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The three nonzero vectors in ${\mathbb{F}}_2^2$ are $e_1$, $e_2$ and $e_1+e_2$. An invertible linear map permutes this three-element set, so part (b) gives a group homomorphismIf a matrix acts trivially on all three vectors, it fixes both basis vectors $e_1,e_2$, so it is the identity matrix. Thus the kernel of a group homomorphism is trivial and $\rho$ is injective.

To count the domain, its first column may be any of the three nonzero vectors. The second column must be outside the two-element span of the first, leaving two choices. HenceAn injection between these finite sets of equal size is surjective, proving

Put $v=P_2-P_1$ and $w=Q_2-Q_1$. They are nonzero vectors with the same Euclidean norm. Choose the rotation matrix $R$ taking the direction $v/\Vert v\Vert$ to $w/\Vert w\Vert$; equal lengths then give $Rv=w$. DefineThis is a translation by $Q_1-P_1$ followed by a rotation about $Q_1$, so it is an isometry. Direct substitution gives $t(P_1)=Q_1$ and $t(P_2)=Q_1+w=Q_2$.

The total derivative of $f$ at $x$ is a linear map $Df(x):{\mathbb{R}}^n\to \mathbb{R}$ satisfyingThe gradient is the vector of partial derivatives, $\nabla f(x)=({\partial }_{1}f(x),\dots ,{\partial }_{n}f(x))$. When $f$ is totally differentiable, these partial derivatives exist andThus the derivative is a linear functional represented by the gradient under the Euclidean inner product. Existence of partial derivatives alone does not imply total differentiability.

Put $v=P_2-P_1$ and $w=Q_2-Q_1$. They are nonzero vectors with the same Euclidean norm. Choose the rotation matrix $R$ taking the direction $v/\Vert v\Vert$ to $w/\Vert w\Vert$; equal lengths then give $Rv=w$. DefineThis is a translation by $Q_1-P_1$ followed by a rotation about $Q_1$, so it is an isometry. Direct substitution gives $t(P_1)=Q_1$ and $t(P_2)=Q_1+w=Q_2$.

The total derivative of $f$ at $x$ is a linear map $Df(x):{\mathbb{R}}^n\to \mathbb{R}$ satisfyingThe gradient is the vector of partial derivatives, $\nabla f(x)=({\partial }_{1}f(x),\dots ,{\partial }_{n}f(x))$. When $f$ is totally differentiable, these partial derivatives exist andThus the derivative is a linear functional represented by the gradient under the Euclidean inner product. Existence of partial derivatives alone does not imply total differentiability.

For a sample of size $n\ge 2$, defineFor the zero-mean Gaussian sample, the standardized vector $(X_1/\sigma ,\dots ,X_n/\sigma )$ has independent standard normal coordinates. Apply an orthogonal transformation whose first row is $(1,\dots ,1)/\sqrt{n}$. The resulting coordinates $Z_1,\dots ,Z_n$ remain independent standard normals: their joint density depends only on squared Euclidean length and the transformation has unit absolute Jacobian. Its first coordinate is $Z_1=\sqrt{n}\overline{X}/\sigma$, while orthogonal decomposition givesThusThe ratio of the first independent standard normal and the square root of the normalized chi-square variable has a Student probability distribution, giving

For the paired height observations, form one cross-minus-self difference per pot. The sampling units for this test are the fifteen differences, not thirty separate plant heights. Under the usual independent normal model for differences, a null mean of zero gives a $t_{14}$ statistic. The observed statistic isFor the two-sided alternative, reject at significance level $\delta$ if $|t_{\mathrm{obs}}|>t_{14,1-\delta /2}$. Its two-sided $p$-value is $2[1-F_{t_{14}}(|t_{\mathrm{obs}}|)]$. This uses the recorded values without performing the numerical calculation forbidden by the question. The normal difference model is what makes the small-sample Student calibration exact.

Let $E(t)$ be the matrix whose columns are the moving orthonormal basis vectors. Then $E{E}^T=I$. Differentiating givesso $W=\dot{E}{E}^T$ is skew-symmetric. Every real skew-symmetric $3\times 3$ matrix has the unique formwhich acts on a vector as $Wv=\mathbf{\omega }\times v$. Since $\dot{E}=WE$, each column satisfiesThe vector $\mathbf{\omega }$ is the instantaneous angular velocity. It may also be written $\mathbf{\omega }=\frac{1}{2}{\sum }_i{\mathbf{e}}_i\times {\dot{\mathbf{e}}}_i$.

Write $r=(x_1,x_2,x_3{)}^T$ for position relative to the centroid. The inertia tensor there isIn particular its diagonal entries integrate the squares of distances from the corresponding axes, and its off-diagonal entries are the negative products of coordinates.

Insert ${\mathbf{v}}_Q={\mathbf{v}}_G+\mathbf{\omega }\times r$ into $T=\frac{1}{2}{\int }_A\rho |{\mathbf{v}}_Q{|}^{2}dV$. The cross term vanishes because ${\int }_A\rho rdV=0$. Using $|\mathbf{\omega }\times r{|}^2=|\mathbf{\omega }{|}^2|r{|}^2-(\mathbf{\omega }\cdot r{)}^2$ givesIf $a$ is the vector from $G$ to $O$, positions relative to $O$ are $r-a$. Expanding the defining tensor integral with this substitution again removes the terms linear in $r$. Thus the tensor parallel axis theorem isSince $O$ is fixed, ${\mathbf{v}}_G=\mathbf{\omega }\times (-a)$, so $m|{\mathbf{v}}_G{|}^2=m{\mathbf{\omega }}^T(|a{|}^{2}I-a{a}^T)\mathbf{\omega }$. Substitution into the energy decomposition yields

The columns outside indices $i,j$ are unchanged coordinate vectors. The remaining two columns have nonzero entries only in rows $i,j$, with the $2\times 2$ block $M(\theta )$. They have squared norm ${\cos }^2\theta +{\sin }^2\theta =1$ and mutual inner product zero; they are also orthogonal to all the unchanged columns. Hence $J^{T}J=I$.

Let $r=\sqrt{x_i^2+x_j^2}$. If $r>0$, choose $\theta \in [-\pi ,\pi ]$ with $\cos \theta =x_i/r$ and $\sin \theta =x_j/r$. ThenEvery other coordinate is unchanged. If $r=0$, choose $\theta =0$; the same conclusions hold. Therefore the procedure can be fixed with the nonnegative pivot conventionThis is a Givens rotation.

Let $N$ be the vector subspace of integrable functions that vanish almost everywhere. The Lebesgue space is the quotient vector space $L^1(\mathbb{R})={\mathcal{L}}^1(\mathbb{R})/N$: its elements are equivalence classes of functions equal almost everywhere. Addition and scalar multiplication are $[f]+[g]=[f+g]$ and $c[f]=[cf]$.

Set $\Vert [f]{\Vert }_1=\int |f|$. This is well-defined because replacing $f$ by an equal-almost-everywhere representative does not change the integral. It is nonnegative, homogeneous, and obeys the triangle inequality by $|f+g|\le |f|+|g|$. Part (i) says its only zero element is the zero class. Hence it is a norm on $L^1$.

For the surface of revolution $r(\eta ,\zeta )=(\eta ,f(\eta )\cos \zeta ,f(\eta )\sin \zeta )$, the coordinate tangent vectors have squared lengths $1+f^{\prime 2}$ and $f^2$, and their dot product is zero. Hence its first fundamental form isUse the energy Lagrangian $L=A{\dot{\eta }}^2+f^2{\dot{\zeta }}^2$. Its Euler-Lagrange equations areFor a meridian $\zeta ={\zeta }_0$, choose arc length as parameter, so $\dot{\eta }=1/\sqrt{A}$. Differentiating gives $2A\ddot{\eta }+A^{\prime }{\dot{\eta }}^2=0$, while the second equation is automatic. Part (b) makes this path an extremal of length. Thus every meridian is a geodesic.

For a nonconstant parallel $\eta ={\eta }_0$, choose constant nonzero $\dot{\zeta }$. The second equation holds, and the first reduces to $f({\eta }_0)f^{\prime }({\eta }_0){\dot{\zeta }}^2=0$. Since $f>0$, this is equivalent toConversely that condition makes both equations hold at constant speed, hence supplies a length extremal. Geometrically, a parallel is geodesic exactly at a stationary radius of the surface.

Differentiating orthonormality gives ${\dot{e}}_i\cdot e_j=-e_i\cdot {\dot{e}}_j$. With $\omega =\frac{1}{2}{\sum }_i{e}_i\times {\dot{e}}_i$, the vector triple product yieldsThe body coordinates $x_i$ of a fixed material point are constant. Differentiating its position $R_G+{\sum }_i{x}_i{e}_i$ therefore givesThis identifies $\omega$ as the angular velocity of the rigid body.

Because $G$ is the center of mass, $\int \rho rdV=0$, so the translational contribution to the angular momentum about $G$ vanishes. The remaining vector triple product givesThis symmetric matrix is the inertia tensor about $G$. The angular-momentum balance about the center of mass is $dL/dt={\tau }_{\mathrm{ext},G}$ in an inertial frame. With no external forces, that torque is zero, so the inertial vector $L$ is constant. Its components in rotating body axes need not individually be constant.

For the tool, let the disc lie at $z=0$ and the rod extend to $z=l=\sqrt{3}a$. The total mass is $3M$ and its center is at $z_G=l/3$. Axial symmetry makes the inertia tensor diagonal with $I_x=I_y$.

The standard disc moments are $M{R}^2/2$ about its normal and $M{R}^2/4$ about a diameter. A thin rod has moment $m{l}^2/12$ about a transverse axis through its center. The parallel axis theorem adds $m{d}^2$ for a parallel shift of distance $d$. ThereforeandThe rod contributes zero about its own longitudinal axis. Thus $\boxed{J=2M{a}^2{I}_3}$, an isotropic inertia tensor even though the shape is not spherical. This gives a nonspherical rigid body with isotropic inertia. Consequently $L=2M{a}^2\omega$. Conservation of $L$ in torque-free motion proves the angular velocity is constant for every initial orientation and spin.

The unrestricted likelihood is maximized at the empirical proportions, while the null likelihood is maximized at ${\pi }_i(\hat{\theta })$. Twice the log of their ratio isLarge values mean the fitted null likelihood is much smaller than the unrestricted maximum, so rejection is in the upper tail.

For a regular identifiable scalar parameter lying in the interior, positive true cell probabilities, and sufficiently large expected counts, Wilks theorem gives an approximate chi-squared distribution with $(k-1)-1=k-2$ degrees of freedom under the null. The unrestricted probability vector has dimension $k-1$ and the fitted null has dimension one. At significance level $\alpha$, reject if $\Lambda (\hat{\theta })>F_{k-2}^{-1}(1-\alpha )$, or report the approximate p-value $1-F_{k-2}(\Lambda )$.

Cells with very small expected counts may require a prespecified pooling scheme or simulation calibration; the asymptotic reference is not an exact finite-sample law. For a regular null with $d$ fitted parameters, the corresponding degrees of freedom are $k-1-d$.

Let $N$ be the vector subspace of integrable functions that vanish almost everywhere. The Lebesgue space is the quotient vector space $L^1(\mathbb{R})={\mathcal{L}}^1(\mathbb{R})/N$: its elements are equivalence classes of functions equal almost everywhere. Addition and scalar multiplication are $[f]+[g]=[f+g]$ and $c[f]=[cf]$.

Set $\Vert [f]{\Vert }_1=\int |f|$. This is well-defined because replacing $f$ by an equal-almost-everywhere representative does not change the integral. It is nonnegative, homogeneous, and obeys the triangle inequality by $|f+g|\le |f|+|g|$. Part (i) says its only zero element is the zero class. Hence it is a norm on $L^1$.

Work in the W boson rest frame, taking the spin-quantization axis along its laboratory motion. Neglect charged-lepton and neutrino masses. The weak charged current produces a negative-helicity ${\ell }^{-}$ and a positive-helicity antineutrino. Their momenta are opposite, so both spin Vectors point opposite to the charged lepton's direction $\hat{\mathbf{n}}$; the final total spin projection along $\hat{\mathbf{n}}$ is $-1$.

Let ${\lambda }_W=+1,0,-1$ denote the parent spin projection along $z$. In the given probability amplitude, use $j=1$ and the rotation element $d_{-1,{\lambda }_W}^1(\theta )$. The supplied spin-one rotation elements giveThe transverse states have spin projections $\pm 1$; the longitudinal state has projection zero, not total spin zero. The original diagram below distinguishes momenta from spins and shows the corresponding normalized angular profiles.

Work in the parent's rest frame, with $M_A>0$. Four-momentum conservation givesso both daughter momenta have magnitude $p$. Their relativistic energy-momentum relations give $E_B^2-p^2{c}^2=M_B^2{c}^4$ and $E_C^2-p^2{c}^2=M_C^2{c}^4$. Subtracting these two relations and using the total energy yieldsSubstitution back gives the relativistic two-body decay momentum:The unit vector $\mathbf{n}$ is not fixed by four-momentum conservation alone. The expression under the square root is the Källén function $\kappa (M_A^2,M_B^2,M_C^2)$. The physical two-body decay threshold is $M_A\ge M_B+M_C$; for massive daughters equality gives zero outgoing momentum. For two massless daughters, each has energy $M_A{c}^2/2$ and momentum magnitude $M_{A}c/2$.

For two particles, let $M=m_1+m_2$ and ${\mathbf{V}}_{\mathrm{cm}}=(m_1{\mathbf{v}}_1+m_2{\mathbf{v}}_2)/M$. The centre-of-mass frame velocities are ${\mathbf{v}}_i^{\ast }={\mathbf{v}}_i-{\mathbf{V}}_{\mathrm{cm}}$, so $m_1{\mathbf{v}}_1^{\ast }+m_2{\mathbf{v}}_2^{\ast }=0$. Thus before and after an isolated collision the two momentum vectors are equal and opposite. For an elastic collision their common magnitude is also unchanged: only their direction rotates.

The kinetic energy decomposition about the center of mass separates unchanged bulk translation from relative motion:The reduced mass $\mu$ reduces the calculation to one relative velocity. After solving the geometry in the centre-of-mass frame, add ${\mathbf{V}}_{\mathrm{cm}}$ to each velocity to return to the laboratory. The four panels below illustrate this for the subsequent example.

The integral and differential forms of Gauss's law in electrostatics areHere $\partial \mathcal{V}$ is a closed surface bounding a volume $\mathcal{V}$, $d\mathbf{A}=\hat{\mathbf{n}}dA$ is its outward area vector, $\mathbf{E}$ is the electric field, $\rho$ is total volume charge density, and $Q_{\mathrm{enc}}={\int }_{\mathcal{V}}\rho d^{3}r$ is enclosed electric charge. The divergence theorem converts one form to the other. For ideal sheets or point charges, interpret the differential equation with their distributional charge densities.

Faraday's law of induction states that the oriented electromotive force around a closed material circuit is minus the rate of change of its linked magnetic flux:The surface orientation and circuit direction obey the right-hand rule. The minus sign expresses Lenz's law: the induced electric current, when allowed to flow, produces a magnetic field opposing the change of linked magnetic flux. For a stationary circuit this reduces to $\oint \mathbf{E}\cdot d\mathbf{\ell }=-d{\Phi }_B/dt$.

The source's “Lorenz force” is the Lorentz force, $q(\mathbf{E}+\mathbf{v}\times \mathbf{B})$. A wire element of cross-section $A_c$ contains $n{A}_{c}d\ell$ carriers of charge $q$ with drift velocity ${\mathbf{v}}_d$. Adding their magnetic forces and using $I=nq{A}_c{v}_d$ givesfor a straight wire in a uniform magnetic field. The length vector follows conventional electric current, irrespective of the carrier sign.

For the unlabelled upward-motion request, define $\hat{\mathbf{t}}=(\cos \varphi ,0,-\sin \varphi )$ down the slope, $\hat{\mathbf{b}}=(0,1,0)$ across the rails and $\hat{\mathbf{n}}=(\sin \varphi ,0,\cos \varphi )$ out of their plane. Choose the vertical magnetic field as $\mathbf{B}=B\hat{\mathbf{z}}$. An electric current $I$ across the bar in direction $-\hat{\mathbf{b}}$ gives a horizontal magnetic force $-IBs\hat{\mathbf{x}}$. The other forces are weight $-mg\hat{\mathbf{z}}$ and the rail normal force $N\hat{\mathbf{n}}$. Uniform upward velocity requiresConsequentlyEach symmetric rail supplies half of $N$. The electric current must make $I\mathbf{L}\times \mathbf{B}$ point horizontally toward the uphill side; for the coordinates just defined it flows along $-\hat{\mathbf{b}}$. Reversing the vertical magnetic field reverses this current direction. The external supply must maintain that electric current against the motional electromotive force during ascent. For $\cos \varphi =0$, no finite horizontal magnetic force can support uniform motion along the vertical rails.

Let the original cylinder axis be $\mathbf{r}=0$, and let the hole axis be at transverse vector $\mathbf{a}$, with $|\mathbf{a}|=a$. The charge density is reproduced by a full cylinder of radius $R_1$ and charge density $+\rho$, superposed with a cylinder of radius $R_2$ centred at $\mathbf{a}$ and charge density $-\rho$. This uses the superposition principle for fixed insulating electric charges; the hole is not a hole in an electrical conductor.

For a complete uniformly charged infinite cylinder of radius $R$, a coaxial Gaussian surface of length $\ell$ gives, by Gauss's law,Consequently defineThe complete electric field, including the exterior if desired, isIn the insulating material, $|\mathbf{r}|<R_1$ and $s=|\mathbf{r}-\mathbf{a}|>R_2$, soInside the hole both contributions use their interior formulas, and all dependence on $\mathbf{r}$ cancels:The electric field in the hole is uniform, points from the original axis towards the hole axis for $\rho >0$, and vanishes for a concentric hole. The two expressions agree at the hole wall, as expected with no imposed surface charge density there. Its uniformity is the useful uniform electric field in an off-axis cylindrical cavity result.