For equal one-second observation windows, the sample mean estimates the Poisson distribution mean:This matches the expected mean count; it is also the maximum-likelihood estimator for independent Poisson distribution observations. With $N=12169$ windows, fitted expected frequencies are $E_j=N{e}^{-\hat{\lambda }}{\hat{\lambda }}^j/j!$ for $j=0,\dots ,4$, and $E_{\ge 5}=N[1-{\sum }_{j=0}^4{e}^{-\hat{\lambda }}{\hat{\lambda }}^j/j!]$.

Each last-column entry is $(O_j-E_j{)}^2/E_j$, and the Pearson statistic is $\boxed{X^2=\sum _{j\in \{0,1,2,3,4,\ge 5\}}(O_j-E_j{)}^2/E_j}$. The displayed expected frequencies and contributions are rounded, so their printed values need not reproduce every final digit of a calculation using unrounded probabilities.

For hard-wall confinement, the stationary spatial modes and energies areTheir stationary time dependence is $e^{-i{E}_{n}t/\hslash }$. The correct triple-angle identity is $\sin 3\theta =3\sin \theta -4{\sin }^3\theta$; the printed hint has its right side reversed in sign. Thus the specified initial state expands asThe coefficients have squared moduli summing to one. The first requested energy is $E_1$, so its probability is $\boxed{9/10}$. The other requested energy is $2E_1$, which is not $n^2{E}_1$ for any positive integer $n$, so its probability is $\boxed{0}$. The occupied excited energy is instead $E_3=9E_1$, with probability $1/10$.

The Time-dependent Schrödinger equation isA normalized quantum state has ${\int }_{{\mathbb{R}}^3}|\psi {|}^2{d}^{3}x=1$; the Hamiltonian operator acts on wave functions satisfying the appropriate domain and boundary conditions.

If $\lambda =0$, only a linear function can solve the equation, and zero endpoint values make it zero. If $\lambda <0$, the corresponding hyperbolic-sine solution also cannot vanish at both endpoints unless it is zero. For $\lambda =k^2>0$, the first endpoint gives $y=C\sin kx$, and the second requires $k=n\pi$ for a positive integer $n$. Thus $\boxed{{\lambda }_{\min }={\pi }^2}$.

The admissible trial function $y=\sin \pi x$ has quotientTherefore the minimum, or infimum before its attainment is established, is $\boxed{\le {\pi }^2}$ as requested. The sharp Poincaré inequality supplies the reverse inequality on the finite-energy zero-trace class, so the exact minimum is also ${\pi }^2$ and is attained by this trial function. The one-sided assertion in the source does not require a full variational existence argument.

Multiply the equation for $y_i$ by $y_j$ and the equation for $y_j$ by $y_i$, then subtract. The $q$ terms cancel and the derivative terms combine into a Wronskian boundary derivative:Integrating over the interval eliminates the boundary term because both eigenfunctions vanish at both endpoints. Since their eigenvalues differ,This is orthogonality in a weighted inner product in the regular Sturm-Liouville setting; the endpoint values must justify the integration by parts.

Here $\exp (I)=eI$. Applying part (a) at that matrix gives $D{s}_{eI}(K)=e(K^t+K)$, while part (b) gives $D{\exp }_I(H)=eH$. ThereforeBoth the outer evaluation point and the inner derivative contribute a factor of $e$.

The requested base point is the identity matrix $I$, not the zero matrix. Since $I$ commutes with every $H$,The matrix power series justifies the quadratic remainder near zero. ConsequentlyAt zero the derivative would instead be $H$; replacing the printed base point would lose the scalar factor.

Expand at a matrix increment $H$:The last term is $O(\Vert H{\Vert }^2)$ in a matrix norm, so it contributes no derivative. HenceThis is a linear operator on matrix increments, not a scalar entrywise derivative.

Since $K\subseteq L$ and $L/F$ is finite, an $F$-linearly independent set in $K$ remains independent in $L$, so $K/F$ is finite. A finite $F$-spanning set of $L$ also spans it over $K$, so $L/K$ is finite.

Take an $F$-basis $a_1,\dots ,a_m$ of $K$ and a $K$-basis $b_1,\dots ,b_n$ of $L$. Expanding coefficients twice shows that all products $a_i{b}_j$ span $L$ over $F$. If ${\sum }_{i,j}{c}_{ij}{a}_i{b}_j=0$ with $c_{ij}\in F$, independence of the $b_j$ forces ${\sum }_i{c}_{ij}{a}_i=0$ for each $j$, and then independence of the $a_i$ forces every coefficient zero. Thus these products form an $F$-basis, proving

The inclusion $F\subseteq K$ makes $K$ a vector space over $F$ by its own addition and multiplication by elements of $F$. The extension is finite when this vector space has a finite basis. Its degree of a field extension isBeing a finite field extension concerns vector-space dimension, not the cardinality of the underlying field.

Replace $L_h$ by $\boxed{{\tilde{L}}_h=L_h+I}$. At a positive interior maximum $v>0$, the neighbour average is no larger than $v$, so $({\tilde{L}}_h\varphi {)}_{jk}\ge v>0$. Thus ${\tilde{L}}_h\varphi \le 0$ cannot have such an interior maximum. The positive-maximum form of the discrete maximum principle therefore remains valid:With $e=u{|}_{{\Omega }_h}-U$ the error equation is ${\tilde{L}}_{h}e=\tau$, where $\tau =O(h^2)$ under the preceding regularity assumptions. Since ${\tilde{L}}_{h}B=1+B\ge 1$ for the same Discrete Poisson barrier on a square, the comparison functions $\pm e-\Vert \tau {\Vert }_{\infty }B$ proveThere is no loss of uniqueness: apply the positive-maximum argument to both signs of a homogeneous discrete solution. The continuous weak formulation also has the coercive bilinear form ${\int }_{\Omega }(|\nabla v{|}^2+v^2)$, so the added positive reaction term strengthens its positive energy.

The five-point equations for the Poisson equation arewith prescribed boundary values. If $L_h\varphi \le 0$, an interior maximal value is at most the average of its four neighbours. Since no neighbour exceeds the maximum, equality forces all four to equal it. Repeating along adjacent nodes propagates that value to the boundary. Thus the discrete maximum principle gives $\max \varphi \le {\max }_{\partial {\Omega }_h}\varphi$, in particular the usual positive-maximum bound with zero included.

For an $O(h^2)$ error estimate assume the exact solution has bounded continuous fourth partial derivatives up to the boundary. The local truncation error $\tau =L_{h}u-f$ then satisfies $\Vert \tau {\Vert }_{\infty }\le C{h}^2$ by the Taylor theorem. Put $e=u{|}_{{\Omega }_h}-U$, so $L_{h}e=\tau$ and $e=0$ on boundary nodes. The Discrete Poisson barrier on a squaresatisfies $L_{h}B=1$ exactly, $B\ge 0$ on the grid boundary, and $B\le 1/8$. Set $\epsilon =\Vert \tau {\Vert }_{\infty }$. Then $L_h(\pm e-\epsilon B)\le 0$, with nonpositive boundary values. Applying the discrete maximum principle twice givesFor nonzero smooth Dirichlet boundary data $g$, impose $U=g$ at the boundary nodes. The error still vanishes there, so the same argument and order apply, under the same fourth-derivative regularity of the exact solution. Smooth boundary values alone do not remove every possible corner-regularity issue.

The four-point midpoint third derivative uses the four values at $x_{j-1},x_j,x_{j+1},x_{j+2}$, centred at $x_{j+1/2}$. The Taylor theorem about that midpoint givesThus, for $n\ge 2$, take the $n-1$ interior equationsImpose the two value conditions exactly, $U_0=\alpha$, $U_{n+1}=\beta$, and approximate the remaining condition with the second-order backward finite difference:This gives $n$ equations for $U_1,\dots ,U_n$. The endpoint derivative formula equals $u^{\prime }(b)-h^2{u}^{\prime \prime \prime }(b)/3+O(h^3)$, so its residual tends to zero along with those of the four-point equations. Every stencil lies within $[a,b]$; no unprescribed ghost value is introduced. The construction is therefore a consistent finite difference method for all three boundary conditions.

For the second-order boundary value problem, introduce an unknown initial slope $s$ and solve the initial value problem $u^{\prime }=v$, $v^{\prime }=g(x,u,v)$ with $u(a)=\alpha$, $v(a)=s$. Its endpoint mismatch is $F(s)=u(b;s)-\beta$. The shooting method seeks a root $F(s)=0$; in general the dependence of the flow on $s$ is nonlinear. A sign-changing bracket for a continuous $F$ allows the bisection method, provided the intervening initial value solutions exist up to $b$.

For Newton's method, solve the parameter sensitivity equation alongside the trajectory. With $z={\partial }_{s}u$ and $w={\partial }_{s}v$,Then $F^{\prime }(s)=z(b)$ and the update is $\boxed{s_{\mathrm{new}}=s-F(s)/z(b)}$ whenever $z(b)\ne 0$.

For the third-order problem, shooting from $a$ leaves both $u^{\prime }(a)$ and $u^{\prime \prime }(a)$ unknown, requiring two shooting parameters to satisfy two right-endpoint conditions. Shooting backwards from $b$ already knows $u(b)=\beta$ and $u^{\prime }(b)=\gamma$; only $u^{\prime \prime }(b)$ is unknown, with the single residual $u(a)-\alpha$. Backward shooting reduces the nonlinear solve from two scalar unknowns to one. This parameter-count advantage does not by itself guarantee numerical stability of backward integration.

For nonnegative initial data, $U_{\min }=0$. The proved Crank-Nicolson maximum principle under a step restriction immediately gives $\boxed{U_{ij}^m\ge 0}$ for all indices. For nonpositive initial data, $U_{\max }=0$ and the same bound gives $\boxed{U_{ij}^m\le 0}$. Equivalently, apply positivity to $-U$, since the finite difference method is linear and has homogeneous Dirichlet boundary conditions. Both conclusions use ${\mu }_x+{\mu }_y\le 1$.

Write $s={\mu }_x+{\mu }_y$. The update can be rearranged asWhen $s\le 1$, the right side is a convex combination of old values: its coefficients are nonnegative and sum to one. Assume all old values lie in $[U_{\min },U_{\max }]$. A new maximum $v>U_{\max }$ is interior because boundary values are zero and $U_{\max }\ge 0$. At its node the left side is at least $(1+s)v-sv=v$, while the right side is at most $U_{\max }$, a contradiction. A new minimum $v<U_{\min }$ similarly makes the left side at most $v$ and the right side at least $U_{\min }$, again a contradiction.

The initial values satisfy the bounds by construction; mathematical induction provesThe new-time matrix is a strictly diagonally dominant matrix with positive diagonal and nonpositive off-diagonal entries, so the linear update is uniquely solvable. This Crank-Nicolson maximum principle under a step restriction concerns preservation of value bounds, which is a stronger requirement than scalar A-stability.