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The inverse function theorem says that a continuously differentiable map on an open subset of ${\mathbb{R}}^n$ whose derivative at a point is invertible has a continuously differentiable inverse between neighborhoods of that point and its image. The derivative of the inverse is the inverse derivative matrix.

For the given map the Jacobian matrix and its determinant areHence every point of the nonempty open sethas a local continuously differentiable inverse. At its image, the derivative of that inverse isAt the omitted points a differentiable inverse with differentiable composition cannot exist, since the chain rule would force the singular derivative to be invertible.

To locate the exceptions on the curve, substitute $x=y^2$ into its defining relation. This gives $y^6-2y^3=y^3(y^3-2)=0$. The two exceptional points are thereforeAt any other point $(a,b)$ of the curve, $f(a,b)=(a,0)$ and the local inverse exists. Because the first coordinate of $f$ is $x$, its inverse at $(t,0)$ has the form $(t,h(t))$. The function $h$ is continuously differentiable for $t$ in a sufficiently small interval $I$ containing $a$. Shrink an interval $J$ containing $b$ so that $I\times J$ lies in the inverse neighborhood and $h(I)\subset J$. A point of the curve in that rectangle maps to $(t,0)$, so injectivity forces it to be $(t,h(t))$. Conversely all these inverse images belong to the curve. This proves the requested local graph description by the inverse function theorem, rather than merely asserting an implicit-function conclusion. Differentiating the relation also gives $h^{\prime }(t)=-(t^2-h(t))/(h(t{)}^2-t)$ there.

Take a finite directed flow network with distinct source $s$, sink $t$ and finite nonnegative flow network edge capacities. A flow assigns $f_e$ to each edge with $0\le f_e\le c_e$ and obeys flow conservation at every vertex except $s,t$. Its value is net outflow from the source, $|f|={\sum }_{s\to v}{f}_e-{\sum }_{v\to s}{f}_e$. A maximal flow here means a maximum flow: one with greatest value among feasible flows. A source-to-sink cut of a flow network is a partition $(S,S^c)$ with $s\in S$, $t\notin S$; its cut capacity is $c(S)={\sum }_{u\in S,v\notin S}{c}_{uv}$, counting directed edges leaving $S$.

The max-flow min-cut theorem statesSumming conservation over $S$ cancels internal edges and givesThis is the upper bound for every flow and every cut. To prove equality, a maximum exists: the feasible edge vectors form a nonempty closed bounded subset of a finite-dimensional real space, and flow value is continuous.

For a maximum $f$, construct its residual network. Each edge has a forward residual edge of capacity $c_e-f_e$ and a backward residual edge of capacity $f_e$; retain those of positive capacity. If a residual source-to-sink path existed, augment by the smallest residual capacity on that finite path. Increasing original forward flows and decreasing backward flows preserves all feasibility conditions and increases value, contradicting maximality. Thus no such augmenting path exists.

Let $S$ be the vertices reachable from $s$ in the residual network. Then $t\notin S$. Every original edge leaving $S$ is saturated, since otherwise its positive forward residual edge would make its endpoint reachable. Every original edge entering $S$ has zero flow, since otherwise its backward residual edge would give the same contradiction. HenceTogether with the universal upper bound this proves equality and identifies a minimum cut. This proof covers real capacities without assuming that an augmenting-path algorithm terminates after finitely many steps. Merely being impossible to increase all edge flows coordinatewise is not the optimization meaning of “maximal” used in the theorem.

With the proton treated as fixed and electron mass $m$, the classical energy and angular momentum areFor a circular orbit, $L=mrv$ and Coulomb's law determines its centripetal acceleration:Eliminating $v$ gives $r=4\pi {ϵ}_0{n}^2{\hslash }^2/(m{e}^2)$. Thus the Bohr radius and orbit radii areThe kinetic energy is half the magnitude of the attractive potential energy. HenceThis is the circular-orbit Bohr model; including proton recoil replaces $m$ by the reduced mass.

Take $S^{\prime }$ to move along $+x$ at velocity $V$, with the origins coincident at zero time. The Lorentz transformation isDifferentiating along the particle trajectory gives the relativistic velocity-addition formulaFor a timelike particle, $|u|<c$ and $|V|<c$. Direct algebra givesso $|u^{\prime }|<c$, in particular $u^{\prime }<c$. Strictness requires a massive, subluminal particle; for a photon with $u=\pm c$, the transformation preserves $u^{\prime }=\pm c$ instead.

Choose positive circulation anticlockwise as viewed from positive $z$. The linked magnetic flux is $\Phi =2\ell XB$, so Faraday law, including the motional contribution, givesThe conducting circuit consists of two arms of length $X$ and two transverse pieces of length $2\ell$. Its resistance is $\mathcal{R}=2R(X+2\ell )$, where $R$ is resistance per unit length. Neglecting self-inductance as in the quasistatic circuit model, the current isOn the moving rod this positive current points along $+y$, so the Lorentz force is $I(2\ell {\mathbf{e}}_y)\times (B{\mathbf{e}}_z)=2\ell IB{\mathbf{e}}_x$. Its mass is $2\ell M$, not the mass of the entire circuit. Newton's equation consequently givesFor constant $B$, the force opposes $\dot{X}$, as expected from Lenz's law. The prescribed time dependence of the external field can also supply energy to the rod.

For an endpoint-fixed variation $\eta =\delta x$,The boundary term vanishes. Thus every solution of the harmonic oscillator equation is stationary for this action functional. For $\sin \omega T\ne 0$, the endpoint-matching solution is the linear combinationwhich directly satisfies $x_c(0)=a$, $x_c(T)=b$ and ${\ddot{x}}_c=-{\omega }^2{x}_c$. This proves stationarity, rather than assuming the explicit path is a minimum.

On that path, $(x_c{\dot{x}}_c{)}^{\prime }={\dot{x}}_c^2-{\omega }^2{x}_c^2$, so integration by parts reduces the entire action to an endpoint term. The endpoint derivatives areTherefore the fixed-endpoint harmonic-oscillator action isThe displayed formula assumes nonresonant endpoints. If $\omega \ne 0$ and $\omega T=k\pi$, a solution exists only for $b=(-1{)}^{k}a$; then $x=a\cos \omega t+B\sin \omega t$ is a family and its action is zero. For $\omega =0$ the limiting stationary path is linear, with action $(b-a{)}^2/(2T)$.

The contour has winding number one around the single pole at zero. By the Cauchy integral formula, or the residue of $1/z$,Rotation by $z\mapsto iz$ carries each oriented edge to the next oriented edge and leaves $dz/z$ unchanged. The four edge integrals are therefore equal, each $\pi i/2$.

On the right edge use $z=1+it$, $-1\le t\le 1$, soThe real integrand $t/(1+t^2)$ is odd and integrates to zero. Dividing by $i$ gives ${\int }_{-1}^1(1+t^2{)}^{-1}dt=\pi /2$ without needing its real antiderivative.

Nonnegativity and symmetry of the maximum product metric follow from those of the two factor metrics. Its value is zero precisely when both coordinates agree. For three points $u=(x,x^{\prime })$, $v=(y,y^{\prime })$, $w=(z,z^{\prime })$, the two triangle inequalities giveAll metric axioms therefore hold.

When the two factors are the same metric space, consider an off-diagonal point $(x,y)$ and set $\delta =d(x,y)>0$. A product-metric ball of radius $\delta /3$ cannot contain any $(z,z)$: such membership would imply $d(x,z),d(y,z)<\delta /3$ and then $d(x,y)<2\delta /3$, a contradiction. The complement of the diagonal is thus open, soThis proves the closed diagonal of a metric space. Equality of the factors here means equality as metric spaces, as intended; merely sharing an underlying set while allowing unrelated metrics would be a different assertion.

The Euler formula for a sphere, applied to the surface of a convex polyhedron, is $V-E+F=2$. In a regular polyhedron let $q\ge 3$ be the common number of edges incident at a vertex. Counting vertex-edge and face-edge incidences, pentagonal faces give $qV=2E$ and $5F=2E$. Hencewhose right-hand side can be positive only if $q<10/3$. Thus $q=3$. Substitution then yields $2=E/15$, provingThese are the counts of the dodecahedron. Here regular polyhedron is understood in the convex, ordinary-face sense used by the stated Euler formula; self-intersecting star constructions do not share these hypotheses.

Write $\tau =(1+\sqrt{-3})/2$. The parity condition gives $R=\mathbb{Z}[\tau ]$, since $(a+b\sqrt{-3})/2=(a-b)/2+b\tau$. For any such element,If $a,b$ are even the numerator is divisible by four; if they are odd it is congruent to $1+3=0$ modulo four. Therefore $N(z)$ is a nonnegative integer. It is zero only at zero, and the Eisenstein-integer norm is multiplicative because complex conjugation respects multiplication.

If $z$ is a unit, $N(z)N(z^{-1})=1$, so $N(z)=1$. Conversely $N(z)=1$ makes $\overline{z}$ its inverse in $R$. The equation $a^2+3b^2=4$ permits exactly $(a,b)=(\pm 2,0)$ or $(\pm 1,\pm 1)$. HenceIndeed ${\tau }^2=\tau -1$ and ${\tau }^3=-1$, so this unit group is cyclic of order six.

For the prime-ideal assertion, ${\tau }^2-\tau +1=0$ and the polynomial $t^2-t+1$ has distinct roots $3,5$ modulo seven. Consequently evaluation at either root defines a surjective ring map $R\to {\mathbb{F}}_7$. Its kernelsare distinct maximal ideals, hence prime ideals. If $A+B\tau$ belongs to both, then $A+3B\equiv A+5B\equiv 0(\mathrm{mod}7)$. Subtraction gives $2B\equiv 0$, so both $A$ and $B$ are divisible by seven. The converse inclusion is immediate, provingEquivalently, the two distinct factors of the reduced polynomial give the Chinese remainder theorem decomposition $R/7R\cong {\mathbb{F}}_7\times {\mathbb{F}}_7$. This proof does not require the permitted unique-factorization assumption.

Use the original PDF: the objective is $z=-x_1+3x_2$, and the second constraint has coefficient $2$ on $x_2$. The TeX transcription corrupts both, as well as the nonnegativity condition. Introduce surplus variables $s_1,s_2$, slack variables $s_3,s_4$, and artificial variables $a_1,a_2$, all nonnegative. The two-phase simplex Phase One problem isThe initial basis $(a_1,a_2,s_3,s_4)$ has values $(3,6,2,5)$, and its objective dictionary is $w=9-3x_2+s_1+s_2$. Enter $x_2$: the ratio test makes $s_3$ leave at $x_2=2$. The resulting dictionary includeswith $w=3-3x_1+s_1+s_2+3s_3$. Enter $x_1$ next; $a_1$ leaves at $x_1=1/2$. Nowand $w=3/2-s_1/2+s_2+3s_3/2+3a_1/2$. Enter $s_1$; the ratio test makes $a_2$ leave at $s_1=3$. This produces $x_1=2$, $x_2=4$, $s_1=3$, $s_4=1$ with both artificial variables zero. Since $w\ge 0$ always, the Phase One minimum is zero, proving feasibility and providing an original-variable basis.

Delete the artificial columns. The starting Phase Two dictionary isEnter $s_2$, whose reduced gain is positive. Only $s_4$ decreases, and it leaves at step one. Eliminating $s_2=1-s_3-s_4$ gives the optimal dictionaryEquivalently, with row equations written as the displayed coefficients times the variables equal to the right-hand side, the optimal simplex tableau isThe objective row represents $z+s_3+2s_4=12$. Nonnegative nonbasic variables cannot improve the objective, so reading them as zero provesBoth nonbasic reduced costs are strictly unfavorable, so this optimal point is unique.