$Question:$ Suppose that $x_n$ is the amount owed on a mortgage after n years, $\$m$ is the monthly repayment and $r$ is the annual percentage interest rate charged on the amount of the mortgage outstanding.

(i) Derive a difference equation satisfied by $x_n$.

(ii) Solve the difference equation derived in (i) for a loan of $\$M$ to be repaid over $N$ years and hence determine what the monthly payment should be.

(iii) If the interest rate is $5\%$, show that the monthly repayment on a loan of $\$50,000$ to be repaid over $25$ years is $\$295.64$.

(iv) What is the total amount paid back on the loan?

$My$ $answer$ $so$ $far:$

(i) Let $\$m=x_0$. We need to derive an equation for $x_n$ which will be the amount owed after $n$ years. Thus, $x_0$ is the loan amount owed after $0$ years. So we know that $x_{n-1}$ is the amount owed from the previous year. Also, $x_n$ (the amount owed for the $n_{th}$ year) is related to $x_{n-1}$ (the amount owed for the previous year). $$x_n = f(x_{n-1}) = x_{n-1} + adjustments$$ We make $12$ monthly payments of $m$, so we substract $12m$: $$x_n =x_{n-1} -12m + adjustments$$ Then we are charged $r$ times the amount owed the previous year: $$x_n = x_{n-1} - 12m + r x_{n-1}$$ and hence, $$x_n=(1+r)x_{n-1} - 12m$$

However, I assume my answer is incorrect as I can't find a way to turn it into a difference equation. Any help would be appreciated.

  • $\begingroup$ Please do not delete questions with good answers. Others have devoted effort to answer your question; deleting the question is disrespectful of their effort and prevents others from benefiting from your question and its answers. $\endgroup$
    – robjohn
    Feb 24, 2014 at 16:37

1 Answer 1


I'll show (i). After the first month, the amount owed, $y_1$, is

$$y_1 = x_n \left (1+\frac{r}{12}\right ) - m$$

$$y_2 = y_1 \left (1+\frac{r}{12}\right ) - m = x_n \left (1+\frac{r}{12}\right )^2 - m \left (1+\frac{r}{12}\right ) - m$$

$$y_3 = y_2 \left (1+\frac{r}{12}\right ) - m = x_n \left (1+\frac{r}{12}\right )^3 - m \left [ \left (1+\frac{r}{12}\right )^2 + \left (1+\frac{r}{12}\right ) +1\right ]$$


$$y_{12} = y_{11} \left (1+\frac{r}{12}\right ) - m = x_n \left (1+\frac{r}{12}\right )^{12} - m \left [ \left (1+\frac{r}{12}\right )^{11}+\cdots + \left (1+\frac{r}{12}\right ) +1\right ]$$

Summing the geometric series, we get

$$x_{n+1} = \left (1+\frac{r}{12}\right )^{12} x_n - 12 m \frac{\left (1+\frac{r}{12}\right )^{12}-1}{r}$$

(ii) Show that a solution to a difference equation of the form $x_{n+1} = a x_n+b$ is

$$x_n = A a^n + \frac{b}{1-a}$$

where $A$ is determined from an initial condition. What is the initial condition here? Then set the expression for $x_n$ to $0$ and solve for $m$.

(iii) Plug in the number into the above.

(iv) Multiply the monthly payment times $25 \cdot 12$.

  • $\begingroup$ Thanks for this. I would just like to ask you how did you get that the solution to a difference equation of the form $x_{n+1} = ax_n + b$ is $$x_n = Aa^n + b/(1-a)$$ I haven't seen an example of how to work these kind of problems. Could you please refer me a relative theorem or the way to work with this? $\endgroup$
    – Belial
    Feb 23, 2014 at 18:28
  • $\begingroup$ @Belial: Note that the first piece satisfies $x_{n+1} = a x_n$; the second piece is what you get from plugging in $x_n = \text{constant}$. The general solution you get from adding these two pieces together. $\endgroup$
    – Ron Gordon
    Feb 23, 2014 at 23:03

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