Condition on the intermediate state at time $m$. The Markov property and time homogeneity giveThese nonnegative sums are valid on a countable state space. Starting from $P(0)=I$, $P(1)=P$, induction gives $\boxed{P(n)=P^n}$. Conditioning on the initial state now yields $\boxed{\mathrm{Pr}(X_n=i)=(\mu P^n{)}_i}$. If $\mu P=\mu$, induction gives $\mu P^n=\mu$ for every $n$, so all one-time distributions equal the stationary initial distribution.

On the line $y=-x$ one has $\overline{iz}=z$. Hence the reflected term in the printed formula is $\overline{w(z)}$, and the total is real on the line. This proves that its stream function is constant there. The nested conjugation is important: it makes the added term analytic as a function of $z$.

For the planar potential dipole its image contribution isThusThe image is the reflected point $-i\bar{a}$, which lies in the opposite half-plane when the original dipole lies strictly inside the fluid half-plane. The wall itself is a limiting boundary, not an admissible position for the singularity.

Across a straight impermeable wall the reflected potential is obtained by conjugating the potential after reflecting its argument. For the real axis the operation is $f(z)\mapsto \overline{f(\bar{z})}$; the sum is real on that axis. For the imaginary axis it is $f(z)\mapsto \overline{f(-\bar{z})}$. Applying both commuting reflections to the dipole producesThe other three poles are outside the first quadrant. The sum is invariant under each reflected-conjugation operation, hence real on each boundary axis. Both walls are therefore streamlines.

For an analytic complex potential $f$ whose singularities lie outside $|z|=a$, the circle theorem adds the reflected potentialThe added term is analytic outside the circle wherever the reflected original is regular; in particular no new exterior singularities occur when $f$ is regular inside the circle. On $|z|=a$, the relation $a^2/\bar{z}=z$ makes $F=f+\bar{f}$ real. Thus its imaginary part, the stream function, is constant on the circle and the normal velocity vanishes. This proves the theorem, with any harmless additive constant left free. Its usual decay condition at infinity selects the surrounding flow inherited from $f$.

Write $a_1=z_1$ for the dipole position to distinguish it from the circle radius, now one. Substitution givesThe image singularity is at $1/{\bar{a}}_1$, strictly inside the solid circle. This exterior potential has the original dipole and an impermeable cylinder boundary.

Measure the axial coordinate $z$ from the vertex. The slice radius is $z\tan \alpha$ and its mass is $dm=3m{z}^2{H}^{-3}dz$, since cross-sectional area is proportional to $z^2$. Its axial moment is $\frac{1}{2}(z\tan \alpha {)}^{2}dm$. ThereforeFor an axis through the vertex perpendicular to the cone axis, a thin disc contributes $[z^2+\frac{1}{4}{z}^2{\tan }^2\alpha ]dm$: the second term is its diametral moment. Integration gives the two equal transverse principal momentsRotational symmetry eliminates the mixed moments and makes these the principal axes. The centre of mass is at $3H/4$, so the parallel-axis theorem would give $A_G=3m{H}^2/80+3m{H}^2{\tan }^2\alpha /20$ and $C_G=C_O$.

If a differentiable function has a constrained local extremum at a point of a smooth surface, its derivative vanishes in every tangent direction; equivalently its gradient is normal to the surface. For a regular equation $F=0$ this means $\nabla V=\lambda \nabla F$ for some real multiplier.

First the box is centred at the origin: holding the other coordinates of a vertex of a polyhedron fixed and comparing the two possible coordinates $u,v$ in the first direction gives $u^2=v^2$, so the distinct extremes are opposites; the other two directions give the same conclusion. Write its positive half-lengths as $ar,bs,ct$, with $r^2+s^2+t^2=1$. Its volume is $8abcrst$. A degenerate box has zero volume, so a positive maximum is interior in $r,s,t>0$. Lagrange multipliers applied to $\log (rst)$ give $1/r=2\lambda r$, and the analogous equations imply $r=s=t=1/\sqrt{3}$. Alternatively the arithmetic–geometric mean inequality applied to $r^2,s^2,t^2$ proves global maximality. Thus

Define $F(x)=x_1^2/a^2+x_2^2/b^2+x_3^2/c^2-1$. Its gradient is $2(x_1/a^2,x_2/b^2,x_3/c^2)$. It cannot vanish on $F^{-1}(0)$, because the only possible zero is the origin and that point has $F=-1$. The regular level-set theorem therefore makes the ellipsoid a two-dimensional embedded submanifold. Its normal space isand the tangent space is the orthogonal complement of that line.

A two-dimensional embedded submanifold is a subset which near every point can be straightened by a smooth local coordinate change in the ambient space to an open piece of the coordinate plane ${\mathbb{R}}^2\times \{0\}$. Equivalently locally it has a regular smooth parametrization of rank two, with the induced local topology.

A tangent vector at $x$ is the derivative ${\gamma }^{\prime }(0)$ of a smooth curve in the submanifold with $\gamma (0)=x$. These derivatives form the two-dimensional tangent space $T_{x}M$. A normal vector is a vector orthogonal to every tangent vector. Its space is $(T_{x}M{)}^{\perp }$, which has $\boxed{\dim (T_{x}M{)}^{\perp }=1}$ in the three-dimensional ambient Euclidean space.

The inclusion $f({\bigcap }_n{K}_n)\subseteq {\bigcap }_{n}f(K_n)$ holds for every map. For the reverse choose $y\in {\bigcap }_{n}f(K_n)$ and let $L_n=K_n\cap f^{-1}(\{y\})$. These sets are nonempty and nested. A Hausdorff space has closed singletons, and continuity makes the fibre closed, so each $L_n$ is compact. Part (b)(i) gives a point $x\in {\bigcap }_n{L}_n$. Then $x\in {\bigcap }_n{K}_n$ and $f(x)=y$. Hence

In the usual real line take $\boxed{F_n=[n,\infty )}$. Each set is nonempty and closed, and $F_{n+1}\subseteq F_n$. No real number lies in them all, because the positive integers are unbounded. Thus their intersection is empty. These sets are not compact.

Let the radius be $r>0$. One inclusion follows from (a). For the reverse, if $\Vert x-c\Vert \le r$, set $x_n=c+(1-1/n)(x-c)$ for $n\ge 2$. Then $\Vert x_n-c\Vert <r$ and $x_n\to x$, so every closed-ball point is in the closure of the open ball. HenceThe positive-radius convention matters: if radius zero is allowed, the open ball is empty whereas the closed ball is $\{c\}$, so the asserted equality fails at that degenerate radius.

The induced-metric balls are intersections with the corresponding ambient balls. On the horizontal arm the distance from the centre is $|x-1/2|$; on the vertical arm it is $\max (1/2,y)$. ConsequentlyThe closure relative to $E$ of the open ball is $[0,1]\times \{0\}$. Every point $(0,y)$ with $y>0$ has positive distance $y$ from that horizontal arm, so none of those points is in the closure. In particular $(0,1/4)$ lies in the closed ball but not in the closure of the open ball, proving strict inclusion.

The condition is $\max (|x-1/2|,|y|)\le 1/2$. Therefore the ball is the closed squareIts four vertices are $(0,\pm 1/2)$ and $(1,\pm 1/2)$.

The distance function $x\mapsto d(x,c)$ is continuous because the triangle inequality gives $|d(x,c)-d(y,c)|\le d(x,y)$. Hence $\overline{B}(c,r)=\{x:d(x,c)\le r\}$ is the inverse image of the closed interval $(-\infty ,r]$, so is closed. Since it contains $B(c,r)$, it also contains the closure of that open ball: