16_hwc1SolnsODDA

16_hwc1SolnsODDA - compute f a = a 2 so that f 2 a = 4 f a...

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x in IR , and neither f nor g are linear. (In fact, neither is homogeneous, and neither is additive). To extend this to IR n , just apply f and g to each entry. 2.29 Define a transformation f from IR 3 to IR 5 as follows. Define f ±² a b c ³´ = r s t u v where r + sx + tx 2 + ux 3 + vx 4 = ( a + bx + cx 2 ) 2 . Is f linear? If so, find the corresponding matrix. If not, explain why not. SOLUTION As we saw in the second solution to Exercise 1.13, 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 The presence of squares indicates that this function is non linear. Since each variable is squared, we should be able to disply the problem with only one variable “active”. So we
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Unformatted text preview: compute f a = a 2 , so that f 2 a = 4 f a and f is not even homogeneous. (It is also not additive.) 2.31 Define a transformation f from IR 3 to IR 2 as follows. Define f ±² a b c ³´ = ² t u v ³ where u + vx + tx 2 = d 2 d x 2 ( x 2 ( a + bx + cx 2 )) . 21 /september/ 2005; 22:09 17...
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