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Conditioning on $N$ and using the moment-generating function of a unit exponential variable,where the geometric series converges precisely when $\theta <p$. This is the moment-generating function of an exponential variable of rate $p$, so the geometric sum of exponential variables satisfies

For any rotation $R$ about the centre, ${\varphi }_R(x)=\varphi (Rx)$ satisfies the same equation and the same Dirichlet data because $V,f,g$ are rotationally invariant. Uniqueness from part (a) gives ${\varphi }_R=\varphi$ for every $R$. Thus $\varphi$ is constant on spheres and is a function of $r$ alone.

For a radial function in three dimensions, the radial Laplacian givesHere regularity at zero selectsA further integration giveswhose apparent singularity is removable, with limit $3+C$ at zero. Since the nonconstant term has value $4e-8$ at $r=1$, the boundary condition $\varphi (1)=3$ gives $C=11-4e$. Therefore

The classification of isolated singularities can be stated through the Laurent expansion of $f$ about $a$:

At every $z_k$, the function $\tan (1/z)$ has a simple pole. The exponential of a pole is an essential singularity, soare essential singularities of $\exp (\tan (1/z))$.

These points accumulate at $0$. Therefore no punctured neighbourhood of zero is a domain of holomorphy for the function: zero is a non-isolated singularity, specifically an accumulation point of essential singularities, rather than an isolated essential singularity. There are no other singularities in $\mathbb{C}$.

The power function of a statistical test for Test 1 is

Put $a=1/\sqrt{10}$. Test 2 rejects whenUsing the two branches of the triangular distribution givesBoth functions equal $0.05$ at the least favourable null value $\theta =0$.

A sufficient statistic $T$ is minimal sufficient if it is a function of every other sufficient statistic: whenever $U$ is sufficient, there is a function $h$ such that $T=h(U)$ almost surely, up to the usual null-set qualification. Thus its level sets give the coarsest sufficient partition of the sample space. In a dominated family with positive densities, the likelihood-ratio criterion for minimal sufficiency says that $T$ is minimal sufficient when

If the urn contains $g$ green balls, it contains $g+2$ red balls: both update rules preserve the difference $R-G=2$. Until absorption at $g=0$, the green count is therefore a birth-death chain withThe functionis harmonic, since

Let ${\tau }_0$ and ${\tau }_N$ be the hitting times of $0$ and $N$. The optional sampling theorem for a supermartingale, applied to the bounded stopped martingale $h(G_{n\wedge {\tau }_0\wedge {\tau }_N})$, givesSolving,Letting $N\to \infty$, the events on the left increase to eventual termination. The harmonic hitting probability for the balanced-difference urn is therefore

Writing the velocity as $(u,v)=(y,ax)$ givesso the flow is incompressible. With the conventiona stream function isThe streamlines are its level sets. For $a>0$, they are the hyperbolaswith separatrices $y=\pm \sqrt{a}x$ and a saddle at the origin. For $a<0$, they are concentric ellipsestraversed clockwise. This is the streamline classification of a planar linear saddle or centre.

The scalar vorticity isHence the flow is irrotational exactly whenThen $(u,v)=(y,x)=\nabla \varphi$, and a velocity potential is

The eigenvalue equation $D_{r}f=\lambda f$ isIf $\lambda =-1$, this forces $f(x+r)=0$ for every $x$, hence $f=0$, so $-1$ is not an eigenvalue. If $c=1+\lambda \ne 0$, choose arbitrary values on representatives of the cosets of $r\mathbb{Z}$ and extend byThis produces nonzero eigenfunctions. Choosing functions supported on distinct cosets gives infinitely many linearly independent eigenfunctions. Thereforeand every eigenspace is infinite-dimensional.

Direct expansion giveswhich is symmetric in $r,s$. Hence

Suppose a degree-$n$ polynomial $p$ were a sum of $n$ periodic functions,Apply the commuting product $D_{r_1}\cdots D_{r_n}$. Every term on the right is killed by its corresponding factor, whereas for leading coefficient $a_n$ the mixed finite difference of a polynomial givesThis contradiction proves

A function $f:(X,d_X)\to (Y,d_Y)$ is uniformly continuous if for every $\epsilon >0$ there is $\delta >0$ such thatfor all $x,x^{\prime }\in X$.

Suppose $f_n\to f$ uniformly and every $f_n$ is uniformly continuous. Given $\epsilon >0$, choose $N$ such thatfor every $x$. Uniform continuity of $f_N$ supplies $\delta >0$ such that $d_X(x,x^{\prime })<\delta$ impliesThe triangle inequality then gives $d_Y(f(x),f(x^{\prime }))<\epsilon$. Thus the uniform limit theorem for uniformly continuous functions proves that $f$ is uniformly continuous.

Pointwise convergence is insufficient. On $[0,1]$, the uniformly continuous functions $f_n(x)=x^n$ converge pointwise towhich is discontinuous and therefore not uniformly continuous.

For a closed piecewise smooth curve $\gamma$ avoiding $w$, its winding number about $w$ is

The argument principle says that if a positively oriented closed curve $\gamma$ bounds a domain $U$, and a meromorphic function $F$ has no zeros or poles on $\gamma$, thenwhere zeros and poles are counted with multiplicity.

Suppose $F$ and $G$ are holomorphic on a neighbourhood of $\overline{U}$ andFor $0\le t\le 1$, $F+tG$ has no boundary zero, because a zero would imply $|F|=t|G|<|F|$. Its argument-principle count is integer-valued and continuous in $t$, hence constant. Thus Rouché's theorem states that

Now let $w_0=f(z_0)$. Since $|w_0|<1\le |f|$ on the unit circle, Rouché's theorem shows that $f$ and $f-w_0$ have the same number of zeros in the unit disc. The latter has the zero $z_0$, so $f$ has at least one zero there. For any $w$ with $|w|<1$, the same boundary inequality shows that $f-w$ has the same positive number of zeros as $f$. ThereforeThis is the unit-disc image from a boundary modulus lower bound.

Finally, take a sufficiently small positively oriented circle $C$ around zero. The residueis the winding number of $g(C)$ around zero, so $k\in \mathbb{Z}$. The pole is simple, hence its residue is nonzero and $k\ne 0$. Forwe haveThe second term cancels the complete principal part at zero, so the integer residue of a logarithmic derivative proves that $h^{\prime }/h$ has a removable singularity there.

Set $u(r,\theta )=R(r)\Theta (\theta )$. Periodicity and separation of variables give the angular modes $1$, $\cos n\theta$, and $\sin n\theta$. The radial equation iswhose solutions are $r^{\pm n}$ for $n\ge 1$; the zero mode has solutions $1$ and $\log r$. Thus a general real harmonic function on the annulus is

The boundary data contain only the $n=2$ cosine mode, so writeThe condition at $r=a$ gives $B=-A{a}^4$, and the condition at $r=b$ fixes $A$. The Dirichlet problem on an annulus for one Fourier mode therefore has solution

Differentiability at $x\in {\mathbb{R}}^2$ means that there is a linear map $Df{|}_x:{\mathbb{R}}^2\to \mathbb{R}$ such thatThe partial derivatives are $D_{i}f(x)=Df{|}_x(e_i)$ when the derivative exists; independently, they may be defined by the corresponding one-variable difference quotients.

Suppose the partial derivatives exist near $x=(x_1,x_2)$ and are continuous at $x$. Apply the one-dimensional mean value theorem along the two coordinate segments from $x$ to $x+h$. It gives intermediate points ${\xi }_h,{\eta }_h\to x$ such thatContinuity of the partials makes the remainder after subtractingequal to $o(\Vert h\Vert )$. Hence $f$ is differentiable and this is its derivative.

For the given function, writing $r=\sqrt{x^2+y^2}$, at $r>0$ we haveAt the origin both partial derivatives are zero, since $f(h,0)/h$ and $f(0,h)/h$ tend to zero. Neither partial is continuous there: along its corresponding coordinate axis the cosine term oscillates without a limit. Neverthelessso $f$ is differentiable at the origin with derivative zero. This illustrates that existence of partial derivatives does not imply their continuity.

The final assertion is false. DefineAgain $|g|\le r^2$, so $g$ is differentiable at the origin and it is smooth elsewhere. But along the $x$-axis,is unbounded near zero, and similarly $D_{2}g(0,y)$ is unbounded. Thus both partial derivatives can be unbounded in every neighbourhood of a point even when the function is differentiable everywhere.

Let $U(s,x)={\mathcal{L}}_t\{u(t,x)\}$. The initial data turn the wave equation intoUsing the Green function from part (ii) with $m=s$ givesSincewe obtain the D'Alembert formula with initial velocity

If $x_k\ge 1$, then positivity and the arithmetic-geometric mean inequality giveThus all iterates remain in $[1,\infty )$.

Consider the strictly convex functionon $[1,\infty )$. Since $f^{\prime }(x)=x^2-a$ and $f^{\prime \prime }(x)=2x$, its Newton minimization step isIts unique minimizer is $x^{\ast }=\sqrt{a}$.