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Zorn lemma states that if every chain in a nonempty partially ordered set has an upper bound in that set, then the set has a maximal element. It is equivalent, over the usual axioms of set theory, to the axiom of choice.

The defining relations for a type-2 generating function giveHenceThe exact differential changes the action only by an endpoint term. Thus the transformed variational principle has canonical one-form $P\cdot dQ-Kdt$ and yields Hamilton's equations in $(Q,P)$ with

For a regular Lagrangian, define the canonical momentum $p_i=\partial L/\partial {\dot{q}}_i$ and invert this relation for $\dot{q}(q,p,t)$. Its Legendre transform is the HamiltonianSince $Ldt=p\cdot dq-Hdt$, the action isFor independent variations with $\delta q(t_1)=\delta q(t_2)=0$,Stationarity for arbitrary $\delta p$ and $\delta q$ gives Hamilton's equations $\dot{q}={\partial }_{p}H$ and $\dot{p}=-{\partial }_{q}H$.

Integrating the density, and using ${\lambda }_{+}+{\lambda }_{-}=A$ and ${\lambda }_{+}{\lambda }_{-}=k_1{k}_2$, givesThe same result follows from first-step analysis: the enzyme waits on average $1/k_1$ to bind, and each complex either produces or returns to the initial enzyme state.

Before the first product forms, the system is in either $E_0$ or $C_0$. Product formation is possible only through $C_0\to E_1$, so the first-passage-time density is the outgoing probability fluxEquivalently, the survival function is $E_0(\tau )+C_0(\tau )$ and its negative derivative is this density.

For zero products, put $A=k_1+k_1^{\prime }+k_2$ andThe initial conditions are $E_0(0)=1$ and $C_0(0)=0$. Eliminating $E_0$ givesThereforeand using $E_0=({\dot{C}}_0+(k_1^{\prime }+k_2)C_0)/k_1$ gives

Write $E_n(t)=P(E,n)$ and $C_n(t)=P(ES,n)$. The master equation balances the transitions $E_n\to C_n$ at rate $k_1$, $C_n\to E_n$ at rate $k_1^{\prime }$, and $C_n\to E_{n+1}$ at rate $k_2$:for $n\ge 0$, with $C_{-1}=0$.

A generalized linear model assumes independent responses from one exponential dispersion family,where ${\sigma }_i^2=a_i{\sigma }^2$, ${\mu }_i=\mathbb{E}[Y_i]=b^{\prime }({\theta }_i)$, and a link function relates the mean to the linear predictor by $g(\mu )=X\beta$. It follows from normalization of the density after shifting the natural parameter thatIndependence therefore gives the joint moment-generating functionwhenever every shifted natural parameter lies in its natural-parameter domain.

Liouville approximation theorem states that if $\alpha$ is a real algebraic number of degree $d\ge 2$, then there is a constant $c(\alpha )>0$ such thatfor every rational $p/q$ with $q>0$. The strict inequality may be replaced by a non-strict one after changing the constant.

If $n={\prod }_{p^{\alpha }\parallel n}{p}^{\alpha }$, multiplicativity and the prime-power formula for the divisor function giveChoose $P_0=2^{1/ϵ}$. For $p\ge P_0$ and $\alpha \ge 1$, the elementary bound $\alpha +1\le 2^{\alpha }$ givesFor each of the finitely many primes $p<P_0$, the sequence $(\alpha +1)p^{-\alpha ϵ}$ tends to zero and therefore has a finite maximum $M_p$ over $\alpha \ge 0$. Consequentlyuniformly in $n$, which proves $\tau (n)\le C(ϵ)n^{ϵ}$.

Let $P=p_1{p}_2\cdots p_{k+1}$ be the product of the first $k+1$ primes, and take $n=P^a$ for positive integers $a$. The prime factorization givesTheir ratio isThus every sufficiently large $a$ gives $\tau (P^a)\ge (\log P^a{)}^k$, producing infinitely many such $n$.

Take both functions to be the constant arithmetic function $1$. Thenso the divisor function is multiplicative by part (a). For $s$ with $\mathrm{Re}s>1$, the relevant Dirichlet series are absolutely convergent, so their Cauchy product may be rearranged:Equivalently, Dirichlet convolution becomes multiplication of absolutely convergent Dirichlet series.

An arithmetic function $f$ is multiplicative when $f(1)=1$ and $f(mn)=f(m)f(n)$ whenever $m$ and $n$ are coprime. If $m$ and $n$ are coprime, every divisor $d$ of $mn$ has a unique factorization $d=d_1{d}_2$ with $d_1\mid m$ and $d_2\mid n$. Hence the Dirichlet convolution of multiplicative functions satisfiesand $(f\star g)(1)=1$, so it is multiplicative.

Let $\epsilon$ be the Identity for Dirichlet convolution, with $\epsilon (1)=1$ and $\epsilon (n)=0$ for $n>1$. The Möbius function satisfies $\mu \star 1=\epsilon$. Therefore, using commutativity and associativity,This is the Möbius inversion formula.