Thomas Calculus 13th [Solutions]

Section 4.7 Newtons Method 305
4.
2
y 2x x 1 y
1 1 5
2 12 12
29
12
2.41667
2
2xn
xn
1
2 2 x xn 1 xn ;
0 0 1 1
1 1
4
x
2 2x
0 0 x1 0
x 1
n
2 0 2 2 2 2 1
25
.41667; x
4 4 1
0 2 x1 2
5
5 1
4
x 5
5 20 25 4 5 1
2 4 2 2 2 2 5 2 12 2 12
1
5.
4 3
y x 2 y 4x
5 113 2500 113
4 2000 2000
2387
2000
4
n
3
4xn
x 2
xn
1 xn
; x
1 2 5
0 1 x1 1
4 4
1.1935
625
256
125
16
5
2
x 5 625 512
2 4 4 2000
6. From Exercise 5,
5 625 512
4 2000
4
n
3
4xn
x 2
xn
1 xn
; x
1 2 1 5
0 1 x1 1 1
4 4 4
5 113
4 2000
1.1935
5 x 2 4
625
2
256
125
16
f ( xn
)
7. f ( x 0 ) 0 and f ( x0 ) 0 xn 1 x n gives x
f ( xn
) 1 x 0 x 2 x 0 x n x 0 for all n 0. That is all, of
the approximations in Newtons method will be the root of f ( x ) 0.
8. It does matter. If you start too far away from x
2 , the calculated values may approach some other root.
Starting with x 0 0.5, for instance, leads to x as the root, not
2
x 2 .
9. If x 0 h
f ( x0
) f ( h)
0 x 1 x0 h
f ( x0
) f ( h)
h
h h h 2 h h;
1
2 h
f ( x ) f ( h)
if x 0 h
0
0 x 1 x0 f ( x0
)
h
f ( h)
h
h h h 2 h h.
1
2 h
10.
1/3 1 2/3
( ) ( )
3
1/3
xn
n 1 n 2 ;
1 2/3 n
x
f x x f x x
x x x x0 1
3
n
x1 2, x2
4, x 3 8, and x 4 16 and so
forth. Since xn
2 x n 1 we may conclude that
n x n .
11. i) is equivalent to solving x 3x 1 0.
ii)
3
is equivalent to solving x 3x 1 0.
3
iii) is equivalent to solving x 3x 1 0.
3
iv) is equivalent to solving x 3x 1 0.
All four equations are equivalent.
3
12. f ( x) x 1 0.5sin x f ( x) 1 0.5cos x
xn
1 0.5sin xn
n 1 n 1 0.5cos xn
x x ; if x 0 1.5, then x 1 1.49870
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