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\begin{equation}f(N)=\frac{\frac{L^2}{y-L}(1-\frac{L}{Nx})+\frac{NL}{N-M}(1-\frac{L}{Nx})\sum\limits_{k=0}^{M}\frac{\binom{N}{k}}{\binom{N}{M}}\left(\frac{L}{Nx-L}\right)^{k-M}}{\frac{L}{N(y-L)}\left[\frac{y}{y-L}(1-\frac{L}{Nx})+M\right]+\frac{1}{N-M}\sum\limits_{k=0}^{M}\frac{\binom{N}{k}}{\binom{N}{M}}k\left(\frac{L}{Nx-L}\right)^{k-M}}\end{equation}

Any suggestion on how to calculate its limit as the positive integer $N\to\infty$? With given positive $M,L,x,y$. Here $Nx>L$, $y>L$, the integer $M\le N$, and $k$ is integer.

I tried the transformation on the combinatorial number as follow, but is didn't seem help.

\begin{equation}f(N)=\frac{\sum\limits_{k=0}^{M}\frac{\binom{M}{k}}{\binom{N-k}{N-M}}\left(\frac{L}{Nx-L}\right)^{k-M}\frac{NL}{N-M}(1-\frac{L}{Nx})+\frac{L^2}{y-L}(1-\frac{L}{Nx})}{\sum\limits_{k=0}^{M}\frac{\binom{M}{k}}{\binom{N-k}{N-M}}\left(\frac{L}{Nx-L}\right)^{k-M}\frac{k}{N-M}+\frac{L}{N(y-L)}\left[\frac{y}{y-L}(1-\frac{L}{Nx})+M\right]}\end{equation}

Numerical result shows it increases sharply as N increases first, and then very slowly, and seems like approaching some constant value. It doesn't seem like the limit will be infinity.

Thanks for any suggestion.

Bob
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    Where does this function come from? – anomaly Aug 19 '15 at 21:09
  • This is the problem I met when I try to do performance analysis with the background of wireless networking. It's like as more resource/station $N$ used, whether there is performance upper bound. – Bob Aug 19 '15 at 21:21
  • In the numerator/denominator- does the summation range over the whole term? In any case, it seems like you should be able to simplify this a lot because the terms on the right hand side seem constant or inverse proportional to $N$ so they should become negligible as $N$ grows large. – Colm Bhandal Aug 19 '15 at 22:39
  • The summation only ranges over the terms related to $k$, and $f(N)$ has been rewritten to avoid the confusion. – Bob Aug 19 '15 at 22:55

1 Answers1

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$$\sum_{k=0}^M\frac{N\choose k}{N\choose M}\left(\frac{L}{Nx-L}\right)^{k-M}\\ \approx\sum_{k=0}^M\frac{N^k/k!}{N^M/M!}\left(\frac L{Nx-L}\right)^{k-M}\\ \approx\sum_{k=0}^M\frac{M!}{k!}\left(\frac L{x-L/N}\right)^{k-M}\\ \approx \sum_{k=0}^M\frac{M!}{k!}\left(\frac Lx\right)^{k-M}\\ = M!\left(\frac xL\right)^M\sum_{k=0}^M\frac {(L/x)^k}{k!}$$ Similarly, in the denominator $$\sum_{k=0}^M\frac{N\choose k}{N\choose M}k\left(\frac L{Nx-L}\right)^{k-M}\approx M!\left(\frac xL\right)^{M-1}\sum_{k=0}^{M-1}\frac {(L/x)^k}{k!}$$ So $$f(N)\approx N\frac{\frac{L^2}{y-L}+LM!\left(\frac xL\right)^M\sum_{k=0}^M\frac {(L/x)^k}{k!}}{\frac{L}{(y-L)}\left[\frac{y}{y-L}+M\right]+M!\left(\frac xL\right)^{M-1}\sum_{k=0}^{M-1}\frac {(L/x)^k}{k!}}$$ If $M$ is much bigger than $L/x$ then you can replace the finite sums in $k$ by $\exp(L/x)$

Empy2
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