Continuously Compounding Interest Rate

Continuous Compounding

In the limit, we compound increasingly frequently to obtain continuously compounded interest rates. In our previous formulation,

(1) $\begin{equation*} \left(1+\frac{r}{m} \right)^m\end{equation*}$

if we increase m to infinity, then we can express it as

(2) $\begin{equation*} \lim_{m\to \infty}\left(1+\frac{r}{m} \right)^m = e^r\end{equation*}$

e^r is an exponential function. e is a mathematical constant, the value of approximately 2.71828…. . Yes, it's ....., which means more digits, and it continues forever. However, we use 2.718 or 2.72 as an approximation. e shows up whenever systems grow exponentially and continuously, for example, interest rate. We can write e as follows:

(3) $\begin{equation*} e=\sum_{n=0}^{\infty}{\frac{1}{n!}}={\frac{1}{0!}}+{\frac{1}{1!}}+{\frac{1}{2!}}+{\frac{1}{3!}}+\dots \end{equation*}$

where, n! is the factorial of a positive integer n, and can be expressed as:

(4) $\begin{equation*} n! = 1 \times 2 \times 3 \times ........ \times (n-2) \times (n-1) \times n\end{equation*}$

For example,

(5) $\begin{equation*} 2! = 2 \times 1=2 \end{equation*}$

(6) $\begin{equation*} 3! = 3 \times 2 \times 1=6 \end{equation*}$

(7) $\begin{equation*} 4! = 4 \times 3 \times 2 \times 1=24 \end{equation*}$

Remember,

(8) $\begin{equation*} 0! = 1 \end{equation*}$

is called empty product. So, we can write,

(9) $\begin{equation*} e=\sum_{n=0}^{\infty}{\frac{1}{n!}}={\frac{1}{1}}+{\frac{1}{1\times 1}}+{\frac{1}{1\times 2}}+{\frac{1}{1\times 2\times 3}}+\dots \end{equation*}$

(10) $\begin{equation*} e=1+\left(1+{\frac{1}{2}}+{\frac{1}{6}}+\dots\right) \end{equation*}$

Let's try to explain the above equation with an example. Imagine, you deposit $1 in the bank with an interest rate of 100% per unit of time. This 'per unit' can be 1 year or 1 month, 1 hour. Here 1 is the unit of measurement. The first 1 outside the bracket in the equation is your $1 deposit or principal. The first 1 in the bracket is the interest earned on the principal as the interest rate is 100%.

Now divide the whole time into two equal lengths. Each length would be 1/2. Then the interest earned on the previous interest of $1 dollar would be 1/2 or $0.5, which is the second element in the bracket. In other words, it is the interest on interest.

Now divide the whole time into three equal lengths, each length would be 1/3. Then the interest earned on the previous interest of $1 dollar and the following interest of 1/2 would be

(11) $\begin{equation*} \frac{1}{6} \left(={\frac{1}{2}}\times{\frac{1}{3}}\right) \end{equation*}$

which is third element in the bracket.

This way it continues forever. As a result,

(12) $\begin{equation*} \left(1+{\frac{1}{2}}+{\frac{1}{6}}+\dots\right) \end{equation*}$

represents (interest + interest on interests + interest on interests on interests + ......). Thus, we find the outcome both exponential and continuous. And we say interest rate, r is continuously compounded per unit of time.

(13) $\begin{equation*} \left (1+{\frac{1}{2}}+{\frac{1}{6}}+\dots)\right \end{equation*}$

can be expressed as

(14) $\begin{equation*} \left({\frac{1}{n})\right)^n \end{equation*}$

e is then the limit of

(15) $\begin{equation*} \left (1+{\frac{1}{n}})\right)^n \end{equation*}$

as n approaches infinity. In other words,

(16) $\begin{equation*} e= \lim_{n\to \infty}\left [1+{\frac{1}{n}\right]}^n \end{equation*}$

Let's r is the interest rate. Then the above equation can be written as

(17) $\begin{equation*} e^r=\sum_{n=0}^{\infty}{\frac{r^n}{n!}}={\frac{1}{0!}}+{\frac{r}{1!}}+{\frac{r^2}{2!}}+{\frac{r^3}{3!}}+\dots= \lim_{n\to \infty}\left [1+{\frac{r}{n}\right]}^n \end{equation*}$

When r = 100% per unit of time, which is continuously compounded, or in decimal format, r = 1, e^r becomes e¹= e, and we end up with equation (5).

What would be the value of $1 if the interest rate, r = 0?

If the interest rate is zero, the principal of $1 will remain $1. This is evident as e^r= e⁰=1.

When r = 25% per unit of time, or in decimal format, r = 0.25, e^r becomes e^0.25, and we end up with:

(18) $\begin{equation*} \lim_{n\to \infty}\left [1+{\frac{0.25}{n}\right]}^n \end{equation*}$

Since the interest rate is normally expressed on a per year basis, and when the rate is also continuously compounded, the above example with a 25% interest rate represents equation (6).

If we continue with 100% interest rate (in decimal, 1) per annum, your $1 becomes

(19) $\begin{equation*} \left(1+r\right) = \left(1+{\frac{1}{1}\right)^1=\$2 \end{equation*}$

with annual compounding (n = 1).

If the bank pays you interest twice a year, then we can say interest rate is compounded semi-annually (n = 2). In other words, the bank is going to pay 50% half-yearly. In such case, your $1 in your account yields

(20) $\begin{equation*} \left(1+{\frac{r}{2}\right)^2 = \left(1+{\frac{1}{2}\right)^2= \$2.25 \end{equation*}$

at the end of the year. Compounding interest rate quarterly (n = 4) yields

(21) $\begin{equation*} \left(1+{\frac{1}{4}\right)^4= \$2.44 \end{equation*}$

and compounding interest rate weekly (n = 52) yields

(22) $\begin{equation*} \left(1+{\frac{1}{52}\right)^{52}= \$2.69 \end{equation*}$

Similarly, compounding daily (n = 365) yields

(23) $\begin{equation*} \left(1+{\frac{1}{365}\right)^{365}= \$2.7145 \end{equation*}$

and compounding daily yields

(24) $\begin{equation*} \left(1+{\frac{1}{365}\right)^{365}= \$2.7145 \end{equation*}$

As the frequency of compounding increases, n grows larger and larger leading to shorter and shorter compounding intervals. Thus, in the limit as n →∞, we get e = 2.71828....

What would be value of $1 in 2 years if the annual interest rate is r with continuously compounding?

It would be

(25) $\begin{equation*} e^{r.2}=\lim_{n\to \infty}\left [1+{\frac{r}{n}\right]}^{n.2} \end{equation*}$

In general, we can write this as:

(26) $\begin{equation*} e^{rt}=\lim_{n\to \infty}\left [1+{\frac{r}{n}\right]}^{nt} \end{equation*}$

Thus, e^rt is what $1 yields in t years with the continuously compounded interest r per annum, where r is expressed in decimal.

Using the first four terms, we get:

(27) $\begin{equation*} e = 1+\left (1+\dfrac{1}{2}+\dfrac{1}{6} \right) = 2.6667 \end{equation*}$

Using the first six terms, we get:

(28) $\begin{equation*} e = 1+\left(1+\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{24}+\dfrac{1}{120} \right)= 2.71667 \end{equation*}$

Similarly, using the first ten terms, we get e = 2.71828. As you can see e converses close to 2.718. Considering t years, we can express it as

(29) $\begin{equation*} \lim_{m\to \infty} \left(1+\frac{r}{m})\right)^{mt} = e^{rt} \end{equation*}$