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For $ϵ\to 0^{+}$, the Taylor expansions needed in the exact energy density areSubstitution yieldsThereforeHere Big O notation describes the neglected small-coupling terms, and the logarithm is natural. The nonanalytic ${ϵ}^2\ln ϵ$ term comes from the formerly degenerate quantum states near the Brillouin zone boundary; a uniform nondegenerate perturbation theory expansion of the integrand would fail there. This is the same mechanism as the Peierls energy gain.

Eliminating the phase velocity from the first two relations yieldsWith $\theta =mh$, putThen $\ell =(c_1/(c_{2}h))\sqrt{a^2{k}^2-{\theta }^2}$. The dispersion relation becomesHere $a>0$ follows from $c_1<c_2$. The parameter $a$ has dimensions of length; $A$ and $ϵ$ are dimensionless.

For the strict mode count for a rigid-backed elastic layer, considerThe right side is negative, so intersections occur only on the negative branches of the tangent,On each such branch $\tan \theta -g(\theta )$ is strictly increasing. If $p_n<A$, its limit at $p_n^{+}$ is negative infinity. At the other end of the available interval, it is positive: either $\theta \to n{\pi }^{-}$ with $n\pi <A$, or $\theta \to A^{-}$ with $A\le n\pi$. The intermediate value theorem and strict monotonicity give exactly one intersection. No such interval exists when $p_n\ge A$.

Thus the exact answer, including all cutoff values, isAway from $A=(n-1/2)\pi$, this equals the printed floor function expression $\lfloor A/\pi +1/2\rfloor$. At $A=(n-1/2)\pi$, the printed expression is one too large: the $n$th interval is empty and the limiting state has $\ell =0$, contradicting exponential decay. This exception is present in the native paper, rather than introduced by transcription.

Now hold $A$ fixed and let $ϵ\to 0^{+}$. Neglecting the right side gives ${\theta }_n\simeq n\pi$ and hence the candidate phase velocitiesThe zero $\theta =0$ is excluded: it would require $c=c_1$, already ruled out. For every $n\pi <A$, the candidate is a valid leading asymptotic expansion. A first correction, obtained by expanding the tangent about $n\pi$, isThe big O notation here assumes a fixed positive gap between $n\pi$ and $A$. In particular, when $A>N\pi$, all $N$ leading approximations are valid. The phase velocity increases strictly with $\theta$, so only the last branch can cause an endpoint difficulty.

If $(N-1/2)\pi <A<N\pi$, the formal last equation root $N\pi$ lies outside the allowed interval. The actual last equation root approaches $A$, where the small factor $ϵ$ is balanced by the vanishing denominator. This is a singular perturbation. For fixed $A$ strictly inside that interval, write ${\theta }_N=A-\delta$. Since $\tan A<0$,The valid endpoint approximation for a rigid-backed shear mode consequently givesHere we used $c^2=c_1^2(1+{\theta }^2/(kh{)}^2)$ and $A^2/(kh{)}^2=c_2^2/c_1^2-1$. Thus this normal mode approaches the upper wave speed from below; it does not approach the inadmissible value obtained from $\theta =N\pi >A$.

There is a further exceptional fixed value $A=N\pi$. Put ${\theta }_N=A-\delta$ again. Now $\tan (A-\delta )\sim -\delta$, so the balance changes toIn the last formula the bracket specifies the first nonzero correction. The zeroth-order candidate $c_N=c_2$ is correct as a limit but misses this correction and the positive exterior exponential decay rate. More generally, close to $A=N\pi$, let $\Delta =A-N\pi$ and ${\theta }_N=N\pi -\delta$. For $|\Delta |,\delta \ll 1$, a transition approximation valid on both sides isIt recovers the ${ϵ}^{2/3}$ balance at $\Delta =0$ and the ordinary expansion above the transition. The fixed-endpoint formula involving $\tan A$ should be used when $A$ is below, and bounded away from, $N\pi$.

Take the ideal electric dipole limit with a fixed smooth electric dipole moment $\mathbf{p}(t)$, so that ${\mathbf{r}}_0(t)=\mathbf{p}(t)/(2q)$; its required time derivatives are bounded on the time interval under consideration. Fix $r>0$ and put $\mathbf{n}=\mathbf{r}/r$, $t_0=t-r/c$. For $s=+1$ or $-1$, the source position is $s\mathbf{p}(t)/(2q)$. A Taylor expansion of the distance givesSubstitute this into the defining equation for retarded time:Boundedness first implies $t_r^s-t_0=O(q^{-1})$. The Taylor expansion $\mathbf{p}(t_r^s)=\mathbf{p}(t_0)+O(q^{-1})$ then changes the displayed correction only by $O(q^{-2})$. HenceThe nearer positive displacement reduces the propagation distance and consequently increases the retarded time. The Big O notation here is for $q\to \infty$ at a fixed observation event away from the origin. Keeping a fixed smooth $\mathbf{p}$ is essential: bounded amplitude alone, with arbitrarily increasing time frequency as $q$ varies, would not justify this Taylor expansion.

For every fixed $x>0$, $\log (x+n)/n\to 0$, so $f_n(x)\to 0$. To exchange the limit with the Lebesgue integral, we need a common integrable bound over the whole half-line.

For $n\ge 1$, $x+n\le n(1+x)$, soTogether with $|\sin x|\le \min (x,1)$, this givesFor $0<x\le 1$, $\log (1+x)\le x\le 1$, hence $G(x)\le 2x^{1-q}$, which is integrable because $q<2$. For $x\ge 1$,Both terms are integrable because $q>1$:The second equality follows from integration by parts, or the substitution $x=e^t$. Consequently $G$ is a Lebesgue integrable function, and every $f_n$ is Lebesgue integrable.

The dominated convergence theorem states that almost everywhere convergence of measurable functions under one integrable absolute-value bound implies L1 convergence, and in particular convergence of their integrals. Here the pointwise limit is zero and the dominator is $G$, soIn fact we obtain the stronger conclusion $\Vert f_n{\Vert }_1\to 0$. A quantitative version follows by retaining the factors $1/n$ in the bound:The preceding endpoint estimates show $A,B<\infty$, so the L1 norm is $O(\log n/n)$ as $n\to \infty$, with $O$ denoting Big O notation. This is logarithmic reweighting under dominated convergence.

Rearranging the outgoing radial null geodesics of the Vaidya metric equation along $r=r_0(v)$ givesFor the prescribed parametrization by $w$, the chain rule yieldsSince $n>1$, $v/w=1+2\beta w^{n-1}\to 1$ as $w\to 0^{+}$. The asymptotic equivalence $w\sim v$ therefore impliesMore precisely $w=v+O(v^n)$ and $M(v)=(\beta /2)v^n[1+O(v^{n-1})]$, using Big O notation. The positivity of $dv/dw$ justifies treating $w$ as a single-valued function of $v$ locally.

Use the positive-exponent quantum Fourier transform convention $F_N|x⟩=N^{-1/2}{\sum }_t{e}^{2\pi ixt/N}|t⟩$. The period of a function here is the least positive translation leaving the function unchanged on the cyclic group ${\mathbb{Z}}_N$. The injective promise means $f(x)=f(x^{\prime })$ exactly when $x\equiv x^{\prime }(\mathrm{mod}r)$. In particular $r\mid N$: the subgroup of period translations is generated by $r$, and wrapping the argument by $N$ preserves its value. This is exact quantum period finding on a finite cyclic group, rather than the truncated, nondividing-period version requiring continued fractions.

Prepare a uniform superposition state in the first quantum register using $F_N|0⟩$, and perform a reversible computation of the function into a second quantum register, cleaning its workspace by uncomputation:A quantum measurement in the computational basis of the second quantum register leaves a periodic coset state in the first:For the quantum Fourier transform of this periodic coset state, the finite geometric series vanishes unless the measured label is $t=mN/r$. On those labels its normalized amplitude is $r^{-1/2}{e}^{2\pi iam/r}$. Thus $m$ has the uniform distribution on a finite set on $\{0,\dots ,r-1\}$, irrespective of $a$.

Compute the greatest common divisor by the Euclidean algorithm, and set $q=N/\gcd (t,N)$, with $\gcd (0,N)=N$. Since $t=mN/r$,The exact period recovery from a Fourier sample is therefore certified by testing $f(q\mathrm{mod}N)=f(0)$. Every candidate $q$ divides $r$ and lies between $1$ and $r$, so the injective promise makes that equality equivalent to $q=r$. This includes $r=N$, where the test uses input $0$, and $r=1$, which always succeeds. The test does not assert that an arbitrary equality of function values certifies an arbitrary period; the Fourier-derived divisor restriction is essential.

The concise result isHere $\phi$ is the Euler totient function, with $\phi (1)=1$. The totient lower bound from the Mertens product gives $\phi (r)/r=\Omega (1/\max \{1,\log \log N\})$ for $r\le N$. Each trial uses polynomial time quantum Fourier transforms, function evaluations and Euclidean algorithm arithmetic in $\log N$. Repeating $O(\max \{1,\log \log N\})$ independent trials gives constant success probability; the herald accepts only correct answers.

The printed asymptotic bound has the wrong direction for this argument. In Big O notation, the stated upper bound on the number of coprime integers supplies no success lower bound and is itself false uniformly: for prime $N$, $\phi (N)=N-1$, not $O(N/\log \log N)$. Likewise the single-trial success need not be $O(1/\log \log N)$, since it tends to one for $r=N$ prime. The meaningful corrected guarantee is the lower bound above. If an upper-bounded success rate is literally required, one may deliberately retain an accepted result with an independent probability proportional to $1/\max \{1,\log \log N\}$; this discards useful successes and is unnecessary for efficient recovery.

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

With $X_1<\cdots <X_n$, it suffices to put the threshold between $X_m$ and $X_{m+1}$ for $m=1,\dots ,n-1$. Form the prefix sumsThey take $O(n)$ operations to compute. The minimized squared errors for the two children areandEach candidate value $L_m+R_m$ therefore takes constant time once the prefix sums are available. Scanning all $n-1$ candidates takes $O(n)$ operations in the sense of Big O notation, and retaining the minimizing $m$ gives the optimal split.

The Big O notation means that there exist constants $C>0$ and $\delta >0$ such thatwithin the domain of $f$. Thus $f(x)/x^2$ stays bounded in a punctured neighbourhood of zero. It need not have a limit: $f(x)=x^2\sin (1/x)$ is one example. The statement imposes no condition on a separately assigned value $f(0)$.

Write $t_j=t^{(j)}(0)$ and $s=1+t_0>0$. The chain rule givesThe condition $t_3{t}_1<0$ ensures $t_1\ne 0$. Since $t_0\ne 0$, also $s\ne 1$ and $\ln s\ne 0$. Hence the first two nonzero terms are the constant and linear terms:The inequality $s>0$, together with continuity of $t$, makes the logarithm defined on a neighbourhood of zero. The remainder is written in Big O notation.

The Maclaurin series $\sinh t=t+t^3/3!+t^5/5!+O(t^7)$, applied with $t=x^2$ and then multiplied by $x$, givesThese are the first three nonzero terms. The remainder is written in Big O notation.

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.

Set $\ell =\ln 2$. Applying the exponential series and the Maclaurin series of the sine givesMultiply the two power series, retaining every contribution through degree four:Since $0<\ln 2<1$, the coefficient of $x^4$ is nonzero, so these are indeed the first two nonzero terms. The remainder is in Big O notation.

Use the exponential series to obtain the Maclaurin series of the hyperbolic sine and hyperbolic cosine:Dividing these power series gives the hyperbolic tangent expansionThe displayed remainder uses Big O notation as $x\to 0$.

Write $F(x)=\ln (e^{ax}-b)$. Direct differentiation, using the chain rule, givesThe stated parameter restrictions ensure that the logarithm is defined near $x=1$, and that $F^{\prime }(1)>0$ and $F^{\prime \prime }(1)<0$. With $h=x-1$, the Taylor series beginsIf $e^a-b\ne 1$, the constant term is nonzero, and the first two nonzero terms areThis covers both $a>\ln (1+b)$ and $0<a<\ln (1+b)$; the constant term is respectively positive or negative.

If $a=\ln (1+b)$, which is possible precisely with $0<a<\ln 2$ and $b=e^a-1$, the constant term vanishes. The first two nonzero terms are thenNo further case is needed, since the linear and quadratic coefficients cannot vanish under the given restrictions. All remainders use Big O notation.

For $n>1$, the Taylor series of the natural logarithm converges absolutely at $1/n$:Multiply this power series by $n+1/2$ and subtract $1$. The constant term cancels the subtracted $1$, and the coefficient of $1/n$ is $-1/2+1/2=0$. For $k\ge 2$, the coefficient of $n^{-k}$ isHenceIn particular $u_n=1/(12n^2)-1/(12n^3)+O(n^{-4})$, in Big O notation.

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 $-1/2$.

Let $K(t)=e^{tR}S{e}^{-tR}$. The product rule and differentiation of the matrix exponential givewhere brackets are the operator commutator. Evaluating at zero gives $K(0)=S$, $K^{\prime }(0)=[R,S]$ and $K^{\prime \prime }(0)=[R,[R,S]]$. The Taylor series at zero is consequentlyIn particular $[R,[R,S]]=R^{2}S-2RSR+S{R}^2$, which also follows by multiplying the two matrix exponential series through quadratic order. For fixed finite matrices the remainder has the stated Big O notation in any matrix norm.

For the intended finite representation, put $m=|S|<\infty$. For each $T\subseteq S$, form the mintermA Boolean valuation satisfies ${\chi }_T$ exactly when its selected subset is $T$. Therefore the weighted formula utility representationrepresents $u$ exactly: at the Boolean valuation for $T$, only that minterm contributes, with weight $u(T)$. This proves completeness and the bound of $2^{|S|}$ rules simultaneously. Zero-weight rules may be omitted. The bound counts rules; writing each minterm also requires $O(m)$ symbols in Big O notation.

For the computational question, assume rule weights are rational numbers encoded in binary. This makes exact equality of sums a finite computational task. With this conventional encoding, equivalence of weighted formula utility representations is co-NP-complete.

For membership, the complement has a short witness: a Boolean valuation on the variables occurring in either rule base. Evaluate each propositional formula, sum the weights of satisfied rules, and accept the witness if the sums differ. Propositional formula evaluation takes polynomial time. Exact sums of binary rational numbers also have polynomially bounded bit lengths: the product of their denominators has bit length at most the sum of the input denominator lengths. Thus non-equivalence lies in NP, and equivalence lies in co-NP.

For hardness, take a Boolean satisfiability problem instance $\theta$ and constructThe first utility function is one at satisfying Boolean valuations and zero elsewhere, while the second is always zero. ConsequentlyThe unsatisfiable Boolean formula problem is co-NP-complete, as the complement of the NP-complete Boolean satisfiability problem. The construction is a polynomial-time reduction, proving hardness even with the single weight $1$.

Two qualifications matter in the literal wording. First, unrestricted real weights have no specified finite encoding, so a standard bit-complexity classification needs the encoding assumption above. Second, the problem says only that $S$ is nonempty. If infinite $S$ is allowed while rule bases remain finite, its completeness claim is false. For $S=\mathbb{N}$, take $u(T)=1$ when $T$ is infinite and $u(T)=0$ when $T$ is finite. A finite rule base mentions only a finite subset $F$ of the variables. The Boolean valuations for $\varnothing$ and $\mathbb{N}\setminus F$ agree on all those variables but have different required values. No such rule base represents this $u$. Thus the construction answers the intended finite-set question and identifies the missing hypothesis in the unrestricted statement.

Adjoin $\dot{\mu }=0$ and put $w=z+\mu$. The extended centre manifold is a graph $w=H(x,y,\mu )$ tangent to $w=0$ at the origin. The equations becomeHere $ϵ>0$ is fixed. The centre-manifold invariance equation, obtained by the chain rule, isWrite $H=H_2+O({\rho }^3)$, where $\rho =|x|+|y|+|\mu |$ and $H_2$ is a homogeneous quadratic polynomial. To quadratic order the centre-manifold invariance equation isThe terms $\mu x,\mu y,{\mu }^2$ have zero coefficients: their equations are homogeneous, and the operator $1+y{\partial }_x-x{\partial }_y$ is invertible on these quadratic polynomials. For $H_2=a{x}^2+bxy+c{y}^2$, comparison of coefficients givesHence $a=3/5$, $b=-2/5$ and $c=2/5$, givingSubstitution into the $y$ equation givesThe constant in the big O notation may depend on fixed $ϵ$. Thus the equations printed in part (b) are the retained terms of the local Taylor series, rather than an exact identity for the full reduced vector field.

For the Hopf bifurcation classification in part (a), set $\mu =0$ and use $x=r\cos \theta$, $y=-r\sin \theta$. The radial derivative is $\dot{r}=(x\dot{x}+y\dot{y})/r$. Its cubic angular mean isHere $⟨\cdot ⟩$ denotes the mean over one revolution. Removing the oscillatory cubic terms by a near-identity transformation gives this same cubic radial coefficient in the Hopf normal form. The linear radial coefficient is $5ϵ\mu /2$.