第5.9节 牛顿方法

5.9 NEWTON’ S METHOD

The Increment Theorem for derivatives shows that when f ′(c) exists and x≈c, f(x) is infinitely close to the tangent line f(c) + f ′(c) (x-c) even compared to x - c. Thus intuitively, when x is real and close to c, f(c) is closely approximated by the tangent line f(c) + f ′(c) (x-c). Newton’s method uses the tangent line to approximate a zero of f(x). It is an iterative method that does not always work but usually gives a very good approximation.

Consider a real function f that crosses the x-axis as in Figure 5.9.1. From the graph we make a first rough approximation x1 to the zero of f(x). To get a better approximation, we take the tangent line at x1 and compute the point x2 where the tangent line intersects the x-axis. At x2, the curve f(x) is very close to zero, so we take x2 as our new approximation. The tangent line has the equation

y = f(x1) + f ′(x1)(x-x1).

We get a formula for x2 by setting y =0 and x = x2 and then solving for x2.

0 = f(x1) + f(x1)(x2 - x1)

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We may then repeat the procedure starting from x2 to get a still better approximation x2 as in Figure 5.9.2,

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NEWTON’S METHOD

When to Use We wish to approximate a zero of f(x), where f ′(x) is continuous and not close to

zero, as in Figure 5.9.1.

Step 1 Sketch the graph of f(x), and choose a point x1 near the zero of f(x). X1

is the first approximation.

Step 2 Compute f ′(x).

Step 3 Compute the second approximation

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Step 4 For a closer approximation repeat Step 3. The (n+1)st approximation is given by

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As a rough check on the accuracy, compute f(xn) and note how close it is to zero.

Steps 3 and 4 can be done conveniently on a hand calculator.

Warning : Since Newton’s method involves division by f ′(x1), avoid starting at a point where the slope is near zero. Figure 5.9.3 shows that when the slope is close to zero, the tangent line is nearly horizontal and the approximation may be poor.

Figure 5.9.3

EXAMPLE 1 Approximate a zero of f(x) = x³ + 2x² - 5 by Newton’s method.

Step 1 The graph is shown in Figure 5.9.4. We choose x1 = 1 as our first approximation.

Step 2 f ′(x) = 3x² + 4x

Step 3 ______________________________

Figure 5.9.4

Step 4 ______________________________

As a check we compute

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One more iteration gives much more accuracy:

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EXAMPLE 2 Approximate the fifth root of 6 by Newton’s method.

Step 1 We must find the zero of f(x) = x5 - 6. The graph is shown in Figure 5.9.5.

Step 2 f(x) =5x4

Step 3 _________________________________

Step 4 _________________________________

As a check we compute

(x3) 5 ~ 6.001

In this example more iterations would be necessary if our first approximation had not been chosen as well. For instance, starting with x1 = 1 we would not reach

The approximation 1.431 until x6, obtaining the successive approximations

x1 = 1, x2 = 2, x3 = 1.675, x4 = 1.49245,

x5 = 1.43583, x6 = 1.43100.

EXAMPLE 3 Approximate the point x where sin x = 1nx.

As one can see from the graphs of sin x and 1n x in Figure 5.9.6, sin x and 1n x cross

at one point x, which is somewhere between x = 1 (where 1n x crosses the x- axis

going up ) and x = π (where sin x crosses the x- axis going down ). To apply Newton’s

method, we let f(x) be the function

f(x) = sin x - 1n x

Shown in Figure 5.9.7. We wish to approximate the zero of f(x).

Figure 5.9.6

Figure 5.9.7

Step 1 Choose x1 = 2 (since the zero of f(x) is between 1 an π).

Step 2 f ′(x) = cos x-1 / x

Step 3 _________________________________

Step 4 Repeat Step 3. The values of xn, f(xn), and f ′(xn) are shown in the table.

The answer is

x ~ 2.219107150.

On a calculator we find that

sin (2.219107150) = 0.797104929

1n( 2.219107150) = 0.797104930.

PROBLEMS FOR SECTION 5.9

Use Newton’s method to find approximate solutions to each of the following equations. (A hand calculator is recommended.)

1 x³ + 5x - 10 = 0 2 2x³ + x + 4 = 0

3 x5 + x³ + x = 1 4 2x5+ 3x = 2

5 x4 = x+1, x > 0 6 x4 = x+1, x < 0

7 x³ - 10x + 4 = 0, x >1 8 x³ - 10x + 4 = 0, 0 < x < 1

9 ________________ 10 __________________

11 ex = 1/x 12 ex + x = 4

13 x + sin x = 2 14 cos x = x², x > 0

15 tan x = ex, 0 < x< π/2 16 ex + 1n x =0