# Analyzing the lower bound of a logarithm of factorials using Stirling's Approximation

I am trying to get the lower bound for:

$f(x) = \ln(\lfloor\frac{x}{4}\rfloor!) - \ln(\lfloor\frac{x}{5}\rfloor!) -\ln(\lfloor\frac{x}{20}\rfloor!) - 2(1.03883)(\sqrt{\frac{x}{4}}) - (1.03883)(\frac{x}{8}) + \ln(\lfloor\frac{x}{10}\rfloor!) - \ln(\lfloor\frac{x}{12}\rfloor!) - \ln(\lfloor(\frac{x}{60}\rfloor!)$

using Stirling's approximation, $n! = \sqrt{2\pi{n}}(\frac{n}{e})^n + r_1(n)$, to show that $f(x)$ is greater than $0$ for all $x \ge 928$?

Let's define $g(x)$ as:

$g(x) = \ln(\sqrt{2\pi\frac{x}{4}}(\frac{x}{4e})^{\frac{x}{4}}) - \ln(\sqrt{2w\pi\frac{x}{5}}(\frac{x}{5e})^{\frac{x}{5}}) -\ln(\sqrt{2w\pi\frac{x}{20}}(\frac{x}{20e})^{\frac{x}{20}}) - 2(1.03883)(\sqrt{\frac{x}{4}}) - (1.03883)(\frac{x}{8}) + \ln(\sqrt{2\pi\frac{x}{10}}(\frac{x}{10e})^{\frac{x}{10}}) - \ln(\sqrt{2w\pi\frac{x}{12}}(\frac{x}{12e})^{\frac{x}{12}}) -\ln(\sqrt{2w\pi\frac{x}{60}}(\frac{x}{60e})^{\frac{x}{60}})$

where $w$ is a very large number.

Would we have established that $f(x) > 0$ for $x \ge 928$ if we show:

(a) $f(928) > 0$

(b) for $x \ge 928$, $g'(x) > 0$?

Attempted Answer: I believe that the answer to my question is no. Derivatives cannot be applied to $f(x)$ because it is not continuous. The floor function is what causes the problem. A better approach is either to analyze the size of the error term based on $x$ or define $f(x)$ in terms of the gamma function which is continuous.

Is this correct?

edit: Changed my question to make my point clearer in light of Marty's point that you need an upper bound of $\ln y$ to establish the lower bound of $\ln x - \ln y$.

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Since you want to show that the expression for $f(x)$ is positive and the expression both adds and subtracts logs of factorials, you need both an upper bound and a lower bound for the log of factorial - lower bound for the terms added and upper bound for terms subtracted.

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So the use of $w$ won't cut it? Would you have any suggestions for establishing an upper bound for $f(x)$? – Larry Freeman Apr 9 '13 at 5:33

After reading up on Stirling's Approximation, I believe that I am now able to answer my own question.

The lower bound of $f(x)$ can be established using the Stirling approximation:

$\ln(\lfloor{x}\rfloor!) = \sqrt{2\pi{x}}(\frac{x}{e})^x + R_k(x)$

where $R_1(x)$ is the error and $|R_k(x)| \le \frac{1}{12x}$

Using the above, we have:

$\ln(\lfloor\frac{x}{4}\rfloor!) - \ln(\lfloor\frac{x}{5}\rfloor!) - \ln(\lfloor\frac{x}{20}\rfloor!) = \lfloor\frac{x}{20}\rfloor\ln 20 + \lfloor\frac{x}{5}\rfloor\ln 5 - \lfloor\frac{x}{4}\rfloor\ln 4-\frac{1}{2}\ln{x} + \ln(5)- \frac{1}{2}\ln(2\pi) + R_1(\frac{x}{4}) - R_2(\frac{x}{5}) - R_3(\frac{x}{20})$

where $|R_1 + R_2 + R_3| < \frac{1}{12*\lfloor\frac{x}{4}\rfloor} + \frac{1}{12*\lfloor\frac{x}{5}\rfloor} + \frac{1}{12*\lfloor\frac{x}{20}\rfloor}$

$\ln(\lfloor\frac{x}{10}\rfloor!) - \ln(\lfloor\frac{x}{12}\rfloor!) - \ln(\lfloor\frac{x}{60}\rfloor!) =$ $\lfloor\frac{x}{60}\rfloor\ln 60 + \lfloor\frac{x}{12}\rfloor\ln 12 - \lfloor\frac{x}{10}\rfloor\ln 10 + \frac{1}{2}\ln(72) -\frac{1}{2}\ln{x} - \frac{1}{2}\ln(2\pi) + R_4(\frac{x}{10}) - R_5(\frac{x}{12}) - R_6(\frac{x}{60})$

where $|R_4 + R_5 + R_6| < \frac{1}{12*\lfloor\frac{x}{10}\rfloor} + \frac{1}{12*\lfloor\frac{x}{12}\rfloor} + \frac{1}{12*\lfloor\frac{x}{60}\rfloor}$

In this way, we have:

$f(x) = \lfloor\frac{x}{20}\rfloor\ln 20 + \lfloor\frac{x}{5}\rfloor\ln 5 - \lfloor\frac{x}{4}\rfloor\ln 4 + \lfloor\frac{x}{60}\rfloor\ln 60 + \lfloor\frac{x}{12}\rfloor\ln 12 - \lfloor\frac{x}{10}\rfloor\ln 10 - (0.12985375)x - (1.03883)\sqrt{x} -\ln(x) + \ln 5 + \frac{1}{2}\ln 72 - \ln(2\pi) + R_1(\frac{x}{4}) - R_2(\frac{x}{5}) - R_3(\frac{x}{20}) + R_4(\frac{x}{10}) - R_5(\frac{x}{12}) - R_6(\frac{x}{60})$

With:

$|R_1(\frac{x}{4}) - R_2(\frac{x}{5}) - R_3(\frac{x}{20}) + R_4(\frac{x}{10}) - R_5(\frac{x}{12}) - R_6(\frac{x}{60})| < \frac{1}{12*\lfloor\frac{x}{4}\rfloor} + \frac{1}{12*\lfloor\frac{x}{5}\rfloor} + \frac{1}{12*\lfloor\frac{x}{20}\rfloor} + \frac{1}{12*\lfloor\frac{x}{10}\rfloor} + \frac{1}{12*\lfloor\frac{x}{12}\rfloor} + \frac{1}{12*\lfloor\frac{x}{60}\rfloor} \le \frac{1}{12*41*\lfloor\frac{x}{60}\rfloor} \le \frac{1}{492*\frac{x-59}{60}}$

So, $f(x) > x(\frac{1}{20}\ln 20 + \frac{1}{5}\ln 5 - \frac{1}{4}\ln 4 + \frac{1}{60}\ln 60 + \frac{1}{12}\ln 12 - \frac{1}{10}\ln 10 - 0.12985375) - (\frac{19}{20}\ln 20 + \frac{4}{5}\ln 5 + \frac{59}{60}\ln 60 + \frac{11}{12}\ln 12) - (1.03883)\sqrt{x} -\ln(x) + \ln 5 + \frac{1}{2}\ln 72 - \ln(2\pi) + R_1(\frac{x}{4}) - R_2(\frac{x}{5}) - R_3(\frac{x}{20}) + R_4(\frac{x}{10}) - R_5(\frac{x}{12}) - R_6(\frac{x}{60})$

So, by my calculations:

$(0.0403029768)x - (1.03883)\sqrt{x} > (\frac{19}{20}\ln 20 + \frac{4}{5}\ln 5 + \frac{59}{60}\ln 60 + \frac{11}{12}\ln 12) + \ln(2\pi) + \ln(x) -\ln(5) -\frac{1}{2}\ln(72) + \frac{1}{492} = 8.529571186 + \ln(x)$

for $x \ge 1331$

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