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Here $0<p<1$, $\theta >0$, and the upward step is either a birth or immigration, the latter having conditional probability $\theta /(k+\theta )$. At zero immigration occurs with probability one. Put $q=1-p$, $a=\theta /p$ and $r=p/q$. The stationary weights of a reflected birth-death chain areTheir successive ratio tends to $r$, so they are summable for $p<1/2$. For $p\ge 1/2$, $u_k>1/2>d_{k+1}$ makes the weights strictly increasing, ruling out a stationary distribution. Thus positive recurrence holds exactly for $p<1/2$. In that case the binomial series gives ${\sum }_k{w}_k=2(1-r{)}^{-a-1}$, andEvery step reverses parity and a two-step return has positive probability, so the period of a state in a Markov chain is two. Starting at zero, $\mathbb{P}(X_{2n}=0)\to 2{\pi }_0=(1-r{)}^{a+1}$, whereas $\mathbb{P}(X_{2n+1}=0)=0$.

For $0\le k\le 1$, define $E(k)={\int }_0^{\pi /2}\sqrt{1-k^2{\sin }^2\theta }d\theta$. An ellipse with semi-major and semi-minor axes $a\ge b>0$ and eccentricity of a conic $e=\sqrt{1-b^2/a^2}$ has perimeter $4aE(e)$. Differentiating its parametrization gives the speed $a\sqrt{1-e^2{\sin }^2\theta }$, and fourfold symmetry reduces the full arc length to a quadrant. The binomial series gives $E(k)=(\pi /2)(1-k^2/4-3k^4/64+O(k^6))$ near zero.

For an ellipse with semiaxes $1,x$, the perimeter integral has integrand $\sqrt{1-t{\cos }^2\theta }$. The uniformly convergent binomial series near $t=0$ permits termwise integration. Integrals of ${\cos }^2\theta$ and ${\cos }^4\theta$ over a full period give the displayed first corrections. The same coefficients follow from the Frobenius method for the transformed perimeter equation.

For a nonnegative integer $L$ and $|z|<1$, this specialization of the binomial series sums the normalizing coefficients of a negative binomial distribution.

The function $F$ is an odd function, passes through $(0,0)$, and has the horizontal asymptote $F=0$ at both infinities. Its derivative isIt decreases for $|x|>1$ and increases for $|x|<1$, with its lowest value $F(-1)=-1/2$ and highest value $F(1)=1/2$. These features determine the graph of a function shown below.

For $0<ϵ<1/2$, the equation is the quadratic equation $ϵx^2-x+ϵ=0$. The quadratic formula gives exactly two real equation roots:The binomial series gives $\sqrt{1-4{ϵ}^2}=1-2{ϵ}^2-2{ϵ}^4-4{ϵ}^6+O({ϵ}^8)$, hence the two requested asymptotic expansions areTheir product is exactly one. Both are positive, with $x_{-}<1<x_{+}$; there are no negative equation roots because $F(x)<0$ for $x<0$. At $ϵ=1/2$ the two real equation roots coincide at one, and for $ϵ>1/2$ there are none. In the stated limit $ϵ\to 0^{+}$, these two branches are therefore exhaustive.

Start with the Maclaurin series $\cos x=1-x^2/2+x^4/24+O(x^6)$. Put $z=-x^2/2+x^4/24+O(x^6)$ into the binomial series $(1+z{)}^{-2}=1-2z+3z^2+O(z^3)$. Keeping terms through $x^4$ givessoSince $d(\tan x)/dx={\sec }^{2}x$ and $\tan 0=0$, term-by-term integration supplies the constant of integration and givesThe expansions hold near zero; the nearest real poles at $\pm \pi /2$ limit their radius of convergence to $\pi /2$. The derivative relation also independently checks all three displayed coefficients of the tangent function.

The binomial series givesThus $p_0=1-\alpha >0$ and, for $k\ge 1$,with the factors after $\beta$ omitted for $k=1$. The signs follow from $0<\beta <1$. As $s\uparrow 1$, the series has nonnegative terms and its sum increases to $G(1)=1$, so monotone convergence gives ${\sum }_{k\ge 0}{p}_k=1$. These are therefore genuine probabilities on the nonnegative integers, proving that $G$ is a probability generating function. Its expected value is infinite, since $G^{\prime }(s)=\alpha \beta (1-s{)}^{\beta -1}\to \infty$; finiteness of the mean is not required for this branching process.

Write $G_n(s)=1-A_n(1-s{)}^{{\beta }^n}$. The composition formula gives $A_{n+1}=A_n{\alpha }^{{\beta }^n}$, with $A_0=1$. Consequently $A_n={\alpha }^{1+\beta +\cdots +{\beta }^{n-1}}$, and summing the geometric series yields the explicit generation generating functionThis is the fractional-power offspring generating function and its iterates.

For a circular orbit the origin of mean anomaly is arbitrary because there is no distinguished pericentre. Choose $M=0$ at the ascending node, so $M$ is also the argument of latitude. If a different periapsis convention is used, replace $M$ below by $M+\omega$. The orbital-frame rotation from inclination and node givesIn particular the crossing at $M=0$ has increasing $z$ for prograde $n>0$ and $0<I<\pi$, confirming the ascending node convention.

Put $ϵ=a/a_{\mathrm{pl}}$ and $c=\cos \psi =\hat{\mathbf{r}}\cdot {\hat{\mathbf{r}}}_{\star }$. The binomial series expansion of the direct disturbing function isThe indirect disturbing function subtracts $G{M}_{\star }ac/a_{\mathrm{pl}}^2$, cancelling the dipole term. The constant $G{M}_{\star }/a_{\mathrm{pl}}$ has no spatial derivative and does not perturb the particle. After omitting that constant, the leading stellar quadrupole disturbing function isThus the PDF's equality is understood modulo a dynamically irrelevant additive constant and higher multipoles.

Define $\ell =M+\Omega$ and let ${\ell }_{\star }$ be the star's orbital longitude as seen from the planet. Its orbital plane has $I_{\mathrm{pl}},{\Omega }_{\mathrm{pl}}$. The vector from planet to star is the negative of the planet's heliocentric vector: if ${\ell }_{\mathrm{pl}}$ describes that latter vector, use ${\ell }_{\star }={\ell }_{\mathrm{pl}}+\pi$. This sign matters in $\cos \psi$ but drops out of ${\cos }^2\psi$. Expanding the two rotated unit vectors givesFor example, the particle's horizontal correction is $-(I^2/2)\sin (\ell -\Omega )(-\sin \Omega ,\cos \Omega )$, while its vertical component is $I\sin (\ell -\Omega )$; their dot products with the corresponding stellar components produce these terms. No cubic terms occur because simultaneous reversal of both small orbital inclinations leaves the dot product unchanged.

The two mean longitudes are uniform for the circular Kepler orbits, so orbit averaging is uniform in both $\ell$ and ${\ell }_{\star }$. Set $c_0=\cos (\ell -{\ell }_{\star })$ and $c_2=\cos \psi -c_0$ through quadratic order. The needed averages follow by integrating sine and cosine products over $[0,2\pi {]}^2$:For the last identity, expand $c_0=\cos \ell \cos {\ell }_{\star }+\sin \ell \sin {\ell }_{\star }$ and factor the two integrations, each using $⟨{\sin }^2\ell ⟩=⟨{\cos }^2\ell ⟩=1/2$. Squaring $c_0+c_2$ now yieldswhere the error includes all fourth-order combinations of the two orbital inclinations. ThereforeMultiplying this by $G{M}_{\star }{a}^2/a_{\mathrm{pl}}^3$ gives the requested secular disturbing function. Equivalently, the exact angular factor for two circular rings is $(3{\cos }^2{I}_m-1)/8$, where $I_m$ is their mutual inclination; its small-angle expansion gives the same result. This is the double-averaged circular tidal potential.

For aligned equatorial and planetary orbital planes, set $I_{\mathrm{pl}}=0$. The secular stellar quadrupole disturbing function then has derivative ${\partial }_I{\overline{\mathcal{R}}}_{\star }=-3G{M}_{\star }{a}^{2}I/(4a_{\mathrm{pl}}^3)$. The supplied Lagrange planetary equations for nodal precession givesThe planetary quadrupole coefficient $J_2>0$ contributes ${\partial }_I{\overline{\mathcal{R}}}_{J_2}=-3G{M}_{\mathrm{pl}}{J}_2{R}_{\mathrm{pl}}^{2}I/(2a^3)$, henceBoth are retrograde nodal precession rates. Equating them cancels $n$ and gives the Laplace radiusThe planetary quadrupole dominates inside $a_L$ and the stellar tide outside it. These are small-inclination rates for planet-bound Kepler orbits in the hierarchical regime $a\ll a_{\mathrm{pl}}$. At exactly $I=0$, the node itself is undefined; the rates are the limiting precession frequencies of an infinitesimal tilt.

Rewrite the integrand as $\sqrt{1-t{\cos }^2\theta }$. The binomial series, uniformly valid in $\theta$ for $|t|$ bounded below one, givesThe integrals of $1,{\cos }^2\theta ,{\cos }^4\theta$ over a full period are respectively $2\pi ,\pi ,3\pi /4$. Termwise integration therefore yields the ellipse perimeter expansion near a circle

The relativistic energy of a particle of rest mass $m$ is $E=\gamma m{c}^2$, where the Lorentz factor is $\gamma =(1-v^2/c^2{)}^{-1/2}$. Using the binomial series at small $v/c$ givesThe second term is the familiar Newtonian kinetic energy. The first is the rest energy, which remains when $v=0$. Thus the new physical idea is mass-energy equivalence: a body's mass represents energy $m{c}^2$ even at rest. The total energy is not merely energy due to motion.

The Taylor theorem with Lagrange remainder says that, if $f$ has $N+1$ continuous derivatives on the interval joining $0$ and $x$, then for some $\xi$ strictly between them,More generally the expansion is about any base point, replacing $x$ by its displacement. The displayed regularity is a sufficient hypothesis for the theorem.

For $f(t)=(1+t{)}^{1/2}$ and $t>-1$, repeated differentiation givesTherefore the requested coefficients are $c_n=f^{(n)}(0)/n!$. For a fixed $0<x<1$, the remainder hasNow $|c_1|=1/2$ and $|c_{n+1}/c_n|=(n-1/2)/(n+1)<1$ for $n\ge 1$, so $|c_{N+1}|\le 1/2$. Hence $|R_N(x)|\le x^{N+1}/2\to 0$. Passing to the limit proves, rather than merely formally suggests, the binomial seriesThe restriction $x>0$ makes the remainder bound especially direct, since $1+\xi \ge 1$.

Let $u_n={\mathbb{P}}_0(X_n=0)$ and $f_n={\mathbb{P}}_0(T=n)$. Decomposing at the first return and using the Markov property gives the renewal equationFor $0<s<1$, define the probability generating functions $U(s)={\sum }_{n\ge 0}{u}_n{s}^n$ and $F(s)={\sum }_{n\ge 1}{f}_n{s}^n={\mathbb{E}}_0[s^T]$. Absolute convergence, or nonnegative summation, justifies the convolution identity $U=1+FU$. The binomial series givesSolving the renewal identity gives the first-return generating function of the simple symmetric random walkAs consistency checks, $F(s)\to 1$ as $s\uparrow 1$, confirming recurrence, while $F^{\prime }(s)=s/\sqrt{1-s^2}\to \infty$. By monotone convergence of ${\sum }_{n}n{f}_n{s}^{n-1}$, this last limit is ${\mathbb{E}}_{0}T=\infty$, confirming null recurrent states.

Put $ϵ=l/R$ and $a=1-\cos {\theta }_0$. ThenThe binomial series givessoSince $v_1^2=2gRaϵ$, a second binomial series expansion yieldsThus the constants in the stated Big O notation expansion are

An analytic branch of a square root determined by the branch $\ell$ isIf ${\psi }_1$ and ${\psi }_2$ are two such branches, their ratio $h={\psi }_1/{\psi }_2$ is analytic and satisfies $h^2=1$. Since a domain is connected and $h$ takes values in the discrete set $\{1,-1\}$, $h$ is constant. Thus ${\psi }_1={\psi }_2$ throughout $D$ or ${\psi }_1=-{\psi }_2$ throughout $D$.

For $z=r{e}^{i\theta }$, the principal square root isOn $D_2$, one may instead takeThe respective removed half-axes are their branch cuts.

On $\mathbb{C}\setminus [-1,1]$, defineIt is analytic there, its square is $1-z^2$, andso it is the required branch. Substitution gives, for $0<|z|<1$,The binomial series yieldshence the first three terms of the Laurent series are

Sinceits residue at zero is $-i$. Under the change of variable $z=1/\zeta$, the two orientation reversals cancel, and the residue theorem gives

For each nonzero period lattice point $\omega$, the binomial series givesPairing $\omega$ with $-\omega$ cancels every odd power. Hence the Laurent series of the Weierstrass elliptic function iswhich is the stated coefficient formula.

For fixed real $a\ne 0$, write the square root as $|a|\sqrt{1+x^2/a^2}$ and use the binomial series with exponent $-1/2$:The absolute values are necessary for negative $a$. This Taylor series is valid for $|x|<|a|$. If $a=0$, the quotient is $x/|x|$ for $x\ne 0$; it has no continuous extension at zero and therefore has no Taylor expansion about zero.

Set $h=k/100$ and $j=m/100$. The exact ratio is $(1+h{)}^p(1+j{)}^q$. Multiplying the two binomial series and retaining total degree at most two gives the second-order relative change of a monomial:Multiply by one hundred to express the change as a percentage:The cross term accounts for the simultaneous changes in both inputs and cannot be recovered by simply adding their separate second-order percentage changes.

Use the fixed exponents $p,q$ in the original PDF. Assume positive $a,b$ so arbitrary real powers and a relative change are defined. Set $h=k/100$. With $b$ fixed, the exact new-to-old ratio is $(1+h{)}^p$. The binomial series, or the one-variable case of the second-order relative change of a monomial, givesTherefore the approximate percentage change through second order isFor negative $k$ this same signed expression describes a decrease.

Use the binomial series and exponential Maclaurin series:The Cauchy product isAdding $\log (1+x)=x-x^2/2+x^3/3-x^4/4+O(x^5)$ cancels the first three powers. Hence the first nonzero term is

Factor out $9$ and use the binomial series with exponent $-1/2$:Thus the expansion through cubic order isThe coefficients of $x$ and $x^3$ vanish because the function is an even function. The underlying binomial expansion converges for $|x|<3$.