Part II: The Answers
336
Answers
501–600
x
x
2
4
3
3
4
2
2
18
4 2
33
16
2 0625
2 0625
2 0625
18
4 2 062
= −
−
=
=
=
−
−
( )
( )
.
.
( .
)
( .
5 5
2 0597725
2 0597725
2 0597725
18
4 2 0597725
2
3
4
4
3
)
.
.
( .
)
( .
)
.
≈
=
−
−
≈
x
0 0597671
2 0597671
2 0597671
18
4 2 0597671
2 059767
5
4
3
x =
−
−
≈
.
( .
)
( .
)
.
1 1
515.
–1.2457342
Use the formula x
x
f x
f x
n
n
n
n
+
( )
′ ( )
1 =
−
with f
(x) = x
5
+ 3, f
'(x) = 5x
4
, and x 1 = –1. Note that
the formula gives you x
x
f x
f x
2
1
1
1
= −
( )
′ ( )
, x
x
f x
f x
3
2
2
2
= −
( )
′ ( )
, and so on. Therefore, you have
x
x
2
5
4
3
5
4
1
1
3
5 1
1 4
1 4
1 4
3
5 1 4
1 27618
= − −
−
−
= −
= − −
−
−
≈ −
( )
( )
.
.
( . )
( . )
.
+
+
4 49
1 2761849
1 2761849
3
5 1 2761849
1 2471501
4
5
4
x
x
= −
−
−
+
−
≈ −
.
( .
)
( .
)
.
5 5
5
4
1 2471501
1 2471501
3
5 1 2471501
1 2457342
= −
−
−
+
−
≈ −
.
( .
)
( .
)
.
516.
0.73909
Newton’s method begins by making one side of the equation zero, labeling the other
side f
(x), and then picking a value of x 1 that’s “close” to a root of f
(x). There’s certainly
a bit of trial and error involved; in this case, you could graph both y = cos x and y = x
and look for a point of intersection to get a rough idea of what the root may be.
Use the formula x
x
f x
f x
n
n
n
n
+
( )
′ ( )
1 =
−
with f
(x) = cos x – x and f
'(x) = –sin x – 1. Note that
the formula gives you x
x
f x
f x
2
1
1
1
= −
( )
′ ( )
, x
x
f x
f x
3
2
2
2
= −
( )
′ ( )
, and so on. You also have to
decide on a value for x 1 . Notice that f
(0) = 1 – 0 = 1, which is close to the desired value
of 0, so you can start with the value x 1 = 0. Therefore, you have
x
x
2
3
0
0
0
0 1
1
1
1
1
1 1
0 75036
= −
−
−
−
=
= −
− ( )
−
−
≈
cos( ) ( )
sin
cos( )
sin( )
.
( )
x x
x
4
0 75036
0 75036
0 75036
0 75036 1
0 73911
=
−
− (
)
−
−
≈
.
cos( .
)
.
sin( .
)
.
5 5
6
0 73911
0 73911
0 73911
0 73911 1
0 73909
=
−
− (
)
−
−
≈
.
cos( .
)
.
sin( .
)
.
x = =
−
− (
)
−
−
≈
0 73909
0 73909
0 73909
0 73909 1
0 73909
.
cos( .
)
.
sin( .
)
.
The approximation is no longer changing, so the solution is 0.73909.
336
Answers
501–600
x
x
2
4
3
3
4
2
2
18
4 2
33
16
2 0625
2 0625
2 0625
18
4 2 062
= −
−
=
=
=
−
−
( )
( )
.
.
( .
)
( .
5 5
2 0597725
2 0597725
2 0597725
18
4 2 0597725
2
3
4
4
3
)
.
.
( .
)
( .
)
.
≈
=
−
−
≈
x
0 0597671
2 0597671
2 0597671
18
4 2 0597671
2 059767
5
4
3
x =
−
−
≈
.
( .
)
( .
)
.
1 1
515.
–1.2457342
Use the formula x
x
f x
f x
n
n
n
n
+
( )
′ ( )
1 =
−
with f
(x) = x
5
+ 3, f
'(x) = 5x
4
, and x 1 = –1. Note that
the formula gives you x
x
f x
f x
2
1
1
1
= −
( )
′ ( )
, x
x
f x
f x
3
2
2
2
= −
( )
′ ( )
, and so on. Therefore, you have
x
x
2
5
4
3
5
4
1
1
3
5 1
1 4
1 4
1 4
3
5 1 4
1 27618
= − −
−
−
= −
= − −
−
−
≈ −
( )
( )
.
.
( . )
( . )
.
+
+
4 49
1 2761849
1 2761849
3
5 1 2761849
1 2471501
4
5
4
x
x
= −
−
−
+
−
≈ −
.
( .
)
( .
)
.
5 5
5
4
1 2471501
1 2471501
3
5 1 2471501
1 2457342
= −
−
−
+
−
≈ −
.
( .
)
( .
)
.
516.
0.73909
Newton’s method begins by making one side of the equation zero, labeling the other
side f
(x), and then picking a value of x 1 that’s “close” to a root of f
(x). There’s certainly
a bit of trial and error involved; in this case, you could graph both y = cos x and y = x
and look for a point of intersection to get a rough idea of what the root may be.
Use the formula x
x
f x
f x
n
n
n
n
+
( )
′ ( )
1 =
−
with f
(x) = cos x – x and f
'(x) = –sin x – 1. Note that
the formula gives you x
x
f x
f x
2
1
1
1
= −
( )
′ ( )
, x
x
f x
f x
3
2
2
2
= −
( )
′ ( )
, and so on. You also have to
decide on a value for x 1 . Notice that f
(0) = 1 – 0 = 1, which is close to the desired value
of 0, so you can start with the value x 1 = 0. Therefore, you have
x
x
2
3
0
0
0
0 1
1
1
1
1
1 1
0 75036
= −
−
−
−
=
= −
− ( )
−
−
≈
cos( ) ( )
sin
cos( )
sin( )
.
( )
x x
x
4
0 75036
0 75036
0 75036
0 75036 1
0 73911
=
−
− (
)
−
−
≈
.
cos( .
)
.
sin( .
)
.
5 5
6
0 73911
0 73911
0 73911
0 73911 1
0 73909
=
−
− (
)
−
−
≈
.
cos( .
)
.
sin( .
)
.
x = =
−
− (
)
−
−
≈
0 73909
0 73909
0 73909
0 73909 1
0 73909
.
cos( .
)
.
sin( .
)
.
The approximation is no longer changing, so the solution is 0.73909.