Taylor Series Error
I) Lagrange Error Bound
f ( n 1) ( z )( x  a)( n 1)
= error bound
(n  1)!
Where
a is where the series is centered
z is a value between a and x (z giving the largest value for f ( n 1) ( z ) .
(z is usually a or x)
-For sin or cos f ( n 1) ( z ) = 1, (even if all z’s give smaller values).
Polynomial value  error bound = range of possible values of the series
Ex 1:
Find the 3rd degree polynomial approximation for e x at 1, centered at 0.
Find the range of possible values for e x at 1, centered at 0. Use the Lagrange error
equation.
ex 
xn
n!
P3 ( x)  1  x 
P3 (1) 
x 2 x3

2 6
8
3
f (31) ( z )(1)(31)
(3  1)!
0<z<1
z=1
f (31) (1)  e
e(1)(4)
e

(4)! 24
Range:
8 e 
  
 3 24 
Ex 2:
Find the 4th degree Maclaurin polynomial approximation for cos(x) where a = 0,
evaluated at 1. Find the Lagrange error bound.
(1)( n ) x 2 n
cos(x) =
(2n)!
P4 ( x)  1 
P4 (1) 
x2 x4

2 4!
37
24
f (2 n2) ( z )( x)(2 n2)
(2n  2)!
f 6 ( z)  1
(1)(1)(6) 1
1
 
(6)!
6! 720
Ex 3:
What degree of Taylor Polynomial for ln(1.2) might have an error < .001
f ( n 1) ( z )( x  a)( n 1)
(n  1)!
(1)( n 1) (n  1)!
f ( x) 
xn
n
n!
n 1
(1.2  1)n1
n ! (.2) n1
1  .2 
z n1


 
(n  1)!
(n  1)! z n1
n 1  z 
z=1

1
n 1
.2   .001
n 1
n = 2.5  n = 3
II) Alternating Series Error

In a series
 (1)
n 1
n 1
an :
The upper bound of the error is found by the n + 1 term.
error < an1
Ex 1:

What degree polynomial for the function

n 0
(1) x n
has an error less than 1/4 for x = 1
n!
n
2
x
2
Third term = 1/6 < 1/4
P2 ( x)  1  x 
III) Actual Error
If the actual value of the function is available, the error of a polynomial estimation can be
found by subtracting the polynomial value from the actual value.
f(x) - Pn ( x) = error
Ex. 1:
What is the error for the fourth degree polynomial approximation of cosx when x =
P4 ( x)  1 
x2 x4

2 24

P4 ( )  .707429
4

cos( )  .707107
4


Error = cos( )  P4 ( )  .000322
4
4

4