> [!NOTE] Theorem (Taylor's theorem for univariate real function, Lagrange remainder) > If $f:I \to \mathbb{R}$ is $n$ times [[Fréchet Differentiation|differentiable]] on the open interval $I$ containing $x_{0}$ then $\begin{align} f(x) =\; &f(x_{0}) + f'(x_{0})(x-x_{0}) + \frac{f''(x_{0})}{2} (x-x_{0})^{2} +\dots \\ &\frac{f^{(n-1)}(x_{0})}{(n-1)!} (x-x_{0})^{n-1} + \frac{f^{(n)}(t)}{n!} (x-x_{0})^{n} \end{align}$for some $t$ between $x$ and $x_{0}.$ ^89aa6a **Proof**. Pick any $a,b \in I$. The function $g$ given by $g(x)=f(x)-\left( f(a)+f'(a)(x-a) +\dots + \frac{f^{(n-1)} (a)}{(n-1)!} (x-a)^{n-1} \right)$satisfies $g(a)=0$, $g'(a)=0$ and so on up to $g^{(n-1)}(a)=0.$ It also satisfies $g^{(n)}(x)=f^{(n)}(x)$ for all $x$ because $f$ and $g$ differ by a polynomial of degree only $n-1.$ If we put $h(x)=g(x)-g(b) \frac{(x-a)^{n}}{(b-a)^{n}}$then $h$ also has its first $n-1$ derivatives vanishing at $a$ but in addition it satisfies $h(b)=0$. We now proceed inductively. Since $h(b)=h(a)=0$ there is a point is a point $t_{1}\in(a,b)$ where $h'(t_{1})=0$ by [[Rolle's Theorem|Rolle's theorem]]. Since $h'(t_{1})=h'(a)=0$ there is a point $t_{2}\in(a,t_{1})$ with $h''(t_{2})=0.$ Continuing this way, we eventually get a point $t=t_{n}$ where $h^{(n)}(t)=0.$ In terms of $g$ this implies $g^{(n)}(t) = g(b) \frac{n!}{(b-a)^{n}}$so $g(b)= \frac{g^{(n)}(t)}{n!} (b-a)^{n} = \frac{f^{(n)}(t)}{n!}(b-a)^{n}$or $f(b) - \left( f(a) +f'(a)(b-a)+\dots+ \frac{f^{(n-1)}(a)}{(n-1)!} (b-a)^{(n-1)}\right)= \frac{f^{(n)}(t)}{n!} (b-a)^{n}$which is exactly the statement of the theorem. $\square$ **Proof**: Note that this Lagrange remainder form of Taylor’s Theorem follows immediately by setting $k = 0$ in the Schlömilch remainder form of Taylor's theorem.