- Calculus, Volume 2: Multi-Variable Calculus and Linear Algebra with Applications to Differential Equations and Probability
- Tom M. Apostol
- Second Edition
- 1991
- 978-1-119-49676-2
2.8 Exercises
29.$\quad$ Let $V$ and $D$ be as in Exercise 28 but let $T$ be the linear transformation that maps $p(x)$ onto $xp(x).$ Prove that $DT - TD = I$ and that $DT^n - T^nD = n T^{n-1}$ for $n \geq 2.$
Proof. $\quad$ With $D$ being the differentiation operation and $T$ being the linear transformation mapping $p(x)$ onto $xp(x),$ we have \begin{align*} (DT - TD)[p(x)] &= DT(p) - TD(p) \\ &= D[xp(x)] - T[p'(x)] \\ &= p(x) + xp'(x) - xp'(x) \\ &= p(x) \end{align*} Thus, we have shown $DT - TD = I.$
$\quad$ If we define the integral powers of $T$ inductively as in Section 2.5, we have $T^0 = I$ and $T^n = TT^{n-1}$ for $n \geq 1.$ But since $T$ is the transformation that maps $p(x)$ onto $xp(x),$ this means that $T$ maps $T[p(x)] = xp(x)$ onto $x^2p(x).$ Now, assume this is true for some $k \geq 2.$ In other words, assume that $T^k(p) = x^kp(x).$ Applying $T$ to $T^k[p(x)]$ gives us \begin{align*} T[T^k(p)] &= xT^k[p(x)] \\ &= x^{k+1}p(x) \end{align*} And since we have proven this for $n = k + 1,$ by induction, we have shown that it is true for each $n \geq 2.$
$\quad$ Now that we have proven that $T^n = x^np$ for each $n\geq 2,$ we can show that \begin{align*} (DT^n - T^nD)[p(x)] &= DT^n[p(x)] - T^nD[p(x)] \\ &= D[x^np(x)] - x^np'(x) \\ &= nx^{n-1}p(x) + x^np'(x) - x^np'(x) \\ &= nT^{n-1}[p(x)] \quad \blacksquare \end{align*}