I'm taking a basic linear algebra/Differential Equations class hybrid (weird right?), and we're currently learning about differential operators. Am I correct in saying that a differential operator is a linear transformation that maps a function to it's derivative n amount of times? So the operator ${D^{(3)}}(f)$ would map from the function f to it's third derivative? Then the book introduced L, which $L(f) = {a_0}{D^{(2)}}(f) + {a_1}D(f) + {a_2}(f)$ where the a's are functions. So is L just another differential operator that can be applied to a function?
Now to the actual book question. It gives me $L_1$ which is $(xD)$, and $L_2$ which is $(xD+1)$, and it wants me to find $L_1L_2$ and $L_2L_2$. Does it just want me to multiply them?
Little workaround about differential operators:
$$\begin{array}{l} {L_1} = xD\\ {L_2} = xD + 1\\ {L_2} = {L_1} + 1\\ {L_2} \circ {L_1} = ({L_1} + 1) \circ {L_1}\\ {L_2} \circ {L_1}(f) = ({L_1} + 1)({L_1}(f))\\ = ({L_1} + 1)(x \cdot D(f))\\ = ({L_1}(x \cdot D(f)) + x \cdot D(f)\\ = (xD)(x \cdot D(f) + x \cdot D(f)\\ = x \cdot D(f) + {x^2} \cdot {D^{(2)}}(f) + x \cdot D(f)\\ = {x^2} \cdot {D^{(2)}}(f) + 2x \cdot D(f)\\ = ({x^2} \cdot {D^{(2)}} + 2x \cdot D)(f)\\ {L_2}{L_1} = {x^2} \cdot {D^{(2)}} + 2x \cdot D\\ \\ {L_2}{L_1} = {L_1}^2 + {L_1} = xD \circ xD + xD = xD + {x^2}{D^{(2)}} + xD = {x^2}{D^{(2)}} + 2xD\\ {L_1}{L_2} = {L_1}^2 + {L_1} = xD \circ xD + xD = xD + {x^2}{D^{(2)}} + xD = {x^2}{D^{(2)}} + 2xD\\ \\ {L_2}{L_2} = ({L_1} + 1) \circ ({L_1} + 1)\\ = {L_1}^2 + 2{L_1} + 1\\ = xD \circ xD + 2xD + 1\\ = xD + {x^2}{D^{(2)}} + 2xD + 1\\ = {x^2}{D^{(2)}} + 3xD + 1 \end{array}$$