> [!theoremb] Theorem > > Let $E, F$ be [[Banach Space|Banach spaces]], $U \subset E$ be [[Open Set|open]], and $f: U \to F$ be of class [$C^p$](Space%20of%20Continuously%20Differentiable%20Functions). Let $x \in U$ and $y \in E$ such that $\bracs{x + ty: t \in [0, 1]} \subset U$. Denote $y^{(k)} = (y, \cdots, y)$ as the $k$-tuple, then > $ > f(x + y) = \sum_{i = 0}^{p - 1}\frac{1}{i!} \cdot D^{i}f(x)y^{(i)} + R_p > $ > where > $ > R_p = \int_0^1 \frac{(1 - t)^{p - 1}}{(p - 1)!}D^pf(x + ty)y^{(p)}dt > $ > *Proof*. The base case of integrating the [[Derivative|derivative]] is just FTC/mean value theorem: > $ > f(x + y) = f(x) + \int_0^1 Df(x + ty)ydt > $ > > Consider the map $t \mapsto Df(x + ty)y$. Let > $ > u = Df(x + ty)y \quad v = -(1 - t) \quad dv = dt > $ > then > $ > \int_0^1Df(x + ty)ydt = \int_{-1}^{0} u dv > $ > Over integration by parts, > $ > \begin{align*} > \int_{-1}^{0}udv &= \braks{uv}_{v = -1}^{0} - \int_{-1}^{0} vdu \\ > &= Df(x)y + \underbrace{\int_{0}^{1} (1-t)D^{2}f(x + ty)y^{(2)}dt}_{R_2} > \end{align*} > $ > Now, let $u = D^{(i)}f(x + ty)y^{(i)}$, and $dv = (1 - t)^{i - 1}/(i - 1)!$, then over integration by parts > $ > \begin{align*} > R_i &= -\int_{-1}^0 udv = \braks{uv}_{v = -1}^{0} + \int_0^1\frac{(1 - t)^i}{i!}D^{(i + 1)}f(x + ty)y^{(i + 1)}dt \\ > &= \frac{1}{i!}D^{(i)}f(x)y^{(i)} + \underbrace{\int_0^1\frac{(1 - t)^i}{i!}D^{(i + 1)}f(x + ty)y^{(i + 1)}dt}_{R_{i + 1}} > \end{align*} > $