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The one-dimensional probability current isAt a finite jump of the potential, both $\psi$ and ${\psi }^{\prime }$ are continuous. A jump in $\psi$ would create a delta derivative in the stationary Schrödinger equation, and integrating that equation across an increasingly small interval then shows continuity of ${\psi }^{\prime }$. Accordingly impose continuity of both quantities at zero and at $a$.

For an incident wave of positive energy, define $k=\sqrt{2mE}/\hslash$. With no wave incident from the right, the left region hasThe interference terms cancel from the imaginary part in the current formula. If $E>V_0$, let $k_0=\sqrt{2m(E-V_0)}/\hslash$. In the right region,For $0<E<V_0$, let ${\kappa }_0=\sqrt{2m(V_0-E)}/\hslash$. Boundedness at positive infinity excludes the growing exponential, soThe decaying solution has zero current even if its amplitude is nonzero. These exterior forms do not require $E$ to lie on a particular side of $V_1$: in the finite middle layer the solution is oscillatory for $E>V_1$, exponential for $E<V_1$, and linear at $E=V_1$; matching determines the amplitudes.

Define the reflection coefficient as reflected current magnitude divided by incident current, and the transmission coefficient as outward right current divided by incident current. ThenThe denominator is $j_{\mathrm{incident}}=\hslash k|A{|}^2/m$. For a real potential, differentiation givesby the stationary Schrödinger equation. Continuity at each interface preserves this conserved current, so $j_{\mathrm{left}}=j_{\mathrm{right}}$. Dividing by incident current yieldsThere can be penetration into the barrier without transmission to infinity, because the entire right half-line remains classically forbidden.

For the real Sturm-Liouville problem, multiply the equations for $y_n$ and $y_m$ by $y_m$ and $y_n$, respectively, and subtract. Integration by parts givesThe boundary term vanishes because both functions are zero at both endpoints. Distinct eigenvalues therefore imply weighted orthogonality. For equal indices the integral is the squared weighted norm, so

Let $c_n={\int }_0^{1}w{y}_n^{2}dx>0$. For $y=\sum a_n{y}_n$ in the operator domain, with convergence sufficient to evaluate its quadratic energy, weighted orthogonality givesFor finite sums these identities follow immediately; the corresponding convergent energy expansion follows by approximation in the operator or quadratic-form domain. For nonzero $y$, the Rayleigh quotient is a weighted average of the eigenvalues, soEquality occurs precisely when only a lowest-eigenvalue component is present.

Assume completeness as permitted and take a sufficiently regular solution of the diffusion equation. It belongs to the weighted expansion space at each positive time; its modal coefficients satisfy ${\dot{a}}_n=-{\lambda }_n{a}_n$. Alternatively differentiate its energy directly and use the preceding inequality:Thus the squared weighted norm is at most its initial value times $e^{-2{\lambda }_{1}t}$. This describes decay when ${\lambda }_1>0$; positivity of $p,w$ alone does not require ${\lambda }_1>0$ if the potential $q$ can be negative. The displayed inequality remains valid regardless of that sign.

Rouché's theorem states that if $f,g$ are holomorphic on a neighborhood of a closed disc with boundary $C$ and $|g|<|f|$ on $C$, then $f$ and $f+g$ have the same number of zeros in the disc, counted with multiplicity. More generally the conclusion holds for a simple closed contour whose interior lies in their domain. To prove it, set $f_t=f+tg$ for $0\le t\le 1$. The strict boundary inequality ensures $f_t\ne 0$ on $C$. By the argument principle,is the integer number of zeros inside. The integrand is continuous in $t$ uniformly on $C$, since its denominator is bounded away from zero on the compact product $C\times [0,1]$. Thus $N(t)$ is continuous and integer-valued, so constant. This proves the theorem.

On $|z|=1$, the term $8z^6$ dominates the remaining terms because $3+1+2+1=7<8$. Thus the polynomial has six zeros inside the unit circle. On $|z|=2$, $3z^9$ dominates because its modulus is $1536$, whereas the sum of the other bounds is $512+32+16+1=561$. There are therefore nine zeros inside radius two. The strict inequalities exclude boundary zeros, so

For the fixed-degree polynomial limit, choose $d+1$ distinct complex nodes ${\zeta }_0,\dots ,{\zeta }_d$. Write $p_n(z)={\sum }_{j=0}^d{a}_{j,n}{z}^j$, inserting zero coefficients if necessary. The vector of values at the nodes equals the fixed Vandermonde matrix times the coefficient vector. Its determinant is ${\prod }_{i<j}({\zeta }_j-{\zeta }_i)\ne 0$. Convergence of the values therefore implies coefficient convergence, say $a_{j,n}\to a_j$. For $p(z)={\sum }_{j=0}^d{a}_j{z}^j$ and a compact set with $|z|\le R$,This proves that the given limit is a polynomial of degree at most $d$ without invoking any theorem about uniform limits of holomorphic functions.

When the limit has $d$ distinct roots, choose pairwise disjoint small closed discs around them, with no limit root on their boundaries. On each boundary $|p|$ has a positive minimum. Uniform convergence makes $|p_n-p|<|p|$ there for all sufficiently large $n$, so Rouché's theorem puts exactly one root $z_{i,n}$ of $p_n$ in each disc. For every smaller disc about $z_i$ the same argument applies eventually. The unique root in the original disc must then lie in the smaller one, provingThe literal demand for a root at every initial index needs this finite-index qualification: $p_1=1$, $p_n=z$ for $n\ge 2$ satisfies the hypotheses with limit $z$, but $p_1$ has no root. If every initial polynomial has a root, arbitrary choices at the finitely many exceptional indices extend the convergent sequences; otherwise the sequences necessarily start at an eventual index.

Let $p=(x_0,y_0)$, $a=f_x(p)$ and $b=f_y(p)$. For sufficiently small $h,k$, decompose the increment into an $x$-segment at height $y_0+k$ and a $y$-segment at $x_0$. The one-variable mean value theorem applies on each segment because existence of its partial derivative gives continuity along the segment. Thus, when both increments are nonzero,for some $0<\theta ,\eta <1$. The same formula with the absent segment omitted covers zero increments. Continuity of both partial derivatives at $p$ makes the remainder after subtracting $ah+bk$ at most $\epsilon (|h|+|k|)$ for small increments, and hence at most $\sqrt{2}\epsilon \sqrt{h^2+k^2}$. Therefore the Fréchet derivative is $(h,k)\mapsto ah+bk$. Continuity throughout the disc is not required; continuity at the point suffices.

For the determinant, expand by multilinearity of the determinant in the columns of $I+H$. Taking no perturbed column gives one; taking exactly one gives $\mathrm{tr}H$; all other terms have degree at least two in the entries of $H$. Since matrix-entry functionals are bounded in any finite-dimensional norm, their sum is $O(\Vert H{\Vert }^2)$. HenceFor invertible $A$, multiplicativity of the determinant gives $\det (A+H)=\det A\det (I+A^{-1}H)$. The bounded linear map $H\mapsto A^{-1}H$ transfers the preceding remainder estimate, giving the derivative of the determinantTo justify an open invertible neighborhood, let $\Vert K\Vert <1$. Completeness and submultiplicativity make the Neumann series $S={\sum }_{j=0}^{\infty }(-K{)}^j$ converge in matrix norm. Multiplying finite partial sums by $I+K$ gives $I-(-K{)}^{N+1}$, which tends to $I$. Continuity of multiplication proves $(I+K)S=S(I+K)=I$. Thus $I+K$ is invertible, and the series gives $(I+H{)}^{-1}=I-H+O(\Vert H{\Vert }^2)$ near zero.

Apply the first-derivative formula at $I+H$ to a test matrix $K$:The last estimate is uniform for $\Vert K\Vert \le 1$, so it proves differentiability of the derivative in the operator norm, rather than merely a directional second derivative. The second derivative of the determinant is the symmetric bilinear mapIts symmetry follows from $\mathrm{tr}(HK)=\mathrm{tr}(KH)$.

A Euclidean domain is an integral domain $R$ with a function $\nu :R\setminus \{0\}\to \mathbb{N}$ such that, for any $a,b\in R$ with $b\ne 0$, there are $q,r\in R$ satisfying $a=bq+r$ and either $r=0$ or $\nu (r)<\nu (b)$. To prove that every ideal is principal, take a nonzero ideal $J$ and choose a nonzero $b\in J$ of minimal Euclidean value. Divide any $a\in J$ by $b$. The remainder belongs to $J$, so minimality forces it to be zero. Thus $J=(b)$; the zero ideal is also principal. Hence every Euclidean domain is a principal ideal domain.

In $R=\mathbb{Z}[\sqrt{-7}]$, consider $J=(2,1+\sqrt{-7})$. It is proper: the map $R\to {\mathbb{F}}_2$ sending $\sqrt{-7}$ to one is a well-defined surjective ring map, and annihilates both generators. Suppose $J=(\eta )$. The positive integer field norm $N(a+b\sqrt{-7})=a^2+7b^2$ is multiplicative, and $\eta$ divides two. Therefore $N(\eta )$ divides four. A proper ideal cannot have a unit generator, so $N(\eta )\ne 1$. There is no element of norm two. The elements of norm four are only $\pm 2$, but neither divides $1+\sqrt{-7}$ in this ring. This is a contradiction. ThusThe argument does not merely show failure of one proposed norm.

Now put $\omega =(1+\sqrt{-7})/2$ and $R^{\prime }=\mathbb{Z}[\omega ]$. Its algebraic field norm issince $\omega \overline{\omega }=2$ and $\omega +\overline{\omega }=1$. To construct Euclidean division, let $z\in \mathbb{C}$. Choose an integer $b$ so that $|\mathrm{Im}z-b\sqrt{7}/2|\le \sqrt{7}/4$, then choose an integer $a$ so that $|\mathrm{Re}z-b/2-a|\le 1/2$. The lattice point $q=a+b\omega$ satisfiesFor $A,B\in R^{\prime }$ with $B\ne 0$, apply this construction to $z=A/B$, and set $r=A-Bq\in R^{\prime }$. Multiplicativity gives $N(r)=N(B)|z-q{|}^2<N(B)$ unless $r=0$. ThereforeThis is the geometric division argument for the norm-Euclidean integers of discriminant minus seven.

A quadratic form over $\mathbb{R}$ or $\mathbb{C}$ is a function $q(x)=B(x,x)$ for a symmetric bilinear form $B$, equivalently a homogeneous polynomial of degree two. Its polarization is $B(x,y)=[q(x+y)-q(x)-q(y)]/2$. If $q$ is not identically zero, choose $v$ with $q(v)\ne 0$. Every $x$ decomposes aswhere the second term lies in $v^{\perp }$ and $kv\cap v^{\perp }=\{0\}$. The quadratic form is consequently a direct sum of its one-dimensional value on $v$ and its restriction to $v^{\perp }$. Induction on dimension diagonalizes it. If $q=0$, any basis suffices. Rescale every nonzero diagonal coefficient: over the reals it becomes $1$ or $-1$, while over the complexes a square root makes it $1$. HenceOver $\mathbb{C}$ the negative coefficients can all be avoided.

For a real quadratic form, let $p$ and $m$ be its positive and negative indices in a diagonalization. Its rank of a quadratic form is $p+m$ and its signature of a quadratic form is $p-m$, or the ordered pair $(p,m)$ in the alternative convention. These indices are intrinsic: the largest dimension of a subspace where the form is positive definite is $p$, since any larger subspace intersects the nonpositive coordinate subspace nontrivially; similarly for $m$. For the form here, completing squares givesThe two displayed linear coordinates, together with $x_1$, form an invertible coordinate change. ThereforeThere is one zero coordinate direction.

The PDF's final dimension condition is $n\ge 2^d$, not the fraction produced by the TeX conversion. We prove the requested common zero of complex quadratic forms by induction on $d$. A complex quadratic form of rank $r$ has canonical expression $x_1^2+\cdots +x_r^2$. Pair its nonzero coordinates into vectors $e_{2j-1}+i{e}_{2j}$ and add the $n-r$ radical coordinate vectors. Their span is a totally isotropic subspace, because the form is zero on every linear combination. This proves that complex quadratic forms have a large isotropic subspace: its dimension isFor $d=0$ there is a nonzero vector whenever $n\ge 1$. For $d\ge 1$, take such a subspace $E$ for $q_1$. If $n\ge 2^d$, then $\dim E\ge 2^{d-1}$. Apply the induction hypothesis to the restrictions of $q_2,\dots ,q_d$ to $E$. It supplies a nonzero $x\in E$ annihilated by those forms, and $q_1(x)=0$ by construction. ThusThe argument makes no nondegeneracy assumption on any of the forms.

Let $x_{ij}\ge 0$ be shipped amounts, with column sums equal to shop demand and row sums at most 15. Total demand is 42, so three units of warehouse stock need not be shipped. A feasible transportation problem allocation isIts row sums are $(15,15,12)$ and its column sums meet every demand. The cost is $18+18+12+3+4+10=65$.

For a transportation dual certificate with capacity inequalities, choose warehouse potentials $u=(0,-1,0)$ and shop potentials $v=(2,3,2,2,1)$. They have $u_i\le 0$ and $u_i+v_j\le c_{ij}$ for every cell. Any feasible allocation therefore hasThe second inequality has this direction because $u_i\le 0$ and warehouse use is at most 15. The exhibited allocation attains the lower bound, so the minimum cost is $\boxed{65}$.