0

How do I expand fractions in a binomial that take one of these three forms:

$$(1)\binom{\frac{p}{q}}{n},(2)\binom{\frac{p}{q}}{\frac{z}{t}}, (3)\binom{n}{\frac{p}{q}}$$ When $t \ne 0, q \ne 0$.

I've tried tried creating an example for myself

$$\binom{-\frac{1}{2}}{2} = \frac{\left(-\frac{1}{2} \right)!}{\left(-\frac{5}{2}\right)! 2!}$$

What's the sequence of multiplication for factorials with the fraction?

I thought it would be $$\frac{\left(-\frac{1}{2} \cdot\frac{1}{2} \right)}{\left(-\frac{5}{2}\cdot-\frac{3}{2}\right) 2}$$

Which gives me the wrong answer, is there a more general way of getting the answer for the first three equations above?

@Gerry Made the distribution clear for me, here's the answer:

  1. When we have the binomial distribution as $$\binom{\frac{p}{q}}{n} = \frac{\frac{p}{q}\left(\frac{p}{q}-1 \right) \cdots \left(\frac{p}{q}-n+1 \right)}{n!}$$

  2. Alternatively, we can use the gamma distribution using the following:
    $(a) \binom{n}{k} = \frac{n!}{k!(n-k)!}$
    $(b) z! = \Gamma(z+1)$
    $(c)\Gamma(z) = (z-1)\Gamma(z-1)$

For the example, this implies $$\binom{-\frac{1}{2}}{2} = \frac{\Gamma(\frac{1}{2})}{\Gamma(3)\cdot \Gamma(-\frac{3}{2})}$$

Taking $\Gamma(\frac{1}{2}) = -\frac{1}{2}\Gamma(-\frac{1}{2}) = (-\frac{1}{2})(-\frac{3}{2})\Gamma(-\frac{3}{2}) $

me.limes
  • 393
  • 1
    ${p/q\choose n}$ is easy, it's just ${p\over q}({p\over q}-1)\cdots({p\over q}-n+1)/n!$. For the others, you'll want to learn about the Gamma function, $\Gamma(z)$, which generalizes the factorial to arguments that are not nonnegative integers. – Gerry Myerson Nov 28 '21 at 23:56
  • Any thoughts??? – Gerry Myerson Nov 30 '21 at 10:02
  • @GerryMyerson I have tried $\frac{-\frac{1}{2}(-\frac{1}{2}-1)}{2}$ which produces the output I needed! As for the gamma function with the others, is this because $\Gamma(z) = (n-1)! \implies (n-1)!=(n-1)n!$ which is pretty much what you showed me. However, $\frac{\Gamma(-\frac{1}{2})}{2!} = -\sqrt{\pi}$ so how would I get it in the right form to answer the above? – me.limes Dec 01 '21 at 10:06
  • 1
    Using ${n\choose k}={n!\over k!(n-k)!}$ and $z!=\Gamma(z+1)$ we get ${-1/2\choose2}={\Gamma(1/2)\over2\Gamma(-3/2)}$ so if you want to use Gamma to evaluate $-1/2\choose2$ you need to know $\Gamma(1/2)$ and $\Gamma(-3/2)$. – Gerry Myerson Dec 01 '21 at 11:50
  • @GerryMyerson That was helpful! I know that $\Gamma(1/2) = \sqrt{\pi}$ and $\Gamma(-3/2) = (-3/2)(-1/2)\Gamma(1/2)$ but the result shows the inverse so $(4/3)\Gamma(1/2)$? – Stackbeans Dec 01 '21 at 12:14
  • 2
    You've got it backwards. $\Gamma(1/2)=(-1/2)\Gamma(-1/2)=(-1/2)(-3/2)\Gamma(-3/2)$. – Gerry Myerson Dec 01 '21 at 12:26
  • So are we OK now? – Gerry Myerson Dec 02 '21 at 22:51
  • Your leading comments were very helpful, thanks! – me.limes Dec 03 '21 at 17:25
  • If you want to be sure I see a comment intended for me, you have to include @Gerry in it. And if you understand how to answer your question now, let me encourage you to write up a solution and post it as an answer. – Gerry Myerson Dec 04 '21 at 23:25
  • You could do that today. – Gerry Myerson Dec 06 '21 at 12:32
  • @GerryMyerson I've updated the post with our comments. I have one final question, the gamma in the numerator, why isn't this $\Gamma(1/2) = -(1/2)\Gamma(-1/2) \implies -2\Gamma(1/2) = \Gamma(-1/2)$. I'm guessing it's because we did $ \Gamma(\frac{1}{2}) =-\frac{1}{2}\Gamma(-\frac{1}{2}) = (-\frac{1}{2})(-\frac{3}{2})\Gamma(-\frac{3}{2})$ – me.limes Dec 06 '21 at 15:30
  • You can avoid the Gamma function in $\displaystyle\binom ab$ whenever $b$ or $a-b$ are integer. –  Dec 06 '21 at 15:35
  • I don't understand your question. $(-1/2)!$ is $\Gamma(1/2)$, which is $-(1/2)\Gamma(-1/2)$. We're trying to evaluate $\Gamma(1/2)\over2\Gamma(-3/2)$, so we'd like to manipulate that expression so the Gammas cancel. We can express $\Gamma(1/2)$ in terms of $\Gamma(-3/2)$, or express $\Gamma(-3/2)$ in terms of $\Gamma(1/2)$, or express both in terms of $\Gamma(-1/2)$. Whichever way we do it, we'll get the same answer. – Gerry Myerson Dec 08 '21 at 12:39

0 Answers0