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Let $L(f)=\mathcal{T}[f]-f^{\prime }(1/3)$. Direct substitution shows $L(1)=L(x)=L(x^2)=0$, so the functional annihilates all quadratic polynomials. Taylor's integral-remainder formula isApplying $L$ and interchanging the finite evaluations and derivative with the continuous integral gives the Peano kernel representation $L(f)={\int }_0^{1}K(t)f^{\prime \prime \prime }(t)dt$, whereAll three expressions are nonnegative; the middle quadratic has positive leading coefficient and discriminant $-1/9$. Therefore the best possible norm bound has constant ${\int }_0^{1}K(t)dt$. Evaluate that integral particularly simply by using $f(x)=x^3/6$, for which $f^{\prime \prime \prime }=1$:It follows thatThe same cubic attains equality, so the constant is sharp, not merely sufficient. This is the sharp Peano bound for an interior three-point first derivative.

Let the fluid be at rest at infinity, and write its velocity potential as $\varphi =f(r)\cos \theta$, the angular dependence selected by translation along the polar axis. The separated Laplace equation iswith radial solutions $r$ and $r^{-2}$. Decay at infinity eliminates the former. The no-penetration condition on the moving sphere is $u_r(a)=U\cos \theta$, giving $f(r)=-U{a}^3/(2r^2)$. Thus the potential flow around a translating sphere isFor density $\rho$, integrate its kinetic energy over the exterior. The angular factor satisfies $\int ({\cos }^2\theta +{\sin }^2\theta /4)d\Omega =2\pi$, and ${\int }_a^{\infty }{r}^{-4}dr=1/(3a^3)$. HenceThe added mass of a sphere is consequently $M_f/2$. For a heavy sphere, $M>M_f$, falling through distance $h$, the net gravitational work after buoyancy is $(M-M_f)gh$. Its own kinetic energy plus that of the fluid is $(M+M_f/2)U^2/2$. Conservation of energy from rest therefore givesThis is the unbounded inviscid-fluid model; walls or viscous dissipation would alter the flow and its energy.

The scattering interpretation assumes $0<ϵ<U$, so $k=\sqrt{ϵ}$ is real and $K=\sqrt{U-ϵ}>0$. The incoming amplitudes are $a$ from the left and $b$ from the right; the outgoing amplitudes are $c,d$. The conserved probability current iswhich gives $j=\hslash k(|a{|}^2-|c{|}^2)/m$ on the left and $j=\hslash k(|d{|}^2-|b{|}^2)/m$ on the right. Equality proves $|c{|}^2+|d{|}^2=|a{|}^2+|b{|}^2$. Equal exterior wavenumbers make the flux weights equal.

Continuity of the wavefunction and its derivative at the finite potential steps yields the four relations, with $C=\cosh KL$ and $S=\sinh KL$,Set $\lambda =K/(ik)$. Substituting $e=a+c$, $f=(a-c)/\lambda$ and subtracting the last two relations after dividing the fourth by $ik$ giveswhere $D=C-(\lambda +{\lambda }^{-1})S/2$. HenceReflection about the barrier midpoint interchanges the left and right incidence problems and, with the chosen phase origins, exchanges $a\leftrightarrow b$ and $c\leftrightarrow d$. ThereforeFor incidence from the left alone, the transmission coefficient is the transmitted-to-incident probability flux ratio, $T=|d/a{|}^2=|D{|}^{-2}$. Since $\lambda =-iK/k$, direct calculation givesFor fixed energy inside the barrier range and $KL\gg 1$, ${\sinh }^{2}KL\sim e^{2KL}/4$, soThis is the thick-barrier quantum tunnelling limit, not a uniform approximation as energy approaches the barrier top while $KL$ remains small.

The dual space is $V^{\ast }={\mathrm{Hom}}_{\mathbb{R}}(V,\mathbb{R})$, the vector space of real-valued linear maps on $V$, with pointwise addition and scalar multiplication. Choose a basis $e_1,\dots ,e_n$ of $V$. Every $v$ has unique coordinates $v={\sum }_i{x}_i{e}_i$, so $e^i(v)=x_i$ defines a linear functional, with $e^i(e_j)={\delta }_{ij}$.

If ${\sum }_i{a}_i{e}^i=0$, evaluation on $e_j$ gives $a_j=0$, proving linear independence. For any $\ell \in V^{\ast }$, linearity gives $\ell (v)={\sum }_i{x}_i\ell (e_i)$, hence $\ell ={\sum }_i\ell (e_i)e^i$. Thus the dual basis spans the dual space as well, proving

Use the Maxwell curl equationsDot the first with $\mathbf{E}/{\mu }_0$ and the second with $\mathbf{B}/{\mu }_0$, then combine them. The identity $\nabla \cdot (\mathbf{E}\times \mathbf{B})=\mathbf{B}\cdot \nabla \times \mathbf{E}-\mathbf{E}\cdot \nabla \times \mathbf{B}$ givesThis proves the Poynting theorem $W_t+\nabla \cdot \mathbf{S}+\mathbf{J}\cdot \mathbf{E}=0$, with the stated energy density and Poynting vector. The last term is work transferred from the electromagnetic field to matter.

For the specified vacuum wave, substitution into the same curl equations gives $k{E}_0=\omega B_0$ and $k{B}_0={ϵ}_0{\mu }_0\omega E_0$. Thus ${\omega }^2=k^2/{ϵ}_0{\mu }_0$, and for propagation along positive $z$,Putting $\vartheta =kz-\omega t$,The equal electric and magnetic contributions account for the factor in $W$. Energy propagates at speed $c$, whose square is $1/({ϵ}_0{\mu }_0)$; the reciprocal product is not itself the speed.

For an integrand $I(y,y^{\prime })$ with no explicit $x$ dependence, the Euler-Lagrange equation is $I_y-d(I_{y^{\prime }})/dx=0$. Differentiate the proposed first integral:Thus $I-y^{\prime }{I}_{y^{\prime }}$ is constant along an extremal, the Beltrami identity.

For the optical path the Fermat principle uses the travel-time integrand $I=e^{-\lambda y}\sqrt{1+y^{\prime 2}}$, in the stated speed units. Its first integral isSet $u=k{e}^{\lambda y}$. Then $u^{\prime 2}={\lambda }^2(1-u^2)$, and integration gives $u=\cos [\lambda (x-x_0)]$ on a positive arch. The symmetric endpoint conditions require $x_0=0$ and $k=\cos \lambda a$. For $\lambda >0$ and $0<\lambda a<\pi /2$, the resulting ray in an exponential-speed medium isIt is above zero inside the endpoints. Substitution also verifies the Euler-Lagrange equation $y^{\prime \prime }=-\lambda (1+y^{\prime 2})$, so the first-integral construction does not accidentally select a spurious constant-height path.

Here $y^{\prime }=-\tan \lambda x$. For $\lambda a$ close to $\pi /2$, the endpoint slopes are steep and the central height $-{\lambda }^{-1}\log \cos \lambda a$ is large. The travel time evaluates directly toIt tends to $2/\lambda$ as the arch height diverges. Positive $\lambda$ is the increasing-speed interpretation needed for a ray arching into $y>0$; for $\lambda <0$ the same algebraic expression bends below that region and is not an admissible interior path. At zero $\lambda$ the limiting straight boundary path has travel time $2a$.

The contraction mapping theorem states: if $(X,d)$ is a nonempty complete metric space and $f:X\to X$ satisfies $d(fx,fy)\le qd(x,y)$ for a fixed $0\le q<1$, then it has exactly one fixed point and the iteration from every starting point converges to it.

To prove existence, choose $x_0$ and let $x_{n+1}=f(x_n)$. Induction gives $d(x_{n+1},x_n)\le q^{n}d(x_1,x_0)$. The triangle inequality and geometric sum then imply, for $m>n$,The iterates are a Cauchy sequence; completeness supplies their limit $x_{\ast }$. The contraction is continuous, so $f(x_{\ast })=\lim f(x_n)=\lim x_{n+1}=x_{\ast }$. If $y_{\ast }$ is another fixed point, $d(x_{\ast },y_{\ast })\le qd(x_{\ast },y_{\ast })$ forces equality of the points. The same estimate gives geometric convergence from every initial point.

For the particular map, the original PDF gives $X=[\sqrt{a/2},\infty )$, which is a closed, complete subset of the real line. For positive $x$, the arithmetic-geometric mean inequality gives $f(x)=(x+a/x)/2\ge \sqrt{a}$, proving $f(X)\subset X$. FurthermoreThe mean value theorem therefore gives $|f(x)-f(y)|\le |x-y|/2$, a genuine contraction. The fixed-point equation is $x^2=a$, and only the positive root belongs to $X$, soThe printed square-root endpoint is essential: the TeX transcription loses it.

Work on the unit sphere, so side lengths are central angles. Place the right-angle vertex at the north pole and choose perpendicular meridians through it. The other vertices can then be writtenTheir dot product is the cosine of the central angle between them, namely the hypotenuse length $c$. Hence the spherical Pythagorean relation isFor the small triangle let its hypotenuse be $c_{\lambda }$. The same formula gives $\cos c_{\lambda }=\cos (\lambda a)\cos (\lambda b)$, so $c_{\lambda }\to 0$. Expanding each cosine about zero yieldsIn particular $c_{\lambda }=O(\lambda )$, and thus $c_{\lambda }^2={\lambda }^2(a^2+b^2)+O({\lambda }^4)$. Consequently $(c_{\lambda }/\lambda {)}^2\to a^2+b^2$, the ordinary Pythagorean theorem in the locally flat limit.

Let $(q_R,q_S,q_P)$ be player two's mixed strategy in Rock paper scissors. The respective expected payoffs from choosing rock, scissors and paper are $1-p$, $-p$ and $2p-1$. Thus player two solves the linear programming problemA linear objective on this probability simplex is maximized by a largest-payoff pure response, or by any mixture of tied maximizers. Scissors is never a maximizing response: rock exceeds it by one. Comparing the other two gives $1-p⋛2p-1$ according as $p⋚2/3$. ThereforeThe optimal expected payoff is $1-p$ below the tie and $2p-1$ above it, with value $1/3$ at the tie.

Let $B$ denote the zero-diagonal matrix with off-diagonal entries appearing in the iteration, and $G=-B/\alpha$ its iteration matrix. Its row sums are three, so $B1=31$. The other eigenvalues are $2\omega +{\omega }^2$ and $2{\omega }^2+\omega$, where $\omega =e^{2\pi i/3}$, namely $-3/2\pm i\sqrt{3}/2$, both of modulus $\sqrt{3}$. Thus the spectral radius is $\rho (G)=3/|\alpha |$.

For $|\alpha |>3$, the infinity-norm estimate $\Vert G{\Vert }_{\infty }=3/|\alpha |<1$ already gives a contraction, so the iterates converge from every initial vector to the unique solution of $(\alpha I+B)x=b$. Conversely, two iterates whose initial difference is a nonzero multiple of $1$ retain the difference $(-3/\alpha {)}^{k}1$. This fails to decay when $|\alpha |\le 3$, so convergence cannot be guaranteed for arbitrary starts. ThereforeAt $\alpha =3$ the offending mode oscillates; at $\alpha =-3$ it is stationary and the coefficient matrix is singular. A specially chosen starting vector can suppress a bad mode, but that is not guaranteed convergence.