# 4.2: The Function e^x

Many of you know the number e as the base of the natural logarithm, which has the value 2.718281828459045. . . . What you may not know is that this number is actually defined as the limit of a sequence of approximating numbers. That is,

$e=\lim_{n→∞}f_n$

$f_n=(1+\frac 1 n)^n\;,\;n=1,2,...$

This means, simply, that the sequence of numbers $$(1+1)^1,(1+\frac 1 2)^2,(1+\frac 1 3)^3, . . .$$, gets arbitrarily close to 2.718281828459045. . . . But why should such a sequence of numbers be so important? In the next several paragraphs we answer this question.

Exercise $$\PageIndex{1}$$

(MATLAB) Write a MATLAB program to evaluate the expression $$f_n=(1+\frac 1 n)^n$$ for $$n=1,2,4,8,16,32,64$$ to show that $$f_n≈e$$ for large $$n$$.

## Derivatives and the Number e

The number $$f_n=(1+\frac 1 n)^n$$ arises in the study of derivatives in the following way. Consider the function

$f(x)=a^x\;,\;a>1$

and ask yourself when the derivative of $$f(x)$$ equals $$f(x)$$. The function $$f(x)$$ is plotted in the Figure for $$a>1$$. The slope of the function at point $$x$$ is

$df(x)dx=\lim_{Δx→0}{\frac {a^{x+Δx}−a^x} {Δx}} = α^x \lim_{Δx→0} {\frac {α^{Δx}−1} {Δx}}$

If there is a special value for $$a$$ such that

$lim_{Δx→0} \frac {a^{Δx}−1} {Δx}=1$

then $$\frac d {dx} f(x)$$ would equal $$f(x)$$. We call this value of a the special (or exceptional) number $$e$$ and write

$f(x)=e^x$

$\frac d {dx}f(x)=e^x$

The number $$e$$ would then be $$e=f(1)$$. Let's write our condition that $$\frac {a^{Δx}−1} {Δx}$$ converges to 1 as

$e^{Δx}−1≅Δx\;,\;Δx\;\mathrm{small}$

or as

$e≅(1+Δx)^{1/Δx}$

Our definition of $$e=\lim_{n→∞}(1+1n)^{1/n}$$ amounts to defining $$Δx=\frac 1 n$$ and allowing $$n→∞$$ in order to make $$Δx→0$$. With this definition for $$e$$, it is clear that the function $$e^x$$ is defined to be $$(e)^x$$:

$e^x=\lim_{Δx→0}(1+Δx)^{x/Δx}$

By letting $$Δx=\frac x n$$ we can write this definition in the more familiar form

$e^x=\lim_{n→∞}(1+\frac x n)^n$

This is our fundamental definition for the function $$e^x$$. When evaluated at $$x=1$$, it produces the definition of $$e$$ given in Equation.

The derivative of $$e^x$$ is, of course,

$\frac d {dx} e^x = \lim_{n→∞}n(1+\frac x n)^{n−1}\frac 1 n = e^x$

This means that Taylor's theorem1 may be used to find another characterization for $$e^x$$:

$e^x=∑^∞_{n=0}[\frac {d^n} {dx^n} e^x]_{x=0} \frac {x^n} {n!} = ∑^∞_{n=0}\frac {x^n} {n!}$

When this series expansion for $$e^x$$ is evaluated at x=1, it produces the following series for e:

$e=∑_{n=0}^∞\frac 1 {n!}$

In this formula, $$n!$$ is the product $$n(n−1)(n−2)⋯(2)1$$. Read $$n!$$ as "n factorial.”

Exercise $$\PageIndex{1}$$

(MATLAB) Write a MATLAB program to evaluate the sum

$S_N=∑_{n=0}^N\frac 1 {n!}$

for $$N=1,2,4,8,16,32,64$$ to show that $$S_N≅e$$ for large $$N$$. Compare $$S_{64}$$ with $$f_{64}$$ from Exercise 1. Which approximation do you prefer?

## Compound Interest and the Function ex

There is an example from your everyday life that shows even more dramatically how the function $$e^x$$ arises. Suppose you invest $$V_0$$ dollars in a savings account that offers 100x% annual interest. (When x=0.01, this is 1%; when x=0.10, this is 10% interest.) If interest is compounded only once per year, you have the simple interest formula for $$V_1$$, the value of your savings account after 1 compound (in this case, 1 year):

$$V_1=(1+x)V_0$$. This result is illustrated in the block diagram of the Figure. In this diagram, your input fortune $$V_0$$ is processed by the “interest block” to produce your output fortune $$V_1$$. If interest is compounded monthly, then the annual interest is divided into 12 equal parts and applied 12 times. The compounding formula for $$V_{12}$$, the value of your savings after 12 compounds (also 1 year) is

$V_{12}=(1+\frac x {12})^{12}V_0$

This result is illustrated in Figure. Can you read the block diagram? The general formula for the value of an account that is compounded n times per year is

$V_n=(1+\frac x n)^nV^0.$

$$V_n$$ is the value of your account after n compounds in a year, when the annual interest rate is 100x%.

Exercise $$\PageIndex{1}$$

Verify in the Equation that a recursion is at work that terminates at $$V_n$$. That is, show that $$V_i+1=(1+\frac x n)V_1$$ for $$i=0,1,...,n−1$$ produces the result $$V_n=(1+\frac x n)^nV_0$$.

Bankers have discovered the (apparent) appeal of infinite, or continuous, compounding:

$V_∞=\lim_{n→∞}(1+\frac x n)^nV_0$

We know that this is just

$V_∞=e^xV_0$

So, when deciding between 100x2% interest compounded daily and 100x2% interest compounded continuously, we need only compare

$(1+\frac {x_1} {365})^{365}\;\;versus\;\;e^{x_2}$

We suggest that daily compounding is about as good as continuous compounding. What do you think? How about monthly compounding?

Exercise $$\PageIndex{1}$$

(MATLAB) Write a MATLAB program to compute and plot simple interest, monthly interest, daily interest, and continuous interest versus interest rate 100x. Use the curves to develop a strategy for saving money.

## Footnotes

Note 1

Taylor's theorem says that a function may be completely characterized by all of its derivatives (provided they all exist)