text:
We study moduli spaces of sheaves over nonprojective K3 surfaces. More precisely, let omega be a Kahler class on a K3 surface S, let rgeq 2 be an integer, and let v = (r,\xi, a) be a Mukai vector on S. We show that if the moduli space M of omega-stable vector bundles with associated Mukai vector v is compact, then M is an irreducible holomorphic symplectic manifold which is deformation equivalent to a Hilbert scheme of points on a K3 surface. Moreover, we show that there is a Hodge isometry between v^{perp} and H^2(M,mathbb{Z}) and that M is projective if and only if S is projective.

### Moduli spaces of bundles over nonprojective K3 surfaces

#### Abstract

We study moduli spaces of sheaves over nonprojective K3 surfaces. More precisely, let omega be a Kahler class on a K3 surface S, let rgeq 2 be an integer, and let v = (r,\xi, a) be a Mukai vector on S. We show that if the moduli space M of omega-stable vector bundles with associated Mukai vector v is compact, then M is an irreducible holomorphic symplectic manifold which is deformation equivalent to a Hilbert scheme of points on a K3 surface. Moreover, we show that there is a Hodge isometry between v^{perp} and H^2(M,mathbb{Z}) and that M is projective if and only if S is projective.
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Utilizza questo identificativo per citare o creare un link a questo documento: http://hdl.handle.net/11567/897031


text:
# Inequality challenge

#### anemone

##### MHB POTW Director
Staff member
Let $x, y$ be real numbers such that $9x^2+8xy+7y^2 \le 6$.

Show that $7x+12xy+5y \le 9$.

#### Opalg

##### MHB Oldtimer
Staff member
[graph]1hbxwofxmo[/graph] [Click on the graph for a larger version.]

The blue curve is the ellipse $9x^2+8xy+7y^2 = 6$. The brown curve is the hyperbola $7x+12xy+5y = 9$. They share a common tangent at the point $(1/2,1/2)$, with gradient $-13/11.$ The interior of the ellipse is the region $9x^2+8xy+7y^2 \leqslant 6$, and the region between the two branches of the hyperbola is given by $7x+12xy+5y \leqslant 9$. From the geometry, it is clear that the first of those regions is contained in the second. I don't see a neat way to prove that algebraically, but I think this is one of those cases where a picture says more than an equation could.

#### Albert

##### Well-known member
Let $x, y$ be real numbers such that $9x^2+8xy+7y^2 \le 6$.

Show that $7x+12xy+5y \le 9$.---(2)

Let $x, y$ be real numbers such that $9x^2+8xy+7y^2 \le 6 ----(1)$.
in fact we only have to discuss :x>0, and y>0
$9x^2+8xy+7y^2 \le 9x^2+4x^2+4y^2+7y^2$ (GM$\le AM)$
the equivalence exists only if x=y
so we have :$24x^2\le 6$
$\therefore x=y\le \dfrac{1}{2}$
$\therefore$ the left side of (2) :$\dfrac{7}{2}+\dfrac{12}{4}+\dfrac{5}{2}\le 9$

#### anemone

##### MHB POTW Director
Staff member
Let $x, y$ be real numbers such that $9x^2+8xy+7y^2 \le 6$.

Show that $7x+12xy+5y \le 9$.
Thank you to both of you, Opalg and Albert for participating. It's good to know there are many ways(graphical/algebraic) to solve a problem.

I think the solution proposed by other that wanted to share here is very similar to the concept of Albert, let's see:

If we let $x=a+\dfrac{1}{2}$ and $y=b+\dfrac{1}{2}$, where $a, b \in R$, the given inequality becomes

$9x^2+8xy+7y^2 \le 6$

$9\left( a+\dfrac{1}{2} \right)^2+8\left( a+\dfrac{1}{2} \right)\left( b+\dfrac{1}{2} \right)+7\left( b+\dfrac{1}{2} \right)^2 \le 6$

$13a+11b+8ab \le -(9a^2+7b^2)$

The inequality that we wanted to prove, its LHS after undergo the transformation turns to

$7x+12xy+5y=7\left( a+\dfrac{1}{2} \right)+12\left( a+\dfrac{1}{2} \right)\left( b+\dfrac{1}{2} \right)+5\left( b+\dfrac{1}{2} \right)=13a+11b+8ab+9 \le 9-(9a^2+7b^2) \le 9$

Adding 9 to both sides of the inequality $13a+11b+8ab \le -(9a^2+7b^2)$ gives

$7x+12xy+5y=13a+11b+8ab+9 \le 9-(9a^2+7b^2) \le 9$

and we are done.

format: plain


text:
## Monday, September 28, 2015

### LeetCode OJ - Unique Paths II

Problem:

Solution:

Unlike Unique Paths, this problem has no closed form solution. Instead, we will use the simple dynamic programming algorithm.

At each cell, we compute the number of ways to reach a certain cell. The initial case is trivial.

ways[0][0] = 1.

Next, we can reach a cell either from above our from left (if they exists), so we have these.

ways[row][col] = ways[row - 1][col] + ways[row][col -1] // Decay the number of zero if array index out of bound.

That's give the following dynamic programming algorithm.

Code:

#include "stdafx.h"

// https://leetcode.com/problems/unique-paths-ii/

#include "LEET_UNIQUE_PATHS_II.h"
#include <map>
#include <iostream>
#include <sstream>
#include <vector>
#include <string>

using namespace std;

namespace _LEET_UNIQUE_PATHS_II
{
class Solution {
public:
int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid)
{
int height = obstacleGrid.size();
if (height > 0)
{
int width = obstacleGrid[0].size();
if (width > 0)
{
for (int h = 0; h < height; h++)
{
}
for (int h = 0; h < height; h++)
{
for (int w = 0; w < width; w++)
{
if (obstacleGrid[h][w] == 1)
{
}
else
{
int above = h == 0 ? 0 : answer[h - 1][w];
int left = w == 0 ? 0 : answer[h][w - 1];
int bonus = (h == 0 && w == 0) ? 1 : 0;
answer[h][w] = above + left + bonus;
}
}
}
return answer[height - 1][width - 1];
}
}
return 0;
}
};
};

using namespace _LEET_UNIQUE_PATHS_II;

int LEET_UNIQUE_PATHS_II()
{
Solution solution;
vector<vector<int>> obstacleGrid;
obstacleGrid.resize(3);
obstacleGrid[0].push_back(0);
obstacleGrid[0].push_back(0);
obstacleGrid[0].push_back(0);
obstacleGrid[1].push_back(0);
obstacleGrid[1].push_back(1);
obstacleGrid[1].push_back(0);
obstacleGrid[2].push_back(0);
obstacleGrid[2].push_back(0);
obstacleGrid[2].push_back(0);
cout << solution.uniquePathsWithObstacles(obstacleGrid) << endl;
return 0;
}


text:
# Isomorphism of R-modules

Does somebody has an example where the left $R$-modules $R^m$ and $R^n$ are isomorphic for all positive integers $m$, and $n$?

#### Solutions Collecting From Web of "Isomorphism of R-modules"

Let $\mathrm{CFM}_\mathbb{N}(R)$ denote the ring of “column finite matrices”, where $R$ is some ring.. Then, one can show that $\mathrm{CFM}_\mathbb{N}(R)\to\mathrm{CFM}_\mathbb{N}(R)^2$ defined by

$$M\mapsto (\text{odd indexed columns of }M,\text{even indexed columns of }M)$$

is an isomorphism of $\mathrm{CFM}_\mathbb{N}(R)$-modules. It clearly then follows that $\mathrm{CFM}_\mathbb{N}(R)^m\cong\mathrm{CFM}_\mathbb{N}(R)^n$ for all $n$ and $m$.

I believe this example is in Dummit and Foote.


text:
# Must have Dover books?

1. Jan 31, 2008

### Shaun Culver

"Must have" Dover books?

Name the field & the book.

2. Jan 31, 2008

### Andy Resnick

Here's some I regularly consult- they are within 5 feet of my computer:

Mathematics Applied to Continuum Mechanics (Segel)
The Principles of Statistical Mechanics (Tolman)
Hydrodynamics (Lamb)
Light Scattering by Small Particles (Van de Hulst)

8. Apr 7, 2010

### Gannet

Re: "Must have" Dover books?

• Aerodynamics - Abbott, Ira H. and Albert E. Von Doenhoff; Theory of Wing Sections; Dover Publications, Inc.; New York, NY; 1959
• Aerodynamics - Ashley, Holt and Marten Landahl; Aerodynamics of Wings and Bodies; Dover Publications, Inc.; New York, NY; 1965
• Aeronautical Engineering - Ashley, Holt; Engineering Analysis of Flight Vehicles; Dover Publications, Inc.; New York, NY; 1974
• Fluid Mechanics - Milne-Thomson, L. M. (Louis Melville); Theoretical Hydrodynamics, Fifth Edition; Dover Publications, Inc.; Mineola, NY; 1968
• Fluid Mechanics - Rosenhead, L.; Laminar Boundary Layers; ; Dover Publications, Inc.; New York, NY; 1963
• Aerodynamics - Thwaites, Bryan; Incompressible Aerodynamics; Dover Publications Inc; New York, NY 10016; 1960
• Fluid Mechanics - Tietjens, O. G.; Fundamentals of Hydro- and Aeromechanics; Dover Publications, Inc.; New York, NY; 1934
• Aeronautical Engineering - Von Mises, Richard; THEORY OF FLIGHT; Dover Publication Inc.; New York, NY; 1995

The ones for aerodynamics and fluid mechanics complements each other.

9. Apr 7, 2010

### turbo

Re: "Must have" Dover books?

Got an interest in observational astronomy? Burnham's 3-volume set is killer. Dated, but still killer. Combine the recommendations in those guides with modern charts, and you've got a lifetime of observing-programs.

Last edited: Apr 7, 2010
10. Apr 7, 2010

### fourier jr

Re: "Must have" Dover books?

i like kline's but maybe not a substitute for spivak's if you're looking for rigour. he writes in the preface to the 2nd edition that "rigour undoubtedly refines the intuition but does not supplant it" and that his approach makes sense because calculus grew out of physical and geometric problems, which makes a lot of sense to me also. if you want something more "hardcore" & with less motivation of its concepts maybe there better books though. what i like about it is its realistic applications. most calculus books have highly contrived, totally unrealistic & unconvincing applications but for example in the sections on polar coordinates he derives kepler's laws; in max/min he does fermat's principle of least time, etc. (i don't know if it's unique in that respect but i still like it)

i guess i could list other must-haves since i'm here
willard - general topology
pfaff/johnsonbaugh - foundations of analysis (not exactly foundations; covers same stuff as little rudin but is more user-friendly)
knopp - theory of functions (complex analysis) & prob book in theory of functions (2 vols in 1)
kamke - theory of sets
suppes - axiomatic set theory
dixon - probs in group theory
tenenbaum/pollard - ODEs
ince - ODEs
schwerdtfeger - geometry of complex numbers
kolmogorov/fomin - intro real analysis & elements of functional analysis
steen/seebach - counterexamples in topology
gelbaum/olmsted - counterexamples in analysis
edwards - riemann's zeta function
dudley - elementary number theory
boyer - history of calculus & history of analytic geometry
crowe - history of vector calculus
ball/coxeter - mathematical recreations & essays
wussing - genesis of the abstract group concept

Last edited: Apr 7, 2010
11. Apr 7, 2010

### turbo

Re: "Must have" Dover books?

Truly, for an amateur astronomer, Burnham's Celestial Handbook (V 1-3) is a treasure. Those of us who learned to find faint objects using Burnhams and Tirion's charts will have a much richer level of experience and knowledge than the generation that relied on the Go-To stuff.

12. Apr 9, 2010

### marcusl

Re: "Must have" Dover books?

By Edmonds? Are you sure that's a Dover book?

Dover books I have used frequently:

Brillouin, Wave Propagation in Solids [a brilliant book]
Harrison, Solid State Theory
Schwartz, Principles of Electrodynamics
Tinkham, Introduction to Superconductivity
(I bought the last two when my hardcover original editions fell apart)
Wainstein and Zubakov, Extraction of Signals From Noise
Grover, Inductance Calculations
Gakhov, Boundary Value Problems

I was sad when Dover switched from paperbacks with sewn bindings to cheap glued ones. Some of the old Dover books are holding up better than hardcover books costing 10 times as much.

13. Apr 9, 2010

### Cod

Re: "Must have" Dover books?

"Essential Calculus with Applications" by Richard A. Silverman

A great calculus book that can be used as a textbook or reference for a Calculus I course. I use it frequently when I get mad at all the textbook fluff in my current textbook. Silverman's book has numbers, proofs, numbers, theorems, and more numbers. Very little sentence structure, but enough to help you understand the concepts. I plan on using it between semesters to keep me fresh for Calculus II this summer.

14. Apr 9, 2010

### Pinu7

Re: "Must have" Dover books?

My favorite Dovers:

Anything by Kolmogorov.

Linear Algebra- Shilov

Themodynamics-Fermi

Differential Geometry- Kreyszig

Calculus of Variations- Gelfand

The Variational Principals of Mechanics- Lanczos(also see his linerar differential operators text)

Theoretical Physics-Joos

Counterexamples in Analysis- Gelbaulm

I think Dover Publications deserves a fields metal.

15. Apr 24, 2010

### Stratosphere

Re: "Must have" Dover books?

How does that compare to https://www.amazon.com/Introduction...d-Griffiths/dp/013805326X/ref=dp_ob_title_bk"

Last edited by a moderator: Apr 25, 2017
16. Apr 27, 2010

### marcusl

Re: "Must have" Dover books?

I can't say enough good things about it, but it's an advanced undergrad book not necessarily the best choice for starting. Griffiths is the most popular there.

Do a PF search on Schwartz electrodynamics and you'll pull up a lot of opinions, mostly positive. Here are two short comments of mine:


text:
# 18th Century Britain. Иностранный язык. Фомина И.В. - 15 стр.

Составители:

Рубрика:

• ## Иностранный язык

15
James Edward had been taken to France and brought up there. His father, the
exiled King James II of England and VII of Scotland, died in 1701, and after that the
Jacobites supporting the king in exile, proclaimed James Edward King James III of
England and VIII of Scotland.
James Edward might call himself a king, but he had to struggle for his father's
domain. In 1708 he attempted to invade Scotland with French soldiers, but failed. He
decided to try his chance again and landed in Britain after Queen Anne's death, but
was forced to flee soon.
Britain was actually entering two decades of relative peace and stability. Local
government was left largely in the hands of country gentlemen owning large estates.
As justices of the peace, they settled the majority of legal disputes. They also
administered roads, bridges, inns, and markets and supervised the local operation of
the Poor Law aid to orphans, paupers, the very old, and those too ill to work. The
Glorious Revolution was not really safe until the failure of the Jacobite rebellion of
1715. It was clear then that the sentiments of the past need not shed blood in the
present, nor imperil the political order on which the future wealth of the nation would
depend. The fact that the counter-attack of the party of church and king had gone
down in ruin in 1714 was even more important.
The party of Church and King was led by Henry St. John, Viscount Bolingbroke.
Bolingbroke may be called one of the most outstanding leaders any political party has
ever had. He was only twenty-two when he became a member of Parliament and
made himself the spokesman of the country gentry against the Whig settlement. Nine
years later, in 1710, he became Secretary of State, determined to "break the body of
the Whigs". . England had been at war with France, and it, was Bolingbroke who
took a leading part in the negotiations which led to the peace of Utrecht in 1713. The
Tories hoped that France would support a Stuart restoration in England, and the
Hanoverian succession was really in danger. But the Tories were divided between the
Hanoverian Tories under Harley who saw the dangers of Jacobitism, and
Bolingbroke, who was prepared to bring back a Stuart king. The Tories were not
united when Queen Anne died and thus lost their opportunity. The Whigs were able
to start the process of the transmutation of limited monarchy into parliamentary
government.
The practice of limited monarchy could not be codified. The system of
government which the British empirically developed has never been completely
explained. The British constitution might be the envy of Europe because it allowed a
hitherto unknown political freedom, but its theory was incomprehensible, and its
working was not as simple as a machine.
The system was monarchical. When political decisions had to be made the king
was the only person known to the law who could make them. William III . secured
the sole power to bestow titles and dignities. He named all offices both civil and
military and the coin bore his effigy. Moreover, he possessed the superintendancy
over all the laws to render them effectual, and justice was administered in his name.
Using these prerogative powers reasonably, William III helped to secure for the
Crown a permanent place in the British constitution. He was his own first minister.
                                          15
James Edward had been taken to France and brought up there. His father, the
exiled King James II of England and VII of Scotland, died in 1701, and after that the
Jacobites supporting the king in exile, proclaimed James Edward King James III of
England and VIII of Scotland.
James Edward might call himself a king, but he had to struggle for his father's
domain. In 1708 he attempted to invade Scotland with French soldiers, but failed. He
decided to try his chance again and landed in Britain after Queen Anne's death, but
was forced to flee soon.
Britain was actually entering two decades of relative peace and stability. Local
government was left largely in the hands of country gentlemen owning large estates.
As justices of the peace, they settled the majority of legal disputes. They also
administered roads, bridges, inns, and markets and supervised the local operation of
the Poor Law aid to orphans, paupers, the very old, and those too ill to work. The
Glorious Revolution was not really safe until the failure of the Jacobite rebellion of
1715. It was clear then that the sentiments of the past need not shed blood in the
present, nor imperil the political order on which the future wealth of the nation would
depend. The fact that the counter-attack of the party of church and king had gone
down in ruin in 1714 was even more important.
The party of Church and King was led by Henry St. John, Viscount Bolingbroke.
Bolingbroke may be called one of the most outstanding leaders any political party has
ever had. He was only twenty-two when he became a member of Parliament and
made himself the spokesman of the country gentry against the Whig settlement. Nine
years later, in 1710, he became Secretary of State, determined to "break the body of
the Whigs". . England had been at war with France, and it, was Bolingbroke who
took a leading part in the negotiations which led to the peace of Utrecht in 1713. The
Tories hoped that France would support a Stuart restoration in England, and the
Hanoverian succession was really in danger. But the Tories were divided between the
Hanoverian Tories under Harley who saw the dangers of Jacobitism, and
Bolingbroke, who was prepared to bring back a Stuart king. The Tories were not
united when Queen Anne died and thus lost their opportunity. The Whigs were able
to start the process of the transmutation of limited monarchy into parliamentary
government.
The practice of limited monarchy could not be codified. The system of
government which the British empirically developed has never been completely
explained. The British constitution might be the envy of Europe because it allowed a
hitherto unknown political freedom, but its theory was incomprehensible, and its
working was not as simple as a machine.
The system was monarchical. When political decisions had to be made the king
was the only person known to the law who could make them. William III . secured


text:
# Inverse fourier transform of $1/(1+s^2)$

Hoi, I want to have the inverse fourier transform $\mathcal{F}^{-1}(\frac{1}{1+s^2})$.

So I thought about using some properties of fourier-transform. But knowing the answer I must make some sort of mistakes in my reasoning, but i dont understand what im doing wrong:

I know the answer is : $$ce^{-|x|}$$ and according to wolframalpha $c= \sqrt{\pi/2}$.

But i got this: Some calculations give:

$$\frac{1}{1+s^2} = \frac{1}{1-is}\cdot \frac{1}{1 +is} = \mathcal{F}(H(t)e^{-t})\cdot \mathcal{F}(H(-t)e^{t}) = \mathcal{F}[H(t)e^{-t}\ast H(-t)e^{t}]$$

That is according some properties of Fourer transform: $F(g \ast f) = F(g)F(f)$

So that would then imply the answer is $$H(t)e^{-t}\ast H(-t)e^{t}$$

But that doesnt give me the right answer...what is my big error. I get calculating this convolution: $\frac{1}{2}e^{-x}$

-
Is it $\frac{1}{1+s^2}$ or $\frac{1}{(1+s)^2}$? – Mhenni Benghorbal Apr 1 '13 at 13:34
$1/(1+s^2)$. Sorry. Otherwise none of what I said wouldve made sense :P – DinkyDoe Apr 1 '13 at 13:47
You can use residue theorem. – Mhenni Benghorbal Apr 1 '13 at 18:56
for what exactly do i use it then. In line of the reasoning above I wouldnt need it, so what is my error? – DinkyDoe Apr 1 '13 at 19:30
The answer should be $\frac{1}{2}e^{-t}H(t) + \frac{1}{2}e^{t}H(-t).$ – Mhenni Benghorbal Apr 2 '13 at 14:15

In your calculation of the convolution I'm sure that at some point you get a $\sqrt{x}^2$, and you're saying that's equal to $x$, when it's actually $|x|$, that is what that answer is telling you. It's a very common mistake, check you're calculations agains to see if that is the problem.

-
ok, do u see a problem with my reasoning here then? Cause im almost sure the convolution itself is the right answer. But see under here, what mistake do I make then... – DinkyDoe Apr 1 '13 at 12:38

There should be a mistake in your computation of convolution, because the map $t\mapsto H(t)e^{-t}\star H(-t)e^{t}$ is even (convolution is commutative).

-
Im sorry: meant to say $H(t)e^{-t}\star H(-t)e^{t}$ – DinkyDoe Apr 1 '13 at 12:28
So, there should be a mistake. I got that. But is my reasoning correct to conclude that $H(t)e^{-t}\star H(-t)e^{t}$ is the right answer? And if so...see under here. What am I misunderstanding here? – DinkyDoe Apr 1 '13 at 12:57

In your Calculation, x shouldn't be the lower border in the integral. It must be H(x)*x. Because, if x < 0, the integral is cut, then the lower border is 0. And in the end it is (x-2H(x)*x) = -|x| ... :)

-


text:
Help w/ Proof in Category Theory

• January 6th 2009, 05:00 PM
mingcai6172
Help w/ Proof in Category Theory
I don't understand this proof, specifically the part in red, I don't understand. Please help me understand this step in the proof. Thanks!

Let Tors be the category whose objects are torsion abelian
groups; if $A$ and $B$ are torsion abelian groups, we deﬁne $\text{Mor}_{\text{\textbf{Tors}}}(A, B)$ to
be the set of all (group) homomorphisms $\phi : A \rightarrow B$. Prove that direct
products exist in Tors; that is, show that given any indexed family ${A_i}_{i \in I}$
where each $A_i$ is a torsion abelian group, there exists a torsion abelian group
which serves as a direct product for this family in Tors.

Proof- Let $T$ be the torsion subgroup (that is, the subgroup of elements of ﬁnite
order) of $P = \prod_{i \in I} A_i = \{f : I \rightarrow A_i : f(i) \in A_i \text{ } \forall i\}$; here of course
$P$ is the direct product of $\{A_i\}_{i \in I}$ in the category Ab. Let $j : T \rightarrow P$ be
the inclusion and for each $i \in I$, let $\pi_i : P \rightarrow A_i$ denote the usual projection
map; that is, $\pi_i(f) = f(i)$. (In coordinate notation, $\pi_i(a_0, a_1, \ldots) = a_i$.)
For each $i \in I$ deﬁne $\tau_i : T \rightarrow A_i$ by $\tau_i = \pi_i * j$. I claim that the group
$T$ together with the maps $\{\tau_i\}_{i \in I}$ constitute a direct product for $A_{i \in I}$ in
Tors. Well, given a torsion group $S$ and maps $\sigma_i : S \rightarrow A_i$ for each $i \in I$,
one deﬁnes $h : S \rightarrow T$ as follows: given $s \in S$, let $h(s) \in T$ be the function
deﬁned by $\color{red}{[h(s)](i) = \sigma_i(s)}$. Then clearly $\tau_i * h = \sigma_i \text{ } \forall i \in I$. Moreover,
if $h' : S \rightarrow T$ is any other map such that $\tau_i * h' = \sigma_i$, then for any $s \in S$ and
$i \in I$, $\color{red}{[h(s)](i) }= (\tau_i * h)(s) = \sigma_i(s) = (\tau_i * h')(s) = \color{red}{[h'(s)](i)}$, so $h = h'$.

I don't understand what $\color{red}{[h(s)](i) = \sigma_i(s)}$ is.
• January 6th 2009, 05:53 PM
GaloisTheory1
Function-valued functions can be confusing; I've always found that stepping through the notation in painstaking detail is very helpful.

By definition, h : S --> T
Therefore, $h(s) \in T$
By definition of T, h(s) is a function $I \to A_i$
h(s)(i) is, therefore, evaluating h(s) at i. (And $h(s)(i) \in A_i$)


text:
• # question_answer Mansi has a bag that contains red, blue and green balls. What fraction of the total balls are blue? A)  $\frac{5}{13}$                      B)           $\frac{2}{10}$             C)  $\frac{4}{13}$                      D)           $\frac{6}{13}$

Total number of balls =13 Number of blue balls = 5 Fraction of blue balls = $\frac{5}{13}$


text:
# How to rewrite these formulas in the opposite form?

#### jenniferruurs

I want to know how I can write from some predicate formula its negative form.

For example
âˆ€xâˆƒy(x R y)
then not(âˆ€xâˆƒy(x R y)) is equal to âˆƒx,not âˆƒy(x R y)))(not exist y)

How do these formula work with conjunction implication and disjunction?

Bellow are examples of some formulas but I also want to know how to rewrite the formulas if there are conjunctions and disjunctions involved.

$\forall a,b,c \in X: (aRb \wedge bRc) \Rightarrow aRc$
what do we have if we have the negation of this formula?

also the opposite of:
$\forall a, b \in X(a R b \Rightarrow \lnot(b R a))$

also for the opposite of:
$\forall a, b, c: a R b \land b R c \Rightarrow \lnot (a R c)$

also for the opposite of
$\forall a, b, c\in X\,(a\,R\, b \land a \,R\, c \to b \,R\, c)$

also for the opposite of:
$\forall a, b \in X(a R b \Leftrightarrow b R a)$

Last edited by a moderator:

#### idontknow

Simply there is another symbol of relation R which is an opposite of the given relation .

#### jenniferruurs

Simply there is another symbol of relation R which is an opposite of the given relation .
What do you mean with this?

1 person

#### idontknow

About the example $$\displaystyle A : \forall x \exists y (x R y)$$ we can have two opposites of A .
$$\displaystyle \neg A =\forall x \not\exists y (xRy)\;$$ , but instead of $$\displaystyle \forall x$$ the $$\displaystyle \exists ! x$$ ( exists only one x) also is a type of opposite of A .
So the best way is to just simply write the opposite of relation R with a symbol .

#### jenniferruurs

$$\displaystyle \neg A =\forall x \not\exists y (xRy)\;$$ , but instead of $$\displaystyle \forall x$$ the $$\displaystyle \exists ! x$$ ( exists only one x) also is a type of opposite of A .
So the best way is to just simply write the opposite of relation R with a symbol .
How would you rewrite this
âˆ€w1w2w3(w1Rw2 âˆ§ w1Rw3 â†’ âˆƒw(w2Rw âˆ§ w3Rw))

If you want to rewrite it like:
not(âˆ€w1w2w3(w1Rw2 âˆ§ w1Rw3 â†’ âˆƒw(w2Rw âˆ§ w3Rw))))

Can this be rewritten as:
âˆƒw1w2w3(w1Rw2 âˆ§ w1Rw3 â†’ (not âˆƒw(not exist w)(w2Rw âˆ§ w3Rw))))?

#### idontknow

Yes but still $$\displaystyle \rightarrow \exists$$ can have the opposite like :
$$\displaystyle \rightarrow \not\exists$$ and $$\displaystyle \not\rightarrow \exists$$ . So you can write the opposite but depending on what relation is .

#### jenniferruurs

Yes but still $$\displaystyle \rightarrow \exists$$ can have the opposite like :
$$\displaystyle \rightarrow \not\exists$$ and $$\displaystyle \not\rightarrow \exists$$ . So you can write the opposite but depending on what relation is .
I think that you and I had it both wrong and that the implication had to be removed.

âˆƒw1w2w3(w1Rw2 âˆ§ w1Rw3 âˆ§ (not âˆƒw(not exist w)(w2Rw âˆ§ w3Rw))))

This is the correct formula do you agree?

1 person

#### idontknow

Yes this way gives the full or true opposite , removing the implication.


text:
Reconstruction algorithms for low-rank tensors and depth-3 multilinear circuits
We give new and efficient black-box reconstruction algorithms for some classes of depth-3 arithmetic circuits. As a consequence, we obtain the first efficient algorithm for computing the tensor rank and for finding the optimal tensor decomposition as a sum of rank-one tensors when then input is a constant-rank tensor. More specifically, we provide efficient learning algorithms that run in randomized polynomial time over general fields and in deterministic polynomial time over and for the following classes: 1) Set-multilinear depth-3 circuits of constant top fan-in ((k) circuits). As a consequence of our algorithm, we obtain the first polynomial time algorithm for tensor rank computation and optimal tensor decomposition of constant-rank tensors. This result holds for d dimensional tensors for any d, but is interesting even for d=3. 2) Sums of powers of constantly many linear forms ((k) circuits). As a consequence we obtain the first polynomial-time algorithm for tensor rank computation and optimal tensor decomposition of constant-rank symmetric tensors. 3) Multilinear depth-3 circuits of constant top fan-in (multilinear (k) circuits). Our algorithm works over all fields of characteristic 0 or large enough characteristic. Prior to our work the only efficient algorithms known were over polynomially-sized finite fields (see. Karnin-Shpilka 09’). Prior more »
Authors:
; ;
Award ID(s):
Publication Date:
NSF-PAR ID:
10292839
Journal Name:
STOC 2021: Proceedings of the 53rd Annual ACM SIGACT Symposium on Theory of Computing
Page Range or eLocation-ID:
809 to 822
3. We study the problem of maximizing a non-monotone submodular function subject to a cardinality constraint in the streaming model. Our main contributions are two single-pass (semi-)streaming algorithms that use $\tilde{O}(k)\cdot\mathrm{poly}(1/\varepsilon)$ memory, where $k$ is the size constraint. At the end of the stream, both our algorithms post-process their data structures using any offline algorithm for submodular maximization, and obtain a solution whose approximation guarantee is $\frac{\alpha}{1+\alpha}-\varepsilon$, where $\alpha$ is the approximation of the offline algorithm. If we use an exact (exponential time) post-processing algorithm, this leads to $\frac{1}{2}-\varepsilon$ approximation (which is nearly optimal). If we post-process with the algorithm of Buchbinder-Feldman '19, that achieves the state-of-the-art offline approximation guarantee of $\alpha=0.385$, we obtain $0.2779$-approximation in polynomial time, improving over the previously best polynomial-time approximation of $0.1715$ due to Feldman'18. One of our algorithms is combinatorial and enjoys fast update and overall running times. Our other algorithm is based on the multilinear extension, enjoys an improved space complexity, and can be made deterministic in some settings of interest.
4. We study an online hypergraph matching problem with delays, motivated by ridesharing applications. In this model, users enter a marketplace sequentially, and are willing to wait up to $d$ timesteps to be matched, after which they will leave the system in favor of an outside option. A platform can match groups of up to $k$ users together, indicating that they will share a ride. Each group of users yields a match value depending on how compatible they are with one another. As an example, in ridesharing, $k$ is the capacity of the service vehicles, and $d$ is the amount of time a user is willing to wait for a driver to be matched to them. We present results for both the utility maximization and cost minimization variants of the problem. In the utility maximization setting, the optimal competitive ratio is $\frac{1}{d}$ whenever $k \geq 3$, and is achievable in polynomial-time for any fixed $k$. In the cost minimization variation, when $k = 2$, the optimal competitive ratio for deterministic algorithms is $\frac{3}{2}$ and is achieved by a polynomial-time thresholding algorithm. When $k>2$, we show that a polynomial-time randomized batching algorithm is $(2 - \frac{1}{d}) \log k$-competitive, and it is NP-hardmore »


text:
# Difference between revisions of "FAQ"

 TODO: This FAQ needs to have a lot of things worked on. To avoid clutter from repeated TODO tags, just note missing answers with "...", and put notes about missing information in parentheses. (See: To-Do List)

### What is ConTeXt?

It is a document production system based on Donald Knuth's [TeX]. It is a set of high-level macros written in the (somewhat arcane) TeX language, and these macros simplify the task of designing layouts, making hyperlinks, etc.

### Is there a hello-world document?

Yes. See First Document and Hello world.

### Is ConTeXt free (open source) software?

Yes. See the Read Me page for details.

See Installation

### Does ConTeXt support Unicode?

Yes. It supports Unicode (utf-8) input, with the \enableregime[utf-8] command. See also Encodings and Regimes (This needs a bit of elaboration!) If you use XeTeX, Unicode should work by default.

### Why does \startalignment[right] or \framed[align=right]{some text} align my text left instead of right and vice versa?

This is for backward compatibility with older versions; see Right and left. Hans was originally thinking of "ragged right" and "ragged left" alignment. Use flushleft and flushright instead.

### What are the differences between ConTeXt and LaTeX?

The main difference between the two, in my opinion, is a philosophical one: LaTeX was created with the idea of separating content and presentation to such an extent that the typical author would write their content and then use a style file created by someone else to provide the visual presentation. Thus, the proliferation of style files and related things.
ConTeXt, on the other hand, retained the idea of separating content and presentation, but was created with the idea of being used for books, where each book tends to have a different layout, and so the expected "end user" is the person doing all the layout. Thus, it's designed to provide a vast amount of flexibility for layout in a way that can be fairly easily defined without needing to write a package (or go find one that's already written).
To some extent, modern LaTeX has become a lot more like ConTeXt's intent, with a large quantity of packages, and classes like "memoir" that are designed to give the end user a large amount of control of the visual presentation. Thus, the difference is more one of focus and arrangement -- ConTeXt is all "of a piece" and most things are built in and designed as a complete set, whereas with LaTeX it's generally the case that when you want to do something different, you need to find the right package, and sometimes find that there are three that almost do what you want and none that are quite right (but, then, that also means that there are usually choices for how to do things, which can be good.)
Another difference is that IMO, ConTeXt is a lot easier to program in. It was designed a fair bit later than LaTeX, for much faster computers, and thus includes a lot of things that make it load and run slower but make programming a lot easier.
Author: Brooks Moses on c.t.t. usenet
(Note that the internal coding of ConTeXt has become much easier, too, with the use of LuaTeX - or perhaps seemingly more complicated, as one has to then learn lua to understand the internal workings of ConTeXt. Most users will never delve into this level, but do indeed benefit from the increased performance of the ConTeXt system. The loading of the ConTeXt macros and the processing of even very large typesetting projects is not slow!)

### What are the advantages of ConTeXt over LaTeX?

ConTeXt is a macro package that’s far more advanced than LaTeX. You can enhance LaTeX with third party packages, but not all macro packages work together with each other. ConTeXt is an integrated, powerful and flexible macro package for which you seldom need third party packages. ConTeXt also has been used to create large and complex on-screen documents, including hyperlinks, on screen buttons, forms, cross-document links, and so on.
Source: LaTeX in proper ConTeXt (by Berend de Boer)

### How can I convert a document into RTF or HTML?

Hot topic on the mailing list!
There's currently no converter such as LaTeX2html available for ConTeXt. If you're ready to pay a price and put some more effort in your document, the best way is to use XML and produce PDF out of it with ConTeXt and HTML using XSLT.
ConTeXt to RTF Conversion

### What about the other way around?

DOC/RTF to ConTeXt via XML

### Can I change to Computer Modern's LaTeX font?

I see diferences in screen between default fonts in ConTeXt (latin modern) and LaTeX (computer modern). Can I switch to latex default font?. I think it's more stetical

No.
(I mean - of course you can; you only need to modify a few type-xxx.tex files and replace lm with cm in the source. But that solution is not/will not be supported officially.)
Is there any tutorial for doing it?
No. There are no tutorials for the things that one is not supposed to mess with. Take a look at how things are done in those files and replace any occurrence of lm with cm.

### Can I use some simpler bibliography method?

I see that the method of [adding references in ConTexT] is like bibtex in latex. Is there any simple methood like build-in latex method (\thebibliography)?

Perhaps [Simple bibliography (in simple bibliography)]?

### How do I make a presentation?

You don't need any special package (like prosper or seminar in LaTeX) to make a presentation. In contrast to LaTeX, changing the document size, changing colors, adding hyperlinks, ... appears to be extremely simple in ConTeXt. See Presentations for more specific topics.

### \setupindenting[medium] doesn't work?

This does work, but the effect is not what you expect. Without yes, you have now set up the default size, but you have not turned paragraph indenting on. You have to execute \setupindenting[yes] before it starts actually indenting your paragraphs.

### Does $$...$$ work for display math, like in (La)TeX?

No; the spacing above and below the equation will be wrong, and maybe other aspects too. Instead use \startformula and \stopformula.

### How can I get a horizontal space with a defined width, so as ex. "\hspace{3em}" or "\hspace*{3em}" in LaTeX?

A horizontal space can be inserted with the TeX-Primitive\hspace〈dimension〉 or with the ConTeXt command \hspace[identifier].

### How can I get the lowercase (medieval, oldstyle figures) numbers in a document?

The mediaeval numbers can be inserted locally by "\os". I would prefer a global setting of them but I don't know how to do it.

### How transcribe this LaTeX macro to ConTEXt?

Could you please advice me, how transcribe this macro, controlling that no last line of a paragraph is shorter than the paragraph indenting?

\parfillskip=\hsize

### Why is \framed{} not centred?

\starttext
\startalignment[middle]
\framed{middle?}
\stopalignment
\startalignment[middle]
middle!
\stopalignment
\stoptext

Answer: \framed switches to vmode, so you need \dontleavehmode:

\starttext
\startalignment[middle]
\dontleavehmode
\framed{middle!}
\stopalignment
\stoptext

### Why is there a line-break in the output after some commands?

\starttext
\externalfigure[hacker][width=1cm] Why a new line here?
\stoptext

\starttext
\dontleavehmode
\externalfigure[hacker][width=1cm] Same line!
\stoptext

### Why the space is gobbled after some commands?

\starttext
X \externalfigure[hacker][width=1cm] X
\stoptext

Answer: When you don't use all of the optional square bracket arguments, the ConTeXt gobbles the space after the last square bracket. You have at least 2 solutions:

\starttext
X \externalfigure[hacker][width=1cm][] X % adding empty square brackets

X \externalfigure[hacker][width=1cm]\ X % adding explicit space command
\stoptext


text:
## Stanford conference

Today I gave a talk at the conference at Stanford, entitled “Embedded contact homology as a (symplectic) field theory”. Very regrettably, I could only come to the conference for this one day. Anyway, my talk was about how to define cobordism maps on ECH induced by strong symplectic cobordisms (for a fixed spin-c structure), and about how the Novikov completion of ECH is a functor on the category of strong symplectic cobordisms, which allows one to recover the Gromov invariant of a closed four-manifold by cutting it along a contact-type hypersurface. This is similar to what I said in two blog posts in May here and here, and I am currently writing up the details (one small change: I have decided to dispense with the formal variable $u$ in the definition of completed ECH).

I also announced one new result: using the cobordism maps on completed ECH, one can make a revised definition of ECH capacities, which agrees with the old definition for the known examples, but which can also be computed for some closed symplectic four-manifolds. In particular, the new ECH capacities of ${\mathbb C}P^2$ are the same as those of a ball of the same volume, and the new ECH capacities of $S^2\times S^2$ are the same as those of a polydisk whose $B^2$ factors have the same areas as the $S^2$ factors. This means that if ECH capacities give an obstruction to embedding something into a ball, then it cannot be embedded into ${\mathbb C}P^2$ either; and likewise for a polydisk and $S^2\times S^2$.  (This is maybe not too surprising because it is already known that if a union of balls can be embedded into ${\mathbb C}P^2$, then it can be embedded into a ball of the same volume, and likewise for $S^2\times S^2$ and a polydisk. I am not proving that if an arbitrary domain can be embedded into ${\mathbb C}P^2$ then it can also be embedded into a ball of the same volume; I am just proving that ECH capacities cannot detect a counterexample.)

Since I didn’t have time to explain the new definition of ECH capacities in my talk, maybe I should do so here.

First, recall the definition of the completed ECH of a contact 3-manifold $(Y,\lambda)$, which we denote by $\overline{ECH}(Y,\lambda)$. This is the homology of a chain complex $\overline{ECC}(Y,\lambda,J)$. The chain complex is a module over the universal Novikov ring $\Lambda$ consisting of formal sums $\sum_{r\in{\mathbb R}}a_rt^r$ such that for all $R\in{\mathbb R}$, there are only finitely many $r such that $a_r\neq 0$. An element of the chain complex is a formal sum $\sum_{\alpha\in\mathcal{O}, r\in{\mathbb R}}a_{\alpha,r}t^r\alpha$ where ${\mathcal O}$ denotes the set of ECH generators, and the coefficients $a_{\alpha,r}\in{\mathbb Z}$. We impose the finiteness condition that for all $R\in{\mathbb R}$, there are only finitely many pairs $(\alpha,r)$ such that $a_{\alpha,r}\neq 0$ and ${\mathcal A}(\alpha)+r, where ${\mathcal A}$ denotes symplectic action. The differential $\overline{\partial}$ is defined in terms of the usual ECH differential $\partial$ by

$\overline{\partial}\sum_{\alpha\in{\mathcal O},r\in{\mathbb R}}a_{\alpha,r}t^r\alpha = \sum_{\alpha,\beta\in{\mathcal O},r\in{\mathbb R}}\langle\partial\alpha,\beta\rangle t^{r+\mathcal{A}(\alpha)-\mathcal{A}(\beta)}\beta.$

Here the guiding principle of the formalism is that whenever we count a holomorphic curve, we multiply by $t$ to the “area” of the holomorphic curve, suitably interpreted. This allows us to define infinite sums of holomorphic curve counts more or less with impunity, since we will get an area bound on the holomorphic curves counted in any coefficient, and we can then use Gromov compactness (Taubes’s version using currents and no genus bound) to show that this is well defined.

With a little work, one can show that $\overline{ECH}(Y,\lambda)$ does not depend on $J$ (basically by “completing” the proof that the usual ECH does not depend on $J$).

Analogously to the way we “completed” the differential, one can also complete the $U$ map to define a map

$\overline{U}:\overline{ECH}(Y,\lambda)\to\overline{ECH}(Y,\lambda),$

which depends only on the choice of a connected component of $Y$.

As I described in the second post referenced above, a strong symplectic cobordism $(X,\omega)$ from $(Y_+,\lambda_+)$ to $(Y_-,\lambda_-)$ induces a map

$\overline{\Phi}(X,\omega): \overline{ECH}(Y_+,\lambda_+) \to \overline{ECH}(Y_-,\lambda_-),$

defined up to sign (I am not sure about some orientation issues). This map is defined using Seiberg-Witten theory, although it is induced by a chain map such that a coefficient of the chain map is nonzero only if there exists a corresponding (possibly broken) holomorphic curve. (It is an interesting open problem to define ECH cobordism maps directly in terms of holomorphic curves). This map satisfies the composition property and so makes ECH into a functor from the category of contact 3-manifolds and strong symplectic cobordisms to the category of $\Lambda$-modules (with maps defined up to sign). If $(X,\omega)$ is closed and connected, then it is a cobordism from the empty set to the empty set, so $\overline{\Phi}(X,\omega)$ is an element of $\Lambda$ (defined up to sign). Because of the way the cobordism map is defined in terms of Seiberg-Witten theory, this element of $\Lambda$ is

$\overline{\Phi}(X,\omega) = \pm\sum_{A\in H_2(X): A\cdot A+c_1(A)=0}t^{\langle\omega,A\rangle}Gr(X,\omega,A),$

where $Gr(X,\omega,A)$ denotes Taubes’s Gromov invariant in the class $A$ (more precisely the Seiberg-Witten invariant of the corresponding spin-c structure; according to a paper of Li-Liu, the definition of Taubes’ Gromov invariant needs to be modified in a manner suggested by McDuff in some cases when $b_2^+(X)=1$).

How can we recover the full Gromov invariant by cutting along a contact type hypersurace? Suppose $(X,\omega)$ is a closed connected symplectic four-manifold, and for a nonnegative integer $d$ define

$Gr_d(X,\omega) = \sum_{A\in H_2(X): A\cdot A+c_1(A)=2d}t^{\langle\omega,A\rangle}Gr(X,\omega,A).$

This is what we would like to compute. Now suppose that $(X,\omega)$ is separated by a contact-type hypersurface $(Y,\lambda)$ into pieces $(X_+,\omega_+)$ and $(X_-,\omega_-)$, where $\omega_\pm$ denotes $\omega|_{X_\pm}$, and our convention is that $Y$ is the concave boundary of $X_+$ and the convex boundary of $X_-$. Then $(X,\omega)$ is the composition of the cobordism $(X_+,\omega_+)$ from the empty set to $(Y,\lambda)$ with the cobordism $(X_-,\omega_-)$ from $(Y,\lambda)$ to the empty set. By the composition property we then have

$Gr_0(X,\omega)= \pm\overline{\Phi}(X_-,\omega_-)\circ\overline{\Phi}(X_+,\omega_+).$

We can recover the rest of the Gromov invariant by inserting the completed $U$ map: If $d$ is a nonnegative integer then

$Gr_d(X,\omega)= \pm\overline{\Phi}(X_-,\omega_-)\circ\overline{U}^d\circ\overline{\Phi}(X_+,\omega_+).$

Here $\overline{U}^d$ can be any composition of $d$ of the completed $U$ maps associated to the components of $Y$. This follows from some basic properties (which I didn’t explain) of the uncompleted cobordism maps with respect to the $U$ map.

I said all of the above in my talk; I will explain the new definition of ECH capacities in the next post.


text:
# Math Help - Derivative of a Trig Function

1. ## Derivative of a Trig Function

Hey guys, could you please explain how to solve this, step by step? Hopefully it's fairly easy but I just can't get it:

f(x) = xe^xcscx

Find f'(x)

Thanks!

2. Originally Posted by Froggdzick
Hey guys, could you please explain how to solve this, step by step? Hopefully it's fairly easy but I just can't get it:

f(x) = xe^xcscx

Find f'(x)
$\frac{d}{dx}[u(x) \cdot v(x) \cdot w(x)] = u'(x) \cdot v(x) \cdot w(x) + u(x) \cdot v'(x) \cdot w(x) + u(x) \cdot v(x) \cdot w'(x)$

3. The way you wrote it could mean many functions. Can you verify which equation you are using (use parentheses or use Latex coding)?

4. Originally Posted by Froggdzick
Hey guys, could you please explain how to solve this, step by step? Hopefully it's fairly easy but I just can't get it:

f(x) = xe^xcscx

Find f'(x)

Thanks!

1(e^xcscx) + ((e^xcscx)(cscx - xcscxcotx))X


text:
Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.

Assume that you have a list and you want to get the n-th element of each component or generally to subset the list. You can use the command sapply(list, "[", c(1,2,3,..))

Let’s see this in practice.

mylist<-list(id<-1:10,
gender<-c("m","m","m","f","f","f","m","f","f","f"),
amt<-c(5,20,30,10,20,50,5,20,10,30)
)

mylist

Output:

> mylist
[[1]]
[1]  1  2  3  4  5  6  7  8  9 10

[[2]]
[1] "m" "m" "m" "f" "f" "f" "m" "f" "f" "f"

[[3]]
[1]  5 20 30 10 20 50  5 20 10 30

Let’s say that we want to get the 3rd and 6th element of the list:

sapply(mylist, "[", c(3,6))

Output:

     [,1] [,2] [,3]
[1,] "3"  "m"  "30"
[2,] "6"  "f"  "50"


text:
Calculate the time from the root to each node of the tree

nodeTimes(tree, tipsOnly = FALSE)

## Arguments

tree An object of class phylo. Logical indicating whether the returned results should be truncated only to the tips of the tree.

## Value

A vector of size the number of nodes in the tree (tips, root, internal) containing the time from the root to the corresponding node in the tree.


text:
Select Page

Question:

True/ False?

The integral $$\int_{0}^{\infty} e^{-x^5}dx$$ is convergent.

Hint:

As x varies from 0 to “infinity”, $$-x^5$$ varies from 1 to “minus infinity”. Basically, the function is “rapidly decreasing”. We can hope that nothing goes wrong in the convergence.

Discussion:

Recall that $$\int_{0}^{\infty} e^{-x}dx$$ is convergent. (In fact the value of this integral is 1 which can be done by simple calculation). We try to use this for our comparison test. For $$1\le x < \infty$$   $$e^{-x^5} \le e^{-x}$$. This is because in this interval, $$x^5 \ge x$$ and because $$e^x$$ is an increasing function. So for $$1\le x < \infty$$, our worries are over.

$$\int_{1}^{\infty} e^{-x^5} \le \int_{1}^{\infty} e^{-x}dx < \infty$$.

Now we only need to check whether the integration is finite on $$[0,1]$$. The intergand is continuous and bounded by 1 (because it is monotonic decreasing and value at 0 is 1) on this interval, hence the integral is bounded. $$\int_{0}^{1} e^{-x^5}dx \le \int_{0}^{1} 1dx = 1$$.

Thus, $$\int_{0}^{\infty} e^{-x^5}dx$$ is convergent.


text:
MathSciNet bibliographic data MR2651848 15A24 (15B48 47A62) Yin, Xiaoyan; Liu, Sanyang Positive definite solutions of the matrix equations $X\pm A^\ast X^{-q}A=Q(q\geq 1)$$X\pm A^\ast X^{-q}A=Q(q\geq 1)$. Comput. Math. Appl. 59 (2010), no. 12, 3727–3739. Article

For users without a MathSciNet license , Relay Station allows linking from MR numbers in online mathematical literature directly to electronic journals and original articles. Subscribers receive the added value of full MathSciNet reviews.


text:
# Pseudocode: “If not” control structure

I use the following packages:

\usepackage{algorithm}
\usepackage{algpseudocode}

to display pseudocode of some algorithms in my latex document. For conditionals I use the follow

\If{success = 0}
\State \Return{$\emptyset$}
\EndIf

Now, this renders the if and endif keywords in bold using the same typeface as the one used for all the other control structures. My question is the following: How can I render if not instead of if? So my code would read something like:

\IfNot{success}
\State \Return{$\emptyset$}
\EndIf

You can take all the details associated with \If (within algpseudocode.sty) and add your own \IfNot:

\documentclass{article}

\usepackage{algpseudocode}

\algnewcommand\algorithmicnot{\textbf{not}}
\algdef{SE}[IF]{IfNot}{EndIf}[1]{\algorithmicif\ \algorithmicnot\ #1\ \algorithmicthen}{\algorithmicend\ \algorithmicif}%

\begin{document}

\begin{algorithmic}
\If{success = 0}
\State \Return $\emptyset$
\EndIf
\IfNot{success}
\State \Return $\emptyset$
\EndIf
\end{algorithmic}

\end{document}

• This did exactly what I was looking for, thanks! – gen Feb 24 at 2:44


text:
# How would you solve the next ODE? (dy)/(dt) =(at + by + m)/( ct + dy + n), where a,b,c,d,m,n are constants and ad=bc.

Separable Differential Equation
How would you solve the next ODE?
$\frac{dy}{dt}=\frac{at+by+m}{ct+dy+n},$
where $a,b,c,d,m,n$ are constants and $ad=bc$
Corrected.
You can still ask an expert for help

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Taniyah Lin
Let $u=at+by$ then $ct+dy=Ku$ for some constant $K$ (as $ad=bc$). Now $\frac{du}{dt}=a+b\frac{dy}{dt}$ and $\frac{dy}{dt}=-\frac{a}{b}+\frac{1}{b}\frac{du}{dt}$. Next we have $\frac{dy}{dt}=\frac{u+m}{Ku+n}$ so $-\frac{a}{b}+\frac{1}{b}\frac{du}{dt}=\frac{u+m}{Ku+n}$. And this can be solved by separation as it is autonomous.


text:
# Geometry Quests by Skanda - Suggestion Box

If you have any suggestions, corrections, doubts regarding the problems of this set, you can freely post it here for me. I would be checking this note very regularly.

Hoping for a positive response........!!!!!!!!!!!

Note by Skanda Prasad
3 years, 7 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

As previously mentioned, this problem currently has 2 possible answers. As we only currently only allow for a unique answer, the options will need to be updated.

When your problem has been deleted, please identify the error and work on fixing them first, before reposting them without any edits.

Please let me know how you want to proceed.

Staff - 3 years, 6 months ago

Awesome questions

- 3 years, 7 months ago

There are issues with question 1, 3, 5. Please fix the problems accordingly.

Also, please refrain from typing your titles in all capital letters, as that is considered very rude on the internet.

Staff - 3 years, 7 months ago

I've done it Sir !! I've deleted the problem 5 and posted it afresh with accurate values.

- 3 years, 7 months ago

Yeah sure why not ?? but could you tell me why is it so considered ??

- 3 years, 7 months ago

Yup! Even in Harry Potter, as you've read it, You see that all the verbal stuff that indicates that a person is shouting is written in Caps.

- 3 years, 7 months ago

Yeah !! I made that relation a few days ago.....

- 3 years, 7 months ago

And, of course, talking about the questions, they are as cool as possible!

- 3 years, 7 months ago

Thanks dude !!

- 3 years, 7 months ago

Good that you've realised it

- 3 years, 7 months ago

It is Believed that whenever you type in capital letters you are "shouting". That is the reason why Sir suggested so.

- 3 years, 7 months ago


text:
# A triangle has sides with lengths of 7 miles, 7 miles, and 9 miles. Is it a right triangle yes or no

1 answer
###### Question:

A triangle has sides with lengths of 7 miles, 7 miles, and 9 miles. Is it a right triangle yes or no

## Answers

1 answer

### In which quadrant or on what axis does each point lie

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### Drag the tiles to the correct boxes to complete the pairs. Match each energy transformation to the correct image.

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2 answers

### PLEASE HELP!! IM REALLY CONFUSED

PLEASE HELP!! IM REALLY CONFUSED...
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### During a medical condition screening lung capacity testing is a standard procedure. True of false

During a medical condition screening lung capacity testing is a standard procedure. True of false...
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### URGENT!!!!!!!!!!!!!!!!! One method for studying opportunity cost is to think in terms of risk and ability. pros and cons. tradeoffs trial and error.

URGENT!!!!!!!!!!!!!!!!! One method for studying opportunity cost is to think in terms of risk and ability. pros and cons. tradeoffs trial and error....
2 answers

### Write the point-slope form of the line that passes through (6, 1) and is perpendicular to a line with a slope of -3. Include all of your work in your final answer. Type your answer in the box provided or use the upload option to submit your solution.

Write the point-slope form of the line that passes through (6, 1) and is perpendicular to a line with a slope of -3. Include all of your work in your final answer. Type your answer in the box provided or use the upload option to submit your solution....
1 answer

### According to the princible of cross-cutting relationships an intrusive rock body is A.always made of the same materials as the rock around it B. younger than the rocks it intrudes into C.older than the rocks it intrudes into D.deposited as sedimentary layers

According to the princible of cross-cutting relationships an intrusive rock body is A.always made of the same materials as the rock around it B. younger than the rocks it intrudes into C.older than the rocks it intrudes into D.deposited as sedimentary layers...
2 answers

### I need a bff plzzzzzzzzzzzzzzzzzzzz

I need a bff plzzzzzzzzzzzzzzzzzzzz...
1 answer

### 22 less than the quotient of an unknown number and 35 is -25. What is the value of the unknown number?

22 less than the quotient of an unknown number and 35 is -25. What is the value of the unknown number?...
1 answer

### How YOU LIKE ME NOW !

how YOU LIKE ME NOW !...
1 answer

### If the Sun suddenly turned off, we would not know it until its light stopped coming. How long would that be, given that the Sun is 1.50x10'' m away?

If the Sun suddenly turned off, we would not know it until its light stopped coming. How long would that be, given that the Sun is 1.50x10'' m away?...
1 answer

### Help plzzThe Earth’s ________________ on its ________ around the _______ takes a year. The Earth can be divided into _______________ or halves. I live in the _______________ ___________________________. It is the _________________________ and its __________________________________ that causes ______________. The Earth’s axis is ______________ (degrees). The Earth’s tilt causes _____________________________________________________. Areas at a __________________________ receive _______________

help plzzThe Earth’s ________________ on its ________ around the _______ takes a year. The Earth can be divided into _______________ or halves. I live in the _______________ ___________________________. It is the _________________________ and its __________________________________ that causes ___...
1 answer

### What could you add too x^2-6+1 to make it a perfect square

what could you add too x^2-6+1 to make it a perfect square...
1 answer

### Suppose that you are going to a conference, and have a set of seminars to select from. Each seminar occurs at a specific time (e.g., 1:00pm - 2:00pm). If two seminars overlap in time, you cannot participate in both of them. Your goal is to go to as many seminars as possible. For example, suppose there is a talk on algorithms from 1pm-2pm, a talk on OS from 1:30pm - 3pm, a talk on programming languages from 2pm-4pm, a talk on data structures from 2:30pm - 3:30pm, and a talk on compilers from 3:30

Suppose that you are going to a conference, and have a set of seminars to select from. Each seminar occurs at a specific time (e.g., 1:00pm - 2:00pm). If two seminars overlap in time, you cannot participate in both of them. Your goal is to go to as many seminars as possible. For example, suppose the...
1 answer

### Petrenko Corporation has outstanding 2,000 $1,000 bonds, each convertible into 50 shares of$10 par value common stock. The bonds are converted on December 31, 2020, when the unamortized discount is $30,000 and the market price of the stock is$21 per share.

Petrenko Corporation has outstanding 2,000 $1,000 bonds, each convertible into 50 shares of$10 par value common stock. The bonds are converted on December 31, 2020, when the unamortized discount is $30,000 and the market price of the stock is$21 per share....
1 answer

### Which research questions would be the most effective in researching the effects of television on children? Check all that apply. What are the most popular television programs among different groups of children? How many hours each day does the average child spend watching television? How does watching television every day impact children's brains? How do different television programs influence children's views about society? Does the majority of children prefer watching movies or television? Is

Which research questions would be the most effective in researching the effects of television on children? Check all that apply. What are the most popular television programs among different groups of children? How many hours each day does the average child spend watching television? How does watchi...
2 answers

### Quickly please Along the coastal plains, what are the three occupations most strongly represented? manufacturing logging farming mining service

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1 answer

1 answer

### What is the power generated by a 3.7 V cell phone battery that provides 2.0 A of current?

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-- 0.010807--


text:
, 04.03.2021 17:50 myaj580

# How many solutions and what is the answer ???

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Questions on the website: 13802536


text:
# How do you solve abs(x-3)> 5?

Jun 14, 2015

x > 8 or x < -2

#### Explanation:

I x-3 I must be zero or positive

Consider I x - 3 I = 5

In this case x = 8 or x = -2

Now think about numbers either side of 8 and -2

If necessary substitute 7, 9 and -3 ,-1 into the original expression to see which side of 8 and -2 you need to be.

Jun 14, 2015

Solve |x - 3| > 5

#### Explanation:

Separate solving in 2 cases:

a. x - 3 > 5 -->$x > 8$

b. -(x - 3) = -x + 3 > 5 -> x < 3 - 5 -->$x < - 2$

Graph:

==========|-2------|0-----------------|8=============


text:
# Relativistic Radiation Mediated Shocks - Astrophysics > High Energy Astrophysical Phenomena

Relativistic Radiation Mediated Shocks - Astrophysics > High Energy Astrophysical Phenomena - Descarga este documento en PDF. Documentación en PDF para descargar gratis. Disponible también para leer online.

Abstract: The structure of relativistic radiation mediated shocks RRMS propagatinginto a cold electron-proton plasma is calculated and analyzed. A qualitativediscussion of the physics of relativistic and non relativistic shocks,including order of magnitude estimates for the relevant temperature and lengthscales, is presented. Detailed numerical solutions are derived for shockLorentz factors $\Gamma u$ in the range $6\le\Gamma u\le30$, using a noveliteration technique solving the hydrodynamics and radiation transport equationsthe protons, electrons and positrons are argued to be coupled by collectiveplasma processes and are treated as a fluid. The shock transitiondeceleration region, where the Lorentz factor $\Gamma$ drops from $\Gamma u$ to $\sim 1$, is characterized by high plasma temperatures $T\sim\Gamma m ec^2$ and highly anisotropic radiation, with characteristicshock-frame energy of upstream and downstream going photons of a few~$\times\,m ec^2$ and $\sim \Gamma^2 m ec^2$, respectively.Photon scattering is dominatedby e$^\pm$ pairs, with pair to proton density ratio reaching$\approx10^2\Gamma u$. The width of the deceleration region, in terms ofThomson optical depths for upstream going photons, is large,$\Delta\tau\sim\Gamma u^2$ $\Delta\tau\sim1$ neglecting the contribution ofpairs due to Klein Nishina suppression of the scattering cross section. A highenergy photon component, narrowly beamed in the downstream direction, with anearly flat power-law like spectrum, $u I u\propto u^0$, and an energycutoff at $\sim \Gamma u^2 m ec^2$ carries a fair fraction of the energy fluxat the end of the deceleration region. An approximate analytic model of RRMS,reproducing the main features of the numerical results, is provided.

Autor: Ran Budnik, Boaz Katz, Amir Sagiv, Eli Waxman

Fuente: https://arxiv.org/


text:
Context.—

Within Medicare's Quality Payment Program, and more specifically the Merit-based Incentive Payment System, pathologists stand to potentially lose or gain approximately $2 billion during the initial 7 years of the program. If you or your group provides services to Medicare beneficiaries, you will likely need to comply with the program. Objective.— To avoid potential reductions in Medicare reimbursement, pathologists need to understand the requirements of these new payment programs. Data Sources.— Each year the Centers for Medicare & Medicaid Services publish a Final Rule detailing the program requirements and updates. 2020 marks the fourth reporting year for the Merit-based Incentive Payment System. Performance this year will impact 2022 Medicare Part B distributions by up to ±9%. Conclusions.— By staying up to date with the ever-evolving Merit-based Incentive Payment System requirements, pathologists will be better equipped to successfully comply with this relatively new payment system, reduce the burden of participating, understand the reporting differences of the various performance categories, and thereby be able to maximize their scoring and incentive potential. In 2015, Congress passed the Medicare Access and CHIP Reauthorization Act (MACRA), which was game-changing for our health care system.1 This legislation replaced the sustainable growth rate system and established in its stead a mandatory new program intended to reward health care providers for higher quality care while reining in costs. This new program bundled prior quality reporting programs: the Physician Quality Reporting System, the Physician Value-Based Modifier, and the Electronic Health Record Incentive Program (Meaningful Use), into a single system of assessments and incentives, which the Centers for Medicare & Medicaid Services (CMS) designated as the Quality Payment Program (QPP). Within the QPP, there are 2 main payment pathways: the Merit-based Incentive Payment System (MIPS) and Advanced Alternative Payment Models (APMs). Reporting for the QPP began January 1, 2017, with the resulting payment adjustments lagging 2 years, starting in 2019. According to CMS, approximately 1 100 000 clinicians are expected to be reimbursed under the QPP in 2022 for 2020 reporting. Most of these clinicians (approximately 880 000) will be considered MIPS eligible, whereas between 210 000 and 270 000 clinicians are expected to be considered Qualifying APM Participants (QP; acronyms list available in the Appendix).2 As stated, most QPP eligible clinicians (ECs) fall into the MIPS pathway, which is the default pathway. An EC is defined by law as any physician, physician assistant, nurse practitioner, clinical nurse specialist, certified registered nurse anesthetist, or any group that includes such professionals and bills CMS. This EC definition can be expanded by CMS in future years. As stated previously, the QPP is a mandatory program; however, an EC can be excluded from MIPS in any of the following circumstances: (1) he or she is a QP; (2) he or she does not exceed the Low Volume Threshold in either of the 2 time segments used to determine eligibility (ie, seeing ≤200 Medicare Part B beneficiaries, providing ≤200 covered professional services to Part B patients, or billing ≤$90,000 for Part B covered professional services); or (3) if he or she is a first-time enrollee in CMS. However, starting in 2019, unless all 3 criteria of the Low Volume Threshold are met, an EC can decide to “opt in” to MIPS. If an EC or group decides to opt in, they will receive a final MIPS score and will receive a payment adjustment based on the data they submit. Once an EC or group decides to opt in, they cannot reverse their decision. An EC or group can also decide to voluntarily report for MIPS. If an EC or group does voluntary reporting, they will not receive a payment adjustment. However, CMS will provide them with a performance feedback report including what their MIPS score would have been based on the data they submit.3

Based on 2017 CMS reporting, 13 409 pathologists were considered MIPS ECs.4  This number of MIPS-eligible pathologists is expected to be lower for the 2018 and 2019 performance years because of the expanded Low Volume Threshold criteria that exempts more clinicians. As we enter year 4 of reporting for the QPP, the program continues to evolve, with scoring becoming more complex, reporting requirements increasing, and potential payment adjustments becoming larger. It is essential that ECs or groups check the status of their MIPS eligibility, using their National Provider Identifier and associated business Tax Identifier Number(s), at the start of each reporting year to understand what actions are needed for them to successfully comply with the program (https://qpp.cms.gov/participation-lookup; accessed October 3, 2019). In addition to providing eligibility status, this tool also informs the EC or group of any special status, such as being deemed facility based, small practice, or non–patient facing. Here, we will review the overarching details of the QPP and specifically MIPS, focusing on what has changed since initial implementation, and describe what an individual pathologist or pathology group needs to do to successfully participate.

## MERIT-BASED INCENTIVE PAYMENT SYSTEM

The 4 categories within MIPS on which performance is assessed are: Quality; Improvement Activities (IAs); Promoting Interoperability or PI (formerly Advancing Care Information); and Cost (formerly Resource Use). Notably, reporting requirements within each category differ depending on whether an EC is considered “Patient Facing” or “Non–Patient Facing.” CMS has defined a Non–Patient-Facing EC as any EC who bills 100 or fewer patient-facing encounters in a calendar year. For those reporting as a group, 75% of the ECs must meet this criterion for the group to be considered Non–Patient Facing. Each year, CMS publishes a list of encounters that qualify as patient facing, which historically has included evaluation and management, apheresis, and certain procedural CPT codes. To date, the vast majority of pathologists are classified as Non–Patient Facing. As Non–Patient-Facing ECs, pathologists will likely not be scored on Promoting Interoperability and Cost because the current performance metrics within those categories do not apply. Although this reduces the burden of reporting, it shifts the weighting of the EC's FS into Quality (85%).

A key point is that the CMS acknowledges that smaller practices are at a disadvantage in reporting MIPS, and accordingly CMS is slowly ramping up the requirements and performance thresholds. For example, during 2017 reporting, CMS only required an EC or group to report on 1 quality measure for 1 patient, 1 IA, or the base measures for the Advancing Care Information (now known as Promoting Interoperability) category in order to avoid the penalty. So by reporting on 1 quality measure, regardless of the number of cases, or attesting to 1 IA in 2017, an EC or group would have avoided the 4% downward payment adjustment in 2019. In 2018, this minimum threshold to avoid the penalty increased to an FS of 15 points and in 2019 to an FS of 30 points. In order to further reduce burdens on small practices, CMS has expanded the Low Volume Threshold criteria to exempt additional solo and small practices from MIPS reporting. CMS also allows small practices of 15 or fewer clinicians to report on quality measures via Medicare Part B claims reporting either as individuals or as a group, a mechanism no longer available starting in 2019 for large practices with more than 15 clinicians. Small practices also receive bonus points that are applied to the Quality category score and receive 3 points for quality measures that do not meet the data completeness criteria (discussed later), whereas large practices receive 0 points.

### Facility-Based Scoring

Starting in 2019 there is a new scoring option for ECs considered to be facility based, where the EC's Quality and Cost category scores will be automatically assigned based on the EC's attributed facility. These scores will be determined by the facility's Hospital Value-Based Purchasing Program performance. For an EC to be classified as facility based, 75% or more of that EC's or group's covered professional services must be billed from the inpatient hospital (Place of Service [POS] 21), on-campus outpatient hospital (POS 22), or emergency department (POS 23), with at least 1 service being billed from POS 21 or POS 23. This alternative scoring methodology is intended to reduce the burden on hospital-based ECs; however, facility-based individuals and groups must still attest to IAs separately in order to maximize their MIPS score. Additionally, facility-based ECs and groups have the option to report their Quality category separately from the facility. Unless the EC's or group's facility has an excellent Value-Based Purchasing Program performance, facility-based ECs and groups will likely find it advantageous to report separately for quality, because CMS will use the highest score (facility based or separately reported) for the MIPS FS.6  Therefore, it is essential that such ECs or groups be aware of their facility's performance within the Hospital Value-Based Purchasing Program. Within CMS's QPP Web site ECs can preview their potential facility-based score based on their facility's historical performance (https://qpp.cms.gov; accessed October 3, 2019).

### Quality Category

As previously mentioned, the Quality performance category is proportionally the most important, accounting for 85% of a Non–Patient-Facing EC's MIPS FS. In order to maximize an EC's or group's quality score, they must either report on at least 6 applicable quality measures, of which 1 must be considered a high priority or outcome measure, or report on a specialty measure set. For CMS to deem the submitted measure data complete, the EC or group must report on at least 70% of the patients eligible for the quality measure. If reporting via claims, that would be 70% of their Medicare patients, but if reporting via any other mechanism, it has to be 70% of all their patients (inclusive of all payers). Additionally, each measure needs to have a minimum of 20 reported cases or patients to receive a performance score.

In 2019, CMS deemed 3 of the College of American Pathologists (CAP's) QPP quality measures as extremely topped out because of high performance (>98%), and retired them from MIPS. This action leaves 5 CAP stewarded QPP quality measures (the Pathology Measure Set; Table 1) to report via Medicare Part B claims. Additionally, starting in 2019 there are 2 American Academy of Dermatology stewarded QPP quality measures in the program that may be applicable to pathologists: QPP 265 and QPP 440. These measures focus on the turnaround time for basal cell carcinoma and squamous cell carcinoma pathology reports (QPP 440), and then the rate of documented communication of biopsy results to the referring physician and the patient (QPP 265; Table 1). In 2020, CMS added QPP 440 to the Qualified Registry Pathology Measure Set, thus making a total of 6 quality measures available for reporting via a Qualified Registry (Table 1). Of those pathology QPP measures that remain in the program, the 2 lung cancer measures and the melanoma measure as well as the 2 American Academy of Dermatology measures were deemed as “high priority” and are eligible for 1 bonus point each once the requirement of reporting on at least 1 high-priority measure is met. In order to receive bonus points, measures must meet case minimum and data completeness requirements and have a performance rate higher than 0%. Seeing the need for alternative mechanisms for pathologists to comply with MIPS, the CAP created the first CMS-approved pathology-specific Qualified Clinical Data Registry (QCDR), the Pathologists Quality Registry. The significance of a QCDR reporting tool will be discussed later in this manuscript, but in short it allows additional CMS-approved quality payment measures to be created for MIPS quality reporting. For 2020, there are 23 total quality payment measures available in CAP's Pathologists Quality Registry (Table 2).7

Table 1

Quality Payment Program (QPP) Pathology-Specific Quality Measures

Table 2

Pathologists Quality Registry Measures

Each reported measure receives a score from 0 to 10 points depending on the EC's performance compared with others who reported on the same measure. The maximum value (10 points) is dependent on whether the measure has a reported benchmark; if CMS cannot score a measure against a benchmark the measure will receive 3 points. Maximum available points are reduced if the measure is considered topped out (7 points), if case minimum requirements are not met (3 points), or if data completeness requirements are not met (0–3 points; Table 3). If 6 measures were applicable to the EC or group that would then equate to a maximum total of 60 possible quality points. If more than 6 measures were reported, CMS will choose the 6 best to determine the quality score. However, the EC would be eligible to receive bonus points for up to 10% of the possible quality points. In this example of 60 possible points, that would equate to up to 6 additional high-priority or outcome measures bonus points even if the measure was not used in calculating the quality score. So it can be advantageous for ECs or groups to report on more than the required 6 measures. Lastly and as previously stated, starting in 2019 CMS gave small group practices, defined as groups with 15 or fewer clinicians, 6 bonus points that apply to the quality category as long as the small practice reports on at least 1 quality measure. An example is presented in the Figure, A.

Table 3

Quality Measure Scoring Rules

A, The top 6 measures were used to calculate the Quality performance score. Most of these are considered topped out and are therefore capped at 7 of 10 points. The 2 new Qualified Clinical Data Registry (QCDR) measures do not have a Centers for Medicare & Medicaid Services (CMS) benchmark and are therefore capped at 3 of 10 points. Quality Payment Program (QPP) measure 396 (#) counted as the 1 required high-priority/outcome measure and therefore does not get the bonus point added. However, the remaining high-priority measures (*) count toward additional bonus points. Also, because this group would be considered a small practice, it receives an additional 6 points to its Quality performance score. B, The small practice received 59 of 60 possible Quality points and attested to 2 medium-weight improvement activities for the full 40 of 40 points. The Cost and Promoting Interoperability categories were not attributed to this group. Therefore, the group received a Final Score (FS) of 98.58 (85% × 59/60 + 15% × 40/40). Abbreviations: BCC, basal cell carcinoma; CAP, College of American Pathologists; HPV, human papillomavirus; MIPS, Merit-based Incentive Payment System; SCC, squamous cell carcinoma.

A, The top 6 measures were used to calculate the Quality performance score. Most of these are considered topped out and are therefore capped at 7 of 10 points. The 2 new Qualified Clinical Data Registry (QCDR) measures do not have a Centers for Medicare & Medicaid Services (CMS) benchmark and are therefore capped at 3 of 10 points. Quality Payment Program (QPP) measure 396 (#) counted as the 1 required high-priority/outcome measure and therefore does not get the bonus point added. However, the remaining high-priority measures (*) count toward additional bonus points. Also, because this group would be considered a small practice, it receives an additional 6 points to its Quality performance score. B, The small practice received 59 of 60 possible Quality points and attested to 2 medium-weight improvement activities for the full 40 of 40 points. The Cost and Promoting Interoperability categories were not attributed to this group. Therefore, the group received a Final Score (FS) of 98.58 (85% × 59/60 + 15% × 40/40). Abbreviations: BCC, basal cell carcinoma; CAP, College of American Pathologists; HPV, human papillomavirus; MIPS, Merit-based Incentive Payment System; SCC, squamous cell carcinoma.

### IA Category

The IA performance category is worth 15% of an EC's MIPS FS. This category aims to recognize activities such as care coordination, patient engagement, and safety. There are more than 100 CMS-defined IAs that are considered medium- or high-weighted activities. An EC or group is able to obtain up to 40 points within this performance category. In order to achieve the full 40 points, a Non–Patient-Facing EC needs to attest to participating in at least 2 medium-weighted or 1 high-weighted activity for a minimum of 90 days in the reporting period. In contrast, Patient-Facing ECs need to attest to 4 medium-weighted or 2 high-weighted activities for full credit in this category. One significant change for this category in 2020 is that although previously a group could attest to an improvement activity if at least 1 clinician in the group participated, starting in 2020 at least 50% of the clinicians in the group must perform the same activity in order for the group to attest to that improvement activity. Clinicians can perform the activity during any continuous 90-day period during the performance year. (Everyone does not need to perform the activity at the same time.) Importantly, CMS recommends keeping documentation of performing the selected activities for 10 years. Through a joint effort with CMS, CAP has created a pathology-specific guide focusing on IAs that are most applicable to pathologists and presenting examples of types of documentation that would comply with performance of the activity. This guide is annually updated as the program evolves, and it is available on the CAP Web site within the Advocacy tab and MIPS Improvement Activities for Pathologists (https://www.cap.org/advocacy/mips-for-pathologists; accessed October 3, 2019). Based on submitted data from the Pathologists Quality Registry, in 2018 the top 3 IAs pathologists attested to were (1) implementation of improvements that contribute to more timely communication of test results, (2) participation in a joint commission evaluation initiative, and (3) implementation of use of specialist reports back to referring clinician or group to close the referral loop.

### Promoting Interoperability Category

The Promoting Interoperability (PI) performance category essentially incorporated the intent of the former Electronic Health Record Incentive Program (Meaningful Use) into the QPP. Historically, pathologists were exempt from the Electronic Health Record Meaningful Use Program because the interoperability and information exchange measures within the program did not apply to laboratory information systems. This continues in MIPS. Non–Patient-Facing clinicians and groups are not scored in this category, and therefore this category is automatically reweighted to zero; this is why so much of the weighting in the Non–Patient-Facing EC FS is on the Quality performance category.

### Cost Category

At the start of the program, the Cost performance category was not scored for any EC or group. That has now changed, so that for most ECs it accounts for 15% of their MIPS FS. However, for pathologists this category will most likely also be reweighted to zero because the current cost measures are based on cost calculations for patients attributed to the EC for either primary care services or because the EC is responsible for providing the plurality of services. Between them, these criteria essentially exclude pathologists. The CMS looks to create more cost metrics, which may in future years change this functionally “exempt” status for pathologists.

## REPORTING MECHANISMS FOR MIPS

Within MIPS, several reporting options are available. For quality measures, these include Medicare Part B claims, Qualified Registry, CMS Web Interface, Electronic Clinical Quality Measures, and QCDR. There are 2 different categories of quality measures: QPP measures and QCDR measures. The QPP measures comprise Medicare Part B claims measures (previously known as claims-based measures) and MIPS Clinical Quality Measures or clinical quality measures (previously known as registry measures). The attributes and limitations of each mechanism are summarized in Table 4. Measure groups do not apply to pathology, and until 2018 using a QCDR was not a viable option. The CMS Web Interface uses a Web interface for transmitting data for prepopulated quality measures and is available to multispecialty group practices of 25 or more clinicians. To date, pathologists have primarily used Medicare Part B claims reporting; however, this option is rapidly being transitioned out of the MIPS. Starting in 2019, only small group practices (≤15 clinicians) can use Medicare Part B claims reporting for either individual or group reporting of quality measures. Of note, those ECs or groups that report via claims will most likely have to find another mechanism for attesting to IAs, such as a QCDR or the CMS QPP Portal. As claims-based reporting continues to be phased out of MIPS, it is felt that a QCDR will allow for the greatest potential success within the program because it allows for a larger menu of quality measure options, timely feedback reports via an interactive dashboard, and ease of reporting for both Quality and IAs within 1 system.

Table 4

Pathologists' Reporting Options for 2020

## MIPS FS CALCULATIONS

As previously mentioned, an overall MIPS FS is calculated based on the EC's or group's performance within each evaluable category. The FS ranges from 0 to 100. For 2020 reporting, the weighting of each category for Patient-Facing ECs and groups will be 45% for Quality performance, 15% for IA, 25% for PI, and 15% for Cost. However, for Non–Patient-Facing ECs and groups, the PI and Cost categories will likely not apply given the current lack of applicability of the current performance metrics within those categories. Therefore, the FS will be based on Quality (85%) and IA (15%). An example is presented in the Figure, B.

Key changes for 2020 FS calculations relate to bonus points attribution and threshold changes to avoid the ±9% penalty or to obtain the exceptional performer bonus. In 2018, small practices received bonus points that were added to their MIPS FS. However, for 2019 and 2020 reporting, the bonus points (6 points) will be added to the Quality performance score and not the MIPS FS. In order to avoid a negative adjustment in 2022, an EC or group must have a 2020 FS above 45 points; in order to receive the additional exceptional performance bonus, they must have a FS score above 85 points.

## SUMMARY AND FUTURE CONSIDERATIONS

As the QPP and specifically the MIPS continue to mature, so will the requirements, performance thresholds, and complexity of the program. The measures available to report on within the various performance categories will change. For example, it is expected that more of the current QPP measures will become topped out and retired from the program. The challenge during this transition is the time it takes for newer QCDR and QPP measures to mature to more meaningfully represent true national performance and thereby obtain a CMS benchmark for maximum scoring potential. Broader uptake and use of the measures by applicable ECs is needed in order for this evolution to occur and to ensure greater longevity of these new measures. In other words, if ECs only chose to report on those measures they know they can perform well on, Quality measures may falsely appear topped out, will be retired prematurely, and may never reach their full point potential. “How can we promote greater quality measure participation and educate around this need?” is a question the CAP is trying to answer.

The CMS continues to explore the addition of new cost metrics, specifically episodes of care measures. Although the current cost measures do not apply to pathology, this may likely change in the coming years. Pathology must stay up to date on the CMS's ongoing pilots and task force efforts in order to effectively advocate for our profession to be viewed fairly. Similarly, ongoing advocacy related to a laboratory information system's ability to be deemed Certified Electronic Health Record Technology may eventually change a pathologist's ability to participate in the Promoting Interoperability category. CAP continues to be involved in advocacy efforts to improve MIPS and to reduce burden on pathologists, including with CMS to highlight the unique challenges pathologists face in MIPS as Non–Patient-Facing clinicians in an inherently Patient-Facing program.

It is incumbent on pathologists to stay abreast of the QPP and annual updates released by the CMS. CAP dedicates a tremendous amount of resources to ensure our pathology community has ready access to the tools and information needed to succeed in MIPS. Getting acquainted with the QPP and specifically MIPS earlier can help ensure an EC or a group's ability to comply with the program, receive maximum scoring, and realize the potential incentive payments.

## References

References
1
Medicare Access and CHIP Reauthorization Act of 2015, HR 2, 114th Cong, 1st Sess (2015)
.
2
Medicare Program; CY 2020 Revisions to Payment Policies Under the Physician Fee Schedule and Other Changes to Part B Payment Policies; Medicare Shared Savings Program Requirements; Medicaid Promoting Interoperability Program Requirements for Eligible Professionals; Establishment of an Ambulance Data Collection System; Updates to the Quality Payment Program; Medicare Enrollment of Opioid Treatment Programs and Enhancements to Provider Enrollment Regulations Concerning Improper Prescribing and Patient Harm; and Amendments to Physician Self-Referral Law Advisory Opinion Regulations Final Rule; and Coding and Payment For Evaluation and Management, Observation and Provision of Self-Administered Esketamine Interim Final Rule: Final Rule and Interim Final Rule 84 FR 62568
.
Fed Regist.
November 15,
2019
. .
3
2019 Merit-Based Incentive Payment System (MIPS) opt-in and voluntary reporting policy fact sheet
.
Centers for Medicare & Medicaid Services Web site
. .
4
2017 Quality Payment Program experience report–appendix: table 5: overall participation rate of eligible clinicians by clinician description
.
Centers for Medicare & Medicaid Services Web site
. .
5
2017 Quality Payment Program performance year data
.
Centers for Medicare & Medicaid Services Web site
. .
6
2019 Merit-based Incentive Payment Program (MIPS) facility-based measurement fact sheet
.
Centers for Medicare & Medicaid Services Web site
. .
7
2020 pathology quality measures
.
College of American Pathologists Web site
. .
Appendix

Glossary of Terms With Definitions

## Author notes

The authors have no relevant financial interest in the products or companies described in this article.


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# Finding the pmf from the distribution function

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• April 22nd 2008, 09:53 PM
cirrus74
Finding the pmf from the distribution function
http://i235.photobucket.com/albums/e...4/untitled.jpg

No idea how to do part (a). Can someone please help me? Thanks!
• April 23rd 2008, 04:42 AM
CaptainBlack
Quote:

Originally Posted by cirrus74
http://i235.photobucket.com/albums/e...4/untitled.jpg

No idea how to do part (a). Can someone please help me? Thanks!

I think we will have to assume that the possible values of $x$ are integers, then:

(a) well obviously we have:

$p_X(x)=0,\ x<-1$

and

$p_X(x)=0,\ x \ge 1$,

so that leaves only $p_X(-1)$ and $p_X(0)$ to be determined, can you do that?

RonL
• April 23rd 2008, 09:25 PM
cirrus74
Well actually, the answer that was given to us was:

http://i235.photobucket.com/albums/e...untitled-1.jpg

But I don't know how they got it?
• April 23rd 2008, 09:36 PM
CaptainBlack
Quote:

Originally Posted by cirrus74
Well actually, the answer that was given to us was:

http://i235.photobucket.com/albums/e...untitled-1.jpg

But I don't know how they got it?

When $F$ jumps by $d_i$ at $x_i$, then $p(x_i)=d_i$

RonL