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Is it already known that ${}_1F_1(a;b;x) \leq \Gamma(b)(1+|x|)^{-a}$ when $a$ is an integer, $a <0,$ and $b>0?$ If it is, what is a reference?

my proof: Since the Kummer function can be written in terms of a generalized Laguerre polynomial, \begin{equation} \label{eqn:Kummer-Laguerre-equality} {}_1F_1(a;b;x) = \frac{\Gamma(1-a)\Gamma(b)}{\Gamma(b-a)} L_{-a}^{(b-1)}(x), \end{equation} when $a < 0,$ we proceed by bounding the generalized Laguerre polynomial on the right hand side.

Let $n = -a$ and $\alpha = b - 1.$ Then $$ L_{n}^{(\alpha)}(x) = \sum_{\ell=0}^n \frac{\Gamma(\alpha + n + 1)}{\Gamma(\alpha + \ell +1)(n-\ell)!\ell!} (-x)^\ell. $$ Our constraints on $a$ and $b$ ensure that each $\Gamma(\cdot)$ term in the above sum is positive. Furthermore, for $\ell=0,1,\ldots,n,$ $$ \Gamma(\alpha + \ell +1) \geq \Gamma(b) \geq \min_{x > 0} \Gamma(x) > 0.88. $$

It follows that $$ L_{n}^{(\alpha)}(x) \leq 1.14 \cdot \Gamma(\alpha + n + 1) \sum_{\ell =0}^n \frac{|x|^\ell}{(n-\ell)!\ell!} = 1.14 \cdot \Gamma(b-a) \frac{1}{(-a)!}(1 + |x|)^{-a}. $$ The last equality is a consequence of the binomial theorem.

The conclusion follows immediately when this estimate is used in the relation expressing ${}_1F_1$ in terms of the Laguerre polynomial.

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Could you give us a little more context? –  Anthony Quas Jan 6 '12 at 3:38
    
Not sure what you meant by context, but I provided a proof. –  Alex Gittens Jan 7 '12 at 0:49

1 Answer 1

Your alleged inequality is false. Here is a counter example.

Let $a=-0.5$, $b=1.5$ and $x=0.1$. Then, a quick calculation shows that the difference between the lhs and rhs is positive = $0.0368....$.

This gap can be made even larger as $a \to 0^-$, but not too much. This suggests that a slightly modified version of the inequality holds.

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Your link to Wolfram Alpha seems not to work (maybe due to the way MathOverflow handles links). Maybe use a URL shortener? –  Benjamin Young Jan 6 '12 at 15:55
    
Thanks Benjamin; I've fixed the URL now. –  Suvrit Jan 6 '12 at 17:10
    
Thanks for pointing that out. I revisited my proof: the rhs of the inequality needs to be divided by the minimum value of the gamma function over the positive real axis, about 0.88. This seems to fix the issue you highlighted. –  Alex Gittens Jan 6 '12 at 23:57

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