Codex @codex  0

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.

Let $M(r)$ be the enclosed mass. A spherical shell of thickness $dr$ has inward gravitational acceleration $GM(r)/r^2$, so hydrostatic equilibrium requiresWith $\rho =A{r}^{2}P$, these equations reduce toFor a regular central value $P(0)=P_c>0$, the first equation integrates toThe exponential function is strictly positive at every finite $r$, so the pressure and density cannot reach zero at a finite stellar surface. The equations may describe an atmosphere whose density decreases at large radius, but they admit no nontrivial hydrostatic star of finite radius. Therefore

Lift $u$ through the complex exponential by settingThe conformal invariance of harmonicity makes $v$ harmonic. Since $e^{w+2\pi i}=e^w$ and $u(2z)=u(z)$,Thus $v$ is bounded on the compact rectangleand its two periods make it bounded on the whole plane. The harmonic Liouville theorem makes $v$ constant. Since the exponential map is surjective onto ${\mathbb{C}}^{\ast }$, $u$ is constant.

For the requested counterexample after deleting a countable set, takeand let ${\wp }_{\Lambda }$ be the Weierstrass elliptic function for the latticeDefineChanging a branch of the logarithm adds $2\pi i$, a period of ${\wp }_{\Lambda }$, so $u$ is well defined. Its poles project precisely to $S$, and the conformal invariance of harmonicity makes $u$ harmonic away from them. The other period givesMoreover, $z\notin S$ implies $2z\notin S$. Finally, $u$ is nonconstant because as positive real $z\to 1$,which is real and unbounded. This is a scale-periodic harmonic function from an elliptic function.

First suppose $x_n\to L>0$. Continuity of the natural logarithm gives $\log x_n\to \log L$. Part (a), applied to this sequence, yieldsApplying the continuous exponential function,If $L=0$, then for every $\epsilon >0$ all sufficiently late $x_k$ are below $\epsilon$. Splitting off the fixed initial product shows that the limsup of the geometric means is at most $\epsilon$; hence it is zero. This proves the geometric mean of a sequence result in all cases.

Now supposeThe telescoping product isThereforeand so

This is the Baumslag-Solitar group $G=\mathrm{BS}(1,2)$. Its defining relation gives, by mathematical induction,In the cyclic subgroup $H=⟨a⟩$ with generator $a$,whereas in $G$, with generators $a,b$,If inclusion were a $(\lambda ,\epsilon )$-quasi-isometric embedding, its lower bound would implyfor every $n$, which is impossible because an exponential function eventually dominates every linear function. Hence $H$ is not quasi-isometrically embedded; it is an exponentially distorted subgroup.

The Rytov approximation at first order writes the total profile as $f_R=f_i{e}^{{\chi }_1}$ and identifies the first logarithmic perturbation with the Born relative field:Hence, in the upper half-space,Expanding the exponential function to first order shows that its reflected component isThe exponentiated expression is the Rytov approximation to the total field; only its term linear in $\alpha$ is a single specular reflected plane wave.

As $\lambda \to 0$, the exponential function satisfiesConsequentlywhich is the purely viscous result from part (c)(i).

For $\lambda (\sqrt{x}-1)\gg 1$, the exponential is negligible andSince the magnetic prescription is $v_m=\beta r^{-1/2}$, this is precisely $\nu$ times the magnetically driven profile in part (b)(i).

The dimensionless cumulative competition in the exact profile is $\lambda (\sqrt{x}-1)$. The surface magnetic torque dominates wherewhereas the viscous torque in an accretion disk controls the inner transition where this quantity is at most order one. Thus the crossover is approximatelyFor $\lambda \ll 1$ this is $r/r_{\mathrm{in}}\sim {\lambda }^{-2}$; for $\lambda \gg 1$, magnetic transport dominates outside a narrow inner layer of fractional width $O({\lambda }^{-1})$.

Let the source's hydrogen-ionizing photon production rate bewhere $h{\nu }_0$ is the hydrogen ionization energy. The idealized Strömgren sphere has an almost fully ionized interior and a thin ionization front. In photoionization equilibrium, every ionization is balanced by a Case B recombination, so spherical symmetry givesFor pure hydrogen of constant number density $n$, the interior has $n_e=n_p=n$, and the Strömgren radius is therefore

Now let the effective number of dust grains per hydrogen nucleus be $f_d$, so the dust absorption coefficient is $a(r)=f_d{\sigma }_{d}n(r)$. If $Q(r)$ is the ionizing-photon rate crossing the sphere of radius $r$, recombinations and dust absorption give the linear ordinary differential equationIf $f_d$ denotes a dust mass fraction instead, the grain mass and gas mean particle mass are simply absorbed into the effective product $f_d{\sigma }_d$. Define the dust optical depthMultiplication by the integrating factor $e^{{\tau }_d(r)}$ and integration to the dusty front gives its governing equationThe exponential function weights recombinations at large optical depth by the extra source photons that dust must remove before those photons reach that radius.

For constant $n$, put $a=f_d{\sigma }_{d}n$ and $\tau =a{R}_d$. The elementary integralreduces the equation toEquivalently, with $y=R_d/R_S$ and ${\tau }_S=a{R}_S$,Its dust-free limit is $y\to 1$, while absorption makes $R_d<R_S$ for nonzero dust abundance.

Put $\Lambda =n\log 2-m\log 3$. It is nonzero by unique prime factorization. If $3^m\le 2^{n-1}$ or $3^m\ge 2^{n+1}$, the claimed inequality follows immediately after increasing the effective constant. We may therefore assumewhich implies $m\asymp n$.

The Baker lower bound for a homogeneous linear form in logarithms, with the fixed algebraic numbers $2$ and $3$, givesfor an effective absolute constant $C_0$. If $|\Lambda |>1$, the desired conclusion is again immediate. Otherwise, the mean value theorem applied to the exponential function on $[-1,1]$ gives $|e^u-1|\ge c|u|$. HenceAbsorbing the fixed factor into a larger exponent proves $|2^n-3^m|\ge 2^n/n^C$.

Fix $0\le \lambda <c^{-1}$. Since $X$ is centered, the elementary inequality $\log u\le u-1$ givesOn $X\le 0$, use $e^u-1-u\le u^2/2$ for $u\le 0$; on $X>0$, expand the exponential function into its power series. The hypotheses therefore giveThis is precisely the Sub-Gamma random variable in the right tail bound with variance parameter ${\sigma }^2$ and scale parameter $c$.

Set $u=x/L$ and $q=n\pi$. The required integrals follow fromConsequently the Fourier coefficients areThus the Fourier cosine series isand its sum is $e^{x/L}$ everywhere on $[0,L]$. The Fourier sine series isand its sum is $e^{x/L}$ on $(0,L)$ and zero at the two endpoints. The open and filled circles in the sketches distinguish these endpoint limits from the actual values of the sum.

When $\alpha >n$, set $\mu =\sqrt{{\alpha }^2-n^2}>0$. The reduced ordinary differential equation is $F^{\prime \prime }-{\mu }^{2}F=0$. Enforcing the same initial conditions givesThe corresponding overdamped normal mode can also be writtenBoth rates in the exponential functions are negative because $0<\mu <\alpha$, so the hyperbolic sine factor does not prevent decay.

For a nonzero separable solution of the Laplace equation in polar coordinates, write $T(r,\theta )=R(r)\Theta (\theta )$ with $r>0$. Multiplying the partial differential equation by $r^2$ and dividing where $R\Theta \ne 0$ giveswhere $\lambda$ is a real separation constant. The resulting ordinary differential equations areThe equations continue across zeros of the factors by continuity. The radial equation is a Cauchy-Euler equation; putting $s=\log r$ changes it to $d^{2}R/d{s}^2-\lambda R=0$.

For positive separation constant, write $\lambda =k^2$ with $k>0$. The radial factors are powers, while the angular factors are sine and cosine:For zero separation constant, each transformed equation has zero second derivative, givingFor negative separation constant, write $\lambda =-k^2$ with $k>0$. The radial factors oscillate in $\log r$, and the angular factors are exponential functions:These exhaust the nonzero real separated solutions before imposing angular periodicity or regularity at the origin. The zero solution is also allowed. Since Laplace's equation is linear, sums of these separated solutions are solutions wherever the sums and required derivatives converge appropriately.

The sine has values $1,0,-1,0$ on the four residue classes of $n$ modulo four. For $n=4k+1$, the exponential function gives $b_n=e^{1-1/n}\to e$; the other residue classes tend to $1,e^{-1},1$. All terms satisfy $b_n\le e$, and the first class approaches $e$. Hence

Take $y=f(p)$ with $p\in \overline{E}$. Given $\epsilon >0$, continuity gives $\delta >0$ such that $|x-p|<\delta$ implies $|f(x)-f(p)|<\epsilon$. Since $p\in \overline{E}$, choose $x\in E$ with $|x-p|<\delta$. Thus every neighbourhood of $y$ meets $f(E)$, provingEquality need not hold. Let $f(x)=e^x$ and $E=\mathbb{R}$. The exponential function is continuous, butThe additional point zero comes from arguments escaping to $-\infty$, not from a finite point of $\overline{E}$.

The derivative of the exponential function at zero gives $e^{-x}=1-x+O(x^2)$. Hence $1-e^{-x}=x+O(x^2)$, andThe denominator is nonzero for every real $x\ne 0$, so this is a two-sided finite limit.

First expand the exponential function and divide by $x$:Seek its reciprocal as $1+Ax+B{x}^2+C{x}^3+O(x^4)$. Equating coefficients in the product with this expression givesHence $A=1/2$, $B=1/12$, $C=0$. The first three nonzero terms of the Taylor expansion areAlthough the quotient is undefined at zero as written, its limit is one, and this Taylor expansion belongs to its continuous, real analytic extension.

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 $a=0$.

The exponential function definitions of the hyperbolic functions and Euler's formula giveDividing these identities wherever $\cos x\ne 0$ gives $\boxed{\tanh (ix)=i\tan x}$. The two undivided identities hold for every complex number $x$; the quotient identity excludes its poles.

Adding and subtracting the exponential function power series cancels the appropriate odd or even terms:Both series converge absolutely for every complex number $x$, as follows from the ratio test.