Principles of Linear Algebra With Maple The NewtonâRaphson ...

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20 Chapter 2. Newton-Raphson Method for Square Systems0 which is really solving simultaneously the square system of two non-linearequations given by f(x,y) = 0 and g(x,y) = 0.In order to do this, we shall have to generalize the one variable Newton-Raphson method iterator formula for solving the equation f(x) = 0 given bythe sequence p k for p 0 the starting guess whereorp k+1 = p k − f(p k)dfdx (p k)p k+1 = g(p k )(2.3)whereg(p k ) = p k − f(p k)(2.4)dfdx (p k)is the (single equation) Newton-Raphson iterator.To see how this can be done, you must realize that now in the two equationcase that[ ]x0p 0 =(2.5)y 0is our starting guess as a point in the xy-plane written as a column vector,[ ]xkp k =(2.6)y kis the kth iteration of our method, and[ ]f(xk ,yF(p k ) = k )g(x k ,y k )are all two component column vectors in R 2 and not numbers as in the singlevariable case. Thus, dividing a two component column vector by a derivativerequires that we be dividing by a 2×2 matrix, or multiplying by the inverseof this matrix. Thus, we need to replace df in our old iterator g(x) by a 2×2dxmatrix consisting of the four partial derivatives of F(x,y), which are ∂f∂x , ∂f∂y ,∂g ∂g, and∂x ∂y .The choiceofthis new derivativematrixisthe 2×2JacobianmatrixJ(x,y),of F(x,y), given by the 2×2 matrix⎡ ⎤∂f ∂f∂x ∂yJ(x,y) =⎢⎣∂g∂x∂g∂y⎥⎦ (2.7)
2.1 Newton-Raphson for Two Equations in Two Unknowns 21Other choices for this 2×2 matrix of partial derivatives are possible and youshould see if they will also work in place of this Jacobian matrix, but thisJacobian matrix seems most logical if you think about the fact that for F(x,y),the first row is f(x,y) and g(x,y) is in the second row while the variables aregiven as x first and y second in all these functions. Thus, the Newton-Raphsonarray (list or sequence) is now for a starting vector p 0 , as defined in equation(2.5), given byp k+1 = p k −(J(p k )) −1 F(p k ) (2.8)where the vector F(p k ) is multiplied on its left by the inverse of the Jacobianmatrix J(x,y) evaluated at p k , i.e., J(p k ) = J(x k ,y k ).Note that finding (J(p k )) −1 in each iteration is a formidable task if thesystem of equations is large, say 25×25 or more, and so at each iteration youcan instead solve for p k+1 by solving the square linear system of equationsJ(p k )p k+1 = J(p k )p k −F(p k )where the components of p k+1 are the unknowns. Of course, it is easy to findp k+1 directly if you are in a small number of equations case such as ours usingthe inverse matrix.Example 2.1.1. Let F : R 2 → R 2 be given in the form of equation (2.1), forf(x,y) = 1 64 (x−11)2 − 1100 (y −7)2 −1, g(x,y) = (x−3) 2 +(y −1) 2 −144[ ] 0We wish to apply the Newton-Raphson method to solve F(x,y) = or0equivalently the square systemf(x,y) = 0, g(x,y) = 0The solutions are the intersection points of these two curves f(x,y) = 0 whichis a hyperbola, and g(x,y) = 0 which is a circle, and is depicted in Figure 2.1.> with(linalg): with(plots):> f:= (x,y) -> (x-11)ˆ2/64 - (y-7)ˆ2/100 - 1:> g:= (x,y) -> (x-3)ˆ2 + (y-1)ˆ2 - 400:> plotf:= implicitplot(f(x,y) = 0, x = -25..25, y = -25..25, color = blue, grid= [50,50]):> plotg:= implicitplot(g(x,y) = 0, x = -25..25, y = -25..25, color = red, grid= [50,50]):
Page 1: Principles of Linear Algebra With M
Page 5 and 6: Chapter 1The Newton-RaphsonMethod f
Page 7 and 8: 1.1 Geometry of the Newton-Raphson
Page 15 and 16: 1.2 Examples of the Newton-Raphson
Page 21 and 22: 1.3 When the Newton-Raphson Method
Page 23: Chapter 2The Newton-RaphsonMethod f
Page 27 and 28: 2.1 Newton-Raphson for Two Equation
Page 33 and 34: 2.1 Newton-Raphson for Three Equati
Page 35 and 36: 2.1 Newton-Raphson for Three Equati
Page 37: 2.1 Newton-Raphson for Three Equati

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