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For $n\ge 1$, the Bernstein polynomial isThe weights sum to one by the binomial theorem. Equivalently, if $K$ has the binomial distribution with parameters $n,x$, then $B_n(f;x)=\mathbb{E}f(K/n)$. This interpretation makes the cancellation of apparent high-degree terms transparent.

For the falling factorial $u^{\underline{r}}=u(u-1)\cdots (u-r+1)$, with $u^{\underline{0}}=1$, direct cancellation in the binomial weights givesThe polynomials $u^{\underline{0}},\dots ,u^{\underline{m}}$ form a monic triangular basis of the degree-at-most-$m$ polynomials. Hence $u^m=u^{\underline{m}}+{\sum }_{r<m}{c}_r{u}^{\underline{r}}$, andFor $m\le n$, its leading coefficient is $n^{\underline{m}}/n^m\ne 0$. If $p$ has actual degree $m$ and leading coefficient $a_m\ne 0$, linearity shows that the leading coefficient of $B_{n}p$ is $a_m{n}^{\underline{m}}/n^m$; lower-degree terms cannot cancel it. Thus the Bernstein polynomial degree preservation property givesThis includes the requested $m<n$, and also explains why the property cannot simply be extended to $m>n$.

The same moments give $\mathbb{E}K=nx$ and $\mathbb{E}{K}^2=n(n-1)x^2+nx$. ThereforeThe first two approximation errors vanish identically, whileAll three convergences are consequently uniform convergence on $[0,1]$.

For real-valued $f\in C[-1,1]$, there is a unique best uniform approximation $p_{\ast }\in {\mathcal{P}}_n$, where ${\mathcal{P}}_n$ means degree at most $n$. If $E=\Vert f-p_{\ast }{\Vert }_{\infty }>0$, the Chebyshev alternation theorem characterizes it by the existence of $n+2$ ordered pointsand a sign $\epsilon \in \{-1,1\}$ such thatConversely, a polynomial in ${\mathcal{P}}_n$ whose error has this alternation is the best one. If the best error is zero, $f$ itself is the exact polynomial approximant. The real-valued qualification matters: this sign-alternation formulation is not the corresponding criterion for arbitrary complex-valued approximation.

The obstruction to a strict improvement is concrete. If $q$ had smaller error at every alternation point, $q-p_{\ast }$ would have alternating strict signs there. The intermediate value theorem would give at least $n+1$ distinct zeros, impossible for a nonzero polynomial of degree at most $n$.

Since $|T_j(x)|\le 1$ on $[-1,1]$ and the coefficients are summable, the Weierstrass M-test gives uniform convergence of this positive lacunary Chebyshev series and makes its sum continuous.

Let $r$ be the largest integer with $3^r\le n$, and set $r=-1$ when $n=0$. The candidate and its tail size areThe empty sum for $n=0$ is zero. The candidate has degree at most $n$. The triangle inequality bounds its error by the proposed tail sum, and equality holds at $x=1$, since every Chebyshev polynomial has $T_j(1)=1$ and all tail coefficients are positive. It remains to prove that another polynomial cannot improve this error.

Put $L=3^{r+1}$, the first omitted degree, so $L>n$. At the points $y_j=\cos (j\pi /L)$ for $j=0,\dots ,L$, every omitted term hasbecause $3^{k-r-1}$ is odd. Uniform convergence therefore givesThese $L+1$ distinct points run in decreasing order; reversing their order still gives alternating signs. Since $L+1\ge n+2$, select $n+2$ consecutive ones and apply the Chebyshev alternation theorem. It proves both optimality and uniqueness of the displayed candidate. Thus the best polynomial and the error stay unchanged between successive powers of three; this exact truncation property depends on the synchronized tail signs, not on a general rule that truncating a Chebyshev expansion is optimal.

The order-$k$ divided difference defining $M_i$ is a finite linear combination of values at its knots. Hence integration can be interchanged with it directly. For a knot value $u\in [a,b]$,It follows thatThe final equality holds because the order-$k$ divided difference of a degree-$k$ monic polynomial is its leading coefficient, one. ThusThis proves the unit-integral normalization of a B-spline. Its support is contained in $[t_i,t_{i+k}]$: below all its knots, the sampled truncated power agrees with a degree-$k-1$ polynomial, annihilated by the order-$k$ divided difference; above all its knots, every sample is zero. No mass is omitted by integration over $[a,b]$.

Since $N_i=(t_{i+k}-t_i)M_i/k$, the corresponding partition-normalized B-spline has integral $(t_{i+k}-t_i)/k$. Its normalization is different from that of $M_i$.

Write the Runge-Kutta method coefficients as $A\in {\mathbb{R}}^{\nu \times \nu }$, $b\in {\mathbb{R}}^{\nu }$, and let $e$ be the all-ones column. On the test equation $y^{\prime }=\lambda y$, with $z=h\lambda$, the stage vector satisfies $Y=e{y}_n+zAY$ and the next value is $y_{n+1}=y_n+z{b}^{T}Y$. Hence its stability function isBoth polynomials have degree at most $\nu$, and the denominator has value one at zero. Order of a Runge-Kutta method $2\nu$ implies that this test-equation approximation agrees with $e^z$ through degree $2\nu$:By uniqueness of the diagonal Padé approximant, $R=[\nu /\nu {]}_{e^z}$. Explicitly,The permitted diagonal Padé approximant property says this rational function has no poles in the closed left half-plane and satisfies $|R(z)|\le 1$ there. ThereforeThere is also no hidden stage-solve singularity from a canceled denominator: the Padé approximant numerator and denominator are coprime and each has degree $\nu$. The original determinant denominator already has degree at most $\nu$, so, with its normalization at zero, it equals $P_{\nu }(-z)$ and cannot contain an additional canceled factor. This proves maximal-order Runge-Kutta methods are A-stable.

Use the Gauss collocation coefficient construction. Find the $\nu$ polynomial roots $c_1,\dots ,c_{\nu }$ of the shifted Legendre polynomial $P_{\nu }(2c-1)$ in $(0,1)$, and formThen setThese are explicit polynomial integrals. For example, if ${\ell }_j(s)={\sum }_{r=0}^{\nu -1}{d}_{jr}{s}^r$, calculate $a_{ij}={\sum }_r{d}_{jr}{c}_i^{r+1}/(r+1)$ and $b_j={\sum }_r{d}_{jr}/(r+1)$. The resulting implicit Gauss collocation method has order $2\nu$; its Gaussian quadrature nodes integrate polynomials through degree $2\nu -1$ exactly. The one-stage example is the implicit midpoint rule, with $c_1=a_{11}=1/2$ and $b_1=1$. No search over nonlinear Runge-Kutta order conditions systems is needed.

The first characteristic polynomial factors asAlso $\rho (1)=0$ and ${\rho }^{\prime }(1)=\sigma (1)=2(1-\alpha )$, so the order conditions for a linear multistep method hold for every $\alpha$. Convergence of the full linear multistep method recurrence requires, in addition, the root condition for a multistep method: all polynomial roots of $\rho$ lie in the closed unit disk and its polynomial roots on the unit circle are simple.

For $-1<\alpha <1$, write $\alpha =\cos \theta$ with $0<\theta <\pi$. The other polynomial roots are $e^{\pm i\theta }$, distinct from each other and from 1, so the root condition for a multistep method holds. For $|\alpha |>1$, the two quadratic polynomial roots are real and reciprocal, and one has modulus greater than one. At $\alpha =1$, $\rho =(w-1{)}^3$; at $\alpha =-1$, $\rho =(w-1)(w+1{)}^2$. Both endpoints violate simplicity. Consistency of a numerical method plus zero-stability therefore givesThe conclusion applies to the recurrence as printed, including its starting data. Canceling common factors is not innocuous: at $\alpha =1$ the polynomials contain $(w-1{)}^2$, and at $\alpha =-13/5$ they contain $w+5$. The reduced recurrences exclude parasitic solutions of the original one. This is the common-factor cancellation defect in a multistep recurrence, so those parameter values are not added to the interval.

For $|\alpha |>1$, an exterior polynomial root of $\rho$ persists by continuity for sufficiently small negative real $z$ in the stability polynomial $\rho (w)-z\sigma (w)$. Hence these parameters cannot be A-stable.

For $-1\le \alpha <1$, the principal polynomial root at $w=1,z=0$ is simple. Let $s(z)=\log w(z)$ denote its analytic logarithm near zero. Using the expansion in part (b), and ${\rho }^{\prime }(1)=2(1-\alpha )$, solve the polynomial root equation to obtainIndeed the residual at $s=z$ is $-(\alpha +5)z^4/12$, and the correction in $s$ cancels it by multiplication with $2(1-\alpha )$. Choose $z=-C{\epsilon }^4/2+i\epsilon$. This lies strictly in the left half-plane, butfor small positive $\epsilon$. Thus $|w(z)|=e^{\mathrm{Re}s(z)}>1$, proving linear instability within the A-stability domain. This is a direct principal-root obstruction to A-stability, rather than an appeal to an order-barrier theorem alone.

Finally, at $\alpha =1$, the full stability polynomial isIt retains a double unit polynomial root for every negative $z$, allowing growing parasitic solutions. ThereforeThe canceled Backward Euler method recurrence at $\alpha =1$ is A-stable, but its admissible solutions and starting relations differ from those of the original method.

Put $a_h^2=k^2+m^2>0$ and $\Lambda =a_h^2+{\pi }^2$. The plane-layer poloidal-toroidal decomposition supplies the assumed poloidal fields asSince $w=-{\Delta }_h\varphi$ and $b_z=-{\Delta }_h\chi$, potentials proportional to $e^{st+ikx+imy}\sin \pi z$ have the required vertical components. Their horizontal components are proportional to $\cos \pi z$, satisfying the stress-free boundary conditions and the horizontal-field boundary conditions. The double-curl representation automatically enforces incompressibility and the solenoidal magnetic-field constraint.

Let $W,\Theta ,C$ be the amplitudes of $w,\theta ,b_z$. The temperature and resistive induction equations immediately giveEliminate the total-pressure perturbation by applying $\Delta$ to the vertical momentum equation and subtracting ${\partial }_z$ times its divergence. The resulting vertical equation isIn particular, $imC=-m^{2}W/(s+\zeta \Lambda )$ shows explicitly that magnetic tension opposes the displacement.

Taking the determinant of these three amplitude equations gives the horizontal-field magnetoconvection dispersion relation:This cubic polynomial includes all roots of the assumed poloidal coupled system; using its determinant form avoids discarding possible roots by dividing by $s+\Lambda$ or $s+\zeta \Lambda$. As a check, $Q{m}^2=0$ factors off magnetic diffusion and leaves the standard stress-free Rayleigh-Bénard convection growth equation.

Write $d{\gamma }_1(x)=(2\pi {)}^{-1/2}{e}^{-x^2/2}dx$. On the dense polynomial algebra in $L^2({\gamma }_1)$, the Gaussian creation and annihilation operators areGaussian integration by parts gives $⟨a^{-}f,g⟩=⟨f,a^{+}g⟩$. Their closed realizations are adjoints. More explicitly, for the orthonormal basis $h_n={\mathrm{He}}_n/\sqrt{n!}$ of normalized Probabilists' Hermite polynomials,Thus the closed annihilation operator has domain $\{\sum c_n{h}_n:\sum n|c_n{|}^2<\infty \}$, and the closed creation operator has domain $\{\sum c_n{h}_n:\sum (n+1)|c_n{|}^2<\infty \}$. These describe the same set of $L^2$ functions, the Gaussian Sobolev space, though the two operators differ. Directly on polynomials, $[a^{-},a^{+}]=I$, and

Use the standard carré du champ operator convention $\Gamma (f,g)=\frac{1}{2}(L(fg)-fLg-gLf)$. The product rule for this semigroup generator givesso the requested squared-gradient and joint energy formulas, for real functions in the standard algebra, areFor complex functions the joint energy is sesquilinear, with $\overline{g^{\prime }}$ in the second factor. If “squared gradient” is defined without the factor $1/2$ in the product-rule expression, its value is instead $2f^{\prime }{g}^{\prime }$ and the joint energy is one half of its integral; the joint energy and logarithmic Sobolev constant used here remain as displayed.

For the Gaussian logarithmic Sobolev inequality, first take smooth bounded $f$ with bounded first and second derivatives, bounded away from zero. The Mehler formula for the Ornstein-Uhlenbeck semigroup isIt preserves ${\gamma }_1$ and differentiating in $x$ gives the Ornstein-Uhlenbeck gradient commutation identityPut $u_t=P_{t}f$. Differentiate its entropy functional. Since $\int u_{t}d{\gamma }_1=\int fd{\gamma }_1$ is constant, and Gaussian integration by parts applies,For such data, Mehler's formula and dominated convergence give $P_{t}f(x)\to \int fd{\gamma }_1$ as $t\to \infty$. The uniform positive upper and lower bounds justify integrating this limit inside the entropy, whose limit is zero. Hence

Apply the allowed weighted Cauchy-Schwarz inequality with $g=f^{\prime }$:After integration, invariance of ${\gamma }_1$ givesIntegrating in $t$ yieldsFor a real smooth compactly supported $g$, take $f=g^2+ϵ$. ThenLetting $ϵ\downarrow 0$ provesSmooth approximation and cutoff in the Gaussian Sobolev norm extend this to the full form domain. The entropy is lower semicontinuous under the resulting $L^2$ convergence: the square densities converge in $L^1$, their masses converge, and, along an almost-everywhere convergent subsequence, Fatou's lemma applies to $h\log h$ after adding the bound $1/e$. The same inequality for complex $g$ follows by applying it to $|g|$ and using the weak-derivative bound $|(|g|{)}^{\prime }|\le |g^{\prime }|$. Thus the constant is at most $2$.

For sharpness of the Gaussian logarithmic Sobolev constant, take $g_s(x)=e^{sx/2}$ with real $s\ne 0$. These functions belong to the Gaussian Sobolev space. The Gaussian moment-generating function givesThereforeThe entropy-to-energy ratio is exactly $2$, so no smaller constant can work. Combining this lower bound with the proved inequality gives

For $f_c(z)=z^2+c$, the Mandelbrot set is the set of parameters for which the finite critical orbit $0,f_c(0),f_c^2(0),\dots$ remains bounded. It is a compact connected parameter set. Its main cardioid consists of parameters with an attracting fixed point, with a period-two bulb attached on the left, smaller bulbs attached recursively and fine filamentary structure. It is symmetric about the real axis and meets that axis in $[-2,1/4]$. The quadratic filled Julia set is connected exactly when $c$ belongs to this parameter set; outside it, the rational critical point escapes and the Julia set is a Cantor set. Thus it organizes the contrasting phase-space dynamics of the quadratic family, rather than itself being a Julia set in the dynamical $z$-plane.

The affine map $h(z)=1/2-z$, its own inverse, givesThis value is unique even among Möbius conjugacies. The finite fixed point of $P$ is zero and its periodic-orbit multiplier is one. Holomorphic conjugacy preserves that periodic-orbit multiplier. Infinity has periodic-orbit multiplier zero for a quadratic polynomial, so a corresponding multiplier-one fixed point of $w^2+c$ must be finite; it obeys $2w=1$ and $w^2+c=w$, forcing $c=1/4$. To locate it explicitly, a quadratic fixed point of periodic-orbit multiplier $\lambda$ gives $c=\lambda /2-{\lambda }^2/4$. The boundary parametrization $|\lambda |=1$ has its cusp at $\lambda =1$, so $c=1/4$ is the right-hand cusp on the Mandelbrot boundary. It is in the set: the critical orbit starts at zero, stays in $[0,1/2]$ and increases to the fixed point $1/2$. For real $c>1/4$, the increment $f_c(x)-x=(x-1/2{)}^2+c-1/4$ is strictly positive; a bounded critical orbit would have a limit solving $x^2+c=x$, which has no real solution. These nearby real parameters escape, proving that the cusp belongs to the boundary of the full Mandelbrot set as well.

The fixed points of $P$ on the sphere are zero and infinity. There is no neighborhood of zero on which all iterates converge to zero, even pointwise: a negative real initial point satisfies $P(x)=x-x^2<x<0$, so its iterates decrease without a finite limit and escape to infinity. Every neighborhood of zero contains such points. At infinity there is a uniformly attracting neighborhood in the chordal metric. In the reciprocal coordinate $w=1/z$,For $|w|\le 1/4$, its modulus is at most $|w|/3$. Thus the reciprocal iterates tend uniformly to zero there, proving uniform convergence to infinity on the corresponding sphere neighborhood.

For the given disc, $|z-1/2|<1/2$ is equivalent to $|z{|}^2<\Re z$, and hence to $\Re (1/z)>1$. Under $w=1/z$ the iteration becomesIf $\Re w>1$, then $\Re (1/(w-1))>0$, so $\Re T(w)>\Re w+1$. Consequently the half-plane is forward invariant andThis proves uniform convergence to zero on the disc and therefore $\Delta \subset F(P)$. It is the explicit parabolic disk for z minus z squared.

Let $U=F_{\Delta }$. The iterate family is normal on the connected open set $U$. Every subsequential limit along $n_j\to \infty$ equals zero on $\Delta$; by the identity theorem it is zero throughout $U$. If the full sequence failed to converge uniformly on some compact subset, a further normal subsequence would contradict that conclusion. Hence $P^n\to 0$ locally uniformly throughout $U$. Complete invariance of the Fatou set puts $P(U)$ in a Fatou component. Since $P(\Delta )\subset \Delta$, that component is $U$, so $P(U)\subseteq U$.

We prove the reverse-image assertion, rather than assuming that a basin has only one component. Each connected component $V$ of $P^{-1}(U)$ maps properly to $U$: for compact $K\subset U$, the sphere preimage $P^{-1}(K)$ is compact, and its intersection with $V$ is closed in it because $V$ is a relatively closed component of $P^{-1}(U)$. Its image is open by holomorphic openness. To prove it is closed in $U$, lift a convergent sequence of image points to preimages in $V$. Their images eventually lie in a compact neighborhood inside $U$; properness gives a convergent preimage subsequence with limit in $V$ and the desired limiting image. Thus the image is also closed, and equals the connected target $U$. Moreover $U$ itself is one such component: it lies in $P^{-1}(U)$ by forward invariance, and any connected inverse-image component containing it lies in $F(P)$ and hence in $U$ by maximality.

Now $1/4\in \Delta$ and its entire preimage is $\{1/2\}$, counted twice. Every component $V$ of $P^{-1}(U)$ must contain a preimage of $1/4$, so every one must be $U$. ThereforeThis is the Fatou component containing an entire fiber is completely invariant argument. The convergence on this component is locally uniform: it cannot be uniform on all of $U$, because $P^n(U)=U$ for every $n$ and $U$ contains nonzero points.

Finally suppose zero belonged to the Fatou set. A normal subsequence on a disc around zero would have limit zero, since that disc meets $\Delta$ in an open set where the entire sequence tends to zero. The identity theorem forces this limit to be zero on the whole disc. Uniform convergence in a smaller finite coordinate disc would then imply derivative convergence at zero by the Cauchy integral formula. Butfor every $n$, whereas the derivative of the zero limit is zero. This contradiction proves

Write $n=km$, with positive integers $k,m$, and put $Q=P^m$. Polynomial substitution respects congruence, so $Q(z)\equiv z(\mathrm{mod}Q(z)-z)$ implies inductively $Q^j(z)\equiv z$ for every $j\ge 1$. Since $Q^k=P^n$,Equivalently one can factor each $Q^{j+1}(z)-Q^j(z)$ by $Q(z)-z$ and telescope. If $Q(z)-z$ is identically zero, then $Q$ is the identity and $P^n-z$ is also zero, so the assertion still holds in the sense that the latter is a polynomial multiple of the former.

For the holomorphic cycle, the chain rule giveswith indices cyclically understood. These are the same product at every point. If its modulus is less than one at one point, it is less than one at every point. Taylor expansion of each return map gives a contracting neighborhood of its fixed point, exactly as in question 1. Hence every point of the holomorphic cycle is attracting for $P^n$ whenever one is; a zero factor makes every return periodic-orbit multiplier zero.

For the quadratic family, factor the second-iterate fixed-point equation:The first factor gives fixed points. A root of the second is also fixed only if subtracting the two factors gives $2z+1=0$, so $z=-1/2$ and $c=-3/4$. If $c\ne -3/4$, the second factor has discriminant $-4c-3\ne 0$ and has two distinct roots, neither fixed. They are exchanged by $P$, since each is fixed by $P^2$ but not by $P$, and form an exact two-cycle. Therefore the unique parameter with no exact two-cycle isAt this parameter,Thus the four finite solutions counted with multiplicity are $3/2$ once and $-1/2$ three times: there are only two distinct finite solutions, both already fixed by $P$. The periodic-orbit multiplier at $3/2$ is three (nine for $P^2$), so it is repelling. At $-1/2$ the periodic-orbit multiplier for $P$ is minus one; for $P^2$ it is one. Indeed, with $u=z+1/2$, the local first-return expression for $P$ is $-u+u^2$, and its square is $u-2u^3+u^4$. This is a parabolic return with a triple fixed-point root. The two-cycle has merged into this fixed point; the triple root is not three distinct periodic points. In the Mandelbrot set picture this is the period-two bulb's attachment to the main cardioid.