@@whataboutwhy2025 I thought it would help you and it did yay, i don't want a channel like this to end up, you have potential and you're great explainer. I hope you'll continue this 🙂👍🏻
WOW! I had no idea Pascal's Triangle was like the davinci code, with all of these hidden things inside of it! My favorite was with the people explanation, like such a great way to figure out combinations!
Pascal's triangle can also be used to make conclusions about higher dimensional geometry. While in 3D, there are obviously 3 different axes that something can rotate around, but in 2D, there's only 1, rather than 2. Now look at the 3rd diagonal, the one with the triangular numbers. 0 (because rotations don't exist in 1D), 1, 3. This is also the diagonal with n choose 2. So any rotation, regardless of dimension can be thought of as going from one axis to another. How many axes of rotation would you have in 4D? It's not 4, but rather 4 choose 2 = 6, the next entry on that diagonal. And this pattern extends to any number of dimensions! The other diagonals also have other meanings in the context of higher dimensional geometry.
Love this. I stumbled on your videos looking for an explanation of the Pythagorean Theorem that I had seen previously from someone else. I also love the channel name, I'm surprised you don't have more subscribers. Keep at it!
This is a fine summary of most of the best-known features of Pascal's Triangle. I can offer a little supplement, but it involves a variation of Pascal's triangle, beginning 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 0 1 8 27 50 55 36 13 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 (i.e., the square pyramidal numbers), 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]. Here are a couple of examples: For even n, F[5n] = F[n] • ((L[n])^4 − 3L[n])^2 + 1). E.g., F[2] = 1; L[2] = 3; F[10] = 1 • (3^4 − 3 • 3^2 + 1) = 81 − 27 + 1) = 55. For odd n, change the minus to a plus. E.g., F[3] = 2; L[3] = 4; F[15] = 2 • (4^4 + 3 • 4^2 + 1) = 2 • (256 + 48 + 1) = 2 • 305 = 610. This enables a vastly accelerated race through the Lucas and Fibonacci numbers, leading to a quickly converging approximation to the golden ratio, as (L[n] / F[n] + 1) / 2 tends to ϕ as n increases. Already, (134 / 610 + 1) / 2 = 1.6180327868852, which is accurate to five decimal places. Technically, it would be possible to produce polynomials based on L[1] and F[1], thus reducing them to simple additions with no need to calculate powers. E.g.: L[5] = L[1] + 5 • L[1] + 5 • L[1] = 1 + 5 + 5 = 11. F[5] = 1 • (L[1] + 3 • L[1] + 1) = 1 + 3 + 1 = 5. Admittedly, these examples are too simple to show the true power of this technique. Here are L[15] and F[15] expressed as sums of coefficients: L[15] = 1 + 15 + 90 + 275 + 450 + 378 + 140 + 15 = 1364. F[15] = 1 + 13 + 66 + 165 + 210 + 126 + 28 + 1 = 610.
Pascal's triangle is not an addition triangle, but a convolution triangle. If you see each new row as the convolution between the latter row with the sequence {1,1} everything shown here is trivial. Another way to trivialize all results is to see that each diagonal is the cumulative sum of the previous diagonal.
Why your channel so underrated.....dont worry....my maths teacher says that the best content are always underrated and it should be...because the one who are curious about maths and all the mathematics terms and stuff will surely find ur videos and I am glad to see ur channels over very viewed channels out there.....❤️❤️😁keep making such interesting videos... Can you please make videos on Ramanujan's number and his theory...cause I am very keen to learn how his theory works...
I am your 890th subscriber. If you know magnificence of this number, please do make a video about that too :) This number is where several fundamental mathematical constants intersect
I remember building Pascals triangle without even knwoing what it was when I was a child. If I think what I have come close to before I withdrew from it I regret it sometimes.
@@whataboutwhy2025 Thanks! is that how you did the cool graphics? I use to teach at a community college and I am just now getting into making youtube videos I would love to learn how you did the animations!!!
Lets try the tetranomial (a+b+c+d)^n. Look in the diagonal and i bet you can guess what is forming. It is very apparent by the time you get to the Octanomial. Coefficients of the binomial 11^n. Tetranomial (11^n)*(101^n). 101^n itself being self similar to 11^n. try(73^n)(137^n). Octanomial coefficients: (11^n)(101^n) (73^n)(137^n) . 1 1 1 1 1 1 2 3 4 3 2 1 1 3 6 10 12 12 10 6 3 1 1 4 10 20 31 40 44 40 31 20 10 41 1 101 10201 1030301 104,060,401
Just wanna say don't stop your channel has quality that's what needed it's just a matter of time your channel blows up!!
Thank you so much, this means a lot!
@@whataboutwhy2025 I thought it would help you and it did yay, i don't want a channel like this to end up, you have potential and you're great explainer. I hope you'll continue this 🙂👍🏻
WOW! I had no idea Pascal's Triangle was like the davinci code, with all of these hidden things inside of it! My favorite was with the people explanation, like such a great way to figure out combinations!
Thank you so much!! Pascal's Triangle is one of my favorite math ideas because there's something for everyone! So glad you enjoyed :)
Pascal's triangle can also be used to make conclusions about higher dimensional geometry. While in 3D, there are obviously 3 different axes that something can rotate around, but in 2D, there's only 1, rather than 2. Now look at the 3rd diagonal, the one with the triangular numbers. 0 (because rotations don't exist in 1D), 1, 3. This is also the diagonal with n choose 2. So any rotation, regardless of dimension can be thought of as going from one axis to another. How many axes of rotation would you have in 4D? It's not 4, but rather 4 choose 2 = 6, the next entry on that diagonal. And this pattern extends to any number of dimensions! The other diagonals also have other meanings in the context of higher dimensional geometry.
Your videos are really underated.
I appreciate your support! :)
Love this. I stumbled on your videos looking for an explanation of the Pythagorean Theorem that I had seen previously from someone else. I also love the channel name, I'm surprised you don't have more subscribers. Keep at it!
This is a fine summary of most of the best-known features of Pascal's Triangle.
I can offer a little supplement, but it involves a variation of Pascal's triangle, beginning 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
0 1 8 27 50 55 36 13 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 (i.e., the square pyramidal numbers), 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]. Here are a couple of examples:
For even n, F[5n] = F[n] • ((L[n])^4 − 3L[n])^2 + 1).
E.g., F[2] = 1; L[2] = 3; F[10] = 1 • (3^4 − 3 • 3^2 + 1) = 81 − 27 + 1) = 55.
For odd n, change the minus to a plus.
E.g., F[3] = 2; L[3] = 4; F[15] = 2 • (4^4 + 3 • 4^2 + 1) = 2 • (256 + 48 + 1) = 2 • 305 = 610.
This enables a vastly accelerated race through the Lucas and Fibonacci numbers, leading to a quickly converging approximation to the golden ratio, as (L[n] / F[n] + 1) / 2 tends to ϕ as n increases.
Already, (134 / 610 + 1) / 2 = 1.6180327868852, which is accurate to five decimal places.
Technically, it would be possible to produce polynomials based on L[1] and F[1], thus reducing them to simple additions with no need to calculate powers.
E.g.:
L[5] = L[1] + 5 • L[1] + 5 • L[1] = 1 + 5 + 5 = 11.
F[5] = 1 • (L[1] + 3 • L[1] + 1) = 1 + 3 + 1 = 5.
Admittedly, these examples are too simple to show the true power of this technique. Here are L[15] and F[15] expressed as sums of coefficients:
L[15] = 1 + 15 + 90 + 275 + 450 + 378 + 140 + 15 = 1364.
F[15] = 1 + 13 + 66 + 165 + 210 + 126 + 28 + 1 = 610.
Awesome! Looking forward for the second part!
Pascal's triangle is not an addition triangle, but a convolution triangle. If you see each new row as the convolution between the latter row with the sequence {1,1} everything shown here is trivial.
Another way to trivialize all results is to see that each diagonal is the cumulative sum of the previous diagonal.
Why your channel so underrated.....dont worry....my maths teacher says that the best content are always underrated and it should be...because the one who are curious about maths and all the mathematics terms and stuff will surely find ur videos and I am glad to see ur channels over very viewed channels out there.....❤️❤️😁keep making such interesting videos...
Can you please make videos on Ramanujan's number and his theory...cause I am very keen to learn how his theory works...
I am your 890th subscriber. If you know magnificence of this number, please do make a video about that too :) This number is where several fundamental mathematical constants intersect
I remember building Pascals triangle without even knwoing what it was when I was a child. If I think what I have come close to before I withdrew from it I regret it sometimes.
This is very well done! what tool did you use to make this video if you dont mind me asking? And thanks for doing this!!!
So glad you enjoyed! :) I used Davinci Resolve to make this video
@@whataboutwhy2025 Thanks! is that how you did the cool graphics? I use to teach at a community college and I am just now getting into making youtube videos I would love to learn how you did the animations!!!
Well done.
Of course: genius profe! Thank you
so helpful
Something I noticed while watching this video is that going straight down the middle of the triangle reveals a pattern of factorials.
Excellent vid, please do part (1+1)=2
... without the music. I just came from another video on that channel. The music there made me leave the video. Same here.
I'm so sorry--I'm fixing it in future videos! I just can't do much now to change it after it's uploaded
There are also Chebyshev polynomials in Pascal's triangle.
Lets try the tetranomial (a+b+c+d)^n. Look in the diagonal and i bet you can guess what is forming. It is very apparent by the time you get to the Octanomial. Coefficients of the binomial 11^n. Tetranomial (11^n)*(101^n). 101^n itself being self similar to 11^n. try(73^n)(137^n). Octanomial coefficients: (11^n)(101^n) (73^n)(137^n) .
1
1 1 1 1
1 2 3 4 3 2 1
1 3 6 10 12 12 10 6 3 1
1 4 10 20 31 40 44 40 31 20 10 41
1
101
10201
1030301
104,060,401