Any ideas on finding a good estimate/approximation for $\frac AB$ where $A = N^L$ and $B = {N+L\choose N}$?
$N^L$ vs. ${N+L\choose N}$
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combinatorics
binomial-coefficients
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2You could try applying Stirling on the factorials implicit in the binomial coefficient... – 2011-05-02
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1I don't understand the notations $N^L$ and $C_{N+L}^N$. What do those mean? – 2011-05-02
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0@Mitch: I am taking $N^L$ as the exponential and $C_{N+L}^N$ as the binomial coefficient of $N+L$ choose $N$ – 2011-05-02
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1In what regime? If $L$ is fixed and $N$ is allowed to grow then the ratio approaches $L!$. – 2011-05-02
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0Wow, just the slightest change in notation (from lower case to upper) made me misunderstand. That's not a problem with the notation, but a problem with my reading ability. – 2011-05-02
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0Well actually $N$ and $L$ are fixed and less than 100. Thanks for pointing to Stirling's approximation, though. – 2011-05-02
3 Answers
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If you expand $B$ as $\frac{(N+L)!}{N!L!}$ and then use Stirling's approximation on the factorials, you will be very close.
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$\log(A/B) = \log(L!) - \sum_{j=1}^L \log(1+j/N)$. You can approximate or bound the sum in various ways, depending on your needs.
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I'm assuming that by $C_{N+L}^N$ you mean the binomial coefficient $(N+L)!/N!L!$. (I would denote this by ${N+L \choose N}$.)
If you're thinking of $L$ as a constant then you can write this as
$$ {(N+L)(N+L-1) \cdots (N+1) \over L!}. $$
And you can expand out the numerator; you get
$$ {(N^L + {L(L+1) \over 2} N^{L-1} + \cdots) \over L!} $$
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0Thank you, this is a good hint as well. – 2011-05-02