694 Appendix C • Answers & Hints for Selected Exercises
00
- (a) L (-x)3k
k=O
on(-1,1)
(c) ~ L., (-1)k k+l xk+3
k=O
on (-1, l]
00
(e) L (-l)k+lkx3k-l
k=O
on (-1, 1)
- These functions are not differentiable at 0, so their Maclaurin coefficients
do not exist.
00 00
- (a) Suppose f(x) = 2:: akxk is even. Then f(x) = f(-x) = 2:: ak(-l)kxk.
k=O k=O
00 00 00 00
Thus, L akxk = L ak(-l)kxk, so L ak[l - (-l)k]xk = 0 = L Oxk. Then
k=O k=O k=O k=O
\;/k EN, ak[l - (-l)k] = 0. :. ak = 0 when k is odd.
- By 8.6.18 (b), tan-^1 x = f (-l)k~:k:; on (-1, 1). Testing endpoints, this
k=O
series converges at ±1 by the alternating series test. :. By Abel's Thm. (8.6.19),
- 00 (-l)k
7r/4 = tan-^11 = hm tan-^1 x = L k+l.
. X-->l - k=O
- (a) Converges absolutely on (-oo,2) U (4,+oo); diverges everywhere else.
(b) Converges absolutely in every interval (2n1f, 2(n + l)7r), n E Z, but
diverges whenever x = 2n7r, n E Z.
EXERCISE SET 8.7
- Assume (b). First, suppose c < x. Since JCn+l) is continuous on [c,x],
the first mean value theorem for integrals applies, so 3z E (c, x) 3 J:(x -
tr J(n+ll(t)dt = (x - z)n J(n+l)(z)(x - c). For c > x , revise this proof slightly.
:. (c).
- Let p(x) be a polynomial of degree n. Then p(x) is analytic at every c ER
The Maclaurin series for p(x) is identical with p(x), while the Taylor series of
p(x) about c =f. 0 is a polynomial of degree n in (x - c) which, when simplified,
is identical with p( x).
- See the solutions of Exercises 6.5.5 and 6.5.13.
00 k+2 00 ( l)k 2k+4
- (a) L x k! on (-oo, oo) (b) L -( 2 k~l)!
k=O k=O
on (-00,00)
( )
x2 x3 x4 x5 x6 x 7
c l+x-~-~+~+~-~-~+··· on (-oo, oo)
oo (-l)kxk+2
( d) 2:: k+ 1 on ( -1, l]
k=O
( ) e 1 - 2!^2 x2 + 4!^2 3x4 - ---m-^2 sx6 + · · · + ( -1 )k^2 2k-1x2k ( 2 k)! + · · · on ( -oo, oo )
(f) 2x2 2! - 23x4 4! + ---m-2sx6 - · · · + ( -1 )k+l 22k-lx2k ( 2 k)! + · · · on ( -oo, oo )
