Riemann's Zeta Function - another approach (RPL)

[Paul Dale]

(06-30-2017 04:50 PM)Dieter Wrote:  Addendum for the 15C and others: The benefit of the e^x–1 method for the resulting accuracy can also be achieved in another way.
2^x/(2^x–2) – 1 = 1/(2^x–1–1)

How about \( e^x - 1 = tanh\left(\frac{x}{2}\right) \left( e^x + 1 \right) \)?

Pauli