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The equation $x^2+ax+b=0$ has two real roots $\alpha$ and $\beta$.

Show that the iterative method given by

$\displaystyle x_{k+1}=-\frac{(ax_k+b)}{x_k}$

is convergent near $x=\alpha$, if $|\alpha|>|\beta|$

prat
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    Ok, what did you do to solve it? – Karl Sep 29 '14 at 12:05
  • I know how to solve an equation using Newton Raphson method. But I have no idea how to proceed in this problem as it asks about where the iterative scheme is going to converge. I took limit as k tends to infinity on both sides and ended up with the original equation – prat Sep 29 '14 at 12:14
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    You have to consider the term $x_k-\alpha$. Did you do this? Showing convergence is to show that this term is approaching zero. You did not show your work, so no possibility to help. – Karl Sep 29 '14 at 17:45

1 Answers1

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You can see this as a fixed point iteration method for find a root of the polynomial function: $$p(x) = x^2 + ax + b,$$ where the iteration function is defined by, $$\phi(x) = - \frac{ax + b}{x}.$$

Since $\phi(x)$ and $\phi'(x)$ are continuous, the iterative method will converge if $|\phi'(x)|<1$, for all $x$ sufficiently near to $\alpha$. We have, $$\phi'(x) = -\frac{b}{x^2} = \frac{\alpha\beta}{x^2}.$$ Then, for $x$ sufficiently near to $\alpha$, $|\phi'(x)|<1$, because $|\alpha|>|\beta|$. So the iterative method is convergent near $x = \alpha$.

Jardel Vieira
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