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Use the Newton-Raphson iteration method (NR) to find one positive root ( Xr > 0 ) of the equation $ X^3 +X -1 =0 $ ??

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closed as off-topic by TZakrevskiy, vonbrand, Andrew D. Hwang, Brian Rushton, Tomás Mar 17 '14 at 19:14

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up vote 1 down vote accepted

Yes. Set $f(x)=x^3+x-1$ and iterate with the Newton Raphson formula:

$$ x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}. $$ You need to guess a value for $x_0$ to start the iteration. Also, $f'(x_n)$ is the derivative of your function $f$ evaluated at $x_n$. If your iterations don't converge, try a different starting value $x_0$.

Hope this helps some.


Paul Safier

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Approach outline:

A plot (courtesy of Wolfram-Alpha) of the function shows:

enter image description here

Choose a starting point, say:

$$x_0 = 1$$

Write the Newton Iteration formula as:

$$x_{n+1} = x_n - \dfrac{f(x)}{f'(x)} = x_n - \dfrac{x_n^3 + x_n -1}{3 x_n^2 + 1}$$

Write out the iterates $x_0, x_1, x_2, \ldots$ and find:

$$x_5 = 0.682328$$

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Thanks for the right answer , i wish i had 15 reputation to give you vote up :( – Ahmed Osama Mar 17 '14 at 18:38
Thank you so much , any way i am student faculty of mechanical engineering , thank you again 7 you are welcome too :D – Ahmed Osama Mar 17 '14 at 18:47

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