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

dan34

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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 \(\displaystyle f\, :\, [a,\, b]\, \rightarrow\, \mathbb{R}\) have the following properties:

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

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

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

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

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

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

In both cases we have for all iterates \(\displaystyle x_n\) the estimate

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




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?
 

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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 \(\displaystyle f\, :\, [a,\, b]\, \rightarrow\, \mathbb{R}\) have the following properties:

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

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

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

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

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

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

In both cases we have for all iterates \(\displaystyle x_n\) the estimate

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




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?

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

Use this approximation as \(\displaystyle x_{1}\) and continue as necessary.
 
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\(\displaystyle x_{0} = 1\)
\(\displaystyle f(x_{0}) = f(1) = \)?
\(\displaystyle min|f'(x)|) = f'(2/3) = \)? How do I know that? I don't get 1/2.
Solve for your first approximation of \(\displaystyle x^{*}\). It is an algebra problem.

Use this approximation as \(\displaystyle x_{1}\) 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)
 
\(\displaystyle f(x) = \dfrac{1}{2}x^{3} - 2x^{2} + 3x + 2\) <== I'm assuming that second term should be x^2

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

\(\displaystyle f''(x) = 3x - 4\) 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.

Okay, now, how about the algorithm? One for free, since I was confusing.

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. \(\displaystyle x_{1} = x_{0} - \dfrac{f(x_{0})}{f'(x_{0})} = -6\) Doesn't seem all that promising, does it? Keep it coming. It will get there. x = 1 was a TERRIBLE place to start.
 
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Finally, calculate the next value. \(\displaystyle x_{1} = x_{0} - \dfrac{f(x_{0})}{f'(x_{0})} = -6\) 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?
 
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Finally, calculate the next value. \(\displaystyle x_{1} = x_{0} - \dfrac{f(x_{0})}{f'(x_{0})} = -6\) 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)
 
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?
 
Never hurts to do another iteration, just for practice.
 
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