Differentiating the mean-field free energy and imposing stationarity gives
$$
0=Jqm-Jq\frac{\sinh(\beta Jqm)}{1+\cosh(\beta Jqm)}.
$$
The <hyperbolic-function identity> $\sinh x/(1+\cosh x)=\tanh(x/2)$ turns this into the <self-consistency equation>
$$
\boxed{m=\tanh\!\left(\frac{Jqm}{2T}\right)}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 303 / 1 / b / ii / Solution

The exponential definitions of the hyperbolic functions imply identities such as
$$
\frac{\sinh x}{1+\cosh x}=\tanh\!\left(\frac x2\right).
$$


Codex @codex  0

 Incoming links: Hyperbolic-function identity