Answer to Question #93978 in Quantitative Methods for ABDULLAHI AA

Question #93978

The interpolating polynominal of degree \\(\\|eqn\\) With the nodes \\(x_{0},x_{1},\\ c dots,x_{n}u) can be written as?


1
Expert's answer
2019-09-09T11:45:11-0400

There are two different kinds of the interpolating polynomials:

1)Lagrange polynomial:


L(x)=i=0nf(xi)0mn,mixxmxixmL(x)=\sum_{i=0}^nf(x_i)\prod_{0\le m \le n, m\ne i}\frac{x-x_m}{x_i-x_m}


2)Newton polynomial:

Forward divided:


N(x)=[y0]+[y0,y1](xx0)++[y0,y1,y2](xx0)(xx1)+...++[y0,...,yn](xx0)(xx1)...(xxn1)N(x)=[y_0]+[y_0,y_1](x-x_0)+ \newline +[y_0,y_1,y_2](x-x_0)(x-x_1)+...+ \newline +[y_0,...,y_n](x-x_0)(x-x_1)...(x-x_{n-1})


Backward divided:


N(x)=[yn]+[yn,yn1](xxn)++[yn,yn1,yn2](xxn)(xxn1)+...++[yn,yn1,...y0](xxn)(xxn1)...(xx1)N(x)=[y_n]+[y_n,y_{n-1}](x-x_n)+ \newline +[y_n,y_{n-1},y_{n-2}](x-x_n)(x-x_{n-1})+...+ \newline +[y_n,y_{n-1},...y_0](x-x_n)(x-x_{n-1})...(x-x_1)


where


yi=f(xi),i=0,1,...,n;[yi]=yi,[yi,yi+1]==yi+1yixi+1xi,[yi,yi+1,yi+2]=[yi+2,yi+1][yi+1,yi]xi+2xi,...,[y0,...,yn]==[yn,...,y1][yn1,...,y0]xnx0.y_i=f(x_i),i=0,1,...,n; [y_i]=y_i,[y_i,y_{i+1}]= \newline =\frac{y_{i+1}-y_i}{x_{i+1}-x_i}, [y_i,y_{i+1},y_{i+2}]= \newline \frac{[y_{i+2},y_{i+1}]-[y_{i+1},y_i]}{x_{i+2}-x_i},...,[y_0,...,y_n]= \newline =\frac{[y_n,...,y_1]-[y_{n-1},...,y_0]}{x_n-x_0}.

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