function that corresponds to Newton's method

**Craka** · November 16th 2008, 11:53 PM

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 $ g(x) = \frac{{3x^4 + 19}}{{4x^3 + 1}} $

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

considering

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

Could someone clarify what I am doing wrong or not interpreting correctly. Thanks

**earboth** · November 17th 2008, 01:11 AM

Originally Posted by Craka

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 $ g(x) = \frac{{3x^4 + 19}}{{4x^3 + 1}} $

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

considering

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

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.

**Craka** · November 17th 2008, 02:18 AM

Thanks, just a little silly algebraic mistake on my behalf

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