$n$th order Taylor polynomial formula varies with the inner product? Let $V$ be the inner product space over $\mathbb{R}$ of infinitely differentiable functions from $\mathbb{R} \to \mathbb{R}$ endowed with inner product $\langle\cdot,\cdot\rangle$ defined by $\langle f,g \rangle = \sum_{k=0}^n\frac{1}{k!^2}f^{(k)}(a)g^{(k)}(a)$. Note $f^{(k)}(a)$ denotes the $k$th derivative of $f:\mathbb{R} \to \mathbb{R}$ evaluated at $x=a \in \mathbb{R}$. Then let $\mathcal{P}_n$ denote the subspace of $V$ containing all polynomials of degree at most $n \in \mathbb{Z}_{\geq 1}$ from $\mathbb{R} \to \mathbb{R}$.
EDIT: $\langle \cdot,\cdot \rangle$ is certainly an inner product over $\mathcal{P}_n$ but I'm unsure if it is actually an inner product over $V$ (perhaps the sum must be an infinite series)? But if we let $V=\mathcal{P}_n$ the question is similar.
If we have a vector $f \in V$ and we want to find the "closest" $p \in \mathcal{P}_n$ to $f$, we can find the orthogonal projection of $f$ onto $\mathcal{P}_n$, $proj_{\mathcal{P}_n}f$.
Notice that the list $\left( (x-a)^0, (x-a)^1, (x-a)^2, \dots, (x-a)^n \right)$ forms an orthonormal basis for $P_n$ and $\langle f,(x-a)^i \rangle = \sum_{k=0}^n\frac{1}{k!^2}f^{(k)}(a)\left( (x-a)^i\right)^{(k)}(a) = \frac{i!}{i!^2}f^{(i)}(a)+\sum_{k \in \{1,...,n \}\setminus\{i\}}0 = \frac{f^{(i)}(a)}{i!}$, so
$$
\begin{aligned}
proj_{\mathcal{P}_n}f &= \sum_{i=0}^n \langle f,(x-a)^i\rangle(x-a)^i \\
&= \sum_{i=0}^n \frac{f^{(i)}(a)}{i!}(x-a)^i = P_n(x) \\
\end{aligned}$$
where $P_n(x)$ denotes the $n$th order Taylor polynomial.
My question: if we had chosen a different inner product for the same vector space $V$, for example $\langle f, g \rangle = 3\sum_{k=0}^nf^{(k)}(a)g^{(k)}(a)$, then $\left( (x-a)^0, (x-a)^1, (x-a)^2, \dots, (x-a)^n \right)$ would no longer being an orthormal basis, so I presume the formula for $proj_{\mathcal{P}_n}f$ would be different? Or perhaps it would be the same using $\frac{\langle f, (x-a)^i\rangle}{\|(x-a)^i\|^2}(x-a)^i$ in the sum... but surely there exists some different choice of inner product that yields a different formula for the projection onto $\mathcal{P}_n$? So the $n$th order Taylor polynomial formula is the arbitrary consequence of choosing/defining an inner product the way we did?
 A: The given bilinear form $$\langle\,\cdot\,,\,\cdot\,\rangle_n = \sum_{k=0}^n \frac{1}{k!^2} f^{(k)}(a) g^{(k)}(a)$$ is not an inner product on the space $V := C^{\infty}(\Bbb R)$ of all infinitely differentiable functions $\Bbb R \to \Bbb R$, for any $a$: The function $h(x) := (x - a)^{n + 1}$ satisfies $h^{(k)}(a) = 0$ for all $0 \leq k \leq n$, so $\langle h, h \rangle_n = 0$ for the nonzero vector $h$, which violates the positive-definiteness axiom.
Even if we extend the summation to be infinite, the resulting bilinear form isn't an inner product on $V$. Showing as much requires a more subtle counterexample, namely, a nonzero, infinitely differentiable function all of whose derivatives at $x = a$ are zero.

If we had chosen a different inner product for the same vector space $V$, for example $$\langle f, g \rangle = 3\sum_{k=0}^nf^{(k)}(a)g^{(k)}(a) ,$$
$\left( (x-a)^0, (x-a)^1, (x-a)^2, \dots, (x-a)^n \right)$ would no longer being an orthormal basis, so I presume the formula for $\operatorname{proj}_{\mathcal{P}_n}f$ would be different?

In any case, yes, different inner products on a vector space typically determine different orthogonal projections. Here's an example that's easier to compute explicitly:
Example On $\Bbb R^2$ consider the inner product $$\langle\!\langle {\bf x}, {\bf y}\rangle\!\rangle_\lambda := {\bf y}^\top \pmatrix{\lambda&\cdot\\\cdot&1} {\bf x} , \qquad \lambda > 0 ,$$ compute the orthogonal projection with respect to $\langle\!\langle\,\cdot\,,\,\cdot\,\rangle\!\rangle_\lambda$ of a general vector $(x, y)$ onto the line spanned by $(1, 1)$, and observe that the result depends with $\lambda$, i.e., on the inner product.
