Number e and compound interest
Suppose you invest some amount [math]\displaystyle{ P }[/math] for a period of time (that can be a week, month, year etc.) at effective rate of compound interest [math]\displaystyle{ i. }[/math] At the end of one such period you have [math]\displaystyle{ P +iP = P(1+i) }[/math] accumulated in your account. If you keep your investment for two such periods, the value in your account will be [math]\displaystyle{ P(1+i) + i[P(1+i)] = P(1+i)^2. }[/math] In general, if the investment is for [math]\displaystyle{ n }[/math] periods, at the end your account will have accumulated [math]\displaystyle{ P(1+i)^n. }[/math]
Suppose, for simplicity, that one period of investment is one year. Suppose further that you invest [math]\displaystyle{ P = 1 }[/math] for the period of one year, and the rate of interest is 100% or, equivalently, [math]\displaystyle{ i = 1 }[/math] (this means that the amount you invested will double at the end of the year, which, of course, rarely occurs in practice). Alternatively, you may consider investing the same amount, [math]\displaystyle{ P =1 }[/math], but so that the effective interest period is a fraction of a year, say, 1 month. In this case, the stated rate of interest, [math]\displaystyle{ i = 1 }[/math] becomes nominal, in the sense that the effective monthly rate of compound interest is computed as [math]\displaystyle{ j= i/12 = 1/12 }[/math]. Thus, at the end of one year you will have
[math]\displaystyle{ (1+j)^{12} = \left (1+\frac{1}{12}\right)^12 }[/math]
in your account. Using the Binomial formula
[math]\displaystyle{ (1+x)^n = \sum_{k=0}^n {n\choose k} x^k }[/math]
we see that
<math> (1+\frac{1}{12}\right)^12