Calculus II, Section 11.10, #24
Taylor and Maclaurin Series
Find the Taylor Series for f(x)centered at the given value of a.[Assume that fhas a power series expansion.
Do not show that Rn(x)→0.]Also find the associated radius of convergence.1
f(x) = cos (x), a =π
2
The general form for a Taylor series is
f(x) =
∞
X
n=0
f(n)(a)
n!(x−a)n
=f(a) + f′(a)
1! (x−a) + f′′ (a)
2! (x−a)2+f′′′(a)
3! (x−a)3+f(4)(a)
4! (x−a)4+···
Let’s organize our work in a table. For this problem, we need enough entries in the table to recognize a
pattern for f(n)π
2in terms of n.
n f(n)(x)f(n)π
2
0 cos (x) 0
1−sin (x)−1
2−cos (x) 0
3 sin (x) 1
4 cos (x) 0
5−sin (x)−1
6−cos (x) 0
7 sin (x) 1
8 cos (x) 0
.
.
..
.
..
.
.
Using the definition of a Taylor series and the values in the table, we get
f(x) = 0 −1
1! x−π
21+ 0 + 1
3! x−π
23+ 0 −1
5! x−π
25+ 0 + 1
7! x−π
27+ 0 + ···
=−1
1! x−π
21+1
3! x−π
23
−1
5! x−π
25+1
7! x−π
27+···
=
∞
X
n=0
(−1)n+1 1
(2n+ 1)! x−π
22n+1
To find the radius and interval of convergence, we use the Ratio Test.
lim
n→∞
an+1
an
= lim
n→∞
(x−π
2)2(n+1)+1
(2(n+1)+1)!
(x−π
2)2n+1
(2n+1)!
1Stewart, Calculus, Early Transcendentals, p. 771, #24.
