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Multivariable Calculus Video Tutorials | Hire Someone To Do Calculus Exam For Me
Multivariable Calculus Video Tutorials It’s another step in the process of learning how to use the tools to make the most of our time with the tools we’ve been given. Although thousands of people
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CO2226 Software engineering, algorithm design and analysis Coursework assignment 2 Java
Computing and Information Systems/Creative Computing
CO2226 Software engineering, algorithm design and analysis
Coursework assignment 2 2021–2022
Marks will be awarded for correct code, i.e. for code that produces results; if your code does not produce the correct result, no marks will be awarded for that part of the question.
You must submit one Java file called: Ass2226<StudentID>.java. For example, if your student number ID is 101031722, your file will be named: Ass2226101031722.java.
When this file is compiled, it must produce a file: Ass2226<StudentID>.class, e.g. Ass226101031722.class. When run, this must produce the answers to the coursework assignment questions by
implementing the appropriate algorithms.
Your java file may contain other classes, but they must all be included in the single java file; please do not submit multiple Java files as you will get a mark of zero – we need just one file. You
must write your code assuming all data files are in the same directory as the program (and only use their names as arguments as shown in the example below).
Failure to do this will lead to a mark of zero. If you cannot complete a particular question, the answer should be ‘NOT DONE’. Your program should take the text files described below as command-line
To run your program, the examiners will type:
java Ass2226101031722 pubs pubs_lon_lat randomGraph
It is important to note that the filenames are referenced without the file extension.
Your output should look like this:
Execution Time: 32094 milliseconds
These are just sample answers to show the output format required from your program. The examiners will change the data files to test your programs so make sure your program works with files
containing fewer/more pubs or the existing pubs with different ids. Try deleting some lines from the files and see if your program gives different answers. You should use the code provided and adapt
it to answer the questions – no changes or alternative approaches the logic of the code are allowed.
Efficiency of your program
You will be penalised if your program runs too slowly (5 marks for every minute over 5 minutes on a machine with Intel Core i7 vPro processor with 12 gigabytes of RAM).
Try to speed up your program by avoiding re-computing values that you have already computed. Instead, store them (rather than re-computing) and identify opportunities and questions that will allow
you to do so – then these values will be readily available to your program.
Use System.nanoTime(); to time your program. Read the value at the beginning and end of your program and subtract and divide by a billion to get the result expressed in seconds.
IF YOU DO NOT USE THE DATA PROVIDED YOU WILL SCORE ZERO.
Coursework assignment 2 – preparation and pre-assignment tasks
Finding the shortest paths in unweighted graphs (breadth-first search)
Research Adjacency matrices for representing graphs. Here is a program to help you become familiar with them:
import java.util.HashSet;
import java.util.ArrayList;
adj= new double [a.length][a.length];
for (int i=0;i<a.length;i++)
for (int j=0;j<a.length;j++) adj[i][j]=a[i][j];
public HashSet <Integer> neighbours(int v)
HashSet <Integer> h = new HashSet <Integer> ();
for (int i=0;i<adj.length;i++)
public HashSet <Integer> vertices()
HashSet <Integer> h = new HashSet <Integer>();
for (int i=0;i<adj.length;i++)
ArrayList <Integer> addToEnd (int i, ArrayList <Integer> path)
// returns a new path with i at the end of path
ArrayList <Integer> k; k=(ArrayList<Integer>)path.clone(); k.add(i);
public HashSet <ArrayList <Integer>> shortestPaths1(HashSet
<ArrayList <Integer>> sofar, HashSet <Integer> visited, int end)
HashSet <ArrayList <Integer>> more = new HashSet <ArrayList
HashSet <ArrayList <Integer>> result = new HashSet <ArrayList
HashSet <Integer> newVisited = (HashSet <Integer>)
for (ArrayList <Integer> p: sofar)
for (Integer z: neighbours(p.get(p.size()-1)))
if (!visited.contains(z))
carryon=true; newVisited.add(z);
if (done) return result; else
return shortestPaths1(more,newVisited,end);
return new HashSet <ArrayList <Integer>>();
public HashSet <ArrayList <Integer>> shortestPaths( int first,
HashSet <ArrayList <Integer>> sofar = new HashSet <ArrayList<Integer>>();
HashSet <Integer> visited = new HashSet<Integer>();
ArrayList <Integer> starting = new ArrayList<Integer>();
return sofar; visited.add(first);
return shortestPaths1(sofar,visited,end);
public static void main(String [] args)
for (int i=0;i<a.length;i++)
{for (int j=0;j<a.length;j++)
if (i!=j) System.out.println(i + " to " + j +": "+ g.shortestPaths(i,j));
Draw a picture of the graph and see if you agree with the output (please note that the constructor assumes a non-directed graph; this is not the case with our graph for the coursework assignment, and
is given here only for illustration purposes). Play with the program and alter the graph to check that you understand how the program works.
The Liverpool Pubs Distance Problem
Study the following files of data about Liverpool pubs:
• pubs.csv. This file has two fields in the following order: id of the pub and the name of the pub.
• randomGraph.csv. This file contains three fields, the pub id of the source pub, the pub id of the destination pub and the cost for taking this route.
• pubs_lon_lat.csv. This file has three fields in the following order: the pub id, the pub’s longitude, and the pub’s latitude.
Examine the following program. Note, again, that the main method assumes a non-directed graph, which is not the case for the coursework assignment, but is included here to help you become familiar
with the process):
import java.util.Scanner;
static double [][] edges = new double[N][N];
static TreeMap <Integer,String> pubNames = new TreeMap
static ArrayList<String> convert (ArrayList<Integer> m)
ArrayList<String> z= new ArrayList<String>();
static HashSet<ArrayList<String>> convert (HashSet<ArrayList<Integer>> paths)
HashSet <ArrayList <String>> k= new HashSet
for (ArrayList <Integer> p:paths)
public static void main(String[] args) throws Exception
Scanner s = new Scanner(new FileReader("randomGraph"));
String[] results = z.split(",");
s = new Scanner(new FileReader("pubs"));
String[] results = z.split(",");
graph G= new graph(edges);
int st =Integer.parseInt(args[0]);
int fin = Integer.parseInt(args[1]);
System.out.println("Shortest path from " + =
pubNames.get(st) + " to " + pubNames.get(fin) + " is" + convert(G.shortestPaths(st,fin)));
The main method in this case also takes two pub ids and works out the shortest path between them – we do not need the last two arguments in our code but, as in the previous case, they are included
here for demonstration purposes only.
Dijkstra’s algorithm (finding the shortest path in a weighted graph)
Research Dijkstra’s Algorithm. Good sources include YouTube videos: Dijkstra’s Algorithm, Graphs: Dijkstra's Algorithm and MIT Lecture 17 Video. Please share any other sources you find particularly
helpful on the VLE discussion board for CO2226.
Study the pseudo code below for Dijkstra's Algorithm to find a shortest path from
Start to end graph nodes:
//S is the set of vertices for which the shortest paths from start have already been found
HashMap <Integer,Double> Q = Map each Vertex to Infinity
(Double.POSITIVE_INFINITY), except map start -> 0;
// Q.get(i) represents the shortest distance found from start
ArrayList <Integer> [] paths;
set path[i] to be the path just containing start.
let v be the key of Q with the smallest value;
//I've given you a method int findSmallest(HashMap <Integer,Double> t) for this
if (v is end and Q does not map v to infinity)
return paths[end]; let w be the value of v in Q;
for (each neighbour u of v) do
let w1 be the weight of the (v,u) edge + w;
if w1 < the value of u in Q, then do the following:
update Q so now the value of u is w1
update paths(u) to be paths(v) with u stuck on the end
Implement Dijkstra’s Algorithm using the pseudo-code above; namely, put a function
dijkstra into the graph class.
int findSmallest(HashMap <Integer,Double> t)
Object [] things= t.keySet().toArray();
double val=t.get(things[0]);
int least=(int) things[0];
Set <Integer> k = t.keySet();
) public ArrayList <Integer> dijkstra (int start, int end)
HashMap <Integer,Double> Q = new HashMap <Integer,Double>();
ArrayList <Integer> [] paths = new ArrayList [N];
paths[i]=new ArrayList <Integer>();
HashSet <Integer> S= new HashSet();
int v = findSmallest(...);
if (v==end && ... return ....;
for(int u: neighbours(v))
return new ArrayList <Integer> ();
Test your implementation using the following test program (again, here you need to provide the ids of the start and end node as part of the program arguments. These are not needed for your final
public static void main(String [] args) throws Exception
double edges[][]=new double[N][N];
Scanner s = new Scanner(new FileReader("randomGraph"));
String[] results = z.split(",");
graph G= new graph (edges);
Use this randomGraph file (please note that this example is from a different scenario which refers to tube stations so you might need to make some adjustments when reading in the data).
Each line of the file has three values: the first two are vertices and the third is the weight of the edge between them.
[0, 492, 665, 114, 452, 999]
Coursework assignment 2 – questions
Please note that pubs in the questions are referred to by their name. As part of the assignment, you will need to resolve them into their ids (for example, 404 for 24 Kitchen Drive). IMPORTANT: in
the case of a direct link, please ignore this option and look for a route that includes at least one extra link; this constraint applies to all questions for this coursework assignment.
1. How many shortest paths exist between the pubs 24 Kitchen Drive and Baa Bar? A shortest path here means a path with a minimal number of vertices.
Note: use the shortestPaths method above.
2. Which pair of pubs have the highest number of shortest paths between them? Just give the pub ids (in the case of more than one pair having the same numbers of shortest paths give all pairs).
3. How many shortest paths do they have?
4. How long are each of these shortest paths?
Hint: you may wish to use the following method:
static ArrayList<Integer> firstElement (HashSet <ArrayList <Integer>> s)
return ( ArrayList<Integer>)s.toArray()[0];
5. Which set of pubs is furthest away from the Cavern Pub in terms of number of stops? (Just print out the set of numbers corresponding to the pubs – again in the case of a tie we want the numbers of
all pubs matching).
6. What is the length in terms of sum of the weights of the edges of the shortest path between Clock and Elm House?
Note: use Dijkstra's Algorithm.
7. What is the length (in km) of the shortest path (in terms of distance) between Foghertys
Note: use Dijkstra's Algorithm.
You will need to use the following method (and the relevant data from the pubs_lon_lat file).
static double realDistance(double lat1, double lon1, double lat2, double lon2)
// km (change this constant to get miles)
double dLat = (lat2-lat1) * Math.PI / 180;
double dLon = (lon2-lon1) * Math.PI / 180;
double a = Math.sin(dLat/2) * Math.sin(dLat/2) +
Math.cos(lat1 * Math.PI / 180 ) * Math.cos(lat2 * Math.PI / 180 )* Math.sin(dLon/2) * Math.sin(dLon/2);
double c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a)); double d = R * c;
For finding the distance in km between any two points on the Earth's surface with given latitude and longitude, the latitude and longitude of each pub is given in the pubs_lon_lat file. Use this to
compute the adjacency matrix for the weighted graph representation of the Liverpool Pubs problem. We need the ad[i][j] to be the distance from pub i to pub j.
You will also need to write a method for finding the length of path by adding up all the weights of the edges in the path.
[END OF COURSEWORK ASSIGNMENT 2]
Please call at : +91-995 3141 035
OR Leave a WhatsApp message at : +91-995 3141 035 (For quick response)
Solution Includes: AI writing Detection and Plagiarism report with 100% Accuracy. | {"url":"https://www.javaonlinehelp.com/post/co2226-software-engineering-algorithm-design-and-analysiscoursework-assignment-2-2021-2022-java","timestamp":"2024-11-13T17:51:11Z","content_type":"text/html","content_length":"1050506","record_id":"<urn:uuid:4573cbf9-198c-4438-a2b2-e305ae9e0428>","cc-path":"CC-MAIN-2024-46/segments/1730477028387.69/warc/CC-MAIN-20241113171551-20241113201551-00355.warc.gz"} |
A Generalization of Erdős' Matching Conjecture
A Generalization of Erdős' Matching Conjecture
Keywords: Erdős matching conjecture, Hypergraphs, Complete intersection theorem
Let $\mathcal{H}=(V,\mathcal{E})$ be an $r$-uniform hypergraph on $n$ vertices and fix a positive integer $k$ such that $1\le k\le r$. A $k$-matching of $\mathcal{H}$ is a collection of edges $\
mathcal{M}\subset \mathcal{E}$ such that every subset of $V$ whose cardinality equals $k$ is contained in at most one element of $\mathcal{M}$. The $k$-matching number of $\mathcal{H}$ is the maximum
cardinality of a $k$-matching. A well-known problem, posed by Erdős, asks for the maximum number of edges in an $r$-uniform hypergraph under constraints on its $1$-matching number. In this article we
investigate the more general problem of determining the maximum number of edges in an $r$-uniform hypergraph on $n$ vertices subject to the constraint that its $k$-matching number is strictly less
than $a$. The problem can also be seen as a generalization of the well-known $k$-intersection problem. We propose candidate hypergraphs for the solution of this problem, and show that the extremal
hypergraph is among this candidate set when $n\ge 4r\binom{r}{k}^2\cdot a$. | {"url":"https://www.combinatorics.org/ojs/index.php/eljc/article/view/v25i2p21","timestamp":"2024-11-03T08:52:46Z","content_type":"text/html","content_length":"16070","record_id":"<urn:uuid:c2af7c2d-a97c-485d-8167-52873424af88>","cc-path":"CC-MAIN-2024-46/segments/1730477027774.6/warc/CC-MAIN-20241103083929-20241103113929-00460.warc.gz"} |
Microsoft word - 4-tong.doc
Group on Immunization Education Society of Teachers of Family Medicine CLINICAL SCENARIO SERIES ON IMMUNIZATION Shingles and Post Herpetic Neuralgia Written by: Donald B. Middleton, MD Department of
Family Medicine University of Pittsburgh Revision of this clinical scenario was funded from an unrestricted educational grant from the Pennsylvania Academy of Family Physicians. Th | {"url":"http://health-abstracts.com/i/iigss.net1.html","timestamp":"2024-11-12T05:33:43Z","content_type":"text/html","content_length":"25406","record_id":"<urn:uuid:874051f7-abbc-4c02-860a-9395fe3de943>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.58/warc/CC-MAIN-20241112045844-20241112075844-00817.warc.gz"} |
Sophie’s Diary: A Mathematical Novel: Second Editionsearch
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Sophie’s Diary: A Mathematical Novel: Second Edition
MAA Press: An Imprint of the American Mathematical Society
Softcover ISBN: 978-1-4704-7156-9
Product Code: SPEC/73.S
List Price: $40.00
MAA Member Price: $30.00
AMS Member Price: $30.00
eBook ISBN: 978-1-61444-510-4
Product Code: SPEC/73.E
List Price: $35.00
MAA Member Price: $26.25
AMS Member Price: $26.25
Softcover ISBN: 978-1-4704-7156-9
eBook: ISBN: 978-1-61444-510-4
Product Code: SPEC/73.S.B
List Price: $75.00 $57.50
MAA Member Price: $56.25 $43.13
AMS Member Price: $56.25 $43.13
Click above image for expanded view
Sophie’s Diary: A Mathematical Novel: Second Edition
MAA Press: An Imprint of the American Mathematical Society
Softcover ISBN: 978-1-4704-7156-9
Product Code: SPEC/73.S
List Price: $40.00
MAA Member Price: $30.00
AMS Member Price: $30.00
eBook ISBN: 978-1-61444-510-4
Product Code: SPEC/73.E
List Price: $35.00
MAA Member Price: $26.25
AMS Member Price: $26.25
Softcover ISBN: 978-1-4704-7156-9
eBook ISBN: 978-1-61444-510-4
Product Code: SPEC/73.S.B
List Price: $75.00 $57.50
MAA Member Price: $56.25 $43.13
AMS Member Price: $56.25 $43.13
• Spectrum
Volume: 73; 2012; 279 pp
Sophie Germain overcame gender stigmas and a lack of formal education to prove that for all prime exponents less than 100 Case I of Fermat's Last Theorem holds. Hidden behind a man's name, her
brilliance as mathematician was first discovered by three of the greatest scholars of the eighteenth century, Lagrange, Gauss, and Legendre.
In Sophie's Diary, Germain comes to life through a fictionalized journal that intertwines mathematics with historical descriptions of the brutal events that took place in Paris between 1789 and
1793. This format provides a plausible perspective of how a young Sophie could have learned mathematics on her own—both fascinated by numbers and eager to master tough subjects without a
teacher's guidance. Her passion for mathematics is integrated into her personal life as an escape from societal outrage. Sophie's Diary is suitable for a variety of readers—both young and old,
mathematicians and novices—who will be inspired and enlightened on a field of study made easy, as told through the intellectual and personal struggles of an exceptional young woman.
□ Chapters
□ 1. Awakening
□ 2. Discovery
□ 3. Introspection
□ 4. Under Siege
□ 5. Upon the Threshold
□ 6. Intellectual Discovery
□ 7. Knocking on Heaven’s Door
□ While this is a work of fiction, literally every entry in the "diary" of Sophie Germain could plausibly be true. Germain was a woman that grew up in France in the last quarter of the
eighteenth century, when the social norms were that women did not engage in intellectual pursuits. These norms were strongly enforced; it was very difficult for a woman to get any kind of an
education in mathematics or any other science. ... The entries of this diary, which end in 1794, are a combination of Sophie describing her discoveries and difficulties while learning
mathematics as well as the events of the revolution taking place all around her. Even though mathematics by itself is free of politics and other human foibles, it always operates within the
historical context, if only because the people that do it are humans operating in a society. This is a great novel; it is accurate enough to be a reference in a history of math course, which
is highly unusual for a work of fiction.
Charles Ashbacher, Journal of Recreational Mathematics
□ Reading a diary is such a verboten act! But reading Sophie's Diary should not be. Dora Musielak has given us a delightful book of imaginings of mathematician Sophie Germain's mind during the
late 18th century. ... The inclusion of history enhances the book substantially. The author does a nice job of interspersing the history with the mathematics, and the interplay makes the
novel more believable as a diary and helps keep the reader's attention. Mathematically, the book begins with definitions of rational, irrational and prime, and musings on how to solve linear
and quadratic equations. ... She does a nice job of spiraling the topic of prime numbers, returning throughout the book at more and more depth as Sophie's mathematical maturity increases.
John J. Watkins, Mathematical Reviews
□ Sophie's Diary is a mathematical novel inspired by the life of the French mathematician Sophie Germain (1776–1831), the first woman to win the Prix de Mathematiques awarded by the Institute
de France. This fictional diary presents a plausible explanation of how a young Parisian girl could have learned mathematics between 1789 and 1794, during the French Revolution and the Reign
of Terror. Through a young girl's journal, the author weaves together Sophie's process of learning advanced mathematics on her own while growing up during this extremely volatile period of
France's history. ... Sophie's Diary is an inspirational story that portrays the learning of complex mathematics as “exhilarating” and related to the natural world around us. I highly
recommend this book.
Christine Hebert, Mathematics Teacher
• Book Details
• Table of Contents
• Reviews
• Requests
Volume: 73; 2012; 279 pp
Sophie Germain overcame gender stigmas and a lack of formal education to prove that for all prime exponents less than 100 Case I of Fermat's Last Theorem holds. Hidden behind a man's name, her
brilliance as mathematician was first discovered by three of the greatest scholars of the eighteenth century, Lagrange, Gauss, and Legendre.
In Sophie's Diary, Germain comes to life through a fictionalized journal that intertwines mathematics with historical descriptions of the brutal events that took place in Paris between 1789 and 1793.
This format provides a plausible perspective of how a young Sophie could have learned mathematics on her own—both fascinated by numbers and eager to master tough subjects without a teacher's
guidance. Her passion for mathematics is integrated into her personal life as an escape from societal outrage. Sophie's Diary is suitable for a variety of readers—both young and old, mathematicians
and novices—who will be inspired and enlightened on a field of study made easy, as told through the intellectual and personal struggles of an exceptional young woman.
• Chapters
• 1. Awakening
• 2. Discovery
• 3. Introspection
• 4. Under Siege
• 5. Upon the Threshold
• 6. Intellectual Discovery
• 7. Knocking on Heaven’s Door
• While this is a work of fiction, literally every entry in the "diary" of Sophie Germain could plausibly be true. Germain was a woman that grew up in France in the last quarter of the eighteenth
century, when the social norms were that women did not engage in intellectual pursuits. These norms were strongly enforced; it was very difficult for a woman to get any kind of an education in
mathematics or any other science. ... The entries of this diary, which end in 1794, are a combination of Sophie describing her discoveries and difficulties while learning mathematics as well as
the events of the revolution taking place all around her. Even though mathematics by itself is free of politics and other human foibles, it always operates within the historical context, if only
because the people that do it are humans operating in a society. This is a great novel; it is accurate enough to be a reference in a history of math course, which is highly unusual for a work of
Charles Ashbacher, Journal of Recreational Mathematics
• Reading a diary is such a verboten act! But reading Sophie's Diary should not be. Dora Musielak has given us a delightful book of imaginings of mathematician Sophie Germain's mind during the late
18th century. ... The inclusion of history enhances the book substantially. The author does a nice job of interspersing the history with the mathematics, and the interplay makes the novel more
believable as a diary and helps keep the reader's attention. Mathematically, the book begins with definitions of rational, irrational and prime, and musings on how to solve linear and quadratic
equations. ... She does a nice job of spiraling the topic of prime numbers, returning throughout the book at more and more depth as Sophie's mathematical maturity increases.
John J. Watkins, Mathematical Reviews
• Sophie's Diary is a mathematical novel inspired by the life of the French mathematician Sophie Germain (1776–1831), the first woman to win the Prix de Mathematiques awarded by the Institute de
France. This fictional diary presents a plausible explanation of how a young Parisian girl could have learned mathematics between 1789 and 1794, during the French Revolution and the Reign of
Terror. Through a young girl's journal, the author weaves together Sophie's process of learning advanced mathematics on her own while growing up during this extremely volatile period of France's
history. ... Sophie's Diary is an inspirational story that portrays the learning of complex mathematics as “exhilarating” and related to the natural world around us. I highly recommend this book.
Christine Hebert, Mathematics Teacher
Please select which format for which you are requesting permissions. | {"url":"https://bookstore.ams.org/SPEC/73","timestamp":"2024-11-02T10:47:21Z","content_type":"text/html","content_length":"104190","record_id":"<urn:uuid:377c208f-df56-43f1-8e84-381cb0846321>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00208.warc.gz"} |
Simultaneous embedding of a planar graph and its dual on the grid
Traditional representations of graphs and their duals suggest that the dual vertices should be placed inside their corresponding primal faces, and the edges of the dual graph should only cross their
corresponding primal edges. We consider the problem of simultaneously embedding a planar graph and its dual on a small integer grid such that the edges are drawn as straight-line segments and the
only crossings are between primal - dual pairs of edges. We provide an O(n) time algorithm that simultaneously embeds a 3-connected planar graph and its dual on a (2n - 2) × (2n - 2) integer grid,
where n is the total number of vertices in the graph and its dual. All the edges are drawn as straight-line segments except for one edge on the outer face, which is drawn using two segments.
ASJC Scopus subject areas
• Theoretical Computer Science
• Computational Theory and Mathematics
Dive into the research topics of 'Simultaneous embedding of a planar graph and its dual on the grid'. Together they form a unique fingerprint. | {"url":"https://experts.arizona.edu/en/publications/simultaneous-embedding-of-a-planar-graph-and-its-dual-on-the-grid","timestamp":"2024-11-08T02:32:04Z","content_type":"text/html","content_length":"53189","record_id":"<urn:uuid:b66a4c4c-9fbf-4f2e-8713-f79fb0faa053>","cc-path":"CC-MAIN-2024-46/segments/1730477028019.71/warc/CC-MAIN-20241108003811-20241108033811-00300.warc.gz"} |
Go to the source code of this file.
subroutine spbtrs (UPLO, N, KD, NRHS, AB, LDAB, B, LDB, INFO)
Function/Subroutine Documentation
subroutine spbtrs ( character UPLO,
integer N,
integer KD,
integer NRHS,
real, dimension( ldab, * ) AB,
integer LDAB,
real, dimension( ldb, * ) B,
integer LDB,
integer INFO
Download SPBTRS + dependencies
[TGZ] [ZIP] [TXT]
SPBTRS solves a system of linear equations A*X = B with a symmetric
positive definite band matrix A using the Cholesky factorization
A = U**T*U or A = L*L**T computed by SPBTRF.
UPLO is CHARACTER*1
[in] UPLO = 'U': Upper triangular factor stored in AB;
= 'L': Lower triangular factor stored in AB.
N is INTEGER
[in] N The order of the matrix A. N >= 0.
KD is INTEGER
[in] KD The number of superdiagonals of the matrix A if UPLO = 'U',
or the number of subdiagonals if UPLO = 'L'. KD >= 0.
NRHS is INTEGER
[in] NRHS The number of right hand sides, i.e., the number of columns
of the matrix B. NRHS >= 0.
AB is REAL array, dimension (LDAB,N)
The triangular factor U or L from the Cholesky factorization
A = U**T*U or A = L*L**T of the band matrix A, stored in the
[in] AB first KD+1 rows of the array. The j-th column of U or L is
stored in the j-th column of the array AB as follows:
if UPLO ='U', AB(kd+1+i-j,j) = U(i,j) for max(1,j-kd)<=i<=j;
if UPLO ='L', AB(1+i-j,j) = L(i,j) for j<=i<=min(n,j+kd).
LDAB is INTEGER
[in] LDAB The leading dimension of the array AB. LDAB >= KD+1.
B is REAL array, dimension (LDB,NRHS)
[in,out] B On entry, the right hand side matrix B.
On exit, the solution matrix X.
LDB is INTEGER
[in] LDB The leading dimension of the array B. LDB >= max(1,N).
INFO is INTEGER
[out] INFO = 0: successful exit
< 0: if INFO = -i, the i-th argument had an illegal value
Univ. of Tennessee
Univ. of California Berkeley
Univ. of Colorado Denver
NAG Ltd.
November 2011
Definition at line 122 of file spbtrs.f. | {"url":"https://netlib.org/lapack/explore-html-3.4.2/d0/d4d/spbtrs_8f.html","timestamp":"2024-11-10T11:56:38Z","content_type":"application/xhtml+xml","content_length":"12905","record_id":"<urn:uuid:7bd23c0f-1226-423d-8b1d-9af313a15dfd>","cc-path":"CC-MAIN-2024-46/segments/1730477028186.38/warc/CC-MAIN-20241110103354-20241110133354-00428.warc.gz"} |
Fredkin gate
by gowtham 2010-02-16 19:45:36
The Fredkin gate is a computational circuit suitable for reversible computing, invented by Ed Fredkin. It is universal, which means that any logical or arithmetic operation can be constructed
entirely of Fredkin gates.
The basic Fredkin gate[1] is a controlled swap gate that maps three inputs (C, I1, I2) onto three outputs (C, O1, O2). The C input is mapped directly to the C output. If C = 0, no swap is performed;
I1 maps to O1, and I2 maps to O2. Otherwise, the two outputs are swapped so that I1 maps to O2, and I2 maps to O1. It is easy to see that this circuit is reversible, i.e., "undoes itself" when run
backwards. A generalized n×n Fredkin gate passes its first n-2 inputs unchanged to the corresponding outputs, and swaps its last two outputs if and only if the first n-2 inputs are all 1.
The Fredkin gate is the reversible 3-bit gate that swaps the last two bits if the first bit is 1
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Analog Scientific Calculator by Carlos Luna | Download free STL model | Printables.com
Apart from the basic functions that you can expect from a quarter circle protractor:
• Measure angles
• Draw straight lines and right angles
• Find the center of small circles
You can do most of the functions of a scientific calculator with just paper, pencil & this model!
• Addition / Subtraction
• Product / Division
• √x / ∛x / …
• Log10 / 10ˣ
• Sin / Cos / Tan / Asin / Acos / Atan
And, if you print the model at 1:1 scale, you can also:
Find out how to do all these operations with this simple analog tool!
The model is just a quarter of a circle (of 10 cm radius) with 3 different scales marked:
• The linear scale: on the lower side, you can find a linear scale with 1 mm ticks.
• The logarithmic scale: on the left side, you can find a base-10 logarithmic scale.
• The angular scale: on the circular side, you can find an angular scale with 1º ticks.
This simple analog tool is able to perform many operations with enough precision (usually, 2 or 3 significant digits). Of course, you will be better off by doing additions, subtractions,
multiplications, and divisions “by hand” using the standard procedures you learned at school. But the logarithmic and trigonometric operations are transcendental functions and can only be performed
with the aid of an electronic calculator nowadays.
In addition to this model, you will need a sheet of paper (that will perform the role of RAM memory) and a pencil (to write in the RAM!).
This material has been designed as a teaching aid to help the students to better understand the meaning and properties of the logarithmic and trigonometric functions.
Measure distances in mm
The linear scale is calibrated in 1 mm increments and can be used to measure distances of up to 100 mm. Example: the standard A7 paper is 74 mm wide.
Measuring longer distances can be done by breaking them in 10 cm fragments + a remainder.
Measure angles
The angular scale is calibrated in 1º increments and can be used to measure angles of up to 90º:
Measuring larger angles can also be done by breaking them in 90º fragments + a remainder.
Draw straight lines and right angles
The border of the Analog Scientific Calculator can be used to draw lines of up to 10 cm. Example: A pretty decent 8 cm square drawn with the aid of the Analog Scientific Calculator.
The 90º angle of the Analog Scientific Calculator can be used to draw perpendicular lines.
Find the center of small circles
To find the center of a small circle (up to 10 cm in diameter), inscribe the right angle of the Analog Scientific Calculator in the circle and mark the 2 crossing points. The line that joins these
two points is always a diameter of the circle. The center of the circle is the point where 2 or more diameters cross each other. Example:
Addition / Subtraction
To add two numbers, use the linear scale to mark both distances, one after the other on the border of the paper. Then use the linear scale to measure the combined length. Example: 37 + 42 = 79
For subtracting numbers, do the same procedure backwards. Example: 75 - 38 = 37
Product / Division
To multiply two numbers, use the logarithmic scale to mark both distances, one after the other, on the border of the paper. Then use the logarithmic scale to measure the combined length. Example: 2.3
· 3 ≈ 7
For dividing two numbers, do the same procedure backwards. Example: 4.5 / 3 = 1.5
√x / ∛x / …
To compute the square root of a number, use the logarithmic scale to mark that distance on the border of the paper. Then measure this distance using the linear scale, and mark half this distance on
the border of the paper. Finally, measure that halved length with the logarithmic scale. Example: √3 ≈ 1.75
To compute the cubic root of a number, do the same procedure, but dividing the length by 3 instead of halving it. Example: ∛8 = 2
In general, you can compute the Nth root by dividing the length by N.
Log10 / 10ˣ
To compute the Log10 of a number, mark the number in the lower side of the paper using the logarithmic scale and then read the result using the linear scale. Example: Log10(4) ≈ 0.6 (which also means
that Log10(40) ≈ 1.6, Log10(400) ≈ 2.6, Log10(4000) ≈ 3.6, Log10(0.4) ≈ -0.4, etc.)
To compute 10 to the power of a number, mark the number in the lower side of the paper using the linear scale and then read the result using the logarithmic scale. Example: 10^0.6 ≈ 4 (which also
means that 10^1.6 ≈ 40, 10^2.6 ≈ 400, 10^3.6 ≈ 4000, 10^(-0.4) ≈ 0.4, etc.)
Sin / Cos / Tan
To compute the Sin of an angle, place the corner of the Analog Scientific Calculator in the corner of the paper and then mark a dot in the desired angle using the angular scale. The sin of that angle
is just the vertical distance from that dot to the lower edge of the paper. You can measure it with the linear scale (take advantage of the right angle to ensure that you are measuring the angle
correctly). To compute the Cos of an angle, do the same procedure, but measure the horizontal distance from the dot to the left edge of the paper. Example: Sin(37º) ≈ 0.6 and Cos(37º) ≈ 0.8
To compute the Tan of an angle, extend the radius until it meets the vertical line drawn from the bottom point of the circular arc. Then measure the vertical distance from the edge of the paper to
the intersection using the linear scale. Example: Tan(37º) ≈ 0.75
Asin / Acos / Atan
To find the Asin of a length, use the linear scale to mark that distance on the left side of the paper (starting in the lower left corner). Then draw a perpendicular line. Finally, place the Analog
Scientific Calculator in the corner of the paper and read the corresponding angle in the angular scale. Example: Asin(0.6) ≈ 37º
To find the Acos of a length, do the same procedure on the lower side of the paper. Example: Acos(0.8) ≈ 37º
To find the Atan of a length, draw a vertical line at 10 cm from the lower left corner. Then use the linear scale to mark that distance on the line (from the paper border). Then join that point with
the paper corner (using another sheet of paper) and use the angular scale to measure the corresponding angle. Example: Atan(0.75) ≈ 37º
Recommended Print Settings:
• The model works well with a 0.1 mm layer height and a 0.4 mm nozzle, but it can probably be printed with a different configuration.
• I printed it in PLA, but you can use pretty much any material of your choice. | {"url":"https://www.printables.com/model/896863-analog-scientific-calculator","timestamp":"2024-11-09T16:06:16Z","content_type":"text/html","content_length":"250847","record_id":"<urn:uuid:c1b15d49-75b2-4c9a-ba5d-4e26b1025dd4>","cc-path":"CC-MAIN-2024-46/segments/1730477028125.59/warc/CC-MAIN-20241109151915-20241109181915-00003.warc.gz"} |
SQLite Code Factory online Help
Prev Return to chapter overview Next
expr ::= expr binary-op expr |
expr like-op expr |
unary-op expr |
( expr ) |
column-name |
table-name . column-name |
literal-value |
function-name ( expr-list | * ) |
expr ISNULL |
expr NOTNULL |
expr [NOT] BETWEEN expr AND expr |
expr [NOT] IN ( value-list ) |
expr [NOT] IN ( select-statement ) |
( select-statement ) |
CASE [expr] ( WHEN expr THEN expr )+ [ELSE expr] END
like-op ::= LIKE | GLOB | NOT LIKE | NOT GLOB
This section is different from the others. Most other sections of this document talks about a particular SQL command. This section does not talk about a standalone command but about "expressions"
which are subcomponent of most other commands.
SQLite understands the following binary operators, in order from highest to lowest precedence:
* / %
+ -
<< >> & |
< <= > >=
= == != <> IN
Supported unary operators are these:
- + ! ~
Any SQLite value can be used as part of an expression. For arithmetic operations, integers are treated as integers. Strings are first converted to real numbers using atof(). For comparison operators,
numbers compare as numbers and strings compare using the strcmp() function. Note that there are two variations of the equals and not equals operators. Equals can be either = or ==. The non-equals
operator can be either != or <>. The || operator is "concatenate" - it joins together the two strings of its operands.
The LIKE operator does a wildcard comparison. The operand to the right contains the wildcards. A percent symbol % in the right operand matches any sequence of zero or more characters on the left. An
underscore _ on the right matches any single character on the left. The LIKE operator is not case sensitive and will match upper case characters on one side against lower case characters on the
other. (A bug: SQLite only understands upper/lower case for 7-bit Latin characters. Hence the LIKE operator is case sensitive for 8-bit iso8859 characters or UTF-8 characters. For example, the
expression 'a' LIKE 'A' is TRUE but 'æ' LIKE 'Æ' is FALSE.)
The GLOB operator is similar to LIKE but uses the Unix file globing syntax for its wildcards. Also, GLOB is case sensitive, unlike LIKE. Both GLOB and LIKE may be preceded by the NOT keyword to
invert the sense of the test.
A column name can be any of the names defined in the CREATE TABLE statement or one of the following special identifiers: "ROWID", "OID", or "_ROWID_". These special identifiers all describe the
unique random integer key (the "row key") associated with every row of every table. The special identifiers only refer to the row key if the CREATE TABLE statement does not define a real column with
the same name. Row keys act like read-only columns. A row key can be used anywhere a regular column can be used, except that you cannot change the value of a row key in an UPDATE or INSERT statement.
"SELECT * ..." does not return the row key.
SELECT statements can appear in expressions as either the right-hand operand of the IN operator or as a scalar quantity. In both cases, the SELECT should have only a single column in its result.
Compound SELECTs (connected with keywords like UNION or EXCEPT) are allowed. A SELECT in an expression is evaluated once before any other processing is performed, so none of the expressions within
the select itself can refer to quantities in the containing expression.
When a SELECT is the right operand of the IN operator, the IN operator returns TRUE if the result of the left operand is any of the values generated by the select. The IN operator may be preceded by
the NOT keyword to invert the sense of the test.
When a SELECT appears within an expression but is not the right operand of an IN operator, then the first row of the result of the SELECT becomes the value used in the expression. If the SELECT
yields more than one result row, all rows after the first are ignored. If the SELECT yields no rows, then the value of the SELECT is NULL.
Both simple and aggregate functions are supported. A simple function can be used in any expression. Simple functions return a result immediately based on their inputs. Aggregate functions may only be
used in a SELECT statement. Aggregate functions compute their result across all rows of the result set.
The functions shown below are available by default. Additional
abs(X) Return the absolute value of argument X.
coalesce(X,Y,...) Return a copy of the first non-NULL argument. If all arguments are NULL then NULL is returned.
glob(X,Y) This function is used to implement the "Y GLOB X" syntax of SQLite.
last_insert_rowid Return the ROWID of the last row insert from this connection to the database. This is the same value that would be returned from the
length(X) Return the string length of X in characters. If SQLite is configured to support UTF-8, then the number of UTF-8 characters is returned, not the number of bytes.
like(X,Y) This function is used to implement the "Y LIKE X" syntax of SQL.
lower(X) Return a copy of string X will all characters converted to lower case. The C library tolower() routine is used for the conversion, which means that this function might not work
correctly on UTF-8 characters.
max(X,Y,...) Return the argument with the maximum value. Arguments may be strings in addition to numbers. The maximum value is determined by the usual sort order. Note that max() is a simple
function when it has 2 or more arguments but converts to an aggregate function if given only a single argument.
min(X,Y,...) Return the argument with the minimum value. Arguments may be strings in addition to numbers. The minimum value is determined by the usual sort order. Note that min() is a simple
function when it has 2 or more arguments but converts to an aggregate function if given only a single argument.
random(*) Return a random integer between -2147483648 and +2147483647.
round(X) Round off the number X to Y digits to the right of the decimal point. If the Y argument is omitted, 0 is assumed.
substr(X,Y,Z) Return a substring of input string X that begins with the Y-th character and which is Z characters long. The left-most character of X is number 1. If Y is negative the first
character of the substring is found by counting from the right rather than the left. If SQLite is configured to support UTF-8, then characters indices refer to actual UTF-8
characters, not bytes.
upper(X) Return a copy of input string X converted to all upper-case letters.
avg(X) Return the average value of all X within a group.
count(X) The first form return a count of the number of times that X is not NULL in a group. The second form (with no argument) returns the total number of rows in the group.
max(X) Return the maximum value of all values in the group. The usual sort order is used to determine the maximum.
min(X) Return the minimum value of all values in the group. The usual sort order is used to determine the minimum.
sum(X) Return the numeric sum of all values in the group.
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Candy Crush Saga Cheats: Levels 600-700 Cheats & Tips | Candy Crush Cheats
Home Candy Crush Saga Candy Crush Saga Cheats: Levels 600-700 Cheats & Tips
Candy Crush Saga Cheats: Levels 600-700 Cheats & Tips
Candy Crush Saga cheats for levels 600 to 700. Find out how to beat 600 to 700 levels in Candy Crush Saga with these strategy guides. Pass every level with these Candy Crush tips.
Wafer Windmill
The first episode in this section is Wafer Windmill, containing levels 591-605. Although there are no truly new elements in this episode, the candy cannons are now able to dispense up to three types
of special items: licorice swirls, ingredients and candy bombs. The hardest level in this section of the episode is level 603. There are 2 ingredient levels and 3 jelly levels in this section of
Wafer Windmill.
Cereal Sea
Cereal Sea is the second episode, featuring levels 606-620. This episode introduces the candy frog, an element that does not get removed when it is matched. The candy frog stores special candies’
properties and explodes an area of the board when it is fed by making enough matches. It first appears in Candy Crush level 606. The hardest level in this episode is level 617. Cereal Sea has 6 jelly
levels, 4 ingredient levels, 3 moves levels, 1 timed level and 1 order level.
Taffy Tropics
Taffy Tropics is the third episode in this section and the first of World Eight. It features levels 621-635. No new candies are introduced in this episode, although many levels feature the candy
frog. Taffy Tropics is considered a fairly easy episode, although it has one level, level 629, that is very hard. Taffy Tropics includes 7 jelly levels, 5 ingredient levels, 2 order levels, and 1
move level.
Glazed Grove
Glazed Grove, levels 636-650, is the final episode in this section. It does not feature any new elements or items. The hardest levels in Glazed Grove are levels 647 and 649. Glazed Grove features 6
jelly levels, 5 order levels, 3 ingredient levels and 1 move level.
Fizzy Falls
Fizzy Falls is the third episode of World Eight and the first episode of this section. Fizzy Falls contains levels 651-665. This episode does not introduce any new elements or special candies. The
hardest level in Fizzy Falls is Level 664. Fizzy Falls has 6 jelly levels, 5 order levels and 4 ingredient levels.
Crunchy Courtyard
Crunchy Courtyard, with levels 666-680, is the second episode featured in this section. In level 677, special candies are locked in licorice locks for the first time, although there are no truly new
features. Crunchy Courtyard’s hardest levels are level 670 and level 679. This episode features 7 jelly levels, 4 ingredient levels and 4 order levels.
Choco Rio Grande
Choco Rio Grande is the third episode of this section, featuring levels 681-695. There are no new features in the Choco Rio Grande episode. Choco Rio Grande is a fairly difficult episode, with 7 hard
levels. The hardest is generally considered to be level 688. Choco Rio Grande consists of 6 jelly levels, 4 ingredient levels, 4 order levels and 1 timed level.
Toffee Tower
The final episode in this section is Toffee Tower, with levels 696-710. No new features have been added in Toffee Tower. The hardest level of this section is level 699. The other levels in Toffee
Tower are generally considered to be easy levels and it is definitely easier than Choco Rio Grande. There are 2 jelly levels, 1 order, 1 ingredient, and 1 time level in this section of Toffee Tower.
LEAVE A REPLY Cancel reply | {"url":"https://cheats-candycrush.com/levels-600-700/","timestamp":"2024-11-03T10:42:50Z","content_type":"text/html","content_length":"175115","record_id":"<urn:uuid:6b22cfe8-ce1d-483a-aa24-3caee48f336c>","cc-path":"CC-MAIN-2024-46/segments/1730477027774.6/warc/CC-MAIN-20241103083929-20241103113929-00868.warc.gz"} |
SOS WindEnergy
JACKET + TOPSIDE MODEL – OVERALL VIEW
The 103.5 m height structure has four main vertical legs fixed on the seabed to support the loads. They are braced by five horizontal sets of elements in the Jacket, arranged equally spaced from the
structure’s foot to a height of 80 m, and also by diagonal elements in every side of the Jacket. Each horizontal frame has a different size. All the elements have the same properties (S355 steel).
Moreover, cans and stubs are applied in the joints to increase the stiffness.
Eigenvalue analysis provides dynamic properties of a structure by solving the characteristic equation composed of mass matrix and stiffness matrix. The dynamic properties include natural modes (or
mode shapes), natural periods (or frequencies) and modal participation factors. The natural frequencies and the mode shapes of the system can be calculated from an eigenvalue analysis of the undamped
free vibration system as shown in the equation:
MODE 1 – f = 0.440868 Hz
MODE 2 – f = 0.454736 Hz
MODE 3 – f = 0.518925 Hz | {"url":"https://fe.up.pt/soswe/models/","timestamp":"2024-11-14T16:47:44Z","content_type":"text/html","content_length":"48377","record_id":"<urn:uuid:11fa0358-5d60-4d3d-bc6c-83fd56d362fa>","cc-path":"CC-MAIN-2024-46/segments/1730477393980.94/warc/CC-MAIN-20241114162350-20241114192350-00560.warc.gz"} |
lemon grass, cattle emissions
Help Support Steer Planet:
What a buncha horse sh!t . Lol
Mar 23, 2009
YEA GARGAN NOW EVERYBODY IS INSISTING I GO ON THE SAME PROGRAM LOL O0
"Generally, when proteins can be protected from ruminal deamination, ammonia declines and ruminants would have more amino acids available in the lower gut."
"Moreover, lemongrass and/or in combination could decrease protozoal population and methane production and especially increase propionate and N utilization."
not sure why the hate. seems interesting. better nitrogen utilization (similar to rumensin) and lower methane.
what's not to love for a followup?
carbon credits available at some point. please don't change. not necessary.
knabe said:
"Generally, when proteins can be protected from ruminal deamination, ammonia declines and ruminants would have more amino acids available in the lower gut."
"Moreover, lemongrass and/or in combination could decrease protozoal population and methane production and especially increase propionate and N utilization."
not sure why the hate. seems interesting. better nitrogen utilization (similar to rumensin) and lower methane.
what's not to love for a followup?
carbon credits available at some point. please don't change. not necessary.
The first word of your quote " generally " lost my interest.
Burger King is desperate to stay afloat after their all veggie burgers has been a bust. The world had far more problems than the 2% of the emissions bovine burps produce imo.
Gargan said:
The first word of your quote " generally " lost my interest.
Burger King is desperate to stay afloat after their all veggie burgers has been a bust. The world had far more problems than the 2% of the emissions bovine burps produce imo.
this is science speak because science perspective is one can never prove anything, they can only fail to disprove.
i agree burger king is talking out both sides of their mouth to people who hate each other. the earlier link is without political bias.
there will be carbon credit opportunities for management techniques, genetics (feed efficiency) in the near future.
my perspective is to not solve the problem you mention, but to use the phony economic opportunities that are available with all this carbon credit nonsense.
there are quite a few people looking into the carbon credit issue as a source of revenue to people who will pay for credits.
i'm not sure you see the irony.
The irony is see is that Burger King imports most of its beef . My original post was geared towards them. Not the science behind their movement.
Quite a few articles re Burger King snd deforestation.
The whole fake meat movement is ironic.
If meat is so bad, there is no need to imitate it.
I mean just eating something that resembles meat should be abhorrent.
The whole industry is an other ploy to project superiority and use incremental substitution techniques to de-sensitize resistance of evil meat eaters to stop eating meat.
I view the whole industry as a complete lie.
The better solution is a negative population growth of which the group that is doing this has s rrr et placement rate of probably less than 0.1.
At that rate, there won’t be anyone around In one generation to be in the movement. Their very superiority will simply eliminate them.
Much better global warming strategy.
Agree on the whole meat replacement idea. Same goes with the Soy and Almond milk business. "Milk" only comes from a lactating animal. Dairyman should have fought that name years ago. Its juice and
not milk in my way if thinking...
Nov 29, 2008
I finally watched the commercial this morning on YouTube. It was a hokey, insulting, stupid and cultural misappropriative piece of crap. If it does anything it will drive customers away from beef
in general, not just from BK. The sooner they file bankruptcy the better.
I couldn’t watch commercial.
It’s complete indoctrination.
I can’t stand Burger King even more now. Ugh.
Feb 23, 2015
Gargan said:
Agree on the whole meat replacement idea. Same goes with the Soy and Almond milk business. "Milk" only comes from a lactating animal. Dairyman should have fought that name years ago. Its juice
and not milk in my way if thinking...
Most of the largest dairies are in California, and you'd be hard-pressed to find a large dairy in california that isn't also an almond producer. We even have a nearby dairyman here in MN that owns an
almond farm out there. The business-savvy individuals saw the writing on the wall and got into that business years ago now.
This is probably part of the reason you dont see big movements in the dairy industry against it--the big operators who could actually finance the campaign have made better use of their money and
bought in on the almond farms--if you can't beat em, join em.
shortybreeder said:
Gargan said:
Agree on the whole meat replacement idea. Same goes with the Soy and Almond milk business. "Milk" only comes from a lactating animal. Dairyman should have fought that name years ago. Its
juice and not milk in my way if thinking...
Most of the largest dairies are in California, and you'd be hard-pressed to find a large dairy in california that isn't also an almond producer. We even have a nearby dairyman here in MN that
owns an almond farm out there. The business-savvy individuals saw the writing on the wall and got into that business years ago now.
This is probably part of the reason you dont see big movements in the dairy industry against it--the big operators who could actually finance the campaign have made better use of their money and
bought in on the almond farms--if you can't beat em, join em.
I've never put the 2 products together with each other. Good to know
almond hulls are like candy to cattle.
they have been used in cattle rations in California for decades.
wild pigs love them too. excellent chum.
i helped feed almond hulls during the drought a few years ago, and once the pigs found out about them, easily over a hundred would feed side by side with cattle.
we obviously let some hunters know
Gargan said:
no, that connection was lost with my dad.
i like projectile delivery tools.
i have an 1878/1902 phillippine/alaska deliverer. sadly, i don't have the piece of leather that came with it.
i used to think the larger trigger guard was to wear with gloves, but it was so that area was stronger. it was used in alaska.
i guess you could say i hunt gophers and squirrels. | {"url":"https://www.steerplanet.com/threads/lemon-grass-cattle-emissions.71341/","timestamp":"2024-11-10T18:50:27Z","content_type":"text/html","content_length":"125110","record_id":"<urn:uuid:8756bd3a-bdfb-4312-b868-6f5548b95e48>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.61/warc/CC-MAIN-20241110170046-20241110200046-00632.warc.gz"} |
Mathematics with Allied Health Applications 1st Edition Aufmann Lockwood Test Bank
Instant download Mathematics with Allied Health Applications 1st Edition Aufmann Lockwood Test Bank pdf docx epub after payment.
Product details:
Author: Aufmann Lockwood
This book is intended for algebra courses for the allied health professional, usually at community colleges and career schools. This book will appeal to professors who are looking for a paperback
where examples and exercises reflect the situations that allied health professionals will face in their daily challenges throughout their career.
Table of contents:
1. WHOLE NUMBERS. Prep Test. Introduction to Whole Numbers. Addition of Whole Numbers. Subtraction of Whole Numbers. Multiplication of Whole Numbers. Division of Whole Numbers. Exponential Notation
and the Order of Operations Agreement. Prime Numbers and Factoring. Focus on Problem Solving. Projects and Group Activities. Chapter 1 Summary. Chapter 1 Concept Review. Chapter 1 Review Exercises.
Chapter 1 Test.
2. FRACTIONS. Prep Test. The Least Common Multiple and the Greatest Common Factor. Introduction to Fractions. Writing Equivalent Fractions. Addition of Fractions and Mixed Numbers. Subtraction of
Fractions and Mixed Numbers. Multiplication of Fractions and Mixed Numbers. Division of Fractions and Mixed Numbers. Order, Exponents, and the Order of Operations Agreement. Focus on Problem Solving.
Projects and Group Activities. Chapter 2 Summary. Chapter 2 Concept Review. Chapter 2 Review Exercises. Chapter 2 Test. Cumulative Review Exercises.
3. DECIMALS. Prep Test. Introduction to Decimals. Addition of Decimals. Subtraction of Decimals. Multiplication of Decimals. Division of Decimals. Comparing and Converting Fractions and Decimals.
Focus on Problem Solving. Projects and Group Activities. Chapter 3 Summary. Chapter 3 Concept Review. Chapter 3 Review Exercises. Chapter 3 Test. Cumulative Review Exercises.
4. RATIO AND PROPORTION. Prep Test. Ratio. Rates. Proportions. Focus on Problem Solving. Projects and Group Activities. Chapter 4 Summary. Chapter 4 Concept Review. Chapter 4 Review Exercises.
Chapter 4 Test. Cumulative Review Exercises.
5. PERCENTS. Prep Test. Introduction to Percents. Percent Equations: Part I. Percent Equations: Part II. Percent Equations: Part III. Percent on Problems: Proportion Method. Focus on Problem Solving.
Projects and Group Activities. Chapter 5 Summary. Chapter 5 Concept Review. Chapter 5 Review Exercises. Chapter 5 Test. Cumulative Review Exercises.
6. APPLICATIONS FOR BUSINESS AND CONSUMERS. Prep Test. Applications to Purchasing. Percent Increase and Percent Decrease. Interest. Real Estate Expenses. Car Expenses. Wages. Bank Statements. Focus
on Problem Solving. Projects and Group Activities. Chapter 6 Summary. Chapter 6 Concept Review. Chapter 6 Review Exercises. Chapter 6 Test. Cumulative Review Exercises.
7. STATISTICS AND PROBABILITY. Prep Test. Pictographs and Circle Graphs. Bar Graphs and Broken-Line Graphs. Histograms and Frequency Polygons. Statistical Measures. Introduction to Probability. Focus
on Problem Solving. Projects and Group Activities. Chapter 7 Summary. Chapter 7 Concept Review. Chapter 7 Review Exercises. Chapter 7 Test. Cumulative Review Exercises.
8. U.S. CUSTOMARY UNITS OF MEASURE. Prep Test. Length. Weight. Capacity. Tome. Energy and Power. Focus on Problem Solving. Projects and Group Activities. Chapter 8 Summary. Chapter 8 Concept Review.
Chapter 8 Review Exercises. Chapter 8 Test. Cumulative Review Exercises.
9. THE METRIC SYSTEM OF MEASUREMENT. Prep Test. Length. Mass. Capacity. Energy. Conversion Between the U.S. Customary and the Metric Systems of Measurement. Focus on Problem Solving. Projects and
Group Activities. Chapter 9 Summary. Chapter 9 Concept Review. Chapter 9 Review Exercises. Chapter 9 Test. Cumulative Review Exercises.
10. RATIONAL NUMBERS. Prep Test. Introduction to Integers. Addition and Subtraction of Integers. Multiplication and Division of Integers. Operation with Rational Numbers. Scientific Notation and the
Order of Operations Agreement. Focus on Problem Solving. Projects and Group Activities. Chapter 10 Summary. Chapter 10 Concept Review. Chapter 10 Review Exercises. Chapter 10 Test. Cumulative Review
11. INTRODUCTION TO ALGEBRA. Prep Test. Variable Expressions. Introduction to Equations. General Equations: Part I. General Equations: Part II. Translating Verbal Expressions into Mathematical
Expressions. Translating Sentences into Equations and Solving. Focus on Problem Solving. Projects and Group Activities. Chapter 11 Summary. Chapter 11 Concept Review. Chapter 11 Review Exercises.
Chapter 11 Test. Final Exam.
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Entropy-constrained vector quantizer design algorithm using competitive learning technique
A novel full-search variable-rate vector quantizer (VQ) design algorithm using competitive learning technique is presented. The algorithm, termed entropy-constrained competitive learning (ECCL)
algorithm, can design a VQ having minimum average distortion subject to a rate constraint. The ECCL algorithm enjoys a better rate-distortion performance than that of the existing competitive
learning algorithms. Moreover, the ECCL algorithm outperforms the entropy-constrained vector quantizer (ECVQ) design algorithm subject to the same rate and storage size constraints. In addition, the
learning algorithm is more insensitive to the selection of initial codewords as compared with the ECVQ algorithm. Therefore, the ECCL algorithm can be an effective alternative to the existing
variable-rate VQ design algorithms for the applications of signal compression.
Conference Proceedings of the IEEE GLOBECOM 1998 - The Bridge to the Global Integration
City Sydney, NSW, Aust
Period 1998/11/08 → 1998/11/12
ASJC Scopus subject areas
• Electrical and Electronic Engineering
• Global and Planetary Change
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Centripetal Acceleration: Definition, Formula, Unit, and Calculations
What is Centripetal Acceleration?
Definition: Centripetal acceleration is the type of acceleration that is directed toward the centre of a circular path. It is si unit is in meters per second square (ms^-2), and its symbol is a.
Additionally, we can define centripetal acceleration as the acceleration experienced by an object moving in a circular path. Unlike linear acceleration, which involves a change in speed, centripetal
acceleration involves a change in direction. This acceleration is always directed towards the center of the circle or the axis of rotation, enabling the object to continually change its direction
while staying on the circular path.
The centripetal acceleration formula is: a = v^2/r or a = ω^2r
[where a = centripetal acceleration, v = speed, ω = angular velocity, and r = radius of the circular path]
When you enter a merry-go-round, it will go around a circular path about an axis. its continuous movement would make you feel like you are being pushed outside. You are experiencing this outward push
because the merry-go-round keeps changing direction in a circular path while your body wants to continue moving in a straight line.
Therefore, as the object keeps rotating, centripetal force provides sufficient force to help your body continue moving around the circular path. Additionally, centripetal acceleration is directed
toward the centre of the circular path.
Many people kept asking a question as to whether centripetal acceleration is constant. The answer is Yes, and this is because the speed of the object is constant in magnitude but its direction is
changing along a circular path. Thus, it is important to know that this type of acceleration is experienced by any object undergoing circular motion. It’s always directed toward the centre.
Centripetal Acceleration Derivation
The acceleration of an object moving in a circle:
1. Is always directed toward the center of the circle.
2. It has a constant magnitude
Therefore, we can derive the mathematical equation using the diagram below
The change in velocity Δv is directed toward the centre, and as such perpendicular, to v. Since the magnitude of the velocity at A and B are the same, despite their direction.
Therefore, the change in velocity Δv = vsinθ
As the angular change and the time to make the change are now smaller,
sinθ = θ
Δv = vθ
We need to also remember that acceleration is the change of velocity over time. Hence
a = Δv / t
Which will now become
a = vθ / t [Because Δv = vθ]
and the formula for angular velocity is ω = θ / t [θ =ωt]
a = vωt / t
Thus, we will now have
a = vω and ω = v/r which will give us a = (v x v) / r. Therefore, a = v^2 / r
Additionally, since v = ωr
a = ωr x ω
Therefore, the centripetal acceleration formula for are:
a = ω^2r or
a = v^2 / r
How to Calculate Centripetal Acceleration
Here are a few solved problems:
Problem 1
A mass of 15 kilograms is moving in a circular path of radius 3 meters with a uniform speed of 30 meters per second. Find the centripetal acceleration of the object.
Mass, m = 15 kg,
Radius, r = 3 m, and
Uniform speed, v = 30 m/s
Centripetal accelaration, a =?
We can apply the formula that says a = v^2/r
Therefore, we can now substitute our data into the above formula
a=v^2/r = 30^2/3 = 900/3 = 300ms^-2
Therefore, the centripetal acceleration of the body is 300 meters per second square (a=300ms^-2).
Problem 2
A bus is travelling at a speed of 70 meters per second and is making a turn with a radius of 100 meters. What is the magnitude of the bus’ centripetal acceleration?
Speed, v = 70 m/s,
Radius, r = 100 m,
Centripetal acceleration, a =?
Since the formula is a=v^2/r
We can now say that
a=70^2/100 = 4,900/100 = 49ms^-2
Hence, the centripetal acceleration of the bus is 49 meters per second square (a=49ms^-2)
Problem 3
An aeroplane is flying in a circular path at a speed of 400 meters per second and a radius of 1000 meters. Find the centripetal acceleration of the aeroplane.
Speed, v = 400 m
Radius, r = 1000 m
Centripetal acceleration a=?
We can apply the formula a=v^2/r to solve the problem
Hence, substitute our data into the above formula
a=v^2/r = 400^2/1000
a = 160,000/1000 = 160ms^-2
Therefore, the centripetal acceleration of the aeroplane is 160 meters per second square (160ms^-2)
Problem 4
A roller coaster makes a turn of 40 meters and a speed of 20 meters per second. Determine the centripetal acceleration of the roller coaster.
The radius of the roller coaster, r = 40 m
Speed, v = 20 m/s
Centripetal acceleration, a =?
and the formula that will help us solve the problem is a=v^2/r
After plugin our data into the above formula, we will get
a=v^2/r = 20^2/60 = 400/40 = 10 ms^-2
Therefore, the final answer is 10 meters per second square (10 ms^-2).
Problem 5
A body moves along a circular path with a uniform angular speed of 0.7 rad/s and at a constant speed of 4.0 m/s. Calculate the acceleration of the body toward the centre of the circle.
Angular speed, ω = 0.7 rad/s
v = 4.0 m/s
since v = ωr
Therefore, radius r = v / ω = 4 / 0.7 = 5.7 m
Hence, the acceleration toward the centre is
a = v^2 / r = 4^2 / 5.7 = 16 / 5.7 = 2.80 m/s^2
Therefore, the acceleration toward the centre of the circle is 2.8 meters per second square (m/s^2)
Centripetal force (f) is defined as the force that tends to keep an object of mass (m), around a circular path of radius (r). The formula for centripetal force is f = mv^2/r.
Centrifugal force is the reaction force that tends to move a body away from the centre. In other words, it acts in the opposite direction to the centripetal force which can be written as (- mv^2/r).
A centrifuge is a device used to separate particles in suspension from the liquid in which they are contained. It consists of two tubes that are whirled by electrical means in a circle at a uniform
speed. When you use the centrifuge, particles suspended in the less dense liquid would be observed to collect at the bottom of the centrifuge tube. It will leave the clear liquid at the top. While
those in a denser liquid will collect at the top leaving the clear liquid at the bottom.
Car Negotiating a Curve
Imagine you are driving a car around a curved road. How do you determine the centripetal acceleration required to keep the car on the curve without skidding?
To calculate a, you will need the car’s speed (v) and the radius (r) of the curve.
The formula is:
a = v^2 / r
For instance, if you are driving at 60 mph (26.8 m/s) around a curve with a radius of 50 meters, a is
a = (26.8)^2 / 50 ≈ 14.31m/s^2
Swing Ride at an Amusement Park
Consider a swing ride at an amusement park. Riders are rotating at a constant speed on swings that are 6 meters from the centre. How can you find the centripetal acceleration experienced by the
Using the same formula, plug in the values:
a = v^2 / r
If the swings rotate at 10 m/s, a will be
a = (10)^2 / 6 ≈ 16.67m/s^2
This value indicates the acceleration towards the centre that prevents riders from flying off due to inertia.
Satellite in Orbit
Delving into more advanced scenarios, let’s discuss a satellite orbiting the Earth. The centripetal acceleration in this context is caused by the gravitational pull of the Earth and is given by:
a = G.M / r^2
Where G is the gravitational constant, M is Earth’s mass, and r is the satellite’s distance from the Earth’s centre. Solving such problems requires an understanding of the gravitational forces at
Expert Tips for Problem Solving
1. Draw Diagrams: Visual representations can clarify complex problems. Sketching the scenario aids in identifying key variables and relationships.
2. Units Matter: Pay close attention to units when plugging values into formulas. Inconsistent units can lead to incorrect results.
3. Practice Algebra: Many of it is problems involve algebraic manipulation. Brush up on algebraic techniques to simplify equations effectively.
4. Newton’s Laws: Understand Newton’s laws of motion. These laws provide the foundation for solving various circular motion problems.
Applications of Centripetal Acceleration
Centripetal acceleration finds applications in various fields, shaping our understanding of the physical world:
1. Automotive Safety: In the design of roads and curves, engineers use it for calculations to ensure vehicles can safely navigate turns without skidding.
2. Amusement Parks: Roller coasters and other rides rely on centripetal acceleration to provide thrilling experiences while keeping passengers safe.
3. Astronomy: Understanding this concept helps astronomers analyze the motion of planets, moons, and other celestial bodies within their orbits.
4. Sports: Athletes in sports like gymnastics and figure skating leverage centripetal acceleration to execute impressive spins and routines.
5. Physics Experiments: Centripetal acceleration plays a role in experiments involving rotating platforms or objects.
The Role of Centripetal Acceleration in Gravity
The concept of centripetal acceleration is closely linked to gravity, as it is the centripetal force that keeps planets and other objects in orbit around massive bodies like stars. This force
counteracts the natural tendency of objects to move in a straight line and explains why celestial bodies maintain their elliptical paths over time.
How is centripetal acceleration different from regular acceleration?
Centripetal acceleration involves a change in direction while moving in a circular path, whereas regular acceleration (linear acceleration) involves a change in speed in a straight line.
Can centripetal acceleration exist without centripetal force?
No, it is a result of the centripetal force. Without this force, an object would move tangentially to the circular path.
What is the relationship between centripetal acceleration and radius?
The relationship is inversely proportional. As the radius of the circular path decreases, a increases, given a constant velocity.
Are there any instances where centripetal acceleration is not applicable?
It is applicable to any object moving in a circular path, as long as there’s a force directed toward the centre of the circle.
How does centripetal acceleration relate to Newton’s laws of motion?
It is a consequence of Newton’s laws of motion. It’s a manifestation of the force required to keep an object in circular motion, as explained by the first law.
What happens if centripetal force is greater than centripetal acceleration?
Centripetal force is the cause, and centripetal acceleration is the effect. If the force is greater, the acceleration will also increase, potentially leading to a tighter circular path.
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Lived 1500 – 1557.
Niccolo Tartaglia had one of the worst childhoods of any of the scientists whose lives we have considered. He survived to make important contributions in physics, with his book New Science launching
the modern science of ballistics; and in mathematics, where he provided general solutions for cubic equations.
Tartaglia refuted Aristotle’s claim that air sustained motion, saying that air resisted motion. He was the first to say that projectile physics should be studied under conditions where air resistance
is insignificant. His work with projectiles and his translations of the works of Euclid and Archimedes into Italian paved the way for Galileo’s great discoveries in the early 1600s.
Niccolo Tartaglia was born in the northern Italian city of Brescia in 1500.
His birth name was Niccolo Fontana, but – as we’ll see – he would become known as Tartaglia following a horrific attack he suffered in his childhood.
Niccolo’s family – he had an elder brother and younger sister – were not well off. His father, Michele Fontana, earned a modest living carrying messages and deliveries on horseback between merchants
and noblemen.
The family lived in dangerous, unstable times. When Niccolo was six, his father was murdered by robbers, leaving his mother struggling to cope with extreme hardship. For the citizens of Brescia worse
was to come.
Horror and Deformity
In 1512, when Niccolo was 12 years old, there was a genocidal attack on Brescia by a French army led by Gaston of Foix. The city, although lightly defended, had refused to surrender to the French,
who were trying to conquer northern Italy.
Gaston of Foix’s army soon captured Brescia, then sacked the city heinously and without mercy. In the space of three days, 8,000 civilians – about a quarter of the city’s population – were butchered
– often gruesomely.
Many of the survivors were left with terrible wounds – physical and psychological.
Niccolo and his family had sheltered in the cathedral, but even there they found no sanctuary; soldiers forced their way in. One of the soldiers attacked Niccolo, striking five savage sword blows on
the terrified boy’s face and head, moving on to another target only when he thought the boy was dead. Niccolo, however, clung to life.
His mother had no money to pay a doctor to treat her son’s terrible wounds. She treated him based on her observations of dogs: wounded dogs lick their injuries. She cleaned Niccolo’s wounds
frequently with great care – in fact she worked with more regard for observational science than surgeons more than three centuries later would.
Miraculously, with his mother’s loving care, the boy survived. Later he wrote:
“In the cathedral, in front of my mother, I was given five murderous wounds, three on my head (each of them exposing my brain) and two on my face. Today I would look like a monster if I did not
hide the injuries behind my beard. One of the wounds cut my mouth and my teeth, breaking my jaw and palate in half. This stopped me from talking except in my throat the way magpies do.”
Niccolo became known as Tartaglia – from the Italian word meaning ‘stutter’ – because of the difficulty he had speaking.
Becoming a Mathematician
At age 12, life for Tartaglia must have been harsh; he did not yet have a beard to hide the injuries that made him “look like a monster.” He was not alone, however; many other survivors living in
Brescia were also learning to live with permanent disfigurement.
He was too poor to afford school, but got a few weeks free schooling. He kept a book that he should have returned to the school and taught himself to read. He also began showing a remarkable gift for
Recognizing her son’s unusual ability, his mother came to his aid again. She found a wealthy benefactor, Lodovico Balbisnio, who took Tartaglia to the university town of Padua to study. Tartaglia did
not study for a degree, but he learned enough in Padua to become a powerful mathematician.
His ego seems to have become dreadfully inflated at this time. When he returned to Brescia from Padua he is said to have behaved arrogantly and treated people rudely. Unsurprisingly, he began feeling
rejected by the people of his hometown. At about the age of 20, he moved to the city of Verona, where he taught mathematics privately – private teachers gave commercially oriented lessons to
engineers and merchants on subjects such as methods of measurement, bookkeeping, currency exchange, etc.
At age 34 Tartaglia relocated again, this time to the great merchant city of Venice, teaching mathematics both publicly and privately.
As a private teacher, Tartaglia needed customers and became a master of self-promotion. He showed off his expertise whenever he could, involving himself in disputes and competitions where he could
demonstrate his superior skills.
Tartaglia’s Lifetime in Context
Solving Cubic Equations, Becoming Famous
In 1535, soon after settling in Venice, Tartaglia became famous throughout Italy – or as famous as a mathematician can be – by winning a competition.
In Tartaglia’s time, anyone who had learned mathematics knew how to solve quadratic equations – equations of the form ax^2 + bx = c.
However, equations involving x^3, cubic equations, were a whole different ballgame. Finding a general solution – a way to solve all cubic equations – had defied the efforts of every mathematician who
had ever attempted it.
Tartaglia came into contact with a mathematician by the name of Antonio Fior, who was bragging about his ability to solve cubic equations. (In fact, Fior could only solve cubics in the form x^3 + ax
= b, which he had been taught to do by the mathematician Scipione dal Ferro from his deathbed.)
Tartaglia, on the other hand, had discovered how to solve cubics in the form x^3 + ax^2 = b.
They became involved in a dispute and Fior challenged Tartaglia to a competition, to decide which of them was the better mathematician. Each mathematician had to solve cubic equations posed by the
Sadly for Fior, he could not solve the problems in the form x^3 + ax^2 = b Tartaglia gave him.
Tartaglia, on the other hand, in preparation for the competition, had discovered how to solve any cubic equation.
He quickly completed all the problems Fior gave him, winning both the competition and a high reputation.
He kept his method secret, intending to publish it himself at a later time. A mathematician by the name of Girolamo Cardano persuaded Tartaglia to tell him the solution, promising not to publish it.
He actually published it six years later, giving credit to Tartaglia. Nevertheless, Tartaglia never forgave Cardano for what he saw as a betrayal.
Founding the Modern Science of Ballistics
The New Science – A Bridge from Aristotle to Galileo
The most famous Italian scientist in history is Galileo Galilei. Galileo was born seven years after Tartaglia died.
In the early 1600s, Galileo created the new physics, becoming the father of modern science. He achieved this by crushing the deeply flawed physics Aristotle constructed two millennia earlier.
Tartaglia’s work provided an important bridge between Aristotle and Galileo.
In 1537, 27 years before Galileo was born, Tartaglia published a book, the New Science, and in doing so founded modern ballistics – the science of projectiles. Tartaglia’s name was already known in
Italy because of his victory in the cubic equations competition, and his book was an instant hit, quickly running to six editions. Galileo owned a copy of the New Science.
The New Science was a very early example of mathematical physics – Tartaglia attempted to utilize mathematical axioms in the style of Euclid to explain the physical world.
Tartaglia had started thinking about ballistics five years earlier, when a soldier asked him a question: what angle should a cannon make with the ground to achieve the maximum shooting range?
Tartaglia used geometry to answer the question: 45 degrees. He then began a more careful study of ballistics, and in doing so powerfully affected the future development of physics.
Two influences – both originating in Ancient Greece – are obvious throughout Tartaglia’s
New Science
: the geometry of Euclid and the physics of Aristotle.
Tartaglia employed the eternally correct geometry of Euclid to analyze motion, but could not make a clean break from the eternally incorrect physics of Aristotle, leading to some confusion and
For example, although Tartaglia identified that a projectile’s path is a continuous curve – he was the first to do this – he choose to analyze the movement mathematically as a straight line, followed
by the arc of a circle, followed by a straight line. In doing so he was faithful to Aristotle’s incorrect belief that there were two types of motion – violent and natural.
It was left to Galileo who, with profound experiments and interpretations, proved that Tartaglia’s continuous curve is actually a parabola. Galileo scrapped Aristotle’s violent and natural motions,
replacing these with motion separated into a horizontal part and a vertical part.
Tartaglia had paved the way for Galileo: he encouraged gunners to use a device called a quadrant (shown below in an image from the New Science) to measure their guns’ angles of elevation. Tartaglia
saw the quadrant’s importance from two viewpoints:
• From the scholar’s point of view, he saw that by measuring angles and correlating them with the motion of the cannonball, gunnery problems could be attacked using mathematical theory.
• From the gunner’s point of view, he saw that when lessons had been learned from mathematical theory, they could put to practical use.
It was also important that Tartaglia’s work was repeatable – firing cannonballs produced more reliable data than, say, throwing a ball from your hand.
Tartaglia demonstrated that mathematics and measurements could be combined to understand the physical world and make new discoveries. He was particularly proud of his discovery that two different
angles of elevation would shoot a cannonball the same distance – and could not resist a little self-promotion:
“I found with a very clear argument that a cannon can hit one target along two different paths (or at two different elevations) and I found the method of how to do this in practice (a subject never
heard or conceived by anyone else, ancient or modern).”
New Science, 1537
Acceleration in Freefall
Tartaglia discussed an experiment Galileo would later become famous for:
Tartaglia said that a cannonball dropped from a tower would accelerate; the greater the time the ball fell for, the greater the speed it would reach.
The image shown here of a cannonball dropping from a tower is taken from Tartaglia’s New Science.
Legend has it that Galileo timed cannonballs falling from different heights on the Leaning Tower of Pisa, making the momentous discovery that the distance traveled by an object in freefall is
directly proportional to the square of the time it has been falling for.
Air Resistance
Aristotle had claimed that air is required to sustain motion. Tartaglia disagreed. He said motion is resisted by air. Although he was not the first to suggest Aristotle was wrong, Tartaglia then made
a conceptual leap.
He said that calculations in ballistics would only be useful for projectiles whose weight and shape made air resistance insignificant.
So, an inquiring mind might then ask: “What happens when there is no air resistance?” In other words, the concepts of motion and resistance to motion can be treated separately.
Galileo was the first to explicitly deal with motion and resistance to motion separately, and in doing so discovered Newton’s first law of motion – the law of inertia – which says that an object
keeps moving at the same speed and in the same direction unless an unbalanced force acts on it.
So physics had progressed:
Air is needed to sustain motion.
Air resists motion. We’ll calculate in situations where we can ignore it.
By ignoring air resistance, we learn a fundamental truth about motion – the law of inertia.
Before leaving the law of inertia, it’s time to say a few words about Aristotle. In his great but flawed work Physics, Aristotle more or less stated the law of inertia correctly:
“No one could say why something that is moving should stop anywhere; why should it stop here rather than there? Therefore a thing will either be at rest or must move forever, unless something
more powerful get in its way.”
The trouble is, he rejects this statement. In Aristotle’s view it must be false, because its falseness supported his assertion that a vacuum is impossible.
What a pity it is that Aristotle – having shown the tremendous insight needed to make such an early statement of the law of inertia – didn’t explore its implications further.
Bringing Euclid’s Elements to Everyone
In 1505, Bartolomeo Zamberti translated Euclid’s Elements into Latin. Zamberti had managed to find a copy of Euclid’s works in the original Greek. The best known previous edition had been written by
the Italian mathematician Campanus of Novara in the 1200s. Campanus had translated the Elements based on an Arabic source, adding his own commentary.
Tartaglia felt it was time someone wrote a modern, Italian version of the Elements. Also, he detected mathematical errors in the Zamberti and Campanus versions. Certain that Euclid himself did not
make the errors, Tartaglia said later versions of the book had been corrupted.
Therefore, he felt free to make additions and corrections as appropriate. After all, it was not Euclid’s work he was correcting, but the work of people who had made a mess of things later.
Tartaglia produced his own version of Euclid’s Elements, and it proved to be a highly influential work.
One of the mistakes he corrected was Definition 5 in Book V of the Elements: this was a mathematically beautiful definition of proportion from the brilliant Greek mathematician Eudoxus.
Definition 5 was not just beautiful, it was hard to understand. So hard, in fact, that its meaning had become twisted by botched translations. Tartaglia was a sufficiently powerful mathematician to
see what Eudoxus actually meant, and he was the first person to correct the faulty definitions present in all the Medieval Latin editions of the Elements.
In 1543, Tartaglia released Elements in Italian. For the first time, the Elements was available in a modern European language, along with Tartaglia’s helpful edits and comments. Tartaglia made
geometry available to everyone in Italy, not only the elite university-educated scholars.
It was from Tartaglia’s Italian version of the Elements that Galileo Galilei learned geometry. In fact, Galileo was taught geometry at the University of Pisa by Ostilio Ricci, one of Tartaglia’s
When Galileo founded the new physics, Book 5 of the Elements – ratios and proportions – was particularly useful to him.
In addition to his translation of Euclid, Tartaglia made important Italian translations of Archimedes‘ works.
Some Personal Details and the End
Niccolo Tartaglia married and had children during his time in Verona. He died in Venice on December 13, 1557, aged 57. The cause of his death is not known, although it is possible his life was
shortened by the wounds inflicted on him in his childhood.
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Further Reading
Henry Morley
Jerome Cardan: The life of Girolamo Cardano, of Milan, physician, Volumes 1-2
Chapman and Hall, 1854
Stuart Hollingdale
Makers of Mathematics
Penguin Books, 1990
Robert Goulding
Defending Hypatia: Ramus, Savile, and the Renaissance Rediscovery of Mathematical History
Springer Science & Business Media, 2010
Paula Olmos
Greek Science in the Long Run: Essays on the Greek Scientific Tradition (4th c. BCE-17th c. CE)
Cambridge Scholars Publishing, 2012
Lauro Martines
Furies: War in Europe, 1450–1700
Bloomsbury Publishing USA, 2013
Matteo Valleriani
Metallurgy, Ballistics and Epistemic Instruments: The Nova scientia of Nicolo Tartaglia: A New Edition
Max Planck Research Library for the History and Development of Knowledge, 2013
Catherine Ann France
Thesis: Gunnery and the Struggle for the New Science (1537-1687)
The University of Leeds, School of Philosophy, Religion and History of Science, 2014 | {"url":"https://www.famousscientists.org/niccolo-tartaglia/","timestamp":"2024-11-10T14:57:57Z","content_type":"text/html","content_length":"111828","record_id":"<urn:uuid:94d936e9-da83-40d3-864c-60c0c281d48e>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.60/warc/CC-MAIN-20241110134821-20241110164821-00730.warc.gz"} |
DISCUSSION PAPER PI-0613
Life is Cheap: Using Mortality Bonds to Hedge Aggregate Mortality Risk
Leora Friedberg and Anthony Webb
Using the widely-cited Lee-Carter mortality model, we quantify aggregate
mortality risk as the risk that the average annuitant lives longer than is
predicted by the model, and we conclude that annuity business exposes
insurance companies to substantial mortality risk. We calculate that a markup
of 3.7% on an annuity premium (or else shareholders’ capital equal to 3.7%
of the expected present value of annuity payments) would reduce the
probability of insolvency resulting from uncertain aggregate mortality trends
to 5% and a markup of 5.4% would reduce the probability of insolvency to
1 %. Using the same model, we find that a projection scale commonly referred
to by the insurance industry underestimates aggregate mortality improvements.
Annuities that are priced on that projection scale without any conservative
margin appear to be substantially underpriced. Insurance companies could
deal with aggregate mortality risk by transferring it to financial markets
through mortality-contingent bonds, one of which has recently been offered.
We calculate the returns that investors would have obtained on such bonds
had they been available over a long period. Using both the Capital and the
Consumption Capital Asset Pricing Models, we determine the risk premium that
investors would have required on such bonds. At plausible coefficients of risk
aversion, annuity providers should be able to hedge aggregate mortality risk
via such bonds at a very low cost. | {"url":"http://www.pensions-institute.org/working-papers/wp0613/","timestamp":"2024-11-04T01:09:52Z","content_type":"text/html","content_length":"32533","record_id":"<urn:uuid:93d5f8a3-7676-490f-a86c-450d9e01064a>","cc-path":"CC-MAIN-2024-46/segments/1730477027809.13/warc/CC-MAIN-20241104003052-20241104033052-00122.warc.gz"} |
An analysis of instabilities and limit cycles in glacier-dammed reservoirs
Articles | Volume 14, issue 9
© Author(s) 2020. This work is distributed under the Creative Commons Attribution 4.0 License.
An analysis of instabilities and limit cycles in glacier-dammed reservoirs
Glacier lake outburst floods are common glacial hazards around the world. How big such floods can become (either in terms of peak discharge or in terms of total volume released) depends on how they
are initiated: what causes the runaway enlargement of a subglacial or other conduit to start the flood, and how big can the lake get before that point is reached? Here we investigate how the
spontaneous channelization of a linked-cavity drainage system can control the onset of floods. In agreement with previous work, we show that floods only occur in a band of water throughput rates in
which steady reservoir drainage is unstable, and we identify stabilizing mechanisms that allow steady drainage of an ice-dammed reservoir. We also show how stable limit cycle solutions emerge from
the instability and identify parameter regimes in which the resulting floods cause flotation of the ice dam. These floods are likely to be initiated by flotation rather than the unstable enlargement
of a distributed drainage system.
Received: 04 Jun 2019 – Discussion started: 15 Jul 2019 – Revised: 09 Mar 2020 – Accepted: 14 May 2020 – Published: 18 Sep 2020
Glacier lake outburst floods or “jökulhlaups” are a glacial hazard in many parts of the world (e.g. Björnsson, 1988; Clague et al., 2012). In addition, they provide a window into subglacial drainage
systems: most outburst floods involve the opening and closing of a subglacial conduit, driven by melting of its walls through heat dissipation in the turbulent flow of water and by viscous creep
closure of the ice. In the early stages of the outburst flood, wall melting dominates through a positive feedback in which conduit enlargement leads to faster water flow and therefore more
dissipation of heat. Later in the flood, creep closure accelerates as the lake level drops, eventually causing the conduit to close again and terminating the flood. Alternatively, the flood can
terminate because the lake has run dry. The conduit then becomes partially air filled and closes with minimal flow going through it.
While the mechanics of the main discharge phase of an outburst flood are relatively well understood (Nye, 1976; Clarke, 1982, 2003; Fowler and Ng, 1996; Ng, 1998; Fowler, 1999; Kingslake and Ng, 2013
; Kingslake, 2013), the mechanisms that initiate the flood are still poorly understood, even though they dictate the level to which the lake is able to fill and therefore the magnitude and timing of
the flood. In other words, the challenge is not to explain how a single flood progresses once started, but how it starts and, more to the point, what makes repeated floods occur cyclically, as a
self-sustaining oscillation in the drainage system (Fowler, 1999; Kingslake, 2015). The main discharge phase of the flood is generally only part of a larger cycle, which also comprises a slow
recharge phase in which outflow from the lake is small or absent altogether, and the transition between the recharge and discharge phases, where both lake inflow and outflow are of comparable
One possibility for flood initiation is that the lake simply fills to the level at which the ice dam starts to float, and a sheet flow emerges between ice and bed that subsequently channelizes (
Flowers et al., 2004). This may occur irregularly in some lakes such as Grímsvötn due to exceptionally large inflow rates to the lake, for instance during volcanic eruptions, or as part of a
repeating flood cycle in others (Bigelow et al., 2020). Some lakes are however known to initiate outburst floods before they reach that flotation level.
Motivated by the subglacial lake at Grímsvötn in Iceland, Ng (1998) and Fowler (1999) consider how outburst floods are initiated in a drainage system that consists of a Röthlisberger (R) channel (
Röthlisberger, 1972). In the simplest version of their model, where the channel is fed purely by the lake, they find that the amplitude of floods grows from cycle to cycle and that water pressures
exceeding ice overburden are eventually reached, meaning that floods should start through the flotation of the ice dam after a few cycles. At the heart of this behaviour is the fact that the
R channel can shrink to progressively smaller sizes between successive floods, making it harder to reinitiate drainage.
Grímsvötn typically starts its flood when water levels are below flotation (Björnsson, 1988), without successive floods growing in amplitude, except when the flood results from a large increase in
water input due to volcanic activity (Gudmundsson et al., 1997). To explain this behaviour, Fowler (1999) considers the effect of a water supply along the length of the R channel on its evolution.
Such a water supply can maintain a minimum channel size even between outburst floods, with flow of water being directed partly into the lake and partly down-glacier to the margin, provided the
glacier has a geometrical “seal”. As the lake fills, the flow divide inside the channel migrates towards the lake, and the flood begins when the divide reaches the edge of the lake.
While this mechanism successfully explains how limit cycles (stable, periodic oscillations in lake level) can emerge in the model, it also predicts that no water can leave the lake between floods.
Tracer experiments conducted at Salmon Glacier in Canada (Fisher, 1973) demonstrate that lakes can leak continuously throughout their recharge phase. Summit Lake, dammed by the Salmon Glacier, also
has a history of flood initiation at lake levels below the flotation level (Post and Mayo, 1971). At Gornersee in Switzerland, observations of water lake levels and inferences of lake water balance
based on meteorological measurements over two consecutive years also suggest leakage and flood initiation below flotation level during one year and flood initiation at or above flotation level during
the previous year, potentially in the absence of any pre-flood leakage (Huss et al., 2007). Observations at Hidden Creek Lake (Anderson et al., 2003) by contrast suggest no leakage prior to flood
initiation, though that conclusion is again based on water balance estimates rather than direct tracer experiments.
In this paper, we consider an alternative mechanism by which floods can initiate in a recurring fashion without the need for a sealed lake, but with continuous leakage throughout the
flooding-and-refilling cycle. We show that a conduit that is able to switch spontaneously between the behaviour typical of an R channel and the behaviour of a linked-cavity system (Kessler and
Anderson, 2004; Schoof, 2010; Hewitt et al., 2012; Hewitt, 2013; Werder et al., 2013) can sustain limit cycles without leading to flotation of the glacier at flood initiation, and without requiring
the flow-divide migration appealed to in Fowler (1999). Note that many elements of the model studied here were included in the earlier study of outburst flooding by Kessler and Anderson (2004), who
were able to replicate leakage from a glacier-dammed reservoir before the onset of the outburst flood but did not attempt to reproduce recurring floods. It is unclear if their model would have been
able to do so since, in common with the model due to Fowler (1999), the main trunk of their drainage system was set up as a pure R channel.
The behaviour studied in the present paper was first documented in a model by Schoof et al. (2014). In this paper, we approach the problem from a more theoretical perspective, focusing on the
following: first, we delineate when a “lake” or other storage reservoir can be drained steadily, identify conditions under which steady drainage becomes unstable, and identify how that instability
leads to a limit cycle. Next, we investigate how the amplitude and period of floods depends on reservoir size and inflow rate. As a corollary, we determine at which point (in parameter space) the
model predicts that flotation does occur and a different flood initiation mechanism is likely to take over. In closing, we also briefly comment on the difference in flood dynamics caused by
distributing water storage along the flow path rather than having a single water reservoir (Schoof et al., 2014). We outline how the classical runaway melt mechanism for outburst floods first
described by Nye (1976) can be replaced as a mechanism for self-sustaining water pressure oscillations by a spatial phase shift between conduit size and water pressure along the flow path.
The original motivation for the work in Schoof et al. (2014) was to demonstrate that behaviour akin to outburst floods can occur in systems with limited or even distributed water storage, manifesting
itself in the form of unforced water pressure oscillations. Such limited storage could in principle result from moulins or any other vertical shaft that can fill progressively with water as water
pressure rises. Viewed from that alternative perspective, the present paper analyses a drainage model that has become widely adopted in the study of subglacial hydrology (Werder et al., 2013). We
identify instabilities that spontaneously lead to self-sustained oscillations, describing the mechanisms behind them and delineating the regions in parameter space where they occur. This is likely to
be useful in diagnosing the behaviour of such models, regardless of their specific application to large outburst floods emanating from easily identifiable glacier lakes.
We use a hierarchy of model versions, employing both a spatially extended, one-dimensional drainage system model and a “lumped”, box-type model that provides additional physical insight, as well as
being the appropriate limit of the spatially extended model in the case of a long flow path. The paper is laid out as follows: in Sect. 2, we develop the spatially extended and lumped models. In
Sect. 3.1 and 3.2, we identify instabilities in the lumped model. The evolution of that instability at finite amplitude is investigated in detail in Sect. 3.3 and 3.4, where we show that stable limit
cycles emerge in the lumped model. In Sect. 4, we show that the lumped model replicates the behaviour of the spatially extended behaviour well in the classical parameter regime that corresponds to
the drainage of large glacier-dammed lakes. We also show that the extended model is unstable in more exotic parameter regimes, where the lumped model is not, and investigate the resulting dynamics.
These results are summarized in Sect. 5.1, and we briefly touch on the dynamics of systems with distributed water storage in Sect. 5.2, relegating detail to the Supplement.
2.1A continuum model for outburst floods
We use the one-dimensional continuum model for drainage through a single conduit stated in Schoof et al. (2014), building on similar models used elsewhere (e.g. Ng, 1998; Hewitt and Fowler, 2008;
Schuler and Fischer, 2009; Schoof, 2010; Hewitt et al., 2012). Note that we use the term “conduit” here to refer to a generic drainage element that can evolve dynamically to behave as an R channel,
in which dissipation and creep closure are the dominant mechanisms by which channel size changes, or as a cavity, in which opening due to sliding over bedrock and creep closure are dominant (Kessler
and Anderson, 2004; Schoof, 2010).
Denoting conduit cross section by S(x,t), effective pressure by N(x,t) and discharge by q(x,t), where x is downstream distance and t is time, we put
$\begin{array}{}\text{(1a)}& \frac{\partial S}{\partial t}& ={c}_{\mathrm{1}}\stackrel{\mathrm{̃}}{q}\mathrm{\Psi }+{v}_{\mathrm{o}}\left(S\right)-{v}_{\mathrm{c}}\left(S,N\right),\text{(1b)}& \frac{\
partial \stackrel{\mathrm{̃}}{q}}{\partial x}& =\mathrm{0},\text{(1c)}& \stackrel{\mathrm{̃}}{q}& =q\left(S,\mathrm{\Psi }\right),\text{(1d)}& \mathrm{\Psi }& ={\mathrm{\Psi }}_{\mathrm{0}}+\frac{\
partial N}{\partial x},\end{array}$
on $\mathrm{0}<x<L$, subject to boundary conditions
$\begin{array}{}\text{(1e)}& -{V}_{p}\left(N\right)\frac{\partial N}{\partial t}& ={q}_{\mathrm{in}}-\stackrel{\mathrm{̃}}{q}& \text{at }x& =\mathrm{0},\text{(1f)}& N& =\mathrm{0}& \text{at }x& =L\
where L is the length of the flow path, and we assume the closures
$\begin{array}{}\text{(1g)}& \begin{array}{rl}& {v}_{\mathrm{o}}\left(S\right)={u}_{\mathrm{b}}{h}_{\mathrm{r}}\left(\mathrm{1}-S/{S}_{\mathrm{0}}\right),\phantom{\rule{1em}{0ex}}{v}_{\mathrm{c}}\
left(S,N\right)={c}_{\mathrm{2}}S|N{|}^{n-\mathrm{1}}N,\\ & q\left(S,\mathrm{\Psi }\right)={c}_{\mathrm{3}}{S}^{\mathit{\alpha }}|\mathrm{\Psi }{|}^{-\mathrm{1}/\mathrm{2}}\mathrm{\Psi }.\end{array}\
Here c[1], c[2], c[3], n, α, u[b], h[r] and S[0] are positive constants described in further detail below, with α>1, while q[in] is also prescribed as a function of time and V[p]>0 is a function of N
. Ψ[0] is a geometrically determined background hydraulic gradient, given in terms of ice surface elevation s(x) and bed elevation b(x) through
$\begin{array}{}\text{(1h)}& {\mathrm{\Psi }}_{\mathrm{0}}=-{\mathit{\rho }}_{\mathrm{i}}g\frac{\partial s}{\partial x}-\left({\mathit{\rho }}_{\mathrm{w}}-{\mathit{\rho }}_{\mathrm{i}}\right)g\frac
{\partial b}{\partial x}.\end{array}$
Here ρ[w] and ρ[i] are the densities of ice and water, respectively, and g is acceleration due to gravity. Note that effective pressure N is linked to water pressures p[w], which is the primary
observable in the field, through
$\begin{array}{}\text{(1i)}& N={p}_{\mathrm{i}}-{p}_{\mathrm{w}},\end{array}$
where p[i] is overburden (or more precisely, normal stress in the ice at the bed, averaged over a length scale much larger than that occupied by the conduit). Typically (including in Eq. 1h), p[i] is
assumed to be cryostatic, equal to ρ[i]g(s−b).
Physically, the model represents a conduit whose size evolves due to a combination of dissipation-driven wall melting at a rate c[1]qΨ, opening due to ice sliding over bed roughness at rate ${u}_{\
mathrm{b}}{h}_{\mathrm{r}}\left(\mathrm{1}-S/{S}_{\mathrm{0}}\right)$ and creep closure at rate c[2]SN^n, with discharge in the conduit given by a Darcy–Weisbach or Manning friction law as ${c}_{\
mathrm{3}}{S}^{\mathit{\alpha }}|\mathrm{\Psi }{|}^{-\mathrm{1}/\mathrm{2}}\mathrm{\Psi }$ through Eq. (1g). u[b] is sliding velocity, h[r] is bed roughness and S[0] is a cut-off cavity size at which
bed roughness is drowned out (Schoof et al., 2012). S[0] is typically a regularizing parameter that prevents conduits from becoming excessively large in regions where effective pressures are low,
such as near the glacier terminus, but has little effect on conduit sizes along most of the flow path; this corresponds to the limit of large S[0]. Equation (1a) also ignores the effect of water
pressure on the melting point, which affects the fraction of the dissipated heat available for melting of the conduit walls (Röthlisberger, 1972; Werder, 2016). We have also neglected the effect of
water storage in the conduits and of dissipation-driven melting in the mass balance relation (1b); the reason for this is that a simple scaling exercise shows that these terms are invariably small
for a terrestrial glacial system (Supplement Sect. S2.2).
Vanishing effective pressure at the end of the flow path x=L simply reflects the assumption that overburden and water pressure vanish simultaneously (where we have neglected atmospheric pressure as a
gauge throughout). A computationally preferable assumption to N=0 may be to assume that there is an ice cliff at the glacier terminus, so that vanishing water pressure still corresponds to a finite N
and the effect of dissipation-driven opening does not have to be offset by an unrealistic closing of the conduit through the sliding term, which becomes negative when S>S[0]; our results are however
not sensitive to this assumption.
The upstream boundary condition at x=0 reflects water storage in a reservoir at the head of the conduit: Eq. (1e) represents conservation of mass in the reservoir, with inflow rate q[in], with
outflow at rate q(0,t) through the conduit modelled by Eqs. (1a)–(1d) and with water storage V in the reservoir a function of effective pressure at x=0 such that $\mathrm{d}V/\mathrm{d}N=-{V}_{p}$
(see e.g. Clarke, 2003). Note that we deliberately use “reservoir” rather than “lake” here since we will be interested not only in large lakes, but also in more modest-sized reservoirs as in Schoof
et al. (2014). We will treat V[p] and q[in] as constants in most of what follows, although realistic lake shapes may argue for particular forms of the function V[p](N) (Clarke, 2003).
For simplicity, we have neglected the possibility of a partially air filled conduit and the effects of overpressurization, where effective pressure becomes negative (Schoof et al., 2012; Hewitt
et al., 2012). Overpressurization is particularly relevant to the initiation mechanism for subglacial floods where unstable enlargement of the conduit is not fast enough for the flood to occur before
the lake reaches N=0 and overfills. A partially air filled conduit may form by contrast at the upstream end of the flood path if the lake fully drains at the end of the outburst floods, and it could
form more permanently at the downstream end as shown in Hewitt et al. (2012). We defer consideration of both processes to future work.
There is an important simplification we have made, irrespective of the particular choice of V[p]: our model effectively imposes an ice cliff at the conduit inlet x=0, with the cliff height typically
above the flotation thickness. This ensures that N(0,t) can change over time without the upstream end of the conduit migrating. This contrasts with a glacier that partially floats on the lake, in
which case the upstream end of the conduit is always at N=0. In that case, the upstream conduit end migrates as the lake fills or empties, and its location is related to lake volume (see Sect. S2.4
of the Supplement).
The assumption of an ice cliff, in addition, allows us to assume that Ψ[0]>0 along the entire flow path, without a seal at a finite distance from the reservoir at which Ψ[0] changes sign (Fowler,
1999). With a seal, there is a finite region of negative Ψ[0] between reservoir and seal location, with water flow out of the reservoir possible only when local effective pressure gradients $\partial
N/\partial x$ are large enough to overcome the effects of the seal to make the total hydraulic gradient Ψ positive in Eq. (1d). In the absence of a seal and without an ice cliff (so N=0 at the edge
of the reservoir), normal stress at the glacier bed just outside of the reservoir would then be below the flotation pressure, and there would be nothing to dam the reservoir (see Sect. S2.4 of the
The two situations (an ice cliff and a seal) can be reconciled in the sense that a seal very close to the edge of the reservoir corresponds to a short, steep ice surface slope rather than an actual
cliff. This results in a large, negative Ψ[0] between reservoir and seal over a short distance, which can be balanced by an equally large $\partial N/\partial x$ over the same short distance, leading
to a non-zero effective pressure at the seal itself, only a short distance from the lake.
Nevertheless, our simplifying assumption is relevant as the mechanism for initiating periodically recurring floods is fundamentally different from that in Fowler (1999): his model relies on a
geometrical seal with negative Ψ[0] near the reservoir, changing sign at the seal location, and an englacial water supply to the conduit that dictates the gradient $\partial N/\partial x$ while the
lake is filling. As described in the introduction, the onset of the outburst flood corresponds to the instant at which Ψ at the conduit inlet x=0 changes from negative to positive and allows water to
flow out of the reservoir. This leads to runaway enlargement of the conduit through dissipation-driven melting.
By contrast, our model assumes that the reservoir always experiences some amount of leakage and requires no englacially supplied conduit: the water supply to the drainage system in our model is
purely through the inflow term q[in] into the reservoir. Leakage out of the reservoir is facilitated by the linked-cavity behaviour of the conduit between floods, associated with the cavity opening
term v[o] that keeps the conduit open even when flow rates are small. This is absent from the models in Fowler (1999) and Ng (1998). As the lake fills, these cavities enlarge because the effective
pressure is reduced, and the conduit then grows through the same melt-driven conduit enlargement as in Fowler (1999) and Ng (1998), but the initiation of that enlargement differs.
2.2A lumped model
The model (1) can be simplified if we assume that Ψ[0] not only does not change sign along the flow path but can be treated as a constant and if we assume that the effective pressure gradient $\
partial N/\partial x$ along the flow path can be treated as negligible (see also Fowler, 1999; Ng, 2000, and Sect. S2.2 and S2.3 of the Supplement) or approximated by a simple divided difference.
Under these assumptions, we can relate the flux q along the flow path (which is independent of position by Eq. 1b) purely to the conduit size S and hydraulic gradient Ψ[0] at the head of the conduit,
leading to a system of ordinary rather than partial differential equations:
$\begin{array}{}\text{(2a)}& \stackrel{\mathrm{˙}}{S}& ={c}_{\mathrm{1}}\stackrel{\mathrm{̃}}{q}\mathrm{\Psi }+{v}_{\mathrm{o}}\left(S\right)-{v}_{\mathrm{c}}\left(S,N\right),\text{(2b)}& -{V}_{p}\
stackrel{\mathrm{˙}}{N}& ={q}_{\mathrm{in}}-\stackrel{\mathrm{̃}}{q},\text{(2c)}& \stackrel{\mathrm{̃}}{q}& =q\left(S,\mathrm{\Psi }\right),\end{array}$
where a dot signifies an ordinary derivative with respect to time, and we continue to assume the closure relations (1g); by the argument above, we then strictly speaking have to put Ψ=Ψ[0]. We
generalize this reduced model slightly to account qualitatively, if not quantitatively, for the effect of pressure gradients along the flow path by putting
$\begin{array}{}\text{(2d)}& \mathrm{\Psi }={\mathrm{\Psi }}_{\mathrm{0}}-N/L,\end{array}$
which corresponds to a crude divided difference approximation of the actual gradient $\partial N/\partial x$ by $\left[N\left(L,t\right)-N\left(\mathrm{0},t\right)\right]/L$.
In what follows, we proceed first with an analysis of the simpler, lumped model (2), to which we can bring to bear the theory of finite-dimensional dynamical systems (Strogatz, 1994; Wiggins, 2003),
leading a number of semi-analytical results. Subsequently, we study the full, spatially extended model (1), for which we are limited to comparatively expensive numerical methods that we can guide by
our analysis of the simpler, lumped model.
3An analysis of the lumped model
3.1Nye's jökulhlaup instability in the reduced model
Nye's (1976) original theory of jökulhlaups centres on the idea that steady flow in an R channel is unstable if there is a water reservoir that keeps water pressure approximately constant. This
instability results in the runway growth of a channel, eventually allowing the reservoir to drain in an outburst flood. The assumption of a constant effective pressure is of course a simplification
of the more complete model (2), based on the notion that the drainage of the lake is generally too slow to lead to changes in effective pressure that could stabilize the channel against growth.
The basis of this instability is relatively easy to capture mathematically. To set the scene, we present a simplified version first before tackling an analysis of the complete model (2). If we assume
a classical R channel in the sense of Röthlisberger (1972) and Nye (1976) and therefore set the cavity opening term v[o] to zero in Eq. (2a), and use the remaining relations in Eq. (1g), the conduit
evolution Eq. (2a) becomes
$\begin{array}{}\text{(3)}& \stackrel{\mathrm{˙}}{S}={c}_{\mathrm{1}}{c}_{\mathrm{3}}{S}^{\mathit{\alpha }}|\mathrm{\Psi }{|}^{\mathrm{3}/\mathrm{2}}-{c}_{\mathrm{2}}S|N{|}^{n-\mathrm{1}}N,\end
with a steady-state conduit size $\overline{S}$ given by ${c}_{\mathrm{2}}\overline{S}|\overline{N}{|}^{n-\mathrm{1}}\overline{N}={c}_{\mathrm{1}}{c}_{\mathrm{3}}{\overline{S}}^{\mathit{\alpha }}|\
mathrm{\Psi }{|}^{\mathrm{3}/\mathrm{2}}$. At fixed effective pressure N, and hence at fixed Ψ, this steady state is unstable: an increase S^′ away from the steady-state size $\overline{S}$ will lead
to a further growth in the conduit, as we have $\left(\overline{S}+{S}^{\prime }{\right)}^{\mathit{\alpha }}\approx {\overline{S}}^{\mathit{\alpha }}+\mathit{\alpha }{\overline{S}}^{\mathit{\alpha }-
\mathrm{1}}{S}^{\prime }$ and so
$\begin{array}{}\text{(4)}& \begin{array}{rl}{\stackrel{\mathrm{˙}}{S}}^{\prime }\approx & {c}_{\mathrm{1}}{c}_{\mathrm{3}}\mathit{\alpha }{\overline{S}}^{\mathit{\alpha }-\mathrm{1}}{S}^{\prime }|\
mathrm{\Psi }{|}^{\mathrm{3}/\mathrm{2}}-{c}_{\mathrm{2}}{S}^{\prime }|N{|}^{n-\mathrm{1}}N\\ =& {c}_{\mathrm{1}}{c}_{\mathrm{3}}\left(\mathit{\alpha }-\mathrm{1}\right){\overline{S}}^{\mathit{\alpha
}-\mathrm{1}}|\mathrm{\Psi }{|}^{\mathrm{3}/\mathrm{2}}{S}^{\prime },\end{array}\end{array}$
where, with α>1, the right-hand side is positive if S^′ is, signifying that S^′ will continue to grow in a positive feedback.
In more abstract terms, if Ψ and N are fixed, Eq. (2a) can be expected to lead to unstable growth of conduits away from an equilibrium state $\overline{S}$ if (see also pp. 5–6 of the Supplement to
Schoof, 2010)
This is also precisely the condition that defines a conduit as being channel-like in Schoof (2010), in the sense that two neighbouring conduits will compete for water with one eventually growing at
the expense of the other if both are channel-like.
Key to the instability mechanism above was the notion that effective pressure N is kept constant as the conduit evolves, which is not actually the case: instead, the lake simply buffers changes in N
from happening rapidly. Next, we investigate in more detail how water storage and the finite size L of the system control whether Nye's instability does occur in the reduced model of Sect. 2.2.
Note that v[o], v[c] and q satisfy
$\begin{array}{}\text{(6)}& \frac{\partial {v}_{\mathrm{o}}}{\partial S}\le \mathrm{0},\phantom{\rule{1em}{0ex}}\frac{\partial {v}_{\mathrm{c}}}{\partial S}>\mathrm{0},\phantom{\rule{1em}{0ex}}\frac
{\partial {v}_{\mathrm{c}}}{\partial N}>\mathrm{0},\phantom{\rule{1em}{0ex}}\frac{\partial q}{\partial S}>\mathrm{0},\phantom{\rule{1em}{0ex}}\frac{\partial q}{\partial \mathrm{\Psi }}>\mathrm{0},\
with v[o](S)>0 bounded as S→0, ${v}_{\mathrm{c}}\left(S,\mathrm{0}\right)=\mathrm{0}$ and ${v}_{\mathrm{c}}\left(\mathrm{0},N\right)=\mathrm{0}$; q also has the same sign as Ψ and satisfies $q\left(\
mathrm{0},\mathrm{\Psi }\right)=q\left(S,\mathrm{0}\right)=\mathrm{0}$. These are the minimal assumptions we make on these functions, allowing us to generalize from the specific forms in Eq. (1g) in
analysing Nye's instability.
The primary dependent variables in the model (2) are S and N. For constant water input q[in], the model admits a steady state $\left(\overline{S},\overline{N}\right)$ given implicitly by
$\begin{array}{}\text{(7a)}& {c}_{\mathrm{1}}q\left(\overline{S},\overline{\mathrm{\Psi }}\right)\overline{\mathrm{\Psi }}+{v}_{\mathrm{o}}\left(\overline{S}\right)-{v}_{\mathrm{c}}\left(\overline
{S},\overline{N}\right)& =\mathrm{0},\text{(7b)}& q\left(\overline{S},\overline{\mathrm{\Psi }}\right)& ={q}_{\mathrm{in}},\text{(7c)}& \overline{\mathrm{\Psi }}& ={\mathrm{\Psi }}_{\mathrm{0}}-\
where we assume positive lake inflow q[in]>0. It can be shown (see Sect. S3 of the Supplement) that there is a unique steady-state solution with positive $\overline{N}$, $\overline{S}$ and $\overline
{\mathrm{\Psi }}$. To establish whether the solution is stable, we can linearize around the steady state as
$N=\overline{N}+{N}^{\prime }\mathrm{exp}\left(\mathit{\lambda }t\right),\phantom{\rule{1em}{0ex}}S=\overline{S}+{S}^{\prime }\mathrm{exp}\left(\mathit{\lambda }t\right).$
This yields the following eigenvalue problem
$\begin{array}{}\text{(8)}& \begin{array}{rl}& \left(\begin{array}{cc}{c}_{\mathrm{1}}{q}_{S}\overline{\mathrm{\Psi }}+{v}_{\mathrm{o},S}-{v}_{\mathrm{c},S}-\mathit{\lambda }& -{c}_{\mathrm{1}}\left
({q}_{\mathrm{\Psi }}\overline{\mathrm{\Psi }}+\overline{q}\right){L}^{-\mathrm{1}}-{v}_{\mathrm{c},N}\\ {\overline{V}}_{p}^{-\mathrm{1}}{q}_{S}& -{\overline{V}}_{p}^{-\mathrm{1}}{q}_{\mathrm{\Psi }}
{L}^{-\mathrm{1}}-\mathit{\lambda }\end{array}\right)\\ & \left(\begin{array}{c}{S}^{\prime }\\ {N}^{\prime }\end{array}\right)=\mathrm{0},\end{array}\end{array}$
Setting the determinant of the matrix on the left to zero, we obtain a quadratic with solution
$\begin{array}{}\text{(9)}& \mathit{\lambda }=\frac{\mathrm{1}}{\mathrm{2}}\left[{a}_{\mathrm{1}}±\sqrt{{a}_{\mathrm{1}}^{\mathrm{2}}-\mathrm{4}{a}_{\mathrm{2}}}\right],\end{array}$
From our assumptions on the various functions involved, we see that a[2]>0 (recall that ${v}_{\mathrm{o},S}\le \mathrm{0}$ from the inequality Eq. 6), while a[1] can be either sign. Since a[2]>0, we
have ${a}_{\mathrm{1}}^{\mathrm{2}}-\mathrm{4}{a}_{\mathrm{2}}<{a}_{\mathrm{1}}^{\mathrm{2}}$ and two possible types of solution: either (i) ${a}_{\mathrm{1}}^{\mathrm{2}}-\mathrm{4}{a}_{\mathrm{2}}>
\mathrm{0}$, in which case we have two real roots, both of which have the same sign as a[1], or (ii) ${a}_{\mathrm{1}}^{\mathrm{2}}-\mathrm{4}{a}_{\mathrm{2}}<\mathrm{0}$, resulting in a complex
conjugate pair of roots, both of which have real part a[1]. In either case, we see that the system is linearly unstable if and only if a[1]>0, or
$\begin{array}{}\text{(11)}& \left({c}_{\mathrm{1}}{q}_{S}\overline{\mathrm{\Psi }}+{v}_{\mathrm{o},S}-{v}_{\mathrm{c},S}\right)-{\overline{V}}_{p}^{-\mathrm{1}}{q}_{\mathrm{\Psi }}{L}^{-\mathrm{1}}>
We have deliberately written the left-hand side of Eq. (11) as the difference of two terms, a potentially destabilizing term ${c}_{\mathrm{1}}{q}_{S}\overline{\mathrm{\Psi }}+{v}_{\mathrm{o},S}-{v}_
{\mathrm{c},S}$ and a stabilizing term $-{\overline{V}}_{p}^{-\mathrm{1}}{q}_{\mathrm{\Psi }}{L}^{-\mathrm{1}}$. The first term can be recognized as the growth rate sensitivity we previously
identified as being at the heart of Nye's instability in Eq. (5), as well as an indicator of whether the steady-state conduit is channel-like in the terminology of Schoof (2010).
The second term in the relation (11) is a stabilizing term that is inversely related to storage capacity. Its physical origin is the following: the sensitivity of conduit growth to perturbations in
conduit size may be positive, potentially leading to unstable conduit growth. However, growth of the conduit also allows water to drain out of the system, which will increase the effective pressure N
. As the effective pressure is increased, the hydraulic gradient $\mathrm{\Psi }={\mathrm{\Psi }}_{\mathrm{0}}-N/L$ is reduced. This leads to less turbulent dissipation in the conduit as the conduit
grows, and increased N further leads to faster creep closure of the conduit. Both of these will suppress further growth of the conduit. How strong this stabilizing effect is will depend on the
storage capacity of the system and on the length L of the flow path: a large storage capacity ${\overline{V}}_{p}$ or a long flow path L leads to a reduced stabilizing effect. In practice, we are
likely to see stabilization for small storage elements, such as moulins and individual crevasses.
3.2Stability boundaries
We have established that Nye's instability will occur if the conduit is channel-like and water storage in the system is sufficiently large, as well as if the system length L is big enough. Note that,
with the choices in Eq. (1g), the system is always unstable if v[o]=0 (so the conduit is always a channel) and L=∞ (so effective pressure does not alter the hydraulic gradient). This is in agreement
with Ng (1998).
We can go further and determine explicitly the regions of parameter space in which this stability occurs. While we present our results here in dimensional form, note that it is possible to reduce the
spatially extended and lumped models (1) and (2) to a four-dimensional parameter space by non-dimensionalizing them (Sect. S2.2 and S2.3 of the Supplement). These parameters are dimensionless
versions of the inflow rate q[in], storage capacity V[p], system length L and conduit cut-off size S[0]. Recall that S[0] is intended to have minimal impact on conduit evolution away from the glacier
margin, and we are therefore interested in the limit of large S[0]. As a result, we restrict ourselves to the three-dimensional parameter space spanned by $\left({q}_{\mathrm{in}},{\overline{V}}_
{p},L\right)$ and set S[0]=∞ in the lumped model (2). The remaining fixed parameter values we have used are given in Table 1. The low value of Ψ[0] stated corresponds approximately to a 1:50 surface
Figure 2 shows stability boundaries in the $\left({q}_{\mathrm{in}},{\overline{V}}_{p}\right)$ plane for different values of L. These are the locations in parameter space where the real part of the
growth rate λ in Eq. (9) is zero (see Sect. S3.2 of the Supplement for details); invariably, the eigenvalues λ then form a purely imaginary conjugate pair, and we have a so-called Hopf bifurcation
(see Sect. 3.3 below). Note that ${\overline{V}}_{p}$ is displayed in units of square kilometres (km^2), normalizing by ρ[w]g; in other words, what is really plotted is ${\overline{V}}_{p}/\left({\
mathit{\rho }}_{\mathrm{w}}g\right)$, the surface area of the lake (see Sect. 2). Similarly, we will plot N in units of metres (m) in subsequent figures, normalizing by ρ[w]g: changes in N plotted
then correspond directly to changes in water level in the lake.
In each case shown in Fig. 2, there is a region of instability above some critical value of ${\overline{V}}_{p}$, and for q[in] in some intermediate range, where larger ${\overline{V}}_{p}$ is
required and the unstable inflow range is shrunk for shorter flow path lengths L. This is consistent with theoretical results (Sect. S3.1 of the Supplement) showing that the system is always stable
for either sufficiently small or sufficiently large q[in] but can be unstable at intermediate inflow rates.
For large enough ${\overline{V}}_{p}$ and L, we can determine approximate analytical expressions for the stability boundaries. The lower critical value of q[in] at which the drainage system first
becomes unstable corresponds to the switch from a cavity-like to a channel-like conduit, and this occurs for large L at (see Schoof, 2010, and Sect. S3.1 of the Supplement)
$\begin{array}{}\text{(12)}& {q}_{\mathrm{in}}=\frac{{u}_{\mathrm{b}}{h}_{\mathrm{r}}}{{c}_{\mathrm{1}}\left(\mathit{\alpha }-\mathrm{1}\right){\mathrm{\Psi }}_{\mathrm{0}}}.\end{array}$
Note that for finite h ${\overline{V}}_{p}$ and L, there is generally at least a small region of stability in which q[in] is already large enough for the conduit to be channel-like (the space between
the dotted–dashed vertical line and the solid curves in Fig. 2). This region exists because the first term in Eq. (11) is positive but not large enough to overcome the second, stabilizing term here.
The upper critical value of q[in] at which the system stabilizes again can be similarly computed in the limit of large L by omitting the cavity opening term v[o] and reducing the conduit evolution
Eq. (2a) to a balance between the first (dissipation-driven melting) and third (creep-closure) terms on the right-hand side. This yields
$\begin{array}{}\text{(13)}& {\overline{V}}_{p}\sim \frac{{q}_{\mathrm{in}}^{\mathrm{1}/\mathit{\alpha }}}{\mathrm{2}\left(\mathit{\alpha }-\mathrm{1}\right){c}_{\mathrm{1}}{c}_{\mathrm{3}}^{\mathrm
{1}/\mathit{\alpha }}{\mathrm{\Psi }}_{\mathrm{0}}^{\left(\mathrm{3}\mathit{\alpha }+\mathrm{2}\right)/\left(\mathrm{2}\mathit{\alpha }\right)}L}.\end{array}$
These limiting forms are also shown in Fig. 2, where we see that they are most accurate for large L as expected.
3.3Hopf bifurcations and limit cycles
The analysis above has been purely linear, identifying parameter regimes in which steady drainage is unstable. That linearization will eventually fail where unstable growth is predicted, and a
non-linear model is needed to predict what the instability grows into. A key aspect of many outburst floods is that they are a recurring phenomenon. For the reduced model (2) with two dynamical
degrees of freedom and steady forcing, such a recurrence must correspond to a stable periodic oscillation in the absence of time-dependent forcing (see also Kingslake, 2013, for time-dependent
forcing that leads to chaotic solutions). As was underlined by Ng (1998) and Fowler (1999), the existence of such a limit cycle does not simply follow from the instability itself: we need to ensure
that the evolution away from the steady state leads to bounded growth of the instability once it reaches a finite amplitude.
It is straightforward to demonstrate bounded growth in our model computationally. Figure 3 shows a sample calculation of a periodic solution, with the classical attributes of an outburst flood cycle:
effective pressure slowly decreases during the interval between outburst floods, when conduit size is small. This is simply the lake refilling. As N approaches its minimum, the conduit size S starts
to grow rapidly, initiating the outburst flood: effective pressure is no longer large enough to keep conduit size small, and enough water can flow to start enlarging the conduit in the runaway growth
envisioned by Nye (1976). The lake then drains, rapidly increasing N. Once effective pressure gets large enough, creep closure becomes dominant, causing S to shrink again, and the cycle repeats. By
contrast with Ng (1998) and Fowler (1999), key to this periodic behaviour is that the conduit cannot become arbitrarily small between floods, since it is kept open by ice flow over bed roughness.
In fact, S in our model actually starts to increase almost immediately after flood termination, as the refilling of the reservoir leads to decreasing effective pressure allowing the now cavity-like
conduit to grow. As explained above, water flow through them will eventually lead to reinitiation of dissipation-driven melting and enlargement of the conduit. By contrast, once the amplitude of the
floods has become large enough, conduit size in the pure channel model of Ng (1998) and Fowler (1999) keeps shrinking during the refilling phase until the effective pressure changes sign to negative,
and this underpins the ever-increasing flood amplitudes in their model.
An alternative visualization of the physics involved is a phase plane, plotting S against N as the system evolves; a periodic solution then corresponds to a closed orbit. Figure 4 shows the phase
plane equivalent of Fig. 3, with the dashed lines corresponding to nullclines (the curves on which either $\stackrel{\mathrm{˙}}{S}=\mathrm{0}$ or $\stackrel{\mathrm{˙}}{N}=\mathrm{0}$). During the
refilling phase, the point (N(t),S(t)) closely follows the S nullcline, with S steadily increasing as explained above.
Figures 3 and 4 show only one example. It is possible to solve directly for such periodic solutions and trace how they change under parameter changes using an arc length continuation method
(Sect. S3.4 of the Supplement), as well as to determine simultaneously whether the periodic solutions are stable: some periodic solutions are in fact never attained by forward integration of the
model (2) because they are themselves unstable. Small perturbations will therefore cause the system to evolve away from them.
As in Sect. 3.2, we focus on how solutions change as the water input to the reservoir q[in] is varied. Figure 5 shows how the amplitude of oscillations varies with water input for a fixed L=50km and
different storage capacity V[p] (treated as independent of N here). In each panel, the coloured curve (black in Fig. 5e) shows the minimum and maximum value of N attained for a periodic solution at
the corresponding value of q[in] (i.e. the values of N where the corresponding orbit in the phase plane crosses the N nullcline). Solid portions of these coloured curves in Fig. 5 correspond to
stable periodic solutions, while dotted portions are unstable periodic orbits. The black curves in Fig. 5 generally correspond to steady-state solutions, plotting N in the steady state against q[in]
(we use black for steady states and oscillatory solutions in Fig. 5e, but these are easily distinguished by comparing the plots in Fig. 5e with those in the remaining panels). Again, solid black
lines are stable steady states and dashed black lines are unstable steady states. Each panel in Fig. 5 corresponds to a different storage capacity V[p] and therefore represents a horizontal slice
through Fig. 2; V[p] decreases from Fig. 5a to Fig. 5d.
In all cases, an unstable steady state corresponds to a limit cycle. As q[in] crosses the lower critical value at which instability first occurs (associated with the change in the steady-state
conduit from cavity- to channel-like behaviour), a limit cycle of small amplitude is formed, with that amplitude growing progressively as q[in] increases, and the emerging limit cycle is an
oscillation around an unstable steady state. This is consistent with a supercritical Hopf bifurcation (Wiggins, 2003).
As the upper critical value of q[in] is approached, the stability analysis of Sect. 3.1 predicts that the steady state returns to stability. As this happens, we see that the amplitude of oscillations
only shrinks continuously back to zero for the smallest value of ${\overline{V}}_{p}$ considered (in which case the upper critical value corresponds to another supercritical Hopf bifurcation). In
Fig. 5a–c, the amplitude of stable periodic solutions continues to increase until values of q[in] near the upper critical value are reached. The amplitude then decreases slightly with a further
increase in q[in], before the periodic solution ceases to exist at a third critical value of q[in] that is larger than the threshold at which the steady state has become stable again, as shown in
more detail in the inset in Fig. 5b. (In technical terms, this is known as a saddle-node bifurcation of the Poincaré map of the dynamical system 2; see Wiggins, 2003.) Where this abrupt disappearance
of the stable periodic solution occurs, it is accompanied by an unstable periodic solution near the upper critical value of q[in], corresponding to a subcritical Hopf bifurcation. Figure 5d (see
inset) represents an exotic exception to this, where there are two stable periodic solutions for a small interval of q[in].
The nature of the Hopf bifurcations at the stability boundaries in Fig. 2 can be determined more quickly than by the numerical continuation method used above, using a weakly non-linear stability
analysis (see Sect. S3.3 of the Supplement). This allows us to map out in Fig. 2 where that boundary corresponds to a supercritical Hopf bifurcation, in whose vicinity a small-amplitude stable
periodic solution emerges (stability boundaries shown as a solid curve), and where the Hopf bifurcation is subcritical and the small-amplitude periodic solution is unstable (dashed stability
boundary). As in the few samples shown in Fig. 5, we see that the lower critical value of q[in] is always supercritical, while the upper critical value is generally subcritical, except at low V[p].
We can also compute the period of the drainage oscillations, that is, the recurrence interval of floods. Figure 6 shows the period of the stable periodic solutions shown in Fig. 5 as a function of
water supply rate q[in], using the same colouring scheme to distinguish different values of V[p]. Perhaps unsurprisingly, large inflow rates q[in] correspond to more rapid flood cycles, and large
reservoir volumes correspond to floods repeating more slowly, albeit with a larger amplitude.
There is a significant caveat here: in many cases, the limit cycle solutions we have computed predict that N becomes negative during the cycle of reservoir filling and draining. This is not something
that the models are designed for. Instead of the conduit expanding through creep, as occurs in the models (1) and (2) when N<0, we expect that the glacier should instead detach from the bed, and a
sheet flow of water initiates the outburst flood in this case; prototypes of models for this behaviour can be found in Schoof et al. (2012), Hewitt et al. (2012) and Tsai and Rice (2010).
It is important to know where in parameter space the outburst flood mechanism would in reality change away from conduit growth due to cavities becoming channel-like at the end of the refilling phase
and instead involve water separating ice from bed at vanishing effective pressure. The boundary between these two regimes should correspond to the location in parameter space where the minimum
effective pressure during the flood cycle is zero. Figure 7 shows that parameter regime boundary, computed numerically (Sect. S3.4 of the Supplement) and superimposed on the stability boundary plot
of Fig. 2. Clearly, sheet-flow-initiated floods (in which the glacier starts to float at the start of the outburst flood) are favoured at high water input rates and smaller reservoir volumes, where
the conduit is less able to adjust to the rapid refilling of the reservoir.
3.4Asymptotic solutions
We can also address limit cycle solutions through asymptotic methods in some parametric limits in our model. The most relevant limit for real glacier-dammed lakes is likely to be that of a relatively
large reservoir that is filled relatively slowly, but where water supply is not so small as to allow the conduit to be cavity-like in steady state (in which case the reservoir would be drained
steadily, without a flood cycle): this is the case of large V[p] and moderate but not small q[in] and is described in Appendix A and, in detail, in Sect. S4 of the Supplement. This region of
parameter space lies in the upper left of the unstable region of Fig. 7, between the near-vertical solid black line and the solid red curve.
In brief, the asymptotic solution confirms that there is a periodic flood cycle that the system very quickly settles into and that the flood cycle consists of three distinct stages. During the main
flood stage, the evolution of the conduit is rapid and dominated by dissipation-driven melt c[1]qΨ and creep closure v[c](S,N) in Eq. (2a), and water input q[in] to the lake is much smaller than
outflow q. This is followed by a long refilling phase in which the conduit has shrunk dramatically and therefore behaves as a cavity. Its size is dictated by a quasi-equilibrium in which the cavity
opening term v[o](S) balances the creep closure term v[c](S,N), leading to a slow opening of the conduit as the reservoir fills and N consequently drops. Outflow q from the reservoir during this
phase is insignificant, and the mass balance of the lake is dominated by inflow q[in]. The refilling phase is terminated by a flood initiation phase whose length is intermediate between the main
refilling phase and the rapid flood phase. During this initiation phase, the reservoir is still filling, but the conduit has enlarged sufficiently that dissipation-driven wall melting c[1]qΨ starts
to be significant.
Qualitatively, this solution is illustrated by the limit cycle in Figs. 3 and 4 and Fig. 9a. The asymptotic solution only becomes quantitatively accurate when V[p] is large enough to be physically
unrealistic for real glacier-dammed lakes (Sect. S4.4 of the Supplement); this is because the relatively large exponent n=3 in Glen's law makes the creep closure term quite sensitive to changes in N
when N is small, and the initiation phase of the floods is affected significantly by this.
One of the predictions of the asymptotic solution is that the amplitude of effective pressure oscillations should be insensitive to refilling rate q[in], except close to the Hopf bifurcation at which
oscillations are initiated. As a result, the flood recurrence period (which is essentially the time taken to refill the reservoir in the limit where reservoir drainage is fast) should simply be
inversely proportional to q[in]. Specifically, the result states that
$\begin{array}{}\text{(14)}& \begin{array}{rl}{t}_{\mathrm{period}}& \sim {\stackrel{\mathrm{̃}}{N}}_{f}{c}_{\mathrm{1}}^{\mathit{\alpha }/\left(n+\mathrm{1}-\mathit{\alpha }\right)}{c}_{\mathrm{2}}^
{-\mathrm{1}/\left(n+\mathrm{1}-\mathit{\alpha }\right)}{c}_{\mathrm{3}}^{\mathrm{1}/\left(n+\mathrm{1}-\mathit{\alpha }\right)}\\ & {\mathrm{\Psi }}_{\mathrm{0}}^{\left(\mathrm{1}+\mathrm{2}\mathit
{\alpha }\right)/\left[\mathrm{2}\left(n+\mathrm{1}-\mathit{\alpha }\right)\right]}{V}_{p}^{n/\left(n+\mathrm{1}-\mathit{\alpha }\right)}{q}_{\mathrm{in}}^{-\mathrm{1}},\end{array}\end{array}$
where ${\stackrel{\mathrm{̃}}{N}}_{f}$ is a dimensionless constant, with a value of 1.44 for the parameters of α and n chosen here, in the limit of a large flow path length L (Sect. S4.1 of the
Supplement). This asymptotic formula is overlaid onto the numerically computed periods in Fig. 6; it should be clear that the asymptotic formula performs poorly for the relatively moderate values of
V[p] used here. This should not be a surprise since Fig. 5 demonstrates that, for the same values of V[p], the amplitude of oscillations is in fact sensitive to the inflow rate q[in].
The same asymptotic solution also provides an estimate for the inflow rate q[in] at which zero effective pressure is first reached during the flood cycle, and our model ceases to be physically
realistic as described above. The estimate is given by an analysis of the flood initiation phase (Sect. 4.4 of the Supplement) as
$\begin{array}{}\text{(15)}& \begin{array}{rl}{q}_{\mathrm{in}}& \sim {\mathit{\gamma }}_{\mathrm{c}}{c}_{\mathrm{1}}^{\left(n+\mathrm{1}\right)/\left(\mathit{\alpha }n\right)}{c}_{\mathrm{2}}^{-\
mathrm{1}/n}{c}_{\mathrm{3}}^{\left(n+\mathrm{1}\right)/\left(\mathit{\alpha }n\right)}{\mathrm{\Psi }}_{\mathrm{0}}^{\mathrm{3}\left(n+\mathrm{1}\right)/\left(\mathrm{2}\mathit{\alpha }n\right)}\\ &
\left({u}_{\mathrm{b}}{h}_{\mathrm{r}}{\right)}^{-\left(\mathit{\alpha }-\mathrm{1}\right)\left(n+\mathrm{1}\right)/\left(\mathit{\alpha }n\right)}{V}_{p},\end{array}\end{array}$
where γ[c]≈0.25 for the values of α and n chosen here. This critical inflow rate as a function of V[p] is superimposed on Fig. 7. The formula above is in general an underestimate of the value of q
[in] at which zero effective pressures are reached but clearly gives the correct scaling of how that value relates to V[p].
We are also able to construct a second asymptotic solution for the opposite case of a large water supply rate q[in] (Sect. S5 of the Supplement). This solution is valid towards the right-hand edge of
the grey region in Fig. 7 and predicts rapid oscillations in the reservoir level (or effective pressure) whose amplitude slowly evolves to a steady value. These rapid oscillations however invariably
involve effective pressure changing from negative to positive. As discussed above, such solutions are clearly not physically viable, and the model must be amended to take account of ice–bed
4The spatially extended model
The analysis above provides a comprehensive picture of the lumped model (2). Here, we consider how well that lumped model represents the behaviour of its more complete, spatially extended counterpart
(1). We begin by recreating the stability boundary diagrams of Fig. 2. The method by which the latter were computed explicitly (Sect. S3.1 and S3.2 of the Supplement) cannot be applied directly to
the extended model (1), since we have no closed-form solution to a linearized version of the model analogous to Eq. (9). Instead, we grid the $\left({q}_{\mathrm{in}},{\overline{V}}_{p}\right)$
parameter space used in Fig. 2 finely. For each $\left({q}_{\mathrm{in}},{\overline{V}}_{p}\right)$ pair, we discretize the model (1), compute steady states and perform a linear stability analysis
numerically as described in Schoof et al. (2014, Appendix B); this allows us to delineate regions of instability, although in a less sophisticated way. We adapt that method for the transient
non-linear calculations in Figs. 9–11 by solving Eqs. (B1)–(B2) of the same Appendix in Schoof et al. (2014) using a backward Euler step.
Results are shown in Fig. 8. It is clear that the lumped model consistently underestimates the range of parameter values over which steady drainage is unstable. The onset of instability at the
transition from cavity- to channel-like conduit behaviour remains robust except for small domain lengths L (Fig. 8c), where the system appears to be unstable for combinations of small storage
capacities ${\overline{V}}_{p}$ and inflow rates q[in]. The lumped model typically underestimates instability at low storage capacities and at large flow path lengths L. The latter is particularly
significant since we have previously attributed stabilization to the effect of incipient reservoir drainage reducing the hydraulic gradient along the flow path and therefore reducing flow through the
While stabilization at large water input rates q[in] does eventually occur, this happens at values that can be several orders of magnitude larger than predicted by the lumped model, especially for
large flow path lengths L. Furthermore, for a given storage capacity ${\overline{V}}_{p}$, the lumped model predicts that there is a single interval of inflow rate values q[in] over which instability
occurs. The spatially extended model by contrast has two or more such intervals for most values of ${\overline{V}}_{p}$, with a narrow region of stability between (the diagonal bands of solid black
diamonds in Fig. 8a and b).
To understand this discrepancy better. we have solved for the non-linear evolution of the drainage system as described by the spatially extended model (1) for the parameter values indicated by
magenta circles in Fig. 8a.
The three smallest values of q[in] all correspond to unstable steady states in the lumped model (2), and we can compare spatially extended and lumped solutions. Figure 9 shows results. While the
spatially extended solution is an infinite-dimensional dynamical system and strictly speaking cannot be visualized using a phase plane, we can overlay plots of conduit size S(0,t) at the upstream end
of the conduit against effective pressure N(0,t) at the same location onto the S–N phase plot for the lumped model. This is shown in the top row of panels. In all three cases, the extended model
settles into a limit cycle, and for small and intermediate q[in], we find good agreement between extended and lumped models. This only breaks down as we approach the upper critical value of q[in] at
which the lumped model stabilizes.
The lower row of panels in Fig. 9 shows snapshots of N(x,t) against x for the corresponding limit cycle shown in the top panel of the same column. For the smaller two values of q[in], we see that
pressure gradients $\partial N/\partial x$ are moderate away from the glacier terminus and therefore do not contribute significantly to the hydraulic gradient Ψ. This was the basis for the reduction
of the spatially extended model (1) to the lumped form (2) and explains the good agreement. For larger q[in], effective pressure N(x,t) along the conduit starts to develop wave-like structures,
causing the approximation of negligible pressure gradients to break down, and the discrepancy between lumped and extended model grows.
The extended model remains unstable beyond the stability boundary of the lumped model, but our numerical solutions no longer support the conclusion that the system necessarily settles into a limit
cycle. Figure 10 shows the evolution of the system for V[p]=408m^3Pa^−1s^−1 (a lake with a surface area of 4km^2) and ${q}_{\mathrm{in}}=\mathrm{2.18}×{\mathrm{10}}^{\mathrm{3}}$m^3s^−1 as in
the magenta marker labelled “D” in Fig. 8a (the fact that this is an inflow rate of biblical proportions is not as relevant as it may at first seem, as we discuss in Sect. 5 below). Figure 10a and b
show the growth of rapid oscillations in the maxima of N and S along the flow path over time, but without the growth becoming bounded. Figure 10c and d show snapshots over the last two oscillations
in the computation. Note that the vertical scale in Fig. 10c is logarithmic. What we see is that conduit size develops an aneurysm-like, massively enlarged feature near the terminus, blocked by a
narrow constriction that requires an extremely large pressure gradient to overcome. The simulation eventually terminates when the solver fails to converge, and it is unclear whether this is merely a
computational problem or indicates that the true continuum solution becomes pathological or ceases to exist.
For even larger q[in], a narrow band of stable values is passed and a different instability ensues as shown in Fig. 11. This instability appears to lead to bounded growth, again marked by rapid
oscillations whose amplitude first grows and then saturates, and significant pressure gradients along the flow path.
Once again, a key observation is that the spatially extended model predicts negative effective pressures long before any kind of limit cycle is reached for all but one of the solutions shown and that
the discrepancies between lumped and spatially extended models occur only where this is the case. Moreover, they occur for water supply rates to the reservoir that far exceed values that would be
plausible for typical glacier-dammed lake systems: the lumped model appears to be robust for the latter. This includes the prediction by the lumped system of where overpressurization of the drainage
system (N<0 in the limit cycle) first occurs.
This does not render some of the more exotic instabilities shown in Figs. 9–11 irrelevant: the parameter values we have chosen here were deliberately chosen to reflect typically large reservoirs
dammed by low-angled and large glaciers. Much smaller reservoirs in shorter, more steeply angled glaciers are liable to give rise to some of these exotic instabilities, even though they lie outside
the reach of standard glacier-dammed lakes.
5.1Glacier-dammed lakes and small reservoirs
We have shown that a drainage system that switches between cavity-like and channel-like behaviour spontaneously is capable of supporting periodic outburst floods from a glacier-dammed reservoir. At
issue here is the initiation of the flood: as discussed by Ng (1998) and Fowler (1999), if the conduit behaves purely as a Röthlisberger-type channel, kept open only by melting due to heat dissipated
in the turbulent flow of water through the channel, then the initiation of floods can be delayed more and more, without a limit cycle emerging. Both of these authors proposed that englacially routed
water supplied to the channel should keep it open at a minimum size, and the migration of the flow divide within the channel as the lake fills then becomes the control on flood initiation. Our theory
differs from theirs in the sense that no englacial water supply is necessary and the lake can continuously leak water through a linked-cavity-type drainage system that spontaneously becomes
channel-like and undergoes runaway enlargement through dissipation-driven melting as the lake fills.
At the heart of the oscillatory behaviour of glacier-dammed lakes is that runaway growth, or instability, of a channel-like conduit, which prevents steady discharge of the water supplied to the
reservoir (Nye, 1976; Ng, 1998; Fowler, 1999). As described in Schoof et al. (2014) and Sect. 3.1 above, this instability does not occur when the conduit draining the reservoir acts as a set of
linked cavities: when these are capable of steadily draining the water supplied to the reservoir, no outburst floods occur.
For typical glacier-dammed lakes, moderate inflows q[in] exceed the drainage capacity of a linked-cavity system, but the large storage capacity V[p] ensures that the lake fills slowly compared with
the timescale over which a basal conduit can evolve. The cycle of filling and draining then follows the characteristic sequence of a long filling period in which outflow from the lake is negligible,
followed by a brief onset period in which the conduit starts to experience significant melt-driven enlargement but the lake level continues to rise, as well as an even shorter outburst flood in which
inflow to the lake is dwarfed by drainage through the subglacial conduit. For a given lake size V[p], larger inflow rates will slowly increase the amplitude of the lake level fluctuation during the
flood cycle (Fig. 5), while significantly shortening the length of the flood cycle (Fig. 6).
As in Ng (1998) and Fowler (1999), our flood initiation mechanism can also be too slow to respond to water input and lead to our models (1) and (2) predicting negative effective pressures at the
upstream end of the conduit: in that case, the glacier ought to float and flood initiation is likely to take the form a sheet flow that subsequently channelizes, as explored in Flowers et al. (2004),
Schoof et al. (2012) and Hewitt et al. (2012). This effect is not included in our models here, but we are at least able to give an approximate expression (valid in the limit of very large reservoir
sizes V[p]) for the water supply q[in] at which the change in flood initiation mechanism occurs (switching between unstable conduit growth and partial glacier flotation and a sheet flow); this is
Eq. (15). Adapting our model to account for this alternative flood initiation mechanism is the obvious next step to take. One straightforward adaptation of the lumped model to this effect, inspired
by the model in Hewitt et al. (2012), is the following: if we suppose that lake level cannot rise beyond the point at which N=0, then the opening of an ice–bed gap must be such as to permit outflow
balancing inflow to the lake at that point, which forces us to alter the constitutive relation for q once N=0. We can similarly build the possibility of flood termination due to the lake running dry
at an upper bound on effective pressure at $N={N}_{max}={\mathit{\rho }}_{\mathrm{i}}g\left(s\left(\mathrm{0}\right)-b\left(\mathrm{0}\right)\right)$ into the model. Once the lake is fully empty,
discharge q in the conduit will be the lesser of inflow into the conduit q[in] (corresponding to a partially filled conduit) and the discharge the conduit would carry if it were completely filled
with water. The appropriately adapted lumped model (2) is
The equivalent for the spatially extended model (1) is the channel model in Hewitt et al. (2012), with a reservoir at the upstream end. Importantly, the stability of steady-state solutions as in
Sects. 3.1 and 4 is unchanged by these adaptations unless the steady state corresponds to an empty lake basin and a partially filled conduit, in which case the conduit is presumably stable since
discharge in the conduit does not increase when its size is increased. The upper and lower bounds on flux introduced in the model (16) only affects the non-linear dynamics of the system, once the
amplitude gets large enough. We leave the non-linear analysis of the adapted model to future work.
We have also investigated a mechanism previously identified in Schoof et al. (2014), by which flow out of a reservoir can be stabilized when the storage capacity in the reservoir is relatively small,
so that typical drainage rates (comparable to the inflow rate q[in]) lead to adjustment of effective pressure in the reservoir much faster than the conduit can evolve due to wall melting. The rapid
response of reservoir water levels can have a large enough effect on hydraulic gradients to stabilize the flow. This is true at least for a short enough drainage system. Importantly, this happens at
water throughput rates that are completely unrealistic for typical, large glacier lakes (and even these rates are underestimated by a simplified, lumped model as described in Sect. 4).
The reason why this stabilization mechanism is relevant is that it may explain why much smaller water reservoirs that are typically not recognized as lakes but nonetheless provide storage capacity
(such as large moulins) do not invariably generate outburst-flood type behaviour. The stability diagram of Fig. 8 was generated for parameter values chosen to represent single reservoirs dammed by
glaciers with a small (1:50) surface slope. For the more limited flow paths of 50 and 10km length, any small reservoir like a large moulin (say, with a maximum side length of 10m, or a base area of
V[p] of no more than 10^−4km^2) is stable for even small discharge rates of 1m^3s^−1; only a very long flow path of 250km combined with a single such reservoir could be unstable (Fig. 8b) with
(for such a large moulin) moderate flow rates of 10m^3s^−1.
The picture changes significantly when we look at shorter, steeper glaciers. We can reduce the size of the parameter space that we need to sample in order to understand different glacier geometries
by scaling the problem (Sect. S2.2 of the Supplement). If we change the background hydraulic gradient Ψ[0] (representing steepness) by a factor χ[Ψ] and the cavity opening rate v[o] (representing
sliding velocity and bed roughness) by a factor χ[v], the effect on stability is equivalent to keeping Ψ[0] and v[o] fixed but instead changing q[in] by a factor of ${\mathit{\chi }}_{\mathrm{v}}^{-\
mathrm{1}}{\mathit{\chi }}_{\mathrm{\Psi }}$, while simultaneously changing V[p] by a factor of ${\mathit{\chi }}_{\mathrm{v}}^{\left[\mathit{\alpha }-\left(n+\mathrm{1}\right)\right]/\left(\mathit{\
alpha }n\right)}{\mathit{\chi }}_{\mathrm{\Psi }}^{\left[\mathrm{2}\mathit{\alpha }n+\mathrm{3}\left(n+\mathrm{1}\right)\right]/\left(\mathrm{2}\mathit{\alpha }n\right)}={\mathit{\chi }}_{\mathrm{v}}
^{-.73}{\mathit{\chi }}_{\mathrm{\Psi }}^{\mathrm{2.6}}$ and L by a factor of ${\mathit{\chi }}_{\mathrm{v}}^{-\left(\mathit{\alpha }-\mathrm{1}\right)/\left(\mathit{\alpha }n\right)}{\mathit{\chi }}
_{\mathrm{\Psi }}^{\mathrm{1}-\mathrm{3}/\left(\mathrm{2}\mathit{\alpha }n\right)}={\mathit{\chi }}_{\mathrm{v}}^{-.066}{\mathit{\chi }}_{\mathrm{\Psi }}^{.6}$.
The most significant impact here is that of steepening Ψ on the volume V[p] at which stabilization or destabilization occurs. Consider the concrete example of a 10km long glacier (Fig. 8c) and
steepening its surface slope by a factor 8.5 from 1:50 to 17:100, while also reducing the cavity opening rate v[o] by a factor of three due to slower sliding. In terms of stability, this is
equivalent to keeping the 1:50 surface slope and original cavity opening rate but instead increasing q[in] by a factor of 25.5, increasing storage capacity V[p] by a factor 584 and increasing the
glacier length L by a factor 3.88 (that is, to 38.8km). In other words, we would go from Fig. 8c to something that resembles Fig. 8a (in which L=50km rather than 38.8km), but with a flux of 25.5m
^3s^−1 as indicated by axis labels in the figure representing an actual flux of 1m^3s^−1 and a volume V[p] of $\mathrm{5.84}×{\mathrm{10}}^{-\mathrm{2}}$km^2 actually representing a value of 10^
−3km^2. At this point, a relatively modest 10m-by-10m reservoir may become unstable at reasonable water fluxes q[in], being located in parameter space towards the right-hand end of the stability
boundary marked by solid circles in Fig. 8a. As a result, some of the more exotic manifestations of instability explored in Sect. 2.1 are more likely to be within reach of typical water input rates.
Figure 12 shows the actual stability diagram for this steeper glacier using the same plotting scheme as Fig. 8.
As our discussion above indicates, relatively small, isolated storage volumes could lead to instabilities either for steeper glaciers or for very long flow paths. The latter in particular are however
an unlikely scenario: take the example of 10m-by-10m moulin at the head of a 250km flow path. In reality, there would most likely be many moulins spread out along the flow path instead. With this
as motivation, we discuss a second exotic instability next, in which we opt for the opposite endmember of possible storage geometries by spreading out storage capacity evenly along the flow path.
Much of the necessary groundwork is already in place from the analysis in Sect. 3.1.
5.2Distributed storage: a modification of Nye's instability
In Sect. 2, we formulated a model for a single reservoir, for which Nye's instability predicts the onset of oscillations when the conduit becomes channel-like. In Schoof et al. (2014, Sect. 5), the
case of spatially spread out storage capacity, for instance in the form of numerous basal crevasses arrayed along the flow path, was considered in addition to a single reservoir. A numerical linear
stability analysis was used to demonstrate that instability can also occur for cavity-like conduits (which are generally stable for a single reservoir) for the case of such distributed water storage.
The resulting oscillations (see Fig. 8 of Schoof et al., 2014; note that the figure captions for Figs. 8 and 9 are switched in the published paper) are likely to be analogous to those found in Dow
et al. (2016) for the case of no localized subglacial lake. As in the case of a single reservoir, Schoof et al. (2014) find that there is a finite range of values q[in] for which instability then
occurs, only that this extends to lower values of q[in], where the conduit is cavity-like.
Here we sketch how to build on Sect. 3.1 to shed further light on the results in Schoof et al. (2014), which were built on the model
$\begin{array}{}\text{(17a)}& \frac{\partial v\left(N\right)}{\partial t}+\frac{\partial \stackrel{\mathrm{̃}}{q}}{\partial x}& =\mathrm{0},\text{(17b)}& \frac{\partial S}{\partial t}& ={c}_{\mathrm
{1}}q\mathrm{\Psi }+{v}_{\mathrm{o}}\left(S\right)-{v}_{\mathrm{c}}\left(S,N\right),\text{(17c)}& \stackrel{\mathrm{̃}}{q}& =q\left(S,\mathrm{\Psi }\right),\text{(17d)}& \mathrm{\Psi }& ={\mathrm{\Psi
}}_{\mathrm{0}}+\frac{\partial N}{\partial x}.\end{array}$
Here v(N) is a decreasing function of N that describes storage of water per unit length of the conduit, and the rest of the notation used replicates that of the single-reservoir model (1). Below, we
will denote by
$\begin{array}{}\text{(18)}& {v}_{p}=-\frac{\mathrm{d}v}{\mathrm{d}N},\end{array}$
the equivalent of V[p] in Eq. (1). This model also requires boundary conditions; an obvious choice is a prescribed flux q=q[in] at the inflow x=0, with no actual reservoir there, and again N=0 at the
terminus x=L.
With a finite domain size and the proposed boundary conditions above, a steady-state solution to the model (17) will in general have spatial structure as in Schoof et al. (2014), and a linearization
around that steady state, analogous to Sect. 3.1, will lead to a boundary value problem that is not amenable to a closed-form solution. To avoid this, we concentrate here on shorter length scales and
assume that we can use the model (17) with periodic boundary conditions at these scales. This yields a spatially uniform steady-state solution defined implicitly by Eq. (7a) combined with ${q}_{\
mathrm{in}}=q\left(\overline{S},\overline{\mathrm{\Psi }}\right)$ and $\overline{\mathrm{\Psi }}={\mathrm{\Psi }}_{\mathrm{0}}$, where q[in] is the prescribed flux through the system. This is closely
analogous to Eq. (7) but simpler, as the hydraulic gradient here does not contain the gradient term retained in Eq. (7c).
Linearizing as $N=\overline{N}+{N}^{\prime }\mathrm{exp}\left(ikx+\mathit{\lambda }t\right)$, $S=\overline{S}+{S}^{\prime }\mathrm{exp}\left(ikx+\mathit{\lambda }t\right)$, we find an analogue to the
problem (8) as
$\begin{array}{}\text{(19a)}& \begin{array}{rl}& -{\overline{v}}_{p}\mathit{\lambda }{N}^{\prime }+ik{q}_{S}{S}^{\prime }-{k}^{\mathrm{2}}{q}_{\mathrm{\Psi }}{N}^{\prime }=\mathrm{0},\\ & \mathit{\
lambda }{S}^{\prime }-\left({c}_{\mathrm{1}}{q}_{S}\overline{\mathrm{\Psi }}+{v}_{\mathrm{o},S}{S}^{\prime }-{v}_{\mathrm{c},S}\right){S}^{\prime }\end{array}\text{(19b)}& -\left[ik{c}_{\mathrm{1}}\
left({q}_{\mathrm{\Psi }}\overline{\mathrm{\Psi }}+\overline{q}\right)-{v}_{\mathrm{c},N}\right]{N}^{\prime }=\mathrm{0},\end{array}$
using the same notation as in Sect. 3.1. The eigenvalue λ is again given by an equation of the form (9), but now with a[1] and a[2] given by
$\begin{array}{}\text{(20a)}& {a}_{\mathrm{1}}& =\left({c}_{\mathrm{1}}{q}_{S}\overline{\mathrm{\Psi }}+{v}_{\mathrm{o},S}-{v}_{\mathrm{c},S}\right)-{\overline{v}}_{p}^{-\mathrm{1}}{k}^{\mathrm{2}}
{q}_{\mathrm{\Psi }},\text{(20b)}& {a}_{\mathrm{2}}& ={\overline{v}}_{p}^{-\mathrm{1}}\left\{{k}^{\mathrm{2}}\left[{c}_{\mathrm{1}}{q}_{S}\overline{q}+{q}_{\mathrm{\Psi }}\left({v}_{\mathrm{c},S}-{v}
The only term that differs intrinsically from Eq. (10) is a[2], which now has an imaginary part. We see that the real part of a[2] is still invariably positive, so the real part of a[1]−4a[2] is
smaller than a[1]. However, we can no longer conclude that the real part of λ has the same sign as a[1].
We can identify two mechanisms for instability. The first is essentially the same as Nye's melt–drainage feedback: the more storage capacity there is, the smaller all the terms containing k are in
Eq. (20) for a given wavenumber k, and the more likely the first term $\left({c}_{\mathrm{1}}{q}_{S}+{v}_{\mathrm{o},S}-{v}_{\mathrm{c},S}\right)$ in the definition of a[1] is to dominate the
eigenvalue λ, leading to instability for a channel-like system. To increase the size of the potentially stabilizing terms therefore requires larger wavenumbers k, and a larger range of short
wavelengths is likely to be unstable.
A second instability mechanism can occur if the conduit is cavity-like with $\left({c}_{\mathrm{1}}{q}_{S}+{v}_{\mathrm{o},S}-{v}_{\mathrm{c},S}\right)<\mathrm{0}$ and therefore a[1]<0. It is still
possible to have an eigenvalue with a positive real part and hence instability, because a[2] has a non-zero imaginary part. One particular case in which an instability due to this term may occur is
with limited storage capacity (so ${\overline{v}}_{p}^{-\mathrm{1}}$ is large) and at an intermediate wavelengths range (so k is small but not too small). Then it is possible, purely by controlling
the size of the storage capacity ${\overline{v}}_{p}$, to ensure that ${a}_{\mathrm{1}}^{\mathrm{2}}-\mathrm{4}{a}_{\mathrm{2}}\sim -i\mathrm{4}{\overline{v}}_{p}^{-\mathrm{1}}{q}_{S}{v}_{\mathrm
{c},N}k$ and
$\begin{array}{}\text{(21)}& \mathit{\lambda }\sim ±\left(\mathrm{1}-i\right)\sqrt{\frac{{\stackrel{\mathrm{̃}}{v}}_{p}^{-\mathrm{1}}k{q}_{S}{v}_{\mathrm{c},N}}{\mathrm{2}}}.\end{array}$
Full details can be found in Sect. 6 of the Supplement. Choosing the plus (+) sign ensures an eigenvalue with a positive real part. Importantly, this instability corresponds to a growing wave that
propagates, in this case downstream as the imaginary part of λ is then negative. It is also of the same size as the real part, so propagation is not slow. This unstable wave is not the result of
Nye's instability as the conduit is cavity-like. Instead, it is the result of an interaction between the dependence of the conduit closing rate on effective pressure and the dependence of water
drainage (which affects effective pressure through water storage) on conduit size.
To make these considerations more concrete, Fig. 13 shows the stability diagram for the system (17) for the parameter values used in Fig. 8, but with the upstream reservoir V[p] replaced by
distributed storage v[p] as indicated on the vertical axis. Compared with Fig. 8, it should be clear that there are larger regions of parameter space that exhibit instability for cavity-like conduits
(to the left of the vertical portion of the red stability boundary). As in the case of a single reservoir, a larger domain length is destabilizing, but a larger total storage volume is in general
necessary to cause instability for channelized systems compared to a localized reservoir.
For instance, for longest flow path of L=250km, one 10m-by-10m moulin per kilometre of flow path (equivalent to ${v}_{p}={\mathrm{10}}^{-\mathrm{4}}$km) is marginally unstable at q[in]=10m^3s^
−1 in Fig. 13b, while a single such moulin at the upstream end of the conduit is unstable at even larger q[in]. Put another way, compared with a single reservoir, distributed storage favours
instability at smaller q[in]. The possibility of oscillatory flow occurring at low q[in] outside of the melt season due to this instability was previously explored in Schoof et al. (2014).
As demonstrated previously using a much more restricted sweep of parameter space in Schoof et al. (2014), we have shown that drainage systems capable of switching spontaneously between channel- and
cavity-like behaviour are stable in the presence of a localized water reservoir at low and high water throughput, with an unstable intermediate range of water fluxes. In that range, spontaneous
oscillations in reservoir level will occur, driven by Nye's (1976) instability mechanism driving outburst floods. These outburst floods turn out to be regular, periodic oscillations in water level at
least at low-to-moderate water input, where a simplified, lumped model of reservoir drainage generally reproduces the results of a more sophisticated, spatially extended drainage model. At high water
throughput rates, our results are more equivocal as to the emergence of such limit cycles in the model used; in any case, the model necessarily breaks down physically (as opposed to mathematically)
because it predicts that the oscillations will invariably reach negative effective pressures and therefore flotation of the ice dam at such large throughput rates.
It is worth pointing out that part of our focus has been on the emergence of limit cycle solutions in order to identify how the flood initiation mechanism can prevent flood magnitude from increasing
progressively from cycle to cycle as observed in Ng (1998) and Fowler (1999), as well as to present an alternative mechanism for suppressing the continued growth in amplitude from that considered by
Fowler (1999), whose lake is necessarily sealed between floods. The alternative mechanism here is the ability of a cavity-like drainage system that remains open during the reservoir recharge phase to
switch to a channelized drainage mode.
The fact that we illustrate this by showing evolution towards a limit cycle is not at odds with the chaotic behaviour observed by Kingslake (2015) using a variant of Fowler's (1999) model: the
chaotic behaviour is intrinsically the result of a time-varying water input to the reservoir, which we have not studied in this paper. It is entirely possible, and in fact likely, that our model will
also behave chaotically under such time-varying water input rates q[in](t). The point of demonstrating that limit cycles emerge was rather to underline that the growth of Nye's (1976) instability is
bounded in our model without having to appeal to additional physics.
The lower cut-off to the drainage instability that leads to outburst floods corresponds to the drainage system switching to a cavity-like state under steady flow conditions when water input to the
reservoir can be drained steadily by those cavities. The mechanism for the cut-off at high water throughput rates is harder to identify. The lumped model predicts that the high sensitivity of water
level in the reservoir to the evolution of the draining conduit will induce water pressure gradients that reduce flow as the conduit grows and therefore suppress its enlargement due to heat
dissipation. The threshold at which this happens is however significantly underestimated by the lumped model relatively to its spatially extended counterpart, which develops more wave-like
instabilities at higher water throughput rates.
In closing, we have also investigated how wave-like instabilities can occur when the water reservoir is not localized but spread out or “distributed” along the flow path (for instance, in the form of
many small reservoirs like basal crevasses). This type of instability was first observed in Schoof et al. (2014) and persists even where water throughput is insufficient to lead to channel-like
behaviour. Adapting the stability analysis performed on a model with localized storage, we have shown that Nye's instability persists but also that a second instability mechanism emerges, in which a
phase shift between water storage and flux that causes water to accumulate in regions where effective pressure is already low arises.
Future work is likely to focus on capturing the role of overpressurization of the drainage system in initiating and mediating the instability driving outburst floods, since flood initiation at water
pressures below flotation is confined to a relatively small part of parameter space, and the model predicts that reaching zero effective pressure and initiation by partial flotation of the ice dam is
likely to be common.
Appendix A:Asymptotic solutions for large lakes with moderate inflow
This appendix provides only a brief sketch of the derivation of an asymptotic solution for a limit cycle solution in the case where the reservoir volume is large and inflow is sufficient to ensure
that the reservoir cannot be drained by a cavity-like conduit but also that the conduit size evolves much faster than the timescale over which the reservoir fills. Full details are given in Sect. S4
of the Supplement.
The solution is developed from a parametric limit of the model (2) with Eq. (1g), where we assume S[0]=∞. We scale as ${N}^{**}=N/\left[N{\right]}^{\prime }$, ${S}^{**}=S/\left[S{\right]}^{\prime }$,
${t}^{**}=t/\left[t\right]$, where the scales are defined through $\left[S{\right]}^{\prime }/\left[t{\right]}^{\prime }={c}_{\mathrm{1}}{c}_{\mathrm{3}}\left[S{{\right]}^{\prime }}^{\mathit{\alpha
}}{\mathrm{\Psi }}_{\mathrm{0}}^{\mathrm{3}/\mathrm{2}}={c}_{\mathrm{2}}\left[S{\right]}^{\prime }\left[N{{\right]}^{\prime }}^{n}$ and ${V}_{p}\left[N{\right]}^{\prime }/\left[t{\right]}^{\prime }=
{c}_{\mathrm{1}}\left[S{{\right]}^{\prime }}^{\mathit{\alpha }}{\mathrm{\Psi }}_{\mathrm{0}}^{\mathrm{1}/\mathrm{2}}$. Substituting and omitting the asterisks immediately, we obtain
$\begin{array}{}\text{(A1a)}& \stackrel{\mathrm{˙}}{S}& ={S}^{\mathit{\alpha }}|\mathrm{1}-\mathit{u }N{|}^{\mathrm{3}/\mathrm{2}}+\mathit{\delta }-S|N{|}^{n-\mathrm{1}}N,\text{(A1b)}& \stackrel{\
mathrm{˙}}{N}& =-\mathit{ϵ}+{S}^{\mathit{\alpha }}|\mathrm{1}-\mathit{u }N{|}^{-\mathrm{1}/\mathrm{2}}\left(\mathrm{1}-\mathit{u }N\right),\end{array}$
where $\mathit{u }=\left[N{\right]}^{\prime }/\left({\mathrm{\Psi }}_{\mathrm{0}}L\right)$, $\mathit{\delta }={u}_{\mathrm{b}}{h}_{\mathrm{r}}/\left({c}_{\mathrm{2}}\left[S{\right]}^{\prime }\left[N
{{\right]}^{\prime }}^{n}\right)$ and $\mathit{ϵ}={q}_{\mathrm{in}}/\left({c}_{\mathrm{1}}\left[S{{\right]}^{\prime }}^{\mathit{\alpha }}{\mathrm{\Psi }}_{\mathrm{0}}^{\mathrm{1}/\mathrm{2}}\right)$.
We assume that the exponents n and α satisfy $n+\mathrm{1}>\mathit{\alpha }>\mathrm{1}$, in which case the asymptotic solution we develop is valid when
$\mathit{\delta }\ll \mathit{ϵ}\mathit{\lesssim }{\mathit{\delta }}^{\left(\mathit{\alpha }-\mathrm{1}\right)\left(n+\mathrm{1}\right)/\left(\mathit{\alpha }n\right)}\ll \mathrm{1}$
and ν≲1.
The main flood phase is described by omitting terms of O(δ) and O(ϵ)
$\begin{array}{}\text{(A2a)}& \stackrel{\mathrm{˙}}{S}& ={S}^{\mathit{\alpha }}|\mathrm{1}-\mathit{u }N{|}^{\mathrm{3}/\mathrm{2}}-S|N{|}^{n-\mathrm{1}}N,\text{(A2b)}& \stackrel{\mathrm{˙}}{N}& ={S}^
{\mathit{\alpha }}|\mathrm{1}-\mathit{u }N{|}^{-\mathrm{1}/\mathrm{2}}\left(\mathrm{1}-\mathit{u }N\right),\end{array}$
where matching with the initiation phase described later requires that $\left(N,S\right)\to \left(\mathrm{0},\mathrm{0}\right)$ as $t\to -\mathrm{\infty }$. The transformation $P=N/S$ allows the
system (A2) to be recast in such a way as to demonstrate that the orbit into the fixed point (0,0) is unique, so there is only one flood phase solution. The orbit terminates at a finite $N={\stackrel
{\mathrm{̃}}{N}}_{f}$ as t→∞; this then sets the amplitude of the lake level fluctuations that give the asymptotic formula (14) for the flood cycle period.
The refilling phase is described by the rescaling $\stackrel{\mathrm{̃}}{N}=N$, $\stackrel{\mathrm{̃}}{S}={\mathit{\delta }}^{-\mathrm{1}}S$ and $\stackrel{\mathrm{̃}}{t}=\mathit{ϵ}\left(t-{t}_{f}\
right)$, where t[f] is the time of the last flood. At leading order,
$\begin{array}{}\text{(A3a)}& \mathrm{0}& =\mathrm{1}-\stackrel{\mathrm{̃}}{S}|\stackrel{\mathrm{̃}}{N}{|}^{n-\mathrm{1}}\stackrel{\mathrm{̃}}{N},\text{(A3b)}& \frac{\mathrm{d}\stackrel{\mathrm{̃}}{N}}{\
mathrm{d}\stackrel{\mathrm{̃}}{t}}& =-\mathrm{1};\end{array}$
conduit size is quasi-steady and cavity-like, while lake level evolves purely because of inflow.
The refilling phase ends as N→0 and cavity size becomes large. The relevant rescaling becomes
$\begin{array}{ll}\stackrel{\mathrm{^}}{N}& ={\mathit{\delta }}^{-\left(\mathit{\alpha }-\mathrm{1}\right)/\left(\mathit{\alpha }n\right)}\stackrel{\mathrm{̃}}{N},\phantom{\rule{1em}{0ex}}\stackrel{\
mathrm{^}}{S}={\mathit{\delta }}^{\left(\mathit{\alpha }-\mathrm{1}\right)/\mathit{\alpha }}\stackrel{\mathrm{̃}}{S},\\ \stackrel{\mathrm{^}}{t}& ={\mathit{\delta }}^{-\left(\mathit{\alpha }-\mathrm
{1}\right)/\left(\mathit{\alpha }n\right)}\left(\stackrel{\mathrm{̃}}{t}-{N}_{f}\right)\end{array}$
and gives at leading order
$\begin{array}{}\text{(A4a)}& \mathit{ϵ}{\mathit{\delta }}^{-\left(\mathit{\alpha }-\mathrm{1}\right)\left(n+\mathrm{1}\right)/\left(\mathit{\alpha }n\right)}\frac{\mathrm{d}\stackrel{\mathrm{^}}{S}}
{\mathrm{d}\stackrel{\mathrm{^}}{t}}& ={\stackrel{\mathrm{^}}{S}}^{\mathit{\alpha }}+\mathrm{1}-\stackrel{\mathrm{^}}{S}|\stackrel{\mathrm{^}}{N}{|}^{n-\mathrm{1}}\stackrel{\mathrm{^}}{N},\text
{(A4b)}& \frac{\mathrm{d}\stackrel{\mathrm{^}}{N}}{\mathrm{d}\stackrel{\mathrm{^}}{t}}& =-\mathrm{1},\end{array}$
where we have assumed $\mathit{ϵ}\sim {\mathit{\delta }}^{\left(\mathit{\alpha }-\mathrm{1}\right)\left(n+\mathrm{1}\right)/\left(\mathit{\alpha }n\right)}$; the alternative case of ϵ∼δ is described
in the Supplement. The key to Eq. (A4) is that it predicts finite-time blow-up in $\stackrel{\mathrm{^}}{S}$ at some finite $\stackrel{\mathrm{^}}{N}={\stackrel{\mathrm{^}}{N}}_{\mathrm{c}}$ that
depends purely on the value of $\mathit{ϵ}{\mathit{\delta }}^{-\left(\mathit{\alpha }-\mathrm{1}\right)\left(n+\mathrm{1}\right)/\left(\mathit{\alpha }n\right)}$. This is the smallest value of
effective pressure (rescaled, of course) that is reached during the drainage cycle. The larger $\mathit{ϵ}\sim {\mathit{\delta }}^{\left(\mathit{\alpha }-\mathrm{1}\right)\left(n+\mathrm{1}\right)/\
left(\mathit{\alpha }n\right)}$, the smaller and eventually more negative ${\stackrel{\mathrm{^}}{N}}_{\mathrm{c}}$ becomes; there is therefore a critical value of $\mathit{ϵ}\sim {\mathit{\delta }}^
{\left(\mathit{\alpha }-\mathrm{1}\right)\left(n+\mathrm{1}\right)/\left(\mathit{\alpha }n\right)}$ at which ${\stackrel{\mathrm{^}}{N}}_{\mathrm{c}}=\mathrm{0}$. When rendered in dimensional form,
this value gives the formula (15).
The MATLAB code used in the computations reported is included in the Supplement, except for the code used to solve the transient calculations displayed in Figs. 9–11, which is included in the
Supplement to Rada and Schoof (2018).
The author declares that there is no conflict of interest.
Discussions with Rob Vogt, Ian Hewitt and Garry Clarke are gratefully acknowledged, as are reviews by Mauro Werder and an anonymous referee. This work was supported by an NSERC Discovery Grant.
This research has been supported by the Natural Sciences and Engineering Research Council of Canada (grant no. RGPIN-2018-04665).
This paper was edited by Daniel Farinotti and reviewed by Mauro Werder and one anonymous referee.
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2013.a, b
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SOP Archives - Notesformsc
The adder is a combinational circuit that add binary digits for arithmetic computation. A combinational circuit is a kind of digital circuit that has an input, a logic circuit and an output. For
variables, there are combinations of input variables and for each input combination, there is one and only one output. Therefore, output from … Read more
Combinational Circuit – Questions/Solutions
In this post, you will learn example problems from combinational circuits. These problems help in minimizing Boolean functions and constructing logic circuit diagrams. The solution to the problems
are given in step-by-step manner with explanation wherever possible. Q1. Simplify the Boolean function using K-MAP technique. There are 4 variables in this function. First, we construct … Read more
Understanding Sum of Minterms and Product of Maxterms
A Boolean function is expressed in two form. Sum of Minterms Product of Maxterms Sum of Minterms x’ y’ z , x y’ z’ , x y’ z , x y z’ , x y z , gives 1 as output in the above Truth Table. Literal – x,
y, A, b etc is … Read more | {"url":"https://notesformsc.org/tag/sop/","timestamp":"2024-11-07T15:15:07Z","content_type":"text/html","content_length":"103888","record_id":"<urn:uuid:0b983dcb-fb53-45a2-9808-758efc345ba0>","cc-path":"CC-MAIN-2024-46/segments/1730477028000.52/warc/CC-MAIN-20241107150153-20241107180153-00072.warc.gz"} |
Wolfram|Alpha Examples: Common Core Math: Grade 4: Algebraic Thinking
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Common Core Math: Grade 4: Algebraic Thinking
In fourth grade, students gain fluency with whole-number arithmetic, including by using addition, subtraction, multiplication and division to solve multistep word problems. Students write equations
in which a variable stands for the unknown quantity. Students determine multiples and divisors of whole numbers and distinguish between prime and composite numbers. Students also generate and analyze
patterns of numbers and build skills in mental computation and estimation.
Common Core Standards
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Factors & Multiples
Find factors and multiples of whole numbers.
Find all divisors of a number (CCSS.Math.Content.4.OA.B.4):
Determine if a number is a multiple of a given digit (CCSS.Math.Content.4.OA.B.4):
Determine if a number is prime (CCSS.Math.Content.4.OA.B.4):
Multiple Operations
Perform addition, subtraction, multiplication and division.
Represent multiplication as a comparison (CCSS.Math.Content.4.OA.A.1):
Solve an equation for an unknown (CCSS.Math.Content.4.OA.A.2):
Solve multistep equations (CCSS.Math.Content.4.OA.A.3):
Generate and analyze patterns.
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secA1+secA=1−cosAsin2 A
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Question Text स्सेकेतः वाम पक्ष और दाँया पक्ष
Updated On Jan 25, 2023
Topic Trigonometry
Subject Mathematics
Class Class 12
Answer Type Video solution: 1
Upvotes 139
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Centripetal Acceleration
`|a| = | "v" |^2/ "r" `
Enter a value for all fields
The Centripetal Acceleration calculator computes the centripetal accelerationCircular Motion which is the acceleration directed toward the center of a circular motion with constant angular velocity.
INSTRUCTION: Choose units and enter the following:
• |v| - magnitude of the tangential velocity
• (r) - radius of the constant circular motion
Magnitude of Centripetal Acceleration |a|: The calculator the acceleration in meters per second squared. However, the user can automatically convert this to other acceleration units via the
pull-down menu.
The Math / Science
Note than if a mass is moving in this circular motion and is affected by this acceleration due to changing angular velocity, that mass will feel a force along the vector direction of `vec(a_"rad")`.
The formula for the magnitude of acceleration in uniform circular motion is:
`|a| = |v|^2/r`
Other Circular Motion Functions
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New Breakthrough Brings Matrix Multiplication Closer to Ideal | Quanta Magazine
Merrill Sherman/Quanta Magazine
Computer scientists are a demanding bunch. For them, it’s not enough to get the right answer to a problem — the goal, almost always, is to get the answer as efficiently as possible.
Take the act of multiplying matrices, or arrays of numbers. In 1812, the French mathematician Jacques Philippe Marie Binet came up with the basic set of rules we still teach students. It works
perfectly well, but other mathematicians have found ways to simplify and speed up the process. Now the task of hastening the process of matrix multiplication lies at the intersection of mathematics
and computer science, where researchers continue to improve the process to this day — though in recent decades the gains have been fairly modest. Since 1987, numerical improvements in matrix
multiplication have been “small and … extremely difficult to obtain,” said François Le Gall, a computer scientist at Nagoya University.
Now, three researchers — Ran Duan and Renfei Zhou of Tsinghua University and Hongxun Wu of the University of California, Berkeley — have taken a major step forward in attacking this perennial
problem. Their new results, presented last November at the Foundations of Computer Science conference, stem from an unexpected new technique, Le Gall said. Although the improvement itself was
relatively small, Le Gall called it “conceptually larger than other previous ones.”
The technique reveals a previously unknown and hence untapped source of potential improvements, and it has already borne fruit: A second paper, published in January, builds upon the first to show how
matrix multiplication can be boosted even further.
“This is a major technical breakthrough,” said William Kuszmaul, a theoretical computer scientist at Harvard University. “It is the biggest improvement in matrix multiplication we’ve seen in more
than a decade.”
Enter the Matrix
It may seem like an obscure problem, but matrix multiplication is a fundamental computational operation. It’s incorporated into a large proportion of the algorithms people use every day for a variety
of tasks, from displaying sharper computer graphics to solving logistical problems in network theory. And as in other areas of computation, speed is paramount. Even slight improvements could
eventually lead to significant savings of time, computational power and money. But for now, theoreticians are mainly interested in figuring out how fast the process can ever be.
The traditional way of multiplying two n-by-n matrices — by multiplying numbers from each row in the first matrix by numbers in the columns in the second — requires n^3 separate multiplications. For
2-by-2 matrices, that means 2^3 or 8 multiplications.
Merrill Sherman/Quanta Magazine
In 1969, the mathematician Volker Strassen revealed a more complicated procedure that could multiply 2-by-2 matrices in just seven multiplicative steps and 18 additions. Two years later, the computer
scientist Shmuel Winograd demonstrated that seven is, indeed, the absolute minimum for 2-by-2 matrices.
Merrill Sherman/Quanta Magazine
Strassen exploited that same idea to show that all larger n-by-n matrices can also be multiplied in fewer than n^3 steps. A key element in this strategy involves a procedure called decomposition —
breaking down a large matrix into successively smaller submatrices, which might end up being as small as 2-by-2 or even 1-by-1 (these are just single numbers).
The rationale for dividing a giant array into tiny pieces is pretty simple, according to Virginia Vassilevska Williams, a computer scientist at the Massachusetts Institute of Technology and co-author
of one of the new papers. “It’s hard for a human to look at a large matrix (say, on the order of 100-by-100) and think of the best possible algorithm,” Vassilevska Williams said. Even 3-by-3 matrices
haven’t been fully solved yet. “Nevertheless, one can use a fast algorithm that one has already developed for small matrices to also obtain a fast algorithm for larger matrices.”
The key to speed, researchers have determined, is to reduce the number of multiplication steps, lowering that exponent from n^3 (for the standard method) as much as they can. The lowest possible
value, n^2, is basically as long as it takes just to write the answer. Computer scientists refer to that exponent as omega, ω, with n^ω being the fewest possible steps needed to successfully multiply
two n-by-n matrices as n grows very large. “The point of this work,” said Zhou, who also co-authored the January 2024 paper, “is to see how close to 2 you can come, and whether it can be achieved in
A Laser Focus
In 1986, Strassen had another big breakthrough when he introduced what’s called the laser method for matrix multiplication. Strassen used it to establish an upper value for omega of 2.48. While the
method is only one step in large matrix multiplications, it’s one of the most important because researchers have continued to improve upon it.
One year later, Winograd and Don Coppersmith introduced a new algorithm that beautifully complemented the laser method. This combination of tools has featured in virtually all subsequent efforts to
speed up matrix multiplication.
Here’s a simplified way of thinking about how these different elements fit together. Let’s start with two large matrices, A and B, that you want to multiply together. First, you decompose them into
many smaller submatrices, or blocks, as they’re sometimes called. Next, you can use Coppersmith and Winograd’s algorithm to serve as a kind of instruction manual for handling, and ultimately
assembling, the blocks. “It tells me what to multiply and what to add and what entries go where” in the product matrix C, Vassilevska Williams said. “It’s just a recipe to build up C from A and B.”
There is a catch, however: You sometimes end up with blocks that have entries in common. Leaving these in the product would be akin to counting those entries twice, so at some point you need to get
rid of those duplicated terms, called overlaps. Researchers do this by “killing” the blocks they’re in — setting their components equal to zero to remove them from the calculation.
That’s where Strassen’s laser method finally comes into play. “The laser method typically works very well and generally finds a good way to kill a subset of blocks to remove the overlap,” Le Gall
said. After the laser has eliminated, or “burned away,” all the overlaps, you can construct the final product matrix, C.
Putting these various techniques together results in an algorithm for multiplying two matrices with a deliberately stingy number of multiplications overall — at least in theory. The laser method is
not intended to be practical; it’s just a way to think about the ideal way to multiply matrices. “We never run the method [on a computer],” Zhou said. “We analyze it.”
And that analysis is what led to the biggest improvement to omega in more than a decade.
A Loss Is Found
Last summer’s paper, by Duan, Zhou and Wu, showed that Strassen’s process could still be sped up significantly. It was all because of a concept they called a hidden loss, buried deep within previous
analyses — “a result of unintentionally killing too many blocks,” Zhou said.
The laser method works by labeling blocks with overlaps as garbage, slated for disposal; other blocks are deemed worthy and will be saved. The selection process, however, is somewhat randomized. A
block rated as garbage may, in fact, turn out to be useful after all. This wasn’t a total surprise, but by examining many of these random choices, Duan’s team determined that the laser method was
systematically undervaluing blocks: More blocks should be saved and fewer thrown out. And, as is usually the case, less waste translates into greater efficiency.
“Being able to keep more blocks without overlap thus leads to … a faster matrix multiplication algorithm,” Le Gall said.
After proving the existence of this loss, Duan’s team modified the way that the laser method labeled blocks, reducing the waste substantially. As a result, they set a new upper bound for omega at
around 2.371866 — an improvement over the previous upper bound of 2.3728596, set in 2020 by Josh Alman and Vassilevska Williams. That may seem like a modest change, lowering the bound by just about
0.001. But it’s the single biggest improvement scientists have seen since 2010. Vassilevska Williams and Alman’s 2020 result, by comparison, only improved upon its predecessor by 0.00001.
But what’s most exciting for researchers isn’t just the new record itself — which didn’t last long. It’s also the fact that the paper revealed a new avenue for improvement that, until then, had gone
totally unnoticed. For nearly four decades, everyone has been relying upon the same laser method, Le Gall said. “Then they found that, well, we can do better.”
The January 2024 paper refined this new approach, enabling Vassilevska Williams, Zhou and their co-authors to further reduce the hidden loss. This led to an additional improvement of omega’s upper
bound, reducing it to 2.371552. The authors also generalized that same technique to improve the multiplication process for rectangular (n-by-m) matrices — a procedure that has applications in graph
theory, machine learning and other areas.
Some further progress along these lines is all but certain, but there are limits. In 2015, Le Gall and two collaborators proved that the current approach — the laser method coupled with the
Coppersmith-Winograd recipe — cannot yield an omega below 2.3078. To make any further improvements, Le Gall said, “you need to improve upon the original [approach] of Coppersmith and Winograd that
has not really changed since 1987.” But so far, nobody has come up with a better way. There may not even be one.
“Improving omega is actually part of understanding this problem,” Zhou said. “If we can understand the problem well, then we can design better algorithms for it. [And] people are still in the very
early stages of understanding this age-old problem.” | {"url":"https://www.quantamagazine.org/new-breakthrough-brings-matrix-multiplication-closer-to-ideal-20240307/","timestamp":"2024-11-06T00:48:24Z","content_type":"text/html","content_length":"211359","record_id":"<urn:uuid:8cfd66d8-b4f1-4c65-aeeb-4446336afa98>","cc-path":"CC-MAIN-2024-46/segments/1730477027906.34/warc/CC-MAIN-20241106003436-20241106033436-00845.warc.gz"} |
The constant objective value property for combinatorial optimization problems
Given a combinatorial optimization problem, we aim at characterizing the set of all instances for which every feasible solution has the same objective value. Our central result deals with
multi-dimensional assignment problems. We show that for the axial and for the planar $d$-dimensional assignment problem instances with constant objective value property are characterized by
sum-decomposable arrays. We provide a counterexample to show that the result does not carry over to general $d$-dimensional assignment problems. Our result for the axial $d$-dimensional assignment
problems can be shown to carry over to the axial $d$-dimensional transportation problem. Moreover, we obtain characterizations when the constant objective value property holds for the minimum
spanning tree problem, the shortest path problem and the minimum weight maximum cardinality matching problem.
View The constant objective value property for combinatorial optimization problems | {"url":"https://optimization-online.org/2014/05/4364/","timestamp":"2024-11-04T13:27:15Z","content_type":"text/html","content_length":"83751","record_id":"<urn:uuid:2e8f5b2a-a044-424b-965b-43ef3bb87fc0>","cc-path":"CC-MAIN-2024-46/segments/1730477027829.31/warc/CC-MAIN-20241104131715-20241104161715-00736.warc.gz"} |
Algebra I Tutors
We provide free access to qualified tutors across the U.S.
Find Greensboro, NC Algebra I Tutors For Lessons, Instruction or Help
A Tutor in the Spotlight!
Chenaya A.
• Asheboro, NC 27205 ( 26.1 mi )
• Algebra I Tutor
• Female Age 30
• Member since: 07/2012
• Will travel up to 10 miles
• Rates from $20 to $40 /hr
• On-line In-Home In-Group
• Also Tutors: Math, Pre-Algebra
Read More About Chenaya A...
In middle school I was a straight A student. Math and Reading come natural to me. I'm excellent at math and I can teach math as to where some people cannot. I comprehend what I read and I can help
those who cannot. In middle school I was a tutor to those who needed help. I am not afraid to help others understand what I do. I am confident in tutoring others.
Patti D.
• Pfafftown, NC 27040 ( 28.6 mi )
• Expert Algebra I Tutor
• Female
• Member Since: 08/2012
• Will travel up to 50 miles
• Tutoring Rate: $50.00 /hr
• I am a Certified Teacher
• On-line In-Group
• Also Tutors: Algebra III, Math, Trigonometry, Pre-Calculus, Pre-Algebra, Geometry, College Algebra, Calculus I, Algebra II
Learning Math can be FUN!
I try to explain concepts by explaining the methods behind them. Once a student understands the concepts, it is easier for them to apply them to other problem situations.
With over 35 years experience teaching math, I have successfully taught students from third grade through college calculus. I enjoy math, and my students are able to understand it in a one-to-one
teaching environment. During this time of closed schools, it is difficult for students to understand and adapt to this new way of learning. Many teachers are overwhelmed as they had to change their
teaching methods with just a couple of months remaining in the school year. I have been teaching online classes at Forsyth Tech. for some time. Transfering this experience to one-on-one tutoring
gives me the opportunity to help students that want to end the school year successfully. Let me help you improve your understanding ... read more
Tziona Liora T.
• Winston-Salem, NC 27104 ( 23.9 mi )
• Expert Algebra I Tutor
• Female Age 27
• Member Since: 11/2014
• Will travel up to 10 miles
• Rates from $20.00 to $45.00 /hr
• On-line In-Home In-Group
• Also Tutors: Pre-Algebra, Algebra II
I will give my all for you succeed.
My teaching style varies from student to student. I like to get to know the student in order to maximize their understanding of the subject. Everybody does learn differently, so I am open to change
for that said student.
To be honest I do not really have any experience of being a tutor. I am a senior at Reagan High school and for the most part I tutor other classmates and family members. Mostly recently I broaden my
range to elementary to high school. I do very well in mathematics except for geometry, Pre-cal, abd calculus because i am not registered for those classes. I'm also good in science.
Shawntese G.
• Greensboro, NC 27403 ( 1 mi )
• Algebra I Tutor
• Female Age 28
• Member Since: 10/2015
• Will travel up to 10 miles
• Rates from $15.00 to $30.00 /hr
• In-Home In-Group
• Also Tutors: Math, Calculus I, Algebra II
A Little Bit About Me...
I have been working at a Child Development Center for 3 years. I enjoyed tutoring math to elementary and middle school aged kids while in High School. I Also enjoy teaching cheerleading because I
have been cheering since age 4. I believe that all children should have access to extra help at all times.
Boukari S.
• Greensboro, NC 27408 ( 1.7 mi )
• Algebra I Tutor
• Male Age 61
• Member Since: 09/2014
• Will travel up to 30 miles
• Rates from $25.00 to $45.00 /hr
• I am a Certified Teacher
• On-line In-Home In-Group
• Also Tutors: Statistics, Pre-Algebra, College Algebra
A Little Bit About Me...
18 years of french teaching in middle in a FRENCH SPEAKING SCHOOL SYSTEM. Ability to help acquire the principles of the language , build vocabulary, sentences formation,learn the verbs and tenses
rules and make you become able to communicate autonomously. French is my primary language and I have both the gift and passion of teaching. I have the capacity to empower my student to understand the
language system and make progress on their own in a teaching by example and by practice on real life situation. No memorization only systematization and mnemotechniques.
Brittany C.
• Greensboro, NC 27401 ( 2 mi )
• Experienced Algebra I Tutor
• Female
• Member Since: 12/2012
• Will travel up to 15 miles
• Rates from $25.00 to $40.00 /hr
• On-line In-Home In-Group
• Also Tutors: Pre-Algebra, College Algebra, Algebra II
creative tutor who believes education is the key to the furu
I have been involved with tutoring other students since 2002 while I was still attending middle school. In high school I excelled in math and science courses and had been requested by my teachers to
help students after school in these subjects. While at Johnson c. Smith university I was a teachers aide for fifth grade students in a weekend program called Saturday acaeemy. My duty was to assist
in math, science and English.
I like to come up wit creative teaching methods to help students grasp information better. I also use a hands on approach to keep the students active and involved with their learning .
Rachel P.
• Greensboro, NC 27405 ( 4.3 mi )
• Experienced Algebra I Tutor
• Female Age 38
• Member Since: 11/2011
• Will travel up to 20 miles
• Rates from $10.00 to $20.00 /hr
• On-line In-Home In-Group
• Also Tutors: Pre-Algebra
My comments on tutoring Algebra I : elementary and high school level
My tutoring strategies: -throughout my experiences, I realized that each person is different and require different teaching methods depending on the individual's preference -I prefer to use study
skills that will benefit the individual; I have a variety of strategies depending on the subject as well. some individuals learn well using flashcards, while others prefer repetitive writing for
memorization - I have had a couple of ADHD clients that prefer learning games and activities because it helps them stay active, but focused at the same time
Hello! My name is Rachel and I live in Greensboro, NC. I am originally from Boston, MA and recently moved to NC about 2 years ago. I graduated from Wheaton College with a major in History and a minor
in French. I have enjoyed tutoring since high school. I have tutored in a variety of subjects and here is some of my experience listed below: ... read more
Y K.
• Greensboro, NC 27405 ( 4.3 mi )
• Algebra I Tutor
• Male
• Member Since: 08/2016
• Will travel up to 20 miles
• Rates from $20.00 to $25.00 /hr
• On-line In-Home In-Group
• Also Tutors: Geometry, College Algebra, Calculus II, Calculus I, Algebra II
Read More About Y K...
Having 10 years of Information Technology and Training experience. Teaching is my passion and take care that every point is explained with right example. With example it makes easy for student to
understand and grasp the concept. .
Brittany W.
• Greensboro, NC 27405 ( 4.3 mi )
• Algebra I Tutor
• Female Age 33
• Member Since: 10/2014
• Will travel up to 20 miles
• Rates from $20.00 to $40.00 /hr
• In-Home In-Group
• Also Tutors: Pre-Algebra, College Algebra, Algebra II
Read More About Brittany W...
Sit down with parents and have them inform me on the different areas their child needs assistance in; based on how the child learns is how I'll tutor.
During my senior year at North Carolina A&T State University, I tutored elementary students. The main focus for my tutors was making sure they completed their homework and also, based off any
information the parents give me about certain areas of study they would like their child working on; I would have all my tutors read at least 2 books to me aloud before ending our session.
Zaron J.
• Greensboro, NC 27405 ( 4.3 mi )
• Algebra I Tutor
• Male Age 42
• Member Since: 05/2015
• Will travel up to 20 miles
• Rates from $25.00 to $40.00 /hr
• On-line In-Home In-Group
• Also Tutors: Algebra III, Math, Pre-Calculus, Pre-Algebra, Geometry, Calculus I, Algebra II
More About This Tutor...
I use the preferences, hobbies, likes, and dislikes of the student to relate the information in which they are learning. I use examples from everyday life that helps them understand information that
is applicable to everyday life.
I have been tutoring for over 8 years. I have also started youth tutoring programs since 2007. I have taught algebra, organic chemistry, physics, and other core mathematics and chemistry classes.
Although I have taught in group settings, I prefer to tutor no more than three students at a time. I have taught grades from 6th grade all the way up to sophomore college level.
Elena P.
• Greensboro, NC 27405 ( 4.3 mi )
• Algebra I Tutor
• Female Age 32
• Member Since: 10/2014
• Will travel up to 30 miles
• Rates from $30.00 to $45.00 /hr
• On-line In-Home In-Group
Specialize in Accounting Principles
Since students learn faster with the right combination of learning styles, All first time students will take a short survey that helps me, as a tutor, to identify my students as visual, auditory,
reading/writing or kinesthetic learners, and align my overall curriculum with these learning styles. Allowing students to access information in terms they are comfortable with will increase their
academic confidence.
My name is Elena P. I recently graduated a 4-year university in Accounting and Finance majors with honor Magna cum laude. Being recommended by the university's Accounting professor, I was an
accounting tutors for my peers and underclass men by creating a one-on-one or group session. I am currently taking CPA exam and scored a 90s for the Financial Reporting section. As a result, I
believe that I have a comprehensive and thorough understanding about the accounting principles an ... read more
Rashonda J.
• Greensboro, NC 27406 ( 5.5 mi )
• Algebra I Tutor
• Female Age 27
• Member Since: 11/2014
• Will travel up to 20 miles
• Rates from $10.00 to $20.00 /hr
• On-line In-Home In-Group
• Also Tutors: Math, Statistics, Pre-Calculus, Pre-Algebra, Geometry, Calculus I, Algebra II
All tutoring sessions MUST be held at a public place!
I do what is best for the child. When giving a example problem , I work it all out first so they can see what I'm doing then i go through and explain in detail why I did what I did.
Worked with children as young as age 5. Have received A's or certification in classes i can tutor. Have helped other high school students as well ..................................................
.................................................. ..........................................l....... ...............
Carla J.
• Greensboro, NC 27406 ( 5.5 mi )
• Expert Algebra I Tutor
• Female Age 56
• Member Since: 11/2012
• Will travel up to 20 miles
• Rates from $10.00 to $30.00 /hr
• On-line In-Home In-Group
• Also Tutors: Math, Pre-Algebra, Geometry, Algebra II
Algebra I is one of my top subjects!
I have worked in the school system for 14 years. I have been a Teacher, Teacher Assistant, Substitute Teacher, a Lead Teacher in a Child Care Center, and a Director of a Center. I have worked with
all ages. I have worked with babies all the way through high school. I have worked with general education and special education. I am pretty easy going when it comes to not knowing, but I do not play
when it comes to school. A mind is a terrible thing to waste. If you don't use it then you will lose.
I enjoy being direct with my students and if they do not understand I will break it up so they can understand it piece by piece.
Cynthia J.
• Greensboro, NC 27406 ( 5.5 mi )
• Algebra I Tutor
• Female Age 32
• Member Since: 05/2015
• Will travel up to 20 miles
• Rates from $10.00 to $40.00 /hr
• On-line In-Home In-Group
• Also Tutors: Math, Pre-Calculus, Pre-Algebra, Geometry, College Algebra, Calculus II, Calculus I, Algebra II
More About This Tutor...
I am a Graduate Assistant at North Carolina A&T State University. I oversee the Bioengineering Lab. I also help my cousin, Daija Coles with her Chemistry II. She feels very confident going into her
final exam next week. I love to teach people especially when they hated that subject in the beginning. I love to help people understand their capability to attack any obstacle.
Edward B.
• Greensboro, NC 27406 ( 5.5 mi )
• Expert Algebra I Tutor
• Male
• Member Since: 07/2013
• Will travel up to 20 miles
• Rates from $25.00 to $50.00 /hr
• In-Home In-Group
• Also Tutors: Pre-Algebra, Geometry, College Algebra, Algebra II
Read More About Edward B...
Taught Basic Math, Pre - Algebra, and Algebra classes Evaluate and grade students' class work, assignments, and papers. Maintain student attendance records, grades, and other required records.
Prepare course materials such as syllabi, homework assignments, and handouts. Compile, administer, and grade examinations or assign this work to others. Keep abreast of developments in the field by
reading current literature, talking with colleagues, and participating in professional conferences. Initiate, facilitate, and moderate classroom discussions. Plan, evaluate, and revise curricula,
course content, and course materials and methods of instruction. Supervise students' laboratory work using NovaNet online program.
Nicole F.
• Whitsett, NC 27377 ( 9.9 mi )
• Algebra I Tutor
• Female Age 40
• Member Since: 04/2014
• Will travel up to 15 miles
• Rates from $25.00 to $50.00 /hr
• On-line In-Home In-Group
• Also Tutors: Math, Pre-Algebra, Geometry, College Algebra, Algebra II
Grad Student w/ extensive Science and Tutoring Background!
One on one, interactive, positive reinforcement
Private one on one tutoring experience for over 10 years. 4 years of group tutoring experience for Princeton Review. Conducted tutoring courses in Algebra, Geometry, and California High School Exit
Exam in class sizes of 5-10 students. Experience working with Elementary, Premedical High School, and delinquent High School seniors. Successful creation and execution of a clinical training manual
at a private medical practice for 3 years. Trained over 30 technicians on physiology and proper medical practices, in group and individual settings. Presented informative Diabetes Lecture to the
community at Bladen County Health Department.
Here is Some Good Information About Algebra I Tutoring
Do you find solving equations and word problems difficult in Algebra class? Are the exponents, proportions, and variables of Algebra keeping you up at night? Taking Algebra courses is a necessary
part of a student’s education. It is very important for a student struggling in Algebra to provide them with the help they need as early as possible. We make finding a qualified and experienced
algebra tutor easy. Our goal is to provide expertly skilled math tutors that can make understanding algebra simple and straightforward.
Cool Facts About Tutoring Near Greensboro, North Carolina
Greensboro is the third largest city in North Carolina. In 2004 the Department of Energy (DOE) awarded Greensboro with entry into the Clean Cities Hall of Fame. Select a tutor today. Results-oriented
and compassionate Greensboro tutors are available now to help your child reach his or her full potential. If you need tutoring, get started with one of our fantastic Greensboro tutors.
Find Greensboro, NC Tutoring Subjects Related to Algebra I
Greensboro, NC Algebra I tutoring may only be the beginning, so searching for other tutoring subjects related to Algebra I will expand your search options. This will ensure having access to
exceptional tutors near Greensboro, NC Or online to help with the skills necessary to succeed.
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Find Local Tutoring Help Near Greensboro, NC
Looking for a tutor near Greensboro, NC? Quickly get matched with expert tutors close to you. Scout out tutors in your community to make learning conveniently fit your availability. Having many
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Explore All Tutoring Locations Within North Carolina
Find Available Online Algebra I Tutors Across the Nation
Do you need homework help tonight or weekly sessions throughout the school year to keep you on track? Find an Online Algebra I tutor to be there when you need them the most.
Search for Online Algebra I Tutors | {"url":"https://www.tutorselect.com/find/greensboro_nc/algebra-1/tutors","timestamp":"2024-11-11T10:51:44Z","content_type":"text/html","content_length":"116008","record_id":"<urn:uuid:bce5a9c7-6de3-471e-a8d1-036ac78d560f>","cc-path":"CC-MAIN-2024-46/segments/1730477028228.41/warc/CC-MAIN-20241111091854-20241111121854-00586.warc.gz"} |
Computational Science and Engineering
The Departments of Chemical Engineering, Computer Science, Earth Science, Ecology, Evolution and Marine Biology, Electrical and Computer Engineering, Mathematics,and Mechanical Engineering each offer
an interdisciplinary master's and Ph.D. degree emphasis in computational science and engineering (CSE).
All students pursuing an emphasis in CSE must complete the requirements outlined below.
• Numerical Methods: (students must take at least three) ECE 210A-B-C-D (cross listed as ME 210A-B-C-D, Math 206A-B-C-D, CS 211A-B-C-D)
• Parallel Computing: Computer Science 240A
• Applied Mathematics: Students whose home department is not Mathematics must take either the Math 214A-B or Math 215A-B sequence or the Chemical Engineering 230A-B sequence (cross-listed as ME
244A-B). Credit will not be given for more than one of these sequences. Advanced courses may be substituted, with approval, as follows: Math 243A-B instead of Math 214A-B and Math 246A-B instead
of Math 215A-B. Students whose home department is Mathematics must take a two course sequence from either Math 243A-B or Math 246A-B.
Requirements for the M.S. in one of the above departments (thesis option only) with the CSE emphasis are as follows:
• Complete the requirements for an M.S. in the chosen department
• Complete the CSE core course sequence
• Write and defend a master's thesis in CSE
The thesis must be written under the supervision of a CSE ladder faculty member from the chosen department. The thesis committee must include a minimum of three permanent ladder faculty members, at
least two from the chosen department and one from CSE (may be CSE faculty member from another department.)
Students pursuing a Ph.D. with an emphasis in CSE must:
• Complete the requirements for a Ph.D. in the chosen department
• Complete the CSE core course sequence
• Write and defend a dissertation in CSE
The student's dissertation must be written under the supervision of a CSE ladder faculty member from the chosen department. The doctoral examination committee must include at least one CSE ladder
faculty member and at least one ladder faculty member from another department.
*** Students must petition with the Graduate Division to add the CSE emphasis to their current degree. Fill out the Change of Degree Status Petition Form and return to the Graduate Division with the
filing fee. | {"url":"https://cse.ucsb.edu/graduation-requirements","timestamp":"2024-11-11T14:53:34Z","content_type":"text/html","content_length":"24455","record_id":"<urn:uuid:2a390b13-8d80-44cb-8767-b850589af96d>","cc-path":"CC-MAIN-2024-46/segments/1730477028230.68/warc/CC-MAIN-20241111123424-20241111153424-00850.warc.gz"} |
Class 10
CBSE Class 10 Maths Syllabus Term 1 Course Structure 2021-22 with Marking Scheme
As students can see in the table below, the CBSE 10th Maths syllabus is divided into 7 units. The marks weightage for each unit is also indicated in the table. The CBSE Syllabus for Class 10 Basic
Maths and Standard Maths with Marking Scheme is also added in the PDF link provided above.
CBSE Syllabus for Class 10 Maths Term 1
One Paper
90 Minutes
Units Unit Name Marks
I Number Systems 06
II Algebra 10
III Coordinate Geometry 06
IV Geometry 06
V Trigonometry 05
VI Mensuration 04
VII Statistics and Probability 03
Total 40
Internal Assessment 10
Grand Total 50
Students can have a look at the CBSE Class 10 Maths Internal Assessment for first term below.
Internal Assessment Marks Total Marks
Periodic Tests 3
Multiple Assessments 2 10 marks for the term
Portfolio 2
Student Enrichment Activities-practical work 3
CBSE Class 10 Maths Syllabus Term 2 for 2021-22
The CBSE Syllabus for Class 10 Maths Term 2 with marks distribution is provided below:
Units Unit Name Marks
I Algebra (Cont.) 10
II Geometry (Cont.) 09
III Trigonometry (Cont.) 07
IV Mensuration (Cont.) 06
V Statistics and Probability (Cont.) 08
Total 40
Internal Assessment 10
Grand Total 50 | {"url":"https://swastiktutorialjeemainneet.in/class-10-duplicate-250/","timestamp":"2024-11-08T21:26:50Z","content_type":"text/html","content_length":"82342","record_id":"<urn:uuid:4f542fa7-b2bb-4a35-9a6c-4f96fd98b07b>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00302.warc.gz"} |
Keras Dense | Complete Guide on Keras Dense in detail
Updated March 15, 2023
Introduction to Keras Dense
Keras dense is one of the widely used layers inside the keras model or neural network where all the connections are made very deeply. In other words, the neurons in the dense layer get their source
of input data from all the other neurons of the previous layer of the network. In this article, we will study keras dense and focus on the pointers like What is keras dense, keras dense network
output, keras dense common methods, keras dense Parameters, Keras dense Dense example, and Conclusion about the same.
What is keras dense?
Keras dense is one of the available layers in keras models, most oftenly added in the neural networks. This layer contains densely connected neurons. Each of the individual neurons of the layer takes
the input data from all the other neurons before a currently existing one.
Internally, the dense layer is where various multiplication of matrix vectors is carried out. We can train the values inside the matrix as they are nothing but the parameters. We can even update
these values using a methodology called backpropagation.
The dense layer produces the resultant output as the vector, which is m dimensional in size. This is why the dense layer is most often used for vector manipulation to change the dimensions of the
vectors. The dense layer can also perform the vectors’ translation, scaling, and rotation operations.
keras dense network output
The dense layer of keras gives the following output after operating activation, as shown by the below equation –
Output of the keras dense = activation (dot (kernel, input) +bias)
The above formula uses a kernel, which is used for the generated weight matrix from the layer, activation helps in carrying out the activation in element-wise format, and the bias value is the vector
of bias generated by the dense layer.
The output of the dense layer is the dot product of the weight matrix or kernel and tensor passed as input. Further, the value of the bias vector is added to the output value, and then it goes
through the activation process carried out element-wise.
keras dense common methods
The dense layer has the following methods that are used for its manipulations and operations –
• Get_weights method – It has the syntax as sampleLayer. Get_weights() helps retrieve the current weights as arrays of NumPy associated with the layers. This method returns the output containing
the weight values in the list format of numPy arrays.
• Set_weights method – This method is used for setting the value of weights for the layer in the form of NumPy arrays. The syntax of the method is sampleEducbaLayer. Set_weights( value_of_weights).
The parameter value_of_weights is the list of NumPy arrays. Note that here the shape and the number of array values specified must match the dimension of the weight. That is, nothing but the size
of the array should match the resultant generated from the get_weights method.
• Get_config method – This method is used for retrieving the configurations of the layer and has its syntax as sampleEducbaLayer.get_config().
• Add_loss method – This method adds the loss tensor or tensors most potentially dependent on the input layers. It has the syntax as sampleEducbaLayer.add_loss (losses, **kwargs)
• Add_metric method – The method proves helpful when adding metric tensor to the existing created layer. The syntax of the method is sampleEducbaLayer.add_metric( value, **kwargs, name = None)
keras dense Parameters
The syntax of the dense layer is as shown below –
Keras. layers. Dense( bias_initializer = ‘zeros’, use_bias = True, activation = None, units, kernel_initializer = ‘glorot_uniform’, bias_constraint = None, activity_regularizer = None,
kernel_regularizer = None, kernel_constraint = None, bias_regularizer = None)
Let us study all the parameters that are passed to the Dense layer and their relevance in detail –
• Activation – It has a key role in applying element-wise activation function execution. When not specified, the default value is linear activation, and the same is used, but it is free for a
change. We can change this activation to any other per requirement by using many available options in keras.
• Units – It is a positive integer and a basic parameter used to specify the size of the output generated from the layer. It has relevance in the weight matrix, which helps specify its size and the
bias vector.
• Use_bias – This parameter is used to decide whether to include the value of the bias vector in the manipulations and calculations that will be done or not. The default value is true when we don’t
specify its value.
• Regularizers – It has three parameters for performing penalty or regularization of the built model. Usually not used, but when specified helps in the model generalization.
• Initializers – It provides the input values for deciding how layer values will be initialized. We have the bias vector and weight matrix in the dense layer, which should be initialized.
• Constraints – These parameters help specify if the bias vector or weight matrix will consider any applied constraints.
Keras Dense example
Let us consider a sample example to demonstrate the creation of the sequential model in which we will add two layers of the dense layer in the model –
sampleEducbaModel = tensorflow.keras.models.Sequential()
sampleEducbaModel.add(tensorflow.keras.layers.Dense(32, activation='relu'))
The output of the code snippet after execution is as shown below –
Keras Dense layer is the layer that contains all the neurons that are deeply connected within themselves. This means that every neuron in the dense layer takes the input from all the other neurons of
the previous layer. We can add as many dense layers as required. It is one of the most commonly used layers.
Recommended Articles
This is a guide to Keras Dense. Here we discuss keras dense network output, keras dense common methods, Parameters, Keras Dense example, and Conclusion. You may also look at the following articles to
learn more – | {"url":"https://www.educba.com/keras-dense/","timestamp":"2024-11-03T12:17:45Z","content_type":"text/html","content_length":"308290","record_id":"<urn:uuid:999a69d8-4ad9-4667-ab66-82c38e457c56>","cc-path":"CC-MAIN-2024-46/segments/1730477027776.9/warc/CC-MAIN-20241103114942-20241103144942-00320.warc.gz"} |
Mathematical Transformations
Clicking the command Tools > Mathematical Transformations... (or the toolbar button
After starting the command, a window is opened which shows the available operators and operands in its upper part and allows the user to enter and edit a mathematical formula in its lower part. The
formula which has been entered most recently is shown as the default formula. This formula can be edited or deleted and replaced by a completely different one.
When the formula is ready for execution, the user has to click Execute. Thereafter the formula is applied to the data matrix. In case of a syntax error, the formula is not executed at all and the
data remain unchanged. If the result of a formula is not defined or infinite, the calculation is aborted and a corresponding error message displayed.
The transformation should be written in standard notation (very much the same as in common programming languages), having the result at the left side and the arguments at the right side of the
equation. Lower case and upper case letters may be mixed arbitrarily and any number of spaces may be inserted into the equation. The assignment can be performed by the sign '=' or alternatively by ':
Hint: In some cases you might consider to use one of the predefined arithmetic conversions as these predefined transformations are executed much faster. Further, while these tools for performing
mathematical transformations are easy to use for quick and simple adjustments, a much better and much more sophisticated alternative is the built-in script language ILabPascal, which allows you
to apply arbitrarily complex operations to the data. | {"url":"http://imagelab.at/help/math_formulas.htm","timestamp":"2024-11-04T07:40:55Z","content_type":"text/html","content_length":"4647","record_id":"<urn:uuid:5e2b7599-9bd4-4daa-9da3-de975c5ad484>","cc-path":"CC-MAIN-2024-46/segments/1730477027819.53/warc/CC-MAIN-20241104065437-20241104095437-00042.warc.gz"} |
axiom holds for all times
## Elucidation This is used when the statement/axiom is assumed to hold true 'eternally' ## How to interpret (informal) First the "atemporal" FOL is derived from the OWL using the
standard interpretation. This axiom is temporalized by embedding the axiom within a for-all-times quantified sentence. The t argument is added to all instantiation predicates and predicates that use
this relation. ## Example Class: nucleus SubClassOf: part_of some cell forall t : forall n : instance_of(n,Nucleus,t) implies exists c : instance_of(c,Cell,t) part_of(n,c,t) ## Notes This
interpretation is *not* the same as an at-all-times relation
## Elucidation This is used when the statement/axiom is assumed to hold true 'eternally' ## How to interpret (informal) First the "atemporal" FOL is derived from the OWL using the standard
interpretation. This axiom is temporalized by embedding the axiom within a for-all-times quantified sentence. The t argument is added to all instantiation predicates and predicates that use this
relation. ## Example Class: nucleus SubClassOf: part_of some cell forall t : forall n : instance_of(n,Nucleus,t) implies exists c : instance_of(c,Cell,t) part_of(n,c,t) ## Notes This interpretation
is *not* the same as an at-all-times relation | {"url":"https://ols.monarchinitiative.org/ontologies/upheno2/individuals?iri=http://purl.obolibrary.org/obo/RO_0001901","timestamp":"2024-11-12T22:37:38Z","content_type":"text/html","content_length":"25287","record_id":"<urn:uuid:90b81878-3895-47ac-ac3e-d59aa49f9eec>","cc-path":"CC-MAIN-2024-46/segments/1730477028290.49/warc/CC-MAIN-20241112212600-20241113002600-00116.warc.gz"} |
N is INTEGER
The order of the matrix. N >= 0.
VL is DOUBLE PRECISION
Lower bound of the interval that contains the desired
eigenvalues. VL < VU. Needed to compute gaps on the left or right
end of the extremal eigenvalues in the desired RANGE.
VU is DOUBLE PRECISION
Upper bound of the interval that contains the desired
eigenvalues. VL < VU.
Note: VU is currently not used by this implementation of DLARRV, VU is
passed to DLARRV because it could be used compute gaps on the right end
of the extremal eigenvalues. However, with not much initial accuracy in
LAMBDA and VU, the formula can lead to an overestimation of the right gap
and thus to inadequately early RQI 'convergence'. This is currently
prevented this by forcing a small right gap. And so it turns out that VU
is currently not used by this implementation of DLARRV.
D is DOUBLE PRECISION array, dimension (N)
On entry, the N diagonal elements of the diagonal matrix D.
On exit, D may be overwritten.
L is DOUBLE PRECISION array, dimension (N)
On entry, the (N-1) subdiagonal elements of the unit
bidiagonal matrix L are in elements 1 to N-1 of L
(if the matrix is not split.) At the end of each block
is stored the corresponding shift as given by DLARRE.
On exit, L is overwritten.
PIVMIN is DOUBLE PRECISION
The minimum pivot allowed in the Sturm sequence.
ISPLIT is INTEGER array, dimension (N)
The splitting points, at which T breaks up into blocks.
The first block consists of rows/columns 1 to
ISPLIT( 1 ), the second of rows/columns ISPLIT( 1 )+1
through ISPLIT( 2 ), etc.
M is INTEGER
The total number of input eigenvalues. 0 <= M <= N.
DOL is INTEGER
DOU is INTEGER
If the user wants to compute only selected eigenvectors from all
the eigenvalues supplied, he can specify an index range DOL:DOU.
Or else the setting DOL=1, DOU=M should be applied.
Note that DOL and DOU refer to the order in which the eigenvalues
are stored in W.
If the user wants to compute only selected eigenpairs, then
the columns DOL-1 to DOU+1 of the eigenvector space Z contain the
computed eigenvectors. All other columns of Z are set to zero.
MINRGP is DOUBLE PRECISION
RTOL1 is DOUBLE PRECISION
RTOL2 is DOUBLE PRECISION
Parameters for bisection.
An interval [LEFT,RIGHT] has converged if
RIGHT-LEFT.LT.MAX( RTOL1*GAP, RTOL2*MAX(|LEFT|,|RIGHT|) )
W is DOUBLE PRECISION array, dimension (N)
The first M elements of W contain the APPROXIMATE eigenvalues for
which eigenvectors are to be computed. The eigenvalues
should be grouped by split-off block and ordered from
smallest to largest within the block ( The output array
W from DLARRE is expected here ). Furthermore, they are with
respect to the shift of the corresponding root representation
for their block. On exit, W holds the eigenvalues of the
UNshifted matrix.
WERR is DOUBLE PRECISION array, dimension (N)
The first M elements contain the semiwidth of the uncertainty
interval of the corresponding eigenvalue in W
WGAP is DOUBLE PRECISION array, dimension (N)
The separation from the right neighbor eigenvalue in W.
IBLOCK is INTEGER array, dimension (N)
The indices of the blocks (submatrices) associated with the
corresponding eigenvalues in W; IBLOCK(i)=1 if eigenvalue
W(i) belongs to the first block from the top, =2 if W(i)
belongs to the second block, etc.
INDEXW is INTEGER array, dimension (N)
The indices of the eigenvalues within each block (submatrix);
for example, INDEXW(i)= 10 and IBLOCK(i)=2 imply that the
i-th eigenvalue W(i) is the 10-th eigenvalue in the second block.
GERS is DOUBLE PRECISION array, dimension (2*N)
The N Gerschgorin intervals (the i-th Gerschgorin interval
is (GERS(2*i-1), GERS(2*i)). The Gerschgorin intervals should
be computed from the original UNshifted matrix.
Z is DOUBLE PRECISION array, dimension (LDZ, max(1,M) )
If INFO = 0, the first M columns of Z contain the
orthonormal eigenvectors of the matrix T
corresponding to the input eigenvalues, with the i-th
column of Z holding the eigenvector associated with W(i).
Note: the user must ensure that at least max(1,M) columns are
supplied in the array Z.
LDZ is INTEGER
The leading dimension of the array Z. LDZ >= 1, and if
JOBZ = 'V', LDZ >= max(1,N).
ISUPPZ is INTEGER array, dimension ( 2*max(1,M) )
The support of the eigenvectors in Z, i.e., the indices
indicating the nonzero elements in Z. The I-th eigenvector
is nonzero only in elements ISUPPZ( 2*I-1 ) through
ISUPPZ( 2*I ).
WORK is DOUBLE PRECISION array, dimension (12*N)
IWORK is INTEGER array, dimension (7*N)
INFO is INTEGER
= 0: successful exit
> 0: A problem occurred in DLARRV.
< 0: One of the called subroutines signaled an internal problem.
Needs inspection of the corresponding parameter IINFO
for further information.
=-1: Problem in DLARRB when refining a child's eigenvalues.
=-2: Problem in DLARRF when computing the RRR of a child.
When a child is inside a tight cluster, it can be difficult
to find an RRR. A partial remedy from the user's point of
view is to make the parameter MINRGP smaller and recompile.
However, as the orthogonality of the computed vectors is
proportional to 1/MINRGP, the user should be aware that
he might be trading in precision when he decreases MINRGP.
=-3: Problem in DLARRB when refining a single eigenvalue
after the Rayleigh correction was rejected.
= 5: The Rayleigh Quotient Iteration failed to converge to
full accuracy in MAXITR steps. | {"url":"https://man.linuxreviews.org/man3/dlarrv.f.3.html","timestamp":"2024-11-14T20:16:06Z","content_type":"text/html","content_length":"13762","record_id":"<urn:uuid:1f47dcc9-d996-4436-8f38-dadc99eeb7b8>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00735.warc.gz"} |
The NUMBERVALUE Function | SumProduct are experts in Excel Training: Financial Modelling, Strategic Data Modelling, Model Auditing, Planning & Strategy, Training Courses, Tips & Online Knowledgebase
A to Z of Excel Functions: The NUMBERVALUE Function
Welcome back to our regular A to Z of Excel Functions blog. Today we look at the NUMBERVALUE function.
The NUMBERVALUE function
The NUMBERVALUE function converts text into a number, in a locale-independent way. This may also be used to convert local-specific values into locale-independent values.
The NUMBERVALUE syntax is as follows:
NUMBERVALUE(text, [decimal_separator], [group_separator])
The NUMBERVALUE function syntax has the following arguments:
• text: this is required. This is the text to convert into a number
• decimal_separator: this argument is optional. This is the character used to separate the integer and fractional part of the result
• group_separator: this argument is also optional. This represents the character used to separate groupings of numbers, such as thousands from hundreds and millions from thousands.
It should be noted that:
• if the decimal_separator and group_separator arguments are not specified, separators from the current locale are used
• if multiple characters are used in the decimal_separator or group_separator arguments, only the first character is used
• if an empty string ("") is specified as the text argument, the result is zero [0]
• empty spaces in the text argument are ignored, even in the middle of the argument. For example, " 3 000 " is returned as 3,000
• if a decimal_separator is used more than once in the text argument, NUMBERVALUE returns the #VALUE! error value
• if the group_separator occurs before the decimal_separator in the text argument , the group_separator is ignored
• if the group_separator occurs after the decimal_separator in the text argument, NUMBERVALUE returns the #VALUE! error value
• if any of the arguments are not valid, NUMBERVALUE returns the #VALUE! error value
• if the text argument ends in one or more percent signs (%), they are used in the calculation of the result. Multiple percent signs are additive if they are used in the text argument just as they
are if they are used in a formula. For example, =NUMBERVALUE("9%%")returns the same result (0.0009) as the formula =9%%.
Please see my example below:
We’ll continue our A to Z of Excel Functions soon. Keep checking back – there’s a new blog post every business day.
A full page of the function articles can be found here. | {"url":"https://www.sumproduct.com/blog/article/a-to-z-of-excel-functions/the-numbervalue-function","timestamp":"2024-11-09T15:35:45Z","content_type":"text/html","content_length":"19387","record_id":"<urn:uuid:f1bddf03-31d5-499b-88a1-70d89f623fe8>","cc-path":"CC-MAIN-2024-46/segments/1730477028125.59/warc/CC-MAIN-20241109151915-20241109181915-00405.warc.gz"} |
1729 is a Magic Number | Riddles360
A bridge is about to collapse. There are four people P, Q, R and S on one of the sides. Before the bridge collapses, they want to cross it. Now since the bridge is too weak, it can only stand the
weight of two people at a time. Also, it is night time and nothing is visible. They have just one torch with them.
Now P takes one minute to cross the bridge, Q takes two minutes to cross, R takes five minutes to cross and S takes ten minutes to cross.
The bridge will collapse in seventeen minutes. How will they be able to cross the bridge before it collapses? | {"url":"https://riddles360.com/riddle/1729-is-a-magic-number","timestamp":"2024-11-14T10:28:43Z","content_type":"text/html","content_length":"43238","record_id":"<urn:uuid:bd92e288-60dc-4f05-a141-9a8e1157159a>","cc-path":"CC-MAIN-2024-46/segments/1730477028558.0/warc/CC-MAIN-20241114094851-20241114124851-00073.warc.gz"} |
How do I use cumulative distribution functions to compute p-values in SPSS, EXCEL and R?
P-values may be computed for an observed sample statistics, x, using the cumulative distribution function (CDF) of a particular distribution.
For x < 0 (1-sided) p-value = CDF(x) = P(X <= x)
For x > 0 (1-sided) p-value = 1- CDF(x) = P(X >= x)
A list of CDFs for various distributions is available in SPSS by clicking transform>compute and choosing CDF and noncentral CDFs.
In EXCEL the normdist(x,mean,standard_dev,cumulative) function will give P(X < -x) + P(X > x) for a z-score, x, coming from a Normal distirbution with given mean and s.d. Placing "False" as the
cumulative gives a two-tailed p-value. For example typing in a spreadsheet cell
and pressing the enter button gives a (two-tailed) p of 0.05.
In R information on the implementation of commonly used cdfs can be obtained by typing at the R prompt help(dt) for the t distribution, help(dnorm) for the Normal distribution, help(dchisq) for the
chi-square distribution and help df for the F distribution.
There also tables (Neave, 1978;1981) which can be photocopied for manual use onto an A4 sheet for specified distributions. The 1981 tables should be in the CBSU library. Howell, (2002) also has
tables giving test statistics at certain specific p-values for a variety of well used distributions. The gaps (to obtain points in the distribution relating to other p-values) in these tables could
be estimated using linear interpolation.
* Programs to obtain p-values using EXCEL and SPSS are available here. | {"url":"https://imaging.mrc-cbu.cam.ac.uk/statswiki/FAQ/pvals","timestamp":"2024-11-08T17:12:36Z","content_type":"text/html","content_length":"11690","record_id":"<urn:uuid:0aa87aff-7811-40ee-bcf7-af97a28539e4>","cc-path":"CC-MAIN-2024-46/segments/1730477028070.17/warc/CC-MAIN-20241108164844-20241108194844-00888.warc.gz"} |
Audiogon Discussion Forum
Hey there, this is pretty typical. Lots of buyers like to run two separate pairs of speakers, but you don't have 2x the circuitry. The jacks are just there for convenience.
The lower the impedance number, the harder to drive.
The top line means "A or B" - each speaker can be rated no less than 4 Ohms.
The bottom line means "A _and_ B " - Each speaker can be no less than 8 Ohms.
Hey thank you so much for your quick response!
So just to understand this better, the notation here is opposite to that in boolean algebra, where A*B would mean A and B, and A+B would mean A or B?
What is the reason behind these different ratings, i.e. what is different between the two configurations (maybe in terms of the wiring or other internal features) that would cause the different
impedance rating?
just to understand this better, the notation here is opposite to that in boolean algebra, where A*B would mean A and B, and A+B would mean A or B?
Forget about boolean algebra.
The top line " A.B-4Ω~16Ω/SPEAKER " means IF playing speaker A OR B only, the amplifier able to safely power 4Ω to 16Ω impedance rating speaker.
The bottom line " A+B-8Ω~16Ω/SPEAKER " means IF playing speaker A AND B together, the amplifier able to safely power 8Ω to 16Ω impedance rating speaker.
What is the reason behind these different ratings, i.e. what is different between the two configurations (maybe in terms of the wiring or other internal features) that would cause the different
impedance rating?
Nikko NA-550 is designed to safely power speaker not less than 4Ω impedance. If playing two sets of speakers together (speaker A and B), both set of speakers are connected in parallel by the speaker
switch and the total impedance apply to the amplifier will be less.
Ah thanks that last paragraph makes perfect sense, thank you.
Yes, I agree that the top line " A.B-4Ω~16Ω/SPEAKER " is confusing, this is much better to understand:
I have a NA-550. I have a 6 ohm pair of speakers on A and an 8 ohm pair on B. I run A+B all the time. I think this is technically out of spec, but is it anything I need to worry about?
Also, the 6 ohm pair is much louder than the 8 ohm pair. Different speakers, obviously. Also different gauge, type, and lengths of speaker wire. Can anyone suggest why there is such a disparity in
the volume of the two pairs? | {"url":"https://d2dve11u4nyc18.cloudfront.net/discussions/understanding-speaker-output-on-nikko-na-550","timestamp":"2024-11-02T13:11:40Z","content_type":"text/html","content_length":"83622","record_id":"<urn:uuid:a0a7aa40-f42d-4206-b1c9-73dc416fa55e>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00729.warc.gz"} |
Extracting digits from an integer
Is there a direct way to extract individual digits from an integer or do you have to loop through to get them.
So if my integer is 48367 how can I return the fourth digit '6' or the second digit '8' - can I do it directly?
Many thanks for your help in advance - I found solutions on the net but could not get them to work - no doubt me and not the solution!
Here are two ways. they might not be the best, but they work.
Here is the link.
Find the power of ten for the digit you want, divide your number by that value, and then modulus the quotient with ten.
unsigned n = 48367;
unsigned x = (n / 10U) % 10;
unsigned y = (n / 1000U) % 10;
After running that, x will be 6 and y will be 8.
1 Like
Wow thank you - I was close but missing the U
For this to work, you have to be counting your digits from the right hand end of the number.
If his number is 48367 and he wants the second digit ( counted from the left, the 8 ), then to implement this scheme, you would need to figure out that there are 5 digits and the 2nd one from the
left is the 4th digit from the right, and then divide by ten to the power of 4-1
This is simple if you are looking at it personally and the number is written down. It's not so straightforward if it is an actual binary number.
I didn't read the question correctly. It seemed to ask that you wanted the digit at a specific position in the number. It didn't say the power associated with it. So, which is it: 1) the digit
character at that position in the number, or 2) the power associated with the digit position?
The easiest way would be to convert the number to an ascii string then return the index that you want.
Granted now instead of looking at each number at "x" place, now you need to look at it as an array.
My first solution in the link I gave, works but it uses pow which is very slow, but again it works and it reads in the "x" place.
No error checking, does test for negative though:
char nth_digit (int val, int n)
char buffer [7] ; // enough space for large negative int and null terminator
sprintf (buffer, "%i", val) ;
return buffer[0] == '-' ? buffer [n+1] : buffer [n] ;
If I wanted to know all of the digits, I'd decompose the number by dividing by ten and taking the remainder, as often as necessary.
If I only wanted one arbitrary digit, I'd figure out if it is a 2,3,4,or 5 digit ( decimal ) number, and then use christop's method.
I wouldn't use sprintf() unless the program was already using it for something else, too much overhead.
What is faster, % or pow? I dont have an Arduino with me to do a speed test.
% by many orders of magnitude, pow is floating point.
So then this:
int ExtractDigit(int V, int P)
return int(V/(pow(10,P-1))) - int(V/(pow(10,P)))*10;
is faster than
int ExtractDigit(int V, int P)
return int(V/(pow(10,P-1))) % 10;
Both product the same result.
Don't use pow at all! Floating point is very slow, its emulated.
unsigned int pow [] = { 1, 10, 100, 1000, 10000 } ;
int ExtractDigit(int V, int P)
return V / pow[P] % 10 ;
But now an array substituting for pow assumes the value will not exceed 5 digits. Once you get to the 6 digits, now you have an array of long variables which uses up more memory.
int ExtractDigit(int V, int P)
unsigned long Pow = 1;
for(byte D= 0; D < (P-1); D++)
Pow *= 10;
return (V / Pow) % 10;
//return (V/(Pow)) - (V/(Pow*10))*10; //Fixed and works
But now an array substituting for pow assumes the value will not exceed 5 digits. Once you get to the 6 digits, now you have an array of long variables which uses up more memory.
int ExtractDigit(int V, int P)
unsigned long Pow = 1;
for(byte D= 0; D < (P-1); D++)
Pow *= 10;
return (V / Pow) % 10;
//return (V/(Pow)) - (V/(Pow*10))*10; //Fixed and works
There's a tradeoff of time vs space. The array method is faster in most cases, and there's some break-even point in the number of digits you want to support where the array is larger than an
iterative multiplication (that point might actually be larger than 6, depending on the size of the code that performs the iterative multiplication).
Sidenote: your functions take slightly different parameters. MarkT's takes P=0 for the unit digit while HazardsMind's takes P=1 for the unit digit. I prefer P=0 for the unit digit because 10^0 = 1.
I started P = 1, because what is at zero place? zero or one?
I started P = 1, because what is at zero place? zero or one?
One. Ten would be at P=1.
Is there a direct way to extract individual digits from an integer or do you have to loop through to get them.
So if my integer is 48367 how can I return the fourth digit '6' or the second digit '8' - can I do it directly?
Many thanks for your help in advance - I found solutions on the net but could not get them to work - no doubt me and not the solution!
How about this:
void setup (void)
uint8_t n;
uint8_t digits[10];
uint32_t value;
uint32_t divider;
char buffer [32];
divider = 1000000000UL;
n = 10;
value = 12345678UL;
while (n--) {
digits[n] = (value / divider);
value -= (digits[n] * divider);
divider /= 10;
n = 10;
while (n--) {
sprintf (buffer, "digit %d is %d\n", n, digits[n]);
Serial.print (buffer);
Serial.print ("\n");
void loop (void)
// nothing
It will allow you to get any digit from up to a 32 bit number. The digits are in [b]digits[n][/b].
For example, if your value is "1234", the zero place digit (digits[0]) is 4.
Hope this helps.
Don't use pow at all! Floating point is very slow, its emulated.
unsigned int pow [] = { 1, 10, 100, 1000, 10000 } ;
int ExtractDigit(int V, int P)
return V / pow[P] % 10 ;
Here's a power function that doesn't need any pre-stored variables and works up to a 64 bit number:
// performs base to the exp power
uint64_t intPower (uint8_t base, uint8_t exp)
uint64_t result = 1;
while (exp--) {
result *= base;
return result; | {"url":"https://forum.arduino.cc/t/extracting-digits-from-an-integer/296749","timestamp":"2024-11-14T08:50:21Z","content_type":"text/html","content_length":"55735","record_id":"<urn:uuid:1aa408be-94b9-44d6-8d2d-4bb9dd9e574f>","cc-path":"CC-MAIN-2024-46/segments/1730477028545.2/warc/CC-MAIN-20241114062951-20241114092951-00065.warc.gz"} |
Estimate of Geothermal Property Profile
The subsurface was divided into three main layers, and the analytical method developed by McDaniel et al. (2018) was applied as a fast method to estimate the thermal conductivity profile. This method
assumes the radial temperature gradients from the U-tube to the subsurface are constant and uniform at the steady state.
Results of two DTRTs are shown in Figure 1 and 2. In the first test, λ for the sediment above 30 ft is 6.2 Btu/hr-ft-℉, and λ for the bedrock from 30 ft to 250 ft is 1.4 Btu/hr-ft-℉; in the second
test, λ for the sediment above 30 ft is 1.2 Btu/hr-ft-℉, and λ for the bedrock from 30 ft to 250 ft is 1.4 Btu/hr-ft-℉. Compared with conventional TRT, DTRT estimates higher thermal conductivities
for bedrock above 250 ft because it considers the low heat transfer efficiency below the depth of 250 ft.
The thermal conductivity of the bedrock (30 – 250 ft) estimated from two DTRTs match well with each other. However, two tests estimate different thermal conductivity for the sediment above the depth
of 30 ft, and in the first test, λ for the sediment above 30 ft is too high. This is due to the spatial fluctuation of the temperature data along the depth, and the thermal conductivity estimated
from the temperature gradient can be easily affected by the fluctuation.
Figure 1 Thermal Conductivity of The First DTRT Using OMNISENS
Figure 2 Thermal Conductivity of The Second DTRT Using ALICIA
The heat capacity profile at the test site is estimated based on typical heat capacity values of soils/rocks listed by Kavanaugh & Rafferty (2014) as shown in Table 1. Thermal diffusivity profile is
calculated and shown in Figure 3. In the first DTRT, the thermal diffusivity of the sediment above 30 ft is 3.5 – 7.5 ft^2/day, and the thermal diffusivity of the bedrock from 30 ft to 250 ft is 0.8
– 1.2 ft^2/day; in the second DTRT, the thermal diffusivity of the sediment above 30 ft is 0.7 – 1.4 ft^2/day, and the thermal diffusivity of the Franciscan Complex from 30 ft to 250 ft is 0.8 – 1.2
Table 1 Layered Heat Capacity
Figure 3 Thermal Diffusivity Profiles | {"url":"https://geomechanics.berkeley.edu/research/berkeleygeothermal/estimate-of-geothermal-property-profile/","timestamp":"2024-11-03T16:12:15Z","content_type":"text/html","content_length":"43885","record_id":"<urn:uuid:ab31524d-7662-49bc-9a1d-1a7cb824aa8f>","cc-path":"CC-MAIN-2024-46/segments/1730477027779.22/warc/CC-MAIN-20241103145859-20241103175859-00654.warc.gz"} |
WSU: Space, Time, and Einstein with Brian Greene
World Science Festival
30 Jul 2020151:26
TLDRThis insightful video script delves into the revolutionary ideas of Albert Einstein's Special Theory of Relativity, which transformed our understanding of space, time, matter, and energy. It
explains how Einstein found that time dilates, lengths contract, and mass and energy are interchangeable, all consequences of the constant speed of light. Through vivid examples, including the
paradoxes of moving poles and aging twins, it demonstrates how these concepts challenge our everyday experiences and intuition. The script emphasizes that these startling revelations all stem from
one groundbreaking insight: the constancy of the speed of light, underscoring the profound impact of Einstein's work on modern physics.
• 💡 Albert Einstein's special theory of relativity revolutionized our understanding of space, time, matter, and energy, fundamentally altering concepts like the flow of time and the structure of
the universe.
• 🚀 Einstein's insights revealed that time passes differently for objects in motion compared to those at rest, leading to phenomena such as time dilation where moving clocks tick slower.
• 🌌 The theory introduced the idea that objects in motion contract along their direction of motion, known as length contraction, challenging our everyday perceptions of distance.
• ⏳ The relativity of simultaneity shows that events perceived as simultaneous from one frame of reference may not be seen as such from another, underscoring the subjective nature of time.
• 🌐 Special relativity suggests that our intuitive understanding of time and space is limited to our direct experiences, which occur within a narrow range of the universe's vast scales.
• 🔬 The speed of light (c) is a cosmic speed limit, and Einstein's equation E=mc² highlights a profound connection between mass and energy, suggesting they are interchangeable.
• 🛰️ Experiments like the muon decay observation from Earth's surface to the upper atmosphere and precise measurements using atomic clocks on airplanes confirm relativity's predictions, showcasing
its applicability to real-world phenomena.
• 🕰️ Time dilation has practical implications, including the need for adjustments in the GPS satellites' clocks to account for differences in time flow compared to Earth's surface.
• 🔄 The twin paradox, involving one twin traveling at high speeds in space and returning younger than their Earth-bound sibling, illustrates the tangible effects of time dilation over significant
• 🧪 Special relativity, built on the constancy of the speed of light, provides a framework that has been confirmed by various experiments, reinforcing its status as a cornerstone of modern physics.
Q & A
• What is the special theory of relativity and who proposed it?
-The special theory of relativity is a theory proposed by Albert Einstein that transforms our understanding of space, time, matter, and energy. It introduces concepts such as time dilation,
length contraction, and the equivalence of mass and energy.
• How does Einstein's theory affect our perception of time in motion?
-Einstein's theory reveals that clocks in motion tick off time at a slower rate compared to those at rest, showing that time dilation occurs as a fundamental aspect of reality missed in everyday
• What is the significance of the equation E=mc² in the special theory of relativity?
-E=mc² is the most famous equation in physics, establishing a deep, hidden connection between mass (m) and energy (E), with c representing the speed of light. It shows that mass can be converted
into energy and vice versa, highlighting the interchangeable nature of the two.
• How does special relativity challenge the traditional notions of space and time?
-Special relativity challenges traditional notions by showing that space and time are not absolute but relative and interconnected. It demonstrates that measurements of space and time vary for
observers in different states of motion, leading to phenomena like time dilation and length contraction.
• What is the relativity of simultaneity, and how does it affect perception of events?
-The relativity of simultaneity is the concept that events perceived as simultaneous by one observer may not be seen as simultaneous by another observer moving relative to the first. This
principle underlines that the concept of 'now' can vary between different frames of reference.
• How does special relativity explain the increase in mass of objects in motion?
-Special relativity explains that as objects move faster, approaching the speed of light, their mass effectively increases when observed from a stationary frame. This mass increase is a
consequence of the energy of motion, further intertwining the concepts of mass and energy.
• Why can't objects with mass reach or exceed the speed of light according to special relativity?
-According to special relativity, as an object's speed approaches the speed of light, its mass increases towards infinity, requiring infinite energy to accelerate further. Since infinite energy
is not attainable, objects with mass cannot reach or exceed the speed of light.
• How does the concept of gamma (γ) factor into time dilation and mass increase?
-The gamma (γ) factor quantifies the degree of time dilation and mass increase for objects in motion relative to an observer. It depends on the object's velocity and shows how time slows down and
mass increases as the object's speed approaches the speed of light.
• What role does acceleration play in distinguishing between observers in special relativity?
-Acceleration plays a crucial role by breaking the symmetry between observers. Unlike constant velocity, where observers can equally claim to be at rest, acceleration provides a clear
distinction, as only non-accelerating observers can claim to be in an inertial frame of reference.
• How does special relativity impact our understanding of the universe?
-Special relativity profoundly impacts our understanding of the universe by showing that our intuitive notions of absolute time and space are incorrect. It reveals a universe where time and space
are fluid, influenced by the observer's motion, fundamentally altering our comprehension of reality.
🌀 Einstein's Revolutionary Theory
In 1905, Albert Einstein's contemplation led to the formulation of the special theory of relativity, fundamentally altering our understanding of space, time, matter, and energy. This theory
introduced revolutionary concepts: the relativity of simultaneity, time dilation, length contraction, and the equivalence of mass and energy (E=mc²). Einstein's insights revealed that our intuitive
grasp of reality, based on everyday experiences, misses the exotic nature of the universe's vast scales. This theory not only challenged conventional wisdom but also unveiled a universe far stranger
and more interconnected than previously imagined.
🔬 The Impact of Extreme Conditions
Einstein's theory revealed that under extreme conditions, such as very small scales, huge masses, or high speeds, new physics emerges. Quantum mechanics governs the realm of the very small, general
relativity the domain of massive objects, and special relativity the behavior of objects moving at high speeds. This exploration into the fabric of the universe shows that our 'normal' experiences
are just a tiny, non-representative sample of reality. The universe operates under rules that are fundamentally different from our everyday observations, especially as we approach the speed of light,
where time and space exhibit bizarre and counterintuitive properties.
🌌 The Relativity of Simultaneity and Time
The constant speed of light, a cornerstone of special relativity, leads to startling implications for simultaneity and time. Observers in relative motion do not agree on the simultaneity of events,
challenging the notion of universal time. This discrepancy becomes evident through thought experiments, demonstrating that clocks in motion tick at a different rate than stationary ones, depending on
the observer's frame of reference. This revelation about time's fluid nature underpins the theory's radical departure from classical physics, emphasizing the subjective experience of time and its
dependency on velocity.
📏 Space and Motion: Length Contraction
Special relativity also profoundly affects our understanding of space. Objects in motion contract in length along the direction of their movement from the perspective of a stationary observer, a
phenomenon known as length contraction. This effect, however, is only appreciable at speeds close to the speed of light. Length contraction, together with time dilation, ensures that the speed of
light remains constant across all frames of reference, further illustrating the intertwined nature of space and time. This concept challenges traditional views of space as a static backdrop,
revealing it as dynamic and relative.
⚛️ The Mass-Energy Equivalence
One of Einstein's most groundbreaking revelations was the mass-energy equivalence, encapsulated in the equation E=mc². This equation signifies that mass and energy are two forms of the same entity,
interchangeable under the right conditions. The theory predicts that as objects accelerate close to the speed of light, their mass effectively increases, necessitating an infinite amount of energy to
surpass the speed of light, thus enforcing it as an ultimate speed limit. This interconversion of mass and energy has profound implications for understanding the universe's workings, from nuclear
reactions to the energy production in stars.
🌐 The Twin Paradox: Time Dilation in Action
The twin paradox delves into the effects of time dilation through a scenario where one twin travels at high speed in space and returns younger than their stay-at-home sibling. This apparent paradox
is resolved by recognizing that only the traveling twin experiences acceleration, breaking the symmetry and validating the time dilation effect predicted by special relativity. This thought
experiment vividly illustrates how velocity and acceleration affect the passage of time, underscoring the non-intuitive nature of the physical universe as described by Einstein's theories.
🚀 Concluding Reflections on Special Relativity
Special relativity, with its counterintuitive notions of time, space, and mass-energy equivalence, represents a monumental shift in our understanding of the universe. It demonstrates that our
everyday experiences are but a narrow slice of reality, bound by the speeds and scales we commonly encounter. Einstein's theory invites us to consider a broader, more exotic cosmos where time slows,
lengths contract, and energy and mass are interchangeable, challenging us to rethink our place in a universe far stranger and more wonderful than we could have imagined.
💡Special Theory of Relativity
The Special Theory of Relativity, proposed by Albert Einstein in 1905, fundamentally changes our understanding of space, time, matter, and energy. It introduces concepts that are counterintuitive to
our everyday experiences, such as time dilation, length contraction, and the equivalence of mass and energy (E=mc^2). The theory is based on two postulates: the laws of physics are the same in all
inertial frames of reference, and the speed of light in a vacuum is the same for all observers, regardless of their relative motion or the motion of the light source. In the script, this theory is
the cornerstone that underpins the entire narrative, exploring its implications on various scales and phenomena.
💡Time Dilation
Time dilation is a phenomenon predicted by the Special Theory of Relativity, where time passes at a slower rate for an object in motion compared to an object at rest, from the perspective of a
stationary observer. This effect becomes significant at speeds close to the speed of light. The script illustrates this concept with examples, such as the hypothetical scenario where clocks in motion
tick off time at a slower rate than stationary clocks, affecting how we perceive the passage of time across different velocities.
💡Length Contraction
Length contraction is another relativistic effect where an object in motion appears shorter in the direction of motion to a stationary observer, with the effect becoming more pronounced as the
object's speed approaches the speed of light. This concept is introduced in the script through the example of measuring the length of a moving train, demonstrating how motion can alter our perception
of spatial dimensions.
E=mc^2 is perhaps the most famous equation in physics, encapsulating the idea that mass and energy are equivalent and interchangeable. This equation, derived from the Special Theory of Relativity,
indicates that a small amount of mass can be converted into a large amount of energy, highlighting a profound connection between these two fundamental concepts. The script discusses this equation as
the culmination of Einstein's insights into the nature of the universe, emphasizing its deep and hidden connection between mass and energy.
💡Relativity of Simultaneity
The relativity of simultaneity is a concept from the Special Theory of Relativity that suggests observers moving relative to each other may disagree on whether two events occur at the same time. This
principle is crucial in understanding how perceptions of time can vary based on the observer's motion. The script uses the example of treaty signing between two nations to illustrate how the timing
of events can appear different to observers in motion relative to one another.
💡Inertial Frame of Reference
An inertial frame of reference is a state of motion where an object either remains at rest or moves at a constant velocity, meaning it's not accelerating. The Special Theory of Relativity applies
within these frames, where the laws of physics are observed to be consistent. The script references this concept to explain scenarios like George and Gracie floating in space, emphasizing the
subjective nature of motion and rest.
💡Speed of Light
The speed of light, denoted as 'c', is a fundamental constant of the universe, representing the maximum speed at which all energy, matter, and information in the universe can travel. It is integral
to the Special Theory of Relativity, which posits that the speed of light in a vacuum is constant and independent of the motion of the source or observer. The script discusses this constancy as a key
insight that led Einstein to develop his revolutionary theory.
💡Gamma Factor
The gamma factor is a mathematical term in the Special Theory of Relativity, denoted by the symbol γ (gamma), which quantifies the amount of time dilation and length contraction experienced by an
object in motion relative to an observer. It's derived from the object's velocity and the speed of light. The script explains how this factor is used to calculate the relativistic effects on time and
space, such as in the Twin Paradox and the pole in the barn paradox.
💡Twin Paradox
The Twin Paradox is a thought experiment in the Special Theory of Relativity, illustrating time dilation's asymmetrical effects. It involves identical twins, where one travels at near-light speed on
a space journey and returns younger than the twin who stayed on Earth. The script addresses this paradox to show the non-intuitive nature of relativistic time, ultimately explaining that the
traveling twin ages less due to experiencing significant acceleration, distinguishing her journey from the inertial frame of reference of the stay-at-home twin.
💡Quantum Mechanics
Quantum Mechanics is mentioned in the context of exploring the universe's fundamental scales, contrasting the macroscopic effects of Relativity with the microscopic phenomena governed by quantum
theory. While not the focus of the script, it's acknowledged as another pillar of modern physics that, together with Relativity, provides a complete picture of the physical universe's workings across
all scales.
The study found significant improvements in reading comprehension scores for students who used the new educational app
Researchers developed a novel algorithm to analyze complex health data and predict disease outbreaks earlier
Professor Smith proposed an innovative theoretical framework to understand social media usage and effects
The experiment resulted in a 75% increase in crop yields, demonstrating the potential of new agricultural techniques
Interview with Dr. Jones, a pioneer in stem cell research, on the ethics of cloning and regenerative medicine
Discussion of study limitations, including small sample size and lack of diversity, affecting generalizability
Researchers developed a low-cost water filtration device to provide clean drinking water in developing regions
Examining the link between social inequality and health outcomes across socioeconomic groups
Overview of the state of climate change science, dire predictions if carbon emissions not reduced
Description of collaborative efforts between neuroscientists and engineers to create advanced brain-computer interfaces
Discussion of challenges in implementing new education policies and suggestions for stakeholder engagement
Presentation of survey results on workforce automation and concerns about impacts on employment
Expert analysis of political and social factors underlying the rise of extremist groups
Professor Patel provides an insightful critique of prevailing economic theory and calls for a new paradigm
Overview of key takeaways and actionable next steps for policymakers based on research findings
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Vaktang Lashkia
According to our database
, Vaktang Lashkia authored at least 2 papers between 1996 and 1998.
Collaborative distances:
Book In proceedings Article PhD thesis Dataset Other
On csauthors.net:
A Finite Basis of the Set of All Monotone Partial Functions Defined over a Finite Poset.
Proceedings of the 28th IEEE International Symposium on Multiple-Valued Logic, 1998
Semirigid sets of diamond orders.
Discret. Math., 1996 | {"url":"https://www.csauthors.net/vaktang-lashkia/","timestamp":"2024-11-13T03:04:26Z","content_type":"text/html","content_length":"16190","record_id":"<urn:uuid:c380cf2c-2f9e-488c-9f16-cf660eca548e>","cc-path":"CC-MAIN-2024-46/segments/1730477028303.91/warc/CC-MAIN-20241113004258-20241113034258-00160.warc.gz"} |
Chemistry 421/821 First Exam Spring 2009 Page 1Name
Chemistry 421/821 – First Exam Spring 2009 page 1Name
1)(15 points)
a. (5 points)Describe the primary difference between a prism and grating monochromator.
A prism is based on the refraction of light and the fact that different wavelength’s of light have different values of refraction index in a medium. A grating monochromator is based on the
diffraction of light (constructive and deconstructive interference).
b. (5 points)Explain how a narrow bandwidth can be both an advantage and disadvantage in UV/vis spectroscopy.
A narrow bandwidth means that a smaller range of wavelengths are transmitted to the sample. Thus, the observed absorbance is more closely correlated to the desired wavelength. This is important since
molar absorptivity is wavelength dependent. You are only getting the wavelength you want. Conversely, a lower bandwidth means the intensity of light reaching the sample has also decreased. This may
be problematic since the sensitivity of measuring an absorbance is directly dependent on the intensity or power of the source wavelength.
c. (5 points) Explain how the additive property of Beer’s law may lead to errors (negative deviation from the expected absorbance)
This is related to the bandwidth question above. The observed total absorbance is the sum of each absorbance at each individual wavelength transmitted to the sample (bandwidth):
AT =A1 +A2 + A3 + … = 3bc + 3bc + 3bc + …
But, the expected absorbance is based on a single wavelength, the center of the bandwidth. If the molar absorptivity is significantly different at each wavelength, the observed absorption at each
wavelength will differ. Thus:
A1≠A2≠A3, so AT is not proportional to A1
Similar effect if the analyte is involved in a chemical equilibrium, where different species exhibit different molar absorptivity at the observed wavelength.
2) (15 points) Given the following calibration curve:
1. (5 points) Which analyte has the lowest sensitivity? Explain.
C, lowest slope
1. (5 points) What are the selectivity coefficients?
kb,a = mb/ma = 3.14/11.714 = 0.27
kc,a = mc/ma = 0.987/11.714 = 0.084
kc,b = mc/mb = 0.987/3.14 = 0.31
1. (5 points) Which analyte has the lowest limit of detection? Explain.
A, highest slope: cm = (Sm-Sbl)/slope
3) (9 points)
a. (3 points)Which compound is likely to have an observable UV/vis spectra?
b. (3 points)The product of a chemical reaction resulted in a red-shift in the UV spectra, which compound is the likely product?
c. (3 points) Which compound is likely to fluoresce?
4) (10 points)Please identify which of the situations described below is a source of systematic or random error:
a. (2 points) Glass cuvette measuring an absorbance at 300 nm.
Systematic (glass absorbs below 350 nm)
b. (2 points) Power fluctuations that affect the stability of the light source.
Random, (random, unpredictable change light power/intensity)
1. (2 points) Rapid decay in the intensity of a deuterium lamp in a Spectronic 20.
Systematic (intensity of light would change by a fixed amount between samples)
1. (2 points) Thermal electron emission in a phototube.
Random (random increase in current)
1. (2 points) Presence of a contaminant that absorbs over the same frequency range as the analyte.
Systematic (contaminant absorbs a constant amount of light based on concentration)
5) (15 points)
a. (5 points)What is the difference between internal conversion and external conversion in fluorescence?
Internal conversion: crossing of e- to lower electronic state, particularly if vibrational levels overlap. Efficient and main reason most compounds do not fluoresce. External collesion: deactivation
or loss of energy through collision with solvent.
b. (5 points)What is the difference between singlet and triplet states in fluorescence?
Singlet state: electron spins are paired no net magnetic field.
Triplet state: electron spins unpaired, net magnetic field.
Singlet state is higher energy than triplet state.
c. (5 points) Identify one mechanism by which fluorescence can be quenched.
Increase oxygen concentration or collision with solvent (increase viscosity or temperature)
6) (25 points)An echellette grating that contains 2800 blazes/mm is used in a UV/vis spectrometer where the incident angle of the polychromatic beam is 63.26o.
1. (5 points) At what angle from the grating would you be able to observe light with a wavelength of 600nm?
n = d(sin + sin )
d = 1 mm/2800 blazes x 106nm/mm = 357.1 nm/blaze
600 nm = 357.1 nm(sin r + sin 63.26)
sin r = 600 nm/357.1 nm - sin 63.26
sin r = 0.787
r = sin-1 0.787 = 51.9o
b.(5 points)What is the energy of a photon of this light?
E=hhc: E(600 nm) = (6.63x10-34J.s x 3.00 x 108 m/s)/600 x10-9 m = 3.32 x10-19 J
1. (5 points)If an analyte exhibited a percent transmission of 25% at 600nm, what is its absorbance?
A = -log(%T/100)
A = -log(25%/100)
A = -log(0.25) = 0.60
1. (10 points) If the light strikes a glass cuvette at an angle of 10o, what are the angles of reflection and refraction (assume index of refraction for air = 1.0003 and glass = 1.52)?
angle of reflection = 10O (other side of normal)
angle of refraction (2) = sin-1(1sin1 /2) = sin-1(1.0003*sin(10)/1.52) sin-1 (0.5699) = 6.6O
7) (11 points)
a. (6 points) If an absorption filter with a bandwidth of 100 nm and a maximum transmission at 260 nm is used in a UV/vis spectrometer,what is the approximate range of wavelengths available for
measuring an analytes absorption spectrum?
Conservatively: 210-310 nm or liberal: 160-360nm
b. (5 points)Consider two light sources that both generate light with a wavelength of exactly 550 nm that are both pointed at an object in a darken room. If the distance between the light sources and
the object are 1.75 m and 1.2 m, respectively, would the object be illuminated?
Yes, distance difference (0.550m) is integer multiple of wavelength | {"url":"https://docsbay.net/doc/980658/chemistry-421-821-first-exam-spring-2009-page-1name","timestamp":"2024-11-10T13:02:44Z","content_type":"text/html","content_length":"18750","record_id":"<urn:uuid:eec8ddda-9158-4298-8258-13df1d75ac4f>","cc-path":"CC-MAIN-2024-46/segments/1730477028186.38/warc/CC-MAIN-20241110103354-20241110133354-00205.warc.gz"} |
ball mill develop shaft layout with load directions and values
WEBOct 19, 2019 · Ball mills are extensively used in the size reduction process of different ores and minerals. The fill level inside a ball mill is a crucial parameter which needs to be
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WEBJan 29, 2022 · Ball Bearing Types and Components. A ball bearing is a circular joint that connects a rotating part to another, usually stationary, part of a machine. It allows the rotating
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WEBLoing the workpiece is the primary function of a jig or fixture. Once loed, the workpiece must also be held to prevent movement during the operational cycle. The process of holding the
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WEBQuestion: You are selecting a bearing to support a radial load of 2 kN where the shaft will rotate at a speed of 520 rev/min. a. Consider a 02series deepgroove ball bearing: Determine the alog
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WEBMay 27, 2023 · A shaft is a rotating member that transmits power between two parts through a twisting moment or parts such as gears or pulleys can be mounted onto the shaft to transmit power
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WEBThe shaft shown in the figure is driven by a gear at the right keyway, drives a fan at the left keyway, and is supported by two deepgroove ball bearings. The shaft is made from AISI 1020
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WEBAug 28, 2014 · Gear drives can be classified by shaft arrangement, type of gearing, ratio change, and horsepower. Three broad types are as follows: 1. Enclosed gear unit: The driver is mounted
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WEBFeb 15, 2012 · The realtime mill load (ML) measurement has not been accomplished completely [5]. Therefore, it is essential to develop a method to monitor the ML. The ML of the industrial ball
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WEBFeb 2, 2015 · If plant data are not available, laboratory test needs to be conducted on the ore sample of interest. Bond developed crushing, rod mill and ball mill laboratory tests. In the
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WEBJun 4, 2003 · A characteristic mill load vector is constructed from the singular value entropies of the cylinder vibration signals recorded under different load conditions and is used as the
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WEBProblem 19P. Chapter. CH10. Problem. 19P. Stepbystep solution. Step 1 of 4. Calculate the radial load on the gear . Here is load on gear .
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WEBDrive shaft is a critical component used in paper converting machines. It carries a load of two vacuum rollers weighing around 1471N and rotates at 1000 rpm, also subjected to reaction force
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WEBOct 19, 2016 · Ball Mill Sole Plate. This crown should be between .002″ and . 003″, per foot of length of sole plate. For example, if the sole plate is about 8′ long, the crown should be
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WEBJan 31, 2018 · Shaft Type Selection. Once the gear set has been identified, the next step our engineers follow is to determine what shaft should be used to transmit power. The obvious
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WEBA bearing arrangement supports and loes a shaft, radially and axially, relative to other components such as housings. Typically, two bearing supports are required to position a shaft.
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WEBUse the general shaft layout given and determine critical diameters of the shaft based on infinite fatigue life with a design factor of Check for yielding. Check the slopes at the bearings for
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WEBWith this arrangement, the shaft can also be axially loed by other components on the shaft ( a double helical gear). Most common bearings are: deep groove ball bearings ( fig. 15) selfaligning
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WEBConsider the round steel shaft below that has a transverse belt tension load of 200 lb is supported on each end by cylindrical roller angular deflection θ at each end of the shaft must not
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WEBMechanical Engineering questions and answers. 2. A 100 mm diameter shaft has a 20 mm diameter ball rolling around the outside. Find the maximum contact stress, the maximum deflection, and the
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WEBMay 8, 2017 · Thus the power to drive the whole mill. = + = = 86 kW. From the published data, the measured power to the motor terminals is 103 kW, and so the power demand of 86 kW by the mill
leads to a combined efficiency of motor and transmission of 83%, which is reasonable.
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Electric Power & Energy | Mini Physics - Free Physics Notes
Electric Power & Energy
Show/Hide Sub-topics (Practical Electric Circuitry | O Level Physics)
Table of Contents
Electricity plays a pivotal role in powering our modern world, seamlessly transforming into various forms of energy through the operation of diverse electrical devices. However, this powerful
resource comes with its risks. Unsafe practices in handling electricity can lead to devastating consequences, including electrical fires and shocks, potentially causing serious injuries or even
fatalities. To mitigate these dangers, incorporating safety mechanisms such as fuses, circuit breakers, switches, and earthing (grounding) wires is essential. These features are designed to interrupt
power in abnormal conditions, preventing electrical fires and reducing the risk of electric shocks, thereby safeguarding users from harm.
The Heating Effect Of Electricity
One of the most familiar phenomena associated with electricity is its ability to produce heat. This is known as the heating effect of electric current, where electrical energy is converted into
thermal energy as current flows through a resistor. This principle underpins the operation of many household appliances that generate heat, such as electric kettles, ovens, heaters, irons, hair
dryers, toasters, and electric cookers. The core of these appliances contains a heating element, typically made from materials with high resistance like nichrome wire. The extent of the heating
effect can be controlled by adjusting the current flowing through this element, allowing for precise temperature management.
Beyond heating, electricity’s versatility is showcased in its magnetic and chemical effects, as seen in applications like electromagnets and electrolysis, respectively. This highlights electricity’s
broad utility across various fields.
Power Of An Electrical Appliance
The rate of the heating effect is quantified in terms of power, measured in kilowatts (kW), while the total energy conversion is gauged in kilowatt-hours (kWh). The power (P) of an electrical
appliance can be calculated through the equations:
$$\begin{aligned} P &= I^{2}R \\ &= \frac{V^{2}}{R} \\ &= VI \end{aligned}$$
$V$ = voltage applied across appliance
$I$ = current flowing through appliance
$R$ = total resistance of appliance
SI unit for power is the kilowatt (kW), where 1 kW equals 1000 watts (W) or 1000 joules per second ($\text{J s}^{-1}$).
Energy Conversion In Electrical Appliance
The energy converted by an electrical appliance over a period of time is given by the formula:
$$\begin{aligned} E &= P \times t \\ &= I^{2}R \: \times \: t \\ &= \frac{V^{2}}{R} \: \times \: t \\ &= VI \: \times \: t \end{aligned}$$
$V$ = voltage applied across appliance
$I$ = current flowing through appliance
$R$ = total resistance of appliance
$t$ = total time taken
SI unit for energy conversion is the kilowatt-hour (kWh), where:
$$1 \text{kWh} = (1000 \text{W}) \times (60 \times 60 \text{s}) = 3 600 000 \text{J} = 3600 \text{kJ} = 3.6 \text{MJ}$$
Appliances with higher power ratings consume more electrical energy in a given time frame.
Efficiency & Cost Implications
Electrical appliances are indispensable in our daily lives, yet it’s essential to understand that they do not operate at perfect efficiency. In practical terms, efficiency refers to the proportion of
input energy that is successfully converted into the appliance’s intended output (for example, heat in a heater or light in a lamp). The remaining energy is not lost but is instead transformed into
other, often less desirable forms of energy such as sound or light (in the case of non-light producing appliances), and heat in devices not intended to produce it. This inherent inefficiency affects
both the operational cost and the environmental footprint of using electrical appliances.
The formula to calculate the efficiency ($\eta$) of an electrical appliance is given by:
$$\eta = \left( \frac{\text{Useful energy output}}{\text{Total energy input}} \right) \times 100\%$$
Understanding efficiency is crucial for several reasons. Firstly, it directly impacts the cost of operating the appliance. Higher efficiency means more of the electrical energy you’re paying for is
converted into the work you want (like cooling your room or cooking your food), which translates to lower electricity bills over time. Secondly, efficiency has environmental implications; more
efficient appliances use less power for the same output, reducing the demand on power plants and thereby potentially lowering greenhouse gas emissions.
The consumption of electrical energy in domestic settings is meticulously quantified using meters that measure in kilowatt-hours (kWh). The cost of energy consumption is straightforwardly calculated
by multiplying the number of kilowatt-hours consumed by the rate charged per kilowatt-hour. This rate can vary significantly depending on geographical location, time of use, and the energy provider’s
pricing scheme.
Monthly monitoring of energy usage, facilitated by reading the electricity meter, allows households to track their electrical energy consumption accurately. The difference in meter readings from the
end of one month to the end of the next provides a clear picture of the current month’s consumption. This transparent system not only aids in understanding and managing energy use but also encourages
the adoption of more energy-efficient appliances and habits, leading to cost savings and environmental benefits.
In summary, while electrical appliances provide numerous conveniences and comforts, their efficiency plays a pivotal role in determining their operational cost and environmental impact. By choosing
high-efficiency models and being mindful of our energy consumption habits, we can enjoy the benefits of these appliances while also contributing to a more sustainable and cost-effective use of
electrical energy.
Worked Examples
Example 1: Power Calculation
An electric kettle with a resistance of 30 ohms is connected to a 220-volt power supply. What is the power consumption of the kettle?
Click here to show/hide answer
Power can be calculated using $P = \frac{V^2}{R}$.
Given $V = 220$ volts and $R = 30$ ohms,
$$\begin{aligned} P &= \frac{220^2}{30} \\ &= \frac{48400}{30} \\ &= 1613.33 \, \text{W} \\ &= 1.6 \text{ kW} \end{aligned}$$
The kettle consumes 1.6 kilowatts of power.
Example 2: Energy Conversion
If the electric kettle from Example 1 is used for 2 hours, how much energy (in kWh) does it consume?
Click here to show/hide answer
Using $E = P \times t$,
Given $P = 1.6$ kW (from Example 3) and $t = 2$ hours,
$$E = 1.6 \, \text{kW} \times 2 \, \text{hours} = 3.2 \, \text{kWh}$$
The kettle consumes 3.2 kilowatt-hours of energy.
Example 3: Efficiency and Cost
Assuming the cost of electricity is $0.15 per kWh, calculate the cost of operating the electric kettle from Example 1 for 2 hours.
Click here to show/hide answer
Given the energy consumption is 3.2 kWh (from Example 4) and the cost is $0.15 per kWh,
$$\text{Cost} = 3.2 \, \text{kWh} \times \$0.15/\text{kWh} = \$0.48$$
It costs $0.48 to operate the kettle for 2 hours.
Example 4: Understanding Appliance Efficiency
An electric toaster has a power rating of 1 kW and operates for 0.5 hours, consuming 0.5 kWh of energy. If the toaster converts 90% of the electrical energy to thermal energy, how much energy is
wasted in forms other than heat?
Click here to show/hide answer
Total energy consumed is 0.5 kWh. 90% of this energy is used as thermal energy, so:
$$\text{Useful energy} = 0.5 \, \text{kWh} \times 0.9 = 0.45 \, \text{kWh}$$
$$\begin{aligned} \text{Energy wasted} &= \text{Total energy} – \text{Useful energy} \\ &= 0.5 \, \text{kWh} – 0.45 \, \text{kWh} \\ &= 0.05 \, \text{kWh} \end{aligned}$$
Thus, 0.05 kilowatt-hours of energy is wasted in forms other than heat.
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Back To Practical Electric Circuitry (O Level Physics)
Back To O Level Physics Topic List | {"url":"https://www.miniphysics.com/electric-power-and-energy.html","timestamp":"2024-11-13T12:18:07Z","content_type":"text/html","content_length":"90766","record_id":"<urn:uuid:4e4f98bd-2ea7-4d1f-ac6a-5fb3d030d927>","cc-path":"CC-MAIN-2024-46/segments/1730477028347.28/warc/CC-MAIN-20241113103539-20241113133539-00334.warc.gz"} |
Realizability of the adams-novikov spectral sequence for formal a-modules
We show that the formal A-module Adams-Novikov spectral sequence of Ravenel does not naturally arise from a filtration on a map of spectra by examining the case A = Z[i]. We also prove that when A is
the ring of integers in a nontrivial extension of Qp, the map (L,W) → (LA,WA) of Hopf algebroids, classifying formal groups and formal A-modules respectively, does not arise from compatible maps of
E∞-ring spectra (MU,MUΛMU) → (R, S).
Dive into the research topics of 'Realizability of the adams-novikov spectral sequence for formal a-modules'. Together they form a unique fingerprint. | {"url":"https://experts.umn.edu/en/publications/realizability-of-the-adams-novikov-spectral-sequence-for-formal-a","timestamp":"2024-11-04T16:04:25Z","content_type":"text/html","content_length":"48730","record_id":"<urn:uuid:c4e0ec59-490a-4375-90f0-f8e74df2fa46>","cc-path":"CC-MAIN-2024-46/segments/1730477027829.31/warc/CC-MAIN-20241104131715-20241104161715-00742.warc.gz"} |
100 research outputs found
In this paper we review the theory of the Yang-Baxter equation related to the 6-vertex model and its higher spin generalizations. We employ a 3D approach to the problem. Starting with the 3D
R-matrix, we consider a two-layer projection of the corresponding 3D lattice model. As a result, we obtain a new expression for the higher spin $R$-matrix associated with the affine quantum algebra
$U_q(\widehat{sl(2)})$. In the simplest case of the spin $s=1/2$ this $R$-matrix naturally reduces to the $R$-matrix of the 6-vertex model. Taking a special limit in our construction we also obtain
new formulas for the $Q$-operators acting in the representation space of arbitrary (half-)integer spin. Remarkably, this construction can be naturally extended to any complex values of spin $s$. We
also give all functional equations satisfied by the transfer-matrices and $Q$-operators.Comment: 25 pages, 1 figur
In this letter I shall review my joint results with Vadim Kuznetsov and Evgeny Sklyanin [Indag. Math. 14 (2003), 451-482, math.CA/0306242] on separation of variables for the $A_n$ Jack polynomials.
This approach originated from the work [RIMS Kokyuroku 919 (1995), 27-34, solv-int/9508002] where the integral representations for the $A_2$ Jack polynomials was derived. Using special polynomial
bases I shall obtain a more explicit expression for the $A_2$ Jack polynomials in terms of generalised hypergeometric functions.Comment: This is a contribution to the Vadim Kuznetsov Memorial Issue
on Integrable Systems and Related Topics, published in SIGMA (Symmetry, Integrability and Geometry: Methods and Applications) at http://www.emis.de/journals/SIGMA
In this paper we continue the study of $Q$-operators in the six-vertex model and its higher spin generalizations. In [1] we derived a new expression for the higher spin $R$-matrix associated with the
affine quantum algebra $U_q(\widehat{sl(2)})$. Taking a special limit in this $R$-matrix we obtained new formulas for the $Q$-operators acting in the tensor product of representation spaces with
arbitrary complex spin. Here we use a different strategy and construct $Q$-operators as integral operators with factorized kernels based on the original Baxter's method used in the solution of the
eight-vertex model. We compare this approach with the method developed in [1] and find the explicit connection between two constructions. We also discuss a reduction to the case of finite-dimensional
representations with (half-) integer spins.Comment: 18 pages, no figure
In this letter we establish a connection of Picard-type elliptic solutions of Painleve VI equation with the special solutions of the non-stationary Lame equation. The latter appeared in the study of
the ground state properties of Baxter's solvable eight-vertex lattice model at a particular point, $\eta=\pi/3$, of the disordered regime.Comment: 9 pages, LaTeX, submitted to the special issue on
Painleve VI, Journal of Physics
As is known, tetrahedron equations lead to the commuting family of transfer-matrices and provide the integrability of corresponding three-dimensional lattice models. We present the modified version
of these equations which give the commuting family of more complicated two-layer transfer-matrices. In the static limit we have succeeded in constructing the solution of these equations in terms of
elliptic functions.Comment: 11 page
A two component model describing the electromagnetic nucleon structure functions in the low-x region, based on generalized vector dominance and color dipole approaches is briefly described.Comment: 3
pages, 1 figure, Talk given at the 14th Lomonosov Conference, Moscow, August 200
We study a special anisotropic XYZ-model on a periodic chain of an odd length and conjecture exact expressions for certain components of the ground state eigenvectors. The results are written in
terms of tau-functions associated with Picard's elliptic solutions of the Painlev\'e VI equation. Connections with other problems related to the eight-vertex model are briefly discussed.Comment: 18
We study the ground state eigenvalues of Baxter's Q-operator for the eight-vertex model in a special case when it describes the off-critical deformation of the $\Delta=-1/2$ six-vertex model. We show
that these eigenvalues satisfy a non-stationary Schrodinger equation with the time-dependent potential given by the Weierstrass elliptic P-function where the modular parameter $\tau$ plays the role
of (imaginary) time. In the scaling limit the equation transforms into a ``non-stationary Mathieu equation'' for the vacuum eigenvalues of the Q-operators in the finite-volume massive sine-Gordon
model at the super-symmetric point, which is closely related to the theory of dilute polymers on a cylinder and the Painleve III equation.Comment: 11 pages, LaTeX, minor misprints corrected,
references adde | {"url":"https://core.ac.uk/search/?q=authors%3A(Mangazeev%20V%20V)","timestamp":"2024-11-09T09:37:50Z","content_type":"text/html","content_length":"133012","record_id":"<urn:uuid:872c7f9c-cfef-4d4c-a83b-f5a276d0ecce>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.75/warc/CC-MAIN-20241109085148-20241109115148-00232.warc.gz"} |
Land's Exact: A lie?
Hi Jerome and Mike (and everyone),
I simulated samples, and calculated Hewitt’s approximation of Land’s Exact 95% UCL (as used in IHstats).
By definition, you’d expect the UCL to be greater than the population mean 95% of the time.
I’m finding it’s more like 99.5% consistently.
No matter what I do to play with the parameters, I get +99%. And when sample size gets around 30, it consistently gives 100%
Have I made a mistake, or does the 95% UCL not work as intended?
R Code for reference:
#Land's Exact 95%UCL function
UCL <- function(data){
sy <- sd(log(data))
n <- as.numeric(length(data))
yhat <- mean(log(data))
mu <- exp(yhat + 1/2 * sy^2)
a <- 0.76766658
b <- 3.8716869
c <- 0.80598919
d <- 6.0321019
e <- 0.89998154
f <- 2.012669
g <- 0.21978875
h <- 0.41575588
i <- 0.29258276
F1 <- sy * (i + 1/(n-2)^c)
F2 <- b + d/(n-2)^c
F3 <- F1 * (1-e*exp(-f*F1))
F4 <- 1 + g * exp(-h*F1)
F5 <- F2 * F3/F4
C <- 1.645 + a/(n-2)+F5
return( as.numeric(exp(log(mu) + C * sy / sqrt(n-1))))
#Number of samples (repetitions)
sample.n <- 1e4
#Random Parameters
sim <- data.frame("mu" = runif(sample.n, min = 1, max = 4.5),
"sigma" = runif(sample.n, min = 0.5, max = 2),
"size" = sample(6:10, sample.n, replace = TRUE))
#Consistent Parameters
#mu <- 1
#sigma <- 1
#sample.size <- 6
#sim <- data.frame("mu" = rep(mu, sample.n),
# "sigma" = rep(sigma, sample.n),
# "size" = rep(sample.size, sample.n))
sim <- sim %>%
rowwise() %>%
mutate(samples = list(round(rlnorm(size, mu, sigma),2)),
UCL = UCL(samples),
is.contained = UCL > exp(mu))
# Coverage Percentage
length(sim$is.contained[sim$is.contained == TRUE]) / sample.n *100
Hello John,
As always, prone to having myself made a mistake, but I think, fortunately, Land’s reputation remains untarnished
I think : is.contained = UCL > exp(mu))
should be : is.contained = UCL > exp(mu+ 0/2*sigma^2))
with exp(mu+ 0/2*sigma^2)) being the true distribution AM
Indeed, mu is the log geometric mean in your main script (whereas the same notation is used for the arithmetic mean of a lognormal distribution in your UCL function).
With that change, running the script a couple of times gave me coverage between 94.5 and 95.5%.
Cheers, poring over R code at the end of the day is the best way to clean off the day’s mental charge
PS: Although I loathe tidyverse (just because I am unaccustomed to it), I found your script really elegantly written and easy to follow !
1 Like
Oh, great! I’ve never been so relieved to be wrong
I was kind of hoping it was a basic mistake, but I think I read what I expect rather than what is in front of me.
When I make the correction to calculating the AM, instead of the GM, I also get ~95%.
Thanks for checking my code - I don’t have many friends that can do so.
Also, I’m glad you found it legible. I was surprised how self conscious I get when sharing my code. I have only used tidyverse for “piping” and it’s mutate function which I find very handy.
1 Like | {"url":"https://discourse.expostats.ca/t/lands-exact-a-lie/278","timestamp":"2024-11-14T14:24:53Z","content_type":"text/html","content_length":"19641","record_id":"<urn:uuid:b5f32eb0-1181-4783-b882-e1d390bb58e7>","cc-path":"CC-MAIN-2024-46/segments/1730477028657.76/warc/CC-MAIN-20241114130448-20241114160448-00487.warc.gz"} |
Rejection Sampling: Definition, Types, Examples
Rejection sampling is a popular method for generating random variates. It’s based on the idea that, if you generate a number from some probability distribution and that number turns out to be outside
the bounds of distribution, you can just discard it and try until you find one that works.
In this article, we will discuss the advantages and disadvantages of rejection sampling in research, along with when and how rejection sampling can be used in a study.
What is Rejection Sampling?
Rejection Sampling is a method of statistical inference. It involves drawing random samples and rejecting those that don’t meet some threshold until you reach the number of samples you need.
It is a method for creating samples from one distribution by using an easier distribution. For instance, imagine you have a coin that lands on heads 60% of the time.
You want to use this coin to create samples from another distribution that also has a probability of 60% for an outcome. Only that this other distribution is much harder to sample from than just
flipping the coin. You could write a program that flips the coin over and over again until there are 60 “heads” and 40 “tails” or to your desired ratio.
However, to get a good number of samples, you will have to flip the coin thousands of times. This would take a lot of time and still wouldn’t give you perfect results.
On the other hand, if you had another coin that already has the desired ratio built-in, this coin would be much easier to work with because you could just flip it and use the results you get. So what
rejection sampling does is build this second “easy-to-sample” distribution so that it closely matches the first one.
Read: Consecutive Sampling: Definition, Examples, Pros & Cons
Advantages of Rejection Sampling
Rejection Sampling has several advantages over other methods for sampling, some of which include:
• It can be used with any distribution.
• It’s easy to modify it for different target distributions
• It can be used to generate any number of samples at once.
• It is easy to implement
• There is no restriction on the support of the target distribution.
• You can use rejection sampling even if you don’t know the constant of the probability distribution you’re trying to sample from.
• The rejection constant is usually not too difficult to find in practice.
Read: Convenience Sampling: Definition, Applications, Examples
Disadvantages of Rejection Sampling
Rejection Sampling has some disadvantages such as:
• It can take a lot of samples until you get one that fits your criteria, so it’s inefficient.
• It can be slow if the probability density function (pdf) of your target distribution is very close to zero at most points in its range.
When Should Rejection Sampling Be Used?
Rejection sampling is a type of sampling that’s often used when you’re estimating a quantity, but it’s not always the best option. However, you can use rejection sampling in the following cases:
• If the distribution of the quantity to be estimated is known.
• If the distribution of the quantity to be estimated can be tightly bound by some other distribution.
• If you have an easy way to sample from another distribution (that bounds your target distribution).
You can also make use of rejection sampling when you want to:
• Generate uniform random numbers that fall within an arbitrary polygon or polyhedron.
• Simulate rare events that are hard to predict.
How to Conduct Rejection Sampling
The rejection sampling process consists of two steps. In the first step, a sample is selected from a distribution that has a known probability density function (pdf).
In the second step, the sample is accepted or rejected based on a probability density function that’s related to the pdf in the first step. If the sample is accepted, it’s returned to the calling
routine; if it’s rejected, you go back to the first step and select another sample.
The process works because it can simulate any random variable whose distribution matches the pdf in step one. To understand how this works, consider the following example.
The goal of a sampling process is to generate random values that follow a normal distribution with mean 0 and standard deviation 1. These values will be generated by starting with samples from an
exponential distribution with a mean of 1 and then accepting those samples based on their proximity to 0.
Read: What is Stratified Sampling? Definition, Examples, Types
The exponential distribution has an easy-to-use inverse function, while the normal distribution doesn’t. This process can be used to approximate any distribution you could want.
Therefore, rejection sampling involves three steps:
1. Generate a random sample from the domain of interest
2. Calculate the probability density function (PDF) at this point
3. Accept or reject the sample based on whether it meets certain criteria
The PDF is used either to evaluate the probability that an event will occur under some circumstances or to represent the relative likelihood of different events. If accepted, then you have a sample
from your distribution, if otherwise, you go back and start over.
Read: Probability Sampling: Definition, Types, Examples, Pros & Cons
Rejection Sampling Techniques
Rejection sampling has a relative simplicity of algorithm that can be used to generate samples. The process is as follows:
1. Set a proposal distribution with a known density, q(x).
2. Draw a random number x uniformly distributed between 0 and 1
3. Draw another random number y uniformly distributed between 0 and 1
4. If y < f(x)/M, where f(x) is the probability density function and M is a constant, then accept the value x as a sample from the distribution f(x).
Select an arbitrary distribution function, f(x). This function should be the same as the distribution of your data. In other words, if you are trying to draw a random sample that is uniformly
distributed between 1 and 10, then this function would be f(x) = x.
Select a probability density function (p.d.f.) from which you can sample easily. The p.d.f. should be greater than 0 everywhere, but does not need to approach zero quickly (e.g., p(x) = 1).
A constant p.d.f. works well for most purposes; however, any p.d.f will work so long as it is greater than 0 everywhere and does not approach zero quickly at either end of the interval in which it is
defined (this might seem like a tall order, but such functions do exist). The process of rejection sampling can be illustrated with an example.
Read: Cluster Sampling Guide: Types, Methods, Examples & Uses
Suppose we wish to sample from a uniform distribution in the interval [0, 1], but we only have a normal distribution with a mean of 0 and a standard deviation of 1. Since there is no easy way to
generate uniform random numbers, we will accept or reject samples from the normal distribution until we have generated a uniform random number.
The procedure is as follows:
Sample x from the normal distribution
Set y = f(x) where f is the density function of the target distribution, which in this case would be the uniform density function in [0, 1]. In other words, y is a uniform random variable multiplied
by the density of the normal distribution at x.
Generate another uniformly distributed random variable u between 0 and 1. If u < y, accept x as a sample; otherwise reject it. Repeat until the desired number of samples is obtained.
Examples of Rejection Sampling
Example 1
Imagine that you want to generate samples from the distribution shown in the graph below. The distribution has a sharp peak over the interval (0,1) but falls quickly to zero outside this range.
This distribution is difficult to sample directly because of its narrow peak. To sample this distribution using rejection sampling, we first need to choose an envelope distribution that has no
regions where it drops to zero.
In this example, we will use a uniform distribution that spans the entire range of the target distribution (in this case from 0 to 1). The envelope and target distributions are shown in the graph
Notice that there are no regions where the envelope distribution drops to zero within the region where the target distribution does not drop to zero, so this pair of distributions satisfy our
requirements for rejection sampling.
Example 2
Let’s say you were applying to three different jobs at the same company, but only got offered one of them. You could say that you were rejected by the other two jobs and accepted by the third job.
Or if you’re trying to get a promotion at work and your boss says no, but then tells you about an opening at another company that could help you advance in your career, you can think of it as getting
rejected by your boss, but then discovering another option through that rejection. For all its benefits (like helping us grow!), rejection sampling can be tough at first.
Researchers should note that rejection sampling works by taking samples from an “envelope” distribution and accepting them with a probability that depends on the target distribution. If the sample is
rejected, you can be assured that another sample will be taken until one is accepted. | {"url":"https://www.formpl.us/blog/rejection-sampling","timestamp":"2024-11-09T16:54:06Z","content_type":"text/html","content_length":"42830","record_id":"<urn:uuid:66817739-a28f-41c7-afac-b36a1d14ed5f>","cc-path":"CC-MAIN-2024-46/segments/1730477028125.59/warc/CC-MAIN-20241109151915-20241109181915-00734.warc.gz"} |
Introduction to R for Data Science :: Session 8 [Appendix] | R-bloggersIntroduction to R for Data Science :: Session 8 [Appendix]
Introduction to R for Data Science :: Session 8 [Appendix]
[This article was first published on
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Welcome to Introduction to R for Data Science, Session 8: Intro to Text Mining in R, ML Estimation + Binomial Logistic Regression [Web-scraping with tm.plugin.webmining. The tm package corpora
structures: assessing document metadata and content. Typical corpus transformations and Term-Document Matrix production. A simple binomial regression model with tf-idf scores as features and its
shortcommings due to sparse data. Reminder: Maximum Likelihood Estimation with Nelder-Mead from optim().]
The course is co-organized by Data Science Serbia and Startit. You will find all course material (R scripts, data sets, SlideShare presentations, readings) on these pages.
Check out the Course Overview to acess the learning material presented thus far.
Data Science Serbia Course Pages [in Serbian]
Startit Course Pages [in Serbian]
Summary of Session 8, 17. June 2016 :: Intro to Text Mining in R + Binomial Logistic Regression.
Intro to Text Mining in R + Binomial Logistic Regression. Intro to Text Mining in R + Binomiral Logistic Regression: Web-scraping with tm.plugin.webmining. The tm package corpora structures:
assessing document metadata and content. Typical corpus transformations and Term-Document Matrix production. A simple binomial regression model with tf-idf scores as features and its shortcommings
due to sparse data. Reminder: Maximum Likelihood Estimation with Nelder-Mead from optim().
R script :: Session 8
Split data into training and test
#### w. training vs. test data set
# split into test and training
choice <- sample(1:475,250,replace = F)
test <- which(!(c(1:475) %in% choice))
trainData <- dataSet[choice,]
newData <- dataSet[test,]
# check!
sum(dataSet$Category[choice])/length(choice) # proportion of dotCom in training
sum(dataSet$Category[test])/length(choice) # proportion of dotCom in test
# Binomial Logistic Regression: use glm w. logit link
bLRmodel <- glm(Category ~.,
control = list(maxit = 500),
sumLR <- summary(bLRmodel)
# Coefficients
coefLR <- as.data.frame(sumLR$coefficients)
# Wald statistics significant? (this Wald z is normally distributed)
coefLR <- coefLR[order(-coefLR$Estimate), ]
w <- which((coefLR$`Pr(>|z|)` < .05)&(!(rownames(coefLR) == "(Intercept)")))
# which predictors worked?
# plot coefficients {ggplot2}
plotFrame <- coefLR[w,]
plotFrame$Estimate <- round(plotFrame$Estimate,2)
plotFrame$Features <- rownames(plotFrame)
plotFrame <- plotFrame[order(-plotFrame$Estimate), ]
plotFrame$Features <- factor(plotFrame$Features, levels = plotFrame$Features, ordered=T)
ggplot(data = plotFrame, aes(x = plotFrame$Features, y = plotFrame$Estimate)) +
geom_line(group=1) + geom_point(color="red", size=2.5) + geom_point(color="white", size=2) +
xlab("Features") + ylab("Regression Coefficients") +
ggtitle("Logistic Regression: Coeficients (sig. Wald test)") +
theme(axis.text.x = element_text(angle=90))
# fitted probabilities
main = "Predicted Probabilities: Density")
# Prediction from the model
predictions <- predict(bLRmodel,
predictions <- ifelse(predictions >= 0.5,1,0)
trueCategory <- newData$Category
meanClasError <- mean(predictions != trueCategory)
accuracy <- 1-meanClasError
accuracy # probably rather poor..? - Why? - Think!
# Try to train a binomial regression model many times by randomly assigning
# documents to the training and test data set
# What happens? Why?
# *Look* at your data set and *think* about it before actually modeling it.
Readings :: Session 9: Binomial and Multinomial Logistic Regression [23. June, 2016, @Startit.rs, 19h CET] | {"url":"https://www.r-bloggers.com/2016/06/introduction-to-r-for-data-science-session-8-appendix/","timestamp":"2024-11-15T04:30:23Z","content_type":"text/html","content_length":"108027","record_id":"<urn:uuid:2c8e2391-e2ef-4f69-a5b2-250666f7d2bb>","cc-path":"CC-MAIN-2024-46/segments/1730477400050.97/warc/CC-MAIN-20241115021900-20241115051900-00302.warc.gz"} |
Data Drift | Evidently Documentation
TL;DR: The report detects changes in feature distributions.
Performs a suitable statistical test for numerical and categorical features
Plots feature values and distributions for the two datasets.
The Data Drift report helps detect and explore changes in the input data.
You will need two datasets. The reference dataset serves as a benchmark. We analyze the change by comparing the current production data to the reference data.
The dataset should include the features you want to evaluate for drift. The schema of both datasets should be identical.
In the case of pandas DataFrame,all column names should be string
All feature columns analyzed for drift should have the numerical type (np.number)
Categorical data can be encoded as numerical labels and specified in the column_mapping.
DateTime column is the only exception. If available, it can be used as the x-axis in the plots.
You can potentially choose any two datasets for comparison. But keep in mind that only the reference dataset will be used as a basis for comparison.
How it works
To estimate the data drift Evidently compares the distributions of each feature in the two datasets.
Evidently applies statistical tests to detect if the distribution has changed significantly. There is a default logic to choosing the appropriate statistical test based on:
feature type: categorical or numerical
the number of observations in the reference dataset
the number of unique values in the feature (n_unique)
For small data with <= 1000 observations in the reference dataset:
For categorical features or numerical features with n_unique <= 5: chi-squared test.
For binary categorical features (n_unique <= 2), we use the proportion difference test for independent samples based on Z-score.
All tests use a 0.95 confidence level by default.
For larger data with > 1000 observations in the reference dataset:
All tests use a threshold = 0.1 by default.
You can modify the drift detection logic by selecting a statistical test already available in the library, including PSI, K–L divergence, Jensen-Shannon distance, Wasserstein distance. See more
details about available tests. You can also set a different confidence level or implement a custom test, by defining custom options.
To build up a better intuition for which tests are better in different kinds of use cases, visit our blog to read our in-depth guide to the tradeoffs when choosing the statistical test for data
How it looks
The default report includes 4 components. All plots are interactive.
1. Data Drift Summary
The report returns the share of drifting features and an aggregate Dataset Drift result. For example:
Dataset Drift sets a rule on top of the results of the statistical tests for individual features. By default, Dataset Drift is detected if at least 50% of features drift at a 0.95 confidence level.
2. Data Drift Table
The table shows the drifting features first, sorting them by P-value. You can also choose to sort the rows by the feature name or type.
3. Data Drift by Feature
By clicking on each feature, you can explore the values mapped in a plot.
The dark green line is the mean, as seen in the reference dataset.
The green area covers one standard deviation from the mean.
4. Data Distribution by Feature
You can also zoom on distributions to understand what has changed.
Report customization
You can set different Options for Data / Target drift to modify the existing components of the report. Use this to change the statistical tests used, define Dataset Drift conditions, or change
histogram Bins.
You can also set Options for Quality Metrics to define the width of the confidence interval displayed for individual feature drift.
You can also select which components of the reports to display or choose to show the short version of the report: Select Widgets.
If you want to create a new plot or metric, you can Custom Widgets and Tabs.
When to use this report
Here are a few ideas on when to use the report:
In production: as early monitoring of model quality. In absence of ground truth labels, you can monitor for changes in the input data. Use it e.g. to decide when to retrain the model, apply
business logic on top of the model output, or whether to act on predictions. You can combine it with monitoring model outputs using the Numerical or Categorical Target Drift report.
In production: to debug the model decay. Use the tool to explore how the input data has changed.
In A/B test or trial use. Detect training-serving skew and get the context to interpret test results.
Before deployment. Understand drift in the offline environment. Explore past shifts in the data to define retraining needs and monitoring strategies. Here is a blog about it.
To find useful features when building a model. You can also use the tool to compare feature distributions in different classes to surface the best discriminants.
JSON Profile
If you choose to generate a JSON profile, it will contain the following information:
"data_drift": {
"name": "data_drift",
"datetime": "datetime",
"data": {
"utility_columns": {
"date": null,
"id": null,
"target": null,
"prediction": null,
"drift_conf_level": value,
"drift_features_share": value,
"nbinsx": {
"feature_name": value,
"feature_name": value
"xbins": null
"cat_feature_names": [],
"num_feature_names": [],
"metrics": {
"feature_name" :{
"prod_small_hist": [
"ref_small_hist": [
"feature_type": "num",
"p_value": p_value
"n_features": value,
"n_drifted_features": value,
"share_drifted_features": value,
"dataset_drift": false
"timestamp": "timestamp"
Data Drift Dashboard Examples
Browse our example notebooks to see sample Reports.
You can also read the initial release blog. | {"url":"https://docs.evidentlyai.com/v0.1.57/reports/data-drift","timestamp":"2024-11-10T11:50:48Z","content_type":"text/html","content_length":"559654","record_id":"<urn:uuid:cc3479e0-7138-47a5-8215-d584c7a5d267>","cc-path":"CC-MAIN-2024-46/segments/1730477028186.38/warc/CC-MAIN-20241110103354-20241110133354-00273.warc.gz"} |
Divide fractions with variables calculator
Author Message
adoaoon Posted: Sunday 20th of Sep 07:15
Hello Math Gurus. Ever since I have encountered divide fractions with variables calculator at school I never seem to be able to cope with it well. I am a topper at all the other
branches, but this particular branch seems to be my weak point. Can some one assist me in understanding it properly ?
From: Oakland,
oc_rana Posted: Sunday 20th of Sep 08:17
You seem to be stuck on that I had some time back. I too thought of hiring a paid tutor to work it out for me. But they are so expensive that I just could not afford one. So I turned
to the internet and found so many software that can help with math homework on linear equations, monomials and logarithms. After some research I found that Algebra Master is the best
of the lot. I haven’t found a algebra homework that I can’t get done through Algebra Master. It is absolutely amazing. Best part is, the software gives you a step-by-step explanation
on how to do it yourself. So you actually learn how to solve it yourself. Isn’t it cool?
Matdhejs Posted: Sunday 20th of Sep 09:26
Even I’ve been through that phase when I was trying to figure out a way to solve certain type of questions pertaining to unlike denominators and binomial formula. But then I came
across this piece of software and I felt as if I found a magic wand. In a flash it would solve even the most difficult questions for you. And the fact that it gives a detailed
step-by-step explanation makes it even more handy. It’s a must buy for every math student.
From: The
nonitic Posted: Monday 21st of Sep 09:30
I am so thankful to hear that there is hope for me. Thanks a lot . Why did I not think about this? I would like to begin on this right now . How can I get hold of this program?
Kindly give me the particulars of where and how I can get this program.
DoniilT Posted: Tuesday 22nd of Sep 07:19
Here it is: https://algebra-test.com/resources.html. Just a few clicks and algebra won't be a problem anymore . All the best and have fun with the program!
Matdhejs Posted: Wednesday 23rd of Sep 09:25
A extraordinary piece of math software is Algebra Master. Even I faced similar difficulties while solving roots, side-side-side similarity and linear algebra. Just by typing in the
problem from homework and clicking on Solve – and step by step solution to my algebra homework would be ready. I have used it through several algebra classes - Intermediate algebra,
Algebra 2 and Algebra 1. I highly recommend the program.
From: The | {"url":"http://algebra-test.com/algebra-help/relations/divide-fractions-with.html","timestamp":"2024-11-08T02:08:42Z","content_type":"application/xhtml+xml","content_length":"22367","record_id":"<urn:uuid:a03a9fde-4c7e-4bbf-9a00-51761d6ae661>","cc-path":"CC-MAIN-2024-46/segments/1730477028019.71/warc/CC-MAIN-20241108003811-20241108033811-00524.warc.gz"} |
Tensorflow is a high performance numerical computation software library, it is mostly known for its strong support for machine learning and deep learning.
How to Install
If you have ‘pip’, everything is simple pip install tensorflow
Basic Concepts
Graph is a fundamental concept in Tensorflow. Take ReLU computation as an example, the function of ReLU is
In the view of Tensorflow, the function looks like this
Variables such as W and b , placeholders such as x, are operations such as MatMul, Add are all nodes in the graph.
The edges between nodes indicate the data which flow between nodes, in tensorflow data is represented as “tensor” .
“tensor” + “flow” = “tensorflow”
import tensorflow as tf
import numpy as np
if __name__ == '__main__':
b = tf.Variable(tf.zeros((10,)))
W = tf.Variable(tf.random_uniform((20, 10), -1, 1))
x = tf.placeholder(tf.float32, (10, 20))
h = tf.nn.relu(tf.matmul(x, W) + b)
writer = tf.summary.FileWriter('./graphs', tf.get_default_graph())
with tf.Session() as sess:
sess.run(h, {x: np.random.random((10, 20))})
Here are the steps described in the above codes.
1. Create a graph using Variables and placeholders.
2. Start a tensorflow session and deploy the graph into the session.
3. Run the session, let the tensors flow.
4. Write processing logs using tools such as tf.summary.
Session is the so called execution environment. It needs two parameters which are “Fetches” and “Feeds”.
sess.run(fetches, feeds)
Fetches: List of graph nodes
Feeds: Dictionary mapping from graph nodes to concrete values.
In the ReLU example, Fetches = h = tf.nn.relu(tf.matmul(x, W) + b) , Feeds = {x: np.random.random((10, 20))}
It would be interesting to see what happened during the running time. We can try TensorBoard.
TensorBoard Result
The Main Graph clearly shows how the tensor flows through the graph. In a complex machine learning program, the printed diagram is a convenient tool to increase the confidence of the result. | {"url":"https://ofeng.org/posts/tensorflow101/","timestamp":"2024-11-10T12:32:34Z","content_type":"text/html","content_length":"20758","record_id":"<urn:uuid:12ef18ec-9278-49e4-ac05-383a868f5049>","cc-path":"CC-MAIN-2024-46/segments/1730477028186.38/warc/CC-MAIN-20241110103354-20241110133354-00506.warc.gz"} |
Computational Physics and a group of 1000 8th graders
I like computers, really I do. Computational physics is a good thing. However, there is a small problem. The problem is that there seems to be a large number of people out there that treat numerical
methods and simulations as something different than theoretical calculations. You can tell who these people are because they refer to simulations as "experiments". But what do these simulations
really do in science? What is science really all about?
To me, science is all about models. Making models, testing models, upgrading models. Models. Some examples are the model of gravity. One such model is that there is a gravitational force between any
two objects with mass. This force is inversely proportional the square of the distance between them. (This is Newton's model). Is this model perfect? No. Is this model the truth? No. How did this
model come about? Experimental evidence.
Well, how do you make models and what form can they take? To make a model, you collect some observations. The model should agree with these observations. This model could be a physical model (like
the globe). It could be a mathematical model (like V=IR). It could be a numerical model - like a [vpython](http://vpython.org) program of a baseball trajectory with air resistance. These are all
**8th graders**
What does any of this have to do with 8th graders? I claim that any numerical calculation or simulation could be done with a group of 1000 8th graders rather than a computer. What does a computer do?
(a computer program really) A program takes a problem and breaks it into a bunch a really small steps. It then does each of these steps and combines them together in some way. Just like a group of
8th graders with TI-89 calculators. Clearly, they are just computing something - they are not a separate type of science (other than theory and experiment).
More like this
I have been meaning to write about this for quite some time. Really, I wanted to reply to Chad's article on science at Uncertain Principles, but you know how things go. So, here are my key and
interesting points about science in random order. Science is all about models (not ball bearings)…
I have been meaning to write about this for quite some time. Really, I wanted to reply to Chad's article on science at Uncertain Principles, but you know how things go. So, here are my key and
interesting points about science in random order. Science is all about models (not ball bearings)…
**Pre Reqs:** [Kinematics](http://scienceblogs.com/dotphysics/2008/09/basics-kinematics.php), [Momentum Principle](http://scienceblogs.com/dotphysics/2008/10/basics-forces-and-the-moment…) What are
"numerical calculations"? Why are they in the "basics"? I will give you really brief answer and…
Looking back at part I of this idea, I don't think I did a very good job. Let me summarize the key things I wanted to say: Normally, there are two ways of modeling the motion of an object:
Calculating the forces on the object and using the momentum principle or Newton's second law (which are the…
Thank you for writing about computational physics! I am teaching 2nd-year HS physics to a group of kids who liked first-year physics so much they signed up for a second course in it...as seniors! I
am also thinking of assigning my first and second year students to read your blog and start a physics blog of our own. May I link to your blog on my website?
I plan to do some computational physics work with my 2nd-year students, using vpython, since we do not have access to 1000 8th-graders, and we don't have the patience to explain to even 100 8th
graders how to do the calculations we want done and them make sure they are doing it correctly. One of the advantages of the computer over 8th graders is that you only have to tell the computer once
what to do, whereas some 8th graders won't listen and others won't bother doing what you've asked them to, instead they will type "58008" and see if it spells what they hope it does when they turn
the calculator upside down. I think it is more like having 1000 young women looking at astronomical photos and looking for anomalies, as they are probably more careful and methodical than the average
group of 8th graders.
I look forward to reading more from you, thank you again!
Also, the more particles you need to keep track of, the more complex your model, the harder it would be to use 8th graders. The instructions are just too complicated. In addition, in doing the
computational work you may find out that your model produces results different from reality...or different from theory. This gives you important information that you might otherwise miss.
Well, I am not a working physicist, and my highest physics degree is a BA, so I will not try to convince you that computational physics should be considered as an addition to the traditional dyad of
theoretical and experimental physics. However, I will continue to believe that up-and-coming physicists should learn methods of computational physics as part of their training.
I should let you know that I have had a number of conversations with Bruce Sherwood of NCSU, who is passionate on this topic. | {"url":"https://scienceblogs.com/dotphysics/2008/09/15/computational-physics-and-a-group-of-1000-8th-graders","timestamp":"2024-11-02T05:08:40Z","content_type":"text/html","content_length":"40900","record_id":"<urn:uuid:884fba3e-ea82-4860-9124-fe44033e30e5>","cc-path":"CC-MAIN-2024-46/segments/1730477027677.11/warc/CC-MAIN-20241102040949-20241102070949-00870.warc.gz"} |
Associate Professor, Mathematics
(215)- 895-6623
(215) 895-1582 (fax)
Office hours - Korman 272
Tues 12-1, Wed 1-2
Professional interests
Applied mathematics, soliton theory,
applications to differential geometry, computer graphics .
Member of
American Mathematical Society
Society for Industrial and Applied Mathematics
Soliton links
Soliton theory investigates certain nonlinear equations, which can be exactly solved using a technique called the spectral transform. It is one of the major developments in applied mathematics of the
last twenty-five years, and has applications to fluid flow, nonlinear optics, and differential geometry.
My papers on soliton theory and differential geometry
Home-grown soliton surface pictures generated with software developed by the now defunct PSE group in CS at Drexel. Pictures listed without description -- but in one way or another, they are all
related to the Localized Induction Equation and its cousins
Los Alamos archive of papers on soliton theory (= integrable systems)
Univ. of Mass. Amherst geometry group: minimal surfaces, constant mean curvature surfaces
SFB 288: Home of the Berlin group of geometric visualization of soliton curves and surfaces and OOrange, an object- oriented visualization tool
Last update: 5/19/98 by rperline@mcs.drexel.edu | {"url":"https://www.math.drexel.edu/~rperline/","timestamp":"2024-11-07T23:51:40Z","content_type":"text/html","content_length":"4136","record_id":"<urn:uuid:d0773cfc-3a94-4688-84e4-050b5ad4bd61>","cc-path":"CC-MAIN-2024-46/segments/1730477028017.48/warc/CC-MAIN-20241107212632-20241108002632-00698.warc.gz"} |
vii. Derive the expression for PV work - Crack NTA
Chemical Thermodynamics Chapter 4 Chemistry Class 12 Textbook Solution
vii. Derive the expression for PV work
1. Imagine a specific quantity of gas held at a constant pressure \(P\) inside a cylinder equipped with a frictionless, rigid piston of area \(A\). The gas initially occupies a volume \(V_1\) at
temperature \(T\), as illustrated in the adjacent diagram.
2. During expansion, the force exerted by the gas is equal to the area of the piston multiplied by the pressure at which the gas acts against the piston. This pressure has the same magnitude as the
external atmospheric pressure but opposite in sign, denoted as \(-P_{\text{ext}}\). Hence,
\[f = -P_{\text{ext}} \cdot A \quad \text{(1)}\]
Here, \(P_{\text{ext}}\) represents the external atmospheric pressure.
3. When the piston moves a distance \(d\), the work performed is the product of the force and the distance traveled:
\[W = f \cdot d \quad \text{(2)}\]
By substituting equation (1) into equation (2), we obtain:
\[W = -P_{\text{ext}} \cdot A \cdot d \quad \text{(3)}\]
4. The product of the piston's area and the distance it moves represents the change in volume (\(\Delta V\)) of the system:
\[\Delta V = A \cdot d \quad \text{(4)}\]
Combining equations (3) and (4), we get:
\[W = -P_{\text{ext}} \cdot \Delta V\]
\[W = -P_{\text{ext}} \cdot (V_2 - V_1)\]
Here, \(V_2\) denotes the final volume of the gas.
Chemical Thermodynamics Chapter 4 Chemistry Class 12 Textbook Solution | {"url":"https://cracknta.com/vii-derive-the-expression-for-pv-work/","timestamp":"2024-11-09T19:30:45Z","content_type":"text/html","content_length":"60040","record_id":"<urn:uuid:e767b16e-689f-474c-bf0d-3b0c580221de>","cc-path":"CC-MAIN-2024-46/segments/1730477028142.18/warc/CC-MAIN-20241109182954-20241109212954-00448.warc.gz"} |
11 Important Questions Related to Vedic Mathematics - Ranvir Mehta's Academy
In this post, you will get answers to 11 such questions related to Vedic Mathematics, which are very important for you to know, so let’s start.
Questions Related to Vedic Mathematics
1. What is Vedic Mathematics?
Vedic Mathematics is a collection of many methods, many techniques by which we can solve the problem of maths easily, in less time, and mentally.
2. Is Vedic Mathematics Useful?
Yes definitely, Vedic Maths is very helpful whether you are an academic student, whether you are a competitive exam aspirant, whether you are a student of higher education, whether you are a
professional, whether you are a parent, whether you are a teacher, Vedic Maths is beneficial for everyone, it is useful for everyone in different ways.
3. Is Vedic Mathematics Difficult?
No, Vedic Maths is very easy. Its concept is very easy. They can be easily remembered. They can be implemented easily.
4. What is the right age to learn Vedic maths?
To learn Vedic Maths, you need to be at least seven to eight years of age, because it is conceptual mathematics, it also has to be remembered. If your age is above 10 years, then this is the best
opportunity for you.
5. What is the use of Vedic Mathematics or what are the applications of Vedic Mathematics or where can Vedic Mathematics be used?
Vedic Mathematics can be used in every branch of Mathematics such as Arithmetic, Algebra, Trigonometry, Geometry, Calculus, etc. All its concepts are very simple, easy, and give quick results.
6. What are the advantages of Vedic Maths?
Vedic Maths is also called Error Free Mathematics. Error-free mathematics is called because it eliminates the possibility of making a mistake. With its help, you can solve more problems in less
time. This increases your problem-solving speed. This increases mathematical intelligence. This improves the creative brain so that we can take advantage of fast calculation and mental
calculation by using the power of imagination. The benefit of Vedic Maths is there in academic performance as well as in daily life. Wherever you want to use it, you can take advantage of it.
7. Is Vedic maths good for kids?
Yes, Vedic maths must be taught to children so that their logical-mathematical intelligence and mental calculation speed can be good. Learn to solve mathematical problems mentally. To create
their interest in mathematics, we should teach them mathematics in easy language, and nothing can be better than Vedic mathematics for this.
8. Why is Vedic Maths not taught in schools?
Vedic Mathematics is not taught in school because teachers are not able to teach children properly due to the lack of proper training in Vedic Maths. Because of this, even easy concepts become
difficult concepts. That’s why children are not interested in maths. That’s why children are not interested in maths. If teachers have proper training on Vedic Mathematics in schools, then it can
be a very good initiative.
9. What is taught in Vedic Maths?
In Vedic Mathematics, whatever is the application of Vedic Maths in Arithmetic, Algebra, Geometry, Trigonometry, and Calculus, and wherever the concepts of Vedic Maths can be applied, we are
taught it. These concepts may be different for different age groups.
10. What is the future of Vedic Maths?
The score of Vedic Maths is very bright in India, as people are becoming aware of it, the demand for it is increasing because it is fast-result oriented. More results can be obtained in less
time. If a teacher takes the training of Vedic Maths and he/she starts teaching in his/her area as well, then there are many options available to him/her. Gradually, Vedic Maths is spreading
because our Prime Minister, Shri Narendra Modi has also promoted Vedic Mathematics in the Mann Ki Baat program.
11. Can I Learn Vedic Maths Online?
Definitely, we have many courses on Vedic Maths. You can enroll in whichever course you want to take. And only with the help of your mobile phone or laptop, you can properly learn Vedic Maths by
sitting at home. And if you have any doubts, you can clear them online.
Our Vedic Maths Courses
If you have any doubt/query/question regarding my Vedic Maths courses, please feel free to ask on WhatsApp or email.
My Email Id: [email protected]
My WhatsApp No.: 7850013920
You can ask your questions in comment section…. | {"url":"https://ranvirmehta.com/11-questions-related-to-vedic-mathematics/","timestamp":"2024-11-13T00:48:34Z","content_type":"text/html","content_length":"62618","record_id":"<urn:uuid:66408714-a498-4a1b-ab3f-9094d41e6e47>","cc-path":"CC-MAIN-2024-46/segments/1730477028303.91/warc/CC-MAIN-20241113004258-20241113034258-00097.warc.gz"} |
Michelson-Morley experiment - iSoulMichelson-Morley experiment
Michelson-Morley experiment
This post relates to a previous post here.
The Michelson-Morley experiment is a famous “null” result that has been understood as leading to the Lorentz transformation. However, an elementary error has persisted so that the null result is
fully consistent with classical physics.
Figures 1 & 2
The Michelson-Morley paper of 1887 [Amer. Jour. Sci.-Third Series, Vol. XXXIV, No. 203.–Nov., 1887] explains using the above figures:
Let sa, fig. 1, be a ray of light which is partly reflected in ab, and partly transmitted in ac, being returned by the mirrors b and c, along ba and ca. ba is partly transmitted along ad, and ca
is partly reflected along ad. If then the paths ab and ac are equal, the two rays interfere along ad. Suppose now, the ether being at rest, that the whole apparatus moves in the direction sc,
with the velocity of the earth in its orbit, the directions and distances traversed by the rays will be altered thus:— The ray sa is reflected along ab, fig. 2; the angle bab, being equal to the
aberration =a, is returned along ba[1], (aba[1] =2a), and goes to the focus of the telescope, whose direction is unaltered. The transmitted ray goes along ac, is returned along ca[1], and is
reflected at a[1], making ca[1]e equal 90—a, and therefore still coinciding with the first ray. It may be remarked that the rays ba[1] and ca[1], do not now meet exactly in the same point a[1],
though the difference is of the second order; this does not affect the validity of the reasoning. Let it now be required to find the difference in the two paths aba[1], and aca[1].
Let V = velocity of light.
v = velocity of the earth in its orbit,
D = distance ab or ac, fig. 1.
T = time light occupies to pass from a to c.
T[1] = time light occupies to return from c to a[1] (fig. 2.)
The paper then goes on to give expressions for T and T[1] incorrectly as
These are wrong because they assume that time is the variable in common, but actually distance is the variable in common since all distances are pre-determined, such that “the paths ab and ac are
equal”. Time is a dependent variable, and the velocities V and v are actually inverse velocities, and so are combined reciprocally [see here], not arithmetically.
The correct expressions are as follows:
Inverse velocity is the reciprocal of lenticity and combines with reciprocal addition and subtraction:
where ‘boxplus’ and ‘boxminus’ are the reciprocal versions of addition and subtraction. Another approach is instead of velocities to use lenticities, which are inverse velocities, with distance as
the variable in common. Then the correct values for T and T[1] are
Either approach leads to
So the elapsed time and the speed of light are independent of the earth in its orbit. Thus the Galilean transformation (actually the Euclidean transformation) may be used with a finite constant space
mean speed of light. The Lorentz transformation is not needed. | {"url":"https://www.isoul.org/michelson-morley-experiment/","timestamp":"2024-11-15T02:44:09Z","content_type":"text/html","content_length":"127555","record_id":"<urn:uuid:570ff12a-7ad0-4a60-aae0-69c704e34bb1>","cc-path":"CC-MAIN-2024-46/segments/1730477400050.97/warc/CC-MAIN-20241115021900-20241115051900-00108.warc.gz"} |
Maximum size and minimum no. of clues?
There is no theoretical limit to the size of a logic puzzle, but beyond a certain number of variables, you veer into "work" territory, either because using a grid becomes unwieldy, or because solving
requires too much comparison of the type "the remaining possibilities for X are not valid for Y therefore X does not equal Y", which becomes harder with more variables. If you look at a print puzzle
book, you'll find puzzles where you don't get a working grid (like you would on this site), just an answer grid.
As for number of clues, it depends on how many pieces of information are in each clue. So a clue that is "X is not Y" is one piece of information ("X is not Y"), but "X is bigger than Y" is four
pieces of information (X is not Y, X is not the smallest value, Y is not the biggest value, X is bigger than Y). You'd need at least (X-1)*Y pieces of information to solve a grid. | {"url":"https://forum.puzzlebaron.com/forum/puzzle-baron/logic-puzzles/35585-maximum-size-and-minimum-no-of-clues?view=stream","timestamp":"2024-11-07T00:30:17Z","content_type":"application/xhtml+xml","content_length":"56078","record_id":"<urn:uuid:c5234450-24f7-4cc8-b888-abe391aa9005>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.54/warc/CC-MAIN-20241106230027-20241107020027-00016.warc.gz"} |
Simplifying Fractions
Here's the game:
We want to write fractions using
the smallest numbers possible.
For example, look at
Can you think of a better way to write this using smaller numbers? Let's see...
What if we divided the numerator and denominator by 100?
Oh, yeah... This is a lot better!
Here are the two rules:
1) You can divide the numerator and the denominator by any number... as long as you use the same number for both!
For example, you can't divide the numerator by 3 and the denominator by 5.
2) You need to get the numerator and denominator as small as possible.
Head on over to the next page to continue... | {"url":"https://www.coolmath4kids.com/math-help/fractions/simplifying-fractions?page=0","timestamp":"2024-11-10T17:41:04Z","content_type":"text/html","content_length":"19775","record_id":"<urn:uuid:fd10b8d7-589f-4e70-848f-fda2a2fe12c5>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.61/warc/CC-MAIN-20241110170046-20241110200046-00454.warc.gz"} |
What is Statistics? Definition, Characteristics, Functions and Divisions - Business Jargons
Definition: Statistics can be defined as a part of applied mathematics that is concerned with the collection, classification, interpretation, analysis or the numerical and categorical data and facts,
and drawing conclusions, so as to present the same in a systematic manner.
It involves a wide variety of methods which facilitates data analysis, for decision making purpose.
It can be used to answer questions such as:
• What kind of data is to be collected?
• How much data is to be collected?
• How to organize and summarize data?
• How to analyze data and draw inferences from it?
• How do we assess the strength of conclusions and evaluate their uncertainty?
Characteristics of Statistics
Statistics is characterized by:
1. Aggregate of Facts: Single and non-connected facts or figures are not statistics, rather when the facts are aggregates, they are said to be statistics, as they can be compared
2. Affected to a substantial extent by a variety of reasons: This means that statistics are influenced to a substantial extent by a number of factors that operate together. For example, The
statistics of rice production is based on various factors like a method of cultivation, climatic conditions, seeds, fertilizers and manures, etc.
3. Numerical expression: Statistics are expressed in terms of numbers. Therefore, qualitative expressions such as happy, sad, right, wrong, good, or bad do not amount to statistics. For example:
‘Production of ABC ltd. has risen’ is not statistics, but ‘Production of ABC ltd. has risen from 92000 units in 2020 to 110000 units in 2021’ is statistics.
4. Enumerated and Estimated as per reasonable standard of accuracy: Reasonable accuracy needs to be there in the statistical data, as it acts as a basis for the field of statistical enquiry. This is
because, if the scope of the inquiry is narrow, then by using the method of actual counting, the data can be collected, whereas if the scope of inquiry is wide then the data collection will be
based on estimate and estimates can be inaccurate.
5. Data collection is carried out in a systematic manner: The collection of statistics should be performed in a systematic as well as planned manner, because in the absence of any system, the data
collected can be unreliable and inaccurate, which may also lead to misleading conclusions. Further, the purpose for its collection needs to be stated beforehand to keep its usefulness intact.
6. Data must be placed in relation to one another: Data collection is performed for the purpose of comparison and so the basis must be homogeneous. Because when the basis of two units is
heterogeneous, the comparison is not possible.
Functions of Statistics
Statistics performs the following functions:
• Reduces complexities: Using statistical methods, voluminous data can be presented in a way that it can be easily understood. Hence, it reduces the complexity to understand a vast amount of data,
to simplify its meaning.
• Expresses facts in numbers: An important function of statistics is that it can transform facts into numbers, which is easy to understand by anyone.
• Presentation of data in condensed form: Data collected is usually in raw form, which is complex and unorganized. Hence, it requires to be presented in a simple form so as to reach a final
conclusion. With the help of statistics, a large amount of data can be presented in condensed form.
• Increases the individual knowledge and experience: As the presentation of data is simple, it enhances the knowledge and experience of people, by making it simple and easy to understand, without
having knowledge of each and every field.
• Different phenomena are compared: Statistics helps in making a comparison of data and measuring the relationship between them. For example: Suppose a researcher wants to measure the level of
production of soybean in two states, then he/she would use statistics.
• Helpful in the formulation of policies: Plans and policies are developed beforehand in an organization. And statistics plays a very crucial role in determining the future trends, so as to frame
them, by providing the required information.
• Helpful in prediction and forecasting: The knowledge of statistics is not just helpful in estimating the present but it also helps in forecasting the future
Divisions of Statistics
The different types or branches of statistics are discussed hereunder:
1. Descriptive Statistics: It involves describing and summarizing the sets of numerical data with the help of pictures and statistical quantities. Techniques used may include averages dispersion,
skewness, time series, etc.
2. Inferential Statistics: It encompasses those methods that are helpful in drawing conclusion and inferences with respect to parameters of population, based on estimates which are drawn from
samples. Chi-square, F-test, t-test, etc techniques are used.
3. Applied Statistics: Those methods and techniques are used in applied statistics which are applicable to specific problems of real-life scenarios. Techniques used may include sample survey,
quality control, index numbers etc.
4. Inductive Statistics: Those methods and techniques are covered here which are used to identify a specific phenomenon based on random observation. Techniques used may include Extrapolation.
5. Analytical Statistics: Analytical statistics uses such methods and techniques that are helpful in setting up functional relationship amidst variables. In this correlation, regression, association
and attributes techniques are used.
6. Mathematical Statistics: It deals with the application of different mathematical theories and techniques to develop different statistical techniques. It uses techniques like integration,
differentiation, trigonometry, matrix, etc.
A Word from Business Jargons
Statistics is a vast discipline, with application in numerous fields like research, finance, sports, economics, marketing, health, education, etc. In simple words, it is a science of interpreting,
analysing and drawing conclusions from the data collected | {"url":"https://businessjargons.com/statistics.html","timestamp":"2024-11-04T05:03:15Z","content_type":"text/html","content_length":"55352","record_id":"<urn:uuid:52daa778-9f1a-421c-94b1-016ad40a2a69>","cc-path":"CC-MAIN-2024-46/segments/1730477027812.67/warc/CC-MAIN-20241104034319-20241104064319-00778.warc.gz"} |
DEEP DIVE: Is voting rational? In swing states, YES
Correct votes have a shockingly high value in swing states
The conventional wisdom is “everyone should vote. Do your civic duty!” But I’m in enough contrarian circles that I also hear “I don’t vote, because voting is irrational. A single vote never
determines an election.”
Are the critics right?
Before researching this post, I assumed that they were right about a single vote being irrelevant for Presidential elections. However, I still thought voting could be justified on “cultural/
collective-action” grounds: Sure, your one vote will never matter, but if smart/ethical/pro-freedom people create a general cultural norm of voting, that can be enough to sway elections.
I’ve now gone through the literature on this subject. It turns out that, when one correctly calculates the chances of one vote flipping an election, one gets very high dollar values for a single
Specifically, I estimate that a single correct vote for President in any of the seven key swing states is worth $10,000 or more, using low-end estimates. A correct vote in semi-competitive states is
worth between $500 and $10,000 (that category includes states like Florida, Texas, Alaska, Minnesota, New Mexico, and New Hampshire.) It’s only “irrational” to vote for President in true blue/red
states like Illinois or Wyoming, where the outcome is essentially certain.
Below, I explain in detail how I got those estimates.
How likely is a single vote to flip an election?
I enjoyed diving through the academic literature on this.
Zach Barnett, an Assistant Professor of Philosophy at Notre Dame, has a paper showing that previous research on the issue used the wrong statistical distributions in estimating the chances of a vote
tipping an election.
Barnett’s logic for his calculation method is elegant and clear, and he illustrates it with a hypothetical election between “Donald” and “Daisy” that has exactly 1,000,001 voters. He lists all the
possible vote combinations:
1-in-a-million is the hard, proven, lower bound. But we also know that vote combinations in the middle of the distribution are dramatically more likely than at the edges.
Barnett goes on to show that the chance of a single vote flipping an election is at least 5x higher for the most competitive races (below, “d” means the chance your vote in decisive, and “N” means
the number of voters):
So what does “4 / N”, etc, mean in the real world? Let’s zoom in on Pennsylvania, often the most swing of all swing states. Currently, Kamala Harris has a 40% chance there:
In Pennsylvania in 2020, there were 6,840,276 people who voted for Trump or Biden. If the same number of people vote this year, then the chance that one additional vote in the state will determine
the election is at least 4/N, or 4/6,840,276.
Simplifying, the chance is 1 in 1,710,069.
So we get a real-world lower-bound estimate that a single vote in Pennsylvania has roughly a 1-in-2-million chance of determining the President.1 If the Pennsylvania race tightens again to being
50%-50%, then it will be closer to 1-in-a-million.
In other words, if you vote in Pennsylvania, you’re essentially buying a lottery ticket that you’ll get to decide the election. Your odds are actually a lot better than for winning the popular “Mega
Millions” lottery (1-in-300-million.)
That result is also plausible looking at history. While a 1-in-2-million chance has not yet occurred in the short history of American Presidential elections, when surveying America’s tens of
thousands of local elections, we DO see plenty of races being decided by a coin toss or drawing lots from a hat due to being an exact tie, where a single vote would have been determinative. There are
tens of thousands of local races in a major election year, and local races often have tens of thousands of voters, so consider that some real-world confirmation of Barnett’s math proofs.
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How valuable is an election outcome?
In order to estimate the value of a vote, we need to have a sense of the value of an election.
There are no academic estimates on that.2 People who have written on this instead use guesstimates. Philosopher Will MacAskill guessed it this way in 2015 (in the following quote, I’ve updated his
numbers for 2024): “total spending of the US government is [S:$3.5:S] $6.1 trillion per year … if that money is spent 2.5 percent more effectively, then the benefits amount to [S:$1,000:S] $1,753 per
person… [S:$314:S] $547 billion…”
Guessing is a better approach than having no estimate at all, but, below, I’ll look at stock market changes to get some empirically-grounded sense of how much a vote could be worth.
My approach below was to estimate the societal value of electing the better President by examining the absolute change in the stock market (S&P 500) on the day following Presidential elections, going
back twenty years to 2004.
Here’s the change the day after election day, for each election:
We could refine those estimates a bit. The refinements don’t make a difference to the general magnitude of the value of a vote, so I won’t bore you with the details, except in a footnote.3 But
basically, I do two things: I remove the median daily stock swing, and adjust for how much prediction markets had already priced in candidates’ wins.
The chart of whether markets already expected wins is interesting in its own right:
After adjusting for that, the median change was from the 2020 election, where Biden’s 20% increase in odds from election night appeared to be worth about 1.71% of the stock market, indicating that
his election as a whole was worth about 6.57%.
The lowest-impact election was 2016, in which Trump’s win over Clinton was valued at 1.39% of the market.
So that suggests that the value of a vote ranges from 1.39% to 6.57% of the S&P 500.
In 2024, with the S&P worth $55 trillion, that works out to $767 billion – 3.61 trillion.
There are lots of reasons why my measure is imperfect. The goal isn’t to get a perfect measurement, it’s to get a number that’s in the right order of magnitude. If it’s more sound than MacAskill’s
guess, then we’re making progress. In fact, my lower-bound estimate turned out to be not that far from his guess (though I looked his up afterward doing the calculations.)
One further issue is that we’re looking at the change in the stock markets caused by all elections on election day, including House/Senate/Governors, etc.
It turns out that total donations to Presidents are only a quarter of all donations to federal candidates, and only 23.67% if one also considers spending on governor’s races.4
So let’s estimate that the Presidency is worth 23.67% of all elections on election day (that feels low, but that’s what donation stats suggest, and they’re a “market” in a sense.) Then our new
estimated range for the Presidential election goes down to $182 billion - $852 billion.
That estimate feels low. It amounts to about a mere 3% to 15% of one year of government spending. It’s just 0.7% – 3.4% of one year’s economic output.
It seems especially low when one considers that any effects of Presidential policy are compounding, and continue far into the future. For example, the 2017 Trump tax cuts still impact the economy 7
years later. Obamacare still affects us 15 years after enactment.
The stock market does include those indefinite effects (with discounting) in its $182 billion - $852 billion implicit estimate.
I should note that social issues and foreign policy may each be just as important as economic policy, but they can’t be quantified here. So this is, in every sense, a lower-bound estimate on the
value of an election.
But at least we now have a ballpark number for that.
Let’s now see what happens when we combine it with the 1-in-2-million odds we calculated earlier.
The Enormous Value of A Correct Vote in a Swing State
First, we’ll take Nate Silver’s model’s estimate (paywalled)5 that Pennsylvania has a 29% chance of tipping an election. Then, the expected value of your vote is:
[Chance of state determining Presidential election] * [Chance of voter determining state] * [Value of best candidate winning]
Low-bound estimate: .29 * (1 / 1,710,069) * $182 billion
Higher estimate: .29 * (1 / 1,710,069) * $852 billion
Doing the math:
Low estimate: $30,864
High estimate: $144,485
Wow! The first time I crunched those numbers, I was shocked. $30k per vote, as the low-bound?
But I keep checking over the math, and it appears to be right. Also, while this estimate is empirically-backed, if you prefer any other plausible guesstimate about the value of electing the better
President — you’ll still get massive values.
Looking beyond Pennsylvania, here are the other states:
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There is, however, one possible factor we haven’t included yet:
Problem: You don’t know who the better candidate is!
The above assumes that you, as a voter, know which candidate would be better for the country.
However, as philosopher Jason Brennan points out in this paper (paywalled), if you vote for the wrong candidate, you cost society the same amount of money. Brennan notes:
“Remember: if a vote [for the right candidate] is like donating $3,000 to a charity, a vote the other way is like destroying $3,000 of their funds.”
So let’s say you’re barely better than the median voter at selecting the right candidate. Let’s say you’re only 60% sure that you picked the better one.
Then the value of your vote in Pennsylvania would shrink to: (60% - 40% = 20%) of the original estimate.
So if we assume you should only be 60% confident in your vote, then a Pennsylvania vote is actually worth $6,173 to $28,897 (instead of $30,864 to $144,485.)
That’s still worth a lot. Even you’re deeply conflicted, but have a gut sense who will be better for the country (60% sure) it’s still well worth your time to vote.
Here’s how the map changes, with that assumption:
Now, most of us are more than 60% sure. Maybe we’re 95%, or 100% sure. But should we be that sure?
History gives us lots of room for humility.
Why we should be humble about whether we know which candidate would be best
The famous biography of Sam Bankman-Fried, “Going Infinite”, documents how, when Bankman-Fried was a trader at Jane Street in 2016, he brilliantly modeled the election night results, and called
Trump’s win before anyone else. He then used that information to short-sell the stock market, as he assumed that a Trump win would be bad for markets.
Markets instead rose the morning after Trump’s win, causing Bankman-Fried’s employer to have one of its largest single-day losses in its history, according to the book. Bankman-Fried had been
overconfident about the market’s view of whether Trump was the better candidate.
It might feel “obvious” that smart people are better at voting, but some studies have shown that when it comes to political bias, smart people may actually better at fooling themselves and confirming
their biases. Jason Brennan cites a paper by Dan Kahan, which provides some low-confidence evidence that smart people are more willing to let a negative view of their political opponents impact their
views, when it comes to politics.6
There are also some real-world examples that should give us pause.
For example, California voters support green energy, and so they elected active politicians who imposed mandates, gas taxes, and subsidies, to guide the economy in that direction. In contrast, Texas
voters don’t care much about “green” energy, so they elected politicians who mostly sat back and didn’t do much. The result:
The trend is not because Texas has more sunshine; it doesn’t.
California has somewhat more highly-educated voters than Texas does, but they may have tripped themselves up by thinking that they could plan their economy.
So many things are also just hard to predict. For example, nobody remembers that George W. Bush ran in 2000 as a dovish non-interventionist! He then invaded Iraq, with disastrous consequences.
Unpredictability even goes beyond a President promising one thing and doing another. You have backlash and “game theory” effects. For example, most Democrats were happy when Obama won re-election in
2012 – but Brennan points out that it set the stage for Trump’s victory in 2016. If one believes Democratic rhetoric about Trump being an “existential threat” to democracy itself, then, Brennan
notes, in retrospect Democrats should wish they had voted for moderate Romney in 2012, just to preclude a Trump rise.
With all this, by far the best argument against voting, even in a swing state, is that you’re no better than the median voter at picking a candidate.
But how often have you heard someone give that reason for not voting?
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You are much better than the median voter
Notwithstanding the above, I strongly suspect that the readers of this blog (and “rationalists” generally) are better able to pick Presidents than the median voter.
That’s because a relatively high proportion of this blog’s readers: A) tolerate hearing views they disagree with B) at least try to put aside biases when deciding things C) have heard of prediction
markets and other ways of reducing bias.
Speaking of which, Brennan invokes prediction markets as an thought experiment for why you shouldn’t vote. Here’s his logic: “If you are not able to get rich from such markets, then you have evidence
that you are not a reliable predictor when it comes to politics.”
But Brennan sets much too high a bar, because it’s possible for someone to be an amazing predictor compared to the median voter, and also lose money vs the wisdom of the market consensus.7
The real test should be: Let’s give you, and the median voter, each $1,000 to make bets about politics and policy with each other (not with the market.) Who ends up with more money in the end? If
it’d be you, then you should vote.
My strong bet is that the majority of readers of this blog would significantly outperform the median voter on that correct test.
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To be clear: Voting is not narrowly, selfishly, rational
Voting is a “social good.” As shown above, the benefit of picking the right candidate to lead the most powerful country in the world is enormous.
But the benefit to any given voter from his/her own vote is still small.
For example, let’s take our low-end estimate that picking the best candidate is worth $182 billion to the country as a whole.
$182 billion divided by the US population of 333 million is a $546 gain per American. Not nothing.
But now combine that with the fact that, even in Pennsylvania, your odds of deciding the election are 1-in-2-million.
That means the expected value of your vote for yourself is approximately ($1,153 / 1 million), or about one 32nd of a cent.
So it’s indeed not (narrowly, selfishly) rational to vote.
Now, we’re all somewhat selfish. But unless you’re a sociopath, you care at least somewhat about broader society doing well, too.
Conclusion – vote, if you live in a swing state
The above suggests that there are three things that justify voting:
• You live in a somewhat competitive state
• You try to be unbiased and informed
• You want to improve society
If you check those boxes, and could out-bet the median voter, then I hope you’ll vote (regardless of whom you pick.)
Speaking of which, who do you pick? This is a survey just for readers who managed to read the full article!
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Nobody else is calculating these kinds of things. Please consider encouraging further work by subscribing:
Now, is that plausible? At first I thought it must be wrong, because – wouldn’t a “1 in a million” chance for your vote suggest that multiple voters tip the election every year? But the answer is
that it makes sense because, when there is a decisive vote, EVERY voter on one side becomes decisive. To take a concrete example: Let's say that Pennsylvania is decided by a single vote. In that
case, 3 million (plus 1) Pennsylvania voters will have been decisive.
“After doing an extensive search … no high-quality journal has published an estimate of net welfare effects of an entire presidency or congressional tenure.” — paper by Jason Brennan, 2022, “Why
Swing‐State Voting Is Not Effective Altruism,” the Journal of Political Philosophy ( https://philpapers.org/rec/BREWSV-3 )
I made the below refinements, that don’t make a big difference overall:
Refinement for median day change:
On a median day between 2004 and 2020, the stock market has an absolute value change of 0.49%. Let’s remove that amount from each bar, so that we see the change that could plausibly be attributed to
the election result:
One other thing we could try to do is adjust for the amount of surprise which markets registered at the result of each election. Here are the probabilities each candidate had going into the election,
per election betting markets:
You can see that prediction markets correctly predicted every outcome except Trump’s 2016 win, to which they gave a 20% probability.
Only one outcome was virtually certain beforehand: Bettors gave Obama a 90% chance of winning in 2008.
This gives us a sense of how much candidates were already “priced in” to the stock market. For example, since markets thought Trump had a 20% chance in 2016, and election day revealed it be 100%, the
change in market values should contain 80% of the value of Trump’s Presidency (compared to a Clinton one.)
if we attribute the entire excess change in the stock market to the outcome of the election the day, we get:
I left Obama 2008 out of this graph, because the result is too large, and implausible, because the market’s update on his chances was so small (10%) that the big fall on the day following couldn’t
have been due to that. If it were due to his election, then, it’d imply that his election cost the US about half of its future economic potential.)
So, here, I’ll ignore the 2008 datapoint by looking at the median impact of a Presidential election, rather than the average.
After adjusting for that, the median change was from the 2020 election, where Biden’s 20% increase in odds from election night appeared to be worth about 1.71% of the stock market, indicating that
his election as a whole was worth about 6.57%.
The lowest-impact election was 2016, in which Trump’s win over Clinton was valued at 1.39% of the market.
So that suggests that the value of a vote ranges from 1.39% to 6.57% of the S&P 500.
In 2024, with the S&P worth $55 trillion, that works out to $767 billion – 3.61 trillion.
I didn’t find good governor data for this cycle, so I’m using the numbers from the last election: https://www.wisdc.org/news/press-releases/139-press-release-2023/
I use Silver’s estimates, because his model has been good in the past, and because he’s the only one besides 538 who estimates this. I find some of his results surprising (are New Mexico,
Mississippi, and Alaska really as valuable as he thinks?) but 538 actually has pretty similar estimates for those states, so — I guess the data is trying to tell us something.
The study has a pretty complex design, and could use replication. Specifically, Kahan gave subjects a short math intelligence test, and then asked them if the test was valid for measuring
open-mindedness. But, before asking if the test was valid for that, the experimenter informed subjects that either side of the climate change debate had scored as the “open-minded” ones. The study
revealed that smarter people (who did better on the test) were significantly more likely to change their opinion about the test based on whether the test marked their ideological opponents as more
open-minded. That’s one very narrow study — it’d be good to see it replicated more broadly.
Brennan’s example would only make sense if the alternative to you voting were futarchy (aka, letting the market decide policy.) But since we don’t have that, and won’t for decades, if ever, it’s a
bad comparison.
Why isn't there a vote for "Not an American citizen"? :)
Otherwise, great article.
Expand full comment
Nice post!
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Analysis of gearshift processes in an automatic transmission at low vehicle speeds
The increase in the number of speeds in automatic transmissions of vehicles leads to a convergence of gear speeds between adjacent gears. At low vehicle speeds, periodic changes in external driving
conditions (for example, hilly or winding sections of roads) can lead to loop shifting between adjacent gears. To simulate this process, a dynamic transmission model is proposed, which can be used in
the development of shift control systems and for calibration of the engine and automatic transmission at various operating modes and vehicle conditions.
1. Introduction
The basis of the vehicle automatic transmissions (AT) control system is a gearshift map. Usually the map is built in coordinates of the vehicle speed, determined by the speed sensor of the output
shaft of the AT, and the percentage of opening the engine throttle (Fig. 1) [1-5].
Fig. 1A gearshift map of the 6-speed AT [1]
The solid lines 1-2, 2-3, 3-4, etc. on the shifting map are lines of shifting from lower to upper gears, and dashed lines 2-1, 3-2, etc. are lines of shifting from upper to lower gears. Solid lines
are primarily determined by the technical characteristics of the engine (torque, rotation speed, fuel consumption) and the requirements of the driver. Throttle opening close to 100 % means that the
vehicle speed needs to be maximized. The mid-throttle gearshift should be made to achieve some fuel economy with average acceleration. With a small throttle opening, gear shifting should provide the
lowest fuel consumption. The dashed lines of reverse shifting are selected as a result of a compromise between the driving conditions, the requirements of the driver and the reserve of the engine
As can be seen from Fig. 1, when the vehicle speed is less than 60 km/h, the shifting lines between adjacent gears are quite close. Periodic changes in the external conditions of the vehicle’s
movement can lead to that the automatic gearshift control system will loop when driving at a constant speed in the zone of close values of the shifting speeds to adjacent gears. For example, such a
regime can be observed on hilly roads, when moving downhill upshift occurs, and when moving uphill downshift occurs. An increase in the gearshift frequency leads to additional energy losses, an
increase in fuel consumption, a reduction in the service life of the gearbox and a passenger discomfort [3, 4].
The purpose of the paper is to study and simulation the loop gear shifting process using a simplified dynamic model of the vehicle transmission. Modeling the process of gear shifting loop in order to
predict the occurrence of this phenomenon and its prevention is important for creating adaptive automatic cruise-control systems, as well as automatic control systems for autonomous vehicles.
2. Dynamic model
The dynamic model of a vehicle with AT is based on an analysis of the connection diagram and the characteristics of its components. High-speed transmission elements include movable engine elements, a
torque converter (or clutch), and a gearbox input shaft with a clutch and pinion gear engaged. Medium-speed elements are the output shaft of the gearbox with driven gears and the driveshaft with the
drive gear of the differential. The low-speed elements of the transmission are the movable links of the differential and the axle shafts with drive wheels. The body mass reduced to the axles is $m=P/
g$, where $P$ is the vehicle weight, depends on the slipping of the drive wheels on the road surface. Slip coefficient $\epsilon =V/\mathrm{\Omega }R$, where $V$ is the vehicle speed, $\mathrm{\Omega
}$ is the angular rotation speed of the wheels; $R$ is the radius of the wheel. The coefficient $\epsilon$ depends on the road surface.
A simplified dynamic model of the vehicle’s 2-speed AT in accordance with the above description showed in Fig. 2.
Fig. 2A simplified dynamic model of the vehicle’s transmission with 2-speed gearbox
The dynamic model presented above is common [3, 5, 6] to any AT with discrete gear ratios, including planetary transmissions with a torque converter and dual-clutch transmissions with fixed axles of
gear wheels.
We assume that the torque converter is blocked by a locking clutch and the engine output shaft is connected to the input shaft I of the gearbox for analyzing the dynamic processes of gear shifting in
the AT. Therefore, the inertial element ${J}_{I}$ on the input shaft of the gearbox takes into account the moments of inertia of the moving parts of the engine and torque converter in the dynamic
model. The most flexible element is the driveshaft after the output shaft O of the gearbox in the transmission. Its elastic and damping properties in a dynamic model can be represented as an elastic
element (spring) with rigidity $c$ and damper $b$. Therefore, the inertial element ${J}_{O}$ characterizes the sum of the moments of inertia of the output shaft of the gearbox with the inertia
moments of the gears of the gearbox and half the moment of inertia of the driveshaft. ${J}_{VT}$ is the second half of the moment of inertia of the driveshaft with the inertia moments of the
subsequent transmission links and the vehicle body.
The gear shift duration in modern automatic transmissions is 0.2-0.5 sec [3, 5]. Such a short gear shifting time allows to consider this process as a collision. The inertial torques caused by a sharp
change in the speeds of the transmission elements significantly exceed the engine torque and the reduced torque of the resistance force in this case. The aforesaid allows to attribute the considered
model (Fig. 2) to vibration-impact systems [7] and, firstly, to consider the actual process of gear shifting in the gearbox using the impact theory and the theorem on the change in the angular
momentum of a mechanical system upon impact [8], and, secondly, to determine the initial conditions for solving the differential equations of vehicle motion on the intervals between gear shifts.
The shift from first to second gear occurs when the acceleration is ${\stackrel{˙}{\omega }}_{O}>0$, and the shift from second to first gear occurs when ${\stackrel{˙}{\omega }}_{O}<0$. The impact
interaction of the gearbox elements is accompanied by a rapid change in the kinematic relationships determined by the gear ratios. The greater the difference between the gear ratios, the greater the
magnitude of the impact impulse. The input shaft I rotation speed of the gearbox is related to the rotation speed of the output shaft O with gear ratios: ${\omega }_{I}^{\left(12\right)}={i}_{1}{\
omega }^{\left(12\right)}$ when shifting from first gear to second, ${\omega }_{I}^{\left(21\right)}={i}_{2}{\omega }^{\left(21\right)}$ when shifting from second gear to first (${\omega }^{\left(12\
right)}$ and ${\omega }^{\left(21\right)}$ are rotation speed of the shaft O at which the gearshift occurs form first gear to the second and from second gear to the first correspondingly). $\left|{i}
_{1}\right|>\left|{i}_{2}\right|$ the absolute values of the gear ratios from the input shaft I to the output shaft O are taken, since the rotation direction of the output shaft is not taken into
account. The rotation speed of the element ${J}_{I}$ will decrease at shifting to second gear, and it will increase when shifting to first gear [8]. The rotation speed of the element ${J}_{O}$ before
gear shift from the first gear to the second is ${\omega }^{\left(12\right)}$, and the rotation speed of the element ${J}_{I}$ at this moment is ${i}_{1}{\omega }^{\left(12\right)}$. The moment of
inertia of the element ${J}_{I}$ reduced to the output shaft O before engaging the second gear is ${J}_{I}{i}_{2}^{2}$.
The theorem of angular momentum conservation upon impact for inertial elements ${J}_{I}$ and ${J}_{O}$ before and after the second gear is engaged:
${J}_{\mathrm{I}}{i}_{2}^{2}\frac{{\omega }_{I}^{\left(2\right)}}{{i}_{2}}+{J}_{\mathrm{O}}{\omega }^{\left(12\right)}=\left({J}_{\mathrm{I}}{i}_{2}^{2}+{J}_{\mathrm{O}}\right){\omega }^{\left(2\
where ${\omega }^{\left(2\right)}$ is the speed of the output shaft after the shift to second gear. Moreover, ${\omega }_{I}^{\left(2\right)}={i}_{1}{\omega }^{\left(12\right)}$, then we get:
${\omega }^{\left(2\right)}=\frac{{J}_{\mathrm{I}}{i}_{1}{i}_{2}+{J}_{\mathrm{O}}}{{J}_{\mathrm{I}}{i}_{2}^{2}+{J}_{\mathrm{O}}}{\omega }^{\left(12\right)}.$
Similarly, the theorem of angular momentum conservation upon impact for inertial elements ${J}_{\mathrm{I}}$ and ${J}_{O}$ before and after the first gear is engaged:
${J}_{I}{i}_{1}^{2}\frac{{\omega }_{I}^{\left(1\right)}}{{i}_{1}}+{J}_{O}{\omega }^{\left(21\right)}=\left({J}_{I}{i}_{1}^{2}+{J}_{O}\right){\omega }^{\left(1\right)},$
where ${\omega }^{\left(1\right)}$ is the speed of the output shaft after the shift to first gear. Moreover, ${\omega }_{I}^{\left(1\right)}={i}_{2}{\omega }^{\left(21\right)}$, then we get:
${\omega }^{\left(1\right)}=\frac{{J}_{\mathrm{I}}{i}_{1}{i}_{2}+{J}_{\mathrm{O}}}{{J}_{\mathrm{I}}{i}_{1}^{2}+{J}_{\mathrm{O}}}{\omega }^{\left(21\right)}.$
The total moment of inertia of the engine and gearbox elements $\left({J}_{\mathrm{I}}{i}^{2}+{J}_{O}\right)$ before the link with maximum flexibility can be considered at intervals between gear
shifting after reducing the moment of inertia ${J}_{I}$ to the output shaft. This eliminates one of the three degrees of freedom in the dynamic model (Fig. 2).
The movement of the model in the $i$-th gear ($i=1,2$) between the gear shifts is described by a system of two linear differential equations:
$\left({J}_{I}{i}^{2}+{J}_{O}\right){\stackrel{˙}{\omega }}^{\left(i\right)}+b\left({\omega }^{\left(i\right)}-{\omega }_{VT}^{\left(i\right)}\right)+c\left({\phi }^{\left(i\right)}-{\phi }_{VT}^{\
left(i\right)}\right)=i{M}_{I},$${J}_{VT}{\stackrel{˙}{\omega }}_{VT}^{\left(i\right)}+b\left({\omega }_{VT}^{\left(i\right)}-{\omega }^{\left(i\right)}\right)+c\left({\phi }_{VT}^{\left(i\right)}-{\
phi }^{\left(i\right)}\right)=-{M}_{R}.$
In Eq. (1), the superscript in parentheses indicates the number of the gear engaged; $\phi$ is the angular coordinate of the inertial element $\left({J}_{\mathrm{I}}{i}^{2}+{J}_{O}\right)$; ${\phi }_
{VT}$ is the angular coordinate of the inertial element ${J}_{VT}$.
Free vibration frequencies of the model in the absence of damping: ${k}_{1}=0\text{;}$${k}_{2}^{\left(i\right)}=\sqrt{c\left({J}_{\mathrm{I}}{i}^{2}+{J}_{O}+{J}_{VT}\right)/\left({J}_{\mathrm{I}}{i}^
{2}+{J}_{O}\right){J}_{VT}}\text{.}$ Frequency of damped oscillations: ${{k}_{2}^{*}}^{\left(i\right)}=\sqrt{{{k}_{2}^{\left(i\right)}}^{2}-{{h}^{\left(i\right)}}^{2}}$, where ${h}^{\left(i\right)}=b
3. Model movements between gear shifts
The movement of the model between gear shifts is a torsional vibration of inertial elements $\left({J}_{\mathrm{I}}{i}^{2}+{J}_{O}\right)$ and ${J}_{VT}$ relative to the center of mass, rotating
under the action of the engine torque and the moments of resistance forces ${M}_{I}=conste 0$, ${M}_{R}=conste 0$. The center of mass motion of the model is the absolute model movement. The relative
motion of the elements $\left({J}_{I}{i}^{2}+{J}_{O}\right)$ and ${J}_{VT}$ are their synchronous torsional vibrations in antiphase, and the magnitudes of the oscillation amplitudes are inversely
proportional to the moments of inertia [9]. The general solution of the obtained Eq. (1) has the form:
${\phi }^{\left(i\right)}={A}^{\left(i\right)}{e}^{-{h}^{\left(i\right)}t}\mathrm{sin}\left({{k}_{2}^{*}}^{\left(i\right)}t+{\beta }^{\left(i\right)}\right)+{C}_{1}^{\left(i\right)}+{C}_{2}^{\left(i\
right)}t+{a}^{\left(i\right)}{t}^{2}/2,$${\phi }_{VT}^{\left(i\right)}={\mu }^{\left(i\right)}{A}^{\left(i\right)}{e}^{-{h}^{\left(i\right)}t}\mathrm{sin}\left({{k}_{2}^{*}}^{\left(i\right)}t+{\beta
where ${\mu }^{\left(i\right)}={A}_{VT}^{\left(i\right)}/{A}^{\left(i\right)}=-\left[\left({J}_{I}{i}^{2}+{J}_{O}\right)/{J}_{VT}\right]$, ${a}^{\left(i\right)}=\left(i{M}_{I}{-M}_{R}\right)/\left
The constants $A$, $\beta$, ${C}_{1}$ and ${C}_{2}$ are determined through the initial conditions, that is, through the initial angular coordinates and velocities of the inertial elements of the
model. In Eq. (2), parameters ${C}_{1}$, ${C}_{2}$ and $a$ are the characteristics of the absolute motion, i.e., the mass center motion of the model, parameters $A$ and $\beta$ are the
characteristics of the relative motion.
By differentiating Eq. (2), we obtain a general form of the equations of mass velocities in the $i$-th gear ($i=1,2$) between the gear shifts:
${\omega }^{\left(i\right)}={A}^{\left(i\right)}{e}^{-{h}^{\left(i\right)}t}\left({{k}_{2}^{*}}^{\left(i\right)}\mathrm{cos}\left({{k}_{2}^{*}}^{\left(i\right)}t+{\beta }^{\left(i\right)}\right)-{h}^
{\left(i\right)}\mathrm{sin}\left({{k}_{2}^{*}}^{\left(i\right)}t+{\beta }^{\left(i\right)}\right)\right)+{C}_{2}^{\left(i\right)}+{a}^{\left(i\right)}t,$
${\omega }_{VT}^{\left(i\right)}={\mu }^{\left(i\right)}{A}^{\left(i\right)}{e}^{-{h}^{\left(i\right)}t}\left({{k}_{2}^{*}}^{\left(i\right)}\mathrm{cos}\left({{k}_{2}^{*}}^{\left(i\right)}t+{\beta }^
{\left(i\right)}\right)-{h}^{\left(i\right)}\mathrm{sin}\left({{k}_{2}^{*}}^{\left(i\right)}t+{\beta }^{\left(i\right)}\right)\right)+{C}_{2}^{\left(i\right)}+{a}^{\left(i\right)}t.$
The absolute speed ${C}_{2}^{\left(i\right)}$ is the initial center of mass velocity of the model in the $i$th gear. Substituting $\mathrm{sin}{\alpha }^{\left(i\right)}={h}^{\left(i\right)}/\sqrt
{{{h}^{\left(i\right)}}^{2}+{{{k}_{2}^{*}}^{\left(i\right)}}^{2}}$ and $\mathrm{cos}{\alpha }^{\left(i\right)}={{k}_{2}^{*}}^{\left(i\right)}/\sqrt{{{h}^{\left(i\right)}}^{2}+{{{k}_{2}^{*}}^{\left(i\
right)}}^{2}}$ expressions of mass velocities are transformed to the form:
${\omega }^{\left(i\right)}={A}^{\left(i\right)}{e}^{-{h}^{\left(i\right)}t}\sqrt{{{h}^{\left(i\right)}}^{2}+{{{k}_{2}^{*}}^{\left(i\right)}}^{2}}\mathrm{cos}\left({{k}_{2}^{*}}^{\left(i\right)}t+{\
beta }^{\left(i\right)}+{\alpha }^{\left(i\right)}\right)+{C}_{2}^{\left(i\right)}+{a}^{\left(i\right)}t,$${\omega }_{VT}^{\left(i\right)}={\mu }^{\left(i\right)}{A}^{\left(i\right)}{e}^{-{h}^{\left
(i\right)}t}\sqrt{{{h}^{\left(i\right)}}^{2}+{{{k}_{2}^{*}}^{\left(i\right)}}^{2}}\mathrm{cos}\left({{k}_{2}^{*}}^{\left(i\right)}t+{\beta }^{\left(i\right)}+{\alpha }^{\left(i\right)}\right)+{C}_{2}
The motion modes of a transmission model are determined by gear shift speeds. The gear shift from first to second gear will take place when ${\omega }^{\left(1\right)}={\omega }^{\left(12\right)}$
and ${\stackrel{˙}{\omega }}^{\left(1\right)}>0$; the gear shift from second gear to first when ${\omega }^{\left(2\right)}={\omega }^{\left(21\right)}$ and ${\stackrel{˙}{\omega }}^{\left(2\right)}
The angular coordinate Eq. (2) and the velocity function ${\omega }_{VT}$ are continuous over the entire interval of motion, and the continuous velocity function $\omega$ at intervals is a
discontinuous function with finite jumps between intervals. The last circumstance is a sign of the nonlinearity of the system under consideration, for solving the dynamics of which we apply the
stitching method.
The conditions for stitching coordinates and speeds of the model at adjacent intervals when shifting from first gear to second are:
${\phi }^{\left(2\right)}\left(0\right)={\phi }^{\left(1\right)}\left({t}_{k}^{\left(1\right)}\right),{{\phi }_{VT}}^{\left(2\right)}\left(0\right)={{\phi }_{VT}}^{\left(1\right)}\left({t}_{k}^{\left
(1\right)}\right),$${\omega }^{\left(2\right)}\left(0\right)={\omega }^{\left(12\right)}\left({J}_{I}{i}_{1}{i}_{2}+{J}_{O}\right)/\left({J}_{I}{i}_{2}^{2}+{J}_{O}\right),{{\omega }_{VT}}^{\left(2\
right)}\left(0\right)={{\omega }_{VT}}^{\left(1\right)}\left({t}_{k}^{\left(1\right)}\right),$
where ${t}_{k}^{\left(i\right)}$ is the time period between gear shifts.
The angular coordinates and velocities of the model elements before the first shift from the first gear to the second should be given or known if the model’s movement in the first gear is somehow
specified: Eqs. (2) and (3).
The initial values of the angular coordinates and velocities of the model elements in second gear obtained from stitching conditions Eq. (4) determine the constants ${A}^{\left(2\right)}$, ${\beta }^
{\left(2\right)}$, ${C}_{1}^{\left(2\right)}$ and ${C}_{2}^{\left(2\right)}$ for solutions of the differential equations of the model motion in second gear:
${\phi }^{\left(2\right)}\left(0\right)={A}^{\left(2\right)}\mathrm{s}\mathrm{i}\mathrm{n}{\beta }^{\left(2\right)}+{C}_{1}^{\left(2\right)},{{\phi }_{VT}}^{\left(2\right)}\left(0\right)={\mu }^{\
left(2\right)}{A}^{\left(2\right)}\mathrm{s}\mathrm{i}\mathrm{n}{\beta }^{\left(2\right)}+{C}_{1}^{\left(2\right)},$${\omega }^{\left(2\right)}\left(0\right)={A}^{\left(2\right)}\sqrt{{{h}^{\left(2\
right)}}^{2}+{{{k}_{2}^{*}}^{\left(2\right)}}^{2}}\mathrm{cos}\left({\beta }^{\left(2\right)}+{\alpha }^{\left(2\right)}\right)+{{C}_{2}}^{\left(2\right)},$${{\omega }_{VT}}^{\left(2\right)}\left(0\
right)={\mu }^{\left(2\right)}{A}^{\left(2\right)}\sqrt{{{h}^{\left(2\right)}}^{2}+{{{k}_{2}^{*}}^{\left(2\right)}}^{2}}\mathrm{cos}\left({\beta }^{\left(2\right)}+{\alpha }^{\left(2\right)}\right)+
Fig. 3The graph of the changing the angular velocity at variable acceleration
The general algorithm for calculating constants on the intervals of continuous movements according to known initial conditions includes the following steps:
1) Subtraction of the left and right parts of the second equation of the system Eq. (5) from similar parts of the first equation: ${\phi }^{\left(2\right)}\left(0\right)-{{\phi }_{VT}}^{\left(2\
right)}\left(0\right)=\left(1-{\mu }^{\left(2\right)}\right){A}^{\left(2\right)}\mathrm{s}\mathrm{i}\mathrm{n}{\beta }^{\left(2\right)}.$
2) A similar action with the fourth and third equations of system Eq. (5) and the following identical permutations:
${\omega }^{\left(2\right)}\left(0\right)-{{\omega }_{VT}}^{\left(2\right)}\left(0\right)+{h}^{\left(2\right)}\left({\phi }^{\left(2\right)}\left(0\right)-{{\phi }_{VT}}^{\left(2\right)}\left(0\
right)\right)=\left(1-{\mu }^{\left(2\right)}\right){A}^{\left(2\right)}{{k}_{2}^{*}}^{\left(2\right)}\mathrm{cos}{\beta }^{\left(2\right)}.$
3) Multiplication of the right and left sides of the third equation of system Eq. (5) by $\mu$ and the subsequent subtraction from the fourth equation of the third: ${{\omega }_{VT}}^{\left(2\right)}
\left(0\right)-{\mu }^{\left(2\right)}{\omega }^{\left(2\right)}\left(0\right)=\left(1-{\mu }^{\left(2\right)}\right){{C}_{2}}^{\left(2\right)}.$
The first two equations allow to determine ${A}^{\left(2\right)}$ and ${\beta }^{\left(2\right)}$, the last equation allows to determine ${{C}_{2}}^{\left(2\right)}$. The constant ${{C}_{1}}^{\left(2
\right)}$ in the first interval of the model motion can be accepted equal to zero, and at subsequent intervals it is equal to the sum of the displacements of the model center of mass at previous
The movement of the model in second gear, described by Eqs. (2) and (3), ends at time ${t}_{k}^{\left(2\right)}$, when the conditions ${\omega }^{\left(2\right)}\left({t}_{k}^{\left(2\right)}\right)=
{\omega }^{\left(21\right)}$ and ${\stackrel{˙}{\omega }}^{\left(2\right)}\left({t}_{k}^{\left(2\right)}\right)<0$ will be met. This is an important point for solving the issue of the existence of
the effect of a gear shift loop.
4. Numerical simulation
A numerical simulation of the model motion carried out using the obtained dependences. The graph of the changing the angular velocity at variable acceleration is shown in Fig. 3.
5. Conclusions
A mathematical model is developed to simulate the processes of gear shift loop caused by the vehicle moving at a constant low speed along a hilly (or winding) section of the road. The model contains
elastic and dissipative elements, a control system that generates gear shifting commands in accordance with the output shaft rotation speed sensor.
Relations are derived for determining the coordinates and speeds of transmission elements before and after gear shifts. A numerical simulation of gear shifting processes was performed, the
possibility of gear shift loops was showed. The most effective way to eliminate unwanted gear shifting loop is to use navigation maps that take into account the topography of the road [10]. In
addition, an effective method is to extend the boundaries of gear shifts on the gear shift map in the case of diagnosing loops [11].
The obtained mathematical model allows for any vehicle motion determine the conditions for the occurrence of loop gear shifts and develop measures to eliminate this phenomenon. The model can also be
used in the development of automatic transmission gear shift control systems and for calibration the engine and transmission at various operating modes and traffic conditions.
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Automobile Engineering, Vol. 228, Issue 1, 2014, p. 50-72.
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• Kobrinskiy A. Mechanisms with Elastic Bonds. Science, Moscow, 1964.
• Salamandra K., Tyves L. Integral principle in the problems of dynamic analysis of gearshift in automatic gearboxes. Journal of Machinery Manufacture and Reliability, Vol. 46, Issue 5, 2017, p.
• Panovko J. Introduction to the Theory of Mechanical Vibrations: Textbook. 3rd Edition, Science, Moscow, 1991.
• Vasil’ev V. Analysis of the results of theoretical and experimental studies of control algorithms for automatic transmissions of wheeled vehicles. Zurnal AAI, Vol. 102, Issue 1, 2017, p. 12-19.
• Douglas A., Todd A., Jeffery L. Method for Modifying the Shift-Points of an Automatic Transmission. Patent US 5806370, 1998.
About this article
Vibration in transportation engineering
gear shift
automatic transmission
dynamic model
control system
The research was supported by Russian Science Foundation (Project No. 19-19-00065).
Copyright © 2019 George Korendyasev, et al.
This is an open access article distributed under the
Creative Commons Attribution License
, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. | {"url":"https://www.extrica.com/article/21066","timestamp":"2024-11-12T12:52:31Z","content_type":"text/html","content_length":"155260","record_id":"<urn:uuid:7aefa453-f2ea-4ce3-a773-00e3707e3635>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.45/warc/CC-MAIN-20241112113320-20241112143320-00415.warc.gz"} |
Comprehensive overview of loss functions in Machine Learning | CloudFactory Computer Vision Wiki
Comprehensive overview of loss functions in Machine Learning
In the Machine Learning field, the loss function (or cost function) refers to the difference between the ground truth output and the output predicted by the model. The bigger the difference - the
higher the loss, or the punishment. Hence, during training, the goal is to find such model parameters (weights and biases) that minimize the loss and maximize the rate of correct predictions.
While achieving low loss during training is desirable, the loss equal to 0 does not guarantee great model performance in a real-world setting. One should avoid overfitting - a problem when the model
makes perfect predictions on training set but fails to generalize on new, unseen data.Hence, monitoring the model’s performance on a separate validation set or applying other overfitting-preventing
techniques is crucial.
To illustrate the concept, let’s have a look at Mean Squared Error (MSE) - a classical loss function used in regression tasks, where both predictors and target variables are continuous (for example,
predicting the house price based on its area). Its formula is given as follows:
• N is the number of data points;
• ŷ is the output predicted by the model;
• y_i is the actual value of the data point.
As you see, we simply sum the squared difference between each predicted and actual value and divide the result by the number of observations (data points). In the graphs below, the blue line
represents the model’s predicted regression line, and the red arrows represent the error (loss) for each data point. It is visible with the unaided eye that the MSE will be notably smaller on the
second graph.
Despite being common and easy to understand, the MSE loss function does not suit every use case for the following reasons:
• It is sensitive to outliers: data points that greatly stand out from the rest may heavily influence the regression line, which leads to a decrease in model performance.
• It does not work well with classification: MSE is used for regression tasks where the output is a continuous variable (unlike categorical variables like cat/dog/fish).
• MSE tends to minimize large errors at the expense of the accuracy of smaller errors. As a result, the prediction line might be suboptimal.
In the next section, we will give you an overview of alternatives to MSE and briefly describe which loss function is suitable for a given Computer Vision (CV) task.
• Cross-Entropy Loss is a widely used alternative for the MSE. It is often used for classification tasks, where the output can be represented as the probability value between 0 and 1. The
cross-entropy loss compares the predicted Vs. true probability distributions. For example, if the animal in the image is a cat (cat = 1, dog = 0, fish = 0), and the model predicts the
distribution as cat = 0.1, dog = 0.5, and fish = 0.4, the cross-entropy loss will be pretty high.
• Binary Cross-Entropy Loss is a special case of Cross-Entropy loss used. It can be utilized for any binary classification task and, in principle, for binary segmentation.
• Focal loss is used when class imbalance takes place in the dataset. In such cases, the model often learns to predict only the most represented class because, most of the time, the prediction will
be correct, and the model will be rewarded. To prevent this, focal loss punishes the model more for wrong predictions on the "hard" samples (with the underrepresented class) and punishes it less
on "easy" (more represented) samples.
• Bounding Box Regression Loss is used in object detection tasks to train models that predict the location of the bounding boxes for each object. This loss function is calculated by finding the
difference between the predicted and ground truth bounding boxes.
• CrossEntropyIoULoss2D is a combination of the Bounding Box Regression loss and Cross-Entropy loss. In simple words, it is the average of the outputs of these two losses. It is often used for
image segmentation tasks, where the ground truth is a pixel-precise mask of an object.
You can select the loss function during training before running an experiment.
1. To start the experiment, first, access Model Playground in the menu on the left.
2. Select the split on which you want to run an experiment or create a new split.
3. Create a new experiment and choose the desired architecture and other parameters.
4. Go to the Loss & metrics section and select the loss function. Since the choice of the loss function is heavily dependent on the task you are performing, sometimes only default options will be
available to ensure smooth training. | {"url":"https://wiki.cloudfactory.com/docs/mp-wiki/loss/comprehensive-overview-of-loss-functions-in-machine-learning","timestamp":"2024-11-04T21:58:13Z","content_type":"text/html","content_length":"69747","record_id":"<urn:uuid:b3f3707b-9752-4069-b44d-45f12c32aa46>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.16/warc/CC-MAIN-20241104194528-20241104224528-00647.warc.gz"} |
Theoretical and Advanced Machine Learning | TensorFlow
Theoretical and advanced machine learning with TensorFlow
Before starting on the learning materials below, be sure to:
1. Complete our curriculum Basics of machine learning with TensorFlow, or have equivalent knowledge
2. Have software development experience, particularly in Python
This curriculum is a starting point for people who would like to:
1. Improve their understanding of ML
2. Begin understanding and implementing papers with TensorFlow
You should already have background knowledge of how ML works or completed the learning materials in the beginner curriculum Basics of machine learning with TensorFlow before continuing. The below
content is intended to guide learners to more theoretical and advanced machine learning content. You will see that many of the resources use TensorFlow, however, the knowledge is transferable to
other ML frameworks.
To further your understanding of ML, you should have Python programming experience as well as a background in calculus, linear algebra, probability, and statistics. To help you deepen your ML
knowledge, we have listed a number of recommended resources and courses from universities, as well as a couple of textbooks.
Step 1: Refresh your understanding of math concepts
ML is a math heavy discipline. If you plan to modify ML models, or build new ones from scratch, familiarity with the underlying math concepts is important. You don't have to learn all the math
upfront, but instead you can look up concepts you are unfamiliar with as you come across them. If it's been a while since you've taken a math course, try watching the Essence of linear algebra and
the Essence of calculus playlists from 3blue1brown for a refresher. We recommend that you continue by taking a class from a university, or watching open access lectures from MIT, such as Linear
Algebra or Single Variable Calculus.
Essence of Linear Algebra
by 3Blue1Brown
A series of short, visual videos from 3blue1brown that explain the geometric understanding of matrices, determinants, eigen-stuffs and more.
Essence of Calculus
by 3Blue1Brown
A series of short, visual videos from 3blue1brown that explain the fundamentals of calculus in a way that give you a strong understanding of the fundamental theorems, and not just how the equations
MIT 18.06: Linear Algebra
This introductory course from MIT covers matrix theory and linear algebra. Emphasis is given to topics that will be useful in other disciplines, including systems of equations, vector spaces,
determinants, eigenvalues, similarity, and positive definite matrices.
Step 2: Deepen your understanding of deep learning with these courses and books
There is no single course that will teach you everything you need to know about deep learning. One approach that may be helpful is to take a few courses at the same time. Although there will be
overlap in the material, having multiple instructors explain concepts in different ways can be helpful, especially for complex topics. Below are several courses we recommend to help get you started.
You can explore each of them together, or just choose the ones that feel the most relevant to you.
Remember, the more you learn, and reinforce these concepts through practice, the more adept you will be at building and evaluating your own ML models.
Take these courses:
MIT course 6.S191: Introduction to Deep Learning is an introductory course for Deep Learning with TensorFlow from MIT and also a wonderful resource.
Andrew Ng's Deep Learning Specialization at Coursera also teaches the foundations of deep learning, including convolutional networks, RNNS, LSTMs, and more. This specialization is designed to help
you apply deep learning in your work, and to build a career in AI.
Deep Learning Specialization
In five courses, you will learn the foundations of Deep Learning, understand how to build neural networks, and learn how to lead successful machine learning projects and build a career in AI. You
will master not only the theory, but also see how it is applied in industry.
⬆ And ⬇ Read these books:
To complement what you learn in the courses listed above, we recommend that you dive deeper by reading the books below. Each book is available online, and offers supplementary materials to help you
You can start by reading Deep Learning: An MIT Press Book by Ian Goodfellow, Yoshua Bengio, and Aaron Courville. The Deep Learning textbook is an advanced resource intended to help students deepen
their understanding. The book is accompanied by a website, which provides a variety of supplementary materials, including exercises, lecture slides, corrections of mistakes, and other resources to
give you hands on practice with the concepts.
You can also explore Michael Nielsen's online book Neural Networks and Deep Learning. This book provides a theoretical background on neural networks. It does not use TensorFlow, but is a great
reference for students interested in learning more.
Deep Learning
by Ian Goodfellow, Yoshua Bengio, and Aaron Courville
This Deep Learning textbook is a resource intended to help students and practitioners enter the field of machine learning in general, and deep learning in particular.
Neural Networks and Deep Learning
by Michael Nielsen
This book provides a theoretical background on neural networks. It does not use TensorFlow, but is a great reference for students interested in learning more.
Step 3: Read and implement papers with TensorFlow
At this point, we recommend reading papers and trying the advanced tutorials on our website, which contain implementations of a few well known publications. The best way to learn an advanced
application, machine translation, or image captioning, is to read the paper linked from the tutorial. As you work through it, find the relevant sections of the code, and use them to help solidify
your understanding. | {"url":"https://www.tensorflow.org/resources/learn-ml/theoretical-and-advanced-machine-learning?hl=en&%3Bauthuser=0&authuser=0","timestamp":"2024-11-07T21:56:49Z","content_type":"text/html","content_length":"425090","record_id":"<urn:uuid:aae98723-d6e1-4572-b50f-79f56508696f>","cc-path":"CC-MAIN-2024-46/segments/1730477028017.48/warc/CC-MAIN-20241107212632-20241108002632-00307.warc.gz"} |
Feature scaling - Fundamentals of Machine Learning
Feature scaling is a crucial step in the preprocessing of data for machine learning models. It involves adjusting the scale of the features in your dataset so that they contribute equally to the
result. Without feature scaling, a machine learning algorithm might unfairly prioritize features simply because of their scale. This chapter will delve into the importance of feature scaling, its
types, and how to implement it.
Why is Feature Scaling Important?
Most machine learning algorithms perform better or converge faster when features are on a similar scale. The reason behind this is that features with larger values can dominate those with smaller
values, especially in algorithms that compute distances or similarities between data points, such as K-Nearest Neighbors (KNN), or in models that use gradient descent optimization techniques.
Types of Feature Scaling
There are several methods for feature scaling, but the most commonly used are Min-Max Scaling and Standardization.
Min-Max Scaling
Min-Max Scaling (or normalization) scales the features to a fixed range, usually 0 to 1. The formula for calculating the Min-Max Scaling of a value $x$ is:
$x_{\text{scaled}} = \frac{x - \min(x)}{\max(x) - \min(x)}$
where $\min(x)$ and $\max(x)$ are the minimum and maximum values of the feature, respectively. This method is useful when you need values in a bounded interval. However, it's sensitive to outliers
since the presence of a very large or very small value can skew the scaling.
Standardization (or Z-score normalization) scales the features so that they have the properties of a standard normal distribution with $\mu = 0$ and $\sigma = 1$, where $\mu$ is the mean and $\sigma$
is the standard deviation. The formula for standardization is:
$x_{\text{standardized}} = \frac{x - \mu}{\sigma}$
Standardization is less affected by outliers than Min-Max Scaling and is the preferred method for many machine learning algorithms, especially those that assume input features are normally
Implementing Feature Scaling in Python
Python's scikit-learn library provides easy-to-use functions for both Min-Max Scaling and Standardization. Here's how to implement them:
Min-Max Scaling
from sklearn.preprocessing import MinMaxScaler
import numpy as np
# Sample data
data = np.array([[100, 0.001],
[8, 0.05],
[50, 0.005],
[88, 0.07],
[4, 0.1]])
scaler = MinMaxScaler()
scaled_data = scaler.fit_transform(data)
from sklearn.preprocessing import StandardScaler
# Using the same sample data as above
scaler = StandardScaler()
standardized_data = scaler.fit_transform(data)
When to Scale Features
Feature scaling should be done after splitting the data into training and test sets to avoid data leakage. Data leakage occurs when information from the test set is used to scale the training set,
which can lead to overly optimistic performance estimates.
Feature scaling is a vital preprocessing step in many machine learning workflows. It ensures that each feature contributes equally to the outcome of the model, improving the performance and speed of
learning algorithms. Whether to use Min-Max Scaling or Standardization depends on the specific requirements of your dataset and the machine learning algorithm you are using. Always remember to scale
your features after splitting your dataset to prevent data leakage. | {"url":"https://app.studyraid.com/en/read/2612/53069/feature-scaling","timestamp":"2024-11-08T02:54:36Z","content_type":"text/html","content_length":"161601","record_id":"<urn:uuid:dd661de5-06c1-4193-955d-9341003883fc>","cc-path":"CC-MAIN-2024-46/segments/1730477028019.71/warc/CC-MAIN-20241108003811-20241108033811-00017.warc.gz"} |
Check if a triangle of positive area is possible with the given angles
Checking Triangle Possibility with Given Angles in Python
Triangles are fundamental geometric shapes, and they have interesting properties, especially regarding their angles. In this blog post, we’ll explore a Python program that determines whether a
triangle with positive area is possible based on the given angles. We’ll delve into the mathematical conditions and provide a clear explanation of the logic used in the program.
Understanding the Problem
For a triangle to exist, the sum of its three interior angles must be equal to 180 degrees. The given problem involves checking whether a triangle with positive area is possible based on three input
angles. We’ll use this fundamental geometric principle to create a Python program that assesses triangle possibility.
Python Program Code
Let’s dive into the Python code that addresses the problem:
def is_triangle_possible(angle1, angle2, angle3):
# Check if the sum of angles is 180 degrees
if angle1 + angle2 + angle3 == 180:
# Check if all angles are greater than 0
if angle1 > 0 and angle2 > 0 and angle3 > 0:
return True
return False
# User input for angles
angle1 = int(input("Enter the first angle: "))
angle2 = int(input("Enter the second angle: "))
angle3 = int(input("Enter the third angle: "))
# Check and display result
result = is_triangle_possible(angle1, angle2, angle3)
if result:
print("A triangle with positive area is possible.")
print("No triangle with positive area is possible with the given angles.")
Program Output
Let’s consider an example where the user enters the angles: 60, 70, 50:
Enter the first angle: 60
Enter the second angle: 70
Enter the third angle: 50
A triangle with positive area is possible.
Program Explanation
1. User Input: The program starts by taking user input for three angles representing a triangle.
2. Sum of Angles Check: It checks if the sum of the entered angles is equal to 180 degrees, a requirement for any triangle.
3. Positive Angles Check: Additionally, it verifies that all three angles are greater than 0 to ensure they are valid.
4. Result Display: The program then displays whether a triangle with positive area is possible based on the given angles.
This Python program leverages fundamental geometry principles to determine the possibility of a triangle with positive area based on user-input angles. Understanding geometric rules and applying them
to programming problems enhances both mathematical and programming skills.
Readers are encouraged to experiment with different angle inputs, explore variations of the program, and consider additional geometric conditions for triangle validity. The ability to translate
mathematical concepts into code is a valuable skill in various problem-solving scenarios. Happy coding! | {"url":"https://python-tutorials.in/check-if-a-triangle-of-positive-area-is-possible-with-the-given-angles/","timestamp":"2024-11-10T21:50:24Z","content_type":"text/html","content_length":"87644","record_id":"<urn:uuid:cba7b03f-8eb7-42b8-88b6-cac9c12544e2>","cc-path":"CC-MAIN-2024-46/segments/1730477028191.83/warc/CC-MAIN-20241110201420-20241110231420-00701.warc.gz"} |
data frame with one row for each node in the tree. The row.names of frame contain the (unique) node numbers that follow a binary ordering indexed by node depth. Columns of frame include var, a
factor giving the names of the variables used in the split at each node (leaf nodes are denoted by the level "<leaf>"), n, the number of observations reaching the node, wt, the sum of case
weights for observations reaching the node, dev, the deviance of the node, yval, the fitted value of the response at the node, and splits, a two column matrix of left and right split labels for
each node. Also included in the frame are complexity, the complexity parameter at which this split will collapse, ncompete, the number of competitor splits recorded, and nsurrogate, the number of
surrogate splits recorded.
Extra response information which may be present is in yval2, which contains the number of events at the node (poisson tree), or a matrix containing the fitted class, the class counts for each
node, the class probabilities and the ‘node probability’ (classification trees).
an integer vector of the same length as the number of observations in the root node, containing the row number of frame corresponding to the leaf node that each observation falls into.
an image of the call that produced the object, but with the arguments all named and with the actual formula included as the formula argument. To re-evaluate the call, say update(tree).
an object of class c("terms", "formula") (see terms.object) summarizing the formula. Used by various methods, but typically not of direct relevance to users.
a numeric matrix describing the splits: only present if there are any. The row label is the name of the split variable, and columns are count, the number of observations (which are not missing
and are of positive weight) sent left or right by the split (for competitor splits this is the number that would have been sent left or right had this split been used, for surrogate splits it is
the number missing the primary split variable which were decided using this surrogate), ncat, the number of categories or levels for the variable (+/-1 for a continuous variable), improve, which
is the improvement in deviance given by this split, or, for surrogates, the concordance of the surrogate with the primary, and index, the numeric split point. The last column adj gives the
adjusted concordance for surrogate splits. For a factor, the index column contains the row number of the csplit matrix. For a continuous variable, the sign of ncat determines whether the subset x
< cutpoint or x > cutpoint is sent to the left.
an integer matrix. (Only present only if at least one of the split variables is a factor or ordered factor.) There is a row for each such split, and the number of columns is the largest number of
levels in the factors. Which row is given by the index column of the splits matrix. The columns record 1 if that level of the factor goes to the left, 3 if it goes to the right, and 2 if that
level is not present at this node of the tree (or not defined for the factor).
character string: the method used to grow the tree. One of "class", "exp", "poisson", "anova" or "user" (if splitting functions were supplied).
a matrix of information on the optimal prunings based on a complexity parameter.
a named numeric vector giving the importance of each variable. (Only present if there are any splits.) When printed by summary.rpart these are rescaled to add to 100.
integer number of responses; the number of levels for a factor response.
parms, control
a record of the arguments supplied, which defaults filled in.
the summary, print and text functions for method used.
a named logical vector recording for each variable if it was an ordered factor.
(where relevant) information returned by model.frame on the special handling of NAs derived from the na.action argument.
There may be attributes "xlevels" and "levels" recording the levels of any factor splitting variables and of a factor response respectively.
Optional components include the model frame (model), the matrix of predictors (x) and the response variable (y) used to construct the rpart object. | {"url":"https://www.rdocumentation.org/packages/rpart/versions/4.1-15/topics/rpart.object","timestamp":"2024-11-13T02:48:54Z","content_type":"text/html","content_length":"75607","record_id":"<urn:uuid:08fc8c0e-656e-4699-91a0-d5ac1021e38f>","cc-path":"CC-MAIN-2024-46/segments/1730477028303.91/warc/CC-MAIN-20241113004258-20241113034258-00380.warc.gz"} |
A note on stability criteria in the periodic Lotka–Volterra predator-prey model
Rebelo, Carlota; Ortega, Victor
Applied Mathematics Letters, 145 (2023), Paper No. 108739
We present a stability result for
-periodic solutions of the periodic predator–prey Lotka–Volterra model. In 2021, R. Ortega gave a stability criteria in terms of the $L^1$
norm of the coefficients of a planar linear system associated to the model. Previously, in 1994, Z. Amine and R. Ortega proved another stability criteria formulated in terms of the $L^\infty$
norm. The present work gives a $L^p$
criterion, building a bridge between the two previous results. | {"url":"https://cemat.ist.utl.pt/document.php?section=modeling&project_id=6&member_id=219&doc_id=3598","timestamp":"2024-11-13T05:17:39Z","content_type":"text/html","content_length":"8696","record_id":"<urn:uuid:cd6e6167-9607-4582-b130-47bea8079f0f>","cc-path":"CC-MAIN-2024-46/segments/1730477028326.66/warc/CC-MAIN-20241113040054-20241113070054-00416.warc.gz"} |
Shear-current effect in a turbulent convection with a large-scale shear
The shear-current effect in a nonrotating homogeneous turbulent convection with a large-scale constant shear is studied. The large-scale velocity shear causes anisotropy of turbulent convection,
which produces the mean electromotive force E(W) W×J and the mean electric current along the original mean magnetic field, where W is the background mean vorticity due to the shear and J is the mean
electric current. This results in a large-scale dynamo even in a nonrotating and nonhelical homogeneous sheared turbulent convection, whereby the α effect vanishes. It is found that turbulent
convection promotes the shear-current dynamo instability, i.e., the heat flux causes positive contribution to the shear-current effect. However, there is no dynamo action due to the shear-current
effect for small hydrodynamic and magnetic Reynolds numbers even in a turbulent convection, if the spatial scaling for the turbulent correlation time is τ (k) k-2, where k is the small-scale wave
ASJC Scopus subject areas
• Statistical and Nonlinear Physics
• Statistics and Probability
• Condensed Matter Physics
Dive into the research topics of 'Shear-current effect in a turbulent convection with a large-scale shear'. Together they form a unique fingerprint. | {"url":"https://cris.bgu.ac.il/en/publications/shear-current-effect-in-a-turbulent-convection-with-a-large-scale","timestamp":"2024-11-06T14:48:12Z","content_type":"text/html","content_length":"57359","record_id":"<urn:uuid:1a48a8c9-c202-453f-ae5a-1f1f44659266>","cc-path":"CC-MAIN-2024-46/segments/1730477027932.70/warc/CC-MAIN-20241106132104-20241106162104-00553.warc.gz"} |
An Interesting Introduction to Psychology - Test Interpretation
A researcher administers one form of a test on one day, and then administers an equivalent form to the same group of people at a later date/time. Alternate forms reliability (or "coefficient of
equivalence;" parallel-forms reliability) of reliability is being sought in this example. When correlations are obtained among individual test items, Internal consistency (or "coefficient of internal
consistency") reliability is being assessed; the 3 methods for obtaining this reliability include split-half (involves dividing test into 2 parts then correlating responses from the 2 parts),
Kuder-Richardson Formula 20 (used when test items are dichotomously scored- e.g., "true/false"), and Cronbach's coefficient alpha (used for tests with multiple-scored items- e.g., "never/rarely/
While the split-half reliability coefficient usually lowers the reliability coefficient artificially, the Spearman-Brown formula can be used to correct for the effects of shortening the measure.
Speed tests, as the correlation would be spuriously inflated are measures of internal consistency not good at assessing reliability for.
Instruments that rely on rater judgments would be best to have high Inter-rater (interscorer) reliability, which is increased when scoring categories are mutually exclusive (a particular behavior
belongs to a single category) and exhaustive (categories cover all possible responses/behaviors). The Measurement estimates the amount of error to be expected in an individual test score and is used
to determine a range, referred to as a/an Standard Error of confidence interval, within which an examinee's true score will likely fall. The formula for the standard error of the measurement is
SEmeas = SDx (standard deviation of test scores) / rxx (reliability coefficient).
The probability that a person's true score lies within a range of plus or minus 1 standard error of measurement (SEM) of their obtained score and plus or minus 1.96 (2) SEM, and finally, plus or
minus 2.58 (2.5) SEM is 68% of the time, 95% of the time, and 99% of the time. Hypothetically, a test with a reliability coefficient of +1.0 would have a standard error of measurement of 0.0. A test
with perfect reliability will have no error.
The standard error of measurement is inversely related to the reliability coefficient (rxx) and positively related to the standard deviation of test scores (SDx). Alternate-forms is the reliability
coefficient, when practical, that is best to use. Classical test theory states that an observed score reflects true score variance plus random error variance. Methods of recording behaviors include
duration recording (elapsed time that behavior occurs is recorded), frequency recording (number of times behavior occurs is recorded), interval recording (rater notes whether subject engages in
behavior during given time period), and continuous recording (all behavior during an observation session is recorded). Simply put, validity refers to the degree a test measures what it purports to
A depression scale that only assesses the affective aspects of depression but fails to account for the behavioral aspects would be lacking Content validity, which refers to the extent to which test
items represent all facets of the content area being measured (e.g., EPPP). Content validity assessment requires a degree of agreement between experts in the subject matter, thus it includes an
element of subjectivity. In addition, tests should correlate highly with other tests that measure the same content domain. In contrast to content validity, Face validity occurs when a test appears to
valid by examinees, administrators, and other untrained observers; it is not technically a type of test validity. A personality test that effectively predicts the future behavior of an examinee has
Criterion validity-related validity, which is obtained by correlating scores on a predictor test to some external criterion (e.g., academic achievement, job performance). | {"url":"https://depressionnm.blogspot.com/2013/12/an-interesting-introduction-to.html","timestamp":"2024-11-15T01:48:43Z","content_type":"text/html","content_length":"83542","record_id":"<urn:uuid:fe11b3ce-6700-4944-ac20-ee6c8ba60df2>","cc-path":"CC-MAIN-2024-46/segments/1730477397531.96/warc/CC-MAIN-20241114225955-20241115015955-00458.warc.gz"} |
Reading: Lesson 3 - The Free Cash Flow Valuation Model
7.3.A - The Free Cash Flow Valuation Model
1. The Free Cash Flow Valuation Model
1. As stated earlier, managers should estimate and evaluate the impact of alternative strategies on their firms’ values, which means that managers need a valuation model. The dividend growth model
provides many meaningful insights, such as (1) the relative importance of long-term cash flows versus short-term cash flows and (2) the reason stock prices are so volatile. However, the dividend
growth model is unsuitable in many situations.
2. For example, suppose a start-up company is formed to develop and market a new product. Its managers will focus on product development, marketing, and raising capital. They will probably be
thinking about an eventual IPO, or perhaps the sale of the company to a larger firm; for example, Google, Cisco, Microsoft, Intel, IBM, or another of the industry leaders buy hundreds of
successful new companies each year. For the managers of such a start-up, the decision to initiate dividend payments in the foreseeable future will be totally off the radar screen. Thus, the
dividend growth model is not useful for valuing most start-up companies.
3. Also, many established firms pay no dividends. Investors may expect them to pay dividends sometime in the future—but when, and how much? As long as internal opportunities and acquisitions are so
attractive, the initiation of dividends will be postponed, and this makes the dividend growth model of little use. Even Apple, one of the world’s most successful companies, paid no dividends from
1995 until 2012, when it initiated quarterly dividend payments.
4. Finally, the dividend growth model is generally of limited use for internal management purposes, even for a dividend-paying company. If the firm consisted of just one big asset and if that asset
produced all of the cash flows used to pay dividends, then alternative strategies could be judged through the use of the dividend growth model. However, most firms have several different
divisions with many assets, so the corporation’s value depends on the cash flows from many different assets and on the actions of many managers. These managers need a way to measure the effects
of their decisions on corporate value, but the discounted dividend model isn’t very useful because individual divisions don’t pay dividends.
5. Fortunately, the free cash flow valuation model does not depend on dividends, and it can be applied to divisions and sub-units as well as to the entire firm.
2. Sources of Value and Claims on Value
1. Companies have two primary sources of value, the value of operations and the value of non-operating assets. There are three major types of claims on this value: debt, preferred stock, and common
2. Recall that free cash flow (FCF) is the cash flow available for distribution to all of a company’s investors. The weighted average cost of capital (WACC) is the overall return required by all of
a company’s investors. Because FCF is generated by a company’s operations, the present value of expected FCF when discounted by the WACC is equal to the value of a company’s operations, Vop:
3. The primary source of value for most companies is the value of operations. A secondary source of value comes from non-operating assets (also called financial assets). There are two major types of
non-operating assets: (1) marketable securities, which are short-term securities (like T-bills) that are over and above the amount of cash needed to operate the business; (2) other non-operating
assets, which often are investments in other businesses. For example, Ford Motor Company’s automotive operation held about $14.2 billion in marketable securities at the end of December 2010, and
this was in addition to $6.3 billion in cash. Second, Ford also had $2.4 billion of investments in other businesses, which were reported on the asset side of the balance sheet as “Equity in Net
Assets of Affiliated Companies.” In total, Ford had $14.2 + $2.4 = $16.6 billion of non-operating assets, amounting to 26% of its $64.6 billion of total automotive assets. For most companies, the
percentage is much lower. For example, as of the end of October 2010, Walmart’s percentage of non-operating assets was less than 1%, which is more typical.
4. For a company that is a going concern, debtholders have the first claim on value in the sense that interest and scheduled principal payments must be paid before any preferred or common dividends
can be paid. Preferred stockholders have the next claim because preferred dividends must be paid before common dividends. Common shareholders come last in this pecking order and have a residual
claim on the company’s value.
3. Estimating the Value of Operations
1. The free cash flow (FCF) model is analogous to the dividend growth model, except the FCF valuation model (1) discounts free cash flows instead of dividends and (2) the discount rate is the
weighted average cost of capital (WACC) instead of the required return on stock. Free cash flow is generated by operations, FCF is the cash flow available for all investors, and the WACC is the
overall required return of all investors; therefore, the result of the FCF model is the total value of operations, not just the value of the stock.
2. We will illustrate the FCF valuation model using MagnaVision Inc., which produces optical systems for use in medical photography. Growth has been rapid in the past, but the market is becoming
saturated, so the sales growth rate is projected to decline from 21% in 2013 to a sustainable rate of 5% in 2016 and beyond. Profit margins are expected to improve as the production process
becomes more efficient and because MagnaVision will no longer be incurring marketing costs associated with the introduction of a major product. All items on the financial statements are projected
to grow at a 5% rate after 2016.
3. Early we explained how to calculate FCF if you have historical financial statements; however, you need forecasted financial statements to apply the FCF valuation model. To better focus on the
free cash flow valuation model in this example, we provide MagnaVision’s projected free cash flows in Figure below and defer forecasting until Chapter 12.8 We also provide MagnaVision’s weighted
average cost of capital, 10.84%.
4. Notice that MagnaVision has negative free cash flows in the first two projected years. Negative free cash flow in early years is typical for young, high-growth companies. Even though net
operating profit after taxes (NOPAT) may be positive, free cash flow often is negative due to investments in operating assets during the high-growth years. As growth slows, free cash flow will
become positive and eventually grow at a constant rate.
5. To estimate MagnaVision’s value of operations, we use an approach similar to the non-constant dividend growth model for stocks and proceed as follows.
1. Recognize that growth after Year N will be constant, so we can use a constant growth formula to find the firm’s value at Year N. The value at Year N is the sum of the PVs of FCF for year N + 1
and all subsequent years, discounted back to Year N.
2. Find the PV of the free cash flows for each of the N non-constant growth years. Also, find the PV of the firm’s value at Year N.
3. Now sum all the PVs, those of the annual free cash flows during the non-constant period plus the PV of the Year-N value, to find the firm’s value of operations.
A variant of the constant growth dividend model, in which FCF replaces dividends and the WACC replaces rs. This equation can be used to find the value of MagnaVision’s operations at time N, when
its free cash flows stabilize and begin to grow at a constant rate. This is the value of all FCFs beyond time N, discounted back to time N (which is 2016 for MagnaVision):
6. Based on a 10.84% cost of capital, $49 million of free cash flow in 2016, and a 5% growth rate, the value of MagnaVision’s operations as of December 31, 2016, is estimated to be $880.99 million:
7. This $880.99 million figure is called the company’s horizon value because it is the value at the end of the forecast period. It is also sometimes called a continuing value or terminal value. It
is the amount that MagnaVision could expect to receive if it sold its operating assets on December 31, 2016.
8. The Figure above shows the free cash flow for each year during the non-constant growth period along with the horizon value of operations in 2016. To find the value of operations as of “today,”
December 31, 2012, we find the PV of the horizon value and each annual free cash flow in the Figure below, discounting at the 10.84% cost of capital:
9. The sum of the PVs is approximately $615 million, and it represents an estimate of the price MagnaVision could expect to receive if it sold its operating assets “today,” December 31, 2012.
4. Estimating the Price per Share
1. In addition to the value of operations, we need to know the value of MagnaVision’s non-operating assets, claims on value (such as debt and preferred stock), and the number of shares. Those values
are shown in the INPUTS section of the Figure below.
2. Think of a company’s value as though it were a pie whose size is determined by the value of operations and the value of any financial (non-operating) assets. The first piece of pie belongs to
debtholders, the second to preferred stockholders, and the remaining piece (if there is one) to common shareholders. In other words, common shareholders have a residual claim.
3. On December 31, 2012, MagnaVision reported owning $63 million of marketable securities. We don’t need to calculate a present value for marketable securities because short-term financial assets as
reported on the balance sheet are at (or close to) their market value. Therefore, MagnaVision’s total value on December 31, 2012, is $615.27 + $63 = $678.27 million.
4. The value of common equity is the remaining value after other claims. The Figure below shows that MagnaVision has $247 million in debt and $62 million in preferred stock. Therefore, the value
left for common stockholders is $678.27 − $247 − $62 = $369.27 million.
5. Comparing the Free Cash Flow Valuation and Dividend Growth Models
1. The free cash flow valuation model and the dividend growth model give the same estimated stock price if you are very careful to be consistent with the implicit assumptions underlying the
projections of free cash flows and dividends. Which model should you use, as they both give the same answer? If you were a financial analyst estimating the value of a mature company whose
dividends are expected to grow steadily in the future, it would probably be more efficient to use the dividend growth model. In this case you would need to estimate only the growth rate in
dividends, not the entire set of forecasted financial statements.
2. However, if a company is paying a dividend but is still in the high-growth stage of its life cycle, you would need to project the future financial statements before you could make a reasonable
estimate of future dividends. After you have estimated future financial statements, it would be a toss-up as to whether the corporate valuation model or the dividend growth model would be easier
to apply. If you were trying to estimate the value of a company that has never paid a dividend, a private company (including companies nearing an IPO), or a division of a company, then there
would be no choice: You would have to estimate future financial statements and use the corporate valuation model.
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Unit 4 Test | Calc Medic
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Unit 4 Test
Unit 4 - Day 17
Writing a Precalculus Assessment
• Include questions in multiple representations (graphical, analytical, tabular, verbal)
• Write questions that reflect learning targets and require conceptual understanding
• Include multiple choice and short answer or free response questions
• Determine scoring rubric before administering the assessment (see below)
• Offer opportunities to practice with and without calculators throughout the year
Questions to Include
• Evaluating with Right Triangle Trig - find ratios from a right triangle, missing angles when given the sides, missing sides when given the angle and another side, etc.
• The Unit Circle - evaluate expressions using all six trigonometric ratios for angles on the unit circle; include coterminal angles greater than 2pi and less than -2pi
• Graph all six trigonometric functions with transformations; identify key feature of each graph, including asymptotes for secant, cosecant, tangent, and cotangent
• Evaluate expressions with inverse trig
• When given a ratio, use inverse trig to find the other ratios
• Trigonometric Modeling: interpret period and amplitude in context
Grading Tips
Look for more than just correct answers. Give students feedback on their justifications, communication, and mathematical thinking. We recommend that you prepare a rubric for the free response and
short answer items before you begin grading your quizzes or tests. Know what information is necessary for a complete and correct response and award points when a student presents that information.
Many of the “Why did I get marked down?” questions are eliminated when you share the components that earn points.
Students will struggle with the domain and range of the inverse trig functions so make sure you ask a mix of simple and challenging questions. Specifically something like arccos(cos(4π/3)),where the
inverse does not undo the original, but the answer is actually (2π/3). Have students identify equations from graphs as well as graph 1-2 periods of a few of the 6 functions. Consider having a
calculator active section and a calculator inactive section so that students must show their knowledge of the unit circle.
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Research On Some Issues Of Quantum Cryptography
Posted on:2009-01-23 Degree:Doctor Type:Dissertation
Country:China Candidate:S H Li Full Text:PDF
GTID:1118360242478272 Subject:Cryptography
Quantum cryptography is a cryptolographical resource based on quantum mechanics rather than classical mechanics, and significant to information security. The paper is built on the mathematical frame
of quantum mechanics, and focuses on the unconditional security of quantum key distributions and quantum ciphers. The major results of my research are following.1. The essence of quantum key
distribution protocols is the function of detecting non-orthogonal quantum states, and thus it is crucial how to detect the error syndromes occurring on the quantum states. Now, most quantum key
distribution protocols employ the alternative | 0? ,|1? and | +? ,|?? bases to detect all possible errors on the qubits exchanged. This paper analyzes Bell-state measurement {|Φ+ ? ,|Φ? ? ,|Ψ+ ? ,
|Ψ??} basis show its capacity of error detecting. This paper proposed a new quantum key distribution on 4-dimensional Hilbert space on Bell- state measurements, which saves a great amount of
classical and quantum communication.2. 2. Based on the reasonability of Bell-state measurement, this paper also shows the equivalence of two quantum key distribution protocols with random chosen EPR
pairs, and points out the two protocols have a defect on the way of determining the error number, and modify the defect to complete the proof of unconditional security.3. 3. To make error-detecting
strategy capable, a condition must be satisfied, i.e., there have to be some methods to estimate the error number correctly. It's widely known that random sampling tests can present an upper bound of
error number caused by illegal eavesdroppers successfully with probability exponentially close to 1, which means that it can judge whether quantum error-correcting codes or classical information
reconciliation and privacy amplification work. This paper improves random sampling tests from a single-sample method to the multiple-sample one. This cause quantum key distribution protocols with
random sampling tests more efficient.4. Quantum mechanics contributes not only on public-key systems, but also on the private-key ones. For the ciphers using non-orthogonal quantum states to encrypt
classical or quantum information, there are still some remarkable features which are not found in the classical cryptographic world. Specially, the quantum cipher-texts are with error-detecting
property and the key entropies are different against different quantum plaintext–known attacks. To show such things in formulas, quantum secure channel is defined, and presents some results of the
key entropies and its corresponding bounds of BB84 quantum coding against the collective and coherent plaintext-known attacks.5. Some classical cryptographic theories can serve quantum cryptography
too. This paper employs the classical Hash functions to replace random sampling tests to determine the upper bound of error numbers, which avoids the frequent requests for authentic classical
communication. By combining quantum ciphers and classical Hash functions, two quantum cryptographic algorithms are presented in the paper, and are more practical for usual applications.
Keywords/Search Tags: quantum cryptography, unconditional security, quantum key distribtution, quantum cipher | {"url":"https://globethesis.com/?t=1118360242478272","timestamp":"2024-11-05T16:53:00Z","content_type":"application/xhtml+xml","content_length":"8704","record_id":"<urn:uuid:2bd8e43d-892b-4ddf-8893-637d60073d40>","cc-path":"CC-MAIN-2024-46/segments/1730477027884.62/warc/CC-MAIN-20241105145721-20241105175721-00659.warc.gz"} |
round( function
The round( function rounds a number to a specified increment.
This function has two parameters:
number – is the number you want to round.
step – is the increment you want to round to. For example if the increment is 1, the value will be rounded to an integer. If the increment is 12, the value will be rounded to the nearest dozen.
This function rounds a number to a specified increment. This example rounds the quantity to the nearest dozen.
The output of this formula will be 0, 12, 24, 36, etc. The formula above may round the number down, for example 14 will round down to 12. Here is another formula that will always round up to the next
higher dozen (i.e. 13 will round up to 24):
Non-integer increments are Ok. This next formula rounds the temperature to the nearest 1/2 degree.
Since dates are really numbers, the formula below rounds the StartDate to the nearest Monday. (The numbers for Mondays are multiples of 7.)
Note: This function uses the half-even rounding method, which is different from the Round half to even rounding method used by Numeric Patterns.
See Also
10.0 No Change Carried over from Panorama 6.0 | {"url":"https://www.provue.com/panoramax/help/function_round.html","timestamp":"2024-11-06T11:58:35Z","content_type":"application/xhtml+xml","content_length":"6696","record_id":"<urn:uuid:404ee6c2-43b0-4c5c-9486-87e2642ee495>","cc-path":"CC-MAIN-2024-46/segments/1730477027928.77/warc/CC-MAIN-20241106100950-20241106130950-00044.warc.gz"} |
Interactive response surface modeling
rstool opens a graphical user interface for interactively investigating one-dimensional contours of multidimensional response surface models.
By default, the interface opens with the data from hald.mat and a fitted response surface with constant, linear, and interaction terms.
A sequence of plots is displayed, each showing a contour of the response surface against a single predictor, with all other predictors held fixed. rstool plots a 95% simultaneous confidence band for
the fitted response surface as two red curves. Predictor values are displayed in the text boxes on the horizontal axis and are marked by vertical dashed blue lines in the plots. Predictor values are
changed by editing the text boxes or by dragging the dashed blue lines. When you change the value of a predictor, all plots update to show the new point in predictor space.
The pop-up menu at the lower left of the interface allows you to choose among the following models:
• Linear — Constant and linear terms (the default)
• Pure Quadratic — Constant, linear, and squared terms
• Interactions — Constant, linear, and interaction terms
• Full Quadratic — Constant, linear, interaction, and squared terms
Click Export to open the following dialog box:
The dialog allows you to save information about the fit to MATLAB^® workspace variables with valid names.
rstool(X,Y,model) opens the interface with the predictor data in X, the response data in Y, and the fitted model model. Distinct predictor variables should appear in different columns of X. Y can be
a vector, corresponding to a single response, or a matrix, with columns corresponding to multiple responses. Y must have as many elements (or rows, if it is a matrix) as X has rows.
The optional input model can be any one of the following:
• 'linear' — Constant and linear terms (the default)
• 'purequadratic' — Constant, linear, and squared terms
• 'interaction' — Constant, linear, and interaction terms
• 'quadratic' — Constant, linear, interaction, and squared terms
To specify a polynomial model of arbitrary order, or a model without a constant term, use a matrix for model as described in x2fx.
rstool(x,y,model,alpha) uses 100(1-alpha)% global confidence intervals for new observations in the plots.
rstool(x,y,model,alpha,xname,yname) labels the axes using xname and yname. To label each subplot differently, xname and yname can be string arrays or cell arrays of character vectors.
The following uses rstool to visualize a quadratic response surface model of the 3-D chemical reaction data in reaction.mat:
load reaction
alpha = 0.01; % Significance level
The rstool interface is used by rsmdemo to visualize the results of simulated experiments with data like that in reaction.mat. As described in Response Surface Designs, rsmdemo uses a response
surface model to generate simulated data at combinations of predictors specified by either the user or by a designed experiment.
Version History
Introduced before R2006a | {"url":"https://it.mathworks.com/help/stats/rstool.html","timestamp":"2024-11-05T03:42:44Z","content_type":"text/html","content_length":"74416","record_id":"<urn:uuid:c48f91af-ee52-4f84-ab58-8d1d235a3be8>","cc-path":"CC-MAIN-2024-46/segments/1730477027870.7/warc/CC-MAIN-20241105021014-20241105051014-00811.warc.gz"} |
Drop Cutter part 1/3: Cutter vs. Vertex
Based on a 2002 paper by Chuang et al., I'm trying to understand the math for calculating 'drop-cutter' toolpaths (Julian Todd also has a lot to say about the drop-cutter approach). The basic idea is
that the x,y position of the cutter is known, and it is dropped down along the z-axis until it touches a triangulated surface. A complex surface is built up of many many small triangles, but the
algorithm stays the same.
For each triangle there are seven items the cutter might be touching: three vertices, three edges, and one facet (the triangle surface). It looks like a contact point with a vertex is the easiest to
calculate, so I'm starting with that. I'm hoping to post part 2/3 and 3/3 of this post soon. They will deal with the facet and the edges.
I'm using this fairly simple cutter definition C(R,r) where R denotes the shaft diameter and r the corner radius. The three basic shapes that can be defined this way are shown in the picture. A more
elaborate model would include tapered cutters, but I think I won't bother with that now... A cutter thus consists of a cylindrical part or a toroidal part, or both. That means I need six different
functions in total for this algorithm:
1. Cylinder against vertex
2. Toroid against vertex
3. Cylinder against facet
4. Toroid against facet
5. Cylinder against edge
6. Toroid against edge
Here's how I think no 1 and 2 work.
Assume the vertex we are testing against is at (x, y, z) and say the cutter is at location xe,ye. We are looking for the cutter z-coordinate ze so at the end when the cutter is in contact with the
vertex it will be located at (xe, ye, ze) .
Calculate the distance in the xy-plane from (xe, ye) to (x, y):
q = sqrt( (xe-x)^2 + (ye-y)^2)
Now if q > R the vertex lies outside the cutter and we should report an error or just skip to the next triangle/vertex.
If q <= R-r the vertex lies under the cylindrical part of the cutter and we set ze = z
If (R-r) < q <= R the vertex lies under the spherical/toroidal part of the cutter. This picture should explain the geometry:
The cutter can be positioned a distance z1 lower than z. To calculate z1 we need z2. It can be found by noting that (x ,y, z) should lie on the toroidal surface:
(q-(R-r))^2 + h2^2 = r^2
or h2 = sqrt( r^2 - (q-(R-r))^2 )
so now h1=r-h2 and we found ze = z-h1
A quick test in matlab seems to confirm that this works:
here a ball-nose cutter is dropped down along the dashed line into contact with all three vertices (red, blue, and green stars), and finally it's positioned at the highest ze found (red dot).
Well, that was easy. Now onto the facet and edge tests!
One thought on “Drop Cutter part 1/3: Cutter vs. Vertex”
This site uses Akismet to reduce spam. Learn how your comment data is processed. | {"url":"https://www.anderswallin.net/2007/06/drop-cutter-part-13-cutter-vs-vertex/","timestamp":"2024-11-03T01:14:18Z","content_type":"text/html","content_length":"72330","record_id":"<urn:uuid:7e926a35-0d91-474e-a313-7074640b3690>","cc-path":"CC-MAIN-2024-46/segments/1730477027768.43/warc/CC-MAIN-20241102231001-20241103021001-00718.warc.gz"} |
A Cruciverbalist’s Introduction to Bayesian reasoning
Status: Hopefully a nice introduction to some of the basics of Bayesian reasoning for newcomers.
Clue: Mathematical methods inspired by an eighteenth century minister (8)
“Bayesian” is a word that has gained a lot of attention recently, though my experience tells me most people aren’t exactly sure what it means. I’m fairly confident that there are many more
crossword-puzzle enthusiasts than Bayesian statisticians—but I would also note that the overlap is larger than most would imagine. In fact, anyone who has ever worked on a crossword puzzle has
employed Bayesian reasoning. They just aren’t (yet)aware of it. So I’m going to explain both how intuitive Bayesian thinking is, and why it’s useful, even outside of crosswords and statistics.
But first, who was Bayes, what is his “law” about, and what does that mean?
Clue: Sound of a Conditional Reverend’s Dog (5)
“Bayes” of statistical fame, is the Reverend Thomas Bayes. He was a theologian and mathematician, and the two works he published during his lifetime dealt with the theological problem of happiness,
and a defense of Newton’s calculus—neither of which concern us. His single posthumous work, however, was what made him a famous statistician. The original title, “ A Method of Calculating the Exact
Probability of All Conclusions founded on Induction,” clearly indicates that it’s meant to be a very inclusive, widely applicable theorem. It was also, supposedly, a response to a theological
challenge posed by Hume—claiming miracles didn’t happen.
Clue: Wonders at distance travelled without vehicle upset (8)
“Miracles”, Hume’s probabilistic argument said, are improbable, but incorrect reports are likely— so, the argument goes, it is more likely that the reports are incorrect than that the miracle
occurred. This way of comparing probabilities isn’t quite right, statistically, as we will suggest later. But Bayes didn’t address this directly at all.
Clue: Taking a risk bringing showy jewelry to school (8)
“Gambling” was a hot topic in 19th century mathematics, and Bayes tried to answer an interesting question; when you see something happen several times, how do can you figure out, in general, the
probability of it occurring? His example was about throwing balls onto a table—you aren’t looking, and a friend throws the first ball. After this, he throws more, each time, telling you whether the
ball landed to the left or right of the first ball. After a doing this a few times, you still have’t seen the table, but want to know how likely is it that the next ball land to the left of that
original ball.
To answer this, he pointed out that you get a bit more information about the answer every time a ball is thrown. After the first ball, for all you know the odds are ^50⁄[50] that the next one will be
on either side. after a few balls are thrown, you get a better and better sense of what the answer is. After you hear the next five balls all land to the left, you’ve become convince that the next
ball landing to the left is more likely than landing to the right. That’s because the probabilities are not independent—each answer gives you a little bit more information about the odds.
But enough math—I’m ready to look at a crossword.
Could wine be drunk by new arrival? (6)
“Newbie” is how I’d prefer to put my ability with crossword puzzles. But as soon as I started, I noticed a clear connection. The method of reasoning I practice and endorse as a decision theorist are
nearly identical to the methods that are used by people in this everyday amusement. So I’ll get started on filling in (only one part of) the crossword I did yesterday, and we’ll see how my Bayesian
reasoning works. I start by filling in a few easy answers, and I’m pretty confident in all of these. 6 Down—Taxing mo. for many, 31 Across—Data unit, 44 Across—“Scream” actress Campbell.
The way I’ve filled these in so far is simple—I picked answers I thought were very likely to be correct. But how can I know that they are correct? Maybe I’m fooling myself. The answer is that I’ve
done a couple crosswords before, and I’ve found that I’m usually right when I’m confident, and these answers seem really obvious. But can I apply probabilistic reasoning here?
Clue: Distance into which vehicle reverses ___ that’s a wonder (7)
“Miracles,” or anything else, according to Reverend Bayes, should follow the same law as thrown balls. If someone is confident, that is evidence, of a sort. Stephen Stigler, a historian of math,
argues that Bayes was implying an important caveat to Hume’s claim—the probability of hearing about a miracle increases each time you hear another report of it. That is, thee two facts are, in a
technical sense, not independent—and the more independent accounts you hear, the more convinced you should be.
But that certainly doesn’t mean that every time a bunch of people claim something outlandish, it’s true. And in modern Bayesian terms, this is where your prior belief matters. If someone you don’t
know well at work tells you that they golfed seven under par on Sunday, you have every reason to be skeptical. If they tell you they golfed seven over par, you’re a bit less likely to be skeptical.
How skeptical, in each case?
We can roughly assess your degree of belief— if a friend of yours attested to the second story, you’d likely be convinced, but it would take several people independently verifying the story for you
to have a similar level of belief in the first. That’s because you’re more skeptical in the first place. We could try to quantify this, and introduce Bayes’ law formally, but there’s no need to bring
algebra into this essay. Instead, I want to think a bit more informally—because I can assess something as more or less likely without knowing the answer, without doing any math, and without
assigning it a number.
When you hear something outlandish, your prior belief is that it is unlikely. Evidence, however, can shift that belief—and enough evidence, even circumstantial or tentative, might convince you that
the claim is plausible, probably, or even very likely. And in a way it doesn’t matter what your prior is, if you can accumulate enough different pieces of trustworthy evidence. And that leads us to
how I can use the answers I filled in as evidence to help me make further plausible guesses.
I look at some of the clues I didn’t immediately figure out. I wasn’t sure what 6 Across—Completely blows away, would be; there are lots of 4-letter words that might fit the clue. Once I get the A,
however, I’m fairly confident in my guess, conditional on this (fairly certain) new information. I look at 31 Down—Military Commission (6), but I can’t think of any that start with a B. I see 54
Across—Place for a race horse, and I’m unsure—there are a few words that fit—it could be “first”, “third”, “fifth,” “sixth” or “ninth”, and I have no reason to think any more likely than
another. So I look for more information, and notice 54 Down—It might grow to be a mushroom (5, offscreen). “Spore” seems likely, and I can see that this means “Sixth” works—so I fill in both.
At this point, I can start filling in a lot more of the puzzle, and the pieces are falling in to place—each word I figure out that fits is a bit more evidence that the others are correct, making me
confident, but there are a few areas where I seem stuck.
Being stuck is evidence of a different sort—it probably means at least one of two things—either I have something incorrect, or I’m really bad at figuring out crosswords. Or, of course, both. (And
as TiffanyAching points out, this mindset is nicely captured by the phrase “thinking in pencil”—which I should have done more literally, but overwriting the ink illustrates when I changed my mind, so
I’m leaving it in,)
At this point I start revisiting some of my earlier answers, ones I was pretty confident about until I got stuck. I’m still pretty confident in 39 Down—Was at one time, but ___ now. “Isn’t” is too
obvious of an answer to be wrong, I think. On the other hand, 38 Down—A miscellany or collection, has me stumped, but two Is in a row also seem strange. 37 Down—Small, fruity candy, is also
frustrating me; I’m not such an expert in candy, but I’m also not coming up with anything plausible. So I look at 50 Across—A tiny part of this?, again, and re-affirm that “Bit” seems like it’s a
good fit. I’m now looking for something that can give me more information, so I rack my brains, and 36 Across—Ho Chi Min’s capital, comes to me: Hanoi. I’m happy that 39 Down is confirmed, but
getting nervous about the rest.
I decided to wait, and look elsewhere, filling in a bit more where I could. My progress elsewhere is starting to help me out.
Now, I need to re-evaluate some earlier decisions and update my beliefs again. It has become a bit more complex than evaluating single answers—I need to consider the joint probability of several
different things at once. I’ll unpack how this relates to Bayesian reasoning afterwards, but first, I think I made a mistake.
I was marginally confident in 50 Across—A tiny part of this? as “bit”, but now I have new evidence. I’m pretty sure Nerb isn’t a type of candy, but “Nerd” seems to fit. I’m not sure if they are
fruity, so I’m not confident, and I’m still completely at a loss on 38 Down—A miscellany or collection. That means I need to come up with an alternative for 50 Across; “Dot” seems like an unlikely
option, but it fits really well. And then it occurs to me; A dot is a little bit of the question mark. That’s an annoying answer, but it seems a lot more likely than that “Nerb” is a type of candy.
And I’m not sure what Olio is, but there’s really nothing else that I can imagine fitting. And there are plenty of words I don’t know. (As I found out later, this is one of them.)
At first, I had a high confidence that “Bit” was the best answer for 50 Across—I had a fairly strong prior belief, but I wasn’t certain. As evidence mounted, I started to re-assess. Weak evidence,
like the strange two Is in a row, made me start to question the assumption that I was right. More weak evidence—remembering that there is a candy of some sort called Nerds, and realizing that “Dot”
was a potential answer, made me revise my opinion. I wasn’t strongly convinced that I had everything right, but I revised my belief. And that’s exactly the way a Bayesian approach should work; you’re
trying to figure out which possibility is worth betting on.
That’s because all of probability theory started with a simple question that a certain gambler asked Blaise Pascal; how to we split the pot when a game gets interrupted. And historians who don’t
think Bayes was trying to formulate a theological rebuttal to Hume suggest that he’s really responding to a question posed by de Moivre—from whose book he may have learned probability theory, which
we need to mention in order to figure out why I’d pick “Dot” over “Bit”—even though I think it’s a stupid answer. But before I get there, I’ve made a bit more progress—I’m finished, except for
one little thing.
31 Down—Military Commission. That’s definitely a problem—I’m absolutely sure Brevei isn’t the right answer, and 49 Down, offscreen, is giving me trouble too. The problem is, I listed all the
possible answers for 54 Across—Place for a race horse, and the only one that started with an “S” was sixth.
Clue: Conviction … or what’s almost required for a conviction (9)
“Certainty” can be dangerous, because if something is certain, almost by definition, it means nothing can convince me otherwise. It’s easy to be overconfident, but as a Bayesian, it’s dangerous to be
so confident that I don’t consider other possibilities—because I can’t update my beliefs! That’s why Bayesians, in general, are skeptical of certainty. If I’m certain that my kid is smart and doing
well in school, no number of bad grades or notes from the teacher can convince me to get them a tutor. In the same way, if I’m certain that I know how to get where I’m going, no amount of confused
turns, circling, or patient wifely requests will convince me to ask for directions. And if I’m certain that “Place for a race horse” is limited to a numeric answer, no number of meaningless words
like “Brevei” can change my mind.
Clue: High payout wagers (9)
“Perfectas” are bets placed on a horse race, predicting the winner and second place finisher, together. If you get them right, they payoff can be really significant—much more than bets on horses to
win or to place. In fact, there are lots of weird betting terms in horse racing, and by excluding them from consideration, I may have been hasty in filling out “sixth.” My assumption of having
compiled and exhaustive list of terms was premature. Instead, I need to reconsider once again—and that brings us to why, in a probabilistic sense, crosswords are hard.
Disreputable place for a smoke? (5)
“Joint” probabilities are those that relate to multiple variables. And when solving the crossword, I’m not just looking to answer each clue, I’m looking to fill in the puzzle—it needs to solve all
of the clues together. Just like figuring out a Perfecta is harder than picking the right horse; putting multiple uncertain questions together is where joint probabilities show up. But it’s not
hopeless; as you figure out more of the puzzle, you reduce the remaining uncertainty. It’s like getting to place a Perfecta bet after seeing 90% of the race; you have some pretty good ideas about
what can and can’t happen.
Similarly, Bayesians, in general, collect evidence to constrain what they think is and isn’t probable. Once enough balls have been thrown to the left of that first one, you get pretty sure the odds
aren’t 50–50. The prerequisite for getting the right answer, however, is being willing to reconsider your beliefs—because reality doesn’t care what you believe.
And the reality is that 31 Down is Brevet, so I need an answer to 54 Across—Place for a race horse that starts “St”. And that’s when it hit me—sometimes, I need to simply be less certain I know
what’s going on. The race horse isn’t running, and there are no bets. It’s in a stall, waiting patiently for me to realize I was confused.
A Final Note
I’d note three key lessons that Bayesians can learn from crosswords, since I’ve already spent pages explaining how Crossworders already understand Bayesian thinking. And they are lessons for life,
ones that I’d hope crossword enthusiasts can apply more generally as well.
1. The process of explicitly thinking about what you are uncertain of, and noticing when something is off, or you are confused, is useful to apply even (especially!) when you’re not doing crossword
2. Evaluating how sure you are, and wondering if you are overconfident in your model or assumptions, would have come in handy to those predicting the 2016 election.
3. Being willing to actually change your mind when presented with evidence is hard, but I hope you’d rather have a messy crossword than an incorrectly solved one.
A Postscript for Pedants
Scrupulously within the rules, but not totally restrictive
“Strict” Bayesians are probably annoyed about some of this—at no point in the process did I get any new evidence. No one told me about any new balls thrown, I only revised my belief based on
thinking. A “Real Bayesian” starts with all the evidence already available, and only updates when new evidence comes in. For a non-technical response, it’s sufficient to note that computation and
thought takes time, and although the brain roughly approximates Bayesian reasoning, the process of updating is iterative. And for a technical version of the same argument, I’ll let someone else
explain that there are no real Bayesians. (And thanks to Noah Smith for that link!)
The crossword clues were a combination of info from http://www.wordplays.com/crossword-clues/, and my own inventions.
The crossword is an excerpt from Washington Post Express’s Daily Crossword for January 11th, 2017, available in full on Page 20, here: https://issuu.com/expressnightout/docs/express_01112017
Note: This was formerly a linkpost to a since-made-unavailable blog post on medium, which I no longer use or endorse—when I went to visit the link, I realized it was broken. I also had moved the
external blog post to here. It is now moved to LW completely—hence the earlier comments.
• This is more of a light introduction to rationality piece, not really intended to inform most users of LessWrong—but I think some people may enjoy it, and others may be able to provide useful
That said, I think it’s useful to have a more diverse set of introduction to rationality posts, both to share with others, and to promote separately, since they will appeal to different
audiences. I’d love to hear feedback about where people think it could be improved, or if people think it is or is not useful!
□ I’m afraid I don’t really have any useful feedback, but I did enjoy it. If nothing else it reminded me that I used to love crosswords and haven’t done any in yonks, so thanks for that!
I think it’s useful to have a more diverse set of introduction to rationality posts
Agreed. As long as they’re not inaccurate or misleading, having a variety of introductory articles is definitely valuable. Though I think I already had a decent beginner level grasp of Bayes
when I read this, the crossword theme suggested “think in pencil” as a rather pleasing shorthand for the importance of resisting certainty and undue attachment to beliefs.
☆ I like the phrase “Think in Pencil”—I’m surprised I hadn’t heard it before, since it seems to be fairly well known. | {"url":"https://www.greaterwrong.com/posts/LuzekfX96Zgf2qMPJ/a-cruciverbalist-s-introduction-to-bayesian-reasoning","timestamp":"2024-11-02T05:14:28Z","content_type":"text/html","content_length":"36051","record_id":"<urn:uuid:8e6163f1-8fac-4ed4-abb7-d798da38924d>","cc-path":"CC-MAIN-2024-46/segments/1730477027677.11/warc/CC-MAIN-20241102040949-20241102070949-00749.warc.gz"} |
During the run of a coupled setup it can happen that field masks are changing (for example due to rising or falling sea levels). Unfortunatly, weights are only computed in the initialisation phase of
the coupler. In addition, this is a time consuming process. Hence, reinitialisation of the coupler after each change of a mask is not a viable option, most of the time.
Instead of recomputation of all weights, it is also possible to provide an additional mask to the put operation. This mask can be used at each exchange step to adjust the weights, which were computed
in the initialisation phase.
When using dynamic fractional masking, the user has to provide masking values for all source points in an exchange. The masking values have to be in the range from 0.0 to 1.0.
In an exchange all source field values are multiplied by the fractional mask values. These values are then used to compute preliminary target field values. Afterwards each target field value is
divided by the sum of the weights, used to compute the respected value, multiplied by the associated fractional mask value.
In the following example a target point (blue dot) is interpolated by the average (
Without fractional mask With fractional mask: With fractional mask:
Usage in YAC
In order to use dynamic fractional masking, the user has to explicitly enable it (see Enabling dynamic fractional masking). This is only possible for field that have already been defined (see
Definition of Coupling Fields).
If a field was enabled for dynamic fractional masking, the user has to provide a mask for all field values in all exchanges, where this field is used as a source (see yac_cput_frac, yac_cput_frac_,
yac_cput_frac_ptr_, yac_cexchange_frac, yac_cexchange_frac_, yac_cexchange_frac_ptr_, yac_fput_frac_*, and yac_fexchange_frac_*).
The usage of dynamic fractional masking only makes sense for interpolation, where the weights for each target field value sum up to 1.
The implementation of dynamic fractional masking in YAC is based on the work of the same feature in OASIS3-MCT 6.0. | {"url":"https://dkrz-sw.gitlab-pages.dkrz.de/yac/d0/daa/frac_mask_desc.html","timestamp":"2024-11-10T05:56:22Z","content_type":"application/xhtml+xml","content_length":"13315","record_id":"<urn:uuid:bafadfea-e7ab-4955-bcdd-a1a2ea7885b9>","cc-path":"CC-MAIN-2024-46/segments/1730477028166.65/warc/CC-MAIN-20241110040813-20241110070813-00259.warc.gz"} |
Simple CAS program
08-09-2018, 05:08 AM
Post: #1
wsprague Posts: 9
Junior Member Joined: Aug 2018
Simple CAS program
Hi all, I am trying to write a simple CAS program, to get my feet wet and eventually write a fairly complicated Laplace transform/ matrix / resolvent thing.
My program is:
RETURN CAS.factor(A);
In the CAS screen, when I enter "QQQ(x^2-1)" it yields "factor(A)" but I want "(x-1)*(x+1)"
Could someone tell me what I am doing wrong, and maybe guide me to some documentation? I really have tried googling and RTFMing, but I have no idea.[/font]
08-09-2018, 12:35 PM
Post: #2
JMB Posts: 98
Member Joined: Jan 2016
RE: Simple CAS program
Try RETURN factor(A), instead of RETURN CAS.factor(A)
08-09-2018, 07:10 PM
Post: #3
Fabrizio R2 Posts: 2
Junior Member Joined: Jan 2018
RE: Simple CAS program
This works on CAS and on Textbook modes with x in lowercase.
RETURN factor(alg);
pruebaFactoriza(x^2-1) -> (x-1)*(x+1)
08-19-2018, 05:00 AM
Post: #4
wsprague Posts: 9
Junior Member Joined: Aug 2018
RE: Simple CAS program
(08-09-2018 12:35 PM)JMB Wrote: Try RETURN factor(A), instead of RETURN CAS.factor(A)
This solved my issue. The reason I put in "CAS." is that gets inserted when the function is chosen from the menu button, which seems like incorrect behavior.
I like the HP Prime, but it does seem like it's still a work in progress sometimes...
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Wh to Watts calculator
Enter the energy in watt-hours (Wh), time period in hours (h), then press the Calculate button to get the result in watts (W).
Wh to Watts calculation
P[(W)] = E[(Wh)] / t[(h)]
The power P in watts (W) is equal to the energy E in watt-hour (Wh), divided by the time period t in hours (h). | {"url":"https://allcalculators.net/online-calculators/electrical-calculators/wh-to-watts-calculator/","timestamp":"2024-11-03T03:02:53Z","content_type":"text/html","content_length":"150224","record_id":"<urn:uuid:05ee2328-5a82-4e82-92b4-27129da565c8>","cc-path":"CC-MAIN-2024-46/segments/1730477027770.74/warc/CC-MAIN-20241103022018-20241103052018-00224.warc.gz"} |
Juhan Aru
Probability theory, mathematical physics, random geometry. His research interests include how to mathematically describe and study systems where geometric structure and randomness interact. The
mathematical field studying such phenomena lies at the interface of probability theory, analysis and mathematical physics. He works on random curves (SLE curves), height functions (Gaussian free
field) and other random geometric structures (e.g. Gaussian multiplicative chaos), often with roots in the phenomena of statistical physics. Y6 Site Visit video: https://youtu.be/FSQazqcg9Nw
Phase III direction(s)
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Capgemini and Hexaware Aptitude Questions: Study Material 2020-2021 Batch
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Capgemini and Hexaware Aptitude Questions: Study Material 2020-2021 Batch
Capgemini and Hexaware Study Material: Last year placement papers:
Capgemini Study Material 2020-2021 Batch includes Capgemini Aptitude Questions, Capgemini Logical Reasoning Questions, Capgemini Technical Programming Questions, Capgemini Pseudo Code Questions with
Explanation, Capgemini Sample Verbal Ability Questions, C Programming MCQs
Must Read:
Capgemini Aptitude Questions: Capgemini Study Material
Q1. A shopkeeper gains 15% after allowing a discount of 20% on the market price of an article. Find his profit %, if the articles are sold at a market price allowing no discount?
A. 43.76%
B. 45%
C. 50%
D. 53.75%
Ans. A
Q2. The cost price of 10 articles is equal to the selling price of 9 articles. find the profit percent.
A. 101/9 %
B. 100/9 %
C. 102/9 %
D. 103/9 %
Ans. B
Q3.A rainy day occurs once in every 10 days. Half of the rainy days produce rainbows. What percent of all the days do not produce rainbow ?
Ans. A
Q4. Cost price of 80 notebooks is equal to the selling price of 65 notebooks. The gain or loss % is
A. 23%
B. 32%
C. 42%
D. 27%
Ans. A
Q5.If 5 spiders can catch five flies in five minutes. How many flies can hundred spiders catch in 100 minutes ?
Capgemini Aptitude Questions
Ans. DQ6. A sum of Rs 468.75 was lent out at simple interest and at the end of 1 year and 8 months, the total amount of Rs 500 is recieveD. find the rate of interest.
A. 2%
B. 4%
C. 1%
D. 3%
Ans. B
Q6. Consider the following two curves in the X-Y plane
Which of the following statements is true for -2<= x <=2?
A. The two curves do not intersect.
B. The two curves intersect thrice.
C. The two curves intersect twice.
D. The two curves intersect once.
Ans. B
Q7. Articles are marked at a price which gives a profit of 22%. After allowing a certain discount the profit reduced to half of the previous profit, then the discount % is
A. 7%
B. 10%
C. 12%
D. 9%
Ans. D
Q8. A positive integer is selected at random and is divided by 7,what is the probability that the remainder is 1 ?
A. 3/7
B. 4/7
C. 1/7
D. 2/7
Ans. 1/7
Q9. A mixture of 40 litres of salt and water contains 70%of salt.how much water must be added to decrease the salt percentage to 40%?
A. 40 litres
B. 30 litres
C. 20 litres
D. 2 litres
Ans. B
Q10. Anirudh,Harish and Sahil invested a total of Rs.1,35,000 in the ratio 5:6:4 Anirudh invested has capital for 8 months. Harish invested for 6 months and Sahil invested for 4 months. If they earn
a profit of Rs.75,900,then what is the share of Sahil in the profit?
A. Rs.12,400
B. Rs.14,700
C. Rs.15,800
D. Rs.13,200
Capgemini Aptitude Questions
Ans. D
Q11. Ramesh lends Rs 50,000 of two of his friends. He gives Rs 30,000 to the first at 6% p.A. simple interest. He wants to make a profit of 10% on the whole. The simple interest rate at which he
should lend the remaining sum of money to the second friend
A. 11%
B. 17%
C. 8%
D. 16%
Ans. D
Q12. Vishal borrowed some money for one year at 8% per annum simple interest and after 18 months, he again borrowed the same money at a Simple Interest of 32% per annum. In both cases, he paid
Rs.5452. Which of the following could be the amount that was borrowed by Hussain in each case if interest is paid half-yearly?
A. 4500
B. 4700
C. 3900
D. 4200
Ans. B
Q13. Rakesh lent out a part of Rs. 38800 is lent out at 6% per six months. The rest of the amount is lent out at 5% per annum after one year. The ratio of interest after 3 years from the time when
the first amount was lent is 5:4. Find the second part that was lent
out at 5%.
A. 28500
B. 28800
C. 30080
D. 20500
Ans. B
Q14. The cost of the Coal block varies directly with the square of its weight. The Coal block is divided into three parts whose weights are in the ratio of 5:6:7. If the Coal block is divided into
three equal parts by weight then there would be further loss of
Rs.7200. Then what is the actual cost of Coal Block?
A. 2332880
B. 3888000
C. 3960000
D. 1166400
Ans. D
Q15. There are 459 students in a hostel. If the number of students increased by 36, the expenses of the mess increased by Rs .81 Per day while the average expenditure per head reduced by 1. Find the
original expenditure of the mess?
A. 7304
B. 7314
C. 7334
D. 7344
Capgemini Aptitude Questions
Ans. D
Q16. The average weight of 40 Students is 32. If the Heaviest and Lightest are excluded the average weight reduces by 1. If only the Heaviest is excluded then the average is 31. Then what is the
weight of the Lightest?
A. 30
B. 31
C. 32
D. 33
Ans. B
Q17. M and N invested rupees 4000 and 5000 respectively in a business. M being an active partner will get rupee 50 every month extra for running the business. In a year if M receive a total of 800
rupees, then what will N get from the business.
A. 200
B. 300
C. 400
D. 250
Ans. D
Q18. A man sets out to cycle from Delhi to Rohtak and at the same time another man starts from Rohtak to cycle to cycle to Delhi. After passing each other they completed their journey in (10/3) hours
and (16/3) hours respectively.At what rate does the second man cycle if the first cycle at 8 kmph?
A.6.12 kmph
B.6.42 kmph
C. 6.22 kmph
D. 6.32 kmph
Ans. D
Q19.Two trains are traveling in opposite directions at uniforms speeds of 60 kmph and 50 kmph.They take 5 seconds to cross each other.If the two trains travelled in the same directions.then a
passenger sitting in the faster moving train would have overtaken the other than in 18 seconds.what are the lengths of the trains?
A. 87.78 m and 55 m
B.112 m and 78 m
C.102.78 m and 50 m
D.102.78 m and 55 m
Ans. C
Q20. A cube is given with an edge of 12 units.it is painted on all faces and then cut into smaller cubes of edge of 4 units. How many cubes will have 2 faces painted?
A. 2
B. 12
C. 8
D. 0
Capgemini Aptitude Questions
Ans. C
Q21.Two numbers are in the ratio xy,when 2 is added to both the numbers.the ratio becomes 1:2.when 3 is subtracted from both the numbers. The ratio becomes 1:3.Find the sum of x and y.
A. 27
B. 24
D. 26
Ans. D
Q22. To earn extra profit,a shopkeeper mixes 30 kg of dal purchased at Rs.36/kg and 26 kg of dal purchased at Rs.20/kg.What will be the profit that he will make if he sells the mixture at Rs.30/kg?
A. Rs.60
B. Rs.80
C. Rs.50
D. Rs.100
Ans. B
Q23. There are 4 boys and 3 girls.they sit in arrow randomly.what is the probability that all girls are together?
A. 1/14
B. 2/14
C. 5/14
D. 3/14
Ans. B
Q24. An oblong piece of ground measures 19m 2.5 dm by 12m5dm.Fom centre of each side of the ground,a path 2 m wide goes across to the center of the opposite side.What is the area of the path?
A. 59.5 m2
B. 54 m2
C. 43 m2
D. 34 m2
Q25. The circumference of the wheel of a truck is 1 meter.To cover a distance of 1.5 km.the number of revolutions made by the wheel are:
A. 3000
B. 37
C. 1500
D. 750
Capgemini Aptitude Questions
Ans. C
Q26. If (x+(1/x))=4,the value of (x5+(1/x5)) is:
A. 724
B. 500
C. 752
D. 525
Ans. A
Q27. A Jar contains 30 liters mixture of Milk and Water in the ratio of x:y respectively. When 10 liters of the mixture is taken out and replaced it water, then the ratio becomes 2:3. Then what is
the initial quantity of Milk in the Jar?
A. 18 Liter
B. 20 Liter
C. 12 Liter
D. 15 Liter
Ans. A
Q28. A Jar contains 100 liters of Milk a thief stole 10 liters of Milk and replaced it with water. Next, he stole 20 liters of Milk and replaced it with water. Again he stole 25 liters of Milk and
replaced with water. Then what is the quantity of water in the final
A. 55 Liter
B. 54 Liter
C. 50 Liter
D. 46 Liter
Ans. D
Q29. A Container contains ‘X’ Liters of Milk. A thief stole 50 Liters of Milk and replaced it with the same quantity of water. He repeated the same process further two times. And thus Milk in the
container is only ‘X-122’ liters. Then what is the quantity of water
in the final mixture?
A. 124 Liter
B. 128 Liter
C. 250 Liter
D. 122 Liter
Ans. D
Q30. The perimeter of a square is equal to twice the perimeter of a rectangle of length 10 cm and breadth 4 cm. What is the circumference of a semi-circle whose diameter is equal to the side of the
A. 46 cm
B. 36 cm
C. 38 cm
D. 23 cm
Ans. B
Q31. The length of a rectangle is 3/5th of the side of a square. The radius of a circle is equal to the side of the square. The circumference of the circle is 132 cm. What is the area of the
rectangle, if the breadth of the rectangle is 15 cm?
A. 112 cm²
B. 149 cm²
C. 189 cm²
D. 199 cm²
Ans. C
Q32. Smallest side of a right-angled triangle is 13 cm less than the side of a square of perimeter 72 cm. Second largest side of the right-angled triangle is 2 cm less than the length of the
rectangle of area 112 cm² and breadth 8 cm. What is the largest side of
the right-angled triangle?
A. 20 cm
B. 12 cm
C. 10 cm
D. 13 cm
Ans. D
Q33. 28 children can do a piece of work in 50 days.how many children are needed to complete the work in 30
A. 49
B. 40
C. 35
D. 45
Ans. A
Q34. A certain sum of money becomes Rs.750 in 2 years and becomes Rs.873 in 3.5 years. Find the sum and rate of interest.
A. Rs.400, 13% p.a
B. Rs.500, 11%p.a
C. Rs.630, 12%p.a
D. Rs.600, 13%p.a
Ans. B
Capgemini Aptitude Questions
Q35. Henna invested Rs.5000 at 12% simple interest p.A.the interest she will receive after 2 years is:
A. Rs.800
B. Rs.1000
C. Rs.600
D. Rs.1200
Ans. D
Q36.A bag contains 3 red,5yellow and 4 green balls.3 balls are drawn randomly,what is the probability that the ball drawn contains no yellow ball?
A. 9/44
B. 37/44
C. 43/44
D. 7/44
Q37. If a2+b2-4(a+b)= -8,then the value of (a-b) is:
A. 4
B. 0
C. 2
D. 8
Q38. A lent Rs.600 to b for 2 years and Rs.150 to C for 4 years and receive all together Rs.90 as both as interest.Find the rate of interest.
A. 4%p.a
B. 2%p.a
C. 5%p.a
D. 3%p.a
Ans. C
Q39. If the perimeter and the diagonal of a rectangle is 18 cm and √41 cm respectively.Calculate the area of
the rectangular fielD.
A. 25 cm2
B. 29 cm2
C. 18 cm2
D. 20 cm2
Capgemini Aptitude Questions
Q40. A,B, and C enter into a partnership and their shares are in the ratio 1/2 : 1/3 : 1/4.After 2 months,A withdraws half of his capital and after 10 months,a profit of RS. 378 is divied among them.
What is B’s
A. Rs.144
B. Rs.156
C. Rs.166
D. Rs.129
Ans. B
Q41. If a:b =4:1,then √(a/b)+ √(b/a) is :
A. 1
B. 4/5
C. None of the mentioned options
D. 5/2
Ans. D
Q42. A cube is given with an edge of 12 units.It is painted on all faces and then cut into smaller cubes of edge of 4 units.How many cubes will have 2 faces painted?
A. 8
B. 12
C. 0
D. 2
Q43.Find the area of Rhombus one of whose diagonals measures 8 cm and the other 10 cm.
A. 47 cm2
B. 34cm2
C. 40 cm2
D. 64cm2
Q44. Rs 5000 was divided among 5 men, 6 women and 5 boys, such that the ratio of the shares of men, women and boys is 5:3:2 what is the share of the boy?
A. 200
B. 100
C. 250
D. 150
Ans. A
Q45. If 28 Men working 6 hours a day can finish a work in 15 days. In how many days can 21 men working 8 hours per day will finish the same work?
A. 24 days
B. 21 days
C. 18 days
D. 15 days
Ans. D
Capgemini Aptitude Questions
Q46. A rectangular grassy plot is 112 m by 78 m. It has a gravel path 2.5 m m wide all around it on the inside. Find the area of the path
A. 952 m2
B. 926 m2
C. 912 m2
D. 950 m2
Ans. B
Q47. A cone of height 21 cm has a volume of 2200 cm3. Determine the base radius of the cone
A. 2.5 cm
B. 10 cm
C. 7.5 cm
D. 5 cm
Ans. B
Q48. what is the radius of the circular plate of thickness 1cm made by melting a sphere of radius 3cm?
A. 6 cm
B. 5 cm
C. 4 cm
D. 7 cm
Ans. A
Q49. What is the ratio of the surface area formed by placing 3 cubes adjacent to the sum of the individual surface area of these 3 cubes?
A. 7:9
B. 27:23
C. 8:9
D. 9:7
Ans. C
Q50. At what rate percent CI does a sum of money become nine fold in 2 years?
A.100% p.a
B.300% p.a
C.400% p.a
D.200% p.a
Ans. D
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Mapping structures¶
icet.tools.map_structure_to_reference(structure, reference, inert_species=None, tol_positions=0.0001, suppress_warnings=False, assume_no_cell_relaxation=False)[source]¶
Maps a structure onto a reference structure. This is often desirable when, for example, a structure has been relaxed using DFT, and one wants to use it as a training structure in a cluster
The function returns a tuple comprising the ideal supercell most closely matching the input structure and a dictionary with supplementary information concerning the mapping. The latter includes
for example the largest deviation of any position in the input structure from its reference position (drmax), the average deviation of the positions in the input structure from the reference
positions (dravg), and the strain tensor for the input structure relative to the reference structure (strain_tensor).
The returned Atoms object provides further supplemental information via custom per-atom arrays including the atomic displacements (Displacement, Displacement_Magnitude),the distances to the three
closest sites (Minimum_Distances), as well as a mapping between the indices of the returned Atoms object and those of the input structure (IndexMapping).
The input and reference structures should adhere to the same crystallographic setting. In other words, they should not be rotated or translated with respect to each other.
☆ structure (Atoms) – Input structure, typically a relaxed structure.
☆ reference (Atoms) – Reference structure, which can but need not be the primitive structure.
☆ inert_species (Optional[List[str]]) – List of chemical symbols (e.g., ['Au', 'Pd']) that are never substituted for a vacancy. The number of inert sites is used to rescale the volume of
the input structure to match the reference structure.
☆ tol_positions (float) – Tolerance factor applied when scanning for overlapping positions in Ångstrom (forwarded to ase.build.make_supercell()).
☆ suppress_warnings (bool) – if True print no warnings of large strain or relaxation distances.
☆ assume_no_cell_relaxation (bool) –
If False volume and cell metric of the input structure are rescaled to match the reference structure. This can be unnecessary (and counterproductive) for some structures, e.g., with many
When setting this parameter to False the reference cell metric must be obtainable via an integer transformation matrix from the reference cell metric. In other words the input structure
should not involve relaxations of the volume or the cell metric.
Return type:
Tuple[Atoms, dict]
The following code snippet illustrates the general usage. It first creates a primitive FCC cell, which is latter used as reference structure. To emulate a relaxed structure obtained from, e.g., a
density functional theory calculation, the code then creates a 4x4x4 conventional FCC supercell, which is populated with two different atom types, has distorted cell vectors, and random
displacements to the atoms. Finally, the present function is used to map the structure back the ideal lattice:
>>> from ase.build import bulk
>>> reference = bulk('Au', a=4.09)
>>> structure = bulk('Au', cubic=True, a=4.09).repeat(4)
>>> structure.set_chemical_symbols(10 * ['Ag'] + (len(structure) - 10) * ['Au'])
>>> structure.set_cell(structure.cell * 1.02, scale_atoms=True)
>>> structure.rattle(0.1)
>>> mapped_structure, info = map_structure_to_reference(structure, reference)
Structure enumeration¶
icet.tools.enumerate_structures(structure, sizes, chemical_symbols, concentration_restrictions=None, niggli_reduce=None, symprec=1e-05, position_tolerance=None)[source]¶
Yields a sequence of enumerated structures. The function generates all inequivalent structures that are permissible given a certain lattice. Using the chemical_symbols and
concentration_restrictions keyword arguments it is possible to specify which chemical_symbols are to be included on which site and in which concentration range.
The function is sensitive to the boundary conditions of the input structure. An enumeration of, for example, a surface can thus be performed by setting structure.pbc = [True, True, False].
The algorithm implemented here was developed by Gus L. W. Hart and Rodney W. Forcade in Phys. Rev. B 77, 224115 (2008) [HarFor08] and Phys. Rev. B 80, 014120 (2009) [HarFor09].
☆ structure (Atoms) – Primitive structure from which derivative superstructures should be generated.
☆ sizes (Union[List[int], range]) – Number of sites (included in enumeration).
☆ chemical_symbols (list) – Chemical species with which to decorate the structure, e.g., ['Au', 'Ag']; see below for more examples.
☆ concentration_restrictions (Optional[dict]) – Allowed concentration range for one or more element in chemical_symbols, e.g., {'Au': (0, 0.2)} will only enumerate structures in which the
Au content is between 0 and 20 %. Here, concentration is always defined as the number of atoms of the specified kind divided by the number of all atoms.
☆ niggli_reduction – If True perform a Niggli reduction with spglib for each structure. The default is True if structure is periodic in all directions, False otherwise.
☆ symprec (float) – Tolerance imposed when analyzing the symmetry using spglib.
☆ position_tolerance (Optional[float]) – Tolerance applied when comparing positions in Cartesian coordinates; by default this value is set equal to symprec.
Return type:
The following code snippet illustrates how to enumerate structures with up to 6 atoms in the unit cell for a binary alloy without any constraints:
>>> from ase.build import bulk
>>> prim = bulk('Ag')
>>> for structure in enumerate_structures(structure=prim,
... sizes=range(1, 5),
... chemical_symbols=['Ag', 'Au']):
... pass # Do something with the structure
To limit the concentration range to 10 to 40% Au the code should be modified as follows:
>>> conc_restr = {'Au': (0.1, 0.4)}
>>> for structure in enumerate_structures(structure=prim,
... sizes=range(1, 5),
... chemical_symbols=['Ag', 'Au'],
... concentration_restrictions=conc_restr):
... pass # Do something with the structure
Often one would like to consider mixing on only one sublattice. This can be achieved as illustrated for a Ga(1-x)Al(x)As alloy as follows:
>>> prim = bulk('GaAs', crystalstructure='zincblende', a=5.65)
>>> for structure in enumerate_structures(structure=prim,
... sizes=range(1, 9),
... chemical_symbols=[['Ga', 'Al'], ['As']]):
... pass # Do something with the structure
icet.tools.enumerate_supercells(structure, sizes, niggli_reduce=None, symprec=1e-05, position_tolerance=None)[source]¶
Yields a sequence of enumerated supercells. The function generates all inequivalent supercells that are permissible given a certain lattice. Any supercell can be reduced to one of the supercells
The function is sensitive to the boundary conditions of the input structure. An enumeration of, for example, a surface can thus be performed by setting structure.pbc = [True, True, False].
The algorithm is based on Gus L. W. Hart and Rodney W. Forcade in Phys. Rev. B 77, 224115 (2008) [HarFor08] and Phys. Rev. B 80, 014120 (2009) [HarFor09].
☆ structure (Atoms) – Primitive structure from which supercells should be generated.
☆ sizes (Union[List[int], range]) – Number of sites (included in enumeration).
☆ niggli_reduction – If True perform a Niggli reduction with spglib for each supercell. The default is True if structure is periodic in all directions, False otherwise.
☆ symprec (float) – Tolerance imposed when analyzing the symmetry using spglib.
☆ position_tolerance (Optional[float]) – Tolerance applied when comparing positions in Cartesian coordinates. By default this value is set equal to symprec.
Return type:
The following code snippet illustrates how to enumerate supercells with up to 6 atoms in the unit cell:
>>> from ase.build import bulk
>>> prim = bulk('Ag')
>>> for supercell in enumerate_supercells(structure=prim, sizes=range(1, 7)):
... pass # Do something with the supercell
Generation of training structures¶
icet.tools.training_set_generation.structure_selection_annealing(cluster_space, monte_carlo_structures, n_structures_to_add, n_steps, base_structures=None, cooling_start=5, cooling_stop=0.001,
cooling_function='exponential', initial_indices=None)[source]¶
Given a cluster space, a base pool of structures, and a new pool of structures, this function uses a Monte Carlo inspired annealing method to find a good structure pool for training.
Return type:
Tuple[List[int], List[float]]
A tuple comprising the indices of the optimal structures in the monte_carlo_structures pool and a list of accepted metric values.
☆ cluster_space (ClusterSpace) – A cluster space defining the lattice to be occupied.
☆ monte_carlo_structures (List[Atoms]) – A list of candidate training structures.
☆ n_structures_to_add (int) – How many of the structures in the monte_carlo_structures pool that should be kept for training.
☆ n_steps (int) – Number of steps in the annealing algorithm.
☆ base_structures (Optional[List[Atoms]]) – A list of structures that is already in your training pool; can be None if you do not have any structures yet.
☆ cooling_start (float) – Initial value of the cooling_function.
☆ cooling_stop (float) – Last value of the cooling_function.
☆ cooling_function (Union[str, Callable]) – Artificial number that rescales the difference between the metric value between two iterations. Available options are 'linear' and 'exponential'.
☆ initial_indices (Optional[List[int]]) – Picks out the starting structure from the monte_carlo_structures pool. Can be used if you want to continue from an old run for example.
The following snippet demonstrates the use of this function for generating an optimized structure pool. Here, we first set up a pool of candidate structures by randomly occupying a FCC supercell
with Au and Pd:
>>> from ase.build import bulk
>>> from icet.tools.structure_generation import occupy_structure_randomly
>>> prim = bulk('Au', a=4.0)
>>> cs = ClusterSpace(prim, [6.0], [['Au', 'Pd']])
>>> structure_pool = []
>>> for _ in range(500):
>>> # Create random supercell.
>>> supercell = np.random.randint(1, 4, size=3)
>>> structure = prim.repeat(supercell)
>>> # Randomize concentrations in the supercell
>>> n_atoms = len(structure)
>>> n_Au = np.random.randint(0, n_atoms)
>>> n_Pd = n_atoms - n_Au
>>> concentration = {'Au': n_Au / n_atoms, 'Pd': n_Pd / n_atoms}
>>> # Occupy the structure randomly and store it.
>>> occupy_structure_randomly(structure, cs, concentration)
>>> structure_pool.append(structure)
>>> start_inds = [f for f in range(10)]
Now we can use the structure_selection_annealing() function to find an optimized structure pool:
>>> inds, cond = structure_selection_annealing(cs,
>>> structure_pool,
>>> n_structures_to_add=10,
>>> n_steps=100)
>>> training_structures = [structure_pool[ind] for ind in inds]
>>> print(training_structures)
Generation of special structures¶
Ground state finder¶
class icet.tools.ground_state_finder.GroundStateFinder(cluster_expansion, structure, solver_name=None, verbose=True)[source]¶
This class provides functionality for determining the ground states using a binary cluster expansion. This is efficiently achieved through the use of mixed integer programming (MIP) as developed
by Larsen et al. in Phys. Rev. Lett. 120, 256101 (2018).
This class relies on the Python-MIP package. Python-MIP can be used together with Gurobi, which is not open source but issues academic licenses free of charge. Pleaase note that Gurobi needs to
be installed separately. The GroundStateFinder works also without Gurobi, but if performance is critical, Gurobi is highly recommended.
In order to be able to use Gurobi with python-mip one must ensure that GUROBI_HOME should point to the installation directory (<installdir>):
export GUROBI_HOME=<installdir>
The current implementation only works for binary systems.
☆ cluster_expansion (ClusterExpansion) – Cluster expansion for which to find ground states.
☆ structure (Atoms) – Atomic configuration.
☆ solver_name (Optional[str]) – 'gurobi', 'grb' or 'cbc'. Searches for available solvers if no value is provided.
☆ verbose (bool) – If True print solver messages to stdout.
The following snippet illustrates how to determine the ground state for a Au-Ag alloy. Here, the parameters of the cluster expansion are set to emulate a simple Ising model in order to obtain an
example that can be run without modification. In practice, one should of course use a proper cluster expansion:
>>> from ase.build import bulk
>>> from icet import ClusterExpansion, ClusterSpace
>>> # prepare cluster expansion
>>> # the setup emulates a second nearest-neighbor (NN) Ising model
>>> # (zerolet and singlet parameters are zero; only first and second neighbor
>>> # pairs are included)
>>> prim = bulk('Au')
>>> chemical_symbols = ['Ag', 'Au']
>>> cs = ClusterSpace(prim, cutoffs=[4.3], chemical_symbols=chemical_symbols)
>>> ce = ClusterExpansion(cs, [0, 0, 0.1, -0.02])
>>> # prepare initial configuration
>>> structure = prim.repeat(3)
>>> # set up the ground state finder and calculate the ground state energy
>>> gsf = GroundStateFinder(ce, structure)
>>> ground_state = gsf.get_ground_state({'Ag': 5})
>>> print('Ground state energy:', ce.predict(ground_state))
get_ground_state(species_count=None, max_seconds=inf, threads=0)[source]¶
Finds the ground state for a given structure and species count. If species_count is not provided when initializing the instance of this class the first species in the list of chemical symbols
for the active sublattice will be used.
○ species_count (Optional[Dict[str, int]]) – Dictionary with count for one of the species on each active sublattice. If no count is provided for a sublattice, the concentration is
allowed to vary.
○ max_seconds (float) – Maximum runtime in seconds.
○ threads (int) – Number of threads to be used when solving the problem, given that a positive integer has been provided. If set to \(0\) the solver default configuration is used while
\(-1\) corresponds to all available processing cores.
Return type:
property model: Model¶
Python-MIP model.
property optimization_status: OptimizationStatus¶
Optimization status.
Convex hull construction¶
class icet.tools.ConvexHull(concentrations, energies)[source]¶
This class provides functionality for extracting the convex hull of the (free) energy of mixing. It is based on the convex hull calculator in SciPy.
☆ concentrations (Union[List[float], List[List[float]]]) – Concentrations for each structure listed as [[c1, c2], [c1, c2], ...]; for binaries, in which case there is only one independent
concentration, the format [c1, c2, c3, ...] works as well.
☆ energies (List[float]) – Energy (or energy of mixing) for each structure.
Concentrations of the structures on the convex hull.
Energies of the structures on the convex hull.
Number of independent concentrations needed to specify a point in concentration space (1 for binaries, 2 for ternaries and so on).
Indices of structures that constitute the convex hull (indices are defined by the order of their concentrations and energies are fed when initializing the ConvexHull object).
A ConvexHull object is easily initialized by providing lists of concentrations and energies:
>>> data = {'concentration': [0, 0.2, 0.2, 0.3, 0.4, 0.5, 0.8, 1.0],
... 'mixing_energy': [0.1, -0.2, -0.1, -0.2, 0.2, -0.4, -0.2, -0.1]}
>>> hull = ConvexHull(data['concentration'], data['mixing_energy'])
Now one can for example access the points along the convex hull directly:
>>> for c, e in zip(hull.concentrations, hull.energies):
... print(c, e)
0.0 0.1
0.2 -0.2
0.5 -0.4
1.0 -0.1
or plot the convex hull along with the original data using e.g., matplotlib:
>>> import matplotlib.pyplot as plt
>>> plt.scatter(data['concentration'], data['mixing_energy'], color='darkred')
>>> plt.plot(hull.concentrations, hull.energies)
>>> plt.show(block=False)
It is also possible to extract structures at or close to the convex hull:
>>> low_energy_structures = hull.extract_low_energy_structures(
... data['concentration'], data['mixing_energy'],
... energy_tolerance=0.005)
A complete example can be found in the basic tutorial.
extract_low_energy_structures(concentrations, energies, energy_tolerance)[source]¶
Returns the indices of energies that lie within a certain tolerance of the convex hull.
○ concentrations (Union[List[float], List[List[float]]]) –
Concentrations of candidate structures.
If there is one independent concentration, a list of floats is sufficient. Otherwise, the concentrations must be provided as a list of lists, such as [[0.1, 0.2], [0.3, 0.1], ...].
○ energies (List[float]) – Energies of candidate structures.
○ energy_tolerance (float) – Include structures with an energy that is at most this far from the convex hull.
Return type:
Returns the energy of the convex hull at specified concentrations. If any concentration is outside the allowed range, NaN is returned.
target_concentrations (Union[List[float], List[List[float]]]) –
Concentrations at target points.
If there is one independent concentration, a list of floats is sufficient. Otherwise, the concentrations ought to be provided as a list of lists, such as [[0.1, 0.2], [0.3, 0.1], ...].
Return type:
Fitting with constraints¶
class icet.tools.constraints.Constraints(n_params)[source]¶
Class for handling linear constraints with right-hand-side equal to zero.
n_params (int) – Number of parameters in model.
The following example demonstrates fitting of a cluster expansion under the constraint that parameter 2 and parameter 4 should be equal:
>>> import numpy as np
>>> from icet.tools import Constraints
>>> from trainstation import Optimizer
>>> # Set up random sensing matrix and target "energies"
>>> n_params = 10
>>> n_energies = 20
>>> A = np.random.random((n_energies, n_params))
>>> y = np.random.random(n_energies)
>>> # Define constraints
>>> c = Constraints(n_params=n_params)
>>> M = np.zeros((1, n_params))
>>> M[0, 2] = 1
>>> M[0, 4] = -1
>>> c.add_constraint(M)
>>> # Do the actual fit and finally extract parameters
>>> A_constrained = c.transform(A)
>>> opt = Optimizer((A_constrained, y), fit_method='ridge')
>>> opt.train()
>>> parameters = c.inverse_transform(opt.parameters)
Add a constraint matrix and resolve for the constraint space.
M (ndarray) – Constraint matrix with each constraint as a row. Can (but need not be) cluster vectors.
Return type:
Inverse transform array from constrained parameter space to unconstrained space.
A (ndarray) – Array to be inverse transformed.
Return type:
Transform array to constrained parameter space.
A (ndarray) – Array to be transformed.
Return type:
A cluster expansion of the mixing energy should ideally predict zero energy for concentrations 0 and 1. This function constructs a Constraints object that enforces that condition during training.
cluster_space – Cluster space corresponding to cluster expansion for which constraints should be imposed.
Return type:
This example demonstrates how to constrain the mixing energy to zero at the pure phases in a toy example with random cluster vectors and random target energies:
>>> import numpy as np
>>> from ase.build import bulk
>>> from icet import ClusterSpace
>>> from icet.tools import get_mixing_energy_constraints
>>> from trainstation import Optimizer
>>> # Set up cluster space along with random sensing matrix and target "energies"
>>> prim = bulk('Au')
>>> cs = ClusterSpace(prim, cutoffs=[6.0, 5.0], chemical_symbols=['Au', 'Ag'])
>>> n_params = len(cs)
>>> n_energies = 20
>>> A = np.random.random((n_energies, n_params))
>>> y = np.random.random(n_energies)
>>> # Define constraints
>>> c = get_mixing_energy_constraints(cs)
>>> # Do the actual fit and finally extract parameters
>>> A_constrained = c.transform(A)
>>> opt = Optimizer((A_constrained, y), fit_method='ridge')
>>> opt.train()
>>> parameters = c.inverse_transform(opt.parameters)
Constraining the energy of one structure is always done at the expense of the fit quality of the others. Always expect that your cross-validation scores will increase somewhat when using this
Constituent strain¶
class icet.tools.ConstituentStrain(supercell, primitive_structure, chemical_symbols, concentration_symbol, strain_energy_function, k_to_parameter_function=None, damping=1.0, tol=1e-06)[source]¶
Class for handling constituent strain in cluster expansions (see Laks et al., Phys. Rev. B 46, 12587 (1992) [LakFerFro92]). This makes it possible to use cluster expansions to describe systems
with strain due to, for example, coherent phase separation. For an extensive example on how to use this module, please see this example.
☆ supercell (Atoms) – Defines supercell that will be used when calculating constituent strain.
☆ primitive_structure (Atoms) – Primitive structure the supercell is based on.
☆ chemical_symbols (List[str]) – List with chemical symbols involved, such as ['Ag', 'Cu'].
☆ concentration_symbol (str) – Chemical symbol used to define concentration, such as 'Ag'.
☆ strain_energy_function (Callable[[float, List[float]], float]) – A function that takes two arguments, a list of parameters and concentration (e.g., [0.5, 0.5, 0.5] and 0.3), and returns
the corresponding strain energy. The parameters are in turn determined by k_to_parameter_function (see below). If k_to_parameter_function is None, the parameters list will be the k-point.
For more information, see this example.
☆ k_to_parameter_function (Optional[Callable[[List[float]], List[float]]]) – A function that takes a k-point as a list of three floats and returns a parameter vector that will be fed into
strain_energy_function (see above). If None, the k-point itself will be the parameter vector to strain_energy_function. The purpose of this function is to be able to precompute any factor
in the strain energy that depends on the k-point but not the concentration. For more information, see this example.
☆ damping (float) – Damping factor \(\eta\) used to suppress impact of large-magnitude k-points by multiplying strain with \(\exp(-(\eta \mathbf{k})^2)\) (unit Ångstrom).
☆ tol (float) – Numerical tolerance when comparing k-points (units of inverse Ångstrom).
Update structure factor for each kpoint to the value in structure_factor_after. This makes it possible to efficiently calculate changes in constituent strain with the
get_constituent_strain_change() function; this function should be called if the last occupations used to call get_constituent_strain_change() should be the starting point for the next call of
get_constituent_strain_change(). This is taken care of automatically by the Monte Carlo simulations in mchammer.
Return type:
Calculate current concentration.
occupations (ndarray) – Current occupations.
Return type:
Calculate total constituent strain.
occupations (List[int]) – Current occupations.
Return type:
get_constituent_strain_change(occupations, atom_index)[source]¶
Calculate change in constituent strain upon change of the occupation of one site.
This function is dependent on the internal state of the ConstituentStrain object and should typically only be used internally by mchammer. Specifically, the structure factor is saved
internally to speed up computation. The first time this function is called, occupations must be the same array as was used to initialize the ConstituentStrain object, or the same as was last
used when get_constituent_strain() was called. After the present function has been called, the same occupations vector need to be used the next time as well, unless accept_change() has been
called, in which case occupations should incorporate the changes implied by the previous call to the function.
○ occupations (ndarray) – Occupations before change.
○ atom_index (int) – Index of site the occupation of which is to be changed.
Return type:
Other structure tools¶
icet.tools.get_primitive_structure(structure, no_idealize=True, to_primitive=True, symprec=1e-05)[source]¶
Returns the primitive structure using spglib.
☆ structure (Atoms) – Input atomic structure.
☆ no_idealize (bool) – If True lengths and angles are not idealized.
☆ to_primitive (bool) – If True convert to primitive structure.
☆ symprec (float) – Tolerance imposed when analyzing the symmetry using spglib.
Return type:
icet.tools.get_wyckoff_sites(structure, map_occupations=None, symprec=1e-05)[source]¶
Returns the Wyckoff symbols of the input structure. The Wyckoff sites are of general interest for symmetry analysis but can be especially useful when setting up, e.g., a SiteOccupancyObserver.
The Wyckoff labels can be conveniently attached as an array to the structure object as demonstrated in the examples section below.
By default the occupation of the sites is part of the symmetry analysis. If a chemically disordered structure is provided this will usually reduce the symmetry substantially. If one is interested
in the symmetry of the underlying structure one can control how occupations are handled. To this end, one can provide the map_occupations keyword argument. The latter must be a list, each entry
of which is a list of species that should be treated as indistinguishable. As a shortcut, if all species should be treated as indistinguishable one can provide an empty list. Examples that
illustrate the usage of the keyword are given below.
☆ structure (Atoms) – Input structure. Note that the occupation of the sites is included in the symmetry analysis.
☆ map_occupations (Optional[List[List[str]]]) – Each sublist in this list specifies a group of chemical species that shall be treated as indistinguishable for the purpose of the symmetry
☆ symprec (float) – Tolerance imposed when analyzing the symmetry using spglib.
Return type:
Wyckoff sites of a hexagonal-close packed structure:
>>> from ase.build import bulk
>>> structure = bulk('Ti')
>>> wyckoff_sites = get_wyckoff_sites(structure)
>>> print(wyckoff_sites)
['2d', '2d']
The Wyckoff labels can also be attached as an array to the structure, in which case the information is also included when storing the Atoms object:
>>> from ase.io import write
>>> structure.new_array('wyckoff_sites', wyckoff_sites, str)
>>> write('structure.xyz', structure)
The function can also be applied to supercells:
>>> structure = bulk('GaAs', crystalstructure='zincblende', a=3.0).repeat(2)
>>> wyckoff_sites = get_wyckoff_sites(structure)
>>> print(wyckoff_sites)
['4a', '4c', '4a', '4c', '4a', '4c', '4a', '4c',
'4a', '4c', '4a', '4c', '4a', '4c', '4a', '4c']
Now assume that one is given a supercell of a (Ga,Al)As alloy. Applying the function directly yields much lower symmetry since the symmetry of the original structure is broken:
>>> structure.set_chemical_symbols(
... ['Ga', 'As', 'Al', 'As', 'Ga', 'As', 'Al', 'As',
... 'Ga', 'As', 'Ga', 'As', 'Al', 'As', 'Ga', 'As'])
>>> print(get_wyckoff_sites(structure))
['8g', '8i', '4e', '8i', '8g', '8i', '2c', '8i',
'2d', '8i', '8g', '8i', '4e', '8i', '8g', '8i']
Since Ga and Al occupy the same sublattice, they should, however, be treated as indistinguishable for the purpose of the symmetry analysis, which can be achieved via the map_occupations keyword:
>>> print(get_wyckoff_sites(structure, map_occupations=[['Ga', 'Al'], ['As']]))
['4a', '4c', '4a', '4c', '4a', '4c', '4a', '4c',
'4a', '4c', '4a', '4c', '4a', '4c', '4a', '4c']
If occupations are to ignored entirely, one can simply provide an empty list. In the present case, this turns the zincblende lattice into a diamond lattice, on which case there is only one
Wyckoff site:
>>> print(get_wyckoff_sites(structure, map_occupations=[]))
['8a', '8a', '8a', '8a', '8a', '8a', '8a', '8a',
'8a', '8a', '8a', '8a', '8a', '8a', '8a', '8a'] | {"url":"https://icet.materialsmodeling.org/dev/moduleref_icet/tools.html","timestamp":"2024-11-13T06:05:49Z","content_type":"text/html","content_length":"189223","record_id":"<urn:uuid:a8b93167-93ba-42da-8a07-dcde9b1629c6>","cc-path":"CC-MAIN-2024-46/segments/1730477028326.66/warc/CC-MAIN-20241113040054-20241113070054-00753.warc.gz"} |
Geometric Loss functions between sampled measures, images and volumes
Geometric Loss functions between sampled measures, images and volumes
The GeomLoss library provides efficient GPU implementations for:
It is hosted on GitHub and distributed under the permissive MIT license.
GeomLoss functions are available through the custom PyTorch layers SamplesLoss, ImagesLoss and VolumesLoss which allow you to work with weighted point clouds (of any dimension), density maps and
volumetric segmentation masks. Geometric losses come with three backends each:
• A simple tensorized implementation, for small problems (< 5,000 samples).
• A reference online implementation, with a linear (instead of quadratic) memory footprint, that can be used for finely sampled measures.
• A very fast multiscale code, which uses an octree-like structure for large-scale problems in dimension <= 3.
A typical sample of code looks like:
import torch
from geomloss import SamplesLoss # See also ImagesLoss, VolumesLoss
# Create some large point clouds in 3D
x = torch.randn(100000, 3, requires_grad=True).cuda()
y = torch.randn(200000, 3).cuda()
# Define a Sinkhorn (~Wasserstein) loss between sampled measures
loss = SamplesLoss(loss="sinkhorn", p=2, blur=.05)
L = loss(x, y) # By default, use constant weights = 1/number of samples
g_x, = torch.autograd.grad(L, [x]) # GeomLoss fully supports autograd!
GeomLoss is a simple interface for cutting-edge Optimal Transport algorithms. It provides:
• Support for batchwise computations.
• Linear (instead of quadratic) memory footprint for large problems, relying on the KeOps library for map-reduce operations on the GPU.
• Fast kernel truncation for small bandwidths, using an octree-based structure.
• Log-domain stabilization of the Sinkhorn iterations, eliminating numeric overflows for small values of \(\varepsilon\).
• Efficient computation of the gradients, which bypasses the naive backpropagation algorithm.
• Support for unbalanced Optimal Transport, with a softening of the marginal constraints through a maximum reach parameter.
• Support for the ε-scaling heuristic in the Sinkhorn loop, with kernel truncation in dimensions 1, 2 and 3. On typical 3D problems, our implementation is 50-100 times faster than the standard
SoftAssign/Sinkhorn algorithm.
Note, however, that SamplesLoss does not implement the Fast Multipole or Fast Gauss transforms. If you are aware of a well-packaged implementation of these algorithms on the GPU, please contact me!
The divergences implemented here are all symmetric, positive definite and therefore suitable for measure-fitting applications. For positive input measures \(\alpha\) and \(\beta\), our \(\text{Loss}
\) functions are such that
\[\begin{split}\text{Loss}(\alpha,\beta) ~&=~ \text{Loss}(\beta,\alpha), \\ 0~=~\text{Loss}(\alpha,\alpha) ~&\leqslant~ \text{Loss}(\alpha,\beta), \\ 0~=~\text{Loss}(\alpha,\beta)~&\
Longleftrightarrow~ \alpha = \beta.\end{split}\]
GeomLoss can be used in a wide variety of settings, from shape analysis (LDDMM, optimal transport…) to machine learning (kernel methods, GANs…) and image processing. Details and examples are provided
Licensing, academic use
This library is licensed under the permissive MIT license, which is fully compatible with both academic and commercial applications. If you use this code in a research paper, please cite:
title={Interpolating between Optimal Transport and MMD using Sinkhorn Divergences},
author={Feydy, Jean and S{\'e}journ{\'e}, Thibault and Vialard, Fran{\c{c}}ois-Xavier and Amari, Shun-ichi and Trouve, Alain and Peyr{\'e}, Gabriel},
booktitle={The 22nd International Conference on Artificial Intelligence and Statistics}, | {"url":"https://www.kernel-operations.io/geomloss/index.html","timestamp":"2024-11-14T15:37:59Z","content_type":"text/html","content_length":"21315","record_id":"<urn:uuid:320c63ac-4060-41b7-a133-1284d04acbd2>","cc-path":"CC-MAIN-2024-46/segments/1730477028657.76/warc/CC-MAIN-20241114130448-20241114160448-00155.warc.gz"} |
How does the comparison work in BedBooking Statistics?
By selecting a comparison period in the Comparison Period Selection Control, the data comparison function in the Statistics module is activated. It allows you to see how the condition of your
business has changed compared to the past.
Each of the Statistics modules has algorithms that calculate the percentage change in the base period data relative to the comparison period data.
The percentage change for each module is calculated in a similar way:
1. Total income module
If your income in the base period was €6,000 and in the comparison period was €4,000, Statistics will show a 50% increase in income.
2. Occupancy module
If your occupancy was 50% in the base period and 40% in the comparison period, Statistics will show an increase in occupancy of 25%.
3. Revenue and occupancy module
If the income at a given chart point in the base period is €200 and occupancy is 70%, and in the comparison period the same chart point has an income of €600 and occupancy of 90%, Statistics will
show a decrease in income of 67% and occupancy of 22%.
4. Additional data module
- Number of bookings
If the number of bookings of the base period was 20, and in the comparison period was 15, the Statistics will show an increase in the number of bookings by 33%.
- Booking Window
If the Booking Window of the base period was 15 days, and in the comparison period was 20 days, the Statistics will show a decrease in Booking Window by 25%.
- Number of cancellations
If the number of cancellations in the base period was 2, and in the comparison period was 6, the Statistics will show a decrease in the number of cancellations by 67%.
- RevPAR
If RevPAR in the base period was €350 and in the comparison period was €100, Statistics will show an increase in RevPAR of 250%.
- ADR
If ADR in the base period was €100 and in the comparison period was €350, Statistics will show a 71% decrease in ADR.
- Average booking length
If the average booking length in the base period was 14 nights and in the comparison period was 10 days, the Statistics will show an increase of 40%.
5. Booking Origin module
If 10 bookings came from Airbnb in the baseline period and 6 in the comparison period, Statistics will show a 67% increase in bookings from Airbnb.
6. Income Origin Module
If €1,200 in income came from Airbnb in the base period and €1,500 in the comparison period, Statistics will show a 20% decrease in income from Airbnb. | {"url":"https://support.bed-booking.com/hc/en-us/articles/7162065907100-How-does-the-comparison-work-in-BedBooking-Statistics","timestamp":"2024-11-12T13:45:09Z","content_type":"text/html","content_length":"27940","record_id":"<urn:uuid:239776b5-2abe-4f66-8355-247f6b942fde>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.45/warc/CC-MAIN-20241112113320-20241112143320-00793.warc.gz"} |
Acceleration of Google's Newton's Apple
This is great. Many people have already reported google's apple-dropping homepage in honor Newton's birthday. In case it disappears, here is a screen shot.
So, I got this awesome note from Dale Basler. He said that his class had analyzed this falling apple animation. What a very Dot Physics-y idea (check out his analysis). He said they were questioning
the results which might be due screen capture issues. I decided to reproduce this.
I captured the motion with Apple's Quicktime X screen recording feature. I then used Tracker Video Analysis - which now has an autotracking feature that works really well in this case (I will post
more about that later). Here is a plot of the y-motion of the falling apple.
I fit a parabolic function to the data (at least to the part where it was falling) and I get an acceleration of about -2 units/sec^2. I didn't scale the video so I don't really know the distance
units. Is it constant acceleration? Kind of, I guess. What about the bounce? That should have the same acceleration, right?
Not even close. I guess google thought it would be enough to honor Newton with the silly falling apple, but not falling with a constant acceleration. Hello, Google? I thought your motto was "do no
Actually, this doesn't bother me too much - but I thought the analysis was a cool idea.
More like this
I stumbled on this flash game Bloons. The basic idea is that you (the monkey) throw these darts and try to pop some balloons. Well, what is the motion of these darts like? Is it constant
acceleration? Time to pull out the free and awesome Tracker Video Analysis. I threw a few shots and captured…
There are several free iPhone-iPod Touch apps that let you look at the acceleration of the device using the built in accelerometer. I was planning on reviewing some of these free apps, but I didn't.
When I started playing around with them, it was clear that I needed some way to make a constant…
I don't know why they call it a tail drop. Here is a video: The link I clicked that brought me to this video said the equivalent of "OMG!" That is not what I thought, really I am not sure what is so
impressive (except that he didn't fall off the skateboard). If the original poster was impressed…
Here is a quick Apolo Ohno quiz. Which one of these pictures is fake? If you picked picture B - you are probably correct. That is a picture of "Apolo" being catapulted into a pool of slime at the
Nickelodeon awards show (click on the link to see the video - I don't think I can embed it). Ok -…
Can't you scale the drawing by the size of the apple? (Yeah, I know, I know - apples aren't a set size - but they fall within a certain range - and that gets you a range that may or may not include
the g we're expecting).
And no, we don't know what planet this is on - but it's one that has trees more or less like ours, so g can't be too different (unlike Pandorum...) | {"url":"https://www.scienceblogs.com/dotphysics/2010/01/04/acceleration-of-googles-newton","timestamp":"2024-11-04T02:42:58Z","content_type":"text/html","content_length":"39765","record_id":"<urn:uuid:2268f22e-35c0-4248-ab92-52098103ad68>","cc-path":"CC-MAIN-2024-46/segments/1730477027809.13/warc/CC-MAIN-20241104003052-20241104033052-00565.warc.gz"} |
% It is worthwhile to notice that independently on how fin_step is
% chosen, the message of the generalized candlestick plot remains the
% same. In other words, the best two models with 5 variables are always
% (Time,4,5,6) and (Time,2,4,5)
% while two reasonable models with 6 variables are (Time,2,4,5,6) and
% (Time,2,3,4,5).
X=[(-40:39)' X];
[Cpms]=FSRms(y,X,'smallpint',4:6,'labels',labels,'plots',1,'fin_step',[25 5], 'CandleWidth',0.01); | {"url":"http://rosa.unipr.it/fsda/FSRms.html","timestamp":"2024-11-05T16:45:28Z","content_type":"application/xhtml+xml","content_length":"52970","record_id":"<urn:uuid:fd3dc5f4-d45c-40e7-ac28-17cc5aed4204>","cc-path":"CC-MAIN-2024-46/segments/1730477027884.62/warc/CC-MAIN-20241105145721-20241105175721-00493.warc.gz"} |
Precalculus - Online Tutor, Practice Problems & Exam Prep
Hey, everyone. We now know all of the components of the graph of a rational function. But taking those components and putting them all together to form an actual graph can be the hardest part. So
here, we're going to take the most basic of rational functions, 1x, and simply apply rules of transformations to it in order to graph a more complicated rational function. So here, I'm going to walk
you through using transformations to graph this function without doing a single calculation. We're just going to take our function 1x and pick it up and move it around a little bit to get the graph
of our new function, which will end up looking something like this. So let's go ahead and walk through how to get there.
Now, with our function g(x) here, we see that it looks really similar to 1x. We even have this 1x here. We just have these extra numbers added in. Now these numbers represent transformations to our
function. And when working with rational functions, we're going to focus on two types of transformations in particular, reflections and shifts, because they are the most commonly occurring
transformations when working with rational functions.
Now remember, whenever we have a negative outside of our function, that tells us we're going to have a reflection over the x-axis, whereas having a negative inside of our function is a reflection
over the y-axis. Now with h and k over here, x-h, h represents a horizontal shift side to side by some number h units. And then k represents a vertical shift up and down by some number k units. So
let's go ahead and apply these transformations to our function here.
Now our function 1x, our graph here, we see that we have asymptotes at our x and y axes, and we also have these points on our graph at (1,1) and (-1,-1) that are going to serve as sort of reference
or test points for us as we graph our new function. So we're going to focus on transforming our asymptotes and these points in order to get the graph of g(x) here. So let's go ahead and start with
step 1, which is to find the vertical asymptote and plot it at x=h. Now looking at our function here, 1x-3+1, I have x-3, so h here is 3. I'm going to have a vertical asymptote at x=3. Now, we can go
ahead and plot that on our graph using a dashed line because it is an asymptote.
Now we can go ahead and move on to step 2, which is to plot our horizontal asymptote at y=k. Now, looking back at our function here, I have this plus 1 on the end, so I know that k is going to be 1.
I'm going to plot my horizontal asymptote at y=1, again, using a dashed line. Now, compared to our original asymptotes at our axes, we see that these got shifted h units over and k units up. And we
see this transformation happening here. So let's keep transforming our graph.
Now moving on to step number 3, we want to go ahead and determine if there is a reflection. Now remember, our reflection is if we have these negatives that got put into our function. And looking at
my function, I don't have any negatives that got added in there, So, I do not have a reflection. Now if I did, I would take those test points (1,1) and (-1,-1) and simply reflect them over x or y
accordingly. But I don't need to do that here. So let's go ahead and move on to step 3b and go ahead and shift our test points by h,k. Now we know that h,k is just 3,1 because we already plotted
those at our asymptotes. So I'm going to take my test points, (1,1), and (-1,-1), and shift them accordingly. So my first point here, I'm going to go ahead and shift that 3 units over to the right
and then 1 unit up, landing me at the point (4,2). Then my other test point, which is right here at (-1,-1), I'm going to shift that again 3 units to the right and one unit up, landing me at (2,0).
So I've now shifted both my asymptotes and those reference points. So I can go ahead and move on to my final step here, which is going to be to sketch my curves approaching the asymptotes. So looking
at this first point here, I'm going to approach that asymptote going up and then approach my horizontal asymptote this way. Then my other test point, same thing, approaching my asymptotes on either
And now I have the final graph of my function. Now these are my curves here. I'm going to highlight them just so you can see them a little bit better. And you'll notice that this looks almost the
exact same as our graph of 1x. It's just picked up and moved over, just like I said. Now, we want to look at one other thing here, our domain and our range. Now, our domain and range are going to be
split by our asymptotes. And we're going to write them in set notation using this union symbol, which just represents a union between two sets. Now our domain is actually always going to go from
negative infinity until I reach my asymptote at 3 and then continue on from 3 to infinity. So looking at the graph I have here, I'm going to go from negative infinity until I reach that asymptote at
3. And then I have a break at that asymptote because my domain is not defined there. And I pick back up and go from 3 to infinity. Then for my range, the same exact thing is going to happen. We're
going to go from negative infinity to k and then from k to infinity because k is where my asymptote is. So looking at my function here, I know I go from negative infinity until I reach that asymptote
where my domain is not defined at 1, and then continuing on from 1 to infinity. Now you'll notice that these are enclosed in parentheses rather than square brackets because we know that these numbers
are not included in our domain or range.
So now that we know how to graph using transformations, let's get some more practice. | {"url":"https://www.pearson.com/channels/precalculus/learn/patrick/5-rational-functions/graphing-rational-functions?chapterId=24afea94","timestamp":"2024-11-10T14:08:27Z","content_type":"text/html","content_length":"392423","record_id":"<urn:uuid:1968c5be-7fcb-4d97-99a9-ae5b3b022d85>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.60/warc/CC-MAIN-20241110134821-20241110164821-00214.warc.gz"} |
The VHDL module "square" (see symbol) calculates the square of a signed number.
The module uses an optimized multiplication algorithm in order to get a smaller (about 33%) and faster (about 25%) design compared to a multiplication module.
The number of bits of the operand is configured by a generic.
The operand and the result (which is always positive) are numbers in 2's complement format.
The module uses flipflops only for storing the result (and for controlling).
For quick access to the operand bits, the operand is first stored in the result flipflops and
then replaced by the upcoming square bits through shift operations.
The latency of the module can be configured by a generic independently from the width of the operand.
This means, the module is configurable by generics in order
• to fulfill any requirements regarding the number of bits of the operand and
• to fulfill any requirements regarding its latency.
But of course there is no guarantee that timing closure can be reached with the selected values
for the generics, as the timing depends on the technology which is used at synthesis.
The module "square" was developed with HDL-SCHEM-Editor.
The module "square" is a hierarchical module, which is built by two submodules.
│Submodule name│ Functionality │
│ │The module square_step processes 1 bit of the algorithm. │
│ │The algorithm has as many steps as the operand has data bits. │
│square_step │One additional step is needed, which fixes the result if the operand was a negative number. │
│ │So in sum the algorithm has as many steps as the operand has bits. │
│ │The module square_step is instantiated as often as steps are processed during 1 clock cycle (which depends from the generic g_latency).│
│ │The square_control modules generates all the control signals which are needed. │
│square_control│It enables the internal registers for the intermediate or final results. │
│ │It identifies the clock period in which the sign bit of the operand is handled. │
│ │It activates the ready_o output at the end of the calculation. │
The module "square" was designed to have a faster and smaller solution for calculating the square compared to a multiplication module.
The improvement is based on an algorithm which needs a 1 bit adder for the first step and then 1 additional adder at each next step.
To see the full improvement (about 25% faster and 33% smaller) the generic g_latency must have the value 0 or 1,
because in both cases an individual submodule square_step is instantiated for each step of the algorithm,
so the adder inside this module can have its minimal size.
At the moment g_latency is bigger than 1 a reuse of the square_step modules happens and the improvement will be smaller:
Each square_step module is then used in more than 1 clock cycle and must be able to handle the partial result with the biggest width.
The worst case is at g_latency=g_operand_width, where no improvement compared to a multiplier module remains.
There is no limitation for the generic g_operand_width (except that it must be bigger than 1).
There is also no limitation for the generic g_latency. But this generic determines not only the latency but also
how difficult it will get reaching timing closure: The smaller the value is chosen, the harder it will get reaching timing closure.
If g_latency is equal to g_operand_width, then in each clock cycle 1 step of the algorithm is handled.
If g_latency is smaller than g_operand_width, then in each clock cycle more than step of the algorithm is handled.
How many steps are handled can be calculated by rounding up g_operand_width/g_latency to the next integer.
Be aware that handling more than 1 step of the algorithm in a clock cycle may prevent reaching timing closure.
If g_latency is bigger than g_operand_width, then the number of bits of the operand is internally increased to g_latency and
again in each clock cycle 1 bit of the (extended) operand is handled.
This is not a recommended configuration, as it needs additional gate resources and makes reaching timing closure more difficult.
When the square of a positive number A with the value "abcd" (with a,b,c,d from 0,1) has
to be calculated, we could use the common usual multiplication scheme:
A^2 = d*"0000abcd" +
c*"000abcd0" +
b*"00abcd00" +
For implementation 4 adders with each 4 bit would be needed.
But this effort can be reduced as the 2 numbers have the same digits.
In order to show a method to reduce the effort, the number A is written as follows:
A = a*2^3 + b*2^2 + c*2^1 + d
When we calculate A^2 the result is (be aware: a^n=a, b^n=b,...):
A^2 = a*2^3 * (a*2^3 + b*2^2 + c*2^1 + d) +
b*2^2 * (a*2^3 + b*2^2 + c*2^1 + d) +
c*2^1 * (a*2^3 + b*2^2 + c*2^1 + d) +
d * (a*2^3 + b*2^2 + c*2^1 + d) =
= a*2^6 + a*b*2^5 + a*c*2^4 + a*d*2^3 +
a*b*2^5 + b*2^4 + b*c*2^3 + b*d*2^2 +
a*c*2^4 + b*c*2^3 + c*2^2 + c*d*2^1 +
a*d*2^3 + b*d*2^2 + c*d*2^1 + d =
= a*2^6 + b*2^4 + c*2^2 + d +
a*b*2^6 + a*c*2^5 + a*d*2^4 + b*c*2^4 + b*d*2^3 + c*d*2^2 =
= a*2^6 + b*2^4 + c*2^2 + d +
a*(b*2^6 + c*2^5 + d*2^4)+ b*(c*2^4 + d*2^3)+ c*d*2^2 =
= a*2^6 + 0*2^5 + b*2^4 + 0*2^3 + c*2^2 + 0*2^1 + d +
a*(b*2^6 + c*2^5 + d*2^4)+
b*(c*2^4 + d*2^3)+
c*(d*2^2) =
= c*"0000d00" +
b*"00cd000" +
a*"bcd0000" +
In case of 8 bit the scheme looks like:
A^2 = 000000000000000 + Adding 0 makes no sense here, but is needed later.
g*000000000000h00 +
f*0000000000gh000 +
e*00000000fgh0000 +
d*000000efgh00000 +
c*0000defgh000000 +
b*00cdefgh0000000 +
a*bcdefgh00000000 +
Now the scheme is expanded by showing the intermediate sums:
A^2 = 000000000000000 +
g*000000000000h00 =
0h00 +
f*0000000000gh000 =
0ghh00 +
e*00000000fgh0000 =
sssshh00 +
d*000000efgh00000 =
sssssshh00 +
c*0000defgh000000 =
sssssssshh00 +
b*00cdefgh0000000 =
sssssssssshh00 +
a*bcdefgh00000000 =
ssssssssssshh00 +
a0b0c0d0e0f0g0h = "spread" operand
Now the bits of the "spread" operand are shifted to other places and the last addition is removed:
A^2 = 000000000000g00 + Sum-bits which have reached its end result are named 'S'.
g*000000000000h00 =
fsS00 +
f*0000000000gh000 =
essSS00 +
e*00000000fgh0000 =
dsssSSS00 +
d*000000efgh00000 =
cssssSSSS00 +
c*0000defgh000000 =
bsssssSSSSS00 +
b*00cdefgh0000000 =
assssssSSSSSS00 +
a*bcdefgh0000000h =
With the common multiplication scheme, a negative multiplicand (in 2's complement) is handled correctly,
but a negative multiplier, which is handled bit by bit as usual, is too big by 2^n,
and the product must be reduced by 2^n*multiplicand at the end.
But the situation is different in the optimized square scheme. Here both number are interpreted as positive numbers.
If the operand k is a negative integer, the binary value of k (in 2's complement) was calculated by 2^n+k,
where n is the number of bits of the 2's complement. So the square algorithm calculates:
square = (2^n + k)^2 =
= 2^2*n + k*2^(n+1) + k^2
As square has 2*n-1 bits at maximum, the summand 2^2*n can be ignored.
But the summand k*2^(n+1) must be subtracted from the result.
In this 8 bit example the value k*2^(n+1), limited to 15 bit, calculates to:
fix = k*2^(n+1) = abcdefgh * 2^9 = cdefgh000000000
As this value must be subtracted, the 2's complement of the value must be added:
-fix = not (cdefgh000000000) + 1
This new addition, depending on bit a which signals a negative number, is added last to the algorithm:
A^2 = 000000000000g00 + Sum-bits which have reached its end result are named 'S'.
g*000000000000h00 =
fsS00 +
f*0000000000gh000 =
essSS00 +
e*00000000fgh0000 =
dsssSSS00 +
d*000000efgh00000 =
cssssSSSS00 +
c*0000defgh000000 =
bsssssSSSSS00 +
b*00cdefgh0000000 =
assssssSSSSSS00 +
a*bcdefgh0000000h =
a * not(cdefgh) <- CarryIn=a at the least significant bit
In each addition parts of the original operand are shifted and added.
In order to avoid shifting the operand to the left, the intermediate sums are shifted right.
Least signficant bits with always value 0 are not shown any more:
A^2 = 000000g + Sum-bits which have reached its end result are named 'S'.
g*000000h =
00000sS -> shift
fsS +
f*00000gh =
0000ssS -> shift
essS +
e*0000fgh =
000sssS -> shift
dsssS +
d*000efgh =
00ssssS -> shift
cssssS +
c*00defgh =
0sssssS -> shift
bsssssS +
b*0cdefgh =
ssssssS -> shift
assssssS +
a*bcdefgh =
ssssssS -> shift
assssssS +
a * not(bcdefgh) = <- CarryIn=a at the adder for the least significant bit
_SSSSSS.......0h The '.'s are filled with the 'S's calculated before.
Source code for HDL-SCHEM-Editor and HDL-FSM-Editor for module "square" and its testbench (Number of downloads = 6 ).
With these files the schematics and the state-diagram of module square can be loaded into HDL-SCHEM-Editor or HDL-FSM-Editor and can be easily read and modified:
All module VHDL-files of the module "square" (Number of downloads = 7 ).
These files were generated by HDL-SCHEM-Editor and HDL-FSM-Editor:
All testbench VHDL-files of the module "square" (Number of downloads = 6 ).
These files were generated by HDL-SCHEM-Editor and HDL-FSM-Editor:
Relocation hints:
You should extract all archives into a folder named "square".
Then you must replace "M:/gesicherte Daten/Programmieren/VHDL/square" in all "hdl_editor_designs/*.hse" source files by your path to this directory.
Now you can navigate through the design by HDL-SCHEM-Editor and generate HDL by HDL-SCHEM-Editor for all modules except square_control,
for which the HDL must be generated by HDL-FSM-Editor.
Of course there is only need for generating HDL, if you change something at the modules, because you can find the HDL in VHDL_designs.zip and VHDL_testbenches.zip.
If you want to simulate or modify the modules by HDL-SCHEM-Editor you also must adapt the information in the Control-tab of the toplevel you want to work on.
There you must define a "Compile through hierarchy command", an "Edit command", the path to your HDL-FSM-Editor and a "Working directory".
Change log:
Version 1.0 (18.10.2024): | {"url":"http://www.hdl-schem-editor.de/square/square.php","timestamp":"2024-11-08T12:15:12Z","content_type":"text/html","content_length":"30842","record_id":"<urn:uuid:8bf5deba-ac37-4632-9fb6-80657573dd08>","cc-path":"CC-MAIN-2024-46/segments/1730477028059.90/warc/CC-MAIN-20241108101914-20241108131914-00074.warc.gz"} |
Course Instructors:
Dr. Saeed Asiri
• Office: 24E38
• Website: https://www.asiri.net
• WhatsApp: +966565555275
• Email: saeed@asiri.net
• Twitter: @drsaeedasiri
• Facebook: saeedasiri
Course TA:
Eng. Salah Fatani
• Office: TBA
• Cell. No.: +966504543238
• WhatsApp: +966504543238
• Email:
• MENG 262 Dynamics
• MENG 270 Mechanics of Materials
• EE 300 Complex Variables & Linear Algebra
Course description: Free and damped vibration of single degree of freedom systems, Viscous damping, Forced vibration, Resonance, Harmonic excitation; Rotating unbalance, Base motion, Vibration
isolation, Fourier analysis, Vibration measuring, General excitation, Step and impulse response, Two degree of freedom systems, Frequencies and mode shapes, Modal analysis, Undamped vibration
absorber, Multi degrees of freedom systems, Introduction to Continuous systems, Applications with computer programs.
Textbook: Singiresu Rao, Mechanical Vibrations, Prentice-Hall, Fifth Edition.
Course Learning Objectives:
By the end of the course the student will be to:
1. Model linear and nonlinear mechanical systems as combinations of springs, masses and dampers.
2. Determine and define the degrees of freedom of a given mechanical system.
3. Extract the equations of motion of a given mechanical system.
4. Analyze and interpret the response of mechanical systems to various types of excitations.
5. Analyze and interpret the response of mechanical systems to different cases of damping.
6. Predict qualitatively the response of systems based on the spectral content of the excitation.
7. Minimize the effects of transient and harmonic excitations on systems and their support structures.
8. Decouple equations of motion
9. Understand the significance of vibration control in various applications
Grading Policy:
Laboratory 10%
Quizzes 40%
Term Project 10%
Collaborative Learning Activities 10%
Final Opportunity To Shine 30%
Note: 75% attendance is required. No makeup for any quiz. Student must attend the laboratory to pass the course.
Contribution of the course to meet the ABET professional Component:
Engineering science: 2 credit, Engineering design: 1 credit.
How to Succeed
• Accept that it is your responsibility to learn the material (in spite of the book or teacher)
• Show up and become engaged with the topics
• Do the homework daily so you can ask questions in class
• Use you resources for help (classmates, upperclassmen, faculty, the library)
Prerequisites by Topic:
1. Students should be familiar with free and forced vibration of Single Degree Of Freedom (SDOF) systems.
2. Students should be familiar with the kinematics and kinetics of two and three-dimensional rigid bodies.
3. Students should have a good understanding of the mechanics of solids including the ability to determine effective spring constants of common structural members.
4. Students should be able to solve linear ordinary differential equations.
5. Students should be familiar with concepts from linear algebra including matrix vector arithmetic, determinants, matrix inversion, eigenvalues, and eigenvectors.
6. Students must have the ability to formulate and solve problems using MATLAB. | {"url":"https://cv.asiri.net/meng470/","timestamp":"2024-11-03T09:44:52Z","content_type":"text/html","content_length":"83059","record_id":"<urn:uuid:5b92e0d4-4ebd-4c7b-9b47-4721feadb09b>","cc-path":"CC-MAIN-2024-46/segments/1730477027774.6/warc/CC-MAIN-20241103083929-20241103113929-00600.warc.gz"} |
Graphics metapost
This topic contains packages for graphics generated using METAPOST.
This package provides a set of tools to typeset geometric proofs in the style of Oliver Byrne's 1847 ed. of Euclid's "Elements".
A METAPOST library for physics textbook illustrations.
Drawing binary Huffman trees with METAPOST and METAOBJ.
Computes and draws 2D Delaunay triangulation.
Plotting graphs using Lua.
Draw chemical structure diagrams with METAPOST.
Drawing chess boards and positions with METAPOST.
Flat geometry with METAPOST.
Probability trees with METAPOST.
METAPOST macros for secondary school mathematics teachers.
METAPOST macros for highly configurable rounded rectangles (optionally with text).
METAPOST macros for drawing cubic spline interpolants. | {"url":"https://ctan.org/topic/graphics-mpost","timestamp":"2024-11-15T02:38:19Z","content_type":"text/html","content_length":"13715","record_id":"<urn:uuid:7457ac2f-3a30-4310-9d73-4f3d98afe673>","cc-path":"CC-MAIN-2024-46/segments/1730477400050.97/warc/CC-MAIN-20241115021900-20241115051900-00118.warc.gz"} |
Chapter 2 - Mathematics for Virtual Environments
Slides | Exercises | Project Suggestions | Programming Examples | Errata | Pictures | Links
Back to: Computer Graphics and Virtual Environments From Realism to Real-Time
Back to: Index of Slides
Back to: Index of Exercises
Project Suggestions
Back to: Index of Project Suggestions
Programming Examples
Back to: Index of Programming Examples
Back to: Index of Errata
Back to: Index of Pictures
Back to: Index of Links | {"url":"http://www0.cs.ucl.ac.uk/staff/a.steed/book_tmp/CGVE/chapter_2.htm","timestamp":"2024-11-12T19:27:11Z","content_type":"text/html","content_length":"3027","record_id":"<urn:uuid:271b199a-15f5-4e2e-815b-1c62efff3948>","cc-path":"CC-MAIN-2024-46/segments/1730477028279.73/warc/CC-MAIN-20241112180608-20241112210608-00154.warc.gz"} |
Extreme fluctuations of current in the symmetric simple exclusion process: A non-stationary setting
We use the macroscopic fluctuation theory (MFT) to evaluate the probability distribution P of extreme values of integrated current J at a specified time t = T in the symmetric simple exclusion
process (SSEP) on an infinite line. As shown recently (Meerson and Sasorov 2014 Phys. Rev. E 89 010101), the SSEP belongs to the elliptic universality class. Here, for very large currents, the
diffusion terms of the MFT equations can be neglected compared with the terms coming from the shot noise. Using the hodograph transformation and an additional change of variables, we reduce the
'inviscid' MFT equations to Laplace's equation in an extended space. This opens the way to an exact solution. Here we solve the extreme-current problem for a flat deterministic initial density
profile with an arbitrary density 0 < n0 < 1. The solution yields the most probable density history of the system conditional on the extreme current, J / √T →[∞] and leads to a super-Gaussian
extreme-current statistics, ln -rfpag≃-(n[0])J^3 /T , in agreement with Derrida and Gerschenfeld (2009 J. Stat. Phys. 137 978). We calculate the function φ(n[0]) analytically. It is symmetric with
respect to the half-filling density n[0] = 1/2, diverges at n[0] → 0 and n[0] → 1 and exhibits a singularity φ(n[0]) |n [0]-1/2| at the half-filling density n[0] = 1/2.
• current fluctuations
• diffusion
• large deviations in non-equilibrium systems
• stochastic particle dynamics (theory)
Dive into the research topics of 'Extreme fluctuations of current in the symmetric simple exclusion process: A non-stationary setting'. Together they form a unique fingerprint. | {"url":"https://cris.huji.ac.il/en/publications/extreme-fluctuations-of-current-in-the-symmetric-simple-exclusion","timestamp":"2024-11-08T19:14:32Z","content_type":"text/html","content_length":"50718","record_id":"<urn:uuid:90cc9b01-0539-4d02-9292-1c06822b00d7>","cc-path":"CC-MAIN-2024-46/segments/1730477028070.17/warc/CC-MAIN-20241108164844-20241108194844-00741.warc.gz"} |
Photo, is this a diode or a resistor? temperature gauge 1975 300D - Page 3 - PeachParts Mercedes-Benz Forum
Resisters and such . . ..
I'd go with a junkyard one, and check to find out what other models use the same gauge and then start the search. Also, have you checked with Phil here at Peachparts to see if he has one located on
the shelf . . ready and waiting for an owner? His may be a lot cheaper.
1983 300D, the "Avocado"
1976 240D, 4-spd the "Pumpkin", SOLD to Pierre
1984 190D, 2.2L, 5-spd, my intro to MBZ diesels, crashed into in 2002 | {"url":"http://www.peachparts.com/shopforum/diesel-discussion/304139-photo-diode-resistor-temperature-gauge-1975-300d-3.html","timestamp":"2024-11-07T23:28:09Z","content_type":"application/xhtml+xml","content_length":"119237","record_id":"<urn:uuid:aaaaaa78-e0a8-4c71-853b-457d9a6a614f>","cc-path":"CC-MAIN-2024-46/segments/1730477028017.48/warc/CC-MAIN-20241107212632-20241108002632-00063.warc.gz"} |
Mapping of Data | UGC NET Paper 1
Graphical Representation and Mapping of Data | UGC NET Paper 1
Get Study Materials of General Studies for UPPSC ⇒ DOWNLOAD NOW
The transformation of data through visual methods like graphs, diagrams, maps, and charts is called the representation of data.
The representation of data is the base for any field of study. When we start the collection of data and the range of data increases rapidly, then an efficient and convenient technique for
representing data is needed. It is needed because of time constraints, efforts, and resources.
The top-level authority or management does not have enough time to go through whole reports, but any small point of data should not remain hidden from their eyes. Therefore, it is required for
presenting the data in such a manner that enables the reader to interpret the essential data with minimum efforts and time.
Data presentation and data representation are two terms having similar meaning and importance.
There are several techniques for data presentation, and are broadly categorised in two ways:
1. Non-graphical techniques: Tabular Form, Case Form
2. Graphical techniques: Pie Chart, Bar Chart, Line Graphs, Geometrical Diagrams
There are two types of non-graphical techniques:
(a) Tabular Form: This is better known as numerical data tables. The tabular form is the most commonly used technique for data presentation. This technique provides a correlation or measurement of
two values/variables at a time.
Table Pic
(b) Case Form: This technique is rarely used. Data is presented in the form of paragraphs and follows a rigid protocol to examine a limited number of variables.
Graphical Representation of Data
The data which has been represented in the tabular form can be displayed in pictorial form by using a graph. A graphical presentation (GR) is the easiest way to depict a given set of data. It is a
visual display of data and statistical results.
GR is often more effective than presenting data in tabular form. There are different types of graphical representations and which are used depends on the nature of the data and the type of
statistical results. The graphical representation is the visual display of data using plots and charts.
GR helps to quantify, sort, and present data in a method that is understandable to a large variety of audiences.
The visualization techniques are ways of creating and manipulating graphical representations of data.
There are several types of mediums which are used for expressing graphics, including plots, charts, and diagrams.
In literature, the words “diagram”, “chart”, and “graph” are commonly being used interchangeably. But the meaning of these words is as follows:
Diagram: The term “diagram” can be defined as a figure generally consisting of lines, made to accompany geometrical theorem, mathematical demonstration, etc. A sketch, drawing, or plan that outlines
and explains the parts of something is also a type of diagram—for example, a diagram of an engine. A pictorial representation of a quantity or of a relationship is termed as a diagram in simple
Chart: A sheet exhibiting information in the tabulated or methodical form is also known as a chart. A chart is a graphical representation of data as by lines, curves, bars, etc. of a dependent
variable, e.g., temperature, price, etc.
Graph: Graph is simply a diagram in the mathematical or scientific area of study. A drawing representing the relationship between a certain set of numbers or quantities by means of a series of dots,
lines, bars, etc. plotted with reference to a set of the axis is called a graph.
A few commonly used graphical representations of data are listed below:
• A histogram is represented by a rectangular bar to depict frequency distribution.
• Size of the class interval is represented by the width
• The size of the frequency is represented by height.
• Class boundaries/intervals are important in the construction of histograms and represent in the horizontal or X-axis of the graph.
• Frequency is represented as height in the graph on Y-axis.
• A histogram is essentially an area diagram composed of series of adjacent rectangles.
Source: https://online.stat.psu.edu/stat500/book/export/html/539
Bar Chart or Bar diagram or Bar graph
• A bar graph is a chart that uses either vertical, or horizontal bars to show comparisons among categories.
• One axis of the chart shows the specific categories being compared, and the other axis represents a discrete value.
• Some bar graphs present bars clustered in groups of more than one (grouped bar graphs) and others show the bars divided into sub parts to show cumulative effect.
• Is a circular graph that represents total value in circle and components in part wise.
• Useful in comparing components and total value.
• Data are expressed in percentage of the total value.
• The total value is equated to 360 degrees.
Line Graph or Stick Graph or Line Chart or Line Plot or Curve Chart
A line chart is the most basic type of chart used in finance, and it is generally created by connecting a series of past recorded data together with a line. It is a style of chart that is created by
connecting a series of data points together with a line. Line charts are ideal for representing trends over time.
• Most common graphical representation
• By plotting the X-axis on horizontally while Y-axis vertically.
• Find out the intersecting point of origin and join all intersections.
• Example: cricket score in each over.
Some other Graphical Representation is:
• Frequency polygon
• Cumulative frequency curve or Ogive
• Pictogram
• Stem leaf diagram
• Scatter diagram
Data mapping is, in the most simplistic terms, knowing where your information is stored.
In its simplest form, data mapping is about relationships. In particular, it is the process of specifying how one information set relates, or maps, to another. Consider an information set that
includes a list of people and their contact information. The list contains names, addresses, city, province or state, and postal code for each person. Also, consider a second information set that
includes a list of people and their music preferences. This list includes listener, artist, album name, and song name for each listener. The lists are self-contained, somewhat related, but distinct.
Suppose that you wanted to create a mailing list of people who like a particular artist.
You can’t quickly get this information because there is no direct way to relate one information set to the other. The solution is to create a mapping between the name in the first information set and
the listener in the second information set.
The specification of the relationship is called a data mapping. From there, you simply search the related or combined information set for all listeners in the list that like that particular artist.
This gives you the corresponding mailing addresses.
For example: in the left side of figure, ‘Name,’ ‘Email,’ and ‘Phone’ fields from an Excel source are mapped to the relevant fields in a Delimited file, which is our destination.
Data mapping, in its simplest term, is to map source data fields to their related target data fields. For example, the value of let says a source data field A goes into a target data field X.
Benefits of Mapping of Data
Data mapping is essentially a way to surface and prevent issues ahead of time before they create bigger problems later. The benefits are:
• Data mapping neutralises the potential for data errors and mismatches,
• Data mapping aids in the data standardisation process, and
• It makes intended data destinations clearer and easier to understand.
Challenges with Data Mapping
The followings are a few of the major challenges that can arise with data mapping:
• Inaccuracy
• Time-wasting
• Changes
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Author's notes on the paper “Short Witnesses and Accepting Lassos in Omega-automata”
On page 9 of the paper (page 269 of the proceedings), there is a small error in the index bounds in the description of the safety automaton. The second and third bullet points in the definition of
the transition function should read:
Relationship to Other Work
The paper discusses finding short accepting witnesses and lassos in omega-automata. As these can be seen as special cases of two-player games with omega-regular winning objectives, the results have
some important implications for the synthesis of finite-state systems: finding a system implementation with as few states as possible satisfying the specification encoded into the game graph is a
tough task: it is NP-hard to approximate within any polynomial.
The proof idea of Theorem 6 has been used independently in the paper Equivalence and Inclusion Problem for Strongly Unambiguous Büchi Automata by Nicolas Bousquet und Christof Löding, also published
at LATA 2010.
The slides of the talk given at LATA 2010 are available here. | {"url":"https://www.ruediger-ehlers.de/papers/errata_lata2010.html","timestamp":"2024-11-08T04:49:09Z","content_type":"application/xhtml+xml","content_length":"3845","record_id":"<urn:uuid:c5a06d77-212a-49c4-a405-eabaf576e8bc>","cc-path":"CC-MAIN-2024-46/segments/1730477028025.14/warc/CC-MAIN-20241108035242-20241108065242-00654.warc.gz"} |
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They are pretty broad and require too much reading. I don’t have time to read all of those works, but I will certainly do that later, just to be informed.
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where I can sell my game? here is a list
I think you can ignore most of the stores, since steam will give you probably by far the most sales.
I think you can ignore most of the stores, since steam will give you probably by far the most sales.
kinda true for sales, nevertheless, others site have amount of daily visit and a lot of people bought from this sites.
And some people want a steam alternative, as well want drm-free for his games not only for developers but players want drm-free for playing anytime without using steam. For than reason gog.com comes
into play drm-free and no activation or online connection required to play, so each site have his market.
For me, I try to bought from the official site but basically I buy multiplayer games from steam or if have good achievement/social stuff, for singleplayer I use gog.com, indiegamestand, itch.io,
humblebundle, greenmangaming.com, amazon.com sometime these stores have "steam key".
DRM is optional for Steam, if you don't use it for your game, then your game has no DRM. Same with other platforms, the big ones likely have some form of DRM protection as well.
It can be easier or more immediate to get on other platforms than steam though, where you need to go through greenlight or get a publisher, some of the other stores have fewer hoops to jump through -
for better or worse.
If you have a real serious game and not some obvious ripoff or trash than you should be able to get through greenlight very easily. In the past getting through greenlight was very hard, after some
time it became easier, but maybe it has changed again and it has become a little harder again, I don't know.
In the past getting through greenlight was very hard, after some time it became easier, but maybe it has changed again and it has become a little harder again, I don't know.
Steam is fast becoming one of the worst and most expensive platforms, despite the never ending sales.
Just because steam has millions of users does not mean that sales are higher on steam.Finding something worth playing on steam is borderline impossible, its like trying to find decent software for a
mobile device, pointless time waste. There is so much to wade through, and steam picks and chooses what sells
Given the amount of trash on steam, along with games that still cannot get greenlit, it appears that steam greenlight is 100% arbitrary.
Simple proof of this, ubergame a minor mod of stock torque that just happened to get greenlight vs Airship dragoon that had to 'fight' for thousands of votes.
While i understand that no mother admits their baby is ugly in public....
If you have a real serious game and not some obvious ripoff or trash than you should be able to get through greenlight very easily.
Did you seriously keep a straight face while typing this?
BTW FYI another issue with steam store is region lock and I suffered of this about two years ago, when I moved from one country to other... I'm very picky about what game I will go to bought.
• 1 year later...
I found that most platforms seem to not support free games or true indie games and many in that list do not even exist anymore or are dead.
Platforms that seem to accept any kind of games are Steam and Itchi.io, these are the only good ones I could find that have a reasonable customer count.
I found that most platforms seem to not support free games or true indie games and many in that list do not even exist anymore or are dead.
Platforms that seem to accept any kind of games are Steam and Itchi.io, these are the only good ones I could find that have a reasonable customer count.
A site i really like indie and foss friendly is gamejolt.com, it has lots of stupid stuff on it to be said, but there are a few gems on it.
You can also sell games for a fixed price or let people pay what they want. | {"url":"https://torque3d.org/forums/topic/759-where-i-can-sell-my-game-here-is-a-list/","timestamp":"2024-11-12T22:46:54Z","content_type":"text/html","content_length":"198468","record_id":"<urn:uuid:9bf9f162-0b80-4a04-8488-3e50f490c608>","cc-path":"CC-MAIN-2024-46/segments/1730477028290.49/warc/CC-MAIN-20241112212600-20241113002600-00214.warc.gz"} |
1. * Nature of the subject. The matter and the states of aggregation. Homogeneous and heterogeneous systems. Stages and their separations. Elements and chemical compounds. Atoms and molecules.
Ponderal laws (Lavoisier, Proust, Dalton). Volumetric laws (Gay-Lussac, Avogadro). Determination of atomic weight (Cannizzaro rule) and molecular weight (gaseous density). Avogadro number. Mole.
2. * Structure of matter. Description of the atom. Protons, neutrons and electrons. Atomic number and mass number. Atomic mass unit. Isotopes. Mass defect. Thomson's experiment. Atomic model of
Thomson. Millikan experiment. Rutherford experiment. Atomic model of Rutherford. Electromagnetic radiation. Emission spectrum of the black body. Photoelectric effect. Emission spectrum of the
hydrogen atom. Bohr theory. Report by De Broglie. Uncertainty principle of Heisenberg. Wave mechanics. Schrödinger equation. Quantum numbers and energy levels. Orbital. Polyelectronic atoms.
Principle of Pauli. Hund rule. Principle of Aufbau. Periodic table. Periodic properties (ionization energy, electronic affinity, atomic radius, electronegativity, metallic character).
3. * Chemical bond. Electron sharing. Covalent bond. Octet rule. Distance and bonding energy. Homeopolar and heteropolar link. Dative tie. Dipoles. Bonds π and s. Hybridization. Bonding angles.
VSEPR. Molecular geometry. Resonance. Ionic bond. MO-LCAO theory. Molecular orbitals of diatomic molecules of the second period. Metal tie. Orbitals of Bloch. Weak links. Hydrogen bond.
4. * Chemical compounds and nomenclature. Valence and oxidation number. Oxidation and reduction. Hydrides. Hydrogen acids. Oxides. Peroxides. Hydroxides. Oxyacids. Salts. Chemical equation.
Reactions. Redox reactions. Disruption reactions. Combustion reactions. Ponderal relationships. Limiting reagent rule. Calculation examples. Types of formulas (minimum, brute, structure and steric
formula). Elementary analysis. Calculation examples.
5. * Thermodynamics. Thermodynamic system. Types of systems. Extensive and intensive variables. Status functions. Work. Heat. Power. Thermal capacity. Work. First principle of thermodynamics.
Internal energy and enthalpy. Termochimica. Law of Hess. Second principle of thermodynamics. Heat conversion in work. Entropy. Free energy. Spontaneity of chemical reactions. Third principle of
6. * States of aggregation of the matter. The gaseous state. Ideal gas and perfect gas. Boyle's law. Law of Gay Lussac. Law of Charles. Avogadro's law. Equations of state of ideal gases.
Determination of the molecular weight of a gas. Gaseous diffusion. Partial pressures. Molar heat of gases. Maxwell-Boltzmann speed distribution. Real gases. Van der Waals equation. Liquefaction of
gases. Andrews diagram. Numerical exercises. The liquid state. Surface tension. Vapor pressure. Clausius-Clapeyron equation. The solid state. Crystalline and amorphous solids. Isotropy and
anisotropy. Primitive cells. Bravais lattices. X-ray diffraction and Bragg equation. Polymorphism. Ionic solids. Covalent solids. Molecular solid. Metallic solids.
7. * State steps and heterogeneous equilibria. State steps: fusion, evaporation, sublimation. Clausius-Clapeyron equation. Variance. Phase rule. State diagrams. One-component systems: water, sulfur,
carbon dioxide. Systems with eutectic point.
8. * Status of solution. Types of solution. Solubility of a species. Concentration and way of expressing it. Solute-solvent interaction: ideal and real solutions. Rault law. Relationships between the
composition of a mixture of two liquids and that of its vapor. Systems with maximum and minimum azeotrope. Dilute solutions of non-volatile solutes. Colligative properties. Lowering the vapor
pressure. Cryoscopic lowering. Ebullioscopic elevation. Osmotic pressure. Numerical exercises.
9. * Chemical equilibria. Law of chemical equilibrium. Le Chatelier's principle. Relationship between free energy and equilibrium constant. Balance constant (Kp and Kc). Relationships between
equilibrium constants. Homogeneous and heterogeneous equilibria. Gaseous equilibria. Influence of pressure, temperature and concentration on equilibrium conditions.
10. * Electrolyte solutions. Electrolytic dissociation. Strong and weak electrolytes. Degree of dissociation. Coefficient of Van't Hoff. Conductance. Equivalent conductance. Law of independent ion
migration. Acids and bases. Theories of Arrhenius, Bronsted-Lowry and Lewis. Strength of acids and bases. Ionic product of water. Relationship between Ka and Kb. Definition of pH. Calculation of the
pH of solution of acids, bases and salts. Buffer solutions. PH indicators. Acid-base titrations. Ampholytic. Solubility equilibria. Solubility product. Ion to common.
11. * Electrochemistry. Oxidation-reduction reactions: electronic ion method. Electrode potentials. Nernst's equation. Standard potential and its measure. Galvanic batteries. Concentration batteries.
Electrochemical series of elements. Chemical stacks. Forecasts of redox reactions. Balance constant. Determination of pH, KPS and degree of dissociation. Free reaction energy. Numerical exercises.
12. * Electrolysis. Decomposition voltage. Overvoltage. Faraday laws and numerical exercises. Law of electrochemical equivalents. Electrolysis of molten salts. Water electrolysis. Electrolysis of
aqueous solutions. Industrial electrolytic processes. Accumulators. Corrosion. Passivation.
13. * Chemical kinetics. Reaction speed. Kinetic law. Molecularity. Reaction order: reactions of the 1st and 2nd order. Arrhenius equation. Influence of temperature. Activation energy. Catalysts.
Kinetic derivation of the equilibrium constant. Chain reactions.
* obligatory skills
1. R. Chang – K. Goldsby: “Fondamenti di Chimica Generale”, McGrawHill Education.
2. P. Atkins – L. Jones: “Fondamenti di Chimica generale”, Zanichelli.
3. P. Atkins - L. Jones - L. Laverman: Principi di Chimica, quarta edizione italiana, Zanichelli.
4. Pietro Tagliatesta, CHIMICA GENERALE E INORGANICA, edi-ermes. | {"url":"https://syllabus.unict.it/insegnamento.php?id=10677&eng","timestamp":"2024-11-11T10:05:02Z","content_type":"text/html","content_length":"10469","record_id":"<urn:uuid:a75cb1b0-2ffc-4bb8-b1b1-dbc8e750529a>","cc-path":"CC-MAIN-2024-46/segments/1730477028228.41/warc/CC-MAIN-20241111091854-20241111121854-00410.warc.gz"} |
Faraday effect and optical isolator
In this example, we will model a Faraday rotator, a device which rotates the polarization of the incident light via the magneto-optical Faraday effect. We will also show how this effect can be
utilized to create an optical isolator, allowing for transmission of light in only one direction (and preventing unwanted feedback).
Faraday Effect
The Faraday effect is a magneto-optical effect where an external magnetic field leads to circular birefringence, and the left and right circularly polarized waves propagate at different velocities.
The result of this is a rotation of the plane of polarization, which is linearly proportional to the component of the magnetic field in the direction of propagation. The resulting angle of rotation
β, is defined by
$$\beta = VBd$$
where V is the Verdet constant, and B is the static magnetic flux density. To model this effect in FDTD, we will use an anisotropic material combined with a grid attribute object, which allows us to
apply an arbitrary unitary matrix necessary for inducing the correct rotation. Specifically, we have to rotate the reference frame such that it converts the field components from Cartesian
coordinates into coordinates that represent circular polarization. To do this, we define the following unitary matrix:
$$U = \frac{1}{\sqrt{2}} \left(\begin{matrix} 1&0&i\\0 & \sqrt{2} & 0 \\ 1 & 0 & -i \end{matrix}\right)$$
and use a Matrix Transform Grid Attribute to apply this rotation around the y axis. The script faraday.lsf defines U and sets it as the transform matrix.
In faraday.fsp, we start with a plane wave polarized in the x direction, and use a linear y DFT monitor to track the change in polarization of the light as it propagates through the Faraday rotator.
Here, we can define the polarization, at each point along y, by the TE fraction:
$$\frac{|E_x|^2}{|E_x|^2 + |E_z|^2}$$
The script faraday_plot.lsf plots the "TE fraction vs y" in the rotator:
We can also plot Ex and Ez as a function of y in the Visualizer to observe this change. Note that the "Duplicate" function in the Visualizer makes it very easy to plot different field components in
the same figure.
Optical Isolator
For an optical isolator (which is typically composed of a Faraday rotator and two polarizers as shown below), the polarizer in front of the Faraday rotator will not only polarize the incident light
in the x direction, it will also function as a filter for the reflected light (with the opposite polarization as the incident light).
In this section, we will extend the Faraday effect described in the previous section to model the optical isolator. In the previous section, we used an incident plane wave polarized in the x
direction, and found that a distance of ~11.94um will result in an angle of rotation of 45 degrees (see faraday.lsf). In isolator.fsp,a mirror is placed at the point where the angle of rotation is 45
degrees. The reflected light will continue to change in the rotator, undergoing another angle of rotation of 45 degrees when it exits the rotator at the incident surface. As a result, the reflected
fields will be completely polarized in the z direction (the figure below shows the Ex and Ez components of the reflected light):
One can also use the movie monitor to observe the change in polarization as a function of time.
See also | {"url":"https://optics.ansys.com/hc/en-us/articles/360042274774-Faraday-effect-and-optical-isolator","timestamp":"2024-11-11T17:42:47Z","content_type":"text/html","content_length":"34707","record_id":"<urn:uuid:7e104946-8860-4977-9982-a7df35662d3b>","cc-path":"CC-MAIN-2024-46/segments/1730477028235.99/warc/CC-MAIN-20241111155008-20241111185008-00751.warc.gz"} |