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Statement (ii) is false. The condition is necessary for an idempotent, but it is not sufficient. For example,satisfies $(I-\alpha {)}^2=0$, hence $\alpha (I-\alpha {)}^2=0$, whileover, for example, $\mathbb{Q}$.

An endomorphism $\alpha$ is a projection onto its image precisely when it is the identity on $\mathrm{im}\alpha$ and kills a complementary subspace. If ${\alpha }^2=\alpha$, thenso $\alpha$ is the identity on its image. Moreover,and $\mathrm{im}\alpha \cap \ker \alpha =\{0\}$. Henceand $\alpha$ is the projection onto its image along its kernel. Conversely, every projection is the identity after one application, so applying it twice has the same effect: ${\alpha }^2=\alpha$.

Statement (i) is false. On a two-dimensional vector space, take the nonzero nilpotent endomorphismThen ${\alpha }^2={\alpha }^3=0$, but ${\alpha }^2\ne \alpha$, so $\alpha$ is not idempotent.

Full column rank of $X$ makes $X^T:{\mathbb{R}}^n\to {\mathbb{R}}^p$ surjective. Given $x\in {\mathbb{R}}^p$, choose $u$ with $x=X^{T}u$. The fitted values for the unrestricted and restricted regressions areThereforeUsing $\mathrm{cov}(Y)={\sigma }^2{I}_n$ and idempotence of the projections,Part (d) now gives

Because $\mathrm{col}Z\subseteq \mathrm{col}X$, the two orthogonal projection matrices satisfyConsequently $P-P_0$ is itself an orthogonal projection, onto the orthogonal complement of $\mathrm{col}Z$ within $\mathrm{col}X$. In particular it is a positive semidefinite matrix, so for every $u\in {\mathbb{R}}^n$,

Write $\tilde{\beta }=({\tilde{\gamma }}^T,0{)}^T$, whereThe final coordinate is deterministic, while the first $p-1$ coordinates have the usual least-squares covariance. Hence

Since $\hat{\beta }={\beta }^0+(X^{T}X{)}^{-1}{X}^T\epsilon$ and $\mathrm{cov}(\epsilon )={\sigma }^2{I}_n$,

Because $X$ has full column rank, $X^{T}X$ is invertible. The ordinary least squares estimator isIt satisfies the normal equationsso the residual is orthogonal to the column space of $X$. For any $\beta$,The two terms are orthogonal, and the Pythagorean theorem in an inner-product space givesThus $\hat{\beta }$ minimizes the least-squares objective.

The first characteristic polynomial of the linear multistep method comes only from its left side:Its roots are zero and one; both lie in the closed unit disc, and the only unit-modulus root is simple. Thus the method satisfies the root condition for a multistep method and is zero-stable. Part (b) showed consistency for every $\alpha ,\beta$. The Dahlquist equivalence theorem states that a consistent linear multistep method is convergent exactly when it is zero-stable. Thereforeunder the usual Lipschitz hypotheses on $f$.

Expand the exact solution about $t_{n+1}=t$. The left side isSince $f(t,y(t))=y^{\prime }(t)$, the right side expands asThe $h{y}^{\prime }$ terms agree for every $\alpha ,\beta$, so the method is always consistent and is generically of order one.

The method is explicit exactly when $1+\alpha =0$. With $\alpha =-1$, its $h^2$ defect vanishes only for $\beta =3/2$, so a generic explicit member has order one; that exceptional explicit member has order two and cannot satisfy the next order condition.

For the unrestricted method to have order at least three, the $h^2$ and $h^3$ coefficients must agree:Solving givesFor these values, $1+2\alpha +\beta =1/2$, so the right-side $h^4$ coefficient is $1/12$, different from the left-side coefficient $1/24$. Hence the order is exactly three.

A one-step or linear multistep method has order $p$ if, when the exact sufficiently smooth solution is substituted for the numerical values, its unscaled local truncation error is $O(h^{p+1})$, and this estimate is not generally $O(h^{p+2})$. For a stable method this corresponds to a global error of order $O(h^p)$ on a fixed time interval.

At $z=0$, the two free-surface conditions for a normal mode giveEliminating $\hat{\zeta }$ yieldsFor the vertical factor found in part (b),Thus the gravity-wave dispersion relation iswhere

For an axisymmetric separated amplitude $\hat{\varphi }(r,z)=R(r)Z(z)$, Laplace equation in cylindrical coordinates becomesThe radial equation isIts solution regular at the axis is the Bessel function $J_0(kr)$. The vertical equation is $Z^{\prime \prime }-k^{2}Z=0$, and the bottom condition $Z^{\prime }(-h)=0$ selectsup to an irrelevant constant multiplier. Finally, the sidewall condition givesTherefore the allowed positive wavenumbers areand

Equation (1) is Laplace equation for the velocity potential. It expresses incompressibility, $\nabla \cdot \mathbf{u}=0$, together with $\mathbf{u}=\nabla \varphi$. Equation (2) is the linearized dynamic free-surface condition: the unsteady Bernoulli equation and constant atmospheric pressure balance the potential acceleration against the gravitational restoring term $g\zeta$. Equation (3), with the printed $c$ understood as $\zeta$, is the kinematic free-surface condition: the surface moves with the vertical fluid velocity. Equations (4) and (5) are no-penetration boundary conditions at the rigid bottom and cylindrical sidewall, respectively.

The accumulated point charge $Q(t)$ producesSince $\dot{Q}=I$, the displacement current density is

Apply the integral Ampère-Maxwell equation to the boundary of a spherical cap of radius $r$ and polar angle $\theta <\pi /2$. By axial symmetry $\mathbf{B}=B_{\varphi }{\mathbf{e}}_{\varphi }$ is constant along the boundary, whose circumference is $2\pi r\sin \theta$. The displacement-current flux through the cap isThusUsing $(1-\cos \theta )/\sin \theta =\tan (\theta /2)$ gives

Translation symmetry in the plane and reflection symmetry across it force the magnetic field to be uniform on each side, parallel to ${\mathbf{e}}_y$, and opposite on the two sides. Apply Ampère's law to a rectangular loop perpendicular to the sheet whose two long sides have length $L$. The enclosed current is $KL$, soThe right-hand rule fixes the directions:Consequentlywhich is the magnetic jump condition for a surface current density.

Write $T=\tanh (ax)$ and $S={\mathrm{sech}}^2(ax)$. Usingtwo differentiations ofgiveThus it solves the time-independent Schrödinger equation in the stated Pöschl-Teller potential, with

As $x\to -\infty$ and $x\to +\infty$, respectively,There is no left-moving term, so this is a reflectionless potential:The incident and transmitted amplitudes have equal modulus becauseHence

Now set $k=i\lambda$ with $\lambda >0$. ThenAt $+\infty$ this decays for every positive $\lambda$. At $-\infty$, however, the factor tends to $\lambda -a$ while $e^{-\lambda x}$ grows. Normalizability therefore requires $\lambda =a$. In that casewhich decays at both ends. The bound-state energy is

For a one-dimensional wavefunction, the probability current isA plane wave $D{e}^{ikx}$ carries current $(\hslash k/m)|D{|}^2$ to the right, while $D{e}^{-ikx}$ carries the negative of this current. Thus the asymptotic terms are interpreted as an incident wave of amplitude $A$, a reflected wave of amplitude $B$, and a transmitted wave of amplitude $C$. Their currents areCurrent conservation for a real potential gives $j_{\mathrm{in}}+j_{\mathrm{ref}}=j_{\mathrm{tr}}$, and thereforewith $P_{\mathrm{ref}}+P_{\mathrm{tr}}=1$.

A normalizable bound state has $E<0$. Writingits decaying asymptotic behaviour is

By separation of variables, the axisymmetric normal modes satisfying regularity at $r=0$ and the fixed-edge condition at $r=1$ arewhere $J_0({\gamma }_k)=0$. The zero initial displacement removes all cosine terms, soDifferentiating at $t=0$ gives the weighted Fourier-Bessel expansionMultiply by $r{J}_0({\gamma }_{\ell }r)$ and integrate from zero to one. The orthogonality and norm from part (i) giveUsingthe remaining integral isHenceand therefore

LetTheir Bessel equations in self-adjoint form areMultiply the first by $v$, the second by $u$, subtract, and integrate. The left side becomes a boundary term:Regularity at zero makes the lower boundary term vanish, whileTherefore

If $\alpha ={\gamma }_k$ and $\beta ={\gamma }_{\ell }$ with $k\ne \ell$, both endpoint values of $J_0$ vanish, so the integral is zero. This is the weighted orthogonality of distinct eigenfunctions in a Sturm-Liouville problem. For the norm, hold $\alpha ={\gamma }_k$ and let $\beta \to {\gamma }_k$. Sincethe quotient tends to

Put $w=z-1$. ThenThusThis is actually a Taylor series, because $z=1$ is a regular point. The point $z=0$ is a simple pole. At infinity,extends analytically to $w=0$, so infinity is a removable singularity of $f$, with extended value zero.

For the real integral, first writeLetWith $z=e^{i\theta }$,The denominator has roots $1/2$ and $2$, so only $1/2$ lies inside the contour. Its residue is $1/(3i)$, and the residue theorem givesConsequently