What Lies Above Pascal's Triangle?

We explore how to extend Pascal's triangle upwards.

Binomial series expansion proof:

00:00 First considerations
01:44 Binomial coefficients
04:35 Extending further upwards
08:38 Binomial theorem
09:40 Proving the pattern continues
15:07 Non-integers
15:48 Alternative series expansion
20:45 Even more values of a & b

[jo]

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.

[slowfreq]

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

[jackkalver]

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

[jbullforg]

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

[jacksonschuetzle]

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

[worldnotworld]

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.

[misraimburgos]

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

[GroundThing]

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

[Aras]

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

[khan]

Never knew this pattern existed, great explanation cheers

[dusaprukiyathan]

"What lies above Pascal's Triangle? Pascal's Triangle."

[alipourzand]

Next time (1+x)^sqrt(2) ☺
Great video

[qiaochuyuan]

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

[sabotagedgamerz]

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!

[BarryRowe-ghyq]

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.

[worldnotworld]

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...

[georgesos]

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

[edmundwoolliams]

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

[oxbmaths]

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?

[AidenOcelot]

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!