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If all members of a set family on $[n]$ have odd size and all distinct pairwise intersections have even size, its characteristic vectors over the finite field ${\mathbb{F}}_2$ are linearly independent. Their pairwise dot products are zero, and each vector has dot product one with itself. Taking the dot product of a linear combination with each vector makes every coefficient vanish, so there are at most $n$ vectors. The set family of all singleton sets attains this bound.

If every member of a set family on $[n]$ and every distinct pairwise intersection has even size, its characteristic vectors span a self-orthogonal binary subspace. The standard dot product is nondegenerate, so $W\subseteq W^{\perp }$ implies $2\dim W\le n$. Thus the set family has at most $2^{\lfloor n/2\rfloor }$ members. All unions of $\lfloor n/2\rfloor$ disjoint pairs of points attain this bound.

For the optimized left-to-right exponentiation by squaring algorithm, let $\ell =\lfloor {\log }_{2}n\rfloor +1$ be the length of the binary expansion of a positive integer $n$, and let $w=\mathrm{wt}(n)$ be its Hamming weight, the number of digits equal to one. There are exactly $\ell -1$ digits after the leading one, and exactly $w-1$ remaining ones. Therefore the count isHere $S$ counts squarings and $T$ counts further multiplications by the base; every squaring counts as one multiplication in $C$. Assignment, digit reading, and any inversion are outside this count. The formula includes $n=1$, which requires zero multiplications; define $C(0)=0$ separately.

For a fixed length $\ell$, the bounds are $\ell -1\le C(n)\le 2\ell -2$. A power of two reaches the lower bound because its Hamming weight is one; $n=2^{\ell }-1$ reaches the upper bound because all digits are one. For $n=13$, $\ell =4$ and $w=3$, giving $C(13)=5$. Thus the arithmetic operation count is $O(\log (n+1))$ in Big O notation, but inputs of similar magnitude can have different counts.

This is a count for the specified algorithm, not a claim that it always minimizes multiplications among all methods. For instance, at $n=15$ the count is six, but the powers with exponents $1,2,3,6,12,15$ can be built in five multiplications: square, multiply by $a$, square, square, then multiply by the already stored $a^3$. Saving operations this way uses a different schedule and stored intermediate powers.

A stretched exponential has the form $e^{-c{x}^{\alpha }}$ with $c>0$ and $0<\alpha <1$. A boundary cost proportional to the square root of an enclosed area naturally produces exponent $\alpha =1/2$.

Methods of detecting exoplanets infer or resolve an exoplanet through its effects on the host star's light or through its own light. The observable determines what can be learned:

An exoplanet secondary eclipse measures the loss of the exoplanet's light when it passes behind its star, complementing exoplanet transit photometry. A detection method also selects which systems are easiest to find: deeper exoplanet transits favour larger radius ratios, and larger stellar radial-velocity semi-amplitudes favour more massive companions, shorter orbital periods, and larger $\sin i$. These preferences must be accounted for when inferring the underlying exoplanet population.

In an economy with a measure space of consumers, a consumption allocation is a measurable, integrable function assigning each consumer a feasible bundle. Exact aggregate feasibility means ${\int }_{A}fd\mu ={\int }_A\omega d\mu$.

A demand correspondence assigns a nonempty set of chosen consumption bundles to each price and wealth pair. Every chosen bundle satisfies the budget constraint. A utility function rationalizes the demand correspondence when the chosen bundles are exactly its maximizers over the budget set.

A price quasi-equilibrium with transfers is an exactly feasible consumption allocation $f$ and a nonzero price vector $p$ for which, almost everywhere, every strictly preferred bundle has cost at least $p\cdot f(a)$. The transfer-adjusted wealth is $p\cdot f(a)$; transfers $p\cdot (f(a)-\omega (a))$ sum to zero. Equal-cost strict improvements may remain, which distinguishes a quasi-equilibrium from a utility-maximizing equilibrium. Nonnegative prices need extra assumptions if goods may be undesirable.

Aggregate excess demand subtracts total endowment from the sum of consumers' demand correspondences. A market-clearing price satisfies $z(p)=0$. Under Walras law, one equation is redundant after price normalization.

An exchange economy is regular when the derivative of its reduced aggregate excess demand has full rank at every equilibrium. For two goods and normalized prices $(p,1)$, this means $Z_p(p)\ne 0$ at every zero of first-good excess demand. Regular equilibria are locally isolated and persist smoothly under small parameter changes.

Write $n=\prod p^{a_p}$ and $d=\prod p^{b_p}$, with $0\le b_p\le a_p$. The prime-power formula for the Euler totient function gives an integer contribution to $\phi (n)/\phi (d)$ from each prime: it is $p^{a_p-b_p}$ when $b_p>0$, and $p^{a_p-1}(p-1)$ when $b_p=0<a_p$. Multiplying proves the divisibility assertion, including $d=1$ with $\phi (1)=1$.

If $n>1$ is odd, its Euler totient function is even and cannot divide $n$. Otherwise write $n=2^c{\prod }_{j=1}^r{p}_j^{a_j}$ with odd primes $p_j$ and $c\ge 1$. Comparing powers of two in $\phi (n)\mid n$ gives ${\sum }_j{v}_2(p_j-1)\le 1$. Thus there is at most one odd prime $p$, with $v_2(p-1)=1$. Since $p-1$ divides $\phi (n)$, it divides $n$, and is coprime to $p$; with only the prime factors $2,p$ available it must be a power of two. Hence $p-1=2$ and $p=3$. Conversely, for $c\ge 1$ the prime-power totient formula verifies divisibility for $2^c{3}^d$, and $n=1$ is immediate.

The map $j\mapsto n-j$ is a bijection of the integers between one and $n-1$ that are coprime to $n$. If their sum is $S$, this bijection gives $S=n\phi (n)-S$, so $S=n\phi (n)/2$. This works even at $n=2$, where the single residue is fixed. The same identity follows from the weighted inclusion-exclusion principle by excluding multiples of the prime factors of $n$.

The divisor sum of Euler's totient function is the identity arithmetic function:Equivalently, $\phi \ast 1=\mathrm{id}$ under Dirichlet convolution, with $1$ the constant-one arithmetic function. ^[ref]

To see why, group the integers $j=1,\dots ,n$ by $h=\gcd (j,n)$. Each $h$ is a positive integer divisor of $n$. Within its group, $j=ha$ with $1\le a\le n/h$ and $\gcd (a,n/h)=1$, by the scaling identity for greatest common divisors. Dividing by $h$ gives a bijection with the integers counted by $\phi (n/h)$. The groups form a partition of a set, soFor $n=12$, the groups with greatest common divisor $h=1,2,3,4,6,12$ are respectively $\{1,5,7,11\},\{2,10\},\{3,9\},\{4,8\},\{6\},\{12\}$. Their sizes are $4,2,2,2,1,1$, totaling $12$. Möbius inversion reverses this divisor sum to recover Euler's totient function from $g(n)=n$.

For a unit modulo n $a$ and a nonnegative integer $k$, division with remainder writes $k=q\phi (n)+r$ with $0\le r<\phi (n)$. Euler's theorem then givesThis replaces a potentially enormous exponent by a small remainder. The exponent itself is reduced modulo Euler's totient function, while the powers are reduced modulo $n$: these are different integer moduli.

For a worked modulo-fifteen calculation, seven is coprime to fifteen and $\phi (15)=8$. ThereforeUse exponentiation by squaring on the remaining exponent $5=(101{)}_2$, reducing after each multiplication:Thus $7^{2029}\mathrm{mod}15=7$. Modular exponentiation supplies the computation; Euler's theorem supplies the optional shortening of its exponent. Exponentiation by squaring also works for nonunit bases, without using this shortening.

The guaranteed exponent $\phi (n)$ is a common return exponent and an upper bound on the shortest period; it need not be that period. The multiplicative order $d={\mathrm{ord}}_n(a)$ is the least positive return exponent for the chosen unit. The Carmichael function $\lambda (n)$ is the least positive return exponent that works for every unit. Their precise relation isEither smaller return exponent can replace $\phi (n)$ in exponent reduction for units. Modulo fifteen, ${\mathrm{ord}}_{15}(7)=\lambda (15)=4<8=\phi (15)$, so $2029=507\cdot 4+1$ gives $7^{2029}\equiv 7$ immediately. For base four the multiplicative order is only two, smaller even than $\lambda (15)$. A return at exponent eight alone does not prove that the multiplicative order is eight.^[ref]

If the reduced exponent is zero, the rule gives $a^k\equiv a^0=1$, and still requires a unit. Applying it to the nonunit three would falsely predict $3^8\equiv 3^0=1(\mathrm{mod}15)$; the actual residue is six. For negative exponents, first interpret the power using a modular multiplicative inverse, as in modular exponentiation.

Choose a reduced residue system $u_1,\dots ,u_{\phi (n)}$. For a unit modulo n $a$, the bijection $u\mapsto au$ on the unit group has inverse $u\mapsto a^{-1}u$. Comparing products before and after this permutation givesThe product $P$ is a unit; its modular inverse cancels it, proving Euler's theorem. Merely being nonzero would not justify cancellation.^[ref]

In the modulo-twelve example, $P=1\cdot 5\cdot 7\cdot 11=385\equiv 1(\mathrm{mod}12)$. The permuted representatives $5,1,11,7$ have the same product, while the unreduced products $(5\cdot 1)(5\cdot 5)(5\cdot 7)(5\cdot 11)$ equal $5^{4}P$. Equating these products modulo twelve exposes the fourth power without calculating it directly. Using all nonzero congruence classes instead would include nonunits, whose product need not permit cancellation.

The Bernoulli principle relates pressure, kinetic energy per unit volume and Newtonian gravitational potential along a streamline of steady flow with inviscid flow. For constant mass density, the Bernoulli equation gives $p/\rho +u^2/2+\Phi$ constant along that streamline. Constancy across different streamlines requires further hypotheses, such as irrotational flow.

Precisely, suppose a fluid of constant positive mass density obeys the Euler equations for an inviscid fluid, its velocity field and pressure are steady, and the body force per unit mass is $-\nabla \Phi$. ThenEvery term has units of energy per unit mass. For a uniform gravitational field, Newtonian gravitational potential is $\Phi =gz$, up to an additive constant. The pressure term represents work by surrounding fluid, so acceleration can exchange pressure work with kinetic energy. Dotting the steady Euler equations with $\mathbf{u}$ gives the displayed directional derivative: this is why constancy follows a streamline. ^[ref]

Continue the horizontal water constriction example from incompressible flow. Assume negligible viscous losses, steady flow and approximately uniform flow at both measurement sections, so the streamline speeds there equal the mean speeds $0.20$ and $0.80\mathrm{m}{\mathrm{s}}^{-1}$. Equal height cancels the Newtonian gravitational potential terms. The ideal pressure drop isThis is the Venturi effect in its ideal limit: the gain in kinetic energy density is supplied by pressure work. Cambridge's treatment uses this relation to connect a measured pressure difference to volumetric flow rate. ^[ref]

The $300\mathrm{Pa}$ result is conditional, rather than a prediction of the actual viscous pipe. Mass conservation fixes $Q$ and the ratio of mean speeds even with viscosity, but nonuniform velocity profiles require an integrated energy balance, and wall friction dissipates mechanical energy. A pump would supply additional work. In particular, constant mean speed in a straight pipe does not imply constant pressure: Poiseuille flow needs a continuing pressure gradient to overcome viscosity. ^[ref]

For a fixed-composition inviscid fluid, specific enthalpy satisfies $dw=Tds+dp/\rho$. Combining this identity with the vorticity cross-product identity gives the displayed equation, with $\mathbf{\omega }=\nabla \times \mathbf{u}$ and $\epsilon =u^2/2+\Phi +w$. It exposes entropy gradients as the thermodynamic source of a mismatch between Bernoulli function gradients and the vorticity term. In a steady adiabatic process, its projection along a streamline makes $\epsilon$ constant there.

For an unmagnetized ideal gas without gravity, the pressure equation gives internal-energy balance ${\partial }_{t}e+{\partial }_x(u_{x}e)=-p{\partial }_x{u}_x$, where $e=p/(\gamma -1)$. The momentum equation gives kinetic-energy balance with source $-u_x{\partial }_{x}p$. Their sum is conservative with the displayed total-energy flux. Tangential velocities contribute to kinetic energy even when all derivatives are in the normal direction. Smooth polytropic flow advects its entropy parameter; a physical shock need not preserve that parameter across the discontinuity.

The linearized Euler equations retain only first-order perturbations about a specified inviscid fluid state. Together with the linearized continuity equation and a pressure-density constitutive relation, they describe small-amplitude acoustic waves. About a uniform resting state, ${\rho }_0{\partial }_t{\mathbf{u}}^{\prime }=-\nabla p^{\prime }$, ${\partial }_t{\rho }^{\prime }+{\rho }_0\nabla \cdot {\mathbf{u}}^{\prime }=0$, and an isentropic pressure perturbation has $p^{\prime }=c_0^2{\rho }^{\prime }$. Gradients and advection of a moving background must be retained when its variation is not negligible.