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For $y=\alpha x+\beta x^2$, the derivatives are $y^{\prime }=\alpha +2\beta x$ and $y^{\prime \prime }=2\beta$. Substituting gives $2\beta x^2-2x(\alpha +2\beta x)+2(\alpha x+\beta x^2)=0$. The independent solutions $x,x^2$ have Wronskian $x^2\ne 0$ on $[1,2]$, so they form a fundamental system of solutions.

Divide the inhomogeneous linear differential equation by $x^2$ and use variation of parameters. Choosing integrals from one gives the particular solutionThus $z=Ax+B{x}^2+z_p$. The first boundary condition gives $A=-B$. The integral endpoint terms cancel when differentiating $z_p$, so the second givesConsequentlyCollect the coefficient of $f(t)$ on either side of $t=x$. A suitable Green function isThe two pieces agree at $x=t$, $G(1,t)=0$, and $G_x(2,t)=0$ for $t<2$. For an interior source point its derivative jump isThus multiplication by the leading coefficient $t^2$ gives unit in a ring source strength. Away from $x=t$, both pieces solve the homogeneous linear differential equation. With continuous $f$, the integral representation has the required second derivative and solves the equation pointwise in $(1,2)$.

For uniqueness, the difference of two solutions is $Ax+B{x}^2$. Its boundary conditions give $A+B=0$ and $A+4B=0$, forcing $A=B=0$. Hence the solution is unique. This is the mixed Green function for an Euler operator with exponents one and two.

A function on the open set $D$ is holomorphic if at every $a\in D$ the complex limitexists. The limit is independent of the direction of approach.

A pole of order $n\ge 1$ at $a$ means that $f(z)=(z-a{)}^{-n}h(z)$ in a punctured neighborhood, where $h$ extends holomorphically across $a$ and $h(a)\ne 0$. Equivalently $(z-a{)}^{n}f(z)$ has a nonzero finite limit at $a$.

For the given quotient, $\cos z=1$ is equivalent, on putting $q=e^{iz}$, to $q+q^{-1}=2$, or $(q-1{)}^2=0$. Thus all possible poles are $a=2\pi k$, $k\in \mathbb{Z}$, with no additional nonreal solutions. At such a point,At $a\ne 0$ the numerator does not vanish, so the pole has order two. At zero the numerator cancels one order:ThereforeThe numerator's cancellation at zero is why not every denominator zero produces the same pole order.

For $z=a+b\sqrt{d}$, let $\bar{z}=a-b\sqrt{d}$ denote the algebraic conjugate, not necessarily ordinary complex conjugation when $d>0$. Conjugation respects multiplication, and $z\bar{z}=a^2-d{b}^2$. ThereforeIf $z$ is a unit in a ring, choose its inverse $w$ in the same ring. Then $N(z)N(w)=N(1)=1$, and both norms are nonnegative integers, forcing $N(z)=1$. Conversely, if $N(z)=1$, then $z\bar{z}=\pm 1$, so $z^{-1}=\pm \bar{z}$ belongs to the ring. Thus $z$ is a unit in a ring exactly when $N(z)=1$.

For negative $d$, the equation becomes $a^2+|d|b^2=1$. If $d=-1$, its integer solutions give the four units $\pm 1,\pm i$. If $d<-1$, the term $|d|b^2$ forces $b=0$, leaving $a=\pm 1$. HenceFor $d=2$, the element $u=1+\sqrt{2}$ has field norm $|1-2|=1$. Its powers $u^n$, $n\ge 0$, are all units by multiplicative function and are distinct, since their positive real values strictly increase with $n$. Thus $\mathbb{Z}[\sqrt{2}]$ has infinitely many units. This argument concerns the stated integer-coefficient ring; it does not replace it by a larger ring of algebraic integers.

Use the monomial basis $e_k=x^{n-k}{y}^k$, $0\le k\le n$. Direct differentiation giveswith the endpoint terms zero. Repeated application therefore gives $P^{n+1}=Q^{n+1}=0$, while $P^n{e}_n=n!e_0\ne 0$ and $Q^n{e}_0=n!e_n\ne 0$. Each nilpotency index is exactly $n+1$, soNextand hence $(PQ-QP)e_k=(n-2k)e_k$. These $n+1$ real eigenvalues are distinct. A polynomial annihilates this operator exactly when it vanishes at all of them, givingThe commutator is diagonal in the displayed monomial basis. Since $n\ge 1$, $P,Q$ are nonzero nilpotent linear map; a diagonalizable nilpotent map would have a diagonal matrix with all eigenvalues zero and thus be the zero map. Therefore only $PQ-QP$ is diagonalizable. The calculation realizes raising and lowering on binary homogeneous polynomials.

The weight is strictly positive and all its polynomial moments are finite. Thus the given integral is a inner product: for a nonzero polynomial, its square is positive on a nonempty interval, so its weighted integral is strictly positive.

In the finite-dimensional vector space $V_n$, subtract the orthogonal projection onto a finite-dimensional subspace of $x^n$ onto $V_{n-1}$:This is monic, since the subtracted polynomial has lower degree, and is orthogonal to every lower-degree polynomial. If ${\varphi }_n,{\psi }_n$ were two such monic polynomials, their difference would lie in $V_{n-1}$ and be orthogonal to that space. Taking its inner product with itself gives zero, hence the difference vanishes. This proves existence and uniqueness for every $n$, with ${\varphi }_0=1$.

Odd moments vanish because the weight is even; the even moments areConsequently ${\varphi }_1=x$, while a monic quadratic orthogonal to $x$ has no linear term, and orthogonality to one gives its constant term $-4/2=-2$. ThusFinally $(-1{)}^n{\varphi }_n(-x)$ is also monic and orthogonal to every lower-degree polynomial, by the change of variable $x\mapsto -x$. Uniqueness implies ${\varphi }_n(-x)=(-1{)}^n{\varphi }_n(x)$. In particular ${\varphi }_{2007}$ is odd andThis is parity of monic orthogonal polynomials for an even weight.

The characteristic equation of a partial differential equation for $u_{xx}+u_{xy}=0$ is $(dy{)}^2-dxdy=0$. Thus the two characteristic curve families are $\boxed{y=\text{constant},y-x=\text{constant}}$. Set $\xi =y$ and $\eta =y-x$. The chain rule gives ${\partial }_x=-{\partial }_{\eta }$ and ${\partial }_y={\partial }_{\xi }+{\partial }_{\eta }$, reducing the equation toIntegrating in the two independent characteristic coordinates gives $U=F(\xi )+G(\eta )$, or $u(x,y)=F(y)+G(y-x)$. At $x=0$, the value condition gives $F(y)+G(y)=0$, while the derivative condition gives $-G^{\prime }(y)=e^y$. Hence $G(y)=-e^y+C$ and $F(y)=e^y-C$, soDirect differentiation gives $u_x=e^{y-x}$, $u_{xx}=-e^{y-x}$ and $u_{xy}=e^{y-x}$, verifying both the equation and the two data. The additive constants cancel in $u$, so they do not create different solutions.

Write the equation as $Ly=f$ with $L=-d^2/d{x}^2+1$. Homogeneous solutions vanishing at the left and right endpoints are $u_L(x)=\sinh (1+x)$ and $u_R(x)=\sinh (1-x)$. Their Wronskian is $u_L{u}_R^{\prime }-u_L^{\prime }{u}_R=-\sinh 2$. The Dirichlet Green kernel for minus second derivative plus one is thereforeIt vanishes at $x=\pm 1$, is continuous at $x=s$, and its first derivative has jump $-1$. Thus $LG(\cdot ,s)={\delta }_s$ in the distributional derivative sense, with the sign appropriate to $L=-d^2/d{x}^2+1$. Integration against $f$ gives the required boundary-value solution. The kernel is positive and symmetric, as is also apparent in the figure.

For the specified step forcing only $0\le s\le 1$ contributes at $x=0$, givingThe discontinuity in $f$ gives a piecewise twice-differentiable solution with continuous value and first derivative, rather than a jump in $y$ itself.

The fundamental theorem of calculus first gives $y^{\prime }(x)=1+{\int }_0^{x}f(y(s))ds$. Integrating again, or differentiating the proposed expression twice and checking its initial values, yields the Volterra integral equationFor $f(y)=-y$, the initial value problem is solved by $\boxed{y(x)=\sin x}$. Its Picard iteration constructs the sine series directly. The base approximation is $y_0(x)=x$. Ifthen elementary integration givesSubstituting this finite sum in the next update, then shifting its index, proves the same formula with $n+1$ in place of $n$. This is the required mathematical induction. The resulting partial sums converge to $\boxed{\sin x}$ for every real $x$, and uniformly on every bounded interval: the absolute terms on $|x|\le R$ are bounded by the convergent series ${\sum }_k{R}^{2k+1}/(2k+1)!$.

The printed update lists $n=1,2,\dots$, which would leave $y_1$ unspecified if read literally. The usual Picard iteration, and the requested induction from $y_0$, use the update starting at $n=0$; that intended convention is used here.

A holomorphic function with $f^{\prime }(z)\ne 0$ at every point is locally a conformal map. Write $a=z_0$. Since $|a|<1$ and $|z|<1$, $1-\overline{a}z\ne 0$, andThus the specified Möbius transformation is conformal throughout the disk. Expanding the two squared absolute values givesso $f$ maps the open unit disk into itself. Solving $w=(a-z)/(1-\overline{a}z)$ gives $z=(a-w)/(1-\overline{a}w)=f(w)$. Equivalently $f(f(z))=z$. Hence $f$ is its own inverse, proving both injectivity and surjectivity on the disk. Nonvanishing derivative establishes the local angle-preserving property; the explicit inverse supplies the separate global bijection claim.

Apply the residue theorem to $F(z)=(z^2+4{)}^{-2}$ on the upper semicircle of radius $R>2$, traversed counterclockwise. Its sole enclosed pole is the double pole at $2i$, andOn the semicircular arc, $|z^2+4|\ge R^2-4$, so the ML inequality bounds its contribution by $\pi R/(R^2-4{)}^2\to 0$. The real integral is absolutely convergent, since its integrand is $O(|x{|}^{-4})$ at infinity. Taking $R\to \infty$ in the contour identity yields

A closed curve has $\gamma (a)=\gamma (b)$. A simple curve has no repeated points apart from the permitted equality of its two endpoints when it is closed. A smooth curve, in the regular contour-integration convention, has a continuous nonzero tangent ${\gamma }^{\prime }$; piecewise smooth contours are treated by adding the integrals over their smooth pieces. For a continuous integrand, its contour integral is defined byFor the positively oriented circle, ${\gamma }^{\prime }(t)=ir{e}^{it}$ and $(\gamma (t)-z_0{)}^n=r^n{e}^{int}$. Direct parameter integration therefore givesFor $n\ne -1$, an elementary antiderivative is $e^{i(n+1)t}/[i(n+1)]$ and its endpoint values agree because $n$ is an integer. For $n=-1$, the integrand of the parameter integral is the constant $i$. No integral theorem is needed.