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ex1solsSummer09

# ex1solsSummer09 - Math 347 C1 HOUR EXAM I 30 June 2009...

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Math 347 C1 HOUR EXAM I 30 June 2009 SOLUTIONS 1. Let a be a non-zero real number. Show that a 2 + 1 a 2 2 . SOLUTION Assume the opposite: a 2 + 1 a 2 < 2 . Then, since a 2 > 0 we can multiply by a 2 without changing the direction of the inequality. Thus a 4 + 1 < 2 a 2 , whence a 4 - 2 a 2 + 1 < 0 . But a 4 - 2 a 2 + 1 = ( a 2 - 1) 2 0 for all values of a , a contradiction. Hence the opposite of our assumption is true, that is, a 2 + 1 a 2 2 . 2. Let p ( x ) be a polynomial of degree r > 0 with real coefficients, p ( x ) = a 0 + a 1 x + a 2 x 2 + . . . + a r x r with a i R . Let A be the sum of the coefficients of p ( x ) with even subscripts and let B be the sum of the coefficients of p ( x ) with odd subscripts. Show that A = 1 2 ( p (1) + p ( - 1)) and B = 1 2 ( p (1) - p ( - 1)) . SOLUTION p (1) = a 0 + a 1 + a 2 + . . . + a r

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and p ( - 1) = a 0 - a 1 + a 2 - . . . ± a r . Adding these two equations we get p (1) + p ( - 1) = 2 a 0 + 2 a 2 + . . . + 2 a r when r is even and p (1) + p ( - 1) = 2 a 0 + 2 a 2 + . . . + 2 a r - 1 when r is odd. In each case, the right hand side is just 2 A and so A = 1 2 ( p (1) + p ( - 1)) .
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