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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?

Pedro
  • 122,002

1 Answers1

10

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

Julián Aguirre
  • 76,354