The error term in Taylor series and convolution.

You can get it by means of the Laplace's transform, although for a restriced class of functions.

If $f\colon[0,\infty)\to\mathbb{R}$ is piecewise continuous such that $|f(x)|\le C\,e^{cx}$ for some constants $C,c\in\mathbb{R}$, its Laplace's transform is defined as $$ \mathcal{L}f(s)=\int_0^\infty e^{-sx}f(x)\,dx,\quad s>c. $$ I will make use of the following properties:

1. Linearity
2. $\mathcal{L}(x^n)=n!/s^{n+1}$
3. $\mathcal{L}(f^{(n)})(s)=s^n\mathcal{L}f(s)-\sum_{k=0}^{n-1}s^{n-k-1}f^{(k)}(0)$
4. $\mathcal{L}(f\ast g)=\mathcal{L}f\cdot\mathcal{L}g$

Now, let $f\colon[0,\infty)\to\mathbb{R}$ be a function with $n+1$ continuous derivatives such that $|f^{(k)}(x)|\le C\,e^{cx}$ for all $x\ge0$, $0\le k\le n+1$ and some constants $C,c\in\mathbb{R}$. Let $$R_n(x)=f(x)-\sum_{k=0}^n\frac{f^{(k)}(0)}{k!}x^n.$$ Then $$\begin{align*} \mathcal{L}R_n(s)&=\mathcal{L}f(s)-\sum_{k=0}^n\frac{f^{(k)}(0)}{k!}\mathcal{L}(x^k)(s)\\ &=\mathcal{L}f(s)-\sum_{k=0}^n\frac{f^{(k)}(0)}{s^{k+1}}\\ &=\frac{1}{s^{n+1}}\Bigl(s^{n+1}\mathcal{L}f(s)-\sum_{k=0}^ns^{n-k}f^{(k)}(0)\Bigr)\\ &=\frac{\mathcal{L}(x^{n})(s)}{n!}\cdot\mathcal{L}(f^{(n+1)})(s)\\ &=\mathcal{L}\Bigl(\frac{x^n}{n!}\ast f^{(n+1)}\Bigr). \end{align*}$$