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This is the second attempt at a proof. My first attempt had a flaw in its logic.

After reviewing the mistake in logic, I believe that with a revised logic, the argument can be saved.

The revision consists of two arguments. The argument presented below covers the condition where $\{\frac{x}{b_2}\} + \{\frac{x}{b_3}\} < 1$. My question here covers the condition where $\{\frac{x}{b_2}\} + \{\frac{x}{b_3}\} \ge 1$

I am very glad to be corrected if someone is able to find a mistake. :-)

This is an attempt to generalize one of the steps in Ramanujan's proof of Bertrand's postulate.

In particular, Ramanujan's does the following comparison in step (8):

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

It occurs to me that this can be generalized to:

$$\ln\Gamma(\frac{x}{b_1}-\frac{3}{16}) - \ln\Gamma(\frac{x}{b_2}+\frac{19}{32}) - \ln\Gamma(\frac{x}{b_3}+\frac{19}{32})\le \ln(\lfloor\frac{x}{b_1}\rfloor!) - \ln(\lfloor\frac{x}{b_2}\rfloor!) - \ln(\lfloor\frac{x}{b_3}\rfloor!)$$

when:

$$\frac{x}{b_1} = \frac{x}{b_2} + \frac{x}{b_3}$$

and:

$$\{\frac{x}{b_2}\} + \{\frac{x}{b_3}\} < 1$$

where:

$$x > 36$$

Here's the argument for this generalization:

Let:

$$\{\frac{x}{b_i}\} = \frac{x}{b_i} - \lfloor\frac{x}{b_i}\rfloor$$

where:

$$0 \le \{\frac{x}{b_i}\} < 1$$

Since:

$$\{\frac{x}{b_1}\} + \lfloor\frac{x}{b_1}\rfloor = \{\frac{x}{b_2}\} + \lfloor\frac{x}{b_2}\rfloor + \{\frac{x}{b_3}\} + \lfloor\frac{x}{b_3}\rfloor$$

We have:

$$\{\frac{x}{b_1}\} = \{\frac{x}{b_2}\} + \{\frac{x}{b_3}\}$$

and

$$\lfloor\frac{x}{b_1}\rfloor = \lfloor\frac{x}{b_2}\rfloor + \lfloor\frac{x}{b_3}\rfloor$$

If $\Delta{t_1} \ge \Delta{t_2} + \Delta{t_3}$ and $x_1 + \Delta{t_1} \ge x_2 + \Delta{t_2} \ge x_3 + \Delta{t_3} > 0$,

Using the logic in the answer here:

$$\frac{\Gamma(x_1 + \Delta{t_1})}{\Gamma(x_1)} \ge \frac{\Gamma(x_2 + \Delta{t_2})}{\Gamma(x_2)}\frac{\Gamma(x_3 + \Delta{t_3})}{\Gamma(x_3)}$$

Let:

$x_1 = \frac{x}{b_1}-\frac{3}{16}$, $\Delta{t_1} = \frac{19}{16}-\{\frac{x}{b_1}\}$,

$x_2 = \frac{x}{b_2}+\frac{19}{32}$, $\Delta{t_2} = \frac{13}{32}-\{\frac{x}{b_2}\}$

$x_3 = \frac{x}{b_3}+\frac{19}{32}$, $\Delta{t_3} = \frac{13}{32}-\{\frac{x}{b_3}\}$

where $\frac{x}{b_2} \ge \frac{x}{b_3}$ (Otherwise, switch the two values).

Then:

$$\frac{\Gamma(\lfloor\frac{x}{b_1}\rfloor+1)}{\Gamma(\frac{x}{b_1}-\frac{3}{16})} \ge \frac{\Gamma(\lfloor\frac{x}{b_2}\rfloor+1)}{\Gamma(\frac{x}{b_2}+\frac{19}{32})}\frac{\Gamma(\lfloor\frac{x}{b_3}\rfloor+1)}{\Gamma(\frac{x}{b_3}+\frac{19}{32})}$$

So then it follows:

$$\ln\Gamma(\lfloor\frac{x}{b_1}\rfloor + 1) - \ln\Gamma(\frac{x}{b_1} - \frac{3}{16}) \ge \ln\Gamma(\lfloor\frac{x}{b_2}\rfloor+1) - \ln\Gamma(\frac{x}{b_2}+\frac{19}{32}) + \ln\Gamma(\lfloor\frac{x}{b_3}\rfloor + 1) - \ln\Gamma(\frac{x}{b_3}+\frac{19}{32})$$

And we have shown:

$$\ln\Gamma(\frac{x}{b_1}-\frac{3}{16}) - \ln\Gamma(\frac{x}{b_2}+\frac{19}{32}) - \ln\Gamma(\frac{x}{b_3}+\frac{19}{32}) \le \ln(\lfloor\frac{x}{b_1}\rfloor!) - \ln(\lfloor\frac{x}{b_2}\rfloor!) - \ln(\lfloor\frac{x}{b_3}\rfloor!)$$

Here's an example:

Let $x=77.1$, $b_1=6$, $b_2=7$, $b_3=42$

So that:

$\{\frac{77.1}{7}\} \approx 0.0143$, $\{\frac{77.1}{42}\} \approx 0.8357$

Then:

$$\ln\Gamma(\frac{77.1}{6}-\frac{3}{16}) - \ln\Gamma(\frac{77.1}{7}+\frac{19}{32}) - \ln\Gamma(\frac{77.1}{42} + \frac{19}{32}) \le \ln(\lfloor\frac{77.1}{6}\rfloor!) - \ln(\lfloor\frac{77.1}{7}\rfloor!) - \ln(\lfloor\frac{77.1}{42}\rfloor!)$$

Please let me know if you see any mistakes.

Thanks,

-Larry


Note 1: I came up with $\frac{x}{b_1}-\frac{3}{16}$ by comparing $\frac{x}{b_1}-1$ and then comparing $\frac{x}{b_1}-\frac{1}{2}$. I found that by taking consistent averages and satisfying these conditions:

If $\Delta{t_1} \ge \Delta{t_2} + \Delta{t_3}$ and $x_1 + \Delta{t_1} \ge x_2 + \Delta{t_2} \ge x_3 + \Delta{t_3} > 0$,

I was able to get a tighter lower bound. I should be able to improve upon $\frac{x}{b_1} - \frac{3}{16}$ if I wanted to.


Note 2: I have not been able to show that this tightening of the lower bound works in all cases. This is a gap in my argument. I still need to show that using $\frac{x}{b_1}-\frac{3}{16}$ is always an improvement to using the more obvious $\frac{x}{b_3}-1$. My argument shows that it will work but not that it is better.

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up vote 1 down vote accepted

Looking at a graph it seems plausible that your result holds for $x>36$, but it does not hold for $35\le x <36$ where the right side is zero.

The flaw in this case is that $\{x/b_3\}>13/16$, allowing $\Delta t_3<0$ and $\Delta t_2>\Delta t_1$, violating the conditions for the $x_i+\Delta t_i$ inequality.

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Once again, thanks very much! Great point. I'll investigate the range of this result and update the question. I'll also work on evaluating when $\frac{x}{b_1} - \frac{3}{16}$ is a tighter lower bound than $\frac{x}{b_1} - 1$. –  Larry Freeman Apr 14 '13 at 16:49
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