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Let $e_n=y^n-y(t_n)$. Subtracting the exact one-step relation from the numerical method and applying the stated Lipschitz continuity givesIf $\Vert {\tau }_{n+1}\Vert \le C{h}^{p+1}$, iteration yields the discrete Gronwall inequalityFor $nh\le t^{\ast }$,andConsequently

The local truncation error is the error made by one numerical step started from the exact solution:A one-step method has local order $p+1$ when ${\tau }_{n+1}=O(h^{p+1})$ uniformly for $t_n$ in each fixed bounded time interval.

The dyed particle satisfies the Lagrangian trajectory equationsFor the proposed approximation, $x=O({\epsilon }^2)$ andAlsoThe initial conditions hold, verifyingthrough order ${\epsilon }^2$.

Over one period, the periodic terms return to their initial values while the secular term changes $x$ by ${\epsilon }^2\pi$. Hence the dyed particle has Stokes driftto this order, despite the zero Eulerian mean velocity.

Both $\sin (x-t)$ and $\cos (x-t)$ have zero average over one period, so $⟨u⟩=0$ at every fixed point.

The material derivative gives the particle accelerationDirect differentiation yieldsTherefore the Eulerian time average at fixed $(x,y)$ is

For a velocity potential, $u=\nabla \varphi$. Writing $\theta =x-t$ givesIts divergence isThus this potential flow is generally compressible.

Substituting the wire potential from part (b) into the flux formula from part (a) gives Neumann's mutual-inductance formulaInterchanging the two curves proves $L_{12}=L_{21}$.

Parametrize the coaxial circles byThen, with $\theta =\varphi -\psi$ and $R=\sqrt{a^2+b^2+c^2}$,where $q=2ab/R^2$. One angular integration contributes $2\pi$, soSince $ab=q{R}^2/2$, this is

The magnetostatic Ampère-Maxwell equation is $\nabla \times B={\mu }_{0}J$. Substituting $B=\nabla \times A$ and using the Coulomb gauge $\nabla \cdot A=0$ givesThe free-space Green function of the Laplacian therefore givesFor the thin wire current stated in the question this becomes

Since the magnetic field is $B=\nabla \times A$, Stokes theorem gives, for any oriented surface $S$ bounded by $C$,Under a gauge transformation $A\mapsto A+\nabla \chi$, the integral changes bybecause $C$ is closed. The flux expression is therefore gauge independent.

Before the change, the normalized ground state of the infinite square well isFor an allowed post-quench energy $E\in (0,U_0)$ satisfying part (i), define the unnormalized eigenfunctionIts normalization factor isThe Born rule therefore givesFor a value of $E$ that is not an eigenvalue, this probability is zero. The sudden change leaves the wavefunction fixed, while the energy eigenbasis changes.

For a normalized wavefunction obeying the infinite-wall Dirichlet boundary condition, integration by parts gives the energy expectationThus every energy eigenvalue is nonnegative.

For $0<E<U_0$, the time-independent Schrödinger equation and the wall conditions givewhere $k=\sqrt{2mE}/\hslash$ and $l=\sqrt{2m(U_0-E)}/\hslash$. Continuity of $\psi$ and ${\psi }^{\prime }$ at the finite potential step givesDividing and rearranging yields the bound-state quantization condition

Extend the forcing oddly across the boundary:The initial displacement $\sin x$ is already odd, so its homogeneous evolution on the line is $\sin x\cos (ct)$. Applying Duhamel's principle to the odd extension givesThe odd-reflection method makes $u(0,t)=0$. At $t=0$ the double integral vanishes together with its first time derivative, so the prescribed initial displacement and velocity are also satisfied.

With $\xi =x+ct$ and $\eta =x-ct$,so the wave equation becomesThus $u=F(\xi )+G(\eta )$. At $t=0$ the initial data giveSolving for $F^{\prime }$ and $G^{\prime }$ and integrating gives d'Alembert's formula>

Forboth $e^z-1$ and $\log (1+z)$ have a simple zero at zero. Hence $g$ has a simple pole, and its residue is

For $h(z)=\sin z\sin (1/z)$, multiplication of the two Laurent series shows that all powers are even:There are infinitely many negative powers, so zero is an essential singularity. There is no $z^{-1}$ term, and therefore

Put $g=1/f$. The hypothesis $|f(z)|\to \infty$ gives $g(z)\to 0$ as $z\to a$, so defining $g(a)=0$ makes $g$ continuous on $U$ and analytic there by the stated assumption. Its zero at $a$ has some finite order $m\ge 1$, and henceThereforehas a pole of order $m$: its Laurent series has $c_{-m}\ne 0$ and $c_n=0$ for $n<-m$.

Now let $f$ be entire and tend to infinity at infinity. The functiontends to infinity as $z\to 0$, so the preceding argument says that $h$ has a pole at zero. If the Taylor series of $f$ is $f(w)={\sum }_{n\ge 0}{a}_n{w}^n$, thenA pole has only finitely many negative powers, so $a_n=0$ for all sufficiently large $n$. Thus $f$ is a polynomial.

The Laurent theorem says that if $f$ is analytic on an annulusthen it has a unique Laurent seriesconverging locally uniformly on that annulus, wherefor any positively oriented circle $C$ in the annulus around $a$.

An isolated singularity at $a$ is a point at which $f$ is not analytic although it is analytic on some punctured neighbourhood. It is removable when every $c_n$ with $n<0$ vanishes; it is a pole of order $m$ when $c_{-m}\ne 0$ and $c_n=0$ for $n<-m$; and it is essential when infinitely many negative-index coefficients are nonzero.

For $0<|z|<1$,For $|z|>1$,The coefficients are unique after the annulus is fixed; these expansions differ because they represent the function on different annuli. At zero the first expansion has principal part $-z^{-1}$, so zero is a simple pole with residue $-1$.

Assume the sequential condition, and let $A\subseteq Y$ be open. If $f^{-1}(A)$ were not open, some $a\in f^{-1}(A)$ would have no ball contained in $f^{-1}(A)$. For each $n$, chooseThen $x_n\to a$, so $f(x_n)\to f(a)\in A$. Because $A$ is open, this forces $f(x_n)\in A$ eventually, a contradiction. Therefore $f^{-1}(A)$ is open. This proves the sequential characterization of continuity in metric spaces.

Assume inverse images of open sets are open. If $x_n\to a$ and $V$ is any open neighbourhood of $f(a)$, then $f^{-1}(V)$ is an open neighbourhood of $a$. By the definition of convergence in a metric space, $x_n\in f^{-1}(V)$ eventually, and hence $f(x_n)\in V$ eventually. Thus $f(x_n)\to f(a)$.

This is always true for metric spaces. Let $x_n\to x$. If $f(x_n)$ did not converge to $f(x)$, some subsequence would remain at least $\epsilon >0$ from $f(x)$. Compactness of $Y$ gives a further subsequenceThen $(x_{n_k},f(x_{n_k}))\to (x,y)$. The graph is closed, so $(x,y)$ belongs to it and $y=f(x)$, contradicting the $\epsilon$ separation. Thus $f(x_n)\to f(x)$ for every convergent sequence, and the equivalence proved above makes $f$ continuous. This is the closed-graph criterion with compact codomain.

This is always true. The continuous image $f(X)$ of a connected space is connected. Its closure is also connected: if the closure were separated into disjoint nonempty relatively open sets, connectedness would put $f(X)$ inside one of them, preventing its closure from meeting the other. Since $f(X)$ is dense in $Y$, its closure is $Y$, so $Y$ is connected.

This may be false because continuity only forces the continuous image of a compact space $f(X)$ to be compact, not the whole codomain. For example, let $X=\{0\}$, let $Y=\mathbb{R}$, and set $f(0)=0$. The domain is compact and $f$ is continuous, but $\mathbb{R}$ is not compact.