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)-e f ' (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+i v=f(x+i y) 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