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

  • A
  • Thread starter rabbed
  • Start date
  • #1
rabbed
243
3
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
 

Answers and Replies

  • #2
14,166
8,148
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.

Code:
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
 
Last edited:
  • #3
rabbed
243
3
Thanks, i'll try that.
If anyone has ideas for an analytic solution, just shoot :)
 

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

  • Last Post
Replies
20
Views
560
  • Last Post
Replies
1
Views
647
Replies
8
Views
753
Replies
34
Views
1K
Replies
12
Views
456
  • Last Post
Replies
6
Views
277
  • Last Post
Replies
5
Views
849
Replies
4
Views
420
  • Last Post
Replies
17
Views
937
  • Last Post
Replies
3
Views
1K
Top