# Thread: Theorem for Newton Method: g(x) = 1/x x^3 - 2x + 3x + 2 on [-1, 1]

1. ## Theorem for Newton Method: g(x) = 1/x x^3 - 2x + 3x + 2 on [-1, 1]

g(x) = 1/2 x3 - 2x + 3x + 2 on the domain [-1, 1]

I have to perform iterations of Newton's method based on this theorem:

Theorem. Let $f\, :\, [a,\, b]\, \rightarrow\, \mathbb{R}$ have the following properties:

. . . . .$\mbox{1. }\, f(a)\, f(b)\, <\, 0,$

. . . . .$\mbox{2. }\, f\, \mbox{ has no critical points in }\, (a,\, b),\, \mbox{ and}$

. . . . .$\mbox{3. }\, f''\, \mbox{ exists, is continuous, and fulfills }\, f''\, \geq\, 0\, \mbox{ or }\, f''\, \leq\, 0\, \mbox{ on whole }\, (a,\, b).$

Then $f(x)\, =\, 0$ has exactly one solution $x^*.$ The Newton sequence $x_n$ converges always towards $x^*$ as $n\, \rightarrow\, \infty$ if the initial guess $x_0$ is chosen according to

. . . . .$\bullet\, \mbox{ if }\, f(a)\, <\, 0,\, f''\, \leq\, 0,\, \mbox{ or }\, f(a)\, >\, 0,\, f''\, \geq\, 0\, \mbox{ then }\, x_0\, \in\, [a,\, x^*],\, \mbox{e.g.}\, x_0\, =\, a$

. . . . .$\bullet\, \mbox{ if }\, f(a)\, <\, 0,\, f''\, \geq\, 0,\, \mbox{ or }\, f(a)\, >\, 0,\, f''\, \leq\, 0\, \mbox{ then }\, x_0\, \in\, [x^*,\, b],\, \mbox{e.g.}\, x_0\, =\, b$

In both cases we have for all iterates $x_n$ the estimate

. . . . .$\displaystyle \big| x_n\, -\, x^{*} \big| \, \leq\, \dfrac{\big|f(x_n)\big|}{\min\limits_{a \leq x \leq b}\, \big|f'(x)\big|}$

a) I have to show that the Conditions 1-3 hold, I proved that.

b) Compute the denominator of that expression in the end, min| g'(x) |, I completed this one also, the value is 1/2

c) choose an initial guess according to the theorem and perform iterations until the error |Xn - X*| lies below 10^-4

I do not understand this part how can I do the interations, if for example my Xo = 1, then how much is my X1, can someone do or explain just one or two interations, so then I am able to continue?

2. Originally Posted by dan34
g(x) = 1/2 x3 - 2x + 3x + 2 on the domain [-1, 1]

I have to perform iterations of Newton's method based on this theorem:

Theorem. Let $f\, :\, [a,\, b]\, \rightarrow\, \mathbb{R}$ have the following properties:

. . . . .$\mbox{1. }\, f(a)\, f(b)\, <\, 0,$

. . . . .$\mbox{2. }\, f\, \mbox{ has no critical points in }\, (a,\, b),\, \mbox{ and}$

. . . . .$\mbox{3. }\, f''\, \mbox{ exists, is continuous, and fulfills }\, f''\, \geq\, 0\, \mbox{ or }\, f''\, \leq\, 0\, \mbox{ on whole }\, (a,\, b).$

Then $f(x)\, =\, 0$ has exactly one solution $x^*.$ The Newton sequence $x_n$ converges always towards $x^*$ as $n\, \rightarrow\, \infty$ if the initial guess $x_0$ is chosen according to

. . . . .$\bullet\, \mbox{ if }\, f(a)\, <\, 0,\, f''\, \leq\, 0,\, \mbox{ or }\, f(a)\, >\, 0,\, f''\, \geq\, 0\, \mbox{ then }\, x_0\, \in\, [a,\, x^*],\, \mbox{e.g.}\, x_0\, =\, a$

. . . . .$\bullet\, \mbox{ if }\, f(a)\, <\, 0,\, f''\, \geq\, 0,\, \mbox{ or }\, f(a)\, >\, 0,\, f''\, \leq\, 0\, \mbox{ then }\, x_0\, \in\, [x^*,\, b],\, \mbox{e.g.}\, x_0\, =\, b$

In both cases we have for all iterates $x_n$ the estimate

. . . . .$\displaystyle \big| x_n\, -\, x^{*} \big| \, \leq\, \dfrac{\big|f(x_n)\big|}{\min\limits_{a \leq x \leq b}\, \big|f'(x)\big|}$

a) I have to show that the Conditions 1-3 hold, I proved that.

b) Compute the denominator of that expression in the end, min| g'(x) |, I completed this one also, the value is 1/2

c) choose an initial guess according to the theorem and perform iterations until the error |Xn - X*| lies below 10^-4

I do not understand this part how can I do the interations, if for example my Xo = 1, then how much is my X1, can someone do or explain just one or two interations, so then I am able to continue?
$x_{0} = 1$
$f(x_{0}) = f(1) =$?
$min|f'(x)|) = f'(2/3) =$? How do I know that? I don't get 1/2.
Solve for your first approximation of $x^{*}$. It is an algebra problem.

Use this approximation as $x_{1}$ and continue as necessary.

3. Originally Posted by tkhunny
$x_{0} = 1$
$f(x_{0}) = f(1) =$?
$min|f'(x)|) = f'(2/3) =$? How do I know that? I don't get 1/2.
Solve for your first approximation of $x^{*}$. It is an algebra problem.

Use this approximation as $x_{1}$ and continue as necessary.

Why f'(2/3), f'(2/3) = | -1/3 | = 1/3, but if you plug f'(1) = 1/2, which is lower than 1/3, therefore f'(1) is the minimum, right? (so the min = 1/2)

4. $f(x) = \dfrac{1}{2}x^{3} - 2x^{2} + 3x + 2$ <== I'm assuming that second term should be x^2

$f'(x) = \dfrac{3}{2}x^{2} - 4x + 3$

$f''(x) = 3x - 4$ 3x - 4 = 0 ==> x = 4/3 ==> min(f'(x)) on [-1,1] is f'(1) = 1/2

Now, I get 1/2. Somehow, 2(3/2) was 6, before. Apologies. That's what I get for typing with one hand.

What we know, then, is that we are no farther away from the solution than f(1)/f'(1) = (7/2)/(1/2) = 7. Doesn't seem very impressive, does it?

Finally, calculate the next value. $x_{1} = x_{0} - \dfrac{f(x_{0})}{f'(x_{0})} = -6$ Doesn't seem all that promising, does it? Keep it coming. It will get there. x = 1 was a TERRIBLE place to start.

5. Finally, calculate the next value. $x_{1} = x_{0} - \dfrac{f(x_{0})}{f'(x_{0})} = -6$ Doesn't seem all that promising, does it? Keep it coming. It will get there. x = 1 was a TERRIBLE place to start.[/QUOTE]

X5 = -0.61125
X6 = -0.49549
X7 = -0.48833

How to know now if I should proceed more or not, how to evaluate that 10-4 ​thing?

6. Originally Posted by dan34
Finally, calculate the next value. $x_{1} = x_{0} - \dfrac{f(x_{0})}{f'(x_{0})} = -6$ Doesn't seem all that promising, does it? Keep it coming. It will get there. x = 1 was a TERRIBLE place to start.

Values are being bigger in every step, they are not going toward 10-4 !?[/QUOTE]Not really!

After (-6), next iteration would be at (-3.58) and f(x) = -57.3
next iteration would be at (-2.01) and f(x) = -16.2 (going towards 0)

If you start with x = 1, it will take about 9 iteration to go to f(x) = -1.9*10^-9

Like TK said, x = 1 is a terrible place to start for this function.

On the other hand, if you start at x = 0, it will take four iteration to get to f(x) = 10^(-4)

7. Originally Posted by Subhotosh Khan
Values are being bigger in every step, they are not going toward 10-4 !?
Not really!

After (-6), next iteration would be at (-3.58) and f(x) = -57.3
next iteration would be at (-2.01) and f(x) = -16.2 (going towards 0)

If you start with x = 1, it will take about 9 iteration to go to f(x) = -1.9*10^-9

Like TK said, x = 1 is a terrible place to start for this function.

On the other hand, if you start at x = 0, it will take four iteration to get to f(x) = 10^(-4)[/QUOTE]

I'm sorry but I did the calculation wrong, now I started again with good calculation and now I am in x7 = -0.48833, and when I try to go further i get the same value for x8, too.

Now f (-0.48833) = -0.00015

Am I done?

8. Never hurts to do another iteration, just for practice.