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.
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)..
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.
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 :)
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.
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?
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.
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
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)
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?
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.
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 !
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!
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.
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?
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.
@@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
@@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)!.
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🙃
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!
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
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.
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.
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.
beautiful how these sums always converge to such nice values
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.
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.
Return of "that's a good place to stop"
OK. Good.
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.
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)..
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.
I think he just said it wrong.
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 :)
@@thomashoffmann8857, good point, I didn't see that, but you are right.
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.
Math is wonderful. People like you show the world reasons why they should pursue math. Just amazing.
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?
I didn't expect such a simple answer given how much work is required to get it.
Seamlessly done!
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?
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.
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
That is basically the same!
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)
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?
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.
@@angelmendez-rivera351 Thank you
@@thomashoffmann8857 Yes thank you
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 !
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 :)
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!
I saw it some where but cannot remember thanks for explaining
Very very very very good channel
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.
Yes... Same question and answer below. One m+1 comes from integration of the exponential function.
you are using the Bose intergral
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?
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.
sum of ln(zeta(n)), from n=2 to infinity, pleasee
@9:40 there is a mistake. You did a good job.
@Michael Penn Tool #1 IS the tool itself.
I made a mistake. You are correct.
@@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
@@axemenace6637 Let me check.
@@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)!.
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🙃
L'Hospital is inherently circular. You have to assume you're dealing with a proper limit.
Excelente resultado muy bueno jajaja
This is true genius....
Power
Fonction Zita, merci professeur
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!
and now we apply the trick, and the trick is... [integrating it]
Why the fuck does wolframalpha give me the answer of 0.497516 ????????
Least view to best thing .
very cute
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
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.
a4 is 4m which means m is a perfect square. (2m)^2
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.
@@angelmendez-rivera351 i read the parantheses wrong. I accidently wrote the - a(n-3) inside the big paranthese. Thanks for pointing it out.
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.