Solve by newtons method x cube minus 2x minus 2 equal to 0
Answers
The Newton-Raphson method approximates the roots of a function. So, we need a function whose root is the cube root we're trying to calculate.
Let's say we're trying to find the cube root of
3
. And let's say that
x
is the cube root of
3
. Therefore,
x
3
=
3
For the Newton-Raphson method to be able to work its magic, we need to set this equation to zero.
x
3
−
3
=
0
Now we will recall the iterative equation for Newton-Raphson.
x
n
+
1
=
x
n
−
f
(
x
n
)
f
'
(
x
n
)
Substituting for
f
(
x
)
=
x
3
−
3
gives us:
x
n
+
1
=
x
n
−
(
x
n
)
3
−
3
3
⋅
(
x
n
)
2
Now, we pick an arbitrary number, (the closer it actually is to
3
√
3
the better) for
x
0
. Let's use
x
0
=
0.5
. Then we substitute each previous number for
x
n
back into the equation to get a closer and closer approximation to a solution of
x
3
−
3
=
0
.
x
1
=
0.5
−
(
0.5
)
3
−
3
3
⋅
(
0.5
)
2
=
4.33333
¯
3
x
2
=
x
1
−
(
x
1
)
3
−
3
3
⋅
(
x
1
)
2
≈
2.94214333
x
3
=
x
2
−
(
x
2
)
3
−
3
3
⋅
(
x
2
)
2
≈
2.07695292
x
4
=
x
3
−
(
x
3
)
3
−
3
3
⋅
(
x
3
)
2
≈
1.61645303
x
5
=
x
4
−
(
x
4
)
3
−
3
3
⋅
(
x
4
)
2
≈
1.46034889
x
6
=
x
5
−
(
x
5
)
3
−
3
3
⋅
(
x
5
)
2
≈
1.44247296
x
7
=
x
6
−
(
x
6
)
3
−
3
3
⋅
(
x
6
)
2
≈
1.4422496
x
8
=
x
7
−
(
x
7
)
3
−
3
3
⋅
(
x
7
)
2
≈
1.44224957
You can see that with only 8 iterations, we've obtained an approximation of
3
√
3
which is correct to 8 decimal places!
You can apply this same logic to whatever cube root you'd like to find, just use
x
3
−
a
=
0
as your equation instead, where
a
is the number whose cube root you're looking for.
Answer:
x= 2+2x/x^2 (also ignore the part that says power of x it couldn't disappear for me)
Step-by-step explanation:
x