[proofplan]
We prove the Cauchy remainder form using [Cauchy's Mean Value Theorem](/theorems/187). Define the auxiliary function $F(t) = f(a+h) - \sum_{k=0}^{n} \frac{(h-t)^k}{k!} f^{(k)}(a+t)$, so that $F(h) = 0$ and $F(0) = R_n$ (the Taylor remainder). A telescoping computation gives $F'(t) = -\frac{(h-t)^n}{n!} f^{(n+1)}(a+t)$. Applying Cauchy's [Mean Value Theorem](/theorems/186) with the comparison function $G(t) = h - t$ (rather than $G(t) = (h-t)^n$ as in the Lagrange case) produces the Cauchy remainder, which retains the factor $(1 - \theta)^n$.
[/proofplan]
[step:Define the auxiliary functions $F$ and $G$ and compute boundary values]
Define
\begin{align*}
F: [0, h] &\to \mathbb{R} \\
t &\mapsto f(a+h) - \sum_{k=0}^{n} \frac{(h - t)^k}{k!} f^{(k)}(a + t),
\end{align*}
and $G: [0, h] \to \mathbb{R}$ by $G(t) = h - t$.
At $t = h$: every term with $k \ge 1$ vanishes because $(h - h)^k = 0$, and the $k = 0$ term is $f^{(0)}(a+h) = f(a+h)$, so $F(h) = 0$.
At $t = 0$: $(h - 0)^k = h^k$, so $F(0) = f(a+h) - \sum_{k=0}^{n} \frac{h^k}{k!} f^{(k)}(a) = R_n$, the Taylor remainder.
The comparison function satisfies $G(0) = h$, $G(h) = 0$, and $G'(t) = -1 \neq 0$ on $(0, h)$.
[/step]
[step:Compute $F'(t)$ by telescoping]
Differentiating $F(t)$ with respect to $t$, we apply the product rule to each term $\frac{(h-t)^k}{k!} f^{(k)}(a+t)$:
\begin{align*}
\frac{d}{dt}\left[\frac{(h-t)^k}{k!} f^{(k)}(a+t)\right] = -\frac{k(h-t)^{k-1}}{k!} f^{(k)}(a+t) + \frac{(h-t)^k}{k!} f^{(k+1)}(a+t).
\end{align*}
The first part simplifies to $-\frac{(h-t)^{k-1}}{(k-1)!} f^{(k)}(a+t)$ for $k \ge 1$.
Summing over $k = 0, \ldots, n$ with the overall minus sign from $F$:
\begin{align*}
F'(t) = \sum_{k=1}^{n} \frac{(h-t)^{k-1}}{(k-1)!} f^{(k)}(a+t) - \sum_{k=0}^{n} \frac{(h-t)^k}{k!} f^{(k+1)}(a+t).
\end{align*}
Re-indexing the first sum with $j = k - 1$ (so $j$ runs from $0$ to $n - 1$):
\begin{align*}
F'(t) = \sum_{j=0}^{n-1} \frac{(h-t)^j}{j!} f^{(j+1)}(a+t) - \sum_{k=0}^{n} \frac{(h-t)^k}{k!} f^{(k+1)}(a+t).
\end{align*}
The terms for $j = 0, \ldots, n-1$ in the first sum cancel with the $k = 0, \ldots, n-1$ terms in the second sum. Only the $k = n$ term from the second sum survives:
\begin{align*}
F'(t) = -\frac{(h-t)^n}{n!} f^{(n+1)}(a+t).
\end{align*}
[/step]
[step:Apply Cauchy's Mean Value Theorem with $G(t) = h - t$]
We verify the hypotheses of [Cauchy's Mean Value Theorem](/theorems/187) on $[0, h]$:
- $F$ is [continuous](/page/Continuity) on $[0, h]$: this follows because $f^{(0)}, \ldots, f^{(n)}$ are continuous by hypothesis.
- $F$ is [differentiable](/page/Derivative) on $(0, h)$: this follows because $f^{(n+1)}$ exists on $(a, a+h)$.
- $G(t) = h - t$ is a polynomial, hence continuous on $[0, h]$ and differentiable on $(0, h)$.
- $G'(t) = -1 \neq 0$ on $(0, h)$.
Cauchy's Mean Value Theorem yields $c \in (0, h)$ such that
\begin{align*}
\frac{F(h) - F(0)}{G(h) - G(0)} = \frac{F'(c)}{G'(c)}.
\end{align*}
[guided]
Note the critical difference from the proof of the Lagrange remainder: there we used $G(t) = (a+h-t)^n$, whose derivative $G'(t) = -n(a+h-t)^{n-1}$ matched the power $(a+h-t)^{n-1}$ in $F'(t)$, allowing cancellation and producing a clean $f^{(n)}(c)/n!$. Here we use $G(t) = h - t$, a linear function with $G'(t) = -1$. Substituting into [Cauchy's Mean Value Theorem](/theorems/187):
\begin{align*}
\frac{F(h) - F(0)}{G(h) - G(0)} = \frac{F'(c)}{G'(c)} = \frac{-\frac{(h-c)^n}{n!}f^{(n+1)}(a+c)}{-1} = \frac{(h-c)^n}{n!}f^{(n+1)}(a+c).
\end{align*}
The power $(h-c)^n$ in $F'(c)$ does not cancel against $G'(c) = -1$, and it survives in the remainder formula. Setting $c = \theta h$ with $\theta \in (0,1)$, this becomes $(h - \theta h)^n = h^n(1-\theta)^n$, which is the characteristic factor of the Cauchy remainder.
The Lagrange form and Cauchy form are both consequences of Cauchy's MVT applied to the same $F$ but with different comparison functions $G$. The Lagrange choice $G = (h-t)^n$ is tuned to simplify the remainder, while the Cauchy choice $G = h - t$ produces a remainder that retains the factor $(1-\theta)^n$ and is more useful for certain convergence arguments (e.g., proving that the Taylor series of $e^x$ converges).
[/guided]
[/step]
[step:Solve for the remainder and write in standard form]
Substituting the computed values:
\begin{align*}
\frac{0 - R_n}{0 - h} = \frac{-\frac{(h-c)^n}{n!} f^{(n+1)}(a+c)}{-1}.
\end{align*}
The left-hand side is $\frac{R_n}{h}$. The right-hand side is $\frac{(h-c)^n}{n!} f^{(n+1)}(a+c)$. Therefore
\begin{align*}
R_n = \frac{h(h-c)^n}{n!} f^{(n+1)}(a+c).
\end{align*}
Setting $c = \theta h$ with $\theta \in (0,1)$ (since $c \in (0,h)$), we have $h - c = h(1 - \theta)$:
\begin{align*}
R_n = \frac{h \cdot h^n(1-\theta)^n}{n!} f^{(n+1)}(a + \theta h) = \frac{h^{n+1}(1-\theta)^n}{n!} f^{(n+1)}(a + \theta h).
\end{align*}
Therefore
\begin{align*}
f(a+h) = \sum_{k=0}^{n} \frac{h^k}{k!} f^{(k)}(a) + \frac{h^{n+1}(1-\theta)^n}{n!} f^{(n+1)}(a + \theta h)
\end{align*}
for some $\theta \in (0,1)$.
[/step]
