Download 2-18-2002, LECTURE 1. The Lagrange Remainder and Applications

2-18-2002, LECTURE
DAVID BEN MCREYNOLDS
1. The Lagrange Remainder and Applications
Let us begin by recalling two definition.
Definition 1.1 (Taylor Polynomial). Let f be a continuous function
with N continuous derivatives. Then we define the N th Taylor Polynomial of f , centered about the point a by
Pa,N,f (x) =
N
X
f (n) (a)
n=0
n!
(x − a)n
For consistency, we denote this simply by PN,a or PN .
Definition 1.2 (Remainder). With f as above, we define the remainder by
RN (x) = f (x) − PN,a,f
Now, if we are going to use the Taylor polynomial as a method or
means of approximation, we must be able to control the accuracy of
our approximation. Historically, these ideas were born early in the development of Calculus born the development of modern mathematical
rigor. The mathematicians of the time felt that the Taylor polynomial
would yield something approximately equal to the function in question. Unfortunately, they were incorrect, since this is not always the
case.1 The Lagrange Remainder theorem does give one the desired
control.
Theorem 1.1 (Lagrange). Let f be continuous in a neighborhood of a
with N + 1 continuous derivatives. Then
f (N +1) (c)
RN (x) =
(x − a)N +1
(N + 1)!
where c is between x and a.
Date: February 17, 2002.
1A classical example of this failure is the function e−1/x2 , except at x = 0, we
define the function to be zero. This is an example of a function with continuous
derivatives of all order, but the Taylor series at x = 0 is identically zero.
1
2
DAVID BEN MCREYNOLDS
Remark 1. As stated in the previous lecture, c depends on x, a, and
N.
Today, we work a few examples utilizing the Lagrange formula.
Example 1.1. Here, we will estimate e using the Taylor polynomials.
Recall that the N th Taylor polynomial of ex is given by
PN (x) =
N
X
xn
n=0
n!
Using the 7th Taylor polynomial, we see that
1
1
1
1
1 1
P7 (1) = 1 + 1 + + +
+
+
+
= 2.71825396825
2 6 24 120 720 5040
on my TI-85 calculator. How close is this to e? We can estimate how
close we are by the Lagrange formula. We know that
ec x8
8!
where c is between 0 and x = 1. Since ex is an increasing function, we
know that the maximum this value could take would occur at c = 1.
Thus
e
|R7 (x)| ≤
8!
Now, since we are trying to approximate e, we cannot exactly use it in
our estimate. We do know that
e − P7 (1) = R7 =
e<3
So
3
≈ 7.44047619048 × 10−5 < 10−4
8!
Thus, our approximation is good to the first four decimal digits. For
comparison, my TI-85 says that
|R7 (x)| <
e ≈ 2.71828182846
and we see that
e − P7 (1) ≈ 2.7860209 × 10−5 < 7.44047619048 × 10−5
So that our calculator is doing a good job.
Question 1. As another (brief example)
P16 (1) = 2.71828182846
so that
e − P16 (1) = 0
on my TI-85. Does this mean that P16 (1) = e?
2-18-2002, LECTURE
3
No, it means that P16 (1) exceeds the accuracy of my calculator in
how well it approximates e. Using an identical argument as above, the
Lagrange formula gives
3
|R16 (x)| <
≈ 8.43437176304 × 10−15 < 10−14
(17!)
So that our approximation is perfect on the first 14 decimal digits of e.
Notice that, for large n, we obtain at least a digit of accuracy for each
Pn (1), since
1
10
<
(n + 1)!
n!
for large n.
Example 1.2. As a second example, we approximate ln 1.1 using the
Taylor polynomials of ln(x + 1). Recall that
PN (x) =
N
X
(−1)n+1 xn
n=1
n
Now, since it will be of use to us later, recall that (we saw this in the
computation of the Taylor polynomials of ln(x + 1))
(ln(x + 1))(n) =
(−1)n+1 (n − 1)!
(x + 1)n
Now, we use P9 to estimate ln 1.1. Here, x = 0.1. My TI-85 gives
P9 (.1) ≈ 0.095310179813
Now, the Lagrange formula says
(10)
f (c)x10 |R9 (x)| = 10!
where c is between 0 and x = 0.1. Now, we notice that the 10th
derivative of ln(x + 1), which is
−9!
(1 + x)10
is increasing on the interval [0, 0.1]. On the other hand, it is also
negative, so that the absolute value is largest at the left end point.
Thus
−(9!)(0.1)10 = 1
|R9 (x)| ≤ 10−9
10!
Now, on my TI-85, I have
ln 1.1 ≈ 0.095310179804
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DAVID BEN MCREYNOLDS
and
|ln 1.1 − P9 (.1)| ≈ 8.674 × 10−12 < 10−11
So that our calculator is doing a good job.
2. Summary
To summarize, we see that the Taylor series can be used in application. The major down side of the Lagrange formula is the imprecision
in the location of the ”magic number”. We work around this by taking the maximum value of the derivative in question on the interval of
consideration. Also, we sometimes have to further estimate our upper
bound when constants appear that we can not compute (or that we are
trying to compute, like in the case of e).
University of Texas at Austin, Department of Mathematics
E-mail address: [email protected]