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Assume $eB>0$; reversing the sign only exchanges the spin labels. Since $\mathbf{B}\cdot \mathbf{\sigma }=B{\sigma }_3$, the Pauli term shifts the spinless energies by $\pm \hslash {\omega }_c/2$. ThusAfter collecting equal energies, the spin splitting of Landau levels isThe level $\ell =0$ contains only $(n,sigma)=(0,-1)$ and has degeneracy $D$. Every level $\ell \ge 1$ contains $(\ell ,-1)$ and $(\ell -1,+1)$ and has degeneracy $2D$.

For $N$ noninteracting electrons, the Pauli exclusion principle fills these states from the bottom. If $0\le N\le D$, every electron fits in the zero-energy level andFor $N>D$, writeThen the first $k$ positive levels are full and $r$ states in level $k+1$ are occupied. The ground-state energy isEquivalently, onThe graph is continuous and piecewise linear: it is flat up to $N=D$, then has slopes $\hslash {\omega }_c,2\hslash {\omega }_c,3\hslash {\omega }_c,\dots$, with kinks at

Take $B>0$ and use the Landau gauge $\mathbf{A}=(0,Bx,0)$. With zero $z$-momentum, the Hamiltonian becomesFor a state $e^{i{k}_{y}y}\chi (x)$, this is a quantum harmonic oscillator in $x$, centred atwith cyclotron frequencyThe Landau levels are therefore

Periodicity in $y$ quantizes $k_y=2\pi \ell /R_y$. Requiring the corresponding oscillator centre to lie in an $x$ interval of length $R_x$ leavesallowed centres for each $n$. This is the degeneracy of a Landau level, equal to the magnetic flux through the rectangle divided by the magnetic flux quantum $h/|e|$. Consistent magnetic periodic boundary conditions make $D$ an integer.

A $j=0$ state exists only when $j_1=j_2=:j$. Since its total magnetic quantum number is zero, every uncoupled basis term must have $m_1+m_2=0$, soApply the total raising operator $J_{+}=J_{1+}+J_{2+}$. The coefficient of
$|j,m+1{⟩}_1|j,-m{⟩}_2$ isThe angular momentum singlet state is annihilated by $J_{+}$, henceAll coefficients consequently have the same modulus and alternating signs. Normalization fixes that modulus to $(2j+1{)}^{-1/2}$, and an arbitrary overall phase may be chosen so thatThus

For $|J,J⟩$, tracing out subsystem 2 gives the reduced density matrixwhich is pure, so its entanglement entropy is zero.

For $|J,J-1⟩$, the two subsystem-2 factors in part (iii) are orthogonal. The partial trace therefore removes the cross terms and givesIts nonzero eigenvalues are $j_1/J$ and $j_2/J$, so the Von Neumann entropy isWhen $j_1=j_2$, both eigenvalues are $1/2$ and $S=\log 2$, the maximal entropy for a rank-two reduced state.

The exact second-iterate equation factors asThe positive nonfixed period-two points come from the final factor. If $y=x^2$, their squared values areThey are real and distinct for $\lambda >2$.

The multiplier of a periodic orbit of an iteration isFor $\lambda =2+\mu$ with $0<\mu \ll 1$,whose modulus is less than one. The new two-cycle is therefore locally asymptotically stable.

Let $x^{\ast }(\lambda )=\sqrt{1-1/\lambda }$ and center the map at this parameter-dependent fixed point:A Taylor polynomial in $X$, with the coefficients expanded at $(x,\lambda )=(1/\sqrt{2},2)$, givesunder the period-doubling scaling $X=O(\sqrt{\mu })$. The coefficients areHere we used $F_x(x^{\ast }(\lambda ),\lambda )=3-2\lambda$, $F_{xx}=-6\lambda x$, and $F_{xxx}=-6\lambda$.

Composing the local map with itself and retaining the terms of order $\mu X$ and $X^3$ givesBesides the fixed solution $X=0$, the two leading solutions therefore satisfyThus the period-two points born at $\lambda =2$ areto leading order. With the coefficients above, the radicand is positive for $\mu >0$.

For $\lambda >1$, the new fixed point in $[0,1]$ isThe origin has multiplier $F_x(0,\lambda )=\lambda >1$ and is therefore unstable. At the new fixed point,HenceThe point $x^{\ast }$ is stable in that interval and unstable for $\lambda >2$. At $\lambda =2$ its multiplier is $-1$, the threshold for a period-doubling bifurcation.

A fixed point satisfiesso either $x=0$ or $x^2=1-1/\lambda$. When $0<\lambda <1$, the latter equation has no real solution, and the origin is the only fixed point. Sincethe criterion for fixed point stability for an iteration shows that the origin is stable for $\lambda <1$.

For $x\in [0,1]$, the nonnegative function $x(1-x^2)$ has derivative $1-3x^2$, so its maximum occurs at $x=1/\sqrt{3}$ and equalsThe map preserves the interval exactly when this maximum is at most one:Thus the upper limit is

For each $m$, the displayed objective is the sum of the minimized linear-regression residual sums of squares on the prefix $1:m$ and suffix $(m+1):n$. Precompute prefix sums ofThis requires $O(n)$ operations. The corresponding five sums over a suffix are obtained by subtracting a prefix sum from the total.

For any block of $k$ observations, these raw sums give in constant timeBy the formula for the residual sum of squares in simple linear regression, its minimized loss isHence both losses for a proposed $m$ take constant time. Evaluating their sum for $m=2,\dots ,n-2$ and keeping the minimum is another $O(n)$ scan, so the complete minimization uses $O(n)$ computations.

WriteThe ordinary least squares estimators givewhereThe denominator is positive because the sorted inputs are distinct and $m\ge 2$.

With $X_1<\cdots <X_n$, it suffices to put the threshold between $X_m$ and $X_{m+1}$ for $m=1,\dots ,n-1$. Form the prefix sumsThey take $O(n)$ operations to compute. The minimized squared errors for the two children areandEach candidate value $L_m+R_m$ therefore takes constant time once the prefix sums are available. Scanning all $n-1$ candidates takes $O(n)$ operations in the sense of Big O notation, and retaining the minimizing $m$ gives the optimal split.

A regression tree chooses a threshold $s$ and constant predictions $c_L,c_R$ to minimize the squared-error criterionwhere $s$ ranges over thresholds producing two nonempty groups. For a fixed split, the least-squares minimizers are the two sample means,Thus the first split is the candidate threshold with the smallest sum of the two within-node residual sums of squares.

No. TakeIt contains more than one point and has the fixed-point property of a closed interval, but it is not homeomorphic to the closed unit disc. Indeed, deleting any point $(t,0)$ with $0<t<1$ disconnects $B$, so such a point is a cut point. Deleting one point from a two-dimensional closed disc leaves a path-connected space. Since being a cut point is preserved by homeomorphisms, the two spaces cannot be homeomorphic. This gives the required counterexample.

The closed unit disc has the fixed-point property by the Brouwer fixed-point theorem. The annulus admits the fixed-point-free rotation of an annulus, for examplewhich is continuous, maps the annulus to itself, and has no fixed point because the origin is not in the annulus. If the disc and annulus were homeomorphic, the homeomorphism invariance of the fixed-point property would give the annulus the fixed-point property, a contradiction. Therefore they are not homeomorphic.

The Brouwer fixed-point theorem for the closed unit disc $\overline{D}\subset {\mathbb{R}}^2$ states that every continuous maphas a fixed point.