Chapter 4 Exponential and Logarithmic
Functions (指数函数和对数函数)
In this Chapter, we will encounter
some important concepts
 Exponential Functions(指数函数)
 Logarithmic Functions(对数函数)
 Differentiation of Logarithmic and
Exponential Functions
1
Section 4.1 Exponential Functions
Exponential function(指数函数): If b is a
positive number other than 1 (b>0, b≠1), there is
a unique function called the exponential function
with base b that is defined by
f(x)=bx for every real number x
NOTE: Such function can be used to describe exponential
and logistic growth and a variety of other important
quantities.
2
Definition of b n for Rational Values of n (and b>0)
Integer powers: If n is a positive integer,
bn  b
b


b



n facters
Fractional powers: If n and m are positive integers,
b
where
m
n/m

 b
m
n
 b
m
n
b denotes the positive mth root.
Negative powers:
b
n
1
 n
b
0
b
1
Zero power:
3
Example 1
Solution:
4
Figure below shows graphs of various members of the
x
family of exponential functions y  b
b
p
(
x
)

x
NOTE: Students often confuse the power function
x
f
(
x
)

b
with the exponential function
5
6
7
Definition: The natural exponential function is
Where
n
10
100
2.59374
2.70481
1000
2.71692
10,000
2.711815
100,000
2.71827
8
Continuous Compounding of Interest(连续利息)
If P is the initial investment (the principal) and r is the interest
rate (expressed as a decimal), the balance B after the interest is
added will be
B=P+Pr=P(1+r)
dollars
9
10
11
12
Example 2
Suppose $1,000 is invested at an annual interest rate of 6%.
Compute the balance after 10 years if the interest is compounded
a. Quarterly
b. Monthly
c. Daily
d. Continuously
Solution:
a. To compute the balance after 10 years if the interest is
compounded quarterly, using the formula
r

B (t )  p1  
 k
kt
with t=10, p=1,000, r=0.06, and k=4:
40
 0.06 
B(10)  1,0001 
  $1,814.02
4 

to be continued
13
b. This time, take t=10, p=1,000, r=0.06, and k=12 to get
120
 0.06 
B(10)  1,0001 

12


 $1,819.40
c. Take t=10, p=1,000, r=0.06, and k=365 to obtain
 0.06 
B(10)  1,0001 

365 

3, 650
 $1,822.03
d. For continuously compounded interest use the formula B(t )  pert
with t=10, p=1,000, and r=0.06:
B(10)  1,000e0.6  $1,822.12
This value, $1,822.12, is an upper bound for the possible balance.
No matter how often interest is compounded, $1,000 invested at an
annual interest rate of 6% can not grow to more than $1,822.12 in
10 years.
14
Present Value(现值)
15
16
Example 3
Sue is about to enter college. When she graduates 4 years from
now, she wants to take a trip to Europe that she estimates will cost
$5,000. How much should she invest now at 7% to have enough
for the trip if interest is compounded:
a. Quarterly
b. Continuously
Solution:
The required future value is F=$5,000 in t=4 years with r=0.07.
a. If the compounding is quarterly, then k=4 and the present value
is
to be continued
17
0.07 

P  5,0001 

4 

4 ( 4 )
 $3,788.08
b. For continuous compounding, the present value is
P  5,000e 0.07( 4)  $3,778.92
Thus, Sue would have to invest about $9 more if interest is
compounded quarterly than if the compounding is continuous.
18
指数增长和指数下降
19
Section 4.2 Logarithmic Functions(对数函数)
20
21
Example 4
Use logarithm rules to rewrite each of the following expressions
in terms of log 5 2 and log. 5 3
5
log
a.
b. log 5 8
c. log 5 36
5 
3
Solution:
a.
5
log 5    log 5 5  log 5 3
3
 1  log 5 3
quotient rule
since log 5 5  1
b. log 5 8  log 5 23  3 log 5 2
c.
power rule
log 5 36  log 5 (2232 )  log 5 22  log 5 32
 2log 5 2  2 log 5 3
product rule
power rule
22
23
24
Example 5
Solution:
25
Exponential function: series representation
Let n=k/r
Expand and take the limit, we get
26
Exponential & Logarithmic functions:
important limits

Lim (x0) (exp(x) – 1 ) / x = 1



Or exp(x) 1 + x as x0 , and
Lim (x0) ln(1+x) / x = 1
From these, we can prove


d/dx ( exp(x) ) = exp(x) ,
d/dx ( ln(x) ) = 1/x
27
Section 4.3 Differentiation of Logarithmic
and Exponential Function
Example 6
Solution:
28
Example 7
29
Solution:
30
Differentiate both sides of the equation
31
Example 8
Solution:
32
33
Example 9
Solution:
34
Taking the derivatives of some complicated functions can
be simplified by using logarithms. This is called
logarithmic differentiation.
Example 10
Solution:
to be continued
35
36
Example 11
Differentiate each of these function
a.
y  2x
b.
y  log 3 x
Solution:
a. To differentiate
follows:
y  2 x , we use logarithmic differentiation as
to be continued
37
b.
38
39
40
The relative rate of change of a quantity Q(x) can be
computed by finding the derivative of lnQ.
d
Q '( x)
(ln Q) 
dx
Q( x)
41
Example 12
Solution:
to be continued
42
43
Summary
 Exponential Functions, Basic Properties of Exponential
Functions, The Natural Exponential Base e.
 Compound Interest, Continuously Compounded Interest,
Present Value.
 Exponential Growth and Decay.
 Logarithmic Functions, The Natural Logarithm.
 Differentiation of logarithmic and Exponential
Functions.
 Optimal holding time.
44
Trigonometric functions: a very
important limit
From this we can prove:
45
Derivatives of trigonometric functions

Using the previous results we can derive




d/dx sin(x) = cos(x)
d/dx cos(x) = - sin(x)
d/dx tan(x) = ?
d/dx sec(x) = ?
46
Hyperbolic functions




sinh (x) = ( exp(x) – exp(-x) )/ 2
cosh (x) = ( exp(x) + exp(-x) )/ 2
d/dx sinh(x) = cosh(x)
d/dx cosh(x) = sinh(x)
47
Taylor’s Series/Expansion
1 2
1 n
e  1 x  x  x 
2!
n!
x
x  (  ,  )
If
then the series is convergent to ex.
Proof:
48
Suppose that the function f(x) has continuous derivatives
of orders 1, 2, ..., k at x=0. Can we approximate f(x) by a
power series?
2
k
f ( x )  c0  c1 x  c2 x   ck x
Expand
We get
Thus,
f
(n)
 n ! cn , n  k ,
(0)  
n  k,
 0,
f (0) 2
f ( x )  f (0)  f (0) x 
x 
2!
 c0  c1 x  c2 x 2 
f ( k ) (0) k

x
k!
 ck x k ,
Suppose f(x) is a function for which you know the value and the value of all
its derivatives at some point x = a. What is the value of f at some nearby
point, say x = a+h, where h is sufficiently small?
For example, you are studying the
motion of an object, and you have
measured its current position and
velocity, and maybe even its
acceleration. If we take f(t) to be the
position of the object at time t, f’ is its
velocity, and f’’ is its acceleration,
then we would like to determine how
much we can know about the motion
a short time later.
Approximation by Increment
50
Taylor's Theorem
51
Taylor Series/Expansion
The remainder can be written in a number of ways. The most useful form is
that derived by J. L. Lagrange (1736-1813), resulting in the Taylor formula with
derivative remainder:
where
is some number between x and a
52
Taylor's formula with integral remainder,
derived by Brook Taylor (1685-1731)
By repeating this integration by parts process on the remaining integral p
times, one has the result:
The last term is referred to as the remainder, Rn(x)
53
Expand f ( x )  e x into Taylor series of x.
Sulution
thus
Since
f
( n)
( x )  e , f ( n ) (0)  1
x
1 2
1 n
e  1 x  x  x 
2!
n!
x
(n  0,1,2,)
x  (  ,  )
Lagrange Residual Term:
y
Obviously,
6
For any real x,
e| x|
lim
| x |n1  0,
n ( n  1)!
y  ex
4
(n  2)
2
(n  0)
thus
Therefore, the expansion is
convergent to f(x)!
(n  3)
1
O
2
1
2
54
x
Useful expansion
55
Euler's formula
named after Leonhard Euler, is a mathematical formula in
complex analysis that establishes the deep relationship
between the trigonometric functions and the complex
exponential function. Euler's formula states that, for any real
number x,
Hence we can get:
56
Proof of Euler's formula
In the last step we have simply recognized the Taylor series for sin(x) and cos(x).
The rearrangement of terms is justified because each series is absolutely
convergent.
57
Indeterminate forms: application of
Taylor Expansion

Expressions of the form 0/0, ∞/∞, 0 ×∞, ∞−∞,
0^∞ and ∞^0 are called indeterminate forms.
58
L’Hospital’s Rule – for indeterminate
forms
f’(x), g’(x) exist,
59
60
Examples:
61
Example:
Find the limit of
as x0
Since
therefore
62
What is Power Series?
Power Series(幂级数)：The expression as

n
2
a
(
x

x
)

a

a
(
x

x
)

a
(
x

x
)
 ...
 n
0
0
1
0
2
0
n 0
Special case：as x0=0 ，i.e., expending at 0, it becomes

n
2
a
x

a

a
x

a
x
 ...
 n
0
1
2
n 0
Convergence (收敛性) of power series：

Power series  an ( x  x0 ) will be absolutely convergent within
n
n 0
interval x  x0  R , and divergent within x  x0  R , i.e., its
convergent interval is [x0-R，x0+R].
Radius of Convergence
63
Absolute convergence and
Conditional convergence
A series
is said to converge absolutely
if the series of absolute values
converges.
A series of real or complex numbers is said to be
conditionally convergent (or semi-convergent) if it is
convergent but not absolutely convergent. A famous example is
the alternating series:
64
Test of convergence: absolute ratio test
Given the series:
and
r
Then, if r < 1, the series converges absolutely.
if r > 1, the series diverges.
if r = 1, no conclusions can be drawn - series
may converge or diverge.
65
Test of convergence: comparison test
# Comparison test 1: If ∑b(n) is an absolutely convergent series such that |a(n) | ≤
C |b(n) | for some number C and for sufficiently large n , then ∑a(n) converges
absolutely as well. If ∑|b(n) | diverges, and |a(n) | ≥ |b(n) | for all sufficiently large n ,
then ∑a(n) also fails to converge absolutely.
# Comparison test 2: If ∑b(n) is an absolutely convergent series such that |an+1 /an
| ≤ |b(n+1) /b(n) | for sufficiently large n , then ∑a(n) converges absolutely as well. If
∑|b(n) | diverges, and |a(n+1) /a(n) | ≥ |b(n+1) /b(n) | for all sufficiently large n , then
∑a(n) also fails to converge absolutely (though it could still be conditionally
convergent, e.g. if the a(n) alternate in sign).
66
Example:
r=
r
67