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Two integral binary quadratic forms are properly equivalent when one is obtained from the other by a change of variables in $S{L}_2(\mathbb{Z})$. For a negative discriminant $d\equiv 0,1(\mathrm{mod}4)$, the class number of binary quadratic forms $h(d)$ is the number of proper equivalence classes of primitive positive-definite forms of discriminant $d$.

If $f$ properly represents $m$, then $f(r,s)=m$ for coprime $r,s$. Extend $(r,s{)}^T$ to a matrix in $S{L}_2(\mathbb{Z})$; after the associated variable change, the coefficient of $x^2$ is $m$, so the transformed form isThe converse follows by evaluating this form at $(1,0)$.

Such a form has discriminant $d$ precisely whenso it exists exactly when $b^2\equiv d(\mathrm{mod}4m)$. Thus $m$ is properly represented by some form of discriminant $d$ exactly when $d$ is a square modulo $4m$.

Now let $d=1-4A$. If $n^2+n+A$ is composite for $0\le n\le A-2$, it has a prime divisor $p<A$, andConversely, if $d$ is a square modulo $4p$ for a prime $p<A$, choose an odd square root $2n+1$ with $0\le n<p$. Then $p\mid n^2+n+A$ and $n\le A-2$, so that value is composite.

Finally, reduction theory says that every nonprincipal positive-definite class of discriminant $1-4A$ has a reduced representative whose leading coefficient has a prime divisor $p<A$ for which $d$ is a square modulo $4p$. Conversely such a prime produces a represented form not equivalent to the principal form $x^2+xy+A{y}^2$. Combining this with the preceding equivalence gives

Without Alice's operation, Bob obtains outcome $b$ with probabilityIf Alice applies $U$ and then measures, but her outcome is not communicated, Bob's probability isbecause ${\sum }_a{\Pi }_a=I_A$. Equivalently, a trace-preserving local operation leaves Bob's reduced density matrix unchanged. This is the no-communication theorem: local operations cannot signal without communication of their outcomes.

Let $R$ be the center of mass and $g$ the uniform gravitational acceleration. The gravitational torque about $R$ is

Choose body-fixed principal axes of inertia. The inertial derivative of the angular momentum isFor torque-free motion $N=0$ and $L_i=I_i{\omega }_i$, giving the Euler equations for a torque-free rigid body

For a symmetric top, let $I_1=I_2=I$ and let $I_3$ be the axial moment. Then ${\omega }_3$ is constant andThus ${\omega }_1^2+{\omega }_2^2$ is constant and the transverse angular-velocity vector rotates uniformly about the body symmetry axis. Its rate is $|\Omega |$; it precesses in the positive axial sense when $(I_3-I){\omega }_3>0$ and in the opposite sense when this product is negative.

Put $t=1/z$ and $y(t)=w(1/t)$. Sincethe transformed equation is

A generalized linear model assumes that the independent responses $Y_i$ have distributions in an exponential family, with means ${\mu }_i$, and thatThe invertible function $g$ is the link function; it connects the mean to the linear predictor.

For binomial regression, write $Y_i\sim \mathrm{Bin}(m_i,p_i)$ independently, so ${\mu }_i=m_i{p}_i$. The logistic regression link iswhile probit regression useswhere $\Phi$ is the standard-normal distribution function. The logit is the canonical link function for the binomial family.

For a deterministic finite automaton the extended transition function of a deterministic finite automaton is recursivelyThe accepted language isFor a nondeterministic finite automaton, $\hat{\delta }(q,w)$ is the set of all states reachable from $q$ while reading $w$; recursively,Thus

The powerset construction has state set $\mathcal{P}(Q)$, initial state $\{q_0\}$, transitionand accepting states $\{S\subseteq Q:S\cap F\ne \varnothing \}$. Induction on $|w|$ givesso the two automata accept exactly the same language. If $|Q|=m$ and $F$ has one state, exactly $2^{m-1}$ of the $2^m$ subset states contain it and are accepting, including states that may be unreachable.