Dot product with which polynomial gives evaluation at $x_0$? I thought of following situation/problem, and was surprised that the solution did not jump out at me. Fix a positive integer $n$ and let $V_n$ be the finite-dimensional vector space of all polynomials of degree $\leq n$ with real coefficents. Make $V_n$ into a inner product space with respect to
$$ \langle p,q \rangle = \int_{-1}^1 p(x) q(x) \ dx.$$
Now, fix some point $x_0 \in \mathbb{R}$ and consider the linear functional 
\begin{align*}
\varphi_{x_0} : V_n \to \mathbb{R} && \varphi_{x_0}(p) = p(x_0).
\end{align*}
Because we are in an inner product space, there is a unique polynomial $p_{x_0,n}$ such that
\begin{align*}
\varphi_{x_0}(p) = \langle p_{x_0,n},p \rangle =\int_{-1}^1 p_{x_0,n}(x) p(x) \ dx &&
\forall p \in V_n.
\end{align*}
Question: What is this polynomial!?
One approach would be to apply the Gramm-Schmidt procedure to $1,x,\ldots,x^n$ to get an orthonormal basis $p_0,p_1,\ldots,p_n$ for $V_n$ and then calculate $\varphi_{x_0}$ on this basis. The required vector should be then $\sum_{i=0}^n \varphi_{x_0}(p_i) p_i$, but I am hoping for a more enlightening representation.
 A: This will be not a very nice and effective answer but gives a more or less direct formula for the polynomial $\varphi_{x_0}$.
We will repeat the proof for Rodriguez's formula for the Legendre polynomials.
Let $q_0,q_1,q_2,\ldots,q_{n+1}$ be the sequence of polynomials such that
$$
q_0 = \varphi_{x_0}; \qquad
q_{k+1}' = q_k \quad\text{and}\quad q_{k+1}(-1)=0 \quad (k=0,1,\ldots,n).
$$
We will compute the polynomial $q_{n+1}$.
The condition, applied to the polynomial $(1-x)^k$ (with $0\le k\le n$) yields
$$
\int_{-1}^1 \varphi_{x_0}(x) \cdot (1-x)^k dx = (1-x_0)^k.
$$
Integrating by parts $k$ times,
\begin{align*}
(1-x_0)^k 
&= \int_{-1}^1 q_0(x) \cdot (1-x)^k dx 
= \underbrace{\Big[q_1(x) \cdot (1-x)^k\Big]_{-1}^1}_0 
+ \int_{-1}^1 q_1(x) \cdot k(1-x)^{k-1} dx = \\
&= k\int_{-1}^1 q_1(x) \cdot (1-x)^{k-1} dx 
= k(k-1)\int_{-1}^1 q_2(x) \cdot (1-x)^{k-2} dx = \dots = \\
&= k! \int_{-1}^1 q_k(x) dx 
= k! \cdot q_{k+1}(1)
= k! \cdot q_{n+1}^{(n-k)}(1). 
\end{align*}
Therefore,
$$
q_{n+1}^{(n-k)}(1) = \frac{(1-x_0)^k}{k!}.
$$
The $n$th Taylor-polynomial of $q_{n+1}$ around $1$ is
$$
\sum_{k=0}^n \frac{q_{n+1}^{(n-k)}(1)}{(n-k)!}(x-1)^{n-k} = 
\sum_{k=0}^n \frac{(1-x_0)^k}{k!}(x-1)^{n-k} = \\ =
\frac1{n!} \sum_{k=0}^n \binom{n}{k}(1-x_0)^k(x-1)^{n-k} = 
\frac{(x-x_0)^n}{n!}
$$
so
$$
q_{n+1}(x) \equiv \frac{(x-x_0)^n}{n!} \pmod{(x-1)^{n+1}}. \tag1
$$
By the condition $q_1(-1)=q_2(-1)=\ldots=q_{n+1}(-1)=0$ we know that 
$$
q_{n+1}(x) \equiv 0 \pmod{(x+1)^{n+1}}. \tag2
$$
Since $\deg q_{n+1}\le 2n+1$, by the Chinese Remainder Theorem, the congruences (1) and (2) uniquely determine $q_{n+1}$. 
A solution of the congruences (1) and (2), but with higher degree is
$$
\left(\bigg(\frac{x-1}2\bigg)^{n+1}-(-1)^{n+1}\right)^{n+1} \cdot \frac{(x-x_0)^n}{n!}.
$$
Dividing with remainders, we can see that 
$$
q_{n+1}(x) = \left(\bigg(\frac{x-1}2\bigg)^{n+1}+(-1)^n\right)^{n+1} \cdot \frac{(x-x_0)^n}{n!} \mod (1-x^2)^{n+1}
$$
and therefore
$$
\varphi_{x_0}(x) =
\left(
\left(\bigg(\frac{x-1}2\bigg)^{n+1}+(-1)^n\right)^{n+1} \cdot \frac{(x-x_0)^n}{n!} \mod (1-x^2)^{n+1}\right)^{(n+1)} .
$$
