Taylor Series
&
Error
n
Series and Iterative methods
Any series ∑xn can be turned into an
iterative method by considering the
sequence of partial sums sn.
sn 
x
k
 x 0  x1  x 2    x n
k 0
s n  s n 1  x n
Functions and Taylor Expansions
From Intermediate Calculus we know any function that has n+1 derivatives at a
point a has a nth Taylor Polynomial expansion centered at a and an expression for
the error.
f x  
n

f
k 
k!
k 0
f  x   f a  
Rn x  
a 
 x  a k
f a 
'
1!
 error
x  a  
f
c 
n 1
x  a 
 n  1 !
f
 n 1 
''
a 
2!
 x  a    
2
f
n 
a 
n!
 x  a n  R n  x 
The value f(k)(a) is the kth derivative
evaluated at a. The function Rn(x)
represents the error where c is a value
between x and a.
Examples of Taylor Series
Because the higher derivatives of some functions repeat in a recognizable pattern
we are able to get a few "standard" Taylor Series expansions. All of the examples
below have the center point of 0 and the error term will go to zero for suitable
values of x. Other more complicated functions can be found using the algebra,
derivatives and integrals.
n
1
1 x

x
2
3

k 0
x
k
k!
 1
x
x

1!
2

2!
  1 k x 2 k 1
x
 2 k  1 !
k 0
x
sin
  1 k x 2 k
x
 2 k !
k 0
 x
n
cos
 1
3
 
3!
n
x
3

2
2!
x

x

 
n2
x
n 1
c
 n  1 !
 
4
4!
e
5
5!
x
1  c 
n
n!
3!
x
1
n
k 0

e 

x  1 x  x  x   x 
k
x
x
n 1
2 n 1

 sin c
 2 n  1 !  2 n  2 !
x
2n

 cos c
 2 n !  2 n  1 !
x
2n2
x
2n2
To find the Taylor series for the function f(x) = arctan(x).

1

1 x
x
 1 x  x  x 
k
2
3
Start with known Taylor Series
k 0

1

1 x
k



x

Substitute –x for x
 1 x  x  x 
2
3
k 0

1
1 x
2

  1
k
x
2k
 1 x  x  x 
2
4
6
k 0
  1 k x 2 k 1

arctan x 

2k  1
k 0
 x
3
x

x
3
5

x
5

cosh x 
x
2k
 2 k !  1 
k 0

x
2

2!
x
k
x
k 0

3
x cosh
k 3
 x  
 x
 2 k !
k 0
x

4!
 x  
 1

 2 k !
2!
x
4
3

x
7
7
To find the Taylor series for the function
cosh
Substitute x2 for x

Integrate both sides.
g  x   x cosh
3
 x
6

Start with known Taylor Series
6!
x
2

4!
x
3
Substitute

6!
4
2!
x

x
5
4!

x
6
6!

Multiply by x3
x for x
A second version of Taylor Polynomials
For any function f(x) consider the function
f(x+h) where we think of x as a constant and h
is the variable. Consider the Taylor expansion
for f(x+h) (with variable h) centered at 0. The
kth derivatives are found to the right.
f x  h  
n

f
k 0
k 
x 
k!
h  f x  
k
f
'
x 
1!
h
f
''
x 
d f x  h 
k
dh
k
h  
2
 f
k 
x  h  h0
 f
k 
h0
k 
f
2!
x 
h 
k
f
 x  c  k 1
h
 k  1 !
 k 1 
k!
This is a more computationally useful way of applying Taylor Series to
approximate functions. We think of the value of h as small (close to zero) and
will represent the distance between computable derivatives and x. Consider the
following example.
Approximate sin(30.01º) with an error of less than 0.00000001.
h

  sin 30

 . 01

. 01 
180
We first need to convert to radians to get the value for h.
sin 30 . 01
x 
  sin 30  .01    sin 


180
6

. 01 
180

We need to figure out the value of k that
 sin
gives the desired accuracy. The exact
Error 
error is given by the expression to the
right.
 6
 c  or  cos  6  c   . 01  


 k  1!
 180 
We know the values for sine and cosine are always less than 1 in absolute
value and  is less than 4. Try different values of k that make this inequality true.
Error 
 sin
 6
 c  or  cos  6  c   . 01 


k  1!
 180 
k
 1 



k  1!  4500 
1
k 1
 . 00000001
When we try k=2 we find the right side is 0.000000000001829. This means
we need go out to the k=2 term.

sin 30 . 01

  sin 

6

. 01 
180
01 
01  2
  sin  6   cos  6  .180
  sin  6  .180

2
3  . 01    1   . 01  


   
  0.50015113 4716148
2
2  180   2   180 
1
sin 30 . 01

  0.50015114
2330817...
k 1
Alternating Series and Error
From intermediate calc remember that a series ∑(-1)kak converges if the
sequence ak are positive terms that decrease to zero, this was called the
alternating series test. The series ∑(-1)kak is called an Alternating Series. The
nice thing about alternating series is that keeping tract of the error is very easy as
given by the theorem below.
Alternating Series Error Theorem
In any convergent alternating series ∑(-1)kak the error is always less than the
first term omitted. In other words:
|Error| < ak+1
Example: Compute 1/e accurate to .0001
1
e
1
 11
e
1
2!
1
1


3!
1
The series for 1/e is alternating so the theorem applies.

4!
We need to go to the k=7 term.
 0.00002480 16  .0001
8!
1
e
 11
1
2!

1
3!

1
4!

1
5!

1
6!

1
7!
 0.36785714 29
1
e
 0.36787944 1 