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
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.
@@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.
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". 🤓
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
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."
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!
Amazing video! At everyone in the comments who is asking why he didn't talk about using 1/2 + 1/2 = 1... he did!! In the end of the video the purpose of generalizing to a and b is to allow you to choose ANY values of a and b that sum to 1, which includes 1/2! So just pause the video at the end and replace a and b both with 1/2 and you have the triangle you're asking for. It's certainly the most aesthetically pleasing and temtping choice, but to everyone insisting that symmetry is the most important property of pascal's triangle that needs to be preserved, I urge you to learn more about Pascal's triangle@ The really neat thing is he showed in the end basically that while there are many other exciting patterns to explore, the only choices for (a,b) that lead to convergent series expansions are (1,0) and (0,1). It's pretty freaking cool. Of course I wanted the "canonical" answer to involve 1/2 as well, that's human instinct, but it's delightfully charming when math gives us an ego check and reminds us there are much more unexpected ways for things to be beautiful. Ultimately I find the general form with a and b to be the most beautiful. It shows the perfect symmetry in its own way and represents an infinite spectrum of pascals triangles, precisely two of which have rows that represent valid convergent series expansions, and one of which is perfectly symmetrical. HOW COULD YOU NOT LOVE IT!!! 😍
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.
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...
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.
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.
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
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.
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 RUclips so amazing.
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.
Great video, I learned a lot through this! Another very interesting thing I see in this expansion is the fact that it simultaneously also displays the binomial expansion for the term (1 - x)^n, so for example n=2: 1 - 2x + x² n=3: 1 - 3x + 3x² + x³ n=4: 1 - 4x + 6x² - 4x³ + x⁴ and so on.
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.
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.
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.
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 :)
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.
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.
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
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.
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!
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.
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.
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?
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.
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 when i was like 10 and super bored i extended it this exact way (without the binomial expansion obviously, i just did it by accident) i thought of it like a rotated, half negated version of it. i also thought it was kinda like being zoomed infinitely into the Sierpinski triangle (i liked fractals back then)
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|
Well he used for the rotated Pascal triangle the binomial expansion of (1+x)^(-n) = sum from k = 0 to infinity of (-1)^k *((n+k-1) choose k)*x^(k) for |x|
The coefficients a and b can be chosen to make the two series converge around an arbitrary point, making this a recipe for a laurent series expansion of the expression
Very cool! I wonder about deferent triangles one might make. I discovered in 5th grade that the difference of the power series for the N-th power eventually yields N! after N iterations.
There are a few ways you could expand it based on a rule. Maybe you are interested in the fibonacci sequence expansion going backwards too. Also valid, since that sequence extends in both directions.
Just noticed also that the sum of elements in a row of Pascal's triangle is the corresponding power of 2. The row at n = -1, you've established should be defined as alternating 1 and -1, and the power of 2 would be 1/2. The agrees with the value of an L-function under analytic continuation that would correspond to 1-1+1-1+... "=" 1/2
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!
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?
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?
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
Hello! Can you define Pascal's triangle as the convergence of the a's, and b's? What other triangles are out there that are also the result of certain a's, and b's convergence other than the Pascal's triangle?
Wow I searched for pascarl's trinagle trivia in and out and this is one of the coolest. I was wondering though what if we apply it to other type of binomial functions like: a(1+1/x)^-1+b(1+x)^-1 which would satisfy the property or something weirder maybe.
Wow this is super interesting!
I agree
woah it's blackpenredpen!!
what i said ong
I thought it would've been the same but upside down or negative or fractions
BPRP?!!!???
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.
Makes sense to learn **pascals** triangle in pascal
You can up with it?
@@saltycucumber2773 no, but they wrote a program for it.
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?
I reckon I need to learn how to program Java's triangle now.
This is so insane how is this the first time I've ever seen this idea?
hi slowfreq i love your music its cool to see you here
@metepure WHAT?!?!??!? Dang dude thanks!!!!
I've seen it before.
I was thinking 1/2 above 1 and alternating -1/2 and 1/2 on either side, since the triangle seemed symmetric.
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.
the question is than, what is above these both? (between)
@@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)
YES! This is the only way the extended Pascal's triangle for negative n carries over the symmetry around the axis of mid coefficients.
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
Mind blown when he showed the original triangle rotated at the top
Truly
Highly suggest you watch the veritasium video on how Newton found a way to calculate pi. Will blow your mind further
Indeed ,that was a key observation I never expected .
I've seen quite some math.
I learned something today.
Math is beautiful.
thanks for suggestion. great video @@bob_kazamakis
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.
Everywhere you look up new thing in Math it seems like beauty always emerge.
as a demon girl, some of the more friendly demons spend our whole lives immersed in the chaos and waterfall of numbers :3
Sociologisst : you shouldn't generalize.
Mathematicians : yes you should !
@@iidoyila_live_is this a maxwell demon’s kind of demon girl or a kink? What does “demon girl” even mean in this context???
The Magical Wife.
@@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.
my smooth brain wanted a and b to both be equal to 1/2 or 0.5
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.
I wonder how that'd look like when extended upwards.
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". 🤓
I once did the expansion while bored in class. It's nice now learning of proofs and further expansions, thanks!
"What lies above Pascal's Triangle? Pascal's Triangle."
It's Pascal's triangle's Wario.
Wascal's triangle.
Pascal's Triangle all the way up
@@maksiiiskam2 RASCAL'S TRIANGLE??!!
A 90-degree rotated Pascal’s triangle.
@@hdthor It'd be 60 degrees
Never knew this pattern existed, great explanation cheers
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
Or you can use generating functions
You should watch veritasium’s video about newton and pi. It’s better and shows how you can calculate things like cube roots using an infinite sum.
How is it obvious in hindsight? And it eoukd just be zeroes and 1s above the first row anyway right since itherwise it wouldnt sum ton2?
@@leif1075 negative numbers is what you have
Also pascal triangle * pascal triangle = cube
cube * pascal triangle = donut
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.
Is there a way to do it without trying to invert (1+x) so that it can grow upwards differently?
@@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."
Wow! I understood none of this lol.
I guess I have some reading to do
Next time (1+x)^sqrt(2) ☺
Great 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!
It is pascals triangle rotated 60 degrees with alternating forms of negative and positive numbers. So fascinating!
I am so happy that you've made this video, I looked this up a few years back and I'm happy that others can enjoy this concept!
Amazing video! At everyone in the comments who is asking why he didn't talk about using 1/2 + 1/2 = 1... he did!! In the end of the video the purpose of generalizing to a and b is to allow you to choose ANY values of a and b that sum to 1, which includes 1/2! So just pause the video at the end and replace a and b both with 1/2 and you have the triangle you're asking for. It's certainly the most aesthetically pleasing and temtping choice, but to everyone insisting that symmetry is the most important property of pascal's triangle that needs to be preserved, I urge you to learn more about Pascal's triangle@ The really neat thing is he showed in the end basically that while there are many other exciting patterns to explore, the only choices for (a,b) that lead to convergent series expansions are (1,0) and (0,1). It's pretty freaking cool. Of course I wanted the "canonical" answer to involve 1/2 as well, that's human instinct, but it's delightfully charming when math gives us an ego check and reminds us there are much more unexpected ways for things to be beautiful. Ultimately I find the general form with a and b to be the most beautiful. It shows the perfect symmetry in its own way and represents an infinite spectrum of pascals triangles, precisely two of which have rows that represent valid convergent series expansions, and one of which is perfectly symmetrical. HOW COULD YOU NOT LOVE IT!!! 😍
This is remarkably well explained. Thank you from just a guy online who is interested in some math
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.
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...
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.
Yes! Perfect way to spend this beautiful Friday morning, accompanying my cup of tea, toast, and bowl of porridge 😊
Tell me you’re from England without telling me you’re from England
@@BlokenArrowexactly xD
I am drinking coffee. We don't do porridge in AZ, as it is already 100 degrees at 6:00 am.
I'm also watching on a Friday morning during breakfast! XD
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.
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
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.
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 RUclips so amazing.
I read the title and think to myself,this is going to be boring.
Then I watch the video.
Wow man! Amazing !
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.
The rotated triangle thing was quite the plot twist!
Great video, I learned a lot through this! Another very interesting thing I see in this expansion is the fact that it simultaneously also displays the binomial expansion for the term (1 - x)^n, so for example
n=2: 1 - 2x + x²
n=3: 1 - 3x + 3x² + x³
n=4: 1 - 4x + 6x² - 4x³ + x⁴
and so on.
Truly interesting! Also you speak so naturally, clearly and fluently. I'm just terrified of people who write x as an inverted c and c put together.
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.
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.
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.
Incredibly clean presentation! I love your style.
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.
How does it do that? Can u elaborate about the -1/2
@@harshlalwani4353 Isn't this the Cesàro summation for the Grandi sequence?
@@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 what CelestinWIDMER said.
@@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.
Been a long time since my math undergrad, and this just reminded me why I fell in love with it. What a cool generalization
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 is insane! I was bored through a couple math classes, so my paper was filled with this exact idea a couple months ago!
What a coincidence! I thought of this like a week ago haha. Very nice explanation!!
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.
Woah math lore just went off the charts.
@voliol8070 Yes, though it should be specified, that is the odd numbers.
うわっすっごいな。目から鱗でした。
I didn't expect this video to be this interesting
Never had this question. But very interesting to see. Thanks a lot!
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.
I don't know why I'm here and why I watched it all, but thanks, i loved it
Never thought about what's above the Pascal's triangle! Cool!
This is absolutely beautiful. Thanks for showing this! ❤
Apparently, the Almighty Algorithm© has decided to finally show your videos to me!
The first one was your recent _Why Is This Almost An Integer_ which I found fascinating as it was talking about this mathematical quirk I'd never heard about.
This one, too, was fascinating. I think I have casually thought about these areas outside of the classical triangle but never followed up on it. It's astounding what turns up exploring Pascal's Triangle which, in the way we know it, is a rather simple, almost trivial, concept.
Now, I'm curious about your other videos!
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
I wish my math classes had been this fascinating!
I'm simply mindblown. Great video man!
This was very fascinating! Great video!
This is eye opening, I thought that Pascal's Triangle *MUST* be symmetrical
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
...
Watch the video, again! Especially the last few minutes where it exactly covers this case.
Feels like a and b could canonically have imaginary parts of +-sqrt(3)/2i if we're considering the imaginary plane.
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.
Very cool, thanks for sharing this!
Wow, that is extremely cool and absolutely not what I expected.
I can't believe this, it's amazing how math can always shock you with new things with just simple things
That's a very beautiful result.
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!
fantastically fascinating! a mathematical masterpiece!
Handsome guy with interesting smart content! Subbed! 😂
Question: can we fill in-between the lines too? As in, can we think of a natural way to fill, say, the 1/2-th line?
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.
Also when you look at the row as a number, it writes out to be 11^n. For example, the 2nd row (1 2 1) is 11^2 which is 121
Awesome! Kp it up with these cool math videos!❤❤❤
superb! Have to reurn to absorb the points made during the last section (regarding a and b). Many thanks,
Beautiful idea! Thanks for sharing!!
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.
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 always wondered this and it's a super interesting result
Amazing! Thanks for the wonderful explanation.
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.
Blaise in Maths Heaven is smiling seeing this presentation!
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
Very amazing thank you very much Dr Barker 😊🎉
literally gasped at the rotated triangle reveal
Very cool stuff on friday 🙂
Wow, I had no idea. Truly a new perspective.
i remember when i was like 10 and super bored i extended it this exact way (without the binomial expansion obviously, i just did it by accident)
i thought of it like a rotated, half negated version of it. i also thought it was kinda like being zoomed infinitely into the Sierpinski triangle (i liked fractals back then)
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|
Well he used for the rotated Pascal triangle the binomial expansion of (1+x)^(-n) = sum from k = 0 to infinity of (-1)^k *((n+k-1) choose k)*x^(k) for |x|
the real question is "why"
because it's cool
Really interesting and amenelly told, thanks!
The coefficients a and b can be chosen to make the two series converge around an arbitrary point, making this a recipe for a laurent series expansion of the expression
Very cool!
I wonder about deferent triangles one might make. I discovered in 5th grade that the difference of the power series for the N-th power eventually yields N! after N iterations.
The case of |x| < 1 and |x| > 1 are both handled here, but what about |x| = 1?
Excellent! Who originally discovered this?
There are a few ways you could expand it based on a rule. Maybe you are interested in the fibonacci sequence expansion going backwards too. Also valid, since that sequence extends in both directions.
is that pascal's triangle but for (a-b)² ?
I never knew what pascals triangle was! Now i know more than i am supposed to know about it 😅
Just noticed also that the sum of elements in a row of Pascal's triangle is the corresponding power of 2. The row at n = -1, you've established should be defined as alternating 1 and -1, and the power of 2 would be 1/2. The agrees with the value of an L-function under analytic continuation that would correspond to 1-1+1-1+... "=" 1/2
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!
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?
this is actually very interesting
This is way above my level of understand of maths but I enjoyed it, still.
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?
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
Hello! Can you define Pascal's triangle as the convergence of the a's, and b's? What other triangles are out there that are also the result of certain a's, and b's convergence other than the Pascal's triangle?
Wow I searched for pascarl's trinagle trivia in and out and this is one of the coolest. I was wondering though what if we apply it to other type of binomial functions like: a(1+1/x)^-1+b(1+x)^-1 which would satisfy the property or something weirder maybe.