For the quadratic polynomial $f(x)=(x-a{)}^2-b^2$ with $b>0$, consider the Newton root-finding iteration towards the polynomial root $x_{\ast }=a+b$. If $x_k\ne a$, its signed error $e_k=x_k-x_{\ast }$ satisfiesIndeed the iteration for $u_k=x_k-a$ is $u_{k+1}=(u_k+b^2/u_k)/2$, and subtracting $b$ yields $(u_k-b{)}^2/(2u_k)$. This exact identity explains quadratic convergence near the simple polynomial root. If an initial error is asymptotic to $C{\epsilon }^m$ and $x_k-a\to b\ne 0$ as $\epsilon \to 0$, one step changes its leading power to $2m$ and its coefficient to $C^2/(2b)$.

Kamiokande, associated with Masatoshi Koshiba, was an underground water Cherenkov detector viewed by photomultipliers. Improved water purity and reduced background enabled solar neutrino observations. A concentration of reconstructed recoil directions toward the Sun distinguished a source signal above an approximately direction-independent background.^[ref]

For solar neutrinos, electron elastic scattering supplies a directional recoil, with different cross sections for active neutrino flavors. Neither the Homestake experiment deficit nor the SN 1987A neutrino burst alone established neutrino oscillation; later evidence compared complementary channels and propagation effects.^[ref]

For a scalar one-dimensional dispersion $E(k)=\hslash v|k|$ and a potential of period $L$, a reciprocal vector $nG$, $G=2\pi /L$, connects degenerate right- and left-moving states at $k=nG/2$. Their unperturbed energy is $E_n=n\pi \hslash v/L$. Thus consecutive allowed Bragg thresholds have spacing $\pi \hslash v/L$, whereas shifting momentum by a full reciprocal period changes a linear branch's energy by $2\pi \hslash v/L$. The potential's Fourier coefficients and higher-order couplings determine which gaps actually open and their widths, so a geometric period alone does not imply an exactly energy-periodic spectrum. A graphene pseudospin selection rule can further suppress scattering; the scalar-chain model omits that internal structure.

Consider a spin-degenerate two-dimensional free electron model with rectangular periods $a>b$ and two electrons per unit cell. Write $K=\pi /a$, $L=\pi /b$, and $k_F^2=4\pi /(ab)$. If $K<k_F<L$, the circular Fermi contour crosses only the vertical boundaries of the first Brillouin zone. Folding translates its protruding caps back by a reciprocal lattice vector $(2K,0)$. A weak periodic potential mixes the intersecting plane waves, reconnecting the contours into two open first-band sheets and a closed second-band pocket. The pocket is centred on a zone boundary: its halves on opposite sides of a reduced-zone plot are identified periodically and belong to one closed sheet.

For a functional $J[y]=\int F(x,y,y^{\prime },y^{\prime \prime })dx$, two applications of integration by parts leave the boundary termIf the endpoint displacement is fixed but its slope is free, $\delta y=0$ while $\delta y^{\prime }$ is arbitrary. The first variation therefore forces $F_{y^{\prime \prime }}=0$ at that endpoint. For the beam energy density $F=\beta (y^{\prime \prime }{)}^2/2-\alpha (y^{\prime }{)}^2/2$ with $\beta >0$, this gives $y^{\prime \prime }=0$, the zero-bending moment condition. A clamped endpoint instead prescribes the slope and has $\delta y^{\prime }=0$.

Coupled inductors are inductors whose magnetic flux linkages depend on each other's electric currents. A two-circuit linear model uses two self-inductances and a mutual inductance.

An integrated circuit technology using complementary MOSFETs. CMOS names a circuit technology; active-pixel sensor names a detector architecture.

An ideal insulating gate blocks particle exchange between the gate and substrate of a MOSFET. Their electrochemical potentials, represented by their band-diagram Fermi levels, therefore need not coincide. An applied gate voltage fixes their difference; with $e>0$ and voltage defined by an electron reservoir, ${\tilde{\mu }}_g-{\tilde{\mu }}_s=-e{V}_g$. Contact work-function differences also affect band bending, so equal Fermi levels do not imply flat bands. Each conducting region can separately have a spatially constant equilibrium electrochemical potential.

In water, D-glucose mainly occupies cyclic forms. The C5 hydroxy group adds to the C1 aldehyde, forming a cyclic hemiacetal; reversible opening restores the aldehyde. Reclosure can give either anomer. Their changing proportions cause mutarotation, the approach of optical rotation to an equilibrium value. This pathway involves changing chemical bonds, rather than simply rotating a hydroxy group about a single chemical bond.^[ref]

In this conventional Haworth projection, C5's $\mathrm{C}{\mathrm{H}}_2\mathrm{OH}$ group points up. The alpha anomer's C1 hydroxy group points down, on the opposite face; the beta anomer's points up, on the same face. C2/C3/C4 retain down/up/down hydroxy groups. These are above/below relationships to a reference ring plane, not a claim that the ring is physically flat.^[ref]

The anomeric carbon is C1, the existing carbonyl group carbon that becomes a new stereocenter; no extra carbon chemical atom appears. The two anomers differ at that centre while retaining the other configurations.^[ref]

Trace the ring as C1–C2–C3–C4–C5–O5–C1, where O5 denotes the oxygen originally in the C5 hydroxy group. It contains five carbon chemical atoms and one oxygen; C6 remains attached outside the ring. The C1 carbonyl group oxygen becomes the C1 hydroxy group oxygen. This accounts for every oxygen: O5 in the ring and five hydroxy groups at C1, C2, C3, C4 and C6. Ring closure preserves ${\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6$. A six-membered ring and a six-carbon sugar are thus different counts.

Recoil is the motion acquired by a body when it exchanges momentum with an emitted, absorbed or scattered particle. For isolated emission from a body initially at rest, momentum conservation makes the remaining body acquire the opposite momentum. If its final mass is $M$ and the recoil is nonrelativistic, its kinetic energy is $E_R\simeq p^2/(2M)$. A photon of energy $E_{\gamma }$ has momentum magnitude $E_{\gamma }/c$, so $E_R\simeq E_{\gamma }^2/(2M{c}^2)$. The transition energy is shared between the photon and the recoiling body; nuclear recoil is an important example.

A uniform square of side $s$ has the displayed moment of inertia about the perpendicular axis through its centre. Integrating $x^2+y^2$ over $[-s/2,s/2{]}^2$ gives $s^4/6$; multiplying by the surface mass density $M/s^2$ gives the result. A square prism of any uniform thickness has the same axial formula when expressed in terms of its total mass. If the side is denoted $2a$ instead, the result is $2M{a}^2/3$, so the side-length convention must be specified before substituting numerical data into a normal mode calculation.

For three mutually perpendicular axes through one point of a nonnegative mass distribution, $I_x+I_y-I_z=2\int z^{2}dm\ge 0$, with cyclic counterparts. Equality holds precisely when the mass distribution is supported in the plane $z=0$, giving the perpendicular axis theorem. The point need not be the centre of mass, and the axes need not be principal axes.

A uniform solid with circular cross-sectional radius $r(z)$ can be sliced into thin discs. A slice of thickness $dz$ has mass $dm=\rho \pi r(z{)}^{2}dz$ and axial moment of inertia $d{I}_z=\frac{1}{2}r(z{)}^{2}dm$. ThusAll slice axes coincide, so no parallel axis theorem correction is needed. For a linearly tapered section of length $l$ with end radii $r_0,r_1$, direct integration gives $M=\rho \pi l(r_0^2+r_0{r}_1+r_1^2)/3$ and $I_z=\rho \pi l(r_0^4+r_0^3{r}_1+r_0^2{r}_1^2+r_0{r}_1^3+r_1^4)/10$.

For a cylinder of uniform mass density $\rho$, length $L$, inner radius $R_i$ and outer radius $R_o$, an annular element has $dm=2\pi \rho Lrdr$. Thus the axial moment of inertia is $I=\int r^{2}dm=\pi \rho L(R_o^4-R_i^4)/2$. Dividing by $M=\pi \rho L(R_o^2-R_i^2)$ yields the displayed result. At fixed mass, the solid-cylinder limit is $M{R}_o^2/2$ and the thin-shell limit is $M{R}^2$.

Set $K_i=K_i^{\prime }$, so that $\alpha ={\alpha }^{\prime }$. The reversible enzyme inhibition equation reduces toAt the chosen $I=30\mu \mathrm{M}$, every reaction rate is one quarter of its uninhibited value. The plateau falls to $25\mu \mathrm{M}\mathrm{mi}{\mathrm{n}}^{-1}$, but half that plateau still occurs at $s=20\mu \mathrm{M}$. Increasing substrate cannot restore the uninhibited plateau. An unchanged apparent Michaelis constant here expresses equality of the two binding effects, rather than an absence of inhibitor binding to the enzyme-substrate complex.