Show that differentiation $D = \frac d{dt}$ is a linear operator on the 3-dimensional space of ‘quasi-polynomials’ $e^t (a_2t^2 + a_1t + a_0)$.

Question

Show that differentiation $D = \frac d{dt}$ is a linear operator on the 3-dimensional space of ‘quasi-polynomials’ $e^t (a_2t^2 + a_1t + a_0)$.

Im a little unsure how to approach this:

Using $(e^t, te^t, t^2e^t)$ as a basis, write out the corresponding matrix.

Answer 1

Let $f(t;a_0,a_1,a_2)=e^t (a_2t^ 2 + a_1t + a_0)$ and $f(t;b_0,b_1,b_2)=e^t (b_2t^ 2 + b_1t + b_0)$ therefore $$\dfrac{d}{dt}(C_1f(t;a_0,a_1,a_2)+C_2f(t;b_0,b_1,b_2)){=\dfrac{d}{dt}f(t;C_1a_0+C_2b_0,C_1a_1+C_2b_1,C_1a_2+C_2b_2)\\=f(t;C_1a_0+C_2b_0,C_1a_1+C_2b_1,C_1a_2+C_2b_2)+f(t;0,2C_1a_2+2C_2b_2,C_1a_1+C_2b_1)\\=C_1(f(t;a_0,a_1,a_2)+f(t;0,2a_2,a_1))+C_2(f(t;b_0,b_1,b_2)+f(t;0,2b_2,b_1))\\=C_1\dfrac{d}{dt}f(t;a_0,a_1,a_2)+C_2\dfrac{d}{dt}f(t;b_0,b_1,b_2)}$$which completes our proof. To represent it in a matrix form we need to find out what happens on the basis under this operator. Let's define $$e_1=f(t;1,0,0)\\e_2=f(t;0,1,0)\\e_3=f(t;0,0,1)$$therefore $$Le_1=e_1\\Le_2=e_1+e_2\\Le_3=e_3+2e_2$$therefore the matrix is $$L=\begin{bmatrix}1&1&0\\0&1&2\\0&0&1\end{bmatrix}$$

Answer 2

If $f$ and $g$ are differentiable functions and $ \alpha \in \mathbb R$, then the elementary rules are: $f+g$ and $ \alpha f$ are differentiable and

$(f+g)'=f'+g'$ and $(\alpha f)'= \alpha f'$.

Can you now show that $d$ is linear ?

Answer 3

Let $V$ be the space of these "quasi polynomials". That $D$ is linear you know already. Therefore we only have to prove $D(V)\subset V$. To this end it suffices to look at $${d\over dt}\bigl(t^n e^t\bigr)= (t^n + nt^{n-1}) e^t\ .$$