![[sol-5.pdf#page=24]] --- $\textbf{Exercise 10.}$ Suppose $m$ is a positive integer. (a) Show that $1, x - 5, \ldots, (x - 5)^m$ is a basis of $\mathcal{P}_m(\mathbb{R})$. (b) What is the dual basis of the basis in (a)? $\textbf{Solution 10.}$ (a) Define $\varphi_0, \varphi_1, \ldots, \varphi_m \in (\mathcal{P}_m(\mathbb{R}))^*$ by $\begin{equation*} \varphi_j(p) = \frac{p^{(j)}(5)}{j!}. \end{equation*}$ Suppose $a_0, a_1, \ldots, a_m \in \mathbb{R}$ and $\begin{equation*} a_0 + a_1(x - 5) + \cdots + a_m(x - 5)^m = 0. \end{equation*}$ Then for each $j = 0, 1, \ldots, m$, we have $\begin{align*} a_j &= \varphi_j(a_0 + a_1(x - 5) + \cdots + a_m(x - 5)^m) \\ &= \varphi_j(0) = 0. \end{align*}$ Thus $a_0 = a_1 = \cdots = a_m = 0$. Hence $1, x - 5, \ldots, (x - 5)^m$ is linearly independent list in $\mathcal{P}_m(\mathbb{R})$ of length $m + 1$, which equals the dimension of $\mathcal{P}_m(\mathbb{R})$. Thus $1, x - 5, \ldots, (x - 5)^m$ is a basis of $\mathcal{P}_m(\mathbb{R})$ (by 2.38). (b) Let $\varphi_0, \varphi_1, \ldots, \varphi_m \in (\mathcal{P}_m(\mathbb{R}))^*$ be defined as in (a). Then $\begin{equation*} \varphi_j((x - 5)^k) = \begin{cases} 1 & \text{if } k = j, \\ 0 & \text{if } k \neq j. \end{cases} \end{equation*}$ Thus $\varphi_0, \varphi_1, \ldots, \varphi_m$ is the dual basis of $1, x - 5, \ldots, (x - 5)^m$. --- $\textbf{Exercise 10.}$ Suppose $m$ is a positive integer. (a) Show that $1, x - 5, \ldots, (x - 5)^m$ is a basis of $\mathcal{P}_m(\mathbb{R})$. (b) What is the dual basis of the basis in (a)? $\textbf{Solution 10.}$ (a) Define $\varphi_0, \varphi_1, \ldots, \varphi_m \in (\mathcal{P}_m(\mathbb{R}))^*$ by $\begin{equation*} \varphi_j(p) = \frac{p^{(j)}(5)}{j!}. \end{equation*}$ Suppose $a_0, a_1, \ldots, a_m \in \mathbb{R}$ and $\begin{equation*} a_0 + a_1(x - 5) + \cdots + a_m(x - 5)^m = 0. \end{equation*}$ Then for each $j = 0, 1, \ldots, m$, we have $\begin{align*} a_j &= \varphi_j(a_0 + a_1(x - 5) + \cdots + a_m(x - 5)^m) \\ &= \varphi_j(0) = 0. \end{align*}$ Thus $a_0 = a_1 = \cdots = a_m = 0$. Hence $1, x - 5, \ldots, (x - 5)^m$ is linearly independent list in $\mathcal{P}_m(\mathbb{R})$ of length $m + 1$, which equals the dimension of $\mathcal{P}_m(\mathbb{R})$. Thus $1, x - 5, \ldots, (x - 5)^m$ is a basis of $\mathcal{P}_m(\mathbb{R})$ (by 2.38). (b) Let $\varphi_0, \varphi_1, \ldots, \varphi_m \in (\mathcal{P}_m(\mathbb{R}))^*$ be defined as in (a). Then $\begin{equation*} \varphi_j((x - 5)^k) = \begin{cases} 1 & \text{if } k = j, \\ 0 & \text{if } k \neq j. \end{cases} \end{equation*}$ Thus $\varphi_0, \varphi_1, \ldots, \varphi_m$ is the dual basis of $1, x - 5, \ldots, (x - 5)^m$. $\textit{Commentary:}$ This exercise demonstrates that the shifted powers $1, x - 5, \ldots, (x - 5)^m$ form a basis for $\mathcal{P}_m(\mathbb{R})$, and it asks for the dual basis. The proof of part (a) uses the same functionals as in Exercise 9, but shifted to the point 5. The linear independence follows from the fact that a polynomial of degree at most $m$ is zero if and only if all its derivatives up to order $m$ vanish at a point. Part (b) is then immediate from the definition of the dual basis. This exercise illustrates the flexibility in choosing bases for polynomial spaces and the corresponding changes in the dual basis. $\textit{Example:}$ Let $m = 2$, so $\{1, x - 5, (x - 5)^2\}$ is a basis of $\mathcal{P}_2(\mathbb{R})$. The dual basis is $\{\varphi_0, \varphi_1, \varphi_2\}$, where \begin{align*} \varphi_0(a + b(x - 5) + c(x - 5)^2) &= a, \\ \varphi_1(a + b(x - 5) + c(x - 5)^2) &= b, \\ \varphi_2(a + b(x - 5) + c(x - 5)^2) &= 2c. \end{align*} You can check that $\varphi_i((x - 5)^j) = \delta_{ij}$ for $i, j = 0, 1, 2$.