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Take fixed $E>0$ and define $k=\sqrt{2mE}/\hslash$. For sufficiently large $V_0$, the barrier region has $\kappa =\sqrt{2m(V_0-E)}/\hslash$ and solves ${\psi }^{\prime \prime }={\kappa }^2\psi$. Solving this constant-coefficient equation gives the exact transfer relationIn the specified limit, $a=U/V_0$ and $\kappa a\to 0$, whileThe transfer matrix therefore tends to $\left(\begin{array}{cc}1& 0\\ \eta & 1\end{array}\right)$. As the two boundaries collapse to the origin, this proves continuity and the derivative jump for a delta potential:The vanishing exterior phase shift $e^{ika}\to 1$ introduces no further factor.

Write the incoming and reflected waves as $e^{ikx}+r{e}^{-ikx}$ on the left and the transmitted wave as $t{e}^{ikx}$ on the right. Matching givesThe denominator is nonzero for real $\eta$ and $k>0$, so the limiting matching system is well defined. The probability currents on either side have the same factor $\hslash k/m$, making the transmission probability $|t{|}^2$. Thus scattering by a delta potential givesThe limit retains a finite reflection probability despite the barrier's vanishing width because its integrated strength remains finite.

Model the two samples as independent multinomial counts with fixed row totals $N_A=N_B=500$ and three category probabilities. The null hypothesis is $p_{Aj}=p_{Bj}=p_j$ for every category $j$. Up to combinatorial factors the likelihood is ${\prod }_{ij}{p}_{ij}^{O_{ij}}$. Without the null restriction its maximizers are $O_{ij}/N_i$; under the null, maximizing ${\sum }_j(O_{Aj}+O_{Bj})\log p_j$ subject to ${\sum }_j{p}_j=1$ givesThe observed column totals are $243,281,476$, giving fitted expected counts $(121.5,140.5,238)$ in each row.

To derive the null distribution, put $D=O_A-O_B$. Under the null its mean is zero and covariance is $1000(\mathrm{diag}p-p{p}^T)$. The multivariate central limit theorem gives, asymptotically, the standardized contrast $Z_j=D_j/\sqrt{1000p_j}$ a centered normal distribution with covarianceThis is the orthogonal projection onto a two-dimensional plane. In an orthonormal basis of that plane, the two nonzero coordinates are independent standard normal variables, so ${\sum }_j{Z}_j^2$ tends to ${\chi }_2^2$. The pooled estimates are consistent; substituting them for $p_j$ leaves this limit unchanged by Slutsky's theorem. Since $O_{Aj}-E_{Aj}=D_j/2$ and $O_{Bj}-E_{Bj}=-D_j/2$, the resulting statistic is precisely the Pearson chi-squared test of homogeneity statisticThis supplies the two degrees of freedom rather than treating all six cells as independent. Every fitted expected count is large, supporting the approximation.

For these data,It exceeds the two-degree-of-freedom 95th percentile $5.99$, but not the 99th percentile $9.21$. Hence reject equal score distributions at the 5% level, but not at the 1% level. The asymptotic p-value is $\mathrm{Pr}({\chi }_2^2\ge 7.56906)=e^{-7.56906/2}\simeq 0.02272$.

For an orthogonal change of Cartesian coordinates $x_i^{\prime }=Q_{ij}{x}_j$, an ordinary rank-$n$ tensor transforms byAn isotropic tensor is invariant under such rotations. Orthogonality gives $Q_{ip}{Q}_{jq}{\delta }_{pq}={\delta }_{ij}$, so transforming any of the three products of two Kronecker deltas leaves it unchanged. Their linear combination with arbitrary scalar coefficients is therefore isotropic.

For completeness, these three products exhaust the rank-four isotropic tensors in three dimensions. Half-turns about coordinate axes force a nonzero component to contain each coordinate label an even number of times. Axis interchanges then leave just the values $X=c_{1111}$, $\alpha =c_{1122}$, $\beta =c_{1212}$ and $\gamma =c_{1221}$. Rotating the first two axes by $\pi /4$ gives $X=\frac{1}{2}X+\frac{1}{2}(\alpha +\beta +\gamma )$, so $X=\alpha +\beta +\gamma$. These component values are exactly those of the three-delta-product expression. Thus isotropy of the elastic medium gives this form for its elasticity tensor.

Contracting with the symmetric strain givesThese are the Lamé parameters. Taking the trace determines the scalar part of the strain:Since ${\sum }_{ij}{e}_{ij}^2=3p^2+{\sum }_{ij}{d}_{ij}^2$, the stored strain energy density isA nonzero traceless test strain proves necessity of $\mu \ge 0$, and a pure scalar strain proves necessity of $3\lambda +2\mu \ge 0$. Conversely these two conditions make both terms nonnegative for every strain. Thus nonnegative isotropic elastic strain energy is equivalent to

First $N(A)$ is finite: by Cauchy-Schwarz, $\Vert Ax{\Vert }_2\le ({\sum }_{ij}{a}_{ij}^2{)}^{1/2}\Vert x{\Vert }_2$. It is nonnegative and absolutely homogeneous. If $N(A)=0$, every $A{e}_j=0$, so every column is zero and $A=0$; the converse is immediate. The triangle inequality for the Euclidean norm gives $N(A+B)\le N(A)+N(B)$ after taking suprema. Thus $N$ is a norm.

For any vector $x$, rescaling gives $\Vert Ax{\Vert }_2\le N(A)\Vert x{\Vert }_2$. Applying this twice givesFurthermore $|a_{ij}|\le \Vert A{e}_j{\Vert }_2\le N(A)$, proving the requested strict entry bound when $N(A)<\epsilon$.

For a polynomial in matrix entries, expand each monomial at $A+H$. The terms involving exactly one entry of $H$ give a linear map $Df(A)[H]$, with coefficients polynomial in the entries of $A$. Every remaining term has at least two entries of $H$ and is $O(N(H{)}^2)$ for $N(H)\le 1$, because $|h_{ij}|\le N(H)$. This proves differentiability. Its derivative is continuous as an operator-valued function: each coefficient is continuous, and for $N(H)\le 1$ every $h_{ij}$ is bounded by one, so the operator-norm difference is bounded by the sum of the coefficient differences. Coordinate continuity follows from the same entry bound. Hence every such polynomial function is continuously differentiable.

In the Leibniz expansion of $\det (I+H)$, the identity permutation supplies $1+{\sum }_i{h}_{ii}$ plus terms of degree at least two. Any nonidentity permutation moves at least two indices, so each of its terms has at least two $H$ entries. ConsequentlyIf $A$ is invertible, multiplicativity of the determinant and the preceding expansion giveThe adjugate identity gives $\mathrm{adj}A=\det (A)A^{-1}$. Thus $d^{\prime }(A)[H]=\mathrm{tr}((\mathrm{adj}A)H)$ for invertible $A$.

For singular $A$, take invertible $A_r\to A$, using the permitted density result. Both $d^{\prime }$ and the adjugate entries are continuous polynomial expressions, so taking limits for every fixed $H$ proves the derivative of the determinant formula on all matrices:Here the standard adjugate is the transpose of the cofactor array, as required by $(\mathrm{adj}A)A=(\det A)I$. In components this derivative is ${\sum }_{ij}{\mathrm{cof}}_{ij}(A)h_{ij}$, also valid at singular matrices.