I remember when I first read the wikipedia entries for the zeta, gamma and digamma functions. Those pages showed all sorts of identities correlating those 3 functions, including several series expansions. At the time I was like, how on earth can all of these identities be found? Your channel is answering that question for me and I thank you for that.
This approximate train of thought is where the rather famous result: sum_{1}^{infinity}{(zeta(2n)-1)/n} = ln(2) comes from! Unfortunately I have never seen any series that use zeta(2n+1). Perhaps you have seen some, though?
Nice, but I want to point out that for the natural even values of the zeta function the is a classic formula involving powers of pi and the Bernouilli numbers (this formula generalizes the Euler's solution of Basel's problem, btw). The fomula reads \zeta(2n)=(-1)^n(1/2)(2\pi)^{2n}B_{2n}(1/(2n)!). This along with your calculation provides a generating formula for either even values of the zeta function or, equivalently, even values of the Bernoilli numbers (btw, all odd values of the Bernouilli numbers, with the exception of the first, which is 1/2, are zero).
For the minute 3:32, the (1-(pi^2*x^2)/(pi^2/x^2)) this leads to 1-(x^2*k^2). I notice that the solution shall be modified. Overall, thank you for this innovative problem.
hii, could you please try this integral, I(α) = \int_0^1 (x^{50}(α-x)^{50}) dx i had this in an exam recently, i tried to use feynman 50 times. i made a mistake but i still got the correct answer 😅
The final result can be further simplified using the reflection formula for digamma:
Final result: 1/x - pi*cot(x*pi)
The answer should instead be (digamma(1+x)-digamma(1-x))/2 as differentiating the power series yields a factor of 2 as noted in 7:35.
Neat derivation!
Hi,
"ok, cool" : 0:20 , 1:14 ,
"terribly sorry about that" : 3:20 , 6:25 , 6:27 , 8:26 .
I remember when I first read the wikipedia entries for the zeta, gamma and digamma functions. Those pages showed all sorts of identities correlating those 3 functions, including several series expansions. At the time I was like, how on earth can all of these identities be found? Your channel is answering that question for me and I thank you for that.
@6:11 it should be 4^k or 2^2k in the denominator since x^2k at x=1/2
This approximate train of thought is where the rather famous result:
sum_{1}^{infinity}{(zeta(2n)-1)/n} = ln(2)
comes from! Unfortunately I have never seen any series that use zeta(2n+1). Perhaps you have seen some, though?
Best place to discover new functions !
This is crazy, I didn't expect the initial solution to be this easy, not to mention all the identities that came of it
Nice, but I want to point out that for the natural even values of the zeta function the is a classic formula involving powers of pi and the Bernouilli numbers (this formula generalizes the Euler's solution of Basel's problem, btw). The fomula reads \zeta(2n)=(-1)^n(1/2)(2\pi)^{2n}B_{2n}(1/(2n)!). This along with your calculation provides a generating formula for either even values of the zeta function or, equivalently, even values of the Bernoilli numbers (btw, all odd values of the Bernouilli numbers, with the exception of the first, which is 1/2, are zero).
For the minute 3:32, the (1-(pi^2*x^2)/(pi^2/x^2)) this leads to 1-(x^2*k^2). I notice that the solution shall be modified. Overall, thank you for this innovative problem.
Are these Kamal special functions?
Biology & Chemistry lover spotted 😂
Do you have a video about a integral of x^2*sech(x)^t ? I'm curious
Well we could have done this by using sinx/x expansion and taking log on both sides but still brilliant
sum k=1 to 10 🔥
hii, could you please try this integral, I(α) = \int_0^1 (x^{50}(α-x)^{50}) dx
i had this in an exam recently, i tried to use feynman 50 times. i made a mistake but i still got the correct answer 😅
Utilizzando la definizione di ξ,e scambiando i simboli di Σ,risulta S=-Σln(1-(x/n)^2)..n=1,2,3...a questo punto....boh...