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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.
Wait! How is the actual error bigger than the error bound for ln 0.5?
Because the series is alternating, we can start with…
Error Bound > Actual Error 0.0417 > 0.03047
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
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:
Does any part of this look familiar?
If our Taylor Series did not have alternating terms:
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
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
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.
Recall that the Taylor Series for ln x centered at x = 1 is given by…
Find the maximum error bound for each approximation.
Wait! How is the actual error bigger than the error bound for ln 0.5?
Because the series is alternating, we can start with…
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
Recall that the Taylor Series for ln x centered at x = 1 is given by…
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…
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
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
which has its largest value when c is smallest.
And therefore…
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
Lagrange Form of the Remainder:
Remainder Estimation Theorem:
If M is the maximum value of on the interval between a and x, then:
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
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.
7
723 0
7!
f
7
733 0
7!
f
7
703 0
7!
f
6
60.5 0
6!
f c 6200 1
0.5 06!
62000.5 0.0043
6!
