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If finite sets satisfy $|A+B|\le K|A|$, then the Plünnecke-Ruzsa inequality bounds iterated sumsets and difference sets by

Write $\Delta ={\theta }_1-{\theta }_2$. Applying the Euler-Lagrange equations to the double pendulum Lagrangian, then dividing by $m{\ell }^2$, givesFor example, the upper generalized momentum is $\partial L/\partial {\dot{\theta }}_1=m{\ell }^2(2{\dot{\theta }}_1+\cos \Delta {\dot{\theta }}_2)$. Its time derivative contains $-m{\ell }^2\sin \Delta ({\dot{\theta }}_1-{\dot{\theta }}_2){\dot{\theta }}_2$. Subtracting $\partial L/\partial {\theta }_1=-m{\ell }^2\sin \Delta {\dot{\theta }}_1{\dot{\theta }}_2-2mg\ell \sin {\theta }_1$ cancels the mixed product and leaves the positive $\sin \Delta {\dot{\theta }}_2^2$ term in the first equation. This cancellation is a useful check on the sign convention.

The matrix multiplying the two angular accelerations isIt is invertible for every configuration, so specified angles and angular velocities determine both angular accelerations. For the same reason $T>0$ whenever the angular velocities are not both zero. The mass cancels from the motion equations, while the timescale is set by $\sqrt{\ell /g}$.

As an exact constructed release, put ${\theta }_1(0)=\alpha$, ${\theta }_2(0)=0$, and both angular velocities zero. Solving the two equations at release yieldsFor $0<\alpha <\pi /2$, the upper rod initially moves back toward vertical, while the lower rod initially moves away from it. The lower pivot's acceleration makes this possible even though the lower rod starts vertical. The small-angle approximation gives ${\ddot{\theta }}_1(0)\simeq -2g\alpha /\ell$ and ${\ddot{\theta }}_2(0)\simeq 2g\alpha /\ell$.

The conserved mechanical energy is $E=T+V$, because the Lagrangian has no explicit time dependence. Directly, $dE/dt$ is $m{\ell }^2$ times the first equation multiplied by ${\dot{\theta }}_1$ plus the second multiplied by ${\dot{\theta }}_2$, so the equations imply $dE/dt=0$. This also supplies a check when computing a trajectory.

Choose Cartesian coordinates with origin at the upper pivot, $x$ to the right and $y$ upward. The two position vectors areThese expressions satisfy the rod holonomic constraints $|{\mathbf{r}}_1|=\ell$ and $|{\mathbf{r}}_2-{\mathbf{r}}_1|=\ell$ automatically. This is the practical advantage of using the two angles as coordinates: the rod tensions need not be unknowns in the motion calculation.^[ref]

Differentiate these position vectors to obtain the velocities. The dot product between their two rod contributions uses $\cos ({\theta }_1-{\theta }_2)=\cos {\theta }_1\cos {\theta }_2+\sin {\theta }_1\sin {\theta }_2$, givingAdding the two kinetic energies and measuring gravitational potential energy from the downward equilibrium givesThe coefficient $2$ has a geometric meaning: the upper rod raises both masses, and its motion contributes to both kinetic energies. The coupling in $T$ is present because the lower mass has a moving pivot. The zero chosen for $V$ changes $L$ by a constant and leaves the Euler-Lagrange equations unchanged.^[ref]

The principle of stationary action applies to $S=\int Ldt$. Its fixed-endpoint first variation, after integration by parts, isIndependent allowed variations therefore give one Euler-Lagrange equation for each angle. Stationary action alone does not assert a minimum.^[ref]

For a smooth projective curve of genus $g$ and a canonical divisor $K$,

This is the full linear system of divisors, obtained by using every global section of the associated line bundle. For a Cartier divisor on an integral projective variety, the complete linear system $|D|$ consists of its effective representatives under linear equivalence of Cartier divisors. A basis of $H^0(X,{\mathcal{O}}_X(D))$ defines a rational map to projective space.

For a nonzero rational function $f$ on a smooth projective curve $C$, its principal divisor isIt records the zeros of $f$ with positive multiplicity and its poles with negative multiplicity. Every principal divisor has degree zero.

Two divisors $D$ and $E$ are linearly equivalent, written $D\sim E$, when $D-E$ is a principal divisor. Multiplication by a rational function whose divisor is $D-E$ gives an isomorphism between their spaces of sections, so $\ell (D)=\ell (E)$.

For $D={\sum }_P{n}_{P}P$ on a curve over an algebraically closed field, its degree is $\mathrm{deg}D={\sum }_P{n}_P$.

A locally integrable function $g$ is the $i$th weak derivative of $u$ whenfor every test function $\phi \in C_c^{\infty }(U)$. Thus a weak derivative is a distributional derivative that is represented by a locally integrable function.

For terminal utility $U$ and independent return innovations,

Let $Q$ be the risk-neutral measure in an $N$-period binomial market. The stock-numeraire measure is defined on terminal events byIf the $Q$-probability of an up move is $q$, its up probability under $\hat{Q}$ isThis change of measure converts discounted expectations containing a factor $S_N$ into probabilities under $\hat{Q}$.

Let $c_0$ be the initial monetary balance in the bank account, negative for borrowing. Matching successor financial payoffs gives the linear systemSubtracting determines the stock holding $\Delta$ in the header; substitution determines $c_0$. The initial cost is $V_0=\Delta S_0+c_0$. For the European call option with $S_u=120$, $S_d=80$, strike price 100 and payoffs 20 and zero,The debt becomes 40 at maturity: half a share minus 40 yields $60-40=20$ in the up state and $40-40=0$ in the down state. Holding these positions until maturity is self-financing. Repeating this two-equation construction at earlier nodes gives backward option pricing and the unique price in the complete market.

The discrete replication method was developed in Cox, Ross and Rubinstein's 1979 paper, whose Sections 2–3 explain borrowing, statewise matching and why physical probability drops out of relative option valuation. Their symbols for physical and pricing probabilities differ from the $p,q$ convention here. ^[ref]

The martingale condition for the discounted stock is $qu{S}_0+(1-q)d{S}_0=R{S}_0$. Therefore the up risk-neutral probability isWith down factor $d=1+a$, up factor $u=1+b$ and $R=1+r$, this is equivalently $(r-a)/(b-a)$. For $d=0.8$, $u=1.2$, $R=1.05$, it gives $q=0.25/0.4=0.625$. Under the risk-neutral measure, the terminal stock price has expected value $(5/8)120+(3/8)80=105$, whose discounted value is 100. The physical probability $p$ need only satisfy $0<p<1$; it does not enter this calculation.

One half-grid sine-transform convention is

The type-II discrete cosine transform of $x_0,\dots ,x_{N-1}$ is

For a primitive fifth root of unity $\zeta$,

For $K=\mathbb{Q}(\sqrt{-3})$ and $\omega =(-1+\sqrt{-3})/2$,

For $K=\mathbb{Q}(\sqrt{10})$,The fundamental unit $3+\sqrt{10}$ has norm minus one.

Choose the real embeddings ${\sigma }_i$ and one embedding ${\tau }_j$ from each complex-conjugate pair. The weighted logarithmic embedding isFor a unit, the sum of these coordinates is $\log |N_{K/\mathbb{Q}}(u)|=0$.

If $K$ is real quadratic and $L/K$ is a number-field extension with ${\mathcal{O}}_L^{\times }/{\mathcal{O}}_K^{\times }$ finite, then the unit ranks agree. If $d=[L:K]$ and $L$ has signature $(r_1,r_2)$, this forcesFor a proper extension the only possibility is $d=2$ and signature $(0,2)$. For $K=\mathbb{Q}(\sqrt{5})$, the example $L=K(i)$ has this signature and a nontrivial finite quotient because $i$ is a new root of unity.