text:
\begin{document}

\maketitle

\begin{abstract}

  We provide results demonstrating the smoothness of some marginal
  log-linear parameterizations for distributions on multi-way
  contingency tables.  First we give an analytical relationship
  between log-linear parameters defined within different margins, and
  use this to prove that some parameterizations are equivalent to ones
  already known to be smooth.  Second we construct an iterative method
  for recovering joint probability distributions from marginal
  log-linear pieces, and prove its correctness in particular cases.
  Finally we use Markov chain theory to prove that certain cyclic
  conditional parameterizations are also smooth.  These results are
  applied to show that certain conditional independence models are
  curved exponential families.
\end{abstract}

\section{Introduction}

Models for multi-way contingency tables may include restrictions on
various marginal or conditional distributions, especially in the
context of longitudinal or causal models \citep[see, for example,][and
references therein]{lang:94, bergsma:09, evans:13}.  Such models can
often be parameterized by combining log-linear parameters from within
different marginal tables.  The resulting \emph{marginal log-linear}
parameterizations, introduced by \citet{br:02}, provide an elegant and
flexible way to parameterize a multivariate discrete probability
distribution.

Setting these marginal log-linear parameters to zero can be used to
define arbitrary conditional independence models \citep{rudas:10,
  forcina:10}, including those corresponding to undirected graphical
models or Bayesian networks.  If these zero parameters can be embedded
into a larger smooth parameterization of the joint distribution, then
the model defined by the conditional independence constraints is a
curved exponential family, and therefore possesses good statistical
properties.  This approach is applied by \citet{rudas:10} and
\citet{evans:13} to classes of graphical models.

Unfortunately, there exist models of conditional independence
which---though believed to be curved exponential families---cannot be
embedded into parameterizations currently known to be smooth.
\citet{forcina:12} studies examples of models defined by `loops' of
conditional independences, such as
\begin{align}
X_1 \indep X_2 \,|\, X_3, \qquad X_1 \indep X_3 \,|\, X_4,
\qquad X_1 \indep X_4 \,|\, X_2, \label{eqn:cyclemod}
\end{align}
which can be defined by constraints on the conditional distributions
$p_{2|13}$, $p_{3|14}$ and $p_{4|12}$ respectively.  However it is not
clear whether a smooth parameterization of the joint distribution can
be constructed using these conditionals.  The model can also be
defined by setting a particular collection of marginal log-linear
parameters to zero (see Section \ref{sec:cycles} for details), but
there is no way to embed these parameters into a smooth
parameterization of the kind studied by \citet{br:02}, so their
results do not apply.  \citet{forcina:12} gives a numerical test for
this model which is highly suggestive of smoothness, but no formal
proof is available.

The contribution of this paper is to show that the class of smooth
discrete parameterizations which can be constructed using marginal
log-linear (MLL) parameters is considerably larger than had previously
been known, and that models such as (\ref{eqn:cyclemod}) can indeed be
embedded into these parameterizations.  We give three different
methods for demonstrating smoothness in this context.  First we
provide an analytical expression for the relationship between
log-linear parameters defined within different marginal distributions;
this allows us to prove the equivalence of various parameterizations.
Second we show that particular fixed point maps relating different
parameters are contractions, and hence can be used to uniquely recover
the joint probability distribution.  Lastly we use Markov chain theory
to show that we can smoothly recover joint probability distributions
from `cyclic' conditional distributions; this is used to show that
certain conditional independence models, including the one above, are
curved exponential families of distributions.

The rest of the paper is organized as follows: Section \ref{sec:mllps}
reviews marginal log-linear parameters and their properties.

Section \ref{sec:smooth} specifies the relationship between log-linear
parameters defined within different margins, enabling certain
parameterizations to be proven equivalent.  Section \ref{sec:fixed}
extends this by constructing fixed point methods that smoothly recover
a joint distribution.  Section \ref{sec:cycles} further extends the
results of Section \ref{sec:smooth} using Markov chain theory, and
demonstrates that certain conditional independence models are curved
exponential families.  Section \ref{sec:discuss} contains
discussion, and a conjecture on the precise characterization of smooth
MLL parameterizations.

\section{Marginal Log-Linear Parameters} \label{sec:mllps}

We consider multivariate distributions over a finite collection of
binary random variables $X_v \in \{0,1\}$, for $v \in V$; we denote
their joint distribution by $p_V \equiv p(x_V) \equiv P(X_V = x_V)$.
All the results herein also hold (or have analogues) in the case of
general finite discrete variables, but the notation becomes more
cumbersome.  For $M \subseteq V$ we denote the marginal distribution
over $X_M = (X_v,\, v \in M)$ by $p_M \equiv p(x_M) \equiv P(X_M =
x_M)$, and for disjoint $A,B \subseteq V$ we denote the relevant
conditional distribution by $p_{A|B} \equiv p(x_A \,|\, x_B) \equiv
P(X_A = x_A \,|\, X_B = x_B)$.  Distributions are assumed to be
strictly positive: $p_V > 0$.

\begin{dfn}
Let $\Delta^k \equiv \{p_V > 0\}$ be the strictly positive probability
simplex of dimension $k = 2^{|V|}-1$.  We say that a homeomorphism
$\theta : \Delta^k \rightarrow \Theta \subseteq \reals^k$ onto an open
set $\Theta$ is a \emph{smooth parameterization} of $\Delta^k$ if
$\theta$ is twice continuously differentiable, and its Jacobian
has full rank $k$ everywhere.
\end{dfn}

The canonical smooth parameterization of $\Delta^k$ is via
\emph{log-linear parameters} $\eta_L$, defined by the M\"obius
expansion
\begin{align*}
\log p_V(x_V) = \sum_{L \subseteq V} (-1)^{|x_L|} \eta_L;
\end{align*}
here $|x_L| = \sum_{v \in L} x_v$ is the number of 1s in $x_L \in
\{0,1\}^L$.  It follows by M\"obius inversion that
\begin{align}
  \eta_L = 2^{-|V|} \sum_{x_V \in \{0,1\}^{|V|}} (-1)^{|x_L|} \log
  p_V(x_V); \label{eqn:llp}
\end{align}
see, for example, \citet{lau:96}.
For example, if $V = \{1,2,3\}$,
\begin{align*}
\eta_{13} = \frac{1}{8} \log \frac{p(0, 0, 0) \, p(0, 1, 0) \, p(1, 0, 1) \, p(1, 1, 1)}{p(1, 0, 0) \, p(1, 1, 0) \, p(0, 0, 1) \, p(0, 1, 1)}.
\end{align*}
It is well known that the collection $\bs\eta \equiv (\eta_L, \,
\emptyset \neq L \subseteq V)$ provides a smooth parameterization of
the joint distribution $p_V$ with $\Theta = \reals^k$.

\begin{dfn}
We define a \emph{marginal log-linear parameter} by analogy with
(\ref{eqn:llp}), as the ordinary log-linear parameter for a particular
marginal distribution.  Let $L \subseteq M \subseteq V$; then
\begin{align*}
\lambda_L^M &= 2^{-|M|} \!\!\!\! \sum_{x_{\sn M} \in \{0,1\}^{|M|}} (-1)^{|x_L|} \log p_M(x_M)
\end{align*}
is the marginal log-linear parameter associated with the \emph{margin}
$M$ and the \emph{effect} $L$.  See \citet{br:02}.
\end{dfn}

Clearly $\lambda_L^V = \eta_L$ and, for example,
\begin{align*}
\lambda_{13}^{13} &= \frac{1}{4} \log \frac{p_{13}(0,0) \, p_{13}(1,1)}{p_{13}(1,0) \, p_{13}(0,1)},
\end{align*}
which is the log-odds ratio between $X_1$ and $X_3$.  In order to fit
a model with the constraint $X_1 \indep X_3$ we could choose a
parameterization that includes $\lambda_{13}^{13}$, and fix it to be
zero.

One way to characterize the main idea of \citet{br:02} is as follows:
given some arbitrary margins $p_{M_1}, \ldots, p_{M_k}$ of a joint
distribution $p_V$, what additional information does one need to
smoothly reconstruct the full joint distribution $p_V$?  They show
that one possibility is to take the collection of log-linear
parameters $\eta_L = \lambda_L^V$ where $L \nsubseteq M_i$ for any $i
= 1,\ldots,k$.

It follows that given any inclusion-respecting sequence of margins
$M_1, \ldots, M_k =V$ (i.e.\ $M_i \subseteq M_j$ only if $i < j$), we
can smoothly parameterize $p_V$ with marginal log-linear parameters of
the form $\lambda_L^{M_i}$, where $L \subseteq M_i$ but $L \nsubseteq
M_j$ for any $j < i$.

\begin{exm} \label{exm:mllp} Take the inclusion-respecting sequence of
  margins $\{1,2\}$, $\{2,3\}$, $\{1,2,3\}$.  This gives us the smooth
  parameterization consisting of the vector $\bs\lambda_{\mathcal{P}}$
  below.  The pairs $(L,M)$ are summarized (grouped by margin) in the
  adjacent table.\footnote{Note that here, and in the sequel, we
    abbreviate sets of integers by omitting the braces and commas in
    order to avoid overburdened notation: so, for example, $23$ means
    $\{2,3\}$.}
\begin{align*}
&\mathcal{P}: \bmllp
12&1, 2, 12\\
23&3, 23\\
123&13, 123.\\
\emllp
&& \bs\lambda_{\mathcal{P}} = (\lambda_1^{12}, \lambda_{2}^{12}, \lambda_{12}^{12},
\lambda_{3}^{23}, \lambda_{23}^{23}, \lambda_{13}^{123}, \lambda_{123}^{123})^T.
\end{align*}
\end{exm}

Now, let $\mathcal{P}$ be an \textbf{arbitrary} collection of
effect-margin pairs $(L,M)$ such that $\emptyset \neq L \subseteq M
\subseteq V$.  Define
\[
\bs\lambda_{\mathcal{P}} = \bs\lambda_{\mathcal{P}}(\bs p) =
(\lambda_L^M : (L,M) \in \mathcal{P})
\]
to be the corresponding vector of marginal log-linear parameters.  The
main question considered by this paper is: under what circumstances
does $\bs\lambda_{\mathcal{P}}$ constitute a smooth parameterization
of $p_V$?

\subsection{Existing Results}

We say that $\mathcal{P}$ is \emph{complete} if every non-empty subset
of $V$ appears as an effect in $\mathcal{P}$ exactly once.  If, in
addition, the margins can be ordered so that each effect appears with
the first margin of which it is a subset, we say that $\mathcal{P}$ is
\emph{hierarchical}.  Parameterizations that can be constructed from
an inclusion-respecting sequence of margins in the manner of Example
\ref{exm:mllp} correspond precisely to hierarchical $\mathcal{P}$.
\citet{br:02} show that if $\mathcal{P}$ is complete and hierarchical
then $\bs\lambda_{\mathcal{P}}$ gives a smooth parameterization of the
joint distribution; in addition, they show that completeness is
necessary for smoothness.  \citet{forcina:12} shows that if
$\mathcal{P}$ is complete and contains only two distinct margins $M$,
then $\bs\lambda_{\mathcal{P}}$ is smooth.

To our knowledge, these are the only existing results on the
smoothness of marginal log-linear parameterizations.  No example has
been provided of a complete parameterization which is non-smooth.  In
Sections \ref{sec:smooth}, \ref{sec:fixed} and \ref{sec:cycles} we
will show that, in fact, many more complete parameterizations are
smooth than had previously been known.

The issue of smoothness in non-hierarchical models was raised by
\citet{forcina:12} in the context of loop models of conditional
independence, and expanded upon by \citet{colombi:14} for models of
context-specific conditional independence; the latter consider a more
general class of models than we do, but there is no overlap in the
theoretical results.  Examples of ordinary conditional independences
models that require non-complete parameterizations (and therefore are
not curved exponential families) are found in \citet{drton:09}.

\section{An Analytical Map between Margins} \label{sec:smooth}

To parameterize a marginal distribution $p_M$ we can use the marginal
log-linear parameters $\{\lambda_L^M : \emptyset \neq L \subseteq
M\}$.  An analogous result holds for conditional distributions: for
disjoint $A,B$ define
\[
\bs\lambda_{A|B} \equiv (\lambda_L^{AB} \,|\, L \subseteq A \cup B, L \cap A \neq \emptyset);
\]
in other words, all the MLL parameters for the margin $A \cup B$ whose
effect contains some element of $A$.  Then $\bs\lambda_{A|B}$
constitutes a smooth parameterization of the conditional distribution
$X_A \,|\, X_B$

A consequence of this is to aid us in understanding the relationship
between log-linear parameters defined within different margins.
Theorem 3 of \citet{br:02} shows that distinct MLL parameters
corresponding to the same effect in different margins (i.e.\
$\lambda_L^M$ and $\lambda_L^N$ with $M \neq N$) are linearly
dependent at certain points in the parameter space, and that therefore
no smooth parameterization can include two such parameters.  The
following theorem elucidates the exact relationship between such
parameters, and will later be used to demonstrate the smoothness of
certain non-hierarchical parameterizations.

\begin{thm} \label{thm:linear} Let $A,M$ be disjoint subsets of $V$.
  The log-linear parameter $\lambda_L^{MA}$ may be decomposed as
\begin{align}
\lambda_L^{MA} = \lambda_L^{M} + f(\bs\lambda_{A|M}), \label{eqn:equal}
\end{align}
for a smooth function $f$, which vanishes whenever $X_A \indep X_v
\,|\, X_{M \setminus \{v\}}$ for some $v \in L$.

In addition, if $K \cap (V \setminus M) \neq \emptyset$
\begin{align}
\frac{\partial \lambda_L^{M}}{\partial \eta_K} =
2^{-|M|} \sum_{x_V} (-1)^{|x_{K \triangle L}|} p(x_{V \setminus M} \,|\, x_M),
\label{eqn:deriv}
\end{align}
(where $(\eta_J : J \neq K)$ are held fixed).
\end{thm}

\begin{proof}
We have
\begin{align*}
\lambda_L^{MA} &= 2^{-|AM|} \sum_{x_{MA}} (-1)^{|x_L|} \log p(x_{MA})\\
  &= 2^{-|AM|} \!\! \sum_{x_{MA}} (-1)^{|x_L|} \left[\log p(x_M) +
		   \log p(x_{A} \,|\, x_M)\right]\\
  &= 2^{-|M|} \sum_{x_{M}} (-1)^{|x_L|} \log p(x_M) +
		  2^{-|AM|} \sum_{x_{MA}} (-1)^{|x_L|} \log p(x_{A} \,|\, x_M)\\
  &= \lambda_L^{M} +
		  2^{-|AM|} \sum_{x_{MA}} (-1)^{|x_L|} \log p(x_{A} \,|\, x_M).
\end{align*}
Since the second term is a smooth function of the conditional
probabilities $p(x_A \,|\, x_M)$, it follows that it is also a smooth
function of the claimed parameters.  The implication of independence
follows from Lemma 2.9 of \citet{evans:13}.

Now,
\begin{align*}
\frac{\partial}{\partial \eta_K} p(x_V) &= \frac{\partial}{\partial \eta_K} \exp\left(\sum_{J \subseteq V} (-1)^{|x_J|} \eta_J \right)
= (-1)^{|x_K|} p(x_V),
\end{align*}
and similarly
\begin{align*}
\frac{\partial}{\partial \eta_K} p(x_M) &= \frac{\partial}{\partial \eta_K} \sum_{y_{V \setminus M}} p(y_{V\setminus M}, x_M) = (-1)^{|x_{K \cap M}|} \sum_{y_{V \setminus M}} (-1)^{|y_{K \setminus M}|} p(y_{V\setminus M}, x_M).
\end{align*}
Hence the derivative of (\ref{eqn:equal}) in the case $A = V \setminus
M$ becomes
\begin{align*}
\frac{\partial f}{\partial \eta_K} &=
2^{-|V|} \sum_{x_{V}} (-1)^{|x_L|} \left\{ (-1)^{|x_K|} \frac{p(x_V)}{p(x_V)} - \frac{(-1)^{|x_{K \cap M}|}}{p(x_M)} \sum_{y_{V \setminus M}} (-1)^{|y_{K \setminus M}|} p(y_{V\setminus M}, x_M) \right\}\\
&= 0 - 2^{-|V|} \sum_{x_{V}} (-1)^{|x_{L}|+|x_{K \cap M}|}  \sum_{y_{V \setminus M}} (-1)^{|y_{K \setminus M}|} p(y_{V \setminus M} \,|\, x_M)
\intertext{and, since there is no dependence upon $x_{V \setminus M}$, this is the same as}
&= - 2^{-|V|} 2^{|V \setminus M|} \sum_{x_{V}} (-1)^{|x_{L}|+|x_{K \cap M}|+|x_{K \setminus M}|} p(x_{V \setminus M} \,|\, x_M).
\end{align*}
Then note that $|x_{L}|+|x_{K \cap M}|+|x_{K \setminus M}| = |x_L| +
|x_K|$ simply counts the number of 1s in $L$ and in $K$, so $|x_{L
  \triangle K}|$ is even if and only if $|x_L| + |x_K|$ is.  Hence
\begin{align*}
\frac{\partial f}{\partial \eta_K}
&= - 2^{-|M|} \sum_{x_{V}} (-1)^{|x_{L \triangle K}|} p(x_{V \setminus M} \,|\, x_M),
\end{align*}
which gives the required result.
\end{proof}

\begin{rmk}
  We have shown that if the conditional distribution of $X_A$ given
  $X_M$ is fixed the relationship between $\lambda_L^M$ and
  $\lambda_L^{MA}$ (and indeed any parameter of the form
  $\lambda_L^{MB}$ for $B \subseteq A$) is linear.  In particular, if
  we \emph{know} $p_{A|M}$, then $\lambda_L^{MA}$ and $\lambda_L^{M}$
  become interchangeable as part of a parameterization, preserving
  smoothness and (when relevant) variation independence.
\end{rmk}

\subsection{Constructing Smooth Parameterizations}

The following example shows how Theorem \ref{thm:linear} can be used
to prove the smoothness of a parameterization.

\begin{exm}
Consider the complete collections $\mathcal{P}$ and $\mathcal{Q}$ below.
\begin{align*}
\mathcal{P}:&\bmllp
3&3\\
23&23\\
123&1, 2, 12, 13, 123.\\
\emllp &&
\mathcal{Q}:
\bmllp
3&3\\
23&2, 23\\
123&1, 12, 13, 123\\
\emllp.
\end{align*}
$\mathcal{P}$ is not hierarchical because in any inclusion-respecting
ordering the margin 23 must precede 123, in which case the effect 2
(contained in the pair $(2,123)$) is not associated with the first
margin of which it is a subset.  Existing results therefore cannot
tell us whether or not $\bs\lambda_{\mathcal{P}}$ is smooth.  However,
by fixing the parameters $\bs\lambda_{1|23} = (\lambda_1^{123},
\lambda_{12}^{123}, \lambda_{13}^{123}, \lambda_{123}^{123})$ Theorem
\ref{thm:linear} shows that $\lambda_{2}^{123}$ and $\lambda_{2}^{23}$
are interchangeable.  Hence $\bs\lambda_{\mathcal{P}}$ is smooth if
and only if $\bs\lambda_{\mathcal{Q}}$ is also smooth which, since
$\mathcal{Q}$ satisfies the conditions of a hierarchical
parameterization, it is.  In addition, $\bs\lambda_{\mathcal{P}}$ and
$\bs\lambda_{\mathcal{Q}}$ are both variation independent
parameterizations (i.e.\ any $\bs\lambda_{\mathcal{P}} \in \reals^7$
corresponds to a valid probability distribution).
\end{exm}

We generalize the approach used in the preceding example with the
following definition and proposition.

\begin{dfn}
Let $\mathcal{P}$ be a collection of MLL parameters, and define
\[
\mathcal{P}_{-v} = \{(L, M \setminus \{v\}) \,|\, (L, M) \in \mathcal{P}, \, v \notin L \}.
\]
That is, all effects involving $v$ are removed, and any margins $M$
containing $v$ are replaced by $M \setminus \{v\}$.
\end{dfn}

\begin{prop} \label{prop:reduce} Let $\mathcal{P}$ be a complete
  collection of marginal log-linear parameters over $V$ such that the
  variable $v$ is not in any margin except $V$.  Then
  $\bs\lambda_{\mathcal{P}}$ is a smooth parameterization of $X_V$ if and
  only if $\bs\lambda_{\mathcal{P}_{-v}}$ is a smooth parameterization of
  $X_{V \setminus v}$.  In addition, $\bs\lambda_{\mathcal{P}}$ is
  variation independent if and only if $\bs\lambda_{\mathcal{P}_{-v}}$ is.
\end{prop}

\begin{proof}
  Since $V$ is the only margin containing $v$ and the parameterization
  is complete, we have the parameters $\bs\lambda_{v|V\setminus v} =
  (\lambda_{L}^V : v \in L)$.  Hence we can smoothly parameterize the
  distribution of $X_v \,|\, X_{V \setminus v}$ with these parameters.

  By Theorem \ref{thm:linear}, any other parameter $\lambda_L^V$ such
  that $v \notin L$ is (having fixed the distribution of $X_v \,|\,
  X_{V \setminus v}$) a smooth function of $\lambda_L^{V \setminus
    v}$.  It follows that we have a smooth map between
  $\bs\lambda_{\mathcal{P}}$ and $(\bs\lambda_{\mathcal{P}_{-v}},
  \bs\lambda_{v|V\setminus v})$.  Since
  $\bs\lambda_{\mathcal{P}_{-v}}$ is a function of $p_{V\setminus v}$,
  and $\bs\lambda_{v|V\setminus v}$ smoothly parameterizes
  $p_{v|V\setminus v}$, it follows that $\bs\lambda_{\mathcal{P}}$
  smoothly parameterizes $p_V$ if and only if
  $\bs\lambda_{\mathcal{P}_{-v}}$ smoothly parameterizes $p_{V
    \setminus v}$.

  Lastly, the two pieces $\bs\lambda_{\mathcal{P}_{-v}}$ and
  $\bs\lambda_{v|V\setminus v}$ are variation independent of one
  another because this is a parameter cut, and parameters within
  $\bs\lambda_{v|V\setminus v}$ are all variation independent since
  they are just ordinary log-linear parameters; therefore
  $\bs\lambda_{\mathcal{P}_{-v}}$ is variation independent if and only
  if $\bs\lambda_{\mathcal{P}}$ is.
\end{proof}

\begin{cor} \label{cor:nested} Any complete parameterization in which
  the margins are strictly nested ($M_1 \subset M_2 \subset \cdots
  \subset M_k = V$) is smooth and variation independent.
\end{cor}

Lemma 6 of \citet{forcina:12} deals with the special case $k=2$, which
to our knowledge was the only prior result showing that a
non-hierarchical MLL parameterization may be smooth.

\begin{exm}
Consider
\begin{align*}
&\mathcal{P}: \bmllp
13&3\\
23&23\\
123&1, 2, 12, 13, 123\\
\emllp &&
\mathcal{Q}: \bmllp
13&3\\
123&1, 2, 12, 23, 13, 123\\
\emllp.
\end{align*}
$\mathcal{P}$ does not satisfy the conditions of Proposition
\ref{prop:reduce}; however, applying Theorem \ref{thm:linear} shows
that $\lambda_{23}^{23}$ is just a linear function of
$\lambda_{23}^{123}$ after fixing the other parameters in the margin
$123$, so $\mathcal{P}$ is smooth if and only if $\mathcal{Q}$ is.
Applying Corollary \ref{cor:nested} shows that
$\bs\lambda_{\mathcal{Q}}$ (and therefore $\bs\lambda_{\mathcal{P}}$) is smooth.
\end{exm}

\begin{prop} \label{prop:cond}
  Let $\mathcal{P}$ be a complete parameterization, and suppose that
  for some $v \in V$, and every $A \subseteq V \setminus \{v\}$, the
  sets $A \cup \{v\}$ and $A$ appear as effects within the same margin
  in $\mathcal{P}$.

  Then $\bs\lambda_{\mathcal{P}}$ is a smooth parameterization of $X_V$
  if and only if $\bs\lambda_{\mathcal{P}_{-v}}$ is a smooth
  parameterization of $X_{V \setminus v}$.  In addition,
  $\bs\lambda_{\mathcal{P}}$ is variation independent if and only if
  $\bs\lambda_{\mathcal{P}_{-v}}$ is variation independent.
\end{prop}

\begin{proof}
  Since $A \subseteq V \setminus \{v\}$ and $A \cup \{v\}$
  appear in the same margin, say $M \cup \{v\}$, set
\begin{align*}
\kappa_A^M(x_v) &= \lambda_A^{Mv} + (-1)^{|x_v|} \lambda_{Av}^{Mv}\\
  &= 2^{-|Mv|} \sum_{y_{Mv}} (-1)^{|y_A|} \left[(-1)^{|y_A|} + (-1)^{|y_{Av}| + |x_v|}\right] \log p_{M|v}(y_M, y_v)
\intertext{which is zero unless $x_v = y_v$, leaving}
  &= 2^{-|M|} \sum_{y_{M}} (-1)^{|y_A|} \log p_{M|v}(y_M, x_v)\\
  &= 2^{-|M|} \sum_{y_{M}} (-1)^{|y_A|} \log p_{M|v}(y_M \,|\, x_v).
\end{align*}

But notice this is of the same form as an MLL parameter for the pair
$(A, M)$ over the conditional distribution $p_{V\setminus v|v}(\cdot
\,|\, x_v)$.  It follows that for fixed $x_v$ the parameters $\left\{
  \kappa_A^M(x_v): (A,M) \in \mathcal{P}, v \notin A \right\}$ form a
complete MLL collection of the form $\mathcal{P}_{-v}$ for the
conditional distribution of $X_{V\setminus v} \,|\, X_v = x_v$.  If
$\bs\lambda_{\mathcal{P}_{-v}}$ is smooth then we can smoothly recover
the conditional distribution $p_{V \setminus v|v}$.  Furthermore, if
the effect $\{v\}$ is in a margin $N\cup \{v\}$, then using
(\ref{eqn:equal}) we obtain
\[
\lambda_v^{Nv} = \lambda_v^v + f(p_{N|v}),
\]
and smoothly recover $\lambda_v^v$.  In addition $\lambda_v^v$ is
variation independent of $p_{N|v}$ (since $p_v$, $p_{N|v}$ constitutes
a parameter cut) and has range $\reals$, so the same is true of
$\lambda_v^{Nv}$.

Conversely if $\bs\lambda_{\mathcal{P}}$ is smooth, then given
parameters $\bs\lambda_{\mathcal{P}_{-v}}$ we can set up a dummy distribution
on $p_V$ in which $\kappa_A^M(x_v)=\lambda_A^M$ for each $x_v$, and
$\lambda_v^N = 0$, thus smoothly recovering $p_{V \setminus v}$.
\end{proof}

\begin{exm} \label{exm:cond}
As an example, consider
\begin{align*}
\mathcal{P}: \bmllp
12&2, 12\\
13&3, 13\\
123&1, 23, 123\\
\emllp
\end{align*}
which is not hierarchical, and nor does it satisfy the conditions of
Proposition \ref{prop:reduce}.  However it does satisfy the the
conditions of Proposition \ref{prop:cond} for $v=1$, and
$\mathcal{P}_{-1}=\{(2,2), (3,3), (23,23)\}$,

which is hierarchical and so certainly smooth.  Hence $\mathcal{P}$
represents a smooth parameterization.
\end{exm}

\section{Fixed Point Mappings} \label{sec:fixed}

The previous section gives analytical maps between some
parameterizations, but Propositions \ref{prop:reduce} and
\ref{prop:cond} only apply directly to a relatively small number of
cases.  In this section we build on these results by presenting
conditions for the existence of a smooth map, even without a closed
form expression.

Given a particular complete MLL parameterization $\bs
\lambda_{\mathcal{P}}$, the identity (\ref{eqn:equal}) in Theorem
\ref{thm:linear} can be written in vector form as
\begin{align*}
\bs\eta = \bs\lambda + \bs f(\bs \eta).
\end{align*}
For a given $\bs\lambda$ this suggests that $\bs\eta$ might be
recovered using fixed point methods; the identity (\ref{eqn:deriv})
gives us information about the Jacobian of $\bs f$.

\begin{exm} \label{exm:psi}
Consider the parameterization based on
\begin{align*}
\mathcal{P}: \bmllp
23&2, 23\\
13&1\\
123&12, 3, 13, 123.\\
\emllp
\end{align*}
If we can smoothly recover $\eta_1$, $\eta_2$ and $\eta_{23}$ from
$\bs\lambda_{\mathcal{P}}$ then it follows that
$\bs\lambda_{\mathcal{P}}$ is a smooth parameterization.  From
(\ref{eqn:equal}) we have
\begin{align*}
\left(\begin{matrix}\eta_2\\\eta_{23} \end{matrix}\right)
&= \left(\begin{matrix}\lambda_2^{23}\\\lambda_{23}^{23} \end{matrix}\right) + \bs f(\eta_{1},\eta_{12},\eta_{13},\eta_{123});\\
\intertext{since $\eta_{12}$, $\eta_{13}$ and $\eta_{123}$ are given in the parameterization
we can assume these to be fixed, so abusing notation slightly}
\left(\begin{matrix}\eta_2\\\eta_{23} \end{matrix}\right) &=  \left(\begin{matrix}\lambda_2^{23}\\\lambda_{23}^{23} \end{matrix}\right) + \bs f(\eta_1) = \left(\begin{matrix}\lambda_2^{23} + f_1(\eta_1)\\\lambda_{23}^{23}  + f_2(\eta_1) \end{matrix}\right).
\end{align*}
Similarly,
$\eta_{1} = \lambda_1^{13} + g(\eta_{2}, \eta_{23})$ for some smooth $g$, so
$\eta_1$ is a solution to the equation
\begin{align*}
x &= \lambda_1^{13} + g(\lambda_2^{23} + f_1(x), \lambda_{23}^{23} + f_2(x))\\
&\equiv \Psi(x).
\end{align*}
If $\Psi$ can be shown to be a contraction mapping, then we are
guaranteed to find a unique solution, and therefore recover the joint
distribution.  In addition, if $\Psi$ is a contraction for all
$\bs\eta$, then since it varies smoothly in $\bs\eta$ we will have
shown that $\bs\lambda_{\mathcal{P}}$ is a smooth parameterization.

\end{exm}

Define $\epsilon = \min_{x_V} p(x_V)$ to be the smallest amount of
probability assigned to any cell in our joint distribution, and
$\Delta_{\epsilon} = \{p : \min_{x_V} p(x_V) \geq \epsilon \}$ to
be the probability simplex consisting of such distributions.  The
Jacobian of an otherwise smooth parameterization can become singular on
the boundary of the probability simplex, so it is useful to have
control over this quantity.

The next result allows us to control the magnitude of the columns (or
rows) of the Jacobian of $\Psi$ in certain examples.  The proof is
given in the appendix.

\begin{lem} \label{lem:mag}
Let $J \subseteq M$, and $\emptyset \neq K \subseteq V \setminus M$.  Then
\begin{align*}
\sum_{\emptyset \neq C \subseteq M} \left| \frac{\partial \lambda_{C}^{M}}{\partial \eta_{JK}} \right|^2 \leq 1 - \epsilon.
\end{align*}
Alternatively, if $\emptyset \neq C \subseteq M$, then
\begin{align*}
&\sum_{\substack{J \subseteq M}} \left| \frac{\partial \lambda_C^{M}}{\partial \eta_{JK}} \right|^2 \leq 1 - \epsilon.
\end{align*}
\end{lem}

\begin{exm} \label{exm:psi2} Returning to the parameterization in
  Example \ref{exm:psi}, the derivative of $\Psi$ is
\begin{align*}
\Psi'(x)
&= \frac{\partial \lambda_1^{13}}{\partial \eta_{2}} \frac{\partial \lambda_2^{23}}{\partial \eta_{1}}
+ \frac{\partial \lambda_1^{13}}{\partial \eta_{23}} \frac{\partial \lambda_{23}^{23}}{\partial \eta_{1}},
\end{align*}
which is the dot product of the vectors
\begin{align*}
&\left( \frac{\partial \lambda_1^{13}}{\partial \eta_{2}}, \frac{\partial \lambda_1^{13}}{\partial \eta_{23}} \right)^T &&
\left( \frac{\partial \lambda_2^{23}}{\partial \eta_{1}}, \frac{\partial \lambda_{23}^{23}}{\partial \eta_{1}} \right)^T.
\end{align*}
By applying the two parts of Lemma \ref{lem:mag}, these vectors each
have magnitude at most $1 - \epsilon$.  Hence $|\Psi'(x)| \leq
1-\epsilon$, and $\Psi$ is a contraction on $\Delta_\epsilon$ for
every $\epsilon > 0$.  It follows that the equation has a unique
solution among all positive probability distributions (and this can be
found by iteratively applying $\Psi$ to any initial distribution), and
by the inverse function theorem it is a smooth function of $\bs
\lambda$.  Hence $\bs\lambda_{\mathcal{P}}$ is indeed smooth.
\end{exm}

\begin{rmk}
  \citet{forcina:12} also uses fixed point methods to recover
  distributions from marginal log-linear parameters, but that approach
  involves computing probabilities directly.  We discuss those methods
  in Section \ref{sec:cycles}.
\end{rmk}

Lemma \ref{lem:mag} enables us to formulate the following
generalization of the idea used in the example above.

\begin{lem} \label{lem:fixedpoint}
  Let $\mathcal{P}$ be complete and such that for any $(L,M) \in
  \mathcal{P}$ with $M \subset V$, there is at most one other
  margin $N \subset V$ in $\mathcal{P}$ with $L \cap (V \setminus
  N) \neq \emptyset$.  Then $\bs\lambda_{\mathcal{P}}$ is smooth.
\end{lem}

\begin{proof}
By Theorem \ref{thm:linear},
\begin{align*}
\eta_{L}  = \lambda_{L}^{M} + f(\bs\lambda_{V\setminus M|M}).
\end{align*}
Since $N$ is the only margin in $\mathcal{P}$ such that $L \cap (V
\setminus N) \neq \emptyset$, it follows that all the parameters in
$\bs\lambda_{V \setminus M|M}$ are known and fixed except for
$(\lambda_{K}^V : K \in \mathbb{L}_N)$, where $\mathbb{L}_N$ is the
set of effects contained in the margin $N$.  Hence
\begin{align}
\eta_{L} = \lambda_{L}^{M}  + f(\bs\eta_{\mathbb{L}_N}). \label{eqn:fp}
\end{align}
Now, consider the vector equation obtained by stacking (\ref{eqn:fp})
over all pairs $(L,M) \in \mathcal{P}$.  This defines a fixed point
equation whose solution is $\bs\eta$, and the column of the Jacobian
corresponding to $L$ has non-zero entries
\begin{align*}
&\frac{\partial \lambda_{C}^{N}}{\partial \eta_{L}}, \qquad C \in
\mathbb{L}_N.
\end{align*}
From Lemma \ref{lem:mag}, each column has magnitude at most
$1-\epsilon$, and therefore the mapping is a contraction on
$\Delta_{\epsilon}$ for each $\epsilon > 0$.  It follows that the
fixed point equation has a unique solution which, by the inverse
function theorem, is a smooth function of $\bs\lambda$.
\end{proof}

From this result we obtain the following corollary, the conditions of
which are easy to verify.

\begin{cor}
Any complete parameterization with at most three margins is smooth.
\end{cor}

\begin{proof}
  Since one of the margins must be $V$, it is clear that the
  conditions of Lemma \ref{lem:fixedpoint} hold.
\end{proof}

\begin{exm} \label{exm:reduce}
Consider $\mathcal{P}$ below.
\begin{align*}
\mathcal{P}:
\bmllp
1&1\\
12&2\\
13&3\\
123&12, 13, 23, 123\\
\emllp
&&
\mathcal{Q}:
\bmllp
1&1\\
123&2, 12, 3, 13, 23, 123\\
\emllp
\end{align*}
Although it does not satisfy the conditions of Lemma
\ref{lem:fixedpoint} directly, one can use the basic idea to set up a
smooth contraction mapping from $\bs\lambda_{\mathcal{P}}$ to
$\bs\lambda_{\mathcal{Q}}$; since $\mathcal{Q}$ is hierarchical, both
parameterizations are smooth.
\end{exm}

\section{Cyclic Parameterizations} \label{sec:cycles}

This section takes a third approach to determining smoothness, by
using Markov chain theory to recover certain marginal distributions.
This method allows us to demonstrate the smoothness of certain
conditional independence models.

\citet[][Example 2]{forcina:12} considers the model defined (up to
some relabelling) by the conditional independences
\begin{align}
&X_1 \indep X_2 \,|\,  X_3, &&X_1 \indep X_3 \,|\, X_4, &&X_1 \indep X_4 \,|\, X_2, \label{eqn:cim}
\end{align}
which is equivalent to setting the parameters
\begin{align}
\bmllp
123&12, 123\\
134&13, 134\\
124&14, 124\\
\emllp \label{eqn:cimp}
\end{align}
to zero.  Note that we cannot embed these parameters into a larger
hierarchical parameterization, because each pairwise effect will
`belong' to a margin preceding it; for example, $12$ is a subset of
$124$, so for hierarchy the margin $123$ must precede $124$; by a
similar argument, $124$ must precede $134$ which must precede $123$.
We therefore have a cyclic parameterization, referred to as a `loop'
by Forcina.  None of the methods used in the previous sections seem
well suited to dealing with this situation.

\citet{forcina:12} presents an algorithm for recovering joint
distributions given parameterizations of this kind, together with a
condition under which it is guaranteed to converge to the unique
solution.  However, this condition is on the spectral radius of a
complicated Jacobian, and is difficult to verify except in a few
special cases: a numerical test is suggested, but this does not
constitute a proof of smoothness.  Here we show that, at least in some
cases, Forcina's algorithm can be recast as a Markov chain whose
stationary distribution is some margin of the relevant probability
distribution.

\begin{thm} \label{thm:markov}
  Let $A_1, \ldots, A_k$ be a disjoint sequence of sets with $k \geq
  2$ such that the conditional distributions $p(x_{A_i} \,|\,
  x_{A_{i-1}}) > 0$ for $i = 2,\ldots,k$ are known, together with
  $p(x_{A_1} \,|\, x_{A_k})$.  Then the marginal distributions $p(x_{A_i})$
  are smoothly recoverable.
\end{thm}

\begin{proof}
Define a $|X_{A_1}| \times |X_{A_1}|$ matrix $M$ with entries
\begin{align*}
M(x_{A_1}', x_{A_1}) = \sum_{x_{A_k}} \cdots \sum_{x_{A_2}} p(x_{A_1} \,|\, x_{A_k}) p(x_{A_k} \,|\, x_{A_{k-1}}) \cdots p(x_{A_2} \,|\, x_{A_1}').
\end{align*}
This is a (right) stochastic matrix with strictly positive entries,
and the marginal distribution $p(x_{A_1})$ satisfies
\begin{align*}
p(x_{A_1}) = \sum_{x_{A_1}'} p(x_{A_1}') M(x_{A_1}', x_{A_1}).
\end{align*}
In other words, $p(x_{A_1})$ is an invariant distribution for the
Markov chain with transition matrix defined by $M$.  Since $M$ has a
finite state-space and all transition probabilities are positive, the
chain is positive recurrent and the equations have a unique solution
\citep[see, e.g.][]{norris:97}.  Hence $p(x_{A_1})$ is defined by the
kernel of the matrix $I - M^T$, and this is a smooth function of the
original conditional probabilities.
\end{proof}

\begin{rmk}
  The Markov chain corresponding to $M$ is that which would be
  obtained by picking some $X_{A_1}$, and then evolving $X_{A_i}$
  using $p(x_{A_i} \,|\, x_{A_{i-1}})$ until we get back to $i=1$.  The
  equations can be solved iteratively by repeatedly right
  multiplying any positive vector by $M$, so that it converges to the
  stationary distribution of the chain; this corresponds precisely to
  Forcina's algorithm.

\end{rmk}

\begin{exm}[\citet{forcina:12}, Example 9] \label{exm:cycle}
Consider the cyclic parameterization $\mathcal{P}$.
\begin{align*}
&\mathcal{P}: \bmllp
12&1, 12\\
23&2, 23\\
13&3, 13\\
123&123\\
\emllp &&
\mathcal{Q}: \bmllp
3&3\\
23&2, 23\\
12&1, 12\\
13&13\\
123&123\\
\emllp
\end{align*}
The parameters corresponding to the first
three margins in $\mathcal{P}$ are equivalent to the conditional
distributions $p_{1|2}$, $p_{2|3}$ and $p_{3|1}$.  Using the
conditionals in the manner suggested by Theorem \ref{thm:markov}, we
can smoothly recover (for example) the margin $p_3$ (or equivalently
$\lambda_3^3$), and consequently $\mathcal{P}$ is equivalent to the
hierarchical parameterization $\mathcal{Q}$.
\end{exm}

\begin{exm} \label{exm:cycle2}
  The parameters (\ref{eqn:cimp}) can be embedded in the
  complete parameterization $\mathcal{P}$ below.
\begin{align*}
&\mathcal{P}:
\bmllp
123&2, 23, 12, 123\\
134&3, 34, 13, 134\\
124&4, 24, 14, 124\\
1234&(other subsets)
\emllp,&&
\mathcal{P}_{-1}:
\bmllp
23&2, 23\\
34&3, 34\\
24&4, 24\\
234&234\\
\emllp.
\end{align*}
$\mathcal{P}$ satisfies Proposition \ref{prop:cond} with $v=1$ and
reduces to $\mathcal{P}_{-1}$, which is isomorphic to the smooth
parameterization in Example \ref{exm:cycle}.  Hence $\mathcal{P}$ is
smooth, and the conditional independence model (\ref{eqn:cim}) is a
curved exponential family.
\end{exm}

\begin{exm} \label{exm:cycle3}
Consider the model defined by
\[
X_1 \indep X_2 \,|\, X_3, \qquad X_2 \indep X_4 \,|\, X_1, \qquad X_1 \indep X_3 \,|\, X_4, \qquad X_3 \indep X_4 \,|\, X_2;
\]
it consists of setting the parameters in $\mathcal{P}$ below to zero.
\begin{align*}
& \mathcal{P}:
\bmllp
123&12, 123\\
124&24, 124\\
134&13, 134\\
234&34, 234\\
\emllp && \mathcal{Q}:
\bmllp
14&1, 4, 14\\
23&2, 3, 23\\
123&12, 123\\
124&24, 124\\
134&13, 134\\
234&34, 234\\
1234&1234\\
\emllp
\end{align*}
We can embed $\mathcal{P}$ in the complete parameterization
$\mathcal{Q}$.  Note that using $\lambda_4^{14}, \lambda_{14}^{14}$
and the fact that $X_4 \indep X_2 \,|\, X_1$, means we can construct
the conditional distribution $p(x_4 \,|\, x_1, x_2)$.  Similarly we
have $p(x_3 \,|\, x_2, x_4)$, $p(x_1 \,|\, x_3, x_4)$ and $p(x_2 \,|\,
x_1, x_3)$.  In a manner analogous to the previous example, we can set
up a Markov chain whose stationary distribution is the marginal
$p(x_1, x_2)$ as follows.  First pick $x_1^{(0)},x_2^{(0)}$.  Now, for $i > 0$
\bitem
\item draw $x_4^{(i)}$ from the distribution $p(x_4 \,|\, x_1^{(i-1)}, x_2^{(i-1)})$;
\item draw $x_3^{(i)}$ from the distribution $p(x_3 \,|\, x_2^{(i-1)}, x_4^{(i)})$;
\item draw $x_1^{(i)}$ from the distribution $p(x_1 \,|\, x_3^{(i)}, x_4^{(i)})$;
\item draw $x_2^{(i)}$ from the distribution $p(x_2 \,|\, x_1^{(i)},
  x_3^{(i)})$.
\eitem
Then the distribution of $(x_1^{(i)}, x_2^{(i)})$ converges to $p_{12}$.
We can therefore smoothly recover a distribution satisfying the conditional
independence constraints from the 7 free parameters.  The dimension of the
model is full, so we have a smooth parameterization of the model, which is
therefore a curved exponential family \citet{lau:96}.
\end{exm}

Note that the construction of the Markov chain in Example
\ref{exm:cycle3} is only possible when the conditional independence
constraints hold, so---unlike in Examples \ref{exm:cycle} and
\ref{exm:cycle2}---we have not actually demonstrated that
$\bs\lambda_{\mathcal{Q}}$ is generally smooth, only that the model
defined by setting $\bs\lambda_{\mathcal{P}} = \bs 0$ is a curved
exponential family.

\begin{rmk}
  Some conditional independence models are non-smooth: e.g.\ the model
  defined by $X_1 \indep X_2, X_4$ and $X_2 \indep X_4 \,|\, X_1, X_3$
  \citep{drton:09}.  This is essentially because it requires that
  $\lambda_{124}^{124} = \lambda_{124}^{1234} = 0$, and setting
  repeated (non-redundant) effects to zero always leads to non-smooth
  parameterizations.

  We remark that all discrete conditional independence models on four
  variables either require repeated effects to be constrained in
  different margins, or can be shown to be smooth using the results of
  this section.  However, the next example shows that for five
  variables the picture is incomplete.
\end{rmk}

\begin{exm} \label{exm:cin}
The conditional independence model defined by
\begin{align*}
& X_1 \indep X_2 \,|\, X_3, && X_1 \indep X_5 \,|\, X_2, && X_1 \indep X_3 \,|\, X_4, \\
& X_3 \indep X_5 \,|\, X_1, &&  X_3 \indep X_4 \,|\, X_2, X_5
\end{align*}
contains no repeated effects, and yet does not appear to be
approachable using the methods outlined above.  Empirically, Forcina's
algorithm seems to converge to the correct solution, which suggests
that the model is indeed smooth.
\end{exm}

\section{Discussion} \label{sec:discuss}

We have presented three new approaches to demonstrating that complete
but non-hierarchical marginal log-linear parameterizations are smooth,
although a general result eludes us.  Note that each of the approaches
provides an explicit algorithm for obtaining the probabilities from
the parameterization, either using the map in Section
\ref{sec:smooth}, the fixed point iteration in Section
\ref{sec:fixed}, or the Markov chain in Section \ref{sec:cycles}.

There are 104 complete MLL
parameterizations on three variables, of which 23 are hierarchical and
a further 4 consist of only two margins, so are smooth by the results
of \citet{br:02} and \citet{forcina:12} respectively.  These 27 were
the only ones known to be smooth prior to this paper.

A further 5 can be shown smooth using Proposition \ref{prop:reduce},
and one using Proposition \ref{prop:cond} (Example \ref{exm:cond}).
Another 26 can be dealt with using Lemma \ref{lem:fixedpoint} in
combination with other methods, and the approach in Example
\ref{exm:reduce} can be applied to three more.  Example
\ref{exm:cycle} brings the total number of known smooth models to 63.

In addition, of the remaining 41 complete parameterizations, there are
smooth mappings between a group of four and a group of three, so it
remains to establish the smoothness (or otherwise) of at most 36
distinct parameterizations.  As an example of a parameterization whose
smoothness is still not established, consider:
\begin{align*}
& \mathcal{P}:
\bmllp
12&1, 2\\
13&3, 13\\
23&23\\
123&12, 123\\
\emllp.
\end{align*}
We conjecture that any complete parameterization is smooth, a result
which would enable us to show that models such as that given in
Example \ref{exm:cin} are curved exponential families of
distributions.

\begin{conj}
Any complete MLL parameterization is smooth.
\end{conj}

\bibliographystyle{plainnat}
\bibliography{mybib}

\newpage

\appendix

\section{Technical Proofs}

\subsection{Proof of Lemma \ref{lem:mag}}

\begin{lem} \label{lem:alt}
  Let $\bs d = (d_A)$ be a vector indexed by subsets $A \subseteq
  \{1,\ldots,k\}$.  Then $\|\bs d\| < 1$ if and only if for any
$B \subseteq [k] \equiv \{1,\ldots,k\}$,
\begin{align*}
\left|\sum_{A \subseteq [k]} (-1)^{|A \cap B|} d_A \right| < 1.
\end{align*}
\end{lem}

\begin{proof}
  The $2^k\times 2^k$-matrix $M$ with $(B,A)$th entry $M_{B,A} =
  2^{-k/2} (-1)^{|A \cap B|}$ is orthogonal, and therefore preserves
  vector lengths.  Then the vector $M \bs d$ has entries with
  magnitude at most $2^{-k/2}$, and therefore has total magnitude at
  most 1.  The same is therefore true of $\bs d$.
\end{proof}

\begin{proof}[Proof of Lemma \ref{lem:mag}]
For $C \subseteq M$, define
\begin{align*}
d_{C} &\equiv -2^{-|M|} \sum_{x_V} (-1)^{|x_{C \triangle {(J \cup K)}}|} p(x_{V\setminus M} \,|\, x_M)\\
&= -2^{-|M|} \sum_{x_V} (-1)^{|x_{C \triangle J}| + |x_K|} p(x_{V\setminus M} \,|\, x_M)
\end{align*}
so that $d_{C} = \frac{\partial \lambda_{C}^{M}}{\partial
  \eta_{JK}}$ for $C \neq \emptyset$.  Given $y_M$,
\begin{align*}
\sum_{C \subseteq M} (-1)^{|y_C|} d_{C} &=  -2^{-|M|} \sum_{C \subseteq M} (-1)^{|y_{C}|} \sum_{x_V} (-1)^{|x_{C \triangle J}| + |x_K|} p(x_{V\setminus M} \,|\, x_M),
\end{align*}
and note that
\begin{align*}
\lefteqn{\sum_{C \subseteq M} (-1)^{|y_{C}|} (-1)^{|x_{C \triangle J}|}}\\
&= \sum_{C \subseteq M \setminus \{v\}} (-1)^{|y_{C}|} (-1)^{|x_{C \triangle J}|} + (-1)^{|y_{Cv}|} (-1)^{|x_{Cv \triangle J}|}\\
&= \sum_{C \subseteq M \setminus \{v\}} (-1)^{|y_{C}|} (-1)^{|x_{C \triangle J}|} \left\{1+(-1)^{|y_v|} (-1)^{|x_{v}|} \right\}\\
\intertext{where the expression in braces is 2 if $x_v=y_v$ or 0 otherwise, so}
&= \sum_{C \subseteq M \setminus \{v\}} 2 (-1)^{|y_{C}|} (-1)^{|x_{C \triangle J}|} \Ind_{\{x_v = y_v\}}\\
&= 2^{|M|} \Ind_{\{x_M = y_M\}}.
\end{align*}
Hence
\begin{align*}
\sum_{C \subseteq M} (-1)^{|y_C|} d_{C} &=  - \sum_{x_{V \setminus M}} (-1)^{|x_K|} p(x_{V\setminus M} \,|\, y_M),
\end{align*}
which is an alternating sum of probabilities which sum to one, so
has absolute value at most $1 - \epsilon$.  The result follows from Lemma
\ref{lem:alt}.
The second result is essentially identical, due to the symmetry between
$L,K$ in (\ref{eqn:deriv}).
\end{proof}

\end{document}


TITLE: Urn balls multinomial with joint probability function
QUESTION [0 upvotes]: An urn contains $w$ white, $b$ black and $r$ red balls. $n$ extractions with replacement are made. $X_w$, $X_b$ and $X_r$ are the random variables representing the number of white, black and red balls extracted. I should find:

domain and probability function for marginal $X_i$, where $i$ is $w$ or $b$ or $r$
domain and joint probability function for random vector $(X_w, X_b, X_r)$
$P(a<X_i<b)$

Question 1
I set $t = w+b+r$ as the sum of the balls inside the urn, so for each colors the probability is $p_w=\frac{w}{w+b+r}$, $p_b=\frac{b}{w+b+r}$ and $p_r=\frac{r}{w+b+r}$
Now the marginal for $X_i$ should be $P(X_i=e)=\binom{n}{e}p_i^e(1-p_i)^{n-e}$ with domain for $X_i$ ${0}\le{e}\le{i}$
Question 2
The domain is the set of 3-tuple $(n_w,n_b,n_r)$ where $n_w + n_b + n_r = n$ and joint probability function is
$P(X_w=n_w, X_b=n_b, X_r=n_r) = \frac{n!}{{n_w!}{n_b!}{n_r!}}{p_w}^{n_w}{p_b}^{n_b}{p_r}^{n_r}$
Question 3
$P(a<X_i<b) = \sum_{k=a-1}^{b-1}\dbinom{n}{k}{p_i}^k(1-p_i)^{n-k}$
Is everything correct?

REPLY [1 votes]: Everything is mostly correct.
In Question $3$, since the inequalities on the left are strict, the bounds shouldn't be included in the summation on the right.
In Question $1$, your dual use of $i$ as an index and as the corresponding variable is slightly unorthodox, but clear enough upon reflection.


TITLE: Clearification on the proof of Goursat's Theorm from Stein's Complex Analysis
QUESTION [2 upvotes]: In the text Complex Analysis by Stein and Shakarchi: Add to Goursat's Theorem is stated as follows: If $\Omega$ is an open set in $C$ and $T \subset \Omega$ a triangle whose interior is also contained in $\Omega$ then the following occurs:
$$ 1.) \, \, \, \, \, \, \, \, \, \, \int_{T}f(z)dz = 0$$
The initial proof of 1.) topologically speaking is done by considering the original triangle, and by giving a bisection of the triangle into it's respective pieces i.e( new tringle are created from the bisection $T_{1}^{(1)}$,$T_{2}^{(2)}$,$T_{3}^{(3)}$,$T_{4}^{(4)}$). It is also important to note the visually the new tringles remain consistent of that of the original triangle. After this I observed that  the author takes the integral of each of the triangles and combines them as follows in 2.)
$$2.) \, \, \, \int_{T_{0}^{(0)}}f(z)dz = \int_{T_{1}^{(1)}}f(z)dz \, + \int_{T_{2}^{(2)}}f(z)dz \, + \int_{T_{3}^{(3)}}f(z)dz + \int_{T_{4}^{(4)}}f(z)dz$$
The following procedure in 2.) as well as the topological basis can be visualized below in the image.What I failed to observe within the proof is why the following is stated within 4.)
$\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $|,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$
$\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$
$\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$  $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ For some j we must have the following:
$$  4.) \, \, \, \, |\int_{T_{0}^{(0)}}f(z)dz | \leq 4 \, |\int_{T_{j}^{(1)}}f(z)dz |$$

REPLY [2 votes]: Okay, I think you're asking why we can justify the equality in (4), so I'll take a stab.
Since $|\int_{T_1^{(1)}} f(z)\ dz + \int_{T_2^{(1)}} f(z)\ dz  + \int_{T_3^{(1)}} f(z)\ dz  + \int_{T_4^{(1)}} f(z)\ dz| = |\int_{T_0} f(z)\ dz|$ the triangle inequality gives us
$$|\int_{T_1^{(1)}} f(z)\ dz| + |\int_{T_2^{(1)}} f(z)\ dz|  + |\int_{T_3^{(1)}} f(z)\ dz|  + |\int_{T_4^{(1)}} f(z)\ dz| \geq |\int_{T_0} f(z)\ dz|$$ so that $\frac{1}{4}|\int_{T_0} f(z)\ dz| \leq |\int_{T_1^{(j)}} f(z)\ dz|$ for some $j$. Multiply by 4 on both sides and you're there!
Obviously, we can repeat this for the nested triangles of $T_j^{(1)}$ and obtain the generalized statement that
$$|\int_{T_0} f(z)\ dz| \leq 4^n|\int_{T_j^{(n)}} f(z)\ dz| $$


TITLE: Is this a Lipschitz function in 2d?
QUESTION [2 upvotes]: I want to show that the function $A(x,y)$ is Lipschitz in two dimensions.
The function is defined as follows
$$A(x,y)=a\frac{\Phi(-y)e^{0.5(y^2-x^2)}+\Phi(-x)e^{0.5(x^2-y^2)}}{(1-a)[1-a(1-\Phi(x))(1-\Phi(y))]+2a\Phi(x)\Phi(y)}$$
where $a\in[-1,1]$ and $\Phi$ is the standard normal cumulative distribution function.
How can I show this is Lipschitz?
Could anyone help me in the right direction?

REPLY [1 votes]: You other question had products like $\Phi(y)\Phi(-y)$, which provide rapid decay in both positive and negative directions.
But here you have things like $\Phi(-y)e^{0.5(y^2-x^2)}$, where $\Phi(-y)\to 1$ as $y\to-\infty$. The exponential factor grows rapidly as $y\to -\infty$.
Let's put $x=0$ and let $y\to -\infty$. Then
$$(1-a)[1-a(1-\Phi(x))(1-\Phi(y))]+2a\Phi(x)\Phi(y) \to (1-a)(1-a/2) $$
and
$$\Phi(-x)e^{0.5(x^2-y^2)}\to 0,$$
so $A(x,y)$ is asymptotic to $\dfrac{a}{ (1-a)(1-a/2)} e^{0.5(y^2-x^2)}$.


TITLE: Proof that these polynomials form a base
QUESTION [2 upvotes]: Prove that the polynomials
$$f'_n=e^{-z}\frac{d^n}{dz^n}(z^{2n}e^z), n=0,1,2,...$$
form a basis in the vector space of all polynomials. Find the expansion coefficients of the $f'_n$ in terms of the basis functions $f_n=z^n$.
I have tried for some time to prove this but I start to believe there is an error with what is being asked. I know that these really look like the Laguerre polynomials, except that the $z^{2n}$ term messes everything up.
Any ideas?

REPLY [3 votes]: The degree of $f_n'$ equals $2n$. So any non trivial linear combination has even degree. This implies that for example $z$ is not a linear combination of the $f_n'$.


TITLE: How to use Bernoulli equation in a rotating frame and is it always true that bernoulli will work?
QUESTION [0 upvotes]: Lets say a fluid is rotating in a cylindrical shape container with every circular fluid streamline rotating with same $w$ by some external mechanism, now imagine asking for pressure difference between two points on a circular cross section of container, how will one apply bernoulli in rotating frame as such ik its validity is here using some pseudo potential energy term, can anyone explain how it will work?

REPLY [1 votes]: The Bernoulli equation is a statement of conservation of energy from: pressure, velocity and elevation. But a specific limitation on the equation is for the effects of "centrifugal force" in curved flow. Also this is not really a case of fluid flow as such because each element of fluid maintains its position in the container (apart from convection). There is no exchange between the three types of energy as the element moves in a circle. Also there is no energy being imparted to the system as a whole, but the velocity of the fluid increases linearly with distance from the centre of rotation, so therefore the kinetic energy of a particle increases with the square of the distance.
The terms $P$ and ${\rho}gh$ account for the linear change in hydrostatic pressure with height (as the elevation decreases the pressure increases), but separately there is an increase in pressure due to velocity: ${\rho}v^2/2 = {\rho}r^2{\omega}^2/2$. If the Bernoulli equation applied, the hydrostatic pressure would have to decrease as the velocity pressure increased away from the centre, which clearly does not happen. So in this case the total energy is not the same for all particles, it is greater, further from the centre.


TITLE: Where can I find a good comprehensive read about the history of Mathematics?
QUESTION [6 upvotes]: I'm doing a Bachelor of Pure Mathematics in Unisversity, and while reading through the book that outlines the course selections, I found one that is listed as "rarely offered", which the department says will likely never be offered again. It is titled "The History of Math", and the synopsis of the course reads as follows:

How did the many powerful theories of modern mathematics develop, and which major mathematicians influenced and shaped this development?
How did the many powerful theories of modern mathematics develop, and which major mathematicians influenced and shaped this development?
In this course, this historical development of mathematics is exemplified by concentrating in some detail on the history of the calculus from its early beginnings through its 18th-century progress to the introduction of mathematical analysis in the 19th century, and the further developments to set theory, the beginnings of topology, and to the structural point of view of the 20th century. Its emphasis is on a deeper understanding of the dynamic nature of mathematics and of the interrelations among various branches of mathematics. This should lead to a better understanding of familiar mathematical topics and also allows the introduction of new mathematical content from a modern point of view.

This is a shame, because I am deeply interested in finding out how some of the popular theories were developed and molded to be taught at lower levels. The one I can think of in particular is how Newton et al came up with the Fundamental Theory of Calculus.
So, aside from haphazardly looking up and reading poorly written (or lacking information) articles on wikipedia, are there any good books that cover this sort of material?

REPLY [1 votes]: Mathematics and Its History by John Stillwell is an excellent survey of the history of mathematics, pitched at undergraduate level. All of Stillwell's books are superb, in my opinion, and this is no exception.


TITLE: Intuition behind the definition of a surface
QUESTION [3 upvotes]: From the book Elementary Differential Geometry by Andrew Pressley, this is the definition of a surface:

A subset $S$ of  $\mathbb{R^3}$ is a surface if, for every point $p \in S$, there is an open set $U$ in $\mathbb{R^2}$ and an open set $W$ in $\mathbb{R^3}$ containing ${P}$ such that $S \cap W$ is homeomorphic to $U$.

I am struggling to understand this definition.
Why do we consider an open set $U$ in $\mathbb{R^2}$
Why do we consider the intersection of $S$ and $W$?
What is the motivation behind this definition?

REPLY [1 votes]: It may help to think about this definition on an example. Let's assume that $S$ is a sphere in $\mathbb{R}^3$ and let $p\in S$. You can think of $W$ as a small portion of space that contains $p$ (a small ball for example). Now $S\cap W$ looks like a small "patch" on the sphere. And we notice that this small patch is approximately flat. If it's small enough (so if $W$ is small enough), you wouldn't distinguish it from some small subset of $\mathbb{R}^2$. Actually, it is homemorphic to a small subset $U$ of $\mathbb{R}^2$. This means that $S\cap W$ can be transformed in a continuous, bijective way, into $U$. So locally, the sphere looks like a plane, topologically speaking.
And the term "locally" is important, because you can't do this for the entire sphere, you won't transform it into a plane with a continuous, bijective transformation. Yet, for every point of the sphere, there is a small neighborhood around this point where you can do it.
More generally, $S$ is a surface if it locally looks like a plane, around every point.
When you draw a circle in $\mathbb{R}^2$ and you look at some small portion of this circle, it looks exactly like a segment, which is a small portion of $\mathbb{R}^2$. If you twist it a little (homeomorphism), it'll be a segment. For surfaces it's exactly the same principle, with one more dimension.


TITLE: Finding when the solutions of $\sin z=u$ with $u\in\mathbb{R}$ are real.
QUESTION [1 upvotes]: Let $z=x+iy$ be a complex number.
I want to find when the solutions of the equation
$$\sin z =u,u\in\mathbb{R}$$
are real.
I used this approach:
Since $\sin z=\sin x\cosh y+i \cos x\sinh y=u$, then
$$\sin x\cosh y=u,\cos x \sinh y=0.$$
From the second equation we have $\cos x=0$ or $y=0$. If $y=0$, the solutions are real, (since $\cosh 0=1$ the solutions would be the solutions of $\sin x=u$). If $\cos x=0$ then $x=\frac{\pi}{2}+n\pi,n\in\mathbb{Z}$ and $\sin x=(-1)^n$.
So, if $-1<u<1$, the case $\cos x=0, y\neq 0$ is impossible since $|\pm\cosh y
|\geq 1$ for all $y$, so in this case all the solutions are real.
For the case $|u|=1$ we get $\pm\cosh(y)=u$ and restricting $x$ appropriately we get $y=0$ too.
In the case that $u>1$ or $u<-1$ then it's clear that $z\notin\mathbb{R}$ so the case $y=0$ is impossible. But since $\cosh y$ doesn't change it's sign, if $u>1$ we must restrict $x=\frac{\pi}{2}+2k\pi$ and we get 2 solutions $y_1=-y_2$ of $\cosh y=u$. So the solutions are given by $y=\pm y_1$ and $x=\frac{\pi}{2}+2k\pi$.
In the case $u<-1$ it's similar, the solutions are $y=\pm y_1',x=-\frac{\pi}{2}+2k\pi$.

I might use another approach:
$$z=\arcsin u=-i\ln(iu\pm\sqrt{1-u^2})$$
If we consider $|u|<1$, then $\sqrt{1-u^2}$ is real, so
$$|\pm\sqrt{1-u^2}+iu|=1$$
and
$$-i\ln(iu\pm\sqrt{1-u^2})=\theta_u+2k\pi$$
where $\theta_u$ satisfies $\cos \theta_u =\pm\sqrt{1-u^2}, \sin \theta_u=u$. So the solutions are real.
In the case $|u|>1$, $\sqrt{1-u^2}=i\sqrt{u^2-1}$ so in this case
$$|i\pm\sqrt{u^2-1}+iu|=|\pm\sqrt{u^2-1}+u|$$
and
$$-i\ln(iu\pm\sqrt{1-u^2})=-i\ln|\pm\sqrt{u^2-1}+u|+2k\pi$$
in this case the solutions are not real, since $u^2-1\neq u^2$ for all $u$.
Is there any easier, shorter, funnier, or even a strange approach to the problem (It doesn't have to be shorter or easier, it may be a large and strange approach, or a geometric one, or something like that)?
Note that we're trying to find out when all the zeroes of $\sin z-u=0$ are real.

REPLY [1 votes]: Since you assume that $u$ is real, there is an alternative method: Use Pythagorean Theorem we get $\cos z=\pm\sqrt{1-u^2}$=$\pm{i}\sqrt{u^2-1}$. Now use $e^{iz}=\cos z+i\sin z$, take the $\ln$ on both sides and you are done. (assuming you know how to take the complex $\ln$). A numerical problem is worked out here, it might be helpful: Finding $z\in \mathbb C$ with $\sin z = 2$?


TITLE: On the methods used to solve an O.D.E.
QUESTION [1 upvotes]: (I'm sorry for the format of the mathematical symbols in advance)
Take the differential equation dy/dx = x * y, for example, the process for solving it goes as follows.
1.) $dy = x * y * dx$
2.) $1/y * dy = x * dx$
3.) $∫ 1/y dy = ∫ x dx$
4.) $ln(y) = (x^2)/2$
5.) e^ln(y) = e^((x^2)/2)
6.) y = e^((x^2)/2)
Now my question is in regards to step 3. When we take the integral of both sides why do we not treat the infinitesimals dy and dx as part of the function that we are integrating? In other words, when taking the integrals why didn't we get $∫ 1/y dy dx = ∫ x dx dx$ instead? What happened to them?
P.S. I am currently taking the regular high school Calculus 2 course so please keep the "advanced" mathematics to a bare minimum when answering the question.

REPLY [3 votes]: $$ y'= xy $$
$$ \frac {y'}{y}= x $$
Take inetgral now on both sides
$$ \color{red}{\int} \frac {y'}{y} \color{red}{dx}=\color{red}{\int} x \color{red}{dx}$$
$$ \color{red}{\int} \frac 1y \frac {dy}{dx}\color{red}{dx}=\color{red}{\int} x \color{red}{dx}$$
$$\int \frac {dy}{y}=\int x dx$$

I prefer this method
$$\frac {y'}{y}= x$$
$$(\ln(y))'= x$$
Then i take integral on both sides
$$\color{red}{\int }(\ln(y))'\color{red}{dx}= \color{red}\int x \color{red}{dx}$$
$$\int \frac {d(\ln(y))}{dx}dx= \int x dx$$
$$\int {d(\ln(y))}= \int x dx$$
$$\ln(y)= \frac  {x^2}2 +K$$


TITLE: What are the closed subschemes of $\mathbb P^2_R$ for a ring $R$? In particular, what are the closed subschemes of $\mathbb P^2_k$ for a field $k$?
QUESTION [1 upvotes]: I am making myself acquainted with the Hilbert Functor $\operatorname{Hilb}_{X/S}$, which is a special case of the $\operatorname{Quot}$ functor. Since $\operatorname{Hilb}_{X/S}$ classifies the isomorphism classes of closed subschemes of $X$ that are isomorphic to $S$, I am interested in the following question:
What are the closed subschemes of $\mathbb P^2_R$ for a ring $R$? More specifically, what are the closed subschemes of $\mathbb P^2_k$ for a field $k$?
I am just asking in order to make myself a little more acquainted with the topic.

REPLY [2 votes]: Question: "What are the closed subschemes of P2R for a ring R? More specifically, what are the closed subschemes of P2k for a field k?"
Answer: For any closed subschemes $Y \subseteq \mathbb{P}^n_A$ there is a homogeneous ideal $I \subseteq A[x_0,..,x_n]$ with $Y \cong V(I)$. This is Hartshorne, Corr.II.5.16.


TITLE: To find order of centre of G
QUESTION [1 upvotes]: Let G be group of order8 generated by a and b such that $a^{4} = b^{2} =1$ and $ ba = a^{3} b $ , then how to found order of centre of G

REPLY [1 votes]: $$Z(G)=\{g\in G: gx=xg ,\forall x\in G\}.$$
Now, $ba=a^3b$ and hence $a,b,a^3$ are not central.
Also,
$$bab^{-1}=a^3 \Rightarrow (bab^{-1})^2=ba^2b^{-1}=a^2,$$
and hence $a^2$ commute with $b$.
Clearly, $a^2$ commute with all powers of $a$ and hence $a^2$ is central.
From here it is easy to check that
$$Z(G)= \langle a^2 \rangle =\{1,a^2\}.$$


TITLE: Relation of $\omega_{\omega_1+1}^{CK}$ to some other ordinals
QUESTION [2 upvotes]: This was posted as a side question in Formal definition of this ordinal? and was split as a separate question based upon suggestion in comments there.
Assume an ordinary ORM model (call it $C_1$). Suppose we add an extra instruction of the form to $u:=u+\omega_1$ (where $u$ is a variable) to the definition of an ordinal register style program (and one or two more basic commands in case they are necessary). Or just consider an OTM with a single parameter $\omega_1$ given to it. Call this model $C_2$. Then define $p$ as the supremum of halting times of these programs (no input). Now as the answer linked below uses $\theta$ to denote $\omega_{\omega_1+1}^{CK}$, I will use the same symbol for uniformity.

Which of the following possibilities can actually occur: (a) $p < \theta$ (b) $p = \theta$ (c) $p > \theta$.

Now given the discussion in comments, assuming the answer given in the linked topic to be correct, it is clear that the possibility (b) above is ruled out. But what about the other two possibilities. Another thing is confinality difference. Consider the possibility $p>\theta$. For that to be true, not only would it have to be true that there exists a program (in model $C_2$) which halts beyond $\theta$ on empty input. But it would also have to be true that there exists some value $\beta<\theta$ such that for all $t$ (where $\beta<t<\theta$) there exists no program which halts at time $t$ on empty input (if this weren't true $cf(\theta)$ would be countable).

Is there a nice/easy way to visualize $\beta$?

The ordinal $q$ was also defined in the previous version of question as follows: One could also consider well-orders of $\omega_1$ that these programs can calculate. For example, say $q$ is the least upper bound of all the ordinals $x$ such that: "there is some program (from model $C_2$) that computes a well-order relation ($\omega_1^2$ to $\{0,1\}$) describing well-order of/on $\omega_1$ with order-type $x$".

One can also ask about the relation of $q$ with $p$,$\theta$ (certainly it seems to me that $p \geq q$ should be true).

REPLY [0 votes]: It seems that the relation between $p$ and $\theta$ should be fairly simple (with $p>\theta$). Because we are talking about sufficiently general programs, $p$ is just the supremum of clocking times given a single parameter $\omega_1$.
To observe why $p>\theta$ is true, we can do the following. First set a counter variable $n$ to $\omega_1+1$. Now just check for the admissibility of $n$. If the check returns false then just increase the value of $n$ by $1$ and check again. The smallest point at which this check will return true will be $\theta$.

Another thing is that if we define $q(\omega_1)$ same as $q$ but by limiting run-times to $\omega_1$, then we have: $q(\omega_1)< \beta < \theta<p$. We can observe that $q(\omega_1)<\theta$ because $q(\omega_1) \leq |\, \omega_1-\mathrm{computable} \,|\,\,$ by its definition. And, since $\omega_1$ is a bad ordinal, we further have $|\, \omega_1-\mathrm{computable} \,|<\theta\,\,$ giving us $q(\omega_1)<\theta$.
The relation of $q$ to the other ordinals isn't clear to me though.


TITLE: When is the periodisation of a function continuous?
QUESTION [3 upvotes]: Consider a function $f\in\mathcal{C}_0(\mathbb{R})$, where $\mathcal{C}_0(\mathbb{R})$ denotes the space of bounded continuous functions vanishing at infinity. I am interested in the $T$-periodisation of such a function, defined as:
$$f_{T}(t)=\sum_{n\in\mathbb{Z}} f(t-nT),\quad \forall t\in \mathbb{R}.$$
As explained in Fischer - On the duality of discrete and periodic functions, $f_{T}$ is a $T$-periodic tempered distribution if $f$ is a rapidly decaying function —i.e. vanishing at infinity faster than any polynomial.
My question concerns the regularity of $f_T$:
For which functions $f\in\mathcal{C}_0(\mathbb{R})$ is the periodised generalised function $f_{T}$ defined above an ordinary, continuous function?
In other words, what should be the assumptions on $f$ so that its periodisation is
continuous?
Any lead would be greatly appreciated. Thank you very much in advance!

REPLY [1 votes]: Short answer: e.g. for Schwartz functions.
Long answer: The Fourier transform of "periodic" is "discrete" and the Fourier transform of "discrete" is "periodic". This is a one-to-one mapping. It is explained in this Fischer - On the duality of discrete and periodic functions.
Analogously, the Fourier transform of "regular" is "local" and the Fourier transform of "local" is "regular". It is another one-to-one mapping. It is explained in Fischer - On the duality of regular and local functions.
The term "regular" refers to ordinary, infinitely differentiable functions which do not grow faster than polynomials. These (regular) functions are so-called multiplication operators for tempered distributions. Their multiplication product with any tempered distribution is again a tempered distribution.
The term "local" refers to tempered distributions which are "local", i.e., they rapidly decay to zero (faster than polynomials). These (generalized) functions are so-called convolution operators for tempered distributions. Their convolution product with any tempered distribution is again a tempered distribution.
The properties of "regular" and "local" fulfill a convolution theorem on tempered distributions.
Now, properties of "periodic", "discrete", "regular" and "local" can be combined. For example, "local+regular" are Schwartz functions and the Fourier transform of Schwartz functions are, again, Schwartz functions ("local+regular"). Moreover, the Fourier transform of "discrete periodic" is again "discrete periodic". It yields the Discrete Fourier Transform (DFT).
Now, the precondition for generalized functions which can be periodized is that they are "local" and the precondition for generalized functions which can be discretized is that they are "regular".
So, back to the original question, in order to periodize an (ordinary or generalized) function, it must be "local" and in order to allow it to be an ordinary function it must be "regular". In other words, Schwartz functions fulfill these two requirements, they are "regular+local".
This property of Schwartz functions of being "regular" and "local" simultaneously, explains their special role as test functions in distribution theory and in quantum physics.
However, there is a difference of "being smooth" in the ordinary and in the generalized functions sense. One may recall, every generalized function is smooth (infinitely differentiable) and, hence, "continuous". To answer this question in the ordinary functions sense, embedded in generalized functions theory, there are more functions beside Schwartz functions. The rectangular function, for example, is smooth in the generalized functions sense but not smooth in the ordinary functions sense. Its periodization, however, yields the function that is constantly 1 for suitable T which is a smooth, ordinary function (in particular continuous). So, obviously, functions which are continuous on an interval [-T/2,+T/2] and such that f(-T/2)=f(+T/2) do also fulfill the requirement.


TITLE: Parametrising a Sphere
QUESTION [2 upvotes]: If parametrising a sphere $r=(\cos{u}\cos{v},\cos{u}\sin{v},\sin{u})$ is it true that limits for $u$ and $v$ will be $-\pi/2$ and $\pi/2$ and $0$ and $2\pi$ respectively? Why is this different to when parametrising a sphere, with parametrization $\rho\sin{\phi}\cos{\theta}, \rho\sin{\phi}\sin{\theta}, \rho\cos{\phi}$ where the limits of $\phi$ are between and $0$ and $\pi$?

REPLY [0 votes]: A useful exercise might be to consider these six points in $(x,y,z)$ coordinates:
$$(A,0,0)_{x,y,z}$$
$$(0,A,0)_{x,y,z}$$
$$(0,0,A)_{x,y,z}$$
$$(-A,0,0)_{x,y,z}$$
$$(0,-A,0)_{x,y,z}$$
$$(0,0,-A)_{x,y,z}$$
Write each of these in standard $(\rho,\phi,\theta)$ spherical coordinates.
Then write the same point in $(r,u,v)$ coordinates,
where $x=r\cos u \cos v,$ $y = r \cos u \sin v,$ and $z = r \sin u.$
Then you'll see how the coordinate systems relate to one another
and why the range of $u$ is different from the range of $\phi$.
The six points have been chosen so that it will be easy to guess the correct
values of the sines and cosines in every case.
This is not meant to be a difficult exercise.


TITLE: planes and lines and circles
QUESTION [1 upvotes]: Consider the points $A(1,2,3)$, $B(-2,3,-1)$ and $C(3,1,-2)$.
$a)$ write an equation of plane $ABC$
$b)$ show that $O$ is on the axis ($\delta$) of circle $S$ circumscribed about triangle $ABC$
I have solved part $a$, and the answer I got was $-9x-23y+z+50=0$.
I just need help solving part $b$.

REPLY [0 votes]: Hint:
1)  Let $r$ the line  passing thorough the origin and orthogonal to the plane:
$$
r) \qquad (x,y,z)^T=t(-9,-23,1)^T
$$
2)  intersect this line with the plane and find the point $O$.
3) verify if the distances from $O$ to the points $A,B,C$ are the same.


TITLE: What would the equation of this graph be?
QUESTION [0 upvotes]: Is this a sine graph? How would I model this graph?

REPLY [3 votes]: It's called interpolating polynomial. There are several methods to find that polynomial based on desired order, number of points to be used, which properties should be conserved and etc. Most straightforward way is to use Lagrange polynomial. For every $n$ distinct points it's possible to construct polynomial of order $n-1$ that passes through those $n$ points.
$$
L(x) = \sum_{k=1}^n \ell_k(x) y_k
$$
where
$$
\ell_k(x) = \prod_{\substack{1 \le m \le n \\m \ne k}}\frac {x-x_m}{x_k-x_m}
$$
In your case, you have dataset
$$
P = \left \{(0, 0.3), (1, 0.8), (2, 0.8), (3, 0.6), (4, 0.7), (5, 1.7) \right \}
$$
You can, and maybe should do it by hands, but I just used Mathematica for that purpose.
pts = {{0, 0.3}, {1, 0.8}, {2, 0.8}, {3, 0.6}, {4, 0.7}, {5, 1.7}};
fLg = InterpolatingPolynomial[pts, x];
pLine = Plot[fLg, {x, 0, 5}, AspectRatio -> Automatic,
GridLines -> {Automatic, Range[0, 2, 0.1]},
PlotRange -> {{0, 5.1}, {0, 2}}, PlotStyle -> {Black, Thick}];
pCircs = Table[Graphics[{Red, Circle[pts[[i]], 0.02]}], {i, 1, 6}];
Show[pLine, pCircs]

which performed `almost' the same graph as you provided

and here's the function itself
$$
L(x) = 0.3 + 0.78 x - 0.266667 x^2 - 0.0291667 x^3 + 0.0166667 x^4 -
 0.000833333 x^5
$$


TITLE: Partial fraction of $\sec(z)$ from $\frac{\pi}{\sin (\pi z)}$
QUESTION [4 upvotes]: Given
$$\frac{\pi}{\sin (\pi z)}=\sum_{n=-\infty}^\infty (-1)^n \frac{1}{z-n},$$
is there a fast way to get
$$\sec(z)=\sum_{n=1}^\infty \frac{(-1)^n(2n-1)\pi}{z^2-(n-1/2)^2\pi^2}?$$
I've tried computing it, but I get a huge mess. Would anybody have some ideas? Thanks in advance.

REPLY [4 votes]: Let's rewrite your first equation as :
$$\frac{\pi}{\cos(\pi/2-\pi z)}=\sum_{n=-\infty}^\infty  \frac{(-1)^n}{z-n}$$
and set $\ x:=\pi\bigl(\frac 12-z\bigr)\ $ then (dividing by $\pi$) :
$$\sec(x)=\sum_{n=-\infty}^\infty  \frac {(-1)^n}{\pi z-\pi/2-(n-1/2)\pi}$$
for $m:=-n$ :
$$\sec(x)=-\sum_{n=1}^\infty  \frac {(-1)^n}{x+(n-1/2)\pi}-\sum_{m=0}^\infty  \frac {(-1)^m}{x+(-m-1/2)\pi}$$
for $n':=m+1$ :
$$\sec(x)=-\sum_{n=1}^\infty  \frac {(-1)^n}{x+(n-1/2)\pi}+\sum_{n'=1}^\infty  \frac {(-1)^{n'}}{x-(n'-1/2)\pi}$$
Putting everything together we conclude :
$$\sec(x)=\sum_{n=1}^\infty \frac{(-1)^n(2n-1)\pi}{x^2-(n-1/2)^2\pi^2}.$$


TITLE: How does one get $u_{y_ky_k}(x_0)\leq 0$ in the proof of the weak maximum principle?
QUESTION [2 upvotes]: The following is the weak maximum principle in Evans's Partial Differential Equations:

Assume $u\in C^2(U)\cap C(\overline{U})$ and $c\equiv0$ in $U$. If $Lu\leq 0$ in $U$, then $\max_{\overline{U}}u=\max_{\partial U}u$. Here

Here is the proof by contradiction. Step 1 is an observation in multivariable calculus:

Then an argument in linear algebra is given as the following

Here is my question:
In the very last line, how does one get $u_{y_ky_k}(x_0)\leq 0$? I can see that $u_{x_kx_k}(x_0)\leq 0$ from (8). Why does change of variables not change the ineqaulity?

REPLY [0 votes]: This is a companion of Jeff's nice answer. First note that
$$
\frac{\partial x_i}{\partial y_k}=o_{ki}.
$$
By chain rule,
$$
u_{y_k}=\sum_iu_{x_i}o_{ki},\quad u_{y_ky_k}=\sum_{i,j}u_{x_i,x_j}o_{ki}o_{kj}.
$$
On the other hand, $\sum_{i,j}u_{x_i,x_j}o_{ki}o_{kj}$ is the $kk$-entry of the matrix $O[D^2u]O^T$.


text:
\begin{definition}[Definition:Partial Derivative/Order]
$u = \map f {x_1, x_2, \ldots, x_n}$ be a [[Definition:Real-Valued Function|function]] of the $n$ [[Definition:Real Independent Variable|independent variables]] $x_1, x_2, \ldots, x_n$.
The '''order''' of a [[Definition:Partial Derivative|partial derivative]] of $u$ is the '''number of times it has been [[Definition:Partial Derivative|(partially) differentiated]]''' by at least one of $x_1, x_2, \ldots, x_n$.
For example:
:a [[Definition:Second Partial Derivative|second partial derivative]] of $u$ is of '''second order''', or '''order $2$'''
:a [[Definition:Third Partial Derivative|third partial derivative]] of $u$ is of '''third order''', or '''order $3$'''
and so on.
\end{definition}

config: wiki


TITLE: Show we can assume algebra of compact operators can be assumed to have trivial nullspace.
QUESTION [1 upvotes]: In the book "Invitation to $C^*$-algebras" by Arveson, I read that:
Let $\mathcal{A}$ be a $C^*$-subalgebra of $B_0(\mathcal{H})$ (= compact operators on the Hilbert space $\mathcal{A}$). The book then reads:

By cutting down to an $\mathcal{A}$-invariant subspace, if necessary,
  we may assume that $\mathcal{A}$ has trivial null-space.

What exactly does this mean and how can we achieve it?
I think it means that we can find a subspace $\mathcal{K} \leq \mathcal{H}$ such that $T\xi \in \mathcal K$ for all $\xi \in \mathcal{K}$ and for all $T \in \mathcal{A}$ and that the restricted operators $T\vert_\mathcal{K}: \mathcal{K} \to \mathcal{K}$ have trivial nullspace, i.e. the common intersection of all their kernels is $0$.
Is this interpretation correct? And how can we prove such subspace exists?

REPLY [1 votes]: The space is $$\mathcal K=\overline{\{A\xi:\ A\in\mathcal A,\ \xi\in \mathcal H\}}.$$
It is obviously invariant for $\mathcal A$. And, for any $A\in\mathcal A$, if $A\ne0$ there exists $\xi$ such that $A\xi\ne0$. Then $A^*(A\xi)\ne0$ (since $\ker A^*A=\ker A$). So the intersection of the kernels of all elements in $\mathcal A$ is $\{0\}$.


TITLE: Maximize $\ x+\frac32 y\ $ subject to...
QUESTION [1 upvotes]: I am stuck on the following problem:

Consider the linear programming problem:
Maximize $x+\frac32 y$
subject to  $$2x+3y \le 16, \\ x+4y \le18,\\ x \ge 0,y \ge0.$$
If $S$ denotes the set of all solutions of the above problem,then  which of the following options is correct?

$S$ is empty
$S$ is a singleton
$S$ is a line segment
$S$ has positive area

My Try:
I have drawn the lines and see that $Z_A$(where $Z$ is $x+\frac32 y)$=Value of $Z$ at $A=8$. Similarly, $Z_B=8,Z_C=\frac{27}{4},Z_D=18$. Now, I can not decide as according to my calculation , option 2 is the right choice but the answer key says ,the answer will be option 3. Can someone enlighten me? Thanks and regards to all.

REPLY [1 votes]: As you noticed it, $(x,y)$ can only be in the pink area.
Obviously, the maximum will be reach on the blue or orange edge.
So you need to calculate the value for both side and you'll find max = 8 is reached on the blue side, which is a line segment.


TITLE: Sum of $34$ Terms Involving Floor Function
QUESTION [0 upvotes]: Find the value of $$\left\lfloor \frac{18}{35}\right\rfloor +\left\lfloor \frac{18\cdot 2}{35}\right\rfloor +\left\lfloor \frac{18\cdot 3}{35}\right\rfloor +\cdots \cdots +\left\lfloor \frac{18\cdot 34}{35}\right\rfloor.$$

I tried the following:
Let $$S=\left\lfloor \frac{18}{35}\right\rfloor +\left\lfloor \frac{18\cdot 2}{35}\right\rfloor +\left\lfloor \frac{18\cdot 3}{35}\right\rfloor +\cdots \cdots +\left\lfloor \frac{18\cdot 34}{35}\right\rfloor.$$
Then  write in the reverse order $$S=\left\lfloor \frac{18\cdot 34}{35}\right\rfloor +\left\lfloor \frac{18\cdot 33}{35}\right\rfloor +\left\lfloor \frac{18\cdot 32}{35}\right\rfloor +\cdots \cdots +\left\lfloor \frac{18}{35}\right\rfloor.$$
Now adding these two equation.  I am not getting what to do; help me please.

REPLY [1 votes]: Hint: Use $$\left\lfloor \frac{18 \times 2n}{35}\right\rfloor = \left\lfloor \frac{(1+35)\times n}{35}\right\rfloor=\left\lfloor n + \frac{n}{35}\right\rfloor=n$$
for $0 \le n<35$.
Also,
$$\left\lfloor \frac{18 \times (2n+1)}{35}\right\rfloor = \left\lfloor \frac{(1+35)\times n +18}{35}\right\rfloor=\left\lfloor n + \frac{n+18}{35}\right\rfloor=n$$
for $0 \le n < 17$.


TITLE: Physical interpretation of Quantum Mechanics Problem in Griffiths
QUESTION [0 upvotes]: In the second edition of "Introduction to Quantum Mechanics" of David J. Griffiths (2016) there is an example that says (p.209, example 5.1)

Suppose we have two noninteracting --they pass right through one another ... -- particles, both of mass $m$, in the infinite square well. The one-particle states are
  $$ \psi_n (x) = \sqrt{\dfrac{2}{a}} \sin \left(\dfrac{n\pi }{a} x \right), \qquad E_n = n^2 K$$
  (where $K\equiv \pi^2 \hbar^2 /2ma^2$, for convenience). If the particles are $\textit{distinguishable}$, with #1 in state $n_1$ and #2 in state $n_2$, the composite wave function is a simple product:
  $$ \psi_{n_1 n_2} (x_1, x_2) = \psi_{n_1} (x_1) \psi (x_2), \qquad E_{n_1 n_2} = (n_1^2+n_2^2) K.$$
  For example, the ground state is
  $$ \psi_{11} = \frac{2}{a} \sin (\pi x_1/a) \sin (\pi x_2/a), \qquad E_{11} = 2K$$
  [...]

and a problem 5.6. (p-214)

Imagine two noninteracting particles, each of mass $m$, in the infinite square well. If one is in the state $\psi_n$, and the other in state $\psi_m (m \neq n)$, calculate $<(x_1-x_2)^2>$, assuming (a) they are distinguishable particles, (b) they are identical bosons, and (c) they are identical fermions.

If they are distinguishable particles:
$$ <(x_1-x_2)^2>\equiv  g_{nm}(a).$$
If they are identical bosons:
$$ <(x_1-x_2)^2> = g_{nm}(a)  -|I_{nm}|^2.$$
If they are identical fermions:
$$ <(x_1-x_2)^2> = g_{nm}(a) + |I_{mn}|^2$$
where
$$ g_{nm}(a) = a^2 \left[ \dfrac{1}{6} -\dfrac{1}{2\pi^2} \left( \dfrac{1}{n^2} +\dfrac{1}{m^2}\right)\right], \qquad n,m \in \mathbb{N}$$
and
$$ I_{mn} =\dfrac{4a}{\pi^2} \dfrac{mn}{(n^2-m^2)^2} \left\{ (-1)^{m+n} -1 \right\}, \qquad n,m \in \mathbb{N}.$$
So, for bosons and fermions if $I_{nm}=0$ they can be considered as distinguishable particles. This occurs when the wave functions have the same parity on $n,m$. What is the physical interpretation of the last sentence?
If the sentence "they can be considered as distinguishable particles" is false, what is the physical interpretation of $|I_{mn}|=0$? What expectation values are needed to prove that the bosons or the fermions can be considered as distinguished particles?

REPLY [2 votes]: Just because $\langle (x_1 - x_2)^2 \rangle$ is unaffected by the particle statistics doesn't mean that "they can be considered as distinguishable particles."  Other expectation values will be affected by the statistics.


TITLE: Poisson Distribution and Median
QUESTION [6 upvotes]: The number of automobile accidents at the corner of Wall and Street is assumed to have Poisson distribution with a mean of five per week.  Let A denote the number of automobile accidents that will occur next week.  Find
(a)  $\Pr[A<3]$
(b)  The median of A
(c) $\sigma_A$

Since this is a Poisson distribution, the probability function is:
$\Pr(A=k)=\cfrac{\lambda^k}{k!} e^{-\lambda}\tag{1}$ where $\lambda=5$ because the rate of occurrence is 5 accidents per week.  Therefore:
(a)  $\Pr[A<3]=e^{-5}(5^0 + 5 ^1 + 5^2/2!)=0.247$
(c)  $\sigma_A=\sqrt{\lambda}=\sqrt{5}$
Although I wasn't sure how to calculate (b).  Equation (1) is valid for $k>0$ so how do you find the median for a probability distribution which can take an infinite number of k values?
Thanks in advance.

REPLY [1 votes]: To complement deinst and Assad's excellent answers, the paper "The median of the Poisson distribution, cited below,
proves explicitly (cf. Theorem 1) that for integer parameter $\lambda$, the median is always just $\lambda$. Certainly, for non-integer parameter, the median must be different from the mean.
Adell, J. A.; Jodrá, P., The median of the Poisson distribution, Metrika 61, No. 3, 337-346 (2005). ZBL1079.62014.


TITLE: Find probability to reject $H_0$ when mean is given
QUESTION [0 upvotes]: The exercise is:

Assume that the population mean is actually 110 grams and that the distribution is normal with standard deviation 4 grams. In a z test of $H_0$: u = 113 against $H_a$: u < 113 with $\alpha = 0.05$, find the probability of rejecting $H_0$ with six observations.

From part (a) in that exercise i found out:
$$\hat x = 112.967 - s = 4.28 - n = 6- t,0.05,5 = -2.015$$
My approach is:
Find the type II error and do $$ 1 - \beta(110) $$
so:
$$P\left( \frac{\hat x - 113}{\frac{4.28}{\sqrt{6}}} > -2.015 | u = 110, \sigma=4\right) = \alpha   $$
$$1 - P\left(\hat x < -2.015\cdot{\frac{4.28}{\sqrt{6}}} + 113\right) $$
$$1 - P\left(\frac{\hat x - 110}{\frac{4}{\sqrt{6}}} < \frac{-2.015\cdot{\frac{4.28}{\sqrt{6}}} + 3)}{\frac{4}{\sqrt{6}}}\right) $$
is:
$$1-P\left(\frac{\hat x - 110}{\frac{4}{\sqrt{6}}} <  -0.32\right)$$
which gives (1-(1-0.3745)) = 0.3745, while the solution is 0.58.

REPLY [1 votes]: The sample mean doesn't enter into this calculation at all. You're answering a question about the probability that (under repeated sampling) you'd reject the null when you're drawing from a population with mean 110 while the hypothesized mean is 113.
Further note that the question specifies a z test, not a t-test - and specifies $\sigma$ - so your critical value is wrong. You need to pay careful attention to the question.
(By comparison, your sample either leads to rejection or it doesn't.)
Here's a calculation (done in the statistical package R) you may find of some value, by considering where the inputs come from, and what they mean:
> pnorm(-1.645-(110-113)/(4/sqrt(6)))
[1] 0.5761748

Note that used with a single argument, pnorm is just the cdf of the standard normal.


TITLE: Solve Algebraical.ly $0.5=\dfrac{365!}{365^{n}(365-n)!} $
QUESTION [1 upvotes]: How does one go about solving this equation? Not sure how to approach this as no factorials will cancel out. Im sorry I meant $\dfrac{365!}{365^{n}(365-n)!}=0.5$.

REPLY [1 votes]: Suppose that you want to solve in the real domain the equation
$$\frac {x!}{x^n (x-y)!}=\frac 12$$ take logarithm to make
$$\log(x!)-y \log(x)-\log[(x-y)!]=-\log(2)$$ Using twice Stirling approximation for large values of $x$, we have
$$-\log(2)=-\frac{(y-1) y}{2 x}+O\left(\frac{1}{x^2}\right)$$ Ignoring the higher order terms and making $x=365$ gives
$$y(y-1)=730\log(2\quad \implies \quad y=\frac{1}{2} \left(1+\sqrt{1+2920 \log (2)}\right)=22.999943$$


TITLE: Simulate random variable RV
QUESTION [1 upvotes]: i need some help to understand what is going on here.I'll be more specific on questions after i give the context.
Suppose we want to simulate $X$ such that:
$P\{X = i \} = p(1-p)^{i-1} , i \geq 1 $
As:
$\sum\limits_{i=1}^{j-1} P\{X = i\} =  1- P\{X > j-1 \} = 1-(1-p)^{j-1}$
we can simulate such a random variable by generating a random number $U$ and setting  $X$ equal to that value $j$ for which:
$1-(1-p)^{j-1} < U < (1-p)^j$
or,equivalently, for which:
$(1-p)^{j} < 1-U < (1-p)^{j-1}$
As, $1-U$ has the same distribution as $U$ , we an thus define $X$ by :
$X = min\{j:(1-p)^j < U \} = min \big \{j:j > \frac{log (U)}{log(1-p)} \big\} = 1 +\big[ \frac{log (U)}{log(1-p)}\big]$
questions:
$\textbf{(1)}$  $\sum\limits_{i=1}^{j-1} P\{X = i\} =  1- P\{X > j-1 \} = 1-(1-p)^{j-1}$ this is nothing more than specifying the CDF (distribution function) right?
$\textbf{2)}$  $1-(1-p)^{j-1} < U < (1-p)^j$ . X gets the value of $j$ if the random number $U$ is between  $1-(1-p)^{j-1}$ and $(1-p)^j$ .Did i get it right?
$\textbf{3)}$
$X = min\{j:(1-p)^j < U \} = min \big \{j:j > \frac{log (U)}{log(1-p)} \big\} = 1 +\big[ \frac{log (U)}{log(1-p)}\big]$
can someone show me or explain the stepwise calculations so that you finally come to $1 +\big[ \frac{log (U)}{log(1-p)}\big]$ , because u don't get the given calculations(in the way they do it)

REPLY [1 votes]: (1) Yes, this is exactly the CDF.
(2) Yes, this is exactly what this means.
(3) This is a simple computation:
$$\begin{align*}
(1-p)^j<U &\Leftrightarrow \log(1-p)^j<\log U\\
&\Leftrightarrow j\log(1-p)<\log U\\
&\Leftrightarrow j>\frac{\log U}{\log(1-p)}\\
\end{align*}$$
Note that we have used two facts: $\log$ is monotone increasing, and therefore it preserves inequalities, and $\log(1-p)$ is negative, and therefore it reverses the inequality when we divide by it.
Finally, we note that the smallest integer $j$ such that $j>\frac{\log U}{\log(1-p)}$ is, by definition, the ceiling of $\frac{\log U}{\log(1-p)}$, which is also equal to $1+\lfloor \frac{\log U}{\log(1-p)}\rfloor$.


TITLE: The derivative evaluated at zero should return back a number, not a function.
QUESTION [0 upvotes]: I got the following feedback on my answer to this problem, and I haven't been able to figure out why: "The derivative evaluated at zero should return back a number, not a function."
Find the tangent line at x = 0 of
y= sin(2x) + 1
sin (0) + 1 = 1, y = 1
y' = cos(2x) * 2
y' = 2cos(2x) = m
y - 1 = 2cos(2x)(x-0)
y = 2xcos(2x) + 1

REPLY [1 votes]: Let's start by understanding why the answer you gave can't be right. What does the graph of "$y=2x\cos(2x)+1$" look like?
Well, one option is to graph it - this will give you something that really doesn't look like a line! But what if we don't have access to a computer, or a graphing calculator?
Well, we can still tell this isn't a line by plugging in a few points:

When $x=0$, we have $y=1$.
When $x={\pi\over 4}$, we have $y=1$.
And when $x=\pi$, we have $y=2\pi+1$.

Draw the points $(0, 1)$, $({\pi\over 4},1)$, and $(\pi, 2\pi+1)$ on a graph; clearly these three points don't lie on a line. So $y=2x\cos(2x)$ can't possibly be the graph of the tangent line.

Alright, now that we know that the answer is wrong, let's try to understand why. In particular, we know that a line is the graph of a function of the form $$y=mx+b$$ for $m$ a constant (not something like "$2\cos(2x)$" in other words), so we're really trying to understand: what constant goes there?
Think back to what the tangent line means. The tangent line to a curve at some point $p$ is the line through $p$ that goes "in the same direction" as the curve, at that point (a phrase you'll hear down the road, if you haven't already, is: the tangent line is the best linear approximation to the curve, at that point). Specifically, we want the tangent line and the curve to "change in the same way" at that point - the rate of change of the tangent line (increasing or decreasing? steep or shallow?) should be the same as the curve.
You probably know where this is going:

The slope of the tangent line to the graph of $y=f(x)$ is the derivative of $f$.

But let's be a bit more specific. The right thing to say is:

The slope of the tangent line to the graph of $y=f(x)$ through the point $(x_0, f(x_0))$ is the derivative of $f$ at the point $x=x_0$.

Remember that the tangent line through one point will probably have a very different slope from the tangent line through a different point! E.g. take $f(x)=x^2$, and look at a point with negative $x$-coordinate versus a point with positive $x$-coordinate; it should be clear if you draw the graph that the tangent line through the former point will be decreasing, while the latter will be increasing. The slope of the tangent line changes with the point: each point yields its own slope. The derivative $f'(x)$ tells us how to find the slope of the tangent line through a given point, but the way we find that slope is to plug the relevant $x$-coordinate into $f'$!
In your example, the derivative is $f'(x)=2\cos(2x)$, and your point is $x=0, y=1$ - so the slope of the tangent line is $f'(0)=2\cos(0)=2$ (note that only the $x$-coordinate $x=0$ is relevant here), and the whole equation of the line is $y=2x+1$ (here the $y$-coordinate $y=1$ comes into play).
Hopefully this helps explain what's going on!


TITLE: Proving well-defineness of function which adds additional variable
QUESTION [2 upvotes]: I'm trying to solve an exercise from my math lecture. The question is:
Let $k \in \mathbb{Z}$, $(G, \circ) := (\mathbb{Z}_{12}, \oplus)$, $(H, *) := (\mathbb{Z}_{8}, \oplus)$ and $\varphi: G \ni [a]_{12} \mapsto [6a + k]_8 \in H$.
Prove that $\varphi$ is well defined.
Based on another exercise which maps $[a]_{12} \mapsto [6a]_8$, I've tried to do this like this:
$$[a]_{12} = [b]_{12} \\
 \iff a \equiv b\pmod {12} \\
 \iff 12 | b-a \\
 \iff (\exists l \in \mathbb{Z})[12l = b-a]$$
Now insert a new variable, $k$.
$$ \iff (\exists l,k \in \mathbb{Z})[12l = (b + k) -(a+k)] \\
\iff (\exists l,k \in \mathbb{Z})[72l = 6(b+k)-6(a+k)] \\
 \iff (\exists l,k \in \mathbb{Z})[8l = 6(b+k)-6(a+k)] \\
 \iff 8 | (6b+k) - (6a+k) \\
 \iff 6a + k \equiv 6b+ k \pmod{8} \\
 \iff [6a +k]_8$$
My problem is actually the introducing $k$ part. This does not look right. I think I can simplify $72l$ to $8l$ because 72 is a multiple of 8, but how should I properly introduce $k$ here?
Also, looked at from a set theorey side, if I see $\varphi$ as a relation, is it correct that the set would be like this?
$$\mathcal{R} = \{([0]_{12}, [0+k]_8), ([0]_{12}, [1+k]_8), \ldots\} $$
i.e. that the $k$ is treated as a constant in the set?

REPLY [0 votes]: From what I can tell from your definition of $\varphi$, $k$ is indeed a "constant", in the sense that you need to prove that for any fixed $k \in \mathbb{Z}$ you can define a map $\varphi_k$.
So it is not an "existence of $k$" matter! You fix $k$ in the beginning, and then argue as
$$[a]_{12} = [b]_{12} \\
 \iff a \equiv b\pmod {12} \\
 \iff 12 | b-a \\
 \iff (\exists l \in \mathbb{Z})[12l = b-a]
12l = b-a \\
72l = 6b-6a \\ $$
so, for any fixed $k \in \mathbb{Z}$
$$72l = 6b-6a +k - k \\
\text{ so }\exists m \in \mathbb{Z} \text{ such that } 8m = (6b+k)-(6a+k) \\
\text{ so } 8 | (6b+k) - (6a+k) \\
\text{ so } 6a + k \equiv 6b+ k \pmod{8} \\
$$
and you are done.


TITLE: On twisted Euler sums
QUESTION [15 upvotes]: An interesting investigation started here and it showed that
$$ \sum_{k\geq 1}\left(\zeta(m)-H_{k}^{(m)}\right)^2 $$
has a closed form in terms of values of the Riemann $\zeta$ function for any integer $m\geq 2$.
I was starting to study the cubic analogue $ \sum_{k\geq 1}\left(\zeta(m)-H_{k}^{(m)}\right)^3 $ and I managed to prove through summation by parts that
$$ \sum_{k\geq 1}\frac{\left(H_k^{(2)}\right)^2}{k^2} =\frac{1}{3}\zeta(2)^3-\frac{2}{3}\zeta(6)+\zeta(3)^2 $$
where the LHS, according to Flajolet and Salvy's notation, is $S_{22,2}$. An explicit evaluation of $\sum_{k\geq 1}\left(\zeta(2)-H_{k}^{(2)}\right)^3 $ is completed by the computation of
$$\boxed{ S_{12,2} = \sum_{k\geq 1}\frac{H_k H_k^{(2)}}{k^2} }$$
which has an odd weight, hence it is not covered by Thm 4.2 of Flajolet and Salvy. On the other hand they suggest that by the kernel $(\psi(-s)+\gamma)^4$ the previous series and the cubic Euler sum $\sum_{n\geq 1}\frac{H_n^3}{(n+1)^2}=\frac{15}{2}\zeta(5)+\zeta(2)\,\zeta(3)$ are strictly related.

Question: can you help me completing this sketch, in order to get an explicit value for $S_{12,2}$ and for $\sum_{k\geq 1}\left(\zeta(2)-H_{k}^{(2)}\right)^3 $? Alternative techniques to summation by parts and residues are equally welcome.

Update: I have just realized this is solved by Mike Spivey's answer to Zaid's question here. On the other hand, Mike Spivey's approach is extremely lengthy, and I would be happy to see a more efficient derivation of $S_{12,2}=\zeta(2)\,\zeta(3)+\zeta(5)$.

REPLY [10 votes]: I am skipping the derivation of Euler Sums $\displaystyle S(1,1;3) = \sum\limits_{n=1}^{\infty} \frac{H_n^2}{n^3}$ and $\displaystyle S(2;3) = \sum\limits_{n=1}^{\infty} \frac{H_n^{(2)}}{n^3}$ (for a derivation see solutions to $\textbf{problem 5406}$ proposed by C.I. Valean, Romania in SSMA).
Now, consider the partial fraction decomposition, \begin{align*}\sum\limits_{k=1}^{n-1}\left(\frac{1}{k(n-k)}\right)^2 = \frac{2}{n^2}\left(H_n^{(2)} + \frac{2H_n}{n}-\frac{3}{n^2}\right) \qquad \cdots (\star)\end{align*}
Multiplying both sides of $(\star)$ with $H_n$ and summing over $n \ge 1$ (and making the change of varible $m = n+k$):
\begin{align}
\sum\limits_{n=1}^{\infty}\frac{H_n}{n^2}\left(H_n^{(2)}+2\frac{H_n}{n} - \frac{3}{n^2}\right) &= \sum\limits_{n=1}^{\infty}\sum\limits_{k=1}^{n-1} \frac{H_n}{k^2(n-k)^2} \\&= \sum\limits_{m,k=1}^{\infty} \frac{H_{m+k}}{m^2k^2} \tag{0}\\&= \sum\limits_{m,k,j=1}^{\infty} \frac{mj+kj}{m^2k^2j^2(m+k+j)} \tag{1} \\&= 2\sum\limits_{m,k,j=1}^{\infty} \frac{jk}{m^2k^2j^2(m+k+j)} \tag{2} \\&= 2\sum\limits_{j,k=1}^{\infty} \frac{1}{jk}\sum\limits_{m=1}^{\infty} \frac{1}{m^2(m+j+k)} \tag{3} \\&= 2\sum\limits_{k,j=1}^{\infty} \frac{1}{kj(j+k)}\left(\zeta(2) - \frac{H_{k+j}}{k+j}\right) \tag{4} \\&= 4\sum\limits_{n=2}^{\infty}\frac{H_{n-1}}{n^2}\left(\zeta(2) - \frac{H_n}{n}\right)\\&= 4\zeta(2)\sum\limits_{n=1}^{\infty}\frac{H_{n-1}}{n^2} + 4\sum\limits_{n=1}^{\infty} \frac{H_n}{n^4}-4\sum\limits_{n=1}^{\infty} \frac{H_n^2}{n^3}\end{align}
Where, in steps $(0)$ and $(3)$ we used the identity, $\displaystyle \frac{H_q}{q} = \sum\limits_{\ell = 1}^{\infty} \frac{1}{\ell (\ell + q)}$. In step $(1)$ we used the symmetry w.r.t. the variables, in step $(2)$ interchanged order of summation and in step $(4)$ made the change of variables $n = j+k$.
Thus, $$ \sum\limits_{n=1}^{\infty} \frac{H_nH_n^{(2)}}{n^2} = 4\zeta(2)\zeta(3) + 7\sum\limits_{n=1}^{\infty} \frac{H_n}{n^4} - 6\sum\limits_{n=1}^{\infty} \frac{H_n^2}{n^3}$$


TITLE: Motivation or Proof of Mechanical Principles
QUESTION [0 upvotes]: I understand both Newtonian and Lagrangian Mechanics are built upon some Principles. Besides Galileo's Principle of Relativity, Newtonian Mechanics counts with Newton's Laws of Motion and Lagrangian Mechanics counts with Hamilton's Principle (AKA the Principle of Least Action).
I am aware you can obtain Hamilton's Principle from Newton's Second Law and, using Lagrangian Mechanics, it is quite easy to prove Newton's First and Third Laws (and I guess it is equally simple with Newtonian Mechanics).
Therefore, my question is: how can one prove Newton's Second Law or what is the motivation for introducing it as a postulate?
I am assuming that the most fundamental block to build Mechanics is Newton's Second Law. If there is another way of building it eliminating any unproven or unmotivated statements, it would also answer my question.  I did not consider building it from Hamilton's Principle because I believe there is no motivation to introduce it as a postulate instead of deriving it from Newton's Laws. If there is a motivation, this motivation would also answer my question.

REPLY [0 votes]: Newton's second law is not a good law unless you already have a definition of "mass". Otherwise saying $\vec{F}=\dot{\vec{p}}$ is just a definition. You should take the 3rd law into account, because that's what "fixes" it a little bit: you can define a standard mass, an so  $m=m_s \frac{a_s}{a}$, and it only works with constant masses in intertial frames. So Newton laws work, but they're not too well formulated, you need something else.
Behind them, there's the principle of "conservation of momentum". Now, if we define what forces are (2nd law), then the 3rd law follows inmediately if momentum is conserved, so the key is conservation of momentum , which is fortunately explained by the Noether's theorems.
On the other hand, you can start saying that "Energy si constant" to get the same results.