[proofplan]
We prove [Taylor's theorem](/theorems/827) with the Lagrange remainder using [Cauchy's Mean Value Theorem](/theorems/187). Define the auxiliary function $F(t) = f(a+h) - \sum_{j=0}^{n-1} \frac{(a+h-t)^j}{j!} f^{(j)}(t)$, which satisfies $F(a+h) = 0$ and $F(a) = f(a+h) - \sum_{j=0}^{n-1} \frac{h^j}{j!} f^{(j)}(a)$ (the remainder we seek). A telescoping computation shows $F'(t) = -\frac{(a+h-t)^{n-1}}{(n-1)!} f^{(n)}(t)$. Applying Cauchy's [Mean Value Theorem](/theorems/186) with the comparison function $G(t) = (a+h-t)^n$ yields the Lagrange form of the remainder.
[/proofplan]
[step:Define the auxiliary function $F$ and compute its boundary values]
Define
\begin{align*}
F: [a, a+h] &\to \mathbb{R} \\
t &\mapsto f(a+h) - \sum_{j=0}^{n-1} \frac{(a+h-t)^j}{j!} f^{(j)}(t).
\end{align*}
At $t = a+h$: every term in the sum with $j \ge 1$ contains the factor $(a+h-(a+h))^j = 0$, and the $j = 0$ term contributes $f^{(0)}(a+h) = f(a+h)$. Therefore
\begin{align*}
F(a+h) = f(a+h) - f(a+h) = 0.
\end{align*}
At $t = a$: we have $(a+h-a)^j = h^j$, so
\begin{align*}
F(a) = f(a+h) - \sum_{j=0}^{n-1} \frac{h^j}{j!} f^{(j)}(a).
\end{align*}
This is precisely the remainder $R_n$ that the theorem claims equals $\frac{h^n}{n!} f^{(n)}(a + \theta h)$.
[/step]
[step:Differentiate $F$ and observe the telescoping cancellation]
We differentiate $F(t)$ with respect to $t$. The constant $f(a+h)$ vanishes, and for the sum we apply the product rule to each term $\frac{(a+h-t)^j}{j!} f^{(j)}(t)$:
\begin{align*}
\frac{d}{dt}\left[\frac{(a+h-t)^j}{j!} f^{(j)}(t)\right] = -\frac{j(a+h-t)^{j-1}}{j!} f^{(j)}(t) + \frac{(a+h-t)^j}{j!} f^{(j+1)}(t).
\end{align*}
The first term simplifies to $-\frac{(a+h-t)^{j-1}}{(j-1)!} f^{(j)}(t)$ for $j \ge 1$, and is absent for $j = 0$. Summing over $j = 0, \ldots, n-1$:
\begin{align*}
-F'(t) = -\sum_{j=1}^{n-1} \frac{(a+h-t)^{j-1}}{(j-1)!} f^{(j)}(t) + \sum_{j=0}^{n-1} \frac{(a+h-t)^j}{j!} f^{(j+1)}(t).
\end{align*}
Re-index the first sum by setting $i = j - 1$ (so $i$ runs from $0$ to $n - 2$):
\begin{align*}
-F'(t) = -\sum_{i=0}^{n-2} \frac{(a+h-t)^i}{i!} f^{(i+1)}(t) + \sum_{j=0}^{n-1} \frac{(a+h-t)^j}{j!} f^{(j+1)}(t).
\end{align*}
The terms with $j = 0, \ldots, n-2$ in the second sum match the terms $i = 0, \ldots, n-2$ in the first sum (with opposite sign), so they cancel. The sole survivor is the $j = n - 1$ term from the second sum:
\begin{align*}
F'(t) = -\frac{(a+h-t)^{n-1}}{(n-1)!} f^{(n)}(t).
\end{align*}
[guided]
The telescoping is the key mechanism. Each term in the sum produces two contributions when differentiated via the product rule:
\begin{align*}
\frac{d}{dt}\left[\frac{(a+h-t)^j}{j!} f^{(j)}(t)\right] = -\frac{(a+h-t)^{j-1}}{(j-1)!} f^{(j)}(t) + \frac{(a+h-t)^j}{j!} f^{(j+1)}(t).
\end{align*}
The first contribution (from differentiating the power $(a+h-t)^j$) lowers the power index by one, while the second (from differentiating $f^{(j)}(t)$) raises the derivative order by one. When we sum over $j = 0, \ldots, n-1$ and collect the two resulting sums, re-indexing the first sum with $i = j - 1$ yields
\begin{align*}
-F'(t) = -\sum_{i=0}^{n-2} \frac{(a+h-t)^i}{i!} f^{(i+1)}(t) + \sum_{j=0}^{n-1} \frac{(a+h-t)^j}{j!} f^{(j+1)}(t).
\end{align*}
The terms for $j = 0, \ldots, n-2$ in the second sum match the terms $i = 0, \ldots, n-2$ in the first sum (both produce $\frac{(a+h-t)^j}{j!} f^{(j+1)}(t)$), but with opposite signs, so they cancel completely. The only unmatched term is $j = n - 1$ in the second sum, since the first sum only runs to $i = n - 2$. The sole survivor is
\begin{align*}
F'(t) = -\frac{(a+h-t)^{n-1}}{(n-1)!} f^{(n)}(t).
\end{align*}
This telescoping is not a coincidence: $F(t)$ is constructed precisely so that its derivative simplifies. The function $F(t)$ is the Taylor remainder viewed as a function of the expansion point, and the telescoping reflects the recursive structure of the Taylor polynomial.
[/guided]
[/step]
[step:Apply Cauchy's Mean Value Theorem with the comparison function $G(t) = (a+h-t)^n$]
Define
\begin{align*}
G: [a, a+h] &\to \mathbb{R} \\
t &\mapsto (a+h-t)^n.
\end{align*}
Then $G'(t) = -n(a+h-t)^{n-1}$, and the boundary values are $G(a) = h^n$ and $G(a+h) = 0$.
We verify the hypotheses of [Cauchy's Mean Value Theorem](/theorems/187) on $[a, a+h]$:
- $F$ is [continuous](/page/Continuity) on $[a, a+h]$: this follows because $f^{(0)}, \ldots, f^{(n-1)}$ are continuous on $[a, a+h]$ by hypothesis.
- $F$ is [differentiable](/page/Derivative) on $(a, a+h)$: this follows because $f^{(n)}$ exists on $(a, a+h)$.
- $G$ is a polynomial, hence continuous on $[a, a+h]$ and differentiable on $(a, a+h)$.
- $G'(t) = -n(a+h-t)^{n-1} \neq 0$ for $t \in (a, a+h)$, since $a+h-t > 0$ on this open interval.
Cauchy's Mean Value Theorem yields a point $c \in (a, a+h)$ such that
\begin{align*}
\frac{F(a+h) - F(a)}{G(a+h) - G(a)} = \frac{F'(c)}{G'(c)}.
\end{align*}
[/step]
[step:Solve for the remainder and obtain the Lagrange form]
Substituting the computed values into the Cauchy MVT equation:
\begin{align*}
\frac{0 - F(a)}{0 - h^n} = \frac{-\frac{(a+h-c)^{n-1}}{(n-1)!} f^{(n)}(c)}{-n(a+h-c)^{n-1}}.
\end{align*}
The left-hand side simplifies to $\frac{F(a)}{h^n}$. On the right-hand side, since $c \in (a, a+h)$ we have $a+h-c > 0$, so $(a+h-c)^{n-1} > 0$ and we cancel this factor together with the minus signs:
\begin{align*}
\frac{F(a)}{h^n} = \frac{f^{(n)}(c)}{n(n-1)!} = \frac{f^{(n)}(c)}{n!}.
\end{align*}
Therefore
\begin{align*}
F(a) = \frac{h^n}{n!} f^{(n)}(c).
\end{align*}
Recalling that $F(a) = f(a+h) - \sum_{j=0}^{n-1} \frac{h^j}{j!} f^{(j)}(a)$, we rearrange:
\begin{align*}
f(a+h) = \sum_{j=0}^{n-1} \frac{h^j}{j!} f^{(j)}(a) + \frac{h^n}{n!} f^{(n)}(c).
\end{align*}
Setting $\theta = \frac{c - a}{h}$, we have $\theta \in (0, 1)$ (since $c \in (a, a+h)$) and $c = a + \theta h$. This gives the Lagrange remainder form:
\begin{align*}
f(a+h) = \sum_{j=0}^{n-1} \frac{h^j}{j!} f^{(j)}(a) + \frac{h^n}{n!} f^{(n)}(a + \theta h).
\end{align*}
[guided]
Let us verify the sign arithmetic carefully. The Cauchy MVT gives
\begin{align*}
\frac{F(a+h) - F(a)}{G(a+h) - G(a)} = \frac{F'(c)}{G'(c)}.
\end{align*}
The numerator on the left is $0 - F(a) = -F(a)$. The denominator is $0 - h^n = -h^n$. So the left-hand side is $\frac{-F(a)}{-h^n} = \frac{F(a)}{h^n}$.
The right-hand side is
\begin{align*}
\frac{F'(c)}{G'(c)} = \frac{-\frac{(a+h-c)^{n-1}}{(n-1)!} f^{(n)}(c)}{-n(a+h-c)^{n-1}} = \frac{(a+h-c)^{n-1} f^{(n)}(c)}{n(n-1)! \cdot (a+h-c)^{n-1}} = \frac{f^{(n)}(c)}{n!}.
\end{align*}
Both minus signs cancel, and the factor $(a+h-c)^{n-1}$ cancels (it is strictly positive since $c < a + h$). Equating left and right:
\begin{align*}
\frac{F(a)}{h^n} = \frac{f^{(n)}(c)}{n!}, \qquad \text{hence} \qquad F(a) = \frac{h^n}{n!} f^{(n)}(c).
\end{align*}
The parametrisation $c = a + \theta h$ with $\theta = (c - a)/h$ maps $c \in (a, a+h)$ bijectively to $\theta \in (0, 1)$, completing the proof.
[/guided]
[/step]
