Take a Diagnostic Exam 37Chapter 14
- Note that
∑∞n= 0(−1)n
2 n=−
1
2
+
1
4
−
1
6
+...is analternating series such thata 1 >a 2 >a 3 ...
i.e.,an>an+ 1 and limn→∞an=nlim→∞1
2 n=0.
Therefore|S−sn|≤an+ 1 and, in this case,
|S−s 5 |≤a 6 , anda 6 =1
12
. Thus
|S−s 5 |≤1
12
is the maximum value.- The series
∑∞n= 03
(n+1)^4is a series with positiveterms, which can be compared to the series
∑∞n= 03
n^4. Also
∑∞n= 03
n^4= 3
∑∞n= 01
n^4
and∑∞n= 01
n^4
is ap-series withp=4, and therefore
convergent.∑∞n= 03
(n+1)^4
is term by termsmaller than∑∞n= 03
n^4
and so∑∞n= 03
(n+1)^4
converges.- The seriesx−
x^2
2+
x^3
3−
x^4
4+···
is an alternating series with general term
(−1)n−^1 xn
n. Using the ratio test for absolute
convergence, we have limn→∞∣∣
∣∣xn+ 1
n+ 1·
n
xn∣∣
∣∣=|x|nlim→∞(
n
n+ 1)
=|x|. The series willconverge absolutely when|x|< 1 ⇒
− 1 <x<1. We do not consider the end
points since the question asks for absolute
convergence.- Investigate the first few derivatives of
f(x)=
1
x. f′(x)=
− 1
x^2
, f′′(x)=
2
x^3
, f′′′(x)=− 6
x^4,
f(4)(x)=24
x^5
and, in general,f(n)(x)=
(−1)nn!
xn+^1.
Evaluate the derivatives atx=2.f(2)=1
2
,
f′(2)=− 1
4
,f′′(2)=2
8
,
f′′′(2)=− 6
16
, f(4)(x)=24
32
and, in general, f(n)(2)=
(−1)nn!
2 n+^1.
The Taylor series isf(x)=1
x=
∑∞n= 0(−1)nn!
2 n+^1
n!
(x−2)n=∑∞n= 0(−1)n
2 n+^1
(x−2)n=
1 / 2
0!
(x−2)^0 +− 1 / 4
1!
(x−2)^1 +2 / 8
2!
(x−2)^2+
− 6 / 16
3!
(x−2)^3 +24 / 32
4!
(x−2)^4 +···=
1
2
−
1
4
(x−2)+1
8
(x−2)^2 −1
16
(x−2)^3+1
32
(x−2)^4 −···- Begin with the MacLaurin series forex.
If f(x)=ex, thenf′(x)=ex,
f′′(x)=ex, and fn(x)=ex.
Thusex= 1 +x+
x^2
2!+
x^3
3!+···.
Replacingxby−x^2 , we havee−x^2 = 1 −x^2 +
x^4
2!−
x^6
3!+
x^8
4!−···.
Thus,e−x^2 =∑∞n= 0(−1)n(x^2 n)
n!