# Paying interest and the number e

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Suppose I borrow a dollar from you and I’ll pay you 100% interest at the end of the year. How much money will you have then?

$1 * (1 + 1) = $2

What happens if instead the interest is calculated as 50% twice in the year?

$1 * (1.5 * 1.5) = $2.25

After 6 months I owe you $1.50 and then at the end of the year I pay 50% interest on that amount.

Or 25% four times per year?

$1 * (1.25 * 1.25 * 1.25 * 1.25) = $2.4414

After 3 months I owe you $1.25. At 6 months I pay you 25% interest on that (which yields $1.5625). At 9 months I pay you 25% interest on that, and so on.

## More generally

The formula is:

(1 + 1/*n*)^*n*

where *n* is the number of periods.

We can use R to look at the more general case. Because of R’s vectorization we can do the formula with lots of different *n*‘s all at once. And we can easily plot the results.

`> eseq <- 1:1000
> plot(eseq, (1+1/eseq)^eseq, type="l", col="blue", lwd=3)`

Figure 1: Resulting amount for compounding frequency.From Figure 1 it becomes believable that as the number of periods increases the amount of money converges to a specific value. That is indeed the case and that number is *e*.

If we put the x-axis on a logarithmic scale, then we can see better what happens in the plot. We’ll also add a horizontal line at the value *e*.

`> plot(eseq, (1+1/eseq)^eseq, type="l", col="blue", lwd=3, log="x")
> abline(h=exp(1), lwd=3, col="gold")`

Figure 2: Resulting amount for compounding frequency with logarithmic x-axis.From Figure 2 we see that daily compounding is virtually the same as continuous compounding.

Natural logarithms are those that use *e* as their base. Note that log returns use natural logarithms.

We’ve just seen why log returns are also called continuously compounded returns.

## Epilogue

Money is the seed of money, and the first franc is sometimes more difficult to acquire than the second million.

Jean-Jacques Rousseau

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