- 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
28.$\quad$ Let $V$ be the linear space of all real polynomials $p(x).$ Let $D$ denote the differentiation operator and let $T$ be the linear transformation that maps $p(x)$ onto $x p'(x).$
(a) $\quad$ Let $p(x) = 2 + 3x - x^2 + 4x^3$ and determine the image of $p$ under each of the following transformations: $D,$ $T,$ $DT,$ $TD,$ $DT - TD,$ $T^2 D^2 - D^2 T^2.$
(b) $\quad$ Determine those $p$ in $V$ for which $T(p) = p.$
(c) $\quad$ Determine those $p$ in $V$ for which $(DT - 2D)(p) = 0.$
(d) $\quad$ Determine those $p$ in $V$ for which $(DT - TD)^n(p) = D^n(p).$
Solution.
(a) $\quad$ Let $p(x) = 2 + 3x - x^2 + 4x^3$ and determine the image of $p$ under each of the following transformations: $D,$ $T,$ $DT,$ $TD,$ $DT - TD,$ $T^2 D^2 - D^2 T^2.$
\begin{align*}
D[p(x)] &= D(2 + 3x - x^2 + 4x^3)
\\
&= 3 - 2x + 12x^2
\\
\\
T[p(x)] &= T(2 + 3x - x^2 + 4x^3)
\\
&= x(3 - 2x + 12x^2)
\\
&= 3x - 2x^2 + 12x^3
\\
\\
DT[p(x)] &= D[T[p(x)]]
\\
&= D(3x - 2x^2 + 12x^3)
\\
&= 3 - 4x + 36x^2
\\
\\
TD[p(x)] &= T(3 - 2x + 12x^2)
\\
&= x(-2 + 24x)
\\
&= -2x + 24x^2
\\
\\
(DT - TD)[p(x)] &= DT[p(x)] - TD[p(x)]
\\
&= (3 - 4x + 36x^2) - (-2x + 24x^2)
\\
&= 3 - 2x + 12x^2
\\
\\
(T^2D^2 - D^2T^2)[p(x)] &= T^2D[3 - 2x + 12x^2] - D^2T[3x - 2x^2 + 12x^3]
\\
&= T^2[-2 + 24x] - D^2[x(3 - 4x + 36x^2)]
\\
&= T[x(24)] - D[3 - 8x + 108x^2]
\\
&= 24x + 8 - 216x
\\
&= 8 - 192x \quad \blacksquare
\end{align*}
(b) $\quad$ Determine those $p$ in $V$ for which $T(p) = p.$
$\quad$ The set of $p$ in $V$ for which $T(p) = p$ is the set of polynomials of the form $cx,$ where $c$ is a scalar. $\quad \blacksquare$
(c) $\quad$ Determine those $p$ in $V$ for which $(DT - 2D)(p) = O.$
$\quad$ We wish to find the set of $p$ satisfying
\begin{align*}
DT[p(x)] - 2D[p(x)] &= D[xp'(x)] - 2p'(x)
\\
&= xp''(x) - p'(x)
\\
&= O
\end{align*}
Setting $x = 1,$ we find that $p''(1) - p'(1) = 0,$ giving us
\begin{align*}
0(c_0) + 0(c_1) + \sum_{k=2}^n k(k-1)c_k - \sum_{k=1}^n kc_k &= 0(c_0) -c_1 + \sum_{k=2}^n k(k-2)c_k
\\
&= 0
\end{align*}
This relation is satisfied by arbitrary $c_0$ and $c_2,$ setting all other $c_k = 0.$ In other words, the set of $p$ satisfying $(DT - 2D)(p) = O$ is the set of polynomials of the form $p(x) = c_0 + c_2x^2,$ where $c_0$ and $c_2$ are arbitrary scalars. $\quad \blacksquare$
(d) $\quad$ Determine those $p$ in $V$ for which $(DT - TD)^n(p) = D^n(p).$
$\quad$ We will prove by induction that $(DT - TD)^n = D^n$ for each $n \geq 1$ and arbitrary $p.$ If $n = 1$ we have
\begin{align*}
(DT - TD)^n(p) &= DT(p) - TD(p)
\\
&= D[xp'(x)] - T[p'(x)]
\\
&= p'(x) + xp''(x) - xp'(x)
\\
&= p'(x)
\\
&= D^n(p)
\end{align*}
Now, assume this is true for some $k \geq 1,$ we will prove it is also true for $k + 1.$ If $n = k + 1,$ we use the inductive definition of integral powers to give us:
\begin{align*}
(DT - TD)^{k+1}(p) &= (DT - TD)(DT - TD)^k(p)
\end{align*}
But by hypothesis, $(DT - TD)^k = D^k(p).$ As a result, we have
\begin{align*}
(DT - TD)(DT - TD)^k(p) &= (DT - TD)[D^k(p)]
\\
&= DT[D^k(p)] - TD[D^k(p)]
\\
&= D[xD^{k+1}(p)] - xD[D^{k+1}(p)]
\\
&= D^{k+1}(p) + xD[D^{k+1}(p)] - xD[D^{k+1}(p)]
\\
&= D^{k+1}(p)
\end{align*}
Thus, we have proven by induction that $(DT - TD)^n(p) = D^n(p)$ for each $n \geq 1.$ But since this is true for arbitrary $p,$ this means that this relation is satisfied by all $p$ in $V. \quad \blacksquare$