Asymptotic approximation of the expression $\sum^{N}_{i=0}\log\Bigg[\binom{\binom{N+1}{i}}{t_i}\Bigg]$

Question

I am wondering about the asymptotic approximation of the following expression: $$S=\sum^{N}_{i=0}\log\Bigg[\binom{\binom{N+1}{i}}{t_i}\Bigg]$$ where $$t_i=\binom{N}{i}-\binom{N-k}{i-k}+\binom{N-k}{i-1}$$ where $k$ is a positive integer. Also we have that $k\ll N$. Also for those binomials that have $i<k$ (namely negative) we count them as zero. I am trying to work out the approximation for $N \rightarrow \infty$.

Answer 1

Here are the asymptotics and a method for $k=1$ through $k=4.$ Unfortunately there does not appear to be a simple pattern developing. Also, I'm using $n=N-1.$

$\textbf{k=1} $ $$S_1:=\sum_{m=0}^n \log{ \Big[ \binom{\binom{n}{m}}{ \binom{n-1}{m}} }\Big] = \sum_{m=1}^{n-1} \log{\Big[ \binom{\binom{n}{m}}{ \binom{n}{m}(1-m/n) } }\Big] $$ The sum starting and ending value has been shifted by 1 towards the center because those end values within the square bracket are 1, and $\log(1)=0.$ Use the 'central value' estimate for the binomials, $$ \binom{n}{m} \sim \binom{n}{n/2} e^{-\tfrac{2}{n}\big(\tfrac{n}{2}-m\big)^2 } .$$ Note: I haven't justified that this is sufficient other than numerically. Use the asymptotic expansion for $\log{\Gamma(1+x)},$ $$ \log{\Gamma(1+x)} \sim x\log{(x/e)} + \log{(\sqrt{2\pi x})} + ...$$ Only the first term of the previous will be used when estimating $S_1.$ In thefollowing let $a=\binom{n}{m}$ and $y=m/n.$ Then $$ S_1 \sim\binom{n}{n/2} \sum_{m=1}^{n-1} e^{-\tfrac{2}{n}\big(\tfrac{n}{2}-m\big)^2 } \big\{ \log{(a/e)}-(1-y)\log{(a\,(1-y)/e)}-y\log{(a \,y /e)} \big\}$$ The $\{\cdot\}$ simplifies so that it is independent of $a,$ $$ \{\cdot\} = H(y):=-\big(y\,\log{y} + (1-y)\log{(1-y)} \big) $$ Therefore $$ S_1 \sim\binom{n}{n/2} \sum_{m=1}^{n-1} e^{-2n\big(\tfrac{m}{n}-\tfrac{1}{2}\big)^2 } H(m/n)$$ In the limit that $n$ is large, interpret the previous expression as a Riemann integral: $$ S_1\sim\binom{n}{n/2} \, n\, \int_0^1 e^{-2n\big(u-\tfrac{1}{2}\big)^2 } H(u) \,du = \binom{n}{n/2} \, n\, \int_{-1/2}^{1/2} e^{-2n\, u^2} H(w_1(u)) \,du$$ where $w_1(u) = 1/2 + u.$ One may expand $H(w_1(u))$ in a power series about $u=0,$ extend the limits on the integral to $ \pm \infty $ so to use Gaussian integrals to get a close form, and expand the leading binomial and $n$ factor to finally get $$S_1 \sim 2^n\big(\log{2} - \frac{\log{2}+2}{4n} + o(1/n) \big) $$
For the other results, a polynomial associated with the index $k$ has been determined. That is, the results will be expressed as $$\quad (A) \quad S_k\sim\binom{n}{n/2} \, n\, \int_{-1/2}^{1/2} e^{-2n\, u^2} H(w_k(u)) \,du$$ $\textbf{k=2:} \quad w_2=\frac{1}{2}+\frac{3}{2}u - 2 u^3 $

$\textbf{k=3:} \quad w_2=\frac{1}{2}+\frac{3}{2}u - 2 u^3 $

$\textbf{k=4:} \quad w_2=\frac{1}{2}+\frac{11}{8}u - u^3 - 2u^5 $

That the $k=2$ is the same as the $k=3$ case is not a typo. There are three sources of error so far unquantified: the central binomial approximation, the dropping of subsequent terms in the $\log{\Gamma}$ expansion, and the error in going from a sum to the integral. Another error will appear when $k$ gets appreciably large with respect to $n:$ One continues to 'pinch' the ends of the integral and the simple Riemann integrals presented will need corrections. This approach may completely fall apart if $k = \alpha n$, with, say, $\alpha \ge 1/10.$

The following is a comparison of values as calculated by the original definition, and by eq. (A). Multiply the entry by the scientific notation designator indicated in the table.

$$ \begin{array}{c|lcr} k & S_{16} & S_{16}^{A} & S_{100} & S_{100}^{A} \\ {} & \quad \cdot 10^4 & & & \cdot 10^{29} \\ \hline 1 & 4.325 & 4.263 & 8.723 & 8.701 \\ 2 & 4.040 & 4.045 & 8.642 & 8.624 \\ 3 & 4.038 & 4.045 & 8.642 & 8.624 \\ 4 & 4.088 & 4.098 & 8.666 & 8.645 \end{array} $$ To me it appears that the values depend weakly on $k$ and an insistence that the asymptotic formula is accurate for these distinctions means that those errors mentioned previously need to be dealt with.