The Exponential Function

If a is a positive number and $a \ne 1$ , the exponential function with base a is

$$y = a^x.$$

You know what this means when x is an integer; for example,

$$a^3 = a\cdot a\cdot a, \quad\quad\hbox{and}\quad\quad a^{-4} = \dfrac{1}{a^4}.$$

You also know what this mean if x is a rational number; for instance,

$$a^{2/3} = {\root 3 \of {a^2}}.$$

What does $a^x$ mean if x is not a rational number? That is, suppose x has a decimal expansion which is infinite and does not repeat itself, such as

$$\pi = 3.14159265358979323846264\ldots.$$

What would $2^\pi$ mean?

There are ways of defining this precisely, but I'll take an intuitive approach which relies on limits. Look at

$$2^3, \quad 2^{3.1}, \quad 2^{3.14}, \ldots.$$

These numbers are

$$2^3 = 8, \quad 2^{3.1} \approx 8.57419, \quad 2^{3.14} \approx 8.81524.$$

If you keep going in this way, the numbers will approach a limit, and that limit is $2^\pi$ :

$$2^\pi = 8.82498\ldots.$$

In a similar way, you can think of $a^x$ as a limit of numbers which you get by using more and more of the decimal expansion of x.

There is a special number which is often used as a base for an exponential function:

$$e = 2.718281828459045\ldots.$$

It can be defined by

$$e = \lim_{x\to \infty} \left(1 + \dfrac{1}{x}\right)^x.$$

e is an irrational number, like $\pi$ ; its decimal expansion is infinite, and does not repeat. It may seem puzzling why anyone would want to use such a weird base for an exponential function, rather than an apparently nicer base such as 2 or 10. It turns out that calculus explains what is special about e; in fact, the exponential function $e^x$ satisfies

$$\der {} x e^x = e^x.$$

No other exponential --- $2^x$ , $10^x$ --- works out to be exactly the same as its derivative. This is why when mathematicians and scientists refer to $e^x$ as the exponential function, and why it is used more than other exponential functions in math and science.

Properties of exponentials. Let a be a positive number, $a \ne 1$ .

  1. $a^x$ is defined for all x, $a^x > 0$ for all x.
  2. $a^x$ increases for all x if $a > 1$ and decreases for all x if $0 < x < 1$ .

$$\matrix{\hbox{\epsfysize=1.75in \epsffile{exp1.eps}} & \hbox{\epsfysize=1.75in \epsffile{exp2.eps}} \cr y = 3^x & y = 0.5^x \cr}$$

  1. $a^0 = 1$ .
  2. $a^pa^q = a^{p+q}$ .
  3. $\dfrac{a^p}{a^q} = a^{p-q}$ .
  4. $(a^p)^q = a^{pq}$ .

Example. Let a be a positive number, $a \ne 1$ . Simplify $\dfrac{a^3a^{-7}}{a^2a^{-9}}$ .

$$\dfrac{a^3a^{-7}}{a^2a^{-9}} = \dfrac{a^{-4}}{a^{-7}} = a^{-4-(-7)} = a^3.\quad\halmos$$


If an amount P (the principal) is invested at an annual interest rate of r% compounded k times a year, then after n years the investment is worth

$$A = P\left(1 + \dfrac{r}{k}\right)^{nk}.$$


Example. $1000 is invested at 6% annual interest, compounded monthly. How much is the investment worth after 10 years?

$$A = 1000\left(1 + \dfrac{0.06}{12}\right)^{10\cdot 12} \approx 1819.40\ {\rm dollars}.\quad\halmos$$


Example. How much money must be invested at 6% annual interest, compounded monthly, so that the investment is worth $2000 after 4 years?

$$2000 = P\left(1 + \dfrac{0.06}{12}\right)^{4\cdot 12}, \quad 2000 \approx 1.27049P, \quad P \approx 1574.20\ {\rm dollars}.$$

$1574.20 is said to be the present value of $2000.


Now suppose there is an imaginary investment where the interest is compounded continuously. You might think that you'd make a lot of money on such an investment!

The interest formula is

$$A = P\left(1 + \dfrac{r}{k}\right)^{nk}.$$

Write this as

$$A = P\left(1 + \dfrac{1}{\dfrac{k}{r}}\right)^{rn(k/r)} = P\left[\left(1 + \dfrac{1}{\dfrac{k}{r}}\right)^{k/r}\right]^{rn}.$$

Letting $k\to \infty$ corresponds to continuous compounding, since k is the number of times you compound each year. But as $k\to \infty$ , $\dfrac{k}{r}\to \infty$ , so

$$\left(1 + \dfrac{1}{\dfrac{k}{r}}\right)^{k/r}\to e.$$

Thus, if a principal P is invested at r% annual interest compounded continuously for n years, the value of the investment is

$$A = Pe^{rn}.$$


Example. $500 is invested for 3 years at 6% interest compounded continuously. Find the value of the investment.

$$A = 500e^{3\cdot 0.06} \approx 598.61\ {\rm dollars}.\quad\halmos$$


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