Don't forget to try the challenge problem at the end! For those of you who saw my approach to floor function integrals last time and want to get straight into the problem, jump to 3:46 where the summing begins. Thanks!
Thanks for your videos! I really appreciate them. Although I've already done my math exams, I like approaching problems of this kind and think your way of explaining your methods is great. Keep it up !
Thank you soooo much this is the coolest question i was ever able to solve and tbh the more i see the sigma functions the more I'm starting to love it i used to hate it😭
@@hokagedattebayo7623 tbh finding them really good so have lots of time to focus on extra stuff which is why i've got this channel! not sure what the AEA is
I tried reversing the order of summation and integration geometrically. With the vertical axis as k, based on the bounds for x and k, the "region" of integration are line segments from (0,k) to (k+1,k). Then the problem becomes sum from k=0 to infty of the integral from 0 to k+1 of the fraction dx. Evaluation then is easy. Not actually sure if this is valid in general, but I do this in programming problems sometimes. 😅
The way I like to think of reindexing double sums such as this is as follows: -- Let S = Sum[n=0 to inf: Sum[k=n to inf: f(n,k)]] -- Consider lattice points (n,k) in the 2D plane that are reached in the summation. In this case these points occupy the triangular region in the top right-hand quadrant on and above the line y=x. -- The inner sum evaluates f(n,k) for points on the line x=n. If instead the inner sum is evaluated for points on the line y=k, it is easily seen that: S = Sum[k=0 to inf: Sum[n=0 to k: f(n,k)]] This is in effect what you end up doing in the video.
@@OscgrMaths I remember a lecturer (when I was doing my maths degree in the 90's) doing such a reindexing without explanation which confused me greatly. I can't remember if another lecturer used the approach above or if I came up with it independently, but I have also now seen it used on other TH-cam videos.
@@franolich3 It's always frustrating when a concept is brushed over like that but great that you came to your own method either way. What was your favourite topic in your degree at the time?
@@OscgrMaths I wasn't very good at pure maths and was more interested in applied maths and theoretical physics anyway. So I enjoyed subjects like vector calculus, Lagrangian/Hamiltonian mechanics, statistical mechanics, quantum field theory, general relativity, numerical solutions of differential equations, differential forms. While this may sound impressive, the truth is I also struggled to gain a solid intuition with the physics. When doing a lot of physics on a maths course, you can get away with not being that good at maths or physics!
@@franolich3 I don't have anywhere as much experience with physics as you but I've always found that intuition is the hardest thing to get when doing physics. That's really interesting thanks for commenting!
there are way easier ways to evaluate that sum by itself, but since i have recently been working with some polylogarithms i saw it as twice the polylogarithm of first degree evaluated at 1/2, which by itself is ln(2),therefore the sum is 2ln(2)
For the challenge at the end, another way of doing it would be to say: S=sum from k=0 to infty of 1/(2^k(k+1)) S/2=sum from k=1 to infty of 1/(k2^k) e^{S/2}=prod from k=1 to infty of e^{1/k2^k} e^{S/2}=prod from k=1 to infty of e^{(1/2)^k/k} e^{S/2}=1/(1-1/2)=2 S/2=ln2 S=2ln2 (that being said I only took e to the power of both sides because I knew the answer would be 2ln2 beforehand)
i recognised the polylogarithm, which was quite surprising as we've not done it in school. im just giving my GCSEs but I'm super into math so I've delved a significant bit ahead of what we learn in school
@@advait8142 yeah i'm the year above and did the same - keep going with it! Have you looked at any contour integration? I'm thinking of doing a video next proving a result for the gamma function (reflection formula) using contour integration.
@@OscgrMaths thanks! i look forward to that video. i absolutely LOVE gamma function. i haven't really delved into contour integration and complex analysis, but i might catch up on that before checking out your video. keep up the great work!
Don't forget to try the challenge problem at the end! For those of you who saw my approach to floor function integrals last time and want to get straight into the problem, jump to 3:46 where the summing begins. Thanks!
For the problem at the end, it looks to be the series expansion of -ln(1-x)/x evaluated at x=1/2, so 2ln2!
Yes that's it - great spot with the -ln(1-x)/x there!
These videos are awesome! I appreciate how well you explain the solutions, making difficult problems look understandable.
Thanks so much! Really glad you enjoyed.
Thanks for your videos! I really appreciate them. Although I've already done my math exams, I like approaching problems of this kind and think your way of explaining your methods is great. Keep it up
!
Will do! Thanks so much for the feedback!
Very nice video! Love to see math lovers on TH-cam! :D
Thank you! Glad you enjoyed.
Thank you soooo much this is the coolest question i was ever able to solve and tbh the more i see the sigma functions the more I'm starting to love it i used to hate it😭
Glad you enjoyed it!! Thanks for the comment.
Great video, what a levels do you do?
maths, further maths, physics and chem, you?
@@OscgrMaths I do the same but Econ instead of chem, how are you finding them honestly? Are you taking the AEA
@@hokagedattebayo7623 tbh finding them really good so have lots of time to focus on extra stuff which is why i've got this channel! not sure what the AEA is
@@hokagedattebayo7623 yeah i do my handle is @oscgr_
I managed to get to the double sum, but couldn't get further. Your solution is absolutely brilliant !
Thank you! I was stumped by the double sum for
a while too.
Great video!
Thanks so much!
I tried reversing the order of summation and integration geometrically.
With the vertical axis as k, based on the bounds for x and k, the "region" of integration are line segments from (0,k) to (k+1,k).
Then the problem becomes sum from k=0 to infty of the integral from 0 to k+1 of the fraction dx. Evaluation then is easy.
Not actually sure if this is valid in general, but I do this in programming problems sometimes. 😅
For the challenge ,the sum is equal to integral from 0 to 1 (1/(1-x/2)) dx which evaluates to ln(4)
Perfect! That's such a nice way to approach it.
The way I like to think of reindexing double sums such as this is as follows:
-- Let S = Sum[n=0 to inf: Sum[k=n to inf: f(n,k)]]
-- Consider lattice points (n,k) in the 2D plane that are reached in the summation. In this case these points occupy the triangular region in the top right-hand quadrant on and above the line y=x.
-- The inner sum evaluates f(n,k) for points on the line x=n. If instead the inner sum is evaluated for points on the line y=k, it is easily seen that:
S = Sum[k=0 to inf: Sum[n=0 to k: f(n,k)]]
This is in effect what you end up doing in the video.
Wow that's so useful!! Thanks so much for the comment, I'll definitely try that approach next time.
@@OscgrMaths I remember a lecturer (when I was doing my maths degree in the 90's) doing such a reindexing without explanation which confused me greatly. I can't remember if another lecturer used the approach above or if I came up with it independently, but I have also now seen it used on other TH-cam videos.
@@franolich3 It's always frustrating when a concept is brushed over like that but great that you came to your own method either way. What was your favourite topic in your degree at the time?
@@OscgrMaths I wasn't very good at pure maths and was more interested in applied maths and theoretical physics anyway. So I enjoyed subjects like vector calculus, Lagrangian/Hamiltonian mechanics, statistical mechanics, quantum field theory, general relativity, numerical solutions of differential equations, differential forms. While this may sound impressive, the truth is I also struggled to gain a solid intuition with the physics. When doing a lot of physics on a maths course, you can get away with not being that good at maths or physics!
@@franolich3 I don't have anywhere as much experience with physics as you but I've always found that intuition is the hardest thing to get when doing physics. That's really interesting thanks for commenting!
there are way easier ways to evaluate that sum by itself, but since i have recently been working with some polylogarithms i saw it as twice the polylogarithm of first degree evaluated at 1/2, which by itself is ln(2),therefore the sum is 2ln(2)
That's it - so great that you recognised it that way!
For the challenge at the end, another way of doing it would be to say:
S=sum from k=0 to infty of 1/(2^k(k+1))
S/2=sum from k=1 to infty of 1/(k2^k)
e^{S/2}=prod from k=1 to infty of e^{1/k2^k}
e^{S/2}=prod from k=1 to infty of e^{(1/2)^k/k}
e^{S/2}=1/(1-1/2)=2
S/2=ln2
S=2ln2
(that being said I only took e to the power of both sides because I knew the answer would be 2ln2 beforehand)
That's such a nice way to do it
Nice problem ❤ from india
@@ayaanmza Thank you so much!
Math is becoming more and more of a programming language xD
is it 2ln(2)?
Yes! How did you do it?
i recognised the polylogarithm, which was quite surprising as we've not done it in school. im just giving my GCSEs but I'm super into math so I've delved a significant bit ahead of what we learn in school
@@advait8142 yeah i'm the year above and did the same - keep going with it! Have you looked at any contour integration? I'm thinking of doing a video next proving a result for the gamma function (reflection formula) using contour integration.
@@OscgrMaths thanks! i look forward to that video. i absolutely LOVE gamma function. i haven't really delved into contour integration and complex analysis, but i might catch up on that before checking out your video. keep up the great work!