Taylor’s Theorem
Section 9.3a
While it is beautiful that certain functions can be
represented exactly by infinite Taylor series, it is the
inexact Taylor series that do all the work…
In practical terms, we would like to be able to use
Taylor polynomials to approximate functions over the
intervals of convergence of the Taylor series, and we
would like to keep the error of the approximation
within specified bounds…
Since the error results from truncating the series down
to a polynomial (that is, cutting it off after some
number of terms), we call it truncation error.
Find a formula for the truncation error if we use
1 x  x  x
interval  1,1.
2
4
6
to approximate
1 1  x
2
 over the
This polynomial is the fourth partial sum of the
geometric series for 1 1  x 2 …


The truncation error is the absolute value of the part
10
2n
that we threw away: 8
x x 
x 
This is the absolute value of a geometric series with:
ax rx
a

1 r
8
Therefore,
x x 
8
10
x 
2n
2
Find a formula for the truncation error if we use
1 x  x  x
interval  1,1.
2
4
Therefore,
6
to approximate
x x 
8
10
1 1  x
x 
2n
8
Truncation error:
x
y
2
1 x
2
 over the
a

1 r
8
8
x
x


2
2
1 x
1 x
Graph this error in [–1, 1] by [–5, 5]
The error is very small near 0, but increases as x gets
closer to 1 or –1.
With this last example, the series (and therefore, the
error) was geometric, but Taylor’s Theorem also works
for nongeometric series…
Every truncation splits a Taylor series into equally
significant pieces: the Taylor polynomial Pn x that
gives us the approximation, and the remainder Rn x
that tells us whether the approximation is any good…
 
 
Taylor’s Theorem with Remainder
If f has derivatives of all orders in an open interval I
containing a, then for each positive integer n and for
each x in I,
f   a 
2
f  x   f  a   f   a  x  a  
 x  a 
2!
 n
f a
n

 x  a   Rn  x  ,
n!
where
Rn  x  
 c  x  a n1


 n  1!
f
 n 1
for some c between a and x.
Taylor’s Theorem with Remainder
The first equation in Taylor’s Theorem is Taylor’s
formula. The function Rn x is the remainder of
order n or the error term for the approximation of
f by Pn x over I. It is also called the Lagrange form
of the remainder, and bounds on Rn x found using
this form are Lagrange error bounds.
 
 
 
 
If Rn x  0 as n   for all x in I, we say that the
Taylor series generated by f at x = a converges to f on
I, and we write
k 

f  x  
k 0
f
a
k!
 x  a
k
Proving Convergence of a Maclaurin Series
Prove that the given series converges to sin(x) for all real x.

2 k 1
x
 1

 2k  1!
k 0
k
By Taylor’s Theorem:
f
 n 1
We need to see what happens
to
Rn  x  as n  
Rn  x  
 c  x  0 n1


 n  1!
f
 n 1
 c  is the (n+1)st derivative of sin(x)…
which lies between –1 and 1 inclusive…
 n 1
So for all x:
Rn  x  
 c  x  0 n1


 n  1!
f
Proving Convergence of a Maclaurin Series
Prove that the given series converges to sin(x) for all real x.
 n 1
 n 1
Rn  x  
 c  x  0 n1


 n  1!
f

c
 n  1!
f
x n 1
n 1
What happens to this last
term as n approaches infinity?
x
1
n 1

x 
 n  1!
 n  1!
The factorial growth in the denominator eventually outstrips
the power growth in the numerator:
n 1
x
Rn  x  
0
 n  1!
This completes the proof!!!
Guided Practice
Find the Taylor polynomial of order four for the function at
x = 0, and use it to approximate the value of the function
at x = 0.2.
f x  cos  x 2
x
f  0   cos
f   0  
2

f   0   
2

sin
 

1
x 0
x
2
2
4

cos
0
x 0
x
2

x 0

2
4
Guided Practice
Find the Taylor polynomial of order four for the function at
x = 0, and use it to approximate the value of the function
at x = 0.2.
f  x   cos  x 2 
3
x
f   0  
sin
0
8
2 x 0
f
 4
 0 

4
16
cos
x
2

x 0

4
16
 4 2
 16 4
3
P4  x   1  0 x 
x  0x 
x
2
4!
2
4
Guided Practice
Find the Taylor polynomial of order four for the function at
x = 0, and use it to approximate the value of the function
at x = 0.2.
f  x   cos  x 2 
 4 2
 16 4
3
P4  x   1  0 x 
x  0x 
x
2
4!
2
 1

2
8
x 
2

4
4
384
x
4
f  0.2   P4  0.2   0.951
How close is this approximation to the actual function value?
Guided Practice
Find the Taylor polynomial of order four for the function at
x = 0, and use it to approximate the value of the function
at x = 0.2.
f x  ln 1  x 2

 

Check the table on p.477:
2
3
x
x
ln 1  x   x   
2 3
So…
ln 1  x
2
x
x  x 



2 2
2
2
  1
2 3
3

n 1
  1
n
x

n
n 1
x 
2 n
n

Guided Practice
Find the Taylor polynomial of order four for the function at
x = 0, and use it to approximate the value of the function
at x = 0.2.
f x  ln 1  x 2

 
ln 1  x
2
x
x  x 



2 2
2
2 3
2
4
3
6
x
x
x   
2 3
2
4
x
P4  x   x 
2
2


  1
  1
n 1
n 1
x 
2 n
n

2n
x

n
f  0.2   P4  0.2   0.0392
Guided Practice
Find the Maclaurin series for the given function.
f  x   xe
2
n

x
x
 x 1  x    
2!
n!

3
n 1
x
x
2
 x x   

2!
n!
x



Guided Practice
Find the Maclaurin series for the given function.
2
x
f  x 
1 2x
 1 
x 

 1 2x 
2

 x 1 2x   2x 
  2x 
 x  2x  4x 
n
2
2
2
3
4
n
2 x
n2

