Discount Rate
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04-10-2022, 04:03 AM
Post: #6
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RE: Discount Rate
(04-09-2022 05:47 PM)Albert Chan Wrote: \(\displaystyle I ≈ \frac{1}{P} - \frac{P}{N^2}\) (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 |
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Messages In This Thread |
Discount Rate - SlideRule - 04-04-2022, 01:54 PM
RE: Discount Rate - Thomas Klemm - 04-08-2022, 07:44 PM
RE: Discount Rate - Albert Chan - 04-08-2022, 10:21 PM
RE: Discount Rate - Albert Chan - 04-09-2022, 12:50 PM
RE: Discount Rate - Eddie W. Shore - 04-10-2022, 11:29 PM
RE: Discount Rate - Albert Chan - 04-09-2022, 05:47 PM
RE: Discount Rate - Albert Chan - 04-10-2022 04:03 AM
RE: Discount Rate - Thomas Klemm - 04-10-2022, 12:19 PM
RE: Discount Rate - Albert Chan - 04-11-2022, 01:18 AM
RE: Discount Rate - Albert Chan - 04-11-2022, 03:11 PM
RE: Discount Rate - Albert Chan - 04-11-2022, 03:57 PM
RE: Discount Rate - Albert Chan - 05-11-2022, 06:22 PM
RE: Discount Rate - Thomas Klemm - 04-11-2022, 02:37 AM
RE: Discount Rate - Thomas Klemm - 04-11-2022, 08:48 PM
RE: Discount Rate - Thomas Klemm - 04-12-2022, 11:11 PM
RE: Discount Rate - Thomas Klemm - 04-12-2022, 11:13 PM
RE: Discount Rate - Thomas Klemm - 04-18-2022, 01:58 PM
RE: Discount Rate - rprosperi - 04-18-2022, 06:41 PM
RE: Discount Rate - Thomas Klemm - 04-18-2022, 07:03 PM
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