>[!Note] Theorem (Taylor's theorem for univariate real valued function, Schlömilch remainder) >If $f:I \to \mathbb{R}$ is $n$ times differentiable on the open interval $I$ containing $x_{0}$ and $0\leq k\leq n-1$ then $\begin{align} f(x) =\; &f(x_{0}) + f'(x_{0})(x-x_{0}) +\dots \frac{f^{(n-1)}(x_{0})}{(n-1)!} (x-x_{0})^{n-1} \\ &+ \frac{f^{(n)}(t)}{(n-1)!(n-k)} (x-x_{0})^{n-k}(x-t)^{k} \end{align}$for some point $t$ between $x$ and $x_{0}.$ >*Proof.* Pick any $a,b\in I.$ Let $R$ be the remainder for which we are trying to find a formula $R = f(b) - \left( f(a) +f'(a) (b-a) +\dots+\frac{f^{(n-1)}(a)}{(n-1)!} (b-a)^{n-1} \right)$Define $h(x) = f(b) - f(x) -f'(x) (b-x) - \dots - \frac{f^{(n-1)}(x)}{(n-1)!}(b-x)^{n-1} - R \frac{(b-x)^{n-k}}{(b-a)^{n-k}}$Clearly $h(b)=0$. Also $h(a)=0$ because of the definition of $R.$ So by Rolle's theorem, there is a number $t$ between $a$ and $b$ where $h'(t) =0.$ So $\begin{align} h'(x) &= -f'(x) + f'(x ) -f''(x ) (b-x) + \frac{f''(x)}{2} 2(b-x) - \frac{f'''(x)}{2} (b-x)^{2} \\ &+ \dots - \frac{f^{(n)}(x)}{(n-1)!}(b-x)^{n-1} + (n-k)R \frac{(b-x)^{n-k-1}}{(b-a)^{n-k}} \\ &= - \frac{f^{(n)}(x)}{(n-1)!}(b-x)^{n-1} + (n-k) R \frac{(b-x)^{n-k-1}}{(b-a)^{n-k}} \end{align}$where for each term we use the product rule, and most of the sum cancels out. If we now substitute $t$ and set the expression equal to $0$ we get $\frac{f^{(n)}(t)}{(n-1)!(n-k)} (x-x_{0})^{n-k}(x-t)^{k}$which is the statement we wanted.