Let $f:\mathbb R\to \mathbb R$ be continuously differentiable. We say that $x_0$ is a simple root of $f$ if $f(x_0)=0$ and $f'(x_0)\neq0$. Show that if $x_0$ is a simple root of $f$, then $x_0$ is a super-attracting fixed point of the Newton's method iteration function $N$ for $f$.

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# Newton’s method yields a super-attracting fixed point!

mark

Levente

Suppose that $x_0$ is a simple root of $f$. Newtons iteration function $N(x) = x - \frac{f(x)}{f'(x)}$, if we evaluate this function at $x_0$ we find $$N(x_0) = x_0 - \frac{f(x_0)}{f'(x_0)} = x_0$$ Since $f(x_0) = 0$ and the derivative of $f$ at $x_0$ is non-zero. This implies that $x_0$ is a fixed point.

Now we show that that $x_0$ is a super attracting fixed point of $N(x)$ namely by showing that $N'(x_0) = 0$. Thus we take the derivative of $N(x)$, $$N'(x) = 1 - \frac{f'(x)f'(x) - f''(x)f(x)}{f'(x)^2}$$ We now evaluate this at $x_0$, since $f(x_0) = 0$ we get $$N'(x_0) = 1 - \frac{f'(x)^2}{f'(x)^2} = 0 $$Thus $x_0$ is a super attracting fixed point of the newton Iteration function of $f(x)$.