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rev1_2011_04_05_03

# rev1_2011_04_05_03 - EE263 S Lall 2011.04.05.03 Problem...

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EE263 S. Lall 2011.04.05.03 Problem Session 1 Solutions 1. Matrix representation of polynomial differentiation. 42035 We can represent a polynomial of degree less than n , p ( x ) = a n - 1 x n - 1 + a n - 2 x n - 2 + · · · + a 1 x + a 0 , as the vector [ a 0 a 1 · · · a n - 1 ] T R n . Consider the linear transformation D that differ- entiates polynomials, i.e. , D p = dp/dx . Find the matrix D that represents D ( i.e. , if the coefficients of p are given by a , then the coefficients of dp/dx are given by Da ). Solution. According to problem ?? it suffices to compute the transformation of the unit vectors e i R n for i = 1 , 2 ,...,n under differentiation. In other words D = bracketleftbig De 1 De 2 · · · De n bracketrightbig . e 1 corresponds to the polynomial p 1 ( x ) = 1, and since D p 1 = 0 we have De 1 = 0 0 0 . . . 0 . e 2 corresponds to p 2 ( x ) = x with D p 2 = 1 and therefore De 2 = 1 0 0 . . . 0 = e 1 . Similarly for p 3 ( x ) = x 2 we get D p 3 = 2 x so De 3 = 0 2 0 . . . 0 = 2 e 2 . It is easy to see that, in general, for i> 1 we have De i = ( i - 1) e i - 1 . Therefore D = 0 1 0 0 · · · 0 0 0 2 0 · · · 0 0 0 0 3

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