Cut the Cards

[Albert Chan]

(08-25-2020 11:41 PM)Gerson W. Barbosa Wrote:  FWIW,

« 1. Psi + EXP 6. OVER INV 12. + / + 1. - »

'H­-¹' STO

Example:

1100 H → 7.58073560027

H­-¹ → 1100

H-­¹ should be accurate to 11 or 12 digits for arguments equal or greater than 7.5

Here is another version, using asymptotic estimate: \(\quad\exp(\psi(x+0.5)) ≈ x + \large{1\over24x} \)

Let k = exp(ψ(x+0.5)), n = x - 0.5 → k = exp(H(n) + ψ(1)) ≈ x + 1/(24x)

If x is big, x ≈ k, we have:                 \(n ≈ k - \large{1\over 24k} - {1\over2}\)

HP50g inverse harmonic function:      « 1. Psi + EXP DUP 24. * INV - .5 - »

Both formulas perform well for large H, but this new version is also good for smaller H

>>> from mpmath import exp, euler, harmonic
>>> f1 = lambda k: k + 6/(1/k+12) - 1
>>> f2 = lambda k: k - 1/(24*k) - .5             # new version
>>> for n in range(50,100,10):
...          h = harmonic(n)
...          k = exp(h - euler)
...          e1, e2 = n - f1(k), n - f2(k)              # under-estimated errors
...          print '%d\t%.10f\t%.10f\t%g' % (n, e1, e2, abs(e1/e2))
...
50       -0.0000013228       0.0000000364       36.3546
60       -0.0000009261       0.0000000212       43.7606
70       -0.0000006844       0.0000000134       51.1669
80       -0.0000005263       0.0000000090       58.5736
90       -0.0000004172       0.0000000063       65.9804