BC 3
Taylor Series
Name: ____________________
In class and on the homework we have determined the following Maclaurin series and intervals of
convergence for the given functions:
e  1 x 
x
x 2 x3
 L 
2! 3!
sin  x   x 
cos  x   1 


n 0
xn
n!
for x  (–, )

x3 x5 x 7
  L 
3! 5! 7 !

x 2 x 4 x6
  L 
2! 4! 6!

x 2n1
 2n  1 !
for x  (–, )
 1 
x2n
 2n  !
for x  (–, )
n
n 0

n 0
x3 x5 x 7
tan  x   x     L 
3 5 7
1
1
2
3
 1 x  x  x L 
1 x
 1 
n

x 2 n1
 1 
 2n  1
n 0

n
for x  [–1, 1]

x
for x  (–1, 1)
n
n 0
2x
Suppose you want to determine the Maclaurin series for the function f  x   e . Clearly you could
find the first few derivatives of f  x  , evaluate each at x = 0, generate the first few terms of the
Maclaurin series, and then look for a pattern in these terms so that you extend these first few terms to an
infinite series. However, one would think there must be an easier way given that we already have a
x
Maclaurin series for g  x   e . And of course there is.
Since f  x   g  2 x  , the Maclaurin series for f  x  can be generated by substituting 2x for x in the
Maclaurin series for g  x  . Thus,
e
2x
 1  2x 
2
3
2x  2x



L
2!
3!
4 x2 8x2
 1 2x 

L 
2!
3!


n 0
 2 x n
n!
x
Also, since the series for g  x   e converges for – < x < ,
2x
the series for f  x   e will converge for – < 2x <  or – < x < .
Taylor Series 1
F12
Use substitution and an appropriate Maclaurin series to determine power series that
represents each of the following functions. (DO NOT use derivatives!) Then determine the
interval of convergence for each of these series.
1.
sin  5x  
interval of convergence:
2.
1
e
x2

interval of convergence:
3.
 x3 
x  cos   
 4 
 
interval of convergence:
4.
1
2 x
3

interval of convergence:
8.
1
1  x2

interval of convergence:
9.
4

2 x
interval of convergence:
Taylor Series 2
Remember that we were also able to find Taylor Series for new functions via differentiation and
integration. For example,
x3 x5 x 7
tan  x   x     L 
3 5 7
1
for f ( x) 
1

x 2 n1
was found by integrating the power series
 1 
2
n

1


n 0

n
 1  x  x  x  L term by term. The radius of convergence for these two
2
4
6
1 x
series will be the same.
2
10. Use differentiation or integration of known series to find MacLaurin series for the
following:
1
a. f ( x) 
(1  x)2
b. g ( x)  ln(1  x)
11.
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(99)(0).
Taylor Series 3

12.
We can show that the series
2
n2
n
converges by the Ratio Test, but to what value does it
n 0
converge? To find out, complete the following steps.
Express
1
as a series:
1 x
1

1 x
Differentiate both sides.
Multiply both sides by x.
Differentiate both sides (again).
Multiply both sides by x (again) and express the right-hand side in simplified -notation
1
for x and simplify both sides of your equation. What result do you get?
2
How cool is that?
Substitute
Taylor Series 4
13. Use ideas similar to those in problem 12 to find the following sums:
2
3
3 3 3 3
a. 1            
4 4 4 4
b.
(1)n 1

3n
n0
c.
2
4
= __________________


n 1
n
1
 n!
= __________________
= __________________
Taylor Series 5