MATH 118, LECTURE 31:
TAYLOR REMAINDERS
1
Taylor Remainders
In applications where we are required to use the Taylor series expansion of a
function, we are not able to compute all of the terms in the series since there
are infinitely many. As we did with standard series a few weeks ago, we will
have to truncate the series at some index and hope that the truncated series
is “close enough” to our desired limit to satisfy our particular application.
Necessarily, when we truncate the series, there is going to be some error
between the true value of the limiting function
f
(
x
) and the truncated Taylor
series expansion.
If we could get a handle on the size of this error, and
show that it is in fact small enough to satisfy us, we would feel confident
in using our truncate Taylor series expansion to approximate the desired
function/solution.
In fact, we have the following result:
Proposition 1.1.
If
f
(
x
)
and its first
n
derivatives are continuous on the
closed interval between
c
and
x
, and if
f
(
x
)
has an
(
n
+ 1)
th
derivative on
the open interval between
c
and
x
, then there exists a point
z
n
between
c
and
x
such that
f
(
x
) =
P
n
(
x
) +
R
n
(
x
)
(1)
where
P
n
(
x
) =
n
summationdisplay
k
=1
f
(
k
)
(
c
)
k
!
(
x

c
)
k
,
R
n
(
x
) =
f
(
n
+1)
(
z
n
)
(
n
+ 1)!
(
x

c
)
n
+1
.
P
n
(
x
) is called the
Taylor polynomial
and
R
n
(
x
) is called the
Taylor
remainder
.
We notice that equation (1) is
exact
—that is to say, there is a point
z
n
in
between
x
and
c
such that
R
n
(
x
) tells us
exactly
what the truncation error
is. The trouble is that we generally cannot easily find which
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '09
 Soares
 Derivative, Taylor Series, Analytic function, Taylor's theorem

Click to edit the document details