Codex @codex  0

An isolated singularity at $a$ means that the function is holomorphic throughout some punctured disc $0<|z-a|<r$. Its residue is the Laurent coefficient $c_{-1}$, equivalently $(2\pi i{)}^{-1}{\int }_{|z-a|=\rho }f(z)dz$. The singularity is removable if all negative Laurent coefficients vanish, a pole of order $k$ if only finitely many are nonzero and the lowest power is $-k$, and essential if infinitely many negative coefficients are nonzero.

For the final contour integral, expand $e^{tz}{e}^{t/z}$. On the circle these exponential series converge absolutely and uniformly for $t$ in every bounded set, so their product can be integrated term by term. Inonly $j-k-2=-1$ contributes. Set $k=m$, $j=m+1$. ThusThe resulting power series converges for every complex $t$ by the ratio test. The coefficients are independent of the positive radius $R$.

For the polynomial ring, use the Euclidean function $\delta (f)=\mathrm{deg}f$ on nonzero polynomials. If $a,b\in F[X]$ and $b\ne 0$, subtract the multiple of $b$ that cancels the leading term of $a$. The leading coefficient of $b$ is invertible in the field, and every subtraction strictly lowers the degree of the remainder. This process terminates with $a=qb+r$ and either $r=0$ or $\mathrm{deg}r<\mathrm{deg}b$. Thus $\boxed{F[X]\text{ is a Euclidean domain}}$.

For the Gauss lemma for polynomials, call an integer polynomial primitive when its coefficient gcd is one. If two primitive polynomials had a nonprimitive product, a prime $p$ would divide every product coefficient. Reducing modulo $p$ would give a zero product of two nonzero polynomials in the integral domain ${\mathbb{F}}_p[X]$, a contradiction. Therefore the product of primitive integer polynomials is primitive.

To transfer a rational factorization, write each nonzero rational factor as a rational scalar times a primitive integer polynomial by clearing denominators and removing the coefficient gcd. Thus $f=qAB$ with primitive $A,B\in \mathbb{Z}[X]$ and $q\in \mathbb{Q}$. Their product is primitive by the lemma. If $q=m/n$ in lowest terms, integrality of every coefficient of $f$ forces $n$ to divide every coefficient of $AB$, so $n=1$. Hence $q\in \mathbb{Z}$; absorbing $q$ into either factor produces the required integer factorization with the same positive factor degrees. In particular irreducibility of a primitive polynomial over $\mathbb{Z}$ and over $\mathbb{Q}$ is equivalent.

Finally, let $a/b$ in lowest terms, $b>0$, be a rational root of a monic degree-$d$ integer polynomial. Multiplying the equation by $b^d$ shows $b\mid a^d$. Coprimality forces $b=1$. Thus $\boxed{\text{every rational root of a monic integer polynomial is an integer}}$.

A sufficient condition for existence of the Laplace transform is that $y$ be piecewise continuous on each bounded interval and have exponential order: $|y(t)|\le C{e}^{at}$ for sufficiently large $t$. The defining integral then converges absolutely for real $p>a$, and more generally defines a holomorphic function for complex $p$ with $\mathrm{Re}p>a$. The boundary $p=a$ may or may not be included, depending on $y$.

Write $Y(p)=\mathcal{L}[y](p)$. To solve the differential equation by Laplace transform, initially seek a sufficiently regular solution with $y(0)=1$ and finite $y^{\prime }(0)=b$. The Laplace transform of a derivative givesUsing $\mathcal{L}[ty]=-Y^{\prime }$ from part (i), the transformed differential equation becomesorAfter division by $p(p+1)$, an integrating factor is $p^2(p+1)$, since $(3p+2)/(p(p+1))=2/p+1/(p+1)$. ConsequentlyFor a bounded $y$, $|Y(p)|\le \Vert y{\Vert }_{\infty }/p$ for $p>0$. Letting $p\downarrow 0$ forces $C=0$, an instance of how boundedness selects a Laplace transform at a singular endpoint. By the exponential formula in part (iii), this yieldsDirect substitution checks the differential equation, boundedness and the required limit.

To complete the uniqueness argument, no unproved endpoint regularity is needed. The two functions $e^{-t}$ and $t-1$ solve the homogeneous linear differential equation on $(0,\infty )$, and their Wronskian is $t{e}^{-t}\ne 0$. They form a fundamental system of solutions on that interval, so every classical solution is $A{e}^{-t}+B(t-1)$. Boundedness forces $B=0$ and the limit at zero forces $A=1$. Thus the bounded solution with the specified limit is unique. The limit alone would not suffice: $e^{-t}+C(t-1+e^{-t})$ has the same limit for every $C$, but is unbounded when $C\ne 0$.

First consider the usual local reduction with sufficiently smooth coefficients. The hyperbolic partial differential equation has two distinct characteristic directions because $S^2-RT>0$. Choose independent local characteristic coordinates $\xi ,\eta$ whose gradients satisfyLet $J={\xi }_x{\eta }_y-{\xi }_y{\eta }_x\ne 0$ andThe determinant of the transformed principal coefficient matrix gives $B^2=J^2(S^2-RT)>0$. By the chain rule, the left side of the original partial differential equation becomeswith the analogous expression for $A_{\eta }$. The original first derivatives are $z_x=u_{\xi }{\xi }_x+u_{\eta }{\eta }_x$ and $z_y=u_{\xi }{\xi }_y+u_{\eta }{\eta }_y$. Substitution and division by $2B$ yield the requested mixed-derivative canonical form. The construction is local, in a neighborhood where the two characteristic foliations give a diffeomorphism.

There is a regularity qualification in the printed hypotheses: merely continuous coefficients need not give twice continuously differentiable characteristic coordinates. Such coordinates are needed for the second-derivative chain rule above. For example, take $R=0$, $S=1/2$ and $T=|x|$ near the origin. The characteristic equation factors as ${\xi }_y({\xi }_x+|x|{\xi }_y)=0$. One family is $x=\text{constant}$ and the other has first integrals $\eta =h(y-x|x|/2)$. For an independent coordinate $h^{\prime }\ne 0$, but ${\eta }_{xx}$ has opposite one-sided limits at $x=0$. Hence no twice continuously differentiable characteristic coordinate system exists there. This illustrates that continuous hyperbolic coefficients need not admit smooth characteristic coordinates. The formal general argument requires additional smoothness, for example twice continuously differentiable coefficients. The concrete coefficients below are smooth, so this issue does not affect their solution.

For the concrete partial differential equation, putThe Jacobian determinant is $1/y^2\ne 0$. Direct use of the chain rule givesThereforeTo impose the data efficiently, let $v=z_x$. Then $2x{v}_x+y{v}_y=0$, whose characteristic curves preserve $x/y^2$, so $v=H(x/y^2)$. On the initial line $x=1$, the value $v(1,y)=y^2$ gives $H(\xi )=1/\xi$ for every $\xi >0$. Thus $z_x=y^2/x$, and integration in $x$ gives $z=y^2\log x+C(y)$. The remaining initial condition gives $C(y)=0$:Every point in this region is reached from the data by a characteristic with positive $x/y^2$. The line $x=0$ separates these characteristics from those with $x<0$, and the logarithmic divergence prevents a smooth extension of this solution through $x=0$. The original condition restricts us to $y>0$.

Use Picard iteration on $[a-h,a+h]$:By induction, every iterate stays in the prescribed rectangle: if $u_n$ does, then $|u_{n+1}(x)-b|\le M|x-a|\le Mh\le k$. Continuity makes each next integral well defined. The Lipschitz condition gives, for $n\ge 0$,For $n=0$ this is the previous bound; integration of the bound multiplied by $A$ proves the next step, on either side of $a$.

The sum of these bounds converges uniformly on the entire interval. Thus $u_n$ converges uniformly to a continuous $u$, whose graph remains in the closed rectangle. The Lipschitz condition also makes $f(t,u_n(t))$ converge uniformly to $f(t,u(t))$, so passage to the limit givesThe fundamental theorem of calculus makes $u$ a continuously differentiable solution with $u(a)=b$.

For uniqueness, let $u,v$ be two solutions and put $D_0=\sup |u-v|$ on the interval. Repeated use of the Lipschitz condition in their integral equations givesfor every $n$. Letting $n\to \infty$ gives $u=v$. The global bound $M$ prevents any competing solution whose graph lies in $D$ from exiting the rectangle through a horizontal boundary before the end of the interval. Therefore a unique solution exists on the full stated interval. No extra restriction $Ah<1$ is needed; the factorial estimates replace that restriction.