Mathematical Series

Chapter 10
Mathematical Series
EXERCISES
Exercise 10.1. Show that in the series of Eq. (10.4) any
term of the series is equal to the sum of all the terms
following it. ( Hint: Factor a factor out of all of the following
terms so that they will equal this factor times the original
series, whose value is now known.)
Let the given term be denoted by
term =
b. 0.001%.
0.001% of 1.64993 is equal to 0.00016449. The
n = 79 term is equal to 0.0001602, so we need the
partial sum S79 . However, this partial sum is equal to
1.6324, so again this approximation does not work very
well.
Exercise 10.3. Find the value of the infinite series
1
2n
∞
The following terms are
[ln (2)]n
n=0
1
1
1
1
+ n+2 + n+3 + n+4 + · · ·
2n+1
2
2
2
1 1 1
1
1
= n+1 1 + + + + · · · + n + · · ·
2
2 4 8
2
1
2
= n+1 = n
2
2
Exercise 10.2. Consider the series
s =1+
1
1
1
1
+ 2 + 2 + ··· + 2 + ···
22
3
4
n
which is known to be convergent and to equal
π 2 6/ = 1.64993 · · ·. Using Eq. (10.5) as an approximation,
determine which partial sum approximates the series to
a. 1%
1% of 1.64993 is equal to 0.016449. The n = 8 term is
equal to 0.015625, so we need the partial sum S8 , which
is equal to 1.2574· · ·. The series is slowly convergent
and S8 is equal to 1.5274, so this approximation does
not work very well.
Mathematics for Physical Chemistry. http://dx.doi.org/10.1016/B978-0-12-415809-2.00056-2
© 2013 Elsevier Inc. All rights reserved.
Determine how well this series is approximated by S2 ,S5 ,
and S10 .
This is a geometric series, so the sum is
s=
1
= 3.25889 · · ·
1 − ln (2)
The partial sums are
S2 = 1 + ln (2) = 1.693 · · ·
S5 = 1 + ln (2) + ln (2)2 + ln (2)3 + ln (2)4
= 2.73746 · · ·
S10 = 3.175461
S20 = 3.256755 · · ·
Exercise 10.4. Evaluate the first 20 partial sums of the
harmonic series.
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Mathematics for Physical Chemistry
Here are the first 20 partial sums, obtained with Excel:
'
Exercise 10.7. Show that the Maclaurin series for e x is
$
ex = 1 +
1
1
1
1
1
x + x2 + x3 + x4 + · · ·
1!
2!
3!
4!
Every derivative of e x is equal to e x
1 dn f
1
an =
=
n! dx n x=0
n!
1.5
1.833333333
2.083333333
2.283333333
Exercise 10.8. Find the Maclaurin series for ln (1 + x).
You can save some work by using the result of the previous
example.
The series is
2.45
2.592857143
2.717857143
2.828968254
ln (1 + x) = a0 + a1 x + a2 x 2 + · · ·
2.928968254
3.019877345
3.103210678
d f dx 3.180133755
3.251562327
The second derivative is
d2 f 1
=
−1
= −1
2 dx 2 x=0
1 + x x=0
3.318228993
3.380728993
3.439552523
3.495108078
3.547739657
3.597739657
&
a0 = ln (1) = 0
1 =
=1
1 + x x=0
x=0
%
Exercise 10.5. Show that the geometric series converges
if r 2 ≺ 1.
If r is positive, we apply the ratio test:
an+1
lim
=r
n→∞ an
If r 2 ≺ 1 and if r is positive, then r ≺ 1, so the series
converges. If r is negative, apply the alternating series test.
Each term is smaller than the previous term and approaches
zero as you go further into the series, so the series converges.
Exercise 10.6. Test the following series for convergence.
∞
1
n2
n=1
Apply the ratio test:
n2
1/(n + 1)2
r = lim
= lim
=1
n→∞
n→∞ (n + 1)2
1/n 2
The ratio test fails. We apply the integral text:
∞
∞
1
1 dx
=
−
=1
2
x
x 1
1
The integral converges, so the series converges.
The derivatives follow a pattern:
n n−1
d f
n−1 (n − 1)! =
(−
1)
= −1
(n −1)!
n
n
dx x=1
(1 + x) x=0
1 dn f
(− 1)n−1
=
n! dx n x=1
n
The series is
ln (1+x) = x−
1
1
1
1
x 2+
x 3−
x 4+
x 5 +· · ·
2
3
4
5
Exercise 10.9. Find the Taylor series for ln (x), expanding
about x = 2, and show that the radius of convergence for
this series is equal to 2, so that the series can represent the
function in the region 0 ≺ x 4.
The first term is determined by letting x = 2 in which
case all of the terms except for a0 vanish:
a0 = ln (2)
The first derivative of ln (x) is 1/x, which equals 1/2 at
x = 2. The second derivative is −1/x 2 , which equals −1/4
at x = 2. The third derivative is 2!/x 3 , which equals 1/4 at
x = 1. The derivatives follow a regular pattern,
n d f
(n − 1)!
n−1 (n − 1)! = (−1)
= ( − 1)n−1
dx n x=2
x n x=2
2n
so that
1
n!
dn f
dx n
=
x=1
(−1)n−1
n2n
CHAPTER | 10 Mathematical Series
and
1
1
ln (x) = ln (2) + (x − 2) − (x − 2)2 + (x − 2)3
8
24
1
− (x − 2)4 + · · ·
64
The function is not analytic at x = 0, so the series is
not valid at x = 0. For positive values of x the series is
alternating, so we apply the alternating series test:
(x − 2)n ( − 1)n−1
tn = an (x − 2)n =
n2n
0 if|x − 2| ≺ 2 or 0 ≺ x 4
lim |tn | =
n→∞
∞ if|x − 2| > 2 or x > 4.
Exercise 10.10. Find the series for 1/(1 − x), expanding
about x = 0. What is the interval of convergence?
1
= a0 + a1 x + a2 x 2 + · · ·
1+x
a0 = 1
1
d
1
=
−
= −1
2
dx 1 + x x=0
(1 + x) x=0
1
1
1 d
2
−
a2 =
=
=1
2
3
2! dx (1 + x)
2! (1 + x) x=0
x=0
a1 =
The pattern continues:
an = (−1)n
1
= 1 − x + x2 − x3 + x4 − x5 + · · ·
1+x
Since the function is not analytic at x = −1, the interval of
convergence is −1 ≺ x ≺ 1
Exercise 10.11. Find the relationship between the
coefficients A3 and B3 .
We begin with the virial equation of state
P=
RT B2
RT B3
RT
+
+
+ ···
2
Vm
Vm
Vm3
We write the pressure virial equation of state:
RT
A2 P
A3 P 2
P=
+
+
+ ···
Vm
Vm
Vm
We replace P and P 2 in this equation with the expression
from the virial equation of state:
A2 RT
RT B2
RT B3
RT
+
+
+
+ ···
P =
Vm
Vm Vm
Vm2
Vm3
2
A3 RT
RT B2
RT B3
+
+
+
+
·
·
·
+ ···
Vm Vm
Vm2
Vm3
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We use the expression for the square of a power series from
Eq. (5) of Appendix C, part 2:
2
RT B2
RT B3
RT
+
+
+ ···
Vm
Vm2
Vm3
4
RT 2
RT
RT B2
1
=
+2
+
O
Vm
Vm
Vm2
Vm
4
2 2 1
RT 2
R T B2
+O
=
+2
3
Vm
Vm
Vm
RT
A2 RT
RT B2
RT B3
P =
+
+
+
+ ···
Vm
Vm Vm
Vm2
Vm3
2 2 4 R T B2
RT 2
1
A3
+2
+O
+
3
Vm
Vm
Vm
Vm
RT
RT A2
RT A2 B2
A3 RT 2
+··· =
+
+
+
Vm
Vm2
Vm3
Vm Vm
=
RT B22
RT
RT B2
A3 R 2 T 2
+
+
+
2
3
Vm
Vm
Vm
Vm3
where we have replaced A2 by B2 . We now equation
2
coefficients of equal powers of 1/Vm in the two series
for P:
RT B3 = RT B22 + A3 R 2 T 2
A3 = B3 − RT B22
Exercise 10.12. Determine how large X 2 can be before
the truncation of Eq. (10.28) that was used in Eq. (10.16)
is inaccurate by more than 1%.
1 2
X + ···
2 2
If the second term is smaller than 1% of the first term
(which was the only one used in the approximation) the
approximation should be adequate. By trial and error, we
find that if X 2 = 0.019, the second term is equal to
0.0095 = 0.95% of the first term.
− ln (X 1 ) = − ln (1 − X 2 ) = X 2 −
Exercise 10.13. From the Maclaurin series for ln (1 + x)
1
1
ln (1 + x) = x − x 2 + x 3 + · · ·
2
3
find the Taylor series for 1/(1 + x), using the fact that
d[ln (1 + x)]
1
=
.
dx
1+x
For what values of x is your series valid?
1
d
1 2 1 3
=
x − x + x ···
1+x
dx
2
3
3x 2
4x 3
2x
+
−
=1−
2
3
4
= 1 − x + x2 − x3 + x4 − x5 + · · ·
which is the series already obtained for 1/(1 + x) in an
earlier example. The series is invalid if x 1.
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Mathematics for Physical Chemistry
Exercise 10.14. Find the formulas for the coefficients in a
Taylor series that expands the function f (x,y) around the
point x = a,y = b.
f (x,y) = a00 + a10 (x − a) + a01 (y − b) + a11 (x − a)
(y − b) + a21 (x − a)2 (y − b) + a12 (x − a)
(y − b)2 + · · ·
a00 = f (a,b)
∂ f a10 =
∂ y x x=a,b=y
∂ f a01 =
∂ y x a,b
2 ∂ f a11 =
∂ y∂ x a,b
n+m ∂
f 1
amn =
m
n
n!m! ∂ y∂ x π
π 2
+ a2 x −
sin (x) = a0 + a1 x −
+ ···
4
4
where x is measured in radians. What is the radius of
convergence of the series?
PROBLEMS
1. Test the following series for convergence.
(( − 1)n (n − 1)/n 2 ).
n=0
We apply the alternating series test:
n
n−1
=
lim
limn→∞ |tn | = limn→∞
n→∞
n2
n2
1
=0
= limn→∞
n
The series is convergent.
3. Test the following series for convergence.
∞
1/n! .
n=0
Try the ratio test
an+1
n!
1
= lim
=0
= lim
n→∞ an
n→∞ (n + 1)!
n→∞ n + 1
The series converges.
5. Find the Taylor series for cos (x), expanding about
x = π/2.
r = lim
cos (x)= a0 + a1 (x − π/2) + a2 (x − π/2)2 + · · ·
a0 = cos π/2 = 0
1 d
1
[cos (x)]
a1 =
= − sin (x)
= −1
1! dx
2!
π/2
π/2
1 d
1
[cos (x)]
= − cos (x)
=0
a2 =
2! dx
2!
π/2
π/2
1 d
1
1
[cos (x)]
sin (x)
=
=
a3 = −
3! dx
3!
3!
π/2
1
1
(x − π/2) + (x − π/2)3
1!
3!
1
− (x − π/2)5 + · · ·
5!
cos (x) = −
7. Find the coefficients of the first few terms of the Taylor
series
a,b
∞
There is a pattern, with all even-numbered coefficients
vanishing and odd-numbered coefficients alternating
between 1/n! and −1/n!.
π/2
1
a0 = sin (π/4) = √ = 0.717107
2
1
1
cos (x)
a1 =
=√
1!
2
π/4
1
1
=− √
a2 = − sin (x)
2!
2! 2
π/4
1
1
cos (x)
= √
a3 =
3!
3! 2
π/4
The coefficients form a regular pattern:
1
√
n! 2
1 1 π
π 2
1
+ √ x−
sin (x) = √ − √ x −
4
4
2
2
2! 2
n 1 π n
π 3
1 − √ x−
+ · · · − −1
√ x−
4
4
3! 2
n! 2
+···
an = −(−1)n
Since the function is analytic everywhere, the radius
of convergence is infinite.
√
2/2 =
9. The sine of π/4 radians (45◦ ) is
0.70710678 . . .. How many terms in the series
sin (x) = x −
x5
x7
x3
+
−
+ ···
3!
5!
7!
must be taken to achieve 1% accuracy at x = π/4?
π
= 0.785398
4
x3
x−
= 0.785398 − 0.080746 = 0.704653
3!
CHAPTER | 10 Mathematical Series
e61
This is accurate to about 0.4%, so only two terms are
needed.
11. Estimate the largest value of x that allows e x to be
approximated to 1% accuracy by the following partial
sum
e x ≈ 1 + x.
The function is not analytic at x = 0, so the interval
of convergence is 0 ≺ x ≺ 2
• 0.20000
1.20000 1.221402758 1.78
• 0.19000
1.19000 1.209249598 1.62
• 0.18000
1.18000 1.197217363 1.46
• 0.17000
1.17000 1.185304851 1.31
1
= b0 + b1 (x − 2) + b2 (x − 2)2 + · · ·
x
1
1
b0 = =
2
2
1 1
b1 = − 2 = −
x 2
4
1 2 1
=
b2 =
2! x 3 2
8
1 6 1
=−
b3 = −
2! x 4 2
16
• 0.15000
1.15000 1.161834243 1.03
The coefficients follow a regular pattern so that
• 0.14000
1.14000 1.150273799 0.90
• 0.14500
1.14500 1.156039570 0.96
• 0.14777
1.14777 1.159246239 0.9999
Here is a table of values:
'
a. • x difference 1 + x
$
ex
%
bn = ( − 1)n
• By trial and error, 1% accuracy is obtained
with x ≺ 0.14777.
&
%
13. Find two different Taylor series to represent the
function
1
f (x) =
x
such that one series is
f (x) = a0 + a1 (x − 1) + a2 (x − 1)2 + · · ·
and the other is
2
f (x) = b0 + b1 (x − 2) + b3 (x − 2) + · · ·
Show that bn = an /2n for any value of n. Find the
interval of convergence for each series (the ratio test
may be used). Which series must you use in the vicinity
of x = 3? Why? Find the Taylor series in powers of
(x − 10) that represents the function ln (x).
1
= a0 + a1 (x − 1) + a2 (x − 1)2 + · · ·
x
1
a0 = = 1
1
1
a1 = − 2 = −1
x 1
1 2 a2 =
=1
2! x 3 1
1 6 a3 = −
=1
2! x 4 1
The coefficients follow a regular pattern so that
1
= 1 − (x − 1) + (x − 1)2
an = ( − 1)
x
−(x − 1)3 + (x − 1)4 + · · ·
n
1
an
= n
2n
2
1
1 1
1
= − (x − 1) + (x − 1)2
x
2 4
8
1
1
3
− (x − 1) + (x − 1)4 + · · ·
16
32
The function is not analytic at x = 0, so the interval
of convergence is 0 ≺ x ≺ 4. The second series must
be used for x = 3, since this value of x is outside the
region of convergence of the first series.
15. Using the Maclaurin series for e x , show that the
derivative of e x is equal to e x .
x3
x4
x2
+
+
+ ···
2!
3!
4!
d x
3x 2
4x 3
2x
e = 0+1+
+
+
+ ···
dx
2!
3!
4!
x2
x3
= 1+x +
+
+ · · · = ex
2!
3!
ex = 1 + x +
17. Find the Taylor series for sin (x), expanding around
π/2.
sin (x) = a0 + a1 (x − π/2) + a2 (x − π/2)2
+a3 (x − π/2)3 + · · ·
a0 = sin π/2 = 1
1 dn
an =
sin
(x)
n
n! d x
π/2
df
= cos (x)
dx
a1 = cos π/2 = 0
d2 f
= − sin (x)
dx 2
1
1
a2 = − sin π/2 = −
2!
2!
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Mathematics for Physical Chemistry
d3 f
= − cos (x)
dx 3
1
a3 = − cos π/2 = 0
3!
d4 f
= sin (x)
dx 4
1
1
sin (π/2) =
a4 =
4!
4!
There is a pattern. Only even values of n occur,and
signs alternate.
1
1
sin (x) = 1 − (x − π/2)2 + (x − π/2)4
2!
4!
1
− (x − π/2)6 + · · ·
6!
19. Find the interval of convergence for the series for
cos (x).
cos (x) = 1 −
x4
x6
x2
+
−
+ ···
2!
4!
6!
Apply the alternating series test. Each term is smaller
than the previous term if x is finite and if you go far
enough into the series. The series converges for all
finite values of x.
21. Using the Maclaurin series, show that
x1
x
es dx = e x 01 = e x1 − 1
0
x
0
x x2
x3
x4
1+x +
+
+
+ · · · dx
2!
3!
4!
0
x1
2
3
4
x
x
x
+
+
+ · · · = x+
2
(3)2! (4)3!
0
es dx =
x3
x4
x2
+
+
+ ··· − 1
= 1+x +
2!
3!
4!
x1
= e −1
23. Find the first few terms of the two-variable Maclaurin
series representing the function
f (x,y) = sin (x + y)
f (0,0)
∂ f ∂x y
0,0
∂ f ∂ y x 0,0
2 ∂ f ∂ y∂ x 0,0
2 ∂ f ∂ x 2 0,0
2 ∂ f ∂ y 2 0,0
3 ∂ f ∂ y∂ x 2 0,0
3 ∂ f ∂ y 2 ∂ x 0,0
4
∂ f
∂ y 2 ∂ x 2 0,0
∂5 f
∂ y2∂ x 3 = sin (0) = 0
= cos (x + y)|0,0 = 1
= cos (x + y)|0,0 = 1
= − sin (x + y)|0,0 = 1
= − sin (x + y)|0,0 = 0
= − sin (x + y)|0,0 = 0
= − cos (x + y) 0,0 = −1
= − cos (x + y) 0,0 = −1
= sin (x + y) 0,0 = 0
= cos (x + y) 0,0 = 1
0,0
There is a pattern. If n + m is even, the derivative
vanishes. If n + m is odd, the derivative has magnitude
1 with alternating signs.
1
(x 2 y + x y 2 )
sin (x + y) = x + y −
1!2!
1
1
(x 2 y 3 + x 3 y 2 ) −
(x 4 y 3 + x 3 y 4 ) + · · ·
+
3!2!
3!4!