Equation system involving InverseDigamma(Digamma(a+1)-b*k)-1

[rabbed]

How would one go about solving this for a and b given N, E and K?

n(0) + n(1) + … + n(K) = N
n(0)*0 + n(1)*1 + … + n(K)*K = E

Where n(k) = InverseDigamma(Digamma(a+1)-b*k)-1

 

[jedishrfu]

I think it would have to be done numerically using Python, Matlab, or Mathematica.

Mathematica has a digamma function:

https://mathworld.wolfram.com/DigammaFunction.html

and Matlab has a digamma function:

https://www.mathworks.com/help/matlab/ref/psi.html

and Python has a version of the digamma function:

https://docs.scipy.org/doc/scipy/reference/generated/scipy.special.digamma.html

Here's some pseudo code to search for the best A, B values across a range of A, B values. It's not the fastest or the best way, and the pseudo-code needs to be written in either Python, Matlab, or Mathematica to find the answer you're looking for.

Also, the means of determining closeness could be better implemented with perhaps some other distance measurement more suited to this search.

given N , E , K

A = 0
B = 0

NDIFF= max_float_value
EDIFF = max_float_value

for a in range(a_min,a_max,a_step)

    for b in range(b_min,b_max,b_step)

        for k = 0 to K:

            X = (a+1)-b*k
            NK = inverse_digamma(digamma(X))
            N2 = N2 + NK
            E2 = E2 +k*NK

        if ((N2-N)^2+(E2-E)^2) < (NDIFF^2+EDIFF^2):
            NDIFF = (N2 - N)
            EDIFF = (E2 - E)
            A = a
            B = b

print "NDIFF = " NDIFF
print "EDIFF = " EDIFF
print "A = ". A, "  B = ", B

 

[rabbed]

Thanks, i'll try that.
If anyone has ideas for an analytic solution, just shoot :)