Chapter IX. Roots of Equations

Transcription

1 14 Chapter IX Roots of Equations Root solving typially onsists of finding values of x whih satisfy a relationship suh as: 3 3 x + x = 1 + e x The problem an be restated as: Find the root of 3 f ( x) = 3 x + x 1 e x In this hapter we will over three methods for root finding: The half-interval method The Newton-Raphson Method The Bairstow method for roots of polynomials 91 The Half-Interval Method Figure 91 Half-Interval Method Find two values that braket a root of the given funtion That is, the root is bounded by ( x, x ) suh that P N f ( x ) > and f ( x ) < P N

2 The half interval method is desribed as follows: Compute x M 3 Compute f ( x M ) = x P + 4 If f ( x ) ε Root = x M x N 143 Example Find 7 using the half-interval method Solution The equation to solve is x 7 = or obtain the root of: f ( x) = x 7 M exit 5 If f ( xm) < xn = xm else xp = xm 6 Repeat steps (1) to (4) g is a small number, typially 1-7, that determines the auray of the root Following the above proedure let xp = 3 f ( 3) = 9 7 = > and xn = f ( ) = 4 7 = 3< x M = = = 5 f ( x M ) = ( 5 ) 7= 75 < 3 f ( x M ) 1 7? no 4 xn = 5 xp = 3 5 Repeat Steps 1 to x M = = 75 f ( x M ) = ( 75) 7 = 565 > 3 f ( x M ) < 1 7? no

3 144 4 xp = 75 xn = 5 et Example Develop a lass Root based on the Half-interval method for root finding Solution The following utilizes pointer to a funtion to allow the user of the lass Root to supply his own funtion without having to make any hanges in the lass Thus leaving the integrity of the lass intat (the way it s supposed to be) #inlude <iostreamh> #inlude <onioh> #inlude <mathh> lass Root private: double xp, xn; double (*f)(double x); //pointer to user supplied funtion publi: Root(double(*F)(double x), double xdp, double xdn) f=f; xp=xdp; xn=xdn; double HLFINT(); ; double Root::HLFINT() double xm; double y; double EPS=1e-7; do xm=(xp+xn)/; y=f(xm); if(y>) xp=xm; else xn=xm; while(fabs(y) > EPS); return xm;

4 145 //User supplied funtion double fun(double x) return x*x-7; // int main() Root sqrt7(fun,3,); out << "Square root of 7 = " << sqrt7hlfint() << endl; geth(); return 1; 9 Newton-Raphson s Method The Taylor series expansion of f(x) about x is given by: f ( x) = f ( x ) + ( x x ) f ( x ) + ( x x)! f ( x ) + If f(x) = then x must be the root of f(x) Negleting terms higher or equal to seond order we an write = f x + x x f x ( ) ( ) ( ) ( x x ) = Writing the above equation in an iterative form we get: xi+ 1 = xi f ( xi ) f ( x ) i f ( x) f ( x ) A graphial interpretation of the Newton-Raphson s method is shown in Fig9

5 146 From Fig 9 we an write: f ( x) x x or 1 x = f ( x ) = x 1 f ( x) f ( x ) Figure 9 The Newton-Raphson s Method After omputing x 1 from x set x = x 1 and repeat the proess In so doing we an approah the root in a few iterations Proedure 1 Selet an initial value x 1 Compute f ( x1) 3 Compute f ( x ) 4 Compute x = x 1 1 f ( x1) f ( x ) 5 If x x1 ε then root x and exit 6 Repeat steps () to (5) 1 In the above proedure we used a different stopping riterion from the one used with the half-interval method You an use any one of the two riteria for terminating the iteration in either methods of

6 147 root finding, however, in some ases where the slope within the viinity of the root is small the seond riterion would be more suitable Example Modify the lass Root to inlude the Newton-Raphson s method Solution To do so we will overload the onstrutor to add the derivative of the funtion and a starting point The modified lass Root is shown next #inlude <iostreamh> #inlude <onioh> #inlude <mathh> lass Root private: double xp, xn; double xst; //start x for newton-raphson double (*f)(double x); //pointer to user supplied funtion double (*fd)(double x); //derivative of (*f)(x) for newton-raphson publi: Root(double(*F)(double x), double xdp, double xdn) f=f; xp=xdp; xn=xdn; Root(double(*F)(double x), double(*fd)(double x), double xi ) f=f; fd=fd; xst=xi; double HLFINT(); //Half interval double NR(); //Newton-Raphson ; double Root::HLFINT() double xm; double y; double EPS=1e-7; do xm=(xp+xn)/; y=f(xm); if(y>) xp=xm;

7 148 else xn=xm; while(fabs(y) > EPS); return xm; double Root::NR() double EPS=1e-7; double x1, x; x1=xst; while(1) x=x1-(*f)(x1)/(*fd)(x1); if(fabs(x-x1)<=eps) break; x1=x; return x; //User supplied funtion double fun(double x) return x*x-7; //User supplied derivative double fund(double x) return *x; // int main() Root sqrt7(fun,3,); Root sqrt7nr(fun,fund,); out << "Square root of 7 using half interval method= " << sqrt7hlfint() << endl; out << "Square root of 7 using Newton Raphson's method= " << sqrt7nrnr() << endl;

8 149 geth(); return 1; Example Find 7 using the Newton-Raphson s method ( = 1) Solution Using Newton-Raphson s method we an derive the iterative equation: x = x1 + a x 1 where a = 7 for this problem We will leave the derivation as well as the hand alulation as an exerise to the reader Exerise Develop a funtion for the square-root of a positive number a Use as initial guess x=a/ if a>1 and x=a if a<1 Exerise The square-root of a omplex number an be obtained as follows: x+ iy = a + ib or x y + ixy = a + ib Equating real to real and imaginary to imaginary we get: x y = a xy = b Solving the above two equations to isolate x we get: 4x 4 4ax b = and y b = x Using the last two equations and Newton-Raphson s method develop an algorithm and funtion for obtaining the square-root of a omplex number 93 The Bairstow s method Bairstow s method is an iterative method whih involves finding quadrati fators, f ( x) = x + ux + v of the polynomial: n n 1 n ax + ax + ax + + a = 1 n

9 15 where a =1 We will first present the proedure, then the C++ ode and then we will follow with the derivation of the method This is going in reverse to the traditional approah of derivation, proedure, and then ode but it helps to keep the fous on development and builds uriosity as to the soure of the approah after one has experiened suess with the method Proedure 1 Selet initial values for u and v Most frequently these are taken as u=v= Calulate b, b,, from 1 b n 3 Calulate 1,, 3,, n 1 from 4 Calulate )u and )v from: bi = ai bi 1u bi v ( i = 3,,, n) where b = 1 and b = a u 1 1 i = bi i 1u i v ( i = 3,,, n 1) where = 1 and = b u 1 1 u = b bn n n n n 1 n 3 1 n n 3 v = n 1 b b n n n 1 n 1 n n n 3 5 Inrement u and v by )u and )v u = u+ u v = v + v 6 Return to step and repeat the proedure until )u and )v approah zero within some preassigned value g suh that u + v < ε An upper limit of iterations should be speified to protet against non-onvergene If

10 151 onvergene to the desired result does not our after the speified number of iterations, then new starting values should be seleted Frequently, suitable starting values for u and v may be determined from u a a and v=a a = n 1 n n n The following C++ ode follows from the above proedure It does not, however, ontain all the bells and whistles For example it does not hek for the number of iterations before onvergene (we leave that part to the reader to modify) The lass RootPoly in the program is tested on two polynomials: x 3x 1x + 1x + 44x + 48 = 3 x 3x x + 9 = #inlude <iostreamh> #inlude <onioh> #inlude <mathh> lass RootPoly private: double *a; double *b; double *; int n; double *rootr; double *rooti; publi: RootPoly(double *oeff, int n); ~RootPoly(); void Bairstow(); double *getrootsr() return rootr; double *getrootsi() return rooti; ; RootPoly::RootPoly(double *oeff, int order)

11 15 n=order; a=new double[n+1]; b=new double[n+1]; =new double[n]; for(int i=; i <= n; i++) a[i]=oeff[i]; rootr = new double[n]; rooti = new double[n]; RootPoly::~RootPoly() delete[] a; delete[] b; delete[] ; delete[] rootr; delete[] rooti; void RootPoly::Bairstow() double u,v; double EPS=1e-7, D, Du, Dv, d; int i,k=; //seleting initial values for u and v if(a[n-]!=) u=a[n-1]/a[n-]; v=a[n]/a[n-]; else u=v=1; //Starting the iterations int iter=; while(n>=) iter++; do b[]=1; b[1]=a[1]-u;

12 153 for(i=; i <= n; i++) b[i]=a[i]-b[i-1]*u-b[i-]*v; []=1; [1]=b[1]-u; for(i=; i < n; i++) [i]=b[i]-[i-1]*u-[i-]*v; if(n==3) D=[n-1]-[n-]*[n-]; Du=(b[n]-b[n-1]*[n-])/D; else D = [n-1]*[n-3]-[n-]*[n-]; Du = (b[n]*[n-3]-b[n-1]*[n-])/d; Dv = (b[n-1]*[n-1]-b[n]*[n-])/d; if((fabs(du)+fabs(dv))<eps) break; u=u+du; v=v+dv; while(1); iter=; d=u*u-4*v; if(d >= ) rootr[k] = (-u + sqrt(d))/; rooti[k] = ; k++; rootr[k] = (-u - sqrt(d))/; rooti[k] = ; k++; else rootr[k] = -u/; rooti[k] = sqrt(-d)/; k++; rootr[k] = -u/; rooti[k] =-sqrt(-d)/; k++; n=n-;

13 154 for(i=; i<=n; i++) a[i] = b[i]; //losing braket for while if(n == 1) rootr[k] = -a[1]; rooti[k] = ; if(n==) d=a[1]*a[1]-4*a[]; if( d < ) rootr[k]=-a[1]/; rooti[k]=sqrt(-d)/; k++; rootr[k]=-a[1]/; rooti[k]=-sqrt(-d)/; else rootr[k]=(-a[1]+sqrt(d))/; rooti[k]=; k++; rootr[k]=(-a[1]-sqrt(d))/; rooti[k]=; // int main() double oeff[]=1,-3,-1,1,44,48; //double oeff[]=1,-3,-1,9; int n=sizeof(oeff)/sizeof(double)-1; out << "n=" << n << endl; RootPoly roots(oeff,n);

14 155 rootsbairstow(); double *RootsReal, *RootsImag; RootsReal=new double[n]; RootsImag=new double[n]; RootsReal=rootsgetRootsr(); RootsImag=rootsgetRootsi(); out << "ROOTS : " << endl; for(int i=; i<n; i++) if( RootsImag[i]>) out << RootsReal[i] << " +i" << RootsImag[i] << endl; else if(rootsimag[i]<) out << RootsReal[i] << " -i" << -RootsImag[i] << endl; else out << RootsReal[i] << endl; geth(); return 1; Those who have utilized proedural languages suh FORTRAN, BASIC or C an realize the great advantage of working with OOP C++ Classes offer the programmer the ability of developing effiient ode Eah member funtion speializes in one speifi task For example memory alloation is arried-out in the onstrutor and dealloation in the destrutor Other member funtions eah is speialized to arry-out a speifi task without the need to inlude data initialization This is arried out by the onstrutor as in the above program along with memory alloation of private pointer variables This a far superior approah to just struture programming used in proedural languages Running the above program for eah polynomial we get the following answers: n=5 ROOTS : i1-1 -i1 4 n=3

15 156 ROOTS : 655 +i i You an hek if the answers are orret by substituting eah oeffiient in the polynomial The program does not arry-out this task but that is something that an be inluded in the version that the reader will develop Anyway, the answers are orret and now we are ready to move on to the derivation of the Bairstows method Dividing the polynomial x n a x n a x n + + a 1 n n 1 n x + a1x + ax + + an n n 1 n = ( x + ux + v)( x + bx + b x + + b x + b ) + b ( x+ u) + b 1 n by a quadrati equation of the form x + ux + v yields a quotient whih is polynomial of order n- and a remainder whih is a polynomial of order one That is: n 3 n n 1 n where bn 1( x + u) + bnis the remainder after division of the polynomial by the quadrati equation The reason the remainder was plaed in this form is to provide simpliity in the solution as we will see later Equating oeffiients of like powers on both sides we get: or we an write a = b + u 1 1 a = b + ub + v 1 a = b + ub + vb a = b + ub + vb i i i 1 i a = b + ub + vb n 1 n 1 n n 3 a = b + ub + vb n n n 1 n

16 157 b = a u 1 1 b = a ub vb 1 b = a ub vb b = a ub vb i i i 1 i b = a ub vb n 1 n 1 n n 3 b = a ub vb n n n 1 n (91) We would like both b n-1 and b n to be zero to ensure that the quadrati x + ux + v is a fator of the polynomial Sine the b s are funtions of u and v and if we an adjust u by )u and v by )v suh that both b n-1 and b n approah zero Of ourse we will have to do this iteratively as we have observed in the proedure we used in developing the C++ ode for Bairstow s method Expressing b n-1 and b n as funtions of two variables u+)u and v+)v and expanding in a Taylor series expansion for two variables We assume that )u and )v are small enough that we an neglet higher-order terms bn bn bn( u+ u, v + v) = bn + u + v v bn 1 bn 1 bn 1( u+ u, v + v) = bn 1 + u + v v (9) where the terms on the right are evaluated for values of u and v Taking partial derivatives of equations (91) with respet to u and defining these in terms of s, we obtain the following:

17 b1 b b3 b4 bi bn 1 bn = 1 = = u b = = b + u+ v = 1 = b + u+ v =! = b + u+ v =! i 1 i i 3 i 1 = b + u+ v = n n 3 n 4 n = b + u+ v = n 1 n n 3 Similarly, taking the partial derivatives of equations (91) with respet to v and relating these with respet to the s, we obtain the following: n 1 (93) b1 v b v b3 v b4 bi v bn 1 v bn v = = 1 = = u b = 1 1 = b + u+ v =! 1 = b + u+ v =! i i 3 i 4 i = b + u+ v = n 3 n 4 n 5 n 3 = b + u+ v = n n 3 n 4 n (94)

18 159 From either Eq93 and 94, we see that i = bi i 1 i v ( i = 3,,, n 1) (95) where = 1 and 1 = b 1 - u From Eqs (93) and (94), Eq(9) an be written in the form: bn = n 1 u+ n v b = u+ v n 1 n n 3 Solving these two simultaneous equations we get: u = b bn n n n n 1 n 3 1 n n 3 v = n 1 b b n n n 1 n 1 n n n 3 whih leads us to the proedure desribed previously for the Bairstow s method Problems 1 Find a root of Fx ( ) = x tanxin the interval [4,45] using the Half-interval method Repeat problem 1 using the Newton-Raphson method with a starting value of 45 3 Find the root of Fx ( ) = xosx sinxin the interval [4,45] using the Newton-Raphson method with a starting value of 4 4 Obtain the square root of the omplex number 7+i6 using the Newton-Raphson Method 5 When the Earth is assumed to be a spheroid, the gravitational attration at a point on its surfae at the latitude angle N is g( φ) = ( sin ( φ) 59sin ( φ ) m / se At what latitude(s), in degrees, is g(n)=98? 6 Obtain the roots of the following polynomial using the Bairstow method:

19 (use the program given in se93) 16 4x 4 + x 3 + 3x + x + 1= 7 Improve the program given in se 93 by inluding onvergene heking and alulation of residual value for eah root, ie the value obtained when substituting eah root in the given polynomial