<p> CHAPTER 3: NUMERICAL ALGORITHMS WITH MATLAB</p><p>Lecture 3.1: Polynomials and polynomial interpolation</p><p>General properties of polynomials:</p><p> A general polynomial of order n: n n-1 2 y = Pn(x) = c1 x + c2 x + … + cn-1 x + cn x + cn+1</p><p> Factorization of polynomials of n-th order by n roots (zeros) x1,x2,…,xn (possibly, complex or multiple): y = Pn(x) = c1 (x – x1)*(x – x2)*…*(x – xn)  Horner's method for nested multiplications of polynomials: y = Pn(x) = ( … ( ( ( c1*x + c2 )*x + c3 )*x + c4) … )*x + cn+1 Example: y = x4 + 2 x3 – 7 x2 – 8 x + 12 y = (x-1)*(x-2)*(x+2)*(x+3) y = ( ( (x + 2)*x – 7)*x – 8)*x + 12</p><p>Polynomials with MATLAB:  representation of polynomials by a row-vector of coefficients p1 = [ 1,2,-7,-8,12], p2 = [ 2,3,5,9,5] p1 = 1 2 -7 -8 12 p2 = 2 3 5 9 5 </p><p> roots: find all zeros of a polynomial from its coefficients r1 = roots(p1)', r2 = roots(p2)' r1 = -3.0000 -2.0000 2.0000 1.0000 r2 = 0.2500 - 1.5612i 0.2500 + 1.5612i -1.0000 -1.0000 </p><p> poly: find coefficients of a polynomial from its zeros q1 = poly(r1), q2 = poly(r2) % NB: coefficients q2 are different from coefficients of p2 by a factor of 2 % The leading-order coefficient c1 is always normalized to 1. q1 = 1.0000 2.0000 -7.0000 -8.0000 12.0000 q2 = 1.0000 1.5000 2.5000 4.5000 2.5000 </p><p>% rounding errors may occur in computations of roots % Wilkinson's example: roots(poly(1:20))' ans = Columns 1 through 11 20.0003 18.9970 18.0118 16.9695 16.0509 14.9319 14.0684 12.9472 12.0345 10.9836 10.0063 Columns 12 through 20 8.9983 8.0003 7.0000 6.0000 5.0000 4.0000 3.0000 2.0000 1.0000 % large rounding error occurs when highly multiple roots are computed % y = (x-1)^6 r = [ 1 1 1 1 1 1 ]; p = poly(r), rr = roots(p)' p = 1 -6 15 -20 15 -6 1 rr = 1.0042 - 0.0025i 1.0042 + 0.0025i 1.0000 - 0.0049i 1.0000 + 0.0049i 0.9958 - 0.0024i 0.9958 + 0.0024i </p><p> polyval: evaluate polynomials at a given point y1 = polyval(p1,2.5) x = [ 0 : 0.2 : 1]; y = polyval(p2,x) y1 = 18.5625 y = 5.0000 7.0272 9.6432 13.1072 17.7552 24.0000 </p><p>% implementation of the Horner algorithm for nested multiplication: function [y] = HornerMultiplication(p,x) [n,m] = size(x); y = p(1)*ones(n,m); for k = 2:length(p) y = y.*x + p(k); end y1 = HornerMultiplication(p1,2.5) y = HornerMultiplication(p2,x) y1 = 18.5625 y = 5.0000 7.0272 9.6432 13.1072 17.7552 24.0000 </p><p> polyder: computes coefficients of the derivative of a given polynomial</p><p>% P(x) = c(1) x^n + c(2) x^(n-1) + … + c(n) x + c(n+1) % P'(x) = n c(1) x^(n-1) + (n-1) c(2) x^(n-2) + … + c(n) Pder1 = polyder(p1), Pder2 = polyder(p2) </p><p>Pder1 = 4 6 -14 -8 Pder2 = 8 9 10 9 </p><p> polyint: computes coefficients of the integral of a given polynomial</p><p>% P(x) = c(1) x^n + c(2) x^(n-1) + … + c(n) x + c(n+1) % P'(x) = c(1) x^(n+1)/(n+1) + c(2) x^n/n + … + c(n) x^2/2 + c(n+1) x + c(n+2) % c(n+2) are constant of integration (to be defined) Pint1 = polyint(p1,10) % the constant of integration is 10 Pint2 = polyint(p2) % the constant of integration is 0 (default) </p><p>Pint1 = 0.2000 0.5000 -2.3333 -4.0000 12.0000 10.0000 Pint2 = 0.4000 0.7500 1.6667 4.5000 5.0000 0 </p><p> conv: computes a product of two polynomials % Let p1 be polynomial of order n, p2 be polynomial of order m, % conv(p1,p2) is polynomial of order (n+m) p = conv(p1,p2) p = 2 7 -3 -18 -12 -57 -47 68 60 </p><p> deconv: computes a division of two polynomials</p><p>% Let p1 be polynomial of order n, p2 be polynomial of order m, % [pq,pr] = deconv(p1,p2) computes the quotient polynomial pq of order (n-m) % and the remainder polynomial pr of order (m-1) such that p1 = p2*pq + pr [pq,pr] = deconv(p1,p2) pq = 0.5000 pr = 0 0.5000 -9.5000 -12.5000 9.5000 </p><p> residue: computes partial-fraction expansion (residues)</p><p>% Let p1 be polynomial of order n and p2 be polynomial of order m % [C,X,R] = residue(p1,p2) finds coefficients C of the residue terms, % locations of poles X and the remainder term of a partial fraction expansion % of the ratio of two polynomials p1(x)/p2(x). % Example: no multiple roots, % p1(x) C(1) C(2) C(n) % ---- = ------+ ------+ ... + ------+ R(x) % p2(x) x - X(1) x - X(2) x - X(n)</p><p>[C,X,R] = residue(p2,p1) </p><p>C = -5.2000 1.2500 4.9500 -2.0000</p><p>X = -3.0000 -2.0000 2.0000 1.0000 R = 2 </p><p>% the inverse operation: % from partial fraction expansion to the ratio of two polynomials [q1,q2] = residue(C,X,R) q1 = 2.0000 3.0000 5.0000 9.0000 5.0000 q2 = 1.0000 2.0000 -7.0000 -8.0000 12.0000 </p><p> polyfit: computes coefficients of interpolating polynomials of order n that passes through a set of (n+1) data points x = [ 1,2,-2,-3,0]; y = [0,0,0,0,12]; p = polyfit(x,y,4) % 4 is the order of the polynomial through 5 data points pp = poly([1,2,-2,-3]) % the same operation if roots of polynomial are known p = 1.0000 2.0000 -7.0000 -8.0000 12.0000 pp = 1 2 -7 -8 12 Polynomial interpolation:</p><p>Problem: Given a set of (n+1) data points: (x1,y1); (x2,y2); …; (xn,yn); (xn+1,yn+1) Find a polynomial of degree n, y = Pn(x), that passes through all (n+1) data points. Assume that the set x1,x2,</p><p>…,xn,xn+1 is ordered in ascendental order, i.e. x1<x2<…<xn<xn+1. The polynomial y = Pn(x) is called the interpolating polynomial for x1<x<xn+1 and extrapolating polynomial for x < x1 and x > xn+1.</p><p>Example: Given a set of five data points for a voltage-current characteristic of a zener diode Voltage -1.00 -0.75 -0.50 -0.25 -0.00 Current -14.58 -6.15 -1.82 -0.23 0.00</p><p>The five points are connected by a polynomial y = P4(x) of order 4. The data points are shown by blue stars, the polynomial is shown by green solid line:</p><p>Methods:  Power series (Vandermonde interpolation)  Lagrange interpolation polynomials  Newton forward difference interpolation  Newton backward difference interpolation</p><p>Vandermonde interpolation:</p><p>The (n+1) coefficients ck of the interpolating polynomial y = Pn(x) are to be matched with (n+1) data points: Pn(xk) = yk. The matching conditions produce a system of (n+1) linear equations for (n+1) unknown coefficients of the interpolating polynomial: n n-1 2 c1 (xk) + c2 (xk) + … + cn-1 (xk) + cn xk + cn+1 = yk The linear system can be solved with the use of MATLAB linear algebra solver. x = [ -1,-0.75,-0.5,-0.25,-0]; A = vander(x) y = [ -14.58;-6.15;-1.82;-0.23;-0.00]; c = A\y; c' xInt = -1 : 0.01 : 0; yInt = HornerMultiplication(c,xInt); plot(xInt,yInt,'g',x,y,'b*');</p><p>A = 1.0000 -1.0000 1.0000 -1.0000 1.0000 0.3164 -0.4219 0.5625 -0.7500 1.0000 0.0625 -0.1250 0.2500 -0.5000 1.0000 0.0039 -0.0156 0.0625 -0.2500 1.0000 0 0 0 0 1.0000 ans = 0.2133 15.0400 0.3067 0.0600 0</p><p>0</p><p>- 5</p><p>- 1 0</p><p>- 1 5 - 1 - 0 . 9 - 0 . 8 - 0 . 7 - 0 . 6 - 0 . 5 - 0 . 4 - 0 . 3 - 0 . 2 - 0 . 1 0</p><p>The Vandermond matrices are ill-conditioned for large n and the linear algerba solvers produce inaccurate numerical results. Lagrange and Newton's interpolations are direct methods of finding the interpolating polynomials, without the necessity to solve linear systems of equations.</p><p>Lagrange interpolation:</p><p>Lagrange polynomials are defined in the factorization form: n1 (x  xi ) Ln,j(x) =  i1 (x  x ) i j j i</p><p> Ln,j(x) is a polynomial of order n</p><p> Ln,j(xi) = 0 for any i  j</p><p> Ln,j(xj) = 1 The interpolating polynomial y = Pn(x) has a simple form in terms of the Lagrange polynomials Ln,j(x): n1 Pn(x) =  y j Ln, j (x) = y1Ln,1(x) + … +yn+1Ln,n+1(x) j1</p><p>The interpolating polynomial y = Pn(x) has order n and it passes through all (n+1) data points. It can be proved that interpolating polynomials are unique, i.e. the polynomials y = Pn(x) obtained by the Vandermonde method and by the Lagrange method are the same polynomials. function [yi] = LagrangeInter(x,y,xi) % Lagrange interpolation algorithm % x,y - row-vectors of (n+1) data values (x,y) % xi - a row-vector of values of x, where the polynomial y = Pn(x) is evaluated % yi - a row-vector of values of y, evaluated with y = Pn(x) n = length(x) - 1; % order of interpolation polynomial y = Pn(x) ni = length(xi); % number of points where the interpolation is to be evaluated L = ones(ni,n+1); % the matrix for Lagrange interpolating polynomials L_(n,j)(x) % L has (n+1) columns for each point j = 1,2,...,n+1 % L has ni rows for each point of xi for j = 1 : (n+1) for i = 1 : (n+1) if (i ~= j) L(:,j) = L(:,j).*(xi' - x(i))/(x(j)-x(i)); end end end yi = y*L'; </p><p>Example of Lagrange interpolation: x = [ -1,-0.75,-0.5,-0.25,-0]; y = [ -14.58,-6.15,-1.82,-0.23,-0.00]; xInt = -1 : 0.01 : 0; yInt = LagrangeInter(x,y,xInt); plot(xInt,yInt,'g',x,y,'b*'); 0</p><p>- 5</p><p>- 1 0</p><p>- 1 5 - 1 - 0 . 9 - 0 . 8 - 0 . 7 - 0 . 6 - 0 . 5 - 0 . 4 - 0 . 3 - 0 . 2 - 0 . 1 0</p>