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

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 2

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 3

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

Answer 4

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 5

$\sqrt{2}$ has such a simple continued fraction that it's so boring to state one defining property of the number, its continued fraction, so I will instead show how to compute another property of that number, its binary notation in time O($n^2)$. I know an easy way to compute the binary notation of $\sqrt{2}$. Start with the ordered pair (1, 2). Compute the square of its average, 1.5. If it's more than 2, make the first coordinate of the next ordered pair be the first coordinate of that ordered pair and the second coordinate of the next ordered pair be the average of the first ordered pair. If it's less than 2, make the first coordinate of the next ordered pair be the average of that ordered pair and the second coordinate of the next ordered pair be the second coordinate of that ordered pair. Repeat the same process again and again. Each time you do it again, it's straight forward to determine another binary digit of $\sqrt{2}$.

Not only that but each of those operations can be done in linear time enabling you to compute the binary digits of $\sqrt{2}$ in only time O($n^2$). That's because for each positive integer $n$, once you have already computed the square of each of the coordinates in the $n^{th}$ ordered pair, you can recognize that the square of their average is just the average of their squares minus $2^{-2n}$ so you can just compute the square of their average in linear time.

Answer 6

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.