DIMACS
DIMACS REU 2013

General Information

Tim
Student: Timothy E. Brown
Office: CoRE 434
School: Morgan State University
E-mail: tibro26@morgan.edu
Project: Experiments with Geometric Modulus Principle for Polynomials

Project Description

Geometric Modulus Principle: Let p(z) be a nonconstant polynomial. If p(z0)=0, then every direction at z0 is an ascent direction for |p(z0)|. If p(z0)≠0 then the cones of ascent and descent direction at z0 partition the unit disc centered at z0 into alternating ascent and descent sectors of equal angle π ⁄ k where k≥1is the smallest index with pk(z0)≠0. (Bahman Kalantari 2011)


Weekly Log

Week 1:
I settled down on Sunday and began working on research in Complex Analysis. After orientation and talking with my advisor and roommates, I became excited to see the result of this program come mid July. On Monday I spoke with Dr. Kalantari and discussed research problems.Friday I gave my presentation.
Week 2:
After meeting my mentor I started look over Polynomial Interploation. To create a smooth uniform graph we agreed to use Hermite Interplotaion. If given zeros and critical points we should be able to construct a polynomial. Now we want to take this polynomial and see if it has the properties of Geometric Modulus Principle
Week 3:
I started to write the program for Hermite Interploation which will generate the polynomials I need for the Geometric Modulus Principle. My next goal is to create a program in Mathmatica which will display the modulus graph and the 2D contour which will highlight direction of ascent and descent.
Week 4:
I got my Hermite Interpolation code to work for complex numbers. Now I am attempting to code Newton's Method so I can visualize the roots that are in the convex hull. By putting all of these programs together I should get a nice 2D contour which highlight gradients and show roots in the convex hull.
Week 5:
This week I constructed a polynomial. In general you are given n and k ∫ (zn - 1)k dz. This creates a polynomial p (z) s.t p'(z)= p''(z) =. ... pk(z) = 0 for roots of zn- 1. Example suppose n=3 and k=1. That is ∫ (z3 - 1)1 dz=z4/4 -z. This generates a polynomial p(z) s.t p'(z)=0 for roots of z3 - 1.
Week 6:
I'm writing a code for the polynomial I created last week. This code will;
  1. Take the modulus of a complex polynomial p(z) (we write p(z) as u(x,y)+iv(x,y) then its modulus is f(x,y) = √(u2 + v2)=|p(z)|. Example: p(z)=z2-1. Set z=x+iy. Then f(x,y)=√((x2-y2-1)2+(2xy)2).
  2. Take a point z0 where p'(z)0)=0 but p''(z)0)≠0, e.g. if p(x)=z2-1, z0=0 is such a point.
  3. Evaluate f(z0)=|p(z0)|, then in a box containing z0for many points evaluate f(x,y). Whenever it is larger than f(z0) color it red and if it is smaller than f(z0) color it black.Then give the image.

Presentations


Additional Information