Hearted Youtube comments on Mathologer (@Mathologer) channel.
-
39
-
39
-
39
-
39
-
39
-
38
-
38
-
38
-
38
-
38
-
38
-
38
-
38
-
38
-
38
-
38
-
38
-
38
-
38
-
37
-
37
-
37
-
37
-
37
-
37
-
37
-
37
-
37
-
37
-
37
-
37
-
37
-
36
-
36
-
36
-
36
-
36
-
36
-
36
-
36
-
36
-
36
-
36
-
The general formula for the sum of 1/x^(2n) is hidden in here too, and only requires a few extra steps. Start from the chapter 2 formula and take logs and derivatives to get the formula at 10:40:
cot(x) = 1/x + 1/(x-pi) + 1/(x+pi) + 1/(x-2pi) + 1/(x+2pi) + ...
Move the 1/x term to the left side, and take 2n-1 more derivatives. The k'th derivative of 1/x is (-1)^k * k!/x^(k+1), so when k = 2n-1 this is -(2n-1)!/x^(2n). So we have
(d/dx)^(2n-1) (cot(x) - 1/x) = -(2n-1)! * (1/(x-pi)^(2n) + 1/(x+pi)^(2n) + ...)
Now divide by -(2n-1)! and take the limit as x -> 0.
1/(2n-1)! * lim x -> 0 [(d/dx)^(2n-1) (1/x - cot(x))] = 1/(-pi)^(2n) + 1/pi^(2n) + 1/(-2pi)^(2n) + 1/(2pi)^(2n) + ...
On the right side, all the negatives are squared away, and we end up with 2 copies of each term. So multiply by pi^(2n)/2, and we get this:
pi^(2n)/2 * 1/(2n-1)! * lim x -> 0 [(d/dx)^(2n-1) (1/x - cot(x))] = 1 + 1/2^(2n) + 1/3^(2n) + ... = zeta(2n)
This formula looks messy, but there's a trick: notice that on the left, we have something of the form 1/k! * k'th derivative of f(x) at x=0. These are just taylor series coefficients! The left side is really just pi^(2n)/2 times the coefficient of x^(2n-1) in the taylor series of 1/x - cot(x).
There are a few other things we can do to make the formula easier to read. We can multiply the function by x to make the powers line up nicely (otherwise the 1/k^8 sum will be related to the coefficient of x^7, instead of x^8). This gives:
zeta(2n) = pi^(2n)/2 * coefficient of x^(2n) in the taylor series of 1 - x cot(x)
The next thing we can do is move the pi^(2n) "inside" the taylor series, by replacing x with pi x. We can also move the factor of 1/2 into the function. Then we get:
zeta(2n) = coefficient of x^(2n) in the taylor series of (1 - pi x cot(pi x))/2, or equivalently,
(1 - pi x cot(pi x))/2 = sum n=1..inf, zeta(2n)x^(2n)
And indeed, if you ask wolframalpha to compute the taylor series of (1 - pi x cot(pi x))/2, you get pi^2/6 x^2 + pi^4/90 x^4 + pi^6/945 x^6 + pi^8/9450 x^8 + ...
Finally, comparing this series to the standard taylor series for cot in terms of Bernoulli numbers gives Euler's general formula for zeta(2n)
35
-
35
-
35
-
Since I made it to the end of this, and since you asked nicely: I'm an enthusiastic amateur, never took much beyond AP calculus formally but I read & study the subject pretty broadly as a hobby.
I was cruising along smoothly here right up until about chapter 4 or so, after which there was a lot of pausing, rewinding, workings-out on paper or Python, and a couple side trips into Wikipedia & Wolfram.
The Pascal/Bernoulli thing was mind-blowing to see in action. It definitely made sense on the surface, though as it got deeper i started to feel like it was going in circles? Something like, "We can easily derive S using B. But how do we get B? Well by deriving it from S of course." At least that was my initial impression.
Really enjoyed it even if it did start to outpace me towards the end (or more accurately, because it did). It'll be a few more viewings before everything clicks for sure, but I'm looking forward to the challenge!
35
-
35
-
35
-
Hi, I did find another way to "build" youre cubes. Instead of adding the hexagons, you can add 1+ (1+2+3), than add 7 + (3+4+5), and 19+ (5+6+7), ...
If you look in the video at 8:39 you see the second cube. To go to the next cube you need to add one side, so 3 pieces, lets say on the left, then you add 4 pieces to the right side (3 pieces for the original cube and one piece for the just added row), and finaly you add the bottom with 5 pieces to get back to the cube.
Now if you look at the numbers you add they are left and right from the yellow marked number in the top row (2 and 4 sandwidch the 3; 5 and 7 the 6,...)
And these yellow numbers are diviceble by 3 (by construction)
So you get a sequens of (3n-1)+(3n+1) and that is the same as 2(3n).
We could write 2(3n) also as 3(2n).
This we can convert in (2n-1)+(2n)+(2n+1).
And this is the secuence you are adding to youre cube.
(Note "n" is the base number from youre last cube, and the counting number of youre tripple (where 3 is the 1st; 6 the 2nd;9 the 3th;12 the 4th,..))
Notice that by every step you go up you start with the same number you stopt with the last time.
(2n+1)=(2n+2-1)=(2(n+1)-1)
So the cube you end up with has a sidelenght of 2n+1. And you start the next step just adding that side to youre cube
I hope you can see how i find this a bit easyer to construct the cube.
Also this is in a way the same you construct youre squares (if you start with "skipping" just one number at the top) if you notice that every odd number is the sum of its position + its position -1. (5 is the 3th odd number and it is 3+2; 9 is the 5th odd number and it is 5+4; ...). Also here you get that the last digit you added the last time you start to add the next time. (but you only get 2 numbers).
Note; this methode does not work in the fourth or higher dimensions. Why? Well I'm not a mathematitian. So I don't know. And I have trouble imagining in four dimentions.
Hope this is helpfull.
Sorry for my bad English, I speak Dutch.
Greetings from Belgium.
35