04_hwc1SolnsODDA

04_hwc1SolnsODDA - f . That leaves the question: Is f one...

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where r + sx + tx 2 + ux 3 + vx 4 = ( a + bx + cx 2 ) 2 . (a) Is f one to one? (b) Is f onto? (c) Is f invertible? SOLUTION (a) Because of the way f is defined in terms of a square of ( a + bx + cx 2 ) 2 , f - a - b - c = f a b c , so f is not one to one. (b) If f were onto, then every quartic polynomial would be the square of a quadratic polynomial. If this were true, every quartic polynomial would have repeated roots, but x ( x - 1)( x - 2)( x - 3) does not. (c) No, f is neither one to one nor onto. SECOND SOLUTION Another way – a bit longer – to solve the problem is to first work out a formula for f . Since ( a + bx + cx 2 ) 2 = a 2 + 2 abx + ( b 2 + 2 ac ) x 2 + 2 bcx 3 + c 2 x 4 , f a b c = a 2 2 ab b 2 + 2 ac 2 bc c 2 We can now see that f is not onto since if f a b c = r s t u v , then r = a 2 0, so no vector on IR 5 whose first entry is negative is an output of
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Unformatted text preview: f . That leaves the question: Is f one to one? The squares are a hint the the answer is no. To prove this is actually the case, we need to find a vector r s t u v for which ( * ) has at least two solutions. To simplify things, let us decrease the number of variables, and see if we can do this with a = b = 0. (Why not? Try simple things first. There is a square in each variable, so keeping one “active” should suffice.) We find f c = c 2 21 /september/ 2005; 22:09 5...
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This note was uploaded on 10/03/2010 for the course MATH 380 taught by Professor Gladue during the Fall '08 term at Roger Williams.

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