best approximation of $\sqrt{2}$

Question

The approximation \begin{align} \sqrt{2} &\approx \frac{1}{8} \operatorname{csch}\left(\frac{3\pi}{2}\right) \operatorname{sech}^3(\pi) \, \left[2+3 \, \sinh\left(\frac{\pi}{2}\right)-\sinh\left(\frac{3\pi}{2}\right) \right. \\ & \hspace{10mm} \left. +3 \, \sinh\left(\frac{5\pi}{2}\right)+\sinh\left(\frac{9\pi}{2}\right)-2 \,\cosh(\pi)+2 \,\cosh(2\pi)+2 \,\cosh(4\pi)\right] \end{align} gives the first $8$ correct digits of $\sqrt{2}$. Is this the best approximation of square root 2 in terms of hyperbolic functions? If not,then please find more examples of this type.

Answer 1

First, replace $\operatorname{csch}$ and $\operatorname{sech}$ with $1 / \sinh$, $1 / \cosh$. Approximation in terms of hyperbolic functions ($\cosh x = \frac{e^x + e^{-x}}{2}, \sinh x = \frac{e^x - e^{-x}}{2}$), is then equivalent to approximation by powers of $e$, since \begin{align*} e^x &= \cosh x + \sinh x \\ e^{-x} &= \cosh x - \sinh x \end{align*} Moreover, you appear to be only considering $\sinh$ and $\cosh r\pi$, where $r$ is rational. This will correspond to $e^{r\pi}$, with $r$ rational. So what you are really asking is: How well can we approximate $\boldsymbol{\sqrt{2}}$ by combining (through addition, multiplication, and division) rational multiples of rational powers of $\boldsymbol{e^{\pi}}$? (If you weren't allowing $\operatorname{csch}$ and $\operatorname{sech}$, it would just be addition and multiplication.)

Answer: $e^\pi$ is known as Gelfond's constant, and is transcendental. Because of this, there is no way to get $\sqrt{2}$ exactly as a sum/product/quotient of powers of it; this would imply that it was algebraic, since $\sqrt{2}$ is algebraic.

However, it is certainly possible to get arbitrarily close to $\sqrt{2}$. In fact, you can get arbitrarily close just by using $r \left(e^\pi\right)^0 = r$, i.e. just with rational numbers. See Diophantine approximation and the continued fraction for $\sqrt{2}$. Since it is possible to get arbitrarily close and impossible to get exactly to $\sqrt{2}$, the approximation you list (nor any other approximation) will be the best approximation for $\sqrt{2}$ using hyperbolic functions.

Answer 2

The continued fraction for $\sqrt{2}$ is a fast rational approximation. Where $\sqrt{2}=1+ \frac{1}{2+ \frac{1}{2+ \frac{1}{\ddots}}}$. The eighth convergent according to wiki is 577/408, read more here.

Answer 3

How about this:

$$\frac35+\frac{\pi}{7-\pi}-\sqrt2=10^{-6}\times6.495680826...$$

This is also interesting:

$$\frac{131836323}{93222358}-\sqrt2=4\times10^{-17}$$

Answer 4

By exploiting the elliptic lambda function, we have: $$ \sqrt{2} = 2\cdot \frac{\theta_2^2(0,e^{-\pi})}{\theta_3^2(0,e^{-\pi})} $$ and by exploting the expansions of the Jacobi theta functions we have: $$ \sqrt{2} \approx 2\,\left(\frac{2-2e^{5\pi}+2e^{8\pi}-e^{9\pi}}{2+2 e^{5 \pi }+2 e^{8 \pi }+e^{9 \pi }}\right)^2$$ that looks way better and is right up to $14$ digits.

Answer 5

$$ \sqrt {2}=-5\,\tanh \left( 1/2 \right) -{\frac {23\,\tanh \left( 3/2 \right) }{5}}+{\frac {21\,\tanh \left( 5/2 \right) }{5}}+{\frac {42\, \tanh \left( 7/2 \right) }{5}}-6\,\tanh \left( 1/3 \right) -{\frac {23 \,\tanh \left( 4/3 \right) }{5}}+\frac{4\,\tanh \left( 5/3 \right)}5 -3\, \tanh \left( 7/3 \right) -\dfrac{3\,\tanh \left( 1/4 \right)}{5} +\frac{9\,\tanh \left( 3/4 \right)}{5} +{\frac {28\,\tanh \left( 5/4 \right) }{5}}-{ \frac {13\,\tanh \left( 7/4 \right) }{5}}+\dfrac{\tanh \left( 1/5 \right)}5 +\dfrac{2\,\tanh \left( 2/5 \right)}{5} $$ with error about $5 \times 10^{-30}$.