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9.3 Taylor’s Theorem: Error Analysis for Series. Tacoma Narrows Bridge: November 7, 1940 . Last time in BC…. 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:. - PowerPoint PPT Presentation
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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
nn
nx
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:
21ln
23ln
2)15.1()15.1(
2
2)15.0()15.0(
2
375.0225.05.0
625.0225.05.0
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.
21ln
23ln 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)n1 (x 1)n
nn1
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 TheoremFor 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
nf 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 cf n 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
nf 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.
21ln
23ln 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
nxxxxxx
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(
21ln
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
nxxxxxx
Recall that the Taylor Series for ln x centered at x = 1 is given by…
1
)5.0(...3
0625.0225.05.0
21ln
n
n
n
xxf 1)( 1
11)1( f
2
1)(x
xf 1
11)1( 2 f
3
2)(x
xf 33 22)( c
ccf
!3cf
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…
21ln 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 Bound 11
)()!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
2c 3 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)
81
616
31
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
2We know that since this is an alternating series, the error bound would be
x 3
3But let’s apply Lagrange (which works on all Taylor Series)…
3
!3)( xcferror
3)1(
2)(x
xf
The third derivative of ln(1+ x) is
error 2(1c) 3
3!x 3
x 3
3(1c)3The value of c that will maximize the error is 0 so…
x 3
3(1c)3 x 3
3Which 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
nMR 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