For the Paczyński-Wiita potential, circular balance givesThereforeThe squared radial epicyclic frequency isCircular orbits change from stable to unstable where ${\kappa }_r^2$ vanishes, so

For $f=1_{[-\pi ,\pi )}$, its $2\pi$-periodization is one almost everywhere. The refinement mask is the periodic function equal to one on $[-\pi /2,\pi /2)$ and zero on the rest of $[-\pi ,\pi )$. Fourier inversion gives the Shannon scaling functionSince $m(t)=\frac{1}{2}{\sum }_n{a}_n{e}^{-int}$, its Fourier coefficients give

Fourier transforming the refinement equation and changing variables givesBy Parseval identity, orthonormality of the translates is equivalent toThese are precisely the Fourier coefficients of the periodization $P(t)={\sum }_k|f(t+2\pi k){|}^2$. Therefore

An orthonormal wavelet is $\psi \in L^2(\mathbb{R})$ such that $\{2^{j/2}\psi (2^{j}x-k){\}}_{j,k\in \mathbb{Z}}$ is an orthonormal basis. A multiresolution analysis is a nested family of closed spaces $V_j$ with trivial intersection, dense union, dyadic scaling, integer-translation invariance of $V_0$, and a generator $\varphi$ whose integer translates form an orthonormal basis of $V_0$. The Meyer-Mallat theorem says every such analysis has an orthonormal wavelet whose translates span $V_{j+1}\ominus V_j$.

The Schoenberg spline operator is positive and $V_{n}1=1$. If ${\xi }_i=a_{1,i}$, then both ${\tau }_i$ and ${\xi }_i$ lie in $[t_i,t_{i+k}]$, soBecause $a_{2,i}$ averages products of knots in the same interval and all points lie in $[0,1]$,The Korovkin theorem now provesfor every $f\in C[0,1]$.

Let $e_m$ denote the $m$th elementary symmetric polynomial of $t_{i+1},\dots ,t_{i+k-1}$. Expanding both sides of the Marsden identity in powers of $x$ and equating the coefficient of $x^{k-1-m}$ givesThus $a_{0,i}=1$, while $a_{1,i}$ and $a_{2,i}$ are respectively the means of the interior knots and of their pairwise products.

The Korovkin theorem states that positive linear operators $L_n:C[0,1]\to C[0,1]$ converge uniformly to the identity on every continuous function if they do so on the three test functions $1,t,t^2$.

Induct on $k$, using the recursive divided-difference formula. After substituting the two induction hypotheses, useand the analogous identity for $g$; adjacent terms telescope. This proves

Apply the Leibniz rule for divided differences to $(x-t{)}_{+}^{k-1}=(x-t)(x-t{)}_{+}^{k-2}$. Only the zeroth and first divided differences of the linear factor survive. After applying the normalization, this gives the Cox-de Boor recursion formula

The explicit divided difference formula makes $M_i(t)$ a finite linear combination of $(t_j-t{)}_{+}^{k-1}$. It is therefore polynomial of degree at most $k-1$ between knots and globally $C^{k-2}$. For $t<t_i$, the nodal data come from a polynomial of degree $k-1$, whose order-$k$ divided difference vanishes; for $t\ge t_{i+k}$ all truncated powers vanish. Thusand normalization does not change the degree, smoothness, knots, or support.

Twice applying the fundamental theorem of calculus giveshence $\boxed{{\omega }_2(f,t)\le t^2\Vert f^{\prime \prime }{\Vert }_{\infty }}$. Part (b) gives $\Vert {\sigma }_n(f)-f{\Vert }_{\infty }=O(n^{-1})$. This cannot be little-$o$ for every $C^2$ function: for $f_0(x)=\cos x$,

The Fejér kernel is nonnegative, and preservation of constants givesBy evenness,Put $\delta =n^{-1/2}$. Use ${\omega }_2(f,|t|)\le (|t|/\delta +1{)}^2{\omega }_2(f,\delta )$ together with $F_n(t)\le n/2$ and $F_n(t)\le C/(n{t}^2)$. Splitting at $1/n$ and $\delta$ shows that the remaining weighted integral is uniformly bounded, so

Substituting the Fourier coefficients into the Fourier partial sum and summing the finite geometric series giveswhere

Since $5^k\equiv 1(\mathrm{mod}4)$,Choose $5^m\le n<5^{m+1}$. For $k\le m$, $\sin (5^k/n)\ge c{5}^k/n$, so each of these $m+1$ same-sign terms has magnitude at least $c/n$. The remaining tail is $O(5^{-m})=O(n^{-1})$. HenceThus $E_n(f)=O(n^{-1})$ does not imply $\omega (f,n^{-1})=O(n^{-1})$; the logarithmic gap prevents a characterization of that approximation class by this first modulus alone.

The inverse theorem for trigonometric approximation givesSumming ${\nu }^{-\alpha }$ provesFor $c_k=a^{-k}$, part (a) gives $E_n(g)\asymp a^{-m}\asymp n^{-{\log }_{5}a}$. If $a<5$, an increment $h=\pi /5^{m+2}\le 1/n$ at $x=0$ has nonnegative summands and its $k=m+2$ term is $\asymp n^{-{\log }_{5}a}$. Part (c) handles $a=5$. Therefore

The trigonometric Chebyshev alternation theorem says that $p_n^{\ast }\in {\mathcal{T}}_n$ is best exactly when its error has at least $2n+2$ cyclically ordered extrema of equal magnitude and alternating sign. Let $5^m\le n<5^{m+1}$ and set $x_j=j\pi /5^{m+1}$. For every $k\ge m+1$,ThusThere are $2\cdot 5^{m+1}\ge 2n+2$ such extrema, while the triangle inequality bounds the tail by their common magnitude. Hence

The body now returns on a bound, highly eccentric orbit whose periapsis crosses more deeply into the planet's neighborhood. Repeated planetary scattering produces a sequence of energy and angular-momentum kicks rather than smooth secular evolution. Possible endpoints include ejection onto an unbound orbit, collision with the planet or star, tidal disruption, or diffusion onto a detached orbit that no longer encounters the planet. Temporary protection by a mean-motion resonance is also possible. The approximate Tisserand parameter constrains weak separated encounters, but the close-encounter sequence is chaotic and does not select a unique final orbit.

The specific orbital energy changes by $\Delta \epsilon =\Delta (v^2)/2<0$, so a marginally bound orbit becomes more tightly bound. Neglecting its tiny initial binding energy,

To obtain the new periapsis, put $c=\sqrt{2}-1$ and $s=\mu /\delta$. Immediately after the encounter,Using the post-encounter energy and angular momentum in the Kepler orbit relations and retaining $O(s^2)$ givesThe inward radial kick therefore lowers the next periapsis as well as shrinking the semi-major axis.

In the planetocentric frame, the incoming relative velocity ${\mathbf{w}}_i$ is prograde and the encounter outside the planet bends it inward through $\theta$, without changing $w=|\mathbf{w}|$. In the inertial frame $\mathbf{v}={\mathbf{v}}_p+\mathbf{w}$, soSubstitution of the previous result yields the negative gravitational assist energy kick

Let $v_K=\sqrt{G{M}_{\ast }/a_p}$. At periapsis the nearly parabolic planetesimal and planet have parallel speeds $\sqrt{2}{v}_K$ and $v_K$, soFor a passage just outside the planet, the impact parameter is $b\simeq \delta a_p$. In the small-angle limit of the supplied gravitational scattering angle,Thus