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A practical family consists of constant-distance parallel surfaces and smooth variable normal offsetsConstants cover wall thickness, cutter radius and clearance. A scalar B-spline or piecewise polynomial displacement field permits thickness variation, local bumps and smooth blending; compact support makes local edits economical. Use the base parameterization or compatible data on its subdivision hierarchy so that evaluating the displacement and its derivatives requires only a small local stencil. Choose displacement size and gradients to avoid unintended folds, and trim global self-intersections where the physical application requires an envelope. A constant displacement smaller than the local curvature radii prevents the corresponding local focal singularities, although global clearance still needs a separate check.

When remaining within a polynomial or rational representation is more useful than prescribing an exact distance, a related practical form is $P+\lambda (u,v)(P_u\times P_v)$ with a low-degree scalar field $\lambda$. Its displacement is normal but has magnitude $|\lambda |\Vert P_u\times P_v\Vert$, so it is not a constant-distance parallel surface. For polynomial $P$ and $\lambda$, it stays polynomial; for rational input it stays rational. More general low-degree vector displacement fields similarly preserve efficient evaluation, while relinquishing the normal-offset condition. These forms make explicit the tradeoff between an exact metric offset, which involves normalization, and an easily represented deformation. Simple scalar normal-displacement fields are useful and directly evaluable; algebraic displacement fields are useful when representation closure is required.

Rolle's theorem states that a continuous function $f:[a,b]\to \mathbb{R}$, differentiable on $(a,b)$ with $a<b$ and $f(a)=f(b)$, has $f^{\prime }(c)=0$ for some $c\in (a,b)$. If $f$ is constant this is immediate. Otherwise the extreme value theorem supplies a maximum or minimum with value different from the common endpoint value, so one such extremum occurs in $(a,b)$. At an interior maximum, the difference quotients for positive increments are nonpositive and those for negative increments are nonnegative. Their common limit, the derivative, must be zero. At a minimum the inequalities reverse, with the same conclusion. This proves Rolle's theorem.

For the two-function identity, write $\Delta f=f(b)-f(a)$ and $\Delta g=g(b)-g(a)$, and set $\varphi (x)=g(x)\Delta f-f(x)\Delta g$. This is a continuous function on the closed interval and is differentiable inside it. Its endpoint values agree:They need not be zero, but Rolle's theorem only requires equality. At the resulting point $c$,which proves the Cauchy mean value theorem identityFor the final endpoint limit, first state the denominator convention explicitly. In the usual zero-over-zero version of L'Hopital rule, $g^{\prime }(x)\ne 0$ on some whole interval $(0,\delta )$; this is understood if the displayed derivative quotient is to be a function defined throughout a deleted right neighborhood of zero. Define $f(0)=g(0)=0$, making both functions continuous at zero. For $0<x<\delta$, Rolle's theorem implies $g(x)\ne 0$, since otherwise $g$ would have equal zero values at $0$ and $x$ and a zero derivative between them. Apply the proved Cauchy mean value theorem on $[0,x]$ to obtain $c_x\in (0,x)$ withGiven $\epsilon >0$, the hypothesis on the derivative quotient supplies $\eta >0$ such that its distance from $\ell$ is less than $\epsilon$ whenever $0<t<\eta$. If $0<x<\min (\eta ,\delta )$ then $0<c_x<x<\eta$, so the same bound holds for $f(x)/g(x)$. ThereforeNo continuity of the derivatives is needed.

The PDF does not explicitly state the nonvanishing condition on $g^{\prime }$. If “the limit exists” permits a limit only along the natural domain where $g^{\prime }\ne 0$, the assertion is false, even with $g>0$ everywhere. The distinction concerns derivative quotient limits on partial domains. Here is a concrete counterexample to that weaker reading. Put $x_n=2^{-n}$ and $h_n=4^{-n}$ for $n\ge 0$. On each interval $[x_n/2,x_n]$, split off the plateau $[3x_n/4,x_n]$. On the plateau setOn the transition interval $[x_n/2,3x_n/4]$ setExtend $g=1,f=0$ for $x\ge 1$. The endpoints fit because $S(0)=0$, $S(1)=1$, while both $S^{\prime }$ and $q^{\prime }$ vanish at zero and one and $q$ also vanishes there. Thus $f,g$ are differentiable for all $x>0$ and tend to zero at zero. The derivative $g^{\prime }$ is nonzero exactly inside the transition intervals, where $f^{\prime }=0$, so the derivative quotient has natural-domain limit zero. But $f(x_n)/g(x_n)=0$, whereas $f(7x_n/8)/g(7x_n/8)=1$. Hence $f/g$ has no limit. The preceding qualified proof is the intended valid form of L'Hopital rule.

The parity operator acts by $(P\psi )(x)=\psi (-x)$. For an even potential, differentiating twice and using $V(-x)=V(x)$ proves $HP=PH$ on the parity-invariant domain of the Hamiltonian operator. Therefore each energy eigenspace is invariant under $P$. The projectors $(I+P)/2$ and $(I-P)/2$ split it into even and odd subspaces, giving a basis of simultaneous eigenstates. An odd state obeys $\psi (0)=-\psi (0)$, hence ${\psi }_{\mathrm{odd}}(0)=0$.

The infinite walls impose $\psi (-a)=\psi (a)=0$, with the wavefunction zero outside. It is continuous at the central delta potential. Integrate the stationary Schrödinger equation over $[-\epsilon ,\epsilon ]$; the energy integral tends to zero and the Dirac delta function contributes $\kappa \psi (0)$. This givesFor the given piecewise positive-energy form, $E={\hslash }^2{\lambda }^2/(2m)$. Replacing $x$ by $-x$ swaps the two pieces and leaves their values unchanged, proving $P\psi =\psi$. The value at zero is $A$, and the one-sided derivatives are $-B\lambda$ and $B\lambda$. The jump condition gives $B=-m\kappa A/({\hslash }^2\lambda )$. The wall at $a$ gives $A\cos (\lambda a)-B\sin (\lambda a)=0$, henceFor $\kappa =0$ the undivided wall equation instead gives the usual free even roots. For $\kappa >0$, set $c=m\kappa /{\hslash }^2>0$ and $k_n=(n+1/2)\pi /a$. In every interval $k_n<\lambda <(n+1)\pi /a$ there is exactly one root: writing $\delta =a(\lambda -k_n)\in (0,\pi /2)$ gives $\cot \delta =\lambda /c$, whose two sides are respectively strictly decreasing and increasing. These are all positive even roots; the quadratic form ${\hslash }^2\int |{\psi }^{\prime }{|}^2/(2m)+\kappa |\psi (0){|}^2$ is nonnegative, so there are no negative eigenvalues. Thus ${\lambda }_n>k_n$ and ${\eta }_n={\hslash }^2({\lambda }_n^2-k_n^2)/(2m)>0$.

The printed limit of this absolute energy difference is false. The exact quantization equation gives ${\delta }_n=\arctan (c/{\lambda }_n)$, so ${\delta }_n\to 0$ but ${\lambda }_n{\delta }_n\to c$. ConsequentlyThus the high-energy shift from a central delta barrier is a finite positive constant, not zero. The corrected vanishing statement is the relative shift ${\eta }_n/E_n^{(0)}\to 0$, or the wave-number difference ${\lambda }_n-k_n\to 0$. Physically, highly energetic particles are weakly affected in relative terms by a fixed point barrier. The free even wavefunction normalized in $[-a,a]$ has density $1/a$ at the origin, so the barrier's first-order expectation is $\kappa /a$, matching the exact limit.

Odd states vanish at zero, so the delta term and the derivative jump both vanish. Their normalized wavefunctions and energies arewith zero wavefunction outside. They do not depend on $\kappa$. These even and odd sectors describe the central delta barrier in a symmetric infinite well.

The charge and its proposed image charge give the exterior electric potentialAt $z=0$ the terms cancel, so $\Phi =0$ everywhere on the plane and its tangential derivatives vanish. Thus the image construction satisfies the earthed-conductor boundary conditions. It has the correct point-charge singularity, is harmonic elsewhere in $z>0$, and decays at infinity; electrostatic uniqueness identifies it with the exterior solution.

Use the steady inviscid, hydrostatic, slowly varying channel approximation, without a hydraulic jump. Conservation of volume flux per unit width gives $Q=u(x)h(x)=u_1{h}_1$. The free-surface Bernoulli equation isDividing by $h_1$ yieldsThe parameter $F$ here is the squared upstream Froude number. For nonzero discharge, regarding $D$ as a function of $h$, its derivative is $dD/dh=-1+Q^2/(g{h}^3)$. It has its unique maximum at $h_c=(Q^2/g{)}^{1/3}=h_1{F}^{1/3}$, where the local Froude number is one. Substitution gives $D^{\ast }=h_1(1-3F^{1/3}/2+F/2)$. This maximum is the hydraulic control over a smooth hump.

For a holomorphic function, compute its complex derivative along the real and imaginary directions. The real direction gives $f^{\prime }=u_x+i{v}_x$, while division by an imaginary increment gives $f^{\prime }=v_y-i{u}_y$. Equating real and imaginary parts proves the Cauchy-Riemann equations:For the specified real part, $u_x=-e^{-x}\cos y$ and $u_y=-e^{-x}\sin y$. Integrating $v_y=u_x$ gives $v=-e^{-x}\sin y+C(x)$; the other relation in the Cauchy-Riemann equations then forces $C^{\prime }(x)=0$. Thus the harmonic conjugate and the holomorphic function are

A subset $U\subseteq {\mathbb{R}}^n$ is an open set when every $x\in U$ has some radius $r>0$ for which $B(x,r)\subseteq U$. If a continuous function takes only the values zero and one, continuity at $x$ with tolerance $1/2$ gives a neighborhood on which $|f(y)-f(x)|<1/2$. The two possible values differ by one, so $f(y)=f(x)$ throughout that neighborhood. Thus $f$ is locally constant, and the zero linear map is its derivative, since the difference $f(x+h)-f(x)$ is exactly zero for all sufficiently small $h$. Hence $D{f}_x=0$ everywhere.

For the bound on $g$, parametrize the segment by $\gamma (t)=b+t(a-b)$, $0\le t\le 1$. The chain rule givesby the definition of operator norm. The mean value theorem applied to the real function $g\circ \gamma$ provesIf the endpoints agree the estimate is immediate. No continuity of the derivative is required.

For the finite-line complement, each removed line ${\ell }_i$ is closed, so $V$ is open. Because $a,b\notin {\ell }_i$, there is a unique affine plane $P_i$ containing $a$ and ${\ell }_i$, and a unique affine plane $Q_i$ containing $b$ and ${\ell }_i$. Choose $c$ outside the union of these finitely many planes, using the permitted fact about finite plane covers. In particular $c\in V$. If $[a,c]$ met ${\ell }_i$ at a point $d$, then $d\ne a$, and the line through $a,d$ lies in $P_i$. Since $c$ is on that same line, it would belong to $P_i$, a contradiction. The same argument using $Q_i$ applies to $[c,b]$. Thus $a$ and $b$ can be joined in $V$ by these two segments, proving polygonal connectivity of a finite-line complement.

Finally a continuous $f:V\to \{0,1\}$ has zero derivative by the first argument. Apply the proved segment estimate with $M=0$ separately on $[a,c]$ and $[c,b]$. It gives $f(a)=f(c)=f(b)$. Since $a,b$ were arbitrary, $f$ is constant on $V$.

HereThis entire function vanishes exactly at $z=inb$, $n\in \mathbb{Z}$, and those points map to $\zeta =(-1{)}^n$. The second derivative there is $(\pi /b{)}^2(-1{)}^n\ne 0$, so each is a simple critical point of the conformal map:Take the half-stripWriting $\pi z/b=u+iv$, its imaginary part maps to $\sinh u\sin v>0$. More precisely, $w=e^{\pi z/b}$ maps $D$ bijectively onto $\{w:\mathrm{Im}w>0,|w|>1\}$, with inverse $z=(b/\pi )\log w$ using arguments in $(0,\pi )$. ThenPart (i) with $b=1$, followed by positive real scaling, proves that this second map is a conformal bijection onto the upper half-plane. Their composition proves the required result and gives the half-strip conformal map by hyperbolic cosine.

For the Joukowski map $F(z)=z+b^2/z$, the derivative is $F^{\prime }(z)=1-b^2/z^2$. Thus it is a conformal map at every finite $z$ except its pole at zero and its critical pointsThe pole maps to $\infty$; as a map of Riemann spheres it is a simple pole and is not a branch point. Neither is the point at infinity, where the map behaves like $z$.

An appropriate domain is the upper exterior half-discFor $z=x+iy$, $\mathrm{Im}F(z)=y(1-b^2/|z{|}^2)>0$ on $D$. To prove surjectivity and injectivity, solve $F(z)=\zeta$ as $z^2-\zeta z+b^2=0$. Its two roots have product $b^2$. For $\mathrm{Im}\zeta >0$, neither root lies on the real axis or on $|z|=b$, since either boundary maps to the real axis. Exactly one root therefore has modulus greater than $b$. The imaginary-part identity forces that root to have positive imaginary part. It is the unique root in $D$, so $F$ maps $D$ conformally and bijectively onto the upper half-plane.

For a one-dimensional wavefunction, the probability density and probability current areThe Time-dependent Schrödinger equation with a real potential is $i\hslash {\psi }_t=-{\hslash }^2{\psi }_{xx}/(2m)+V\psi$. Multiply it and its complex conjugate by ${\psi }^{\ast }$ and $\psi$, respectively. Subtracting cancels the real-potential terms and givesThis derives the continuity equation ${\rho }_t+j_x=0$.

For $k>0$, the specified free-region wavefunction combines a right-moving incoming wave of unit amplitude and a left-moving reflected wave with complex reflection amplitude $R$. Applying the free Schrödinger operator to either plane wave gives $E={\hslash }^2{k}^2/(2m)$. Its time factor has modulus one, soThe density includes the stationary interference pattern; $|R{|}^2$, rather than $R$, is the reflection probability.

For the finite square well, let $k=\sqrt{2mE}/\hslash$ outside and $q=\sqrt{2m(E+V_0)}/\hslash$ inside. Write the stationary parts as $e^{ikx}+R{e}^{-ikx}$ to the left, $A{e}^{iqx}+B{e}^{-iqx}$ inside, and $T{e}^{ikx}$ to the right. At both finite potential steps, the wavefunction and its first spatial derivative are continuous. Integrating the stationary Schrödinger equation across a shrinking step proves derivative continuity; a jump of the wavefunction would create an unmatched distributional derivative and is excluded.

Propagate the boundary values at $a$ back through the well. The values at zero areMatching them to $1+R$ and $ik(1-R)$, and adding the two equations after division of the second by $ik$, givesSince the exterior wave numbers agree, the transmission probability is the current ratio $|T{|}^2$, soWhen $V_0=3E$, $q=2k=\sqrt{8mE}/\hslash$, yieldingClassically an incoming particle of positive energy crosses the attractive well and emerges with its original energy, so transmission is one. The quantum result agrees exactly when $qa$ is an integer multiple of $\pi$:These are widths with resonant transmission through a square well; the degenerate width zero also has unit transmission.

The contraction mapping theorem states that a contraction mapping $T:X\to X$ on a nonempty complete metric space has a unique fixed point. Here contraction means that some $0\le q<1$ satisfies $d(Tx,Ty)\le qd(x,y)$ for all $x,y$.

For a proof, choose $x_0$ and iterate $x_{r+1}=T{x}_r$. Induction gives $d(x_{r+1},x_r)\le q^{r}d(x_1,x_0)$. For $m>n$, the triangle inequality and a geometric series giveThus the iterates form a Cauchy sequence, and completeness supplies a limit $a\in X$. A contraction mapping is continuous, so $Ta=\lim T{x}_n=\lim x_{n+1}=a$. If $b$ is another fixed point, $d(a,b)=d(Ta,Tb)\le qd(a,b)$, forcing $d(a,b)=0$. This proves existence and uniqueness.

For the exponential problem, use $h=f\circ f$ on the complete metric space $\mathbb{R}$. Its derivative isPutting $t=e^{-x}>0$, the function $t{e}^{-t}$ has maximum $1/e$ at $t=1$, as its derivative is $(1-t)e^{-t}$. By the mean value theorem, $|h(x)-h(y)|\le e^{-1}|x-y|$, so $h$ is a contraction mapping on all of $\mathbb{R}$. Let $a$ be its unique fixed point. The compositions commute, so $h(f(a))=f(h(a))=f(a)$; uniqueness of the fixed point of $h$ forces $f(a)=a$. Conversely every fixed point of $f$ is fixed by $h$. Hence there is exactly one real solution of $x=e^{-x}$. It lies in $(0,1)$ because $a=e^{-a}>0$ and then $e^{-a}<1$. This uses the iterated contraction argument without claiming that $f$ itself contracts on all of $\mathbb{R}$.

A function is Fréchet differentiable at $x$ when there is a linear map $A:{\mathbb{R}}^n\to {\mathbb{R}}^m$ such thatThis linear map, which is unique, is the derivative $Df(x)$.

For the chain rule, if $f$ is differentiable at $x$ and $g$ is differentiable at $f(x)$, then $g\circ f$ is differentiable at $x$ and $D(g\circ f)(x)=Dg(f(x))Df(x)$. Indeed write $f(x+h)-f(x)=Ah+r(h)=v(h)$ and $g(f(x)+v)=g(f(x))+Bv+s(v)$, where $r(h)=o(|h|)$, $s(v)=o(|v|)$. Since $v(h)=O(|h|)$, substitution giveswhich proves the assertion, including paths on which $v(h)=0$.

For $g_1$, write $q=x^2-y^2$. Away from the two diagonal lines, $q\ne 0$ and the formula is a composition of smooth functions, hence differentiable. At the origin, $|g_1(x,y)|\le |x^2-y^2|\le x^2+y^2$, so its ratio to the displacement length tends to zero. Its derivative there is therefore zero: this is quadratic damping restores differentiability at the origin.

At a nonzero point $(a,b)$ with $a=\pm b$, necessarily $a\ne 0$. Along the $x$ direction,The denominator inside the sine is locally an invertible function of $t$, with nonzero derivative $2a$. Choose sequences tending to zero for which its reciprocal equals $2\pi n+\pi /2$ and $2\pi n+3\pi /2$. The quotients tend to $2a$ and $-2a$. Thus even this directional derivative fails to exist, and the function is not differentiable there. ConsequentlyFor $g_2$, the formula is smooth away from zero. At zero, $|g_2(x,y)|\le x^2+y^2$, so the same remainder estimate gives zero derivative. HenceNo limit of the sine factor itself is needed for either derivative at the origin.