Mathc complexes/071
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c00f.c |
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/* ------------------------------------ */
/* Save as : c00f.c */
/* ------------------------------------ */
#include "w_a.h"
/* ------------------------------------ */
#define RA R5
#define CA C5
#define Cb C1
/* ------------------------------------ */
/* ------------------------------------ */
int main(void)
{
double xy[10] ={
1, -2,
2, -2,
3, 3,
4, -9,
5, 4, };
double ab[RA*(CA+Cb)]={
/* x**4 x**3 x**2 x**1 x**0 y */
+1, +1, +1, +1, +1, -2,
+16, +8, +4, +2, +1, -2,
+81, +27, +9, +3, +1, +3,
+256, +64, +16, +4, +1, -9,
+625, +125, +25, +5, +1, +4,
};
double **XY = ca_A_mRZ(xy,i_mZ(R5,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 of the curve \n\n");
printf(" y = ax**4 + bx**3 + cx**2 + dx + e \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**4 x**3 x**2 x**1 x**0 y");
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**4 %+.3fx**3 %+.3fx**2 %+.3fx %+.3f\n\n"
,x[R1][C1],x[R2][C1],x[R3][C1],x[R4][C1],x[R5][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 of the curve
y = ax**4 + bx**3 + cx**2 + dx + e
that passes through the points.
x y
+1 -2
+2 -2
+3 +3
+4 -9
+5 +4
Using the given XY, we obtain this matrix.
x**4 x**3 x**2 x**1 x**0 y
+1.00 +1.00 +1.00 +1.00 +1.00 -2.00
+16.00 +8.00 +4.00 +2.00 +1.00 -2.00
+81.00 +27.00 +9.00 +3.00 +1.00 +3.00
+256.00 +64.00 +16.00 +4.00 +1.00 -9.00
+625.00 +125.00 +25.00 +5.00 +1.00 +4.00
Press return to continue.
Q :
+0.0015 +0.0485 +0.4216 +0.8487 +0.3156
+0.0235 +0.2856 +0.7083 -0.1335 -0.6312
+0.1190 +0.6174 +0.2365 -0.3877 +0.6312
+0.3762 +0.6420 -0.4915 +0.3242 -0.3156
+0.9185 -0.3504 +0.1519 -0.0805 +0.0631
R :
+680.4256 +142.3006 +30.1502 +6.5033 +1.4388
+0.0000 +16.2950 +8.2602 +3.2880 +1.2431
-0.0000 -0.0000 +1.3158 +1.3410 +1.0268
+0.0000 +0.0000 +0.0000 +0.3128 +0.5711
-0.0000 -0.0000 -0.0000 -0.0000 +0.0631
Press return to continue.
Q_T :
+1.4697e-03 +2.3515e-02 +1.1904e-01 +3.7624e-01 +9.1854e-01
+4.8534e-02 +2.8560e-01 +6.1737e-01 +6.4201e-01 -3.5037e-01
+4.2165e-01 +7.0827e-01 +2.3651e-01 -4.9150e-01 +1.5186e-01
+8.4868e-01 -1.3354e-01 -3.8774e-01 +3.2419e-01 -8.0475e-02
+3.1560e-01 -6.3119e-01 +6.3119e-01 -3.1560e-01 +6.3119e-02
invR :
+1.4697e-03 -1.2834e-02 +4.6895e-02 -9.6700e-02 +3.3138e-01
+0.0000e+00 +6.1368e-02 -3.8527e-01 +1.0067e+00 -4.0502e+00
+0.0000e+00 -0.0000e+00 +7.6002e-01 -3.2586e+00 +1.7121e+01
+0.0000e+00 +0.0000e+00 +0.0000e+00 +3.1973e+00 -2.8930e+01
+0.0000e+00 -0.0000e+00 +0.0000e+00 +0.0000e+00 +1.5843e+01
Press return to continue.
Solving this system yields a unique
least squares solution, namely
x = invR * Q_T * b :
+2.667
-30.333
+117.833
-181.167
+89.000
The coefficients a, b, c, d of the curve are :
y = +2.667x**4 -30.333x**3 +117.833x**2 -181.167x +89.000