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MIT18_014F10_pset2sols

MIT18_014F10_pset2sols - 18.014 Problem Set 2 Solutions...

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18.014 Problem Set 2 Solutions Total: 24 points Problem 1: Let f ( x ) = k n =0 c k x k be a polynomial of degree n . (d) If f ( x ) = 0 for n + 1 distinct real values of x , then every coefficient c k of f is zero and f ( x ) = 0 for all real x . (e) Let g ( x ) = m b k x k be a polynomial of degree m where m n . If g ( x ) = f ( x ) k =0 for m + 1 distinct real values of x , then m = n , b k = c k for each k , and g ( x ) = f ( x ) for all real x . Solution (4 points) We prove ( d ) by induction on n . Since the statement is true for all integers n 0, our base case is n = 0. If f is a polynomial of degree 0, then f = c is a constant. If f has n + 1 = 1 real roots, then f ( x ) = 0 for some x ; hence, c = 0 and f ( x ) = 0 for all x . Assume the statement is true for all polynomials of degree n ; we prove the statement for a polynomial f of degree n + 1. By hypothesis, f has n + 2 distinct real roots, { a 1 , . . . , a n +2 } . Using part ( c ) of this problem (which we did together in recitation), f ( x ) = ( x a n +2 ) f n ( x ) where f n is a polynomial of degree n . But, the roots { a 1 , . . . , a n +2 } are distinct; hence, a i a n +2 = 0 if i < n + 2 and f n ( a i ) = 0 for i = 1 , . . . , n + 1. Thus, by the in- duction hypothesis, f n = 0 because f n is a polynomial of degree n with n +1 distinct real roots, { a l , . . . , a n +1 } . Since f n = 0, we conclude f ( x ) = ( x a n +2 ) f n ( x ) = 0 for every real x . Moreover, since every coefficient of f n is zero, every coefficient of f is zero. This is what we wanted to show. (e). If g ( x ) = f ( x ) for m + 1 distinct values of x , then g ( x ) f ( x ) has m + 1 distinct real roots. Moreover, since deg f = n , deg g = m , and m n , we observe deg( g f ) m . Thus, by part (d), g ( x ) f ( x ) = 0 and every coefficient of g ( x ) f ( x ) is zero. This implies g ( x ) = f ( x ) for all real x ; moreover, since the coefficients of g ( x ) f ( x ) are b k a k , it implies b k = a k for all k and m = deg g = deg f = n . Problem 2: Let A = { 1 , 2 , 3 , 4 , 5 } , and let M denote the set of all subsets of A . For each set S in M , let n ( S ) defnote the number of elements of S . If S = { 1 , 2 , 3 , 4 } 1
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and T = { 3 , 4 , 5 } , compute n ( S T ), n ( S T ), n ( S T ), and
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