Efficient Formula for $\zeta(3)$

Question

I was messing around with Desmos to find an exact form for $\zeta(3)$ and was using the following triple integral form: $$ \zeta(3)=\int_0^1{\frac 1 z\int_0^z{\frac 1 y\int_0^y{\frac 1 {1-x}}}}dxdydz $$ I set it to $a$ in Desmos, and evaluated:

$$ \frac {\ln{\big(\frac \pi 2\big)}}{a^3} $$

To my surprise, Desmos actually gave me a fraction option for the value, which was $\frac {238072} {915685}$. Skeptical, I searched Wolfram Alpha for: $$ \sqrt[3]{\frac {915685} {238072}\ln{\big(\frac \pi 2\big)}}-\zeta(3) $$ and it gave me an error of $\approx2.098\times10^{-16}$. So I only found an approximation, not an exact form, which doesn't surprise me.

This begs the question: How can I more accurately represent $\zeta(3)$ in Desmos? Any thoughts would be much appreciated.

Answer 1

Desmos probably utilized a continued fraction expansion of the machine precision value of $\frac{\log \pi/2}{\zeta(3)^3}$ to obtain an efficient rational approximation. Specifically, the sequence of convergents is:

$$\left\{0,\frac{1}{3},\frac{1}{4},\frac{6}{23},\frac{7}{27},\frac{13}{50},\frac{774}{2977},\frac{787}{3027},\frac{1561}{6004},\frac{78837}{303227}, \color{red}{\frac{238072}{915685}}, \frac{2086065701}{8023535197}, \ldots \right\}$$

The next convergent in the sequence yields an absolute error of about $1.2 \times 10^{-20}$.

For whatever reason, you chose the function $$f(a) = \frac{\log \frac{\pi}{2}}{a^3}$$

from which to compute a rational approximation, but other functions can also be chosen. For instance, zeta functions evaluated at positive even arguments are rational multiples of a corresponding even power of $\pi$; e.g.,

$$\zeta(2) = \frac{\pi^2}{6}, \quad \zeta(4) = \frac{\pi^4}{90}, \ldots.$$

I believe it remains an open question if odd arguments of the zeta function show similar phenomena, i.e., if they are rational multiples of an odd power of $\pi$. To investigate, we could compute the convergents for the choice of function $$g(a) = \frac{a}{\pi^3}$$ at $a = \zeta(3)$ which gives $$\left\{0,\frac{1}{25},\frac{1}{26},\frac{4}{103},\frac{5}{129},\frac{34}{877},\frac{107}{2760},\frac{141}{3637},\frac{248}{6397},\frac{637}{16431},\frac{885}{22828},\frac{9487}{244711},\frac{29346}{756961}, \ldots \right\}$$ but computing additional terms requires increasingly high numerical precision of $\zeta(3)$ and $\pi$. If such a rational number existed, the sequence of convergents would terminate.

---

I should also point out that the convergent $\frac{238072}{915685}$ happens to be an "unusually good" approximation in the sense that the size (number of digits) of the denominator is small compared to the number of decimal places the convergent matches the precise value. This is observable from the continued fraction expansion

$$\frac{\log \pi/2}{\zeta(3)^3} = \frac{1}{3 + \frac{1}{1 + \frac{1}{5 + \frac{1}{1 + \frac{1}{1 + \frac{1}{59 + \frac{1}{1 + \frac{1}{1 + \frac{1}{50 + \frac{1}{3 + \frac{1}{\color{red}{8762} + \frac{1}{\ddots}}}}}}}}}}}}$$

When the expansion term has a large value, this means it gets "close" to being a rational number. So the aforementioned convergent--the one that Desmos found--is the one that stops just before $8762$; i.e., it simply lets the part $$\frac{1}{8762 + \frac{1}{2 + \frac{1}{7 + \frac{1}{\ddots}}}} \approx 0.$$

To give you an idea of just how good this approximation is, $8762$ is the $11^{\rm th}$ term in the continued fraction expansion (not counting the initial $0$ since we are dealing with a number that is less than $1$). The next time we see a number in the expansion sequence that is at least as big is at the $9231^{\rm st}$ term, when it is $18858$. I am not sure how much precision is attained for the corresponding convergent.

Answer 2

We grab the bull by the horns and work with the continued fraction for $\zeta(3)$ itself.

The partial quotients are available from OEIS A013631, and from these we get the convergents below.

$\dfrac11,\dfrac54,\dfrac65\dfrac{113}{94},\color{blue}{\dfrac{119}{99}},\dfrac{232}{193},\dfrac{351}{292},\dfrac{1636}{1361},\dfrac{1987}{1653},\color{blue}{\dfrac{19519}{16238}},\dfrac{177658}{147795},\dfrac{374835}{311828},\dfrac{552493}{459623},\dfrac{927328}{771451},\dfrac{1479821}{1231074},\color{blue}{\dfrac{3886970}{3233599}},...$

Certain of these are colored blue. If we regard an integer in a computer as being represented by $k$ bytes and one of the bits is used for the sign, then with one-byte representation we can use numbers up to $2^7-1$, wuth two-byte representation we can go up to $2^{15}-1$, and with three bytes available we can go to $2^{23}-1$. The blue numbers represent the maxium numerator and denominator within these respective bounds.

So how did we do? Below are the values of the blue fractions, rendered to as many decimal places as would match the exact value of $\zeta(3)$. Thus if we allow $15$ bits plus the sign bit (two bytes), the accuracy to eight decimal places actually corresponds to about $25$ bits in base two. The three-byte limit gives about $40$ bits of accuracy.

$119/99 \approx 1.202$ (the next decimal place is correctly $0$, but the rounded number to four decimal places is $1.2020$ and $\zeta(3)$ rounds to $1.2021$)

$19519/16238\approx 1.20205690$

$3886970/3233599\approx 1.2020569031596$