How Can Newton-Raphson Method Solve These Complex Nonlinear Equations?

[wel]

The three non-linear equations are given by
\begin{equation}
c[(6.7 * 10^8) + (1.2 * 10^8)s+(1-q)(2.6*10^8)]-0.00114532=0
\end{equation}
\begin{equation}
s[2.001 *c + 835(1-q)]-2.001*c =0
\end{equation}
\begin{equation}
q[2.73 + (5.98*10^{10})c]-(5.98 *10^{10})c =0
\end{equation}
Using the Newton-Raphson Method solve these equations in terms of [itex]c,s[/itex] and [itex]q[/itex].

=> It is really difficult question for me because i don't know very much about the Newton-Raphson Method and also these non-linear equations contain 3 variables.

I have try by applying the Newton-Raphson method to each equations:-
\begin{equation}
f(c,s,q)=0= c[(6.7 * 10^8) + (1.2 * 10^8)s+(1-q)(2.6*10^8)]-0.00114532
\end{equation}
\begin{equation}
g(c,s,q)=0= s[2.001 *c + 835(1-q)]-2.001*c
\end{equation}
\begin{equation}
h(c,s,q)=0= q[2.73 + (5.98*10^{10})c]-(5.98 *10^{10})c
\end{equation}
now i guess i need to work out [itex]f'(c,s,q), g'(c,s,q), h'(c,s,q)[/itex] but i don't know how?

and after working out [itex]f'(c,s,q), g'(c,s,q), h'(c,s,q)[/itex] . After that i think i need to use Newton-raphson iteration:

[itex]c_{n+1}= c_n - \frac{f(c,s,q)}{f'(c,s,q)}[/itex]

but the [itex]f(c,s,q)[/itex] and [itex]f'(c,s,q)[/itex] contains the [itex]s[/itex] and [itex]q[/itex].

Similarly, for

[itex]s_{n+1}= s_n - \frac{g(c,s,q)}{g'(c,s,q)}[/itex]

will have [itex]g(c,s,q)[/itex] and [itex]g'(c,s,q)[/itex] containing the [itex]c[/itex] and [itex]q[/itex[].

[itex]q_{n+1}= q_n - \frac{h(c,s,q)}{h'(c,s,q)}[/itex]

will have [itex]h(c,s,q)[/itex] and [itex]h'(c,s,q)[/itex] containing the [itex]c[/itex].

so am i not sure what to do please help me. to find the values of [itex]c,s,q[/itex].

 

[LCKurtz]

Perhaps you should read how the Newton Raphson method works for multi variable problems. Here's one link:

http://fourier.eng.hmc.edu/e161/lectures/ica/node13.html

[Edit, added] Why don't you simplify the equations a bit first? Also, are you using something like Maple or Mathematica? If so, why not let the package solve the system? I'm not a computer scientist, but I would be wary of a system with huge numbers and little bitty numbers.

 

[wel]

I need to calculate by hand without any mathematical software. that's the reason i am having difficulty. please help me.

 

[Ray Vickson]

'

I need to calculate by hand without any mathematical software. that's the reason i am having difficulty. please help me.

You have already been told what you need to do; have you tried it?

Personally, I would be skeptical about the use of a standard method on your problem as it stands, because your problem is very badly scaled. You can get a much better-behaved system by using scaled variables: let ##s = 10^{-8} s'## and ##c = 10^{-9} c'##. Then the system becomes
[tex] .93 c'-.26 c' q-.00114532 +.7065655170 \times 10^{-11} c' s' = 0\\
835 s'-835 s'q-.2001 c' + .2001 \times 10^{-8} c' s'= 0 \\
2.73 q+59.8 c' q-59.8 c' = 0
[/tex]
In this new system the bad scaling is limited to the terms in ##c' s'## in the first two equations. It is very probable that much better accuracy can be obtained from solving the new system.

 

[LCKurtz]

@Ray: Interestingly enough, Maple will give an answer to that system. Then if you solve the same system without the ##c's'## terms, you get the same answer. Not too surprising, I guess. I'm thinking those two terms would also raise havoc when calculating the inverse of the Jacobian for Newton-Rhapson.

 

[Ray Vickson]

@Ray: Interestingly enough, Maple will give an answer to that system. Then if you solve the same system without the ##c's'## terms, you get the same answer. Not too surprising, I guess. I'm thinking those two terms would also raise havoc when calculating the inverse of the Jacobian for Newton-Rhapson.

Yes, I did everything in Maple.

If you take the original system but first convert the numbers from floats to rationals, the maple command 'solve' produces a solution in terms of some 'Root_Of' expressions; then applying the 'allvalues' command produces 4 symbolic solutions, but which are huge, multi-page monsters that involve enormous rational numbers, etc. Applying 'evalf[n]' for n-digit evaluation produces two complex and two real solutions. Applying 'solve' on the original system, with no rational conversion, produces essentially the same 4 floating-point solutions. Using 'fsolve' instead produces one of the 4 solutions, and if I am not mistaken, is done essentially by implementing Newton's method. I suspect that the implementation was done carefully enough to get around bad scaling, perhaps by using a self-scaling version of Newton.

 

[LCKurtz]

@wel: Where did this problem come from? If it is just a problem where you are supposed to learn how Newton-Rhapson method works, then it is a ridiculous teaching example. On the other hand, if you are studying numerical analysis and learning about how to handle systems that aren't well behaved, it would be a more appropriate problem. Please give us the context in which this problem arose.