Added appropriate form of Taylor's formula.

src/dg/derivative/taylor.tex

@@ -62,16 +62,17 @@
 
 \begin{theorem}[Taylor's Formula, Peano Remainder]
 \label{theorem:taylor-peano}
-    Let $E$ be a topological vector space, $\sigma \subset \mathfrak{B}(E)$ be an ideal that includes bounded sets contained in finite-dimensional subspaces, $F$ be a separated locally convex space, $U \subset E$ be open, and $f: U \to F$ be $n$-fold $\tilde \sigma$-differentiable at $x_0 \in U$, then there exists $r \in \mathcal{R}_\sigma^n(E; F)$ such that
+    Let $E$ be a topological vector space, $\sigma \subset \mathfrak{B}(E)$ be an ideal containing bounded subsets in finite-dimensional subspaces, $F$ be a separated locally convex space, $U \subset E$ be open, and $f: U \to F$ be $n$-fold $\tilde \sigma$-differentiable at $x_0 \in U$, then there exists $r \in \mathcal{R}_\sigma^n(E; F)$ such that
     \[
-    g(x_0 + h) = g(x_0) + \sum_{k = 1}^n \frac{1}{k!}D^k_\sigma f(x_0)(h^{(k)}) + r(h)
+    f(x_0 + h) = f(x_0) + \sum_{k = 1}^n \frac{1}{k!}D^k_\sigma f(x_0)(h^{(k)}) + r(h)
     \]
 
+    for sufficiently small $h$. 
 \end{theorem}
 \begin{proof}[Proof {{\cite[Theorem 4.7.3]{Bogachev}}}. ]
     Let
     \[
-    r(h) = g(x_0 + h) - g(x) - \sum_{k = 1}^n \frac{1}{k!}D^k_\sigma f(x_0)(h^{(k)})
+    r(h) = f(x_0 + h) - f(x) - \sum_{k = 1}^n \frac{1}{k!}D^k_\sigma f(x_0)(h^{(k)})
     \]
 
     For any $1 \le k \le n$, $D^k_\sigma(x_0) \in B^k_\sigma(E; F)$ is symmetric by \autoref{theorem:derivative-symmetric}. Let $T_k(h) = \frac{1}{k!}D^k_\sigma f(x_0)(h^{(k)})$, then by \autoref{theorem:power-rule}, for any $\bracs{t_j}_1^\ell \in E$,
@@ -85,7 +86,7 @@
 
     so
     \[
-    D^k_\sigma r(0) = D^k_\sigma g(x_0) - D^k_\sigma(x_0) = 0
+    D^k_\sigma r(0) = D^k_\sigma g(x_0) - D^k_\sigma f(x_0) = 0
     \]
 
 
@@ -121,3 +122,41 @@
 
     Therefore $r \in \mathcal{R}_\sigma^{n+1}(E; F)$.
 \end{proof}
+
+
+\begin{theorem}[Taylor's Formula, Integral Remainder]
+\label{theorem:taylor-integral}
+    Let $E$ be a topological vector space, $\sigma \subset \mathfrak{B}(E)$ be a covering ideal, $F$ be a separated locally convex space, $U \subset E$ be open, and $f \in C^{n+1}_\sigma(E; F)$, then for any $x_0 \in U$ and $h \in E$ such that $[x_0, x_0 + h] \subset U$, then
+    \[
+    f(x_0 + h) = f(x_0) + \sum_{k = 1}^n \frac{1}{k!}D^k_\sigma f(x_0)(h^{(k)}) + r(h)
+    \]
+
+    where
+    \[
+    r(h) = \int_0^1 \frac{(1 - t)^{n}}{n!}D^{n+1}_\sigma f(x_0 + th)(h^{(n+1)}) dt
+    \]
+
+    In particular, for any continuous seminorm $[\cdot]_F: F \to [0, \infty)$, 
+    \[
+    [r(h)]_F \le \frac{1}{(n+1)!} \cdot \sup_{t \in [0, 1]}[D^{n+1}_\sigma f(x_0 + th)(h^{n+1})]
+    \]
+\end{theorem}
+\begin{proof}{Proof, {{\cite[Section XIII.6]{Lang}}}. }
+    Firstly, if $n = 0$, then by the \hyperref[Fundamental Theorem of Calculus]{theorem:ftc-riemann}, 
+    \[
+    f(x_0 + h) - f(x_0) = \int_0^1 D_\sigma f(x_0 + th)(h) dt
+    \]
+
+    Assume inductively that the theorem holds for $n \in \natz$. Let
+    \[
+    u(t) =  D^{n+1}_\sigma f(x + ty)(h^{(n+1)}) \quad v(t) = -\frac{(1 - t)^{n+1}}{(n+1)!}
+    \]
+
+    then $Dv(t) = (1 - t)^n/n!$, and using the \hyperref[change of variables formula]{theorem:rs-change-of-variables} and \hyperref[integration by parts]{theorem:rs-ibp},
+    \begin{align*}
+        r(h) &= \frac{1}{n!}\int_0^1 (1 - t)^{n}D^{n+1}_\sigma f(x_0 + th)(h^{(n+1)}) dt \\
+        &= \int_0^1 udv = u(1)v(1) - u(0)v(0) - \int_0^1 vdu \\
+        &= D_\sigma^{n+1}(x_0)(h^{(n+1)}) \\
+        &+ \int_0^1 \frac{(1 - t)^{n+1}}{(n+1)!}D^{n+2}_\sigma f(x_0 + th)(h^{(n+2)}) dt
+    \end{align*}
+\end{proof}