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Put $W=1+{\rho }^{\prime 2}+{\rho }^2{\varphi }^{\prime 2}$. The Euler–Lagrange equations areWhen $n=n(\rho )$, the helical ansatz $\rho =R$, $\varphi ={\varphi }_0+\omega z$ automatically satisfies the second equation. The first becomesso $a=1$.

The endpoints have the same cylindrical angle exactly when $\omega L=2\pi k$ for some integer $k$. Thus a positive $L$ with $L=2\pi k/\omega (R)$ admits the two special solutions of opposite handedness, using $\omega$ and $-\omega$.

The product topology has basis $U\times V$ with $U$ open in $X$ and $V$ open in $Y$. Since ${\Pi }_X^{-1}(U)=U\times Y$ and similarly for ${\Pi }_Y$, both projections are continuous.

If $Y$ is Hausdorff, a point $(y_1,y_2)$ off the diagonal has disjoint product neighborhoods, so the diagonal is closed. Conversely, if the diagonal is closed, its complement supplies disjoint neighborhoods of any two distinct points.

The graph is the inverse image of the diagonal under $(x,y)\mapsto (f(x),y)$, so a continuous map into a Hausdorff space has closed graph. Conversely, if $X,Y$ are compact Hausdorff and $C\subseteq Y$ is closed, then ${\Gamma }_f\cap (X\times C)$ is compact. Its projection onto $X$ is $f^{-1}(C)$, compact and hence closed. Thus $f$ is continuous.

The function $f(0)=0$, $f(x)=1/x$ for $x\ne 0$ is discontinuous but has closed graph: a convergent graph sequence approaching $x=0$ cannot have $x\ne 0$ and bounded second coordinate unless it eventually reaches the isolated graph point $(0,0)$.

For a linear map $T:V\to W$ with $V$ finite-dimensional,Nowwhich is the required inequality after rank-nullity for $\alpha$.

If $X$ and $Y$ represent the same map, then $Y=C^{-1}XB$, where $B$ changes new domain coordinates to old ones and $C$ does the same in the codomain.

Invertible block row and column operations reduceto $\mathrm{diag}(P,S-R{P}^{-1}Q)$, proving the rank formula. Apply it to $\left(\begin{array}{cc}{I}_n& Q\\ R& I_m\end{array}\right)$ first with the upper-left block and then with the lower-right block. Equating the results gives

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