More Taylor Problems
BC3

1) Find the interval of convergence of
( x  3)n
. Show all work.
n
n 1
4
n 0
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 ƒ'(0).
b. Find ƒ(60)(0).
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].
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.
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).
b) If the number of terms you found in part a) are used, will the estimate for f(4) be too large or
too small?
c) Will the series converge to f(4) exactly? How do you know?

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) .
b. If the radius of convergence of this series is 3 then the interval of convergence is
( 6, 0) .
c. If the series converges at x = 5 , then the series converges at x = 1 .
d. There are no values of x for which this series converges.
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
2
4
6
4
=
8
1 1 1 1
1             
3  3  3  3
1
1
1
1
 2
 3
 4

2 2  2! 2  3! 2  4!
=
=