Suppose we initially have $$ 100$ in our bank account and the bank has a generous interest rate of $50 %$ compounded annually. This means that every year, the bank will increase however much money we have in our bank account by $50 %$ . Let's see how our money grows year after year:

Today	1st year	2nd year	3rd year
$$ 100$	$100 (1 + 0.5) = $ 150$	$150 (1 + 0.5) = $ 225$	$225 (1 + 0.5) = $ 337.5$

In general, we can see that the new balance for any year is computed by:

\begin{matrix} (1) & P_{t} = P_{t - 1} (1 + r) \end{matrix}

Where:

$P_{t}$ is the bank balance of the $t$ -th year.
$P_{t - 1}$ is the bank balance of the previous ( $t - 1$ -th) year.
$r$ is the interest rate in decimals.

But the bank balance of the previous year $P_{n - 1}$ is computed as follows:

\begin{matrix} (2) & P_{t - 1} = P_{t - 2} (1 + r) \end{matrix}

Substituting $(2)$ into $(1)$

\begin{matrix} (3) & P_{t} = P_{t - 2} (1 + r)^{2} \end{matrix}

Again $P_{t - 2}$ is computed using $P_{t - 3}$ :

\begin{matrix} (4) & P_{t - 2} = P_{t - 3} (1 + r) \end{matrix}

Substituting $(4)$ into $(3)$ gives:

\begin{matrix} (5) & P_{t} = P_{t - 3} (1 + r)^{3} \end{matrix}

We can keep repeating this until we get to the present day, that is, when we have our initial bank balance $P_{0}$ :

\begin{matrix} (6) & P_{t} = P_{0} (1 + r)^{t} \end{matrix}

This equation gives us the bank balance after $n$ years when the interest $r$ is compounded annually. What if the interest is not compounded annually, but instead semi-annually, that is, twice a year? By definition of compound interest, this would mean that we would receive the interest twice as frequently, but the interest rate will also be cut in half.

Once again, let's assume that we initially have $$ 100$ in our bank account and the bank has an interest rate of $50 %$ compounded semi-annually:

Today	0.5 year	1 year	1.5 year
$$ 100$	$100 (1 + 0.25) = $ 125$	$125 (1 + 0.25) = $ 156.25$	$156.25 (1 + 0.25) = $ 195.3125$

We can see that the formula to compute the bank account after $t$ years is:

\begin{matrix} (7) & P_{t} = P_{t - (1 / 2)} (1 + \frac{r}{2}) \end{matrix}

Here, $P_{t}$ is dependent on $P_{t - (1 / 2)}$ , that is, the bank balance half a year ago. Similarly, $P_{t - (1 / 2)}$ is computed by:

\begin{matrix} (8) & P_{t - (1 / 2)} = P_{t - 1} (1 + \frac{r}{2}) \end{matrix}

Substituting $(8)$ into $(7)$ gives:

\begin{matrix} (9) & P_{t} = P_{t - 1} (1 + \frac{r}{2})^{2} \end{matrix}

If we keep repeating the same process until we reach the present day, we should end up with:

\begin{matrix} (10) & P_{t} = P_{0} (1 + \frac{r}{2})^{2 t} \end{matrix}

Notice the following:

the exponent is $2 t$ instead of $t$ because the frequency in which we receive the interest is doubled.
we are dividing $r$ by $2$ because the interest rate is cut in half.

If the interest is compounded quarterly, then we can guess that the formula is:

\begin{matrix} (11) & P_{t} = P_{0} (1 + \frac{r}{4})^{4 t} \end{matrix}

This leads us to the general formula for compound interest.

Theorem.

Formula to compute compound interest

The compound interest is calculated by:

\begin{array}{r} A = P (1 + \frac{r}{n})^{n t} \end{array}

Where:

$A$ is the final amount in the bank account after $t$ years.
$P$ is the principal, that is, the starting amount in the bank account.
$r$ is the interest rate (e.g. $0.1$ ).
$n$ is the number of compounds per year.
$t$ is the number of years elapsed.

Example.

Computing compound interest (1)

Suppose we have $$ 100$ in our bank account and the bank has an interest rate of $40 %$ compounded quarterly. What would our balance be at the end of $2$ years?

Solution. We are given the following information:

P = $ 100, r = 0.4, n = 4, t = 2

Let's directly apply our formula:

\begin{aligned} A & = P (1 + \frac{r}{n})^{n t} \\ = (100) (1 + \frac{0.4}{4})^{(4) (2)} \\ = (100) (1.1)^{8} \\ \approx $ 214 \end{aligned}

Therefore, we should expect to have roughly $$ 214$ in our bank account after $2$ years!

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Example.

Computing compound interest (2)

Suppose we have $$ 150$ stored in our bank account. The bank has an interest rate of $10 %$ compounded annually. What would our bank balance be after $3$ years?

Solution. We are given the following information:

P = $ 150, r = 0.1, n = 1, t = 3

We apply our formula for compound interest:

\begin{aligned} A & = P (1 + \frac{r}{n})^{n t} \\ = (150) (1 + \frac{0.1}{1})^{1 \cdot 3} \\ = (150) (1.1)^{3} \\ = $ 199.65 \end{aligned}

Therefore, after $3$ years, we would end up with roughly $200$ dollars in our bank account!

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Published by Isshin Inada

Edited by 0 others

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