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shouldnt (1+x)^-1 = 1/(1+x)?

Out of bounds nonsense probability distributions. Glitches in quantum mechanics

Some 20 Years ago I played around with this Idea. You do not have to be a Mathematician to have Fun with it.

Sry but what a weird way of writing “x”?! 😅

@blackpenredpen - I’ve also expanded the standard triangle into higher dimensions. The problem I’m having is understanding Python enough to generalize the multinomial distribution.

I never knew what pascals triangle was! Now i know more than i am supposed to know about it 😅

Why would we continue the constant of 1's for the B but not the same constant of 1's for the A? Not seeing the connection.

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.

if u can move the ruler u only need the 0 and 1 cm marks to measure any integer distance

made me think that if the ancient mathematicians had a CZcams channel & modern dress, it wouldn't look too different from this

So I see some people choosing a=b=½, but you can have even more fun with any conjugate complex numbers with real part ½. Interestingly, the negative 3rd roots of unity have this property.

could we apply these principals to the complex plane? ie could a be (0.5 + 0.5i) and b could be (0.5-0.5i)? what would we see if we iterated on these values? 😊

This is only for (1+x)^n but what about (x+y)^n = 1x² + 2xy + 1y² (if n=2)

So in fact Pascal triangle is a Pascal half-plane.

Incredibly clean presentation! I love your style.

Awesome! Kp it up with these cool math videos!❤❤❤

Wow, I had no idea. Truly a new perspective.

I can't believe this, it's amazing how math can always shock you with new things with just simple things

1) Pascal triangle is about INTEGERS, more specifically- NATURAL NUMBERS. 2) What sense, say- geometrical sense it makes to extend it upwards?

My guess is, it's completely arbitrary

But what's above the rotated triangle now?

The existence of 3π/2 expanding Pascal's triangle and π/4 expanding Pascal's triangle implies the existence of 3π/4 expanding Pascal's triangle

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This is so fucking cool

square root of -0. I thought everyone knew this

As Below, So Above

lascaP triangle ?

Pascal would be rotating in his grave.😊

Care is needed here. There must be more to be un-earthed though.

Function keys. What they do depends on your operating system.

.. it's a reflection about the 0 wedge.....

Handsome guy with interesting smart content! Subbed! 😂

This is really cool and all, but am I the only one distracted by the fact that when he writes an x on the board, because of the way he draws it it always looks like a cartoon butt with the cheeks facing each other? 😂

Wow, that is extremely cool and absolutely not what I expected.

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.

Is there a pretty way to define a and b to make it so when abs(x) > 1 then a would be 0 and b would be 1 and vice versa as in to fix the possible values of a and b depending only on x? Any of them can be defined like a = 1 - b (swapping them works just as well, in this case if we restrict b to be 1 or 0, it would work). But how would we calculate b like that using just the value of x? Probably it would be using limits to make sure things go to 0

I always wondered this and it's a super interesting result

👏

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. 🤓

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?

Very cool, thanks for sharing this!

Pascal's Trefoil

Did I miss something or did your first proof only works for X = 1?

I've investigated this myself, but it didn't occur to me that there could be infinitely many solutions. There is, however, as far as I can tell, only one symmetrical solution: −1½ ½ 0 0 ½ −1½ −1 ½ 0 ½ −1 ½ −½ ½ ½ −½ ½ 0 0 0 1 0 0 0 0 0 0 1 1 0 0 0 0 0 0 1 2 1 0 0 0 I have a related question. Have you ever come across variations of Pascal's triangle, such as one with 0, 1, 2, 0 in a row? 0 1 2 0 0 1 3 2 0 0 1 4 5 2 0 0 1 5 9 7 2 0 0 1 6 14 16 9 2 0 0 1 7 20 30 25 11 2 0 This particular variation has some fine properties, such as encompassing the Fibonacci numbers (descending shallow diagonals), the Lucas numbers (ascending shallow diagonals), the natural numbers, the odd numbers, the natural squares, the sums of the natural squares, and a host of other sequences. Perhaps more significantly, though, the individual elements of the shallow ascending diagonals provide the coefficients for polynomials that define L[2n], L[3n], L[4n], etc., in terms of L[n], the nth Lucas number. (Please excuse the formatting. Subscripts are fiddly and almost illegible in YT comments.) I'll give you a couple of illustrative examples: For even n, L[5n] = (L[n])^5 − 5(L[n])^3 + 5(L[n]). E.g., L[2] = 3; L[10] = 3^5 − 5 • 3^3 + 5 • 3 = 243 − 135 + 15 = 123. For odd n, change the minus to a plus. E.g.; L[3] = 4; L[15] = 4^5 + 5 • 4^3 + 5 • 4 = 1024 + 320 + 15 = 1364. Pascal's triangle can do similar things for Fibonacci numbers, but the polynomials are in terms of Lucas numbers, L[n], with all coefficients multiplied by F[n]. I won't bore you with the details.

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.

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

My mouth dropped when the triangle showed up again but rotated 😮

Fascinating! 😮

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.

From any number, the two numbers above are exactly half the original number. So what is above the 1? 0.5 and 0.5 What's above the 0 (next to the starting 1)? 0.5 and -0.5