For an $m$-quasirandom group, uniform expectations, and the normalized convolution on a finite group, a scalar mean-zero function $f$ satisfiesThe Fourier analysis on a finite group proof bounds each nontrivial matrix component of $f$ in operator norm using its weighted Hilbert-Schmidt norm. For subsets of subset density values $a,b,c$, the error in their normalized product count is at most $\sqrt{a(1-a)b(1-b)c(1-c)/m}$. In particular $abc>1/m$ guarantees a solution of $xy=z$ in the three subsets.

The top quark is the up-type member of the third quark generation. Its mass exceeds the W boson mass, so a real W boson cannot decay into an on-shell top quark. A massless approximation for accessible decay products must not silently remove this channel threshold.

Apply a Controlled-Z gate to input $a|0⟩+b|1⟩$ and a fresh $|+⟩$ quantum ancilla. Measuring the input in the equatorial qubit measurement basis at angle $\alpha$, with outcome $s$, leaves the unnormalized outputEach branch has probability $1/2$, independently of the input. The logical output is $J(\alpha )|\psi ⟩$ with known Pauli frame $X^s$; the input qubit has been measured, so this does not clone it.

For $0\le A<I$ and normalized $|b⟩={\sum }_j{\beta }_j|u_j⟩$, filtering by $h(\lambda )=\lambda$ succeeds withThe inequality follows by replacing each nonnegative ${\lambda }_j^2$ by its minimum and using ${\sum }_j|{\beta }_j{|}^2=1$. Equality holds exactly when all occupied eigenvectors belong to the minimum-eigenvalue eigenspace. If ${\lambda }_{\min }=0$, success is positive exactly for inputs outside the kernel of $A$.

If $|b⟩$ lies in the kernel of $A$, then $A|b⟩=0$ has no normalized quantum state. A request to produce that normalized vector with positive probability is consequently undefined. For example, $A=\mathrm{diag}(0,1/2)$ on one qubit has distinct dyadic eigenvalues, but annihilates $|0⟩$. The corrected filtering assertion requires $A|b⟩\ne 0$; it cannot be repaired by assigning a positive success probability to this input.

Query a known set of $m$ inputs and store its output table. If no function collision occurs there, each of its $m$ distinct outputs has exactly one partner in the complement, for a two-to-one function. A compute-phase-uncompute construction marks those partners using two function queries and reversible table comparison. Known-subset Grover search then uses $O(\sqrt{(N-m)/m})$ such phase queries. Including table preparation and a final verification query givesChoosing $m\asymp N^{1/3}$ balances both terms and gives $T(m)=O(N^{1/3})$. The Grover rotation angle rounding bound gives success tending to one. This is a query bound; it does not make reversible table lookup free in a gate or physical-memory cost model.

If $q$ is the quantile function of an absolutely continuous probability distribution with probability density function $f$, then $q^{\prime }(u)=1/f(q(u))$ for almost every $u\in (0,1)$. Gaps in the support may give jumps in $q$, so the identity concerns its ordinary almost-everywhere derivative. It follows by differentiating the inverse relation to the cumulative distribution function.

For $Q(u)=\inf \{x:F(x)\ge u\}$, $0<u<1$, the quantile function is a monotone function and is left-continuous. The distribution function is right-continuous, giving $Q(u)\le x$ exactly when $u\le F(x)$. If ${\sup }_{v<u}Q(v)<Q(u)$, a point between them has $F(x)<u$ and contradicts this equivalence for some $v\in (F(x),u)$. A jump just to the right of $u$ is possible when $F$ is flat on an interval.

For a Newtonian Kepler orbit, let $M=M_1+M_2$ be the total mass and $\mu =M_1{M}_2/M$ the reduced mass; put $k=GM$, $\mathbf{n}=\mathbf{r}/r$, $u=\dot{r}$ and ${\mathbf{v}}_{\perp }=\dot{\mathbf{r}}-u\mathbf{n}$, with $v^2=\dot{\mathbf{r}}\cdot \dot{\mathbf{r}}$. In the convention $q_{ij}=\mu (3r_i{r}_j-r^2{\delta }_{ij})/2$, differentiating the acceleration $\ddot{\mathbf{r}}=-k\mathbf{r}/r^3$ givesThe two terms have zero cross tensor contraction, with squared norms $6u^2$ and $72v_{\perp }^2$. Therefore ${\stackrel{...}{q}}_{ij}{\stackrel{...}{q}}_{ij}=72{\mu }^2{k}^2(v^2-11u^2/12)/r^4$. Substitution into the quadrupole formula in this convention, $L=4G{\stackrel{...}{q}}_{ij}{\stackrel{...}{q}}_{ij}/(45c^5)$, proves the luminosity. It is nonnegative since $v^2-11u^2/12=v_{\perp }^2+u^2/12$. The formula uses a weak gravitational field, slow motion and leading radiation order; orbital averaging gives a secular luminosity.

For a slowly evolving circular binary star with separation $a$, reduced mass $\mu$ and orbital angular frequency $\Omega$, the leading gravitational wave losses satisfy $\dot{E}=\Omega \dot{J}$. One direct check uses the standard quadrupole angular-momentum loss lawwhere $Q$ is the mass quadrupole moment and ${ϵ}_{ijk}$ is the Levi-Civita symbol. For $\mathbf{r}=a(\cos \Omega t,\sin \Omega t,0)$, put $A=\mu a^2$; then $Q_{xx}=A(1/6+\cos (2\Omega t)/2)$, $Q_{yy}=A(1/6-\cos (2\Omega t)/2)$, $Q_{xy}=A\sin (2\Omega t)/2$, and $Q_{zz}=-A/3$. Substitution gives ${\dot{J}}_z=-32G{A}^2{\Omega }^5/(5c^5)$ and the quadrupole formula gives $\dot{E}=-32G{A}^2{\Omega }^6/(5c^5)$. Their ratio proves the balance relation. The same relation follows from the rotating wave's angular harmonic number two and frequency $2\Omega$. The additional angular-momentum flux law, not the energy flux alone, is needed to justify preservation of secular circularity.

For a point mass moving as $z=L\cos \omega t$ along one fixed axis, the trace-free mass quadrupole moment is $m{L}^2{\cos }^2\omega t\mathrm{diag}(-1/3,-1/3,2/3)$. Its radiating component has twice the source angular frequency. The quadrupole formula gives $⟨P⟩=16G{m}^2{L}^4{\omega }^6/(15c^5)$ for this contribution in the leading slow-motion weak-field approximation. Any physical driver or support has its own quadrupole contribution; this expression is for the prescribed mass alone.

For two equal point masses at $\pm z$ falling from rest at $\pm z_0$ under Newtonian gravity, in geometrized units one has $\ddot{z}=-m/(4z^2)$ and $\dot{z}=-\sqrt{m(1/z-1/z_0)/2}$. The only nonzero component of the second mass moment tensor is $I_{zz}=2m{z}^2$, and its third derivative is $-m^2\dot{z}/z^2$. The trace-free mass quadrupole moment and quadrupole formula give the displayed power. From infinity, integrating down to $z=z_c$ gives $E=2\sqrt{2}{m}^{9/2}/(105z_c^{7/2})$. Relative to the total initial mass $2m$ the fraction is $\sqrt{2}(m/z_c{)}^{7/2}/105$. The weak-field slow-motion approximation must be distinguished from its extrapolation to a compact endpoint.

On a compact interval, positive quadrature rule weights whose sum is the interval length define uniformly bounded functionals on continuous functions. If the rules eventually integrate every fixed polynomial exactly, the Weierstrass approximation theorem proves convergence for every continuous function.

For a quadratic potential with the end set by ${ϵ}_V=1$, a mode leaving $N$ e-folds earlier has ${\varphi }_{\ast }^2=(4N+2)M_{\mathrm{pl}}^2$. Its curvature amplitude is $(m/M_{\mathrm{pl}}{)}^2(2N+1{)}^2/(24{\pi }^2)$. Thus the scalar amplitude fixes the inflaton mass while the tilt and tensor ratio depend only on the exit e-fold count at this order.

The quadratic potential has equal potential slow-roll parameters, ${ϵ}_V={\eta }_V=1/(2N+1)$. Consequently $n_s=1-4/(2N+1)$ and $r=16/(2N+1)$ at leading order. The relation $r=4(1-n_s)$ is independent of the inflaton mass, so a satisfactory scalar tilt does not by itself guarantee a satisfactory tensor prediction.

Conditioning on selecting item $i$ leaves capacity $B-w_i$. Its fractional knapsack problem bounds that item's total interaction with the other selected items. Thus the ordered-pair quadratic objective is at most ${\sum }_i(v_i+q_i)x_i$. Remove overweight items before defining their residual-capacity problems.

For a real quadratic $a{s}^2+bs+c$ with $a>0$, strict negativity occurs between its two distinct real roots, if they exist. Completing the square gives discriminant $b^2-4ac>0$ as the existence condition. A physical domain restriction must be intersected with that interval.

For a quadratic equation $a{x}^2+bx+c=0$ with $a\ne 0$, completing the square givesOver the complex numbers both roots exist, with a repeated root of a polynomial when the discriminant $b^2-4ac$ is zero. Over the real numbers the discriminant determines whether there are two distinct roots, one repeated root, or no real root.

For $\lambda \to +\infty$ and $0\le \theta \le \pi$, consider the Gaussian integral ${\int }_{-1/\sqrt{2}}^{e^{i\theta }}{e}^{-\lambda z^2}dz$. Here, a saddle point contribution dominates for $0\le \theta \le \pi /4$. The moving endpoint contribution in steepest descent dominates for $\pi /4<\theta <5\pi /6$. At $5\pi /6$ the two endpoints have equal exponential rates and both are needed; for larger angles the fixed endpoint dominates. Saddle connectivity switches at $\pi /2$, but that change is subdominant in the leading endpoint regime.

The product preparations $00,0+,+0,++$ are excluded respectively by the orthonormal vectors $(01+10)/\sqrt{2}$, $(00-01+10+11)/2$, $(00+01-10+11)/2$ and $(00-11)/\sqrt{2}$, written in the computational basis. Each vector has zero inner product with its labelled preparation. This projective measurement gives antidistinguishable quantum states and drives the two-copy PBR theorem contradiction.