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.)

The matrix representation of the linear transformation is

,

which is a matrix of dimension .

Step-by-step explanation:

First, as we are working with vector spaces, in particular with 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: .

The differential operator is such that . So, in the image we will use the canonical base too.

Recall that, if we have a polynomial , its derivative is , i.e. . 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

.

It is no difficult to notice that

The next step is to write the image of each element of the base as a linear combination of the base of . Then,

and so on, until

.

The coefficients of this linear combinations are written as columns of matrix, which is the representation we are looking for:

.

In this case we have a matrix of dimension , because we are considering a linear transformation from to itself, rather to .

c

step-by-step explanation:

it isn't d, because the interquartile range is from 27-57. and the interquartile range is basically half the data, it's pretty obvious.

it isn't b, because like i said earlier, the interquartile range is from 27-57. 57-27 is 30. the real range has to be longer

not a either. although it looks like it, 3/4 of the data is not greater than 27. but if c is wrong than this should be correct

the correct answer is c, because 13 is the point in which the plot starts

d

step-by-step explanation: