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Write $z=x+iy$ and $w=u+iv$. Multiplying the numerator and denominator of the Möbius transformation by $1+\bar{z}$ givesPull back the given upper-half-plane solution to obtain the harmonic measure of a semicircle in a disk:The principal real arctangent is appropriate because $v>0$. By conformal invariance of harmonicity, composing a harmonic function with this analytic conformal map is harmonic. More explicitly, ${\Delta }_z(f\circ w)=|w^{\prime }(z){|}^2({\Delta }_{w}f)\circ w=0$. At a boundary point on the open upper semicircle the ratio tends to $+\infty$, yielding one; on the lower semicircle it tends to $-\infty$, yielding zero.

At $z=\pm 1$ the two boundary prescriptions jump, so no continuous boundary value can be imposed there and the interior limit depends on the approach. The bounded harmonic solution is interpreted with the specified values away from those endpoints. Boundedness also supplies the usual uniqueness for this Dirichlet problem; without a boundedness or comparable regularity restriction, singular harmonic functions at the omitted endpoints would allow extra solutions.

The function is Fréchet differentiable at $(a,b)$ when there is a linear map $L:{\mathbb{R}}^2\to \mathbb{R}$ such thatIf it exists, $L(h,k)=f_x(a,b)h+f_y(a,b)k$, as is seen by taking increments along each coordinate axis.

To prove differentiability from continuous partial derivatives, take a small rectangle in the given neighbourhood and split the increment along its sides. The one-variable mean value theorem givesfor some $0<\theta ,\eta <1$ when the corresponding increment is nonzero. If an increment is zero, its term is simply omitted. The required one-variable continuity follows from existence of the corresponding partial derivative on the segment. Subtracting $L(h,k)$ gives a remainder whose absolute value, by continuity of both partial derivatives at $(a,b)$, is at mostfor sufficiently small increments. Since $ϵ$ is arbitrary, the normalized remainder tends to zero. Thus the function is differentiable at $(a,b)$, with the displayed derivative.

Parametrize the surface by $r(x,y)=(x,y,xy)$. Its tangent vectors are $r_x=(1,0,y)$ and $r_y=(0,1,x)$. Choose the upward unit normalThe first fundamental form coefficients areSince $r_{xx}=r_{yy}=0$ and $r_{xy}=(0,0,1)$, the second fundamental form coefficients are $e=g=0$ and $f=(1+x^2+y^2{)}^{-1/2}$. The determinant of the shape operator gives the Gaussian curvature:Changing the normal's sign changes both second-form factors and leaves this determinant unchanged. Although the question calls the surface a hyperboloid, $z=xy$ is a hyperbolic paraboloid; the calculation uses the stated surface equation.

In a zero-sum game, an entry $a_{ij}$ is the net amount received by the row player when the players choose actions $i,j$; the column player receives its negative. A mixed strategy $p$ has nonnegative entries summing to one. Against column strategy $q$, expected row payoff is $p^{T}Aq$. An optimal row strategy maximizes its guaranteed payoff ${\min }_q{p}^{T}Aq$; an optimal column strategy minimizes ${\max }_p{p}^{T}Aq$.

The betting payoff must subtract the player's own stake: a winner receives both stakes but has already paid its own. Its net gain is the opponent's stake, and a loser loses its own. Ties give zero. Thus the row-payoff matrix isIt is antisymmetric. Interchanging players therefore changes the sign of the game value while leaving the game unchanged, so its value is zero. An optimal column strategy $p$ must prevent any pure row choice from gaining positively, namely $(Ap{)}_i\le 0$. Antisymmetry then also gives $p^{T}A\ge 0$, so the same vector is an optimal row strategy. This is the support certificate for an antisymmetric matrix game.

Seek a mixture supported on actions $1,3,4$ making those three opposing choices indifferent. Their zero-payoff equations areWith $p_1+p_3+p_4=1$, these giveCheck every opponent action, including the omitted second one:This verifies optimality rather than simply assuming that a support choice is valid.

It is also unique. Let $r$ be any other optimal strategy, so $Ar\le 0$. For the displayed $p$, antisymmetry gives $p^{T}Ar=-(Ap{)}^{T}r=(7/8)r_2$. The left side is nonpositive and the right side nonnegative, forcing $r_2=0$. Since the other three entries of $p$ are positive, $p^{T}Ar=0$ also forces $(Ar{)}_1=(Ar{)}_3=(Ar{)}_4=0$. Their equations and normalization recover the same probabilities. Thus both players' unique optimal strategy uses bets one, three and four in the proportions $3:4:1$.

For an inviscid flow of constant density $\rho$, driven by a conservative body force $-\nabla \Phi$, the Euler equation isThe vector identity $(u\cdot \nabla )u=\nabla (|u{|}^2/2)-u\times (\nabla \times u)$ and irrotationality give the Unsteady Bernoulli equation. Specifically, wherever a single-valued potential $u=\nabla \varphi$ is available,On a connected potential-flow region this meansThe integration function is spatially constant but need not be constant in time; it can be removed by changing the potential's time-dependent gauge. A simply connected irrotational region provides the requisite global potential. For a barotropic fluid the pressure term is replaced by $\int dp/\rho (p)$; here the liquid density is constant.

For the tube, let $U(t)$ be its approximately uniform axial velocity. The large reservoir's surface velocity is negligible and gravity is neglected as stipulated. Entrance-region inertia contributes an effective length of order $a$, negligible against $L$. The potential difference between the outlet and the reservoir is consequently $LU$ to leading order. Subtract Bernoulli at the reservoir surface, where pressure is $P+p_a$, from its value at the outlet, where pressure is $p_a$ and speed is $U$. This givesThe inertial term cannot be dropped during startup. With $U_{\infty }=\sqrt{2P/\rho }$, separate variables:Hence the inviscid startup in a pressure-driven tube givesInitially $U\sim Pt/(\rho L)$, the acceleration of the stationary liquid column. At long times $U\to \sqrt{2P/\rho }$, the inviscid pressure-to-kinetic-energy limit. The result applies while the tube remains full and reservoir dimensions are large enough that surface motion and the neglected entrance inertia remain small.

In SI units, the source-free Maxwell equations in vacuum areUse complex amplitudes with phase $e^{i(\omega t-k\cdot x)}$ and take real parts at the end. The electric divergence equation gives $k\cdot E_0=0$. Faraday's equation gives $k\times E_0=\omega B_0$, soThe magnetic divergence equation follows because $k\cdot (k\times E_0)=0$. Ampere's equation becomes $k\times B_0=-{\mu }_0{ϵ}_0\omega E_0$. Since $k\times (k\times E_0)=-|k{|}^2{E}_0$, a nonzero wave satisfiesThus both fields are transverse, mutually perpendicular for linear polarization, and obey the vacuum plane-wave dispersion relation.

For the interface let $n=e_z$ point from the minus side to the plus side and write $[E]=E_{+}-E_{-}$, similarly for $B$. Applying the integral Maxwell equations to a thin pillbox and a thin loop gives the electromagnetic boundary conditionsHere $j$ is the tangential surface current. The jump notation fixes the signs, including the direction of the normal.

For the perfect conductor, take the time-varying interior fields to vanish. Choose positive $k$ with $\omega =ck$. The incident electric field has complex form $E_0{e}_x{e}^{i\omega t+ikz}$. To cancel its tangential value at the surface, use the reflected field $-E_0{e}_x{e}^{i\omega t-ikz}$, whose wavevector points upward. The total vacuum fields are thereforeThe sign of each magnetic contribution follows from $B_0=k\times E_0/\omega$. At $z=0$, the electric field is zero and the magnetic field is tangential. The jump conditions yieldFor a real incident amplitude this is $j=(2E_0/{\mu }_{0}c)\cos (\omega t)e_x$. It supplies the reflected wave while satisfying the normal-field and tangential-electric conditions. Any independently imposed static interior field is separate from this oscillatory response.

For a normalized state and an observable $A$, define $⟨A⟩=⟨\psi ,A\psi ⟩$ and ${\Delta }_{\psi }A=\Vert (A-⟨A⟩)\psi \Vert$. Suppose first that the position and momentum variances are finite and the state has the regularity needed for the usual position-momentum quadratic forms. Put $a=⟨x⟩$, $b=⟨p⟩$, $u=(x-a)\psi$ and $v=(-i\hslash {\partial }_x-b)\psi$. The Cauchy-Schwarz inequality givesThe term involving $b$ has zero imaginary part. Integration by parts givesConsequently $\Im ⟨u,v⟩=\hslash /2$, andThis proves the Heisenberg uncertainty relation. For smooth rapidly decaying states the boundary terms vanish directly. The same proof extends to $x\psi \in L^2$ and weak derivative ${\psi }^{\prime }\in L^2$ using smooth cutoffs and taking their radius to infinity; the necessary mixed products are integrable by Cauchy-Schwarz. Normalization alone need not give finite variances or an operator-domain momentum expectation. A state with infinite spread is interpreted through the corresponding extended bound, rather than assigning a finite uncertainty to an undefined expression.

For the specified Gaussian wave packet, write $s=\hslash t$ and $a_t=1+is$. Its probability density isIt is a normalized centered normal distribution with variance $1+s^2$. Symmetry gives $⟨x⟩=0$, and henceAlso ${\partial }_x\psi =-x\psi /(2a_t)$, so $p\psi =i\hslash x\psi /(2a_t)$ and $⟨p⟩=0$ by oddness. Integration by parts, or directly taking the squared norm, givesThusThe packet saturates the bound at $t=0$ and has a larger position-momentum uncertainty product at every nonzero time. Its spatial spreading does not change the momentum distribution.

Let $p$ first be an arbitrary degree-$n$ polynomial with leading coefficient $a_n\ne 0$. Choose $R$ so large that on $|z|=R$,The homotopy $a_n{z}^n+s[p(z)-a_n{z}^n]$, $0\le s\le 1$, never meets zero on that circle. Hence the image curve $p(R{e}^{i\theta })$ has winding number $n$, the same as $a_n{R}^n{e}^{in\theta }$. The argument principle counts precisely $n$ zeros inside, with multiplicity, since there are no poles. Choose $R$ large enough that the same leading-term domination holds for every $|z|\ge R$; the lower-term ratios decrease with radius, so this also excludes all exterior zeros. This proves the fundamental theorem of algebra with its multiplicity count. A degree-zero nonzero polynomial has no zeros.

Now use the specified bound for the monic quartic. Compare $\Gamma$ with the closed curveIts first piece travels once counterclockwise around a circle, and its second stays at the endpoint, so its winding number is one. On the first piece, $|\Gamma -{\gamma }_0|<R^4$. On the second, both $p(R)$ and $p(iR)$ differ from $R^4$ by less than $R^4$; every convex combination of those errors also has modulus less than $R^4$. Thus the quoted winding number stability result gives

For real coefficients define $q(y)=y^4-b{y}^2+d$. It has no real zeros and positive leading coefficient, so $q(y)>0$ for every real $y$, including $d=q(0)>0$. Thereforelies strictly in the right half-plane for every real $y$. Also $p(x)>0$ for $x\ge 0$: there are no zeros there, and the polynomial is positive for sufficiently large positive $x$. These facts exclude roots on the positive real and imaginary axes.

Consider the positively oriented quarter-disk boundary: first the arc from $R$ to $iR$, then the imaginary axis to zero, then the real axis back to $R$. Under $p$, its last two pieces form a path from $p(iR)$ to $p(R)$ entirely within the right half-plane. The chord closing $\Gamma$ also lies there. Since this half-plane is convex and excludes zero, deforming one closing path into the other changes no winding number. Thus the full image of the quarter-disk boundary has winding number one. By the argument principle there is exactly one root inside that quarter-disk, counted with multiplicity.

Finally, all polynomial roots lie in $|z|<R$ under the given strict coefficient bound: for $|z|\ge R$, the sum of lower terms divided by $|z{|}^4$ is smaller than one. HenceIts multiplicity is one. This is the quarter-sector winding test for a real quartic; it accounts for the axis exclusions and does not replace the sector boundary by a chord without justifying the deformation.

The inverse function theorem says: if $U\subseteq {\mathbb{R}}^2$ is open, $F:U\to {\mathbb{R}}^2$ is smooth, $a\in U$, and $DF(a)$ is invertible, then there are open neighbourhoods $V\subseteq U$ of $a$ and $W$ of $F(a)$ for which $F:V\to W$ is a bijection with a smooth inverse. Its derivative is $D(F^{-1})(F(x))=[DF(x){]}^{-1}$ after shrinking $V$ if necessary. A smooth function here has derivatives of every order.

For the given map,Thus its local smooth-invertibility set isA differentiable local inverse is impossible at either excluded line by the chain rule. Even a merely continuous local inverse is impossible there: at fixed $y$, the cubic first component has a strict local extremum at these $x$ values, giving equal values at two nearby $x$ coordinates and preventing local injectivity.

On the curve, $y^2=x^3-x=x(x-1)(x+1)\ge 0$. Its sign intervals give exactly $x\in [-1,0]\cup [1,\infty )$, so the two prescribed subsets cover the curve and are disjoint. For fixed $y$, the function $x^3-x$ is strictly increasing on $[1,\infty )$, starts at zero and tends to infinity. The intermediate value theorem therefore gives a unique $p(y)\ge 1$ with $p(y{)}^3-p(y)=y^2$.

At every point of this second component, $3x^2-1\ge 2$, so the inverse function theorem applies. Locally its smooth inverse evaluated at $(0,y)$ has the form $(p(y),y)$, by the proved uniqueness. These local descriptions agree on overlaps, proving $p$ is smooth on all of $\mathbb{R}$. For example,whose nonzero denominator also makes the smoothness mechanism transparent.

Use the unit sphere $X^2+Y^2+Z^2=1$, its north pole $(0,0,1)$, and the equatorial complex plane $Z=0$. The line from the north pole to a sphere point meets that plane atThe north pole maps to $\infty$. The inverse stereographic projection isThe antipodal point has coordinate $w=-(X+iY)/(1+Z)$. Since $X^2+Y^2=(1-Z)(1+Z)$, this is $-1/\bar{z}$ when both sides are finite and nonzero. Zero and infinity correspond to the two poles and are exchanged by the same extended convention. The inverse formula also proves the converse, giving the antipodal stereographic coordinate relation

The requested rotation result is that sphere rotations correspond exactly to Möbius transformationsThe matrices $\left(\begin{array}{cc}a& b\\ -\bar{b}& \bar{a}\end{array}\right)$ form $SU(2)$, and opposite matrices induce the same transformation. Thus the rotation group $SO(3)$ is identified with $SU(2)/\{\pm I\}$. This is the sphere rotations as special-unitary Möbius transformations result, which need not be proved here.

For a planar circle $|z-c|=r$, $c=a+ib$ and $r>0$, substitute $|z{|}^2=(1+Z)/(1-Z)$ and $\Re (\bar{c}z)=(aX+bY)/(1-Z)$. Multiplication by $1-Z$ gives the circle-plane relation under stereographic projection:The lift lies in this plane, and conversely its intersection with the sphere projects onto the circle. It is a nondegenerate circle on the sphere. The north pole is not on the displayed plane, since substitution there gives two rather than zero.

A great circle is the intersection with a plane through the sphere's centre. Here this holds exactly whenThe unit circle lifts to the equator. If a different circle lifts to a great circle, its plane and the equatorial plane are distinct planes through the origin, so their intersection cuts the sphere at two antipodal equatorial points. Their planar coordinates are $z$ and $-z$ on $|z|=1$.

Conversely, if the planar circle meets the unit circle in $z$ and $-z$, their lifts are antipodal equatorial points. The plane of the lifted circle contains their midpoint, the origin, so that circle is great. Equivalently, adding $|z-c{|}^2=r^2$ and $|-z-c{|}^2=r^2$ gives $r^2=1+|c{|}^2$. The exclusion of the unit circle ensures the two intersection points are a genuine pair rather than coincident circles. This proves the antipodal equatorial intersection criterion for a great circle.

Use the usual commutative unital-ring convention for Noetherian rings. A ring is Noetherian exactly when every ideal is finitely generated, equivalently when ascending chains of ideals stabilize. Indeed, the union of an ascending chain is an ideal; finitely many generators all lie in a single chain member. Conversely, if an ideal cannot be finitely generated, adjoining a new element at each stage constructs a strictly ascending chain.

First assume $R[X]$ is Noetherian. Evaluation at zero gives a surjective ring homomorphism $R[X]\to R$ with kernel $(X)$. A quotient of a Noetherian ring is Noetherian: lift an ideal to the source, take finitely many generators there, and map those generators down. Thus $R\cong R[X]/(X)$ is Noetherian.

For the converse, assume $R$ is Noetherian and take any ideal $I\subseteq R[X]$. For $d\ge 0$, let $J_d$ be the set of coefficients of $X^d$ in elements of $I$ of degree at most $d$. It is an ideal of $R$, since sums and scalar multiples of such polynomials stay in $I$. Multiplication by $X$ shows $J_d\subseteq J_{d+1}$. The ascending chain stabilizes, say at $N$.

For each $0\le d\le N$, choose finitely many generators $a_{d,j}$ of $J_d$ and corresponding $f_{d,j}\in I$ of degree at most $d$ having coefficient $a_{d,j}$ at $X^d$. Discard zero generators. We claim these finitely many polynomials generate $I$. If a nonzero $f\in I$ has degree $m\le N$, express its leading coefficient as ${\sum }_j{r}_j{a}_{m,j}$ and subtract ${\sum }_j{r}_j{f}_{m,j}$. The degree decreases. If $m>N$, use $J_m=J_N$, express the leading coefficient using the $a_{N,j}$, and subtractAgain the leading term cancels and the degree decreases. Induction on degree, including degree zero, expresses every $f$ in the claimed ideal. Therefore every ideal of $R[X]$ is finitely generated, proving the Hilbert basis theorem andThe argument is the bounded-degree coefficient proof of Hilbert basis theorem; it never divides by a leading coefficient and therefore does not require $R$ to be an integral domain. With the analogous left-ideal convention the same proof applies to left-Noetherian rings with a central indeterminate.