I saw your comment before I hit start, so I was waiting for him to say this... but then he said "one of my favorite *special* functions" and I knew he meant something different. Because why the floor function is indeed awesome, I knew it wasn't "special" in the mathematical sense of the term!
It is super interesting to see how Michael adjusts his approach as he goes, but also super confusing when he suddenly rewinds without warning. Would it be possible, even if slightly janky, to insert a "rewind" effect to demonstrate that the backwards jumps are deliberate rather than an editing error?
Here is a another approach to the second way , notice that the integral of dilog(x) from 0 to 1 = the integral of dilog(1-x) form 0 to 1 and use the remarkable identitiy dilog(x)+dilog(1-x)+ log(x) log(1-x)=zeta(2) so it suffices to evaluate the integral of log(x)log(1-x) from 0 to 1 , this can be easily calculated by using a little trick.
Interesting and instructive problem and solutions as usual, Professor Penn, but it seems there were some problems with the editing today...a couple of redundant bits left in possibly by mistake.
Christmas time at Michael's place. "Help yourself to a nut. There's a nutcracker there, or you can use one of my favourite tools - that steamroller outside."
This was fun; I've been looking for perspective on sums and (for me at least) this has just the right level of progression to help me pause and analyze what's familiar to others. Thanks 😊
Partial fractions is actually not a theorem about integration, but rather a theorem about the structure of rational functions. As such it comes up in many areas.
What about following sum sum((n choose 2m)(m choose k),m=k..floor(n/2)) Result should depend on n and k Wolfram Alpha does not calculate this sum correctly
Starting off saying he's going to use "one of my favorite functions", I was confused as to how the floor function was going to come into play.
I saw your comment before I hit start, so I was waiting for him to say this... but then he said "one of my favorite *special* functions" and I knew he meant something different. Because why the floor function is indeed awesome, I knew it wasn't "special" in the mathematical sense of the term!
Generating function is another one of his. I can attest to that.
@@wesleydeng71
Sure ,i was waiting for Generating function
For the interchange of integration and summation, the monotone convergence theorem can also be used since everything is positive.
It is super interesting to see how Michael adjusts his approach as he goes, but also super confusing when he suddenly rewinds without warning.
Would it be possible, even if slightly janky, to insert a "rewind" effect to demonstrate that the backwards jumps are deliberate rather than an editing error?
It would be so on brand with the humor of Stephanie - make it happen, please?
13:27
Here is a another approach to the second way , notice that the integral of dilog(x) from 0 to 1 = the integral of dilog(1-x) form 0 to 1 and use the remarkable identitiy dilog(x)+dilog(1-x)+ log(x) log(1-x)=zeta(2) so it suffices to evaluate the integral of log(x)log(1-x) from 0 to 1 , this can be easily calculated by using a little trick.
Interesting and instructive problem and solutions as usual, Professor Penn, but it seems there were some problems with the editing today...a couple of redundant bits left in possibly by mistake.
Christmas time at Michael's place. "Help yourself to a nut. There's a nutcracker there, or you can use one of my favourite tools - that steamroller outside."
That was very cool!
Thank you, professor!
Great, now do the same for sum 1/n³
Thank you MP for doing so many great math vids!
Some problem with the video cutting? Thatś the second time that I noticed a start-over in between😊
This was my favourite explanation of new bounds thank you so much
This was fun; I've been looking for perspective on sums and (for me at least) this has just the right level of progression to help me pause and analyze what's familiar to others. Thanks 😊
Partial fractions is actually not a theorem about integration, but rather a theorem about the structure of rational functions. As such it comes up in many areas.
Great to see my favorite math professor at large but I prefer the first solution 😅 Many thanks Michael for Your hard and inspiring work
The video was a little jumpy, but the ride was enjoyable.
What about following sum
sum((n choose 2m)(m choose k),m=k..floor(n/2))
Result should depend on n and k
Wolfram Alpha does not calculate this sum correctly
分かりやすいです
But it's in English though 😂
5:58 - 6:04 ?? Is that ment to be there
Absolutely! It was a little spoiler in case you were about losing attention ;-)
Harkens back to when he used to edit his own videos :)
* meant
Some heave time travelling this time around ;-)
it's looks like ζ(2)+ζ(3) !
but isn't
Amazing!
Hi,
And so, compared to the sum of the inverses of the squares, it only subtracts 1? not very glorious 😆
Editing problems.
I'm happy I got the solution ! This is not so easy on this channel !! Thanks for your cool problems.
I recently saw a similar video about calculating using the integral of the dilogarithm ua-cam.com/video/pmuFPRiNgjc/v-deo.html
I pushed the 2^8th like
there was no need to bring this horrible calculus into a simple problem
Michael's favorite trick is fucking awesome
Stop your major cursing, especially in a mathematics forum. It is rude, ignorant, and needless.