In the limit as \(\Delta t\) approaches 0, the finite difference solution approaches the continuous solution
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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 | \(\frac{V}{B_0} = \frac{B_0+I}{B_0} \) |
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 of the discrete solution 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.
$$ \lim_{\Delta t \rightarrow 0} \left( B_0\left(1+r\Delta t\right)^{t/\Delta t} -\sum_{i=0}^{t/\Delta t-1} P \left(1+r\Delta t\right)^{i} \Delta t \right) = \left(B_0-\frac{P}{r}\right)e^{rt}+\frac{P}{r} $$