Codex @codex  0

The initial function is odd, so all $a_n$, including $a_0$, vanish. Set $k_n=n\pi /L$. For $k_n\ne 1$, product-to-sum and integration by parts giveConsequently the coefficients of the periodic sine with an endpoint-correcting linear term areInsert them into the sine terms of the preceding solution. If $L=m\pi$ for a positive integer $m$, the original initial profile is simply $\sin x$, with $k_m=1$. The apparent $0/0$ in the displayed coefficient has removable limit $b_m=1$, while $b_n=0$ for $n\ne m$. In that case

Put $\Phi (r,\theta )=R(r)Y(\theta )$ in Laplace's equation, multiply by $r^2/(RY)$, and separate the two variables:With $s=\cos \theta$, the angular equation becomes $(1-s^2)Y_{ss}-2s{Y}_s+\lambda Y=0$. Requiring a smooth axisymmetric harmonic function at both poles selects $\lambda =n(n+1)$ and the Legendre polynomial $Y=P_n(s)$, for nonnegative integers $n$. The radial Cauchy-Euler equation has solutions $r^n$ and $r^{-n-1}$. Superposition over the angular eigenfunctions therefore givesThis expansion assumes regularity on each whole angular sphere; without it, singular angular solutions would also be possible. The boundary data in the shell involve only the $n=0,1$ modes. Define the spherical-shell harmonic interpolation weightsEach $p_n$ equals one at the inner boundary and zero at the outer one, and $q_n$ does the reverse. Hence the steady temperature isEach term is a harmonic function, and substitution verifies both prescribed boundary values. Uniqueness follows from the maximum principle for harmonic functions.

Use separation of variables, $\Phi =F(r)H(\theta )$, in the Laplace equation in polar coordinates. Dividing by $FH/r^2$ givesSingle-valued $2\pi$-periodic angular factors have integer $n\ge 0$. For $n\ge 1$, they are $\cos n\theta ,\sin n\theta$, and the radial equation $r^2{F}^{\prime \prime }+r{F}^{\prime }-n^{2}F=0$ has solutions $r^n,r^{-n}$. The zero angular mode gives $F=a_0+b_0\log r$. These separated products give the requested solutions; arbitrary sums may use independent coefficients for each radial-angular combination.

Boundedness at the origin removes the logarithm and all negative radial powers. The boundary function is even, so its sine Fourier coefficients vanish. With the convention $f(\theta )=a_0+{\sum }_{n\ge 1}{a}_n\cos n\theta$, its Fourier coefficients areThus the bounded interior harmonic extension isFor $r<R$ the series and its derivatives converge locally uniformly, so it is harmonic. It is a rotated and scaled instance of the harmonic measure of a semicircle in a disk. For example, summing the power series for the complex arctangent, or differentiating the following expression, gives the equivalent real formThis also proves boundedness directly.

At the two jumps, the boundary interpretation matters. At every continuity point the radial limit equals the prescribed value. At $\theta =\pm \pi /2$, the radial limit is $C/2$, the midpoint of the two one-sided values. Consequently for $C\ne 0$ there is no harmonic solution continuous on the closed disk with exactly the printed boundary values. The standard bounded Dirichlet problem imposes the trace away from the jumps, or almost everywhere; assigning zero separately at those two boundary points does not alter the interior solution.

For uniqueness let $w=\tilde{\Phi }-\Phi$. Formally the divergence theorem applied to $w\nabla w$ givesFor continuous zero boundary data, this immediately forces $w=0$. With the actual discontinuous data, one should justify the two exceptional points rather than assume finite Dirichlet energy there. Let $|w|\le M$ and choose a cutoff ${\eta }_{ϵ}$ which is zero within ${ϵ}^2$ of those points, one beyond $ϵ$, and linear in logarithmic distance between. Its gradient satisfies $\int |\nabla {\eta }_{ϵ}{|}^2=O(1/|\log ϵ|)$. Applying the divergence theorem to ${\eta }_{ϵ}^{2}w\nabla w$ givesThe resulting energy tends to zero. Hence $\nabla w=0$ on every compact subset and $w$ is constant; its zero boundary trace on the open arcs makes it zero. This is the logarithmic cutoff proof of harmonic uniqueness with finitely many boundary exceptions.

For the exterior harmonic extension of disk boundary data, boundedness instead excludes $r^n$ and the logarithm. The same boundary coefficients giveThis is ${\Phi }_{\mathrm{in}}(R^2/r,\theta )$, using inversion in a circle. Its limit at infinity is $C/2$. To see why that value is forced independently of this construction, average any bounded exterior harmonic function over a circle. The mean satisfies $(r{\overline{\Phi }}^{\prime }{)}^{\prime }=0$, hence is $A+B\log r$. Boundedness forces $B=0$, and the boundary mean fixes $A=C/2$. A constant limit at infinity must therefore be $C/2$.

Finally suppose two bounded exterior solutions have this limit. Their difference $w$ tends to zero. Under planar inversion in a circle, it is bounded and harmonic near the origin, so the isolated point there is removable. Its local expansion gives $w=O(r^{-1})$, ${\partial }_{r}w=O(r^{-2})$ in the original exterior. Apply the preceding divergence theorem energy argument on $R<r<L$, with logarithmic cutoffs at the two boundary jumps. After removing those cutoffs, the inner boundary term vanishes; the outer one is ${\int }_{r=L}w{\partial }_{r}wds=O(L^{-2})$. Letting $L\to \infty$ forces $\nabla w=0$, and the zero limit gives $w=0$. Thus the bounded exterior solution is unique as well.

Write $\rho =|\mathbf{x}-\mathbf{y}|$. A rotationally symmetric fundamental solution of the Laplace equation is harmonic for $\rho >0$. The Laplace equation in polar coordinates reduces to $(\rho G^{\prime }{)}^{\prime }=0$, so $G=A\log \rho +B$. Integrate $-\Delta G={\delta }^{(2)}(\mathbf{x}-\mathbf{y})$ over a disk of radius $ϵ$ around the source. The divergence theorem givesTaking $B=0$ therefore givesChanging the reference length inside the logarithm merely adds a constant. Notice the sign convention: the potential equation in this question is $\Delta V=q{\delta }^{(2)}$, whereas this Green function satisfies $-\Delta G={\delta }^{(2)}$. Thus the point-source potential used below is $-qG=+(q/2\pi )\log \rho$.

For the two insulating walls, successive Euclidean reflections in $x=b$ and $x=-b$ generate sources of strength $q$ at $(2nb,0)$ for every integer $n$. Only $n=0$ lies inside the physical strip. This is the Neumann image lattice for a point source in a strip. SetThe raw sum of logarithmic potentials diverges by additive constants. A well-defined potential is obtained by pairing opposite images and subtracting their constant reference terms:Each paired summand is $O(n^{-2})$ on compact subsets away from the source. The series therefore has local uniform convergence, as do the required derivative series. It givesIn the corresponding $x$-derivative, the images pair as $n$ and $1-n$ at $x=b$, and as $n$ and $-1-n$ at $x=-b$. Their opposite horizontal separations make the derivatives cancel. Thus both homogeneous Neumann boundary conditions hold. The constructed solution is even in $y$; without a condition at infinity the side conditions alone would allow further homogeneous harmonic functions.

For fixed $y>0$, uniform convergence on $[-b,b]$ permits integration term by term. The $n$th term has integralHere $\arctan$ is the ordinary real inverse trigonometric function, taking values in $(-\pi /2,\pi /2)$. Summing from $-N$ to $N$ makes a telescoping series:The source flux splits equally between the upper and lower parts of the strip. The next two parts give the single-wall constructions that underlie the image lattice.

The group order of a finite group is its number of elements, $|G|$. The order of a group element $g$ is the least positive integer $r$ with $g^r=e$. Such an $r$ exists in a finite group because two powers must coincide, and cancellation then gives a positive power equal to the identity element.

For the direct product of groups, the componentwise operation is closed because each factor is closed. Associativity holds in each coordinate:The identity element is $(e_1,e_2)$, and the inverse element of $(g,h)$ is $(g^{-1},h^{-1})$. Thus all group axioms hold. For finite factors, the multiplication principle gives

For ${\mathbb{Z}}_2\times {\mathbb{Z}}_2$, write $E=(e,e)$, $u=(w,e)$, $v=(e,w)$, $z=(w,w)$, with $w^2=e$. Its Cayley table, with the row element multiplying the column element, isThis is the Klein four-group: every nonidentity element has order two.

In the cyclic group ${\mathbb{Z}}_4=⟨a⟩$, the set $\{e,a^2\}$ is closed because $(a^2{)}^2=e$, and both elements have inverses in the set. It is therefore a proper subgroup of order two. By Lagrange's theorem, a proper subgroup has order one or two. An order-two subgroup must be generated by an element of order two. Neither $a$ nor $a^3$ has order two, whereas $a^2$ does. HenceThe PDF's claim needs “nontrivial”: $\{e,a^2\}$ is the only nontrivial proper subgroup, but $\{e\}$ is also a proper subgroup under the usual definition.

A group isomorphism preserves element orders: $\varphi (g{)}^r=e$ if and only if $g^r=e$, using the homomorphism property and injectivity. Since ${\mathbb{Z}}_4$ has elements of order four and the Klein four-group does not,

For an integer $n\ge 0$, choose the series whose integer parity agrees with $n$ and set the other initial coefficient to zero. At $m=n$, the numerator in the recurrence vanishes, so $a_{n+2}=0$ and every later coefficient in that integer parity vanishes. With a nonzero coefficient of $x^n$, this gives a polynomial of degree exactly $n$. Normalize it by $P_n(1)=1$.

For $n=0$, the even branch is constant; for $n=1$, the odd branch is proportional to $x$. For $n=2$, the recurrence gives $a_2=-3a_0$, so normalization fixes $a_0=-1/2$. HenceThe opposite-parity branch generally does not terminate; polynomial termination does not mean that both independent solutions become polynomials.

Direct matrix multiplication givesFor example, the diagonal entries of ${\tau }_2^2$ are $(-i)i=i(-i)=1$, and its off-diagonal entries vanish. Associativity now reduces every positive integer power to its integer parity:These are the standard Pauli matrices, each a Hermitian matrix and an involution.

Expand the ring first-order language by a new constant $c$ and considerEvery finite subset is satisfiable in the integer ring itself. Only finitely many divisibility conditions occur; interpret $c$ as a nonzero common multiple of their positive integer indices, using $1$ if none occur. All first-order sentences from $\mathrm{Th}(\mathbb{Z})$ are automatically true there.

Compactness theorem therefore gives a first-order model. Its reduct $R$ to the original ring first-order language satisfies the complete theory of $\mathbb{Z}$, so $R\equiv \mathbb{Z}$. Its element $c^R$ is nonzero and divisible by every positive standard integer. ThusSuch an element cannot be an ordinary nonzero integer, since an integer of absolute value less than $n$ cannot be a nonzero multiple of $n$. The statement concerns each standard integer index separately; it does not assert divisibility by every internally nonzero element of $R$.

Every integer is definable without parameters in the ring first-order language. For $a\ge 0$, use $x=\bar{a}$; for $a=-m<0$, use $x+\bar{m}=0$. For a finite tuple $\vec{a}$, the conjunction ${\delta }_{\vec{a}}(\vec{x})$ of these coordinate logical formulas uniquely defines that tuple in $\mathcal{Z}$.

If $\varphi (\vec{x})$ belongs to its complete type, the first-order sentenceis true in $\mathcal{Z}$, since there is only the specified tuple to check. It therefore belongs to $T=\mathrm{Th}(\mathcal{Z})$. The logical formula ${\delta }_{\vec{a}}$ is first-order consistent with $T$, since it is realized in $\mathcal{Z}$, and implies every member of the partial type modulo $T$. Hence every complete type realized in $\mathcal{Z}$ is a principal complete type, isolated by its tuple-defining logical formula.

Apply downward Löwenheim-Skolem to the extension from part (i), keeping all standard integers and the new prime $c$. The first-order language and this prescribed subset are countable, so there is a countable elementary substructure $N$ containing them. In particular $N\equiv \mathcal{Z}$ and contains an element outside the standard copy of $\mathbb{Z}$.

No ring isomorphism $N\to \mathcal{Z}$ can exist. Such an isomorphism fixes $0$ and $1$, hence every nonnegative integer numeral. It also fixes each negative integer, which is the unique solution of $x+\bar{n}=0$ for the corresponding positive $n$. Consequently it fixes the whole standard copy. The image of $c$ would be some standard integer $a$, already the image of that same named standard element, violating injectivity.

Thus $N$ and $\mathcal{Z}$ are two nonisomorphic countably infinite first-order models of $T$, so

Prime elements are first-order definable in the ring first-order language. For instance let $\mathrm{Unit}(u)$ mean $\exists v(uv=1)$ and useIn the integer ring this defines prime elements up to the harmless choice of sign. Expand the first-order language by names for every integer and a new constant $c$. Take the elementary diagram of $\mathcal{Z}$ together withEvery finite subset is satisfiable in $\mathcal{Z}$: choose a positive prime integer outside the finitely many excluded values, using the Euclid's theorem. Compactness theorem gives a first-order model of the entire set. The complete-diagram argument embeds $\mathcal{Z}$ elementarily into its reduct; identify that copy with the original integer ring.

The resulting elementary extension contains $c$, a prime element unequal to every named standard integer. ThereforeThe construction rules out all standard integers, not merely the finite list appearing in any one approximation.

Take the parameter-free complete typeof the nonstandard prime constructed in part (i). It is first-order consistent and complete because it consists of exactly the logical formulas true of an actual element in a first-order model of $T$. It contains $x\ne \bar{a}$ for every integer $a$, and consequently no element of $\mathcal{Z}$ realizes it. ThusIt is necessarily a nonprincipal complete type. If a first-order consistent logical formula isolated it, completeness of $T$ would force the existential closure of that logical formula to hold in every first-order model of $T$, including $\mathcal{Z}$; a witness there would realize the partial type, a contradiction. This connects the explicit omission with the hypothesis of the omitting types theorem.

Take the first-order theory of an infinite set in the first-order language of logical equality alone, with an axiom asserting at least $n$ distinct elements for each positive integer $n$. It is complete and eliminates quantifiers: a new logical variable either equals a specified old logical variable or is required to avoid only finitely many existing values, which an infinite domain permits.

For every infinite cardinal $\kappa$, any two first-order models of size $\kappa$ are isomorphic because every bijection between their underlying sets preserves logical equality, the whole first-order language. ThusIts complete $n$-types record only the partition of coordinate indices according to logical equality. There are finitely many such partitions, and each is isolated by the corresponding logical equalities and inequalities. Ryll-Nardzewski therefore recovers countable categoricity; there are no nonprincipal complete finite-arity partial types to omit in this example.

Use the first-order theory $T_{\mathbb{Z}}$ of a bijection $f$ with named inverse $g$ and no finite cycles. Its axioms areEvery first-order model is a disjoint union of free two-sided integer orbits. Write $f^k$ for positive powers of $f$ and for powers of $g$ when $k$ is negative.

This first-order theory is complete and eliminates quantifiers. Every first-order term reduces to $f^k(x_i)$, and atomic logical formulas reduce to statements $x_j=f^k(x_i)$. To eliminate an existential logical variable in a conjunction of literals, first remove self-equalities using the no-finite-cycle axioms. If an equation fixes that logical variable as an integer translate of another logical variable, substitute that value. If no such equation remains, only finitely many points are excluded by the remaining inequalities, so some value exists in the infinite domain. Apply this to disjunctive normal form and then induct on logical formulas. The resulting parameter-free first-order sentences are decided, proving completeness.

For each fixed finite $n$, a complete type records a partition of the logical variables into common orbits and finitely many integer offsets within each block. There are countably many possibilities. Countable first-order models with one orbit and with two orbits are not isomorphic, so the first-order theory is not ${\aleph }_0$-categorical.

Thus this example has the source's explicitly defined countable-type property. In fact it also satisfies ordinary omega-stability: over a countable parameter set, an element is either a specified integer translate of a parameter, giving countably many possibilities, or lies in an entirely new orbit, giving one further complete one-type. The same finite-orbit-pattern description handles finite tuples. ThereforeThe source prints the label “$0$-stable”; the construction meets its stated definition as well as the usual stronger omega-stability condition. The difference between countably many and finitely many partial types is exactly why Ryll-Nardzewski does not yield categoricity here.

Reuse the complete free-bijection theory from part (e). Quantifier elimination shows that every one-variable atomic statement is already decided by the no-finite-cycle axioms. Thus there is exactly one complete one-type, isolated by $x=x$.

For two logical variables, each integer $k$ gives a different complete type isolated byThere is also the complete typesaying that the two elements lie in different orbits. These exhaust the possibilities by quantifier elimination. In particularThe last complete type is a nonprincipal complete type. Any logical formula uses only finitely many integer offsets after quantifier elimination; a pair in a single orbit with a sufficiently different offset agrees with a different-orbit pair on those tests. Hence no single logical formula can isolate $p_{\infty }$.

The one-orbit first-order model $(\mathbb{Z},f(n)=n+1,g(n)=n-1)$ omits $p_{\infty }$, while a two-orbit first-order model realizes it. This illustrates the omitting types theorem and shows concretely why finitely many complete one-types alone cannot replace the all-arities condition in Ryll-Nardzewski.

The Ryll-Nardzewski theorem says that for a complete theory $T$ in a countable first-order language with infinite first-order models, the following are equivalent: $T$ is ${\aleph }_0$-categorical; for every positive integer $n$, the set $S_n(T)$ of complete parameter-free $n$-types is finite; every such finite-arity complete type is isolated by a logical formula. The countable-language hypothesis and the requirement for every finite arity are important.

The omitting types theorem says that, for a first-order consistent complete theory in a countable first-order language, a countable family of nonprincipal partial types in finite tuples of logical variables can all be omitted in a countable first-order model. Here nonprincipal means that no logical formula first-order consistent with the first-order theory implies every logical formula of the partial type. Omission means that no tuple in the constructed first-order model satisfies all logical formulas of that partial type. The complete one-type version is the special case of one partial type in one logical variable.

The natural-number successor chain embeds into the integer successor chain by ordinary inclusion. Thus $\mathbb{Z}$ witnesses the existential joint satisfiability in the first assertion. The two chains are not elementarily equivalent: every integer has a predecessor, whereas zero has none in the natural-number chain.

For the universal assertion, construct an elementary extension of the natural-number chain containing a two-sided successor chain. To its elementary diagram add distinct constants $d_j$ for $j\in \mathbb{Z}$ with $s(d_j)=d_{j+1}$ and require them to differ from all named natural numbers. Every finite subset is realized in the natural-number chain: place the finitely many required consecutive blocks sufficiently far to the right, away from the finitely many forbidden named Elements. Compactness yields an elementary extension $A^{\ast }$ with the required two-sided chain. That chain is a substructure isomorphic to the integer successor structure and hence is a model of its complete theory.

Consequently both joint theories in the two assertions are satisfiable, though the integer chain cannot embed into the natural-number chain itself: repeated predecessors would give an infinite descending sequence of natural numbers. The second assertion correctly uses an elementarily equivalent ambient structure rather than the original one.

By Vieta's formulas, the elementary symmetric polynomials in the roots giveFor each root, ${\alpha }_j^3+c_1{\alpha }_j^2+c_2{\alpha }_j+c_3=0$. Multiplying by ${\alpha }_j^k$ and summing gives the power sum recurrenceIts initial values are $s_0=3$, $s_1=-c_1$ and $s_2=c_1^2-2c_2$. They are polynomials with integer coefficients in $c_1,c_2,c_3$. The recurrence uses only addition and multiplication by those coefficients, so induction proves the same for every $s_k$. This is the cubic case of Newton's identities.

For $c_1=0$, put ${\alpha }_1=u$, ${\alpha }_2=v$, ${\alpha }_3=-u-v$. Then $c_2=-(u^2+uv+v^2)$ and $c_3=uv(u+v)$. Expanding the product of root differences givesThis is the discriminant of a depressed cubic in depressed form, and the displayed identity also proves its expression in integer-coefficient symmetric polynomials.

Put $\theta =\sqrt{p+\sqrt{p}}$. Then $({\theta }^2-p{)}^2=p$, so $\theta$ is a root ofThis polynomial satisfies Eisenstein's criterion at $p$: every nonleading coefficient is divisible by $p$, while the constant coefficient $p(p-1)$ is not divisible by $p^2$. ThereforeNow suppose $p=n^2+1$, taking $n>0$ without loss of generality. Let $\beta =\sqrt{p-\sqrt{p}}$. Since $\theta \beta =n\sqrt{p}$ and $\sqrt{p}={\theta }^2-p$,Consequently all four distinct roots $\pm \theta ,\pm \beta$ lie in this field. This gives a cyclic quartic field from a nested quadratic radical. It is a splitting field in characteristic zero, hence a normal extension and a Galois extension.

There is a Galois automorphism $\sigma$ sending $\theta$ to $\beta$: a root image defines an embedding, and normality makes it an automorphism of the splitting field. It sends $\sqrt{p}={\theta }^2-p$ to $-\sqrt{p}$, soThus $\theta \mapsto \beta \mapsto -\theta \mapsto -\beta \mapsto \theta$ is a four-cycle. The Galois group has order four and contains an element of order four, givingFor the requested lower degrees, $(q,r)=(3,4)$ gives $\sqrt{q+\sqrt{r}}=\sqrt{5}$ and degree two; $(q,r)=(3,1)$ gives $2$ and degree one. Both pairs use positive integers.

For the $n$ distinct complex embeddings ${\sigma }_1,\dots ,{\sigma }_n$ of the number field, define the discriminant of elements of a number field byThe equality follows by multiplying the transpose of the embedding matrix by itself. Each field trace is rational: it is also the trace of the rational matrix of multiplication by $w_i{w}_j$ on $K$. Thus the determinant lies in $\mathbb{Q}$. If the $w_i$ are algebraic integers, the products and their conjugates are algebraic integers, and so are their sums. A rational algebraic integer is an integer, hence all trace entries and their determinant are integers.

Write $u_j={\sum }_i{a}_{ij}{v}_i$. The inclusion of the generated abelian groups makes $A=(a_{ij})$ an integer matrix. Its embedding matrix is the embedding matrix of the $v_i$ multiplied by $A$. Taking squared determinants givesFor independent algebraic integers $v_i$, choose an integral basis ${\omega }_i$ of ${\mathcal{O}}_K$. The same formula gives $D(v)=[{\mathcal{O}}_K:{\sum }_i\mathbb{Z}{v}_i{]}^{2}D(\omega )$. If $D(v)$ is nonzero and a square-free integer, the integer index must be one, proving that the $v_i$ form an integral basis.

Now put $\beta =3\alpha +4$. Since $\mathbb{Q}(\beta )=\mathbb{Q}(\alpha )$ has degree three, the monic polynomialis its minimal polynomial of an algebraic element. The field norm is the negative of its constant term:For the discriminant calculation, $f^{\prime }(\alpha )=3{\alpha }^2+4\alpha$. In the basis $1,\alpha ,{\alpha }^2$, using ${\alpha }^3=-2{\alpha }^2-1$, multiplication by this element has matrixThus the given derivative-norm formula gives $D(1,\alpha ,{\alpha }^2)=-59$. Since $\alpha$ satisfies a monic polynomial with integer coefficients, $1,\alpha ,{\alpha }^2$ are algebraic integers; the fact that a squarefree discriminant certifies an integral basis proves