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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.

Take the scalene triangle vertex set $T=\{(0,0),(1,0),(0,2)\}$. Its edge lengths are $1,2,\sqrt{5}$. Each vertex has a different unordered pair of incident edge lengths. A Euclidean isometry preserving the set must preserve these pairs, and therefore fixes every vertex.

To justify rigidity of a scalene triangle fully, Euclidean distances from an arbitrary plane point to these three fixed points determine that point uniquely: subtracting its squared Euclidean distance to the origin from the other two squared Euclidean distances determines its two coordinates. Thus an isometry fixing the three vertices fixes every plane point. Only the identity preserves $T$.

For the general union of group translates, let $h\in G$. Left multiplication by $h$ permutes $G$, soTherefore $G\subseteq H$. This containment uses only the group action; the absence of symmetries of $T$ is not needed for the containment itself.

Place the cube vertices at $(\pm 1,\pm 1,\pm 1)$. The four vertices whose coordinate product is $+1$ form $T$; those with product $-1$ form $T^{\prime }$. Within either set, two distinct vertices differ in exactly two coordinates, so every edge has length $2\sqrt{2}$ and each set is a regular tetrahedron.

A member of the cube rotation group is a signed permutation matrix of determinant one. There are $3!$ permutations of the coordinate axes and four choices of signs for each permutation that give determinant one, so $|G|=24$. Such a matrix maps a coordinate-product class to a coordinate-product class. Moreover, the rotation through a quarter-turn $(x,y,z)\mapsto (-y,x,z)$ interchanges the classes. Thus the orbit of $T$ consists of exactly $T,T^{\prime }$. By the orbit-stabilizer theorem, the tetrahedron stabilizer in the cube rotation group has order

For a finite group $G$ acting on a set $X$, write $G_x=\{g\in G:gx=x\}$ for the stabilizer subgroup and $Gx=\{gx:g\in G\}$ for the orbit of $x$. The orbit-stabilizer theorem givesIndeed, $g{G}_x\mapsto gx$ is a bijection from the left cosets of $G_x$ onto $Gx$: two images agree precisely when $h^{-1}g\in G_x$. Each coset has $|G_x|$ elements, proving the formula.

Use the reduced-denominator function unbounded on every interval:Every value is a finite real number. In any nondegenerate interval $[a,b]\subseteq [0,1]$, its interior contains infinitely many rational numbers, whereas there are only finitely many reduced fractions in $[0,1]$ with denominator at most any fixed $M$. Some rational number in $(a,b)$ therefore has denominator exceeding $M$. Hence the function is unbounded on every such interval. It reverses the positive values of the Thomae function rather than requiring an infinite value at any point.

Take the step functionIt is bounded between zero and one, and is discontinuous at $x=1/2$.

Take $f_2(x)=x$. It is continuous and bounded on $(0,1)$, with infimum zero and supremum one, neither attained on this open interval.

Take $f_1(x)=1/x$. It is continuous on $(0,1)$ and unbounded as $x\to 0$ from the right.