Section 9.7 – Taylor’s
Theorem
Taylor’s Theorem
If 𝑔 is a function with 𝑛 + 1 continuous
derivatives on 𝑎, 𝑥 (or 𝑥, 𝑎 ), then there is
a number 𝑐 in (𝑎, 𝑥) (or 𝑥, 𝑎 ) such that
The approximate value of 𝑔 𝑥 using the
Taylor Polynomial centered at 𝑥 = 𝑎
g  x   g  a   g   a  x  a  
The actual
value of 𝑔 𝑥
g   a  x  a 
2!
2
 ... 
g
(n)
 a  x  a 
n!
n

g
( n 1)
 c  x  a 
 n  1!
n 1
An unknown function of
𝑥 because a derivative
is evaluated at some
point 𝑐 in the interval.
Like all of the “Value Theorems,” this is an existence theorem.
Taylor’s Theorem Re-Worded
If 𝑇𝑛 𝑥 is the nth degree Taylor Polynomial
for 𝑔 at 𝑥 = 𝑎, then
The approximate
value.
𝑔 𝑥 = 𝑇𝑛 𝑥 + 𝑅𝑛 𝑥
The actual
value.
The amount needed for the approximate
to be equal to the actual.
where 𝑅𝑛 is the “remainder.”
Technical Note: 𝑬𝒓𝒓𝒐𝒓 = 𝑹𝒏 𝒙
Example 1
Let 𝑔 𝑥 = 𝑒 𝑥 , express 𝑔 2 in terms of a thirddegree Taylor polynomial centered at 𝑎 = 0 and
the remainder.
2
3
𝑥
𝑥
Find the third degree
𝑇3 𝑥 = 1 + 𝑥 + +
Taylor polynomial:
2! 3!
Use the Taylor Theorem Equation:
𝑔 2 = 𝑇3 2 + 𝑅3 2
2
3
2
2
𝑒 2 = 1 + 2 + + + 𝑅3 2
2! 3!
7.389 = 6.333 + 𝑅3 2
We can calculate this value since we know the function. But what
if we did not know the function?
The Remainder in Taylor’s Theorem
Also known as the Lagrange Form of the
Remainder, for some 𝑐 in the interval 𝑎, 𝑥 .
Rn  x  
g
( n 1)
 c  x  a 
 n  1!
n 1
According to Taylor’s Theorem, the remainder (or error) is one degree
higher than the Taylor Polynomial used to approximate the actual value.
Remember, this is the same result from power zooming!
Example 1 Continued
Let 𝑔 𝑥 = 𝑒 𝑥 , express 𝑔 2 in terms of a thirddegree Taylor polynomial centered at 𝑎 = 0 and
the remainder.
𝑔 𝑛+1 𝑐 𝑥 − 𝑎
𝑛+1 !
7.389 = 6.333 + 𝑅 2
What would
happen if we
could find the 7.389
Taylor
polynomial but
didn’t know the 7.389
function and
could not
1.0557
calculate this
value?
= 6.333 +
3
𝑔 3+1
𝑐 2−0
3+1 !
𝑛+1
3+1
This equation is
𝑒 𝑐 24
= 6.333 +
true for some
4!
𝑐
4
value 𝑐 on 0,2
𝑒 2
=
4!
Notice 0 < 𝑐 < 2
𝑐 = 0.45969
Example 1 Continued
Let 𝑔 𝑥 = ? , express 𝑔 2 in terms of a thirddegree Taylor polynomial centered at 𝑎 = 0 and
the remainder.
𝑔 𝑛+1 𝑐 𝑥 − 𝑎
𝑛+1 !
𝑔 2 = 6.333 + 𝑅 2
3
𝑔 3+1
We can
usually NOT 𝑔 2 = 6.333 +
find the
value for 𝑐.
𝑒 𝑐 24
𝑔 2 = 6.333 +
4!
𝑐 2−0
3+1 !
𝑛+1
3+1
All Taylor’s Theorem
says is this equation
MUST be true for
SOME value 𝑐 on 0,2
Seek Comfort in the Familiar: Taylor’s Theorem, like the Intermediate Value,
Extreme Value, and Mean Value Theorem is an existence theorem.
Example 1 Continued
Let 𝑔 𝑥 = ? , express 𝑔 2 in terms of a thirddegree Taylor polynomial centered at 𝑎 = 0 and
the remainder.
Instead of actually calculating the remainder (or error). The best we can
usually do is find a bound for the error. In other words, something the error
is always less than. The worst case scenario for your approximation.
𝑒 𝑐 24
𝑔 2 = 6.333 +
4!
Try to find the maximum value of the error
𝑒 𝑐 24
is strictly
4!
increasing on 0,2 .
𝑒 𝑐 24
4!
Thus
reaches
a maximum when
𝑐 = 2.
𝑒 𝑐 24
4!
on 0,2
Therefore
𝐸𝑟𝑟𝑜𝑟 ≤
𝑒 2 24
4!
OR
𝐸𝑟𝑟𝑜𝑟 ≤ 4.926
The Lagrange Error Bound
One useful consequence of Taylor’s
Theorem is that:
The ABSOLUTE VALUE of
the maximum value of
𝑔 𝑛+1 𝑥 on 𝑎, 𝑥
The Error
Rn  x  
M xa
n 1
 n  1!
Use the
absolute
value to
keep things
positive
when
calculating
error.
Example 2
Determine the degree of the Maclaurin polynomial
in order to approximate the square root of 𝑒 to
within 0.0001.
We want to use 𝑔 𝑥 = 𝑒 𝑥 to approximate 𝑒 0.5 .
Investigate the error function on the interval 0,0.5 :
The derivative
is always 𝑒 𝑥
The series is centered at 0 so 𝑎 = 0
𝑔
𝑛+1
𝑐 𝑥−𝑎
𝑅𝑛 𝑥 =
𝑛+1 !
Worst case scenario: 𝑐 = 0.5
𝑒 𝑐 0.5 − 0 𝑛+1
𝑅𝑛 0.5 =
𝑛+1 !
The error has to be
less than 0.0001
𝑒 0.5 0.5 𝑛+1
0.0001 ≥
𝑛+1 !
𝑛+1
We are trying
to approximate
𝑒 0.5 , so 𝑥 = 0.5.
Use a calculator to
find a value for
𝑛 that satisfies the
inequality.
n
Max Error
0
0.82436
1
0.20609
2
0.03435
3
0.00429
4
0.00043
5
0.00004
5th Degree Polynomial
Example 3
If the third degree Maclaurin polynomial for sin 𝑥 is used to
approximate sin 0.1 , determine the accuracy of the
approximation.
We want to use 𝑔 𝑥 = sin 𝑥 to approximate sin 0.1 .
Investigate the error function on the interval 0,0.1 :
We are trying to
approximate sin 0.1, so
𝑥 = 0.1.
𝑅𝑛 𝑥 =
Worst case scenario is that
the derivative equals 1
𝑅3 0.1 =
𝑅3 0.1
𝑔
𝑛+1
𝑔
3+1
𝑐 𝑥 − 𝑎 𝑛+1
𝑛+1 !
𝑐 0.1 − 0 3+1
3+1 !
1 ∙ 0.14
<
4!
Or 𝑅3 0.1
Using the 3rd degree
polynomial, so 𝑛 = 3.
< 0.000004
Extension: Taylor Theorem and The Mean
Value Theorem
Let 𝑛 = 0 in Taylor’s Theorem for some 𝑐 in 𝑎, 𝑏 :
Taylor
Series is 1
term long
f  b   f  a   f '  c  b  a 
Remainder
Solve for the
derivative
f b  f  a 
f 'c 
b  a 
Mean Value Theorem
is a special case of
Taylors Theorem.