Interchanging $x$ and $y$ in Taylor's Theorem for $f(x+y)$; is there a deeper reason for equality? Let $f$ be analytic on some interval containing $x,y,x+y$. Taylor's Theorem gives
$$
f(x+y) = \sum_{n=0}^{\infty} \frac{f^{(n)}(y)}{n!}x^n;
$$
$$
f(y+x) =\sum_{n=0}^{\infty} \frac{f^{(n)}(x)}{n!}y^n 
$$From the perspective of $f$, the interchange of $x,y$ is obvious: clearly $f(x+y)=f(y+x)$. From the perspective of the series, I don't think it's immediate that
$$f(y) + f^{(1)}(y) x+ \frac{f^{(2)}(y)}{2}x^2+\frac{f^{(3)}(y)}{6}x^3+\cdots = 
$$
$$f(x) + f^{(1)}(x) y+ \frac{f^{(2)}(x)}{2}y^2+\frac{f^{(3)}(x)}{6}y^3+\cdots.$$ Is there a deeper reason why the two series are equal; namely, is it possible to transform one into the other without going through $f$?
 A: The question is mainly about formal power series.
For any commutative ring $R$, we have the ring of formal power series $R[[x]]$, and we can take formal derivative of any element in $R[[x]]$.
Now for two variables, we can define the ring $R[[x, y]]$ either as $(R[[x]])[[y]]$ or as $(R[[y]])[[x]]$. It turns out that the two definitions are canonically isomorphic to each other, hence can be identified.
However, an important remark is that the natural topology on $R[[x, y]]$ is neither the $x$-adic topology nor the $y$-adic topology: it's the $(x, y)$-adic topology, i.e. we view $R[[x, y]]$ as the inverse limit $\varprojlim\limits_kR[x, y]/(x, y)^k$.
This is important, because e.g. the expression $f(x + y)$ doesn't make sense if we equip $R[[x, y]]$ with the $x$-adic topology, since $x + y$ has $x$-adic valuation $0$ and hence the sum $\sum a_n(x + y)^n$ doesn't converge. On the other hand, it converges for the $(x, y)$-adic topology.

With these in mind, we now understand that $f(x + y) = \sum_{n=0}^{\infty} \frac{f^{(n)}(y)}{n!}x^n$ is a valid identity in the ring $R[[x, y]]$, when $R$ is a $\Bbb Q$-algebra (i.e. every nonzero integer is invertible in $R$).
Same applies to the other identity where we exchange $x$ and $y$, and hence we get the wanted equality as an identity of formal power series.

It is also possible (and not very complicated) to prove it directly. Write $f(y) = \sum a_m y^m$. We then have:
\begin{eqnarray}
\sum_n\frac{f^{(n)}(y)}{n!}x^n &=& \sum_n\sum_m a_m \binom m n y^{m - n} x^n \\
&=& \sum_n \sum_m a_{m + n} \binom {m + n} n x^n y^m
\end{eqnarray}
which is obviously symmetrical in $x$ and $y$.
