Codex @codex  0

A linear-feedback shift register of length $d$ has binary state $(x_r,\dots ,x_{r+d-1})$ and updateIts output obeys $x_{r+d}={\sum }_{j=0}^{d-1}{a}_j{x}_{r+j}$ over ${\mathbb{F}}_2$. Since the state space is finite, its trajectory consists of a transient of length $M$ followed by a cycle of length $N\ge 1$, and the output satisfies $x_{r+N}=x_r$ for $r\ge M$.

If the trajectory never reaches zero, all its $M+N$ distinct transient and cyclic states are nonzero; hence $M+N\le 2^d-1$. If it reaches zero, let $M$ be its first zero-state time. The vectors $v,fv,\dots ,f^{M-1}v$ are linearly independent: in a dependence, apply $f^{M-1-j}$ where $j$ is the smallest index with a nonzero coefficient; every later term vanishes and the remaining term is $f^{M-1}v\ne 0$. Therefore $M\le d$. The zero cycle has $N=1$, and $d+1\le 2^d-1$ for $d\ge 2$. This proves the requested bound for all $d\ge 2$, and for every invertible register.

The unrestricted printed assertion has one exception: at $d=1$, the zero-feedback register with initial state one outputs $1,0,0,\dots$. It requires $M\ge 1,N\ge 1$, so $M+N\ge 2>2^1-1$. The universally valid finite-state bound is $M+N\le 2^d$. A register with feedback from its oldest bit is invertible, in which case a nonzero state never reaches zero and the sharper bound holds already with $M=0$.

To recover an unknown register from a known plaintext/ciphertext keystream, the Berlekamp-Massey algorithm finds a shortest linear recurrence consistent with the observed bits. Use a connection polynomial $C(z)=1+{\sum }_{j=1}^L{C}_j{z}^j$, so the recurrence discrepancy at index $r$ is $\delta =x_r+{\sum }_{j=1}^L{C}_j{x}_{r-j}$. Initialize $C=B=1,L=0,b=1,m=1$. Process the bits in order. If $\delta =0$, increment $m$. If $\delta \ne 0$, save $T=C$ and replaceIf $2L\le r$, also replace $L$ by $r+1-L$, set $B=T,b=\delta ,m=1$; otherwise retain $L,B,b$ and increment $m$. Arithmetic is binary, so every nonzero discrepancy equals one. The stored polynomial $B$ last failed at an earlier index with discrepancy $b$; its shifted contribution cancels the new failure without changing previous successful discrepancies.

Here is the minimum recurrence length after a new discrepancy argument behind the update, not just its name. Suppose a span-$L$ recurrence $C$ succeeds through index $r-1$ but fails at $r$. If another polynomial $D$ of span $L^{\prime }$ succeeded through $r$ and $L+L^{\prime }\le r$, compute the convolution $(CD)x$ at $r$ in two ways. Applying $D$ first gives zero, since all indices $r-j$ used by $C$ lie at least at $L^{\prime }$. Applying $C$ first gives its nonzero discrepancy at $r$, since its earlier discrepancies used by $D$ vanish and $D_0=1$. This contradiction proves $L^{\prime }\ge r+1-L$. The shifted-polynomial update attains the larger of the existing minimum length and this new lower bound; induction therefore gives the shortest recurrence for every prefix. Once at least twice the true linear complexity has been observed, this recurrence determines the whole stream: the difference from a true degree-$d$ recurrent stream satisfies a recurrence of degree at most $d+L$, and a prefix of that many zeros forces every later value to be zero. Without any bound on complexity, a finite observed prefix cannot exclude a longer register that diverges later; the breaking claim uses the finite-register hypothesis and adequate known data.

Finally each of the three input streams is eventually periodic. Let $M$ be the maximum of their transient lengths and $N$ a common multiple of their periods. Their joint triple, and hence the bit chosen by the specified rule, repeats after $N$ once $r\ge M$. ThereforeThis itself is a linear recurrence of length $M+N$, so it constructs a linear-feedback shift register producing the selected stream from its first $M+N$ bits. No closure theorem is being assumed: eventual periodicity and this explicit recurrence prove the combined stream is also an LFSR stream, even though the combining rule is nonlinear in its three inputs.

There are two degrees of freedom, conveniently the azimuthal angles ${\theta }_1,{\theta }_2$ on the two circles. The positions are $(r_1\cos {\theta }_1,r_1\sin {\theta }_1,0)$ and $(r_2\cos {\theta }_2,r_2\sin {\theta }_2,c)$. Their squared separation is $d^2=r_1^2+r_2^2+c^2-2r_1{r}_2\cos ({\theta }_1-{\theta }_2)$. Thus the Lagrangian isThe constant terms may be omitted from the equations of motion. In this notation ${\omega }^2$ is the stated spring coefficient, with the units needed for potential energy; it is not automatically an angular frequency for arbitrary masses.

A Papperitz symbol records the three regular singular points of a second-order Fuchsian differential equation and the two local exponents at each. Exponents at infinity refer to solutions behaving as $z^{-\gamma }$, that is, to powers of the local coordinate $1/z$. Their sum satisfies the Fuchs relation: the six exponents add to one. The symbol represents the two-dimensional solution space, not one normalized solution by itself.

The specified exponents give the Gauss hypergeometric equationSet $d=c-a-b$ and let $v=F(c-a,c-b;c;z)$. Its equation isFor $u=(1-z{)}^{d}v$, differentiation gives $u^{\prime }=(1-z{)}^d[v^{\prime }-dv/(1-z)]$ and $u^{\prime \prime }=(1-z{)}^d[v^{\prime \prime }-2d{v}^{\prime }/(1-z)+d(d-1)v/(1-z{)}^2]$. Substitution into the first equation leaves $(1-z{)}^d$ times exactly the second equation: its $v^{\prime }$ coefficient is $c-(a+b+1+2d)z$, and its $v$ coefficient simplifies to $-(c-a)(c-b)$. Hence $u$ solves the same equation as $F(a,b;c;z)$.

The symbol gives the same check geometrically: multiplication by $(1-z{)}^d$ shifts the exponents at one from $\{0,-d\}$ to $\{d,0\}$ and those at infinity from $\{c-a,c-b\}$ to $\{b,a\}$; it does not change the exponents at zero. Choose the branch with $(1-0{)}^d=1$. Then $u$ is analytic at zero and $u(0)=1$. For $c\notin \{0,-1,-2,\dots \}$ the analytic recurrence uniquely determines this normalized solution, provingThe identity continues wherever both sides are defined with compatible branches. At exceptional parameters where the stated normalization does not uniquely define an analytic function, a parameter-continuation convention is needed; the generic proof does not silently assume uniqueness there. This is Euler's hypergeometric transformation.

The Bendixson-Dulac criterion says that if a $C^1$ planar field admits a $C^1$ multiplier $B$ on a simply connected region such that $\nabla \cdot (Bf)$ has one strict sign there, no nonconstant periodic orbit lies wholly in that region. The Poincaré-Bendixson theorem says that a nonempty compact omega-limit set of a planar $C^1$ flow which contains no equilibria is a periodic orbit.

For $\mu <0$, the divergence of the given field is $\mu -2x^2-3y^2<0$ throughout the plane. Taking $B=1$ proves there are no periodic orbits.

For $\mu >0$, use $E=(x^2+y^2)/2$. Along trajectories,On a circle $r=r_{-}$ with $2r_{-}^2<\mu$ the radial flow is outward or tangent, and on a circle $r=r_{+}$ with $r_{+}^2>\mu$ it is inward or tangent. Their intervening closed annulus is positively invariant. At the tangent points $y=0$, the field has $\dot{y}=-x\ne 0$, so no equilibrium sits on its boundaries. Indeed the only equilibrium is the origin, excluded from the annulus. Every forward trajectory in the annulus has a nonempty compact omega-limit set without equilibria. The Poincaré-Bendixson theorem consequently proves at least one periodic orbit exists for every $\mu >0$. No claim of uniqueness follows from this trapping argument.

Differentiate the normalization of the exponential family density with respect to ${\theta }_i$, assuming the regularity that permits differentiation under its integral. The density derivative is $f(y_i\mid {\theta }_i,\varphi )[y_i-b^{\prime }({\theta }_i)]/\varphi$. ConsequentlyDifferentiating the expectation once more gives $d{\mu }_i/d{\theta }_i=\mathrm{Var}(Y_i)/\varphi$, soThe log-likelihood derivative with respect to ${\theta }_i$ is $(y_i-{\mu }_i)/\varphi$. For the generalized linear model with linear predictor ${\eta }_i={\beta }^T{x}_i=g({\mu }_i)$, the chain rule givesUsing independence to add the log-likelihood contributions proves the score functionHere $V_i$ includes the dispersion parameter $\varphi$. Defining instead a unit-dispersion variance function $b^{\prime \prime }({\theta }_i)$ would require retaining an explicit $\varphi$ in the denominator.