An incredible integral from the 2023 MIT integration bee semi finals

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Күн бұрын

Пікірлер: 41

@manstuckinabox3679 Жыл бұрын

me watching maths 505 take 20 mins to drill down a wall of an integral while I wait for some series expansion to pop of of nowhere

@MrWael1970 Жыл бұрын

Thank you for your great effort

@sahilharidas7172 Жыл бұрын

where can i find all the MIT integration bee questions?

@supercellyt9449 Жыл бұрын

16:57 -> Since the integration is from 0 to pi/2, the integration of cos terms individually give rise to sine terms, whose upper limit doesn't simplify to 0, I think we would get a sum like 2022*pi + 1 + 2*((-1/3)+(1/5)-(1/7)+(1/9)...+(1/1009)) - 1/1011 Correct me if I'm wrong

@mavenfromheaven1190 Жыл бұрын

very nice observation, you are right. on 15:36 he used incorrect formula, it should be 1/2*(cos(diff) + cos(sum)), not 1/2*(cos(diff/2) + cos(sum/2)) tbh the integral was soooo hard i don't blame him for this small mistake

@supercellyt9449 Жыл бұрын

Yes, that makes more sense

@paulor.r.correia1789 Жыл бұрын

Thanks for your work

@RiteshKumar-sy9sp Жыл бұрын

Hey I used an approach taught by my teacher We declare I as a function of n We find I(n+1) - I(n) Then we consider this as another function J(n) now we find J (n+1) - J(n) Which turns out to be zero which means J(n ) is independent of n which means difference of I(n+1) - I(n) a constant so we calculate I(1) - I(0) which turns to be π/2 which means I(n ) is an arithmetic progression of common difference of π/2 so we get I(n) = n*π/2 putting the values we get 2022π this approach is more easy to me, tell me if there are any mistakes

@maths_505 Жыл бұрын

A wonderfully creative approach indeed

@maths_505 Жыл бұрын

I'll note this for future videos

@manstuckinabox3679 Жыл бұрын

why is n= 4044? how do you define I(n)? where 's the n? can you explain more lol this is very interesting! I think this hints on the fact that you can use somewhat of a Feynman's technique, instead of a discrete n we use a real s and differentiate under the integral sign.

@RiteshKumar-sy9sp Жыл бұрын

@@manstuckinabox3679 n is 1011 , 4 is multiplied outside the integral of (1 + cos(2022x))/(1 + cos2x), we know 1 + cos 2y = 2 cos^2y Put that in integral You get cos^2(1011x)/cos^2(x) After applying the phase shift we get sin^2 1011x / sin^2 x now follow the steps above and use some trigonometric identities

@BridgeBum Жыл бұрын

Trying to understand what you are saying. How are I and J defined?

@Brainiac_69 11 ай бұрын

A beautiful question! Btw I proceeded simplifying with kings rule and it reduces really fast as the cosine terms get cancelled on adding the same integral. But the problem is, I get a different answer :’). I’ve tried cross checking for any mistakes but end up getting the same result.

@Joellie859 Жыл бұрын

Hi I used complex functions to solve it .first we have to take a look at de moivre’s theorem which says that ^n =r^n *

@ikarienator Жыл бұрын

I wonder how this can be done in 3 minutes

@lagnugg 5 ай бұрын

maybe they gave points for parts of solutions? i solved this integral a bit differently from the video, but my solution is long as well, and i have no idea what shortcuts i could use

@benwhitaker9888 Жыл бұрын

9:20 you didn’t change the limits. I’m going to assume you realised and just used the -dx to flip the limits back to 0 -> pi/2

@maths_505 Жыл бұрын

Yup That's exactly what happened

@jonasdaverio9369 9 ай бұрын

Hi, there is a much much easier way. First do u = pi - x, it changes the sign of cos(nx) with odd n, but for even n. Then add to the original, it cancels out most of the numerator. You arrive with I = 2 int (1 - cos 2022x)/(1 - cos 2x) = 2 int sin^2(1011x)/sin^2(x). Now use the Dirichlet kernel formula on this, and get (1 + 2 sum cos 2nx)^2. You expand the sum as you did, and get 1 + 4 sum cos 2nx + 4 sum_nm cos(2n x) cos(2m x). The second term integrates to 0. The third term is 2 sum_nm (cos((n+m)x) + cos((n-m)x), from which the first term integrates to 0, and the second integrates to 0 except when n=m. You thus get 2pi + 4 int sum_nm delta_nm (Kronecker delta) which gives 2022 pi.

@maths_505 9 ай бұрын

Damnnnnn.....nice one

@jonasdaverio9369 9 ай бұрын

@@maths_505I'm surprised you didn't think of it, because the King property comes everywhere in MIT problems, and also you used a changed of variable which I would never have thought about (well, I'm hyperbolizing but you get what I mean)