Function for newton method  y=(9-x)^1/3 - 3x -2
X value= 1

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Request for Question Clarification by mathtalk-ga on 27 Jul 2004 05:35 PDT

Hi, kevswife-ga:

cunoodle-ga has _evaluated_ y at x = 1 (obtaining y = -3, correctly).

However the reference to Newton's method suggests that you are looking
for something quite different, likely a root of the equation y(x) = 0.

Note that y(0) > 0 while y(1) < 0, so by the intermediate value
theorem there is a root x between 0 and 1.

One can use x = 1 as the initial value for such an iteration.  It
remains to be clarified, if a root is what you are seeking, how many
iterations or how much precision in the root is required.

regards, mathtalk-ga

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Clarification of Question by kevswife-ga on 27 Jul 2004 10:31 PDT

I need a root using Newtons Method. Starting with a seed of -100

y=(9-x)^1/3 - 3x -2

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Request for Question Clarification by mathtalk-ga on 27 Jul 2004 11:48 PDT

Since Newton's method does not give an exact root, I would think you
need to specify either the maximum number of iterations to use or the
"precision" you are willing to accept in a root (or both!).

The choice of initial "seed" x = -100, far from the actual root, seems
possibly to have been chosen to demonstrate a difficulty in obtaining
convergence with Newton's method.  I've not done the computations for
this function, but it is possible with functions related to the
cube-root used here to have the Newton corrections overshoot the mark
by larger and larger margins, illustrating that there is no guarantee
of convergence for an arbitrary starting value.  [When you have an
isolated root of a twice-continuously differentiable function, there
is a "basin of attraction" neighborhood of the root in which starting
values will converge to the root.]

regards, mathtalk-ga

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Clarification of Question by kevswife-ga on 27 Jul 2004 12:09 PDT

find a root using Newton's Method to solve the problem y=(9-x)^1/3 -3x-2

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Clarification of Question by kevswife-ga on 27 Jul 2004 12:13 PDT

10 iterations
precision of .0001

This question can be broken up into 3 parts

1. "(9-x)^1/3"
2. "- 3x"
3. "- 2"

Explanation of each step
1.  (9-x) = (9-1) = (8)
When rasing to the power '^' of (1/3) you are in essence asking for
the "cube root" of an answer.  The cube root of 8 is 2.
2.  "- 3x"  = "- 3 * 1" = "- 3"

So your answer would be figured like this..
2 - 3 - 1 = -2
y = -2

Here it is all at once...

y=(9-x)^1/3 - 3x -2

now sub 1 for X...
y=(9-1)^1/3 - 3 * 1 - 2

basic subtraction...
y=(8)^1/3 - 3 * 1 - 2

multiply the 3 and the 1
y=(8)^1/3 - 3 - 2

take the cube root of 8
y=2 - 3 - 2

do the final subtraction...
y = -2


(9-1) = 8
8^(1/3) = 2
3x = 3 * 1 = 3



2-3-2 = -3

I MADE A MISTAKE!!!

I am VERY sorry but I made a basic mistake on the subtraction section...

should be 
2 - 3 - 2
the final CORRECT answer should be "y = -3"

Again here it is written out...

Here it is all at once...

y=(9-x)^1/3 - 3x -2

now sub 1 for X...
y=(9-1)^1/3 - 3 * 1 - 2

basic subtraction...
y=(8)^1/3 - 3 * 1 - 2

multiply the 3 and the 1
y=(8)^1/3 - 3 - 2

take the cube root of 8
y=2 - 3 - 2

do the final subtraction...
y = -3  <<<< THIS is the CORRECT answer

Using Newton's method in finding a root of f(x).

x1 = x0 -[f(x0)]/[(f'(x0)]  <---***
where
x0, read as x-sub-0, is initial guess as the root.
x1, read as x-sub-1, is the root nearer to the correct root than x0.
f(x0) is the value of f(x) when x=x0.
f'(x0) is the value of the first derivative of f(x) when x=x0.

Two or three iterations is enough to get very closde to the correct
root. But I will use only one iteration here. Just to show the method.

f(x) = (9-x)^(1/3) -3x -2  <---***
So,
f'(x) = (1/3)[(9-x)^(-2/3)](-1) -3
f'(x) = -(1/3)[(9-x)^(-2/3)] -3  <----***

I will use x0 = -1.
So,
x1 = -1 -{[(9-(-1))^(1/3) -3(-1) -2]/[-(1/3)(9 -(-1))^(-2/3) -3]}
x1 = -1 -{[(10)^(1/3) +3 -2]/[-(1/3)(10)^(-2/3) -3]}
x1 = -1 -{[2.154434 +3 -2]/[-0.071814 -3]}
x1 = -1 +(3.154434)/(3.071814)
x1 = 0.026896  <----that should be close to the correct root.

If you like, get an x2, where, by Newton's method,
x2 = x1 -{[f(x1)]/[f'(x1)]}  <---***

Let us check that x1 = 0.026896,
f(0.026896) should be close to zero if x=0.026896 is a root of f(x).
f(0.026896) = (9 -0.026896)^(1/3) -3(0.026896) -2
= 2.0781 -0.080688 -2
= -0.002588  <---very close to zero. Not bad.

A second iteration, or x2, should come up really close to the correct root of f(x).