I want to prove, that $\binom{a/m}{k}$, $a \in \mathbb{Z}, k \in \mathbb{N}, m \in \mathbb{N}_{\geq 1}$, can be rewritten as $\frac{b}{m^n}$ with $b \in \mathbb{Z}, n \in \mathbb{N}$. I thought of proving it by induction but failed at transforming $\binom{a/m}{k+1}$. An other idea was to use the binomial series as $\sum_{k=0}^\infty \binom{a/m}{k} x^k = \left(\sum_{k=0}^\infty \binom{a}{k} x^k\right)^\frac{1}{m}$ but I'm not sure how to continue at that point.

  • $\begingroup$ Where does the $n$ comes from? $\endgroup$ – mrtaurho Jun 27 at 16:02
  • $\begingroup$ Do you mean $n$ in $\frac{b}{m^n}$? This part of the identity was given in the task. $\endgroup$ – Tim Jun 27 at 16:03
  • $\begingroup$ But there is only a $k$ on the LHS (besides the $a$ and the $m$); or are you missing a summation sign? $\endgroup$ – mrtaurho Jun 27 at 16:05
  • $\begingroup$ No, there is no summation sign missing, i have to prove, that $\binom{a/m}{k}$ can be rewritten as $\frac{b}{m^n}$. $\endgroup$ – Tim Jun 27 at 16:08
  • 1
    $\begingroup$ And how are $k$ and $n$ related? $\endgroup$ – mrtaurho Jun 27 at 16:09

The given task lacks several clarifications which makes it overall impossible to solve; at least to solve at it was intented. However, I want to expand my commentary a little bit.

First, note that you are given an expression of the form $f(m,a,k)=g(m,b,n)$ but we are lacking precise informations about the nature of the relation between the unknows, especially we do not know how to rewrite $b$ and $n$ in terms of $m,a$ and $k$ (assuming that at least the LHS- $m$ and the RHS-$m$ coincide).
As I mentioned within one of my comments it may be interpretable as a given form rather than an actual closed-form, e.g. $36$ is of the form $3^n2^m$ for some positive integers $n,m$. We are not given which integers, i.e. $3^n2^m$ is not the solution but $3^22^2$ is, only the hint how the form should look like (and $3^22^2$ certainly is of the given form).

Note that there is a natural extension of the binomial coefficient using the Gamma Function (if you never heard of this function either, just think of it as an extension of the factorial, which is restricted to natural numbers, to the complex plan) given by

$$\binom xy~=~\frac{\Gamma(x+1)}{\Gamma(y+1)\Gamma(x-y+1)}~~~~~x,y\in\Bbb C\tag1$$

Also, there is a generalization of the concept of the factorial, known as Multifactorial. Where the standard factorial can be written (for natural numbers) as $n!=n(n-1)\cdots1$, the double factorial may be written as $n!!=n!^{(2)}=n(n-2)\cdots 2$ for even $n$ and $n!!=n!^{(2)}=n(n-2)\cdots 1$ for odd $n$, and so on. I guess the idea is clear now, and yet again we can extend the domain of those functions to the complex plane by invoking, again, the Gamma Function (seriously, read about this one!) to obtain

$$n!^{(\alpha)}~=~\alpha^{\frac{n-1}\alpha}\frac{\Gamma\left(\frac n\alpha+1\right)}{\Gamma\left(\frac1\alpha+1\right)}~~~~~n,\alpha\in\Bbb C\tag2$$

So, to solve your given task we may combine $(1)$ and $(2)$ which yields to

\begin{align*} \binom{\frac am}k&=\frac{\Gamma\left(\frac am+1\right)}{\Gamma(k+1)\Gamma\left(\frac am-k+1\right)}\\ &=\frac1{k!}\frac{a!^{(m)}\Gamma\left(\frac1m+1\right)m^{-\frac{a-1}m}}{(a-km)!^{(m)}\Gamma\left(\frac1m+1\right)m^{-\frac{a-km-1}m}}\\ &=\frac1{k!m^k}\frac{a!^{(m)}}{(a-km)!^{(m)}}\\ &=\frac1{m^k}\frac{a(a-m)\cdots(a-(k-1)m)}{k!} \end{align*}

$$\therefore~\binom{\frac am}k~=~\frac1{m^k}\frac{a(a-m)\cdots(a-(k-1)m)}{k!}\tag{$\star$}$$

Well, $(\star)$ is a closed-form I would say. Note that we got the $\frac1{m^n}$ factor as of the form $\frac1{m^k}$ which consolidates my theory that $k$ and $n$ are meant to be the same. However, it remains to show that the occuring fraction takes integer values (this is the mysterious $b$) for all combinations of $a,k$ and $m$ and I have absolutely no idea where to get started here (okay, to be honest I guess you can attack this one by some kind of division argument but I do not know how to execute this precisely).

As a side note: we do not need to use the Gamma Function here (even though it is a pretty nice function) but simply utilize a slithly weaker generalization of the binomial coefficient which is given by

$$\binom\alpha k~=~\frac{\alpha(\alpha-1)\cdots(\alpha-k+1)}{k!}~~~~~\alpha\in\Bbb C,~k\in\Bbb N\tag3$$

Since the denominator cosists of precisely $k$-factors we get $(\star)$ way faster by plugging in. However, I still prefer formula $(1)$ and $(2)$ since they are way more general than $(3)$ (but quite an overkill for this task).

  • $\begingroup$ Thank you very much for the explanation! I think I can prove by induction, that $\frac{\prod_{i=0}^{k-1}(a-im)}{k!} \in \mathbb{Z}$ (as $a \in \mathbb{Z}$ and $k,m \in \mathbb{N}$). $\endgroup$ – Tim Jun 27 at 19:43
  • $\begingroup$ @Tim Glad to help! Sounds like a good idea to me (and seems doable). If there is something left which you like to have explained: tell me! Otherwise, feel free to accept the answer (and maybe upvote it too) if you are satisfied :) $\endgroup$ – mrtaurho Jun 27 at 19:49

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