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Shift the absorbing interval to $\{0,\dots ,a+1\}$ and start at $1$. The gambler’s-ruin harmonic functions give $\mathbb{P}({\tau }_a<{\tau }_{-1})=1/(a+1)$ and $\mathbb{E}T=1(a+1-1)=a$.

The extinction probability is the smallest fixed point of the convex generating function $F_1$ in $[0,1]$. Since $F_1\prime (1)=\mu \le 1$ and $p_0,p_2>0$ make strict convexity relevant, the only fixed point is $1$.

The required harmonic function isOn the upper and lower boundary its argument is respectively $\pi /2$ and $-\pi /2$, while at infinity the logarithm tends to $0$.

For fixed $x$ add $K\Pi (t)$ to $p(t)-f(t)$ so the resulting function also vanishes at $t=x$. Repeated Rolle gives a point $\xi$ where its $(n+1)$st derivative vanishes, yielding the formula.

The inverse function theorem says that a $C^1$ map with invertible derivative at $x$ restricts to a $C^1$ diffeomorphism between neighbourhoods of $x$ and $f(x)$. It makes every zero isolated. Compactness of $\bar{\Omega }$, together with separation from zero on the boundary, then makes the zero set finite and interior. Continuity of $Df$ gives the displayed radii. Choose disjoint balls around the zeros and a positive lower bound for $|f|$ on their complement and boundary. A sufficiently small $C^1$ perturbation has no zeros outside, while the quoted local lemma gives exactly one zero in each ball with the same determinant sign. Summing proves $N(g)=N(f)$.

Expanding the derivative limit along real and imaginary increments gives $u_x=v_y$ and $u_y=-v_x$. Since $\mathrm{Re}(z{e}^z)=e^x(x\cos y-y\sin y)$, the required function is $f(z)=z{e}^z$, and $f(0)=0$. Two such functions differ by a holomorphic function with zero real part; the open mapping theorem makes it constant, and the value at zero makes that constant zero.

Bounded means finite diameter; sequential compactness means every sequence has a convergent subsequence. A failure of uniform continuity supplies two close sequences whose image distances stay apart; a convergent subsequence contradicts continuity. Into a discrete metric, continuity is local constancy, and compactness upgrades the radii uniformly. The Cantor set is closed in compact $[0,1]$, hence sequentially compact. Uniform local constancy gives a positive separation scale, so finitely many initial Cantor-code (binary) digits determine $f$. Removing zero destroys compactness: the hinted function reading the digit after the first $1$ is continuous at every remaining sequence but depends on arbitrarily late digits.

A choice function on a family $\mathcal{A}$ of nonempty sets satisfies $c(A)\in A$. Zermelo's theorem says every set has a total order under which every nonempty subset has a least element.

$(g,V)$ continues $(f,U)$ when they agree on a nonempty connected component reached by an overlapping chain of function elements. A complete analytic function contains every continuation. Its Riemann surface is the set of its germs; germ neighbourhoods are charts, projection records the base point, and germ evaluation gives a holomorphic map to $\mathbb{C}$.

The characteristic function of $S_n/n$ isThus every $S_n/n$ already has the standard Cauchy law, and convergence in law follows from characteristic-function convergence theorem.

The marginal utility price isThe value function is concave; its tangent at zero lies above its graph. Evaluating that supporting inequality at the utility-indifference displacement $Y-(1+r)\pi (Y)$ gives
$E[U^{\prime }(X_0)(Y-(1+r)\pi (Y))]\ge 0$, hence ${\pi }_0(Y)\ge \pi (Y)$.

The basic functions are zero $Z(\mathbf{x})=0$, successor $S(n)=n+1$, and projections $P_i^k(\mathbf{x})=x_i$. Close these under composition and primitive recursion:

No. Point evaluation is continuous on $H^1(\mathbb{R})$, and every limit of functions supported away from zero has trace $u(0)=0$. A smooth function with nonzero value at zero belongs to $H^1$ but not to this $H_0^1$.

For $V=Z(f_1,\dots ,f_r)$, $T_{V,p}$ is the common kernel of $d{f}_i(p)$; $\dim V$ is the transcendence degree of its function field, and $p$ is smooth when $\dim T_{V,p}=\dim V$. A nonzero maximal Jacobian minor defines a nonempty open smooth locus (generic rank gives nonemptiness).

A line $ux+vy+wz=0$ corresponds to $[u:v:w]\in ({\mathbb{P}}^2{)}^{\ast }$. Lines through $P=[a:b:c]$ satisfy $au+bv+cw=0$, a projective line. For a smooth curve $F=0$, the dual morphism is $P\mapsto [F_x(P):F_y(P):F_z(P)]$.

For $F=x^3+y^3+z^3$, eliminating $x,y,z$ from $[u:v:w]=[x^2:y^2:z^2]$ givesThe dual sextic is singular: its nine cusps are the tangent lines at the nine flexes of the cubic. Thus points of $X$ mapping to singularities are precisely inflection points, where the tangent has higher contact.

Runge theorem says that if $K\subset \mathbb{C}$ is compact and $\mathbb{C}\setminus K$ is connected, every function holomorphic near $K$ is uniformly approximable on $K$ by polynomials.

Choose a compact exhaustion $K_j$ of the slit disc $\Delta \setminus L^{+}$, rounding the two sides of the slit so that $0\notin K_j$ and $\mathbb{C}\setminus K_j$ is connected. Runge applied to $1/z$ gives a polynomial $p_j$ with ${\sup }_{K_j}|p_j-1/z|<1/j$. Every compact subset of the slit disc lies in all sufficiently large $K_j$, proving locally uniform convergence.

For perfect balance, $k=2^{n-1}$ and the all-zero probability is zero. For other balances the one-shot outcomes overlap the constant case. Repeat Deutsch-Jozsa algorithm and declare nonconstant upon any nonzero result. Constants are never misclassified; a $k$-balanced function fails after $R$ trials with probability $q^R$, $q=(1-2k/2^n{)}^2$. It suffices that $R\ge \log ϵ/\log q$, using that many oracle calls.

For machine index $m$, $f_{m,k}$ is the partial $k$-ary function computed by that machine under the fixed coding, and $W_m$ is its halting domain (equivalently the r.e. set enumerated by machine $m$, under the course convention).

Kleene's recursion theorem says that for every total recursive map $f$ on program indices there is an index $e$ with the same partial function as program $f(e)$.