That was great! There was definitely some work in it, but any time you can use Feynman's trick and Laplace transforms in one integral you know it's going to be fun! I probably shouldn't even bring this up because every time I do calculus professors tell my I am vile and despicable and a conspiracy theorist, but I really think it is a shame that these "advanced" techniques are kept secret by these power hungry professors! ; )
Beautiful! One tiny comment - I would explicitly mention Laplace Transform (evaluated at S=1) already at 02:50. Think it would give some intuition as to where we're going😊
@@owl3math Already at 02:50 you have f(S) the Laplace of (u-sin u) /u^2, so it's helpful to mention it (actually you can simply use the properties of Laplace and avoid Feynman's) 🍻
Great question but I'm not sure. I checked the website and they mention: "30 problems in 2 hours for a team of 3". This is like another math problem: if it takes 3 students to bake 3 cakes in 3 hours how long does it take them to do an integral?
Yea but at 3:25 you need to show that your function inside your integral is uniformly convergent in order to put the limit inside the integral and calculate it to 0
@@EvTheBadConlanger by using the definition but its not that easy sometimes. Maybe in this case it would be easier to use the dominated convergence theorem. I am self though on these subjects so i dont know a lot but there are good guides on both internet and yt
What happens in the situation where when you plug in f(infinity)=0 you get C=infinity, then it can be any value involving s that cancels with the part that diverges right, or is this situation impossible.
That's a good point. I was working on UK #14 and I think it is that exact scenario. I get the right answer but I don't know that I have the full justification because at that point we are using Feynman's trick when the integral is divergent. Kind of troubling :(
A beauty of process
thank you :)
Nice problem ❤
thanks!
That was great! There was definitely some work in it, but any time you can use Feynman's trick and Laplace transforms in one integral you know it's going to be fun!
I probably shouldn't even bring this up because every time I do calculus professors tell my I am vile and despicable and a conspiracy theorist, but I really think it is a shame that these "advanced" techniques are kept secret by these power hungry professors! ; )
Ha!!! Good point! These are the tricks they don’t want you to know 😂
Beautiful! One tiny comment - I would explicitly mention Laplace Transform (evaluated at S=1) already at 02:50. Think it would give some intuition as to where we're going😊
Thanks! At that point I say our goal is f(1). You mean like make it more clear that s = 1?
@@owl3math Already at 02:50 you have f(S) the Laplace of (u-sin u) /u^2, so it's helpful to mention it (actually you can simply use the properties of Laplace and avoid Feynman's) 🍻
Nice problem. What was the time limit on problems in this competition?
Great question but I'm not sure. I checked the website and they mention: "30 problems in 2 hours for a team of 3". This is like another math problem: if it takes 3 students to bake 3 cakes in 3 hours how long does it take them to do an integral?
further complicating it the document I have has only 20 problems 🤨
Yea but at 3:25 you need to show that your function inside your integral is uniformly convergent in order to put the limit inside the integral and calculate it to 0
How would you do that? /genuine
@@EvTheBadConlanger by using the definition but its not that easy sometimes. Maybe in this case it would be easier to use the dominated convergence theorem. I am self though on these subjects so i dont know a lot but there are good guides on both internet and yt
@@alexkaralekas4060Your point on dominated convergence as a sufficient condition to differentiate under the integral sign is fair.
What happens in the situation where when you plug in f(infinity)=0 you get C=infinity, then it can be any value involving s that cancels with the part that diverges right, or is this situation impossible.
That's a good point. I was working on UK #14 and I think it is that exact scenario. I get the right answer but I don't know that I have the full justification because at that point we are using Feynman's trick when the integral is divergent. Kind of troubling :(
@@owl3math ah you make a fair point
The Feynman's trick would only apply on integrals that are convergent. Otherwise differentiating under integral sign would be invalid.