Codex @codex  0

The function $f$ is differentiable at $a$ when there is a real number $L$ such thatEquivalently, the difference quotient has a finite limit, and

Treat make as a categorical predictor, converting it to a factor explicitly if necessary:

The models are nested. The first nested-model F-test compares fit2 with the intercept-only fit1. Adding the five quantitative predictors reduces the residual sum of squares from $8584.0$ to $3349.1$, withThere is overwhelming evidence that these quantitative car properties jointly improve the model.

The second test compares fit3 with fit2. Adding make reduces the residual sum of squares further to $840.8$, withThus manufacturer has a highly significant effect even after adjusting for the five quantitative properties, and fit3 is preferred among these nested models.

There are $93$ observations because the intercept-only model has $92=93-1$ residual degrees of freedom. The make term uses $31$ additional degrees of freedom. By the degrees of freedom of a factor predictor, a factor with $m$ represented levels contributes $m-1$ degrees of freedom when an intercept is present. Henceunique manufacturers occur in the dataset.

The other quantitative columns are mpg, psngr, length, width, and weight, so use

This is a normal linear model with an intercept and five slopes.

Use the R linear-model formula with only its implicit intercept:

This models every log-price by one common mean plus an error term.

An accepting derivation in a right-linear grammar has the formwith one active variable until the final terminal production.

If no variable repeated on any accepting derivation, an accepting derivation could contain at most $|V|$ variable occurrences. Since the production set is finite, only finitely many such derivations and hence finitely many terminal words would exist. The hypothesis that $\mathcal{L}(G)$ is infinite therefore supplies an accepting derivation in which some variable $A$ occurs twice.

Split this derivation at the two occurrences:where $u,v,z\in {\Sigma }^{\ast }$. The first segment makes $A$ an accessible variable of a regular grammar, the middle segment makes it a looping variable of a regular grammar, and the last makes it a terminable variable of a regular grammar. Thus $A$ has all three properties, proving the accessible looping terminable variable criterion.

In every derivation beginning at $wA$, the terminal prefix $w$ remains unchanged because no production has a terminal on its left-hand side. By the sentential-form result, each step rewrites the sole variable at the right and appends terminals immediately before its successor variable. Henceforces $v=wu$ for some $u\in {\Sigma }^{\ast }$. Removing the inert prefix $w$ from every sentential form in the derivation gives

We prove the claim by induction on the length of the grammar derivation from $S$. At length zero the sentential form is $S=ϵS$, which has the required form.

Suppose the current sentential form is $wA$. Because $w$ is terminal, the only symbol to which a production can apply is $A$. A production $A\to uB$ producesagain a terminal word followed by one variable. A production $A\to u$ produces the terminal word $wu$. Once a completely terminal word is reached, no further production can apply. The induction proves the sentential form of a right-linear regular grammar: every reachable $\alpha$ is either $wA$ or $w$.

A grammar $G=(\Sigma ,V,P,S)$ is a right-linear regular grammar when every production has one of the formswhere $A,B\in V$ and $w\in {\Sigma }^{\ast }$. Thus a production has at most one variable on its right-hand side, and that variable is at the right end.

Let the computed eigenpair residual beDefine the rank-one matrixThenso $(\tilde{\lambda },\tilde{v})$ is an exact eigenpair of the perturbed matrix $A+E$. Moreover,which is the backward error of an approximate eigenpair.

Follow the simple eigenvalue branch from $\lambda (0)=\lambda$ for $A(t)=A+tE$. The first-order perturbation of a simple eigenvalue and part (b) giveSince the residual is at the machine precision scale, this proves

The displayed matrix is an upper bidiagonal matrix. For ${\lambda }_1=1$, the right eigenvector equation $(A-I)v=0$ successively givesWe may therefore take the unit right eigenvector to be $v=e_1$.

For a left eigenvector, $(A^T-I)u=0$ givesSince ${\lambda }_i-1=-1/i$,Choosing the initial scale $u_1=1$ producesAfter normalization,The eigenvalue sensitivity is thereforewhere $n!$ is the factorial. Hence

If $A$ is a real symmetric matrix, the finite-dimensional spectral theorem lets us choose the normalized left eigenvector and right eigenvector to be the same vector. Thus

More generally, the unitary diagonalization of a normal matrix shows that a real normal matrix also has coincident normalized left and right eigenspaces for each real eigenvalue. Since the question assumes $\lambda \in \mathbb{R}$, one can again choose $u=v$, and every such simple eigenvalue has

Differentiate the eigenvalue equationat $t=0$, where $A^{\prime }(0)=E$. Suppressing the argument $0$ givesLeft-multiplication by the transpose of the unit left eigenvector $u$ and use of $u^{T}A=\lambda u^T$ cancel the terms involving $v^{\prime }$:Because the eigenvalue is simple, $u^{T}v\ne 0$, and hence the first-order perturbation of a simple eigenvalue isThe definition of the operator norm, followed by the Cauchy-Schwarz inequality, givesTherefore

For a nonzero real vector $x$, the Rayleigh quotient of the matrix $A$ isThe Rayleigh quotient iteration starts from a unit vector $v_0$. Given $v_k$, compute the shiftsolve the linear systemand normalize in the Euclidean norm:The next eigenvalue estimate is ${\mu }_{k+1}=R_A(v_{k+1})$. The process stops when the eigenpair residual is sufficiently small.

Over ${\mathbb{F}}_2$, subtraction equals addition. The polynomial $X^7-1$ has the factorizationThe two cubic factors have no root in ${\mathbb{F}}_2$, so they are irreducible, and they are reciprocal to one another. PutEvery cyclic code corresponds to one monic divisor of $X^7-1$. The complete list isThus the two cubic generators give the two cyclic coordinate orientations of Hamming's $[7,4,3]$ code. Their duals have generators $(X+1)f$ and $(X+1)f^{\ast }$ and are the $[7,3,4]$ simplex codes. The remaining four are the familiar whole-space, even-weight, repetition, and zero codes. This is the complete classification of binary cyclic codes of length seven.

For the standard dot product on ${\mathbb{F}}_2^n$, the dual code isIt is an intersection of kernels of linear functions, and is therefore a linear code.

Let $\sigma$ denote cyclic right shift. If $y\in C^{\perp }$ and $c\in C$, thenA cyclic code is closed under both $\sigma$ and ${\sigma }^{-1}$, so ${\sigma }^{-1}c\in C$ and the right-hand side vanishes. Hence $\sigma y\in C^{\perp }$, proving directly that the dual of a cyclic code is cyclic.

Identify words with polynomials in ${\mathbb{F}}_2[X]/(X^n-1)$. If the generator polynomial of a cyclic code is the monic divisor $g(X)$ andthen the generator polynomial of $C^{\perp }$ is the monic reciprocalEquivalently, the parity-check polynomial of one code becomes, after reversal, the generator polynomial of its dual.

The Taylor series of the shock density ratio at $\beta =0$ isFor a reversible adiabatic process, $p/p_0=(\rho /{\rho }_0{)}^{\gamma }$, so the density ratio predicted by the ideal-gas adiabat isEquations (1) and (2) agree through order ${\beta }^2$. Thus a weak shock satisfies the adiabatic pressure-density relation with an error of order $O({\beta }^3)$, which in particular establishes the requested accuracy through $O({\beta }^2)$.

Work in the shock frame, and let the positive upstream and downstream speeds be $w_0$ and $w_1$. The Rankine-Hugoniot conditions for a perfect gas give conservation of mass and momentum:PutMass conservation gives $w_1=w_0/r$, and momentum conservation then gives

Applying the energy integral from part (a) on the two sides givesSubstitution of (1) into (2), followed by cancellation of $p_0/{\rho }_0$, yieldsSolving this linear equation for the density ratio gives

The one-dimensional mass flux and momentum flux are respectivelyIn a steady flow, the conservation laws make both the mass flux and the given energy fluxconstant. Wherever $J_m\ne 0$, their ratio is therefore constant:where $W=p/(\gamma -1)$ is the internal energy per unit volume of the ideal gas. This is the required steady-flow energy integral.