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What makes some things difficult or easy to understand?

I am thinking really deep about this topic with no answer. For example while learning set theory first time you find it very difficult to get your head around concepts like Union, Intersection etc but after some practice it becomes quite obvious, but as we go further there comes concepts like Group theory, Abstract algebra etc which even after reading 100 times doesn't makes any sense. On the other side you see some people playing with those concepts very easily.

Has anyone any idea on this topic?

• Abstract concepts are more difficult to understand. Also, domains with many concepts connected with many complex relations. – DesignerAnalyst Oct 26 '15 at 16:19
• @DesignerAnalyst your comment seems the most abstract comment to me :) – CodeYogi Oct 26 '15 at 16:44
• The complete answer probably involves some reasonably complex Person X Stimulus type X Context interactions (i.e., it depends). – mrt Oct 28 '15 at 2:08
• @mrt Please explain Stimulus and Context. – CodeYogi Oct 29 '15 at 3:11
• This question is pretty broad as it covers a bunch of different domains of cognitive psychology. Are you asking about mathematical concepts specifically? In what domain of mathematics? Are you talking about crossing between domains of mathematics? Is the question specific to adults or to developing children? – Seanny123 Oct 31 '15 at 19:37

Firstly lets define criteria for what difficult learning material is. My interpretation of the matter states that the task of developing mastery or competency in understanding and applying difficult content:

• Takes an increased amount of time
• or Demands increased cognitive effort

It's relevant to keep in mind The 2 systems that occur in the brain

The brain is known to have a limited amount of cognitive capacity on the amount of calculations or work able to be performed, at once. This capacity is constant, unless brain development due to certain hard activities occur, in which, overtime the capacity of the brain will increase, but this happens over a long period of time(weeks months etc).

Thinking Fast and Slow by Daniel Kahneman references an experiment where demand was found to correlate to pupil dilation:

We recorded strings of digits on beats of the metronome and instructed the subject to repeat or transform the digits one by one, maintaining the same rhythm. We soon discovered that the size of the pupil varied second by second, reflecting the changing demands of the task.

At a certain point though, the capacity of the brain is met:

reaches an almost intolerable peak as you rush to produce a transformed string during and immediately after the pause, and relaxes gradually as you “unload” your short-term memory. The pupil data corresponded precisely to subjective experience: longer strings reliably caused larger dilations, the transformation task compounded the effort, and the peak of pupil size coincided with maximum effort.

The evidence provided strongly suggests that the theory put forward has some accuracy

According to Thinking Fast and Slow there are several factors which use up this cognitive capacity/energy, but the most important ones to consider are:

• Multitasking Mentally computing a series of tasks, while keeping the state of other tasks in memory. For example when multiplying 15 * 34 without a calculator, you must keep the value of 5 * 34 in your head while doing 10* 34
• Resisting subconscious processes For example unlearning habits, or the Stroop effect
• Memory how much memory you are actively storing

These factors, which tax cognitive capacity, are ultimately the root cause of what determines the difficulty of concept, the more the learning material provokes any of these tasks, the more difficult the content is to learn.

These factors are mainly affected by two elements of knowledge: the complexity of the knowledge, and how well it relates to information currently in your head.

There are many theories on what knowledge is, and how it is interpreted by humans, including Bloom's taxonomy but the most relevant one for this question is Concept Learning which I will briefly state the 3 levels of concepts:

Not a concept simple recall or discrimination of knowledge

Concrete or Perceptual Concepts knowledge that bears physical similarity to one another from the learner.

Defined, or relational and associated concepts where the stimuli bears no obvious physical similarities, but cohere due to functional properties.

from not a concept -> concrete or perceptual concepts, concepts basically grow in amount of similar or distinct properties, and abstractions; distinct physical properties, but similar functional properties.

The more properties equates to more demand for memory. The more distinct and similar properties present, the more demand for multitasking, which therefore makes complex concepts harder to understand.

In Make it Stick, a book on learning strategies suggests that humans learn content through understanding it in short term memory, but after practice, the information goes through a process of:

consolidation, in which memory traces (the brain’s representations of the new learning) are strengthened, given meaning, and connected to prior knowledge

This point relates to Resisting Subconscious Processes which is described in "thinking fast and slow" tax cognitive capacity.

Also the more foreign the information is to you;the information already stored in your head, the harder consolidation will be, taking up more mental capacity.

Another consideration is elaboration. Knowledge is suggested to trigger the thought of other knowledge when the brain understands the relationship between knowledge. If concepts are not understood completely, and instead its information is just memorized, then this process will not occur and competency and mastery in understanding and applying the concept will be reduced. This is discussed in A Mind Of Numbers. This process not triggering will then require the logical part of the mind to figure out the connection, where the subconscious mind does not supply it, and demand more mental capacity.

So basically, the more complex information is, and the more foreign the information is, the more taxing the information will be in learning it, which makes it harder to understand.

• Well written! but I would need couple to readings more to understand it completely since its a foreign knowledge to me :) – CodeYogi Oct 28 '15 at 4:24
• That said! I think only hard work doesn't pay off completely and you feel cheated, lose motivation and neglected to work harder next time. It feels that your mind resists you to think further, you just need to finish the study material as soon as possible. I know setting possible goals might help, but in this case your mind becomes your own enemy. There have been countless times when given a puzzle / problem I find that my mind is thinking nothing I am just gazing over it sweating, feel my heart pumped up and fell headache. Feels like my mind has become too lazy to do anything. – CodeYogi Oct 28 '15 at 4:30
• Also, I forgot to mention that I even feel too distracted to read long posts like these in one sitting. Really I find it too disturbing like I mentioned in my first comment. While reading your post I switched tab many times, changed songs, had an urge to watch some cool video. Is this normal? – CodeYogi Oct 28 '15 at 4:34
• Its completely normal, I am probably worse than you!!, I've got finals tomorrow and haven't studied all day. Its called procrastination, though I do not believe your problem directly relates to difficult concepts, it is more to do with your emotional intelligence, how well you can control your thoughts. Its basically procrastination, as the work seems to hard or boring to do. I suggest you apply the stuff in this video:youtube.com/watch?v=wmx_35rQIRg – tristo Oct 28 '15 at 6:14

Part of it has to do with the ability to construct a model of reality. Every developer can relate to that. Visible, concrete things are easy to deconstruct in pragmatic terms. But once you get into abstract concepts, it gets harder and harder to bind it with your existing view of the world.

Some concepts are also very difficult to grasp if you can't connect parts of the concept with things you already understand. This process is called chunking, and it can be understood as dividing the problem until each part already makes sense to you, or is small enough to tackle the question on your own.

On top of this, you have to know what you're really looking for in order to ask the right questions. Some students just ask "How does this work? I don't understand", but the teacher can't answer something so vague. Richard Feynman make a very good statement about this problem in this interview.

• Ya, for example I tried to visualize tree while learning counting but it becomes too cumbersome when you start too difficult problems. – CodeYogi Oct 28 '15 at 4:12
• Please check my second comment on @tristo's answer. – CodeYogi Oct 28 '15 at 4:31


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# Physics of guitar strings

Guitarists normally press down hard on the frets and then pluck a string to obtain a note. However, one can also create notes by just touching the string above a particular fret and plucking.

For example, if I just press on a string about two thirds of the way up the guitar (from the bridge), a note of one octave higher in pitch is created than when I press down firmly at the same place. I know that an increase of one octave is the same as doubling the pitch.

How does this happen? What's the physics behind it? I assumed it was something to do with nodes, but I wasn't sure.

• Hint: the note you describe, one octave higher, is obtained exactly at half the length (12th fret), not 2/3. – fqq Jan 20 '15 at 21:26
• @fqq "one octave higher in pitch than when I press down firmly at the same place", i.e. the 3rd harmonic of the open string, compared to fretting the string to 2/3 of its open length. – Nathan Reed Jan 20 '15 at 21:35
• And to the OP, the phenomenon is called guitar harmonics and Wikipedia has a good explanation. – Nathan Reed Jan 20 '15 at 21:38
• @NathanReed you are right, I didn't read well, sorry about that. That's correct. – fqq Jan 20 '15 at 22:49

How does this happen? What's the physics behind it?

If you fret the string and pluck it, the string will vibrate strongest at the middle point (between the bridge and the fret). This is the fundamental vibrational mode of the string.

By placing your finger lightly at, e.g., the middle of the string and plucking (while quickly removing your finger), you force the string to vibrate such that the middle is a node.

In such a case, the string is vibrating in a higher order (2nd, 3rd, etc.) mode where there are one or more nodes along the string instead of just at the ends.

Image credit

The physics requires that, in the higher order modes, the frequency of the amplitude variations are higher which corresponds to higher pitch.

For example, in the 2nd mode, the pitch is twice as high as in the fundamental mode.

Animated gif credit

• Nice visuals! Did you do the animation in $\LaTeX$? – Bill N Jan 20 '15 at 22:36
• @BillN, I was adding the credits just as your comment came in. – Alfred Centauri Jan 20 '15 at 22:39
• The actual waveshape differs from what is shown above. If you look at slow-mo videos of guitar strings the waveforms look more like trapezoids than sine waves. – ja72 Jun 6 '17 at 18:54
• @ja72, if you pluck the guitar string near the bridge, far from the anti-node, it will be 'bright' sounding indicating the presence of several harmonics as would be expected of say, a trapezoidal wave (however, the higher harmonics will decay faster than the fundamental and eventually, the string will vibrate more or less in the fundamental mode). If you pluck the guitar string at the anti-node, it will vibrate nearly sinusoidally. If, as the OP asks about, you lightly touch the string above say, the 12 fret and let go while plucking, you'll excited the string into the 2nd harmonic. – Alfred Centauri Jun 6 '17 at 23:41
• @ja72, see for example, this – Alfred Centauri Jun 6 '17 at 23:46

A string fixed at both ends, as on a guitar, can vibrate in a "standing wave" mode at several different frequencies. The lowest frequency, the fundamental, is such that the length of the string matches half of a wavelength, $L=\lambda/2$. The middle of the string has the maximum displacement from rest and the two ends don't move. In this case, the mid-point of the string is called an antinode and the ends are (always) nodes. The frequency of the string, and hence the frequency of the sound wave in the air generated by the string vibrations is $$f_1 = \frac{v}{\lambda} = \frac{v}{2L}$$ where $v$ is the speed of the wave in the string. This speed depends on the string material, thickness, and the tension (tightness) of the string. Thicker and more dense string means slower speed and tighter means higher speed. That's why higher pitched strings are generally thin and lower pitched are thick. It helps for the guitar strings to have a nearly uniform tension so that the neck isn't pulled sideways.

This fundamental isn't the only frequency present when a string is plucked. Another standing wave has a node in the middle and antinodes at 1/4 and 3/4. This means that the length of the string equals the wavelength of this wave, $L=\lambda_2$. So the frequency of this wave is $$f_2=\frac{v}{L} = 2f_1.$$ Aha! This second wave is one octave above the fundamental. It's called the 2nd partial, the 2nd (musical) harmonic, and the 1st overtone of a string (fixed at both ends). You can get this note to sound strongly in two ways: 1) press at the 12th fret, which forces the length of the string to $L_{new}=L/2$ so now the new fundamental is $v/L$ or 2) touch the string just over the 12th fret (at $L/2$) which forces a node where there would normally be an antinode of the fundamental. This prevents the fundamental from sounding, so you can hear the overtone.

You see, when you pluck a guitar string, you add energy to the string at hundreds of different frequencies. This is due to something called the Fourier Theorem. Only the frequencies in this pluck which match possible standing waves in the string will persist longer than a few milliseconds. All the others a quickly dissipated into the internal energy of the string and guitar components, never to be heard again. So a pluck will result in the string having the fundamental, the 1st overtone, the 2nd, the 3rd and so forth.

Let's look at the 2nd overtone. This requires another node, with the nodes evenly spaced on the string, so now we have 3 antinodes and the length of the string is split into 3 half wavelengths, $L=3\frac{\lambda_3}{2}$. The resulting frequency is $$f_3=\frac{v}{L(2/3)}=\frac{3v}{2L} = 3f_1$$ This frequency is also $3/2f_2$. A frequency ratio of 3/2 is a musical fifth, for example, G up to D.

If you consider the position of playing D on the G string, that's at the 7th fret, and it's also 1/3 of the way from the nut to the bridge (your 2/3 position). If you press there, the new length is the string is 2/3L, so the fundamental of this fretted string would be $$f_{fret7}=\frac{v}{2(L)(2/3)} = \frac{3v}{4L}$$. This is exactly half of the frequency of the 2nd overtone (also called the 3rd harmonic). Merely touching the string just over the 7th fret kills the original fundamental and 1st overtone, allowing the 2nd overtone to be heard with a frequency an octave higher than the 7th fret pressed note.

And that's how it happens. If you touch just above the 5th fret you will get a double octave because now you have killed the vibrations of the first three harmonics and forced a node at the nodal location for the 4th harmonic/3rd overtone.

• Small nitpick. The Fourier Theorem doesn't cause the multiple frequencies ("due to"), it helps describe them. – Floris Jan 21 '15 at 0:19
• @Floris Fourier Theorem allows you to show that the pluck waveform is equivalent to multiple frequency sine/cosine inputs. "Due to" is about "how we explain" the multiple f's. Never meant to imply cause. – Bill N Jan 21 '15 at 4:45

## Harmonic Response

Consider an elastic string stretched between two fixed points a distance $L$ apart. The harmonic response is a standing wave with speed $c=\sqrt{E/\rho}$

\begin{align} y(x,t) &=\sum_{i=1}^{\infty}\sin\left(\frac{i\pi x}{L}\right)\left(A_{i}\sin\left(\frac{i\pi ct}{L}\right)+B_{i}\cos\left(\frac{i\pi ct}{L}\right)\right) \\ \dot{y}(x,t) &=\sum_{i=1}^{\infty}\sin\left(\frac{i\pi x}{L}\right)\frac{i\pi c}{L}\left(A_{i}\cos\left(\frac{i\pi ct}{L}\right)-B_{i}\sin\left(\frac{i\pi ct}{L}\right)\right) \end{align}

with unknown coefficients $A_{i}$ and $B_{i}$ depending on the initial conditions. This is a result of the equation of motion

$$T\,\frac{\partial^{2}y(x,t)}{\partial x^{2}}-\rho\,S\,\frac{\partial^{2}y(x,t)}{\partial t^{2}}=0$$

where $T$ is the elastic tension in $\rm{[N]}$, $E$ the elastic modulus in $[{\rm N/m^2}]$, $S$ is the string area in $[{\rm m^2}]$ and $\rho$ the mass density in $[\rm{kg}/{m^{3}}]$.

## Initial Conditions

Now consider at $t=0$ the shape and speed of the string to be $y(x,0)=U(x)$ and $\dot{y}(x,o)=V(x)$.

For example, for a slow pluck at point $x_p$ the initial speed is zero $V(x)=0$ and the shape is $$U(x)=\begin{cases} U_{0}\left(\frac{x}{x_{p}}\right) & 0\geq x\geq x_{p}\\ U_{0}\left(1-\frac{x-x_{p}}{L-x_{p}}\right) & x_{p}>x\geq L \end{cases}$$ where $U_0$ is the amplitude of the pluck

## Fourier Analysis

By expanding out the following frequency decomposition we get the coefficients $A_i$ and $B_i$.

\left. \begin{aligned} \int\limits _{0}^{L}\sin\left(\frac{j\pi x}{L}\right)U(x)\,{\rm d}x&=\frac{L}{2}\sum_{i=1}^{\infty}\left[B_{i}\delta_{ij}\right]=\frac{L}{2}B_{j} \\ \int\limits _{0}^{L}\sin\left(\frac{j\pi x}{L}\right)V(x)\,{\rm d}x&=\frac{L}{2}\sum_{i=1}^{\infty}\left[\frac{i\pi c}{L}A_{i}\delta_{ij}\right]=\frac{i\pi c}{2}A_{j} \end{aligned} \right\} \boxed{ \begin{aligned} A_{i}&=\frac{2}{i\pi c}\int\limits _{0}^{L}\sin\left(\frac{i\pi x}{L}\right)V(x)\,{\rm d}x \\ B_{i}&=\frac{2}{L}\int\limits _{0}^{L}\sin\left(\frac{i\pi x}{L}\right)U(x)\,{\rm d}x \end{aligned} }

Again for the slow plucked string we get \begin{aligned} A_{i}&=0 \\ B_{i}&=U_{0}\frac{2L^{2}\sin\left(\frac{\pi ix_{p}}{L}\right)}{\pi^{2}i^{2}x_{p}\left(L-x_{p}\right)} \end{aligned}

## Example

A slow plucked string in the middle $x_p = \frac{1}{2} L$ has solution

$$y(x,t) = \sum_{i=1}^{\infty} \frac{8 U_0}{i^2 \pi^2} \sin\left( \frac{i \pi x}{L} \right) \sin\left( \frac{i \pi}{2} \right) \cos\left( \frac{\pi c i t}{L} \right)$$ $$y(x,t) \approx \frac{8 U_0}{\pi^2} \left[ \cos(\theta) \sin\left( \frac{\pi x}{L} \right) - \frac{1}{9} \cos(3\theta) \sin\left( \frac{3 \pi x}{L} \right) +\frac{1}{25} \cos(5\theta) \sin\left( \frac{5 \pi x}{L} \right) + \ldots \right]$$

where $\theta = \frac{\pi c t}{L}$ is a new independent variable replacing time.

Additionally to the "usual" frequency of the string, it always produces tones with integer multiple frequencies with decreasing volume due to an effect called standing waves. When you damp say in the middle of the string only multiples of 2 remain because only those standing waves have a node there. What you hear will be an octave because the lowest of the frequencies will be the loudest. If you damp at a third of the string, only multiples of 3 times the ground frequency remain etc.

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Math Help - taylor series

1. taylor series

how would i develop a 3rd order taylor series for sqrt(x) @ x=1 and use it to estimate sqrt(2)??

2. Originally Posted by thedoge
how would i develop a 3rd order taylor series for sqrt(x) @ x=1 and use it to estimate sqrt(2)??
Here

3. Hear here!


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# A question on almost disjoint collection

Here is a theorem:

Let $E$ be an uncountable infinite set. Then there is a collection $\mathcal{A}$ of subsets of $E$ such that $|\mathcal{A}|=|E|^\omega$, $|A|=\omega$ for each $A\in \mathcal{A}$, and the intersection of any two distinct elements of $\mathcal{A}$ is finite.

How to show it? I hope the proof a little simple and clear, because it is too difficult in my textbook. Thanks.

ADD: Thanks Brain. In my textbook, it said, it suffices to construct the desired collection on the set of all finite subset of $\omega \times E$. For each $f\in {}^\omega E$, let $A(f)=\{(f|n):n<\omega\}$. One can easily check $A(f)\cap A(g)$ is finite whenever $f$ and $g$ are distinct elements of ${}^\omega E$, so $\{A(f): f\in {}^\omega E\}$ is the desired collection of sets. I cannot check this sentence: One can easily check $A(f)\cap A(g)$ is finite whenever $f$ and $g$ are distinct elements of ${}^\omega E$.

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For a regular cardinal $\kappa$, by a result of Ulam, it is a disjoint union of $\kappa$ stationary sets. For singular cardinals, use the same argument but on the cofinality $\kappa$. –  Eran Mar 12 '13 at 13:03

## 1 Answer

Let $T={}^{<\omega}E=\bigcup_{n\in\omega}{}^nE$, the set of functions from finite ordinals into $E$. Then $\langle T,\subseteq\rangle$ is a tree of height $\omega$. Note that for $s\in{}^mE$ and $t\in{}^nE$, $s\subseteq t$ iff $t\upharpoonright m=s$. Each $\sigma\in{}^\omega E$ defines a branch through $T$; for $\sigma\in{}^\omega E$ let $B_\sigma=\{\sigma\upharpoonright n:n\in\omega\}$, the set of nodes on that branch. Each $B_\sigma$ is a countably infinite subset of $T$. $|T|=\omega\cdot|E|=|E|$, so there is a bijection $\varphi:T\to E$; for each $\sigma\in{}^\omega E$ let $A_\sigma=\varphi[B_\sigma]$, and let $\mathscr{A}=\left\{A_\sigma:\sigma\in{}^\omega E\right\}$. I leave it to you to verify that $\mathscr{A}$ has the desired properties.

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Thanks Brain for your answer. I have a small question in the proof of my textbook. Could you help me? see the added part. –  Simple Mar 13 '13 at 5:31
@Simple: If $f,g\in{}^\omega E$ and $f\ne g$, then there is an $n\in\omega$ such that $f(n)\ne g(n)$. Then for every $k>n$ we have $f\upharpoonright k\ne g\upharpoonright k$, since $(f\upharpoonright k)(n)\ne(g\upharpoonright k)(n)$. This means that $A(f)\cap A(g)\subseteq\{f\upharpoonright k:k\le n\}$, which is finite. –  Brian M. Scott Mar 13 '13 at 8:10
Maybe i'm not clear to the $A(f)$, however, by the definition, $A(f)$ seems that it is not $\subset E$, it is a collection of some subsets of $E$. Is it right? –  Simple Mar 13 '13 at 8:58
@Simple: The proof in your book says right up front that it’s constructing subsets of $\omega\times E$, not of $E$, just as I constructed subsets $B_\sigma$ of $T$, not of $E$. But $|\omega\times E|=|E|$, so there is a bijection $\varphi:\omega\times E\to E$, and the sets $\varphi[A(f)]$, which are subsets of $E$, are still almost disjoint. All of the properties that you want $\mathscr{A}$ to have are preserved by bijections, so you can transfer such a family from one set of a given cardinality to any other set of the same cardinality. –  Brian M. Scott Mar 13 '13 at 14:52


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# A geometry problem by karan tatariya

Find the total number of solutions of the equation

$2x=3\pi(1-\cos x).$

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15 views

In the signal flow graph of Figure, the gain $c / r$ will be

1. $\frac{11}{9}$
2. $\frac{22}{15}$
3. $\frac{24}{23}$
4. $\frac{44}{23}$


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# Compiling libigl As A Static Library¶

Warning

Compiling libigl as a static library is considerably more difficult than using it as a header-only library (see installation instruction instead). Do it only if you are experienced with C++, cmake and make, and you want to improve your compilation times.

Libigl is developed most often on Mac OS X, though has current users in Linux and Windows.

### Linux/Mac OS X/Cygwin¶

Libigl may also be compiled to a static library. This is advantageous when building a project with libigl, since when used as an header-only library can slow down compile times.

To build the entire libigl library producing at least lib/libigl.a and possible other (automatically detected) extras, e.g. lib/libiglcgal.a issue:

cmake ../ -DCMAKE_BUILD_TYPE=Release\
-DLIBIGL_USE_STATIC_LIBRARY=ON\
-DCMAKE_INSTALL_PREFIX=/path/to/custom/installation

make
make install

This will install libigl into the directory specified by CMAKE_INSTALL_PREFIX, which is set to /usr/local by deafault on macOS and Linux. The installation process does not only create ${CMAKE_INSTALL_PREFIX}/lib/libigl.a, it copies the libigl headers into ${CMAKE_INSTALL_PREFIX}/include/igl/ too.

Or if you base your project on the libigl-example-project you can add

option(LIBIGL_USE_STATIC_LIBRARY     "Use libIGL as static librarie" ON)

before the following line

find_package(LIBIGL REQUIRED QUIET)

in the CMakeLists.txt to always build libigl as static library for your project. See example-project aswell.

Tip

If you have changed the value for LIBIGL_USE_STATIC_LIBRARY in your CMakeLists.txt make sure to remove or update the CMakeCache.txt in your build directory.

Warning

You should expect to see a few linker warnings of the form:

/opt/local/bin/ranlib: file: libigl.a(*.cpp.o) has no symbols

These are (admittedly unpopular) functions that have never been used by us statically so we haven’t explicit instantiations (yet).

## Development¶

Further documentation for developers is listed in style-guidelines.

See LICENSE.txt

## Zipping¶

Zip this directory without .git litter and binaries using:

git archive -prefix=libigl/ -o libigl.zip master

## Explicit Instantiations Of Templated Functions¶

Special care must be taken by the developers of each function and class in the libigl library that uses C++ templates. If this function is intended to be compiled into the statically linked libigl library then function is only compiled for each explicitly instantiated declaration. These should be added at the bottom of the corresponding .cpp file surrounded by a

#ifdef IGL_STATIC_LIBRARY

Of course, a developer may not know ahead of time which instantiations should be explicitly included in the igl static lib. One way to find out is to add one explicit instantiation for each call in one’s own project. This only ever needs to be done once for each template.

The process is somewhat mechanical using a linker with reasonable error output.

Supposed for example we have compiled the igl static lib, including the cat.h and cat.cpp functions, without any explicit instantiation. Say using the makefile in the libigl directory:

cd $LIBIGL make  Now if we try to compile a project and link against it we may get an error like: Undefined symbols for architecture x86_64: "Eigen::Matrix<int, -1, -1, 0, -1, -1> igl::cat<Eigen::Matrix<int, -1, -1, 0, -1, -1> >(int, Eigen::Matrix<int, -1, -1, 0, -1, -1> const&, Eigen::Matrix<int, -1, -1, 0, -1, -1> const&)", referenced from: uniform_sample(Eigen::Matrix<double, -1, -1, 0, -1, -1> const&, Eigen::Matrix<int, -1, -1, 0, -1, -1> const&, int, double, Eigen::Matrix<double, -1, -1, 0, -1, -1>&)in Skinning.o "Eigen::SparseMatrix<double, 0, int> igl::cat<Eigen::SparseMatrix<double, 0, int> >(int, Eigen::SparseMatrix<double, 0, int> const&, Eigen::SparseMatrix<double, 0, int> const&)", referenced from: covariance_scatter_matrix(Eigen::Matrix<double, -1, -1, 0, -1, -1> const&, Eigen::Matrix<int, -1, -1, 0, -1, -1> const&, ArapEnergy, Eigen::SparseMatrix<double, 0, int>&)in arap_dof.o arap_rhs(Eigen::Matrix<double, -1, -1, 0, -1, -1> const&, Eigen::Matrix<int, -1, -1, 0, -1, -1> const&, ArapEnergy, Eigen::SparseMatrix<double, 0, int>&)in arap_dof.o  This looks like a mess, but luckily we don’t really need to read it all. Just copy the first part in quotes Eigen::Matrix<int, -1, -1, 0, -1, -1> igl::cat<Eigen::Matrix<int, -1, -1, 0, -1, -1> >(int, Eigen::Matrix<int, -1, -1, 0, -1, -1> const&, Eigen::Matrix<int, -1, -1, 0, -1, -1> const&)  , then append it to the list of explicit template instantiations at the end of cat.cpp after the word template and followed by a semi-colon. Like this: #ifdef IGL_STATIC_LIBRARY // Explicit template instantiation template Eigen::Matrix<int, -1, -1, 0, -1, -1> igl::cat<Eigen::Matrix<int, -1, -1, 0, -1, -1> >(int, Eigen::Matrix<int, -1, -1, 0, -1, -1> const&, Eigen::Matrix<int, -1, -1, 0, -1, -1> const&); #endif  Then you must recompile the IGL static library. cd$LIBIGL
make

And try to compile your project again, potentially repeating this process until no more symbols are undefined.

Note

It may be useful to check that you code compiles with no errors first using the headers-only version to be sure that all errors are from missing template instantiations.

If you’re using make then the following command will reveal each missing symbol on its own line:

make 2>&1 | grep "referenced from" | sed -e "s/, referenced from.*//"

### Benefits Of Static Library¶

• Faster compile time: Because the libigl library is already compiled, only the new code in ones project must be compiled and then linked to IGL. This means compile times are generally faster.

• Debug or optimized: The IGL static library may be compiled in debug mode or optimized release mode regardless of whether one’s project is being optimized or debugged.

### Drawbacks Of Static Library¶

• Hard to use templates: Special care (by the developers of the library) needs to be taken when exposing templated functions.


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# Properties

 Label 1225.1.q.a Level $1225$ Weight $1$ Character orbit 1225.q Analytic conductor $0.611$ Analytic rank $0$ Dimension $4$ Projective image $D_{2}$ CM/RM discs -7, -35, 5 Inner twists $16$

# Related objects

## Newspace parameters

 Level: $$N$$ $$=$$ $$1225 = 5^{2} \cdot 7^{2}$$ Weight: $$k$$ $$=$$ $$1$$ Character orbit: $$[\chi]$$ $$=$$ 1225.q (of order $$12$$, degree $$4$$, not minimal)

## Newform invariants

 Self dual: no Analytic conductor: $$0.611354640475$$ Analytic rank: $$0$$ Dimension: $$4$$ Coefficient field: $$\Q(\zeta_{12})$$ Defining polynomial: $$x^{4} - x^{2} + 1$$ Coefficient ring: $$\Z[a_1, \ldots, a_{4}]$$ Coefficient ring index: $$1$$ Twist minimal: yes Projective image $$D_{2}$$ Projective field Galois closure of $$\Q(\sqrt{5}, \sqrt{-7})$$ Artin image $C_3\times OD_{16}$ Artin field Galois closure of $$\mathbb{Q}[x]/(x^{24} - \cdots)$$

## $q$-expansion

The $$q$$-expansion and trace form are shown below.

 $$f(q)$$ $$=$$ $$q + \zeta_{12} q^{4} -\zeta_{12}^{5} q^{9} +O(q^{10})$$ $$q + \zeta_{12} q^{4} -\zeta_{12}^{5} q^{9} + 2 \zeta_{12}^{4} q^{11} + \zeta_{12}^{2} q^{16} -2 \zeta_{12}^{3} q^{29} + q^{36} + 2 \zeta_{12}^{5} q^{44} + \zeta_{12}^{3} q^{64} -2 q^{71} -2 \zeta_{12}^{5} q^{79} -\zeta_{12}^{4} q^{81} + 2 \zeta_{12}^{3} q^{99} +O(q^{100})$$ $$\operatorname{Tr}(f)(q)$$ $$=$$ $$4q + O(q^{10})$$ $$4q - 4q^{11} + 2q^{16} + 4q^{36} - 8q^{71} + 2q^{81} + O(q^{100})$$

## Character values

We give the values of $$\chi$$ on generators for $$\left(\mathbb{Z}/1225\mathbb{Z}\right)^\times$$.

 $$n$$ $$101$$ $$1177$$ $$\chi(n)$$ $$-\zeta_{12}^{2}$$ $$\zeta_{12}^{3}$$

## Embeddings

For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.

For more information on an embedded modular form you can click on its label.

Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
18.1
 −0.866025 − 0.500000i 0.866025 + 0.500000i 0.866025 − 0.500000i −0.866025 + 0.500000i
0 0 −0.866025 0.500000i 0 0 0 0 −0.866025 + 0.500000i 0
557.1 0 0 0.866025 + 0.500000i 0 0 0 0 0.866025 0.500000i 0
618.1 0 0 0.866025 0.500000i 0 0 0 0 0.866025 + 0.500000i 0
1157.1 0 0 −0.866025 + 0.500000i 0 0 0 0 −0.866025 0.500000i 0
 $$n$$: e.g. 2-40 or 990-1000 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles

## Inner twists

Char Parity Ord Mult Type
1.a even 1 1 trivial
5.b even 2 1 RM by $$\Q(\sqrt{5})$$
7.b odd 2 1 CM by $$\Q(\sqrt{-7})$$
35.c odd 2 1 CM by $$\Q(\sqrt{-35})$$
5.c odd 4 2 inner
7.c even 3 1 inner
7.d odd 6 1 inner
35.f even 4 2 inner
35.i odd 6 1 inner
35.j even 6 1 inner
35.k even 12 2 inner
35.l odd 12 2 inner

## Twists

By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 1225.1.q.a 4
5.b even 2 1 RM 1225.1.q.a 4
5.c odd 4 2 inner 1225.1.q.a 4
7.b odd 2 1 CM 1225.1.q.a 4
7.c even 3 1 1225.1.g.a 2
7.c even 3 1 inner 1225.1.q.a 4
7.d odd 6 1 1225.1.g.a 2
7.d odd 6 1 inner 1225.1.q.a 4
35.c odd 2 1 CM 1225.1.q.a 4
35.f even 4 2 inner 1225.1.q.a 4
35.i odd 6 1 1225.1.g.a 2
35.i odd 6 1 inner 1225.1.q.a 4
35.j even 6 1 1225.1.g.a 2
35.j even 6 1 inner 1225.1.q.a 4
35.k even 12 2 1225.1.g.a 2
35.k even 12 2 inner 1225.1.q.a 4
35.l odd 12 2 1225.1.g.a 2
35.l odd 12 2 inner 1225.1.q.a 4

By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
1225.1.g.a 2 7.c even 3 1
1225.1.g.a 2 7.d odd 6 1
1225.1.g.a 2 35.i odd 6 1
1225.1.g.a 2 35.j even 6 1
1225.1.g.a 2 35.k even 12 2
1225.1.g.a 2 35.l odd 12 2
1225.1.q.a 4 1.a even 1 1 trivial
1225.1.q.a 4 5.b even 2 1 RM
1225.1.q.a 4 5.c odd 4 2 inner
1225.1.q.a 4 7.b odd 2 1 CM
1225.1.q.a 4 7.c even 3 1 inner
1225.1.q.a 4 7.d odd 6 1 inner
1225.1.q.a 4 35.c odd 2 1 CM
1225.1.q.a 4 35.f even 4 2 inner
1225.1.q.a 4 35.i odd 6 1 inner
1225.1.q.a 4 35.j even 6 1 inner
1225.1.q.a 4 35.k even 12 2 inner
1225.1.q.a 4 35.l odd 12 2 inner

## Hecke kernels

This newform subspace can be constructed as the kernel of the linear operator $$T_{2}$$ acting on $$S_{1}^{\mathrm{new}}(1225, [\chi])$$.


text:
Countability of eigenvalues of a linear operator

Is it true that every closed operator on a separable Hilbert H space only has countably many eigenvalues?

Or put the other way around, if I want to ensure that a (not necessarily bounded) linear operator on a separable Hilbert space only has countably many eigenvalues, is closedness (or better said, closability) a sufficient condition?

(By the term eigenvalue, I do not only mean a point in the spectrum of course, but one that actually fulfills $Tx = \lambda x$.)

-
So are there any other natural conditions that ensure that the number of eigenvalues is countable apart from normality of the operator? –  Kofi Oct 15 '11 at 21:26
Compact operators have countably many eigenvalues. If you think about why normal operators do too, you'll see that it's essentially because eigenvectors corresponding to different eigenvalues are orthogonal. So any operator satisfying this condition will have countably many eigenvalues. For example subnormal operators--or more generally, hyponormal operators--fit the bill. –  Faisal Oct 15 '11 at 23:14

Let $T:\ell^2\rightarrow\ell^2$ be the backwards shift operator, $T(a_n) = (a_2,a_3,\cdots)$. This is a contraction. For any $\lambda\in\mathbb C$, consider the sequence given by $a_n = \lambda^n$. Thus $(a_{n+1}) = (\lambda^2,\lambda^3,\cdots) = \lambda(\lambda,\lambda^2,\cdots)$ and so, if $(a_n)\in\ell^2$, then $(a_n)$ is an eigenvector of $T$, for eigenvalue $\lambda$. Of course, $\sum_n |\lambda^n|^2 = \sum_n |\lambda^2|^n <\infty$ if and only if $|\lambda|<1$.

So even a bounded operator can have a continuum of eigenvectors.

-
How this can be possible if we work in a separate Hilbert space ? Yes we may have a continuum of spectral values but certainly not a continuum of eigenvectors ! –  user36539 Sep 27 at 18:16
Separable = countable dense subset... –  Matthew Daws Sep 27 at 20:57
in a separable HS the basis is countable. you probably mean a continuum of "eigenvectors" which belong to a space wider than HS (Dirac) –  user36539 Sep 28 at 10:20
The example shows that you are wrong, user36539. It is an operator with uncountably many ACTUAL eigenvectors to differnt eigenvalues in the SAME Hilbert Space! –  Kofi Sep 29 at 17:11
Certainly not. In fact there are bounded operators with uncountably many eigenvalues. For example, the left shift $S^\ast$ defined on $\ell^2$ by $S^\ast(x_1,x_2,\ldots)=(x_2,x_3,\ldots)$ has point spectrum equal to the open unit disk.
Laplacian given on the sphere S, its domain is a dense subset of $L^2(S)$ but its resolvent is compact...Obviously the Laplacian acts on function belonging to $L^2(S)$ which take values on the sphere –  user36539 Sep 29 at 20:32
@Kofi Computer scientists prefer high level language (C, Pascal) than Assembler contrary to the situation for (pure) mathematicians which preffer "assembler-like" language...If you preffer Laplacian given on sections of TS viewed as a Hilbert space of half-densities $L^2(S)$ –  user36539 Sep 29 at 20:40


text:
# Wydawnictwa / Czasopisma IMPAN / Studia Mathematica / Wszystkie zeszyty

## Triebel–Lizorkin spaces with non-doubling measures

### Tom 162 / 2004

Studia Mathematica 162 (2004), 105-140 MSC: Primary 42B35; Secondary 46E35, 42B25, 47B06, 46B10, 43A99. DOI: 10.4064/sm162-2-2

#### Streszczenie

Suppose that $\mu$ is a Radon measure on ${{{{\mathbb R}}}^d},$ which may be non-doubling. The only condition assumed on $\mu$ is a growth condition, namely, there is a constant $C_0>0$ such that for all $x\in \mathop {\rm supp}(\mu )$ and $r>0,$ $$\mu (B(x, r))\le C_0r^n,$$ where $0< n\leq d.$ The authors provide a theory of Triebel–Lizorkin spaces ${\dot F^s_{pq}(\mu )}$ for $1< p< \infty$, $1\le q\le \infty$ and $|s|< \theta$, where $\theta >0$ is a real number which depends on the non-doubling measure $\mu$, $C_0$, $n$ and $d$. The method does not use the vector-valued maximal function inequality of Fefferman and Stein and is new even for the classical case. As applications, the lifting properties of these spaces by using the Riesz potential operators and the dual spaces are given.

#### Autorzy

• Yongsheng HanDepartment of Mathematics
Auburn University
Auburn, AL 36849-5310, U.S.A.
e-mail
• Dachun YangDepartment of Mathematics
Beijing Normal University
Beijing 100875
People's Republic of China
e-mail

## Przeszukaj wydawnictwa IMPAN

Zbyt krótkie zapytanie. Wpisz co najmniej 4 znaki.

Odśwież obrazek


text:
XXXVIII Workshop on Geometric Methods in Physics 30.06-6.07.2019 VIII School on Geometry and Physics 24-28.06.2019
 Home First Announcement Second Announcement Travel and local information Participants of Workshop Participants of School Registration Photos Proceedings Previous Workshops Email

# Stephen B. Sontz

## Coherent States for a Toeplitz Quantization of the Manin Plane

In the theory of Toeplitz quantization of algebras coherent states are defined as eigenvectors of a Toeplitz annihilation operator. These coherent states are studied in the case when the algebra is the generically non-commutative Manin plane. We introduce the resolution of the identity, upper and lower symbols as well as a coherent state quantization, which in turn quantizes the Toeplitz quantization. We thereby have a curious combination of quantization schemes which might be a novelty. We proceed by identifying a generalized Segal–Bargmann space $\mathcal{SB}$ of square-integrable, anti-holomorphic functions as the image of a coherent state transform. Then $\mathcal{SB}$ has a reproducing kernel function which allows us to define a secondary Toeplitz quantization, whose symbols are functions.

 Event sponsored by: University of Bialystok

Webpage by: Tomasz Golinski


text:
lilypond-user
[Top][All Lists]

## Re: How do I make all measures have the same width?

 From: James Lowe Subject: Re: How do I make all measures have the same width? Date: Wed, 09 Jun 2010 13:58:32 +0100 User-agent: Thunderbird 2.0.0.24 (Windows/20100228)

I think this maybe what you are looking for

http://lilypond.org/doc/v2.13///Documentation/notation/line-breaking

The second example.

James

snozpacker wrote:
I am working on a Lilypond template to format jazz lead sheets so that they
resemble the formatting of the (hand written) lead sheets in the Chuck Sher
New Real Books.
I know that Lilypond is programmed to carefully place bar lines to make
classical scores look beautiful.  But for the many jazz tunes that are
organized into 4-bar phrases, lead sheets are usually written so that each
line has four bars, and each bar is the same width.  This way, all phrases
start at the beginning of the line, and when you scan down the page, all of
the bar lines are right on top of each other.
In the example below, you can see that the first bar of the first line is
much bigger than the first bar of the second line.  If one were writing this
snippet out by hand, one would probably first draw equally spaced bar lines
on the first two lines of staff paper and then write in the notes, to ensure
that the bar lines line up

I haven't been able to find a way to do this in Lilypond. How do I do it? Thanks.

Example code:

\paper {
indent = 0\mm
}

title = "Angst"
}

\score{
<<

\new ChordNames \chords { bes2.:m7 bes:maj7 e:maj7/bes b:m7/bes
e:m7 a:m7 c:maj7 fis:m7
}

\new Staff {
\time 3/4

\set Staff.minimumVerticalExtent = #'(-5.5 . 5.5)
\clef treble

\relative c'' {

\override Score.BarNumber  #'break-visibility = #end-of-line-invisible
\set Score.barNumberVisibility = #(every-nth-bar-number-visible 40)

r8 f, r f'~ \times 2/3 {f ees des}
c2.
r8 e, r dis'~ \times 2/3 {dis cis b}
\times 3/4 {a8 cis r4 a fis} \break
g2.~ g r8 d r b'~ \times 2/3{b a g}
\times 3/4 {fis4 a cis e} \break

}
}

}


text:
Language:   Search:   Contact
Zentralblatt MATH has released its new interface!
For an improved author identification, see the new author database of ZBMATH.

Query:
Fill in the form and click »Search«...
Format:
Display: entries per page entries
Zbl 1081.34056
Preda, Petre; Pogan, Alin; Preda, Ciprian
Schäffer spaces and uniform exponential stability of linear skew-product semiflows.
(English)
[J] J. Differ. Equations 212, No. 1, 191-207 (2005). ISSN 0022-0396

Let $\Theta$ be a locally compact metric space and $X$ a Banach space. A mapping $\sigma: \Theta\times\bbfR_+\to\Theta$ is called a semiflow if it is continuous and such that $$\sigma(\theta, 0)= \theta\text{ for }\theta\in\Theta,\tag1$$ $$\sigma(\theta, t+s)= \sigma(\sigma(\theta, s),t)\text{ for all }t,s\ge 0\text{ and }\theta\in \Theta.\tag2$$ Let $\sigma$ be a semiflow on $\Theta$ and let $\phi:\Theta\times\bbfR_+\to B(X)$ be a continuous mapping satisfying some extra conditions (for example, $\phi(\theta, 0)= I$ -- the identity operator on $X$; $\phi(\theta, t+s)= \phi(\sigma(\theta, t), s)\phi(\theta, t)$ for $t,s\ge 0$ and $\theta\in\Theta)$. The pair $\pi= (\phi,\sigma)$ is called linear strongly continuous skew-product semiflow on $X\times\Theta$. A strongly continuous skew-product semiflow $\pi= (\phi,\sigma)$ is called uniformly exponentially stable if there exist $N> 0$ and $\nu> 0$ such that $\Vert\phi(\theta, t)\Vert\le N\exp(-\nu t)$ for $t\ge 0$ and $\theta\in\Theta$. Moreover, the authors used the concept that a pair $(E,F)$ of Schäffer spaces is admissible to $\pi$. The main result of the paper asserts that $\pi$ is uniformly exponentially stable if and only if there exists a pair $(E,F)$ of Schäffer spaces admissible to $\pi$ and satisfying some additional condition. This nice paper contains also some other results and interesting examples explaining the basic concepts and illustrating the obtained results.
[J. Banaś (Rzeszów)]
MSC 2000:
*34G10 Linear ODE in abstract spaces
34D20 Lyapunov stability of ODE

Keywords: Linear skew-product semiflow; Uniform exponential stability; Admissibility

Highlights
Master Server


text:
# Signal Ground and ground in general

Hello Forum,

If we connect one of the terminals of an AC voltage source to ground, the earthed terminal will maintain the same electric potential as planet earth. Earth can be assumed to be a huge spherical conductor whose electric potential does not change (much) when connected to a time-varying source: the little positive or negative charge it acquires does not change things much...
so the terminal connected to ground will remain at that stable earth potential (which we arbitrarily call zero-volt potential) while the other AC source terminal change electric potential (larger than zero, smaller than zero) in reference to to grounded terminal...

If we did not ground either terminals of the AC source, the potential difference would still be there but we would not know the instantaneous electric potential of either terminals. All we would know is that their difference is oscillating in time sinusoidally.

Do I understand things correctly?

thanks
fog37


text:
Next Optimal Move in Tic Tac Toe
Hard Accuracy: 64.89% Submissions: 111 Points: 8

You are given a middle game situation of the game Tic Tac Toe. It is given that it is player "X's" turn and you need to give to most optimal position for the turn. The situation is given as a 3 x 3 character matrix board. '_' refers to the place is empty. 'o' refers that player O marked it in his turn at some time and 'x' refers that player X marked it in his turn at some time. It is player X's turn. Tell him the most optimal solution.(Assume player O played first).

Example 1:

Input: board = {{o, _, _},
{_, _, _},
{_, _, _}}
Output: 1 1
Explaination: Placing a 'x' in the (1, 1)
that is the center of the board is the most
optimal approach for x.

You do not need to read input or print anything. Your task is to complete the function findBestMove() which takes board as input parameter and returns the best optimal move in a list where the first one is the row index and the second one is the column index.

Expected Time Complexity: O(29)
Expected Auxiliary Space: O(1)

Constraints:
board[i][j] = 'o' / 'x' / '_'

We are replacing the old Disqus forum with the new Discussions section given below.


text:
## Benedikt Brütsch ; Wolfgang Thomas - Solving Infinite Games in the Baire Space

fi:8743 - Fundamenta Informaticae, October 21, 2022, Volume 186, Issues 1-4: Trakhtenbrot's centenary
Solving Infinite Games in the Baire Space

Authors: Benedikt Brütsch ; Wolfgang Thomas

Infinite games (in the form of Gale-Stewart games) are studied where a play is a sequence of natural numbers chosen by two players in alternation, the winning condition being a subset of the Baire space $\omega^\omega$. We consider such games defined by a natural kind of parity automata over the alphabet $\mathbb{N}$, called $\mathbb{N}$-MSO-automata, where transitions are specified by monadic second-order formulas over the successor structure of the natural numbers. We show that the classical Büchi-Landweber Theorem (for finite-state games in the Cantor space $2^\omega$) holds again for the present games: A game defined by a deterministic parity $\mathbb{N}$-MSO-automaton is determined, the winner can be computed, and an $\mathbb{N}$-MSO-transducer realizing a winning strategy for the winner can be constructed.

Volume: Volume 186, Issues 1-4: Trakhtenbrot's centenary
Published on: October 21, 2022
Accepted on: June 21, 2022
Submitted on: November 23, 2021
Keywords: Computer Science - Computer Science and Game Theory,Computer Science - Logic in Computer Science

## Consultation statistics

This page has been seen 14 times.


text:
How can I calculate the Cumulant-Generating Function in Matlab?

Let $M(h)$ be the moment-generating function, then the cumulant generating function is given by

$$K(h)=\text{ln}M(h)=\\ =\kappa_1h+\frac{1}{2!}h^2\kappa_2+\frac{1}{3!}h^3\kappa_3+\ldots$$ where $\kappa_1, \kappa_2, \ldots$, are the cumulants.

If $L=\sum_{j=1}^Nc_jx_j$ is a function of $N$ independent variables, then the cumulant-generating function for $L$ is given by $$K(h)=\sum_{j=1}^NK_j(c_jh).$$

-
You should use the equation environment and latex. Do you know how to calculate the moment generating function $M(h)$ then the cumulant is just $\ln M(h)$ just look at the defintions (or e.g. wikipedia). –  Richard Jan 9 at 11:45
Then for a normal distribution as in M(y)= E(e A*τ(t)) where A= e(β*mean+(1/2) β^2*sigma^2) Thus I need to compute the Taylor expansion around the mean (τ (t)) –  Shikamekika SK Jan 9 at 11:58
Please yous latex, I can not read your formula. Whatever: you just take the log that's it! No expansion needed. If you know the moment generating function, then evaluate it and take the log. –  Richard Jan 9 at 11:59
The issue is I did not find any function that allows the evaluation of the moment generating function so I can’ compute the ‘Cumulant moment generating function’ –  Shikamekika SK Jan 9 at 12:08
For the mgf you usually use the analytic expression if it is known. In the case of the normal distribution you write a function yourself (if it is not built in somewhwere in a package). But you really don't need the series expansion. –  Richard Jan 9 at 13:05

I am trying to make things clear with this answer. In the case of the normal distribution it holds that the moment generating function (mgf) is given by $$M(h) = \exp(\mu h + \frac12 \sigma^2 h^2),$$ where $\mu$ is the mean and $\sigma^2$ is the variance. Thus the cumulant generating function $C(h)$ which is given by $C(h) = \ln (M(h))$ reduces to $$C(h) = \mu h + \frac12 \sigma^2 h^2.$$ I am sure you can evaluate this in Matlab.

In the case of log-normal the mgf is not defined.

-


text:
Notice: The contact form functionality has been restored. Apologies for any inconvenience.

### Clone hba-107d15

 Clone and library
Clone
Clone name
hba-107d15
LE_HBa0107D15
C09HBa0107D15
P107D15
Clone type
BAC
Contamination
none
Estimated length
132,154
Library
Name
Tomato HindIII BAC Library
Cloning host
E. coli
Organism
Solanum lycopersicum (formerly Lycopersicon esculentum)
Cloning vector
Accession
Heinz 1706
Restriction enzyme
HindIII
Ordering Information
BAC Clones
BAC clones can be ordered from the clone ordering page at TFGD
 Physical mapping
 Fingerprint Contig Builds (FPC)
Tomato FPC (AGI 2005)
ctg221 : [coherent], 11 members:
C12HBa0004J02, C12HBa0008G21, C12HBa0046F10, C12HBa0063E16, C09HBa0107D15, LE_HBa0120L04, C12HBa0121A09, LE_HBa0171J11, LE_HBa0190J18, LE_HBa0298O16, LE_HBa0321C23
 Marker matches - overgo what's this?
Conflicted matches
None

 Marker matches - BLAST None
 Marker matches - manual None
 Physical map (from marker matches)
 FISH map
 FISH images
(percentage of arm length)
138-04 1CompositeFluorescencePhase32.1%
138-05 1CompositeFluorescencePhase27.4%
138-07b 1CompositeFluorescencePhase32.5%
138-08 1CompositeFluorescencePhase28.9%
157-08a 1CompositeFluorescencePhase33.8%
157-08b 1CompositeFluorescencePhase28.2%
157-09 1CompositeFluorescencePhase36.1%
138-12 1CompositeFluorescencePhase30.1%
157-13 1CompositeFluorescencePhase28.1%
157-14 1CompositeFluorescencePhase36.3%
138-18 1CompositeFluorescencePhase30.1%
157-24 1CompositeFluorescencePhase30.1%
157-30 1CompositeFluorescencePhase33.5%
Mean±sd31.3%±3.0%

Given an estimated length of 9.7µm for chromosome 9Q and the mean percentage above, this BAC is approximately 3.0µm±0.3µm from the chromosome centromere.

1Experiment conducted in the lab of Stephen Stack at Colorado State University.
 IL Mapping None
 Sequencing
This clone is being sequenced by the Chromosome 9 Sequencing Project. (View projects)
Log in as a curator or sequencer to edit this clone's registry information.
Sequencing Project
Chromosome 9
HTGS phase
1 - fragmented, unordered
Sequencing Status
complete
Sequencing Organization
not recorded
Full Sequence
End Sequences
Warnings
• sequence is not finished to HTGS3
FTP
Chromosome Assembly
Restriction Fragments
 HindIII in vitro (36) 1011, 1084, 1155, 1291, 1354, 1473, 1646, 1646, 1710, 1948, 2009, 2134, 2356, 2500, 2755, 2861, 2980, 3053, 3133, 3205, 3314, 3408, 3506, 3506, 3702, 4377, 4542, 4748, 5222, 5678, 6015, 6521, 6742, 8510, 9275, 11784 HindIII in silico (34) 1001, 1070, 1146, 1280, 1350, 1640, 1650, 1653, 1713, 1922, 2126, 2376, 2511, 2765, 2885, 2981, 3066, 3132, 3335, 3436, 3506, 3553, 3678, 3738, 4374, 4647, 5214, 5767, 6411, 6719, 7281, 8481, 11663, 12729 Fragment Lengths Match Score 0.96
 Annotations
Browse
LE_HBa0107D15 has no browsable sequence annotations


text:
# An integral involving ζ(2) (And Euler's first proof for the Basel Problem)

Posted on Wed 29 April 2020 in Mathematics

Evaluate the integral $$\int_0^1 \frac{\log(x)}{x-1}dx$$

There are a few methods of doing this: the first one uses the taylor series expansion of $$\log(1-x)$$: $$I = -\int_0^1 \frac{\log(1-x)}{x}dx$$ $$I = \int_0^1 \frac 1x \left( x + \frac{x^2}{2} + \frac{x^3}{3} + ... \right)$$

This expression nicely integrates to give $$I = 1 + \frac{1}{2^2} + \frac{1}{3^2} + ... = \boxed{\frac{\pi^2}{6}}$$

The last step here involves the summation $$\zeta(2) = \sum_{k=1}^{\infty} \frac 1 {k^2} = \frac{\pi^2}{6}$$. This summation is known as the Basel Problem (more on this later). For now, we'll use this as a fact and move forward.

Another solution that does not involve the taylor series requires some substitutions and limit evaluation.

$$\begin{gather} I = \int_0^1\frac{\log x}{x-1} dx= -\int_0^1\log x(1+x+x^2+...)dx \\ = -\int_0^1 \log x \sum_{r=0}^\infty x^r dx \\ = -\sum_{r=0}^\infty\int_0^1x^r \log x dx \end{gather}$$

let $$\log x = t \implies dx = e^t dt$$: $$\begin{gather}I = -\sum_{r=0}^\infty \int_{-\infty}^0te^{(r+1)t}dt \\ =-\sum_{r=0}^\infty \left[\frac{te^{(r+1)t}}{(r+1)} - \frac{e^{(r+1)t}}{(r+1)^2}\right]_{-\infty}^0\\ = \sum_{r=0}^\infty \left[\frac{1}{(r+1)^2} + \lim_{t \to -\infty} \frac{(t(r+1)-1)e^{(r+1)t}}{(r+1)^2} \right]\\ = \sum_{r=0}^\infty \left[\frac{1}{(r+1)^2} - \lim_{t \to \infty} \frac{(t(r+1)+1)}{e^{(r+1)t}(r+1)^2} \right] \\ = \sum_{r=0}^\infty \left[\frac{1}{(r+1)^2} - \lim_{t \to \infty} \frac{r+1}{e^{(r+1)t}(r+1)^3} \right] \\ = \sum_{r=0}^\infty \frac{1}{(r+1)^2}\\ = \sum_{k=1}^\infty \frac{1}{k^2}\\ \boxed{I = \frac{\pi^2}{6}} \end{gather}$$

This solution also involves the same summation. There is, however, a solution that does not involve the summation: The general form of this integral is known as Spence's Function or Dilogarithm. It's a transcendental function which has the values $$Li_2(0) = 1$$ and $$Li_2(1) = \zeta(2) = \frac{\pi^2}{6}$$. If you know these values beforehand, you can cut out the lengthy calculation.

Back to the Basel Problem, and a bit of a primer first. The Riemann-Zeta function is a mathematical function that is defined as $$\zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s}$$. Thus, a convenient way of referring to the sum $$\sum_{k=1}^{\infty} \frac{1}{k^2}$$ would be $$\zeta(2)$$, which I will use hereon. The quest for the value of $$\zeta(2)$$ far predates the Riemann-zeta function (and Bernhard Riemann himself for that matter). This problem was first posed by Pietro Mengoli, an Italian mathematician who had proved that the harmonic series $$\left( \sum_{n=1}^{\infty} \frac 1n \right)$$ diverges and also found the value of the sum $$\sum_{r=1}^\infty (-1)^{r-1} \frac 1r$$ to be $$\log(2)$$. The problem is called the Basel problem as Basel in Switzerland is the hometown of Euler, who solved the problem and the Bernoulli family, who unsuccessfully attacked the problem.

The story goes that the first proof Euler gave was not accepted as it involved an unproven factorization of $$\sin(x)$$. Euler being Euler, gave a second proof and a third proof as well, which were accepted. The initial proof is as follows: From the taylor series expansion of $$\sin(x)$$, we have $$\sin(x) = x + \frac{x^3}{3!} + \frac{x^5}{5!} + ...$$

$$\sin(x)$$ can also be written as a function of it's roots, according to the Weierstrass Factorization Theorem. Note that this theorem was published a 100 years after Euler's proof, which is why Euler's proof was initially not accepted. Euler factorized $$\sin(x)$$ as follows: $$\sin(x) = x(1 - \frac x\pi)(1 + \frac x\pi)(1 - \frac{x}{2\pi})(1+\frac{x}{2\pi})...$$ $$\sin(x) = x(1-\frac{x^2}{\pi^2})(1-\frac{x^2}{4\pi^2})... \tag{1}$$

Why did Euler choose to factorize $$\sin(x)$$ as such? He could also have chosen the function $$x(x-\pi)(x+\pi)(x-2\pi)(x+2\pi)...$$. A simple trick I used to explain this is that the function chosen by Euler is bounded and also converges to a common limit. Dividing both sides by $$x$$ and evaluating the limit $$lim_{x \to 0} \frac{\sin(x)}{x}$$ in equation $$(1)$$, we see that both sides converge to 1.

Back to the proof, since we have two equivalent representations of $$\sin(x)$$, all we have to do is compare the coefficients of $$x^3$$ in both of them to find the value of $$\zeta(2)$$ $$\frac{1}{3!} = \frac{1}{\pi^2} \sum_{k=1}^{\infty} \frac{1}{k^2}$$ $$\boxed{\sum_{k=1}^{\infty} \frac{1}{k^2} = \frac{\pi^2}{6}}$$

This completes the proof. A really beautiful and simple proof, this was the one of the two proofs (Along with Cauchy's proof) that I was able to understand. For more proofs involving higher order mathematics, you can check out the links here and here.


text:
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• CommentRowNumber1.
• CommentAuthorMike Shulman
• CommentTimeMar 24th 2012

Started a general page about beta reduction, and a stub for eta reduction but I have to run now.

• CommentRowNumber2.
• CommentAuthorMike Shulman
• CommentTimeMar 25th 2012

Now a bit more at eta reduction.

• CommentRowNumber3.
• CommentAuthorTobyBartels
• CommentTimeMar 25th 2012
• (edited Mar 25th 2012)

Since the version with symbols is definitely ‘$\beta$-reduction’, I’ve moved the name without symbols to beta-reduction (with a hyphen). This triggers the cache bug, so I mention it here. The same with eta-reduction … but can I try to convince you that $\eta$-reduction is wrong and $\eta$-expansion is correct? ETA: No, the cache bug did not appear!

• CommentRowNumber4.
• CommentAuthorTobyBartels
• CommentTimeMar 25th 2012

I put in a stub for alpha-equivalence.

• CommentRowNumber5.
• CommentAuthorMike Shulman
• CommentTimeMar 26th 2012

Thanks. I’m glad you brought up the reduction/expansion issue. What about calling the page eta-equivalence or eta-conversion? I know there are situations in which $\eta$-expansion seems to match better with $\beta$-reduction than $\eta$-reduction does (looking like the unit and multiplication of a monad or something, I don’t recall the details), but I don’t recall being convinced that that is a universal phenomenon. Is there a reason to believe that $\eta$-conversion has a well-defined “correct” directionality at all?

Also, by the “trivial type” do you mean the unit type? If so, I don’t understand your remark about it. Are you thinking of a positive or a negative definition of the trivial type?

• CommentRowNumber6.
• CommentAuthorTobyBartels
• CommentTimeMar 26th 2012

I like eta-conversion so moved the page. However, I understand that term a bit differently from you; see my edits.

Yes, I meant the unit type, defined negatively. I altered your material on the product type to produce material on the unit type, so you can see for yourself.

• CommentRowNumber7.
• CommentAuthorMike Shulman
• CommentTimeMar 26th 2012

Thanks; your meaning of $\eta$-conversion is better. And I see what you mean about the unit type. Interestingly, the positive definition of the unit type does have a well-defined $\eta$-reduction; I added some details to unit type.

Now I’m wondering about the empty type. Its positive presentation is “dual” to the negative unit type in that there are no constructors, and a single eliminator

$e:\varnothing ⊢{\mathrm{abort}}_{C}\left(e\right):C.$

So there can again be no $\beta$-reduction, but it seems like the $\eta$-conversion should be $e\phantom{\rule{thickmathspace}{0ex}}{↔}_{\eta }\phantom{\rule{thickmathspace}{0ex}}{\mathrm{abort}}_{\varnothing }\left(e\right)$, which makes sense both as an expansion and as a reduction. Am I wrong? If not, why the asymmetry, I wonder? Is it because we’ve already broken the symmetry by formulating everything with variables and terms instead of covariables and coterms?

• CommentRowNumber8.
• CommentAuthorMike Shulman
• CommentTimeMar 26th 2012

Ah, no, I’m wrong. The $\eta$-conversion for the positive empty type says that (in the context of a term $e:\varnothing$), any term $c:C$ of any type is convertible to $\mathrm{abort}\left(e\right)$. This also only makes sense as an expansion.

• CommentRowNumber9.
• CommentAuthorMike Shulman
• CommentTimeMar 26th 2012

I’ve updated empty type with this, and also corrected the $\eta$-rule at sum type in the corresponding way.

• CommentRowNumber10.
• CommentAuthorMike Shulman
• CommentTimeMar 27th 2012

I added some comments to eta-conversion about propositional $\eta$-conversions and how they are related to dependent eliminators. I also expanded function type and dependent product type to describe not only the usual negative presentations, but also the “higher-order” positive presentations and their relationship.

• CommentRowNumber11.
• CommentAuthorTobyBartels
• CommentTimeMar 30th 2012

I’ve never seen the positive approach to dependent product types seriously considered, although it’s implicit (in my opinion) in Martin-Löf’s earliest work. So I’m glad to see that people have worked it out (even though I don’t really like it for requiring a stronger metatheory).

• CommentRowNumber12.
• CommentAuthorMike Shulman
• CommentTimeMar 30th 2012

Yeah, the stronger metatheory is a shame. But as Richard’s paper shows, the resulting dependent eliminator sheds a useful light on the propositional eta conversion rule.


text:
## Abstract and Applied Analysis

### On Input-to-State Stability of Impulsive Stochastic Systems with Time Delays

#### Abstract

This paper is concerned with pth moment input-to-state stability (p-ISS) and stochastic input-to-state stability (SISS) of impulsive stochastic systems with time delays. Razumikhin-type theorems ensuring p-ISS/SISS are established for the mentioned systems with external input affecting both the continuous and the discrete dynamics. It is shown that when the impulse-free delayed stochastic dynamics are p-ISS/SISS but the impulses are destabilizing, the p-ISS/SISS property of the impulsive stochastic systems can be preserved if the length of the impulsive interval is large enough. In particular, if the impulse-free delayed stochastic dynamics are p-ISS/SISS and the discrete dynamics are marginally stable for the zero input, the impulsive stochastic system is p-ISS/SISS regardless of how often or how seldom the impulses occur. To overcome the difficulties caused by the coexistence of time delays, impulses, and stochastic effects, Razumikhin techniques and piecewise continuous Lyapunov functions as well as stochastic analysis techniques are involved together. An example is provided to illustrate the effectiveness and advantages of our results.

#### Article information

Source
Abstr. Appl. Anal., Volume 2014, Special Issue (2013), Article ID 589562, 10 pages.

Dates
First available in Project Euclid: 6 October 2014

https://projecteuclid.org/euclid.aaa/1412605759

Digital Object Identifier
doi:10.1155/2014/589562

Mathematical Reviews number (MathSciNet)
MR3193526

Zentralblatt MATH identifier
07022665

#### Citation

Yao, Fengqi; Cheng, Pei; Shen, Hao; Qiu, Li. On Input-to-State Stability of Impulsive Stochastic Systems with Time Delays. Abstr. Appl. Anal. 2014, Special Issue (2013), Article ID 589562, 10 pages. doi:10.1155/2014/589562. https://projecteuclid.org/euclid.aaa/1412605759

#### References

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• M. Lazar, W. P. M. H. Heemels, and A. R. Teel, “Further input-to-state stability subtleties for discrete-time systems,” IEEE Transactions on Automatic Control, vol. 58, pp. 1609–1613, 2013.
• J. Liu, X. Liu, and W.-C. Xie, “Input-to-state stability of impulsive and switching hybrid systems with time-delay,” Automatica, vol. 47, no. 5, pp. 899–908, 2011.
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• F. L. Sun, Z. H. Guan, X. H. Zhang, and J. C. Chen, “Exponential-weighted input-to-state stability of hybrid impulsive switched systems,” IET Control Theory and Applications, vol. 6, pp. 430–436, 2012.
• Y. E. Wang, R. Wang, and J. Zhao, “Input-to-state stability of non-linear impulsive and switched delay systems,” IET Control Theory and Applications, vol. 7, pp. 1179–1185, 2013.
• W.-H. Chen and W. X. Zheng, “Input-to-state stability for networked control systems via an improved impulsive system approach,” Automatica, vol. 47, no. 4, pp. 789–796, 2011.
• Z. Liu, R. E. Torres, N. Patel, and Q. Wang, “Further development of input-to-state stabilizing control for dynamic neural network systems,” IEEE Transactions on Systems, Man, and Cybernetics Part A: Systems and Humans, vol. 38, no. 6, pp. 1425–1433, 2008.
• E. N. Sanchez and J. P. Perez, “Input-to-state stabilization of dynamic neural networks,” IEEE Transactions on Systems, Man, and Cybernetics Part A: Systems and Humans, vol. 33, no. 4, pp. 532–536, 2003.
• S. Zhu and Y. Shen, “Two algebraic criteria for input-to-state stability of recurrent neural networks with time-varying delays,” Neural Computing and Applications, vol. 22, pp. 1163–1169, 2013.
• V. Lakshmikantham, D. D. Bainov, and P. S. Simenov, Theory of Impulsive Differential Equations, World Scientific, Singapore, 1989.
• J. P. Hespanha, D. Liberzon, and A. R. Teel, “Lyapunov conditions for input-to-state stability of impulsive systems,” Automatica, vol. 44, no. 11, pp. 2735–2744, 2008.
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• S. Dashkovskiy, M. Kosmykov, A. Mironchenko, and L. Naujok, “Stability of interconnected impulsive systems with and without time delays, using Lyapunov methods,” Nonlinear Analysis: Hybrid Systems, vol. 6, no. 3, pp. 899–915, 2012.
• X. Mao, Stochastic Differential Equations and Applications, Horwood, Chichester, UK, 2nd edition, 2007.
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• Z.-G. Wu, P. Shi, H. Su, and J. Chu, “Passivity analysis for discrete-time stochastic markovian jump neural networks with mixed time delays,” IEEE Transactions on Neural Networks, vol. 22, no. 10, pp. 1566–1575, 2011.
• H. Shen, S. Xu, J. Lu, and J. Zhou, “Passivity-based control for uncertain stochastic jumping systems with mode-dependent round-trip time delays,” Journal of the Franklin Institute, vol. 349, no. 5, pp. 1665–1680, 2012.
• S. Peng and Y. Zhang, “Razumikhin-type theorems on pth moment exponential stability of impulsive stochastic delay differential equations,” IEEE Transactions on Automatic Control, vol. 55, no. 8, pp. 1917–1922, 2010.
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• H. Khalil, Nonlinear Systems, Prentice Hall, New Jersey, NJ, USA, 3rd edition, Edition, 2002. \endinput


text:
## anonymous one year ago find centroid of the region found in first quadrant bounded by y=sqrt(1-x^2),y=0,x=0 does this just basically make it not balance so that I have to find xbar and ybar? using their formulas

1. dm91

this question is confusing

2. anonymous

are xbar and ybar anything other than 0?

3. dm91

what do you think it is?

4. anonymous

well I think none are going to be zero. I was looking at my notes and don't know how my teacher was able to determine xbar to equal zero just by looking at the problem.

5. ganeshie8

You need to find the coordinates of the center of mass : |dw:1442386634593:dw|

6. ganeshie8

$\overline{x} = \dfrac{1}{A} \iint xdA =\dfrac{1}{\pi/4} \int\limits_0^1 x\sqrt{1-x^2}\, dx = ?$ $\overline{y} = \dfrac{1}{A} \iint ydA = \dfrac{1}{\pi/4}\int\limits_0^1 \frac{1-x^2}{2}dx=?$

7. anonymous

I understand that, what I need help with is , how do I know if xbar or ybar will be 0 just by looking at it?

8. ganeshie8

neither of them could be 0, forget math. look at the region, where do you expect the center of mass to lie ?

9. ganeshie8

|dw:1442387249476:dw|

10. ganeshie8

somewhere around here : |dw:1442387453637:dw|


text:
## STP

https://doi.org/10.1351/goldbook.S06036
Abbreviation for standard temperature ($$273.15\ \text{K}$$ or $$0\ ^{\,\unicode{x26ac}}\text{C}$$) and pressure ($$10^{5}\ \text{Pa}$$); usually employed in reporting gas volumes. Note that flow meters calibrated in standard gas volumes per unit time often refer to volumes at $$25\ ^{\,\unicode{x26ac}}\text{C}$$, not $$0\ ^{\,\unicode{x26ac}}\text{C}$$.


text:
# Looking for star database

I am looking for a particular set of information to essentially make a star map for a game.

What I need:
All stars within 11,462 light years of Earth.
Their stellar classification.
If they have any planets around them (hypothetical or otherwise)
Their position using an x, y, z coordinate relative to earth or galaxy center (I don't really get declination, etc)

I need the extrapolated position/distance data, not the catalogs themselves.

The best one I can find is HYG, HabCat, and HYGHag is limited to to like 120,000 of 1,2 billion estimated stars... Are there any better?

• Are you sure that you want a volume-limited catalog ("all stars within X light years"), and not a flux-limited catalog ("all stars which appear brighter than X as seen from Earth") ? Google for the Hipparcos catalog to get 3D positions of brighter stars. – mtewes Sep 2 '16 at 9:19
• All is what I "want"... Cuz I like to keep as close to "reality" as possible, but minimally I need all stars within 100ly of Earth for anothe reason too. – Durakken Sep 2 '16 at 9:28
• This is going to be a lot of work on your part to compile something like this. All this information, in this format, is not going to be compiled easily. Likely you'll need to use several sources and do a lot of converting. A good place to get exoplanet data is exoplanet.eu where you can download all their data. However, it will require significant work to match up their star names with stars from other catalogues. Often a single star will have many different names, depending on what catalogue its in. – zephyr Sep 2 '16 at 12:53
• FYI, I'm not sure if you know, but the general naming format for exoplanets is STAR_NAME X where X is some lower-case letter starting at b for the closest planet to the star and proceeding alphabetically for every planet thereafter. An example would be HD 1160 b which is the closest planet around the star HD 1160 (the HD referencing the Henry Draper Catalogue). – zephyr Sep 2 '16 at 12:58
• @zephyr Yes I knew that. Though thanks ^.^ The exoplanet database looks nice – Durakken Sep 3 '16 at 4:33

Best chance would be the Hipparcos catalogue. The first set of Gaia data will be released Mid-September 2016, but I don't know if it will be more accurate than Hipparcos already.

All stars within 11,462 light years of Earth.

That won't be easy. These catalogues have a magnitude (brightness) cutoff. Brighter, more massive stars can be seen further away than smaller, darker ones.

Their stellar classification.

Spectral type should be in there.

If they have any planets around them (hypothetical or otherwise)

Searching for Exoplanets is an ongoing project, and there is no complete catalogue in any sense of the word yet for any radius. Most Exoplanets were found with the Kepler mission, but that one looked in a fairly small part of the sky. What exactly do you mean by hypothetical planets?

Their position using an x, y, z coordinate relative to earth or galaxy center (I don't really get declination, etc)

You'll have to convert that yourself. If you know how to program, there are libraries for coordinate conversion. I would argue, though, that an Earth-centred spherical coordinate system is preferential for most applications.

Distances are hard to measure. Hipparcos and Gaia use the parallax method. This gets less accurate with increasing distance. Also, distances are naturally measured in parsec, but the conversion to light years is trivial.

The Gaia catalogue will have $10^9$ stars with positions and velocities. It will also be more accurate than Hipparcos. But the measurements are ongoing, and the final data release is planned for 2022.

• Isn't there a compendium/compilation catalog? That while one catalog might not have it another does so it's all included in 1 catalog or no? What I mean with the planets is that I'm going to have to generate the majority of those anyways, but if there are possible planets there if it's noted I'd find it useful to have it there to see what it's saying a star has. As to the coordinate system... it doesn't hurt to ask ^.^ I hate that system, but yeah I'm going to have to convert anyways, I was just hoping not to have to figure that out too. – Durakken Sep 2 '16 at 9:34
• As far as lightyear vs parsec that's fine... I consider both bleh measures outside of average human practical use ^.^ I have my own unit that is equivalent to 24 Terameters so technically if they measured in terameters that would be ideal for me ^.^ ---- Also is there a rough guess at how many stars are in that volume? I need to know that to know how many I need to generate, even with a catalog. – Durakken Sep 2 '16 at 9:37
• @Durakken: Well, length conversion is simple, and Parsec is what the actual measurement is, so that's what's used. As for a compendium catalogue of stars and planets, to my knowledge there isn't one. When the Hipparcos catalogue was made, there were no / very few exoplanets. Also, it's mostly used by people interested in stellar evolution and galactic dynamics, and these people don't need planets. Exoplanet researchers are better off with a specialized catalogue that doesn't consist of 99.x% irrelevant entries. Best chance is to find an exoplanet catalogue and match positions with Hipparcos. – Alex Sep 2 '16 at 9:53
• To find sunlike stars at that distance, abs mag 4.83, your catalog will need a cutoff of at least magnitude 17.6. A decent calculator: calctool.org/CALC/phys/astronomy/star_magnitude You might be able to find suns that far out in a catalog, if not hidden behind nebula, but you're going to have to just make up the distant K and M type star locations. – Wayfaring Stranger Sep 2 '16 at 13:28
• Well dl the HIP and Tycho Catalog... They're useless to me. i need the position data calculated out, not the arc seconds and such cuz I don't understand the math to get the position out of that... – Durakken Sep 3 '16 at 4:13


text:
# Estimating mutual information of sine time series using 2D kernel density estimators

This is an attempt to reproduce Moon et al. 1995, and the author's copy can be obtained through here.

As a benchmark, we estimate the time-lagged mutual information of a simple sine signal $\sin(0.02\pi t)$ using Gaussian kernels: $$K(y)=\dfrac{\exp\left(-\dfrac{(y-y_i)^TS^{-1}(y-y_i)}{2h^2}\right)}{(2\pi)^{d/2}h^d\det(S)^{1/2}},$$ where $h$ is chosen to be $$h=\left\{\dfrac{4}{(d+2)}\right\}^{1/d+4}n^{-1/(d+4)},$$ where $d$ is the dimension of the multivariate kernel, and $n$ is the number of data points.

I wrote a Matlab function (the code is at the end of the post).

Suppose $x_t$ is the original time series $\sin(0.02\pi t)$ with 400 data points $t=1,2,\cdots,400$, and $x_{t-\tau}$ is $\sin[0.02\pi (t-\tau)]$. The author plotted the estimated mutual information $I_{x_t,x_{t-\tau}}$ as a function of $\tau$:

However, what I got from my code is

While the qualitative features (like the diverging mutual information at lag 0, 50 and 100), but the magnitude is far off and the overall shape is not right.

My 1st suspicion is the fact that I used different window widths for joint pdf and marginal pdf, and they may not be consistent. So my question is: how to consistently construct joint and marginal pdf estimate in the context of mutual information estimation?

Second question is: is there anything wrong in the code besides the possibly inconsistent pdf KDEs?

Edit 1:

The reason why I choose KDE over other methods is that I think KDE is simple and intuitive in concept and possibly also simple in implementation. Besides MI, I also want to estimate transfer entropy and possibly other information-theoretic measures as well. I feel that estimating joint probability is general enough to apply on various information-theoretic measures. Please let me know if you have comment on my argument above.

I didn't use anything that can be found only in Matlab. Code in R, Python, etc. is welcome.

Edit 2: the Matlab code

The Matlab code:

function [MI] = get_MI(xt, xt_lag)
% xt is original time series, xt_lag is the lagged one, both are column
% vectors
x_pair = [xt' xt_lag']; n=length(xt);

d=1; h_d1=(4/(d+2))^(1/(d+4)) * n^(-1/(d+4));

d=2; h_d2=(4/(d+2))^(1/(d+4)) * n^(-1/(d+4));

MI = 0.;

Sxy_pair = cov(x_pair); invS_pair = Sxy_pair\eye(2); detS_pair = det(Sxy_pair);

Sxy_xt = cov(xt'); invS_xt = Sxy_xt\eye(1); detS_xt = det(Sxy_xt);

Sxy_lag = cov(xt_lag'); invS_lag = Sxy_lag\eye(1); detS_lag = det(Sxy_lag);

for ix=1:n
Psq = p_mkde(x_pair(ix,:)', x_pair', h_d2, invS_pair, detS_pair);
Ps = p_mkde(x_pair(ix,1), xt, h_d1, invS_xt, detS_xt);
Pq = p_mkde(x_pair(ix,2), xt_lag, h_d1, invS_lag, detS_lag);

MI = MI + (log2(Psq) - log2(Ps) - log2(Pq));
end

MI = MI/n;

end

function [pxy]=p_mkde(x,X,h,invS,detS)

s1=size(X);
d=s1(1);
N=s1(2);

pxy_sum=0;
for ix=1:N
p2=(x-X(:,ix))'*invS*(x-X(:,ix));
pxy_sum=pxy_sum+1/sqrt((2*pi)^d*detS)*exp(-p2/(2*h^2));
end
pxy=1/(N*h^d)*pxy_sum;
end

• It may be the case that the original picture was simplified because it's three in one. In particular, the sine has to be below 1, but it's not. – James Oct 1 '14 at 18:59
• A general comment: it's much easier to estimate MI than to estimate the joint density. If MI is all you need, you might want to consider directly estimating MI (or entropies. E.g. by using a nearest neighbor based estimator). – Memming Oct 1 '14 at 22:25
• @James, what did you mean by "simplified"? I don't understand your argument. – wdg Oct 2 '14 at 2:11
• @Memming, actually my purpose is to learn about KDE technique. Besides MI, I also need transfer entropy and possibly other information-theoretic measures so that I need an estimate method that is general enough like KDE. – wdg Oct 2 '14 at 2:13
• @wdg if you can estimate MI, KL-div and entropy, most of the other information-theoretic quantities including transfer entropy can be easily estimated. I recommend you NOT to estimate the joint density unless you have a low (<5) dimensional problem and A LOT of data. – Memming Oct 3 '14 at 15:20

Let me start by restating that this is a terrible MI estimation procedure. Please don't use this in any serious business, unless you are absolutely sure you know what you are doing.

You are making two critical mistakes in your code:

1. Your scale doesn't match because you didn't do the integration properly. Mutual information is defined as in the paper is for probability, not probability density. But your kernel density estimator gives you back probability density which integrates to one, but does not sum to one on the points on a grid. If you want to use the sum formula as in the paper, you need to convert the density back to probability.

2. You are adding the probabilities only at the points where your data live, and counting multiple times on those regions. You should be integrating over the support of the entire distribution! An easy way to do this would be to make a uniform grid to integrate over. I think you should carefully review the definition of mutual information. (You need to add the probabilities, not the point wise MI estimates over the data.)

This is not a code-review site, but a couple of small things on your code.

• never use reserved matlab function names as variables: sum is one of them
• You don't need inv, so don't use it. Use \ and / instead
• try not to use i as a looping variable. i is for imaginary numbers in MATLAB (so is j)
• hint: MATLAB has a KDE for 1D which you can verify your code against: ksdensity

EDIT: I misinterpreted the code. I was wrong. OP is replacing expectation with sum over samples, so there's no problem using density nor having a grid.

• Thanks for the comments on the code. Regarding your point 1.: I understand your point. How can I convert the density back to probability in the context of kernel density estimate? – wdg Oct 5 '14 at 11:15
• In response to your point 2., I add only the probabilities of the data point, because I'm using resubstitution estimate: $H_n=-\frac{1}{n}\sum_{i=1}^n\log f_n(X_i)$. If the underlying process of the time series is ergodic, then time average is as good as ensemble average. – wdg Oct 5 '14 at 12:23
• @wdg Aha. Sorry for not understanding your code. If you are replacing expectation by sum over samples, then it's hard to correct for point 1. But, the units in the denominator should cancel the units in the numerator, so I actually think I was totally wrong now. :'( – Memming Oct 6 '14 at 13:34
• Why is this a terrible MI estimation procedure? – user76284 Nov 12 '17 at 18:00


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# Altered images

In a recent post I described the following problem which I encountered while sitting in a dentist waiting room:

Images are randomly selected from a library of images and shown on a screen. After watching the screen for a while, I notice one or more of the images is a repeat showing of an earlier image. How can I use information on the number of images observed and the number of repeats to estimate how many images there are in the entire library?

I had two great replies suggesting solutions to this problem. The first was from Nity.Raj@smartodds.co.uk

Surely the efficient thing to do is to hack the database of images so you just find out how many there are in fact, rather than estimating?

It’s the perfect answer, but I just need to run it past someone with a legal background who’s connected to Smartodds to check it’s compliant with relevant internet communication laws. Can anyone suggest somebody suitable?

The other idea was from Ian.Rutherford@smartbapps.co.uk who suggested this:

I would take the total of all the images seen and divide it by the number of times I spotted the 23 to Leigh Park to give an estimation of the number of different images

You’ll have to read the original post to understand the ’23 to Leigh Park’ bit of this answer, but you can take it as a reference to any one of the images that you’ve seen. So, let’s suppose I’ve seen 100 images, and I’ve seen one particular image that I’m interested in 4 times. Then Ian’s suggestion is to estimate the total number of images as

$100/4=25$

Ian didn’t explain his answer, so I hope I’m not doing him a disservice, but I think the reasoning for this solution is as follows. Suppose the population size is N and I observe v images. Then since the images occur at random, the probability I will see any particular image when a random image is shown is 1/N. So the average, or expected, number of times I will see a particular image in a sequence of v images is v/N. If I end up seeing the image t times, this means I should estimate v/N with t. But rearranging this, it means I estimate N with v/t.

It’s a really smart answer, but I think there are two slight drawbacks.

1. Suppose, in the sequence of 100 images, I’d already seen 26 (or more) different images. In that case I’d know the estimate of 25 was bound to be an under-estimate.
2. This estimate uses information based on the number of repeats of just one image. Clearly, the number of repeats of each of the different images I observe is equally relevant, and it must be wasteful not to use the information they contain as well.

That said, the simplicity and logic of the answer are both extremely appealing.

But before receiving these answers, and actually while waiting at the dentist, I had my own idea. I’m not sure it’s better than Nity’s or Ian’s, and it has its own drawbacks. But it tells a nice story of how methods from one area of Statistics can be relevant for something apparently unrelated.

So, imagine you’re an ecologist and there’s concern that pollution levels have led to a reduction in the number of fish in a lake. To assess this possibility you need to get an estimate of how many fish there are in the lake.  The lake is large and deep, so surface observations are not useful. And you don’t have equipment to make sub-surface measurements.

What are you going to do?

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One standard statistical approach to this problem is a technique called mark and recapture. There are many variations on this method, some quite sophisticated, but we’ll discuss just the simplest, which works as follows.

A number of fish are caught (unharmed), marked and released back into the lake. Let this number of fish be n, say.

Some time later, a second sample of fish – let’s say of size K – is taken from the lake. We observe that k fish of this second sample have the mark that we applied in the first sample. So k/K is the proportion of fish in the second sample that have been marked. But since this is just a random sample from the lake, we’d expect this proportion to be similar to the proportion of marked fish in the entire lake, which will be n/N.

Expressing this mathematically, we have an approximation

$k/K \approx n/N$

But we can rearrange this to get:

$N \approx nK/k$

In other words, we could use

$\hat{N}= nK/k$

as an estimate for the number of fish, since we’d expect this to be a reasonable approximation to the actual number N.

So, let’s suppose I originally caught, marked and released 100 fish. I subsequently catch a further 50 fish, of which 5 are marked. Then, n=100, K=50, k=5 and so

$\hat{N} = nK/k = 100 \times 50 /5 =1000$

and I’d estimate that the lake contains 1000 fish.

Now, maybe you can see where this is going. Suppose instead of a lake of fish, we have a library of images. This method would allow me to estimate the size of the population of images, just as it does a population of fish. But there’s a slight catch (if you’ll pardon the pun). When I take a sample of fish from a lake, each of the fish in the sample is unique. But when I look at a selection of images at the dentist, some of them may be repeats. So I can’t quite treat my sample of images in exactly the same way as I would a sample of fish. To get round this problem I have to ignore the repeated images within each sample. So, my strategy is this:

1. Observe a number of the images, ignoring any repeats. Call the number of unique images n.
2. Observe a second set of images. Let the number of unique images in this set be K, but keeping count of repeats with the first set. Let’s say the number of repeats with the first sample is k.

The estimate of the population size – for the same reasons as estimating fish population sizes – is then

$\hat{N} = nK/k$.

So, suppose I chose to look at images for 10 minutes. In that period there were 85 images, but 5 of these were repeats. So, n=80. I then watch for another 5 minutes and observe 30 unique images, 4 of which were also observed in the first sample. So, n=80, K=30, m=4 and my estimate of the number of images in the database is

$\hat{N} = nK/k = 80 \times 30 /4 =600$

Is this answer any better than Ian’s? I believe it uses more information available in the data, since it doesn’t focus on just one image. It’s also less likely to give an answer that is inconsistent with the data that I’ve already seen. But it does have drawbacks and limitations:

1. Ignoring the information on repeats within each sample must also be wasteful of relevant information.
2. The distinction between the first sample and second sample is arbitrary, and it might be that different choices lead to different answers.
3. Keeping track of repeats within and across the two samples might be difficult in practice.

In a subsequent post I’ll do a more detailed study of the performance of the two methods. In the meantime, let me summarise what I think are the main points from this discussion:

1. Statistical problems can occur in the most surprising places
2. There’s usually no right or wrong way of tackling a statistical problem. One approach might be best from one point of view, while another is better from a different point of view.
3. Statistics is a very connected subject: a technique that has been developed for one type of problem may be transferable to a completely different type of problem.
4. Simple answers are not always be the best – though sometimes they are – but simplicity is a virtue in itself.

Having said all that, there are various conventional ways of judging the performance of a statistical procedure, and I’ll use some of these to compare my solution with Iain’s in the follow-up post. Meantime, I’d still be happy to receive alternative solutions to the problem, whose performance I can also compare against mine and Ian’s.

## 2 thoughts on “Altered images”

1. Ian Rutherford says:

Stu your solution is much better from a statistical point of view and Nity’s preferable in a lawless society!

1. Stuart Coles says:

Thanks Ian. I honestly don’t know how our solutions would actually compare in practice – I definitely wouldn’t be confident enough to say that mine is ‘much better’, but, I’ll do some simulations to try to find out. Yeah, Nity though 🙂


text:
formula | res |

is formula.

(label) (l).

# formula

## English

### Noun

(wikipedia formula)
• (mathematics) Any mathematical rule expressed symbolically.
• $x = \frac \left\{-b \pm \sqrt\left\{b^2 - 4ac\right\}\right\}\left\{2a\right\}$ is a formula for finding the roots of the quadratic equation ax2 + bx + c = 0 .
• (chemistry) A symbolic expression of the structure of a compound.
• H2O is the formula for water (Dihydrogen monoxide)
• a plan or method for dealing with a problem or for achieving a result
• The company's winning formula includes excellent service and quality products.
• A formulation; a prescription; a mixture or solution made in a prescribed manner; the identity and quantities of ingredients of such a mixture.
• The formula of the rocket fuel has not been revealed.
• Drink given to babies to substitute for mother's milk.
• (logic) A syntactic expression of a proposition, built up from quantifiers, logical connectives, variables, relation and operation symbols, and, depending on the type of logic, possibly other operators such as modal, temporal, deontic or epistemic ones.
• #### Synonyms

* (in mathematics) mathematical formula * (in chemistry) chemical formula

#### Hyponyms

* (logic) sentence

#### Derived terms

* (l) * formulate * formulation * Formula One * formula racing * multiformula

* (l)