text:
\section{Proof of the fundamental sum rule~\eqref{eq:interpolation:main2}--\eqref{eq:interpolation:remainder}}
\label{proof:interpolation:main2}

Similar computations go back to \cite{FranzLeone} and were applied in Nishimori symmetric situations in \cite{Mon:2005,Mac:2007a,KudM:2009}, so we will
be relatively brief.
We compute $h_{T,1; \epsilon,\delta}$ and $\frac{dh_{t,s; \epsilon,\delta}}{ds}$ in \eqref{eq:interpolation:main1}.
From the definitions \eqref{eq:Hamiltonian_tse}, \eqref{eq:partitionFunction_tse}, \eqref{eq:htse}, and the identity $e^{\sigma x}=(1+\sigma\tanh x)\cosh x$ for $\sigma \in\{-1,+1\}$, we can expand $h_{T,1;\epsilon,\delta}$ as
\begin{align}
h_{T,1; \epsilon,\delta}

	= \mathbb{E}\Big[ & \ln\Big( \prod_{t'=1}^{T} \prod_{B=1}^{e_{i}^{(t')}} (1 + \tanh U_{B}^{(t')})
	+ e^{-2(H+\tilde{H})} \prod_{t'=1}^{T} \prod_{B=1}^{e_{i}^{(t')}} (1 - \tanh U_{B}^{(t')})\Big)
	\nonumber \\ &
	- \frac{\alpha K}{T} \sum_{t'=1}^{T} \ln(1+\tanh U^{(t')})\Big].
		\label{eq:h_T1}
\end{align}
Note that the first term is part of~\eqref{eq:RSFreeEntropy_coupled1}.
For $\frac{dh_{t,s; \epsilon,\delta}}{ds}$ we use the following property of the Poisson distribution: for any function $f(X)$ of a r.v. $X$ with Poisson distribution and mean $\nu$ we have
\begin{align}
\frac{d \,\mathbb{E}f(X)}{d \nu} = \mathbb{E} f(X+1) - \mathbb{E} f(X).
\label{eq:dPoissonMean}
\end{align}
This allows us to write
\begin{align}
\frac{dh_{t,s; \epsilon,\delta}}{ds} \-
	& = - \frac{\alpha }{T} \mathbb{E}_{t,s; \epsilon,\delta}\mathbb{E}_{B, \tilde{J}_B }  \ln \< e^{\tilde{J}_{B} (\sigma_{B}-1)} \>_{t,s; \epsilon,\delta}  \-
		+ \frac{\alpha K}{nT} \sum_{i=1}^{n} \mathbb{E}_{t,s; \epsilon,\delta}\mathbb{E}_{U_i^{(t)}} \ln \< e^{U_{i}^{(t)} (\sigma_{i}-1)} \>_{t,s; \epsilon,\delta}
		\label{eq:dh_ts}
\end{align}
where we distinguish the expectation $\mathbb{E}_{t,s; \epsilon,\delta}$ with respect to the original interpolating model with Hamiltonian \eqref{eq:Hamiltonian_tse}
and the expectation with respect to
an ``extra measurement'' and its neighborood $\mathbb{E}_{B, \tilde{J}_B }$ and an ``extra field'' $\mathbb{E}_{U_i^{(t)}}$.
Standard algebra, using again the identity $e^{\pm x}=(1\pm\tanh x)\cosh x$, leads to

\begin{align}
\mathbb{E}  \ln \< e^{\tilde{J}_{B} (\sigma_{B} -1)} \>_{t,s; \epsilon,\delta}
	& = \mathbb{E}_{t,s; \epsilon,\delta}\mathbb{E}_{B,\tilde{J}_B}
	 \ln \big( 1 + \langle \sigma_B \rangle_{t,s; \epsilon,\delta} \tanh \tilde{J}_B \big )
	- \mathbb{E}_{\tilde{J}_B} \ln(1+ \tanh \tilde J_B) \nonumber \\
	& =\sum_{p=1}^{\infty} \frac{(-1)^{p+1}}{p} \frac{1}{n^K} \sum_{i_1,\dots,i_K}
	\mathbb{E}[\langle \sigma_{i_1} \cdots \sigma_{i_K} \rangle_{t,s; \epsilon,\delta}^p]
	\mathbb{E}[(\tanh \tilde{J})^p]
	-\mathbb{E} \ln(1+ \tanh \tilde J)
	\nonumber
\end{align}
and similarly, using \eqref{eq:BP-update},
\begin{align}
	& \frac{1}{n} \sum_{i=1}^{n} \mathbb{E} \ln \< e^{U_{i}^{(t)} (\sigma_{i}-1)} \>_{t,s; \epsilon,\delta}  \nonumber \\
	 = ~&\frac{1}{n}\sum_{i=1}^{n} \mathbb{E}_{t,s; \epsilon,\delta}\mathbb{E}_{U_i^{(t)}}
	\ln \big ( 1 + \langle \sigma_i \rangle_{t,s; \epsilon,\delta} \tanh U_i^{(t)}  \big )
	-
	\frac{1}{n}\sum_{i=1}^{n} \mathbb{E}\ln(1+ \tanh U_i^{(t)})
	\nonumber \\
	=~& \frac{1}{n}\sum_{i=1}^{n} \mathbb{E}_{t,s; \epsilon,\delta}\mathbb{E}_{\tilde{J}, \underline{V}^{(t)} }
	 \ln\big( 1 + \langle \sigma_i \rangle_{t,s; \epsilon,\delta} \tanh \tilde{J} \prod_{j=1}^{K-1} \tanh V_j^{(t)} \big)
	-  \mathbb{E}\ln(1+ \tanh U^{(t)}) \nonumber \\
	=~& \sum_{p=1}^{\infty} \frac{(-1)^{p+1}}{p}  \mathbb{E}[(\tanh \tilde{J})^p]
	\mathbb{E} [ (\tanh V^{(t)})^p]^{K-1}\frac{1}{n}
	\sum_{i=1}^n \mathbb{E}[\langle \sigma_i \rangle_{t,s; \epsilon,\delta}^p]
	- \mathbb{E} \ln(1+ \tanh U^{(t)}). \nonumber
\end{align}
Recall $Q_{p} \equiv \frac{1}{n} \sum_{i} \sigma_i^{(1)} \cdots \sigma_i^{(p)}$ and thus
$$
\< Q_p \>_{t,s; \epsilon,\delta} = \frac{1}{n} \sum_{i} \langle \sigma_{i}\rangle_{t,s; \epsilon,\delta}^p,
\quad \< Q_p^K \>_{t,s; \epsilon,\delta} = \frac{1}{n^K} \sum_{i_1,\dots,i_K} \langle \sigma_{i_1} \ldots \sigma_{i_K} \rangle_{t,s; \epsilon,\delta}^p\,.
$$
Recall also $q_p^{(t)} \equiv \mathbb{E} [(\tanh V^{(t)})^p]$. Then \eqref{eq:dh_ts} becomes
\begin{align}
\frac{dh_{t,s; \epsilon,\delta}}{ds} \-
	 & =
	 - \frac{\alpha}{T} \sum_{p=1}^{\infty} \frac{(-1)^{p+1}}{p}
	\mathbb{E}[(\tanh \tilde{J})^p]
	\mathbb{E}[\langle Q_p^K \rangle_{t,s; \epsilon,\delta} - K (q_p^{(t)})^{K-1} \langle Q_p \rangle_{t,s; \epsilon,\delta}]
	\nonumber \\
	& \ + \frac{\alpha}{T}\mathbb{E}\ln (1 + \tanh\tilde J)
	- \frac{\alpha K}{T}\mathbb{E}\ln (1 + \tanh U^{(t)})
	\nonumber \\
	& = - \frac{\alpha}{T} \sum_{p=1}^{\infty} \frac{(-1)^{p+1}}{p} \mathbb{E}[( \tanh \tilde{J})^p]
	\mathbb{E} \big\langle Q_{p}^{K} - K (q_{p}^{(t)})^{K-1} ( Q_{p} - q_{p}^{(t)}) - (q_{p}^{(t)})^K \big\rangle_{t,s; \epsilon,\delta} \nonumber \\
	& \ + \frac{\alpha(K-1)}{T} \mathbb{E}\ln\big( 1 + \tanh \tilde{J} \prod_{j=1}^{K} \tanh V_j^{(t)}\big)
	+ \frac{\alpha}{T} \mathbb{E}\ln(1+ \tanh \tilde J)
	- \frac{\alpha K}{T} \mathbb{E}\ln(1+ \tanh U^{(t)}) .
	\label{eq:dh_st _final}
\end{align}

Substituting \eqref{eq:h_T1} and \eqref{eq:dh_st _final} into \eqref{eq:interpolation:main1} gives~\eqref{eq:interpolation:main2}, where
\begin{align}
\sR_{t,s; \epsilon,\delta} = \sum_{p=1}^{\infty} \frac{(-1)^{p+1}}{p} \mathbb{E}[ (\tanh \tilde{J})^{p} ]
\mathbb{E} \big\langle Q_{p}^{K} - K (q_{p}^{(t)})^{K-1} ( Q_{p} - q_{p}^{(t)}) - (q_{p}^{(t)})^K
\big\rangle_{t,s; \epsilon,\delta}.
\label{eq:interpolation:remainder:unpaired}
\end{align}

An application of \eqref{eq:Nishimori} yields

$$
\mathbb{E}\langle Q_{2p-1}^m\rangle_{t, s;\epsilon,\delta} = \mathbb{E}\langle Q_{2p}^m\rangle_{t, s;\epsilon,\delta}
$$
for all $m\in \mathbb{N}$ and $p\geq 1$.
Similarly
an application of \eqref{eq:channelSymmetry} yields $$q_{2p-1}^{(t)} = q_{2p}^{(t)} \quad \text{as well as} \quad \mathbb{E}[ (\tanh \tilde{J})^{2p-1} ]= \mathbb{E}[ (\tanh \tilde{J})^{2p} ]$$ for $p\geq 1$. Therefore
combining the odd and even terms of \eqref{eq:interpolation:remainder:unpaired} we obtain the form in \eqref{eq:interpolation:remainder}.


TITLE: Closed Forms of Certain Zeta constants?
QUESTION [0 upvotes]: The Riemann Zeta function $\zeta(s)=\sum_{n=1}^\infty \frac{1}{n^s}$ converges for $\operatorname{Re}(s)>1$. I am interested in some particular "irrational " values of it such as:

$\zeta(\pi)=1.176241738\ldots$,
$\zeta(e)=1.2690096043\ldots$,
$\zeta(\sqrt2)=3.020737679\ldots$,
$\ldots$

Are there closed form representations for these and constants? Are there formulas which consists of these constants?

REPLY [5 votes]: There is no reason to suspect that these have a "closed form". There isn't even a known closed form for $\zeta(3)$...


TITLE: There's no surjective ring homomorphism from $\mathbb{Z}[x_1,\dots,x_n]$ onto $\mathbb{Q}$.
QUESTION [1 upvotes]: I'm trying to prove that, if a field $A$ is also a finitely generated $\mathbb{Z}$-algebra, then $A$ is finite. The proof I found for this depends on the fact that $\mathbb{Q}$ cannot be a finitely generated $\mathbb{Z}$-algebra.
So, my question (which is equivalent to the title of the post, right?) is how to prove that $\mathbb{Q}$ cannot be a finitely generated $\mathbb{Z}$-algebra?
Thanks.

REPLY [3 votes]: Hint  $\,\ \Bbb Q = \Bbb Z\left[\dfrac{a_1}{b_1},\cdots,\dfrac{a_n}{b_n}\right]\,\Rightarrow\,\Bbb Q = \Bbb Z\left[\dfrac{1}b\right],\ b = b_1\cdots b_n\,\Rightarrow\,$ every prime $\,p\mid b\ $ since
$$\frac{1}p \in \Bbb Z\left[\frac{1}b\right]\,\Rightarrow\, \frac{1}p =  \frac{a}{b^n}\,\Rightarrow\, ap = b^n\,\Rightarrow\, p\mid b^n\,\Rightarrow\, p\mid b$$
Remark $\ $ Kaplansky calls G-domains those domains whose quotient field is so finitely generated (equivalently, as above, singly-generated), see section $1$-$3$ of his Commutative Rings.
The above yields that a PID is a G-domain iff it has finitely many primes. More generally one can show that a Noetherian domain is a G-domain iff it has finitely many nonzero prime ideals all of which are maximal, i.e. the domain is $1$-dimensional and semilocal.


TITLE: Is $(0,0)$ a solution to $x^y-y^x=0$?
QUESTION [4 upvotes]: I am trying to determine if $(0,0)$ a solution to $x^y-y^x=0$.  My hunch is that it is undefined since $0^0$ is an indeterminate form.  To attempt to prove this, I have tried the usual "different paths give different limits" trick with
$(x,ax^n)\rightarrow(0,0)$
$(x,\sin(x))\rightarrow(0,0)$
$(x,e^x-1)\rightarrow(0,0)$
None of the above accomplished my goal.
I did find this older post ($x^y = y^x$ for integers $x$ and $y$) which included an answer by "Yuval Filmus" which states $y=x=0$ is $\it{trivially}$ a solution.  He accomplished this by defining $0^0$ to some value and moves on.  I would like to see something more rigorous, if it exists.
Any hints on how to proceed?
Edited for wording.
Edit:  Stefan Smith confirms my suspicion that the comment of Yuval Filmus cannot be made rigorous.

REPLY [4 votes]: This might sound unsatisfying, but if you define $0^0$ to be any finite number, then $(0,0)$ is a solution, otherwise it isn't. I personally like $0^0=1$, but some people disagree. This site has a lot to say about the issue of $0^0$, and you can search for that. It is ultimately a matter of definition.


TITLE: Evaluating $\sum_{r=1}^{89} \frac{1}{1+\tan^3 r}$
QUESTION [5 upvotes]: $$\sum_{r=1}^{89} \frac{1}{1+\tan^3 r }$$
where $r$ is in degrees

I tried this a lot using the $a^3+b^3$ identity but I don't seem to be getting anything fruitful :(
Can someone please give me a hint? I don't think the $V_n$ method is working here. ($T_r+T_{r-1}$)
Is it related to the $\tan(A+B)$ identity? However I'm not getting anything through that too.
I would be grateful if someone helped. Thanks!

REPLY [10 votes]: Let $S=\displaystyle\sum_{r=1}^{89} \cfrac{1}{1+\tan^3 r }$
$S=\displaystyle\sum_{r=1}^{89} \cfrac{\cos^3r}{\cos^3r+\sin^3 r }$
Now Since  $\displaystyle\sum_{r=a}^{b}f(r)=\displaystyle\sum_{r=a}^{b}f(a+b-r)$
Using which we get $S=\displaystyle\sum_{r=1}^{89} \cfrac{\cos^3(90-r)}{\cos^3(90-r)+\sin^3(90-r) }=\displaystyle\sum_{r=1}^{89} \cfrac{\sin^3r}{\cos^3r+\sin^3 r }$
Therefore $S=\displaystyle\sum_{r=1}^{89} \frac{\cos^3r}{\cos^3r+\sin^3 r} =\displaystyle\sum_{r=1}^{89} \frac{\sin^3r}{\cos^3r+\sin^3 r }$
Therefore $2S=\displaystyle\sum_{r=1}^{89} \cfrac{\sin^3r+\cos^3r}{\cos^3r+\sin^3 r }=\displaystyle\sum_{r=1}^{89} 1=89$
Therefore $S=\displaystyle \boxed{{\frac{89}{2}}}$


TITLE: Riemann surfaces with an atlas all of whose open sets are biholomorphic to $\mathbb{C}$?
QUESTION [10 upvotes]: Is there a compact Riemann surface other than the sphere with an atlas consisting of open subsets biholomorphic to $\mathbb{C}$? Is there a  compact Riemann surface other than the sphere which possesses an open subset biholomorphic to $\mathbb{C}$?

REPLY [19 votes]: Using the uniformization theorem, we can prove that the only compact connected Riemann surface admitting an open set biholomorphic to $\mathbb{C}$ is the Riemann sphere $\mathbb{P}^1(\mathbb{C})$. This answers negatively both questions of the OP.
Indeed, let $X$ be a compact connected Riemann surface with an open set $U$ biholomorphic to $\mathbb{C}$. Choose a base point in $U$, and let $\tilde{X}$ be the universal cover of $X$. By the uniformization theorem, $\tilde{X}$ is isomorphic either to $\mathbb{P}^1(\mathbb{C})$, or to $\mathbb{C}$, or to the open unit disk $D$. In the first case we have necessarily $X \cong \mathbb{P}^1(\mathbb{C})$, so let us assume we are not in this case. Since $\mathbb{C}$ is simply connected, the inclusion map $f : \mathbb{C} \to X$ lifts to a map $\tilde{f} : \mathbb{C} \to \tilde{X}$ which is necessarily injective. If $\tilde{X}=D$ then we get a nonconstant bounded entire function, which is impossible. Finally if $\tilde{X}=\mathbb{C}$ then $\tilde{f}$ is a nonconstant entire function, so by Picard's theorem misses at most one value. But $X \cong \mathbb{C}/\Lambda$ and $\tilde{X} \to X$ is far from being injective, which gives a contradiction. (Alternatively, we can use the fact that any injective entire function $\tilde{f}$ is of the form $\tilde{f}(z)=az+b$ with $a \neq 0$, see this question at MSE.)
EDIT. Here is a more elementary argument avoiding the use of the uniformization theorem. The idea was suggested by an attempt of a student of mine.
Let $X$ be a compact connected Riemann surface with an open set $U$ biholomorphic to $\mathbb{C}$. Let $\omega \in \Omega^1(X)$ be a holomorphic $1$-form. The $2$-form $\sigma = i \cdot \omega \wedge \overline{\omega}$ is everywhere non-negative, and is integrable on $X$ since $X$ is compact. It follows that $\sigma$ is integrable on $U$. Let us write $\omega |_U = f(z) dz$ with $f(z) = \sum_{n \geq 0} a_n z^n$ an entire function. Then
\begin{equation*}
\int_U \sigma = \int_{\mathbb{C}} |f(z)|^2 i dz \wedge d\overline{z} = \int_0^\infty \int_0^{2\pi} |f(re^{i \theta})|^2 2r dr d\theta.
\end{equation*}
Now we have
\begin{equation*}
\int_0^{2\pi} |f(re^{i\theta})|^2 d\theta = \int_0^{2\pi} \sum_{m,n \geq 0} a_m \overline{a_n} r^{m+n} e^{i(m-n)\theta} d\theta = 2\pi \sum_{n \geq 0} |a_n|^2 r^{2n}.
\end{equation*}
Assume that $f$ is nonzero, so that there exists an integer $k \geq 0$ such that $a_k \neq 0$. Then
\begin{equation*}
\int_U \sigma \geq 4\pi |a_k|^2 \int_0^\infty r^{2k+1} dr
\end{equation*}
and the last integral is infinite, contradicting the integrability of $\sigma$ on $U$. It follows that $f=0$ and thus $\Omega^1(X)=\{0\}$, so that $X$ has genus $0$ and $X \cong \mathbb{P}^1(\mathbb{C})$.

REPLY [11 votes]: Compact Riemann surfaces of genus at least $2$ are uniformized by the unit disk, hence they
do not admit any non-constant holomorphic map from $\mathbb{C}$.


TITLE: How will the compression ratio increase thermal effeciency of otto cycle while Win is higher?
QUESTION [0 upvotes]: Why the ideal otto cycle thermal effeciency increase with compression ratio although that will increase work input? While the temperature difference in the combustion chamber could be low,the overall effect is decrease in nth otto. Regardless of the compression ratio.

REPLY [2 votes]: the reason that efficiency increases with compression ratio is it raises the temperature at which the heat is added to the working fluid. The work input is recovered during the expansion phase of the cycle- it is not lost.