Constructing Newton iteration converging to non-root

Question

Is it possible to construct a Newton sequence $x_{n+1} := x_{n} - f(x_n)/f'(x_{n})$ such that $\{x_{n}\}$ is a Cauchy sequence converging to $x^*$, but $x^{*}$ is not a root of $f$? (Perhaps because $f$ has no roots?)

Answer 1

It is not possible if we assume that $x^\star$ lies in the domain of definition of $f$ and that $f$ and $f'$ are continuous at that point. For a counterexample where $f'$ is not continuous at $x^\star$ see the nice answer by Oscar Lanzi.

In my setting we use $$ f(x_n) = f'(x_n) (x_n - x_{n+1}) $$ and take the limit (I assume $f'$ to be continuous at $x^*$).

If we take the limit in the equation above we get (using the continuity of $f'$ and $f$ at $x^\star$) $$ f(x^\star) = f(\lim_{n\rightarrow \infty} x_n) = \lim_{n\rightarrow \infty} f(x_n) = \lim_{n\rightarrow \infty} f'(x_n) (x_n - x_{n+1}) = \lim_{n\rightarrow \infty} f'(x_n) \cdot \lim_{n\rightarrow \infty} (x_n - x_{n+1}) = f'(x^\star) \cdot (x^\star - x^\star) = 0.$$

Answer 2

Try this function:

$f(x)=\max(1-\sqrt{|x|},|x|)$

For most initial guesses, and for all initial guesses more than one-half in absolute value, you converge to zero. But the function has no real zeroes.

Answer 3

An idea to construct such a function $f(x)$:

First, find a sequence $\{ x_n \}$ increasingly convergent to $x^*$. Let $f(x_n)=1$ and $f(x^*)=1$. (This $\{x_n\}$ will be the sequence found by the newton method's iteration, and the point $x^*$ will be the converging point which is not a root of $f(x)$)

Second, linearly connect point $(x_n,f(x_n))$ and $(x_{n+1},0)$ and take a small enough neighbor for each $x_n$ as a part of function $f(x)$. (This step makes newton method iterate from $x_n$ to $x_{n+1}$)

Third, smoothly connect the neighbors of $x_n$s.

Note: when the neighbors are small enough the value of $\frac{f(x^*)-f(x)}{x^*-x}$ in the neighbor can be arbitrary small and therefore we can construct the function $f(x)$ be differentiable at $x^*$ as well.