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Evaluation of the probability generating function at zero selects the constant coefficient, soTo identify this limit with ultimate extinction, let $E_n=\{X_n=0\}$. The absorbing-zero property of the branching process gives $E_n\subseteq E_{n+1}$. Ultimate extinction means that some finite generation is empty, namely $E={\bigcup }_{n\ge 0}{E}_n$. Continuity of probability on increasing events therefore givesThis set argument is essential: it uses absorption, rather than inferring extinction merely from a numerical limit of generation probabilities.

The binomial series givesThus $p_0=1-\alpha >0$ and, for $k\ge 1$,with the factors after $\beta$ omitted for $k=1$. The signs follow from $0<\beta <1$. As $s\uparrow 1$, the series has nonnegative terms and its sum increases to $G(1)=1$, so monotone convergence gives ${\sum }_{k\ge 0}{p}_k=1$. These are therefore genuine probabilities on the nonnegative integers, proving that $G$ is a probability generating function. Its expected value is infinite, since $G^{\prime }(s)=\alpha \beta (1-s{)}^{\beta -1}\to \infty$; finiteness of the mean is not required for this branching process.

Write $G_n(s)=1-A_n(1-s{)}^{{\beta }^n}$. The composition formula gives $A_{n+1}=A_n{\alpha }^{{\beta }^n}$, with $A_0=1$. Consequently $A_n={\alpha }^{1+\beta +\cdots +{\beta }^{n-1}}$, and summing the geometric series yields the explicit generation generating functionThis is the fractional-power offspring generating function and its iterates.

Given $X_n=k$, the next size is a sum of $k$ independent reproduction counts. Multiplication of their probability generating functions gives $\mathbb{E}[s^{X_{n+1}}\mid X_n=k]=G(s{)}^k$. Averaging this conditional expectation provesWith one initial individual, $G_0(s)=s$ and $G_1(s)=G(s)$, so $G_n$ is the $n$-fold composition of $G$ with itself.

A branching process models reproduction through successive generations. In the Galton-Watson process used here, every individual independently produces a nonnegative integer number of offspring with a common offspring distribution; these reproduction counts are independent across individuals and generations. If $X_n$ is the generation size, thenwhere the ${\xi }_{n,j}$ have that distribution and the empty sum is zero. Thus zero is absorbing: once a generation is empty, every subsequent generation is empty.

Another convolution of probability densities gives, for $0\le s\le 1$,Only this part of the density is needed because the fourth length is at most one.

Four positive lengths $a,b,c,d$ can form a nondegenerate quadrilateral if and only if every length is less than the sum of the other three. Necessity follows by following the other three sides between the endpoints of the longest side and applying the triangle inequality. For sufficiency, select a diagonal length $r$ in the intersectionBoth intervals are nonempty, and their intersection is nonempty exactly when $|a-b|<c+d$ and $|c-d|<a+b$. These are precisely the four strict length inequalities. Construct triangles with sides $(a,b,r)$ and $(c,d,r)$ on opposite sides of their common diagonal. Their union has the required simple, nondegenerate quadrilateral as boundary.

The failure events, in which one length is at least the sum of the others, are disjoint except on sets of probability zero. For the specified length $W$, independence and the density above giveThere are four symmetric choices of the overly long side, so the probability isThis is the four-sided case of the polygon probability for independent uniform rods.

Three positive lengths form a nondegenerate triangle precisely when each is less than the sum of the other two. Necessity is the triangle inequality. For sufficiency, two circles centered at the ends of one side intersect off that side whenever the remaining lengths satisfy these strict inequalities, giving the third vertex.

Failure thus occurs when $X\ge Y+Z$, $Y\ge X+Z$, or $Z\ge X+Y$. Two such inequalities cannot hold together for strictly positive lengths. Zero lengths and equality cases have probability zero because the variables have continuous probability density functions. By symmetry each failure event has the probability $1/6$ just calculated. Hence the probability of forming a triangle is

Conditioning on $U=X+Y$, independence of $Z$ gives $\mathbb{P}(Z>u)=1-u$ for $0\le u\le 1$ and zero for $u>1$. Using the preceding probability density function,

For independent random variables, the probability density function of a sum is the convolution of probability densities. Since each density is one on $[0,1]$ and zero elsewhere,This is the length of $[0,1]\cap [u-1,u]$. Therefore the density isThe two triangular areas sum to one, as a probability density function must.

With equal type probabilities, $F$ and $S$ have the same marginal probability generating function, namely $G_N((1+s)/2)$. Their assumed independence, together with $N=F+S$, givesSet $H(t)=G_N(1-t)$ for $0\le t\le 1$. The identity becomes $H(t)=H(t/2{)}^2$ and, after $k$ iterations,The finite expected value gives $G_N(1-h)=1-\mu h+o(h)$ as $h\downarrow 0$. To justify this without any analyticity assumption at the endpoint, note that $(1-(1-h{)}^N)/h\to N$ and is bounded above by $N$; dominated convergence gives the derivative $G_N^{\prime }(1-)=\mu$. Substituting the expansion into the iterated identity yieldsThe left side is independent of $k$, so it equals the limit. Thus $G_N(s)=e^{\mu (s-1)}$ on $[0,1]$ and uniqueness of the power-series coefficients proves $N\sim \mathrm{Poisson}(\mu )$. Only finite expected value was needed; the given finite variance is stronger. This proves the Poisson characterization by independent binomial splitting rather than assuming the original count distribution.

A Poisson distribution with parameter $\mu$ has probability generating function $G_N(s)=e^{\mu (s-1)}$. The preceding part gives $G_F(s)=e^{\mu p(s-1)}$, the probability generating function of a Poisson distribution with parameter $\mu p$. To prove independence, retain both counts:Consequently the joint probability generating function factors asComparing the coefficients of $s^f{t}^k$ proves that the joint probabilities are the products of the marginal probabilities. Hence $F$ and $S$ are independent, with distributions $\mathrm{Poisson}(\mu p)$ and $\mathrm{Poisson}(\mu (1-p))$. This is Poisson thinning, including the degenerate counts when $p=0$ or $p=1$.

Conditional on $N=n$, the $n$ type indicators are independent random variables with the Bernoulli distribution of parameter $p$. Thus $F\mid N=n$ has the binomial distribution and its probability generating function is $(1-p+ps{)}^n$. The general rule used here is that for a nonnegative integer-valued random variable $X$, $G_X(s)=\mathbb{E}[s^X]$; for a sum of independent such variables, their probability generating functions multiply, since the expected value of the product $s^{X_1}\cdots s^{X_n}$ factors. The law of total expectation then averages the conditional result:This calculation uses the given independent labeling of objects, including conditional on their number.

Let $S_N={\sum }_{k=1}^N{a}_k$, with $S_0=0$. The indices in the $n$th dyadic block have one sign, soA convergent sequence of partial sums must be a Cauchy sequence: for every $\epsilon >0$, all sufficiently late pairs of partial sums differ by less than $\epsilon$. The displayed differences violate this necessary condition with $\epsilon =1/4$, at arbitrarily late endpoints. Thus the series diverges, even though its individual terms tend to zero. This dyadic-block alternation of reciprocal terms has growing block lengths; the fixed-length block argument of the previous part does not apply.

Group three consecutive terms. Their block sum isSince $2n-1\ge n$ and $4n-1\ge 3n$ for $n\ge 1$, we have $|b_n|\le 1/(2n^2)$. The comparison test for series and the preceding proof of convergence of $\sum n^{-2}$ show that $\sum b_n$ converges absolutely.

This establishes convergence of the original partial sums at indices $3N$. Each intermediate partial sum differs from one of these by at most two of the original terms; those terms tend to zero. Thus all partial sums have the same limit of a sequence, which is the mechanism of convergence from fixed-length series blocks. The original series is not absolutely convergent, since its absolute-value series contains ${\sum }_n(2n-1{)}^{-1}\ge \frac{1}{2}{\sum }_n{n}^{-1}$. Therefore the original series converges conditionally.

The absolute-value series can be grouped in pairs of nonnegative terms:Indeed, $1/n^2\le {\int }_{n-1}^n{x}^{-2}dx$ for $n\ge 2$, so these positive partial sums are bounded by two. Completeness of the real numbers proves their convergence. The triangle inequality bounds every late signed sum by the corresponding late absolute-value sum, so absolute convergence implies convergence. Thus the series converges absolutely.

The positive terms $b_n=1/\log (n+1)$ decrease to zero. The alternating series test just proved therefore gives convergence of the signed series. It is conditional convergence: $\log (n+1)\le n$, so $b_n\ge 1/n$, and the absolute-value series dominates the divergent harmonic series.

The comparison test for series used here says that if $0\le c_n\le d_n$, convergence of $\sum d_n$ implies convergence of $\sum c_n$, while divergence of $\sum c_n$ implies divergence of $\sum d_n$. This follows by comparing increasing partial sums. For completeness, the harmonic series diverges because every block from $2^k$ to $2^{k+1}-1$ has sum at least $1/2$.

Orient the axis towards $n=(3/5,4/5,0)$ and measure positive angles by the right-hand rule about $n$. The vectors $n$, $m=(-4/5,3/5,0)$ and $e_3=(0,0,1)$ form a positively oriented orthonormal basis, since $n\times m=e_3$. Its change-of-basis matrix isRelative to this basis, the rotation fixes the first coordinate and rotates the last two. Writing $c=\cos \alpha$ and $s=\sin \alpha$, its matrix in the standard basis is thereforeIn particular $Rn=n$, $Rm=cm+s{e}_3$, and $R{e}_3=-sm+c{e}_3$, verifying the axis and the direction of rotation. Reversing the chosen orientation of the axis replaces $\alpha$ by $-\alpha$.

Take positive angles counterclockwise. The rotation sends the standard basis vectors to $(\cos \alpha ,\sin \alpha )$ and $(-\sin \alpha ,\cos \alpha )$. These images form the columns of its matrix, soThe columns are an orthonormal basis and the determinant is one, as required for a plane rotation.

Complex conjugation fixes $0,1,\infty$, but sends $i$ to $-i$, so it is not the identity. The preceding classification of fixed points of a Möbius transformation says that a nonidentity Möbius transformation has at most two fixed points. Since complex conjugation has at least three, it cannot be a Möbius transformation.

Write a Möbius transformation as $g(z)=(az+b)/(cz+d)$, with $ad-bc\ne 0$, on the Riemann sphere. If $c\ne 0$, infinity is not fixed, and its finite fixed points solveThe pole $-d/c$ cannot solve this equation, since substitution gives $(ad-bc)/c\ne 0$. The quadratic has one or two distinct complex roots, so these give exactly one or two fixed points.

If $c=0$, then $a,d\ne 0$ and infinity is fixed. If $a\ne d$, there is one additional finite fixed point, namely $b/(d-a)$. If $a=d$ and $b\ne 0$, the transformation is a nonzero translation, with no finite fixed point. If $a=d$ and $b=0$, it is the identity and fixes the entire sphere.

Thus a nonidentity Möbius transformation has one or two fixed points; the identity has infinitely many. The examples $z\mapsto z+1$, $z\mapsto 2z$, and $z\mapsto z$ realize the three possibilities respectively.

Use the same triangle and let $G$ consist of translations $x\mapsto x+m$ for $m\in {\mathbb{Z}}^2$. All its vertices have integer coordinates, and one vertex is the origin, soThe rotation through a quarter-turn $(x,y)\mapsto (-y,x)$ preserves this integer lattice, and hence belongs to $H$. It fixes the origin and is not the identity, whereas the only translation fixing the origin is the identity. It is therefore outside $G$, proving $G⊊H$. This is symmetry enlargement under group translates even though $T$ itself has trivial stabilizer subgroup.