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I am still struggling to build my intuition as far as reasoning with ratios of gamma functions.

Reasoning with factorials is significantly clearer.

Consider this example. I would appreciate if anyone could help me to understand how to complete the following with regard to gamma functions.

Let $n > 1$ be any integer.

Clearly:

$$\frac{(2n + 2)!}{(2n)!} = (2n+2)(2n+1) > (n+1)^2 = n^2+2n+1$$

So, changing this to a ratio of Gamma functions, the equivalent is:

$$\frac{\Gamma(2n + 3)}{\Gamma(2n+1)} = (2n+2)(2n+1) > (n+1)^2 = n^2+2n+1$$

So far, so good.

My problem comes down to evaluating when a fraction less than 1 gets applied.

For example, consider the value of $\frac{1.25506}{\ln n}$ which is less than $1$ for $n > e^{1.25506}$

While it is easy to figure out any given value and it is straight forward to generate a graph, how do I show that this value is true for $n > 800$ for example. How would I determine the derivative and show that is increasing (which I suspect it is)?

$$\frac{\Gamma(2n+ 3 - \frac{1.25506}{\ln n})}{\Gamma(2n+1)} > n^2+2n+1$$

In other words, as I leave the safety of factorials, I am at a loss for how to prove or disprove the inequality for all $n > k$ where $k > 800$ for example.


Edit: I think that the inequality may not be true for $\dfrac{5n}{3}$.

I am switching from $\dfrac{5n}{3}$ to $2n$. I believe that this inequality might be true for a reasonably sized $n$.

I believe that the inequality is true for $n=800$

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    $\begingroup$ Numerical evidence doesn't suggest the inequality is true. $\endgroup$ – FearfulSymmetry May 14 '20 at 22:10
  • $\begingroup$ Thanks Integrand. I will change it from $\frac{5n}{3}$ to $2n$ which I believe may be true for a lower $n$. For example, I believe that is true for $n=930$ $\endgroup$ – Larry Freeman May 14 '20 at 22:38
  • $\begingroup$ I start being fascinated by your problems ! $\endgroup$ – Claude Leibovici May 23 '20 at 8:08
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Suppose that we consider the function $$f(n)=\log \left(\Gamma \left(2n+3-\frac{a}{\log (n)}\right)\right)-\log (\Gamma (2 n+1))-2 \log (n+1)$$

Using Stirling approxiamtion followed by Taylor series, we have $$f(n)=-\left(\frac{a \log (2)}{\log (n)}+a-2\log (2)\right)+\frac{a^2-5 a \log (n)-2 \log ^2(n)}{4 n \log ^2(n)}+\cdots$$

Ignoring the second term leads to a lower bound $$n_{\text{low}}=2^{-\frac{a}{a-2 \log (2)}}$$ which, for the value of $a$, gives $n_{\text{low}}= 756.660$. Including the second term, Newton method converges immediatly at $n=792.720$.

Using Newton method for $f(n)=0$ with $n_0=n_{\text{low}}$, the iterates are $$\left( \begin{array}{cc} k & n_k \\ 0 & 756.6600 \\ 1 & 791.6120 \\ 2 & 792.7187 \\ 3 & 792.7197 \end{array} \right)$$

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