Taylor Series
From Functions to Series and Back, again!
Chapter 1
FROM FUNCTIONS TO SERIES
Before: Given a power series can we find a formula for it?
Sometimes we could . . .

1
f ( x)   x 
for all x in  1,1 .
1 x
n 0
n

Sometimes not . . .
1
k
(
x

3)

k 2
2
k
k 1
Chapter 1
FROM FUNCTIONS TO SERIES
Now we want to turn this around:
Given a particular function, can we find a power series for it?
•The book suggests that sometimes we can (chart you used
for homework!)
•We also suggested something similar in our discussion of
Taylor Polynomials!
Approximating f ( x)  ( x  1) e
The third order Maclaurin
Polynomial is the best
3rd degree polynomial
approximation for f at
x = 0.
And, as we added more
and more terms the
approximation got better
and better.
 x2
Approximating f ( x)  ( x  1) e
The hundredth Maclaurin
Polynomial is the best
100th degree polynomial
approximation for f at
x = 0.
What if we take this to its
logical conclusion? We
get Taylor Series!
 x2
Taylor Series for cos(x)
1. Find the Maclaurin series for f (x) = cos(x).
2. Use Taylor’s Theorem to show that this series
converges to f for all values of x.
Maclaurin Series for f (x) = cos(x)
n
f (n)(x)
f (n)(0)
0
cos( x)
cos  0  1
a0  1
1
 sin( x)
 sin  0  0
a1  0
2
 cos( x)
 cos  0  1
a2  
3
sin( x)
sin(0)  0
a3  0
cos( x)
cos(0)  1
a4 
4
an= f (n)(0)/n!
1
2!
1
4!
Can we deduce the rest of the pattern? What happens, in
general, with the coefficients?
Maclaurin Series for f (x) = cos(x)
n
f (n)(x)
f (n)(0)
0
cos( x)
cos  0  1
a0  1
1
 sin( x)
 sin  0  0
a1  0
2
 cos( x)
 cos  0  1
a2  
3
sin( x)
sin(0)  0
a3  0
cos( x)
cos(0)  1
a4 
4
x 2 x 4 x6
1   
2! 4! 6!
an= f (n)(0)/n!
(1)n x2 n

 2n  !
k 0

1
2!
1
4!
Show that this converges to cos(x)
We start with the general set-up for Taylor’s Theorem.
K n1
n 1
cos( x)  Pn ( x) 
x
 n  1!
What is Kn+1?
d n1
cos( x)  1 for all n and all x.
n1
dx
It follows that
n 1
1
n 1
x
cos( x)  Pn ( x) 
x

 n  1!
 n  1!
This quantity goes
to 0 as n→∞!
Show that this converges to cos(x)
We start with the general set-up for Taylor’s Theorem.
K n1
n 1
cos( x)  Pn ( x) 
x
 n  1!
What is Kn+1?
d n1
cos( x)  1 for all n and all x.
n1
dx
It follows that
n 1
1
n 1
x
cos( x)  Pn ( x) 
x

 n  1!
 n  1!
So what can
we conclude?
Show that this converges to cos(x)
For all real x, cos( x)  Pn ( x) 

0.
n
What (on Earth!) does this mean?
•The Pn’s are the
?
.
Show that this converges to cos(x)
For all real x, cos( x)  Pn ( x) 

0.
n
What (on Earth!) does this mean?
•The Pn’s are the partial sums of the Maclaurin Series
•So cos( x)  Pn ( x)
converges to cos(x)!
0
means that the series
.
Chapter 2
FROM SERIES TO FUNCTIONS
What is the relationship between
Taylor Series and other power series?
Coefficients of a Power Series
Suppose that we have function f given by a power series
f ( x)  a0  a1 ( x  x0 )  a2 ( x  x0 )2  a3 ( x  x0 )3 
What can we say about the relationship between f and
the coefficients a0, a1, a2, a3, a4, a5, . . ?
Answer: Quite a bit, and the reasoning should look
somewhat familiar to you.
Here’s how it goes . . .
If
f ( x)  a0  a1 ( x  x0 )  a2 ( x  x0 )2  a3 ( x  x0 )3 
Then
f ( x0 )  a0  a1 ( x0  x0 )  a2 ( x0  x0 )2  a3 ( x0  x0 )3 
 a0
Furthermore, Theorem 13 (pg. 591 in OZ), says that
f ( x)  a1  2a2 ( x  x0 )  3a3 ( x  x0 )2  4a4 ( x  x0 )3
Thus f ( x0 )  a1.
Is this beginning to look familiar?
It should remind you of the process by which we computed the
coefficients of the Taylor polynomial approximations
Continuing to take derivatives and evaluate at x0,
we have . . .
f ( x)  2a2  3  2a3 ( x  x0 )  4  3a4 ( x  x0 )2 
 f ( x0 )  2a2
f ( x0 )
 a2 
2
f ( x)  3!a3  4!a4 ( x  x0 )  5  4  3a5 ( x  x0 ) 2 
 f ( x0 )  3!a3
f ( x0 )
 a3 
3!
In general, we have:
(n)
f
( x0 )
( n)
f ( x0 )  n!an which tells us that an 
n!
In other words, if a function f is given by a power series that is
centered at x0, that power series must be the Taylor series for f
based at x0.
If, we have
f ( x)  a0  a1 ( x  x0 )  a2 ( x  x0 )2  a3 ( x  x0 )3 
Then
(3)
f ( x0 )
f
( x0 )
f ( x)  f ( x0 )  f ( x0 )( x  x0 ) 
( x  x0 ) 2 
( x  x0 )3 
2
3!
It is easy to see that Taylor series are just a special kind
of power series. Our discovery tells us that they are
really the only kind of power series there is.
To reiterate: If a function f is given
by a power series, that power series
must be the Taylor Series for f at the
same base point.
Chapter 3
TAYLOR SERIES CAUTIONS
Weird stuff can happen if we are not careful.
Now where were we?
If a function f is given by a power series, that power
series must be the Taylor Series for f at the same base
point.
Notice what this does not say.
•It does not say that every function is given by its
Taylor Series.
•It does not even say that every function that has a
Taylor series is given by its Taylor Series.
For a Function f, Some
Questions Arise
1. If f has a Taylor Series, does the series
converge?
Answer: Often, but not always, and certainly not
always on the whole domain of the function.

Consider the familiar case of
f ( x) 
1
.
2
1  3x
What is the Taylor Series for this function? What can
we say about its convergence?
Next Question . . .
2. If the Taylor Series for f converges, is it equal
to the f on its interval of convergence?
Answer: Often, but not always.

Consider the absolute value function
f ( x)  x .
We know that we cannot expand it in a Taylor
series about x = 0. (Why?)
But f (x) = | x | has derivatives of all orders at all
other points.
What if we consider a Taylor series expansion
about x = 1?
Taylor Series for f (x)= |x| based
at x = 1.
What about the derivatives of f
at x = 1?
What do we get for the Taylor
series expansion at x = 1?
The Taylor Series expansion
for f (x) = | x | converges on
the entire real line, but is equal
to f only on the interval [0,∞)!
Things can get really weird
e  1 x2
f ( x)  
0
if x  0
if x  0
Facts:
• f is continuous and has
derivatives of all orders at x = 0.
• f (n)(0)=0 for all n.
What does all this tell us about the Maclaurin Series for f ?
The Maclaurin Series for f converges everywhere, but is equal
to f only at x = 0!
So where does this leave us?
To Summarize:
Even if we can compute the Taylor Series for a function,
•the Taylor Series does not always converge.
•If the Taylor Series converges, the Taylor Series is not
necessarily equal to the function, even on its interval of
convergence.
We know how to determine whether (and where) the
Taylor series converges---Ratio test!
But how do we know if the Taylor Series of the function
is equal to the function on the interval where it converges?
The answer is . . . Taylor’s Theorem.
Ostebee and Zorn assure us that
Though a Taylor series for f “might converge at all x
but perhaps to a limit other than f, . . .
“Taylor’s theorem guarantees that this
unfortunate event seldom occurs.”
Chapter 4
TAYLOR SERIES TO THE
RESCUE
Most of our familiar calculus functions ARE
given by their Taylor Series, and Taylor’s
Theorem helps us to establish this.
More examples.
Taylor Series
f ( x)  sin( x)
1. Find the Taylor series for f that is based at x = p/4.
2. Show that this Taylor series converges to f for all values of x.
Taylor Series for f (x) = sin(x) at

0
sin( x)
1
cos( x)
2
 sin( x)
3
 cos( x)
4
sin( x)


an= f (n)( p 4)/n!
f (n)( p 4 )
f (n)(x)
n
 4 2
cos p   1
4
2
 sin p    1
4
2
 cos p    1
4
2
sin p   1
4
2
sin p


 1


/4
1 1

2 0!
1 1
a1 

2 1!
1 1
a2  
2 2!
1 1
a3  
2 3!
1 1
a4 
2 4!
a0 


1
2
1
2

2
3
4
5
1
1
1 1
1 1
1 1
1 1
p
p
p
p
p

x

x

x

x

x

4
4
4
4
4
2
2
22
2 3!
2 4!
2 5!
Show that this converges to sin(x)
We start with the general set-up for Taylor’s Theorem.
n 1
K n1
p
sin( x)  Pn ( x) 
x  Notice that I didn’t have to
4 what P was in order to
 n  1! know
n
gather this information. (In
other words, our second
question is independent of our
first.)
What is Kn+1?
It follows that
sin( x)  Pn ( x) 
x p
n 1
4
for all x.
 n  1!
What happens
to this quantity
As n→∞?
Epilogue---Two points of view
Power series as functions
First a
series . . .

n
a
(
x

x
)
n
0
n 0
. . . Then a
function
First a
function . . .
f ( x)
. . . Then a
series

f ( x)   an ( x  x0 )
Taylor Series

n
n 0
Guarantees that f is equal to the
power series where the power
series converges.

n 0
f ( n ) ( x0 )
( x  x0 )n
n!
No a priori guarantee that f is
equal to its Taylor series.
Why the Taylor series, then?
Power series as functions
First a
series . . .

n
a
(
x

x
)
n
0
n 0
. . . Then a
function
First a
function . . .
f ( x)
. . . Then a
series

f ( x)   an ( x  x0 )
Taylor Series

n
n 0
Guarantees that the power series
we started with is, in fact, the
TAYLOR SERIES FOR f .

n 0
f ( n ) ( x0 )
( x  x0 )n
n!
If f is equal to any power series at all,
that power series must be the Taylor
series for f.
That’swhere we look! Taylor’s
Theorem helps!