368 STEP 4. Review the Knowledge You Need to Score High
Since∑∞
n= 11
n
diverges, this series diverges as
well. Whenx= 2 a, the series becomes
∑∞
n= 1(−1)n−^1 (2a−a)n
nan=
∑∞
n= 1(−1)n−^1
n. This
alternating series converges; therefore, the
interval of convergence is (0, 2a].
14. Calculate the derivatives and evaluate at
x=1.f(x)=ex^2 and f(1)=e.
f′(x)= 2 xex^2 and f′(1)= 2 e.
f′′(x)= 4 x^2 ex^2 + 2 ex^2 andf′′(1)= 6 e.
f′′′(x)= 8 x^3 ex^2 + 12 xex^2 and f′′′(1)= 20 e.
f(4)(x)= 16 x^4 ex^2 + 48 x^2 ex^2 + 12 ex^2 and
f(4)(1)= 76 e. Then the function
f(x)=ex^2 can be approximated by
e
0!
(x−1)^0 +
2 e
1!
(x−1)^1 +
6 e
2!
(x−1)^2 +
20 e
3!
(x−1)^3 +
76 e
4!
(x−1)^4. Simplifying
f(x)=ex^2 ≈e+ 2 e(x−1)+ 3 e(x−1)^2 +
10 e
3(x−1)^3 +
19 e
6(x−1)^4.- f
(
1
2)
=cos(
π
2)
=0. Find thederivatives and evaluate atx=1
2
.
f′(
1
2)
=−πsinπx|x= 1 / 2 =−π,f′′(
1
2)
=−π^2 cosπx|x= 1 / 2 =0,f′′′(
1
2)
=π^3 sinπx|x= 1 / 2 =π^3 , andf(4)(
1
2)
=π^4 cosπx|x= 1 / 2 =0. Thenf(x)=cosπxaroundx=1
2
can beapproximated by0
0!
(
x−1
2
) 0
+−π
1!(
x−1
2
) 1
+0
2!
(
x−1
2
) 2
+
π^3
3!(
x−1
2
) 3
+0
4!
(
x−1
2
) 4
or−π(
x−1
2
)
+
π^3
6(
x−1
2
) 3
.- Atx=e,f(e)=lne=1,
f′(e)=
1
x∣∣
∣
∣x=e=1
e
, f′′(e)=− 1
x^2∣∣
∣
∣x=e=− 1
e^2,
f′′′(e)=2
x^3∣∣
∣∣
x=e=
2
e^3, andf(4)(e)=− 6
x^4∣∣
∣∣
x=e=
− 6
e^4
.f(x)=lnxcan beapproximated by1
0!
(x−e)^0 +
1 /e
1!(x−e)^1 +
− 1 /e^2
2!(x−e)^2 +
2 /e^3
3!
(x−e)^3 +
− 6 /e^4
4!
(x−e)^4 = 1 +
x−e
e
−
(x−e)^2
2 e^2+
(x−e)^3
3 e^3−
(x−e)^4
4 e^4.
- Calculate the derivatives and evaluate at
x=0.f(x)=
1
1 −x
, f(0)=1,f′(x)=1
(1−x)^2
,f′(0)=1,f′′(x)=2
(1−x)^3
,f′′(x)=2,f′′′(x)=6
(1−x)^4
,f′′′(x)=6,f(4)(x)=24
(1−x)^5
,f(4)(x)=24. In general,
f(n)(0)=n!, so the MacLaurin seriesf(x)=1
1 −x=
∑∞
n= 0f(n)(x)
n!
xn=
∑∞
n= 0n!
n!
xn=∑∞
n= 0xn. The series converges tof(x)=1
1 −x
when limn→∞∣∣
∣∣xn+ 1
xn∣∣
∣∣=nlim→∞|x|<1. The series converges on
(−1, 1). Whenx=1, the series becomes
∑∞
n= 01 n, which diverges. Whenx=−1, theseries becomes
∑∞
n= 0(−1)n, which diverges.
Therefore, the interval of convergence
is (−1, 1).