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A binary linear-feedback shift register of length $L$ stores $L$ bits. At each clock it outputs one bit, shifts the state, and inserts a fixed linear combination over ${\mathbb{F}}_2$ of the stored bits. Equivalently, its output obeys a recurrence

The Berlekamp-Massey algorithm processes the observed symbols from left to right. It maintains the shortest current connection polynomial $C(D)$, its degree $L$, and a saved polynomial from the most recent degree increase. At step $n$ it computes the discrepancyIf $d_n=0$, the current recurrence already predicts the new symbol. If $d_n\ne 0$, a shifted multiple of the saved polynomial is added to $C$ to cancel the discrepancy; when $2L\le n$, the algorithm also replaces $L$ by $n+1-L$ and saves the old polynomial.

For the observed binary sequence, the nonzero-discrepancy updates areAll discrepancies through $n=11$ then vanish. Thus the shortest recurrence isor, equivalently, $s_{n+4}=s_{n+1}+s_n$. The connection polynomial is $1+D^3+D^4$, and with the reciprocal convention the feedback polynomial is

For a Fourier mode $e^{i(kx-\omega t)}$, the wave equation givesTaking the positive-frequency branch,For $k>0$, the phase velocity and group velocity areThe frequency rises from zero and approaches $1$; $c_p$ reaches its maximum at $k=1$, whileshows that the front velocity isThus the wave at the front hasMoreover $c_g/c_p=2/(1+k^2)$, so the packet outruns its crests for $0<k<1$, the speeds agree at $k=1$, and the crests outrun the packet for $k>1$.

The initial conditions give the exact superpositionThe condition $A^{\ast }(-k)=A(k)$ makes this solution real. For $V=x/t$ with $0<V<V_m$, the equationhas two positive roots $0<{\alpha }_1<1/\sqrt{3}<{\alpha }_2$. The second exponential has the corresponding stationary points $-{\alpha }_1,-{\alpha }_2$. Sincethe stationary phase method and pairing of complex-conjugate contributions giveFor $-V_m<V<0$, the same picture is reflected and the stationary points have negative wavenumber. For $|V|>V_m$ there are no real stationary points, so a smooth localized spectrum gives a much smaller nonstationary tail outside the wave front.

The exact gap between the plane and the circle isIn the narrow region relevant to lubrication theory,so changes occur over the streamwise scale $x=O(\sqrt{a{h}_0})$; below use $x=\sqrt{2a{h}_0}X$.

Choose the positive $x$ direction to be the direction of the cylinder surface velocity $U=a\Omega$ in the gap, and write $\mu =\rho \nu$ for the dynamic viscosity. The lubrication velocity profile and volume flux per unit axial length areSince $Q$ is constant and the pressure returns to the same ambient value as $x\to \pm \infty$, ${\int }_{-\infty }^{\infty }{p}_{x}dx=0$. ThereforeUsing $h=h_0(1+X^2)$ and the supplied integrals givesin the direction of the cylinder surface motion.

The shear stress exerted by the fluid on the cylinder is opposite toEliminating $p_x$ and inserting $Q=2U{h}_0/3$ giveswhere the sign is relative to $U$. Sincethe resisting torque per unit axial length is

Finally, the ratio of streamwise fluid inertia $U^2/\sqrt{a{h}_0}$ to transverse viscous acceleration $\nu U/h_0^2$ is of orderThe stated restriction makes this lubrication-scale Reynolds number small, justifying neglect of inertia.

A three-dimensional Bravais lattice is the set of all integer combinations of three linearly independent primitive vectors. Its reciprocal lattice isand the first Brillouin zone is the Wigner-Seitz cell of the origin in reciprocal space.

For the stated FCC lattice, the primitive-cell volume isThe reciprocal-basis formula givesA reciprocal primitive cell, and hence the Brillouin zone, has volume

In the first Born approximation, with momentum transfer $q=k-k^{\prime }$, the scattering amplitude isFor the spherical shell, align the polar axis with $q$ and use its radial Dirac delta function:Therefore

Translation of each atomic potential multiplies its transform by a phase, so the crystal lattice structure factor givesFor an infinite lattice, the lattice Poisson summation formula turns the sum intoThus the amplitude vanishes away from reciprocal-lattice momentum transfers.

In elastic scattering, $|k|=|k^{\prime }|$ and henceThe two shortest relevant reciprocal-vector lengths of this FCC lattice areBoth are kinematically available when $k>2\pi /a$. Assigning their scattering angles ${\theta }_2$ and ${\theta }_1$, respectively, gives

The angular momentum commutation relations implyThus $S_{\pm }|s,\sigma ⟩$ has the same total-spin eigenvalue and magnetic quantum number $\sigma \pm 1$. Its squared norm follows fromChoosing the conventional positive phases gives the spin ladder operator formula

For $s=1$, in the ordered basis $|1,1⟩,|1,0⟩,|1,-1⟩$,Since $S_x=(S_{+}+S_{-})/2$,

The incident state is the spin coherent state obtained by rotating $|1,1⟩$ through $\theta$ about the $y$ axis. Its amplitudes in the $S_z$ basis areup to an irrelevant phase convention. The Born rule therefore gives

An interval map has a horseshoe for an interval map when two disjoint subintervals are each mapped across a common interval. It is Glendinning-chaotic when one of its positive iterates has such a horseshoe.

Let the points of the three-cycle be $x_1<x_2<x_3$ and put $I_0=[x_1,x_2]$, $I_1=[x_2,x_3]$. There are two possible cyclic orders. If $x_1\mapsto x_2\mapsto x_3\mapsto x_1$, the intermediate value theorem givesfor the interval covering relation. Thus there are two closed covering walks of length two based at $I_1$. In the reverse cyclic order there are instead the arrows $I_0\to I_0$, $I_0\to I_1$ and $I_1\to I_0$, again giving two closed walks of length two. The horseshoe from two closed covering walks therefore shows that $F^2$ has a horseshoe in either case, so $F$ is chaotic.

The Sharkovsky theorem orders the positive integers asand says that a cycle of a given period forces cycles of every period to its right.

Suppose $N=2^{r}q$ with $q>1$ odd. Then $F^{2^r}$ has a $q$-cycle. The Sharkovsky theorem gives that map a six-cycle, and squaring it produces a three-cycle. Hence $F^{2^{r+1}}$ has a three-cycle and an iterate of $F$ has a horseshoe. Thus

A horseshoe contains the symbolic dynamics of the Bernoulli shift. Periodic binary words give cycles of every symbolic period; after translating from an iterate back to $F$, this supplies cycles of many periods that are not powers of two.

For the logistic map, a ten-cycle would imply chaos because $10$ is not a power of two. Hence there is no ten-cycle for $\mu <{\mu }_{\infty }$. A three-cycle forces every period, so a ten-cycle exists for $\mu >1+\sqrt{8}$. The stated facts alone give no conclusion for the intermediate range:

The intended integral representation of the modified Bessel function isDifferentiation under the integral sign givesSinceintegration by parts and the vanishing endpoint term yield $x{I}_0^{\prime \prime }+I_0^{\prime }-x{I}_0=0$.

For large positive $x$, Laplace's method localizes the integral near the endpoint $\theta =0$. Put $\theta =u/\sqrt{x}$ and useExtending the exponentially small tail to infinity and evaluating the resulting Gaussian integrals givesand therefore

Now write $y=x^{-1/2}w$. Direct differentiation givesso substitution into the ordinary differential equation givesFor the Liouville–Green approximation, put $Q(x)=1-1/(4x^2)$. The two approximate solutions areso at leading order $y_{\pm }(x)\sim x^{-1/2}{e}^{\pm x}$. The growing solution agrees with the preceding asymptotic expansion, including its leading normalization $I_0(x)\sim e^x/\sqrt{2\pi x}$.

LetAssume for a contradiction that $S=a/b$ is rational. For every $N\ge b$, both $N!S$ andare integers, because the $m_n$ are integers. Their positive differencemust therefore be a positive integer.

On the other hand, concavity of $x^{\alpha }$ for $0\le \alpha <1$ gives $(N+k{)}^{\alpha }\le N^{\alpha }+k^{\alpha }\le N^{\alpha }+k$. Since $(N+1)\cdots (N+k)\ge (N+1{)}^k$,Eventually $0<R_N<1$, contradicting its integrality. Hence

The integrality assumption is essential. For example, take $\alpha =0$, $A=2$, andThen $1<m_n<2$, but the exponential series givesThus the result does not remain true for arbitrary real $m_n$.