12.8 Taylor’s Theorem: Error Analysis for Series
Tacoma Narrows Bridge: November 7, 1940
Previously 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:
Recall that the Taylor Series for ln x centered at x = 1 is given by…
Find the maximum error bound for each approximation.
Because the series is alternating, we can start with…
Error Bound > Actual Error
0.0417 > 0.03047
Wait! How is the actual error bigger than the error bound for ln 0.5?
And now, the exciting
conclusion of Chapter 12…
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:
Lagrange Error Bound
In this case, c is the number between x and a that will give us the largest result for
Does any part of this
look familiar?
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:
If our Taylor Series did not have
alternating terms:
This is just the next term of the series Note that working with
which is all we need if it is an
is the part that makes the Lagrange Error
Alternating Series
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:
Lagrange Error Bound
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…
Find the maximum error bound for each approximation.
Because the series is alternating, we can start with…
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…
First of all, when plugging in ½ for x, what happens to your series?
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 = 2, we will need to go to n = 3 to find our Remainder
Term(error bound):
The Taylor Series for
centered at x = 1
The third derivative gives us this
coefficient:
This is the part of the
error bound formula
that we need
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
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 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…
which has its largest value when c is smallest.
c = 0.5
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
We know that since this is an alternating series, the error
bound would be
But let’s apply Lagrange (which works on all Taylor Series)…
The third derivative of
ln(1+ x) is
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:
and
If M is the maximum value of
then:
on the interval between a and x,
Remainder Estimation Theorem:
Note from the way that it is described above that M is just another way of saying that
you have to maximize
Remember that the only difference you need to worry about between Alternating Series
error and La Grange is finding
We are done.
We are done. Just Kidding.
f
7
 2
7!
3  0
7
f
7
 3
7!
3  0
7
f
7
 0
7!
3  0
7
f
 6
c
6!
 0.5  0 
6
200 1
6

 0.5  0 
6!
200
6
 0.5  0.0043
6!