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Independence gives $S_n\sim N(0,n)$. For every integer $m$,since subtraction of an integer does not change the exponential. These are precisely the Fourier coefficients of the distribution on the circle.

One can also obtain a direct density proof on $[0,1]$, avoiding an endpoint issue in transferring circle convergence. Periodize the normal density:Its coefficients are the displayed Gaussian values, so its absolutely convergent Fourier series isThe sum of the absolute values of the nonconstant coefficients tends to zero, for example by domination by the summable sequence at $n=1$. Therefore $f_n\to 1$ uniformly, and integrating any bounded continuous function on $[0,1]$ givesIn fact the convergence is in total variation.

A bounded linear operator is compact if it maps the unit ball to a set with compact closure; equivalently, every bounded sequence has a subsequence whose images converge.

If $S,T$ are compact operators, select a subsequence on which $S{x}_n$ converges and then a further subsequence on which $T{x}_n$ converges. This proves compactness of any linear combination. For norm closure, suppose compact $T_j$ converge to $T$ in operator norm. Given $\epsilon >0$, choose $j$ with $\Vert T-T_j\Vert <\epsilon /3$. A finite $\epsilon /3$-net for $T_j$ of the unit ball is then a finite $2\epsilon /3$-net for its image under $T$. This image is totally bounded, and its closure is compact because $X$ is complete. Thus $B_0(X)$ is a closed linear subspace.

For bounded $S$ and compact $T$, the image of a bounded sequence under $S$ is bounded, so $TS$ is compact; and a convergent subsequence of $T{x}_n$ remains convergent after applying the continuous map $S$, so $ST$ is compact. Therefore the compact operators form a closed two-sided ideal in $B(X)$.

For the weighted backward shift, truncate its output after coordinate $N$, obtaining a finite-rank operator $T_N$. The exact norm estimate isHence $T$ is compact. Its iterates areThus for every nonzero complex $\lambda$ the Neumann seriesconverges absolutely in operator norm. Zero is an eigenvalue, since $T{e}_1=0$, whereas every nonzero value is in the resolvent set. Consequently

The Snake lemma applies to a commutative diagram of modules with exact rows $0\to A^{\prime }\stackrel{i^{\prime }}{\to }{B}^{\prime }\stackrel{j^{\prime }}{\to }{C}^{\prime }\to 0$ and $0\to A\stackrel{i}{\to }B\stackrel{j}{\to }C\to 0$, and vertical maps $f,g,h$. It supplies the exact sequenceFor $c^{\prime }\in \ker h$, choose $b^{\prime }$ with $j^{\prime }{b}^{\prime }=c^{\prime }$. Commutativity gives $jg{b}^{\prime }=0$, so $g{b}^{\prime }=i(a)$ for a unique $a\in A$. Define $\delta (c^{\prime })=[a]$ modulo $f(A^{\prime })$. Changing $b^{\prime }$ by $i^{\prime }(a^{\prime })$ changes $a$ by $f(a^{\prime })$, proving well-definedness. This construction is additive. Exactness follows by the same lift-and-chase: $\delta (c^{\prime })=0$ exactly when the lift can be adjusted into $\ker g$; a class in $\mathrm{coker}f$ maps to zero in $\mathrm{coker}g$ exactly when it is obtained from such a lift. The remaining kernel and cokernel positions follow directly from row exactness.

For an open cover $X=U\cup V$, use the short exact sequence of singular chain complexeswhere the last complex consists of chains whose simplices lie in one member of the cover. This is exact because a simplex present in both summands lies in their intersection. Barycentric subdivision and its chain homotopy to the identity show that this small-chain inclusion induces an isomorphism on homology: compactness of each simplex and a Lebesgue-number argument place sufficiently subdivided simplices inside the cover.

Applying the Snake lemma to a short exact sequence of chain complexes gives the connecting homomorphism $[c]\mapsto [da]$: lift a quotient cycle $c$ to the middle complex, take its differential, and identify that differential with an element of the subcomplex. Its independence of the lift and representative follows by the same chase. Repeating this degree by degree, or chasing the cycles and boundaries directly, givesThis is the Mayer–Vietoris sequence. Specifically, split a small cycle as $a+b$; then $da=-db$ lies in the intersection, and $\delta [a+b]=[da]$ with the sign convention above.

For the final truncation, use homological grading $d:C_j\to C_{j-1}$. The span $B$ in degrees below $n$ is a subcomplex. The span called $A$ in degrees at least $n$ need not be a subcomplex with the inherited differential: $d{C}_n$ may lie in $C_{n-1}$. Give $A$ the quotient differential, with $d:A_n\to A_{n-1}=0$ zero and all higher differentials inherited. Thenis a short exact sequence of chain complexes. Its only potentially nonzero connecting map isIndeed $H_n(A)=C_n/d{C}_{n+1}$; $da$ is a cycle in $B$, and changing $a$ by $db$ changes $da$ by zero. In other degrees either the source or the relevant target of the connecting map vanishes.

A rational root of the monic cubic would be one of $\pm 1,\pm 3$, and substitution excludes all four. A reducible cubic over $\mathbb{Q}$ would have a rational root, so the polynomial is irreducible andIts polynomial discriminant is $-4(-1{)}^3-27(3{)}^2=-239$. The integer 239 is prime: it has no prime divisor among $2,3,5,7,11,13$, the primes not exceeding its square root.

The discriminant-index formula for an integral lattice for an integral primitive element isBoth discriminants are integers, and the left side is $-239$. Its squarefreeness forces the positive index to be one. Since $\alpha$ is an algebraic integer by its monic equation,

Since $a=kp\ge 2$, we have $a^p\equiv 1(\mathrm{mod}{a}^p-1)$ and $\gcd (a,a^p-1)=1$. Its multiplicative order therefore divides the prime number $p$. It cannot be one, because $0<a-1<a^p-1$. Thus the order is exactly $p$.

To find the desired prime divisor, use the cyclotomic polynomialChoose a prime factor $q$ of this integer. Since $p\mid a$, we have ${\Phi }_p(a)\equiv 1(\mathrm{mod}p)$, so $q\ne p$. Also $q\nmid a$. If $a\equiv 1(\mathrm{mod}q)$, then ${\Phi }_p(a)\equiv p(\mathrm{mod}q)$, contradicting $q\ne p$. Consequently $a$ has multiplicative order $p$ in the multiplicative group of a finite field ${\mathbb{F}}_q^{\times }$. Lagrange theorem givesThis $q$ divides $a^p-1$ as required. Notice that order $p$ modulo a composite integer alone would not justify selecting an arbitrary prime divisor; the cyclotomic polynomial selects a suitable one.

If there were only finitely many such primes $q_1,\dots ,q_r$, choose $k=q_1\cdots q_r$, with the empty product interpreted as one. Each $q_i$ divides $a=kp$ and hence does not divide $a^p-1$. The preceding argument supplies a further prime $q\equiv 1(\mathrm{mod}p)$ dividing that integer, a contradiction. There are infinitely many primes in this residue class.

The circle group is $U(1)=\{z\in \mathbb{C}:|z|=1\}$ under multiplication. For continuous finite-dimensional complex representations, its irreducibles are the one-dimensional charactersA group representation is a continuous homomorphism to the invertible linear maps; it is irreducible when it has no proper nonzero invariant subspace. Compactness permits an invariant Hermitian inner product by averaging. Since the resulting unitary operators commute for $U(1)$, simultaneous diagonalization reduces an irreducible to a character, and the continuous characters of the circle have the stated integer winding number.

The group $SU(2)$ consists of unitary two-by-two complex matrices of determinant one. Each has the unique formidentifying it homeomorphically with the unit 3-sphere in ${\mathbb{R}}^4$. This is the Spin group $\mathrm{Spin}(3)$. A unitary matrix is conjugate to $\mathrm{diag}(e^{i\theta },e^{-i\theta })$, and its conjugacy class is determined by $0\le \theta \le \pi$, or equivalently its trace $2\cos \theta$.

The irreducible representations of SU2 are $V_n={\mathrm{Sym}}^n({\mathbb{C}}^2)$ for integers $n\ge 0$, with dimension $n+1$. The symmetric power can be realized as homogeneous degree-$n$ polynomials in two variables, with the induced linear action. On the diagonal torus its eigenvalues are $e^{i(n-2j)\theta }$ for $j=0,\dots ,n$, so its character, the trace of the representing operator, isAt the endpoints use the continuous values ${\chi }_n(0)=n+1$ and ${\chi }_n(\pi )=(-1{)}^n(n+1)$. The highest-weight classification says this list is complete.

For the final action, multiplication of block matrices gives $(AB{)}_1=A_1{B}_1$, so conjugation is a linear group representation. Decompose the underlying three-dimensional space as $V_1\oplus V_0$. Its endomorphism space isThe two off-diagonal blocks are each $V_1$, since the invariant alternating form identifies $V_1^{\ast }$ with $V_1$. The scalar part of $\mathrm{End}(V_1)$ is $V_0$ and its traceless part is $V_2$: the same alternating form identifies traceless endomorphisms with symmetric tensors of degree two. The lower scalar block is a second $V_0$. ThusThe dimensions add to $2+4+3=9$, and its character is $2{\chi }_0+2{\chi }_1+{\chi }_2$.

For a connected plane drawing, Euler formula for a connected planar graph isFor a simple planar graph with $v\ge 3$, counting edge sides around faces gives $3f\le 2e$ and hence $e\le 3v-6$. Thus some vertex has degree at most five. A tree or a graph with bridges obeys the same bound; a face boundary is counted as a closed walk with edge multiplicity, and the small cases are immediate.

Induct on the number of vertices to prove the five colour theorem. Remove a vertex $v$ of degree at most five and five-colour the remainder. If fewer than five colors appear at its neighbours, extend the coloring. Otherwise it has five neighbours in their cyclic order, with colors $1,2,3,4,5$. If the color-1 neighbour is not connected to the color-3 neighbour by a path using only colors 1 and 3, swap these colors in the component containing the former, freeing color 1 at $v$. If they are connected, that path together with its two edges to $v$ is a simple closed curve. The colour-2 and colour-4 neighbours lie on opposite sides, so cannot be joined by a path using only colors 2 and 4, which would have to cross it. Swap colors in the component containing the colour-2 neighbour and give $v$ color 2. This establishes the induction without the four colour theorem.

For a triangle-free simple planar graph with $v\ge 3$, the analogous bound is $e\le 2v-4$, so some vertex has degree at most three. One can prove the bound for connected graphs with no bridges by $4f\le 2e$; bridges and isolated components are then joined or split off, with the small tree cases checked directly. Every subgraph is still triangle-free and planar. Successively remove a vertex of degree at most three, and color in reverse order with four colors. Thus triangle-free planar graphs have chromatic number at most four.

Finally, let $G$ be minimal 5-chromatic. Removing any vertex leaves a graph colorable with four colors. If its degree is at most three, that coloring extends immediately. If its degree is four, either a colour is unused at its neighbours, or all four appear once in cyclic order. The same two-path argument, now using opposite pairs 1,3 and 2,4, permits a Kempe chain interchange freeing one colour. Therefore a planar minimal 5-chromatic graph must satisfyWithout planarity only $\delta (G)\ge 4$ follows from the elementary extension argument. The complete graph $K_5$ is a counterexample to the stronger assertion: it is minimal 5-chromatic and every vertex has degree four. Deleting an edge allows its endpoints to share a color; deleting a vertex also reduces the required number of colors.

The completeness theorem for propositional logic states that $\Gamma \models \varphi$ implies $\Gamma ⊢\varphi$: every semantic consequence is formally derivable. Equivalently, every syntactically consistent set of formulas has a truth assignment satisfying it.

First prove the equivalent formulation. With countably many primitive propositions there are countably many formulas; enumerate them as ${\varphi }_1,{\varphi }_2,\dots$. Starting from a consistent $\Gamma$, construct increasing consistent sets ${\Gamma }_n$ by adjoining ${\varphi }_n$ if this remains consistent, and otherwise adjoining $\neg {\varphi }_n$. The second choice is consistent: if adjoining either choice were inconsistent, the deduction theorem for propositional logic would give both $\neg {\varphi }_n$ and $\neg \neg {\varphi }_n$ as consequences of ${\Gamma }_{n-1}$, contradicting consistency.

The union $\Delta ={\bigcup }_n{\Gamma }_n$ is consistent, since a finite proof of a contradiction would already use premises in one stage. It decides every formula. It is also deductively closed: if $\Delta ⊢\psi$ and $\psi \notin \Delta$, then $\neg \psi \in \Delta$, violating consistency.

Give an atomic proposition $P$ the truth value true exactly when $P\in \Delta$. Induction proves the truth lemma. For negation, exactly one of $\psi ,\neg \psi$ lies in $\Delta$. For implication, if $\psi \to \chi \in \Delta$ and $\psi \in \Delta$, modus ponens gives $\chi \in \Delta$. Conversely, if $\psi \notin \Delta$, then $\neg \psi \in \Delta$ entails $\psi \to \chi$ by a classical propositional tautology; if $\chi \in \Delta$, it too entails $\psi \to \chi$. Hence membership of an implication agrees with its truth table. Other connectives follow from their definitions or the corresponding tautologies. Thus the assignment satisfies $\Delta$, and therefore $\Gamma$.

Now if $\Gamma ⊬\varphi$, the set $\Gamma \cup \{\neg \varphi \}$ is consistent, by the deduction theorem for propositional logic and classical double-negation elimination. Its satisfying assignment is a model of $\Gamma$ in which $\varphi$ is false. Contraposition provesThe reverse implication is soundness theorem for propositional logic, proved by checking axioms and inference rules preserve truth.

For uncountably many primitive propositions, replace the enumeration by a maximal-consistent extension using Zorn lemma. The union of a chain of consistent extensions is consistent for the same finite-proof reason. Maximality then decides each formula: if adjoining $\psi$ creates inconsistency, the deduction argument gives $\neg \psi$ already in the extension. The same truth lemma applies with no change. Alternatively, well-order the formulas and run the same recursion with unions at limit stages. This modification uses a choice principle, unlike the countable enumeration argument.

For a canonical coordinate and momentum, the Poisson bracket isThe chain rule and Hamilton equations $\dot{q}=H_p$, $\dot{p}=-H_q$ giveFor the stated complex variables, direct differentiation givesSince $H=\omega a{a}^{\ast }$, the product rule for the Poisson bracket yieldsThe linear change of variables is invertible, so treating $a,a^{\ast }$ as independent coordinates when differentiating givesIn particular $a(t)=a(0)e^{-i\omega t}$ and $a^{\ast }(t)=a^{\ast }(0)e^{i\omega t}$. On a continuously chosen branch along a nonzero trajectory,are constant. The logarithm is undefined at the zero-energy equilibrium, and a fixed principal branch need not be continuous across its cut; the branch qualification is necessary.

A fixed point is locally attracting when its multiplier satisfies $|F^{\prime }(x_{\ast })|<1$ and repelling when $|F^{\prime }(x_{\ast })|>1$. A multiplier crossing $+1$ can create or destroy fixed points in a fold bifurcation, or exchange their stability when branches cross, as in a transcritical bifurcation. A multiplier crossing $-1$ gives a flip, or period-doubling bifurcation, in which a two-cycle can emerge. These are the two multiplier mechanisms; additional symmetry can distinguish special cases at $+1$.

For the logistic map, the fixed points areThe second lies in $[0,1]$ only for $\mu \ge 1$. Their multipliers are $\mu$ and $2-\mu$. Thus zero attracts for $0<\mu <1$; the branches meet at $\mu =1$ and exchange stability, with the nonzero branch attracting for $1<\mu <3$. At $\mu =1$, positive orbits decrease to zero since $F(x)=x-x^2$. At $\mu =3$, put $x=2/3+z$. Then $z\mapsto -z-3z^2$ and its second iterate isso the fixed point is still weakly attracting at the bifurcation, although linear attraction has disappeared.

The supplied factorization gives the nonfixed two-cycle points for $\mu >3$:They are interchanged by $F$. Using their sum and product in $F^{\prime }(x_{+})F^{\prime }(x_{-})$ givesThe two-cycle is attracting exactly forIt is born from the fixed point at multiplier $+1$ for the second iterate, and at $\mu =1+\sqrt{6}$ its multiplier reaches $-1$, giving the next period doubling. The boundary cycles are nonhyperbolic; the strict multiplier test is not a proof of attraction there.

Beyond this threshold the logistic map undergoes further period doublings, with attracting cycles of periods $4,8,16,\dots$ and then chaotic parameter ranges, interspersed with periodic windows. It is not correct to say that every parameter above this threshold is chaotic. At $\mu =4$, the substitution $x={\sin }^2(\pi \theta )$ gives $F(x)={\sin }^2(2\pi \theta )$, illustrating its chaotic doubling-map behavior.