Calculating e^x-1 on classic HPs

[Albert Chan]

(01-13-2016 03:46 PM)Gerson W. Barbosa Wrote:  \[e^{x}-1=\frac{2}{\coth \frac{x}{2}-1}\]
... this will require two additional steps to handle x=0

Is there any reason to use a possible divide-by-0 coth(x/2) ?

If t = tanh(x/2), we have exp(x) - 1 = 2 t / (1 - t), avoided the singularity.