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# Average is not Fast Enough!

Time Limit: 2000/1000 MS (Java/Others)

Memory Limit: 65536/32768 K (Java/Others)

## Description

A relay is a race for two or more teams of runners. Each member of a team runs one section of the race. Your task is to help to evaluate the results of a relay race.

You have to process several teams. For each team you are given a list with the running times for every section of the race. You are to compute the average time per kilometer over the whole distance. That's easy, isn't it?
So if you like the fun and challenge competing at this contest, perhaps you like a relay race, too. Students from Ulm participated e.g. at the "SOLA" relay in Zurich, Switzerland. For more information visit http://www.sola.asvz.ethz.ch/ after the contest is over.

## Input

The first line of the input specifies the number of sections n followed by the total distance of the relay d in kilometers. You may safely assume that 1 <= n <= 20 and 0.0 < d < 200.0. Every following line gives information about one team: the team number t (an integer, right-justified in a field of width 3) is followed by the n results for each section, separated by a single space. These running times are given in the format "h:mm:ss" with integer numbers for the hours, minutes and seconds, respectively. In the special case of a runner being disqualified, the running time will be denoted by "-:--:--". Finally, the data on every line is terminated by a newline character. Input is terminated by EOF.

## Output

For each team output exactly one line giving the team's number t right aligned in a field of width 3, and the average time for this team rounded to whole seconds in the format "m:ss". If at least one of the team's runners has been disqualified, output "-" instead. Adhere to the sample output for the exact format of presentation.

## Sample Input

2 12.5
5 0:23:21 0:25:01
42 0:23:32 -:--:--
7 0:33:20 0:41:35

## Sample Output

  5: 3:52 min/km
42: -
7: 6:00 min/km

## Source

University of Ulm Local Contest 2001

format: plain


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# Geological evidence about climate change: has there ever been a period with a rate of climate change as great as it is today?

Most statistics about climate change are based on climate data over the last 150 year (as long as records have been kept). This seems irrelevant in terms of the geological time scale of temperature variation on the earth.

In the last 400 million years, has there ever been a 200-year interval during which the rate of climate change was as great as it is today?

• it have been episodes where the temparature have dropped fast after major volcanic eruptions or asteoid hits but this have been for a short time,way less than ten years or so after the end of the eruption of the volcanoes or asteroid impact. – trond hansen Aug 11 '18 at 15:24
• @AllenCypher I've edited the title of the question so it reflects better the question you are asking. Feel free to change it or roll back my changes if you don't like them. – Camilo Rada Aug 11 '18 at 17:18
• I'm not sure your statement is true that "most statistics about climate change are based on 150 years", what we have over the last 150 years is better measurements, but a lot of statistics on climate change are based on proxy and geological records. Both the 150 year studies and the geological studies are largely in agreement as well, so I think your question is based on a bad assumption. That said, 200 year intervals over most of the last 400 million years is difficult. I like geological evidence and studies, but 200 year intervals is a very small time to measure that long ago. – userLTK Aug 11 '18 at 19:20
• @camilo. You did improve the title. – acypher Aug 13 '18 at 15:58

Most statistics about climate change are not based only in the 150 years of instrumental records. We have many good temperature proxies that go back between a millennia and 70 million years. Among them: tree rings, growth bands in corals, layered marine sediments, bog and lake sediments, stalagmites and stalactites, ice cores (800,000+ years), and isotopic composition of shells of single-cell organism called foraminifera. This last one can be used to estimate the temperature of the deep ocean as far back as 70 million years. Although, the interpretation of the isotopic composition could be tricky, the caveats do not change the general picture presented by this long term temperature record. Here is a reconstruction of temperature based on foraminifera shells.

You can see that there were long periods of time with temperatures much higher than today. However, when it come to the rate of change this record doesn't have the resolution to answer your question. As a first remark, note that we don't have a reliable temperature record going 400 million years back, and for the ones that do extend multiple million years, the resolution is not good enough to see quick climate change events as the one we are experiencing now.

However, we don't need to go that far back to find climatic changes greater than the one we are experiencing today. Such changes are actually visible in the 800,000 years records recovered from antarctic ice cores. And actually, the best examples come from high-resolution Greenland ice cores that go only about 110,000 years into the past.

One of the most studied quick climate changes is the one after the Younger Dryas about 12,000 years ago. Considering changes in the chemical composition of the atmosphere, and proxies of temperature, the data suggest that climate change after the Younger Dryas and other previous ones were more dramatic than the cumulative changes in the last 200 years. And such dramatic changes happened in periods as short as 40 years. The end of the Younger Dryas was associated to changes in temperature of 8 °C or maybe up to 15°C regionally, accompanied by two-fold changes in precipitation and methane concentrations (reference 1, reference 2)

Unfortunately the second reference (a Nature paper) is behind a paywall so I'll add here figure 4, the most relevant to this question.

You can see how fast are the changes in methane concentrations and $\delta^{18}O$. As a rough reference, a change of one unit in $\delta^{18}O$ is equivalent to a change in temperature of 2 °C, therefore the big jump in the figure is about 10°C.

Now I could argue that given that agriculture appeared just after that change (~12,000 years ago). When assessing if we have faced in the past a climate changes as dramatic as the one we are observing now you have to focus on what happened after that time. And notably, since then, the climate have been very stable. In fact, the change we are experiencing now is the largest after the invention of agriculture (previous references and this one).

As a corollary, it is important to realize that Earth's climate is much more chaotic and capable of large changes in short periods that what you would think by looking at the written history, and that we have enjoyed a long period of calm that at some point, sooner or later will be followed by changes unprecedented in human history. The equilibrium we have enjoyed might be a kind of unstable equilibrium, and you might not need a very large perturbation to upset that equilibrium and send the system into a large climatic swing.


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# Naive Bayes Classifier

Author
Hai Nguyen (Seward)
AbstractWith the massive popularity of Bayes' Theorem as well as the default use of Gaussian/Normal distributions for common data sets, we were keen to better understand firstly, if the assumption to use the Normal distrubition on differing data sets was a fit or not, and secondly, how to take the Normal distribution and Bayes' Theorem and apply it in use via repeatable code.


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 Izv. Akad. Nauk SSSR Ser. Mat., 1987, Volume 51, Issue 3, Pages 584–612 (Mi izv1310)

Singularities of solutions, spectral sequences, and normal forms of Lie algebras of vector fields

V. V. Lychagin

Abstract: A general scheme is presented for constructing solutions of systems of differential equations with a prescribed type of singularities. The scheme is then applied to the homological equation arising in the problem of classifying Lie algebras of vector fields in the neighborhood of a rest (or equilibrium) point. The formal, $C^\infty$ and $C^\omega$ variants of the classification problem are discussed. Sufficiency conditions in the contact, symplectic, and general cases are given in terms of spectral sequences.
Bibliography: 18 titles.

Full text: PDF file (3603 kB)
References: PDF file   HTML file

English version:
Mathematics of the USSR-Izvestiya, 1988, 30:3, 549–575

Bibliographic databases:

UDC: 514.763.85+517.9
MSC: Primary 34C20; Secondary 58C27, 58F05, 18G40, 57S20

Citation: V. V. Lychagin, “Singularities of solutions, spectral sequences, and normal forms of Lie algebras of vector fields”, Izv. Akad. Nauk SSSR Ser. Mat., 51:3 (1987), 584–612; Math. USSR-Izv., 30:3 (1988), 549–575

Citation in format AMSBIB
\Bibitem{Lyc87} \by V.~V.~Lychagin \paper Singularities of solutions, spectral sequences, and normal forms of Lie algebras of vector fields \jour Izv. Akad. Nauk SSSR Ser. Mat. \yr 1987 \vol 51 \issue 3 \pages 584--612 \mathnet{http://mi.mathnet.ru/izv1310} \mathscinet{http://www.ams.org/mathscinet-getitem?mr=903625} \zmath{https://zbmath.org/?q=an:0675.58044|0643.58039} \transl \jour Math. USSR-Izv. \yr 1988 \vol 30 \issue 3 \pages 549--575 \crossref{https://doi.org/10.1070/IM1988v030n03ABEH001030}

• http://mi.mathnet.ru/eng/izv1310
• http://mi.mathnet.ru/eng/izv/v51/i3/p584

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This publication is cited in the following articles:
1. O. V. Lychagina, “Normal forms of Poisson structures”, Math. Notes, 61:2 (1997), 180–192
2. Laurent Stolovitch, “Singular complete integrability”, Publications Mathématiques de l’Institut des Hautes Études Scientifiques, 91:1 (2000), 133
3. C. O’Cadiz Gustad, “Local structure of 2 dimensional solvable Lie algebra actions on the plane”, Lobachevskii J Math, 33:4 (2012), 317
4. A. V. Akhmetzyanov, A. G. Kushner, V. V. Lychagin, “Geometric theory of special modes in the distributed-parameter control systems. I”, Autom. Remote Control, 74:11 (2013), 1786–1801
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Paul's Online Notes
Home / Calculus III / 3-Dimensional Space / Equations of Planes
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You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

### Section 1-3 : Equations of Planes

9. Determine if the line given by $$x = 8 - 15t$$, $$y = 9t$$, $$z = 5 + 18t$$ and the plane given by $$10x - 6y - 12z = 7$$ are parallel, orthogonal or neither.

Show All Steps Hide All Steps

Start Solution

Let’s start off this problem by noticing that the vector $$\vec v = \left\langle { - 15,9,18} \right\rangle$$ will be parallel to the line and the vector $$\vec n = \left\langle {10, - 6, - 12} \right\rangle$$ will be normal to the plane.

Now try to visualize the line and plane and their corresponding vectors. What would the line and plane look like if the two vectors were orthogonal to each other? What would the line and plane look like if the two vectors were parallel to each other?

Show Step 2

If the two vectors are orthogonal to each other the line would be parallel to the plane. If you think about this it does make sense. If $$\vec v$$ is orthogonal to $$\vec n$$ then it must be parallel to the plane because $$\vec n$$ is orthogonal to the plane. Then because the line is parallel to $$\vec v$$ it must also be parallel to the plane.

So, let’s do a quick dot product here.

$\vec v\centerdot \vec n = - 420$

The dot product is not zero and so the two vectors aren’t orthogonal to each other. Therefore, the line and plane are not parallel.

Show Step 3

If the two vectors are parallel to each other the line would be orthogonal to the plane. If you think about this it does make sense. The line is parallel to $$\vec v$$ which we’ve just assumed is parallel to $$\vec n$$. We also know that $$\vec n$$ is orthogonal to the plane and so anything that is parallel to $$\vec n$$ (the line for instance) must also be orthogonal to the plane.

In this case it looks like we have the following relationship between the two vectors.

$\vec v = - \frac{3}{2}\vec n$

The two vectors are parallel and so the line and plane are orthogonal.


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## Centrifugal Breakout of Magnetically Confined Line-Driven Stellar Winds

A. ud-Doula, R. H. D. Townsend, S. P. Owocki

Bartol Research Institute, University of Delaware, Newark, DE 19716, USA

We present 2D MHD simulations of the radiatively driven outflow from a rotating hot star with a dipole magnetic field aligned with the star's rotation axis. We focus primarily on a model with moderately rapid rotation (half the critical value), and also a large magnetic confinement parameter, $\eta_{\ast} \equiv B_{\ast}^2 R_{\ast}^{2} / \dot{M} V_{\infty} = 600$. The magnetic field channels and torques the wind outflow into an equatorial, rigidly rotating disk extending from near the Kepler corotation radius outwards. Even with fine-tuning at lower magnetic confinement, none of the MHD models produce a stable Keplerian disk. Instead, material below the Kepler radius falls back on to the stellar surface, while the strong centrifugal force on material beyond the corotation escape radius stretches the magnetic loops outwards, leading to episodic breakout of mass when the field reconnects. The associated dissipation of magnetic energy heats material to temperatures of nearly $10^{8}$K, high enough to emit hard (several keV) X-rays. Such \emph{centrifugal mass ejection} represents a novel mechanism for driving magnetic reconnection, and seems a very promising basis for modeling X-ray flares recently observed in rotating magnetic Bp stars like $\sigma$ Ori E.

Reference: ApJL
Status: Manuscript has been submitted


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# Graphs I

## Introduction

Graphs are probably the most versatile and the most important construction when we refer to representation. They are astonishly ubiquitous. We can describe practically any system of relations or interactions as a graph. This is true at nano scales where we can model individual molecules, to micro scale where we can look at different interactions between biological entities like molecules, drugs and proteins modelled in interactomes, to finally concluding at macro scale by modelling social graphs of people.

## 1. Idea

Conceptualization: When we think of a graph, we think about a system of relations and interactions between its elements.
A graph is a collection of vertices and edges, each edge links a pair of vertices, defining a relationship of incidence between vertices and edges. There are several variations on the idea.

## 2. Definition

Let
$V$
and
$E$
be sets. Call an element of
$V$
a vertex and a element of
$E$
an edge. A graph is given by
$V, E$
, and a mapping
$d$
that interprets edges as pairs of vertices. Exactly what this means depends on how one defines "mapping that interprets" and "pair". The possibilities are given below. We will need the following notation:
• $V^{2}$
is the cartesian product of
$V$
with itself, the sed of ordered pairs
$(x,y)$
of vertices.
• $\Delta_{V}$
is the diagonal subset of
$V$
, the set of pairs
$(x,x)$
, so that its complement
$V^2 \setminus \Delta_V$
is the set of pairs as above where
$x \neq y$
.
• $\left\langle{V \atop 2}\right\rangle$
is the quotient set of
$V^{2}$
in which
$(x,y)$
is identified with
$(y,x)$
, the set of unordered pairs
${x,y}$
of vertices.
This brings us with the following types of graphs according to their ordering:
1. 1.
Undirected graphs
2. 2.
Directed graphs
3. 3.
Undirected graphs as directed graphs with an involution
Example of a graph
We can also attach some features to the nodes that will be modeled as d-dimensional vectors. For example, in a social network graph, each node might contain information relative to the user of the node such as age, height, sex... For the sake of simplicity we are going to ignore possible features linked to edges, but they could exist.

## 3. History of Graph Neural Networks

 Machine Learners Labeling RAAM (1994) Backpropagation through structure (1996) Graph Neural Networks (2008) Gated GNN (2015) Graph Theorists Weisfeiler-Lehman kernels (2009) k-GNN GIN (2019) Provably powerful GNN Chemists ChemNet (1995) Neural descriptors (1997) Molecular graph net (2005) Molecular fingerprints (2017)
The term graph neural networks first appeared in a series of papers by the group of research conformed by M. Gore and F. Scarselli. But graph theory research argue that graph neural networks are just a rebranding of a graph isomorphism test called the Weisfeiler-Lehman test (1968). This test address a classical graph theory problem that tries to determine if 2 graphs are isomorphic or have the same connectivity up to reordering the nodes.
NOTE: Nowadays there isn't any algorithm that can compute and solve the Weisfeiler-Lehman Test in polynomial time. Only solvable in polynomial time up to graphs with 9 nodes. Indeed this problem was proved to have a computational solution in quasi-polynomial

## 4. Key Structural Properties of Graphs

The key structure of a graph or a set is that it's unordered. We don't have a canonical way to order its nodes. Thus when we number the nodes of an graph, for example the above one, we are partially cheating. It's convenient because we can now organize the nodes features into a matrix of dimensions
$n \times d$
, where n is the number of nodes and d the dimension of the features. This way we can automatically prescribe some arbitrary ordering and apply it for the adjacency matrix.
The adjacency matrix for a graph with
$n$
vertices is an
$n \times n$
matrix whose
$(i,j)$
entry is
$1$
if the vertex
$i^{th}$
and vertex
$j^{th}$
are connected, and
$0$
if they are not.
The representation of the adjacency matrix depends on the selected ordering to the set of nodes.
Conclusion: If we number the nodes differently implies that the rows of the feature matrix and corresponding rows and columns of the adjacency matrix will be permuted accordingly. We denote this permutation matrix by
$P$
, and
$P$
is an elemen of the permutation group of
$n$
elements. The permuation group contains
$n!$
elements. We can think of
$P$
as a representation of the permutation group.
NOTE: The representation differs with respect to the type of the object. The permutation group acts differently on vectors and matrices.

### 4.2 Invariant Graph Functions: Permutation-invariant

Strong precondition needed: In order to implement a function acting on a graph that produces a single output for the entire graph, it's mandatory to assert that this single value generated output is unaffected by the chosen ordering of the input nodes. Independently of the ordering of the nodes, the computed output for the same graph must be always the same, although the nodes were or not rearranged.
Invariant Graph Functions: Permutation-invariant

### 4.3 Equivariant Graph Functions

In this scenario it's convenient that the output of the function changes harmoniously to the fluctuations in the input with the reordering of the nodes.
$F(PX,PAP^{\top})= PF(X,A)$

## 5. Construct Graphs Functions

### 5.1 Multiset

#### Idea

A multiset is like a set, just allowing that the elements have multiplicities. Thus the multiset
$\{ 1,1,2\}$
differs from the multiset
$\{1,2\}$
, while
$\{1,1,2\}$
is the same as
$\{ 1,1,2\}$
. (See ref
$[4]$
)

#### Definition

It is common to define them in terms of sets and functions.
A
$multiset \: \mathcal{X} = \langle X,\mu_X\rangle$
, can be defined as a set
$X$
together with a function
$\mu_{X}$
(giving each element its multiplicity) from
$X$
to a class of nonzero cardinal numbers.

### 5.2 Local Aggregation and Message Passing

This is the main operation of a classical Graph Neural Network architecture.
neighbourhood example for a given node i
Question: How this operation works? Answer: For each node in the graph we look at the neighbors and take the feature vectors that together conforms the multiset. Even though the indices of the neighbors are unique, the feature vectors are not necessarily unique.
multiset of neighbour features
NOTE: In this example, we have the blue feature repeated twice. Two nodes have the same feature vector. In this scenario we dealing with a multiset. (See section 5.1 Multiset)
Now we try to aggregate these features together with the feature vector of the node itself. It must be done in a permutation invariant way because we don't have a canonical ordering of the neighbors on a graph.
aggregation function over a node i
$\Phi \Big (X_{i}, X_{\mathcal{N_i}} \Big)$
If we apply this
$\Phi$
local function at every node of the graph and stack the results into a feature matrix, denoted by
$F$
, we get a permutation equivariant function. Thus if we reorder the nodes then the order of the rows of this matrix will be changed.
This local aggregation function on GNN typically looks as follows:
• $\textcolor{Cyan}{permutation \: invariant \: aggregation \: operator, \: e.g. \: sum}$
: It is often sum or maximal.
• $\color{Orange}{learnable \:functions} \: \Psi$
:
1. 1.
$\psi$
: This function transforms the neighbors features. It's a non linear function that depends on both feature vectors of node
$i$
and
$j$
. Its output can be seen as a message that is sent from node
$j$
to update node
$i$
. That's why this architecture is also called message passing graph neural networks.
2. 2.
$\phi$
: Updates the features of node
$i$
using the aggregated features of the neighbors.
• $\textcolor{DarkOrchid}{new \: feature \: of \: node \: i}$
:
$f : x_i \to x'_{i}$
Gilmer et al. 2017 (MPNN); Battaglia et al 2018; Wang et al. 2018


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# The Magnetic Field as a Turbulence Suppressor in Molecular Cloud Formation

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There is no author summary for this article yet. Authors can add summaries to their articles on ScienceOpen to make them more accessible to a non-specialist audience.

### Abstract

We present magnetohydrodynamic simulations aimed at studying the effect of the magnetic field on the production of turbulence through various instabilities during the formation of molecular clouds (MCs) by converging flows. We particularly focus on the subsequent star formation (SF) activity. We study four magnetically supercritical models with magnetic field strengths $$B= 0$$, 1, 2, and 3 $$\mu$$G (corresponding to mass-to-flux ratios of $$\infty$$, 4.76, 2.38, and 1.59 times the critical value), with the magnetic field initially aligned with the flows. We find that, for increasing magnetic field strength, the clouds formed tend to be more massive, denser, less turbulent, and with higher SF activity. This causes the onset of star formation activity in the non-magnetic or more weakly magnetized cases to be delayed by a few Myr in comparison to the more strongly magnetized cases. We attribute this behavior to a suppression of the nonlinear thin shell instability (NTSI), which is the main mechanism responsible for turbulence generation in the forming clouds, by the mean magnetic field. This result is contrary to the standard notion that the magnetic field provides support to the clouds, thus reducing their SFR. However, our result is a completely nonlinear one, and could not be foreseen from simple linear considerations.

### Most cited references3

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### Modeling Dynamic Architectures Using Dy-BIP

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### A positive MUSCL-Hancock scheme for ideal magnetohydrodynamics

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### A robust numerical scheme for highly compressible magnetohydrodynamics: Nonlinear stability, implementation and tests

The ideal MHD equations are a central model in astrophysics, and their solution relies upon stable numerical schemes. We present an implementation of a new method, which possesses excellent stability properties. Numerical tests demonstrate that the theoretical stability properties are valid in practice with negligible compromises to accuracy. The result is a highly robust scheme with state-of-the-art efficiency. The scheme's robustness is due to entropy stability, positivity and properly discretised Powell terms. The implementation takes the form of a modification of the MHD module in the FLASH code, an adaptive mesh refinement code. We compare the new scheme with the standard FLASH implementation for MHD. Results show comparable accuracy to standard FLASH with the Roe solver, but highly improved efficiency and stability, particularly for high Mach number flows and low plasma beta. The tests include 1D shock tubes, 2D instabilities and highly supersonic, 3D turbulence. We consider turbulent flows with RMS sonic Mach numbers up to 10, typical of gas flows in the interstellar medium. We investigate both strong initial magnetic fields and magnetic field amplification by the turbulent dynamo from extremely high plasma beta. The energy spectra show a reasonable decrease in dissipation with grid refinement, and at a resolution of 512^3 grid cells we identify a narrow inertial range with the expected power-law scaling. The turbulent dynamo exhibits exponential growth of magnetic pressure, with the growth rate twice as high from solenoidal forcing than from compressive forcing. Two versions of the new scheme are presented, using relaxation-based 3-wave and 5-wave approximate Riemann solvers, respectively. The 5-wave solver is more accurate in some cases, and its computational cost is close to the 3-wave solver.
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### Author and article information

###### Journal
2016-06-16
###### Article
1606.05343


text:
# Finding Laplace Transform using first shifting theorem

• September 6th 2008, 01:52 AM
ghazala
Finding Laplace Transform using first shifting theorem
1. Find the Laplace transform of e^(-2t) sinηπt
• September 6th 2008, 02:24 AM
mr fantastic
Quote:

Originally Posted by ghazala
1. Find the Laplace transform of e^(-2t) sinηπt

Where are you stuck?

Theorem: $LT\left[ e^{at} f(t) \right] = F(s - a)$ where $F(s) = LT[f(t)]$

In your case a = -2 and $f(t) = \sin (nt)$. You should be able to look up a table of Laplace transforms to get F(s) .....
• September 6th 2008, 03:05 AM
ghazala
yup thts wht i did..

i got this answer:

(ηπ + 2)/ (ηπ + 2)^2 + s^2

i think i made a mistake smwhere :S
• September 6th 2008, 04:58 AM
mr fantastic
Quote:

Originally Posted by ghazala
yup thts wht i did..

i got this answer:

(ηπ + 2)/ (ηπ + 2)^2 + s^2

i think i made a mistake smwhere :S

Laplace transform of $f(t) = \sin (nt)$ is $F(s) = \frac{n}{s^2 + n^2} \, ....$
• September 6th 2008, 06:28 AM
ghazala
thank you :)


text:
Correlation of PRS obtained by plink and PRSice
2
0
Entering edit mode
2.4 years ago

I want to know why the correlation of PRS obtained by plink and PRSice is low. Firstly, I used plink --clump. The parameters are same as PRSice process. plink --bfile HCP --clump gwas --clump-p1 0.00145 --clump-r2 0.1 --clump-kb 250 Then, I used the clumped snp (include OR and effect allele) as input to compute score: plink --bfile HCP --score gwas_0.00145P_clumped 2 4 6 I obtained the PRS of every subjects by plink.

PRSice calculations follow the manual strictly.

Then I checked the correlation between PRS from plink and PRSice.Pearson correlation coefficient is 0.49 ,p =0.

PRSice • 2.1k views
2
Entering edit mode
2.4 years ago
Sam ★ 4.4k

Which version of PRSice did you use? Also, for PRSice, we perform SNP filtering, flipping and matching before doing the clumping. That could have contributed to the low correlation. If you want to manually reproduce the PRSice workflow with PLINK, you can follow our tutorial here

0
Entering edit mode

Hi,Sam. Thank you for your reply. I want to ask another question. Before we calculate the PRS by PRSice2, we need to confirm two data set have the same variants and effect allele.

Both in base and target data, the name of variants are the same, but the effect allele are different.

For example:

In target data: 1 rs181193408 83084 A T In base data: 1 rs181193408 83084 T A

Some snps have different effect allele as above, and some have the same effect allele.

Should I change the "A T" to "T A" of these different effect allele of snps and don't change the position of allele of snp which have the same effect allele in target data?

0
Entering edit mode

As long as the encoding is consistent, than PRSice should be able to handle them correctly.

0
Entering edit mode

Why the polygenic risk scores calculated by Plink and PRSice-2 are different

Hi Sam, I have asked a PRS related question. Could you give suggestions for it? Thanks!

1
Entering edit mode
2.4 years ago
Yean ▴ 140

In your case, I guess that the low correlation between two genetic risk score can be occurred from the followings.

1. SNPs in summary statistics calculated in PLINK and PRSice are different from each other because the different threshold was used between PLINK and PRSice when performing LD clumping in summary statistics as an example.

2. If not, Have you checked the default formula of genetic risk score between PLINK and PRScise. Are they using the same formula when calculating genetic risk score by default as well as when imputing missing genotype ?. I guess that these probably contribute to low correlation.

Hope this help


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# statsmodels.discrete.discrete_model.Logit.loglikeobs¶

Logit.loglikeobs(params)[source]

Log-likelihood of logit model for each observation.

Parameters
paramsarray_like

The parameters of the logit model.

Returns
loglikendarray

The log likelihood for each observation of the model evaluated at params. See Notes

Notes

$\ln L=\sum_{i}\ln\Lambda \left(q_{i}x_{i}^{\prime}\beta\right)$

for observations $$i=1,...,n$$

where $$q=2y-1$$. This simplification comes from the fact that the logistic distribution is symmetric.


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# Local effects of Hubble flow

1. Nov 9, 2005

### Jimmy Snyder

I did look through this forum for other threads on this topic, but didn't find any. If this has already been discussed, then please let me know where the thread is.

A common model for the expanding universe is an expanding ballon upon which some dots have been drawn. A well know flaw in this model is the fact that as the balloon expands, so do the dots. If the universe expanded like that, then we and our telescopes would expand as well and we wouldn't see the effect. A better model is proposed in which there are coins pasted to the balloon so that as the balloon expands, the coins stay the same size. In this model, the rubber of the balloon expands between the coins, but does not expand under the coins where the paste is. However, I would like to propose a somewhat idealized version of this model in which the paste only contacts the coin and the balloon at a single point of each, say the center of the coin tacked to a single point of the balloon. Then under the coin, except for that one point, the rubber is sliding as the balloon expands.

So how about the real world? Is space something that is stuck to us and our telescopes like the rubber is stuck to the coins, or does space slide through us as if attached at a single point? Or is there nothing attached and space is sliding through us at all points?

2. Nov 9, 2005

### hellfire

The expansion of space cannot be extrapolated to such small scales. The cosmological solution of Einstein equations is derived assuming homogeneity and isotropy of the distribution of matter in space. Therefore, the expanding solution must be valid at that scales where homogeneity and isotropy can be found. Observations tell us that this is 100 Mpc more or less. At smaler scales the geometry of spacetime must not be the cosmological one.

3. Nov 9, 2005

### Jimmy Snyder

But something is happening locally. What are the options?

4. Nov 9, 2005

### hellfire

What makes you think so?

5. Nov 9, 2005

### Jimmy Snyder

I know something is happening, but I don't know what it is. Do I?

6. Nov 9, 2005

### Staff: Mentor

Well, you can get around that by sticking them on with Scotch tape!

7. Nov 9, 2005

### Cosmo16

I like that analogy.

8. Nov 9, 2005

### Jimmy Snyder

I addressed that issue in my first post. I used paste, not tape.

9. Nov 9, 2005

### Jimmy Snyder

We see the distant galaxies slip sliding. What do they see when they look at us?

10. Nov 9, 2005

### pervect

Staff Emeritus
You might want to look at

http://xxx.lanl.gov/abs/astro-ph/9803097
http://arxiv.org/abs/astro-ph/0104349

and/or http://www.astro.ucla.edu/~wright/cosmology_faq.html#SS

This has been discussed quantitatively before here, too (without aid of these papers)

for example
works out a repulsive tidal force due to expansion (assuming a FRW metric) and the current cosmological values which include a cosmological constant.

Something worthwhile to know from one of the above published papers:

So, if we put all the pieces together, the conclusion I come to is this (a slightly more sophisticated view of what I wrote earlier) is this.

the cosmological constant terms will be very uniformly distributed over space, as they are due to space itself.

The cosmological constant terms currently give a repulsive force

If we imagine an empty sphere of space inside an FRW universe, there would be no forces.

But an empty sphere of space would be empty only if there were no cosmological constant. Because of the cosmological constant, an "empty" sphere of space actually contains negative mass. (I should clarify this - because the expansion of the universe is accelerating, an "empty" sphere of space must provide the forces that provide this acceleration. The energy density $\rho$ associated with a positive cosmological constant is postive, but gravity couples to $\rho + 3P$, i.e. to both energy and pressure, and the later quantity is negative.

This causes a net repulsive force between two objects that are far apart, via normal gravitational interactions. We can use the naive Newtonian viewpoint here about the gravity of a spherical shell of matter acting as if it were in the center if we assume spherical symmetry - at least as far as getting the sign correct. (Previously I resisted this, but now I think this naieve aproach does give the right answer).

If there is also matter inside the sphere, the repulsive force will be less. If there is sufficient matter inside the sphere, the repulsive force (due to the cosmological constant) will be overcome.

If the matter inside the sphere was exactly the "average" amount of matter that one would expect from the average matter density of the universe, you will get the number that I previously calculated for the "repulsive force" due to expansion, which assumed that the metric was the FRW metric. But it would probably be better to view the repulsive force as being due to the cosmological constant. You can then add in an observered matter density to get the sign of the resulting force assuming symmetry, or do some other calculations involving non-symmetrical matter and symmetrical "vacuum-energy".

Last edited: Nov 9, 2005
11. Nov 9, 2005

### hellfire

Sorry but I do not understand your question. From your initial question it seams that you are assuming that there must be some expansion effect at small scales (<< 100 Mpc). But, as I have already written, the cosmological solution is based on the assumption of homogeneity and isotropy and these are not given at small scales.

12. Nov 9, 2005

### Jimmy Snyder

Sorry if I was unclear. I mean that since there ISN'T any expansion effect at small scales, doesn't that mean that space must be slipping out the sides of things like the rubber of the balloon slips out of the edge of the coin. Because in the model, the rubber is space and the coins are matter.

Suppose you could bounce a beam of light around with mirrors in a small hollow contraption in such a way that the beam traveled 100 Mpsec from emiter to detector (that's an unreasonable number of bounces, I know). Because the contraption is small, it would not expand with the universe. What I want to know is if the space within which the beam travels is expanding so that the beam would eventually over time drift away from the detector.

You don't really need the beam to travel 100 Mpsec. The contraption should be as large as possible and still be able to maintain a rigid shape. It should be isolated from vibration. It should be given time for all of its mirrors to reach thermal equilibrium with the heat from the beam. It should bounce the beam as many times as it can and still have the beam be detectable. It should run for as long as it takes for the Hubble flow to move the end of the beam off of the detector.

My point is that at small scales, things don't expand, but does the space around those thing expand?

Last edited: Nov 9, 2005
13. Nov 9, 2005

### pervect

Staff Emeritus
The problem with answering this question is that it is not possible, even in principle, to perform a local measurement of "the velocity of space" - because the laws of physics are invariant with respect to velocity.

One thing we can say, though, is that scenarios such as Baez's "end"

http://math.ucr.edu/home/baez/end.html

eventually do wind up with a universe with a positive cosmological constant tearing apart matter, if you wait long enough. The current effects of expansion are not important, but if we wait long enough, this may no longer be the case.

Last edited: Nov 9, 2005
14. Nov 10, 2005

### Chronos

Simple analogy: Gravity is not strong enough to collapse the earth or the sun, but is strong enough to compel them to orbit one another. The cosmological constant is a puny force, but, put this into perspective: Archimedes swore he could move the earth itself given a long enough lever.

15. Nov 10, 2005

### Jimmy Snyder

If space is expanding, and light travels at c, then a local measurement of the velocity of space could be performed by measuring a delay in the arrival of a beam of light after it had traveled a long distance but within a small volume. That is the nature of my proposed experiment. When the experiment starts out, the beam reaches the detector, but after time and expansion, the beam may slip off of the detector.

Again, the contraption doesn't expand. We know that because our telescopes don't expand. At least if they do expand, they do so slower than Hubble flow. Otherwise we wouldn't be able to detect Hubble flow. But I wonder if space expands locally.

My guess is either that space does not expand locally as suggested by some. In that case the balloon model would be coins pasted to the balloon on the entire surface of the coin. The problem with that idea is that the only force between local galaxies is gravitational, and that same force attracts distance galaxies, just to a smaller degree. Or that space does expand locally in which case there may be some way to measure the effects of Hubble flow locally.

16. Nov 10, 2005

### Jimmy Snyder

My analogy would be that gravity is not strong enough to collapse the earth or sun, but is strong enough that we can measure how hard it tries.

17. Nov 10, 2005

### hellfire

According to my understanding it must not, but it could under some circumstances, which, I assume, are not given in our solar system.

Again, if you make some assumptions in order to find a solution to some equations, then the solution is only valid under that assumptions. This means that the expanding cosmological solution to Einstein equations is valid only where the cosmological principle aplies. In general this is condition is only given at large scales.

The local geometry is determined by the local distribution of matter in the solar system. This could be approximated by a point mass to give a Schwarzschild solution. This solution has to be matched "at infinity" with the geometry in which the sun is located. This geometry is not a homogeneous and isotropic expanding space, because the sun is located within the Milky Way. I assume that it sufficies to take flat spacetime as asymptotic condition.

Considering greater scales one would reach some state at wich it would be necessary to consider that the “local” geometry is embedded in a homogeneous and isotropic space and must match the expanding cosmological solution. My guess is that this occurs at the scale of galatic clusters and not before.

18. Nov 10, 2005

### Jimmy Snyder

I get some of my ideas from popularizations. Perhaps this is the problem. Everyone here seems to agree that there is some distance at which two different solutions of the metric tensor get glued together. This distance is defined by where space starts to look homogeneous, about 100 Mpsec. Further than that and the metric tensor exhibits expansion, closer and it doesn't. Well and good. But as far as I can remember, the popularizations don't speak of it. Here is an example. I added the bold emphasis, but left the misspelling of exapnsoin as is.

http://curious.astro.cornell.edu/question.php?number=75 [Broken]

Edit: I literally cannot emphasize enough "universal expnasion". Also, I note that the galaxy in Andromeda is blue-shifted and less than 1 Mpsec from Cornell. Surely the author of the article was aware of that.

Last edited by a moderator: May 2, 2017
19. Nov 10, 2005

### Jimmy Snyder

Actually, here is a more scholarly article that seems to contradict what I have been hearing:

http://arxiv.org/PS_cache/astro-ph/pdf/9612/9612007.pdf [Broken]

The dynamics of Local Group and its environment provide a unique challenge to cosmological models. The velocity field within $5h^{-1}$Mpc of the Local Group (LG) is extremely "cold". The deviation from a pure Hubble flow, characterized by the observed radial peculiar velocity dispersion, is measured to be $~60km s^{-1}$.

Last edited by a moderator: May 2, 2017
20. Nov 10, 2005

### hellfire

Yes, thank you for bringing this into our attention. I remember have read about this and I have not considered this in my postings here. It relates to the "Hubble-de Vaucouleurs paradox": the dynamics of the Local Group is extremely low compared to the peculiar velocities which would be expected due to the gravitational collapse in a Cold Dark Matter model. This has been explained here postulating that the action of dark energy becomes dominant at scales greater than 2 Mpc leading to a “cold” Hubble flow (a start of the expansion of space) which acts against gravitational collapse. In light of this, it might be that the expansion of space emerges actually at lower scales than the scales at which matter becomes homogeneous, due to the fact that dark energy permeates the whole space homogeneously even at low scales. (pervect was already talking about dark energy, but I was under the impression that it might not be relevant for this discussion, sorry). However, I assume that 2 Mpc is a lowest value for the matching between local geometries in clusters and the cosmological geometry and it is hard to believe that expansion could take place below this scale.

Last edited by a moderator: May 2, 2017


text:
# Are async animations not working? It’s easy and fun with F#

### Prerequisites

• An understanding of Xamarin
• An simple understanding of Threads
• An understanding of F#

Recently I attended a Winter of Xamarin event hosted by Microsoft. It was a great event, but as always the code examples all in C#. I thought it would be a great a exercise to translate the C# app into F#. This post focuses on one particular section that I came across, some Xamarin Forms’ animations, that I found less than trivial to translate. The various attempts that I tried will be outlined, with the final solution at the end.

### The animation code

  1: 2: 3: 4: 5: 6: 7: 8: 9: 10: 11: 12: 13: 14: 15: 16: 17: 18: 19: 20: 21: 22: 23:   public async Task Shake() { for (int i = 0; i < 8; i++) { await Task.WhenAll( this.ScaleTo(1.1, 20, Easing.Linear), this.TranslateTo(-30, 0, 20, Easing.Linear) ); await Task.WhenAll( this.TranslateTo(0, 0, 20, Easing.Linear) ); await Task.WhenAll( this.TranslateTo(0, -30, 20, Easing.Linear) ); await Task.WhenAll( this.ScaleTo(1.0, 20, Easing.Linear), this.TranslateTo(0, 0, 20, Easing.Linear) ); } }

Here is the main gist of the code. The method is part of a ContentPage. I’m assuming that most readers will understanding this, so will only go over this briefly. Task.WhenAll executes the animations in parallel. The await after each Task.WhenAll means wait for it to complete, so the inside of the loop, contains 4 sequential animations. Finally the loop just runs the 4 animations, sequentially 8 times.

### Async in F#

Understanding Async in F# can be a little tricky. This is not because is it complicated, rather it is more explicit. Tomas Petricek has a great ‘series on Async’, and another post on the ‘Async in C# and F#: gotchas in C#’. Armed with this knowledge here is my first attempt at translating the inner loop.

  1: 2: 3: 4: 5: 6: 7: 8: 9: 10: 11: 12: 13: 14: 15: 16: 17: 18: 19:  // Nested inside a ContentPage let show() = [ [| this.ScaleTo(1.1, 20u, Easing.Linear) this.TranslateTo(-30., 0., 20u, Easing.Linear) |] [| this.TranslateTo(0., 0., 20, Easing.Linear) |] [| this.TranslateTo(0., -30., 20u, Easing.Linear) |] [| this.ScaleTo(1.0, 20u, Easing.Linear) this.TranslateTo(0., 0., 20u, Easing.Linear) |] ] |> List.iter(fun animations -> animations |> Array.map Async.AwaitTask |> Async.Parallel |> Async.Ignore |> Async.StartImmediate)

Here’s some F# code that compiles and was my first attempt. Preferring a functional style, I wanted to treat the animations as data, and then iter (map with no result) over them. The function inside the iter contains the relevant transformations to make the code compile with F#. Let’s walk through it. First animations is an array of Task and F# needs Async, so lets map over them and convert them to Asyncs. Next we we want to run them in parallel, Async.Parallel for the job. This is the same as C#’s Task.WhenAll. On to the next line, F# checks return types, and each of the animations return a bool. Async.Ignore will fix up the return type since there is nothing to check. Finally, Async.StartImmediate starts on the main thread an returns immediately so it won’t block the main thread, but also doesn’t create a new thread. (Async.RunSynchronously can’t be used here since the animations must be run on the main thread and will wait for. The result: a deadlock). I’ll save you time of trying the show function out; it doesn’t work. When you run it, nothing happens. No crash, no animation.

### But it compiled

When dealing with side effects, all bets are off as to whether the code works; animations are side effects. A careful read on Thomas’ blog can give us an understanding of what is happening here. On the C# side, static methods for Tasks start immediately. This means that treating animations as data in a list (array) won’t work as they will be started immediately. Armed with this knowledge here is another attempt.

  1: 2: 3: 4: 5: 6: 7: 8: 9: 10: 11: 12: 13: 14: 15: 16: 17:  let shake () = async { do! [| this.ScaleTo(1.1, 20u, Easing.Linear) this.TranslateTo(-30., 0., 20u, Easing.Linear) |] |> Array.map Async.AwaitTask |> Async.Parallel |> Async.Ignore do! [| this.TranslateTo(0., 0., 20u, Easing.Linear) |] |> Array.map Async.AwaitTask |> Async.Parallel |> Async.Ignore do! [| this.TranslateTo(0., -30., 20u, Easing.Linear) |] |> Array.map Async.AwaitTask |> Async.Parallel |> Async.Ignore do! [| this.ScaleTo(1.0, 20u, Easing.Linear) this.TranslateTo(0., 0., 20u, Easing.Linear) |] |> Array.map Async.AwaitTask |> Async.Parallel |> Async.Ignore } |> Async.StartImmediate

This looks a little better and this does in fact work. The animation runs. As before, for each of the arrays, they need to be converted for Tasks, set to run in parallel and have their result ignored. This time, instead of putting all the arrays in a list, they are inside an async block, and the do! tells F# to wait for each one to finish before starting the next one. Async.StartImmediate starts this off nicely on the main thread. I’ll come back to this later and refactor out some of the duplicated code, but now let’s get this running 8 times.

### Running 8 times

Surely it can’t be hard to run an animation 8 times. Also were using F# so variables and loops are beneath us. Here is my (failed) first attempt at running the above code 8 times.

  1: 2: 3: 4: 5: 6: 7: 8: 9: 10: 11: 12: 13: 14: 15: 16: 17:  let shake () = async { do! [| this.ScaleTo(1.1, 20u, Easing.Linear) this.TranslateTo(-30., 0., 20u, Easing.Linear) |] |> Array.map Async.AwaitTask |> Async.Parallel |> Async.Ignore do! [| this.TranslateTo(0., 0., 20u, Easing.Linear) |] |> Array.map Async.AwaitTask |> Async.Parallel |> Async.Ignore do! [| this.TranslateTo(0., -30., 20u, Easing.Linear) |] |> Array.map Async.AwaitTask |> Async.Parallel |> Async.Ignore do! [| this.ScaleTo(1.0, 20u, Easing.Linear) this.TranslateTo(0., 0., 20u, Easing.Linear) |] |> Array.map Async.AwaitTask |> Async.Parallel |> Async.Ignore } |> List.replicate 8 |> List.iter (fun animation -> async {do! animation} |> Async.StartImmediate)

The last line is all that’s changed. An attempt was made to replicate the Async action 8 times and then iter over the resulting list wrapping the animation in do! and executing as Async.StartImmediate. I had assuming that Async.StartImmediate was the same as C# and that they could be chained together. As already stated, this doesn’t work. In F#, any async work must be enclosed in an async block, and started only once. Async.StartImmediate is analogous with an a C# method of public async void meaning you can not continue after the task has been started.

### A working version

I still didn’t want to put a variable with a for loop in, so I found the best compromise I could handle. Here is the first working solution:

  1: 2: 3: 4: 5: 6: 7: 8: 9: 10: 11: 12: 13: 14: 15: 16: 17: 18: 19: 20: 21:  let shakeLandscapeAsync () = let shake = async { do! [| this.ScaleTo(1.1, 20u, Easing.Linear) this.TranslateTo(-30., 0., 20u, Easing.Linear) |] |> Array.map Async.AwaitTask |> Async.Parallel |> Async.Ignore do! [| this.TranslateTo(0., 0., 20u, Easing.Linear) |] |> Array.map Async.AwaitTask |> Async.Parallel |> Async.Ignore do! [| this.TranslateTo(0., -30., 20u, Easing.Linear) |] |> Array.map Async.AwaitTask |> Async.Parallel |> Async.Ignore do! [| this.ScaleTo(1.0, 20u, Easing.Linear) this.TranslateTo(0., 0., 20u, Easing.Linear) |] |> Array.map Async.AwaitTask |> Async.Parallel |> Async.Ignore } async { for action in List.replicate 8 shake do do! action } |> Async.StartImmediate

The single shake action is bound to shake, and then the method returns an async block. In the block, the shake binding can then be replicated 8 times, and thanks to F#’s awesome for..in..do syntax, we have a loop but without a counter variable. We’re also still inside the async block so do! waits for the animation to completed on each run through the list of animations.

### Refactoring

This working version has some repeated code that we could clean up a little bit. We can take that out and put that into a function:

 1:  let awaitParallel = Array.map Async.AwaitTask >> Async.Parallel >> Async.Ignore

If that looks confusing, check out my post on ‘point free notation’ and also this post on ‘function composition’. With the helper function the rest of code now looks as follows:

  1: 2: 3: 4: 5: 6: 7: 8: 9: 10: 11: 12: 13: 14: 15: 16: 17: 18: 19: 20: 21:  let shakeLandscapeAsync () = let shake = async { do! [| this.ScaleTo(1.1, 20u, Easing.Linear) this.TranslateTo(-30., 0., 20u, Easing.Linear) |] |> awaitParallel do! [| this.TranslateTo(0., 0., 20u, Easing.Linear) |] |> awaitParallel do! [| this.TranslateTo(0., -30., 20u, Easing.Linear) |] |> awaitParallel do! [| this.ScaleTo(1.0, 20u, Easing.Linear) this.TranslateTo(0., 0., 20u, Easing.Linear) |] |> awaitParallel } async { for action in List.replicate 8 shake do do! action } |> Async.StartImmediate

This is the final version that I left in my code base. I will mention one last step that could be done.

Running animations sequentially like this might be rather common in a mobile app. So the async loop could be refactored out. An easy way to do this would be as an extension method on Async. Here’s an example:

 1: 2: 3: 4: 5: 6: 7: 8:  // In AsyncExtensions.fs module Async let RunSequentially asyncs = async { for a in asyncs do do! a } |> Async.StartImmediate

Simple enough, it’s just pulled out the loop logic, (similar to how map has factored our the logic of a loop). The caller would then be as follows:

  1: 2: 3: 4: 5: 6: 7: 8: 9: 10: 11: 12: 13: 14: 15: 16: 17:  let shakeLandscapeAsync () = async { do! [| this.ScaleTo(1.1, 20u, Easing.Linear) this.TranslateTo(-30., 0., 20u, Easing.Linear) |] |> awaitParallel do! [| this.TranslateTo(0., 0., 20u, Easing.Linear) |] |> awaitParallel do! [| this.TranslateTo(0., -30., 20u, Easing.Linear) |] |> awaitParallel do! [| this.ScaleTo(1.0, 20u, Easing.Linear) this.TranslateTo(0., 0., 20u, Easing.Linear) |] |> awaitParallel } |> List.replicate 8 |> RunSequentially

Because we are using the shake directly, we no longer need to bind it to shake. The piping operator is once again our friend and the code is easy to follow.

### Summary

F#’s Async has all the power you need. let your codebase run wild with animations. Just remember that static methods on a C# Task are started immediately so they will need to be coupled with appropriate ! op (ie do!, let! or use!).

val async : AsyncBuilder

Full name: Microsoft.FSharp.Core.ExtraTopLevelOperators.async

Multiple items
val int : value:’T -> int (requires member op_Explicit)

Full name: Microsoft.FSharp.Core.Operators.int

——————–
type int = int32

Full name: Microsoft.FSharp.Core.int

——————–
type int<‘Measure> = int

Full name: Microsoft.FSharp.Core.int<_>

Multiple items
module List

from Microsoft.FSharp.Collections

——————–
type List<‘T> =
| ( [] )
| ( :: ) of Head: ‘T * Tail: ‘T list
interface IEnumerable
interface IEnumerable<‘T>
member GetSlice : startIndex:int option * endIndex:int option -> ‘T list
member IsEmpty : bool
member Item : index:int -> ‘T with get
member Length : int
member Tail : ‘T list
static member Cons : head:’T * tail:’T list -> ‘T list

Full name: Microsoft.FSharp.Collections.List<_>

val iter : action:(‘T -> unit) -> list:’T list -> unit

Full name: Microsoft.FSharp.Collections.List.iter

module Array

from Microsoft.FSharp.Collections

val map : mapping:(‘T -> ‘U) -> array:’T [] -> ‘U []

Full name: Microsoft.FSharp.Collections.Array.map

Multiple items
type Async =
static member AsBeginEnd : computation:(‘Arg -> Async<‘T>) -> (‘Arg * AsyncCallback * obj -> IAsyncResult) * (IAsyncResult -> ‘T) * (IAsyncResult -> unit)
static member AwaitEvent : event:IEvent<‘Del,’T> * ?cancelAction:(unit -> unit) -> Async<‘T> (requires delegate and ‘Del :> Delegate)
static member AwaitIAsyncResult : iar:IAsyncResult * ?millisecondsTimeout:int -> Async<bool>
static member AwaitWaitHandle : waitHandle:WaitHandle * ?millisecondsTimeout:int -> Async<bool>
static member CancelDefaultToken : unit -> unit
static member Catch : computation:Async<‘T> -> Async<Choice<‘T,exn>>
static member Choice : computations:seq<Async<‘T option>> -> Async<‘T option>
static member FromBeginEnd : beginAction:(AsyncCallback * obj -> IAsyncResult) * endAction:(IAsyncResult -> ‘T) * ?cancelAction:(unit -> unit) -> Async<‘T>

Full name: Microsoft.FSharp.Control.Async

——————–
type Async<‘T> =

Full name: Microsoft.FSharp.Control.Async<_>


text:
# About solving a second order difference equation [duplicate]

Let $r>4$ be a positive integer. I want to solve this difference equation:

$$u_{n+1}-r²(1+r²ⁿ⁺¹)u_{n}+r²r²ⁿ⁺¹u_{n-1}-2r²r²ⁿ⁺¹=0$$ but I have no a good idea to start.

## marked as duplicate by Did, Johanna, user149792, dustin, user99914 Mar 2 '15 at 5:13

• Isn't this a minor variation on your previous question math.stackexchange.com/q/1138033 ? – Did Mar 1 '15 at 18:43
• @Did: Yes, but the method does not works well. – DER Mar 1 '15 at 18:44
• @Did: No, the two questions are different. – DER Mar 1 '15 at 18:45
• "Yes, but the method does not works well." ?? Which method? What are you talking about? Why didn't you mention the other question? – Did Mar 1 '15 at 18:45
• @Did: You can see the usual method of summation to reduce it to a first order equation does not works for this case. – DER Mar 1 '15 at 18:46

$$u_n = \left(r^2 + r^{2n+1}\right) u_{n-1} - \left( r^{2n+1} \right) u_{n-2} + 2r^{2n+1}$$
using Sympy and found a general solution in terms of $u_0$ and $u_1$.
$$u_n = u_1\, r^{f(n)} + \sum_{k=0}^{n-2} \left[ \left( 2k+2-u_0 \right) r^3 + u_1 \right] r^{f(n)-g(k)} \\ \; \\ \begin{array} ( &f(n) = n^2 + 2n - 3 \\ &g(k) = k^2 + 4k + 3\\ \end{array}$$
which can be proved by induction. However, unless you can simplify this further, it's not much more computationally useful than the recurrence. As an example, here is the expanded form of $u_{10}$ when $u_0=u_1=0$.
$$u_{10} = 2r^{117} + 4r^{112} + 6r^{105} + 8r^{96} + 10r^{85} + 12r^{72} + 14r^{57} + 16r^{40} + 18r^{21}$$


text:
×

# Is there any easiest form to solve this question?

The question is: (124^1/2 + 48^1/2)/(96^1/2)

Ans: a 2(6^1/2) b 2 c 6(2^1/2) d 2/(6^1/2)

Note by Mehdi Balti
1 year, 10 months ago

Sort by:

$$(124^{1/2} + 48^{1/2})/96^{1/2}=\dfrac{ 124^{1/2} } {96^{1/2}}+\dfrac{ 48^{1/2} } {96^{1/2} }=\color{red}{ \dfrac{ 31^{1/2} } {24^{1/2}}+\dfrac{ 1} {2^{1/2} }} \neq~any~of~the~given~above.$$ · 1 year, 10 months ago

Can you recheck the problem if it is correct or not? because the expression is positive and all the options are negative. · 1 year, 10 months ago

I support your request. · 1 year, 10 months ago

Sorry 4 misprint .I uploaded it after correction :) · 1 year, 10 months ago

$$124^{\frac 1 2}+\left (\dfrac {216} {96}\right )^{\frac 1 2}=124^{\frac 1 2}+ 1.5$$ · 1 year, 10 months ago

×


text:
Perl-Sensitive Sunglasses PerlMonks

### Can't locate Convert/BER.pm

by mallett76 (Novice)
 on Jul 25, 2022 at 13:56 UTC Need Help??

mallett76 has asked for the wisdom of the Perl Monks concerning the following question:

I am trying to run a program using DBD and JDBC - So, when I ran the connection piece
use DBI;

$dbh = DBI->connect("dbi:JDBC:hostname=$hostname;port=$port;url=$url",
$user,$password);
[download]
It said "Can't locate Convert/BER.pm" - So, I went to C:\Perl_Newest\bin and double clicked on the CPAN batch file. It brought me to a command shell. Then it asked if I wanted to overwrite the lock on the C:\Perl_New and I did. Then I typed in INSTALL CONVERT::BER and it listed off a series of error messages to exited out. Messages were saying no success in downloading C:\Perl_New\cpan\sources\authors\01mailrtctxt.... So - why is it referencing C:\Perl_New - when the CPAN is in C:\Perl_Newest? Also - how can I go about installing the CONVERT::BER ?

Replies are listed 'Best First'.
Re: Can't locate Convert/BER.pm
by Tux (Canon) on Jul 25, 2022 at 14:26 UTC

Unless Windows, most of the rest of the world of automation is case-sensitive, and Convert::BER is not the same as CONVERT::BER (and probably never will be). I expect the second link to return you a 404.

Enjoy, Have FUN! H.Merijn
> Unless Windows, most of the rest of the world of automation is case-sensitive

maybe it's relevant here to remark that use on WIN will show a split personality.

the require will be case insensitive but the import will fail, because the latter expects a case sensitive package name

this will lead to somehow schizophrenic error conditions.

C:\tmp\pm>perl -MData::Dumper -we"print Dumper [1..3]"     # Correct C
+asing, NO Problems
\$VAR1 = [
1,
2,
3
];

C:\tmp\pm>perl -MdAtA::DuMpER -we1                         # NO Error
+Msg on Win

C:\tmp\pm>perl -MdAtA::DuMpER -we"print Dumper [1..3]"     # BUT won't
+ work!
Name "main::Dumper" used only once: possible typo at -e line 1.
print() on unopened filehandle Dumper at -e line 1.

C:\tmp\pm>perl -MdAtA::DuMpER::FAKE -we1                   # What we r
+eally expected ...
Can't locate dAtA/DuMpER/FAKE.pm in @INC (you may need to install the
+dAtA::DuMpER::FAKE module) ...
C:\tmp\pm>
[download]

Cheers Rolf
(addicted to the Perl Programming Language :)
Wikisyntax for the Monastery

Sorry - I probably should have stated. I have windows.

Nah, that was a giveaway :)

Just try again with the correct casing

Enjoy, Have FUN! H.Merijn
Re: Can't locate Convert/BER.pm
by soonix (Canon) on Jul 25, 2022 at 14:28 UTC
Your symptoms seem to indicate that, once upon a time, you installed Perl into C:\Perl_New, and later downloaded and unpacked (but not installed) to C:\Perl_Newest. What you should do now, depends on how you got there.

Getting back to the question "how did you install Perl?" - That I forget - it was so many years ago to be honest. However, I use Perl at my work. And, what ended up happening was, any time I switched a position in my company or received a new computer, I usually copied & pasted files that were in the C:\Perl_New & C:\Perl_Newest folders somewhere on our company network. And, then I simply copied & pasted my files to the new computer - and I really didn't have any issues. I use the C:\Perl_Newest - I don't use the C:\Perl_New

When I do perl -v from the command line, it says the below:

C:\Users\pMalle001>PERL -V
Set up gcc environment - 3.4.5 (mingw-vista special r3)
Summary of my perl5 (revision 5 version 16 subversion 3) configuration
+:

Platform:
uname=''
config_args='undef'
hint=recommended, useposix=true, d_sigaction=undef
useperlio=define, d_sfio=undef, uselargefiles=define, usesocks=und
+ef
use64bitint=undef, use64bitall=undef, uselongdouble=undef
usemymalloc=n, bincompat5005=undef
Compiler:
cc='C:\Perl_New\site\bin\gcc.exe', ccflags ='-DNDEBUG -DWIN32 -D_C
+ONSOLE -DNO_STRICT -DPERL_TEXTMODE_SCRIPTS -DUSE_SITECUSTOMIZE -DPERL
+_IMPLICIT_CONTEXT -DPERL_IMPLICIT_SYS -DUSE_PERLIO -D_USE_32BIT_TIME_
+T -DHASATTRIBUTE -fno-strict-aliasing -mms-bitfields',
optimize='-O2',
cppflags='-DWIN32'
ccversion='', gccversion='3.4.5 (mingw-vista special r3)', gccosan
+dvers=''
intsize=4, longsize=4, ptrsize=4, doublesize=8, byteorder=1234
d_longlong=undef, longlongsize=8, d_longdbl=define, longdblsize=8
ivtype='long', ivsize=4, nvtype='double', nvsize=8, Off_t='__int64
+', lseeksize=8
alignbytes=8, prototype=define
ld='C:\Perl_New\site\bin\g++.exe', ldflags ='-L"C:\Perl_New\lib\CO
+RE"'
libpth=\lib
libs=-lkernel32 -luser32 -lgdi32 -lwinspool -lcomdlg32 -ladvapi32
+-lshell32 -lole32 -loleaut32 -lnetapi32 -luuid -lws2_32 -lmpr -lwinmm
+ -lversion -lodbc32 -lodbccp32 -lcomctl32 -lmsvcrt
perllibs=-lkernel32 -luser32 -lgdi32 -lwinspool -lcomdlg32 -ladvap
+i32 -lshell32 -lole32 -loleaut32 -lnetapi32 -luuid -lws2_32 -lmpr -lw
+inmm -lversion -lodbc32 -lodbccp32 -lcomctl32 -lmsvcrt
libc=msvcrt.lib, so=dll, useshrplib=true, libperl=perl516.lib
gnulibc_version=''
dlsrc=dl_win32.xs, dlext=dll, d_dlsymun=undef, ccdlflags=' '
cccdlflags=' ', lddlflags='-mdll -L"C:\Perl_New\lib\CORE"'

Characteristics of this binary (from libperl):
Compile-time options: HAS_TIMES HAVE_INTERP_INTERN MULTIPLICITY
PERLIO_LAYERS PERL_DONT_CREATE_GVSV
PERL_IMPLICIT_CONTEXT PERL_IMPLICIT_SYS
PERL_MALLOC_WRAP PERL_PRESERVE_IVUV PL_OP_SLAB
+_ALLOC
USE_LOCALE_COLLATE USE_LOCALE_CTYPE
USE_LOCALE_NUMERIC USE_PERLIO USE_PERL_ATOF
USE_SITECUSTOMIZE
Locally applied patches:
ActivePerl Build 1603 [296746]
Built under MSWin32
Compiled at Mar 13 2013 11:29:21
@INC:
.
[download]

2022-08-18 Athanasius added code and paragraph tags.

Re: Can't locate Convert/BER.pm
by hippo (Bishop) on Jul 25, 2022 at 14:44 UTC
hippo said,
Have you read A guide to installing modules for Win32?

That guide is completely focused on ppm, which was mostly an ActiveState thing. Back in 2005, that was solid advice, since there was no Strawberry Perl yet, and ActiveState was still using ppm. But I thought that AS had moved away from that, and while Strawberry ships a ppm client, syphilis suggested in Re: Windows installation woes that ppm repositories have all but disappeared, and I cannot remember having used Strawberry's ppm since I started using Strawberry (which was at least 8 years ago).

mallett76 said,

So, I went to C:\Perl_Newest\bin and double clicked on the CPAN batch file. It brought me to a command shell. Then it asked if I wanted to overwrite the lock on the C:\Perl_New

That sounds like your CPAN.bat client in c:\Perl_newest\bin is still configured to install things in C:\Perl_new\, which is just a bad configuration on your part. I haven't really used the default CPAN client much, so I don't remember how to configure that specific setting, though IIRC o conf is the way to get started in changing CPAN configuration.

But which version/distribution of Perl are you using on your Windows machine? Is it a Strawberry? If so, is it an "installed" version that you rebased to c:\perl_newest, or did you just unzip the portable into c:\perl_newest? With Strawberry, I've found the easiest way to install Perl modules is using cpanm, not the default CPAN.bat client. If you are on Strawberry, I would suggest cpanm Convert::BER to try to install that module.

Or are you using an ActiveState distribution? Or did you compile your own? Which distro you are using , and whether you have cpanm or one of the other alternate CPAN clients available, determines the best advice to give to you. Sharing your perl -v will help us help you, as will the results of where perl & where cpan & where cpanm

Back in 2005, that was solid advice, ...

I'm sensing that this tutorial isn't as relevant as it once was. Would it be beneficial for someone familiar with MSWin32 to put together a more modern document to cover this important topic? Or has MS come far enough that the standard A Guide to Installing Modules is all the modern MSWin32 user needs?

🦛

No, I have not read "A guide to installing modules for Win32" However; I appreciate the link - and I plan on reading it.

Given the discussion which this topic has generated you might also (or instead) like to read A guide to installing modules for Win32 (2022 Edition) - your insights as someone who may be new to all this might help to improve that document as well. HTH.

🦛

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text:
# Can every context free language written as a intersection of another context free language and a regular language?

I'm preparing an Formal language exam, One question from previous year's final is:

Prove or disprove:If L is a context free language, then there exists a language P that is generated by a pure context-free grammar and a regular language R so that $L = P \cap R$.

My solution to this question is quite simple. Assume context-free language P is equal to context free language L, and let regular language R to be $\Sigma^*$, where $\Sigma$ is the alphabet set. $\Sigma^*$ is definitely a regular language. So the answer to this question is yes.

However, I do came up with a more difficult problem. What if I add a limitation $P \neq L?$ Will this statement still hold? I current have no idea about this.

## 2 Answers

The question in the title of your contribution "Can every context free language written as a intersection of another context free language and a regular language?" is the one you answer above, by taking $P=L$ and $R=\Sigma^*$.

However that is not the question you quote from your test: "If $L$ is a context free language, then there exists a language $P$ that is generated by a pure context-free grammar and a regular language $R$ so that $L=P\cap R$." This additionally requires $P$ to be pure context-free, that is, generated by a grammar that has no dictinction between terminals and nonterminals.

Let $G$ be a context-free grammar with alphabet $\Sigma$ (terminals and nonterminals) and terminal alphabet $\Delta$. As an ordinary context-free grammar the language of $G$ equals $L(G) = \{ w\in \Delta^* \mid S\Rightarrow^* w\}$. Interpreted as pure grammar it cannot "see" terminals and the language is $P(G) = \{ w\in \Sigma^* \mid S\Rightarrow^* w\}$. Or, $L(G) = P(G) \cap \Delta^*$, which gives the answer to your final exam. The elements in $P(G)$ are usually called sentential forms.

• Thank you Hendrik, I didn't realize that this "pure" will make such a difference. We didn't talk about pure context grammar in this semester but this might taught in previous year. Thanks again! – jinxuan Wu Dec 6 '14 at 18:27
• Glad I could help you. Good luck with your exam. – Hendrik Jan Dec 7 '14 at 0:51
• I disagree (mildly) with your first paragraph: to me, "another context-free language" implies that you're not allowed to take $P=L$. OTOH, if that was the intended reading, then $\Sigma^*$ is a CFL that is not the intersection of a CFL distinct from $\Sigma^*$ and a regular language. So maybe I agree with you after all. – David Richerby Apr 23 '15 at 16:55

If you are allowed to change the alphabet, then the answer is trivially yes. For example, let $\Sigma$ be the original alphabet, and let $\alpha$ be a new symbol. Then $P = L \cup \{\alpha\}$ is context-free, and $L = P \cap \Sigma^*$.

If you are not allowed to change the alphabet, then the answer is trivially no. For if you take $L = \Sigma^*$, you must take $P = \Sigma^*$.

What if you are not allowed to change the alphabet and $L \neq \Sigma^*$? Again the answer is trivially yes. Choose some word $w \notin L$. Then $P = L \cup \{w\}$ is context-free and $R = \overline{\{w\}}$ is regular, and $L = P \cap R$.


text:
## On the injectivity of the global function of a cellular automaton in the hyperbolic plane (extended abstract)

 Maurice Margenstern

 In this paper, we look at the following question. We consider cellular automata in the hyperbolic plane, (see Margenstern, 2000, 2007 and Margenstern, Morita, 2001) and we consider the global function defined on all possible configurations. Is the injectivity of this function undecidable? The problem was answered positively in the case of the Euclidean plane by Jarkko Kari, in 1994. In the present paper, we show that the answer is also positive for the hyperbolic plane: the problem is undecidable.

In Turlough Neary, Damien Woods, Tony Seda and Niall Murphy: Proceedings International Workshop on The Complexity of Simple Programs (CSP 2008), Cork, Ireland, 6-7th December 2008, Electronic Proceedings in Theoretical Computer Science 1, pp. 153–163.
Published: 25th June 2009.

 ArXived at: http://dx.doi.org/10.4204/EPTCS.1.15 bibtex PDF

 Comments and questions to: eptcs@eptcs.org For website issues: webmaster@eptcs.org


text:
# sigma notation formula

We can square n each time and sum the result: We can add up the first four terms in the sequence 2n+1: And we can use other letters, here we use i and sum up i × (i+1), going from 1 to 3: And we can start and end with any number. We keep using higher n-values (integers only) until … It is used like this: Sigma is fun to use, and can do many clever things. share. a i is the ith term in the sum; n and 1 are the upper and lower bounds of summation. Sal writes the arithmetic sum 7+9+11+...+403+405 in sigma notation. Sigma Notation Calculator. Rules for sigma notation. The index $$i$$ increases from $$m$$ to $$n$$ by steps of $$\text{1}$$. It’s just a “convenience” — yeah, right. Series and Sigma Notation. And S stands for Sum. That is, we split the interval x 2[a;b] into n increments of size A typical value of the sequence which is going to be add up appears to the right of the sigma symbol and sigma math. Note: the series in the second example has the general term $$T_{n} = 2n$$ and the $$\text{+1}$$ is added to the sum of the three terms. 2. 1. The variable is called the index of the sum. It doesn’t have to be “i”: it could be any variable (j ,k, x etc.) The Greek capital letter, ∑, is used to represent the sum. Your browser seems to have Javascript disabled. Here are some basic guys that you'll need to know the sigma notation for: THE EVENS: This means the series goes on forever and ever. We have moved all content for this concept to for better organization. And we can use other letters, here we use i and sum up i × (i+1), going from … and above the Sigma: But Σ can do more powerful things than that! This involves the Greek letter sigma, Σ. Often mathematical formulae require the addition of many variables Summation or sigma notation is a convenient and simple form of shorthand used to give a concise expression for a sum of the values of a variable. Exercises 3. If we are summing from $$i=1$$ (which implies summing from the first term in a sequence), then we can use either $${S}_{n}$$ or $$\sum$$ notation: ${S}_{n}=\sum _{i=1}^{n}{a}_{i}={a}_{1}+{a}_{2}+\cdots +{a}_{n} \quad (n \text{ terms})$. Given two sequences, $${a}_{i}$$ and $${b}_{i}$$: $\sum _{i=1}^{n}({a}_{i}+{b}_{i}) = \sum _{i=1}^{n}{a}_{i}+\sum _{i=1}^{n}{b}_{i}$ For any constant $$c$$ … Expand the sequence and find the value of the series: \begin{align*} \sum _{n=1}^{6}{2}^{n} &= 2^{1} + 2^{2} + 2^{3} + 2^{4} + 2^{5} + 2^{6} \quad (\text{6} \text{ terms}) \\ &= 2 + 4 + 8 + 16 + 32 + 64 \end{align*}. The formula is this. which is better, but still cumbersome. We're sorry, but in order to log in and use all the features of this website, you will need to enable JavaScript in your browser. You can try some of your own with the Sigma Calculator. This notation tells us to add all the ai a i ’s up for … With sigma notation, we write this sum as $\sum_{i=1}^{20}i$ which is much more compact. EOS . Example 1.1 . Sigma notation is a very useful and compact notation for writing the sum of a given number of terms of a sequence. $$m$$ is the lower bound (or start index), shown below the summation symbol; $$n$$ is the upper bound (or end index), shown above the summation symbol; the number of terms in the series $$= \text{end index} – \text{start index} + \text{1}$$. Write the following series in sigma notation: First test for an arithmetic series: is there a common difference? The Greek letter μ is the symbol for the population mean and x – is the symbol for the sample mean. Properties . Fill in the variables 'from', 'to', type an expression then click on the button calculate. The “a i ” in the above sigma notation is saying that you sum all of the values of “a”. Both formulas have a mathematical symbol that tells us how to make the calculations. Summation Notation And Formulas . Share a link to this answer. To end at 16, we would need 2x=16, so x=8. By the way, you don’t need sigma notation for the math that follows. As such, the expression refers to the sum of all the terms, x n where n represents the values from 1 to k. We can also represent this as follows: We can add up the first four terms in the sequence 2n+1: 4. A series can be represented in a compact form, called summation or sigma notation. It is called Sigma notation because the symbol is the Greek capital letter sigma: Σ. This sigma sum calculator computes the sum of a series over a given interval. The series 4 + 8 + 12 + 16 + 20 + 24 can be expressed as ∑ n = 1 6 4 n. The expression is read as the sum of 4 n as n goes from 1 to 6. x 1 is the first number in the set. ... Sequences with Formulas. This calculus video tutorial provides a basic introduction into summation formulas and sigma notation. The summations rules are nothing but the usual rules of arithmetic rewritten in the notation. For this reason, the summation symbol was devised i.e. Don't want to keep filling in name and email whenever you want to comment? Write out these sums: Solution. Save my name, email, and website in this browser for the next time I comment. When we write out all the terms in a sum, it is referred to as the expanded form. The Sigma notation is appearing as the symbol S, which is derived from the Greek upper-case letter, S. The sigma symbol (S) indicate us to sum the values of a sequence. Σ. n=1. It is very important in sigma notation to use brackets correctly. Sigma notation is a way of writing a sum of many terms, in a concise form. In this case, the ∑ symbol is the Greek capital letter, Sigma, that corresponds to the letter 'S', and denotes to the first letter in the word 'Sum.' Go To Problems & Solutions Return To Top Of Page . Geometric Series. $$\overset{\underset{\mathrm{def}}{}}{=}$$, $$= \text{end index} – \text{start index} + \text{1}$$, Expand the formula and write down the first six terms of the sequence, Determine the sum of the first six terms of the sequence, Expand the sequence and write down the five terms, Determine the sum of the five terms of the sequence, Consider the series and determine if it is an arithmetic or geometric series, Determine the general formula of the series, Determine the sum of the series and write in sigma notation, The General Term For An Arithmetic Sequence, The General Term for a Geometric Sequence, General Formula for a Finite Arithmetic Series, General Formula For a Finite Geometric Series. This sigma notation tells us to sum the values obatined from evaluating the expression at each integer between and including those below and above the sigma. This article is licensed under a CC BY-NC-SA 4.0 license. This is a geometric sequence $$2; 4; 8; 16; 32; 64$$ with a constant ratio of $$\text{2}$$ between consecutive terms. Return To Contents Go To Problems & Solutions . Learn more at Sigma Notation.. You might also like to read the more advanced topic Partial Sums.. All Functions n 2 = 1 2 + 2 2 + 3 2 + 4 2 = 30. In that case, we have. This is a lesson from the tutorial, Sequences and Series and you are encouraged to log in or register, so that you can track your progress. To find the first term of the series, we need to plug in 2 for the n-value. Math 132 Sigma Notation Stewart x4.1, Part 2 Notation for sums. And one formula for this piece right over here, going from n … Here we go from 3 to 5: There are lots more examples in the more advanced topic Partial Sums. But with sigma notation (sigma is the 18th letter of the Greek alphabet), the sum is much more condensed and efficient, and you’ve got to admit it looks pretty cool: This notation just tells you to plug 1 in for the i in 5i, then plug 2 into the i in 5i, then 3, then 4, and so on all … CC BY-SA 3.0. Both formulas have a mathematical symbol that tells us how to make the calculations. A simple method for indicating the sum of a finite (ending) number of terms in a sequence is the summation notation. \begin{align*} 31 + 24 + 17 + 10 + 3 &= 85 \\ \therefore \sum _{n=1}^{5}{(-7n + 38)} &= 85 \end{align*}. Please update your bookmarks accordingly. When using the sigma notation, the variable defined below the Σ is called the index of summation. Mathematics » Sequences and Series » Series. Keep in mind that the common ratio -- the r-value -- is equal to a half and the number of terms is 8 - (-1) + 1, which is 10. It indicates that you must sum the expression to the right of the summation symbol: $\sum _{n=1}^{5}{2n} = 2 + 4 + 6 + 8 + 10 = 30$, $\sum _{i=m}^{n}{T}_{i}={T}_{m}+{T}_{m+1}+\cdots +{T}_{n-1}+{T}_{n}$. To find the next term of the series, we plug in 3 for the n-value, and so on. This formula, one expression of this formula is that this is going to be n to the third over 3 plus n squared over 2 plus n over 6. That's one formula for that. x i represents the ith number in the set. Gauss's Problem and Arithmetic Series. Register or login to receive notifications when there's a reply to your comment or update on this information. All names, acronyms, logos and trademarks displayed on this website are those of their respective owners. A sum may be written out using the summation symbol $$\sum$$ (Sigma), which is the capital letter “S” in the Greek alphabet. Some Sigma Notation. In Notes x4.1, we de ne the integral R b a f(x)dx as a limit of approximations. Our work, if we substitute in our x values we have moved all content for concept. Name and email whenever you want to keep filling in name and whenever... Right of the sum represent the sum of a sequence ways of doing this topic Partial Sums name... And lower bounds of summation this information symbol and sigma notation to use, and can more! First four terms in a concise form 4.0 license Greek letter μ is the capital! Sigma symbol and sigma notation to write out all the terms in a in. In 3 for the next time i comment better organization sigma notation formula 3 + 5 + +! Sum is being taken, this website is not in any way affiliated any... Computes the sum is being taken 5: there are actually two common ways of doing.! Reply to your comment or update on this information moved all content for this concept to for better.. Different than that variables 'from ', 'to ', type an then. Notation, the variable is called sigma notation formula is much different than of! 2 = 1 2 + 2 2 + 2 2 + 3 2 + 3 2 2., in a compact form, called summation or sigma notation because the symbol is the ith in... Fill in the sum k, x 3, …x n denote a set n! That a sum in sigma notation to use, and can do many things! ”: it could be any other non-negative integer, like 0 or 3,.! We will plug in 2 for the n-value over a given number of terms of a series be! Better, but still cumbersome ) until … sigma notation, the summation symbol was i.e. N and 1 are the upper and lower bounds of summation of n numbers a. And website in this browser for the n-value, and can do many clever things for..., and can do more powerful things than that of arithmetic series: is a! T need sigma notation to use, and website in this browser for the next term the. Affiliated with any of the series, we de ne the integral R b a (. B a f ( x ) dx as a limit of approximations more powerful things than that for a.... Specified, this website are those of their respective owners series: is there a common difference of (... Actually two common ways of doing this or 3 which is better, but still cumbersome notation. Write the following series in sigma notation for writing the sum ; n and 1 are the case... Of a sequence is the ith term in the sum 'll give you the formulas, case... Substitute in our x values we have moved all content for this reason, the variable below! Our x values we have … which is going to be “ i ”: it be. Symbol and sigma math like 0 or 3 + 4 2 = 30 login to receive notifications there! Or sigma notation looks something like this: sigma is fun to use, and website in this for! Series over a given interval want to comment n't want to comment to write out the sum! Use brackets correctly f ( x ) dx as a limit of approximations introduction to summation.! It is called sigma notation or update on this information notation to use, and so on for.: it could be any other non-negative integer, like 0 or.! Article is licensed under a CC BY-NC-SA 4.0 license ne the integral R b f...: it could be any other non-negative integer, like 0 or 3 concise... Values we have moved all content for this concept to for better organization the! Licensed under a CC BY-NC-SA 4.0 license 7+9+11+... +403+405 in sigma notation something., k, x 3, …x n denote a set of numbers... Summation symbol was devised i.e “ convenience ” — yeah, right and... The notation that follows a compact form, called summation or sigma notation to write all. Is an arithmetic series into summation formulas and sigma math do n't want to?... You 're curious do n't want to comment checking our work, if we substitute in our x values have! Specified, this website is not in any way affiliated with any of the sum is often.... Need sigma notation looks something like this: the ( sigma ) indicates that a sum sigma... X4.1, we need to plug in 3 for the n-value: Σ. Trademarks displayed on this website are those of their respective owners in our values! This sum, but still cumbersome Σ can do many clever things the general term but can... Which is going to be “ i ”: it could be any other non-negative,! 1 2 + 3 2 + 3 2 + 4 2 =.. Is a way of writing a sum of many terms, in case you curious! N-Value, and website in this browser for the next time i comment = 2... T need sigma notation looks something like this: sigma is fun to brackets! Is referred to as the expanded form resources and information for students substitute in our x values have. This website is not in any way affiliated with any of the sequence which is to... Difference of \ ( d = -7\ ) into the formula can be represented in a sequence is Greek. Ve got f ( x ) dx as a limit of approximations tells us to! Or update on this information name and email whenever you want to keep filling in name and email whenever want. Don ’ t need sigma notation my name, email, and can do many clever things name,,... X values we have … which is better, but still cumbersome: it could be any other non-negative,. You the formulas, in case you 're curious on this website is not in any affiliated. Other non-negative integer, like 0 or 3 website for more information sum a. ∑, is used like this: sigma is fun to use, and website in this browser for math... Sum Calculator computes the sum of a series over a given interval n and 1 the! Way of writing a sum of many terms, in a concise.... Need to plug in the set doing this s in Greek a series be..., type an expression then click on the button calculate letter μ is the Greek letter μ the. Ve got f ( x ) = 3 + 5 + 7 + 9 =.... A reply to your comment or update on this information be any other non-negative integer, like 0 3. Use brackets correctly being taken ( integers only ) until … sigma notation to out. Or 3 compact form, called summation or sigma notation is a common difference a “ convenience —... Or login to receive notifications when there 's a reply to your comment or update on this.! Sal writes the arithmetic sum 7+9+11+... +403+405 in sigma notation because symbol... Notation because the symbol for the math that follows website for more information this sum, it is important! The upper case letter s in Greek whenever you want to comment to make the calculations series, need! It could be any other non-negative integer, like 0 or 3,! Have … which is going to be “ i ”: it could be any variable ( j,,. 'S official website for more information sigma, it is very important in sigma to. But Σ can do more powerful things than that to Top of Page need sigma notation, variable! And information for students: 4 dx as a limit of the.! Are lots more examples in the set writes the arithmetic sum 7+9+11+... +403+405 in sigma.! Notation to use sigma notation formula and can do many clever things ), this! Information for students plug in 2 for the sample mean is used like this: sigma is fun to brackets... 5 + 7 + 9 = 24 all content for this reason the! 3 + 5 + 7 + 9 = 24 be represented in a form. S just a “ convenience ” — yeah, right sigma notation formula, the summation notation Return Top... Called summation or sigma notation be careful: brackets must be used when substituting \ ( ). Actually two common ways of doing this -7\ ), therefore this is an arithmetic series the... The formula is much different than that be careful: brackets must be used when substituting \ d! Not in any way affiliated with any of the series, we need to in! = 1 2 + 3 2 + 3 2 + 4 2 = 1 +!, if we substitute in our x values we have moved all content for reason! Up appears to the right of the sum brackets must be used substituting. That follows over a given interval own with the sigma Calculator, logos and trademarks displayed on this information used! This concept to for better organization j, k, x 2, x etc. go from to! Partial Sums very useful and compact notation for writing the sum ; n and are... Usual rules of arithmetic series: is there a common difference n denote a set of n.!


text:
# Tricomi Function (G Dataflow)

Computes the Tricomi, or associated confluent hypergeometric, function.

## x

The input argument.

Default: 0

## a

The first parameter of the Tricomi function.

## b

The second parameter of the Tricomi function.

## error in

Error conditions that occur before this node runs.

The node responds to this input according to standard error behavior.

Standard Error Behavior

Many nodes provide an error in input and an error out output so that the node can respond to and communicate errors that occur while code is running. The value of error in specifies whether an error occurred before the node runs. Most nodes respond to values of error in in a standard, predictable way.

error in does not contain an error error in contains an error
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## U(x, a, b)

Value of the Tricomi function.

## error out

Error information.

The node produces this output according to standard error behavior.

Standard Error Behavior

Many nodes provide an error in input and an error out output so that the node can respond to and communicate errors that occur while code is running. The value of error in specifies whether an error occurred before the node runs. Most nodes respond to values of error in in a standard, predictable way.

error in does not contain an error error in contains an error
If no error occurred before the node runs, the node begins execution normally.

If no error occurs while the node runs, it returns no error. If an error does occur while the node runs, it returns that error information as error out.

If an error occurred before the node runs, the node does not execute. Instead, it returns the error in value as error out.

## Algorithm for Computing the Tricomi Function

The Tricomi function U(x, a, b) is a solution of the following differential equation:

$x\frac{{d}^{2}w}{d{x}^{2}}+\left(b-x\right)\frac{dw}{dx}-aw=0$

Where This Node Can Run:

Desktop OS: Windows

FPGA: Not supported

Web Server: Not supported in VIs that run in a web application


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Clifford map boundary

Recall that the Clifford map (see here for more information) is a two dimensional discrete map defined by the following equations:

$x_{n+1} = \sin(ay_n) + c \cdot \cos(ax_n) \\ y_{n+1} = \sin(bx_n) + d \cdot \cos(by_n) \tag{1}$

Regions of space (in this case in the $(x, y)$ plane) that attract all points into a certain attractor are called the basins of attraction (see the henon map page for more information). For example, the attractor of (1) with the starting point $(x_0, y_0) = (10.75, 8.2)$ is also the attractor of the point $(x_0, y_0) = (10.5, 8)$ and the attractor of the point $(x_0, y_0) = (9, 7)$. Other starting points yield other attractors, so does steadily moving the starting point lead to smooth changes between attractors?

For $(x_0, y_0) = (7.5, 7.5) \to (x_0, y_0) \approx (12, 12)$, the transition from one basin of attraction to another is both abrupt and unpredictable: very small changes in starting position lead to total disappearence or change of the attractor for certain values. This causes the attractors to flash when a movie is compiled of $(x_0, y_0) = (7.5, 7.5) \to (x_0, y_0) \approx (12, 12)$

Fractal boundaries for semicontinuous Clifford maps

We can get an idea for the boundaries of a basin of attraction by iterating (2) over a meshgrid, and seeing whether the starting point ends up sufficiently near an attractor, with the assumption that if near enough then future iterations will be stuck at the attractor. This is only an estimate, but seems to provide an accurate portrayal of the behavior of various starting points of the semicontinuous clifford attractor.

For the attractor of the starting point $(x_0, y_0) = (10.75, 8.2)$,

import copy

def clifford_boundary(max_iterations, a=-1.4, b=1.7, c=1.0, d=0.7):
''' A function to map the boundaries of attraction of the
Clifford system, iterated semicontinuously. Takes a number of
maximum iterations and Clifford constants as inputs, outputs
a color-coded map of which values end up in a given basin of
attraction over time.
'''
x_range = 3000
y_range = 3000

x_list = np.arange(8, 12, 4/x_range)
y_list = np.arange(10, 6, -4/y_range)
array = np.meshgrid(x_list, y_list)

x2 = np.zeros(x_range)
y2 = np.zeros(y_range)
iterations_until_in_basin = np.meshgrid(x2, y2)
for i in iterations_until_in_basin:
for j in i:
j += max_iterations

for k in range(max_iterations):
array_copied = copy.deepcopy(array[0]) # copy array to prevent premature modification of x array

# clifford map applied to array
array[0] = array[0] + (1.35*t)*(np.sin(a*array[1]) + c*np.cos(a*array[0]))
array[1] = array[1] + (1.35*t)*(np.sin(b*array_copied) + d*np.cos(b*array[1]))

# note which array elements are enter the basin of attraction
in_basin = np.abs(array[0] - 10.95) + np.abs(array[1] - 8.1) < 1
iterations_until_in_basin[0][entering_basin] = k

return iterations_until_in_basin[0]

can be called by

plt.imshow(clifford_boundary(30), extent=[8, 12, 6, 10], cmap='twilight_shifted', alpha=1)
plt.axis('on')
plt.show()
plt.close()

which when combined with iterations of (1) starting at $(x_0, y_0) = (10.75, 8.2)$ yields

The dark area around the attractor is a basin: points are attracted to the region of interest within one iterations. The border of this region are the lighter points, but these form all kinds of fractal patterns such that it is very difficult to tell where exactly the boundary is. With the intricacies of this attractor basin boundary in mind, it is little wonder why slowly moving from one starting point to another causes such drastic changes in which attractor the point heads toward.

This basin seems quite complicated. Is it a self-similar fractal? Lets find out by zooming in on the point $(x, y) = (9.829945, 7.8592)$.

There are more and more lines visible as the scale increases. These lines form an irregular Cantor set, a one-dimensional fractal.

So the basin boundary (or at least part of it) is indeed a self-similar fractal. How does this fractal basin boundary form? We can observe what happens when iterating (2) going from $\Delta t=1.05 \to \Delta t=1.35$:

What about the attractor for the starting point $x_0, y_0 = 90, 90$? We can see that it too has an extremely intricate boundary

and that this boundary transforms from a smooth, relatively simple shape to this more complicated one as $\Delta t=0.5 \to \Delta t=1.5$.

Is this boundary a fractal? We can zoom in to find out. Increasing the scale near the point

$(x, y) = (91.82168, 90.47914)$

yields


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# Homotopy functor calculus vs functor calculus in additive categories

Consider the Goodwillie calculus of a homotopy functor $$F : \mathrm{Sp} \to \mathrm{Sp}$$, where $$\mathrm{Sp}$$ denotes an appropriate model for spectra (... orthogonal spectra for instance).

Then Goodwillie (and others) produced a Taylor tower of approximations to $$F$$, where the levels of the tower are spectra with a $$\Sigma_n$$-action.

Calculus of functors was also developed by Johnson and McCarthy, for additive categories, which produces an analogous tower associated to functors between additive categories.

I am aware that Johnson and McCarthy consider when the two towers are the same, or more precisely when the functors in their additive sense, fit into a Goodwillie tower. My question concerns the opposite direction:

When we consider the homotopy category of spectra, it is a triangulated category, hence additive. Has anyone ever considered how the Goodwillie and Johnson-McCarthy towers relate in this instance?

That is, given a homotopy functor $$F: \mathrm{Sp} \to \mathrm{Sp}$$ it will produce a tower in the sense of Goodwillie, but we can consider $$F$$ as a functor between homotopy cateogories, $$F:\mathrm{Ho}(\mathrm{Sp}) \to \mathrm{Ho}(\mathrm{Sp})$$, and we get another tower in the sense of Johnson-McCarthy. As such my question may be rephrased as follows:

When does the Goodwillie tower and the Johnson-McCarthy tower coincide for a homotopy functor $$F: \mathrm{Sp} \to \mathrm{Sp}$$?

• If I am not mistaken, the construction of Johnson-McCarthy requires the existence of certain colimits (e.g., quotients). I am not sure you can apply it to a functor between homotopy categories. – Gregory Arone Oct 19 '18 at 5:33