I came across these numbers in my work some time ago. This type of expressions do not exist in closed form (not to confuse with Vandermonde convolution), I already know that. To simplify I denote $$P(r;s;k)=\binom{\frac{n}{2}+r}{k} \binom{\frac{n}{2}-r}{k+s}$$

Here is why I think these numbers are interesting: anyone can see combinatorial structures in algorithms or elsewhere, but these expressions converge.

Experiment 1, $r=0, s=0:$ There are $n$ fair coins, of them half are heads $H$, half are tails $T$. Each coin is flipped w.p. $\frac{1}{n}$, i.e. proportional to size of the sample. The question is, after the experiment is over, what is the probability that the ratio of $H$/$T$ remains the same, i.e. half and half. The tricky thing is, in order to preserve the ratio, we need to flip an $\mathit{even}$ number of coins, or, more specifically, an equal number of $H$ and $T$ (obviously two $H$ or two $T$ won't do): either 0 $H$ and 0 $T$, or 1 $H$ and 1 $T$ and so on till $\frac{n}{2} \ H, \frac{n}{2} \ T$. The probability to preserve the ratio (and the first such number) is therefore $$ h_{0,0}= \lim _{n \to \infty} \sum_{k=0}^{\frac{n}{2}} P(0;0;k) \frac{1}{n^{2k}} \bigg(1-\frac{1}{n}\bigg)^{n-2k} \approx 0.465267 $$

Experiment 2, $r=0,s=1$: same setup as before, but now I need to obtain $\textit{exactly}$ one more $H$ coin as a result of the experiment (or $T$, it does not matter by symmetry). To do this, I need to flip $exactly$ one more $T$ than $H$, hence we need to flip an $odd$ number of coins:

$$ h_{0,1}= \lim _{n \to \infty} \sum_{k=0}^{\frac{n}{2}-1} P(0;1;k) \frac{1}{n^{2k+1}} \bigg(1-\frac{1}{n}\bigg)^{n-2k-1} \approx 0.208912 $$

Experiment 3, $r=1,s=0$. This comes directly after experiment 2: we have $\frac{n}{2}+1 \ H$ and $\frac{n}{2}-1 \ T$. We want the same result as in Experiment 1: maintain the current proportion of $H$ and $T$ (i.e. flip an even number of coins), but clearly this time around the upper bound on the summation is $\frac{n}{2}-1$.

$$ h_{1,0}= \lim _{n \to \infty} \sum_{k=0}^{\frac{n}{2}-1} P(1;0;k) \frac{1}{n^{2k}} \bigg(1-\frac{1}{n}\bigg)^{n-2k} \approx 0.465225 $$

and so on for $h_{r,s}, 0 \leq r,s \leq n$. So here is what I'd like to know:

1) Have these numbers arisen in some other context (I never ended up publishing a paper)

2) Is there some rigorous way to prove these numbers are irrational? I've shown the $\lim$ part simply by taking large $n$, so I guess this not rigorous enough.

3) If there is something I've missed, I'd like to know too.

Thanks for the suggestions.

  • $\begingroup$ I'm getting $\approx 0.46576$ for $h_{0,0}$. That's close to your value but not exactly the same. Could my number be correct, or did I make a mistake somewhere? If it's correct, I'll type up my answer sometime in the next day or two. Unfortunately, I don't have time right now. $\endgroup$ – Mike Spivey Jan 13 '13 at 7:08
  • $\begingroup$ I mentioned that my proof is purely numerical and is probably not rigorous, hence different values for $h$. Is the question too silly/unreasonable as it didn't generate too much interest? $\endgroup$ – Alex Jan 13 '13 at 16:07
  • 2
    $\begingroup$ It's really hard to make sense sometimes of how this community responds to questions and answers. I've been surprised many times at which of my questions and answers get a lot of attention and which don't. At any rate, I think your question is interesting (and hard). Also, I will post my answer at some point. We'll see if I have time this afternoon to type it up. $\endgroup$ – Mike Spivey Jan 13 '13 at 20:53
  • $\begingroup$ Thanks Mike. Please do give your answer. Do you think posting this question on MO would make sense? $\endgroup$ – Alex Jan 13 '13 at 20:55
  • $\begingroup$ I would wait a few more days. I think there are some really good mathematicians who browse the unanswered questions here every few days or so, so you might yet get a complete answer. $\endgroup$ – Mike Spivey Jan 13 '13 at 20:59

I find $$h_{0,0} = \frac{I_0(1)}{e} \approx 0.4657596,$$ where $I_0(x)$ is a modified Bessel function. That answers your question about $h_{0,0}$ arising in some other context.

As far as the irrationality of $h_{0,0}$, of course $1/e$ is transcendental. The value of $I_0(1)$ is irrational as well (see argument at end of post). Since irrational times irrational might be rational, though, this doesn't completely answer your question about $h_{0,0}$.

I'm not sure about $h_{0,1}$ and $h_{1,0}$. Perhaps my argument below could be modified.

Proof that $h_{0,0} = I_0(1)/e$.

Since the total number of coins must be even, let's consider $2n$ coins. Then $$h_{0,0} = \lim_{n \to \infty} \sum_{k=0}^n \binom{n}{k} \binom{n}{k} \left(\frac{1}{2n}\right)^{2k} \left(1 - \frac{1}{2n}\right)^{2n-2k}.$$ With a little algebra, the summation can be simplified to $$\left(1 - \frac{1}{2n}\right)^{2n} \sum_{k=0}^n \binom{n}{k}^2 \frac{1}{(2n-1)^{2k}}.$$ Now, $\sum_{k=0}^n \binom{n}{k}^2 x^k$ can be expressed in terms of Legendre polynomials $P_n$. (See, for instance, my answer here.) With $x = 1/(2n-1)^2$, we have $$\sum_{k=0}^n \binom{n}{k}^2 x^k = (1-x)^n P_n\left(\frac{1+x}{1-x}\right) = \left(1-\frac{1}{(2n-1)^2}\right)^n P_n\left(1 + \frac{1}{2n^2-2n}\right).$$

Since $\left(1 - \frac{1}{2n}\right)^{2n} \to 1/e$ and $\left(1-\frac{1}{(2n-1)^2}\right)^n \to 1$ as $n \to \infty$, we're left with determining the asymptotics of the Legendre polynomial expression.

According to the DLMF, for $x > 1$, and $n \to \infty$, $$P_n(x) \sim \frac{\cosh^{-1} x}{\sinh(\cosh^{-1}x)} I_0\left((n+1/2)\cosh^{-1}(x)\right).$$

The asymptotic calculations from here are a bit painful. Essentially we need to use $\cosh^{-1}(x) = \log(x+ \sqrt{x^2-1})$ (see, for instance, here) and $\sinh(\cosh^{-1}x) = \sqrt{x^2-1}$. We get $\displaystyle \frac{\cosh^{-1} y}{\sinh(\cosh^{-1}y)} \to 1$ and $(n+1/2)\cosh^{-1}(y) \to 1$ as $n \to \infty$ for $y = 1 + \frac{1}{2n^2-2n}$.

This establishes $$h_{0,0} = \frac{I_0(1)}{e},$$ as I mention as the top of the post.

Proof that $I_0(1)$ is irrational.

The modified Bessel function $I_0(x)$ has the nice representation $$I_0(x) = \sum_{k=0}^{\infty} \frac{x^{2k}}{k!k!4^k}.$$ Thus $$I_0(1) = \sum_{k=0}^{\infty} \frac{1}{k!k!4^k}.$$ To prove that the value of this infinite sum is irrational, you can mimic Sondow's proof of the irrationality of $e$. Start with the interval $[1,2]$. At step $k$, divide the current interval into $4k^2$ equal subintervals and take the second subinterval each time. This creates a sequence of nested intervals such that $\displaystyle \frac{a_k}{k!k!4^k} < I_0(1) < \frac{a_k+1}{k!k!4^k}$ and $a_k$ is an integer. This proves that $I_0(1)$ cannot be represented in the form $\displaystyle \frac{r}{k!k!4^k}$ for any integer $r$. However, every rational number can be expressed in the form $\displaystyle \frac{r}{k!k!4^k}$ for some integer $r$. For instance, $\displaystyle \frac{p}{q} = \frac{p (q-1)! q! 4^q}{q!q!4^q}$. Therefore, $I_0(1)$ is irrational.

  • $\begingroup$ You might also note that according to Maple $$I_0(x)/e^x = \sum _{k=0}^{\infty }{\frac { \left( -1 \right) ^{k} \left( 2\,k \right) !\,{2}^{-k}{x}^{k}}{ \left( k! \right) ^{3}}} $$ I don't know if that will help in proving that this is irrational for $x=1$. $\endgroup$ – Robert Israel Jan 14 '13 at 4:21
  • $\begingroup$ Wow, that's a great answer! I knew about the representation using Legendre polynomial, but the rest is new. One more thing I'm thinking is something with the general expression for $h(r,s)$. I still may put it on MO in case someone gets interested. $\endgroup$ – Alex Jan 14 '13 at 4:22
  • $\begingroup$ @RobertIsrael: Thanks! I'll have to take a look at that. $\endgroup$ – Mike Spivey Jan 14 '13 at 4:47

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