- #1
- 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
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