0
$\begingroup$

Let's consider a Newton expansion over consecutive derivatives of a function:

$$F(x)=\sum_{m=0}^{\infty} \binom {-1}m \sum_{k=0}^m\binom mk(-1)^{m-k}f^{(k)}(x)$$

Can it be proven that such expansion, when converges, has the properties of an antiderivative? What are the conditions for that if it is not always the case?

Improve this question
$\endgroup$
7
  • $\begingroup$ was the earlier answer unsatisfactory? mathoverflow.net/questions/130886 $\endgroup$
    – Carlo Beenakker
    Commented Nov 7, 2013 at 11:32
  • 1
    $\begingroup$ @Carlo Beenakker I think it does not answer the question because it does not mention the other types of differintegral, which I was asking for comparison with. From what I understand, it says that the continuity is difficult to prove. This question is more narrow, it asks about just antiderivative. $\endgroup$
    – Anixx
    Commented Nov 7, 2013 at 11:42
  • $\begingroup$ Why do you write $\binom{-1}m$ rather than $\frac{(-1)\cdots(-m)}{m!} = (-1)^m$? By "properties of an antiderivative", do you mean more than that $\partial F = f$? $\endgroup$
    – Theo Johnson-Freyd
    Commented Jan 9, 2014 at 18:23
  • $\begingroup$ @Theo Johnson-Freyd no, I mean exactly that property. $\endgroup$
    – Anixx
    Commented Jan 9, 2014 at 20:23
  • $\begingroup$ Well, the answer is clearly "yes" for $f(x) = \exp(\lambda x)$ for $\lambda \in \mathbb C$ with $|\lambda-1|<1$, and therefore for linear combinations of such functions (more generally, e.g., if your Fourier transform has compact support in the interior of the disk). But I'm sure there are cases where F converges in some not-great way to some not-very-regular function. $\endgroup$
    – Theo Johnson-Freyd
    Commented Jan 9, 2014 at 20:48

0

Reset to default