The coordinate $x_1^{\prime }$ does not occur. The perpendicular cross-section isThus the surface is an elliptic cylinder extending in both directions along the zero-eigenvalue axis. Its cross-sectional semiaxes are $\sqrt{k/{\lambda }_2}$ and $\sqrt{k/{\lambda }_3}$. It becomes a circular cylinder when the two positive eigenvalues coincide.

Let ${\lambda }_1={\lambda }_2=\lambda >0$ and ${\lambda }_3=-\gamma <0$. ThenThis is a one-sheet hyperboloid of revolution around the $x_3^{\prime }$ axis. Its circular waist has radius $\sqrt{k/\lambda }$ at $x_3^{\prime }=0$, and its radius grows with $|x_3^{\prime }|$. In the original coordinates the axis is the eigenvector direction belonging to the negative eigenvalue.

Writing the common positive eigenvalue as $\lambda$, the equation is $|x^{\prime }{|}^2=k/\lambda$. Hence it is a sphere withEvery orthonormal set of axes is a principal-axis set in this fully degenerate case.

Along a differentiable curve, the chain rule givesby the Euler-Lagrange equation. Thus when $F_x=0$ the Beltrami identity is

When $F_y=0$, the Euler-Lagrange equation reduces immediately to $d{F}_{y^{\prime }}/dx=0$. ThereforeThis is conservation of the generalized momentum associated with a cyclic coordinate.

For $c>1$, the catenary has a positive minimum height $x_0=(c-1)/\beta$ above the road. Its derivatives areThus the light ray is symmetric about $y=y_0$, descends towards a horizontal tangent at its minimum, and then rises. It never intersects the road. This smooth upward bending in the refractive index gradient produces an inferior mirage, without reflection at the road.

For the displayed form take $c>0$ and $1+\beta x\ge c$. Substitute $s=(1+\beta x)/c$ in the first integral:Inverting the hyperbolic cosine gives the catenary light ray in a linear refractive indexThe minimum occurs at $y=y_0$ and $x=x_0=(c-1)/\beta$, so the integration constant has the specified meaning. The integral above describes the $y\ge y_0$ graph branch; the other sign of the square root continues it smoothly to the other side of the turning point in $x(y)$.

The signed-constant qualification matters: for $c<0$ in part (a), integration reverses horizontal direction, and the positive prefactor in the full catenary is $|c|/\beta$, with minimum height $(|c|-1)/\beta$. The printed expression presupposes $c>0$. The separate case $c=0$ gives $y$ constant, a vertical ray. At the turning point $dy/dx$ diverges, so $y(x)$ is not a single regular graph across it, although $x(y)$ is smooth.

The integrand $f=n(x)\sqrt{1+(y^{\prime }{)}^2}$ has no explicit dependence on $y$, so part (i) givesFor positive refractive index and a regular graph with $n(x)>|c|$, squaring and then restoring the sign of $y^{\prime }$ yieldsIt is the continuous-medium version of Snell's law, expressing conserved horizontal optical momentum. The Euler-Lagrange equation establishes stationarity; on a monotone-height segment, convexity of the integrand in $y^{\prime }$ also gives a minimum among graphs with fixed endpoints.

If $f$ is independent of $y$, then $f_y=0$ and the Euler-Lagrange equation reduces to $d{f}_{y^{\prime }}/dx=0$. Therefore the first integral isThis is the conserved generalized momentum for a cyclic coordinate.

The boundary term $[f_{y^{\prime }}\eta {]}_a^b$ vanishes, so stationarity and the fundamental lemma of the calculus of variations give the Euler-Lagrange equation

Iterate the Gamma function recurrence to obtain, near $z=-m$ for $m\ge 0$,The numerator is holomorphic and equals $\Gamma (1)=1$ at $z=-m$. Exactly one denominator factor, $z+m$, vanishes; the product of the remaining factors there is $(-m)(-m+1)\cdots (-1)=(-1{)}^{m}m!$. Thus there is a simple pole, with Gamma function residue at a nonpositive integerFor $m=0$ the empty product is one. The same recurrence with a sufficiently large shift defines the continuation at every point away from these integers, so no further poles occur. At the poles, the recurrence is an identity of meromorphic functions, rather than an equality of finite values.

For $\mathrm{Re}z>0$, integration by parts givesThe boundary term vanishes because $t^{\mathrm{Re}z}\to 0$ at zero and exponential decay dominates any fixed power at infinity. ThusThis also gives $\Gamma (n+1)=n!$ for nonnegative integers, linking the Gamma function to the factorial.

The derivative of a holomorphic function can be evaluated using an increment along the real axis:Identifying the components of the gradient $\mathbf{V}=\nabla u$ gives

Since $\mathbf{V}=\nabla u$, its divergence is the Laplacian of the velocity potential. Using the Cauchy-Riemann equations and equality of mixed partial derivatives givesThus this is an incompressible flow, and $u$ is a harmonic function.

Expand $v={\sum }_j{c}_j{x}_j$, where $c_j=x_j^{†}v$. By orthonormality,Since the spectrum avoids $[a,b]$, each ${\lambda }_j$ is either below $a$ or above $b$, making every product strictly positive. This proves positivity of a quadratic polynomial across a spectral gap:In particular $\beta >0$ for every nonzero $v$. For $v=0$, $\beta =0$; the printed strict claim for an arbitrary vector must exclude this case.

For $B{x}_i={\lambda }_i{x}_i$, Hermitian symmetry givesDistinct eigenvalues therefore imply $x_i^{†}{x}_j=0$ for $i\ne j$. Together with the unit normalizations,The $n$ mutually orthogonal nonzero vectors are linearly independent and span the $n$-dimensional space, so they form an orthonormal basis. Here ${\delta }_{ij}$ is the Kronecker delta.

For any Hermitian matrix $K$, the scalar $x^{†}Kx$ is real because its complex conjugate equals $x^{†}{K}^{†}x=x^{†}Kx$. Applying this to $H,S$ in part (a) givesThese are the real and imaginary quadratic forms associated with the numerical range of $A$.

Multiply the eigenvalue equation $Ax=\lambda x$ on the left by $x^{†}$. The unit vector normalization is $x^{†}x=1$, soNo normal matrix assumption on $A$ is needed: this is an identity for each normalized eigenvector.

Applying $A^2=A$ to an eigenvector gives ${\lambda }_j^2={\lambda }_j$, so each eigenvalue is either zero or one. If all are one, the spectral theorem gives $A=I$, hence $\det A=1$ and $\mathrm{tr}A=n$. Otherwise, if $m$ eigenvalues are one, with $0\le m<n$, thenThus the trace of this orthogonal projection equals its rank, including $m=0$ for the zero matrix.

The convergent power series defining the matrix exponential gives $e^A=U{e}^D{U}^{†}$, with $e^D=\mathrm{diag}(e^{{\lambda }_1},\dots ,e^{{\lambda }_n})$. Hence the matrix exponential determinant identity isThe diagonalization proof uses the given Hermitian hypothesis; the identity itself holds for every square real or complex matrix.