Finding a closed form for ζ(4)

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[manstuckinabox]

And that's a good what? AND THAT'S A GOOD WHAT PROFESSOR?!?

[michel_dutch]

You know you're a math nerd when you know this value by heart.

[nHans]

1:45 "Little Twiddle" — thanks! Until now, I did not know the technical term for the "~" symbol. Always a good day when you learn something new!

[MikeOxmol_]

You can also calculate ζ(4) using Fourier series of f(x) = π^2 - x^2 on [-π, π], and then using Parseval's theorem which relates sum of squares of coefficients with the square of the original function. Parseval might be a cool idea for a future video.

[emilwrisberg]

I have used Michael's approach to find a recursive formula for zeta(2*m), where m is any positive integer. Thank you Michael!

[robertlunderwood]

The good place to stop was obviously before "a good place to stop".

[JustinWilsonPhysics]

I'll never know if that was a good place to stop or not. The agony.

[lexinwonderland]

I love the animation, it's definitely a nice fresh take! Great vid as always:)

[Woah]

Intresting think is that zeta of a even number is always equal to a rational number(it's pi to the power of the imput divided by a number) but it's odd form has no closed form

[nickruffmath]

I paused at 4:20 and did the calculation myself. The difference was I took x=0 instead of x=pi at the evaluation step. It still works and I got the same answer in the end! However, there was some extra computation...

Choosing x=0 turns the cos(nx) into 1 but it doesn't eliminate the (-1)^n . It's easy to fix by splitting the sum into pairs of adjacent odd and even values of n.

This led me to a general formula for the alternating version,

Sum(n = 1 to inf) of (-1)^n / n^(2k)

splitting and re-indexing means it's equal to:

Sum(n = 1 to inf) of ( -1/(2n-1)^(2k) + 1/(2n)^(2k) )

In other words, negative of the odd powers, plus the even powers.

Set S = zeta(2k)

E = just the even powers
Factor out the 1/2^(2k) and you just get
E = S / 2^(2k)

Then the odd powers are S - E

So the alternating version = -(S-E) + E = 2E - S = S ( 2 / 2^(2k) - 1)

= S * (2^(1-2k) - 1)

In other words
AlternatingZeta(2k) = Zeta(2k) * (-1 + 1/2^(2k-1))

For example,
The Basel problem k=1 is zeta(2) is pi^2 over 6

The alternating version (-1)^n / n^2
Sums to: pi^2 / 6 times (-1 + 1/2^1) or -pi^2 over 12

The alternating version of zeta(4) is -7/8 times pi^4 over 90

Because -1 + 1/8

[dj-maxus]

Awesome edits! The resulting pace of the video feels great

[Noam_.Menashe]

The easiest way I found to find zeta(2n) without some sort of formula are the polygamma functions and their reflection formulae. It does require you to take many derivatives though.

[bjornfeuerbacher]

@Michael: Could you please do a video on the connection between the values of zeta(2n) and the Bernoulli numbers?

[Kurtlane]

Wonderful.

Now could you please do zeta (3)?

[shyaamganesh]

Another way: we can equate the coefficient of x^4 in the traditional method of finding zeta(2) using the idea: roots of sin x/x. So now, zeta(4)=(zeta(2))^2-2*<the sum that we calculated above, it will come out to be pi^4/120>. So, zeta(4)=pi^4(1/36-1/60)= pi^4/90.

[CamiKite]

It's also possible to obtain ζ(4) Euler's way by "expanding" the 5th order coefficient of the product wich is also x^5/120 (taylor serie)

[nirajmehta]

no idea why, but blackpenredpen's picture popping up made me chuckle. great video as always!

[Ensivion]

this method can be used to find higher even numbers of the zeta function.

[jacob]

I love the new animations throughout the video.

[skb]

By using repeated integration by parts, this can be generalized to the Fourier series of x^(2n) on [-pi, pi] and it yields a recursive formula for zeta(2n) in terms of all previous zeta(2k) values. If we define z_n to be the coefficient of pi^(2n) in zeta(2n), we get a recursive formula for a sequence of rational numbers related to the Bernoulli numbers.