Taylor and Maclaurin Series
Lesson 9.10
Convergent Power Series Form
• Consider representing f(x) by a power series
• For all x in open interval I
• Containing c
• Then
f ( x)   an  x  c 
f ( n ) (c )
an 
n!
f "(c)
f ( x)  f (c)  f '(c)( x  c) 
( x  c) 2  ...
2!
n
Taylor Series
• If a function f(x) has derivatives of all orders
at x = c, then the series


n 0
f ( n ) (c)
f "(c)
n
( x  c)  f (c)  f '(c)( x  c) 
( x  c)2  ...
n!
2!
is called the Taylor series for f(x) at c.
• If c = 0, the series is the
Maclaurin series for f .
Taylor Series
• This is an extension of the Taylor polynomials
from section 9.7
• We said for f(x) = sin x, Taylor Polynomial of
degree 7
2
3
x
x
sin x  M 7 ( x)  0  x  0   
2! 3!
x 4 x5
x6 x7
0   0 
4! 5!
6! 7!
Guidelines for Finding Taylor
Series
1. Differentiate f(x) several times
f (c), f "(c), f '''(c),..., f
•
(c)
Evaluate each derivative at c
2. Use the sequence to form
the Taylor coefficients
•
(n)
f ( n ) (c )
an 
n!
Determine the interval of convergence
3. Within this interval of convergence,
determine whether or not the series
converges to f(x)
Series for Composite Function
• What about when f(x) = cos(x2)?
• Note the series for cos x
x2 x4 x6
cos x  1     ...
2! 4! 6!
• Now substitute x2 in for the x's
4
8
12
x
x x
cos x  1   
 ...
2! 4! 6!
2
Binomial Series
• Consider the function
f ( x)  1  x 
k
• This produces the binomial series
• We seek a Maclaurin series for this function
• Generate the successive derivatives
(n)
f
(0)  ?
• Determine
• Now create the series using the pattern
f "(c) 2
f ( x)  f (0)  f '(0)  x 
 x  ...
2!
Binomial Series
•
2
n
k

(
k

1)
x
k

(
k

1)

...

(
k

n

1)

x
f ( x)  (1  x) k  1  k  
 ... 
2
n!
• We note that
an1
lim
1
n  a
n
• Thus Ratio Test tells us radius of convergence
R=1
• Series converges to some function in interval
-1 < x < 1
Combining Power Series
• Consider f ( x)  e x  arctan x
x 2 x3
e  1  x    ...
2! 3!
and
x
• We know
x3 x5
arctan x  x    ...
3 5
• So we could multiply and collect like terms
1 3
e  arctan x  x  x  x  ...
6
x
2
Assignment
• Lesson 9.10
• Page 685
• 1 – 29 odd