hw5 104.docx - 1(a Let’s think of a random polynomial of...

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1. (a) Let’s think of a random polynomial of degree n r(x), such that r(x) : (x-x 0 )(x-x 1 )(x-x 2 )……(x-x n-1 ) After defining r(x), we remember that we can define the difference between a polynomial of degree at most n and a polynomial of degree at most n-1 by defining an error e e(xj) = p n (x j ) – p n-1 (x j ) , whenever p n (x) interpolates as x j points. Given that both r(x) and e(x) are polynomials of degree at most n, lets assume there is a case in which r(x)=e(x), and re-arrange our error equality above to define p n (x), p n (x)= p n-1 (x) +C n r(x), where C is just a given constant. If we assume our polynomial p n (x) interpolates at given nodes, we can then say f(x j )=p n (x j ). We use this to say f(x j )=p n-1 (x j ) + C n r(x j ), and so we can define our constant as C n = ( f(x j ) – p n-1 (x j ) ) / r(x j ) Now that our Cn is defined, we can go back and redefine p n (x) as : P n (x) = ∑C k r k (x), from k=0 to n. If we take Lagrange’s definition, f[x 0 ,x 1 ,…x k ] = ( [x 1 ,x 2 ….x n ] – [x 0 ,x 1 ….x n-1 ] ) / x k -x 0 we can combine it with our Cn definition and say write p n (x) as p n (x) = p n-1 + f[x
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  • Fall '08
  • Staff
  • CN, x-xn-1, = pn

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