I'm having a little bit of trouble trying to get my head around the following problem.
I have this main function
$$S_n(\alpha, \beta) = \dfrac{dV}{V} \dfrac{\alpha}{d\alpha} = \dfrac{4 \alpha \beta^n}{(1 + \alpha)^2 - (1-\alpha)^2\beta^{2 n}}$$
The author explains that "For all values of $\alpha$, the logarithm of $S$ is, asymptotically for small $\beta$, linearly related to the logarithm of $\beta$. Thus, the slope can be determined by:
$$\dfrac{d \ln S}{d ln \beta}$$
From my understanding, I approached this problem by substituting $ln(x)$ using
$$\dfrac{d \ln(x)}{dx} = 1/ x \Rightarrow d \ln (x) = \dfrac{dx}{x} $$
Therefore, the slope would be obtained using:
$$ \dfrac{dS}{d \beta} \dfrac{\beta}{S}$$
But I'm not sure if this approach is correct, as so far it hasn't got me close to the results of the author.
Any suggestion is highly appreciated.
For $\beta$ small we have
$$S_n(\alpha, \beta) = \dfrac{4 \alpha \beta^n}{ (1 + \alpha)^2 - (1-\alpha)^2\beta^{2 n}} \sim\frac{4\alpha}{(1 + \alpha)^2}\beta^n $$
then
$$\log S\sim n\log \beta+\log \left(\frac{4\alpha}{(1 + \alpha)^2 }\right)$$
and the slope should be $n$.