Supplemental Notes for Calculus II
Ellen Ziliak and Alexander Hulpke
Fall 2011
Alexander Hulpke
Department of Mathematics
Colorado State University
1874 Campus Delivery
Fort Collins, CO, 80523
© 2009 by the authors. This work is licensed under the Creative Commons Attribution-NoncommercialShare Alike 3.0 United States License. To view a copy of this license, visit http://creativecommons.org/
licenses/by-nc-sa/3.0/us/ or send a letter to Creative Commons, 171 Second Street, Suite 300, San
Francisco, California, 94105, USA.
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1
Convergence of Improper Integrals
In this chapter we will look at whether or not improper integrals converge or diverge.
DEFINITION 1
An improper integral converges if when it is evaluated the result is a finite value. An improper integral diverges if when it is evaluated the result is an infinite or undefined value.
We begin with an example. Suppose a scientist is observing a particle that is moving in a force field. He has
been able to determine from several measurements he made that the particle’s acceleration is proportional to
its velocity, resulting in a differential equation
dv
= −2 ⋅ v
dt
where v is the velocity of the particle. He also measured the initial velocity to be 100m/s. Solving this initial
value problem1 we get v(t) = 100e −2t .
The scientists now asks whether the particle will move arbitrarily far. To answer this question we need to
determine whether or not
∞
∫0
100e −2t dt converges. In this case the value of the integral is not difficult to
calculate using the methods previously described in this section:
lim
∫
b
100e −2t dt =
b→∞ 0
=
lim −50e −2t ∣
b→∞
lim −50e
b→∞
−2b
b
0
+ 50
= 50
So the total distance traveled by the particle at the end of time is 50m, which is finite.
1
This is the equation for exponential growth:
dv
= −2 ⋅ v ⇒
dt
∫
dv
=
v
∫ −2dt
After evaluating this integral we get ln ∣v∣ = −2t + C which we can then solve for v(t) to get v(t) = Ae −2t . The initial condition for
v(0) then gives v(t) = 100e −2t .
1
Now suppose instead of such a simple function for the velocity we got instead that the velocity of the
sin2 (t)
. If we want to know, whether such a particle travels a finite distance, we
particle is given by v(t) =
t2
∞ sin2 (t)
would have to determine whether
dt converges or diverges. We cannot just evaluate the integral,
t2
0
so we must discover some other techniques to answer this question.
∫
Direct Comparison
∞
∫a
Our goal is to determine if
f (x)dx converges without having to compute the value of the integral. One
way to do this is to compare with an integral we know something about. Let us first consider an example
where the limits of integration are finite to develop intuition about how we compare two functions.
EXAMPLE 1
Solution
1
1
∫−1 f (x)dx compared to ∫−1 g(x)dx?
1
First we look at the graphs of f (x) and g(x) below: We know geometrically ∫ f (x)dx can
−1
If f (x) = x 2 + 2 and g(x) = x 2 , what can we say about
y
f(x)=x2+2
4
3
3
2. 5
2
2
1. 5
f(x)=x2
-1
-0.8 -0.6 -0.4
1
1
0. 5
-0.2
0 .2
0. 4
0 .6
0. 8
x
1
-1
0
-0.8 -0.6 -0.4 -0. 2
be interpreted as the area under the curve of f (x) on [−1, 1] and
0 .2
0. 4
0 .6
0. 8
x
1
1
∫−1 g(x)dx can be interpreted as the area
under the curve of g(x) on [−1, 1]. And we notice on the interval from [−1, 1] that f (x) > g(x) (in fact this is
always true but we only care about this interval) the area under g clearly must be smaller than the area under
f.
We have seen that if f (x) > g(x) then
1
∫−1
f (x)dx >
1
∫−1 g(x)dx.
This result holds of course for any other interval:
THEOREM 1
If on the entire interval [a, b] we have f (x) > g(x) ≥ 0 then
Proof is given by picture on p348
We now want to consider what happens in the limit case:
2
b
∫a
f (x)dx >
b
∫a
g(x)dx.
EXAMPLE
2
∞
Does
∫1
Solution
1
dx converge?
ex
Consider f (x) = e1x and g(x) = x12 on the interval [1, ∞) Let us first notice that e1x is smaller
1
0. 8
0. 6
g(x)=1 ∕ x2
0. 4
0. 2
f(x)=1 ∕ e x
0
2
4
6
8
10
12
14
16
18
x
20
∞
1
1 ) when p > 1,
dx converged (to p−1
xp
1 = 1. Therefore, extrapolating from the finite case, we would
so the area under x12 from [1, ∞) is equal to 2−1
∞ 1
dx < 1 which would mean that this improper integral converges.
assume that
ex
1
than x12 on the entire interval from [1, ∞). We also know that
∫1
∫
To verify this, let us calculate this integral directly, so we can verify that this guess was correct.
∫1
∞
b
b
1
−x
−x
dx
=
lim
e
dx
=
lim
−e
∣
ex
b→∞ 1
b→∞
1
1
= lim −e −b + e −1 = = 0.367879 < 1
e
b→∞
∫
This comparison property holds for any other pair of functions. We get the following generalization of
Theorem 1:
THEOREM 2
Let f(x) and g(x) be continuous on [a, ∞) with f (x) > g(x) ≥ 0 on the whole interval. Then if
converges, then
∞
∫a
∞
∫a
f (x)dx
g(x)dx converges.
With this theorem we can answer the question from the start of the chapter:
EXAMPLE 3
The Particle Revisited
∞
sin2 (t)
dt converges.
t2
1
∞ 1
sin2 (t)
1
Solution
Since sin2 (t) < 1 always, if we compare
<
.
Since
dt converges we can cont2
t2
t2
1
∞ sin2 (t)
clude that
dt converges. We therefore know that the distance traveled by the particle is bounded
t2
1
∞ 1
by
dt = 1 m, however we cannot give a concrete value for this distance.
t2
1
A similar argument can be used to show that an improper integral diverges.
In the example we wanted to know whether or not
∫
∫
∫
∫
3
EXAMPLE
4
∞
Does
∫2
√
Solution
1
dx converge?
x −1
Consider f (x) = √1 and g(x) = √ 1
x
x−1
on the interval [2, ∞). We see in the graph that
12
10
8
6
f(x)=1 ∕ √x
g(x)=1 ∕ √x-1
4
2
0
1
2
4
8
6
x
10
∞
2 1
1
√ dx diverges. since
√ dx is a
x
1
x
x
finite value (namely 0.82842112) throwing this part out won’t change the conclusion that the value of the
∞ 1
√ dx diverges.
integral is infinite, so
2
x
Since √ 1 > √1 on the entire interval [2, ∞), the area under √ 1 on [2, ∞) must be larger than the
x
x−1
x−1
∞
1
1
√
dx should also diverge. Again we check the conclusion by actually
area under √ on [2, ∞), so
x
2
x −1
calculating the value of this integral:
√1
x−1
> √1 for x > 1. We also know from previous work that
∫1
∫
∫
∫
∫2
∞
√
b
1
1
√
dx = lim
dx
b→∞ 2
x −1
x −1
∫
du = 1 and substituting the limits of integration in we have u(2) = 1 and u(b) = b −1,
let u = x −1 then we have dx
yielding:
∫
b→∞ 1
lim
∞
b−1
−1
1
u 2 du = lim 2u 2 ∣
b→∞
b−1
1
1
= lim 2(b − 1) 2 − 2 = ∞
b→∞
1
dx diverges.
x −1
Again we state this observation as a general result:
Therefore
∫2
√
THEOREM 3
Let f(x) and g(x) be continuous on [a, ∞) with f (x) < g(x) on the whole interval. Then if
diverges, then
∞
∫a
∞
∫a
f (x)dx
g(x)dx diverges.
In all of the examples we have seen so far we have had f (x) ≥ g(x) on the entire interval we have considered. In some situations this makes it unnecessaily hard to find a function to compare with. We will see in
the next example that one can in fact weaken this condition:
4
EXAMPLE 5
∫1
Does
∞
Solution
sin2 (x) 1
+ dx converge?
x
4
sin2 (x)
Let f (x) = x + 41 and g(x) = x1 on the interval [1, ∞). Then initially we have f (x) oscil-
1
0. 8
0. 6
f(x)=sin2(x)/x+1∕4
0. 4
0. 2
g(x)=1 ∕ x
5
15
10
20
x
25
sin2 (x)
lating over g(x) from [1, 3.5021575], then after that point we have x1 < x + 41 from [3.5021575, ∞). Since
3.5021575 sin2 (x)
3.5021575 1
1
+ dx = 1.425137 is finite and
dx = 1.2533763 is also finite, we know that
x
4
x
1
1
the behavior on the interval from 1 to 3.5021575 will not affect the convergence or divergence of the improper
integral.
∞ 1
We also know that
dx diverges; similarly by splitting up the interval we have
x
1
∫
∫
∫
∫1
letting us conclude that
∞
∞
∞
3.5021575 1
∞
1
1
dx −
dx =
dx
x
x
x
1
3.5021575
∫
∫
1
∫3.5021575 x dx diverges as well. Then since x1 < sin x(x) + 41 from [3.5021575, ∞) we
2
sin2 (x) 1
+ dx also diverges.
x
4
3.5021575
∞
sin2 (x) 1
Comparison now tells us that
+ dx diverges, from which we conclude by the same
x
4
3.5021575
∞ sin 2 (x)
1
argument as before that
+ dx diverges,
x
4
1
Again we state this observation in a theorem which generalizes the theorems stated before in this section.
can conclude that
∫
∫
∫
5
THEOREM 4
Direct Comparison Test for Improper Integrals
Let f(x) and g(x) be piecewise-continuous on [a, ∞)
a) If there exists a value b ≥ a such that f (x) ≥ g(x) for all x ≥ b and
∞
∫a
∫a
∞
about
∞
∫a
∫a
∞
∫a
f (x)dx diverges then
g(x)dx diverges.
Notice if we happen to choose a f (x) ≥ g(x) and
∞
f (x)dx converges then
g(x)dx converges.
b) If there exists a value b ≥ a such that g(x) ≥ f (x) for all x ≥ b and
∫a
∞
g(x)dx. Similarly, if g(x) ≥ f (x) and
∞
∫a
∞
∫a
f (x)dx diverges we cannot conclude anything
f (x)dx converges we cannot say anything about
g(x)dx.
Limit Comparison Test
Finding
a point b in Theorem 4 for which f (x) ≥ g(x) can be hard. It also leaves the case that f and g
1
1
repeatedly are “oscillating around each other” but “stay close”. For example consider f (x) = and g(x) =
x
1 +0.sin(x)/2
8
:
x
0.6
y
g(x)=(1+sin(x)/2)/x
0.4
0.2
0
f(x)=1/x
10
20
30
40
x
50
One might hope that the functions are “close enough” to let us compare the convergence behavior.
The proper context for describing “close enough” is that of growth, which we studied in section 7.6. We
recall the relevant definition:
DEFINITION 2
Let f (x) and g(x) be real valued functions which are positive for sufficiently large values of x.
f (x)
g(x)
= ∞, or — equivalently — if lim
= 0.
x→∞
g(x)
f (x)
We also say that g(x) grows slower than f (x) as x → ∞.
a) f grows faster than g as x tends to infinity if lim
x→∞
6
b) f and g grow at the same rate as x → ∞ if lim
x→∞
f (x)
= L where L is finite and positive.
g(x)
For improper integrals to converge, of course we need functions to converge to zero, so we don’t compare
“growth” but “decay”. We get this by simply considering the reciprocal function 1/ f (x) and 1/g(x). We get
the following analogous definition:
DEFINITION 3
Let f (x) and g(x) be real valued functions which are positive for sufficiently large values of x.
f (x)
g(x)
= 0, or — equivalently — if lim
= ∞.
x→∞
g(x)
f (x)
We also say that g(x) decays slower than f (x) as x → ∞.
a) f decays faster than g as x tends to infinity if lim
x→∞
g(x)
= L where L is finite and positive.
x→∞ f (x)
b) f and g decay at the same rate as x → ∞ if lim
Similar to the case of growth, if f decays faster then g there must be a value b such that f (x) < g(x) for
all x ≥ b. This means that we can apply Theorem 4 and make deductions about convergence:
EXAMPLE
6
∞
∫1
1
dx converge?
1
1
Solution
Let f (x) = 3
and g(x) = 2
on the interval from [1, ∞).
x +1
x +1
We find that
1
x3 + 1
g(x)
2
lim
= lim x 1+1 = lim 2
=∞
x→∞ x + 1
x→∞ f (x) x→∞
3
Does
x3 + 1
x +1
Therefore f (x) decays faster than g(x), which means at some point b, for all x ≥ b we have f (x) ≥ g(x). We
can see in the graph of the function that this is true.
1
f (x)=1/(x3+1)
0. 8
0. 6
0. 4
0. 2
0
g(x)=1/(x2+1)
2
4
Now we study if
directly.
6
∫1
∞
8
10
12
14
16
18
x
20
1
dx converges. There are two ways to do this. The first is to compute the integral
x2 + 1
∫1
∞
b
b
1
1
dx
=
lim
dx
=
lim
atan(x)∣
x2 + 1
b→∞ 1 x 2 + 1
b→∞
1
π π π
= lim atan(b) − atan(1) = − =
2 4 4
b→∞
∫
7
Therefore the integral converges.
∞ 1
1
1
(We could also have use the direct comparison test, and compared to 2 ≥ 2
so since
dx
x
x +1
x2
1
∞
1
converges we can conclude that
dx converges.)
2
x +1
1
∞
1
1
1
decays faster than 2 , we therefore know that
dx converges.
As 3
x +1
x +1
x3 + 1
1
∫
∫
∫
Again we get an analogue condition for divergence.
If f (x) and g(x) decay at the same rate, convergence of ∫ f (x)dx implies convergence of ∫ g(x) and vice
versa. (This is not simply a reduction to Theorem 4, but a result that has to be proven separately.) Collecting
the behavior for all three cases, we get the following theorem:
THEOREM 5
Limit Comparison Test for Improper Integrals
Let f (x) and g(x) be piecewise-continuous on [a, ∞) and suppose that f (x) ≥ 0 and g(x) ≥ 0 for all x ≥ a
∞
∞
f (x)
= L > 0 then
f (x)dx and
g(x)dx both converge or both diverge.
x→∞ g(x)
a
a
∫
a) If lim
∫
∞
∞
f (x)
= 0 and
g(x)dx converges then
f (x)dx converges.
x→∞ g(x)
a
a
b) If lim
∫
∫
∞
∞
f (x)
= ∞ and
g(x)dx diverges then
f (x)dx diverges.
x→∞ g(x)
a
a
c) If lim
∫
∫
One can show that all previous theorems in this chapter, including the direct comparison test, are special
cases of this theorem. This means that any problem which can be done by direct comparison can also be done
by limit comparison. (Still, often direct comparison is easier, as no limit needs to be computed.)
∞
∞
f (x)
Also notice again that if lim
= 0 and
f (x)dx converges, we get no conclusion about
g(x)dx
x→∞ g(x)
a
a
∞
∞
f (x)
and similarly if lim
= ∞ and
f (x)dx diverges, we get no conclusion about
g(x)dx.
x→∞ g(x)
a
a
∫
∫
∫
Let us look at an example for comparison of same decay order:
EXAMPLE 7
Does
∫2
∞
Solution
x3 + x2 + 1
dx converge?
x4 − x2
x3 + x2 + 1
1
Let f (x) =
and g(x) = on the interval [2, ∞).
x
x4 − x2
Direct comparison is difficult, so we look at
f (x)
lim
x→∞ g(x)
=
=
x 3 +x 2 +1
4
2
lim x −x
= lim
1
x→∞
x→∞
x
x4 + x3 + x
lim
x→∞
x4 − x2
8
=1
x(x 3 + x 2 + 1)
x4 − x2
∫
Therefore we can conclude that these two functions decay at the same rate. Since
∞
∫2
∞
1
dx diverges we
x
x3 + x2 + 1
dx also diverges.
x4 − x2
2
In this example we could calculate the integral directly, using partial fractions2 , giving (of course) the
same result, albeit with far more effort.
thus know that
∫
Finding Comparison Candidates
The final question we want to study in this chapter is how to choose good comparison functions. For this we
can consider growth order of the reciprocal functions. In the previous example, the reciprocal function would
1
x4 − x2
be
= 3
. From chapter 7.6, we know that this is of the same growth order as x 4−3 = x, which
f (x) x + x 2 + 1
1
leads us to a comparison with g(x) = .
x
∞ 1
Consideration of the growth order will often lead to integrals of the form
dx as comparison
a xp
candidates3 . For such integrals we have explicitly determined the criterion that they converge if and only if
∫
2
We have
∫
2
∞
x3 + x2 + 1
dx
x4 − x2
=
=
lim
b→∞
lim
b→∞
∫
b
∫
b
2
2
x3 + x2 + 1
dx = lim
b→∞
x4 − x2
x3 + x2 + 1
dx
2
x (x − 1)(x + 1)
∫
b
2
x3 + x2 + 1
dx
x 2 (x 2 − 1)
The method for solving the integral is partial fractions, we get
x3 + x2 + 1
x4 − x2
=
=
=
C
D
A B
+
+ +
x2 x x − 1 x + 1
A(x 2 − 1) + B(x 3 − x) + C(x 3 + x 2 ) + D(x 3 − x 2 )
x 2 (x − 1)(x + 1)
Ax 2 − A + Bx 3 − Bx + Cx 3 + Cx 2 + Dx 3 − Dx 2
x 2 (x − 1)(x + 1)
Comparing coefficients for powers of x in the numerator, we get the following equations
B+C+D
=
B
=
A+C − D
−A
We can then solve for the variables and get A = 1, B = 0, C =
lim
b→∞
∫
2
b
1
1
1
+
+
dx
x 2 2(x − 1) 2(x + 1)
=
=
=
1
,
2
D=
=
=
1
.
2
1
1
0
−1
Substituting in the original integral, we get
b
1
1
lim x −1 + ln∣x − 1∣ + ln∣x + 1∣∣
b→∞
2
2
2
1 1
1
1 1
1
lim + ln∣b − 1∣ + ln∣b + 1∣ − − ln∣1∣ − ln∣3∣
b→∞ b
2
2
2 2
2
1 1
0 + ∞ + ∞ − − ln(3) = ∞
2 2
Thus the integral diverges, as we found already by the limit comparison test with far less effort.
3
which is the main reason we bothered with stating a theorem for such integrals
9
p > 1. This not only indicates comparison candidates, but also whether the integral will converge or diverge.
If we cannot find a function of the same growth order as 1/ f (x), parts b) and c) of Theorem 5 often can be
used to find functions of strictly faster decay for which the improper integral diverges, respectively functions
of slower decay for which the integral converges.
Going back to the example starting this chapter, we find for example that for f (t) = 100e −2t the reciprocal
1/ f (t) = e 2t /100 is of larger growth than any power of t, thus we could compare to g(t) = 1/t 2 to show
convergence of
∞
∫0
f (t)dt without explicitly calculating the antiderivative.
EXAMPLE 8
Determine, by finding suitable comparisons, whether the following improper integrals converge:
1.
2.
3.
4.
5.
6.
∫1
∫1
∫1
∫1
∫1
∫1
∞
3
√ dx
x+ x
∞
ln x
x 3/2
∞
dx
x2
dx
2x
∞
1
dx
(ln x)2
∞
x2 + 1
√ dx
x3 x
∞
1
sin ( ) dx
x
Solution
√
3
√ , then 1/ f (x) = 1/3(x + x). This is of the same growth order as x, thus the integral
x+ x
∞ 1
diverges by limit comparison with
dx.
x
1
1. If f (x) =
∫
x 3/2
. We know that ln x grows slower than any power of x, thus we see
ln x
x 3/2
that x 3/2−ε for arbitrary small ε, for example x 5/4 , is a function of smaller growth. We get that
2. If f (x) =
ln x
, then 1/ f (x) =
ln x
x 3/2
lim
x→∞ 1
x 5/4
As
∫1
∞
1
x 5/4
= lim
ln x ⋅ x 5/4
x 3/2
x→∞
= lim
ln x
x→∞ x 1/4
dx converges, we conclude by limit comparison that
∫1
=0
∞
ln x
x 3/2
dx converges.
x2
2x
3. If f (x) = x , then 1/ f (x) = 2 . This is of growth larger than any power of x, for example x 2 . Therefore
2
x
x2
4
∞ 1
x
x
dx.
lim 21 = lim x = 0, and the integral converges by limit comparison with
x→∞
x→∞ 2
x2
1
2
∫
x
10
1
, then 1/ f (x) = (ln x)2 . This grows slower than x, thus the integral diverges by
(ln x)2
∞ 1
dx. (Of course we also have that (ln x)2 grows less than x 2 , but limit
limit comparison with
x
1
1
comparison with 2 is a case in which we cannot draw any conclusion.)
x
√
√
x3 x
x2 + 1
x3 x
, which is of the same growth order as x 2 = x 3/2 . The integral
5. If f (x) = 3 √ , then 1/ f (x) = 2
x +1
x x
∞ 1
dx.
converges by limit comparison with
1
x 3/2
4. If f (x) =
∫
∫
6. This one is tricky, as we don’t know the growth order for 1/ sin(1/x). The 1/x suggests limit comparison
with 1/x. We get
L’Hopital
−x −2
1
=
lim
= lim
=1
1
1
x→∞ sin ( )
x→∞ −x −2 cos ( ) x→∞ cos ( 1 )
x
x
x
lim
∞
1
x
∞
1
1
dx, diverges, so does
sin ( ) dx. (Once we know power series we’ll see another explax
x
1
nation for this behavior: For small values of z, sin(z) behaves like z. Thus sin ( x1 ) will behave like x1 for
large x.)
As
∫1
∫
We close this chapter with the remark that we considered only criteria for convergence tests for the case
of
∞
∫a
f (x)dx. Obviously similar tests are possible for the other types of improper integrals.
EXERCISES
Determine whether the following improper integrals
converge or diverge. You may use direct comparison
or limit comparison (or explicit computation of an
antiderivative)
1.
2.
3.
∫1
∫3
∫1
4.
∫1
5.
∫2
∞
6.
1 + cos(x)
dx
x2
∫1
7.
∫1
8.
∫1
∫1
∞
1
dx
ln(ln(x))
∞
2x
dx
3x − 1
9.
x + 2x
dx
x 2 2x
10.
∫1
1
dx
(ln(x))2
11.
∫1
∞
∞
11
∞
x3
dx
3x
∞ 3x
x3
dx
∞
−2
√ dx
x x
∞
ln(x)
dx
x 1.001
∞ 3x−1 + 1
3x
∞
√
dx
1
x2 + 2
dx
-
2
Geometric Series
In this chapter we will discuss one of the simplest and most useful types of infinite series, the geometric series.
They are derived from instances where geometric growth occurs, this is growth that is proportional to the
current value, i.e the growth rate is constant. (Here we are looking at the discrete situation, in a continuous
situation a differential equation for constant growth rate produces exponential functions as solutions.)
A simple example of geometric growth is interest accrued annually on a bank account. Each year we take
the amount of money in the account the value of the account and multiply it by the interest rate and add it to
the value to get a new value. Therefore the constant of proportionality is one plus the interest rate.
EXAMPLE 1
If you put $100 into a savings account that earned 3% interest accrued annually, how much would you have
after one year? two years? n years?
Solution
Let p n be the value of our account at time n, so p0 = 100 since we begin initially with $100.
Next at the end of one year we earn 3% interest on this account so p1 = (1.03) ⋅ p0 = (1.03) ⋅ 100, hence
p1 is proportional to p0 with the constant of proportionality being 1.03. This process repeats so that p2 =
(1.03) ⋅ p1 = (1.03)2 ⋅ 100. At this point we notice a pattern p n = (1.03)n ⋅ 100.
As a second example of geometric growth we consider university scholarships which come from endowments. In this case, an endowment (a fixed sum) is given to the university. A scholarship then is funded from
the interest of this money (without using up the money, the scholarship therefore exists forever).
EXAMPLE 2
Endowments
Delighted by the calculus course he took at CSU, an alumnus wants to endow a scholarship for mathematics
students that will pay $2000 each year. The university guarantees an interest rate of 5% per year. How much
must the alumnus donate to guarantee this scholarship will be available forever?
Solution
If we let x=the value of the endowment, then after one year the endowment is worth 1.05 ⋅ x
and we want to give a $2000 scholarship, so we want 1.05x = x + 2000. Therefore the endowment will never
be touched and the scholarship is funded by the interest earned. Solving this equation we get 0.05x = 2000
which gives us that x = 40, 000. Therefore the alumni must give an endowment of $40, 000 to guarantee the
scholarship is funded forever.
A geometric series now is a series whose terms show geometric growth, i.e. s n = r ⋅ s n−1 for a constant r
independent of n. This type of series has applications in several fields including physics, biology, economics
and finance. We will look at a few examples of where geometric series occur in our everyday lives including:
repeating decimals and calculating dosage of medicine.
13
DEFINITION 1
Let r be a ratio and a any nonzero constant. A geometric series is a series of the form
∞
∞
∞
n=1
n=1
n=0
a + ar + ar 2 + . . . + ar n−1 + . . . = ∑ ar n−1 = a ∑ r n−1 = a ∑ r n
As an example of a geometric series we look at the “Race course paradox” of Zeno1 . Suppose a runner
wanted to travel a given distance, say one kilometer. Then he must first travel the first half kilometer, and the
next half kilometer remains. Next the runner must travel half of the next half kilometer – a quarter kilometer
– and the next quarter kilometer remains. It might seem (so claimed Zeno) as if the runner never reaches the
goal.
However if we look at the distance traveled after k iterations, we get a geometric sum with ratio 21 :
1 1 1
1
1
+ + + +⋯+ k
2 4 8 16
2
If the process continues infinitely many times, we thus get get the following formula for the distance traveled:
∞ 1
1
1 1 1
+ + +...+ n +... = ∑ n
2 4 8
2
n=1 2
We will see that this sums up to 1 (i.e. the full distance is traveled, resolving the paradox).
Notice that there is a slight shift in this formula from the one given in the definition of a geometric series
in that we start not with r 0 = 1 but with r 1 = 21 while in the definition this power is zero. This can be fixed by
letting the constant a be 21 , This sort of index change arises often enough that we will first look at how we can
manipulate such infinite sums.
Reindexing a Series
k
We begin this discussion with a brief reminder about the Σ notation for sums. Given ∑ a n we call n the
n=1
index of summation, a n the n th term of the sum, and the upper and lower bounds of summation are k and
1 respectively.
Now we consider reindexing our infinite series, notice in the following examples we are working with
geometric series, however this algorithm will be useful with other types of series that will be discussed later.
1
Zeno of Elea, Greek philosopher, ca. 490 BC - ca. 430 BC
14
Most of the infinite series we have seen so far have a lower bound of summation of one, however their are
times when we are given an infinite series where the lower bound of summation will be a value other than
one. We consider first a geometric series which does not have a lower bound of one:
∞ −1
−1 1
1
−1
+ −
+ . . . + ( )n−1 + . . . = ∑ ( )n−1
27 81 243
3
n=4 3
∞
This is still a geometric series, however our definition requires the series to have the form ∑ ar n−1 , so reinn=1
dexing is required. To reindex a series we follow the following basic algorithm:
Algorithm 1
Reindexing a series
1 Choose a new letter to be the index of summation (m)
2 Relate m to the original index of summation (n) such that when n=4 we have m=1. So we have the
equation m=n-3.
3 Solve for original index of summation (n) and substitute into the n th term of the sum:
−1 m+3−1 ∞ −1 m+2
)
= ∑( )
m=1 3
m=1 3
∞
n = m+3 ⇒ ∑(
Notice if the upper bound of summation were finite we would have to decrease its value by 3 as well,
however since we are summing to infinity our upper bound of summation will remain infinite.
4 Solve for an a such that the power on r is m-1.
∞
−1 m+2 ∞ −1 3 −1 m−1
)
= ∑( ) ⋅( )
3
m=1 3
m=1 3
∑(
since m-1+3=m+2.
∞
m−1
−1
−1
3
Therefore a = ( −1
. Notice that
3 ) = 27 and r = 3 so we have a geometric series of the form ∑ ar
m=1
in this example the ratio was negative, therefore a geometric series can have negative or positive r values. We
now look at another example of this reindexing algorithm before we develop an equation for the value of a
geometric series.
EXAMPLE 3
∞
4
, rewrite this series so that the lower bound of summation is n=1.
n=9 2n + 3
∞
∞
4
4
Let m = n − 8, then n = m + 8 so we have ∑
=∑
2(m
+
8)
+
3
2m
+ 19
m=1
n=1
Consider the series ∑
Solution
Using this algorithm we can always manipulate a given geometric series so that it can be written in the
∞
∞
n=1
n=0
form ∑ ar n−1 or equivalently ∑ ar n next we consider how to calculate the value of this series.
15
The Formula for the Geometric Series
k
In this section we will derive a general formula for the value of ∑ ar n . We can apply this in the limit case
n=0
∞ 1
k = ∞ to get the value of an infinite series, for example to show that the series ∑ n from Zeno’s paradox
n=1 2
has value 1.
DEFINITION 2
∞
The k th partial sum s k of a Geometric series ∑ ar n is the (finite) sum of the first k terms:
n=0
k−1
s k = a + ar + ar 2 + . . . + ar k−1 = ∑ ar n
n=0
Using this definition we can consider the sequence s1 , s2 , s3 , . . . , s k , . . . of partial sums, which converges
∞
k−1
∞
to ∑ ar n , since lim s k = lim ∑ ar n = ∑ ar n . Therefore, once we know a formula for the value of s k we
n=0
k→∞
k→∞ n=0
∞
n
n=0
can compute the value of ∑ ar as the limit.
n=0
We begin by considering
and
s k = a + ar + ar 2 + . . . + ar k−1
rs k = ar + ar 2 + ar 3 + . . . + ar k .
Notice that s k and rs k share several terms in common, in fact
s k − rs k = (a + ar + ar 2 + . . . + ar k−1 ) − (ar + ar 2 + ar 3 + . . . + ar k ) = a − ar k
Now if we factor both sides of this equation, we get
s k (1 − r) = a(1 − r k ).
As long as r ≠ 1 we can divide both sides by (1 − r) to get
k−1
1 − rk
s k = ∑ ar = a
.
1−r
n=0
n
(What happens in the case where r=1? We get s k = a + a(1) + a(1)2 + . . . + a(1)k−1 = ka.)
Before we go on to get a formula for the infinite sum, we see how the formula for s k might be used on its
own:
EXAMPLE 4
The magnitude of repeated growth
Suppose we are given a chessboard and place one grain of rice on the first square, two grains of rice on the
second square, four grains of rice on the third square continuing on so that there are 2n−1 grains of rice on the
n th square. How many grains of rice are on the chess board? (There are 8 ⋅ 8 = 64 squares on a chess board.)
16
Solution
Since there are 64 squares on the chess board, we want to compute s64 . Our ratio is 2 and a is
one, hence we are asked to find
s64 = 1 + 2 + 4 + . . . + 263 =
1(1 − 264 )
= 1.84467x1019
1−2
A similar calculation underlies the repayment of loans or mortgages:
EXAMPLE 5
Paying back a loan
Suppose we take out a loan of L dollars which is paid back periodically (typically monthly). The periodic
payment is a dollars, the fixed interest rate per period is i. If b k is the loan sum outstanding after k time
periods, we have that
b k+1 = b k ⋅ (1 + i) − a.
Using b0 = L and setting r = 1 + i, in the first step we have
b1 = Lr − a
we then continue on to the next step where we get
b2 = b1 (r) − a = (Lr − a)(r) − a
= Lr 2 − (a + ar)
We might be starting to notice a pattern, however it becomes obvious in the next step
b3 = b2 (r) − a = (Lr 2 − (a + ar))(r) − a
= Lr 3 − (a + ar + ar 2 )
At this point we can solve this recursion to
k−1
b k = L ⋅ r k − (a + ar + ar 2 + ⋯ + ar k−1 ) = L ⋅ r k − ∑ ar n
n=0
Sum formula
=
L ⋅ rk − a
1 − rk
.
1−r
A bank now would set b m = 0 (where m is the time after which the loan should be paid off, e.g. m = 12⋅30 = 360
for a 30 year mortgage) and solve for a to determine the necessary monthly repayment, given the loan sum
and interest rate.
For example, if we have a mortgage of L = $200, 000, an annual interest rate of 6% (leading to a monthly
rate of i = .06/12 = 0.005, i.e. r = 1.005) and a monthly repayment sum2 of a = $1, 200, we find that the
outstanding sum after k months is
b k = 200, 000 ⋅ 1.005k − 1, 200
1.005k − 1
= 200, 000 ⋅ 1.005k − 240, 000 (1.005k − 1) .
0.005
After 10 years (120 months) this leaves an outstanding amount of $167, 224.13, after 20 years $107, 591.82,
roughly half3 , after 30 years $−903.00 (i.e. the loan is paid off after 30 years less one month4 ).
2
Moralistic remark: Incidentally, initial interest amounts to $1000 per month at the start of the loan. An interest only loan thus
does not save much and is a very bad deal!
3
In general, this means that a loan is paid off half after roughly 23 of its planned life time
4
The total cost of the loan then will have been $430, 800, more than double the loan amount. (Though inflation means that the
actual value will be less.)
17
If the interest rate instead was 7% annually (i = .07/12 = 0.005833 monthly), we get with same repayment
sum a remaining loan amount of $159, 256.18 after 30 years, which is not even halfway paid off. A monthly
repayment sum of $1330 would be needed5 to have the loan paid off after 30 years.
∞
Now since we saw earlier that the sequence s1 , s2 , s3 , . . . , s k , . . . of partial sums converges to ∑ ar n−1 , we
can use the formula just derived to compute a value for the limit of a geometric series.
THEOREM 1
∞
The geometric series a + ar + ar 2 + . . . + ar n−1 + . . . = ∑ ar n−1 converges to
n=1
otherwise.
n=1
a
when ∣r∣ < 1, and diverges
1−r
Proof
Case 1: If ∣r∣ < 1 then ∣r∣k < 1, and as k gets larger this will cause ∣r k ∣ to get smaller and yield
lim r k = 0. Thus, by the rules for limits,
k→∞
a(1 − r k ) a(1 − 0)
a
=
=
.
1−r
1−r
k→∞ 1 − r
lim s k = lim
k→∞
Case 2: When ∣r∣ ≥ 1 then ∣r∣k ≥ 1. Therefore when k tends to infinity the numerator a(1 − r k ) tends to plus
a(1 − r k )
or minus infinity, and the denominator 1 − r is a fixed value. So lim
diverges.
k→∞ 1 − r
Now that we have a formula for the value of a geometric series let us verify the solution to the Zeno’s
∞ 1
paradox example. We are looking at ∑ n . First notice that this sum is not in the correct form, we must
n=1 2
∞ 1
∞ 1
∞ 1
1
1
factor ( 21 ) out of the n th term. We get ∑ ( )n = ∑ ( ) ⋅ ( )n−1 = ∑ ( ) ⋅ ( )n , hence a = 21 and r = 21 < 1
2
2
n=1 2
n=1 2
n=0 2
so this series converges to
1
1
a
= 2 1 = 21 = 1.
1−r 1− 2 2
Next we will look at some interesting examples where we can use this formula.
Examples
In this section we will look at how people may use geometric series in real life applications.
In our first example we look at how much medication will remain in a persons body after taking medication at regular intervals
EXAMPLE 6
Repeated Drug Dosage
A CSU student comes down with bacterial laryngitis, the student goes to the Healthcenter to see a doctor who
prescribes 250mg doses of antibiotics that the student must take every six hours over the next several days.
5
I.e. a change of one percentage point in the interest rate increased the monthly payment (and thus the total loan cost) by 10%!
No wonder people go bonkers about interest rates.
18
Since the body will digest the antibiotics between doses and it is known that 5 percent of the drug remains
in the body at the end of six hours, how can we calculate the quantity of the drug that remains in the student
after the fifth dose? At the end of five full days on the drug?
Solution
Let us begin by looking at the quantity of the drug in the body after each dose. Let q n represent
the quantity of the drug in the students body after the n th dose. Initially
q1 = 250
at the next dose we will have 5% of that 250mg remaining and another 250mg is taken, so
q2 = 250(0.05) + 250
At the third dose we have 5% of q2 remaining and another 250mg is taken, so therefore
q3 = 0.05 ⋅ q2 + 250 = 0.05(250(0.05) + 250) + 250
= 250(0.05)2 + 250(0.05) + 250.
Now on the fourth dose q4 we have 0.05 ⋅ q3 remaining in the body and another 250mg is taken.
q4 = 0.05 ⋅ q3 + 250 = 0.05(250(0.05)2 + 250(0.05) + 250) + 250
= 250(0.05)3 + 250(0.05)2 + 250(0.05) + 250.
Notice at this point that a pattern has developed, we have
n
q n = s n = ∑ 250(0.05)n−1
k=1
Therefore we can use our formula s n =
calculate q5 =
250(1−0.055 )
1−0.05
a(1−r n )
1−r
to calculate q n =
250(1−0.05 n )
. With this information we can
1−0.05
= 263.1578125mg in the students body after 5 doses. If the student continues to
take the antibiotic for 5 full days or 20 doses they will have q20 =
250(1−0.0520 )
1−0.05
= 263.1578947mg.
Suppose the student continued to take this drug forever, how much medication would there be in the
students body at each dose?
∞
This question is asking us to compute ∑ 250(0.05)n−1 which luckily for the student 0.05 < 1 will converge
n=1
to a finite number of
= 263.1578947mg. Since 0.0520 = 9.536743x10−27 which is practically zero, we
get the same value up to the ten-millionth place if the student took the medication forever as we got when
the student took the medication for five days.
250
1−0.05
EXAMPLE 7
Repeating Decimal
Express the repeating decimal 0.08 as a fraction of two integers.
Solution
To express 0.08 as a fraction we first notice that 08 is the repeated value, so we want to break
up 0.08 into a sum of the repeated values. Therefore we have 0.08 = 0.08 + 0.0008 + 0.000008 + . . .. Once we
8 ,
have this written as a sum we can now write each term in the sum as a fraction, hence we have 0.08 = 100
8 = 8 , and 0.000008 = 8 . Using this information we should recognize a pattern
0.0008 = 10,000
1002
1003
0.08 =
∞ 8
8
8
8
+
+
+
.
.
.
=
∑
n
100 1002 1003
n=1 100
19
However at this point we notice that we have a geometric series which is not written in the correct form for
∞ 8
1 n−1
8 ) we get
us to apply our formula. If we factor out ( 100
)(
)
which is now in the correct form.
∑(
n=1 100 100
1 < 1 so our geometric series converges and since a = 8 the series converges to
This gives r = 100
100
8
100
1
1 − 100
8
8
= 100
=
= 0.08
99
99
100
Being able to express a repeating decimal as a fraction is useful when we are calculating with a large number
of values where the precision of the answer is vital since a fraction will not have any round off error.
Another paradox due to Zeno initially seems even more puzzling, which is the reason we consider it last.
Achilles, the fastest runner in ancient Greece, runs 100 times as fast as the Tortoise. But — so the paradox
claims — if the Tortoise is given an advance, Achilles will never be able to pass the Tortoise.
EXAMPLE 8
Achilles and the Tortoise
Suppose that Achilles runs 10m/s and the Tortoise only 0.1m/s, furthermore the Tortoise is given a head start
of 100m. After 10 seconds, Achilles has reached the place where the Tortoise started. But in this time, the
Tortoise has run 1m ahead. Achilles will reach this distance in 0.1 seconds. But then the Tortoise has moved
another 0.01m. Achilles will take 0.001 seconds to reach this, and so on. He will never reach the Tortoise.
Where is the error in this argument?
Solution
The paradox is resolved, if we realize what time period is considered: All events take place within
∞
1
10
1000
=
=
= 10.10
n
1
99
1 − 100
n=0 100
10 + 0.1 + 0.001 + ⋯ = 10 ∑
seconds. This is exactly the time when Achilles reaches (and overtakes!) the Tortoise. The paradox arises from
the implicit (and wrong, as we have calculated!) suggestion, that it describes the events for all time.
EXERCISES
1. What monthly payment is needed to pay a car 3. The goverment – due to an election year in
loan of $ 15, 000 off after 48 months if the interest rate spending mode – enacts a stimulus package that puts
is 7%?
$150 billion back into the economy. We assume that
all the people who have extra money to spend would
spend 80% of it and save 20%. Thus, of the extra in2. A standard method for valuing a business (ex- come generated by the package, $150⋅ 45 billion = $120
cluding any inventory) is the “discounted cash flow” billion would be spent again and so become extra inmethod: The value of the business itself is the present come to someone else. Assume that these people also
day value of its future earnings, discounted by an as- spend 80% of their additional income (which would
sumed interest rate, i.e. the amount of money that be $96 = 120 ⋅ 45 = 150( 45 )2 billion) and so on.
would have to be invested at an assumed interest rate a) Let C = 150 be the amount of the stimulus package
to get the same income as the business.
(in billion dollars) and r = 4/5 the spending factor.
Calculate the present day value of a business Write down an infinite sum, involving C and r, for
whose annual earnings are $50, 000, assuming an in- the amount of money added to the economy.
terest rate of 5%.
b) Calculate the value for the infinite sum in a).
20
-
3
Series as Discrete Analogues of Improper
Integrals
The geometric series, studied in the previous chapter, are special in that we can completely describe under
what condition it converges, and also have an explicit formula for its limit. In general finding the limit of
a series can be very hard, many of the techniques1 needed are far beyond this course. Instead we want to
study the easier question, whether a given series converges. This is very similar to what we did with improper
integrals, and indeed we will see in this chapter that the convergence of series and the convergence of improper
integrals are intimately related: Once we know how to test convergence of improper integrals we can use
essentially the same methods to test convergence for series.
The Integral Test
∞
If we are given a series of the form ∑ a n , the only condition for convergence we know is the n-th term test,
n=1
which requires that a n be non-increasing. This condition however is not sufficient in general, as for example
∞ 1
the divergent series ∑ shows: Its terms are decreasing, but the series does not converge.
n=1 n
We now want to see, that we can interpret the partial sums of a series as approximation of an integral. The
infinite sum of the series then becomes an approximation of an improper integral (of type I), and convergence
of the series implies convergence of the improper integral and vice versa.
To see why this should hold, we need to go back to the definition of integrals which we did using Riemann
sums.
EXAMPLE 1
∞ 1
Does ∑ 2 converge?
n=1 n
Solution
1
k 1
Consider a n = n12 . Then any partial sum of the form ∑ 2 can be interpreted as (right endn=1 n
for example using methods from complex analysis, a 400-level course
21
point) Riemann sum2 for 1 +
∫1
k
1
dx (the blue boxes).
x2
1.0
y
0.5
0
0
We see that
2
4
6
8
x
10
k 1
k 1
dx
∑ 2 ≤1+
1 x2
n=1 n
∫
We now consider the limit k → ∞. The integral on the right will converge (as integral of x −p for p > 1), thus
the sequence of partial sums is bounded from above. On the other hand (we are summing up positive terms)
the sequence of partial sums is increasing, thus this sequence (which is the series) must converge.
Let us consider the same example once more, but as Riemann sum for left endpoints (green boxes), apk+1 1
proximating
dx:
x2
1
∫
1.0
y
0.5
0
0
2
4
6
8
x
10
The Riemann sum now is an upper approximation, thus we get that
∫1
2
k+1
k 1
k 1
1
dx
≤
≤
1
+
dx
∑
2
x2
1 x2
n=1 n
∫
1
As ∫0 x −2 diverges, we need to consider the leftmost blue box separately
22
We therefore see that we can also consider the series as upper approximation of the integral, by a similar
1
argument as in the previous example we get that convergence of the series ∑∞
n=1 n 2 implies convergence of
∞ 1
the improper integral. ∫1 x 2 dx.
This observation is not specific to x12 , it holds for any decreasing function. We thus get the following
theorem3 :
THEOREM 1
Integral Test
Let a n be a sequence of positive terms. Suppose that there is a function f defined for all positive real numbers
such that
1. a n = f (n) for all n,
2. f is piecewise-continuousa ,
3. f is decreasing for x > N (where N is some integer, e.g. N = 0 if f is decreasing over the positive real
axis)
Then
a
∫1
∞
∞
f (x)dx converges if and only if ∑ a n converges.
n=1
otherwise we cannot guarantee that integrals exist even in the case of finite limits
Proof
Convergence of an infinite sum or an improper integral is not affected by the behavior over a finite
interval. Thus we can consider without loss of generality the case N = 0. The argument then is essentially the
same as in the discussion for x12 .
EXAMPLE 2
∞ 1
Does ∑ n converge?
n=1 e
Solution
The function f (x) = e1x is continuous and decreasing for x > 0. We have seen in the chapter on
∞ 1
∞ 1
improper integrals that
dx
converges.
Thus
by
theorem
1
also
converges.
∑
n
ex
1
n=1 e
∫
EXAMPLE 3
p-series
∞ 1
For which values of p does ∑ p converge?
n=1 n
∫
The function f (x) = x1p is continuous and decreasing for x ≥ 1. We know that
1
∞ 1
verges, if and only if p > 1. Thus, by theorem 1, ∑ p converges if and only if p > 1.
n=1 n
∞ 1
∞ 1
Thus (we have seen this previously) ∑ 2 converges and ∑ diverges.
n=1 n
n=1 n
Solution
3
In the book this is theorem 9 on p. 773
23
∞
1
conxp
Direct Comparison Test for Series
In the chapter on improper integrals we learned about powerful comparison tests that can be used to determine convergence. If we wanted to test convergence for a series, we therefore could consider the corresponding improper integral, and resolve its convergence via a comparison test. Doing so however it turns out to be
unnecessarily formal; because theorem 1 has an if and only if condition, we can (by defining for a series ∑ a n
a function f (x) with f (x) = a n if n ≤ x < n + 1) translate the comparison test for improper integrals to a
comparison tests for series. (Compare the statement to theorem 4.)
THEOREM 2
Direct Comparison Test for Series
Let a n and b n be non-increasing sequences of non-negative numbers.
∞
∞
n=1
n=1
a) If there exists an integer value N such that a n ≥ b n for all n ≥ N and ∑ a n converges then ∑ b n
converges.
∞
∞
n=1
n=1
b) If there exists an integer value N such that a n ≤ b n for all n ≥ N and ∑ a n diverges then ∑ b n
diverges.
As with improper integrals, this theorem does not reach any conclusion if a n ≤ c n and ∑ c n diverges or if
a n ≥ d n and ∑ d n diverges.
EXAMPLE 4
∞
sin2 (n)
converge?
n2
n=1
Does ∑
∞ 1
≤ n12 for all values of n. We also know that ∑ 2 dx converges. By theon=1 n
∞ sin2 (n)
dx converges.
rem 2, part a), we can conclude that ∑
n2
n=1
Solution
We know that
sin2 (n)
n2
EXAMPLE 5
∞
1
Determine if ∑ 3
converges.
n
+3
n=1
∞ 1
Solution
Let us consider ∑ 3 , which is a p-series for p = 3 > 1 and therefore converges. Since n 31+3 ≤ n13
n=1 n
∞
1
for all n ≥ 1 the direct comparison test shows ∑ 3
converges.
n
+3
n=1
Limit Comparison Test for Series
With the same argument as used for the ordinary comparison test, the limit comparison test can be adapted
to series. (Again, compare to theorem 5.)
24
THEOREM 3
Limit Comparison Test for Series
Suppose that a n > 0 and b n > 0 for all n ≥ N where N is an integer
an
= L > 0, then ∑ a n and ∑ b n both converge or both diverge.
n→∞ b n
a) If lim
an
n→∞ b n
= 0 and ∑ b n converges then ∑ a n converges.
an
n→∞ b n
= ∞ and ∑ b n diverges then ∑ a n diverges.
b) If lim
c) If lim
an
an
= 0 and ∑ b n diverges or lim
= ∞ and ∑ b n converges the theorem
n→∞ b n
n→∞ b n
Again notice that if lim
gives us no conclusion.
EXAMPLE 6
∞
1
, does this series converge or diverge?
n=1 n + 2
∞ 1
Solution
We know that ∑ √ diverges (it is a p-series for p = 21 < 1).
n=1 n
We calculate the limit of the quotient:
Consider ∑ √
√1
n+2
lim
n→∞ √1
n
√
√
n( √1 )
n
n
= lim √
= lim √
n→∞ n + 2 n→∞ n + 2( √1 )
n
√
1
= lim √
=1
n→∞
1 + n2
∞
So by the limit comparison test we conclude that ∑ √
n=1
1
diverges. (Note that direct comparison will not
n+2
1
1
work in this example, as √
< √ .)
n
n+2
As with improper integrals, candidates for comparison are often determined by growth classes.
EXAMPLE 7
∞
(ln(n))2
converge or diverge?
n3
n=1
Solution
We know that ln(x) grows less than any power of x, thus (ln(x))2 should also grow less than
any power of x. Thus we assume that the sum converges.
2
To show this, we just need to pick a suitable exponent x ε in relation to ln(x) to make (x ε ) x −3 decay
Does ∑
1
faster than x1 . We pick x 8 . Then
∞
and ∑
n=1
1
11
n4
1
(n 8 )2
1
= 11
3
n
n4
converges, as 114 > 1.
25
We now use limit comparison:
(ln(n))2
n3
lim
1
n→∞
11
n4
11
(ln(n))2
n 4 (ln(n))2
=
lim
= lim
1
n→∞
n→∞
n3
n4
To apply L’Hopital’s rule, requiring derivatives, we rewrite this expression in the variable x instead of n:
=
lim
∞
1
11
lim
L’Hopital
=
1
x4
x→∞
L’Hopital
=
Since ∑
(ln(x))2
8( x1 )
x→∞ 1 −3
4
4x
= lim
x→∞
32
1
x4
lim
x→∞
2(ln(x))( x1 )
1
4
−3
x4
= lim
x→∞
8(ln(x))
1
x4
= 0.
∞
(ln(n))2
converges.
n3
n=1
converges, by the limit comparison test we can conclude that ∑
n4
As in this example (and with improper integrals), comparison to a p-series is often useful.
Another fruitful comparison candidate is given by geometric series, the method for finding a suitable
geometric factor is given by the ratio test, which we shall study next:
n=1
The Ratio Test
In this section we want to see how the geometric series can be used along with the comparison test to reach
∞
conclusions about the convergence of series. Recall that a geometric series is a series of the form ∑ ar n ,
n=0
which we have shown will converge if ∣r∣ < 1.
Suppose we want to test a given series ∑ a n (with nonnegative values a n ) for convergence by comparing
n
with a geometric series ∑ ar n . For comparison to work, we need that the terms a n of our series fall as quickly
n
as the terms ar n of the geometric series.
We can ensure this by looking at the ratio of subsequent terms: Assume for the moment that
a n+1 ar n+1
≤
=r
an
ar n
and that ∣r∣ < 1, so we know that ∑ ar n converges.
n
a
Because of the inequality just assumed, we get — setting b ∶= 0 — that
a
a1
a a
= 1 ⋅ 0
ar 1 a0 a
a2
a
a
= 2⋅ 1
ar 2 a1 ar 1
1
r
1
⋅
r
⋅
a n+1
a
an 1
= n+1 ⋅ n ⋅
a n ar r
ar n+1
26
1
=b
r
1
≤ r⋅b⋅ =b
r
⋮
1
≤ r⋅b⋅ =b
r
≤ r⋅b⋅
and therefore
an
≤b<∞
ar n
By the limit comparison test (Theorem 3) we therefore know that ∑ a n also converges.
lim
n→∞
n
To do this in a concrete example, of course we would have to determine r. However in the end we don’t
really care about the value of r, but whether ∑ a n converges. This is guaranteed if (eventually — think of the
n
a
possibility of a positive number N for the comparison test) there is a number r < 1 such that n+1 ≤ r. But
an
this is guaranteed, if
a
lim → ∞ n+1 = L < 1
n
an
1−L
(Think of the ε-criterion for the limit of the sequence: If the limit L is smaller than 1, we set ε =
and can
2
a
find an N such that n+1 ≤ L + ε < 1 for n ≥ N. We simply set r ∶= L + ε.)
an
a
We have thus seen that ∑ a n converges, if lim → ∞ n+1 < 1. This is the first version of the Ratio Test.
n
an
n
To state this test in general, we note that an analogous argument can be given to show divergence (comparison with a diverging geometric series), if the ratio has limit > 1.
THEOREM 4
Ratio Test for Series with positive terms
Let ∑ a n be a series with positive terms and suppose that
a n+1
=r
n→∞ a n
lim
then
(a) the series converges if r < 1
(b) the series diverges if r > 1 or if r is infinite
(c) the test is inconclusive if r = 1.
∞
1
which
n=1 n
To see why we really cannot deduce anything in the case of r = 1, consider two examples: ∑
∞ 1
diverges by the p-series test and ∑ 2 which converges by the p-series test. Let us look at the limits of the
n=1 n
ratios in each of these cases:
1
n( n1 )
n
1
= lim
= lim
=1
1
n→∞ n + 1 n→∞ (n + 1)( ) n→∞ 1 + 1
n
n
lim n+1
= lim
1
n→∞
Similarly
n
1
n 2 +1
lim
n→∞ 1
n2
n2 ( n12 )
n2
1
= lim 2
= lim
= lim
=1
n→∞ n + 1 n→∞ (n 2 + 1)( 1 ) n→∞ 1 + 1
n2
n2
Therefore we see that in the case where ∣r∣ = 1 the test is inconclusive.
27
EXAMPLE 8
∞ n2
Consider the series ∑ n . We want to determine if this series converges or diverges.
n=1 2
Solution
Consider the ratio
a n+1
an
=
=
(n+1)2
2 n+1
n2
2n
=
(n + 1)2 2n
2n+1 n2
(n + 1)2 2n (n + 1)2
=
2 n 21 n 2
2n2
Now consider the behavior as n gets large (in other words take the limit as n → ∞), we get
(n + 1)2 ( n12 )
(1 + n1 )2 1
(n + 1)2
lim
=
lim
=
lim
= < 1.
n→∞ 2n 2
n→∞ 2n 2 ( 1 )
n→∞
2
2
n2
∞ n2
By the ratio test we therefore conclude that the series ∑ n converges.
n=1 2
EXAMPLE 9
∞ n!
Determine whether ∑ n converges or diverges?
n=1 e
Solution
We begin by considering the ratio
(n+1)!
(n + 1)n!e n (n + 1)
a n+1
n+1
= e n! =
=
an
e
e n e 1 n!
n
e
Now we look at the behavior as n tends toward infinity lim
n→∞
∞ n!
∑ n diverges.
n=1 e
n+1
= ∞ therefore we can conclude the series
e
Sometimes the limit in the ratio test requires L’Hopital’s rule for evaluation:
EXAMPLE 10
∞
ln(n)
converges or diverges?
n
n=1 2 n
Determine whether ∑
Solution
We begin by considering the ratio
ln(n+1)
ln(n + 1)n2n
ln(n + 1)n
a n+1 2n+1 (n+1)
=
= n
=
ln(n)
an
2 2(n + 1) ln(n) 2(n + 1) ln(n)
n
2 n
ln(n + 1)n
∞ , therefore we
. It has the indeterminant form ∞
n→∞ 2(n + 1) ln(n)
Next we take the limit of this function lim
have to use L’Hopital’s rule.
Writing the function in x (to indicate that we now consider it as a continuous, not a discrete function),
we get
28
x
+ ln(x + 1)
ln(x + 1)x
= lim x+1
x→∞ 2(x + 1) ln(x) x→∞ 2((1) ln(x) + x+1 )
x
lim
Again we have the indeterminant form ∞
∞ , so we apply L’Hopital’s rule once more, and get.
=
=
(x+1)(1)−x(1)
1
+ x+1
(x+1)2
lim
(x)(1)−(x+1)(1)
x→∞
2( x1 +
)
x2
2
(x + 2)(x )
lim
x→∞
(x + 1)2 (2x − 2)
=
= lim
(x+1)
1
+ (x+1)2
(x+1)2
lim
x→∞ 2( x − 1 )
x2
x2
3
2
x + 2x
2x 3 + 2x 2 − 2x − 2
x→∞
=
=
x+2
(x+1)2
lim
x→∞ 2x−2
x2
x 3 + 2x 2
1
lim
2 x→∞ x 3 + x 2 − x − 1
At this point we have only polynomials left. We know that the behaviour of such a quotient is determined
x 3 + 2x 2
by the leading terms (formally, one applies L’Hopital’s rule multiple times), thus lim 3
= 1 and
x→∞ x + x 2 − x − 1
1
a
lim n+1 = < 1.
x→∞ a n
2
∞ ln(n)
Therefore we can conclude by the ratio test that the series ∑ n converges.
n=1 2 n
EXAMPLE 11
∞ n!
Determine whether ∑ n converges or diverges?
n=1 n
Solution
We begin by considering the ratio
(n + 1)!n n
(n + 1)n!n n
nn
a n+1
=
=
=
an
(n + 1)n+1 n! (n + 1)n (n + 1)n! (n + 1)n
n n
) has indeterminant form 1∞ . Using the fact that logarithm and exponential
n+1
function are continuous, we therefore use that
We note that lim (
n→∞
n n
n n
) = exp ( lim ln ((
) ))
x→∞ n + 1
x→∞
n+1
lim (
We now consider the inner limit
lim ln(
n→∞
n
ln( n+1 )
n
n n
) = lim n ln(
) = lim
.
1
n→∞
n→∞
n+1
n+1
n
This has the indeterminant form 00 which again we will resolve by applying L’Hopital’s rule. (Again, we write
this as a function in x, to indicate that this is using a continuous function.) By L’Hopital’s rule we get
=
=
=
lim
x )
ln( x+1
1
x
1
x(x+1)
lim
x→∞ −1
x2
x→∞
lim
x→∞
1
= lim
x→∞
⋅
(x+1)(1)−x(1)
(x+1)2
−1
x2
=
x+1
1
x (x+1)2
lim
−1
x→∞
x2
−x 2
−x
= lim
x→∞ x(x + 1) x→∞ (x + 1)
= lim
−x( x1 )
(x
x
x+1
+ 1)( x1 )
= lim
x→∞
−1
= −1.
(1 + 1/x)
29
We thus know that
lim
n→∞
n n
a n+1
== exp ( lim ln ((
) )) = exp(−1) = e −1 < 1
x→∞
an
n+1
∞ n!
and therefore we can conclude that ∑ n converges.
n=1 n
So far all the terms of the series we considered were positive. Once we get to power series, however also
negative terms will occur. We therefore want to establish that — by considering absolute values — the ratio
test also applies to these series.
THEOREM 5
Ratio Test for Series
Let ∑ a n be a series with positive terms and suppose that
a
lim ∣ n+1 ∣ = r
n→∞ a n
then
(a) the series converges if r < 1
(b) the series diverges if r > 1 or if r is infinite
(c) the test is inconclusive if r = 1.
EXAMPLE 12
∞
(−2)n
does it converge or diverge?
n=1 n!
Consider the series ∑
Solution
As not all terms are positive, we cannot immediately apply the ratio test. Therefore we begin by
looking at how this series compares to the corresponding series of positive terms
∞
∞ 2n
(−2)n
∣=∑ .
n!
n=1
n=1 n!
∑∣
(−2)n 2n
≤
.
n!
n!
However, the comparison test as given only holds for positive series, so this is not enough to make a conclusion.
∞
2n
Therefore, we also consider the corresponding series of negative terms: ∑ (−1) . In this series each
n!
n=1
Each term in our original series is smaller than or equal to the terms of this new series, since
n
term is smaller than or equal to the terms of our original series, since − 2n! ≤
30
(−2) n
n! .
n
n!
n
n!
n
n!
∞
∞ 2n
∞ (−2)n
2n
and − ∑
converge, we can conclude that ∑
n=1 n!
n=1 n!
n=1 n!
converges since the value of its terms are squeezed between the two other series terms and partial sums of the
original series therefore are bounded in terms of the partial sums of the positive and negative series, both of
which are converging.
Let us look at this in the context of the ratio test. By taking the absolute value of quotients we treat all
∞ (−2)n
three series “simultaneously”: For ∑
we thus get the ratio
n=1 n!
Therefore, it seems plausible that if both ∑
RRR (−2)n+1 RRR ∣−2∣n+1
2 n+1
R
R
R
R
a
2n+1 n!
2n ⋅ 2 ⋅ n!
2
R (n+1)! RRR (n+1)!
(n+1)!
=
= 2n =
=
=
.
∣ n+1 ∣ = RRRR
R
n
n
R
∣−2∣
RRR (−2) RRR
an
(n + 1)! 2n (n + 1)n! ⋅ 2n n + 1
n!
RRR n! RRR
n!
2
= 0 < 1, we conclude by the ratio test that this series converges.
n+1
∞
∞ 2n
∞ 2n
2n
Since ∑ (−1) = (−1) ∑
this is just a scalar multiple of ∑
and scalar multiples do not affect
n!
n=1
n=1 n!
n=1 n!
As lim
n→∞
31
∞
convergence. Therefore, we can conclude ∑ (−1)
n=1
∞
2n
converges. Hence by the argument above we conclude
n!
(−2)n
must converge as well. Therefore if we have a series with negative terms, we can determine the
∑
n=1 n!
convergence using the ratio test on the absolute value of the series. In fact the proof of this statement is the
same as the proof of the ratio test, but using ∣a n ∣ instead of a n .
EXERCISES
∞
√
n
6. ∑ 2
n=1 n + 1
Determine whether the following series converge or
diverge. You may use the integral test, direct comparison, limit comparison, or the ratio test.
∞
1
7. ∑ sin ( )
n
n=1
∞
1
n
n=1 10
1. ∑
∞
∞
8. ∑
ln(n)
2. ∑
n=1 n
n=1
n
1
√
n
∞
n
1−n
n
n=1 n2
∞
2n
3. ∑ n
n=1 3
9. ∑
∞ 2n
4. ∑ 2
n=1 n
∞ n3
10. ∑ n
n=1 3
∞
n
5. ∑ 3
n=1 n + 5
11. ∑
∞
n10
n=1 n!
32
-
4
Working with Power Series
In this chapter we finally see the reason why we were bothering with series. A power series is a series involving
powers of x, thus as long as it converges we obtain a series value for every x-value. In other words: the series
describes a function. Such a function might not look as nice as the functions you have seen so far, but in fact
has many advantages. By calculating partial sums we can approximate the value of the function. (This is in
fact how your calculator evaluates functions such as sine or logarithm. When you press sine there is no little
man in the calculator measuring the length among the circle; instead power series is evaluated.)
We have seen in the main textbook that Power series have an (possibly infinite) interval, centered around
the center of the series (that’s why its called “center”), in which they are converging (absolute value of ratio
< 1). They diverge outside the closed interval. (In this course we ignore the behavior at the end points.)
Power Series as Polynomials
Power series look a lot like polynomials. We therefore would like to treat them like polynomials — in fact this
is one of the main reasons for using power series. Before we can do so, however we will have to establish that
this is a valid thing to do, for example we have to show that adding two power series gives the same result as
adding the coefficients off the power series:
∞
∞
∞
n=1
n=1
n=1
∑ a n + ∑ b n = ∑ (a n + b n )
If we only had finite sums this would be immediately clear, the law of commutativity tells us that the result is
the same. This is something you are probably using without actually thinking about it because you learned it
as soon as you learn the addition of numbers.
Surprisingly (though you will see more examples of such behavior in advanced calculus), this is far more
difficult for infinite sums. The problem is basically that the law of commutativity only lets us move numbers in
addition by finitely many places, but we need to move them infinitely many places. That can cause difficulties,
because it potentially allows us to keep a particular summand for “later addition” in perpetuity, thus changing
the value of any finite approximation (and thus also the value of the sum). The following example illustrates
this in a particular case.
EXAMPLE 1
What value should the infinite sum
1−1+1−1+1−1+1−1+1−1+⋯
33
have?
Solution
Let us first note that it is not clear whether this infinite series actually converges. What we will
be doing now therefore is also an illustration of the need to prove convergence.
What we will be doing looks rather harmless at first sight. We will simply change the order of addition,
by introducing pairs of parentheses in the sum. This changes the order of addition and is equivalent to rearranging the terms of this series. In each of the cases we can then ”evaluate” the sum easily:
1 − 1 + 1 − 1 + 1 − 1 + ⋯ = (1 − 1) + (1 − 1) + (1 − 1) + ⋯ = 0 + 0 + 0 + 0 + ⋯ = 0
= 1 + (−1 + 1) + (−1 + 1) + (−1 + 1) + ⋯ = 1 + 0 + 0 + 0 + ⋯ = 1
This cannot be right (or the world is insane, a possibility which we do not want to consider here). In fact one
could do even worse things. We could group each −1 with the +1 following three positions later and therefore
get the value of 2.
The root of the problem turns out to be the negative summands in between. If all were positive, we had
the sum 1 + 1 + 1 + ⋯ which clearly diverges. In such a case we say, that this series is not absolutely convergent.
The example shows, that rearrangement is not a valid operation for such a series.1
Fortunately a theorem from advanced calculus tells us that this obstacle we observed here is the only one.
If the series converges absolutely, i.e. it would converge even if all minuses were replaced by pluses, we are
permitted to rearrange.
THEOREM 1
∞
The Rearrangement Theorem for Absolutely Convergent Series
∞
Suppose that ∑ a n converges absolutely, i.e. ∑ ∣a n ∣ converges as well, and b1 , b2 , . . . , b n , . . . is any arrangen=1
n=1
ment of the sequence {a n }, then ∑ b n converges absolutely, and
∞
∞
n=1
n=1
∑ bn = ∑ an
This theorem is stated as Theorem 17 in the textbook on p. 790, a sketch of the proof is outlined on pg 794
exercise 60. We will be using this theorem in this lecture only to establish Theorems 2-5 below.
For power series we are now in luck: because of the ratio test (which we also could do with absolute values)
a power series converges absolutely inside its interval of convergence. We therefore are permitted to rearrange
its terms. In particular, when adding to a power series (centered at the same point) we can collect their terms
together in the same way as we would do for finite sums. We get the following important theorem:
1
Modern Physics has a method called Renormalization, which is essentially about such reordering.
34
THEOREM 2
Term-by-Term Summation for Power Series
∞
∞
n=l
n=0
Suppose that f (x) = ∑ c n (x − a)n and g(x) = ∑ d n (x − a)n are two power series, both centered at the
same point a, and that b is chosen inside the interval of convergence of both f and g. (Thus f (b) and g(b)
are both absolutely convergent series.) Then
∞
f (b) + g(b) = ∑ (c n + d n )(b − a)n
n=0
We write that
∞
f (x) + g(x) = ∑ (c n + d n )(x − a)n
n=0
for x inside the interval of convergence for both series.
EXAMPLE 2
xn
xn
and g(x) = ∑ . Determine the common interval of convergence for both series and
n n!
n n
write for x inside this interval f (x) + g(x) as a single sum.
Solution
Let f (x) = ∑
The same holds for products. Again, once we are permitted to rearrange terms, we can take the same
process as used for multiplying polynomials, and generalize it to the case of power series, as long as long as
we are inside the interval of convergence. The resulting formula is in fact the same as holds for the general
multiplication of polynomials. (In the textbook this is theorem 21 on p. 803.)
THEOREM 3
Term-by-Term Multiplication for Power Series
∞
∞
n=l
n=0
Suppose that f (x) = ∑ c n (x − a)n and g(x) = ∑ d n (x − a)n are two power series, both centered at the
same point a, and that b is chosen inside the interval of convergence of both f and g. Then
∞
⎛ n
⎞
n
∑ c k ⋅ d n−k (b − a)
⎠
n=0 ⎝k=0
f (b) ⋅ g(b) = ∑
We write that
∞
⎛ n
⎞
n
∑ c k ⋅ d n−k (x − a)
⎝
⎠
n=0 k=0
f (x) ⋅ g(x) = ∑
for x inside the interval of convergence for both series.
Proof
To simplify notation we assume that a = 0. (The general proof works the same way by replacing x
35
by x − a, but this is more messy to write down.) We multiply out:
(c0 x 0 + c1 x 1 + c2 x 2 + ⋯)
(d0 x 0 + d1 x 1 + d2 x 2 + ⋯)
⋅
= c0 x 0 (d0 x 0 + d1 x 1 + d2 x 2 + ⋯) + c1 x 1 (d0 x 0 + d1 x 1 + d2 x 2 + ⋯)
+c2 x 2 (d0 x 0 + d1 x 1 + d2 x 2 + ⋯) + ⋯
= c0 d0 x 0 x 0 + c0 d1 x 0 x 1 + c0 d2 x 0 x 2 + ⋯ + c1 d0 x 1 x 0 + c1 d1 x 1 x 1 + ⋯
c2 d0 x 2 x 0 + c2 d1 x 2 x 1 + ⋯
= c0 d0 x 0+0 + c0 d1 x 0+1 + c0 d2 x 0+2 + ⋯ + c1 d0 x 1+0 + c1 d1 x 1+1 + ⋯
c2 d0 x 2+0 + c2 d1 x 2+1 + ⋯
Now we collect the summands according to the power of x. This is where we need to rearrange the terms,
which is permitted because of the absolute convergence.
=
(c0 d0 ) x 0 + (c0 d1 + c1 d0 ) x 1 + (c0 d2 + c1 d1 + c2 d0 ) x 2 + ⋯
´¹¹ ¹ ¸¹ ¹ ¹ ¶
´¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¸ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹¶
´¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹¸¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹¶
n=0,
k=0
n=1,
k=0,1
n=2,
k=0,1,2
n
+ (c0 d n + c1 d n−1 + c2 d n−2 + ⋯ + c n d0 ) x + ⋯
´¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¸ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹¶
k=0,1,2,...,n
EXAMPLE 3
xn
xn
and g(x) = ∑ . For x inside the common interval of convergence, write f (x) ⋅ g(x) as
n n!
n n
Let f (x) = ∑
a sum.
Solution
Since we are doing calculus, we don’t only want to do arithmetic with power series, but also differentiate
and integrate. Two theorems from advanced calculus state, that this also can be done, as one would do it for
polynomials:
THEOREM 4
∞
The Term-by-Term Differentiation Theorem
Let f (x) = ∑ c n (x − a)n . Then f (x) is differentiable for x inside the interval of convergence, and we have
that
n=0
f ′ (x) =
∞
d
f (x) = ∑ nc n (x − a)n−1 ,
dx
n=1
and this derived series converges (absolutely).
Notice that the series now begins with n = 1 because the derivative of the constant term is zero.
36
THEOREM 5
∞
The Term-by-Term Integration Theorem
Let f (x) = ∑ c n (x − a)n . Then for any x inside the interval of convergence of this series also the series
n=0
∞
F(x) = ∑ c n
n=0
converges (absolutely), and we have that
∫
(x − a)n+1
n+1
d
F(x) = f (x). Therefore f (x) has an antiderivative, and
dx
∞
f (x)dx = ∑ c n
n=0
(x − a)n+1
+C
n+1
This establishes that inside the interval of convergence, we can treat power series just like polynomials.
A power series as solution to a differential equation
To give you a real world example consider the Bessel functions2 . These functions first arose in solving Kepler’s
equations for planetary motion. Since that time these functions have been applied in many different physical
situations — for example the temperature distribution in a circular plate.
EXAMPLE 4
The Bessel function
We define the Bessel function of order 0 as
∞ (−1)n x 2n
J0 (x) = ∑ 2n
2
n=0 2 (n!)
What is its interval of convergence?
Solution
To find the interval of convergence, set a n = (−1)n x 2n /(22n (n!)2 ). Then
R
RRR
∣x∣2
(−1)n+1 x 2(n+1)
a
22n (n!)2 RRRR
=
→0<1
⋅
R
∣ n+1 ∣ = RRRRR
R
an
RRR 22(n+1) ((n + 1)!)2 (−1)n x 2n RRRR 4(n + 1)2
Thus the series converges for all real x, the Bessel function is defined on the whole real axis.
If we want to see how the Bessel function looks we can start looking at the approximations given by partial
sums:
s2 (x) = 1 −
x2
,
4
s3 (x) = 1 −
x2 x4
+ ,
4 64
s4 (x) = 1 −
x2 x4
x6
+ −
,
4 64 2304
s6 (x) = 1 −
x2 x4
x6
x8
+ −
+
4 64 2304 147456
as shown below. The second graph is the full Bessel function J0 (x).
As we have absolute convergence within the interval of convergence (and thus may reorder terms!) we
can treat arithmetic with power series in the same way as arithmetic with polynomials (i.e. sum up termwise or form the product from partial sums), which anyhow looks like the natural thing to do. We also see
2
Friedrich Bessel, German astronomer, 1784-1846
37
Degree=4
Deg=8
Deg=6
Degree=2
that we can calculate derivative and anti-derivative of a power series very easily while in general calculating
anti-derivatives is hard.
Later we will see that we can express solutions to differential equations easily in terms of power series.
EXAMPLE 5
The Bessel function as solution for a differential equation
In continuation of the previous example, we want to see that the Bessel function J0 (x) is a solution to the
differential equation
x 2 y′′ (x) + x y′ (x) + x 2 y(x) = 0
Solution
For this, we calculate the first two derivatives (note that we can leave out the n = 0 term in the
derivatives):
∞ (−1)n x 2n
J0 (x) = ∑ 2n
,
2
n=0 2 (n!)
∞
∞
(−1)n ⋅ 2n ⋅ x 2n−1
,
22n (n!)2
n=1
(−1)n 2n(2n − 1)x 2n−2
22n (n!)2
n=1
J ′ 0 (x) = ∑
J0′′ (x) = ∑
and multiply appropriately (doing an index-shift to have all three series start at n = 1) to get
(−1)n x 2n+2 ∞
−(−1)(n+1) x 2(n+1)
=
∑
2n
2
2((n+1)−1) ((n + 1) − 1)!)2
n=0 2 (n!)
n=0 2
∞
x 2 J0 (x) = ∑
m = n+1
=
∞
−(−1)m x 2m
,
∑ 2(m−1)
((m − 1)!)2
m=1 2
as well as:
∞
(−1)n ⋅ 2n ⋅ x 2n
,
22n (n!)2
n=1
x J0 ′ (x) = ∑
∞
(−1)n 2n(2n − 1)x 2n
22n (n!)2
n=1
x 2 J0′′ (x) = ∑
38
and therefore (collecting according to powers of x 2n ):
∞
1
2n(2n − 1) 2n
2n
x 2 J0′′ + x J0′ + x 2 J0 = ∑ (−1)n (−
+ 2n
+ 2n
)x
2
2(n−1)
2
2
(n!)
2 (n!)2
2
((n − 1)!)
n=1
∞
= ∑ (−1)n
n=1
∞
−4n2 + 2n + 2n(2n − 1) 2n
x
22n (n!)2
0
x 2n = 0
= ∑ (−1)n 2n
2
2
(n!)
n=1
This verifies that the Bessel function is a solution to the given differential equation.
There are many other so-called special functions, defined as power series, which occur in applications.
Expressing known functions as power series
To be able to work with power series and “ordinary” functions, the next big question then is how to translate
between “traditional” functions, given by a formula, and power series.
We can do some cases using partial fractions and geometric series.
EXAMPLE 6
Power series for rational functions
9
Determine a power series for 3
.
x + 3x 2 − 4
Solution
In a partial fraction decomposition, we write
9
x 3 + 3x 2 − 4
=
1
1
3
9
=
−
−
2
x − 1 x + 2 (x + 2)2
(x − 1)(x + 2)
Then we note that (by geometric series):
1
x −1
1
x +2
For the power series for
= −1 − x − x 2 − x 3 − ⋯
=
∞
1 1
1
1
1
− x + x 2 − x 3 + ⋯ = ∑ (−1)n n+1 x n
2 4
8
16
2
n=0
1
, remember that
(x + 2)2
1
−1
∫ (x + 2)2 dx = x + 2 , we can thus get a series for
1
−1
by term-by-term differentiation of a series for
(which we know already):
x +2
(x + 2)2
1
(x + 2)2
∞ −1
d −1
d ∞
n 1
n
n+1 n−1
=
−(−1)
x
=
∑
∑ ( ) nx
dx x + 2 dx n=0
2
2n+1
n=1
−1 2
−1 3
−1 4
−1
= ( ) + ( ) ⋅ 2x + ( ) ⋅ 3x 2 + ( )5 ⋅ 4x 3 + ⋯
2
2
2
2
1 1
3 2 1 3
=
− x + x − x +⋯
4 4
16
8
=
39
Note that one could get this result also by multiplying the series for
1
(x + 2)2
1
x+2
with itself, using the formula for products:
1 1
1 1 1 1
1 1 1 1 1 1
= ( ⋅ ) + (− ⋅ − ⋅ ) x + ( ⋅ + ⋅ + ⋅ ) x 2
2 2
2 4 4 2
2 8 4 4 8 2
1 1
1 1 1 1
1 1
+ (− ⋅ − ⋅ − ⋅ − ⋅ ) x 3 + ⋯
2 16 4 8 8 4 16 2
1 1
3
1
=
− x + x2 − x3 + ⋯
4 4
16
8
by geometric series and multiplication (remember reindexing!)
In summary we get that
9
9 27
9
= − − x 2 − x 3 + ⋯.
4 16
16
The general case will have to wait for the next chapters.
x 3 + 3x 2 − 4
EXERCISES
1. Determine a power series for
1
.
(x + 2)(x − 2)
3. Determine the first four terms of a power series
1
for
, by multiplication. Compare with the re(x − 2)2
sult of problem 2.
d
1
ln(1 + x) =
. Determine a
dx
1+x
1
from a power series for 1 , and use term-wise integration
2. Determine a power series for
2
(x − 2)
1+x
power series for its antiderivative.
to obtain a power series for ln(1 + x).
4.
40
We know that
-
5
Taylor polynomials as function
approximations
One of the principal uses of Taylor polynomials is to approximate functions. The key to this is the formula
for the error term. Let us recall the formula:
Let f be a function that is infinitely often differentiable and M be a positive constant such that
∣ f (n+1) (t)∣ ≤ M for all t between x and a, then the remainder term R n (x) = f (x) = Tn (x) in
Taylor’s Theorem satisfies
∣R n (x)∣ ≤ M
∣x − a∣n+1
.
(n + 1)!
If f is well-behaved (essentially any function you will consider in calculus is), then for each finite interval
there will be one (possibly very large) value of M, such that ∣ f (n+1) (t)∣ ≤ M for every n. The reason for this is,
that otherwise higher derivatives of f must have larger and larger values, and therefore oscillate rather wildly.
∣x − a∣n
. From the study of series we remember that
The remainder estimate however then is essentially
(n + 1)!
n! grows faster1 than x n . Thus the error becomes smaller and smaller, the larger n gets.
Vice versa, for a fixed n we can make the error arbitrary small, by choosing x to be close to the center a.
This principle is fundamental to applications in Physics and Engineering:
Approximation principle for well-behaved functions
If f is a well-behaved function, and Tn the Taylor polynomial for f of degree n centered at a, then the function
Tn (x) is an approximation for f (x). The approximation quality gets better, the larger n is and the closer x is
to a.
For values of x very close to a the linear approximation T1 (x) is often sufficiently good.
We ilustrate this principle in the case of deciding which function is largest:
1
Once n > x the subsequent factors in n! are larger than the factors in x n
41
EXAMPLE 1
Using Power Series to Approximate Functions
x , sin(x), and e x − 1 for small (positive) values of x.
Suppose we wish to compare the values of functions 1−x
Which function is largest?
Solution
We begin expressing each of these functions as power series, centered at a = 0:
∞
∞
1
) = x ⋅ ∑ x n = ∑ x n+1 = x + x 2 + x 3 + ⋯
1−x
n=0
n=0
∞ (−1)n x 2n+1
x3 x5
=x−
+
−⋯
sin(x) = ∑
3! 5!
n=0 (2n + 1)!
x
1−x
= x(
ex − 1 =
∞
∞ xn
xn
x2 x3
−1= ∑
=x+
+
+⋯
2! 3!
n=0 n!
n=1 n!
∑
We see that all three functions agree on the Taylor approximations T0 (x) and T1 (x), thus we need to consider
at least T2 (x). We get the following approximations:
x
≃ x + x2
1−x
sin(x) ≃ x
ex − 1 ≃ x +
For positive values of x, we then note that x + x 2 > x +
x2
2
x2
> x, thus for small values of x we have that
2
x
> e x − 1 > sin(x).
1−x
A look at the graph verifies this result: The graph for x ∈ [0..0.01] shows that for even smaller values of x
x
x
sin x
x
e
x
x
sin x
ex
x
x
sin x
ex
the graphs are almost identical, this is a consequence of the fact that the linear approximation of all three
functions is the same. Of course, for larger values of x the behaviour can be quite different, as the third graph
shows.
42
EXERCISES
1. By looking at their Taylor series, decide which of a) Determine the interval (open interval: you do not
the following functions is largest, and which smallest need to decide on the behavior at the end points) of
convergence for W(x). You may use without proof
for x near 0:
n n+1 1
that
lim
=
(
)
1
1
n→∞ n + 1
e
√
ex ,
,
1−x
1 − 2x
b) Calculate a power series for W ′ (x).
c) Let I = [0, 0.1]. You are given the information, that
for the second derivative ∣W (2) (x)∣ ≤ 2 for x ∈ I.
2. Consider the power series
Using the formula for the error term of a Taylor poly∞ (−n)n−1
nomial, show that ∣W(x) − x∣ ≤ x 2 (and thus x − x 2 ≤
xn.
W(x) = ∑
W(x) ≤ x + x 2 ) for x ∈ I.
n!
n=1
43
-
Homework Assignments
EXERCISES
8.2.A
Evaluate the integrals
7.2.B
∫ sin3 x cos3 xdx
∫ sin2 x cos2 xdx
Evaluate the integrals
(ln x)2
dx
x
(3x − 1)
dx
9 − 2x + 3x 2
∫
∫
7.1.A Let f (x) = 31 x + 53 . Find the inverse f −1 (x) 7.3.A Evaluate the integrals
√
and identify the domain and range of f −1 . As a check,
t
−1
−1
e
e t − 1dt
show that f ( f (x)) = f ( f (x)) = x.
∫
∫ t ⋅ 3t dt
2
2)3 .
7.1.B Let f (x) = (x −
Show that f is one-toone on R. Find a formula for d f −1 /dx at x = 8 using
7.4.A Verify that the function e −x (x − 2)−2 is a sotheorem 1.
lution to the initial value problem
7.6.A Given that α = arccos(4/9), find sin α, tan α
1
(x − 2)y′ + x y = 0,
y(0) =
and sec α.
4
7.6.B
Simplify the expression tan(arcsin(x/7)).
7.4.B
7.6.C Evaluate the integrals
1
∫ √−x 2 − 6x − 8 dx
4r
∫ √1 − r4 dr
7.2.A Differentiate:
d
cos x
ln ( √
)
2
dx
x + 1(x − 1)5
Solve the separable differential equation
√
y′ ⋅ 1 − x 2 = x y
7.4.C When cane sugar is dissolved in water, it converts to invert sugar over time. The amount f (t) of
unconverted cane sugar at time t decreases, following the differential equation f ′ = −0.2 f . Suppose we
start with a solution containing 100g of cane sugar.
How much cane sugar remains after 5 hours? after 10
hours?
How long will it take, until only 1g of cane sugar remains?
45
8.0.A
10.7.A Determine the open (i.e. ignoring the behavior at the endpoints) interval of convergence for
Evaluate the integrals
x −5
∫ 3 − x dx
1
∫
8.1.A
x 2 − 14x
+ 50
∞
x 2n+1
n=10 3n + 1
∑
dx
∞
(x − 3)n
7
n=1
∑
Evaluate the integrals
∫ x −9 ⋅ ln xdx
∫ arccos tdt
8.3.A Evaluate the integrals
x2
∫ (1 + x 2)3/2 dx
x2
∫ (1 − x 2)3/2 dx
10.7.B
f (x) = x −
Expand the quotient
2x + 4
(x − 2)(x 2 + 4)
8.4.B Evaluate the integral
1
∫ (x − 4)2(x − 1) dx
Evaluate the improper integral
∫2
Determine a power series for
1
.
(2 + x 3 )2
d
1
ln(1 + x) =
. Deterdx
1+x
1
mine a power series for
, and use term-wise in1+x
tegration to obtain a power series for ln(1 + x).
10.7.D We know that
by partial fractions.
8.7.A
x2 x3 x4
+
−
+⋯
2
3
4
i) determine the (open, ignoring endpoints) interval,
in which f (x) converges.
ii) Determine its term-by-teem derivative f ′ (x).
iii) Determine the value of f ′ (x) as a fraction (Hint:
Geometric series).
iv) Give a formula for the value of f (x), using an antiderivative of the function found in iii)
10.7.C
8.4.A
Consider the power series
∞
1
dx
(x + 4)3
10.8.A Find the first four terms of a Taylor series
about a = 0 for
sin(x)
1−x
10.8.B
Determine the Taylor series for f (x) =
√
4 − x about a = 0.
10.1.A Find a formula for the terms of the following
sequences:
10.9.A Determine a Power series for the following
1, 4, 7, 10, 13, . . .
functions about a = 0:
2
1 2 6 24
f (x) = x 2 ⋅ e x
,− , ,− ,...
2 4 8 16
√
f (x) = cos( x)
10.1.B Give an N/ε-proof, that the sequence a n =
1 − cos(2x)
1 − 2n
f (x) = sin2 (x) =
converges to −1.
2
1 + 2n
46
10.9.B Identify the following functions from their 11.2.A Calculate the arc length of ln(cos(x)) from
0 to π4 .
Taylor series about a = 0
1 + x3 +
x 6 x 9 x 12
+
+
+⋯
2! 3!
4!
1 − 4x + 42 x 2 − 43 x 3 + 44 x 4 − 45 x 5 + ⋯
x 12 x 20 x 28
x4 −
+
−
⋯
3
5
7
x3
6
11.3.A
Convert the equation of the curve
r=
1
2 − cos θ
into cartesian coordinates
10.9.C We approximate sin(x) by x − on the interval [−1, 1]. What estimate can you make for the er- 11.3.B Convert the equation of the curve
ror?
Suppose we want to approximate sin(x) on the interxy = 1
val [−1, 1] with an error of at most 10−5 , what degree
Taylor polynomial do we need to choose?
into polar coordinates
10.9.D By looking at their Taylor series, decide
which of the following functions is largest, and which 11.4.A Find the points of intersection of the polar
curves
smallest for x near 0:
r = 2 + sin θ, r = 2 − sin θ
1
1
√
ex ,
,
1−x
1 − 2x
11.5.A Sketch the curves r1 = 2 + sin 2θ and r2 =
10.10.A Find a power series solving the differential 2 + cos 2θ.
equation
Determine the area inside the curve r1 = 2 + sin 2θ
y′ − 2y = 4x
but outside the curve r2 = 2 + cos 2θ.
under the initial condition y(0) = 1. Identify the solution as a function.
A.7.A Determine all complex fifth roots of 32.
10.10.B Find a power series solving the differential
A.7.B Calculate 1√
equation
2+i 3
y ′′ + y = 0
under the initial conditions y(0) = 2, y′ (0) = 1. Iden- A.7.C Determine all complex roots of the equation
tify the solution as a function.
x4 + 3 ∗ x2 + 9 = 0
47