Consider the vector space P5 consisting of all polynomial functions of degree 5
ID: 2943025 • Letter: C
Question
Consider the vector space P5 consisting of all polynomial functions of degree 5 or less?
Consider the vector space P5 consisting of all polynomial functions of degree 5 or less.
Then any vector u = f(x) in P5 is a polynomial of the form
u = a0 + a1x + a2x^2 + a3x^3 + a4x^4 + a5x^5:
The defining traits of vectors in P5 is the coefficients of the vector (polynomial function). Therefore,
we could just as easily associate any vector u with a (column) vector in R6 as
u =
(open bracket)
a0
a1
a2
a3
a4
a5
(close bracket)
:
This allows us to represent differentiation of a polynomial function using matrix multiplication
Du =(d/dx)u
where D is a matrix of appropriate size for the matrix-vector product AND the resulting vector (out-
put) has precisely the correct components that represent the coeficients of the derivative u0 = f0(x).
(a) Determine the exact matrix form for D and show how Du = u0 for the polynomial functions
u = 2x5 - 3x4 + 7x2 - 5x - 9 and v = (pie)x4 - 6x3 + 3x2 + 2x + 8.
(b) In the general case of u = a0 + a1x + a2x^2 + a3x^3 + a4x^4 + a5x^5, find Du.
(c) Calculate D^2 and show that D^2u is indeed the second derivative of u.
(d) Explain why D^ku = 0 for any integer k > 5.
(e) Describe the nullspace of D as well as the nullspace of D^2.
Explanation / Answer
a) D has to be a 6x6 matrix, and Du has to equal the derivative of the polynomial, which is a1 + 2a2x + 3a3x^2 + 4a4x^3 + 5a5x^4 Du = a1 2a2 3a3 4a4 5a5 0 The matrix that does this is D = 0.....1.....0.....0.....0.....0 0.....0.....2.....0.....0.....0 0.....0.....0.....3.....0.....0 0.....0.....0.....0.....4.....0 0.....0.....0.....0.....0.....5 0.....0.....0.....0.....0.....0 I'll leave it to you to show that it works. b) This is no different than (a). The matrix D that I've given you works for ALL 5th degree polynomials. c) I've shown you D. Hopefully you know how to multiply matrices. Just multiply D by itself. You'll see that the result is D^2 = 0.....0.....2.....0.....0.....0 0.....0.....0.....6.....0.....0 0.....0.....0.....0....12....0 0.....0.....0.....0.....0....20 0.....0.....0.....0.....0.....0 0.....0.....0.....0.....0.....0 d) Because it is a matrix representation of the differential operator, and the 6th derivative of a 5th degree polynomial is 0. e) The null space of D consists of vectors of the form a0 0 0 0 0 0 In other words, any constant polynomial is mapped to 0. The null space of D^2 is all polynomials of the form a0 a1 0 0 0 0 This is because when we take two derivatives, the constant term and the a1 term vanish.