TITLE: Signs of solutions to a quadratic equation
QUESTION [0 upvotes]: If we have two solutions to a quadratic equation which are
$L_1=-a+\sqrt{a^2-b^2}$ and $L_2=-a-\sqrt{a^2-b^2}$ and the only fact we are given is that $a,b>0$,
Is it possible to know if $L_1$ and $L_2$ can be of opposite signs if
$\sqrt{a^2-b^2}>0$?
We obviously know $L_2$ would be negative but what sign would $L_1$ be?
If we assume $-a+\sqrt{a^2-b^2}>0$ then $\sqrt{a^2-b^2}>a$ then square both sides you get $a^2-b^2>a^2$ so $b^2<0$ which is a contradiction so $-a+\sqrt{a^2-b^2}<0$ right?
But if we at the beginning just took $-a>-\sqrt{a^2-b^2}$ and then square both sides, we have $b^2>0$ which contradicts the contradiction, so is it possible to know the bold question?

REPLY [2 votes]: Since $L_1 L_2=b^2>0$, they cannot have opposite signs.


TITLE: short exact sequences and functors
QUESTION [1 upvotes]: Suppose
$$0 \longrightarrow A \longrightarrow B\longrightarrow C \longrightarrow 0$$
is a short exact sequence and $F$ is a functor, so we get maps $F(A)\longrightarrow F(B)$ and $F(B) \longrightarrow F(C)$. Is the sequence $$F(A)\longrightarrow F(B)\longrightarrow F(C)$$ exact in general?

REPLY [4 votes]: You can find a couple of examples at MO/46019.
Note that since you didn't assume that $F$ is additive, there are lots of trivial examples (for example non-zero constant functors).


TITLE: Eigenvalues are roots of cyclotomic polynomial
QUESTION [1 upvotes]: I am reading Lyndon and Shupp's 'combinatorial group theory'. At page 25 it is stated that

if $g$ is an element of finite order $n$ in $\mathbb{GL}(2, \mathbb{Z})$, its eigenvalues must be roots of the cyclotomic polynomial $\Phi_n(x)$,$\ldots$

I can see why the eigenvalues are $n$th roots of unity, but I do not see why they should be primitive roots.

REPLY [0 votes]: This is false. Take$$g=\begin{pmatrix}1&0\\0&-1\end{pmatrix}.$$Then $g$ has order $2$, but only one of its eigenvalues is a root of $\Phi_2$.


TITLE: Prove the following inequality for all $n \in \mathbb{N}$
QUESTION [1 upvotes]: Let $f(x)$ be a derivable function on the interval $(0,1)$ and continuous on $[0,1]$.
Assume that $|f'(x)| \leq M$ for every $x \in (a,b)$.
Prove the following inequality for all $n\in\mathbb{N}$,
$$
  \left| \int_0^1 f(x) dx - \frac{1}{n} \sum_{k=1}^n f(k/n) \right| \leq \frac{M}{n}.
$$

REPLY [3 votes]: If I'm not mistaken, we can get a slightly sharper bound.
Write
$$\int_0^1\,f(x)\,dx = \sum_{k=1}^n\int_{(k-1)/n}^{k/n}\,f(x)\,dx$$
so then we need to estimate
\begin{align}
&\left| \sum_{k=1}^n\left[\left(\int_{(k-1)/n}^{k/n}\,f(x)\,dx\right)-\frac1n f(k/n)\right]  \right|
\\=\,\,&
\left| \sum_{k=1}^n\left(\int_{(k-1)/n}^{k/n}\,f(x) - f(k/n)\,dx\right)  \right|.
\end{align}
Now, use the mean value theorem.
For each $x\in\Big((k-1)/n,\, k/n\Big)$, there is some $\eta_x \in (x,\,k/n)$ with
$$f(x) - f(k/n) = f'(\eta_x) \big(x-k/n\big).$$
Then our estimate becomes
\begin{align}
&\left| \sum_{k=1}^n\left(\int_{(k-1)/n}^{k/n}\,f(x) - f(k/n)\,dx\right)  \right|
\\\leqslant \,\,&
\sum_{k=1}^n\int_{(k-1)/n}^{k/n}\,|f'(\eta_x)| \big(k/n - x\big)\,dx
\\\leqslant \,\,&
M\,\sum_{k=1}^n\int_{(k-1)/n}^{k/n}\,\big(k/n - x\big)\,dx
\\= \,\,&
M\,\sum_{k=1}^n\left[\frac{kx}n - \frac{x^2}2\right]_{(k-1)/n}^{k/n}
\\= \,\,&
M\,\sum_{k=1}^n\frac1{2n^2} = \frac M{2n}.
\end{align}

As a bonus, it's easy to verify that the bound is sharp when $f$ is affine (that is, when $f'$ is constant).


TITLE: Are these properties for independence false?
QUESTION [1 upvotes]: I have tried to prove that if:

If $X$ is independent of $Y$, and $Y=Y_1+Y_2$, and $Y_1$ and $Y_2$ are
  independent. Then $X$ and $Y_1$ is independent.

But I am not able to prove this, but I can't find a counterexample either. I almost had it if the range of $Y_1$ was countable(because then I could split it up in a sum).
But in proving this, I used that:

If $A$ and $B$ are independent, so that $P(A \cap B)=P(A)P(B)$, then if $C$
  is en event with $P(C)>0$, then $P (A\cap B\mid C)=P(A\mid C)P(B\mid C)$.

But this seems wrong by looking at the case:

So this last one is wrong? Do you know anything about the first one? I am not able to prove the first one, or find a counterexample. What I end up with is this:
I try to prove that $X$ and $Y_1$ is independent:
$P(X\in A, Y_1\in B)=P(X \in A, Y-Y_2 \in B)$. And then I try to use some sort of trick learned in elementary statistics where I in some way integrate over all possible values $Y_2$ can take. But I can't get the argument wo work correctly. Any hints or tips?

REPLY [2 votes]: Here is a counterexample, although not in $\mathbb{R}$: take $X$ and $Y_2$ to be independent uniform random variables on $\{0,1\}$, and set $Y_1=X$.
Then $Y\stackrel{\rm def}{=}Y_1+Y_2 \mod 2$ is uniform on $\{0,1\}$ and independent of $X$, but clearly we do not have $X=Y_1$ independent of each other.
(If you want to remove the $\operatorname{mod} 2$ thing, just work in $\mathbb{Z}_2$.)


text:
\section{Existing stability results and their limitations}\label{sec:existing_limitations}

As already mentioned, the classical stability results all involve the bottleneck distance between persistence diagrams. A complete overview of these results is beyond the scope of this paper and we direct the reader to \cite{chazal2014persistence} for the stability of geometric constructions and ~\cite{Bubenik_2014} for the categorical foundations of $\infty$-Wasserstein stability. This should  not be considered as a complete list as there is a large body of work on stability which we do not address review here.

\subsection{Lipschitz functions on compact manifolds}

The most relevant related work to the results presented in this paper can be found in  ~\cite{cohen2010lipschitz}.
To the best of our knowledge, this paper contains the main existing stability result for bounding the ($p\neq \infty$)-Wasserstein distance between two persistence diagrams. It is for the setting of sub-level set filtrations of Lipschitz functions.

\begin{theorem}[Wasserstein Stability Theorem~\cite{cohen2010lipschitz}]
Let $X$ be a triangulable, compact metric space that implies bounded degree-$k$ total persistence, for $k\geq 1$ and let $f,g:X\to \R$ be two tame Lipschitz functions. Then $$W_p(f,g)\leq C^{1/p}\|f-g\|_\infty^{1-\frac{p}{k}}$$ for all $p\geq k$, where $C=C_X \max\{Lip(f)^k, Lip(g)^k\}$ and $C_X$ is a constant dependent on $X$.
\end{theorem}
To put our results into context, it is worthwhile understanding the limitations of this theorem. We will find lower bounds on $C_X$ and $k$, restricting ourselves  to the case where $X$ is a compact $d$-dimensional manifold.  An important aspect is the bounded degree-$k$ total persistence which will force this stability result to only hold for sufficiently large $p$.
\begin{definition}
A metric space $X$ \emph{implies bounded degree $k$-total persistence} if there exists a constant $C_X$ that depends only on $X$ such that
$$||\Dgm(X(f))||^k_k  <C_X $$
for every tame function $f$ with Lipschitz constant $Lip(f)\leq 1$.
\end{definition}

To construct a counterexample we will use a function which is the sum of functions with supports over disjoint balls of small radius.

\begin{lemma}
Given an $d$-dimensional compact Riemannian manifold $X$ and $r>0$ small enough, there exists a packing of $\left\lfloor \frac{\vol(X)}{\kappa \omega_d 2^d r^d}\right\rfloor$ disjoint balls of radius $r$ in $X$, where $\omega_d$ is the volume of the $d$-dimensional Euclidean ball and $\kappa$ is a constant which depends on the infimum of the scalar curvature of $X$.
\end{lemma}
We note that since we restrict ourselves to compact manifolds, the curvature must be bounded.  The proof then follows from standard arguments involving packing and covering numbers.

\begin{lemma}\label{lem:teepee}
Let $X$ an $d$-dimensional compact manifold. If $X$ has bounded degree-$k$ total persistence then $k\geq d$.
\end{lemma}
\begin{proof}
We will prove this via a counterexample of the contrapositive.
Let $P=\{p_1, \ldots p_N\}$ be the centers of a packing of $N=\left\lfloor \frac{\vol(X)}{\kappa\omega_d 2^d r^d}\right\rfloor$ disjoint balls of radius $r$ in $X$. Set $T_{r,p}$ to be a teepee shaped function about $p$ with height $r$, with $T_{r,p}(x)=\max\{r-d(x,p), 0\}$.
We then consider functions $f_r=\sum_{i=1}^N T_{r,p_i}$(see Figure~\ref{fig:teepee}).  Observe that $f$ is 1-Lipschitz. Then,
$$||\Dgm(X(f))||^k_k  = \sum_{i=1}^N r^k = \left\lfloor \frac{\vol(X)}{\kappa\omega_d 2^d r^d}\right\rfloor r^k = O(r^{d-k})$$
For this to be uniformly bounded for all small enough $r>0$, we require $d\geq k$.
\end{proof}
\begin{figure}[th]
  \centering
  \begin{tikzpicture}[scale=0.6,every node/.style={scale=0.8}]

    \draw [<->,thick] (-2.4,0) -- (2.4,0) node (xaxis) [at end, below] {$x$};
    \draw [->,thick] (0,-0.1) -- (0,2) node [at start, below] {$p$};
    \draw (-0.1,1.5) -- (0.1,1.5) node (peak) [at start,left] {$r$};
    \draw (-1.5,0) -- (0,1.5) -- (1.5,0);
    \draw (-1.5,-0.1) -- (-1.5,0.1) node (l) [at start, below] {$p-r$};
    \draw (1.5,-0.1) -- (1.5,0.1) node (r) [at start, below] {$p+r$};

  \end{tikzpicture}\hspace{1cm}
  \begin{tikzpicture}[scale=0.6,every node/.style={scale=0.8}]

    \draw [<->,thick] (-8,0) -- (8,0) node (xaxis) [at end, below] {$x$};
    \draw [->,thick] (0,-0.1) -- (0,2) node [at start, below] {$p_i$};
    \draw (-0.1,1.5) -- (0.1,1.5) node (peak) [at start,left] {$r$};
    \draw (-1.5,0) -- (0,1.5) -- (1.5,0);
    \draw (-1.5,-0.1) -- (-1.5,0.1) node [at start, below] {$p_i-r$};
    \draw (1.5,-0.1) -- (1.5,0.1) node  [at start, below] {$p_i+r$};

    \draw  (5,-0.1) -- (5,0.1) node [at start, below] {$p_{i+1}$};
    \draw (3.5,0) -- (5,1.5) -- (6.5,0);
    \draw (3.5,-0.1) -- (3.5,0.1) node  [at start, below] {$p_{i+1}-r$};
    \draw (6.5,-0.1) -- (6.5,0.1) node  [at start, below] {$p_{i+1}+r$};

    \draw  (-5,-0.1) -- (-5,0.1) node [at start, below] {$p_{i-1}$};
    \draw (-3.5,0) -- (-5,1.5) -- (-6.5,0);
    \draw (-3.5,-0.1) -- (-3.5,0.1) node  [at start, below] {$p_{i-1}+r$};
    \draw (-6.5,-0.1) -- (-6.5,0.1) node  [at start, below] {$p_{i-1}-r$};

    \node (ld) at (-7,0.75) {$\mathbf{\cdots}$};
    \node (rd) at (7,0.75) {$\mathbf{\cdots}$};
  \end{tikzpicture}

  \caption{\label{fig:teepee} (Left) The teepee function and (right) sum of teepee functions from Lemma~\ref{lem:teepee}.}
\end{figure}
The same example used in the above lemma, coupled with $g$ the zero function provides a lower bound on $C_X$.  We see that $C_X$  grows linearly as a function of the volume of $X$.

Other papers that prove Wasserstein stability results are few and far between. In \cite{chen2011diffusion}, Chen and Edelsbrunner consider non-Lipschitz functions on non-compact spaces, using scale-space diffusion. They focus on convergence properties as opposed to stability but also attain some Wasserstein stability results. Crucially, just as in the Lipschitz case, this $p$-Wasserstein stability only holds for $p > d$ where $d$ is the dimension of the domain. The condition $p > d$ also appears in stability results for \v Cech filtrations, or equivalently distance filtrations, for point clouds in $\R^d$. We partially address the case of Vietoris-Rips filtrations in Section~\ref{sec:rips}.

\subsection{Erroneous appeals to previous $p$-Wasserstein stability results}
Unfortunately, the Lipschitz Wasserstein stability theorem in \cite{cohen2010lipschitz} appears to be one of the most misunderstood and miscited results within the field of topological data analysis. Common errors include using a small $p$ (often $1$ or $2$) for high dimensional data, assertions that the persistence diagrams depend Lipschitz-continuously on data and applying the theorem to Vietoris-Rips filtrations. Luckily, many of the erroneous applications can now be covered by the stability results in this paper. Rather than discuss individual examples, in Section~\ref{sec:consequences}, we examine the consequences for topological summaries, which are increasingly the most common way to apply persistence diagrams to data.

config: arxiv


TITLE: Definition of quotient of sigma algebras
QUESTION [3 upvotes]: Let $(\Omega,\mathcal{A},\mu)$ a measure space. Let $I$ be some $\sigma$-ideal in $(\Omega,\mathcal{A})$, i.e. it contains the empty set and contains subsets and countable unions of its elements.

How do you define $\Omega/ I$ ? And what are the equivalence classes defined as?

I am just familiar with quotient space of vector spaces, but not with that general case. Any literature is welcome aswell.

REPLY [5 votes]: In the standard group theoretic context, $G/H$ is the equivalence classes of the relation $a\cdot b^{-1}\in H$ (consider them as right cosets here). If the group is abelian, it simply means that $a-b\in H$, and the cosets are exactly $a+H=\{a+b\mid b\in H\}$.
In the ring theoretic case, this is also how we define the equivalence relation defined by the ideal.
Note that a $\sigma$-algebra, indeed any algebra of sets, is a ring of characteristics $2$ with $\triangle$, symmetric difference, as the addition and $\cap$ as multiplication.
So naturally, $A\sim_I B$ if and only if $A\mathbin{\triangle} B\in I$. And the equivalence class of $A$ is $[A]_I = \{A\mathbin{\triangle} X\mid X\in I\}$.


TITLE: Verifying Solution of PDE
QUESTION [0 upvotes]: Suppose that w(x,t; τ) is the solution of the following initial value problem of the
wave equation:
$w_{tt} −c^2 w_{xx} = 0, −∞< x < +∞,  t > τ $
$w = 0$ and $w_t = f(x,τ), $ when $t = τ$.
Show that $u(x,t) = \int_0 ^t w(x,t,τ) dτ$ is the solution of the initial value problem of the inhomogeneous wave equation
$u_{tt} −c^2 u_{xx} = f(x,t), −∞< x < +∞, t > 0 $
$u(x,0) = 0, u_t(x,0) = 0$
I computed $u_t , u_{tt}, u_x, u_{xx}$ (by making use of the Leibniz Rule) as follows:
$u_t = w(x,t,t) + \int_0 ^t w_t (x,t,τ) dτ$
$u_{tt} = 2w_t(x,t,t) +\int_0 ^t w_{tt} (x,t,τ)dτ$
$u_x = \int_0 ^t w_x(x,t,τ)dτ $
$u_{xx}=\int_0 ^t w_{xx}(x,t,τ)dτ  $
but this gives $u_t (x,0) = w(x,0)$ which I'm not sure should be zero. I also couldn't show $u_{tt}-c^2 u_{xx} = f(x,t)$ and the usage of $\tau$ only made me more confused... maybe I'm not computing the derivatives correctly?
Any help would be greatly appreciated.

REPLY [1 votes]: Consider
$$u(x,t) = \int_0^t w(x,t;\tau)d\tau.$$
We have that $w(x,t;t)=0$ and $w_t(x,t;t)=f(x,t).$ (A note - don't get confused by the dummy variables).
The first time derivative is therefore
$$u_t(x,t) = w(x,t;t) + \int_0^t w_t(x,t;\tau)d\tau,$$
which simplifies to
$$u_t(x,t) = \int_0^t w_t(x,t;\tau)d\tau.$$
The second is then
$$u_{tt}(x,t) = w_t(x,t,t) + \int_0^t w_{tt}(x,t;\tau)d\tau$$
which simplifies to
$$u_{tt}(x,t) = f(x,t) + \int_0^t w_{tt}(x,t;\tau)d\tau.$$
Plugging these into the original PDE therefore gives
$$f(x,t) + \int_0^t \Big[w_{tt}(x,t;\tau)-c^2w_{xx}(x,t;\tau)\Big]d\tau = f(x,t),$$
which works out to $0=0$.
Notice this depends on @Dan's correct application of Leibnitz's rule.


TITLE: Unconventional way to solve trig question
QUESTION [2 upvotes]: I was working on a trig question and got stuck, but then I noticed a possible way to solve the problem. However, this way seemed to be slightly unconventional and possibly not what the book was looking for.
The question was: "Find $k$ and $b$ when $\sqrt3 \sin x + \cos x=k\sin(x+b)$"
The first thing I did was to expand $k\sin(x+b)$ to $k\sin x\cos b+k\cos x\sin b$.
Here is where I got stuck. I tried several different things, but I hit dead ends. Then I tried to equate coefficients and got $k\cos b=\sqrt3$ and $k\sin b=1$ which simplified to:
$\cos b=\displaystyle{\sqrt3 \over k}$ and $\sin b=\displaystyle{1\over k}$. The answer from there is fairly simple to get which is $k=2$ and $b=\displaystyle{\pi\over6}$.
However this method seems rather dubious and so I was wondering if someone new a better way of solving the problem using more rigorous mathematical methods.

REPLY [1 votes]: Suppose that $a\sin(x)+b\cos(x)=c\sin(x)+d\cos(x)$ holds for all $x$ in some neighborhood of $x_0$. Then we have the following equation and its derivative:
$$
\begin{align}
0&=(a-c)\sin(x_0)+(b-d)\cos(x_0)\tag{1}\\
0&=(a-c)\cos(x_0)-(b-d)\sin(x_0)\tag{2}
\end{align}
$$
Adding the squares of $(1)$ and $(2)$ gives
$$
0=(a-c)^2+(b-d)^2\tag{3}
$$
Thus, $a=c$ and $b=d$.
So if we have
$$
\begin{align}
\sqrt{3}\sin(x)+\cos(x)
&=k\sin(x+b)\\
&=k\cos(b)\sin(x)+k\sin(b)\cos(x)\tag{4}
\end{align}
$$
for all $x$ in some neighborhood of $x_0$, then we must have
$$
k\cos(b)=\sqrt{3}\quad\text{ and }\quad k\sin(b)=1\tag{5}
$$
and $(5)$ was solved above.