I don’t know how much you’ll enjoy this post, but calculating values of the zeta function is one of those things that struck high-school me as being incredibly difficult and requiring mastery over the arcane arts of mathematics, so it’s pretty cool to know that I can do this now 1. Without further ado, let’s find a formula for $\zeta(2k)\dots$ 2
Let’s start with Parseval. Recall that we have a nice theory of Fourier analysis on the circle $S^1$ for the space $C^2(S^1)$ of twice differentiable (with continuous second derivative) functions. Note furthermore that viewing circle functions as functions $f:[0,1]\to\C$ with $f(0)=f(1)$ gives an embedding $C^2(S^1)\into C^2([0,1])$. Note that the set $\bracks{e_n}_{n\in\Z}$ where $e_n(x)=e^{-2\pi inx}$ is orthonormal with respect to the following inner product on $C^2([0,1])$:
We claim (without proof) that the (linear) span of $\bracks{e_n}$ is dense in $C^2([0,1])$, letting us make a limit argument to show that for any $f\in C^2([0,1])$, one has
The comment about embedding $C^2(S^1)$ into $C^2([0,1])$ is meant to help you recognize that $\angles{f,e_n}$ is exactly the $n$th Fourier coefficient of $f$. Our strategy for calculating $\zeta(2k)$ will be to find some function $f\in C^2([0,1])$ whose Fourier series has coefficients (roughly) of the form $c_n=\frac1{n^k}$ and then apply Parseval’s above identity.
The functions we’ll use are the (Jacob) Bernoulli polynomials $B_k(x)$ defined by the following identity 3
In particular, $\int_0^1B_k(x)\dx=0^k$ (Here, $0^0=1$). Note that one can show that $B_k’(x)=kB_{k-1}(x)$ 4, and for good measure, one can calculate
and so on. Note that this derivative relation gives
for $k>1$ whereas $B_1(1)-B_1(0)=1$. Let the $k$th Bernoulli number $B_k$ be $B_k(0)$ (some authors use $B_k(1)$ instead). With this, we’ve done enough set up, so let’s get to the zeta stuff. 5
Define $c_k(n):=\int_0^1B_k(x)e^{-2\pi inx}\dx$, the $n$th Fourier coefficient of $B_k(x)$, and calculate
for $k>1$ (since the product vanishes) while $c_1(n)=-1/(2\pi in)$ (since the integral vanishes). This gives
where the $\cdots$ indicates that an induction argument is taking place behind the scenes. Now, the careful reader should be up in arms with the above equalities because they only hold for $n\neq0$ (this is needed for the integration by parts to work as claimed). By definition of the Bernoulli polynomials, we have $c_k(0)=0$. At this point, Parseval says
The left hand side simplifies as
so our fabled $\zeta$ finally appears. To calculate the right hand side, integrate by parts (and induct) some more:
Putting these two sides together gives
In particular, plugging in $k=1$ recovers the classic
which is a good sign 6. Furthermore, I haven’t actually checked this myself yet, but I’m pretty sure that using the functional equation for the $\zeta$ function along with this calculation let’s you show that $\zeta(1-2k)\in\Q$ for all $k$ (and so $\zeta(1-k)\in\Q$ for all $k$ since it’s 0 when $k$ is odd) which is quite surprising. To end, I’ll really indulge a high-school fantasy by “proving” everyone’s favorite “theorem.”
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And also nice to know that I didn’t have to master any arcane arts to do so ↩
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using an argument that’s rigorous modulo me playing fast-and-loose with the definition of Bernoulli numbers/polynomials and with proving Parseval’s identity ↩
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Exercise: prove that this defines a unique (degree k) polynomial ↩
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A perhaps suggestive observation is that this relation is held by the polynomials B_k(x)=x^k ↩
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Read: Let’s do some GRE prep by integrating by parts ↩
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A better sign is that the general formula we calculated here is the same one appearing in Wikipedia ↩