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The Jensen zero-count bound gives $O(T\log T)$ zeros up to ordinate $T$. Since their real parts lie in $(0,1)$, a fixed-point real logarithmic derivative summand is $O(|\Im \rho {|}^{-2})$. Dyadic ordinate bands then contribute $O(j{2}^{-j})$. This proves absolute convergence of the real sum, while the unpaired complex sum of reciprocals need not converge absolutely.

The logarithmic derivative of the modular discriminant gives $E_2={\Delta }^{\prime }/(2\pi i\Delta )$. Differentiating its weight-twelve transformation yields the anomalous term $6c(c\tau +d)/(\pi i)$. Subtracting the displayed inverse-height term cancels that anomaly. The result transforms with weight two under the modular group but is not holomorphic, so it is not a holomorphic modular form.

The argument principle turns a boundary journey into an interior count: follow the values of a meromorphic function around a contour, and count their net turns around zero. Every enclosed zero of a function contributes its multiplicity of a zero; every enclosed pole subtracts its order. There is no need to find each zero of a function separately.

Let $\gamma$ be a piecewise smooth simple closed curve with positive orientation of a contour. Suppose $f$ is a meromorphic function on an open neighborhood of the curve and its interior, with no zeros of a function or poles on $\gamma$. ThenHere $N$ is the sum of the zero multiplicities inside $\gamma$, and $P$ is the sum of the orders of its enclosed poles. The image contour $f\circ \gamma$ may cross itself or revisit a point. Its signed winding number, rather than the number of visually bounded regions, is the quantity that counts.^[ref]

For the polynomial $f(z)=z^3$ and the unit circle $\gamma (t)=e^{it}$, $0\le t\le 2\pi$, the image is $e^{3it}$. It goes around zero three times although its drawn outline is just one circle. The interior contains one distinct zero of a function, at zero, with multiplicity of a zero three: the count is three. Reversing the orientation of a contour reverses the winding number and the sign of the contour integral.

An argument must be followed continuously along the journey. Subtracting the principal arguments at the endpoints loses whole turns: a closed image contour has the same endpoint value even when its winding number is nonzero. Nor does winding number zero prove that there are no zeros of a function: one enclosed simple zero of a function and one enclosed simple pole cancel in $N-P$. Known poles must be added back to recover $N$.

The mechanism is short. Near a zero of a function of order $m$, factor $f(z)=(z-a{)}^{m}g(z)$ with $g$ a nonvanishing holomorphic function. Its logarithmic derivative is $f^{\prime }/f=m/(z-a)+g^{\prime }/g$, so its residue is $m$. A pole of order $m$ gives residue $-m$. The residue theorem adds those contributions. Changing variables to $w=f(z)$ identifies the resulting contour integral with the winding number. This is a local algebraic count joined to a global geometric count.^[ref]

The usefulness goes beyond explicit polynomials. Rouché's theorem keeps a count unchanged when a boundary perturbation cannot push the image contour through zero. The Nyquist stability criterion applies the same counting idea to poles governing a system under feedback control.^[ref]

Take the even primitive Dirichlet character ${\chi }_{12}$ whose values at $1,5,7,11$ modulo twelve are $1,-1,-1,1$. Its Gauss sum of a Dirichlet character is $\sqrt{12}$. The displayed series $A$ has $A(-1/\tau )=\sqrt{-i\tau }A(\tau )$ and $A(\tau +1)=e^{\pi i/12}A(\tau )$. Thus $A^{24}$ is the normalized weight-twelve modular discriminant. It has no interior zeros by the valence formula for the modular group. Its logarithmic derivative equals $E_2/24$, so integration gives the Dedekind eta function product. The odd character modulo four similarly produces a series $B$ with $B^8=A^{24}$ and leading term $q^{1/8}$; consequently $B=A^3$.

The Floquet growth rate measures the exponential trend of a periodic linear differential equation over many coefficient periods. A Floquet multiplier $m$ measures the factor per period $T$; the corresponding rate per unit time iswhere ${\rho }_m$ is a Floquet exponent. Taking the real part removes the ambiguity in the complex logarithm.

For the scalar example under periodic solution, $m=e^{2\pi a}$ and the Floquet growth rate is $a$, independently of $b$. Indeed,The instantaneous logarithmic derivative keeps oscillating even though its average over complete periods is exactly $a$. Sampling at the same phase in each period removes the modulation altogether: $y(nT)=m^{n}y(0)$.

For the general finite-dimensional homogeneous linear differential equation $x^{\prime }=A(t)x$, assume that $A$ is continuous on the real line and $T$-periodic, with $T>0$. Normalize its fundamental matrix by $\Phi (0)=I$ and set $M=\Phi (T)$. The maximal asymptotic exponential rate, using any induced matrix norm, isHere $\rho (M)$ is the spectral radius, $m_j$ are the Floquet multipliers, and invertibility gives $\rho (M)>0$. This concerns growth of the solution operator; it does not say that every initial condition attains the maximum.^[ref]

To prove the formula, write $t=nT+s$ with $0\le s<T$. Periodicity and uniqueness giveBoth $\Phi (s)$ and $\Phi (s{)}^{-1}$ have uniformly bounded matrix norms on a period. Consequently $\Vert \Phi (t)\Vert$ differs from $\Vert M^n\Vert$ by at most fixed multiplicative factors. The spectral radius formula gives $\Vert M^n{\Vert }^{1/n}\to \rho (M)$, proving the claimed limit without choosing a matrix logarithm.

For a particular nonzero initial condition $v$, decompose $v$ into the generalized eigenspaces of $M$ over the complex numbers. The rate of $\Phi (t)v$ is the largest $T^{-1}\log |m_j|$ among components that are nonzero. Generic data have a component in the dominant generalized eigenspaces and attain $\gamma$; data contained in the sum of the subdominant generalized eigenspaces have a smaller rate. For example, with $A=\mathrm{diag}(0.2,-0.3)$, the maximal rate is $0.2$, while $v=(0,1)$ decays at rate $-0.3$. These invariant subspaces describe initial data at phase zero. Their transported versions $E(t)=\Phi (t)E(0)$ can vary with time and satisfy $E(t+T)=E(t)$.

A negative Floquet growth rate allows temporary amplification. For the constant non-normal matrixthe initial condition $v=(0,1)$ has Euclidean norm $e^{-t}\sqrt{100t^2+1}$, which rises from $1$ to about $3.697$ at $t=1$ before eventually decaying. Its exponential rate is still $-1$: the Jordan block contributes a polynomial factor, whose logarithm divided by $t$ tends to zero. At a unit-modulus Floquet multiplier, such polynomial growth can even be unbounded with zero exponential rate; $A=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$ gives $e^{tA}v=(t,1)$. Thus zero Floquet growth rate alone proves neither boundedness nor a periodic solution.

For a primitive nonprincipal Dirichlet character of conductor $q$ and parity $a$, the completed Dirichlet L-function is entire of order one. Its genus-one Hadamard factorization gives the displayed logarithmic derivative, with $B_{\chi }$ the logarithmic derivative of the completion at zero and zeros counted with multiplicity. Combined summands converge locally normally away from zeros. For an imprimitive character induced by ${\chi }^{\ast }$, add ${\sum }_{p\mid Q,p\nmid q}{\chi }^{\ast }(p)\log p/(p^s-{\chi }^{\ast }(p))$ for the removed Euler factors. The primitive principal case is the global partial-fraction expansion of the zeta logarithmic derivative.

// Target: algebraic-number-theory.bigb

A laboratory determination of the gravitational constant must connect a sensor's reading to a gravitational signal from independently characterized masses. For an ideal two-point mass force measurement, invert the model:These are signed small changes obtained by taking the logarithmic derivative, not an instruction to add standard uncertainties with signs. For independent small input uncertainties, linear propagation of uncertainty instead givesCorrelated inputs require their covariances too. In particular, common geometric calibration errors must not be treated as independent simply because several distances were recorded.

For a concrete model check, hold the actual force and masses fixed but overestimate a true $r=0.100\mathrm{m}$ by $1\mu \mathrm{m}$. The exact inferred ratio isThe resulting bias is $20.0001$ parts per million, already comparable with the relative standard uncertainty of the recommended gravitational constant. A sensor can resolve very small changes while its inferred answer remains limited by source geometry. The example concerns one idealized separation; it is not an uncertainty estimate for any actual experiment.

Making the source mass larger or bringing it closer strengthens the signal, but also makes detailed mass density and geometry important. Nearby apparatus supplies its own Newtonian gravitational field; electromagnetic shielding does not remove that attraction. Sensor offsets, changing temperature, vibration and source motion can imitate or alter a signal. The Cavendish experiment illustrates calibration of a torsion constant; a different sensor requires its own calibrated response and a full source model.

Repeating readings can reduce independent measurement noise, but a fixed scale error survives averaging. A useful experiment deliberately changes the source configuration, checks whether the signal follows the predicted change, and characterizes what moves with the source. A systematic error that changes with that configuration also survives background subtraction. Comparing methods with different sensors and different dominant errors is therefore valuable.

In a thin ideal-gas grey atmosphere with $T^4=T_e^4(1/2+3\tau /4)$, constant enclosed mass $M$ and luminosity $L$, and $\kappa ={\kappa }_0\rho T^{13}$, divide the stellar hydrostatic equation by radiative diffusion in a star. This gives $PdP/dT=16\pi acGM\mathcal{R}{T}^{-9}/(3{\kappa }_{0}L\mu )$. Integrating from $P=0$, $T^4=T_e^4/2$, yieldsLogarithmic derivative gives $\nabla =T^8/T_e^8-1/4$. For a monatomic gas gas, the Schwarzschild criterion reaches equality at $\nabla =2/5$, hence $T/T_e=(13/20{)}^{1/8}$. The purely radiative profile cannot be continued past this point as a convectively stable model.

For constant nonzero $c$, putting $y=x^{\prime }/(cx)$ in $y^{\prime }+c{y}^2+a(t)y+b(t)=0$ cancels the quadratic derivative terms and gives $x^{\prime \prime }+a(t)x^{\prime }+cb(t)x=0$. The logarithmic derivative is defined on intervals where $x\ne 0$, and multiplying $x$ by a nonzero scalar does not change $y$. For variable $c(t)\ne 0$, the derivative coefficient becomes $a-c^{\prime }/c$.

For a monic polynomial $p(x)={\prod }_i(x-a_i)$ with distinct ordered real roots of a polynomial, each critical point $\xi$ lies between two consecutive roots by Rolle's theorem. Its logarithmic derivative gives ${\sum }_i(\xi -a_i{)}^{-1}=0$. Implicit differentiation gives the displayed formula, so moving any root to the right moves every ordered critical point to the right, while the roots remain distinct. Iterating this statement for derivatives preserves componentwise ordering of their roots. Weak inequalities extend by continuity to repeated roots.

For a nonconstant meromorphic function on the plane, the displayed estimate holds outside an exceptional set of radii of finite linear measure, for large $r$. The logarithmic derivative has only simple principal parts at zeros and poles; Poisson-Jensen formula estimates and control of increasing auxiliary growth functions produce the logarithmic error. It supplies the error term in the Nevanlinna second main theorem.

The Painlevé property excludes movable critical singularities, such as branch points, of the analytically continued general solution of a complex differential equation. A common stronger formulation requires every movable singularity to be a pole, so local solutions are meromorphic there. Fixed singularities determined by the coefficients are distinguished from movable ones whose locations depend on initial data. Riccati equations with meromorphic coefficients have only movable poles at ordinary points: linearization expresses their solutions as logarithmic derivatives of linear-system solutions. More generally, set $w=f/g$ and solve the holomorphic linear system $f^{\prime }=bf+cg$, $g^{\prime }=-af$. Its quotient satisfies $w^{\prime }=a{w}^2+bw+c$ directly. Initial data $g(z_0)=1$, $f(z_0)=w(z_0)$ recover every local solution, and holomorphic $f,g$ make the quotient a meromorphic function, even at zeros of $a$. At a zero of $g$ with $a\ne 0$, uniqueness prevents $f=0$ there, so $g^{\prime }=-af\ne 0$ and the pole is simple. This proves the absence of movable branching without assuming that the quadratic coefficient never vanishes.

First alternative. The Euler product of the Riemann zeta function is absolutely convergent for $\sigma >1$. Its logarithm and the nonnegative trigonometric polynomialgiveConsequently $\zeta (\sigma {)}^3|\zeta (\sigma +it){|}^4|\zeta (\sigma +2it)|\ge 1$. Suppose $t\ne 0$ and the Riemann zeta function had a zero of order $m\ge 1$ at $1+it$. As $\sigma \downarrow 1$, its simple pole at one gives $\zeta (\sigma )=O((\sigma -1{)}^{-1})$, the alleged zero gives $\zeta (\sigma +it)=O((\sigma -1{)}^m)$, and $\zeta (\sigma +2it)$ remains bounded. The product would be $O((\sigma -1{)}^{4m-3})\to 0$, contradicting its lower bound. Thus there are no zeta zeros on $\Re s=1$; the point $s=1$ itself is a pole, not a zero.

For the first integral of the Chebyshev function, defineThe logarithmic derivative identity $-{\zeta }^{\prime }/\zeta (s)={\sum }_n\Lambda (n)n^{-s}$, valid for $\Re s>1$, leads toTo verify the kernel rather than assume an inversion formula, integrate $y^s/(s(s+1))$ on the vertical line $\Re s=c$. Closing to the left when $y>1$ gives the residues at zero and minus one, namely $1-y^{-1}$; closing to the right when $0<y<1$ gives zero. At $y=1$ the integral is zero as well. The horizontal integrals tend to zero because of the quadratic denominator, and the distant vertical integrals vanish in the indicated direction. Multiplying this kernel by $x$ with $y=x/n$ gives $(x-n{)}^{+}$. The absolute convergence of the Dirichlet series on $\Re s=c$, together with the integrable denominator, permits interchange with the integral and proves the formula.

Here is a quantitative contour route from this relation to the Prime number theorem. Question 3 establishes a zero-free region of the Riemann zeta function and the accompanying bounds for its logarithmic derivative, independently of the Prime number theorem. In a sufficiently narrow half-width region they implyon the left and horizontal sides of the rectangle with heights $\pm T$ and left edge ${\sigma }_L=1-b/(2\log (T+3))$, after making $b>0$ small enough also for bounded heights. The explanation of this bound is included at the end of Question 3. Take $c=1+1/\log x$ and $T=\exp (\sqrt{\log x})$. On the original line, the omitted tails aresince the Euler product bounds $|{\zeta }^{\prime }/\zeta (c+it)|$ by $-{\zeta }^{\prime }/\zeta (c)=O(\log x)$. Shift the truncated integral to the left edge. The only enclosed pole is at $s=1$, with residue $x^2/2$; the kernel poles at zero and minus one lie outside the rectangle. The left edge contributes $O(x^{2-b/(2\log (T+3))}{\log }^2(T+3))$, since $|s(s+1){|}^{-1}\ll (1+t^2{)}^{-1}$ there. The two horizontal edges contribute $O(x^2{\log }^2(T+3)/T^2)$. In particular, for some $a>0$,

It remains to remove the smoothing; differentiating an asymptotic formula would not justify this step. The monotonicity of the Second Chebyshev function gives, for $h=\epsilon x$ with $0<\epsilon <1$,First let $x\to \infty$ with $\epsilon$ fixed. After division by $x$, the lower and upper bounds tend to $1-\epsilon /2$ and $1+\epsilon /2$. Let $\epsilon \downarrow 0$ to obtain $\psi (x)\sim x$. Higher prime powers contribute $O(\sqrt{x}\log x)$ by the Chebyshev estimate, so $\theta (x)\sim x$. Finally partial summation givesThe integral is $O(x/{\log }^{2}x)$, by splitting at $\sqrt{x}$ and using $\theta (t)=O(t)$. This proves the Prime number theorem with $\pi (x)\sim x/\log x$.

Second alternative. We prove the functional equation of the Riemann zeta function from a theta function. Set $\Theta (u)={\sum }_{n\in \mathbb{Z}}{e}^{-\pi n^{2}u}$ for $u>0$. With Fourier transform convention $\hat{f}(\xi )={\int }_{\mathbb{R}}f(v)e^{-2\pi iv\xi }dv$, the Gaussian function $f(v)=e^{-\pi u{v}^2}$ hasIndeed, differentiation under the integral and integration by parts give ${\hat{f}}^{\prime }(\xi )=-2\pi \xi \hat{f}(\xi )/u$, while the Gaussian integral gives $\hat{f}(0)=u^{-1/2}$. Periodize $f$: the smooth period-one function ${\sum }_{n}f(v+n)$ has Fourier coefficients $\hat{f}(k)$, by integrating on $[0,1]$ and interchanging its absolutely convergent sum with the integral. Its absolutely convergent Fourier series, evaluated at zero, proves the Poisson summation formula in this case. HenceFor $\Re s>1$, termwise integration and the defining integral of the Gamma function give the Mellin representation of the completed Riemann zeta function:Split at one and substitute $u=1/v$ in the lower integral. The elementary part isand the remaining part becomes an integral over $[1,\infty )$. ThusThe integral is an entire function of $s$, because $\Theta (u)-1$ decays exponentially and uniform convergence holds on every compact subset of the complex plane. This proves meromorphic continuation of the completed Riemann zeta function, with only the displayed simple poles, and the right side is invariant under $s\mapsto 1-s$. ThereforeThe Gamma reflection formula and Gamma duplication formula give the equivalent form $\zeta (s)=2^s{\pi }^{s-1}\sin (\pi s/2)\Gamma (1-s)\zeta (1-s)$, understood by meromorphic continuation. Multiplication by $s(s-1)/2$ gives the entire Riemann xi function $\xi (s)=\xi (1-s)$.

Let $N(T)$ count positive ordinates of Nontrivial zeros of the Riemann zeta function up to height $T$, including multiplicity. The Riemann–von Mangoldt formula isMore precisely, away from ordinates of zeros, the constant term is $7/8$, the additional term is $S(T)={\pi }^{-1}\arg \zeta (1/2+iT)$, and the remaining error is $O(T^{-1})$. Here the argument is continued along $2\to 2+iT\to 1/2+iT$, and $S(T)=O(\log T)$. The coarse formula displayed above holds at all large heights with the usual right-continuous counting convention. Subtract it at $T+1$ and $T$: the main term changes by $O(\log T)$ and each error is $O(\log T)$. Hence

For infinitely many reciprocal-logarithmic gaps between zeta zeros, fix any $0<c<2\pi$, for example $c=\pi$. Suppose all sufficiently late gaps satisfied ${\gamma }_{n+1}-{\gamma }_n\le c/\log n$. Summing and usinggives ${\gamma }_n\le (c+o(1))n/\log n$. The last elementary estimate follows by integral comparison with $1/\log t$ and one integration by parts. The Riemann–von Mangoldt formula would implya contradiction. This reasoning is valid whether repeated ordinates are listed according to multiplicity or only distinct ordinates are listed, since in either case $n\le N({\gamma }_n)$. Consequently ${\gamma }_{n+1}-{\gamma }_n>c/\log n$ infinitely often, for any fixed $0<c<2\pi$.

Write $s=\sigma +it$ and let $\rho =\beta +i\gamma$ run over Nontrivial zeros of the Riemann zeta function, including multiplicity. The bounds involving $\log |t|$ below are for $|t|\ge 3$. At bounded heights their uniform versions use $\log (|t|+3)$; $\log |t|$ itself cannot be the right bound near zero or $|t|=1$.

Start with the Hadamard factorization of the Riemann xi function, an entire function of order one. Its logarithmic derivative has the convergent expansionDifferentiating $\xi (s)=\frac{1}{2}s(s-1){\pi }^{-s/2}\Gamma (s/2)\zeta (s)$ givesThe real sum of $1/\rho$ converges absolutely: $0\le \beta \le 1$ makes its terms $O({\gamma }^{-2})$ at large height, and the order-one Hadamard factorization ensures the required summability. Thus these terms and $\Re B$ form a fixed constant. The Stirling formula for the digamma function gives, uniformly for $1<\sigma \le 2$ and $|t|\ge 3$,This is the useful real form of the global partial-fraction expansion of the zeta logarithmic derivative. Every term subtracted is nonnegative because the Euler product excludes zeros with $\beta >1$. At the simple pole $s=1$, one also has $F(\sigma )=1/(\sigma -1)+O(1)$ as $\sigma \downarrow 1$.

The three-four-one zero-free-region argument now detects a zero too close to one. The Von Mangoldt function is nonnegative and the absolutely convergent Dirichlet series givesbecause the corresponding coefficient at $n$ is $\Lambda (n)n^{-\sigma }[3+4\cos (t\log n)+\cos (2t\log n)]$. If $\beta +it$ is a zero, retain its term in the expansion at height $t$ and discard the other nonpositive terms. The expansion at height $2t$ gives an upper bound even without retaining a particular zero. We obtain, for an absolute constant $C_0>0$,Choose $0<a<1$ so small that $5/(9a)>C_0$, put $b=a/8$, and take $\sigma =1+a/\log |t|$. If $1-\beta \le b/\log |t|$, the last inequality would requirea contradiction. Thereforefor some fixed $b>0$. This proves the required zero-free region of the Riemann zeta function. Nonvanishing on $\Re s=1$ at bounded nonzero heights, proved in Question 2, and the isolation of zeros allow a decrease of $b$ giving the uniform formulation with $\log (|t|+3)$ as well. Near $s=1$, apply this observation to the nonzero holomorphic function $(s-1)\zeta (s)$.

Next evaluate the real expansion at $\sigma =2$. The absolutely convergent Dirichlet series bounds $|{\zeta }^{\prime }/\zeta (2+it)|\le {\sum }_n\Lambda (n)n^{-2}$ uniformly in $t$, soSince $1\le 2-\beta \le 2$, each term on the left is at least $1/[4+(t-\gamma {)}^2]$. The smoothed zeta zero-count bound follows:Each zero satisfying $|t-\gamma |<1$ contributes more than $1/5$, so this also gives the local zero count for the Riemann zeta function:The same kernel sum is bounded at compact heights, since its tail converges uniformly there; the bounds with $\log (|t|+3)$ are consequently valid for all real $t$.

For completeness, the contour estimate used in Question 2 follows from these very formulas. Subtract the complex logarithmic derivative expansion at $2+it$ from that at $\sigma +it$, with $1-b/(2\log (T+3))\le \sigma \le 2$ and $|t|\le T$. Terms with $|t-\gamma |>1$ have differences $O((t-\gamma {)}^{-2})$, whose sum is $O(\log (|t|+3))$ by the kernel bound. There are $O(\log (|t|+3))$ remaining terms. After reducing $b$, the zero-free region of the Riemann zeta function separates every such zero horizontally from the half-width boundary by $\gg 1/\log (T+3)$: their heights satisfy $|\gamma |\le T+1$. Thus on the left edge, and on the horizontal edges at $t=\pm T$, each reciprocal is $O(\log (T+3))$. The Gamma function and pole terms satisfy the same required bound; at small heights on the left edge the pole distance is $\gg 1/\log (T+3)$. This proves $|{\zeta }^{\prime }/\zeta |\ll {\log }^2(T+3)$ on those edges and justifies the contour shift without choosing heights that dodge unknown zeros.

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

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 $|q|<1$.

The Painlevé property says that the general solution has no movable critical singularities: analytic continuation around a singularity whose location depends on the initial data must not create a different branch. A fixed singularity of a complex differential equation has its location prescribed by the coefficients; a movable singularity of a complex differential equation has a location which varies with the initial data. A commonly used stronger form requires movable singularities to be poles, so that the solution is locally a meromorphic function there. Movable branch points are excluded; a movable pole is permitted. We will prove the stronger form for this equation by finding its complete general solution family as rational functions.

Use the linearization of a Riccati equation. If $w^{\prime }=a{w}^2+bw+c$, put $w=-v^{\prime }/(av)$ wherever $a,v\ne 0$. Differentiating this expression and cancelling the quadratic terms yieldsConversely a nonzero solution $v$ of this linear ordinary differential equation gives a solution $w$, wherever the quotient is defined. Every local finite solution $w$ at an ordinary point arises this way: solve $v^{\prime }=-awv$ with a nonzero initial value and differentiate to get the second-order equation. Multiplying $v$ by a nonzero constant leaves $w$ unchanged, giving the one effective constant of integration expected for a first-order Riccati equation.

For the present coefficients, $a^{\prime }/a=1/z+1/(z+1)$, so the linear ordinary differential equation simplifies toTwo direct substitutions give solutions $v_1=z-1$ and $v_2=1/(z+1)$. Their Wronskian iswhich is nonzero at every ordinary point $z\notin \{0,-1\}$. They therefore form a fundamental pair there, and all local solutions have the formTaking its logarithmic derivative givesThus, for $C_1\ne 0$, writing $k=(C_2-C_1)/C_1$, the general family isThe remaining member, corresponding to $C_1=0$, isThe completeness of the linearizing pair proves that no other local solution is omitted. These are rational functions, single-valued meromorphic functions; hence their continuation has no branch points. Their only possible moving singularities are zeros of $z^2+k$. A root $z_0$ outside $\{0,-1\}$ is simple, and its residue isEquivalently, poles of a Riccati solution from zeros of its linearizing solution are simple at ordinary points, because a nontrivial solution of a linear ordinary differential equation cannot have both $v(z_0)=0$ and $v^{\prime }(z_0)=0$. The exceptional collisions $k=0$ or $k=-1$ occur at the fixed coefficient singularities, and still give rational functions. Consequently every movable singularity is a pole, establishing the Painlevé property in its stronger meromorphic sense.

A precise endpoint version of the Perron formula is as follows. If $F(s)={\sum }_{n\ge 1}{a}_n{n}^{-s}$ converges absolutely on $\Re s=c>0$, then, for $x>0$,When $x$ is not an integer, the left side is simply the sum over $n\le x$.