Linear Algebra and the C Language/a0bk

__NOTOC__ Install and compile this file in your working directory. /* ------------------------------------ */ /* Save as : c00c.c */ /* ------------------------------------ */ #include "v_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_mR(xy,i_mR(R4,C2)); double **Ab = ca_A_mR(ab,i_Abr_Ac_bc_mR(RA,CA,Cb)); double **A = c_Ab_A_mR(Ab,i_mR(RA,CA)); double **b = c_Ab_b_mR(Ab,i_mR(RA,Cb)); double **Q = i_mR(RA,CA); double **R = i_mR(CA,CA); double **invR = i_mR(CA,CA); double **Q_T = i_mR(CA,RA); double **invR_Q_T = i_mR(CA,RA); double **x = i_mR(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**3 + bx**2 + cx + d \n\n"); printf(" that passes through the points. \n\n"); printf(" x y"); p_mR(XY,S5,P0,C6); printf(" Using the given points, we obtain this matrix.\n"); printf(" x**3 x**2 x**1 x**0 y\n"); p_mR(Ab,S7,P2,C6); stop(); clrscrn(); QR_mR(A,Q,R); printf(" Q :"); p_mR(Q,S10,P4,C6); printf(" R :"); p_mR(R,S10,P4,C6); stop(); clrscrn(); transpose_mR(Q,Q_T); printf(" Q_T :"); pE_mR(Q_T,S12,P4,C6); inv_mR(R,invR); printf(" invR :"); pE_mR(invR,S12,P4,C6); stop(); clrscrn(); printf(" Solving this system yields a unique\n" " least squares solution, namely \n\n"); mul_mR(invR,Q_T,invR_Q_T); mul_mR(invR_Q_T,b,x); printf(" x = invR * Q_T * b :"); p_mR(x,S10,P2,C6); printf("\n The coefficients a, b, c of the curve are : \n\n" " y = %+.2fx**3 %+.2fx**2 %+.2fx %+.2f\n\n" ,x[R1][C1],x[R2][C1],x[R3][C1],x[R4][C1]); stop(); f_mR(XY); f_mR(A); f_mR(b); f_mR(Ab); f_mR(Q); f_mR(Q_T); f_mR(R); f_mR(invR); f_mR(invR_Q_T); f_mR(x); return 0; } /* ------------------------------------ */ /* ------------------------------------ */ Presentation : Let's calculate the coefficients of a polynomial. y = ax**3 + bx**2 + cx + d Which passes through these four points. x[1], y[1] x[2], y[2] x[3], y[3] x[4], y[4] Using the points we obtain the matrix: x**3 x**2 x**1 x**0 y x[1]**3 x[1]**2 x[1]**1 x[1]**0 y[1] x[2]**3 x[2]**2 x[2]**1 x[2]**0 y[2] x[3]**3 x[3]**2 x[3]**1 x[3]**0 y[3] x[4]**3 x[4]**2 x[4]**1 x[4]**0 y[4] That we can write: x**3 x**2 x 1 y x[1]**3 x[1]**2 x[1] 1 y[1] x[2]**3 x[2]**2 x[2] 1 y[2] x[3]**3 x[3]**2 x[3] 1 y[3] x[4]**3 x[4]**2 x[4] 1 y[4] Let's use the QR_mR() function to solve the system that will give us the coefficients a, b, c, d Screen output example: Find the coefficients a, b, c 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 points, 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 : -0.14 -0.73 +1.31 +4.43 The coefficients a, b, c of the curve are : y = -0.14x**3 -0.73x**2 +1.31x +4.43 Press return to continue. Copy and paste in Octave: function y = f (x) y = -0.139285714*x^3 -0.732142857*x^2 +1.307142857*x +4.428571429; endfunction f (-5) f (-2) f (+2) f (+3)