CHALLENGE PROBLEMS
■
1
CHALLENGE PROBLEMS:
CHAPTER 8
A Click here for answers.
S
Click here for solutions.
1. If f 共x兲 苷 sin共x 3 兲, find f 共15兲共0兲.
2. A function f is defined by
f 共x兲 苷 lim
nl⬁
x 2n ⫺ 1
x 2n ⫹ 1
Where is f continuous?
3. (a) Show that tan 2 x 苷 cot 2 x ⫺ 2 cot x.
1
1
(b) Find the sum of the series
⬁
兺
n苷1
P¢
4
P∞
FIGURE FOR PROBLEM 4
ⱍ
ⱍ
ⱍ
ⱍ
4. Let 兵Pn 其 be a sequence of points determined as in the figure. Thus AP1 苷 1, Pn Pn⫹1 苷 2 n⫺1,
P£
2
8
1
x
tan n
2n
2
P™
1
A 1 P¡
and angle APn Pn⫹1 is a right angle. Find lim n l ⬁ ⬔Pn APn⫹1 .
5. To construct the snowflake curve, start with an equilateral triangle with sides of length 1.
Step 1 in the construction is to divide each side into three equal parts, construct an equilateral
triangle on the middle part, and then delete the middle part (see the figure). Step 2 is to repeat Step
1 for each side of the resulting polygon. This process is repeated at each succeeding step. The
snowflake curve is the curve that results from repeating this process indefinitely.
(a) Let sn , ln , and pn represent the number of sides, the length of a side, and the total length of the
n th approximating curve (the curve obtained after Step n of the construction), respectively.
Find formulas for sn , ln , and pn .
(b) Show that pn l ⬁ as n l ⬁.
(c) Sum an infinite series to find the area enclosed by the snowflake curve.
Parts (b) and (c) show that the snowflake curve is infinitely long but encloses only a finite area.
1
2
3
6. Find the sum of the series
1⫹
1
1
1
1
1
1
1
⫹ ⫹ ⫹ ⫹ ⫹ ⫹
⫹ ⭈⭈⭈
2
3
4
6
8
9
12
where the terms are the reciprocals of the positive integers whose only prime factors are 2s and 3s.
7. (a) Show that for xy 苷 ⫺1,
arctan x ⫺ arctan y 苷 arctan
x⫺y
1 ⫹ xy
Thomson Brooks-Cole copyright 2007
if the left side lies between ⫺␲兾2 and ␲兾2.
(b) Show that
120
1
arctan 119 ⫺ arctan 239 苷
␲
4
(c) Deduce the following formula of John Machin (1680–1751):
1
1
4 arctan 5 ⫺ arctan 239 苷
␲
4
2
■
CHALLENGE PROBLEMS
(d) Use the Maclaurin series for arctan to show that
0.197395560 ⬍ arctan 15 ⬍ 0.197395562
(e) Show that
0.004184075 ⬍ arctan 239 ⬍ 0.004184077
1
(f) Deduce that, correct to seven decimal places,
␲ ⬇ 3.1415927
Machin used this method in 1706 to find ␲ correct to 100 decimal places. Recently, with the aid of
computers, the value of ␲ has been computed to increasingly greater accuracy. In 1999, Takahashi
and Kanada, using methods of Borwein and Brent兾Salamin, calculated the value of ␲ to
206,158,430,000 decimal places!
8. (a) Prove a formula similar to the one in Problem 7(a) but involving arccot instead of arctan.
(b) Find the sum of the series
⬁
兺 arccot共n
2
⫹ n ⫹ 1兲
n苷0
9. Find the interval of convergence of 冘⬁n苷1 n 3x n and find its sum.
10. If a 0 ⫹ a 1 ⫹ a 2 ⫹ ⭈ ⭈ ⭈ ⫹ a k 苷 0, show that
lim (a0 sn ⫹ a1 sn ⫹ 1 ⫹ a2 sn ⫹ 2 ⫹ ⭈ ⭈ ⭈ ⫹ ak sn ⫹ k ) 苷 0
nl⬁
Hint: Try the special cases k 苷 1 and k 苷 2 first. If you can see how to prove the assertion for
these cases, then you will probably see how to prove it in general.
⬁
11. Find the sum of the series
冉
兺 ln
n苷2
冊
1
.
n2
1⫺
12. Suppose you have a large supply of books, all the same size, and you stack them at the edge of a
1
1 6
8
1
4
FIGURE FOR PROBLEM 12
1
2
table, with each book extending farther beyond the edge of the table than the one beneath it. Show
that it is possible to do this so that the top book extends entirely beyond the table. In fact, show that
the top book can extend any distance at all beyond the edge of the table if the stack is high enough.
Use the following method of stacking: The top book extends half its length beyond the second
book. The second book extends a quarter of its length beyond the third. The third extends one-sixth
of its length beyond the fourth, and so on. (Try it yourself with a deck of cards.) Consider centers
of mass.
13. Let
u苷1⫹
x6
x9
x3
⫹
⫹
⫹ ⭈⭈⭈
3!
6!
9!
v苷x⫹
x4
x7
x 10
⫹
⫹
⫹ ⭈⭈⭈
4!
7!
10!
w苷
x2
x5
x8
⫹
⫹
⫹ ⭈⭈⭈
2!
5!
8!
Show that u 3 ⫹ v 3 ⫹ w 3 ⫺ 3u vw 苷 1.
14. If p ⬎ 1, evaluate the expression
1
1
1
⫹ p ⫹ p ⫹ ⭈⭈⭈
2p
3
4
1
1
1
1 ⫺ p ⫹ p ⫺ p ⫹ ⭈⭈⭈
2
3
4
1⫹
Thomson Brooks-Cole copyright 2007
15. Suppose that circles of equal diameter are packed tightly in n rows inside an equilateral tri-
angle. (The figure illustrates the case n 苷 4.) If A is the area of the triangle and An is the total area
occupied by the n rows of circles, show that
FIGURE FOR PROBLEM 15
lim
nl⬁
An
␲
苷
A
2 s3
CHALLENGE PROBLEMS
■
3
16. A sequence 兵a n 其 is defined recursively by the equations
a0 苷 a1 苷 1
n共n ⫺ 1兲a n 苷 共n ⫺ 1兲共n ⫺ 2兲a n⫺1 ⫺ 共n ⫺ 3兲a n⫺2
Find the sum of the series 冘⬁n苷0 an.
17. Taking the value of x x at 0 to be 1 and integrating a series term-by-term, show that
P¡
P∞
y
P™
0
⬁
兺
x x dx 苷
n苷1
共⫺1兲n⫺1
nn
18. Starting with the vertices P1共0, 1兲, P2共1, 1兲, P3共1, 0兲, P4共0, 0兲 of a square, we construct further
P˜
Pˆ
points as shown in the figure: P5 is the midpoint of P1 P2, P6 is the midpoint of P2 P3, P7 is the midpoint of P3 P4, and so on. The polygonal spiral path P1 P2 P3 P4 P5 P6 P7 . . . approaches a point P
inside the square.
(a) If the coordinates of Pn are 共x n, yn 兲, show that 12 x n ⫹ x n⫹1 ⫹ x n⫹2 ⫹ x n⫹3 苷 2 and find a similar
equation for the y-coordinates.
(b) Find the coordinates of P.
Pß
P¡¸
P¢
1
P£
P¶
19. If f 共x兲 苷
冘⬁m苷0 cm x m has positive radius of convergence and e f 共x兲 苷 冘⬁n苷0 dn x n, show that
FIGURE FOR PROBLEM 18
n
ndn 苷
兺 ic d
i
n艌1
n⫺i
i苷1
1
20. Right-angled triangles are constructed as in the figure. Each triangle has height 1 and its base is the
1
hypotenuse of the preceding triangle. Show that this sequence of triangles makes indefinitely many
turns around P by showing that 冘 ␪ n is a divergent series.
1
1
21. Consider the series whose terms are the reciprocals of the positive integers that can be written in
¨£
¨™
¨¡
P
FIGURE FOR PROBLEM 20
1
1
base 10 notation without using the digit 0. Show that this series is convergent and the sum is less
than 90.
22. (a) Show that the Maclaurin series of the function
f 共x兲 苷
x
1 ⫺ x ⫺ x2
⬁
is
兺
fn x n
n苷1
Thomson Brooks-Cole copyright 2007
where fn is the nth Fibonacci number, that is, f1 苷 1, f2 苷 1, and fn 苷 fn⫺1 ⫹ fn⫺2
for n 艌 3. [Hint: Write x兾共1 ⫺ x ⫺ x 2兲 苷 c0 ⫹ c1 x ⫹ c2 x 2 ⫹ . . . and multiply both sides of
this equation by 1 ⫺ x ⫺ x 2.]
(b) By writing f 共x兲 as a sum of partial fractions and thereby obtaining the Maclaurin series in a
different way, find an explicit formula for the nth Fibonacci number.
4
■
CHALLENGE PROBLEMS
ANSWERS
S
Solutions
1. 15!兾5! 苷 10,897,286,400
3. (b) 0 if x 苷 0, 共1兾x兲 ⫺ cot x if x 苷 k␲, k an integer
5. (a) sn 苷 3 ⴢ 4 , ln 苷 1兾3 , pn 苷 4 n兾3 n⫺1
n
n
Thomson Brooks-Cole copyright 2007
9. 共⫺1, 1兲, 共x 3 ⫹ 4x 2 ⫹ x兲兾共1 ⫺ x兲4
(c) 2 s3兾5
1
11. ln 2
CHALLENGE PROBLEMS
SOLUTIONS
E
Exercises
1. It would be far too much work to compute 15 derivatives of f . The key idea is to remember that f (n) (0) occurs in
the coefficient of xn in the Maclaurin series of f . We start with the Maclaurin series for sin:
sin x = x −
x5
x15
x9
x3
+
− · · · . Then sin(x3 ) = x3 −
+
− · · · , and so the coefficient of x15 is
3!
5!
3!
5!
f (15) (0)
1
15!
= . Therefore, f (15) (0) =
= 6 · 7 · 8 · 9 · 10 · 11 · 12 · 13 · 14 · 15 = 10,897,286,400.
15!
5!
5!
3. (a) From Formula 14a in Appendix A, with x = y = θ, we get tan 2θ =
2 cot 2θ =
2 tan θ
1 − tan2 θ
,
so
cot
2θ
=
1 − tan2 θ
2 tan θ
1 − tan2 θ
= cot θ − tan θ. Replacing θ by 12 x, we get 2 cot x = cot 12 x − tan 12 x, or
tan θ
tan 12 x = cot 12 x − 2 cot x
(b) From part (a) with
∞ 1
S
x
tan n is
n
2
n=1 2
x
x
x
x
in place of x, tan n = cot n − 2 cot n−1 , so the nth partial sum of
2n−1
2
2
2
tan(x/4)
tan(x/8)
tan(x/2n )
tan(x/2)
+
+
+··· +
2
4
8
2n
cot(x/4)
cot(x/2)
cot(x/4)
cot(x/2)
cot(x/8)
− cot x +
−
−
=
+
+ ···
2
4
2
8
4
cot(x/2n )
cot(x/2n−1 )
cot(x/2n )
−
(telescoping sum)
=
−
cot
x
+
+
2n
2n−1
2n
sn =
cos(x/2n )
x/2n
1
1
cos(x/2n )
cot(x/2n )
=
·
→ · 1 = as n → ∞ since x/2n → 0
= n
n
n
2
2 sin(x/2 )
x
sin(x/2n )
x
x
for x 6= 0. Therefore, if x 6= 0 and x 6= kπ where k is any integer, then
Now
∞ 1
S
1
1
x
x
−
cot
x
+
= − cot x +
tan
=
lim
s
=
lim
cot
n
n
n
n
n
n→∞
n→∞
2
2
2
2
x
n=1
If x = 0, then all terms in the series are 0, so the sum is 0.
5. (a) At each stage, each side is replaced by four shorter sides, each of
length
1
3
of the side length at the preceding stage. Writing s0 and
s0 = 3
0
Thomson Brooks-Cole copyright 2007
for the number of sides and the length of the side of the initial
triangle, we generate the table at right. In general, we have
n
sn = 3 · 4n and n = 13 , so the length of the perimeter at the nth
n
n
stage of construction is pn = sn n = 3 · 4n · 13 = 3 · 43 .
(b) pn =
4n
3n−1
=4
n−1
4
. Since
3
4
3
> 1, pn → ∞ as n → ∞.
s1 = 3 · 4
2
s2 = 3 · 4
3
s3 = 3 · 4
···
0
=1
1
= 1/3
2
= 1/32
3
= 1/33
···
⇒
■
5
6
■
CHALLENGE PROBLEMS
(c) The area of each of the small triangles added at a given stage is one-ninth of the area of the triangle added at the
preceding stage. Let a be the area of the original triangle. Then the area an of each of the small triangles added
1
a
= n . Since a small triangle is added to each side at every
9n
9
at stage n is an = a ·
stage, it follows that the total area An added to the figure at the nth stage is
An = sn−1 · an = 3 · 4n−1 ·
a
4n−1
= a · 2n−1 . Then the total area enclosed by the snowflake curve is
n
9
3
A = a + A1 + A2 + A3 + · · · = a + a ·
4
1
42
43
+ a · 3 + a · 5 + a · 7 + · · · . After the first term, this is a
3
3
3
3
geometric series with common ratio 49 , so A = a +
equilateral triangle with side 1 is a =
8
5
·
√
3
4
=
√
2 3
5 .
1
2
8a
a/3
a 9
. But the area of the original
=a+ · =
3 5
5
1 − 49
· 1 · sin π3 =
√
3
4 .
So the area enclosed by the snowflake curve is
7. (a) Let a = arctan x and b = arctan y. Then, from Formula 14b in Appendix A,
tan a − tan b
tan(arctan x) − tan(arctan y)
x−y
=
=
.
1 + tan a tan b
1 + tan(arctan x) tan(arctan y)
1 + xy
tan(a − b) =
Now arctan x − arctan y = a − b = arctan(tan(a − b)) = arctan
x−y
since − π2 < a − b <
1 + xy
π
.
2
(b) From part (a) we have
1
− arctan 239
= arctan
arctan 120
119
1
120
1
− 239
119
1
+ 120
· 239
119
= arctan
28,561
28,441
28,561
28,441
= arctan 1 =
(c) Replacing y by −y in the formula of part (a), we get arctan x + arctan y = arctan
4 arctan 15 = 2 arctan 15 + arctan 15 = 2 arctan
= arctan
5
12
+
5
12
1−
5
12
·
5
12
x+y
. So
1 − xy
1
5
+ 15
5
5
5
= 2 arctan 12
= arctan 12
+ arctan 12
1 − 15 · 15
= arctan 120
119
1
1
Thus, from part (b), we have 4 arctan 15 − arctan 239
= arctan 120
− arctan 239
=
119
(d) From Example 7 in Section 8.6 we have arctan x = x −
arctan
π
4
π
.
4
x5
x7
x9
x11
x3
+
−
+
−
+ · · · , so
3
5
7
9
11
1
1
1
1
1
1
1
= −
+
−
+
−
+ ···
5
5
3 · 53
5 · 55
7 · 57
9 · 59
11 · 511
This is an alternating series and the size of the terms decreases to 0, so by the Alternating Series Estimation
Theorem, the sum lies between s5 and s6 , that is, 0.197395560 < arctan 15 < 0.197395562.
Thomson Brooks-Cole copyright 2007
(e) From the series in part (d) we get arctan
1
1
1
1
=
−
+
− · · · . The third term is less than
239
239
3 · 2393
5 · 2395
2.6 × 10−13 , so by the Alternating Series Estimation Theorem, we have, to nine decimal places,
1
1
arctan 239
≈ s2 ≈ 0.004184076. Thus, 0.004184075 < arctan 239
< 0.004184077.
CHALLENGE PROBLEMS
1
(f ) From part (c) we have π = 16 arctan 15 − 4 arctan 239
, so from parts (d) and (e) we have
16(0.197395560) − 4(0.004184077) < π < 16(0.197395562) − 4(0.004184075) ⇒
3.141592652 < π < 3.141592692. So, to 7 decimal places, π ≈ 3.1415927.
∞
S
1
, |x| < 1, and differentiate:
1−x
∞
S n
d
d
1
1
=
x
for |x| < 1 ⇒
=
=
dx n=0
dx 1 − x
(1 − x)2
9. We start with the geometric series
xn =
n=0
∞
S
nxn−1
n=1
∞
S
nxn = x
n=1
∞
S
nxn−1 =
n=1
x
for |x| < 1. Differentiate again:
(1 − x)2
∞
S
x
d
(1 − x)2 − x · 2 (1 − x) (−1)
x+1
x2 + x
=
⇒
n2 xn =
2 =
4
3
dx (1 − x)
(1 − x)
(1 − x)
(1 − x)3
n=1
(1 − x)3 (2x + 1) − x2 + x 3 (1 − x)2 (−1)
d x2 + x
x2 + 4x + 1
=
=
⇒
3 =
6
dx (1 − x)
(1 − x)
(1 − x)4
n2 xn−1 =
n=1
∞
S
∞
S
n3 xn−1
n=1
⇒
x3 + 4x2 + x
, |x| < 1. The radius of convergence is 1 because that is the radius of convergence for
(1 − x)4
n=1
S
the geometric series we started with. If x = ±1, the series is n3 (±1)n , which diverges by the Test For
∞
S
n3 xn =
Divergence, so the interval of convergence is (−1, 1).
2
1
n −1
(n + 1)(n − 1)
11. ln 1 − 2 = ln
= ln
= ln[(n + 1)(n − 1)] − ln n2
n
n2
n2
= ln(n + 1) + ln(n − 1) − 2 ln n
= ln(n − 1) − ln n − ln n + ln(n + 1)
= ln
[
k n
1
n−1
− ln
for k ≥ 2. Then
ln 1 − 2 =
ln
n
n
n+1
n=2
n=2
k
[
Let sk =
sk =
∞
[
n=2
n−1
n−1
n
− [ln n − ln(n + 1)] = ln
− ln
.
n
n
n+1
2
2
3
k−1
k
1
k
1
+ ln − ln
+ · · · + ln
− ln
= ln − ln
, so
ln − ln
2
3
3
4
k
k+1
2
k+1
1
1
k
1
ln 1 − 2 = lim sk = lim ln − ln
= ln − ln 1 = ln 1 − ln 2 − ln 1 = − ln 2.
k→∞
k→∞
n
2
k+1
2
13. u = 1 +
x6
x9
x4
x7
x10
x2
x5
x8
x3
+
+
+ ···, v = x +
+
+
+ ···, w =
+
+
+ ···.
3!
6!
9!
4!
7!
10!
2!
5!
8!
Use the Ratio Test to show that the series for u, v, and w have positive radii of convergence (∞ in each case), so
Theorem 8.6.2 applies, and hence, we may differentiate each of these series:
Thomson Brooks-Cole copyright 2007
3x2
6x5
9x8
x2
x5
x8
du
=
+
+
+··· =
+
+
+ ··· = w
dx
3!
6!
9!
2!
5!
8!
Similarly,
dv
x3
x6
x9
dw
x4
x7
x10
=1+
+
+
+ · · · = u, and
=x+
+
+
+ · · · = v.
dx
3!
6!
9!
dx
4!
7!
10!
■
7
8
■
CHALLENGE PROBLEMS
So u0 = w, v 0 = u, and w0 = v. Now differentiate the left hand side of the desired equation:
d 3
u + v 3 + w3 − 3uvw = 3u2 u0 + 3v2 v 0 + 3w2 w0 − 3(u0 vw + uv 0 w + uvw0 )
dx
= 3u2 w + 3v 2 u + 3w2 v − 3(vw2 + u2 w + uv 2 ) = 0
⇒
u3 + v 3 + w3 − 3uvw = C. To find the value of the constant C, we put x = 0 in the last equation and get
13 + 03 + 03 − 3(1 · 0 · 0) = C
⇒ C = 1, so u3 + v 3 + w3 − 3uvw = 1.
15. If L is the length of a side of the equilateral triangle, then the area is A = 12 L ·
√
3
2 L
=
√
3 2
4 L
and so L2 =
√4 A.
3
Let r be the radius of one of the circles. When there are n rows of circles, the figure shows that
L=
√
√
√ L
√
3r + r + (n − 2)(2r) + r + 3r = r 2n − 2 + 2 3 , so r = .
2 n+ 3−1
The number of circles is 1 + 2 + · · · + n =
n(n + 1)
, and so the total area of the circles is
2
L2
n(n + 1) 2
n(n + 1)
πr =
π √
2
2
2
4 n+ 3−1
√
4A/ 3
n(n + 1)
π =
√
2
2
4 n+ 3−1
An =
πA
n(n + 1)
= √
2 √
2
3
n+ 3−1
⇒
π
n(n + 1)
An
= √
2 √
A
2
3
n+ 3−1
π
π
1 + 1/n
= √
2 √ → √ as n → ∞
2 3
1+
3 − 1 /n 2 3
17. As in Section 8.6 we have to integrate the function xx by integrating series. Writing xx = (eln x )x = ex ln x and
using the Maclaurin series for ex , we have xx = (eln x )x = ex ln x =
∞
∞
[
[
(x ln x)n
xn (ln x)n
=
.
n!
n!
n=0
n=0
As with power series, we can integrate this series term-by-term:
]
1
xx dx =
0
∞ ]
[
n=0
1
0
]
∞
[
xn (ln x)n
1
dx =
n!
n!
n=0
xn (ln x)n dx
0
xn+1
n(ln x)n−1
dx and v =
:
x
n+1
n+1
1
] 1
] 1
] 1
n
x
(ln x)n − lim
xn (ln x)n−1 dx
xn (ln x)n dx = lim
xn (ln x)n dx = lim
+
n
+
1
t→0+
t→0+ n + 1
t→0
0
t
t
t
] 1
n
n
n−1
=0−
x (ln x)
dx
n+1 0
We integrate by parts with u = (ln x)n , dv = xn dx, so du =
Thomson Brooks-Cole copyright 2007
1
(where l’Hospital’s Rule was used to help evaluate the first limit).
CHALLENGE PROBLEMS
Further integration by parts gives
these steps, we get
]
1
xx dx =
0
19. Let f (x) =
]
m=0
1
0
xn (ln x)k dx = −
1
xn (ln x)n dx =
0
]
∞
[
1
n!
n=0
S∞
]
(−1)n n!
(n + 1)n
xn (ln x)n dx =
0
xn dx =
0
n=0
coefficient of xn−1 . But also
n=0
1
S∞
cm xm and g(x) = ef (x) =
S∞
]
1
xn (ln x)k−1 dx and, combining
0
(−1)n n!
(n + 1)n+1
⇒
∞
∞
∞
[
[
[
1 (−1)n n!
(−1)n
(−1)n−1
=
=
.
n+1
n+1
n! (n + 1)
(n + 1)
nn
n=0
n=0
n=1
1
g0 (x) = ef (x) f 0 (x) =
]
k
n+1
dn xn
S∞
m=1
dn xn . Then g 0 (x) =
mcm xm−1
S∞
n=0
ndn xn−1 , so ndn occurs as the
= d0 + d1 x + d2 x2 + · · · + dn−1 xn−1 + · · · c1 + 2c2 x + 3c3 x2 + · · · + ncn xn−1 + · · ·
so the coefficient of xn−1 is c1 dn−1 + 2c2 dn−2 + 3c3 dn−3 + · · · + ncn d0 =
S
ndn = n
i=1 ici dn−i .
Sn
i=1
ici dn−i . Therefore,
21. Call the series S. We group the terms according to the number of digits in their denominators:
S=
1
1
+
1
2
+ ··· +
~}
g1
1
8
+
1
9
€
+
1
 11
+··· +
~}
g2
1
99
€
+
1
 111
+ ··· +
~}
g3
1
999
€
+···
Now in the group gn , since we have 9 choices for each of the n digits in the denominator, there are 9n terms.
Furthermore, each term in gn is less than
Now
S∞
n=1
[except for the first term in g1 ]. So gn < 9n ·
9 n−1
9 10
is a geometric series with a = 9 and r =
S∞ 9 n−1
S
=
S= ∞
n=1 gn <
n=1 9 10
Thomson Brooks-Cole copyright 2007
1
10n−1
9
1 − 9/10
= 90.
9
10
1
10n−1
9 n−1
= 9 10
.
< 1. Therefore, by the Comparison Test,
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9