Transcript erroranalysis - Mr. Murphy's Website

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…
( x  1) ( x  1) ( x  1)
f ( x)  ln x  ( x  1) 


...
2
3
4
( x  1)
  (1)
n
2

3
4
n
n 1
n 1
Use the first two terms of the Taylor Series for ln x centered at x = 1 to
approximate:
3
0.25
(1.5  1) 2
ln  (1.5  1) 
 0.5 
 0.375
2
2
2
1
(0.5  1) 2
0.25
ln  (0.5  1) 
 0.5 
 0.625
2
2
2
Recall that the Taylor Series for ln
x centered at x = 1 is given by…

n
(x 1)2 (x 1)3 (x 1)4
(x
1)
f (x)  ln x  (x 1) 


... (1)n 1
2
3
4
n
n 1
Find the maximum error bound for each approximation.
Because the series is alternating, we can start with…
3
(
1
.
5

1
)
3  error 
 0.0417
ln
3
2
actualerror | ln(1.5)  0.625| 0.03047
3
(
0
.
5

1
)
1  error 
 0.0417
ln
3
2
actualerror | ln(0.5)  (0.625) | 0.068147
Wait! How is the actual error bigger than the error bound for ln 0.5?
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:
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.
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!
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:
f   a 
f n  a 
2
f  x   f  a   f   a  x  a  
 x  a   
 x  a n  Rn  x 
2!
n!
Lagrange Error Bound
Rn  x  
 c  x  a n1


 n  1!
f
 n 1
In this case, c is the number between x and a that will give
us the largest result for Rn (x)
Does any part of
this look familiar?
f  n 1  c 
Rn  x  
 n  1!
 x  a
n 1
This remainder term is just like the Alternating Series error (note
that it uses the n + 1 term) except for the
If our Taylor Series had
alternating terms:
f n1 (a)
Rn ( x) 
( x  a) n
(n  1)!
This is just the next term of the
series which is all we need if it is
an Alternating Series
f n1 c 
If our Taylor Series did not
have alternating terms:
Rn  x  
f  n 1  c 
 n  1!
 x  a
n 1
n1
c
f
Note that working with
is the part that makes the Lagrange
Error Bound more complicated.
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:
f   a 
f n  a 
2
f  x   f  a   f   a  x  a  
 x  a   
 x  a n  Rn  x 
2!
n!
Lagrange Error Bound
Rn  x  
 c  x  a n1


 n  1!
f
 n 1
Why this is the case involves a mind-bending proof so we just won’t do
it here.
Now let’s go back to our last problem…
Recall that the Taylor Series for ln
x centered at x = 1 is given by…
n

( x  1) 2 ( x  1)3 ( x  1) 4
(
x

1
)
ln x  ( x  1) 


...   (1) n1
2
3
4
n
n 1
Find the maximum error bound for each approximation.
Because the series is alternating, we can start with…
3
(
1
.
5

1
)
3  error 
 0.0417
ln
3
2
actualerror | ln(1.5)  0.625| 0.03047
3
(
0
.
5

1
)
1  error 
 0.0417
ln
3
2
actualerror | ln(0.5)  (0.625) | 0.068147
Wait! How is the actual error bigger than the error bound for ln 0.5?
Recall that the Taylor Series for ln
x centered at x = 1 is given by…
n

( x  1) 2 ( x  1)3 ( x  1) 4
(
x

1
)
ln x  ( x  1) 


...   (1) n1
2
3
4
n
n 1
First of all, when plugging in ½ for x, what happens to your series?
n

1
(0.5  1) 2 (0.5  1)3
(

0
.
5
)
ln  (0.5  1) 

...   (1) n1
2
2
3
n
n 1

1
0.25 0.0625
(0.5) n
ln  0.5 

...  
2
2
3
n
n 1
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
Since we used terms up through n
Remainder Term(error bound):
The Taylor Series for
f ( x)  ln x
1
f ( x) 
x
1
f ( x)  2
x
2
f ( x)  3
x
= 2, we will need to go to n = 3 to find our
f ( x)  ln x
centered at x = 1
f (1)  ln1  0
The third derivative
1
f (1)   1 gives us this
1
coefficient:
1
f (1)   2  1
3
f c  2c
1

3!
3!
2
3
f (c )  3  2c
c
This is the part

of the error
bound formula
that we need
3
(
0
.
5

1
)
1  error 
 0.0417
ln
3
2
actualerror | ln(0.5)  (0.625) | 0.068147
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…
The Lagrange Error Bound
f n1 (c)
( x  a) n1
(n  1)!
3



2
c
f (c )
(0.125)
error 
(0.5  1)3 
3!
3!
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.
3
2
c
f (c)
(0.125)
error 
(0.5  1)3 
3!
3!
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
And therefore…
2
c3
which has its largest value when c is smallest.
c = 0.5
3
3
(c)
f
2(0.5)
error 
(0.5  1)3  2c (0.125) 
(0.125)
3!
3!
3!
16 1 1


6 8 3
Which is larger than
the actual error!
And we always want the error bound
to be larger than the actual error
actualerror 
| ln(0.5)  (0.625) | 0.068147
Let’s try using Lagrange on an alternating series
x2
ln(1 x)  x 
2
We know that since this is an alternating
series, the error bound would be x 3
3
But let’s apply Lagrange (which works on all Taylor Series)…
f (c) 3
error 
x
3!
The third derivative
of ln(1+ x) is
2(1 c) 3 3
x3
error 
x 
3!
3(1 c) 3
x3
x3
3 
3(1 c)
3

2
( x) 
f 
(1  x) 3
The value of c that will
maximize the error is 0 so…
Which is the same as the
Alternating Series error bound
Most text books will describe the error bound two ways:
Lagrange Form of the Remainder: Rn  x  
f  n 1  c 
 n  1!
 x  a
n 1
and
If M is the maximum value of
between a and x, then:
Remainder Estimation Theorem:
f
 n 1
 x
on the interval
M
n 1
Rn  x  
xa
 n  1!
Note from the way that it is described above that M is just
another way of saying that you have to maximize f n1 c

Remember that the only difference you need to worry
about between Alternating Series error and La Grange is
finding f n1 c
