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

  1. #1
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    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 [tex]f\, :\, [a,\, b]\, \rightarrow\, \mathbb{R}[/tex] have the following properties:

    . . . . .[tex]\mbox{1. }\, f(a)\, f(b)\, <\, 0,[/tex]

    . . . . .[tex]\mbox{2. }\, f\, \mbox{ has no critical points in }\, (a,\, b),\, \mbox{ and}[/tex]

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

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

    . . . . .[tex]\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[/tex]

    . . . . .[tex]\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 [/tex]

    In both cases we have for all iterates [tex]x_n[/tex] the estimate

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




    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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    Last edited by stapel; 11-10-2017 at 04:34 PM. Reason: Typing out the text in the graphic; creating useful subject line.

  2. #2
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    Quote Originally Posted by dan34 View Post
    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 [tex]f\, :\, [a,\, b]\, \rightarrow\, \mathbb{R}[/tex] have the following properties:

    . . . . .[tex]\mbox{1. }\, f(a)\, f(b)\, <\, 0,[/tex]

    . . . . .[tex]\mbox{2. }\, f\, \mbox{ has no critical points in }\, (a,\, b),\, \mbox{ and}[/tex]

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

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

    . . . . .[tex]\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[/tex]

    . . . . .[tex]\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 [/tex]

    In both cases we have for all iterates [tex]x_n[/tex] the estimate

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




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

    Use this approximation as [tex]x_{1}[/tex] and continue as necessary.
    Last edited by stapel; 11-10-2017 at 04:34 PM. Reason: Copying typed-out graphical content into reply.
    "Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.

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

    Use this approximation as [tex]x_{1}[/tex] 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)

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    [tex]f(x) = \dfrac{1}{2}x^{3} - 2x^{2} + 3x + 2[/tex] <== I'm assuming that second term should be x^2

    [tex]f'(x) = \dfrac{3}{2}x^{2} - 4x + 3[/tex]

    [tex]f''(x) = 3x - 4[/tex] 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. [tex]x_{1} = x_{0} - \dfrac{f(x_{0})}{f'(x_{0})} = -6[/tex] Doesn't seem all that promising, does it? Keep it coming. It will get there. x = 1 was a TERRIBLE place to start.
    Last edited by tkhunny; 11-03-2017 at 06:06 PM.
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    Finally, calculate the next value. [tex]x_{1} = x_{0} - \dfrac{f(x_{0})}{f'(x_{0})} = -6[/tex] 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?
    Last edited by dan34; 11-04-2017 at 10:09 AM.

  6. #6
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    Quote Originally Posted by dan34 View Post
    Finally, calculate the next value. [tex]x_{1} = x_{0} - \dfrac{f(x_{0})}{f'(x_{0})} = -6[/tex] 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)
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

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    Quote Originally Posted by Subhotosh Khan View Post
    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?

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    Never hurts to do another iteration, just for practice.
    "Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.

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