Difference between revisions of "Number e and compound interest"
m |
m |
||
| (5 intermediate revisions by the same user not shown) | |||
| Line 3: | Line 3: | ||
Suppose, for simplicity, that one period of investment is one year. Suppose further that you invest <math> P = 1 </math> for the period of one year, and the rate of interest is 100% or, equivalently, <math> 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> 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> i = 1 </math> becomes nominal, in the sense that the effective monthly rate of compound interest is computed as <math> j= i/12 = 1/12 </math>. Thus, at the end of one year you will have | Suppose, for simplicity, that one period of investment is one year. Suppose further that you invest <math> P = 1 </math> for the period of one year, and the rate of interest is 100% or, equivalently, <math> 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> 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> i = 1 </math> becomes nominal, in the sense that the effective monthly rate of compound interest is computed as <math> j= i/12 = 1/12 </math>. Thus, at the end of one year you will have | ||
| − | <math> (1+j)^{12} = \left (1+\frac{1}{12}\right)^12 </math> | + | <math> (1+j)^{12} = \left (1+\frac{1}{12}\right)^{12} </math> |
in your account. Using the Binomial formula | in your account. Using the Binomial formula | ||
| Line 11: | Line 11: | ||
we see that | we see that | ||
| − | <math> \left(1+\frac{1}{12}\right)^12 \ge 1 + 12\cdot \frac{1}{12} = 2 </math> | + | <math> \left(1+\frac{1}{12}\right)^{12} \ge 1 + 12\cdot \frac{1}{12} = 2 </math> |
and thus investing at nominal rate of interest 1 is more profitable than at effective annual rate equal to 1. | and thus investing at nominal rate of interest 1 is more profitable than at effective annual rate equal to 1. | ||
| Line 21: | Line 21: | ||
For very large values of <math> n </math> we therefore can consider the limit | For very large values of <math> n </math> we therefore can consider the limit | ||
| − | <math> \lim _{n\to\infty \left(1+\frac{1}{n}\right)^n </math> | + | <math> \lim _{n\to\infty} \left(1+\frac{1}{n}\right)^n </math> |
| − | which corresponds to the interest compounded "continuously". | + | which corresponds to the interest compounded "continuously". It can be shown that the above limit exists (and is approximately equal to <math> 2.7.</math>) This observation leads to the definition of famous mathematical constant - number <math> e</math>: |
| + | |||
| + | <math> e = \lim _{n\to\infty} \left(1+\frac{1}{n}\right)^n.</math> | ||
Latest revision as of 16:03, 3 April 2022
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]\displaystyle{ \left(1+\frac{1}{12}\right)^{12} \ge 1 + 12\cdot \frac{1}{12} = 2 }[/math]
and thus investing at nominal rate of interest 1 is more profitable than at effective annual rate equal to 1.
Further, we may consider scenarios in which the interest is compounded at higher frequencies, e.g. weekly or daily. In general, if [math]\displaystyle{ n }[/math] is the number of times compounding occurs per year, and 1 is the corresponding nominal rate of interest, the value accumulated after one year will be
[math]\displaystyle{ \left(1+\frac{1}{n}\right)^n. }[/math]
For very large values of [math]\displaystyle{ n }[/math] we therefore can consider the limit
[math]\displaystyle{ \lim _{n\to\infty} \left(1+\frac{1}{n}\right)^n }[/math]
which corresponds to the interest compounded "continuously". It can be shown that the above limit exists (and is approximately equal to [math]\displaystyle{ 2.7. }[/math]) This observation leads to the definition of famous mathematical constant - number [math]\displaystyle{ e }[/math]:
[math]\displaystyle{ e = \lim _{n\to\infty} \left(1+\frac{1}{n}\right)^n. }[/math]