New Sum of Powers Log Function

[Albert Chan]

(03-30-2021 01:44 AM)Albert Chan Wrote:  If we have LambertW, we can get a much better guess, without solver.

s ≈ ∫(t^x, t=1/2 .. n+1/2) = (n+1/2)^(x+1)/(x+1) - (1/2)^(x+1)/(x+1)

If we drop the last term, and let N=n+1/2, X=x+1, we have ...

We could estimate the dropped term, and add it back.

Code:

def guess2(n,s):
    t = -log(n+.5)
    p = lambertw(t/s,-1) / t
    s += 0.5**p/p
    return lambertw(t/s,-1) / t - 1

Above estimated x has similar error as roughx(n,s), but twice as fast. (on my machine)

(03-30-2021 01:44 AM)Albert Chan Wrote:  Turns out, LambertW -1 branch is the one we need.

Illustrate the reason, by example:

>>> n, s = 100, 1000
>>> t = -log(n+.5)
>>> p = lambertw(t/s,-1) / t       # W1 resulted p
>>> p, (n+.5)**p/p, .5**p/p
(1.60037803179321, 1000.0, 0.2060704060225)
>>>
>>> p = lambertw(t/s, 0) / t       # W0 resulted p
>>> p, (n+.5)**p/p, .5**p/p
(0.00100464230172027, 1000.0, 994.686243766351)

Both solved p, for s = (n+.5)**p/p.
But, the assumption that last term can be dropped is false for W0