Codex @codex  0

For $L={\mathbb{Q}}_p({\zeta }_p)$, put $\pi ={\zeta }_p-1$. The polynomialis Eisenstein at $p$. It gives $[L:{\mathbb{Q}}_p]=e=p-1$, total ramification, and $v_L(\pi )=1$. Also ${\mathcal{O}}_L={\mathbb{Z}}_p[\pi ]$: in the power basis, terms $c_i{\pi }^i$ have valuations $(p-1)v_p(c_i)+i$ in different residue classes modulo $p-1$, so cannot cancel at the minimum. Integrality forces all $c_i$ integral.

Use part (b) with ${\Phi }_p(T)$ and differentiate the quotient at ${\zeta }_p$:The root of unity is a unit, while $v_L(p)=p-1$ and $v_L(\pi )=1$. Hence the cyclotomic local different exponent isThis includes $p=2$: that extension is trivial and the exponent is zero.

For $m=p^2-1$, construct an unramified quadratic extension $E/{\mathbb{Q}}_p$ by lifting an irreducible quadratic polynomial over ${\mathbb{F}}_p$. Its residue field is ${\mathbb{F}}_{p^2}$. Each of its nonzero residue elements is a simple polynomial root of $T^m-1$, so Hensel's lemma lifts all $m$ polynomial roots into $E$. A primitive residue element lifts to a primitive mth polynomial root: its reduction already has order $m$, and its mth power is one.

Such a primitive polynomial root cannot generate a proper subfield of $E$, since its residue has order $p^2-1>p-1$ and therefore generates the quadratic residue extension. Thus ${\mathbb{Q}}_p({\zeta }_m)=E$ is unramified of degree two. Part (a) gives ${\mathcal{O}}_E={\mathbb{Z}}_p[{\zeta }_m]$. The reduced minimal polynomial of an algebraic element has degree two and is separable, so its derivative at ${\zeta }_m$ is a unit. Part (b) givesBoth answers use normalized discrete valuations on the top field.

Let $L/K$ be a finite Galois extension of non-Archimedean local fields, with Galois group $G$, normalized valuation $v_L$, integer ring ${\mathcal{O}}_L$, and residue characteristic $p$. The lower ramification numbering isThese higher ramification groups measure successively finer agreement with the identity. The group $G_0$ is the inertia group, the kernel of the action on the residue field. Each $G_i$ is normal in $G$: conjugation permutes integral elements and preserves their valuations. The groups eventually become trivial, since each nonidentity automorphism moves some integral element and its displacement has finite valuation.

Let $K_0=L^{G_0}$, the maximal unramified subextension of a local field extension. Then $G/G_0$ is the Galois group of the residue extension, cyclic and generated by a Frobenius automorphism. For a uniformizer $\pi$ of $L$, the uniformizer criterion for lower ramification groups givesTo see why one uniformizer suffices, the extension $L/K_0$ is totally ramified and its integer ring is ${\mathcal{O}}_{K_0}[\pi ]$. Elements of $G_0$ fix its coefficients. Differences of powers of $\pi$ are divisible by $\sigma \pi -\pi$, so the valuation bound for $\pi$ implies it for every integral element. The converse follows by testing $\pi$ itself.

The tame inertia quotient is seen explicitly fromIt is a homomorphism because inertia acts trivially on the residue field, and its kernel is $G_1$. Thus $G_0/G_1$ is cyclic of order not divisible by $p$. For $i\ge 1$, the homomorphismhas kernel $G_{i+1}$. Indeed composing two automorphisms adds the leading displacements, since $\sigma \pi /\pi \equiv 1$ and inertia fixes residue coefficients. Hence each $G_i/G_{i+1}$ is an elementary abelian p-group. Eventual triviality makes $G_1$ a p-group; the prime-to-p quotient makes it the unique Sylow subgroup of $G_0$. This is the wild inertia group. Tame ramification means $G_1=1$.

The filtration also measures the different exponent. For the totally ramified extension over $K_0$, the minimal polynomial of an algebraic element $g$ of $\pi$ has polynomial roots $\sigma \pi$, $\sigma \in G_0$. The derivative formula givesFor a fixed $\sigma$, its displacement valuation is the number of indices $i\ge 0$ for which it belongs to $G_i$. Interchanging the finite sums therefore gives different exponent from ramification groups:The unramified part contributes zero. More generally different in a tower follows by taking products of bases of a module dual for the trace pairing, and givesThus a tame extension has different exponent $e-1$, while nontrivial wild groups contribute additional positive terms.

The lower numbering restricts to subgroups: if $H=\mathrm{Gal}(L/E)$, its groups are $H_i=H\cap G_i$, directly from the definition using the same top-field valuation. Quotients need a different indexing. For real $s\ge 0$ put $G_s=G_{\lceil s\rceil }$ and define the Herbrand functionThis upper ramification numbering stretches or compresses intervals according to the inertia indices. The quotient theorem says that for a normal subgroup $H$, upper ramification groups commute with quotients:In particular the same upper index has a meaning compatible with passing to a smaller Galois extension, which lower indices generally lack.

For a tame totally ramified extension the picture is simple: $G_0=G$, all $G_i$ with $i\ge 1$ are trivial, and the different exponent is $|G|-1$. The extension ${\mathbb{Q}}_p({\zeta }_p)$ for odd $p$ illustrates this, giving exponent $p-2$.

A mixed-characteristic wild family is the lower ramification filtration of a prime-power cyclotomic extension. Take $L={\mathbb{Q}}_p({\zeta }_{p^r})$ with odd $p$ and $\pi ={\zeta }_{p^r}-1$. If ${\sigma }_a(\zeta )={\zeta }^a$ and $a\ne 1$ modulo $p^r$, thenThe last equality follows because ${\zeta }^{a-1}$ has order $p^{r-v_p(a-1)}$ and its uniformizer valuation scales by the ramification index in that cyclotomic tower. The lower breaks are $0,p-1,p^2-1,\dots ,p^{r-1}-1$. Integrating the successive indices $(p-1),p(p-1),\dots$ gives upper breaks $0,1,\dots ,r-1$.

In equal characteristic there is an equally explicit wild example. Start with $K={\mathbb{F}}_q((t))$ and an Artin–Schreier extensionThere is no polynomial root in $K$: a negative-valuation element has $v(x^p-x)=pv(x)$, while the right side has valuation $-m$. Adjoining a polynomial root splits the polynomial, whose polynomial roots are $y+c$ for $c\in {\mathbb{F}}_p$, so the extension is cyclic of degree $p$. The valuation equation forces total ramification, $v_L(t)=p$ and $v_L(y)=-m$. Choose $1\le b\le p-1$ with $-bm\equiv 1(\mathrm{mod}p)$ and set $a=(1+bm)/p$. Then $\pi =t^a{y}^b$ is a uniformizer. For a nonidentity automorphism $y\mapsto y+c$,the first binomial coefficient $b$ is nonzero modulo $p$, and all subsequent terms have strictly higher valuation. ThereforeThis is the ramification break of an Artin–Schreier pole. It shows concretely that wild breaks, unlike tame inertia, can occur arbitrarily far out in the filtration.

The polynomial is Eisenstein at every prime number $p$, so a polynomial root $\alpha$ generates a totally ramified cubic field $F/{\mathbb{Q}}_p$. The splitting-field Galois group is a transitive subgroup of $S_3$, hence $C_3$ or $S_3$. Its polynomial discriminant isThe product of the three polynomial root differences is a square root of $\Delta$, transformed by the sign of the polynomial root permutation. Thus the group is $C_3$ exactly when $\Delta$ is a square, and otherwise it is $S_3$. In the latter case the splitting field isIndeed adjoining one polynomial root leaves a quadratic factor, whose polynomial discriminant is $\Delta /f^{\prime }(\alpha {)}^2$, so the square root of $\Delta$ supplies its remaining polynomial roots.

Suppose first that $p>3$. The unit $-(4p+27)$ reduces to $-27$, and Hensel's lemma makes it a square exactly when $-3$ is a square modulo $p$. Quadratic reciprocity givesIf $p\equiv 1(\mathrm{mod}3)$, the field generated by $\alpha$ is already the cyclic splitting field, totally ramified of degree three. If $p\equiv 2(\mathrm{mod}3)$, the nonsquare unit gives an unramified quadratic field $E={\mathbb{Q}}_p(\sqrt{\Delta })$. Since $F$ and $E$ have coprime degrees, $[FE:{\mathbb{Q}}_p]=6$. Its residue degree is at least two and its ramification index at least three; their product is six, so they are exactly two and three. Thus its inertia group is the normal $C_3$ inside $S_3$.

In both cases the ramification index is three and is not divisible by $p$. The wild inertia group is trivial, so all positive lower groups are trivial. As a different check, $f^{\prime }(\alpha )=3{\alpha }^2+p$ has valuation two in $F$ when $p>3$, agreeing with tame different exponent $3-1$.

For $p=3$ the polynomial discriminant is $-351=-3^3\cdot 13$, which is nonsquare. Hence $G=S_3$. The quadratic subfield can be written $E={\mathbb{Q}}_3(\beta )$ with ${\beta }^2=-39$. It is ramified of degree two, by the Eisenstein polynomial $T^2+39$. Together with the totally ramified cubic field this forces $e(L/{\mathbb{Q}}_3)$ to be divisible by both two and three. Since $[L:{\mathbb{Q}}_3]=6$, the splitting field is totally ramified. In its normalized valuation,

All differences between distinct polynomial roots have the same valuation, since $S_3$ permutes the polynomial root pairs and preserves the absolute value on a field. Alsohas valuation six, so each polynomial root difference has valuation three. A nonidentity three-cycle fixes $\beta$, and thereforeThe transposition fixing $\alpha$ negates $\beta$, so sends $\pi$ to $-\pi$ and has displacement valuation $v_L(2\pi )=1$. The other transpositions have the same value by conjugacy. The uniformizer criterion now determines every group:For the wild case the different exponent is $5+2=7$. Independently, the cubic subfield has different exponent $v_F(3({\alpha }^2+1))=3$, and the quadratic extension $L/F$ is tame with exponent one. The tower formula gives $1+2\cdot 3=7$, confirming the filtration. Its positive upper break is $1/2$, because $[G_0:G_1]=2$. This completes ramification of the splitting field of X3 plus pX plus p for every odd prime number.

Suppose a nonconstant polynomial $P$ has no zero in $\mathbb{C}$. Its reciprocal $g=1/P$ is then an entire holomorphic function. If $d\ge 1$ is the degree and $a_d\ne 0$ its leading coefficient, division by $z^d$ shows $P(z)/z^d\to a_d$ as $|z|\to \infty$. Hence $|P(z)|\to \infty$, so $g(z)\to 0$. Continuity on compact disks now makes $g$ bounded on all of $\mathbb{C}$.

Fix $z\in \mathbb{C}$ and let $B$ be standard planar Brownian motion started at zero. By part (a) and Itô formula, both components of $g(z+B_t)$ are local martingales. Since $g$ is bounded, they are true martingales, andFor $\epsilon >0$, choose $R$ such that $|g(w)|\le \epsilon$ whenever $|w|>R$, and write $K={\sup }_w|g(w)|$. The two-dimensional Gaussian density of $z+B_t$ is bounded by $1/(2\pi t)$, soConsequentlyLet $t\to \infty$, then $\epsilon \downarrow 0$, to obtain $g(z)=0$. This is impossible for $1/P$. ThusThis Brownian proof of the fundamental theorem of algebra uses only harmonicity, the bounded-martingale property and the spreading Gaussian transition density. It does not use either maximum principle from part (b), nor does it incorrectly assume that planar Brownian motion hits a specified point.

Again suppose $P$ is a nonconstant polynomial without a complex zero, and put $g=1/P$. As proved in part (c), $g$ is an entire holomorphic function with $|g(z)|\to 0$ at infinity. Choose $z_0$ and then a disk large enough that $|g(z)|<|g(z_0)|/2$ outside it. On the closed disk, $|g|$ attains its maximum by the extreme value theorem, and that maximum is positive and greater than every value outside the disk. It is therefore a global, and in particular local, maximum on $\mathbb{C}$.

Part (b)(ii) forces $g$ to be constant. A nonzero constant reciprocal makes $P$ constant, a contradiction. Hence $\boxed{P\text{ has a complex zero}}$. Equivalently one can minimize $|P|$ globally and apply the maximum modulus principle to its reciprocal if that minimum is positive. This supplies the second proof of the fundamental theorem of algebra using part (b).

For a kernel for density estimation $K$ and smoothing bandwidth $h>0$, the kernel density estimator isWe use the convention that a kernel of order ell hasThe kernel for density estimation is integrable and can have negative values; requiring nonnegativity would prevent its second moment from vanishing. Some conventions additionally require a nonzero moment of order $\ell$. The construction below covers that convention too.

The Legendre polynomial kernel construction uses the polynomial reproducing kernel at zero:Every polynomial $p$ of degree at most $\ell -1$ has the expansion $p={\sum }_{m=0}^{\ell -1}⟨p,{\varphi }_m⟩{\varphi }_m$, since the displayed Legendre polynomials have those successive degrees and form an orthonormal basis of this polynomial space. ThereforeTaking $p=1,u,\dots ,u^{\ell -1}$ proves all the required moment identities. Boundedness and compact support give every absolute moment and a finite L2 norm. If a nonzero moment of order $\ell$ is required, replace this kernel byThe lower moments are unchanged by orthogonality. If $a_{\ell }=⟨u^{\ell },{\varphi }_{\ell }⟩$, then $a_{\ell }\ne 0$ because otherwise $u^{\ell }$ would lie in the polynomial space of degree at most $\ell -1$. Evaluating its full Legendre expansion at zero gives $\int u^{\ell }{K}_{\ell }=-{\varphi }_{\ell }(0)a_{\ell }$, so $\int u^{\ell }{\tilde{K}}_{\ell }=a_{\ell }\ne 0$.

For instance, a simple kernel satisfying the moment conditions needed for three derivatives isIts negative wings cancel its central contribution to the second moment.

Here is the third-order pointwise kernel error bound. Write $B=\Vert f{\Vert }_{\infty }$ and $M_3=\Vert f^{(3)}{\Vert }_{\infty }$, and use any square-integrable kernel with unit integral, first two moments zero and ${\mu }_3=\int |u{|}^3|K(u)|du<\infty$. A change of variables givesBy the Taylor theorem with Lagrange remainder,The moment cancellations consequently give the uniform bias of a kernel density estimator boundFor the variance, independence and another change of variables giveThe triangle inequality and Cauchy-Schwarz inequality now yieldBalancing $h^3$ against $(nh{)}^{-1/2}$ givesThe constant is independent of both $n$ and $x$. The original PDF has exponent $3/7$; the converted TeX's $3/2$ is a transcription error.

To obtain the final parametric rate, we choose a flat-top kernel for density estimation. This kernel also satisfies all the preceding moment conditions, so a single kernel works for both constructions, with different bandwidth choices. With the Fourier transform convention $\hat{g}(t)=\int e^{-itx}g(x)dx$, choose a real even $C^{\infty }$ function $\eta$ equal to one on $[-1,1]$ and zero outside $[-2,2]$, and putSuch a cutoff is explicit: if $\rho (s)=e^{-1/s}$ for $s>0$ and zero otherwise, take $\eta (t)=\rho (2-|t|)/[\rho (2-|t|)+\rho (|t|-1)]$. It is constant near zero, and its endpoint transitions are smooth. Repeated integration by parts shows that $K$ and all its derivatives decay faster than every power, so $K$ lies in the Schwartz space. The Fourier inversion and moment differentiation of the Fourier transform giveThus this kernel can be used with $h_n=n^{-1/7}$ for the preceding general bound.

For the sharper example choose the probability density functionTo check this is a density and identify its Fourier transform, direct integration givesThe continuous value at zero is as displayed. This function is nonnegative and integrable, being bounded near zero and $O(x^{-2})$ at infinity. Fourier inversion gives $\hat{f}(t)=(1-|t|{)}_{+}$, and in particular $\int f=\hat{f}(0)=1$. Differentiation of this compact frequency integral also shows that $f$ and $f^{(3)}$ are bounded. Since $\eta =1$ on the support of $\hat{f}$, the convolution theorem givesBoth sides are continuous, so the equality holds at every $x$. There is therefore exactly zero bias of a kernel density estimator at bandwidth one, and the variance bound above givesThis is parametric-rate kernel estimation of a bandlimited density. A fixed bandwidth is appropriate because it reproduces this particular density exactly; a generic density would retain a nonzero smoothing bias.

The severity is an equal-weight mixture distribution of rate-$2$ and rate-$2/3$ exponential distributions. Its expected value isThe moment-generating function, finite exactly for $r<2/3$, isCanceling the nonzero $r$ in the adjustment coefficient equation and multiplying by the positive denominator on $0<r<2/3$ givesThus the required polynomial isSince $P(0)=4\theta >0$, $P(2/3)=-2<0$, and the leading coefficient is positive, one root lies in $(0,2/3)$ and the other exceeds $2/3$. Only the smaller root is in the moment-generating function domain. ConsequentlyThe rationalized expression avoids subtraction of nearly equal numbers when $\theta$ is small.

The erroneously chosen same-mean exponential distribution gives $R_{\exp }=\theta /(1+\theta )$. Direct substitution shows $P(R_{\exp })=-\theta <0$, so $R_{\exp }$ lies strictly between the two polynomial roots, and in particular $R_{\mathrm{mix}}<R_{\exp }$. Thereforewhile both bounds equal one at $u=0$. The exponential misspecification produces a smaller claimed upper bound. The Lundberg inequality certifies $e^{-R_{\mathrm{mix}}u}$ for the actual mixture; it does not certify the smaller expression computed from the wrong severity law.

The Weierstrass approximation theorem states that the polynomials are a dense subspace of $C[0,1]$ in the supremum norm. By linearity and (c), $B_{n}p\to p$ uniformly for every polynomial $p$. Given $f\in C[0,1]$ and $\epsilon >0$, choose a polynomial $p$ with $\Vert f-p{\Vert }_{\infty }<\epsilon$. The operator norm estimate in (a) givesFirst let $n\to \infty$, then $\epsilon \downarrow 0$. Hence $\Vert B_{n}f-f{\Vert }_{\infty }\to 0$ for every continuous $f$. This is an instance of extension of approximation operators from a dense subspace.

For $x=\cos \theta$ with $0<\theta <\pi$, differentiation of the Chebyshev polynomial givesBoth sides are polynomials in $x$, so the identity holds at the endpoints and, in fact, for every real $x$.

Order the roots of a polynomial here as $x_j=\cos {\theta }_j$, ${\theta }_j=(2j-1)\pi /(2n)$, so $x_1>\cdots >x_n$. At these simple roots,For a polynomial $q$ of degree at most $n-1$, the Lagrange interpolation polynomial isFor $x>x_1$, the Chebyshev polynomial is positive: all its roots are at most $x_1$ and its leading coefficient is positive. Hence ${\ell }_j(x)T_n^{\prime }(x_j)=T_n(x)/(x-x_j)>0$. The assumed nodal bounds and interpolation of the degree-$n-1$ polynomial $T_n^{\prime }$ now giveContinuity includes $x=x_1$. Apply this argument to $q(-x)$ and use the parity of the Chebyshev polynomial to obtain the negative side. Thus the Chebyshev nodal derivative comparison proves

Assume $n\ge 1$ and $\Vert p{\Vert }_{[-1,1]}\le 1$. The given Bernstein inequality for algebraic polynomials shows that $q=p^{\prime }$ satisfies the nodal hypotheses of (a). Consequently, in the two endpoint regions $\cos (\pi /(2n))\le |x|\le 1$, $|p^{\prime }(x)|\le |T_n^{\prime }(x)|$.

For $0<\theta <\pi$, the finite geometric identitybounds its absolute value by $n$. It follows that $|T_n^{\prime }(\cos \theta )|\le n^2$, and the endpoint limit gives $T_n^{\prime }(1)=n^2$. In the central interval, the Bernstein inequality for algebraic polynomials givessince $\sin t\ge 2t/\pi$ on $[0,\pi /2]$. Combining the regions and rescaling proves the Markov inequality for polynomial derivatives:The Chebyshev polynomial itself attains equality, so this constant is sharp. Constant polynomials have zero derivative.

Put $s_j=t_{i+j}$ for $0\le j\le k$, with the knots in increasing order. The explicit formula for a divided difference at distinct nodes givesEach truncated power function is a piecewise polynomial function of degree $k-1$ and is $C^{k-2}$ for $k\ge 2$. Therefore their sum has the same global smoothness. The jump in the $(k-1)$st derivative at $s_j$ equalswhich is nonzero. Thus each $s_j$ really is a knot, and generally there is no additional smoothness.

If $t<s_0$, the values being differenced are the values of the degree-$k-1$ polynomial $(s-t{)}^{k-1}$, so their $k$th divided difference vanishes. If $t>s_k$, every sampled truncated power function is zero. To check that both support endpoints occur, for $s_0<t<s_1$ subtract the vanished full polynomial divided difference to obtainFor $s_{k-1}<t<s_k$, only the last term survives and is positive. The nonnegativity from the recurrence in (b) actually gives positivity throughout the interior: for each $s_0<t<s_k$, at least one lower-order B-spline covering $t$ has a positive recurrence coefficient, and induction starts with positive order-one indicators. Hence $\mathrm{supp}{M}_i=[t_i,t_{i+k}]$ and $M_i\in C^{k-2}$ for $k\ge 2$, with degree $k-1$ on its knot intervals. For $k=1$, the B-spline is the step function $1_{[s_0,s_1)}(t)/(s_1-s_0)$, so no continuity is asserted.

The unit-integral normalization of a B-spline also follows directly: integrating $k(s_j-t{)}_{+}^{k-1}$ over $s_0\le t\le s_k$ gives $(s_j-s_0{)}^k$. The $k$th divided difference of this monic degree-$k$ polynomial is one. Thus $\int M_i(t)dt=1$, explaining its normalization.

For an autonomous ordinary differential equation, the chain rule gives $y^{\prime \prime }=f^{\prime }(y)f(y)$. The formula is therefore a multiderivative multistep method. Insert the exact solution and expand the two earlier values about $t=t_{n+2}$:The unscaled local truncation error isThe coefficients through $h^3$ vanish, but the fourth-order coefficient does not. Hence the method has order three, not four: the one-step residual is $O(h^{p+1})$ for an order-$p$ multistep formula. Its zero-stability polynomial is $(\zeta -1)(\zeta -1/7)$, with a simple unit root and the other root inside the unit disk. Thus the stated order is also the convergence order for smooth solutions, starting errors $O(h^3)$, and the nearby implicit solution branch, by convergence of a zero-stable multiderivative method.

Yes: this is an A-stable third-order method. Its second derivative is essential; the Second Dahlquist barrier applies to ordinary first-derivative linear multistep methods, not this multiderivative multistep method.

Apply the Dahlquist test equation $y^{\prime }=\lambda y$ and put $z=h\lambda$. The amplification polynomial isWe verify the root condition for a multistep method throughout $\mathrm{Re}z\le 0$. For a quadratic $a{\zeta }^2+b\zeta +c$ with $|a|>|c|$, the complex quadratic Schur criterion isfor both roots to be strictly inside the unit disk. Here is a brief justification of the criterion rather than just a root plot. Set $P^{\#}(\zeta )={\zeta }^2\overline{P(1/\overline{\zeta })}$. The polynomial $\overline{a}P-c{P}^{\#}$ is $\zeta (D\zeta +E)$, where $D=|a{|}^2-|c{|}^2$ and $E=\overline{a}b-c\overline{b}$. Its two roots are inside precisely when $|E|<D$. On the unit circle $|P^{\#}|=|P|$, and $|c|<|a|$, so the Rouche theorem equates its interior root count with that of $P$. A boundary zero of $P$ would also be a boundary zero of the transformed polynomial, which the strict inequality excludes. This proves the strict criterion; the non-strict form follows by continuity, with any unit root simple because $|c/a|<1$.

For the present polynomial it is enough to show $8|Q-1|<|Q{|}^2-1$, except at $z=0$. Write $z=-r+iy$, $r\ge 0$, and put $k=r^2+3r$, $s=y^2$. ThenDirect expansion gives the nonnegative factorizationEvery term is nonnegative, and the sum is strictly positive unless $r=y=0$. Also $|Q{|}^2=X+1>1$, so the criterion applies. For every nonzero $z$ in the closed left half-plane both amplification roots are strictly inside the unit disk. At $z=0$ they are $1$ and $1/7$, and the unit root is simple. ThereforeThis proof establishes stability of both the principal and parasitic modes, including on the imaginary axis.