Exponential growth and decay (Part 7): Paying off credit-card debt via recurrence relations

The following problem in differential equations has a very practical application for anyone who has either (1) taken out a loan to buy a house or a car or (2) is trying to pay off credit card debt. To my surprise, most math majors haven’t thought through the obvious applications of exponential functions as a means of engaging their future students, even though it is directly pertinent to their lives (both the students’ and the teachers’).

You have a balance of $2,000 on your credit card. Interest is compounded continuously with a rate of growth of 25% per year. If you pay the minimum amount of $50 per month (or $600 per year), how long will it take for the balance to be paid?

In previous posts, I approached this problem using differential equations. There’s another way to approach this problem that avoids using calculus that, hypothetically, is within the grasp of talented Precalculus students. Instead of treating this problem as a differential equation, we instead treat it as a first-order difference equation (also called a recurrence relation):

A_{n+1} = r A_n - k

The idea is that the amount owed is multiplied by a factor r (which is greater than 1), and from this product the amount paid is deducted. With this approach — and unlike the approach using calculus — the payment period would be each month and not per year. Therefore, we can write

A_{n+1} = \displaystyle \left( 1 + \frac{0.25}{12} \right) A_n - 50

Notice that the meaning of the 25% has changed somewhat… it’s no longer the relative rate of growth, as the 25% has been equally divided for the 12 months.

green line

A full treatment of the solution of difference equations belongs to a proper course in discrete mathematics. In the previous posts, I demonstrated how this difference equation could be solved by directly finding A_1, A_2, A_3, \dots and looking for a pattern.

In this post, I’d like to present an alternative method for deriving the solution. I’ll let the reader decide for him/herself as to whether this technique is pedagogically superior to the previous method. We will attempt to find a solution of the form

A_n = a r^n + b,

where a and b are unknown constants.Why do we guess the solution to have this form? I won’t dive into the details, but this is entirely analogous to constructing the characteristic equation of a linear differential equation with constant coefficients as well as using the method of undetermined coefficients to find a particular solution to a inhomogeneous linear differential equation with constant coefficients.

Substituting n+1 instead of n, we find that

A_{n+1} = a r^{n+1} + b.

So we plug both of these into the difference equation:

A_{n+1} = r A_n - k

a r^{n+1} + b = r \left( a r^n + b \right) - k

a r^{n+1} + b = a r^{n+1} + r b - k

b = r b - k

k = (r-1) b

\displaystyle \frac{k}{r-1} = b

We also use the fact that A_0 = P:

A_0 = a r^0 + b

P = a + b

P - b = a

\displaystyle P - \frac{k}{r-1} = a

Combining these, we obtain the solution of the difference equation:

A_n = \displaystyle \left( P - \frac{k}{r-1} \right) r^n +\frac{k}{r-1}

Unsurprisingly, this matches the solution that was obtained in the previous two posts (though the terms have been rearranged).

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  1. Exponential growth and decay: Index | Mean Green Math

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