The $d$ distinct powers $1,a,\dots ,a^{d-1}$ all solve $X^d=1$. By the allowed root bound they are all its solutions. A residue $b$ of order $d$ must therefore equal $a^j$; conversely $\mathrm{ord}(a^j)=d/\gcd (d,j)$, because $d\mid jt$ characterizes the exponents $t$ with $(a^j{)}^t=1$. This proves the equivalence with $\gcd (d,j)=1$ and gives exactly $\phi (d)$ elements of order $d$ whenever any such element exists.

To establish existence at the full order too, put $m=p-1$ and let $N_d$ count elements of order $d$. By Lagrange's theorem for finite groups, these orders divide $m$. The first result gives $N_d=0$ or $\phi (d)$. Thereforeusing the standard divisor-sum identity for Euler's function. Equality forces every deficit to vanish, so $N_m=\boxed{\phi (p-1)}$.

Finally let $h=g^{(p-1)/8}$, which has order eight. Its fourth power is the unique nonidentity square root of one, namely $-1$. Since $h^7=h^{-1}$,because $h^{-2}=h^6=-h^2$. This proves the stated residue squares to two.

The characteristic of a field is the least positive integer $p$ with $p{1}_F=0$, or zero if none exists. Finiteness forces a positive characteristic, since the sequence of integer multiples of one must repeat. If this least integer factored into two smaller positive factors, their nonzero field multiples would multiply to zero, contradicting the absence of zero divisors. Hence it is prime. The field is a finite-dimensional vector space over its prime subfield ${\mathbb{F}}_p$, so choosing a basis gives $|F|=p^a$.

A nonzero polynomial of degree $n$ over a field has at most $n$ roots: each root supplies a linear factor by polynomial division, and induction applies to the quotient. In particular $X^n-1$ has at most $n$ roots. The elements of $F^{\times }$ whose orders divide $n$ are precisely those roots. The permitted finite-group criterion therefore proves $F^{\times }$ is cyclic.

For the final assertion take an odd Fermat prime $p=2^m+1$, $m\ge 1$. Primality forces $m$ to be a power of two: an odd factor $s>1$ in $m=2^{r}s$ would factor $(2^{2^r}{)}^s+1$ nontrivially. Since $2^m\equiv -1(\mathrm{mod}p)$, the order of two divides $2m$ but not $m$. As $2m$ is a power of two, its only divisor not dividing $m$ is $2m$ itself. ThusIt generates the full group exactly when $2m=2^m$, which holds for $m=1,2$ and fails for $m\ge 3$. Therefore the possibilities are $p=3,5$, equivalently $p\le 5$ among odd Fermat primes. The oddness is implicit in referring to two as a unit; $p=2$ is not an admissible case for that claim.

Let $C$ be the clump size and put $q=e^{-\lambda t}$. The probability that all its members die, conditional on $C=n$, is $(1-q{)}^n$. Summing the geometric series givesSubtracting from one yields $\boxed{p(t,\lambda )=q/(1-\theta +\theta q)}$, for $0\le \theta <1$.

Assuming independent clumps and independent repeats, the combined likelihood is proportional to $p^R(1-p{)}^{N-R}$. For $0<R<N$ its unique maximizer is $\hat{p}=R/N$. The relation between $p$ and $q$ is one-to-one, with $q=(1-\theta )p/(1-\theta p)$. For $t>0$ this givesIf $R=N$, the estimate is the boundary value zero; if strictly positive rates are required, it is only a limiting supremum. If $R=0$, the likelihood approaches its supremum as $\lambda \to \infty$ and there is no finite estimate. These boundary cases explain when the printed logarithmic expression is finite.

Apart from its multinomial coefficient, the likelihood is ${\prod }_{ij}{\theta }_{ij}^{n_{ij}}$. Maximizing its logarithm subject to $\sum {\theta }_{ij}=1$ gives $n_{ij}/{\theta }_{ij}$ equal to a common multiplier, which is $n$. Hence ${\hat{\theta }}_{ij}=n_{ij}/n$.

Under independence the log likelihood separates as ${\sum }_i{n}_{i\cdot }\log {\varphi }_i+{\sum }_j{n}_{\cdot j}\log {\rho }_j$. The two normalization constraints give ${\hat{\varphi }}_i=n_{i\cdot }/n$ and ${\hat{\rho }}_j=n_{\cdot j}/n$, so the fitted probabilities are $n_{i\cdot }{n}_{\cdot j}/n^2$. Therefore twice the difference of the maximized log likelihoods isThe unrestricted model has $rc-1$ free parameters and the independence model has $(r-1)+(c-1)$, leaving $(r-1)(c-1)=4$ degrees of freedom for the table. Under the null, the likelihood-ratio statistic is asymptotically chi-squared with four degrees of freedom. Since $25.192>18.467$, reject independence at the 0.1% significance level. This is evidence of association in the sampled variables; it does not by itself identify a causal direction.

The central limit theorem states that for independent identically distributed variables of finite mean $\mu$ and positive finite variance ${\sigma }^2$,For a continuous strictly increasing cumulative distribution function, its inverse is defined on $(0,1)$. Thus $\mathbb{P}(F(X)\le y)=\mathbb{P}(X\le F^{-1}(y))=y$ for $0<y<1$, proving the uniform law.

If $T=-\log (1-Y)$, then for $t\ge 0$, $\mathbb{P}(T\le t)=\mathbb{P}(Y\le 1-e^{-t})=1-e^{-t}$. Hence $f_T(t)=e^{-t}{1}_{(0,\infty )}(t)$, an exponential density of rate one. Its first two moments are one and two, by integration by parts, so $\mathbb{E}T=1$ and $\mathrm{Var}T=1$.

Although the original variables may have different distributions, their separate transforms $Y_k=F_k(X_k)$ are independent and all uniform. Consequently $T_k=-\log (1-Y_k)$ are independent identically distributed exponentials of rate one, and the statistic is exactlyThe stated central limit theorem therefore yields $\boxed{Z_n\stackrel{d}{\to }N(0,1)}$.

A material interface obeys ${\eta }_t+{\varphi }_{j,x}{\eta }_x={\varphi }_{j,y}$ at its actual height. Keeping only first-order terms and evaluating at the undisturbed interface gives ${\varphi }_{1,y}={\varphi }_{2,y}={\eta }_t$ at $y=0$. Linearized unsteady Bernoulli gives the pressure perturbation $-{\rho }_j({\varphi }_{j,t}+g\eta )$, after absorbing time-only constants into the potentials. Pressure continuity therefore givesFor $\vartheta =kx-\omega t$, choose potentials decaying away from the interface:They are harmonic and their normal derivatives equal $A\omega \sin \vartheta ={\eta }_t$ at zero. The dynamic condition reduces toIf the denser fluid is above, the squared frequency is negative. The corresponding perturbations grow or decay exponentially instead of oscillating: this is Rayleigh-Taylor instability, with growing rate $\sqrt{gk({\rho }_2-{\rho }_1)/({\rho }_1+{\rho }_2)}$ in the ideal model without surface tension.