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The terms $u$ and $\alpha v$ represent intrinsic exponential population growth. The losses $-uv$ and $-\alpha uv$ describe competition between the species; $-{ϵ}_1{u}^2$ and $-\alpha {ϵ}_2{v}^2$ describe crowding within each species. The positive constant $\alpha$ sets the relative time scale of the second population.

The four equilibrium points in the nonnegative quadrant areThe last follows by solving ${ϵ}_{1}u+v=1$ and $u+{ϵ}_{2}v=1$ and is strictly positive under the assumptions. The Jacobian matrix isAt the origin its eigenvalues are $1,\alpha >0$, so it is unstable. At the first single-species equilibrium they are $-1$ and $\alpha (1-1/{ϵ}_1)<0$; at the second they are $1-1/{ϵ}_2<0$ and $-\alpha$. Both are asymptotically stable. At coexistence,Thus coexistence is a saddle; either species-only equilibrium is stable, and extinction is unstable. The saddle's stable manifold separates the competing basins of attraction.

The normal-model log-likelihood is $-\frac{n}{2}\log {\sigma }^2-\Vert Y-X\beta {\Vert }^2/(2{\sigma }^2)$ up to a constant. Differentiating gives the maximum-likelihood estimatorsLet $P=X(X^{T}X{)}^{-1}{X}^T$. The fitted component $P\epsilon$ and residual $(I-P)\epsilon$ are orthogonal jointly normal projections and hence independent. ThereforeThe prediction error $\tilde{y}-y^{\ast }=x^{\ast T}(\hat{\beta }-\beta )-{\epsilon }^{\ast }$ is normal with mean zero and variance ${\sigma }^2{\tau }^2$, and is independent of the residual variance estimate. Dividing by the independent estimated standard deviation yieldsThus the prediction interval in a normal linear model isThe extra $1$ includes the new observation's independent noise; omitting it would give a confidence interval for its conditional mean instead.

A binary linear-feedback shift register stores $L$ bits, shifts them at each clock step, and inserts a fixed linear combination over ${\mathbb{F}}_2$ of the stored bits. Its output obeys a recurrence $s_n+c_1{s}_{n-1}+\cdots +c_L{s}_{n-L}=0$. Write its connection polynomial as $C(D)=1+c_{1}D+\cdots +c_L{D}^L$; the corresponding monic feedback polynomial is $z^{L}C(z^{-1})$.

The Berlekamp-Massey algorithm maintains the shortest recurrence fitting the observed prefix. Initialize $C=B=1$, length $L=0$, displacement $m=1$, and previous nonzero discrepancy $b=1$. At symbol $n$, calculate $d=s_n+{\sum }_{j=1}^L{c}_j{s}_{n-j}$. If $d=0$, increase $m$. If $d\ne 0$, save $T=C$ and replace $C$ by $C-(d/b)D^{m}B$. If $2L\le n$, replace $L$ by $n+1-L$, $B$ by $T$, $b$ by $d$, and $m$ by $1$; otherwise only increase $m$. In the binary field subtraction equals addition. The shifted previous discrepancy polynomial cancels the current discrepancy while preserving earlier equations. A failed recurrence of length $L$ at position $n$ requires a corrected length at least $n+1-L$ when this exceeds $L$, explaining the length update. Thus the algorithm recovers the shortest compatible register; a known bound $L$ and at least $2L$ output symbols suffice to identify its minimal recurrence.

For the printed twelve-bit prefix, the nonzero discrepancies occur at positions $n=0,1,2,5,6,7$ (indexing from zero). The successive updated pairs $(L,C)$ areAll remaining discrepancies vanish. ThereforeThe first four observed bits are the fill. Length four is necessary: a length-three recurrence would need to hold already at $n=3$ and no such recurrence fits the prefix. Indeed the equations at $n=3,4,5$ force $c_2=1$, $c_1+c_3=0$, $c_1+c_3=1$, a contradiction. Finite observations cannot rule out longer nonminimal registers.

For a Euclidean circle with centre $c$ and radius $r$, two distinct finite points are inverse when they lie on the same ray from $c$ and their distances from $c$ have product $r^2$. Points on the circle are fixed, and $c$ and infinity are inverse to each other. In complex coordinates this uniquely defines inversion in a circle byUniqueness follows from the ray and product conditions, including the specified centre–infinity pair; the formula also gives $I_{\Gamma }^2=\mathrm{id}$. A circle through infinity is a straight line in the complex plane; its inversion is reflection in that line, fixing infinity. For a line through $c$ with direction angle $\theta$, its formula is $c+e^{2i\theta }\overline{z-c}$.

Thus every inversion is an anti-Möbius transformation, of the form $(a\overline{z}+b)/(c\overline{z}+d)$ with nonzero determinant. If $J(z)=\overline{z}$, write such an inversion as $M\circ J$ with $M$ a Möbius transformation. Since $J\circ M\circ J$ is the Möbius transformation with conjugated coefficients, the product of two inversions is $M_1\circ (J\circ M_2\circ J)$, hence Möbius. Pairing consecutive inversions proves every composition of an even number of inversions is a Möbius transformation.

For a plane acoustic wave propagating along $\hat{\mathbf{k}}$, the linear momentum equation and dispersion $\omega =c_{0}k$ give $\tilde{p}={\rho }_0{c}_0{u}_{\parallel }$ and $\mathbf{u}=u_{\parallel }\hat{\mathbf{k}}$. Therefore the instantaneous acoustic intensity isFor harmonic complex amplitudes the time average instead has the factor one half, $⟨\mathbf{I}⟩=\frac{1}{2}{\rho }_0{c}_0|\hat{\mathbf{u}}{|}^2\hat{\mathbf{k}}$.

For a radial velocity potential, set $v=r\varphi$. The spherical wave equation reduces to $v_{tt}=c_0^2{v}_{rr}$, soNo incoming radiation from infinity means $G=0$. Linearizing the moving-surface condition at $r=a$ gives ${\varphi }_r(a,t)=i\omega aϵe^{i\omega t}$, with real parts understood. Write the outgoing solution as $\varphi =A{e}^{i\omega (t-(r-a)/c_0)}/r$. Its radial derivative at $a$ is $-A(1+i\omega a/c_0)e^{i\omega t}/a^2$. Therefore the outgoing acoustic field of a pulsating sphere isIn the far field the velocity amplitude has magnitude $(\omega /c_0)|A|/r$. Multiply its average plane-wave intensity by the spherical area to obtain the mean acoustic power of a pulsating sphere:This is positive outward radiated power, evaluated to leading quadratic order in the small real displacement amplitude.