Proof by intuition done by Leonhard Euler, sum of 1/n^2, (feat. Max)

Sum of 1/n^2,
pi^2/6,

blackpenredpen,
math for fun,

[lucanalon]

To show b=1, just compare the constant terms in the two infinite-nomials

[funny_monke]

5:38 to skip the intro to polynomials and roots

[dsfdsfsfdsfdsfd]

Take limit as x->0 of both sides. Sinx/x approaches 1 (well known limit) and right side approaches beta*(1*1*1*1*...)=beta. So beta=1.

[Czeckie]

when you fiddle with the coefficient by x^4, you'll get sum of 1/n^4 = pi^4/90

[OonHan]

sin(x) isnt a polynomial but the Taylor series of it is a polynomial

[samferrer]

That was very elegant!!! It bridges algebra with analysis ... I like that ... He was all over the place ... but he will get better for sure ...

[tulliusagrippa]

It it does not terminate, , it is not a polynomial — it is a power series!

[zazkegirotron]

Recommendation: dude, if you are making a video about complex numbers, nth term sums and taylor series. obviously the audience knows about polynomials. explain everything or not explain.

nice video!

[SartajKhan-jgnz]

I watched this video so much that I just love the 'nice' at the end of the video.

[abdelmoulamsaddaq]

I like this démonstration
I like your team

[MrRyanroberson]

Multiply everything by pi, and then the sum is a bunch of circles with radius 1/n, and for n is all integers you get a cone like shape, whose area is proportional to pi*h/x, for h is the largest value of n, and x is something dependant on n... It ends up as h/x = pi^2/6

[Romulo_Cunha]

I just love your channel, thx so much for sharing all these cool videos. Cheers from Brazil

[copperfield]

I like this version more, as I don't have deal with some fancy theorem that I don't understand

[acerovalderas]

Beautiful proof! The three of you are grea

[YorangeJuice]

“Proof by intuition” gives the same vibe as using “this was once revealed to me in a dream” as a citation

[Handelsbilanzdefizit]

Some years ago, I found a proof that Pi²/6 is the same as:
e^(Σ_n Σ_p 1/(n*p^2n)) where 'n' are integers 1 to ∞, and 'p' are primes.
I think Euler did the same, but he had a lot more time ^^

[Prasen]

At 8:10, sinx = infinite factors it is correct, but a constant must be mutliplied like insteas of x(x-pi)(x+pi) ... it should be beta.x.(x-pi)(x+pi)... etc

[วิระพิทักษ์ถิร]

Very very very great and nice solutions that I have ever seen. thank you very much for your presentation.

[brianasgarian]

For the issue looking at the constant B does this make sense to anyone?

Note I’m using xn in place of n^2Pi^2

Keeping in mind that a polynomial can be factored as P(x)=B(x - x1)(x - x2)…(x - xn)

But in this form THE INFINITE PRODUCT does not converge. In the video the constant is left out for the infinite product for sin(x) when the instructor forms the polynomial after the Taylor series.

Bx(x − x1)(x − x2)(x − x3) . . . (x - xn) n → ∞.

The answer is no, because (x − xn) → ∞.

However we can modify it in the following way: factoring each xn and modifying B to B′
That is just dividing each factor on xn, and adjusting the constant B in front accordingly.

The first infinite product does not converge but the second does:

B’x(1 - x/x1)(1 - x/x2)…(1 - x/xn) n → ∞ converges

So the first B is undefined but B’ can be found by dividing both sides by x and looking at the limit of sin(x)/x where x -> 0.

[JuanDeLaCruz-wxpf]

I'd like to have friends like that some day, nice video <3