Post on 02-Aug-2020
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9.3 Taylor’s Theorem: Error Analysis for Series
Tacoma Narrows Bridge: November 7, 1940
Last time in BC…
...4
)1(
3
)1(
2
)1()1(ln)(
432 −−
−+
−−−==
xxxxxxf
∑−
−=∞
=
+
1
1 )1()1(
n
n
n
n
x
So the Taylor Series for ln x centered at x = 1 is given by…
Use the first two terms of the Taylor Series for ln x centered at x = 1 to
approximate:
2
1ln
2
3ln
2
)15.1()15.1(
2−−−≈
2
)15.0()15.0(
2−−−≈
375.02
25.05.0 =−=
625.02
25.05.0 −=−−=
068147.0|)625.0()5.0ln(| =−−=erroractual
0417.03
)15.0( 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.
2
1ln
2
3ln 0417.0
3
)15.1( 3
=−
≤→ error
03047.0|625.0)5.1ln(| =−=erroractual
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 −1) −(x −1)
2
2+
(x −1)3
3−
(x −1)4
4... = (−1)n +1 (x −1)
n
nn =1
∞
∑
And now, the exciting
conclusion of Chapter 9…
Since each term of a convergent alternating series
moves the partial sum a little closer to the limit:
This is also a good tool to remember because it is
easier than the Lagrange Error Bound…which you’ll
find out about soon enough…
Muhahahahahahaaa!
Alternating Series Estimation Theorem
For a convergent alternating series, the truncation
error is less than the first missing term, and is the
same sign as that term.
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! !
nn
n
f fa af f f Rx a a x a x a x a x
n
′′′= + + + ⋅⋅⋅ +− − −
Lagrange Error Bound ( )( ) ( )
( )( )
11
1 !
nn
n
f cR x x a
n
++
= −+
In this case, c is the number between x and a that will give
us the largest result for )(xRn
( )( ) ( )
( )( )
11
1 !
nn
n
f cR 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 ( )( )cfn 1+
If our Taylor Series had
alternating terms:
Does any part of
this look familiar?
If our Taylor Series did not
have alternating terms:
nn
n axn
afxR )(
)!1(
)()(
1
−+
=+
( )( ) ( )
( )( )
11
1 !
nn
n
f cR x x a
n
++
= −+
This is just the next term of the
series which is all we need if it is
an Alternating Series
is the part that makes the Lagrange
Error Bound more complicated.
Note that working with( )( )cf n 1+
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! !
nn
n
f fa af f f Rx a a x a x a x a x
n
′′′= + + + ⋅⋅⋅ +− − −
Lagrange Error Bound ( )( ) ( )
( )( )
11
1 !
nn
n
f cR x x a
n
++
= −+
Now let’s go back to our last problem…
Why this is the case involves a mind-bending proof so we just won’t do
it here.
068147.0|)625.0()5.0ln(| =−−=erroractual
0417.03
)15.0( 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.
2
1ln
2
3ln 0417.0
3
)15.1( 3
=−
≤→ error
03047.0|625.0)5.1ln(| =−=erroractual
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
1432 )1(
)1(...4
)1(
3
)1(
2
)1()1(ln
n
nn
n
xxxxxx
First of all, when plugging in ½ for x, what happens to your series?
∑∞
=
+ −−=
−+
−−−=
1
132 )5.0(
)1(...3
)15.0(
2
)15.0()15.0(
2
1ln
n
nn
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 have
to use it to find our error bound in this case
∑∞
=
+ −−=
−−
−+
−−−=
1
1432 )1(
)1(...4
)1(
3
)1(
2
)1()1(ln
n
nn
n
xxxxxx
Recall that the Taylor Series for ln x centered at x = 1 is given by…
∑∞
=
−=−−−=1
)5.0(...
3
0625.0
2
25.05.0
2
1ln
n
n
n
xxf
1)( =′ 1
1
1)1( ==′f
2
1)(
xxf
−=′′ 1
1
1)1(
2−=−=′′f
3
2)(
xxf =′′′ 3
32
2)( −==′′′ c
ccf
( )=
′′′
!3
cf 2c−3
3!
xxf ln)( = 01ln)1( ==f
xxf ln)( =The Taylor Series for centered at x = 1
Since we used terms up through n = 2, we will need to go to n = 3 to find our
Remainder Term(error bound):
This is the part
of the error
bound formula
that we need
The third derivative
gives us this
coefficient:
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 the
Remainder Term which is also called…
2
1ln 0417.0
3
)15.0( 3
=−
≤→ error
068147.0|)625.0()5.0ln(| =−−=erroractual
3)15.0(!3
)(−
′′′=
cferror )125.0(
!3
2 3−
=c The third derivative
of 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 2c–3.
The Lagrange Error Bound1
1
)()!1(
)( ++
−+
nn
axn
cf
3)15.0(!3
)(−
′′′=
cferror )125.0(
!3
2 3−
=c The third derivative
of 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 = 0.5
2
c3 which has its largest value when c is smallest.
And therefore…
3)15.0(!3
)(−
′′′=
cferror )125.0(
!3
2 3−
=c
=2(0.5)−3
3!(0.125)
==8
1
6
16
3
1
068147.0|)625.0()5.0ln(| =−−=erroractual
Which is larger than the actual error!
And we always want the error bound to be larger than the actual error
Let’s try using Lagrange on an alternating series
ln(1+ x) ≈ x −x
2
2
We know that since this is an alternating
series, the error bound would be x3
3
But let’s apply Lagrange (which works on all Taylor Series)…
3
!3
)(x
cferror
′′′≤ 3)1(
2)(
xxf
+=′′′The third derivative
of ln(1+ x) is
error ≤2(1+ c)−3
3!x
3=
x3
3(1+ c)3
The value of c that will
maximize the error is 0 so…
x3
3(1+ c)3=
x3
3
Which is the same as the
Alternating Series error bound
Lagrange Form of the Remainder: ( )( ) ( )
( )( )
11
1 !
nn
n
f cR x x a
n
++
= −+
Remainder Estimation Theorem:
If M is the maximum value of on the interval
between a and x, then:
( ) ( )1nf x
+
( )( )
1
1 !
n
n
MR 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 ( )( )cf n 1+
Remember that the only difference you need to worry
about between Alternating Series error and La Grange is
finding ( )( )cf n 1+