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After division by $z$, the coefficients are $P=2/z-1$ and $Q=-1/z$. At $0$, $zP$ and $z^{2}Q$ are analytic functions, so $0$ is a regular singular point. Its indicial equation is $r(r-1)+2r=r(r+1)=0$. There are no other finite singular points of a differential equation. At infinity the regular singular point criterion for a second-order equation would require $P=O(z^{-1})$, but $P\to -1$; hence infinity is an irregular singular point.

Substituting $y=f/z$ reduces the equation to $f^{\prime \prime }-f^{\prime }=0$, soA linearly independent pair is $1/z$ and $(e^z-1)/z$. Their Wronskian is $e^z/z^2\ne 0$. The second has a removable singularity at $0$ and value $1$, whereas the first has a simple pole, agreeing with the two characteristic exponents at a regular singular point, $0,-1$. The integer difference does not require a logarithm. At infinity $e^z/z$ has an essential singularity, consistent with the irregular singular point; the existence of the simpler solution $1/z$ does not make infinity regular.

Expand the finite Euler product into geometric series. By unique prime factorization, it is the sum of $1/n$ over positive integers whose prime factors do not exceed $x$, and hence contains every term up to $m=\lfloor x\rfloor$. The harmonic sum satisfiesThe last inequality uses $m+1>x$, including when $x$ is an integer. For each prime number $p$,The telescoping series over all integers $n\ge 2$ bounds the sum of these remainders by $\frac{1}{2}{\sum }_{n\ge 2}[n(n-1){]}^{-1}=1/2$. Taking logarithms of the product bound therefore givesThis explicit constant suffices; no asymptotic theorem about the distribution of prime numbers is needed.

Let $g=a{U}_0\gg 1$ and $z=ka$. A denominator zero away from the removable zero at $z=0$ obeysNear $z=\pi$, write $z=\pi +\delta$ and choose the logarithm branch with $\delta \to 0$. Expanding givesConsequentlyThe lower-half-plane pole is a scattering resonance: an odd standing wave is nearly confined between the two high repulsive barriers but leaks out slowly. If $E=E_R-i\Gamma /2={\hslash }^2{k}^2/(2m)$, then $\Gamma \simeq 2{\hslash }^2\pi \gamma /(m{a}^2)$ and its probability lifetime is $\hslash /\Gamma$. It is not a genuine bound state of the repulsive potential.

All entropies and information rates below use base-two logarithms. The quantum fidelity uses the unsquared convention.

Use base-two logarithms. The weakly typical sequence definition isEquivalently, $2^{-n(H(X)+\epsilon )}\le p(x^n)\le 2^{-n(H(X)-\epsilon )}$. By the weak law of large numbers applied to the independent and identically distributed random variables $-{\log }_{2}p(X_j)$, the probability of this typical set tends to one.

This captures the usual probability scale of a long sample, but it need not make every symbol frequency representative. For a fair binary source, every word has probability $2^{-n}$ and $H(X)=1$, so even the all-zero word is weakly typical. The strongly typical sequence definition additionally controls empirical symbol frequencies and excludes this word for small tolerance. Thus weak typicality agrees with the high-probability-set intuition, while allowing individually unrepresentative members.

There are two positive roots for $u>1$: the derivative of $x{e}^{1/x}$ is $e^{1/x}(1-1/x)$, so the function decreases to its minimum $e$ at $x=1$ and then increases. Negative $x$ cannot solve the equation. For the larger root, expansion of the exponential function in $1/x$ givesso series reversion yieldsFor the smaller root, set $y=1/x$ and take a logarithm. Then $y-\log y=u$, so successive substitution gives $y=u+\log u+O(\log u/u)$ andThese are the first two nonzero terms of the two asymptotic expansions. Equivalently, the roots are $-1/W_0(-e^{-u})$ and $-1/W_{-1}(-e^{-u})$, from the real branches of Lambert W.

Apply the Jensen inequality to the convex exponential function under the trial equilibrium measure:The given identity therefore implies $e^{-F}\ge e^{-F_0-⟨H-H_0{⟩}_0}$. Taking minus the logarithm reverses the inequality and proves the Feynman-Bogoliubov inequalityThe units $\beta =1$ make $H$ and $F$ dimensionless here. Equality holds when $H-H_0$ is constant on the support of the trial measure. Optimizing the trial parameters tightens an upper bound, which is the basis of the Gaussian variational approximation.

On either open half-line, part (i) says $\hat{u}=-v^{\prime }$. Differentiating the formula in part (iii) makes all elementary terms cancel except the logarithm and the bracket. ThereforewhereandwhereBoth functions are smooth on their respective half-lines, proving the required assertion.

Put $L=L^{\ast }(X)$ and $\mu =\mathbb{E}L$. Since $L$ is a function of $X$, the chain rule for information entropy and part (c) givePart (a), applied to the nonnegative integer-valued random variable $L$, yieldsThe logarithm inequality $\log (1+t)\le t\log e$ implies $\mu \log (1+1/\mu )\le \log e$. ThusPart (c) also gives $\mu \le H(X)$, so monotonicity of the logarithm lets us replace $\log (1+\mu )$ by $\log (H(X)+1)$. Rearranging proves

The function $y_1=x$ solves the linear ordinary differential equation. Apply reduction of order by writing $y=xv(x)$. Substituting $y^{\prime }=v+x{v}^{\prime }$ and $y^{\prime \prime }=2v^{\prime }+x{v}^{\prime \prime }$ and cancelling the terms in $v$ givesFor $w=v^{\prime }$, this becomes a first-order linear ordinary differential equation:A partial fraction decomposition provides the required antiderivative:ThereforeThe logarithm has positive argument on the required interval. The Wronskian of $x$ and $1+x\log ((x-1)/x)$ is $1/(x-1)\ne 0$, proving their linear independence and hence completeness of this general solution.

Put $u_n=11^n\log n/(2^{2n+1}n!)$ for $n\ge 2$. The factorial givesThe logarithm identityfollows from continuity of the logarithm at one and $\log n\to +\infty$. The product rule for limits of sequences therefore makes the displayed ratio tend to zero. In particular, some $N$ satisfies $0<u_{n+1}\le u_n/2$ for all $n\ge N$. By mathematical induction, $0<u_{N+j}\le u_N{2}^{-j}$. A geometric sequence with ratio in $(0,1)$ tends to zero, so the squeeze theorem proves

The endpoint radii are $5$ and $13$, but substituting those two numbers in the finite conical frustum formula is invalid. A rotation about the $z$-axis preserves the horizontal angle between the endpoint position vectors; here $(3,4)$ and $(12,5)$ are not parallel. The radius along the generating segment is not linear in height.

Parametrize the actual segment byIts rotated circle at height $1+6t$ has radiusThus the rotated image is the surface of revolution $\mathbf{R}(t,\theta )=(\rho (t)\cos \theta ,\rho (t)\sin \theta ,1+6t)$. Applying the conical frustum formula to infinitesimal pieces of this radial profile, or directly using the cross product as in ii, gives the surface area of a surface of revolutionTo evaluate it explicitly, put $u=82t+31$. Then $9676t^2+7316t+1861=(59u^2+19602)/41$. DefineThe chain rule gives $G^{\prime }(u)=2\sqrt{59u^2+19602}$, so the answer isThe positive logarithm arguments make this a real-valued expression. Rotating the segment's endpoints independently into one vertical plane before joining them would instead create a different surface, with surface area $180\pi \simeq 565.4867$.

Separation of variables givesHere $x>1$ makes the argument of the second logarithm positive. Substituting the initial condition yields $3y^2+1=4(x^2-1{)}^3$. The positive branch is thereforeThis is the maximal connected interval containing $\sqrt{2}$ on which this branch solves the original differential equation. At its left endpoint $y=0$, the original right-hand side has a zero denominator, so one cannot extend the solution through that point merely by continuing the squared equation.

Use the real logarithm near zero and subtract its two Taylor series:Thus the first three nonzero terms are

Divide by $\cos x$ on an interval where it is nonzero. For the resulting first-order linear differential equation, use the integrating factor $M=\sec x+\tan x$, since $M^{\prime }=\sec xM$. The right-hand side after division is also $M^{\prime }$, soThe identity $(\sec x+\tan x)(\sec x-\tan x)=1$ shows $M$ never vanishes on the interval, givingAn integrating factor may have either fixed sign; if obtained by integrating $\sec x$, the real logarithm is $\ln |\sec x+\tan x|$.

For fixed real $a$, the base is positive for all sufficiently large positive $x$. Take its logarithm and use the Taylor expansion $\ln (1-u)=-u+O(u^2)$:Continuity of the exponential function then gives $\boxed{\underset{x\to \infty }{\lim }(1-a/x{)}^x=e^{-a}}$. This includes $a=0$.

Put $u=\ln x$, so $du=dx/x$. The integral becomes $\int du/u$, givingFor the real logarithm, choose an interval in $0<x<1$ or $x>1$; the original integrand is singular at one and undefined for nonpositive $x$.

This is a separable differential equation. Multiply by $e^{-y}$ and integrate:The initial condition gives $C=-1$. HenceFor a real solution the outer logarithm requires a positive argument. The maximal interval containing the initial point is therefore $\boxed{|x|<\sqrt{e^2-1}}$. At either endpoint $y$ diverges, so the real solution cannot continue through it with a finite value.

The even-power formula reduces every summand to a scalar multiple of $I$:Both series have absolute convergence. Apply the logarithm expansion ${\sum }_{n\ge 1}{z}^n/n=-\log (1-z)$ for $|z|<1$ to obtainThus the off-diagonal entries vanish and both diagonal entries have the required value.

The argument $3x^3+2x$ is odd, so its squared sine is even. Replacing $x$ by $-x$ interchanges the numerator and denominator inside the logarithm, changing that factor's sign. Thus the full integrand is an odd function. It is continuous on the symmetric interval because $|x|\le 1/\pi <1$. Therefore