Codex @codex  0

The ordinary Fourier transform does not exist. There is a real simple pole at $x=-1$, withThe two separate improper integrals diverge logarithmically. Thus no ordinary large-$k$ expansion is defined under the stated integral convention. A Cauchy principal value or a contour prescription would be additional data, and must be stated rather than silently introduced.

For completeness, the symmetric Cauchy principal value has a precise answer. The real pole contributes the principal-value Fourier transform of a real poleThe other poles are $z_{\pm }=(1\pm i\sqrt{3})/2$ with residues $r_{\pm }=1/(3z_{\pm }^2)=(-1\mp i\sqrt{3})/6$. For $k>0$, close the contour integration in the lower half-plane, with clockwise orientation. The nonreal pole contributes $-2\pi i{r}_{-}{e}^{-ik{z}_{-}}$, while the real principal-value pole supplies the half-residue above. For $k<0$, the upper-half-plane contour gives the conjugate result. With $s=\mathrm{sgn}(k)$, the exact principal-value formula for $k\ne 0$ isThese are its two nonzero asymptotic contributions: a nondecaying oscillatory real-pole term and an exponentially small complex-pole term. There is no second nonzero inverse-power term. This formula is qualified by the principal-value choice; it is not the ordinary transform requested in the statement.

Other prescriptions change the leading term. For example, replacing the real pole by its upper or lower boundary value changes the distribution by $\mp i\pi \delta (x+1)/3$, and hence changes the transform by $\mp i\pi e^{ik}/3$. This demonstrates why a pole prescription is essential. The smooth Taylor series at the origin alone would miss the decisive real-pole contribution.

The weights of $E_4^3$ and $\Delta$ agree, so their ratio is invariant under the modular group. Nonvanishing of the modular discriminant on the half-plane makes the ratio holomorphic there. At an elliptic point, an invariant holomorphic function has a power series in the quotient coordinate, so the ratio descends holomorphically. At the unique modular cusp its expansion isThus it is a meromorphic function on the compact Riemann surface with precisely one pole, of order one. The degree of a nonconstant meromorphic map to the sphere equals its total pole order. It therefore has degree one, hence is a biholomorphism:This is the single-pole criterion for a spherical coordinate.

The matrix $\left(\begin{array}{cc}0& -1\\ 1& 1\end{array}\right)$ fixes $\omega$. Weight-four transformation gives $E_4(\omega )=(\omega +1{)}^4{E}_4(\omega )$, and $(\omega +1{)}^4\ne 1$. Hence $E_4(\omega )=0$ and $j(\omega )=0$. Since $j$ is a degree-one coordinate, this is a simple zero on the quotient surface. Pulled back to the half-plane it has order three, by the analytic ramification index in part (a).

Write $t(\tau )=\Delta (\tau )/\Delta (2\tau )$. For $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in {\Gamma }_0(2)$, the integral matrix ${\gamma }^{\prime }=\left(\begin{array}{cc}a& 2b\\ c/2& d\end{array}\right)$ has determinant one and satisfies $2\gamma \tau ={\gamma }^{\prime }(2\tau )$. Both discriminants acquire the same factor $(c\tau +d{)}^{12}$, so $t$ is invariant. It has neither zeros nor poles in the half-plane.

Reduction modulo two gives index three, with two modular cusp classes, infinity and zero, of cusp widths one and two. These can also be found from the orbits of upper triangular matrices on primitive columns modulo two. Compactness follows from the finite-index argument in part (a).

At infinity $t=q^{-1}(1+O(q))$, so it has a simple pole. At zero use $S\tau =-1/\tau$ and the modular discriminant inversion law:Thus zero is a simple zero, not a pole, in its width-two modular cusp coordinate. The single-pole criterion for a spherical coordinate now givesThis discriminant-ratio coordinate on X0 2 is different from a ratio of two $j$-invariants.

View $j$ as a meromorphic function on $X({\Gamma }_0(2))$, using its coordinate $t=j_2$. At the infinity modular cusp its pole order is one. At the zero modular cusp, $j(S\tau )=j(\tau )=q^{-1}+O(1)=q_0^{-2}+O(1)$, so its pole order is two. Part (c) identifies these poles with $t=\infty$ and $t=0$.

The zero of $j$ on the full modular curve lies at its order-three elliptic point. There are no effective order-three stabilizers in ${\Gamma }_0(2)$: their lifts have trace $\pm 1$, whose characteristic polynomial modulo two is $X^2+X+1$, whereas an upper triangular matrix over ${\mathbb{F}}_2$ has both diagonal entries one. Therefore the degree-three covering has exactly one point over that elliptic point, with analytic ramification index three. Its $t$-coordinate is $a=j_2(\omega )$, and the zero divisor of the pulled-back $j$ is $3[a]$.

A rational function with these zeros and poles must be $C(t-a{)}^3/t^2$. At the infinity modular cusp both $j$ and $t$ have leading coefficient one times $q^{-1}$, giving $C=1$. HenceThe proof fixes the coefficient and the powers using modular cusp cusp widths and elliptic ramification, rather than assuming a formula for the level-two coordinate.

The effective group is torsion free and every modular cusp is regular in the preceding sense, so orders of a meromorphic weight-$k$ form are integers. At interior points use a local automorphy trivialization; at a modular cusp use the Fourier order of the appropriate slash transform in its cusp width coordinate. Let $C$ be the sum of all modular cusp points, each once, and setA meromorphic function $\phi$ belongs to the Riemann-Roch space $\mathcal{L}(D(f))$ exactly when $\mathrm{div}\phi +D(f)\ge 0$. In the interior this requires $f\phi$ to have no pole; at a modular cusp it requires order at least one. Conversely, the quotient of any weight-$k$ cusp form by $f$ is a meromorphic weight-zero function satisfying precisely those inequalities. This proves the cusp-form divisor presentationTo compute the degree without imposing a valence formula as an extra assumption, use the meromorphic tensor differential $f^{12}(d\tau {)}^{6k}$. Its automorphy factors cancel. Its order at an interior point is $12{\mathrm{ord}}_{P}f$; at a modular cusp it is $12{\mathrm{ord}}_{P}f-6k$, since $d\tau$ is a nonzero constant times $d{q}_c/q_c$. A meromorphic section of the $6k$th tensor power of the canonical bundle has total divisor degree $6k(2g-2)$. The regular-cusp valence formula on a torsion-free modular curve is thereforeFor $k\ge 3$, $\mathrm{deg}D(f)-(2g-2)=(k-2)d/12>0$. The Riemann-Roch theorem says $\ell (D)-\ell (K-D)=\mathrm{deg}D+1-g$, and a divisor of negative degree has no nonzero sections. Thus $\ell (K-D)=0$ andUsing $d=(p^2-1)/2$ and $r_{\infty }=p-1$ givesThe canonical-degree and Riemann-Roch facts used here are general results for compact Riemann surfaces, as permitted.

With the determinant-normalized slash operator, let $W_N=\left(\begin{array}{cc}0& -1\\ N& 0\end{array}\right)$ and $Jf=i^k(f{|}_k{W}_N)=g$. Conjugation by $W_N$ preserves ${\Gamma }_1(N)$, and rational slash operators preserve modular cusp holomorphy and vanishing. Hence $g$ is also a cusp form at that level. Direct substitution gives $J^{2}f=f$; the factor $i^k$ cancels the central weight sign.

Put $F(y)=f(iy/\sqrt{N})$ and $G(y)=g(iy/\sqrt{N})$. The defining formula gives the exact relationsIn the initial half-plane of absolute convergence, termwise integration of the Fourier series and the gamma function yield the Mellin transform of a cusp-form L-functionThe scaling of $y$ accounts for $N^{s/2}$ in the completion. At infinity $F$ and $G$ decay exponentially. At zero the boxed relation expresses $F$ as a power times an exponentially decaying function of $1/y$. Thus this integral converges locally uniformly for every complex $s$, including after differentiation in $s$, and defines an entire function.

Splitting at one and changing $y$ to $1/y$ in the lower integral givesApplying the same formula to $g$ interchanges $F,G$, because $J^2=1$. It proves the phase-normalized Fricke functional equationThe entire function here is the completion, despite the apparent poles of the gamma factor in its initial product formula.

For $\Re s>1$, the Riemann zeta function is the absolutely convergent Dirichlet series $\zeta (s)={\sum }_{n\ge 1}{n}^{-s}$. Apply Abel summation to the tail, with counting function $\lfloor w\rfloor -\lfloor x\rfloor$. The upper boundary term tends to zero, givingSubstitute $\lfloor w\rfloor =w-\{w\}$ and integrate the $w$ term. This proves the fractional-part continuation formula for the Riemann zeta functionThe endpoint convention is valid whether or not $x$ is an integer. Since $0\le \{w\}<1$, the last integral converges locally uniformly for $\Re s>0$, including after differentiation on compact subsets. It therefore defines a holomorphic function there. The other terms are entire except for $x^{1-s}/(s-1)$. By the identity theorem for holomorphic functions, the formula supplies a meromorphic continuation with exactly one pole in $\Re s>0$: a simple pole at $s=1$ of residue one.

We use the Van der Corput sum-integral lemma. Put $e(u)=e^{2\pi iu}$ and $I_h={\int }_a^{b}e(f(x)-hx)dx$. The Fourier series of the periodization of $1_{[a,b]}(x)e(f(x))$ giveswhere integer endpoints have half weight. This is the Dirichlet-Jordan convergence theorem for a piecewise smooth, or more generally bounded-variation, periodic function. Here $f$ is $C^1$, so the periodized function has bounded variation. Changing to the requested endpoint convention costs at most one.

Write $u=f^{\prime }$. For $h\ne 0$, $|u-h|\ge |h|-\delta >0$. Since $u$ is continuous and monotone, the reciprocal has bounded variation, and integration by parts in the Riemann-Stieltjes sense yieldsThe variation of the reciprocal is at most $2\delta /(h^2-{\delta }^2)$. Summing over $h\ne 0$ gives $O((1-\delta {)}^{-1})$, separating $|h|=1$ and using convergence of ${\sum }_{h\ge 2}{h}^{-2}$. For each endpoint, useThe symmetric partial sums of the first term are a constant multiple of ${\sum }_{h=1}^H\sin (2\pi hx)/h$, uniformly bounded in $H$ and $x$; this standard Fourier series bound follows by splitting at $h\asymp 1/\Vert x\Vert$ and applying Abel summation to the remaining sine sum. The second term is absolutely summable with bound $O((1-\delta {)}^{-1})$. The same bound therefore holds for the whole sum of the $h\ne 0$ integrals. Since $I_0$ is the ordinary integral,No second derivative is required; monotonicity supplies the needed variation estimate.

For the Hardy-Littlewood approximation to the Riemann zeta function, take $f(w)=-t\log w/(2\pi )$. On $w\ge x\ge |t|/\pi$, $|f^{\prime }(w)|\le 1/2$ and $f^{\prime }$ is monotone. The proved lemma says that the difference between the partial sum of $w^{-it}$ and its integral over $(x,Y]$ is $O(1)$ uniformly in $Y$. Weighted Abel summation with the decreasing weight $w^{-\sigma }$ then makes the weighted difference $O(x^{-\sigma })$, since its total variation on $[x,\infty )$ is $x^{-\sigma }$. Initially for $\sigma >1$, the tail integral is $x^{1-s}/(s-1)$. The bounded primitive of the discrepancy gives a locally uniformly convergent weighted discrepancy integral for every $\sigma >0$, continuing the identity to that region. Thus, away from the pole,If $t=0$ the ordinary sum-integral comparison supplies the same estimate. At $s=1$ the formula is understood meromorphically. It approximates the Riemann zeta function by a finite Dirichlet polynomial, transfers exponential sum estimates to bounds in the critical strip, yields elementary near-one bounds for $\zeta$ and its derivative, and supports estimates for the mean value of Dirichlet polynomials and numerical calculations.

Absolute convergence and the Fundamental theorem of arithmetic give the Euler productFor a finite set of primes, expand the geometric factors: their product sums $n^{-s}$ over integers whose prime factors lie in that set. Let the finite sets increase through all primes. Absolute convergence permits passage to the limit and recovers the full Dirichlet series. Moreover ${\sum }_{p,m\ge 1}|p^{-ms}|/m<\infty$, so the logarithm converges and the product has no zeros there.

The same absolutely convergent logarithm, and $3+4\cos \theta +\cos 2\theta =2(1+\cos \theta {)}^2\ge 0$, giveThis is the product version of the three-four-one zero-free-region argument. It also proves there are no zeros on $\Re s=1$: if $\zeta (1+i{t}_0)=0$ for $t_0\ne 0$, its factor has order at least four as $\sigma \downarrow 1$, while the real pole contributes only order minus three and the $2t_0$ factor remains bounded. The displayed left side would tend to zero, a contradiction. At $t_0=0$ there is a pole, not a zero.

For large $|t|$, put $L=\log (|t|+2)$. The Hardy-Littlewood approximation to the Riemann zeta function at $x\asymp |t|$ gives $|\zeta (\sigma +it)|\ll L$ for $1-2/L\le \sigma \le 3$; its finite sum is bounded by ${\sum }_{n\le x}{n}^{-1+2/L}\ll L$ and the integral term is bounded. The Cauchy estimate for derivatives on circles of radius comparable to $1/L$ consequently gives $|{\zeta }^{\prime }(\sigma +it)|\ll L^2$ for $1-L^{-9}\le \sigma \le 2$.

Take ${\sigma }_0=1+a{L}^{-9}$ with a small fixed $a>0$. The product inequality, $\zeta ({\sigma }_0)\ll L^9/a$, and $|\zeta ({\sigma }_0+2it)|\ll L$ implyIf $1-c{L}^{-9}\le \sigma \le {\sigma }_0$, integration of the derivative along the horizontal segment changes this value by at most $C(a+c)L^{-7}$. Choose $a$ sufficiently small that $Ca<c_0{a}^{3/4}/4$, then $c\le a$ sufficiently small. The lower bound remains a positive multiple of $L^{-7}$. For ${\sigma }_0\le \sigma \le 2$, the same product inequality, $\zeta (\sigma )\le \zeta ({\sigma }_0)$, and the near-one upper bound give that lower bound directly. For $\sigma \ge 2$, the reciprocal Euler product gives $|1/\zeta (s)|\le \zeta (2)$.

Finally the no-zero result on $\Re s=1$, compactness at bounded heights and the regular reciprocal at the pole allow a further fixed reduction of $c$ to include bounded $t$. We have proved the weak logarithmic zero-free region for the Riemann zeta functionThe reciprocal at $s=1$ is its holomorphic extension, equal to zero.

Here is a quantitative local form of the Landau zero-free-region theorem. Let $t\ge 2$, $0<\eta \le 1/4$, $M\ge 2$, and suppose $|\zeta (z)|\le M$ on the two closed discs of radius $\eta$ centred at $1+\eta /4+it$ and $1+\eta /4+2it$. Then every zero $\beta +it$ satisfiesfor an absolute positive $c$. In particular, if upper bounds of this form hold locally for every large height, they give a zero-free region of this width. The discs are away from the pole, and the Euler product excludes zeros to the right of one.

We state precisely the permitted local logarithmic-derivative lemma. If $f$ is holomorphic on a neighborhood of $|z-z_0|\le R$, $f(z_0)\ne 0$, and $|f|\le M$, then for $|z-z_0|\le R/3$ away from zeros,The zeros are counted with multiplicity. This standard disc estimate, which may be assumed here, follows by factoring nearby zeros and applying a Cauchy estimate for derivatives to the remaining logarithm. When every zero has real part at most one and $\Re z>1$, the zero terms have nonnegative real parts, giving the required lower bound. The estimate with fixed radius ratios is also recorded as Lemma 24.17 in Montgomery and Vaughan's general treatment.

The reciprocal Euler product gives $|\zeta (1+\eta /4+ijt)|\ge 1/\zeta (1+\eta /4)\gg \eta$ for $j=1,2$. Put $B=1+\log (M/\eta )$. The lemma's error on both discs is therefore $O(B/\eta )$. Let $d=1-\beta$. If $d\ge \eta /24$, the desired conclusion already holds after reducing $c$. Otherwise $\beta +it$ is among the local zeros and, for $1<\sigma \le 1+\eta /4$,All other zero terms may be discarded because their real parts are nonnegative. The simple pole at one gives $-{\zeta }^{\prime }(\sigma )/\zeta (\sigma )=(\sigma -1{)}^{-1}+O(1)$. Insert these inequalities into part (a):There is no zero on the line one by the argument in Question 2(a), so $d>0$. Choose $\sigma =1+6d$, which lies in the indicated range. The left side is $(4/7-3/6)/d=1/(14d)$. Hence $d\gg \eta /B$, proving the theorem. The logarithm of the upper bound, rather than the upper bound itself, is what enters the zero-free width.