672 Appendix C • Answers & Hints for Selected Exercises
Since the sequence { ~~ } is decreasing when n ::::: 2, I.Rn ( x) I < 0.005 when
n::::: 10. Thus, T 10 (x) is the desired Taylor polynomial about 0.
(b) We want IRn(x)I < 0.005 for all x E [-1, l]. Since (n~l)! lxln+l <
e' ·ln+I (n+l)! , we can accomp l" is h t h" is b y ma k" mg (n+l)! e <. 0 005.n 4 5
e .023 .0038
(n+l)!Since the sequence{~} is decreasing, l.Rn(x)I < 0 .0 05 when n::::: 5. Use Ts(x).
- For the Taylor polynomials Tn(x) about 0 for the functions sinx or cosx,
Hn D( X ) - ±(sin(n+l)! e or cosc) X n+l ) SO IR n ( X )I < (n+l)! lxln+I • B Y C Or.^2 • •^3 11 ) n~1! l" (n+l)J lxln+I _ -
0 for all x ER:. Rn(x)--> 0.
- Let x E I, x -=/:- a. By Taylor's thm., :Jc between a and x 3 Rn(x) =
f(n+l)(c) (n+l)! ( x-a )n+l , so I Rn ( x )I ::::; Mn+(n+l)^1 (x-a)n+I!. We shall apply Thm. 2.3.10. Let- Mn+'(x-a)n+1. I Yn+I I -. I Mn+2(x-a)''+2 (n+l)! 1-
Y - (n+l)!. Then J~1! Yn - J~1! (n+2)!. Mn+'(x-a)n+1 -
- Mn+'(x-a)n+1. I Yn+I I -. I Mn+2(x-a)''+2 (n+l)! 1-
lim Mlx+- 2 al = 0. Thus, by Thm. 2.3.10, lim Rn(x) = 0.
n-+oo n n-+oo
- Calculating J(k)(x) by hand fork= 1, 2, · · · , 6 requires some hard work,
but a computer algebra system will make it easy. You will find that J(k) (0) = 0
for k = 1, 2,-· · , 5, but JC^6 l(O) = 720. :. By the nth derivative test, f has a
local minimum at 0. - See the footnote.
EXERCISE SET 6.6- Modify the proof of Case 2 by changing inequalities appropriately.
- (a) lim f(x) = L {::}Ve> 0, 35 > 0 3 xo - 5 < x < xo =? lf(x) - LI < e
x-+xQ
{::}Ve> 0, 35 > 0 3 - xo < -x < -xo + 5 =? lf(x) - LI< e
{::}Ve> 0, 35 > 0 3 - xo < x < -xo + 5 =? lf(-x) - LI < e
{::} lim f(-x) = L.
x--+-xci - We prove Case 4; Cases 5 and 6 are proved similarly.
Given: (i) J, g : I--> JR, where I is an interval with right end-point a= x 0.
(ii) f , g are differentiable on I ; (iii) Vx E I, g(x )g' (x) -=f. O;
(iv) lim f(x) = lim g(x) = O; (v) lim_ ~;~~? = L.
X--JoXo X-+Xo X-+Xo
Let -I= (-xo,-a), for some a< xo in I. Define F,G: -I--> JR by
F(x) = J(-x) and G(x) = g( -x). Then Vx E -I, F'(x) = -f'(-x) and