(15C) Halley's Method

[Albert Chan]

(06-14-2022 01:20 AM)Gil Wrote:  6500*R^30+50000=(R^30-1)/(R-1)*1000,
with R=1+interest_rate/100,

Above setup were solving NFV = 0, for R

NFV = 6500*R^30 + 50000 - (R^30-1)/(R-1)*1000

For comparison, I was solving for f = NPMT/n = 0

f = NFV * (R-1)/(R^30-1) = (56500 / (R^30-1) + 6500) * (R-1) - 1000

Solving for f=0 for R is more stable, and, likely converge faster.

From the edges, solve f=0 with Newton's method, we have 1-sided convergence.
In other words, starting from 2 edges, we can get both roots (if existed).

Newton's method for NFV=0 starting from edges, for this example, we get only 1 root.
Interestingly, Halley's Irrational formula for NFV=0 work for small edge, R = 1+i = 0.98

0.98
1.0810815846
1.08936323285
1.08960584869
1.08960585626

For R < 1.12175, argument inside square root are positive.
Thus, it will not work on the other edge, with R = 1+i = 1+2/13 ≈ 1.15385

We have only very small guess window to get the other root, R ≈ 1.11100

R(where NFV'>0) to R(where (NFV')^2 - 2*NFV*(NFV'') ≥ 0) = (1.10110, 1.12175)

I would stay away using Halley's Irrational formula.
To avoid issues, it required very close guess, more trouble than it is worth.