exercicios resolvidos 11.10
DESCRIPTION
Exercicios de calculo 3 resolvidos.TRANSCRIPT
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SECTION 11.10 TAYLOR AND MACLAURIN SERIES 489
39. By Example 7, tan31 { ="Sq=0
(31)q {2q+1
2q + 1 for |{| ? 1. In particular, for { =1I3, we
have6
= tan31
1I3
="Sq=0
(31)q1@I32q+1
2q + 1 ="Sq=0
(31)q1
3
q1I3
1
2q + 1 , so
= 6I3
"Sq=0
(31)q(2q + 1)3q = 2
I3"Sq=0
(31)q(2q + 1)3q .
11.10 Taylor and Maclaurin Series
1. Using Theorem 5 with"Sq=0
eq({3 5)q, eq = i(q)(d)q! , so e8 =
i (8)(5)8!
.
3. Since i (q)(0) = (q + 1)!, Equation 7 gives the Maclaurin series
"Sq=0
i (q)(0)q! {
q ="Sq=0
(q + 1)!q! {
q ="Sq=0
(q + 1){q. Applying the Ratio Test with dq = (q + 1){q gives us
limq
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490 CHAPTER 11 INFINITE SEQUENCES AND SERIES
9.q i (q)({) i (q)(0)0 h5{ 11 5h5{ 52 52h5{ 253 53h5{ 1254 54h5{ 625...
......
h5{ ="Sq=0
i (q)(0)q! {
q ="Sq=0
5q
q! {q.
limq