Let K be any field, and let a1,...,a, be pairwise distinct elements of K (that is, a; +a; for all i + j). For each i = 1,...,n, define Pi = (x– a1).. (x – a;-1)(x– a;+1)-…· (x- a,) E K[x]. Note that...

Let K be any field, and let a1,...,a, be pairwise distinct elements of K (that is, a; +a;<br>for all i + j). For each i =<br>1,...,n, define<br>Pi = (x– a1).. (x – a;-1)(x– a;+1)-…· (x- a,) E K[x].<br>Note that the (x– a;) factor has been left out of p;, so deg p; =n– 1.<br>(a) Prove that p:(a;) #0 if and only if i = j.<br>(b) Let b1,...,bz be elements of K (some of them maybe equal). Using part (a),<br>explain how to find a polynomial q E K[x], with degq <n (or q = 0), such that<br>q(a;) = b; for each i = 1,...,n.<br>You don't have to include a proof.<br>(c) Prove that there cannot exist two different polynomials q,r EK[x], both of degree<br>less than n, such that q(a;) = r(a;) for each i = 1,...,n.<br>

Extracted text: Let K be any field, and let a1,...,a, be pairwise distinct elements of K (that is, a; +a; for all i + j). For each i = 1,...,n, define Pi = (x– a1).. (x – a;-1)(x– a;+1)-…· (x- a,) E K[x]. Note that the (x– a;) factor has been left out of p;, so deg p; =n– 1. (a) Prove that p:(a;) #0 if and only if i = j. (b) Let b1,...,bz be elements of K (some of them maybe equal). Using part (a), explain how to find a polynomial q E K[x], with degq

Jun 04, 2022

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Let K be any field, and let a1,...,a, be pairwise distinct elements of K (that is, a; +a; for all i + j). For each i = 1,...,n, define Pi = (x– a1).. (x – a;-1)(x– a;+1)-…· (x- a,) E K[x]. Note that...

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