BC 3
Taylor Quiz
Name:
Calculator Allowed
Clearly show all appropriate work for full credit. NO magically calculator "leaps of faith," please.
1 (1 pt each) The function y  f ( x) is approximated near x = 0 by the third degree Taylor Polynomial
1
f (0) 2 f (0) 3
P3 ( x)  3x  x 2  x 3  f (0)  f (0) x 
x 
x 
3
2!
3!
f (0)  0
f (0)  3
f (0)  2
f (0)  2
2 (5 pts) Determine all values of x (exact) for which x 2  x3  x 4  x5  x6  L L L  1 .
x  x  x  x  x L L L  1
2
3
4
5
6
2
x
1
1 x
 x  1 x
2
 x2  x  1  0
x
1 5
2
1 5
 1, this value is not in the interval of converence.
2
1 5
So, x 
is the only solution.
2
Since, x 
3(3 pts each) Find the value of each series by recognizing the function and the point at which it is
evaluated. [Exact answer please.] Indicate clearly the function and the value of x you use.
32 34 36 38
   

2! 4! 6! 8!
cos(3)  1 
32 34 36 38
32 34 36 38
  
  cos(3)  1     

2! 4! 6! 8!
2! 4! 6! 8!
1 1 1 1
1      
3 5 7 9
IMSA
= cos(3)  1 , since
= tan 1 (1) 

4
Suppose g  x   xe
4(6 pts)
 x2
.
a. Find the Maclaurin series for g  x  .
   x   x 

 x2

2
g  x   x 1   x 
2!

 
2
2 3
3!
2 4
4!


L L 

2 n 1
x5 x 7
n x
 x  x    L L  (1)
 L L , x  (, )
2! 3!
n!
3
b. Evaluate g
g
(100)
 0  and g (101)  0  , the 100th and 101st derivatives of g at
x = 0.
 0   0 , since the coefficient on x100 is 0 in the Maclaurin series,
g (101)  0  (1)50
101!

 g (101)  0  
(100)
and
101!
50!
5(8 pts) Let f  x  be a function such that f
50!
 n
 0  (1)n n2
for all n ≥ 0. Thus, f  0  0,
f     1, f   2  4 , and so on.
a. Find the MacLaurin series for f  x  . Write out the first 4 non-zero terms along with the
general term.
f (0)  f (0) x 
f (0) 2
4
9
16
( 1) n  n 2 n
x  L L L   x  x 2  x3  x 4  L L L 
x L L
2!
2!
3!
4!
n!
b. Given that the MacLaurin series converges at x  1 , how many non-zero terms are needed to
approximate f 1 with an error of at most 0.001? Clearly show analysis.
2
n
4 9 16
( 1) n  n 2
L . By A.S.T., f 1  S n  an 1 
f 1 = 1     L L L 
.
n!
2! 3! 4!
n!
92
1
1


 .001 .
From calculator,
9! 4480 1000
So we need eight non-zero terms.
IMSA

6. (2 pts) Evaluate the series
4
 (4n)! .
n0
Work on this. I will give you some help if you want.
IMSA