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Proof of the Newton Raphson method for finding complex roots
By Amars Birdi (P2769) on Saturday, March 24, 2001 - 01:27 pm:

We all know the graphical proof of the Newton Raphson method that draws a tangent to the graph in question and then finds where this intersects the x-axis, giving better approximations. How could I adapt this proof to explain why the Newton Raphson method works for complex roots (provided that you put complex starting points in).
Thanks!!!

By David Loeffler (P865) on Saturday, March 24, 2001 - 10:59 pm:

Well, obviously the graphical proof doesn't work unless you are lucky enough to be able to visualise four dimensions, so you have to use Taylor series.

Suppose that you are trying to find a root of a function f(x), and that root is a. Let your initial guess be X = a + e, where e is quite small.

Then we can expand f(x) in a Taylor series at X, so we have f(a) = f(X - e) = f(X) - ef'(X) + something very small.

However, since a is a root, f(a) = 0. So we have f(X) - ef'(X) = 0, or e = f(X)/f'(X).

Thus a better estimate for a is X - e = X - f(X)/f'(X), which is the Newton - Raphson formula.

(However, we have assumed that differentiation still exists for complex functions. This is not trivial, but relatively safe for well behaved functions like polynomials and rational functions.)

David

By Arun Iyer (P4587) on Wednesday, June 13, 2001 - 08:47 pm:

Well,
if we can represent complex numbers on a graph then why can we not give the same graphical explanation for complex numbers as well.
Arun

By Kerwin Hui (Kwkh2) on Thursday, June 14, 2001 - 01:17 am:

I don't quite understand your question. We can graph a complex function in 4-D, and we will obtain a surface as our graph. The reason why we need 4-D is that we need 2-D for a complex number (since we have not experience something like complex lines, but only real lines).

Now, assuming you can visualise 4-D (don't ask me how to do this!), we can find a tangent plane to our surface u+iv=f(x+iy) in our 4-D space x,y,u,v and work through the exact detail of the proof in the real case, except now we have to set two of the variables to zero, viz u=v=0.

Unlike the real case, we can guarantee quadratic convergence in the complex case (for sufficiently small e).

Kerwin