Remainder Estimation Theorem (Lagrange) 9.3
By Collette M. and McKinsey W.
The Taylor Series allows you to approximate a function. An infinite series allows you to get a series
almost identical to the original function. In the first half of 9.3, you learn how to solve for a Taylor Series.
Now, we take it one step further. The longer the Taylor Series, or in other words the greater the nth
term, the closer it gets to the function. If we stop the Taylor Series at n=5, for example, all the other
terms for n > 5 will be an error from the original function. And we want to find that error! Why? Because
we’re cool! We have a master equation that we can plug everything in to find the error. It is Remainder
Estimation Theorem (Lagrange)!
Remainder Estimation Theorem:
The Remainder
(Answer or Error)
The interval
of lxl given
Rn x
f
Constant of the next term evaluated
at the interval given to find a max
(Also known as M)
n 1
c
n 1!
x a
The next term in the Taylor series
without its constant
The last nth term given in the
Taylor Series.
Rn x
n 1
f
n 1
c
n 1!
x a
n 1
Examples:
1. If the function is centered at zero and f(x) = x +
error when lxl < 0.1.
+
approximates
, find the maximum
A) Find the next derivative in the Taylor Series. In this problem, the next derivative would be
(x).
f(x) =
f’(x) =
f”(x) =
f”’(x) =
(x) =
| x=0 = 1
| x=0 = 1
|x=0 = 1
|x=0 = 1
| x=0 = 1
B) Find the constant of the next term evaluated at lxl = 0.1.
We now know
(-0.1) =
(x) =
Now all we have to do is plug in ± 0.1 for x.
= 1.105
= 0.905
*Since 1.105 is > than 0.905, we know we need to
0.1 for x instead of -0.1.
1.105 is our constant for the next term!
C) Plug into the formula!
(0.1) ≤ 1.105
= 4.604 X
2. If the function is centered at x=0 for f(x) = 1 maximum error when lxl < 0.2.
+
is the approximation for cos(x). Find
A) Find the next derivative in the Taylor Series.
f(x) = cos(x)
f’(x) = -sin(x)
f”(x) = -cos(x)
f”’(x) = sin(x)
(x) = cox(x)
(x) = -sin(x)
|x=0 = 1
|x=0 = 0
|x=0 = -1
|x=0 = 0
|x=0 = 1
|x=0 = 0
*This problem is a little different, because
(x)=0. So the next derivative is
(x) = -cos(x) |x=0 = -1
B) Find the constant of the next term evaluated at lxl = 0.2
= -cos(0.2) = -0.980
(-0.2) = -cos(-0.2) = -0.980
*Why are both answers the same? Cos(x) is an
even function so f(x) and f(-x) both= f(x). So you
could use either x. We’ll use 0.2.
Our constant of the next term is -0.980.
C) Plug into equation
(0.2) ≤ -0.980
= -8.711 X
*Notice we used
instead of .
still exists, it is just = 0. If you did use
technically wouldn’t be the right term to use.
it
3. Given the function f(x) = sin(5x). Find the Taylor Series to the fourth term when x is centered
around . Find the maximum error when lxl < 0.2
A) Find the Taylor series.
f(x) = sin(5x)
|x
=1
f’(x)= 5cos(5x)
|x=
=0
f”(x) = -25sin(5x)
|x=
= -25
f”’(x)= -125cos(5x) |x=
(x)= 625sin(5x)
1–
|x=
=0
= 625
+
B) Find the next term. Which if (x) since
(x)= 3,125cos(5x) |x= = 0
(x)= -15,625sin(5x) |x=
(x)= 0.
= -15,625
C) Find the constant of the next term evaluated at lxl < 0.2
(0.2) = -15,625sin(5(0.2)) = -13,147.984
(-0.2) = -15,625sin(5(-0.2)) = 13,147.984
Our constant for the next term is 13,147.948 at x=-0.2
D) Plug it into the equation
(-0.2) ≤ 13,147.948
= 0.337
4. The approximation of
= 1 + is used when x is small. The function is about x=0. Estimate
the MINIMUM error when lxl< 0.01.
A) Find the next term
f(x) =
or
f’(x) =
f”(x)=
| x=0 = 1
|x=0 =
|x=0 =
B) Find the constant of the next term evaluated at lxl<0.01
f”(0.01)=
= -0.246
f”(-0.01)=
= -0.254
Our constant for the next term is -0.254 at x=-0.01
C) Plug into equation
(-0.01) = -0.254
= 8.467 X
5. The approximation ln(1-x) = -x - is used when x is small. The function is about x=0. Use the
Remainder Estimation Theorem to get a bound for the maximum error when lxl<0.1.
A) Find the next term.
f(x)= ln(1-x)
f’(x)=
or –
|x=0 = 0
|x=0 = -1
f”(x)=
f”’(x) =
|x=0 = -1
|x=0 = -1
B) Find the constant of the next term.
f”’(0.1)=
= 1.372
f”’(-0.1)=
= 0.751
The constant of the next term is 1.372 at x= 0.1
C) Plug into the equation
(0.1) = 1.372
= 4.573 X
The reason why there is not (!) factorial is because the Maclaurin Series of ln(1+x) doesn’t
have a factorial (!).
INTERNET SITE:
www.online.math.uh.edu/HoustonACT/Greg_Kelly.../Calc09_3.ppt
Free Response Question
Let h be a function having derivatives of all orders for x>0. Selected values of h and its first four
derivatives are indicated in the table below. The function h and these four derivatives are increasing on
the interval 1 ≤x ≤3
x
h(x)
h’(x)
h”(x)
h”’(x)
1
11
30
42
99
2
80
128
3
317
(x)
18
1. Use Lagrange error bound to show that the third-degree Taylor Polynomial for h about x=1
approximates h(0.9) with error less than 8 x
.
ANSWER:
A) Find the constant of the next term.
h(x)= 11
h’(x)= 30
h”(x)= 42
h”’(x)= 99
(x)= 18
The constant of the next term is 18.
*We are going to use (1) (the constant of the next term) because it is a close
approximation of (0.9)
B) Plug into the equation.
(0.9) ≤ 18
= 7.5 X
7.5 X
< 8.0 X