Let V = span{e2x, xe2x, x2e2x}.
(a) Show that d/dx (a1e^2x + a2xe^2x + a3x^2e^2x) ∈ V for any a1, a2, a3 ∈ R
(b) let [1 0 0], [0 1 0], and [0 0 1] represent the functions e^2x, xe^2x and x^2e^2x, respectively.
For example, [3 4 5] represents 3e^2x + 4xe^2x + 5x^2e^2x .
Find the matrix of differentiation as a linear transformation on V .
(a) To show that the derivative of any function in V belongs to V, we need to show that it can be written as a linear combination of the given basis functions.
Let f(x) = a1e^(2x) + a2xe^(2x) + a3x^2e^(2x), where a1, a2, a3 are real constants.
Now, let's compute the derivative of f(x):
f'(x) = d/dx (a1e^(2x) + a2xe^(2x) + a3x^2e^(2x))
= a1(2e^(2x)) + a2(e^(2x) + 2xe^(2x)) + a3(2xe^(2x) + 2x^2e^(2x))
= 2a1e^(2x) + a2e^(2x) + 2a2xe^(2x) + 2a3xe^(2x) + 2a3x^2e^(2x).
Notice that the derivative f'(x) can be written as a linear combination of the basis functions:
f'(x) = (2a1 + a2)e^(2x) + (2a2 + 2a3)xe^(2x) + 2a3x^2e^(2x).
Since f'(x) can be expressed as a linear combination of the basis functions xe^(2x), e^(2x), and x^2e^(2x), we can conclude that d/dx (a1e^2x + a2xe^2x + a3x^2e^2x) belongs to V for any a1, a2, a3 ∈ R.
(b) To find the matrix of differentiation as a linear transformation on V, we need to determine how the linear transformation maps the basis vectors [1 0 0], [0 1 0], and [0 0 1].
Let's compute the derivatives of the basis functions:
d/dx (e^(2x)) = 2e^(2x), so the derivative of [1 0 0] is [2 0 0].
d/dx (xe^(2x)) = e^(2x) + 2xe^(2x), so the derivative of [0 1 0] is [1 2 0].
d/dx (x^2e^(2x)) = 2xe^(2x) + 2x^2e^(2x), so the derivative of [0 0 1] is [0 2 2].
Therefore, the matrix of differentiation as a linear transformation on V is:
[2 0 0]
[1 2 0]
[0 2 2]
To show that the derivative of any function in V is also in V, we need to differentiate the function and show that the resulting expression can be written as a linear combination of the basis vectors.
(a) Let's start by differentiating the function f(x) = a1e^2x + a2xe^2x + a3x^2e^2x:
f'(x) = d/dx(a1e^2x + a2xe^2x + a3x^2e^2x)
= a1 * d/dx(e^2x) + a2 * d/dx(xe^2x) + a3 * d/dx(x^2e^2x)
Now, let's calculate the derivative of each of the terms:
d/dx(e^2x) = 2e^2x
d/dx(xe^2x) = e^2x + 2xe^2x
d/dx(x^2e^2x) = 2xe^2x + 2x^2e^2x
Substituting these derivatives back into the expression for f'(x), we get:
f'(x) = a1 * 2e^2x + a2 * (e^2x + 2xe^2x) + a3 * (2xe^2x + 2x^2e^2x)
= (2a1 + a2)e^2x + (a2 + 2a3)xe^2x + (2a3)x^2e^2x
We can see that f'(x) is a linear combination of the basis vectors e^2x, xe^2x, and x^2e^2x, with coefficients (2a1 + a2), (a2 + 2a3), and (2a3) respectively. Hence, f'(x) belongs to V.
(b) To find the matrix of differentiation as a linear transformation on V, we need to determine how each basis vector in V is transformed under differentiation and represent those transformations in matrix form.
Let's differentiate each basis vector and represent them as linear combinations of the basis vectors:
d/dx(e^2x) = 2e^2x = 2[1 0 0]
d/dx(xe^2x) = e^2x + 2xe^2x = 1[1 0 0] + 2[0 1 0] = [1 2 0]
d/dx(x^2e^2x) = 2xe^2x + 2x^2e^2x = 2[0 1 0] + 2[0 0 1] = [0 2 2]
Now, we can represent these transformations as a matrix by putting the coefficients together:
Matrix of differentiation = [2 1 0]
[0 2 2]
[0 0 0]
Hence, the matrix of differentiation as a linear transformation on V is given by:
[2 1 0]
[0 2 2]
[0 0 0]