Financial Mathematics: Continuous Compounding Interest

Earlier we set up the basic compounding (discounting) factor for when interest is compounding at smaller intervals than 1 year:
$$\left(1+\frac{i}{m}\right)^{nt}$$
where $i$ is the interest rate, $n$ is the number of divisions in a year and $t$ is time expressed in years.

Next we're going to suppose that compounding occurs, instead of annually, quarterly, monthly or even daily, but rather in a continuous manner. This means that our time interval goes to 0 or the number of intervals in the year is infinite.

But before that I want to introduce a quick concept:

Accumulator Function

The accumulator function shows how, say, money in account earning interest changes over time. The fundamental relationship looks like this:

$$A(t) = A(0)a(t)$$
where $ A(0) $ is the initial amount invested (or the present value) and $ A(t) $ describes the value of that investment at some future time. Therefore $ a(t) $ is the factor at which interest is accrued or discounted.

For simple interest we have:
$$a(t) = (1+it)$$
For annual compounding interest we have:
$$a(t)= (1+i)^t$$
And lastly, for periodic compounding interest we have (as stated above):
$$a(t)=\left(1+\frac{i}{n}\right)^{nt}$$

Continuous Compounding Interest

There are two techniques I will use to show this and both require calculus. For those of you not mathematically inclined, you may wish to skip to the result and hope I haven't made any errors.

Derivation 1

So to begin we will look at it periodic compounding interest but we're going to do is take the limit as $\lim n \to \infty$:
$$a(t)= \lim_{n \to \infty} \left(1+\frac{i}{n}\right)^{nt} $$
We will make the following substitutions: $x = \frac{i}{n}$ and, consequently, $nt = \frac{it}{x}$. As a result we will look at the $\lim x \to 0$.
$$a(t)=\lim_{x \to 0} (1+x)^{it/x}$$
Next, we'll take the natural log and then exponentiate this expression:
$$a(t)=\lim_{x \to 0}\exp\left(\frac{it}{x} \ln(1+x)\right)$$
It can be shown that by l'Hôpital's rule that:
$$\lim_{x \to 0}\frac{\ln(1+x)}{x}=1$$
As a result, the entire expression converges to:
$$a(t)=e^{it}$$
Derivation 2

The second approach requires that we set up a differential equation. Imagine I have an account whose value is a function of time, $A(t)$. We want to know how this account changes in value over time. We'll assume no money is deposited to this account except for the interest earned on it.

Now for simplicity, we'll initially assume that interest is compounded annually. So if I have $ \$100 $ and I save it in an account earning $5 \%$ interest, how much as my account changed? If $ \Delta A $ is the change in the value of the account and $ \Delta t $ is the interval of time, then the amount that the account changes is:
$$\Delta A = A(t) i \Delta t = \$ 100 \times 5 \% \times 1 (\text{year}) = \$ 5 $$
Now if we allow our time interval, $\Delta t$, to get infinitely small, we set up this basic differential equation:
$$dA = A(t) i dt$$
Another way to view this (and this may help those of you who are not mathematically inclined) is as follows:
$$\frac{dA}{dt} =A(t) i $$
The left-hand side of the equation is called a derivative which is just a generalization of slope. It tells you how much your account value will change as time progresses. The right side of the equation says that it will increase by the interest paid expressed as an interest rate multiplied by the current value of your account.

To solve this equation we're going to rewrite it again and integrate both sides (notice my hands waving while I yell "calculus!"):
$$\int{\frac{dA}{A(t)}} = \int{idt} $$
This turns out to be a separable ordinary differential equation which, given an initial value of $A(0)$, has a solution:
$$\ln |A(t)| =it + c$$
$$A(t) = A(0)e^{it}$$
As such, $ a(t)=e^{it} $ just as we showed above.

Use of Continuous Compounding Interest

Apart from instances in which interest is compounded continuously, what use does it have? Well, it serves as a decent approximation for compounding at less frequent intervals (monthly, daily, etc).

Suppose you have $ \$ 100 $ and you're compounding that at a rate of  $6 \% $ interest. How much money will you have after $t$ number of years depending on how frequently you compound it?

After 20 years, there's a clear advantage to having interest compounding at more frequent intervals. You would have $ \$ 11.30 $ more ($ 11.3 \% $ more of the initial value) by compounding continuously instead of annually.

Nonetheless, continuous compounding is a pretty good approximation for daily, monthly or even quarterly compounding and so it often gets used as the expression can be much easier to work with.

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