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📄 tlmdif.c

📁 该程序实现了非线性最小二乘问题和非线性方程组的解法
💻 C
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/*      driver for lmdif example. */#include <stdio.h>#include <math.h>#include <cminpack.h>int fcn(void *p, int m, int n, const double *x, double *fvec, int iflag);int main(){  int i, j, m, n, maxfev, mode, nprint, info, nfev, ldfjac;  int ipvt[3];  double ftol, xtol, gtol, epsfcn, factor, fnorm;  double x[3], fvec[15], diag[3], fjac[15*3], qtf[3],     wa1[3], wa2[3], wa3[3], wa4[15];  double covfac;  m = 15;  n = 3;/*      the following starting values provide a rough fit. */  x[1-1] = 1.;  x[2-1] = 1.;  x[3-1] = 1.;  ldfjac = 15;  /*      set ftol and xtol to the square root of the machine */  /*      and gtol to zero. unless high solutions are */  /*      required, these are the recommended settings. */  ftol = sqrt(dpmpar(1));  xtol = sqrt(dpmpar(1));  gtol = 0.;  maxfev = 800;  epsfcn = 0.;  mode = 1;  factor = 1.e2;  nprint = 0;  info = lmdif(fcn, 0, m, n, x, fvec, ftol, xtol, gtol, maxfev, epsfcn, 	 diag, mode, factor, nprint, &nfev, fjac, ldfjac, 	 ipvt, qtf, wa1, wa2, wa3, wa4);  fnorm = enorm(m, fvec);  printf("      final l2 norm of the residuals%15.7g\n\n", fnorm);  printf("      number of function evaluations%10i\n\n", nfev);  printf("      exit parameter                %10i\n\n", info);  printf("      final approximate solution\n");  for (j=1; j<=n; j++) printf("%s%15.7g", j%3==1?"\n     ":"", x[j-1]);  printf("\n");  ftol = dpmpar(1);  covfac = fnorm*fnorm/(m-n);  covar(n, fjac, ldfjac, ipvt, ftol, wa1);  printf("      covariance\n");  for (i=1; i<=n; i++) {    for (j=1; j<=n; j++)      printf("%s%15.7g", j%3==1?"\n     ":"", fjac[(i-1)*ldfjac+j-1]*covfac);  }  printf("\n");  return 0;}int fcn(void *p, int m, int n, const double *x, double *fvec, int iflag){/*      subroutine fcn for lmdif example. */  int i;  double tmp1, tmp2, tmp3;  double y[15]={1.4e-1, 1.8e-1, 2.2e-1, 2.5e-1, 2.9e-1, 3.2e-1, 3.5e-1,		3.9e-1, 3.7e-1, 5.8e-1, 7.3e-1, 9.6e-1, 1.34, 2.1, 4.39};  if (iflag == 0)    {      /*      insert print statements here when nprint is positive. */      return 0;    }  for (i = 1; i <= 15; i++)    {      tmp1 = i;      tmp2 = 16 - i;      tmp3 = tmp1;      if (i > 8) tmp3 = tmp2;      fvec[i-1] = y[i-1] - (x[1-1] + tmp1/(x[2-1]*tmp2 + x[3-1]*tmp3));    }  return 0;}

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