e

Bernoulli’s compound interest problem

The discovery of e emerged from work Swiss mathematician Jakob Bernoulli did on compound interest. Bernoulli considered the following problem.

If one deposits $1 at a bank that gives 100 percent interest per year, after one year the final value in the account is $2. However, if the same investment gives 50 percent interest but pays twice a year (still 100 percent interest in the year), one has $1.50 after six months and then $2.25 at the end of the year.

Clearly, paying interest more often within a year offers an investment that pays more in total interest.

If the bank gives 100 percent annual interest compounded monthly (100/12 percent each month), the final value in dollars is: $1{\left(1+\frac{1}{12}\right)}^{12}\approx 2.61$.

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If the interest is paid n times through the year, the interest is(1 + 1/n )ⁿ.Bernoulli saw that the amount of the return on the investment continued to get larger, but there was a decreasing rate of increase. As the interest is paid more often, a smaller amount is paid each time, and the value in the account tends to a specific limit. Bernoulli did not calculate the limit as n goes to infinity but determined that it was between 2 and 3.

Series expansion of e

In 1748 Euler published Introductio in analysin infinitorum in which he calculated e to 18 decimal places. He showed that e was the limit in Bernoulli’s compound interest problem and that e could also be calculated from an infinite series: $e=1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...$For the generalized exponential function:$e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...$.

Euler’s identity

In the Euler identity,$e^{ix}=cosx+isinx$where i is the square root of −1 (see imaginary number).

When x is equal to π or 2π, the formula yields two elegant expressions relating π, e, and i:$e^{i\pi }=-1$$e^{2i\pi }=1$