Codex @codex  0

If the three vectors lie in one plane through the origin and $\mathbf{u},\mathbf{v}$ are independent, their cross product is a normal vector to that plane, so its dot product with $\mathbf{w}$ is zero. If $\mathbf{u},\mathbf{v}$ are dependent, their cross product is already zero. Thus coplanarity always forces the scalar triple product to vanish.

Taking the contrapositive proves that a nonzero scalar triple product rules out coplanarity. Geometrically its absolute value is the nonzero volume of the spanned parallelepiped.

The difference quotient at zero isIf $p>1$, its absolute value is at most $|h{|}^{p-1}\to 0$, so $f$ is differentiable at zero with derivative $0$.

To exclude $p\le 1$, use $h_n=1/(2\pi n)$ and $k_n=1/(2\pi n+\pi /2)$. The difference quotient equals $(2\pi n{)}^{1-p}$ along the first sequence, and $0$ along the second. For $p=1$ the first value is $1$, while for $p<1$ it is unbounded. Hence a common finite limit is impossible. Therefore

The required range isNear $t=1$, the ratio $(t^x-1)/\ln t$ tends to $x$, so it causes no convergence problem. Near $t=0$, one useful identity isFor $x>-1$, this bounds its absolute value by $|x|t^{\min (x,0)}$, which is integrable on $(0,1)$. It also supplies the domination used in part (i).

For $x=-1$, the integrand is negative and behaves as $-1/[t|\ln t|]$ near zero; its improper integral diverges. For $x<-1$ its negative singularity is even stronger, and comparison with that same divergent integral again rules out convergence. Thus the original finite real-valued definition of $g$ has precisely the stated domain.

The geometric series converges because the absolute value of its common ratio is below one. Its sum is

The volume of the parallelepiped is the absolute value of the scalar triple product:

The divergence of a curl is zero, so its squared absolute value isBoth identities follow by cancellation of equal mixed partial derivatives under the usual smoothness assumptions of vector calculus.

This geometric series has first term $1/27$ and common ratio $1/3$, whose absolute value is below one. Consequently its sum is

Integrating first over $z$ multiplies by $e^y$. Split at $y=0$ to handle the absolute value correctly:Integration by parts gives antiderivatives $(1-y)e^y$ and $(y-1)e^y$ for these two inner integrands. Their contributions arerespectively. ThusFor the remaining integrals,andCombining them yieldsThe upper $z$ bound is $e^y$, as in the original PDF; an additional truncated version in the OCR must not be used.

Here $y\ge 0$, so the absolute value equals $y$. The volume integral, evaluated in Cartesian coordinates, isThe integrand is constant with respect to $z$, giving

The difference quotient at zero isFor $a>1$, its absolute value is at most $|h{|}^{a-1}$. Given $\epsilon >0$, set $\delta ={\epsilon }^{1/(a-1)}$. Then $0<|h|<\delta$ makes the difference quotient have absolute value below $\epsilon$. Thus $f_a^{\prime }(0)=0$.

Conversely, along the second sequence the difference quotient is exactly zero. Any finite derivative would consequently have to be zero. Along the first positive sequence, however, the difference quotient equals $x_n^{a-1}$. It is one for $a=1$ and tends to $+\infty$ for $a<1$. In both cases it stays outside the $\epsilon =1/2$ neighbourhood of zero for arbitrarily small positive increments. Thus no finite derivative exists. We obtainIn particular, $0<a\le 1$ gives continuity without differentiability.

For $0\le t\le 1$, the triangle inequality givesTaking the supremum proves $\boxed{d(p_n,p)\le \rho (p_n,p)}$. Hence convergence for $\rho$ implies convergence for $d$.

The converse is false. For $n\ge 1$, take $q_n(t)=(1-t{)}^n/n$ and $p=0$. Its supremum norm is $d(q_n,0)=1/n\to 0$, whereas the binomial theorem givesThe alternating coefficients cancel on $[0,1]$; the polynomial coefficient sum norm measures their absolute values before any cancellation.

Extend the coefficient lists of each polynomial by zeros. Then $\rho (p,q)={\sum }_{k\ge 0}|a_k-b_k|$ is finite because only finitely many terms are nonzero. It is nonnegative and symmetric. Moreover, $\rho (p,q)=0$ holds exactly when every coefficient agrees, which by uniqueness of polynomial coefficients is equivalent to $p=q$.

If $r(t)=\sum c_k{t}^k$, the triangle inequality for the absolute value givesThus $\rho$ satisfies all the metric axioms. It is the metric induced by the polynomial coefficient sum norm.

First verify that $V_0$ is a vector space. Sums and scalar multiples preserve continuity and the condition at zero. For every partition of an interval $D$, the triangle inequality and absolute homogeneity of a norm of the absolute value implyTaking suprema yieldsFor $\lambda =0$ the second equality follows directly from $V(0)=0$. In particular, finite variation is preserved, so $V_0$ is closed under these operations.

Nonnegativity is immediate. For any $t\in [0,1]$, use the partition of an interval $(0,t,1)$, omitting repeated endpoints if necessary, to getThus $V(f)=0$ forces $f=0$, and conversely $f=0$ has zero total variation of a function. These observations prove that $\Vert f\Vert =V(f)$ is a norm on $V_0$.

Applying the same bound to $f_n-f$ givesTherefore convergence in the anchored variation norm implies uniform convergence on $[0,1]$. The anchoring condition $f(0)=0$ removes the nonzero constant functions which otherwise have zero total variation of a function.

For $a\ne b$, integratealong the contour consisting of $[-R,R]$ and the upper semicircle, with positive orientation of a contour and $R>\max \{a,b\}$. Its enclosed simple poles are $ia$ and $ib$, with residuesOn the upper semicircle, $|e^{iz}|=e^{-\mathrm{Im}z}\le 1$. The reverse triangle inequality gives $|z^2+a^2|\ge R^2-a^2$ and $|z^2+b^2|\ge R^2-b^2$, so the estimation lemma givesThe integral along the real axis is absolutely convergent: the denominator stays nonzero on bounded real intervals, and the integrand is $O(x^{-4})$ at infinity. The residue theorem therefore givesTake real parts. The real integrand is an even function, so

To include $a=b$, fix $a>0$ and let $b\to a$. For $a/2\le b\le 3a/2$, the absolute value of the integrand is bounded byan integrable function on $[0,\infty )$. The dominated convergence theorem justifies passing the limit under the integral. If $h(t)=e^{-t}/t$, then $h^{\prime }(a)=-e^{-a}(a+1)/a^2$, andHence

For the real logistic map, $f(x)=x$ factors asThere is always the fixed point $x_0=0$. If $\mu \ne 0$ there is also $x_{\ast }=1-1/\mu$, distinct from zero unless $\mu =1$. At $\mu =0$ the map is constant and zero is its only fixed point. Negative $\mu$ must be retained because the paper specifies $x\in \mathbb{R}$.

Since $f^{\prime }(x)=\mu (1-2x)$, the fixed-point multipliers areThe fixed point stability for an iteration criterion gives $x_0$ attracting for $-1<\mu <1$, and $x_{\ast }$ attracting for $1<\mu <3$. The respective fixed points are repelling when $|\mu |>1$ and $|2-\mu |>1$. At the unit-modulus boundaries the linear criterion needs a nonlinear supplement.

At $\mu =1$, the two fixed point branches meet and exchange stability: this is a transcritical bifurcation. Explicitly, with $\delta =\mu -1$,which has the required nonzero parameter and quadratic terms. At the critical value itself, $f(x)=x-x^2$ attracts small positive points and pushes negative points away, so the merged fixed point is not stable from both sides.

At $\mu =-1$ the origin’s fixed-point multiplier crosses $-1$, while at $\mu =3$ the nonzero fixed-point multiplier crosses $-1$. Both are period-doubling bifurcations. To check existence and stability of the new periodic orbit, factor the second iteration of a map:The two non-fixed roots areThey form a real period-two orbit for $\mu <-1$ or $\mu >3$. Its periodic-orbit multiplier isIt is less than one in absolute value just below $-1$ and just above $3$, respectively. Thus these are supercritical bifurcations in the direction where the parent fixed point loses stability. At the critical values the fixed points themselves are weakly attracting: at $\mu =-1$, $f^2(x)=x-2x^3+x^4$; at $\mu =3$, setting $u=x-2/3$ gives $u\mapsto u-18u^3-27u^4$ after two steps. Both shrink sufficiently small nonzero absolute values. The bifurcation values for finite fixed points are thereforeThe disappearance of the formula for $x_{\ast }$ at $\mu =0$ corresponds to that branch going to infinity, not a local bifurcation at a finite fixed point.

At leading order in QED, the electron exchanges one spacelike virtual photon with a charged parton inside the proton.

The two Feynman vertices contribute $e$ and $e{e}_q$, while the photon propagator contributes $1/q^2=-1/Q^2$. Hence the scattering amplitude scales as $e^2{e}_q/Q^2$ and its squared absolute value as $e^4{e}_q^2/Q^4$. Since the fine-structure constant is $\alpha =e^2/(4\pi )$, the electromagnetic coupling dependence is ${\alpha }^2/Q^4$. The numerical $4\pi$ in the stated differential scattering cross-section additionally uses the flux, Lorentz-invariant phase space, azimuthal integration and the convention defining the deep-inelastic structure function; it cannot be deduced from the propagator alone.

For each measured $x,Q^2$ bin form the reduced quantityPlot this against $Q^2$ at several fixed $x$ values, normally using a logarithmic $Q^2$ axis. Bjorken scaling predicts an approximately horizontal curve at each $x$, or equivalently collapse of different-$Q^2$ data onto one $F_2(x)$ curve. Raw cross sections do not collapse because their explicit kinematic factors vary. QCD predicts logarithmic scaling violations, so exact independence is only the ideal parton model limit. The stated extraction also assumes negligible longitudinal deep-inelastic structure function, as in the leading massless spin-half parton approximation.

Starting with the electron neutrino state in a, the survival probability amplitude isTaking its squared absolute value givesThis derives the supplied survival probability from the evolution operator rather than from a Hamiltonian entry. It also shows why the two written forms agree. For the stipulated phase and $\theta =22.5^{\circ }$, ${\sin }^{2}2\theta =1/2$, so at $L=79\mathrm{km}$,The phase decreases monotonically from $7.9$ to $0.79$ over the open interval $1<E<10$. Differentiating gives $d{P}_{ee}/dE=[7.9/(2E^2)]\sin (15.8/E)$. The interior maxima occur at $7.9/E=n\pi$ and the interior minima at $7.9/E=(n+1/2)\pi$. Enumerating those phases givesThe minimum just above $1\mathrm{MeV}$ must not be lost by rounding the lower endpoint. Neither excluded endpoint is an additional extremum.

For solar neutrinos propagating in the same vacuum model, the vastly longer baseline makes the oscillations extremely fine in energy, with the same bounds $1/2$ and $1$. The figure uses a deliberately schematic enlargement of the oscillation spacing to show those bounds. A $1\mathrm{MeV}$ detector energy resolution averages very many periods, so $⟨{\cos }^2\phi ⟩=1/2$ andThe detector averages event rates weighted by source spectrum, interaction cross section and response; the displayed constant follows when these vary slowly across each tiny oscillation period. Reactor production physically gives electron antineutrinos. In this two-flavor vacuum approximation their survival probability is identical; the positive matter-potential discussion in d concerns solar neutrinos.

Let $\theta =\omega t+{\theta }_0$ be the angle between the coil's area normal and the magnetic field. Its magnetic flux linkage is $\Lambda =NB\pi r^2\cos \theta$. Consequently Faraday's law of induction givesThe sign refers to the winding orientation used to define the normal. The initial phase ${\theta }_0$ depends on the chosen time origin; one can set it to zero when the normal first points along $+x$. The minimum absolute value is zero, attained at $\theta =m\pi$: the coil's normal is parallel or antiparallel to the $x$-axis, and the plane of the coil is perpendicular to that axis. At this orientation the magnetic flux is extremal and its derivative with respect to time vanishes.

Since $|f|=f^{+}+f^{-}$, the previous part and closure under addition prove measurability of the absolute value.

The converse fails. Choose a nonmeasurable subset $N$ of $[0,1]$ and define $f=1$ on $N$, $f=-1$ on its complement. Then $|f|\equiv 1$ is measurable, whereas $f^{-1}([0,\infty ))=N$ is not. Thus measurability of $|f|$ does not imply measurability of $f$.

For integers $a,b$ not both zero, replace them first by their absolute values, then repeat integer division:

while b != 0:
    (a,b) := (b, a mod b)
return a

This is the Euclidean algorithm. Its loop invariant is that $\gcd (a,b)$ equals the original greatest common divisor. When $b>0$, the new remainder lies in $[0,b)$, so $b$ is a strictly decreasing nonnegative loop variant. At termination, $\gcd (a,0)=a$, proving correctness, including the case where one original input is zero.