sas可否解方程组，求教高手

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求教高手

sas可否解方程组，非线性的，如何操作？

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关键词：解方程组 方程组 解方程 如何操作 非线性 高手 SAS 求教 解方程

Newton's Method for Solving Nonlinear Systems of Equations

This example solves a nonlinear system of equations by Newton's method. Let the nonlinear system be represented by

F(x) = 0

where x is a vector and F is a vector-valued, possibly nonlinear, function.

In order to find x such that F goes to 0, an initial estimate x₀ is chosen, and Newton's iterative method for converging to the solution is used:

x_n+1 = x_n - J^-1(x_n) F(x_n)

where J(x) is the Jacobian matrix of partial derivatives of F with respect to x.

For optimization problems, the same method is used, where F(x) is the gradient of the objective function and J(x) becomes the Hessian (Newton-Raphson).

In this example, the system to be solved is

The code is organized into three modules: NEWTON, FUN, and DERIV.

 /* Newton's Method to Solve a Nonlinear Function */ /* The user must supply initial values, */ /* and the FUN and DERIV functions. */ /* on entry: FUN evaluates the function f in terms of x */ /* initial values are given to x */ /* DERIV evaluates jacobian j */ /* tuning variables: CONVERGE, MAXITER. */ /* on exit: solution in x, function value in f close to 0 */ /* ITER has number of iterations. */ start newton; run fun; /* evaluate function at starting values */ do iter=1 to maxiter /* iterate until maxiter iterations */ while(max(abs(f))>converge); /* or convergence */ run deriv; /* evaluate derivatives in j */ delta=-solve(j,f); /* solve for correction vector */ x=x+delta; /* the new approximation */ run fun; /* evaluate the function */ end; finish newton; maxiter=15; /* default maximum iterations */ converge=.000001; /* default convergence criterion */ /* User-supplied function evaluation */ start fun; x1=x[1] ; x2=x[2] ; /* extract the values */ f= (x1+x2-x1*x2+2)// (x1*exp(-x2)-1); /* evaluate the function */ finish fun; /* User-supplied derivatives of the function */ start deriv; /* evaluate jacobian */ j=((1-x2)||(1-x1) )//(exp(-x2)||(-x1*exp(-x2))); finish deriv; do; print "Solving the system: X1+X2-X1*X2+2=0, X1*EXP(-X2)-1=0" ,; x={.1, -2}; /* starting values */ run newton; print x f; end; Newton-Raphson是其中的一种方法

多谢，可否帮我解一个实例好让我明白？