Download Calculus Math 1710.200 Fall 2012 (Cohen) Lecture Notes

Calculus Math 1710.200 Fall 2012 (Cohen) Lecture Notes
For the purposes of this class, we will regard calculus as the study of limits
and limit processes. Without yet formally recalling the definition of a limit, let’s
recall the main ways in which one applies the concept in an introductory Calculus
I course (which the student should have recently taken).
Super-Brief Calculus I Review.
The Derivative. Given a function f and a point x, we wish to compute the
instantaneous rate of change of f at x. Geometrically, we may identify this
instantaneous rate of change with the slope of the line tangent to the graph
of f at the point x. How can we compute such a thing? The process is achieved in
just two major steps: a “geometric approximation” step and then a “limit” step.
y2 − y1
to compute the average rate of change
x2 − x1
between x and any other nearby point, which we denote x + h. Since the
function f determines the y-values at x and x + h, the formula we obtain
is the famous difference quotient below:
(1) Use the slope formula
f (x + h) − f (x)
h
If we take the distance h to be a very small number, then this slope should
be a very close approximation to the slope we hope to compute. Indeed, in
general a smaller choice of h should lead to a closer approximation.
(2) Since step (1) gives us an infinite family of arbitrarily close approximations
to the slope of the tangent line, the natural final step is to let h shrink to
0, i.e. compute the limit as h → 0 of our approximations. So we obtain the
formula
f 0 (x) = lim
h→0
f (x + h) − f (x)
h
We call this value f 0 (x) the derivative of f at x, and it is precisely the
slope of the tangent line we seek.
The Definite Integral. Now, given a function f and two points a and b with
a < b, we wish to compute the area under the curve of f , bounded by the lines
x = a and x = b. Again we proceed by a “geometric approximation” step, and then
a “limit” step. The student should certainly observe the clear analogy between this
process and the one above for computing the derivative!
(1) First we consider any positive integer n, and we subdivide the interval [a, b]
into n equal pieces, indexed by the endpoints a = x0 < x1 < x2 < ... < xk <
... < xn = b. The width of each piece [xk−1 , xk ] is ∆x = xk − xk−1 = b−a
n .
From each piece [xk−1 , xk ], we select any test point xk , and we imagine
1
2
drawing a rectangle with base [xk−1 , xk ], and height determined by the
test point f (xk ). The area of this rectangle is obviously (height)·(width)=
f (xk ) · ∆x. Adding together the area of each of the n rectangles, we obtain
the Riemann sum:
n
X
f (xk ) · ∆x
k=1
This Riemann sum gives the approximate area of the region bounded by
the curve of f . It is easy to see that, in most cases, choosing a finer mesh,
i.e. subdividing [a, b] into more, smaller pieces, will yield a closer approximation to the actual area. In other words, choosing larger values of n
(equivalently, smaller values of ∆x) should lead to a closer approximation
to the area we hope to compute.
(2) Now, again, step (1 yields for us an infinite family of arbitrarily close approximations to the true area under the curve. So again we take a limit,
this time as n → ∞, or equivalently, as ∆x → 0. We get the formula for
the definite integral of f from a to b:
Rb
a
f (x)dx = lim
n→∞
n
X
f (xk ) · ∆x
k=1
This integral, when it exists, is the area under the curve of f bounded by
x = a and x = b, as desired.
The two processes above should have been the main focus of the reader’s Calculus I course, together with the following remarkable theorem:
Theorem 1 (The Fundamental Theorem of Calculus). Let f be any continuous
function and let a be any point. Define the area function A of f (centered at a)
by the rule
A(x) =
Rb
a
f (t)dt
Stated informally, A is the function which takes a real number x for input, and
for output computes the definite integral of f between the (non-variable) point a
and the input x, i.e. gives the area under the curve of f between a and x. Then
the following two statements hold:
(1) A0 (x) = f (x); and
(2)
Rb
a
f (x)dx = F (b) − F (a), where F is any antiderivative of f .
Informally, the FTC (Fundamental Theorem of Calculus) may be read as: (1)
The definite integral function of f is an antiderivative of f ; and (2) any antiderivative of f also computes the definite integral of f . In other words, integration and
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antidifferentiation are essentially the same thing. So the seemingly unrelated
geometric processes of taking the derivative and computing the definite integral are
in a vital sense, inverse processes of one another!
Are Limits a New Technology?
The student will probably not have seen the definition of a limit prior to his or
her Calculus I course. Thus the “limit process” may seem at first glance contrived,
i.e. a notion invented solely for the purposes of differentiation and integration.
While this impression may not be totally wrong, here we pose a few simple mathematical questions to help motivate our line of inquiry regarding limits.
Recall: A real number x is called rational if x can be written as a fraction, i.e.
if x = pq for some whole numbers p and q. Examples of rational numbers are 81 ,
11
9 1092
we call x irrational. Some famous
6 , 12, − 2 , 37 , etc. If x is not rational,√then
√
examples of irrational numbers are π, e, 2, 3, etc.
The student should also recall the following alternative characterization of
rationals in terms of their decimal expansions:
Theorem 2.
(1) If x is a rational number, then the decimal expansion of x is
either finite or infinitely repeating.
(2) If x is an irrational number, then the decimal expansion of x is infinite
and non-repeating.
Example 1. Consider the number x = 0.101001000100001000001..., where the
pattern of 0’s and 1’s continues in the obvious way. The above theorem implies that
x must be irrational.
In the assertions above, we have taken the notion of an “infinite decimal expansion” for granted. But now we pose the following question to the student:
Question 1. Just what does it mean to have an infinite decimal expansion?
For example, finite decimal expansions are easily computed using the place-value
system. Consider the expansion x = 0.125. By our theorem, we know x must be
rational, but what is the value of x? We compute below:
1
1
1
+2·
+5·
10
100
1000
100
20
5
+
+
1000 1000 1000
125
1000
1 · 125
8 · 125
1
8
x = 0.125 = 0 · 1 + 1 ·
=
=
=
=
4
So x = 18 , and the decimal expansion 0.125 has a clear meaning. The same
should be true for any such finite expansion.
But how can we compute the value of the decimal expansion of say, π = 3.14159...
? The computation should look something like this:
π = 3.14159... = 3 · 1 + 1 ·
1
1
1
1
1
+4·
+1·
+5·
+9·
+ ...
10
100
1000
10000
100000
What exactly does it mean to add together infinitely many numbers? How can
an infinite collection of fractions sum to a finite number π? Or does the infinite
decimal expansion represent something else entirely?
Question 2. Does every real number have one and only one decimal expansion?
To understand why this is a good question, recall that the rational numbers are
dense in the real number line, that is, whenever x and y are real numbers with
x < y, there always exists a rational number r such that x < r < y. (In fact, there
must exist infinitely many- why?)
Example 2. Let x = 121.3791 and let y = 121.3792. Find a rational number r
such that x < r < y.
Example 3. Let x = π = 3.14159 and let y = π + .01 = 3.15159.... Find a rational
number r such that x < r < y.
The above two examples can be easily solved because the order of the real number
line is determined lexicographically, i.e. “alphabetically on the digits,” of the
decimal expansions. So consider the following problem.
Example 4. Let x = 0.99999... and let y = 1 = 1.00000... Find a rational number
r such that x < r < y.
This new problem seems to have no easy solution; certainly there is no decimal
expansion which lies lexicographically between those of x and y. How can we resolve this dilemma?
Possible explanations for the strange example above:
(1) There are “ghost rationals” between 0.9999... and 1, i.e. rational numbers
which do not have decimal expansion representations.
(2) Not every decimal expansion represents a real number, i.e. the expression
0.9999... is not actually a point on the real number line.
(3) 0.9999... is in fact NOT less than 1, i.e. we have 0.9999... = 1 and hence
no rationals need lie between them.
Which of the above explanations does the reader think is true? Or none of them?
Question 3. What does the expression 2x mean in general?
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For instance, for any positive integer n, we expect that 2n = |2 · 2 · {z
2 · ... · 2}, i.e.
n
the product of 2 multiplied with itself n times.
Now notice that if n and m are positive integers, then we have
2n · 2m = |2 · 2 {z
· ... · 2} · 2| · 2 {z
· ... · 2} = 2n+m .
n
m
This rule gives us a natural extension of the definition of exponentiation to
negative integers. Specifically, we would like for the definition of 2−n to satisfy
2−n · 2n = 2−n+n = 20 = 1; this only happens if 2−n = 21n . So we adopt this
convention.
We may also extend the definition of exponentiation to rational numbers in a
similar way. Notice that for any two integers n and m, we have
· ... · 2} ·... · 2| · 2 {z
· ... · 2} = 2n·m .
(2n )m = |2 · 2 {z
· ... · 2} · 2| · 2 {z
n
n
n
|
{z
}
m
Since (2n )m = 2n·m holds for integers, we would like the definition
of 21/n to
√
1
n
satisfy (21/n )n = 2 n ·n = 21 = 2. This only happens if 21/n = 2, so we adopt
this convention as well. With this rule established, for all rational numbers pq , we
√
√
may write 2p/q = q 2p = ( q 2)p , which is the definition we all know and understand.
So how does one extend√the definition to irrational numbers? Just what is meant
by the expressions 2π or 2 2 ?
Question 4 (Zeno’s Paradox). Does Achilles ever catch the tortoise?
Here is a variation of one of Zeno of Elea’s paradoxes (5th century BC): Achilles
is chasing a tortoise. The tortoise is 1000 meters away, but Achilles runs 10 times
as fast as the tortoise, so he expects to do well. Unfortunately, by the time Achilles
runs the first 1000 meters, the tortoise has advanced another 100 meters. By the
time Achilles runs this additional 100 meters, the tortoise has advanced 10 more
meters. By the time Achilles goes the next 10, the tortoise has gone another 1
meter. The tortoise continually moves away from Achilles: next .1 meter, then .01
meter, then .001 meter, etc., ad infinitum. So Achilles never catches the tortoise.
Can this paradox be resolved? Our intuition says that surely Achilles catches
the tortoise, but where and when? It seems that the distance Achilles must travel
is given by the sum
1000 + 100 + 10 + 1 + .1 + .01 + .001 + ...
But is it possible to compute such an “infinite sum”?
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One of our goals by the end of this class will be to have solid answers for these
questions and more. Questions 1, 2, and 4 will have to get answered later in the
course, but we can turn to a rigorous answer for Question 3 almost immediately,
as we will do in the next section.
The Logarithmic Function and Exponential Function.
First let’s recall the college algebra definition of the logarithm. For real numbers
b and x, we write that
y = logb x if and only if by = x.
In other words, we define logb x to be the unique exponent y for which by =
x. We also define the natural logarithm ln to be the logarithm of base e, i.e.
ln x = loge x. In light of Question 3 above, the student should wonder how well they
understand these definitions. Still, we are able to compute a few simple examples.
Example 5. Compute the following logarithms.
(1) log2 32
(2) ln e11
(3) log5
(4) log7
1
25
√
8
7
(5) log81 3
Now let’s use our calculus techniques to give a more rigorous definition.
Definition 1. The natural logarithm of a real number x > 0, denoted ln x, is
given by
Z x
1
ln x =
dt.
1 t
Domain of the logarithm. The domain of the natural logarithm is (0, ∞) by
definition.
Range of the logarithm. Consider the graph of y = 1t and visualize the corRx 1
responding areas computed by ln x = 1 dt. (A picture helps here.) Since the
t
graph of 1t lies entirely above the x-axis, we can see that ln x must be an increasing
function. Clearly by our Calculus I integration rules, we must have ln 1 = 0, and if
x > 1 then ln x is positive while if x < 1 then ln x is negative.
Judging from the picture, it appears that we may have lim ln x = ∞, and since
x→∞
1
lim = ∞, it appears that we may have lim ln x = −∞. Neither of these facts is
x→0
t→0 t
immediately obvious, but they turn out to be true: the range of ln x is (−∞, ∞) as
we expect! (This will become clear much later in the course when we discuss the
7
harmonic series.)
Derivative of the logarithm. The derivative of the natural log follows immediately from the FTC.
d
d
(ln x) =
dx
dx
Z
1
x
1
1
dt = , for x > 0.
t
x
By differentiating ln(−x) using the chain rule for x < 0, we can also see that
d
1
ln |x| = .
dx
x
Logarithm of a product. From our college algebra class, we may hope that
the rule ln xy = ln x + ln y holds. Let’s check that this is true using our definition:
Z
xy
1
dt
t
Z x
Z xy
1
1
=
dt +
dt
t
t
Z1 x
Zxy
1
1
=
dt +
du (where u = t/x; check with substitution rule)
t
u
1
1
= ln x + ln y
ln(xy) =
1
From this rule it is easy to derive that ln xy = ln y − ln x.
Integral of x−1 . Recall that x−1 is the only power of x which we are not able
to integrate using the power rule, for obvious reasons. Now we can find its integral!
d
Since dx
ln |x| = x1 , we have
Z
Example 6. Find
dy
dx
(1) y = ln(4x)
(2) y = x ln x
(3) y = ln | sec x|
(4) y =
ln x2
x2
1
dx = ln |x| + C.
x
for the following functions.
8
4
Z
Example 7. Evaluate
0
x
dx.
x2 + 9
HOMEWORK (Due 9/14/12): Section 7.2 #8, 9, 10, 13, 15, 16, 18, 20, 23,
24, 26, 29, 30, 31, 32, 34, 39, 41
We have established that f (x) = ln x is a continuous increasing function on
(0, ∞), with range (−∞, ∞). Hence it admits some unique inverse function f −1 (x);
denote this function by exp(x) = f −1 (x). Recall that since exp is the inverse
function of ln, it has the following properties:
(1) The domain of exp is (−∞, ∞), the range of ln;
(2) The range of exp is (0, ∞), the domain of ln;
(3) ln(exp(x)) = x for all x in (−∞, ∞); and
(4) exp(ln(x)) = x for all x in (0, ∞).
Definition 2. The exponential function is the unique function exp(x) which is
the inverse of ln(x), and which has all the properties listed above.
Now let x1 , x2 be any two real numbers, and set y1 = exp(x1 ), y2 = exp(x2 ).
Then x1 = ln(y1 ) and x2 = ln(y2 ) and hence, using the properties of ln, the following must be true:
exp(x1 + x2 ) = exp(ln y1 + ln y2 )
= exp(ln(y1 · y2 ))
= y1 · y2
= exp(x1 ) · exp(x2 )
We refer to the above as the multiplicative property of exp.
Definition 3. Define e = exp(1). Notice it follows immediately from our definitions that ln e = 1.
Theorem 3. For any positive whole number n, exp(n) = en .
Proof. Using the fact that ln x + ln y = ln(xy), we have
9
exp(n) = exp(1 + 1 + ... + 1)
|
{z
}
n
= exp(ln e + ln e + ... + ln e)
{z
}
|
n
= exp(ln(e| · e {z
· ... · e}))
n
= exp(ln(en ))
= en .
Theorem 4. For any integer n, we have exp(n) = en .
Proof. We have already proved this for any nonnegative n. So suppose n is negative and let m = −n, so m is positive. Then, using the previous theorem, and the
multiplicative property of exp, we have
1 = exp(0)
= exp(n − n)
= exp(n + m)
= exp(n) · exp(m)
= exp(n) · em .
It follows that exp(n) =
1
= e−m = en , and the theorem is proved.
em
Theorem 5. For any rational number x, we have exp(x) = ex .
Proof. Let x = pq , where p and q are integers. Then using our previous theorems,
and the multiplicative property again, we have
q
p
p
p
p
exp
= exp
· exp
· ... · exp
q
q
q
q
|
{z
}
q


p p

= exp  + + ... +
q
q
|
{z
q
= exp(p)
= ep .
p


q
}
10
Hence we must have exp
p
q
√
q
=
ep = ep/q , as desired!
Answer (Answer 3). Astoundingly, the above theorems imply that exp(x) = ex
for all rational numbers x, and hence the function exp(x) is an extension of the
natural definition of ex to the entire real number line! Thus from now on we will
regard the two functions as being exactly the same. This gives us a rigorous answer
to Question 3: we may set 2x = (eln 2 )x = ex ln 2 and we have a continuous
exponentiation function which is defined on the entire number line, and which has
all the nice properties we expect of the function 2x .
Definition 4. Define ex = exp(x) for all real numbers x. This justifies our use of
the term exponential function, and we have the properties eln x = x and ln ex = x
for all reasonable inputs x. (As a consequence, our function ln is also an extension
of the usual natural logarithm from college algebra to the entire positive real number
line.) More generally, for any real number b, define bx = ex ln b = exp(x ln b).
Derivative and Integral of the Exponential Function. Let’s first compute
the derivative of ex . Notice that since ln ex = 1, we must have
d
d
ln ex =
x
dx
dx
= 1.
On the other hand, by the chain rule we must have
d
dx
So we get
1
ex
·
d x
dx e
ln ex =
1
ex
·
d x
dx e
.
= 1 and hence
d x
dx e
= ex
It then follows immediately that
R
ex dx = ex + C
Remark. The above properties are very striking! If the reader ever takes a real
analysis course, they may learn the following alternative definition of the exponential function: exp(x) = ex is the UNIQUE function which (1) satisfies exp(0) = 1
and (2) is the derivative of itself!
Example 8. Compute the following derivatives.
− 4ex + e−3x )
t
(1)
d
2x
dx (3e
(2)
d
dt
e
e2t−1
11
(3)
d
cos πx
)
dx (e
Example 9. Find an equation for the line tangent to the graph of f (x) = 2x −
at the point (0, − 12 ).
Example 10. Evaluate
R
Example 11. Let f (x) =
ex
2
ex
dx.
1 + ex
√
(x3 − 1)4 3x − 1
. Find f 0 (x). [Hint: Use logarithmic
x2 + 4
differentiation!]
One more logarithm property. Notice that for any real number p, we have
ln xp = ln ep ln x = p ln x
So exponents may still by shifted to the front of logarithms, as we expect from
our previous education in logarithms.
Logarithms and Exponentials in Other Bases
We have already defined the function bx = ex ln b ; now let us define its inverse.
Definition 5. For any base b > 0, with b 6= 1, define the logarithm of base b,
denoted logb x, to be the inverse of the exponential function bx . In other words, let
logb x be the unique function for which blogb x = x and logb bx = x, for all reasonable
inputs x.
Derivative of bx . We compute the derivative using the chain rule below:
d x
d x ln b
b =
e
dx
dx
d
x ln b
=e
·
dx
= bx · ln b
x ln b
So we have a new derivative rule.
d x
dx b
= bx ln b
Example 12. Find the derivatives of the following functions.
(1) f (x) = 3x
(2) g(t) = 108 · 2t/12
Integral of bx . Since
d x
dx b
= bx ln b, the following is also true:
12
R
bx dx =
1
ln b
· bx + C
Example 13. Evaluate the following integrals.
R
2
(1) x · 3x dx
Z
(2)
1
4
√
6−
√
x
x
dx
Example 14. Determine whether the graph of f (x) = xx has any horizontal tangent lines.
Derivative of logb x. First notice that since blogb (x) = x, we have
d
dx x = 1. On the other hand by the chain rule, we have
d logb x
dx b
Thus we have x · ln b ·
d
dx
= blogb x · ln b ·
d
dx
logb x
d logb x
dx b
=
logb x = 1, and hence
d
dx
logb x =
1
x ln b
Example 15. Compute the derivatives of the following functions.
(1) f (x) = log5 (2x + 1)
(2) T (n) = n log2 n
Exponential Models.
Definition 6. An exponential growth model is a function of the form y(t) =
y0 · ekt , where y0 is some initial value constant, and k is some positive growth
constant.
Exponential growth models are typically used to model things like population
growth, interest accumulation, cellular growth, etc.
Definition 7. If y(t) represents some quantity (like population or dollars) which
is changing with respect to time, then we call y 0 (t) the growth rate of y. We call
y 0 (t)
y(t) the relative growth rate of y, i.e. the growth rate of the quantity divided by
the size of the quantity.
Notice that if y(t) is an exponential growth model, then y 0 (t) = k·y0 ·ekt = k·y(t).
In other words, the growth rate of an exponential growth model is proportional to
0
(t)
= k, we can say that the relative growth rate of
its value. Equivalently, since yy(t)
an exponential growth model is constant.
Example 16. The population of Town A is given by P (t) = 1500 + 125t, while
the population of Town B is given by Q(t) = 1500e0.1t , where t is time measured in
years. Find the growth rates and the relative growth rates of the two towns.
13
Example 17. Assume that the population of the world in 1999 was 6.0 billion, and
that the population in 2009 was 6.9 billion.
(1) Find an exponential growth model for the world population that fits these
two data points.
(2) From 1999, how long will it take the world’s population to double?
(3) From 2009, how long will it take the world’s population to double?
(4) Give a general formula for doubling time of the world’s population at time t.
(5) Find the growth rate y 0 (t).
(6) How fast was the population growing in 2010 according to our model?
The above example illustrates a property of exponential growth models: that
doubling time is a constant. Let’s compute it in general. If y(t) = y0 · ekt is any
exponential growth model, then we wish to find the particular time t at which y(t)
is twice the inital population y(0) = y0 . In other words, we wish to solve the equation
2y0 = y0 · ekt
Dividing out the y0 from both sides and taking the natural logarithm of both
sides, we get ln 2 = kt and hence t = lnk2 .
Definition 8. If y(t) is an exponential growth model, then the doubling time of
y is the quantity T2 = lnk2 .
Example 18. Suppose you invest 500 in a savings account which pays 6.18% interest each year.
(1) How long will it take for the account to accumulate 1000?
(2) How long will it take for the account to accumulate 2500?
Definition 9. An exponential decay model is a function of the form y(t) =
y0 · e−kt , where y0 is an initial value constant and k is a positive decay constant.
The half-life of y(t) is T1/2 = lnk2 .
Example 19. An initial dose of y0 = 100 mg of a drug is administered to a patient.
The amount of drug remaining in the patient’s blood after t hours is given by an
exponential decay model y(t) = y0 · e−kt . The half-life of this particular drug is 16
hours.
(1) Find the exponential decay function which models the amount of the drug
remaining in the blood.
(2) How much time is required for the drug to reach 1% of the initial dose?
14
(3) If a second 100 mg dose is given 12 hours after the first dose, how much
time is required for the drug level to reach 1mg?
HOMEWORK (Due 9/21/12): Section 7.3 #21, 22, 23, 25, 26, 30, 31, 32, 41,
44, 47, 51; Section 7.4 #11, 13, 14, 18, 21
Inverse Trigonometric Functions.
First recall the following definitions from trigonometry.
Definition 10. The arcsine function, denoted by arcsin x or sin−1 x, is the partial inverse of sin with domain [−1, 1] and range [− π2 , π2 ]. In other words, for
−1 ≤ x ≤ 1, we define arcsine by the rule
y = arcsin x if and only if sin y = x and − π2 ≤ y ≤
π
2.
We also define the arccosine function, denoted arccos x or cos−1 x, to be the
partial inverse of cos with domain [−1, 1] and range [0, π]. So for −1 ≤ x ≤ 1, we
have
y = arccos x if and only if cos y = x and 0 ≤ y ≤ π.
Example 20. Evaluate the following expressions.
√
(1) arcsin(
3
2 )
√
(2) arccos(−
3
2 )
(3) arccos(cos 3π)
(4) sin(arcsin( 21 ))
Derivative of arcsin x. We can now find the derivative of y = arcsin x by
using implicit differentiation. First recall that for the expression y = arcsin x to be
reasonable, we must require that −1 ≤ x ≤ 1 and − π2 ≤ y ≤ π2 . Now we do a few
computations:
y = arcsin x
sin y = x
d
d
sin y =
x
dx
dx
y 0 · cos y = 1
1
y0 =
cos y
We wish to express the derivative y 0 = cos1 y above in terms of our input variable
x. To do this, let’s recall the Pythagorean identity:
15
sin2 y + cos2 y = 1
Solving for cos y in the above, we get
q
cos y = ± 1 − sin2 y
p
= ± 1 − x2
But since y lies in the interval [− π2 , π2 , we must have cos y ≥ 0. This means we
√
can drop the ± sign in the above equality, i.e. we definitely have cos y = 1 − x2
and hence
y0 =
d
dx
arcsin x = √
1
1 − x2
Example 21. Compute the following derivatives.
(1)
d
2
dx [arcsin(x
(2)
d
dx [cos(arcsin x)]
− 1)]
Derivative of arccos x. First observe that the following identity holds:
arcsin x + arccos x =
π
2
(A picture helps to see the above.) Now differentiate with respect to x on both
sides of the equation:
√ 1
1−x2
+
d
dx
arccos x = 0
For the above to be true, we must have
d
dx
arccos x = − √
1
1 − x2
Other inverse trigonometric functions and their derivatives. We now
recall our other four inverse trigonometric functions, together with their domain
and range:
(1) The arctangent function, denoted arctan x or tan−1 x, with domain (−∞, ∞)
and range (− π2 , π2 ).
(2) The arccotangent function, denoted arccot x or cot−1 x, with domain
(−∞, ∞) and range (0, π).
16
(3) The arcsecant function, denoted arcsec x or sec−1 x, with domain (−∞, −1]∪
[1, ∞) and range [0, π2 ) ∪ ( π2 , π].
(4) The arccosecant function, denoted arccsc x or csc−1 , with domain (−∞, −1]∪
[1, ∞) and range [− π2 , 0) ∪ (0, π2 ].
Now we compute the derivatives of arctan x and arcsec x below. The former is
easy but the latter presents some minor technical difficulty.
y = arctan x
tan y = x
d
d
tan y =
x
dx
dx
y 0 sec2 y = 1
1
y0 =
sec2 y
1
y0 =
1 + tan2 y
1
y0 =
1 + x2
This gives us
d
dx
arctan x =
1
1 + x2
Now let’s turn to the arcsecant function. We have:
y = arcsec x
sec y = x
d
d
sec y =
x
dx
dx
y 0 sec y tan y = 1
y0 =
1
sec y tan y
Now we wish to write the right-hand side of the last equation abovepin terms of
x. We may use the identity 1 + tan2 y = sec2 y to obtain tan y = ± sec2 y − 1.
Now consider the value of the ± sign in the previous equality. If x ≥ 1, then
since y = arcsec x, we will have 0 ≤ y < π2 , in which case tan y > 0. So in
this case we take the ± to be +. On the other hand if x ≤ 1, then we will have
17
− π2 < y ≤ 0 and hence tan y < 0, so we will take ± to be −. In other words we have
1
d
dx arcsec
x =
d
dx arcsec
x =−
sec y
p
sec2
1
= √
x x2 − 1
1
=− √
x x2 − 1
y−1
1
sec y
p
sec2 y − 1
if x ≥ 1;
if x ≤ −1.
We may easily write the above piecewise function in the following compact form:
d
dx arcsec
x=
1
√
|x| x2 − 1
Now it is not too hard to see that the identities arctan x + arccot x = π2 and
arcsec x + arccsc x = π2 also hold. So, taking derivates and solving (just like in the
arccos x case), we obtain the last two derivatives we seek:
d
dx arccot
d
dx arccsc
Example 22.
x=−
x=−
1
1 + x2
1
|x| x2 − 1
√
√
(1) Evaluate f 0 (2 3), where f (x) = x arctan( x2 ).
(2) Find an equation of the line tangent to the graph of g(x) = arcsec (2x) at
the point 1, π3 ).
Integrals involving inverse trigonometric functions. Using our new derivative rules together with the chain rule, it is easy to verify that the following
indefinite integral formulas hold:
Z Z Z √
1
a2 − x2
1
2
a + x2
√
dx = arcsin
1
x x2 − a2
dx =
x
a
+ C, a > 0
x
1
arctan
+ C, a 6= 0
a
a
dx =
1
arcsec
a
x
+ C, a > 0
a
HOMEWORK (Due 9/26/12): Section 7.5 #41-52, #65-72
Example 23. Determine the following integrals.
18
(1)
R
√ 4
dx
9−x2
(2)
R
dx
16x2 +1
(3)
R5
√
√ dx
5/ 3 x 4x2 −25
L’Hospital’s rule and growth rates.
∞
∞
L’Hospital’s rule is a powerful tool for computing limits of “indeterminate form
or 00 .” We recall the statement of L’Hospital’s rule below.
Theorem 6 (L’Hospital’s Rule). Suppose f and g are functions differentiable at
(x)
is of indeterminate form, i.e.
x = a, and limx→a fg(x)
limx→a f (x) = limx→a g(x) = 0, or
limx→a f (x) = ±∞ and limx→a g(x) = ±∞.
Then limx→a
ists).
f (x)
g(x)
= limx→a
f 0 (x)
g 0 (x)
(provided the limit on the right-hand side ex-
Now that we have the natural logarithm at our disposal, we can apply L’Hospital’s
rule to an even larger class of limits- specifically, those limits of “indeterminate form
1∞ , 00 , and ∞0 .”
Example 24. Evaluate the following limits.
(1) lim xx
x→0+
x
1
(2) lim 1 +
x→∞
x
(3) lim+ (csc x)x
x→0
Solution. We’ll solve part (a) and leave the rest to the reader. First notice that we
may rewrite the problem as follows:
lim xx = lim ex ln x
x→0+
x→0+
Set L = limx→0+ x ln x. Now since et is a continuous function, we may carry the
limit above up into the exponent and write the following:
lim x ln x
lim xx = lim ex ln x = ex→0+
= eL
x→0+
x→0+
So if we can just compute L then we will have the limit we seek. This is where
L’Hospital’s rule comes in:
19
L = lim+ x ln x
x→0
ln x
∞
← Indetermine form
1
∞
x
1
= lim+ x1 ← by L’Hospital’s rule!
x→0
− x2
= lim −x
= lim+
x→0
x→0+
= 0.
It follows that we have limx→0+ xx = eL = e0 = 1.
The reader should notice when she attempts part (b), that if computed correctly,
the limit should be e!
Example 25. A particular theory for modeling the spread of an epidemic predicts
that the number of infected people t days after the start of the epidemic is given
t2
approximately by the function N (t) = 2.5 · 0.01t . In the long run (as t → ∞), does
e
the epidemic spread or does it die out?
Notice that the above example confirms our intuition that exponential functions
“grow faster” than polynomial ones. Let us make this notion precise.
Definition 11. Suppose f and g are functions with lim f (x) = lim g(x) = ∞.
x→∞
x→∞
We say that f grows faster than g if
lim
x→∞
g(x)
f (x)
= 0, or, equivalently, lim
= ∞.
x→∞ g(x)
f (x)
We say that f and g have comparable growth rates if
lim
x→∞
f (x)
=M
g(x)
for some fixed number M with 0 < M < ∞.
Example 26. Rank the following families of functions in order of how fast they
grow in comparison to one another.
(1) mx where m > 0 (linear growth)
(2) xp where p > 0 (polynomial growth)
(3) xx (superexponential growth)
(4) ln x (logarithmic growth)
20
(5) lnq x where q > 0
(6) xp lnq x where p, q > 0
(7) bx where b > 1 (exponential growth)
Integration by parts.
At this point in our development of the calculus, the student should recognize
that our differentiation techniques (chain rule, product rule, quotient rule, etc.)
give us a tremendous amount of power to compute derivatives, but that our integration techniques are comparatively lacking. So far the only major integration tool
we have developed is the substitution rule, which one should think of as a “reverse
chain rule.” We now wish to expand the number of integration tools at our disposal
by introducing a “reverse product rule.”
So consider a pair of differentiable functions u(x) and v(x). The product rule
tells us that
d
dx u(x)v(x)
= u0 (x)v(x) + u(x)v 0 (x)
and hence
u(x)v 0 (x) =
d
dx u(x)v(x)
− v(x)u0 (x)
Now we antidifferentiate both sides (and invoke the Fundamental Theorem of
Calculus) to obtain an integration rule, which we call integration by parts:
R
u(x)v 0 (x)dx = u(x)v(x) =
R
v(x)u0 (x)dx
Note that in the above expression we need not include any +C term, since there
is an indefinite integral on both sides of the equation. We may also write the rule
for integration by parts in the following memorable compact fashion, by using just
a slight abuse of notation:
R
udv = uv −
R
vdu
Example 27. Compute the following antiderivatives using integration by parts.
(1)
R
xex dx
(2)
R
x sin xdx
It may be necessary to use integration by parts more than one time, as the
following example illustrates.
Example 28. Compute the following antiderivatives.
21
(1)
R
x2 ex dx
(2)
R
e2x sin xdx
The Fundamental Theorem of Calculus also implies that we may use integration
by parts to compute definition integrals, using the following rule.
Rb
a
u(x)v 0 (x)dx = [u(x)v(x)]ba −
Rb
a
v(x)u0 (x)dx
Example 29. Compute the following definite integrals.
(1)
R2
(2)
R π/2
1
0
ln xdx
x cos 2xdx
HOMEWORK (Due 8/3/12): Section 7.6 #7, 8, 10, 12, 13, 19, 20, 23, 24, 28;
Section 8.1 #7, 8, 9, 10, 15, 17, 25, 29, 32, 33
Integration of Rational Functions by the Method of Partial Fractions.
Z
Example 30. Evaluate
3x
dx.
x2 + 2x − 8
Solution. The key to evaluating this integral is knowing that every rational function
may be rewritten as a partial fraction expansion. Note that the denominator
x2 + 2x − 8 in the integrand above factors into (x + 4)(x − 2). Then it is possible
to find two real numbers A and B for which
x2
3x
3x
A
B
=
=
+
.
+ 2x − 8
(x + 4)(x − 2)
x+4 x−2
The only trick is to find just what the values of A and B should be. To find the
values of A and B, multiply both sides of the equation above by (x + 4)(x − 2) to
clear all denominators. Then combine like terms:
3x = A(x − 2) + B(x + 4)
= Ax − 2A + Bx + 4B
= (A + B)x + (−2A + 4B)
Now we have obtained the equality 3x = (A + B)x + (−2A + 4B); it follows that
we must have 3 = A + B and 0 = −2A + 4B. So we have a system of two equations
in two variables; now we may resort to our algebra techniques! If we carefully solve
the given system, we should get A = 2 and B = 1. Thus we may convert our
original integral into the following, much easier problem:
22
Z
Z
Z
3x
2
1
dx =
dx +
dx
x2 + 2x − 8
x+4
x−2
= 2 ln |x + 4| + ln |x − 2| + C.
3x + 7x − 2
dx.
x3 − x2 − 2x
Z
5x2 − 3x + 2
dx.
x3 − 2x2
Example 31. Evaluate
Example 32. Evaluate
2
Z
Note: The same techniques we have used above will work for the last example above, if we keep track of one important detail: Notice that the denominator
x3 − 2x2 factors into x2 (x − 2). Since x appears as a factor with multiplicity 2;
we must represent x twice in the partial fraction expansion: once as an x term,
and once as an x2 term. In other words, we will be able to find a partial fraction
expansion of the form
A
B
C
5x2 − 3x + 2
= + 2+
,
2
x (x − 2)
x
x
x−2
for some real numbers A, B, and C. Otherwise we may proceed as usual.
Z
7x2 − 13x + 13
Example 33. Evaluate
dx.
(x − 2)(x2 − 2x + 3)
Note: To solve the above, we must again keep track of an important detail.
Notice that the polynomial x2 − 2x + 3 which appears in the denominator is an
irreducible quadratic, i.e. it cannot be factored into binomials with real coefficients. We may still find a partial fraction expansion for the integrand, but it will
take the form
7x2 − 13x + 13
A
Bx + C
=
+ 2
,
2
(x − 2)(x − 2x + 3)
x − 2 x − 2x + 3
where A, B, and C are some real numbers. A similar principle holds whenever an
irreducible quadratic appears in the denominator of a rational expression. These integrals can in general be challenging to compute, and may involve both the method
of completing the square (from the student’s previous algebra course) and the use
of the arctangent function.
Z
z+1
dz.
Example 34. Evaluate
z(z 2 + 4)
HOMEWORK (Due 10/8/12): Section 8.4 #10, 11, 15, 16, 19, 20, 30, 31
Trigonometric Integrals.
In the next few examples we will make use of the following well-known trigonometric identities, called respectively the Pythagorean identity and the halfangle formulas.
23
sin2 x + cos2 x = 1
sin2 x = 21 (1 − cos 2x)
cos2 x = 12 (1 + cos 2x)
Example 35. Evaluate the following integrals.
R
(1) R cos5 xdx
(2) sin4 xdx
Example 36. Evaluate the following integrals.
(1)
R
sin4 x cos2 xdx
(2)
R
sin3 x cos−2 xdx
The student can see from the above examples that evaluating integrals involving powers of trigonometric functions often involves using identities to reduce the
exponents involved to lower, more manageable levels. Of course, this process can
tend to become tedious when dealing with large exponents. For this purpose, we
introduce the following reduction formulas which the student may use at will
from now on:
R
sinn xdx = −
R
cosn xdx =
R
R
sinn−1 x cos x n − 1 R
+
sinn−2 xdx
n
n
cosn−1 x sin x n − 1 R
+
cosn−2 xdx
n
n
tann xdx =
secn xdx =
tann−1 x R
− tann−2 xdx
n−1
secn−2 x tan x n − 2 R
+
secn−2 xdx
n−1
n−1
Example 37. Evaluate tan4 xdx.
R
Example 38. Find sec xdx.
Trigonometric Substitutions
The main trick we will develop in this section is based on the following idea:
Suppose
we are trying to evaluate an integral which involves a term of the form
√
a2 − x2 , where a is some positive real number. For this term to make sense,
we must have x2 ≤ a2 and hence −a ≤ x ≤ a. So −1 ≤ xa ≤ 1, and hence we
24
may set θ = arcsin xa . In other words, we may write x = a sin θ for some angle θ.
p
√
2 − x2 =
a
a2 − a2 sin2 θ =
Then
if
we
make
a
change
of
variable,
we
may
write
p
2
a 1 − sin θ = a| cos θ|. In other words, by substituting a trigonometric function
into some expression, it may become possible to simplify the original expression
using our familiar trigonometric identities. We’ll use this idea for the next few
examples.
R
Example 39. Evaluate (16−x12 )3/2 dx.
Example 40. Use integration to compute the area of a perfect circle of radius r.
R
1
Example 41. Evaluate (1+x
2 )2 dx.
HOMEWORK (Due 10/24/12): Section 8.2 #10, 11, 13, 14, 22, 23; Section
8.3 #8, 9, 12, 13, 21, 22; Section 8.7 #5, 6, 7, 9, 10, 14, 28, 33
Improper Integrals.
Example 42. Let b be any real number greater than 1. Compute the following
integrals.
Rb
(1) 1 1dx
Rb 1
(2) 1 x2 dx
Solution. We have
Z
b
1dx = [x]b1
1
=b−1
and
Z
1
b
1
1
dx = [− ]b1
2
x
x
1
=1− .
b
Let us consider the geometric meanings of the expressions we have computed
Rb
above. If we sketch the picture, the first integral 1 1dx may be interpreted as the
area of the box with the x-axis for a base, the line x = 1 for a left side, the line
y = 1 for a top side, and the line x = b for a right side. If we allow b to grow
large, i.e. we allow the right side of the box to slide further right, the integral
confirms our intuition that the area of the box becomes larger and larger. It grows
Rb
unboundedly, i.e. we have limb→∞ 1 1dx = limb→∞ (b − 1) = ∞.
25
Rb
On the other hand, the second integral 1 x12 dx = 1 − 1b behaves remarkably
differently. If we compute the area bounded above by the graph of x12 between, say,
1 and 2, we get 1 − 21 = 12 . If we compute from 1 to 3 we get 23 ; if we compute
from 1 to 4 we get 34 , and so on. In general the area is increasing, which confirms
our intuition. But what happens as b becomes very large? Notice that no matter
how large b is, the area 1 − 1b will never exceed 1, i.e. the area does not grow
unboundedly as in the previous example.
Thus the region bounded by the curve x12 , the x-axis, and the line x = 1 is
Rb
infinitely long but has finite area. Since limb→∞ 1 x12 dx = limb→∞ 1 − 1b = 1,
we say that the region has area 1.
This is our first glimpse at a phenomenon that will become increasingly common
for us: that of an infinite “process” which yields a finite result.
Definition 12. Let f be a continuous function. We define improper integrals
with infinite bounds of integration as below, provided the given limits exist:
Z ∞
Z b
f (x)dx = lim
f (x)dx for any real number a;
b→∞
a
Z
b
a
Z
b
f (x)dx = lim
a→−∞
−∞
Z
∞
Z
f (x)dx = lim
a→−∞
−∞
f (x)dx for any real number b;
a
c
Z
f (x)dx + lim
b→∞
a
b
f (x)dx for any real number c.
c
If the limit exists and is finite, then we say the integral converges. If the limit
does not exist, or is equal to ∞ or −∞, then we say the integral diverges.
Example 43. Evaluate each integral.
R∞
(1) 0 e−3x dx
(2)
R∞
0
1
1+x2 dx
Example 44. Let R be the region bounded by the graph of y = x−1 and the x-axis,
to the right of the line x = 1.
(1) What is the volume of the solid generated when R is revolved about the
x-axis?
(2) What is the volume of the solid generated when R is revolved about the
y-axis?
We may also define improper integrals for functions with vertical asymptotes, as
we see in the next example.
R1
Example 45. What should the value of 0 √1x dx be?
Definition 13. Let f be a function continuous at x 6= p and with a vertical asymptote at x = p. Define the improper integrals (with unbounded integrand) as follows:
Z p
Z c
f (x)dx = lim−
f (x)dx for a < p;
a
c→p
a
26
b
Z
p
Z
b
a
b
Z
f (x)dx = lim+
d→p
Z
f (x)dx = lim−
c→p
a
f (x)dx for p < b;
d
c
Z
f (x)dx + lim+
d→p
a
f (x)dx for a < p < b.
d
Again, we say the integral converges if the limit exists and is finite; otherwise we
say the integral diverges.
R3
1
Example 46. Compute −3 √9−x
dx.
2
Example 47. State whether the following integrals converge or diverge. If they
converge, give the value of the integral. (Be careful!)
R1 1
dx
(1) −1 x1/3
(2)
R1
1
dx
−1 x3
Sequences.
Definition 14. A sequence is an infinite list of numbers, enumerated by the natural numbers starting from 1. We use the following notations to refer to the sequence
whose n-th term is the number an :
(a1 , a2 , a3 , ..., an , ...) = (an )∞
n=1 = (an ).
We should start with a few examples of sequences.
Example 48. For each sequence defined below, write out the first 5 terms of the
sequence.
(1) Let a1 = 1, and for each integer n ≥ 1, let an+1 = 3 + an .
(2) Let f (x) = x3 − 1. For each integer n ≥ 1, let an = f (n).
(3) For each integer n ≥ 1, let an =
1
2n .
(4) For each integer n ≥ 1, let an be the n-th digit after the decimal point in
the decimal expansion of the number π − 3.
Each of the examples above clearly defines a sequence, but they are all defined
by different means. For the first example, we say that the sequence (an ) is defined
by a recurrence relation, or we say that it is defined implicitly. For the second
and third examples, the sequence (an ) is defined explicitly, i.e. each term an may
be computed exactly as a well-defined function f of n.
We have not defined the fourth example via a recurrence relation nor by an
explicit formula, but it is still a valid sequence nonetheless.
Example 49. Let (an ) = (−2, 5, 12, 19, ...) and let (bn ) = (3, 6, 12, 24, 48, ...).
27
(1) Find recurrence relations that could generate the terms of the sequences
(an ) and (bn ).
(2) Using the relations you found, compute what the next two terms could be
in each sequence.
(3) For each sequence, find an explicit formula for the n-th term generated by
your recurrence relation.
Definition 15. If the terms of a sequence (an )∞
n=1 become arbitrarily close to some
fixed number L as n grows sufficiently large, then we say that L is the limit of (an ),
or that (an ) converges to L. We also write
lim an = L, or (an ) → L,
n→∞
so long as the latter notation does not introduce confusion about what limit we are
taking.
If there is no such number L then we say the sequence diverges.
Example 50. Write the first four terms of each sequence. Do you believe the limit
converges or diverges? If you believe it converges, then to what value?
2
∞
2n − 1
(1)
2n2
∞n=1
(−1)n
(2)
n2 + 1 n=1
(3) (cos(πn))∞
n=1
(4) (cos(2πn))∞
n=1
(5) (n)∞
n=1
(6) (an )∞
n=1 , where a1 = 1 and an+1 = −2an for all integers n ≥ 1.
(7) (an )∞
n=1 , where an is the n-th digit after the decimal point in the decimal
expansion of π − 3
Let’s make it easier on ourselves to compute limits of explicitly-defined sequences.
Theorem 7. Suppose f is a function for which f (n) = an for all positive integers
n. If lim f (x) = L, then lim an = L.
x→∞
n→∞
Question 5. Does the converse of the above theorem hold? That is, is it true that
if f (n) = an for some function f , and lim an = L, then lim f (x) = L?
n→∞
x→∞
In addition to the above theorem, the following nice properties all hold for any
convergent sequences (an ), (bn ) with lim an = A and lim bn = B..
n→∞
(1) lim (an ± bn ) = A ± B;
n→∞
n→∞
28
(2) lim can = cA, where c is any constant;
n→∞
(3) lim an bn = AB; and
n→∞
(4) lim
n→∞
an
A
= , provided B 6= 0.
bn
B
Example 51. Compute the limits of the following sequences, if they exist.
3n3
(1) (an ) where an = 3
.
n +1
n
5+n
.
(2) (bn ) where bn =
n
(3) (cn ) where cn = e−n n10 .
Definition 16. If (an ) is a sequence for which an+1 ≥ an for every n, then we say
(an ) is nondecreasing. If (an ) satisfies an+1 ≤ an for every n, then we say (an )
is nonincreasing. If (an ) is either nonincreasing or nondecreasing, we say (an )
is monotone. If there exists any real number M for which |an | ≤ M for every
integer n, then we say the sequence (an ) is bounded.
Example 52. Are the following sequences monotone? Bounded?
(1) (1 − n1 )∞
n=1
(2) (15 − 3n)∞
n=1
Theorem 8. Every bounded monotone sequence converges to some limit.
HOMEWORK (Due 10/31/12): Section 9.1 #10, 11, 14, 15, 25, 26, 27, 29;
Section 9.2 #9, 10, 11, 12, 13, 14, 18, 21, 27, 28
Infinite Series.
Now we arrive at one of the major developments of the course- the extension of
the basic arithmetical operation of addition, which up until now in our education
has formally applied only to finite collections of numbers, to encompass infinite
collections of numbers. We have already seen during our study of improper integrals (see, for example, the comments following Example 42) that when we involve
limits, it is possible for “infinite processes” to yield “finite results.”
Let’s look at some explicit examples.
Example 53. Find a geometric way to visualize the “infinite sum”
1
1
1
1
1
2 + 4 + 8 + 16 + ... + 2n + ...
What should the value of this sum be? Consider the area of a unit square in the
xy-plane.
Example 54. Does it make sense to compute the infinite sum 1 + 1 + 1 + 1 + ... ?
Now we will formalize the concept of “infinite addition” by applying the technology of sequences which we developed in the previous section.
29
Definition 17. Let (an )∞
n=1 be an infinite sequence. We refer to the infinite sum
a1 + a2 + a3 + ...
as an infinite series. We denote this series by
∞
X
an
n=1
which is a natural extension of our sigma notation for finite sums.
Note here that sequences and series are intimately connected, but that they
represent different concepts. A sequence is always just an infinite list of numbers
without any additional structure; a series always refers to an infinite addition of
numbers.
Definition 18. Let
∞
X
an be an infinite series. Define a new sequence (Sn )∞
n=1 as
n=0
follows:
S1 = a1
S2 = a1 + a2
S3 = a1 + a2 + a3
S4 = a1 + a2 + a3 + a4
...
In general, define Sn by Sn = a1 + a2 + ... + an for every integer n ≥ 1.
We call this sequence (Sn ) the sequence of partial sums of
converges to some limit L, then we say that the series
∞
X
∞
X
an . If (Sn )
n=0
an converges to L, and
n=0
we write
∞
X
an = L.
n=0
If the sequence (Sn ) diverges, then we say that the series
∞
X
n=1
Example 55.
(2) Compute
(1) Compute
∞
X
n=1
∞
X
1
, if the series converges.
n
2
n=1
1, if the series converges.
an diverges.
30
Example 56. Consider the infinite series
∞
X
1
.
n(n
+ 1)
n=1
(1) Find the first four terms in the sequence of partial sums.
(2) Do you think the series converges? If so, to what sum?
P∞
Example 57. Compute k=1 (−1)n n, if the sum converges. (Be careful!)
Definition 19. A geometric series is a series of the form
∞
X
arn = a + ar + ar2 + ar3 + ...
n=0
where a and r are some fixed real numbers.
Example 58. Consider the series
3+
3
2
+
3
4
+
3
8
+
3
16
+ ...
Is the series geometric? Does it converge, and if so to what value?
It is not hard to see that if
∞
X
arn is a geometric series where |r| ≥ 1, then
n=0
the series diverges. On the other hand, if we take |r| < 1, then we can prove the
following theorem.
Theorem 9. Let
∞
X
arn be a geometric series with |r| < 1. Then the series always
n=0
converges, and moreover we can compute the sum by the following formula:
∞
X
n=0
arn =
a
.
1−r
Proof. For each n set
Sn = a + ar + ... + arn−1 + arn
So (Sn ) is the sequence of partial sums. We must check if (Sn ) converges.
We can do this by making the following clever observation. Multiply both sides
of the equation above by r. Then we get
rSn = r(a + ar + ... + arn )
= ar + ar2 + ... + arn + arn+1 .
31
Subtracting the two equalities we’ve obtained, and then solving for Sn , we get
the following.
Sn − rSn = (a + ar + ... + arn ) − (ar + ar2 + ... + arn + arn+1 )
Sn − rSn = a − arn+1
(1 − r)Sn = a(1 − rn+1 )
Sn = a ·
1 − rn+1
.
1−r
n+1
So we have an explicit formula for the value of each partial sum Sn = a · 1−r
1−r .
All that remains to do is take the limit. Since |r| < 1, we have lim rn+1 = 0, and
n→∞
hence
∞
X
n=0
an = lim Sn
n→∞
= lim a ·
n→∞
1 − rn+1
1−r
1
1−r
a
.
=
1−r
=a·
So the series converges to
a
as we hoped.
1−r
Example 59. Compute the following geometric sums, if they converge.
∞
X
(1)
1.1n
(2)
n=0
∞
X
n=0
∞
X
e−k
n
3
4
n=0
n
∞
X
3
(4)
3
4
n=2
n
∞
X
3
(5)
3 −
4
n=0
(3)
3
Question 6. Suppose (an ) is a sequence for which P
limn→∞ an = L, where L > 0 is
∞
some positive number. Is it possible for the series n=1 an to converge?
32
Theorem 10 (Divergence Test). If
∞
n=1 an
lently, if (an ) does not converge to 0, then
converges, then lim an = 0. Equiva-
∞
X
n→∞
an diverges.
n=1
HOMEWORK (Due 11/7/12): Section 9.1 #50, 51, 53, 54; Section 9.3 #19,
20, 21, 24, 26, 30, 37, 48, 50, 51, 52, 53
Telescoping Series.
Example 60. Evaluate the following series.
∞ X
1
1
− k+1
(1)
3k
3
k=1
∞
X
1
(2)
(k + 1)(k + 2)
k=1
The Harmonic Series.
Consider the infinite series:
∞
X
1
1 1 1 1
= 1 + + + + + ...
n
2 3 4 5
n=1
The series above is called the harmonic series. Does the harmonic series converge or diverge?
Since the sequence ( n1 )∞
n=1 converges to 0, it seems plausible that the harmonic
series might converge. Computing the first few terms in the sequence of partial
sums, we may obtain:
S1 = 1
3
S2 =
2
11
S3 =
6
25
S4 =
12
...
The trend of convergence or divergence may not be clear from the first few terms.
What about if we consider very large n-th partial sums? It can be shown that the
sum of the first 1, 000, 000 terms of the harmonic series is less than 15. The number
of atoms comprising the planet Earth is (very roughly) 1050 ; the sum of the first
1050 terms of the harmonic series is approximately 115.6. The number of atoms in
the observable universe is (very roughly) 1080 ; the sum of the first 1080 terms of
33
the harmonic series is approximately 184.7.
It may be unclear from this information whether the series in fact converges or
diverges. In fact, we can show through an elementary argument that the harmonic
series diverges. Simply make the following observation:
∞
X
1
1 1 1 1 1 1
1
+ + + + + + +...
= |{z}
1 +
n
2
|{z}
|3 {z 4} |5 6 {z 7 8}
n=1
≥1/2
≥1/2
Since computing the infinite sum
≥1/2
≥1/2
∞
X
1
must involve adding terms whose sum is
n
n=1
≥ 21 infinitely many times, the sum must diverge to infinity. It just diverges very,
very slowly!
Now let’s go back and rigorously observe something which we claimed at the
beginning of the course (See second paragraph of “Range of the Logarithm” after
Definition 1).
Example 61. Show that lim ln x = ∞.
x→∞
Theorem 11 (Integral Test). Suppose f is a continuous, positive, decreasing function for x ≥ 1 and suppose an = f (n) for every positive integer n ≥ 1. Then the
∞
X
R∞
infinite sum
an and the improper integral 1 f (x)dx either both converge, or
n=1
else both diverge. However, if they both converge, their values need not in general
be the same.
Example 62. Determine whether the following series converge or diverge.
∞
X
n
(1)
n+1
n=0
∞
X
k
(2)
k2 + 1
(3)
(4)
k=1
∞
X
n=3
∞
X
√
1
2k − 5
1
, where p > 1 is some fixed real number.
np
n=1
The last example above yields the following fact.
∞
X
1
Theorem 12.
always converges if p > 1.
p
n
n=1
In the next few sections we will develop many methods for determining whether
a given series converges or diverges. In general it will not always be completely
clear which test, if any, will be the appropriate one to apply when faced with any
particular series. It is up to the student to develop the intuition and problemsolving skills for attacking such problems. However, as a general rule, it will be
34
extremely helpful to recall the notion of “relative growth rates” which we developed
in Example 26, and think of them in terms of sequences (Listen carefully in lecture
for more details!). Let us recall now the relative growth rates of some common
sequences, in order from slowest to fastest:
(lnq n) << (np ) << (np lnr n) << (np+s ) << bn << (n!) << (nn ) ,
where p, q, r, s are any real numbers and b > 1.
The Ratio Test.
Theorem 13 (Ratio Test). Let
∞
X
an be a series with positive terms. Let
n=1
r = lim
n→∞
an+1
.
an
(1) If 0 ≤ r < 1, the series converges.
(2) If r > 1 (including r = ∞), the series diverges.
(3) if r = 1, the test is inconclusive.
Example 63. Determine whether the following series converge.
(1)
(2)
∞
X
10n
n!
n=1
∞
X
nn
n=1
n!
The Root Test.
Theorem 14 (Root Test). Let
∞
X
an be a series with nonnegative terms. Let
n=1
p = lim
√
n
n→∞
an .
(1) If 0 ≤ p < 1, the series converges.
(2) If p > 1 (including p = ∞), the series diverges.
(3) If p = 1, the test is inconclusive.
Example 64. Determine whether the following series converge.
(1)
(2)
n
∞ X
4n2 − 3
n=1
∞
X
k=1
7n2 + 6
2k
k 10
The Direct Comparison Test.
Theorem 15 (Direct Comparison Test). Let
∞
X
n=1
an and
∞
X
bn be series with pos-
n=1
itive terms, and such that 0 < an ≤ bn for every integer n ≥ 1.
35
(1) If
(2) If
∞
X
n=1
∞
X
bn converges, then
an diverges, then
n=1
∞
X
an converges.
n=1
∞
X
bn diverges.
n=1
Example 65. Determine whether the following series converge.
(1)
(2)
∞
X
k=1
∞
X
k=1
k3
−1
2k 4
ln k
k3
The Limit Comparison Test.
Theorem 16 (Limit Comparison Test). Let
∞
X
n=1
an and
∞
X
bn be series with pos-
n=1
itive terms, and let
lim
n→∞
(1) If 0 < L < ∞, then
(2) If L = 0 and
∞
X
∞
X
n=1
an and
an
= L.
bn
∞
X
bn either both converge or both diverge.
n=1
bn converges, then
n=1
∞
X
(3) If L = ∞ and
n=1
bn diverges, then
∞
X
n=1
∞
X
an converges.
an diverges.
n=1
Example 66. Determine whether the following series converge.
(1)
(2)
∞
X
k 4 − 2k 2 + 3
k=1
∞
X
k=1
2k 6 − k + 5
ln k
k2
HOMEWORK (Due 11/14/12): Section 9.4 #18, 20, 25, 26, 44, 45, 47, 48;
Section 9.5 #11, 13, 14, 20, 22, 24, 28, 33, 35, 43, 44, 48
Decimal Expansions and Zeno’s Paradox.
Now we are finally ready to rigorously define the concept of an infinite decimal
expansion for a real number. Since every positive real number may be written as
the sum of an integer and a number between 0 and 1 (and its decimal expansion is
easy to compute from this decomposition), we will simplify things by only defining
infinite decimal expansions for numbers x in the interval [0, 1].
Definition 20. Let x be a real number in [0, 1]. Let (dn )∞
n=1 be a sequence such
that for every positive integer n, we have dn = 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9. In
36
other words, (dn ) is a sequence of digits. Suppose x and (dn ) satisfy the following
equality:
x=
∞
X
1
10n
dn ·
n=1
Then we say that (dn ) is an infinite decimal expansion for x, and we write
that
x = 0.d1 d2 d3 d4 ...
Answer (Answer 1). The above definition is the standard definition of a decimal representation, and it coincides with our most of our intuition regarding the meaning
of such a representation- so we have an answer to Question 1 from the beginning
of the course. Notice that if (dn ) is any sequence of digits, then 0.d1 d2 d3 ... really
does define some real number in [0, 1]. This is because dn · 101n ≤ 9 · 101n for every
n, and the sum
∞
X
9·
n=1
1
10n
always converges since it is just a geometric series. So the sum
0.d1 d2 d3 ... =
∞
X
dn ·
n=1
1
10n
also converges by the Direct Comparison Test.
Answer (Answer 2). Now we turn to Question 2, which is really two questions:
Does every real number have one, and only one, infinite decimal expansion?
We have already seen that every expansion defines a real number; now we will
see that every real number x in [0, 1] really does admit a decimal expansion. This
expansion is actually easy to construct recursively, by just considering finite decimal expansions. Let d1 be the greatest digit for which 0.d1 ≤ x. Then let d2
be the greatest digit for which 0.d1 d2 ≤ x. Let d3 be the greatest digit for which
0.d1 d2 d3 ≤ x, etc. Continue this process to obtain an infinite sequence (dn ); it is
not overly difficult to check that in the end we have x = 0.d1 d2 d3 ....
But is it true that each number x admits only one decimal expansion? Try computing the following two real numbers using the definition of the decimal expansion
and the geometric series formula.
1.0000... = 1 +
∞
X
0·
n=1
0.9999... =
∞
X
n=1
9·
1
10n
1
10n
37
This computation should yield insight as to why there appear to be no rational
numbers between 0.999... and 1, despite our claim that the rational numbers are
dense in the real number line.
It is a fact that any number which has a decimal expansion which ends in an
infinite tail of 0’s, also has a decimal expansion which ends in an infinite tail of 9’s.
However, no real number admits more than two distinct decimal expansions.
Answer (Answer 4). Our last question, Question 4, is now the easiest for us to
answer, given what we have already done. Our computation from the beginning of
the semester implies that the number of meters Achilles must travel to catch the
tortoise is given by
1000 + 100 + 10 + 1 + .1 + .01 + .001 + ...
Now we know it is possible for an infinite collection of summands to accumulate
to a finite sum, which resolves Zeno’s paradox. So Achilles need only run 1111.111...
meters before he catches the slow-moving reptile, or 1111 and 19 meters.
Alternating Series.
Example 67. Determine if the series
∞
X
(−1)k+1
k
this series the alternating harmonic series.)
converges or diverges. (We call
k=1
k
X
(−1)k+1
1 1 1
(−1)k+1
= 1 − + − + ... +
,
k
2 3 4
k
n=1
so Sk is the sequence of partial sums of the given series.
Solution. For each k ≥ 1 let Sk =
First consider the sequence which consists only of the even terms (S2 , S4 , S6 , ..., S2k , ...).
Notice that since 13 > 41 , we have 31 − 14 > 0, and hence S4 = S2 + 13 − 14 > S2 .
1
1
In fact for any even number 2k, we have 2k+1
− 2k+2
> 0 and hence S2k+2 =
1
1
S2k + 2k+1 − 2k+2 > S2k . It follows that the sequence of even terms (S2 , S4 , ...) is
strictly increasing.
In addition, the sequence (S2 , S4 , ...) is bounded above by 1 = S1 . So the sequence (S2 , S4 , ...) is a bounded monotone sequence, and hence it converges to some
limit L by our theorem from earlier.
If we consider the sequence (S1 , S3 , S5 , ..., S2k−1 , ...) of odd partial sums, then
arguments which are very similar to the above will show that the sequence is strictly
decreasing and bounded below by S2 ; so the sequence of odd partial sums also converges to some limit, say M , by our same theorem from earlier.
If we can check that L = M , then we will have shown the series converges;
otherwise we will have shown it diverges. To see the relationship between the two
1
for every positive integer k. Hence
limits, simply observe that S2k = S2k−1 − 2k
38
we have
L = lim S2k
k→∞
1
= lim S2k−1 −
k→∞
2k
= lim S2k−1 − lim
k→∞
k→∞
1
2k
=M −0
= M.
So the series converges!
Question 7. To what value does the alternating harmonic series converge? We
don’t have the tools to compute this right now, but we will see the answer in the
future, in Example 72.
Notice that the argument above depends only on three facts: that the sequence of
terms in the series alternates signs; that the terms of the (non-alternating) harmonic
series are a decreasing sequence; and that the terms converge to 0. So the solution
above may be easily modified to prove the following theorem (which we leave to
the reader):
Theorem 17 (Alternating Series Test). Suppose (an )∞
n=1 is a sequence of positive
terms for which
(1) an+1 ≤ an for every positive integer n, and
(2) lim an = 0.
n→∞
Then the alternating series
∞
X
(−1)n+1 an converges.
n=1
Example 68. Determine whether the following series converge or diverge.
(1)
∞
X
(−1)k+1
k2
k=1
(2) 2 −
(3)
3
2
+
4
3
−
5
4
+
6
5
− ...
∞
X
(−1)k ln k
k=2
k
A Note On Polynomials.
Recall that a polynomial is a function of the form
39
p(x) = c0 + c1 x + c2 x2 + ... + cn xn =
n
X
ck xk ,
k=0
where c0 , c1 , ..., cn are some real numbers, which we call coefficients. The positive
integer n in the above is called the degree of the polynomial p(x).
Now let a be any real number, and consider the function q(x) defined below:
q(x) = c0 + c1 (x − a) + c2 (x − a)2 + ... + cn (x − a)n =
n
X
ck (x − a)k .
k=0
It is obvious that q(x) is another polynomial of degree n, and that q(x) = p(x−a).
The reader should recall from a previous course that if q(x) = p(x − a), then the
graph of q(x) is just a shift of the graph of p(x); in particular, a horizontal shift
to the right, of magnitude a. If this is the case we say that the polynomial q(x) is
centered at a.
Now that we have defined infinite sums, and recovered some geometric intuition
for finite polynomials, it is not too much of a leap to our next topic: infinite polynomials.
Power Series.
Let us start off with a concrete example. Recall from our section on geometric
series that the following equation holds:
∞
X
rk = 1 + r + r2 + r3 + ... =
k=0
1
1−r
whenever r is a real number with |r| < 1. Let us now define a function f by the
following rule:
f (x) =
∞
X
xk = 1 + x + x2 + x3 + ...
k=0
Notice that f (x) only makes sense for numbers x which satisfy |x| < 1, since
the series above only converges for such x. So the domain of f is the open interval
1
(−1, 1). Moreover, by our geometric series formula, we have f (x) = 1−x
for all x
in (−1, 1); so f is continuous and infinitely differentiable on this restricted domain.
This function f is the simplest example of an “infinite polynomial.” We refer to
such functions as power series.
Definition 21. A power series has the general form
∞
X
k=0
ck (x − a)k ,
40
where a is a real number called the center of the series, and c0 , c1 , c2 , ... are real
numbers called the coefficients of the series. The set of values x for which the
series converges is called the interval of convergence of the series. The radius
of convergence of the series, denoted R, is the distance from the center of the
series to the boundary of the interval of convergence.
Example 69. Find the interval and radius of convergence for each power series.
∞
X
(−1)k (x − 2)k
(1)
4k
(2)
(3)
k=0
∞
X
k=0
∞
X
xk
k!
k!xk
k=1
Theorem 18. A power series
∞
X
ck (x − a)k centered at a converges in one of three
k=0
ways:
(1) The series converges for all x, in which case the interval of convergence is
(−∞, ∞) and the radius of convergence is R = ∞.
(2) There is a real number R > 0 such that the series converges for |x − a| < R,
and diverges for |x − a| > R, in which case the radius of convergence is R.
(3) There series converges only at a, in which case the radius of convergence is
R = 0.
Example 70. Use the Ratio Test to find the radius and interval of convergence of
∞
X
(x − 2)k
√
.
k
k=1
Theorem 19. Suppose f (x) =
∞
X
k=0
ck xk and g(x) =
∞
X
dk xk are two power series
k=0
with the same interval of convergence I. Then the following hold.
∞
X
(ck + dk )xk converges to the function (f + g)(x) on I.
(1)
k=0
(2) xm
∞
X
k=0
ck xk =
∞
X
ck xk+m converges to xm f (x) on I whenever m is a non-
k=0
negative integer.
(3) If h(x) = bxm where m is a positive integer and b is a real number, then
∞
X
ck (h(x))k converges to the composite function f (h(x)) for all x such
k=0
that h(x) is in I.
Example 71. Find the power series and interval of convergence for the following
functions.
41
x5
1−x
1
(2)
1 − 2x
1
(3)
1 + x2
(1)
Differentiating and Integrating Power Series.
Theorem 20. Let f (x) =
∞
X
ck xk be a power series with interval of convergence
k=0
I.
(1) f is a continuous function on I.
(2) The power series may be differentiated or integrated
term by term, and the
R
resulting power series converges to f 0 (x) or f (x)dx, respectively, at all
points in the interior of I.
∞
X
1
=
xk for |x| < 1.
Example 72. Let f (x) =
1−x
k=0
(1) Differentiate the series above to find a power series for f 0 (x), and identify
the function it represents.
(2) Integrate the series above term by term and identify the function it represents.
(3) To what value does the alternating harmonic series converge?
Example 73. Find power series representations centered at 0 for the following
functions and give their intervals of convergence.
(1) arctan x
1+x
(2) ln
1−x
Taylor Series.
We have shown that power series always represent continuous functions. Now
we will consider the converse: given an arbitrary continuous function, may it be
represented as some convergent power series? We will attack this problem in the
usual way: given a function f , we will attempt to find some polynomials which
closely approximate it; if we can do such a thing, then we will pass to a limit.
Example 74. Let f (x) = ln x.
(1) Find a linear function p1 which approximates f at x = 1, in the sense that
p1 (1) = f (1) and p01 (1) = f 0 (1).
42
(2) Find a quadratic function p2 which approximates f at x = 1, in the sense
that p2 (1) = f (1), p02 (1) = f 0 (1), and p002 (1) = f 0 (1).
(3) Find a cubic function p3 which approximates f at x = 1, in the sense that
(3)
p3 (1) = f (1), p03 (1) = f 0 (1), p003 (1) = f 00 (1), and p3 (1) = f (3) (1).
(4) Find an n-th degree polynomial pn which satisfies p(k) (1) = f (k) (1) for every
non-negative integer k ≤ n.
Definition 22. Let f be a function with at least n derivatives at a point a. The
nth-order Taylor Polynomial for f centered at a, denoted pn , is given by
pn (x) = f (a) + f 0 (a)(x − a) +
f 0 (a)
f (n) (a)
(x − a)2 + ... +
(x − a)n
2
n!
or
pn (x) =
n
X
f (k) (a)
k=0
k!
(x − a)k
(k)
This function pn has the property that pn (a) = f (k) (a) for all non-negative integers
k ≤ n.
Definition 23. Suppose f is a function with derivatives of all orders on an interval
containing the point a. The Taylor series for f centered at a is
∞
X
f (k) (a)
k=0
k!
(x − a)k
A Taylor series centered at 0 is called a Maclaurin series.
Example 75. Compute the Maclaurin series for the following functions, and give
the interval of convergence.
(1) f (x) = sin x
(2) f (x) =
1
1−x
(3) f (x) = ex
(4) f (x) = e3x
HOMEWORK (Due 12/5/12): Section 9.6 #15, 16, 17; Section 10.2 #10, 11,
12, 14, 21, 22, 33, 34, 40; Section 10.3 #10, 12, 13, 17, 19, 20