Mathematics Asked by hallowdance on December 13, 2021
I’m working through a statistical physics book and one of the problems makes the claim that the quantity:
$$H =frac{1}{N}sum_{n=1}^Nfrac{1}{frac{a}{N} + frac{b}{N^{5/3}}(n^{5/3}-(n-1)^{5/3})}$$
can be expressed in the large $N$ limit as the integral:
$$H = Nint_0^1 frac{text{d}u}{a+frac{5}{3}b u^{2/3}}.$$
However, I am unable to arrive at that result. I’ve tried to use LRAM but I got a different result. Any tips will be appreciated.
Observe that, for $n$ large, $$ n^{5/3} - (n-1)^{5/3} = n^{5/3} left[1- left(1 - frac{1}{n}right)^{5/3}right] sim n^{5/3} frac{5}{3n} = frac{5}{3} n^{2/3} $$ Using the above approximation (even if, in my opinion, in this case it is not fully justifiable), you can write $H$ as $$ H sim N cdot sum_{n=1}^N frac{1}{N} cdot frac{1}{a + frac{5}{3} b left(frac{n}{N}right)^{2/3}},. $$ Finally, you can use the fact that, if $f$ is a continuous function, then $$ sum_{n=1}^N frac{1}{N} fleft(frac{n}{N}right) to int_0^1 f(u), du, $$ as $Nto +infty$.
Answered by Rigel on December 13, 2021
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