Equivalence of Taylor Expansions Let $E \subset \mathbb{R}^n$ be open, $f: E \rightarrow \mathbb{R}$, and $x \in E$. Assume that for $y$ in a neighborhood of $x$ we have $$f(x+y) = \sum_{|\alpha| \leq k} c_{\alpha}y^{\alpha} + o(\lvert\lvert y \rvert\rvert^k)$$ as $y \rightarrow 0$ and $$f(x+y) = \sum_{|\alpha| \leq k} \tilde{c}_{\alpha}y^{\alpha} + o(\lvert\lvert y \rvert\rvert^k)$$ as $y \rightarrow 0$. Show that $c_{\alpha} = \tilde{c}_{\alpha}$ for all $|\alpha| \leq k$.
Is this trivally easy by setting the two equations equal to one another, or am I missing something?
 A: The generic hazard is: Until you have this result, you do not know whether a disagreement at one multiindex can be corrected by some carefully constructed set of contributions from higher weight multiindices.  After you have this result, you know that all (truncated) power series for this domain/range pair have the same leading coefficients.  If you change the domain or range, you should re-verify that you have (essentially) unique series representations.  This isn't automatic.
Far safer to compare with the representation of $0$.  So subtract the one equation from the other.  On the one side, you get $0$, which has a very easy to compute power series expansion.  
Suppose that there is some multiindex where $c_\alpha - \tilde{c}_\alpha \neq 0$.  The set of multiindices of such coefficients has a minimal weight element, so let $\alpha$ be one of minimal weight.  Show that the truncated series on the right-hand side is not $0 - o(||y||^k)$.  (Equiv.:  the truncated series is not $o(||y||^k)$.)  (You know a lowest order derivative which is nonzero, so you know a path to look along.)
