In Problems 15 through 18, use a second degree Taylor polynomial centered appropri-ately to approximate the expression given.. , 5 Compare the size of the difference between the actual v
Trang 1Alternating signs:
(−1)kand (−1)k+1can be used to indicate alternating signs Which is needed to do the job is determined by the notational system you happen to have chosen The simplest way of determining which you need is by trial and error Try (−1)kand check it with
a particular k-value If it doesn’t work, switch to (−1)k+1
34 using the appropriate second degree Taylor polynomial
SOLUTION Let f (x) = x1 We must center the Taylor polynomial at a point near 34 at which the values
of f , f, and fcan be readily computed
An off-the-cuff approximation of√5
34 is√5
34 ≈√532 = 2; we know that√534 is a bit more than 2 Center the Taylor polynomial at x = 32
P2(x) = f (32) + f(32)(x − 32) + f
(32) 2! (x − 32)2
f (x) = x1 f (32) = 2
f(x) = 1
5x
−4 f(32) =1
5
1
324 =1 5
1
24= 1 80
f(x) = −4
25x
−9 f(32) =−4
25
1
29= −1
25 · 27= −1
3200 Therefore,
P2(x) = 2 + 1
80(x − 32) − 1
6400(x − 32)2 5
√
34 ≈ P2(34) = 2 + 1
80(2) − 4
6400= 2 + 1
40− 1
1600= 2.024375 This agrees with the actual value of√5
34 to four decimal places
If you study closely the numerical data in this section you can start to get a sense of the magnitude of the error involved in a Taylor polynomial approximation The size of the error can be estimated by graphing f (x) − Pn(x)using a calculator or computer In the next section we will state Taylor’s Theorem, which will provide not only a method of estimating errors independent of a calculator, but also an invaluable theoretical tool
P R O B L E M S F O R S E C T I O N 3 0 1
For Problems 1 through 7, do the following.
(a) Compute the fourth degree Taylor polynomial for f (x) at x = 0.
(b) On the same set of axes, graph f (x), P1(x), P2(x), P3(x), and P4(x).
(c) Use P1(x), P2(x), P3(x), and P4(x) to approximate f (0.1) and f (0.3) Compare
these approximations to those given by a calculator.
1 f (x) = e−x
2 f (x) = ln(1 + x)
3 f (x) = tan−1x
Trang 2932 CHAPTER 30 Series
4 f (x) = (1 + x)4
5 f (x) =√1 + x
6 f (x) = 2x4− 3x2+ x − 1
7 f (x) = (1 + x)−2
8 Below is a graph of f (x) For each quadratic given, explain why the quadratic could
notbe the second degree Taylor polynomial for f (x) at x = 0
(a) 2 + 3x − 12x2 (b) −1 − 5x + 2x2 (c) −2 + 2x −13x2
y
x
y = f(x)
9 Let P2(x) = a0+ a1x + a2x2be the second degree Taylor polynomial generated by
f (x)at x = 0, where the graph of f is the one given in Problem 8 above Use the graph
to determine the signs of a0, a1, and a2
10 (a) Find the second degree Taylor polynomial generated by sec x at x = 0
(b) Graph P2(x)and sec x on the same set of axes
11 (a) Compute the third degree Taylor polynomial for tan x about x = 0
(b) Why is it reasonable to expect the coefficient of the x2term to be zero?
12 The graph of a differentiable function f (x) is given Use the graph to determine the signs of the coefficients of the second degree Taylor polynomials indicated f has a minimum at x = 0 and a point of inflection at x = 2
(a) P2(x) = a0+ a1x + a2x2 (b) P2(x) = a0+ a1(x − 1) + a2(x − 1)2 (c) P2(x) = a0+ a1(x − 2) + a2(x − 2)2 (d) P2(x) = a0+ a1(x − 3) + a2(x − 3)2
Trang 3x
y = f(x)
13 Let f (x) = ln(1 + x) Find the nth degree Taylor polynomial generated by f about
x = 0
14 Compute the nth degree Taylor polynomial expansion of f (x) =1xabout x = 1 Graph
f and P1, P2, P3, and P4on a common set of axes
In Problems 15 through 18, use a second degree Taylor polynomial centered appropri-ately to approximate the expression given.
15.√3
8.3
16.√
103
17 tan−1(0.75)
18.√3
29
19 Compute the third degree Taylor polynomial generated by sin x at x =π4
20 Find the fifth degree Taylor polynomial for√x
centered at x = 9
21 Write the third degree Taylor polynomial centered about x = 0 for f (x) = (1+x)1 p, where p is constant
Introduction to Error Analysis: Problems 22 and 23.
22 Let f (x) = ex Use the data given in the table on page 928 to compute the following (a) f (0.1) − Pk(0.1) for k = 1, 2, , 5
(b) f (0.2) − Pk(0.2) for k = 1, 2, , 5
(c) f (0.5) − Pk(0.5) for k = 1, 2, , 5
(d) f (1) − Pk(1) for k = 1, 2, , 5
Compare the size of the difference between the actual value of the function and the poly-nomial approximation with that of the first “unused” term of the Taylor polypoly-nomial— that is, the last term of the next higher degree polynomial—and observe that they have the same order of magnitude
Trang 4934 CHAPTER 30 Series
23 Let f (x) = exand let Pk(x)be its kth degree Taylor polynomial about x = 0 Graph
Rk(x) = f (x) − Pk(x)for k = 1, 2, , 5
24 Use a third degree Taylor polynomial to approximate ln 0.9
25 f (x) = 12 + 3(x − 1) + 5(x − 1)2+ 7(x − 1)3 Find the following
(a) f(1) (b) f(1) (c) f(1) (d) f (1)
26 f (x) =√3 + 12(x − 5)3+ 17(x − 5)6 Find the following
(a) f (5) (b) f(5) (c) f(5) (d) f(6)(5)
27 Compute the sixth degree Taylor polynomial generated by sin x about x = π
28 Compute the sixth degree Taylor polynomial generated by cos x about x = −π2
29 Let f (x) = (1 + x)p (a) Compute the third degree Taylor polynomial for f (x) around x = 0
(b) Compute the fifth degree Taylor polynomial for f (x) around x = 0
30 Using the results of Problem 29(a), approximate the following Compare your results with the numerical approximations given by a calculator
(a) √ 1.002 (b) 1.031 (c) √3
1.001
An approximation is of limited use unless we have a notion of the magnitude of the error involved Every Taylor polynomial Pn(x)has an associated error function, Rn(x), defined by
function = polynomial
approximation
+associated error
Rn(x)is referred to as the Taylor remainder; Rn(x) = f (x) − Pn(x)
For a Taylor polynomial centered at x = b we expect the magnitude of the remainder
to decrease as n increases and as x approaches b Because each successive refinement
of a Taylor polynomial involves a higher derivative, we might expect Rn(x)to involve the (n + 1)st derivative of f While Taylor’s Theorem does not pin down the remainder precisely, it provides a means of putting an upper bound on the magnitude of the error
Trang 5T a y l o r ’ s T h e o r e m
Suppose f and all its derivatives exist in an open interval I centered at x = b Then for each x in I
f (x) = f (b) + f(b)(x − b) +f
(b) 2! (x − b)2+f
(b) 3! (x − b)3+ · · · +
f(n)(b) n! (x − b)n+ Rn(x), where
Rn(x) =f
(n+1)(c) (n + 1)! (x − b)
n+1 for some number c in I , c between x and b
Note that Rn(x)has the same form as the next term of a Taylor polynomial except that the (n + 1)st derivative is evaluated at some c between x and b instead of at b itself Its form agrees with the expectations laid out before the statement of Taylor’s Theorem When applying the theorem we do not expect to be able to find c; if we could, an approximation wouldn’t have been needed
In practice, we look for a bound, M, such that |f(n+1)(c)| ≤ M for all c between x and
band use the inequality
|Rn(x)| ≤ M
(n + 1)!|x − b|
n+1
This is referred to as Taylor’s Inequality
A sketch of the proof of Taylor’s Theorem is given in Appendix H
Let’s revisit some of the problems from the previous section and see what information Taylor’s remainder provides about the accuracy of approximations
EXAMPLE 30.6 Give a good5upper bound for the error involved in estimating sin 0.1 using the
approxima-tion sin x ≈ x −x3!3
SOLUTION We can call x −x3!3 either P3(x)or P4(x), the two being equal We’ll call it P4(x)as this
will give a better bound on the error
f (0.1) = P4(0.1) + R4(0.1) sin(0.1) =
0.1 −(0.1)
3
3!
+ R4(0.1)
Taylor’s Theorem says Rn(x) =f(n+1)!(n+1)(c)(x − b)n+1 for some c between x and b In this example n = 4, f (x) = sin x, b = 0, and x = 0.1
R4(0.1) = f
(5)(c) 5! (0.1)
5
5
Trang 6936 CHAPTER 30 Series
The derivatives of sin x are ± sin x and ± cos x, so |f(5)(c)| ≤ 1
0 ≤ |R4(0.1)| ≤ 1
5!
1
105= 1
120 · 105= 1
1.2 × 107= 8.3 × 10−8
EXAMPLE 30.7 We want to use an nth degree Taylor polynomial for excentered at x = 0 to approximate e
How large must n be to assure that the answer differs from e by no more than 10−7? Assume
we know e < 3.
SOLUTION |Rn(x)| =|f(n+1)!(n+1)(c)||x − b|n+1 for some c between x and b In this example f (x) = ex,
b = 0, and x = 1
|f(n+1)(c)| = |ec| = ecfor c between 0 and 1
ecincreases with c, so ec< e1<3
|Rn(1)| ≤ 3
(n + 1)!· 1 =
3 (n + 1)!
We must find an integer n such that(n+1)!3 ≤1017, or, equivalently,
(n + 1)! ≥ 3 · 107
We find n by trial and error 11! = 39,916,800 > 3 · 107, whereas 10! is not large enough
n + 1 = 11, so n = 10
We must use the 10th degree Taylor polynomial: P10(x) =10k=0xk!n Checking, we see that10k=0k!1 = 1 + 1 +2!1 +3!1 + · · · +10!1 ≈ 2.718281801, which differs from e by less than 10−7 In fact, we solved the problem efficiently; had we used one less term of the expansion, the error would have been more than 10−7
34 using a second degree Taylor polynomial centered
at x = 32 Find a reasonable upper bound for the magnitude of the error
SOLUTION |Rn(x)| =|f(n+1)!(n+1)(c)||x − b|n+1for some c between x and b In this example n = 2, f (x) =
x1, b = 32, and x = 34
|R2(34)| =|f
(c)|
3! |34 − 32|3
|R2(34)| =|f
(c)|
6 · 8 for some c between 32 and 34
We must find M such that |f(c)| ≤ M
f(x) =1
5x
−4; f(x) = −4
25x
−9; f(x) = 36
125x
−14 5
|f(c)| = 36
125 1 (√5c)14 for some c between 32 and 34
Trang 7The smaller c, the larger |f(c)|, so
0 ≤ |f(c)| ≤ 36
125
1 (√5 32)14 = 36
125
1
214 = 9
125 · 212
0 ≤ |R2(34)| ≤ 9
125 · 212
1
6(8) = 3
125 · 210 = 3
128000≈ 2.344 × 10−5 The error is less than 2.4 × 10−5
Taylor’s Theorem gave a good estimate of the error; the actual error involved in Example 30.5 is approximately 2.24 × 10−5
If a computer or graphing calculator is at our disposal, error estimates can be readily available Suppose, for example, that we plan to use the third degree Taylor polynomial for
ln x centered at x = 1 in order to approximate ln x for x ∈ [0.3, 1.7] We want an upper bound for the error involved in doing so In other words, for x ∈ [0.3, 1.7] we use the approximation
ln x ≈ (x − 1) −(x − 1)
2
3
3 and want an estimate of R3(x) = ln x −
(x − 1) − (x−1)2 2 +(x−1)3 3
We can simply graph
|R3(x)| on [0.3, 1.7], obtaining the graph shown in Figure 30.6 Using the tracer we estimate that the magnitude of the error is less than 0.145
As an exercise, use Taylor’s Remainder to estimate the error
x y
0.1 0.1 0.2
0.3
(.3, ≈ 14464)
Graph of |R3(x)| = ln x – (x–1) – (x–1)
2
2
(x–1)3
3 + ] |
Figure 30.6
EXAMPLE 30.9 Use graphical methods to find an upper bound for the error involved in using the tangent
line approximation 1 −12xto approximate √1
1+x for |x| < 0.001
SOLUTION Graph R2(x) = (1 + x)−1 −
1 − 12xon the domain [−0.001, 001] (Play around with the range to obtain a useful graph.) The graph is given in Figure 30.7 on the following page
Trang 8938 CHAPTER 30 Series
3.8 × 10 –7
.001 –.001
y
x
(–.001, ≈ 3.8 × 10 –7 ) (.001, ≈ 3.8 × 10 –7 )
R2(x) = (1 + x) – [1 – x]12
1
–
Figure 30.7
For |x| < 0.001, the approximation √1
1+x ≈ 1 − 12x produces an error of less than
4 × 10−7
Any physicist will attest to the fact that physicists often use Taylor polynomials to simplify mathematical expressions In fact, they often use only first or second degree polynomials While this may at first strike you as a dubious strategy, the following example will demonstrate that in certain situations the error introduced is minimal
EXAMPLE 30.10 According to Newtonian physics, an object’s kinetic energy, K, is given by
K =1
2m0v
2
, where m0is the mass of the object at rest and v is its velocity
Einstein’s theory of special relativity produces a more involved expression for K According to Einstein, the mass of an object is a function of its velocity, m =√ m0
1−v 2 /c 2 Einstein’s theory says energy, E, equals mc2, where c is the speed of light He concludes that an object’s kinetic energy is given by the difference mc2− m0c2 Using the expression for m, Einstein’s theory says
K = m0c2
1
1 − v2/c2− 1
= m0c2
1 − v
2
c2
−1
− 1
(30.3)
Our goal in this example is to show that if an object is traveling much slower than the speed
of light, then according to Einstein’s theory, the error involved in using the Newtonian expression for K is small
SOLUTION We begin by noting that if v is substantially less than c, then vc is small, andvc is even
smaller From Example 30.9 we know that (1 + x)−1 can be well approximated by its first degree Taylor polynomial, 1 −12x, for |x| small Let x =−vc22 Using the approximation
1 − v
2
c2
−1
=
1 +
−v
2
c2
−1
≈ 1 − 1 2
−v
2
c2
in Equation (30.3) we obtain
Trang 9K = m0c2
1 − v
2
c2
−1
− 1
≈ m0c2
1 + 1 2
v2
c2 − 1
= m0c21 2
v2
c2 =1
2m0v
2
Let’s estimate the size of the error introduced by using the Newtonian expression for K for
an object traveling at speeds of 300 m/s or less c = 3 · 108m/s
We’ll find an upper bound for the error in replacing (1 + x)−1by 1 −12xfor |x| ≤300c22 and multiply the answer by m0c2
|R1(x)| =|f
(a)|
2! |x|2 for some a between 0 and 300
f (x) = (1 + x)−1; f(x) = −1
2(1 + x)−3; f(x) = 3
4(1 + x)−5
|R1(x)| = 3
2 · 4(√1 + a)5x2≤ 3
8
1 −300c22
5
3004
c4 ≈ 3.75 × 10−25
Multiplying by m0c2gives m03.375 × 10−8
Therefore, for speeds of up to 300 m/s, the error incurred in computing K using Newtonian physics is less than 3.4 × 10−8m0, where m0is the mass of the body at rest
P R O B L E M S F O R S E C T I O N 3 0 2
1 Find a good upper bound for the magnitude of the error involved in approximating cos x by 1 −x2!2 +x4!4 for |x| ≤ 0.2 Do this using Taylor’s Inequality; then check your answer by graphing the remainder function
2 Use the third degree Taylor polynomial for exat x = 0 to estimate√e Then use Taylor’s Theorem to get a reasonable upper bound for the remainder
3 We will use the nth degree Taylor polynomial for ex,nk=0 xk!k, to approximate √1
e What should n be in order to guarantee that the approximation is off by less than 10−5?
4 Use the third degree Taylor polynomial for ln x centered at x = 1, (x − 1) −(x−1)2 2 +
(x−1) 3
3 , to approximate ln(1.5) Then give an upper bound for the remainder using Taylor’s Theorem
5 The second degree Taylor polynomial for f (x) = (1 + x)pis 1 + px +p(p−1)2! x2 If the second degree Taylor polynomial is used to approximate√
1 + x for |x| ≤ 0.2, find
an upper bound for the magnitude of the error Use the Taylor Inequality; then check your answer by graphing R (x)
Trang 10940 CHAPTER 30 Series
6 For x near zero, cos x ≈ 1 − x2!2 +x4!4 − · · · + (−1)n x(2n)!2n What degree Taylor polyno-mial must be used to approximate cos(0.2) with error less than 1018?
7 Approximate √3
27.5 using an appropriate second degree Taylor polynomial Find a good upper bound for the error by using Taylor’s Inequality
8 The second degree Taylor polynomial generated by ln(1 + x) about x = 0 is x − x22 Use Taylor’s Theorem to find a good upper bound on the error involved in using this polynomial to approximate the following
(a) ln(1.2) (b) ln(0.8)
9 By graphing R2(x), estimate the values of x for which the approximation
ln x ≈ (x − 1) − (x−1)2! 2 can be used without producing an error of magnitude greater than 10−3
10 For x near zero, ex≈ 1 + x +x2!2 +x3!3 Find a reasonable upper bound for the magni-tude of the error involved in using this approximation for |x| < 0.5 Use Taylor’s Inequality and check your answer by graphing R3(x)
11 A hyena is loping down a straight path away from a stream The hyena is 6 m from the stream, moving at a rate of 2 m/s and decelerating at a rate of 0.1 m/s2 Use a second degree Taylor polynomial to estimate its distance from the stream 1 second later
12 What degree Taylor polynomial for exabout x = 0 must be used to approximate e0.3 with error less than 10−5?
13 (a) Find the nth degree Taylor polynomial for f (x) =1−x1 centered at x = 0 (b) How many nonzero terms of the polynomial in part (a) must be used to approximate f
1 2
with error less than 10−5?
14 According to Einstein’s theory of special relativity, the mass of an object moving with velocity v m/s is given by
m =m0
1 − vc22 ,
where m0is the mass of the object at rest and c is the speed of light, c = 3 × 108m/s (a) Use the first degree Taylor polynomial for √1
1+x to arrive at the estimate
m ≈ m0+m0
2
v2
c2 (b) If an object is moving at 100 m/s, find an upper bound for the error involved in using the approximation given in part (a)