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For an autonomous Lagrangian, the canonical momentum and energy areAlong a solution of the Euler-Lagrange equation,The absence of explicit time dependence is essential to this conservation law.

The classical action is $S_c(q_I,q_F,T)={\int }_0^{T}L(q_c,{\dot{q}}_c)dt$. If the final endpoint is moved along one fixed classical trajectory, while the initial endpoint is held fixed, the fundamental theorem of calculus givesThis total derivative is different from the partial derivative with fixed final position.

For a family of nearby classical trajectories, integration by parts gives the endpoint variationTo obtain the final term, the endpoint change in the variation at fixed parameter time is $\delta q(T)=\delta q_F-{\dot{q}}_F\delta T$, while the changing upper integration limit contributes $L_F\delta T$. The interior integral vanishes by the Euler-Lagrange equation. Therefore, on any smooth branch of classical boundary-value solutions,The chain rule along the endpoint trajectory now reads $d{S}_c/dT=p_F{\dot{q}}_F-E=L_F$, consistently with the direct calculation.

Choose $z=e^{it}$ on the right semicircle, with the continuous branch $\log z=it$. ThenHere the integral is along that semicircle. Since $|A|>1$, the pole lies outside the unit disc and the geometric series converges uniformly on the contour.

PutThe Euler integral for the hypergeometric function, initially for $\mathrm{Re}u>-1$, givesThe prefactor follows from $\Gamma (u+2)/\Gamma (u+1)=u+1$. Hence an antiderivative of the contour integrand before multiplication by $1/i$ isEvaluating at the endpoints with the chosen branch yieldsThis is first obtained where the individual expressions are defined. The original finite-interval integral is entire in $u$, so analytic continuation gives the result elsewhere, taking removable limits at the apparent exceptional negative integers. That convention matters: the two separate hypergeometric terms may be singular even though their combination is finite.

A useful version of the Poincaré-Bendixson theorem says that a forward orbit of a continuously differentiable planar autonomous vector field which stays in a compact set has a nonempty compact omega-limit set; if that omega-limit set contains no equilibrium, it is a periodic orbit.

Let $r^2=x^2+y^2$. Direct calculation givesFor $r>0$, $|xy|\le r^2/2$, so $-9/8\le \dot{\theta }\le -7/8$. In particular there is no equilibrium away from the origin.

On $r=1/2$, both coefficients in the radial expression are strictly positive, so the flow crosses outward. On $r=2$, both are strictly negative, so the flow crosses inward. Consequently the compact annulusis forward invariant. Any orbit starting there stays there, and its omega-limit set avoids the only equilibrium, the origin. The Poincaré-Bendixson theorem therefore supplies at least one periodic orbit in this annulus.

Put $u=\epsilon v+O({\epsilon }^2)$. The linearization at the zero solution isThe appropriate Fourier sine series isOrthogonality of the modes givesThus each mode grows or decays exponentially with its own eigenvalue, and the largest growth rate is the $n=1$ value. Strict linear stability analysis, meaning decay of every mode, is equivalent toAt equality the first mode is neutral in the linearization; the nonlinear terms decide its subsequent behavior. Above this length the first mode grows, giving linear instability.

Write $Y_i=\log ({\mathrm{price}}_i)$ and fit the normal linear modelThe basic linear regression analysis assumes centered independent normal random variables ${\epsilon }_i$ of a common variance ${\sigma }^2$, conditional on the predictors. Independence between makes does not by itself establish independence between different models of the same make; that stronger assumption should be checked, or within-make dependence modelled. The requested R fit is

fit <- lm(log(price) ~ mpg + psngr + length + width + weight, data = cars)

The estimate column contains the ordinary least squares coefficients. The standard error estimates the standard deviation of each coefficient estimator. The t value is its estimate divided by its standard error. The last column gives the two-sided p-value for a zero coefficient, using a Student t-distribution with $93-6=87$ residual degrees of freedom under the stated normal linear model; the stars summarize significance thresholds.

Conditional on the other predictors, passenger capacity and width have negative fitted coefficients, while length and weight have positive ones. All four are significant at the conventional 5% level; length is borderline. There is little evidence for an additional mpg effect once the others are included. These are conditional associations, not causal claims. For example, increasing weight by 100 pounds multiplies the fitted price by approximately $\exp (0.08373)$, about 1.087.

A reasonable candidate simplification removes mpg and compares the nested models by an F-test, rather than treating a nonsignificant coefficient as proof of no effect:

fit2 <- update(fit, . ~ . - mpg)
anova(fit2, fit)

Inspect regression residuals, normal quantiles, residual-versus-fitted plots, leverage and influential observations. The correlated size predictors may produce multicollinearity; missing make effects, nonlinear terms or unequal residual variances may also matter. A random make effect or a covariance adjustment addresses the possible dependence noted above.

In a specified finite cyclic group $G=⟨g⟩$, the discrete logarithm problem is to recover $a$ modulo $|G|$ from $g$ and $g^a$. A common choice is a large prime-order subgroup of a multiplicative group of a finite field.

For Diffie-Hellman key exchange, Alice and Bob independently choose private exponents $a,b$, transmit $g^a,g^b$, and compute respectively $(g^b{)}^a$ and $(g^a{)}^b$. Both obtainSolving a discrete logarithm problem reveals a private exponent and breaks this exchange. The converse implication is not known in general: recovering $g^{ab}$ from $g^a,g^b$ is the computational Diffie–Hellman problem, rather than literally the same task as computing a logarithm.

The advantage is that parties without an already shared secret can establish a fresh session key, which can then be used with efficient symmetric encryption. A classical secret-key system requires an earlier secure key distribution; ordinary public-key encryption can instead transmit a chosen key, but ephemeral Diffie-Hellman key exchange also permits forward secrecy when private session exponents are erased and the exchange is authenticated. Authentication is essential: an unauthenticated exchange is vulnerable to a man-in-the-middle attack.

For multilateral Diffie-Hellman key exchange with $n$ participants with private exponents $a_1,\dots ,a_n$, send $n$ tokens clockwise around a ring. Participant $j$ starts the token $g^{a_j}$ and sends it to participant $j+1$. On each subsequent receipt, the receiver raises the token to their own private exponent and sends it onwards, until that token has made $n-1$ transmissions and reaches participant $j-1$. Its exponent then contains every exponent except $a_{j-1}$. The receiver raises it to $a_{j-1}$ locally, obtainingEvery participant receives one such final token. There are $n$ tokens with exactly $n-1$ transmitted group elements per token, giving $n(n-1)$ transmissions. No private exponent is transmitted.

There are two conventions for loxodromic Möbius transformations. In the inclusive convention, a transformation conjugate to $z\mapsto \lambda z$ with $|\lambda |\ne 1$ is loxodromic, including a hyperbolic Möbius transformation with $\lambda >0$. Under that convention the assertion about discs is false: $z\mapsto 2z$ preserves the upper half-plane, and a Cayley conjugation gives an invariant ordinary disc.

The question uses the classification in which hyperbolic Möbius transformations and genuinely spiralling loxodromic Möbius transformations are separate. Thus here the multiplier satisfies $|\lambda |\ne 1$ and $\lambda \notin (0,\infty )$. Such a Möbius transformation has two distinct fixed points; after sending them to zero and infinity it acts by the stated dilation and rotation.

For a determinant-one representative, put $\tau =a+d$. Its eigenvalues are $\eta ,{\eta }^{-1}$, with $\eta +{\eta }^{-1}=\tau$, and the normalized multiplier is ${\eta }^2$ or its reciprocal. If $\tau$ is real, the eigenvalues are either real or conjugate unit-modulus numbers. These give the hyperbolic, elliptic, parabolic or identity cases, not a genuine spiral. Conversely, a real eigenvalue or a unit-modulus eigenvalue makes $\tau$ real. Hence the practical criterion in the question's convention isChanging the determinant-one representative by its sign does not affect this criterion.

Finally, conjugate an invariant disc to the upper half-plane. A Möbius transformation preserving that half-plane has a representative in $S{L}_2(\mathbb{R})$: preservation of the boundary gives real coefficients up to a common scalar, and preservation of the chosen side makes the real determinant positive, allowing normalization to one. Its trace is real. Trace is invariant under conjugacy, up to the harmless sign of the representative, so a disc-preserving transformation cannot be genuinely loxodromic. This proves the intended assertion while specifying its necessary terminology.

For one-dimensional isentropic compressible flow, continuity and momentum balance areSet $Q(\rho )={\int }_{{\rho }_0}^{\rho }c(s)ds/s$. Combining the equations gives the Riemann invariantsFor the power-law equation of state, $c=c_0(\rho /{\rho }_0{)}^{(\gamma -1)/2}$ and direct integration gives $Q=2(c-c_0)/(\gamma -1)$.

In the expansion into the region left of the undisturbed gas, the invariant $u-Q$ remains zero. Hence $c=c_0+(\gamma -1)u/2$. In the centred simple wave, the other characteristic speed equals $\xi =x/t=u+c$, givingWhen the gas follows the piston, $u=-V$ at the piston and $c_p=c_0-(\gamma -1)V/2$. This requires $V<2c_0/(\gamma -1)$ for positive density. The three gas regions areAt the threshold the uniform region collapses. If $V$ exceeds the threshold, a vacuum gap separates the piston from the gas, whose free edge moves at $-2c_0/(\gamma -1)$.

A particle initially at $x_0>0$ stays at rest until the fan head arrives at $t_0=x_0/c_0$. Afterwards it lies in the fan and satisfiesMultiplication by $t^{-2/(\gamma +1)}$ and integration give $x=C{t}^{2/(\gamma +1)}-2c_{0}t/(\gamma -1)$. Matching $x(t_0)=x_0$ determines $C$, yieldingSince $2/(\gamma +1)<1$, differentiating provesThus each fixed particle approaches the velocity of the vacuum edge.