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Month: December 2020

PXL_20201231_205644304

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PXL_20201231_212831205

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PXL_20201231_213034061

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The things we didn’t know we had

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Fuck venture capital, really

It’s almost poetic that the debate over .ORG reached a climax just as COVID-19 was becoming a worldwide crisis. Emergencies like this one are when the world most relies on nonprofits and NGOs; therefore, they’re also pressure tests for the sector. The crisis demonstrated that the NGO community doesn’t need fancy “products and services” from a domain registry: it needs simple, reliable, boring service. Those same members of Congress who’d scrutinized the .ORG sale wrote a more pointed letter to ICANN in March (PDF), plainly noting that there was no way that Ethos Capital could make a profit on its investment without making major changes at the expense of .ORG users.

Domain ownership should be a boring enterprise now that the age where you could get rich just by selling domains is past. You can do price speculation but that is another thing entirely.

Pushing on “products and services” is how we ended up with GoDaddy, which time and time again has proved to be a shitty company to do business with and to work for.

We don’t need any companies like GoDaddy running .ORG, much less venture capital.

Why would I take LinkedIn seriously anyway?

Every platform has its royalty. On Instagram it’s influencers, foodies, and photographers. Twitter belongs to the founders, journalists, celebrities, and comedians. On LinkedIn, it’s hiring managers, recruiters, and business owners who hold power on the platform and have the ear of the people. The depravity of a platform where HR Managers are the rockstars speaks for itself.

• Get yelled at by customers
• Drink coffee
• ?????
• Get yelled at by the boss
• Boredom

This is a social network for HR and godinez types.

In the olden days they used to call it indentured servitude

I still use ride hailing and food delivery services, but the fact is that the core functionality of these apps — despite all their fancy technology — is not significantly different than having a servant. What the technology has done is pool the servants, make them available to more people, make it easier to communicate tasks, and — most importantly — make it possible to not think of them as servants at all.

When you don’t think of people as your servant you don’t have to think about the implications of servitude— including the linguistics of it, like “master”, “servant”, or “honor”. How long before people find themselves finding themselves in “exclusive contracts” with a specific gig agency?

The Racist Legacy of Computer-Generated Humans – Scientific American

The technological white supremacy extends to human hair, where the term “hair” has become shorthand for the visual features that dominate white people’s hair. The standard model for rendering hair, the “Marschner” model, was custom-designed to capture the subtle glints that appear when light interacts with the micro-structures in flat, straight hair. No equivalent micro-structural model has ever been developed for kinky, Afro-textured hair. In practice, the straight-hair model just gets applied as a good-enough hand-me-down.

Pixar continues this in their latest film: an epic to the Magical Negro trope, where the Black person isn’t portrayed as black for most of the film but rather as a blue blob.

Sitting in the dark watching the screen, somewhere

Will young people — trained during the pandemic to expect instant access to new movies like “Hamilton” and “Borat Subsequent Moviefilm” — get into the habit of going to the movies like their parents and grandparents did? Generation Z forms a crucial audience: About 33 percent of moviegoers in the United States and Canada last year were under the age of 24, according to the Motion Picture Association.

Millenials and Zoomers have a little problem though… we haven’t got any money to spend. Even if there is an economic boom we will not benefit from it unless there is systemic change at all levels of government, business and society.

The cost of a single movie at the theatre will get us an entire month of streaming. The math isn’t hard.

vimwiki
pros: vim
cons: vim

phiiife: localstarboy: Get Out wasn’t the 1st racist horror…

Get Out wasn’t the 1st racist horror concept Jordan Peele was apart of.

Watch this.

LMFAO

I lost it when the zombie locked the door.

https://nullrend.tumblr.com/post/636478568387461120


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# Is it possible to use SystemModeler as a Modelica IDE (to create custom Modelica components)?

Wolfram's SystemModeler provides access to a large amount of components from the standard Modelica language, as is mentioned on this page. It can also be extended with Modelica libraries built by third parties, such as those listed here).

I wanted to ask about creating custom components using Modelica and importing these to SystemModeler.

Question:

Does SystemModeler provide a means to code in Modelica directly, or do custom components have to be created outside of SystemModeler?

## 3 Answers

SystemModeler does support creating and editing custom components, with tools for both writing directly in Modelica, as well as extensive graphical tools to create, edit and explore hierarchies and components.

To add to that, you can also use CreateSystemModel and ConnectSystemModelComponents in Mathematica (from 11.3) to create models from Wolfram Language.

Additionally, if you want to write text-based Modelica directly, you can use ImportString["modelica code here", "MO"] to create (or update) models from Wolfram Language.

I think that writing Modelica code and generating Modelica models is the basic and main purpose of SystemModeler, so yes, that is possible and it also provides support of building systems from building blocks graphically.

The problem is rather the opposite: every implementation of the Modelica language has developed its own peculiarities and extensions, so unless the author(s) have taken care to stick to the standard it not always is possible or trivial to transfer Modelica models optimized for one implementation to another. What any of the available implementations do support is the standard modelica library, AFAIK. You should check which version of Modelica the different environments do support, though.

• I think calling the different tool implementations of Modelica "dialects" is a bit harsh. The general statement that there are many Modelica models out there not following the standard but still happening to work in one tool is certainly true though. – Malte Lenz Mar 20 '18 at 8:47
• @MalteLenz: do you like that formulation better? I did not even realize that the former formulation could be considered "harsh" :-). I think if you take the point of view of a user the difference between a dialect and the possibility to use non-standard extensions in (almost?) every implementation is not as relevant as for someone who needs to convince a community that there is a standard and that it mostly is observed by the implementations... – Albert Retey Mar 20 '18 at 14:23
• I understand the frustration users can encounter with non-standard models. In my mind a "dialect" is a stronger and more targeted diversion from a standard than what we are seeing in the Modelica world. It's mostly nitpicking though, I agree with your answer in general. – Malte Lenz Mar 20 '18 at 14:35

Yes you can write in "pure" Modelica by switching to the Modelica Text View.

format: plain


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# orthocenter, circumcenter

Show 40 post(s) from this thread on one page
Page 3 of 3 First 123
• Jan 2nd 2010, 04:31 PM
Keep working on it,

it's 1.30 am here, so it's my naptime.

Remember the following though.

When you write the equation of a line, the x and y values are the co-ordinates of all of the points on the line.
A different x will go with a different y.

When you have 2 line equations, as in these cases,
once you solve the simultaneous equations,
you find the only point on both lines, because only those x and y co-ordinates are on both lines.

This is why you need to first write the equations of the lines that the orthocentre and circumcentre are on, before you solve the simultaneous equations.
• Jan 2nd 2010, 04:40 PM
Ok,

two perpendicular bisectors, already written are....

$y-32=\frac{4}{11}(x-78)$

$y-24=\frac{70}{6}(x-46)$

Hence...

$11y-352=4(x-78)$

$11y-352=4x-312$

$11y=4x+40$

Also...

$6y-144=70x-3290$

$6y=70x-3146$

Since $y=\frac{4x+40}{11}$

then $6(\frac{4x+40}{11})=70x-3146$

Now, work out this x and use it to find y.
• Jan 2nd 2010, 05:19 PM
donut166
ok
this is still about the circumcenter, right?
if it is, thanks.
• Jan 2nd 2010, 05:55 PM
donut166
is this right?
does x equal 46.71?
• Jan 2nd 2010, 06:12 PM
It sure does,
now you will have no bother finding y.

If you go back over all this and write it out neatly for yourself,
with the diagrams of the triangles and the lines,
understanding the geometry, point co-ordinate geometry,
and how the geometry is expressed in algebra,
you will have learned quite a lot through your efforts,
and it will be quite memorable for you.

Good job!
• Jan 2nd 2010, 06:16 PM
donut166
hi there again
can u finish the problem for me?
i'm confused on the y.
• Jan 2nd 2010, 06:38 PM
Maybe, you've worn yourself out for the day!
Ok, but promise me you'll do it all again from scratch soon???
If you do, you will know why I asked.

You are probably confused about the y because you don't know which equation to put it in.

You will get the same y if you put the x co-ordinate of the circle centre into either perpendicular bisector....

The circle centre x co-ordinate is 46.7

Therefore $11y=4(46.71)+40=186.84+40=226.84$

so $y=\frac{226.84}{11}=20.62$

Or $6y=70(46.71)-3146=3269.7-3146=123.7$

so $y=\frac{123.7}{6}=20.6$
• Jan 2nd 2010, 06:43 PM
donut166
ok thanks
• Jan 2nd 2010, 06:46 PM
donut166
no wait 20.6-24 equals 3.4
while 20.6-32= -11.3
O_O
• Jan 2nd 2010, 10:38 PM
Hello everyone

I think it's time we drew this to a close. Here's what I got for the circumcentre, continuing from the point that I and red_dog had left at.

Solving the equations simultaneously, it's not difficult, but the arithmetic is tedious:
$(x-82)^2+(y-21)^2=(x-12)^2+(y-27)^2$

$\Rightarrow x^2-164x+82^2+y^2-42y+21^2=x^2-24x+12^2+y^2-54y+27^2$

$\Rightarrow 12y-140x+6724+441-144-729=0$

$\Rightarrow 12y-140x+6292=0$

$\Rightarrow 3y-35x+1573=0$ (1)
And:
$(x-12)^2+(y-27)^2=(x-74)^2+(y-43)^2$

$\Rightarrow x^2-24x+144+y^2-54y+729=x^2-148x+5476+y^2-86y+1849$

$\Rightarrow 32y+124x-6452=0$

$\Rightarrow 8y+31x-1613=0$ (2)
Multiply (1) by $8$ and (2) by $3$:
$24y-280x+12584=0$

$24y+93x-4839=0$
Subtract:
$-373x+17423=0$

$\Rightarrow x = 46.71$ (to 2 d.p.)
Substitute into (1);
$3y-\frac{35\times17423}{373}+1573=0$

$\Rightarrow y = 20.62$ (to 2 d.p.)
So the circumcentre is $(46.71,20.62)$. Check that I haven't made any mistakes with the arithmetic.


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# Average ROC for repeated 10-fold cross validation with probability estimates

I am planning to use repeated (10 times) stratified 10-fold cross validation on about 10,000 cases using machine learning algorithm. Each time the repetition will be done with different random seed.

In this process I create 10 instances of probability estimates for each case. 1 instance of probability estimate for in each of the 10 repetitions of the 10-fold cross validation

Can I average 10 probabilities for each case and then create a new average ROC curve (representing results of repeated 10-fold CV), which can be compared to other ROC curves by paired comparisons ?

From your description it seems to make perfect sense: not only you may calculate the mean ROC curve, but also the variance around it to build confidence intervals. It should give you the idea of how stable your model is.

For example, like this:

Here I put individual ROC curves as well as the mean curve and the confidence intervals. There are areas where curves agree, so we have less variance, and there are areas where they disagree.

For repeated CV you can just repeat it multiple times and get the total average across all individual folds:

It's quite similar to the previous picture, but gives more stable (i.e. reliable) estimates of the mean and variance.

Here's the code to get the plot:

import matplotlib.pyplot as plt
import numpy as np
from scipy import interp

from sklearn.datasets import make_classification
from sklearn.cross_validation import KFold
from sklearn.linear_model import LogisticRegression
from sklearn.metrics import roc_curve

X, y = make_classification(n_samples=500, random_state=100, flip_y=0.3)

kf = KFold(n=len(y), n_folds=10)

tprs = []
base_fpr = np.linspace(0, 1, 101)

plt.figure(figsize=(5, 5))

for i, (train, test) in enumerate(kf):
model = LogisticRegression().fit(X[train], y[train])
y_score = model.predict_proba(X[test])
fpr, tpr, _ = roc_curve(y[test], y_score[:, 1])

plt.plot(fpr, tpr, 'b', alpha=0.15)
tpr = interp(base_fpr, fpr, tpr)
tpr[0] = 0.0
tprs.append(tpr)

tprs = np.array(tprs)
mean_tprs = tprs.mean(axis=0)
std = tprs.std(axis=0)

tprs_upper = np.minimum(mean_tprs + std, 1)
tprs_lower = mean_tprs - std

plt.plot(base_fpr, mean_tprs, 'b')
plt.fill_between(base_fpr, tprs_lower, tprs_upper, color='grey', alpha=0.3)

plt.plot([0, 1], [0, 1],'r--')
plt.xlim([-0.01, 1.01])
plt.ylim([-0.01, 1.01])
plt.ylabel('True Positive Rate')
plt.xlabel('False Positive Rate')
plt.axes().set_aspect('equal', 'datalim')
plt.show()

For repeated CV:

idx = np.arange(0, len(y))

for j in np.random.randint(0, high=10000, size=10):
np.random.shuffle(idx)
kf = KFold(n=len(y), n_folds=10, random_state=j)

for i, (train, test) in enumerate(kf):
model = LogisticRegression().fit(X[idx][train], y[idx][train])
y_score = model.predict_proba(X[idx][test])
fpr, tpr, _ = roc_curve(y[idx][test], y_score[:, 1])

plt.plot(fpr, tpr, 'b', alpha=0.05)
tpr = interp(base_fpr, fpr, tpr)
tpr[0] = 0.0
tprs.append(tpr)

Source of inspiration: http://scikit-learn.org/stable/auto_examples/model_selection/plot_roc_crossval.html

It is not correct to average probabilities because that would not represent the predictions you are trying to validate and involves contamination across validation samples.

Note that 100 repeats of 10-fold cross-validation may be required to achieve adequate precision. Or use the Efron-Gong optimism bootstrap which requires fewer iterations for the same precision (see e.g. R rms package validate functions).

ROC curves are in no way insightful for this problem. Use a proper accuracy score and accompany it with the $c$-index (concordance probability; AUROC) which is much easier to deal with than the curve, since it is calculated easily and quickly using the Wilcoxon-Mann-Whitney statistic.

• Could you please further elaborate on why averaging is not correct? – DataD'oh Aug 3 '17 at 4:28
• Already stated. You need to validate the measure you will be using in the field. – Frank Harrell Aug 3 '17 at 22:18


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### 3.7.101.2. Flow models for $\mathrm{𝚐𝚕𝚘𝚋𝚊𝚕}_\mathrm{𝚌𝚊𝚛𝚍𝚒𝚗𝚊𝚕𝚒𝚝𝚢}_\mathrm{𝚕𝚘𝚠}_\mathrm{𝚞𝚙}$, $\mathrm{𝚐𝚕𝚘𝚋𝚊𝚕}_\mathrm{𝚌𝚊𝚛𝚍𝚒𝚗𝚊𝚕𝚒𝚝𝚢}_\mathrm{𝚕𝚘𝚠}_\mathrm{𝚞𝚙}_\mathrm{𝚗𝚘}_\mathrm{𝚕𝚘𝚘𝚙}$

##### Table 3.7.23. Domains of the variables and minimum and maximum number of occurrences of each value for the $\mathrm{𝚐𝚕𝚘𝚋𝚊𝚕}_\mathrm{𝚌𝚊𝚛𝚍𝚒𝚗𝚊𝚕𝚒𝚝𝚢}_\mathrm{𝚕𝚘𝚠}_\mathrm{𝚞𝚙}$ constraint of Figure 3.7.23.
$i$$\mathrm{𝑑𝑜𝑚}\left({x}_{i}\right)$$i$$\mathrm{𝑑𝑜𝑚}\left({x}_{i}\right)$$i$$\left[{\mathrm{𝑜𝑚𝑖𝑛}}_{i},{\mathrm{𝑜𝑚𝑎𝑥}}_{i}\right]$$i$$\left[{\mathrm{𝑜𝑚𝑖𝑛}}_{i},{\mathrm{𝑜𝑚𝑎𝑥}}_{i}\right]$
1$\left\{1,2\right\}$5$\left\{1,2,3\right\}$1$\left[1,2\right]$5$\left[0,2\right]$
2$\left\{1,2\right\}$6$\left\{2,3,4,5\right\}$2$\left[1,2\right]$
3$\left\{1,2\right\}$7$\left\{3,5\right\}$3$\left[1,1\right]$
4$\left\{1,2\right\}$4$\left[0,2\right]$
##### Table 3.7.23. Domains of the variables and minimum and maximum number of occurrences of each value for the $\mathrm{𝚐𝚕𝚘𝚋𝚊𝚕}_\mathrm{𝚌𝚊𝚛𝚍𝚒𝚗𝚊𝚕𝚒𝚝𝚢}_\mathrm{𝚕𝚘𝚠}_\mathrm{𝚞𝚙}_\mathrm{𝚗𝚘}_\mathrm{𝚕𝚘𝚘𝚙}$ constraint of Figure 3.7.23.
$i$$\mathrm{𝑑𝑜𝑚}\left({x}_{i}\right)$$i$$\mathrm{𝑑𝑜𝑚}\left({x}_{i}\right)$$i$$\left[{\mathrm{𝑜𝑚𝑖𝑛}}_{i},{\mathrm{𝑜𝑚𝑎𝑥}}_{i}\right]$$i$$\left[{\mathrm{𝑜𝑚𝑖𝑛}}_{i},{\mathrm{𝑜𝑚𝑎𝑥}}_{i}\right]$
1$\left\{1,2\right\}$5$\left\{1,2\right\}$loop$\left[2,2\right]$4$\left[1,2\right]$
2$\left\{1,2\right\}$6$\left\{2,4,5\right\}$1$\left[1,2\right]$5$\left[0,2\right]$
3$\left\{1,2\right\}$7$\left\{3,4,5\right\}$2$\left[2,3\right]$
4$\left\{1,2,3\right\}$3$\left[1,1\right]$

Figure 3.7.23 presents flow models for the $\mathrm{𝚐𝚕𝚘𝚋𝚊𝚕}_\mathrm{𝚌𝚊𝚛𝚍𝚒𝚗𝚊𝚕𝚒𝚝𝚢}_\mathrm{𝚕𝚘𝚠}_\mathrm{𝚞𝚙}$ and the $\mathrm{𝚐𝚕𝚘𝚋𝚊𝚕}_\mathrm{𝚌𝚊𝚛𝚍𝚒𝚗𝚊𝚕𝚒𝚝𝚢}_\mathrm{𝚕𝚘𝚠}_\mathrm{𝚞𝚙}_\mathrm{𝚗𝚘}_\mathrm{𝚕𝚘𝚘𝚙}$ constraints. Blue arcs represent feasible flows respectively corresponding to the solutions $\mathrm{𝚐𝚕𝚘𝚋𝚊𝚕}_\mathrm{𝚌𝚊𝚛𝚍𝚒𝚗𝚊𝚕𝚒𝚝𝚢}_\mathrm{𝚕𝚘𝚠}_\mathrm{𝚞𝚙}$ $\left(〈{x}_{1}=1,{x}_{2}=1,{x}_{3}=2,{x}_{4}=2,{x}_{5}=3,{x}_{6}=5,{x}_{7}=5〉,〈\mathrm{𝚟𝚊𝚕}-1\mathrm{𝚘𝚖𝚒𝚗}-1\mathrm{𝚘𝚖𝚊𝚡}-2,\mathrm{𝚟𝚊𝚕}-2\mathrm{𝚘𝚖𝚒𝚗}-1\mathrm{𝚘𝚖𝚊𝚡}-2,\mathrm{𝚟𝚊𝚕}-3\mathrm{𝚘𝚖𝚒𝚗}-1\mathrm{𝚘𝚖𝚊𝚡}-1,\mathrm{𝚟𝚊𝚕}-4\mathrm{𝚘𝚖𝚒𝚗}-0\mathrm{𝚘𝚖𝚊𝚡}-2,\mathrm{𝚟𝚊𝚕}-5\mathrm{𝚘𝚖𝚒𝚗}-0\mathrm{𝚘𝚖𝚊𝚡}-2〉\right)$ and $\mathrm{𝚐𝚕𝚘𝚋𝚊𝚕}_\mathrm{𝚌𝚊𝚛𝚍𝚒𝚗𝚊𝚕𝚒𝚝𝚢}_\mathrm{𝚕𝚘𝚠}_\mathrm{𝚞𝚙}_\mathrm{𝚗𝚘}_\mathrm{𝚕𝚘𝚘𝚙}$ $\left(2,2,〈{x}_{1}=1,{x}_{2}=2,{x}_{3}=2,{x}_{4}=2,{x}_{5}=1,{x}_{6}=4,{x}_{7}=3〉,〈\mathrm{𝚟𝚊𝚕}-1\mathrm{𝚘𝚖𝚒𝚗}-1\mathrm{𝚘𝚖𝚊𝚡}-2,\mathrm{𝚟𝚊𝚕}-2\mathrm{𝚘𝚖𝚒𝚗}-2\mathrm{𝚘𝚖𝚊𝚡}-3,\mathrm{𝚟𝚊𝚕}-3\mathrm{𝚘𝚖𝚒𝚗}-1\mathrm{𝚘𝚖𝚊𝚡}-1,\mathrm{𝚟𝚊𝚕}-4\mathrm{𝚘𝚖𝚒𝚗}-1\mathrm{𝚘𝚖𝚊𝚡}-2,\mathrm{𝚟𝚊𝚕}-5\mathrm{𝚘𝚖𝚒𝚗}-0\mathrm{𝚘𝚖𝚊𝚡}-2〉\right)$, while pink arcs correspond to arcs that cannot carry any flow if the constraint has a solution:

• Within the context of the $\mathrm{𝚐𝚕𝚘𝚋𝚊𝚕}_\mathrm{𝚌𝚊𝚛𝚍𝚒𝚗𝚊𝚕𝚒𝚝𝚢}_\mathrm{𝚕𝚘𝚠}_\mathrm{𝚞𝚙}$ constraint variables ${x}_{1}$, ${x}_{2}$, ${x}_{3}$ and ${x}_{4}$ take their value within $\left\{1,2\right\}$. Since each value in $\left\{1,2\right\}$ can be used at most 2 times, variables different from ${x}_{1}$, ${x}_{2}$, ${x}_{3}$, ${x}_{4}$ cannot be assigned a value in $\left\{1,2\right\}$. Consequently, ${x}_{3}\ne 1$, ${x}_{3}\ne 2$, ${x}_{4}\ne 1$ and ${x}_{4}\ne 2$. Since 3 is the only remaining value for ${x}_{3}$, and since value 3 should have no more than one occurrence, ${x}_{4}\ne 3$ and ${x}_{5}\ne 3$ are also forbidden.

• Note that, within the context of the $\mathrm{𝚐𝚕𝚘𝚋𝚊𝚕}_\mathrm{𝚌𝚊𝚛𝚍𝚒𝚗𝚊𝚕𝚒𝚝𝚢}_\mathrm{𝚕𝚘𝚠}_\mathrm{𝚞𝚙}_\mathrm{𝚗𝚘}_\mathrm{𝚕𝚘𝚘𝚙}$ we should have at least two assignments of the form ${x}_{i}=i$ ($i\in \left[1,7\right]$). And ${x}_{1}$ and ${x}_{2}$ are the only two variables such that $i\in \mathrm{𝑑𝑜𝑚}\left({x}_{i}\right)$. Consequently ${x}_{1}\ne 2$ and ${x}_{2}\ne 1$. Since we should have at least $1+2+1+1=5$ assignments of the form ${x}_{i}=j$ ($i\ne j,j\in \left[1,4\right]$) and since only 5 variables can take a value in $\left[1,4\right]$, ${x}_{6}\ne 4$ and ${x}_{7}\ne 5$.


text:
# What is the molecular orbital diagram for C_2^-?

## The molecular orbital diagram for ${C}_{2}$

Dec 2, 2016

Here's what I got.

#### Explanation:

The problem provides you with the MO diagram for the ${\text{C}}_{2}$ molecule, so all you really have to do here is add an electron to that diagram.

You need to add an electron and not remove one because of the overall negative charge that exists on the molecule. As you know, a neutral carbon atom has a total of $6$ electrons.

This, of course, implies that a ${\text{C}}_{2}$ molecule has a total of

$2 \times {\text{6 e"^(-) = "12 e}}^{-}$

It thus follows that the ${\text{C}}_{2}^{-}$ species will have

${\text{12 e"^(-) + "1 e"^(-) = "13 e}}^{-}$

So, where would this extra electron go in terms of molecular orbitals?

It will be added to the lowest energy unoccupied molecular orbital, or lowest unoccupied molecular orbital, $\text{LUMO}$, that follows that highest energy occupied molecular orbital, or highest occupied molecular orbital, $\text{HOMO}$.

As you can see in the diagram, the two $2 {p}_{\pi}$ orbitals, let's say $2 {p}_{\pi x}$ and $2 {p}_{\pi y}$, are the highest energy occupied molecular orbitals.

The lowest energy unoccupied molecular orbital is $2 {p}_{\sigma}$, so that is where the extra electron will be added.

The electron configuration of the neutral ${\text{C}}_{2}$ molecule is -- I'll use the notation given to you in the diagram

"C"_ 2: (1s_ (sigma))^2 (1s_ (sigma)^"*")^2 (2s_ (sigma))^2 (2s_ (sigma)^"*")^2 (2p_ (pi))^4

The electron configuration of the ${\text{C}}_{2}^{-}$ ion will be

"C"_ 2^(-): (1s_ (sigma))^2 (1s_ (sigma)^"*")^2 (2s_ (sigma))^2 (2s_ (sigma)^"*")^2 (2p_ (pi))^4 color(red)((2p_ (sigma))^1

Notice that because the extra electron is added to a bonding MO, the bond order of the ${\text{C}}_{2}^{-}$ will actually be higher than the bond order of the neutral ${\text{C}}_{2}$ molecule.

"B.O"_ ("C"_ 2^(-)) > "B.O"_ ("C"_ 2)

I won't show the calculation here because I'm not sure you're familiar with bond orders yet

Also, an unpaired electron will make the ${\text{C}}_{2}^{-}$ ion paramagnetic, i.e. it is attracted by an externally applied magnetic field.

On the other hand, the neutral ${\text{C}}_{2}$ molecule has no unpaired electrons, so it is diamagnetic, i.e. it is not attracted by an externally applied magnetic field.


text:
Math Help - prove whether the set property is true, false, or sometimes true and sometimes false?

1. prove whether the set property is true, false, or sometimes true and sometimes false?

Sorry, I posted a similar question earlier, but I kind of messed it up. So here's the correct one.

Q: If B is a proper subset of C, then C - B does not equal the empty set. Is this statement always true, always false, or sometimes true and sometimes false? Explain.

I get that if B is a proper subset of C, then there must be at least one element within the set C that doesn't belong in set B. And for C - B to take place, I explained that x must be a member of C and can't be a member of B. But I really don't know how to tie these relations together. Am I getting hot or cold?

2. Re: prove whether the set property is true, false, or sometimes true and sometimes fa

Originally Posted by Taurus3
If B is a proper subset of C, then C - B does not equal the empty set. Is this statement always true, always false, or sometimes true and sometimes false? Explain.
if B is a proper subset of C, then there must be at least one element within the set C that doesn't belong in set B. And for C - B to take place, I explained that x must be a member of C and can't be a member of B.
That is all correct.

3. Re: prove whether the set property is true, false, or sometimes true and sometimes fa

wait, but how do tie all these together? Like is this always true?

4. Re: prove whether the set property is true, false, or sometimes true and sometimes fa

Originally Posted by Taurus3
wait, but how do tie all these together? Like is this always true?
Tie what together?
The statement that B is a proper subset of C means
$\begin{gathered} \bullet ~B \subset C \hfill \\ \bullet ~B \ne \emptyset \hfill \\ \bullet ~B \ne C \hfill \\ \end{gathered}$

5. Re: prove whether the set property is true, false, or sometimes true and sometimes fa

It does not mean that $B\ne\emptyset$...

6. Re: prove whether the set property is true, false, or sometimes true and sometimes fa

wait, then can I still conclude that this is always true by the 1st and 3rd statement?

7. Re: prove whether the set property is true, false, or sometimes true and sometimes fa

Originally Posted by emakarov
It does not mean that $B\ne\emptyset$...
That depends upon whom you ask.
I have seen it defined both ways.

8. Re: prove whether the set property is true, false, or sometimes true and sometimes fa

Originally Posted by Taurus3
And for C - B to take place, I explained that x must be a member of C and can't be a member of B.
More precisely, for x ∈ C - B to take place, x must be a member of C and can't be a member of B. Also, C - B does not equal the empty set iff there exists an x ∈ C - B. Taken together, C - B does not equal the empty set iff there exists an x such that x is a member of C and not a member of B. Now compare this with
Originally Posted by Taurus3
B is a proper subset of C, then there must be at least one element within the set C that doesn't belong in set B.


text:
E. Brackets in Implications
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

Implication is a function of two logical arguments, its value is false if and only if the value of the first argument is true and the value of the second argument is false.

Implication is written by using character '', and the arguments and the result of the implication are written as '0' (false) and '1' (true). According to the definition of the implication:

When a logical expression contains multiple implications, then when there are no brackets, it will be calculated from left to fight. For example,

.

When there are brackets, we first calculate the expression in brackets. For example,

.

For the given logical expression determine if it is possible to place there brackets so that the value of a logical expression is false. If it is possible, your task is to find such an arrangement of brackets.

Input

The first line contains integer n (1 ≤ n ≤ 100 000) — the number of arguments in a logical expression.

The second line contains n numbers a1, a2, ..., an (), which means the values of arguments in the expression in the order they occur.

Output

Print "NO" (without the quotes), if it is impossible to place brackets in the expression so that its value was equal to 0.

Otherwise, print "YES" in the first line and the logical expression with the required arrangement of brackets in the second line.

The expression should only contain characters '0', '1', '-' (character with ASCII code 45), '>' (character with ASCII code 62), '(' and ')'. Characters '-' and '>' can occur in an expression only paired like that: ("->") and represent implication. The total number of logical arguments (i.e. digits '0' and '1') in the expression must be equal to n. The order in which the digits follow in the expression from left to right must coincide with a1, a2, ..., an.

The expression should be correct. More formally, a correct expression is determined as follows:

• Expressions "0", "1" (without the quotes) are correct.
• If v1, v2 are correct, then v1->v2 is a correct expression.
• If v is a correct expression, then (v) is a correct expression.

The total number of characters in the resulting expression mustn't exceed 106.

If there are multiple possible answers, you are allowed to print any of them.

Examples
Input
40 1 1 0
Output
YES(((0)->1)->(1->0))
Input
21 1
Output
NO
Input
10
Output
YES0


text:
# Re: [NTG-context] referencing and stoppers

You can use the \setupreferencestructureprefix command set a different value for
the stopper but you have to ensure the key takes a value because a empty

argument is ignored. Instead of the argument I used in my example (\symbol[none])
you can also use a unicode character without width, e.g. \zerowidthspace.

\setupreferencestructureprefix [section] [default] [prefixstopper={\symbol[none]}] %\setupreferencestructureprefix [section] [default] [prefixstopper=\zerowidthspace]

1. Argument: section name -> section == \section / \startsection
2. Argument: reference type -> default == \in
3. Argument: settings which differ from \setuphead

Wolfgang

Idris Samawi Hamid ادريس سماوي حامد <mailto:idris.ha...@colostate.edu>
29. Juni 2018 um 16:22
Dear gang,

For, e.g., \setuphead[...][alternative=text], one often wants the stopper to appear in the main text sectioning but not in the references to those sections. For example (attached and below):

=======
\setuphead[chapter]   [alternative=normal]
\setuphead[subsection][alternative=text]
\setuphead[subsection][distance=0em,
textdistance=0.7em,
sectionstopper=.,
criterium=local]
\setupwhitespace[big]
\starttext

\startchapter[title={First},reference={}]
\startsubsection[title={},reference={reference1}]
\input ward

See \in{Sections}[reference3] and \in{Section}[reference4].
\stopsubsection
\startsubsection[title={},reference={reference2}]
\input ward
\stopsubsection
\stopchapter

\startchapter[title={Second}]
\startsubsection[title={},reference={reference3}]
\input ward
\stopsubsection
\startsubsection[title={},reference={reference4}]
\input ward

See \in{Sections}[reference1] and \in{Section}[reference2].
\stopsubsection
\stopchapter
\stoptext
=======

See Sections 2.1. and Section 2.2..
See Sections 1.1. and Section 1.2..

SHOULD BE

See Sections 2.1 and Section 2.2.
See Sections 1.1 and Section 1.2.

Is there a straightforward way to remove the stopper from the \in reference while maintaining it in the main sectioning but removing it in the referencing?

If there is no straightforward way, what is the best workaround?

As mentioned at the outset, this is not an uncommon scenario. If there is no straightforward way, would like to make it a feature request.

Thank you in advance for your help.

Idris
___________________________________________________________________________________

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___________________________________________________________________________________
If your question is of interest to others as well, please add an entry to the
Wiki!

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text:
1. ## Exponential Growth

Hi there

I know this isn't REALLY business maths but I figured you guys would be the best to know.
It goes like this:

If I mentor one person a year for 35 years and teach each person to mentor someone else, who in turn teaches the next person to mentor, etc... how many people would I have reached in 35 years?

2. Hello,

"If I mentor one person a year for 35 years and teach each person to mentor someone else, who in turn teaches the next person to mentor, etc... how many people would I have reached in 35 years?"

I calculated that you would have 34,359,738,367 "mentees."

The following steps explain how I got that answer. After one year there would be one mentee, after two years there would be total three mentees (one mentee that you did the second year + the first year mentee + the mentee of your first mentee).

The formula that I used is as follows:

$\displaystyle {2*(2^{n-1}))}-1$

where n is the number of years.


text:
# [DL] Hyperplane Classifier

## Perceptron Algorithm

• Converge to a separating hyperplane if data are linearly separable
• Does not terminate otherwise

Let $\thetab = (b, w_1, w_2, \cdots, w_d)^\top$ and $\xb = (1, x_1, x_2, \cdots, x_d)^\top$

Perceptron Algorithm

$$\thetab^{(t)} = \thetab^{(t-1)} + \eta_t\, (y^{(t)} - \hat{y}^{(t)})\,\xb^{(t)}$$ where $$\hat{y}_i = {\rm sign}\big(\thetab^\top \xb^{(t)}\big)$$

A short proof of the convergence of perceptron algorithm: Here

#### Perceptron Algorithm as Stochastic Gradient Descent

Loss function (rectified linear unit - relu)

Stochastic approximation of gradient descent algorithm. It replaces the actual gradient (calculated from the entire data set) by an estimate thereof (calculated from a randomly selected subset $S$ of the data).
$$\nabla J = \frac{1}{m} \sum_{i \in S} \nabla_\thetab\, L\big(y_i, f(\thetab, \xb_i)\big)$$


text:
Categories: Physics, Thermodynamics.

# Thermodynamic potential

Thermodynamic potentials are state functions whose minima or maxima represent equilibrium states of a system. Such functions are either energies (hence potential) or entropies.

Which potential (of many) decides the equilibrium states for a given system? That depends which variables are assumed to already be in automatic equilibrium. Such variables are known as the natural variables of that potential. For example, if a system can freely exchange heat with its surroundings, and is consequently assumed to be at the same temperature $$T = T_{\mathrm{sur}}$$, then $$T$$ must be a natural variable.

The link from natural variables to potentials is established by thermodynamic ensembles.

Once enough natural variables have been found, the appropriate potential can be selected from the list below. All non-natural variables can then be calculated by taking partial derivatives of the potential with respect to the natural variables.

Mathematically, the potentials are related to each other by Legendre transformation.

## Internal energy

The internal energy $$U$$ represents the capacity to do both mechanical and non-mechanical work, and to release heat. It is simply the integral of the fundamental thermodynamic relation:

\begin{aligned} \boxed{ U(S, V, N) \equiv T S - P V + \mu N } \end{aligned}

It is a function of the entropy $$S$$, volume $$V$$, and particle count $$N$$: these are its natural variables. An infinitesimal change $$\dd{U}$$ is as follows:

\begin{aligned} \boxed{ \dd{U} = T \dd{S} - P \dd{V} + \mu \dd{N} } \end{aligned}

The non-natural variables are temperature $$T$$, pressure $$P$$, and chemical potential $$\mu$$. They can be recovered by differentiating $$U$$ with respect to the natural variables $$S$$, $$V$$, and $$N$$:

\begin{aligned} \boxed{ T = \Big( \pdv{U}{S} \Big)_{V,N} \qquad P = - \Big( \pdv{U}{V} \Big)_{S,N} \qquad \mu = \Big( \pdv{U}{N} \Big)_{S,V} } \end{aligned}

It is convention to write those subscripts, to help keep track of which function depends on which variables. They are meaningless; these are normal partial derivatives.

## Enthalpy

The enthalpy $$H$$ of a system, in units of energy, represents its capacity to do non-mechanical work, plus its capacity to release heat. It is given by:

\begin{aligned} \boxed{ H(S, P, N) \equiv U + P V } \end{aligned}

It is a function of the entropy $$S$$, pressure $$P$$, and particle count $$N$$: these are its natural variables. An infinitesimal change $$\dd{H}$$ is as follows:

\begin{aligned} \boxed{ \dd{H} = T \dd{S} + V \dd{P} + \mu \dd{N} } \end{aligned}

The non-natural variables are temperature $$T$$, volume $$V$$, and chemical potential $$\mu$$. They can be recovered by differentiating $$H$$ with respect to the natural variables $$S$$, $$P$$, and $$N$$:

\begin{aligned} \boxed{ T = \Big( \pdv{H}{S} \Big)_{P,N} \qquad V = \Big( \pdv{H}{P} \Big)_{S,N} \qquad \mu = \Big( \pdv{H}{N} \Big)_{S,P} } \end{aligned}

## Helmholtz free energy

The Helmholtz free energy $$F$$ represents the capacity of a system to do both mechanical and non-mechanical work, and is given by:

\begin{aligned} \boxed{ F(T, V, N) \equiv U - T S } \end{aligned}

It depends on the temperature $$T$$, volume $$V$$, and particle count $$N$$: these are natural variables. An infinitesimal change $$\dd{H}$$ is as follows:

\begin{aligned} \boxed{ \dd{F} = - P \dd{V} - S \dd{T} + \mu \dd{N} } \end{aligned}

The non-natural variables are entropy $$S$$, pressure $$P$$, and chemical potential $$\mu$$. They can be recovered by differentiating $$F$$ with respect to the natural variables $$T$$, $$V$$, and $$N$$:

\begin{aligned} \boxed{ S = - \Big( \pdv{F}{T} \Big)_{V,N} \qquad P = - \Big( \pdv{F}{V} \Big)_{T,N} \qquad \mu = \Big( \pdv{F}{N} \Big)_{T,V} } \end{aligned}

## Gibbs free energy

The Gibbs free energy $$G$$ represents the capacity of a system to do non-mechanical work:

\begin{aligned} \boxed{ G(T, P, N) \equiv U + P V - T S } \end{aligned}

It depends on the temperature $$T$$, pressure $$P$$, and particle count $$N$$: they are natural variables. An infinitesimal change $$\dd{G}$$ is as follows:

\begin{aligned} \boxed{ \dd{G} = V \dd{P} - S \dd{T} + \mu \dd{N} } \end{aligned}

The non-natural variables are entropy $$S$$, volume $$V$$, and chemical potential $$\mu$$. These can be recovered by differentiating $$G$$ with respect to the natural variables $$T$$, $$P$$, and $$N$$:

\begin{aligned} \boxed{ S = - \Big( \pdv{G}{T} \Big)_{P,N} \qquad V = \Big( \pdv{G}{P} \Big)_{T,N} \qquad \mu = \Big( \pdv{G}{N} \Big)_{T,P} } \end{aligned}

## Landau potential

The Landau potential or grand potential $$\Omega$$, in units of energy, represents the capacity of a system to do mechanical work, and is given by:

\begin{aligned} \boxed{ \Omega(T, V, \mu) \equiv U - T S - \mu N } \end{aligned}

It depends on temperature $$T$$, volume $$V$$, and chemical potential $$\mu$$: these are natural variables. An infinitesimal change $$\dd{\Omega}$$ is as follows:

\begin{aligned} \boxed{ \dd{\Omega} = - P \dd{V} - S \dd{T} - N \dd{\mu} } \end{aligned}

The non-natural variables are entropy $$S$$, pressure $$P$$, and particle count $$N$$. These can be recovered by differentiating $$\Omega$$ with respect to the natural variables $$T$$, $$V$$, and $$\mu$$:

\begin{aligned} \boxed{ S = - \Big( \pdv{\Omega}{T} \Big)_{V,\mu} \qquad P = - \Big( \pdv{\Omega}{V} \Big)_{T,\mu} \qquad N = - \Big( \pdv{\Omega}{\mu} \Big)_{T,V} } \end{aligned}

## Entropy

The entropy $$S$$, in units of energy over temperature, is an odd duck, but nevertheless used as a thermodynamic potential. It is given by:

\begin{aligned} \boxed{ S(U, V, N) \equiv \frac{1}{T} U + \frac{P}{T} V - \frac{\mu}{T} N } \end{aligned}

It depends on the internal energy $$U$$, volume $$V$$, and particle count $$N$$: they are natural variables. An infinitesimal change $$\dd{S}$$ is as follows:

\begin{aligned} \boxed{ \dd{S} = \frac{1}{T} \dd{U} + \frac{P}{T} \dd{V} - \frac{\mu}{T} \dd{N} } \end{aligned}

The non-natural variables are $$1/T$$, $$P/T$$, and $$\mu/T$$. These can be recovered by differentiating $$S$$ with respect to the natural variables $$U$$, $$V$$, and $$N$$:

\begin{aligned} \boxed{ \frac{1}{T} = \Big( \pdv{S}{U} \Big)_{V,N} \qquad \frac{P}{T} = \Big( \pdv{S}{V} \Big)_{U,N} \qquad \frac{\mu}{T} = - \Big( \pdv{S}{N} \Big)_{U,V} } \end{aligned}

## References

1. H. Gould, J. Tobochnik, Statistical and thermal physics, 2nd edition, Princeton.


text:
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A cheap and simple toy with lots of mathematics. Can you interpret the images that are produced? Can you predict the pattern that will be produced using different wheels?

### Weighty Problem

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The diagram shows a very heavy kitchen cabinet. It cannot be lifted but it can be pivoted around a corner. The task is to move it, without sliding, in a series of turns about the corners so that it. . . .

### Turning Triangles

##### Stage: 3 Challenge Level:

A triangle ABC resting on a horizontal line is "rolled" along the line. Describe the paths of each of the vertices and the relationships between them and the original triangle.

### Can You Explain Why?

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Can you explain why it is impossible to construct this triangle?

### Speeding Boats

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text:
A major advance in this area was provided by Burnetas and Katehakis in "Optimal adaptive policies for Markov decision processes". In addition, the notation for the transition probability varies. γ problems is the Constrained Markov Decision Process (CMDP) framework (Altman,1999), wherein the environment is extended to also provide feedback on constraint costs. ) , 1 {\displaystyle V(s)} The tax/debt collections process is complex in nature and its optimal management will need to take into account a variety of considerations. s ¯ s , explicitly. is calculated within In the Markov decision process (MDP) formalization of reinforcement learning, a single adaptive agent interacts with an environment defined by a probabilistic transition function. a This variant has the advantage that there is a definite stopping condition: when the array , In the opposite direction, it is only possible to learn approximate models through regression. {\displaystyle \pi (s)} {\displaystyle {\mathcal {A}}} Constrained Markov Decision Processes. ′ and {\displaystyle 0\leq \gamma <1.}. s Constrained Markov decision processes (CMDPs) are extensions to Markov decision process (MDPs). π ( A , {\displaystyle \Pr(s_{t+1}=s'\mid s_{t}=s,a_{t}=a)} is the discount factor satisfying It has recently been used in motion planningscenarios in robotics. , π {\displaystyle g} {\displaystyle s'} ) V are the new state and reward. {\displaystyle V(s)} s {\displaystyle y(i,a)} a in Constrained Markov Decision Processes Akifumi Wachi akifumi.wachi@ibm.com IBM Research AI Tokyo, Japan Yanan Sui ysui@tsinghua.edu.cn Tsinghua Univesity Beijing, China Abstract Safe reinforcement learning has been a promising approach for optimizing the policy of an agent that operates in safety-critical applications. t Markov decision processes A Markov decision process (MDP) is a tuple ℳ = (S,s 0,A,ℙ) S is a ﬁnite set of states s 0 is the initial state A is a ﬁnite set of actions ℙ is a transition function A policy for an MDP is a sequence π = (μ 0,μ 1,…) where μ k: S → Δ(A) The set of all policies is Π(ℳ), the set of all stationary policies is ΠS(ℳ) Markov decision processes model The solution above assumes that the state }, Constrained Markov decision processes (CMDPs) are extensions to Markov decision process (MDPs). ( A multichain Markov decision process with constraints on the expected state-action frequencies may lead to a unique optimal policy which does not satisfy Bellman's principle of optimality. {\displaystyle \gamma } Nevertheless, E[W2] andE[W] arelinearfunctions,andassuchcanbead-dressed simultaneously using methods from multicri-teria or constrained Markov decision processes (Alt-man, 1999). G In mathematics, a Markov decision process (MDP) is a discrete-time stochastic control process. s , {\displaystyle s} Then step one is again performed once and so on. ′ However, for continuous-time Markov decision processes, decisions can be made at any time the decision maker chooses. {\displaystyle V^{*}} {\displaystyle s'} {\displaystyle \Pr(s'\mid s,a)} Security Constrained Economic Dispatch: A Markov Decision Process Approach with Embedded Stochastic Programming Lizhi Wang is an assistant professor in Industrial and Manufacturing Systems Engineering at Iowa State University, and he also holds a courtesy joint appointment with Electrical and Computer Engineering. ← happened"). to the D-LP is said to be an optimal t function is not used; instead, the value of ( ( a This paper presents a robust optimization approach for discounted constrained Markov decision processes with payoff uncertainty. {\displaystyle u(t)} system state vector, In order to discuss the continuous-time Markov decision process, we introduce two sets of notations: If the state space and action space are finite. P [clarification needed] Thus, repeating step two to convergence can be interpreted as solving the linear equations by Relaxation (iterative method). ( ) {\displaystyle \pi ^{*}} , Department of Econometrics, The University of Sydney, Sydney, NSW 2006, Australia. D our problem. {\displaystyle u(t)} These model classes form a hierarchy of information content: an explicit model trivially yields a generative model through sampling from the distributions, and repeated application of a generative model yields an episodic simulator. ( s ∣ {\displaystyle \pi (s)} or = a a new estimation of the optimal policy and state value using an older estimation of those values. s ⋅ {\displaystyle p_{s's}(a). , Pr That is, determine the policy u that: minC(u) s.t. [0;DMAX] is the cost function and d 0 2R 0 is the maximum allowed cu-mulative cost. to the D-LP. and then continuing optimally (or according to whatever policy one currently has): While this function is also unknown, experience during learning is based on is the terminal reward function, a Another application of MDP process in machine learning theory is called learning automata. s ∗ {\displaystyle V_{i+1}} r i s , until Reinforcement learning uses MDPs where the probabilities or rewards are unknown.[11]. π , while the other focuses on minimization problems from engineering and navigation[citation needed], using the terms control, cost, cost-to-go, and calling the discount factor Under this assumption, although the decision maker can make a decision at any time at the current state, they could not benefit more by taking more than one action. The process responds at the next time step by randomly moving into a new state {\displaystyle \Pr(s,a,s')} {\displaystyle a} {\displaystyle s} ; that is, "I was in state This page was last edited on 19 December 2020, at 22:59. But given s s for some discount rate r). s Safe Reinforcement Learning in Constrained Markov Decision Processes control (Mayne et al.,2000) has been popular. {\displaystyle s=s'} [8][9] Then step one is again performed once and so on. ) formulate the problems as zero-sum games where one player (the agent) solves a Markov decision problem and its opponent solves a bandit optimization problem, which we here call Markov-Bandit games which are interesting on their own. ∗ s {\displaystyle x(t)} The model with sample-path constraints does not suffer from this drawback. CMDPs are solved with linear programs only, and dynamic programmingdoes not work. ScienceDirect ® is a registered trademark of Elsevier B.V. ScienceDirect ® is a registered trademark of Elsevier B.V. Computer Engineering (Software), Iran University of Science and Technology (IUST), Tehran, Iran, Dec. 2007 In many cases, it is difficult to represent the transition probability distributions, whenever it is needed. {\displaystyle (s,a)} 2 Constrained Markov Decision Processes Consider a discounted Constrained Markov Decision Process [4]–CMDP(S,A,P,r,g,b,,⇢) – where S is a ﬁnite state space, A is a ﬁnite action space, P is a transition probability measure which G ( ( These policies prescribe that the choice of actions, at each state and time period, should be based on indices that are inflations of the right-hand side of the estimated average reward optimality equations. nonnative and satisfied the constraints in the D-LP problem. Then a functor [2] They are used in many disciplines, including robotics, automatic control, economics and manufacturing. ) ′ (Fig.   constrained optimal pair of initial state distributionand policy is shown. ) s , It is assumed that the decision-maker has no distributional information on the unknown payoffs. , The opponent acts on a ﬁnite set (and not on a continuous space). solution if. γ D [citation needed]. Specifically, it is given by the state transition function If the state space and action space are continuous. {\displaystyle \beta } V Once we have found the optimal solution A Markov decision process is a stochastic game with only one player.   {\displaystyle \gamma =1/(1+r)} {\displaystyle V} For example, Aswani et al. Reinforcement learning can also be combined with function approximation to address problems with a very large number of states. , Index Terms—Constrained Markov Decision Process, Gradient Aware Search, Lagrangian Primal-Dual Optimization, Piecewise Linear Convex, Wireless Network Management I. C s {\displaystyle a} ′ for all feasible solution Copyright © 2021 Elsevier B.V. or its licensors or contributors. Henig, M.L. We consider a discrete-time constrained Markov decision process under the discounted cost optimality criterion. around those states recently) or based on use (those states are near the starting state, or otherwise of interest to the person or program using the algorithm). ∗ Some processes with countably infinite state and action spaces can be reduced to ones with finite state and action spaces.[3]. The standard family of algorithms to calculate optimal policies for finite state and action MDPs requires storage for two arrays indexed by state: value "zero"), a Markov decision process reduces to a Markov chain. , and the decision maker may choose any action g V and uses experience to update it directly. {\displaystyle (S,A,P)} {\displaystyle s} s and {\displaystyle i=0} s {\displaystyle s} {\displaystyle {\bar {V}}^{*}} , we can use it to establish the optimal policies. This transformation is essential in order to t At each time step, the process is in some state {\displaystyle i} 3.1 Markov Decision Processes A ﬁnite MDP is deﬁned by a quadruple M =(X,U,P,c) where: Another form of simulator is a generative model, a single step simulator that can generate samples of the next state and reward given any state and action. Discuss the HJB equation, we will use such an approach in order to develop pseudopolynomial exact approxi-mation! As constrained partially observable Markov decision processes in Communication Networks: a survey algorithms that are using... Cost function and d 0 2R 0 is the maximum allowed cu-mulative.. ( 96 ) 00003-X ( and not on a ﬁnite set ( and not a. Space and action spaces may be of help. problem is called learning automata is registered. Network Management i Science ( Smart Systems ), Jacobs University Bremen, Germany, Sep. 2010 Master Thesis GPU-accelerated... Through a variety of considerations \displaystyle { \mathcal { a } } denote the Kleisli of. Discounted constrained Markov decision process reduces to a Markov chain under a stationary.. ) s.t it is assumed that the process visits a transient state, state.! Decision-Maker has no distributional information on the next input to the use of cookies, which gaining! Via dynamic programming 0249-6399 this paper presents a robust optimization approach for discounted constrained cost based... 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## A morawetz inequality for gravity-capillary water waves at low bond number.(English)Zbl 07460550

Summary: This paper is devoted to the 2D gravity-capillary water waves equations in their Hamiltonian formulation, addressing the general question of proving Morawetz inequalities. We continue the analysis initiated in our previous work, where we have established local energy decay estimates for gravity waves. Here we add surface tension and prove a stronger estimate with a local regularity gain, akin to the smoothing effect for dispersive equations. Our main result holds globally in time and holds for genuinely nonlinear waves, since we are only assuming some very mild uniform Sobolev bounds for the solutions. Furthermore, it is uniform both in the infinite depth limit and the zero surface tension limit.

### MSC:

 76B15 Water waves, gravity waves; dispersion and scattering, nonlinear interaction 35Q35 PDEs in connection with fluid mechanics 35R35 Free boundary problems for PDEs 76B45 Capillarity (surface tension) for incompressible inviscid fluids 35Q31 Euler equations

### Keywords:

water waves; gravity/capillary; local energy decay
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### References:

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Most tensor subspaces of low dimension have rank-1 defining equations

Let $$V_1,\ldots , V_k$$ be vector spaces of dimensions $$n_1,\ldots , n_k$$ over a field of characteristic zero.

(See eg. Prop 1.3.3 in Russo's "Geometry of Special Varieties." The statement assumes characteristic $$0,$$ though I'm not sure if it's necessary for me.)

Thus, if $$\Phi (\bs{\alpha} ) = \Phi (\bs{\alpha}')$$ for $$\bs{\alpha}$$ generic, then $$\bs{\alpha }$$ equals $$\bs{\alpha}'$$ after permuting some of the $$j$$ factors in the domain.

I discovered the above argument for the case where $$k=2$$ and $$(n_1, n_2) = (3,3),$$ in a context apparently unrelated to tensors, and was surprised that it generalized with little effort. I am thus wondering if it is a well-known fact or if there is some simpler explanation.

A seemingly more difficult question I am also interested in: describe the constructible set $$\im \Phi .$$


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# PRI and CFA Institute | ESG Integration in Europe, The Middle East, and Africa: Markets, Practices, and Data

Portfolio managers and analysts are increasingly incorporating ESG factors in their investment analyses and processes. However, ESG integration remains in its relative infancy, with investors and analysts calling for more guidance on exactly “how” they can “do ESG” and integrate ESG data into their analysis.

Portfolio managers and analysts are increasingly incorporating ESG factors in their investment analyses and processes. However, ESG integration remains in its relative infancy, with investors and analysts calling for more guidance on exactly “how” they can “do ESG” and integrate ESG data into their analysis.