Taylor Theorem and Remainder Estimation
One of the questions you should have with the Taylor Series and its approximation of
function f, is “how accurate is the approximation”.
Taylor’s Theorem with Remainder
If function f has derivatives of all orders in an open interval I containing a, then for
each positive integer n and for each value x in the interval I
n
f
a
f 
a

2

n
f 
x f 
a f 

a 
x a 
x a ... 
x a Rn 
x
2!
n!
f n 1
c
n
1
Where Rn 
x 
x a for some c between a and x
!
n 1
That is
Rn 
x f 
x Pn 
x
Rn 
x is often called the remainder term. This remainder term serves two purposes;
it enables us to obtain an estimate of the error in using a Taylor polynomial to
approximate a given function. Also
Rn 
x provides the means to prove that a Taylor
series for a given function f actually converges to f.
Example
Use a Taylor polynomial with
P4 
x to approximate the value of ln 
1.2 and estimate
the error in the value.
Solution:
k
f 

c
k
P4 
x 
x c 
k!
k 0
f 
0
f
0
f
0
f

0



0
1
2
3

x 0
x 0 
x 0
x 0 ...
0!
1!
2!
3!

4
2
3


f
0 
x
f
0 
x
f 
0 
x4


f 
0 f 
0 
x



2!
3!
4!
1
2
6

x2

x3
2
3
4
1 0
1 0 
1 0  4
1



ln 
1 0 

x


x
1 0
2!
3!
4!
x 2 x3 x4
0 x   
2
3 4
2
3
x
x
x4
x   
2
3 4
4
To estimate
ln 
1.2 , we apply the series ln 
1 x when x=0.2
ln 
1.2 ln 
1 0.2
0.2 0.2 0.2
2

0.2 
3
4


2
3
4
0.2 0.02 0.0026666 0.0004
0.1822666
We can use the remainder term to estimate the error in this approximation.
error ln 
1.2 P4 
1.2 R4 
1.2 

c

4
1

1.2 1 
4 1!
f
41
4! c

5!
5

0.2 
5
Where c is between 1 and 1.2. This gives us the following bound on the error:
5
0.2
error 
5c 5
5
0.2

0.0000397
5
5

1.1
The approximation of
ln 
1.2 0.1822666 0.00004
Example
Use the fifth degree Maclaurin polynomial to approximate
sin 
0.3and determine the
accuracy of the approximation.
Solution:
The Maclaurin series for
sin 
x is:
x3 x5 x7
n 1
x 2n 1
x    ... 
1
...
3! 5! 7!
2 n 1
!

Then
x3 x5
P5 
x x  
3! 5!
x3 x5
x  
6 120
Therefore
3
5

0.3 
0.3
f
0.3P5 
0.3
0.3
6

120
0.29552025
6
f 

c 
0.3
The remainder term R5 
0.3
where c is between 0 and 0.3
6!
6
c 
0.3
sin 

0.000001013
R5 
0.3
6
6!
Therefore we obtained an answer accurate to 6 decimal places.