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A cubic over a field is reducible exactly when it has a root in that field: a nontrivial factorization must contain a degree-one factor. By the rational root theorem, a rational root of this monic integer polynomial must be an integer dividing its constant coefficient, hence either $1$ or $-1$. Substitution gives $-2$ and $4$, respectively. Neither is zero. Thus the cubic is irreducible over $\mathbb{Q}$.

Use the Eisenstein criterion with prime $5$. The leading coefficient is one; every lower coefficient, including the zero coefficients, is divisible by $5$; and the constant coefficient $5$ is not divisible by $25$. The polynomial is monic and hence primitive. Therefore it is irreducible over $\mathbb{Q}$.

This relative absolute-error loss weights each parameter by $1/\theta$. Its posterior expected loss isThe effective density for the Bayes estimator under weighted absolute loss is proportional to $\pi (\theta \mid x)/\theta$, hence an exponential distribution of rate $b$. Its median solves $1-e^{-ba}=1/2$.

More explicitly, for $a\ge 0$, differentiating the split integrals at $\theta =a$ gives $R_x^{\prime }(a)=b(1-2e^{-ba})$ and $R_x^{\prime \prime }(a)=2b^2{e}^{-ba}>0$. For $a<0$, the loss has derivative $-b$, so its minimum cannot lie there. The unique global minimum is thereforeThis is the weighted posterior median; the ordinary median of the Gamma posterior would minimize unweighted absolute loss and would give a different answer.

Under squared-error loss, the posterior expected loss decomposes asIt is uniquely minimized by the posterior mean. Integrating the Gamma posterior gives $\mathbb{E}[\theta \mid x]=b^2{\int }_0^{\infty }{\theta }^2{e}^{-b\theta }d\theta =2/b$. Hence the Bayes estimator is

Take the Fourier transform in $x$ of the Laplace equation. Writing $U(s,y)$ for the transformed solution gives $U_{yy}-s^{2}U=0$. Its boundary values givewith the multiplier at $s=0$ understood as $1-y$. The particular data simplify this to $U(s,y)=2\sinh (s(1-y))$ for $|s|<1$, and zero outside. Thus, with $a=1-y$, the Fourier solution of the strip Dirichlet problem isThe displayed closed expression applies in the open strip. The integral defines its continuous boundary values, including the removable point $(x,a)=(0,0)$. It gives $u(x,0)=g(x)$ and $u(x,1)=0$. Differentiation under this finite integral verifies $u_{xx}+u_{yy}=0$ directly, since each integrand has opposite second derivatives in $x$ and $y$.

The boundary conditions alone do not specify a unique solution without a growth restriction: adding $C{e}^{\pi x}\sin (\pi y)$ preserves both boundary values and harmonicity. The constructed solution is bounded, and it is unique among bounded solutions continuous on the closed strip. To see this, apply the maximum principle for harmonic functions to a bounded real or imaginary component $w$ of the difference. For $0<\delta <\pi$, the positive harmonic function $B(x,y)=\cosh (\delta x)\cos (\delta (y-1/2))$ has a positive minimum factor in $y\in [0,1]$. For any $\epsilon >0$, the functions $\pm w-\epsilon B$ are nonpositive on the horizontal sides of a sufficiently wide rectangle and on its vertical sides, because $w$ is bounded and $B$ grows there. The maximum principle gives $|w|\le \epsilon B$ at every fixed point. Letting $\epsilon \downarrow 0$ proves uniqueness in this natural Fourier solution class.

With the Fourier transform convention of 5A, use the printed Fourier inversion formula directly. The possible choice of values at $s=\pm 1$ has no effect on the integral. Integration over the support givesCombining the two fractions yieldsIn particular $g$ is imaginary-valued and odd, as expected from the real odd transformed function with this inverse-transform sign.

At $c=0$ the left-hand root is $-2$, and the derivative of the polynomial with respect to $z$ is $3(-2{)}^2=12\ne 0$. The holomorphic implicit function theorem states that if $P(z,c)$ is holomorphic near $(z_0,c_0)$, $P(z_0,c_0)=0$ and $P_z(z_0,c_0)\ne 0$, there is a unique holomorphic function $z=\gamma (c)$ near $c_0$, with $\gamma (c_0)=z_0$ and $P(\gamma (c),c)=0$. It therefore supplies an analytic root branch $\gamma (c)$ near zero. Differentiate its equation:It follows that ${\gamma }^{\prime }(0)=i/6$ andFor sufficiently small positive $c$ its imaginary part is $c/6+O(c^2)>0$. By the unique-root localization from the preceding part, this is the root in the left half-plane. Thus that root lies above the real axis for all sufficiently small positive $c$.

On $|z|=3$, the perturbation has modulus $|icz+8|\le 3c+8<11<27=|z^3|$. Thus Rouché's theorem gives three roots inside $|z|<3$, counted with multiplicity.

For the quadrant information, compare instead with $P_0(z)=z^3+8$, whose roots are $r_1=-2$, $r_2=1+i\sqrt{3}$, $r_3=1-i\sqrt{3}$. Their pairwise distances are $2\sqrt{3}$. On the circle $|z-r_j|=1$,The three unit discs are disjoint, and Rouché localization of polynomial roots gives one root in each, counted with multiplicity. The disc around $1+i\sqrt{3}$ lies strictly in the first quadrant in its interior; that around $1-i\sqrt{3}$ lies in the fourth quadrant; and that around $-2$ lies in $\mathrm{Re}z<-1$. Consequently there is exactly one root in each of those two right-hand quadrants, and exactly one root in the left half-plane. The same count also shows that each root is simple.

The form of Rouché's theorem used here is as follows. Suppose $f$ and $g$ are holomorphic on a neighborhood of the closure of a bounded Jordan domain, and $|g-f|<|f|$ on its positively oriented boundary. Then $f$ and $g$ have the same number of zeros inside, counted with multiplicity.

To prove this form, put $f_t=f+t(g-f)$ for $0\le t\le 1$. The strict inequality prevents $f_t$ from vanishing on the boundary. The argument principle counts its interior zeros by $(2\pi i{)}^{-1}\oint f_t^{\prime }/f_t$. This integral varies continuously with $t$: the integrand is continuous and its denominator stays uniformly away from zero on the compact boundary times $[0,1]$. Its values are integers, so the integral is constant. Its values at $t=0,1$ are the respective zero counts. This proves Rouché's theorem with the multiplicities and boundary hypotheses required below.

Use the usual convention that a ring homomorphism preserves the identity. Such a map from the quotient ring to $\mathbb{C}$ fixes $\mathbb{Q}$: it fixes integers, and preservation of inverses fixes rational numbers. It is therefore determined by the image $\zeta$ of the class of $x$. The defining relation forces ${\zeta }^3=1$. Conversely, evaluation at any such $\zeta$ annihilates $(x^3-1)$ and hence factors through the quotient. Thus there are exactly three unital homomorphisms, namelyIf the convention permits nonunital ring homomorphisms, add the zero map. These are the only extra possibilities: the image of $1$ is an idempotent in $\mathbb{C}$, hence either zero or one, and image zero forces the whole map to vanish.

In the quotient ring, the classes of $x-1$ and $x^2+x+1$ are nonzero: neither nonzero polynomial of degree less than three belongs to the ideal generated by $x^3-1$. But their product is the zero class, sinceThe quotient thus contains zero divisors, so it is not an integral domain.

The coefficient identityholds over every field. Both factors have positive degree, including in characteristics $2$ and $3$, because their leading coefficients remain one. Therefore there is no field over which this polynomial is irreducible.

Consider the ideal $I=(2,x)$ in $\mathbb{Z}[x]$. It is proper: evaluation at zero followed by reduction modulo $2$ maps the ring onto ${\mathbb{F}}_2$ and has kernel $I$. Suppose $I=(f)$. Then $f$ divides the constant polynomial $2$, so $f$ is a constant integer dividing $2$. Up to sign, the only possibilities are $1$ and $2$. The first would make the ideal the whole ring; the second cannot divide $x$ in $\mathbb{Z}[x]$. Both are impossible. This proves the integer polynomial ring is not a principal ideal domain result: $\mathbb{Z}[x]$ is not a principal ideal domain.

In the polynomial ring $\mathbb{Z}[x]$, the ideal $(x)$ is nonzero and prime: the quotient ring $\mathbb{Z}[x]/(x)$ is isomorphic to the integral domain $\mathbb{Z}$, by evaluation at zero. It is not maximal, since that quotient is not a field. Equivalently $(x)⊊(2,x)⊊\mathbb{Z}[x]$. Thus $(x)$ is a nonzero prime ideal which is not maximal.

Choose an orthonormal basis of eigenvectors of the positive-definite operator $\alpha$ using the proved spectral theorem. Its eigenvalues ${\lambda }_j$ are strictly positive, because $⟨\alpha e_j,e_j⟩={\lambda }_j>0$. Define $\beta e_j=\sqrt{{\lambda }_j}{e}_j$. It is a positive-definite Hermitian operator and ${\beta }^2=\alpha$, establishing existence of the positive square root of an operator.

For uniqueness, let $B$ be any positive-definite Hermitian square root. The identity $B^2=\alpha$ implies $B\alpha =B^3=\alpha B$, so $B$ preserves each eigenspace $E_{\lambda }$ of $\alpha$. Its restriction to that space is Hermitian and positive-definite. Diagonalize that restriction by the same spectral theorem. Every resulting eigenvalue $\mu$ satisfies $\mu >0$ and ${\mu }^2=\lambda$, hence $\mu =\sqrt{\lambda }$. Therefore $B$ acts as $\sqrt{\lambda }I$ on the entire eigenspace, even when $\lambda$ is repeated. These eigenspaces span the space, so the positive-definite Hermitian square root is unique, with eigenvalues $\sqrt{{\lambda }_j}$.

The adjoint reverses products, so for the two Hermitian operatorsThus $\alpha \beta$ is Hermitian if and only if $\alpha \beta =\beta \alpha$. No positivity hypothesis is needed for this equivalence.

For a Hermitian operator $A$, every eigenvalue is real. Indeed, if $Av=\lambda v$ with $v\ne 0$, self-adjointness gives $\lambda ⟨v,v⟩=⟨Av,v⟩=⟨v,Av⟩=\overline{\lambda }⟨v,v⟩$. To prove the finite-dimensional spectral theorem, proceed by induction on the dimension. The fundamental theorem of algebra supplies a root of the characteristic polynomial, hence a unit eigenvector $v$. Its orthogonal complement is invariant: for $w\perp v$, $⟨Aw,v⟩=⟨w,Av⟩=0$. The restriction of $A$ to that complement is again Hermitian. Induction supplies an orthonormal basis of eigenvectors there; adjoining $v$ gives an orthonormal eigenbasis of the whole space. The zero-dimensional case is immediate. Consequently a Hermitian operator is diagonal in an orthonormal basis, with real diagonal entries.

Near zero the $n=0$ summand supplies the pole and the remaining series is holomorphic, so $f(w)=w^{-2}+F(w)$ with $F$ holomorphic near zero. Choose the local logarithm with $L(1)=0$. Writing $h=z-1$ gives $L(z)=h-h^2/2+O(h^3)$ and henceTherefore $z=1$ is a double pole, with leading coefficient ${\lim }_{z\to 1}(z-1{)}^{2}g(z)=-4{\pi }^2$. Its residue, if desired, is $-4{\pi }^2$.

Fix $z_0\ne 0,1$ and take a small disk about it that avoids zero and one. There is a holomorphic logarithm $L$ on this disk. The function $L(z)/(2\pi i)$ cannot be an integer there: if it were an integer, exponentiation would give $z=1$. Thus $f(L(z)/(2\pi i))$ is holomorphic on the disk.

Where the prescribed principal logarithm and $L$ are both evaluated, they differ by $2\pi ik$ for an integer $k$. By the period-one identity, their compositions with $f$ coincide. This remains true on either side of the principal branch cut, and at a point on that cut one may instead choose the logarithm with argument between zero and $2\pi$. The local holomorphic expressions therefore agree with the stated pointwise definition and with each other. This is periodic holomorphic descent through the exponential map, provingThe principal logarithm itself is not analytic across its cut; periodicity of the outer function removes that obstruction.

Let $K\subset \mathbb{C}\setminus \mathbb{Z}$ be compact and contained in $|z|\le R$. For $|n|>2R$, $|z-n|\ge |n|/2$, hence $|(z-n{)}^{-2}|\le 4/n^2$ uniformly on $K$. The tail is therefore absolutely and uniformly convergent by the Weierstrass M-test. Its finitely many remaining summands are holomorphic on a neighborhood of $K$.

The theorem that a locally uniform limit of holomorphic functions is holomorphic now proves that the sum is analytic on $\mathbb{C}\setminus \mathbb{Z}$. Absolute convergence also permits a shift of the integer index:Thus the function has period one.