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The ideal pressure-forced response becomes resonant whenequivalently $\tan \alpha =-\alpha$. Damping regularizes the divergent linear response and introduces a nonzero phase lag and mean supplied power.

For a finite source alphabet with probabilities $p_i>0$ and binary prefix code lengths ${\ell }_i$, the expected length satisfiesTo see this, let $K={\sum }_i{2}^{-{\ell }_i}\le 1$ by Kraft inequality and put $q_i=2^{-{\ell }_i}/K$. Direct calculation givesusing nonnegativity of relative entropy. Thus the information entropy measures the unavoidable average description length, rather than the length of every individual message.

If at least two symbols have positive probability, the Shannon lengths ${\ell }_i=\lceil -{\log }_2{p}_i\rceil$ are positive, satisfy Kraft inequality, and obey $H(p)\le \overline{\ell }<H(p)+1$. For a one-symbol source, $H=0$; if codewords must have positive length, a one-bit codeword has expected length $1$, so the strict upper bound requires that separate convention or a nontrivial alphabet.

Order source probabilities as $p_1\ge \cdots \ge p_N$, putand take the first $l_i$ binary digits of $b_i$ as the codeword for symbol $i$. Since $b_j-b_i\ge p_i\ge 2^{-l_i}$ for $j>i$, two cumulative probabilities cannot lie in the same dyadic interval selected by the earlier codeword. The resulting code is prefix-free.

Shannon-Fano coding comprises two related prefix code constructions. Shannon coding assigns lengths $\lceil -{\log }_2{p}_i\rceil$ to symbols of probability $p_i$; Fano coding recursively partitions the ordered symbols into groups of nearly equal total probability and assigns a bit at each partition.

The increments $\Delta A_{n+1}=V_n-\mathbb{E}[V_{n+1}\mid {\mathcal{F}}_n]$ of a supermartingale are predictable and nonnegative.

If $M$ is a martingale and $A$ is predictable, then, subject to integrability,is a martingale.

For symmetric time-independent constitutive tensors,and the macroscopic Maxwell equations give ${\partial }_{t}u+\nabla \cdot S=-E\cdot J$.

For a single traveling vacuum plane electromagnetic wave, $\mathbf{B}=\hat{\mathbf{k}}\times \mathbf{E}/c$ gives equal electric and magnetic contributions to electromagnetic energy:For the fields with linear polarization along $x$ and magnetic orientation along $y$ specified above,Angle brackets mean an average over one period at fixed position. The factor $1/2$ comes from averaging squared cosine; $E_0$ is the peak amplitude, whereas the root mean square field is $E_{\mathrm{rms}}=E_0/\sqrt{2}$. The equal energy contributions oscillate together, rather than trading energy as in an RLC circuit. ^[ref].

For $E_0=100V/m$, using $c=299792458m/s$ and ${ϵ}_0\simeq 8.854\times 10^{-12}F/m$,A $0.010{\mathrm{m}}^2$ surface normal to $+z$ therefore has average crossing power $0.133\mathrm{W}$. This is incident power; the amount absorbed depends on the surface. Doubling the field amplitude quadruples the average flux.

Suppose an analytic function near $c$ is not identically zero there. Its Taylor series has a first nonzero coefficient $a_m$, andThe term $a_m(z-c{)}^m$ is its leading Taylor term. Dividing by $(z-c{)}^m$ gives a limit of $a_m\ne 0$ as $z\to c$, so this term determines the leading local magnitude and angular behaviour. The remainder uses big O notation; it is smaller by one power near the centre.

For a real-valued analytic function of a real variable, an even integer $m$ gives the same nearby sign on both sides of $c$, while an odd integer $m$ gives opposite signs. For example, $f(x)=x^2(1+x)$ has leading Taylor term $x^2$ and is positive on both sides of zero sufficiently nearby; $g(x)=x^3(1+x)$ has leading Taylor term $x^3$ and changes sign there. The finite factor $1+x$ stays positive for $|x|<1$. The ordinary constant term is the leading one when $f(c)\ne 0$.

Inside its radius of convergence, a power series $f(z)={\sum }_{n\ge 0}{a}_n(z-c{)}^n$ can be differentiated term by term:The differentiated series has the same radius of convergence. The conclusion concerns the interior, including uniform convergence on smaller closed disks; it does not assert that differentiated boundary series converge wherever the original did. For example, ${\sum }_{n\ge 1}{z}^n/n$ converges conditionally at $z=-1$, but its differentiated series ${\sum }_{n\ge 1}{z}^{n-1}$ does not.

Applied to the geometric series, it gives the useful identityDifferentiation turns the plain coefficients into integer weights. This is one way generating functions translate a weighted counting question into an operation on a known series. A proof controls the original and differentiated tails on smaller disks; the Weierstrass M-test provides the needed uniform bounds. See Lebl, Basic Analysis, §11.3: power-series differentiation and analyticity.

The Cauchy product has coefficients $c_n={\sum }_{k=0}^n{a}_k{b}_{n-k}$. Absolute convergence permits regrouping and makes its sum the product of the two sums.

The radius of convergence of ${\sum }_{n\ge 0}{a}_n(z-c{)}^n$ is the unique $R\in [0,\infty ]$ separating interior absolute convergence from exterior divergence. It converges absolutely when $|z-c|<R$ and diverges when $|z-c|>R$. At the centre it always has the finite value $a_0$, since all positive powers vanish. The boundary convergence of a power series is a separate question.

For $0<R<\infty$, the interior is a disk in the complex plane and an open interval $(c-R,c+R)$ on the real line. A power series converges uniformly on every smaller closed disk $|z-c|\le r<R$. This gives stable polynomial approximation over a whole compact region, rather than a separate limiting calculation at each point. See Jiří Lebl, Basic Analysis, §11.3.2: convergence of power series.

Four examples reveal what $R$ measures:

The Cauchy-Hadamard theorem computes $R$ directly from the coefficients. The attractive feature is that one growth statistic decides convergence throughout the interior and exterior, including sparse series with many zero coefficients. Finite changes to the coefficients do not alter the radius, though they can change the sum.

Subspace iteration repeatedly applies a matrix to several independent vectors and orthonormalizes them, converging to a dominant invariant subspace when a spectral gap separates the selected eigenvalue cluster.

Inverse iteration repeatedly solves $(A-sI)y_{k+1}=q_k$ and applies vector normalization. It is the power method for $(A-sI{)}^{-1}$, so $s$ must not be an eigenvalue. For a real symmetric matrix, if one eigenvalue is uniquely closest to $s$ and the starting vector has a nonzero component in its eigenspace, the normalized directions converge to that eigenspace. Ties or an absent initial component can prevent that conclusion.

For a diagonalizable matrix, the active spectrum consists of eigenvalues whose eigenspaces have nonzero components of the starting vector in its spectral decomposition. These components are orthogonal projections for a self-adjoint operator, but need not be orthogonal projections for a general diagonalizable matrix. With a unique largest modulus in this active spectrum, the power method selects its eigendirection; an exactly absent dominant component is never created in exact arithmetic.

For a real symmetric matrix with $|{\lambda }_1|>|{\lambda }_2|$ and a starting vector having nonzero component in the leading eigendirection, power-method direction errors arewhereas the Rayleigh quotient error isThe improvement occurs because orthogonality cancels terms linear in the direction error.

Shallow-water potential vorticity is materially conserved by inviscid, unforced shallow water equations. Linearizing about rest with depth $H_0$ gives an anomaly proportional to

The second potential slow-roll parameter isSlow roll usually requires both ${ϵ}_V\ll 1$ and $|{\eta }_V|\ll 1$.

If $A=Q\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_n)Q^T$ is real symmetric and positive semidefinite, its principal square root isIt is symmetric and positive semidefinite, and is positive definite exactly when $A$ is.

When $b_j=1$ and $a_j$ are positive integers, adjacent convergents satisfyEven and odd convergents are monotone from opposite sides, their separation is $1/(q_n{q}_{n+1})$, and $q_n\to \infty$. They therefore have a common limit $x$, with