5. Nonlinear Functions of Several Variables Jieun Yu (jieunyu@korea.ac.kr) Kyungsun Kwon (sadokwon@korea.ac.kr) Kyunghwi Kim (kyunghwi@korea.ac.kr) Sungjin Kim (sungsink@korea.ac.kr) Computer Networks Research Lab. Dept. of Computer Science and Engineering Korea University

Contents Introduction Newton’s Method for Nonlinear Systems Fixed-Point Iteration for Nonlinear Systems Minimum of a Nonlinear Function of Several Variables MATLAB’s Methods Case Study

Introduction Chapter 1 Chapter 2 Chapter 3,4 Chapter 5

Introduction A zero of a nonlinear function of a single variable Nonlinear functions of several variables The latter problem is much more difficult Newton’s method is applicable to this problem as well as that of a single variable

5.1 Newton’s Method for Nonlinear Systems

Newton’s Method Tangent line function Iteration form Tangent plane function Iteration using r, s

Newton’s Method (cont’) Step1 Step2

5.1.1 Matrix-Vextor Notation Step1 Step2

Ex5.1 Intersection of a Circle and a Parabola

Ex5.1 Intersection of a Circle and a Parabola Step1 Partial derivative Initial estimate (x0,y0) = (1/2, 1/2)

Ex5.1 Intersection of a Circle and a Parabola Step2 The second approximate solution (x1,y1)

Ex5.1 Intersection of a Circle and a Parabola Initial estimate and 2 iterations of Newton’s method iteration x y | | 0.5 1 0.875 0.625 0.39528 2 0.79067 0.61806 0.084611 The true solution is (x, y) = (0.78615, 0.61803)

5.1.2 MATLAB Function for Newton’s Method for Nonlinear Systems \ operation -> 100p

Example 5.2 a System of Three Equations

Example 5.2 a System of Three Equations

Example 5.3 Intersection of a Circle and an Ellipse y 1 0.5 x -0.5 -1 -1 -0.5 0.5 1

Example 5.3 Intersection of a Circle and an Ellipse

Example 5.4 Positioning a Robot Arm the end of the first link(x1,y1) the end of the second link(x2,y2) We need to solve 10 (x2,y2) 8 6 β 4 (10,4) (x1,y1) 2 α 2 4 6 8 10

Example 5.4 Positioning a Robot Arm Find the angles so that the arm will move to the point(10,4) The length of link (d1,d2) = (5, 6) initial angles (α , β) = (0.7,0.7) The system of equations

5.2 Fixed-Point Iteration For Nonlinear Systems

5.2 Fixed-Point Iteration For Nonlinear Systems Ex 5.5 Fixed-Point Iteration for a System of two Nonlinear Function Consider the problem of finding a zero of the system Coverting form The graph of the equations and

5.2 Fixed-Point Iteration For Nonlinear Systems 0.2 0.4 0.6 0.8 1 Iteration 0.60000 1 0.53840 0.18160 0.42291 2 0.50255 0.15444 0.044975 3 0.50275 0.15062 0.0038204 4 0.50235 0.00039661 5 0.50238 0.15058 4.9381e-05

5.2 Fixed-Point Iteration For Nonlinear Systems

5.2 Fixed-Point Iteration For Nonlinear Systems Using MATLAB function G = ex5_5(x) G = [ (-0.1*x(1)^3 + 0.1*x(2) + 0.5) (0.1*x(1) + 0.1*x(2)^3 + 0.1)]; % Actually using function : Fixed_pt_sys >> x=Fixed_pt_sys(@ex5_5, [0.6 0.6], 0.00001, 5)

5.2 Fixed-Point Iteration For Nonlinear Systems Result

5.2 Fixed-Point Iteration For Nonlinear Systems Ex 5.5 Fixed-Point Iteration for a System of Three Nonlinear Function Coverting this system to a fixed point iteration form

5.2 Fixed-Point Iteration For Nonlinear Systems Using the preceding MATLAB function and a starting estimate of (2, 2, 2) Iteration 2.0000 1 3.7600 2.1000 -1.6750 4.0759 2 3.5729 1.6528 -1.4113 0.5518 3 3.6502 1.7621 -1.4876 0.15403 4 3.6272 1.7232 -1.4643 0.050903 5 3.6346 1.7350 -1.4719 0.0115899 6 3.6323 1.7312 -1.4695 0.0050696 7 3.6330 1.7324 -1.4702 0.0016031 8 3.6328 1.7320 -1.4700 0.00050866 9 1.7321 -1.4701 0.00016117 10 5.1098e-05

5.2 Fixed-Point Iteration For Nonlinear Systems Using MATLAB function G = ex5_6(x) G = [ (-0.02*x(1)^2 - 0.02*x(2)^2 - 0.02*x(3)^2 + 4) (-0.05*x(1)^2 - 0.05*x(3)^2 + 2.5) (0.025*x(1)^2 + 0.025*x(2)^2 - 1.875)]; % Actually using function : Fixed_pt_sys >> x=Fixed_pt_sys(@ex5_6, [2 2 2], 0.00001, 10)

5.2 Fixed-Point Iteration For Nonlinear Systems Result

5.2 Fixed-Point Iteration For Nonlinear Systems If g(x) maps D into D, then g has a fixed point in D. In other words, if g(x) is in D whenever x is in D, then there is some point p in D such that p=g(p) If sequence of approximations to the fixed point intial point is sufficiently close to the fixed point p If there is a constant K<1 such that for every x in D For each i= 1,….,n and each j=1,…,n, A bound on the error at the mth step is given by

5.2 Fixed-Point Iteration For Nonlinear Systems Example 5.5 First check to make sure that g(x) maps the retangle That is, for we have In fact and Which are all less than 0.5 (for ), as is required for the corollary

5.3 Minimum of a Nonlinear Function of Several Variables

Example 5.7 and , we define The minimum value of h(x,y) is 0, which occurs when f(x,y)=0 and g(x,y)=0. The derivatives for the gradient are

Table 5.8 Result of ffmin Table 5.8 Approximate zeros at each step of iteration Step x y Change 0.5 1 0.875 0.625 -0.26831 2 0.74512 0.61133 -0.035987 3 0.79251 0.61902 -0.0079939 4 0.78446 0.6178 -0.00019414 5 0.78657 0.6181 -1.3631e-05

5.3.1 MATLAB Function for Minimization by Gradient Descent

5.3.1 MATLAB Function for Minimization by Gradient Descent function h = ex_min(x) h = (x(1)^2 + x(2)^2 - 1)^2 + (x(1)^2 - x(2) )^2; function dh = ex_min_g(x) dh = [ -(4*(x(1)^2 + x(2)^2 - 1)*x(1) + 4*(x(1)^2 - x(2))*x(1)) -(4*(x(1)^2 + x(2)^2 - 1)*x(2) - 2*(x(1)^2 - x(2)))]'; xmin = ffmin(@ex_min, @ex_min_g, [0.5 0.5], 0, 5) 0 0.5000 0.5000 1.0000 0.8750 0.6250 -0.2683 2.0000 0.7451 0.6113 -0.0360 3.0000 0.7925 0.6190 -0.0080 4.0000 0.7845 0.6178 -0.0002 5.0000 0.7866 0.6181 -0.0000

Example 5.8 Given three points in the plane, we wish to find the location of the point P=(x,y) sp that the sum of the squares of the distances from P to the three given points, is as small as possible.

Figure 5.7

5.4 MATLAB’s Methods

MATLAB’s METHODS FMINS finds the minimum of a scalar function of several variables, starting at an initial estimate Note The fmins function was replaced by fminsearch in Release 11 (MATLAB 5.3). In Release 12 (MATLAB 6.0), fmins displays a warning message and calls fminsearch Syntax x = fmins('fun',x0) x = fminsearch(fun,x0) starts at the point x0 and finds a local minimum x of the function described in fun. x0 can be a scalar, vector, or matrix.

MATLAB’s METHODS Examples a A classic test example for multidimensional minimization is the Rosenbrock banana function The traditional starting point is (-1.2,1). The M-file banana.m defines the function. The minimum is at (1,1) and has the value 0. a

5.5 Nonlinear system: Case Study

Nonlinear system: Case Study The analytical model to compute the 802.11 DCF throughput

Nonlinear system: Case Study The stationary probability τ The station transmits a packet in a generic slot time The conditional collision probability p aa

Nonlinear system: Case Study Two Equations represent a nonlinear system in the two unknowns τ and p W = Cwmin m = Maximum backoff stage N = The number of stations

Nonlinear system: Case Study Solve the nonlinear problem of two variables Assumptions W = 32 m = 3 N = 3, 10, 50 Newton’s Method and fminsearch(fun,x0) Problem1 by Newton’s Method (W=32, m = 3, N =3) 64τ p^4 + 34pτ - 33τ - 4p +2 =0 (1) p = 1-(1- τ)^2 (2) *Equations (1) and (2) is nonlinear system

Nonlinear system: Case Study function f = newton_case1(x) f = [ (64*x(2)*x(1)^4 + 34*x(1)*x(2) -33*x(2) -4*x(1) +2) (x(2)^2 - 2*x(2) + x(1)) ]; function df = newton_case1_j(x) df = [ (256*x(2)*x(1)^3 + 34*x(2) - 4) (64*x(1)^4 + 34*x(1)-33) 1 ( 2*x(2) -2) ]; τ = 0.4165 p = 0.6595 aa

Nonlinear system: Case Study Problem1 by fminsearch(fun,x0) (W=32, m = 3, N =3) function h = fminsearch_case1(x) h = (64*x(2)*x(1)^4 + 34*x(1)*x(2) -33*x(2) -4*x(1) +2)^2 + ((1-x(2))^2 -1 +x(1))^2;

Nonlinear system: Case Study Problem2 by Newton’s Method (W=32, m = 3, N =10) 64τ p^4 + 34pτ - 33τ - 4p +2 =0 (1) p = 1-(1- τ)^9 (2) *Equations (1) and (2) is nonlinear system τ = 0.1259 p = 0.7021 dd

Nonlinear system: Case Study Problem2 by fminsearch(fun,x0) (W=32, m = 3, N =10) function h = fminsearch_case1(x) h = (64*x(2)*x(1)^4 + 34*x(1)*x(2) -33*x(2) -4*x(1) +2)^2 + ((1-x(2))^2 -1 +x(1))^9;

Nonlinear system: Case Study Problem3 by Newton’s Method (W=32, m = 3, N =50) 64τ p^4 + 34pτ - 33τ - 4p +2 =0 (1) p = 1-(1- τ)^49 (2) *Equations (1) and (2) is nonlinear system τ = 0.0427 p = 0.8823 dd

Nonlinear system: Case Study Problem2 by fminsearch(fun,x0) (W=32, m = 3, N =50) function h = fminsearch_case1(x) h = (64*x(2)*x(1)^4 + 34*x(1)*x(2) -33*x(2) -4*x(1) +2)^2 + ((1-x(2))^2 -1 +x(1))^49;