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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.

In local coordinates $u=(u^1,u^2)$ with metric $g_{ij}={\partial }_i\sigma \cdot {\partial }_j\sigma$, the energy isThe Euler–Lagrange equations are equivalentlywhere ${\Gamma }_{ij}^k=\frac{1}{2}{g}^{k\ell }({\partial }_i{g}_{j\ell }+{\partial }_j{g}_{i\ell }-{\partial }_{\ell }{g}_{ij})$.

For the plane-section curve $\eta$, constant speed gives ${\eta }^{\prime \prime }\perp {\eta }^{\prime }$. Both ${\eta }^{\prime \prime }$ and the surface normal lie in $P$, while $P$ is spanned by ${\eta }^{\prime }$ and that normal along the intersection. Hence ${\eta }^{\prime \prime }$ is normal to the surface. Its tangential acceleration vanishes, which is the geodesic equation.

The Leibniz definition isThe adjugate is the transpose of the cofactor matrix. Cofactor expansion givesFor the displayed tridiagonal matrix, expansion along the last row gives $D_n=2D_{n-1}-D_{n-2}$, with $D_1=2,D_2=3$. Induction yields