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# HIN1100 User Guide

## Part 1: Setup

Next, you will need to connect the pieces:

1. Connect the VINT Hub to your computer with a USB cable.
2. Connect the HIN1100 to the VINT Hub using the Phidget Cable.

Now that you have everything together, let's start using the HIN1100!.

### Phidget Control Panel

In order to demonstrate the functionality of the HIN1100, the Phidget Control Panel running on a Windows machine will be used.

The Phidget Control Panel is available for use on both macOS and Windows machines.

#### Windows

To open the Phidget Control Panel on Windows, find the icon in the taskbar. If it is not there, open up the start menu and search for Phidget Control Panel

#### macOS

To open the Phidget Control Panel on macOS, open Finder and navigate to the Phidget Control Panel in the Applications list. Double click on the icon to bring up the Phidget Control Panel.

### First Look

After plugging the HIN1100 into your computer and opening the Phidget Control Panel, you will see something like this:

The Phidget Control Panel will list all connected Phidgets and associated objects, as well as the following information:

• Serial number: allows you to differentiate between similar Phidgets.
• Channel: allows you to differentiate between similar objects on a Phidget.
• Version number: corresponds to the firmware version your Phidget is running. If your Phidget is listed in red, your firmware is out of date. Update the firmware by double-clicking the entry.

The Phidget Control Panel can also be used to test your device. Double-clicking on an object will open an example.

## Part 2: Using Your Phidget

The 2-axis Thumbstick Phidget tracks the horizontal and vertical position of the stick between -1 and 1. The stick will snap back to the center when released. The thumbstick can also be clicked down to activate a push-button, similar to a video game controller.

### Explore Your Phidget Channels Using The Control Panel

1. Open your Control Panel, and you will find the following channels:

2. Double click on a channel to open an example program.

Joystick Axis: Reports the tilt of the joystick (Channel 0 for up/down and channel 1 for left/right)

In your Control Panel, double click on "Joystick Axis":

Joystick Button: Reports the state of the pushbutton

In your Control Panel, double click on "Joystick Button":

## Part 4: Advanced Topics and Troubleshooting

How do I know what channel, serial number, or hub port to use in my program?

Before you open a Phidget channel in your program, you can set these properties to specify which channel to open. You can find this information through the Control Panel.

1. Open the Control Panel and double-click on the red map pin icon:

2. The Addressing Information window will open. Here you will find all the information you need to address your Phidget in your program.

See the Phidget22 API for your language to determine exact syntax for each property.

Setting the Change Trigger and Data Interval

The Change Trigger is the minimum change in the sensor data needed to trigger a new data event. The Data Interval is the time (in ms) between data events sent out from your Phidget. You can modify one or both of these values to achieve different data outputs. You can learn more about these two properties here.

Converting to Polar Coordinates

For some applications, you may want to convert the 2-axis data of the Thumbstick Phidget into an angle and magnitude. This can be helpful if you want to use it like a video game controller. You can use the following formulas:

${\displaystyle {\text{r}}={\sqrt {x^{2}+y^{2}}}}$
${\displaystyle \phi =atan2(x,y)}$

Where r is the magnitude, ϕ is the angle, and x and y are the two axis values. Atan2 is the 2-argument arctangent function, which you can find in most math libraries. For example, in C# you could use Math.Atan2(x,y). Depending on your language, ϕ may be in radians or degrees. Since x and y range from -1.0 to 1.0, the resulting r will range from -√2 to √2, so you may also want to modify the first formula like so:

${\displaystyle {\text{r}}={\frac {\sqrt {x^{2}+y^{2}}}{\sqrt {2}}}}$

This way, your r will range from 0 to 1, which is more intuitive than -√2 to √2.


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# Most effective approach when creating % chance of something

When dealing with % chance in a game what are the most effective ways to achieve this? When I use random.random()[python] or Math.random()[javascript] and use an if/elif/else conditional it seems like the things with a very low chance of happening, happened far more often than they should have.

For example, a python game I worked on used something like this:

qaul_eq = random.random()
if qual_eq <= .5:
quality = common
elif qual_eq <= .75:
quality = uncommon
elif qual_eq <=.9:
quality = rare
elif qual_eq <=.97:
quality = epic
else:
quality = legendary

Javascript DEMO

There is a 2% chance that quality will be legendary yet when printed 20 times every time it is run at least 1 item in almost every test run is legendary. In my python game it isn't uncommon for 2 or 3 of the items to be legendary. I'd chalk it up to good luck if not for the large amount of test runs I have conducted.

• To use an aphorism commonly credited to Einstein, "everything should be as simple as possible, but not simpler" – DMGregory Apr 28 '16 at 21:31
• It's important to think about the user's mental model: can they understand the algorithms? Ultimately, in RPGs, knowing (and sometimes mastering) algorithms can be a critical part of the game. – ashes999 Apr 28 '16 at 21:49
• @Philipp I looked around at other questions that were broad and unrelated to code that had + ratings and assumed it was okay to ask a question such as this. I'll take note that it isn't. I'll also edit my question to be focused on the second one. – Shniper Apr 28 '16 at 21:53
• With the code you show (Legendary has about a 3% chance, not 2%, since a roll of 0.970001 still goes to legendary), we would expect a batch of 20 drops to include at least 1 Legendary almost half the time (45.6%). We'd expect 2-3 Legendaries in a batch about 12% of the time, which matches your description that it "isn't uncommon." So, rather than a mysterious code glitch, this sounds like you might just need to recalibrate your expectations for how low you need to set an outcome's probability to get the desired result over repeated trials. – DMGregory Apr 28 '16 at 23:00
• Not a duplicate, but related: How can I make a “random” generator that is biased by prior events?. The techniques described in that question might be an inspiration for alternative methods than just rolling a random number whenever an event occurs. – Philipp Apr 29 '16 at 8:27

Your algorithm generates a "legendary" rarity whenever a random number in the range 0..1 rolls above 0.97. That's a 3% chance or a chance of one in 33. It's not a 2% chance because random() generates double-precision floating-point numbers. 0.970000000023 is just as likely as 0.9999998735.
If you want an event to happen every nth times on average, do if (random() < 1.0 / n). So to get something to happen once in 50, do random() < 0.02. Or random() > 0.98 when you prefer that form. Checking randomly-generated floating-point numbers for equality is quite pointless because of their high presision (by the way: this also applies to floating-point numbers in most other situations).

format: plain


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# Shortest walk through a given subset of edges

Given an undirected weighted graph $G = (V, \{E,F\})$, how to find the shortest walk that passes through all edges $e \in E$ exactly once?

I'd like to know if there is a general approach to this problem. However, additional constraint may be added for my specific scenario.

The graph $G_0 = (V, E)$ initially has only edges that must be crossed, but it's disconnected:

• Each connected component is formed by 2 vertices and 1 edge.
• Each vertex $v$ is the endpoint of exactly one edge $\in E$. This means that any two edges $e_i, e_j \in E$ don't have a vertex $v$ in common.
• The cost of each edge $e \in E$ is $0$.
• The number of edges in $E$ is at most 1000, usually between 10 and 100.

Then other edges $f \in F$ are added with to link any two vertices (except the one already linked by $e$), so $G = (V, \{E,F\})$.

• The cost of each edge $f \in F$ is $> 0$

Related questions

I found a few related questions, but they doesn't seem to help:

Not working solutions

I tried a few approaches so far, but I'm unable to make them work properly. E.g.:

• Convert $G$ to its line graph $L$. I can't figure out how to assign a proper weight to edges in $L$
• If I can find a proper way to construct $L$, adding a dummy vertex $s$ on $L$ allows to solve this with TSP. Then I'll know the order of the edges $e$, but not the direction.
• I really don't want to use brute-force. It may become impractical very soon.

Almost-Working solutions for the specific case

Since by construction every walk will be of the form: $e_1, f_1, e_2, ... f_n, e_n$:

1. Compute the weight of the path for each permutation of edges in $E$. The edge $f$ between two edges $e$ will be the mininmum weight one. Pick up the minimum weight path. This solution is brute-force, so not really practical for a large number of edges in E.

2. Start from an edge $e$. Follow the min-cost edge $f$ starting from the end-vertex in $e$. Repeat until all edges in $E$ are reached. This solution is not guaranteed to find the shortest path.

This is NP-hard, so it's very unlikely that a polynomial-time algorithm exists.

Given any instance $G=(V, E)$ of Hamiltonian Path, create a new graph $G'=(V', E')$ in which every vertex $v \in V$ becomes a pair of vertices $v_+, v_-$ connected by an edge in $G'$. All of these edges should also be added to $F$. Then for each $(u, v) \in E$, add the corresponding 4 edges $(u_-, v_-), (u_-, v_+), (u_+, v_-), (u_+, v_+)$ to $E'$ (but not to $F$). All edges in $E'$ get weight 1.

Now if and only if there is a Hamiltonian Path in $G$, there is a path (and thus also a walk) of length $2|V|-1$ in $G'$ that passes through all the edges in $F$. (Exercise: Prove that if a walk having length $2|V|-1$ and passing through all edges in $F$ exists in $G'$, then $G$ has a Hamiltonian Path; the other direction is easy.) IOW, if your problem could be solved efficiently by some algorithm, then that algorithm could be used as a subroutine to solve Hamiltonian Path efficiently too, and people don't think that's very likely.

• My answer doesn't describe an algorithm for solving your problem; it describes an "algorithm" for solving the Hamiltonian Path problem by converting it into an instance of your problem. This is one way to show that your problem is NP-hard -- that is, that it's very unlikely that any algorithm exists that can solve your problem. – j_random_hacker Dec 15 '16 at 15:01
• "Also, I understand that this problem may probably be NP. This isn't a major concern right now." You mean "NP-hard", BTW ("NP" is a larger class of problems). It is a major concern that your problem is NP-hard if you "really don't want to use brute-force", since being NP-hard effectively means that brute force is the only way that guarantees an optimal solution. – j_random_hacker Dec 15 '16 at 15:03
• It's unfortunate that the problem is NP-hard, but that's not my fault -- I'm just the messenger here. What you perhaps don't seem to appreciate is that the proof sketch above saves you time hunting for something that doesn't exist. You can now pick a brute-force algorithm with your conscience clear that nothing better exists "out there, somewhere". (Not quite true; for many NP-hard problems there exist algorithms that save time over the most naive brute force, but they are all still exponential in the input size -- at best, the exponent decreases a bit.) – j_random_hacker Dec 15 '16 at 15:43
• You can turn an instance of your problem into a TSP instance with a simple transformation, and then apply one of the advanced (exact or heuristic) solvers to it: Just subdivide each edge in $E$ (earlier I had $E$ and $F$ mixed up, BTW) by adding a new vertex in the middle. Now any TSP tour can only visit such a vertex by visiting one of its 2 neighbours, then the vertex, then the other neighbour -- equivalent to visiting the original "edge". Finally add a single extra vertex at a huge distance from every vertex in your original graph (and not adjacent to the other extra vertices). – j_random_hacker Dec 15 '16 at 16:08
• You're welcome. The tour you get from the TSP solver will necessarily include exactly 2 edges to this "far-away" vertex -- cut them to find the path you want. (Actually this will find the shortest path that visits all edges in $E$ exactly once; to allow a vertex to be revisited multiple times, you can turn each original vertex into $n$ copies, all connected to each other by 0-length edges.) – j_random_hacker Dec 15 '16 at 16:13

If you want a walk from $v$ to $u$, and it has to cross every edge of $F$ once, you can do the next: -Duplicate al the nodes from G |F| times (you put every copy in levels) -A node $v_{i}$ (node in level i) has and edge to a node $v_{i+1}$ iff this two had an edge in the original graph and this edge is in F. Then, you want the min walk from $v_0$ to $u_{|F|}$. You can use Dijkstra for this. I think this works but I'm not sure.

If this is your original graph, the red edges are the ones I want for my path

Then I say to copy it so you have the next graph

Then you want a path from u_o to v_2 since there are the paths that uses two red edges. I added weight 1 to the edges because you don't want to cross it more than once. (If the weight were 0, then the min path can go through this edge many times without cost)

• Thanks, let me think about it. However, I need to go trough edges in E exactly once, not F. It's just notation, doesn't change much – Miki Dec 15 '16 at 12:29
• I can't figure out how to make this work. Can you please elaborate? – Miki Dec 15 '16 at 13:03
• I need to verify this, but thanks for your effort – Miki Dec 15 '16 at 15:43
• I'm afraid this won't work reliably for $|F| \ge 2$, since nothing stops Dijkstra from choosing several layer-traversing edges that correspond to the same original edge in $F$, and leaving other edges from $F$ unvisited. You need to somehow record which edges from $F$ have already been visited by the path so far, so that they can be forbidden from later layers... You can do this by turning each vertex in $G$ into $2^{|F|}$ vertices, each corresponding to a different set of already-visited $F$-edges, but of course this makes the problem exponential-time (and -space). – j_random_hacker Dec 15 '16 at 16:45
• Another way is to fix a particular ordering of $F$-edges. You can then apply your poly-time and -space algorithm directly -- but you'll need to rerun it for every possible ordering (i.e., $|F|!$ times) and take the best overall. – j_random_hacker Dec 15 '16 at 16:48


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# All Questions

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### ASN.1 OID of bcrypt

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### How does DES decryption work? Is it the same as encryption or the reverse? [duplicate]

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### salting with password hash to improve security?

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### Finite fields in elliptic curve

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242 views

### How to Compute C^2 in AES MixColumns Matrix?

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374 views

### Ed25519 choice of private key implementation

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176 views

### Are there any good examples of Contemporary Mechanical Cryptography?

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I know: Never do your own crypto. But this question is just theoretical. Assume you do a Diffie-Hellman Key Exchange with a server to produce a shared secret $x$. Then use a cryptographic hash ...


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# Solving Recurrence Relation $a_n=6a_{n-1} - 9a_{n-2}$ for $n\geqslant2$

So the problem is to solve this recurrence relation with the initial conditions $a_0 = 2, a_1 = 21$.

$a_n=6a_{n-1} - 9a_{n-2}$ for $n\geqslant2$

And also find the value of $a_{2016}.$

Here is my solution but I'm not entirely sure if it's correct. Was wondering if anyone can confirm what I did is valid or perhaps I made a mistake somewhere? Thanks in advance.

• Sounds about right. Apr 12 '17 at 21:44
• @IvanNeretin thanks! Apr 12 '17 at 21:46
• Very good. continue. Apr 12 '17 at 22:11
• @Jamie here is an other method. Apr 12 '17 at 22:26

Put $a_n=3^nb_n$.

then

$$3^nb_n=2.3^nb_{n-1}-3^nb_{n-2}$$

or

$b_n-b_{n-1}=b_{n-1}-b_{n-2}$ =constante $$=b_1-b_0=7-2=5$$

thus

$$b_n=5n+2$$

and

$$\boxed {a_n=3^n (5n+2)}$$

Generatingfunctionologically, define $$A(z) = \sum_{n \ge 0} a_n z^n$$, shift the recurrence by 2, multiply by $$z^n$$, sum over $$n \ge 0$$ and recognize resulting sums:

\begin{align*} \sum_{n \ge 0} a_{n + 2} z^n &= 6 \sum_{n \ge 0} a_{n + 1} z^n - 9 \sum_{n \ge 0} a_n z^n \\ \frac{A(z) - a_0 - a_1 z}{z^2} &= 6 \frac{A(z) - a_0}{z} - 9 A(z) \end{align*}

Solve for $$A(z)$$ with the given values of $$a_0. a_1$$, write as partial fractions:

\begin{align*} A(z) &= \frac{2 + 9 z}{1 - 6 z + 9 z^2} \\ &= \frac{5}{(1 - 3 z)^2} - \frac{3}{1 - 3 z} \end{align*}

Use the generalized binomial theorem:

\begin{align*} (1 + x)^{-m} &= \sum_{n \ge 0} (-1)^n \binom{-m}{n} x^n \\ &= \sum_{n \ge 0} \binom{n + m - 1}{m - 1} x^n \end{align*}

extract the desired coefficient:

\begin{align*} [z^n] A(z) &= 5 \cdot \binom{n + 2 - 1}{2 - 1} \cdot 3^n - 3 \cdot 3^n \\ &= (5 n + 2) \cdot 3^n \end{align*}


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What is the difference between Gradient Descent and Stochastic Gradient Descent?

What is the difference between Gradient Descent and Stochastic Gradient Descent?

I am not very familiar with these, can you describe the difference with a short example?

5 Answers

For a quick simple explanation:

In both gradient descent (GD) and stochastic gradient descent (SGD), you update a set of parameters in an iterative manner to minimize an error function.

While in GD, you have to run through ALL the samples in your training set to do a single update for a parameter in a particular iteration, in SGD, on the other hand, you use ONLY ONE or SUBSET of training sample from your training set to do the update for a parameter in a particular iteration. If you use SUBSET, it is called Minibatch Stochastic gradient Descent.

Thus, if the number of training samples are large, in fact very large, then using gradient descent may take too long because in every iteration when you are updating the values of the parameters, you are running through the complete training set. On the other hand, using SGD will be faster because you use only one training sample and it starts improving itself right away from the first sample.

SGD often converges much faster compared to GD but the error function is not as well minimized as in the case of GD. Often in most cases, the close approximation that you get in SGD for the parameter values are enough because they reach the optimal values and keep oscillating there.

If you need an example of this with a practical case, check Andrew NG's notes here where he clearly shows you the steps involved in both the cases. cs229-notes

Source: Quora Thread

• thanks, Briefly like this? There are three variants of the Gradient Descent: Batch, Stochastic and Minibatch: Batch updates the weights after all training samples have been evaluated. Stochastic, weights are updated after each training sample. The Minibatch combines the best of both worlds. We do not use the full data set, but we do not use the single data point. We use a randomly selected set of data from our data set. In this way, we reduce the calculation cost and achieve a lower variance than the stochastic version. Aug 7, 2018 at 15:50
• Note that the above link to cs229-notes is down. However, Wayback Machine, aligned with date of post, delivers - yay! web.archive.org/web/20180618211933/http://cs229.stanford.edu/… Jan 17, 2021 at 18:17
• @Sociopath Great explanation! I've tried both GD and SGD and GD performs better however I am not really understanding the reason behind it. do you know why? and if yes, is moving from SGD -> GD a way to mitigate underfitting ? May 11, 2021 at 23:32
• So I'm sure I get this right. The stochastic part just refers to "randomly" selecting a value out of all values to be representative for the update? Sep 11, 2021 at 20:45
• @JoshuaOlson: Correct! Dec 13, 2021 at 12:08

The inclusion of the word stochastic simply means the random samples from the training data are chosen in each run to update parameter during optimisation, within the framework of gradient descent.

Doing so not only computed errors and updates weights in faster iterations (because we only process a small selection of samples in one go), it also often helps to move towards an optimum more quickly. Have a look at the answers here, for more information as to why using stochastic minibatches for training offers advantages.

One perhaps downside, is that the path to the optimum (assuming it would always be the same optimum) can be much noisier. So instead of a nice smooth loss curve, showing how the error descreases in each iteration of gradient descent, you might see something like this:

We clearly see the loss decreasing over time, however there are large variations from epoch to epoch (training batch to training batch), so the curve is noisy.

This is simply because we compute the mean error over our stochastically/randomly selected subset, from the entire dataset, in each iteration. Some samples will produce high error, some low. So the average can vary, depending on which samples we randomly used for one iteration of gradient descent.

• thanks, Briefly like this? There are three variants of the Gradient Descent: Batch, Stochastic and Minibatch: Batch updates the weights after all training samples have been evaluated. Stochastic, weights are updated after each training sample. The Minibatch combines the best of both worlds. We do not use the full data set, but we do not use the single data point. We use a randomly selected set of data from our data set. In this way, we reduce the calculation cost and achieve a lower variance than the stochastic version. Aug 7, 2018 at 15:51
• I'd say there is batch, where a batch is the entire training set (so basically one epoch), then there is mini-batch, where a subset is used (so any number less than the entire set $N$) - this subset is chosen at random, so it is stochastic. Using a single sample would be referred to as online learning, and is a subset of mini-batch... Or simply mini-batch with n=1. Aug 7, 2018 at 19:34
• tks, this is clear!
– Dee
Sep 28, 2019 at 2:20

In Gradient Descent or Batch Gradient Descent, we use the whole training data per epoch whereas, in Stochastic Gradient Descent, we use only single training example per epoch and Mini-batch Gradient Descent lies in between of these two extremes, in which we can use a mini-batch(small portion) of training data per epoch, thumb rule for selecting the size of mini-batch is in power of 2 like 32, 64, 128 etc.
For more details: cs231n lecture notes

Gradient Descent is an algorithm to minimize the $$J(\Theta)$$!

Idea: For current value of theta, calculate the $$J(\Theta)$$, then take small step in direction of negative gradient. Repeat.

Update Equation =

Algorithm:

while True:
theta_grad = evaluate_gradient(J,corpus,theta)
theta = theta - alpha * theta_grad

But the problem is $$J(\Theta)$$ is the function of all corpus in windows, so very expensive to compute.

Stochastic Gradient Descent repeatedly sample the window and update after each one

Stochastic Gradient Descent Algorithm:

while True:
window = sample_window(corpus)
theta_grad = evaluate_gradient(J,window,theta)
theta = theta - alpha * theta_grad

Usually the sample window size is the power of 2 say 32, 64 as mini batch.

Both algorithms are quite similar. The only difference comes while iterating. In Gradient Descent, we consider all the points in calculating loss and derivative, while in Stochastic gradient descent, we use single point in loss function and its derivative randomly. Check out these two articles, both are inter-related and well explained. I hope it helps.

• Please update the links and/or give the exact titles of the articles. thx Jun 11, 2021 at 13:14


text:
# Synopsis: Insulating behavior in topological insulators

A new topological insulator that approaches true insulating behavior in the bulk is synthesized.

3D topological insulators represent a unique quantum state for matter that is supposed to show insulating behavior in the bulk and spin-dependent metallic conduction on the surface. In practice, the best-known exemplars of materials that show a topologically protected metallic surface state, such as ${\text{Bi}}_{2}{\text{Se}}_{3}$ and ${\text{Bi}}_{2}{\text{Te}}_{3}$, are also conducting in the bulk due to the presence of vacancies. Significant efforts in trying to find a topological insulator that is truly insulating in the bulk have met with little success.

Presenting their results as a Rapid Communication in Physical Review B, Zhi Ren and colleagues from Osaka University, Japan, have synthesized a new topological insulator, ${\text{Bi}}_{2}{\text{Te}}_{2}\text{Se}$, that approaches insulating behavior in the bulk with a high resistivity. Ren et al. demonstrate variable-range hopping that is the hallmark of an insulator in high-quality single crystals of ${\text{Bi}}_{2}{\text{Te}}_{2}\text{Se}$ and Shubnikov-de Haas oscillations coming from the 2D surface metallic state. Surface contribution to the total conductance of the crystal at $6%$ is the largest ever achieved in a topological insulator. From a detailed study of the Hall effect, the authors also determine the transport mechanism in the bulk that reveals an impurity band in the band gap along with hopping conduction of localized electrons. These results pave the way for exploiting the unique surface conduction properties of topological insulators. – Sarma Kancharla

More Features »

### Announcements

More Announcements »

## Previous Synopsis

Particles and Fields

Fluid Dynamics

## Related Articles

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### Synopsis: Quantum Circulator on a Chip

A circulator that routes microwave signals is suitable for scaling up quantum-computing architectures. Read More »

Spintronics

### Synopsis: Flip-Flopping the Bands

A pair of semiconductor quantum wells with an inverted band structure hosts electrons whose spins are almost all in the same quantum state.   Read More »

Nanophysics

### Focus: Negative Resistance with a Single Atom

Current flowing through a single silicon atom can be made to decrease with increasing voltage, potentially allowing the integration of a new type of component into microelectronic circuits. Read More »


text:
## anonymous 5 years ago Show that a linear transformation T:V-->W is injective if and only if it has the property of mapping linearly independent subsets of V to linearly independent subsets of W.

Let $$x\in \ker T$$. If $$x\neq 0$$ then $$\{x\}$$ is linearly independent. But $$\{T(x)\}=\{0\}$$ is linearly dependent. Hence $$x=0$$ and therefore $$T$$ is injective.


text:
QpiddConfig - file /etc/qpid/qpidd.conf¶

class insights.parsers.qpidd_conf.QpiddConf(context)[source]

Parse the qpidd configuration file.

Produces a simple dictionary of keys and values from the configuration file contents , stored in the data attribute. The object also functions as a dictionary itself thanks to the insights.core.LegacyItemAccess mixin class.

Sample configuration file:

# Configuration file for qpidd. Entries are of the form:
# name=value
#
# (Note: no spaces on either side of '='). Using default settings:
# "qpidd --help" or "man qpidd" for more details.
#cluster-mechanism=ANONYMOUS
log-enable=error+
log-to-syslog=yes
auth=no
require-encryption=yes
ssl-require-client-authentication=yes
ssl-port=5672
ssl-cert-db=/etc/pki/katello/nssdb
ssl-cert-name=broker

interface=lo

Examples

>>> qpidd_conf['auth']
'no'
>>> 'require-encryption' in qpidd_conf
True

parse_content(content)[source]

This method must be implemented by classes based on this class.


text:
# Loop-generated LaTeX table using tokens

I am trying to create a template sort of LaTeX document for myself where I can set a number near the beginning (like a variable) and later in the document there shall be a table with as many rows as that number. Also, in the first column there shall be an increasing list of numbers. So if I set the 'variable' to 5, I would get a table with numbers from 1 to 5 in the first column.

I have followed this professional's answer to a similar question to create this table using tokens. My problem is that I can't get the numbers in my first column like I desire. Here is an MWE

\documentclass{article}

\def\numberoflines{10}

\newcounter{i}
\setcounter{i}{0}
\newtoks\tabtoks
\newcommand*\resettabtoks{\tabtoks{}}
\newcommand*\printtabtoks{\the\tabtoks}

\begin{document}

\resettabtoks
\loop\ifnum\thei<\numberoflines
\stepcounter{i}
\repeat

\begin{tabular}{| c | c |}
\hline
number & empty \\
\hline
\printtabtoks
\end{tabular}

\end{document}

yes, I know vertical lines are ugly; this is a special use case, not for professional typesetting, please ignore

Output:

Is there some fix to get 1-10 instead of 10 everywhere?

The main problem is that you aren't expanding \thei, so the token register will contain

\thei & \\ \hline
\thei & \\ \hline
\thei & \\ \hline
\thei & \\ \hline
\thei & \\ \hline
\thei & \\ \hline
\thei & \\ \hline
\thei & \\ \hline
\thei & \\ \hline

However, the idea of doing

\expandafter\addtabtoks\expandafter{\thei & \\ \hline}

doesn't work, because the first level expansion of \thei is \arabic{i}; a further expansion gives \number\c@i and a third expansion would be needed, leading to a whopping 14 repetitions of \expandafter. Not ideal.

The simplest trick is to force expansion all the way with

\expandafter\addtabtoks\expandafter{\romannumeral-Q\thei & \\ \hline}

Also \stepcounter{i} should be moved before this instruction.

\documentclass{article}

\def\numberoflines{10}

\newcounter{i}
\setcounter{i}{0}
\newtoks\tabtoks
\newcommand*\resettabtoks{\tabtoks{}}
\newcommand*\printtabtoks{\the\tabtoks}

\begin{document}

\resettabtoks
\loop\ifnum\thei<\numberoflines
\stepcounter{i}
\repeat
\showthe\tabtoks

\begin{tabular}{| c | c |}
\hline
number & empty \\
\hline
\printtabtoks
\end{tabular}

\end{document}

You may find the following solution better. Any number of rows can be specified at runtime.

\documentclass{article}
\usepackage{xparse}

\ExplSyntaxOn
\NewDocumentCommand{\makerows}{m}
{
\int_step_function:nN { #1 } \__bertalanp_mr_row:n
}
\cs_new:Nn \__bertalanp_mr_row:n { #1 & \\ \hline }
\ExplSyntaxOff

\begin{document}

\begin{tabular}{| c | c |}
\hline
number & empty \\
\hline
\makerows{10}
\end{tabular}

\end{document}

• The alternative solution you provided clearly seems superior, thank you. I am not even sure if using LaTeX for things like this is appropriate (is it?), but this xparse solution is still elegant.
– bp99
Oct 27, 2018 at 20:36
• In my opinion (not that you asked for it), doing things like this is where LaTeX really shines Oct 28, 2018 at 8:16

Three Four ways to do this on-the-fly without pre-accumulating tokens, using xinttools. The first method only uses a counter.

### expandable replication with normal LaTeX counter

Sorry for \romannumeral but this is the syntax for using \xintreplicate. If you prefer, define \newcommand{\xintReplicate}{\romannumeral\xintreplicate} to hide it and use \xintReplicate.

\documentclass{article}
\usepackage{xinttools}

\newcounter{i}

\newcommand{\makerows}[1]{%
\romannumeral\xintreplicate{#1}{\stepcounter{i}\thei&\\\hline}%
}

\begin{document}

\setcounter{i}{0}
\begin{tabular}{| c | c |}
\hline
number & empty \\
\hline
\makerows{10}
\end{tabular}

\end{document}

No counter is used in the next three methods.

### expandable loop with magic loop indexing

This is completely expandable loop; a somewhat delicate beast to use. The parameter #1 is assumed to be explicit digits (the end of line space terminates them in the \ifnum test).

\documentclass{article}
\usepackage{xinttools}

\newcommand{\makerows}[1]{%
\xintiloop[1+1]
\xintiloopindex & \\ \hline
\unless\ifnum\xintiloopindex=#1
\repeat
}
\begin{document}

\begin{tabular}{| c | c |}
\hline
number & empty \\
\hline
\makerows{10}
\end{tabular}

\end{document}

### xintFor loop over integers (starts at 1 per default) with a Break

Here the ##1 is encapsulated in a \numexpr, hence the need for \number.

\documentclass{article}
\usepackage{xinttools}

\newcommand{\makerows}[1]{%
\xintFor ##1 in \xintintegers
\do
{\number##1 & \\ \hline
\ifnum#1=##1\expandafter\xintBreakFor\fi}%
}

\begin{document}

\begin{tabular}{| c | c |}
\hline
number & empty \\
\hline
\makerows{10}
\end{tabular}

\end{document}

### xintFor loop with pre-computed list of integers

Here the ##1 are digit tokens.

\documentclass{article}
\usepackage{xinttools}

\newcommand{\makerows}[1]{%
\xintFor* ##1 in {\xintSeq{1}{#1}}
\do
{##1 & \\ \hline}%
}

\begin{document}

\begin{tabular}{| c | c |}
\hline
number & empty \\
\hline
\makerows{10}
\end{tabular}

\end{document}

### output (same for all four methods, of course)

• for the first method (\xintreplicate) one only needs the smaller package xintkernel, not needed to load xinttools.
– user4686
Oct 28, 2018 at 13:08

You need some more expansion. As it is only the macro \thei is stored and it will be executed when typesetting it (i.e. at the end when i=10).

\documentclass{article}

\def\numberoflines{10}

\newcounter{i}
\setcounter{i}{0}
\newtoks\tabtoks
\newcommand*\resettabtoks{\tabtoks{}}
\newcommand*\printtabtoks{\the\tabtoks}

\begin{document}

\resettabtoks
\loop\ifnum\value{i}<\numberoflines
\stepcounter{i}
\repeat

\begin{tabular}{| c | c |}
\hline
number & empty \\
\hline
\printtabtoks
\end{tabular}

\end{document}

Edit: \expanded is currently LuaLaTeX-only. Here's a more general version:

\documentclass{article}

\def\numberoflines{10}

\newcounter{i}
\setcounter{i}{0}
\newtoks\tabtoks
\newcommand*\resettabtoks{\tabtoks{}}
\newcommand*\printtabtoks{\the\tabtoks}

\begin{document}

\resettabtoks
\loop\ifnum\value{i}<\numberoflines
\edef\tmpi{\thei}
\stepcounter{i}
\repeat

\begin{tabular}{| c | c |}
\hline
number & empty \\
\hline
\printtabtoks
\end{tabular}

\end{document}

• What is expanded? xelatex gives me undefined control sequence for that.
– bp99
Oct 27, 2018 at 20:01
• At present, only LuaTeX features \expanded. Oct 27, 2018 at 20:03
• @bertalanp99 It is LuaTeX only, but I have added a more general approach. Oct 27, 2018 at 20:06
• This prints the numbers from 0 to 9. For 1 to 10, \stepcounter{i} must be moved just after \numberoflines. Oct 27, 2018 at 20:33


text:
# Limit Inferior/Examples/Sequence of Reciprocals

## Example of Limit Inferior

Let $\sequence {a_n}$ be the sequence defined as:

$\forall n \in \N_{>0}: a_n = \dfrac 1 n$

The limit inferior of $\sequence {a_n}$ is given by:

$\displaystyle \map {\liminf_{n \mathop \to \infty} } {a_n} = 0$

## Proof

From Sequence of Reciprocals is Null Sequence, $\sequence {a_n}$ is convergent:

$\displaystyle \lim_{n \mathop \to \infty} \dfrac 1 n = 0$

Let $L$ be the set of all real numbers which are the limit of some subsequence of $\sequence {a_n}$.

By Limit of Subsequence equals Limit of Real Sequence, all such subsequences have limit $0$.

Hence by definition of limit inferior:

$\displaystyle \map {\liminf_{n \mathop \to \infty} } {a_n} = \inf \set 0 = 0$

$\blacksquare$


text:
# example of fundamental theorem of demography

Assume a population with a (age, sex, etc.) that is described by a vector $x(t)$, where $x_{1}(t),\ldots,x_{n}(t)$ represents the number of individuals in the population who possess the characteristic at a level $1,\ldots,n$, at time $t$.

For example, consider age-groups, and assume $x_{0}(t)$ is the number of individuals in the population that are aged 0 to 1 year, $x_{1}(t)$ is the number of individuals aged 1 to 2 years, etc.

Suppose that the transition from one class to another is described by a matrix $A(t)$. In the case of age-groups, this matrix will for example describe mortality in a given age-group. This matrix, in the case of non deterministic modelling, will define a Markov chain.

The fundamental theorem of demography then states that if the matrix $A(t)$ satisfies the required properties, then the distribution vector $x(t)$ converges to the eigenvector associated to the dominant eigenvalue, regardless of the behavior of the total population $\|x(t)\|$.

Hence, in the case of age-groups, the proportion of individuals aged, say, 1 to 2 years, tends to a fixed value, even if the total population increases.

Title example of fundamental theorem of demography ExampleOfFundamentalTheoremOfDemography 2013-03-22 14:15:59 2013-03-22 14:15:59 mathcam (2727) mathcam (2727) 6 mathcam (2727) Example msc 92D25 msc 37A30


text:
Income is the consumption and saving opportunity gained by an entity within a specified timeframe, which is generally expressed in monetary terms.[1] Income is difficult to define conceptually and the definition may be different across fields.[2][page needed] For example, a person's income in an economic sense may be different from their income as defined by law.[2]

An extremely important definition of income is Haig–Simons income, which defines income as Consumption + Change in net worth and is widely used in economics.[2]

For households and individuals in the United States, income is defined by tax law as a sum that includes any wage, salary, profit, interest payment, rent, or other form of earnings received in a calendar year.[3] Discretionary income is often defined as gross income minus taxes and other deductions (e.g., mandatory pension contributions), and is widely used as a basis to compare the welfare of taxpayers.

In the field of public economics, the concept may comprise the accumulation of both monetary and non-monetary consumption ability, with the former (monetary) being used as a proxy for total income.

For a firm, gross income can be defined as sum of all revenue minus the cost of goods sold. Net income nets out expenses: net income equals revenue minus cost of goods sold, expenses, depreciation, interest, and taxes.[1]

Economic definitions

Full and Haig–Simons income

 Main article: Haig–Simons income

"Full income" refers to the accumulation of both the monetary and the non-monetary consumption-ability of any given entity, such as a person or a household. According to what the economist Nicholas Barr describes as the "classical definition of income" (the 1938 Haig–Simons definition): "income may be defined as the... sum of (1) the market value of rights exercised in consumption and (2) the change in the value of the store of property rights..." Since the consumption potential of non-monetary goods, such as leisure, cannot be measured, monetary income may be thought of as a proxy for full income.[1] As such, however, it is criticized[by whom?] for being unreliable, i.e. failing to accurately reflect affluence (and thus the consumption opportunities) of any given agent. It omits the utility a person may derive from non-monetary income and, on a macroeconomic level, fails to accurately chart social welfare. According to Barr, "in practice money income as a proportion of total income varies widely and unsystematically. Non-observability of full-income prevent a complete characterization of the individual opportunity set, forcing us to use the unreliable yardstick of money income.

Factor income

In economics, "factor income" is the return accruing for a person, or a nation, derived from the "factors of production": rental income, wages generated by labor, the interest created by capital, and profits from entrepreneurial ventures.[4]

In consumer theory 'income' is another name for the "budget constraint," an amount ${\displaystyle Y}$ to be spent on different goods x and y in quantities ${\displaystyle x}$ and ${\displaystyle y}$ at prices ${\displaystyle P_{x))$ and ${\displaystyle P_{y))$. The basic equation for this is

${\displaystyle Y=P_{x}\cdot x+P_{y}\cdot y}$

This equation implies two things. First buying one more unit of good x implies buying ${\displaystyle {\frac {P_{x)){P_{y))))$ less units of good y. So, ${\displaystyle {\frac {P_{x)){P_{y))))$ is the relative price of a unit of x as to the number of units given up in y. Second, if the price of x falls for a fixed ${\displaystyle Y}$ and fixed ${\displaystyle P_{y},}$ then its relative price falls. The usual hypothesis, the law of demand, is that the quantity demanded of x would increase at the lower price. The analysis can be generalized to more than two goods.

The theoretical generalization to more than one period is a multi-period wealth and income constraint. For example, the same person can gain more productive skills or acquire more productive income-earning assets to earn a higher income. In the multi-period case, something might also happen to the economy beyond the control of the individual to reduce (or increase) the flow of income. Changing measured income and its relation to consumption over time might be modeled accordingly, such as in the permanent income hypothesis.

Legal definitions

Definitions under the Internal Revenue Code

 Main article: Internal Revenue Code section 61

Except as otherwise provided in this subtitle, gross income means all income from whatever source derived, including (but not limited to) the following items: (1) Compensation for services, including fees, commissions, fringe benefits, and similar items; (2) Gross income derived from business; (3) Gains derived from dealings in property; (4) Interest; (5) Rents; (6) Royalties; (7) Dividends; (8) Annuities; (9) Income from life insurance and endowment contracts; (10) Pensions; (11) Income from discharge of indebtedness; (12) Distributive share of partnership gross income; (13) Income in respect of a decedent; and (14) Income from an interest in an estate or trust.

26 U.S. Code § 61 - Gross income defined. There are also some statutory exclusions from income.[2]

Definition under US Case law

 Main article: Commissioner v. Glenshaw Glass Co.

Income is an "undeniable accessions to wealth, clearly realized, and over which the taxpayer has complete dominion." Commentators say that this is a pretty good definition of income.[2]

Taxable income is usually lower than Haig-Simons income.[2] This is because unrealized appreciation (e.g., the increase in the value of stock over the course of a year) is economic income but not taxable income, and because there are many statutory exclusions from taxable income, including workman's compensation, SSI, gifts, child support, and in-kind government transfers.[5]

Accounting definitions

 Main article: Income (United States legal definitions)

The International Accounting Standards Board (IASB) uses the following definition: "Income is increases in economic benefits during the accounting period in the form of inflows or enhancements of assets or decreases of liabilities that result in increases in equity, other than those relating to contributions from equity participants." [F.70] (IFRS Framework).

Previously the IFRS conceptual framework (4.29) stated: "The definition of income encompasses both revenue and gains. Revenue arises in the course of the ordinary activities of an entity and is referred to by a variety of different names including sales, fees, interest, dividends, royalties and rent. 4.30: Gains represent other items that meet the definition of income and may, or may not, arise in the course of the ordinary activities of an entity. Gains represent increases in economic benefits and as such are no different in nature from revenue. Hence, they are not regarded as constituting a separate element in this Conceptual Framework."[6]

The current IFRS conceptual framework [7] (4.68) no longer draws a distinction between revenue and gains. Nevertheless, the distinction continues to be drawn at the standard and reporting levels. For example, IFRS 9.5.7.1 states: "A gain or loss on a financial asset or financial liability that is measured at fair value shall be recognised in profit or loss ..." while the IASB defined IFRS XBRL taxonomy [8] includes OtherGainsLosses, GainsLossesOnNetMonetaryPosition and similar items.

US GAAP does not define income but does define comprehensive income (CON 8.4.E75): Comprehensive income is the change in equity of a business entity during a period from transactions and other events and circumstances from nonowner sources. It includes all changes in equity during a period except those resulting from investments by owners and distributions to owners.

According to John Hicks' definitions, income "is the maximum amount which can be spent during a period if there is to be an expectation of maintaining intact, the capital value of prospective receipts (in money terms)”.[9]

"Nonincome"

Debt

Borrowing or repaying money is not income under any definition, for either the borrower or the lender.[2] Interest and forgiveness of debt are income.

Psychic income

"Nonmonetary joy," such as watching a sunset or having sex, simply is not income.[2] Similarly, nonmonetary suffering, such as heartbreak or labor, are not negative income. This may seem trivial, but the noninclusion of psychic income has important effects in economics and tax policy.[2] It encourages people to find happiness in nonmonetary, nontaxable ways, and means that reported income may overstate or understate the wellbeing of a given individual.[2]

Income growth

Income per capita has been increasing steadily in most countries.[10] Many factors contribute to people having a higher income, including education,[11] globalisation and favorable political circumstances such as economic freedom and peace. Increases in income also tend to lead to people choosing to work fewer hours. Developed countries (defined as countries with a "developed economy") have higher incomes as opposed to developing countries tending to have lower incomes.

Income inequality

Income inequality is the extent to which income is distributed in an uneven manner. It can be measured by various methods, including the Lorenz curve and the Gini coefficient. Many economists argue that certain amounts of inequality are necessary and desirable but that excessive inequality leads to efficiency problems and social injustice.[1] Thereby necessitating initiatives like the United Nations Sustainable Development Goal 10 aimed at reducing inequality.[12]

National income, measured by statistics such as net national income (NNI), measures the total income of individuals, corporations, and government in the economy. For more information see Measures of national income and output.

Income in philosophy and ethics

Throughout history, many have written about the impact of income on morality and society. Saint Paul wrote 'For the love of money is a root of all kinds of evil:' (1 Timothy 6:10 (ASV)).

Some scholars have come to the conclusion that material progress and prosperity, as manifested in continuous income growth at both the individual and the national level, provide the indispensable foundation for sustaining any kind of morality. This argument was explicitly given by Adam Smith in his Theory of Moral Sentiments,[13] and has more recently been developed by Harvard economist Benjamin Friedman in his book The Moral Consequences of Economic Growth.[14]

Income and health

A landmark systematic review from Harvard University researchers in the Cochrane Collaboration found that income given in the form of unconditional cash transfers leads to reductions in disease, improvements in food security and dietary diversity, increases in children's school attendance, decreases in extreme poverty, and higher health care spending.[15][16]

History

Income is conventionally denoted by "Y" in economics. John Hicks used "I" for income, but Keynes wrote to him in 1937, "after trying both, I believe it is easier to use Y for income and I for investment." Some consider Y as an alternative letter for the phoneme I in languages like Spanish,[17] although Y as the "Greek I" was actually pronounced like the modern German ü or the phonetic /y/.

References

1. ^ a b c d Barr, N. (2004). Problems and definition of measurement. In Economics of the welfare state. New York: Oxford University Press. pp. 121–124
2. McCaffery, Edward (2012). The Oxford Introductions to U.S. Law: Income Tax Law 1st Edition. Oxford University Press.
3. ^ Case, K. & Fair, R. (2007). Principles of Economics. Upper Saddle River, NJ: Pearson Education. p. 54.
4. ^ Staff (2012). "factor income". BusinessDictionary.com. WebFinance, Inc. Archived from the original on 18 June 2012. Retrieved 20 June 2012.
5. ^ Brooks, John R., "The Definitions of Income" (2018). Georgetown Law Faculty Publications and Other Works. 1952. https://scholarship.law.georgetown.edu/facpub/1952/
6. ^ admin. "Conceptual Framework for Financial Reporting 2018". www.iasplus.com. Retrieved 2022-06-28.
7. ^
8. ^
9. ^ "Oxbridge Notes". Retrieved 18 August 2016.
10. ^
11. ^
12. ^ "Goal 10 targets". UNDP. Retrieved 2020-09-23.
13. ^ Smith, Adam (2009). The theory of moral sentiments. Oxford: Clarendon. OCLC 1017407319.
14. ^ Friedman, Benjamin M (2006). The moral consequences of economic growth. New York, NY: Vintage Books. ISBN 978-1-4000-9571-1. OCLC 71353264.
15. ^ Pega, Frank; Liu, Sze; Walter, Stefan; Pabayo, Roman; Saith, Ruhi; Lhachimi, Stefan (2017). "Unconditional cash transfers for reducing poverty and vulnerabilities: effect on use of health services and health outcomes in low- and middle-income countries". Cochrane Database of Systematic Reviews. 11 (4): CD011135. doi:10.1002/14651858.CD011135.pub2. PMC 6486161. PMID 29139110.
16. ^ Pega, Frank; Pabayo, Roman; Benny, Claire; Lee, Eun-Young; Lhachimi, Stefan; Liu, Sze (2022). "Unconditional cash transfers for reducing poverty and vulnerabilities: effect on use of health services and health outcomes in low- and middle-income countries". Cochrane Database of Systematic Reviews. 2022 (3): CD011135. doi:10.1002/14651858.CD011135.pub3. PMC 8962215. PMID 35348196.
17. ^ "Why Y?". Greg Mankiw's Blog. December 21, 2016.


text:
# The sales

The sales tax rate is 4.447% for the city and 4% for the state. Find the total amount paid for 2 boxes of chocolates at $17.96 each. ### Correct answer: t = 38.9542 ### Step-by-step explanation: Did you find an error or inaccuracy? Feel free to write us. Thank you! Tips to related online calculators Our percentage calculator will help you quickly calculate various typical tasks with percentages. #### You need to know the following knowledge to solve this word math problem: ## Related math problems and questions: • Sale A camera has a listed price of$751.98 before tax. If the sales tax rate is 9.25%, find the total cost of the camera with sales tax included.
• Four pupils
Four pupils divided $1485 so that the second received 50% less than the first, the third 1/2 less than a fourth, and fourth$ 154 less than the first. How much money had each of them?
• Find sales tax
What is 8.25% sales tax out of 1607.00?
• Saving 9
An amount of $2000 is invested at an interest of 5% per month. If$ 200 is added at the beginning of each successive month but no withdrawals. Give an expression for the value accumulated after n months. After how many months will the amount has accumula
• Simple interest 4
Find the simple interest if $x USD at$p% for $m days. Assume a$r-day year.
• Loan
If you take a bank loan $10000 and we want to repay after the year, we have to pay the total amount$ 10320/ What is the annual interest rate on this loan?
• Sales tax
A sales tax on a Php 10,800 appliance is Php.1,620. What is the rate of sales tax?
• Simple interest 2
Find the simple interest if $x USD at$p% for $m months. • Sales off If a sweater sells for$ 19 after a 5% markdown, what was its original price?
• Exchange rates
If the Canadian dollar appreciated by C$0.005 relative to the US dollar, what would be the new value of the Canadian dollar per US dollar? Assume the current exchange rate was US$1 = C$0.907. • Loan 4 Mr. Santos borrowed Php.18,000 at 1.5% interest for one year. What was the total amount he paid for the end of the 1-year term? • A man 2 A man divides$10,000 into two investments, one at 10% and the other at 30%. Find how much is invested at each rate so that the two investments produce the same income annually.
• A fisherman
A fisherman buys carnivores to fish. He could buy either 6 larvae and 4 worms for $132 or 4 larvae and 7 worms per$ 127. What is the price of larvae and worms? Argue the answer.
• Associative law multiplication
In a warehouse, you obtain a 20% discount but you must pay a 15% sales tax. Which would you prefer to have calculated first: discount or tax? Explain. (result write as: 1 = first discount, 2 = first tax, 3 = no matter what first)
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Two accounts in the bank, one per year interest 2%, the second 3%. Total interest income 1900 USD. If interest rates were reversed, the yield would be USD 200 higher. What are the amounts on each account?
• Loan
Apply for a $59000 loan, the loan repayment period is 8 years, the interest rate 7%. How much should I pay for every month (or every year if paid yearly). Example is for practise geometric progression and/or periodic payment for an annuity. • Tabitha Tabitha manufactures a product that sells very well. The capacity of her facility is 241,000 units per year. The fixed costs are$122,000 per year, and the variable costs are $11 per unit. The product currently sells for$17. a. What total revenue is requ


text:
# Synopsis: Cross-Country Time Keeping

A new distance record is set in the fiber transmission of stable frequency signals capable of synchronizing atomic clocks.

Next-generation atomic clocks are so precise they can’t be synchronized remotely with traditional communication pathways. Researchers are therefore investigating novel synchronization methods. A new milestone in this development is presented in Physical Review Letters, with the longest distance transmission of a highly stable optical frequency. The signal was sent back and forth across Germany on optical fibers, while keeping a fixed frequency to within a few parts in ${10}^{19}$.

Several applications, such as navigation and fundamental physics, require the comparison of clocks at large physical separation. In geodesy, for example, the time difference between two distant clocks can provide relative elevation measurements with centimeter precision. Currently, clock signals are relayed by satellite communication, but the frequency of these radio signals drifts over time by as much as a few parts per ${10}^{16}$. Higher stability is needed to compare recently developed optical atomic clocks that have precisions on the order of one part in ${10}^{17}$.

Several past experiments have shown that optical fibers can faithfully transmit a clock-synchronizing frequency signal over hundreds of kilometers. Stefan Droste of Max Planck Institute of Quantum Optics, Germany, and his colleagues have now sent a highly stable $194$ terahertz ($1542$ nanometer) frequency over a distance of $1840$ kilometers, doubling their previous record. The team achieved this result by equipping the dedicated optical fiber connecting two German research institutions with active stabilization to overcome frequency shifts from thermal noise and acoustic noise. The method might one day link together optical clocks around the world. – Michael Schirber

### Announcements

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## Previous Synopsis

Particles and Fields

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Atomic and Molecular Physics

## Related Articles

Atomic and Molecular Physics

### Viewpoint: Journey from Classical to Quantum in Two Dimensions

Two separate groups have extracted the thermodynamic equation of state for a two-dimensional gas of fermionic atoms, revealing its peculiar quantum features. Read More »

Quantum Physics

### Synopsis: Tiny Oscillator Works as Photon Changing Room

A new device converts the frequency of a photon using the vibrations in a mechanical oscillator. Read More »

Quantum Information

### Synopsis: All-Around Single-Photon Source

A quantum dot embedded in a micropillar is an efficient source of pure and indistinguishable single photons. Read More »


text:
# Integration: Trigonometric substitution

• Sep 18th 2009, 09:56 PM
Sam1111
Integration: Trigonometric substitution
Hey

The problem i am stuck on is: find the integral of (x^2)/(sqrt(9-x^2))

I have managed to complete the problem to the second to last step where i get (9 theta)/2-9/4 sin(2 theta),

since i substituted at the beginning x=3sin(theta) i can rearrange for theta and get the first part = 9/2sin^-1(x/3).
What I am struggling with is to complete the 9/4sin(2theta) by putting this back in terms of x.
The answer i am looking for is 1/2 x sqrt(9-x^2) so if someone could explain how to get 9/4 sin(2 theta) to 1/2 x sqrt(9-x^2) that be be extremely helpful.

Thanks
• Sep 18th 2009, 10:10 PM
Chris L T521
Quote:

Originally Posted by Sam1111
Hey

The problem i am stuck on is: find the integral of (x^2)/(sqrt(9-x^2))

I have managed to complete the problem to the second to last step where i get (9 theta)/2-9/4 sin(2 theta),

since i substituted at the beginning x=3sin(theta) i can rearrange for theta and get the first part = 9/2sin^-1(x/3).
What I am struggling with is to complete the 9/4sin(2theta) by putting this back in terms of x.
The answer i am looking for is 1/2 x sqrt(9-x^2) so if someone could explain how to get 9/4 sin(2 theta) to 1/2 x sqrt(9-x^2) that be be extremely helpful.

Thanks

$\tfrac{9}{4}\sin\!\left(2\theta\right)=\tfrac{9}{4 }\left(2\sin\theta\cos\theta\right)=\tfrac{9}{2}\s in\theta\cos\theta$. From your substitution, we know that $x=3\sin\theta\implies \sin\theta=\frac{x}{3}$. With this, it is easily shown that $\cos\theta=\frac{\sqrt{9-x^2}}{3}$.

Thus, $\tfrac{9}{2}\sin\theta\cos\theta=\tfrac{9}{2}\left (\tfrac{1}{3}x\cdot\tfrac{1}{3}\sqrt{9-x^2}\right)=\tfrac{1}{2}x\sqrt{9-x^2}$

Does this make sense?


text:
### unt311's blog

By unt311, history, 5 weeks ago,

I've been trying to do this question since many hours and am unable to visualize the authors solution. The question involves maintaining a 1d dp from a 2d dp.

I'm unable to understand what does dp[j] represent in the authors solution after ignoring (i) from dp states ?

• +2

 » 5 weeks ago, # | ← Rev. 2 →   +1 It's not a very clear tutorial, in fairness.The idea is that you have an array of counts (cnt in the author's solution), which after 0 indices has cnt[0] = 1, and cnt[x] = 0 for all other x (up to 10^6). You iterate through each index of the array a, and update the cnt array each time (this is like the first dimension of the DP, except that you aren't storing all the old cnt arrays, you're just updating the new one, to avoid MLE). The second dimension of the DP is handled by looking at all the factors of a[i], and these are stored in a temporary vector (cur). At the end of each iteration, cur is used to update all of the indices of cnt which could possibly have changed, thus avoiding iterating through every index and risking a TLE.
•  » » 5 weeks ago, # ^ | ← Rev. 2 →   0 while updating the dp[j] why do we need to iterate through the divisors(j) in the descending order?
•  » » » 5 weeks ago, # ^ |   0 The reason for that is that when we have a factor j, and we need to update cnt[j], we must add to j the value of cnt[j-1] from the previous iteration since we effectively want to do dp[i][j] += dp[i-1][j-1]. By iterating in descending order, we ensure that we have dp[i-1][j-1], not dp[i][j]
 » 5 weeks ago, # |   0 The basic idea is that $dp_{i}$ = # of good subsequences of length $i$. Then you can iterate through the array, find all of the divisors of $a[i]$ and then update the answer accordingly. Don't forget to increase $dp[1]$ after each iteration! My code: https://codeforces.com/contest/1061/submission/104371163


text:
Corpus ID: 199452940

# Functional probabilistic programming for scalable Bayesian modelling

@article{Law2019FunctionalPP,
title={Functional probabilistic programming for scalable Bayesian modelling},
author={Jonathan Law and Darren J. Wilkinson},
journal={arXiv: Computation},
year={2019}
}
• Published 6 August 2019
• Computer Science, Mathematics
• arXiv: Computation
Bayesian inference involves the specification of a statistical model by a statistician or practitioner, with careful thought about what each parameter represents. This results in particularly interpretable models which can be used to explain relationships present in the observed data. Bayesian models are useful when an experiment has only a small number of observations and in applications where transparency of data driven decisions is important. Traditionally, parameter inference in Bayesian… Expand
1 Citations

#### Figures from this paper

Automatic Backward Filtering Forward Guiding for Markov processes and graphical models
• Mathematics, Computer Science
• 2020
This work backpropagate the information provided by observations through the model to transform the generative (forward) model into a pre-conditional model guided by the data, which approximates the actual conditional model with known likelihood-ratio between the two. Expand

#### References

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An architectural design of a library for Bayesian modelling and inference in modern functional programming languages that enables deterministic testing of inherently stochastic Monte Carlo algorithms is presented and it is demonstrated using OCaml that an expressive module system can also implement the design. Expand
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JAGS is a program for Bayesian Graphical modelling which aims for compatibility with Classic BUGS. The program could eventually be developed as an R package. This article explains the motivations forExpand
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The No-U-Turn Sampler (NUTS), an extension to HMC that eliminates the need to set a number of steps L, and derives a method for adapting the step size parameter {\epsilon} on the fly based on primal-dual averaging. Expand


text:
# Morgantown Algebra Days

## Speakers

### Jürgen Herzog

University of Duisburg-Essen, Germany

### Haydee Lindo

Harvey Mudd College, U.S.
Personal Website

### Linquan Ma

Purdue University, U.S.
Personal Website

### Sarasij Maitra

University of Utah, U.S.
Sarasij Maitra Personal Website

### Takumi Murayama

Purdue University, U.S.
Personal Website

### Rebecca R.G.

George Mason University, U.S.
Personal Website

### Liana Şega

University of Missouri-Kansas City, U.S.
Personal Website

## TALK TITLES AND ABSTRACTS

Talks will be at G15 Life Sciences Building (First Floor). View the Full Schedule.

### JÜrgen herzog

#### Rings of Teter type and the trace of the canonical module and the Teter number of determinantal rings

A 0-dimensional local ring $(R, \mathfrak{m})$ is called a Teter ring, if there exists a local Gorenstein ring $G$ such that $R\simeq G/(0:\mathfrak{m}_G)$. This class of rings was introduced in 1974 by William Teter. It has been shown by Huneke, Vraciu and others that $R$ is a Teter ring if and only if there exists an epimorphism $\varphi: \omega_R\to \mathfrak{m}$. We call $R$ of Teter type, if there exists an epimorphism $\omega_R\to \mbox{tr}(\omega_R)$, and we define the {\em Teter number} of $R$ to be the smallest number $s$ for which there exist $R$-module homomorphisms $\varphi_i: \omega_R\to R$ such that $\mbox{tr}(\omega_R)=\sum_{i=1}^s\varphi_i(\omega_R)$.

Various characterizations of rings of Teter type are given and families of rings of Teter type are presented. We determine the canonical trace of generic determinantal rings and compute its Teter numbers. As an application we determine the canonical trace of a Cohen-Macaulay ring $R$ of codimension two, which is generically Gorenstein. It is shown that if the defining ideal $I$ of $R$ is generated by $n$ elements, then $\mbox{tr}(\omega_R)$ is generated by the (n-2)-minors of the Hilbert-Burch matrix of $I$.

This is joint work with Gasanova, Hibi and Moradi, and with Ficarra, Stamate and Trivedi.

Coming soon...

### Linquan Ma

#### Vanishing and non-negativity of the first normal Hilbert coefficient

Let $(R,m)$ be a Noetherian local ring such that $\widehat{R}$ is reduced. We prove that, when $\widehat{R}$ is $S_2$, if there exists a parameter ideal $Q\subseteq R$ such that the first normal Hilbert-coefficient $\overline{e_1}(Q)$ vanishes, then $R$ is regular and $\nu(m/Q)\leq 1$. This leads to an affirmative answer to a problem raised by Goto-Hong-Mandal. We also give a new proof of the non-negativity of the first normal Hilbert-coefficient (which also shows the non-negativity of the first tight Hilbert-coefficient in positive characteristic).

This is joint work with Pham Hung Quy.

Coming soon...

Coming soon...

Coming soon...

Coming soon...


text:
# Solution of the α-potential mystery in the γ process and its impact on the Nd/Sm ratio in meteorites.

@article{Rauscher2013SolutionOT,
title={Solution of the $\alpha$-potential mystery in the $\gamma$ process and its impact on the Nd/Sm ratio in meteorites.},
author={Thomas Rauscher},
journal={Physical review letters},
year={2013},
volume={111 6},
pages={
061104
}
}
• T. Rauscher
• Published 18 July 2013
• Physics
• Physical review letters
The 146Sm/144Sm ratio in the early solar system has been constrained by Nd/Sm isotope ratios in meteoritic material. Predictions of 146Sm and 144Sm production in the γ process in massive stars are at odds with these constraints, and this is partly due to deficiencies in the prediction of the reaction rates involved. The production ratio depends almost exclusively on the (γ,n)/(γ,α) branching at 148Gd. A measurement of 144Sm(α,γ)148Gd at low energy had discovered considerable discrepancies…
17 Citations

## Figures and Tables from this paper

Measurement of (α,n) reaction crosssections of erbium isotopes for testingastrophysical rate predictions
• Physics
• 2015
The γ-process in core-collapse and/or type Ia supernova explosions is thought to explain the origin of the majority of the so-called p nuclei (the 35 protonrich isotopes between Se and Hg).
Origin of the p-process radionuclides 92Nb and 146Sm in the early solar system and inferences on the birth of the Sun
• Physics
Proceedings of the National Academy of Sciences
• 2016
Radioactive nuclei present in the early solar system are used to derive that at least 10 million years passed between the last of these explosions and the birth of the Sun, lending further support to the idea that the Sun was born in a massive star-forming region together with many thousands of stellar siblings.
Challenges in nucleosynthesis of trans-iron elements
Nucleosynthesis beyond Fe poses additional challenges not encountered when studying astrophysical processes involving light nuclei. Astrophysical sites and conditions are not well known for some of
The production of proton-rich isotopes beyond iron: The γ-process in stars
• Physics
• 2016
Beyond iron, a small fraction of the total abundances in the Solar System is made of proton-rich isotopes, the p-nuclei. The clear understanding of their production is a fundamental challenge for
Determination of the photodisintegration reaction rates involving charged particles: Systematic calculations and proposed measurements based on the facility for Extreme Light Infrastructure–Nuclear Physics
• H. Lan
• Physics
Physical Review C
• 2018
Photodisintegration reaction rates involving charged particles are of relevance to the p-process nucleosynthesis that aims at explaining the production of the stable neutron-deficient nuclides
Accelerator Mass Spectrometry in Laboratory Nuclear Astrophysics
• Physics
• 2016
The extreme sensitivity and discrimination power of accelerator mass spectrometry (AMS) allows for the search and the detection of rare nuclides either in natural samples or produced in the
Uncertainties of α-particle optical potential assessment around and below the Coulomb barrier
• Physics
• 2017
A competition of the low-energy Coulomb excitation (CE) with the compound nucleus (CN) formation in α-induced reactions below the Coulomb barrier has recently been assumed in order to make possible

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Phd by thesis
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text:
# zbMATH — the first resource for mathematics

##### Examples
 Geometry Search for the term Geometry in any field. Queries are case-independent. Funct* Wildcard queries are specified by * (e.g. functions, functorial, etc.). Otherwise the search is exact. "Topological group" Phrases (multi-words) should be set in "straight quotation marks". au: Bourbaki & ti: Algebra Search for author and title. The and-operator & is default and can be omitted. Chebyshev | Tschebyscheff The or-operator | allows to search for Chebyshev or Tschebyscheff. "Quasi* map*" py: 1989 The resulting documents have publication year 1989. so: Eur* J* Mat* Soc* cc: 14 Search for publications in a particular source with a Mathematics Subject Classification code (cc) in 14. "Partial diff* eq*" ! elliptic The not-operator ! eliminates all results containing the word elliptic. dt: b & au: Hilbert The document type is set to books; alternatively: j for journal articles, a for book articles. py: 2000-2015 cc: (94A | 11T) Number ranges are accepted. Terms can be grouped within (parentheses). la: chinese Find documents in a given language. ISO 639-1 language codes can also be used.

##### Operators
 a & b logic and a | b logic or !ab logic not abc* right wildcard "ab c" phrase (ab c) parentheses
##### Fields
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Combinatorics in pure mathematics. (English) Zbl 0851.52017
Graham, R. L. (ed.) et al., Handbook of combinatorics. Vol. 1-2. Amsterdam: Elsevier (North-Holland). 2039-2082 (1995).

The authors present a wide variety of applications of combinatorial reasoning in fields ranging from metric geometry to knot theory. The article consists of seven chapters, described in detail below. The topics covered give just a sampling of the applications of combinatorics in pure mathematics, yet provide a good impression of the breadth of combinatorial techniques that have proved useful in other fields. Definitions, proofs and examples are given sufficient to make the material easily accessible to non-specialists.

The first chapter concerns finite dimensional normed spaces. Arguments from extremal combinatorics are used to prove the existence of subspaces on which the norm is well-approximated by or isometric to the ${\ell }_{1}$ or ${\ell }_{\infty }$ norms, and whose dimensions exceed prescribed bounds. The second treats the problem of characterizing those $n$ by $n$ matrices which arise as matrices of mutual distances between finite sets of points in ${L}_{1}$ spaces. The set of such matrices forms a polyhedral cone ${H}_{n}$, called the Hamming cone, and satisfy so-called hypermetric inequalities. The geometry of the hypermetric cone is related to multi-commodity flows on graphs and lattice-point-free ellipsoids. These connections provide non-trivial results toward a description of ${H}_{n}$. The third section gives an application of matching theory to measure theory, presenting a simple proof of the existence of the Haar measure on a compact topological group using P. Hall’s Marriage Theorem.

The minimal degree of a group action of $G$ on ${\Omega }$ is the smallest number of points moved by a non-trivial element of the group. This is the subject of the fourth chapter. The action $G$ on ${\Omega }×{\Omega }$ is modelled by a collection of digraphs called a coherent configuration. Elementary combinatorial arguments then lead to good lower bounds on the minimal degree.

The fifth section provides a nice introduction to combinatorial methods in commutative algebra. An algebra with straightening law is an algebra $A$ with a specified set of generators and a special kind of normal form algorithm. The algorithm is encoded by a partially-ordered set $P$. Combinatorial properties of $P$, such as shellability, can be used to prove that the algebra is Cohen-Macauley and to construct a particularly nice decomposition of $A$, via the Stanley-Reisner ring of $P$. The method is demonstrated with an analysis of the Grassmann variety in its Plücker embedding.

The next section outlines some of the applications of combinatorial methods in the theory of complex hyperplane arrangements. A finite set of hyperplanes or arrangement in ${C}^{\ell }$ is seen to be a realization of combinatorial structure in the form of a matroid. The matroid controls much of the topology and analytic geometry of the arrangement, allowing application of combinatorial tools such as the broken circuit complex. Included in this section are discussions of the cohomology algebra of the complement of a union of hyperplanes, the homotopy type of the complement and of the singularity link determined by an arrangement, arrangements whose matroids are supersolvable, uniform, or “2-generic”, and an extensive treatment of free arrangements.

The final section exposes the connection between the recently discovered polynomial invariants of knots and links and the classical Tutte polynomial of graph theory. A definition of the Jones polynomial is given in terms of Kauffman’s bracket polynomial on link diagrams, and is proven to be an isotopy invariant of links. In the case of an alternating knot or link, the Jones polynomial is shown to be a specialization of the Tutte polynomial of the graph obtained by 2-coloring the regions of the knot diagram, placing a vertex in each region, and connecting regions of the same color with an edge. The authors present a well-known application of the Jones polynomial in the proof of Tait’s conjecture on the number of crossings in alternating projections of an alternating link.

##### MSC:
 52B40 Matroids (convex geometry) 52C35 Arrangements of points, flats, hyperplanes 57Q45 Knots and links in high dimensions (PL-topology) 05B35 Matroids, geometric lattices (combinatorics)


text:
## Abstract and Applied Analysis

### Positive Solutions for Third-Order $p$-Laplacian Functional Dynamic Equations on Time Scales

#### Abstract

We study the following third-order $p$-Laplacian functional dynamic equation on time scales: ${[{\mathrm{\Phi }}_{p}({u}^{\Delta \nabla }(t))]}^{\nabla }+a(t)f(u(t),u(\mu (t)))=0$, $t\in {(0,T)}_{\mathbf{T}}$,  $u(t)=\phi (t)$,  $t\in {[-r,0]}_{\mathbf{T}}$,  ${u}^{\Delta }(0)={u}^{\Delta \nabla }(T)=0$, and $u(T)+{B}_{0}({u}^{\Delta }(\eta ))=0$. By applying the Five-Functional Fixed Point Theorem, the existence criteria of three positive solutions are established.

#### Article information

Source
Abstr. Appl. Anal., Volume 2014, Special Issue (2014), Article ID 676052, 6 pages.

Dates
First available in Project Euclid: 2 October 2014

https://projecteuclid.org/euclid.aaa/1412278108

Digital Object Identifier
doi:10.1155/2014/676052

Mathematical Reviews number (MathSciNet)
MR3248871

Zentralblatt MATH identifier
1343.34209

#### Citation

Guan, Wen; Wang, Da-Bin. Positive Solutions for Third-Order $p$ -Laplacian Functional Dynamic Equations on Time Scales. Abstr. Appl. Anal. 2014, Special Issue (2014), Article ID 676052, 6 pages. doi:10.1155/2014/676052. https://projecteuclid.org/euclid.aaa/1412278108

#### References

• D. Anderson, R. Avery, and J. Henderson, “Existence of solutions for a one dimensional $p$-Laplacian on time-scales,” Journal of Difference Equations and Applications, vol. 10, no. 10, pp. 889–896, 2004.
• D. R. Anderson, “Existence of solutions for a first-order $p$-Laplacian BVP on time scales,” Nonlinear Analysis: Theory, Methods & Applications, vol. 69, no. 12, pp. 4521–4525, 2008.
• L. Bian, X. He, and H. Sun, “Multiple positive solutions of $m$-point BVPs for third-order $p$-Laplacian dynamic equations on time scales,” Advances in Difference Equations, vol. 2009, Article ID 262857, 12 pages, 2009.
• A. Cabada, “Existence results for $p$-Laplacian boundary value problems on time scales,” Advances in Difference Equations, vol. 11, Article ID 21819, 2006.
• A. Cabada, “Discontinuous functional $p$-Laplacian boundary value problems on time scales,” International Journal of Difference Equations, vol. 2, no. 1, pp. 51–60, 2007.
• W. Guan, “Three positive solutions for $p$-Laplacian functional dynamic equations on time scales,” Electronic Journal of Qualitative Theory of Differential Equations, No. 28, 7 pages, 2008.
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# Can we construct a Hilbert space where the operator $A_u v := -\frac{1}{2} v'' + (vF + v\int_\mathbb{R} Su + u\int_\mathbb{R} Sv )'$ is symmetric?

It seems not to be a easy problem.

I'd like to know if one can define a pertinent Hilbert space where the operator $$A_p v := -\frac{1}{2} v^{\prime\prime} + (vF + v\int_\mathbb{R} Sp + p\int_\mathbb{R} Sv )^\prime$$ is symmetric. Here, $p$ satisfies the following differential equation with $p \in L^2_1(\mathbb{S})\cap C^\infty(\mathbb{S})$ $$\frac{p^\prime}{2p} = F + \int Sp$$ and $u,v$ and $F,S$ can be taken to satisfy at least the following weak regularity conditions:

• $u,v \in L^2_0(\mathbb{R})\cap C^2(\mathbb{R})$
• $F,S \in C^\infty(\mathbb{R})$ and odd

but the regularity will probably be restricted much more by choosing the nice Hilbert space that we wish. We note that $\int_\mathbb{R} Sp= \int_\mathbb{R} S(x) p(x) dx$.

My motivation for a possible positive answer for this question comes from the following similar problem. Given a fixed $p \in L^2_1(\mathbb{S})\cap C^\infty(\mathbb{S})$ ($\mathbb{S} = \mathbb{R}/(2\pi\mathbb{Z})$ and $p \neq 0$), solution of $\frac{p^\prime}{2p}= S*p$ ($S(\theta) = \sin(\theta)$) for, the operator

$$B_p v := -\frac{1}{2} v^{\prime\prime} + ( v\int_\mathbb{S} S*p + p\int_\mathbb{S} S*v )^\prime$$

a such Hilbert space is $H_ {-1,1/p}$ provided with $\left\langle f,g\right\rangle_{H_ {-1,1/p}} := \int \frac{\mathcal{F}\mathcal{G}}{p}$ where $\mathcal{F}$ and $\mathcal{G}$ are respectively the primitives in $L^2(\mathbb{R})$ of $f$ and $g$ such that $\int \frac{\mathcal{F}} {p} = \int \frac{\mathcal{G}} {p} =0$ . Here $H_ {-1,1/p}:= H_{1,p}^\prime$ (notation:$V^\prime$ is the dual space of $V$), $H_{1,p}:=${$\overline{ f \in C^1(\mathbb{S}): \int f =0 } ^H$} (which is also a Hilbert space with scalar product $\left\langle f,g\right\rangle_{H_ {1,p}} := \int (f^\prime g^\prime p)$ ) and $H =L^2 _0 (\mathbb{S})$.

We can prove for this case that $$\left\langle f,B_pg\right\rangle_{H_ {-1,1/p}} = \left\langle B_pf,g\right\rangle_{H_ {-1,1/p}}=-\frac{1}{2}\int\frac{fg}{p}+\int fS^\prime *g$$ then $B_p$ is clearly symetric in $H_ {-1,1/p}$

I expected had well motivated my question. I've tried several integral and derivative restrictions under spaces similar to $H_ {-1,1/p}$ but I've not been successful yet. I'd be glad for some advice. If you visit the topic please leave a message with your opinion.

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Please clarify the ambiguity in the last integrals: Do you mean $\int(Su)$ or $(\int S)u$? –  Dirk Jun 18 '12 at 6:31
There's no more ambiguity in the text. I mean $\int (Su)$. –  Paul Jun 18 '12 at 21:35
Crossposted to MO: mathoverflow.net/questions/99907/… –  user16299 Jun 19 '12 at 2:34