Weight-two discriminant root at level eleven ($f_{11}(\tau )=(\Delta (\tau )\Delta (11\tau ){)}^{1/12}$)

The effective group ${\Gamma }_0(11)$ is torsion-free, has index twelve and has two regular cusps. If its weight-two cusp form space is one-dimensional, any nonzero member has order one at each cusp and no interior zeros by the regular-cusp valence formula on a torsion-free modular curve. Its twelfth power and $\Delta (\tau )\Delta (11\tau )$ have the same weight and divisor. Their ratio is a holomorphic function without poles on the compactified modular curve, hence a nonzero constant. Normalizing the leading Fourier coefficient to one proves the displayed formula, with the holomorphic root chosen to start with $q$. It equals $q{\prod }_{n\ge 1}(1-q^n{)}^2(1-q^{11n}{)}^2$.