Recursive Function for Loan ProblemDate: 11/13/97 at 08:14:15 From: dennis hotetz Subject: Recursive function for loan problem I do not understand this question. My college math teacher wants me to show him the recursive function formula for a loan of $100,000, payable at 6 percent, over 20 years. Can you help? Date: 11/13/97 at 10:15:50 From: Doctor Mitteldorf Subject: Re: Recursive function for loan problem Dear Dennis, If the balance of the loan in one month is x, then the next month it will go up by 1/2 percent to 1.005x, but then a payment of p dollars will be made against it, so the following month balance will be 1.005 x - p Let's write 1.005 as f for short, and the original 100,000 we'll write as x0. Then we can make the following chart: Time, months Balance 0 x0 1 f*x0-p 2 f*(f*x0-p)-p 3 f*(f*(f*x0-p)-p 4 f*(f*(f*(f*x0-p)-p)-p At the end of the 20 years, we have the 240th payment, and the balance is zero. What will the 240th line look like? First, we'll have f^(240)*x0. Then we'll have a succession of p's, each with one more f multiplying it p*(1+f+f^2+...+f^239) The sum must come out to zero, since we're done paying off the loan at this point: *** f^(240)*x0 - p*(1+f+f^2+...+f^239) = 0 From this we can get a formula for p. We just need an old trick from HS algebra that's used for summing 1+f+f^2+...+f^239. Let's write 240 as n for generality, and we'll define S as the value of the sum. Look what happens if we multiply S by f: S = 1+f+f^2+...+f^n-1 fS = f+f^2++f^3...+f^n Notice that after we've multiplied S by f, we get back most of the terms in the original S, just shifted over by one. The only differences are at the endpoints: there's one extra term on the right and we're missing one on the left. Expressing this algebraically, we have: fS = S-1+f^n We can solve this equation for S: S = (f^n-1)/(f-1) Let's go back now to the line marked ***. We can rewrite this as f^n*x0 - p*S = 0 Solve for p: p = x0 * (f^n/S) Using our formula for S, we have p = x0 * f^n * (f-1) / (f^n-1) We can simplify slightly by dividing numerator and denominator by f^n: p = x0 * (f-1) / (1-f^(-n)) In the case you cite, where f = 1.005, x0 = 100,000 and n = 240, I get p = 716.43 -Doctor Mitteldorf, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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