text:
\begin{document}

\begin{abstract}
Given a sequence of Boolean functions \( (f_n)_{n \geq 1} \),  \( f_n \colon \{ 0,1 \}^{n}  \to \{ 0,1 \}\), and a sequence \( (X^{(n)})_{n\geq 1} \) of continuous time \( p_n \)-biased random walks \( X^{(n)} =  (X_t^{(n)})_{t \geq 0}\) on \( \{ 0,1 \}^{n} \), let \( C_n \) be the (random) number of times in \( (0,1) \) at which the process  \( (f_n(X_t))_{t \geq 0} \) changes its value. In~\cite{js2006}, the authors conjectured that if \( (f_n)_{n \geq 1} \) is    non-degenerate, transitive and satisfies \( \lim_{n \to \infty}  \mathbb{E}[C_n] = \infty\), then \( (C_n)_{n \geq 1} \) is not tight. We give an explicit example of a sequence of Boolean functions which disproves   this conjecture.
\end{abstract}

\maketitle

\section{Introduction}

The aim of this paper is to present an example of a sequence of Boolean functions, which show that a conjecture made in~\cite{js2006}, in its full generality, is false. To be able to present this conjecture, we first give some background.

For each \( n \geq 1 \), fix some \( p_n \in (0,1) \) and let \(X^{(n)} = (X_t^{(n)})_{t \geq 0}\) be the continuous time random walk on the \( n \)-dimensional hypercube defined as follows.  For each \( i \in [n] \coloneqq \{1,2, \ldots, n \} \) independently, let \( (X_t^{(n)}(i))_{t \geq 0} \) be the continuous time Markov chain on \( \{ 0,1 \} \) which at random times, distributed according to a rate one Poisson process, is assigned a new value, chosen according to   \(  (1-p_n) \delta_0  + p_n \delta_1   \), independently of the Poisson process. The unique stationary distribution of $(X^{(n)}_t)_{t\geq  0}$, denoted by  \( \pi_n\), is the   measure \( ((1-p_n) \delta_0  + p_n \delta )^{\otimes n} \) on \( \{0,1 \}^n \). Throughout this paper, we will always assume that  \( X_0^{(n)}   \) is chosen with respect to this measure. When  \( t > 0 \) is small, the difference between \( X_0^{(n)} \) and \( X_t^{(n)} \) is often thought of as noise, describing a small proportion \( 1-e^{-t} \approx t\)  of the bits being miscounted or corrupted.

A function \( f_n \colon \{ 0,1 \}^{n} \mapsto \{0,1\} \) will be referred to as a Boolean function.  Some classical examples of Boolean functions are the so called \emph{Dictator function} \( f_n(x) = x(1) \), the \emph{Majority function} \( f_n(x) = \sgn \bigl(\sum_{i=1}^{n} (x(i) - 1/2) \bigr) \) and the \emph{Parity function} \( \sgn  \bigl( \prod_{i=1}^{n} (x(i)-1/2 ) \bigr)\)  (see e.g.~\cite{odonnell, gs}).

 Since it is sometimes not natural to require that a sequence of Boolean functions is defined for each \( n \in \mathbb{N} \), we only require that a sequence of Boolean functions is defined for \( n \) in an infinite sub-sequence of \( \mathbb{N} \). Such sub-sequences of \( \mathbb{N} \) will be denoted by \( \mathcal{N} = \{ n_1,n_2, \ldots \} \), where \( 1 \leq n_1 < n_2 < \ldots \). To simplify notation, whenever we consider the limit of a sequence \( (x_{n_i})_{i \geq 1} \) and the dependency on \( \mathcal{N} \) is clear, we will abuse notation and write \( \lim_{n \to \infty} x_n \) instead of \( \lim_{i \to \infty} x_{n_i}\).  Also, we will write \( (x_n)_{n \in \mathcal{N}} \) instead of \( (x_{n_i})_{i \geq 1} \).

One of the main objectives of~\cite{js2006} was to introduce notation which describes possible behaviours of \( (f_n(X_t^{(n)}))_{t \geq 0} \). Some of these definitions which will be relevant for this paper is given in the following definition.

\begin{definition}
Let \( (f_n)_{n \in \mathcal{N}}\), \( f_n \colon \{ 0,1 \}^{n} \to \{ 0,1 \} \), be a sequence of Boolean functions. For \( n \in \mathcal{N} \), let \( C_n = C_n(f_n) \) denote the (random) number of times in \( (0,1) \) at which \( (f_n(X_t^{(n)}))_{t \geq 0} \) has changed its value, i.e.\ let   \( C_n \coloneqq  \lim_{N \to \infty}  \sum_{i=0}^{N-1} \mathbf{1}({f_n(X^{(n)}_{i/N}) \neq f_n(X^{(n)}_{(i+1)/N})}) \).

The sequence \( (f_n)_{n \in \mathcal{N}} \) is said to be
\begin{enumerate}[(i)]
	\item \emph{lame} if
	\begin{equation*}
		\lim_{n \to \infty} P(C_n=0) = 1,
	\end{equation*}
	\item \emph{tame} if   \( (C_n)_{n\geq 1} \) is tight, that is for every \( \varepsilon > 0 \) there is \( k \geq 1 \) and \( n_0 \geq 1 \) such that
	\begin{equation*}
		   P(C_n\geq k) < \varepsilon \qquad \forall n \in \mathcal{N} \colon n \geq n_0,
	\end{equation*}
	\item \emph{volatile} if \( C_n \Rightarrow \infty \) in distribution.
\end{enumerate}

\end{definition}

In~\cite{js2006}, the authors showed that a sequence of Dictator functions is tame and that a sequence of Parity functions is volatile, while a sequence of Majority functions is neither tame nor volatile. More generally, the authors also showed that any noise sensitive sequence of Boolean functions (see e.g.~\cite{bks, gs, odonnell}) is volatile, while any sequence of Boolean functions which is lame or tame is noise stable~\cite{bks, gs, odonnell}. As noted in~\cite{f}~and~\cite{pg}, there are many sequences of functions which are both noise stable and volatile, and hence the opposite does not hold.

Given a sequence \( (p_n)_{n \geq 1} \), \( p_n \in (0,1) \), a sequence of Boolean functions \( (f_n)_{n \in \mathcal{N}} \) is said to be \emph{non-degenerate} if \( 0 < \liminf_{n \to \infty } P(f_n(X_0^{(n)}) = 1) \leq \liminf_{n \to \infty } P(f_n(X_0^{(n)}) = 1)  < 1 \). A Boolean function \( f_n \colon \{ 0,1 \}^{n} \to \{ 0,1 \} \) is said to be transitive if for all \( i,j \in [n] \coloneqq \{ 1,2, \ldots, n \} \) there is a permutation \( \sigma \) of \( [n] \) which is such that (i) \( \sigma(i) = j \) and (ii) for all \( x \in \{ 0,1 \}^{n} \), if we define \( \sigma(x) \coloneqq (x(\sigma(k)))_{k \in [n]} \), then \( f_n(x) = f_n(\sigma(x)) \). To simplify notation, we will abuse notation slightly and say that a sequence of Boolean functions \( (f_n)_{n\geq 1} \) is transitive if \( f_n \) is transitive for each \( n \geq 1 \).

In~\cite{js2006}, the authors show that a sufficient, but not necessary, condition for a non-degenerate sequence \( (f_n)_{n \in \mathcal{N}} \) of Boolean functions to be tame, is that \( \sup_n \mathbb{E}[C_n]< \infty \). It is natural to ask if this condition is also necessary for some natural subset of the set of all sequences of Boolean functions. This is the motivation for the following conjecture.

\begin{conjecture}[Conjecture~1.21 in~\cite{js2006}]\label{the conjecture}
For any sequences \( (p_n)_{n \geq 1} \) and \( \mathcal{N} \), if \( (f_n)_{n \in \mathcal{N}} \) is transitive, non-degenerate  and \( \lim_{n \to \infty} \mathbb{E} [C_n] = \infty \), then \( (f_n)_{n \in \mathcal{N}} \) is not tame.
\end{conjecture}

The main objective of this paper is to show that this conjecture, in its full generality, is false. This result will follow as an immediate consequence of the following theorem, which is our main result.

\begin{theorem}\label{theorem: first result}
Let  \( (p_n)_{n \geq 1} \) be a decreasing sequence which is such that
\begin{enumerate}[(A)]
	\item \( \lim_{n \to \infty} n p_n = \infty \)
	\item \( \lim_{n \to \infty} n p_n^r = 0 \) for some \( r \geq 2 \), and
	\item for any mapping \( \phi \colon \mathbb{N} \to \mathbb{N}  \) which satisfies \( \lim_{n \to \infty} |n - \phi( n)|/(n \land \phi(n))  = 0 \), we have \( \lim_{n \to \infty} p_n / p_{\phi(n)} = 1\).
	\end{enumerate}
Then there is a sequence of positive integers \( \mathcal{N} \) and  a sequence \( (f_n)_{n \in \mathcal{N}} \) of Boolean functions, \( f_n \colon \{ 0,1 \}^{n} \to \{ 0,1 \} \), which (w.r.t\ \( (p_n)_{n \geq 1}\)) is
\begin{enumerate}[(a)]
    \item  non-degenerate,
    \item transitive,
    \item tame, and
    \item \( \lim_{n \to \infty}\mathbb{E}[C_n] = \infty \).
\end{enumerate}
\end{theorem}

In the proof of Theorem~\ref{theorem: first result} we give  explicit examples of sequences of functions which satisfies the above conditions, and hence contradicts Conjecture~\ref{the conjecture}, for sequences \( (p_n)_{n \geq 1} \) which satisfies the assumptions above.
We remark however that this example cannot directly be extended to the case  \( 0 \ll p_n \ll 1 \). In particular, the conjecture might hold with some additional restriction on the sequence \( (p_n)_{n \geq 1} \).

\begin{remark}
The assumption that \( \lim_{n \to \infty} np_n = \infty \) is very natural. To see this, note that the number of jumps of \( (X_t)_{t \geq 0} \) in \( (0,1) \) is given by \( 2np_n(1-p_n) \), and hence if \( \limsup_{n \to \infty} np_n < \infty \), then any sequence \( (f_n)_{n \geq 1} \) of Boolean functions is tame and satisfies    \( \limsup_{n \to \infty} \mathbb{E}[C_n] < \infty \).
\end{remark}

\begin{remark}
With slightly more work, one can modify the example given in the proof of Theorem~\ref{theorem: first result} to get a sequence of functions which, in addition to satisfying (a), (b), (c) and (d) of Theorem~\ref{theorem: first result}, is also monotone in the sense that if \( x,x' \in \{ 0,1 \}^n \), \( n \in \mathcal{N} \), if \( x(i) \leq x'(i) \) for all \( i \in [n] \), then \( f(x) \leq f(x') \).
\end{remark}

 \begin{remark}
 The same idea which is used in the proof of Theorem~\ref{theorem: first result} work in general to disprove Conjecture~\ref{the conjecture} whenever one can find a non-degenerate, tame and transitive sequence of Boolean functions. In particular, this implies that the assumption that \( \mathbb{E}[C_n] = \infty \) can be dropped from Conjecture~\ref{the conjecture}.
 \end{remark}

 With the previous remark in mind, we suggest the following modified conjecture.
 \begin{conjecture}\label{the conjecture II}
If \( (p_n)_{n \geq 1} \) satisfies \( \lim_{n \to \infty }np_n^r = \infty \) for all \( r > 0 \), and \( (f_n)_{n \geq 1} \) is transitive and  non-degenerate (w.r.t.\ \((p_n)_{n\geq 1}\)), then \( (f_n)_{n \geq 1} \) is not tame.
\end{conjecture}

\section{Proof of the main result}

\begin{definition}[Easily convinced tribes]\label{def: easily convinced tribes}
Fix \( r \geq 2 \) and \( n \geq 1 \), and let \( \ell_n \geq 2\) and \( k_n \) be positive integers with the property that \( \ell_n k_n = n \). Partition \( [n] \) into \( k_n \) sets \( S_1^{(n)}, S_2^{(n)}, \ldots, S_{k_n}^{(n)}\), each of size \( \ell_n \), and for \( x \in \{ 0,1 \}^n \), let \( g_n(x) = g_n^{(k_n,\ell_n,r)}(x) \) be equal to one exactly when there is some \( j \in \{ 1,2, \ldots, k_n \} \) such that \( \sum_{i \in S_j^{(n)}} x(i) \geq r \).
\end{definition}

Since Definition~\ref{def: easily convinced tribes} requires that \( k_n\ell_n = n \) and that \( \ell_n \geq 2 \), \( g_n^{(k_n, \ell_n, r)} \) is only well defined when \( n \) is not a prime, and we will in general only want to consider sub-sequences \( \mathcal{N} \) of \( \mathbb{N} \) which have the property that \( k_n \) and \( \ell_n \) can be chosen such that they satisfy certain growth conditions.

We will now show that we can choose \( r  \), \( (p_n)_{n \geq 1} \),  \( \mathcal{N}   \), \( (\ell_n)_{n \in \mathcal{N}} \) and  \( (k_n)_{n\in \mathcal{N}} \)  so that (a), (b) and (d) of Theorem~\ref{theorem: first result} hold.

\begin{lemma}\label{lemma: gn}
For any \( r \geq 2 \) and any decreasing sequence \( (p_n)_{n \geq 1} \) which satisfies the assumptions of Theorem~\ref{theorem: first result}, there is \( \mathcal{N} \) and sequences  \( (\ell_n)_{n \in \mathcal{N}} \), \( (k_n)_{n \in \mathcal{N}} \) of positive integers such that \( (g_n)_{n \in  \mathcal{N}} \) is
\begin{enumerate}[(a)]
    \item non-degenerate,
    \item transitive, and
    \item tame.
\end{enumerate}
\end{lemma}

\begin{proof}[Proof of Lemma~\ref{lemma: gn}]
Assume that there are sequences \( \mathcal{L} \), \( (\ell_n)_{n \in \mathcal{L}} \) and \( (k_n)_{n \in \mathcal{N}} \) such that
\begin{enumerate}[(i)]
    \item \( 2r < \inf \ell_n  \),
    \item \( \lim_{n \to \infty}p_n \ell_n = 0 \), and
    \item \( p_n^r \ell_n^r k_n \asymp 1 \).
\end{enumerate}
We first show that this assumption implies that the conclusions of the lemma hold, and then show that we can find sequences \( \mathcal{L} \), \( (\ell_n)_{n \in \mathcal{L}} \) and \( (k_n)_{n \in \mathcal{N}} \) with these properties.

\begin{proof}[Proof of (a)]
    \renewcommand{\qedsymbol}{}
Note first that
  \begin{equation}\label{eq: proof eq 1}
    P(g_n(X_0^{(n)})=0) = \biggl( \sum_{i=0}^{r-1} \binom{\ell_n}{i} p_n^i (1-p_n)^{\ell_n-i}  \biggr)^{k_n}.
    \end{equation}

    For integers \( 0 < r < \ell \) such that \( 2 r < \ell \), define  \( T \colon \mathbb{R} \to \mathbb{R}\) by \( T(x) \coloneqq \sum_{i=0}^{r-1} \binom{\ell}{i} x^i (1-x)^{l-i} \). Then
        \begin{equation*}
        \begin{cases}
        T(x) = \sum_{i=0}^{r-1} \binom{\ell}{i} x^i (1-x)^{l-i} \cr
        T'(x) = -\binom{\ell}{r} \cdot rx^{r-1} (1-x)^{\ell-r} \cr
        T''(x) = -\binom{\ell}{r} \cdot r(r-1)x^{r-2} (1-x)^{\ell-r}+\binom{\ell}{r} \cdot r(\ell-r)x^{r-1} (1-x)^{\ell-r-1} \cr
        \ldots  \cr
        T^{(m)}(x) = -\binom{\ell}{r} \sum_{i=1}^{m \land r\land \ell-r}  \binom{r}{i} (r)_i x^{r-i} (\ell-r)_{m-i}(1-x)^{\ell-r-i}(-1)^{m-i}
        \end{cases}
    \end{equation*}
    and hence
    \begin{equation*}
        \begin{cases}
        T(0) = 1 \cr
        T^{(m)}(0) = 0 \text{ if } j = 1,2, \ldots, r-1  \cr
        T^{(r)}(0) = -(\ell)_r    .
        \end{cases}
    \end{equation*}
    Moreover, if we assume that  \( x \in (0,1) \) and that \( \ell x < 1 \), then for all \( \xi \in (0,x) \) we have that
    \begin{align*}

        &|T^{(r+1)}(\xi)| = \Bigl| -\binom{\ell}{r} \sum_{i=1}^{r}  \binom{r}{i} (r)_i \xi^{r-i} (\ell-r)_{r+1-i}(1-\xi)^{\ell-r-i}   (-1)^{r+1-i} \Bigr|
        \\&\qquad \leq \binom{\ell}{r}    \sum_{i=1}^{r}  \binom{r}{i} (r)_i \xi^{r-i} (\ell-r)_{r+1-i}
        \leq \binom{\ell}{r}    \sum_{i=1}^{r}  \binom{r}{i} (r)_i \xi^{r-i} \ell^{r-i+1}
        \\&\qquad \leq \binom{\ell}{r}    \sum_{i=1}^{r}  \binom{r}{i} (r)_i \cdot \ell
        \leq  \ell^{r+1} 2^r .
    \end{align*}
    Applying Taylor's theorem to the right hand side of~\eqref{eq: proof eq 1}, and noting that \( 2^r < (r+1)! \), we obtain
  \begin{align*}
    &P\bigl(f_n(X_0^{(n)})=0 \bigr) =  \Bigl(1 - (\ell_n)_r p_n^r + C_n p_n^{r+1} \ell_n^{r+1}\Bigr)^{k_n}
    \end{align*}
    where \( |C_n| < 1 \) for all \( n \). By using the inequalities \( e^{-2 x} \leq 1-x \leq e^{-x} \), valid for all \( x \in [0,1/2) \), the desired conclusion follows by applying (iii).
\end{proof}

\begin{proof}[Proof of (b)]
    \renewcommand{\qedsymbol}{}

Fix some \( n \in \mathcal{L}\) and \( i,i' \in [n] \). We need to show that there is a permutation \( \sigma \) which is such that \( \sigma(i) = i' \) and \( f_n(\sigma(x)) = f_n(x) \) for all \( x \in \{ 0,1 \}^n \). We now divide into two cases.

First, if \( i,i' \in S_m^{(n)} \) for some \( m \in [k_n] \), then we can set \( \sigma = (ii') \).

On the other hand, if there are distinct \( m,m' \in [k_n] \) such that  \( i \in S_m^{(n)} = \{ i, i_2, \ldots, i_{\ell_n}\} \) and \( i' \in S_{m'}^{(n)} = \{ i', i_2', \ldots, i_{\ell_n}'\}\), then we can set \( \sigma = (ii') \prod_{j=2}^{\ell_n} (i_j i_j')\).

This concludes the proof of (b).
\end{proof}

 \begin{proof}[Proof of (c)]
    \renewcommand{\qedsymbol}{}
Fix some \( n \in \mathcal{L} \) and note that whenever \( g_n(X_0^{(n)}) =1\), the distribution of the smallest time \( t > 0 \) at which \( g_n(X_t^{(n)})=0 \) stochastically dominates an exponential distribution with rate \( r \). From this the desired conclusion follows.
\end{proof}

To complete the proof of Lemma~\ref{lemma: gn} it now remains only to show that there are sequences \( \mathcal{N} \),  \( (\ell_n)_{n \in \mathcal{L}} \) and \( (k_n)_{n \in \mathcal{N}} \) such that (i), (ii) and (iii) hold.

To this end, for each  \( n \geq 1 \) let \(  \ell_n  \coloneqq  (np_n^r)^{-1/(r-1)}  \) and \(  k_n \coloneqq n/\ell_n = (np_n)^{r/(r-1)}\). Then one easily verifies that \( 2r < \inf \ell_n  \), \( \lim_{n \to \infty}p_n \ell_n = 0 \), and \( p_n^r \ell_n^r k_n \asymp 1 \).
However, in general, neither \(  \ell_n \) nor \(  k_n \) need to be integers.

To fix this problem, define
\begin{equation*}
    \begin{cases}
        \hat n \coloneqq \lceil  \ell_n \rceil \, \lceil  k_n \rceil, \cr
        \hat \ell_{\hat n} \coloneqq \lceil  \ell_n \rceil, \text{ and}\cr
        \hat k_{\hat n} \coloneqq \lceil  k_n \rceil .
    \end{cases}
\end{equation*}
Let \( { \mathcal{N}}\subseteq \mathbb{N} \) be an infinite sequence on which the mapping \( n \mapsto \hat n \) it is a bijection, and let \( \hat {\mathcal{N}} \) be its image.

We will show that the desired properties hold for \( \hat{\mathcal{N}}\), \( (p_{\hat n})_{\hat n \in \hat {\mathcal{N}}} \), \( (\hat \ell_{\hat n})_{\hat n \in \hat {\mathcal{N}}} \) and \( (\hat k_{\hat n})_{\hat n \in \hat {\mathcal{N}}} \). To this end, note first that
\begin{equation*}
    \inf_{\hat n \in \hat {\mathcal{N}}} \hat \ell_{\hat n}  = \inf_{ n \in  {\mathcal{N}}} \lceil \ell_n \rceil > \inf_{ n \in \mathbb{N}} \lceil \ell_n \rceil > \inf_{ n \in \mathbb{N}}  \ell_n   > 2r,
\end{equation*}
and hence (i) holds.

Next, since \( \hat n \geq n \) for each \( n \in \mathbb{N} \) and \( (p_n)_{n \geq 1} \) is decreasing, we have that \( p_{\hat n} \leq p_n \) for all \( n \in \mathbb{N} \). Using this observation, we obtain
\begin{equation*}
    \lim_{\hat n \to \infty} \hat \ell_{\hat n} p_{\hat n} \leq \lim_{\hat n \to \infty} \hat \ell_{\hat n} p_{n} = \lim_{n \to \infty}   \lceil \ell_{n} \rceil p_{n}
    = \lim_{n \to \infty}   \bigl(\ell_np_n + (\lceil \ell_{n} \rceil - \ell_n) p_{n}\bigr)=0,
\end{equation*}
and hence (ii) holds.

Finally, to see that (iii) holds, note that for any \( n \in \mathbb{N}\),
\begin{align*}
    &| n - \hat n| =  \hat \ell_{\hat n} - \hat k_{\hat n} -\ell_nk_n
    = \lceil k_n \rceil  \lceil \ell_n \rceil  - \ell_nk_n
    \\&\qquad = (\lceil k_n \rceil - k_n)(\lceil \ell_n \rceil - \ell_n) + \ell_n(\lceil k_n \rceil - k_n) +  k_n(\lceil \ell_n \rceil - \ell_n) < \ell_n + k_n +1.
\end{align*}
Since both \( \ell_n \to \infty \) and \( k_n \to \infty \) by definition, we have \( \lim_{n \to \infty} |n - \hat n|/\hat n  = 0 \), and hence by assumption, \( \lim_{n \to \infty} p_n / p_{\hat n} = 1\). This implies in particular that for \( \hat n \in \hat {\mathcal{N}} \), we have
\begin{equation*}
    p_{\hat n}^r \hat \ell_{\hat n}^r \hat k_{\hat n}
    \sim p_{n}^r \hat \ell_{\hat n}^r \hat k_{\hat n}
    = p_{n}^r  \lceil \ell_{ n}\rceil^r \lceil  k_{ n}  \rceil
     = p_{n}^r  \bigl(\ell_n - (\ell_n - \lceil \ell_{ n}\rceil)\bigr)^r \bigl(k_n - (k_n-\lceil  k_{ n}  \rceil)\bigr)
\end{equation*}
Using the assumption that \( p_n\ell_n \to 0 \), it follows that
\begin{equation*}
    p_{\hat n}^r \hat \ell_{\hat n}^r \hat k_{\hat n}
    \sim  p_{n}^r  \bigl(\ell_n - (\ell_n - \lceil \ell_{ n}\rceil)\bigr)^r k_n
    =   p_{n}^r \ell_n^{r} k_n \sum_{i=0}^{r} \binom{r}{i}     \biggl( \frac{\lceil \ell_{ n}\rceil - \ell_n}{\ell_n}\biggr)^{r-i}
    \sim    p_{n}^r \ell_n^{r} k_n
    \asymp 1.
\end{equation*}

This concludes the proof.
\end{proof}

\begin{remarks}
Using essentially the same argument as in the proof of Theorem~\ref{theorem: first result}(c), one can show that for any \( r \geq 2 \), \( \mathcal{N} \), \( (\ell_n)_{n \in \mathcal{N}} \) and \( (k_n)_{n\in \mathcal N} \), we have \( \lim_{n \to\infty }\mathbb{E}[C_n(g_n)] < \infty \). This has two interesting consequences.
	\begin{enumerate}
		\item For any choice of parameters, the sequence of Boolean functions defined in~\ref{def: easily convinced tribes} does not satisfy (c) in Theorem~\ref{theorem: first result}, and hence does not provides a counter-example to Conjecture~\ref{the conjecture}.
		\item Given a Boolean function \( f_n \colon \{ 0,1 \}^n \to \{ 0, 1 \}^n\) and \( i \in \{ 1,2 \ldots, n \} \), we let \( I_i^{(p_n)}(f_n) \) denote the \emph{influence of the \( i \)th bit} on \( f_n \) at \( p_n \), defined as the probability that resampling the \( i\)th bit of \( X_0^{(n)} \) according to \( (1-p_n)\delta_0 + p_n \delta_1 \) changes the value of \( f_n(X_0^{(n)}) \). Note that this definition agrees with the definition of influence given in~\cite{js2006}, but differs with a factor \( 2p_n(1-p_n) \) from the definition of influence used in e.g.\ \cite{bkkkl} and \cite{odonnell}.

			We let \( I^{(p_n)}(f_n) \coloneqq \sum_{i=1}^n I_i^{(p_n)}(f_n) \) and call this the \emph{total influence} of \( f_n \) at \( p_n \).
		By Proposition~1.19 in~\cite{js2006} the total influence of \( g_n \) is equal to  \(  \mathbb{E}[C_n(g_n)]  \). It thus follows from Lemma~\ref{lemma: gn} that when \( \lim_{n \to \infty}np_n^r = 0 \) for some \( r \geq 2 \),  there is a sequence of Boolean functions which is non-degenerate, transitive and have bounded total influence.  Using Proposition~1.19 in~\cite{js2006} together with the proof of Lemma~\ref{lemma: hn} below, it in fact follows that any such sequence could be used to create a counter-example to Conjecture~\ref{the conjecture} as in the proof of Theorem~\ref{theorem: first result}. By contrast, by Theorem~1 in~\cite{bkkkl}, the total influence of any non-degenerate and transitive Boolean function \( f_n \colon \{ 0,1 \}^n \to \{ 0,1 \} \) is  of order at least \( p_n^2 \log n \). In particular, this implies that when \( p_n \gg (\log n)^{-1/2}   \) there can be no sequence of non-degenerate and transitive Boolean functions with bounded total influence. This does however not exclude the possibility of a counter-example to Conjecture~\ref{the conjecture} in this regime.
	\end{enumerate}
\end{remarks}

We now want to modify the sequence \( (g_n)_{n \in \mathcal{N}} \) slightly to obtain a sequence \( (f_n)_{n \in \mathcal{N}} \) of Boolean functions which in addition to satisfying (a), (b) and (c) of Lemma~\ref{lemma: gn} also satisfies \( \lim_{n \to \infty} \mathbb{E}[C_n(f_n)] = \infty \). To this end, we first define a degenerate sequence of Boolean functions with this property.

\begin{definition}
For each \( n \geq 1 \), let \( a_n > 0 \) and \( H_n \coloneqq np_n + a_n \sqrt{np_n(1-p_n)} \).
For \( x \in \{ 0,1 \}^n \), let \( \|x \|\coloneqq \sum_{i=1}^n x_i \) and define \( h_n(x) \coloneqq I(\| x \|  \geq H_n ) \).
\end{definition}

\begin{lemma}\label{lemma: hn}
If \( \lim_{n \to \infty} np_n = \infty \) and \( a_n = \sqrt{\log(np_n)}/2\), then \( (h_n)_{n \geq 1} \) is
\begin{enumerate}[(a)]
    \item  degenerate,
    \item transitive
    \item lame, and satisfies
    \item \( \lim_{n \to \infty}\mathbb{E}\bigl[C_n(h_n) \bigr] = \infty \).
\end{enumerate}
\end{lemma}

\begin{proof}
Note first that the assumptions on \( (p_n)_{n \geq 1} \) and \( (a_n)_{n \geq 1} \) together imply that \( \lim_{n \to \infty} a_n = \infty \) and \( \sqrt{np_n}e^{-a_n^2} \to \infty\).

\begin{proof}[Proof of (a)]
    \renewcommand{\qedsymbol}{}
By definition, we have  \(  \mathbb{E}\bigl[\|X_0^{(n)}\|\bigr] = np_n \) and \( \Var \bigl(\|X_0^{(n)}\| \bigr) = np_n(1-p_n) \). Using Chebyshev's inequality, we thus obtain
\begin{align*}
    P\bigl(h_n(X_0^{(n)}) = 1\bigr) = P\Bigl(\| X_0^{(n)}\| \geq \mathbb{E}\bigl[\| X_0^{(n)}\|\bigr] + a_n \sqrt{\Var\bigl(\| X_0^{(n)}\|\bigr)}\Bigr) \leq a_n^{-2}.
\end{align*}
Since \( a_n \to \infty \), this implies that \((h_n)_{n \geq 1} \) is degenerate, which is the desired conclusion.
\end{proof}

\begin{proof}[Proof of (b)]
    \renewcommand{\qedsymbol}{}
Since for any \( x \in \{ 0,1 \}^n \), \( h_n(x) \) depends on \( x \) only through \( \| x \|\), \( (h_n)_{n \geq 1} \) is transitive.
\end{proof}

\begin{proof}[Proof of (c)]
    \renewcommand{\qedsymbol}{}
Recall that whenever \( np_n(1-p_n) \to \infty \),
\begin{equation*}
   \biggl( \frac{\|X_t^{(n)}\|-np_n}{\sqrt{2np_n(1-p_n)}} \biggr)_{t \geq 0} \overset{\mathcal{D}}{\Rightarrow} (Z_t)_{t \geq 0},
\end{equation*}
where \( (Z_t)_{t \geq 0} \) is a so-called Ornstein-Uhlenbeck process with infinitesimal mean and variance given by \( \mu(z) = -z \) and \( \sigma^2(x) = 1 \) respectively (see e.g.\ pp.\ 170--173 in~\cite{kt}). Given \( z \in \mathbb{R} \), let \( \tau_z \) denote the first time \( t \geq 0 \) at which \( Z_t = z\), given that \( Z_0 \) is chosen according to the stationary distribution of \( (Z_t)_{t \geq 0} \).

By Corollary 1 in~\cite{rs} (see also~\cite{crs}), when \( z>0 \) is large, we have \(  \mathbb{E} [\tau_z] \sim 1/\hat h(z)\) and \( \Var (\tau_z) \sim 1/\hat h(z)^2\), where \( \hat h(z) = z \exp(-z^2/2) /\sqrt{2 \pi}\). By the Paley-Zygmund inequality, this implies that for any finite time \( t > 0 \),  \( \lim_{z \to \infty} P(\tau_{z} > t) = 1 \). This implies in particular that \( (h_n)_{n \geq 1} \) is lame whenever \( a_n \to \infty \), completing the proof of (c).
\end{proof}

\begin{proof}[Proof of (d)]
    \renewcommand{\qedsymbol}{}
By Proposition 1.19 in~\cite{js2006}, for each \( n \geq 1 \) we have
\begin{equation*}
    \mathbb{E}[C_n(h_n)] = \sum_{i=1}^n I_i^{(p_n)}(h_n)
\end{equation*}
where \( I_i^{(p_n)}(h_n) \) is the so-called influence of the \( i \)th bit on \( h_n \) at \( p_n \), defined as the probability that resampling the \( i\)th bit of \( X_0^{(n)} \) according to \( (1-p_n)\delta_0 + p_n \delta_1 \) changes the value of \( h_n(X_0^{(n)}) \). Using this result, we obtain
\begin{align*}
    &\mathbb{E}[C_n(h_n)] = nI_1^{(p_n)}(h_n) = n \Biggl( P\bigl(\|X_0^{(n)} \|=H_n-1\bigr) \cdot \frac{n-(H_n-1)}{n} \cdot p_n + P\bigl(\|X_0^{(n)} \|=H_n\bigr) \cdot \frac{H_n}{n} \cdot (1-p_n) \Biggr)
    \\&\qquad =n \Biggl( \binom{n}{H_n-1} p_n^{H_n-1}(1-p_n)^{n-H_n+1}  \cdot \frac{n-(H_n-1)}{n} \cdot p_n + \binom{n}{H_n} p_n^{H_n}(1-p_n)^{n-H_n}  \cdot \frac{H_n}{n} \cdot (1-p_n) \Biggr)
    \\&\qquad= 2H_n   \binom{n}{H_n} p_n^{H_n}(1-p_n)^{n-H_n+1}.
\end{align*}
Using Stirling's formula, it follows that
\begin{align*}
    &\mathbb{E}[C_n(h_n)]
    \sim 2 \sqrt{np_n} \cdot    \frac{e^{-a_n^2}}{\sqrt{2 \pi}}.
\end{align*}
In particular, if \( \sqrt{np_n}e^{-a_n^2} \to \infty \), then \(\lim_{n \to \infty}\mathbb{E}[C_n(h_n)] = \infty \). This completes the proof of (d).
\end{proof}
\end{proof}

We are now ready to give a proof of our main result.

\begin{proof}[Proof of Theorem~\ref{theorem: first result}]
Fix some \( r \geq 2 \) and sequences \( \mathcal{N} \), \( (\ell_n)_{n \in \mathcal{N}} \), \( (k_n)_{n \in \mathcal{N}} \) and \( (a_n)_{n \geq 1} \) so that the assumptions of Lemmas~\ref{lemma: gn} and Lemma~\ref{lemma: hn} both hold. For \( n \in \mathcal{N} \) and \( x \in \{ 0,1 \}^n \), let  \( H_n \coloneqq np_n + a_n\sqrt{np_n(1-p_n)} \) and define
\begin{equation}
    f_n(x) \coloneqq  g_n(x)I(\|x\| < H_n-1) + I(\|x \| \geq H_n) = g_n(x)I(\|x\| < H_n-1) + h_n(x).
\end{equation}
Note that \( f_n (x)\) and \( h_n (x)\) agree whenever  \( \| x \| \geq H_n-1 \).
Combining Lemma~\ref{lemma: gn} and Lemma~\ref{lemma: hn},  the desired conclusion now immediately follows.
\end{proof}

\section*{Acknowledgements}

The author would like to thank the anonymous referee for several useful comments.
The author acknowledges support from the European Research Council, Grant Agreement No. 682537.


TITLE: Why the fire is blue when there is enough oxygen?
QUESTION [2 upvotes]: In the kitchen the flame is blue when there is enough oxygen. Is it true that the amount of oxygen has effect in fire color, why blue, why yellow sometimes?

REPLY [7 votes]: Yellow flames are due to glowing particles of soot that emit a black-body spectrum. These will only form when there is not enough oxygen in the gas mixture (incomplete combustion).
The blue emission comes from excited molecules that are produced in the chemical reactions.


TITLE: What is the fundamental group of the octahedral symmetry group Oh?
QUESTION [0 upvotes]: Do discrete groups have a well defined topology? If so, what is the fundamental group of the octahedral symmetry group Oh? In other words, are all the maps of one-spheres on the symmetry elements of Oh continuously deformable into each other?

REPLY [2 votes]: Of course it has a well-defined topology...the discrete one (all subsets are open subsets).
The fundamental group requires a choice of base point.  Given that, of course all the base-point-preserving maps are continuously deformable to each other, since there's only one of them!  So the fundamental group is trivial.


TITLE: Determine size or number of jordan blocks
QUESTION [1 upvotes]: I have done most of the work but I struggle to put this matrix into Jordan Normal Form.
$$C=\begin{bmatrix}1 & 0 & 0 & 0&0\\1 & -1 & 0 & 0&-1 \\1 & -1 & 0 & 0&-1 \\ 0 & 0 & 0 & 0&-1\\-1&1&0&0&1\end{bmatrix}$$
The characteristic polynomial is $(x-1)x^4$, so the eigenvalues are $1,0,0,0,0$.
The rank of $C$ is 3. The rank of $C^2=2$.
I know there is only $1$ Jordan block of size $1$ for the eigenvalue $1$, $J_1(1)$.
But I am stuck to determine the blocks for the eigenvalue $0$. I know the block sizes must add up to $4$. Can someone help me finish finding the Jordan blocks for $0$?

REPLY [1 votes]: As at Jordan normal form (Basis)
we see here the method of finding some column vector $r_5,$ such that $C^3 r_5 = 0$ but $C^2 r_5 \neq 0.$ Once we choose that, we are forced to take $r_4 = C r_5,$ then column $r_3 = C r_4.$ This becomes an eigenvector because $Cr_3 = C^2 r_4 = C^3 r_5 = 0.$
I have picked $r_5 = (0,0,0,0,-1)^T$ and finished up. The other null vector $r_2$ can be anything in the two- dimensional subspace of genuine zero eigenvectors, so more choices are possible.
$$
\left(
\begin{array}{ccccc}
1&0&0&0&0 \\
0&-1&1&0&0 \\
0&-1&0&1&0 \\
-1&1&0&0&0 \\
0&-1&0&0&-1 \\
\end{array}
\right)
\left(
\begin{array}{ccccc}
1&0&0&0&0 \\
1&0&0&1&0 \\
1&1&0&1&0 \\
1&0&1&1&0 \\
-1&0&0&-1&-1 \\
\end{array}
\right) =
\left(
\begin{array}{ccccc}
1&0&0&0&0 \\
0&1&0&0&0 \\
0&0&1&0&0 \\
0&0&0&1& 0\\
0&0&0&0&1 \\
\end{array}
\right)
$$
$$
\left(
\begin{array}{ccccc}
1&0&0&0&0 \\
0&-1&1&0&0 \\
0&-1&0&1&0 \\
-1&1&0&0&0 \\
0&-1&0&0&-1 \\
\end{array}
\right)
\left(
\begin{array}{ccccc}
1&0&0&0&0 \\
1&-1&0&0&-1 \\
1&-1&0&0&-1 \\
0&0&0&0&-1 \\
-1&1&0&0&1 \\
\end{array}
\right)
\left(
\begin{array}{ccccc}
1&0&0&0&0 \\
1&0&0&1&0 \\
1&1&0&1&0 \\
1&0&1&1&0 \\
-1&0&0&-1&-1 \\
\end{array}
\right) =
\left(
\begin{array}{c|c|ccc}
1&0&0&0&0 \\ \hline
0&0&0&0&0 \\ \hline
0&0&0&1&0 \\
0&0&0&0& 1\\
0&0&0&0&0 \\
\end{array}
\right)
$$
$$
\left(
\begin{array}{ccccc}
1&0&0&0&0 \\
1&0&0&1&0 \\
1&1&0&1&0 \\
1&0&1&1&0 \\
-1&0&0&-1&-1 \\
\end{array}
\right)
\left(
\begin{array}{c|c|ccc}
1&0&0&0&0 \\ \hline
0&0&0&0&0 \\ \hline
0&0&0&1&0 \\
0&0&0&0& 1\\
0&0&0&0&0 \\
\end{array}
\right)
\left(
\begin{array}{ccccc}
1&0&0&0&0 \\
0&-1&1&0&0 \\
0&-1&0&1&0 \\
-1&1&0&0&0 \\
0&-1&0&0&-1 \\
\end{array}
\right) =
\left(
\begin{array}{ccccc}
1&0&0&0&0 \\
1&-1&0&0&-1 \\
1&-1&0&0&-1 \\
0&0&0&0&-1 \\
-1&1&0&0&1 \\
\end{array}
\right)
$$


TITLE: Let $G$ be an arbitrary group.Let $S$ and $H$ be subgroups of $G$ which are not equal with $G$. Show that $G\setminus H$ is not subset of $S$.
QUESTION [1 upvotes]: Let $G$ be an arbitrary group.
Let $S$ and $H$ be subgroups of $G$ which are not equal with $G$. Show that $G\setminus H$ is not subset of $S$.
$S$ subgroup of $G$, then for all $s_1, s_2 \in S, s_1s_2^{-1} \in S$ and $H$ subgroup of $G$, then for all $h_1,h_2 \in H, h_1h_2^{-1} \in H$. Let $G = \{e_G,g,g^2,\dots, g^{n-1}\}$.
What next?
Any idea? Thanks in advance.

REPLY [1 votes]: See also the answer of @JCCA: a group cannot be written as the union of two of its proper (:= being not equal to the whole group) subgroups. Hence it follows:

**Proposition** Let $H$ be a proper subgroup of a group $G$. Then $\langle G \setminus H \rangle=G$.
**Proof** Obviously $G=H \cup G \setminus H \subseteq H \cup \langle G \setminus H \rangle$. Since $H$ is proper the proposition follows.

So in your case, if $G \setminus H \subseteq S$. then, since $S$ is a subgroup, also $\langle G \setminus H \rangle \subseteq S$, a contradiction to $S$ being proper.


TITLE: Projective varieties with isomorphic coordinate rings
QUESTION [2 upvotes]: Let $X$, $Y$ be projective varieties with isomorphic homogeneous coordinate rings (as graded rings). What can we say about $X$ and $Y$? Are they necessarily isomorphic?

REPLY [0 votes]: Question: "Let X, Y be projective varieties with isomorphic homogeneous coordinate rings (as graded rings). What can we say about X and Y? Are they necessarily isomorphic?"
Answer: Yes: If $X:=Proj(T), Y:=Proj(S)$ with $T:=\oplus_{d \geq 0}T_d, S:=\oplus_{d \geq 0}S_d$ and $T \cong S$ as graded rings, it follows $Proj(S) \cong Proj(T)$ holds.


TITLE: The effect of tension force on a pendulum swinging in simple harmonic motion
QUESTION [2 upvotes]: In physics class, I learned that the reason a pendulum moves about the equilibrium position is due to the force acting towards the center, which is the resultant force of tension and weight.
However, since tension force is orthogonal to the direction, shouldn't only gravity act as the object's restoring force about the equilibrium position? What is the role of the string's tension in simple harmonic motion, and does the horizontal component of tension have anything to do with the restoring force mgsinθ?

REPLY [0 votes]: Look at this pendulum diagram. the pendulum can rotate about the point A. taking the sum of the torques about point A, you get:
$$\sum_\tau=I_p\,\ddot{\theta}+m\,g\,h=0\quad,(h=L\,\sin(\theta))$$
where dot mean time derivative and $I_p=m\,L^2$ is the pendulum inertia.
so the tension force T doesn't affect   the equation of motion.


text:
/-
Copyright (c) 2014 Parikshit Khanna. All rights reserved.
Released under Apache 2.0 license as described in the file LICENSE.
Authors: Parikshit Khanna, Jeremy Avigad, Leonardo de Moura, Floris van Doorn, Mario Carneiro,
Scott Morrison
-/
import data.list.count
import data.list.infix

/-!
# Lattice structure of lists

This files prove basic properties about `list.disjoint`, `list.union`, `list.inter` and
`list.bag_inter`, which are defined in core Lean and `data.list.defs`.

`l₁ ∪ l₂` is the list where all elements of `l₁` have been inserted in `l₂` in order. For example,
`[0, 0, 1, 2, 2, 3] ∪ [4, 3, 3, 0] = [1, 2, 4, 3, 3, 0]`

`l₁ ∩ l₂` is the list of elements of `l₁` in order which are in `l₂`. For example,
`[0, 0, 1, 2, 2, 3] ∪ [4, 3, 3, 0] = [0, 0, 3]`

`bag_inter l₁ l₂` is the list of elements that are in both `l₁` and `l₂`, counted with multiplicity
and in the order they appear in `l₁`. As opposed to `list.inter`, `list.bag_inter` copes well with
multiplicity. For example,
`bag_inter [0, 1, 2, 3, 2, 1, 0] [1, 0, 1, 4, 3] = [0, 1, 3, 1]`
-/

open nat

namespace list
variables {α : Type*} {l l₁ l₂ : list α} {p : α → Prop} {a : α}

/-! ### `disjoint` -/

section disjoint

lemma disjoint.symm (d : disjoint l₁ l₂) : disjoint l₂ l₁ := λ a i₂ i₁, d i₁ i₂

lemma disjoint_comm : disjoint l₁ l₂ ↔ disjoint l₂ l₁ := ⟨disjoint.symm, disjoint.symm⟩

lemma disjoint_left : disjoint l₁ l₂ ↔ ∀ ⦃a⦄, a ∈ l₁ → a ∉ l₂ := iff.rfl

lemma disjoint_right : disjoint l₁ l₂ ↔ ∀ ⦃a⦄, a ∈ l₂ → a ∉ l₁ := disjoint_comm

lemma disjoint_iff_ne : disjoint l₁ l₂ ↔ ∀ a ∈ l₁, ∀ b ∈ l₂, a ≠ b :=
by simp only [disjoint_left, imp_not_comm, forall_eq']

lemma disjoint_of_subset_left (ss : l₁ ⊆ l) (d : disjoint l l₂) : disjoint l₁ l₂ := λ x m, d (ss m)

lemma disjoint_of_subset_right (ss : l₂ ⊆ l) (d : disjoint l₁ l) : disjoint l₁ l₂ :=
λ x m m₁, d m (ss m₁)

lemma disjoint_of_disjoint_cons_left {l₁ l₂} : disjoint (a :: l₁) l₂ → disjoint l₁ l₂ :=
disjoint_of_subset_left (list.subset_cons _ _)

lemma disjoint_of_disjoint_cons_right {l₁ l₂} : disjoint l₁ (a :: l₂) → disjoint l₁ l₂ :=
disjoint_of_subset_right (list.subset_cons _ _)

@[simp] lemma disjoint_nil_left (l : list α) : disjoint [] l := λ a, (not_mem_nil a).elim

@[simp] lemma disjoint_nil_right (l : list α) : disjoint l [] :=
by { rw disjoint_comm, exact disjoint_nil_left _ }

@[simp, priority 1100] lemma singleton_disjoint : disjoint [a] l ↔ a ∉ l :=
by { simp only [disjoint, mem_singleton, forall_eq], refl }

@[simp, priority 1100] lemma disjoint_singleton : disjoint l [a] ↔ a ∉ l :=
by rw [disjoint_comm, singleton_disjoint]

@[simp] lemma disjoint_append_left : disjoint (l₁ ++ l₂) l ↔ disjoint l₁ l ∧ disjoint l₂ l :=
by simp only [disjoint, mem_append, or_imp_distrib, forall_and_distrib]

@[simp] lemma disjoint_append_right : disjoint l (l₁ ++ l₂) ↔ disjoint l l₁ ∧ disjoint l l₂ :=
disjoint_comm.trans $ by simp only [disjoint_comm, disjoint_append_left]

@[simp] lemma disjoint_cons_left : disjoint (a :: l₁) l₂ ↔ a ∉ l₂ ∧ disjoint l₁ l₂ :=
(@disjoint_append_left _ l₂ [a] l₁).trans $ by simp only [singleton_disjoint]

@[simp] lemma disjoint_cons_right : disjoint l₁ (a :: l₂) ↔ a ∉ l₁ ∧ disjoint l₁ l₂ :=
disjoint_comm.trans $ by simp only [disjoint_comm, disjoint_cons_left]

lemma disjoint_of_disjoint_append_left_left (d : disjoint (l₁ ++ l₂) l) : disjoint l₁ l :=
(disjoint_append_left.1 d).1

lemma disjoint_of_disjoint_append_left_right (d : disjoint (l₁ ++ l₂) l) : disjoint l₂ l :=
(disjoint_append_left.1 d).2

lemma disjoint_of_disjoint_append_right_left (d : disjoint l (l₁ ++ l₂)) : disjoint l l₁ :=
(disjoint_append_right.1 d).1

lemma disjoint_of_disjoint_append_right_right (d : disjoint l (l₁ ++ l₂)) : disjoint l l₂ :=
(disjoint_append_right.1 d).2

lemma disjoint_take_drop {m n : ℕ} (hl : l.nodup) (h : m ≤ n) : disjoint (l.take m) (l.drop n) :=
begin
  induction l generalizing m n,
  case list.nil : m n
  { simp },
  case list.cons : x xs xs_ih m n
  { cases m; cases n; simp only [disjoint_cons_left, mem_cons_iff, disjoint_cons_right, drop,
                                 true_or, eq_self_iff_true, not_true, false_and,
                                 disjoint_nil_left, take],
    { cases h },
    cases hl with _ _ h₀ h₁, split,
    { intro h, exact h₀ _ (mem_of_mem_drop h) rfl, },
    solve_by_elim [le_of_succ_le_succ] { max_depth := 4 } },
end

end disjoint

variable [decidable_eq α]

/-! ### `union` -/

section union

@[simp] lemma nil_union (l : list α) : [] ∪ l = l := rfl

@[simp] lemma cons_union (l₁ l₂ : list α) (a : α) : a :: l₁ ∪ l₂ = insert a (l₁ ∪ l₂) := rfl

@[simp] lemma mem_union : a ∈ l₁ ∪ l₂ ↔ a ∈ l₁ ∨ a ∈ l₂ :=
by induction l₁; simp only [nil_union, not_mem_nil, false_or, cons_union, mem_insert_iff,
  mem_cons_iff, or_assoc, *]

lemma mem_union_left (h : a ∈ l₁) (l₂ : list α) : a ∈ l₁ ∪ l₂ := mem_union.2 (or.inl h)

lemma mem_union_right (l₁ : list α) (h : a ∈ l₂) : a ∈ l₁ ∪ l₂ := mem_union.2 (or.inr h)

lemma sublist_suffix_of_union : ∀ l₁ l₂ : list α, ∃ t, t <+ l₁ ∧ t ++ l₂ = l₁ ∪ l₂
| [] l₂ := ⟨[], by refl, rfl⟩
| (a :: l₁) l₂ := let ⟨t, s, e⟩ := sublist_suffix_of_union l₁ l₂ in
  if h : a ∈ l₁ ∪ l₂
  then ⟨t, sublist_cons_of_sublist _ s, by simp only [e, cons_union, insert_of_mem h]⟩
  else ⟨a::t, s.cons_cons _, by simp only [cons_append, cons_union, e, insert_of_not_mem h];
    split; refl⟩

lemma suffix_union_right (l₁ l₂ : list α) : l₂ <:+ l₁ ∪ l₂ :=
(sublist_suffix_of_union l₁ l₂).imp (λ a, and.right)

lemma union_sublist_append (l₁ l₂ : list α) : l₁ ∪ l₂ <+ l₁ ++ l₂ :=
let ⟨t, s, e⟩ := sublist_suffix_of_union l₁ l₂ in
e ▸ (append_sublist_append_right _).2 s

lemma forall_mem_union : (∀ x ∈ l₁ ∪ l₂, p x) ↔ (∀ x ∈ l₁, p x) ∧ (∀ x ∈ l₂, p x) :=
by simp only [mem_union, or_imp_distrib, forall_and_distrib]

lemma forall_mem_of_forall_mem_union_left (h : ∀ x ∈ l₁ ∪ l₂, p x) : ∀ x ∈ l₁, p x :=
(forall_mem_union.1 h).1

lemma forall_mem_of_forall_mem_union_right
   (h : ∀ x ∈ l₁ ∪ l₂, p x) : ∀ x ∈ l₂, p x :=
(forall_mem_union.1 h).2

end union

/-! ### `inter` -/

section inter

@[simp] lemma inter_nil (l : list α) : [] ∩ l = [] := rfl

@[simp] lemma inter_cons_of_mem (l₁ : list α) (h : a ∈ l₂) :
  (a :: l₁) ∩ l₂ = a :: (l₁ ∩ l₂) :=
if_pos h

@[simp] lemma inter_cons_of_not_mem (l₁ : list α) (h : a ∉ l₂) :
  (a :: l₁) ∩ l₂ = l₁ ∩ l₂ :=
if_neg h

lemma mem_of_mem_inter_left : a ∈ l₁ ∩ l₂ → a ∈ l₁ := mem_of_mem_filter

lemma mem_of_mem_inter_right : a ∈ l₁ ∩ l₂ → a ∈ l₂ := of_mem_filter

lemma mem_inter_of_mem_of_mem : a ∈ l₁ → a ∈ l₂ → a ∈ l₁ ∩ l₂ :=
mem_filter_of_mem

@[simp] lemma mem_inter : a ∈ l₁ ∩ l₂ ↔ a ∈ l₁ ∧ a ∈ l₂ := mem_filter

lemma inter_subset_left (l₁ l₂ : list α) : l₁ ∩ l₂ ⊆ l₁ := filter_subset _

lemma inter_subset_right (l₁ l₂ : list α) : l₁ ∩ l₂ ⊆ l₂ := λ a, mem_of_mem_inter_right

lemma subset_inter {l l₁ l₂ : list α} (h₁ : l ⊆ l₁) (h₂ : l ⊆ l₂) : l ⊆ l₁ ∩ l₂ :=
λ a h, mem_inter.2 ⟨h₁ h, h₂ h⟩

lemma inter_eq_nil_iff_disjoint : l₁ ∩ l₂ = [] ↔ disjoint l₁ l₂ :=
by { simp only [eq_nil_iff_forall_not_mem, mem_inter, not_and], refl }

lemma forall_mem_inter_of_forall_left (h : ∀ x ∈ l₁, p x)
  (l₂ : list α) :
  ∀ x, x ∈ l₁ ∩ l₂ → p x :=
ball.imp_left (λ x, mem_of_mem_inter_left) h

lemma forall_mem_inter_of_forall_right (l₁ : list α)
  (h : ∀ x ∈ l₂, p x) :
  ∀ x, x ∈ l₁ ∩ l₂ → p x :=
ball.imp_left (λ x, mem_of_mem_inter_right) h

@[simp] lemma inter_reverse {xs ys : list α} : xs.inter ys.reverse = xs.inter ys :=
by simp only [list.inter, mem_reverse]

end inter

/-! ### `bag_inter` -/

section bag_inter

@[simp] lemma nil_bag_inter (l : list α) : [].bag_inter l = [] :=
by cases l; refl

@[simp] lemma bag_inter_nil (l : list α) : l.bag_inter [] = [] :=
by cases l; refl

@[simp] lemma cons_bag_inter_of_pos (l₁ : list α) (h : a ∈ l₂) :
  (a :: l₁).bag_inter l₂ = a :: l₁.bag_inter (l₂.erase a) :=
by cases l₂; exact if_pos h

@[simp] lemma cons_bag_inter_of_neg (l₁ : list α) (h : a ∉ l₂) :
  (a :: l₁).bag_inter l₂ = l₁.bag_inter l₂ :=
begin
  cases l₂, {simp only [bag_inter_nil]},
  simp only [erase_of_not_mem h, list.bag_inter, if_neg h]
end

@[simp] lemma mem_bag_inter {a : α} : ∀ {l₁ l₂ : list α}, a ∈ l₁.bag_inter l₂ ↔ a ∈ l₁ ∧ a ∈ l₂
| []        l₂ := by simp only [nil_bag_inter, not_mem_nil, false_and]
| (b :: l₁) l₂ := begin
    by_cases b ∈ l₂,
    { rw [cons_bag_inter_of_pos _ h, mem_cons_iff, mem_cons_iff, mem_bag_inter],
      by_cases ba : a = b,
      { simp only [ba, h, eq_self_iff_true, true_or, true_and] },
      { simp only [mem_erase_of_ne ba, ba, false_or] } },
    { rw [cons_bag_inter_of_neg _ h, mem_bag_inter, mem_cons_iff, or_and_distrib_right],
      symmetry, apply or_iff_right_of_imp,
      rintro ⟨rfl, h'⟩, exact h.elim h' }
  end

@[simp] lemma count_bag_inter {a : α} :
  ∀ {l₁ l₂ : list α}, count a (l₁.bag_inter l₂) = min (count a l₁) (count a l₂)
| []         l₂         := by simp
| l₁         []         := by simp
| (h₁ :: l₁) (h₂ :: l₂) :=
begin
  simp only [list.bag_inter, list.mem_cons_iff],
  by_cases p₁ : h₂ = h₁; by_cases p₂ : h₁ = a,
  { simp only [p₁, p₂, count_bag_inter, min_succ_succ, erase_cons_head, if_true, mem_cons_iff,
               count_cons_self, true_or, eq_self_iff_true] },
  { simp only [p₁, ne.symm p₂, count_bag_inter, count_cons, erase_cons_head, if_true, mem_cons_iff,
               true_or, eq_self_iff_true, if_false] },
  { rw p₂ at p₁,
    by_cases p₃ : a ∈ l₂,
    { simp only [p₁, ne.symm p₁, p₂, p₃, erase_cons, count_bag_inter, eq.symm (min_succ_succ _ _),
                 succ_pred_eq_of_pos (count_pos.2 p₃), if_true, mem_cons_iff, false_or,
                 count_cons_self, eq_self_iff_true, if_false, ne.def, not_false_iff,
                 count_erase_self, list.count_cons_of_ne] },
    { simp [ne.symm p₁, p₂, p₃] } },
  { by_cases p₄ : h₁ ∈ l₂; simp only [ne.symm p₁, ne.symm p₂, p₄, count_bag_inter, if_true,
      if_false, mem_cons_iff, false_or, eq_self_iff_true, ne.def, not_false_iff,count_erase_of_ne,
      count_cons_of_ne] }
end

lemma bag_inter_sublist_left : ∀ l₁ l₂ : list α, l₁.bag_inter l₂ <+ l₁
| []        l₂ := by simp [nil_sublist]
| (b :: l₁) l₂ := begin
  by_cases b ∈ l₂; simp [h],
  { exact (bag_inter_sublist_left _ _).cons_cons _ },
  { apply sublist_cons_of_sublist, apply bag_inter_sublist_left }
end

lemma bag_inter_nil_iff_inter_nil : ∀ l₁ l₂ : list α, l₁.bag_inter l₂ = [] ↔ l₁ ∩ l₂ = []
| []        l₂ := by simp
| (b :: l₁) l₂ :=
begin
  by_cases h : b ∈ l₂; simp [h],
  exact bag_inter_nil_iff_inter_nil l₁ l₂
end

end bag_inter
end list

subset_name: curated


text:
\section{Conclusions}
\label{sec:conc}
\noindent This paper presents a highly accurate boundary integral method to simulate deforming droplets in Stokes flow. The method can be used for both droplets and inviscid bubbles, both which may be covered by insoluble surfactants. The boundary integral method is coupled with a pseudo-spectral method for surfactant concentration, and together they are spectrally accurate in space.

The errors introduced by the numerical evaluation of the integrals in the boundary integral equation when droplets get close to each other are accurately estimated using contour integrals. These errors are removed using a special quadrature, which enables the method to simulate close drop-drop interactions very accurately.

Given a sufficient spatial resolution, the accuracy of the method is limited only by the set tolerance of the adaptive time-stepping scheme. To the knowledge of the authors, this method is more accurate than other methods to simulate Stokes flow in 2D currently available.

By presenting a set of easily accessible algorithms based on exact and semi-analytical solutions, the hope of the authors is that this will set a standard of validation for any proposed new method for surfactant-laden drops in 2D.

config: arxiv


TITLE: Stability of planet-enclosing shell supported by atmosphere
QUESTION [3 upvotes]: I understand that a hollow spherical shell of uniform density gives net zero gravitational attraction to objects inside it. So in vacuum, a ball at the centre of a spherical shell would not be stable in its position. Any slight deviation would continue and the ball would collide with the inside of the shell.
Would this be different in the presence of an atmosphere? Specifically, if the inner planet is massive enough to have an atmosphere that extends beyond the shell, and the shell is not sealed so that the atmosphere flows through in places.
I am imagining that the air pressure on the surface of the shell would be greater on the inside than the outside due to the thickness of the shell. This would have no net effect when the planet is at the centre of the shell, but any deviation would lead to the part of the shell nearest the planet being in higher pressure atmosphere. Would this lead to the difference in pressure between inside and outside being greater at that point than others, pushing the shell and planet apart to restore the planet to the centre?
My thinking is that the net outwards force on the shell at the point furthest from the planet will be less than the net outwards force on the shell at the point closest to the planet, due to there being a greater pressure difference across the thickness of the shell. Am I missing any effects that would prevent or counteract this?

This idea came to me as an extension of one of the suggestions in this worldbuilding answer.

REPLY [2 votes]: There would be a slight restorative force from the gravitational pull of the atmosphere itself.
First, imagine an ocean world with a very small core. If the core was more dense than the ocean it will be "sink" to the center.
Similarly, if your "core" was a shell, it would also "sink" to the center.
Thus, if you had both a sphere and a shell they would both tend towards the center, aligning to each other through the gravity of the ocean. The same restorative forces would exist with a very light "ocean" of gas; they would just be much much weaker.
This restorative force may not be enough to overcome instability that may come from rotating.


TITLE: If a multiple integral is zero over some region, can I say the integrand is zero?
QUESTION [3 upvotes]: Consider the following problem
Let the integral of a real function of 3 real variables $F(x,y,z)$ over some volume $V$ of $R^{3}$ vanish,
$\int$$\int$$\int$$dxdydz$ $F(x,y,z)$$=0$
Now assume this continues to hold for any finite volume $V$ contained in a larger volume $V'$ of $R^{3}$.
Can I say the integrand is necessarily zero over $V'$?
This seems to be valid for definite integrals on a line (although I'm also not sure), and I feel the function would have to vanish, but I don't see how to prove this.

REPLY [5 votes]: Sure, the function must be zero almost everywhere, under the minimal assumption that it is measurable. Note that these are very reasonable and non-restrictive conditions:

The almost everywhere condition is required since you can modify a function on a measure zero (i.e., zero volume) set to be anything you like, without changing any integrals.
The measurability condition is required for the integrals to even be well-defined.

Although the question is stated in $\mathbb{R}^3$, the result actually holds on measure spaces more generally, and in my opinion the reasoning behind the result is easier to understand in this more general setting.
The argument is straightforward. Suppose for the sake of contradiction that the integral of your function over any measurable set is zero, yet the function itself is strictly nonzero on some set $V$ with positive measure.
Decompose $V$ into the union of a piece where the function is strictly positive and another piece where the function is strictly negative. Since our function is measurable, these pieces are measurable (they are the intersection of a measurable set with the pullback of a measurable set under a measurable function). Furthermore, at least one of these pieces must have positive measure, since their union is $V$, which has positive measure. The integral over that piece must then be nonzero*, a contradiction.

*For a proof that the integral of a strictly positive (or strictly negative) function over a set of positive measure is nonzero, see the answer to the question here: Lebesgue integral of a positive function on a set of positive measure . The basic premise is to decompose your set into a countable union of subsets where your function is bounded away from zero by successively smaller and smaller tolerances. Since the countable union of measure zero sets has measure zero, at least one of these subsets must have positive measure, and so the integral on that subset is strictly nonzero.

For some further intuition about this result and why it should be true physically, imagine that your function represents temperature in a room. If you measure the temperature in the room with a thermometer, you don't get the exact temperature at a point - rather, you get a local average temperature over a small region of space around the thermometer tip. This local "average temperature" is essentially the integral of your function over that set of space, divided by its volume. The more spatially accurate the thermometer, the smaller the volume of integration.
If the integral of the temperature function is zero over all volumes of space, that is like saying that any thermometer you use to measure it, no matter how accurate, will always say that the room has temperature zero. There is no physical measurement possible that can distinguish the temperature field from the zero field.


TITLE: Why is the Math-tea argument wrong?
QUESTION [1 upvotes]: The Math-tea argument says there must be some real numbers that cannot be specified because we only have countably many definitions.
Why is that wrong?
Is there a simple explanation?
I don't understand this blog.
https://alexanderpruss.blogspot.com/2020/11/the-math-tea-argument.html

I don't understand what does this mean:

For instance, there is no guarantee that there is a set of all the definable real numbers.

Why does it matter that there's no set that contains the enumeration?
There is no set of all sets, yet we can enumerate through every set of sets that has a finite description by considering all combinations of the following characters from least length to greatest length: { },,
{}
{{}}
{{{}}}
{{{{}}}}
{{},{{}}}
{{{{{}}}}}
{{{},{{}}}}
{{{{{{}}}}}}
{{{{},{{}}}}}
{{},{{},{{}}}}

etc.

The axioms of set theory tell us that for any predicate F in the language of set theory there is a set of all the numbers that satisfy F.

I think this makes sense - we can specify a predicate that corresponds to a subset of real numbers that satisfy the predicate
I don't follow the logic here:

Elegant as this argument is, it has crucial set-theoretic flaws. For instance, there is no guarantee that there is a set of all the definable real numbers. The axioms of set theory tell us that for any predicate F in the language of set theory there is a set of all the numbers that satisfy F. But the predicate "is definable" is in English, not in set theory.

REPLY [7 votes]: I would describe the situation as follows: it's not that the math tea argument is wrong, so much as that it is easy to misapply it. Specifically, here's a theorem which on the face of it may seem to contradict the math tea argument:

$(@)$ Assuming $\mathsf{ZFC}$ is consistent, there is a model $M$ of $\mathsf{ZFC}$ such that every element of $M$ is parameter-freely definable in $M$. In particular, every real number in $M$ is parameter-freely definable in $M$.

(Really it's more convenient to talk about $\omega$-models here, but I'll ignore this since it makes things more technical and isn't the key point.)
The issue is that we have to be very careful about what system the math tea argument takes place in. There's a version of the math tea argument which we can run inside $M$, but it is substantially weaker than one might expect.
Specifically, for $n\in\mathbb{N}$ consider the sentence

$(*)_n$: "There is a real which is not parameter-freely $\Sigma_n$-definable over $M$.

Each $(*)_n$ is outright provable in $\mathsf{ZFC}$ via essentially the math tea argument, and so holds in $M$ in particular. However, the following sentence

$(\dagger)$: "There is a real which is not definable over the universe"

is not even expressible in the language of set theory - this is a consequence of Tarski's undefinability theorem - and so we can't talk about whether or not $M$ thinks it's true.
We can (express and) prove in $\mathsf{ZFC}$ the statement

$(\dagger)_{set}$: "For every model $A$ of $\mathsf{ZFC}$, there is a real $r$ which is not definable over $A$,"

again using the math tea argument. This is because we can talk about truth within a set, rather than over the universe. However, note that the real $r$ is not guaranteed to live in $A$. So e.g. the fact that $(@)$ is a $\mathsf{ZFC}$-theorem doesn't contradict the fact that $\mathsf{ZFC}$ proves each $(*)_n$ and $(\dagger)_{set}$.


TITLE: Nearest Neighbor Global matching
QUESTION [2 upvotes]: I'm trying to compare two datasets; a satellite image and some known points. In the satellite image I've got 20 to 50 detections that should match up to a set of lat/lon points that I have. There is generally a unknown uniform offset between the two datasets (this offset is a function of the position in the image, speed/direction of the object and size). This problem isn't too difficult when the points are far apart from each other but when they get close I start running into mismatching problems.

So in this dummy pic I've got some satellite image detections (red circles) that correspond to some ground truth samples (green squares). I generally end up with some false red circles (bad detections, artifacts etc) and some unmatchable green squares (resolution too bad to detect them, obscured by terrain). There is often a uniform offset that's not predictable (top left corner) that causes some mismatches. This can't usually be corrected for since it's a function of a bunch of variables and not just a geo-coding error. If I used shortest distance matching, for example, the samples in the dotted ellipse would be matched.
What's the correct algorithm to be using here? Some kind of global nearest neighbor matching with the input features being straight up x,y coords? Is there something like this in sci-kit learn or some other python lib? My maths background isn't great so I'm struggling to find the right keywords to search for.

REPLY [2 votes]: I already mentioned that this is called the "simultaneous pose and correspondence registration" problem in a comment. Additionally, I want to add something on the mathematical side:
First, you may want to run a point set registration algorithm, for example Iterative Closets Points (ICP) to match the two point clouds (red and green) to match as close as possible.
Now assume for a moment that there are no outliers at all. Mathematically speaking, what you are hoping for is a minimum-weight perfect matching on the complete bipartite graph, where on the one side the vertices are the green squares, and on the other side the vertices are the red circles, and the weight of an edge is given by the square of the euclidian distance between the corresponding objects. In your initial post, you correctly observed that the nearest neighbor assignment not necessarily gives a minimum cost perfect matching. (It is both possible that neither a minimum cost matching, nor a matching at all emerge). There are fast algorithms for the minimum weight perfect matching problem, but you should definitely use a library, because these algorithms are very complicated and error-prone to implement by yourself.
Finally, you have to deal with outliers. One way to do this, is the following: First introduce artificial dummy vertices, in such a way that the number of red circles and green squares is equal, and let the distance between a dummy vertex and any other vertex always be equal to $C$, where $C$ is some fixed large number. Then run the minimum weight perfect matching algorithm and at the end discard all edges from the matching which have an unnaturally high weight.
Note: Again, I am not an expert, but I think this will lead to an acceptable solution. If you can find libraries for your problem online, you should prefer them to my solution.


TITLE: Where does the energy used to rotate the axis of a flywheel orthogonally go?
QUESTION [1 upvotes]: I have a wheel spinning away from me in the vertical plane. The angular momentum is pointing to the left. The axis of rotation is horizontal, and it is supported on either side by a bearing and the whole mount is on a turntable constrained to rotate about the Z-axis. Think of a free spinning bike wheel. It requires force to rotate the assembly due to the gyroscopic action of the flywheel but lets say I rotate it 360 degrees so that the direction of the angular momentum is brought back to where it was initially. Now, a force has been applied through an entire 360 degrees of rotation, reversing the direction of the wheels angular momentum twice.

My question is WHERE has the energy gone ? What has it been converted to ? ie What evidence is there of work having been done ? (It seems that this energy just disappears, but we know that can't be.)

REPLY [3 votes]: Ok, this question was very well posed and I think it deserves as high-quality an answer as we can manage. Here is my feeble attempt...
Most people who have manipulated a gyroscope observe that it seems to take more force to orthogonally rotate the gyroscope when it is spinning. What actually happens is that the original force they apply is redirected. This causes the person to instinctively apply a second corrective force to counter the redirection of the original force. This second force seems, to a person, to constitute additional work. In reality, the second force is only a static force, not a force through a distance, so it does not constitute actual “work”.
It is much like two people trying to move a car where the first person pushes on the rear bumper (the original force) and a second person tries to help by pushing on the car sideways (the second force). It seems like more total force is needed to move the car in this way, but so long as the car doesn’t move in the sideways direction that the second person is pushing it in, the second person does no additional work.
Human experience with manipulating spinning gyroscopes trains us to believe that more work must be done to manipulate a spinning gyroscope because we feel that second force and allow it to distort our estimate of the amount of work that is really being done. In fact, the amount of "real" work that needs to be done to orthogonally rotate a spinning gyroscope IS THE SAME as the amount of work that needs to be done to rotate a non-spinning gyroscope. (Queue small mind-blowing hand explosion)
If one were to built the rig depicted in the excellent figure above, one would observe that the friction-less bearings supporting the shaft along the z-axis would apply the secondary forces, and that the effort needed to rotate the spinning gyroscope around the z-axis would not seem to be more significant.


TITLE: Is there a continuous function $f$ satisfying $f(x)^2=f\left(x^2\right)$, $f(0)=1$ and $ f(1)=0$?
QUESTION [6 upvotes]: I was wondering if there is a continuous function $f:[0,1]\to\mathbb R$ satisfying
$$f(x)^2=f\left(x^2\right)\text,$$
for all $x\in[0,1]$, $f(0)=1$ and $ f(1)=0$.
Clearly, some easy functions like polynomials are not satisfied. I guess there is a way to construct an example since it only needs a continuous function.

REPLY [9 votes]: There is no such function. Let $g(x)=|f(x)|$. Note that $g(x^{2})=(g(x))^{2}$. By iteration we get $g(x^{2^{n}})=(g(x))^{2^{n}}$ for every positive integer $n$. By continuity $g(x^{2^{n}}) \to g(0)=1$ for $|x| <1$. So $(g(x))^{2^{n}} \to 1$ for such $x$. But this implies $g(x)=1$. [Indeed $g(x) >1$ implies $(g(x))^{2^{n}} \to \infty$ and $g(x)<1$ implies $(g(x))^{2^{n}} \to 0$]. We have proved that $|f(x)|=1$ for $|x|<1$. By continuity $|f(1)|=1$ contradicting the hypothesis.


TITLE: Why is a number field always of the form $\mathbb Q(\alpha)$ for $\alpha$ algebraic?
QUESTION [7 upvotes]: My definition of a number field is "a finite extension of $\mathbb Q$". I want to prove that if $L$ is a finite field extension of $\mathbb Q$, then $L = \mathbb Q(\alpha)$ for some $\alpha$ algebraic over $\mathbb Q$.
I can prove things that look helpful. I know that if $L/K$ is a finite extension with $\mathrm{char}K = 0$  and with finitely many intermediate fields, then $L = K(\alpha)$ for some algebraic $\alpha$. I also know that if $\alpha$ is algebraic over $K$, then there are only finitely many intermediate fields between $K$ and $K(\alpha)$ (where those fields are determined by the factors of the minimal polynomial of $\alpha$ over $K$).
I can't pull a proof together though. I'd obviously be done if I could prove that a finite extension of $\mathbb Q$ has finitely many intermediate extensions. Any hints would be greatly appreciated.

REPLY [2 votes]: Use the primitive element theorem to get the existence of a generator. Then the fact that a number field forms a finite extension of $\mathbb{Q}$ tells you that any such generator has to be algebraic (since finite extensions are automatically algebraic).


TITLE: Basic primal and dual solution. Feasible? Why?
QUESTION [0 upvotes]: Questions

Attempt
I managed to solve (c) but I have trouble understanding (d) any directions would be incredibly useful.

REPLY [0 votes]: $$
\begin{array}{cc}
\begin{array}{rr}
\min\quad c^Tx & \\
\text{s.t.} \quad Ax&=b \\
x &\ge 0
\end{array} & \qquad
\begin{array}{rr}
\max\quad b^Ty & \\
\text{s.t.} \; A^Ty&\le c \\
y & \text{free}
\end{array} \\
{\cal (P)} & \qquad {\cal (D)}
\end{array}
$$
The last four columns of $A$ is in the current basis.
$$A=\begin{pmatrix}
2&0&0&2&1&1\\
-1&2&1&-2&2&3\\
-1&2&2&0&1&-2\\
1&-2&-1&1&0&1
\end{pmatrix},\;
B=\begin{pmatrix}
0&2&1&1\\
1&-2&2&3\\
2&0&1&-2\\
-1&1&0&1
\end{pmatrix},\\
B^{-1}=\begin{pmatrix}
9/5 & -1/5 & -7/5 & -4 \\
4/5 & -1/5 & -2/5 & -1 \\
-8/5 & 2/5 & 9/5 & 4 \\
1 & 0 & -1 & -2
\end{pmatrix},\;
b=\begin{pmatrix}3\\-1\\2\\0\end{pmatrix},\;
c=\begin{pmatrix}0 \\ 4 \\ 10 \\ 5 \\ 5 \\ 0 \end{pmatrix}
$$
The primal basic solution $x_B$ satisfies $Bx_B=b$, so
$$x_B=B^{-1}b=
\begin{pmatrix}
9/5 & -1/5 & -7/5 & -4 \\
4/5 & -1/5 & -2/5 & -1 \\
-8/5 & 2/5 & 9/5 & 4 \\
1 & 0 & -1 & -2
\end{pmatrix}
\begin{pmatrix}3\\-1\\2\\0\end{pmatrix}
= \begin{pmatrix} 14/5 \\ 9/5 \\ -8/5 \\ 1 \end{pmatrix}.
$$
The primal basic solution is $(0,0,14/5 , 9/5 , -8/5 , 1)^T$.  By complementary slackness, the 3rd-6th dual constraints are equalities:
$B^T y = c_B := (c_3,\dots,c_6)^T$, so
$$y = (B^{-1})^T c_B
=\begin{pmatrix}
9/5  & 4/5  & -8/5 & 1 \\
-1/5 & -1/5 & 2/5  & 0 \\
-7/5& -2/5& 9/5 & -1 \\
-4 & -1 & 4  & -2
\end{pmatrix}
\begin{pmatrix} 10 \\ 5 \\ 5 \\ 0 \end{pmatrix}
= \begin{pmatrix} 14 \\ -1 \\ -7 \\ -25 \end{pmatrix}
$$
$$c^T x
= \begin{pmatrix} 0 \\ 4 \\ 10 \\ 5 \\ 5 \\ 0 \end{pmatrix}^T
\begin{pmatrix} 0 \\ 0 \\ 14/5 \\ 9/5 \\ -8/5 \\ 1 \end{pmatrix} = 29 \\
b^T y
= \begin{pmatrix} 3\\-1\\2\\0 \end{pmatrix}^T
\begin{pmatrix} 14 \\ -1 \\ -7 \\ -25 \end{pmatrix} = 29
$$
The corresponding dual solution is $y=(14, -1, -7, -25)^T$.
The primal basic solution not feasible as $x_5 =-8/5 < 0$.