Euler's brilliant solution to the Basel problem vs Cauchy's cool residue theorem approach

[@maths_505]

You can follow me on Instagram for write ups that come in handy for my videos: instagram.com/maths.505?igshid=MzRlODBiNWFlZA== If you like the videos and would like to support the channel: www.patreon.com/Maths505

[@uxydna]

Euler's idea to use coefficients is simply marvelous and so intuitive... deservedly a classic solution

[@ΙΗΣΟΥΣΧριστος-θ2γ]

Suiiii.He did it.Great presentation plus u found a way to combine the basel problem with complex analysis. At this point i may as well say thank you for ur service man ur a legend😂

[@quentinrenon9876]

The second one is so wild!

[@realcirno1750]

the fact that f(z)cot(pi z) has a residue of f(n) at integers where f has no pole is a genius insight that paves the way for so many summation problems (along with the fact that the sum of the residues is 0 if f is O(1/z^2) which this vid seems to leave out) i think cauchy wins, his method is more rigorous and generalizable, the class of series that can be evaluated with hadamard product is much smaller than those that can be evaluated with cotangent summation

[@Galileosays]

Great. Nice to see how the the residue theorem can be applied to Basel's problem.

[@mokhtarmougai5088]

that "It's Euler" reason can shut every hater mouth 💀

[@anthony9656]

Great presentation, thanks! On the Basil problem, there is a different more generalized approach due to Euler in the “spectacular sums” section of the book “The Calculus Gallery”. Definitely worth checking out.

[@herbertdiazmoraga7258]

FUCKING LOVE COMPLEX VARIABLE MAAAAN

[@Ghaith7702]

this is verry intresting but a bit too much for me unfortunatly awesome videos btw

[@timemasterdm2462]

A beautiful pair of solutions.

[@danielrosado3213]

epic

[@MrWael1970]

Very nice proof. But when you substitute by k = 1 at the first method, you forget to raise the power of (-1) to 1. Thank you very much for your amazing effort.

[@giuseppemalaguti435]

Ok coool

[@maths_505]

grazie per aver digitato in inglese questa volta😂

[@giuseppemalaguti435]

@@maths_505 quando scrivo in italiano, scrivo numeri e formule...

[@zeggwaghismail827]

Try to solve it using polynomials in cot^2.

[@PotentialEnergy-ew1js]

Make a video how to learn maths... you're really impressive 🙏

[@maths_505]

Well the best way to learn it is to actually do it. Nd in case you're looking for resources then go for MIT and Harvard lectures on KZbin

[@thewolverine7516]

​@@maths_505What about your promise? When are your full theory lectures gonna come?

[@maths_505]

@@thewolverine7516 probably September. Thing is I really want them to be good so everytime I write up a lecture I end up redoing the whole thing cuz I feel it's not good enough. So its gonna take me a while to get em ready. Plus alot of work goes into making content for this channel so the routine is pretty tough these days.

[@maths_505]

@@thewolverine7516 I'll probably just start with one or two courses. Specifically complex analysis and differential equations and I'll build up from there

[@thewolverine7516]

@@maths_505 Take your time, I believe when you launch it, it's gonna be extremely resourceful.

[@Predaking4ever]

Basel Wars (in the voice of Beast Wars)!!!

[@anupamamehra6068]

hi math 505 i have a question for you: solve the indefinite integral: ∫ x^2 squareroot( 36 x^2 - 1) dx

[@maths_505]

Use IBP

[@aravindakannank.s.]

all videos are released at midnight in India

[@عليعقل-ص6ث]

Second one best

[@saulmendoza1652]

Very neat!!

[@Super-gt9lk]

Hello, is it possible to integrate x/(e^x-1) from x=0 to x=infty WITHOUT using series expansion? Context: I am collecting the ways to compute sum(1/n^2). After Laplace transform and MCT, I encounter this integral, but it would make me a clown if I use series expansion since it just goes all the way back sum(1/n^2).

[@tzebengng9722]

There are many ways to determine the sums. You might like to check out my calculus web at firebase for a telescopic series with simple trigonometric integrals. It is much more elegant.

[@Super-gt9lk]

@@tzebengng9722 Although I couldn't find the desired answer there (since I am trying the technique in the paper I recently read, which is about evaluating infinite sum via Laplace transform), your website is extremely informative and would be helpful for my further studying. So...thanks a lot, I surely would check your website from time to time in the future.

[@tzebengng9722]

Yes, it is possible. You can re-express the integral in terms of log and find a suitable function to apply the differentiation under the integral sign. Check my calculus web again, I will the answer soon. @@Super-gt9lk

[@tzebengng9722]

With a change of variable, you can show that the integral is equal to 2/3 of the integral of ln(x)/(x^2-1) from 0 to infinity. Then you can apply differentiation under the integration sign to evaluate this integral. Use the obvious function 1/2 of ln(1+t^2(x^2-1))/(x^2-1). For details see my calculus web.

[@oom_boudewijns6920]

3:22 why not multiply the pix^3 term with the x^2 terms??? u are ignoring the x^5 terms?

[@maths_505]

Multiply out the first few terms....you'll see why....

[@jurgensand9201]

I still do not understand: all x^5 terms are positive and do not sum to 0; correspondingly all x^7 terms are negativ and do not sum to 0, etc. So why do you ignore them?

[@jurgensand9201]

I understand now: the higher potential terms of x do not cancel out, you simply ignore them in your calculation. Finally you only compare the x^3 term with the x^3 term of Fourier expansion of the sin function. So, the corresponding comparations of the x^5, x^7 and so on terms could give more results ...

[@cycklist]

Zed

[@ngc-fo5te]

Zee

[@saulmendoza1652]

The link to the paper is dead

[@maths_505]

Fixed it