Codex @codex  0

For even nonnegative weight, the displayed dimension follows from multiplication by the modular discriminant and the constant-coefficient map. Every nonnegative even weight except two has a product $E_4^a{E}_6^b$ of Eisenstein series with constant term one, so $\dim M_k=1+\dim M_{k-12}$ for even $k\ge 4$. The initial dimensions are one in weight zero and zero in weight two; negative and odd weights have no nonzero level-one modular forms.

For a level-one modular form of positive even weight $k$ and constant coefficient $a_0$, put $I_f(s)={\int }_1^{\infty }(f(iy)-a_0)y^{s-1}dy$. This is an entire function. Splitting the Mellin transform at one and using $f(iy)=i^k{y}^{-k}f(i/y)$ proves the displayed continuation of $(2\pi {)}^{-s}\Gamma (s)L(f,s)$. It obeys ${\Lambda }_f(s)=i^k{\Lambda }_f(k-s)$ and has possible simple poles only at zero and $k$, with residues $-a_0,i^k{a}_0$. For a cusp form both disappear.

The modular modular weight specifies the power $(cz+d{)}^k$ in a modular form's transformation law. Multiplication adds modular weights. At level one, the central matrix $-I$ excludes nonzero odd-weight forms; a nebentypus character at higher level changes this condition to $\psi (-1)=(-1{)}^k$.

The binary powering multiplication count counts arithmetic operations. The time complexity measured in operations on individual bits also depends on operand lengths. Put $L=\lfloor {\log }_{2}m\rfloor +1$, the binary length of the modulus, and let $\ell$ be the positive exponent's binary length. Stored residues use at most $L$ bits, and an unreduced product uses at most $2L$ bits.

With ordinary long multiplication and division with remainder, one modular product costs $O(L^2)$ operations on bits. Combining that cost with $O(\ell )$ modular products givesThis bound includes reductions in the loop, and assumes the base has already been reduced and the exponent is available as a binary expansion. Reading the exponent takes $O(\ell )$ additional work; initial reduction of a very long base depends on its input length, and an inverse for a negative exponent adds further work. More generally, if a modular product costs $R(L)$ operations on bits, the powering loop costs $O(\ell R(L))$. Faster multiplication and reduction methods can improve the quadratic bound.^[ref]

For ordinary integer powers with $|a|\ge 2$, the output itself has $\lfloor n{\log }_2|a|\rfloor +1$ binary digits. Even writing it takes work proportional to that length. Consequently $O(\log (n+1))$ multiplications does not imply $O(\log (n+1))$ operations on bits: the operands grow during ordinary exponentiation by squaring, whereas modular exponentiation keeps them bounded by the modulus. A squaring or multiplication by a small fixed base can also cost less than a general product, so equal counts need not mean equal running times.

A modular symbol is the relative homology class of a path between two modular cusps. Oriented paths satisfy $\{\alpha ,\beta \}+\{\beta ,\gamma \}=\{\alpha ,\gamma \}$. Continued fractions subdivide rational-endpoint paths into unimodular edges, giving a finite presentation through Manin symbols. The Hecke operators act by finite sums of transformed paths, so their matrices can be computed by rational linear algebra.

An elliptic point of a modular curve is the image of a point of the complex upper half-plane with a nontrivial stabilizer subgroup in the effective modular subgroup. For the full modular group the stabilizer subgroup orders are two and three, represented by $i$ and $e^{2\pi i/3}$. A local disk coordinate $u$ upstairs descends as $u^e$ when the stabilizer subgroup order is $e$. The central matrices $\pm I$ are factored out before counting this order.

Direct differentiation givesConsequently this transformation takes solutions of the modified Korteweg-De Vries equation to solutions of the Korteweg-De Vries equation in these sign conventions. Both summands in $u$ have the same scaling weight, so the modified-equation scaling induces $(x,t,u)\mapsto (e^{s}x,e^{3s}t,e^{-2s}u)$.

For the three corner-normalized Lax pair eigenfunctions, let $s={\mu }_3(0,0,k)$ and $S=e^{4i{k}^{3}T{\hat{\sigma }}_3}{\mu }_2(0,T,k{)}^{-1}$. Their constant connection matrices give the displayed global relation. If their second columns are $(b,a{)}^T$ and $(B,A{)}^T$, its upper-right entry is $Ab-Ba=e^{8i{k}^{3}T}{\mu }_{3,12}(0,T,k)$. The right side is analytic in the lower half-plane. In the linear limit this is the backward-sign Airy half-line global relation at Fourier parameter $2k$, coupling all three boundary traces rather than allowing three independent boundary conditions.

For $k\ne 0$, differentiation gives $dW=-2e^{-ikz-\lambda \bar{z}/(ik)}(q_{z\bar{z}}-\lambda q)dz\wedge d\bar{z}$. Thus the differential form is closed precisely for the modified Helmholtz equation $\Delta q-4\lambda q=0$. Integrating around an oriented domain gives a global relation. The primitive $\mu e^{-ikz-\lambda \bar{z}/(ik)}$ also produces compatible first-order equations whose spectral jumps reconstruct $q$.

Here $B(p)=\sqrt{p^2+4\lambda }$ and $S_j$ are the Fourier sine transforms of the boundary values. The modified Helmholtz closed spectral one-form and its reflected global relations remove the unknown normal traces. Equivalently, four reflected fundamental solutions give a zero-boundary Green function in the quadrant. Its inward derivatives give the displayed formula. The two terms recover the bottom and left traces respectively; matching corner values ensure continuity at the corner.

A scalar-tensor theory uses a metric tensor and an additional scalar field to mediate gravity. A scalar multiplier of the Ricci scalar makes the curvature coupling depend on that field. The Brans-Dicke theory is a standard example; the choice of scalar potential and kinetic coupling distinguishes different members of the class.

With fixed budget $s^T\pi =x$, positive definite covariance $V$, and mean vector $\mu$ independent of $s$, set $A=s^T{V}^{-1}s$, $B=s^T{V}^{-1}\mu$, $C={\mu }^T{V}^{-1}\mu$ and $D=AC-B^2>0$. The minimum variance at mean $m$ is the displayed parabola. Its upper-mean branch $m\ge Bx/A$ is efficient: the lower branch has a competitor with equal variance and higher mean. If $\mu$ is proportional to $s$, only one mean is feasible and this two-dimensional parabola degenerates.

The efficient frontier is the set of portfolios for which no other feasible portfolio has at least as much expected return and no greater variance, with at least one strict improvement.

Let terminal asset prices $S$ have mean $\mu$ and positive definite covariance matrix $V$, with a Gaussian distribution. Consider a finite optimizer of $\mathbb{E}[U({\pi }^{T}S)]$ subject to the linear budget $s^T\pi =w$. Assume $U\in C^2(\mathbb{R})$ is increasing, concave and non-affine, and its derivatives satisfy the integrability needed for differentiation and Gaussian integration by parts. Then every optimizer lies in the linear span of $V^{-1}s$ and $V^{-1}\mu$.

The first-order condition and Gaussian identity giveFor $\pi \ne 0$, the terminal wealth has full Gaussian support. Continuity and non-affine concavity make $U^{\prime \prime }<0$ on some interval, so $\mathbb{E}{U}^{\prime \prime }({\pi }^{T}S)<0$. Rearranging proves the two-fund representation. The zero portfolio belongs to the same span automatically.

The regularity conditions justify the displayed derivative proof. Affine utility is an exception to its conclusion: if means are proportional to initial prices, every budget-feasible portfolio can be optimal, including portfolios outside this span.

The set of attainable return standard deviations and expected returns for budget-feasible risky investment portfolios. Negative weights are allowed when short selling is permitted. With positive definite covariance matrix and nonconstant means, the minimum-risk boundary is a hyperbola in standard-deviation/mean coordinates; its upper branch is the mean-variance efficient frontier. Adding a risk-free asset changes the budget constraint on risky holdings and produces a different attainable set.

Removes a mapping from an operating-system process's address space. Mapping lifetime is separate from file descriptor lifetime; removal also does not supply the completed writeback promised by memory-mapped file synchronization.

A mixing-length closure uses a prescribed mixing length $\ell$ to approximate turbulent transport. For mean shear ${\partial }_{z}U$, a standard model gives eddy viscosity ${\nu }_t={\ell }^2|{\partial }_{z}U|$ and down-gradient Reynolds stress. Scalar-diffusion versions replace momentum transport by an appropriate scalar flux. The choice of length and correlations is a modelling assumption rather than an exact closure of the Navier-Stokes equation.

The support of a mixed strategy consists of the pure strategies used with positive probability. At a Nash equilibrium, every supported pure strategy is a best response, while unused strategies may give no higher payoff.

// Target: computer-science.bigb

In mitotic anaphase, sister chromatids separate into daughter chromosomes moving toward opposite poles.

Release of cohesin-mediated association permits the sister centromere regions to separate. Each former chromatid is then counted as an independent daughter chromosome. This changes the counting convention without making additional DNA copies. For the $2n=4$ example, the whole undivided biological cell transiently contains eight daughter chromosomes, four moving toward each pole. Movement toward poles and separation of the poles both contribute to partition in the illustrated animal setting.^[ref]

Model the joint distribution by multiplying a response model by a sequential dropout model. For example, the logit of dropout before visit $j$, conditional on remaining observed, can depend on the currently unobserved $Y_{ij}$ as well as previous responses and baseline covariates. The observed-data likelihood sums over the missing binary responses. Such a dependence parameter is usually weakly identified or unidentified from observed responses alone; sensitivity analysis must vary it over scientifically justified values.

Missingness depends only on fully observed covariates $X$, not on the response after conditioning on them. A responder-only marginal estimate may still be biased when response rates and outcome probabilities both vary across covariate strata. MAR standardization over a fully observed covariate weights within-stratum respondent estimates using covariate frequencies from the whole target cohort.