Does there exist a function such that

[epkid08]

If we define a finite difference operator as $$\Delta a_n = a_{n+1}-a_n$$

Can we prove or disprove the existence of a function F, $$F:\mathbb{Z}\rightarrow\mathbb{Z}$$, such that $$\Delta F(g_n)=\frac{\Delta g_n}{ g_n}$$, where g is some arbitrary function?
Edit: fixed Big typo

 

[CRGreathouse]

Is that $\exists g\exists F$, $\forall g\exists F$, or $\exists F\forall g$?

 

[epkid08]

$$\exists F\forall g$$

 

[willem2]

suppose $g_0 = 1, g_1 = 2$ and $g_2 = 1$

Then $$f(2) - f(1) = \frac {g_2 - g_1} {g_1} = 1$$ and

$$f(1) - f(2) = \frac {g_3 - g_2} {g_2} = -1/2$$

so there can't be any F for this g

 

[epkid08]

Can you generalize a non-piecewise function for g that has the values $g_0 = 1, g_1 = 2$, and $g_2 = 1$?

 

[willem2]

Can you generalize a non-piecewise function for g that has the values $g_0 = 1, g_1 = 2$, and $g_2 = 1$?

Why? Of course there is a quadratic going through these points.

Is the range of F really $\mathbb{Z}$?. That is a problem with the division by $g_n$

 

[epkid08]

Is the range of F really $\mathbb{Z}$?. That is a problem with the division by $g_n$

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Now that I think about it, it shouldn't be.