Proof of Taylor's theorem with the Lagrange remainder by the Cauchy mean value theorem

Proof

We expand $f(x)$ around a moving point $t$ and apply the Cauchy mean value theorem in $t$.

1. Let $\alpha < \beta$ be the endpoints of $J$, so $\set{\alpha, \beta} = \set{a, x}$. The hypothesis means that $f, f', \dots, f^{(k)}$ exist on $J$, with one-sided derivatives at $\alpha$ and $\beta$.
2. For $t \in J$ define $F(t) = \sum_{j=0}^{k} \dfrac{f^{(j)}(t)}{j!} (x - t)^j$ and $G(t) = (x - t)^{k+1}$.
3. For $j < k$, $f^{(j)}$ is differentiable on $J$, so it is continuous on $J$. Inside $J$ this is differentiability implies continuity.
4. At an endpoint, write $f^{(j)}(s) - f^{(j)}(\alpha) = (s - \alpha) \cdot \dfrac{f^{(j)}(s) - f^{(j)}(\alpha)}{s - \alpha}$. The quotient has a finite limit, so the product tends to $0$. The endpoint $\beta$ is similar.
5. Also $f^{(k)}$ is continuous on $J$ by hypothesis. So $F$ and $G$ are continuous on $J$, as sums of products of continuous functions.
6. By the power rule and the chain rule, $\dfrac{d}{dt} \dfrac{(x - t)^j}{j!} = -\dfrac{(x - t)^{j-1}}{(j-1)!}$ for $j \ge 1$. Also $G'(t) = -(k + 1)(x - t)^k$.
7. For $0 \le j \le k$ and $t \in (\alpha, \beta)$ let $A_j(t) = \dfrac{f^{(j+1)}(t)}{j!} (x - t)^j$. This uses the hypothesis that $f^{(k+1)}$ exists on $(\alpha, \beta)$.
8. By the product rule and step 6, the $j$-th term of $F$ has derivative $A_j - A_{j-1}$ for $j \ge 1$. The term with $j = 0$ is $f$, with derivative $A_0$.
9. Adding, the sum telescopes: $F'(t) = A_0 + \sum_{j=1}^{k} (A_j - A_{j-1}) = A_k = \dfrac{f^{(k+1)}(t)}{k!} (x - t)^k$ on $(\alpha, \beta)$.
10. By the Cauchy mean value theorem, there is $\xi \in (\alpha, \beta)$ with $(F(\beta) - F(\alpha)) G'(\xi) = (G(\beta) - G(\alpha)) F'(\xi)$.
11. Exchanging $\alpha$ and $\beta$ multiplies both sides by $-1$. So $(F(x) - F(a)) G'(\xi) = (G(x) - G(a)) F'(\xi)$, and $\xi$ lies strictly between $a$ and $x$.
12. At $t = x$, every term of $F$ with $j \ge 1$ vanishes, so $F(x) = f(x)$. At $t = a$, $F(a) = P_k(x)$, the value at $x$ of the Taylor polynomial of $f$ at $a$.
13. Also $G(x) = 0$ and $G(a) = (x - a)^{k+1}$.
14. Substituting steps 6, 9, 12 and 13 into step 11 gives $-(k + 1)(x - \xi)^k (f(x) - P_k(x)) = -(x - a)^{k+1} \dfrac{f^{(k+1)}(\xi)}{k!} (x - \xi)^k$.
15. Since $\xi \ne x$, we have $(x - \xi)^k \ne 0$. Dividing by $-(k + 1)(x - \xi)^k$ gives $f(x) = P_k(x) + \dfrac{f^{(k+1)}(\xi)}{(k + 1)!} (x - a)^{k+1}$.