Yikes, I worked on this problem, too! I thought it was not a problem at first, but it just kept getting more problematic as I proceeded. I could not fathom doing this under a time constraint.
For an intuitive method of tackling the last step (the infinite product), and most likely the method they used in the competition, you can separate it into product[(n+1)/(n-1)] * product[(n^2-n+1)/(n^2+n+1)] and make it the product from 2 to K where K -> infinity. Then, you can see that the first part of product will be ((K+1)!/2)/(K-1)! = (K^2+K)/2. For the second part of product, Define the sequence a_n = n^2 - n + 1 and you’ll see that a_(n+1) = n^2 + n + 1 so the second part equals product[ a_n / a_(n+1) ] and telescoping the product yields (a_2)/(a_(K+1)) = (4-2+1)/(K^2 + K + 1) = 3/(K^2+K+1). Putting the two together yields Product = lim K -> +inf [ 3(K^2 + K)/2(K^2 + K + 1) = 3/2
@@ronbannon I’m still in high school so I don’t have a good deal of experience using mathematical notation over the internet / in forums such as stack exchange
Absolutely the best integral video and explanation I've watched this year
Haha thanks a lot my friend for your support👍👍👍
Yikes, I worked on this problem, too! I thought it was not a problem at first, but it just kept getting more problematic as I proceeded. I could not fathom doing this under a time constraint.
Yikes, I tried this with an estimated (and anticipated time) of like 30 minutes but took me more than 2 hours uh
@@domedebali632 I followed a lot of deadends myself. Finally figured it out.
Ohhh haha yes this took longer than I thought, too👍👍👍
For an intuitive method of tackling the last step (the infinite product), and most likely the method they used in the competition, you can separate it into product[(n+1)/(n-1)] * product[(n^2-n+1)/(n^2+n+1)] and make it the product from 2 to K where K -> infinity. Then, you can see that the first part of product will be ((K+1)!/2)/(K-1)! = (K^2+K)/2. For the second part of product, Define the sequence a_n = n^2 - n + 1 and you’ll see that a_(n+1) = n^2 + n + 1 so the second part equals product[ a_n / a_(n+1) ] and telescoping the product yields (a_2)/(a_(K+1)) = (4-2+1)/(K^2 + K + 1) = 3/(K^2+K+1). Putting the two together yields Product = lim K -> +inf [ 3(K^2 + K)/2(K^2 + K + 1) = 3/2
I'm not sure why RUclips doesn't support LaTeX-like input, as hints like yours would be way easier to read if it did. In any case, thanks for sharing.
@@ronbannon that's right. RUclips should support LaTeX or similar inputs
@@ronbannon yeah, sorry if it’s a bit messy. I hope it was still comprehensible!
@@ronbannon I’m still in high school so I don’t have a good deal of experience using mathematical notation over the internet / in forums such as stack exchange
@filipeoliveira7001 Wow, you're a great student. I was not complaining about your comment, I was noting that RUclips needs to support LaTeX!
Awesome video professor
Thank you so much my friend haha👍👍👍
You are the best professor
Thanks a lot my friend haha 👍👍👍
You are like integration God
Haha thanks a lot my friend👍👍👍
awesome. I love using sigma notations when doing integrals. Thanks Doc PK
Same here bro
Thanks a lot my friend👍👍👍
Sooo cooolll🎉
Thank you my friend👍👍👍
Ok good solution sir
When will you upload integral advanced techniques and method to solve
Thank you my friend for your comments?m! Pulling those up now👍👍👍