= Weight-two discriminant root at level eleven
{title2=$f_{11}(\tau)=(\Delta(\tau)\Delta(11\tau))^{1/12}$}
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 $q\prod_{n\geq1}(1-q^n)^2(1-q^{11n})^2$.
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