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Write $E=-{\hslash }^2{\kappa }^2/(2m)$. An even bound state is proportional to $\cos (kx)$ inside the well and to $e^{-\kappa |x|}$ outside, whereContinuity of the logarithmic derivative at $a$ givesFor sufficiently small $a$, $ka$ lies in the first monotone branch, so this equation has exactly one positive root and the even state is unique up to scale.

As $a\downarrow 0$, $\kappa \sim k^{2}a\to m/{\hslash }^2$. HenceThe wells converge distributionally to $V_0(x)=-\delta (x)$, whose normalized even bound state isIts derivative jump is precisely the one imposed by the delta potential.

An allowable parametrisation is a smooth homeomorphism from an open subset of ${\mathbb{R}}^2$ onto an open subset of $S$, with derivative of rank two. Away from the axis, the rotation orbit has nonzero tangent. A transverse curve supplied by the submanifold theorem, followed by the rotation action, giveswith $|v|<\pi$ and $(f^{\prime },g^{\prime })\ne (0,0)$.

For a ruled parametrisation, regularity is exactly

Rotate and translate along the axis so the specified ruling isHere $s\ne 0$ because the line is not parallel to the axis, and $c\ne 0$: if $c=0$, rotating the horizontal tangent line makes the ruled parametrisation singular at its closest point. Rotating $L$ givesa one-sheet hyperboloid. Rotation invariance puts this whole surface in $\Sigma$. Connectedness and the fact that a complete embedded hyperboloid cannot be a proper subset of another connected embedded surface force equality. Rescaling radial and axial coordinates gives a diffeomorphism with $x^2+y^2=1+z^2$.

Uniform continuity means that for every $\epsilon >0$ there is a single $\delta >0$ such that $d(x,y)<\delta$ implies $|f(x)-f(y)|<\epsilon$ for all $x,y$. A $d^{\prime }$-Cauchy sequence converges uniformly pointwise to a bounded function; passing a uniform-continuity estimate through a sufficiently close member proves that the limit is uniformly continuous. Thus $C_{b,u}(X)$ is complete.

If $f\in C_0({\mathbb{R}}^n)$, decay at infinity and compactness of a ball make it bounded. Uniform continuity on a sufficiently large compact ball, together with small values outside it, proves global uniform continuity. Hence $C_0\subset C_{b,u}$.

A uniform limit of functions vanishing at infinity also vanishes at infinity, so $C_0$ is closed. It is not compact: translate a fixed compactly supported bump of height one so that the supports are disjoint. The resulting sequence has pairwise sup distance one and no convergent subsequence.

With $\omega =B/m$, magnetic force does no work, so $|\dot{x}|$ is constant; its zero $z$ component also makes $\dot{z}$ constant. For the stated data,This is a circle of radius $u/\omega$ centred at $(0,-u/\omega )$.

With $\alpha =\mu /m$, the complex velocity is $\dot{x}+i\dot{y}=u{e}^{-(\alpha +i\omega )t}$. HenceThe path spirals into $u(\alpha ,-\omega )/({\alpha }^2+{\omega }^2)$. As $u$ varies, this lies on $y=-(\omega /\alpha )x$, at distancefrom the origin.

A set is countable when it is finite or admits a bijection with a subset of $\mathbb{N}$. For each degree, the coefficient tuples for polynomials in $A[x]$ form a finite Cartesian power of the countable set $A$; a countable union over degrees is countable. In particular the algebraic numbers, being roots of countably many integer polynomials with finitely many roots each, are countable. Since $\mathbb{R}$ is uncountable, uncountably many real numbers are transcendental.

The set $Q$ is uncountable. Partition $N$ into four-element blocks and, independently on each block, choose either of two fixed-point-free permutations. Binary sequences then inject into $Q$.

Finally choose a line $L$ distinct from and nonparallel to every $L_i$, possible because only countably many directions are excluded. Each $L\cap L_i$ has at most one point, so the union covers only countably many points of the uncountable line $L$. It cannot cover the plane.

If $\sqrt{2}+\sqrt{3}$ were rational, squaring would make $\sqrt{6}$ rational, a contradiction. Alsoso it is irrational but algebraic, satisfying $x^2+2x-1=0$.

For lattice vertices $u,v,w$, the triangle area isa half-integer. Triangulating a convex lattice polygon from one vertex makes its area rational.

A regular octagon of side length $s$ has area $2(1+\sqrt{2})s^2$. Lattice endpoints make $s^2$ a positive integer, so this area would be irrational, contradicting the preceding result. Hence no such octagon exists.