$$\langle X \rangle_f := f^{-1} \left(\frac{1}{n} \sum_i f(x_i)\right)$$
$$\langle X + Y \rangle_\gamma = \langle X \rangle_\gamma + \langle Y \rangle_\gamma \Leftrightarrow X \perp \!\!\! \perp Y$$
$$ \frac{1}{\gamma} \frac{\pi_i^{-\gamma}}{\sum_k \pi_k^{1-\gamma}} - \alpha_0 - \frac{\alpha_1}{\beta \gamma} \frac{e^{\gamma \beta \epsilon_i}}{\sum_k \pi_k e^{\gamma \beta \epsilon_k}} = 0$$
$$\mathcal{U}_\gamma^\beta = U_\gamma^\beta(\pi)$$
$$\mathcal{R}_\gamma^\beta = R_\gamma(\pi)$$
$$\mathcal{F}_\gamma^\beta = \mathcal{U}_\gamma^\beta - \frac{1}{\beta} \mathcal{R}_\gamma^\beta$$
$$ \mathcal{R}_\gamma^\beta = \beta^2 \, \frac{\mathcal{F}^{\beta'} - \mathcal{F}^\beta}{\beta'-\beta} $$
which is the \(\beta\) rescaling of the free energy difference.
$$ \mathcal{U}^\beta = - \left(\frac{\partial \Psi^\beta}{\partial \beta}\right)\, $$
$$\rho(\alpha,s) \mathrm{d} \alpha = c(\alpha) s^{-f(\alpha)} \mathrm{d} \alpha$$
where \(f(\alpha)\) is called as multifractal spectrum
Finally, the scaling exponents of thermodynamic potentials are
$$\beta U_\beta^{\gamma}(s) = \frac{\tau_\gamma - \tau_1}{\gamma}\ln s + \mathcal{O}(1)$$
$$\beta F_\beta^{\gamma}(s) = \frac{\tau_\gamma}{\gamma-1} \ln s + \mathcal{O}(1) $$
$$M_\gamma = \gamma U_\gamma^\beta(s) = \tau_1 - \tau_\gamma + o(1) $$
Other topics in the paper: