Using $E=-\nabla \varphi$ and the electrostatic Poisson equation, $\rho =-{ϵ}_0{\nabla }^2\varphi$, integration by parts givesThe boundary potential is zero, so the surface term vanishes. Hence $\boxed{W=({ϵ}_0/2){\int }_D{E}^{2}dV}$.

For the capacitor use the usual parallel-plate approximation, neglecting edge fields. The two gaps have widths $H+a$ and $H-a$, and their field magnitudes are $V/(H+a)$ and $V/(H-a)$. The field-energy expression givesThe two faces of the middle plate carry charges $Q_{-}={ϵ}_{0}AV/(H+a)$ and $Q_{+}={ϵ}_{0}AV/(H-a)$. Its total charge is their sum; the grounded plates have zero potential and contribute zero to $\frac{1}{2}\sum Q_j{\varphi }_j$. The charge-potential expression therefore gives $W=\frac{1}{2}V(Q_{-}+Q_{+})$, exactly the same result. Since $H^2-a^2$ is largest at $a=0$, the energy at fixed $V$ is minimized when the middle plate is centered, with value ${ϵ}_{0}A{V}^2/H$.

This is the electrostatic energy of a three-plate capacitor in the intended uniform-field model. Finite circular plates have fringing corrections; no small-gap-to-radius ratio is explicitly stated in the source, so the simple formula should not be described as an exact finite-disc solution.

The fixed-end wave equation has spatial modes $\sin (n\pi x)$ and temporal factors $\cos (n\pi t),\sin (n\pi t)$. The zero initial velocity removes every temporal sine term. Expand the initial displacement in a Fourier sine series, whose coefficients areTwo integrations by parts give the last equality; the endpoint terms from $x(1-x)$ vanish and the remaining derivative endpoints distinguish even from odd $n$. ConsequentlyThis solves the initial-boundary problem in the finite-energy sense, with convergent first-derivative series. At $t=0$ the energy in the question is ${\int }_0^1(1-2x{)}^{2}dx=1/3$. At any time, sine/cosine orthogonality gives the contribution of mode $n$ as $(n\pi {)}^2{b}_n^2/2$: its kinetic and elastic pieces have factors ${\sin }^2(n\pi t)$ and ${\cos }^2(n\pi t)$ summing to one. ThereforeThe square summability of the derivative coefficients justifies the orthogonality calculation and its limit.

Let $F(z)=p(z)/q(z)$. Its only possible finite poles are the distinct nonreal roots ${\alpha }_j$, and the residue of $F(z)e^{iz}$ there is $p({\alpha }_j)e^{i{\alpha }_j}/q^{\prime }({\alpha }_j)$; this is zero if the numerator cancels that pole. Close the real segment by a positively oriented semicircle in the upper half-plane, where $|e^{iz}|=e^{-\mathrm{Im}z}$ decays.

For sufficiently large radius $R$, all roots lie strictly inside or outside the contour as appropriate, and the degree bound gives $|F(z)|\le C/R$ uniformly on the arc. Parametrizing by $z=R{e}^{i\theta }$ bounds the arc integral byHere $\sin \theta \ge 2\theta /\pi$ on $[0,\pi /2]$, and symmetry handles the other half. The residue theorem therefore givesTo justify that this is the ordinary improper integral rather than just its symmetric principal value, note $F(x)=O(|x{|}^{-1})$ and $F^{\prime }(x)=O(|x{|}^{-2})$. On either tail, integration by parts writes the integral as a vanishing endpoint term $F(x)e^{ix}/i$ minus an absolutely convergent integral involving $F^{\prime }$. Thus both one-sided limits exist. When the degree bound is stricter the original integral is already absolutely convergent. This proves the Fourier integral of a rational function with simple nonreal poles formula with all tail and arc limits accounted for.

A norm $p$ is nonnegative with $p(x)=0$ exactly when $x=0$, satisfies $p(ax)=|a|p(x)$, and obeys $p(x+y)\le p(x)+p(y)$. The Euclidean norm has positivity and absolute homogeneity immediately. For the triangle inequality, positivity of $\Vert x-ty{\Vert }_2^2$ for all real $t$ gives the Cauchy-Schwarz inequality $|x\cdot y|\le \Vert x{\Vert }_2\Vert y{\Vert }_2$ (with $y=0$ immediate). Hencewhich proves the triangle inequality after taking square roots.

For the prescribed set, use the Minkowski functionalSince $U$ is open and contains zero, some Euclidean ball of radius $r>0$ lies inside it. Since it is bounded, it lies inside a ball of radius $R$. These inclusions imply $\Vert x{\Vert }_2/R\le p_U(x)\le \Vert x{\Vert }_2/r$, so the functional is finite and positive for nonzero $x$; it is zero at zero. Symmetry gives $p_U(-x)=p_U(x)$, and substitution in the infimum gives $p_U(ax)=a{p}_U(x)$ for $a>0$, hence absolute homogeneity for all real scalars.

Convexity and $0\in U$ imply $sU\subseteq U$ for $0\le s\le 1$. Consequently, for $a>p_U(x)$ and $b>p_U(y)$, one has $x/a,y/b\in U$. Convexity then givesso $p_U(x+y)\le a+b$. Letting $a\downarrow p_U(x)$ and $b\downarrow p_U(y)$ proves the triangle inequality.

Finally, if $p_U(x)<1$, there is a $t<1$ with $x\in tU\subseteq U$. Conversely, if $x\in U$ and $x\ne 0$, openness permits $(1+\delta )x\in U$ for some $\delta >0$, so $p_U(x)\le 1/(1+\delta )<1$; zero is immediate. Thus $\boxed{U=\{x:p_U(x)<1\}}$, completing the norm from a bounded symmetric convex neighbourhood construction.

In the upper half-plane model, a hyperbolic line is either a vertical line or a Euclidean semicircle perpendicular to the real axis. For the vertical line $x=a$, set $R_L(z)=2a-\bar{z}$. For the circle $|z-a|=r$ with real $a$, setThe vertical formula preserves $y$ and Euclidean speed. For the circular formula, $\mathrm{Im}{R}_L(z)=r^2\mathrm{Im}z/|z-a{|}^2$, while its local Euclidean length multiplier is $r^2/|z-a{|}^2$. Thus both preserve $|dz|/\mathrm{Im}z$, the hyperbolic metric, and consequently all curve lengths and distances. Each is an involution preserving the half-plane; each fixes its specified line pointwise and moves points off that line. This establishes the required hyperbolic reflection explicitly.

To factor an arbitrary isometry $g$, first recall why a bisector reflection exchanges any distinct points $P,Q$. The hyperbolic distance formula isEquating the distances to $P$ and $Q$ yields a vertical line or a circle with centre on the real axis, hence a hyperbolic line. Substitution in the reflection formulas above shows that its reflection exchanges $P$ and $Q$. Choose a base point $P$. If $gP\ne P$, compose $g$ on the left with their bisector reflection, obtaining $h$ that fixes $P$; otherwise take $h=g$.

Choose $Q\ne P$. The points $hQ,Q$ have the same distance from $P$, so their bisector contains $P$. If they differ, its reflection corrects $hQ$ while retaining $P$. We obtain an isometry $k$ fixing both $P,Q$ after at most two reflections.

An isometry fixing two points is either the identity or reflection in their joining line. To see this directly, a point with prescribed distances from $P,Q$ lies at the intersection of two hyperbolic circles; these are Euclidean circles in this model and have at most two intersections. When there are two, reflection in the line $PQ$ interchanges them. Pick a third point off that line. Either $k$ fixes it, or composing $k$ with that line reflection fixes it. An isometry fixing three noncollinear points is the identity: the first two distances leave only the reflected pair, and the distance to the third point distinguishes that pair. Thus $k$ is the identity or one additional reflection. Reversing the compositions proves every isometry is a product of at most three hyperbolic reflections.

The Fundamental theorem of finitely generated abelian groups states that every such group is isomorphic towith uniquely determined rank and invariant factors, allowing an empty finite part. Modulo $pA$, the free summand contributes $(\mathbb{Z}/p\mathbb{Z}{)}^r$. If $A/pA=0$ for some prime, this forces $r=0$, so $A$ is finite. Conversely, if $A$ is finite, choose a prime $p$ coprime to $|A|$. Bezout's identity gives an integer $b$ with $bp\equiv 1(\mathrm{mod}|A|)$. Since $|A|$ kills every element, multiplication by $p$ is onto, with inverse multiplication by $b$. Hence $pA=A$. This proves $\boxed{A\text{ finite}⟺A/pA=0\text{ for some prime }p}$ under finite generation.

For the requested nonzero example use the additive divisible group $A=\mathbb{Q}$. Each rational $q$ equals $p(q/p)$, so $p\mathbb{Q}=\mathbb{Q}$ for every prime. It is not finitely generated: a finite list of rational generators has a common denominator $D$, and every integral combination then lies in $D^{-1}\mathbb{Z}$. The rational $1/(2D)$ does not lie there. This proves non-finite-generation directly and shows why the hypothesis in finite generation detected by a prime quotient is essential.

Both candidate sets contain zero. The sum of subspaces is closed under linear combinations because $a(u_1+w_1)+b(u_2+w_2)=(a{u}_1+b{u}_2)+(a{w}_1+b{w}_2)$, with the two terms in $U$ and $W$. The intersection is closed because any linear combination of vectors lying in both subspaces still lies in both. Thus each is a vector subspace.

Take a basis $s_1,\dots ,s_r$ of the intersection, extend it by $u_1,\dots ,u_a$ to a basis of $U$, and by $w_1,\dots ,w_b$ to a basis of $W$. The combined list $s,u,w$ spans $U+W$. To prove independence, a zero combination makes the $u$ combination equal a vector in both $U$ and $W$, so it is a combination of the $s$'s. Independence of the basis of $U$ makes all $u$ coefficients zero; independence of the basis of $W$ then makes all remaining coefficients zero. Hence $\dim (U+W)=r+a+b$, giving the dimension formula for a sum of subspaces

For the concrete kernels, the equations $Ax=0$ giveApplying the two rows of $B$ to this expression gives $4(s-t)$ and $5(s-t)/3$. Thus the intersection has basis $v_0=(0,2,1,1)$. Choose $v_1=(5,-1,3,0)$ in $U$ and $v_2=(-4,-2,1,0)$ in $W$. ThenIndeed $U$ has dimension two because $A$ has rank two, and $v_0,v_1$ are independent. Likewise $W$ has dimension two and $v_0,v_2$ are independent, so the last list spans the sum. Its first three coordinates have determinant $-48$, confirming independence and dimension three.

Assume independent samples $X_1,\dots ,X_n$ and $Y_1,\dots ,Y_m$, each consisting of independent normal random variables, with respective means ${\mu }_1,{\mu }_2$ and a common unknown variance ${\sigma }^2>0$. Take $n,m\ge 2$. Normality, independence and equal variances are the assumptions giving the exact pooled Student's t-test; unequal variances would require a different test. We test $H_0:{\mu }_1={\mu }_2$ against an unrestricted difference.

Put $N=n+m$ andThe normal likelihood is proportional to $({\sigma }^2{)}^{-N/2}\exp [-S/(2{\sigma }^2)]$, where $S$ is the residual sum of squares. Under the unrestricted model, the maximizing means are $\bar{X},\bar{Y}$ and $S=W$. Under $H_0$, the maximizing common mean is $(n\bar{X}+m\bar{Y})/N$ and decomposition about the sample means gives $S=W+B$. For either model, maximizing over variance gives ${\hat{\sigma }}^2=S/N$. Thus the generalized likelihood-ratio test statistic isSmall $\Lambda$ is therefore equivalent to large $|T|$.

Under the null, $Z=(\bar{X}-\bar{Y})/[\sigma \sqrt{1/n+1/m}]$ is standard normal. In each sample an orthogonal change of Gaussian coordinates separates the sample-mean coordinate from the centered residual coordinates; those coordinates are independent. Consequently $V=W/{\sigma }^2$ has a chi-squared distribution with $N-2$ degrees of freedom and is independent of $Z$. Therefore $T=Z/\sqrt{V/(N-2)}$ has the Student's t-distribution with $N-2$ degrees of freedom. At significance level $\alpha$, reject exactly when $\boxed{|T|>t_{N-2,1-\alpha /2}}$, the two-sample pooled t-test.

Use a translating accelerated reference frame; its axes do not rotate. If $\mathbf{R}=(\xi ,\eta ,\zeta )$ is the block's position relative to the table, then its inertial position is $\mathbf{r}=\mathbf{R}+\mathbf{A}\sin \omega t$. Its relative velocity is $\dot{\mathbf{R}}$ and its inertial acceleration is $\ddot{\mathbf{R}}-{\omega }^2\mathbf{A}\sin \omega t$. Let $N$ be the upward normal force. During contact $\zeta =\dot{\zeta }=\ddot{\zeta }=0$, and the linear tangential resistance givesThese are Newton's second law transformed to the table's reference frame; the extra terms arise from the table's acceleration. The last equation gives $N=m(g-A_z{\omega }^2\sin \omega t)$ while contact persists. If contact is lost, both the contact resistance and $N$ vanish, and free flight must be described by the corresponding gravity-only equations.

For $A_z=0$, assume $\lambda >0$ and put $\gamma =\lambda /m$. Seek a periodic particular solution ${\xi }_p=b\sin \omega t+c\cos \omega t$. Substitution into the first differential equation givesSolving yieldsThe homogeneous solutions are a constant and a multiple of $e^{-\gamma t}$. Choose the phase $0<\theta <\pi /2$ with $\tan \theta =\omega /\gamma =m\omega /\lambda$. Then $b=-A_x{\sin }^2\theta$ and $c=-A_x\sin \theta \cos \theta$, soAlso $\eta (t)={\eta }_{\infty }+C_{\eta }{e}^{-\lambda t/m}$. The decaying transients leave the stated steady orbit. Positive damping is essential to this late-time assertion; for $\lambda =0$ an additional constant drift may persist.

No attractive contact force is available, so $N$ cannot be negative. Over a whole oscillation its minimum is $m(g-{\omega }^2|A_z|)$. Thus strictly positive contact loading throughout requires $\boxed{{\omega }^2|A_z|<g}$, or the printed form with $A_z$ taken as a nonnegative vertical amplitude. If ${\omega }^2|A_z|>g$, contact cannot persist: it would require $N<0$ during part of the cycle. At equality, $N$ vanishes only at isolated instants and the ideal constrained motion can still remain in contact. Hence the strict inequality characterizes positive loading; mere nonnegative contact permits the marginal equality case.

For each $x\in X$, the product ${\prod }_{i=1}^m(1-g_i(x))$ is $1$ precisely outside every $X_i$, and is $0$ otherwise. It is therefore the indicator function of $Y$. Expanding the finite product givesMultiply by $f(x)$ and sum over the finite set $X$. The product of indicator functions is the indicator function of the corresponding intersection, soHere the empty intersection is $X$, and the empty product is $1$. Grouping the subsets $S$ by their cardinality gives the requested weighted inclusion-exclusion principle, with no contribution for $r>m$.

Let $\mathcal{P}$ be the finite set of distinct prime factors of $n$, and work in $\{0,\dots ,n-1\}$. For $S\subseteq \mathcal{P}$ put $d_S={\prod }_{p\in S}p$, including $d_{\varnothing }=1$. Then $d_S\mid n$, and an integer belongs to every $X_p$ for $p\in S$ exactly when it is a multiple of $d_S$. Thus the intersection has size $n/d_S$. The complement consists precisely of the residues coprime to $n$. Using weight $1$ givesThis proves the product formula for the Euler totient function without requiring $n$ to be squarefree.

For the weight $f(x)=x$, the sum over the multiples of any divisor $d$ in this interval isThe weighted inclusion-exclusion principle therefore givesSince $n>1$, $\mathcal{P}$ is nonempty and the last sum is $(1-1{)}^{|\mathcal{P}|}=0$. The first factors as before, givingThis sum of reduced residues also follows by pairing the bijection $x\mapsto n-x$; at $n=2$ its sole fixed residue causes no problem because the sum of both copies still equals $n\phi (n)$.

The binomial polynomial $P_r$ equals $1$ when $r=0$. For $n\ge 0$, it has value $\left(\genfrac{}{}{0ex}{}{n}{r}\right)$ if $r\le n$ and value zero otherwise. These are integers. If $n=-k<0$, thenHence each $P_r$ is an integer-valued polynomial on all of $\mathbb{Z}$, including negative arguments.

For $r\ge 1$, factor the common product in the two polynomials:For $r=1$ the common product is empty and equals $1$, so this argument also covers that boundary case.

The polynomial $P_r$ has polynomial degree $r$ and leading coefficient $1/r!$. For a polynomial $F$ of polynomial degree $d$ with leading coefficient $a_d\in \mathbb{Q}$, subtract $d!a_d{P}_d$; the remainder has smaller polynomial degree. Repeating gives the required expansion with rational numbers $c_r$. For uniqueness, suppose two expansions differ and let $s$ be the largest index with different coefficients. Their difference then has a nonzero term of polynomial degree $s$, with leading coefficient equal to that coefficient difference divided by $s!$. It cannot be the zero polynomial. Thus the expansion is unique.

Replacing $X$ by $X+1$ in the earlier identity gives $\Delta P_r(X)=P_{r-1}(X)$ for $r\ge 1$, while $\Delta P_0=0$. By linearity of the forward difference operator,An empty sum is zero when $d=0$. If $H=F-G$ were nonconstant, with polynomial degree $s\ge 1$ and leading coefficient $a_s$, then $\Delta H$ would have polynomial degree $s-1$ and leading coefficient $s{a}_s\ne 0$. Therefore $\Delta F=\Delta G$ forces $H$ to be constant.

There is a direct way to recover every coefficient. Since $P_0(0)=1$ and $P_j(0)=0$ for $j\ge 1$, repeated application of the forward difference operator givesThe last equality follows by expanding $(T-I{)}^r$, where $TF(X)=F(X+1)$; the commuting shifts obey the binomial theorem. If $F$ is an integer-valued polynomial, every summand on the right is an integer, proving that all $c_r(F)$ are integers. Together with the first paragraph this gives both directions of the Newton series for an integer-valued polynomial sequence description.

For a prime number $p$ and an integer $x$ coprime to $p$, multiplication by $x$ permutes the nonzero residue classes modulo $p$. Indeed, $ax\equiv bx(\mathrm{mod}p)$ implies $a\equiv b(\mathrm{mod}p)$, and none of the products is zero. Taking the product over the $p-1$ classes givesThe factorial is coprime to $p$ and can be cancelled, proving Fermat little theorem: $\boxed{x^{p-1}\equiv 1(\mathrm{mod}p)}$.

Now $n=mp$ and $x$ is coprime to $n$, so it is coprime to $p$. Since $(n-1)-(m-1)=m(p-1)$, the proved theorem givesThus the two specified integer congruences modulo $p$ are equivalent. In this particular step the additional hypothesis $(m,p)=1$ is not needed; it will hold for the squarefree factorization used next.

If $n$ is a squarefree integer, its distinct prime factors are pairwise coprime. Consequently $n$ divides $x^{n-1}-1$ exactly when every prime factor does. Applying the preceding equivalence with $m=n/p$ provesFinally retain this squarefree hypothesis. If $p-1$ divides $n-1$, write $n-1=k_p(p-1)$. Fermat little theorem gives $x^{n-1}=(x^{p-1}{)}^{k_p}\equiv 1(\mathrm{mod}p)$ for every prime factor. Their pairwise coprimality then gives $\boxed{x^{n-1}\equiv 1(\mathrm{mod}n)}$. For composite $n$ this is the sufficient direction of the Korselt criterion for a Carmichael number; primality is not excluded by the requested implication.

Number the natural numbers from one. If the power set could be enumerated as $A_1,A_2,\dots$, form $D=\{j:j\notin A_j\}$. For each $j$, membership of $j$ in $D$ is opposite to its membership in $A_j$, so $D\ne A_j$. This contradicts completeness of the enumeration. Thus the Cantor diagonal argument proves that the power set is an uncountable set.

For a finite subset $A$, define $b(A)={\sum }_{j\in A}{2}^{j-1}$. This is a nonnegative integer and the map is injective: if two finite subsets differ, let $k$ be their largest differing element. Its contribution $2^{k-1}$ exceeds the sum $1+2+\cdots +2^{k-2}$ of all smaller possible contributions, so their codes differ. There are infinitely many singleton subsets. Hence the countability of finite subsets of a countable set here is countable infinitude.

To treat all bijections, associate to every subset $A\subseteq \mathbb{N}$ the permutation that swaps $2j-1$ and $2j$ when $j\in A$, and fixes both when $j\notin A$. These disjoint pair swaps define a bijection of $\mathbb{N}$. Different subsets give different permutations, so the map from the power set into $X$ is injective. Therefore $X$ is an uncountable set.

Every finitely supported permutation fixes all elements above some $k$. A bijection fixing that tail must permute $\{1,\dots ,k\}$: it cannot send a smaller element into the fixed tail without violating injectivity. There are exactly $k!$ such permutations. Thusis a countable union of finite sets. Enumerate first by $k$, then by one of the finitely many permutations at that $k$; duplicates can be omitted. Infinitely many distinct pair swaps belong to $Y$, so $Y$ is countably infinite. The full family is uncountable, while the finitely supported family is countably infinite.

Let $z$ measure height above the release point, and use $\mathcal{V}=4\pi a^3/3$ for the bubble volume to avoid confusing it with a speed. For constant radius, Newton's second law, the specified buoyancy and the given Stokes law giveWith positive $\alpha ,\rho ,\mu ,a$, defineThe differential equation becomes $\dot{u}+(u/\tau )=U/\tau$. Its solution with $u(0)=0$ is $u(t)=U(1-e^{-t/\tau })$. Integration using $z(0)=0$ yieldsAn arbitrary initial height is added to this expression. The effective inertia changes the relaxation time $\tau$, but not the terminal velocity obtained by balancing buoyancy and drag.

For the dissolving bubble assume $\beta >0$ and use the specified instantaneous terminal velocity, rather than the constant-radius transient formula. The bubble disappears at $t_{\ast }=a_0^2/\beta$, andIts total rise is therefore $\boxed{z(t_{\ast })=\rho g{a}_0^4/(9\mu \beta )}$. This second model neglects the initial acceleration and uses exactly the terminal-speed approximation stipulated for the shrinking bubble.

Choose the positive direction along the incoming car's velocity. Treat each car as a translating body, take $m_1,m_2>0$, and neglect external impulse, rotation and energy losses during the short elastic collision. If $v_1,v_2$ are the final signed velocities, conservation of momentum and conservation of energy giveFactoring the second relation and using the first gives $U_1+v_1=v_2$ on the collision branch. The other algebraic branch, $v_1=U_1,v_2=0$, describes no impulse and cannot describe separating bodies after contact. Solving the collision branch yieldsFor $0<R<1$ the first car rebounds and the second moves forwards. At $R=1$ the first stops and the second takes its original velocity. For $R>1$ both move forwards, with $v_2-v_1=U_1>0$, so they separate. In the limits $R\to 0$ and $R\to \infty$, the first car's velocity approaches $-U_1$ and $U_1$, respectively.

For the deformable car, take $y=0$ at first contact. The resisting force does negative work. The work-energy theorem therefore givesAt the first stop $v=0$, whence $\sqrt{L-y}=\sqrt{L}-M{V}^2/4$. The speed bound ensures that this right-hand side is positive, so squaring introduces no extraneous solution. The stopping compression isFor $0<V<2L^{1/4}/M^{1/2}$ this lies strictly between $0$ and $L$, as required before complete crushing. The coefficient in the stipulated force law is taken in its stated normalization.

For the odd-square sum, the initial case $n=1$ gives $1=(4-1)/3$. Assume the formula at $n$. Adding the next square givesThe initial case and this induction step prove $\boxed{\sum _{j=1}^n(2j-1{)}^2=(4n^3-n)/3}$ for every positive integer $n$.

For the divisibility assertion, $1^3+5\cdot 1=6$. The change when $n$ is replaced by $n+1$ isOne of the consecutive integers $n,n+1$ is even, so the displayed change is divisible by $6$. If $n^3+5n$ is divisible by $6$, its successor is too. Thus induction proves $\boxed{6\mid n^3+5n}$ for every $n\ge 1$.