In numerous cases throughout this book, as in the real world, we
find problems where an exact numerical answer is either impossible or impractical
to express as a decimal number. Examples of these situations include calculation
of , p,
e, and ln(2). As we have seen there are many ways to characterize and define
these numbers and use these characterizations to find an estimate. For
our discussion in this chapter it is important to notice that each of these
numbers can be characterized as arising from the evaluation of a function
which is the solution to a differential equation. [See Table 1 below.]
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We have discussed approximations from the very beginning of our work on the derivative, with Newton's method for estimating the roots of equations, Euler's method for estimating particular solutions to differential equations, and numerous methods for estimating definite integrals. In the following sections we will delve further into the nature and applications of polynomial approximations based on imformation about the derivatives of a function.
To provide further motivation for what we shall refer to as Taylor's approximation theory, let's consider first a simple situation where an exact value for a function is attainable but perhaps not worth the effort.
EXAMPLE IX.A.1: Suppose
and we wish to find f(.5).
Of course the exact answer is .25 + .125 + (.5)100, but if we only need an estimate of the answer it is easy to settle for .375. The error we make in using this estimate is relatively small [namely (.5)100]. In this example we should note two things.
Taylor's Theory-Objective and Key Ideas: The main concerns
of Taylor's theory for estimating function values are
Taylor's theory generalizes these two ideas that use the derivative to estimate.
IX.A. Estimating ex with Polynomials: An Introduction to Taylor's Theory.
To focus our discussion more specifially, in this section we will consider
only and
begin by trying to estimate f(1)=e illustrating the key ideas of
Taylor's
theory.
At x = 0, we find that f(0) = 1, f '(0) =1, and since
f
'(x)
= f(x) we have that f (n)(0) = 1 for all
natural numbers
n. We now consider our first result, typical of
the Taylor's theory approach to estimation using polynomials and derivative
information.
PROPOSITION IX.A.1: If
then Pn(x) is a polynomial of degree n
so that
and Pn(b) is approximately equal to eb. In
fact, if we let Rn = eb - Pn(b)
then
for
some c between 0 and b.
Estimating e: Before we proceed to justify this result, we'll
apply
this result using n = 5 to estimate the value of e [= e1].
First,,
so according to the proposition, e is approximately equal to
(
about 2.716667) and R5=e-P5(1) where
for some c beween 0 and 1. Thus, e is approximately 2.716667 and the error
in using this estimate is R5. Since c is between 0 and 1, ec
is between 1 and e, so we can deduce that R5, the difference
between e and the estimate, P5(1), is no greater than
e/720. Using estimates of e from our earlier work in Chapter VI we know
that e < 3, so our errror R5 is no larger than 3/720
= 1/240.
A more accurate estimate can be obtained by using a larger value for n. Try using n=6 and 7 to see the improvement. [This can be done progressively using a spreadsheet which you can examine now or later.]
Proof of Proposition: We'll begin our proof by finding Pn'(x).
From this it follows easily that .
Now suppose that b is not 0 and let Rn = eb
-
Pn(b).
We need only justify the formula for evaluating Rn.
For convenience we'll write R=Rn and let
Furthermore
But
, so
.
Now we apply the Mean Value (or Rolle's) Theorem to the function g,
we can say that there is a number c between 0 and b where g'(c)=0.
Thus
and
.
Solving this last equation for R gives
.
EXAMPLE IX.A.2: Estimate .
Use this to estimate e.
Solution: Using Proposition IX.1 with n=4 for each x between 0 and 1 we have
where
for some c with 0<c<x<1.
Thus
[Remember ec<3.] and so
for all x between 0 and 1. Now we use the monotone property of the
definite integral [Ch. V . ** ] to obtain the estimate:
,
so
Therefore e is approximately
as in the first estimate we saw in the note after the statement of the
theorem, and the error in using this estimate is less than
as we also saw previously.
The next example follows a similar analysis but applied to a more difficult yet still important integral.
EXAMPLE IX.A.3: Estimate
.
Solution: For each t between 0 and 1, -t2 is between -1 and 0. Let x = -t2. By Proposition IX.1 with n = 4 for each x between -1 and 0 we have
Since c < 0 , 0 < ec < 1, and since x < 0 we
have that .
Substituting -t2 for x ,we see that
Now since
we have
.
By evaluating these integrals we obtain
or .
Therefore,
is approximately equal to
and this is an overestimate by no more than .
Comment: In both of these examples we have been able to estimate a definite integral involving the exponential function by using a Taylor polynomial of degree 4. It should be apparent that by using a higher degree for the estimating polynomial, the error term will become smaller and we will obtain a more precise estimate. The systematic pattern in these polynomials should allow you to find more precise estimates for the last example without much difficulty.
Exercises IX.A.: