Solutions to Homework I
Y. Zhou
Section 1.1: 5.a. In order to find the Taylor series of f (x) = ln(x + 1) at 0, we need to
find its derivatives:
f 0 (x) = (1 + x)−1 ,
f 00 (x) = −(1 + x)−2 ,
f (3) (x) = 2(1 + x)−3 ,
f (k) (x) = (k − 1)!(−1)(k+1) (1 + x)−k .
Thus the Taylor series of f (x) at 0 is
∞
X
1 (k)
f (0)xk + Rn (x)
f (x) =
k=0
∞
X
k!
1
(k − 1)!(−1)(k+1) xk
k!
k=0
=
∞
X
(−1)(k+1) k
x ,
=
k
k=0
or
n
X
(−1)(k+1) k
x + Rn (x)
f (x) =
k
k=0
when truncated. Here the reminder Rn (x) can be a Lagrange reminder
Rn (x) =
=
1
f (n+1) (ξ)x(n+1)
(n + 1)!
(−1)(n+2)
(1 + ξ)−(n+1) xn+1
n+1
or an integral reminder
Rn (x) =
1
n!
Z x
f (n+1) (t)(x − t)n dt
0
= (−1)(n+2)
Z x
(1 + t)−(n+1) (x − t)n dt.
0
5.b. We need to determine the smallest n such that
|Rn (0.5)| ≤ 10−8 .
Since
0.5n ,
n + 1
|Rn (0.5)| ≤ it is sufficient to find the smallest n such that
0.5n −8
n + 1 ≤ 10 ,
1
which requires that n ≥ 23.
5.c. Following 5.b we need to find the smallest n such that
0.6n −10
,
n + 1 ≤ 10
which requires that n ≥ 38.
24. Recall the Taylor series of cos(x) at 0 is
cos(x) =
∞
X
(−1)k
k=0
x2k
,
(2k)!
and thus the full reminder for cos(x) ≈ 1 − 0.5x2 is
1 4
1
1
x − x6 + x8 · · ·
4!
6!
8!
1 4
4! 2 4! 4
x (1 − x + x · · ·)
4!
6!
8!
1 4
x
4!
Rn (x) =
=
<
if x < 1. Thus if Rn (x) is required to be less than 1.0 × 10−3 , it is sufficient to ask
1 4
x ≤ 10−3 ,
4!
√
suggesting that |x| ≤ 4 4! × 10−3 = 0.3936. Note: Other estimates also apply. For example,
Scott and Heather use the Lagrange reminder at n = 2 and estimate the derivative, which
could give a shaper bound of x.
35. Suppose we are truncating the series at the nth term, i.e.,
e≈
n
X
1
k=0
k!
.
But than the full reminder will be
R =
∞
X
1
k!
k=n+1
1
(n + 1)! (n + 1)!
1+
+
+ ···
(n + 1)!
(n + 2)! (n + 3)!
1
(1 + 1/2 + 1/4 + 1/8 + · · ·)
(n + 1)!
2
.
(n + 1)!
=
≤
≤
It is straightforward to find that n has to be at least 21 to make R ≤ 0.6 × 10−20 . A little
shaper estimate can be found by using the Lagrange remainder at n and estimating the
derivative.
Section1.2 I only do the first three problems.
7.a. False, since
(n + 1)/n2
n+1
= lim
=1
n→∞
n→∞
1/n
n
lim
2
7.b. False, since
√
(n + 1)/ n
=∞
n→∞
1
lim
7.c. False, since
1/ ln n
= ∞.
n→∞ 1/n
lim
9. From the hints we know that the limit is −1/2. If the rate of convergence is p, then
there exists a nonzero constant C such that
lim
h→0
1
[(1
h2
+ h) − eh ] − (− 12 )
= C.
hp
We notice that
lim
1
[(1
h2
h→0
+ h) − eh ] − (− 12 )
hp
2
=
=
=
=
(1 + h) − eh + h2
lim
h→0
h(p+2)
1 − eh + h
lim
h→0 (p + 2)h(p+1)
−eh + 1
,
lim
h→0 (p + 2) · (p + 1) · hp
−eh
lim
,
h→0 (p + 2) · (p + 1) · p · hp−1
which equals −1/6 if p = 1, ∞ is p > 1 and 0 if p < 1. Hence p = 1 is required, and the
convergence rate is 1. In other words,
1
1
1
[(1 + h) − eh ] = − + O(h) = − + o(hβ )
2
h
2
2
for any β < 1.
22. By the definition of linear convergence it is sufficient to show that there is a nonzero
constant C such that
(1 + aθn+1 )/(1 + aθn ) − 1
= C.
n→∞ (1 + aθ n )/(1 + aθ n−1 ) − 1
lim
We notice that
θ(1 + aθn−1 )
(1 + aθn+1 )/(1 + aθn ) − 1
=
lim
,
n→∞
n→∞ (1 + aθ n )/(1 + aθ n−1 ) − 1
1 + aθn
lim
which convergences to θ if θ ∈ (0, 1).
30. Sine the left-hand side of the equation is a series, we might resort to the Taylor
expansion. Notice the Taylor series of 1/(1 − x) at x = 0 is
1/(1 − x) =
∞
X
k=0
3
xk ,
thus for any fixed n,
Pn
lim
k
k=0 x
x→0
− 1/(1 − x)
xn
=
∞
X
xk
lim
x→0
xn
k=n+1
2
3
= x + x + x + · · · = 0.
This proves the argument.
14. Taking derivative with respect to x on the both sides we obtain
6x2 y 2 + 2x2 · 2y ·
dy
dy
+ 4xy + x2
+ ex = 0,
dx
dx
from which we can solve dy/dx to be
dy
6x2 y 2 + 2xy + ex
=−
dx
4x3 y + x2
Section 2.3 1. It can be shown that
α + 0.1β = f (0.1) × 1012 ,
α + 0.9β = f (0.9)/0.912 ,
from which (α, β) is solved to be (0.6659, −0.5991).
On the other hand, it is seen that
"
α
β
#
"
=
"
=
"
=
1 0.1
1 0.9
#−1 "
·
1.125 −0.125
1.25
1.25
1012
0
0
3.5407
# "
·
# "
·
1012
0
0
3.5407
1.125 × 1012 −0.4426
1.25 × 1012
4.4259
# "
·
f (0.1)
f (0.9)
# "
f (0.1)
f (0.9)
·
#
f (0.1)
f (0.9)
#
#
.
The condition number of this coefficient matrix is 6.30 × 1011 , indicating that (α, β) is
extremely sensitive to the slight change of (f (0.1), f (0.9)).
5.c The condition number of the function is
cond(f (x)) =
x sin0 (x)
= x cot(x),
sin(x)
which will be large wherever x is close to multiples of π.
5.e The condition number of the function is
cond(f (x)) =
x(ex /x)0
= x − 1,
ex /x
which will be large only if x is large.
4