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Write the holomorphic function as $f=u+iv$. Analyticity gives smooth real and imaginary parts satisfying the Cauchy-Riemann equations $u_x=v_y$, $u_y=-v_x$. Differentiating these identities showsThus both parts are harmonic functions. The product rule then givesSince $f^{\prime }=u_x+i{v}_x$, this proves the Laplacian identity for the squared modulus of a holomorphic function,

In the upper half-plane model $H=\{(x,y):y>0\}$ the hyperbolic metric isThe length of a differentiable curve is $\int \sqrt{{\dot{x}}^2+{\dot{y}}^2}dt/y$. The area element is the square root of the metric determinant times $dxdy$, so the hyperbolic area of a measurable region $S$ is ${\int }_S{y}^{-2}dxdy$.

The Gauss-Bonnet theorem for a geodesic hyperbolic triangle of angles $\alpha ,\beta ,\gamma$ in curvature $-1$ gives area $\pi -\alpha -\beta -\gamma$; an ideal vertex has angle zero. The horizontal edge in the present region is not a geodesic, so it is simpler to use the area element directly:The open boundary does not affect the integral because its area is zero.

The minimal polynomial $m_A$ is the monic polynomial of smallest degree satisfying $m_A(A)=0$. An annihilating polynomial exists, for example because $I,A,\dots ,A^{n^2}$ are linearly dependent in the $n^2$-dimensional space of matrices. A nonzero constant cannot annihilate $A$, so the minimal degree is positive.

If $p$ and $q$ are two such monic polynomials of minimal degree $d$, then $(p-q)(A)=0$ and $\mathrm{deg}(p-q)<d$ unless $p=q$. Minimality therefore forces $p=q$. Polynomial division also shows that $m_A$ divides every annihilating polynomial: the remainder after division would otherwise be an annihilator of smaller degree. This proves uniqueness rather than merely choosing one of several minimal-degree polynomials.

If $A$ is real, coefficientwise complex conjugation of $m_A(A)=0$ gives $\overline{m_A}(A)=0$. The conjugated polynomial is monic of the same degree, hence uniqueness gives $\overline{m_A}=m_A$. The minimal polynomial of a real matrix has real coefficients.

For $n\ge 3$, use the block diagonal matrixThe two-dimensional Jordan block is annihilated by $(t-1{)}^2$ but not by $t-1$, while the scalar block requires the factor $t+1$. A polynomial annihilates the direct sum exactly when it annihilates every block. Thus its minimal polynomial is the least common multiple $(t-1{)}^2(t+1)$. For $n=3$ the final identity block is omitted.

For the natural filtration ${\mathcal{F}}_n=\sigma (X_0,\dots ,X_n)$, a stopping time $T$ takes values in $\{0,1,\dots \}\cup \{\infty \}$ and satisfies $\{T\le n\}\in {\mathcal{F}}_n$ for each $n$. Whether it has occurred by time $n$ can be decided from the observations through that time.

The Strong Markov property for a time-homogeneous Markov chain says that, on $\{T<\infty \}$, conditional on ${\mathcal{F}}_T$, the process $(X_{T+r}{)}_{r\ge 0}$ has the law of a fresh chain started at $X_T$. In particular its future conditional law depends on the past only through $X_T$. For example $\mathbb{P}(X_{T+r}=j\mid {\mathcal{F}}_T)=(P^r{)}_{X_{T}j}$ on that event, and the corresponding statement holds for every finite future path.

Start at $i$ and let $T_0=0$, $T_{r+1}=\inf \{n>T_r:X_n=i\}$, with $\inf \varnothing =\infty$. These are stopping times. Put $f={\mathbb{P}}_i(T_1<\infty )$. Each finite return restarts the chain at $i$, so the Strong Markov property gives inductivelyThe events decrease as $r$ increases; infinitely many visits occur exactly when every return time is finite. Continuity of probability therefore givesNo irreducibility assumption is needed for this zero-one conclusion at the starting state.

Let $N={\sum }_{n\ge 0}{1}_{\{X_n=i\}}$, counting the initial visit. Summing nonnegative random variables and then using the tail-sum identity givesIf $f<1$, this is $1/(1-f)<\infty$. Thus divergence of the given sum forces $f=1$, and infinitely many visits then occur with probability one. This proves the recurrence criterion by return probabilities directly from the regeneration at return times; divergent probability sums alone would not suffice for arbitrary dependent events.

A sufficient statistic $T(X)$ for $\theta$ is a function of the observations whose conditional distribution of $X$ given $T(X)$ does not depend on $\theta$, whenever that conditioning event is possible. It retains all parameter dependence of the likelihood.

In the discrete case the Fisher-Neyman factorization theorem provides a direct criterion: $f(x\mid \theta )=g(T(x),\theta )h(x)$, where $h$ is independent of the parameter. To justify sufficiency, for a value $t$ with positive probability, sum this expression over its fiber and divide:The parameter-dependent factor cancels. Conversely, parameter-independent conditional probabilities $h_t(x)$ give the factorization $f(x\mid \theta )={\mathbb{P}}_{\theta }(T=t)h_t(x)$ with $t=T(x)$. Thus the criterion is both necessary and sufficient, with conditional laws only required on fibers of positive probability.

For the shifted exponential sample the likelihood isIt factors through $T=X_{(1)}={\min }_i{X}_i$, so the sample minimum is a sufficient statistic. Writing $Y_i=X_i-\theta$, independence gives for $t\ge 0$Therefore $T-\theta$ is exponential with rate $n\lambda$. Its mean is $(n\lambda {)}^{-1}$ and its variance is $(n\lambda {)}^{-2}$. The unbiased endpoint estimator for a shifted exponential sample is consequentlyAlthough the parameter is nonnegative, this unbiased estimator can be negative. Truncating it at zero would change its expectation and lose the requested property of an unbiased estimator.

Define source strength $m$ here as the volume flux through a small sphere around the source. For incompressible potential flow, $\mathbf{u}=\nabla \varphi$ and the governing conditions areEquivalently the Laplace equation holds away from the source, with local behavior $\varphi \sim -m/(4\pi |\mathbf{x}-{\mathbf{x}}_s|)$. The boundary condition expresses impermeability. The method of images uses an equal source below the plane:The image singularity lies outside the fluid, the two vertical derivatives cancel on the wall, and the source has the specified outward flux.

Let $s=\sqrt{x^2+y^2}$. On the wall $u_z=0$ and the tangential radial velocity is $u_s=ms/[2\pi (s^2+a^2{)}^{3/2}]$. With ambient pressure $p_{\infty }$, the steady Bernoulli equation in this irrotational flow givesThe fluid pressure is reduced everywhere on the wall except at its axis and in the infinite-distance limit. The finite hydrodynamic force is the excess over the uniform ambient-pressure force. Integrating its magnitude over the plane yieldsPutting $u=s^2$ gives the last integral as $\frac{1}{2}{\int }_0^{\infty }u(u+a^2{)}^{-3}du=1/(4a^2)$. HenceRelative to ambient pressure on the other side, the boundary is attracted toward the source. This is the hydrodynamic attraction of a plane wall to a point source.

There is an alternative source-strength convention: if $m$ denotes the coefficient in $u_r=m/r^2$ rather than total volume flux, replace the flux above by $4\pi m$. Then $\varphi =-m(1/r_{+}+1/r_{-})$, $p-p_{\infty }=-2\rho m^2{s}^2/(s^2+a^2{)}^3$, and $F=\pi \rho m^2/a^2$. The physical result is identical once the normalization is fixed.

Take zero electric potential at infinity and put $K=(4\pi {\epsilon }_0{)}^{-1}$. By spherical symmetry and Gauss's law, the electric field is radial, with magnitude determined by the total enclosed electric charge. Thus, away from the idealized shell surfaces,Integrating $E_r=-d\Phi /dr$ from infinity, or adding the three spherical-shell potentials, givesThe expressions agree at each interface, so the potential is continuous everywhere, including the origin. The electric field has different interior and exterior limits at each infinitely thin charged surface, with jump equal to surface charge density divided by ${\epsilon }_0$. An ideal sheet has no single classical field value on the sheet itself; the displayed adjacent limits specify the field there. In conducting material of nonzero thickness the internal field is zero.

The electrostatic energy is the integral of ${\epsilon }_0|\mathbf{E}{|}^2/2$. There is no contribution from $r<a$, and spherical integration givesThe dependence on $b$ cancels because the magnitude of enclosed electric charge is $|q|$ on both sides of the middle shell, even though the field direction changes.

No: ordering the attractive potentials does not impose the proposed ordering of reflection probabilities. Take $k=\pi /(4a)$ and let $M$ denote the particle mass. Choose $V_2=0$ everywhere andThese satisfy every stated ordering and support condition. The incoming energy is ${\hslash }^2{k}^2/(2M)$; the interior wavenumber in the finite square well is $q=\sqrt{k^2-2M{V}_1/{\hslash }^2}=3k$.

Here is an explicit matching calculation. For a general well of width $L=2a$, take exterior waves $e^{ik(x+a)}+r{e}^{-ik(x+a)}$ on the left and $t{e}^{ik(x-a)}$ on the right. Shifting the incident phase in this way does not change the reflection probability. Continuity of the wavefunction and its first derivative gives the propagation relationInverting it and adding the two equations after dividing the derivative equation by $ik$ yieldsThe incident and reflected exterior waves have the same speed, so their flux ratio isFor the chosen parameters $qL=3\pi /2$ and $(q^2-k^2{)}^2/(4k^2{q}^2)=16/9$. ThusThe zero potential has no reflected component. This proves the failure with explicit probabilities. More generally the same expression vanishes at $qL\in \pi \mathbb{Z}$, explaining resonant transmission through a square well and why increasing attractive well depth does not monotonically control reflection.

In orthonormal Cartesian coordinates, a second-rank tensor has components transforming under a change of axes $Q$ as $M_{ij}^{\prime }=Q_{ik}{Q}_{j\ell }{M}_{k\ell }$, where $Q{Q}^T=I$. Its contraction is invariant:Thus $M_{ii}$ is a scalar. This is tensor contraction with the Euclidean metric. In general nonorthonormal coordinates the invariant expression is $g^{ij}{M}_{ij}$, not an unweighted sum of diagonal covariant components.

For the planar plate, every mass element has $z=0$. The inertia tensor is consequentlyIt follows directly that ${\mathbf{e}}_z$ is an eigenvector andThe real symmetric matrix has three real eigenvalues. Its trace is $2{\int }_D(x^2+y^2)\rho dS=2M_{\perp }$, so $M_1+M_2+M_{\perp }=2M_{\perp }$. Hence the perpendicular axis theorem gives $M_{\perp }=M_1+M_2$.

For a disc of constant areal density ${\rho }_0$, symmetry gives ${\int }_{D}xydS=0$, and polar integration gives ${\int }_D{x}^{2}dS={\int }_D{y}^{2}dS=\pi a^4/4$. Thereforewhere the scalar $M$ in the last expression denotes the total mass rather than the tensor.

Let $F(z)=e^{az}/(1+e^z)$ and integrate counterclockwise around the rectangle with vertices $-R,R,R+2\pi i,-R+2\pi i$. It contains a single simple pole at $z=\pi i$, with residueOn the top edge $F(x+2\pi i)=e^{2\pi ia}F(x)$, and its orientation reverses that of the bottom edge. On the right edge $|F|\le e^{aR}/(e^R-1)$, so its integral tends to zero for $a<1$. On the left $|F|\le e^{-aR}/(1-e^{-R})$, which tends to zero for $a>0$. The residue theorem therefore givesSince $1-e^{2\pi ia}=-2i{e}^{\pi ia}\sin (\pi a)$,The restriction is also necessary for ordinary convergence: at $-\infty$ the integrand behaves as $e^{ax}$, requiring $a>0$, while at $+\infty$ it behaves as $e^{(a-1)x}$, requiring $a<1$. At either endpoint one tail approaches a nonzero constant; outside this range one tail grows. There is no cancellation because the real integrand is positive.

Let $S_N(x)={\sum }_{n=1}^N[a_n\cos (nx)+b_n\sin (nx)]$. Under the first coefficient bounds,This proves the uniform Cauchy condition on the whole real line and hence uniform convergence to $f$. It is the Weierstrass M-test applied with the summable majorant $2c{n}^{-1-\epsilon }$. Every partial sum is continuous. To see continuity of the limit explicitly, at a chosen $x_0$ first choose $N$ with small uniform error; then make $|S_N(x)-S_N(x_0)|$ small by continuity. The inequalityproves continuity. Also $S_N(x+2\pi )=S_N(x)$ for every $N$, so taking limits gives $f(x+2\pi )=f(x)$.

Under the stronger coefficient bounds, the derivative series has summable uniform majorant $2c{n}^{-1-\epsilon }$. Thus $S_N^{\prime }$ converges uniformly to a continuous function $g$ given by that series. The original series also converges uniformly. On any finite interval, the fundamental theorem of calculus gives $S_N(x)-S_N(0)={\int }_0^x{S}_N^{\prime }(t)dt$. Uniform convergence permits passage to the limit inside this finite integral, since the error is at most $|x|\Vert S_N^{\prime }-g{\Vert }_{\infty }$. Therefore $f(x)-f(0)={\int }_0^{x}g(t)dt$, and continuity of $g$ gives $f^{\prime }=g$. This justifies Termwise differentiation of a Fourier series rather than assuming it:

Use the following standard facts: $A_5$ has order $60$ and is a nonabelian simple group; the coset action on the cosets of an index-$j$ subgroup is a transitive homomorphism into $S_j$. If $H<A_5$ has index $j>1$, that action is nontrivial. Its kernel is a normal subgroup, hence is trivial by simplicity. It would therefore embed $A_5$ in $S_j$. For $j=2,3,4$ this is impossible since $j!<60$. There are no subgroups of indices 2, 3 or 4.

An index-5 subgroup has order $12$ by Lagrange's theorem. Consider its natural action on the five letters. It cannot be transitive, since an orbit of size 5 would divide its order by the orbit-stabilizer theorem. If it had no fixed letter, its orbits would have sizes 2 and 3. It would then be a subgroup of the permutations preserving that partition, namely $S_2\times S_3$. Exactly half of those 12 permutations are even, so $(S_2\times S_3)\cap A_5$ has order 6, too small to contain $H$.

Thus $H$ fixes a letter $i$. The stabilizer of $i$ in $A_5$ is the copy of $A_4$ consisting of even permutations of the other four letters; it has order 12. Containment and equality of orders force $H$ to equal that stabilizer. Conversely each such stabilizer has index 5. ConsequentlyThese five subgroups are distinct: for $i\ne j$, a three-cycle on letters other than $i$ can move $j$, so it belongs to the stabilizer of $i$ but not that of $j$.