9.10 Taylor and
Maclaurin Series
Colin Maclaurin
1698-1746
In the previous lesson we were able to take certain functions that could be
written “in the form” of
a
1 r
and write them as a geometric power series.
Today we are going to make this more general so that
we can write any function as a power series!
Suppose we have a function represented by the polynomial:
f ( x)  ao  a1 ( x  c)  a2 ( x  c)2  a3 ( x  c)3  ...
How could we determine what the coefficients
an
are in terms of f ?
If we let x = c, all the terms after the first become 0 and we have:
f (c)  ao
If we differentiate term by term, we get:
f '( x)  a1  2a2 ( x  c)  3a3 ( x  c)2  4a4 ( x  c)3  ...
Again, if we let x = c, all the terms after the first become 0 and we have:
f '(c)  a1
f '( x)  a1  2a2 ( x  c)  3a3 ( x  c)2  4a4 ( x  c)3  ...
If we differentiate again we get:
f "( x)  2a2  2  3a3 ( x  c)  3  4a4 ( x  c)  ...
2
Again, if we let x = c, all the terms after the first become 0 and we have:
f "(c)  2a2
Let’s differentiate one more time to get:
f "'( x)  2  3a3  2  3  4a4 ( x  c)  3  4  5a5 ( x  c)  ...
2
Finally when x = c:
f "'(c)  2  3a3  3!a3
So what is the pattern??
In each case when x = c, we got:
f ( n) (c)  n!an
Solving this equation for an we get:
an 
f
(n)
(c )
n!
Where have we seen this before?
These are the coefficients of the Taylor polynomial that represents f(x).
Definition of Taylor Series


n 0
(n)
f ( n ) (c)
f
'(
c
)
f
"(
c
)
f
(c)
( x  c)n  f (c) 
( x  c) 
( x  c) 2  ...
( x  c) n  ...
n!
1!
2!
n!
is the Taylor series for f(x) at c.
If c = 0, then what would we call this??
The Maclaurin series for f(x)!
This is useful because now we can find a series representation
for any function for which we can find its derivatives at c!
Ex. 1 (Together):
f

n 0
f
f ( x)  sin x
f (0)  sin 0  0
f '( x)  cos x
f '(0)  cos 0  1
f "( x)   sin x
f "(0)   sin 0  0
( x)   cos x
f (3) (0)   cos 0  1
(3)
f (4) ( x)  sin x
f (4) (0)  sin 0  0
( x)  cos x
f (5) (0)  cos 0  1
f

Form the Maclaurin series forf ( x)  sin x
and find its interval of convergence.
(5)
(0) n
0 2 (1) 3 0 4 1 5
x  0  (1) x  x 
x  x  x  ...
n!
2!
3!
4!
5!
(n)


n 0
f ( n ) (0) n
0 2 (1) 3 0 4 1 5
x  0  (1) x  x 
x  x  x  ...
n!
2!
3!
4!
5!

x3 x5 x 7
(1)n x 2 n1
 x    ...  
3! 5! 7!
n  0 (2n  1)!
To find the interval of
convergence, use the Ratio Test:
x 2 n 3 (2n  3)!
lim 2 n 1
n  x
(2n  1)!
x2
 lim
 0 1
n  (2 n  3)(2 n  2)
Converges for all x, so the interval of convergence is:
(, )
We know this series converges for all x, but what does it converge to??
Does it necessarily converge to
sin x
?
In otherwords, the series could agree with sinx at its derivatives but
maybe not at points in between the derivatives.
How could we tell if the series really does converge to sinx everywhere
(and in the long run), and not just at its derivatives?
Recall that a Taylor polynomial of n terms has a remainder:
(n)
f '(c)
f "(c)
f
(c )
2
f ( x )  f (c ) 
( x  c) 
( x  c)  ... 
( x  c) n  Rn ( x)
1!
2!
n!
What is the remainder equal to??
f ( n1) ( z )
Rn ( x) 
( x  c)n1
(n  1)!
( n 1)
f
( z)
We need lim Rn ( x) 
( x  c)n1  0
(n  1)!
n

To say that f ( x)  
n 0
f n (c )
( x  c) n
n!
Convergence of a Taylor Series
If
lim Rn ( x)  0
n 
for all x in the interval I centered at c, then
the Taylor series for f converges and equals f(x).
Ex. 2
We already showed that the Maclaurin series for
for all x. Now show that it in fact converges to
Since
f ( n1) ( x)   sin x
or
sin x
sin x
converges
for all x.
f ( n1) ( x)   cos x
Taylor’s inequality tells us that our remainder is bounded
by the maximum value for the (n+1)th derivative. What is that?
( n 1)
n 1
x
f
( z)
n 1
0  Rn ( x) 
( x  c) 
(n  1)!
(n  1)!
0  Rn ( x) 
lim
n 
x
x
n 1
(n  1)!
n 1
(n  1)!
Rn ( x)  0
 0  lim
n 
Therefore the Maclaurin series for
function
sin x
sin x converges to the
for all x.
2 n 1
(1) x
sin x  
n 0 (2n  1)!

n
For certain functions (categorized as transcendental functions)
if the Taylor Polynomial is infinite, becoming a
Taylor Series, then the polynomial is actually
equivalent to the original function.
WOW!!!
http://calculusapplets.com/taylor.html
Taylor Series and Maclaurin Series
The diagram visually illustrates the convergence of the Maclaurin series
for sin x by comparing the graphs of the Maclaurin polynomials P1(x),
P3(x), P5(x), and P7(x) with the graph of the sine function. Notice that as
the degree of the polynomial increases, its graph more closely
resembles that of the sine function.
Guidelines for Finding a Taylor Series
1) Differentiate f(x) several times and evaluate each derivative at c.
Try to recognize a pattern in these numbers.
f (c), f '(c), f "(c), f "'(c),..., f ( n) (c),...
2)
Use the sequence developed in the first step to form the Taylor
coefficients
and determine the interval of
f ( n ) (c )
an 
n!
convergence for the resulting power series.
3)
Within this interval of convergence, determine whether or not
the series converges to f(x).
Ex. 3 (You try) Find the Taylor series for
f ( x)  ex
Since e is always its own derivative,


n 0
f ( n ) (2)
( x  2)n 
n!
centered at c = 2.
f ( n) (2)  e2

e2
n
(
x

2)

n 0 n !
Find the interval of convergence using the Ratio Test:
e2 ( x  2)n1
n!
(
x

2)
lim
 2
0 1
n  lim
n 
(n  1)! e ( x  2)
n  n  1
Interval of convergence:
(, )

2
e
e   ( x  2)n ?
n 0 n !
x
x
e
0  Rn ( x) 
( x  2)n1
(n  1)!
To show convergence for all x, we need numbers between 2 and x such that:
x
e
lim Rn ( x)  lim
( x  2)n1  0?
n 
n (n  1)!
n 1
(
x

2)
 e x lim
n  (n  1)!
Ratio Test?
lim Rn ( x)  0
n 
Therefore the Taylor series for f ( x)  e x at x  2
converges to the function f ( x)  e x for all x.
Deriving Taylor Series from a Basic List
Ex. 4
Find the Maclaurin series for
f ( x)  sin x2
Since we already know that Maclaurin series for sin
that to build our series:
x3 x5 x 7
sin x  x     ...
3! 5! 7!
6
10
14
x
x
x
2
2
sin( x )  x  

 ...
3! 5! 7!


n 0
f
(n)

2(2 n 1)
(0) n
n x
x   (1)
n!
(2n  1)!
n 0
x
we can use
Ex. 5
x2 x4 x6
Given that cos x  1 
   ... find the power series for
2! 4! 6!
f ( x)  cos x
No need to start from scratch…
x  x  x

x  1


2
cos
2!
4
4!
x x 2 x3
 1     ...
2! 4! 6!
6!
6
 ...
Group-work
Series FRQ Practice Problem
Homework
Pg. 685 1-11 odd, 21-25 odd, 31, 43-46
Day 2 pg. 685 15-19 odd, MMM 220