# function that corresponds to Newton's method

• Nov 16th 2008, 11:53 PM
Craka
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 $
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
• Nov 17th 2008, 01:11 AM
earboth
Quote:

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

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