9.3 Taylor’s Theorem: Error Analysis for Series
Tacoma Narrows Bridge: November 7, 1940
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9.3 Taylor's Theorem: Error Analysis for Series, Lecture notes of Signals and Systems
Recall that the Taylor Series for ln x centered at x = 1 is given by… Find the maximum error bound for each approximation. 2. 1 ln. 2. 3.
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Taylor’s Theorem: Error Analysis for Series
Tacoma Narrows Bridge: November 7, 1940
Last time in BC…
ln( |^
error
actual
3
error
Recall that the Taylor Series for
ln
x
centered at
x
= 1 is given by…
Find the maximum error bound for each approximation.
ln
ln
3
error
ln( |^
error
actual
Wait!
How is the actual error bigger than the error bound for ln 0.5?
Because the series is alternating, we can start with…
f
(
x
)
=
ln
x
=
(
x
−
−
(
x
−
2
2
(
x
−
3
3
−
(
x
−
4
4
...
=
(
−
n
1
(
x
−
n
n
n
=
1 ∞ ∑
And now, the excitingconclusion of Chapter 9…
Taylor’s Theorem with Remainder
If
f
has derivatives of all orders in an open interval
I
containing
a
, then for each positive integer
n
and for each
x
in
I
:
(
)
(
)
(
)(
)
(
)
(
)
(^
)^
(
)
(
)
(
)
2
2!
! n
n
n
f^
f
a
a
f^
f
f^
R
x
a
a
x
a
x
a
x
a
x
n
′′
′
=
⋅⋅⋅
−
−
−
Lagrange Error Bound
(
)
(^
)^
(
)
(
)
(
)
1
1
n
n
n
f
c
R
x
x
a
n
In this case,
c
is the number between
x
and
a
that will give
us the largest result for
)
(
x
R
n
(
)
(^
)^
(
)
(
)
(
)
1
1
n
n
n
f
c
R
x
x
a
n
This remainder term is just like the Alternating Series error
(note
that it uses the
n
- 1 term)
except for the
(^
)^
(
) c
f
n
1
Does any part ofthis look familiar? If our Taylor Series hadalternating terms:
If our Taylor Series did not have alternating terms:
n
n
n
a
x
n
a
f
x
R
1
(
)
(^
)^
(
)
(
)
(
)
1
1
n
n
n
f
c
R
x
x
a
n
This is just the next term of theseries which is all we need if it isan Alternating Series
Note that working withis the part that makes the LagrangeError Bound more complicated.
(^
)^
(
) c
f
n
1
ln( |^
error
actual
(^0417). 0
3
) 1
(^5). 0 (
3
error
Recall that the Taylor Series for
ln
x
centered at
x
= 1 is given by…
Find the maximum error bound for each approximation.
ln
ln
3
error
ln( |^
error
actual
Wait!
How is the actual error bigger than the error bound for ln 0.5?
Because the series is alternating, we can start with…
∑
∞ =
1
1
4
3
2
ln
n
n
n
n x x x x x x
First of all, when plugging in
½
for
x
, what happens to your series?
∑
∞ =
1
1
3
2
ln
n
n
n
n
Note that when
x
=
½
,^
the series is no longer alternating.
So now what do we do?Since the Remainder Term will work for any Taylor Series, we’ll haveto use it to find our error bound in this case
∑
∞ =
1
1
4
3
2
ln
n
n
n
n x x x x x
Recall that the Taylor Series for x
ln
x
centered at
x
= 1 is given by…
∑
∞ =
− = − − − =
1
ln
n
n
n
We saw that plugging in ½ for
x
makes each term of the series positive and
therefore it is no longer an alternating series. So we need to use theRemainder Term which is also called…
ln
3
error
ln( |^
error
actual
c
f
error
3 −
c
The third derivativeof ln
x
at
x
=
c
What value of
c
will give us the maximum error?
Normally, we wouldn’t care about the actual value of
c
but in this case, we
need to find out what value of
c
will give us the maximum value for 2
c
.
The Lagrange Error Bound
1
1
n
n
a
x
n
c
f
c
f
error
3 −
c
The third derivativeof ln
x
at
x
=
c
The question is what value of
c
between
x
and
a
will give us the maximum error?
So we are looking for a number for
c
between 0.5 and 1.
Let’s rewrite it as
c
(^2) c
3
which has its largest value when
c
is smallest.
And therefore…
c
f
error
3 −
c
−
3
ln( |^
error
Which is larger thanthe actual error! actual
And we always want the error boundto be larger than the actual error
Lagrange Form of the Remainder:
(^
)^
1
1
n
n
n
f
c
R
x
x
a
n
Remainder Estimation Theorem:
If
M
is the maximum value of
on the interval
between
a
and
x
, then:
(^
)^
1 n
f
x
1
n
n
M
R
x
x
a
n
Most text books will describe the error bound two ways:
and
Note from the way that it is described above that
M
is just
another way of saying that you have to maximize
(^
)^
c
f
n
1
Remember that the only difference you need to worryabout between Alternating Series error and La Grange isfinding
(^
)^
c
f
n
1