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

In the frame rotating with the planet, put the planet at fixed radius $a_p$. At periapsis the planetesimal moves prograde faster than the planet and passes just outside it. Its relative longitude initially increases, reaches a turning point near $r=2^{1/4}{a}_p$ at a lead of about $10^{\circ }$, and then decreases because its angular speed has fallen below the frame rate. Meanwhile its radius continues to grow along the outgoing parabolic branch. A sketch should therefore show an exterior cusp-like closest passage, a short prograde sweep to maximum lead, and a long outward trail that bends retrograde in the rotating frame.

Treat the just-bound orbit as a parabolic Kepler orbit with $q\simeq a_p$. Its specific angular momentum is $h=\sqrt{2G{M}_{\ast }q}$, so its instantaneous angular speed is $h/r^2$. Equating this to the planet's mean motion $n_p=\sqrt{G{M}_{\ast }/a_p^3}$ givesThe parabolic orbit equation $r=2q/(1+\cos f)$ then gives $f\simeq 47^{\circ }$, or $50^{\circ }$ to the nearest ten degrees.

If the bodies were in conjunction at periapsis, Barker equation givesThe planetesimal is therefore at its maximum lead, approximatelyahead of the planet as seen from the star.

Very near one planet, that planet dominates both the particle's free-precession coefficient and its forcing coefficient, with the leading terms arranged approximately as $\dot{z}\simeq iA(z-z_p)-z/\tau$. Since $|A|\tau \gg 1$ close to the planet, the late-time forced solution approachesThe particle's orbit therefore tends toward the planet's eccentricity and apsidal direction. Extremely close to the planet, close encounters, co-orbital dynamics, and individual mean-motion resonances invalidate the orbit-averaged linear approximation.

For $t\gg \tau$, the free term has disappeared. Exactly at $A=g_1$, the resonant contribution tends towith finite eccentricity $|{\nu }_1|\tau$ and a quarter-cycle phase shift. Without damping, exact resonance instead giveswhose secular resonance amplitude grows linearly in the ideal linear theory.

Linear eccentricity damping adds $-z/\tau$ to the complex equation:Its solution isUnlike the undamped free eccentricity, the homogeneous term decays exponentially. The forced response survives, acquires a phase lag, and has finite amplitude