Here is a simple polynomial equation:
b^2 + 2b + 1 = 0
I could easily solve this as:
import numpy as np from scipy.optimize import fsolve eq = lambda b : np.power(b,2) + 2*b + 1 fsolve(eq, np.linspace(0,1,2))
Similarly I could solve any equation, that has finite number of terms. But how do I solve an equation with infinite number of terms which is given as :
The above equation could be written as :
(1 - l) * (5.5 + 4.0*l + 4*l^2 + 6*l^3 + 5*l^4 + 5*l^5 + 5*l^6 + 5*l^7 + 5*l^8 + 5*l^9 + 5*l^10 ) = 5
when n goes from 1 to 10. But I want to solve this for sufficiently large value of
n such that LHS ~= RHS.
I know the values of LHS and
G1 -> Ginf but cannot understand how could I compute the value of lambda here.
I tried looking at numpy polynomial functions but could not find a function that is relevant here.