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 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
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.
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.
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
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.
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.
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.
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!
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.
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.
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. :)
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!
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.)
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.
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!!
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 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.
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.
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.
Pascal's triangle describes the coefficients of the resulting polynomial from (a + b)^n A 3 dimensional Pascal's pyramid made to represent the coefficients of the resulting polynomial from (a + b + c)^n is related in really cool ways to the standard Pascal's triangle and I would have liked some more details on that.
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.
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.
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.
For a function f(x)=u(x)v(x), the nth-derivative denoted by f{n} is given by: f{n}=∑(nCi u{i}v{n-i}) for i going from 1 to n, and nCi is the binomial coefficient. Yes one can generalise the product rule for differentiation to any nth derivative where it turns out that the coefficients are from Pascal's triangle.
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.
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.
write pascals triangle and use it to calculate powers of 11 11^3= 1331 11^4=14641 11^5= 1 5 (10) (10) 5 1 start from right last dt =1, 2nd last =5, 3rd last but digit is 10 so we take only unit digit =0, 4th last digit is (10+1 carry from last ) 11 ,so unit dt of 11 =1, 5th last dt is 5+1 carry from last last digit =6 6th last digit is same as no carry is done = 1 so 11^5=161051 11^6=1 6 (15)(20)(15) 6 1 =1771561 it works like a basic addition trick (with diffrent rules ) when you know pascals triangle
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.
It can also be used as the coefficients of terms of the product of the nth power of a binomial, (a+b)^n. As of row 2, (a+b)^2 = *1* a^2 + *2* ab + *1* b^2, and the powers of a decreases in order, and the powers of b increases in order. Also works for row 3. (a+b)^3 = *1* a^3 + *3* a^2 b + *3* a b^2 + *1* b^3 Thanx for reading my comment! Edit: I like this video!
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.
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
Lovely video 🙂. As a comment, I would say that complete information about the prime factorization of any positive integer n can be obtained from the entries in the n-th row of Pascal's triangle. If frac( x ) represents the fractional part of x, and v_p(n) denotes the exponent of the prime p in the factorization of n, then we can apply the formula: v_p(n) = p * Sum_[ j = 1..floor(log_p(n)) ] frac( binomial(n, p^j) * p^(j - 1) / n ) ref: journal INTEGERS (22) _paper number: 61
If you take a row, put each number in binary and write each number vertically each ending at the same line then the shape of that will be a parabola which gets better and better the further down the triangle you go. There's a nice animation on wikipedia
Pascal's triangle is amazing! one year ago I used it in order to find tan(n*arctan(x)) in terms of f/g where n is a whole number and f and g are polinomials
Another neat part of the triangle is how it so immediately simplifies the (a+b)^x formula and just gives you the constant factor in front. No need to remember, nor do manual multiplying to remember! (a+b)^0 = 1 (a+b)^1 = 1a+1b (a+b)^2 = 1a^2 +2ab +1b^2 (a+b)^3 = 1a^3 + 3ba^2 + 3ab^2 + 1b^3 (a+b)^4 = 1a^4 + 4ba^3 + 6(ab)^2 + 4ab^3 + 1b^4 And so on :D
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
Some may have already said this but if you use each slot as a single digit number and carry the remainder over dont have to do that huge calculation at 4:45 I hope that makes sense
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 :)
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.
This is a nice pattern derived from the Pascal's triangle, when one of the edges (1 1 1..) is cut off and alternation between positive/negative values is added, and the numbers are then used as coefficients in polynomials. The resulting polynomials are the gnomons of the perfect powers. 1 2 -1 3 -3 1 4 -6 4 -1 etc n = sum(k=1..n)(1) n² = sum(k=1..n)(2k - 1) n³ = sum(k=1..n)(3k² - 3k + 1) n⁴ = sum(k=1..n)(4k³ - 6k² + 4k - 1) etc
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!
It's also related to the long-hand method for taking whole-number roots. I don't have time to go into detail, but I stumbled across the relationship when I was trying to figure out how to take third roots by hand. Square roots by hand is related to the 1-2 line as you multiple through and third roots to the 1-3-3 line, and so on, 1-4-6-4 for 4th roots, etc. :-)
Finally!! One of my least understood mathematical construct which appears everywhere and is extremely useful is finally here!! Expecting more explanations on its relations with permutation and arrangements. Also, Greeting From China!
I gave thumb up for a Autobot logo t-shirt before even watching :) I love Pascal's traingle, I remember trying to get more and more rows during boring mathematics classes in high school ;) Just one minor nit-pick: you forgot to mention (or show if I missed it) most common way of using the triangle, that is the n-row numbers are coefficients in (x + y)^n formula, i.e. row 2 means (x + y)^2, which is x^2+2xy+y^2
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.
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.
Lol, fermat’s last theorem reference. I like it.
Jeo okay fermat
It'd probably all fit now a days lol
I love how you guys constantly interview people who have a genuine love of mathematics! It makes these videos so amazing.
yes they just go around Europe asking random people in the streets.
I will never forget klein bottle man
instablaster
??.
same!!! love this channel sm
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.
You can verfy if a number n is prime by looking in the n-th row and checking if every number(beside the 1´s) in that line is 0 modulo n.
really? wow
So n is a prime factor of every number in the nth row (except the ones). Cool pattern.
Pascal's triangle never stops amazing me.
i was super excited when i first discovered this pattern myself
That makes me wonder...is testing pascal's triangle as efficient as simply testing factors?
00:50 "We can go on as long as we want and for me it is when the first double digit shows up because that is when adding gets hard." lol
You explained that better than an hour of lecture in 2 minutes, well done!
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
Please ramble on more about this triangle!
Did you know that if you put a point at every edge of Pascal's triangle you get a triangle?
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.
Except you don't, because it's infinite and only has one corner and two edges.
jenecomprends exactly. more like please don't
Zardo Dhieldor an edge if defined by two vectors, so maybe its a vector and two rays?
I used to watch you guys when I was like 6 or 7 and now I can finally understand what you are talking about
Fibonacci numbers in Pascal triangle..... JUST AWESOME
My triangle is amazing huh?
You didn't discover it you thief!
Justin Ward but it has my name :c
You stole the triangle from Chinese mathematicians. Thief!
yee
A more illustrative way to write out the part at 4:47 would be:
100,000
050,000
010,000
001,000
000,050
000,001
-------------
161,051
I love it when mathematicians can't hide the hype while talking about numbers!!!!
All of this is adorable!!!! XD
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.
So cool to see an undergrad on numberphile
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
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.
Yosef MacGruber Therefore, your counterargument is incorrect.
New conjecture: Pascal's triangle encodes literally everything that can be encoded.
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.
@@samuelking1624 a hundred years later, its a millenium prize problem, and we all were here to witness history right before our eyes
Finally, a video about Pascal's triangle, and it's amazing.
honestly what I really like it how excited she is to explain the triangle, her enthusiasm had me interested
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
I first was introduced to Pascal's Triangle in the third grade... when I was eight.
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.
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!
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.
What are dimensional triangle numbers?
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.
11:09 Brady's little exhale when he gets what's going on is the exact same reaction I made :D Very cool!
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. :)
6:26 is my favorite thing. All the mods. Had a great maths teacher who showed this pattern to us.
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!
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.)
hey enlong chiou! i remember you from g+ ! ur awesome!!
enlong chiou give us the nontrivial zeros!
A favorite of mine. If this is confusing just google 'Faulhaber's formula.'
Let x/(exp(x) - 1) = a + (b/1!)x + (c/2!)x^2 + (d/3!)x^3 + (e/4!)x^4 + (f/5!)x^5 + ... [and let mx^n = (m)*(x^n)]
Then
1^0 + 2^0 + 3^0 + ... + n^0 = (1/1)(an)
1^1 + 2^1 + 3^1 + ... + n^1 = (1/2)(an^2 - 2bn)
1^2 + 2^2 + 3^2 + ... + n^2 = (1/3)(an^3 - 3bn^2 + 3cn)
1^3 + 2^3 + 3^3 + ... + n^3 = (1/4)(an^4 - 4bn^3 + 6cn^2 - 4dn)
1^4 + 2^4 + 3^4 + ... + n^4 = (1/5)(an^5 - 5bn^4 + 10cn^3 - 10dn^2 + 5en)
1^5 + 2^5 + 3^5 + ... + n^5 = (1/6)(an^6 - 6bn^5 + 15cn^4 - 20dn^3 + 15en^2 - 6fn)
etc.
Notice the 'truncated' version of Pascal's triangle.
enlong chiou Amazing!!!
Says the anonymous guy called "Fish Bones"
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.
what if you extended the triangle to a tetrahedron or any simplex of n dimensions?
wow this is mind blowing stuff and i have some ideas.... thank you man.
I've heard they've done that for a trinomial distribution pattern, pretty neato
Imagine a programm that represent this pattern as tetrahedron in a 3d world and highlight some numbers.
Look up Pascal's Pyramid
You get the multinomial coefficients.
Casandra Monroe's enthusiasm is inspiring.
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!!
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!
Thank you -- your enthusiasm is infectious!
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.
she is so cool
Are you ok?
@@yvesnyfelerph.d.8297 yes
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.
This is like magic. I didn't realize Pascal's Triangle had so much in it!
Its magic :D
This is made by 1st grader additioning
Wow I dont regret watching this. well done, I'm quite suprised to find this out. Thank you
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.
you could download pari gp and try it with high precision. though you may be relating it to pascal's simplex. en.wikipedia.org/wiki/Pascal%27s_simplex
It's great to have those kind of videos of brilliant people for free available for everyone, Thank you!
im so happy numberphile finally did a video on pascal's triangle
Pascal's triangle describes the coefficients of the resulting polynomial from (a + b)^n
A 3 dimensional Pascal's pyramid made to represent the coefficients of the resulting polynomial from (a + b + c)^n is related in really cool ways to the standard Pascal's triangle and I would have liked some more details on that.
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.
Uncanny and completely amazing that what at first blush feels like a children's exercise contains all these other relationships and qualities,
12:15 'there are still other things i can ramble on about....'
Me: TELL ME NOWWWW
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.
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.
Casandra Monroe was amazing in this video!!! Bring her back for more please!!
She's wearing an Autobot shirt, she is immediately the best guest you've ever had. :D
For a function f(x)=u(x)v(x), the nth-derivative denoted by f{n} is given by:
f{n}=∑(nCi u{i}v{n-i}) for i going from 1 to n, and nCi is the binomial coefficient.
Yes one can generalise the product rule for differentiation to any nth derivative where it turns out that the coefficients are from Pascal's triangle.
When I saw the Fibonacci Sequence in that I actually said, "It's so beautiful!"
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.
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.
The 8th row is 255 and is not prime. The next prime (257) is the 9th row
yeah, the primes are at (assuming the top is the 0'th row) locations 1, 2, 4, 8, 16, ...
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.
write pascals triangle and use it to calculate powers of 11
11^3= 1331
11^4=14641
11^5= 1 5 (10) (10) 5 1
start from right last dt =1,
2nd last =5,
3rd last but digit is 10 so we take only unit digit =0,
4th last digit is (10+1 carry from last ) 11 ,so unit dt of 11 =1,
5th last dt is 5+1 carry from last last digit =6
6th last digit is same as no carry is done = 1
so 11^5=161051
11^6=1 6 (15)(20)(15) 6 1 =1771561
it works like a basic addition trick (with diffrent rules ) when you know pascals triangle
Smart mathematician doing beautiful stuff!!!
Yea Maths is fantastic! You can always be amazed by learning new ways to interpreter Pascal's Triangle!
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.
It can also be used as the coefficients of terms of the product of the nth power of a binomial, (a+b)^n.
As of row 2,
(a+b)^2 = *1* a^2 + *2* ab + *1* b^2,
and the powers of a decreases in order, and the powers of b increases in order.
Also works for row 3.
(a+b)^3 = *1* a^3 + *3* a^2 b + *3* a b^2 + *1* b^3
Thanx for reading my comment!
Edit: I like this video!
I've never seen the Fibonacci pattern before. Mind blowing.
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.
I always thought pascal's triangle was kind of boring. My mistake! hehe XD
Has anyone ever tried to apply the same methods to different bases?
I'd love to see what those look like, out of silly curiosity
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
I love her passion and energy.
Can you show a Parker Triangle next?
Lovely video 🙂. As a comment, I would say that complete information about the prime factorization of any positive integer n can be obtained from the entries in the n-th row of Pascal's triangle. If frac( x ) represents the fractional part of x, and v_p(n) denotes the exponent of the prime p in the factorization of n, then we can apply the formula:
v_p(n) = p * Sum_[ j = 1..floor(log_p(n)) ] frac( binomial(n, p^j) * p^(j - 1) / n )
ref: journal INTEGERS (22) _paper number: 61
nth
Nth + 1
suc(suc(Nth))
N shell!
n^mth
Zeroth
If you take a row, put each number in binary and write each number vertically each ending at the same line then the shape of that will be a parabola which gets better and better the further down the triangle you go. There's a nice animation on wikipedia
It's cool to see an undergraduate on here. :) Gotta love the enthusiasm!
When you do it in mod 2, then the number of 1's in each row is always a power of two. Great video!
Illuminati confirmed
Get back to being a meme on Etho's channel! :P
We take the Triangle and push it somewhere else
lol who'd have known you'd leave Etho's channel to find your way to a top comment here?
Yes. Once again.
After millions of confirmations.
This confirms it again.
Are you following me? I just came from Project Ozone #29... :)
I love so many things about this video!
there is also the binomial thing
Pascal's triangle is amazing! one year ago I used it in order to find tan(n*arctan(x)) in terms of f/g where n is a whole number and f and g are polinomials
in Italy we call it "Tartaglia's triangle"
Thanks to you i instantly solved a circuits problem involving infinite resistances organized as a pascal triangle. Thanks!
*Claimed by Nintendo LLC
I love this, quite useful for my discrete math course
Parker triangle - the new Parker square
Another neat part of the triangle is how it so immediately simplifies the (a+b)^x formula and just gives you the constant factor in front. No need to remember, nor do manual multiplying to remember!
(a+b)^0 = 1
(a+b)^1 = 1a+1b
(a+b)^2 = 1a^2 +2ab +1b^2
(a+b)^3 = 1a^3 + 3ba^2 + 3ab^2 + 1b^3
(a+b)^4 = 1a^4 + 4ba^3 + 6(ab)^2 + 4ab^3 + 1b^4
And so on :D
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
Prateek Gurjar I don't get it.
units of pressure is pascal..which is Newton per meter square
That One Guy "Newtons per meter squared" is a unit of pressure which is also known as Pascal
exactly !
I think in english we say "square meter" and not "meter sqaure". I could be wrong
Some may have already said this but if you use each slot as a single digit number and carry the remainder over dont have to do that huge calculation at 4:45
I hope that makes sense
I'm in love.
A little bit of both.
Mojos Bigstick A little bit of both what?
There was previous reply asking if I was in love with Ms Monroe or with Pascal's triangle. I'm a little bit smitten with both.
I was worried you fell in love with the triangle only.
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 :)
420 Blaise it
Phort won tea, Belize it
Angry Seal nice
Angry Seal +
wow just wow
Where? Why? Tell me I *need* to blaze it!
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.
+Andrew Gutsuleac you you also used the little bell to get notifications!
Cassandra, it wasn't nice to tease us with three or four Fermat primes without mentioning that the run of primes is interrupted after 2^2^4 + 1.
This is a nice pattern derived from the Pascal's triangle, when one of the edges (1 1 1..) is cut off and alternation between positive/negative values is added, and the numbers are then used as coefficients in polynomials. The resulting polynomials are the gnomons of the perfect powers.
1
2 -1
3 -3 1
4 -6 4 -1
etc
n = sum(k=1..n)(1)
n² = sum(k=1..n)(2k - 1)
n³ = sum(k=1..n)(3k² - 3k + 1)
n⁴ = sum(k=1..n)(4k³ - 6k² + 4k - 1)
etc
I wouldn't be surprised if Pi showed up in that triangle in a way… somewhere. 🤔
HyperQbeMusic u just gave me something to work on... thanks
Excellent explanations!
Mathimus Prime...
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!
The Legend of Zelda reference \(°-°)/
It's also related to the long-hand method for taking whole-number roots. I don't have time to go into detail, but I stumbled across the relationship when I was trying to figure out how to take third roots by hand. Square roots by hand is related to the 1-2 line as you multiple through and third roots to the 1-3-3 line, and so on, 1-4-6-4 for 4th roots, etc. :-)
The right name is Tartaglia's Triangle :-)
Facciamoci sentire ahaha
Marco Beri quindi non sono l'unico italiano qui
Ehi, Triangolo di Yang Hui, ci siamo arrivati prima noi :-)
stavo pensando l'esatta stessa cosa... perché questi tizi lo stanno chiamando "triangolo di pascal".....?
Volevo proprio scriverlo..
Finally!! One of my least understood mathematical construct which appears everywhere and is extremely useful is finally here!! Expecting more explanations on its relations with permutation and arrangements. Also, Greeting From China!
Didn't expect to see Fibonacci sequence AT ALL!
damnnn Casandra Monroe back at it again with
a transformers shirt... .rly?
Someone My exact sentiments.
Would be appropriate if it was a video about primes.
yeah not the best shirt for a mathematician to wear... it takes your focus of the actual video
I didn't even notice it until the 8 minute mark or so.
pure mathematics joke...
I gave thumb up for a Autobot logo t-shirt before even watching :)
I love Pascal's traingle, I remember trying to get more and more rows during boring mathematics classes in high school ;)
Just one minor nit-pick: you forgot to mention (or show if I missed it) most common way of using the triangle, that is the n-row numbers are coefficients in (x + y)^n formula, i.e. row 2 means (x + y)^2, which is x^2+2xy+y^2
It's inevitable to find patterns if you look for them.
search for pattern is like a hidden background door in maths.you could find something useful for physiks or other topics.
especially when they're actually there
There is a world of difference between seeing patterns from random noise and being able to _prove_ there's patterns in something like this.
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.
Sometimes you don't even have to focus on finding a pattern. People find patterns that don't really exist everyday almost.
Fantastic episode! Bloody brilliant. Love the presenter's enthusiasm too!