The Power Of Recursion: An UNSOLVABLE Integral
Vložit
- čas přidán 18. 06. 2024
- Hope everyone enjoyed! I thought this integral was particularly unique and love how theres so many approaches - please comment all of yours down below!! Please comment with any questions or suggestions for new topics, and as always, subscribe to stay updated. ~ Thanks for watching!Still looking for your answer to my last two video challenges @blackpenredpen ... Thanks again for all the recent support and please keep sharing and interacting!!#maths #mathematics #integrals #MIT #Cambridge #JEE #recursion #problemsolving #whoknew #fascinating #functions #euler #funproblems #proofs #functions #physics #sums #series #limits #whiteboard #math505 #blackpenredpen #integral #trig #trigonometry
WOOOOOOOOOOO 1K!!!!
I KNOW!!! Thanks so much for everything - couldn't have got anywhere near without the help or the inspiration to start a channel in the first place.
Wow, it’s neat how many different kinds of mathematical tools we used to solve this one
That's why I thought it was such a good one to share!
I’d just like to answer to the title of the video: in France, we are being taught the general method for integrals of 1/(ax^2+bx+c)^n where a, b, c are the coefficients of an irreducible polynomial considered in the real space. If the polynomial is of an order above 2, or it’s reducible, you can use partial fraction decomposition. Actually that last method is how you would start solving any rational fraction integration problem!
Thanks! That's really interesting to hear and a great problem to solve.
prepa?
this very specific class of problem was standard in all textbooks at least through the 90s- i definitely did this problem in high school. i feel like it may have been dropped from some curricula as not being worth the time but i wouldn't call this an arcane topic.
Using residue theorem is also nice for this one
Definitely agree!
Nice one, I remember solving the trig case as a question on my cal2 finals. A wonderful exercise indeed!
Thank you!!
Damn son, where'd you find this?
inf*0 is not necessarily zero. its indeterminate without further work
Yeah I was thinking of leaving it as a challenge at the end... you can gauge roughly it ends up at 0 because the power of 2n for cosine scales more rapidly than tan but of course there is a longer rigorous proof. I didn't want to include it in the video because it was already a bit long but feel free to share it here!
One can avoid the issue of the indeterminate form by using a hyperbolic substitution instead:
x = sinh(u)
dx = cosh(u).du
x=0 => u=0
x-->inf => u-->inf
Letting sinh(u,n) = (sinh(u))^n etc:
J[n] = Integral[0 to inf: 1 / (1+x^2)^n]
= Integral[0 to inf: cosh(u) / (1+sinh(u,2))^n]
= Integral[0 to inf: cosh(u) / cosh(u,2n)]
= Integral[0 to inf: 1 / cosh(u,2n-1)]
J[n] = Integral[0 to inf: cosh(u,2) / cosh(u,2n+1)]
= Integral[0 to inf: (1+sinh(u,2)) / cosh(u,2n+1)]
= Integral[0 to inf: 1 / cosh(u,2n+1)] + Integral[0 to inf: sinh(u,2) / cosh(u,2n+1)]
= J[n+1] + Integral[0 to inf: sinh(u) . sinh(u) / cosh(u,2n+1)]
= J[n+1] + (-1/2n).[0 to inf: sinh(u) / cosh(u,2n)] - (-1/2n).Integral[0 to inf: cosh(u) / cosh(u,2n)]
= J[n+1] - (1/2n).[0 to inf: tanh(u) / cosh(u,2n-1)] + (1/2n).Integral[0 to inf: 1 / cosh(u,2n-1)]
= J[n+1] - (1/2n).[(1/inf) - (0/1)] + (1/2n).J[n]
= J[n+1] + (1/2n).J[n]
=> J[n+1] = J[n].(2n-1)/2n
@@franolich3 Nicely done!!
You can rewrite tan x = sinx/cos x. If n > 0 then the cos x in the denominator is cancelled by a cos x in the numerator and you end up with a positive power of cos x multiplied by sin x. Then cos x->0 as x->pi/2, and sin x->0 as x->0, and there are no infinities.
@@RamblingMaths Yes this is what I was thinking too... Thanks for sharing, great solution!
wow this is so fancy and elegant
Thank you!
This one is well known in BG as well, it's the notorious "14000th integral" ('coz ~14k freshmen got F on their Calculus 1 exam in a single day, one of the problems being the general form of this one, $I_n = \int \frac{Px + Q}{(ax^2 + bx + c)^n} dx$) 😃.
Great video ! Would love to see more series in the channel
Absolutely - series are some of my favourites too, thanks so much the comment!
Beta function enters the room
I had never heard of recursive integrals. That's amazing.
Thanks so much!
very nice
Thank you!!
Beta(n+1/2,1/2) sneaking up
Bold of you to assume anyone learns integrals from school (unless they want to mess up their thinking ability and hamper their originality)
Let n be a positive integer. The integral converges by majoration with the case when n = 1.
For all x in the set of real numbers:
0 ≤ 1 / (1 + x²)^(n + 1) ≤ 1 / (1 + x²)^n
We can integrate this relation because we have shown that these integrals converge. Thus, the sequence (Iₙ) is positive and decreasing, which means it is convergent.
Next, we can integrate by parts on the interval [0, A] where A > 0, to establish the recurrence relation between Iₙ and Iₙ₊₁, leading to the same result!
Using trigonometry, as demonstrated in the video, is also a nice approach
Great video on what would be a classic prep class exercise in France! :)
Nice! Love that method, great approach.
using beta function, the integral is trivial after subing x=tan(theta)
Yeah I mentioned at the start - beta function is definitely a shortcut after one substitution, but I wanted to offer a more accessible second method. Great spot though! That's part of what makes the beta function so useful.
Can you work on complex numbers next time, I really like to see a problem with that.
Definitely - do you mean a complex integral (like contour integration) or just anything complex in general?
@@OscgrMaths love to see complex integral
I didn't know that this simple technique, that I have found by experimenting(things they don't teach at school, but also subtract from students) with unfamiliar integral by myself for 3 mins btw, is SO HARD that it deserves a whole video of explanation :)))). Lol, we learn something everyday don't we?
Glad you enjoy this kind of maths! Maybe check out some of the other videos on my channel like the one from the BMO if you want a different kind of question.
I have a textbook, you know? Absolutely no need for videos, bruh
bro stfu like how can you be this much of a virgin. lol ig we learn something new every day
can you make more logarithmic integrals
Yeah definitely, thanks for the suggestion!
Another “simplification” is making the (2n)!/(n!)(n!) => 2nCn, neat
Wow that's a great point, makes the answer even more satisfying!
You should teach the chain rule differently lol
Sorry! I was hoping everyone watching would know it already. Would you like me to do a separate video teaching it?
The derivative of the outside function evaluated at the inside function times the derivative of the inside function
@@OscgrMaths That’s one of the better things to explain, if you do try to explain it, I wouldn’t do it the way you did
@@coreymonsta7505 Yeah I think given the full time I'd use the fact that substitution u for inner function allows you to find dy/du and du/dx, then show the cancellation comes after that giving dy/dx. But sometimes with these longer videos I try and rush over the basic stuff - I'll keep it in mind next time!
@@OscgrMaths I mean just explain how the rule works during the video in a different way
Don't use hand to wipe the board, oil on the skin ruins the pens and the board
Thanks for the tip!! I'll make sure to always use my rubber.
The integral is just pi*i*residue at z=i. (complete a contour in the UHP), and using the evenness of the function.
The residue is pretty straight forward as 1/(1+z^2)^n = 1/(z+i)^n(z-i)^n. So we have a pole of order n. So we just need to do 1/(n-1)! * the n-1'th derivative of (z+i)^-n evaluated at z = i. This is 1/(n-1)! * (-1)^(n-1) n(n+1)(n+2)...(2n-2) (2i)^(2n-1). Multiplying by pi*i and a bit of cleaning up gives the answer.
Yes!! I was hoping somebody would take a contour approach... This is a beautiful solution, very nicely done.
asnwer=1 oo sin cos tan
Fantastic. [[ Technical comment: is it possible to direct the microphone more directly towrad your voice? You're picking up an awful lot of marker-squeak, which, though not as bad as fingernails on a chalkboard, is somewhat annoying...
Yes, I'm thinking of getting a lapel mic which I'm hoping will pick up more voice and less board. Thanks for the comment!