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Math Help - Help- Newton - Raphson Method

  1. #1
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    Help- Newton - Raphson Method

    Consider the function f(x) = x- 4 In x

    Show that the equation f(x) = 0 has a solution in the interval (1,2)

    Show that for this function f the Newton - raphson method formula can be expressed as xn +1 = 4x n (1-4 In x)/4 - x n (n = 0,1,2....).
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  2. #2
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    Hello, fair_lady0072002!

    I'll do the second part . . . it's trickier.
    And I believe there's an extra "4" is the answer . . .


    Consider the function: f(x) \:= \:x- 4\ln x

    (b) Show that for this function f, the Newton- Raphson method formula
    can be expressed as: . x_{n +1} \;= \;\frac{4x_n(1 - 4\ln x_n)}{4 - x_n}? for n = 0,\,1,\,2\,\hdots

    Formula: . x_{n+1}\;=\;x_n - \frac{f(x_n)}{f'(x_n)}

    I'm going to drop the subscripts (temporarily) . . .

    We have: . f'(x) \:=\:1 - \frac{4}{x}

    Then: . x_{n+1}\;=\;x - \frac{x - 4\ln x}{1 - \frac{4}{x}}


    Multiply top and bottom of the fraction by x:\;\;x_{n+1}\;=\;x - \frac{x^2 - 4x\ln x}{x - 4}

    Get a common denominator: . x_{n+1} \;= \;\frac{x}{1}\cdot\frac{x-4}{x-4} - \frac{x^2 - 4x\ln x}{x - 4}


    x_{x+1}\;= \;\frac{x^2 - 4x - (x^2 - 4x\ln x)}{x - 4} \;= \;\frac{x^2 - 4x - x^2 + 4x\ln x}{x - 4} \;= \frac{-4x + 4x\ln x}{x - 4}


    Factor: . x_{n+1}\;=\;\frac{-4x(1 - \ln x)}{-(4 - x)}\quad\Rightarrow\quad \boxed{x_{n+1} \;= \;\frac{4x_n(1 - \ln x_n)}{4 - x_n}}

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  3. #3
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    My mistake....sorry and thanx

    second part should be

    x n+1 = 4x n (1- Inx n)/4-x n
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  4. #4
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    Hello again, fair_lady0072002!

    I hope this suffices as a solution . . .


    Consider the function: f(x) \:= \:x- 4\ln x

    Show that the equation f(x) = 0 has a solution in the interval (1,2)

    It can shown that f(x) is continuous for all x > 0.

    We find that: . f(1) \;= \;1 - 4\ln1 \;= \;1
    . . . . . . and: . f(2) \;= \;2 - 4\ln 2 \;\approx \;\text{-}0.773

    Since the continuous function changes from positive to negative over the interval,
    . . it must achieve f(x) = 0 somewhere on the interval.

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