 Solution

At the tricritical point, $a_4=0$ and the zero-field Landau free energy is $f=a_2{m}^2+a_6{m}^6$. Below $T_c$, minimization givesSince $a_2\sim T-T_c$, the order-parameter critical exponent isAt this minimum, $f_{\mathrm{sing}}=a_2{m}^2+a_6{m}^6$ is proportional to $-|T-T_c{|}^{3/2}$. Comparing this with $f_{\mathrm{sing}}\sim |T-T_c{|}^{2-\alpha }$ gives the heat-capacity critical exponentAfter adding the magnetic contribution $-Bm$, the inverse zero-field magnetic susceptibility is the curvature $f^{\prime \prime }(m)$. Above $T_c$ it is $2a_2$, while below $T_c$ it is $-8a_2$, so $\chi \sim |T-T_c{|}^{-1}$ and the magnetic-susceptibility critical exponent isFinally, exactly at $T_c$ the equation of state is $B=6a_6{m}^5$. Hence $m\sim B^{1/5}$ and the critical-isotherm exponent is