This was fantastic, thank you! The way you convey those crazy (& tedious) book-keeping steps in a neat and comprehensible manner while still maintaining the rigor is truly exceptional.
At around the 11:00 mark, when you where trimming away the edges of the sum, is there a reason you didn’t also trim off the j=1 term? This would increase the coefficient of the c_(k-1) to 7 and raise the lower limit of the sum to j=2. (Maybe you address this later; I’m currently at the 11:00 mark and wanted to comment this while it was on my mind).
Ah, I thought about discussing that in the video, but decided against it! The problem with doing that is it only works when k >= 3. If k = 2 then the sum is already empty, so the coefficient "5" doesn't actually become a "7". But then it would become "7" when k >= 3. So you can do it, but it leads to a slightly uglier (in my opinion) situation where you need explicit values for c0, c1, and c2, and then you have a recurrence involving "7c_{k-1}". Whereas mine you need explicit values for c0 and c1, and then you have a recurrence involving "5c_{k-1}". Either way works though. (There were a lot of choices like this that I made throughout the video. Most of them are arbitrary, but I made my choices to try to make the math work out as cleanly as possible.)
At the very beginning when you are defining c_k, you define -b as(1-3x) but then inside the sqrt you define b^2 as (1-3x)^2 . Should you not define it as (3x+1) then instead distribute the -1 to the other term (4ac)? Genuinely asking bc im not sure haha
(-b)^2 = b^2, so I didn't worry about the negative sign on b in the b^2 - 4ac part. To perhaps make it a bit clearer though: since -b = 1-3x, we have b = 3x-1 (not 3x+1). Then b^2 = 9x^2 - 6x + 1 (and then the rest of the calculation is the same as what I showed).
This was fantastic, thank you! The way you convey those crazy (& tedious) book-keeping steps in a neat and comprehensible manner while still maintaining the rigor is truly exceptional.
Your applied calculus set of videos is really helping me understand math better sending thanks from the U.S
At around the 11:00 mark, when you where trimming away the edges of the sum, is there a reason you didn’t also trim off the j=1 term? This would increase the coefficient of the c_(k-1) to 7 and raise the lower limit of the sum to j=2. (Maybe you address this later; I’m currently at the 11:00 mark and wanted to comment this while it was on my mind).
Ah, I thought about discussing that in the video, but decided against it! The problem with doing that is it only works when k >= 3. If k = 2 then the sum is already empty, so the coefficient "5" doesn't actually become a "7". But then it would become "7" when k >= 3. So you can do it, but it leads to a slightly uglier (in my opinion) situation where you need explicit values for c0, c1, and c2, and then you have a recurrence involving "7c_{k-1}". Whereas mine you need explicit values for c0 and c1, and then you have a recurrence involving "5c_{k-1}". Either way works though.
(There were a lot of choices like this that I made throughout the video. Most of them are arbitrary, but I made my choices to try to make the math work out as cleanly as possible.)
At the very beginning when you are defining c_k, you define -b as(1-3x) but then inside the sqrt you define b^2 as (1-3x)^2 . Should you not define it as (3x+1) then instead distribute the -1 to the other term (4ac)? Genuinely asking bc im not sure haha
(-b)^2 = b^2, so I didn't worry about the negative sign on b in the b^2 - 4ac part. To perhaps make it a bit clearer though: since -b = 1-3x, we have b = 3x-1 (not 3x+1). Then b^2 = 9x^2 - 6x + 1 (and then the rest of the calculation is the same as what I showed).
@@NathanielMath Makes sense, thank you! Loved watching the video, outlined the process and made it super easy to follow