# Newtonian Method, just need algebra help

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• October 21st 2007, 04:54 PM
mistykz
Newtonian Method, just need algebra help
f(x) = 4x^4 - 4x^2
x0 = sqrt(21)/7
x1 = -sqrt(21)/7

i need to show, by hand, that you can calculate x1 from x0. i tried, but the answers didn't match up. exact square roots (no decimals) are to be used.
the equation i used was x1 = x0 - f(x0)/f '(x0), is that correct?
• October 21st 2007, 06:38 PM
topsquark
Quote:

Originally Posted by mistykz
f(x) = 4x^4 - 4x^2
x0 = sqrt(21)/7
x1 = -sqrt(21)/7

i need to show, by hand, that you can calculate x1 from x0. i tried, but the answers didn't match up. exact square roots (no decimals) are to be used.
the equation i used was x1 = x0 - f(x0)/f '(x0), is that correct?

Yes, you have the correct equation.
$f(x) = 4x^4 - 4x^2$

Thus
$f^{\prime}(x) = 16x^3 - 8x$

Thus
$x_{n + 1} = x_n - \frac{4x_n^4 - 4x_n^2}{16x_n^3 - 8x_n}$

So with $x_0 = \frac{\sqrt{21}}{7}$ we get
$x_1 = \frac{\sqrt{21}}{7} - \frac{4 \left ( \frac{\sqrt{21}}{7} \right )^4 - 4 \left ( \frac{\sqrt{21}}{7} \right ) ^2}{16 \left ( \frac{\sqrt{21}}{7} \right ) ^3 - 8 \left ( \frac{\sqrt{21}}{7} \right )}$

$= \frac{\sqrt{21}}{7} - \frac{4 \left ( \frac{21^2}{7^4} \right ) - 4 \left ( \frac{21}{7^2} \right ) }{16 \left ( \frac{21 \sqrt{21}}{7^3} \right ) - 8 \left ( \frac{\sqrt{21}}{7} \right )}$

$= \frac{\sqrt{21}}{7} - \frac{4 \left ( \frac{21^2}{7^4} \right ) - 4 \left ( \frac{21}{7^2} \right ) }{16 \left ( \frac{21 \sqrt{21}}{7^3} \right ) - 8 \left ( \frac{\sqrt{21}}{7} \right )} \cdot \frac{ \frac{ \sqrt{21}}{7^4} }{ \frac{ \sqrt{21}}{7^4} }$

$= \frac{\sqrt{21}}{7} - \frac{4 \cdot 21^2 \sqrt{21} - 4 \cdot 21 \sqrt{21} \cdot 7^2 }{16 \cdot 21^2 \cdot 7 - 8 \cdot 21 \cdot 7^3}$

I'm going to factor that last fraction a bit...
$= \frac{\sqrt{21}}{7} - \frac{21 - 7^2}{2 \cdot 21 - 7^2} \cdot \frac{4 \cdot 21 \sqrt{21}}{8 \cdot 21 \cdot 7}$

$= \frac{\sqrt{21}}{7} - \frac{-28}{-7} \cdot \frac{\sqrt{21}}{2 \cdot 7}$

$= \frac{\sqrt{21}}{7} - 2 \cdot \frac{\sqrt{21}}{7}$

$= -\frac{\sqrt{21}}{7}$

(whew! (Sweating) )

-Dan
• October 21st 2007, 09:22 PM
mistykz
wow, i was nowhere near close. thanks a ton!