Math 106a Solutions
Quiz 5
10/29/10
1. Let f(x) = ln(1 −x).
(a) The 5th-order Taylor polynomial of f(x) centered at x= 0 may be found as follows:
f(x) = ln(1 −x) =⇒f(0) = ln(1) = 0
f0(x) = −1
1−x=−(1 −x)−1=⇒f0(0) = −1
1−0=−1
f00(x) = −(1 −x)−2=⇒f00(0) = −1
(1 −0)2=−1
f000(x) = −2(1 −x)−3=⇒f000(0) = −2
(1 −0)3=−2
f(4)(x) = −(3!)(1 −x)−4=⇒f(4)(0) = −3!
(1 −0)4=−3!
f(5)(x) = −(4!)(1 −x)−5=⇒f(5)(0) = −4!
(1 −0)5=−4!
P5(x) = f(0) + f0(0)(x−0) + f00(0)
2! (x−0)2+f000(0)
3! (x−0)3+f(4)(0)
4! (x−0)4+f(5)(0)
5! (x−0)5
=−x−1
2x2−2
3!x3−3!
4!x4−4!
5!x5
Therefore, P5(x) = −x−x2
2−x3
3−x4
4−x5
5.
(b) Use the Taylor polynomial P5(x) to find an estimate for ln(0.5).
First notice that f(0.5) = ln(1 −0.5) = ln(0.5).
Therefore, to estimate ln(0.5) we need to evaluate P5(x) when x= 0.5.
P5(0.5) = −0.5−(0.5)2
2−(0.5)3
3−(0.5)4
4−(0.5)5
5≈ −0.689.
In other words, ln(0.5) ≈ −0.689 .
2. To determine whether Zπ/2
0
sin x
√cos xdx converges or diverges we first need to recognize that
f(x) = sin x
√cos xis undefined at x=π
2. Therefore, the integral is improper and must be evaluated as
shown below.
Zπ/2
0
sin x
√cos xdx = lim
b→π
2
−Zb
0
sin x
√cos xdx
= lim
b→π
2
−Zcos b
1
−du
√u(where u= cos xand du =−sin x dx)
= lim
b→π
2
−Zcos b
1
(−u−1/2)du = lim
b→π
2
−h−2u1/2icos b
1
= lim
b→π
2
−h−2√cos b+ 2√1i= 2 (since √cos b→0 as b→π
2
−)
Therefore, the improper integral converges, in fact, Zπ/2
0
sin x
√cos xdx = 2 .
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