text:
## Foundations of randomness, Day 2

Back to the fundamentals of randomness and our workshop in Stellenbosch – I want to discuss the remaining two days of talks!

The second day started off with a whirlwind tour of the TCS approach to pseudo-randomness by David Zuckerman. David surveyed known constructions of pseudo-random generators and extractors; he also gave a very comprehensive overview of the different types of sources that have been considered in this area and what kind of extraction (deterministic, seeded, two-source) is known to be possible (or not) for each type of source.

I’ll refer you to David’s slides for details. Two particular sources caught my attention. First David mentioned small-space sources, for which there are good deterministic extractors but so far only in the regime of high (polynomial) min-entropy. This type of source seems quite relevant in practice, and it is natural to ask whether it is possible to do better using device-independent quantum extraction. How could we leverage the fact that the source is generated using bounded space? A second interesting category of sources are non-oblivious bit-fixing sources, which are sources such that some of the bits are uniformly random, and others deterministic, but such that which bits are of which type can be chosen adversarially by the source based on the values taken by previous bits. Together with Santha-Vazirani sources these are the two examples David gave of natural sources from which deterministic extraction is not possible. But we do know that it is possible to extract from a single Santha-Vazirani source, of bias arbitrarily close to ${1/2}$, using a device-independent quantum procedure. So what about non-oblivious bit-fixing sources — can it be done there as well?

Work on device-independent randomness extraction from the quantum information community has so far focused on a small set of sources — uniform seeds of limited length, arbitrary Santha-Vazirani sources, and more recently general min-entropy sources — but I see no reason why these are the only cases of interest; the restricted focus seems to be mostly for historical reasons. In pseudo-randomness different types of sources are often motivated by applications to derandomization, lower bounds, or combinatorial constructions. In quantum information, rather than “beating the classical” we should focus on the scenario that are best motivated by the relevant applications, which so far come mostly from cryptography. Here device-independent randomness certification procedures seem most relevant in their role as “decouplers” (see the discussion on “random to whom” from Valerio’s talk). What kind of structure are we willing to assume on the kind of information an adversary may have kept on a particular source? This question seems particularly relevant in light of recent works on device-independent two-source extraction.

David’s talk was followed with a talk by one of his students, Eshan Chattopadhyay, on their recent breakthrough construction of a two-source extractor for poly-logarithmic min-entropy. It is hard to do any justice to such an intricate (and beautiful!) result, and instead I will point you to Eshan’s slides or the great talk that David gave on the same topic on TCS+ just a few weeks ago.

For the last talk of the morning session Henry Yuen treated us to an (arguably :-)) even more formidable treat: infinite randomness expansion! Henry presented the main steps that led him and Matthew Coudron to their beautiful result showing how, starting from just a few uniformly random bits, one could bootstrap a quantum device-independent expansion procedure and generate as many uniformly random bits as one desires — yes, that’s an ${\infty}$! (I realize a bit late that I never properly explained what “device-independent” even means… if I have any “classical” readers left, for context for Henry’s talk see a blog post of Henry’s for all the required background; for the quantum crypto motivation see the viewpoint by Roger Colbeck; for the computer scientist perhaps the introduction to this paper is readable.)

In Henry and Matt’s protocol the length of the initial seed only needs to be at least some universal constant. The number of random bits available to start with will govern the security parameter (the distance from uniform) of the bits produced, but aside from that any small initial seed will do. The protocol uses two pairs of devices (those can be arbitrarily correlated, e.g. share randomness or quantum entanglement, but are not allowed to communicate directly) which take turns in expanding the current string of bits by an exponential factor. The key observation required to argue that the protocol works is that each expansion step, not only increases the number of bits available, but simultaneously “purifies” them, in the sense that the bits produced will have high entropy even conditioned on any information, classical or quantum, in the possession of the other pair of devices. Given that this other pair generated the input bits used for the expansion in the first place this is not a trivial observation, and is  essentially the content of the powerful “equivalence lemma” of Chung, Shi and Wu discussed in the post describing Valerio Scarani’s talk. The lemma guarantees that the bits produced by a device-independent procedure are random even from the point of view of an adversary who shares arbitrary correlations with the seed (including knowing it perfectly).

The result is quite impressive: if a few uniformly random bits are available, then it is automatically guaranteed that infinitely many just-as-good (in some respects even better) random bits can be generated! So much for our derandomization problems… The next step is to ask if uniformly random bits are even needed: how about weaker sources of randomness, such as Santha-Vazirani sources or general min-entropy sources? The task of device-independent extraction from the former was taken on by Colbeck and Renner (with much follow-up work); for the latter there is the work of Chung, Shi and Wu, which unfortunately still requires a number of devices that scales with the length of the source as well as the security parameter. More to be done!

The afternoon brought two more talks, with a very different focus but equally stimulating. The first was delivered by Ruediger Schack, who presented “The QBist perspective on randomness and laws of nature”. Ruediger’s talk followed the one by Carl Hoefer on the previous day in challenging our attitudes towards randomness and the meaning of probabilities, but Ruediger took us in quite a different direction. One of the points he made is that Bell’s theorem can be interpreted as giving us the following mutually exclusive alternatives about the world: either locality does not hold, and we must accept that far-away objects can have a nonlocal (instantaneous) influences on one another, or we insist on keeping locality — as Einstein did — but ascribe a different “meaning” to probabilistic statements, or predictions, that arise from the Born rule. (Note that in the first option in “nonlocal influence” I am including the kind of non-signaling “influence” which arises from measuring half of an entangled pair.)

This is where QBism — for “Quantum Bayenism” — comes in. According to Ruediger, the solution it offers to this conundrum goes through the assertion that, contrary to e.g. the Copenhagen interpretation, “probabilities are not determined by real properties”. This allows one to “restore locality” while not having recourse to local hidden variables (as Bohmian mechanics would) either. Thus in QBism “there is no Born rule”, in the sense that the rule does not describe an intrinsic probabilistic fact about the world; rather, probabilities are seen as the reflection of an agent’s subjective belief and have no objective existence. The fact that, if Alice measures her half of an EPR pair in the computational basis and obtains a ${|0\rangle}$, then Bob will with certainty observe a ${|0\rangle}$ as well, if he measures in the same basis, is not a “fact” about their systems; rather it is a prescription for how a rational Bob should update his belief as to the outcomes of an experiment he could perform, were he to be informed of the situation at Alice’s side.

The main conceptual tool used to formulate this interpretation of probabilities is the Dutch book used as a guide to how an agent (observer) should update its probabilities. I admit I find this interpretation a bit hard to digest — I cannot help but think of betting as a highly irrational procedure, and in general I would be hard-pressed to bet on any precise number or probability; the Born rule makes very precise predictions and ascribing them to an agent’s subjective belief feels like a bit of a stretch. Quite possibly it is also a problem of language, and I am not an expert on QBism (even the “usual” Bayesianism is far from obvious to me). Ruediger’s slides are very clear and give a good introduction for anyone interested in digging further.

The last talk of the day was given by Carl Miller, on “The extremes of quantum random number generation”. Carl explained how two very different principles enabled him and Yaoyun Shi to derive two independent lower bounds, each of interest in a separate regime, on the amount of randomness that can be generated from a pair of quantum devices in the sequential scenario: the devices are used repeatedly in sequence and one is interested in the min-entropy rate of one of the device’s outputs, as a function of the average observed violation of (in this case) the CHSH inequality: as the violation increases to the maximum the rate will improve to 1, and the question is how high a rate can be maintained is the violation is be bounded away from the optimum.

Carl’s first principle of measurement disturbance leads to a good rate in the high error regime, i.e. when the observed violation deviates significantly from optimum, say it is around ${0.8}$ (for an optimum of ${\approx .85}$). The general idea is that a violation of the CHSH inequality necessarily implies the use of “non-classical” measurements, which do not commute (in fact a maximal violation requires anti-commuting measurements), by Alice’s device. Two non-commuting measurements do not have a common eigenstate, hence whatever the state of the device at least one of Alice’s measurements will perturb it — the outcome of that measurement on the state cannot be deterministic, and randomness is produced.

Of course this is just the intuition, and making it quantitative is more challenging. In their paper Carl and Yaoyun formalize it by introducing a new uncertainty relation (see this very recent survey on the topic for much more) expressed in terms of Renyi ${\alpha}$-entropy for ${\alpha > 1}$. The proof relies on strong convexity of the underlying Schatten-normed space of matrices, and Carl gave some geometric intuition for the uncertainty relation. The use of ${\alpha}$-Renyi entropy is important for the inductive step; it allows to bypass the lack of good chain-rule like arguments for the min-entropy.

The second principle introduced in Carl’s talk is self-testing. This is the idea that a device demonstrating a close-to-optimal violation of the CHSH inequality must have a certain rigid structure: not only does it generate random bits, but the device’s outcomes must be produced in a very specific way. Specifically the device’s state and measurements must be equivalent, up to local isometries, to those used in the “canonical” optimal strategy for the CHSH game. This principle is well-known (see e.g. the paper by McKague for a proof), but usually one only expects it to be of use in the “tiny error” regime, i.e. when the observed violation is of order ${opt - \varepsilon}$ for small epsilon. Carl manages to squeeze the techniques further, and by carefully fine-tuning the argument is able to get good bounds on the randomness produced when the violation is large but still a constant away from optimum, say ${0.83}$ or so. In that regime the bound is better than the one obtained from the first principle; I’ll refer you to Carl’s slides for the precise curves.

After Carl’s talk, the mind full of new ideas we left for a little walk to the Lanzerac wine estate, where we were shown around the wine cellar — including the mandatory wine
tasting of course, with the highlight being the local pinotage — and finished off the day with a nice dinner on the terrasse: it’s late October in South Africa, summer is coming!


text:
# Coordinate spaces and transformation matrices

I'm trying to get an object from object space, into projected space using these intermediate matrices:

The first matrix (I) is the one that transforms from object space into inertial space, but since my object is not rotated or translated in any way inside the object space, this matrix is the 4x4 identity matrix.

The second matrix (W) is the one that transforms from inertial space into world space, which is just a scale transform matrix of factor a = 14.1 on all coordinates, since the inertial space origin coincides with the world space origin.

    /a 0 0 0\
W = |0 a 0 0|
|0 0 a 0|
\0 0 0 1/

The third matrix (C) is the one that transforms from world space, into camera space. This matrix is a translation matrix with a translation of (0, 0, 10), because I want the camera to be located behind the object, so the object must be positioned 10 units into the z axis.

    /1 0 0  0\
C = |0 1 0  0|
|0 0 1 10|
\0 0 0  1/

And finally, the fourth matrix is the projection matrix (P). Bearing in mind that the eye is at the origin of the world space and the projection plane is defined by z = 1, the projection matrix is:

    /1 0 0   0\
P = |0 1 0   0|
|0 0 1   0|
\0 0 1/d 0/

where d is the distance from the eye to the projection plane, so d = 1.

I'm multiplying them like this: (((P x C) x W) x I) x V, where V is the vertex' coordinates in column vector form:

    /x\
V = |y|
|z|
\1/

After I get the result, I divide x and y coordinates by w to get the actual screen coordinates.

Apparenly, I'm doing something wrong or missing something completely here, because it's not rendering properly. Here's a picture of what is supposed to be the bottom side of the Stanford Dragon:

Also, I should add that this is a software renderer so no DirectX or OpenGL stuff here.

• Your matrices look reasonable to me. You might try taking a few example points and doing the calculations by hand / drawing diagrams to see what the results should be at each stage, then putting those points into your code and stepping through it in a debugger to ensure all the results match. – Nathan Reed Dec 4 '12 at 21:29
• @NathanReed I tried several points just now and the results coincide perfectly. – toby Dec 4 '12 at 22:10
• Then, are you sure there is not something wrong with your triangle clipper or rasterizer? Perhaps trying a simpler model, like a cube or just a single triangle, would help illuminate things. – Nathan Reed Dec 4 '12 at 23:03

format: plain


text:
"

BENJAMIN BLUNDELL'S  WEBSITE
----------------------------

A Mosaic of snaffled images

13-01-2012

## A Mosaic of snaffled images

13th of January, 2012

As is my way, I decided to embark on some intelligent vandalism, only this time, in the digital realm. We have several machines in our lab at Central St Martins and access to these is completely open. There are no student passwords and as such, students here have become lazy and left all their data on the desktop or trash. I decided to write a script to pull off all the data from these machines, find the images and mash them all together in a mosaic.

At first, I had to grab the images, rename them, resize them and put them in the right sort of place. This is actually a tricky problem because files have spaces, many directory levels and all the rest.

!/bin/bash
c=0
find . -type f -name "*.jpg" -o -name "*.jpeg"
-o -name "*.JPG" -o -name "*.png" -o -name "*.PNG"
-o -name "*.tif" -o -name "*.tiff"
-o -name "*.bmp" -o -name "*.BMP"
-o -name "*.TIFF" -o -name "*.TIF"
-o -name "*.JPEG" | while read FILE
do
fn="$(basename "$FILE")"
echo "$FILE" en=${fn*.}
dn="$(dirname "$FILE")"
cp -- "$FILE"$1/$c.$en
convert "$1/$c.$en" -resize 64x64 "$1/\$c.jpg"
let c=c+1
done

This worked more or less. There are probably improvements we can make to this but for now, I had around 300,000 images already! Quite a lot to pick from.

Originally, I wanted to use an off-the-shelf solution. I found MacOSaiX, Mazaika and Mozodojo. The image below is created with Mazaika.

The problem with all of these is they are limited to certain sizes, despite giving good results. I wanted this to be A0 at 300DPI when printed. The problem of course, is that the image sizes are huge. Most programs just can't cope with it. Mazaika is limited in how many you can use unless you buy it (and its too expensive for what it is) and the others can't cope. So, in the end, I decided to write my own.

Using OpenCV I came up with two programs that you can get from my github page. The first goes through the set of images and creates a text file of average RGB values. The second performs the matching based on this database. OpenCV is used throughout to process the images. The result was a program that took a while to run, but wouldn't crash.

You can see that although it has worked, it's not brilliant. It matches well but I noticed that some areas were just too similar. I adjusted with a small amount of random jitter in the source colour.

After looking around the web, I finally found OpenMP which is a lovely concept, baked right into GCC. It allows you to take control of many cores on your machine, and create threads easily.

pragma omp parallel for

This one line basically sped up one of the steps by a factor of 7. Essentially, this task, without local neighbourhood refinement, is all parallel and so, OpenMP was a wonderful find and a free win. Great!

There is more to be done for sure - this sort of large scale processing could be useful for video, or maybe transformed into Scala or similar for parallel processing over many machines?


text:
# Ring is a Filed if and only if the Zero Ideal is a Maximal Ideal

## Problem 172

Let $R$ be a commutative ring.

Then prove that $R$ is a field if and only if $\{0\}$ is a maximal ideal of $R$.

## Proof.

### $(\implies)$: If $R$ is a field, then $\{0\}$ is a maximal ideal

Suppose that $R$ is a field and let $I$ be a non zero ideal:
$\{0\} \subsetneq I \subset R.$

Then the ideal $I$ contains a nonzero element $x \neq 0$. Since $R$ is a field, we have the inverse $x^{-1}\in R$.
Then it follows that $1=x^{-1}x \in I$ since $x$ is in the ideal $I$.

Since $1\in I$, any element $r \in R$ is in $I$ as $r=r\cdot 1 \in I$.
Thus we have $I=R$ and this proves that $\{0\}$ is a maximal ideal of $R$.

### $(\impliedby)$: If $\{0\}$ is a maximal ideal, then $R$ is a field

Let us now suppose that $\{0\}$ is a maximal ideal of $R$.
Let $x$ be any nonzero element in $R$.

Then the ideal $(x)$ generated by the element $x$ properly contains the ideal $\{0\}$.
Since $\{0\}$ is a maximal ideal, we must have $(x)=R$.

Since $1\in R=(x)$, there exists $y\in R$ such that $1=xy$.
This implies that the element $x$ is invertible. Therefore any nonzero element of $R$ is invertible, and hence $R$ is a field.

##### Nilpotent Element a in a Ring and Unit Element $1-ab$
Let $R$ be a commutative ring with $1 \neq 0$. An element $a\in R$ is called nilpotent if $a^n=0$ for...


text:
# URR Calculator - Urea Reduction Ratio

Created by Łucja Zaborowska, MD, PhD candidate
Reviewed by Steven Wooding
Based on research by
Owen, W. F., Jr, Lew, N. L., Liu, Y., Lowrie, E. G., Lazarus, J. M. The urea reduction ratio and serum albumin concentration as predictors of mortality in patients undergoing hemodialysis The New England journal of medicine (September 1993)
Last updated: Nov 16, 2022

The URR calculator allows you to check whether the performed dialysis was of good quality - adequate to the patient's state - using the urea reduction ratio (URR) calculations.

Follow the short text below to remind yourself about the most important issues connected to the URR lab tests, the use of URR in dialysis, and its basic formula.

We try our best to make our Omni Calculators as precise and reliable as possible. However, this tool can never replace professional medical advice.

## What does URR stand for, and how to use it?

The URR is an abbreviation for the urea reduction ratio - we can also call it the urea reduction percentage (PRU). You should measure it every month or on every 12th dialysis session.

URR is a simple method for assessing the quality of dialysis (a life-saving treatment used in the final stage of kidney failure). It is directly connected to the patient's outcome. The URR assesses the ability of the dialysis to eliminate the waste products from the blood. Because of its simplicity, this ratio can also be used for various comparisons, within one patient's results or the entire nation's dialysis standards.

URR is also closely related to the Kt/V equation- both of these values can be used as dialysis adequacy tests (you can read more about Kt/V in the paragraph below, or in a separate Kt/V dialysis calculator).

❗ The dialysis is considered to be adequate, only if the URR ratio is greater than or equal to 65% (≥65%).

The studies have confirmed that patients with lower than adequate URR values had more health problems and a greater risk of death.

What can we do to increase the URR?

• Increase time on dialysis;
• Increase blood flow through the dialyzer - it's the case of both the dialyzer and the quality of the patient's blood vessels;
• Find and eliminate any circulation problems.

Interested in the topic? Omni has plenty of useful tools to offer. For instance, you can use GFR calculator for detecting a kidney disease at its early stage and urine output calculator for computing fluid balance and urine output rate (ml/kg) per hour.

## How to calculate URR in dialysis?

The URR calculations for dialysis are straightforward, and you can easily perform them single-handed! In this section, we'd like to present two different versions of the urea reduction ratio formula - choose the one that suits you best.

\footnotesize \begin{align*} &1.\ \text{URR} = [(\text{U}_{pre}- \text{U}_{post})/\text{U}_{pre}] \cdot 100\%\\ \\ &2.\ \text{URR} = [1 - (\text{U}_{post}/\text{U}_{pre})] \cdot 100%\\ \end{align*}

💡 The choice of urea units is wide (mg/dL, mg%, mg/L, g/L, g/dL); you only need to remember to use the same unit for both urea values.

For example, if a patient's urea level before the dialysis is equal to 40 mg/dL, and afterward is lower than 20 mg/dL, we can perform the following URR calculations:

URR = [(40 mg/dL - 20 mg/dL)/ 40 mg/dL] × 100%

URR = [20/ 40] × 100%

URR = ½ × 100% = 50%

Our URR is lower than 65%, which means that the performed dialysis was not adequate.

Still hungry for knowledge? 🍰=📚
Check one of our brilliant, topic-related tools, such as FEUrea for Acute Kidney Injury (AKI) and BUN/ creatinine ratio for prerenal causes of AKI.

## URR versus Kt/V

URR can also be calculated using the Kt/V equation for dialysis; we can put the relationship between these two values into the following equation:

(K × t)/V = -ln(1 - URR)

where:

• K means the dialyzer clearance; describes the blood flow rate in milliliters per minute (mL/min);
• t is the time of dialysis in minutes;
• V stands for total body fluid volume;
• URR is the short version of Urea Reduction Ratio, given in %; and
• ln is used in mathematics to indicate the use of natural logarithm.
Łucja Zaborowska, MD, PhD candidate
Pre-dialysis urea level
mg/dL
Post dialysis urea level
mg/dL
Urea reduction ratio (URR)
%
💡 If you're looking for the Kt/V vs. URR relationship, try the Kt/V calculator.
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The grams to cups calculator converts between cups and grams. You can choose between 20 different popular kitchen ingredients or directly type in the product density.

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Find out how much plastic you use throughout the year with this plastic footprint calculator. Rethink your habits, reduce your plastic waste, and make your life a little greener.


text:
# Disjoint paths

## Kriesell's Conjecture ★★

Author(s): Kriesell

\begin{conjecture} Let $G$ be a graph and let $T\subseteq V(G)$ such that for any pair $u,v\in T$ there are $2k$ edge-disjoint paths from $u$ to $v$ in $G$. Then $G$ contains $k$ edge-disjoint trees, each of which contains $T$. \end{conjecture}

## Approximation Ratio for Maximum Edge Disjoint Paths problem ★★

Author(s): Bentz

\begin{conjecture} Can the approximation ratio $O(\sqrt{n})$ be improved for the Maximum Edge Disjoint Paths problem (MaxEDP) in planar graphs or can an inapproximability result stronger than $\mathcal{APX}$-hardness? \end{conjecture}


text:
# Adjunction between Pro and Ind Completions

Hi,

Let $\mathcal{C}$ be a small category with finite limits and finite colimits.

The Ind-completion $Ind(\mathcal{C})$ is a locally finitely presentable category which is the free cocompletion of $\mathcal{C}$ under directed colimits. That is, there is a full embedding $I : \mathcal{C} \hookrightarrow Ind(\mathcal{C})$ and for any functor $H : \mathcal{C} \to \mathcal{D}$ where $\mathcal{D}$ has directed colimits, there is an essentially unique directed colimit preserving extension $H^* : Ind(\mathcal{C}) \to \mathcal{D}$ with $H^* \circ I = H$.

The latter uses the fact that $\mathcal{C}$ has finite colimits. We can also construct the Pro-completion $Pro(\mathcal{C})$. Since $\mathcal{C}$ has finite limits it arises as the free completion of $\mathcal{C}$ under inverse limits and is dual to the locally finitely presentable category $Ind(\mathcal{C}^{op})$.

Then there is an adjunction $F \dashv U$ where $U : Pro(\mathcal{C}) \to Ind(\mathcal{C})$ is the unique inverse limit preserving extension of $\mathcal{C} \hookrightarrow Ind(\mathcal{C})$, likewise $F$ uniquely extends $\mathcal{C} \hookrightarrow Pro(\mathcal{C})$. For example if $\mathcal{C} = \mathsf{FinSet}$ then essentially $U : \mathsf{Stone} \to \mathsf{Set}$ is the usual forgetful functor and $F$ is the Stone-Cech compactification.

Questions:

• Is it known whether $U$ and $F$ are always faithful and preserve both epis and monos?
• Are any other general preservation properties known?
• Aside from "Stone Spaces", what is a good source for the connection between Pro and Ind completions?

Any help much appreciated.

Here are some examples where $U$ and $F$ are both faithful and both preserve epis and monos.

1. If $\mathcal{C}$ is complete or cocomplete then it is essentially a complete lattice (since $C$ is small), so $Ind(\mathcal{C}) \cong Pro(\mathcal{C}) \cong C$ and $U$, $F$ define an equivalence.
2. If $\mathcal{C} = \mathsf{FinSet}$ then $U : \mathsf{Stone} \to \mathsf{Set}$ is the forgetful functor, whose left adjoint is the Stone-Cech compactification.
3. If $\mathcal{C} = \mathsf{FinBA}$ then $U : \mathsf{CABA} \to \mathsf{BA}$ is the faithful forgetful functor, whose left adjoint could be described as the canonical extension of a boolean algebra.
4. If $\mathcal{C} = \mathsf{FinDL}$ then $U : \mathsf{Priestley} \to \mathsf{Poset}$ is the forgetful functor, whose left adjoint is the ordered Stone-Cech compactification.
5. If $\mathcal{C} = \mathsf{FinPoset}$ then $U : \mathsf{DADL} \to \mathsf{DL}$ is the faithful forgetful functor where $\mathsf{DADL}$ is the category of doubly algebraic distributive lattices with complete lattice morphisms. Its left adjoint could be described as the canonical extension of a distributive lattice.
6. If $\mathcal{C} = \mathsf{FinJSL}_\bot$ (finite join semilattices with bottom) then $U : \mathsf{JSL}_\bot^{op} \to \mathsf{JSL}_\bot$ sends semilattices to their ideal completion and morphisms to the right adjoint of their extension.
7. If $\mathcal{C} = \mathsf{FDVect}(\mathbb{F})$ (finite dimensional vector spaces) then $U : \mathsf{Vect}(\mathbb{F})^{op} \to \mathsf{Vect}(\mathbb{F})$ is the usual functor sending a vector space to its dual space.
8. If $\mathcal{C} = \mathsf{FinAb}$ (finite abelian groups) then $U : \mathsf{Ab}(\mathsf{Stone}) \to \mathsf{TorAb}$ is the forgetful functor from the Stone topological abelian groups to abelian groups with torsion.

Admittedly I haven't checked all the details, but I'd also be interested in other such examples.

-
What is the Stone-Cech compactification of a finite set? –  René Jul 24 '12 at 11:45
@René: The Stone-Cech compactification of a finite set is the same set again endowed with the discrete topology. More generally, given any set Z then FZ is the Stone space dual to the complete and atomic boolean algebra Pow(Z) of subsets. –  Rob Myers Jul 24 '12 at 12:17

No a answere but a specification and description of the issue (too long for a comment).

The free cocompletion property of $Ind(\mathcal{C})$ dont use the hypothesis "$\mathcal{C}$ has finite colimits". ANyway $Ind(\mathcal{C})$ is equivalent to the full subcategory $P_1(\mathcal{C})\subset \mathcal{C}^>$ of presheves that are coimits of a small, filtred diagram of representable. In this way the inclusion $i$ corresponds to the yoneda inclusion $h_-: \mathcal{C}\to P_1(\mathcal{C})$ and it preserves limit, and finite colimits. The inclusion $P_1(\mathcal{C})\subset \mathcal{C}^>$ create (small) filter colimits (if $P=\varinjlim_{i\in I}P_i$ is a filtrant colimit with $O_i\in P_1(P_1(\mathcal{C})$ then combining the comma categories $P_1(\mathcal{C})\downarrow P_i$ we make a filtrant (small) diagram of $P_1(\mathcal{C})\downarrow P_i$

Is $F: \mathcal{C}\to \mathcal{D}$ where $\mathcal{D}$ as filter colimits we have a (iso)unique extentions $F': P_1(\mathcal{C})\to \mathcal{D}$ with $F'(P):=\varinjlim_{\ (X, x)\in \mathcal{C}\downarrow P} F(X)$, if $P$ has the Ind-representation $(X_i)_{i\in I}$ i.e. $P=\varinjlim_i h_{X_i}$ then the diagram of the $h_{X_i}$ is a final diagram on the comma category $\mathcal{C}\downarrow P$ then $F'(P):=\varinjlim_i F(X_i)$

If $\mathcal{C}$ has finite colimits $P_1(\mathcal{C})\cong Cart(\mathcal{C}^{op}, Set)$ the latter is the the category cartesians presheaves i.e. that maps finite colimits of $\mathcal{C}$ to finite limits in $Set$, of course the embedding $Cart(\mathcal{C}^{op}, Set)\subset \mathcal{C}^>$ create limits .

All above as a dual version for $Proj(\mathcal{C}):=(Ind(\mathcal{C}^{op}))^{op}$, it is a free completion of $\mathcal{C}$, and if $\mathcal{C}$ has finite limits $Proj(\mathcal{C})$ is equivalent to $Cart(\mathcal{C}, Set)^{op}$ (dual to the category of copresheaves that preserving finite limits) it has (small) colimits and we have the embedding $\iota:=(h^-)^{op}: \mathcal{C}\to Cart(\mathcal{C}, Set)^{op}$.

Now the inclusion $h_-: \mathcal{C}\to Cart(\mathcal{C}^{op}, Set)$ and $\iota: \mathcal{C}\to Cart(\mathcal{C}, Set)^{op}$ induce for the universal properties of these two completions the functors $U: Cart(\mathcal{C}, Set)^{op}\to Cart(\mathcal{C}^{op}, Set)$ with $U(Q)=\varprojlim_{\ h^Y\to Q} h_Y$, and $F: Cart(\mathcal{C}^{op}, Set)\to Cart(\mathcal{C}, Set)^{op}$ with

$F(P)=\varprojlim_{\ h_X\to P} h^X$ (the limit is in $\mathcal{C}^{<}$ the copresheaves category)

we have the natural isomorphisms:

$Cart(\mathcal{C}, Set)^{op}(F(P), Q)=Cart(\mathcal{C}, Set)(Q, F(P))=$

$\mathcal{C}^{<}(\varinjlim_{\ h^Y\to Q}h^Y, \varprojlim_{\ h_X\to P}h^X)=$

$\varprojlim_{\ h_X\to P} \varprojlim_{h^Y\to Q}\mathcal{C}^<(h^Y, h^X)\cong$

$\varprojlim_{\ h_X\to P} \varprojlim_{h^Y\to Q}\mathcal{C}^>(h_X, h_Y)\cong$

$\mathcal{C}^>(\varinjlim_{\ h_X\to P} h_X, \varprojlim_{h^Y\to Q}h_Y) \cong$

$Cart(\mathcal{C}^{op}, Set)(P, U(Q))$

then $U$ is a adjoint to $F$.

-
Well, thanks for writing this. An advantage of C having finite colimits and finite limits is that Ind(C) is lfp and Pro(C) is co-lfp i.e. they can be approximated in some sense. It also holds for the examples I have in mind, which I will now add. –  Rob Myers Jul 24 '12 at 21:18


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# Systems of equations with elimination: 4x-2y=5 & 2x-y=2.5

CCSS Math: 8.EE.C.8b

## Video transcript

We're told to solve and graph the solution for the system of equations right here. And the first thing that jumps out at me, is that we might be able to eliminate one of the variables. And if we just focus on the x, we have a 4x here and we have a 2x right here. If we were to just add them right now, we would get a 6x. So that wouldn't eliminate it. But if we can multiply this 2x by negative 2, it'll become a negative 4x, and then when you add it, they would cancel out. So let's multiply this equation, this second equation, by negative 2. So I'm going to multiply both sides of this equation by negative 2. And the whole motivation is so that this 2x becomes a negative 4x. And, of course, I can't just multiply only the 2x. Anything I do to the left-hand side of the equation I have to do to every term, and I have to do to both sides of the equation. So the second equation becomes negative 4x-- that's negative 2 times 2x-- plus-- we have negative 2 times negative y-- which is plus 2y is equal to 2.5 times negative 2, is equal to negative 5. I just rewrote the second equation, multiplying both sides by negative 2. Now, this top equation-- I'll write it on the bottom now-- we have 4x minus 2y is equal to positive 5. And now we can eliminate it. We can say, hey, look, the negative 4x and the positive 4x should cancel out, or they will cancel out. So let's add these two equations. Let's add the left side to the left side, the right side to the right side, and we can do that because these two things are equal. We're doing the same thing to both sides of the equation. So what do we get? If we take our negative 4x plus our 4x, well, those cancel out. So you're left with nothing. Maybe I could write a 0 there. 0x if you want. And then you have your plus 2y and your negative 2y. Those also cancel out. So you're also left with 0y. And then that equals negative 5 plus 5 is equal to 0. So this just simplifies to 0 equals 0, which is true, but it's kind of bizarre. We had all these x's and y's. Everything canceled out. So let's explore this a little bit more. Let's graph it and see what this 0 equals 0 is telling us when we try to solve this system of equations. So let me graph this top guy. I'll do it in blue. So right now it's in standard form. Let's put it in slope-intercept form. So we have 4x minus 2y is equal to 5. Let's subtract 4x from both sides. I want the x terms on the right-hand side. So then I'm left with negative 2y is equal to negative 4x plus 5. Now we can divide both sides by negative 2. And we are left with y is equal to positive 2x, right, that's positive 2x, minus 2.5. So let's graph that. The y-intercept is negative 2.5. So negative 2.5 right there, and then it has a slope of 2. So if we move up 1, if we move up in the x-direction, if we move to the right 1 in the positive x-direction, we will move up 2. So 1, 2. Right there. And if we were to do it again, we move up 1, 2. Just like that. So the line's going to look something like this. I'll try my best to draw a straight line. This is the hardest part about a lot of these problems. There you go. So that's the top equation. Now, let me draw the bottom equation. Let me draw and I'll do it in this green color. So this bottom equation was 2x minus y is equal to 2.5. And we can subtract 2x from both sides. The left-hand side becomes negative y is equal to 2x plus-- or is equal to negative 2x plus 2.5. Now let's multiply or divide both sides by negative 1. And you get y is equal to positive 2x minus 2.5. And let's try to graph this, and you already might notice something interesting about these two equations. You try to graph this, the y-intercept is at negative 2.5, right there. The slope is 2. So it's going to be this exact same line. And you saw that algebraically. I didn't have to graph it. These two lines have the exact same equation when you put them in slope-intercept form. That's the first equation. That's the second equation. So what this 0 equals 0 is telling us is actually that these are the same line. That these actually have an infinite number of solutions. Any point on this line, which is both of those lines, will satisfy both of these equations. You give me an arbitrary y, solve for x in the top equation, that x and y will also satisfy the bottom equation. So this actually has an infinite number of solutions. These are the same line.


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# Analytical Approximation of A Convolution

Consider the function $f(x)=\frac{1}{\sqrt{2 \pi} s |x|} e^{-\frac{(a x +b )^2}{2 (s x)^2}}$, where $s>0$. I am interested in the convolution of $f$ with the Gaussian kernel $g(x)=\frac{1}{\sqrt{2 \pi}}e^{-\frac{x^2}{2}}$. Let's refer to this convolution as $h(x)$,

$h(x)=\int_\mathbb{R} g(x-t) \, f(t) \, dt$.

I first tried to obtain a closed form expression for $h(x)$, but I failed. So my next goal is find an approximation to $h(x)$. However, that turns out to be challenging too. I appreciate any idea how to obtain an "analytical" approximation to $h(x)$.

Best

Golabi

• Thank you John. Integrating $\frac{1}{|t|}$ does not converge around zero as you said, but I do not understand why you eliminated the exp in front of $\frac{1}{|t|}$. More precisely it seems the full $f$ is well-behaved because $\lim_{t \rightarrow 0} f(t) = 0$. – Golabi Feb 9 '17 at 19:56
• Ok. I made a mistake previously. – John Feb 10 '17 at 1:43


text: Answer. As Example:, 8x2 + 5x – 10 = 0 is a quadratic equation. Quadratic functions make a parabolic U-shape on a graph. x 1 = (-b … The quadratic function f (x) = a (x - h) 2 + k, a not equal to zero, is said to be in standard form . Quadratic functions are symmetric about a vertical axis of symmetry. In this example we are considering two … A(L) = −2L. (x + 2) (x + 5) = x 2 + 5x + 2x + 10 = x 2 + 7x + 10. where a, b, c are real numbers and the important thing is a must be not equal to zero. Khan Academy is a 501(c)(3) nonprofit organization. f(x) = -x 2 + 2x + 3. Use the quadratic formula to find the roots of x 2 -5x+6 = 0. Standard Form. The factors of the quadratic equation are: (x + 2) (x + 5) Equating each factor to zero gives; x + 2 = 0 x= -2. x + 5 = 0 x = -5. The quadratic formula, an example. Substitute the values in the quadratic formula. Our mission is to provide a free, world-class education to anyone, anywhere. Verify the factors using the distributive property of multiplication. The quadratic function f(x) = a x 2 + b x + c can be written in vertex form as follows: f(x) = a (x - h) 2 + k The discriminant D of the quadratic equation: a x 2 + b x + c = 0 is given by D = b 2 - 4 a c . Example 2 f(x) = -4 + 5x -x 2 . The quadratic has a negative leading coefficient, so the graph will open downward, and the vertex will be the maximum value for the area. The functions in parts (a) and (b) of Exercise 1 are examples of quadratic functions in standard form . Now, let us find sum and product of roots of the quadratic equation. It is represented in terms of variable “x” as ax2 + bx + c = 0. A quadratic equation is an equation that can be written as ax ² + bx + c where a ≠ 0. Example 5. Then, the two factors of -15 are. Examples of quadratic equations $$y = 5x^2 + 2x + 5 \\ y = 11x^2 + 22 \\ y = x^2 - 4x +5 \\ y = -x^2 + + 5$$ Non Examples x2 + 2x - 15 = 0. In this unit, we learn how to solve quadratic equations, and how to analyze and graph quadratic functions. Solution. x 2 - (α + β)x + α β = 0. A ( L) = − 2 L 2 + 8 0 L. \displaystyle A\left (L\right)=-2 {L}^ {2}+80L. Quadratic functions follow the standard form: f(x) = ax 2 + bx + c. If ax 2 is not present, the function will be linear and not quadratic. +5 and … Therefore, the solution is x = – 2, x = – 5. If a is negative, the parabola is flipped upside down. The maximum revenue is the value of the quadratic function (1) at z = 2" R = = -200 + 400 + 1600 = 1800 dollars. Solution : In the given quadratic equation, the coefficient of x2 is 1. x 2 - (1/α + 1/β)x + (1/α) (1/β) = 0. x 2 - ( (α + β)/α β)x + (1/αβ) = 0. x 2 - ( ( - √2 )/3)x + (1/3) = 0. Graphing Quadratic Functions in Factored Form. Example 1. Graphing Parabolas in Factored Form y = a ( x − r ) ( x − s ) Show Step-by-step Solutions. The general form of a quadratic equation is y = a ( x + b ) ( x + c) where a, b and c are real numbers and a is not equal. Example. The market for the commodity is in equilibrium when supply equals demand. The function, written in general form, is. 2. . α β = 3/1 = 3. here α = 1/α and β = 1/β. + 80L. In other words, a quadratic equation must have a squared term as its highest power. Graphing Parabolas in Factored Form y=a (x-r) (x-s) - … (The attendance then is 200 + 50*2 = 300 and (for the check purpose) $6*300 =$1800). This form of representation is called standard form of quadratic equation. x2 + √2x + 3 = 0. α + β = -√2/1 = - √2. The revenue is maximal $1800 at the ticket price$6. Comparing the equation with the general form ax 2 + bx + c = 0 gives, a = 1, b = -5 and c = 6. b 2 – 4ac = (-5)2 – 4×1×6 = 1. When a quadratic function is in standard form, then it is easy to sketch its graph by reflecting, shifting, and stretching/shrinking the parabola y = x 2. 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text:
# interrupt with delay in main loop

#### denison

Joined Oct 13, 2018
283
Just wondering in Arduino programming. If you have a 5 second delay in the main loop does an interrupt still work? Should think it would but not sure. Does anybody know for sure?

#### Ian0

Joined Aug 7, 2020
4,906
If the delay is simply executing NOP instuctions, then the interrupt should work. The 5 second delay will then be extended by the length of time it takes to execute the interrupt.
If the delay is created by a counter, then the main loop will be just checking the counter for completion over and over again, and the interrupt should work, but the delay will remain at 5 seconds.

#### BobTPH

Joined Jun 5, 2013
4,785
That is the whole point of interrupts, they interrupt the execution of normal code, and possibly even other interrupts.

Bob

#### Papabravo

Joined Feb 24, 2006
18,161
This all assumes that you are not disabling interrupts inside the delay loop.

#### John P

Joined Oct 14, 2008
1,971
An interrupt should work. But I suggest that a delay as long as 5 seconds is a bad idea. It's time when your processor will be doing nothing, and you'd be giving up the opportunity to do anything else. I'd be happier with a regular interrupt at (let's say) 1000 times a second, and if you want a delay, increment a counter to reach it. Like:
Code:
  five_sec_counter = 5000;       // This line is in main(); five_sec_counter is a global variable

// Do this in the interrupt
if (five_sec_counter != 0)
{
if (--five_sec_counter == 0)
;      // Action when the delay ends. It could be setting a flag which is periodically checked in main()
}

#### dl324

Joined Mar 30, 2015
14,357
If you have a 5 second delay in the main loop does an interrupt still work? Should think it would but not sure. Does anybody know for sure?
As long as you're not disabling interrupts.

Using delays is usually a bad idea. Setting up ISR's is usually the way to go.

#### djsfantasi

Joined Apr 11, 2010
8,263
An ISR should do one simple task and then return. That’s it.

It shouldn’t be used to perform tasks asynchronously. It CAN set a flag and the main loop can execute process a more complex task when it has the time.

An example. An ISR reads a character from a serial device and adds it to a string or char array. Return. The main loop determines if all data is received and processed it if ready.

A bad example. Read a character. Check if all data is received. If so process the data. Otherwise, add the character to a string or char array. Return. The main loop does none of these tasks.

The worst, absolute worst, thing you can do is add a delay to your ISR. For at least one very good reason.

ISRs should disable interrupts. Otherwise, you’re inviting data corruption when an interrupt occurs during an interrupt. But if interrupts are disabled and contains a delay - you will (I almost said might) lose data and the execution of your code will become erratic.

The use of interrupts is very powerful… if used correctly.


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# Russell's paradox

Lowp tae: navigation, rake

In the foondations o mathematics, Russell's paradox (an aa kent as Russell's antinomy), discovered bi Bertrand Russell in 1901, shawed that some attemptit formalisations o the naive set theory creatit by Georg Cantor led tae a contradiction. The same paradox haed been discovered a year afore bi Ernst Zermelo but he did nae publish the idea, which remained kent anly tae Hilbert, Husserl an ither members o the Varsity o Göttingen.

Accordin tae naive set theory, any definable collection is a set. Let R be the set o aw sets that are nae members o themselves. If R is nae a member o itsel, then its definition dictates that it must contain itsel, an if it contains itsel, then it contradicts its ain definition as the set o aw sets that are nae members o themsels. This contradiction is Russell's paradox. Seembolically:

${\displaystyle {\text{Let }}R=\{x\mid x\not \in x\}{\text{, then }}R\in R\iff R\not \in R}$


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# Hacking Tube

## CSAW CTF 2015 -- FTP & FTP2

Category: Reversing (FTP) & Exploitable (FTP2)
Points: 300 (FTP) & 300 (FTP2)

## FTP

64 bit ELF. It's a FTP-like service, we can list all the acceptable command by sending the HELP command. Here are some important commands that we'll need to pass the challenges:

USER [username]: enter username to login
PASS [password]: enter password after sending the USER command to login as the [username]
PASV: open a port for passive mode
LIST: list the files in the directory
STOR: upload a file
RETR: download a file
RDF: read the reversing solution's flag

So after some static analysis with the help of IDA Pro, I figure out that in order to pass the reversing challenge, we'll need to login as the user blankwall. The password checking function's at 0x401540, let's take a look at it:

At first I was like "Ah, that's easy!", since we have the constraint system, we can just set it up and leave the rest to Z3. But after I have the solution and enter the password, the service respond it with a frustrating "Invalid login credentials". Knowing that Hex-Rays' decompiler might have the incorrect decompiling result, I re-check the password checking logic by reversing directly from the x64 assembly, not the pseudo code, and finally found the root of the problem:

What really matters is the assembly eax, [rbp+var_4] at line 4. Notice that when the program move v3 to the register eax, it doesn't clear the highest 32 bits of the register rax. So when it runs to the line ecx, [rdx+rax], rax isn't just simply v3 & 0xFFFFFFFF, it's actually ((v3 << 5) & 0xFFFFFFFF00000000) | (v3 & 0xFFFFFFFF), and that's where the Hex-Rays decompiler made the mistake.

So now we have the correct constraint system. Wrote a Z3 python script and retrieve the password:

Since I don't know the password length, I just brute force it from 1 ~ 10. We can found a solution at length 6:

Now we get the login password, time to capture the flag :)

The flag is: flag{n0_c0ok1e_ju\$t_a_f1ag_f0r_you}

## FTP2

So now we're logged in as a valid user, we can finally do some other stuff. After sending PASV and the LIST command, I found that there's a flag.txt in the directory. At first I try to download the file, but the service response "Invalid character specified". Well that's strange :/ so I went to the RETR function and start analyzing.

So...to sum it up, the program will detect whether if the filename has the character store in 0x604408, and if it does, it will refuse to let us download the file. Remeber the function that does the password checking?

So apparently we can't have 'f' in our filename, we'll need to find another way to bypass the filter. By checking other functions, I finally found a way to bypass it.

Here in the STOR function, if we upload a file that is big enough, we can overwrite the data at 0x604408. So it's quite simple: just create a file that is larger than 512 bytes, then upload it to the server. After that, we can download the flag.txt and get the flag.

The flag is: flag{exploiting_ftp_servers_in_2015}


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407 493 6601

# 6th Grade - Ratios and Proportions

## Ratios

• A comparison of two numbers by division is called a Ratio.
• A ratio can be expressed in three different ways: .
• A ratio $\frac{a}{b}$ is read as "a ratio of a to b".
• A ratio is a fraction that can be simplified.
• Two ratios are called equal ratios or equivalent ratios if are equal.
• $\frac{2}{5}$ and $\frac{4}{10}$are equal ratios as $\frac{4}{10}=\frac{2}{5}×\frac{2}{2}$ (numerator and denominator are multiplied by the same number).
• A table of equivalent ratios is called a ratio table.
•  2 4 6 8 10 5 10 15 20 25
• All ratios in the ratio table are equivalent.

## Proportion

• Two ratios are said to be in proportion if their cross-product are equal.
•   are in proportion if .

## Solved Examples

1. Find the ratio of 7 to 10.

The ratio of 7 to 10 is $\frac{7}{10}$

2. Find the ratio of 16 oranges to 21 apples.

The ratio of 16 to 21 is $\frac{16}{21}$

3. Complete the following ratio table:

 Number of Lemons 5 10 ? Glasses of Lemonade 10 ? 30

To keep the ratio table equal, the top and the bottom numbers should be multiplied by the same number.

 Number of Lemons 5 10 ($5×2$) 15 ($5×3$) Glasses of Lemonade 10 20 ($10×2$) 30 ($10×3$)

4. Do the following ratios form a proportion: ?

The cross-product should be the same for the ratios to form a proportion. Since , the ratios are not in proportion.

5. 6 pastries cost $5. What should 18 pastries cost? Put the provided information as a proportion: $\frac{6}{5}=\frac{18}{?}$. Since 6 is multiplied by 3 to get 18, 5 must be multiplied by 3. The answer is$15.

## Cheat Sheet

• The quantity that comes first in the ratio statement becomes the numerator in the ratio fraction.
• A proportion consists of 4 terms. It is a comparison of two ratios.
• If the numerator and denominator of a ratio are multiplied by the same number, the resulting ratio is equivalent to the first one.
• Their cross-product should be the same for the ratios to be in proportion.

## Blunder Area

• In the case of word problems, pay attention to which value goes as a numerator and which one goes as a denominator.
• Not all ratios are in proportion.


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# Tag Info

60

Imagine you have a very heavy book and a bookcase, and your goal is to put the book on the top shelf of the bookcase. How much time would you spend doing that? Maybe five seconds, maybe fifteen. Would going much slower help you? No, it would not, because simply carrying the book is exhausting to you. You would never be able to hold the book up for an entire ...

43

Wouldn't i inevitably spiral to sun surface even if i was faster than 0km/s ? No. On reasonable timescales, an orbit will have a fixed distance of closest approach, called "periapsis." (These timescales shorten if you're close enough to what you're orbiting that an atmosphere can drag you down). You don't really need to "drop in straight line" (which ...

43

A very good question! The reason is essentially to do with tides. And a slightly over-simplified summary is: If the moon orbits more slowly than the rotation of the parent body (as our Moon does, 12 degrees per day while the Earth rotates about 360 degrees per day) then the moon will gradually orbit further and further away. If the moon orbits faster than ...

38

Primarily, locations of spaceports would change. California, not Florida would host the NASA's main launch site. Russia would be in slightly better position, able to send rockets over the Black Sea, nicer inclinations than currently available from Baikonur - although Vostochny wouldn't happen or would be closer to Chita. ESA could forget about French Guiana, ...

35

To answer your title question: By using its engines. However you seems to be quite puzzled by the fact that velocity of an object can decrease and increase over the course of an orbit. If the orbit is perfectly circular, the speed will always remain the same (until thrusters are used). However, as is the case with Chandrayaan-2, most orbits are ...

31

The Answer is on a page by Sven Grahn. No ocean going tracking ships were used. Only ground stations on the territory of the USSR. In the other answers some russian sources about the use of tracking ships were found, so the information from Sven Grahn may be partialy wrong. Short waves were used for long distance (5000 km) transmission even beyond line-of-...

28

I feel the need to correct some issues that were brought up in the other answers. Yes, gravity is an attraction-only force. But due to its relative weakness, objects in space can attain large velocities, before getting the chance to collide with any single target. In physics we would speak of excess angular momentum, which is hard to get rid of in space, ...

25

No, unless your structure is located directly on the equator and your satellite follows a perfectly circular orbit, atmospheric "orbits" aren't possible, even in a vacuum tunnel. Because the Earth is on an axis of ~23 degrees and rotates every day, it is not possible to create an orbit which has no ground track precession except for equatorial ...

24

What you're describing is (more or less) the StarTram "gen 1" design. The reference design has: 40 tonne unmanned cargo projectile, 25 tonnes of payload, ~2 m wide, ~13 m long. A 130 km maglev acceleration tunnel, evacuated. An exit point 6000 m up, on a mountain. A plasma window to allow projectile egress into atmosphere without repressurising the entire ...

23

A polar orbit can be geosynchronous and always follow the same path, but that path cannot be straight along a meridian. If the orbit passes on top of both poles, then the orbit lies on a plane containing both poles, which cuts the earth longitudinally like two opposite meridians do. However, as the satellite travels along that orbit, the Earth moves so its ...

23

The thing your're missing is that the Hohmann Transfer orbit takes time, and both Mars and Earth are moving around the sun. For the Hohmann Transfer orbit to work, the position of Mars at arrival has to be opposite the point of of Earth at Departure. The following image depicts Earth's and Mars' orbit as circular, rather than elliptical to simplify ...

19

Many novel launch schemes need some amount of help from rockets. What kills a lot of them is doing a tradeoff study of just enlarging the rocket part and getting rid of the non-rocket part. Surprisingly often, that works out to be better and cheaper. --Henry Spencer This is a system that needs a rocket part, as one of these two cases would necessary ...

18

You need below 2866 m/s of orbital velocity at 1 AU to crash into the Sun. You technically don't need to slow down exactly to 0 m/s relative to the Sun in order to crash into it. Let's calculate the approximate velocity required to graze the "surface" of the Sun. This is an excellent answer on how to calculate apoapsis and periapsis of an orbit. So first, ...

17

Ships were used to receive telemetry. Work in the ocean took place on February 12, 1961. At this time, preparations for the launch of the world's first manned spacecraft Vostok were completed. At the Control Center, a decision is made to use a tracking ships for telemetric monitoring of the onboard equipment of the spacecraft. On April 12, 1961, ...

16

And note that if you want to hit the sun the cheaper (but slow!) way to do it is to head out. 12.32km/sec will take you to infinity, at infinity a burn of 0m/sec will kill your orbital velocity and you'll come straight in. Of course this will take infinite time, but even going only as far as Jupiter's orbit means you use less energy to drop your periapsis ...

15

According to this source (in Russian; includes communication log and post-flight debriefing) the only ground stations that are listed as those that were communicating with Gagarin were located in USSR. This leads to conclusion that there were no other tracking stations located elsewhere. The list of the stations: Located at launch site (Baikonur). ...

15

I've found that language barrier problem exists. There are many info in Russian that have no proper translations in English. There were space tracking ships too. Information that USSR had not them at the time is not correct, the first ships were equipped in 1959-60. Sources: 1. http://niskgd.ru/pages/qa.htm quotes: Первые 6 судов, что начали работу до ...

15

There are two ways in which a massive orbiting body, such as a planet, can clear a smaller object from the vicinity of its orbit. One, obviously, is by colliding with it. The other, more common way is called the gravitational slingshot effect.* This is a trick that many space probes have used to gain (or lose) extra speed and thus get further away from (...

15

Such a tunnel is not plausible for a number of reasons. 1. Problems with orbits First of all, as other people have said it would only work for equatorial orbits which were either circular (very long tunnel) or had a period which is some rational multiple of the Earth's rotational period. And, again as other people have said, the real Earth is nothing like ...

10

Gravity does not cancel out at any of the Lagrange points, not even L1. As you point out, at L2 both the central body and orbiting body are pulling the same direction. But there is a third character in this tug of war, centrifugal acceleration. It's not truly an acceleration but inertia in a rotating frame. But when you're in the rotating frame it sure ...

10

Orbits beneath synchronous orbits have a higher angular velocity than their planets rotation, orbits above have a slower angular velocity. Drag (atmospheric or tidal) would try to match the angular velocity to the planets rotation. So below a synchronous orbit objects get slower, above it they would speed up (and slow down the rotation of the body they are ...

8

No, such an orbit is not possible (or: it's not possible unless the satellite is accelerated most of the time, which is not plausible in terms of fuel). In particular it's not possible for the path of the orbit to trace out a great circle over surface the Earth (in a coordinate system rotating with the Earth). Your picture shows a great circle. For a ...

8

What you are looking for is called orbit determination, and in particular batch least squares orbit determination. To learn about it I can recommend Statistical Orbit Determination by Bob Schutz, Byron Tapley and George H. Born. I understand that you want to try to do it yourself. Nevertheless, if at some point you are looking to do it with software, it is ...

8

If you define perfection as absolutely zero eccentricity then perfection is impossible. There will always be eccentricity in an orbit, even if it is very small. Orbits vary due to: Spacecraft system inaccuracy: no spacecraft is perfect, no matter how accurate Changes in the density of the orbited planet Gravitational influence of other celestial bodies: my ...

8

This would be a tunnel a couple thousand kilometers long, that extends from the surface of the Earth to approximately low earth orbit altitude at both ends, strong enough to keep vacuum inside and the atmosphere out at sea level, such that its openings are in place for a space station in a highly-elliptical geosynchronous orbit comes flying through at about ...

7

In this supplementary answer I've crudely processed all TLEs for Vanguard 1 and plotted the trends. I used the mean motion (revs/day) to get a period, divided by 0.9975 (estimating from this answer) to undo the effects of $J_2$, then used $a^3=GM (T/2 \pi)^2$ to estimate a semimajor axis, periapsis and apoapsis based on the TLE's eccentricity value. This is ...

7

The general concept of doing things slowly generally allowing better efficiency in a design is certainly correct. The difficulty with applying this to rocket launches is the concept of gravity loses that in general mean getting to orbit as quickly as possible is better. Slow/efficient space access via space elevator and related designs is the logical ...

6

In addition to Russell Borogove's good answer there is another factor here to keep in mind: When you are dealing with planets you can't just add together separate burns. If you add up the energy needed to escape Earth, the energy needed to go from Earth's orbit to Mars' orbit and the energy needed to enter Mars' orbit you will get more than 16 km/sec. In ...

6

16 km/s is about right for Earth surface to low Mars orbit, summing up a few entries in this table. I am not sure as how he did it though, and was unsure if he used the escape velocity of Earth, the Hohmann Transfer, and also if that was only that velocity computed needed to be obtained once At a minimum, this would be split into ascent to LEO (~9.4 km/...

6

After reading all your answers I'd like to summarise the situation. The black circles are the circular orbits and the red ellipse is the transfer orbit. Consider a spacecraft in the elliptical orbit. At the point P the velocity is greater than the circular orbital velocity, and that's why the distance from the centre increases. And at the point A the ...

Only top voted, non community-wiki answers of a minimum length are eligible


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## Introductory Chemistry (5th Edition)

a) $2.6 moles NO_{2}$ b) $11.6 moles NO_{2}$ c) $8900 moles NO_{2}$ d) $2.012\times10^{-3} moles NO_{2}$
We can solve this by looking at the balanced chemical equation and using the mole ratios: $2N_{2}O_{5} --> 4NO_{2} + O_{2}$ a) $1.3 moles N_{2}O_{5} \times\frac{4 moles NO_{2}}{2 moles N_{2}O_{5}} = 2.6 moles NO_{2}$ b) $5.8 moles N_{2}O_{5} \times\frac{4 moles NO_{2}}{2 moles N_{2}O_{5}} = 11.6 moles NO_{2}$ c) $4.45\times10^{3} moles N_{2}O_{5} \times\frac{4 moles NO_{2}}{2 moles N_{2}O_{5}} = 8900 moles NO_{2}$ d) $1.006\times10^{-3} moles N_{2}O_{5} \times\frac{4 moles NO_{2}}{2 moles N_{2}O_{5}} = 2.012\times10^{-3} moles NO_{2}$


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<meta http-equiv="refresh" content="1; url=/nojavascript/"> Proofs with CPCTC | CK-12 Foundation
You are reading an older version of this FlexBook® textbook: CK-12 Foundation and Leadership Public Schools, College Access Reader: Geometry Go to the latest version.

# 4.8: Proofs with CPCTC

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Apply various triangle congruence postulates and theorems.
• Know the ways in which you can prove parts of a triangle congruent.

As you can see, there are many different ways to prove that two triangles are congruent. It is important to know all of the different ways that can prove congruence, and it is important to know which combinations of sides and angles do not prove congruence.

## Congruence Theorem Review

As you have studied in the previous lessons, there are five theorems and postulates that provide different ways in which you can prove two triangles congruent without checking all of the angles and all of the sides. It is important to know these five rules well so that you can use them in practical applications.

Name Corresponding congruent parts Picture Does it prove congruence?
SSS Three sides Yes
SAS Two sides and the angle between them Yes
ASA Two angles and the side between them Yes
AAS Two angles and a side not between them Yes
HL The hypotenuse and a leg in a right triangle Yes

Example 1

What rule can prove that the triangles below are congruent?

A. SSS

B. SAS

C. ASA

D. AAS

The two triangles in the picture have two pairs of congruent angles ($\angle DCE \cong \angle ORN$ and $\angle CED \cong \angle RNO$) and one pair of corresponding congruent sides $(\overline{DC} \cong \overline{OR})$.

So, the triangle congruence postulate you choose must have two A’s (for the angles) and one $S$ (for the side). You can eliminate choices A and B for this reason.

Now that you are deciding between choices C and D, you need to identify where the side is located in relation to the given angles. It is adjacent to one angle, but it is not in between them.

Therefore, you can prove congruence using AAS.

The correct answer is D.

1. The two triangles below are congruent because of the SAS postulate. Mark the SAS congruent parts with tic marks and arcs:

2. True/False: SSS, SAS, ASA, AAS, and HL (in a right triangle) are 5 different ways to prove that triangles are congruent.

${\;}$

## Proving Parts Congruent

It is one thing to identify congruence when all of the important identifying information is provided, but sometimes you will have to identify congruent parts on your own. This may take a bit of thought, and you must use some deductive reasoning (finding conclusions based on facts) to find the missing parts.

When you were creating proofs, you also used the reflexive property of congruence. This property states that any segment or angle is congruent to itself. While this may sound obvious, it can be very helpful in proofs, as you saw in those examples.

Recall that, if two triangles are congruent, then all pairs of corresponding sides and all pairs of corresponding angles are congruent.

We say that the Corresponding Parts (sides and angles) of Congruent Triangles are Congruent, or CPCTC.

1. What is a part of a triangle?

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${\;}$

${\;}$

2. What does it mean for parts to be corresponding?

${\;}$

${\;}$

${\;}$

3. Explain in your own words what CPCTC means.

${\;}$

${\;}$

${\;}$

How do we use the concept of CPCTC in a proof?

Check out the example on the following page…

Example 2

How could you prove that segment $AB$ is congruent to segment $DE$ in the diagram below?

We can see that $\overline{BC} \cong \overline{CE}$ and $\overline{AC} \cong \overline{CD}$ because of the tic marks in the figure.

Think of all of the postulates and theorems (in the chart at the beginning of Lesson 7 and in your graphic organizer) that have 2 S’s (SSS and SAS). We may be able to use SSS or SAS to show the triangles are congruent. However, to use SSS, we would need $\overline{AB} \cong \overline{DE}$ and we cannot yet make this assumption.

Can we show that two of the angles are congruent?

Notice that $\angle BCA$ and $\angle ECD$ are vertical angles (non-adjacent angles made by the intersection of two lines—i.e., angles on the opposite sides of the intersection).

The Vertical Angle Theorem states that all vertical angles are also congruent.

So, this tells us that $\angle BCA \cong \angle ECD$.

We now have two congruent sides and a congruent angle between the sides!

• The congruent sides are:

___________ $\cong$ ___________ and ___________ $\cong$ ___________

• The congruent angles are:

_______________ $\cong$ _______________

By putting all of this information together, you can confirm that $\Delta ABC \cong \Delta DEC$ by the SAS Postulate.

Finally, if the two triangles are congruent, then their corresponding parts are congruent.

Therefore segment $AB$ is congruent to segment $DE$ by CPCTC.

1. In the space below, draw a picture of two triangles that are congruent because of the ASA postulate. Make sure to mark the congruent parts with tic marks and arcs!

${\;}$

${\;}$

${\;}$

${\;}$

2. If you know that 2 pairs of angles and 1 pair of sides are congruent in your picture above, what other parts of the triangles are congruent? (Hint: there are 3 answers!)

${\;}$

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3. Your answers to question #2 above are true because of CPCTC. The letters CPCTC stand for: (Fill in the blanks) ______________________ ______________________ of ________________________ ________________________ are _________________________ .

## Finding Distances (a real-word application)

One way to use congruent triangles is to help you find distances in real life—usually using a map or a diagram as a model.

When using congruent triangles to identify distances, be sure you always match up corresponding sides. The most common error on this type of problem involves matching two sides that are not corresponding.

Example 3

The map below shows five different towns. The town of Meridian is exactly halfway between two pairs of cities: it is halfway between Camden and Grenata AND it is halfway between Lowell and Morsetown.

Using the information in the map, what is the distance between Camden and Lowell?

The first step in this problem is to identify whether or not the triangles drawn on the map are congruent.

Since you know that the distance from Camden to Meridian is the same as Meridian to Grenata (since Meridian is halfway between those cities), those two sides of the triangles are congruent. You can make a tic mark on these sides.

Similarly, since the distance from Lowell to Meridian is the same as Meridian to Morsetown (again because Meridian is halfway between), those two sides are also a congruent pair and you can give them congruent tic marks.

Finally, you can tell that the angles between these lines (at the intersection where Meridian is) are also congruent because they are vertical angles.

With all of your congruent sides and angles marked, your map will look like this:

So, by the SAS postulate, these two triangles are congruent.

This allows us to find the distance between Camden and Lowell by identifying its corresponding side on the other triangle. Because they are both opposite the vertical angle, the side connecting Camden and Lowell corresponds to the side connecting Morsetown and Grenata.

Since the triangles are congruent, these corresponding sides will also be congruent to each other. Therefore, the distance between Camden and Lowell is 5 miles.

This use of the definition of congruent triangles is one of the most powerful tools you will use in geometry class! It is often abbreviated as CPCTC, meaning Corresponding Parts of Congruent Triangles are Congruent.

8 , 9 , 10

## Date Created:

Feb 23, 2012

May 12, 2014
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# Formation of stars

How are stars formed? I read somewhere that when dust, gas, rocks etc come close to each other due to gravity, slowly and gradually build up a lot of pressure and form stars. If what I read is right, then from where did these materials come from?

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# Computational Aeroacoustics (CAA) part1

This is the first in a series of posts about the computational aeroacoustics that is abbreviated as CAA. Aeroacoustics is mainly concerned with the generation of sound or noise through a fluid flow. Familiar examples are

They can be studied by both experimental and computational methods. In an experimental approach, the aerodynamically generated sound is measured in the anechoic chamber that is designed to prevent reflections of sound on the wall, making the room echo-free. The following video created by Microsoft will help us to understand the structure of it.

The computational techniques for simulating flow-generated noise can be classified into two broad categories:

 Direct Approaches

The governing equations of the compressible fluid flow is the compressible Navier-Stokes equations and they also describe the generation and propagation of acoustic noise, so we can solve the computational aeroacoustics problems by solving the transient compressible Navier-Stokes equations both in the source region where the flow disturbances generate noise and propagation region where the generated acoustic waves propagate. Acoustic waves have to be resolved in both regions so that the noise can be accurately simulated at observation locations. However, the solution of the Navier–Stokes equations with fine mesh over large domains to determine farfield noise is computationally very expensive.

As is the case in the experimental approaches, it is essential to prevent the reflection of the acoustic waves on the artificially truncated boundary (i.e. it does not stretch to infinity) of the computational domain in order to obtain an accurate result. A variety of numerical techniques have been developed for this purpose:

1. Navier-Stokes Characteristic Boundary Conditions (NSCBC)
2. Artificial dissipation and damping in an absorbing zone
3. Grid stretching and numerical filtering in a “sponge layer” or “exit zone”
4. Perfectly matched layer (PML)

In OpenFOAM, acousticDampingSource is categorized into Ⅱ and the waveTransmissive boundary condition is an approximation of NSCBC.

 Hybrid Approaches (Acoustic Analogy)

David P. Lockard and Jay H. Casper [1] states that:

The physics-based, airframe noise prediction methodology under investigation is a hybrid of aeroacoustic theory and
computational fluid dynamics (CFD). The near-field aerodynamics associated with an airframe component are simulated to obtain the source input to an acoustic analogy that propagates sound to the far field. The acoustic analogy employed within this current framework is that of Ffowcs Williams and Hawkings, who extended the analogies of Lighthill and Curle to the formulation of aerodynamic sound generated by a surface in arbitrary motion.

• Lighthill’s analogy

Lighthill derived the following wave equation \eqref{eq:Lighthill} from the compressible Navier-Stokes equations:
\begin{align}
\frac{\partial^2 \rho}{\partial t^2} – c_0^2 \nabla^2 \rho = \frac{\partial^2 T_{ij}}{\partial x_i \partial x_j}, \tag{1} \label{eq:Lighthill}
\end{align}
where the so-called Lighthill (turbulence) stress tensor is expressed as

\begin{align}
T_{ij} = \rho u_i u_j + \left( p-c_0^2 \rho \right)\delta_{ij} -\mu \left(\frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i} – \frac{2}{3}\delta_{ij}\frac{u_k}{x_k} \right), \tag{2} \label{eq:Tij}
\end{align}
$$\delta_{ij}$$ is the Kronecker delta and $$c_0$$ is the speed of sound in the medium in its equilibrium state.

• Curle’s analogy

The existence of the objects (solid walls) is not considered in the Lighthill’s theory and Curle developed the theory so that it can be dealt with. The density variation at the observer location $$\boldsymbol{x}$$ is calculated from the following equation \eqref{eq:Curle}
\begin{align}
\acute{\rho}(\boldsymbol{x}, t) &= \rho(\boldsymbol{x}, t)\;- \rho_0 \\
&=\frac{1}{4 \pi c_0^2} \frac{\partial^2}{\partial x_i \partial x_j} \int_{V}\frac{[T_{ij}]}{r}dV – \frac{1}{4 \pi c_0^2} \frac{\partial}{\partial x_i}\int_{S}\frac{[P_i]}{r}dS, \tag{3} \label{eq:Curle}
\end{align}
where
\begin{align}
P_i = -n_j \{ \delta_{ij}p -\mu \left(\frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i} – \frac{2}{3}\delta_{ij}\frac{u_k}{x_k} \right) \}, \tag{4} \label{eq:Pi}
\end{align}
$$r$$ is the distance between the receiver $$\boldsymbol{x}$$ and the source position $$\boldsymbol{y}$$ in $$V$$ (or on $$S$$) and the operator [] denotes evaluation at the retarded time $$t-r/c_0$$.

• Ffowcs-Williams and Hawkings (FW-H) analogy

In the following video, we can watch the lecture by James Lighthill and John Ffowcs Williams !

 References (English)
 References (Japanese)

## Author: fumiya

CFD engineer in Japan


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## Introduction#

This course covers research topics related to economic slack. So what is economic slack? Economic slack describes the amount of productive resources in the economy that are unused. There are many different forms of slack: people who cannot find a job and remain unemployed; machines left idle in a factory; employed workers left idle on the job; hotel rooms, restaurants table, airplane seats that remain vacant; durable goods that cannot be sold and depreciate; perishable goods that cannot be sold and perish.

Economic slack represents a waste of productive resources, and it therefore is something that should be limited. Moreover, in addition to its wastefulness, unemployment generates other, large costs to society. People who are unemployed suffer from lower mental and physical health than employed workers. Even employed people in areas with high unemployment report lower well-being. Accordingly, good economic policy should stabilize economic slack at a desirable level and avoid periods of elevated slack.

The course is centered around formal modeling, but it also presents evidence supporting the assumptions introduced in the models. In the course we cover several models of economic slack and use them to answer several questions:

• Why does slack exist?
• How does slack affect economic life?
• Why does slack vary over time?
• How are slack fluctuations related to price and wage rigidities?
• What is the socially optimal amount of slack?
• How should monetary policy respond to fluctuations in slack over the business cycle?
• How should fiscal policy respond to fluctuations in slack over the business cycle?
• What happens at the zero lower bound?

#### Introductory video#

1. What is slack and why do we care about it?

• Krugman (2000) – This paper argues that uncomplicated models can be helpful. In this course we will develop models that are as uncomplicated as possible—and that are much simpler than models typically studied in graduate macroeconomic courses.
• Frey, Stutzer (2002) – This survey provides evidence that personal unemployment and aggregate unemployment significantly reduce happiness. The evidence refers to the pure effect of unemployment, which controls for the income loss and other inderect effects of unemployment.

This section first reviews the Kuhnian model of science. It then uses the Kuhnian perspective to understand how, over the past century, business-cycle macroeconomics evolved from the IS-LM model (inspired by Keynes’s General Theory) to the General Disequilibrium model (with nonclearing markets) to the Real Business-Cycle model (with perfectly competitive markets) and finally to the New Keynesian model (with monopolistically competitive markets).

#### Lecture videos#

• Summers (1986) – This paper discusses the origins and limitations of the Real Business-Cycle model.
• Benassy (1993) – This survey reviews the General Disequilibrium literature.
• Blanchard, Kiyotaki (1987) – This paper develops a business-cycle model with monopolistic competition that is a direct precursor to the New Keynesian model.

## Prevalence of slack and matching function#

This section documents the presence of economic slack on the labor market (unemployed workers) and on the product market (idle labor and capital). It also documents that such slack always coexists with vacant jobs and unfulfilled consumption. Then, we introduce the matching function, which is the tool that we will use to model the coexistence of unemployed workers and vacant jobs, and of idle labor and capital and unfulfilled consumption. The matching function summarizes the complex process through which workers searching for jobs meet firms searching for employees, and firms searching for customers meet consumers searching for sellers.

#### Lecture videos#

• Petrongolo, Pissarides (2001) – This survey reviews the microfoundations of the matching function, its empirical properties, and its applications.
• Shimer (2007) – This paper generates an aggregate matching function from mismatch in local labor markets.
• Montgomery (1991) – This paper generates an aggregate matching function from wage competition between firms.

## Basic model of slack#

This section develops a basic macroeconomic model of slack. The model is static. It is built around a matching function. Because of the matching function, self-employed workers are not able to sell all their services: there is always some slack. Wealth (in the form of real money balances) enters the utility function. Thanks to this assumption, and although the model is static, the aggregate demand is nondegenerate.

## Model of slack with income and wealth inequality#

This section introduces income and wealth inequality in the basic model of slack. We compute the aggregate demand and aggregate supply curves with inequality, and show how the model with inequality can be solved. In the model the marginal propensity to spend varies with slack, and the deviation from Say’s Law appears clearly.

#### Lecture videos#

• Saez, Zucman (2019) – This paper documents the rise of income and wealth inequality in the United States. The data come from distributional macroeconomic accounts.

## Discussion of the solution concept#

This section provides additional discussions of the solution concept used in the basic model of slack, and discusses an interesting special case. It also shows how the model solution is the equilibrium (in the sense from physics not economics) of a dynamical model in which households slowly learn the market tightness.

## Price and wage rigidities#

This section argues that prices and wages are not fully flexible but instead somewhat rigid. (Fairness seems to be a key reason behind price and wage rigidities.) We see how realistic pricing norms can be inserted into the basic model of slack. We also compute comparative statics in response to aggregate demand and aggregate supply shocks under fixed prices and rigid prices. We contrast these results to those obtained under bargained prices.

#### Lecture videos#

• Blinder (1994) – This paper argues, based on a survey of 200 US firms, that prices are sticky and that firms do not change prices more often by fear of antagonizing customers, who might find such price changes unfair.
• Bewley (2004) – This paper provides evidence of wage rigidity. It then argues that firms avoids pay cuts because they damage morale, which eventually reduces productivity, increases turnover, and complicates recruiting.
• Eyster, Madarasz, Michaillat (2021) – This paper develops a model of pricing in which buyers care about the fairness of markups, and firms take these concerns into account when setting prices. The model yields price rigidity and realistic Phillips curves.

## Model of slack with labor and product markets#

This section presents a model with two markets and two types of slack: a labor market with unemployment and a product market with idleness. Each market is organized around a matching function. Unemployment and idleness interact with each other. For instance, after an increase in aggregate demand, firms find more customers. This reduces the idle time of firms’ employees and thus increases firms’ labor demand. This in turn reduces unemployment.

#### Lecture videos#

• Michaillat, Saez (2015, sections 3–6) – These sections develop the model of slack with labor and product markets, and assess the sources of unemployment fluctuations in the United States.
• Diamond (2011) – This is Peter Diamond’s Nobel lecture. It discusses the applications of the matching framework to the product market and other markets.
• Wasmer, Weil (2004) – This paper develops a model of slack with labor and financial markets—each organized around a matching function.

## Dynamic model of slack#

This section presents a dynamic version of the basic model of slack. An advantage of moving to a dynamic environment is that interest rates appear into the model. Indeed, the real interest rate is a key determinant of aggregate demand. By setting a nominal interest rate, the central bank can stabilize the economy. The model is useful to quantify the effect of monetary policy on unemployment—for instance to assess the possibility of a soft landing in the aftermath of the pandemic inflation spike.

#### Lecture videos#

• Michaillat, Saez (2022, sections 1–4) – These sections develop the dynamic model of slack and perform various comparative statics.
• Michaillat, Saez (2021) – The dynamic model of slack assumes that wealth enters households’ utility function. This paper shows that introducing wealth in the utility is also helpful in the New Keynesian model. Indeed, it resolves all the anomalies of the New Keynesian model at the zero lower bound.
• Ball, Leigh, Loungani (2017) – This paper documents the prevalence of Okun’s law—the negative correlation between output and unemployment rate—in the United States since 1948. Okun’s law implies that output and market tightness are negatively correlated over the business cycle, which in turn implies that aggregate demand shocks are the main source of cyclical fluctuations.

## Social welfare, efficiency, and inefficiency#

The model of slack is generally inefficient. Except in a knife-edge case, there is too much or too little slack. This section therefore develops a simple formula for the efficient amount of slack. It shows that the US economy is generally inefficient, and is especially exceedingly slack in slumps.

#### Lecture videos#

• Michaillat, Saez (2022) – This paper derives the formula $u^\ast = \sqrt{uv}$.
• Michaillat, Saez (2021) – This paper derives a sufficient-statistic formula for the efficient unemployment rate that generalizes the formula $u^\ast = \sqrt{uv}$. The general formula involves the Beveridge elasticity, cost of unemployment, and cost of recruiting.
• Chetty (2009) – This survey describes the sufficient-statistic method for welfare and policy analysis.

## Optimal monetary policy over the business cycle#

This section describes optimal monetary policy over the business cycle. Monetary policy influences the aggregate-demand curve, so it can be used to shrink the unemployment gap. In fact, the optimal monetary policy is to adjust interest rates to eliminate the unemployment gap entirely. So the central bank should lower rates in bad times, when unemployment is inefficiently high, and raise rates in good times, when unemployment is inefficiently low.

#### Lecture videos#

• Michaillat, Saez (2022, sections 5–6) – These sections obtain the sufficient-statistic formula for optimal monetary policy and apply it to the US economy.
• Bernanke, Blinder (1993) – This paper estimates the response of the federal funds rate to unemployment, and the effect of the federal funds rate on unemployment.
• Coibion (2012) – This paper blends the narrative and VAR approaches to estimate the monetary multiplier—the effect of an increase in the federal funds rate on the unemployment rate.

## Optimal government spending over the business cycle#

This section studies optimal public expenditure over the business cycle. It shows that that optimal public expenditure deviates from the Samuelson rule to reduce, but not eliminate, the unemployment gap. The amplitude of the deviation depends on the unemployment gap, fiscal multiplier, and elasticity of substitution between public and private goods. Since the unemployment gap is countercyclical, optimal public expenditure is also countercyclical. That is, the government should spend more in bad times and less in good times.


text:
# Asymptotically Polynomial Type Solutions for Some 2-dimensional Coupled Nonlinear Odes

Volume 14, Issue 3, pp 211-221 Publication Date: April 28, 2015
• 804 Views

### Authors

B. V. K. Bharadwaj - Department of Mathematics and Computer Science Sri Sathya Sai Institute of Higher Learning Prasanthinilayam – 515134, INDIA. Pallav Kumar Baruah - Department of Mathematics and Computer Science Sri Sathya Sai Institute of Higher Learning Prasanthinilayam – 515134, INDIA.

### Abstract

In this paper we have considered the following coupled system of nonlinear ordinary differential equations. $x^{n_1}_1(t)=f_1(t,x_2(t))$ $x^{n_2}_2(t)=f_2(t,x_1(t))$ where $f_1,f_2$ are real valued functions on $[t_0,\infty)×R, \quad t\geq t_0>0$. We have given sufficient conditions on the nonlinear functions $f_1,f_2$, such that the solutions pair $x_1,x_2$ asymptotically behaves like a pair of real polynomials.

### Keywords

• Nonlinear Coupled Ordinary Differential Equations
• Fixed-point Theorem
• Assymptotically Polynomial like solutions

•  34D05
•  47H10
•  34A34

### References

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