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The product defining the modular discriminant converges locally uniformly for $|q|<1$ and has integral Fourier coefficients. Its logarithmic derivative isTherefore $D\Delta =E_2\Delta$. Comparing the coefficient of $q^n$ gives $n\tau (n)=\tau (n)-24{\sum }_{j=1}^{n-1}{\sigma }_1(j)\tau (n-j)$, orThis also fixes the initial value $\tau (1)=1$.

For the Ramanujan tau congruence modulo five, first justify the link between this product and $E_4,E_6$. Question 1 constructs ${\Delta }_0=(E_4^3-E_6^2)/1728\in S_{12}$ with first coefficient one. Its Serre derivative is in $S_{14}$, which vanishes by the valence formula for the modular group: a nonzero such cusp form would contribute at least $1+1/2>14/12$, since it vanishes at infinity and at $i$. Hence $D{\Delta }_0=E_2{\Delta }_0$. The product and ${\Delta }_0$ have the same differential equation and leading coefficient. Equivalently their quotient has derivative zero near $q=0$ and limit one; the identity theorem givesAlso $D_6{E}_6\in M_8=\mathbb{C}{E}_4^2$. Its constant coefficient is $-1/2$, so the Serre derivative gives $2D{E}_6=E_2{E}_6-E_4^2$.

Work now in the formal power series ring ${\mathbb{F}}_5[[q]]$. The integral expansions give $E_4\equiv 1$ and $E_2\equiv E_6(\mathrm{mod}5)$, because $d^5\equiv d(\mathrm{mod}5)$, $-24\equiv -504\equiv 1$, and $240\equiv 0$. The two identities reduce toThus $3\Delta \equiv -2D{E}_6\equiv 3D{E}_6$, and division by the nonzero element $3$ gives $\Delta \equiv D{E}_6$. Since the coefficient of $q^n$ in $D{E}_6$ is $-504n{\sigma }_5(n)$, we conclude

The required anomaly cancellation depends on the exact Eisenstein series of weight two lawHere is a way to establish it without assuming that $E_2$ is a modular form. The lattice definition of the completed nonholomorphic Eisenstein series in Question 6 is modular invariant. Its Poisson summation calculation, proved there, gives the finite Laurent coefficient at $s=1$ asThe residue is constant, so $C$ is also invariant. This calculation uses only convergent Gaussian integrals and the analytic infinite product, not modularity of that product. Using the Wirtinger derivative ${\partial }_{\tau }$, with ${\partial }_{\tau }y=1/(2i)$, differentiation of the locally uniformly convergent series givesDifferentiate $C(\gamma \tau )=C(\tau )$ to see that $E_2^{\ast }=E_2-3/(\pi y)$ transforms with weight two. Substituting $\Im (\gamma \tau )=y/|j{|}^2$ yieldssince $\overline{j}-j=-2icy$. This is the weight-two transformation from an invariant Eisenstein limit.

Combine this law with the derivative transformation of a weak modular form proved in part (i). The anomalous terms in $Df-(k/12)E_{2}f$ cancel exactly, leaving the weight-$k+2$ transformation law. Both functions are holomorphic on the complex upper half-plane, and their Fourier expansions contain no negative powers. Hence the Serre derivative satisfiesIts constant coefficient is $-k{a}_0(f)/12$; it need not be a cusp form when $f$ is not cuspidal.

Extend $\chi$ by zero to integers not coprime to $N$, and write $g(\chi )={\sum }_{a\mathrm{mod}N}\chi (a)e^{2\pi ia/N}$ for its Gauss sum of a Dirichlet character. We use the analytic branch of $\sqrt{-i\tau }$ that is positive when $\tau$ is positive imaginary. With the convention $\hat{F}(t)={\int }_{\mathbb{R}}F(x)e^{-2\pi ixt}dx$, the Gaussian Fourier transform and its differentiated version areThe first formula follows by completing the square, first on the imaginary axis and then by analytic continuation; the second follows by differentiating the Fourier transform in $t$. All the functions are Schwartz functions on the real axis.

For completeness, the finite Fourier transform of a primitive Dirichlet character satisfiesFor a unit $r$ this is substitution by its inverse. If $p\mid (r,N)$, primitivity supplies a unit $u\equiv 1(\mathrm{mod}N/p)$ with $\chi (u)\ne 1$: otherwise the character would factor through modulus $N/p$. Multiplication by $u$ fixes the additive exponential but multiplies the sum by $\chi (u)$, so the sum vanishes. This proves the identity also on nonunits. Finite Parseval identity now gives $|g(\chi ){|}^2=N$, since exactly $\phi (N)$ frequencies have nonzero character values. In particular this Gauss sum of a Dirichlet character is nonzero.

Apply Poisson summation to each residue class: ${\sum }_{j}F(a+Nj)=N^{-1}{\sum }_r{e}^{2\pi ira/N}\hat{F}(r/N)$. Multiply by $\chi (a)$ and sum over $a$. The formulas above give the two character theta transformationsandIn the odd case the prefactor originally reads $g(\chi )/(N^2\tau \sqrt{-i\tau })$; using $\tau =i(-i\tau )$ explains the factor $i$. Opposite parity makes the corresponding series identically zero by pairing $n,-n$. This is why the weight-three-halves series is needed for odd Dirichlet characters.

To deduce the products from these transformations, use two specific primitive Dirichlet characters. The even character ${\chi }_{12}$ takes values $1,-1,-1,1$ on $1,5,7,11$ modulo twelve. It is primitive: its values do not factor through six or four, hence through no proper divisor of twelve. Its Gauss sum of a Dirichlet character is $2\sqrt{3}=\sqrt{12}$. PutThe even transformation gives $A(-1/\tau )=\sqrt{-i\tau }A(\tau )$; also $A(\tau +1)=e^{\pi i/12}A(\tau )$, since each nonzero term has $n^2\equiv 1(\mathrm{mod}24)$. Thus $A^{24}$ transforms as a weight-twelve modular form under the generators $S,T$ of the modular group. It is holomorphic and starts with $q$, so it is a normalized cusp form. The proof in Question 1 shows $S_{12}=\mathbb{C}{\Delta }_0$, giving $A^{24}={\Delta }_0$. In particular $A$ has no interior zeros, by the valence formula for the modular group.

The Serre derivative argument in Question 2 gives $D{\Delta }_0=E_2{\Delta }_0$. Hence $DA/A=E_2/24$. The analytic product $P(\tau )=q^{1/24}{\prod }_{n\ge 1}(1-q^n)$ is nonzero and has the same logarithmic derivative, since $DP/P=1/24-\sum {\sigma }_1(n)q^n$. Therefore $A/P$ is constant on the connected complex upper half-plane; its limit at infinity is one. This proves $A=P$, without assuming either of the product identities being deduced. Pair the terms $n,-n$ in $A$, choosing representatives $n=6r+1$. Since ${\chi }_{12}(6r+1)=(-1{)}^r$, we getThis is the Euler pentagonal identity, derived by the character theta proof of the eta product.

For the second product use the odd primitive character ${\chi }_4(1)=1$, ${\chi }_4(3)=-1$, with Gauss sum of a Dirichlet character $2i$. SetThe odd transformation gives $B(-1/\tau )=(-i\tau {)}^{3/2}B(\tau )$ and the series gives $B(\tau +1)=e^{\pi i/4}B(\tau )$. Thus $B^8$ is another normalized weight-twelve cusp form, and $B^8={\Delta }_0=A^{24}$. Because $A$ is nonzero, $B/A^3$ has eighth power one, so it is constant; its leading coefficient fixes it to one. Pairing positive and negative odd integers now givesSince $B=A^3=q^{1/8}{\prod }_{n\ge 1}(1-q^n{)}^3$, the Jacobi cubic product identity follows:All pairings and differentiations take place in locally uniformly convergent theta functions or products on $|q|<1$.

Use $\Theta =(2\pi i{)}^{-1}d/d\tau$. For $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$, differentiating the weight-$k$ law givesThe transformations of $E_2$ under the generators $S,T$, and the chain rule for their compositions, giveEquivalently, one can check just $S,T$ in the next cancellation, since they generate the group. Substituting these formulas intocancels the extra term and gives $(D_{k}f)(\gamma \tau )=(c\tau +d{)}^{k+2}{D}_{k}f(\tau )$. Both constituents are holomorphic on $\mathbb{H}$ and have Fourier expansions of a modular form with nonnegative exponents, so $D_{k}f$ is holomorphic at the modular cusp. ThusThis operator is the Serre derivative.

If $f={\sum }_{n\ge 0}{a}_n{q}^n$, then its derivative has constant coefficient zero, while $E_2$ has constant coefficient one. ThereforeFor $k\ne 0$ this vanishes exactly when $a_0(f)=0$, proving the claimed cusp form equivalence in the intended positive-weight setting. At weight zero the printed equivalence is false: $f=1$ is a noncuspidal weight-zero modular form, but $D_{0}f=0$ belongs to $S_2$. Indeed all weight-zero forms are constants by compactness of $X(1)$. Negative-weight level-one modular forms are zero by the valence formula for the modular group, so there is no further nonzero exception. The transformation and holomorphy conclusion for $D_{k}f$ remains true at weight zero.

Write $D=(2\pi i{)}^{-1}d/dz=qd/dq$ and use the Serre derivative $D_{k}f=Df-(k/12)E_{2}f$. Translation invariance of $f$ and the Eisenstein series of weight two gives $D_{k}f(z+1)=D_{k}f(z)$.

Differentiate the weight-$k$ transformation $f(-1/z)=z^{k}f(z)$. The chain rule yieldsThe assumed transformation of the Eisenstein series of weight two givesSubtracting cancels the extra term. Hence $D_{k}f(-1/z)=z^{k+2}{D}_{k}f(z)$. Since $S$ and $T=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$ generate the modular group, these two transformations prove the weight-$k+2$ law for all its elements. Holomorphy on the complex upper half-plane follows from the formula.

For the Fourier expansion of a modular form $f={\sum }_{n\ge 0}{a}_n{q}^n$, differentiation gives $Df={\sum }_{n\ge 1}n{a}_n{q}^n$. The product of the convergent series for $E_2$ and $f$ likewise has no negative powers. Thus $D_{k}f$ is holomorphic at a cusp at infinity, and therefore at every cusp of the modular group. Its constant coefficient is $-k{a}_0/12$. Since $k>0$, it vanishes exactly when $a_0=0$. Consequently

Apply the Serre derivative in weight twelve to the modular discriminant. The preceding result makes $D_{12}\Delta =D\Delta -E_2\Delta$ a weight-fourteen cusp form. The vanishing of weight-fourteen level-one cusp forms follows directly from the valence formula for the modular group: a nonzero such form has order at infinity at least one, and $f(i)=i^{14}f(i)=-f(i)$ forces order at $i$ at least one. Thus its weighted number of zeros would be at least $1+1/2>14/12$, a contradiction. Therefore $D\Delta =E_2\Delta$.

Compare the Fourier coefficients at positive indices using $\Delta ={\sum }_{n\ge 1}\tau (n)q^n$ and $E_2=1-24{\sum }_{r\ge 1}{\sigma }_1(r)q^r$. This gives $n\tau (n)=\tau (n)-24{\sum }_{r=1}^{n-1}{\sigma }_1(r)\tau (n-r)$. For the Ramanujan tau function, the concise recurrence isAt $n=1$ the sum is empty and both sides are zero; normalization supplies $\tau (1)=1$. For instance the next coefficient is $\tau (2)=-24$.

The modular discriminant satisfies $1728\Delta =E_4^3-E_6^2$, while the Serre derivative gives $2\Theta E_6=E_2{E}_6-E_4^2$. Their integral Fourier expansions have $E_4\equiv 1$ and $E_2\equiv E_6(\mathrm{mod}5)$, because $d^5\equiv d(\mathrm{mod}5)$. Reducing the two identities gives $\Delta \equiv \Theta E_6(\mathrm{mod}5)$. The coefficient of $q^n$ in $\Theta E_6$ is $-504n{\sigma }_5(n)\equiv n{\sigma }_5(n)$.

The Ramanujan tau function gives the Fourier coefficients at positive indices of the modular discriminant: $\Delta (z)={\sum }_{n\ge 1}\tau (n)q^n$, with $\tau (1)=1$. The Serre derivative and vanishing of weight-fourteen level-one cusp forms give $qd\Delta /dq=E_2\Delta$ and hence the convolution recurrence