Відео

As someone who has a degree in data science, done far too many modules that involve complex multivariable calculus and i have never in my life understood e so much. So many insane integrations and distributions now make sense, why dont they just start with this?

Well done! The visuals were crisp and the explanation was very clear. *SUBSCRIBED*

The fact that x⁴ + y⁴ + z⁴ is agonizingly close to 4 but not 4 is killing me

I'm so happy that you're back :)

Thank you for such an amazing video!!

I have a better and easiest solution , anyone want?

=(1+2+3+4+5+....+100)² =5050² =(5000+50)² =5000²+2(5000)(50)+50² =(50•100)²+10000(50)+2500 =2500•10000+500000+2500 =25000000+500000+2500 =25002500+500000 =25502500

one week ago watched the 1st 20s of the video so only the first problem and found he answer and didn't check the answer, then went into a rabbit hole exploring symetric ploynomial expressions, what coefficient resulting product of any two basic elementary expressions, and wanted to find out if you always can find solution of sums of powers if you have enough symetrix expressions and if you can determine if you what you have, and got stuck hinking ithas to do with rings complicated combinatorics and partition numbers relted topics. i came back today and had a lot adun watching this, from a different solution to the first problem , to the variabkes being roots of single polynomial, to them being written in terms of elementary forms, similar to bases on vector spaxes, and a reduction strategy similar to the shaun substitution lemma for base conversion in vector spaces. also the quality and pacing and the right amount of examples make this even better. keep up the goodwork. i might try to code the algorithm at the end

Nice vid , understood it completely

Wym?

Use vieta relation and voilá get the answer

this is outrageously overcomplicated. Simply note Sn = x^n + y^n +z^n; P1 = x+y+z; P2 = xy + yz + xz; P3 = xyz. We know S1 S2 S3, lets go for S4! We have P1 = S1 = 1; P2 = (1/2)(S1²-S2) = -1/2; P3 = (1/6)(S1^3 + 2S3 - 3S1S2) = 1/6. Then let P(X) = (X-x)(X-y)(X-z) = X^3 - P1X^2 + P2X - P3. We have P(x) = P(y) = P(z)=0. with x : x^3 = P1x^2-P2x+P3. Multiply by x^n : x^(3+n) = P1x^(2+n) - P2x^(1+n) + P3x^n. Sum with y and z same way, you get the recurrence S(n+3) = S(n+2) + (1/2)S(n+1) +(1/6) Sn. Apply n = 1 : S4 = 3+1+1/6 = 25/6

Congratulations on getting honorable mention this year! I hope you keep participating in SoME because your videos are great 🫶

2:13 THIS IS RIDICULOUS.

This may not be the point of the video, but I really want to track along with all of the algebra to simplify the initial systems and I am absolutely lost at the 2:30 time-stamp. How does +/- (1-z)*sqrt(-3z^2+2z-3) simplify to 0?!? What am I missing in that step?

complex numbers just killed me

Allen Donald Lewis Ruth Smith Michelle

Johnson Angela Lewis Patricia Lewis Mary

Oh, I watch you a year ago)

The mistake is that 1/2= 0 not pi, lol it's like basic math.

I enjoyed your videos. But, I don't see the benefit I would have by subscribing, with only 1 video published a year.

Those sound effects are divine

dude I did not expect to just stumble into some cool math I've never heard of. are there any books recommended for going deeper into this?

also mildly interesting is that the zeroth diagonal can be used to calculate sum of numbers to zeroth power. don't know why you would want that thou

n²(n+1)²/4

So, his true love gave him 364 gifts for Christmas?

Hi! Nice video! Refreshing to listen to something original using only elementary math! I have been playing around with symmetrical variables my self. I came up with my own notation called symmetric exponentiation and it’s exactly like the s-terms. So when ever I have indices for variables in any problem where the indices are interchangeable without changing the system (thus the variables are symmetric) I used the notation s_n=x^(n) instead where x is the variable and using () to denote the symmetry just like one does in the Einstein notation sometimes. For instance, the problem of find the inscribed circle has the solution: r^2=x^(3)/x^(1) Have a very neat formula regarding inscribed and subscribed circles for polygons that would be fun to discuss with you!

......... it is, at time of writing, precisely 6 minutes to 3am. maybe i should sleep

Compound interest turns x2 into xe

okay fishboy

I love the sounds the numbers make when they are placed! It sounds like a game.

Basically, computing the summation of a pascal diagonal type polynomial is really easy. In a linear algebra sense, if you use them as a basis for polynomials, summation amounts to shifting your vector. EG the polynomial 1+n+n(n+1)/2 can be represented as (1,1,1,0,,,0), the summation is just (0,1,1,1,0000)

The sum of integer powers can be written down in a closed form, found it in hs. It’s a double sum however. Sum from k=0 to a of (N+1 choose k) sum from j=0 to k of (-1)^(j*k)(k choose j)*k^a. a is the integer power.

Fun little math fact: The sum from 1 to n plus the sum from 1 to n+1 will always give a perfect square.

This video is nicely done. I have made a video about this approach last year: ua-cam.com/video/k-9YCo_C4J4/v-deo.html, using the so-called Hockey stick identity in Pascal's triangle. The coefficients how they are combined in general is not super nice. Related to Stirling numbers if someone is interested.

6:00 y'all ate fr

I always look forward to your SoME videos; can't believe it's already been a year since the last one!

Absolutely love the video! It's clear that a lot of work went into it, as the animations are crisp and clean, while the presentation is both clear and enjoyable. The approach that you have outlined for finding sums of powers is very cool, especially because it is not one of the most conventional/standard methods. However, a natural question arises: what are the constants c_1, c_2, c_3, ... etc? Sure, they can be computed by brute force, but is there a deeper pattern or significance to them of which you are aware? Also, is there a combinatorial interpretation of this method beyond mere algebraic equivalence? These seem to be missing pieces to an otherwise compelling story. But as a youtube video on this topic, it is definitely miles beyond most of them out there, so I look forward to your next one. Thanks!

What is the upper number for the Ck in the general formula?

Nice singing!

2:17 I never knew an induction proof would be explained so sneakily without making it feel like an induction proof

That was incredible! The animations were soooo clean and well thought through. Such a great explanation and y’all made the alternating talking parts feel very natural which I’m sure is not easy to do. Hope to see more videos!

it's a small detail, but I appreciate that your rendering of Pascal's triangle uses hexagons. it makes it a lot easier to view diagonal paths than when the square visualization is used.

I couldt understand this metod, but now i can

But how can i find a general form for the constants?

Great video! Pascal's triangle is my favorite mathematical object, due to all the little secrets it holds, like this one. And the presentation was top-notch. Really like how you took the time to explain the relation between n choose k and the triangle; even if it's apparent the triangle is combinatoric-y (you combine the upper terms to get the lower), the relation between the formula and the triangle is often pulled through thin air. If I may have one criticism though, it is the video felt a little open-ended. Or rather, I was fully expecting there to be a *second* trick coming up, to somehow extract those coefficients directly from the triangle. Like, surely there must be a more elegant way of getting those? And then the video just ended.

ok fishboy

This was one of the first indie math videos I have ever encountered, looking at it again, it's so nicely explained!

epic as usual

I love how on the outro you talk about e being natural and all I could think of is "e is about as natural as you two are in front of the camera" 😂 Sorry, great video. I did spot some problems with it (as someone who entirely understands the topic at hand), it could have been slightly improved, but it was almost perfect at what it wanted to and needed to achieve, so it's great!