# The maths behind Compound Interest

So we have all heard of the power of compounding, so I thought I would do a brief post on the maths behind compound interest.

# What is compound interest?

Compound interest is where you earn interest on the interest you have earned. Simple interest is interest only on the principle balance.

Lets look at a very simple example. Lets assume you have \$100, and the annual rate of interest is 10%, and lets also assume the interest is payable once a year.

So at year 0 (i.e. the start) you have \$100
In Year 1, you earn 10% of \$100, which is \$10. So now the balance is \$110.
In Year 2, you earn 10% on \$110, which is \$11. So now the balance is \$110+\$11 = \$121.
In Year 3, you earn 10% on \$121, which is \$12.10. So now the balance is \$121+12.10 = \$132.10

This pattern repeats year after year. If you haven't picked it up already, each year we are multiplying by 1.1 form the previous value.

So the balance at time n, i.e. B(n) = 100 * 1.1^n

In the more general sense, B(n) = B(0) * (1+r)^n
where
B(n) = The balance at year n
r = is the interest rate expressed as a decimal

# How about more frequent compounding

Lets look at another example, how much the balance will be at the end of one year, at different compounding frequencies.

So if you are paid interest at the end of the year, the balance would be 100*1.1 = \$110, same as above.

But what if you are paid interest twice in the year, so 5% in the first half and 5% in the second half.

at the end of the first 6 months, the balance would be \$105, and the interest in the second 6 months would be 5% of \$105, which is \$5.25. Therefore the balance at the end of the year is \$110.25, and the effective annual yield is 10.25%

so we can generalise the formula to be
B(e) = B(s) * (1+r/n)^n

Where
B(e) = Balance at the end of the year
B(s) = Balance at the start of the year
r = interest rate (p.a.)
n = number of times the interest is paid per year

So if we compounded 12 times a year (i.e. monthly) the answer would be 100*(1+0.1/12)^12 = 110.47

or 10.47% effective annual yield.

So as you can see the yield increases as we compound more frequently

# What happens if we compound continuously

So if we decide we wanted to compound every second, or every millisecond, eventually getting to infinitely small intervals, what happens?

so the equation (1+r/n)^n as n approaches infinity tends to e^r
where,
r = interest rate
e = Euler's constant - 2.718

So if the interest rate is 10%, the continuously compounded balance is 100*e^0.1 = \$110.52

The effective rate is e^n-1 = 10.52%