$\textbf{Exercise 9.}$ Suppose $m$ is a positive integer. Show that the dual basis of the basis $1, x, \ldots, x^m$ of $\mathcal{P}_m(\mathbb{R})$ is $\varphi_0, \varphi_1, \ldots, \varphi_m$, where $\begin{equation*} \varphi_k(p) = \frac{p^{(k)}(0)}{k!}. \end{equation*}$ Here $p^{(k)}$ denotes the $k$th derivative of $p$, with the understanding that the $0$th derivative of $p$ is $p$. $\textbf{Solution 9.}$ Note that $\begin{equation*} \varphi_k(x^j) = \frac{(x^j)^{(k)}(0)}{k!} = \begin{cases} 1 & \text{if } j = k, \\ 0 & \text{if } j \neq k. \end{cases} \end{equation*}$ Thus $\varphi_0, \varphi_1, \ldots, \varphi_m$ is indeed the dual basis of $1, x, \ldots, x^m$. --- ![[sol-5.pdf#page=23]] --- $\textbf{Exercise 9.}$ Suppose $m$ is a positive integer. Show that the dual basis of the basis $1, x, \ldots, x^m$ of $\mathcal{P}_m(\mathbb{R})$ is $\varphi_0, \varphi_1, \ldots, \varphi_m$, where $\begin{equation*} \varphi_k(p) = \frac{p^{(k)}(0)}{k!}. \end{equation*}$ Here $p^{(k)}$ denotes the $k$th derivative of $p$, with the understanding that the $0$th derivative of $p$ is $p$. $\textbf{Solution 9.}$ Note that $\begin{equation*} \varphi_k(x^j) = \frac{(x^j)^{(k)}(0)}{k!} = \begin{cases} 1 & \text{if } j = k, \\ 0 & \text{if } j \neq k. \end{cases} \end{equation*}$ Thus $\varphi_0, \varphi_1, \ldots, \varphi_m$ is indeed the dual basis of $1, x, \ldots, x^m$. $\textit{Commentary:}$ This exercise gives an explicit description of the dual basis for the standard basis of the vector space of polynomials of degree at most $m$. The dual basis functionals are defined in terms of the derivatives of a polynomial at 0, scaled by a factorial factor. The verification that this is indeed the dual basis is a straightforward calculation using the properties of derivatives. This dual basis is important in many applications, including interpolation and approximation theory. $\textit{Example:}$ Let $m = 2$, so $\mathcal{P}_2(\mathbb{R})$ has the basis $\{1, x, x^2\}$. The dual basis is $\{\varphi_0, \varphi_1, \varphi_2\}$, where $\begin{align*} \varphi_0(a + bx + cx^2) &= a, \\ \varphi_1(a + bx + cx^2) &= b, \\ \varphi_2(a + bx + cx^2) &= 2c. \end{align*}$ You can check that $\varphi_i(x^j) = \delta_{ij}$ (the Kronecker delta) for $i, j = 0, 1, 2$. ---