The Most Intimidating Integral I've Ever Seen
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- Äas pĆidĂĄn 27. 06. 2021
- This Putnam Series was given on the Putnam exam in 1997 (Problem A3). Let's figure out a way to deal with this Putnam problem!
For those that don't know, the Putnam math competition features some super interesting and challenging problems!
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Disclaimer: This video is for entertainment purposes only and should not be considered academic. Though all information is provided in good faith, no warranty of any kind, expressed or implied, is made with regards to the accuracy, validity, reliability, consistency, adequacy, or completeness of this information.
Putnam 1997 A3
Putnam Exam 1997 A3
#math #brithemathguy #putnam
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Beautiful presentation! Love it!
Thank you so much!
Wonderful! Simply wonderful!
This I'm placing on my journal to sleep.
Yup
Outstanding. Sometimes I wonder who's more impressive: the student who solved the integral or the person who conjured it.
Arms getting bigger, so is the channel!
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The echo is a little jarring but nonetheless still a beautiful solution to such an intimidating integral! Good stuff
And i thought i was the one who felt something was different.
Sorry about that! It should be fixed in the future.
@@BriTheMathGuy no problem the math was great as always. Love you and your content. đ.
That's a very very beautiful way of solving a particularly intimidating integral, you just won a suscriber
Thanks so much!
That was so insightful. I have never dealt with an integral like that, but now I am confident that if I ever see one, not to panic. Thank you! I really enjoyed this video.
Wonderful!
That feeling when n factorial cancels
Watching these videos makes me realize that my hunger for scientific knowledge is still stronger and bigger than my fatigue after a full-time, warehouse-assistant working day.
We all crave it! Thanks for watching after your tough day!
Very nice presentation! To be absolutely rigorous though, it'd be nice to mention that each of the series converge for all positive x (ratio test) and that the sum and integral can be interchanged (e.g. tonelli's theorem)
Watched to the end, liked, saved to favorite math playlist, already subscribed, there isn't just anything left to do.
Become a Putnam fellow
Wow thank you so much!
@@BriTheMathGuy me too!!!!!!!!!!!!
Phenomenal!!
Your way of presenting a problem is mind-blowing.
Discussing the possible methods in a step, how to start solving it, best approach ...
Everything illustrates how good you are in math and throws light on the beauty of math
5:11 i think swapping the integral and the infinite sum there requires using the dominated convergence theorem(if we think about it rigorously), very good presentation overall
What a tremendous exposition! New subscriber! Thank you for your material! đčđ„
Subscribed!! Brilliant way of solving the integral as well as presenting it. Loved the video!
Awesome, thank you!
Your videos kick ass man, I want to make ones just like them! I love this fast paced but concise format
Thanks so much! Best of luck!!
The format of black screen, the math in all the details and the clean process with all the steps makes these series of tutorial useful.
Wow!
What a great way with words! I love your channel.
Thanks so much! Have a great day!
This was amazing; thank you so much for sharing!
Glad you enjoyed it!
That is a suprisingly beautiful result! Thank you for covering this in a video. :D
My pleasure!
That was amazingđđ Congratulations
Thanks so much!!
The second power series (1 + xÂČ/2ÂČ + ...) equals the the Bessel function of the first kind J_0 evaluated at ix, although I don't know how that would be helpful in this problem.
It would be helpful if you are familiar with the Bessel functions, since they satisfy many integral equations.
You release that you're good at math when u start watching the contents in x2
Beautiful problem, and very beautiful answer.
Using the sum representation of the exponential function and the Gamma function⊠what a ride haha.
Love your channel!!
Many thanks!
Such a great video!
Glad you liked it!!
ăăăăăăBravo!!
ăăŁăĄăăăĄăăăăăăăăŁăă§ă!!!đđđđđđ
Thanks so much!
Your videos are so fun to watch.đ
Glad you like them!
2:18
My dirty brain just hears a curse word
i appreciate this hope that maths will be fun and famous like nothing before once
Wow!! Absolutely marvelous!!
Thank you! Cheers!
The way you explain, makes these intimidating integrals seem easier
Glad you think so! Have a great day!
5:05 why can we do this ? Permute the sum and the integral?
Is it because the sum is converging uniformly on [0,+infinity] ?
I'd say dominated convergence theorem, with something like exp(-u+u/2) = exp(-u/2) being the integrable dominant.
The integral of the sum is the sum of the integrals because the integral is a linear function. Then you just put out of the integral the terms that don't have 'u', which means they are constants.
@@Rzko U can do this when Everything is finite , I mean when the sum is finite , but when It's a series (infinite sum) , you need more argument : you need to know if the sum is converging , how it's converging in order to switch the sum with the integral
@@Rzko And yeah Thank you , I did understand the following steps
@@tueur2squall973 are you sure about that? An infinite sum is just the limit of a partial sum (idk if we say like that in english)
Your presentation of the solution always gets me. My best wishes to you and please please continue
Thank you, I will!
So good video !
Thanks so much!
I really enjoy these videos! Can't wait to start taking higher level maths in uni
I'm so glad!
4:15 When he said "we still have some exes lingering about' , I felt that
By looking at that integral, I instantly understood that I would not be able to solve it if I try.
*And I was not disappointed*
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Beautiful. I am so proud of myself that I solved it on my own.
Edit: Okay maybe I didn't solve it completely correct lol I messed up a 2^r and got the answer e instead of sqrt(e)........ that is fine right!?!?!
no
yes
Yesnât
Yes, making mistakes is good for the growth of math skills.
Great video, thanks Bri. What program are you using for the text?
I like that you get into the math immediately
Nice proof ! Now you just need to justify swapping the sum and the integral.. as it cannot always be done .
The sum converges to less than e^(u/2)
Great u make maths lucid
Such the one of the best teacher ever
Incredible explanation!
Glad you think so!
Beautiful!
Glad you enjoyed it!
That worked out so perfectly lmao
i have no clue what he is talking about but i still love it
cool integral, great video:D
Glad you liked it!
Amazing!
You are!
Very nice. Thanks.
Most welcome!
I can explain this integral just one word.
WOW
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The awkward moment when a solution is as pretty as the one presenting it.
You really should become a math professor....
Currently an instructor (no Phd)đ
@@BriTheMathGuy great!!!
This is awesome
This video is sooooo Good â€â€
from Ethiopia, Africa
Incredible!
Glad you thought so!
bro pls make more videos on putnam integrals .They are really interesting. Thank you in advance
Is there a Taylor series expansion that expands to the factor of (n!)^2?
Are there any other places where i could find an integral like this with two sums multiplied together in the integrand?
đ€©, it was a crazy integral, it involved power series gamma u sub,I want more integrals like this
Does the second series converge on 0 to INF though? I can't really see it because I suck at evaluating them functional series.
The second series absolutely converges.
If I call U(n) = {x^(2n)}/{[2^(2n)]*(n!)^2}, then you have:
U(n+1)/U(n) = {x^2}/{4(n+1)^2}, which converges to 0 as n -> infinity, for all x in (0, infinity).
By the ratio test, that series converges.
Always awesome like you are :-)
You're the best!
You say that to all of them.
For the first factor I did the following
pulled out x,
substituted u=-x^2/2
For the second factor i have got second order linear differential equation but not with constant coefficients
xy''+y'-xy=0
Second factor will probably be Bessel function
but when we get first factor Gamma function will be helpful
Nice, I was able to do this one! Really awesome integral
Great job!
This is so good
Glad you thought so!
can you cancel the x, in the same quick step where you cancel the 2^n?? since one of the limits of integration is 0 you would have a 0/0 in x, I'm not sure if it is allowed to cancel out the x there
this goes way beyond advanced level of that pesky JEE
Really? I thought the advanced JEE was the hardest test
hmmmm.. that integral can simplify like
Ê (1-x*exp(-x^2))*BesselI(0,x) dx
and BesselI(0,x) is modified Bessel Function of the First kind
Woah!!! Mind blown...
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Super!
Thank you! Cheers!
That was so enjoyable
I lost track for the first few times but I'm glad I understood this in the end :)
it's a tricky one! :) thanks for watching!
@@BriTheMathGuy yea your vids are quite interesting... Who knew a bio nerd like me would binge math questions some day... Thanks for ur efforts
I hear so much stuff about the Putnam being ridiculously hard, but every step here was the most obvious thing to do given the current stage. Like it's not something you just scribble down in a hurry, but it's something I imagine most mathematically experienced people could do. Lovely presentation though
That was quite an aesthetic one
Use double factorials. These are useful.
I love watching *other* people do integrals :)
Incredible
Glad you thought so!
Amazing.
Thank you! Cheers!
Awesome
at 3:30 u had a chance to turn that sum into e^2x*sum(n=0,infinity,1/2^2n)
The second part can also be written as (x^n)^2 / ((2^n)^2 * (n!)^2) and we can take the entire term into square like (x^n / 2^n * n!) ^2 which we can write as ((x^n/2^n)/n!)^2 = ((x/2)^n /n!) ^2 so we can put it into e^x form like (e^(x/2))^2 which basically is e^x.
You made a mistake. In general, given a sequence a_n, the sum of (a_n)^2 is not equal to (the sum of a_n)^2
that was pretty cool
Glad you thought so!
You did a HARD putnum problem in 6 minutes!
So impressed !
I think you are a genius !
He explained the solution in 6 minutes. No telling how long it took him to find the solution.
@@magicmulder no but at least he is a genius .........
@@aashsyed1277 He is very good, but most math students could solve that one. Genius is rare. Very rare.
I like your videos very much. One tiny suggestion though- can you slow down your speed while explaining such problems. You go very fast, which is problematic to understand what you are saying. I mean, before even I understand the concept you told, you move to another concept.
why echo?
Sorry about that! It should be fixed in the future.
Why do we always end up at gamma in these types of problems đ,
I don't know! đ
Intimidating â€ïž
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I got the first sum. But was clueless about what to do with (n!)^2...
Subbing x^2/2=u was brilliant bruh
This one was very beautiful
Glad you thought so!
Is the echo on purpose ?
Had an issue while recording, Sorry about that! It should be fixed in the future.
Simplified results are beauty gives extraterrestrial vibes.
awesome
What's with the audio issues?
Sorry about that! It should be fixed in the future.
Nice answer
Glad you thought so!
it's crazy how something that looks absolutely nasty like this can simplify down into âe at the end
Mathematics always blows up my mind
đ€Ż
I liked this
Niceđ
Thanks â