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For the chain with $0$ absorbing, equal-probability nearest-neighbour moves between $1,2,3$, and a probability-$1/2$ self-loop at $3$, starting from $1$ the expected absorption time is $6$, the probability of visiting $3$ before $0$ is $1/3$, and the expected number of visits to $3$ before absorption is $2$.

If extinction scales as $\sigma \propto {\lambda }^{\gamma }$, an isothermal hydrostatic atmosphere has $d{R}_p/d\log \lambda =\gamma H$. The measured continuum slope can therefore estimate atmospheric temperature when gravity and mean molecular mass are known.

This is the atmospheric instance of a scale height.

For an isothermal ideal-gas atmosphere, $H=k_{B}T/(\mu m_{H}g)$. A strong transmission feature spanning $N_H$ scale heights has approximate transit-depth amplitude $2R_p{N}_{H}H/R_{\ast }^2$.

For blackbody star and planet, define the planet-star contrastA complete exoplanet secondary eclipse removes the planet's light from the combined out-of-eclipse light. Its exact normalized depth is $d_{\lambda }=r_{\lambda }/(1+r_{\lambda })$. The common approximation $d_{\lambda }\simeq r_{\lambda }$ requires $r_{\lambda }\ll 1$. This thermal model omits reflected light and spatial variations.

Geometric albedo compares a body's full-phase brightness with that of a flat, perfectly diffusing reference disc of the same cross-section.

Suppose a function on ${\mathbb{Z}}_N$ has least period $r\mid N$ and is injective on each period. A Fourier sample $c=sN/r$ gives the candidate denominator $q=r/\gcd (s,r)$. If equality of function values can be tested efficiently, then $q$ is the full period exactly when $f(q)=f(0)$, so each successful run is certified. A sample succeeds with probability $\phi (r)/r$, and an inverse-polylogarithmic lower bound for this ratio permits amplification to constant success probability with polynomially many repetitions.

For the smooth curve $y^2=f(x)$ with square-free $f$ of degree $2n$, the differential $dx/y$ has divisorThus the canonical degree is $2n-4$, the genus is $n-1$, and $L(K)$ has basis $1,x,\dots ,x^{n-2}$.

The affine germ surface over the plane with the eight roots of unity removed is compactified by adding the eight simple branch points above those roots and two unbranched points above infinity. The degree-two map to the Riemann sphere has total ramification eight, so the Riemann-Hurwitz formula gives genus three.

A forward-start call fixes its strike at a future time as a multiple of the stock price then. In a homogeneous binomial model, its time-zero price reduces to that of an ordinary call over the remaining periods.

Conditioning on the first binomial move expresses a call with $N+1$ periods as a positive weighted sum of two $N$-period calls with rescaled strikes.

Put-call parity fixes the price difference between a European call option and a European put option on the same underlying stock, with the same strike price $K$ and expiry $T$. Assume no dividends or other payments from the stock before or at expiry, a known constant continuously compounded interest rate $r$, and a frictionless market. Borrowing and lending are available at that same interest rate; short selling of the stock, European call option, European put option, and zero-coupon bond is permitted. Settlement is certain, and the buy-and-hold portfolios below may be held until $T$ without interim funding restrictions or forced liquidation. These assumptions exclude transaction costs, taxes, borrowing fees, and execution or settlement risks from the arbitrage argument.

At expiry, the two positive parts cancel their corners:Thus a European call option and a short European put option together have the same terminal financial payoff as one stock share with a debt of $K$. A particularly readable claim replication uses two portfolios with nonnegative terminal amounts: portfolio A holds one European call option and a risk-free zero-coupon bond paying $K$ at $T$; portfolio B holds one European put option and one stock share. The zero-coupon bond costs $K{e}^{-rT}$ today, since the discount factor is $e^{-rT}$. There are no intermediate cash flows. For every terminal stock price,For the strike price $K=100$, even the following very different outcomes leave the two portfolios equal:

The law of one price forces $A_0=B_0$ in this setting: otherwise buy the cheaper portfolio and sell the more expensive portfolio using short selling. Their terminal obligations cancel in every scenario, while the initial price difference can grow in the bank account. Hence absence of arbitrage givesWith a constant per-period interest rate $i$ over $N$ discrete periods, the same reasoning gives $C_0-P_0=S_0-K(1+i{)}^{-N}$. This is a different compounding convention; $i$ is not the continuously compounded interest rate $r$.

For an illustrative stock price $S_0=100$, strike price $K=100$, expiry one year away, interest rate $r=0.05$ per year, and given European call option price $C_0=10$, put-call parity requiresBoth portfolios therefore cost approximately $105.12294245$. The inputs specify no stock price dynamics: $C_0=10$ is supplied for illustration, not calculated from the Black-Scholes model. Put-call parity determines the European put option price conditional on that European call option price; it does not determine both prices by itself. At zero interest rate, the same $S_0=K=100$ and $C_0=10$ give the exact result $P_0=10$.

To compute both directions of arbitrage, keep all the preceding inputs except the quoted European put option price. If it is $6$, portfolio B costs $106$ while portfolio A costs $105.12294245$. Buy A, sell B, and place the initial surplus $6-(10+100e^{-0.05}-100)\approx 0.87705755$ in the bank account. At expiry, A and the obligations from selling B cancel exactly, leaving $96e^{0.05}-100\approx 0.92202525$ in every scenario. If the European put option instead costs $4$, buy B for $104$ and sell A for $105.12294245$; the initial surplus is approximately $1.12294245$, growing to $100-94e^{0.05}\approx 1.18051694$. Both constructions start with zero external capital. They are model calculations under the stated frictionless market assumptions.

The beauty of put-call parity is that uncertain individual financial payoffs produce an exact relation between today's prices. No forecast of the terminal stock price and no probability distribution are needed. It provides a consistency check and lets claim replication exchange one contract for a combination of others. Dividends introduce extra cash flows, and American put options can be exercised early, so the displayed European equality cannot simply be transferred to those contracts. K. Foster's ^[ref] explain the matching portfolios; Jun Pan's ^[ref] discuss put-call parity and its dividend adjustment.

An original modern research source is Hans R. Stoll's ^[ref], The Journal of Finance 24(5), 801-824 (December 1969). Stoll studied the relation between contract prices and tested it against market data; this was a relation between prices rather than a formula fixing each price individually. Robert C. Merton's ^[ref], 28(1), 183-184, highlighted the importance of excluding premature exercise for an exact arbitrage equality. This distinction explains why the exercise assumption belongs in the statement of put-call parity.

For every complex number $z$ with $|z|\le 1$,Applying this to each prime-power term in logarithms of Euler products proves thatfor $\sigma >1$ and every completely multiplicative $f$ bounded by one.

A field configuration or identity is on shell when the equations of motion hold.

If every face of a connected planar graph has size at least $g>2$, double-counting edge-face incidences and using Euler's formula gives

A finite simple graph that is a connected graph and a planar graph, with $n\ge 3$ vertices and $m$ edges, satisfies $m\le 3n-6$. Choose a plane graph: each facial boundary walk has length at least three, so $3f\le 2m$. Substitute $f=2-n+m$ from the Euler formula for a connected planar graph.^[ref]

The complete graph $K_4$ attains equality: $6=3\cdot 4-6$. The bound also holds for a disconnected finite simple graph that is a planar graph with $n\ge 3$: place its connected components of a graph in disjoint regions and join them with new bridges in a graph, then apply the connected bound.

For an integer matrix $A$ and prime $p$, the commuting-matrix binomial theorem and divisibility of $\left(\genfrac{}{}{0ex}{}{p}{j}\right)$ for $0<j<p$ give

For every odd prime,

Euler's criterion at $a=-1$ gives

This is a ball for the Euclidean metric.

The closed Euclidean ball of radius $R$ about $a\in {\mathbb{R}}^n$ is $\{x:\Vert x-a{\Vert }_2\le R\}$; replacing $\le$ by $<$ gives the open ball.

A proper Euclidean motion has the form $E(x)=Ax+b$ with $A\in SO(3)$. It preserves lengths, dot products, cross products, orientation, curvature, and torsion.

This is the orientation-preserving subgroup of the three-dimensional Euclidean group.