Log-Arcsine Algorithm

[Thomas Klemm]

With a minor change we can also calculate the inverse hyperbolic sine function.
We just have to start with:

\(
x \sqrt{1 + x^2}
\)

and

\(
x
\)

Example

2 SQRT 1

XEQ 00

0.88137358702

---

To find the recurrence relation we can define the following functions with somewhat suggestive names:

\(
\begin{align}
s(z) &= \frac{z - \frac{1}{z}}{2} \\
\\
c(z) &= \frac{z + \frac{1}{z}}{2} \\
\end{align}
\)

It's easy to show (see below) that:

\(
\begin{align}
2 s(z) c(z) &= s(z^2) \\
\\
2 c(z)^2 &= 1 + c(z^2) \\
\end{align}
\)

From the latter we conclude:

\(
\begin{align}
c(z) = \sqrt{\frac{1 + c(z^2)}{2}}
\end{align}
\)

Using the substitution \(z \to z^{\tfrac{1}{2}}\) we get:

\(
\begin{align}
c(z^{\frac{1}{2}}) &= \sqrt{\frac{1 + c(z)}{2}} \\
\\
s(z)
&= 2 \, s(z^{\frac{1}{2}}) \, c(z^{\frac{1}{2}}) \\
\\
&= 2 \, s(z^{\frac{1}{2}}) \, \sqrt{\frac{1 + c(z)}{2}} \\
\\
&= 2 \, s(z^{\frac{1}{2}}) \, \sqrt{\frac{s(z) + s(z)c(z)}{2s(z)}} \\
\\
&= 2 \, s(z^{\frac{1}{2}}) \, \sqrt{\frac{s(z) + \frac{1}{2}s(z^2)}{2s(z)}} \\
\end{align}
\)

Or then after rearranging it:

\(
\begin{align}
2 \, s(z^{\frac{1}{2}}) = s(z) \, \sqrt{\frac{2s(z)}{s(z) + \frac{1}{2}s(z^2)}}
\end{align}
\)

With:

\(
\begin{align}
\ell(h) = \frac{1}{2h}(x^h - x^{-h}) = \frac{s(x^h)}{h}
\end{align}
\)

Or then with \(z = x^h = x^{2^{-n}}\) we have:

\(
\begin{align}
\ell_n = \ell(2^{-n}) = 2^n s(x^{2^{-n}}) = 2^n s(z) = s_n
\end{align}
\)

This leads to the recurrence relation:

\(
\begin{align}
s_{n+1}
&= \ell(2^{-n-1}) \\
&= 2^{n+1} s(x^{2^{-n-1}}) \\
&= 2^n \cdot 2 s(z^{\tfrac{1}{2}}) \\
\\
&= 2^n s(z) \sqrt{\frac{2s(z)}{s(z) + \tfrac{1}{2}s(z^2)}} \\
\\
&= 2^n s(z) \sqrt{\frac{2 \cdot 2^n s(z)}{2^n s(z) + \tfrac{1}{2} \cdot 2^n s(z^2)}} \\
\\
&= s_n \sqrt{\frac{2 s_n}{s_n + 2^{n-1} s(z^2)}} \\
\\
&= s_n \sqrt{\frac{2 s_n}{s_n + s_{n-1}}} \\
\end{align}
\)

We can plug in \(z = e^x\) or \(z = e^{ix}\) to get related formulas for \(\sinh(x)\) and \(\sin(x)\):

\(
\begin{align}
2 \sinh(\frac{1}{2} z) &= \sinh(z) \sqrt{\frac{2\sinh(z)}{\sinh(z) + \frac{1}{2}\sinh(2z)}} \\
\\
2 \sin(\frac{1}{2} z) &= \sin(z) \sqrt{\frac{2\sin(z)}{\sin(z) + \frac{1}{2}\sin(2z)}} \\
\end{align}
\)

That's due to the fact that:

\(
\begin{align}
s(e^x) = \frac{e^x - e^{-x}}{2} = \sinh(x) \\
\\
s(e^{ix}) = \frac{e^{ix} - e^{-ix}}{2} = \sin(x) \\
\end{align}
\)

---

\(
\begin{align}
2 s(z) c(z)
&= 2 \frac{z - \frac{1}{z}}{2} \frac{z + \frac{1}{z}}{2} \\
&= \frac{\left(z - \frac{1}{z} \right) \left(z + \frac{1}{z} \right)}{2} \\
&= \frac{z^2 - \frac{1}{z^2}}{2} \\
&= s(z^2) \\
\\
2 c(z)^2
&= 2 \left( \frac{z + \frac{1}{z}}{2} \right)^2 \\
&= \frac{\left(z + \frac{1}{z} \right)^2}{2} \\
&= \frac{z^2 + 2 + \frac{1}{z^2}}{2} \\
&= 1 + \frac{z^2 + \frac{1}{z^2}}{2} \\
&= 1 + c(z^2) \\
\end{align}
\)