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A procedure associated with Pierre de Fermat for extrema and tangents. Compare expressions at $a$ and $a+e$, remove common terms, divide by a remaining power of $e$, and suppress terms still containing $e$. Its interpretation must be distinguished from ordinary equality at finite $e$ and from a modern limit definition.^[ref]

A submerged object looks shallower because its emerging light rays bend away from the surface normal vector. Their backward extensions give an approximate virtual image nearer the surface. For an object at perpendicular depth $d$ in a medium of refractive index $n_{\mathrm{obj}}$, viewed through a flat boundary from a medium of refractive index $n_{\mathrm{obs}}$, the near-normal paraxial approximation gives $d_{\mathrm{par}}=(n_{\mathrm{obs}}/n_{\mathrm{obj}})d$. Thus the model water depth $12\mathrm{cm}$ appears approximately $9\mathrm{cm}$ from air. This is a bundle-of-light rays statement, not a deflection of the axial light ray.^[ref]

For general transmitted light rays, specify what “depth” means. Put the object at $(0,-d)$ below a horizontal boundary $z=0$. A light ray reaches the boundary at $(x,0)$, with object-side angle ${\theta }_{\mathrm{obj}}$ and observer-side angle ${\theta }_{\mathrm{obs}}$, both measured from the upward normal vector. For $x>0$, its backward extension meets the object's normal line $x=0$ at $z=-d_{\mathrm{axis}}$, whereThis axis-intercept depth is exact for that one transmitted light ray in the stated model. It tends to $d_{\mathrm{par}}$ as the angles tend to zero; at exactly zero the quotient is interpreted as a limit. These elementary tangent, sine and cosine relations show precisely where the small-angle approximation enters.

The original calculations below use $d=12\mathrm{cm}$, $n_{\mathrm{obj}}=4/3$ and $n_{\mathrm{obs}}=1$:

At appreciable obliquity these depths differ, so the full outgoing bundle has no single exact virtual image on the object's normal line. An oblique eye samples a narrow bundle; its local apparent position also depends on that bundle and cannot be inferred by declaring $d_{\mathrm{axis}}$ to be a universal image depth. A plane refracting boundary gives approximate image formation near normal incidence, with departures at larger angles. The construction assumes a flat surface, uniform transparent media with isotropy, and a propagating transmitted light ray; it does not extend to total internal reflection.

The map $C(z)={\tan }^2(\pi \sqrt{z}/4)$ is a conformal map from $\{x+iy:y^2<4(1-x)\}$ onto the unit disc. A principal square root first maps the slit parabolic region to a vertical strip, and a scaled tangent followed by squaring maps the strip to a slit disc. Evenness in the square-root variable gives a holomorphic extension across the removed axis, filling the target slit.

The angle-addition identities give $\sin 2x=2\sin x\cos x$, $\cos 2x={\cos }^{2}x-{\sin }^{2}x=1-2{\sin }^{2}x=2{\cos }^{2}x-1$, and $\tan 2x=2\tan x/(1-{\tan }^{2}x)$ wherever the tangent expressions are defined. The sine formula converts ${\sin }^{2}x{\cos }^{2}x$ into ${\sin }^2(2x)/4$, which is useful for trigonometric inequalities and integration.

The inverse of the tangent restricted to $(-\pi /2,\pi /2)$, taking real values in that interval. Its derivative is $(1+x^2{)}^{-1}$. A formula involving an inverse tangent of a ratio may need a continuous branch of a multivalued function to describe a smooth field through a pole of that ratio.

A conformal map preserves oriented angles between regular smooth curves through each point. For a holomorphic map with nonzero derivative, the derivative also describes the local scale and rotation.

Indeed, if two curves ${\gamma }_1,{\gamma }_2$ meet at $z$ and have nonzero tangent vectors $v_1,v_2$, the real chain rule and complex differentiability giveBecause $f^{\prime }(z)\ne 0$, the image tangents are nonzero andTaking arguments proves preservation of oriented angles. Multiplication by $f^{\prime }(z)$ rotates every tangent by $\arg f^{\prime }(z)$ and scales every length by $|f^{\prime }(z)|$. Thus the hypothesis gives conformality at every point of $U$; global injectivity is not needed for this local conclusion.

For a nonsingular curve $y^2=x^3+Ax+B$ with integers $A,B$, the Nagell–Lutz theorem states that a finite rational torsion point has integer coordinates and satisfies either $y=0$ orThe point at infinity is the identity and is treated separately. Integrality is the result allowed in the source. To prove the remaining divisibility, suppose $y\ne 0$. Then the doubled point is finite and also torsion, so its coordinates are integral. Its tangent slope is $m=(3x^2+A)/(2y)$ and its abscissa is $m^2-2x$. Thus $m^2$ is an integer. A rational number whose square is an integer is itself an integer, so $2y$ divides $3x^2+A$. In particular $y^2$ divides ${\psi }_1(x)=(3x^2+A{)}^2$. It also divides ${\psi }_2(x)=x^3+Ax+B=y^2$. Evaluating the permitted polynomial identity at the integer $x$ therefore proves the divisibility of $4A^3+27B^2$ by $y^2$.

For $y=0$, the point equals its negative and has order two. The theorem gives a finite set of torsion candidates, but the divisibility condition by itself does not assert that every candidate is torsion.

Nonsingularity with $b=a^2$ forces $a\ne 0$. TakeIt lies on the curve. Its tangent slope is $a/(2a)=1/2$, so the doubling formula gives $x(2P)=1/4$. If $P$ were torsion, $2P$ would be torsion as well, contradicting the integrality just proved. Therefore $P$ has infinite order.

For $y^2=x^3+x+2$, the point $P=(1,2)$ doubles to $(-1,0)$: its tangent slope is $(3+1)/(2\cdot 2)=1$, giving $x(2P)=-1$ and $y(2P)=0$. Thus $P$ has order four.

The curve has good reduction at $3$, since $4+27\cdot 4=112$ is nonzero modulo three. Its reduced points are the point at infinity, two points above $x=1$, and one above $x=2$, so there are four in total. Injectivity of torsion reduction bounds the rational torsion order by four. The known four points exhaust it:

Eliminating the phase velocity from the first two relations yieldsWith $\theta =mh$, putThen $\ell =(c_1/(c_{2}h))\sqrt{a^2{k}^2-{\theta }^2}$. The dispersion relation becomesHere $a>0$ follows from $c_1<c_2$. The parameter $a$ has dimensions of length; $A$ and $ϵ$ are dimensionless.

For the strict mode count for a rigid-backed elastic layer, considerThe right side is negative, so intersections occur only on the negative branches of the tangent,On each such branch $\tan \theta -g(\theta )$ is strictly increasing. If $p_n<A$, its limit at $p_n^{+}$ is negative infinity. At the other end of the available interval, it is positive: either $\theta \to n{\pi }^{-}$ with $n\pi <A$, or $\theta \to A^{-}$ with $A\le n\pi$. The intermediate value theorem and strict monotonicity give exactly one intersection. No such interval exists when $p_n\ge A$.

Thus the exact answer, including all cutoff values, isAway from $A=(n-1/2)\pi$, this equals the printed floor function expression $\lfloor A/\pi +1/2\rfloor$. At $A=(n-1/2)\pi$, the printed expression is one too large: the $n$th interval is empty and the limiting state has $\ell =0$, contradicting exponential decay. This exception is present in the native paper, rather than introduced by transcription.

Now hold $A$ fixed and let $ϵ\to 0^{+}$. Neglecting the right side gives ${\theta }_n\simeq n\pi$ and hence the candidate phase velocitiesThe zero $\theta =0$ is excluded: it would require $c=c_1$, already ruled out. For every $n\pi <A$, the candidate is a valid leading asymptotic expansion. A first correction, obtained by expanding the tangent about $n\pi$, isThe big O notation here assumes a fixed positive gap between $n\pi$ and $A$. In particular, when $A>N\pi$, all $N$ leading approximations are valid. The phase velocity increases strictly with $\theta$, so only the last branch can cause an endpoint difficulty.

If $(N-1/2)\pi <A<N\pi$, the formal last equation root $N\pi$ lies outside the allowed interval. The actual last equation root approaches $A$, where the small factor $ϵ$ is balanced by the vanishing denominator. This is a singular perturbation. For fixed $A$ strictly inside that interval, write ${\theta }_N=A-\delta$. Since $\tan A<0$,The valid endpoint approximation for a rigid-backed shear mode consequently givesHere we used $c^2=c_1^2(1+{\theta }^2/(kh{)}^2)$ and $A^2/(kh{)}^2=c_2^2/c_1^2-1$. Thus this normal mode approaches the upper wave speed from below; it does not approach the inadmissible value obtained from $\theta =N\pi >A$.

There is a further exceptional fixed value $A=N\pi$. Put ${\theta }_N=A-\delta$ again. Now $\tan (A-\delta )\sim -\delta$, so the balance changes toIn the last formula the bracket specifies the first nonzero correction. The zeroth-order candidate $c_N=c_2$ is correct as a limit but misses this correction and the positive exterior exponential decay rate. More generally, close to $A=N\pi$, let $\Delta =A-N\pi$ and ${\theta }_N=N\pi -\delta$. For $|\Delta |,\delta \ll 1$, a transition approximation valid on both sides isIt recovers the ${ϵ}^{2/3}$ balance at $\Delta =0$ and the ordinary expansion above the transition. The fixed-endpoint formula involving $\tan A$ should be used when $A$ is below, and bounded away from, $N\pi$.

The figure illustrates Pierre de Fermat's use of adequality to construct a tangent to a parabola. Reading the local exam image is decisive: $D$ is the vertex of a parabola, $DC$ lies on its axis of symmetry, $B$ lies on the parabola, and the proposed tangent $BE$ meets the extended axis at $E$. The horizontal through the auxiliary point $O$ meets the axis at $I$; $O$ lies on $BE$, outside the parabola. In particular, $O$ must not be treated as a second point of the parabola. The target is the subtangent $CE$. The diagram and this construction occur in the tangent discussion associated with Varia opera mathematica by Pierre de Fermat. The 1679 book was posthumous: its date does not date the invention of the method.^[ref][ref]

Write $d=CD>0$, $a=CE>0$, and $e=CI$, with $I$ between $C$ and $D$. These are lengths, and the increment $e$ is distinct from the labelled point $E$. Then $DI=d-e$ and $EI=a-e$. The parabola's characteristic relation makes the square of an ordinate proportional to its distance from the vertex of a parabola. At height $I$, the actual ordinate of the parabola is shorter than $OI$. Hencewhere the equality follows from similar triangles $BCE$ and $OIE$.

The method of adequality replaces the comparison by a relation used to extract the condition of contact:Expanding and cancelling $d{a}^2$ gives $-2dae+d{e}^2\sim -a^{2}e$. Divide by the nonzero increment $e$, then suppress the remaining term containing $e$. The result is $a^2=2da$, so, since $a>0$,Thus $DE=DC$: the intersection with the axis lies as far above the vertex of a parabola as $C$ lies below it. This is a concrete construction, not just a rule for computing a numerical slope.

A modern check clarifies why the calculation succeeds without pretending that Pierre de Fermat used a modern limit. Put $D=(0,0)$, let $y$ increase down the axis, and write the parabola as $x^2=ky$, $k>0$. For $B=(b,d)$, with $b^2=kd$, the proposed tangent through $E=(0,-d)$ is $y=2bx/k-d$. Substitution givesIt meets the parabola with a double root at $B$, and for every other point its $y$ coordinate is less than that of the parabola. This also verifies the strict comparison used for $O$.

The significance is a systematic algebraic condition for geometric contact. Subtracting neighbouring expressions, dividing by an increment, and discarding its remaining terms anticipates differentiation. But the historical question is precisely what licenses the last step. The symbol $\sim$ above marks adequality, not ordinary equality at a finite displacement. Treating the construction as an already stated epsilon-delta definition of a limit would conceal the methodological innovation and its unresolved explanatory burden.

The substitution $u=x+y$ turns the equation into $u^{\prime }=1+u^2$. Separating and integrating gives $\arctan u=x+C$, henceEach real solution is defined on an interval containing no pole of the tangent. There is no real constant-$u$ solution discarded by division, since $1+u^2>0$.

Let $D=y(3y-2)$. Implicit differentiation of the two constraints gives the linear systemAdding them, then substituting back, gives at regular pointsThe relevant Jacobian determinant is $2(1+a)yz(3y-2)$. Thus the special parameter value is $\boxed{a=-1}$, assuming the other factors are nonzero. At that parameter, adding the original constraints gives $(x+1{)}^2=0$: the intersection curve lies wholly in the plane $x=-1$ and cannot be parametrized by $x$ on an interval. Calling its derivatives “infinite” describes a vertical projection, not a finite-valued differentiable pair $y(x),z(x)$ at $a=-1$; the displayed ratios then have simultaneous zero numerators on the intersection and are undefined.

There are also point-dependent failures when $y=0$, $y=2/3$ or $z=0$. These can produce vertical tangents or require a different local parameter, and simultaneous numerator zeros need separate analysis. The implicit function theorem guarantees finite $y^{\prime },z^{\prime }$ only where the stated Jacobian determinant is nonzero. This distinction is the geometry of a singular projection of an implicit curve.

The Schwarz-Pick lemma says that every holomorphic self-map $F$ of the unit disc contracts pseudohyperbolic distance and therefore hyperbolic distance:Its infinitesimal form isEquality for distinct points in the distance inequality, or at one point in the derivative inequality, holds exactly when $F$ is a disc Möbius transformation.

For completeness, first prove the Schwarz lemma. If $H:\mathbb{D}\to \mathbb{D}$ is holomorphic and $H(0)=0$, the quotient $H(z)/z$ extends holomorphically at zero. On $|z|=r<1$ its modulus is at most $1/r$. The maximum modulus principle, followed by $r\uparrow 1$, gives $|H(z)|\le |z|$ and $|H^{\prime }(0)|\le 1$. Equality at a nonzero point or in the derivative forces that quotient to be a constant of modulus one, again by the maximum modulus principle; hence $H(z)=e^{i\theta }z$.

Now let ${\varphi }_a(z)=(z-a)/(1-\overline{a}z)$ and apply this result to $H={\varphi }_{F(w)}\circ F\circ {\varphi }_w^{-1}$, which fixes zero. The modulus inequality is exactly the displayed distance inequality, and $\rho =\log ((1+\delta )/(1-\delta ))$ is increasing in $\delta$, giving hyperbolic contraction. Also $|{\varphi }_a^{\prime }(a)|=(1-|a{|}^2{)}^{-1}$ and $|({\varphi }_w^{-1}{)}^{\prime }(0)|=1-|w{|}^2$, so the chain rule proves the derivative inequality. The equality statement follows from the rotation case of the Schwarz lemma after undoing the two automorphisms. Conversely a disc automorphism preserves the distance and attains equality.

There is a maximum of the derivative modulus in the given annulus family, which is nonempty because it contains the constant function one. Indeed the functions are bounded by $e$, and Cauchy estimates bound their derivatives at zero. Take a sequence whose derivative moduli approach the finite supremum. By Montel theorem it has a subsequence converging locally uniformly to a holomorphic function $f$, with $f(0)=1$ and $e^{-1}\le |f|\le e$. The upper equality cannot occur at an interior point, because the maximum modulus principle would force a constant of modulus $e$, contrary to $f(0)=1$. The lower equality is excluded by applying the same argument to $1/f$, which is holomorphic because $|f|\ge e^{-1}$. Thus the limit still maps into the open annulus. Locally uniform convergence gives convergence of derivatives by the Cauchy integral formula, so this limit attains the supremum.

To compute it and all extremizers, the nonvanishing function $f$ on the simply connected disc has a holomorphic logarithm $h$ with $h(0)=0$. One can construct it by integrating $f^{\prime }/f$ from zero; differentiation shows $e^h/f$ is constant, hence one. Its real part obeys $-1<\mathrm{Re}h<1$. The mapis a biholomorphism from this vertical strip onto the disc. To check the inverse explicitly, for $|z|<1$ the ratio $(1+iz)/(1-iz)$ has positive real part, so its logarithm with value zero at zero has imaginary part in $(-\pi /2,\pi /2)$. Thereforehas real part between minus one and one, and direct substitution gives $T(s(z))=z$. The strip restriction removes the period ambiguity of the tangent, so these are inverse maps.

Applying the Schwarz lemma to $T\circ h$ gives $(\pi /4)|h^{\prime }(0)|\le 1$. Since $f^{\prime }(0)=h^{\prime }(0)$, the exact maximum isIt is attained by $f(z)=\exp ((4/\pi )\arctan z)$. The equality case requires $T(h(z))=e^{i\theta }z$, and hence all maximizing functions areTheir derivatives are $(4/\pi )e^{i\theta }$, so the extremizer is not unique: there is exactly this circle of distinct functions, with the parameter considered modulo $2\pi$. This is the extremal derivative of a disc map into a symmetric annulus for $L=1$.

It helps to separate the tangent and the squaring map. Put $t=\tan (z/2)$, with $z=x+iy$. The identityshows that $0<x<\pi /2$ gives $\mathrm{Re}t>0$ and $|t|<1$. To see that every point of this right half-disc occurs, setThen $z=-i\mathrm{Log}R$ has $0<\mathrm{Re}z<\pi /2$ and satisfies $\tan (z/2)=t$. The principal complex logarithm is unambiguous because $R$ is in the first quadrant.

Squaring maps this right half-disc bijectively onto the unit disc with the interval $(-1,0]$ removed: the argument of $t$ lies between $-\pi /2$ and $\pi /2$, and doubles into $(-\pi ,\pi )$. Thus the desired image isNeither derivative vanishes in the domain, so the composition is also a conformal map onto that slit disc.

The linear dispersion relation for a mode $e^{i(kx-\omega t)}$ is $5{\omega }^2=k^4+4$. The proposed modes share phase velocity $c$, so $5c^2{k}_j^2=k_j^4+4$. A quadratic wave interaction generates the second harmonic and the difference harmonic. With only two positive wavenumbers, phase matching for a quadratic wave interaction requires $k_2=2k_1$. Equating their phase velocities givesand thereforeThis is a two-to-one resonance of dispersive waves. Its amplitude changes accumulate on $t=O({ϵ}^{-1})$. Choose the slow time $T=ϵt/5$; this harmless constant rescaling makes the later amplitude formulas simple. With $\xi =x-t$, the leading real field is $A_0{e}^{i\xi }+B_0{e}^{2i\xi }+{\overline{A}}_0{e}^{-i\xi }+{\overline{B}}_0{e}^{-2i\xi }$.

For the fundamental, the order-$ϵ$ cross derivative in $5{\psi }_{tt}$ is $-2i(A_0{)}_T{e}^{i\xi }$, while the fundamental coefficient in ${\psi }_0{\psi }_{0x}$ is $i{\overline{A}}_0{B}_0$. For the second harmonic the corresponding terms are $-4i(B_0{)}_T{e}^{2i\xi }$ and $i{A}_0^2$. The solvability condition in the method of multiple scales removes the resonant forcing and givesThe nonresonant third and fourth harmonics enter the correction. They do not change these leading amplitude equations.

Write $A_0=R{e}^{i\theta }$, $B_0=P{e}^{i\varphi }$ and $\delta =\varphi -2\theta$. Taking real and imaginary parts gives the explosive two-to-one amplitude system in polar form:The polar phases are used where their corresponding amplitudes are nonzero. These equations implyFor the second identity, differentiate using ${\delta }_T=(R^2/(4P)+P)\sin \delta$: the terms proportional to $\sin \delta \cos \delta$ cancel. Thereforeare first integrals, and in particular

For constant phases with $\varphi =2\theta +\pi$, these real equations reduce to $R_T=RP/2$ and $P_T=R^2/4$. The initial data give $R^2-2P^2=8$, so the Riccati equation for $P$ is $2P_T=P^2+4$. Integrating and using $P(0)=2$ givesThe tangent and secant function have a pole at $T=\pi /4$, corresponding to $t=5\pi /(4ϵ)$. Both modes grow through their locked resonant interaction. This is formal finite-time blowup of the reduced amplitude equations; the weakly nonlinear expansion loses validity as the amplitudes become large, so it cannot establish a singularity of the full partial differential equation.

Source: released examination PDF, bundle pages 230–233.

Solutions to all seven questions, including every optional question and Section C. In Question 6(b), the printed transcendental equation excludes valid separated solutions at common poles of the tangents; the complete characteristic equation and an explicit counterexample are given below.

For $\omega \ne 0$, substitution of the separated solutions into the heat equations givesThe zero exterior boundary values are incorporated byTo obtain a eigenvalue equation for heat conduction in two joined slabs, write $u=\omega {\alpha }_{1}L$ and $v=\omega {\alpha }_{2}L$. Continuity of temperature and heat flux at zero gives the homogeneous linear systemA nonzero amplitude pair exists precisely when its determinant vanishes, equivalentlyIf $\cos u\cos v\ne 0$, division by $k_1{k}_2{\alpha }_1{\alpha }_2\cos u\cos v$ gives the printed transcendental equationThe printed necessity claim needs the qualification that both tangents are defined. If just one of $\cos u,\cos v$ vanishes, $D(\omega )\ne 0$. If both cosines vanish, however, $D(\omega )=0$ and legitimate normal modes remain. Their temperature at the thermal interface need not be zero; the temperature condition fixes the amplitude ratio, while each heat flux is zero there.

For a concrete counterexample, take identical slabs: ${\rho }_1={\rho }_2=\rho$, $c_1=c_2=c$, $k_1=k_2=k$, and $\alpha =\sqrt{\rho c/k}$. Thenwith each formula restricted to its own slab, satisfies both heat equations and all four boundary conditions. But $\omega \alpha L=\pi /2$, so neither tangent is defined. This directly disproves the unqualified printed assertion. The determinant condition retains these missing normal modes.

Finally, $\omega =0$ must be checked before dividing by it. Then $G_1=A(x+L)$, $G_2=B(L-x)$, and interface continuity gives $A=B$, while heat flux continuity gives $k_{1}A=-k_{2}B$. Positivity of the thermal conductivities forces $A=B=0$. Thus $\boxed{\omega =0\text{ gives only the trivial solution}.}$ In the determinant formula, its apparent zero at $\omega =0$ is spurious because the chosen sine basis degenerates there.

The Bolzano-Weierstrass theorem states that every bounded sequence of real numbers has a convergent subsequence. It is enough to select infinitely many terms of the tangent sequence that stay uniformly bounded.

For each positive integer $k$, choose $n_k=\lfloor k\pi +1/2\rfloor$, using the floor function. Then $|n_k-k\pi |\le 1/2$. Moreover $n_{k+1}-n_k\ge \pi -1>0$, so these are strictly increasing indices. By the period and monotonicity of the tangent on $(-\pi /2,\pi /2)$,Thus $(\tan n_k)$ is a bounded subsequence. Applying the Bolzano-Weierstrass theorem to it gives a convergent subsequence, which is also a subsequence of the original sequence. Such a convergent subsequence therefore exists.

The tangent addition formula givesHence $\boxed{f(v)=\tan v}$. The restriction $xy\ne 1$ excludes the poles of tangent that can arise in the possible range $-\pi <v<\pi$. One should not replace this identity by $v=\arctan u$ globally: the principal inverse tangent can differ from $v$ by an integer multiple of $\pi$.