Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 4.8 NEWTON’S METHOD ¤ 205
11. To approximate x = 5√ 20 (so that x 5 =20),wecantakef(x) =x 5 − 20. Sof 0 (x) =5x 4 , and thus,
x n+1 = x n − x5 n − 20
.Since 5√ 32 = 2 and 32 is reasonably close to 20, we’ll use x
5x 4 1 =2. We need to find approximations
n
until they agree to eight decimal places. x 1 =2 ⇒ x 2 =1.85, x 3 ≈ 1.82148614, x 4 ≈ 1.82056514,
x 5 ≈ 1.82056420 ≈ x 6 .So 5√ 20 ≈ 1.82056420, to eight decimal places.
Here is a quick and easy method for finding the iterations for Newton’s method on a programmable calculator.
(The screens shown are from the TI-84 Plus, but the method is similar on other calculators.) Assign f(x) =x 5 − 20
to Y 1 ,andf 0 (x) =5x 4 to Y 2 .Nowstorex 1 =2in X andthenenterX − Y 1 /Y 2 → X to get x 2 =1.85. By successively
pressing the ENTER key, you get the approximations x 3, x 4, ....
In Derive, load the utility file SOLVE. EnterNEWTON(xˆ5-20,x,2) and then APPROXIMATE to get
[2, 1.85, 1.82148614, 1.82056514, 1.82056420]. You can request a specific iteration by adding a fourth argument. For
example, NEWTON(xˆ5-20,x,2,2) gives [2, 1.85, 1.82148614].
In Maple, make the assignments f := x → xˆ5 − 20;, g := x → x − f(x)/D(f)(x);,andx := 2.;. Repeatedly execute
the command x := g(x); to generate successive approximations.
In Mathematica, make the assignments f[x_ ]:=xˆ5 − 20, g[x_ ]:=x − f[x]/f 0 [x],andx =2. Repeatedly execute the
command x = g[x] to generate successive approximations.
13. f(x) =x 4 − 2x 3 +5x 2 − 6 ⇒ f 0 (x) =4x 3 − 6x 2 +10x ⇒ x n+1 = x n − x4 n − 2x 3 n +5x 2 n − 6
. We need to find
4x 3 n − 6x 2 n +10x n
approximations until they agree to six decimal places. We’ll let x 1 equal the midpoint of the given interval, [1, 2].
x 1 =1.5 ⇒ x 2 =1.2625, x 3 ≈ 1.218808, x 4 ≈ 1.217563, x 5 ≈ 1.217562 ≈ x 6. So the root is 1.217562 to six decimal
places.
15. sin x = x 2 ,sof(x) =sinx − x 2 ⇒ f 0 (x) =cosx − 2x ⇒
x n+1 = x n − sin xn − x2 n
cos x n − 2x n
.Fromthefigure, the positive root of sin x = x 2 is
near 1. x 1 =1 ⇒ x 2 ≈ 0.891396, x 3 ≈ 0.876985, x 4 ≈ 0.876726 ≈ x 5.So
thepositiverootis0.876726, to six decimal places.