(15C) Halley's Method

[Albert Chan]

(06-14-2022 02:59 AM)Albert Chan Wrote:  2 sign changes, thus 0 or 2 positive roots (for R = 1+i)
Based on rate edges, (PMT/FV , PMT/-PV) = (-1/50, 2/13)

2 Rates (if exist at all): -1/50 < i < 2/13

If we can show we have 1 root, we must have 2.

I was thinking of a fast way to locate extremum, check f sign, to decide number of roots.

Let decay factor, g = n*x / ((1+x)^n - 1)

f = (pv+fv)/n*g + pv*x + pmt

f' = (pv+fv)/n*g' + pv

f' = 0      → g' = -n*pv/(pv+fv)

g' is too complex to be invertible back for x
Instead, we approximate g with simple exponential decay function.
Matching g value and slope at x=0, we have g ≈ exp((1-n)/2*x)
exp is invertible with ln, we have:

Extremum: x_m ≈ m * ln(-m*n*pv / (pv+fv))      , where m = 2/(1-n)

Reusing previous example, n=30, pv=6500, pmt=-1000, fv=50000

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

x_m ≈ -2/29 * ln(2/29 * 30 * 6500/56500) ≈ 0.09899

f(x = 0.09899) ≈ -6.46                        // true extremum, f(x = 0.1002) ≈ -6.52
f(x = 0) = (pv+fv)/n + pmt = 883.33

We have locked in 1 roots ⇒ we have 2 roots.