1
$\begingroup$

I am trying to find an upper and lower bound for the following function:

$$f(x) = \ln(\lfloor\frac{x}{b_1}\rfloor!) - \ln(\lfloor\frac{x}{b_2}\rfloor!) - \ln(\lfloor\frac{x}{b_3}\rfloor!)$$

where

$$\frac{1}{b_1} = \frac{1}{b_2} + \frac{1}{b_3}$$

Is it always the case:

$\ln\Gamma(\frac{x}{b_1}) - \ln\Gamma(\frac{x}{b_2}+\frac{1}{2}) - \ln\Gamma(\frac{x }{b_3}+\frac{1}{2}) \le \ln(\lfloor\frac{x}{b_1}\rfloor!) - \ln(\lfloor\frac{x}{b_2}\rfloor!) - \ln(\lfloor\frac{x}{b_3}\rfloor!) \le \ln\Gamma(\frac{x}{b_1}+1) - \ln\Gamma(\frac{x}{b_2} + \frac{1}{2}) - \ln\Gamma(\frac{x}{b_3}+\frac{1}{2})$

If you can show analysis behind your answer, that will really help.

My main interest is the analysis behind the answer.


Edit: I removed the attempt at a partial answer which was incorrect. I posted the partial answer as a separate question here.

The answer to my question above is no. It is true when $\{\frac{x}{b_2}\} + \{\frac{x}{b_3}\} \ge 1$.

$\endgroup$
  • $\begingroup$ $\ln(\Gamma(x+1))=-\gamma x+\int_0^x\int_0^1\frac{1-r^\phi}{1-r}dr\ d\phi$ $\endgroup$ – Simply Beautiful Art Oct 15 '16 at 0:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.