MHB How to Solve Non-Linear Equations of 3 Variables Using Newton-Raphson Method?

AI Thread Summary
The discussion focuses on solving a system of three non-linear equations using the Newton-Raphson method, specifically for the variables c, s, and q. The user expresses difficulty in calculating the derivatives of the functions involved and applying the Newton-Raphson iteration due to the interdependence of the variables. They seek guidance on how to derive the necessary derivatives and proceed with the iteration process. The conversation emphasizes the importance of forming a vector equation and utilizing the Jacobian matrix for the Newton-Raphson method. Overall, the thread highlights the challenges of applying this numerical method to multi-variable non-linear equations.
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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 $c$,$s$ and $q$.=> 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 $f'(c,s,q), g'(c,s,q), h'(c,s,q)$ but i don't know how?and after working out $f'(c,s,q), g'(c,s,q), h'(c,s,q)$ . After that i think i need to use Newton-raphson iteration:$c_{n+1}= c_n - \frac{f(c,s,q)}{f'(c,s,q)}$ but the $f(c,s,q)$ and $f'(c,s,q)$ contains the $s$ and $q$.

Similarly, for $s_{n+1}= s_n - \frac{g(c,s,q)}{g'(c,s,q)}$ will have $g(c,s,q)$ and $g'(c,s,q)$ containing the $c$ and $q$.$q_{n+1}= q_n - \frac{h(c,s,q)}{h'(c,s,q)}$ will have $h(c,s,q)$ and $h'(c,s,q)$ containing the $c$.so am i not sure what to do please help me. to find the values of $c,s,q$.
 
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grandy said:
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 $c$,$s$ and $q$.=> 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 $f'(c,s,q), g'(c,s,q), h'(c,s,q)$ but i don't know how?and after working out $f'(c,s,q), g'(c,s,q), h'(c,s,q)$ . After that i think i need to use Newton-raphson iteration:$c_{n+1}= c_n - \frac{f(c,s,q)}{f'(c,s,q)}$ but the $f(c,s,q)$ and $f'(c,s,q)$ contains the $s$ and $q$.

Similarly, for $s_{n+1}= s_n - \frac{g(c,s,q)}{g'(c,s,q)}$ will have $g(c,s,q)$ and $g'(c,s,q)$ containing the $c$ and $q$.$q_{n+1}= q_n - \frac{h(c,s,q)}{h'(c,s,q)}$ will have $h(c,s,q)$ and $h'(c,s,q)$ containing the $c$.so am i not sure what to do please help me. to find the values of $c,s,q$.

Write your system as a vector equation in a vector variable, you want a solution of the system:
$${\bf{f}}({\bf{x}})={\bf{0}}$$

The NR iteration for this is:

$${\bf{x}}_{n+1}={\bf{x}}_n-[{\rm{J}}({\bf{x}}_n)]^{-1}{\bf{f}}({\bf{x_n}})$$

where ${\rm{J}}({\bf{x}})$ is the matrix with $i.j$ th element equal to $ \dfrac{\partial {\bf{f}}_i}{\partial{\bf{x}}_j}$
 
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