For the linearized system a centre occurs when $\boxed{{\lambda }_{1,2}=\pm i\beta ,\beta \ne 0}$. Its real solutions are periodic ellipses after an invertible linear coordinate change.

As explained by a centre in the linearization need not be a nonlinear centre, for a nonlinear system these eigenvalues alone do not guarantee a centre. For example $\dot{X}=-Y-X(X^2+Y^2)$, $\dot{Y}=X-Y(X^2+Y^2)$ has eigenvalues $\pm i$ at zero but polar equations $\dot{r}=-r^3$, $\dot{\theta }=1$, so trajectories spiral inward. Removing the cubic terms gives a genuine centre with the same Jacobian. Thus there is no eigenvalue-only sufficient condition for a nonlinear centre in the stated generality.

The eigenvalues are real and have opposite signs: $\boxed{{\lambda }_1{\lambda }_2<0}$. One eigendirection contracts and the other expands, giving a saddle.

Both eigenvalues are real and positive: $\boxed{{\lambda }_1>0,{\lambda }_2>0}$. The real eigendirections carry exponentially growing factors, giving an unstable node.

Both eigenvalues are real and negative: $\boxed{{\lambda }_1<0,{\lambda }_2<0}$, with distinctness as assumed. The two independent real eigendirections carry exponentially decaying factors $e^{{\lambda }_{j}t}$, giving a stable node.

The same integration by parts with $p$ replaced by $p+c$ givesFor the remaining unnumbered requests, the derivative and convolution formulas, under the usual convergence hypotheses, areApplying them to the general equation yields $p\bar{x}-a+\bar{k}\bar{x}=\bar{y}$ and thereforeFor the particular data, $\bar{y}=1/p$, $\bar{k}=1/(p+2)$ and $a=0$, soThe four transforms above giveAs a direct check, $x(0)=0$, $x^{\prime }=(1+t)e^{-t}$, andAdding the last two expressions gives one, verifying the original equation as well as the required limit.

Combining the exponential factors gives

Integration by parts, whose boundary term vanishes for $\mathrm{Re}p>0$, gives

For a function for which the integral converges, the Laplace transform isFor a continuous function of exponential order this exists in a sufficiently far right half-plane. For the first input,This follows by evaluating the exponential antiderivative and using its decay at infinity.

Write $r=\sqrt{x^2+y^2}>0$. For $u=x/(1+r)$, $v=y/(1+r)$,The two Cauchy-Riemann equations would require $x^2=y^2$ and $xy=0$ simultaneously, impossible when $r>0$. Thus $f$ is not complex differentiable at any nonzero point.

At zero, however,So $f$ is complex differentiable at zero, with derivative one, but is not holomorphic on $\mathbb{C}$ or on any nonempty open subset. Complex differentiability at a single point is weaker than holomorphy in a neighborhood.

A function is holomorphic on an open set if it is complex differentiable at every point of that set. At $z_0=x_0+i{y}_0$, take difference quotients first along real increments. Their limit is $f^{\prime }(z_0)$, so $u_x=\mathrm{Re}{f}^{\prime }(z_0)$ and $v_x=\mathrm{Im}{f}^{\prime }(z_0)$. Along imaginary increments,so $u_y=-\mathrm{Im}{f}^{\prime }(z_0)$ and $v_y=\mathrm{Re}{f}^{\prime }(z_0)$. These limits prove existence of all four partial derivatives and the Cauchy-Riemann equations $u_x=v_y$, $u_y=-v_x$.

For the first example, the identity function and the reciprocal function are holomorphic away from zero, so $f$ is holomorphic on the given domain, with

An element $a$ is a prime element when it is nonzero and not a unit, and $a\mid bc$ implies $a\mid b$ or $a\mid c$. If $a=bc$ and $a$ is a prime element, assume for example $b=ad$. Cancellation in $a=adc$ gives $dc=1$, so $c$ is a unit. This proves every prime element is an irreducible element.

In a Euclidean domain, each nonzero ideal has a generator of least Euclidean value: division of another member by that generator leaves a remainder in the ideal of smaller value, so the remainder must vanish. Let $a$ be an irreducible element and suppose $a\mid bc$ but $a\nmid b$. Write $(a,b)=(d)$. Since $d\mid a$, irreducibility says $d$ is a unit or is associate to $a$. The second case would give $a\mid b$, so $d$ is a unit. Hence $1=ra+sb$ for some $r,s$. Multiplying by $c$ shows $a\mid c$. Thus irreducibles in a Euclidean domain are prime.

For $R=\mathbb{Z}[i\sqrt{7}]$, use the multiplicative integer field norm $N(m+ni\sqrt{7})=m^2+7n^2$. Units have field norm one and are exactly $\pm 1$. If $2=uv$ were a product of nonunits, their norms would be integers greater than one multiplying to four, hence both would have field norm two. But $m^2+7n^2=2$ has no integer solution: $n\ne 0$ makes the left side at least seven, and $n=0$ would give $m^2=2$. Therefore 2 is an irreducible element.

On the other hand,is divisible by 2, while neither factor is, because divisibility by 2 requires both integer coefficients to be even. Thus 2 is not a prime element in this ring. It is also not a prime element in $\mathbb{Z}[i]$: $2\mid (1+i{)}^2=2i$, but $2\nmid 1+i$. In fact $2=(1+i)(1-i)$ is already a nontrivial factorization there.

An element $a$ of an integral domain is an irreducible element when it is nonzero and not a unit, and any factorization $a=bc$ has at least one unit factor. The exclusions of zero and units are part of the definition.

Let $\omega =e^{2\pi i/3}$ and integrate $1/(1+z^3)$ around the positively oriented sector from argument zero to $2\pi /3$, with outer radius $R>1$. The single pole inside is $z_0=e^{i\pi /3}$, with residue $1/(3z_0^2)=1/(3\omega )$. The outward real radial segment contributes $I_R={\int }_0^{R}dx/(1+x^3)$; the return radial segment contributes $-\omega I_R$ because $(\omega x{)}^3=x^3$.

On the circular arc, $|1+z^3|\ge R^3-1$, so its integral has modulus at most $(2\pi R/3)/(R^3-1)\to 0$. Thus the residue theorem yieldsSince $\omega (1-\omega )=\omega -{\omega }^2=i\sqrt{3}$,The ordinary improper integral converges at both endpoints, so no principal-value interpretation is needed here.

Integrate $e^{iz}/z$ around the upper half-disc with the upper half-disc of radius $\epsilon$ about zero removed. The real segments run from $-R$ to $-\epsilon$ and $\epsilon$ to $R$; the small semicircle is clockwise. There are no poles inside this indented contour.

The large arc contribution tends to zero: its modulus is at most ${\int }_0^{\pi }{e}^{-R\sin \theta }d\theta$, which tends to zero by dominated convergence. On the small arc $z=\epsilon e^{i\theta }$, $\theta$ decreases from $\pi$ to zero, so its integral tends to ${\int }_{\pi }^{0}id\theta =-i\pi$. The contour theorem consequently givesTaking imaginary parts is legitimate for each truncated integral. The imaginary integrand $\sin x/x$ is continuous at zero after assigning its limiting value one, so the inner indentation can be removed. Therefore

For a bounded entire function, the preceding result with $n=0$ gives Liouville's theorem. If a nonconstant polynomial $p$ had no complex root, $1/p$ would be entire. The leading term gives $|p(z)|\to \infty$ at infinity, so $1/p$ is bounded outside a large disc and is also bounded on the remaining compact disc. Liouville's theorem would make $1/p$, and hence $p$, constant. This proves the fundamental theorem of algebra.

Finally, away from the zeros of $\cos z$, the holomorphic quotient $h=f/\cos z$ has modulus at most $C$. At each zero it therefore has a removable singularity. These isolated points have no finite accumulation, so removing all of them gives a bounded entire function. Liouville's theorem gives a constant $c$ with $|c|\le C$, andConversely every such multiple satisfies the assumed bound.

Take a circle of radius $R\ge S$ about zero. The Cauchy estimate gives, for each integer $k>n$,Therefore all these derivatives vanish. The Taylor series of an entire function converges to it everywhere, and henceAs usual $n$ is a nonnegative integer; if a real exponent is allowed, the degree is at most its integer part. A negative growth exponent forces $f=0$.

If $f$ is holomorphic on a neighborhood of the closed disc, the differentiated Cauchy integral formula iswith positive orientation. On the circle the integrand has modulus at most $M/r^{k+1}$ and the contour has length $2\pi r$. Thus

Let $J={\bigcup }_n{I}_n$. It contains zero. If $a,b\in J$, both belong to $I_N$ for some common $N$, so $a-b\in I_N\subseteq J$. If $r\in R$, then $ra\in I_N\subseteq J$. These are the ideal axioms, proving $J$ is an ideal.

In a principal ideal domain write $J=(g)$. Since $g\in J$, there is $n_0$ with $g\in I_{n_0}$. For $n\ge n_0$,All inclusions are equalities. Hence the chain stabilizes at $I_{n_0}$. This includes $J=(0)$ and is the ascending chain condition for ideals.

An integral domain is a nonzero commutative ring with identity and no zero divisors. A principal ideal domain is an integral domain in which every ideal is generated by one element.

The leading coefficient of a product of nonzero real polynomials is nonzero, so $\mathbb{R}[x]$ is an integral domain. The zero ideal is principal. In any nonzero ideal $J$, choose a nonzero polynomial $g$ of least degree. Divide an arbitrary $f\in J$ by $g$: $f=qg+r$ with $r=0$ or $\mathrm{deg}r<\mathrm{deg}g$. Since $r\in J$, minimality forces $r=0$. Thus $J=(g)$ and $\mathbb{R}[x]$ is a principal ideal domain.

In $\mathbb{Z}[x]$ consider $J=(2,x)$. If $J=(h)$, then $h$ divides the constant polynomial 2, so $h$ is constant, and divisibility of $x$ forces $h=\pm 1$. That would give $J=\mathbb{Z}[x]$. But the constant coefficient of every element $2p+xq$ is even, so $1\notin J$. This contradiction proves $\mathbb{Z}[x]$ is not a principal ideal domain.

The sum in question is $⟨Pu,Pv⟩$. Applying Cauchy-Schwarz and then Bessel's inequality givesIf either original vector is zero, equality holds. If both are nonzero, equality forces both projection norms to equal their original norms and also forces dependence of the projected vectors. Thus apart from the zero-vector cases, equality holds exactly when $u,v\in W$ and $u,v$ are linearly dependent.

For the final unnumbered request, apply Cauchy-Schwarz in ${\mathbb{R}}^k$ to $p_i=a^{i/2}$ and $q_i=a^{-i/2}$. Their inner product is $k$, soEquality requires the positive vectors $p,q$ to be proportional, equivalently $a^i$ constant for $i=1,\dots ,k$. For $k\ge 2$ this is $a=1$. If $k=1$, equality holds for every $a>0$.