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\( B(t) \quad \text{or} \quad B(\tau) \) Loan balance vs time
\( B_i \) Loan balance after the ith payment
\( B_0\) principal or loan balance at \(t=0\)
\( \phi_i \) Fraction of payment to interest during the ith payment \(\phi = \frac{rB_i}{P} \)
\( \phi \) Fraction of payment to interest \(\phi = \frac{rB}{P} \)
\( r \) Interest rate
\( t_\text{term} \) Loan term
\( n \) Number of loan payments
\( \Delta t \) Time between loan payments\( \Delta t = \frac{t_\text{term}}{n} \)
\( rt_\text{term} \) Loan product, important parameter which fully specifies a loan
\( R \) Helpful collection of variables\( R = 1+r\Delta t = 1+\frac{rt_\text{term}}{n} \)
\( P \) Repayment rate (dollars per time)
\( \tau \) Fraction of loan term \(\tau= \frac{t}{t_\text{term}}\)
\( I \) Total payment to interest
\( V \) Sum of all payments \(V = B_0+I \)
\( \frac{V}{B_0} \) Overpay ratio \(\frac{V}{B_0} = \frac{B_0+I}{B_0} \)

In the limit as \(\Delta t\) approaches 0, the finite difference solution approaches the continuous solution

Goals of this article

This article is purely a matter of mathematical masturbation. Having solved an ODE with a continuous and finite difference method, we are going to show that by taking the limit as \(\Delta t \rightarrow\) 0 we can recover the continuous solution.

Recovering the continuous solution by limits

Let us review the ODE solutions we found and then show how to recover the continuous solution by taking a limit. We began with an ordinary differential equation.

$$ \frac{\partial B}{\partial t} = Br - P $$

The two solutions, discrete and continuous, look different but they should be equivalent in the limit where the discretization parameter \(\Delta t\) approaches 0.

Continuous solution Discrete solution
\( \displaystyle B(t) = \left(B_0-\frac{P}{r}\right)e^{rt}+\frac{P}{r} \) \( \displaystyle B_{i} = B_0\left(1+r\Delta t\right)^{t/\Delta t} -\sum_{i=0}^{t/\Delta t-1} P \left(1+r\Delta t\right)^{i} \Delta t \)

In the limit that \(\Delta t \rightarrow\) 0, we must show two things.

$$\begin{align} B_0\left(1+r\Delta t\right)^{t/\Delta t} &= B_0 e^{rt} \\ \sum_{i=0}^{t/\Delta t-1} P \left(1+r\Delta t\right)^{i} \Delta t &= \frac{P}{r}\left( e^{rt}-1\right) \end{align}$$

Begin with the simple case.

$$ \begin{align} \lim_{\Delta t \rightarrow 0}& B_0 \left(1+r\Delta t\right)^{t/\Delta t} \\ \lim_{\Delta t \rightarrow 0}& B_0 e^{t\log(1+r\Delta t)/\Delta t} \\ \lim_{\Delta t \rightarrow 0}& B_0 e^{rt \frac{1}{1+r\Delta t}} \qquad \text{L'Hopital's rule}\\ & B_0 e^{rt} \end{align} $$

Notice that the more complicated case is the same problem but wrapped in a definition of a definite integral.

$$ \lim_{\Delta t \rightarrow 0} \sum_{i=0}^{t/\Delta t-1} f(i) \Delta t = \int_0^{t}f(\tau) d\tau $$

The limit of a sum is the sum of the limits.

$$ \begin{align} \lim_{\Delta t \rightarrow 0} \sum_{i=0}^{t/\Delta t-1} & P \left(1+r\Delta t\right)^{i} \Delta t \\ \lim_{\Delta t \rightarrow 0} \sum_{i=0}^{t/\Delta t-1} & Pe^{rt} \Delta t\\ \int_0^{t} & Pe^{r\tau} d\tau\\ & \frac{P}{r}\left( e^{rt}-1\right) \end{align} $$

This section just serves to show there is no contradiction between the two methods - the limit of the discrete solution as \(\Delta t \rightarrow\) 0 is the continuous solution.

$$ \lim_{\Delta t \rightarrow 0} \left( B_0\left(1+r\Delta t\right)^{t/\Delta t} -\sum_{i=0}^{t/\Delta t-1} P \left(1+r\Delta t\right)^{i} \Delta t \right) = \left(B_0-\frac{P}{r}\right)e^{rt}+\frac{P}{r} $$
© MC Byington