Power sums with Calculus.

🌟Support the channel🌟

🌟my other channels🌟

🌟My Links🌟

🌟How I make Thumbnails🌟

🌟Suggest a problem🌟

[ianfowler]

This the first time I have seen such a development. Bloody marvelous! It's hard to express in words, but one can only be struck by a sense of "wonder" or "awe" that when you "follow the math" correctly, it often leads you a known result from a seemingly unrelated and oblique direction. But in the end, what this really demonstrates is that mathematics (calculus included) is all connected in such a basic fundamental way so that final result naturally emerges from the chaos. Very cool.

It's the same sense of "awe" I got when I applied 1) Newton's Second Law 2) The Universal Law of Gravitation 3) Radial Force and 4) Calculus and out pops (almost magically) Conservation of Angular Momentum and Energy, Kepler's 2nd Law and finally the orbital path being a conic section. Thanks Michael.

[mustafaunal]

WOW!! Thank you so much! I am sooo happy now! 😄😄

[Noam_.Menashe]

If instead of a sum of exponentials, we use an alternating sum of inverse exponentials (whose geometric sum converges then) we could evaluate the Dirichlet Eta function in negative integers as derivatives of it, and we could then get the zeta function at negative integers using the formulas.


Also, does this make it easier to get the general formula? It seems like you can get the Bernoulli numbers off it but I can't just find out how.

[Kapomafioso]

Huh, this is a really neat trick I never seen or thought of!

[holyshit]

These sums can be calculated with generating function but we have to express n^k in Newton's divided difference polynomial form

[channalbert]

I need to ask this question, otherwise I'll implode. I cannot recall where I saw this. I cannot explain why I find it uncomfortable. However, it just works. It works amazingly so.

What I'm talking is a way yo treat the Bernoulli numbers as if they were some sort of generalized element, ฿ (for lack of a better term), which can be exponenciated to a power and have its correspondence with the Bernoulli numbers: ฿^n := B_n. [This is where I start to dislike the lack of formality. What does it *mean* for an abstract element that to the k-th power has a different form everytime?]

You can easily find the Bernoulli terms with the pseudo-equation ฿=฿-1, and its powers:
— ฿^0 = B_0 = 1
— ฿^1 = ฿^1 -1 → ฿^1 = B_1 = 1/2,
— ฿^3 = (฿-1)^3 → ฿^2 = B_2 = 1/6
and so on (considering all B^(2k+1) for k>0 are zero). *Why* does it work? Is there some truth behind this seemingly childish way of looking at the element ฿ as more than just a dummy variable that when exponenciated gives you the k-th value of the Bernoulli numbers?

Here's the kicker: the expression at the end of the video is elegant and there are so many important numbers and famous coefficients (binomial, bernoulli). But how (*how?*) does it scream so much to turn the end result into a form like 1^{k+1} + 2^{k+1} + ... + n^{k+1} = (฿ + n)^k/k? *It makes no sense that it makes sense.*

Can somebody more knowledgeable in this area tell me? Or maybe Michael would be up for exploring what truth is behind this seemingly child's play, hokus-pukus piece of mathematics yet so consistently effective?

[johannmeier]

Why could you compare the coefficients of x^k in the sums in the end? After re-indexing, both sums had x^k, but they still started at different points (0 and -1). What did the reindexing get you, then? Without reindexing you could have done the same. You compared the coefficients of "x^0" with the ones from "x^-1" (and so on). Why was this allowed?

[marouaniAymen]

I really loved this video, but where the Bernoulli Numbers came from ? is there any simple explanation for the idea and derivation behind ?

[assilsiahmed]

Is it true to say that the sum of (e^x)^n is the same as the sum of a geometric sequence because here e^x isn't a constant ?!

[thes]

Here's an interesting summation problem I've been working on for a while: Given a positive integer p, what is the polynomial over indeterminate x given by the sum of k^p from k=1 to k=x? Is there a general formula for the coefficients of that polynomial?
I've found some patterns. Let Q(p, i) be defined such that the sum of k^p from k=1 to k=x is equal to the sum of Q(p, i)x^i from i=0 to i=x+1. I've found that Q(p, i)=1/2 for p=i, that Q(p, i)=1/i for p+1=i, and a few other patterns, but I haven't found a formula.

[VinodKumar-rhds]

Fantastic but i had already lerant it by Bernoulli thearom

[krisbrandenberger]

@ 14:35 Should be the kth derivative of f_n evaluated at 0.

[robshaw]

the triangular number formula doesnt really need induction - just reorder the sum 1 + n + 2 + (n-1) + 3 + (n-2) + .... all the pairs add to n+1, so there's floor(n/2) * (n+1) + <the middle term, (n+1)/2 if n is odd>. just consider the odd/even cases for a direct proof.

[Jack_Callcott_AU]

Gee Mike .. Thank you for a brilliant video. I watched a Mathologer video about power sums, but yours is better. I will watch this video a few times because the result is so nicely developed. I'd like to see more about Bernoulli numbers and Bernoulli polynomials.

[leif]

Desr.GOD didn't everyone else think he would just take the derivative of the expression and get k(1)^k-1) plus..k(n)^k-1) tomsolve this..WHY IS E THE NATURAL NHMBER HERE AT ALL..? Not clear.. Surely everyone else was wondering this??

[farfa]

Wouldn't you want a closed form to be easier to calculate than the thing you started with? It kinda looks like we started from the closed form and got to something much more complex lol

[AnisYousuf-wu]

Enkuwan calculus tewegnna medemer meqenesum azaye new wedaje m8naqejaberegn midre tequway ytenquwy follower hula atMetam mariamullahudan