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The stated bound is Lipschitz continuity with constant $1$. For every $x$ and every $\epsilon >0$, taking $\delta =\epsilon$ ensures $|f(x)-f(y)|<\epsilon$ whenever $|x-y|<\delta$. The same $\delta$ works at every point, so $f$ is both continuous and uniformly continuous.

For an integer $N\ge 1$, partition $[0,1]$ into $N$ equal intervals and define a step functionAt the remaining endpoint set $g_N(1)=f((N-1)/N)$, so the final interval may be taken closed on the right. The Lipschitz bound givesChoosing $N\ge 1/\epsilon$ proves the requested uniform step approximation on a compact interval:A sufficiently fine partition gives a piecewise constant approximation uniformly over the whole interval.

The continuous function $f$ and every finite step function $g_N$ are Riemann integrable. Put $c_j=f(j/N)$. Integrating each constant piece explicitly givesSince the sine differences have absolute value at most $2$, the triangle inequality and uniform approximation giveTo prove convergence, fix $\epsilon >0$ and first choose $N$ with $1/N<\epsilon /2$. With this $N$ held fixed, the finite number ${\sum }_j|c_j|$ is independent of $n$, so the second term is also less than $\epsilon /2$ for sufficiently large $n$. ThereforeFirst make the approximation error small, then let the oscillation frequency grow. This is the step-function proof of the Riemann-Lebesgue lemma; it does not require assuming that a Lipschitz function has an everywhere-defined derivative.

The Rolle theorem states that if a real function is continuous on $[u,v]$, differentiable on $(u,v)$, and has equal endpoint values, then its derivative vanishes somewhere in $(u,v)$. Order the $k$ distinct real roots as $r_1<\cdots <r_k$. The theorem gives a root of $f^{\prime }$ in each interval $(r_j,r_{j+1})$. These intervals are disjoint, so the obtained roots are distinct:

We prove the requested bound for positive roots of a polynomial by mathematical induction on the number of nonzero terms. In fact, the proof also works with roots counted according to their multiplicity. Write $V(p)$ for the number of sign changes and $N_{+}(p)$ for the positive-root count.

A one-term polynomial $a_1{x}^{d_1}$ has no positive roots and no sign changes. For the induction step, factor out the smallest power of $x$:For $x>0$ this factor is nonzero, so $N_{+}(q)=N_{+}(p)$, including multiplicities, and $V(q)=V(p)$. Differentiation removes exactly the constant term:The exponents remain strictly increasing, and the positive multipliers $e_j$ do not change coefficient signs. The induction hypothesis therefore gives

First suppose $a_1{a}_2<0$. Removing $a_1$ removes exactly one sign change, so $V(q)=V(q^{\prime })+1$. If $q$ has positive roots $r_1<\cdots <r_k$ of multiplicities $m_1,\dots ,m_k$, its derivative has multiplicity $m_j-1$ at $r_j$ and, by the Rolle theorem, an additional root between each pair. Thus the Rolle root count with multiplicities giveswhen there is a positive root. ConsequentlyIf there is no positive root, the desired bound is automatic.

Now suppose $a_1{a}_2>0$, so $V(q)=V(q^{\prime })$. Again the zero-root case is immediate. Otherwise, multiplying $q$ by $-1$ if necessary lets us assume $a_1>0$ and $a_2>0$ without changing roots or sign changes. For sufficiently small $x>0$,since $q(0)=a_1$ and the lowest-power term of $q^{\prime }$ has positive coefficient. Choose such a point $u$ before the first positive root $r_1$. The mean value theorem on $[u,r_1]$ gives a point $v\in (u,r_1)$ withBut $q^{\prime }(u)>0$. The intermediate value theorem applied to the polynomial $q^{\prime }$ therefore gives an extra root in $(u,v)$, before $r_1$. It is distinct from the roots at $r_j$ and those between consecutive roots. HenceBoth sign cases complete the induction:An initial sign change allows one extra positive root; without that sign change, the derivative must already turn before the first positive root. This is the upper-bound part of the Descartes' rule of signs. The multiplicity claim used above follows by differentiating $q(x)=(x-r{)}^{m}h(x)$ with $h(r)\ne 0$: the derivative has a factor $(x-r{)}^{m-1}$ and its remaining factor is nonzero at $r$.

The effective number of relativistic energy degrees of freedom count two photon polarizations, four electron/positron spin states, and six neutrino/antineutrino states, with the Fermi-Dirac distribution factor $7/8$ for fermions. Treating these species as relativistic and at a common temperature gives $\boxed{g_{\ast }=2+(7/8)(4+6)=10.75}$.

The mass-equivalent radiation mass density is $\rho =({\pi }^2/30)g_{\ast }(k_{B}T{)}^4/({\hslash }^3{c}^5)$. Therefore the Friedmann equation givesAt $T_0=10^{10}\mathrm{K}$, $H_0\simeq 0.503{\mathrm{s}}^{-1}$. Equating the specified weak interaction rate to $H_0(T/T_0{)}^2$ gives $(T_{\ast }/T_0{)}^3=H_0/(1{\mathrm{s}}^{-1})$ and henceThe approximate equilibrium neutron-to-proton ratio, neglecting chemical potentials, is $r_{\ast }\simeq \exp [-(m_n-m_p)c^2/(k_B{T}_{\ast })]$, about $0.152$ using the mass difference $1.293\mathrm{MeV}$. This is a freeze-out estimate rather than a detailed reaction-network computation.

After weak cosmological weak freeze-out, free neutrons undergo beta-minus decay; neutrinos decouple, and electron-positron annihilation heats the photons relative to the neutrinos. Initially the abundant energetic photons destroy deuterium almost as soon as it forms. Once the deuterium bottleneck ends, nuclear reactions rapidly pass through deuterium, tritium and helium-three to helium-4. Reactions eventually become too slow compared with cosmic expansion; nearly all surviving neutrons are bound in helium-four. Its mass fraction is approximately $Y_4=2r/(1+r)$, where $r$ is evaluated when nuclear assembly begins.

Increasing $g_{\ast }$ increases the Hubble parameter, raises $T_{\ast }$, increases the freeze-out neutron-to-proton ratio, and leaves less time for free-neutron decay after freeze-out. Both effects increase $Y_4$; decreasing $g_{\ast }$ reverses them. Most deuterium is only an intermediate product and is burned into helium, so its residual abundance and its absolute change are tiny compared with the helium mass fraction. This does not imply a negligible fractional change in residual deuterium. Faster expansion also shortens the time available to burn it, and its relative abundance can be appreciably sensitive to extra radiation. The distinction matters when interpreting the final wording of the question; joint deuterium and helium abundances constrain extra relativistic species, as discussed in the Particle Data Group nucleosynthesis review.

The Papperitz symbol describes a homogeneous second-order Fuchsian differential equation on the Riemann sphere, with three regular singular points and the two indicial exponents at each. At a finite point $z_j$, an exponent $r$ corresponds locally to a factor $(z-z_j{)}^r$; at infinity it corresponds to $z^{-r}$. Repeated or resonant exponents can require logarithms. The six exponents obey the Fuchs relation that their sum is one. With three singularities there is no independent accessory parameter, so these data determine the normalized equation.

A Möbius transformation of a Papperitz symbol moves the singular points without changing their exponents. Multiplication of the dependent variable by a local power shifts the two exponents by that power, using the reciprocal local coordinate at infinity.

Set $w=z/(z-1)$ and $v(w)=F(a,c-b,c;w)$. At $w=0,1,\infty$, the exponent pairs for $v$ are respectively $(0,1-c)$, $(0,b-a)$, and $(a,c-b)$. The Möbius transformation sends $z=0,1,\infty$ to $w=0,\infty ,1$. Before rescaling, $v(w(z))$ therefore has exponent pairs $(0,1-c)$ at zero, $(a,c-b)$ at one, and $(0,b-a)$ at infinity. Multiplication by $(1-z{)}^{-a}$ subtracts $a$ at one and adds $a$ at infinity. The resulting pairs areexactly those of the original Gauss hypergeometric equation. Near zero choose the branch with $(1-z{)}^{-a}=1$ at zero. The resulting solution is analytic there and has value one. Uniqueness of the normalized analytic solution yieldsThis is the Pfaff transformation. The normalization presupposes the usual admissible parameters, in particular $c\notin \{0,-1,-2,\dots \}$. Elsewhere the identity holds by analytic continuation with compatible branches, or by parameter limits wherever those are defined.

This is a generalized linear model with independent Bernoulli distributions for the responses and the logit link. Under treatment coding with colour one as the reference level, the linear predictor isColour is a categorical variable, not a numerical trend. Width has the same slope for every colour, since no interaction term is fitted. The parameters are estimated by maximum likelihood estimation.

The approximate Wald confidence interval for the width coefficient is $\boxed{0.46\pm 1.96(0.106)}$, or approximately $(0.2522,0.6678)$. This uses the approximate normal distribution of the maximum-likelihood estimator, rather than a confidence interval for an individual binary response.

For the requested crab, the fitted linear predictor is $\hat{\eta }=-11.38-0.22+0.46(20)=-2.40$. Applying the inverse logit link gives $\boxed{\hat{p}=(1+e^{2.40}{)}^{-1}\simeq 0.0832}$. The rounded displayed coefficients need not reproduce the reported test statistics exactly.

Work over the finite field ${\mathbb{F}}_2$. Reduce polynomials modulo $X_i^2-X_i$, so their monomials are square-free. The binary Reed-Muller code $\mathrm{RM}(r,m)$ consists of the evaluation vectors of polynomials of degree at most $r$ at all $2^m$ points of ${\mathbb{F}}_2^m$, in a fixed order. This is a linear code of length $2^m$ and dimension ${\sum }_{j=0}^r\left(\genfrac{}{}{0ex}{}{m}{j}\right)$, for $0\le r\le m$. Square-free monomials give independent evaluation functions, as follows inductively by splitting according to the last variable.

The minimum Hamming distance of a linear code equals its least nonzero Hamming weight. The result isHere $r=0$ is the repetition code, while $r=m$ is the full binary vector space and has distance one. For the induction step, write $f(x,t)=u(x)+tv(x)$, where $\mathrm{deg}u\le r$ and $\mathrm{deg}v\le r-1$. Its evaluation vector is $(u,u+v)$, the Reed-Muller bar-product recursion. If $v=0$, its nonzero Hamming weight is at least $2d(r,m-1)=2^{m-r}$, with the full-space endpoint treated as above. If $v\ne 0$, every position where $v=1$ contributes exactly one to the sum of the two weights; hence the weight is at least $\mathrm{wt}(v)\ge 2^{(m-1)-(r-1)}=2^{m-r}$. The monomial $X_1\cdots X_r$ is one exactly on $2^{m-r}$ points, attaining the bound. Degrees greater than $m$ give the same full code; the zero code at negative degree has no nonzero word whose weight could define a finite minimum.