1
$\begingroup$

When I have $$f(a,N,x)={}_1F_1\left(1;N;a\frac{x}{1+x}\right)$$ where $N$ positive integer, $a>0$ and $x\in[0,\infty)$, I may have a bound $$f(a,N,x)\leq{}_1F_1\left(1;N;a\right)$$ as $\frac{x}{1+x}\in[0,1]$. This is good enough for my work.

However, now, I have $$f(a,N,x)={}_1F_1\left(1;N;a x\right)$$ which cannot be bounded as previous case.

Does anyone have idea of getting some bound such as $$f(a,N,x)\leq g(a,N) h(a,N,x)$$ where parameter $x$ in $h(a,N,x)$ may be only in polynomial or exponential functions, e.g., $x^pe^{-qx}$, $x^pe^{-qx^2}$.

$\endgroup$

1 Answer 1

3
$\begingroup$

I'm unsure of what is 'good enough' for your work, but one such bound arises as follows:

First, we rewrite this particular special case of Kummer's (confluent hypergeometric) function in terms of the lower incomplete gamma function:

$$_1F_1(1;N;ax) = (N-1)(ax)^{1-N}e^{ax}\gamma(N-1,ax).$$

This is a well-known identity, e.g., lower incomplete gamma

Next, we can use the following bound

$$(N-1)(ax)^{1-N}\gamma(N-1,ax) \leq \frac{1}{N}(1 + (N-1)e^{-ax}).$$

(this comes from the paper Inequalities and Bounds for the Incomplete Gamma Function)

Combining these results in the bound:

$$_1F_1(1;N;ax) \leq \frac{e^{ax}}{N}(1 + (N-1)e^{-ax}) = \frac{e^{ax}-1}{N} + 1.$$

Hopefully this will suffice for your purposes.

$\endgroup$
3
  • $\begingroup$ Is it possible to use $e^a$ instead $e^{ax}$? Thanks ! $\endgroup$
    – Frey
    May 11, 2017 at 9:13
  • $\begingroup$ Because i really need $e^{-px}$ form or else $x^p$ only which is the most preferred one. $\endgroup$
    – Frey
    May 11, 2017 at 9:23
  • $\begingroup$ @frey What ranged of values do you have for your inputs? $\endgroup$
    – David
    May 11, 2017 at 11:20

You must log in to answer this question.

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