No. The propertyis an index set: it depends only on the domain $W_n$ of the partial function computed, not on the syntax of its program. It is nontrivial. For example, an index whose domain is a singleton belongs to $\mathcal{S}$, whereas an index whose domain has two elements does not. The Rice theorem therefore shows that $\mathcal{S}$ is not a computable set, so no algorithm can decide the requested property for arbitrary $n$.

Invoking the Church–Turing thesis, this also rules out any effective procedure in the informal sense, because such a procedure would be represented by a computable function.

For each $x$, let $h(x)$ be an effectively obtained index of the partial functionThe existence of the total computable index transformation $h$ follows from the S-m-n theorem. By the recursion theorem, some $m$ satisfies $f_{m,1}\simeq f_{h(m),1}$. Therefore its domain isand hence

Apply the S-m-n theorem to $g$. It gives a total computable function $h$ such thatThe recursion theorem supplies an index $e$ satisfying $f_{e,1}\simeq f_{h(e),1}$. Consequently

Rice theorem says that every nontrivial extensional property of programs computing partial functions is undecidable. Equivalently, a nontrivial index set cannot be a computable set.

Let $A$ be such an index set. Replacing $A$ by its complement if necessary, assume that the nowhere-defined function has no index in $A$. Since the property is nontrivial, choose an index $a\in A$. If membership in $A$ were decidable, we could decide the diagonal halting problem as follows. For each $x$, defineBy the S-m-n theorem, a total computable function $q$ produces an index $q(x)$ for ${\psi }_x$. If $f_{x,1}(x)$ diverges, then ${\psi }_x$ is nowhere defined and $q(x)\notin A$. If it halts, then ${\psi }_x\simeq f_{a,1}$ and extensionality gives $q(x)\in A$. A decider for $A$ would therefore decide whether $f_{x,1}(x)$ halts, a contradiction. Hence $A$ is undecidable.

Fix an acceptable effective enumeration $f_{e,n}$ of the $n$-ary partial computable functions. The S-m-n theorem states that for every $m,n\ge 1$ there is a total computable function $s_n^m$ such thatwhere $\simeq$ means that both sides have the same value when defined and are otherwise both undefined.

The recursion theorem states that every total computable transformation $h$ of program indices has a semantic fixed point: there is an $e$ such that

By Fermat's little theorem,Thus the sum of $g$ over ${\mathbb{F}}_p^3$ counts, modulo $p$, the triples on which the ternary quadratic form $f$ vanishes.

Expand $f^{p-1}$ with the multinomial theorem. A term indexed by $r+s+t=p-1$ contributes a constant multiple ofAt least one of $r,s,t$ is strictly less than $(p-1)/2$. Its corresponding exponent is therefore in $\{0,1,\dots ,p-2\}$, so part (b), including its $k=0$ case, makes that power sum zero modulo $p$. Every term in the expansion consequently vanishes after summation. The constant term contributes $p^3\equiv 0$, and therefore

The zero triple is one solution of $f=0$. If it were the only solution, the displayed sum would be congruent to $1$ rather than $0$. Hence there is another triple $(x,y,z)$, not all divisible by $p$, satisfying

For $k=0$, the convention $0^0=1$ gives $S_0=p\equiv 0(\mathrm{mod}p)$. Now suppose $1\le k<p-1$ and choose a primitive root $g$ modulo $p$. The nonzero residues are $g^j$ for $0\le j\le p-2$, while the $x=0$ term vanishes. HenceThis is a finite geometric series. Since $g^k\not\equiv 1(\mathrm{mod}p)$ but $g^{k(p-1)}\equiv 1(\mathrm{mod}p)$ by Fermat's little theorem, multiplication by the nonzero residue $g^k-1$ givesThe factor $g^k-1$ is invertible modulo the prime number $p$, so

A primitive root modulo $p$ is an integer $g$ whose residue class generates the multiplicative group of a finite field ${\mathbb{F}}_p^{\times }$. Equivalently, its multiplicative order modulo $p$ is $p-1$, so

When $k=N/4$,One Grover search algorithm iteration therefore givesThe selective sign change $I_G$ is implemented with one application of the Boolean quantum oracle: prepare its answer qubit in $|-⟩=(|0⟩-|1⟩)/\sqrt{2}$ and use quantum phase kickback, obtaining $(-1{)}^{f(x)}$ on each $|x⟩$. The remaining operations $H_n$, $I_0$, and the final computational-basis measurement are independent of $f$. Measurement of $|{\psi }_G⟩$ returns some $x\in G$ with probability one, so exactly one query to $U_f$ suffices.

Because $H_n|0^n⟩=|{\psi }_0⟩$,which is the reflection operator in the line spanned by $|{\psi }_0⟩$. Meanwhile $I_G$ changes the sign of every good basis state and fixes every bad basis state; onit is the reflection in the $|{\psi }_B⟩$ axis. Both reflections preserve $\mathcal{P}$, so their product $Q$ does too.

WriteIn Euclidean geometry, the composition of reflections in two lines through the origin has a rotation matrix through twice the oriented angle between the lines. Hence $Q$ rotates the good-bad plane by $2\theta$ from $|{\psi }_B⟩$ toward $|{\psi }_G⟩$. In particular, the Grover rotation angle formula is

The cosmological constant energy is constant, whereas the matter density scales as $a^{-3}$. HenceThe transition occurs when $\rho +3P={\rho }_{\mathrm{dust}}-2{\rho }_{\mathrm{cc}}=0$, so

At equality, let ${\rho }_{\mathrm{cc}}={\rho }_{\mathrm{dust}}={\rho }_{\ast }$. The cosmological constant has $P_{\mathrm{cc}}=-{\rho }_{\ast }$, while nonrelativistic matter has $P_{\mathrm{dust}}=0$. ThereforeThe scale factor consequently has $\ddot{a}>0$: the expansion is accelerated at $a=a_{\mathrm{eq}}$.

For the equation of state $P=w\rho$ with positive energy density,The expansion is accelerated for $w<-1/3$, coasting for $w=-1/3$, and decelerated for $w>-1/3$.

Differentiate the Friedmann equation $H^2=8\pi G\rho /3$ and use the continuity equation $\dot{\rho }=-3H(\rho +P)$:so $\dot{H}=-4\pi G(\rho +P)$. Since $H=\dot{a}/a$,Hence the Raychaudhuri equation is

Let $f$ and $g$ be conserved quantities, allowing explicit time dependence. ThenUsing bilinearity of the Poisson bracket, commutation of partial differentiation with it, and the Jacobi identity for the Poisson bracket,Thus $\{f,g\}$ is also conserved.

By the multivariable chain rule and Hamilton's equations,The sum is the Poisson bracket $\{f,H\}$, giving

The canonical Poisson-bracket relations are

The Poisson bracket of two phase-space functions is

For canonical position and momentum coordinates $(q_i,p_i)$, Hamilton's equations are

The first model alone would suggest that more cycling is beneficial, but cycling was not randomized in that model: its coefficient can be distorted by confounding or reverse causality. The randomized incentive both substantially increased cycling and substantially increased asthma episodes.

If the incentive affects asthma only through cycling, it is an instrumental variable, and the corresponding Wald estimator of the effect per additional weekly mile isadditional episodes. Under that exclusion assumption the evidence therefore advises against cycling to work more often for this population. Without the exclusion assumption, the experiment establishes the harmful effect of the incentive itself but cannot by itself distinguish an effect of cycling from a direct effect of the monetary incentive.