Series 361Step 2: Differentiate:
− 1
(x+1)^2
=− 1 + 2 x− 3 x^2 + 4 x^3 −···=∑∞n= 0(−1)n+^1 (n+1)xn.Step 3: Multiply by−1.
1
(x+1)^2
= 1 − 2 x+ 3 x^2 − 4 x^3 +···=
∑∞n= 0(−1)n(n+1)xnExample 3
Use a MacLaurin series to approximate the integral
∫ 10sin(x^2 )dxto three decimal place
accuracy.
Step 1: Substitutex^2 forxin the MacLaurin series representing sinx.
sin(x^2 )=(x^2 )−
(x^2 )^3
3!+
(x^2 )^5
5!−
(x^2 )^7
7!
···=x^2 −
x^6
3!+
x^10
5!−
x^14
7!+···
Step 2:
∫ 10sin(x^2 )dx=∫ 10(
x^2 −
x^6
3!+
x^10
5!−
x^14
7!...
)
dx=
x^3
3−
x^7
7 ·3!+
x^11
11 ·5!−
x^15
15 ·7!+···
∣
∣∣
∣10=
1
3
−
1
7 ·3!
+
1
11 ·5!
−
1
15 ·7!
+···=
∑∞n= 01
(4n+3)(2n+1)!Step 3: For this alternating series, the absolute error for thenth partial sum is less than the
n+1 term so|S−sn|<1
(4n+4)(2n+2)!. We want three decimal place accuracy,
so we need|S−sn|<1
(4n+4)(2n+2)!
≤ 0 .0005 or 2000≤(4n+4)(2n+2)!
This occurs forn≥2.Step 4: Taking the suma 0 +a 1 +a 2 =
1
3
−
1
7 .3!
+
1
11 ·5!
= 0 .3103,
∫ 10sin(x^2 )dx≈ 0 .3103.Error Bounds
The remainder,Rn(x), for a Taylor series is the difference between the actual value of the
functionf(x) and thenth partial sum that approximates the function. If the function f(x)
can be differentiatedn+1 times on an interval containingx 0 , and if|f(n+1)(x)|≤Mfor all
xin that interval, then|Rn(x)|≤
M
(n+1)!
|x−x 0 |n+^1 for allxin the interval.Example 1
Approximate
√
eaccurate to three decimal places.Step 1: Substitute
1
2
forxin the MacLaurin series representation forex.e^1 /^2 =∑∞n= 0(1/2)n
n!= 1 +
1
2
+
(1/2)^2
2
+
(1/2)^3
6
+···
= 1 +
1
2
+
1
8
+
1
48
+···=
∑∞n= 01
2 n·n!