Taylor Series
The Coefficients of a 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.
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 already familiar . . . Taylor’s Theorem.
Recall
Taylor’s Theorem
Suppose that f is repeatedly differentiable on an interval I
containing x0 and that
Pn ( x)  a0  a1 ( x  x0 )  a2 ( x  x0 )2  a3 ( x  x0 )3 
 an ( x  x0 ) n
is the nth order Taylor polynomial based at x0.
Suppose that Kn+1 is a number such that for all z in I,
| f ( n1) ( z ) |  K n1.
Then for x in I,
| f ( x)  Pn ( x) | 
K n1
n 1
x  x0
 n  1!
What does this tell us?
Pointing out that a Taylor series for f “might converge at
all x but perhaps to a limit other than f,” Ostebee and
Zorn assure us that
“Taylor’s theorem guarantees that this
unfortunate event seldom occurs.”
Pinning this down
 Recall that Pn is the nth partial sum of theTaylor Series of f
based at x0.
 And thus
| f ( x)  Pn ( x) |
Measures the error made by Pn(x) in approximating f (x).
 Taylor’s theorem gives us an upper bound on this error!
The Taylor series for f will converge to f if and only if for all x
| f (x) - Pn(x) |
goes to zero as n →∞. Taylor’s theorem can help us establish this.
Using Taylor’s Theorem
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.
1. Taylor Series for f (x) = sin(x)

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


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
3!
4!
5!
2
2
2
2
2
2
Show that this converges to sin(x)
We start with the general set-up for Taylor’s Theorem.
n 1
K n1
sin( x)  Pn ( x) 
x p
4
 n  1!
What is Kn+1?
d n1
sin( x)  1 for all n and all x.
n 1
dx
It follows that
sin( x)  Pn ( x) 
x p
n 1
4
for all x.
 n  1!
What happens
to this quantity
As n→∞?
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!
Now it’s your turn
Repeat this exercise with the Maclaurin series for f (x) = cos(2x) .
1. Find the Maclaurin series for f (x) = cos(2x).
2. Show that this series converges to f for all values of x.
1. Taylor Series for f (x) = cos(2x)
n
f (n)(x)
0
cos(2 x)
cos  0  1
a0  1
1
2sin(2 x)
2sin  0  0
a1  0
2
2 cos(2 x)
2 cos  0  2
3
23 sin(2 x)
4
2
4
2 cos(2 x)
f (n)(0)
an= f (n)(0)/n!
2
2
22
a2 
2!
23 sin(0)  0
a3  0
2 cos(0)  2
24
a4 
4!
4
2 2 2 4 4 26 6
1 x  x  x 
2!
4!
6!
4
(1) n 22 n 2 n

x
 2n !
n 0

Show that this converges to
cos(2x)
We start with the general set-up for Taylor’s Theorem.
K n1
n 1
cos(2 x)  Pn ( x) 
x
 n  1!
d n1
n 1
What is Kn+1?
cos(2
x
)

2
for all n and all x.
n 1
dx
It follows that
n 1
2x
cos(2 x)  Pn ( x) 
for all x.
 n  1!
This quantity goes
to 0 as n→∞!