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Math Help - newtons method to locate t

  1. #1
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    newtons method to locate t

    use newtons method to locate t starting with b as an initial guess.
    let u=b - f(b)f'(b) denote the first approximation supplied by newtons method
    which of the followin are true
    1.u>b
    2.u<b
    3.u>a
    4.u<a
    5.u>t
    6.u<t
    7.impossible to determine

    so i had

    x^2=612
    f(x)=x^2-612
    f'(x)=2x
    b=10
    t=sqrt(612)
    10-(10^2-612)/(2*10)=35.6
    35.6-(35.6^2-612/(2*35.6)=26.395

    1.u>b true
    2.u<b false

    im not ure what a is supposed to be
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  2. #2
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    Hi there, this is quite hard to follow, maybe its the order on how is it written.

    I get given x_0 = 10 with f(x) = x^2-216 \implies f'(x) = 2x

    x_1 = x_0-\frac{f(x_0)}{f'(x_0)} = 10-\frac{f(10)}{f'(10)} = 10-\frac{10^2-216}{2\times 10} =10-\frac{-116}{20} =4.2

    x_2 = x_1-\frac{f(x_1)}{f'(x_1)} = 4.2-\frac{f(4.2)}{f'(4.2)} = \dots

    not sure where a or t comes from? Are you confusing this with the bi-section method?
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  3. #3
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    Quote Originally Posted by dat1611 View Post
    use newtons method to locate t starting with b as an initial guess.
    let u=b - f(b)f'(b) denote the first approximation supplied by newtons method
    which of the followin are true
    1.u>b
    2.u<b
    3.u>a
    4.u<a
    5.u>t
    6.u<t
    7.impossible to determine
    As written, that question is meaningless.
    The Newton’s method is not even correct.
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  4. #4
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    suppose you use newtons method to locate t starting with b as an initial guess.
    let u=b - f(b)/f'(b) denote the first approximation supplied by newtons method
    ii) which of the followin are true?
    1.u>b
    2.u<b
    3.u>a
    4.u<a
    5.u>t
    6.u<t
    7.impossible to determine

    thats excactly how its written
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  5. #5
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    Quote Originally Posted by dat1611 View Post

    thats excactly how its written
    The second time you have supplied this you have divided the function and derivative as required. Additional information is required to answer the question as is so i'm picking option 7.
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  6. #6
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    I agree with option 7.
    Consider: f(x)=x^3-3x+6,~\7~b=1
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  7. #7
    MHF Contributor chisigma's Avatar
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    The Newton's method find the solutions of an equation of the type f(x)=0 starting fron an x_{1} and generating the sequence defined by...

    \displaystyle x_{n+1} = x_{n} - \frac{f(x_{n})}{f^{'} (x_{n})} (1)

    A simple question: how to select x_{1} in order to have convergence?...

    The simple answer : it depends from the specific f(x) ...

    As easy example lets consider the equation x^{2} - 4 =0. In that case the (1) becomes...

    \displaystyle \Delta_{n} = x_{n+1} - x_{n} = \frac{4-x^{2}_{n}}{2\ x_{n}} = \varphi (x_{n}) (2)

    The function \varphi(*) is illustrated in the figure...



    The solution is x_{0} = 2 and the 'red line' is l(x) , i.e. the line with slope -1 crossing the x axis in x=x_{0} . Because for x>0 is \varphi (x) \ge l(x) any initial value x_{1} > 0 will produce convergence and that is easy to be verified. Not for all f(x) however the matter is so easy ...

    Kind regards

    \chi \sigma
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  8. #8
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    Quote Originally Posted by dat1611 View Post
    suppose you use newtons method to locate t starting with b as an initial guess.
    let u=b - f(b)/f'(b) denote the first approximation supplied by newtons method
    ii) which of the followin are true?
    1.u>b
    2.u<b
    If f(b) and f'(b) have the same sign, the u< b. If f(b) and f'(b) have different sign, then u> b. Since it is possible to give examples of both, neither of these can be true.

    3.u>a
    4.u<a
    These don't make sense. There was no "a" in your problem.

    5.u>t
    6.u<t
    These don't make sense. "t" is a variable and can have any value.

    7.impossible to determine

    thats excactly how its written
    I guess 7 is the only one left.
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  9. #9
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by Plato View Post
    I agree with option 7.
    Consider: f(x)=x^3-3x+6,~\7~b=1
    It is a very interesting case ... the equation is...

    \displaystyle  x^{3} - 3\ x + 6 = 0 (1)

    ...and its only real solution is x_{0} = -2.35530139761.... The corresponding \varphi(*) [see my previous post...] is...

    \displaystyle \varphi(x)= - \frac{1}{3}\ \frac{x^{3} - 3\ x + 6}{x^{2}-1} (2)

    ... and it is represented here...



    The function has two singularities in x=1 and x=-1 and it is evident [see my previous post...] that any initial value x_{1} < -1 will produce a sequence converging to x_{0}. For x_{1}>-1 and x_{1} \ne 1 the analysis is quite complex and the convergence is not guaranted...

    Kind regards

    \chi \sigma
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