//Example 5.23 //Newton's Forward Difference Formula //Page no. 147 clc;close;clear; printf(' x\t y\t d1\td2\td3\td4\t') printf('\n--------------------------------------------------------------------') h=5; z=[80,5026;85,5674;90,6362;95,7088;100,7854] deff('y=f(x,p)','y=z(x,2)+p*z(x-1,3)+p*(p+1)*z(x-2,4)/2+p*(p+1)*(p+2)*z(x-3,5)/6+p*(p+1)*(p+2)*(p+3)*z(x-4,6)/24') x01=100;x11=105; for i=3:7 for j=1:7-i z(j,i)=z(j+1,i-1)-z(j,i-1) end end printf('\n') for i=1:5 for j=1:6 if z(i,j)==0 then printf(' \t') else if j==1 then printf(' %i\t',z(i,j)) else printf('%i\t',z(i,j)) end end end printf('\n') end x=poly(0,'x') l=z(1,2)+x*z(1,3)+x*(x-1)*z(1,4)/2+x*(x-1)*(x-2)*z(1,5)/6 disp(l,"The required equation is :") p=(x11-x01)/h; disp(f(5,p),"fp (105) =");