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.
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.
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)
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
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!?!?!
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
The second power series (1 + x²/2² + ...) 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.
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
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!!
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
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.
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
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.
This is really dangerous from a rigour perspective. The 2nd power series leads to dominated convergence issues that most will just hand wave away, but they are there.
3:35 HOW CAN THIS APPEAR ON MY RECOMMENDED AFTER I HAVE JUST MISTAKEN EXACTLY THAT THING AT THE TUESDAY TEST!??!?!?! THAT EXACT du=x*dx IS where u=(x^2)/2 IS THE EXACT THING I MISSED! I didn't realise x can be written as the derivative of (x^2)/2 at the test and i magically passed it even with that mistake.
