1. The Solution of a Difference Equation for a Compound Interest Account

2. Basic Compound Interest Formula Recall that the new balance of an account that earns compound interest can be found by using the formula B new = (1 + i) B previous. Recall the following: B new represents the new balance (which can also be thought of as the next balance) B previous represents the previous balance i represents the interest rate per compound period (i.e. i = r/m)

3. Formula for a Future Balance of a Compound Interest Account A future balance of a compound interest account can be found by using the formula F = (1 + i) n P. Recall the following: F represents the future balance n represents the number of compound periods i represents the interest rate per compound period (i.e. i = r/m) P represents the principal deposited into the account The above formula allows us to find a future balance without having to find any of the previous balances.

4. The formula, F = (1 + i) n P, is a solution of the difference equation B new = (1 + i) B previous. The following slides are a proof of the above statement.

5. Recall the General Form of a Difference Equation In chapter 11, a generalized difference equation (with an initial value) is written as: y n = a∙y n-1 + b, y 0 where y n represents the next value in the list y n-1 represents the previous value in the list a is some value that is the coefficient of y n-1 b is some constant value y 0 is the initial value

6. The formula B new = (1 + i)B previous is a difference equation since it is in the form of y n = a∙y n-1 + b. The next slide will show this, but first we will rewrite B new = (1 + i)B previous as B new = (1 + i)∙B previous + 0.

7. B new = (1 + i) ∙ B previous + 0 y n = a ∙ y n-1 + b where B new is equivalent to the y n B previous is equivalent to the y n-1 a = (1 + i) b = 0 Note that b/(1 – a) = 0 [Proof: b/(1 – a) = 0/(1 – (1 + i)) = 0 / (– i) = 0. ]

8. The Initial Value The initial value, y 0, for a account that earns compound interest is the principal, P (which is the initial deposit). Thus y 0 = P.

9. Recall When Solving a Difference Equation Since a ≠ 1 (the reason will be, or was, discussed in the lecture), then the solution of the difference equation can be found using y n = b/(1 – a) + ( y 0 – b/(1 – a) ) a n. Now substituting the chapter 10 notations into the above formula.

10. The Substitutions Substituting 0 for b/(1 – a), P for y 0, and (1+i) for a, the solution of y n = b/(1 – a) + ( y 0 – b/(1 – a) ) a n becomes y n = 0 + ( P – 0) (1+i) n. Note that we did not change the notation of y n ; we will address this later.

11. Simplification Using Algebra The formula y n = 0 + ( P – 0) (1+i) n can be simplified to y n = P (1 + i) n ; which is equivalent to y n = (1 + i) n P. We used the additive property of 0 and the commutative property of multiplication.

12. The y n Notation Change Now since y n, in y n = (1 + i) n P, represents a specified value in the list (mainly the n th value in the list after the initial value) without using any of the previous values (with the exception of the initial value P), the y n can be changed to an F to represent the future value. So y n = (1 + i) n P is changed to F = (1 + i) n P.

13. Therefore F = (1 + i) n P is the solution of the difference equation B new = (1 + i)B previous. Q.E.D.