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I am studying Newton-Raphson Method but I am facing questions in my head. As far as I know Newton Raphson Method works on real values, but what if Newton Raphson Method faces an imaginary number when it is find real values on x-axis? and What to do when I have them when using Newton Raphson Method?

Please let me have your advice. I am sorry if I have placed my questions in the wrong places. Please do not answer if you are not going to explain in a plain language.

Thank you.

user122358
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    Can you elaborate on "How do I know if I have an imaginary number"? – hrkrshnn Jul 09 '14 at 17:08
  • I just have edited my question. – user122358 Jul 09 '14 at 17:10
  • It is still not clear... – Surb Jul 09 '14 at 17:10
  • Which part? Pinpoint which part is not clear? – user122358 Jul 09 '14 at 17:11
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    How could you possibly 'face an imaginary number'? The update is of the form $x_{n+1} = x_n -\alpha_n d_n$ where $d_n$ is obtained from solving a linear system of equations. – copper.hat Jul 09 '14 at 17:11
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    Why not face an imaginary number? – user122358 Jul 09 '14 at 17:12
  • Give an example of what you are talking about. – copper.hat Jul 09 '14 at 17:12
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    Ok, then let me ask you that if I am ever going to face an imaginary number? – user122358 Jul 09 '14 at 17:13
  • No you won't because you only consider real numbers and real operations between them... – Surb Jul 09 '14 at 17:14
  • I guess you're wondering what happens if you're trying to solve an equation that contains complex values, or from a complex starting point. The answer is that it gets complicated. Seriously complicated: http://www.chiark.greenend.org.uk/~sgtatham/newton/ – Dan Piponi Jul 09 '14 at 17:17
  • How is this different from http://math.stackexchange.com/q/861105/27978? – copper.hat Jul 09 '14 at 17:19
  • Your question/concern is unclear. You need to elaborate what your issue is. Are you asking if you can apply Newton's method to complex equations (yes, you can) or are you asking if complex numbers can ever arise when solving real systems using Newton's method (no, you can't, for reasons that have been pointed out twice above)? – copper.hat Jul 09 '14 at 17:21
  • The situation is like this. I am currently studying on Newton Raphson Method, a friend of mine told that what if I meet an imaginary root while performing Newton Raphson Method. I couldn't answer what his question was. I have been think about that if it is possible that I will face imaginary roots. – user122358 Jul 09 '14 at 17:43

1 Answers1

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I am not entirely sure what your issue is, I will address what I think the possible issues are.

The iteration is defined by $x_{n+1} = x_n - { f(x_n) \over f'(x_n) }$.

You can do computations in $\mathbb{R}$ or $\mathbb{C}$. In either case, you must have $f'(x_n) \neq 0$ for the iteration to be well-defined.

If the function maps reals to reals (that is, $f(\mathbb{R}) \subset \mathbb{R}$), then if $x_0 \in \mathbb{R}$, and the iterates are well-defined, then the above equation shows that $x_n \in \mathbb{R}$ for all $n$. So, $x_n$ can never 'turn' complex using this scheme.

If you are doing computations in $\mathbb{C}$, then it is entirely possible that the iterates will be complex.

It is entirely possible that the system $f(x) = 0$ has no solutions in either the real or complex numbers. For example, take $f(x) = 1$.

Here is a more illustrative example: Let $f(x) = x^2+1$. Note that if $x$ is real, then $f(x)$ will be real.

As a function $f: \mathbb{C} \to \mathbb{C}$, the equation $f(x)=0$ has two solutions ($\pm i$, of course).

As a function $f: \mathbb{R} \to \mathbb{R}$, the equation $f(x)=0$ has no solutions (since $f(x) \ge 1$ for all real $x$).

If you start with $x_0$ 'near' $\pm i$, and use complex arithmetic, then $x_n$ will converge very quickly to $\pm i$.

If you start with $x_0 \in \mathbb{R}$, then either $f'(x_n) = 0$ for some $n$ (in which case we must have $x_n = 0$, and subsequent iterates are not defined) or the iterates do not converge (if they did we would get a contradiction for this function). However, if they are defined, then the iterates are all real.

copper.hat
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