Are you lucky enough to have some money to invest? If yes then you might have been struggling with compound interest and this would put you in the exalted company of the great seventeenth century mathematician Jacob Bernoulli. Suppose you have $1 and invest it for a year at an (admittedly unrealistic) interest rate of 100%. If the interest is credited once at the end of the year, you will walk away with
$1+$1 = $2.
If interest is compounded twice a year, at 50% after six months and again at the end of the year, you will walk away with
$1+$0.5 + $0.5(1+0.5) = $(1+1/2)2 = $2.25
In a similar vein, if it is credited monthly, you will get
$(1+1/12)12= $1.0816 = $2.613.
What, so Bernoulli wondered, happens as interest is charged more and more frequently: weekly, daily, hourly, once a minute, and so on? This corresponds to substituting larger and larger values for k into the formula
The numbers that result don’t jump about erratically, neither do they grow beyond all bounds. Instead, as Bernoulli worked out, they approach a specific value:
usually denoted by the letter e. If interest is compounded continuously, at every instant in time, your return after a year will be exactly e dollars. More generally, the value of an investment of P after t years at an interest rate of r, compounded continuously, is $ Pert.
Why would you do such a crazy thing as continuous compounding? Because it makes the maths easier. For example, if the interest rate is 6% for the first year and 7% for the second, then after two years with continuous compounding you have
$ Pe0.06 e0.07.
This can be worked out by simply adding the powers:
Pe0.06+0.07 = Pe0.13 = 1.14P,
rounded to two decimal places.
Of course your bank doesn’t always talk to you in terms of continuous compounding, so before you can use the handy number e in your calculations you need to convert the rate given to you by your bank into a continuously compounded rate. For example suppose the rate is 6% compounded once a year. We need to find the number r which gives
Per = P(1+0.06).