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In reviewing Ramanujan's proof of Bertrand's postulate, Ramanujan observes that:

$$\ln\Gamma(x) - 2\ln\Gamma(\frac{x+1}{2}) \le \ln(\lfloor{x}\rfloor!) - 2\ln(\lfloor\frac{x}{2}\rfloor!)$$

I have noticed that under some circumstances, the following lower bound is better.

$$\ln\Gamma(x) - 2\ln\Gamma(\frac{x+1}{2}) \le \ln\Gamma(x+1) - 2\ln\Gamma(\frac{x+2}{2}) \le \ln(\lfloor{x}\rfloor!) - 2\ln(\lfloor\frac{x}{2}\rfloor!)$$

For example, this is true when $x=34$. I'm assuming that it is not true in all cases or Ramanujan would have used the tighter lower bound in his proof.

How can I show that $\ln\Gamma(x) - 2\ln\Gamma(\frac{x+1}{2})$ is guaranteed to be lower in all cases $x \ge 2$ and identify the conditions when $\ln\Gamma(x+1) - 2\ln\Gamma(\frac{x+2}{2})$ is the better lower bound?

Here's my thinking on this question:

  • $\ln\Gamma(x-1) - 2\ln\Gamma(\frac{x}{2})$ is an increasing function (see here)
  • There are some known conditions when $\ln\Gamma(x-1) - 2\ln\Gamma(\frac{x}{2}) \le \ln\Gamma(\lfloor{x}\rfloor+1) - 2\ln\Gamma(\lfloor\frac{x}{2}\rfloor+1)$
  • My goal with this question is to show the conditions when $\ln\Gamma(x+1) - 2\ln\Gamma(\frac{x+2}{2}) \le \ln(\lfloor{x}\rfloor!) - 2\ln\Gamma(\lfloor\frac{x}{2}\rfloor!)$ and when it is not.

Thanks very much,

-Larry

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Larry Freeman
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    "I'm assuming that it is not true in all cases or Ramanujan would have used the tighter lower bound in his proof." I don't know about your example, but this isn't true in general. If a simpler approximation is sufficient for a given situation then people often don't clutter up their work with extra terms. Years ago when I suggested some miniscule corrections to an approximation, my supervisor responded "We're not trying to build a high-precision bomb, we're trying to understand something." – Douglas B. Staple Apr 15 '13 at 17:12

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You want to prove $f(x)\leq g(x) \leq h(x)$, where: \begin{align} f(x) &= \ln\Gamma(x) - 2\ln\Gamma\left(\frac{x+1}{2}\right),\\ g(x) &= \ln\Gamma(x+1) - 2\ln\Gamma\left(\frac{x+2}{2}\right),\\ h(x) &= \ln\left( \lfloor{x}\rfloor! \right) - 2\ln\left( \left\lfloor\frac{x}{2} \right\rfloor!\right). \end{align}

Firstly, this is only true for $x\in \mathbb{N}$ with $x\geq 2$. For $x=1$ we have: \begin{align} g(1) &= \ln\Gamma(2) - 2\ln\Gamma\left(3/2\right)\\ &= 2\ln(2) - \ln(\pi)\\ &= 0.24\ldots\\ h(1) &= \ln\left( 1 \right) - 2\ln\left( 1 \right)\\ &= 0, \end{align} so g(1) > h(1). For $x\in \mathbb{R}$, $g(x)>h(x)$ infinitely often.

Secondly, note that for even $x$, $g(x)=h(x)$.

Finally, for odd $x\geq 3$, you can prove $h(x)>g(x)$ by comparing term-by term, and noting that $\Gamma(x/2 + 1)$ is monotonic in this regime ($\Gamma(x)$ has a single minimum at $x=1.46\ldots$).

Comparing $f(x)$ and $g(x)$ is a bit simpler and I'll leave it out for now.

Douglas B. Staple
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    Hi Douglas, I am interested in $x \in \mathbb{R}$ where $x \ge 2$. I am especially interested under what conditions $g(x) > h(x)$ and also the reasoning to show that for all $x \in \mathbb{R}$ where $x \ge 2$, $f(x) \le h(x)$. Thanks. – Larry Freeman Apr 16 '13 at 00:39
  • @Larry In your question you say (in terms of my notation) that Ramanujan observed $f(x) \leq h(x)$ and that you want to prove $g(x) \leq h(x)$. So I'm surprised to hear you say now that you want to prove $f(x) \leq h(x)$! – Douglas B. Staple Apr 16 '13 at 00:49
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    @Larry Regarding the sign of $h(x)-g(x)$, we have $h(x)<g(x)\ \forall x \in (n,n+1)$, for every $n\geq 2$ an even integer, and $h(x)\geq g(x)\ \forall x \in [n,n+1]$ for every $n\geq 3$ an odd integer. You should plot the difference of these two functions. – Douglas B. Staple Apr 16 '13 at 00:55
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    Thanks, Douglas! That's exactly what I was looking for. I'm sorry if my interest in $f(x) \le h(x)$ wasn't clear. I tried to make this second question clear when I wrote: "How can I show that $\ln\Gamma(x)-2\ln\Gamma(\frac{x+1}{2})$ is guaranteed to be lower in all cases $x \ge 2$." – Larry Freeman Apr 16 '13 at 02:40