Theorem: Let $f$ be $(n+1)$ times differentiable on an open interval containing $a$ and $x$. Then $f(x) = \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x-a)^k + \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}$ for some $c$ between $a$ and $x$.

Step 1: Setup.

Define $T_n(x) = \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x-a)^k$ and $R_n(x) = f(x) - T_n(x)$. We need to show that $R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}$ for some $c$ between $a$ and $x$.

Fix $x \neq a$ and define the constant $M$ by the equation $R_n(x) = M \cdot \frac{(x-a)^{n+1}}{(n+1)!}$

We need to show $M = f^{(n+1)}(c)$ for some $c$ between $a$ and $x$.

Step 2: Construct an auxiliary function.

Define $g(t) = f(t) - T_n(t) - M \cdot \frac{(t-a)^{n+1}}{(n+1)!}$

where $T_n(t) = \sum_{k=0}^n \frac{f^{(k)}(a)}{k!}(t-a)^k$ is viewed as a function of $t$.

We observe that $g(a) = 0$ because $T_n(a) = f(a)$ and the last term vanishes at $t = a$. Also $g(x) = 0$ by our choice of $M$.

Step 3: Apply the generalized Rolle's theorem.

Since $g(a) = g(x) = 0$ and $g$ is differentiable, Rolle's theorem gives $c_1$ between $a$ and $x$ with $g'(c_1) = 0$.

Now compute $g'(t)$: $g'(t) = f'(t) - \sum_{k=1}^n \frac{f^{(k)}(a)}{(k-1)!}(t-a)^{k-1} - M \cdot \frac{(t-a)^n}{n!}$

We have $g'(a) = f'(a) - f'(a) = 0$ (the sum at $t=a$ gives $f'(a)$ from the $k=1$ term). So $g'(a) = g'(c_1) = 0$, and Rolle's theorem gives $c_2$ between $a$ and $c_1$ with $g''(c_2) = 0$.

Continuing this process: at each stage, $g^{(j)}(a) = 0$ (since the Taylor polynomial was chosen to match $f$ up to order $n$) and $g^{(j)}(c_j) = 0$ from the previous step. By Rolle's theorem, there exists $c_{j+1}$ between $a$ and $c_j$ with $g^{(j+1)}(c_{j+1}) = 0$.

Step 4: The $(n+1)$-th derivative.

After $n+1$ applications of Rolle's theorem, we obtain $c_{n+1}$ (call it $c$) between $a$ and $x$ with $g^{(n+1)}(c) = 0$.

Now compute $g^{(n+1)}(t)$:

• $\frac{d^{n+1}}{dt^{n+1}} f(t) = f^{(n+1)}(t)$
• $\frac{d^{n+1}}{dt^{n+1}} T_n(t) = 0$ (since $T_n$ is a polynomial of degree $\leq n$)
• $\frac{d^{n+1}}{dt^{n+1}} \left[M \cdot \frac{(t-a)^{n+1}}{(n+1)!}\right] = M$

Therefore $g^{(n+1)}(t) = f^{(n+1)}(t) - M$.

Setting $g^{(n+1)}(c) = 0$ gives $f^{(n+1)}(c) = M$.

Step 5: Conclusion.

Substituting back: $R_n(x) = M \cdot \frac{(x-a)^{n+1}}{(n+1)!} = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}$. $\square$