MIT18_014F10_pset2sols

MIT18_014F10_pset2sols - 18.014 Problem Set 2 Solutions...

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Unformatted text preview: 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 ) = for n + 1 distinct real values of x , then every coecient c k of f is zero and f ( x ) = 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 ) = for some x ; hence, c = and f ( x ) = 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 = if i < n + 2 and f n ( a i ) = for i = 1 ,...,n + 1. Thus, by the in- duction hypothesis, f n = 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 ) = for every real x . Moreover, since every coecient of f n is zero, every coecient 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 ) = and every coecient of g ( x ) f ( x ) is zero. This implies g ( x ) = f ( x ) for all real...
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MIT18_014F10_pset2sols - 18.014 Problem Set 2 Solutions...

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