TVM solve for interest rate, revisited

[Albert Chan]

With Q-method proven, I was able to improve its accuracy.

\((1+i)^Q \,≈\, (1+i_0)^Q - 1\)


\(\displaystyle
i \,=\, i_0 - \frac{i_0}{(1+i)^n}
\,≈\, i_0 - \frac{i_0}{((1+i_0)^Q-1)^\frac{n}{Q}}
\)

This is suitable for C = n*i0 > 1, because denominator has size around C^2

(06-27-2024 09:21 PM)Albert Chan Wrote:  lua> n, i, pv, pmt, fv = 360, 0.0065, 225e3, -1619.71, 0
lua> i0 = pmt/-pv
lua> i0
0.007198711111111112
lua> q = log1p(1/n) / log(2)     -- slightly better than q = 1/n / log(2)
lua> EFF(APR(i0,q)-1,q)
0.006464576864221504

lua> i0 - i0/APR(i0,q)^(n*q)
0.006491096452188186

lua> Q = (n+.5)*log(2)        -- my preference is big Q
lua> i0 - i0/EFF(i0,Q)^(n/Q)
0.006491096805454773