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+//Eg-10.2
+//pg-432
+
+clear
+clc
+close()
+// A = (x-x0)/h = 0.1*(x-30)
+//A is the greek alphabet 'alpha'
+
+Y = [31.8;55.3;92.5;149.4];
+
+X = [30;40;50;60];
+
+T = zeros(4,4);
+
+T(:,1) = Y;
+
+for(j = 2:4)
+    for(i = 1:4+1-j)
+        T(i,j) = T(i+1,j-1) - T(i,j-1);
+    end
+end
+
+//disp(T)
+
+//from equation [10], the interpolating polunomial is 
+
+//p3 = f(x0) + A*Df(x0) + A(A-1)/2!*D2f(x0) + A(A-1)(A-2)/3!*D3f(x0)
+
+//note that A is used in place of 'alpha' and D in place of 'delta'
+
+// The above expression p3 can also be written as 
+
+//p3 = f(x0) + A * [ Df(x0) - D2f(x0)/2 + 1/3*D3f(x0) ] + A^2 * [ D2f(x0)/2 - 1/2*D3f(x0)] + A^3/6 * D3f(x0)..............call this expression 1
+
+f = T(1,1);
+Df = T(1,2);
+D2f = T(1,3);
+D3f = T(1,4);
+
+//Substituting the values of D,D2,D3 in the expression 1 we finally get
+
+// p3 = a0 + a1*A + a2*A^2 + a3*A^3
+
+a0 = f;
+a1 = Df - D2f/2 + 1/3*D3f;
+a2 = D2f/2 - 1/2*D3f;
+a3 = 1/6*D3f;
+
+//disp(a0,a1,a2,a3)
+
+//Now taking A = 0.1*(x-30)
+
+//p3 = b0 + b1*x + b2*x^2 + b3*x^3
+
+b0 = a0 -3*a1 + 9*a2 - 27*a3;
+b1 = 0.1*a1 - 0.6*a2 + 2.7*a3;
+b2 = 0.01*a2 - 0.09*a3;
+b3 =  0.001*a3;
+
+//disp(b3,b2,b1,b0)
+
+printf('The polynomial is  p(T) = (%f)*T^3 + (%f)*T^2 + (%f)*T + (%f)\n',b3,b2,b1,b0)
+
+deff('out = func(in)','out = b3*in^3 + b2*in^2 + b1*in + b0')
+
+
+x = 30:60;
+y = func(x);
+
+plot(x,y)
+plot(X,Y,'db')
+
+legend('Interpolated polynomial','Experimental data points')
+
+xlabel('Temperature')
+ylabel('Vapour Pressure of Water')