2 particle measured to have spin-5/2 in some direction will be measured to have spin-3/2 in an orthogonal direction. Also the probability that a spin-7/2 particle with spin 3/2 will be measured to have spin-5/2 in an orthogonal direction. However, it is easy to see that the rows do not add to unity, and that is because i have removed a normalization constant to make them all whole numbers. the normalization constant for the n-th layer is simply 2^(-n). Any more questions i would be happy to answer.
I love Pascal's Triangle and all the possibilities hidden in it. I'm from India and I am proud to say that it was originally found in the works of Pingala, the Indian Mathematician, thousands of years ago and his student brought it to Arabia and it travelled West where Pascal made it popular. This is called Meru Prasthara in Sanskrit and it means "Ladder to Mt Meru". In Vedic literature, Mt Meru is so tall that the Himalaya is like a small stone on it. :) Indeed, this triangle is infinite!
I first learnt about Pascal's Triangle because of how you can use it in the expansion of brackets with a high power. So if you had (x+a)^5, you would go to Row 5 of the triangle, and each value is a coefficient, in order, so you add that to each product, and list the powers in descending order for x and ascending order for a. Will definitely save a lot of time in exams. In January, I had a summer school for my math this year (to prepare us for senior highschool math) and we looked at other ways it can be used then as well (mainly combinations and permutations - I don't remember which). It's such an amazing mathematical tool.
Pacals triangle patterns make more sense if you think of each diaganol as a cumulative frequency of the diagonal before it 0+0+0+0+0+0+0+0+0... 0+1 +1 +1 +1 +1 +1... 0+1 +2 +3+4 +5... 0+1+3+6+10... 0+1+4+10... 0+1+5... 0+1... 0... One of the patterns I found was that the sum of squares up to n=((nC1)x(n+1C2))-(n+1C3) so for example, sum of squares up to 4=(4x10)-10 They're are patterns for individual square numbers and cube numbers too, but they're way to complicated for me to explain. It makes me wonder if they're are patterns for every exponent, and they're just too complicated to find. Also pascals triangle can be used to find the constants terms for any binomial expansion, a pascals pyramid can be used to find the constant terms for a nominal expansion. Pascals triangle is real neato.
oh, also the diagonals are numbers which are needed to construct a triangle based shape, so the third row is triangular numbers, the fourth row is tetrahedral numbers, and then whatever the fourth dimentional equivalent is is the fourth row I think.
I've loved Pascal's Triangle since I first learned about it in 7th grade. I had so much fun playing with it in mod 2, mod 3, etc. Different pattern each time!
My friend and I discovered the Serpenski triangle that's hidden inside in a bit of a different way. Instead of applying mod2, we found the absolute value of the difference of the two numbers above. In the same way that you add to produce Pascal's Triangle, we subtracted to create this Difference Triangle, and it turned out to be identical to the mod2 serpenski fractal thing.
Here's one of the patterns that I've used to teach multiplication to elementary school students: 0 9, 1 8,2 7, 3 6, 4 5, 5 4, 6 3, 7 2, 8 1, 9 0 In the first column on the left, write the numbers in ascending order from zero up to nine and in the second column in descending order from nine to zero and, presto!, you've got the nine's multiplication table. I have patterns for the fours, sixes, sevens, eights and also division.
9:00 So... the primes' locations on the Pascal triangle are located exactly on the Fibonacci sequence's locations, ignoring 1. They occur at locations 2, then 3, then 5, and the next one (not imaged) would occur at the 8th row. So that's another way to get Fibonacci from Pascal's triangle.
They are Fermat primes, and so far we only know 5 Fermat primes. Which means the rest of the 2^n rows doesn't create a prime number as long as we know.
I've been watching Numberphile for quite a while and I really enjoy the videos. But today I discovered I wasn't even subscribed.... It got me by surprise so I immediately clicked the button.
Indicotherium its happened on every base... if u put a number n on the formula, you get (n+1)^x... x means row.. for examples.. n = 3, x = 2, u'll get 1(3^2) + 2 (3) + 1 = 9+6+1 = 16 = 4^2 n = 4, x = 3, u'll get 1(4^3) + 3(4^2) + 3(4) +1 = 64 + 48 + 12 + 1 = 125 = 5^3 and so on
it goes waaay back into Hindu, Buddhist, and Jain mathematics as "Mount Meru" long before pascal it was known to Pingala in or before the 2nd century BC much to explore in it's relationship to cellular automata.
I think that the reason why there are so many things in pascal's triangle is because it encompasses the one thing almost all of our math is based on, addition. This means that since most of our math has it's roots at addition, we'll find many things inside of the summitive nature of this triangle.
There's something about the way she's presented this topic that's really _transformed_ my views on Pascal's Triangle. Now I think there might be more than meets the eye.
I love the geometric implications in Pascal's triangle. The first row is a point. The second row is a line segment. The 3rd row is a square with a diagonal line (1 point, 2 adjacent points, 1 point). 4th row is a cube with opposite triangular vertex figures (1 point, 3 adjacent points, 3 more adjacent points, 1 point). Beyond that it's hard to explain but it still works and it's BEAUTIFUL!!! What's more, these geometric figures can be used as simple diagrams to understand why Pascal's triangle works for binomial expansion. The third row, for example, is an expansion of (a+b)^2. So pick a corner and label it a^2. The two adjacent corners are ab, and the last corner is b^2.
Extremely interesting. I got curious and searched online for PI and e, and no surprises, they are hidden in this triangle as well. Amazing!. Thanks for this.
This video just gave me even more things i did not know about Pascal's triangle. Like, i knew about the diagonal sequences, the powers of two and the factorial coefficient, but i did not knew about the rest. Truly amazing how many things this triangle hides :D Thank you very much brady :)
TheTruthSentMe Sure you can choose to select and omit things at your choosing and then find any pattern. but when ever single number in a sequence is used, in a systematic way...it's not a made up pattern. it's really encoded in the numbers. just like everything in this video.
I don't know but I was really happy that this was a lady doing the video. it was fantastic. And even better was that she fucking loves Maths and the mysteries that it holds. What a video man. Kudos to you
Every side of a triangle can be a base, but a vertex cannot. So when a vertex is on the bottom and a side is on the top, you can legitimately say the triangle is upside down.
lowlize but you can't legitimately call it "upside down" or "rightside up" or anything like that because those aren't terms nor are the definitions in geometry. Just because in human society when a triangle is on its base it looks "correct" doesn't mean it is
I know, but if a side can be called a base there must be a reason, right? Even in abstract geometry we work using our physical intuition of a space with a privileged direction (that defined by gravity), so if it can be useful to construct an immediate visual representation (upside down triangle) without ambiguity, why not use it?
