Analysis by SBC

392

Name

안선주

Subject

연습문제 4.1_7

File

hwk4_7.jpg (15 KB) - Download : 111

%% 7. {(-2,-15),(-1,-8),(0,-3)}을 보간하는 2차 다항식을 만드시오. 그리고 p2(x)의 근 중에서 x = 0의 우측으로 향한 첫 번째 근을 찾으시오. 다항식 p2(x)가 최대값을 갖는가?
% x0=-2; x1=-1; x2=0;
% y0=-15; y1=-8; y2=-3;

% L2,0(x) = (x-x1)*(x-x2)/(x0-x1)*(x0-x2);
% L2,1(x) = (x-x0)*(x-x2)/(x1-x0)*(x1-x2);
% L2,2(x) = (x-x0)*(x-x1)/(x2-x0)*(x2-x1);

% p2(x) = L2,0(x)*(-15) + L2,1(x)*(-8) + L2,2(x)*(-3);
% p2(x) = (-20/3)*x^2 + (37/3)*x - 3;

f=@(x) (-20/3)*x.^2 + (37/3)*x - 3;
h = 1; xj = -2:h:0; yj = f(xj); xv = -2:0.01:0; fv=f(xv);
yv = interp_dd(xj,yj,xv);
plot (xv,fv)
title('f(x) = (-20/3)*x^2 + (37/3)*x - 3')
hold on
plot(xj,yj,'ro')

%%%%%%%%%%%실행결과%%%%%%%%%%%%%%%%%
------ Divided Differences of f --------
i       x_i     f(x_i)       D_i
----------------------------------------
   0    -2.00   -54.3333 -5.4333e+001
   1    -1.00   -22.0000 3.2333e+001
   2     0.00    -3.0000 -6.6667e+000
----------------------------------------

>> solve('(-20/3)*x^2 + (37/3)*x - 3=0')

ans =

37/40-1/40*649^(1/2)
37/40+1/40*649^(1/2)

최대값 = 37/40 = 0.9250

DATE: 2013.05.06 - 15:52

연습문제 4.1_18

4.33