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Wei Tsang

Mat 351

Prof PrestonMay 7, 2006

Newtons Method for Complex PolynomialMany important questions in mathematics involve the solution of equations or

systems of equations. Unfortunately, many of them cannot be solved by algebraic

manipulation. We therefore need to find a way to solve the equations numerically, in thehope of making accuracy as much as possible. There is a simple technique of elementary

calculus known as Newtons method. Newtons method is a method for iteratively

approximating the root of a function f(x) =0 using the derivative (under appropriate

hypothesis). In other words, we want to find a value x* such that f(x*) =0, and weproceed it by giving an educated guess and then refining our guess over and over again.

If the function is quite simple, we may be able to do some algebra to find a root x

exactly. For instance, if f(x) is a quadratic polynomial, we can use the quadratic formula.

If f(x) is a degree 3 or 4 polynomial, there are messier formulas which work as well. Butif f(x) is a higher degree polynomial or an even more complicated function, there is no

analog to the quadratic formula, i.e. there is no systematic process to algebraicallydetermine the roots exactly. We have to approximate, and Newton's method is just one

way of doing this.

We will first start our discussion on Newton's method over the real number line, i.e.

when x is always real and so is f(x). Later then we will see that the situation becomesmuch more interesting when we enlarge our scope, letting x and f(x) be complex numbers.

For simplicity, we will use the function f(z) =z3-1 in this case. Of course, we all know

that z=1 is the only real root, and there are also two other complex rootsz= -1/2 + (3

1/2/2) i = -0.5 + 0.866 i and z = -1/2 - (3

1/2/2) i = -0.5 - 0.866 i. (1)

, which can be calculated by formula to the k-th root of a complex number, and so we do

not really need Newton's method in this example. However, since many physical scienceequation has no real solution, we introduce complex numbers and are much more

interested in how a given complex number z converging to the root of the equation, for

example, f(z)=z3-1 under the sequence given by Newtons method. We will later see that

even in this simple case we already get pictures which are quite complicated enough than

on the real number line.

The Newton's method formula

The idea behind Newton's method is simple. Suppose that a mapping f: [a, b]R is a

differentiable mapping and f(x) changes sign, then there is a unique x* in the closedinterval [a, b] such that f(x*)=0. The existence of root x* and the convergence to the root

are guaranteed by Contraction Mapping Theorem. So if we take a starting "guess" for a

root locally close to the x* and call it x0, draw the tangent line to f(x) at x0, and figure outwhere this tangent line crosses the x-axis. This is the next guess x1 for a root. If f(x) were

equal to its tangent line, x1 would actually be a root of f(x), of course. In symbol, by

linear approximation,

f(x*)-f(x0 )= f '(x0 ) (x*-x0 )

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Then x*= x1 - f(x0 )/ f '(x0 )

In general, this is not the case. However, the tangent line approximates f(x) for x near

x0, and so we hope that x1 is not "too far off" from a root of f(x), in particular hopefully

closer than x0. Repeating the process with x1 instead of x0, we will get the nextapproximation x2, and so on. Thus each xn is determined by f(xn-1) and the slope of the

tangent line at xn, which is f '(xn-1). The appropriate formula is

xn = xn-1 - f(xn-1) / f '(xn-1) (2)

Demonstration of Newtons method:

Examples for f(x) =x3-1 over the real numbers

In this case, formula (2) becomes

xn = xn-1 - (xn-13

-1) / 3xn-12

. (3)Let's start with the guess x0=0.5.Iterating formula (3) gives the sequence

shown at right. We are getting closer and

closer to the root x=1, so things are

working well. In fact, you can see fromthe picture that any guess x0>1 will

quickly converge to the root x=1. In fact,if you randomly pick almost any x0,chances are xn will converge to x=1.

Notice, however, that if x0 is close to 0,

then the tangent line there is almost flatand x1 will be blasted way out to theright on the x axis. This is not much of a

problem, though, since x2 will be pushed

rapidly back, as the graph shows forx1=1.7 (approx.).

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If we choose x0=0, we are in trouble,

since the tangent is horizontal, f '(x0) =0and formula (*) crashes. Furthermore, if

we choose a different x0 which

eventually "lands" on xn=0, we have the

same problem. The graph at right showsthis in red for x0=-1.434, which crashes

in 2 steps in this manner. There is an

infinite sequence of "bad" negativepoints like this. Note that if we choose an

initial guess very close to such a bad

point, xn will eventually become close-but-not-equal-to 0, and then xn+1 will get

blasted way out to the right and from

there eventually converge to x=1. Thecloser we began to a bad point, the more

steps it will take to get "reasonablyclose" to x=1. An example of this is

shown in blue.

Using complex numbers

The question may arise what is going to happen if we now allow x (and f(x)) to becomplex numbers rather than just real numbers. So now our domain of a function is on a

complex plane. Can we get the other complex roots? This question makes perfect sense

since a polynomial of degree n has n roots, which is the Fundamental Theorem ofAlgebra. It is also possible to define the derivative over complex numbers just as for real

numbers, so formula (2) makes sense. It turns out that polynomials are all complex-

differentiable, and the complex-derivative is (surprise!) the same as the real-derivative.So formula (3) still holds, even though the geometry is much more complicated. (In part

since x is now in some sense 2-dimensional and likewise f(x), so we need to think in 4

dimensions!)

We change notation and call the variable z instead of x, and write x and y for its realand imaginary parts, so that z = x + i y. Starting with some (complex) guess z0, we can

apply the formula and get a sequence zn, which hopefully converges to a (complex) root.

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Here is what happens for three values of z0 close to 0.5 + 0.5 i.

The picture shows that one choice (blue) converges to the real root, but another (green)converges to one of the other complex roots. But if we choose precisely z0 = 0.539 +

0.417 i (determined with some effort!), then z1= 0 .539 - 0.417 i, and z2=z0! So we getinto an infinite cycle back and forth between these two values and do not converge to any

root. This is a different sort of "bad point" than the ones we found on the negative real

axis, when f '(x)=0.

Incidentally, the grey shading in the graph represents |f(z)|. The dark squares are nearthe roots, when |f(z)| is 0, and the squares are progressively lighter as |f(z)| gets bigger,

and hence f(z) gets "further from 0". So the darken square is the complex root of the

equation f(z)=0.Recall that the complex-valued polynomial of degree n has n roots. From the graph

above, we might question which root is being converged if we guess a z0 somewherebetween two roots. The question leads to an interesting phenomenon called Newton basin,which is fractal.

Using the same function f(z)=z3- 1to give a fractal, since the polynomial has the three

toots of unity as its roots. The three roots are 1, -0.5+0.866i, and -0.5-0.866i.

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We see the three white small circles are the three roots of the f(z)=0. To answer to whichroot is converge, notice that if we pick z0 in the pink region, the sequence defined by

Newtons method would converge to 1, likewise, z0 would converge to -0.5+0.866i if it is

in blue and to -0.5-0.866i if in light blue. (See reference)

Reference:

http://aleph0.clarku.edu/~djoyce/newton/examples.html