# Difference equation - compound interest

#### Peytonator

Joined Jun 30, 2008
105
Hello,

A car is purchased on hire purchase for $a. The debt is to be paid by a sequence of N payments over a total time of NT time periods. The interest charges are compounded at a rate r% per time period T. What is the difference eqation representing this system? My attempt: Let y(k) = outstanding debt after the kth payment of p y(k+1) = y(k) + r*y(k) - p Is this correct? Where does the "a" come in? Thanks #### Kermit2 Joined Feb 5, 2010 4,162 #### Charles Colbert Joined May 5, 2019 1 Σ Hello, A car is purchased on hire purchase for$a. The debt is to be paid by a sequence of N payments over a total time of NT time periods. The interest charges are compounded at a rate r% per time period T.

What is the difference eqation representing this system?

My attempt:

Let y(k) = outstanding debt after the kth payment of p

y(k+1) = y(k) + r*y(k) - p

Is this correct? Where does the "a" come in? General equation is y[n] = (1+r) y[n-1] +x[n]; where Y[n] is current amount at n months; 1+r is keep the 1 + interest; y[n-1] is the previous amount;x[n] >0 money put adds to y[n] and x[n] <0 money taken out subtracts from y[n]. By zero-state (zs) methods, will give the following relationships. x[n] >0, X[n]=βu[n].

Given h[n] = (1+r)^n*u[n]=α^n*u[n], y[-1]=0 and α = 1+r and yzs = Σ(-∞ to ∞) x* h[n-i] = Σ (i=0 to n) x[n-i]* h = Σ(i=0 to n) βα^i ; configuring the equation for negative payments and a present value end up with A=P[r(1+r)^n/(1+r)^n-1].

#### jpanhalt

Joined Jan 18, 2008
10,461
@Charles Colbert

Welcome to AAC.

You are responding to a thread that is 9 years old. In all probability, the TS is now a professor at that same school using the same old questions. 