Polynomials. Computer Programming for Engineers (2014 Spring)

Transcription

1 Computer Programmig for Egieers (014 Sprig) Polyomials Hyougshick Kim Departmet of Computer Sciece ad Egieerig College of Iformatio ad Commuicatio Egieerig Sugkyukwa Uiversity

2 Polyomials A th order polyomial i variable is writte as 1 p ) a a a a ( 1 1 It is atural to associate a row vector A p with p, amely A few eamples: Row vector represetatios of are >> Ap = [ ]; >> Aq = [ 3-4 ]; leadig zeros But [ ] also represets q

3 Fuctios operatig o polyomials Today lecture s goal is to write several useful fuctios for polyomial maipulatio, icludig Polyomial additio Polyomial subtractio Polyomial multiplicatio Polyomial evaluatio (use polyval) Plottig the graph of a polyomial Roots of a polyomial (use roots) The iputs to the fuctios will be row vectors, represetig polyomials.

4 Polyomial Additio Addig two polyomials requires addig their coefficiets. The sum of a a a a b b b b is simply ) ( ) ( ) ( ) ( b a b a b a b a I terms of the row-vector represetatio of the polyomials, we simply add them, elemet-by-elemet. But the row vectors may have differet legths, ad we eed to alig them.

5 Polyomial Additio The row vector represetatios of 3 a( ) are b( ) 3 4 >> A = [ ]; >> B = [ 3-4 ]; but obviously >> C = A + B will give a error. We eed to pad B with a zero >> C = A + [ 0 B ]; or add B oly to the correct elemets of A. >> C = A; >> C(:ed) = C(:ed) + B;

6 Addig Polyomials TASK: Write a MATLAB fuctio to calculate the additio of two polyomials. The iputs are two polyomials A ad B. The output is the sum of two polyomials. fuctio C = addpoly(a,b) % This adds two row vectors, iterpretig % them as polyomials. % Should check that A ad B are row vectors

7 Aswer fuctio C = addpoly(a,b) % This adds two row vectors, iterpretig % them as polyomials. % Should check that A ad B are row vectors lea = legth(a); leb = legth(b); if lea==leb C = A+B; elseif lea<leb C = [zeros(1,leb-lea) A] + B; else C = A + [zeros(1,lea-leb) B]; ed

8 Subtractig Polyomials TASK: Write a MATLAB fuctio to calculate the subtractio of two polyomials. The iputs are two polyomials A ad B. The output is the differece of two polyomials. fuctio C = subpoly(a,b) % This subtracts two row vectors, % iterpretig them as polyomials. % Should check that A ad B are row vectors

9 Multiplyig Polyomial Multiplicatio a() = b() = Epress the product as ( )(3 + 4) = ( )*3 add A*B(1) to C(1:4) + ( )* add A*B() to C(:5) + ( )*(-4) A = [ ] B = [3-4] The result is 5 th order, so we eed a 1-by-6 array for the result C = [ ] Do this for every elemet of B (for loop) add A*B(3) to C(3:6) A*B(1) A*B(i) A*B() A*B(3) + C(i:i+legth(A)-1) C(1:1+legth(A)-1) C(:+legth(A)-1) C(3:3+legth(A)-1)

10 Multiplyig Polyomials TASK: Write a MATLAB fuctio to calculate the multiplicatio of two polyomials. The iputs are two polyomials A ad B. The output is the multiplicatio of two polyomials. fuctio C = multpoly(a,b) % This computes the product of the iputs, % iterpretig them as polyomials.

11 Aswer fuctio C = multpoly(a,b) % This computes the product of the iputs, % iterpretig them as polyomials. lea = legth(a); leb = legth(b); lec = lea + leb - 1; C = zeros(1,lec); for i=1:leb id = i:i+lea-1; C(id) = A*B(i) + C(id); ed

12 Polyomial Evaluatio Use the Matlab fuctio polyval >> A = [ ]; >> =.6; >> y = polyval(a,) polyval works for vectors too >> = lispace(-3,3,00); >> y = polyval(a,); >> plot(,y)

13 Roots of Polyomials A th order polyomial (with ozero leadig coefficiet) p 1 ( ) a1 a a a 1 It is a fudametal theorem of algebra that the equatio p( ) has solutios, called the roots of p. Label these roots r 1, r,, r. Also, p ca be factored ito the for Eve if the coefficiets of p are real umbers, the roots may be comple. 0 p( ) a1( 1 r )( r ) ( r )

14 The roots commad The commad roots computes the roots of a polyomial, ad returs them as a colum vector. Compute the roots of >> roots([1 1 -]) >> roots([1 3 5]) >> roots([ ]) 5 3 ) ( q 3 4 ) ( 3 4 v ) ( p

15 Solvig Algebraic Equatios From liearity, it is easy to solve the equatio or eve the system of equatios 5 y 1 6 y 3 But what about a equatio like ta( e ) 1 0

16 Solvig Algebraic Equatios I geeral, give a cotiuous fuctio f, how do you fid the (ay, all) solutios to f ( ) 0 f() Graph of f() versus roots of f

17 Iterative Methods Iterative methods start with a guess, labeled 0 ad through easy calculatios, geerate a sequece 1,, 3, Goal is that sequece satisfies lim f 0 ad covergece to a limit, * * is a solutio

18 Rate of Covergece Suppose { } is a sequece, covergig to a limit *. Let e deote the differece *- If there is a costat C, ad positive epoet such that for large e 1 C e the the sequece { } coverges with order to *. liear covergece is =1 The sequece 3,.5,.5,.15,.065, coverges liearly to (here C = ½) The sequece 3,.1,.01,.001,.0001, coverges liearly to (here C=0.1) quadratic covergece is = The sequece, 1, 1, 1, 1, 1,... coverges quadratically to (here C=1)

19 Stoppig Criteria Suppose { } is a sequece, covergig to a limit *. The limit * has the property f(*) = 0. Let tol be a positive umber But we do t kow * Absolute (i ) * < tol Relative (i ) * < tol * Absolute -1 < tol Relative -1 < tol Absolute i f f( ) < tol Sometimes (i bisectio, for eample) we ca boud this, eve without kowig *.

20 Bisectio Method, basic idea Suppose f is cotiuous, f 0 f 0 L R L f L M f R Itermediate value theorem R There must be a root * betwee L ad R. Let M be the midpoit. Based o sig of f M replace either L or R with M, ad repeat.

21 Simple pseudo-code: Bisectio % Start with L, R, fl = f(l); fr = f(r); while StoppigCriteriaNotMet M = 0.5*(L+R); fm = f(m); if fl*fm<0 % check the sig of fm R = M; else L = M; ed ed

22 Questios?