This video is absolutely captivating! The zeta function has always held a special place in my heart. The way these series expansions gracefully unfold truly captures the essence of its mesmerizing properties. Thank you for sharing this intriguing exploration of the zeta function and its series expansions.
Another derivative will give me something related to Apery's constant and for n^3 we'll have something related to zeta(4) which has a nice closed form.
Check out the playlist for Feynman's trick. There you'll find the integral of e^(-x^2)lnx from zero to infinity. I've evaluated a few derivatives of the gamma function there
You have to make sure they are converging first, by fubini’s theorem, but he doesn’t mention it in the video cause we assume the sums are converging. A way you can show it is we know that zeta(n)=2. So we can write out the same sum S, bounded by a second summation S_1, replacing the zeta function with 2, and showing that the new upper bound sum S_1 exists, hence the original sum S does too.
@@gabriel_talih ah i see. what about the summation index? i was thinking that the order of summation can only be interchanged if the indices are independent; an index can't be the limit of summation of another one
Thank you for the nice problems and their solutions. I found it easier to start with letting 1/k^(n+1) = 1/n! int_{0}^{oo} t^n e^(-t k) dt. Then doing the double sum gives the integrals I1 = int_{0..oo} (e^(t/2)-1)/(e^t - 1) dt = ln(4), I2 = int_{0..oo} (1-e^(-t/2)-1)/(e^t - 1) dt = 2-ln(4), and I3 = \frac{1}{2} \int_0^{\infty } \frac{t e^{t/2}}{e^t-1} \, dt = pi^2/4, respectively. Now I1 and I2 were elementary (after an obvious substitution), I3 leads to the sum of the inverse of odd squares. No mention of the the polygamma function and Euler gamma is necessary.
@@mars_titanI see exactly what you mean now. Zeta(n+1) is bounded above by 2, allowing the series with zeta(n+1) coefficients to be in turn bounded above by twice the geometric series with common ratio x, which evidently converges since |x|
This video is absolutely captivating! The zeta function has always held a special place in my heart. The way these series expansions gracefully unfold truly captures the essence of its mesmerizing properties. Thank you for sharing this intriguing exploration of the zeta function and its series expansions.
Nice method and video! Can you evaluate the last sum when replacing n by n^2 or n^3 or n^4? I did so, using Maple‘s identify routine.
Another derivative will give me something related to Apery's constant and for n^3 we'll have something related to zeta(4) which has a nice closed form.
@@maths_505 if I understand that properly, you can get zeta of m+1 by computing n^m or am I wrong?
Maths 505 reminding us that dealing with sums can be fun and not always cumbersome.
Very good tricks and smart solutions. Again you are really talented.
Thank you Professor
Fascinating video :-) Could you tell please the link to the gamma'(1/2) proof?
Check out the playlist for Feynman's trick. There you'll find the integral of e^(-x^2)lnx from zero to infinity. I've evaluated a few derivatives of the gamma function there
@@maths_505 Great! Many thanks 🙂
wait, can we always switch the order of two sums like that?
You have to make sure they are converging first, by fubini’s theorem, but he doesn’t mention it in the video cause we assume the sums are converging.
A way you can show it is we know that zeta(n)=2. So we can write out the same sum S, bounded by a second summation S_1, replacing the zeta function with 2, and showing that the new upper bound sum S_1 exists, hence the original sum S does too.
@@gabriel_talih ah i see. what about the summation index? i was thinking that the order of summation can only be interchanged if the indices are independent; an index can't be the limit of summation of another one
@@GeoffryGifari I think the indices are independent of one another? We range over all k>=1 and n>=1.
@@aldomorelli6379 yes, it is in this case
2:46 What happened to the exponent k? I don't understand that.
It’s not an exponent he’s multiplying the top and bottom by k
@@rebel2358 thanks, I didn't understand what "expanding" meant.
Hay can you please suggest me a good book to practice calculus? Foreign author would be great...
What level of calculus?
Cal 1, 2 or 3
@@maths_505 for integration and vector calculus... I guess it is calculus 2&3
Thank you for the nice problems and their solutions. I found it easier to start with letting 1/k^(n+1) = 1/n! int_{0}^{oo} t^n e^(-t k) dt. Then doing the double sum gives the integrals I1 = int_{0..oo} (e^(t/2)-1)/(e^t - 1) dt = ln(4), I2 = int_{0..oo} (1-e^(-t/2)-1)/(e^t - 1) dt = 2-ln(4), and I3 = \frac{1}{2} \int_0^{\infty } \frac{t e^{t/2}}{e^t-1} \, dt = pi^2/4, respectively. Now I1 and I2 were elementary (after an obvious substitution), I3 leads to the sum of the inverse of odd squares. No mention of the the polygamma function and Euler gamma is necessary.
Hello! Can you please tell what app you use to write on?
Samsung notes😊
in this period he used his phone
yeah I know he is a madman😅
My request has been accepted 🎉😊
Excellent request
@@jkid1134 🗿👍
How can we be sure that the sum of zeta(n + 1) * x ^ n converges?
|x|
@@mars_titanCan you please elaborate? What convergence rule are you referring to?
@@mars_titanI see exactly what you mean now. Zeta(n+1) is bounded above by 2, allowing the series with zeta(n+1) coefficients to be in turn bounded above by twice the geometric series with common ratio x, which evidently converges since |x|
@@slavinojunepri7648 exactly!