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Let $k, n ,m \in N$ and such that $0\leq k \leq n \leq m$. When the following ineuality is true? $$ \frac{2^{m-k}\Gamma(n+1)\Gamma\left(\left[\frac{m+1-k}{2}\right]\right)\Gamma(m+1-n)}{\Gamma(m+1)\Gamma\left(\left[\frac{m+2-k}{2}\right]\right)}\leq \sqrt{\pi} $$

Thank you for your help.

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    $\begingroup$ If $m-k$ is odd then that inequality is equivalent to $$k \geq m - \log_2 \left[\sqrt{\pi} \binom{m}{n} \right].$$ Is this the kind of thing you're looking for? $\endgroup$ – Antonio Vargas Dec 11 '12 at 0:56
  • $\begingroup$ Thank you, but I wanted the result for any $k, n$ and $m$. And also I would like to get condition only in terms of $n$ and $m$. $\endgroup$ – Michael Dec 11 '12 at 1:13
  • $\begingroup$ The other case, when $m-k$ is even, is equivalent to $$k \geq m+1 - \log_2\left[\sqrt{\pi}(m-k)\binom{m}{n}\right].$$ This second case is satisfied when, e.g., $$k \geq m+1 - \log_2\left[\sqrt{\pi}\binom{m}{n}\right],$$ which removes the $k$ dependence in the right-hand side. I do not see a way to completely remove the $k$ dependence, and I'm interested to see if someone else does. $\endgroup$ – Antonio Vargas Dec 11 '12 at 3:07
  • $\begingroup$ @Antonio Vargas : Could you please share your solution, thank you. $\endgroup$ – Michael Dec 13 '12 at 3:28
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I'm assuming that "$[\cdots]$" is the floor function, sometimes written "$\lfloor \cdots \rfloor$".

Since $m$ and $n$ are integers we have

$$\Gamma(n+1) = n! \qquad \Gamma(m+1) = m! \qquad \Gamma(m+1-n) = (m-n)!$$

This allows us to rewrite the left-hand side of the inequality as

$$ 2^{m-k} \frac{\Gamma\left(\left[\frac{m+1-k}{2}\right]\right)}{\Gamma\left(\left[\frac{m+2-k}{2}\right]\right)} \cdot \frac{n!(m-n)!}{m!} = 2^{m-k} \frac{\Gamma\left(\left[\frac{m+1-k}{2}\right]\right)}{\Gamma\left(\left[\frac{m+2-k}{2}\right]\right)} \cdot \frac{1}{\binom{m}{n}}. $$

Now if $m-k$ is even then $m+2-k$ is even, which implies that $(m+2-k)/2$ is an integer and

$$ \left[\frac{m+2-k}{2}\right] = \frac{m+2-k}{2} = \frac{m-k}{2} + 1. $$

This tells us that

$$ \Gamma\left(\left[\frac{m+2-k}{2}\right]\right) = \Gamma\left(\frac{m-k}{2} + 1\right) = \frac{m-k}{2} \,\Gamma\left(\frac{m-k}{2}\right). $$

Then since

$$ \frac{m+1-k}{2} = \frac{m+2-k}{2} - \frac{1}{2}, $$

we have

$$ \left[\frac{m+1-k}{2}\right] = \frac{m+2-k}{2} - 1 = \frac{m-k}{2}, $$

which gives us

$$ \Gamma\left(\left[\frac{m+1-k}{2}\right]\right) = \Gamma\left(\frac{m-k}{2}\right). $$

Some cancellation happens in the inequality and we are left with

$$ \frac{2^{m+1-k}}{(m-k)\binom{m}{n}} \leq \sqrt{\pi}. $$

After multiplying by the denominator and taking logarithms this is equivalent to

$$ k \geq m+1 - \log_2\left[\sqrt{\pi}(m-k)\binom{m}{n}\right] \qquad \text{if } m-k \text{ is even.} $$

If $m-k$ is odd then similar considerations reveal that

$$ \Gamma\left(\left[\frac{m+1-k}{2}\right]\right) = \Gamma\left(\left[\frac{m+2-k}{2}\right]\right), $$

which leaves the inequality in the form

$$ \frac{2^{m-k}}{\binom{m}{n}} \leq \sqrt{\pi}. $$

By again multiplying by the denominator and taking logarithms we see that this is equivalent to

$$ k \geq m - \log_2\left[\sqrt{\pi}\binom{m}{n}\right] \qquad \text{if } m-k \text{ is odd.} $$

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