How to Solve nonlinear system by newton method?

Solve the system of non-linear equation by newton raphson method x squared plus y equals 11 and y squared plus x equals 7?

• Answer:

  I don't have access to maple at the moment so I can't give you the answer as I'm not going to spend 30mins doing an algorithm a computer could do in 1 second, but I will tell you what the algorithm is/how to solve it. In order to solve a non-linear system numerically by Newton's method, you need to reduce it into a linear system. Recall the newton's method for single equations of one variable, xn+1=xn-f(xn)/f'(xn) the essential theory behind the derivation of this equation is by expressing f(x) as a Taylor polynomial, and assuming that the error (x-x0) is so small that errors of higher terms can be ignored, giving 0≈f(x0)+(x-x0)f'(x0) and solving this equation for x gives us our iterative equation above. The theory for non-linear systems is similar to this in that instead of having g(x)=x-f(x)/f'(x), we have G(x)=x-A(x)-1F(x) where G(x) and F(x) are vector functions and A(x) is some invertible matrix whose entries are functions from Rn->R i.e. from non-linear to linear. By dividing our non-linear vector function F(x) by the matrix A(x) we are essentially linearising our system. I won't go over the theory of it but as it happens, our matrix A(x) is the Jacobian matrix of our system so we have G(x)=x-J(x)-1F(x) This gives us an iterative formula x(n+1)=x(n)-J(x(n))-1F(x(n)) however for our algorithm we modify this slightly so that we do not have to calculate the Jacobian and the system at each point. We do this by setting y(n)=-J(x(n))-1F(x(n)) in other words before we start our algorithm we find out what -J(x)F(x) is and set that to y (I am omitting the (x) for simplicity's sake) so that instead of having to calculate the Jacobian and the system separately and then multiply them together, we just calculate y, turning our formula into x(n+1)=x(n)+y(n). This also has the effect of making our formula more accurate by reducing round off errors, as instead of having to do three calculations, finding the value of the Jacobian and the system then multiplying them together, we only have one calculation, finding the value of y. This means our algorithm is: INPUT: y(x), initial approximation x=(x1,...,xn), tolerance TOL, maximum number of iterations N. step 1: set k=1 step 2: if k≤N then do steps 3-6, otherwise do step 7 step 3: calculate y(x) step 4: set x=x+y(x) step 5: if y(x)<TOL then OUTPUT ('Procedure successful' print (k,x)); STOP #comment, this is not meant to be the norm of y but rather the norm between the old x and the x just calculated i.e. we're calculating the difference between approximations as an absolute error value and comparing it against an acceptable degree of error, the tolerance. If it is less than the tolerance the approximation is deemed to be 'good enough' and is given as the solution to the system along with the number of iterations it took to get it. step 6: set k=k+1, return to step 2. step 7: OUTPUT ('Procedure unsuccessful, can not find solution within tolerance'); STOP. And there's how to solve a non-linear system by Newton's method. For your problem this means that F(x)=(x2+y-11, y2+x-7), J(x)=([2x,1],[1,2y]) i.e. 2x and 1 on the top row of the matrix and 1 and 2y on the bottom, J(x)-1=1/(4xy-1)*([2y,-1],[-1,2x]) and y=-J(x)-1F(x)=1/(4xy-1)*(2x2y+y2-22y-x+7, 2xy2+x2-14x-y+11). Note that I have used bold parentheses to indicate a vector, or a matrix with rows separated by [], and that for your initial approximation x=(x1, x2), x1 equates to the coordinate of x, and x2 to the coordinate of y. Important note: as with most iterative methods, the success of the algorithm is highly dependent on what initial value you choose and, particularly in this case, how close it is to the real solution. If it is too far away it can have a tendency to 'fly away' from the solution and not give you an answer.