What Lies Above Pascal's Triangle?

Wow this is super interesting!

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.

Mind blown when he showed the original triangle rotated at the top

This is so insane how is this the first time I've ever seen this idea?

Everywhere you look up new thing in Math it seems like beauty always emerge.

I was thinking 1/2 above 1 and alternating -1/2 and 1/2 on either side, since the triangle seemed symmetric.

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.

Never knew this pattern existed, great explanation cheers

my smooth brain wanted a and b to both be equal to 1/2 or 0.5

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!

Yes! Perfect way to spend this beautiful Friday morning, accompanying my cup of tea, toast, and bowl of porridge 😊

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

Next time (1+x)^sqrt(2) ☺ Great video

I once did the expansion while bored in class. It's nice now learning of proofs and further expansions, thanks!

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 read the title and think to myself,this is going to be boring. Then I watch the video. Wow man! Amazing !

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

This is actually more interesting than I initially thought

This is remarkably well explained. Thank you from just a guy online who is interested in some math

This is absolutely beautiful. Thanks for showing this! ❤