Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Posting-Version: version B 2.10.1 6/24/83; site imsvax.UUCP Path: utzoo!watmath!clyde!burl!ulysses!gamma!exodus!mhtsa!mh3bs!eagle!harpo!seismo!rlgvax!cvl!elsie!imsvax!leung From: leung@imsvax.UUCP Newsgroups: net.sources Subject: Re: loan analysis program (Proof) Message-ID: <156@imsvax.UUCP> Date: Tue, 7-Feb-84 13:42:49 EST Article-I.D.: imsvax.156 Posted: Tue Feb 7 13:42:49 1984 Date-Received: Fri, 10-Feb-84 02:37:19 EST Organization: IMS Inc, Rockville MD Lines: 109 Posted Feb 2 stevens@hsi.UUCP in net.source has a program for loan analysis. There is no analytic proof of the formula he used. The following is a proof of the formula. The following lemma will be useful. Lemma. n _ \ i n+1 1 + a / (1 + a) = (1 + a) (*) - i=0 Proof: For i=0, lhs = 1 + a = rhs Suppose for i=n, (*) holds For i=n+1, n+2 lhs = (1 + a) n+1 = (1 + a) (1 + a) n _ \ i = (1 + a) (1 + a / (1 + a) ) - i=0 n+1 _ \ i = 1 + a + a / (1 + a) - i=1 n+1 _ \ = 1 + a / (1 + a) = rhs - i=0 Q.E.D. Back to the Loan Equation: Theorem: n AI (1 + I) P = ------------ n (1 + I) - 1 where P = Monthly Payment A = Loan Amount I = Monthly Interest Rate (Annual Interest Rate / 12) n = Number of Payments Proof: For n=1, the final balance is A(1 + I) - P = 0 For n=2, the final balance is (A(1 + I) - P)(1 + I) - P = 0 2 A(1 + I) - P(1 + (1 + I)) = 0 For n=3, the final balance is 2 (A(1 + I) - P(1 + (1 + I)))(1 + I) - P = 0 3 2 A(1 + I) - P(1 + (1 + I) + (1 + I) ) = 0 For n=n, the final balance is n-1 _ n \ j A(1 + I) - p / (1 + I) = 0 - j=0 then, n-1 _ n \ j AI(1 + I) - pI / (1 + I) = 0 - j=0 Using the above Lemma, n n AI(1 + I) - p((1 + I) - 1) = 0 This completes the proof.