- 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
2.12 Exercises
$\quad$ In the linear space of all real-valued functions, each of the following sets is independent and spans a finite-dimensional subspace $V.$ Use the given set as a basis for $V$ and let $D: V \rightarrow V$ be the differentiation operator. In each case, find the matrix of $D$ and of $D^2$ relative to this choice of basis.
16. $\quad$ $(\sin x, \cos x, x \sin x, x \cos x).$
Solution. $\quad$ Given the basis $(\sin x, \cos x, x \sin x, x \cos x)$ for $V,$ $D(V),$ and $D^2(V),$ we get the following: \begin{align*} D(\sin x) &= \cos x \\ &= 0(\sin x) + 1(\cos x) + 0(x \sin x) + 0(x \cos x) \\ D(\cos x) &= -\sin x \\ &= -1(\sin x) + 0(\cos x) + 0(x \sin x) + 0(x \cos x) \\ D(x \sin x) &= \sin x + x\cos x \\ &= 1(\sin x) + 0(\cos x) + 0(x \sin x) + 1(x \cos x) \\ D(x \cos x) &= \cos x - x\sin x \\ &= 0(\sin x) + 1(\cos x) - 1(x \sin x) + 0(x \cos x) \\ \\ D^2(\sin x) &= D(\cos x) \\ &= -\sin x \\ &= -1(\sin x) + 0(\cos x) + 0(x \sin x) + 0(x \cos x) \\ D^2(\cos x) &= D(-\sin x) \\ &= -\cos x \\ &= 0(\sin x) -1(\cos x) + 0(x \sin x) + 0(x \cos x) \\ D^2(x \sin x) &= D(\sin x + x\cos x) \\ &= \cos x + (\cos x - x\sin x) \\ &= 0(\sin x) + 2(\cos x) - 1(x \sin x) + 1(x \cos x) \\ D^2(x \cos x) &= D(\cos x - x\sin x) \\ &= -\sin x - (\sin x + x\cos x) \\ &= -2(\sin x) + 0(\cos x) + 0(x \sin x) - 1(x \cos x) \end{align*} Which gives us the following matrices of $D$ and $D^2$ relative to the basis $(\sin x, \cos x, x \sin x, x \cos x)$ \begin{align*} (d'_{ik}) &= \begin{bmatrix} 0 & -1 & 1 & 0 \\ 1 & 0 & 0 & 1 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0 \end{bmatrix} \qquad (d''_{ik}) = \begin{bmatrix} -1 & 0 & 0 & -2 \\ 0 & -1 & 2 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 1 & -1 \end{bmatrix} \quad \blacksquare \end{align*}