Difference Equations to Differential Equations — Section 6.1: The Exponential Function
Section 6.1
Difference Equations
to
Differential Equations
The Exponential Function
At this point we have seen all the major concepts of calculus: derivatives, integrals, and
power series. For the rest of the book we will be concerned with how these ideas apply
in various circumstances. In particular, in this chapter we will introduce the remaining
elementary functions of calculus: the exponential function, the natural logarithm function,
the inverse trigonometric functions, and the hyperbolic trigonometric functions. As they
are introduced, we will discuss related issues involving derivatives, integrals, and power
series, as well as applications to the physical world.
We will begin by considering the exponential function. We first saw this function in
Section 5.7, but we will redefine it here for completeness.
Definition
The exponential function, with value at x denoted by exp(x), is defined by
∞
X
xn
x2
x3
=1+x+
+
+ ···.
exp(x) =
n!
2!
3!
n=0
(6.1.1)
We saw in Section 5.7 that this series converges absolutely for all values of x; hence
the domain of the exponential function is (−∞, ∞) . We should also note that exp(0) = 1.
Using the properties of power series, it is an easy matter to compute the derivative of
the exponential function:
!
X
∞
∞
∞
∞
X
X
d
d X xn
d xn
xn−1
xn
exp(x) =
=
=
=
= exp(x).
dx
dx n=0 n!
dx n!
(n − 1)! n=0 n!
n=0
n=1
Proposition
d
exp(x) = exp(x).
dx
Example
(6.1.2)
Using the chain rule, we have
d
exp(4x) = 4 exp(x).
dx
Example
Similarly,
d
exp(x2 ) = 2x exp(x2 ).
dx
In fact, the exponential function is the only function f for which both f (0) = 1 and
0
f (x) = f (x) for all x. To see this, we first demonstrate a more general property. Suppose
1
Copyright c by Dan Sloughter 2000
2
The Exponential Function
Section 6.1
f is any function for which f (0) = c and f 0 (x) = kf (x) for all x, where c and k are
constants. Then it follows that
d
(kf (x)) = kf 0 (x) = k 2 f (x),
dx
d 2
f 000 (x) =
(k f (x)) = k 2 f 0 (x) = k 3 f (x),
dx
f 00 (x) =
and, in general,
f (n) (x) = k n f (x)
(6.1.3)
f (n) (0) = k n c
(6.1.4)
for n = 0, 1, 2, . . .. Hence
for n = 0, 1, 2, . . .. Thus the Taylor series for f about 0 is given by
∞
X
f (n) (0)
n!
n=0
n
x =
∞
X
ck n
n!
n=0
n
x =c
∞
X
(kx)n
n=0
n!
= c exp(kx),
(6.1.5)
where the final equality follows from the definition of the exponential function. Now, as a
consequence of Taylor’s theorem, if Pn is the nth order Taylor polynomial for f at 0, then
|f (x) − Pn (x)| ≤
M
|x|n+1 ,
(n + 1)!
(6.1.6)
where M is the maximum value of |f (n+1) | on the closed interval from 0 to x. But
f (n+1) (x) = k n+1 f (x),
so
M = |k|n+1 L
where L is the maximum value of |f | on the closed interval from 0 to x. Hence
|f (x) − Pn (x)| ≤
L|kx|n+1
|k|n+1 L n+1
|x|
=
.
(n + 1)!
(n + 1)!
(6.1.7)
As we have seen before,
L|kx|n+1
=0
n→∞ (n + 1)!
lim
(6.1.8)
for any value of x, so it follows that
f (x) = lim Pn (x) =
n→∞
∞
X
f (n) (0)
n=0
n!
xn
(6.1.9)
for all x. In other words, f has a Taylor series representation, and so, using (6.1.5), we
have
f (x) = c exp(x).
Section 6.1
The Exponential Function
3
Proposition If f is a function for which f (0) = c and f 0 (x) = kf (x) for all x, where c
and k are constants, then
f (x) = c exp(x)
(6.1.10)
for all x.
In particular, if we let c = 1 and k = 1 in this proposition, then f (x) = exp(x).
In many ways, it is this property that makes the exponential function one of the most
important functions in mathematics.
Now consider a function f defined by f (x) = exp(x + b) for some constant b. Then
f 0 (x) = exp(x + b) = f (x),
so, by the previous proposition, we must have f (x) = c exp(x) for all x, where c = f (0) =
exp(b). That is, for all values of x,
exp(x + b) = f (x) = exp(b) exp(x).
This demonstrates a fundamental algebraic property of the exponential function: For any
numbers a and b,
exp(a + b) = exp(a) exp(b).
(6.1.11)
It follows from (6.1.11) that for any number a,
exp(a) exp(−a) = exp(a − a) = exp(0) = 1.
That is,
exp(−a) =
1
exp(a)
(6.1.12)
More generally, using both (6.1.11) and (6.1.12), we have
exp(a − b) = exp(a) exp(−b) =
exp(a)
.
exp(b)
(6.1.13)
for any numbers a and b, another important algebraic property of the exponential function.
We shall soon see that the number exp(1) plays a special role in this discussion.
Definition
The value of the exponential function at 1 is denoted by e. That is,
e = exp(1) = 1 + 1 +
1
1
+ + ···.
2 3!
(6.1.14)
It may be shown, although not easily, that e is an irrational number. Much more easily
(see Problem 5), it may be shown that, to 5 decimal places, e is given by 2.71828. The use
of the letter e to denote this number originates with Leonhard Euler (1707-1783), one of
the most prolific mathematicians of all time.
4
The Exponential Function
Section 6.1
Notice that for any positive integer n,
exp(n) = exp(1 + 1 + · · · + 1) = exp(1) exp(1) · · · exp(1) = (exp(1))n = en
|
{z
}
|
{z
}
n times
n times
and
exp(−n) =
1
1
= n = e−n .
exp(n)
e
(6.1.15)
(6.1.16)
Combining this with exp(0) = 1, we have
exp(n) = en
(6.1.17)
for all integers n. Moreover, for any integer n 6= 0,
n
1
1
1
1
= exp
exp
· · · exp
exp
n
n
n
n
|
{z
}
n times
1
1
1
= exp
+ + ··· +
n n
n
|
{z
}
n times
= exp(1) = e,
showing that
1
1
= en .
exp
n
(6.1.18)
Hence if m and n are integers with n 6= 0, then
exp
m
n
m
m
1
m
1
1
1
1
= exp
+ + ··· +
= exp
= en
= en.
n n
n
n
|
{z
}
m times
(6.1.19)
The next proposition summarizes these facts.
Proposition
For any rational number r,
exp(r) = er .
(6.1.20)
That is, evaluating the exponential function at a rational number r is equivalent to
raising e to the rth power. A natural question at this point is to ask whether the same result
holds for irrational numbers. A little thought shows that this question is not meaningful;
although we know what it means to raise a number to a rational power (namely, for integers
m and n,
√
m
a n = n am ,
Section 6.1
The Exponential Function
5
m
that is, a n is the nth root of the mth power of a), we have never defined what it means to
raise a number to an irrational power. For example, at this point we do not have a meaning
to associate with the symbol 2π . We will now take the first step toward remedying this
situation by defining es for an irrational number s.
Definition
If s is an irrational number, then we define
es = exp(s).
(6.1.21)
With this definition we can now say that
exp(x) = ex
(6.1.22)
for any real number x. The properties of the exponential function stated in (6.1.11) and
(6.1.13) may be restated as
ex+y = ex ey
(6.1.23)
and
ex
(6.1.24)
ey
for any real numbers x and y. Hence exponents behave in this new situation exactly the
way we should expect them to behave.
ex−y =
From our previous result that
d
exp(x) = exp(x),
dx
it now follows that
d x
e = ex .
dx
From this differentiation rule we obtain the indefinite integral
Z
ex dx = ex + c.
Example
(6.1.25)
(6.1.26)
Using the chain rule, we have
d 2x
e = 2e2x .
dx
Example
Using the product and chain rules,
2
2 d
d
d 4x2
3xe4x = 3x
e
+ e4x
(3x)
dx
dx
dx
2
2
= (3x) 8xe4x
+ e4x
2
= (3 + 24x2 )e4x .
(3)
6
The Exponential Function
Example
Section 6.1
Since
d −4x
e
= −4e−4x ,
dx
it follows that
Z
1
e−4x dx = − e−4x + c.
4
Notice the similarity between the evaluation of the integral in the last example and
the evaluation of the integral
Z
1
cos(−4x)dx = − sin(−4x) + c.
4
In fact, just as, for any a 6= 0,
Z
cos(ax)dx =
we have
Z
Z
Example
eax dx =
1
sin(ax) + c,
a
1 ax
e + c.
a
(6.1.27)
2
3xe2x dx, we use the substitution
To evaluate
u = 2x2
du = 4xdx.
Then
1
du = xdx, so
4
Z
2x2
3xe
Z
Example
To evaluate
3
dx =
4
Z
eu du =
3 u
3 2
e + c = e2x + c.
4
4
2xex dx, we use integration by parts with
u = 2x
dv = ex dx
du = 2dx
v = ex .
Then
Z
x
x
2xe dx = 2xe −
Z
2ex dx = 2xex − 2ex + c.
Notice the similarity between the technique
for evaluating the integral in the last
Z
example and the technique for evaluating
2x sin(x)dx.
Section 6.1
The Exponential Function
7
30
25
20
15
10
5
-4
-2
2
4
x
Figure 6.1.1 Graph of y = e
Z
Example
ex sin(x)dx may also be handled by integration by parts, al-
The integral
though with a little more work than in the previous example. Here we will let
u = sin(x)
dv = ex dx
du = cos(x)dx
v = ex .
Then
Z
x
Z
x
e sin(x)dx = e sin(x) −
ex cos(x)dx.
We now perform another integration by parts by choosing
u = cos(x)
dv = ex dx
du = − sin(x)dx
v = ex .
Then
Z
x
x
x
Z
e sin(x)dx
x
e sin(x) = e sin(x) − e cos(x) +
Z
x
x
= e sin(x) − e cos(x) − ex sin(x)dx.
At first glance it may
Z seem that we are back to where we started; however, all we need to
do now is solve for
Z
2
so
ex sin(x)dx. That is, we have
ex sin(x)dx = ex sin(x) − ex cos(x) = ex (sin(x) − cos(x)),
Z
ex sin(x)dx =
1 x
e (sin(x) − cos(x)) + c.
2
Note that we have added an arbitrary constant c since we are seeking the general antiderivative.
8
The Exponential Function
Section 6.1
30
25
20
15
10
5
-4
-2
2
4
−x
Figure 6.1.2 Graph of y = e
We now have sufficient information about the exponential function to understand the
geometry of its graph. Since e > 0 , we know that ex > 0 for all rational values of x, and
hence, by continuity, for all values of x. Since e > 1, it follows that
lim ex = ∞
x→∞
and
1
= 0.
u→∞ eu
lim ex = lim e−u = lim
x→−∞
u→∞
(6.1.28)
(6.1.29)
Moreover, since
d x
e = ex > 0
dx
(6.1.30)
and
d2 x
e = ex > 0
(6.1.31)
2
dx
for all x, the graph of y = ex is always increasing and always concave up. Moreover,
(6.1.30) and (6.1.31) indicate that as x increases, the graph is not only increasing, but its
slope is increasing at the same rate that y is increasing. Thus we should expect y to grow
at a very rapid rate, as we see in Figure 6.1.1. This rate of growth is characterized as
exponential growth. Figure 6.1.2 shows the graph of y = e−x , which is the graph of y = ex
reflected about the y-axis. In this case y decreases asymptotically toward 0 as x increases;
this is known as exponential decay
We will close this section with an application to the problem of uninhibited population
growth, a problem we first considered in Section 1.4.
Uninhibited population growth
Recall from Section 1.4 that if xn represents the size of a population after n units of time
and the population grows at a constant rate of α100% per unit of time, then the sequence
{xn } must satisfy the linear difference equation
xn+1 − xn = αxn
(6.1.32)
Section 6.1
The Exponential Function
9
for n = 0, 1, 2, . . .. At that time we saw that the solution of this equation is given by
xn = (1 + α)n x0 .
(6.1.33)
The crucial aspect of (6.1.32) is the statement that amount of change in the size of the
population over any unit of time is proportional to the current size of the population.
Hence if x(t) represents the size of a population at time t, where the population can
change continuously over time, then the continuous time analog of (6.1.32) is the differential
equation
ẋ(t) = αx(t)
(6.1.34)
for all time t. If x0 is the size of the population at time t = 0, then we know from our
work in this section that the only solution to this equation is the function
x(t) = x0 eαt .
(6.1.35)
Hence if the size of a population is growing at a rate which is proportional to itself, an
assumption which, as we noted in Section 1.4, is often reasonable over short periods of
time, then the population will grow exponentially. As in Section 1.4, we refer to such
growth as uninhibited population growth.
Example In 1970 the population of the United States was 203.3 million and in 1980
the population was 226.5 million. Assuming an uninhibited growth model and letting x(t)
represent the population t years after 1970, by (6.1.35) we should have
x(t) = 203.3eαt
for some constant α. Since x(10) = 226.5, we can find α by solving
226.5 = 203.3e10α .
That is, we need to find a value for α such that
e10α =
226.5
= 1.114.
203.3
Unfortunately, solving this equation exactly requires being able to reverse the process of
applying the exponential function. In other words, we need an inverse for the exponential
function. We shall take up that problem in the next section; for now we may use a
numerical approximation. You should verify that α = 0.0108 satisfies the equation. Thus
this model would predict the population of the United States t years after 1970 to be
x(t) = 203.3e0.0108t .
For example, this model would predict a 1990 population of
x(20) = 203.3e(0.0108)(20) = 252.3
10
The Exponential Function
Section 6.1
1000
800
600
400
200
0
20
40
60
80
100
120
140
Figure 6.1.3 Uninhibited growth model for the United States (1970-2120)
and a population in the year 2000 of
x(30) = 203.3e(0.0108)(30) = 281.1.
While the prediction for 1990 is fairly accurate (the actual population was approximately
249.6 million), the second prediction differs significantly from the Census Bureau’s own
prediction of a population of 268.3 million for the year 2000. As we discussed in Sections
1.4 and 1.5, an uninhibited growth model is a simple model which cannot be expected to
be accurate for predictions too far into the future.
We shall have more to say about population models in Section 6.3, where we will also
see another example of a differential equation. We will have a much fuller discussion of
differential equations in Chapter 8.
Problems
1. Find the derivative of each of the following functions.
(a) f (x) = 3e2x
(c) h(z) = (3z 2 − 6)e5z
3x
(e) g(x) = x
2e
3s − 1
(g) f (s) = −2s
e
+2
(b) g(t) = 4t2 e3t
3
2. Evaluate each of the following integrals.
Z
(a)
3e2x dx
Z
(c)
4te3t dt
(d) f (x) = e3x sin(2x)
(f) h(t) = e−6t cos(4t)
(h) g(θ) = 5θe6θ sin(2θ)
Z
(b)
Z
(d)
2
4xe3x dx
5ye−y dy
Section 6.1
Z
(e)
z 2 ez dz
Z
(g)
ex cos(x)dx
The Exponential Function
Z
(f)
x3 e−2x dx
Z
(h)
e−2t sin(3t)dt
11
3. Find the maximum value of f (x) = x2 e−x on the interval (0, ∞).
4. (a) Use the Taylor series for e−x to show that e−1 > 13 . Hence conclude that e < 3.
(b) Show that if Pn is the nth order Taylor polynomial for ex at 0, then
3x
|x|n+1
(n + 1)!
|ex − Pn (x)| ≤
for any value of x.
(c) Use (b) to find an approximation for e with an error of less than 0.000005.
5. Find the following limits.
ex − 1
x→0
x
(a) lim xe−x
(b) lim
x→∞
e−t − 1
(d) lim x2 e−2x
x→∞
t→0
t
6. (a) Show that lim xn e−x = 0 for any positive integer n.
(c) lim
x→∞
(b) Use (a) to show that if p is any polynomial, then lim p(x)e−x = 0. This shows
x→∞
that ex grows faster as x → ∞ than any polynomial function.
7. Graph the following functions on the specified intervals.
(a) f (t) = 3e2t on [−3, 3]
et − e−t
(c) g(t) = t
on [−10, 10]
e + e−t
8. Evaluate the following improper integrals.
Z ∞
(a)
e−x dx
Z0 ∞
(c)
xe−x dx
Z0 ∞
(e)
x2 e−x xdx
(b) g(x) = 4x2 e−2x on [0, 5]
t
(d) f (t) = e− 4 sin(3t) on [0, 10]
Z ∞
(b)
Z0 ∞
(d)
Z0 ∞
(f)
0
3e−2x dx
3xe−2x dx
2
xe−x dx
0
9. Use the integral test to show that the infinite series
∞
X
e−n converges.
n=0
2
10. (a) Find the Taylor series for f (x) = e−x .
(b) Use (a) to find the the Taylor series for
2
erf(x) = √
π
Z x
0
2
e−t dt,
12
The Exponential Function
Section 6.1
known as the error function.
(c) Use your result in (b) to approximate erf(1) with an error less than 0.0001.
11. Suppose x(t) is the population of a certain country t years after 1985, x(0) = 23.4
million, and
ẋ(t) = 0.008x(t).
(a) What will the population of the country be in the year 2000?
(b) In what year will the population be twice what it was in 1985?
12. Let
f (x) =
1
e− x2 , if x 6= 0
0,
if x = 0.
(a) Graph f on the interval [−5, 5].
(b) Show that f 0 (0) = 0.
(c) Show that f (n) (0) = 0 for n = 0, 1, 2, . . ..
(d) Show that f is C ∞ on (−∞, ∞).
(e) Note that the Taylor series for f about 0 converges for all x in (−∞, ∞), but does
not converge to f (x) except at 0. Thus f is C ∞ on (−∞, ∞), but not analytic at
0.