Power sums with Calculus.

[@ianfowler9340]

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.

[@samueldeandrade8535]

"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". This statement actually makes no sense.

[@ianfowler9340]

@@samueldeandrade8535 I was trying to say that you can start with a set of true and well known results, apply the mathematics correctly (sometimes blindly), and be lead to an unexpected or new result or a known result but in an unexpected and seemingly unrelated way.

[@samueldeandrade8535]

@@ianfowler9340 well, ok. But what you said was too much. You made an invalid generalization. Your talking is very imprecise. Look, "apply the mathematics ... (sometimes blindly) and be lead to an unexpected or new result". What evidence you have for that case? And what, more precisely, do you mean by "applying math blindly"? Man, be mathematical. Mean a lot by saying little. You are doing kinda the opposite.

[@letis2madeo995]

​@@samueldeandrade8535the video is an example of that.... we don't know where we're going with the derivativesof the closed expression. That statement is definitely correct.

[@samueldeandrade8535]

@@letis2madeo995 man, let the guy talk. You are just adding more wrong statements. "the video is an example of that...." Of "that" what? "We don't know where we're going with the derivatives of the closed expression"? Wtf do you mean? What you want to know? What do you think you should know? We defined a function, we realised its derivatives at 0 are the sums of power, we calculate its derivatives by some other expression, we obtain a formula that must equal the other one. What else do you need to know? "That statement is definitely correct". What statement? You are not mathematical too.

[@mustafaunal1834]

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!

[@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?

[@johannmeier6707]

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?

[@bjornfeuerbacher5514]

No, he compared the coefficients of x^0 with the ones from x^0 and so on. But nevertheless, you have a point. I think the formula given at 13:40 is slightly wrong: The sum probably should start at k = 1, not at k = 0. Because otherwise that sum would start with 1/x, i. e. it would diverge for x going to zero, which obviously contradicts fn(0) = n.

[@holyshit922]

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

[@marouaniAymen]

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

[@shrayanpramanik8985]

Cute observation!

[@assilsiahmed6905]

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 ?!

[@tuanvuhoang4853]

it is fantastic!

[@goodplacetostop2973]

17:18

[@thes7274473]

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.

[@leif1075]

Is x an integer also or any real number?

[@leif1075]

And "polynomial over" indeter.inate x is such bizarre unclear language wouldn't yiu agree..why don't mathematicians use more intelligible terminology?

[@GreenMeansGOF]

I assume B_0(n)=0?

[@dodgsonlluis]

No, it isn't, it's 1. I am confused, the expressions doesn't equate at all. I think that, simply, the term for k = -1 have not to appear.

[@bjornfeuerbacher5514]

Actually, it's B_0(n) = 1, as one could look up, or calculate from the formula given at the end. I think the sum given at 13:40 is slightly wrong, probably it should start at k = 1, not at k = 0.

[@krisbrandenberger544]

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

[@VinodKumar-rh7ds]

Fantastic but i had already lerant it by Bernoulli thearom

[@robshaw2639]

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) + . just consider the odd/even cases for a direct proof.

[@idjles]

Hallo Gauss? Nice to see you again after 250 years!

[@samueldeandrade8535]

Impressive. You manage do say the obvious, avoiding the simplest way to do it. Man, bring this guy a Fields Medal.

[@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.

[@farfa2937]

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

[@leif1075]

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??

[@AnisYousuf-w8u]

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