# Finding initial conditions while solving system of non-linear equations by Newton Raphson method.

Friends

I was solving a question on system of non-linear equations by Newton-Raphson method and i came by a problem where there was no initial conditions given ,so we have to find it out! ,

but it uses a method[use of approximations] which i am not getting any intuition about and also since the convergence of the N-R method depends on the choice of initial conditions , so there must be a reason of doing this procedure below to obtain the initial conditions .

Any idea guys , why it is so ?

Thank you!

• Yes but Why $\sin(x+y)=0 , \cos(z-y)=1,\sin(z)=0$ was taken ?.. , – BAYMAX Feb 25 '17 at 4:42
• We also have no idea that by doing so, the solutions may converge or not! – BAYMAX Feb 25 '17 at 4:43
• Note that $$\max \sin(x+y)=\max \cos(z-y)=\max\sin z=1$$ So a good starting point is to replace the sines and cosines by their maximum or minimum and see the approximate values of $x,y,z$ are near zero – polfosol Feb 25 '17 at 6:13

There is no real good reason. From the coefficients alone these are the dominant terms of the equations, i.e., they change most with changes in the variables. An equally justified (or not justified) initial point would be $(x,y,z)=(0,0,0)$ where then for the next point you would linearize $\sin(x+y)=x+y+h.o.t.$ (higher order terms), $\cos(z-y)^2=1+h.o.t.$ and $\sin(z)=z+h.o.t.$
If you want to experiment, compute the solutions for the homotopy \begin{align} 10x+t\sin(x+y)&=1\\ 8y-t\cos^2(z-y)&=1\\ 12z+t\sin(z)&=1 \end{align} using the given trivial solution at $t=0$ as initial point and incrementing $t$ in steps of, say, $0.1$ towards 1, in each step using Newton to compute the solution and using the approximation of the last step as initial point for the next $t$.
This homotopy continuation or watershed method (simulated annealing if one replaces $t$ with $(1-t)$) is one way to globalize the Newton method.