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Suppose $I$ is prime. Ifthen $ab\in I$, so $a\in I$ or $b\in I$. Thus one factor is zero and $R/I$ is an integral domain.

Conversely, if $R/I$ is an integral domain and $ab\in I$, thenso $a+I=0+I$ or $b+I=0+I$. Therefore $a\in I$ or $b\in I$, and $I$ is prime. This proves the prime ideal quotient criterion.

Reduce coefficients modulo $I$:This is a surjective ring homomorphism and its kernel consists exactly of polynomials all of whose coefficients lie in $I$, namely $I[X]$. The first isomorphism theorem gives the coefficientwise quotient of a polynomial ringIf $I$ is prime, then $R/I$ is an integral domain, so $(R/I)[X]$ is an integral domain. The quotient criterion therefore shows that $I[X]$ is prime in $R[X]$.

If the bound is to hold with finite $c$, the error functional must vanish on every polynomial whose fourth derivative is zero. Thus the scheme must be exact for degrees zero through three. Applying it to powers of $t=x+1$ givesSolving,This is the four-point one-sided second-derivative formula.

The Peano kernel theorem says that if a linear functional $L$ annihilates all polynomials of degree below $r$, then for $f\in C^r[a,b]$,

Here $r=4$ and $L=e$. Under the stated nonnegativity assumption,The integral equals $e(q)$ for $q(x)=(x+1{)}^4/24$, since $q^{(4)}=1$. Now $q^{\prime \prime }(-1)=0$ andThereforeEquality is attained by $q$, so the sharp Peano-kernel constant for the four-point endpoint second derivative is

The Cauchy integral theorem states that if $f$ is holomorphic on a simply connected domain, then its integral around every closed piecewise smooth contour in that domain is zero.

Put $q=b-k$. Completing the square givesAfter the change of variable $z=x-iq/(2a)$, the integral runs along the horizontal line $\mathrm{Im}z=-q/(2a)$. The integrand $e^{-a{z}^2}$ is entire. Apply Cauchy's theorem to a rectangle joining this line to the real axis; the two vertical integrals tend to zero as their real parts tend to $\pm \infty$, because $a>0$. The contour may therefore be shifted to the real line. Hence the Fourier transform of a Gaussian gives

Normalizability requires $\mathrm{Re}B>0$. Sincethe given Gaussian integral givesAlso ${\psi }_x=-2Bx\psi$, so integration by parts yieldsThereforeThese are the second moments of a complex Gaussian wave packet.

Newton's method for minimization usesFor a quantitative local bound, suppose on a convex neighbourhood containing the iterates thatwith $m>0$, and let $x^{\ast }$ be the minimizer. The integral form of the gradient and the Hessian Lipschitz bound give the quadratic convergence bound for Newton's methodIf $q=M\Vert x_0-x^{\ast }\Vert /(2m)<1$, induction yieldsSince $f(x)-f(x^{\ast })\le L\Vert x-x^{\ast }{\Vert }^2/2$,

A function $f:U\to \mathbb{C}$ is holomorphic when it is complex differentiable at every point of the open connected set $U$. Morera's theorem states that a continuous function on a domain is holomorphic if its integral around every triangle whose interior lies in the domain is zero.

The integrand $e^{tz}/(1+t^2)$ is continuous jointly in $(t,z)$ on $[0,1]\times \mathbb{C}$, so the displayed integral defines a continuous function. For any triangle $T$, Fubini's theorem and the Cauchy integral theorem giveMorera's theorem therefore proves that $f$ is entire.

The function $1/z$ is holomorphic on $\mathbb{C}\setminus \{0\}$ but has no antiderivative there, because an antiderivative would integrate to zero around every closed curve whereasThis is the standard period obstruction to a holomorphic antiderivative.

Define the two shell integralsFor an external ${\varphi }_1$, the ${\varphi }_1^4$ part of the interaction gives the scalar tadpole coefficient $12g_0{I}_1$, while $2g_0{\varphi }_1^2{\varphi }_2^2$ gives $4g_0{I}_2$. Interchanging the components givesWhen the masses agree, $I_1=I_2$ and both formulas reduce to $4(N+2)g_0{I}_1$ with $N=2$.

Let the source's hydrogen-ionizing photon production rate bewhere $h{\nu }_0$ is the hydrogen ionization energy. The idealized Strömgren sphere has an almost fully ionized interior and a thin ionization front. In photoionization equilibrium, every ionization is balanced by a Case B recombination, so spherical symmetry givesFor pure hydrogen of constant number density $n$, the interior has $n_e=n_p=n$, and the Strömgren radius is therefore

Now let the effective number of dust grains per hydrogen nucleus be $f_d$, so the dust absorption coefficient is $a(r)=f_d{\sigma }_{d}n(r)$. If $Q(r)$ is the ionizing-photon rate crossing the sphere of radius $r$, recombinations and dust absorption give the linear ordinary differential equationIf $f_d$ denotes a dust mass fraction instead, the grain mass and gas mean particle mass are simply absorbed into the effective product $f_d{\sigma }_d$. Define the dust optical depthMultiplication by the integrating factor $e^{{\tau }_d(r)}$ and integration to the dusty front gives its governing equationThe exponential function weights recombinations at large optical depth by the extra source photons that dust must remove before those photons reach that radius.

For constant $n$, put $a=f_d{\sigma }_{d}n$ and $\tau =a{R}_d$. The elementary integralreduces the equation toEquivalently, with $y=R_d/R_S$ and ${\tau }_S=a{R}_S$,Its dust-free limit is $y\to 1$, while absorption makes $R_d<R_S$ for nonzero dust abundance.

An algebraic number is an algebraic integer when it is a root of a monic polynomial in $\mathbb{Z}[X]$.

If $\alpha$ and $\beta$ are algebraic integers, the ring $\mathbb{Z}[\alpha ,\beta ]$ is a finitely generated $\mathbb{Z}$-module. Multiplication by $\alpha \beta$ preserves this module, so the algebraic integer module criterion shows that $\alpha \beta$ is an algebraic integer.

The ring of integers of a quadratic field givessince $2\equiv 2(\mathrm{mod}4)$. For a direct verification, if $x=a+b\sqrt{2}$ with $a,b\in \mathbb{Q}$ is integral, then its trace $2a$ and norm $a^2-2b^2$ are integers. Writing $2a=m\in \mathbb{Z}$ and reducing the norm condition in lowest terms shows first that $b$ can have denominator at most $2$; a denominator $2$ with odd numerator would require $m^2\equiv 2(\mathrm{mod}4)$, which is impossible. Thus $b\in \mathbb{Z}$, and then the norm condition forces $m$ even, so $a\in \mathbb{Z}$. Conversely, every $a+b\sqrt{2}$ with $a,b\in \mathbb{Z}$ is integral because $\sqrt{2}$ is integral and algebraic integers form a ring.

Now let $f$ have degree $d$ and roots ${\alpha }_1,\dots ,{\alpha }_d$. Sinceand $|{\alpha }_1|>1$, the integer $|a_d|$ cannot be $2$; hence $|a_d|=1$. Thus $f$ is monic up to sign and ${\alpha }_1$ is an algebraic integer.

Let $B$ be the product of all roots with modulus greater than one, counting multiplicity. Complex roots occur in conjugate pairs, so $B$ is real up to the signs of the real roots andAs a product of algebraic integers, $B$ is an algebraic integer; the equality forces $B=\pm 2$. Thereforeis the absolute value of the product of the remaining conjugates, is rational, and is an algebraic integer. It is consequently a nonzero rational integer. Every remaining factor has modulus at most one, so this integer is at most one. It must equal one, and henceThis is the mahler-measure norm argument at measure two.

For $f_z(y)=f(yz)$, the substitution $u=yz$ givesHomogeneity of $K$ and the substitution $x=yz$ giveApply part (a), or equivalently the Minkowski integral inequality, to this integral and use the preceding scaling identity:

Conditioned on the jump-chain states, the successive holding times are independent exponentials with rates $q_{i_0},\dots ,q_{i_n}$. Henceequals the integral over $s_0,\dots ,s_{n-1}\ge 0$ with ${\sum }_{r<n}{s}_r\le t$ ofThis simplex integral is unchanged by reversing the $n+1$ time portions and the rates attached to them. It is thereforeas required. This is holding-time reversal along a fixed CTMC jump path.

The Markov property at deterministic times givesThus $(Z_n)$ is a discrete-time Markov chain with transition matrixwhen the minimal chain is nonexplosive, $P(h)=e^{hQ}$.

For an irreducible chain, state $i$ is recurrent in continuous time exactly whenand it is recurrent for the skeleton exactly when ${\sum }_{n\ge 0}{p}_{ii}(nh)=\infty$. If $t\in [nh,(n+1)h]$, the event of staying at $i$ suppliesIntegrating over each interval shows that the integral diverges exactly when the series does. Irreducibility then makes recurrence equivalent for the two chains. This is recurrence equivalence for a CTMC and its fixed-time skeleton.

On the finite state space, $P_t=e^{tQ}$ andThereforeThe matrix semigroup satisfies $\frac{d}{dt}{P}_{t}f=P_{t}Qf$. Integrating from $0$ to $t$ gives the backward integral equation

The classical one-particle phase-space integral isThe momentum integral isHence

For singularities $a<c_1<\cdots <c_N<b$, the Cauchy principal value iswhere the assumed tail integrals make the two unbounded pieces meaningful. Equivalently, remove a symmetric interval of radius $\epsilon$ about each $c_j$ and take the limit of the remaining integral.

Fix $w\in \mathbb{C}$ and apply the stated clockwise argument principle to $h=f-w$. Its logarithmic derivative is periodic, so the integrals over opposite sides of a fundamental cell cancel. Hence $P-Z=0$. The poles of $f-w$ are precisely the poles of $f$, with unchanged multiplicities, while its zeros are the solutions of $f(z)=w$. Thus the value multiplicity of an elliptic function is the same fixed pole count for every finite $w$, counting multiplicity.

A reduced positive definite binary quadratic form $[a,b,c]$ satisfieswith $b\ge 0$ when either $|b|=a$ or $a=c$. For $d\equiv 0$ or $3(\mathrm{mod}4)$, the class number of a negative discriminant $h(-d)$ is the number of proper equivalence classes of primitive positive definite integral forms of discriminant $-d$, equivalently the number of their reduced representatives.

WritePartitioning the $k$ complete prime-power factors between $r$ and $s$ gives $2^k$ ordered factorizations $d=rs$ with $\gcd (r,s)=1$. After identifying $(r,s)$ with $(s,r)$, there are $2^{k-1}$ choices with $r\le s$. Each gives the primitive reduced formwhose discriminant is $-4rs=-4d$. The uniqueness of reduced representatives makes these classes distinct, proving the coprime-factorization lower bound for a quadratic-form class number

The inequality can be strict. For $d=5$, there is one distinct prime factor, while the reduced primitive forms of discriminant $-20$ areThus $h(-20)=2>1=2^{1-1}$.

The Dirichlet unit theorem states that for a number field of signature $(r_1,r_2)$,

For $K=\mathbb{Q}(\sqrt{5})$,The field is real quadratic, so the unit rank is one and its only roots of unity are $\pm 1$. To see that $\varphi$ is fundamental, suppose a positive unit satisfies $1<u<\varphi$. If its norm is $1$, its integral trace $u+u^{-1}$ lies strictly between $2$ and $\sqrt{5}$; if its norm is $-1$, its trace $u-u^{-1}$ lies strictly between $0$ and $1$. Neither interval contains an integer. Reducing any positive unit by a suitable power of $\varphi$ now proves the units of the quadratic field Q square root of five formula

If ${\mathcal{O}}_L^{\times }/{\mathcal{O}}_K^{\times }$ is finite, the two unit groups have the same rank, namely one. Put $d=[L:K]$ and let $(r_1,r_2)$ be the signature of $L$. ThenTherefore $r_1=4-2d$ and $r_2=2d-2$. For a proper extension, nonnegativity forceswith signature $(0,2)$.

This degree occurs: take $L=K(i)$. It is a totally imaginary quadratic extension of $K$, so the unit ranks agree and the quotient is finite. It is nontrivial because $i\in {\mathcal{O}}_L^{\times }$ but $i\notin {\mathcal{O}}_K^{\times }$ (and its coset has order two). This is the finite relative unit quotient over a real quadratic field example.

The Hahn-Banach theorem says that if $Y$ is a linear subspace of a real normed space $X$ and $g:Y\to \mathbb{R}$ is bounded and linear, then there is a bounded linear extension $G:X\to \mathbb{R}$ with $\Vert G\Vert =\Vert g\Vert$.

For $x\in X$, the map $\hat{x}(f)=f(x)$ is linear on $X^{\prime }$ andso $\hat{x}\in X^{\prime \prime }$ and $\Vert \hat{x}\Vert \le \Vert x\Vert$. The assignment $Jx=\hat{x}$ is visibly linear. If $x\ne 0$, define $g$ on $\mathrm{span}\{x\}$ byIt has norm one, and Hahn--Banach extends it to an $f\in X^{\prime }$ of norm one with $f(x)=\Vert x\Vert$. ThereforeThus the canonical embedding into the bidual is an isometry and

On $C([0,1])$, point evaluation $\ell (f)=f(0)$ has norm one for the $L^{\infty }$ norm, because a continuous function's essential supremum equals its supremum. Hahn--Banach extends it to $\Lambda \in (L^{\infty }{)}^{\prime }$. If $\Lambda$ were represented by some $g\in L^1$, thenfor every continuous $f$. Choose continuous $0\le f_n\le 1$ with $f_n(0)=1$ and support in $[0,1/n]$. Absolute continuity of the integral makes the right side tend to zero, while the left side is always one, a contradiction. This singular functional on L infinity proves