# The maths behind Compound Interest

in STEMGeeks2 months ago

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%

# Concluding comments

Compound interest is a powerful tool people use to generate wealth, the maths isn't that hard when you think about it, its just multiplying the interest rate on your balance several times. Most scientific calculators (and excel) can do this pretty simply, its a very useful tool to have in your repertoire of knowledge, especially those interested in investing. But the other interesting fact was, that the benefits of compounding more frequently aren't really large, at normal rates of returns (if you are in a defi farm earning 1000% p.a. that's a different story)

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It would be really cool if Hive frontends supported proper LaTeX equations in their markdown. I'm often put off from writing mathematical posts on hive because the equations look so ugly.

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Agree, it would be nice

2 months ago (edited)

Consider Quicklatex.com. That is what I use often as a workaround solution (for now).

That's a pretty neat tool, thanks for letting me know about it! I'll use it next post.

Though we get interest in compounded format, i never had the chance to look at the formulae and the mathematics behind the same. This was really a great calculation. Thank you for the post.

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