Maxima を利用しよう -------------------------------------- 区間数2 -------------------------------------- p0(x)=a0*x^3+b0*x^2+c0*x+d0 p1(x)=a1*x^3+b1*x^2+c1*x+d1 に関する条件式: (0) 関数値 d0=y0 a0+b0+c0+d0=y1 a1+b1+c1+d1=y1 8*a1+4*b1+2*c1+d1=y2 (1) 1階微分 3*a0+2*b0+c0=3*a1+2*b1+c1 (2) 2階微分 6*a0+2*b0=6*a1+2*b1 (3) 自然境界 b0=0 12*a1+2*b1=0 -- Maxima への入力: reset(y0); reset(y1); reset(y2); a:matrix([0,0,0,1,0,0,0,0],[1,1,1,1,0,0,0,0],[0,0,0,0,1,1,1,1],[0,0,0,0,8,4,2,1],[3,2,1,0,-3,-2,-1,0],[3,1,0,0,-3,-1,0,0],[0,1,0,0,0,0,0,0],[0,0,0,0,6,1,0,0]); v:transpose(matrix([y0,y1,y1,y2,0,0,0,0])); b:invert(a); b.v; -- Maxima の出力 ( 出力を Copy as Text ): matrix([y2/4-y1/2+y0/4],[0],[-y2/4+(3*y1)/2-(5*y0)/4],[y0],[-y2/4+y1/2-y0/4],[(3*y2)/2-3*y1+(3*y0)/2],[-(7*y2)/4+(9*y1)/2-(11*y0)/4],[y2/2-y1+(3*y0)/2]) -- Excel の計算式: y2, y1, y0 を観測値の絶対参照に置換して =$b$22/4-$b$12/2+$b$2/4 =0 =-$b$22/4+(3*$b$12)/2-(5*$b$2)/4 =$b$2 =-$b$22/4+$b$12/2-$b$2/4 =(3*$b$22)/2-3*$b$12+(3*$b$2)/2 =-(7*$b$22)/4+(9*$b$12)/2-(11*$b$2)/4 =$b$22/2-$b$12+(3*$b$2)/2 -------------------------------------- 区間数3 -------------------------------------- p0(x)=a0*x^3+b0*x^2+c0*x+d0 p1(x)=a1*x^3+b1*x^2+c1*x+d1 p2(x)=a2*x^3+b2*x^2+c2*x+d2 に関する条件式: (0) 関数値 d0=y0 a0+b0+c0+d0=y1 a1+b1+c1+d1=y1 8*a1+4*b1+2*c1+d1=y2 8*a2+4*b2+2*c2+d2=y2 27*a2+9*b2+3*c2+d2=y3 (1) 1階微分 3*a0+2*b0+c0=3*a1+2*b1+c1 12*a1+4*b1+c1=12*a2+4*b2+c2 (2) 2階微分 6*a0+2*b0=6*a1+2*b1 12*a1+2*b1=12*a2+2*b2 (3) 自然境界 b0=0 18*a2+2*b2=0 -- Maxima への入力: reset(y0); reset(y1); reset(y2); reset(y3); a:matrix([0,0,0,1,0,0,0,0,0,0,0,0],[1,1,1,1,0,0,0,0,0,0,0,0],[0,0,0,0,1,1,1,1,0,0,0,0],[0,0,0,0,8,4,2,1,0,0,0,0],[0,0,0,0,0,0,0,0,8,4,2,1],[0,0,0,0,0,0,0,0,27,9,3,1],[3,2,1,0,-3,-2,-1,0,0,0,0,0],[0,0,0,0,12,4,1,0,-12,-4,-1,0],[3,1,0,0,-3,-1,0,0,0,0,0,0],[0,0,0,0,6,1,0,0,-6,-1,0,0],[0,1,0,0,0,0,0,0,0,0,0,0],[0,0,0,0,0,0,0,0,9,1,0,0]); v:transpose(matrix([y0,y1,y1,y2,y2,y3,0,0,0,0,0,0])); b:invert(a); b.v; -- Maxima の出力 ( 出力を Copy as Text ): matrix([-y3/15+(2*y2)/5-(3*y1)/5+(4*y0)/15],[0],[y3/15-(2*y2)/5+(8*y1)/5-(19*y0)/15],[y0],[y3/3-y2+y1-y0/3],[-(6*y3)/5+(21*y2)/5-(24*y1)/5+(9*y0)/5],[(19*y3)/15-(23*y2)/5+(32*y1)/5-(46*y0)/15],[-(2*y3)/5+(7*y2)/5-(8*y1)/5+(8*y0)/5],[-(4*y3)/15+(3*y2)/5-(2*y1)/5+y0/15],[(12*y3)/5-(27*y2)/5+(18*y1)/5-(3*y0)/5],[-(89*y3)/15+(73*y2)/5-(52*y1)/5+(26*y0)/15],[(22*y3)/5-(57*y2)/5+(48*y1)/5-(8*y0)/5]) -- Excel の計算式: y3, y2, y1, y0 を観測値の絶対参照に置換して =-$b$32/15+(2*$b$22)/5-(3*$b$12)/5+(4*$b$2)/15 =0 =$b$32/15-(2*$b$22)/5+(8*$b$12)/5-(19*$b$2)/15 =$b$2 =$b$32/3-$b$22+$b$12-$b$2/3 =-(6*$b$32)/5+(21*$b$22)/5-(24*$b$12)/5+(9*$b$2)/5 =(19*$b$32)/15-(23*$b$22)/5+(32*$b$12)/5-(46*$b$2)/15 =-(2*$b$32)/5+(7*$b$22)/5-(8*$b$12)/5+(8*$b$2)/5 =-(4*$b$32)/15+(3*$b$22)/5-(2*$b$12)/5+$b$2/15 =(12*$b$32)/5-(27*$b$22)/5+(18*$b$12)/5-(3*$b$2)/5 =-(89*$b$32)/15+(73*$b$22)/5-(52*$b$12)/5+(26*$b$2)/15 =(22*$b$32)/5-(57*$b$22)/5+(48*$b$12)/5-(8*$b$2)/5 -------------------------------------- 参考までに -------------------------------------- eq:[d0=y0,a0+b0+c0+d0=y1,a1+b1+c1+d1=y1,8*a1+4*b1+2*c1+d1=y2,3*a0+2*b0+c0=3*a1+2*b1+c1,6*a0+2*b0=6*a1+2*b1,b0=0,12*a1+2*b1=0]; var:[a0,b0,c0,d0,a1,b1,c1,d1]; solve[eq,var]; では失敗 --------------------------------------