Riemann's Zeta Function - another approach (RPL)

[Dieter]

(06-30-2017 11:23 PM)Paul Dale Wrote:  How about \( e^x - 1 = tanh\left(\frac{x}{2}\right) \left( e^x + 1 \right) \)?

Ah, this reminds me of an earlier thread on e^x–1.

In this special case the function may be coded this way, cf. Gerson's post #19 in the mentioned thread:

Code:

 .. ...         ...
 90 2           1
 91 RCL 0       STO+ 0
 92 1           2
 93 +           RCL 0
 94 y^x         y^x
 95 LastX       +
 96 2           /
 97 LN          2
 98 *           LN
 99 2           2
100 /           /
101 TANH        RCL* 0
102 *           TANH
103 Last        /
104 +           CHS
105 /
106 CHS

Do not use ln √2 instead of ½ ln 2 here as on the 10 digit calculators this causes an error of several ULPs.

Dieter