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For small $e$, the binomial series gives the uniform expansion $\sqrt{1-e^2{\sin }^2\theta }=1-(e^2/2){\sin }^2\theta +O(e^4)$. Since ${\int }_0^{2\pi }{\sin }^2\theta d\theta =\pi$, integration yieldsAlso $b=a\sqrt{1-e^2}=a(1-e^2/2+O(e^4))$, so the familiar symmetric approximation $\boxed{L=\pi (a+b)+O(a{e}^4)}$ is equivalent to this order. This is the first correction in the ellipse perimeter expansion near a circle; the two approximate formulas need not agree at order $e^4$.

Multiply the Maclaurin series $e^{-x}=1-x+x^2/2+O(x^3)$ by the binomial series $(1+x{)}^{-1/2}=1-x/2+3x^2/8+O(x^3)$. The constant, linear and quadratic coefficients are therefore

The generalized binomial series givesIn particular,The recurrence $u_{n+1}=u_n(1/3-n)/(n+1)$ follows directly from the coefficient product. The power series has radius of convergence one and represents the real cube root for $|x|<1$.

Use the binomial series with exponent $1/2$ and argument $-3x$:Thus the first three nonzero terms are

The usual interior condition for the binomial series is $|3x|<1$, giving $-1/3<x<1/3$ as its open interval of convergence. In this particular square-root case the full series also converges absolutely at both endpoints, so the maximal real range on which its sum equals the original function isFor an endpoint check, write $\sqrt{1-u}=1-{\sum }_{n\ge 1}{b}_n{u}^n$, where $b_n=a_{n-1}-a_n>0$ and $a_n=\left(\genfrac{}{}{0ex}{}{2n}{n}\right)/4^n\to 0$. Then $\sum b_n=1$ by telescoping, establishing absolute convergence at $u=\pm 1$. The square-root binomial series at its endpoints proves continuity of the sum there. The endpoint values are $\sqrt{2}$ at $x=-1/3$ and $0$ at $x=1/3$. Outside this range the power-series terms do not tend to zero. A three-term truncation remains an approximation, rather than an exact formula throughout the interval.

The binomial series gives $(1+ϵ{)}^{-\alpha }=1-\alpha ϵ+O({ϵ}^2)$. ThereforeTo justify the limiting approximation over the infinite interval, also use the right-endpoint sum. It is $L_{ϵ}=(1+ϵ{)}^{-1-\alpha }{U}_{ϵ}$, and monotonicity gives $L_{ϵ}\le I\le U_{ϵ}$. Both bounds have the same limit. The geometric partition bounds for a power-law improper integral and the squeeze theorem thus proveThis bounding argument matters because the geometric partition has unbounded widths far along the half-line; a finite-interval maximum-mesh argument cannot simply be applied to the entire infinite partition at fixed $ϵ$.

For $a/|b|\ll 1$, apply the binomial series to the result in (c):ThereforeFor $\alpha =2\pi$, the leading term is $\pi b^2$, the area of the base disk. A very shallow cap is nearly planar; its first area excess is $\pi a^2$.

Multiply the Maclaurin series of hyperbolic cosine by the binomial series for the reciprocal square root:The quadratic terms cancel, while the quartic coefficient is $1/24-1/4+3/8=1/6$. HenceThe absence of odd powers follows from the function's even symmetry.

For $a\ne 0$, factor out $|a|$, not $a$, and apply the binomial series with exponent $1/2$:For each fixed nonzero $a$, these are the first three nonzero terms of the Taylor series about zero; the full binomial series converges when $|x|<|a|$.

For $a=0$, the function is $|x|$. Its right derivative at zero is $1$ and its left derivative is $-1$, so no Taylor series about zero exists when $a=0$. The one-sided formulas are exactly $x$ for $x\ge 0$ and $-x$ for $x\le 0$; they do not form a single two-sided Taylor expansion, nor can they supply three nonzero terms. The parameter case explicitly included in the question must therefore be treated as this exception.

The Maclaurin series of the hyperbolic cosine is $\cosh x=1+x^2/2+x^4/24+O(x^6)$. The binomial series with exponent $-1/2$ gives $(1+u{)}^{-1/2}=1-u/2+3u^2/8+O(u^3)$. Substituting $u=x^2/2+x^4/24+O(x^6)$ yieldsThese are the first three nonzero terms; the remainder is written in Big O notation.

The binomial series beginsEquating coefficients gives $na=-2$ and $n(n-1)a^2/2=3$. Since $na\ne 0$, substitute $a=-2/n$ into the second equation to obtain $2(n-1)/n=3$, so $n=-2$ and $a=1$. ThereforeIndeed $(1+x{)}^{-2}=1-2x+3x^2-\cdots$ for $|x|<1$.

It is enough to choose an admissible parameter, without solving the minimizing equation. Take $a=V_0^2>0$ in the Gaussian variational bound for an attractive Gaussian well. The Rayleigh-Ritz variational principle giveswhere an explicit remainder isThe denominator tends to $2$ as $V_0\to 0$, proving $\epsilon =O(V_0^4)$ in Big O notation. More precisely the binomial series givesand thereforeThis is an upper bound on the exact ground state energy, rather than an assertion that the leading coefficient of its exact weak-potential expansion is $-1/2$.

The relativistic kinetic energy is the total energy minus the rest energy:Here $\gamma$ is the Lorentz factor and $c$ the speed of light. The binomial series, with $b=v/c$, givesFor $v\ll c$, the leading term is the classical kinetic energy $M{v}^2/2$. The ratio of the fourth-order term to that leading term is $3v^2/(4c^2)$. ConsequentlyThe condition concerns one term of the binomial series; it is not the condition that the entire relativistic correction be exactly $1\%$.

To calculate magnetic-field cancellation of an arc and tangent leads away from the plane, put $a^2=R^2+h^2$. Each point on the arc is distance $a$ from $(0,0,h)$, and the axial component of $d\mathbf{l}\times \mathbf{s}$ still has magnitude $R^{2}d\varphi$. Thus its axial magnetic field isFor either tangent lead the axial numerator remains $Rds$, while the cubed distance is $(R^2+h^2+s^2{)}^{3/2}$. HenceAdding both leads givesFor $ϵ=h/R\ll 1$, a binomial series yieldsThe exact condition for cancellation of the axial component isFor example, keeping $\theta =2$ would leave a small positive axial magnetic field proportional to $h^2/R^2$. Both contributions weaken with height, but the arc factor is $(1+{ϵ}^2{)}^{-3/2}$ whereas the lead factor is $(1+{ϵ}^2{)}^{-1}$. The arc therefore loses a greater fraction of its axial magnetic field. Increasing its angle compensates for this loss. Physically, the arc is all at the same finite radius, whereas distant elements of the semi-infinite leads undergo a smaller fractional change in distance on moving a little out of the plane.

The cancellation here concerns $B_z$ only; off the plane it does not assert that every component of the magnetic field vanishes. The small-height result also stays within the tangent-lead geometry shown in the original paper.

Assume $p<1/2$, and put $a=\theta /p$ and $r=p/q\in (0,1)$. The detailed balance equations giveMultiplying from $0$ to $k-1$, the last factor telescopes, while the product involving $a$ is evaluated by the Gamma function:Consequently the requested invariant distribution can be expressed using the paper's power series asThe assumed convergence of $g_{p,\theta }(r)$ ensures that this normalization is finite and positive.

For an explicit normalizing constant, the binomial series givesThe termwise differentiation of a power series is valid for $|r|<1$, soThe last equality uses $r=p/q$ and $a=\theta /p$: $ar+\theta (1-r)=\theta /q+\theta (q-p)/q=2\theta$. ThereforeThese positive terms sum to one and satisfy the detailed balance equations, so part (a)(ii) verifies stationarity. Irreducibility makes the invariant distribution unique.

For a free particle of rest mass $m$, subtract its rest energy $m{c}^2$ from its total energy $\gamma m{c}^2$, where $\gamma$ is the Lorentz factor. The low-speed binomial series gives $K=m{v}^2/2+3m{v}^4/(8c^2)+O(m{v}^6/c^4)$, recovering classical kinetic energy.

For positive $a,b$, fixed real $p,q$, and $f(a,b)=a^p{b}^q$, change the inputs to $a(1+h),b(1+j)$ with $h,j$ small. The ratio of the new value to the old one is exactly $(1+h{)}^p(1+j{)}^q$. Applying the binomial series to each factor and multiplying proves the displayed Taylor expansion. Positive inputs and $1+h,1+j>0$ make arbitrary real exponents well defined. A percentage change of $k$ in $a$ means $h=k/100$, not $h=k$. The mixed term $pqhj$ is required even though there is no such term in either one-variable expansion separately.

Write $a_n=\left(\genfrac{}{}{0ex}{}{2n}{n}\right)/4^n$, $a_0=1$. The binomial series for $\sqrt{1-u}$ has positive numbers $b_n=a_{n-1}-a_n=\left(\genfrac{}{}{0ex}{}{2n}{n}\right)/[4^n(2n-1)]$ as the magnitudes of its nonconstant coefficients. The product $a_n={\prod }_{j=1}^n[1-1/(2j)]$ tends to zero, since $\ln a_n\le -\frac{1}{2}{\sum }_{j=1}^{n}1/j\to -\infty$. Thus $\sum b_n=1$ by telescoping. The Weierstrass M-test gives uniform convergence and absolute convergence on $[-1,1]$. Continuity extends the interior identity to both endpoints. Its radius of convergence is one; outside this interval its terms do not tend to zero.

Over the complex numbers, if $N^m=0$, thenHere the generalized binomial coefficient is $\left(\genfrac{}{}{0ex}{}{\alpha }{j}\right)=\alpha (\alpha -1)\cdots (\alpha -j+1)/j!$, with $\left(\genfrac{}{}{0ex}{}{\alpha }{0}\right)=1$. This is an exact finite identity, even when an infinite binomial series would not converge.

One can construct the required polynomial $q(z)=1+{\sum }_{j=1}^{m-1}{c}_j{z}^j$ without any infinite series. At polynomial degree $j$, the coefficient of $q(z{)}^3$ is $3c_j$ plus terms involving only earlier coefficients. Recursively choose it to agree with $1+z$ through polynomial degree $m-1$; the first coefficients are $1/3,-1/9,5/81$. The stated generalized binomial coefficients solve this recursion: near $z=0$, the binomial series for $(1+z{)}^{1/3}$ cubes to $1+z$, and truncating it changes no coefficient below polynomial degree $m$. Thus $q(z{)}^3-(1+z)$ is divisible by $z^m$ as a polynomial identity. Substituting the nilpotent matrix $N$ kills the remaining terms, all divisible by $N^m$. For a Jordan block $\lambda I+N$ with $\lambda \ne 0$, replace $N$ by $N/\lambda$ and multiply $Q$ by a scalar cube root of $\lambda$.

For a forward photon and a positive collinear Lorentz transformation, the Doppler factor is $D=\gamma (1+\beta )$. The identity $\gamma \beta =\sqrt{{\gamma }^2-1}$ gives the exact expression in the title. Expanding the square root by the binomial series gives $D=2\gamma -1/(2\gamma )-1/(8{\gamma }^3)+O({\gamma }^{-5})$. A leading approximation $D\sim 2\gamma$ has relative error of order ${\gamma }^{-2}$; retaining a purported subleading term requires expanding all dependent quantities consistently. For example, $\sqrt{4{\gamma }^2-1}$ has subleading term $-1/(4\gamma )$ and is not the exact longitudinal Doppler factor.