1
$\begingroup$

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 dont 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 $c,s,q$.

$\endgroup$
  • $\begingroup$ You need to treat the three functions as the components of a single vector-valued function of three variables. The Newton iteration is then formally the same as in the one variable case, but with the inverse of the differential of the multivariable function (a 3 x 3 matrix) in place of the reciprocal of the derivative of the single variable function. $\endgroup$ – Dan Fox Jun 3 '14 at 15:57
2
$\begingroup$

The Newton-Raphson method is based on considering the tangent line. We need some linear algebra to understand and implement the substitute of "tangent line" in multiple dimensions.

A nonlinear system of $n$ equations with $n$ unknowns can be written in vector form as $\vec F(\vec x)=0$. The first order partial derivatives of $\vec F$ form the Jacobian matrix $J$: put the components of $\vec F$ in a column, then take derivatives in each variable. For example if $$ \vec F(\vec x) = \begin{pmatrix} x_1^2e^{3x_2}-30 \\ x_1x_2-\sin(x_1+x_2^2) \end{pmatrix} $$ then $$ J = \begin{pmatrix} 2x_1e^{3x_2} & 3x_1^2e^{3x_2} \\ x_2-\cos(x_1+x_2^2) & x_1-2x_2\cos(x_1+x_2^2) \end{pmatrix} $$ The Jacobian matrix provides a linear approximation to $\vec F$: near a point $\vec x_0$ we have $$\vec F(\vec x) \approx \vec F(\vec x_0) + J(\vec x-\vec x_0) \tag1$$ This is the analog of tangent line.

Following the idea of single variable method, we equate the right hand side of (1) to $\vec 0$ and solve: $$\vec x - \vec x_0 = -J^{-1} \vec F(\vec x_0) \tag2$$ Note that (2) is merely for writing down the theoretical approach, in practice we do not invert the matrix $J$. Rather, we let the software (Matlab or whatever) solve the system with matrix $J$ and right hand side $-\vec F(\vec x_0)$, which it can do efficiently.

Having solved the system, you obtain the new point $\vec x$, which takes the role of $\vec x_0$ at the next step of iteration. Continue until the norm $|\vec x - \vec x_0|$ becomes small... or until the allowed number of iterations runs out (indicating the method fails to converge).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.