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The flatness problem is that $|\Omega -1|$ grows during ordinary decelerating expansion, so its small present value requires the early density to have been extraordinarily close to critical. The horizon problem is the near uniformity of regions whose past light cones do not meet in a noninflationary hot Big Bang model.

Inflation requires accelerated expansion,For a constant equation of state $P=w\rho$, this means $w<-1/3$; representative scale factors are $a\propto t^p$ with $p>1$ and $a\propto e^{Ht}$. Equivalently, the comoving Hubble radius $(aH{)}^{-1}$ decreases. Sincemany e-folds drive the universe toward spatial flatness. A small region that was initially in causal contact is stretched to a size far larger than the later Hubble radius, allowing the observed universe to originate from one equilibrated patch.

By the primitive-root modulus classification, $P(n)>0$ only forwith $p$ an odd prime, and thenFix $k\ge 1$. The finite fibres of the Euler totient function imply that there are only finitely many possible values $m=\phi (n)$ satisfying $\phi (m)=k$. For each such $m$, the same result gives only finitely many $n$ satisfying $\phi (n)=m$. The union of these finitely many finite sets is finite, proving the claim.

By lifting a primitive root to odd prime powers, it is enough to test primitivity modulo $25$. The positive integers below eight either fail to be units or have orderswhereasTherefore the three smallest required integers are

An integer $b$ is a primitive root modulo $n$ when it is coprime to $n$ and its residue class generates the multiplicative group of integers modulo n. Equivalently,

With the restricted operations, begin in $|0⟩|0⟩$ and apply $F$ to the first register:Now applyFor every basis state this compute-phase-uncompute construction acts asThe first register is therefore the same phase state used in part (a). Apply $F^{-1}$ and measure. Again, outcome $0$ occurs with probability one for a constant string and probability zero for a balanced string. The construction uses exactly two queries to $O_x$ and only the permitted operations.

Use the available unitaries to preparein the first register andin the answer qubit. The quantum phase kickback identity givesChoose a unitary $F$ with $F|0⟩=|s⟩$, apply $F^{-1}$ to the first register, and measure it. Its amplitude at $|0⟩$ isThis equals $+1$ or $-1$ for a constant string and zero for a balanced string. Thus outcome $0$ means constant, while every other outcome means balanced, with certainty after one oracle query. This is the Deutsch-Jozsa test with an arbitrary uniform-state unitary.

There is no stable positive coexistence state: the only one is a saddle for every $\rho ,\alpha >0$ when $\beta >1$. Moreover, the Bendixson-Dulac criterion with Dulac function $B(x,y)=1/(xy)$ applies in the positive quadrant, sinceThus no periodic orbit can support long-term coexistence. Except on the saddle's stable manifold, one species is excluded: species 2 becomes extinct for initial data in the basin of $(1,0)$, ecologically when species 1 is sufficiently abundant and species 2 sufficiently scarce. On the other side, species 1 becomes extinct and the unregulated second species grows.

The Jacobian isFor $\beta >1$, the origin has eigenvalues $\rho ,1$ and is an unstable node. At $(1,0)$ the eigenvalues are $-\rho$ and $1-\beta$, so it is a stable node. At the coexistence state,so it is a saddle.

The nullclines are $x=0$, $x+\alpha y=1$, $y=0$, and $x=1/\beta$. The stable manifold of the coexistence saddle separates trajectories tending to $(1,0)$ from trajectories on which $x\to 0$ and $y$ grows without bound. The coordinate axes are invariant.

SetThenThe steady states arewhere the coexistence state is biologically relevant exactly when $\beta >1$.

The log-likelihood has derivativeDifferentiating the normalizing identity for the family givesHence an interior maximum satisfies, and under nondegeneracy uniquely satisfies,

Independence gives the joint densityBy the factorization criterion, its dependence on the sample relevant to $\theta$ is only through ${\sum }_i{Y}_i$, equivalently through $\overline{Y}$. Thus $\overline{Y}$ is a sufficient statistic for $\theta$.

A one-parameter exponential family has densitieswith support independent of the natural parameter $\theta$. In the present canonical setting the natural statistic is $T(y)=y$.

This language is not regular. If it were regular, its intersection with the regular language $0^{\ast }{1}^{\ast }$ would be regular because regular languages are closed under intersection. That intersection is exactlywhich part (ii) proved nonregular.

The language $\{0^n{1}^n:n\ge 0\}$ is not regular. For pumping length $p$, take $0^p{1}^p$. Any permitted pumped block lies entirely among the initial zeros. Pumping it down changes the number of zeros without changing the number of ones, producing a word outside the language and contradicting the pumping lemma.

The pumping lemma for regular languages states that if $L$ is regular, then some $p\ge 1$ has the following property: every $w\in L$ with $|w|\ge p$ can be written $w=xyz$ with $|xy|\le p$, $|y|\ge 1$, and $x{y}^{i}z\in L$ for every $i\ge 0$. Indeed, while a deterministic finite automaton with $p$ states reads the first $p$ symbols, two of the first $p+1$ visited states coincide; the intervening nonempty loop may be traversed any number of times.

The language $\{0^{n^2}1:n\ge 0\}$ is not regular. If its pumping length were $p$, apply the lemma to $w=0^{p^2}1$. The pumped block is $y=0^r$ for some $1\le r\le p$. Pumping once more gives $p^2+r$ zeros, butso their number is not a square.