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//Example 7.13
//Newtons Backward Formula
//Page no. 248
clc;close;clear;
printf(' x\t y\t d\t d2\t d3\t d4\t d5\n')
printf('--------------------------------------------------------')
h=0.5;
deff('y=f2(x)','y=(z(x,4)-z(x,5)+z(x,6))/h^2')
z=[1.5,3.375;2,7;2.5,13.625;3,24;3.5,38.875;4,59];
for i=1:6
for j=3:7
z(i,j)=-1
end
end
for i=3:7
for j=1:8-i
z(j,i)=z(j+1,i-1)-z(j,i-1)
end
end
printf('\n')
for i=1:6
for j=1:7
if z(i,j)==-1 then
printf(' \t')
elseif j==1
printf(' %.1f\t',z(i,j))
else
printf('%.3f\t',z(i,j))
end
end
printf('\n')
end
j=1;y1=0;
for i=3:6
y1=y1+(-1)^(i-1)*z(j,i)/(i-2)
end
y1=y1/h;
y2(7)=f2(1);
printf('\n\n f`(1.5)= %g',y1)
printf('\n\n f``(1.5) = %g',y2(7))
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