Past exam of the mathematics course of the University of Cambridge / 2004 / ib / Paper 4

For the given orthonormal basis, the orthogonal projection isIt belongs to $W$, and $⟨v-\pi v,e_j⟩=0$ for every basis vector. Thus $v-\pi v\in W^{\perp }$. For any $w\in W$, decompose $v-w=(v-\pi v)+(\pi v-w)$. The summands are orthogonal, so the Pythagorean theorem in an inner-product space for the inner product givesEquality holds exactly when $w=\pi v$. Hence $\pi v$ is the unique closest point of $W$ to $v$. This proves the minimizing property as well as the formula, including $W=\{0\}$ and $W=V$.

A polynomial in $\mathbb{Z}[x]$ is a primitive polynomial when the greatest common divisor of its coefficients is one. The Gauss lemma for polynomials states that a product of primitive integer polynomials is primitive, and that a primitive positive-degree integer polynomial is reducible over $\mathbb{Q}$ exactly when it factors into positive-degree integer polynomials. More generally, polynomial content is multiplicative up to sign.

The Eisenstein criterion states that a primitive polynomial $a_n{x}^n+\cdots +a_0\in \mathbb{Z}[x]$ is irreducible over $\mathbb{Q}$ if there is a prime $p$ such that $p\nmid a_n$, $p\mid a_j$ for every $j<n$, and $p^2\nmid a_0$. These criteria permit integer coefficient tests for rational irreducibility; the following parts apply them with their exact divisibility hypotheses.

Use the Eisenstein criterion with prime $5$. The leading coefficient is one; every lower coefficient, including the zero coefficients, is divisible by $5$; and the constant coefficient $5$ is not divisible by $25$. The polynomial is monic and hence primitive. Therefore it is irreducible over $\mathbb{Q}$.

A cubic over a field is reducible exactly when it has a root in that field: a nontrivial factorization must contain a degree-one factor. By the rational root theorem, a rational root of this monic integer polynomial must be an integer dividing its constant coefficient, hence either $1$ or $-1$. Substitution gives $-2$ and $4$, respectively. Neither is zero. Thus the cubic is irreducible over $\mathbb{Q}$.

Let $P(x)$ denote the polynomial under consideration. The change $x\mapsto x+1$ is an automorphism of $\mathbb{Q}[x]$ and preserves degrees, so it preserves reducibility. Its transformed polynomial isIt is monic; every other coefficient is divisible by $p$, because $p$ is prime and $1\le j<p$; and its constant coefficient is $p$, not divisible by $p^2$. The Eisenstein criterion proves that $P(x+1)$ is irreducible, and therefore $P$ is irreducible. The argument also includes $p=2$, when the polynomial is linear. This is the prime-index case of an irreducible polynomial arising as a cyclotomic polynomial.

The map is differentiable at $u$ when there exists a linear map $A:{\mathbb{R}}^n\to {\mathbb{R}}^m$ such thatThe increments are taken sufficiently small to stay in the open domain; $A=Df(u)$ is the derivative. This is a joint approximation in all directions, not just the existence of coordinate partial derivatives.

At $(a,b)$ with $ab\ne 0$, choose $\delta <\min (|a|,|b|)$. If $\Vert (h,k)\Vert <\delta$, neither coordinate changes sign, andThus the remainder in the differentiability definition is identically zero throughout that neighborhood. Consequentlywhich proves the requested differentiability directly.

The disc lies in the upper-right half-plane and avoids zero and the negative real axis. On the slit plane take the holomorphic logarithm $\mathrm{Log}z=\log |z|+i\arg z$ with $-\pi <\arg z<\pi$. Its derivative is $1/z$, and $e^{\mathrm{Log}z}=z$. Therefore the holomorphic square rootis analytic and satisfies $f^2=z$. One can also construct it without a global branch theorem: fix either scalar square root $a$ of $3+3i$ and expand $a(1+(z-3-3i)/(3+3i){)}^{1/2}$ by its convergent binomial series. The series converges on the entire disc because $|z-3-3i|/|3+3i|<1/\sqrt{18}$. Either sign gives an admissible choice.

The slit plane is connected and contains $D$. If two holomorphic functions on it extend the same $f$, their difference vanishes on the nonempty open disc $D$. The identity theorem forces their difference to vanish on the entire slit plane. Thus there is at most one analytic extension of the chosen $f$. Choosing the opposite square root initially changes that chosen function; it does not produce two extensions of the same function.

Suppose an analytic extension $F$ existed on the punctured plane. The identity theorem would extend $F^2=z$ from $D$ to that connected domain, so $F$ would never vanish and differentiation would give $2F^{\prime }/F=1/z$. Integrate around the positively oriented unit circle. The argument principle says $\int F^{\prime }/F$ is $2\pi i$ times the integer winding number of the closed curve $F(e^{it})$ about zero, whereasThis would make that winding number $1/2$, a contradiction. Hence no such analytic extension exists. Equivalently, continuation of a square root once around zero reverses its sign, preventing a single-valued holomorphic function on the punctured plane.

Use the Fourier transform conventionThis convention also gives the sign of the inverse transform in 15A. The Parseval identity, in its bilinear form, isIn particular $\int |f{|}^2=(2\pi {)}^{-1}\int |\hat{f}{|}^2$. Initially take $f,g$ to be Schwartz functions; the formula then extends to $L^2$ with Fourier transforms interpreted by completion.

Here is a proof which also justifies the needed limit operations. For $\epsilon >0$, formAbsolute integrability allows Fubini's theorem to replace the transformed factors by their defining integrals. The elementary Gaussian integral givesso $I_{\epsilon }=\int f(x)\overline{(K_{\epsilon }\ast g)(x)}dx$, with convolution $(h\ast g)(x)=\int h(x-y)g(y)dy$. The kernel is nonnegative, has integral one, and its mass outside any fixed neighborhood of zero tends to zero. Since a Schwartz function is bounded and uniformly continuous, splitting the convolution difference into small and large $|x-y|$ proves $K_{\epsilon }\ast g\to g$ uniformly. As $f\in L^1$, the last integral consequently tends to $\int f\overline{g}$.

On the spectral side, repeated integration by parts makes the transforms rapidly decreasing; their product is integrable. The dominated convergence theorem therefore gives $I_{\epsilon }\to (2\pi {)}^{-1}\int \hat{f}\overline{\hat{g}}$. Equating the two limits proves the displayed identity. For general $L^2$ functions, approximate by Schwartz functions. The norm identity makes their transforms Cauchy in $L^2$ with the prescribed factor; the limit defines the $L^2$ Fourier transform independently of the approximation. Taking limits in the bilinear identity proves it for all $L^2$ pairs. Thus the proof includes the hypotheses and avoids unjustified exchanges of nonabsolutely convergent integrals.

Divide the differential equation by $\sqrt{1-x^2}$. Its Sturm-Liouville form isHere $p(x)=\sqrt{1-x^2}$ and weight $w(x)=(1-x^2{)}^{-1/2}$. Multiply the equations for $T_n,T_m$ by the other polynomial and subtract. Integration by parts gives a boundary term $p(T_m{T}_n^{\prime }-T_n{T}_m^{\prime })$, which vanishes at both endpoints because polynomial derivatives are finite and $p\to 0$. HenceWith the standard nonnegative degree indices $n,m$, this proves orthogonality for $n\ne m$. If signed integer indices are used literally, the condition is $|n|\ne |m|$: $T_{-n}=T_n$, so merely $n\ne m$ would not suffice.

Set $x=\cos \theta$ and $Y(\theta )=y(\cos \theta )$. The chain rule gives $(1-x^2)y^{\prime \prime }-x{y}^{\prime }=Y^{\prime \prime }$, so $Y^{\prime \prime }+n^{2}Y=0$. For $n>0$, $Y=A\cos (n\theta )+B\sin (n\theta )$. A polynomial $y$ has $Y^{\prime }(0)=0$, forcing $B=0$. At $n=0$, $Y=A+B\theta$ and the same endpoint condition gives $B=0$. Therefore every polynomial solution is proportional toThe cosine recurrence yields $T_{n+1}=2x{T}_n-T_{n-1}$, $T_0=1,T_1=x$, confirming that these functions are polynomials. In this normalization the Chebyshev polynomial weight identity isIt is simply cosine orthogonality after $x=\cos \theta$.

The four-momentum is $(E/c,p_x,p_y,p_z)$. It transforms by the same Lorentz boost as other four-vectors. For an observer moving at $v$ in the positive $x$ direction, its time and longitudinal components transform asSubstituting the given momentum components proves all three displayed transformation relations.

A photon has zero rest mass, so the relativistic energy-momentum relation gives $E=cp$ on the positive-energy branch. Lorentz invariance preserves $E^2-c^2{p}^2=0$, and $E^{\prime }=\gamma E(1-(v/c)\cos \theta )>0$ for $|v|<c$. Therefore $E^{\prime }=c{p}^{\prime }$. Dividing the transformed longitudinal momentum by $p^{\prime }=E^{\prime }/c$ gives the relativistic aberration formulaAs $v\to c^{-}$, for every fixed $\theta \ne 0$ the ratio tends to $-1$: photon directions are concentrated along the negative $x$ direction in the observer's frame. The exactly forward photon is exceptional: if $\theta =0$, then $\cos {\theta }^{\prime }=1$ for every $v<c$, and its energy tends to zero. For other fixed angles $E^{\prime }$ diverges. The limiting angular behavior is not uniform near $\theta =0$, and $v=c$ itself is not an admissible inertial observer velocity.

Taking the curl of the Euler equation gives the vorticity equation for an inviscid incompressible flow without body forces:The given velocity has divergence $-1+0+1=0$ and curl $(0,0,\omega (t))$. Its vorticity is spatially uniform, so its advective derivative vanishes. The stretching term is $\omega {\partial }_{z}u=(0,0,\omega )$. ThusThe signed vertical component is ${\omega }_0{e}^t$; its magnitude is $|{\omega }_0|e^t$, with the source's magnitude convention taking ${\omega }_0\ge 0$. As an additional consistency check, after imposing this equation the material acceleration is $(x,0,z-1)$, so a pressure $p=-\rho [x^2+(z-1{)}^2]/2+p_0(t)$ satisfies the full Euler equation.

Write $S_{xx}={\sum }_i{x}_i^2$. When $S_{xx}>0$, minimizing ${\sum }_i(Y_i-b{x}_i{)}^2$ gives the strictly convex quadratic normal equation $b{S}_{xx}={\sum }_i{x}_i{Y}_i$. The least-squares estimator is thereforeThis is linear regression through the origin, with no fitted intercept. For $n\ge 2$, let $\mathrm{SSE}={\sum }_i(Y_i-\hat{\beta }{x}_i{)}^2$ and $s^2=\mathrm{SSE}/(n-1)$. An orthonormal change of coordinates in the Gaussian vector $Y$ separates the unit design direction $x/\sqrt{S_{xx}}$ from its $n-1$ orthogonal residual directions. The first coordinate is normal with mean $\beta \sqrt{S_{xx}}$ and variance ${\sigma }^2$; the remaining coordinates are independent centered normals with that variance. Consequentlyand these variables are independent. Thus under the null, the Student t test for regression through the origin statistic isFor a prescribed significance level $\eta$, reject the two-sided null when $|T|>t_{n-1,1-\eta /2}$. The two-sided p-value is $2[1-F_{t_{n-1}}(|T|)]$. There is one fitted coefficient, so the variance degrees of freedom are $n-1$, not $n-2$.

The assumptions $S_{xx}>0$ and $n\ge 2$ matter. If all $x_i=0$, the distribution contains no information about $\beta$ and no informative test is possible. With one nonzero design observation there is no residual variance degree of freedom, so the stated unknown-variance t test cannot be constructed.

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.

For a Hermitian operator $A$, every eigenvalue is real. Indeed, if $Av=\lambda v$ with $v\ne 0$, self-adjointness gives $\lambda ⟨v,v⟩=⟨Av,v⟩=⟨v,Av⟩=\overline{\lambda }⟨v,v⟩$. To prove the finite-dimensional spectral theorem, proceed by induction on the dimension. The fundamental theorem of algebra supplies a root of the characteristic polynomial, hence a unit eigenvector $v$. Its orthogonal complement is invariant: for $w\perp v$, $⟨Aw,v⟩=⟨w,Av⟩=0$. The restriction of $A$ to that complement is again Hermitian. Induction supplies an orthonormal basis of eigenvectors there; adjoining $v$ gives an orthonormal eigenbasis of the whole space. The zero-dimensional case is immediate. Consequently a Hermitian operator is diagonal in an orthonormal basis, with real diagonal entries.

The adjoint reverses products, so for the two Hermitian operatorsThus $\alpha \beta$ is Hermitian if and only if $\alpha \beta =\beta \alpha$. No positivity hypothesis is needed for this equivalence.

Choose an orthonormal basis of eigenvectors of the positive-definite operator $\alpha$ using the proved spectral theorem. Its eigenvalues ${\lambda }_j$ are strictly positive, because $⟨\alpha e_j,e_j⟩={\lambda }_j>0$. Define $\beta e_j=\sqrt{{\lambda }_j}{e}_j$. It is a positive-definite Hermitian operator and ${\beta }^2=\alpha$, establishing existence of the positive square root of an operator.

For uniqueness, let $B$ be any positive-definite Hermitian square root. The identity $B^2=\alpha$ implies $B\alpha =B^3=\alpha B$, so $B$ preserves each eigenspace $E_{\lambda }$ of $\alpha$. Its restriction to that space is Hermitian and positive-definite. Diagonalize that restriction by the same spectral theorem. Every resulting eigenvalue $\mu$ satisfies $\mu >0$ and ${\mu }^2=\lambda$, hence $\mu =\sqrt{\lambda }$. Therefore $B$ acts as $\sqrt{\lambda }I$ on the entire eigenspace, even when $\lambda$ is repeated. These eigenspaces span the space, so the positive-definite Hermitian square root is unique, with eigenvalues $\sqrt{{\lambda }_j}$.

In the polynomial ring $\mathbb{Z}[x]$, the ideal $(x)$ is nonzero and prime: the quotient ring $\mathbb{Z}[x]/(x)$ is isomorphic to the integral domain $\mathbb{Z}$, by evaluation at zero. It is not maximal, since that quotient is not a field. Equivalently $(x)⊊(2,x)⊊\mathbb{Z}[x]$. Thus $(x)$ is a nonzero prime ideal which is not maximal.

Consider the ideal $I=(2,x)$ in $\mathbb{Z}[x]$. It is proper: evaluation at zero followed by reduction modulo $2$ maps the ring onto ${\mathbb{F}}_2$ and has kernel $I$. Suppose $I=(f)$. Then $f$ divides the constant polynomial $2$, so $f$ is a constant integer dividing $2$. Up to sign, the only possibilities are $1$ and $2$. The first would make the ideal the whole ring; the second cannot divide $x$ in $\mathbb{Z}[x]$. Both are impossible. This proves the integer polynomial ring is not a principal ideal domain result: $\mathbb{Z}[x]$ is not a principal ideal domain.

The coefficient identityholds over every field. Both factors have positive degree, including in characteristics $2$ and $3$, because their leading coefficients remain one. Therefore there is no field over which this polynomial is irreducible.

In the quotient ring, the classes of $x-1$ and $x^2+x+1$ are nonzero: neither nonzero polynomial of degree less than three belongs to the ideal generated by $x^3-1$. But their product is the zero class, sinceThe quotient thus contains zero divisors, so it is not an integral domain.

Use the usual convention that a ring homomorphism preserves the identity. Such a map from the quotient ring to $\mathbb{C}$ fixes $\mathbb{Q}$: it fixes integers, and preservation of inverses fixes rational numbers. It is therefore determined by the image $\zeta$ of the class of $x$. The defining relation forces ${\zeta }^3=1$. Conversely, evaluation at any such $\zeta$ annihilates $(x^3-1)$ and hence factors through the quotient. Thus there are exactly three unital homomorphisms, namelyIf the convention permits nonunital ring homomorphisms, add the zero map. These are the only extra possibilities: the image of $1$ is an idempotent in $\mathbb{C}$, hence either zero or one, and image zero forces the whole map to vanish.

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.

The form of Rouché's theorem used here is as follows. Suppose $f$ and $g$ are holomorphic on a neighborhood of the closure of a bounded Jordan domain, and $|g-f|<|f|$ on its positively oriented boundary. Then $f$ and $g$ have the same number of zeros inside, counted with multiplicity.

To prove this form, put $f_t=f+t(g-f)$ for $0\le t\le 1$. The strict inequality prevents $f_t$ from vanishing on the boundary. The argument principle counts its interior zeros by $(2\pi i{)}^{-1}\oint f_t^{\prime }/f_t$. This integral varies continuously with $t$: the integrand is continuous and its denominator stays uniformly away from zero on the compact boundary times $[0,1]$. Its values are integers, so the integral is constant. Its values at $t=0,1$ are the respective zero counts. This proves Rouché's theorem with the multiplicities and boundary hypotheses required below.

On $|z|=3$, the perturbation has modulus $|icz+8|\le 3c+8<11<27=|z^3|$. Thus Rouché's theorem gives three roots inside $|z|<3$, counted with multiplicity.

For the quadrant information, compare instead with $P_0(z)=z^3+8$, whose roots are $r_1=-2$, $r_2=1+i\sqrt{3}$, $r_3=1-i\sqrt{3}$. Their pairwise distances are $2\sqrt{3}$. On the circle $|z-r_j|=1$,The three unit discs are disjoint, and Rouché localization of polynomial roots gives one root in each, counted with multiplicity. The disc around $1+i\sqrt{3}$ lies strictly in the first quadrant in its interior; that around $1-i\sqrt{3}$ lies in the fourth quadrant; and that around $-2$ lies in $\mathrm{Re}z<-1$. Consequently there is exactly one root in each of those two right-hand quadrants, and exactly one root in the left half-plane. The same count also shows that each root is simple.

At $c=0$ the left-hand root is $-2$, and the derivative of the polynomial with respect to $z$ is $3(-2{)}^2=12\ne 0$. The holomorphic implicit function theorem states that if $P(z,c)$ is holomorphic near $(z_0,c_0)$, $P(z_0,c_0)=0$ and $P_z(z_0,c_0)\ne 0$, there is a unique holomorphic function $z=\gamma (c)$ near $c_0$, with $\gamma (c_0)=z_0$ and $P(\gamma (c),c)=0$. It therefore supplies an analytic root branch $\gamma (c)$ near zero. Differentiate its equation:It follows that ${\gamma }^{\prime }(0)=i/6$ andFor sufficiently small positive $c$ its imaginary part is $c/6+O(c^2)>0$. By the unique-root localization from the preceding part, this is the root in the left half-plane. Thus that root lies above the real axis for all sufficiently small positive $c$.

With the Fourier transform convention of 5A, use the printed Fourier inversion formula directly. The possible choice of values at $s=\pm 1$ has no effect on the integral. Integration over the support givesCombining the two fractions yieldsIn particular $g$ is imaginary-valued and odd, as expected from the real odd transformed function with this inverse-transform sign.

Take the Fourier transform in $x$ of the Laplace equation. Writing $U(s,y)$ for the transformed solution gives $U_{yy}-s^{2}U=0$. Its boundary values givewith the multiplier at $s=0$ understood as $1-y$. The particular data simplify this to $U(s,y)=2\sinh (s(1-y))$ for $|s|<1$, and zero outside. Thus, with $a=1-y$, the Fourier solution of the strip Dirichlet problem isThe displayed closed expression applies in the open strip. The integral defines its continuous boundary values, including the removable point $(x,a)=(0,0)$. It gives $u(x,0)=g(x)$ and $u(x,1)=0$. Differentiation under this finite integral verifies $u_{xx}+u_{yy}=0$ directly, since each integrand has opposite second derivatives in $x$ and $y$.

The boundary conditions alone do not specify a unique solution without a growth restriction: adding $C{e}^{\pi x}\sin (\pi y)$ preserves both boundary values and harmonicity. The constructed solution is bounded, and it is unique among bounded solutions continuous on the closed strip. To see this, apply the maximum principle for harmonic functions to a bounded real or imaginary component $w$ of the difference. For $0<\delta <\pi$, the positive harmonic function $B(x,y)=\cosh (\delta x)\cos (\delta (y-1/2))$ has a positive minimum factor in $y\in [0,1]$. For any $\epsilon >0$, the functions $\pm w-\epsilon B$ are nonpositive on the horizontal sides of a sufficiently wide rectangle and on its vertical sides, because $w$ is bounded and $B$ grows there. The maximum principle gives $|w|\le \epsilon B$ at every fixed point. Letting $\epsilon \downarrow 0$ proves uniqueness in this natural Fourier solution class.

For the radial Helmholtz Dirichlet Green function, set $H(x,\xi )=xG(x,\xi )$. Multiplication of the radial equation by $x$ givesThe first endpoint follows from the required boundedness of $G$ at zero. For $k\ne 0$ and $\sin k\ne 0$, use the left solution $\sin kx$ and right solution $\sin k(1-x)$. Their Wronskian is $-k\sin k$. Matching continuity and the derivative jump therefore gives, with $x_{<}=\min (x,\xi )$ and $x_{>}=\max (x,\xi )$,This is finite at zero, vanishes at one, is continuous at $x=\xi$, and has $G_x(\xi +,\xi )-G_x(\xi -,\xi )=1$. Thus it has exactly the prescribed delta source, not its negative. Its weighted reciprocity is $x^{2}G(x,\xi )={\xi }^{2}G(\xi ,x)$, as expected for the radial weight.

Integrating the Green function against the constant forcing givesThe integrals are respectively $-x\cos (kx)/k+\sin (kx)/k^2$ and $(1-x\cos (k(1-x)))/k-\sin (k(1-x))/k^2$. After cancellation their weighted sum is $(\sin (kx)-x\sin k)/k$. HenceThe values at zero are interpreted by continuity. At $k=0$ the limit exists: $G(x,\xi )=-\xi x_{<}(1-x_{>})/x$ and $y(x)=(x^2-1)/6$.

The exceptional nonzero values $k\in \pi \mathbb{Z}$ need a genuine qualification. There is then a nonzero bounded homogeneous solution $\sin (kx)/x$ satisfying both endpoint conditions, so an inverse Green function does not exist. Moreover the stated forcing is incompatible, not merely nonunique. If $h=xy$ solved the problem, multiply $h^{\prime \prime }+k^{2}h=x$ by $\sin (kx)$ and integrate by parts using $h(0)=h(1)=0$. The left side is zero, whereasThus there is no solution for nonzero $k\in \pi \mathbb{Z}$. This explains exactly when the generic formula is applicable.

With metric signature $(+,-,-,-)$, four-momentum is $p^{\mu }=(E/c,\mathbf{p})$; for a particle of rest mass $m$ it satisfies $p^{\mu }{p}_{\mu }=m^2{c}^2$. For a massive particle $p^{\mu }=m{u}^{\mu }$, where $u^{\mu }$ is its four-velocity. The total four-momentum is the sum of the individual four-momenta. If that sum is timelike, its center-of-momentum frame exists and the total energy in that frame isThis follows because the spatial momentum is zero in that frame and the Minkowski norm is invariant under a Lorentz boost. A null total momentum, such as parallel photons, would not have such a rest frame.

For the photon striking an electron initially at rest, the initial invariant gives $W^2=m_e^2{c}^4+2m_e{c}^2{E}_{\gamma }$. The three final particles have total rest energy $3m_e{c}^2$. Their center-of-momentum energy is at least this value, with equality when all have zero relative momentum. Energy-momentum conservation therefore requiresThe photon-induced triplet pair-production threshold is consequently four electron rest-energy units. This is attainable as a kinematic threshold: at equality the center-of-momentum speed in the laboratory is $4c/5$, and boosting the three resting final particles gives each energy $5m_e{c}^2/3$ and momentum $4m_{e}c/3$. Their totals are the initial $5m_e{c}^2$ and $4m_{e}c$. Recoil of the original electron is why twice the rest energy is insufficient.

For steady incompressible flow of constant density without a body force, the Euler equation and mass conservation giveApply the divergence theorem componentwise to a fluid volume. With $\mathbf{n}$ its outward normal, the integral momentum flux balance isIf a solid bounds part of the volume, retain its pressure contribution; that contribution supplies the force on the fluid. This derivation includes pressure as well as transported momentum.

Choose the axis in the incoming jet direction. The printed diagram measures $\alpha$ between that axis and the downstream sheet: it is the cone's half-angle. Along a streamline, the steady Euler equation integrates to the Bernoulli equation $p/\rho +|\mathbf{u}{|}^2/2=\text{constant}$. At the incoming free jet and the outgoing free sheet the pressure is atmospheric. With gravity neglected, the outgoing speed is therefore $u$. This argument uses far free surfaces; it does not assert atmospheric pressure on the wetted sphere.

At radial distance $r$, the sheet's circumference is $2\pi r\sin \alpha$. Its cross-section normal to its velocity is this circumference times its small thickness $d$. Conservation of mass gives $Au=(2\pi r\sin \alpha )du$, henceFor the axial momentum balance, transverse contributions cancel by axial symmetry. The outgoing axial momentum rate is $\rho A{u}^2\cos \alpha$ and the incoming rate is $\rho A{u}^2$. Uniform atmospheric pressure cancels on the closed control surface when expressed as gauge pressure. The sphere's force on the fluid is therefore $\rho A{u}^2(\cos \alpha -1)$ in the axis direction. By action and reaction, the fluid force on the sphere is downstream, with magnitudeThe downstream direction in the original diagram fixes this sign and the factor $1-\cos \alpha$. If one calls the full cone opening its vertex angle, that full opening is $2\alpha$; using $\alpha$ as the full opening would conflict with the printed thickness formula.

A Bayes estimator minimizes the integrated loss function under the prior and sampling distribution. For a nonnegative loss, Tonelli's theorem and the posterior distribution give the risk decompositionwhere $m$ is the marginal density. Thus choosing, for each observed $x$, a measurable minimizer of the posterior expected loss minimizes the integrated risk. Observations of marginal probability zero do not affect the rule.