Linear Algebra and the C Language/a0bi

__NOTOC__ Install and compile this file in your working directory. /* ------------------------------------ */ /* Save as : c00a.c */ /* ------------------------------------ */ #include "v_a.h" /* ------------------------------------ */ /* ------------------------------------ */ #define RA R3 #define CA C3 #define Cb C1 /* ------------------------------------ */ /* ------------------------------------ */ int main(void) { double xy[6] ={1, 6, 2, 3, 3, 5 }; double ab[RA*(CA+Cb)]={ /* x**2 x**1 x**0 y */ +1, +1, +1, +6, +4, +2, +1, +3, +9, +3, +1, +5, }; double **XY = ca_A_mR(xy,i_mR(R3,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**2 + bx + c (x**0 = 1) \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**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); invgj_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**2 %+.2fx %+.2f\n\n" ,x[R1][C1],x[R2][C1],x[R3][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**2 + bx + c Which passes through these three points. x[1], y[1] x[2], y[2] x[3], y[3] Using the points we obtain the matrix: x**2 x**1 x**0 y x[1]**2 x[1]**1 x[1]**0 y[1] x[2]**2 x[2]**1 x[2]**0 y[2] x[3]**2 x[3]**1 x[3]**0 y[3] That we can write: x**2 x 1 y x[1]**2 x[1] 1 y[1] x[2]**2 x[2] 1 y[2] x[3]**2 x[3] 1 y[3] Let's use the QR_mR(); function to solve the system that will give us the coefficients a, b, c Screen output example: Find the coefficients a, b, c of the curve y = ax**2 + bx + c (x**0 = 1) that passes through the points. x y +1 +6 +2 +3 +3 +5 Using the given points, we obtain this matrix. x**2 x**1 x**0 y +1.00 +1.00 +1.00 +6.00 +4.00 +2.00 +1.00 +3.00 +9.00 +3.00 +1.00 +5.00 Press return to continue. Q : +0.1010 +0.7184 +0.6882 +0.4041 +0.6025 -0.6882 +0.9091 -0.3476 +0.2294 R : +9.8995 +3.6365 +1.4142 -0.0000 +0.8806 +0.9733 +0.0000 +0.0000 +0.2294 Press return to continue. Q_T : +1.0102e-01 +4.0406e-01 +9.0914e-01 +7.1841e-01 +6.0254e-01 -3.4762e-01 +6.8825e-01 -6.8825e-01 +2.2942e-01 invR : +1.0102e-01 -4.1714e-01 +1.1471e+00 +0.0000e+00 +1.1355e+00 -4.8177e+00 -0.0000e+00 -0.0000e+00 +4.3589e+00 Press return to continue. Solving this system yields a unique least squares solution, namely x = invR * Q_T * b : +2.50 -10.50 +14.00 The coefficients a, b, c of the curve are : y = +2.50x**2 -10.50x +14.00 Press return to continue. Copy and paste in Octave: function y = f (x) y = +2.50*x^2 -10.50*x +14.00; endfunction f (+1) f (+2) f (+3)