TITLE: Online high quality colloquium talks
QUESTION [21 upvotes]: In my department we're thinking about showing online lectures one day per week at lunch, as sort of a virtual colloquium appropriate to mathematics undergraduates as well as faculty. To start with we'd probably not want to launch into a lecture series on a single topic, but instead show high quality colloquium talks. I'm looking for links to things like Voevodsky's lecture "An Intuitive Introduction to Motivic Homotopy Theory", for example. (We give only the title here because links to YouTube aren't allowed on MO. If you intend to answer this question with a YouTube link, instead please type the title of the talk so it can be googled.)
We intend to start with MSRI talks and ICM lectures, but would like to know which are the best quality talks in these collections!
So by now you've probably figured out that the purpose of the current question is to collect other high-quality online colloquium talks:

Question: What is a link to your favorite online talk suitable for (reasonably) general audiences, that is not part of a lecture series?

It would even be useful for us if you provided a recommendation for a particularly high quality talk found in the MSRI or ICM archives, as all of these talks are not created equal.
Thanks in advance for allowing us to benefit from your experience!
To clarify how this question is different from, for example this one, I'm looking for specific lecture recommendations and not only general collections of lectures. The hope is that the community can share its good taste in order to benefit our virtual colloquium (and help other departments who might want to try this).

REPLY [2 votes]: For specific topics, I found very high quality and worth watching videos at the Arizona Winter School. Those are yearly 5 days schools around arithmetic geometry, providing 4 lectures with one daily session, videos and notes, so it is a bit of a lecture series, however they are short, willing to be pedagogical and introductive, hence the first talk of a topic is probably suitable for a lunch colloquium.


text:
\subsection{1-bit Compressed Sensing}
\label{sec:onebitcs}

One of the confounding factors regarding the practical implementation
of CS in hardware devices is the treatment of measurement quantization.
The original CS analysis provides recovery bounds based
upon the assumption of real-valued measurements. However, in practice,
hardware devices cannot capture such values with infinite precision, and
so some kind of quantization on the measurements must be implemented.
Specifically, if $Q(\cdot)$ is
a uniform scalar quantizer, then $\vy = Q(\Phi \vx, B),$
where $B$ is the number of bits used to represent the measurement.

If signal recoverability is
significantly impacted by small $B$, then the dimensionality reduction
provided by CS may be lost by the requirement for many bits to encode
each measurement.

Thankfully, recent works have shown CS recovery to be robust
to quantization and the non-linear error it introduces.
In fact, CS has been shown \cite{BB2008,JLB2011} to be robust even in
the extreme case $B=1$ known as 1-bit CS.
In this case, the quantized measurements are given by
\begin{equation}
	\vy = \text{sign}(\Phi \vx).
    \label{eq:1bitcs}
\end{equation}
The non-linearity and severity of 1-bit CS requires special
treatment from the CS recovery procedure. In \cite{BB2008},
a renormalized fixed-point continuation (RFPC) algorithm was proposed.

Later, \cite{JLB2011} analyzed the sensitivity of 1-bit CS to sign
flips and proposed a noise-robust recovery algorithm, binary
iterative hard thresholding (BIHT).

Recognizing the capability of GAMP to handle non-linear output
channels, \cite{KGR2012} proposed the use of GAMP for signal recovery
from quantized CS measurements.
Further analysis of message-passing approaches to the
1-bit CS problem from the perspective of statistical mechanics was
given in \cite{XK2013} where a modified fixed-point iteration was derived
via the cavity method which provided both improved recovery accuracy and reconstruction time as compared to the RFPC.
Additionally, the authors used replica analysis to
estimate the optimal MSE performance of $\ell_1$-minimization based
1-bit CS reconstruction. Finally, this analysis is extended in \cite{XKZ2014}
to include the theoretical Bayesian optimal performance, which we will use
as a baseline of comparison in Fig.~\ref{fig:1bit_all}.

Both methods \cite{KGR2012} and \cite{XK2013} show the
effectiveness of algorithms grounded in statistical mechanics for
quantized CS reconstruction. However, both assume an amenable set of
projections. Even projections possessing small mean can cause large
degradations in performance.
While mean removal is occasionally effective in the usual CS setting,
it cannot be used for 1-bit CS due to the nature of the sign operation
in \eqref{eq:1bitcs}. An algorithm that can handle troublesome
projectors can therefore be of great use.  In Sec. \ref{sec:amp}, we
show how the \swamp~can be modified to the general-channel setting, as
was done in GAMP. This generalization allows for 1-bit CS recovery
with \swamp~under much more relaxed requirements for $\Phi$.

In Fig.~\ref{fig:1bit_all},
we see Generalized \swamp~(G-\swamp) results for $\Phi_{\mu i} \sim \mathcal{N}
\left( \frac{20}{N}, \frac{1}{N} \right)$.
We observe that G-\swamp~performs admirably even for this non-neglible
mean on the projectors. In terms of recovery performance, it does not quite
meet the theoretical Bayes optimal performance \cite{XKZ2014}, however, this
is expected as the Bayes optimal performance is calculated for $\gamma = 0$.
Additionally, we see that even for this non-zero mean, G-\swamp~outperforms both
the BIHT's empirical performance for the same mean,
as well as the best-case theoretical $\ell_1$ performance for zero mean \cite{XK2013}.
Finally, in Fig.~\ref{fig:1bit_vsgamp}, we see that
GAMP fails to provide any meaningful signal recovery for $\gamma$ small, while
G-\swamp~continues to converge to low-MSE even for large values of $\gamma$.

\begin{figure}[ht]
    \centering
    \begin{subfigure}[b]{0.4\figurewidth}
        \includegraphics[width=\textwidth]{1bit}
        \caption{}
        \label{fig:1bit_all}
    \end{subfigure} \hspace{2em}
    \begin{subfigure}[b]{0.5\figurewidth}
        \includegraphics[width=\textwidth]{compar_1bit}
        \caption{}
        \label{fig:1bit_vsgamp}
    \end{subfigure}
    \caption{Results for 1-bit CS. (a) Top: Comparison between the
      Bayes optimal MSE for zero-mean projectors
      \cite{XKZ2014} (dashed lines) and that obtained by \swamp~for
      projectors with $\gamma = 20$ (markers) for three
      different levels of signal sparsity. The reported empirical
      results were obtained by averaging over 200 instances of size $N
      = 512$. Bottom: Comparison of \swamp~and BIHT for $\rho = 1/8$
      for experiment conditions identical to the figure above; theoretical
      results for zero-mean projectors are also presented for completeness,
      including theoretical $\ell_1$ performance \cite{XK2013}. (b)
      Single instance comparison between GAMP and G-\swamp~ for 1-bit
      CS with $N = 2048$, $\rho = 1/8$, and $\alpha = 3$.}
\end{figure}

config: arxiv


TITLE: Why use the letter "k" in the function transformation formula $f(x - h) + k$?
QUESTION [1 upvotes]: This is strictly a historical "why is it the letter k rather than say v for vertical" question -- is it the initial letter of something from a specific language? Is it arbitrary?
While we're at it, did "h" originally stand for "horizontal" or was it something else?

REPLY [0 votes]: 1.There are a lot of words that can be used but we use letter "k". It just to show that the word represents a constant,or in German konstant. So we can get an idea of use of k mostly for a constant.
2.Its is obvious that "h" doesn't stand for horizontal.It can be any thing but one can think it as a coincidence.


TITLE: $u(x)$ satisfies the integral equation $u(x) = \int_0^x \sin(u(t))u(t)^p dt$ on $0 \leq x \leq 1$. show that $u(x) = 0$ on this interval
QUESTION [3 upvotes]: Suppose that $u(x)$ is continuous and satisfies the integral equation
\begin{equation}\label{1.4.1}
        u(x) = \int_0^x \sin(u(t))u(t)^p dt
    \end{equation}
on the interval $0 \leq x \leq 1$. show that $u(x) = 0$ on this interval if $p \geq 0$.
This is what I have:
Since $\sin(u(t))u(t)^p$ is continuous, it follows from the integral definition that $u(x)$ is differentiable. Let us differentiate both sides of equation above with respect to $x$. This yields:
\begin{equation}
    u'(x) = \sin(u(x))u(x)^p
    \end{equation}
This ODE is separabale and becomes:
\begin{equation}
        \frac{1}{\sin(u(x))u(x)^p}du(x) = dx
    \end{equation}
However, this doesn't seem easily solvable so I'm not sure how to show that $u(x) = 0$ from this.

REPLY [0 votes]: Use @Matthew Pilling's setting. Assume there exists $t\in(0,1)$ such that $u(t)>0$. Because $u$ is continuous and $u(0)=0$, we can identify $a,b\in [0,1]$ such that $u(a)=0,a<t<b,$ and $u>0$ on $(a,b)$. Since $\sin\big(u(t)\big)$ is also continuous, we can find $c\in (a,b)$ such that $\sin \big(u(t)\big)>0$ on $(a,c)$. Now consider the IVP:
$$\frac{dy}{dx}=y^p\sin(y),x\in(a,c).$$
So
$$ y'=y^p\sin(y)\le y^{p+1}, x\in(a,c)$$
and hence
$$ y^{-p-1}y'\le 1. \tag1$$
Integrating (1) from $x_0$ to $x$ ($x_0,x\in(a,c)$, $x_0<x$), one has
$$ -\frac1p\bigg[y^{-p}(x)-y^{-p}(x_0)\bigg]\le x-x_0\le1 $$
from which
$$ y^{-p}(x)\ge y^{-p}(x_0)-p. \tag2$$
Choose $x_0$ close to $a$ such that
$$ y^{-p}(x_0)> p $$
and then from (2), one has
$$ 0< y^p(x)\le \frac{y^p(x_0)}{1-py^p(x_0)}. \tag3$$
Letting $x_0\to a^+$ in (3) gives
$$ 0<y^p(x)\le0, x\in(a,c)$$
which is absurd for $x\in(a,c)$.


TITLE: Work done when you bring an bringing an object down from a height
QUESTION [1 upvotes]: I am going to explain this question through an example.
Suppose I lift an object I apply a force $mg $ then I apply additional force, that would be $ma $ so total force would be $m(g+a)$. My doubt is that the work done by a person in lifting a box by applying more force than the weight that should be $m(g+a) *h $ right?
I would love to know if my answer is right or if someone could kindly correct me

REPLY [1 votes]: Normally in lifting, you apply a momentary extra force to get the object moving, and then proceed at a constant speed.  As Dale points out, the extra work goes into kinetic energy.