Parametrize one ellipse as $\gamma (t)=c+A(\cos t,\sin t)$ and substitute it into the quadratic equation $Q=0$ of the other. At a transverse intersection, $g(t)=Q(\gamma (t))$ has a simple zero, so its sign changes. Periodicity makes the number of zeros even. The function is a trigonometric polynomial of degree at most two, so multiplication by $z^2$, with $z=e^{it}$, gives a nonzero polynomial of degree at most four. Hence there are at most four distinct intersection points. Coincident ellipses are excluded by transversality.

For $Q\sim {\chi }_m^2(\lambda )$ with $m>2$, completing the square in its Gaussian representation gives $\mathbb{E}{e}^{-tQ}=(1+2t{)}^{-m/2}{e}^{-\lambda t/(1+2t)}$. Integrate the identity $1/q={\int }_0^{\infty }{e}^{-tq}dt$ and use Tonelli theorem to obtain the displayed reciprocal moment. It is strictly decreasing in $\lambda \ge 0$, since its positive integrand is strictly decreasing for every $t>0$. At $\lambda =0$ the integral equals $1/(m-2)$, proving finiteness and the sharp maximum over noncentrality parameters.

If an autoregressive model has white noise of positive variance, a unique weakly stationary process solving its equation on all integer times exists precisely when its autoregressive polynomial has no root on the unit circle. In that case $1/A(z)$ has an absolutely summable Laurent series on an annulus containing the unit circle; its coefficients define a bilateral linear filter of a stationary time series. Applying that inverse also proves uniqueness among bounded second-moment solutions. A root on the unit circle forces the candidate spectral density of a stationary process to have a nonintegrable singularity. The stronger causality root criterion for an autoregressive model requires every root outside the closed unit disk.

The real matrix $B_M=\left(\begin{array}{cc}1& M\\ 0& 1\end{array}\right)$ has both eigenvalues equal to one, whereas $x^T{B}_{M}x/(x^{T}x)=1+M/2$ for $x=(1,1{)}^T$. Hence no universal constant bounds this quotient by the maximum eigenvalue of every real matrix with real eigenvalues.

Extend the interior grid vector by zero. With axial weights $2/(3h^2)$ and diagonal weights $1/(6h^2)$, the negative central coefficient equals the total neighbour weight. Expanding the edge-square sum gives the displayed quadratic form. Equality forces all axial differences to vanish, and connection to the zero exterior forces the vector to be zero. Simultaneous row and column permutations preserve symmetry and negative definiteness.

The Jacobi method has axial neighbour weights $1/5$ and diagonal weights $1/20$. Interior row sums are one, boundary-adjacent sums are smaller, and the positive-weight graph is connected. If an eigenvalue had modulus one, a maximal-modulus eigenvector component would force all its neighbours to have the same modulus. Propagation reaches a deficient row, a contradiction. This proves strict spectral-radius control even though the infinity norm of the iteration matrix can equal one.

Let ${\Gamma }_9$ have center weight $-10/3$, axial weights $2/3$ and diagonal weights $1/6$, and let $M_h$ have center weight $2/3$ and axial weights $1/12$. Taylor expansion gives $h^{-2}{\Gamma }_{9}u=\Delta u+h^2{\Delta }^{2}u/12+O(h^4)$ and $M_{h}u=u+h^2\Delta u/12+O(h^4)$. On a smooth solution of the constant-coefficient Helmholtz equation, the second-order term becomes $h^2\Delta (\Delta u+\lambda u)/12=0$. Thus the normalized PDE residual is fourth order, while the original unscaled stencil residual is sixth order. This local cancellation alone does not prove uniform global accuracy at or near a resonance.

For the square-grid nine-point finite-difference stencil for the Laplacian,If $\Delta u=0$, both displayed correction terms vanish: the mixed sixth derivatives sum to $\Delta u_{xxyy}=0$, and the pure sixth derivatives sum to the negative of that sum. Expanding two orders further yields $D_{9}u=h^6{u}_{xxxxxxxx}/3024+O(h^8)$ for a smooth harmonic function. Thus the compact harmonic scheme has sixth-order normalized consistency of a numerical method and, with the discrete maximum principle inverse bound, sixth-order nodal convergence on a square with exact smooth Dirichlet boundary data. For example $\mathrm{Re}(x+iy{)}^8$ shows that the sixth-order term need not vanish.

For the Dirichlet discrete Laplacian on a square, let $T=h^{-2}\mathrm{tridiag}(1,-2,1)$. The standard axial five-point operator is $T\otimes I+I\otimes T$, and the axial-weight $2/3$, corner-weight $1/6$ nine-point operator additionally contains $h^{2}T\otimes T/6$. Its eigenvalues are therefore strictly larger than the five-point values for the same sine modes. Both lie above the corresponding negative continuum Laplacian eigenvalues, so the five-point operator has smaller spectral error. For fixed mode numbers the leading errors are respectively ${\pi }^4{h}^2(k^4+l^4)/12$ and ${\pi }^4{h}^2(k^2+l^2{)}^2/12$.

There are $q^{n^2-n}$ nilpotent matrices of size n over $F_q$. Every matrix has a unique invariant direct sum on which it is nilpotent and invertible. Counting those ordered decompositions gives $q^{n^2}/|G{L}_n(q)|={\sum }_{k=0}^n{a}_k/|G{L}_k(q)|$. Subtraction at n and n−1 yields $a_n=q^{n^2-n}$. Translation by the identity counts unipotent matrices.

For a real or complex two-by-two matrix $A$, the Cayley-Hamilton theorem reads $A^2-(\mathrm{tr}A)A+(\det A)I=0$. Hence $A^2=0$ exactly when $\mathrm{tr}A=\det A=0$. Conversely, if $A^m=0$ for any positive integer $m$, every eigenvalue is zero and the same identity gives $A^2=0$. Thus a nonzero nilpotent matrix of this size has nilpotency index two.

For a nilpotent ideal $I$, its nilpotence index of an ideal is the least positive integer $n$ for which $I^n=0$. In $k[X]/(g^e)$, with $g$ an irreducible polynomial, the unique maximal ideal $(g)$ has index $e$. In a discrete valuation ring with $p=u{\pi }^e$, the maximal ideal of its reduction modulo $p$ has the same index. Comparing these indices under a quotient ring isomorphism recovers the ramification exponent in the Kummer-Dedekind theorem.

Group commutators are taken coordinatewise in a direct product of groups, giving ${\Gamma }_j(A\times B)={\Gamma }_j(A)\times {\Gamma }_j(B)$. A finite direct product of nilpotent groups is nilpotent, with class the maximum of the factor classes.

// Target: algebra.bigb

When $|\Phi |\to 1$ and its gauge covariant derivative decays fast enough, on a large circle write $\Phi =\rho e^{i\chi }$ locally. The tangential derivative gives $A=d\chi -\mathrm{Im}(\overline{\Phi }D\Phi )/{\rho }^2$. The integrated error tends to zero, so Stokes theorem turns total magnetic flux into the phase change $2\pi N$. This integer is the winding number of the normalized Higgs field on the circle. For positive Bogomolny vortex equations it is the total zero multiplicity.

For a unit Abelian Higgs vortex, the piecewise profiles $h=r/R$ and $a=(r/R{)}^2$ inside $r=R$, with both equal to one outside, are continuous finite-energy trial fields. In the normalization $E=\int [B^2/2+|D\varphi {|}^2/2+(1-|\varphi {|}^2{)}^2/8]$, their radial energy is the displayed function. Its minimum has $R^4=48$ and $E=\pi (2/3+1/\sqrt{3})$, an upper-bound estimate above the exact Bogomolny bound $\pi$. Derivative jumps at the matching radius are allowed in the variational energy space; the trial profile is not an exact solution.

For independent exponential distribution observations with rate $\lambda$, the likelihood ratio for a larger simple rate is decreasing in their sum. The Neyman-Pearson lemma therefore selects a lower-tail sum threshold $c$. The sum has a gamma distribution, giving the displayed power; its derivative is $c{e}^{-\lambda c}(\lambda c{)}^{n-1}/(n-1)!>0$. Calibrating at ${\lambda }_0$ gives size $\alpha$ for the composite null $\lambda \le {\lambda }_0$ and a uniformly most powerful test against larger rates.

For two simple hypotheses with probability densities $f_0,f_1$, a critical region $C$ has sum of Type I error and Type II error probabilities ${\int }_C{f}_0+{\int }_{C^c}{f}_1$. Pointwise minimization chooses $C=\{f_1>f_0\}$, with either decision allowed on the equality set. Thus the optimal likelihood ratio threshold is $1$, and the minimum is $\int \min (f_0,f_1)=1-\frac{1}{2}\int |f_1-f_0|$. This criterion weights the two error probabilities equally; unequal costs or prior probabilities change the threshold.

For two unit-scale Laplace distributions with locations ${\theta }_0<{\theta }_1$, the likelihood ratio is nondecreasing in $X$, with constant tails and an increasing middle interval. Therefore an upper-tail statistical test of exact size of a statistical test $\alpha$ is a most powerful test. Its cutoff is $c_{\alpha }={\theta }_0-\log (2\alpha )$ for $\alpha \le 1/2$, and $c_{\alpha }={\theta }_0+\log (2(1-\alpha ))$ otherwise. Under location ${\theta }_1$, its statistical power is $\frac{1}{2}{e}^{{\theta }_1-c_{\alpha }}$ if $c_{\alpha }\ge {\theta }_1$, and $1-\frac{1}{2}{e}^{c_{\alpha }-{\theta }_1}$ otherwise. Constant likelihood ratio tails can make the most powerful test nonunique.

For a Newtonian central configuration, set ${\mathbf{x}}_i(t)=a(t){\mathbf{a}}_i$ with $a>0$. The gravitational force scales as $a^{-2}$, reducing all equations to the displayed scalar ordinary differential equation. Its first integral is ${\dot{a}}^2/2-\Lambda /a=k/2$. The total energy is $kI/2$, so the scalar energy constant exactly tracks the many-body energy for the fixed shape.

For equal initial temperatures and the same Newton's law of cooling coefficient $a>0$, compare a unit Dirac delta function cooling impulse at time one with unit-rate cooling on the interval from one to two. The impulse produces a temperature jump of minus one. The finite-duration forcing gives a continuous temperature with change $-(1-e^{-a(t-1)})/a$ while it is active. Relative to the common unforced cooling curve, the impulse contributes $-e^{-a(t-1)}$. Consequently the temperatures cross at $t=1+\log (1+a)/a$, strictly between one and two. After time two the impulse-cooled body is warmer: the later cooling receives less subsequent exponential attenuation.