Series expansion of monotonic functions on positive numbers?

Question

I'm interested in a series expansion of 1d functions of a positive number which are monotonic decreasing and go to 0 at infinity. I need this for a physics problem where I'm trying to fit interaction energy. I could write the energy as an arbitrary series, but depending on the coeffcients that can easily end up non-monotonic which is known to not be possible physically. I'd like to have some guarantee that any function can be represented. The series doesn't have to consist of orthogonal functions. Ideally, I'd like to use a rather small number of terms, let's say 3-5.

A couple of ideas:

• sum of $A_k/(r+r_0)^{p_k}$ (all $A_k$ positive, all $p_k$ > 1)
• sum of $A_k e^{(-B_k r)}$ (all $A_k$, $B_k$ positive)

Would something like that work? Are there standard approaches to this problem?

Answer 1

Not sure this will work:

Assume wlg that the range of the function $f(x)$ is $[0,\infty)$ and $f(0)=1$

Since you want monotonically decreasing and positive, the derivative is negative. Lets look at minus the derivative $P(x) \equiv -f'(x)$ where $f(x)$ is the function you want to approximate. Due to monitonicity, $P > 0$ and goes to zero at infinity. Also, $$\int_0^\infty P(x)dx = f(0) = 1> 0$$ from the definition.

So we can consider $P$ as a probability distribution function (PDF).

There exists expansions for PDF's that are "almost" always positive definite, two examples being:

• Edgeworth_series

• random pdf I found on the internet

After we have a positive definite approximation for $P$ we can reconstruct $$f_{approx}(x)=1-\int_0^xP_{approx}(y)dy$$

Answer 2

Since you want a good approximation with few terms, then I'll suppose that $(p_k)_k$ and $(B_k)_k$ are monotone increasing sequences. In this case, the ideas you proposed cannot work for any positive monotone decreasing function that tends to zero, because we can always propose a function with a slower decay than the family of functions can represent.

Let us analyze your first family of functions $a_k/(r_0+r)^{p_k}$ (the second family is analogous). Suppose we are interested in representing $f(r) = 1/\ln(2+r)$ as $$ f(r) = \sum_{k=1}^\infty \frac{a_k}{(r_0+r)^{p_k}}. $$ Then, $$ f((\alpha-1)r_0) = \sum_{k=1}^\infty \frac{a_k}{(\alpha r_0)^{p_k}} \le \frac{1}{\alpha^{p_1}} \sum_{k=1}^\infty \frac{a_k}{r_0^{p_k}} = \frac{f(0)}{\alpha^{p_1}}. $$ However, this is a contradiction, because $\dfrac{\ln(2+(\alpha-1)r_0)}{\alpha^{p_1}} \to 0$ as $\alpha\to\infty$ for any $p_1>0$.