Taylor Problem Solutions
BC3
( x  3)n
1) Find the interval of convergence of  n
. Show all work.
n 1
n 0 4

L  lim
( x  3) n 1 4n n  1
( x  3) n  1
( x  3)

 lim


n
n 1
n 
4
4
4
n  2 ( x  3)
n2
L 1
( x  3)
 1  4  x  3  4  1  x  7
4
n 
Endpoints:

  1
( x  3)n
(4)n
x  1   n
 n

which converges by AST.
n  1 n 0 4 n  1 n 0 n  1
n 0 4

n


( x  3)n
(4)n
1


which diverges by p  test.


n
n
n  1 n 0 4 n  1 n 0 n  1
n 0 4

x7
Therefore, interval of convergence = [1, 7)
2)The Maclaurin series for some function is given by
1 2 x 3x 2 4 x3 5 x 4 6 x5 7 x 6 8 x 7
f ( x)  






..........
2! 3! 4!
5!
6!
7!
8!
9!
a. Find f (0) .
b. Find f (60) (0) .
f (0) 2
x 
2!
2 1
So, f (0) is the coefficient of x in our series for f, i.e., f (0)   .
3! 3
(60)
f (0) 61
61 60! 1

 f (60) (0) 

Similarly,
. Note that the coefficient of x 60 in the series
60!
62!
62!
62
for f is positive – all powers of x that are multiples of 4 have positive coefficients.
Remember that by Taylor’s Theorem: f ( x)  f (0)  f (0) x 
1
sin x
dx with an error of less than 0.001. (This integral is not
3) Use series to approximate 

0 x
improper at the origin, so don’t worry about that.) Explain how you know your answer has the
desired accuracy. Is your approximation an overestimate or an underestimate? How do you
know? [Do not add up the terms of the series].
1
1
3
5
7
 sin x dx   1  x  x  x  x

 x
0 x
3! 5! 7!
0 
Integrating term by term, we get
1


 dx  

0
 x2 x4 x6
1  3!  5!  7!


 dx

1
3
5
7
 sin x dx  x  x  x  x

0 x
3  3! 5  5! 7  7!
1
1
1
1
 1


0
3  3! 5  5! 7  7!
1
Since this is an alternating series with an 
, an is decreasing, positive, and
(2n  1)  (2n  1)!
1
sin x
1

tending toward 0. Hence, 
. If
dx  Sn  an1 
0 x
(2n  1)  (2n  1)!
1
1
1
n  3,


 .001 .
(2n  1)  (2n  1)! (7)  (7)! 1000
1
sin x
1
1
dx  1 

So, 
with error less than .001

0 x
3  3! 5  5!
4) Write Taylor’s formula with remainder for g(x) = x3/2, with x0 = 16 and n = 3, by finding the
appropriate derivative of g and using Taylor’s Theorem. Be detailed about the error term.
So,
3
g ( x)  x 2 
3 1
g ( x)  x 2 
2
3 1
g ( x)  x 2 
4
3 3
g ( x)   x 2 
8
9 5
g (4) ( x)  x 2
16
Therefore,
g (16)  64
g (16)  6
3
g (16) 
16
g (16)  
g (4) (c ) 
3
256
9
5
16  c 2
9
g ( x)  64  6( x  16) 
16 and x.
5
2
3
3
( x  16)2 
( x  16)3  16  c ( x  16)4 , where c is between
16  2!
1256  3!
4!
5) Let f(x) be a function whose nth derivative lies in the range [0, 1/n]. That is, 0  f(n)(x)  1/n
for all n and for all x.
a) If the Taylor series for f(x) about x0 = 6 is used to estimate f(4), how many terms are necessary
to achieve an error of less than 0.01? (It’s at least 5, so don’t check small numbers).
f ( n1) (c)
f ( n1) (c)
2n1
Error =
.
( x  6)n1 
(4  6)n1 
(n  1)!
(n  1)!
(n  1)  (n  1)!
So we need
2n1
1
2n1
8
1
. If n = 6,
, So we need n  6 .



(n  1)  (n  1)! 100
(n  1)  (n  1)! 2205 100
b) If the number of terms you found in part a) are used, will the estimate for f(4) be too large or
f 7 (6)
too small? Since a7 
(4  6)7  0 , the approximation is too high. This follows because the
7!
series used to approximate f(4) is an alternating series.
c) Will the series converge to f(4) exactly? Yes. How do you know? Because the remainder
term is approaching zero.

6) Consider the power series

n  0
an (x  3) n . Indicate whether the following statements must
be true, which may be true, and which cannot be true. Briefly justify your answers.
a. The interval of convergence of this power series is ( 4, 2) .
This may be true. The center of this series is 3 , as is the center of the interval
( 4, 2) . Without knowing an , we cannot know the exact interval (nor radius) of
convergence.
b. If the radius of convergence of this series is 3 then the interval of convergence is
( 6, 0) .
This may be true. Given the radius of convergence does not tell us whether the series
converges at the endpoints x  6 or x  0 .
c. If the series converges at x = 5 , then the series converges at x = 1 .
This may be true. If x = 1 is an endpoint, the series may converge or diverge at the
other endpoint.
d. There are no values of x for which this series converges.
This cannot be true since the series will converge at x  3
7).
Find the value of each series by recognizing the function and the point at which it is
evaluated. [Exact answer please.]
2
3
3 3 3 3
1       
4 4 4 4
4
1
4

 3 7
1   
 4
=
3
[geometric series with a  1 , r   ]
4
2
4
6
8
1 1 1 1
1             
3  3  3  3
=
1
[geometric series w/ a  1 , r   ]
9
1
1
1
1
 2
 3
 4

2 2  2! 2  3! 2  4!
1
= e 2 1
1
9

10
 1
1   
 9