05_hwc1SolnsODDA

# 05_hwc1SolnsODDA - and now it is clear that 0 0 0 0 f 0 = f...

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Unformatted text preview: and now it is clear that 0 0 0 0 f 0 = f 0 = 0 , −1 1 0 1 so f is not one to one. 1.15 Deﬁne a transformation f from IR3 to IR2 as follows. Deﬁne a b c t u v f = where u + vx + tx2 = d2 (x2 (a + bx + cx2 )) . dx2 (The polynomial transformation is multiply by x2 , then diﬀerentiate twice). (a) Is f one to one? (b) Is f onto? (c) Is f invertible? SOLUTION There are several things going on in the deﬁnition of f , so it is probably wise to begin by simply writing down a formula for f . Since d2 2 (x (a + bx + cx2 )) = 2a + 6bx + 12cx2 , dx2 the formula is very simple: It is a 2a f b = 6b . c 12c Since it is clear how to undo this transformation, it is clear that the inverse is t t/2 f −1 u = u/6 . v/12 v Since we have found, it is therefore both one–to-one and onto. Section 2 2.1 Let a = 1 −1 and let b = 2 . 1 21/september/2005; 22:09 6 ...
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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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