Adv Alegbra HW Solutions 94

Adv Alegbra HW Solutions 94 - 94 (iii) We prove that ( f g...

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Unformatted text preview: 94 (iii) We prove that ( f g ) = f g + f g by induction on deg( f ) (of course, the equation holds when f = 0). The base step deg( f ) = 0 has been done in part (ii). For the inductive step, write f (x ) = ak x k + h (x ), where h (x ) = 0 or deg(h ) < k = deg( f ). If h (x ) = 0, then ( f g ) = ak x k bn x n ak bn x n +k = = (n + k )ak bn x n +k −1 = k ak bn x n +k −1 + = kak x k −1 n ak bn x n +k −1 bn x n + ak x k n bn x n −1 = (ak x k ) g (x ) + ak x k g . If h (x ) = 0, then ( f g ) = (ak x k + h (x ))g (x ) = ak x k g ( x ) + h ( x ) g ( x ) = ak x k g (x ) + D (hg ) = ak x k g (x ) + ak x k g + h g + hg = ak x k + h g + (ak x k + h )g = f g + fg . (iv) We prove that ( f n ) = n f n −1 f by induction on n ≥ 1. If we define f 0 to be the constant function 1, then the base step holds. For the inductive step, f n +1 = fn f = fn f + fn f = (n f n −1 f ) f + f n f = (n + 1) f n f . 3.37 Assume that (x − a ) | f (x ) in R [x ]. Prove that (x − a )2 | f (x ) if and only if (x − a ) | f in R [x ]. ...
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