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Math Help - function that corresponds to Newton's method

  1. #1
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    Question function that corresponds to Newton's method

    Question is if f(x) = x^4  + x - 19 then which of the following functions correspond to Newton's method x_{n + 1}  = g(x_n)

    Answer is given as <br />
g(x) = \frac{{3x^4  + 19}}{{4x^3  + 1}}<br />

    But I would have thought it would be
    <br />
g(x) = \frac{{ - x^4  + 4x^3  - x + 20}}{{4x^3  + 1}}<br />

    considering

    <br />
\begin{array}{l}<br />
 x_{n + 1}  = x_n  - \frac{{f(x)}}{{f'(x)}} \\ <br />
 x_{n + 1}  = x_n  - \frac{{x^4  + x - 19}}{{4x^3  + 1}} \\ <br />
 x_{n + 1}  = \frac{{4x^3  + 1}}{{4x^3  + 1}} - \frac{{x^4  + x - 19}}{{4x^3  + 1}} \\ <br />
 x_{n + 1}  = \frac{{ - x^4  + 4x^3  - x + 20}}{{4x^3  + 1}} \\ <br />
 \end{array}<br />

    Could someone clarify what I am doing wrong or not interpreting correctly. Thanks
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  2. #2
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    earboth's Avatar
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    Quote Originally Posted by Craka View Post
    Question is if f(x) = x^4  + x - 19 then which of the following functions correspond to Newton's method x_{n + 1}  = g(x_n)

    Answer is given as <br />
g(x) = \frac{{3x^4  + 19}}{{4x^3  + 1}}<br />

    But I would have thought it would be
    <br />
g(x) = \frac{{ - x^4  + 4x^3  - x + 20}}{{4x^3  + 1}}<br />

    considering

    <br />
\begin{array}{l}<br />
 x_{n + 1}  = x_n  - \frac{{f(x)}}{{f'(x)}} \\ <br />
 x_{n + 1}  = x_n  - \frac{{x^4  + x - 19}}{{4x^3  + 1}} \\ <br />
 x_{n + 1}  = \frac{{4x^3  + 1}}{{4x^3  + 1}} - \frac{{x^4  + x - 19}}{{4x^3  + 1}} \\ <br />
 x_{n + 1}  = \frac{{ - x^4  + 4x^3  - x + 20}}{{4x^3  + 1}} \\ <br />
 \end{array}<br />

    Could someone clarify what I am doing wrong or not interpreting correctly. Thanks
    Your mistake is hidden in the third row of your transformations:

    According to Newtons formula you are looking for:

    g(x)=x-\dfrac{x^4+x-19}{4x^3+1}=\dfrac{x(4x^3+1)}{4x^3+1}-\dfrac{x^4+x-19}{4x^3+1}=\dfrac{4x^4+x-(x^4+x-19)}{4x^3+1} =  \dfrac{4x^4+x-x^4-x+19}{4x^3+1}

    Collect like terms and you'll get the given answer.
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  3. #3
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    Thanks, just a little silly algebraic mistake on my behalf
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