Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 23 / 5 / c

Use the product printed in the original PDF, with factors $(1-q^n{)}^2$ and $(1-q^{11n}{)}^2$. The Fricke involution normalizes ${\Gamma }_0(11)$ and preserves its modular cusp space. Therefore $g=-11^{-1}{\tau }^{-2}f(-1/(11\tau ))$ lies in the given one-dimensional space, so $g=cf$ for a constant $c$.

At the Fricke fixed point ${\tau }_{\ast }=i/\sqrt{11}$, the prefactor $-11^{-1}{\tau }_{\ast }^{-2}$ is one. Thus $g({\tau }_{\ast })=f({\tau }_{\ast })$. The product has $0<q_{\ast }<1$, every factor is positive, and its limit is nonzero since ${\sum }_n{q}_{\ast }^n+{\sum }_n{q}_{\ast }^{11n}<\infty$. Hence $f({\tau }_{\ast })>0$, forcing $c=1$. This Fricke sign from a nonvanishing fixed-point value provesPart (b) now gives $\Lambda (f,s)=\Lambda (f,2-s)$. Its Taylor series at one contains only even powers, so its order of vanishing is even. The function is not identically zero, since its first Fourier coefficient is one. In this example the product is positive on the entire positive imaginary axis, and its Mellin integral at $s=1$ is positive. Thus the stronger conclusion isThe TeX aid duplicates and corrupts the product in this part; neither corrupted expression is used.