Newton's method with no real roots So as the title would suggest I'm currently reading about Newton's method for finding roots. I'm having trouble understanding the reasoning for a function without a root.
It reads as following:
"Consider the function $f(x) =1+x^2$. Clearly f has no real roots though it does have complexroots $x\pm i$.The Newton method formula for f is:
$x_{n+1} = x_{n} - \frac{1+x^2}{2x_{n}}=\frac{x^2-1}{2x_{n}}$"
What is happening here? 
Many thanks to whomever might expand this a little for me!
 A: Newton's Method works only if the sequence of iterates converges. This is not always the case. 
For example if you choose $f(x)=\sqrt[3] x$ and try Newton's Method to find the root $x=0$ 
We get $$ x_{n+1}=x_n - \frac{f(x_n)}{f'(x_n)} = x_n - \frac{x_n^{1/3}}{(1/3) x_n^{-2/3}}=-2x_n $$ The sequence of iterates starting at $x=1$ is $$1,-2,4,-8,...$$ which does not converge to $x=0$ 
A: There's a typo in your displayed equation. The $x$ in the numerator should be $x_n$.  Then one just writes
$$x_{n+1} = x_n -\frac{1+x_n^2}{2x_n} = x_n\frac{2x_n}{2x_n} - \frac{1+x_n^2}{2x_n} =\frac{2x_n^2-x_n^2-1}{2x_n}. $$
A: In some cases, the Newton's Method does not work. For example, if you encounter a stationary point in the process (division by zero). 
Your example is another one in which Newton's Method does not work, because starting with a real number you only go through real numbers in the process. BUT you could start with a complex number, and (with a bit of luck) you will converge to the correct complex root.
