New PRNG for calculators

[Albert Chan]

(06-08-2024 03:44 PM)ttw Wrote:  After more reading and a night's sleep, it's the entire line from -inf to +inf for which frac(x) is uniform.

I would have guessed FP(exp(x)) would skew to tiny values, not uniform.

Negative side, x<ln(0.1), FP(exp(x)) = exp(x) < 0.1
Positive side, numerically FP(exp(x)) = 0, if x goes big. (assume it did not overflow)

The other two exp() is needed to keep x in reasonable domain.

e^e^0 .. e^e^1 = e .. e^e ≈ 2.71828 .. 15.1543