Discount Rate

[Albert Chan]

(04-09-2022 05:47 PM)Albert Chan Wrote:  \(\displaystyle I ≈ \frac{1}{P} - \frac{P}{N^2}\)

This is even better, and work well with big N.
...
Does anyone know how this is derived ?

(10-16-2020 04:02 PM)Albert Chan Wrote:  XCas> C := I*N / (1 - (1+I)^-N)       // C = |N*PMT/PV|, "compounding factor"

This is my guess on how formula is derived, using continuous compounding.
With small I, (1+I)^-N = exp(ln(1+I)*-N) ≈ exp(-I*N)

Let x = I*N, and solve for x

C = N/P ≈ x / (1 - e^-x)
x = I*N = C + W(-C*exp(-C))            // W has 2 branches, but it does not matter here

Let y = -C*exp(-C), solve for I

I ≈ 1/P + W(y)/N

If we can show -W(y) ≈ 1/C, we have I ≈ 1/P - 1/(CN) = 1/P - P/N^2, and prove is done.

There are 2 LambertW branches, but we only want estimate when C close to 1
Both branches behave about the same, because y is very close to -1/e

C = 1 + ε      → y = 1/e * (-1 + ε^2/2 - ε^3/3 + ε^4/8 - ...)

I borrowed lyuka e^W estimation formula, replace 0.3 by 1/3, for Puiseux series.

e^W(y) ≈ 1/e + sqrt ((2/e)*(y+1/e)) + (y+1/e)/3

-W(y)
= 1 - ln(e * e^W(y))
= 1 - ln(1 + sqrt(2*(e*y+1)) + (e*y+1)/3)
= 1 - ln(1 + ε - 1/6*ε^2 + ...)
= 1 - 2*atanh(ε/2 - 1/3*ε^2 + ...)
≈ 1 - ε + 2/3*ε^2
≈ 1/(1+ε)
= 1/C            QED

To confirm math is correct, at least numerically.

lua> z = 1e-4
lua> c = 1 + z
lua> y = -c*exp(-c)
lua> -W(y), 1 - z + 2/3*z^2, 1/c
0.9999000066665401      0.9999000066666667      0.9999000099990001