The Basel Problem Part 1: Euler-Maclaurin Approximation

This is the first video in a two part series explaining how Euler discovered that the sum of the reciprocals of the square numbers is π^2/6, leading him to define the zeta function, and how Riemann discovered the surprising connection between the zeroes of the zeta function and the distribution of the primes, leading ultimately to his statement of the Riemann Hypothesis. This video focuses on how Euler developed a method to approximate this sum to 17 decimal places, as well as how the Bernoulli numbers naturally appear as part of this problem.

Theme music by Keith Welker.

[ericbischoff]

32:39 the Bernouilli coefficients Bk(0) and Bk(1) are wrong for k = 1 (signs are inverted).

Thanks for the great contents.

[KarlFredrik]

It's hard to understand how smart Euler was. Always breathtaking when his derivations are worked through

[maxdodson]

I remember that a year or so ago I stumbled onto this video and, although it was amazingly put together, I was completely lost. Returning to this after completing an introductory calculus course and understanding the arguments is a positive experience like no other. Thank you so much for this resource!

[JakeFace]

5:40 Why did they calculate the decimal expansion of pi^2/6? Why did Euler know that decimal expansion so well as to recognize it?

[loosebet]

How the heck Euler figured out that 1.644... is pi^2/6?

[AntonBourbon]

Thank you very much for an interesting video digging deeper than the others!
48:11 : You " _can in no way imagine this sum to be calculated by hand to 20 decimal places_ "? I just timed how much it would take me to calculate it (using a pen and paper). Took 1 hour 13 minutes, including going to the kitchen to heat my tea in a microwave 3 times. I have never been fast, and I stopped calculating on paper some 30 years ago when calculators became ubiquitous. My point is, I would not be surprised if this can be done in 20 minutes if you practice calculations by hand all the time. (You could miss the fact that all the fractions involved have relatively "easy" denominators, that is, the decimal fractions have a relatively short period:
1/6 = 0.1(6)
1/30 = 0.0(3)
1/42 = 0.0(238095)
1/30 = already calculated above
5/66 = 0.0(75)
691/2730 = 0.2(531135)
7/6 = 1 + already calculated above
3617/510 = 7.0(9215686274509803) -- the most difficult division here, as 1/51=1/(3*17), and 1/17 has a period of whole 16 digits
43867/798 ~= 54.971... -- I never figured out the period, as this is divided by 10^19, and you only need 4 digits (54.97) to have 21 digits in the result.)

[Mizziri]

The triangle area proof is pretty simple with a telescoping series, but that visual demonstration is so much more elegant.

[littlenarwhal]

Very interesting video: the length is not at all a deterrent considering how well explained and entertaining this is.

[jaeimp]

@6:35 As the triangles slide to the left against the y-axis it becomes clear that all triangles have a base of 1, and therefore the sum of the area of all infinite triangles is 1/2 times 1 times the sum of the heights of each one of the infinite triangles. Now the heights of all infinite triangles add up to 1, as geometrically shown on @6:45.

[ytkerfuffles]

On 35:50, basically everything must be symmetrical by induction. even cases are like a U shape always and odd cases are like a ^u kind of up down up shape, and symmetry carries over to the next case when you integrate, because the second half and first half of the interval alternates between being the mirror image horizontally and the mirror image both horizontally and vertically of the second half of the interval. By symmetry and the fact that the start and end values of the polynomials at 0 and 1 are the same, if said value is not 0 then the odd cases will not have an integral of 0 as the left and right half will not have areas that cancel.

[revaldonkwinika]

Most underrated youtube channel. Math does not have to be difficult.

[a.e.]

This is real great stuff. Many thanks for it from Germany!!

[richardfredlund]

wow... that was really fascinating. I'm not a professional mathematician but like maths so actually did an investigation into sum of 1/x^n .. a few years ago and that was indeed the first time I came across the Bernoulli numbers. (they popped out when I inverted this matrix ( very closely related to pascals triangle)) . So then I found them on wikipedia but didn't make much sense of it. Really this video makes it much more clear what these things are (than any other i've seen). Thank you.

[xyzct]

That was wonderfully entertaining! Thank you, sir!!

[tricycle]

I just happened to see this today. I have wanted to give a talk on this for over ten years but could not make it clear enough for a math circle. Now I don't have to. Thanks

[radonkule]

pleaaaasee make more of these or anything that has nice. mathematical objects like infinite series and stuff like that

[wdyuyi]

great video!
25:18 the "special A" was written identical as "area A", be careful to distinguish between them.

[seanoneill]

Way to go, thanks, you create access to “next level” ideas

[rdbury]

2:40 - There is some missing logic here. By the same argument you get 1 + 1 + 1 + 1 + ... = (-1 + 2) + (-2 +3) + (-3 + 4) + ... = -1 + (2 - 2) + (3 - 3) + ... = -1. Telescoping sums only work if the final term approaches zero. Nowadays you could compute the sum in a spreadsheet, look up the the first few digits in OEIS (A013661 btw) and boom, pi^2/6. Sometime I wonder how much math is missed because we have computers for computations. But I guess without computers no one would have ever heard of the Mandelbrot set so it more than balances out.

[qsfrankfurt]

Please can we get a lot, lot more videos on this channel?