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)=\\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)=[y_0]+[y_0,y_1](x-x_0)+\n\\newline\n+[y_0,y_1,y_2](x-x_0)(x-x_1)+...+\n\\newline\n+[y_0,...,y_n](x-x_0)(x-x_1)...(x-x_{n-1})"


Backward divided:


"N(x)=[y_n]+[y_n,y_{n-1}](x-x_n)+\n\\newline\n+[y_n,y_{n-1},y_{n-2}](x-x_n)(x-x_{n-1})+...+\n\\newline\n+[y_n,y_{n-1},...y_0](x-x_n)(x-x_{n-1})...(x-x_1)"


where


"y_i=f(x_i),i=0,1,...,n; \n[y_i]=y_i,[y_i,y_{i+1}]=\n\\newline\n=\\frac{y_{i+1}-y_i}{x_{i+1}-x_i}, [y_i,y_{i+1},y_{i+2}]=\n\\newline\n\\frac{[y_{i+2},y_{i+1}]-[y_{i+1},y_i]}{x_{i+2}-x_i},...,[y_0,...,y_n]=\n\\newline\n=\\frac{[y_n,...,y_1]-[y_{n-1},...,y_0]}{x_n-x_0}."

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