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Here the target $C$ of $g$ is the unit circle. The boundary loop $g(e^{2\pi it})$ extends through the disk, so the radial homotopy $g(s{e}^{2\pi it})$ contracts it to a constant loop. Thus its winding number is zero. On the boundary put $h(z)=g(z)/z^n$, also circle-valued. Adding argument lifts gives $w(h)=0-n=-n$. If $z^n+g(z)$ never vanished, $h$ would avoid $-1$, which is the only negative real point of the unit circle. Part (iii) would give $w(h)=0$, contradicting $n\ne 0$. Therefore there is a boundary point with $z^n+g(z)=0$. The circle-valued hypothesis is essential; it is not a claim about arbitrary complex-valued $g$.

The principal argument is continuous and single-valued on $\mathbb{C}\setminus (-\infty ,0]$. The given loop stays in this slit plane, and its initial and final values have the same principal argument. The definition therefore gives $\boxed{w(\gamma ;0)=0}$.

A loop homotopy is a continuous $H:[0,1{]}^2\to \mathbb{C}\setminus \{0\}$ with $H(0,t)={\gamma }_0(t)$, $H(1,t)={\gamma }_1(t)$ and $H(s,0)=H(s,1)$ for every $s$. Compactness gives a positive lower bound $r$ for $|H|$ and uniform continuity. Partition the $s$ interval finely enough that neighboring loops $\eta ,\delta$ satisfy $|\eta (t)-\delta (t)|<r$. Their quotient lies in $|z-1|<1$, and has a single-valued continuous argument with equal endpoint values. Its winding number is zero. Arguments of a product add, so $w(\eta )=w(\delta )+w(\eta /\delta )=w(\delta )$. Applying this at every step provesThis proves homotopy invariance of winding number directly, including continuous curves without differentiability.

For a continuous loop in $\mathbb{C}\setminus \{0\}$, choose a continuous argument lift $\theta :[0,1]\to \mathbb{R}$ with $\gamma (t)/|\gamma (t)|=e^{i\theta (t)}$. Such a lift is constructed by subdividing the interval into pieces whose images lie in an argument chart and matching the arguments at their endpoints. The winding number of a continuous closed path isChanging the initial lift adds one constant multiple of $2\pi$ and does not change this integer.

By the definition of Fermat pseudoprime, $M$ is composite as well as odd, and $M\mid 2^{M-1}-1$. Put $L=2^M-1$. A proper factorization $M=ab$ makes $2^a-1$ a proper divisor of $L$, so $L$ is composite. NowSince $2^M\equiv 1(\mathrm{mod}L)$, $2^d\equiv 1(\mathrm{mod}L)$. This is the first passing condition of the Miller-Rabin primality test, with $s=1$. Thus $2^M-1$ is a base-two strong pseudoprime. If “pseudoprime” were used to include primes, the compositeness conclusion would fail, for instance $M=3$ gives $L=7$; the standard composite definition matters.

Since $p\ge 5$ is odd, $2^p\equiv 2(\mathrm{mod}3)$ andis a product of two odd integers greater than one. Thus $N$ is odd and composite. Also $2^{2p}=4^p\equiv 1(\mathrm{mod}N)$. Fermat's theorem modulo $p$ gives $p\mid 4^{p-1}-1$, and $p\ne 3$, so $p\mid N-1=4(4^{p-1}-1)/3$. This number is even; hence $2p\mid N-1$. Thereforeproving that $N$ is a base-two Fermat pseudoprime.

Testing a weak Neumann Poisson problem with the constant function one gives ${\int }_{\Omega }f=0$, proving necessity. For sufficiency, use the closed Hilbert space of mean-zero $H^1$ functions. The given Poincaré inequality makes $\Vert \nabla v{\Vert }_{L^2}$ an equivalent norm there. SetThe negative sign corresponds to $\Delta u=f$. Poincare's inequality gives boundedness of $\ell$ and coercivity of $a$. The Lax-Milgram theorem supplies a unique mean-zero solution. Every $v\in H^1$ is the sum of its mean-zero part and a constant; since $\int f=0$, the weak identity extends to every $v\in H^1$. This is precisely the weak equation with zero normal derivative as its natural boundary condition. Existence and uniqueness in the mean-zero space hold exactly when $\bar{f}=0$.

The Lax-Milgram theorem states that on a real Hilbert space, a bounded bilinear form $a$ satisfying $a(v,v)\ge c\Vert v{\Vert }^2$ for some $c>0$ represents every bounded linear functional uniquely: there is a unique $u$ with $a(u,v)=\ell (v)$ for all $v$.

The clamped trace conditions define a closed subspace $H_{\partial }^2(\Omega )$ of the Sobolev space $H^2(\Omega )$. On it take $a(u,v)={\int }_{\Omega }\Delta u\Delta v$ and $\ell (v)={\int }_{\Omega }fv$. Both are bounded by the Cauchy-Schwarz inequality. The Dirichlet elliptic regularity estimate on a smooth bounded domain gives $\Vert v{\Vert }_{H^2}\le C\Vert \Delta v{\Vert }_{L^2}$ for $v\in H^2\cap H_0^1$, so $a(v,v)\ge C^{-2}\Vert v{\Vert }_{H^2}^2$. The Lax-Milgram theorem now gives exactly one $u$ satisfyingTesting with compactly supported smooth $v$ and integrating by parts twice gives ${\Delta }^{2}u=f$ in distributions; the two boundary conditions hold as traces by the space's definition. This is the unique weak clamped biharmonic solution.

The Weierstrass approximation theorem says that for every real continuous function $f$ on $[0,1]$ and every $\epsilon >0$, there is a real polynomial $P$ with ${\sup }_{[0,1]}|f-P|<\epsilon$. Equivalently, polynomials are dense in $C[0,1]$ in the uniform norm.

Write ${\sigma }^2={\Sigma }_{\eta }>0$ and assume $V_0>0$. The scalar Kalman filter reduces toTaking reciprocal variance and dividing the mean by variance givesThusSince the process noise is zero and $A=1$, $X_t=X_0$ and $Y_j=X_0+{\eta }_j$. The strong law of large numbers gives $t^{-1}{\sum }_j{Y}_j\to X_0$ almost surely. The prior's finite weight vanishes in the displayed formula, so both limits equal $X_0$ almost surely. A zero prior variance instead means $X_0$ is known, and the assertion follows directly without reciprocal-variance formulas.

Use the usual state-space assumption that the prior $X_0$ is independent of future process and observation noises. Suppose inductively $X_t\mid {\mathcal{Y}}_t\sim N({\hat{X}}_t,V_t)$. Prediction through the linear dynamics gives $X_{t+1}\mid {\mathcal{Y}}_t\sim N(A{\hat{X}}_t,M_t)$, where $M_t=A{V}_t{A}^T+{\Sigma }_{\epsilon }$. The pair $(X_{t+1},Y_{t+1})$ is conditionally multivariate normal, with means $A{\hat{X}}_t,CA{\hat{X}}_t$, cross covariance $M_t{C}^T$, and observation covariance $R_t={\Sigma }_{\eta }+C{M}_t{C}^T$. Applying part (a) gives the Kalman filterThis proves the inductive Gaussian assertion. The inverse requires nonsingular innovation covariance, as in the printed formula. Importantly, prediction uses $A{\hat{X}}_t$: the TeX conversion drops this $A$ in one place, whereas the original PDF includes it.

Assume $V_{YY}$ is invertible. PutThis and $Y$ are jointly multivariate normal, with $\mathrm{Cov}(R,Y)=0$ and $\mathrm{Cov}(R)=V_{XX}-V_{XY}{V}_{YY}^{-1}{V}_{YX}$. Their joint Gaussian characteristic function factors when the cross covariance vanishes, so they are independent. Conditioning on $Y=y$ therefore leaves the law of $R$ unchanged, proving

Passing to the deterministic-service limit in part (c), near zero, givesDifferentiation at zero gives $\mathbb{E}B=(1+\lambda \mathbb{E}B)/\mu$. Alternatively, the same branching expectation equation gives $\mathbb{E}{B}_k=1/(\mu -\lambda )$ for every $k$, because $\mathbb{E}{S}_k=1/\mu$. The assumed convergence of finite moment-generating functions at a positive point near zero bounds exponential moments and makes these means uniformly integrable, so the means pass to the limit. Since $T=A+B$, the answers areThey correspond to an M/D/1 queue with deterministic service time $1/\mu$. The calculation only needs finite transforms near zero; the literal assumption of finite transforms for every $\theta <\lambda$ is stronger than necessary and is not generally true for positive busy-period arguments. For instance, with $k=1$ and $\lambda =\mu /2$, part (c) gives $\lambda b^2-(\mu +\lambda -\theta )b+\mu =0$. Its discriminant is negative for $(\sqrt{\mu }-\sqrt{\lambda }{)}^2<\theta <\lambda$, so no finite busy-period transform can exist there.

The first arrival time $A$ has an exponential distribution of rate $\lambda$ and is independent of the following busy period. ThusSubstitute ${\varphi }_{B_k}=(\lambda -\theta ){\varphi }_{T_k}/\lambda$ into part (c). Its service-transform argument becomes $\theta +\lambda ({\varphi }_{B_k}-1)=(\lambda -\theta )({\varphi }_{T_k}-1)$. ThereforeAs before, positive arguments must lie in the domain of finite moment-generating functions.

For the busy-period branching equation, during the first service of a busy period, every arriving customer initiates a subtree of further work. The independence and stationary increments of the Poisson process, together with independent services, give the distributional recursionwhere conditional on $S_k=s$, $N(S_k)$ is Poisson with mean $\lambda s$, and the descendant busy periods are independent copies of $B_k$. One may explore these subtrees in order; the server's total workload is unchanged by that ordering. Consequently, wherever the moment-generating functions are finite,For $\lambda <\mu$, this identity in particular holds in a sufficiently small neighbourhood of zero.

An M/G/1 queue has Poisson process arrivals of rate $\lambda$, independent identically distributed service times of law $G$, one server, and here infinite waiting capacity. Use first-come-first-served service. Let $X_n$ be the number left just after the $n$th departure. During the next service, let $A_n$ be the number of arrivals. ThenConditional on the next service time $S_n=s$, $A_n$ is Poisson with mean $\lambda s$. If $X_{n-1}=0$, the server waits for one arrival to start service; that arrival is the customer being served, and is not counted in $A_n$. This constructs the embedded Markov chain.

Independence and the moment-generating function of an exponential distribution giveFor each fixed $\theta$, taking logarithms and using $-\log (1-x)\sim x$ yields ${\varphi }_{S_k}(\theta )\to e^{\theta /\mu }$. The limiting service time is the constant $S=1/\mu$. Its moment-generating function is exactly this limit.