The error term in Taylor series and convolution. I've been wondering a lot why is the remainder of the Taylor expansion of a function, $R_n(x)$, expressed (in one of the many forms) as something very similar to aconvolution. Precisely:
$$R_n(x) = \int_a^x \frac{(x-t)^n}{n!}f^{(n+1)}(t)dt$$
and if we're dealing with the Mc Laurin series, then it is a convolution:
$$R_n(x) = \int_0^x \frac{(x-t)^n}{n!}f^{(n+1)}(t)dt$$
I studied this particular error formula because I find it very elegant, and it isn't hard to manipulate.
I know the proof of this formula is simply made by induction on $n$, we start of by the linear approximation:
$$f(x)=f(a)+f'(a)(x-a)+R_1(x)$$, so that
$$R_1(x) = f(x)-f(a) - f'(a) (x-a)$$
$${R_1}(x) = \int\limits_a^x {f'\left( t \right)dt}  - \int\limits_a^x {f'(a)dt} $$
$${R_1}(x) = \int\limits_a^x {f'\left( t \right) - f'\left( a \right)dt} $$
So now we integrate by parts with
$$f'\left( t \right) - f'\left( a \right) = u$$
$$t - x = v$$
to get
$${R_1}(x) = \int\limits_a^x {\left( {x - t} \right)f''\left( t \right)dt} $$
We can similarily do this with $R_2(x)$, since
$${R_2}(x) = {R_1}(x) - f''\left( a \right)\frac{{{{\left( {x - a} \right)}^2}}}{{2!}}$$
$${R_2}(x) = \int\limits_a^x {\left( {x - t} \right)f''\left( t \right)dt}  - \int\limits_a^x {\left( {x - t} \right)f''\left( a \right)dt} $$
$${R_2}(x) = \int\limits_a^x {\left( {x - t} \right)\left( {f''\left( t \right) - f''\left( a \right)} \right)dt} $$
So again integrating by parts gives
$${R_2}(x) = \int\limits_a^x {\frac{{{{\left( {x - t} \right)}^2}}}{{2!}}f'''\left( t \right)dt} $$
Q1: Can this be proved in an alternative way, noticing that the error is a 
convolution between $\dfrac{{{x^n}}}{{n!}}$ and $f^{(n+1)}(x)$?
Q2: How can this be interpreted in the scope of convolution "theory" and similar ideas?
 A: 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:


*

*Linearity

*$\mathcal{L}(x^n)=n!/s^{n+1}$

*$\mathcal{L}(f^{(n)})(s)=s^n\mathcal{L}f(s)-\sum_{k=0}^{n-1}s^{n-k-1}f^{(k)}(0)$

*$\mathcal{L}(f\ast g)=\mathcal{L}f\cdot\mathcal{L}g$


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
$$\begin{align*}
\mathcal{L}R_n(s)&=\mathcal{L}f(s)-\sum_{k=0}^n\frac{f^{(k)}(0)}{k!}\mathcal{L}(x^k)(s)\\
&=\mathcal{L}f(s)-\sum_{k=0}^n\frac{f^{(k)}(0)}{s^{k+1}}\\
&=\frac{1}{s^{n+1}}\Bigl(s^{n+1}\mathcal{L}f(s)-\sum_{k=0}^ns^{n-k}f^{(k)}(0)\Bigr)\\
&=\frac{\mathcal{L}(x^{n})(s)}{n!}\cdot\mathcal{L}(f^{(n+1)})(s)\\
&=\mathcal{L}\Bigl(\frac{x^n}{n!}\ast f^{(n+1)}\Bigr).
\end{align*}$$
