I Is long division for a non-polynomic function possible?

1. Mar 4, 2016

ecneicScience

My question is: suppose you have a function $F(x)$ which has an asymptote at $x=x*$. Can you decompose $F(x)$ so that
$$F(x) = G(x) + H(x)$$
where $G(x)$ is defined at $x=x*$ and $H(x)$ contains the asymptotic behaviour at $x=x*$ and goes to $0$ at plus or minus $\inf$? This is commonly done when $F(x)$ is a polynomial in order to approximate $F(x)$'s behaviour with the function $G(x)$ when $x$ is sufficiently far away from $x^{*}$.

I have a very complicated function $F(x)$ that I would like to extract $G(x)$ from. If you're curious here it is:

$$F(x) = \frac{(2 a_1 z^{-a_2} -2 a_1 z^{-a_1} + z^{-a_1} - z^{-a_2}+ a_1^2 z^{-a_1} - a_2^2 z^{-a_2}) ln(z)^2 + (2 a_2 z^{-a_2} -2 a_1 z^{-a_1}+2 z^{-a_2} - 2 z^{-a_1}) ln(z) + 2 z^{-a_1} - 2 z^{-a_2}} {(a_1 z^{-a_1} - a_2 z^{-a_2} + z^{-a_2} - z^{-a_1}) ln(z)^2 + (z^{-a_1}-z^{-a_2}) ln(z)}$$

where
$$z = 10^{x-x^{*}}$$
and $a_2>a_1>0$ and $x^*>0$. I've attached a graph for an instance of the function where $x^{*}=2$. As you can see there is an asymptote at $x=2$. I want either $G(x)$ or the slope of $G(x)$ at $x=x^{*}$. Has anyone encountered a similar problem?

Last edited: Mar 5, 2016
2. Mar 5, 2016

Staff: Mentor

Such a decomposition is always possible and never unique (e.g. for every solution you can add a Gaussian to G and subtract it from H). That also means that the slope of G(x) is not fixed at x=x*, it can have any value you like.

To find a possible decomposition, it is sufficient to describe the behavior either for x around x* or for x to +- infinity. The latter is probably easier here as it does not involve finding x*.

3. Apr 15, 2016

ecneicScience

Thank you very much for your insight. You made the arbitrariness of G(x) very clear to me. Sorry it took so long to get back to you.