BC Calculus – FDW 3rd Edition – Section 9.3 Even Answers
Quick Review 9.3
2. 7 4. ½ 6. yes 8. yes 10. no
Section 9.3
 x 4 , f (.2)  .9511 4. x 2  x , f (.2)  .0392
2. 1  8 x 2  384
2
2
x5
5!
6.

x7
7!
4

x9
9!
4
2 n 5
 ...  ( 1)n (2xn 5)!  ... 8. 1  x 2 
x4
3
2n
x)
 ...  (1)n (2
 ...
2(2 n )!
10. x 2  2 x3  4 x 4  ...  2n  2 x n , n  2 12. P12(x) 14.
16. Lagrange form of the remainder: Rn ( x ) 
x10
1 x
n 1
f   (c)
( x  a )n 1
( n 1)!
for c   a, x  or  x, a 
Using #6, for n > 2, f ( n 1) (c)  1 and for a Maclaurin Series, a  0 , so our remainder
is: Rn  ( n 11)! x n 1 . The limit as n   is 0 which implies convergence.
*The tricky part of this problem is to argue that the factorial growth of the denominator
is greater than the power growth of the numerator. See example 3 p.497.
18. Remainder Estimation Theorem (see Theorem 4 p.498), we need to find M & r
such that f ( n 1) (t )  Mr n 1 for all t between a and x , then Rn ( x )  M
n 1
r n1 x  a
( n 1)!
.
2 x , we find the derivatives of cos2x ,
Using #8 ‘double angle formula hint’ cos2 x  1cos
2
f ( x )  cos2 x , f '( x )  2sin 2 x , f ''( x )  4cos2 x , f '''( x )  8sin 2 x and notice that
since sin 2 x or cos2 x  1, we let M = 1, and although our coefficient keeps increasing
we can let r = 2 so that for R3(x), f (4) (t )  Mr n 1 =1 24 for all t in interval, I .
*Again we’ll use the argument that the factorial denominator will grow faster than the
power growth of the numerator. Rn ( x ) 
n 1
2x
.
( n 1)!
(Don’t worry about the initial ½ term
or the division by 2.) The limit as n   is 0 which implies convergence.
2
20. |error| < .0026, 1  x2 is small since the next term (in the alternating series) is positive.
22. 1.27  10 5
24. sinh and cosh series are like sine and cosine (respectively) without the alternating
negative signs: sinh x  x 
x3
3!

x5
5!
2 n 1
 ...  (2xn 1)!  ... cosh x  1 
x2
2!

x4
4!
We can find them by combining appropriate exponential series. For cosh x  e

1 x 
 1 x 

2 1
x2
2!
x2
2!
 ... 
xn
n!
x2
2!
 ...   1

x4
4!
n xn
n!
2n
 ...  (2x n )!


2n
 ...  (2x n )!  ...
x
 e x
2
Then dividing by ‘2’, we get the boxed series above.
,
BC Calculus – Section 9.3 Even Answers (continued)
f ( b) f ( a )
 f '(c ) as f (b)  f (a)  f '(c)(b  a) . Now consider b  x and let
26. Rewrite
b a
‘a’ = a , then we have: f ( x )  f (a)  f '(c)  x  a   P0 ( x )  R0 ( x ) , the zero polynomial
with remainder term for f(x) about x  a .
2
2
28.(a) 1  x (b) 1  x  x2 (c)
30.(a) 1 (b) 1  x2 (c)
32. P2 ( x )  1  kx  k ( k2!1) x 2 , using a graphing calculator, I found the intersection of |f(x) –
P2(x)| and the horizontal line, y = .01 to get 0  x  .215 .
34. P3 ( x )  1  x  x 2  x 3 vs f ( x )  11x and using a graphing calculator…
Max error = 1.111  104
2
3
n
3
2 n1
5
36.(a)  x  x2  x3 ...  xn ... (b) 2 x  23x  25x ...  22xn1 ...
38. False. f '(a)  0  0 degree
40. D 42. B
44. cos2 x  1 
(a) sin x 
2
 2 x 2
 2 x 2
22!
(b) f '( x )  2 x 
(c) sin 2 x
2!


 2 x 4
 2 x 4
24!
23 x3
3!
4!
 ... 
 ... 
 ... 
 1n 22 n x 2 n
(2 n )!
 1n1 22 n1 x2 n
(2 n )!
 1n1 22 n1 x2 n1
3
5
 2 x  (23!x )  (25!x )
i
i
1
(2n 1)!
 ... 
 ... Subtract from ‘1’ and divide by ‘2’ to get:
 ... = f(x)
 ... differentiating term by term (for n = 1, 2, …)
 1n1 22 n1 x2 n1
(2n 1)!
substitution method (for n = 0, 1, 2, …)
e  e    cis  cis      2cos   cos
(b) 21i  ei  ei   21i   cos  i sin    cos  i sin(    21i  2i sin   sin
46.(a)
1
2
2
1
2
48.(a) Differentiate the right-side and note the complex conjugate multiplication.
(b) Breaking up part (a) and using Euler’s formula:  e abi  x dx   eax ebxi dx =
e
ax
cos bx dx  i  eax sin bx dx 
a ib
a2 b2
e
a bi  x

a ib
a2 b2
eax  cos bx  i sin bx 
Now expand and group the real and complex terms to get:
ax
ax
ax
ax
 e cos bx dx  2e 2  a cos bx  b sin bx  &  e sin bx dx  2e 2  a sin bx  b cos bx 
a b
a b
Quick Quiz – Sections 9.1 – 9.3
2. A 4.(a) Geometric Series converge iff |r| < 1 or -5 < x < 1 (b) a = 2, r =
x 2
3
,
a
1 r
 16x