This was one of my HW problems in my Fourier analysis class. It's a neat little result that I find really charming. We had to find the generating function for this Fourier series and then work from there.

Each step seemes intuvative but it takes a lot of nerve to perform those steps without second thought. Actually I am 14 years and i dont have much knowledge so first i see the solution and try it myself and if i am stuck i start the problem again But one day I'll surpass you its my challange 😤 I will solve those questions without seeing the solution and i will do it no matter what it takes❤

@kirbyharbinger6985

Жыл бұрын

im 14 too no wayy

I'm not sure if one day i will reach that level. But for now i'm really appreciating to follow your reasoning. Thanks a lot for showing us the way.

Again complex analysis makes surprising appearance. Thank you very much.

Thank you very much. I love your utilisation of special fonctions.

I really enjoyed this video and pretty much all your videos. Great way to pass the time and reflect to younger days! Keep ‘em coming! Thanks for your efforts.

5:36 you can also prove the recurrence relation for the digamma function by starting from that of the gamma function and taking the log of both sides, splitting up the log on the right, and differentiating. It’s the same thing just done much faster.

Awesome video and very elegant solution! Also, the derivation of side steps is very neat! Don't stop!

I think this can also be done using complex analysis, typical residue theorem problem (I remember doing this or a similar one in my complex variables exam). But seeing it done here using only real analysis is incredibly satisfying! Subscribed! 😊 Edit: Well, I suppose it still uses complex numbers, but doesn’t rely on the residue theorem, which is incredibly satisfying!

@maths_505

Жыл бұрын

The gamma and digamma functions are both subjects of complex analysis so yes the video is solved using complex analysis rather than real analysis.

@edmundwoolliams1240

4 ай бұрын

​@@maths_505 But can't gamma and digamma just be restricted to a real domain, in that sense they can be treated with real analysis?

Very Smart Proof. I like it really. Thank you for your nice video

We had to compute this exact result using a contour integral over the complex plane and using the residue theorem as an example sheet question

Thank U so much ❤❤❤👍👍👍👍👍

Loved it! Though each and every step is simple and clear, putting them into that sequence is a matter for the genius! ما شاء الله عليك يا عبقري.

@maths_505

2 ай бұрын

Thank you my friend ❤️ All praise to Allah ❤️

Use residue theorem. Consider the function f(z) = cos(πz) /((z^2+1)*sin(πz)). Consider the sequence of square contours centered at the origin with sides of length (2n+1). It has resides at +-i, 0, +-1, +-2,…+-n. Integrate f(z) on this sequence of contours and take n to infinity. It can be proven that this sequence of integrals goes to zero as n approaches infinity, which takes some effort (you can prove it goes to zero on the vertical and horizontal, respectively). On the right hand side of the equation, we break down the big square into smaller squares centered at all the integers and +-i and use Taylor expansion centered at each of the integers and +-i on each of the cos and sin. Apply the residue theorem on each. Note the symmetry of z= n and z = -n for any nonzero integer n. We calculate the residue at 0, +i, -i, and then we are left with twice the sum of residues at 1,2,3,4… and some coefficients in the Taylor series. This part of the sum also happens to be the infinite sum of 1/(n^2+1) multiplied by some coefficient. Adding the residue at 0, i, -i, this right hand side is equal to the left hand side which could be shown to be zero as n goes to infinity. Bonus: if you consider f(z) = cos(πz) /((z^2)*sin(πz)), we have the Basel problem’s solution. And instead of cos (πz), if you use 1, then we have solutions for alternating series. For example, the infinite sum of (-1)^(n+1) / n^2 and so on and so forth.

@nakhleasmar9175

Жыл бұрын

or see the book by Asmar and Grafakos, Complex Analysis with Applications, page 347.

"cancel out" is my favorite combination of words 👍 :) I hope the cold got better 👌

@maths_505

Жыл бұрын

Yeah it's been quite a while since this video came out so the cold did subside 😂

@maths_505

Жыл бұрын

And yes the magic words 'cancel' and 'out' >>>> 'please' and 'thank you'

@phenixorbitall3917

Жыл бұрын

:)

I saw the computation of this sum using Fourier Series of the function e^x defined on (-pi, pi] with period 2*pi, then we just set x = pi

@nakhleasmar9175

Жыл бұрын

See my book "Partial Differential Equations with Fourier Series and Boundary Value Problems," page p. 63 and p.66 # 13.

I am smelling something big comming up next is it ??❤

@maths_505

Жыл бұрын

Well its actually unrelated to this result but yes it's something to do with the digamma boi

Get well soon!

You could have just derive the formula of sinx/x that consist of infintie product after plug in the natural log function and then evaluate at x=i

Theres a fast method of finding this result by taking the euler product for sine, taking the natural log, and taking the derivative to find cot(x).

This qn came to my mind in high school 😂

1:30 After split into 2 sum I decided to transform n>>-n Then I use the cotangent expansion

Digamma function is the second loveliest non-elementary function for me after of course, the Gamma function.

@maths_505

Жыл бұрын

Same here

@user-uh9bo2im1h

10 ай бұрын

Beta on top

I actually posted this proof on my IG page to solve an integral recently. Exact same solution

@gagadaddy8713

Жыл бұрын

I am mesmerized for every step 505 applied here. It is even a bit difficult to reason the meaning of digamma function for me, not talking about how he can come a connection between a real number summation series with the digamma function. Anyway, if you said that you had figure out the solution all by yourself, please have my salute, mate! 👨‍🎓

Interesting

So I solved this problem the wrong way on high school. First we need to find a function that has roots on n²+1 that can be done with sin(pi *sqrt(x-1)) this will have roots at 1, 2, 5, ..., n²+1 beyond. Ok now the idea was to make a polinomial function like a taylor series on that, if we do so, if we approximate enough we will have a polinomial function with many of these roots if we derive the logarithmic ln of a polinomial with x at 0 we will get the negative of the sum of the inverse of the roots, sadly our function doesn't behave like we want and we can't make a nice taylor approximation of it, but we can do of sin(pi*sqrt(x-1))/(x-1) , this will exclude the root 1 so we need to add it at the end, we take the log of it, derive at x=0 and beng we get the same result. Shame that it doesn't work so well for 1/n⁴ alike scenarios, we end up getting some poles on some derivatives ):

Just today I found the value of a very familiar sum The sum k=1to∞ of 1/(1+π²k²) I used the cot infinite sum over all integers in your proof off it the step where k is form 1to∞ And then substituted I as an input I made the sum equal to cot(i) and used its complex form With a little simplifying you get that the sum equal to 1/(e²-1)

@maths_505

Жыл бұрын

Beautiful!

This is how I solved it once I found it. Reflection formulae ftw. I later solved it by differentiating the sine product but it's less beautiful.

How did you learn the series for all these functions?

@Obotron7

Жыл бұрын

We usually use wikipedia to find series representations of special functions

@maths_505

Жыл бұрын

Google em

@mohandaskandukuri4625

Жыл бұрын

​@@Obotron7 not only calculus aslo start series on trigonometry algebra. Etc😊😊❤❤

You appeal without proof to properties of the psi and gamma functions. So how is this easier than the traditional way of using the residues of cot (pi z) to sum the original series?

@maths_505

Жыл бұрын

Dude I literally proved both properties of the digamm function in the video. I have separate videos proving the properties of the gamma function but the ones here are just too trivial.

Cool, the digamma function! But I was expecting a solution development more along the lines of kzread.info/dash/bejne/n4BnyNamnMezn9o.html#t=3m25s. 😜

@maths_505

Жыл бұрын

Dude I see your point but there was no chance I was letting go of a chance to use the digamma function and prove its properties 😂