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For an Einstein-de Sitter universe, $H=\dot{a}/a=2/(3t)$. The flat matter-only Friedmann equation $H^2=8\pi G{\rho }_m/3$ gives ${\rho }_m=1/(6\pi G{t}^2)$. With zero sound speed the perturbation equation is thereforeSubstituting $t^s$ yields $s(s-1)+4s/3-2/3=0$, or $(3s-2)(s+1)=0$. Hencewith growing mode proportional to the scale factor and decaying mode proportional to $t^{-1}$.

The physical Jeans length is the wavelength for which pressure and gravity balance:The physical wavelength of the comoving mode is $2\pi a/k$. Modes with wavelength larger than ${\lambda }_J$ have a gravitationally destabilizing coefficient and can grow; shorter wavelengths are pressure-supported and oscillate, with expansion also damping the motion. A time-dependent background makes the exact growth history depend on the epoch, not just an instantaneous sign. For pressureless matter the Jeans length vanishes.

Write $ϵ=1-q$ and $b/a=1-s$. Then $\log (1-s)=-s+O(s^2)$ and $\log q=-ϵ+O({ϵ}^2)$, givingThe largest permitted integer is still selected by the strict inequality from part (b). This says that longer master sequences need a proportionately larger selective advantage or smaller per-digit copying error.

For $0<q<1$, a finite positive equilibrium exists exactly when $a{q}^L>b$, and thenThe derivative of the ratio's vector field at $z_{\ast }$ is $-(a{q}^L-b)<0$, so this equilibrium is locally asymptotically stable; every positive initial ratio tends to it. The critical real length is $L_c=\log (b/a)/\log q$, and the admissible integer lengths satisfy $L<L_c$. Thus the largest positive integer length, if one exists, is $\boxed{L_{\max }=\lceil L_c\rceil -1}$. At equality the equilibrium is zero, so rounding down without accounting for the strict inequality can be wrong. If $q=1$, mutants are never produced from masters and the ratio grows without a finite positive equilibrium; the finite threshold formula concerns imperfect copying.

The probability of an exact copy is $Q=q^L$. Incorrect copies contribute mutants at rate $a(1-Q)x$, giving $\boxed{\dot{y}=by+a(1-Q)x}$. For $y>0$, differentiating the ratio givesThis is a logistic-type equation for relative concentration; the absolute concentrations need not have a stationary value.

Define the hat matrix $H=X(X^{T}X{)}^{-1}{X}^T$. It is symmetric and idempotent, the orthogonal projection onto the column space of $X$. Since $HX=X$,The covariance is ${\sigma }^{2}H{H}^T={\sigma }^{2}H$. This normal distribution can be singular when $p<n$, since all fitted vectors lie in the $p$-dimensional model subspace.

The least-squares estimator minimizes $\Vert Y-Xb{\Vert }^2$. Differentiating in $b$ gives $-2X^T(Y-Xb)=0$, hence $\boxed{X^{T}X\hat{\beta }=X^{T}Y}$. Full column rank makes $X^{T}X$ positive definite, soThis is a linear transformation of a multivariate normal distribution. Its mean is $\beta$ and its covariance is ${\sigma }^2(X^{T}X{)}^{-1}$. Therefore $\boxed{\hat{\beta }\sim N_p(\beta ,{\sigma }^2(X^{T}X{)}^{-1})}$.

For the linear test equation $f(y)=\lambda y$ put $z=h\lambda$. The second stage solves $(1-az)k_2=\lambda y[1+(1-a)z]$, and substitution into the update gives the stability functionA-stability requires no poles and $|R(z)|\le 1$ throughout the closed left half-plane. If $a\ne 1/2$, the quadratic numerator gives unbounded magnitude as $z\to -\infty$; when $a=0$ the denominator is constant and the same conclusion holds. Thus $a=1/2$ is necessary. At this value $R(z)=(1+z/2)/(1-z/2)$, whose only pole is at positive $z=2$. Moreoveron the left half-plane. Hence the method is A-stable exactly for $\boxed{a=1/2}$. Its limit at negative infinity is $-1$, so it is not L-stable.

Use the stage solution continuous from $k_2=f(y)$ as $h\to 0$. Its displacement is $\delta =h[(1-a)f+a{k}_2]=hf+a{h}^2{f}^{\prime }f+O(h^3)$. Taylor expansion of $f(y+\delta )$ givesConsequently the Runge-Kutta method advances byThe exact solution has third-order term $h^3(f^{\prime 2}f+f^{\prime \prime }{f}^2)/6$. The first two orders agree for every real $a$, but the $f^{\prime \prime }{f}^2$ coefficient is always $1/4$ instead of $1/6$. Therefore the general nonlinear method has $\boxed{\text{order }2}$ for every fixed real $a$. The special value $a=1/3$ matches the linear test equation at third order but does not raise the nonlinear order. This is the two-stage Runge-Kutta family with an implicit second stage.

Put $\Psi =\mu e^{-i(kx+2k^{2}t)\sigma }$. The Lax pair becomes ${\Psi }_x=U\Psi$, ${\Psi }_t=V\Psi$, withEquality of mixed derivatives, using invertibility, gives the zero-curvature condition $U_t-V_x+[U,V]=0$. The terms quadratic in $k$ cancel. The terms linear in $k$ cancel because $\sigma Q=-Q\sigma$, $\sigma Q_x=-Q_x\sigma$ and $Q^2$ commutes with $\sigma$. For the constant term, the identitiescancel the differentiated $Q^2$ term, leaving $Q_t+i{Q}_{xx}\sigma -2i{Q}^3\sigma =0$. Multiplying by $i$ provesOnce $S$ is known, its time dependence enters through the explicit exponential conjugation. The boundary relation and normalization form a linear problem for $\mu$, and $Q=i[\sigma ,{\mu }_1]$ reconstructs the nonlinear solution. This is the requested linearization. Its disadvantage as an initial-value method is that it does not itself determine $S$ from a prescribed $Q(x,0)$. The inverse scattering transform supplies that direct-scattering step, including any discrete spectral data and their pole conditions, before reconstruction. Solvability cannot be inferred merely from writing an arbitrary jump.

The $k^{-1}$ coefficient of the first equation is ${\mu }_{1x}+i[\sigma ,{\mu }_2]=Q{\mu }_1$. Substitution in the expression for $B$ gives $B=-2{\mu }_{1x}$. Decompose ${\mu }_1$ into diagonal and off-diagonal parts. Since $Q=i[\sigma ,{\mu }_1]$, its off-diagonal part is $iQ\sigma /2$. ThusThe commutator in the $k^{-1}$ equation has zero diagonal; multiplication of two off-diagonal two-by-two matrices is diagonal. Combining the two parts proves

There is a literal inconsistency in the stated Riemann-Hilbert problem: a jump matrix $S(k)$ tending to zero is incompatible with ${\mu }_{\pm }\to I$. Along the real axis the conjugating exponentials are unitary, so the right side of the jump tends to zero while the left side tends to $I$. No normalized solution satisfies the printed decay condition. The intended condition is $S(k)\to I$, or replacement of $S$ in the jump by $I+S$ when $S$ denotes its decaying part.

For this corrected problem assume a regular, invertible, pole-free solution, differentiable in $x,t$, with the required expansion uniformly at infinity. These conditions matter: arbitrary jump data need not be solvable, and poles would contribute extra terms. Write $\sigma ={\sigma }_3$ and $\mu =I+{\mu }_1/k+{\mu }_2/k^2+\cdots$. Matching the nondecaying coefficients in the two differential expressions yieldsThe first expression makes $Q$ off-diagonal.

These matches really establish the differential equations, rather than just formal identities. Let $G=e^{-i(kx+2k^{2}t)\sigma }S(k)e^{i(kx+2k^{2}t)\sigma }$. It satisfies $G_x=-ik[\sigma ,G]$ and $G_t=-2i{k}^2[\sigma ,G]$. Differentiating ${\mu }_{+}={\mu }_{-}G$ consequently shows thathave matching boundary values. They therefore extend holomorphically across the contour. The chosen coefficients make both residuals $O(k^{-1})$ at infinity. Their absence of poles and Liouville theorem imply $R_x=R_t=0$, proving the required Lax pair under the corrected hypotheses. This is matrix NLS reconstruction from a normalized Riemann-Hilbert problem.

Put $P(t)={\prod }_i(t-x_i)$. For every polynomial $q$ of degree at most $n-1$, $Pq$ has degree at most $2n-1$ and vanishes at all nodes; thus ${\int }_{-1}^{1}Pq=0$. The monic degree-$n$ orthogonal polynomial for this integral is unique: the difference of two such monic polynomials has degree below $n$ and is orthogonal to itself, hence is zero. Consequently $P$ is the monic normalization of the degree-$n$ Legendre polynomial, proving the given nodes are exactly its roots. Together the three results characterize this Gaussian quadrature rule.

Apply exactness to the constant polynomial one. Its integral is two and its value at every node is one, so $\boxed{\sum _{i=1}^n{a}_i=2}$.

For each node define its Lagrange interpolation cardinal polynomial ${\ell }_i(t)={\prod }_{j\ne i}(t-x_j)/(x_i-x_j)$. Its square has degree $2n-2<2n$ and nodal values ${\ell }_i(x_j{)}^2={\delta }_{ij}$. Exactness therefore gives $a_i={\int }_{-1}^1{\ell }_i(t{)}^{2}dt$. The polynomial is nonzero, so continuity makes this integral strictly positive. Thus $\boxed{a_i>0}$ for every node.

If the first move is bad but the buyer misses exercise, the remaining European claim is worth $2/3$, whereas the existing hedge is worth three. At that node rehedge the continuation claim: its stock holding is $(0-5)/(16-10)=-5/6$ and its bank holding is $2/3+(5/6)12=32/3$.

Starting from the original short $1/6$ and bank balance five, short another $2/3$ share for eight, giving bank cash thirteen. Retain $32/3$ for the new hedge and withdraw $\boxed{7/3}$ as guaranteed time-one profit. At time two the retained bank grows to $40/3$. If price is sixteen, covering the short costs $40/3$ and the put expires worthless. If price is ten, covering costs $25/3$ and the remaining five pays the put. Thus the profit does not depend on the second move. Investing the withdrawn profit at the riskless rate makes it $35/12$ at time two.

The terminal European put option payoffs are zero at forty-five, thirty-six and sixteen, and five at ten. At the good first node its value is zero. At the bad node its continuation value is $(4/5)[(5/6)0+(1/6)5]=2/3$, below the exercise payoff three, so optimal exercise occurs there. At time zero the continuation value is $(4/5)[(3/8)0+(5/8)3]=\boxed{3/2}$, exceeding the immediate payoff zero.

To hedge one sold American put option, short $1/6$ of a share and put four in the bank at time zero. The portfolio costs $-15/6+4=3/2$, exactly the premium. At time one the bank holds five. After a good move the short position costs five to close, exhausting the portfolio and liability. After a bad move it costs two to close, leaving three for the exercised put. Thus after two bad periods the rational buyer already exercised at time one, and the hedge pays three and closes then; there is no remaining time-two obligation. Physical delivery gives the same accounting by immediately selling the stock received on exercise.

At gross rate $R=5/4$, the conditional up probabilities of the equivalent martingale measure areAll lie strictly between zero and one. The four terminal path probabilities are $1/16,5/16,25/48,5/48$ in the order good-good, good-bad, bad-good, bad-bad.

At the node with stock price thirty, both possible next prices exceed thirty-three, the value of thirty invested at gross rate $11/10$. Take no position at time zero. If the first move is good, borrow thirty at time one and buy one share; if it is bad, do nothing. At time two sell the share for forty-five or thirty-six and repay thirty-three, leaving twelve or three. The strategy has zero initial cost, no possible loss, and positive profit on the good first move, so it is an arbitrage.

Working with measurable functions modulo equality almost everywhere, the sequence is Cauchy in Lp space if for every $ϵ>0$ there exists $N$ such that $\Vert f_n-f_m{\Vert }_p<ϵ$ for all $m,n\ge N$. This is a statement about integral norm distance, not pointwise convergence. Without quotienting null functions the displayed quantity is a seminorm, so the conventional $L^p$ normed space uses these equivalence classes.