beautiful how these sums always converge to such nice values
@samharper5881
4 жыл бұрын
I figured out a few of these identities a while ago just by playing around with Mathematica; it's been cool to see them shown in these videos lately.
@emanuellandeholm5657
4 жыл бұрын
This identity, and its proof, was most satisfying to me.Thanks for this exercise, Michael Penn! 27:00 is the exact moment my jaw dropped.
@sumukhhegde6677
4 жыл бұрын
Return of "that's a good place to stop"
@peterbrough2461
4 жыл бұрын
OK. Good.
@jayantimajumder4151
2 жыл бұрын
In first tool, the occured integral is the Laplace transform of x^(2n-1) in terms of (m+1).. So, we get:(2n-1)factorial/(m+1)^2n.The factorials will be cancel out and the rest is infinite sum from m=1to inf.1/m^(2n) which is Riemann zetaof (2n)..
@joshuagreenberg1301
4 жыл бұрын
When doing the repeated integration by parts to establish the first of the "tools", I'm confused. He says we end up doing it 2n times, but why isn't it 2n-1 times? It seems like 2n-1 iterations is enough to get the power of x down to zero.
@leanansidhe6332
4 жыл бұрын
I think he just said it wrong.
@thomashoffmann8857
4 жыл бұрын
You need to do 2n-1 times the partial integration. Another (m+1) comes from integration of the exponential function. Together you will get 2n factors of (m+1). The explanation was wrong but the calculation was right in the end :)
@sergiokorochinsky49
4 жыл бұрын
@@thomashoffmann8857, good point, I didn't see that, but you are right.
@alainbrizard4719
4 жыл бұрын
At 4:12, introduce the transformation t = (m+1) x and use properties of the Gamma function. In one line, you get zeta(2n). No need for integration by parts.
@pharaohgarmar5611
4 жыл бұрын
Brilliant. Two suggestions: (1) explain why other routes may not work or are more difficult (2) have a numbering system for your videos so that when you refer back, it’s easier to find.
@riadsouissi
4 жыл бұрын
Another way was to replace 1/2n+1 by integral x^2n from 0 to 1 and 1/n+1 by integral of y^n. After expanding the sum (over k) of the zeta function, and swapping the sum with the double integral, I end up with a sum of a geometric series, then with some substitution, I get: Sum(-k * integral from 0 to 1/k of (log(1-t^2)/t^2 + 1) dt) Developing the integral, I get: Sum(k(k-1)log(k-1)+k(k+1)log(k+1)-2k^2log(k)-1). Notice the -1 at the end. This series has some nice telescoping factors (across 3 terms) so I end up with terms left over at infinity and sum of (-1). I get: Lim k to infinity of (k(k+1)log(1+1/k) -k). Using Taylor expansion of the logarithm to the second degree, the limit gives me exactly 1/2. By the way, using similar method, I can also calculate sum(zeta(2n)/(2n+1)(n)) = log(2*pi)-1 (in this case, the telescoping series leave a log of k factorial, which can be approximated using striling formula)
@michuosas
4 жыл бұрын
Math is wonderful. People like you show the world reasons why they should pursue math. Just amazing.
@The1RandomFool
3 жыл бұрын
I didn't expect such a simple answer given how much work is required to get it.
@Kapomafioso
Жыл бұрын
4:30 easier proof for general n (not just integer), without using integration by parts: substitute (m+1) x = u, replace 2n = a integrand = x^(a-1) / (a-1)! e^(-(m+1)x) = u^(a-1) / ((m+1)^(a-1) * Gamma(a)) e^(-u) * du / (m+1) = (1/(m+1)^a * Gamma(a)) * u^(a-1) e^(-u) du Integrating over u yields Gamma (a) by definition, so (summing over m) zeta(a) = sum_m = 0^infty 1/(m+1)^a = sum_m = 1^infty 1/m^a This holds for Re(a) > 1.
@Bayerwaldler
4 жыл бұрын
11:25 Instead of L'Hopital: e^(-x) = 1 - x + x^2/2 + h.o.t. (higher order terms) as x->0 by Taylor expansion. So the term in paranthesis equals -(x^2/2) / x^2 + h.o.t = -1/2 + h.o.t -> -1/2 as x->0
@digxx
4 жыл бұрын
That is basically the same!
@bndrcr82a08e349g
2 жыл бұрын
Very very very very good channel
@radhab8761
4 жыл бұрын
I saw it some where but cannot remember thanks for explaining
@user-yt8xc8zw6v
4 жыл бұрын
Hi your work is amazingly stunning for the first tool , we may use a well-known functinal equation to prove it, namely, integral from 0 to infinity of (u^(x-1))*(e^u -1) du = gamma(x) * zeta(x) and plugging x= 2n Have a great day!
@davidgoett8655
4 жыл бұрын
At 2:31, could you please give a little more explanation of how to use the dominated convergence theorem? What is the dominating function here?
@martinyang5596
4 жыл бұрын
Seamlessly done!
@chritophergaafele8922
4 жыл бұрын
you are using the Bose intergral
@nicolascamargo8339
Жыл бұрын
Excelente resultado muy bueno jajaja
@alainbarnier1995
4 жыл бұрын
Someone could explain the e^(x) + e^(-x) sum please, at around 18:50 ? Or the link to the "previous video" that explain it ? And a big up to Pr Penn for introduce us to Riemann-Zeta function and all intersting stuff !
@mosesose8814
2 жыл бұрын
Maybe a bit late but the sum is the taylor expansion of cosh(x) which is equal to (e^{x}+e^{-x})/2 where he takes the factor of 2 into the brackets :)
@LouisEmery
4 жыл бұрын
That was wonderful. It's like watching my kids building minecraft structures, i.e. moving things (factors) up and down. BTW, who asked to evaluate the sum of the zeta function in the first place? Is there an application?
@Re-lx1md
2 жыл бұрын
Power
@txikitofandango
4 жыл бұрын
9:30 If you did the integration-by-parts process (2n) times, you'd still end up with an integral of a constant times e^(-(m+1)x), right? So where did that part go?
@axemenace6637
4 жыл бұрын
Actually, you only take the antiderivative 2n-1 times. Then, you take the integral of e^-(m+1)x, and find that it equals 1/m+1. This gives you 1/(m+1)^2n.
@tucantipack
4 жыл бұрын
sum of ln(zeta(n)), from n=2 to infinity, pleasee
@hanibahout
4 жыл бұрын
Question : at 9.30 shouldn't the iteration be repeated 2n-1 times and the integrale calculated generating an additionnal (m+1) in the denominator and resulting in a (m+1)to the power 2n ? thank you for your answer.
@thomashoffmann8857
4 жыл бұрын
Yes... Same question and answer below. One m+1 comes from integration of the exponential function.
@chritophergaafele8922
4 жыл бұрын
This is true genius....
@FisicoNuclearCuantico
4 жыл бұрын
@9:40 there is a mistake. You did a good job.
@FisicoNuclearCuantico
4 жыл бұрын
@Michael Penn Tool #1 IS the tool itself.
@FisicoNuclearCuantico
4 жыл бұрын
I made a mistake. You are correct.
@axemenace6637
4 жыл бұрын
@@FisicoNuclearCuantico his explanation was wrong, but the result is correct. You only do integration by parts 2n-1 times, and the other 1/m+1 term comes from the antiderivative of e^-(m+1)x
@FisicoNuclearCuantico
4 жыл бұрын
@@axemenace6637 Let me check.
@FisicoNuclearCuantico
4 жыл бұрын
@@axemenace6637 The proof is correct, but rather incomplete. I prefer considering the integral of e^(-nx)*x^(s - 1) from zero to infinity and then integrating by parts. This will give (1/n^s)*Pi(s - 1), where Pi(x) is the extended Gamma function. We then introduce a Summation on both sides of the equation from n = 1 to infinity. This will give equation three on Riemann's 1859 paper www.claymath.org/sites/default/files/ezeta.pdf. You then divide both sides by Pi(s - 1) and substitute s equals 2n. Recall Pi(s - 1) = (s - 1)!.
@bhanusri3732
4 жыл бұрын
At 7:22 l'hospital rule have to be applied couple of times but when i tried i need to apply 2n - 1 times to cancel out the x term in numerator? What is the way to do it in couple of times?
@thomashoffmann8857
4 жыл бұрын
Each time you apply l'hospitals rule the exponential function stays the same but the power of x is reduced by one. So in the end the x-power vanishes and e^-(m+1)x stays and wins.
@bhanusri3732
4 жыл бұрын
@@angelmendez-rivera351 Thank you
@bhanusri3732
4 жыл бұрын
@@thomashoffmann8857 Yes thank you
@orangehokage2065
3 жыл бұрын
Least view to best thing .
@boukaddid
4 жыл бұрын
Fonction Zita, merci professeur
@advaithnair8152
4 жыл бұрын
Sir pls. Solve this for me A sequence a(n) is defined by a(1) = a(2) = 1, a(3) = 4 and for all other n a(n) = m(a(n-1) + a(n-2)) - a(n-3) Here m is a positive integer. Determine all m such that every term of the sequence is a whole square
@yakov9ify
4 жыл бұрын
We know that all squares are either 1 or 0 in mod 4 so a(n) will also follow that then if m is 0 in mod 4 then term a(4) will never be a square. Now the first three terms are 0 1 and 1 mod 4 so in order for a(4) to be a square m cannot be 3 mod 4 either which narrows this down to 1,2 mod 4. Now if it is 1 mod 4 then a(4) will be 0 mod 4 and thus a(5) will be 3 mod 4 which isn't a square so now m can't be 1 mod 4 either. The only option is thus 2 mod 4. So that narrows down the options but I haven't thought about the rest.
@michuosas
4 жыл бұрын
a4 is 4m which means m is a perfect square. (2m)^2
@michuosas
4 жыл бұрын
I have also checked other terms of the sequence and for many there is no m really that would reveal them as perfect squares. If you meant that m is strictly positive you my friend have no m to satisfy the propriety.
@michuosas
4 жыл бұрын
@@angelmendez-rivera351 i read the parantheses wrong. I accidently wrote the - a(n-3) inside the big paranthese. Thanks for pointing it out.
@michuosas
4 жыл бұрын
I believe that for m=2 it may be provable by induction that the general term is a(n) =f(n) ^2 where f(n) is the sequence of Fibonacci numbers. It sort of implies that this maybe the only solution for m but this is just a guess.
@zygoloid
4 жыл бұрын
and now we apply the trick, and the trick is... [integrating it]
@wallythe-k4036
4 жыл бұрын
Almost perfect except at 12:00 that limit using L’H the second term will result in circular reasoning, remind of how you derive the limit of e^-x using the definition🙃
@emanuellandeholm5657
4 жыл бұрын
L'Hospital is inherently circular. You have to assume you're dealing with a proper limit.
@hybmnzz2658
4 жыл бұрын
Pretty much what Angel said. I have to say its a very satisfying exercise starting with lim n->inf (1+x/n)^n and using the binomial theorem to arrive at the taylor series of e^x. You should try it really. I think its really e^x to blame cause there are so many equivalent definitions for it.
@Czeckie
4 жыл бұрын
very cute
@neeraj399
4 жыл бұрын
I need some help regarding something I ended up deriving by chance. I have posted it as a question here: math.stackexchange.com/questions/3738442/identity-involving-a-relation-between-zetas-and-zetas1-for-integers-s . I’ll be glad if you can do a video explaining this identity. Thanks!
@georgekh541
2 жыл бұрын
Why the fuck does wolframalpha give me the answer of 0.497516 ????????
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