Set $\Theta =e^{-ikx+i{k}^{2}t}$. Expanding the proposed local relation givesThus its vanishing is exactly $i{q}_t+q_{xx}=0$. Integrating over $x>0$, using decay at infinity, yieldsIntegrate in time and use the initial data to obtain the global relationHere ${\tilde{g}}_j(z,t)={\int }_0^t{e}^{iz\tau }{g}_j(\tau )d\tau$, with $g_0=q(0,\cdot )$ and $g_1=q_x(0,\cdot )$. In particular these finite-time transforms are entire in $z$.

Write $F(k)=e^{ikx-i{k}^{2}t}$ and $B_{\pm }(k)=k{\tilde{g}}_0(k^2,t)\pm i{\tilde{g}}_1(k^2,t)$. Substitution into part (i), including replacing $k$ by $-k$ in the reflected transform, givesFor the boundary integrands, rotate the negative real ray to the positive imaginary ray through the second quadrant. There are no singularities, and for $k=a+ib$ with $a\le 0,b\ge 0$,After writing $F{B}_{+}$ as its finite time integral, this estimate and the usual smooth-data arc bounds justify the rotation. The positive real ray stays fixed and the new imaginary ray is directed downwards, givingThe same argument applies separately to the ${\tilde{g}}_0$ and ${\tilde{g}}_1$ terms. Thus the net boundary integral in the inversion formula isFor Dirichlet boundary conditions choose $c=-1$, eliminating the unknown normal derivative. For Neumann boundary conditions choose $c=+1$, eliminating the unknown boundary value. The resulting unified transform representations areEach contains only the initial transform and the prescribed boundary transform. The signs depend on the PDF's orientation of $L$, checked above.

Extend $q$ by zero to $x<0$ and apply the assumed Fourier inversion theorem. For sufficiently regular, decaying data and $x>0$ this gives $q(x)=(2\pi {)}^{-1}{\int }_{\mathbb{R}}{e}^{ikx}\hat{q}(k)dk$, with $\hat{q}(k)={\int }_0^{\infty }{e}^{-ikx}q(x)dx$. If $\mathrm{Im}k\le 0$, the exponential in this last integral is bounded in magnitude by one; it defines a holomorphic transform in the lower half-plane, with its boundary values on the real axis.

The PDF orients $L$ from $i\infty$ down to zero, then from zero to $+\infty$. The function $e^{ikx}\hat{q}(-k)$ is holomorphic in the upper half-plane, and for $x>0$ its exponential decays there. Close the two rays through a large first-quadrant arc. Cauchy integral theorem and Jordan lemma giveFor example, if $q,q^{\prime }$ are integrable, integration by parts gives $\hat{q}(-k)=O(1/k)$ in that quadrant, which suffices for the arc estimate. Standard limiting arguments extend the identity to the admissible Fourier-inversion class. Adding any constant multiple of this zero integral to ordinary inversion provesThe extra integral is a reflected, zero-supported contribution, not an extra datum.

With $\psi =0$, the angular momentum becomes $\mathbf{L}=m\mathbf{r}\times \dot{\mathbf{r}}$ and is constant. The identity $\mathbf{L}\cdot \mathbf{r}=0$ confines the trajectory to the plane through the origin perpendicular to $\mathbf{L}$ when $\mathbf{L}\ne 0$. If $\mathbf{L}=0$, the velocity is radial and a central force preserves the radial line; this degenerate motion is also planar.

For the proposed Laplace-Runge-Lenz vector convention, differentiate $\mathbf{K}=\mathbf{L}\times \dot{\mathbf{r}}-\varphi (r)\mathbf{r}$. Since $\ddot{\mathbf{r}}=-{\varphi }^{\prime }(r)\hat{\mathbf{r}}/m$ and $\mathbf{L}$ is constant, the vector triple product givesThereforeIt vanishes for general motion precisely when $(r\varphi {)}^{\prime }=0$, givingThis is an inverse-square potential, attractive when $C<0$ and repulsive when $C>0$; $C=0$ is the free-particle case. An additive constant in $\varphi$ leaves the force unchanged but would change this particular $\mathbf{K}$ by a constant multiple of $\mathbf{r}$, so its definition requires the displayed choice of energy zero.

For $\psi =\mu /r$, the radial gradient givesThe force $-\nabla \varphi$ is radial and has no torque about the origin. For the remaining force, use the vector triple product:Meanwhile the radial unit vector obeysSince $\psi \mathbf{r}=\mu \hat{\mathbf{r}}$, these derivatives cancel. Thus the generalized angular momentum in a radial monopole field isThe argument applies away from $r=0$, where the field is defined, for any radial $\varphi$. In contrast to ordinary angular momentum, this conserved vector has $\mathbf{L}\cdot \hat{\mathbf{r}}=-\mu$; it fixes a cone rather than generally fixing a plane.

Take the scalar product of Newton's second law with the velocity $\dot{\mathbf{r}}$. The cross product is perpendicular to its first factor, soConsequentlyThis is conservation of energy: the sum of kinetic energy and potential energy stays fixed. The magnetic-type Lorentz force $\dot{\mathbf{r}}\times \mathbf{B}$ does no work, because it is perpendicular to the instantaneous velocity. It may turn the trajectory but cannot directly change its speed, regardless of the spatial variation of $\mathbf{B}$.

Factor the equation asFor the nontrivial roots, the quadratic formula over residues givesThe supplied square root is $14$, and $2^{-1}\equiv 100(\mathrm{mod}199)$. ThereforeIndeed $107^2-107+1=11343=57\cdot 199$, so $107^3+1$ is divisible by $199$, and $107\not\equiv -1$. The other nontrivial root is $(1-14)100\equiv 93$; it satisfies $93^2-93+1=43\cdot 199$. Together with $198\equiv -1$, these are the three cube roots of minus one. The complex-number hint reflects the same quadratic factor: the two roots there are $(1\pm \sqrt{-3})/2$.

Write $p=3k+2$. A solution has $p\nmid x$, so Fermat's little theorem givesThus $x\equiv (-1{)}^k(\mathrm{mod}p)$. If $p$ is odd, then $k$ is odd and $x\equiv -1$. The remaining prime case $p=2$ has $k=0$ and $x\equiv 1\equiv -1(\mathrm{mod}2)$ as well. Since $-1$ itself always satisfies the equation, the unique solution is $x\equiv -1(\mathrm{mod}p)$.

Set $m=(p-1)/2$. Pairing each of $1,\dots ,m$ with its negative residue givesHere $p\equiv 1(\mathrm{mod}4)$ makes $m$ even. By Wilson theorem,satisfies $x^2\equiv -1(\mathrm{mod}p)$. Its negative is the other root. This proves existence by an explicit factorial construction.

Suppose $x^2\equiv -1(\mathrm{mod}p)$. Then $p\nmid x$, so Fermat's little theorem gives $x^{p-1}\equiv 1$. But $p\equiv 3(\mathrm{mod}4)$ makes $(p-1)/2$ odd, and henceSince this $p$ is odd, $1\not\equiv -1(\mathrm{mod}p)$, a contradiction. There is no square root of minus one in this case.

Yes. A concrete example of nonunique factorization in a congruence submonoid isThe four factors are distinct and each is congruent to one modulo ten. The primes $3,7,13,17$ can be checked by trial division by two and three. Their elementary prime decompositions areFor any one of these products $pq$, where $p,q$ are the indicated distinct primes, its positive divisors are $1,p,q,pq$. This divisor assertion does not require unique factorization: if $d\mid pq$ and $p\nmid d$, then $p\mid pq/d$ by the Euclid lemma proved in (a), so $d\mid q$ and $d$ is either one or $q$. If $p\mid d$, write $d=pe$ and obtain $e\mid q$, giving $d=p$ or $pq$.

None of $3,7,13,17$ is congruent to one modulo ten. Consequently no listed product has a proper nonunit divisor in $S$. Thus each is an irreducible element of a commutative monoid, and the displayed equality proves the claim. Ordinary prime decompositions and irreducibility relative to the smaller commutative monoid $S$ are different notions.

The greatest common divisor $d=\gcd (a,b)$ is the largest positive integer dividing both $a$ and $b$. To prove Bezout identity without assuming unique factorization, let $d_0$ be the least positive member ofThis set has positive members, for example $a$, so well-ordering principle supplies $d_0$. Divide $a$ by $d_0$ using Euclidean division: $a=q{d}_0+r$, $0\le r<d_0$. The remainder $r=a-q{d}_0$ is another integer linear combination of $a,b$. Minimality forces $r=0$, so $d_0\mid a$. The same argument gives $d_0\mid b$. Every common divisor divides every integer linear combination, hence divides $d_0$. Therefore $d_0$ is the greatest common divisor, andThis argument also proves that every common divisor divides the greatest common divisor.

The positive integers with the stated product-divisibility property areThe case $n=1$ is immediate. If $p$ is prime and $p\nmid a$, its only possible common positive divisor with $a$ is one. By Bezout identity, $\lambda p+\mu a=1$. Multiplying by $b$ shows that $p\mid ab$ implies $p\mid b$. This proves Euclid lemma directly. Conversely, if $n>1$ is not prime, it has a divisor $r$ with $1<r<n$. Put $s=n/r$; then $1<s<n$. Although $n\mid rs$, neither $r$ nor $s$ is divisible by $n$, disproving the property for a composite $n$.

Finally, suppose $ab=cd$ for four distinct primes. Then $a\mid cd$, so Euclid lemma gives $a\mid c$ or $a\mid d$. A prime has no positive divisor other than one and itself, so this would mean $a=c$ or $a=d$, a contradiction. Products of these two disjoint pairs cannot be equal. No form of the fundamental theorem of arithmetic was used: the required prime-divisibility fact was proved from Bezout identity.

Can be false. Take $X=\{0,1,2,\dots \}$ and defineThen $(f\circ g)(n)=n$, so their function composition is the identity function and hence a bijection. However, $g$ is not surjective, since zero has no preimage, while $f$ is not injective, since $f(0)=f(1)=0$.

The valid implications are only that $g$ must be injective and $f$ must be surjective: equality of two $g$ values implies equality after composition, and every output of the composition is an output of $f$. On a finite $X$, either of these properties implies a bijection, so an infinite set is essential to this counterexample.

Always true. If $(f\circ g)(x)=(f\circ g)(x^{\prime })$, injectivity of $f$ gives $g(x)=g(x^{\prime })$, and injectivity of $g$ gives $x=x^{\prime }$. Thus the function composition is injective. For any $y\in X$, since $f$ is surjective, choose $z$ with $f(z)=y$. Since $g$ is surjective, choose $x$ with $g(x)=z$. Then $(f\circ g)(x)=y$, proving that the composition is surjective. Consequently $f\circ g$ is a bijection; its inverse function is $g^{-1}\circ f^{-1}$.

The incident four-momentum has energy $E$ and spatial momentum $p=\gamma mv$, while the stationary target has energy $m{c}^2$ and zero spatial momentum. By the relativistic energy-momentum relation, $c^2{p}^2=E^2-m^2{c}^4$. For the fixed-target threshold for production of equal-mass particles, the squared total centre-of-momentum energy isIn the centre-of-momentum frame, each of the final $N$ particles has energy at least its rest energy $m{c}^2$. Four-momentum conservation therefore requires $W\ge Nm{c}^2$, and thusFor the physically relevant $N\ge 2$, this is equivalent to $\gamma \ge (N^2-2)/2$. The speed is an increasing function of $\gamma$, soConsequentlyEquality is the kinematic threshold at which all final particles are at rest in their common centre-of-momentum frame. It shows why a collision with a stationary target needs more incident energy than merely the additional rest energy: some laboratory energy must remain as collective motion. A one-particle final state of the specified mass, with no other outgoing energy, is impossible because its invariant energy would be $m{c}^2$, whereas the two-particle initial state has invariant energy at least $2m{c}^2$.

The relativistic energy of a particle of rest mass $m$ is $E=\gamma m{c}^2$, where the Lorentz factor is $\gamma =(1-v^2/c^2{)}^{-1/2}$. Using the binomial series at small $v/c$ givesThe second term is the familiar Newtonian kinetic energy. The first is the rest energy, which remains when $v=0$. Thus the new physical idea is mass-energy equivalence: a body's mass represents energy $m{c}^2$ even at rest. The total energy is not merely energy due to motion.

Now $\dot{m}=\beta$ and $\dot{M}=\alpha -\beta$, soFor $\alpha >0$, set $w=u\beta -\alpha v$. Then $\dot{w}=-\alpha w/M$ and $w(0)=u\beta$. For $\alpha \ne \beta$, integration givesConsequentlyThis solution of the rocket equation with stationary mass accretion covers increasing or decreasing rocket mass, as long as $M>0$. For equal-rate rocket burning and accretion, in the case $\alpha =\beta >0$, the mass is constant and the linear equation instead givesIt is also the continuous equal-rate limit of the preceding expression when written in terms of $t$.

For the no-accretion limit, expand the power as an exponential:Substitution givesrecovering (a). At a fixed time with $M_0-\beta t>0$, the limiting mass is $M=M_0-\beta t$, so the same calculation recovers the time-dependent no-dust solution.

Without dust collection, $dM=-dm$. The rocket equation reduces toIntegrating from $v=0$, $M=M_0$ givesThis is the usual rocket equation with constant relative exhaust speed, valid while the mass remains positive and the prescribed ejection continues.

Neglect the centrifugal acceleration, which is of order ${\omega }^2$, and use the local flat-surface approximation. To zeroth order in $\omega$, the height above $P$ and flight time areInserting this vertical velocity into the Coriolis acceleration gives the eastward componentWith $x(0)=\dot{x}(0)=0$, two integrations yieldto first order. Horizontal velocity is already of order $\omega$, so its effect on vertical acceleration is of order ${\omega }^2$. Thus the flight time remains $T_0$ at this order. The Coriolis deflection of a vertical projectile at landing isThe displacement is westward, and its magnitude is $4\omega v^3\sin \theta /(3g^2)$. The horizontal speed remains westward during the flight: $\dot{x}=-2\omega \sin \theta z_0(t)$. This explains why the eastward acceleration on descent does not cancel the earlier displacement.

At $P$, use the local right-handed basis ${\mathbf{e}}_E,{\mathbf{e}}_N,{\mathbf{e}}_U$ pointing east, north and vertically outward. The Earth's rotation vector has componentsInitially $\dot{\mathbf{r}}=0$ in the rotating reference frame, so the Coriolis acceleration is zero. Evaluating the centrifugal acceleration givesThe gravitational acceleration is $-g{\mathbf{e}}_U$. HenceThe small angle $\delta$ from the downward vertical therefore satisfiesFor the northern hemisphere, $0\le \theta \le \pi /2$, this is the printed expression without absolute values. More generally, the printed signed meridional expression specifies the direction of the tilt; an unsigned angle uses its absolute value. The horizontal component is toward the equator and vanishes at both poles and at the equator.

A simple group has no normal subgroup except the identity subgroup and the whole group. To prove the Simplicity of the alternating group A5, we compute its conjugacy classes. Since the sign homomorphism is onto, $|A_5|=5!/2=60$. Its possible nonidentity cycle types are a three-cycle, a product of two disjoint transpositions, and a five-cycle.

There are $\left(\genfrac{}{}{0ex}{}{5}{3}\right)\cdot 2=20$ three-cycles. The centralizer in $S_5$ of $(123)$ consists of its three powers and an independent interchange of the two unused letters. Only the three powers are even. Its centralizer in $A_5$ therefore has order three, giving one conjugacy class of size $60/3=20$ by the orbit-stabilizer theorem.

There are $5\cdot 3=15$ products of two disjoint transpositions: choose the fixed letter and then a pairing of the other four letters. For $(12)(34)$, a commuting permutation fixes $5$ and may interchange the two pairs and swap within them, giving a centralizer of order eight in $S_5$. It contains an odd element, $(12)$, so exactly half its elements are even. The centralizer in $A_5$ has order four, and these fifteen elements form one conjugacy class of size $60/4=15$.

Finally, there are $4!=24$ five-cycles. A permutation commuting with a five-cycle is determined by the image of one letter and must be a power of that cycle. All five powers are even, so the centralizer in $A_5$ has order five. Each such conjugacy class has size $60/5=12$, giving two classes of size twelve.

A normal subgroup must be a union of conjugacy classes and must include the identity. Its order therefore has the formThe resulting distinct possibilities areOf these, only $1$ and $60$ divide $60$. Lagrange's theorem rules out every other possible order. Hence