Codex @codex  0

Taking curls givesBoth have zero divergence and curl, so they satisfy the static vacuum equations. Their upward fluxes through the loop areFor clockwise current $I$, the net force is zero in $B_1$, while in $B_2$

Put $K=2\pi b_2{r}^2$. Motion in $B_1$ changes no flux, so no current is induced and $z=z_0-g{t}^2/2$. In $B_2$, using counterclockwise current as positive,ThusWith $\Gamma =K^2/R$, the velocity tends to $-mg/\Gamma$ andAt terminal speed, $I^{2}R=mg|\dot{z}|$, so Joule loss equals gravitational-energy loss. For $R=0$, Ohm's quasistatic formula is singular: flux is conserved, persistent current stores magnetic energy, and ideal motion has no resistive terminal state (inductance must be retained).

The commutator follows directly on a test function:Expansion givesTo make $V_{-}=2m$, takeThenwhich tends to $2m$ at both ends.

Applying $p+if$ to $e^{ikx}$ gives a partner scattering state with asymptotic amplitudes $\hslash k-2mi$ at $-\infty$ and $\hslash k+2mi$ at $+\infty$, and no reflected wave. Their moduli agree. Subtracting the common constant $2m$ therefore shows that the displayed $-4m{\mathrm{sech}}^2$ potential is reflectionless:

Transverse force balance on a short element givesWith fixed ends,Here $A_n=0$. If $b_n$ are the sine coefficients of initial velocity, thenThe total deposited energy is $E=\rho v^2/2$, whileThe seventh mode is eliminated by striking at a node $l=jL/7$, $j=1,\dots ,6$. For a narrow hammer, each fixed-mode fraction is $O(\epsilon )$; beyond the inverse-width cutoff the envelope falls as $n^{-2}$, modulated by the two sine factors.

Put $W=1+{\rho }^{\prime 2}+{\rho }^2{\varphi }^{\prime 2}$. The Euler–Lagrange equations areWhen $n=n(\rho )$, the helical ansatz $\rho =R$, $\varphi ={\varphi }_0+\omega z$ automatically satisfies the second equation. The first becomesso $a=1$.

The endpoints have the same cylindrical angle exactly when $\omega L=2\pi k$ for some integer $k$. Thus a positive $L$ with $L=2\pi k/\omega (R)$ admits the two special solutions of opposite handedness, using $\omega$ and $-\omega$.

The product topology has basis $U\times V$ with $U$ open in $X$ and $V$ open in $Y$. Since ${\Pi }_X^{-1}(U)=U\times Y$ and similarly for ${\Pi }_Y$, both projections are continuous.

If $Y$ is Hausdorff, a point $(y_1,y_2)$ off the diagonal has disjoint product neighborhoods, so the diagonal is closed. Conversely, if the diagonal is closed, its complement supplies disjoint neighborhoods of any two distinct points.

The graph is the inverse image of the diagonal under $(x,y)\mapsto (f(x),y)$, so a continuous map into a Hausdorff space has closed graph. Conversely, if $X,Y$ are compact Hausdorff and $C\subseteq Y$ is closed, then ${\Gamma }_f\cap (X\times C)$ is compact. Its projection onto $X$ is $f^{-1}(C)$, compact and hence closed. Thus $f$ is continuous.

The function $f(0)=0$, $f(x)=1/x$ for $x\ne 0$ is discontinuous but has closed graph: a convergent graph sequence approaching $x=0$ cannot have $x\ne 0$ and bounded second coordinate unless it eventually reaches the isolated graph point $(0,0)$.

For a linear map $T:V\to W$ with $V$ finite-dimensional,Nowwhich is the required inequality after rank-nullity for $\alpha$.

If $X$ and $Y$ represent the same map, then $Y=C^{-1}XB$, where $B$ changes new domain coordinates to old ones and $C$ does the same in the codomain.

Invertible block row and column operations reduceto $\mathrm{diag}(P,S-R{P}^{-1}Q)$, proving the rank formula. Apply it to $\left(\begin{array}{cc}{I}_n& Q\\ R& I_m\end{array}\right)$ first with the upper-left block and then with the lower-right block. Equating the results gives

The Neyman–Pearson lemma says that among tests of size at most $\alpha$ for two simple hypotheses, a likelihood-ratio test rejecting where $p(x;{\theta }_1)/p(x;{\theta }_0)>k$ is most powerful, with boundary randomisation if required. If $\phi$ is that test and $\psi$ any competing test, choose $k$ so the sizes agree. Pointwise,Integration and the size inequality yield $E_1\phi \ge E_1\psi$.

Herewhich decreases with $|x|$. Thus reject for $|X|\le c$, where

For every tangent vector $v\perp x_0$, stationarity of $Q$ on the sphere gives $2v^{T}A{x}_0=0$. Hence $A{x}_0$ is parallel to $x_0$, say $A{x}_0=E{x}_0$. Taking the inner product with $x_0$ gives $E=Q(x_0)$. The spectral theorem shows that this is the largest eigenvalue.

For $A=\left(\begin{array}{cc}1& t\\ t& 1\end{array}\right)$ the eigenvalues are $1\pm t$, soIts graph is a V translated upward and is convex.