Mathc complexes/06y
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c00c.c |
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/* ------------------------------------ */
/* Save as : c00c.c */
/* ------------------------------------ */
#include "w_a.h"
/* ------------------------------------ */
#define RA R4
#define CA C4
#define Cb C1
/* ------------------------------------ */
/* ------------------------------------ */
int main(void)
{
double xy[8] ={ -5, -3,
-2, 0,
2, 3,
3, -2 };
double ab[RA*(CA+Cb)]={
/* x**3 x**2 x**1 x**0 y */
-125, +25, -5, +1, -3,
-8, +4, -2, +1, +0,
+8, +4, +2, +1, +3,
+27, +9, +3, +1, -2,
};
double **XY = ca_A_mRZ(xy,i_mZ(R4,C2));
double **Ab = ca_A_mRZ(ab,i_Abr_Ac_bc_mZ(RA,CA,Cb));
double **A = c_Ab_A_mZ(Ab,i_mZ(RA,CA));
double **b = c_Ab_b_mZ(Ab,i_mZ(RA,Cb));
double **Q = i_mZ(RA,CA);
double **R = i_mZ(CA,CA);
double **invR = i_mZ(CA,CA);
double **Q_T = i_mZ(CA,RA);
double **invR_Q_T = i_mZ(CA,RA);
double **x = i_mZ(CA,Cb); // x invR * Q_T * b
clrscrn();
printf("\n");
printf(" Find the coefficients a, b, c, d of the curve \n\n");
printf(" y = ax**3 + bx**2 + cx + d \n\n");
printf(" that passes through the points. \n\n");
printf(" x y");
p_mRZ(XY,S5,P0,C6);
printf("\n");
printf(" Using the given XY, we obtain this matrix.\n");
printf(" x**3 x**2 x**1 x**0 y\n");
p_mRZ(Ab,S7,P2,C6);
stop();
clrscrn();
QR_mZ(A,Q,R);
printf(" Q :");
p_mRZ(Q,S10,P4,C6);
printf(" R :");
p_mRZ(R,S10,P4,C6);
stop();
clrscrn();
transpose_mZ(Q,Q_T);
printf(" Q_T :");
pE_mRZ(Q_T,S12,P4,C6);
inv_mZ(R,invR);
printf(" invR :");
pE_mRZ(invR,S12,P4,C6);
stop();
clrscrn();
printf(" Solving this system yields a unique\n"
" least squares solution, namely \n\n");
mul_mZ(invR,Q_T,invR_Q_T);
mul_mZ(invR_Q_T,b,x);
printf(" x = invR * Q_T * b :");
p_mRZ(x,S10,P3,C6);
printf(" The coefficients a, b, c, d of the curve are : \n\n"
" y = %+.3fx**3 %+.3fx**2 %+.3fx %+.3f\n\n"
,x[R1][C1],x[R2][C1],x[R3][C1],x[R4][C1]);
stop();
f_mZ(XY);
f_mZ(A);
f_mZ(b);
f_mZ(Ab);
f_mZ(Q);
f_mZ(Q_T);
f_mZ(R);
f_mZ(invR);
f_mZ(invR_Q_T);
f_mZ(x);
return 0;
}
/* ------------------------------------ */
/* ------------------------------------ */
Exemple de sortie écran :
Find the coefficients a, b, c, d of the curve
y = ax**3 + bx**2 + cx + d
that passes through the points.
x y
-5 -3
-2 +0
+2 +3
+3 -2
Using the given XY, we obtain this matrix.
x**3 x**2 x**1 x**0 y
-125.00 +25.00 -5.00 +1.00 -3.00
-8.00 +4.00 -2.00 +1.00 +0.00
+8.00 +4.00 +2.00 +1.00 +3.00
+27.00 +9.00 +3.00 +1.00 -2.00
Press return to continue.
Q :
-0.9737 +0.2054 +0.0819 -0.0556
-0.0623 +0.1700 -0.9033 +0.3889
+0.0623 +0.3529 +0.4209 +0.8333
+0.2103 +0.8969 -0.0131 -0.3889
R :
+128.3822 -22.4486 +5.7485 -0.7633
+0.0000 +15.2990 +2.0292 +1.6252
-0.0000 +0.0000 +2.1995 -0.4136
-0.0000 +0.0000 -0.0000 +0.7778
Press return to continue.
Q_T :
-9.7365e-01 -6.2314e-02 +6.2314e-02 +2.1031e-01
+2.0543e-01 +1.7002e-01 +3.5289e-01 +8.9686e-01
+8.1914e-02 -9.0331e-01 +4.2088e-01 -1.3124e-02
-5.5556e-02 +3.8889e-01 +8.3333e-01 -3.8889e-01
invR :
+7.7892e-03 +1.1429e-02 -3.0902e-02 -3.2672e-02
+0.0000e+00 +6.5364e-02 -6.0304e-02 -1.6865e-01
+0.0000e+00 -0.0000e+00 +4.5466e-01 +2.4180e-01
-0.0000e+00 +0.0000e+00 -0.0000e+00 +1.2857e+00
Press return to continue.
Solving this system yields a unique
least squares solution, namely
x = invR * Q_T * b :
-0.139
-0.732
+1.307
+4.429
The coefficients a, b, c, d of the curve are :
y = -0.139x**3 -0.732x**2 +1.307x +4.429
Press return to continue.