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Factor the expression using the complex conjugate:It is real, so

Since $z\overline{z}=|z{|}^2=x^2+y^2$, the complex conjugate givesTherefore

The complex sine function obeysThus

Using Euler's formula, the complex exponential function satisfiesHence

Since $Y_1=X_1^2$ and a standard normal variable has second and fourth moments $1$ and $3$,Thus $T_n$ has the chi-squared distribution with $n$ degrees of freedom and

Convergence in distribution $W_n\Rightarrow W$ means that the cumulative distribution functions satisfy $F_{W_n}(x)\to F_W(x)$ at every continuity point of $F_W$. The Lévy continuity theorem states that this is equivalent to pointwise convergence of characteristic functions to the characteristic function of $W$, provided the limiting function is continuous at zero.

For fixed $t$, independence and a Gaussian integral giveUsing $\log (1-z)=-z-z^2/2+O(z^3)$,Therefore ${\phi }_{V_n}(t)\to e^{-t^2/2}$, the characteristic function of the standard normal distribution. Lévy's theorem now givesThis derives the limit directly, without assuming the central limit theorem.

The standard normal density isIt is nonnegative, and the Gaussian integral gives ${\int }_{-\infty }^{\infty }\varphi (z)dz=1$, so it is a probability density function. Completing the square gives the moment-generating function of a standard normal variableThus $\mathbb{E}Z=M_Z^{\prime }(0)=0$ and

Boundedness of $(z_n)$ makes both real sequences $(x_n)$ and $(y_n)$ bounded. The Bolzano-Weierstrass theorem first gives a subsequence $(z_{n_j})$ for which $(x_{n_j})$ converges. Applying it again to $(y_{n_j})$ gives a subsubsequence for which the imaginary parts also converge. Since convergence of complex numbers is equivalent to convergence of their real parts and imaginary parts, this subsubsequence converges in $\mathbb{C}$.

Choose a closed interval containing the bounded sequence. Bisect it and retain a half containing infinitely many terms; repeat. The resulting intervals are nested and their lengths tend to zero. By the nested interval theorem they have one common point $x$, and choosing successively indexed terms from these intervals produces a subsequence converging to $x$. This proves the Bolzano-Weierstrass theorem.

First consider rotation predominantly about the third principal axis, writing ${\omega }_3=\Omega +O({\omega }_1^2+{\omega }_2^2)$. To first order the third Euler equation gives $\dot{\Omega }=0$, while the transverse components satisfyThereforeand the same equation holds for ${\omega }_2$.

Order the principal moments as $I_1<I_2<I_3$. The coefficient above is negative, so perturbations about axis three oscillate and remain bounded. Cyclically applying the same linear stability of principal-axis rotation calculation about axis one givesso that rotation is also stable. About axis two, however,which has an exponentially growing solution. Thus the intermediate axis theorem gives

Multiply the three Euler equations for a torque-free rigid body by ${\omega }_1,{\omega }_2,{\omega }_3$ and add. The cubic terms cancel, giving conservation of the rotational kinetic energyInstead multiply the equations by $I_1{\omega }_1,I_2{\omega }_2,I_3{\omega }_3$ and add. The cubic terms again cancel, givingthe squared magnitude of the conserved angular momentum. For an asymmetric body these two quadratic forms are not proportional, so the torque-free rigid-body invariants are independent.

Let $e_1,e_2,e_3$ be a principal body frame, let $I_1,I_2,I_3$ be the corresponding principal moments of inertia, and write the angular velocity as $\mathbf{\omega }={\sum }_i{\omega }_i{e}_i$. The body-frame components of the angular momentum areNo external torque acts, so $\mathbf{L}$ is constant in the fixed space frame. Since a body-fixed basis vector satisfies ${\dot{e}}_i=\mathbf{\omega }\times e_i$,Taking body-frame components gives the Euler equations for a torque-free rigid body:

Let the argument of $z$ increase by $2\pi$ along a counterclockwise path enclosing both branch points. On a large circle, the branch chosen in part (a) satisfiesThe change in the primitive is consequentlyEquivalently, this is the composition of the two reflections described by the monodromy of the complex inverse sine. Thus, for this orientation,The reverse orientation gives $\mathrm{Arcsin}(z)+2\pi$. Both continued values invert the same $z$, so $\sin (w-2\pi )=\sin w$ on an open set of inverse values. The identity theorem extends this equality to every $w\in \mathbb{C}$, proving the periodicity of the complex sine:

Start with the germ $w=\mathrm{Arcsin}(z)$ selected in part (a) and continue it along paths in $\mathbb{C}\setminus \{-1,1\}$. At the two branch points,The monodromy reflection at a square-root branch point therefore shows that continuation once around $1$ sends $w$ to $\pi -w$, while continuation once around $-1$ sends it to $-\pi -w$. Composing these reflections produces every translation by an even multiple of $\pi$. Hence the branches of the complex inverse sine are

Putand take the branch cut $[-1,1]$. On $\mathbb{C}\setminus [-1,1]$, choose the analytic branch of a square root whose upper boundary value isIts lower boundary value is $s(x-i0)=-\sqrt{1-x^2}$. This fixes the integrand to be $+1$ at $0^{+}$ as required.

For $0\le \arg z\le \pi$, start at $0^{+}$ and use any path to $z$ lying above the cut; path deformations within that region do not alter the integral by the Cauchy integral theorem. For $\pi <\arg z<2\pi$, start in the same way, continue counterclockwise around the branch point $-1$ from the upper bank to the lower bank, and then continue to $z$ through the lower half-plane. This second prescription is the continuation obtained as $argz$ increases through $\pi$.

Again put $x=1+u$. ThenThe linearized equation is thereforewhose sole characteristic exponent is $-\alpha <0$ and is independent of $T$. By delay-independent stability from an undelayed linearization,

Put $x(t)=1+u(t)$. Retaining terms linear in $u$ givesThe exponential ansatz $u=e^{\lambda t}$ gives the characteristic equation of a delay differential equationAt $T=0$ its root is $\lambda =-\alpha$, and the remaining delay roots lie far in the left half-plane, so the equilibrium is stable for sufficiently small $T$.

A stability change can first occur when $\lambda =i\omega$. Separating real and imaginary parts givesFor the first positive crossing, $\omega T=\pi /3$ and $\omega =\sqrt{3}\alpha$. ThusAll positive-frequency crossings have $\omega T=\pi /3+2\pi k$, so there is no earlier imaginary-axis crossing. Implicit differentiation ofgives, with $q=\alpha T_c$,so the conjugate pair crosses into the right half-plane as $T$ increases. Therefore the stability threshold for delayed quadratic crowding is

Let $K\subseteq \mathbb{N}$ be the undecidable set denoted $\mathbb{K}$ in the question. The language satisfies the pumping lemma for regular languages with pumping length one. For any nonempty $s\in L$, write $s=xyz$ with $x=\epsilon$, $y$ its first symbol, and $z$ the remaining suffix.

If $s\in b^{\ast }$, then $x{y}^{i}z\in b^{\ast }$ for every $i\ge 0$. Otherwise $s=wa{b}^n$ for some $n\in K$. If the chosen first symbol lies in $w$, pumping it preserves the same final suffix $a{b}^n$. If $w=\epsilon$, then $s=a{b}^n$: pumping down gives $b^n\in b^{\ast }$, while every positive pump gives $a^i{b}^n$ and all but its last $a$ can be absorbed into $w$. Thus every sufficiently long word has a valid pumping decomposition.

Nevertheless, if $L$ were regular language, closure under intersection would makeregular. A finite automaton deciding this unary language would decide whether $n\in K$ by running it on $a{b}^n$, contradicting the halting problem. ThereforeThis is a nonregular language satisfying the pumping lemma.