Is this reasonable closed form approximation for the value of $\zeta(3)$ using trigonometric functions just a coincidence?

Question

I have found this expression for an approximate value of $\zeta(3)$:

$$\zeta(3)≅\sqrt{\zeta(6)}\left(\frac{1}{\cos \left(\frac{\pi}{18}\right)}+\tan \left(\frac{\pi}{18}\right) \right)= \frac{\pi^3}{3\sqrt{105}}\left(\frac{1}{\cos \left(\frac{\pi}{18}\right)}+\tan \left(\frac{\pi}{18}\right) \right)=1.2020...$$ using $\zeta(6)=\frac{\pi^6}{945}$

It is correct in the 5 digits of the approximate numerical value $1.2020...$ displayed above.

This expression has no explicit infinite series or products or obviously tweaked constants. Admittedly, the constant 18 was tweaked, but is does not look tweaked (I would prefer to say it is reverse engineered).

The expression was derived as follows:

The starting formula involves the Liouville function $\lambda(n)$. It is either $+1$ or $-1$, depending on whether the number of factors in the prime factorization of $n$ is even or odd ($\lambda(1)=+1$, as $1$ has $0$ prime factors).

$$\frac{\zeta(2s)}{\zeta(s)}=\sum_{n=1}^\infty \frac{\lambda(n)}{n^s}$$ where $s>1$

It can be derived from Euler's product formula for $\zeta(2s)$.

Now the terms of $\zeta(s)=\sum_{n=1}^\infty \frac{1}{n^s}$ can be separated on the basis of the value of $\lambda(n)$.

$$\zeta_+(s)=\frac 12 \left(\zeta(s)+\frac{\zeta(2s)}{\zeta(s)}\right)$$ is the sum of the terms having $\lambda(n)=+1$.

$$\zeta_-(s)=\frac 12 \left(\zeta(s)-\frac{\zeta(2s)}{\zeta(s)}\right)$$ is the (positive) sum of the (absolute value of the) terms having $\lambda(n)=-1$.

The following three relations hold:

$$\zeta_+(s)+\zeta_-(s)=\zeta(s)$$ $$\zeta_+(s)-\zeta_-(s)=\frac{\zeta(2s)}{\zeta(s)}$$ and, multiplying the two relations above: $$\zeta_+(s)^2-\zeta_-(s)^2=\zeta(2s)=\left(\sqrt{\zeta(2s)}\right)^2$$ or $$\zeta_+(s)^2=\left(\sqrt{\zeta(2s)}\right)^2+\zeta_-(s)^2$$

The latter is the the Pythagorean equation for a right triangle with hypotenuse length $\zeta_+(s)$ and right angle sides lengths $\sqrt{\zeta(2s)}$ and $\zeta_-(s)$.

If we call the angle between the hypotenuse and the $\sqrt{\zeta(2s)}$ side $\alpha(s)$, then $$\sin(\alpha(s))=\frac{\zeta_-(s)}{\zeta_+(s)}=\frac{\zeta(s)-{\zeta(2s)\over \zeta(s)}}{\zeta(s)+{\zeta(2s)\over \zeta(s)}}=\frac{1-{\zeta(2s)\over \zeta(s)^2}}{1+{\zeta(2s)\over \zeta(s)^2}}=q(s)$$ The rightmost fraction is also defined in the limit $\lim_{s\to 1}$.

So $\alpha(s)=\arcsin(q(s))$, $\cos(\alpha(s))=\frac{\sqrt{\zeta(2s)}}{\zeta_+(s)}$, $\tan(\alpha(s))=\frac{\zeta_-(s)}{\sqrt{\zeta(2s)}}$, which gives

$$\zeta(s)=\zeta_+(s)+\zeta_-(s)=\sqrt{\zeta(2s)}\left(\frac{1}{\cos(\alpha(s))}+\tan(\alpha(s))\right)$$

Plugging in s=3, with numerical values values for $\zeta(3)$ and $\zeta(6)$ from WolframAlpha, one finds

$\alpha(3)=\arcsin(q(3))=$ 0.1745439... $=$ π/17.99886... in radians, or 10.0006...° in degrees.

This is very close to $\frac{\pi}{18}$ radians i.e. $10°$.

This link has the WolframAlpha calculation for $\zeta(3)$ using $\alpha(3)=\frac{\pi}{18}$.

Is the reasonably close approximation of $\zeta(3)$ using the round value $\frac{\pi}{18}$ for $\alpha(3)$ just a coincidence, or is there an explanation for it?

Answer 1

If we write $$\zeta(2n+1)=\sqrt{\zeta(4n+2)}\left(\frac{1}{\cos(\alpha_n)}+\tan(\alpha_n)\right)$$ then $$\alpha_n=2 \tan ^{-1}\left(\frac{\zeta (2 n+1)-\sqrt{\zeta (4 n+2)}}{\zeta (2n+1)+\sqrt{\zeta (4 n+2)}}\right)$$ Defining $$\beta_n=2\log_{\color{red}{18}} \left(\frac{\pi }{\alpha_n }\right)$$ which, at least for the first, are quite "close" to consecutive integer numbers $$\left( \begin{array}{cc} n & \beta_n \\ 1 & 1.99996 \\ 2 & 3.09699 \\ 3 & 4.10893 \\ 4 & 5.09077 \\ 5 & 6.05999 \\ 6 & 7.02367 \\ 7 & 7.98489 \\ 8 & 8.94502 \\ 9 & 9.90466 \\ \end{array} \right)$$

Answer 2

The value $q(3)=\frac{\zeta_-(3)}{\zeta_+(3)}$ is very close to being a root of the function $f(x)=8x^3-6x+1$, as $|f(q(3))|<6\times 10^{-5}$ in WolframAlpha. The corresponding exact root of $f$ is $\sin(\frac{\pi}{18})$, which explains why $q(3)$ is so close to $\sin(\frac{\pi}{18})$.

Here the determination of the proximity of $q(3)=\frac{\zeta_-(3)}{\zeta_+(3)}$ to a root of $f$ is based on the numerical values of $\zeta_+(3)$ and $\zeta_-(3)$. The derivation of this proximity from the definitions of $\zeta_+(3)$ and $\zeta_-(3)$ remains to be found.

To get an expression for an approximate value of $\zeta(5)$ using the same procedure, note that $q(5)$ is very close to being a root of the function $f_5(x)=22x^3-28x+1$. The corresponding exact root of $f_5$ is: $$2\sqrt{\frac{14}{33}}\sin\left(\frac{1}{3}\arcsin\left(\frac{3}{56}\sqrt{\frac{33}{14}}\right)\right)$$

Note that $$q(s)=\frac{1-{\zeta(2s)\over \zeta(s)^2}}{1+{\zeta(2s)\over \zeta(s)^2}} \iff {\zeta(2s)\over \zeta(s)^2}=\frac{1-q(s)}{1+q(s)}$$ so $$\zeta(s)=\sqrt{\zeta(2s) \cdot \frac{1+q(s)}{1-q(s)}}=\sqrt{\zeta(2s)\left(1+\frac{2}{{1\over q(s)}-1} \right)}$$ Plugging in $s=5$, and using $\zeta(10)=\frac{\pi^{10}}{93555}$ and the above approximation for $q(5)$, one obtains: $$\zeta(5)≅1.036927\dots$$ which is correct in the decimals displayed.

Alternatively, the equally accurate function $g_5(x)=\frac{22}{28}x^2-28x+1$ can be used. $q(5)$ is very close to this root of $g_5$: $$\frac{14^2-\sqrt{42 \times 911}}{11}=\frac{14}{14^2+\sqrt{42 \times 911}}$$ This gives $\zeta(5)≅1.036927\dots$, having the same precision.

For a less complicated, and less accurate, approximate expression for $q(5)$, one can use $h_5(x)=x^2-28x+1$, with root $$14-\sqrt{13 \times15}={1 \over {14+\sqrt{13 \times15}}}$$ which gives $\zeta(5)≅1.0369\dots$.

When going from an approximation for $\zeta(3)$ using the polynomial $8x^3-6x+1$, to an approximation for $\zeta(5)$, we adjusted the coefficients of $x^3$ and $x$. We can also adjust (parameterise) the constant $1$. The middle solution, that corresponds to $q(s)$, of the equation $$8x^3-6x+c(s)=0 , -2 \le c(s) \le 2$$ is $$x=\sin\left(\frac{1}{3}\arcsin\left(\frac{c(s)}{2}\right)\right)$$

If we require $q(s)$ to be equal to this solution (for $0 \lt q(s) \le \frac{1}{2}$), i.e. $\alpha(s)=\frac{1}{3}\arcsin\left(\frac{c(s)}{2}\right)$ with $0 \lt \alpha(s) \le \frac{\pi}{6}$, then $$c(s)=2\sin(3\arcsin(q(s))=2\sin(3\alpha(s))$$

If we choose $c(5)=\frac{303}{1415}$ to get an approximation of $q(5)$ and hence $\zeta(5)$, i.e. we choose $\alpha(5)≅\frac{1}{3}\arcsin\left(\frac{303}{2830}\right)$, we get: $$\zeta(5)≅1.0369277\dots$$ which is even more precise than the approximations above.

For the case where $\frac{1}{2} \le q(s) \lt 1$ the largest solution of the equation $8x^3-6x+c(s)=0$ corresponds to $q(s)$. That solution is: $$x=\sin\left(\frac{1}{3}\left(\pi-\arcsin\left(\frac{c(s)}{2}\right)\right)\right)$$

With $q(s)$ equal to this solution, $\alpha(s)=\frac{1}{3}\left(\pi-\arcsin\left(\frac{c(s)}{2}\right)\right)$ with $\frac{\pi}{6} \le \alpha(s) \lt \frac{\pi}{2}$. Then $$c(s)=2\sin(\pi-3\arcsin(q(s))=2\sin(3\arcsin(q(s))=2\sin(3\alpha(s))$$ (as $\sin(\pi-a)=\sin(a))$, which is the same formula for $c(s)$ as above.

Note that $\frac{1}{2} \le q(s) \lt 1$ implies $1 \lt s \le 1.8416497\dots$, so this interval does not contain any whole number values for $s$.

Returning to the original approximation for $\zeta(3)$, note that $$\frac{1}{\cos x}+\tan x=\sqrt{\frac{1+\sin x}{1-\sin x}}=\sqrt{\frac{2}{1-\sin x}-1} \;,\; 0 \le x \lt \frac{\pi}{2}$$ Also note that $\sin\left(\frac{\pi}{18}\right)=\sin\left(\frac{1}{3}\arcsin\left(\frac{1}{2}\right)\right)$, since $\sin(\frac{\pi}{6})=\frac{1}{2}$.

Hence the approximation for $\zeta(3)$ can also be written as $$\zeta(3)≅\frac{\pi^3}{3}\sqrt{\frac{1}{105} \left(\frac{2}{1-\sin\left(\frac{1}{3}\arcsin\left(\frac{1}{2}\right)\right)}-1\right)}=1.2020\dots$$

Let $c(s)$ now be determined by: $q(s)=\sin\left(\frac{1}{s}\arcsin(c(s)\right)$, so that $c(s)=\sin(s\cdot\arcsin(q(s)))$, where, as before, $s>1$ and $$q(s)=\frac{1-{\zeta(2s)\over \zeta(s)^2}}{1+{\zeta(2s)\over \zeta(s)^2}}$$ For $s=3$ we get $c(3)=\sin(3\arcsin(q(3)))=3q(3)-4q(3)^3$, since $$\sin3x=3\sin x-4\sin^3x$$ In the above approximation for $\zeta(3)$, $\frac{1}{2}≅c(3)$ and $\sin\left(\frac{1}{3}\arcsin\left(\frac{1}{2}\right)\right)≅q(3)$.

If the value of $c(3)$ is approximated by the first two terms of the (greedy) Egyptian fraction expansion of its numeric value, we get: $$c(3)≅\frac{1}{2}+\frac{1}{35047}$$

With this approximation for the value of $c(3)$ the approximation for $\zeta(3)$ becomes $$\zeta(3)≅\frac{\pi^3}{3}\sqrt{\frac{1}{105} \left(\frac{2}{1-\sin\left(\frac{1}{3}\arcsin\left(\frac{1}{2}+\frac{1}{35047}\right)\right)}-1\right)}=1.2020569031595\dots$$

Adding the reciprocal of a 5-digit number results in a 9-digit increase in the precision of the $\zeta(3)$ approximation.

Answer 3

As soon as the discussion comes on the subject of approximates of remarkable mathematical constants one thinks to ISC (Inverse Symbolic Calculator). Of course this isn't the present subject but the methods and softwares developed in this field can be distorted to find a lot of approximate formulas.

For example the ISC of the Canadian Centre for Experimental and Constructive Mathematics gives : $$\frac{\sqrt{6+4^{5/6}}}{4^{2/3}}=\simeq 1.202056898...$$ which is an appoximate of $\zeta(3)$ with deviation $<0.000\,000\,005$ $$\zeta(3)\simeq 1.202056903...$$ One can easily compute this value with a simple scientific pocket calculator.

Just for the fun I tried with my own ISC. I got a lot of formulas as approximates of $\zeta(3)$. One of them for example : $$\frac{5^{1/3}}{\sin(1)\sqrt{\pi-\cos(5)}}\simeq 1.20205690297... \qquad \text{Deviation } <0.000\,000\,000\,2$$

Another one : $$\sqrt{\frac{\pi}{2}}-\frac{e^{-\pi}}{\ln\left(\sin(2)+\sqrt{2}\right)} \simeq 1.20205690305... \qquad \text{Deviation } <0.000\,000\,000\,1$$

For French readers : https://fr.scribd.com/doc/14161596/Mathematiques-experimentales