Looking for TVM formulas

[Dieter]

(04-06-2014 07:46 PM)Dieter Wrote:  Since we're at it: hyperbolics can also be used for ln1+x.

\(ln(1+x) = 2 \cdot artanh(\frac{x}{x+2})\)

Seems to work well if x is not too large so that the artanh-argument does not get too close to 1.

This one seems even better and does not have problems with very large x:

\(ln(1+x) = arsinh(\frac{x^2+2x}{2+2x})\)

For better numeric accuracy this can be written as

\(y = \frac{x}{2};   z = y + \frac{y}{2y+1}\)

\(ln(1+x) = arsinh z\)

With \(z \rightarrow \infty,   arsinh z \rightarrow ln 2z\)  and  \(z \rightarrow \frac{x+1}{2}\),  so that \(arsinh z \rightarrow ln(1+x)\).

Dieter