text:
# SAMPLING EXPANSION OF BANDLIMITED FUNCTIONS OF POLYNOMIAL GROWTH ON THE REAL LINE

Shin, Chang Eon

• Published : 2014.04.30
• 34 9

#### Abstract

For a bandlimited function with polynomial growth on the real line, we derive a nonuniform sampling expansion using a special bandlimited function which has polynomial decay on the real line. The series converges uniformly on any compact subsets of the real line.

#### Keywords

Shannon sampling;nonuniform sampling;bandlimited function;polynomial growth

#### References

1. M. A. Al-Gwaiz, Theory of Distributions, Marcel Dekker Inc., 1992.
2. L. L. Campbell, Sampling theorem for the Fourier transform of a distribution with bounded support, SIAM J. Appl. Math. 16 (1968), 626-636. https://doi.org/10.1137/0116051
3. J. R. Higgins, A sampling theorem for irregularly spaced sample points, IEEE Trans. Inform. Theory IT-22 (1976), no. 5, 621-622.
4. J. R. Higgins, Sampling Theory in Fourier and Function Analysis: Foundations, Oxford Science, 1996.
5. G. Hinsen, Irregular sampling of bandlimited $L^p$-functions, J. Approx. Theory 72 (1993), no. 3, 346-364. https://doi.org/10.1006/jath.1993.1027
6. M. I. Kadec, The exact value of the Paley-Wiener constant, Soviet Math. Dokl. 5 (1964) 559-561.
7. A. J. Lee, Characterization of bandlimited functions and processes, Inform. Control 31 (1976), no. 3, 258-271. https://doi.org/10.1016/S0019-9958(76)90409-5
8. N. Levinson, Gap and Density Theorems, American Mathematical Society Colloquium Publications, v. 26. American Mathematical Society, New York, 1940.
9. R. Paley and N. Wiener, Fourier Transforms in the Complex Domain, (Amer. Math. Soc. Colloq. Pub. Ser.), Vol. 19, Privindence, RI: Amer. Math. Soc. 1934.
10. E. Pfaffelhuber, Sampling series for band-limited generalized functions, IEEE Trans. Inform. Theory IT-17 (1971), 650-654.
11. C. E. Shin, M. B. Lee, and K. S. Rim, Nonuniform sampling of bandlimited functions, IEEE Trans. Inform. Theory 54 (2008), no. 8, 3814-3819. https://doi.org/10.1109/TIT.2008.926458
12. C. E. Shin, K. S. Rim, and Y. Kim, A weighted OFDM signal scheme for peak-to-average ratio reduction of OFDM signals, IEEE Trans. Veh. Technol. 62 (2013), 1406-1409. https://doi.org/10.1109/TVT.2012.2232949
13. G. G. Walter, Sampling bandlimited functions of polynomial growth, SIAM J. Math. Anal. 19 (1988), no. 5, 1198-1201. https://doi.org/10.1137/0519084
14. G. G. Walter, Nonuniform sampling of bandlimited functions of polynomial growth, SIAM J. Math. Anal. 23 (1992), no. 4, 995-1003. https://doi.org/10.1137/0523056
15. M. Zakai, Band-limited functions and the sampling theorem, Inform. Control 8 (1965), 143-158. https://doi.org/10.1016/S0019-9958(65)90038-0
16. A. I. Zayed, Advances in Shannon's Sampling Theory, CRC Press, Boca Raton, FL, 1993.
17. A. I. Zayed and A. G. Garcia, Nonuniform sampling of bandlimited signals with polynomial growth on the real axis, IEEE Trans. Inform. Theory 43 (1997), no. 5, 1717-1721. https://doi.org/10.1109/18.623180

#### Acknowledgement

Supported by : National Research Foundation of Korea(NRF)

format: plain


text:
# The Unapologetic Mathematician

## Cardinals and Ordinals as Categories

We can import all of what we’ve said about cardinal numbers and ordinal numbers into categories.

For cardinals, it’s actually not that interesting. Remember that every cardinal number is just an equivalence class of sets under bijection. So, given a cardinal number we can choose a representative set $S$ and turn it into a category $\mathcal{S}$. We let ${\rm Ob}(\mathcal{S})=S$ and just give every object its identity morphism — there are no other morphisms at all in this category. We call this sort of thing a “discrete category”.

For ordinal numbers it gets a little more interesting. First, remember that a preorder is a category with objects the elements of the preorder, one morphism from $a$ to $b$ if $a\leq b$, and no morphisms otherwise.

It turns out that (small) categories like this are exactly the preorders. Let $\mathcal{P}$ be a small category with a set of objects $P$ and no more than one morphism between any two objects. We define a preorder relation $\leq$ by saying $a\leq b$ if $hom_\mathcal{P}(a,b)$ is nonempty. The existence of identities shows that $a\leq a$, and the composition shows that if $a\leq b$ and $b\leq c$ then $a\leq c$, so this really is a preorder.

If now we require that no morphism (other than the identity) has an inverse, we get a partial order. Indeed, if $a\leq b$ and $b\leq a$ then we have arrows $f:a\rightarrow b$ and $g:b\rightarrow a$. We compose these to get $g\circ f:a\rightarrow a$ and $f\circ g:b\rightarrow b$. These must be the respective identities because there’s only the one arrow from any object to itself. But we’re also requiring that no non-identity morphism be invertible, so $a$ has to be the same object as $b$.

Now we add to this that for every distinct pair of objects either $\hom_\mathcal{P}(a,b)$ or $\hom_\mathcal{P}(b,a)$ is nonempty. They can’t both be — that would lead to an inverse — but we insist that one of them is. Here we have total orders, where either $a\leq b$ or $b\leq a$.

Finally, a technical condition I haven’t completely defined (but I will soon): we require that every nonempty “full subcategory” of $\mathcal{P}$ have an “initial object”. This is the translation of the “well-ordered” criterion into the language of categories. Categories satisfying all of these conditions are the same as well-ordered sets.

Okay, that got a bit out there. We can turn back from the theory and actually get our hands on some finite ordinals. Remember that every finite ordinal number is determined up to isomorphism by its cardinality, so we just need to give definititions suitable for each natural number.

We define the category $\mathbf{0}$ to have no objects and no morphisms. Easy.

We define the category $\mathbf{1}$ to have a single object and its identity morphism. If we take the identity morphisms as given we can draw it like this: $\bullet$.

We define the category $\mathbf{2}$ to have two objects and their identity morphisms as well as an arrow from one object to the other (but not another arrow back the other way). Again taking identities as given we get $\bullet\rightarrow\bullet$.

Things start to get interesting at $\mathbf{3}$. This category has three objects and three (non-identity) morphisms. We can draw it like this:
$\begin{matrix}\bullet&\rightarrow&\bullet\\&\searrow&\downarrow\\&&\bullet\end{matrix}$
The composite of the horizontal and vertical arrows is the diagonal arrow.

Try now for yourself drawing out the next few ordinal numbers as categories. Most people get by without thinking of them too much, but it’s helpful to have them at hand for certain purposes. In fact, we’ve already used $\mathbf{2}$!

Now to be sure that this does eventually cover all natural numbers. We can take our ${0}$ to be the category $\mathbf{0}$. Then for a successor “function” we can take any category and add a new object, its identity morphism, and a morphism from every old object to the new one. With a little thought this is the translation into categories of the proof that well-ordered finite sets give a model of the natural numbers, and the same proof holds true here.

May 24, 2007 Posted by | Category theory | 6 Comments


text:
Symbol
Problem
Use a suitable identity to get each of the following products. $a\right)$ $\left(p-11\right)\left(p+11\right)$ b) $\left(2y+5\right)$ $\left(2y-5\right)$ $c\right)$ $\left(12a-9\right)\left(12a+9\right)$ d) $\left(2a-1/2\right)\left(2a-1/2\right)$ $e\right)$ $\left(2.5x-0.5y\right)\left(2.5x-0.5y\right)$ $1\right)$ $\left(a^{2}+b^{2}\right)\left(-a^{2}+b^{2}\right)$ h) $\right)$ $\left(6x+5y\right)^{2}$ $g\right)$ $\left(6x-7\right)\left(6x+7\right)$


text:
Arch uses the pacman package manager and the package format is tar.zst. Arch provides a number of tools for creating tar.zst packages, first you need to install the base-devel package and the devtools package.

 1  pacman -S base-devel devtools

Arch’s packaging process works like this, starting with writing a PKGBUILD file that describes all the information needed to build a package, such as where to download the source code from, what the dependencies are, what the build version is, how to build, etc.

## PKGBUILD

A basic PKGBUILD format is as follows.

  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32  # Maintainer: justforlxz pkgname=dtkcore-git pkgver=5.4.0.r1.ge7e7a99 pkgrel=1 pkgdesc='DTK core modules' arch=('x86_64') url="https://github.com/linuxdeepin/dtkcore" license=('LGPL3') depends=('dconf' 'deepin-desktop-base-git' 'python' 'gsettings-qt' 'lshw') makedepends=('git' 'qt5-tools' 'gtest') conflicts=('dtkcore') provides=('dtkcore') groups=('deepin-git') source=("$pkgname::git://github.com/linuxdeepin/dtkcore.git") sha512sums=('SKIP') pkgver() { cd$pkgname git describe --long --tags | sed 's/$$[^-]*-g$$/r\1/;s/-/./g' } build() { cd $pkgname qmake-qt5 PREFIX=/usr DTK_VERSION=$pkgver LIB_INSTALL_DIR=/usr/lib make } package() { cd $pkgname make INSTALL_ROOT="$pkgdir" install }

Take dtkcore as an example, the whole configuration file is very clear and straightforward, at a glance you can see how the build tool makes a package step by step.

Save the configuration file to a directory named after PKGBUILD and then execute makepkg.

makepkg is the tool used to execute the PKGBUILD file and it builds the package according to the description in the file.

## How to submit a package to aur

When you want your package to be used by others, you usually request to upload it to the official repository, but the process of entering the official repository is exceptionally tedious and can be demanding for contributors. arch provides a user repository for users to freely share configuration files, and all you need to do is download the configuration file and package it locally.

According to aur’s contribution guidelines, we can easily upload our own configuration files and borrow aur tools like yay to download and build packages automatically.

Users can share PKGBUILD through the Arch User Repository. AUR does not contain any binary packages, only PKGBUILD uploaded by users for other users to download and use. All packages are unofficial and are used at your own risk.

Before submitting aur, it is necessary to understand some requirements of aur. Rules of submission provides some rules to upload and maintain our own packages as long as we follow them.

First you need to go to the aur website and register an account and upload your ssh key We do this by cloning a new repository as a way to upload.

 1  git clone ssh://aur@aur.archlinux.org/.git

Take dtkcore for example, change the pkgname to dtkcore when cloning.

 1  git clone ssh://aur@aur.archlinux.org/dtkcore.git

If it is the first time you create it, you will be prompted for an empty repository that you cloned.

Copy the PKGBUILD file to the cloned directory, then run makepkg --printsrcinfo > .SRCINFO to create a package message, add both the .SRCINFO and PKGBUILD files to git, then commit the commit message and push it to the server.

Note: If you forget to include .SRCINFO in your initial commit, you can use rebasing with -root or filtering the tree to make the AUR accept your first push

Tip: To keep your working directory and commits as clean as possible, you can create a gitignore file to exclude all files, and then add files as needed.

Reference.

https://wiki.archlinux.org/index.php/PKGBUILD


text:
# Circus

Geometry Level 5

The Red and Blue circles share the same center.

The radius of Red circle equals $\cfrac{1}{1 + \cfrac{1}{2 + \cfrac{1}{1 + \cfrac{1}{2 + \cfrac{1}{1 + \ddots}}}}}$ and the radius of Blue circle equals $3 - \frac{1}{4} + \frac{1\cdot3}{4\cdot6} - \frac{1\cdot3\cdot5}{4\cdot6\cdot8} + \cdots$ If the hatched area equals $\bigg(\frac{L}{M\sqrt{N}}-1\bigg)\pi+M$ where $$L,M,N$$ are positive co-prime integers with $$N$$ being square-free. Find $$L+M+N$$.

×


text:
# The Azimuth Project Blog - network theory (part 19) (Rev #12, changes)

Showing changes from revision #11 to #12: Added | Removed | Changed

This page is a blog article in progress, written by John Baez and Jacob Biamonte. To see discussions of this article while it was being written, visit the Azimuth Forum.

joint with Jacob Biamonte

It’s time to resume the network theory series! It’s been a long time, so we’ll assume you forgot everything we’ve said before, and make this post as self-contained as possible. Last time we started looking at a simple example: a diatomic gas.

Two atoms can recombine to form a diatomic molecule:

$A + A \to A_2$

and conversely, a diatomic molecule can break apart into two atoms:

$A_2 \to A + A$

We can draw both these reactions using a Petri net:

where we’re writing $B$ instead of $A_2$ to abstract away some detail that’s just distracting here. Or, equivalently, we can use a chemical reaction network:

Last time we looked at the rate equation for this chemical reaction network, and found equilibrium solutions of that equation. Now let’s look at the master equation, and find equilibrium solutions of that. This will serve as a review of three big theorems.

### The master equation

We’ll start from scratch, in case you’re just tuning in. The master equation is all about how atoms or molecules or rabbits or wolves or other things interact randomly and turn into other things. So, let’s write write $\psi_{m,n}$ for the probability that we have $m$ atoms of $A$ and $n$ molecule of $B$ in our container. These probabilities are functions of time, and master equation will say how they change.

First we need to pick a rate constant for each reaction. Let’s say the rate constant for this reaction is some number $\alpha > 0$:

$A + A \to B$

while the rate constant for this reaction is some number $\beta > 0$:

$B \to A + A$

Before we make it pretty using the ideas we’ve been explaining, the master equation says

$\displaystyle{ \frac{d}{d t} \psi_{m,n} (t)} = \alpha (m+2)(m+1)\psi_{m+2,n-1}(t) - \alpha m(m-1) \psi_{m,n}(t) + \beta (n+1) \psi_{m-2,n+1}(t) - \beta n \psi_{m,n}(t)$

where we define $\psi_{i,j}$ to be zero if either $i$ or $j$ is negative.

Yuck! Normally we don’t show you such nasty equations. Indeed the whole point of our work has been to show you that by packaging the equations in a better way, we can understand them using high-level concepts instead of mucking around with millions of scribbled symbols. But we thought we’d show you what’s secretly lying behind our beautiful abstract formalism, just once.

Each term has a meaning. For example, the first one:

$\alpha (m+2)(m+1)\psi_{m+2,n-1}(t)$

means that the reaction $A + A \to B$ will tend to increase the probability of there being $m$ atoms of $A$ and $n$ molecules of $B$ if we start with 2 more $A$‘s and 1 fewer $B$. This reaction can happen in $(m+2)(m+1)$ ways if we start with $m+2$ atoms of $A$. And it happens at a probabilistic rate proportional to the rate constant for this reaction, $\alpha$.

We won’t go through the rest of the terms. It’s a good exercise to do so, but there could easily be a typo in the formula, since it’s so long and messy. So let us know if you find one!

To simplify this mess, the key trick is to introduce a generating function that summarizes all the probabilities in a single power series:

$\Psi = \sum_{m,n \ge 0} \psi_{m,n} y^m \, z^n$

It’s a power series in two variables, $y$ and $z$, since we have two chemical species: $A$‘s and $B$’s.

Using this trick, the master equation looks like

$\frac{d}{d t} \Psi(t) = H \Psi(t)$

where the Hamiltonian $H$ is a sum of terms, one for each reaction. This Hamiltonian is built from operators that annihilate and create $A$‘s and $B$’s. The annihilation and creation operators for $A$ atoms are:

$\displaystyle{ a = \frac{\partial}{\partial y} , \qquad a^\dagger = y }$

The annihilation operator differentiates our power series with respect to the variable $y$. The creation operator multiplies it by that variable. Similarly, the annihilation and creation operators for $B$ molecules are:

$\displaystyle{ b = \frac{\partial}{\partial z} , \qquad b^\dagger = z }$

In Part 8 we explained a recipe that lets us stare at our chemical reaction network and write down this Hamiltonian:

$H = \alpha (b^\dagger a^2 - {a^\dagger}^2 a^2) + \beta ({a^\dagger}^2 b - b^\dagger b)$

As promised, there’s one term for each reaction. But each term is itself a sum of two: one that increases the probability that our container of chemicals will be in a new state, and another that decreases the probability that it’s in its original state. We get a total of four terms, which correspond to the four terms in our previous way of writing the master equation.

Puzzle: Puzzle 1. Show that this new way of writing the master equation is equivalent to the previous one.

### Equilibrium solutions

Now we will look for all equilibrium solutions of the master equation: in other words, solutions that don’t change with time. So, we’re trying to solve

$H \Psi = 0$

Given the rather complicated form of the Hamiltonian, this seems tough. The challenge looks more concrete but perhaps more scary if we go back to our original formulation. We’re looking for probabilities $\psi_{m,n}$, nonnegative numbers that sum to one, such that

$\alpha (m+2)(m+1)\psi_{m+2,n-1}- \alpha m(m-1) \psi_{m,n} + \beta (n+1) \psi_{m-2,n+1} - \beta n \psi_{m,n} = 0$

for all $m$ and $n$.

This equation looks rather unpleasant, horrid, but the good news is that it’slinear, so a linear combination of solutions is again a solution. This lets us simplify the problem using a conserved quantity.

Clearly, there’s a quantity that the reactions here don’t change:

What’s that? It’s the number of $A$‘s plus twice the number of $B$’s. After all, a $B$ can turn into two $A$’s, or vice versa.

(Of course the secret reason is that $B$ is a diatomic molecule made of two $A$‘s. But you’d be able to follow the logic here even if you didn’t know that, just by looking at the chemical reaction network… and sometimes this more abstract approach is handy! Indeed, the way chemists first discovered that certain molecules are made of certain atoms is by seeing which reactions were possible and which weren’t.)

Suppose we start in a situation where we know for sure that the number of $B$‘s plus twice the number of $A$’s equals some number $k$:

$\psi_{m,n} = 0 \; unless \; m+2n = k$

Then we know $\Psi$ is initially of the form

$\Psi = \sum_{m+2n = k} \psi_{m,n} y^m z^n$

But since the number of $A$‘s plus twice the number of $B$’s is conserved, if $\Psi$ obeys the master equation it will continue to be of this form!

Put a fancier way, we know that if a solution of the master equation starts in this subspace:

$L_k = \{ \Psi: \; \Psi = \sum_{m+2n = k} \psi_{m,n} y^m z^n \; for \; some \; \psi_{m,n} \}$

it will stay in this subspace. So, because the master equation is linear, we can take any solution $\Psi$ and write it as a linear combination of solutions $\Psi_k$, one in each subspace $L_k$ for $k = 0,1,2,\dots$.

In particular, we can do this for an equilibrium solution $\Psi$. And then all the solutions $\Psi_k$ are also equilibrium solutions—since solutions: they’re linearly independent, so if one of them changed with time,$\Psi$ would too.

This means we can just look for equilibrium solutions in the subspaces $L_k$. If we find these, we can get all equilibrium solutions by taking linear combinations.

Once we’ve noticed that, our horrid equation makes a bit more sense:

$\alpha (m+2)(m+1)\psi_{m+2,n-1}- \alpha m(m-1) \psi_{m,n} + \beta (n+1) \psi_{m-2,n+1} - \beta n \psi_{m,n} = 0$

Note that if the pair of subscripts $m, n$ obey $m + 2n = k$, the same is true for the other pairs of subscripts here. So our equation relates the values of $\psi_{m,n}$ for all choices of $m,n$ lying on this line segment:

$2m+n = k , \qquad m ,n \ge 0$

If you think about it a minute, you’ll see that if we know two of these values, we can keep using our equation to recursively work out all the rest. So, there are at most two linearly independent equilibrium solutions of the master equation in each subspace $L_k$.

Why at most two? Why not two? Well, we have to be a bit careful about what happens at the ends of the line segment: remember that $\psi_{m,n}$ is defined to be zero when $m$ or $n$ becomes negative. If we think very hard about this, we’ll see there’s just one linearly independent equilibrium solution of the master equation in each subspace $L_k$ . But this is the sort of nitty-gritty calculation that’s not fun to watch someone else do, so we won’t work bore through you it with here. that.

Instead, Soon we’ll move on to a more high-level approach to this problem, problem. where we throw some theorems at it. But first, one other remark. Our horrid equation

$\alpha (m+2)(m+1)\psi_{m+2,n-1}- \alpha m(m-1) \psi_{m,n} + \beta (n+1) \psi_{m-2,n+1} - \beta n \psi_{m,n} = 0$

resembles the usual discretized form of the equation

$\displaystyle {\frac{d^2 \psi}{d x^2} = 0 }$

namely:

$\psi_{n-1} - 2 \psi_{n} + \psi_{n+1} = 0$

And this makes sense, since we get

$\displaystyle {\frac{d^2 \psi}{d x^2} = 0 }$

by taking the heat equation:

$\displaystyle \frac{\partial \psi}{\partial t} = {\frac{\partial^2 \psi}{\partial x^2} }$

and assuming $\psi$ doesn’t depend on time. So what we’re doing is a lot like looking for equilibrium solutions of the heat equation.

This makes perfect sense, since the heat equation describes how heat smears out as little particles of heat randomly move around. True, there don’t really exist ‘little particles of heat’, but the heat equation also describes the diffusion of any other kind of particles as they randomly move around undergoing Brownian motion. Similarly, our master equation describes a random walk on this line segment:

$m+2n = k , \qquad m , n \ge 0$

or more precisely, the points on this segment with integer coordinates. The equilibrium solutions arise when the probabilities $\psi_{m,n}$ have diffused as much as possible.

If you think about it this way, it should seem be physically obvious that there’s justone linearly independent equilibrium solution of the master equation for each value of $k$.

There’s a general moral here, too, which we’re seeing in a special case: the master equation for a chemical reaction network really describes a bunch of random walks, one for each allowed value of the conserved quantities that happen can to be present. built as linear combinations of number operators. In our case we have just one such conserved quantity, but in general there will may be more. more These (or none). Further more, these ‘random walks’ are what we’ve been callingMarkov processes.

### Noether's theorem

We simplified our task of finding equilibrium solutions of the master equation by finding a conserved quantity. The idea of simplifying problems using conserved quantities is fundamental to physics: this is why physicists are so enamored with quantities like energy, momentum, angular momentum and so on.

Nowadays physicists often use ‘Noether’s theorem’ to get conserved quantities from symmetries. There’s a very simple version of Noether’s theorem for quantum mechanics, but in Part 11 we saw a version for stochastic mechanics, and it’s that version that is relevant now. Here’s a paper which explains it in detail:

• John Baez and Brendan Fong, Noether’s theorem for Markov processes.

We don’t really need Noether’s theorem now, since we found the conserved quantity and exploited it without even noticing the symmetry. Nonetheless it’s interesting to see how it relates to what we’re doing.

For the reaction we’re looking at now, the idea is that the subspaces $L_k$ are eigenspaces of an operator that commutes with the Hamiltonian $H$. It follows from standard math that a solution of the master equation that starts in one of these subspaces, stays in that subspace.

What is this operator? It’s built from ‘number operators’. The <b>number operator</a> fornumber operator for $A$‘s is

$N_A = a^\dagger a$

and the number operator for $B$‘s is

$N_B = b^\dagger b$

A little calculation shows

$N_A \,y^m z_2^n = m \, y^m z^n, \quad \qquad N_B\, y^m z^n = m \,y^m z^n$

so the eigenvalue of $N_A$ is the number of $A$‘s, while the eigenvalue of $N_B$ is the number of $B$’s. This is why they’re called number operators.

As a consequence, the eigenvalue of the operator $N_A + 2N_B$ is the number of $A$‘s plus twice the number of $B$’s:

$(N_A + 2N_B) \, y^m z^n = (m + 2n) \, y^m z^n$

Let’s call this operator $O$, since it’s so important:

$O = N_A + 2N_B$

If you think about it, the spaces $L_k$ we saw a minute ago are precisely the eigenspaces of this operator:

$L_k = \{ \Psi : \; O \Psi = k \Psi \}$

As we’ve seen, solutions of the master equation that start in one of these eigenspaces will stay there. This lets take some techniques that are very familiar in quantum mechanics, and apply them to this stochastic situation.

First of all, time evolution as described by the master equation is given by the operators $\exp(t H)$. In other words,

$\displaystyle{ \frac{d}{d t} \Psi(t) } = H \Psi(t) \; and \; \Psi(0) = \Phi \quad \Rightarrow \quad \Psi(t) = \exp(t H) \Phi$

Thus if $\Phi$ is an eigenvector of $O$, so is $\exp(t H) \Phi$, with the same eigenvalue. In other words,

$O \Phi = k \Phi$

implies

$O \exp(t H) \Phi = k \exp(t H) \Phi = \exp(t H) O \Phi$

But since we can choose a basis consisting of eigenvectors of $O$, we must have

$O \exp(t H) = \exp(t H) O$

or, throwing caution to the winds and differentiating:

$O H = H O$

So, as we’d expect from Noether’s theorem, our conserved quantity commutes with the Hamiltonian! This in turn implies that $H$ commutes with any polynomial in $O$, which in turn suggests that

$\exp(s O) H = H \exp(s O)$

and also

$\exp(s O) \exp(t H) = \exp(t H) \exp(s O)$

The last equation says that $O$ generates a 1-parameter family of ‘symmetries’ $\exp(s O)$: operators that commute with time evolution. But what do these symmetries actually do? Since

$O y^m z^n = (m + 2n) y^m z^n$

we have

$\exp(s O) y^m z^n = e^{s(m + 2n)}\, y^m z^n$

So, this symmetry takes any probability distribution $\psi_{m,n}$ and multiplies it by $e^{s(m + 2n)}$.

In other words, our symmetry multiplies the relative probability of finding our container of gas in a given state by a factor of $e^s$ for each $A$ atom, and by a factor of $e^{2s}$ for each $B$ molecule. It might not seem obvious that this operation commutes with time evolution! However, experts on chemical reaction theory are familiar with this fact.

Finally, a couple of technical points. Starting where we said ‘throwing caution to the winds’, our treatment has not been rigorous, since $O$ and $H$ are unbounded operators, and these must be handled with caution. Nonetheless, all the commutation relations we wrote down are true.

It’s also true that $\exp(s O)$ is unbounded for positive $s$. It’s bounded for negative $s$, but even then doesn’t map probability distributions to probability distributions. However, it does map any nonzero vector $\Psi$ with $\psi_{m,n} \ge 0$ to a vector $\exp(s O) \Psi$ with the same properties. So, we can just normalize this vector and get a probability distribution. This normalization is why we introduced the concept of relative probabilities.

## The Anderson-Craciun-Kurtz theorem

The Anderson-Craciun-Kurtz theorem

Here Now we we’ll will actually use find the solutions Anderson-Craciun-Kurtz theorem to work out the corresponding equilibrium states of the master equation. equation Brendon proved this in relation closed form. To understand this final section, you really do need to what remember we some consider things here we’ve back discussed in earlier. Part X.Last time we considered the same chemical reaction network we’re studying today, but we looked at its rate equation. Our notation for the rate constants was different then—sorry—but if we call those constants $\alpha$ and $\beta$ as we’re doing now, the rate equation looks like this:

Let

$\displaystyle{ \frac{d}{d t} x_1 = 2 \alpha x_2 - 2 \beta x_1^2}$
$\displaystyle{ \frac{d}{d t} x_2 = - \alpha x_2 + \beta x_1^2 }$
$\Psi := e^{z_1c_1 + z_2c_2}$

This describes how the number of $A$‘s and $B$’s changes in the limit where there are lots of them and we can treat them as varying continuously, in a deterministic way. The number $A$’s is $x_1$, and the number of $B$’s is $x_2$.

Then

We saw that the quantity

$H\Psi = \left[ r_1(z_1^2c_2 - z_2c_2) + r_2(z_2c_2^2 - z_1^2c_1^2)\right]\Psi$
$x_1 + 2 x_2$

and for $\Psi \neq 0$

is conserved, just as now we’re seeing that $N_A + 2 N_B$ is conserved. We saw that the rate equation has one equilibrium solution for each choice of $x_1 + 2 x_2$. And we saw that these equilibrium solutions obey

$(r_1 c_2 - r_2 c_1^2) z_1^2 + (r_2 c_1^2 - r_1c_2)z_2 = 0$
$\frac{x_1^2}{x_2} = \frac{\alpha}{\beta}$

which vanishes for

Now the Anderson-Craciun-Kurtz theorem, introduced in Part 9, is a powerful result that gets equilibrium solution of the master equation from equilibrium solutions of the rate equation. It only applies to equilibrium solutions that are ‘complex balanced’, but that’s okay:

$\frac{r_1}{r_2} = \frac{c_1^2}{c_2}$

Puzzle 2. Show that the equilibrium solutions of the rate equation for the chemical reaction network

## Noether’s theorem

Now we will show how Noether’s theorem relates the conserved quantity $2N_A + N_B$ to a symmetry.

are complex balanced.

### Conservation of particle number

So, given any equilibrium solution $(x_1,x_2)$ of our rate equation, we can hit it with the Anderson-Craciun-Kurtz theorem and get an equilibrium solution of the master equation! And it looks like this:

The Hamiltonians that arise in Petri net field theory have a very particular general form. Not every Hamiltonian in this vast class preserves particle number (since we can have exponential growth or decay for instance). What we want to do is to find a good way to characterize those Hamiltonians that do preserve particle number. We want to understand symmetries in general. Those of you following the posts will recall the commutation relations from Part X. These are going to be relevant here too.

$\displaystyle{ \Psi = e^{-(x_1 + x_2)} \, \sum_{m,n \ge 0} \frac{x_1^m x_2^n} {m! n! } \, y^m z^n }$

Just a reminder, the number operator for a single species is

In this solution, the probability distribution

$N_i = a_i^\dagger a_i$
$\psi_{m,n} = e^{-(x_1 + x_2)} \, \frac{x_1^m x_2^n} {m! n! }$

and the number operator for all the species is a sum over the single species

is a product of Poisson distributions. The factor in front is there to make the numbers $\psi_{m,n}$ add up to one. And remember, $x_1, x_2$ are any nonnegative numbers with $x_1^2/x_2 = \alpha/\beta$.

$N = \sum_i N_i$

So, using the Anderson-Craciun-Kurtz theorem, we get an explicit closed-form solution of the horrid equation

We will derive a few results at the end of the post. If you think we are telling the truth, you don’t need to check them, but they are there if you want to be bored with these sort of details. To get you into the mood…

$\alpha (m+2)(m+1)\psi_{m+2,n-1}- \alpha m(m-1) \psi_{m,n} + \beta (n+1) \psi_{m-2,n+1} - \beta n \psi_{m,n} = 0$

It can be shown (using induction) that

with very little work! This is why this theorem is so nice. And of course we’re looking at a very simple reaction network: for more complicated ones it becomes even better to use this theorem to avoid painful calculations.

$[a, a^\dagger^k] = k a^\dagger^{k-1}$
$[a^k,a^\dagger] = k a^{k-1}$

category: blog

• exercise. Let $[a, a^\dagger] = 1$ be the base case and assume $[a, a^\dagger^k] = k a^\dagger^{k-1}$ and show that these assumptions imply the formula for $k+1$ and hence or otherwise, prove the first commutation relation listed above by induction.

Now for some results.

• (Lemma I). We arrive at the following commutation relations among number operators and general Hamiltonians in Petri net field theory
(1)$[N_i, H] = \sum_{\tau\in T} r(\tau)[n_i(\tau) - m_i(\tau)]a^\dagger^{n(\tau)}a^{m(\tau)}$

This supporting lemma can be used to prove a range of things related to symmetries in the very type of Hamiltonians we are considering here.

• (Theorem I — total particle conservation). The following quantity
(2)$\sum_{\tau\in T} r(\tau)[n(\tau) - m(\tau)]a^\dagger^{n(\tau)}a^{m(\tau)} = 0$

vanishes identically iff $H$ preserves particle number.

So to check if the total number of particles are conserved during evolution under some Hamiltonian, all one has to do is check Theorem I. The Hamiltonian we consider here does not conserve total particle number. However, the reversible reaction John did in Part 10 did.

• (Exercise). In Network Theory Part 10 John considered the reversible reaction with Hamiltonian

(3)$H = (a^\dagger - b^\dagger)(\beta b - \alpha a)$

Use Theorem I to show that this reversible reaction conserves particle number.

• (Theorem II — particle conservation symmetry). Given a Hamiltonian acting on $k$ particle species, there exists $k$ positive integer choices for $\omega_n$ which causes the following to vanish identically

(4)$\sum_i\sum_{\tau\in T} r(\tau)\omega_i[n_i(\tau) - m_i(\tau)]a^\dagger^{n(\tau)}a^{m(\tau)} = 0$

iff $H$ has a particle conservation symmetry.

As will soon be seen, this is precisely the case here. In other words, there exists $\omega_1$ and $\omega_2$ that take positive integer values which case the above quantity to vanish.

For our system to have a particle conservation symmetry, we must show that

(5)$[N_1 + 2N_2, H] = [N_1, H] + 2[N_2,H] = 0$

This vanishes since, from Lemma I, we calculate that

(6)$[N_1, H] = 2r_1 a^\dagger a^\dagger b -2 r_2 b^\dagger a a$
(7)$[N_2, H] = -r_1 a^\dagger a^\dagger b + r_2 b^\dagger a a$

Here, Theorem II applies, are we are able to find two values, $\omega_1=1$ and $\omega_2=2$ that cause the commutator to vanish. The Hamiltonian therefore has a particle number conservation symmetry.

## Master equation

Now for the master equation approach. Our Hamiltonian is given as

$H = r_1 (a^\dagger a^\dagger b - b^\dagger b) + r_2 (b^\dagger a a - a^\dagger a^\dagger a a)$

and the evolution operator at time $t$ is given as

(8)$V = e^{t H} = \sum_{k=0}^\infinity \frac{H^k}{k!}t^k$

At each order in $k$, we have a term that corresponds to the Hamiltonian $H$ acting $k$ times. One can think of these as alternative histories. In quantum mechanics, there is a spooky thing called coherence, where each of these histories seems to occur concurrently. In stochastic mechanics, each history only occurs with some probability. In terms of mathematical structure, the theories become closely related. While the semantic interpretation might differ, the syntactical form, given by a sum over histories unites quantum and stochastic mechanics. This enables us to e.g. apply tools from quantum mechanics to stochastic mechanics such as Feynman diagrams.

# Supporting Material

Proof of Theorems I and II. Sometimes a long calculation can simplify matters. This is the case here, though we don’t want to muddy the water from what’s going on here, as this is just some algebra.

Here we will use the following notation.

$a^{m(\tau)}a^\dagger_i = a^{m_i(\tau)}_i a^\dagger_i a^{m'(\tau)}$
$a_i a^\dagger^{n(\tau)} = a^\dagger^{n'(\tau)}a_i a_i^\dagger^{n_i(\tau)}$

where the vector $m'(\tau)$ has its $i$th component zero which is why we are able to move term(s) $a^{m'(\tau)}$ to the right. This enables us to express the more general commutation relations,

$[a^{m(\tau)},a^\dagger_i] = m_i(\tau)a_i^{m_i(\tau)-1}a^{m'(\tau)}$
$[a_i, a^\dagger^{n(\tau)}] = n_i(\tau) (a_i^\dagger)^{n_i(\tau)-1}a^\dagger^{n'(\tau)}$

Using these relations, it follows that (is better notation possible?)

$[H, a_i^\dagger] = \sum_{\tau\in T} r(\tau)(a^\dagger^{n(\tau)} - a^\dagger^{m(\tau)})a_i^{m_i(\tau)-1} a^{m'(\tau)} m_i(\tau)$

and also

$[a_i, H] = \sum_{\tau\in T} r(\tau)[n_i(\tau)a^\dagger^{n_i(\tau)-1}a^\dagger^{n'(\tau)} - m_i(\tau)a^\dagger^{m_i(\tau)-1}a^\dagger^{m'(\tau)}]a^{m(\tau)}$

These will simplify the calculation of the commutation of the Hamiltonian and the number operator.

$[\sum_i a_i^\dagger a_i, H] = \sum_i [a_i^\dagger a_i, H] = \sum_i(a_i^\dagger [a_i, H] - [H, a_i^\dagger]a_i)$

For this to vanish, the following quantity must vanish identically.

$\sum_i (a_i^\dagger K_2 - K_1 a_i)$

where

$a_i^\dagger K_2 = a_i^\dagger[a_i, H] = \sum_i \sum_{\tau\in T} r(\tau)[n_i(\tau)a^\dagger^{n(\tau)} - m_i(\tau)a^\dagger^{m(\tau)}]a^{m(\tau)}$
$K_1 a_i = [H, a_i^\dagger]a_i = \sum_i \sum_{\tau\in T} r(\tau)(a^\dagger^{n(\tau)} - a^\dagger^{m(\tau)})a^{m(\tau)}m_i(\tau)$

Both of the theorems then follow from applications of the above.

# Solution to exercises

Particle Conservation of the simple reversible reaction. In Network Theory Part 10 John considered the reversible reaction with Hamiltonian

$H = (a^\dagger - b^\dagger)(\beta b - \alpha a)$

We find its commutation relations with the creation (destruction) operators of both species to be

$[H, a^\dagger] = \alpha (b^\dagger - a^\dagger)$
$[H, a] = \alpha a - \beta b$
$[b^\dagger, H] = \beta (b^\dagger - a^\dagger )$
$[b, H] = \alpha a - \beta b$

Now we see that the particle number is conserved for this Hamiltonian by calculating

$[N, H] = [a^\dagger a, H] + [b^\dagger b, H] = a^\dagger[a, H] + [a^\dagger, H]a + b^\dagger[b, H] + [b^\dagger, H]b = 0$

This could have also been shown, using Theorem I directly.

## New stuff

We’ve been working hard to understand the parallels and differences between quantum and stochastic mechanics. Last time we showed how the methods developed in prior posts can be used to model chemical reaction networks. This time, we are going to report the details of a battle.

This is the same quantum vs. stochastic battle we’ve talked about in prior posts, but this time we are going talk in detail about the odd nature of eigenstates in quantum mechanics, and how we can’t expect this structure in stochastic mechanics.


text:
# Transistor power dissipation, current, and voltage

Correct me if I am wrong, but from my understanding a transistor is current controlled in that the current flowing from Collector to Emitter is varied by the amount of current going from Base to Emitter. I initial though that a transistor was controlling the current flow based upon B->E and not changing voltage.

I recently was looking into transistors further and read the following:

The power dissipated in the transistor is the voltage drop across the collector emitter junction times the collector current

http://www.physics.unlv.edu/~bill/PHYS483/transbas.pdf

I thought that there wouldn't be a voltage drop when using a transistor since B->E is controlling current flow, not voltage.

• In order for the transistor to control the current, $V_{ce}$ has to adjust to whatever is needed to make the current flow in the external circuit. – The Photon Mar 31 '16 at 18:22

Assume this schematic:

simulate this circuit – Schematic created using CircuitLab

What I1 does and why isn't important, we just know it sources exactly 10μA.

The transistor Q1 is a type that, in this set-up happens to have an exact current gain factor of 100. Since this is a theoretical question, we can just assume that.

Now, that means that 1mA will flow into the collector right?

We also know that Ohm's Law teaches us that 1mA through 100 Ohm will only cause:

V = I * R = 0.001A * 100 Ohm = 0.1V

Add to that that the Battery BAT1 is 12V, we'll assume it ideal, so that we can say that across the resistor and transistor there is a total of 12V. If Q1 and R1 have 12V across them and R1 has only 0.1V across it, where does the rest go?

The only place it can: It falls across the Transistor.

Now, of course in the real world this voltage will slightly influence the current gain through some more complex formulas, but this is just to sketch the idea.

In the real world, there are quite a few other similar situations where such things happen.

And even if you use it as a hard-switch, the transistor will have a saturation voltage, the voltage that will always fall across the Emitter-Collector region, even if the path is fully turned on.

There are plenty transistors where this is only 0.1V in many use-cases, but if you'd put 100A through it, that's still 10W, which can be enough to kill any transistor. (Though types that can't handle that usually also don't handle 100A too well to begin with).

• In your example, would the power dissipation be 11.9v * .001A = 0.0119 watts? – Justin Mar 31 '16 at 18:41
• @Justin That is very nearly correct. The difference is ignorable, but since you didn't round away the microwatts: The base current times the base-emitter voltage will also cause a few microwatts of heating. – Asmyldof Mar 31 '16 at 18:43
• And now for a practical question, If I have a 12v 5 watt (therefore 0.5 amps) LED strip that I would like to control the brightness of, can I do this safely with a transistor? – Justin Mar 31 '16 at 18:47
• @Justin I think for that it'd be smart if you look up "PWM" for "Pulse Width Modulation" on teh interwebz. Or just this site. You can do it with a properly heat-sinked transistor and current-control, but PWM will be much more energy efficient. (ps: 12V, 5W is not exactly 0.5A, but 5W/12V =~ 0.42A) – Asmyldof Mar 31 '16 at 18:58
• Thanks for the reply, yes I was thinking about doing PWM, but the arduino outputs at 5v and doing PWM on a relay would a bit noisy. – Justin Mar 31 '16 at 19:00

Ahem.

There is no contradiction in the collector current being controlled by Ibe and the existence of a voltage drop over CE. In fact it is a basic effect of KVL that there will be a voltage drop.

If an ideal transistor had 0V over CE all the time it was no controlled current source. Look up a V-I diagram of a (controlled) current source, which is the CE part of a BJT in not saturated mode, and you will see, that the voltage can be any positive or negative voltage for any given current the source is designed (or controlled) to deliver. For a NPN-BJT, of course, voltage has to be positive to prevent the transistor from smoking.

More precisely spoken: That the current over CE is controlled means that the voltage isn't. By no means the voltage is zero. If it was 0 regardless the behaviour and design of the surronding circuitry it was somehow a controlled voltage.

That quote is specifically related to transistors operating in the linear region. In the linear region of operation, the transistor behaves a bit like a resistor. Since the collector current is controlled by the base current, then there must be some voltage component from collector to emitter. Ohm's law tells us that if we have a voltage and a current flowing, there must be a resistance (or effective resistance in this case).

A very good explanation of BJT operation can be seen here: https://electronics.stackexchange.com/a/13082/82922

• What is a linear region? – Justin Mar 31 '16 at 18:34
• Take a look at this. In this case, the linear region is before it reaches saturation. In the graph to the right, you can see how prior to the device reaching saturation (when collector current will not increase despite an increase in base current), the device behaves almost linearly. – Brendan Simpson Mar 31 '16 at 18:35

Up here at 300 degrees above absolute zero (normal room temperature) EVERYTHING has resistance (=voltage drop), even semiconductors operating as digital (on/off) switches. Remember to consider the difference between theoretical explanations and what we experience here in the Real World.

• The non-zero ohmic resistance of a semiconductor ist not the primary cause of the voltage drop on CE on a NPN-BJT. The primary cause is the current imprinted into the whole circuitry by the controlled current source. Given a circuitry of resistors, voltage sources and similar building a loop around CE the voltage drop can be derived using KVL. The resulting resistance between CE is purely virtual then. – Ariser Mar 31 '16 at 18:40


text:
# Section8.4Exercises

##### 1

Let $G$ be a group and let $H\leq G$ have index 2. Prove that $H\unlhd G\text{.}$

##### 2

Let $G$ be an abelian group with $N\unlhd G\text{.}$ Prove that $G/N$ is abelian.

##### 3

Find the following.

1. $|2\Z/6\Z|$
2. $|H|\text{,}$ for $H=2+\langle 6\rangle \subseteq \Z_{12}$
3. $o(2+\langle 6\rangle)$ in $\Z_{12}/\langle 6\rangle$
4. $\langle fH\rangle$ in $D_4/H\text{,}$ where $H=\{e,r^2\}$
5. $|(\Z_6\times \Z_8)/(\langle 3\rangle\times \langle 2\rangle)|$
6. $|(\Z_{15} \times \Z_{24})/\langle (5,4)\rangle|$
##### 4

For each of the following, find a familiar group to which the given group is isomorphic. (Hint: Consider the group order, properties such as abelianness and cyclicity, group tables, orders of elements, etc.)

1. $\Z/14\Z$
2. $3\Z/12\Z$
3. $S_8/A_8$
4. $(\Z_4 \times \Z_{15})/(\langle 2 \rangle \times \langle 3 \rangle )$
5. $D_4/\langle r^2 \rangle$
##### 5

Let $H\unlhd G$ with index $k\text{,}$ and let $a\in G\text{.}$ Prove that $a^k\in H\text{.}$


text:
# Vector space bases without axiom of choice

I want to find an example of a vector space with no base if we assume that axiom of choice is incorrect. This question might be duplicate so please alert me. Thanks.

• I can't prove it, but I'd be surprised if one could establish a basis for the space of all (real- or complex-valued) sequences without the axiom of choice. – Daniel Fischer Aug 6 '13 at 22:06
• @DanielFischer DERP. – Pedro Tamaroff Aug 6 '13 at 22:09
• @PeterTamaroff Nope, COFFEE! – Daniel Fischer Aug 6 '13 at 22:09
• @DanielFischer Why no both? – Pedro Tamaroff Aug 6 '13 at 22:10
• Inspired by the comment of @ZhenLin below: If the axiom of choice is incorrect, the only thing we know is that there exists a setting where it fails. But we don't know which one, which probably makes it impossible to pin down a concrete vector space without a base. So probably, you should reformulate this part to something like "if we cannot invoke the axiom of choice". – azimut Aug 6 '13 at 22:24

The proof is not as constructive as one would expect.

I suggest that you take a look at Andreas Blass' paper in which he proved that the existence of bases implies the axiom of choice. From the proof it is easy to construct a counterexample.

The proof idea is to take a family of non-empty sets, and to show that there exists a choice function (not quite exactly, Blass goes through another equivalence first). Assuming the axiom of choice fails, there is a family of non-empty sets which doesn't have a choice function. From this you can construct a counterexample to the principle that Blass is using in his equivalence, and then you can easily construct the vector space which doesn't have a basis.

On the other hand, it is not very hard to construct specific models in which there are specific vector spaces without bases. $\ell_2$ doesn't have a basis in Solovay's model, and generally in models of $\sf ZF+DC+BP$ (where $\sf BP$ is the statement that every set of real numbers have the Baire property).

I think for any field $F$, the $F$-vector space $F^{\mathbb{N}}$ of all sequences over $F$ provides a suitable example. (Note that the unit vectors only generate the subspace consisting of all vectors with finitely many non-zero entries.)
• Just because we can't describe the basis doesn't mean it doesn't exist! For this particular example, if the axiom of choice holds up to a sufficiently large cardinal, then $F^{\mathbb{N}}$ would have a basis. But the axiom of choice could still fail higher up. – Zhen Lin Aug 6 '13 at 22:21


text:
# The impact of the stretching exponent on fragility of glass-forming liquids

Authors
Type
Published Article
Journal
Journal of Thermal Analysis and Calorimetry
Publisher
Springer Netherlands
Publication Date
Sep 17, 2016
Volume
127
Issue
3
Pages
1975–1981
Identifiers
DOI: 10.1007/s10973-016-5828-5
Source
Springer Nature
Keywords
The paper compared two factors governing fragility. The first one is thermodynamic, depending on the change in configurational entropy. The second one is kinetic and is determined by the stretching exponent β. Using a simple mathematical development, we show that the kinetic term mK=f(βTg)∂lnβT/∂lnT|T=Tg\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$m_{\text{K}} = f( {\beta_{{T_{\text{g}} }} } )\partial \ln \beta \left( T \right)/\partial \ln T|_{{T = T_{\text{g}} }}$$\end{document} is always smaller than 1, where Tg is the glass transition temperature and f(βTg)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$f( {\beta_{{T_{\text{g}} }} } )$$\end{document} is the given functions of the stretching exponent at T = Tg. As a result, the contribution of the kinetic term on the fragility of strong glasses is <6 %. Fragile glasses have a larger value of fragility index m, and for them, the influence of the kinetic term is significantly smaller. In any case, the thermodynamic term plays the main role in fragility and the kinetic one can be neglected.


text: Equations can be formatted inline or as displayed formulas. More importantly, markdown documents are easy to read. 4 & 5 & 6 \\ 6.12 For hardcore LaTeX users (*) | R Markdown Cookbook. For kables(), a list with each element being a returned value from kable().. format: A character string. If the subscript or superscript has just one character, there is no need to delimit with braces. This post was produced with R Markdown. \end{align} You can download the latest version of RStudio at https://www.rstudio.com/products/rstudio/download/. As you can see, the code of the table does not need to represent the spacing of the table - that is accomplished within the markdown. Useful Latex Equations used in R Markdown for Statistics - stats_equations.Rmd The source code is available here as a gist. \], http://rmarkdown.rstudio.com/authoring_basics.html. In proving a result, it is often useful to include comments. R Markdown supports a reproducible workflow for dozens of static and dynamic output formats including HTML, PDF, MS … To create an R Markdown document that uses the Distill format, first install the distillR package from CRAN: Using Distill for R Markdown requires Pandoc v2.0 or higher. Here is an example of how you format. I include basic syntax to R Markdown and a minimal working example of how the analysis itself can be conducted within R with the knitr package. " The great thing about using a standard R Markdown, instead of the fully $$\LaTeX$$ approach with the .Rnw file, is that you can use $$\LaTeX$$ syntax as much or little as you like, still get the same results, and, using the same $$\TeX$$ installation, produce a great looking PDF document. LaTeX can certainly produce any type of table you’d like to create, but it does so in a way that can be very difficult to follow visually. As the commands are the ones already used in latex syntax, this works as expected in a tex output document, and thus with pdf. In the example below, we use the code {rrr} to indicate that each column is right justified. 1 & 2 & 3 \\ PDF via R Markdown. abstract: " This document provides an introduction to R Markdown, argues for its benefits, and presents a sample manuscript template intended for an academic audience. 1Introduction This document gives a gallery of tables which can be made using the xtable package to create LATEX output.It doubles as a regression check for the package. \begin{align} Itâs always better to give than to receive. Much of the Greek alphabet is implemented in LaTeX, as indicated in the LaTeX cheat sheet available at the course website. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% In the following example, I compute the sum of the integers from 2 to 19 in a hidden code block. You can embed an R code chunk like this: Note that the echo = FALSE parameter was added to the code chunk to prevent printing of the R code that generated the plot. Now Iâm going to create a series of sections using secondary headings. \end{array}\right] Click … Many statistical expressions use Greek letters. using 666 words. Notice how I define new symbols \Xbar and \sumn to make things much simpler! This creates a new LaTeX command that defines \mx to be the equivalent of \mathbf. Displayed fractions are typeset using the \frac operator. There are both upper and lower case versions available for some letters. latex font-lock syntax-highlighting markdown-mode polymode. 1 & 2 & 3 \\ Here is how your format it. ... Here’s a straightforward example where, if you have title: in your .Rmd YAML header, the \maketitle command will be executed in your LaTeX … Markdown file extension is .md These functions, such as $$\sin$$, $$\cos$$, etc. Method 1: Start a new R Markdown file in R Studio. (Notice that, in the third example, I use the tilde character for a forced space. You can send custom arguments to your .tex doc by simply defining new variables in the YAML header (relevant info in the pandoc docs here). When you click the Knit button in RStudio, a document will be generated that includes both content as well as the output of any embedded R code chunks within the document. Here is an example of one way you can do that. Here’s an excellent guide to LaTeX, and here is an example of some beautifully aligned math derivations: Using LaTeX Math features in R Markdown (Image by Author) 10. This book showcases short, practical examples of lesser-known tips and tricks to helps users get the most out of these tools. I often need to write short reports which are not full blown manuscripts, e.g. The biggest challenge is that it is not possible to translate LaTeX to Markdown perfectly: the translation is almost surely lossy (because Markdown is simple but LaTeX can be complicated). Markdown is an easy-to-use plain text formatting syntax. In the latter case, they are centered and set off from the main text. Note: this can be confusing, because the R Markdown language delimits superscripts with two carets. Markdown is a tool used to create formatted documents. on R markdownis a particular kind of markdown document. Markdown. There are lots of ways you can exploit this capability. However, if there is more than one character, braces must be used. Modifying R Markdown's LaTeX styles How to make a simple LaTeX template for R Markdown from scratch 4 minute read Published: 25 Apr, 2019. )}\\ consists of text (especially no floating environments), go with Markdown. Here’s a minimal example of what the defaults within a .Rmd will give you: Now, two specific things I’d like to change are: A solution to these two problems easily generalizes to the broader question of “How do I format the title and H1-H6 specs in the context of LaTeX rendering from .Rmd files?”. In the former case, the mathematical material occurs smoothly in the line of text. Tables printed with {gtsummary} can be seamlessly integrated into R markdown documents. Turn your analyses into high quality documents, reports, presentations and dashboards with R Markdown. x: For kable(), x is an R object, which is typically a matrix or data frame. In LaTeX, you can create parentheses, brackets, and braces which size themselves automatically to contain large expressions. R Markdown is a free, open source tool that is installed like any other R package. {rrr} Also notice the double-backslash newline character at the end of every line of the equation except the last, \[ I cannot think of a reason why you must use LaTeX to write a novel. 7 & 8 & 9 By studying the document source code file, compiling it, and observing the result, side-by-side with the source, you’ll learn a lot about the R Markdown and LaTeX mathematical typesetting language, and you’ll be able to produce nice-looking documents with R input and output neatly formatted. This book showcases short, practical examples of lesser-known tips and tricks to helps users get the most out of these tools. These are indicated with the â operator, followed by a subscript for the material appearing below the summation sign, and a superscript for any material appearing above the summation sign. For figure cross-referencing to work, you’ll also need to add a figure caption to the same code chunk using the knitr chunk option fig.cap , such as fig.cap="A good plot" . R Markdown files have the file extension “.Rmd”. and tagged R, Script and Tips Here is an example, \[\mathbf{X} = \left[\begin{array} \end{align} \end{array} After reading this book, you will understand how R Markdown documents are transformed from plain text and how you may customize nearly every step of this processing. Those new to R Markdown will appreciate the short, practical examples that address the most common issues users encounter. Adding a pagebreak was already possible with rmarkdown when output is pdf_document() or latex_document(), without any restriction about the version of pandoc. This is an R Markdown document. Here’s a straightforward example where, if you have title: in your .Rmd YAML header, the \maketitle command will be executed in your LaTeX render: Now the problem feels more tractable: all we have to do is modify the \maketitle defaults in the usual LaTeX manner within the template.tex document. This summation expression is in display form. Converting knitr/LaTeX to PDF RStudio. markdown documentation: Creating a table. This is an R Markdown document. It’s a very simple language that allows you to write HTML in a shortened way. The following examples, followed by their typeset versions, should make this clear. Open R Studio, then go to File $$\rightarrow$$ New file $$\rightarrow$$ R Markdown. annual grant progress reports. See more ideas about latex, article template, latex math. &= \sumn X_i - \sumn X_i \\ We indicate a square root using the \sqrt operator. © Text CC-BY and Code MIT unless otherwise specified. The number of #s at the beginning of the line indicates whether it is treated as a section, sub-section, sub-sub-section, etc. Posted on February 10, 2016 by steve in R Markdown The frontmatter to an R Markdown document. &= 0 )}\\ In LaTeX equations, a single caret indicates the superscript. Generally LaTeX does spacing for you automatically, and unless you use the tilde character, R will ignore your attempts to add spaces. This repository holds my working template for such purposes. 4 & 5 & 6 \\ You should consider no longer using LaTeX as a front-end for your manuscripts. can be a bit mysterious. ), you’ll see familiar LaTeX commands that are often surrounded by if(X) statements that are triggered if X appears in your .Rmd YAML. \begin{align} It will require substantial knowledge about LaTeX and Pandoc templates. Let’s start with the following: And don’t forget to include template.tex in your .Rmd YAML header like so: Next, we clearly need to fix the fact that section titles are now larger than the document title! Creating a new CV with vitae can be done using the RStudio R Markdown template selector: . Each row is then entered, with cells separated by the & symbol, and each line (except the last) terminated by \\., In math textbooks, matrices are often surrounded by brackets, and are assigned to a boldface letter. The relevant remote repo is here, and you can copy the local version you’re using into your working directory with this line: If you look past the pandoc nastiness in this template file (I at least find it nasty, being that I was mostly unfamiliar with pandoc scripting! Sometimes, you need to include an R name or command inline in a sentence. To indicate a subscript, use the underscore _ character. The sum of the integers from 2 to 19 is 189. Markdown source files, file type .md, contain text and formatting commands. Inline mathematical material is set off by the use of single dollar-sign characters. Possible values are latex, html, pipe (Pandoc's pipe tables), simple (Pandoc's simple tables), and rst.The value of this argument will be automatically determined if the function is called within a knitr document. You can do more advanced calculations in a hidden code block, assign the results to variables, and then simply use the variable names to insert the results in a sentence. I really want to use the double-stroke letters. are indicated in LaTeX with a backslash. I guessed I need something like latex as a second major mode or a markdown-latex-r-polymode but didn't get very far. Thankfully, RStudio will render a LaTeX pdf, but formatting beyond the defaults (which are still nice!) NOTE: if you need a void column you must add a space between the pipes. Ultimately though, I don’t want to abandon the LaTeX look in the compiled document. The distinguishing feature of R markdownis that it cooperates with R. Like LATEX with Sweave, code chunks can be included. LaTeX typesets special functions in a different font from mathematical variables. Patchwork. Authors should be cautious about following formatting advice for other types of markdown when working on R markdown. This creates a new LaTeX command that defines \mx to be the equivalent of \mathbf Once youâve done this, you can use your new command. 2. x &=4-3 && \text{(Subtract 3 from both sides. In this next example, however, failure to use braces creates an error, as LaTeX sets only the first character as a subscript, Here is an expression that uses both subscripts and superscripts. Currently, {gt} supports HTML output, with LaTeX and RTF planned for the future. Here’s the full .Rmd: Here’s a bonus I learned along the way. In some situations, you want to display R code but not evaluate it. In order to fit neatly in a line, summation expressions (and similar constructs) are formatted slightly differently in their inline and display versions. All common mathematical symbols are implemented, and you can find a listing on the LaTeX cheat sheet. Defining a new LaTeX command in a R Markdown document is also quite straight forward. The latex part should also be highlighted. \def\Xbar{\overline{X}_\bullet} Markdown and LaTeX Intro. Choose a title and author, but leave the rest as it is, and press the OK button: The R markdown file you opened, already contains example code. Matrics are presented in the array environment. For example, Suppose you are asked to prove something that requires several lines of development. Use a wrapper for LaTeX instead, like R Markdown. Some examples. Sometimes, you want a result without showing the user that you used R to get it. ), $a \pm b$ $x \ge 15$ $a_i \ge 0~~~\forall i$. Using with PDF/ latex documents. {rrr} But the default in RStudio is still to use Sweave, so you first need to change that default.Go to the RStudio (on menu bar) → Preferences and select Sweave on the left. The table below summarizes the default print engine utilized for {gtsummary} tables for various R Markdown output formats. The sqrt function computes the square root of a number. In this section, we show you some rudiments of the LaTeX typesetting language. Markdown and R Markdown. After reading this book, you will understand how R Markdown documents … \]. This option is ultimately included in the template.tex provided in the repository, and here’s the relevant YAML and output: Published If you don't or can't work on this by yourself, you can wait for my next package. Though such documents don’t need to adhere to a strict template, I still want them to look nice. Markdown is a simple formatting syntax for authoring HTML, PDF, and MS Word documents. To indicate a superscript, use a single caret character ^. Headers. If you are using RStudio then you should use RStudio v1.2.718 or higher (which comes bundled with Pandoc v2.0). (5) discusses the implications of R Markdown. Then I display the result inline. So I can't for the life of me including this dsfont package in R Markdown. Latex to render mathematical and scientific writing. It can be used on some websites like Stack Overflow or to write documentations (essentially on GitHub). Here is an example, $\sum_{i=1}^{n}\left( \frac{X_i}{Y_i} \right)$. by Travis Gerke There are a number of different ways to edit and produce LaTeX documents, which we’ll revist in a future post. in LaTeX, R and R markdown An R Markdown template using the bookdown package for preparing a PhD thesis at the University of Washington . Here is an example that also illustrates how to typeset an integral. In raw markdown, you would for example write a cross-reference to a figure like this: \@ref(fig:label), where the label is the name of the code chunk used to make the figure. The post may be most useful if the source code and displayed post are viewed side by side. One begins with the statement \begin{array} and ends with the statement \end{array}. 7 & 8 & 9 Download and Install R. Before installing RStudio we will need to download and install R itself. For more details on using R Markdown see http://rmarkdown.rstudio.com. Suppose I want an optional subtitle parameter. 3+x &=4 && \text{(Solve for } x \text{. Example. LaTeX \newcommand in R Markdown. x &=1 && \text{(Yielding the solution.)} For example, suppose you are proving that the sum of deviation scores is always equal to zero in any list of numbers. \], If you are going to use a lot of matrices, you should place the following early in your document. Here is an example. For example to produce the following table, we’d use this code: which produces output like this… \[\begin{array} You do this using the \left and \right operators. Apr 9, 2019 - Explore Andrew Zieffler's board "Rmarkdown and LaTeX" on Pinterest. %% Comment -- define some macros In this post, I want to deal with one of the major issues that people find difficult with LaTeX: tables. Let’s do this with the LaTeX sectsty package – you can basically stuff this code anywhere in the preamble, like so: Let’s also reduce the overall margins a touch via the geometry argument in the YAML while we’re at it. Following the opening statement, a format code is used to indicate the formatting of each column. \def\sumn{\sum_{i=1}^{n}} I write my class notes in markdown (.md file extension). This summation expression $$\sum_{i=1}^n X_i$$ appears inline. You can use RStudio to convert a .Rnw file to PDF and preview the result, in the same way you worked with R Markdown.. An R Markdown Template for Academic Manuscripts. Markdown which is a markup language that is a superset of HTML. This is accomplished like so: With this trick, you can start to do even fancier things (literally), such as include fancyhdr options. This new book written by the developers of R Markdown is an essential reference that will help users learn and make full use of the software. Consequently, if you wish to use a dollar sign (for example, to indicate currency), you need to preface it with a back-slash. Notice that in the above case, braces were not actually needed. The key formatting constructs are discussed at http://rmarkdown.rstudio.com/authoring_basics.html. \sumn \left(X_i - \Xbar\right) &= \sumn X_i - \sumn \Xbar \\ Thursday, April 25, 2019 LaTeX is a fantastic way to create and display print-ready scientific documents. Getting started. The first official book authored by the core R Markdown developers that provides a comprehensive and accurate reference to the R Markdown ecosystem. With R Markdown, you can easily create reproducible data analysis reports, presentations, dashboards, interactive applications, books, dissertations, websites, and journal articles, while enjoying the simplicity of Markdown and the great … The character # at the beginning of a line means that the rest of the line is interpreted as a section header. Define the command as you would in LaTeX… &= \sumn X_i - n \Xbar \\ Use the following command to install R Markdown: install.packages("rmarkdown") Now that R Markdown is installed, open a new R Markdown file in RStudio by navigating to File > New File > R Markdown…. Introduction. Notice the key role that the alignment tab character & plays in telling LaTeX where to align the equations. this is bold in markdown \center{this centers text} in latex. Use a productive notebook interface to weave together narrative text and code to produce elegantly formatted output. I’ve accomplished this for years by writing directly in LaTeX, but I want to align my process with my recent transition to composing most docs in RStudio/Rmd. By studying the document source code file, compiling it, and observing the result, side-by-side with the source, youâll learn a lot about the R Markdown and LaTeX mathematical typesetting language, and youâll be able to produce nice-looking documents with R input and output neatly formatted. To start, we will borrow the LaTeX template R Markdown is currently using (h/t SO). They're meant not to look like they've been marked up with tags or formatting instructions.. Import a Directory of CSV Files at Once Using {purrr} and {readr}, Conditional recoding of values in a data frame with recode_if(), Left-justify the title/author/date section, Modify the font specs used in section titles. You can see above how I constructed main section headings. The vitae package currently supports 5 popular CV templates, and adding more is a relatively simple process (details in the creating vitae templates vignette).. Markdown's formatting commands are simpler than most other formatting languages, such as LaTeX or HTML, because it has a smaller number of features. You can align the equations like this. Use multiple languages including R, Python, and SQL. Using RStudio then you should consider no longer using LaTeX as a section header forced! Are easy to read name or command inline in a hidden code block Markdown Cookbook v2.0! Download and Install R. Before installing RStudio we will need to write short reports which are still nice! mathematical. Statement \end { array }: //rmarkdown.rstudio.com/authoring_basics.html if you are proving that the sum of the integers from to. And produce LaTeX documents, which is typically a matrix or data frame to be the of. Latex: tables a series of sections using secondary headings the future the integers from 2 to 19 is.! ) new file \ ( \sum_ { i=1 } ^n X_i\ ) appears inline { i=1 } ^n ). That allows you to write documentations ( essentially on GitHub ) inline or as displayed formulas this a. Write documentations ( essentially on GitHub ) formatting constructs are discussed at http: //rmarkdown.rstudio.com character # at course! & \text { ( Subtract 3 from both sides why you must use LaTeX to a... Available at the course website get the most out of these tools the major issues that find! Together narrative text and formatting commands use the tilde character, R will your! B\ ] \ [ a_i \ge 0~~~\forall i\ ] single caret character ^ sheet available the... Such as \ ( \cos\ ), go with Markdown nice! template, compute! For the future RTF planned for the future third example, I want to display R code not. Practical examples of lesser-known tips and tricks to helps users get the most out of these tools by core! Line means that the alignment tab character & plays in telling LaTeX where to the! For the life of me including this dsfont package in R Studio, then go to file \ \sum_. To 19 is 189 superscript, use a wrapper for LaTeX instead like... Example that also illustrates how to typeset an integral R Markdown language delimits superscripts with carets. About latex in r markdown example formatting advice for other types of Markdown when working on R Markdown Andrew Zieffler 's board Rmarkdown! Braces were not actually needed the square root using the \left and operators. Beginning of a line means that the rest of the line is interpreted as a front-end your... Not full blown manuscripts, e.g the LaTeX cheat sheet and accurate reference to the R file. With the statement \begin { align } 3+x & =4 & & \text { ( Subtract 3 from both.! Begins with the statement \begin { array } and ends with the \end... Scores is always equal to zero in any list of numbers I can not think a... Extension “.Rmd ” lots of ways you can wait for my next package seamlessly! Output formats are centered and set off from the main text & plays in telling where! Bookdown package for preparing a PhD thesis at the beginning of a line means that the tab! Create a series of sections using secondary headings a markdown-latex-r-polymode but did n't get very.. Side by side Install R. Before installing RStudio we will need to delimit with.. Install R. Before installing RStudio we will borrow the LaTeX cheat sheet and Install R. Before installing RStudio we borrow. Deal with one of the integers from 2 to 19 in a R Markdown ecosystem currently! & =1 & & \text { ) R Markdown documents for you automatically, and SQL MS Word.. I want to display R code but not evaluate it character for a forced space both sides interpreted a. Several lines of development though such documents don ’ t want to deal one. ( Subtract 3 from both sides telling LaTeX where to align the equations of development to typeset integral!, go with Markdown contain text and code MIT unless otherwise specified for { gtsummary } can included... This using the bookdown package for preparing a PhD thesis at the beginning of a why. Make this clear apr 9, 2019 - Explore Andrew Zieffler 's board ` Rmarkdown LaTeX... We indicate a superscript, use a productive notebook interface to weave together text. Be most useful if the source code is used to indicate that each is!, as indicated in the example latex in r markdown example, we show you some rudiments of major! Look like they 've been marked up with tags or formatting instructions other! 2016 by steve in R Markdown output formats inline or as displayed formulas other R package align the.! Element being a returned value from kable ( ), etc posted on February 10, 2016 by steve R... Types of Markdown when working on R Markdown template selector: compute the sum of the Greek alphabet implemented. Than one character, there is no need to delimit with braces the! Install R. Before installing RStudio we will borrow the LaTeX template R Markdown extension. Include comments much latex in r markdown example may be most useful if the subscript or has. Secondary headings narrative text and code to produce elegantly formatted output one begins the. N'T get very far x \ge 15\ ] \ [ a \pm b\ ] [. The default print engine utilized for { gtsummary } tables for various R file! Latex look in the third example, Suppose you are asked to prove something that requires lines! Of text ( especially no floating environments ), x is an example of one way you can find listing... Going to create formatted documents to create a series of sections using secondary headings: tables Markdown working! Way you can create parentheses, brackets, and braces which size themselves automatically to contain large.. Or data frame single caret character ^ implemented in LaTeX, x is an example of one you. } tables for various R Markdown developers that provides a comprehensive and accurate reference to the R Markdown strict. Code is used to create formatted documents LaTeX pdf, but formatting beyond the (. & =4 & & \text { ( Yielding the solution. ) \\. And Install R. Before installing RStudio we will borrow the LaTeX look in the line text. The \left and \right operators I still want them to look like they 've marked. Wrapper for LaTeX instead, like R Markdown floating environments ), go with Markdown 2016 by steve in Markdown! And braces which size themselves automatically to contain large expressions case versions available for some letters to zero in list... Available at the beginning of a number of different ways to edit produce... Latex to write documentations ( essentially on GitHub ) in Markdown (.md file extension “.Rmd ” planned the... May be most useful if the source code is used to indicate a superscript, use the code { }... X \text { ( Solve for } x \text { ( Solve for } x \text.! Off by the core R Markdown will appreciate the short, practical examples that address the most out of tools. Formatting syntax for authoring HTML, pdf, and MS Word documents and LaTeX '' on Pinterest including dsfont. Formatting advice for other types of Markdown when working on R Markdown document is quite... At http: //rmarkdown.rstudio.com the R Markdown file extension is.md Method 1: Start a new Markdown! This dsfont package in R Markdown documents … Markdown documentation: Creating a new CV with vitae can be.... To look like they 've been marked up with tags or formatting..... IâM going to create a series of sections latex in r markdown example secondary headings produce elegantly formatted output of the LaTeX look the... N'T or ca n't for the future be cautious about following formatting advice other... Next package.Rmd ” installing RStudio we will need to include an R Markdown formats! Will render a LaTeX pdf, and SQL, like R Markdown developers that provides a comprehensive and accurate to! We indicate a square root of a number of different ways to and. Integrated into R Markdown centers text } in LaTeX equations, latex in r markdown example single caret indicates the.... Line means that the sum of the Greek alphabet is implemented in LaTeX, as indicated in the case... Comprehensive and accurate reference to the R Markdown template selector: dsfont package in R Studio, then go file... Learned along the way: for kable ( ), go with Markdown sqrt function computes the root... The source code is used to create a series of sections using secondary headings a listing on the LaTeX language... Tool used to indicate a square root using the RStudio R Markdown document also... Define new symbols \Xbar and \sumn to make things much simpler different ways to edit and produce LaTeX documents which! ).. format: a character string short reports which are not full blown manuscripts e.g. And lower case versions available for some letters formatted inline or as displayed formulas a... Other types of Markdown when working on R Markdown developers that provides a comprehensive and accurate reference to the Markdown... Code chunks can be done using the RStudio R Markdown template using the bookdown package for a! That defines \mx to be the equivalent of \mathbf summation expression \ ( \sin\ ), \ ( )! Code block the key formatting constructs are discussed at http: //rmarkdown.rstudio.com/authoring_basics.html for you,... Posted on February 10, 2016 by steve in R Studio, then go to file \ ( ). First official book authored by the use of single dollar-sign characters result without showing the user that you R... Selector: is right justified a table and formatting commands n't work on this yourself! To prove something that requires several lines of development © text CC-BY and code to produce elegantly output... Other types of Markdown when working on R Markdown template using the bookdown package for preparing PhD! A matrix or data frame how R Markdown file extension is.md Method 1: Start new!


text:
# Halogen Oxyacid Trend (acid strength)

Why is $\ce{HClO4}$ more acidic than $\ce{HBrO4}$? But $\ce{HCl}$ is less acidic than $\ce{HBr}$? What determines the acidity? Is it the concentration of $\ce{H+}$? If so, how do these trends develop along the halogen group?

• Would you mind to give the $pK_a$ values and the solvent in which they were measured for the acids you are asking about? – Klaus-Dieter Warzecha Mar 23 '14 at 17:42

In the determination of the $\mathrm{pK_a}$ values of really strong acids, water as a medium is irrelevant at a certain point.

Ivo Leito and coworkers have examined the acidity of various compounds in 1,2-dichloroethane (DCE).

In this medium the $\mathrm{pK_a}$ values for $\ce{HCl}$, $\ce{HNO3}$, $\ce{HBr}$, $\ce{HI}$, $\ce{HBF4}$, an $\ce{HClO4}$ have been determined as -0.4, -1.7, -4.9, -7.7, -10.3 and -13.0.

In this survey, 1,1,2,3,3-pentacyanopropene (1), was found to be even more acidic with a $\mathrm{pK_a}$ of -15.0!

EDIT It is important to note that these values are likely to be correct in their relative order. But the values are based on picric acid, for which the $\mathrm{pK_a}$ was arbitrarily set to zero. This initially escaped my attention!

Thanks to Martin for pointing out this important detail in his comment below!

Being an oddball, the rather unstable perbromic acid ($\ce{HBrO4}$) was not in the list!

It was synthesized by Evan Appelman (even on a larger scale) by bubbling fluorine gas through an alkaline solution of 1M $\ce{NaBrO3}$ and 5M $\ce{NaOH}$ until the solution becomes acid.

Amazingly, the author further states in a footnote:

The reaction is not smooth, and small explosions may take place in the vapor above the solution.

• Could you clarify/ prove your first statement. From my understanding $\mathrm{pK_a}$ values always involve a solvent. For superacids they are measured relative to each other. Your cited paper uses 1,2-dichloroethane as solvent and the $\mathrm{pK_a}$ values have to be anchored to an acid with a known value. For the cited case this was not possible as exact measurements in DCE were not accessible. So your posted numbers are quite arbitrary. – Martin - マーチン Mar 24 '14 at 8:45
• @Martin Good observation! One does indeed need a reference. In the paper cited, pikric acid was arbitrarily set to 0 an I forgot to mention that. This does however not invalidate the relative order. – Klaus-Dieter Warzecha Mar 24 '14 at 9:29

Assuming you are talking about acids in water, and taking into account that the acids you mention are all regarded as strong acids, I would focus on the relative stability of the corresponding conjugated bases.

$\ce{Br-}$ being larger than $\ce{Cl-}$ can better allocate the negative charge. Then, $\ce{HBr}$ ($pK_{a}$= −8.7) is stronger acid than $\ce{HCl}$ ($pK_{a}$=-6.3).

On the other hand $\ce{ClO4-}$ is more stable than $\ce{BrO4-}$. Here, another factor plays a role and should explain the difference in stability: the charge dispersal associated to resonance. Being the structure similar, we would expect the same effect: however, the resonance stabilization that characterizes such kind of ions has to be more effective for $\ce{ClO4-}$, due to a better overlap between $p$ orbitals ($2p$ of oxygen with $3p$ of chlorine, better than with $4p$ of bromine). As a consequence, $\ce{HClO4}$ ($pK_{a}$= −8) is stronger acid than $\ce{HBrO4}$ ($pK_{a}$= ? haven't found it, but here $\ce{HBrO4}$ is said to be an "almost strong acid", therefore not stronger than $\ce{H3O+}$).

These considerations tentatively explain the difference in $pK_{a}$ for such strong acids.

• Perbromic acid: the Wikipedia article has been heavily edited. You sure it still says perbromic acid is almost strong? – Oscar Lanzi Aug 26 '18 at 20:59

There are two different kinds of acids mentioned. First, binary acids such as $\ce{HCl}$, $\ce{HBr}$, $\ce{HI}$, etc. The other acids mentioned were inorganic oxyacids or oxoacids.

The most important factor in oxoacid strength is the ability to donate a proton which is influenced by electronegativity and bond polarity.

Oxoacids are usually designated with X as the central atom, and thus, their structure can be written as:

$$-\text{X}-\text{O}-\text{H}$$

X can be bonded to other things. i.e. hydrogens, more -OH groups, etc.

If X was an alkali or alkaline earth metal, and thus electropositive, the corresponding compound would be a base. This is not an acid. The reason for this is the $X-O$ bond is extremely covalent, almost to the point where the strength is equivalent to bond in ionic strength. The oxygen atom thus has a large partial negative charge $\delta^-$ and the electron density accumulates on the $-O-H$ while the hydrogen, $\ce{H^+}$, binds tightly with the oxygen, preventing deprotonation.

As the number of lone oxygens on $-X$ increases, the strength of the oxoacid increases as well.

Taken from Principles of Modern Chemistry by Oxtoby et. al 7th Edition:

If the formula of these acids is written as $XO_nOH_m$, the corresponding acid strengths fall into distinct classes according to the value of n, the number of lone oxygen atoms Each increase of 1 in n increases the acid ionization constant $K_a$ by a factor of about $10^5$.

Another way to examine the strength of acids is to look at the stability of their conjugate base which can be written as $XO_{n+1}(OH)^-_{m-1}$. As more oxygens are added on the central atom, the net negative charge is distributed more evenly around the base. As bases become more stable, they are less prone to accepting a proton. Remember that conjugate bases of strong acids are weak so intuitively, this makes sense.

Put briefly, as $-X$ becomes more electronegative, electron density is pulled towards the central atom. As the electron density is pulled from the $-H$ bond to $-X-O$, the $O-H$ bond breaks easily, and a proton is produced.

If we discuss organic acids and not binary nor inorganic acids, we not only have to talk about electronegativies, but steric hinderance, and resonance structures as well. If you would like, I can explain those as well.


text:
# Create Financial Dashboards Using Excel PowerPivot

Microsoft Excel has always been the spreadsheet software of choice for organizations around the world to perform financial analysis and generate all kinds of reports. In fact, you can safely say that most banks and financial institutions will stop functioning if excel is taken away from them for even one day. Companies use excel both as a database and as a tool to analyze their data. However, when it comes to large data sets and complex analysis, excel falls short. It also gets difficult to maintain and update these spreadsheets; they are highly labor intensive and prone to errors. That’s the reason why many companies are moving towards real business intelligence tools such as SAS and Cognos. If you are on a low budget, these expensive BI tools may not be suitable for you.

There are many of us who would like to continue with excel, if only it was more powerful. Thanks to Microsoft, we now have PowerPivot, an Excel add-in, that allows you to work with massive amounts of data within Excel, perform a variety of analysis and create powerful dashboards that are second to none.

The best part is that it’s a free add-in that you can download from Microsoft’s website and start using right away. PowerPivot Download Link. If you have Microsoft Office 2016, the good news is that PowerPivot comes pre-bundled with it so you don't have to download anything.

In the rest of the article, I will highlight some of the features and capabilities of PowerPivot that can help you create powerful dashboards for financial analysis.

• Work with a massive amount of data. Excel allows you to have up to 1 million rows in a single sheet. This is a problem when you have larger datasets. With PowerPivot, you can create special sheets that have no limits in terms of how many rows of data you have. It is not uncommon for people to have over 100 million records in this special sheet. PowerPivot uses an in-memory technology called xVelocity that provides the unmatched analytical performance to process billions of rows within seconds. Since it’s an in-memory technology, how much data you can process depends on the RAM you have in your computer.
• Integrate data from different sources. Excel has always allowed us to integrate data from various sources such as XML, MS Access, and SQL Server. Even though you could get the data, there was no easy way to form relationships between these data sets and perform a comprehensive analysis. Of course, you could use LOOKUP functions to reference and join data from different sheets; however, it’s not really practical when working with large datasets. PowerPivot makes this easy. You can import data from any data source, and use its data model capacities to easily form relationships between different datasets. For example, you can take a data set from another excel sheet and another data set from SQL server and perform an inner join to combine the data.
• Create formulas using DAX. DAX stands for Data Analysis Expressions. This is a formula and query language for PowerPivot which is similar to excel formulas and is used to perform calculations. It includes some of the functions that are used in Excel formulas and additional functions that are designed to work with relational data and perform dynamic aggregation. Unlike excel formulas that work with ranges, DAX functions work with tables and columns. You can perform advanced lookups to related values and related tables. You can also easily perform sophisticated aggregations that generally require knowledge of relational database and OLAP concepts.
• Create professional dashboards. With the help of PowerPivot, you can create powerful dashboards using just excel. You can perform a whole lot of visual analysis and also use the slicer to visually filter the data. These are self-contained apps that can also be published to SharePoint 2010 and accessed by users from a browser. Once an app/dashboard is ready, you can schedule it to refresh itself with the latest data. The dashboard is always updated without even opening it.

This summarizes some of the important features of PowerPivot. If you have any questions, feel free to leave a comment below or start a new thread in the discussion forum.

# R Programming Bundle: 25% OFF

Get our R Programming - Data Science for Finance Bundle for just $29$39.
Get it now for just \$29


text:
# Thread: There are 100 photos in the envelope, among whose is the wanted one?

1. ## There are 100 photos in the envelope, among whose is the wanted one?

There are There are 100 photos in the envelope, among whose is the wanted one. Tim picks up 10 photos randomly. What is the probability that the wanted one is among these ten photos?

My Solution: I would say, since we have ONLY 1 wanted photo and we pick 10 photos therefore the probability that the photo Tim wants is one of those 10 is
C(1,100) / C(10,100)

My friend says : No! It's (C(1,1) * C(9,99)) / C(10,100)

which one is correct and WHY? Thank you

2. If we group the 100 in groups of 10, each group has a 10% chance so 10% is the probability that 10 contains the 1.

3. Therefore, non of mine and my friend's solutions are correct?

4. Hello, Narek!

Therefore, neither mine and my friend's solutions are correct?

Your friend's solution, which should be written: $\dfrac{C(1,1)\cdot C(99,9)}{C(100,10)}$
. . is equal to $\dfrac{1}{10}$ . . . and is correct!

His/her reasoning is correct.

There are $C(100,10)$ possible outcomes; that's the denominator.

There is 1 special photo and 99 Others.

There is $C(1,1)$ way to get the special photo.
And there are $C(99,9)$ ways to choose 9 Others.

Hence, there are: . $C(1,1)\cdot C(99,9)$ ways to select 10 photos,
. . including the special one; that's the numerator.

Please pass this on to your friend: . Good work!

5. Originally Posted by Narek
My Solution: I would say, since we have ONLY 1 wanted photo and we pick 10 photos therefore the probability that the photo Tim wants is one of those 10 is C(1,100) / C(10,100)
In addition to what has been written already, maybe the following helps you seeing why this is wrong:

Suppose there are only 10 or 11 photos instead of 100 (and again, there's 1 you want, and you pick 10). Then according to your method, what are the probabilities of the right photo being in your pick?

(condsidering that with 10 photos the probability should be 1, and with 11 it should be less than 1, obviously)


text:
# Count of Subsets

How many ways can you choose a subset from the set $P = \{1, 2, 3, .. n\}$? The subset $S$ should meet the following condition:

When you choose $x \in P$ to create $S$, you cannot choose $a * x$ and $b * x$ for S.

Example :

n = 3 , a = 2, b = 3
so total subsets are 5 ,i.e, {}, {1}, {2}, {3}, {2, 3}
as if in a particular subset there is 1 so 1*2 = 2 and 1*3 cant be there.
so {1,2}, {1,3} and {1,2,3} can't be there

-
Is there a particular context that inspired your question? I suspect any formula will be a little ugly. –  Lepidopterist Mar 16 '13 at 22:14
Since you're choosing $x$, the only cases $x = a * x$ or $x = b * x$ is for $x, a, b \in \{0, 1\}$. A formula to describe this should simply include such restraint. Either you didn't mean what you wanted, or the solution should be something hereby. –  Rubens Mar 16 '13 at 22:16
Look.. we have to form different subsets and the condition is if u choose x for your subset then you can't choose ax and bx for that subset. –  Shweta Mar 16 '13 at 22:20
Ah, I must have misunderstood your post, then. –  Rubens Mar 16 '13 at 22:22
It probably will be a complicated formula. There seem to be more than $2^{\pi(n)}$ such subsets, where $\pi(n)$ is the number of primes less than or equal to $n,$ because if I understand your definition correctly, any subset $S$ which consists entirely of primes will qualify. I would think that this size is of the right order of magnitude for the answer. –  Geoff Robinson Mar 16 '13 at 23:06

If I understand the question correctly, then you're seeking a formula for $I_{a,b,n}$, the number of subsets $S \subseteq \{1,2,\ldots,n\}$ that satisfy the properties

• $x \in S \implies ax \not\in S$ and
• $x \in S \implies bx \not\in S$.

I think I have to agree with the comments thus far, you're unlikely to encounter a slick exact formula.

We can rephrase this problem as a graph-theoretic problem. Let $G=G(a,b,n)$ be the graph with vertex set $\{1,2,\ldots,n\}$ and edge set $$\big\{\{x,y\}:x<y \text{ and either } y=ax \text{ or } y=bx\big\}.$$ Then the problem asks to count the number $I_{a,b,n}$ of independent sets in this graph. (Alternatively, the number of cliques in the complement of this graph.) The graph $G(2,3,30)$ is drawn below.

We observe that the maximum degree of this graph is at most $4$ (degree $4$ occurs at vertex $v$ when $\max(a,b) v \leq n$ and $v/a \in \mathbb{Z}$ and $v/b \in \mathbb{Z}$). So, while there's lots of theoretical results discussing the difficulty of finding, enumerating, and approximating the number of independent sets, there might be some hope for an algorithmic answer for such a low bounded degree.

Extending Geoff Robinson's comment, not only does the set of primes up to $n$ form an independent set in $G$, but the set $$S_{a,b}=\{i \in \{1,2,\ldots,n\}:i \not\equiv 0 \pmod a \text{ and } i \not\equiv 0 \pmod b\}$$ forms an independent set of $G$ (since every edge has an endpoint whose index is divisible by $a$ or $b$).

Provided $2 \leq a \leq b$, we see $|S_{a,b}|$ is minimised when $a=2$ and $b=3$, so $|S_{a,b}| \geq \lfloor \tfrac{1}{3} n \rfloor$ (we include anything congruent to $\pm 1 \pmod 6$). Any subset of $S_{a,b}$ is an independent set in the graph, so there are at least $2^{|S_{a,b}|}$ independent sets in $G(a,b,n)$. So...

The number of independent sets in $G$ is at least $2^{\lfloor n/3 \rfloor}$.

The true value is probably not much more than this. As a rough guide, in a $4$-regular graph, the number of independent sets is at most $\approx 2^{0.62 n}$ (see Zhao Yufei "The Number of Independent Sets in a Regular Graph", Comb. Prob. Comp. (2010) 19, 315-320).

The reader might also be interested to note that $G$ is a subgraph of the complement of the coprime graph on $\{1,2,\ldots,n\}$, where edges $ij$ occur whenever $\gcd(i,j)=1$.

-


text:
Stalks

## Stalks

Let $X$ be a topological space, let $x\in X$ and let $ℱ:𝒯\to 𝒜$ be a functor.

Let ${𝒩}_{X}$ be the neighourhood filter at $x$.

The stalk of $𝒰$ at $x$ is $ℱ x = lim U ∈ 𝒩 x ℱ U (direct limit).$

Let $E = ∏ x ∈ X ℱ x$ with topology given by setting $W$ open if it satisfies the following constraint. $If U ∈ 𝒯 and s ∈ ℱ U then s ~ -1 W is open,$ where $s ~ : U ⟶ E x ⟼ s x is given by ℱ U ⟶ ℱ x s ⟼ s x$ is the natural homomorphism.

The sheafification of $ℱ$ is the sheaf $\stackrel{~}{ℱ}$ given by $ℱ ~ U = s ′ = s ′ x ∈ ∏ x ∈ U ℱ x | if x ∈ U then there exists V ∈ 𝒯 , and s ∈ ℱ V such that x ∈ V ⊆ U , and s ′ y y ∈ V = s y y ∈ V = s ′ : U → E | s ′ is continuous and p ∘ s ~ = id U$ where $p:E\to X$ sends ${ℱ}_{x}$ to $x$.

We have homomorphisms $ℱ U ↪ ℱ ~ U s ↦ s x x ∈ U .$