Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let

$$L(x)=Q\left(\frac{x}{2},\frac{a}{a+f(x)/\sqrt{x}}Q^{-1}\left(\frac{x}{2},1-b^{1/g(x)}\right)\right)$$

where $Q(s,x)=\frac{\Gamma(s,x)}{\Gamma(s)}$ is the upper incomplete gamma function $\Gamma(s,x)=\int_x^\infty t^{s-1}e^{-t}dt$ regularized by gamma function $\Gamma(s)$, $Q^{-1}(s,z)$ is the solution for $x$ in $z=Q(s,x)$, $a>0$ is a constant, $\frac{1}{2}<b<1$ is also a constant, and $g(x)=\omega(x)$ is a function that is asymptotically greater than $x$.

I am interested in the behavior of the limit $\lim_{x\rightarrow\infty}L(x)$ for varying asymptotics of $f(x)$. Specifically, numerical evaluations seem to indicate that when $f(x)=\omega(\sqrt{\log(x)})$, $\lim_{x\rightarrow\infty}L(x)=0$ and when $f(x)=\mathcal{O}(\sqrt{\log(x)})$, $\lim_{x\rightarrow\infty}L(x)=1$. However, I am having trouble proving this analytically (my usual Taylor expansion tricks don't seem to work here). Any help?

share|improve this question

1 Answer 1

up vote 1 down vote accepted

Below I give a sketch of a proof.

First an approximation for the inverse incomplete gamma function, $Q^{-1}$, is needed. Henceforth I assume that

$$ g(x)\sim \gamma x^s $$

for large $x$ with $s>0$. Then ($0<b<1$)

$$ 1-b^{1/g(x)}\sim (\gamma x^s)^{-1}\ln\left(\frac{1}{b}\right), $$

which is small for large $x$.

Now we use a formula due to Tricomi, which can be found here

$$ Q(\alpha+1,\alpha+\sqrt{2 \alpha} y) = \frac{1}{2} \text{erfc}(y), $$

for $\alpha\rightarrow \infty$.

Setting $\alpha=x/2$ one gets for large $x$ (dropping constant and lower terms) the following approximation

$$ Q\left(\frac{x}{2},\frac{x}{2}\left(1+\frac{2 y}{\sqrt{x}}\right)\right) \sim \frac{1}{2} \text{erfc}(y) . $$

Now we choose $y$ such that the right side is approximately equal to $(\gamma x^s)^{-1}\ln\left(\frac{1}{b}\right)$. With the help of the expansion of the inverse erfc, $\text{inverfc}$ (see e.g. here .)

$$ \text{inverfc} (y) \sim \frac{1}{\sqrt{2}} \sqrt{-\ln\left(\pi y^2 \ln \frac{1}{y}\right)} . $$

Again using the fact that $x$ is large, we find $y\sim \sqrt{s} \sqrt{\ln x}$, keeping only the highest order in $\ln x$.

This gives

$$ Q^{-1}\left(\frac{x}{2},1-b^{1/g(x)} \right) \sim \frac{x}{2}\left(1+2\sqrt{ s}\sqrt{ \frac{\ln x}{x}}\right), $$

independent of $\gamma$ and $b$.

Insertion and once more dropping terms of lower order in $x$ (assuming $f(x)/\sqrt{x}\rightarrow 0$ for $x\rightarrow \infty$) $$ L(x) \sim Q\left(\frac{x}{2},\frac{x}{2}\left[1+\frac{1}{\sqrt{x}} \left(2 \sqrt{s \ln x}- \frac{f(x)}{a}\right)\right]\right). $$

Another asymptotic formula for the incomplete gamma functions comes into play (see, e.g., here):

$$ \Gamma(z,\lambda z) \sim (\lambda z)^z e^{-\lambda z}\frac{1}{z(\lambda - 1)}, $$

For this formula to be valid $\lambda > 1$.

Using the above and Stirling's formula for the $\Gamma(z)$ with $z=x/2$ and setting

$$ \lambda = \lambda(x) := 1+\frac{1}{\sqrt{x}} \left(2 \sqrt{s \ln x}- \frac{f(x)}{a}\right). $$

after a lot of cancellations the result is

$$ L(x) \sim (2 \sqrt{s \ln x}- f(x)/a)^{-1}\rightarrow 0 $$

for $x \rightarrow \infty$.

Note that

$$ \lambda(x)^{x/2} \sim e^{x(\lambda(x)-1)/2}, $$

which cancels exactly with the exponential term in the approximation, thus leaving only the $\sqrt{x}/(\lambda(x)-1)$.

If we have $\lambda = 1$, i.e. $f(x) = 2 a \sqrt{s \ln x}$, another formula might be used (see, e.g. here)

$$ \Gamma(z,z) \sim \sqrt{\frac{\pi}{2}} z^{z-\frac{1}{2}}e^{-z} . $$

With this I find

$$ L(x)\sim Q \left(\frac{x}{2},\frac{x}{2}\right)\sim \frac{1}{2}, $$

which is in disagreement with the numerical results. One has instead to take into account next order terms.

share|improve this answer
    
Thanks for the derivation! I haven't gotten a chance to completely go through it yet, however, there are two things that I noticed right away in the beginning: 1) I am fine with the assumption of a $g(x)$ being a power function of $x$, but I think you meant $s>1$ (you have $s>0$ which I think is just a typo); 2) Are you sure that $Q(\alpha+1,\alpha+\sqrt{2\alpha}y)\approx\frac{1}{2}\operatorname{erfc}(y)$? The DLMF link you posted states that $Q(\alpha+1,\alpha+\sqrt{2\alpha}y)\approx\frac{1}{2}\operatorname{erfc}(-y)$... Does that impact the result? –  Bullmoose Aug 22 '13 at 22:31
    
Back to going through your proof and noticed that I made a mistake in my comment: indeed $Q(\alpha+1,\alpha+\sqrt{2\alpha}y)\approx \frac{1}{2}\operatorname{erfc}(y)$ since Tricomi's formula in DLMF is for the complement $P(\alpha+1,\alpha+\sqrt{2\alpha}y)=1-Q(\alpha+1,\alpha+\sqrt{2\alpha}y)$... my mistake, though point #1 in my previous comment still stands. :) Continuing reading the proof... –  Bullmoose Aug 23 '13 at 2:03
    
I finished reading the proof -- I think it's correct (that was quite a tour de force through DLMF). I think there is one more typo in the final approximation of $Q^{-1}$ (after "This gives"): I think you want the $\ln x$ under the square root sign (it's there for the remainder of the proof). Regarding your last statement, my numerical evaluations weren't perfect and indeed $L(x)$ might be going to $\frac{1}{2}$ instead of 1. Still, it looks like that there is indeed a threshold on the asymptotics of $f(x)$. Thanks! –  Bullmoose Aug 23 '13 at 2:29
    
Thank you Bullmoose for going through the proof ! I had indeed to flip a lot of DLMF pages. I corrected one typo (the ln is now under the square root). Thank you for that. . $s>0$ is no typo. The proof works perfect for all $s>0$, I guess. One just needs that $g(x)$ grows infinitely as $x \rightarrow \infty$. –  Johannes Trost Aug 23 '13 at 7:08
    
Thank you for the proof! I really like how you inverted the $Q$ function -- that's a neat trick that I'll use in the future. While the proof works as is, after a brief search I found this paper by Temme. I haven't had a chance to read it in detail, but it seems to apply to a very general range of scenarios, so I am now wondering whether your derivation of $Q^{-1}$ matches the form given in that paper (I'll try to read it carefully this coming week.) –  Bullmoose Aug 25 '13 at 5:01

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.