MATH SOLVE

4 months ago

Q:
# stanford Polynomial differentiation. Suppose p is a polynomial of degree n − 1 or less, given by p(t) = c1 + c2t + · · · + cntn−1. Its derivative (with respect to t) p′(t) is a polynomial of degree n−2 or less, given by p′(t) = d1 +d2t+···+dn−1tn−2. Find a matrix D for which d = Dc. (Give the entries of D, and be sure to specify its dimensions.)

Accepted Solution

A:

Answer:The matrix representation of the linear transformation is[tex] \begin{pmatrix} 0 & 1 & 0 & 0 & \cdots & 0\\ 0 & 0 & 2 & 0 & \cdots & 0\\ 0 & 0 & 0 & 3 & \cdots & 0\\ \cdot & \cdot & \cdot & \cdot & \cdots & \cdot\\ 0 & 0 & 0 & 0 & \cdots & n-1\\ 0 & 0 & 0 & 0 & \cdots & 0\end{pmatrix}[/tex],which is a matrix of dimension [tex]n\times n[/tex].Step-by-step explanation:First, as we are working with vector spaces, in particular with [tex]P_{n-1}[/tex] the space of real polynomials of degree at most n-1, we need to set a base. As usual we will use the canonical base: [tex]\{1,t,t^2,\ldots,t^{n-1}\}[/tex]. The differential operator [tex]D[/tex] is such that [tex]D:P_{n-1}\rightarrow P_{n-1}[/tex]. So, in the image we will use the canonical base too.Recall that, if we have a polynomial [tex]p(t) = c_0 + c_1t + \cdots +c_{n-1}t^{n-1}[/tex], its derivative is [tex]p'(t) = c_1 + 2c_2t +\cdots +(n-1)c_{n-1}t^{n-2}[/tex], i.e. [tex]p' = Dp[/tex]. In order to construct the matrix representation of a linear transformation we must evaluate the transformation at the elements of the base. In this case we must find [tex]\{D1, Dt, Dt^2,\ldots, Dt^{n-1}\}[/tex].It is no difficult to notice that[tex]\{D1, Dt, Dt^2,\ldots, Dt^{n-1}\} = \{0, 1, 2t, \ldtos, (n-1)t^{n-2}\}.[/tex]The next step is to write the image of each element of the base as a linear combination of the base of [tex]P_{n-2}[/tex]. Then,[tex] 0 = 0*1+0*t+\cdots + 0*t^{n-2}[/tex][tex] 1 = 1*1+0*t+\cdots + 0*t^{n-2}[/tex][tex] 2t = 0*1+2*t+\cdots + 0*t^{n-2}[/tex][tex] 3t^2 = 0*1+0*t+3t^2+\cdots + 0*t^{n-2}[/tex]and so on, until[tex] 0 = 0*1+0*t+\cdots + (n-1)*t^{n-2}[/tex].The coefficients of this linear combinations are written as columns of matrix, which is the representation we are looking for:[tex] \begin{pmatrix} 0 & 1 & 0 & 0 & \cdots & 0\\ 0 & 0 & 2 & 0 & \cdots & 0\\ 0 & 0 & 0 & 3 & \cdots & 0\\ \cdot & \cdot & \cdot & \cdot & \cdots & \cdot\\ 0 & 0 & 0 & 0 & \cdots & n-1\\ 0 & 0 & 0 & 0 & \cdots & 0\end{pmatrix}[/tex].In this case we have a matrix of dimension [tex]n\times n[/tex], because we are considering a linear transformation from [tex]P_{n-1}[/tex] to itself, rather to [tex]P_{n-2}[/tex].