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I have been working with ratios of gamma functions and I am surprised how difficult it is to make even elementary conclusions. I am hoping it is just the learning curve.

Consider the following problem. Find a positive real $c$ for an integer $x > 1$ such that:

$$\frac{\Gamma(2x+3-c)}{\Gamma(2x+1)} = x^2$$

Here's my reasoning for why I am confident that a solution exists for each integer $x$.

$\dfrac{\Gamma(2x+3)}{\Gamma(2x+1)} = \dfrac{(2x+2)!}{(2x)!}=(2x+2)(2x+1) = 4x^2 + 6x + 2 > x^2$

I am completely at a loss how to tackle what appears to me to be such a simple use of the Gamma function.

I would greatly appreciate if someone can either solve this problem for some $x > 1$ or help me to understand the methods that could be applied to this problem to provide an estimate for $c$ with, ideally, an upper and lower bound.

For the estimate, I am looking for something more interesting than the trivial:

$$0 < c < 1$$

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  • $\begingroup$ Wikipedia gives the identity $$ \Gamma(\varepsilon-n) = (-1)^{n-1} \frac{\Gamma(-\varepsilon)\Gamma(1+\varepsilon)}{\Gamma(n+1-\varepsilon)} $$In particular, this gives $$ \Gamma(2n+3-\varepsilon)= - \frac{\Gamma(-\varepsilon)\Gamma(1+\varepsilon)}{\Gamma(\varepsilon-2n-2) }, $$but I wasn't able to get anything from that. $\endgroup$ – FearfulSymmetry May 18 '20 at 0:18
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One may use Newton's method to solve $\ln \Gamma(2x+3-c) = 2\ln x + \ln \Gamma(2x+1)$ to find $c$ (see [2]):

Choose the initial $c_0$, $$c_{k+1} = c_k - \frac{\ln \Gamma(2x+3-c_k) - 2\ln x - \ln \Gamma(2x+1)}{-\Psi(2x+3-c_k)}, \ k=0, 1, 2, \cdots$$ where $\Psi(x) = (\ln \Gamma(x))'$ is the digamma function.

(Remark: $c \mapsto \ln \Gamma(2x+3-c)$ is a convex function.)

For bounds, one may use good bounds for the gamma function. For example, in the following, we give some simple bounds.

First, we deal with the case when $x > 2$. Clearly, $0 < c < 1$.

The equation is written as $$\frac{\Gamma(2x+3-c)}{\Gamma(2x+3)} = \frac{x^2}{(2x+1)(2x+2)}$$ or $$\frac{\Gamma(2x+3)}{\Gamma(2x+3-c)} = \frac{(2x+1)(2x+2)}{x^2}. \tag{1}$$

Recall Gautschi's inequality [1]: for $y > 0$ and $s \in (0, 1)$, $$y^{1-s} < \frac{\Gamma(y+1)}{\Gamma(y+s)} < (y+1)^{1-s}.$$ By letting $y = 2x + 2$ and $s = 1-c$ in Gautschi's inequality, we have $$(2x+2)^c < \frac{\Gamma(2x+3)}{\Gamma(2x+3-c)} < (2x+3)^c.\tag{2}$$ From (1) and (2), we have $$(2x+2)^c < \frac{(2x+1)(2x+2)}{x^2} < (2x+3)^c$$ which gives $$\frac{\ln (2x+1) + \ln (2x+2) - 2\ln x}{\ln (2x+3)} < c < \frac{\ln (2x+1) + \ln (2x+2) - 2\ln x}{\ln (2x+2)}.$$

Second, we deal with the case when $x = 2$. Clearly, $1 < c < 2$.

The equation is written as $$\frac{\Gamma(2x+3-c)}{\Gamma(2x+2)} = \frac{x^2}{2x+1}$$ or $$\frac{\Gamma(2x+2)}{\Gamma(2x+3-c)} = \frac{2x+1}{x^2}. \tag{3}$$

Using Gautschi's inequality, we get $$1 + \frac{\ln 5 - 2\ln 2}{\ln 6} < c < 2 - \frac{2\ln 2}{\ln 5}.$$

Reference

[1] https://en.wikipedia.org/wiki/Gautschi%27s_inequality

[2] Folitse Komla Amenyou, "Properties and Computation of the Inverse of the Gamma function".

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The unknown parameter $c$ is a function of $x$ and we can almost get its exact value.

Writing $${\Gamma(2x+3-c(x))} = x^2\,{\Gamma(2x+1)}\implies c(x)=2x+3-\Gamma^{-1}\Big[x^2\,{\Gamma(2x+1)} \Big]$$ Using this, we then have the almost exact solution $$c(x)\sim 2 x+\frac{5}{2}-\frac{e A}{W(A)}\qquad \text{with}\qquad A=\frac 1 e \log \left(\frac{x^2\, \Gamma (2 x+1)}{\sqrt{2 \pi }}\right)$$ where appears Lambert function.

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Using only numerical methods, for a given $x$, it does not make any problem to solve almost exactly for $c$ the equation $${\Gamma(2x+3-c(x))} = x^2\,{\Gamma(2x+1)}$$

A quite good empirical model seems to be $$c(x)=\alpha +\frac \beta {\gamma+\log^\delta(x)}$$ With $R^2=0.999949$ $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ a & 0.007276 & 0.001186 & \{0.004898,0.009654\} \\ b & 1.447447 & 0.021832 & \{1.403676,1.491217\} \\ c & 0.565313 & 0.026577 & \{0.512028,0.618598\} \\ d & 1.056799 & 0.010748 & \{1.035250,1.078347\} \\ \end{array}$$ which leads to very narrow bounds.

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