Pascal's triangle was one of the first programs I wrote in 1982 learning Pascal. It was just simple enough for a 13 years old child. And here we are, a PhD talks 25 minutes about it. In Math there is always so much more in every problem. I love it, thank you.
That sounds really boring. Whybwould abyoen want to sit innfront of a dsmn co outer all day or even for a few bours? And isnt math in general tedious and frustrating?
That's what I thought too since generalising something something involves conserving many properties such as symmetry. He probably just used other generalisations because they were more interesting.
@@schwingedeshaehers so there's 1, above that is ½ ½, above that is ¼ ¼ ¼, aboce that is ⅛ ⅛ ⅛ ⅛ etc. This breaks a pattern though, which is that the sum of the numbers in the nth row is 2ⁿ⁻¹, but if n = 0, the sum would be 1 instead of 2⁻¹. We could create a new formula for nonpositive values of n which is 2ⁿ⁻¹(2-n)
Well I mean if it was wouldn’t just be half of Pascal’s triangle (ignoring the negatives)? Edit: wait nvm but what would it be? Edit: wait nvm it would be that Edit: wait nvm it could also be other things
That feels like a canonical choice to me. As pointed out in another comment yesterday, generalizing something sometimes involves conserving many properties, such as symmetry.
Here is another way to exclude 1/2. The sum of the n-th row of the Pascal triangle is 2^n (expansion of (1+1)^2). But 1/2-1/2+1/2..=1/4. Because 1-1+1..=1/2. 😉Indeed, we can see this sum as the sum of a geometrical sequence of reason r=-1 which is 1/(1-r) for r>0.. If we apply it for - 1 we get sum=1/2. It is a bit of a stretch, as mathematically the sum is just a non converging serie alternating between 0 and 1,but it makes sense somehow..
Can you really do 1/2-1/2+1/2… though? In the middle we have a +1/2+1/2 which means this series isn’t exactly alternating and order matters with infinite series
@@PedroCristian how (and why) are you assigning values to non-convergent series? :-s PS it's a ratio not a "reason". ;-) Also -1 is not >0 so you can't really "apply it for". 🤓
@@cyrilmeynier5688Matgematicians have means to prove you can generalise, and a keenness to be careful about whether they can. People who generalise other people have neither.
It's also fascinating that the symmetry of the bottom (classic) part of the triangle is broken -- arbitrarily, really -- by a decision about whether we read the coefficients right-to-left or conversely.
This is honestly the first place I've seen the series expansion of (1+x)^-n for |x|>1 by leveraging the fact that (1+x)^-n = x^-n*(1+1/x)^-n, and it's one of those things that feels so obvious in hindsight, that it's hard to believe it never occurred to me before seeing this video
The final diagram with 3 separate scalar multiples of Pascal’s Triangles with 0s in between was truly beautiful. The proof (for the binomial coefficients doing that) was also extremely satisfying and cleanly done! I’ve also never seen expansions of real binomials for abs(x) > 1. Just wow!
Hi, great video! What's happening at the end there in terms of the doubly infinite series can actually be made sense of rigorously, if you're willing to ignore convergence. Just like you can write down formal power series with arbitrary coefficients while ignoring convergence, you can write down "doubly infinite formal Laurent series" with arbitrary coefficients in both directions while ignoring convergence. These things are no longer closed under multiplication so they don't form a ring, but you can still multiply such a series by a polynomial (even a Laurent polynomial), so they are still a *module* over polynomials (or even Laurent polynomials), and what your calculation is doing is repeatedly attempting to invert the operation of multiplication by (1 + x). The reason you get this 1-parameter family of choices when you try to extend upward is that multiplication by (1 + x) is not invertible (unlike in formal power series where it is invertible) - there's a series p(x) such that (1 + x) p(x) = 0, namely the series p(x) = sum_{n in Z} (-1)^n x^n, which has the property that if you multiply it by x you get -p(x)! This series can be interpreted as the "Dirac delta at x = -1," since it has the more general property that if f(x) is any polynomial then f(x) q(x) = f(-1) q(x). The fact that it has coefficients going off infinitely in both directions is related to what happens when you take the Fourier transform of the Dirac delta, and the fact that doubly infinite formal Laurent series aren't closed under multiplication is related to the fact that you can't multiply Dirac deltas together. So, every time you try to invert (1 + x) you end up with another 1-parameter family of choices you can make, because every time you have the freedom to add another multiple of p(x). (And it's exactly 1 parameter, not more; it's not hard to show the kernel of multiplication by (1 + x) is 1-dimensional with basis p(x).) Cheers, Qiaochu
Oh wow! I didn't get most of that due to a lacking math background, but it's incredible! I'm definitely going to look into the Fourier transform of the Dirac delta function.
@@levivanveen6568 Well, if by "grow upwards differently" you mean "in a way that still satisfies Pascal's rule" then this is the only option: what I mean by "trying to invert (1 + x)" is just an abstract way of saying "trying to extend upwards while keeping Pascal's rule."
Not just the English. Im Irish and we love our tea and porridge. Also I think porridge is mostly a Scottish thing rather than English. But then again everyone can enjoy good food with good maths.
Whats interesting is that if you add up all the numbers in the nth row in Pascal's Triangle, you get 2^n For example, 1 3 3 1 is the third row and 1+3+3+1 = 2^3 = 8. But at the -1st row at 4:18, you get zeros on the left and 1-1+1-1+... on the right, which is Grandi's series which "evaluates" to 1/2, which is 2^-1, and I can assume that the other negative rows "evaluate" to 2^n as well.
This is utterly remarkable. How do such things work so well? It reminds me of extrapolating the notion of exponent from integer "counts" of multiplication out to negative and fractional exponents: just extend the additive arithmetic for whole exponents to any number, "pretending" that it still holds, and voila...
Veritasium also made a video about this and it was about Newton and π. He figured out that you could squeeze fractional values between the integers. He had invented integrals and he could combine these ideas to calculate π as an infinite sum
When I did this in undergrad, i found that a = b = 1. Instead of extending with the addition rule, I used the n choose k formula for the coefficients and made the assumption that -1! would be some kind of undefined infinite value, but 1/-1! would be 0, and the factorial relation would be preserved. The pattern fills out like you would expect with a=b=1, but the reason why the addition rule breaks down at that spot is because the proof of the addition rule for n choose k would do a division by 0 at that spot. After that, I played around with higher dimensional pascal's triangles, which helps when you have the assumption that the outsides of the triangles/tetrahedrons are 0.
Legend says that way off in the distance to the right, there is another triangle. The whole thing is a triforce, if you will. The bottom left, the original, is the triforce of power. The top, above the first, is wisdom. The third is only legendary, yet hypothetically possible with the oddity that is named infinity. Its name? Courage.
This is an piece of poetry of course, but note if you take pascal's triangle (well, the normal positive parts), you actually get a sierpinski triangle. A mega-triforce, if you will.
Nice video and explanation :) How about doing a sequel on "What Lies Between Numbers in Pascal's Triangle"? Is there a continuous function that smoothly interpolates all the discrete values of Pascal's triangle, like how the gamma function interpolates discrete values of the factorial function?
The kth term of the nth row can be found as n!/((n-k)!k!). Switching to the continuous version of the factorial gives Π(n)/(Π(n-k)Π(k)), (though you might be more familiar with Γ(n+1)/(Γ(n-k+1)Γ(k+1)) ) This version would give smooth intermediate values for Pascal's triangle, however since the Pi function diverges for negative integers (corresponding to the Gamma function diverging for negative integers and zero), it wouldn't be able to give the discrete values of the extended Pascal's triangle without some limits. There is another way to extend Pascal's triangle as Π(a + b + c + ...)/(Π(a)Π(b)Π(c)...) for multinomial expansions with arbitrary numbers of terms.
I was led to what lied above the Pascal Triangle by extending the Fibonacci number backwards from zero. (Be aware that the Fibonacci numbers are also obtained from Pascal Triangle) and then filled the missing upper layers. It turned out that if the bottom triangle is the expansion of (x+1) to the powers of n then the upper parts are the expansion of (x-1) to the powers of n. And also whereas the sum of the coefficients of every row is 2 to the power of n, those of the above sum to zero.
Very cool. Pascal’s simplices are some of my favorite patterns in mathematics, so this is very cool. It also gives an argument for why the sum 1-1+1-1+… “converges” to 1/2 in some contexts.
@@harshlalwani4353 the sum of each line in Pascals triangle is a power of two, for example 1+2+1=4=2² or 1+3+3+1=8=2³ if we take the -1th line, which is 1-1+1-1+1-1+... it should sum up to 2^(-1) = 1/2 and it actually kinda does
@@harshlalwani4353 This is an interesting example of generalization, interesting by the fact isn't 'settled' yet. You know, math expands when operations are generalized. For example, take substraction. With natural numbers, some substractions are impossible. You can't calculate 3-4 (with natural numbers), unless you generalize substraction, and that's accomplished by introducing negative numbers and defining substraction as a type of addition that uses those negative numbers. Same with division: with integers, you can't divide 5/2, but you can if you introduce rational numbers. So the pattern is: you find something you can't do with the current axioms you have, so you expand those axioms so you now can, and probably you are also redefining what the operation means. With infinite sums, you have the same issue. Convergent sums can be solved, but divergent sums can't. So, is there a way to generalize maths so we can redefine the operation of sum so we can solve them? The answer is 'yes', there are several ways to redefine the sum operation so you can assign a numerical value to a divergent series. The problem here is that there is more than one way to accomplish that, and another problem is that the redifinition of the 'sum' concept makes him a lot less intuitive. The issue would be resolved, probably, when real life applications appear (and they always do, but sometimes it can take a couple of hundreds of years) and they use one specific way of making the sum.
What if we start with a = b = 0.5? I started to explore this myself. All of the numbers in the row have a fixed power of two as the denominator, e.g., 1/2 for row -1, 1/4 for row -2, etc.. But each numerator row is a series of numbers I have not seen before: Each series is symmetrical, extending to infinity in both directions. Looking only at the right side, I see: Row 0: 1 0 0 0 0 ... Row -1: 1 -1 1 -1 ... Row -2: 1 -3 5 -6 ... Row -3: 1 -7 17 -29 ... Row -4: 1 -15 49 -107 ... What are these numbers? They are derived from b = 2*c-a, where c is the number below and to the right of a we can simplify things by turning the isosceles triangle into a right triangle: 1 1 1 1 2 1 1 3 3 1 ...
Your -6 on row -2 should be -7 (leading to different values in the next rows as well). The pattern is clearer in the diagonals: focus on the top right part, consider the diagonal going to top right, again omitting the powers of 2 in the denominator. 1,1,1,1,1,1... 1,3,7,15,31,63... 1,5,17,49,129,321... 1,7,31,111,351... First diagonal: 1 Second diagonal: 2^n - 1 Third diagonal: (n-1)2^n + 1 Fourth diagonal: (n²+n+2)2^n - 1 Fifth diagonal: (n³/3+n²+8n/3)2^n + 1 So the pattern seems to be alternating +1 and -1, added to 2^n times a polynomial in n. More interestingly, they are the coefficients of the series expansion of: 1/(1-x) = 1 + x + x² + ... (first diagonal) 1/((1-x)(1-2x)) = 1 + 3x + 7x² + 15x³ + ... (second diagonal) 1/((1-x)(1-2x)²) = 1 + 5x + 17x² + 49x³ + ... (third diagonal) 1/((1-x)(1-2x)³) = 1 + 7x + 31x² + 111x³ + ... (fourth diagonal) and so on. If you don't know yet, oeis.org is a great tool to recognize this kind of sequences.
Math is so beautiful. Somethimes I wish I had studied Math in university. But then again, I can just enjoy the beauty of Math at home through videos like this.
This is one of the two options, yes If you follow the / diagonal, a is 0 and b is 1 If you follow the \ diagonal, a is 1 and b is zero They're both the same, just flipped over the y axis
You are sofa-king good at this I really hope you do this for the rest of your life. Recorded videos like this are forever. You're so good at this and there's no overhead but the upside to society in perpetuity is exponentially great ROI on this investment. Keep at it, please. This kind of content is what makes RU-vid so amazing.
Caught your video in recommended, didn't watch it in full but skipped ahead to 20:37 to get a sneak preview, I never thought about this, very interesting.
I am interested but not well educated in math and this sort of succinct explanation of an interesting (to me emergent) phenomenon always blows my mind. I don't know whether I'd use any takeaways in life but am pleasantly surprised by youtube suggesting this :)
this shit crazy, you rotate it 120 degrees either side, resulting in 6 hexants (6 triangles in a hexagon) 1. the OG Pascal triangle 2, 4, 6. the Zeroes triangles 3&5. the As and Bs Pascal triangles, similar to the og triangle except for the alternative change between positive and negative values, depending on if it's seperated by even or odd number of zeroes
I remember extending the Pascal triangle upwards on a whim in high school math, but I did not at all understand that the numbers (after arbitrarily going with 0 1 though I don't remember which direction I went in) correspond to these coefficients. It's really very beautiful.
Many many thanks for this very nice and exhaustive video! The "expansion to the left" shown by you has also another possible interpretation, sliding the rows to make it look like an infinite square matrix. Then Pascal's original triangle is equal to the exponential of a particular "subdiagonal" matrix, i.e. a matrix having all zeroes on the main diagonal and {0, 1, 2, 3, ...} on the parallel diagonal situated one place lower. If we expand this diagonal also above with negative values, and exponentiate, we get Pascal's triangle expanded to the upper left.
22:53 I think that’s a brilliant maths quote: ‘bearing in mind that this isn’t actually going to be valid for any values’ but we’ll just keep going anyway cus it’s interesting!
Excellent video! This is the first time I’ve seen the rest of Pascal’s Triangle explored. I got out the popcorn when I saw the 1, -2, 3, -4, …. appear. This gon b gud. 🤓
I would like to note that this generalization preserves in a certain way the property that the sum of the terms of the n'th row is 2^n (starting to count from the 0'th row). If you consider Ramanujan sumation, then this property actually holds for negative n.
The upper extension of Pascal's can be rewritten very much like the usual triangle, except that the numbers down the left hand side oscillate between 1 and -1, the numbers down the right hand side are all 1 as usual, and all the other numbers within the triangle are found by subtracting the number above to the right from the number above on the left.
To be honest i think the topic could be explored with much more fun. With excel (or google sheets) you can just experiment directly with the number. It's obvious that you can select one number in each row arbitrary (useful to set them at the center or directly around it). I was trying to preserve another property of triangle - symmetry (selected 0.5+0.5 to construct top most 1). Experimenting around it's quickly become obvious that if you set all other rows to zero at the center you'll get a nice pattern of inverted triangle of zeros. And all non-zero values forms two Pascal's triangles with all values divided by 2 and oscillating signs. Diagonals of "1" continues as diagonals of "0.5". And if you change one 0.5 to 1 (or to 0) then you'll lose symmetry but you'll get copy of original triangle on one side an just zeros on the other, which is kinda neat... Why bother with binominals at all?
Together with the relationship between Pascals and Sirpinskis triangle, that gives every odd number in the Pascal Triangle a certain color and every even number another certain color, resulting in a Sirpinski-Triangle-Looking-Pattern, this expansion to "what lies above" gives the Sirpinski a somewhat zero-th iteration or just an miror image pattern. But in general, its cool, Pascals Triangle is 1 of my favorite math topics and now thanks for making it into an Hourglass. So if above Pascal triangle is just a fliped version of the triangle, its more like Pascals Hourglass. Or a rather distorted Hourglass as there are now infinite nonzero values to the right. Proof is 15:04 for the Hourglass. And as 18:49 says, the uper part of the Hourglass is bent to the side coresponding to if |x|>1 or |x|1 or |x|1 or |x|
You made an arbitrary choice in expanding 1/(1 + x) as 1 - x + x² - · · · (|x| < 1). It could just as well have been expanded as 1/x - 1/x² + 1/x³ - · · · (|x| > 1), pushing all the non-zero coefficients to the left rather than the right. More natural would be to invoke the symmetry of the original table so that diagonally above, and each side of, the solitary 1 in the zeroth line is a ½. Then ±½ will alternate in both directions along line -1.
Very nice, but since it appears the choice of a and b is arbitrary, provided they sum up to 1, I would have chosen a = b = 1/2, in order to preserve the triangle simmetry
There is no 'correct way' to do this. You can do what the guy did, preserving the property of Pascal's triangle representing the coeficients of functions, or preserving the symmetry, that wouldn't preserve that property of Pascal's triangle but maybe it can serve other purposes.
In short: another Pascal's triangle, but rotated 120 degrees counterclockwise, with the previously "left", now bottom chain of 1's alternating between 1 and -1, producing a Pascal triangle with alternating positive and negative values. And a whole bunch of zeroes
That is wonderful! Is that original with you? Whether it is or not, you've made a video that tickles my math brain in a lovely way, Thank you very very. This is also a cellular automaton. Have you tried complex numbers for a and b?
I've always extended Pascal's Triangle in my mind as if there was an extremely tiny, infinitesimal chance that there was a glitch in the summation. So you start with an infinite hexagonal grid of 0s, each 0 the sum of two other zeros. Except randomly, at some point in the infinite expanse...there's a "glitch" and 0+0=1. And from that, the entire Pascal's Triangle is generated. Similarly: the Fibonacci series.
A fun thing to do is to replace the outer rows of 1s with different patterns. Similarly, you can use the rules of pascal's triangle to reconstruct the upper layers given a "seed" cluster of numbers from some arbitrary layer.
The similarity of the two triangles indicates that you could extend the nCr operation to negative values of n, by mapping them to the positive domain. Formally, (-n C k) = (-1)^k * ((n - k) C k) I vaguely recognize that, might have learned it in combinatorics class! It could be useful for some applications of generating functions
Crazy to think this all started because people a long time ago wanted to communicate quantity with one another. Modern math is so out of the box when you think about it. Something as simple as the decimal point, negative numbers, all the way to complex numbers, calculus, relativity, and quantum physics are all such abstract concepts. Negatives exist because someone subtracted too much from a number. Complex numbers were 'discovered' because people needed a way to compute equations with square roots of negatives. Relativity exists because someone noticed clocks tick slightly different in motion, which allowed us to calculate the curvature of spacetime inside a rotating black hole. That's like giving someone one day of stock market data and them successfully reconstructing the entire history of that market. Quantum physics exists because someone noticed some particles aren't sure where they are. Our understanding of numbers is so out of bounds it's crazy. Humans are good at taking what they have, and pushing them so far beyond their limits they break reality.
If you look at the diagonal that starts with all 1's then it's just f(i,j) = sgn(i)^(j+1)*nCk(i+j-2,j-1)). That is, this is the "square form" of the triangle. Each column then is simply a diagonal in the triangle. The diagonal of this is just a row in the P.T.. The same idea holds: f(i,j) + f(i-1,j+1) = f(i,j+1)
Well I'm amazed. What an incredible finish! A and B as coefficients that satisfy (a+b = 1), but for which X is undefined. Could there be complex values of X where there's a valid solution?
Nice! Now I wonder also what happens when two Pascal's triangles (separated by N zeroes) collide. In fact this perspective remembers me of cellular automata (I had to watch without audio so I don't know if it was mentioned)
Problem: given a natural number n that does not occur in the standard Pascal's triangle , find the position of the other triangle's tip that will cause n to appear in the collision.
@@tylerduncan5908 On second thought, the question as stated makes no sense because all natural numbers appear in Pascal's triangle just under the 1s. Note to self: don't come up with brilliant math ideas after 10PM.
@Axacqk what do you mean? I was thinking about what would happen if you had, say, two pascal's triangles separated horizontally by a string of 0's in the uppermost row. For example ------1----------1------ -----1-1--------1-1----- ----1-2-1-----1-2-1---- ---1-3-3-1---1-3-3-1--- --1-4-6-4-1-1-4-6-4-1-- 1.5.10.10.5.2.5.10.10.5.1 And so on. I wonder what that would represent, and what the pattern would become as the 2 triangles merge more and more.
The extension above the "standard: Pascals triangle where the sum of the a and b values do not converge as you show, reminds me of the analytic extension "trick" that is used in, say the Reimann Zeta function. I'm sure that is not accurate, but it does seem as if there is an analogous process going on here.
Also, as you suggest there are ways of using any number presumably the reals, as coefficients, do this mean you are generalising Pascals Triangle in a similar way to generalising the factorials, as teh Gamma function? Intriguing stuff!
That's really cool! So...instead of Pascal's Triangle, it's more like...Pascal's Hexagon, with 3 triangles of 0's and 3 Pascal's Triangles with 60º between them.. :)
