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False. If uniform convergence held, some polynomial $P_n$ would satisfy ${\sup }_{t\in \mathbb{R}}|P_n(t)-\sin t|<1$. Such a polynomial would be bounded on $\mathbb{R}$, hence constant: a nonconstant polynomial grows without bound in absolute value along a real half-line. For any constant $c$, however,a contradiction.

Write $|F|=p^n$, where $n=[F:{\mathbb{F}}_p]$ since $F$ is a finite-dimensional vector space over its prime field. Every nonzero element satisfies $x^{p^n-1}=1$ by Lagrange theorem, so $F$ is exactly the splitting field of $X^{p^n}-X$ over ${\mathbb{F}}_p$. The derivative is $-1$, making it a separable polynomial. Thus the extension is a normal field extension and a separable field extension, hence a Galois extension.

The finite-field Frobenius automorphism $\sigma (x)=x^p$ fixes ${\mathbb{F}}_p$ and has ${\sigma }^n=1$. If ${\sigma }^d=1$ with $0<d<n$, all $p^n$ elements would be roots of the degree-$p^d$ polynomial $X^{p^d}-X$, impossible. It has order $n$, equal to the degree of the Galois extension. Therefore

Over ${\mathbb{F}}_7$, the polynomial $t^4-1$ factors as $(t-1)(t+1)(t^2+1)$. Since $-1$ is not a square modulo $7$, the last factor is an irreducible polynomial. The splitting field is ${\mathbb{F}}_{49}$, so $\boxed{\mathrm{Gal}(t^4-1/{\mathbb{F}}_7)\cong C_2}$. The finite-field Frobenius automorphism fixes $1,-1$ and swaps the other two roots.

The polynomial $t^3-t-2$ has values $3,3,4,2,3$ at $0,1,2,3,4$ respectively, modulo $5$. Thus $t^3-t-2$ has no root and is an irreducible polynomial over ${\mathbb{F}}_5$. One root generates ${\mathbb{F}}_{125}$; all its Frobenius conjugates are there, so this is its splitting field. Hence $\boxed{\mathrm{Gal}(t^3-t-2/{\mathbb{F}}_5)\cong C_3}$, generated by $x\mapsto x^5$, a 3-cycle on the roots.

Conjugation by $P$ partitions $H$ into orbits. Each non-singleton orbit has size divisible by $p$, by the orbit-stabilizer theorem. The singleton orbits are exactly $H\cap Z(P)$. Since $|H|$ is divisible by $p$, $|H\cap Z(P)|\equiv 0(\mathrm{mod}p)$. This set contains $1$, so $\boxed{|H\cap Z(P)|\ge p}$.

For the deduction we also give the representation-theoretic ingredient behind Burnside's theorem: a nonabelian finite simple group has no nontrivial conjugacy class of prime power size. Here is a proof, so no stronger theorem is being assumed without justification. Suppose $g\ne 1$ has class size $p^b$. For an irreducible character $\chi$ of degree $d$ coprime to $p$, the class sum acts as the scalar $p^b\chi (g)/d$, which is an algebraic integer: the class sum has an integer-entry matrix in the regular representation, which contains every irreducible representation, so this scalar is an eigenvalue of an integer-entry matrix. Also $\chi (g)$ is an algebraic integer. Bezout's identity then makes $\chi (g)/d$ an algebraic integer. All its algebraic conjugates have modulus at most $1$, because character values of a representation are sums of $d$ roots of unity. An algebraic integer with this property is zero or a root of unity: the monic polynomials of all its powers have uniformly bounded integer coefficients, so two powers must coincide unless it is zero.

Every nontrivial irreducible representation of a simple group is faithful. If $\chi (g)/d$ were a root of unity, equality in the triangle inequality would force its representing matrix to be scalar, hence $g$ central, impossible. Thus $\chi (g)=0$ whenever $p\nmid d$, except for the trivial character of a representation. Orthogonality of the identity column and the $g$ column in the character table now givesThis would make $1/p$ an algebraic integer, contradicting the fact that rational algebraic integers are integers. The same proof covers class size $1$ directly through triviality of the center.

If an abelian subgroup $A$ had index $p^a$, every $g\in A\setminus \{1\}$ would have $A\subseteq C_G(g)$, making its class size a divisor of $p^a$, a contradiction. Such a $g$ exists: if $A=1$, then $G$ is a nontrivial p-group, whose center is nontrivial by the first argument with $H=P$. Therefore a nonabelian simple group has no abelian subgroup of prime power index.

Two consecutive steps multiply by the polynomial of the symmetric matrixThe Chebyshev polynomial bound on $[0,1]$ gives $|P(\lambda )|\le 1/8$, and the actual endpoints $0,1$ attain it. At the dominant eigenvalue, $P(1+ϵ)=1/8+ϵ+{ϵ}^2$. ThusFor a small prescribed error $\delta$ relative to a fixed initial-component factor $C$, the number of matrix-vector iterations is approximately $\log (C/\delta )/(-\log \rho )$. The three leading denominators are $ϵ,2ϵ,4ϵ$. Therefore the iteration counts are in the ratio $1:1/2:1/4$: the single shift halves and the alternating double shift quarters the leading count. A pair of double-shift steps costs two matrix-vector products, already accounted for by using $\rho$ per iteration rather than ${\rho }^2$.

Identify a binary word with its coefficient polynomial in $R={\mathbb{F}}_2[X]/(X^7-1)$. A cyclic code is exactly an ideal of $R$: closure under multiplication by $X$ is closure under cyclic shift. Its inverse image in the principal ideal domain ${\mathbb{F}}_2[X]$ has a unique monic generator polynomial of a cyclic code $g$ dividing $X^7-1$. PutThe cubics have no root in ${\mathbb{F}}_2$, hence are irreducible polynomials. The eight monic divisors yield all binary cyclic codes of length seven. A code generated by $g$ has basis $g,Xg,\dots ,X^{6-\mathrm{deg}g}g$ and dimension $7-\mathrm{deg}g$, with the zero code treated separately.

For the dual of a cyclic code, let $h=(X^7-1)/g$. Orthogonality of coefficient vectors to all cyclic shifts gives the generator $h^{\ast }(X)=X^{\mathrm{deg}h}h(X^{-1})$. Here $a^{\ast }=a,b^{\ast }=c,c^{\ast }=b$. Thus the complete pairing, with the respective dimensions, isThese are respectively the whole-space/zero, even-weight/repetition, and two Hamming code/binary simplex code pairs. The reciprocal formula follows because the product of a word polynomial and the reversed polynomial of an orthogonal word has zero coefficients in all cyclic positions; the resulting inclusion has the correct complementary dimension, hence is equality.

A cyclotomic field over $\mathbb{Q}$ is a field $\mathbb{Q}({\zeta }_m)$ generated by an $m$th root of unity of exact order $m$. It contains all roots of $X^m-1$, and is its splitting field. In characteristic zero this polynomial is separable, so the extension is Galois.

Every automorphism sends ${\zeta }_m$ to ${\zeta }_m^a$ with $a\in (\mathbb{Z}/m\mathbb{Z}{)}^{\times }$, and is determined by that image. Composition multiplies the exponents modulo $m$, yielding an injective homomorphism into an abelian group. Thus the Galois group is abelian. Using irreducibility of the cyclotomic polynomial, every unit exponent occurs andIf “cyclotomic” is used more broadly for a subfield of such a field, it is still Galois and abelian, because every subgroup of an abelian Galois group is normal.

The Clebsch-Gordan decomposition for SU2 statesRealize the tensor product as polynomials of bidegree $(m,n)$ in $(x_1,y_1)$ and $(x_2,y_2)$. The determinant $D=x_1{y}_2-y_1{x}_2$ is invariant under the simultaneous SU2 action. For each $r$, the vectorhas torus weight $q=m+n-2r$ and is killed by $E=x_1{\partial }_{y_1}+x_2{\partial }_{y_2}$. Applying $F=y_1{\partial }_{x_1}+y_2{\partial }_{x_2}$ produces $q+1$ nonzero vectors of distinct weights, with $F^{q+1}{v}_r=0$. The relations $[E,F]=H$ and $H{v}_r=q{v}_r$ give $E{F}^j{v}_r=j(q-j+1)F^{j-1}{v}_r$, so their span is an irreducible copy of $V_q$.

These irreducible submodules have distinct highest weights and form a direct sum: a sum of earlier ones cannot contain an irreducible module inequivalent to all its summands. Equivalently use complete reducibility, obtained by averaging an inner product over the compact group. Assuming $m\le n$, their total dimension iswhich equals the dimension of the tensor product. Hence they exhaust it, proving the theorem.

The Poisson bracket is $\{f,g\}=\mathbf{r}\cdot (\nabla f\times \nabla g)$. Since $\nabla {\rho }^2=2\mathbf{r}$, the scalar triple product vanishes, andEquivalently, check the identity on $x,y,z$ and extend to every polynomial using the Leibniz rule. Thus ${\rho }^2$ is a Casimir function of a Poisson manifold of this rotational Lie-Poisson structure on R3.

The Hamilton's equations areLinearization about $(1,0,0)$ gives $\dot{\alpha }=0$, $\dot{\beta }=(C-A)\gamma$, $\dot{\gamma }=(A-B)\beta$. The two nonzero exponents satisfy ${\lambda }^2=(C-A)(A-B)$. ConsequentlyWhen $A$ is outside this interval the transverse motion is oscillatory. At equality the linearization is nonhyperbolic; if just one transverse coupling vanishes, the other permits secular linear growth, rather than an exponential instability. This is the familiar intermediate-axis instability described by the Euler equations for a torque-free rigid body.

Let the nonconstant polynomial have degree $d$ and leading coefficient $c$. If $d$ is odd, its limits at $+\infty$ and $-\infty$ have opposite signs, so the intermediate value theorem shows that its image is $\mathbb{R}$. If $d$ is even and $c>0$, it tends to $+\infty$ at both ends. Outside a sufficiently large compact interval it exceeds $f(0)$, while on that interval the extreme value theorem gives a global minimum $a$. Continuity and the intermediate value theorem then give image $[a,\infty )$. For $c<0$, applying this argument to $-f$ gives $(-\infty ,a]$. These are exactly the three possible images.

Work in the prime field ${\mathbb{F}}_p$ and set $r=p-1$. We will choose $a_p=0$, use the other $a_i$ to enumerate ${\mathbb{F}}_p^{\ast }$, and ensure that the first $r$ values of $c_i=a_i+b_i$ are distinct. The zero-sum condition will then force the last value to be the missing field element.

Consider the polynomialIt has total degree of a polynomial $r(r-1)$. Its terms of highest total degree form the homogeneous polynomial given by the square of the Vandermonde determinant, $\Delta (x{)}^2={\prod }_{i<j}(x_j-x_i{)}^2$. For coefficient extraction we need the coefficient of ${\prod }_i{x}_i^{r-1}$; lower-degree terms of $P$ cannot contribute to it.

Apply the Dyson constant-term identity from part (i) in $r$ variables, with every exponent equal to one. Pairing its factors for $i,j$ givesThe constant term is $r!$. Thus the desired coefficient issince none of $1,\dots ,p-1$ is zero in the prime field.

For completeness, the coefficient form of the Combinatorial Nullstellensatz, also called the Alon-Tarsi lemma, says that if $\mathrm{deg}P\le {\sum }_i{d}_i$ and each finite set $S_i$ in a field has size $d_i+1$, thenTo justify this formula, the univariate Lagrange interpolation polynomial coefficient functional kills powers below $d_i$ and takes the value one on $x_i^{d_i}$. Apply the product of these functionals to each monomial of $P$. Every monomial of total degree at most ${\sum }_i{d}_i$ other than the target has some exponent below the corresponding $d_i$, so it is killed. The target survives with coefficient one. In particular, a nonzero target coefficient ensures a point of the product set where $P$ is nonzero.

Use $S_i={\mathbb{F}}_p^{\ast }$ and $d_i=r-1$ for every $i$. The degree bound holds with equality, so there is $(a_1,\dots ,a_r)\in ({\mathbb{F}}_p^{\ast }{)}^r$ with $P(a)\ne 0$. Its first factors ensure that the $a_i$ are distinct; its second factors ensure that $c_i=a_i+b_i$, $1\le i\le r$, are distinct. With $a_p=0$, the $a_i$ enumerate the whole prime field.

Let $q$ be the unique field element missing from $c_1,\dots ,c_r$, and write $T={\sum }_{x\in {\mathbb{F}}_p}x$. Since ${\sum }_i{b}_i=0$,The same sum is $T-q$, hence $q=b_p$. Taking $c_p=b_p=a_p+b_p$ completes the enumeration. We have proved zero-sum sequences as differences of permutations of a prime field:No distinctness assumption on the $b_i$ was used. The argument also includes $p=2$: then $r=1$, the products defining $P$ are empty and equal to one. Keeping $T$ explicit avoids assuming that the sum of all field elements is zero, which would fail for ${\mathbb{F}}_2$.

The tautological bundle $\mathcal{O}(-1)$ over Complex projective space has fibre over $[\ell ]$ equal to the line $\ell \subset {\mathbb{C}}^{r+1}$. Define $\mathcal{O}(1)=\mathcal{O}(-1{)}^{\ast }$ and the twisting sheaf on projective space, viewed analytically as a holomorphic line bundle, by

For ${\mathbb{CP}}^1$, use coordinates $z=Z_1/Z_0$ and $w=Z_0/Z_1=1/z$. The tautological frames $(1,z)$ and $(w,1)$ differ by $z^{-1}$, so frames $e_0,e_1$ for $\mathcal{O}(d)$ satisfy $e_1=z^d{e}_0$. A global holomorphic section is therefore described by entire functions $s_0(z),s_1(w)$ withWrite $s_0(z)={\sum }_{k\ge 0}{a}_k{z}^k$. Holomorphicity of $s_1$ at zero permits only the powers $w^{d-k}$ with $d-k\ge 0$. For $d\ge 0$, this says $s_0$ is a polynomial of degree at most $d$; for $d<0$, every coefficient must vanish. Equivalently, when $d\ge 0$ the sections are homogeneous polynomials of degree $d$ in $Z_0,Z_1$. Thus

The fiber monodromy is periodic when isotopy is allowed to move the boundary, in the sense of a periodic surface homeomorphism. Here is a justification from the Seifert fibered space structure, rather than a special formula about torus knots. Its base is a disk with cone-point orders $p,q$. The fiber Seifert surface is an incompressible surface because its inclusion in the mapping torus is injective on fundamental groups. It is also a boundary-incompressible surface: lift a proposed boundary-compressing disk to the infinite cyclic cover $\Sigma \times \mathbb{R}$ and project to $\Sigma$. This would homotope an essential arc of $\Sigma$ into $\partial \Sigma$, contradicting its essentiality. The classification of incompressible surfaces in Seifert fibered spaces therefore applies. A vertical surface in a Seifert fibered space has zero intersection with a regular Seifert fiber, whereas the fibration class evaluates on $f=pqm$ as $pq$. Thus the Seifert surface is isotopic to a horizontal surface in a Seifert fibered space, with $pq$ intersections per regular orbit.

Following the oriented Seifert fibers from one intersection to the next gives a first-return homeomorphism of the horizontal topological surface. It is a representative of the fiber monodromy, and its $pq$-th power is the identity. It has order $pq$: a regular orbit meets the connected horizontal surface in a Seifert fibered space in $pq$ cyclically ordered points. This realizes the periodic case of the Nielsen–Thurston classification theorem. It is neither an Anosov homeomorphism nor a pseudo-Anosov homeomorphism.

There is a boundary convention here. A representative fixed pointwise on the boundary is not periodic relative to the boundary: its $pq$-th power is a boundary Dehn twist (with sign determined by the return-map convention). Capping the boundary, or allowing it to rotate during isotopy, gives the finite-order representative intended by “periodic”. This distinguishes periodic surface type from literal finite order in the mapping class group relative to the boundary.

For any fibered knot, its Alexander module is $H_1(\Sigma ;\mathbb{Z})$ with the deck transformation given by homological monodromy, so its Alexander polynomial of a knot is $\det (tI-{\varphi }_{\ast })$ up to a unit. The general Turaev torsion relation $\Delta (t)\doteq (1-t)\tau (X)$ and part (b) therefore giveThis rational expression is a polynomial: in its factorization into cyclotomic polynomials, $\gcd (p,q)=1$ cancels all denominator factors. It is monic, has constant coefficient one, and has polynomial degree $pq+1-p-q=(p-1)(q-1)$. The characteristic polynomial has polynomial degree $2g$, is monic, and has constant coefficient $\det {\varphi }_{\ast }=1$, because homological monodromy preserves the intersection form. These facts remove the unit ambiguity and giveThe sign change between $\det ({\varphi }_{\ast }-tI)$ and $\det (tI-{\varphi }_{\ast })$ is trivial because $\mathrm{rank}{H}_1(\Sigma )=2g$ is even. As a separate geometric check, the orbifold Euler characteristic of the base is $1/p+1/q-1$, and its $pq$-sheeted horizontal surface in a Seifert fibered space has Euler characteristic $\chi (\Sigma )=p+q-pq=1-2g$.

The space-time Hermite polynomials obtained by scaling the Probabilists' Hermite polynomials are genuine polynomials in both variables, including at $t=0$:This expansion follows by differentiating the defining Gaussian exponential, and shows that $H_n(x,0)=x^n$ extends smoothly through $t=0$. Consequently the Itô formula is applicable from time zero. The given backward heat equation cancels the drift, and the derivative identity givesThe integrand is continuous and adapted, hence previsible, and stopping the Brownian motion and time makes it bounded. Thus $H_n(B_t,t)$ is a local martingale for every $n\ge 0$. The $n=0$ case is the constant $H_0=1$; in fact Gaussian moments make the displayed Itô integral square-integrable on every finite horizon, so these are true martingales as well.

The first three Probabilists' Hermite polynomials are $h_1(x)=x$, $h_2(x)=x^2-1$ and $h_3(x)=x^3-3x$. Hence

Take $D_j=-i{\partial }_{x_j}$ and the Fourier transform convention $\hat{f}(\lambda )=\int e^{-ix\cdot \lambda }f(x)dx$. Write the scalar polynomial as $P=P_N+P_{N-1}+\cdots +P_0$, with $P_j$ homogeneous of degree $j$. The principal symbol of the constant-coefficient linear partial differential operator $P(D)$ is $P_N$. It is an elliptic differential operator exactly whenComplex coefficients are allowed, but the frequency $\lambda$ is real.

By compactness of the unit sphere, $c={\min }_{|\omega |=1}|P_N(\omega )|>0$. Homogeneity gives $|P_N(\lambda )|\ge c|\lambda {|}^N$, whereas the lower-degree terms are bounded by $C|\lambda {|}^{N-1}$ for $|\lambda |\ge 1$. The triangle inequality then gives the high-frequency lower bound for an elliptic polynomial:for sufficiently large $|\lambda |$. ThusThe Japanese bracket is $⟨\lambda ⟩=(1+|\lambda {|}^2{)}^{1/2}$. If $N=0$, a nonzero constant $P$ satisfies the same assertion directly.

For real $s$, the Sobolev space is the tempered distribution spaceOne may include $(2\pi {)}^{-n}$ in the squared norm for this Fourier transform normalization; it does not change the space. The Local Sobolev space iswhere multiplication of a distribution by a smooth function is followed by extension by zero.

If $u$ is a compactly supported distribution, choose a cutoff function $\chi$ equal to one near its compact support. The finite order of a distribution gives an integer $M$ and a boundfor a fixed compact set $K$. This also makes $u$ a tempered distribution. Its Fourier transform of a compactly supported distribution is the smooth functionThe Leibniz rule gives the last estimate, and parameter differentiation under the finite-order pairing proves smoothness of this Fourier transform. Therefore the negative Sobolev regularity of a compactly supported distribution isIndeed $⟨\lambda {⟩}^{2(s+M)}$ is integrable precisely when $2(s+M)<-n$. This proves the claimed existence of a sufficiently negative Sobolev space index.

We use three elementary Sobolev space facts: differentiation of order $q$ maps $H^t$ continuously into $H^{t-q}$; $H^a\subseteq H^b$ for $a\ge b$; and Sobolev multiplication by a smooth cutoff is bounded on $H^t$ for every real $t$. The first two follow directly from the frequency weights. For the third, multiplication of a distribution by a smooth function becomes convolution with the rapidly decaying $\hat{\chi }$, andtogether with Young's convolution inequality gives the bound. These facts apply after localization as well.

The high-frequency lower bound for an elliptic polynomial gives a useful global implication. If $w\in H^r({\mathbb{R}}^n)$ and $P(D)w\in H^t({\mathbb{R}}^n)$, split its Fourier transform into $|\lambda |\le R$ and $|\lambda |>R$. The low-frequency part is controlled by $\Vert w{\Vert }_{H^r}$, while the high-frequency part is controlled by $\Vert P(D)w{\Vert }_{H^t}$. Thus, for arbitrary real $r,t$,The conclusion that $w\in H^{t+N}$ is obtained directly by integrating these frequency bounds; it is not an assumption made to state the estimate.

For the cutoff bootstrap for local elliptic regularity, fix $U⋐X$. A cutoff function equal to one near $\overline{U}$ makes $u$ a compactly supported distribution after multiplication, so the preceding negative-index argument gives $u\in H_{\mathrm{loc}}^r(U)$ for some finite $r$. Suppose inductively that $u\in H_{\mathrm{loc}}^q(U)$. For any test function $\chi \in C_c^{\infty }(U)$,The Leibniz rule shows that this commutator has order at most $N-1$, with smooth functions as coefficients, all with compact support:A second cutoff function equal to one near $\mathrm{supp}\chi$ lets us apply the stated Sobolev space bounds to every term. Hence $[P(D),\chi ]u\in H^{q-N+1}({\mathbb{R}}^n)$, while $\chi P(D)u\in H^s({\mathbb{R}}^n)$. For $N\ge 1$, the global estimate yieldsThis holds for every such $\chi$, so it improves the Local Sobolev space index by one until $s+N$ is reached. A finite number of iterations starting at $r$ provesSince $U⋐X$ was arbitrary, the conclusion holds throughout $X$. For $N=0$, division by the nonzero constant $P$ proves it immediately. This is the asserted elliptic regularity, proved without assuming an initial nonnegative Sobolev space index.

A first-order example on ${\mathbb{R}}^2$ is the first-order Cauchy-Riemann operatorIt is an elliptic differential operator with complex coefficients. In one real variable, $D=-id/dx$ is already a first-order elliptic differential operator.

There is no scalar elliptic differential operator of odd order in three variables. If its order $N$ were odd, its principal symbol would satisfy $P_N(-\omega )=-P_N(\omega )$ on $S^2$. Regard the continuous mapThe Borsuk-Ulam theorem gives $f(\omega )=f(-\omega )$ for some $\omega$. Oddness then gives $f(\omega )=0$, contradicting ellipticity. This is the odd-order obstruction for scalar elliptic operators in at least three dimensions; restricting to a three-dimensional subspace proves the higher-dimensional case too. For real coefficients alone, the same obstruction follows from the intermediate value theorem along a path between antipodal points.

The scalar qualification matters. An elliptic system of differential equations can be first order in three variables: with the Pauli matrices, the matrix symbol $A(\lambda )={\sum }_{j=1}^3{\sigma }_j{\lambda }_j$ satisfies $A(\lambda {)}^2=|\lambda {|}^{2}1$ and is invertible for $\lambda \ne 0$. This does not contradict the scalar polynomial obstruction.

With $N$ vanishing moments, a wavelet annihilates every polynomial of degree below $N$. A Taylor polynomial then explains small wavelet coefficients on smooth parts of a signal, and efficient best N-term approximation. Compact support localizes coefficients near a feature, limits the number of boundary interactions, and permits a finite filter implementation. Increasing the number of vanishing moments while keeping an orthonormal basis generally requires a larger support of a function: a finite orthonormal filter with $N$ vanishing moments needs at least $2N$ taps, and the minimal-support Daubechies wavelet has support length $2N-1$ in the standard normalization.

The Haar wavelet, $\psi ={\chi }_{[0,1/2)}-{\chi }_{[1/2,1)}$, has one vanishing moment, unit support length and discontinuities. It is inexpensive and particularly suitable for piecewise constant data with sharp jumps. A Daubechies wavelet of a larger order has more vanishing moments and a longer finite filter; sufficiently large orders also provide greater regularity. It is useful when smooth trends should yield small coefficients, although the wider support of a function can spread a jump across more coefficients. Choose Haar for compact jump localization; choose a higher-order Daubechies wavelet for smooth polynomial structure. More vanishing moments alone does not make every low-order wavelet highly differentiable.