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+//inverse intrpolation

+//example 3.25

+//page  118

+clc;clear;close;

+x=[2 3 4 5];

+y=[8 27 64 125];

+for i=1:3

+    d1(i)=y(i+1)-y(i);

+end

+for i=1:2

+    d2(i)=d1(i+1)-d1(i);

+end

+for i=1:1

+    d3(i)=d2(i+1)-d2(i);

+end

+yu=10;//square rooot of 10

+y0=y(1);

+d=[d1(1) d2(1) d3(1)];

+u1=(yu-y0)/d1(1);

+u2=((yu-y0-u1*(u1-1)*d2(1)/2)/d1(1));

+u3=(yu-y0-u2*(u2-1)*d2(1)/2-u2*(u2-1)*(u2-2)*d3(1)/6)/d1(1);

+u4=(yu-y0-u3*(u3-1)*d2(1)/2-u3*(u3-1)*(u3-2)*d3(1)/6)/d1(1);

+u5=(yu-y0-u4*(u4-1)*d2(1)/2-u4*(u4-1)*(u4-2)*d3(1)/6)/d1(1);

+printf(' %f \n %f \n %f \n %f \n %f \n ',u1,u2,u3,u4,u5);

+printf(' the approximate square root of %d is: %0.3f',yu,x(1)+u5)
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