AP Calculus BC
Taylor's Polynomials and Error Bounds
Name:
Generalized Mean Value Theorem:
Let f be a differentiable function on an open interval ( a , x ) and continuous on a closed interval [a, x] .
Then there exists a point c  (a, x) such that
f '(c) 
f ( x)  f (a )
xa
If we let
n
Pn ( x)  
k 0
f ( k ) (a)
( x  a) k
k!
be the nth degree Taylor Polynomial evaluated around x = a, then Taylor’s theorem says there exists a
c  (a, x) such that
f ( n1) (c) f ( x)  Pn ( x)

(n  1)!
( x  a)n1
The Original Mean Value Theorem is the case where n=0 . Are the statements the same for n=0?
The idea is to proceed by induction, but as you can imagine, it gets a little complicated. So we won't
prove it here.
Taylor's Theorem:
If f is n -differentiable on an interval containing a, (a  ò, a  ò) ,then there exists a number c  ( a, b)
such that
f (a )
f ( x)  f (a )  f '(a )( x  a ) 
( x  a)2 
2!
 Pn ( x)  Rn ( x)
f ( n ) (a )

( x  a ) n  Rn ( x)
n!
f ( n1) (c)
We call Rn ( x) 
( x  a) n1 the remainder of the nth degree Taylor Polynomial, Pn ( x) .
(n  1)!
Taylor Series and Polynomials, L. Marizza A. Bailey
AP Calculus BC
Taylor's Polynomials and Error Bounds
Name:
The purpose of Taylor’s Theorem
ln(x) is differentiable at x = 1
The Taylor Polynomial of
degree 5 is a good estimation,
but the error increases as x
gets further away from 1.
The error at 2 is given by
The error at x = 2, is .109814. How
can we estimate the error at each
point?
Taylor Series and Polynomials, L. Marizza A. Bailey
AP Calculus BC
Taylor's Polynomials and Error Bounds
Name:
LaGrange’s Error Bound
The idea behind this uses Taylor’s Theorem, but the statement is as follows:
If f ( x) has n  1 derivatives and | f ( n 1) (c) | M for all c  (a, x),
then the remainder |R n ( x) | M
| x  a |n 1
(n  1)!
Practice Problem: [1999 BC 2]
The function f has derivatives of all orders for all real numbers x.
Assume f (2)  3, f '(2)  5, f ''(2)  3, and f '''(2)  8.
a) Write the third degree Taylor Polynomial for f about x = 2 and use it to approximate f ( 1.5) .
b) The fourth derivative of f satisfies the inequality f (4) ( x)  3 for all x in the closed interval [1.5, 2] .
Use the Lagrange Error Bound on the approximation to f (1.5) found in part (a ) to explain why
f (1.5)  5.
c) Wrote the fourth degree Taylor Polynomial, P ( x ) for g ( x)  f ( x 2  2) about x = 0.
Taylor Series and Polynomials, L. Marizza A. Bailey