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Complete the square: $z^2-2iz=(z-i{)}^2+1$. The pole of $(1-iz{)}^{-1}$ is at $z=-i$, so the contour may be shifted upward through the saddle point $z=i$ without crossing a singularity. Setting $z=t+i$ gives the exact steepest-descent representationNear the saddle,Odd terms integrate to zero, whileThe Watson lemma therefore yields the full asymptotic expansion

One form of Watson lemma is this. If $f(t)$ has the asymptotic expansion $f(t)\sim {\sum }_{n\ge 0}{a}_n{t}^{{\lambda }_n-1}$ as $t\downarrow 0$, with $0<{\lambda }_0<{\lambda }_1<\cdots$, and the Laplace integral has suitable growth control away from zero, thenas $x\to \infty$, with truncation after any fixed number of terms justified by the corresponding remainder estimate.

True. Exhaust the quadrant by compact sets $K_n$ with connected complements and with every compact subset of $\Omega$ eventually contained in the interior of $K_n$. Runge's theorem gives a polynomial $P_n$ satisfyingThus $P_n\to f$ locally uniformly. For a fixed $z\in \Omega$, choose a small circle about $z$ eventually contained in every $K_n$. The Cauchy integral formula for derivatives applied to $P_n-f$ givesfor every $r\ge 0$.

True when “uniform” means on all of the unbounded quadrant $\Omega$. A uniformly convergent sequence $(P_n)$ is uniformly Cauchy. For sufficiently large $m,n$, the polynomial $P_n-P_m$ is bounded on $\Omega$. Every nonconstant polynomial is unbounded along some ray contained in $\Omega$, so $P_n-P_m$ is constant. Fixing one large $m$, the limit is therefore $P_m$ plus the limit of constants, hence is a polynomial.

Condition on the information through time $n-1$. Independence of ${\xi }_n$ and the definition of the Bellman equation for terminal-wealth utility implybecause the actual holding is one candidate in the supremum. Thus $V(n,X_n)$ is a supermartingale for every strategy.

For the stated strategy it is a martingale, soEvery competing strategy has expected terminal value at most $V(0,X_0)$ by the supermartingale inequality from part (a), evaluated at the bounded time $N$. Hence ${\theta }^{\ast }$ is optimal.

If ${\theta }_n$ shares are held during period $n$, the remaining wealth $X_{n-1}-{\theta }_n{S}_{n-1}$ is in the bank account. Self-financing therefore giveswhich rearranges to the required equation.

For a bounded stopping time $T\le N$, writeThe indicator is measurable at time $n-1$. Taking conditional expectations term by term and using the supermartingale inequality gives $\mathbb{E}{\hat{M}}_T\le \mathbb{E}{\hat{M}}_0$. The same calculation gives equality for the martingale $M$. Since their initial values agree, $\mathbb{E}{M}_T\ge \mathbb{E}{\hat{M}}_T$.

For bootstrapping, repeatedly sample $X_1^{\ast },\dots ,X_n^{\ast }$ with replacement from the empirical distribution of the observations and compute ${\hat{\beta }}^{\ast }=\alpha /{\overline{X}}^{\ast }$. Conditional on the data, estimate the $\gamma /2$ and $1-\gamma /2$ quantiles $q_l,q_u$ of $\sqrt{n}({\hat{\beta }}^{\ast }-\hat{\beta })$. The basic bootstrap interval isAs the number of resamples tends to infinity it estimates the conditional quantiles; bootstrap consistency and the continuous mapping theorem then give asymptotic coverage $1-\gamma$.

Let $z_{1-\gamma /2}$ be the corresponding standard normal distribution quantile. Slutsky theorem permits replacing $\beta$ by the consistent estimator $\hat{\beta }$, giving the confidence intervalAfter standardization, the preceding convergence in distribution and continuity of the normal distribution show that its coverage tends to $1-\gamma$.

The log-likelihood is, up to constants, $n\alpha \log \beta -\beta {\sum }_i{X}_i$, so the maximum-likelihood estimator is $\hat{\beta }=\alpha /\overline{X}$. By the central limit theorem,Applying the delta method to $g(x)=\alpha /x$ gives

In equilibrium the chain is reversible by detailed balance. Reverse time. Forward departures become reverse accepted arrivals; reversibility identifies the reversed chain with the original one, whose attempted arrivals form a Poisson process. The accepted points are exactly those occurring while the state is below $N$. The standard quasi-reversibility argument therefore makes the equilibrium departure process Poisson, with rate equal to the mean departure rate

On states $0,\dots ,N$, births have rate $1$ below $N$ and deaths have rate $\mu$ above $0$. For $\mu >0$ this finite continuous-time Markov chain is irreducible, so its unique stationary law isFor $\mu =1$ this is uniform. If the parameter value $\mu =0$ is admitted, state $N$ is absorbing and the unique stationary law is instead the point mass at $N$.

The queue length is a birth-death process with birth rate $\lambda r^n$ in state $n$ and death rate $\mu \min (n,2)$. Its reversible weights satisfyFor $0<r<1$, the factor $r^{n(n-1)/2}$ makes their sum finite for every $\lambda ,\mu >0$. For $r=1$, the tail is geometric with ratio $\lambda /(2\mu )$, so a stationary distribution exists exactly when $\lambda <2\mu$.

Write a conic as $x^{T}Ax=0$ for a symmetric $3\times 3$ matrix $A$, with off-diagonal entries equal to half the corresponding coefficients. A union of two distinct lines has matrix rank two. Thus $R^{\circ }$ is the rank-two part of the determinant hypersurface $\det A=0$. Its closure also contains rank-one matrices, representing double lines, so it is not Zariski closed. Since the determinant is one nonzero homogeneous equation in ${\mathbb{P}}^5$, its dense rank-two locus has dimension $4$.

The final sentence of the converted question writes $R^c$, but the natural intended set is $R^{\circ }$. The homogeneous ideal of its closure is the principal ideal generated by the cubic polynomial $\det A$: the determinant hypersurface is irreducible, and $R^{\circ }$ is dense in it.

A projective variety is smooth when every local ring is regular, equivalently when its Jacobian has the expected rank at every point. The affine curve $y=x^3$ is smooth because the derivative of $y-x^3$ with respect to $y$ is $1$, while its projective closure $Y{Z}^2=X^3$ is singular at $[0:1:0]$.

The genus is ${\dim }_k{H}^1(C,{\mathcal{O}}_C)$. Intersect the Fermat cubic threefold with the plane $X_3=X_4=0$. This gives the smooth plane cubic $X_0^3+X_1^3+X_2^3=0$, whose genus of a smooth plane curve is $(3-1)(3-2)/2=1$.

An irreducible plane conic is smooth: a singular quadratic form in three variables has rank at most two and factors over the algebraically closed field, contradicting irreducibility. Projection from any point of the conic, or its degree-two complete linear system, then identifies it with the projective line.

Finally, a nonzero ternary quadratic form is determined up to nonzero scalar by its six coefficients. Thus generalized conics are parametrized bijectively by ${\mathbb{P}}^5$.

The monodromy around both $0$ and $1$ must be a product of two disjoint transpositions, hence a $(2,2)$ cycle type in $S_4$. The product relation for loops around the three punctures says ${\sigma }_0{\sigma }_1{\sigma }_{\infty }=1$. By the stated fact, ${\sigma }_0{\sigma }_1$ is either the identity or another $(2,2)$ permutation. Consequently ${\sigma }_{\infty }=({\sigma }_0{\sigma }_1{)}^{-1}$ cannot contain a $3$-cycle. But ramification index $3$ over infinity would force exactly such a cycle, a contradiction.

Outside the finite set of branch points, the derivative is nonzero. The inverse function theorem supplies evenly covered neighbourhoods, and compactness of the Riemann sphere makes every fibre finite, so the restriction is a covering map.

Lift a loop $\gamma$ from each point of the fibre over $z_0$; its endpoints define a permutation ${\sigma }_{\gamma }$. The homotopy lifting property shows that this permutation depends only on the based homotopy class. The punctured source is connected, so a path between any two points in the fibre projects to a loop whose lift joins them. Hence the monodromy group acts transitively.

A point is ramified with multiplicity $m_i$ exactly when it is a zero of $f^{\prime }$ of multiplicity $m_i-1$. Since $f^{\prime }$ has degree $d-1$, the fundamental theorem of algebra gives ${\sum }_i(m_i-1)=d-1$.

Conversely, choose distinct complex numbers $a_i$ and put $g(z)=c{\prod }_i(z-a_i{)}^{m_i-1}$ with $c\ne 0$. The assumption makes $g$ a polynomial of degree $d-1$. Any antiderivative $f$ of $g$ has degree $d$ and precisely the prescribed ramification multiplicities.

Taking the Fourier transform of a derivative givesThus, away from $\xi =0$,This formula determines the $L^2$ solution uniquely; a distribution supported only at $\xi =0$ cannot belong to $L^2$ unless it is zero.

For $|\xi |\le 1$, the Fourier transform estimate $|\hat{f}(\xi )|\le C\Vert f{\Vert }_1$ and local integrability of $|\xi {|}^{-2}$ in three dimensions control the squared norm. For $|\xi |>1$,The Plancherel theorem therefore yields $\Vert B_j{\Vert }_{H^{s+1}}\le C(\Vert f{\Vert }_1+\Vert f{\Vert }_{H^s})$. Finally the Sobolev embedding theorem in three dimensions gives $H^{s+1}\subset C^1$ when $s+1>1+3/2$, namely when $s>3/2$.

The Riemann-Lebesgue lemma states that the Fourier transform of an $L^1$ function is continuous and tends to zero at infinity.

For a Schwartz function $f$, differentiation under the integral and integration by parts giveThe Riemann-Lebesgue lemma bounds the right side by an $L^1$ seminorm of $D^{\alpha }(x^{\beta }f)$. Every Schwartz seminorm of $\hat{f}$ is therefore bounded by finitely many Schwartz seminorms of $f$. Hence the Fourier transform maps $\mathcal{S}({\mathbb{R}}^n)$ continuously into itself.