Chapter 4 | Applications of Derivatives
475
3 −3 x +1 over the
4.45 Letting x 0 =0, let’s use Newton’s method to approximate the root of f ( x ) = x
interval [0, 1] by calculating x 1 and x 2 .
Newton’s method can also be used to approximate square roots. Here we show how to approximate 2. This method can be modified to approximate the square root of any positive number. Example 4.47 Finding a Square Root
Use Newton’s method to approximate 2 ( Figure 4.79 ). Let f ( x ) = x 2 −2, let x
0 =2, and calculate
2 −2 has a zero at 2, the initial value x
x 1 , x 2 , x 3 , x 4 , x 5 . (We note that since f ( x ) = x
0 =2 is a
reasonable choice to approximate 2.)
Solution For f ( x ) = x 2 −2, f ′( x ) =2 x . From Equation 4.8 , we know that x n = x n −1 −
f ( x n −1 ) f ′( x n −1 )
x 2
n −1 −2 2 x n −1
= x n −1 −
= 1 2 = 1 2
x n −1 + 1 x ⎛ ⎝ x n −1 + 2 x n −1
⎞ ⎠ .
n −1
Therefore,
⎛ ⎝ x 0 + 2 x 0 ⎛ ⎝ x 1 + 2 x 1
⎞ ⎠ = 1 2 ⎞ ⎠ = 1 2
⎛ ⎝ 2+ 2 2
⎞ ⎠ =1.5
x 1 = 1 2 x 2 = 1 2
⎛ ⎝ 1.5+ 2 1.5
⎞ ⎠ ≈ 1.416666667.
Continuing in this way, we find that
x 1 =1.5 x 2 ≈ 1.416666667 x 3 ≈ 1.414215686 x 4 ≈ 1.414213562 x 5 ≈ 1.414213562. Since we obtained the same value for x 4 and x 5 , it is unlikely that the value x n will change on any subsequent application of Newton’s method. We conclude that 2 ≈ 1.414213562.
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