Thread: newtons method initial valve (Xo)

1. newtons method initial valve (Xo)

hi guys, just a question on newtons method. I want to know how to do determine the first value we input into the equation? So for Xo ? i know you could guess this and just start the iterations from there but might take longer than knowing an initial value to start with so its shorter and you get more accurate results

for example

$\displaystyle x^3 = 4x + 1$
$\displaystyle x^3 - 4x - 1 = 0$

$\displaystyle f(x) = x^3 - 4x - 1$
$\displaystyle f ' (x) = 3x^2 - 4$

then we have $\displaystyle x - \frac {f(x)}{f'(x)}$

Xo = ?

$\displaystyle X1 = Xo - \frac {(Xo)^3 - 4(Xo) - 1}{3(Xo)^2 - 4}$

2. Originally Posted by jvignacio
hi guys, just a question on newtons method. I want to know how to do determine the first value we input into the equation? So for Xo ? i know you could guess this and just start the iterations from there but might take longer than knowing an initial value to start with so its shorter and you get more accurate results

for example

$\displaystyle x^3 = 4x + 1$
$\displaystyle x^3 - 4x - 1 = 0$

$\displaystyle f(x) = x^3 - 4x - 1$
$\displaystyle f ' (x) = 3x^2 - 4$

then we have $\displaystyle x - \frac {f(x)}{f'(x)}$

Xo = ?

$\displaystyle X1 = Xo - \frac {(Xo)^3 - 4(Xo) - 1}{3(Xo)^2 - 4}$

You guess, though often an educated guess. Do a little sketch of y=x^3 and y=4x+1, or try some values, or tabulate some values.

If you compute f(-10), f(-1), f(0), f(1), f(10) you will see that f changes sign between -10 and -x and between -1 and 0 and again between 1 and 10. Now as f(x) is continuous it has a root in each of the intervals (-10,-1), (-1,0) and (1, 10).

Choose the mid point of the interval containing the root you want might be a good method of choosing an initial value. As f(x) is a cubic these are all the roots that it has.

CB