Proofs from Taylor's theorem

We prove the integral formula by induction. When $n=0$, the Fundamental Theorem of Calculus gives

Assume that

For the next remainder, take $u=(x-t)^k$ and $dv=f^{(k+1)}(t)\,dt$, so $du=-k(x-t)^{k-1}\,dt$ and $v=f^{(k)}(t)$. We obtain

The $t=x$ endpoint vanishes because $(x-x)^k=0$ and the penultimate line uses the induction hypothesis $R_k(x)=f(x)-P_{k-1}(x)$. Each induction step removes one Taylor term, which proves the integral form for every $n$.

To obtain the Lagrange form, apply the mean value theorem for integrals to the continuous function $f^{(n+1)}$ with the non-negative weight $|x-t|^n$ between $a$ and $x$. Accounting for the orientation when $x<a$ gives $\int_a^x f^{(n+1)}(t)(x-t)^n\,dt =f^{(n+1)}(c)(x-a)^{n+1}/(n+1)$ for some intermediate $c$.

Taylor's theorem gives

because the lower-order terms vanish. Since $f^{(k+1)}$ is continuous near $a$, it is bounded on a smaller neighbourhood, so the remainder estimate gives $R_{k+1}(x)=O(|x-a|^{k+1})$ and hence $R_{k+1}(x)=o(|x-a|^k)$. Therefore the first nonzero term determines the sign of $f(x)-f(a)$ sufficiently close to $a$.

When $k$ is even, $(x-a)^k$ is positive on both sides of $a$, so the sign of $f^{(k)}(a)$ determines a minimum or maximum. When $k$ is odd, then $k\ge3$ and the same expansion applied to $f''$ gives

Since $k-2$ is odd, $f''$ changes sign across $a$; together with $f'(a)=0$, this gives a horizontal point of inflection.