0
$\begingroup$

In the theory of indefinite sums, anti-differences and finite calculus, ‎the following ‎difference ‎functional ‎equation ‎and ‎its ‎solutions ‎are ‎very ‎important:

‎$$‎\bigtriangleup ‎F(x):=F(x+1)-F(x)=f(x) ‎\quad‎\quad‎(1),‎$$

where‎ ‎$‎\bigtriangleup‎$ ‎is ‎the ‎forward ‎difference ‎operator when $f$ is given and $F$ is unknown. ‎ Also‎, if ‎$‎‎D_f=\mathbb{R}$‎, then there exists a special solution ‎$‎‎F_0(x)$ for equation (1) and ‎ ‎we ‎have ‎the ‎general ‎solutions of ‎it ‎as ‎follows‎: ‎‎‎$‎‎F=F_0+‎\lambda$,‎‎ ‎which ‎‎$‎‎‎\lambda$ ‎is a‎ ‎one-periodic ‎function.‎

Now‎, in my research I deal to the following functional equation ‎ $$F(x+1)+F(x)=f(x)‎‎$$

‎but I don't have any knowledge about it and its solution. ‎ What is the name of this equation and where can I learn more about it?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
5
$\begingroup$

Denote $g(x)=F(x) (-1)^{\lfloor x\rfloor}$ and get the difference equation for $g$. Multiplying by $\sin \pi x$ also works.

Improve this answer
$\endgroup$
7
  • $\begingroup$ Do you mean that by applying this $g$, the new functional equation turns into the functional equation (1)? $\endgroup$
    – soodehMehboodi
    Commented Sep 17, 2018 at 15:19
  • $\begingroup$ Or you give just an special solution. $\endgroup$
    – soodehMehboodi
    Commented Sep 17, 2018 at 15:31
  • 1
    $\begingroup$ Yes, it turns out into (1) $\endgroup$
    – Fedor Petrov
    Commented Sep 17, 2018 at 15:36
  • 1
    $\begingroup$ Henec, if ‎we denote ‎‎$h(x)=F(x)‎\varepsilon(x)‎ (-1)^{\lfloor x\rfloor} ‎\quad‎\text{where}‎\quad‎ \varepsilon(x) = \begin{cases} 1 & \text{if } \lfloor x\rfloor\text{is odd}, \\ -1 & \text{if }\lfloor x\rfloor \text{is even}. \end{cases} $‎‎‎‎ then ‎$‎‎h$ ‎works ‎for ‎all ‎‎$‎‎x$.‎ ‎ $\endgroup$
    – soodehMehboodi
    Commented Sep 18, 2018 at 13:00
  • 1
    $\begingroup$ Your $h$ equals $F$, remove one of multiples (any of two). $\endgroup$
    – Fedor Petrov
    Commented Sep 18, 2018 at 13:26

Not the answer you're looking for? Browse other questions tagged .