Linear Algebra and the C Language/a0bn

__NOTOC__ Install and compile this file in your working directory. /* ------------------------------------ */ /* Save as : c00f.c */ /* ------------------------------------ */ #include "v_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_mR(xy,i_mR(R5,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**4 + bx**3 + cx**2 + dx + e \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**4 x**3 x**2 x**1 x**0 y"); 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**4 %+.2fx**3 %+.2fx**2 %+.2fx %+.2f\n\n" ,x[R1][C1],x[R2][C1],x[R3][C1],x[R4][C1],x[R5][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**4 + bx**3 + cx**2 + dx + e Which passes through these five points. x[1], y[1] x[2], y[2] x[3], y[3] x[4], y[4] x[5], y[5] Using the points we obtain the matrix: x**4 x**3 x**2 x**1 x**0 y x[1]**4 x[1]**3 x[1]**2 x[1]**1 x[1]**0 y[1] x[2]**4 x[2]**3 x[2]**2 x[2]**1 x[2]**0 y[2] x[3]**4 x[3]**3 x[3]**2 x[3]**1 x[3]**0 y[3] x[4]**4 x[4]**3 x[4]**2 x[4]**1 x[4]**0 y[4] x[5]**4 x[5]**3 x[5]**2 x[5]**1 x[5]**0 y[5] That we can write: x**4 x**3 x**2 x 1 y x[1]**4 x[1]**3 x[1]**2 x[1] 1 y[1] x[2]**4 x[2]**3 x[2]**2 x[2] 1 y[2] x[3]**4 x[3]**3 x[3]**2 x[3] 1 y[3] x[4]**4 x[4]**3 x[4]**2 x[4] 1 y[4] x[5]**4 x[5]**3 x[5]**2 x[5] 1 y[5] Let's use the QR_mR() function to solve the system that will give us the coefficients a, b, c, d, e Screen output example: 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 points, 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 : +2.67 -30.33 +117.83 -181.17 +89.00 The coefficients a, b, c of the curve are : y = +2.67x**4 -30.33x**3 +117.83x**2 -181.17x +89.00 Press return to continue. Copy and paste in Octave: function y = f (x) y = +2.666665810*x^4 -30.333323719*x^3 +117.833298608*x^2 -181.166625268*x +88.999995106; endfunction f (+1) f (+2) f (+3) f (+4) f (+5)