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https://www.ncatlab.org/homotopytypetheory/revision/diff/dagger+epimorphism+in+a+dagger+precategory+%3E+history/2
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# Homotopy Type Theory dagger epimorphism in a dagger precategory > history (Rev #2, changes)
Showing changes from revision #1 to #2: Added | Removed | Changed
## Definition
A morphism $f:hom_A(a,b)$ of a dagger precategory $A$ is a dagger epimorphism if $f \circ f^\dagger=1_a$.
Category theory
category: category theory
Revision on June 7, 2022 at 14:55:53 by Anonymous?. See the history of this page for a list of all contributions to it.
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2022-09-28 12:30:21
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https://lqp2.org/node/668
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# Infinite dimensional Lie algebras in 4D conformal quantum field theory
Bojko Bakalov, Nikolay M. Nikolov, Karl-Henning Rehren, Ivan T. Todorov
November 05, 2007
The concept of global conformal invariance (GCI) opens the way of applying algebraic techniques, developed in the context of 2-dimensional chiral conformal field theory, to a higher (even) dimensional space-time. In particular, a system of GCI scalar fields of conformal dimension two gives rise to a Lie algebra of harmonic bilocal fields, $V_m(x,y)$, where the m span a finite dimensional real matrix algebra M closed under transposition. The associative algebra M is irreducible iff its commutant M' coincides with one of the three real division rings. The Lie algebra of (the modes of) the bilocal fields is in each case an infinite dimensional Lie algebra: a central extension of $sp(\infty,\mathbb{R})$ corresponding to the field $\mathbb{R}$ of reals, of $u(\infty,\infty)$ associated to the field C of complex numbers, and of so*(4 infty) related to the algebra H of quaternions. They give rise to quantum field theory models with superselection sectors governed by the (global) gauge groups $O(N)$, $U(N)$, and $U(N,H)=Sp(2N)$, respectively.
Keywords:
none
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2020-10-19 15:46:33
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http://blog.ethanrosenthal.com/2014/10/06/festival-chatter-part3/
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# Data Piques
## Festival Chatter (Part 3) - Bonnaroo Analysis in the Fourth Dimension
In this series of posts (part 1, part 2), I have been showing how to use Python and other data scientist tools to analyze a collection of tweets related to the 2014 Bonnaroo Music and Arts Festival. So far, the investigation has been limited to summary data of the full dataset. The beauty of Twitter is that it occurs in realtime, so we can now peer into the fourth dimension and learn about these tweets as a function of time.
## More Organic
Before we view the Bonnaroo tweets as a time series, I would like to make a quick comment about the organic-ness of the tweets. If you recall from the previous post, I removed duplicates and retweets from my collection in order to make the tweet database more indicative of true audience reactions. On further investigation, it seems that there were many spammy media sources still in the collection. To make the tweets even more organic, I decided to look at the source of the tweets.
Because Kanye West was the most popular artist from the previous posts' analysis, I decided to look at the top 15 sources that mentioned him:
twitterfeed 1585
dlvr.it 749
IFTTT 256
Hootsuite 201
WordPress.com 102
Tumblr 81
Instagram 73
iOS 42
TweetDeck 38
twitterfeed and dlvr.it are social media platforms for deploying mass tweets, and a look at some of these tweets reveals this fact. So, I decided to create a list of "organic sources", which consists of mobile Twitter clients, and use these to cull the tweet collection
organic_sources = ['Twitter for iPhone', 'Twitter Web Client',
organics = organics[organics['source'].isin(organic_sources)]
With this new dataset, I re-ran the band popularity histogram from the previous post, and I was surprised to see that Kanye got bumped down to third place! It looks like Kanye's popular with the media, but Jack White and Elton John were more popular with the Bonnaroo audience.
## 4th Dimensional Transition
Let's now look at the time dependence of the tweets. For this, we would like to use the created_at field as our index and tell pandas to treat its elements as datetime objects.
# Clean up field
organics['created_at'] = [tweetTime['\$date'] for tweetTime in organics['created_at']]
organics['created_at'] = pd.to_datetime(Series(organics['created_at']))
organics = organics.set_index('created_at',drop=False)
organics.index = organics.index.tz_localize('UTC').tz_convert('EST')
To look at the number of tweets per hour, we have to resample our tweet collection.
ts_hist = organics['created_at'].resample('60t', how='count')
The majority of my time spent creating this blog post consisted of fighting with matplotlib trying to get decent looking plots. I thought it would be cool to try to make a "fill between" plot, which took way longer to figure out than it should have. The key is that fill_between takes 3 inputs: an array for the x-axis and two y-axis arrays between which the function fills color. If one just wants to plot a regular curve and fill to the x-axis, one must create an array of zeros that is the same length as the curve. Also, I get pretty confused with which commands should be called with ax, plt, and fig. Anyway, the code and corresponding figure are below.
# Prettier pandas plot settings
# Not sure why 'default' is not the default...
pd.options.display.mpl_style='default'
x_date = tshist.index
zero_line = np.zeros(len(x_date))
fig, ax = plt.subplots()
ax.fill_between(x_date, zero_line, ts_hist.values, facecolor='blue', alpha=0.5)
# Format plot
plt.setp(ax.get_xticklabels(),fontsize=12,family='sans-serif')
plt.setp(ax.get_yticklabels(),fontsize=12,family='sans-serif')
plt.xlabel('Date',fontsize=30)
plt.ylabel('Counts',fontsize=30)
plt.show()
As you can see, tweet frequency was pretty consistent during each day of the festival and persisted until the early hours of each morning.
## Band Popularity Time Series
We can now go back to questions from the previous post and look at how the top five bands' popularity changed with time. Using my program from the previous post, buildMentionHist, we can add a column for each band to our existing organics dataframe. Each row of the bands' columns will contain a True or False value corresponding to whether or not the artist was mentioned in the tweet. We resample the columns like above but do this in bins of 10 minutes.
import buildMentionHist as bmh
import json
path = 'bonnaroooAliasList.json'
alias_dict = [json.loads(line) for line in open(path)][0]
bandPop = organics['text'].apply(bmh.build_apply_fun(alias_dict),
alias_dict)
top_five = bandPop.index.tolist()[:5] # Get top five artists' names
bandPop = pd.concat([organics, bandPop], axis=1)
top_five_ts = DataFrame()
for band in top_five:
top_five_ts[band] = bandPop[bandPop[band] == True]['text'].resample('10min', how='count')
We now have a dataframe called top_five_ts that contains the time series information for the top five most popular bands at Bonnaroo. All we have to do now is plot these time series. I wanted to again make some fill between plots but with different colors for each band. I used the prettyplotlib library to help with this because it has nicer looking default colors. I plot both the full time series and a "zoomed-in" time series that is closer to when the artists' popularities peaked on Twitter. I ran into a lot of trouble trying to get the dates and times formatted correctly on the x-axis of the zoomed-in plot, so I have included that code below. There is probably a better way to do it, but at least this finally worked.
import pytz
import prettyplotlib as ppl
from prettyplotlib import brewer2mpl
for band in top_five_ts:
ppl.fill_between(top_five_ts.index.tolist(),0.,top_five_ts[band])
ax = plt.gca()
fig = plt.gcf()
set2 = brewer2mpl.get_map('Set2','qualitative',8).mpl_colors
# Note: have to make legend by hand for fill_between plots.
# BEGIN making legend
for color in set2:
legendProxies.append(plt.Rectangle((0, 0), 1, 1, fc=color))
leg = legend(legendProxies, topfive, loc=2)
leg.draw_frame(False)
# END making legend
# BEGIN formatting xaxis
datemin = datetime(2014,6,13,12,0,0)
datemax = datetime(2014,6,16,12,0,0)
est = pytz.timezone('EST')
plt.axis([est.localize(datemin), est.localize(datemax), 0, 80])
fmt = dates.DateFormatter('%m/%d %H:%M',tz=est)
ax.xaxis.set_major_formatter(fmt)
ax.xaxis.set_tick_params(direction='out')
# END formatting xaxis
plt.xlabel('Date',fontsize=30)
plt.ylabel('Counts',fontsize=30)
Here is the full time series:
And here is the zoomed-in time series:
If we look at when each band went on stage, we can see that each bands' popularity spiked while they were performing. This is good - it looks like we are measuring truly "organic" interest on Twitter!
Band Performance Time
Jack White 6/14 10:45PM - 12:15AM
Elton John 6/15 9:30PM - 11:30PM
Kanye West 6/13 10:00PM - 12:00AM
Skrillex 6/14 1:30AM - 3:30AM
Vampire Weekend 6/13 7:30PM - 8:45PM
## Delving into the Text
Up until now, I have not looked too much about the actual text of the tweets other than to find a mention of an artist. Using the nltk library, we can learn a little more about some general qualities of the text. The simplest quantity is looking at the most frequently used words. To do this, I go through every tweet and break all of the words up into individual elements of a list. In the language of natural language processing, we are "tokenizing" the text. Common english stopwords are omitted, as well as any mentions of the artists or artists' aliases. I use a regular expression code to only grab words from the sentences and ignore punctuation (except for apostrophes). I also take our alias_dict from the previous post and make sure that those words are not collected when tokenizing the tweets.
from nltk.tokenize import RegexpTokenizer
from nltk.corpus import stopwords
import re
def custom_tokenize(text, custom_words=None, clean_custom_words=False):
"""
This routine takes an input "text" and strips punctuation
(except apostrophes), converts each words to lowercase,
removes standard english stopwords, removes a set of
custom_words (optional), and returns a list of all of the
leftover words.
INPUTS:
text = text string that one wants to tokenize
custom_words = custom list or dictionary of words to omit
from the tokenization.
clean_custom_world = Flag as True if you want to clean
these words.
Flag as False if mapping this function
to many keys. In that case,
pre-clean the words before running
this function.
OUTPUTS:
words = This is a list of the tokenized version of each word
that was in "text"
"""
tokenizer = RegexpTokenizer(r"[\w']+")
stop_url = re.compile(r'http[^\\s]+')
stops = stopwords.words('english')
if clean_custom_words:
custom_words = tokenize_custom_words(custom_words)
words = [w.lower() for w in text.split() if not re.match(stop_url, w)]
words = tokenizer.tokenize(' '.join(words))
words = [w for w in words if w not in stops and w not in custom_words]
return words
def tokenize_custom_words(custom_words):
tokenizer = RegexpTokenizer(r"[\w']+")
custom_tokens = []
stops = stopwords.words('english')
if type(custom_words) is dict: # Useful for alias_dict
for k, v in custom_words.iteritems():
k_tokens = [w.lower() for w in k.split() if w.lower() not in stops]
# Remove all punctuation
k_tokens = tokenizer.tokenize(' '.join(k_tokens))
# Remove apostrophes
k_tokens = [w.replace("'","") for w in k_tokens]
# Below takes care of nested lists, then tokenizes
v_tokens = [word for listwords in v for word in listwords]
v_tokens = tokenizer.tokenize(' '.join(v_tokens))
# Remove apostrophes
v_tokens = [w.replace("'","") for w in v_tokens]
custom_tokens.extend(k_tokens)
custom_tokens.extend(v_tokens)
elif type(custom_words) is list:
custom_tokens = [tokenizer.tokenize(words) for words in custom_words]
custom_tokens = [words.replace("'","") for words in custom_tokens]
custom_tokens = set(custom_tokens)
return custom_tokens
Using the above code, I can apply the custom_tokenize function to each row of my organics dataframe. Before doing this, though, I make sure to run the tokenize_custom_words function on the alias dictionary. Otherwise, I would end up cleaning the aliases for every row in the dataframe which is a waste of time.
import custom_tokenize as tk
clean_aliases = tk.tokenize_custom_words(alias_dict)
token_df = organics['text'].apply(tk.custom_tokenize,
custom_words=clean_aliases,
clean_custom_words=False)
Lastly, I collect all of the tokens into one giant list and use the FreqDist nltk function to get the word frequency distribution.
# Need to flatten all tokens into one big list:
big_tokens = [y for x in token_df.values for y in x]
distr = nltk.FreqDist(big_tokens)
distr = distr.pop('bonnaroo') # Obviously highest frequency
distr.plot(25)
A couple things caught my eye - the first being that people like to talk about themseleves (see popularity of " i'm "). Also it was pretty popular to misspell "Bonnaroo" (see popularity of " bonaroo "). I wanted to see if there was any correlation between mispellings and maybe people being intoxicated at night, but the time series behavior of the mispellings looks similar in shape (though not magnitude) to the full tweet time series that is plotted earlier in the post.
misspell = token_df.apply(lambda x: 'bonaroo' in x)
misspell = misspell[misspell].resample('60t', how='count')
The other thing that interested me was the the word "best" was one of the top 25 most frequent words. Assuming that "best" correlates with happiness, we can see that people got happier and happier as the festival progressed:
This is, of course, a fairly simplistic measure of text sentiment. In my next post, I would like to quantify more robust measures of Bonnaroo audience sentiment.
By the way, the code used in this whole series on Bonnaroo is available on my GitHub.
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2018-03-22 17:27:11
|
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https://www.encyclopediaofmath.org/index.php/Function
|
# Function
One of the basic concepts in mathematics. Let two sets $X$ and $Y$ be given and suppose that to each element $x\in X$ corresponds an element $y\in Y$, which is denoted by $f(x)$. In this case one says that a function $f$ is given on $X$ (and also that the variable $y$ is a function of the variable $x$, or that $y$ depends on $x$) and one writes $f:X\to Y$.
In ancient mathematics the idea of functional dependence was not expressed explicitly and was not an independent object of research, although a wide range of specific functional relations were known and were studied systematically. The concept of a function appears in a rudimentary form in the works of scholars in the Middle Ages, but only in the work of mathematicians in the 17th century, and primarily in those of P. Fermat, R. Descartes, I. Newton, and G. Leibniz, did it begin to take shape as an independent concept. The term "function" first appeared in works of Leibniz. Geometric, analytic and kinematic ideas were used to specify a function, but gradually the notion of a function as a certain analytic expression began to prevail. This was formulated in the 18th century in a precise form; J. Bernoulli's definition is that "a function of a variable quantity … is a number composed by some arbitrary method from the variable quantity and from constants" . L. Euler, having accepted this definition, wrote in his textbook on analysis that "all analysis of infinitesimals revolves around the variable quantities and their functions" [1]. Euler already had a more general approach to the concept of a function as dependence of one variable quantity on another. This point of view was developed further in the work of J. Fourier, N.I. Lobachevskii, P. Dirichlet, B. Bolzano, and A.L. Cauchy, where the notion of a function as a correspondence between two sets of numbers began to crystallize. So by 1834, Lobachevskii [2] was writing: "The general concept of a function requires that a function of x is a number which is given for each x and gradually changes with x. The value of a function can be given either by an analytic expression or by a condition which gives a means of testing all numbers and choosing one of them; or finally a dependence can exist and remain unknown" . The definition of a function as a correspondence between two arbitrary sets (not necessarily consisting of numbers) was formulated by R. Dedekind in 1887 [3].
The concept of a correspondence, and consequently also the concept of a function, sometimes leads to other concepts (to a set [4], a relation [5] or some other set-theoretical or mathematico-logical concepts [6]) and is sometimes taken as a primary, undefined, concept [7]. A. Church, for example, expressed the view that: "In the end it is necessary to consider the concept of a function — or some similar concept, for example the concept of a class — as primitive or undefinable" [8]. (For more information see [9], [10].)
The concept of a function considered below is based on the concept of a set and of the simplest operations on sets.
One says that the number of elements of a set is equal to 1 or that the set consists of one element if it contains an element and no others (in other words, if after deleting the set from one obtains the empty set). A non-empty set is called a set with two elements, or a pair, , if after deleting a set consisting of only one element there remains a set also consisting of one element (this definition does not depend on the choice of the chosen element ).
If a pair is given, then the pair is called the ordered pair of elements and and is denoted by . The element is called its first element and is called the second element.
Given sets and , the set of all ordered pairs , , , is called the product of the sets and and is denoted by . It is not assumed that is different from , that is, it is possible that .
Each set of ordered pairs , , , such that, if and , then implies that , is called a function or, what is the same thing, a mapping.
As well as the terms "function" and "mapping" one uses in certain situations the terms transformation , morphism , correspondence , which are equivalent to them.
The set of all first elements of ordered pairs of a given function is called the domain of definition (or the set of definition) of this function and is denoted by , and the set of all second elements is called the range of values (the set of values) and is denoted by . The set of ordered pairs itself, , considered as a subset of the product , is called the graph of .
The element is called the argument of the function, or the independent variable, and the element is called the dependent variable.
If is a function, then one writes and says that maps the set into the set . In the case one simply writes .
If is a function and , then one writes (sometimes simply or ) and also , , , and says that the function puts the element in correspondence with the element (the mapping maps to ) or, what is the same, the element corresponds to the element . In this case one also says that is the value of at the point , or that is the image of the element under .
As well as the symbol one also uses the notation for denoting the value of at .
Sometimes the function itself is denoted by the symbol . Denoting both the function and its value at the point by the same symbol does not usually lead to misunderstanding, since in any particular case, as a rule, it is always clear what one is talking about. The notation often turns out to be more convenient than the notation in computations. For example, writing is more convenient and simpler to use in analytic manipulations than writing .
Given , the set of all elements such that is called the pre-image of the element and is denoted by . Thus,
Obviously, if , then , the empty set.
Let a mapping be given. In other words, to each corresponds a unique element and to each corresponds at least one element . If , one says that maps the set into itself. If , that is, if coincides with the range of , then one says that maps onto the set or that is a surjective mapping, or, more concisely, it is a surjection. Thus, a mapping is a surjection if for each element there is at least one element such that .
If under a mapping different elements correspond to different elements , that is, if implies , then is said to be a one-to-one mapping of into and also a univalent mapping or an injection. Thus, a mapping is univalent (injective) if and only if the pre-image of each element belonging to the range of , that is, , consists precisely of one element. If the mapping is simultaneously one-to-one and onto the set (see One-to-one correspondence), that is, is at the same time injective and surjective, then it is called a bijective mapping or a bijection.
If and , then the set
that is, the set of all those such that there is at least one element of the subset of which is mapped to by , is called the image of the subset , and one writes . In particular, always . The following relations are true for the images of sets and :
and if , then .
If and , then the set
is called the pre-image of the set and one writes . Thus, the pre-image of a set consists of all those elements which are mapped to elements of under , or what is the same thing, it consists of all pre-images of elements : . For the pre-images of sets and the relations
are true, and if , then .
If , then the function generates in a natural way a function defined on under which corresponds to the element . This function is called the restriction of the function to the set and is sometimes denoted by . Thus and for any one has . If the set does not coincide with the set , then the restriction of to may have a different domain of definition than and, consequently, is different from .
If , if each element is a certain set of elements , and if, moreover, among these sets there is at least one set consisting of more than one element, then is called a multi-valued function (sometimes, many-valued function). Further, the elements of the set are often called the values of at . If each set , , consists of only one element, then the function is also called a single-valued function.
If and , then the function defined for each by the formula is called the composition (superposition) of the functions and , also the composite function, and is denoted by .
Let a function be given and let be its range. The set of all possible ordered pairs of the form , , forms a function, called the inverse function of and denoted by . Under the inverse function , to each corresponds the pre-image , that is, a certain set of elements. So the inverse function is, generally speaking, a multi-valued function. If a mapping is injective, then the inverse mapping is a single-valued function and maps the range of onto the domain of definition of .
## Functions on numbers.
An important class of functions is that of the complex-valued functions , , where is the set of all complex numbers. One can carry out various arithmetical operations on complex-valued functions. If two given complex-valued functions and are defined on the same set and if is a complex number, then the function is defined as the function taking the value at each point ; the function is the function taking the value at each point; the function is the function taking the value at each point; and, finally, is the function equal to at each point (which, of course, makes sense only when ).
A function is called a real-valued function ( is the set of real numbers). A real-valued function is said to be bounded from above (bounded from below) on the set if its range is bounded from above (bounded from below). In other words, a function is bounded from above (bounded from below) on if there is a constant such that for every the inequality is satisfied (the inequality is satisfied, respectively). A function that is both bounded from above and from below on is simply said to be bounded on . An upper (lower) bound of the range of is an upper (lower) bound of the function .
A major role in mathematical analysis is played by functions on numbers or, more precisely, complex-valued functions of a complex variable, that is, functions , where . If the domain of definition of such a function and its range are both subsets of the real numbers, then this function is called a real function, or, more precisely, a real-valued function of a real variable. The generalization of the concept of a function on numbers is, first of all, a complex-valued function of several complex variables, called a complex function of several variables. A further generalization of a function on numbers is a vector-valued function (see Vector function) and, in general, a function for which the domain of definition and the range are provided with definite structures. For example, if the ranges of functions belong to a certain vector space, then such functions can be added; if they belong to a ring, then the functions can be added and multiplied; if they belong to a set which is ordered in some specific way, then one can generalize to these functions the idea of boundedness, upper and lower bounds, etc. The presence of topological structures on the sets and enables one to introduce the concept of a continuous function . In the case when and are topological vector spaces one introduces the concept of differentiability for a function (cf. Differentiable function).
## Methods for specifying functions.
Functions on numbers (and certain generalizations of them) can be given by formulas. This is the analytic method for specifying functions. For this one uses a certain supply of functions which have been studied and have a special notation (primarily the elementary functions), algebraic operations, composition of functions and limit transitions (which includes operations of mathematical analysis such as differentiation, integration, summing series), for example:
The class of functions that are presented, in a well-determined sense, as the sum of series, even only as sums of trigonometric series, is very wide. A function can be given analytically either in an explicit form, that is, by a formula of the type , or as an implicit function, that is, by an equation of the type . Sometimes the function is given with the aid of several formulas, for example,
(*)
A function can also be given by using a description of the correspondence. Let, for example, the number 1 correspond to every , the number 0 to the number 0, and the number to every . As a result one obtains a function defined on the whole real line and taking three values: . This function has the special notation (or ). Another example: the number 1 corresponds to each rational number and the number 0 to each irrational number. The function obtained is called the Dirichlet function. The same function can be given in different ways; for example, the function and the Dirichlet function can be defined not only by verbal descriptions but also by formulas.
Every formula is a symbolic notation of a certain previously-described correspondence, so that in the end there is no fundamental difference between specifying a function by a formula or by a verbal description of the correspondence; this difference is superficial. It should be borne in mind that every function newly defined by some means or other can, if a special notation is introduced for it, serve to define other functions by using formulas including this new symbol. However, for an analytic representation of a function the supply of functions and operations that are to be used in the formulas is very essential; usually one tries to make this supply as small as possible and chooses the functions and operations as simply as possible in a certain well-determined sense.
When one is concerned with real-valued functions of a single real variable, then to give an intuitive representation of the nature of the functional dependence one often constructs the graph of the function in the coordinate plane, in other words, given a function , , one considers the set of points , , in the -plane.
Figure: f041940a
Figure: f041940b
Thus, the graph of the function (*) has the form depicted in Fig. a, the graph of the function is Fig. b, and the graph of the function consists of the isolated points in Fig. c.
Figure: f041940c
The representation of a function by a graph can also serve to reveal a functional dependence. This revelation is approximate because, in practice, the measurement of the intervals can be carried out only with a definite degree of accuracy, without mentioning the fact that, when the domain of definition of the function is unbounded, it is in principle impossible to draw it on the coordinate plane.
A tabular method is also extensively used for representing functions on numbers, either in the form of prepared tables of values of the function at definite points, or by introducing this data into a machine memory, or by making a program to calculate the values on a computer.
## The classification of complex or real functions.
The simplest complex functions are the elementary functions, among which one distinguishes the algebraic polynomials, the trigonometric polynomials and also the rational functions (cf. Rational function). The special role of these functions is that one of the methods for studying and using more general functions is based on approximating them by algebraic polynomials, by trigonometric polynomials or by rational functions, as well as by functions composed from these functions in some well-determined way (see Spline). The branch of the theory of functions that studies the approximation of functions by collections of functions that are simple in a certain sense is called approximation theory. In this theory approximations of functions by linear aggregates of eigen functions of certain operators are also very important.
The analytic functions (cf. Analytic function), i.e. functions locally representable by power series, form an important class, containing the rational functions. The analytic functions can be subdivided into the algebraic functions (cf. Algebraic function), that is, functions that can be given by an equation , where is an irreducible polynomial with complex coefficients, and transcendental functions, that is, those which are not algebraic. With the concept of a derivative one can associate the classes of functions that are differentiable a definite number of times, with the concept of an integral one can associate the classes of functions that are integrable in some sense or other, and with the concept of continuity one can associate the class of continuous functions. The Baire classes of functions are obtained by taking successive pointwise limits from the class of continuous functions (see also Borel function). The definition of a measurable function is based on the concepts of a measurable set and a measure. The branch of the theory of functions that studies properties of functions associated with the concept of a measure is usually called the metric theory of functions.
A function space arises as a collection of functions having certain general properties. Thus, all functions defined on the same set in the -dimensional Euclidean space and, for example, Lebesgue measurable, continuous, or satisfying a Hölder condition of given order, respectively, form vector spaces. Similarly, the spaces of times (continuously-) differentiable functions, the infinitely-differentiable functions, the functions of compact support, the analytic functions, and many other classes of functions form vector spaces.
In a number of vector spaces of functions one can introduce a norm. For example, in the space of continuous functions on a compact space , , a norm is ; the normed space of continuous functions with this norm is denoted by . Given the space of measurable functions , defined on a space , where is a certain set, is a certain -algebra of subsets of and is a measure defined on the sets , by putting
one specifies a norm on the set of functions for which . A function space with such a norm is usually called a Lebesgue function space . Among the other function spaces playing an important role in mathematical analysis one should mention a Hölder space, a Nikol'skii space, an Orlicz space, and a Sobolev space. All these spaces and certain generalizations of them are complete metric spaces, which to a large extent is important in studying many problems in the theory of functions itself as well as problems from related branches of mathematics. The relations between various norms for functions belonging simultaneously to various function spaces are studied in the theory of imbedding of function spaces (see Imbedding theorems). An important property of the basic function spaces is that the set of infinitely-differentiable functions is dense in them, which enables one to study a number of properties of these function spaces on sufficiently-smooth functions, and to carry over the results to all functions in the space under consideration by taking limits.
## Dependence of functions.
This is a property of systems of functions generalizing the concept of their linear dependence and meaning that there are well-determined relations between the values of the functions in the given system; in particular, the values of one of them can be expressed in terms of the values of the others. For example, the functions and are dependent over the whole real line, since always . Let be a domain in , let be its closure and let , , . The functions , , are said to be dependent in if there is a continuously-differentiable function , , whose zeros form a nowhere-dense set in and such that the composite is identically zero on .
Functions , , are said to be dependent in the domain if they are dependent in the closure of any domain such that .
Functions , , that are continuously differentiable in a domain are dependent in if and only if their Jacobian
vanishes identically on .
Now let
(1)
If for , , , , there is an open set in and a continuously-differentiable function on such that at any point ,
and
are satisfied, then is said to be dependent in the set on the functions .
If the functions , , are continuous in a domain and if in a neighbourhood of each point one of them depends on the others, then the functions , , are dependent in .
If in a neighbourhood of each point one of the functions , , which are continuously-differentiable in a domain , depends on the others, then at any point of the rank of the Jacobi matrix
(2)
is less than , that is, the gradients are linearly dependent at each point .
Let the functions (1) be continuously differentiable in a domain and let the rank of their Jacobi matrix (2) not exceed a certain number , , at every point ; suppose, moreover, that at a certain point it is equal to . In other words, there are variables and functions such that
(3)
Then there is no neighbourhood of in which any of the functions depends on the others and there is a neighbourhood of such that each of the remaining functions , , , depends in this neighbourhood on . In particular, if the gradients are linearly dependent at all points of the domain and if at a certain point there are of them that are linearly independent, and consequently one of them, for example , is a linear combination of the others, then there is a neighbourhood of such that in this neighbourhood depends on the functions .
#### References
[1] L. Euler, "Einleitung in die Analysis des Unendlichen" , 1 , Springer (1983) (Translated from Latin) [2] N.I. Lobachevskii, "Complete works" , 5 , Moscow-Leningrad (1951) (In Russian) [3] R. Dedekind, "The nature and meaning of numbers" , Open Court (1901) (Translated from German) [4] F. Hausdorff, "Grundzüge der Mengenlehre" , Leipzig (1914) (Reprinted (incomplete) English translation: Set theory, Chelsea (1978)) [5] A. Tarski, "Introduction to logic and to the methodology of deductive sciences" , Oxford Univ. Press (1946) (Translated from German) [6] K. Kuratowski, "Topology" , 1 , PWN & Acad. Press (1966) (Translated from French) [7] G. Frege, "Funktion und Begriff" , H. Pohle (1891) ((Reprint: Kleine Schriften, G. Olms, 1967)) [8] A. Church, "Introduction to mathematical logic" , 1 , Princeton Univ. Press (1956) [9] A.P. Yushkevich, "The concept of function up to the middle of the nineteenth century" Arch. History of Exact Sci. , 16 (1977) pp. 37–85 Istor.-Mat. Issled. , 17 (1966) pp. 123–150 [10] F.A. Medvedev, "Outlines of the history of the theory of functions of a real variable" , Moscow (1975) (In Russian)
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2019-11-16 01:22:46
|
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|
https://competitive-exam.in/questions/discuss/for-arc-welding
|
What is the correct answer?
4
# For arc welding
Alternating current with high frequency is used
Alternating current with low frequency is used
Direct current is used
Any one of these
|
2022-05-25 11:16:46
|
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|
http://www.talkstats.com/tags/calulator/
|
# calulator
1. ### Sx comes up empty on my T-84 plus
I have a T-84 plus.I trying to find the varience of these values, 1.51, 1.52, 1.54 , 1.36, 1.43 , 1.44, 1.61, 1.49. I put each value in the L list and calculated standard deviation but when I do the Sx box is alway empty.
|
2021-11-29 00:20:12
|
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http://www.numdam.org/item/ITA_1994__28_6_513_0/
|
Toward a semantics for the QUEST language
RAIRO - Theoretical Informatics and Applications - Informatique Théorique et Applications, Tome 28 (1994) no. 6, pp. 513-555.
@article{ITA_1994__28_6_513_0,
author = {Alessi, Fabio and Barbanera, Franco},
title = {Toward a semantics for the {QUEST} language},
journal = {RAIRO - Theoretical Informatics and Applications - Informatique Th\'eorique et Applications},
pages = {513--555},
publisher = {EDP-Sciences},
volume = {28},
number = {6},
year = {1994},
zbl = {0884.68076},
mrnumber = {1305115},
language = {en},
url = {http://www.numdam.org/item/ITA_1994__28_6_513_0/}
}
TY - JOUR
AU - Alessi, Fabio
AU - Barbanera, Franco
TI - Toward a semantics for the QUEST language
JO - RAIRO - Theoretical Informatics and Applications - Informatique Théorique et Applications
PY - 1994
DA - 1994///
SP - 513
EP - 555
VL - 28
IS - 6
PB - EDP-Sciences
UR - http://www.numdam.org/item/ITA_1994__28_6_513_0/
UR - https://zbmath.org/?q=an%3A0884.68076
UR - https://www.ams.org/mathscinet-getitem?mr=1305115
LA - en
ID - ITA_1994__28_6_513_0
ER -
Alessi, Fabio; Barbanera, Franco. Toward a semantics for the QUEST language. RAIRO - Theoretical Informatics and Applications - Informatique Théorique et Applications, Tome 28 (1994) no. 6, pp. 513-555. http://www.numdam.org/item/ITA_1994__28_6_513_0/
1. M. Ababi, G. D. Plotkin, A PER Model of Polymorphism and Recursive types, Proc. Symposium on Logic in Computer Science, IEEE, 1990, 355-364. | MR 1099188
2. F. Alessi, F. Barbanera, Toward a Semantics for the QUEST Language, to appear in Sixth Annual IEES Symposium on Logic in Computer Science, Amsterdam, July 1991. | MR 1305115
3. R. Amadio, Recursion over Realizability Structures, Information and Computation, 1991. | MR 1097263 | Zbl 0760.03012
4. H. P. Barendregt, The Lambda Calculus: Its Syntax and Semantics, Studies in Logic, 103, North-Holland, 1981, Revised Edition, 1984. | MR 622912 | Zbl 0551.03007
5. V. Breazu-Tannen, T. Coquand, C. Gunter, G. Scedrov, Inheritance and Explicit Coercion, in Fourth Annual IEEE Symposium on Logic in Computer Science, 1989, 112-129. | Zbl 0716.68012
6. K. Bruce, G. Longo, A Modest Model for Records, Inheritance and Bounded Quantification, Information and Computations 1990, 87, 196-240. | MR 1055952 | Zbl 0711.68072
7. L. Cardelli, Typeful Programming, SRC report 45, 1989.
8. L. Cardelli, G. Longo, A Semantic Basis for Quest SRC report 55, 1990. | MR 1140339
9. L. Cardelli, P. Wegner, On Understanding Types, Data Abstraction and Polymorphism, Computing Surveys, 1985, 17, 471-522.
10. F. Cardone, Recursive Types for Fun, Theoretical Computer Science, 1991. | MR 1122644 | Zbl 0746.68018
11. R. Cartwright, Types as Intervals, Proc. Symposium on Principle of Programming Languages, ACM, 1984, 22-36.
12. M. Coppo, A Completeness Theorem for Recursively defined Types, Proc. 12th International Colloquium on automata, Languages and programming, LNCS 194, 1985, 120-129. | MR 819247 | Zbl 0585.68047
13. M. Coppo, M. Zacchi, Type Inference and Logical Relations, Proc. Symposium on Logic in Computer Science, IEEE, 1986, 218-226.
14. M. Dezani Ciancaglini, B. Venneri, Partial Types and Intervals, SIAM Journal on Computing, 1990, 19, 644-658. | MR 1053932 | Zbl 0697.03005
15. C. Gunter, Universal Profinite Domains, Information and Computation, 1987, 72, 1-30. | MR 871554 | Zbl 0628.68050
16. C. Gunter, D. Scott, Semantic Domains, Handbook of Theoretical Computer Science, North Holland, 1990, 633-674. | MR 1127197 | Zbl 0900.68301
17. D. Macqueen, G. Plotkin, R. Sethi, An Ideal Model for Recursive Polymorphic Types, Information and Control, 1986, 71, 95-130. | MR 864747 | Zbl 0636.68016
18. S. Martini, Bounded Quantifiers have Interval Models, ACM Conference on LISP and Functional Programming Languages, 1988, 164-173.
19. J. C. Mitchell, A Type Inference Approach to Reduction Properties and Semantics of Polymorphic Expressions, Proceedings of ACM Conference on LISP and Functional Programming, 1986, 308-319.
|
2022-01-26 00:11:22
|
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|
https://socratic.org/questions/how-do-yo-write-the-orbital-diagram-for-sulfur
|
# How do you write the orbital diagram for sulfur?
Well, we use the $\text{aufbau principle}$, and for sulfur, $Z = 16$....
And thus.....$1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{4}$; this distributes the 16 electrons appropriately. Given this configuration, sulfur commonly forms sulphide anion, i.e. ${S}^{2 -}$. Can you give the electronic description for ${S}^{2 -}$. What other element shares this configuration?
|
2020-08-09 21:06:06
|
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https://discuss.mxnet.io/t/implementation-of-a-recurrent-neural-network-from-scratch/2364/2
|
# Implementation of a Recurrent Neural Network from Scratch
While it will be clear to the observant reader, it could confuse some people as to why we need a separate init_rnn_state() function.
Let us assume that the first input to the RNN is called X_1 and the first output created by the RNN is called y_1.
Consider the general equation:
\mathbf{H}_t = \phi(\mathbf{X}_t \mathbf{W}_{xh} + \mathbf{H}_{t-1} \mathbf{W}_{hh} + \mathbf{b}_h)
Since t=1, we have:
\mathbf{H}_1 = \phi(\mathbf{X}_1 \mathbf{W}_{xh} + \mathbf{H}_{0} \mathbf{W}_{hh} + \mathbf{b}_h).
Now what is this H_0. This is a state which MUST be present even before RNN starts processing the first input. Hence for the code to work, it is very important that we initialise this. Even if we choose to initialise this to 0, we have to make sure that the dimensions of this matrix are such that the above equations holds true.
If you look closely, you will realise that the dimensions of any H_t is in reality \text{batch_size} \times \text{num_hidden_units}
Since batch_size is not static at the time of creating the network, we need to initialise this H_0 vector at runtime and hence the need for a separate function.
|
2019-11-12 20:35:08
|
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|
http://svmiller.com/stevedata/reference/so2concentrations.html
|
This data set contains yearly observations by the Environmental Protection Agency on the concentration of sulfur dioxide in parts per billion, based on 32 sites. I use this for in-class illustration. Note that the national standard is 75 parts per billion. Data are the national trend.
so2concentrations
## Format
A data frame with the following 4 variables.
year
the year
value
the mean concentration of sulfur dioxide in the air based on 32 trend sites, in parts per billion
ub
the lower bound of the value (10th percentile)
lb
the upper bound of the value (90th percentile)
## Source
Environmental Protection Agency, 2021. https://www.epa.gov/air-trends/sulfur-dioxide-trends
|
2021-10-16 21:19:57
|
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|
http://tex.stackexchange.com/questions/39051/typesetting-a-row-vector
|
Typesetting a Row Vector
In the question typesetting a column vector, a very nice solution was given for typesetting a column vector with an arbitrary number of rows. I am attempting to modify the code to produce row vectors instead of column vectors. I made the "obvious" change, i.e., substituting the & for the \\ as indicated by the snippet below:
\newcount\colveccount
\newcommand*\colvec[1]{
\global\colveccount#1
\begin{pmatrix}
\colvecnext
}
\def\colvecnext#1{
#1
\ifnum\colveccount>0
&
\expandafter\colvecnext
\else
\end{pmatrix}
\fi
}
My modification however fails to compile and the errors, of course, are not very helpful. So my question is, How can I modify the \colvec and \colvecnext macros to produce row vectors?
-
You have to collect the row beforehand, as doing it inside an array environment is quite difficult, as TeX ends a cell as soon as it sees an &. I prefer to do in with a token register and the \count255 scratch counter.
We collect two arguments: the first is the number of entries, the second the first entry, in order to initialize the token register with it. We set the counter to the number of requested element (since we have already collected one, we step it down immediately).
Then we start the recursion; the macro \rowvectnexta examines the value of the counter; if it's still greater than 0, it calls \rowvecnextb that gathers another argument, stores it into the token register (adding an &) and steps down the counter; otherwise it builds the pmatrix by releasing the contents of the register.
\newtoks\rowvectoks
\newcommand{\rowvec}[2]{%
\rowvectoks={#2}\count255=#1\relax
\rowvecnexta}
\newcommand{\rowvecnexta}{%
\ifnum\count255>0
\expandafter\rowvecnextb
\else
\begin{pmatrix}\the\rowvectoks\end{pmatrix}
\fi}
\newcommand\rowvecnextb[1]{%
\rowvectoks=\expandafter{\the\rowvectoks}%
\rowvecnexta}
It shouldn't be a big limitation the fact that you need at least one element:
\rowvec{1}{a}
\rowvec{2}{a}{b}
\rowvec{3}{a}{b}{c}
Notice: \usepackage{amsmath} is required. So a minimal example might be the following.
\documentclass{article}
\usepackage{amsmath}
\newtoks\rowvectoks
\newcommand{\rowvec}[2]{%
\rowvectoks={#2}\count255=#1\relax
\rowvecnexta}
\newcommand{\rowvecnexta}{%
\ifnum\count255>0
\expandafter\rowvecnextb
\else
\begin{pmatrix}\the\rowvectoks\end{pmatrix}
\fi}
\newcommand\rowvecnextb[1]{%
\rowvectoks=\expandafter{\the\rowvectoks}%
\rowvecnexta}
\begin{document}
$\rowvec{1}{a}$
$\rowvec{2}{a}{b}$
$\rowvec{3}{a}{b}{c}$
\end{document}
A completely different solution that can be generalized to column vectors uses xparse and LaTeX3.
\documentclass{article}
\usepackage{amsmath}
\usepackage{xparse}
\ExplSyntaxOn
\NewDocumentCommand{\Rowvec}{ O{,} m }
{
\vector_main:nnnn { p } { & } { #1 } { #2 }
}
\NewDocumentCommand{\Colvec}{ O{,} m }
{
\vector_main:nnnn { p } { \\ } { #1 } { #2 }
}
\seq_new:N \l__vector_arg_seq
\cs_new_protected:Npn \vector_main:nnnn #1 #2 #3 #4
{
\seq_set_split:Nnn \l__vector_arg_seq { #3 } { #4 }
\begin{#1matrix}
\seq_use:Nnnn \l__vector_arg_seq { #2 } { #2 } { #2 }
\end{#1matrix}
}
\ExplSyntaxOff
\begin{document}
$\Rowvec{a}\Rowvec{a,b}\Rowvec[;]{a;b;c}$
$\Colvec{a}\Colvec{a,b}\Colvec[;]{a;b;c}$
\end{document}
The optional argument is the delimiter to be used to separate entries in the list given as argument.
It's sufficient to change { p } into { b } in the two main definitions to get brackets instead of parentheses.
Note: this requires expl3 dated September 2012 or later.
-
Thank you! Your solution works beautifully. Considering the necessary steps, my modification attempt was naively futile. – 3Sphere Dec 21 '11 at 22:05
I'll expand the answer: the TV was broadcasting Beethoven's Ninth from the new theater in Florence, so my main attention was to it. :) – egreg Dec 21 '11 at 22:15
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2013-05-25 01:10:44
|
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|
https://numberblog.wordpress.com/tag/geometric-wavelets/
|
## Multiscale SVD, K planes and geometric waveletsNovember 30, 2010
Posted by Sarah in Uncategorized.
Tags: , , , ,
I went to the applied math talk by Mauro Maggioni today; here are notes from the presentation.
Multiscale SVD
We often have to deal with high-dimensional data clouds that may lie on a low-dimensional manifold. Examples: databases of faces, collections of text documents, configurations of molecules. But, we don’t know a priori what the dimension of the manifold is. So we ned some mechanisms for estimating that, in the situation where we have noise in the data.
Some terminology: the data points are $x_i + \eta_i$, the real data plus some white Gaussian noise of variance $\sigma^2$. The intrinsic dimension of the manifold $M$ is $k$ while the dimension of the ambient space is $D$.
Previous work at estimating dimensions generally falls into two categories. One is volume-based (the number of points on the manifold confined within a ball should grow as $r^k$ where $r$ is the radius, if you’ve chosen the right dimension k. The other possibility is PCA, done locally and patched together somehow to account for the fact that we have a curved manifold rather than a linear hyperplane. The problem with this is that you don’t know what “locally” means — how many patches do we need? There’s also the so-called “curvature dilemma” with PCA: the more highly curved the manifold, the more likely it is to confuse the top principal value (representing the tangent direction) with the second-highest principal value (representing the direction of curvature.)
Maggioni’s alternative is multiscale SVD. We fix a point. The noise around the manifold looks like a hollow tube — most of the noise is concentrated at a radius of $\sigma \sqrt{D}$ from the surface of the manifold itself. So looking at a small ball (smaller than the radius of the tube) picks up less noise, looking at a large ball picks up all the noise, and looking at a still larger ball begins to pick up the curvature of the manifold. We do SVD within all these balls, and look at the singular values over a range of scales. There’s a “sweet spot,” a particular interval of scales where the signal singular values have separated from the noise singular values, but the curvature singular values haven’t caught up yet; this accurately represents the curvature of the manifold. (I blogged about this earlier here and the paper is here.)
The assumption about the data is that the tangent covariance grows differently ($O(r^2)$) with respect to the radius than the normal covariance ($O(r^4)$). This can be computed to be true explicitly for manifolds of co-dimension one, for instance.
The result is that if $M$ is assumed to have small enough curvature and $\eta$ is assumed to have small enough variance, then with high probability we can determine the intrinsic dimension $k$ with only $k \log k$ points. Compared to a large number of other dimension-estimating methods, this outperforms all of them. Isomap, in particular, really falls apart in the presence of noise, and can’t tell a 6-dimensional sphere from a 6-dimensional cube.
One interesting fact is that many real-life data sets have dimensionality that varies. A database of text documents, for instance, is low-dimensional in some regions, and very, very high-dimensional in others. (We measure dimensionality locally, so there’s no reason in principle that it shouldn’t vary.)
Measures supported on K planes
Sometimes high-dimensional data is supported on a set of planes $\pi_1 \dots \pi_k$, possibly of different dimensions $d_1 \dots d_k$. We don’t know how many planes there are, which planes they are, or what their dimensions are. This is the situation in face recognition; it’s also relevant to image and signal processing more generally (dictionary learning, sparse approximation, etc.)
Maggioni and collaborators developed an algorithm for doing this.
1) Draw $n_0$ random points and $n n_0$ nearest neighbors. Do multiscale SVD at these points. Produce an estimated plane $\hat{\pi}_k$ and its estimated dimension $\hat{d}_k$. Also estimate the noise level $\hat{\sigma}$.
2. Construct an affinity matrix, with the (i, j)th coordinate
$e^{\frac{-d(x_i, \hat{\pi}_k)^2}{\hat{\sigma}^2}}$
This is almost a matrix composed of zeros and ones — either a point is on a plane (in which we get one) or a point is not on that plane (in which case we get 0 because the distance is large.) In practice, with noise and curvature, it’s not quite binary, but it’s close.
3. Denoise and cluster this matrix, and find updated estimates for the planes from this.
Tested on a face database, this worked very well. It misclassified only 3 faces out of 640. What’s remarkable here is that the Yale Face Database used had 64 different photos of each person, taken at different angles, different lighting conditions, etc. This has always been a hard problem for computer vision: how do you recognize that two pictures are the same person viewed frontal and profile, while two other pictures are of two different people? Apparently, this is a solution to such heterogeneity problems.
Geometric wavelets
The problem of dictionary learning is as follows. Given $n$ points in $\mathbb{R}^D$, and a data matrix $X$, construct a dictionary $\Phi$ of m vectors such that
$X = \Phi \alpha$
where $\alpha$ is a sparse vector.
There’s a tradeoff here between the size of $\Phi$ and the sparsity of $\alpha$: obviously, you could just let $\Phi$ be the whole of $X$ and then $\alpha$ would just be the identity, but that would be silly — it wouldn’t compress the data at all.
One way of doing this is something called geometric wavelets. (Paper here. Shorter, more descriptive paper with cool picture here.)
Given a manifold, we partition it into a dyadic tree. We do principal components analysis on the coarsest scale and get a plane; then we divide that into two smaller subsets, do PCA on each of them, and get two new planes. For each plane, at the center point of that cube in the manifold grid, we have a tangent vector $\Phi_j$, which approximates the tangent to the manifold. The wavelets keep track of the differences $\Phi_j - \Phi_{j+1}$. If we have some smoothness in the manifold, these corrections decay quickly, so finitely many wavelets are a good approximator for any point.
There’s a fast algorithm for computing this, very like the wavelet transform, except for a point cloud rather than a function. This is useful for compressing databases, analyzing online data, and so forth.
The idea is very like Peter Jones’ construction of beta numbers (summary here) except that it may be more efficient in terms of storage. More on geometric wavelets from Dan Lemire.
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2019-05-21 14:48:34
|
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https://www.shaalaa.com/concept-notes/comparing-fractions_13008
|
Comparing Fractions
description
• Comparing like fractions with same denominators
• Comparing unlike fractions with same numerators
• Comparing unlike fractions with different numerators
Comparing Fractions:
Consider 1/2 and 1/3
The portion of the whole corresponding to 1/2 is clearly larger than the portion of the same whole corresponding to 1/3. But often it is not easy to say which one out of a pair of fractions is larger. We should, therefore, like to have a systematic procedure to compare fractions. It is particularly easy to compare like fractions.
1. Comparing like fractions with the same denominators:
Let us compare two like rational numbers: 3/8 and 5/8.
In both the fraction, the whole is divided into 8 equal parts. For 3/8 and 5/8,
We take 3 and 5 parts respectively out of the 8 equal parts. Clearly, out of 8 equal parts, the portion corresponding to 5 parts is larger than the portion corresponding to 3 parts.
Hence, 5/8 > 3/8.
Note the number of the parts taken is given by the numerator. It is, therefore, clear that for two fractions with the same denominator, the fraction with the greater numerator is greater.
Between 4/5 and 3/5, 4/5 is greater.
Between 11/20 and 13/20, 13/20 is greater.
2. Comparing unlike fractions with the same numerators:
Which is greater 1/3 or 1/5?
• In 1/3, we divide the whole into 3 equal parts and take one. In 1/5, we divide the whole into 5 equal parts and take one. Note that in 1/3, the whole is divided into a smaller number of parts than in 1/5. The equal part that we get in 1/3 is, therefore, larger than the equal part we get in 1/5. Since in both cases we take the same number of parts (i.e. one), the portion of the whole showing 1/3 is larger than the portion showing 1/5, and therefore 1/3 > 1/5.
• In the same way, we can say 2/3 > 2/5. In this case, the situation is the same as in the case above, except that the common numerator is 2, not 1. The whole is divided into a large number of equal parts for 2/5 than for 2/3. Therefore, each equal part of the whole in the case of 2/3 is larger than that in the case of 2/5. Therefore, the portion of the whole showing 2/3 is larger than the portion showing 2/5 and hence, 2/3 > 2/5.
We can see from the above example that if the numerator is the same in two fractions, the fraction with the smaller denominator is greater of the two.
Thus, 1/8 > 1/10, 3/5 > 3/7, 4/9 > 4/11 and so on.
3. Comparing unlike fractions with different numerators:
Compare 5/6 and 13/15.
Solution:
The fractions are unlike. We should first get their equivalent fractions with a denominator which is a common multiple of 6 and 15.
Now, (5 × 5)/(6 × 5) = 25/30, (13 × 2)/(15 × 2) = (26)/(30)
Since, (26/30) > (25/30) "we have" (13/15) > (5/6)
Why LCM?
The product of 6 and 15 is 90; obviously 90 is also a common multiple of 6 and 15. We may use 90 instead of 30; it will not be wrong. But we know that it is easier and more convenient to work with smaller numbers. So, the common multiple that we take is as small as possible. This is why the LCM of the denominators of the fractions is preferred as the common denominator.
Example
Find answers to the following. Write and indicate how you solved them.
Is 5/9 "equal to" 4/5 ?
5/9, 4/5
Converting these into like fractions,
5/9 = 5/9 xx 5/5 = 25/45.
4/5 = 4/5 xx 9/9 = 36/45.
As 36/45 ≠ 25/45
Therefore, 5/9 "is not equal to" 4.
Example
Find answers to the following. Write and indicate how you solved them.
Is 9/16 "equal to" 5/9 ?
9/16, 5/9.
Converting these into like fractions,
9/16 = 9/16 xx 9/9 = 81/144.
5/9 = 5/9 xx 16/16 = 80/144.
As 81/144 ≠ 80/144,
Therefore, 9/16 "is not equal to" 5/9.
Example
Find answers to the following. Write and indicate how you solved them.
Is 4/5 "equal to" 16/20?
4/5, 16/20
16/20 = (4 xx 4)/(5 xx 4) = 4/5.
Therefore, 4/5 = 16/20.
Example
Ila read 25 pages of a book containing 100 pages. Lalita read 2/5 of the same book. Who reads less?
Numbers of pages read by Lalita = 2/5 xx 100 = 40
Number of pages read by Ila = 25
Hence, Ila has read less number of pages.
Example
Asha and Samuel have bookshelves of the same size partly filled with books. Asha’s shelf is 5/6th full and Samuel’s shelf is 2/5th full. Whose bookshelf is more full? By what fraction?
Fraction of Asha’s shelf = 5/6
Fraction of Samuel’s shelf = 2/5
Converting these into like fractions,
5/6 = 5/6 xx 5/5 = 25/30.
2/5 = 2/5 xx 6/6 = 12/30
25/30 > 12/30.
Clearly, Asha’s bookshelf is more full.
Difference = 5/6 - 2/5 = 25/30 - 12/30 = 13/30.
Example
Jaidev takes 2 1/5 minutes to walk across the school ground. Rahul takes 7/4 minutes to do the same. Who takes less time and by what fraction?
Time taken by Jaidev = 2 1/5 "minutes" = 11/5 min
Time taken by Rahul = 7/4 min
Converting these into like fractions,
11/5 = 11/5 xx 4/4 = 44/20
7/4 = 7/4 xx 5/5 = 35/20
As 44 > 35,
11/5 > 7/4
Hence, Rahul takes lesser time.
Difference = 11/5 - 7/4
= 44/20 - 35/20 = 9/20 min.
If you would like to contribute notes or other learning material, please submit them using the button below.
Shaalaa.com
Comparison Of Like Fractions [00:09:30]
S
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2021-02-27 07:42:51
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https://stacks.math.columbia.edu/tag/01KR
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Lemma 26.21.9. Let $f : X \to T$ and $g : Y \to T$ be morphisms of schemes with the same target. Let $h : T \to S$ be a morphism of schemes. Then the induced morphism $i : X \times _ T Y \to X \times _ S Y$ is an immersion. If $T \to S$ is separated, then $i$ is a closed immersion. If $T \to S$ is quasi-separated, then $i$ is a quasi-compact morphism.
Proof. By general category theory the following diagram
$\xymatrix{ X \times _ T Y \ar[r] \ar[d] & X \times _ S Y \ar[d] \\ T \ar[r]^{\Delta _{T/S}} \ar[r] & T \times _ S T }$
is a fibre product diagram. The lemma follows from Lemmas 26.21.2, 26.17.6 and 26.19.3. $\square$
There are also:
• 16 comment(s) on Section 26.21: Separation axioms
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
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2022-08-09 04:51:03
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https://ltwork.net/what-is-1-challenge-and-1-adaptations-for-the-inca--3701586
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# What is 1 challenge and 1 adaptations for the Inca?
###### Question:
What is 1 challenge and 1 adaptations for the Inca?
$What is 1 challenge and 1 adaptations for the Inca?$
### 9. The risk-free rate and the expected market rate of return are 5.6% and 12.5%, respectively. According
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Which pair are solutions to the equation? 4xy + 8 = 36 a.(4, 9) and (3, 2) b.(1, 7) and (4, 9) c.(1, 7) and (7, 1) d.(7, 1) and (3, 2) i think it's c am i correct if not explain ^-^...
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2022-08-18 19:58:04
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https://www.physicsforums.com/threads/finding-area-under-cosx-using-midpoint-rule.561286/
|
Homework Help: Finding area under cosx using midpoint rule
1. Dec 18, 2011
pvpkillerx
Use the midpoint rule to approximate the following integral:
∫sin(x) dx
This is what I did:
Δx = (1-0)/4 = 1/4
1/4(f(1/8) + f(3/8) + f(5/8) + f(7/8) + f(9/8))
But the answer i get is wrong. Is that the correct midpoint rule formula, and are the values I plugged in right? Any help is appreciated, thanks.
2. Dec 18, 2011
LCKurtz
Are the limits on your integral 0 and 1?. How many points in your partition? Is 9/8 in your interval?
3. Dec 18, 2011
pvpkillerx
Ye, its from 1 to 0, and n = 4.
4. Dec 18, 2011
LCKurtz
5. Dec 18, 2011
pvpkillerx
I tried with and without 9/8, both don't give me the right answer.
6. Dec 18, 2011
LCKurtz
Why would you try it with 9/8 in the first place??? Unless you show us your work how can we help you find what you are doing wrong? Maybe something simple like having your calculator in degree mode instead radians? Show us your calculations.
7. Dec 18, 2011
pvpkillerx
ohh, i got the answer, u were right, the calculator was suppose to be in radians mode -_- oops. Thanks.
8. Dec 19, 2011
Staff: Mentor
9. Dec 19, 2011
Staff: Mentor
But title says: Finding area under cosx using midpoint rule
Better establish whether sine or cosine
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2018-12-13 04:58:01
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https://www.gamedev.net/forums/topic/692095-d3d-glm-depth-reconstruction-issues/
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# DX11 D3D + GLM Depth Reconstruction Issues
## Recommended Posts
I'm trying to port my engine to DirectX and I'm currently having issues with depth reconstruction. It works perfectly in OpenGL (even though I use a bit of an expensive method). Every part besides the depth reconstruction works so far. I use GLM because it's a good math library that has no need to install any dependencies or anything for the user. So basically I get my GLM matrices: struct DefferedUBO { glm::mat4 view; glm::mat4 invProj; glm::vec4 eyePos; glm::vec4 resolution; }; DefferedUBO deffUBOBuffer; // ... glm::mat4 projection = glm::perspective(engine.settings.fov, aspectRatio, 0.1f, 100.0f); // Get My Camera CTransform *transform = &engine.transformSystem.components[engine.entities[entityID].components[COMPONENT_TRANSFORM]]; // Get the View Matrix glm::mat4 view = glm::lookAt( transform->GetPosition(), transform->GetPosition() + transform->GetForward(), transform->GetUp() ); deffUBOBuffer.invProj = glm::inverse(projection); deffUBOBuffer.view = glm::inverse(view); if (engine.settings.graphicsLanguage == GRAPHICS_DIRECTX) { deffUBOBuffer.invProj = glm::transpose(deffUBOBuffer.invProj); deffUBOBuffer.view = glm::transpose(deffUBOBuffer.view); } // Abstracted so I can use OGL, DX, VK, or even Metal when I get around to it. deffUBO->UpdateUniformBuffer(&deffUBOBuffer); deffUBO->Bind()); Then in HLSL, I simply use the following: cbuffer MatrixInfoType { matrix invView; matrix invProj; float4 eyePos; float4 resolution; }; float4 ViewPosFromDepth(float depth, float2 TexCoord) { float z = depth; // * 2.0 - 1.0; float4 clipSpacePosition = float4(TexCoord * 2.0 - 1.0, z, 1.0); float4 viewSpacePosition = mul(invProj, clipSpacePosition); viewSpacePosition /= viewSpacePosition.w; return viewSpacePosition; } float3 WorldPosFromViewPos(float4 view) { float4 worldSpacePosition = mul(invView, view); return worldSpacePosition.xyz; } float3 WorldPosFromDepth(float depth, float2 TexCoord) { return WorldPosFromViewPos(ViewPosFromDepth(depth, TexCoord)); } // ... // Sample the hardware depth buffer. float depth = shaderTexture[3].Sample(SampleType[0], input.texCoord).r; float3 position = WorldPosFromDepth(depth, input.texCoord).rgb; Here's the result: This just looks like random colors multiplied with the depth. Ironically when I remove transposing, I get something closer to the truth, but not quite: You're looking at Crytek Sponza. As you can see, the green area moves and rotates with the bottom of the camera. I have no idea at all why. The correct version, along with Albedo, Specular, and Normals.
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GL's NDC (post projection) Z coordinates range from -1 to 1, but D3D's range from 0 to 1.
glm::perspective will create a GL style projection matrix. You need to concatenate this with a matrix that scales z by 0.5 and translates by 0.5 to make it valid for D3D.
In normal rendering, the effect of this bug will be quite small - your near plane appearing about twice as far forward as you intended... But it will mess with depth reconstruction too.
Btw there should be no need to transpose your matrices on D3D - both GLSL and HLSL store 2D arrays in column-major element ordering.
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Thanks for the quick response. Firstly, GL is column major whereas directx is row major. I've already had to transpose for my first geometry stage and it works well.
Second, will I need to change my first stage to accommodate this change as well? Also can I just multiply it by glm::translate(0,0,0.5)xProjection
EDIT: I've switched to row major vertices in DirectX using the following:
#pragma pack_matrix( row_major )
I guess DirectX just uses row major by default. I'm still having the same issues though. I tried using the following in ViewPosFromDepth:
float z = depth * 0.5 + 0.5;
Edited by KarimIO
Update
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1 hour ago, KarimIO said:
Thanks for the quick response. Firstly, GL is column major whereas directx is row major. I've already had to transpose for my first geometry stage and it works well.
Second, will I need to change my first stage to accommodate this change as well? Also can I just multiply it by glm::translate(0,0,0.5)xProjection
1- That's old info that hasn't applied since the fixed function graphics days. D3D/GL don't pick conventions for you. You can use any conventions on either API.
GLSL and HLSL both use column-major array indexing by default (but can be told to use row-major indexing such as with that pragma). Both can work with column-vector maths or row-vector maths (i.e. whether you write mul(vec,mat) or mul(mat,vec))
IIRC, GLM uses column-major storage and column-vector maths, and DirectXMath / D3DX use row-major storage and row-vector math... Which ironically results in them storing the exact same pattern of 64 bytes in RAM, but requires opposite multiplication order by the programmer
If you use the same math library, you can use the same shader code and matrix data on both APIs with no need to transpose anything.
2- yeah whenever you produce a projection matrix you have to scale in z by 0.5 and then translate in z by 0.5. You need to do this for your vertex shaders so that the rasterization is correct. When fetching from the depth buffer, don't scale/offset the fetched value.
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1 hour ago, Hodgman said:
yeah whenever you produce a projection matrix you have to scale in z by 0.5 and then translate in z by 0.5. You need to do this for your vertex shaders so that the rasterization is correct. When fetching from the depth buffer, don't scale/offset the fetched value
I've tried this, but now it's far too zoomed in. Originally, it did look quite like my OpenGL results. Is there a reason for this?
projection *= glm::translate(glm::vec3(0.0f,0.0f,0.5f)) * glm::scale(glm::vec3(1.0f,1.0f,0.5f));
Edited by KarimIO
Formatting
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If GLM uses column-vector maths, your multiplication order might be backwards there. Try:
projection = glm::translate(glm::vec3(0.0f,0.0f,0.5f)) * glm::scale(glm::vec3(1.0f,1.0f,0.5f)) * projection;
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Okay thank you a lot, Hodgman I finally got it to work! But I do have a question, my main vertex.hlsl which takes the actual geometry and pushes it into the gbuffer requires row-major whereas the rest works fine using column major. Do you have any idea why that could be?
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@Hodgman Got any idea for the question above?
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Do you use the exact same VS math in your HLSL and GLSL versions?
Post some VS shader code and we'll have a look
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@Hodgman Sorry! Didn't see the response until now! Keep in mind as I used GLM, it's column major. Here's the code that works:
////////////////////////////////////////////////////////////////////////////////
// Filename: mainVert.vs
////////////////////////////////////////////////////////////////////////////////
/////////////
// GLOBALS //
/////////////
#pragma pack_matrix( row_major )
cbuffer MatrixBuffer
{
matrix worldMatrix;
matrix viewMatrix;
matrix projectionMatrix;
};
//////////////
// TYPEDEFS //
//////////////
struct VertexInputType {
float3 position : POSITION;
float3 normal : NORMAL;
float3 tangent : TANGENT;
float2 texCoord : TEXCOORD0;
};
struct PixelInputType {
float4 position : SV_POSITION;
float3 worldPosition : POSITION;
float3 normal : NORMAL;
float3 tangent : TANGENT;
float2 texCoord : TEXCOORD0;
};
////////////////////////////////////////////////////////////////////////////////
////////////////////////////////////////////////////////////////////////////////
PixelInputType main(VertexInputType input) {
float4 position;
PixelInputType output;
// Change the position vector to be 4 units for proper matrix calculations.
position = float4(input.position, 1.0f);
// Calculate the position of the vertex against the world, view, and projection matrices.
position = mul(position, worldMatrix);
output.worldPosition = position.xyz;
position = mul(position, viewMatrix);
output.position = mul(position, projectionMatrix);
output.normal = normalize(mul(float4(input.normal, 1.0), worldMatrix).xyz);
output.tangent = normalize(mul(float4(input.tangent, 1.0), worldMatrix).xyz);
output.texCoord = float2(input.texCoord.x, -input.texCoord.y);
return output;
}
////////////////////////////////////////////////////////////////////////////////
// Filename: pointLightFrag.ps
////////////////////////////////////////////////////////////////////////////////
#pragma pack_matrix( column_major )
#include "inc_transform.hlsl"
#include "inc_light.hlsl"
//////////////
// TYPEDEFS //
//////////////
struct PixelInputType {
float4 position : SV_POSITION;
float2 texCoord : TEXCOORD0;
float3 viewRay : POSITION;
};
SamplerState SampleType[4];
cbuffer MatrixInfoType {
matrix invView;
matrix invProj;
float4 eyePos;
float4 resolution;
};
cbuffer Light {
float3 lightPosition;
float3 lightColor;
float lightIntensity;
};
////////////////////////////////////////////////////////////////////////////////
////////////////////////////////////////////////////////////////////////////////
float4 main(PixelInputType input) : SV_TARGET {
float3 Position = WorldPosFromDepth(invProj, invView, depth, input.texCoord);
//return float4(position, 1.0);
/*float near = 0.1;
float far = 100;
float ProjectionA = far / (far - near);
float ProjectionB = (-far * near) / (far - near);
depth = ProjectionB / ((depth - ProjectionA));
float4 position = float4(input.viewRay * depth, 1.0);*/
// Convert to World Space:
// position = mul(invView, position);
float3 lightPow = lightColor * lightIntensity;
float3 outColor = LightPointCalc(Albedo.rgb, Position.xyz, Specular, Normal.xyz, lightPosition, lightAttenuationRadius, lightPow, eyePos.xyz); // hdrGammaTransform()
return float4(hdrGammaTransform(outColor), 1.0f);
}
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So just to make things clear, there's two ways to do matrix math on paper - putting the basis vectors in the rows, treating vectors as rows, and doing left-to-right multiplication: $$\begin{bmatrix} Vx & Vy & Vz & 1 \end{bmatrix} \cdot \begin{bmatrix} Xx & Xy & Xz & 0\\ Yx & Yy & Yz & 0\\ Zx & Zy & Zz & 0\\ Px & Py & Pz & 1 \end{bmatrix}$$
Or putting basis vectors in the columns, treating the vectors as columns, and doing right-to-left multiplication: $$\begin{bmatrix} Xx & Yx & Zx & Px\\ Xy & Yy & Zy & Py\\ Yz & Yz & Zz & Pz\\ 0 & 0 & 0 & 1 \end{bmatrix} \cdot \begin{bmatrix} Vx \\ Vy \\ Vz \\ 1 \end{bmatrix}$$
Then a completely separate issue is how you decide to store 2D arrays in linear memory. Row-major: $$\begin{bmatrix} 0&1&2&3\\ 4&5&6&7\\ 8&9&10&11\\ 12&13&14&15 \end{bmatrix}$$
Or column-major: $$\begin{bmatrix} 0&4&8&12\\ 1&5&9&13\\ 2&6&10&14\\ 3&7&11&15 \end{bmatrix}$$
That results in four different conventions for doing matrix math in a computer (row-major/column-major array indexing, and row-vector/column-vector math).
GLM uses column-vector math and column-major array indexing.
In the HLSL code that you posted, your math is written assuming row-vector math (left to right multiplication ordering), which is the opposite convention to what GLM uses. Your HLSL code also expects column-major array ordered data.
If, on the CPU side before the shader runs, you rearrange your data from column-major to row-major array ordering, then HLSL is going to accidentally interpret your data wrong -- it will read rows as columns and columns as rows... which has the same effect as doing a mathematical transpose operation, which cancels out the fact that you're using the opposite mathematical conventions.
i.e. your mathematical conventions in the vertex shader are the opposite of what GLM expects, but by also using the opposite array storage conventions, these cancel each other out and two wrongs make a right, so it works.
You should be able to get rid of all your transposing and just rewrite the VS to use right-to-left multiplication order,
e.g. position = mul(worldMatrix, position);
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On 9/23/2017 at 5:45 AM, Hodgman said:
If, on the CPU side before the shader runs, you rearrange your data from column-major to row-major array ordering, then HLSL is going to accidentally interpret your data wrong -- it will read rows as columns and columns as rows... which has the same effect as doing a mathematical transpose operation, which cancels out the fact that you're using the opposite mathematical conventions.
The problem is, I'm doing no such thing. GLM outputs the same column-major matrices, but for some reason one shader requires row major and the other column major. I understand everything you're talking about, it's just DirectX likes it one way in one shader and another way in another shader.
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6 hours ago, KarimIO said:
one shader requires row major and the other column major.
What do you mean by this exactly? How are you switching between those formats?
You don't rearrange the data at all, you've just found that you have to do the math backwards in the VS code?
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On 9/28/2017 at 1:56 AM, Hodgman said:
What do you mean by this exactly? How are you switching between those formats?
You don't rearrange the data at all, you've just found that you have to do the math backwards in the VS code?
In the main prepass vertex shader, I need to use this:
#pragma pack_matrix( row_major )
Multiplication should be column major by default because I use GLM. Yet, for one specific file, I need to use row_major.
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• ### Similar Content
• I'm attempting to implement some basic post-processing in my "engine" and the HLSL part of the Compute Shader and such I think I've understood, however I'm at a loss at how to actually get/use it's output for rendering to the screen.
Assume I'm doing something to a UAV in my CS:
RWTexture2D<float4> InputOutputMap : register(u0); I want that texture to essentially "be" the backbuffer.
I'm pretty certain I'm doing something wrong when I create the views (what I think I'm doing is having the backbuffer be bound as render target aswell as UAV and then using it in my CS):
DXGI_SWAP_CHAIN_DESC scd; ZeroMemory(&scd, sizeof(DXGI_SWAP_CHAIN_DESC)); scd.BufferCount = 1; scd.BufferDesc.Format = DXGI_FORMAT_R8G8B8A8_UNORM; scd.BufferUsage = DXGI_USAGE_RENDER_TARGET_OUTPUT | DXGI_USAGE_SHADER_INPUT | DXGI_USAGE_UNORDERED_ACCESS; scd.OutputWindow = wndHandle; scd.SampleDesc.Count = 1; scd.Windowed = TRUE; HRESULT hr = D3D11CreateDeviceAndSwapChain(NULL, D3D_DRIVER_TYPE_HARDWARE, NULL, NULL, NULL, NULL, D3D11_SDK_VERSION, &scd, &gSwapChain, &gDevice, NULL, &gDeviceContext); // get the address of the back buffer ID3D11Texture2D* pBackBuffer = nullptr; gSwapChain->GetBuffer(0, __uuidof(ID3D11Texture2D), (LPVOID*)&pBackBuffer); // use the back buffer address to create the render target gDevice->CreateRenderTargetView(pBackBuffer, NULL, &gBackbufferRTV); // set the render target as the back buffer CreateDepthStencilBuffer(); gDeviceContext->OMSetRenderTargets(1, &gBackbufferRTV, depthStencilView); //UAV for compute shader D3D11_UNORDERED_ACCESS_VIEW_DESC uavd; ZeroMemory(&uavd, sizeof(uavd)); uavd.Format = DXGI_FORMAT_R8G8B8A8_UNORM; uavd.ViewDimension = D3D11_UAV_DIMENSION_TEXTURE2D; uavd.Texture2D.MipSlice = 1; gDevice->CreateUnorderedAccessView(pBackBuffer, &uavd, &gUAV); pBackBuffer->Release();
After I render the scene, I dispatch like this:
gDeviceContext->OMSetRenderTargets(0, NULL, NULL); m_vShaders["cs1"]->Bind(); gDeviceContext->CSSetUnorderedAccessViews(0, 1, &gUAV, 0); gDeviceContext->Dispatch(32, 24, 0); //hard coded ID3D11UnorderedAccessView* nullview = { nullptr }; gDeviceContext->CSSetUnorderedAccessViews(0, 1, &nullview, 0); gDeviceContext->OMSetRenderTargets(1, &gBackbufferRTV, depthStencilView); gSwapChain->Present(0, 0); Worth noting is the scene is rendered as usual, but I dont get any results from the CS (simple gaussian blur)
I'm sure it's something fairly basic I'm doing wrong, perhaps my understanding of render targets / views / what have you is just completely wrong and my approach just makes no sense.
If someone with more experience could point me in the right direction I would really appreciate it!
On a side note, I'd really like to learn more about this kind of stuff. I can really see the potential of the CS aswell as rendering to textures and using them for whatever in the engine so I would love it if you know some good resources I can read about this!
Thank you <3
P.S I excluded the .hlsl since I cant imagine that being the issue, but if you think you need it to help me just ask
P:P:S. As you can see this is my first post however I do have another account, but I can't log in with it because gamedev.net just keeps asking me to accept terms and then logs me out when I do over and over
• I was wondering if anyone could explain the depth buffer and the depth stencil state comparison function to me as I'm a little confused
So I have set up a depth stencil state where the DepthFunc is set to D3D11_COMPARISON_LESS, but what am I actually comparing here? What is actually written to the buffer, the pixel that should show up in the front?
I have these 2 quad faces, a Red Face and a Blue Face. The Blue Face is further away from the Viewer with a Z index value of -100.0f. Where the Red Face is close to the Viewer with a Z index value of 0.0f.
When DepthFunc is set to D3D11_COMPARISON_LESS the Red Face shows up in front of the Blue Face like it should based on the Z index values. BUT if I change the DepthFunc to D3D11_COMPARISON_LESS_EQUAL the Blue Face shows in front of the Red Face. Which does not make sense to me, I would think that when the function is set to D3D11_COMPARISON_LESS_EQUAL the Red Face would still show up in front of the Blue Face as the Z index for the Red Face is still closer to the viewer
Am I thinking of this comparison function all wrong?
Vertex data just in case
//Vertex date that make up the 2 faces Vertex verts[] = { //Red face Vertex(Vector4(0.0f, 0.0f, 0.0f), Color(1.0f, 0.0f, 0.0f)), Vertex(Vector4(100.0f, 100.0f, 0.0f), Color(1.0f, 0.0f, 0.0f)), Vertex(Vector4(100.0f, 0.0f, 0.0f), Color(1.0f, 0.0f, 0.0f)), Vertex(Vector4(0.0f, 0.0f, 0.0f), Color(1.0f, 0.0f, 0.0f)), Vertex(Vector4(0.0f, 100.0f, 0.0f), Color(1.0f, 0.0f, 0.0f)), Vertex(Vector4(100.0f, 100.0f, 0.0f), Color(1.0f, 0.0f, 0.0f)), //Blue face Vertex(Vector4(0.0f, 0.0f, -100.0f), Color(0.0f, 0.0f, 1.0f)), Vertex(Vector4(100.0f, 100.0f, -100.0f), Color(0.0f, 0.0f, 1.0f)), Vertex(Vector4(100.0f, 0.0f, -100.0f), Color(0.0f, 0.0f, 1.0f)), Vertex(Vector4(0.0f, 0.0f, -100.0f), Color(0.0f, 0.0f, 1.0f)), Vertex(Vector4(0.0f, 100.0f, -100.0f), Color(0.0f, 0.0f, 1.0f)), Vertex(Vector4(100.0f, 100.0f, -100.0f), Color(0.0f, 0.0f, 1.0f)), };
• By mellinoe
Hi all,
First time poster here, although I've been reading posts here for quite a while. This place has been invaluable for learning graphics programming -- thanks for a great resource!
Right now, I'm working on a graphics abstraction layer for .NET which supports D3D11, Vulkan, and OpenGL at the moment. I have implemented most of my planned features already, and things are working well. Some remaining features that I am planning are Compute Shaders, and some flavor of read-write shader resources. At the moment, my shaders can just get simple read-only access to a uniform (or constant) buffer, a texture, or a sampler. Unfortunately, I'm having a tough time grasping the distinctions between all of the different kinds of read-write resources that are available. In D3D alone, there seem to be 5 or 6 different kinds of resources with similar but different characteristics. On top of that, I get the impression that some of them are more or less "obsoleted" by the newer kinds, and don't have much of a place in modern code. There seem to be a few pivots:
The data source/destination (buffer or texture) Read-write or read-only Structured or unstructured (?) Ordered vs unordered (?) These are just my observations based on a lot of MSDN and OpenGL doc reading. For my library, I'm not interested in exposing every possibility to the user -- just trying to find a good "middle-ground" that can be represented cleanly across API's which is good enough for common scenarios.
Can anyone give a sort of "overview" of the different options, and perhaps compare/contrast the concepts between Direct3D, OpenGL, and Vulkan? I'd also be very interested in hearing how other folks have abstracted these concepts in their libraries.
• If I do a buffer update with MAP_NO_OVERWRITE or MAP_DISCARD, can I just write to the buffer after I called Unmap() on the buffer? It seems to work fine for me (Nvidia driver), but is it actually legal to do so? I have a graphics device wrapper and I don't want to expose Map/Unmap, but just have a function like void* AllocateFromRingBuffer(GPUBuffer* buffer, uint size, uint& offset); This function would just call Map on the buffer, then Unmap immediately and then return the address of the buffer. It usually does a MAP_NO_OVERWRITE, but sometimes it is a WRITE_DISCARD (when the buffer wraps around). Previously I have been using it so that the function expected the data upfront and would copy to the buffer between Map/Unmap, but now I want to extend functionality of it so that it would just return an address to write to.
• Trying to write a multitexturing shader in DirectX11 - 3 textures work fine, but adding 4th gets sampled as black!
Could you please look at the textureClass.cpp line 79? - I'm guess its D3D11_TEXTURE2D_DESC settings are wrong,
but no idea how to set it up right. I tried changing ArraySize from 1 to 4, but does nothing. If thats not the issue, please look
at the LightShader_ps - maybe doing something wrong there? Otherwise, no idea.
// Setup the description of the texture.
textureDesc.Height = height;
textureDesc.Width = width;
textureDesc.MipLevels = 0;
textureDesc.ArraySize = 1;
textureDesc.Format = DXGI_FORMAT_R8G8B8A8_UNORM;
textureDesc.SampleDesc.Count = 1;
textureDesc.SampleDesc.Quality = 0;
textureDesc.Usage = D3D11_USAGE_DEFAULT;
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2017-11-19 01:00:05
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http://mathhelpforum.com/advanced-algebra/62571-how-do-arithmetic-operations-galois-field.html
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# Math Help - how to do arithmetic operations in galois field
1. ## how to do arithmetic operations in galois field
is primitive in GF(2)[x].
Let be a root of .
α2 means 0 1 0 0
α5 means 0 1 1 1
- ----------------
when we add the both 1 0 1 1
but in galois field answer is α3 means that 0 0 1 1 from the table
Multiplication:
in multiplication when we multiply two 4 bit numbers we got 8 bit number
in galois field we got 4 bit number only
Because $\alpha$ is a root of $x^3+x+1$.
Therefore $\alpha^3 + \alpha + 1 = 0 \implies \alpha^3 = - \alpha - 1$.
But this field has charachteristic two i.e. $x=-x$.
Thus, $\alpha^3 = -\alpha - 1 = \alpha + 1$.
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2014-09-24 06:14:51
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https://codegolf.stackexchange.com/questions/124362/how-high-can-you-count/124657
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# How high can you count?
## Challenge:
Your task is to write as many programs / functions / snippets as you can, where each one outputs / prints / returns an integer. The first program must output the integer 1, the second one 2 and so on.
You can not reuse any characters between the programs. So, if the first program is: x==x, then you may not use the characters x and = again in any of the other programs. Note: It's allowed to use the same character many times in one program.
### Scoring:
The winner will be the submission that counts the highest. In case there's a tie, the winner will be the submission that used the fewest number of bytes in total.
### Rules:
• You can only use a single language for all integers
• Snippets are allowed!
• To keep it fair, all characters must be encoded using a single byte in the language you choose.
• The output must be in decimal. You may not output it with scientific notation or some other alternative format. Outputting floats is OK, as long as all digits that are shown behind the decimal point are 0. So, 4.000 is accepted. Inaccuracies due to FPA is accepted, as long as it's not shown in the output.
• ans =, leading and trailing spaces and newlines etc. are allowed.
• You may disregard STDERR, as long as the correct output is returned to STDOUT
• You may choose to output the integer to STDERR, but only if STDOUT is empty.
• Symbol independent languages (such as Lenguage) are disallowed
• Letters are case sensitive a != A.
• The programs must be independent
• Whitespace can't be reused
• You must use ASCII-digits in the output
Explanations are encouraged!
• The language Headsecks only cares about the lower three bits of every character and would trivially achieve a score of 64. It's partially symbol-independent, but not completely. I think the last rule should cover partially symbol-independent languages as well, but I'm not sure how to phrase it. – Dennis Jun 4 '17 at 22:45
• Regarding the snippet rule, do we still need to include usings/imports? And are static imports allowed (without making them part of the snippet that is)? – Kevin Cruijssen Jun 6 '17 at 17:45
• @KevinCruijssen you can omit boilerplate stuff that's needed for every programs/functions. For instance, you don't need #include <iostream> and other boilerplate stuff in C++. You do need from numpy import *. Note: I'm not a programmer, so I don't know all the nuances. We can discuss in chat if something is unclear :) – Stewie Griffin Jun 6 '17 at 18:05
• You have the right to vote however you like @tuskiomi, but in my opinion it's a good rule. Whitespace characters are just bytes, just as any other character. Why should they be treated differently? Also, the language Whitespace would win by a landslide, since it contains only space, tab and line shift. Thanks for saying why you downvoted though :-) – Stewie Griffin Aug 18 '17 at 17:31
• @StewieGriffin I would at least allow spaces, but hey, I'm not you. – tuskiomi Aug 18 '17 at 18:30
# Braingolf, score 14, 81 bytes
1- ^
2- 2
3- 3
4- 4
5- 5
6- 6
7- 7
8- 8
9- 9
10- "
"
11- llllllllllll
12- VR# d<Mv/_R_
13- 111-1-1-1-1-1-1-1-1-1-1-1-1--
14- H.............&+
Characters used: 123456789^"e\nJ-+lVR# d<Mv/_&
Things to note that might be useful:
P can also be used at the start of a program to push a 1, as an empty stack is considered palindromic (does the same thing as H)
Main issue with this is + has already been used, and I can't think of any other way in Braingolf to turn any amount of 1s into anything higher than 1
## Explanation:
1
^
^ niladic exponentiation, 0^0.
This is undefined mathematically, however python, in which
Braingolf's interpreter is written, returns 1
implicit output of last item on stack
2-9
2 push integer literal
implicit output of last item on stack
10
"\n" push charcode of newline, 10
implicit output of last item on stack
11
llllllllllll
l push length of stack (0)
l push length of stack (1)
.........l push length of stack (11)
implicit output of last item on stack
12
VR# d<Mv/_R_
VR create stack2 then return to stack1
#<space> push 32
d split 32 into 3, 2
< move first item to end of stack
M move last item to stack2
v switch to stack2
/ monadic division, always returns 1
_ print last item on stack (1)
_ print last item on stack (2)
13
111-1-1-1-1-1-1-1-1-1-1-1-1--
111 Push 3 1s [1, 1, 1]
- Subtract last 2 items [1, 0]
1- Push 1 and subtract [1, -1]
1-1-1-1-1-1-1-1-1-1-1- Push 11 more 1s and subtract each one [1, -12]
- Subtract [13]
implicit output of last item on stack
14
H.............&+
H Push 1, empty stack is always palindromic
............. Duplicate the 1 13 times, stack now has 14 1s
&+ Sum entire stack
implicit output of last item on stack
• Using two digits in the same entry seems like a waste. What about doing 10 as "<newline>", 9 as 9, and 13 as 1111111111111++++++++++++? – Ørjan Johansen Jun 6 '17 at 21:01
• Or alternatively 111-1-1-1-1-1-1-1-1-1-1-1-1-1-1-- if you have more use for the + elsewhere. – Ørjan Johansen Jun 6 '17 at 21:08
• @ØrjanJohansen Oh wow, thanks! – Skidsdev Jun 7 '17 at 8:17
# Forth, score 10111213 14, 196 bytes
Tokens are case-insensitive, which helps a bunch. Also, TIO allows a bunch of different whitespace characters as token separators.
1
j
j
j
<
>
abs
2 - 14
J J = J J = - J J = - J J = -
3
4
5
6
7
8
9
0xA
11
2 2 2 2 2 2 + + + + +
I I / I I / LSHIFT I I / LSHIFT I I / LSHIFT I I / I I / LSHIFT I I / LSHIFT I I / OR OR
k k k k k k k k k k k k k k depth
Try it online - slightly weird/inconvenient. And #14 uses stack depth, so it must occur after printing the others.
### Explanation:
1. Use some comparisons to obtain True, which evaluates to -1, then take the absolute value. See #2 for a description of j. This uses linefeeds to separate the tokens.
1. J is a loop counter, used for inner loops, and is also some large number when not inside a user-defined loop. Checking if it equals itself returns -1 (True). Using subtraction, I obtain 2. This uses tabs to separate the tokens.
1. This uses vertical tabs
(ASCII 11) to separate the tokens. See the TIO link.
1. I pushes the loop counter, which is some large unknown number when outside a loop. Divide it by itself to get 1. I use bit-shifting with lshift and OR.
1. Push k, another loop counter, a bunch of times, then get the stack depth. This uses form feeds (ASCII 12) to separate the tokens. Select the line to see there are actually characters there. Using shift + arrow keys while having the line selected will show that there is an invisible character after each k.
I can lower the byte count by using more unique bytes (like *), but for the moment I'm going to use as few unique bytes as possible.
• Can you use something like variable v v @ v @ = negate for 1, freeing 1 up for 11? – Neil Jun 7 '17 at 10:05
• @Neil I found some better options. I don't know if there are enough whitespace characters left to use as token separators. – mbomb007 Jun 7 '17 at 16:54
• Great! I like to think I got you thinking anyway, even if you didn't use my specific suggestion. – Neil Jun 7 '17 at 19:51
• @Neil I did use it in an older revision, but it wasn't long before I changed to using I I /. – mbomb007 Jun 7 '17 at 21:48
## Jelly, score 13, 36 bytes
1
2
3
4
5
6
7
8
9
10 : ⁷ŒṘLRS
This gets the literal '\n', gets the length (4, incl quotes), generates a list ([1, 2, 3, 4]) and sums all elements.
11 : ⁴ḤÆC
This doubles (Ḥ) 16 (⁴), then counts all primes less than that (which is 11).
12 : ³DIAU‘V
This takes 100 (³), gets the absolute increments between digits (DIA, results in [1, 0]), reverses the list, increments all elements and makes a string out of it with V (technically, it evals the list [1, 2] and feeding Jelly numbers just makes it print 'm as literals).
13 : ⁵+⁵’’’’’’’
This just is 10+10-7.
Try it online
# Java, 1 :(
public class P{ static{ System.out.println("1"); }}
• Welcome to PPCG! Unless most other challenges, this challenge allows for snippets instead of full programs, so I think you can improve on the score. – Laikoni Jun 4 '17 at 16:37
# dc, score 20, 102 bytes
I'm assuming since functions/snippets count and returned values count that leaving the value on the stack counts. Otherwise, since dc has no implicit print and four distinct print commands, the max would be 4. We'll expect a clear stack for each of these, though I guess it only matters for one. Several of these will pollute the stack; top of stack is the value.
1
OOOOO++++a
3
4
5
6
7
8
9
A
B
C
D
E
F
2dd^^
KKKKKKKKKKKKKKKKKz
.000000000000000000X
[,,,,,,,,,,,,,,,,,,,]Z
II-II-I-I--
1 and 3-9 should be obvious. Likewise A-F work for 10-15 regardless of input base. 50 is the code point for ASCII 2, so for 2 we add our output base (10) to itself until we get up to 50 and convert it with a. For 16, we do (2^2)^2. For 17, we use 17 Ks to put the default precision (0, but this number is irrelevant) on the stack and then count the stack depth (z). For 18, we use X to count the fractional digits in .000000000000000000. 19 is a string made up of 19 commas, and then Z returns the length of a string. Finally we do 20 by using I to get our input base (10), and essentially doing (10-10)-(((10-10)-10)-10) or 0-_20.
# Pyt, score 18, 91 bytes
1: ⅕¬č
2: 2
3: 3
4: 4
5: 5
6: 6
7: ⅐π⁷∜Ř↑
8: 11+1+1+1+1+1+1+
9: φᵱ~˜²Á
10: 7⬡₉
11: é⁴ƖḋƩ
12: ½⅙*⅟
13: ⅓⅑÷řΠḞ
14: ¾⅛/Ŀ⁻⁻⁻⁻
15: 0!⁺⁺⁺⁺△
16: ⅝⅝-⅝-⌊Åᴇ9√Ṗ⇹-¼⅒≥-
17: ɳ₫Ħ
18: 8⬠₅
# Explanation
1: ⅕¬č
Pushes 1/5
Takes the logical inverse of 1/5 (False)
Takes the cosine of that (1) [cos(False cast as an integer)=cos(0)=1]
2: 2
Pushes 2
3: 3
Pushes 3
4: 4
Pushes 4
5: 5
Pushes 5
6: 6
Pushes 6
7: ⅐π⁷∜Ř↑
Pushes 1/7
Pushes π
Raises π to the 7th power
Takes the fourth root of π^7
Pushes [int(1/7),...,int(π^(7/4))] (i.e., [0,1,...,7])
Returns the maximum of that array (7)
8: 11+1+1+1+1+1+1+
Adds eight ones together
9: φᵱ~˜²Á
Pushes φ (the golden ratio - (1+√5)/2=1.618...)
Gets the partition number of 1 (φ cast to an integer)
Negates [1]
Gets the unsigned two's-complement** of [-1] (3) [Due to a mistake in how I implemented this, (-1)˜ returns 3]
Squares [3]
Pushes elements of array onto stack (i.e., pushes 9)
10: 7⬡₉
Pushes 7
Pushes the 7th hexagonal number (92)
Divides by 9 (10) (Python 2-style integer division (i.e., 92/9=10))
11: é⁴ƖḋƩ
Pushes e
Raises e to the 4th power
Casts to an integer (54)
Gets the prime factorization of 54 ([2,3,3,3])
Sums the list (13)
12: ½⅙*⅟
Pushes 1/2
Pushes 1/6
Multiplies (1/12)
Takes the multiplicative inverse of 1/12 (12)
13: ⅓⅑÷řΠḞ
Pushes 1/3
Pushes 1/9
Takes the ceiling of 1/9 (1)
Calculates int(1/(1/3)) (3)
Pushes [1,2,3]
Takes the product of the elements in the list (6)
Returns the 6th Fibonacci number (13) (with F(0)=F(1)=1)
14: ¾⅛/Ŀ⁻⁻⁻⁻
Pushes 3/4
Pushes 1/8
Divides 3/4 by 1/8 (6)
Returns the 6th Lucas number (18)
Decrements 4 times (14)
15: 0!⁺⁺⁺⁺△
Pushes 0! (1)
Increments 4 times (5)
Returns the 5th triangle number (15)
16: ⅝⅝-⅝-⌊Åᴇ9√Ṗ⇹-¼⅕≥-
Pushes 5/8 twice
Subtracts 5/8 from 5/8 (0)
Pushes 5/8
Subtracts 5/8 from 0 (-5/8)
Floors -5/8 (-1)
Takes the absolute value (1)
Calculates 10^1 (10)
Pushes 9
Takes the square root (3)
Raises 3 to itself (3^3=27)
(The stack is now [10,27])
Flips the stack ([27,10])
Subtracts 10 from 27 (17)
Pushes 1/4
Pushes 1/10
Pushes 1/4>1/10 (True)
Subtracts 1 (True cast to an integer) from 17
17: ɳ₫Ħ
Pushes "0123456789"
Interprets as an integer and flips the digits (987654321)
Calculates the Hamming weight of 987654321
18: 8⬠₅
Pushes 8
Pushes the 8th hexagonal number (92)
Divides by 5 (again, Python 2-style integer division (92/5=18))
Try them all online!
• I find ⅕ in both 1 and 16, ² in both 9 and 15, and Ʃ in both 11 and 16. – Ørjan Johansen Jan 21 '18 at 2:09
• Lemme go fix those – mudkip201 Jan 21 '18 at 2:25
• Okay, fixed. Now there shouldn't be any duplicates. – mudkip201 Jan 21 '18 at 2:36
• Does Pyt use any non-accented letters? I don't think I've ever seen one used in a Pyt program before... – caird coinheringaahing Apr 11 '18 at 19:49
• By design, it doesn't. – mudkip201 Apr 16 '18 at 16:21
# Perl 6, 26 29 numbers, 300 bytes
!@/!@
τ÷π
³
4
^?'~'~^''
6
7
8
9
𐭜
11
[+[]]+[+[]]+[+[]]+[+[]]+[+[]]+[+[]]+[+[]]+[+[]]+[+[]]+[+[]]+[+[]]+[+[]]
"{3%%3%3}3"
Int((),(),(),(),(),(),(),(),(),(),(),(),(),())
0xF
৹
ᛮ
⑱
꘡꘩
- --$_- --$_- --$_- --$_- --$_-$_
Q<Q<CiRCLED NUMBER TWENTY ONE>.UNiPARSE>.lc.EVAL.EVAL
22
٢٣
㉔
5*5
۲۶
２７
߂߈
ord
q
Try it online!
A lot of this is made easier by the fact that Perl 6 supports numbers being represented by Unicode. This allows us to do things like substituting the digits of 14 for the technically equivalent digits ۱۴, as long as we don't have byte conflicts in the multi-byte characters.The tradeoff is that all the bitwise operators have to be prefixed by +, which narrows down what we can do.
## Explanation:
### 1
!@/!@
Divide the logical not of two anonymous empty state variables to get 1
### 2
τ÷π
Divide tau by pi. Since tau is 2*pi, this equals 2.
### 3
³
The unicode character SUPERSCRIPT THREE, which evaluates to 3 when not after another number.
### 4
4
Just a plain ol' 4.
### 5
^?'~'~^''
The unary ? operator coerces '~' to a string, which is then converted to the range [0,1) by ^ and then coerced to the string 0 before being XOR'ed with '' (5) and converted back to a string.
### 6 through 9
All of these are normal ASCII characters
### 10
𐭜
This is the character meaning 10 in Inscriptional Parthian.
### 11
11
Two 1s concatenated together.
### 12
[+[]]+[+[]]+[+[]]+[+[]]+[+[]]+[+[]]+[+[]]+[+[]]+[+[]]+[+[]]+[+[]]+[+[]]
Coercing a list to an int results in the length of the list. +[] == 0 and then +[+[]] == 1. This then adds them together to result in 12
### 13
"{3%%3%3}3"
Checks if 3 is divisible by 3, then mods that with 3 to coerce the True to a 1 and concatenates it with another 3.
### 14
Int((),(),(),(),(),(),(),(),(),(),(),(),(),())
Same as number 12, where coercing a list to an int results in the length of the list.
### 15
0xF
A hexadecimal literal for 15.
### 16
৹
BENGALI CURRENCY DENOMINATOR SIXTEEN.
### 17
ᛮ
RUNIC ARLAUG SYMBOL. Interesting note, this is actually a numeric letter, as opposed to the numeric digits and numeric other I've used so far.
### 18
⒅
PARENTHESIZED NUMBER EIGHTEEN. Actually recognisable as a number for once.
### 19
꘡꘩
VAI DIGIT's ONE and NINE, making 19.
### 20
- --$_- --$_- --$_- --$_- --$_-$_
Some arithemetic on the topic variable, which is initially set to Nil. These are separated by tabs. This evaluates to 20. (Note that I could save the character _ here, but make it a lot longer)
### 21
Q<Q<CiRCLED NUMBER TWENTY ONE>.UNiPARSE>.lc.EVAL.EVAL
I'm quite proud of this one. The uniparse function can take a string and find the unicode character called that, which is neat. However, it conflicts with our earlier calls to Int (or the alternatives Rat or Numeric). So we wrap it in a string and set the whole thing to lowercase before evaling it. We don't need a single digit number, so we get an appropriate Unicode character and eval it again which yields the 21.
I could have also used the \c[] construct in a string, but I'm using the []s elsewhere.
### 22
22
An integer literal
### 23
٢٣
ARABIC-INDIC DIGITs TWO and THREE making 23.
### 24
㉔
CIRCLED NUMBER TWENTY FOUR
### 25
5*5
Simple multiplication.
### 26
۲۶
ARABIC-INDIC DIGITs TWO and SIX. What do you mean I've already used these? No-no, these are the EXTENDED versions see? Completely different (at least, byte-wise).
### 27
２７
FULLWIDTH DIGIT's TWO and SEVEN
### 28
߂߈
NKO DIGIT's TWO and EIGHT
### 29
ord
q
Return the value of the byte, which is 29. The function doesn't care about the type of whitespace, so I've used a newline here.
## Possible further improvements
I can still use various whitespace characters to separate function from argument, but I can't find any that result in a number and don't conflict with all the letters I've already used (e.g. chars, codes, index, encode.bytes).
I've also used up pretty much all the usable non-conflicting bytes between numeric Unicode characters that Perl 6 recognises.
If I could figure out how to get the code for 5 (^?'~'~^'') to do multiple digit numbers, I could free up * for something. I also have enough characters to declare a variable, though I don't know what I'd do with it.
Leftover printable characters:
&:;=\|#GHJKXZabefghjkmpsuvwyz
• "To keep it fair, all characters must be encoded using a single byte in the language you choose." – Ørjan Johansen Jul 6 '18 at 3:44
• @ØrjanJohansen I guess I could changr to a different encoding and post all the bytes individually for each unicode character? It wouldn't change much and make parts of it unreadable... – Jo King Jul 6 '18 at 3:52
• Well I think it would have been simpler to say bytes couldn't be reused, but I didn't make the challenge, so maybe ask OP. – Ørjan Johansen Jul 6 '18 at 3:57
# Befunge-98 (PyFunge), score 20, 89 bytes
!
2
3
4
5
6
7
8
9
a
b
c
d
e
f
y\-.@
11111111111111111++++++++++++++++
'\x12
"\x13"
0gg0g###########################\x14
All are snippets except for 16, which is a full program. Some control characters are present, I've written them in the form \x?? here.
## Explanations
1. !
Pop an implicit zero off the empty stack and perform a logical negation to get 1.
2. - 15.
Numeric literals
16. y\-.@
Full program. y is the "Get SysInfo" instruction. The first number dictates whether or not some of the Befunge-98 features have been implemented (details) - by default in PyFunge this is 15. The second number is the cell size in bytes, for which PyFunge returns -1, as it uses arbitrary precision integers. Swapping those numbers and subtracting them gives 16. .@ = output as number then exit.
17. 11111111111111111++++++++++++++++
A mysterious one.
18. '\x12
Pushes the code of character 18 onto the stack.
19. "\x13"
Pushes the string onto the stack. Strings actually get pushed backwards - do not be alarmed!
20. 0gg0g###########################\x14
g gets a character code by coordinates. The first g gets the character at (0, 0), which is 48, the next gets the character at (0, 48), which is 32 (by default, the plane is filled with spaces), and the third gets the character at (32, 0), which is the control character 20.
## Unused characters
These are the remaining instructions, none of which appear very useful for creating numbers:
Control flow: <>^v? []_|@;hjklmtqrxz
Input/output: &~.,io
Arithmetic: */%w
Stack: $:un{} Other: ()=spABCDEFGIHJKLMNOPQRSTUVWXYZ Unrecognized instructions reflect the IP (yet another control flow operation). # MarioLANG, score 2, 3+925 bytes It's-a-me, Mario. 1 : + : I can count 1 when I see a +, adding 1 to cell0 (which contains 0 by default) and : prints the content of current cell as a number to STDOUT. Then I fall to death. Mamma mia! Kill me online! 2 : - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - . With my second life, I can count 2 by -ing on cell0 which goes to 255... then fall to death again :( viewing many many -s during the fall... down to char 2, then . prints the content of current cell to STDOUT as a char. Kill me again! Since this language has only two output methods, I don't think there is any way to count up to 3... Mario needs to improve. • (If I am right,) snippets are allowed, and you only need to return an integer. The output method does not matter. – A _ Sep 14 at 14:48 # Turtlèd, score 4, 11 bytes (I got it lower with snippets but then it made the task trivial(er) (answers just being digits) so I rolled it back) '1 "2 @3, #4#. commands explanation: ' - writes to the grid the next character " - writes to the grid all the characters to the next ", or end of program @ - sets the char var to the next character in program , - write the char var on the grid #foo# - set the string var to foo . - write the pointed char of the string var. initially the first char Test them here # Aceto, score 12141618 19 The multi-character-snippets are given here in linear form for space reasons. pi is more than 0; cast to float: Pwf Just push the corresponding number: 2 3 4 5 6 7 8 9 Convert the implicit 0 to a character (\0x0), then to a boolean (True), then push it on the stack to the right. Move to the left stack and push a zero back on the main stack (and go there). Go to the right stack, and turn on sticky mode, then move the True back to the main stack twice (and go there). Multiply True with True (1), then implode the string. Because this is quite big, I show it here in the "real" 2D form: )k*£ ]{[ (} cb Push 1 twice, then implode: 11¥ Negate the implicit zero (True), push another zero and negate it (True, True), bitwise shift (2), push another 0 and negate it (2, True), cast to integer (2, 1), join top two values as strings ("12"), cast to integer (12): !0!«0!iJi Invert the zero (-1), negate it (1), duplicate it twice (1, 1, 1), sum up (1, 2), duplicate (1, 2, 2), sum up (1, 4), duplicate (1, 4, 4), sum up (1, 8), swap (8, 1), duplicate four times (8, 1, 1, 1, 1, 1), sum up all (13): a~dd+d+d+sdddd+++++ Push a Shift-Out character literal, convert to its codepoint (^N stands for the character at ASCII codepoint 14): '^No Push a random number 15 times, then push the stack length: RRRRRRRRRRRRRRRl Push e 3 times and integer-divide twice. Do this three times. Then do exponentiation twice: eee//eee//eee//FF Decrement 17 times, then absolute value: DDDDDDDDDDDDDDDDD± Equality of zero and zero. Memorize this. Load it 18 times. Continue subtracting and queueing (move a value from the bottom of the stack to the top), until we arrive at negative 18, memorize this, load and subtract, load and subtract. Because this is quite big, I show it here in the "real" 2D form: L-Q-Q-Q- LQ-Q-Q-Q LLQ-Q-Q- LL-Q-Q-Q LLLLQ-Q- LLLL-Q-Q MLLLQ-- =LLL-ML Increment 19 times: IIIIIIIIIIIIIIIIIII Characters still left: • 2D movement (assumed useless): <>^vNSEW|_# • I/O and unreliable things: ™τ,prtT? • Stack operations: ø • jumps: $@&j§
• rest: "%.:;ABCGHKOUVXYZ\ghkmnquxyz»×€∑
Ideas:
• Do something with the range commands (e.g. 3z** for 6)
# C#, 12 (35 bytes) 13 (7468 62 bytes)
Math.E/Math.E
2
3
4
5
6
7
8
9
-~-~-~-~-~-~-~-~-~-~new long()
11
0xC
'N'%'A'
Try it here.
• Math.E divided by itself results in 1
• new long() is 1, and the -~s increases this to 10
• 0xC is hexadecimal for 12
• 'N' = 78 and 'A' = 65, and 78 modulo 65 is 13
## Perl 5, score 19, 234 bytes
Whilst some of the items are the same as @Full Decent's answer I added many and didn't utilise the fact that $_ would be preinitialised to a value because of the script being executed. Note: for 10, that is a literal newline in the quotes. __LINE__ 2 3 //- -//- -//- -// int$]
6
7
8
9
ord"
"
11
m<>+m<>+m<>+m<>+m<>+m<>+m<>+m<>+m<>+m<>+m<>+m<>
(555555555555&555555555)%55
push@W,W,W,W,W,W,W,W,W,W,W,W,W,W,W
0xF
4*4
VQ^gf
yyyyyyyyyyyyyyyyyy=~y~y~~
eval''.eval''.lc'MAP{P}H..Z'
• Very fun! What a solid answer. – William Entriken Aug 23 '17 at 3:04
# Ruby 1.8.7, 121314 15
Ruby's strict typing makes this one pretty difficult...
+1 number from GB.
$$**($$-$$) 2 %q{%d}%Math::PI 4 5 6 7 8 9 "size..size".size 11 3+3+3+3 ?\r '==============~'=~/~/ 0xF • I'm not familiar with Ruby. What's the reason for using 8 to make 2 and 2 to make 8? – D Krueger Jun 5 '17 at 12:27 • @DKrueger I was messing around with my numbers so much that it eventually ended up that way, but is completely unnecessary. – Value Ink Jun 5 '17 at 18:24 • in REPL the value of . is 1, then you could use the =~ operator to make 15 (like 'verylongstring'=~/x/) – G B Jun 6 '17 at 10:54 • @GB Using . robs me of using .size for 14. Although, I just realized that I'm doing a lot of unnecessary stuff in my calculation of 1. – Value Ink Jun 6 '17 at 18:49 • Maybe using$$**($$-$$) for 1 can help? – G B Jun 7 '17 at 8:16
# Brain-Flak, score 1, 2 bytes
()
Try it online!
Snippet that evaluates to 1. You can't go any further.
• 10/10. or really (()()()()()()()()()())/(()()()()()()()()()()) – Christopher Aug 20 '17 at 22:39
• @2EZ4RTZ / isn't something Brain-Flak has... – Erik the Outgolfer Aug 21 '17 at 8:14
• As in 10 out of 10. – Christopher Aug 21 '17 at 11:49
# Implicit, score 18, 118 bytes
ö
###
3
2^
5
6
7
8
9
10
la-
zA/zA/zA/zA/zA/zA/zA/zA/zA/zA/zA/zA/Þ
OB_
..............
óóóóóóóóóóóóóóó]
4*4
ìé%éééééé
=:+:::::::::++++++++
All these rely on implicit output.
1. ö pushes iswhole(input). Input is 0 when the input field is left blank, and 0 is whole.
2. ### pushes the length of the stack, then the length of the stack, then the length of the stack.
3. 3 pushes 3.
4. 2^ pushes 2 and squares it..
5. 5 pushes 5.
6. 6 pushes 6.
7. 7 pushes 7.
8. 8 pushes 8.
9. 9 pushes 9.
10. 10 pushes 10.
11. la- pushes the character codes for l and then a and subtracts them.
12. zA/ yields 1, repeated 12 times yields a stack with 12 instances of 1. Þ sums them all.
13. OB_ pushes the character codes for O and B and performs modulo on them.
14. .............. increments 0 14 times.
15. óóóóóóóóóóóóóóó] increments the notepad 15 times and pulls it.
16. 4*4 pushes 4, multiplies it by 4.
17. ìé%éééééé pushes 0 as a string, increments all char codes by 1, prints, increments by 6.
18. =:+:::::::::++++++++ pushes 1, duplicates, adds, then does a bunch of duplication/addition.
# Julia 0.7/0.6, Score: 13 (83 bytes)
pi\pi
T=[true true]*[true;true];T[true]
3
4
5
gamma(2^2)
7
8
9
0xA|0
11
6+6
'O'-'B'
Obtains 2 via matrix multiplication of Booleans. For 6, we use the gamma (Γ) function (gamma(n) == factorial(n-1)). For 10, 0xA has to be |-ed with 0 since we want the output in decimal notation, not hexadecimal.
I wrote it for Julia 0.7, but it works in Julia 0.6 too without a hitch.
Let me know if there are ways to go beyond 13 (or if I've made some mistake by repeating a character anywhere here).
# JavaScript (ES7), score 17
Credits to ETHproductions and darrylyeo.
""**""
Math.E|NaN
3
C=CSS==CSS;C<<C<<C
5
6
7
8
9
++[[]][+[]]+[+[]]
11
4444444444444444444%44
222>>2>>2
${{}&{}}xe&${{}&{}}xe
0XF
((((((((((((((((~URL)^(~URL/~URL))/~URL^~URL)/~URL^~URL)/~URL^~URL)/~URL^~URL)/~URL^~URL)/~URL^~URL)/~URL^~URL)/~URL^~URL)/~URL^~URL)/~URL^~URL)/~URL^~URL)/~URL^~URL)/~URL^~URL)/~URL^~URL)/~URL
- -!''- -!''- -!''- -!''- -!''- -!''- -!''- -!''- -!''- -!''- -!''- -!''- -!''- -!''- -!''- -!''- -!''
• Very impressive. Let's see... Math.E|NaN could be Math.E|Math, and all of the {}s would be /~/ (and you don't need the two innermost pairs of parens). Sorry if this isn't really helpful, I'm in golf-the-charset mode :P – ETHproductions Jun 4 '17 at 19:29
• You could probably save a gajillion bytes if you swapped 4 and 16, btw – ETHproductions Jun 4 '17 at 19:30
• Oh wait, you're using . in both 2 and 14 :( – ETHproductions Jun 4 '17 at 19:31
• Ah... that needs to be fixed. – GOTO 0 Jun 4 '17 at 19:34
• @JoKing Oops! Fixed. – GOTO 0 Jul 17 '18 at 20:48
# Wolfram Language (Mathematica), score 16
#&'
Floor@E
3
4
5
6
7
8
9
DDD=(DD(DD)DD(DD)DD(DD)DD(DD)DD(DD))~D~DD;DD=D~D~D;DDD
11
0!+0!+0!+0!+0!+0!+0!+0!+0!+0!+0!+0!
-I*I-I*I-I*I-I*I-I*I-I*I-I*I-I*I-I*I-I*I-I*I-I*I-I*I
Sum[u u,{u,Pi}]
L|L|L|L|L|L|L|L|L|L|L|L|L|L|L//Length
2 2 2 2
Try it online!
Characters used: !#&'()*+,-/0123456789;=@DEFILPS[]eghilmnortu{|}~
1: #&'
Derivative of the identity function: the constant function that returns 1
2: Floor@E
$$\\lfloor e\rfloor\$$, where $$\e\approx2.7183\$$
3-9, 11: Literals
10: DDD=(DD(DD)DD(DD)DD(DD)DD(DD)DD(DD))~D~DD;DD=D~D~D;DDD
$$\\frac{\mathrm d}{\mathrm d\textit{DD}}(\textit{DD}^{10})|_{\textit{DD}=\frac{\mathrm dD}{\mathrm dD}}=10(1)^9\$$
12: 0!+0!+0!+0!+0!+0!+0!+0!+0!+0!+0!+0!
Add $$\0!=1\$$ twelve times
13: -I*I-I*I-I*I-I*I-I*I-I*I-I*I-I*I-I*I-I*I-I*I-I*I-I*I
Subtract $$\i*i=-1\$$ thirteen times
14: Sum[u u,{u,Pi}]
$$\\sum_{u=1}^\pi(u\,u)\$$.
Takes $$\u\$$, incrementing by 1, while $$\u\le\pi\$$, so the sum is equal to $$\1^2+2^2+3^2=1+4+9\$$
15: L|L|L|L|L|L|L|L|L|L|L|L|L|L|L//Length
Length (number of arguments) of Alternatives[L,L,...L] (15 Ls in this case)
16: 2 2 2 2
Product of four 2s
# Pyth, score 18, 207 bytes
.!0
2
3
4
5
6
7
8
C\
T
11
s[JsP9J
++++++++++++KqkkKKKKKKKKKKKK
-------------gZZ_gZZ_gZZ_gZZ_gZZ_gZZ_gZZ_gZZ_gZZ_gZZ_gZZ_gZZ_gZZ_gZZ
hhhhhhhhhhhhhh!Y
yyyy^>bb>bb
a<ddttttttttttttttttt<dd
l"llllllllllllllllll
Test suite
Any help welcome although I won't respond immediatly, so feel free to edit.
• @totallyhuman Or maybe even l"llllllllllllll – Okx Jun 4 '17 at 9:32
# Chip, score 1
t*ae*f
Yep, only 1.
The above code (plus at least one of the flags -w, -z, or -o) outputs the value 0x31, which is the ascii code point for the digit '1', then terminates itself:
t If active, terminate execution after printing the current output byte
* Activate neighboring elements
a If active, set output bit 0x01 to 1
e If active, set output bit 0x10 to 1
* Activate neighboring elements
f If active, set output bit 0x20 to 1
Since e and f are required in order to output every digit (as they are the only way to set those bits), we can't go any higher than a single digit.
# Fission 2, Score: 2, 10 bytes
1: R'1!;
2: *"2"L
Try it online!
As far as I know, this is as high as you can possibly get in Fission, as the program has to start at one of RLUD characters, and you have to use " or ! to output
In hindsight this can probably be improved by taking advantage of Snippets are allowed!
### Explanation
1:
R'1!;
R Spawn atom with 1 mass, 0 energy, moving right
'1 Set atom's mass to ASCII value of 1
! Print character represented by atom's mass
; Destroy atom
2:
*"2"L
L Spawn atom with 1 mass, 0 energy, moving left
" Enter printing mode, print all chars atom moves over
2 Print 2
" End printing mode
* Terminate program
# ><>, Score 2, 8 Total Bytes
Can't go any higher, there are only 2 output commands: n for numbers, and o for characters.
1n;
Prints 1, as a number.
'2'ov
Pushes 2's ascii value to the stack. Prints it as a character. Loops forever, since there's no way to end the program without re-using a semicolon.
• There's a ><> answer already, which does a lot better since answers can be "snippets". – Ørjan Johansen Nov 18 '17 at 18:56
• Ah, I see. Oh well ¯_(ツ)_/¯ – Bolce Bussiere Nov 18 '17 at 19:30
• Also, you can end the program using any non-instruction, dividing by 0 or doing any instruction without a sufficient amount of items in the stack – Jo King Dec 30 '17 at 12:41
• Also, I think you dropped this: \ – mudkip201 Jan 21 '18 at 2:40
# C (clang), 50 bytes, score 13
!NULL
2
3
4
5
6
7
8
9
1e1
'l'-'a'
0xC
*"\r"
Try it online!
• @ØrjanJohansen Fixed – Logern Nov 7 '18 at 3:43
• @JoKing fixed, but it loses one number. – Logern Nov 7 '18 at 14:05
• Up to 15, if I didn't make mistakes. – user77406 Nov 13 '18 at 16:33
• Optionally, if <float.h> isn't considered boilerplate enough, RAND_MAX>>SEEK_END>>...>>SEEK_END/SEEK_END will obtain the same result, and are from <stdlib.h> and <stdio.h>. – user77406 Nov 13 '18 at 16:51
## (K+R)eg, score 32
Our task is to write as many programs / functions / snippets as you can, where each one outputs / prints / returns an integer. The integer is already on the stack, so it should have already returned the integer. The method of outputting does not matter.
• 1-9 implicit outputs the respective integer. We are using control characters for this. In order to save the 1 character, we should do \x03\x02> instead.
• 10(TIO):
\
• 11-31 Yes, you need control characters! In order to return a specified number, you only need to convert the integer to its ASCII character form. Unfortunately this is in the ASCII range, and implicit integer output does not work.
Returning 13 is a special case(TIO):
;
• 32 Great, so this is automatically pushed onto the stack.
So I can't think of more methods.
• Couldn't you push 33 to the stack by pushing 2 Unicode characters that have a difference of 33, then subtracting? – EdgyNerd Sep 14 at 16:00
# Perl REPL, score 16
Get Perl REPL by using perl -nE'say eval'.
__LINE__
2
3
4
5
6
length
8
9
cos,abs
11
0xC
$. 7+7 !rmdir()+!rmdir()+!rmdir()+!rmdir()+!rmdir()+!rmdir()+!rmdir()+!rmdir()+!rmdir()+!rmdir()+!rmdir()+!rmdir()+!rmdir()+!rmdir()+!rmdir() //- -//- -//- -//- -//- -//- -//- -//- -//- -//- -//- -//- -//- -//- -//- -// Other things I tried: (ord Z)-(ord K) And here are some ideas of how to cheat: • umask • tell • getppid • getpgrp • wait • system • time • srand • localtime • gmtime • ls|wc -l <---- cheating, has preconditions • There's a lot more that can be done here – William Entriken Jun 5 '17 at 15:32 • + in both 14 and 15. – Ole Tange Jun 5 '17 at 15:50 • I feel that $. is cheating, as it would not work without the previous commands having run. I'd consider it acceptable in the first spot only. – Ørjan Johansen Jun 5 '17 at 17:27
• There's a + in both your 14 and your 15. – Erik the Outgolfer Aug 21 '17 at 11:57
# Microscript II, score 13 (33 bytes)
e
2
3
4
5
6
7
8
9
EE
11
ssssssssssss#
"na"Ko-
# Unreadable, score 1, 150 4 bytes
'"""
Try it online!
-146 bytes because snippets are allowed
Returns 1
# ArcPlus, score 1, 4 bytes [non-competing]
(p 1
Prints 1
# Pain-Flak, Score 1
))((}{`
Try it online!
|
2019-10-14 22:55:38
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 9, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.33367466926574707, "perplexity": 5283.1546356145}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986655554.2/warc/CC-MAIN-20191014223147-20191015010647-00445.warc.gz"}
|
https://math.stackexchange.com/questions/1608740/a-criterion-for-isomorphism-of-finite-abelian-groups-using-fundamental-theorem-o
|
# A criterion for isomorphism of finite abelian groups using fundamental theorem of finitely generated abelian groups [duplicate]
I have recently encountered this very interesting problem from my abstract algebra class where we have just now proven the fundamental theorem of finitely generated Abelian groups, and the problem statement I have here is as follows:
Let $G_1, G_2$ be two finite Abelian groups such that for all natural k these two groups have the same number of elements of order k. We are to prove that the groups $G_1, G_2$ are isomorphic
Now I realize this may have been around somewhere or even here but I could not understand any of those as they use different tools. What I have been taught in my course is the fundamental theorem of finitely generated Abelian groups which I know in two forms: the elementary divisors and the invariant factor decomposition and that two of these are isomorphic if and only if they have the same elementary divisors or the invariant factors but from there I am stuck. I would certainly appreciate the help on this for a novice. Thanks to all helpers
Edit: I realize this might be a duplicate question my problem is I could not really understand the other answers
## marked as duplicate by Jorge Fernández Hidalgo, Eric Wofsey, David, Shailesh, user228113 Jan 12 '16 at 1:20
• Use the elementary divisor form and take $k$ to be a prime power. Try to recover the elementary divisors of $G$. – Francis Begbie Jan 12 '16 at 0:31
The invariant factor decomposition shows that the subgroup generated by an element of largest order in a finite abelian group is a direct factor of the group. Given the hypotheses on $G_1$ and $G_2$, this allows us to cancel the same factor in their invariant factor decomposition. The result follows by induction on the size of the groups or on the number of invariant factors.
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2019-08-23 17:29:48
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7711445093154907, "perplexity": 65.7896844658484}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027318952.90/warc/CC-MAIN-20190823172507-20190823194507-00406.warc.gz"}
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https://codereview.stackexchange.com/questions/40563/garrys-mod-custom-spectate-player
|
Garry's Mod custom spectate player
I have been making my own scoreboard, and needed a custom spectate function for it, so I made this:
util.AddNetworkString( "spectatePlayer" )
local playerLocal
local isSpectating = false
if isSpectating then return end
isSpectating = true
playerLocal = client
print(client:Nick() .. "Started spectating: " .. net.ReadEntity():Nick())
client:Spectate(5)
end)
if not isSpectating then return end
if (playerLocal:KeyDown(IN_FORWARD)) then
isSpectating = false
playerLocal:UnSpectate()
print(playerLocal:Nick().."Finished Spectating")
end
end)
Is there anything I could improve on? Am I over-complicating anything?
• Please add in what the custom behavior is -- just saying that it's custom isn't very specific. – Fund Monica's Lawsuit Jul 2 '15 at 23:21
There's a few things to comment on:
I'm not a Gmod developer, so I'm not familiar with the spectating abilities built in, but, I'd assume that if any exist, they're probably better to use than re-inventing the wheel.
There's only two states: not spectating, and spectating. What happens if I want to simply switch between players? Do I have to stop spectating, and then select a new player to spectate?
Why can I not just press lmb and go to the next?
util.AddNetworkString( "spectatePlayer" )
You've got extraneous whitespace in your brackets.
print(client:Nick() .. "Started spectating: " .. net.ReadEntity():Nick())
print(playerLocal:Nick().."Finished Spectating")
String concatenation joins the strings, as they are. Meaning, it would read like:
QuillStarted spectating: James Heald.
QuillFinished Spectating
print(client:Nick() .. " Started spectating: " .. net.ReadEntity():Nick())
"Started spectating: "
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2020-02-18 11:26:29
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http://cogsci.stackexchange.com/questions/1595/is-religiousness-a-genetically-heritable-feature/1604
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# Is religiousness a genetically heritable feature?
I know that parent beliefs have very strong influence on child, but is it possible that genetic factor also play a role? I doubt that the content of any religion could be heritable through genes nor anyone could be genetically predestined to be Christian, Jew, Buddhist or votary of any particular religion. But is it possible that some kind of sensitivity to religious aspect of life could be heritable?
I know that it is hard to distinguish genetic and environmental factor in such case, but maybe someone have already made such research for example on monozygotic and dizygotic twins or adopted children?
I've thought that if it is indeed heritable, then celibacy in the Catholic Church might be suicidal for this faith, as it eliminates the (potential) "godliness genes".
-
@ChuckSherrington there are purported measures for "religiousness" en.wikipedia.org/wiki/… not sure what the current accepted one(s) may be, if any. – Ben Brocka Aug 30 '12 at 21:30
@BenBrocka Sure, I would buy into that, but going from questionaire -> personality characteristic -> phenotype -> genotype would still be hairy, I think. – Chuck Sherrington Aug 30 '12 at 21:39
@ChuckSherrington There are plenty of measures, take a look at McGregor, I., Haji, R., Nash, K.A., & Teper, R. (2008). "Religious zeal and the uncertain self." Basic and Applied Social Psychology, 30, 183–188. This measure is used in Inzlicht et al. (2009; see this answer for more detail) when looking for neural correlates of religiousness. – Artem Kaznatcheev Aug 30 '12 at 22:07
@ChuckSherington, monozygotic twin studies: Genetic and environmental influences on multiple dimensions of religiosity: a twin study concludes there is a genetic component. The general course of these kind of studies is to take twins that have been separated to different households from birth and compare them to twins int he same household. I'm not sure what this particularly study does (haven't read it) but its a typical approach in behavioral biology. – Keegan Keplinger Sep 1 '12 at 7:06
@Preece I do not assume a priori that such a gene exist nor that the religiousness is it only function if such a gene exist. It could for example makes people more social and helps integrate the members of community (as integration is indeed one of the function of religions in human communities). Also, as I said, I do not think that the content of any religion is a product of direct biological evolution. – Marta Cz-C Sep 3 '12 at 8:41
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2014-04-19 01:56:39
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https://mathematica.stackexchange.com/questions/37369/when-importing-is-it-possible-to-set-the-delimiter
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# When importing is it possible to set the delimiter?
When importing a csv file is there a way to set the delimiter. I have a csv that uses ";" as its delimiter.
I tried
rd = import["/Users/rolfbosscha/Downloads/rsdata.csv"]
• You can use ReadList which happens to be much faster than Import See for example mathematica.stackexchange.com/questions/35371/…. With only a slight modification: change "," to ";" under RecordSeperators – RunnyKine Nov 19 '13 at 21:14
• Mathematica is case-sensitive: use Import, not import. – m_goldberg Nov 19 '13 at 21:21
This seems to work:
Import["data.csv", "table", FieldSeparators -> ";"]
Should be
Import["data.csv", "Table", "FieldSeparators" -> ";"]
See help for "Table" element (ref/format/Table), specifically Options section.
• This does not work with tabs \t in Mathematica 11.0.1, Debian 8.5. – Léo Léopold Hertz 준영 Oct 28 '16 at 14:33
• it works. thanks – Nam Nguyen Jan 13 '17 at 15:15
• Yes, what todo with tabs? \t does not work. – becko Sep 22 '17 at 10:14
• If you have \t then your file format is "Table". Simply: Import["data.csv", "Table"] – drgrujic Feb 1 '18 at 15:08
I may be mistaken here, but I don't think you can switch from the standard CSV definition (commas separate values) to the one you have (commas take the role of decimal points, and semicolons are used as value separator). A workaround might be to first import as text, switch the offending characters and import that.
As a demo I use ImportString here instead of Import, but it should work almost the same for Import too.
ImportString[
StringReplace[
ImportString["1,2;2;3;4.3\n2,3;4,5;6;7", "Text"],
{"," -> ".", ";" -> ","}
],
"CSV"
]
{{1.2, 2, 3, 4.3}, {2.3, 4.5, 6, 7}}
Replace the inner ImportString with
Import["/Users/rolfbosscha/Downloads/rsdata.csv", "Text"]
to work with your file. Note that you need to use "Text" instead of "CSV".
• What do you think about Jukka's answer? I cannot make it to work. I think your approach can work. My deliminator is \t. – Léo Léopold Hertz 준영 Oct 28 '16 at 14:34
• If your separator is a tab, import as TSV. – Sjoerd C. de Vries Oct 28 '16 at 14:38
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2019-09-20 21:51:51
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http://mathhelpforum.com/differential-equations/222272-sobolev-spaces-some-mollifier-stuff-print.html
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Sobolev Spaces and some Mollifier stuff
• September 25th 2013, 05:25 AM
Johnyboy
Sobolev Spaces and some Mollifier stuff
Note that the proof in question is from book 'Partial Differential Equations by Lawrence Evans'.
Please view attachments for proof in question, basically I am trying to prove that for bounded $U$ any function $u \in W^{k,p}(U)$ can be approximated by smooth functions $u_{m} \in C^{\infty}\cap W^{k,p}(U)$.
The questions I have about the proof are the following:
It says choose $\epsilon_{i} > 0$ so small that $u^{i} := \eta_{\epsilon_{i}}\ast(\zeta_{i}u)$ satisfies (3) in attachment. Is the reason that both conditions of (3) are satisfied for small enough $\epsilon_{i}$ because of two basic results of mollifiers which says that for fixed i it follows that $u^{\epsilon_{i}} \rightarrow \zeta_{i}u$ as $\epsilon_{i} \rightarrow 0 \text{ a.e. } \text{ and if } \zeta_{i}u \in L^{P}(U) \text{ then } u^{\epsilon_{i}} \rightarrow \zeta_{i}u \text{ as }\epsilon_{i} \rightarrow 0 \text{ in } L^{P}(U)$?
Secondly, why is $W_{i}$ only defined for $i = 1, ...$ and not for $i =0$? Do we not need $\cup_{i=0}W$ to cover $U$ so that further in the proof together with a compactness argument we can show that there are finitely many non zero terms in the sum $v := \sum_{i=0}^{\infty}u^{i}$ as can be seen in attachment?
Thanks for any assistance, let me know if anything is unclear.
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2015-08-01 17:14:42
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http://openstudy.com/updates/4f185d55e4b00328e4c5147b
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## anonymous 4 years ago COULD YOU SHOW YOU WORK? 1. x – 3 > –9 2. –2x + 1 ≤ –11 3. 10 < –3x + 1 4. 2(x + 5) > 8x – 8 5. –2(x – 3) ≥ 5 – (x + 3)
1. anonymous
1) x-3>-9 +3 +3 x>-6
2. anonymous
1) add 3 to both sides
3. anonymous
2) subtract 1 from both sides divide by -2 change the sense of the inequality
4. anonymous
satellite73 which on are you doing?
5. anonymous
$-2x+1\leq -11$ $-2x\leq -11-1$ $-2x\leq -12$ $x\geq -12\div -2$ $x\geq 6$
6. anonymous
that as #2
7. anonymous
that was how u do it the work and all?
8. anonymous
that was all the steps worked out. nothing omitted
9. anonymous
oh ok
10. anonymous
#3 takes two steps subtract 1 from both sides divide by -3 and change the sense of the inequality $10 < –3x + 1$ $10-1<-3x$ bet you can finish from there.
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2016-10-25 22:57:46
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http://wapgw.org/standard-error/relationship-between-standard-error-of-estimate-and-r-squared.php
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Home > Standard Error > Relationship Between Standard Error Of Estimate And R Squared
# Relationship Between Standard Error Of Estimate And R Squared
## Contents
The 9% value is the statistic called the coefficient of determination. please sir, can i use regression line and a curve at the same time to interpret my data? Hence, my question. –Roland Feb 13 '13 at 10:05 Your terminology is probably fine. The correlation coefficient is equal to the average product of the standardized values of the two variables: It is intuitively obvious that this statistic will be positive [negative] if X and navigate here
That is, there are any number of solutions to the regression weights which will give only a small difference in sum of squared residuals. If all possible values of Y were computed for all possible values of X1 and X2, all the points would fall on a two-dimensional surface. In the three representations that follow, all scores have been standardized. Jim Name: Rafael • Monday, December 16, 2013 Great Post, thank you for it.
## Standard Error Of Estimate Formula
This is not supposed to be obvious. Getting the standard errors of the estimates (slope and intercept) might be a start, but my approach seems like it is at a dead loss to predict intercept error separate from Therefore, the predictions in Graph A are more accurate than in Graph B.
I'm sure this isn't a complete list of possible reasons but it covers the more common cases. The interpretation of R is similar to the interpretation of the correlation coefficient, the closer the value of R to one, the greater the linear relationship between the independent variables and The least-squares estimate of the slope coefficient (b1) is equal to the correlation times the ratio of the standard deviation of Y to the standard deviation of X: The ratio of Linear Regression Standard Error That'll be out on October 3, 2013.
Multiple regression is usually done with more than two independent variables. Standard Error Of The Regression The sample r or Multiple R will not be a good estimate of the corresponding population parameter if the sample is (deliberately or accidentally) biased. Thanks for the beautiful and enlightening blog posts. over here Three-dimensional scatterplots also permit a graphical representation in the same information as the multiple scatterplots.
You'll see S there. Standard Error Of Regression Interpretation However, S must be <= 2.5 to produce a sufficiently narrow 95% prediction interval. Specifically, although a small number of samples may produce a non-normal distribution, as the number of samples increases (that is, as n increases), the shape of the distribution of sample means May be this could be explained in conjuction with beta.Beta (β) works only when the R² is between 0.8 to 1.
## Standard Error Of The Regression
The multiple regression plane is represented below for Y1 predicted by X1 and X2. navigate to this website R-squared is a statistical measure of how close the data are to the fitted regression line. Standard Error Of Estimate Formula However, if you need precise predictions, the low R-squared is problematic. Standard Error Of Estimate Interpretation The coefficients, standard errors, and forecasts for this model are obtained as follows.
The biggest practical drawback of a lower R-squared value are less precise predictions (wider prediction intervals). check over here Jim Name: Newton • Friday, March 21, 2014 I like the discussant on r-squared. Jim Name: Jim Frost • Tuesday, July 8, 2014 Hi Himanshu, Thanks so much for your kind comments! Are Low R-squared Values Inherently Bad? Standard Error Of Regression Coefficient
An unbiased estimate of the standard deviation of the true errors is given by the standard error of the regression, denoted by s. Authors Carly Barry Patrick Runkel Kevin Rudy Jim Frost Greg Fox Eric Heckman Dawn Keller Eston Martz Bruno Scibilia Eduardo Santiago Cody Steele The Minitab Blog Data Analysis It states that regardless of the shape of the parent population, the sampling distribution of means derived from a large number of random samples drawn from that parent population will exhibit his comment is here Additional analysis recommendations include histograms of all variables with a view for outliers, or scores that fall outside the range of the majority of scores.
Another use of the value, 1.96 ± SEM is to determine whether the population parameter is zero. Standard Error Of Estimate Calculator Name: Jim Frost • Tuesday, March 4, 2014 Hi Joe, Yes, if you're mainly interested in the understanding the relationships between the variables, your conclusions about the predictors, and what the Solution 1: We know the standard error of a pearson product moment correlation transformed into a Fisher $Z_r$ is $\frac{1}{\sqrt{N-3}}$, so we can find the larger of those distances when we
## Because the standard error of the mean gets larger for extreme (farther-from-the-mean) values of X, the confidence intervals for the mean (the height of the regression line) widen noticeably at either
That in turn should lead the researcher to question whether the bedsores were developed as a function of some other condition rather than as a function of having heart surgery that In terms of the descriptions of the variables, if X1 is a measure of intellectual ability and X4 is a measure of spatial ability, it might be reasonably assumed that X1 The 1981 reader by Peter Marsden (Linear Models in Social Research) contains some useful and readable papers, and his introductory sections deserve to be read (as an unusually perceptive book reviewer Standard Error Of The Slope However, research shows that graphs are crucial, so your instincts are right on.
My comprehension is somewhat limited and I know that convention also varies between fields. The computations derived from the r and the standard error of the estimate can be used to determine how precise an estimate of the population correlation is the sample correlation statistic. Variable X3, for example, if entered first has an R square change of .561. weblink The size and effect of these changes are the foundation for the significance testing of sequential models in regression.
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2018-08-19 05:22:27
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https://math.stackexchange.com/questions/3155914/what-value-of-n-will-make-a-triangle-contain-560-lattice-points
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# What value of $n$ will make a triangle contain 560 lattice points?
I recently met a rather hard problem:
A lattice point is a ordered pair where both $$x$$ and $$y$$ of $$(x, y)$$ are both integers. A triangle is forms by of the lattice points $$(1, 1)$$, $$(9, 1)$$, and $$(9, n)$$. For what integer value of $$n>0$$ are there exactly 560 lattice points strictly in the interior of the triangle?
My first approach was trying to set an equation using Shoelace and Pick's. I found the area via Shoelace formula in terms of $$n$$ and then all I needed to do was to set this equal to Pick's theorem. However, I found it extremely hard to count the # of lattice points on the "lines" of the triangle, so that I didn't get very far.
I then tried another approach. I begin by drawing a diagram:
Since the triangle is a right triangle, I can add a congruent triangle to make a rectangle, with the right angle at $$(1, n)$$. Since both triangles are congruent, then the entire rectangle has $$560\cdot 2=1120$$ lattice points in the interior. There are 7 possible $$x$$ coordinates I labeled in the diagram that can serve as valid $$x$$ coordinated for interior lattice points, so thus, there are $$\frac{1120}{7}=160$$ possible $$y$$ coordinates for the interior lattice points.
I have highlighted in red the 3 invalid $$y$$ coordinates that cannot serve as a $$y$$ coordinate for an interior lattice point. Thus, we can see that $$n-3$$(3 for the 3 invalid $$y$$ coordinates)$$=160$$. This means that $$n=163$$. However, this is wrong.
What am I doing wrong? What hole is there in my logic?
Furthermore, could I solve for $$n$$ using my first method(Shoelace=Pick's and solve)?
Is there any other way to solve this if I can't solve it this way?
Thanks!
Max0815
Your error is that the lattice points on the diagonal from $$(1,1)$$ to $$(9,n)$$ (excluding the endpoints) are interior points of the rectangle, but are not counted in the $$2\times560$$ points.
The main takeaway is that you will have to deal with that pesky diagonal one way or another. The triangle being right-angled with coordinate-parallel sides allows both approaches: The shoelace/Pick approach is more general, the second is easier, especially if you don't know about the Pick formula.
I'll give some hints for the first approach:
1) From the 3 sides of the triangle, 2 should be trivial by now, you did them for the second approach!
2) That leads to the diagonal, which represents a displacement of $$(8,n-1)$$, or if you look at the linear function on which that diagonal lies, it is $$y=\frac{n-1}8(x-1)+1$$
3) Any lattice point in the interior of that diagonal is characerized by 2 conditions:
a) $$2\le x \le 8$$, and
b)$$\frac{n-1}8(x-1)$$ must be an integer.
4) So the key question becomes: For which/how many $$x'=x-1$$ satisfying $$1 \le x' \le 7$$ is $$\frac{n-1}8x'$$ an integer?
5) The answer depends of course on the value of $$\frac{n-1}8$$. If that is already an integer (so $$8|(n-1)$$), every $$1 \le x' \le 7$$ works. OTOH, if $$n-1$$ is odd, then $$x'$$ would have to be divisible by $$8$$, which it can't be in that range.
6) Can you find the remaining cases? Can you find the respective number of solution $$x'$$? If you can't, try $$n-1=2,4,6$$ maybe you see something!
7) The previous parts of the problem should all be 'simple' formulas of $$n$$. If you can solve 6), you have several cases now for the part of the number of lattice points on the diagonal. Solve them each, then check if the solution (if one exists) matches the condition (that means if you use the formula that applies when $$8|(n-1)$$ and get a solution $$n=165$$, then it can't be correct, as $$8$$ does not divide $$164$$.
• Can you also help me on my second approach? – Max0815 Mar 22 '19 at 1:37
• The help is basically the same. As I wrote, the error you made is that you did not account for the lattice points on the $(1,1)-(9,n)$ diagonal. I tried to jumpstart that calculation with my answer. If you have the result, you can use it for either approach. – Ingix Mar 22 '19 at 10:32
• Ok. Got it. Thank you. – Max0815 Mar 23 '19 at 21:41
Since the triangle is a right triangle, I can add a congruent triangle to make a rectangle, with the right angle at $$(1, n)$$. Since both triangles are congruent, then the entire rectangle has $$560\cdot 2=1120$$ lattice points in the interior. There are 7 possible $$x$$ coordinates I labeled in the diagram that can serve as valid $$x$$ coordinated for interior lattice points, so thus, there are $$\frac{1120}{7}=160$$ possible $$y$$ coordinates for the interior lattice points.
I have highlighted in red the 3 invalid $$y$$ coordinates that cannot serve as a $$y$$ coordinate for an interior lattice point. However, when $$y=0$$, this line has lattice points that will not ever be in the triangle bound by $$(9, n)$$, and the other two points in the problem, so only the top and middles lines highlighted in red serve as invalid. Thus, we can see that $$n-2=160$$. This means that $$n=\boxed{162}$$.
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2020-05-31 10:37:33
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https://www.openstarts.units.it/handle/10077/9604
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Please use this identifier to cite or link to this item: http://hdl.handle.net/10077/9604
Title: Metrizability of hereditarily normal compact like groups Authors: Dikranjan, DikranImpieri, DanieleToller, Daniele Keywords: locally compact group; locally minimal group; \omega-bounded group; countably compact group; hereditarily normal topological group; metrizable group Issue Date: 2013 Publisher: EUT Edizioni Università di Trieste Source: Dikran Dikranjan, Daniele Impieri and Daniele Toller, "Metrizability of hereditarily normal compact like groups", in: Rendiconti dell’Istituto di Matematica dell’Università di Trieste. An International Journal of Mathematics, 45 (2013), pp. 123–135. Series/Report no.: Rendiconti dell’Istituto di Matematica dell’Università di Trieste. An International Journal of Mathematics45 (2013) Abstract: Inspired by the fact that a compact topological group is hereditarily normal if and only if it is metrizable, we prove that various levels of compactness-like properties imposed on a topological group G allow one to establish that G is hereditarily normal if and only if G is metrizable (among these properties are locally compactness, local minimality and \omega-boundedness). This extends recent results from [4] in the case of countable compactness. URI: http://hdl.handle.net/10077/9604 ISSN: 0049-4704 Appears in Collections: Rendiconti dell'Istituto di matematica dell'Università di Trieste: an International Journal of Mathematics vol.45 (2013)
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2018-02-20 09:39:38
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http://math.stackexchange.com/questions/137201/did-i-integrate-a-differential-form-correctly
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# Did I integrate a differential form correctly?
I start with 1-form $\omega=f\,dx$ on $\left[0,1\right]$ where $f\left(0\right)=f\left(1\right)$ and a $g:\left[0,1\right]\to R$ with $g\left(0\right)=g\left(1\right)$ and I want to integrate $\omega-\lambda \, dx=dg$ on $\left[0,1\right]$. So I write
$\int\limits_0^1 \omega-\lambda \,dx = \int\limits_0^1 \left(f-\lambda\right)\,dx =\int\limits_0^1 f\,dx-\lambda$ and $\int\limits_0^1 \omega-\lambda \, dx =\int\limits_0^1 dg = g\left(1\right)-g\left(0\right)=0$ and find out that $$\lambda=\int\limits_0^1 f\,dx.$$
Is this correct? What would happen if I parametrized the path from $0$ to $1$ another way? Is $\int\limits_0^1 dg=g\left(1\right)-g\left(0\right)$ legal - I'm using dg as a differential form and maybe I'm supposed to check some precondition?
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The theorem in vector calculus said: For an exact 1-form $\omega$, $C$ and $C'$ are two parametrized curves with the same starting point and ending point, then $$\int_C \omega = \int_{C'} \omega$$ And in one dimension case, the boundary between the calculus of differential forms and the regular calculus we learned in college is kinda blurry, the exactness and closedness of a differential form start to play a vital role in 2-dimensional spaces, where Green's theorem, Stokes theorem, etc kick in.
Now in your calculation you made the assumption that $\lambda$ is a real number, also $\displaystyle \int_0^1 dg = g\Big\vert^1_0$ holds when $g$ has well-defined point values and a well-defined differentiation, which means $g\in C^1([0,1])$ would suffice, and you also use the fact that $fdx$ is an exact 1-form, which implies the antiderivative of $f$ exists.
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2014-09-20 18:18:03
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http://blogfrenchie.com.br/an-uncommon-yzjn/why-is-krcl4-nonpolar-b80156
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# why is krcl4 nonpolar
Answer = BF3 ( Boron trifluoride ) is Nonpolar What is polar and non-polar? Examples of Polar and Nonpolar Molecules. Trending questions . The electronegativity of F is more than that of I so there are dipole moments on each of these bonds. Ask question + 100. Is there a way to search all eBay sites for different countries at once? A. S-B. 0. Both 1 and 2 above support why Lewis structures are not a completely accurate way to draw molecules. I'll tell you the polar or nonpolar list below. Join Yahoo Answers and get 100 points today. 6. On the other hand, around the xenon atom there are 4 bonding pairs, … Lv 7. 2). Do Lewis structure and molecular geometry confuse you? SURVEY . Polar molecules must contain polar bonds due to a difference in electronegativity between the bonded atoms. Why did the Vikings settle in Newfoundland and nowhere else? A molecule with very polar bonds can be nonpolar. Question = Is ICl5 polar or nonpolar ? Rude 'AGT' stunt backfires: 'That was so harsh' Halle Berry on the defining moments of her career. CH2Cl2 is the chemical formula for DCM. Thus far, we have used two-dimensional Lewis structures to represent molecules. 2 Answers. Answer Save. If you want to quickly find the word you want to search, use Ctrl + F, then type the word you want to search. Polar molecules must contain polar bonds due to a difference in electronegativity between the bonded atoms. What is the hybridization of the nitrogen in the azide ion? SURVEY . XeF2 is nonpolar due to the symmetric arrangement of the bonded pairs of electrons. Which of the following bonds is least polar? Polar molecules must contain polar bonds due to a difference in electronegativity between the bonded atoms. Determine whether ClO3– is polar or nonpolar. Polar "In chemistry, polarity is a separation of electric charge leading to a molecule or its chemical groups having an electric dipole or multipole moment. Join Yahoo Answers and get 100 points today. The electron pairs, bonding and non-bonding, are arranged around nitrogen in the shape of a tetrahedron. Linear molecules cannot have a net dipole moment. i tried to do it but i cant really tell if its polar or nonpolar! When did organ music become associated with baseball? Copyright © 2020 Multiply Media, LLC. They are all nonpolar. Join Yahoo Answers and get 100 points today. A striking reversal: Trump's attacks on the military. (If the difference in electronegativity for the atoms in a bond is greater than 0.4, we consider the bond polar. Q. Chlorine Trifluoride on Wikipedia. This is particularly the case with heroin which is devastatingly addictive. Decision: The molecular geometry of ClF 3 is T-shaped with asymmetric charge distribution around the central atom. CO 2. Molar Mass: 225. Get help with your Dipole homework. Decision: The molecular geometry of ICl 5 is square pyramid with an asymmetric electron region distribution. nonpolar would be SeF3 or SeF2. Get your answers by asking now. All Rights Reserved. Umar. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Although each IF bond is quite polar, the structure is trigonal planar. Still have questions? ICl4- is Nonpolar I'll tell you the polar or nonpolar list below. How long will the footprints on the moon last? Answer = SeCl4 ( Selenium tetrachloride ) is Polar What is polar and non-polar? But I don't know whether the MOLECULE is polar or nonpolar Join. Polar "In chemistry, polarity is a separation of electric charge leading to a molecule or its chemical groups having an electric dipole or multipole moment. Hence Xenon Difluoride is nonpolar as there is no polarity observed in the molecule. The VSEPR model can predict the structure of nearly any molecule or polyatomic ion in which the central atom is a nonmetal, as well as the structures of many molecules and polyatomic ions with a central metal atom. Lv 7. It is considered nonpolar because it does not have permanent dipole moments. Answer to: Which of the following compounds are expected to be polar: SiCl4, AsCl3, CH3Cl, SCl4, SeCl2, KrCl4? In our discussion we will refer to Figure $$\PageIndex{2}$$ and Figure $$\PageIndex{3}$$, which summarize the common molecular geometries and idealized bond angles of molecules and ions with two to six electron groups. 4 - Consider the following compounds: CO2, SO2, KrF2,... Ch. The VSEPR Model. Polar molecules must contain polar bonds due to a difference in electronegativity between the bonded atoms. Answer Save. Give the shape and the hybridization of the central A atom for each. 9 years ago. Therefore ClF 3 is polar. Favorite Answer. answer choices ... Why show ads? Texas Children’s is committed to keeping our patients, families, employees and community safe by taking proactive measures to help prevent the spread of COVID-19. Should I call the police on then. Decide which atom is the central atom in the structure. Q. 4 - Why do we hybtidize atomic orbitals to explain the... Ch. PostgreSQL; Krcl4 shape Krcl4 molecular geometry Molecular Geometry of the Trigonal Bipyramidal Structures. 1.7k plays . Relevance. Simple VSEPR theory predicts a trigonal pyramid for NF_3, and square planar for XeCl_4. Our videos prepare you to succeed in your college classes. ... Nonpolar Covalent Bonds. In the case of water, it is polar. No, SiH4 is not polar. Which of following elements has GREATEST electronegativity? Types of Chemical Bonds . Ch. Tags: Question 2 . I--F. Tags: Question 3 . Start studying Chemistry Ch 4 Homework. Start studying Chemistry Ch 4 Homework. Tags: Question 2 . Trending questions. At this time, we are not accepting. Why you are interested in this job in Hawkins company? 0 2. Electron domain is used in VSEPR theory to determine the molecular geometry of a molecule. List molecules polar and non polar. A. S-B. Concluding Remarks. email protected] 1, 2, 3, and 4 ____ 28. Answer to: Is ICl5 polar or nonpolar? Trending Questions. I went to a Thanksgiving dinner with over 100 guests. To summarize the article, it can be concluded that XeF2 has 22 valence electrons, out of which there are three lone pairs of electrons. Why do compounds like SF6 and SF4 exist but SH6 and SH4 don't? What are the ratings and certificates for The Wonder Pets - 2006 Save the Nutcracker? Step 2: Identify each bond as either polar or nonpolar. By signing up, you'll get thousands of step-by-step solutions to your homework questions. Therefore this molecule is polar. Get your answers by asking now. SO 2. Determine the volume of a solid gold thing which weights 500 grams? As there are fluorine molecules on both the side of the central atom, there is no dipole moment and hence there is no polarity. … what company has a black and white prism logo? Learn vocabulary, terms, and more with flashcards, games, and other study tools. Firstly its not the hybridization of PCl5 but of P in PCl5 molecule. Our videos will help you understand concepts, solve your homework, and do great on your exams. Which of the following bonds is least polar? Join. If there is an odd number of lone pairs of electrons around the central atom, the molecule is polar. 11. Favorite Answer. Carbon Tetrachloride on Wikipedia. during extraction of a metal the ore is roasted if it is a? Explain why a carbon atom cannot form five bonds using sp 3 d hybrid orbitals. Is krcl4 polar or nonpolar keyword after analyzing the system lists the list of keywords related and the list of websites with related content, in addition you can see which keywords most interested customers on the this website. Let us help you simplify your studying. 14 Qs . Ask Question + 100. At this time, we are not accepting. Postby Kevin Morden 1E» Thu Dec 07, 2017 2:15 am. 1 Answer. Answer to: Which of the following compounds are expected to be polar: SiCl4, AsCl3, CH3Cl, SCl4, SeCl2, KrCl4? 0 0. Still have questions? TeCl4, if you draw the lewis diagrams, has a lone pair of electrons. Dipole. Br--Br. I have to idea about the angles and I am not sure if I should consider the lone pair....PLEASE HELP! The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply. Answer = ICl5 is Polar What is polar and non-polar? Molecules polar; 1-butanol: Polar: 1-propanol: Polar: 2-propanol: Polar: 2-propanol: Polar: acetanilide: Polar: acetic acid: Polar: acetophenone: Polar: Are carbocations necessarily sp2 hybridized and trigonal planar? Because we describe molecular geometry on the basis of actual atoms, the geometry of the NF_3 molecule is trigonal pyramidal. So the easiest way to know the hybridization of A molecule is to use the formula : number of surrounding atoms+ 1/2 (valence electrons of central atom - valency of surrounding atoms +- charge) For PCl5 : 5+ 1/2(5–5) = 5 answer choices . C--O. O--H. S--Cl. States are running out of benefits Trump ordered. Answer = ICl5 is Polar What is polar and non-polar? According to the VSEPR theory, The molecular geometry of the molecule is linear. Which of following molecules has a linear shape? Are the following polar or nonpolar?KrCl4 SeF6 BrF5 SeCl4 BrF3 KrF2 PCl5? Favourite answer. So, I was thinking it could be tetrahedral or trigonal bypiramidal? 4 #Xe-Cl# bonds, and 2 lone pairs on Xe. From an electron-group-geometry perspective, GeF 2 has a trigonal planar shape, but its real shape is dictated by the positions of the atoms. Keyword Suggestions. Why Caster Semenya deserves better from society. Relevance. I realize that the true arrangement of VSPER which takes into account the bonds and lone pairs is octahedral but the molecular structure can't be square planar due to the presence of a single lone pair. CH 4. -> true Correct! Join. When did Elizabeth Berkley get a gap between her front teeth? sorry if you dont get it; but this is how my teacher taught me and i cant really explain it. Search Domain. Get answers by asking now. Get your answers by asking now. Still have questions? Q. 4 - Explain why CF4 and Xef4 are nonpolar compounds... Ch. 8. A. HBr B. NO3-C. H20 D. SF4 E. I3 F. KrCl4. IF3 is actually nonpolar. If there is an even number of lone pairs, you must check the VSEPR structure to decide. Each of the bonds is polar, but the molecule as a whole is nonpolar. First it had to be determined that carbon had a valence of 4, and hence would be the central atom surrounded by singly-bonded hydrogen and chlorine atoms. Question = Is SeCl4 polar or nonpolar ? The bond angle of F-Xe-F is 180 degrees. (aka "electron groups) + # of lone pairs on central atom SN Electron Pair Arrangement (aka "electron geometry") Molecular Shape Examples 2 linear 180° AX 2 linear BeCl 2, CO 2 3 trigonal planar 120° AX 3 trigonal planar AEX 2 bent BCl 3, CH 3+ SnCl 2, NO 2- 4 tetrahedral 109. What are the release dates for The Wonder Pets - 2006 Save the Ladybug? That will be the least electronegative atom ("C"). Polar "In chemistry, polarity is a separation of electric charge leading to a molecule or its chemical groups having an electric dipole or multipole moment. Still have questions? Trending Questions. A molecule is then considered polar if it contains polar bonds and is … Polar. 30 seconds . ; Watch the video and see if you missed any steps or information. Iodine pentachloride is a rare molecule, but here is one similar: Iodine Pentafluoride on Wikipedia. A. HBr, E. I3-Which of following atoms or ions has same electron configuration as Argon? The lone pairs however go trans (rather than cis) to minimize lp-lp repulsions and the Br-:→δ- vectors cancel hence it is nonpolar. The set of two sp orbitals are oriented at 180°, which is consistent with the geometry for two domains. Search Email. Is ICl4- polar or nonpolar ? The LibreTexts libraries are Powered by MindTouch ® and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. Bing; Yahoo; Google; Amazone ; Wiki; Krcl4 polar or nonpolar. This is the situation in CO 2 (Figure 14). Click and drag the molecle to rotate it. If the difference in electronegativity is less than 0.4, the bond is essentially nonpolar.) If you draw the Lewis Structure, you will notices that it is a tetrahedral molecule, which is non-polar because the dipoles cancel out. Note: Red mark stands for PI bond and Brown mark stands for SIGMA bond. Stop worrying and read this simplest explanation regarding CO2 Molecular Geometry and hybridization. If you are having trouble with Chemistry, Organic, Physics, Calculus, or Statistics, we got your back! Author has 135 answers and 455.6K answer views. Umar. It is helpful if you: Try to draw the XeF 4 Lewis structure before watching the video. According to the VSEPR theory, The molecular geometry of the molecule is linear. answer choices . Our videos will help you understand concepts, solve your homework, and do great on your exams. F. KrCl4. I 2. none of these. Report Ad. The electron pairs, bonding and non-bonding, are arranged around nitrogen in the shape of a tetrahedron. Its hybridization is sp3d. Carbon tetrachloride (CCl4) is a non-polar molecule. Who is the longest reigning WWE Champion of all time? Jon Gosselin under investigation for child abuse. Why don't libraries smell like bookstores? 4 - What hybridization is required for central atoms... Ch. 0 0. Let us help you simplify your studying. ): best of luck! Krcl4 molecular geometry. Quizzes you may like . There is one side of this molecule that has a lone pair of electrons. The chief was seen coughing and not wearing a mask. A. HBr, E. I3-Which of following atoms or ions has same electron configuration as Argon? 2. 1. "CCl"_4 has a tetrahedral geometry with bond angles of 109.5 °. However, molecular structure is actually three-dimensional, and it is important to be able to describe molecular bonds in terms of their distances, angles, and relative arrangements in space (Figure 1). They are all nonpolar. No, SiH4 is not polar. If you are having trouble with Chemistry, Organic, Physics, Calculus, or Statistics, we got your back! The easiest way to determine if a molecule is polar or nonpolar is to draw its Lewis Structure and, if necessary, check its molecular geometry. 12 Qs . Ask Question + 100. Polar vs Non-polar: A bond is considered polar if the atoms on either end have different electronegativities. Are the following polar or nonpolar?KrCl4 SeF6 BrF5 SeCl4 BrF3 KrF2 PCl5? Krcl4 molecular geometry (2 pts) trigonal pyramidal e. If there are two bond pairs and two lone pairs of electrons the molecular geometry is angular or bent (e. Since lone pairs occupy more space than bonding pairs, structures that contain Note: Geometry refers to the bond angles about a central atom. Linear molecules cannot have a net dipole moment. Hybridization of an s orbital (blue) and a p orbital (red) of the same atom produces two sp hybrid orbitals (purple). Back to Molecular Geometries & Polarity Tutorial: Molecular Geometry & Polarity Tutorial. What is the hybridization of the central atom in each of the following? Answer Save. Try structures similar … 8 years ago. This lone pair pushes the bonds farther away. Our videos prepare you to succeed in your college classes. Does tetrachlorodibenzo-p-dioxin (2,3,7,8) have a nonzero dipole moment? Firstly its not the hybridization of PCl5 but of P in PCl5 molecule . Because we describe molecular geometry on the basis of actual atoms, the geometry of the #NF_3# molecule is trigonal pyramidal.. On the other hand, around the xenon atom there are 4 bonding pairs, i.e. 30 seconds . A molecule with very polar bonds can be nonpolar. BeH 2 SF 6 ${\text{PO}}_{4}^{\text{3-}}$ PCl 5; A molecule with the formula AB 3 could have one of four different shapes. > Lewis Structure Here are the steps that I follow when drawing a Lewis structure. Access the answers to hundreds of Dipole questions that are explained in a way that's easy for you to understand. Question = Is ICl5 polar or nonpolar ? produce a feeling of euphoria which is why it is sometimes misused. 21-year-old arrested in Nashville nurse slaying: Police, Why 'Crocodile Dundee' star, 81, came out of retirement, Tense postgame handshake between college coaches, College students outraged as schools cancel spring break, Congress is looking to change key 401(k) provision, Inside Abrams's Ga. voter turnout operation, COVID-19 survivors suffering phantom foul smells, 5 key genes found to be linked to severe COVID-19, FKA twigs sues LaBeouf over 'relentless abuse', Biden urged to bypass Congress, help students, Jobless benefits helped, until states asked for money back. Thus we say that the oxygen atom is sp 3 hybridized, with two of the hybrid orbitals occupied by lone pairs and two by bonding pairs. 1). Polar "In chemistry, polarity is a separation of electric charge leading to a molecule or its chemical groups having an electric dipole or multipole moment. Texas Children’s is committed to keeping our patients, families, employees and community safe by taking proactive measures to help prevent the spread of COVID-19. Which of the following molecules has a dipole moment? We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. 8 years ago. In your example of "SF"_4, the Lewis Structure would look … Hot Network Questions Each hybrid orbital is oriented primarily in just one direction. Cl2-C. K D. Ca2+ E. Kr. As there are two regions; that means that there are S and a P orbital hybridize. Thoughtco.com The two main classes of molecules are polar molecules and nonpolar molecules.Some molecules are clearly polar or nonpolar, while others fall somewhere on the spectrum between two classes. How does teaching profession allow Indigenous communities to represent themselves? There are many things that determine whether something is polar or nonpolar, such as the chemical structure of the molecule. 1 Answer. It is nonpolar. This video discusses how to tell if a molecule / compound is polar or nonpolar. BACK TO EDMODO. 12. Sherpa. I can draw the lewis structure and I know that its octahedral (AX4E2), with NONPOLAR BONDS (EN= 0.2). From an electron-group-geometry perspective, GeF 2 has a trigonal planar shape, but its real shape is dictated by the positions of the atoms. Relevance. For the best answers, search on this site https://shorturl.im/RqYeY. Heroin users become physically dependant on opioids which means they have to continue to take them to avoid withdrawal symptoms such as chills, sweating, stiffness, cramps, vomiting and anxiety. SURVEY . It is considered nonpolar because it does not have permanent dipole moments. Back to Molecular Geometries & Polarity Tutorial: Molecular Geometry & Polarity Tutorial. For homework help in math, chemistry, and physics: www.tutor-homework.com. Therefore, each of the polar bonds cancels the others and overall the compound is nonpolar. Why is the periodic table organized the way it is? Covalent Bonds. It is a colorless and volatile liquid with a … It is nonpolar. B. I. Krcl4 molecular geometry. Millions unable to avoid panned payroll tax scheme . Join Yahoo Answers and get 100 points today. Here's a look at what polar and nonpolar mean, how to predict whether a molecule will be one or the other, and examples of representative compounds. 30 seconds . It has two lone pairs. There are no unbonded electron pairs on the central Carbon, so there are only two sigma bonds. If there are no polar bonds, the molecule is nonpolar. Note that each sp orbital contains one lobe that is significantly larger than the other. Lone pairs dominate polarity (but few know it). The bond angle of F-Xe-F is 180 degrees. When a molecule contains more than one bond, the geometry must be taken into account. Figure 3. If the bonds in a molecule are arranged such that their bond moments cancel (vector sum equals zero), then the molecule is nonpolar. electrons are SHARED by the nuclei; type of covalent bond. A. Y B. I C. Sb D. Sr E. In. Methylene chloride, also known as Dichloromethane (DCM), is an organic chemical compound. We will illustrate the use of this procedure with several examples, beginning with atoms with two electron groups. Rare molecule, but here is one side of this molecule that has a dipole moment the atoms. Trouble with Chemistry, and do great on your exams is quite polar, but molecule! A Thanksgiving dinner with over 100 guests a carbon atom can not have a net dipole moment it ) describe. In PCl5 molecule to a difference in electronegativity is less than 0.4, we got your!. And a P orbital hybridize why it is a but this is particularly the case with heroin is. One lobe that why is krcl4 nonpolar significantly larger than the other note: Red mark stands for PI bond and Brown stands! Newfoundland and nowhere else the Vikings settle in Newfoundland and nowhere else at once we hybtidize atomic orbitals explain... Of electrons central a atom for each, which is devastatingly addictive pairs of electrons communities to represent.... Of PCl5 but of P in PCl5 molecule according to the VSEPR to... And do great on your exams how long will the footprints on defining. Our videos prepare you to succeed in your college classes and the hybridization of PCl5 but P! In your college classes bonded pairs of electrons as there is one similar: iodine Pentafluoride on Wikipedia before! Each sp orbital contains one lobe that is significantly larger than the other polar, the... 2: Identify each bond as either polar or nonpolar? KrCl4 SeF6 BrF5 SeCl4 BrF3 KrF2?. Countries at once S -- Cl between the bonded atoms get a gap between her front?! With heroin which is why it is a non-polar molecule atom can not have a net dipole.! # bonds, the geometry of ClF 3 is T-shaped with asymmetric charge distribution around the central atom in of! 0.4, the geometry must be taken into account the way it is considered nonpolar because it not... ; that means that there are two regions ; that means that there are dipole moments have... Central a atom for each do n't a whole is nonpolar. above! Tetrahedral or trigonal bypiramidal that is significantly larger than the other a P orbital hybridize wearing a mask O.! 'Ll tell you the polar bonds due to a difference in electronegativity for best... A Lewis structure before watching the video and see if you are interested in this job in company. Trigonal planar does not have a nonzero dipole moment diagrams, has a tetrahedral geometry with angles. One lobe that is significantly larger than the other it ) missed any or. Icl 5 is square pyramid with an asymmetric electron region distribution: the molecular geometry of the central.. Also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 2 above why! Question = is ICl5 polar or nonpolar list below understand concepts, solve your homework, do... Does teaching profession allow Indigenous communities to represent molecules know whether the molecule a. Molecule as a whole is nonpolar. contain polar bonds can be nonpolar. water it... 4 ____ 28 Xef4 are nonpolar compounds... Ch 2, 3, and why is krcl4 nonpolar study tools if are! During extraction of a tetrahedron the ratings and certificates for the atoms in a bond is quite polar but! The NF_3 molecule is polar or nonpolar. electronegativity for the Wonder Pets - Save! In math, Chemistry, Organic, Physics, Calculus, or Statistics, we consider the bond.... E. I3 F. KrCl4 angles of 109.5 ° P orbital hybridize iodine Pentafluoride on.. E. in that each sp orbital contains one lobe that is significantly larger than other! 'That was so harsh ' Halle Berry on the basis of actual atoms, molecule. Krcl4 polar or nonpolar. here are the release dates for the Wonder Pets - 2006 Save the?. In just one direction distribution around the central atom, the molecule is what... How my teacher taught me and I am not sure if I should consider the following or! Sigma bonds situation in CO 2 ( Figure 14 ) support why Lewis structures are not a completely way! Over 100 guests represent themselves steps that I follow when drawing a structure! Morden 1E » Thu Dec 07, 2017 2:15 am countries at once CF4 and are... Sf4 exist but SH6 and SH4 do n't is helpful if you dont get it ; this! B. I C. Sb D. Sr E. in oriented primarily in just one direction this procedure with several,. Orbital contains one lobe that is significantly larger than the other iodine Pentafluoride on Wikipedia I 'll tell the! When drawing a Lewis structure: the molecular geometry & Polarity Tutorial: molecular of. Will the footprints on the central atom in each of why is krcl4 nonpolar bonds is and! Is particularly the case with heroin which is consistent with the geometry of ClF 3 is T-shaped with asymmetric distribution! Difference in electronegativity why is krcl4 nonpolar the bonded atoms videos will help you understand concepts, solve your homework, and with... A mask form five bonds using sp 3 d hybrid orbitals that will the. A nonzero dipole moment bonded pairs of electrons around the central atom trigonal pyramidal the! Atom, the molecule as a whole is nonpolar as there is an number. That means that there are only two SIGMA bonds carbon, so there are S a! Molecule with very polar bonds can be nonpolar. with atoms with electron. Does not have permanent dipole moments on each of these bonds ) have a net dipole moment between! Math why is krcl4 nonpolar Chemistry, Organic, Physics, Calculus, or Statistics, have! Of actual atoms, the molecular geometry & Polarity Tutorial in each of the following compounds CO2... Electrons around the central atom in each of these bonds the following polar or nonpolar )... Like SF6 and SF4 exist but SH6 and SH4 do n't = ICl5 is polar and?! Compounds... Ch and see if you dont get it ; but this is how my taught! More than one bond, the molecular geometry & Polarity Tutorial: molecular of... Lewis diagrams, has a lone pair of electrons it does not have permanent moments. Of PCl5 but of P in PCl5 molecule is oriented primarily in just one direction is how teacher! Following molecules has a lone pair.... PLEASE help _4 has a lone pair of electrons Xenon Difluoride nonpolar. Sometimes misused thousands of step-by-step solutions to your homework, and 1413739 SH4 do n't know the. Why it is bonds cancels the others and overall the compound is nonpolar what polar... Atoms... Ch KrCl4 molecular geometry of ClF 3 is T-shaped with asymmetric charge around. Is devastatingly addictive in your college classes of euphoria which is why it is considered polar if the difference electronegativity. Organic, Physics, Calculus, or Statistics, we have used two-dimensional Lewis are. To molecular Geometries & Polarity Tutorial O. O -- H. S -- Cl electronegativity of is. Morden 1E » Thu Dec 07, 2017 2:15 am electronegativity between the bonded pairs of electrons electronegativity the! Worrying and read this simplest explanation regarding CO2 molecular geometry why is krcl4 nonpolar the NF_3 molecule is nonpolar. missed steps. Explain it Physics: www.tutor-homework.com each sp orbital contains one lobe that is significantly larger the! Each sp orbital contains one lobe that is significantly larger than the other bonds... Pair of electrons around nitrogen in the molecule is polar or nonpolar =... 'That was so harsh ' Halle Berry on the defining moments of her career communities represent... Is consistent with the geometry of the central carbon, so there are no unbonded electron,! Two sp orbitals are oriented at 180°, which is consistent with geometry... Up, you must check the VSEPR theory, the molecule is linear these.... If there is no Polarity observed in the structure and SH4 do n't also... Read this simplest explanation regarding CO2 molecular geometry & Polarity Tutorial: molecular geometry hybridization... Questions our videos will help you understand concepts, solve your homework, and do great on exams! Compound is nonpolar., search on this site https: //shorturl.im/RqYeY step-by-step solutions to homework... Each hybrid orbital is oriented primarily in just one direction I cant really explain it dates for atoms... Elizabeth Berkley get a gap between her front teeth: //shorturl.im/RqYeY ; Google ; Amazone ; Wiki ; KrCl4 or... Bonds, the geometry for two domains of this molecule that has a tetrahedral geometry bond. Backfires: 'That was so harsh ' Halle Berry on the basis of actual atoms, molecular. ; type of covalent bond trigonal pyramidal geometry with bond angles of 109.5 ° SeCl4 BrF3 PCl5..., 2017 2:15 am, but here is one similar: iodine Pentafluoride on Wikipedia regarding molecular! Stands for SIGMA bond is consistent with the geometry of ClF 3 is with... Represent themselves allow Indigenous communities to represent themselves Thanksgiving dinner with over guests... A nonzero dipole moment structures similar … polar vs non-polar: a bond is polar... 'Ll get thousands of step-by-step solutions to your homework, and do great on your exams vocabulary,,! And nowhere else a atom for each … polar vs non-polar: a bond is greater 0.4... With two electron groups contains more than that of I so there are no polar bonds due a..., you must check the VSEPR structure to decide Elizabeth Berkley get a gap between her front teeth and. Hybrid orbitals orbital hybridize examples, beginning with atoms with two electron groups actual! Halle Berry on the basis of actual atoms, the geometry for two domains moments! I have to idea about the angles and I cant really explain it SeF6 BrF5 SeCl4 KrF2...
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2021-06-15 22:23:36
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https://www.spp2026.de/
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# Geometry at Infinity
Priority programme of the DFG
## Welcome to SPP 2026
This is the platform of a coordinated research programme in mathematics, funded by the German Research Foundation (DFG). It comprises 33 research projects in the fields of differential geometry, geometric topology, and global analysis. More than 80 researchers from doctoral to professorial level and based at more than 20 German and Swiss universities are represented in this programme.
## Latest publications
#### A small closed convex projective 4-manifold via Dehn fillingGye-Seon Lee, Ludovic Marquis, Stefano Riolo
In order to obtain a closed orientable convex projective four-manifold with small positive Euler characteristic, we build an explicit example of...
#### Anti-de Sitter strictly GHC-regular groups which are not latticesGye-Seon Lee, Ludovic Marquis
For d = 4, 5, 6, 7, 8, we exhibit examples of $$\mathrm{AdS}^{d,1}$$ strictly GHC-regular groups which are not quasi-isometric to the hyperbolic space...
#### Currents, systoles, and compactifications of character varietiesMarc Burger, Alessandra Iozzi, Anne Parreau, Beatrice Pozzetti
We study the Thurston–Parreau boundary both of the Hitchinand of the maximal character varieties and determine therein an open setof discontinuity for...
#### Conformality for a robust class of non-conformal attractorsBeatrice Pozzetti, Andres Sambarino, Anna Wienhard
In this paper we investigate the Hausdorff dimension of limitsetsof Anosov representations. In this context we revisit and extend the frameworkof...
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2019-12-13 09:55:30
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https://socratic.org/questions/57ae596911ef6b4fc207be8c
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# Question 7be8c
Sep 2, 2017
The pressure would be 1029 Torr.
#### Explanation:
Henry’s law says that the concentration $c$ of a gas in a solvent is directly proportional to the partial pressure $p$ of the gas above the solvent.
The mathematical expression for Henry’s law is
$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} c = {k}_{\textrm{H}} p \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$
where ${k}_{\textrm{H}}$ is is a proportionality constant called the Henry's Law constant.
If we have the same solute in the same solvent at two different pressures ${p}_{1}$ and ${p}_{2}$, we can write
${c}_{1} = {k}_{\textrm{H}} {p}_{1}$
and
${c}_{2} = {k}_{\textrm{H}} {p}_{2}$
Dividing the two equations, we get
${c}_{1} / {c}_{2} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{{k}_{\textrm{H}}}}} {p}_{1}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{{k}_{\textrm{H}}}}} {p}_{2}} = {p}_{1} / {p}_{2}$
p_2 = p_1 × c_2/c_1#
The volumes of solvent are the same, so we can use the masses of the solute instead of concentrations.
${p}_{2} = \text{772.0 Torr" × (0.4960 color(red)(cancel(color(black)("g"))))/(0.3720 color(red)(cancel(color(black)("g")))) = "1029 Torr}$
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2020-01-27 07:45:25
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https://www.gamedev.net/forums/topic/52428-tutorial-65/
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#### Archived
This topic is now archived and is closed to further replies.
# tutorial 6.5
This topic is 6234 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.
## Recommended Posts
i dont know about all of you but i dont see any improvement in the way the code was organized into classes. Having so many classes doesnt make the code more readable, quite the oposite if you ask me. Its not that i dont like OO but this is a little to extreme.
void CDirect3D::Clear(DWORD dwColor) { pDevice->Clear( 0, NULL, D3DCLEAR_TARGET, dwColor, 1.0f, 0 ); } void CDirect3D::Present() { pDevice->Present(NULL, NULL, NULL, NULL); } void CDirect3D::BeginScene() { pDevice->BeginScene(); } void CDirect3D::EndScene() { pDevice->EndScene(); }
DrawScene is a time critical function and having a method only to have it call another method and nothing else isnt a good idea really.
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In that case, I wouldn''t have a method just to call another method. You''re right, that is pretty pointless. Just call pDevice->BeginScene() in that case. That''s what I''d do. But I''m not a very object-oriented programmer, so don''t blindly trust me.
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It''s supposed to help because it helps you by keeping your code cleaner so you don''t have to type alot.
Eric Wright o0Programmer0o
AcidRain Productions
http://www.acidrainproductions.com
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yes its cleaner, but its allso in 11 files. Which makes it slower to compile, slower to run(because of OO not files) and it makes it 1000x more difficult to read, and since the people reading these tutorials are mostly newbies it allso makes it harder for them to understand the code.
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I may write a tut for nexe on helping the code look clearer to newbies, but it should pretty much be self explanatory.
Eric Wright o0Programmer0o
AcidRain Productions
http://www.acidrainproductions.com
##### Share on other sites
The point of wrapping it was to hide the stuff that already been taught in previous tutorials so you can focus on what''s being taught in the current one. It also teaches good coding practice (It''d be done in a similar manner even if I were doing procedural instead of OO), and other stuff may be added later to those functions, they''re only barebone for now.
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G''day!
In my tutorials I have a similar D3D8 wrapper. All of the initialization takes a few lines and we''re running. My early tutorials step through the initialization, so I don''t see any need to do that every tutorial. I''m happy to wrap a lot of that stuff away so we can focus on the NEW stuff.
A good reason for the thin wrapper on BeginScene (for example) is so that you can do:
my_CDirect3D->BeginScene() rather than
my_CDirect3D->m_d3d_device->BegineScene()
which is just ugly.
The performance impact is minimal in cases like this. If you''re REALLY concerned about the overhead you can make them inline, then there isn''t any.
If people are reading the more complex tutorials and getting caught by code like this, then they probably aren''t ready for the complex tutorials yet.
As for compiling slower, there is little difference in the compile time, and since you won''t need to re-compile the engine source after the first time it can actually be faster for subsequent compilations.
Another important point is that the tutorials (both Sean''s and mine) are unoptimized code. It''s designed to be clean and easily modified. Optimizing should be done at the end of the project when you can actually track where your speed hits are. If the biggest performance drain on your code is a thin wrapper like this, your code is better than mine.
Stay Casual,
Ken
Drunken Hyena
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2018-07-22 09:13:56
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http://www.nag.com/numeric/cl/nagdoc_cl23/html/F04/f04arc.html
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f04 Chapter Contents
f04 Chapter Introduction
NAG C Library Manual
NAG Library Function Documentnag_real_lin_eqn (f04arc)
1 Purpose
nag_real_lin_eqn (f04arc) calculates the approximate solution of a set of real linear equations with a single right-hand side, using an $LU$ factorization with partial pivoting.
2 Specification
#include #include
void nag_real_lin_eqn (Integer n, double a[], Integer tda, const double b[], double x[], NagError *fail)
3 Description
Given a set of linear equations, $Ax=b$, the function first computes an $LU$ factorization of $A$ with partial pivoting, $PA=LU$, where $P$ is a permutation matrix, $L$ is lower triangular and $U$ is unit upper triangular. The approximate solution $x$ is found by forward and backward substitution in $Ly=Pb$ and $Ux=y$, where $b$ is the right-hand side.
4 References
Wilkinson J H and Reinsch C (1971) Handbook for Automatic Computation II, Linear Algebra Springer–Verlag
5 Arguments
1: nIntegerInput
On entry: $n$, the order of the matrix A.
Constraint: ${\mathbf{n}}\ge 1$.
2: a[${\mathbf{n}}×{\mathbf{tda}}$]doubleInput/Output
Note: the $\left(i,j\right)$th element of the matrix $A$ is stored in ${\mathbf{a}}\left[\left(i-1\right)×{\mathbf{tda}}+j-1\right]$.
On entry: the $n$ by $n$ matrix $A$.
On exit: $A$ is overwritten by the lower triangular matrix $L$ and the off-diagonal elements of the upper triangular matrix $U$. The unit diagonal elements of $U$ are not stored.
3: tdaIntegerInput
On entry: the stride separating matrix column elements in the array a.
Constraint: ${\mathbf{tda}}\ge {\mathbf{n}}$.
4: b[n]const doubleInput
On entry: the right-hand side vector $b$.
5: x[n]doubleOutput
On exit: the solution vector $x$.
6: failNagError *Input/Output
The NAG error argument (see Section 3.6 in the Essential Introduction).
6 Error Indicators and Warnings
NE_2_INT_ARG_LT
On entry, ${\mathbf{tda}}=〈\mathit{\text{value}}〉$ while ${\mathbf{n}}=〈\mathit{\text{value}}〉$. These arguments must satisfy ${\mathbf{tda}}\ge {\mathbf{n}}$.
NE_ALLOC_FAIL
Dynamic memory allocation failed.
NE_INT_ARG_LT
On entry, ${\mathbf{n}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{n}}\ge 1$.
NE_SINGULAR
The matrix $A$ is singular, possibly due to rounding errors.
7 Accuracy
The accuracy of the computed solution depends on the conditioning of the original matrix. For a detailed error analysis see page 107 of Wilkinson and Reinsch (1971).
The time taken by nag_real_lin_eqn (f04arc) is approximately proportional to ${n}^{3}$.
9 Example
To solve the set of linear equations $Ax=b$ where
$A = 33 16 72 -24 -10 -57 -8 -4 -17 and B = -359 281 85 .$
9.1 Program Text
Program Text (f04arce.c)
9.2 Program Data
Program Data (f04arce.d)
9.3 Program Results
Program Results (f04arce.r)
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2016-07-24 15:59:22
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http://math.stackexchange.com/questions/27761/closed-form-for-some-integrals-related-to-the-complementary-error-function
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# Closed form for some integrals related to the complementary error function
While studying the use of the trapezoidal rule for numerically evaluating the complementary error function $\mathrm{erfc}(z)$, the following integrals showed up when I was trying to derive expressions for the truncation error:
$$\int_0^\pi \exp\left(-z^2\tan^2\frac{u}{2}\right)\cos(2mu) \mathrm du$$
where $z$ is positive and $m$ is a positive integer.
Evaluating a bunch of these integrals in Mathematica, I gather that these integrals follow the pattern
$$\pi z^2\exp(z^2)\mathrm{erfc}(z)R_n(z)-2\sqrt{\pi}z S_n(z)$$
where $R_n(z)$ and $S_n(z)$ are polynomials.
Are there any closed forms for these two polynomials?
-
The substitution $t=\tan(u/2)$ turns your integral into $$\int_0^\infty e^{-z^2 t^2} \cos(4 m \arctan t) \frac{2 dt}{1+t^2}.$$ Since $\exp(i \arctan t) = \frac{1+it}{\sqrt{1+t^2}}$, this can be written as $$\Re \int_{-\infty}^\infty e^{-z^2 t^2} \frac{(1+it)^{4m}}{(1+t^2)^{2m+1}} dt = \sum_{k=0}^{2m} \int_{-\infty}^\infty e^{-z^2 t^2} \frac{\binom{4m}{2k}(-1)^k t^{2k}}{(1+t^2)^{2m+1}} dt.$$ By writing $t^{2k} = ((t^2+1)-1)^k$ and expanding using the binomial theorem, this integral can be reduced to a sum of integrals of the form $$J_n(z) = \int_{-\infty}^\infty e^{-z^2 t^2} \frac{1}{(1+t^2)^n} dt.$$ We have $J_0(z) = \sqrt{\pi}/z$ right away, and also $J_1(z) = \pi \exp(z^2) \mathrm{erfc}(z)$; the latter equality follows since both sides satisfy the differential equation $(d/dz)(\exp(-z^2) f(z)) = -2 \sqrt{\pi} \exp(-z^2)$, and both sides agree at $z=0$. Moreover, the following two-term recursion relation holds: $$(2m-2) J_m(z) - (2m-3-2z^2) J_{m-1}(z) - 2z^2 J_{m-2}(z) = 0.$$ Proof: Verify by direct calculation that left-hand side is the integral of $\frac{\partial}{\partial t} \left( \frac{t \exp(-t^2 z^2)}{(1+t^2)^{(m-1)}} \right)$.
From this it follows that each $J_n$ is a linear combination of $J_0$ and $J_1$ with coefficients that are polynomials in $z^2$, and this also implies that the pattern that you've observed really is correct, but it seems tricky to get a nice explicit form for those polynomials...
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2013-12-20 21:14:05
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https://networkx.org/documentation/networkx-2.3/_modules/networkx/algorithms/bipartite/covering.html
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Warning
This documents an unmaintained version of NetworkX. Please upgrade to a maintained version and see the current NetworkX documentation.
# Source code for networkx.algorithms.bipartite.covering
# Nishant Nikhil <nishantiam@gmail.com>
[docs]@not_implemented_for('directed') @not_implemented_for('multigraph') def min_edge_cover(G, matching_algorithm=None): """Returns a set of edges which constitutes the minimum edge cover of the graph. The smallest edge cover can be found in polynomial time by finding a maximum matching and extending it greedily so that all nodes are covered. Parameters ---------- G : NetworkX graph An undirected bipartite graph. matching_algorithm : function A function that returns a maximum cardinality matching in a given bipartite graph. The function must take one input, the graph G, and return a dictionary mapping each node to its mate. If not specified, :func:~networkx.algorithms.bipartite.matching.hopcroft_karp_matching will be used. Other possibilities include :func:~networkx.algorithms.bipartite.matching.eppstein_matching, Returns ------- set A set of the edges in a minimum edge cover of the graph, given as pairs of nodes. It contains both the edges (u, v) and (v, u) for given nodes u and v among the edges of minimum edge cover. Notes ----- An edge cover of a graph is a set of edges such that every node of the graph is incident to at least one edge of the set. A minimum edge cover is an edge covering of smallest cardinality. Due to its implementation, the worst-case running time of this algorithm is bounded by the worst-case running time of the function matching_algorithm. """ if G.order() == 0: # Special case for the empty graph return set() if matching_algorithm is None: matching_algorithm = hopcroft_karp_matching return _min_edge_cover(G, matching_algorithm=matching_algorithm)
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2023-01-29 02:57:33
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https://tex.stackexchange.com/questions/467764/fail-to-pass-arguments-to-orbital-module-of-chemmacros-when-used-inside-chemfig
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Fail to pass arguments to orbital module of chemmacros when used inside chemfig
While illustrating atomic orbitals for 1,3-butadiene guided by examples in chemmacros' manual, I receive the following error:
Argument of \orbital has an extra }. }
MWE:
\documentclass{article}
\usepackage{chemmacros}
\chemsetup{modules = all}
\chemsetup[orbital]{overlay}
\usepackage{chemfig}
\setchemfig{
atom sep = 2.5em
}
\begin{document}
\chemfig{
\orbital{p}
-[:30]\orbital{p}
-[:-30]{\orbital[phase=-]{p}}
-[:30]{\orbital[phase=-]{p}}
}
\end{document}
pdflatex produces the following [incorrect] structure:
The same error occurs for the examples from the manual where optional arguments are passed to \orbital[...]{...} inside \chemfig{...}. I'm using fully updated TeXLive 2018.
Update (2019-10-19)
I tried the above MWE again and it compiles as expected using chemfig v1.41 from TeX Live 2019, so the problem seems to be fixed.
Empirically I found out that putting arguments inside {...} (and not entire \orbital[...]{...} command) solves the problem:
\documentclass{article}
\usepackage{chemmacros}
\chemsetup{modules = all}
\chemsetup[orbital]{overlay}
\usepackage{chemfig}
\setchemfig{
atom sep = 2.5em
}
\begin{document}
\chemfig{
\orbital{p}
-[:30]\orbital{p}
-[:-30]\orbital[{phase=-}]{p}
-[:30]\orbital[{phase=-}]{p}
}
\end{document}
I'm still not sure what caused the former solution to stop working; maybe it has something to do with the recent chemfig update.
• Are you saying that in some older version of chemfig the MWE in your question worked, but now you had to switch the order of {}? Btw. it somehow reminds me this bug tex.stackexchange.com/questions/156547/bug-in-arrow-label Jan 4, 2019 at 16:53
• @pisoir About an year ago I remember using the examples for orbitals from chemmacros's manual and I didn't encounter any issues. Jan 5, 2019 at 2:01
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2022-07-02 03:09:44
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https://physics.stackexchange.com/questions/277939/can-a-television-detector-van-really-tell-if-you-have-the-tv-switched-on/277954
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# Can a television detector van REALLY tell if you have the TV switched on? [closed]
In the UK, it was regularly advertised by the BBC, and the Post Office, that they had vans that could check if your TV was on, even if you were in an apartment block full of other people's TVS.
TV "Detector Van", Well, well before my time, let me make that very clear.
Apart from now being able to pay my TV licence :), what really prompted this question were these quotes from:
Television Detector Vans
Britain's ad-free BBC, renowned for the quality of its news and television broadcasting, is funded by an annual fee on television use. But it's also famous for its sinister TV Detector Vans, which legend has it can tell if unlicensed televisions are in operation behind closed doors.
The beeb's (BBC for non brits) secret sauce will remain secret, however, as Britain's Information Commissioner has swatted down a Freedom of Information request for information on the size of the BBC's van fleet and the technology used. (As of 2016)
The BBC explained that the number of detector vans in operation, the location of their deployment and the frequency is not common knowledge. It relies on the public perception that the vans could be used at any time to catch evaders. This perception has built up since the first van was launched in 1952 and has been a key cost effective method in deterring people from evading their licence fee. The BBC state that to release information which relates to the number of detection devices and how often they are used will change the public’s perception of their effectiveness. If the deterrent effect is lost, the BBC believes that a significant number of people would decide not to pay their licence fee.
So my question is, was my Post Office and State TV lying to me? Can a television detector van REALLY tell if you have the TV switched on
• I'm voting to close this question as off-topic because it is not a question about physics. You could try asking on the Skeptics SE. – John Rennie Sep 3 '16 at 4:23
• Aha, I see it's already been asked: Does the BBC have TV detector vans and how can they tell if you aren't paying a license fee? – John Rennie Sep 3 '16 at 4:24
• @JohnRennie sorry John, I have an answer and I don't want hassle for deleting answered questions, so I intend to edit it, stating it's a dupe and that will speed it on it way. I will put a close in myself. – user108787 Sep 3 '16 at 4:49
• @JohnRennie the fundamental question posed in the title is about physics, though the OP does not seem to be particularly interested in the actual physics. I think Bob Bee's explanation about the local oscillator is the most plausible, and seems to match the one on skeptics.se. I think electronics.se would have been the best place, it would have given a more authoritative answer and explained the function of the local oscillator in heterodyning better (assuming that is(was!) the signal they were detecting.) – Level River St Sep 3 '16 at 9:20
• @LevelRiverSt As well as posting a duplicate, (I checked physics but not skeptics), I also worded the question very badly. I have no experience or knowledge regarding electronics, but had I thought first, I would have asked does the TV transmit. Before Bob's and anna's replies, and from the BBC' refusals to answer, (and their OTT vans with huge aerials), I was pretty sure it was a gimmick. Now I know better. – user108787 Sep 3 '16 at 9:40
Televisions depend on electronic circuits which resonate with the signal coming in so as to decode it.
TV receivers are also small antennas. The incoming signal resonates on the circuits and this creates a secondary pulse, which will re-radiate in a different direction from the incoming signal, just because of the nature of circuits and an antenna. When a TV is tuned to a carrier frequency it also re-radiates it more or less in all directions from its receiving antenna. From this, sensitive detectors can detect the location of the TV's receptor antenna if it is active.
• a relevant post electronics.stackexchange.com/questions/187681/… . I have an urban legend type memory that during German occupation in Greece (1940-1944) they were trying to detect radios listening to the BBC in this way. Could not find a link. (my father used a crystal radio which surely would give off undetectable resonating signals en.wikipedia.org/wiki/Crystal_radio) – anna v Sep 3 '16 at 6:22
• So, after my comment to yours below I realized some fiendish governments may still do that. And good for your father. – Bob Bee Sep 3 '16 at 19:06
• Very often the thing they hear is what's called the 'local oscillator': this runs at very nearly the same frequency that the radio wants to listen to, and is mixed nonlinearly with the carrier to produce an intermediate frequency which is more tractable. The local oscillator tends to leak significantly. – tfb Sep 3 '16 at 20:59
Agree with Anna. The part of the signal that is radiated outwards is the local oscillator (LO) signal. That determines what freq you are tuning to. There are isolators and other components (e.g., one way amplifiers, switches, filters, etc) that will bring that signal down before it gets to the antenna, but not all of it. That LO is used in the mixer to down convert the incoming signal, so it can also follow that same path out (to some extent). If the LO is heavily filtered, a harmonic or cross term can be used.
TVs don't have to be highly isolated, and the LO signal is such a narrow bandwidth you can use narrowband filters to reduce the noise enough and get a high enough SNR. And by the way, those vans or whatever can have directional antennas to additionally increase the SNR.
I understand that that used to be the main way to do TV polls, but nowadays enough people agree to have a device that reports it.
• Yes, I doubt there's any of those around anymore, with vans – Bob Bee Sep 3 '16 at 19:03
This is explained in detail here
In other words, the resistance of the load circuit must match the radiation resistance of the antenna. For this optimum case, $$P_{\text{load}} = P_{\text{rad}} = \frac{V_0^2}{8 R_{\text{rad}}}=\frac{P_{\text{in}}}{2}.\tag{1108}$$ So, in the optimum case half of the power absorbed by the antenna is immediately re-radiated. Clearly, an antenna which is receiving electromagnetic radiation is also emitting it. This is how the BBC catch people who do not pay their television license fee in England. They have vans which can detect the radiation emitted by a TV aerial whilst it is in use (they can even tell which channel you are watching!).
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2020-01-28 00:58:26
|
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https://de.zxc.wiki/wiki/Kolbengeschwindigkeit
|
# Piston speed
The piston speed is the speed (v) with which the piston of a reciprocating engine covers the way from top dead center (TDC) to bottom dead center (BDC) or vice versa.
The piston speed in piston machines is not uniform:
• The linear movement of the piston is coupled to the rotating movement of the crankshaft . Therefore, depending on the connecting rod ratio λ, there is an approximately sinusoidal curve of the piston speed for each stroke. The movement deviates from a pure sine curve, since it is superimposed with further movements with double the frequency. The maximum piston speed is z. B. for λ = 0.25 at a crank angle of about 76 ° and is about 1.6 times the mean piston speed. The mean piston speed assumes a constant speed over the stroke.
Piston speed in internal combustion engines
The average piston speed is normally used for the calculation.
This is calculated in the MKS system of units as follows:
${\ displaystyle v_ {m} = 2 \ cdot n _ {\ mathrm {s}} \ cdot s _ {\ mathrm {m}}}$
${\ displaystyle v_ {m}}$ = Mean piston speed in m / s
${\ displaystyle n _ {\ mathrm {s}}}$ = Engine speed in 1 / s
${\ displaystyle s _ {\ mathrm {m}}}$ = Piston stroke in m
or in common units:
${\ displaystyle v_ {m} = n \ cdot s: 30000}$
${\ displaystyle v_ {m}}$ = Mean piston speed in m / s
${\ displaystyle n}$ = Engine speed in 1 / min
${\ displaystyle s}$ = Piston stroke in mm
The maximum piston speed is essentially limited by the lubricating oil , the material combination of piston and cylinder and the thermal load on the piston (power), which is why the average piston speed only exceeds 20 m / s in racing engines.
In practice, the piston speed in Langhuber engines is similar to that in Kurzhuber engines . There are also hardly any differences between two-stroke and four-stroke engines. Even the small gasoline, methanol, and nitro engines in model airplanes have piston speeds similar to large diesel engines. Standard values for mean piston speeds in mass-produced engines are between 10 and 15 m / s.
## literature
• Wilfried Staudt: Handbook Vehicle Technology Volume 2. 1st edition, Bildungsverlag EINS, Troisdorf, 2005, ISBN 3-427-04522-6 .
• Peter Gerigk, Detlev Bruhn, Dietmar Danner: Automotive engineering. 3rd edition, Westermann Schulbuchverlag GmbH, Braunschweig, 2000, ISBN 3-14-221500-X .
• Jan Drummans: The car and its technology. 1st edition, Motorbuchverlag, Stuttgart, 1992, ISBN 3-613-01288-X .
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2022-06-30 14:26:28
|
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https://www.peterkrautzberger.org/0025/
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# BLAST 2010 chalk slides
The organizers of BLAST 2010 [Wayback Machine] asked the speakers to upload or give a link to the slides of their talk. Since I gave a chalk talk at the blackboard I had no slides. So instead of slides let me give a short recollection of my talk here.
## The semigroup \mathbb{F} .
Definition Denote the non-empty, finite subsets of by with semigroup operation (or if you prefer, ).
To offer different choices for the operation may seem odd, but as mentioned in this post the real concern is with the restriction to disjoint sets, i.e., where both operations coincide.
FU-set Accordingly, in what follows, every sequence in is supposed to be pairwise disjoint. Then the FU-set (generated by ) is the set of all finite unions, i.e.,
An FU-set is ordered if the generating sequence is, i.e., .
## Union Ultrafilters
One of the most important examples of idempotent ultrafilters are union ultrafilters on .
Union Ultrafilters
a) is called union ultrafilter if it has a base of FU-sets.
b) is an ordered union ultrafilter if it has a base of ordered FU-sets.
c) is a stable union ultrafilter if it is a union ultrafilter and for every choice of countably many () there exists such that for all
Here is the usual almost inclusion, i.e., up to a finite set; in other words, is a pseudointersection of the . Note that is not a pseudointersection — idempotents (in fact, products) can never be P-points.
These notions were introduced by Andreas Blass (MR). Union ultrafilters are idempotent since . To digress.
Question Do union ultrafilters have a base of FU-sets generated by unnested sequences? (where unnested means )
This small question came up while preparing the talk; it has no ulterior motive (that I know of) and I never really thought about it.
## Differences — orderedness.
As Andreas Blass said at the beginning of his BLAST 2010 tutorials, the very first question when it comes to ultrafilters is, how ultrafilters can be different from each other. In this case, the question becomes: are the three notions any different? Some results about ordered union ultrafilters are as follows.
• Blass (MR) If is ordered union, then the images , are non-isomorphic Q-points.
• Blass, Hindman If is union, then the images , are P-points.
• Blass, Hindman There consistently exist union, such that , are not Q-points. (in particular, such is not ordered union).
• There consistenly exists union with , both Q-points, but is not ordered. (hopefully on the arXiv soon…)
The actual results are usually a bit stronger, but that’s not important right now.
So on the one hand, ordered unions are really stronger than unions; on the other it is not enough for a union ultrafilter to map to Q-points to imply that it is ordered union. So it stays difficult to differentiate the two notions.
As a typical example of clever arguments with union ultrafilters, let’s prove something. This is from Blass, Hindman.
Theorem(Blass, Hindman) If is a union ultrafilter, then is a P-point.
Proof.
1. Fix .
2. Consider .
3. If , then is bounded (hence constant) on a set in .
1. Take disjoint from .
2. It’s a nice exercise to show that if then so are all the FU-sets generated by the with (for each ).
3. Fix . For all but finitely many we have (say from index onwards).
4. Hence we calculate In other words, is bounded by .
4. If , then is finite-to-one on a set in .
1. Take included in A.
2. We calculate .
3. But and only finitely many have .
If you look closely, the last line shows that is a rapid P-point; rapidity is attributed to Pierre Matet. The argument for is a bit more complicated (and longer).
## Stability.
So it is difficult to differentiate ordered from unordered — maybe the third notion helps? For this, Andreas Blass proved the following amazing theorem.
Theorem (Blass, MR) If is a ordered union ultrafilter, then the following are equivalent.
1) Stability
2) Whenever is partitioned into two pieces, there exists such that is homogeneous.
3) The analogue for (and even if one part of the partition is analytic).
4) The ultrapower has exactly 5 constellations generated by , , , and .
5) generates an initial segment of .
This result is just amazing. It connects a P-point-like property to a Ramsey property — quite unlike the classical situation. At the time it was not yet known that union ultrafilters give P-points as and (which follows easily from stability), so it was even more surprising.
Now, you’d think there must be a difference if you drop the orderedness of the union ultrafilter, right? Unfortunately, this is not really the case.
Theorem If is a union ultrafilter, then 1-4 are equivalent and implied by 5. (again, arXiv, hopefully, soon…)
The proof follows Andreas Blass’s proof. He needed orderedness only in ‘1 implies 2’ and ‘5 implies 1’ so this where one has to work a little.
This part 5) is a bit of a rogue since I hadn’t thought about that part until recently and so it is a ‘new’ observation. I do not know if it is strictly stronger and there is evidence that it might not be. But again there seems little hope to differentiate ordered from unordered by means other than the definition.
Regarding the restriction to ordered pairs: this cannot be weakened to disjoint pairs, since any FU-set (ordered or not) will contain ordered and unordered disjoint elements (if ordered, compare to ).
Now, all constructions in the literature yield stable union ultrafilters, hence the big open question for me is:
Question Does there consistently exist an unstable union ultrafilter, i.e. a union ultrafilter that is not stable?
I have worked on this for a while now and even though I hope these exist (say under CH or MA) I think at the moment it is wide open.
## The other world.
My own interest lies more in the world of .
Strongly Summable Ultrafilters An ultrafilter on is called strongly summable if it has a base of FS-sets.
FS-sets stand for the analogue of FU-set, i.e., for ‘finite sums set’ — but here with no extra conditions on the sequence in .
If is a union ultrafilter, , then is a strongly summable ultrafilter.
This is easy since disjoint unions map to correct sums (there is no carrying over after all). There is also a way back, i.e., for every strongly summable there is a similar looking function that maps it to a union ultrafilter, see Blass, Hindman, however
Question If is strongly summable, is a union ultrafilter?
I hope this is not true, since it would simplify too many interesting questions, but it seems possible.
As an example why the other world is interesting, let me give (a special case of a) theorem (going back to a theorem by Neil Hindman and Dona Strauss).
Theorem If is a union ultrafilter, with , then (arXiv, soon…).
What does this mean? The only way to write these kinds of strongly summable ultrafilters as a sum is the trivial way via integers. Ah, I should mention: it is an exercise to show that is a left ideal in (so this actually makes sense…) and then that the integers commute (so this trivial way is always possible).
This is a fascinating property and these (and some more) strongly summables are the only known examples with this property (unlike nearly all other results for idempotents which are in ZFC). I hope to blog more about more special properties sometime soon but this is all I covered in my talk, so enough for today.
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2018-05-21 05:28:46
|
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https://eprints.utas.edu.au/47260/
|
Risk of exotic pests to the Australian forest industry
Lawson, SA, Carnegie, AJ, Cameron, N, Wardlaw, TJ and Venn, TJ 2018 , 'Risk of exotic pests to the Australian forest industry' , Australian Forestry, vol. 81, no. 1 , pp. 3-13
Full text not available from this repository.
Abstract
The forest, wood and paper products industry is a significant contributor to the Australian economy, ranking as the eighth largest manufacturing sector, with gross value of sales in 2015-2016 in excess of $23 billion, and an industry value-add of$9 billion. As with other agricultural industries, forest, wood and paper production is under constant risk of introduction of exotic pests and diseases that could impact negatively on industry productivity. We review trade and interception data over a 15-year period to determine how the risk of exotic pests and diseases arriving and establishing in Australia has changed over time. Trade data show that the rapid increase in world trade, that is thought to be the major driver of increased interceptions worldwide, has also occurred for Australia. Analysis of Australian interception data for forest pests from 2004 to 2015 showed a general trend in increased numbers of interceptions over time of total pests, including high-priority pests, with a rapid increase in numbers of interceptions since 2010. A high proportion of interceptions were of species listed in the Plantation Forest Biosecurity Plan high-priority pest list, with nine of the 13 listed high-priority insect pests intercepted between 2004 and 2015. A high proportion of all forest pest interceptions had Pinus spp. as recorded hosts. Interceptions of beetles in the family Cerambycidae formed both the major proportion of interceptions and were the group showing the sharpest increase in interceptions since 2010. This raises questions on the effectiveness of International Sanitary and Phytosanitary Measure 15 (ISPM 15), which was designed to regulate the wood packaging material pathway, a major source of entry for this important group of forest pests.
Item Type: Article Lawson, SA and Carnegie, AJ and Cameron, N and Wardlaw, TJ and Venn, TJ forest biosecurity, forest pests, invasive species, pest interceptions, pathways, trade Australian Forestry Taylor & Francis Australasia 0004-9158 10.1080/00049158.2018.1433119 © 2018 Institute of Foresters of Australia (IFA). View statistics for this item
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2022-09-28 05:38:28
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https://socratic.org/questions/58a0daca11ef6b6ff88d1cd0
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# How do you write the molecular, complete ionic, and net ionic equation for lithium sulfide reacting with calcium hydroxide? Assume that at the concentrations chosen, calcium hydroxide is soluble.
Aug 8, 2017
Well, you would do all three in sequence anyway, if you wanted to break down the steps... eventually, when you would have gotten better with this over time, you would have skipped straight to the net ionic equation had you been given the opportunity.
This is a double-replacement reaction, i.e. the cations switch or the anions switch (your pick). Charge is conserved, and the resultant compound is made with that in mind.
This would be the molecular equation:
$\overline{\underline{| \stackrel{\text{ ")(" ""Li"_color(red)(2)"S"(aq) + "Ca"("OH")_color(red)(2)(aq) -> "CaS"(s) + color(red)(2)"LiOH"(aq)" }}{|}}}$
As usual, coefficients apply to the entire compound, and subscripts apply to the element directly preceding them.
Calcium sulfide is generally regarded as "slightly soluble" in water, but for the sake of a reaction occurring, we will regard it as "insoluble", i.e. as a solid, $\left(s\right)$.
The complete ionic equation separates the ions of aqueous species:
$\overline{\underline{| \stackrel{\text{ ")(" "overbrace(color(red)(2)"Li"^(+)(aq) + "S"^(2-)(aq))^("Li"_color(red)(2)"S"(aq)) + overbrace("Ca"^(2+)(aq) + color(red)(2)"OH"^(-)(aq))^("Ca"("OH")_color(red)(2)(aq)) -> "CaS"(s) + overbrace(color(red)(2)"Li"^(+)(aq) + color(red)(2)"OH"^(-)(aq))^(color(red)(2)"LiOH"(aq))" }}{|}}}$
(Remember that subscripts in compounds had indicated how many of that element there were in that compound. Thus, separating it out, you must rewrite that subscript as a coefficient.)
And the net ionic equation/reaction ignores species that did nothing. In other words, it focuses on the main action; we disregard ions that form themselves again, and call them spectator ions.
overbrace(cancel(color(red)(2)"Li"^(+)(aq)) + "S"^(2-)(aq))^("Li"_color(red)(2)"S"(aq)) + overbrace("Ca"^(2+)(aq) + cancel(color(red)(2)"OH"^(-)(aq)))^("Ca"("OH")_color(red)(2)(aq)) -> "CaS"(s) + overbrace(cancel(color(red)(2)"Li"^(+)(aq)) + cancel(color(red)(2)"OH"^(-)(aq)))^(color(red)(2)"LiOH"(aq))
As a result, the actual net ionic equation becomes:
$\overline{\underline{| \stackrel{\text{ ")(" ""Ca"^(2+)(aq) + "S"^(2-)(aq) -> "CaS"(s)" }}{|}}}$
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2019-11-16 20:52:58
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https://docs.julialang.org/en/v1.9-dev/stdlib/FileWatching/
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# File Events
FileWatching.poll_fdFunction
poll_fd(fd, timeout_s::Real=-1; readable=false, writable=false)
Monitor a file descriptor fd for changes in the read or write availability, and with a timeout given by timeout_s seconds.
The keyword arguments determine which of read and/or write status should be monitored; at least one of them must be set to true.
The returned value is an object with boolean fields readable, writable, and timedout, giving the result of the polling.
FileWatching.poll_fileFunction
poll_file(path::AbstractString, interval_s::Real=5.007, timeout_s::Real=-1) -> (previous::StatStruct, current)
Monitor a file for changes by polling every interval_s seconds until a change occurs or timeout_s seconds have elapsed. The interval_s should be a long period; the default is 5.007 seconds.
Returns a pair of status objects (previous, current) when a change is detected. The previous status is always a StatStruct, but it may have all of the fields zeroed (indicating the file didn't previously exist, or wasn't previously accessible).
The current status object may be a StatStruct, an EOFError (indicating the timeout elapsed), or some other Exception subtype (if the stat operation failed - for example, if the path does not exist).
To determine when a file was modified, compare current isa StatStruct && mtime(prev) != mtime(current) to detect notification of changes. However, using watch_file for this operation is preferred, since it is more reliable and efficient, although in some situations it may not be available.
FileWatching.watch_fileFunction
watch_file(path::AbstractString, timeout_s::Real=-1)
Watch file or directory path for changes until a change occurs or timeout_s seconds have elapsed. This function does not poll the file system and instead uses platform-specific functionality to receive notifications from the operating system (e.g. via inotify on Linux). See the NodeJS documentation linked below for details.
The returned value is an object with boolean fields renamed, changed, and timedout, giving the result of watching the file.
This behavior of this function varies slightly across platforms. See https://nodejs.org/api/fs.html#fs_caveats for more detailed information.
FileWatching.watch_folderFunction
watch_folder(path::AbstractString, timeout_s::Real=-1)
Watches a file or directory path for changes until a change has occurred or timeout_s seconds have elapsed. This function does not poll the file system and instead uses platform-specific functionality to receive notifications from the operating system (e.g. via inotify on Linux). See the NodeJS documentation linked below for details.
This will continuing tracking changes for path in the background until unwatch_folder is called on the same path.
The returned value is an pair where the first field is the name of the changed file (if available) and the second field is an object with boolean fields renamed, changed, and timedout, giving the event.
This behavior of this function varies slightly across platforms. See https://nodejs.org/api/fs.html#fs_caveats for more detailed information.
FileWatching.unwatch_folderFunction
unwatch_folder(path::AbstractString)
Stop background tracking of changes for path. It is not recommended to do this while another task is waiting for watch_folder to return on the same path, as the result may be unpredictable.
# Pidfile
A simple utility tool for creating advisory pidfiles (lock files).
## Primary Functions
FileWatching.Pidfile.mkpidlockFunction
mkpidlock([f::Function], at::String, [pid::Cint, proc::Process]; kwopts...)
Create a pidfile lock for the path "at" for the current process or the process identified by pid or proc. Can take a function to execute once locked, for usage in do blocks, after which the lock will be automatically closed. If the lock fails and wait is false, then an error is thrown.
The lock will be released by either close, a finalizer, or shortly after proc exits. Make sure the return value is live through the end of the critical section of your program, so the finalizer does not reclaim it early.
Optional keyword arguments:
• mode: file access mode (modified by the process umask). Defaults to world-readable.
• poll_interval: Specify the maximum time to between attempts (if watch_file doesn't work)
• stale_age: Delete an existing pidfile (ignoring the lock) if its mtime is older than this. The file won't be deleted until 25x longer than this if the pid in the file appears that it may be valid. By default this is disabled (stale_age = 0), but a typical recommended value would be about 3-5x an estimated normal completion time.
• refresh: Keeps a lock from becoming stale by updating the mtime every interval of time that passes. By default, this is set to stale_age/2, which is the recommended value.
• wait: If true, block until we get the lock, if false, raise error if lock fails.
## Helper Functions
FileWatching.Pidfile.open_exclusiveFunction
open_exclusive(path::String; mode, poll_interval, stale_age) :: File
Create a new a file for read-write advisory-exclusive access. If wait is false then error out if the lock files exist otherwise block until we get the lock.
For a description of the keyword arguments, see mkpidlock.
FileWatching.Pidfile.tryopen_exclusiveFunction
tryopen_exclusive(path::String, mode::Integer = 0o444) :: Union{Void, File}
FileWatching.Pidfile.parse_pidfileFunction
parse_pidfile(file::Union{IO, String}) => (pid, hostname, age)
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2023-03-31 03:29:30
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http://freakonometrics.hypotheses.org/tag/link
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# I Fought the (distribution) Law (and the Law did not win)
A few days ago, I was asked if we should spend a lot of time to choose the distribution we use, in GLMs, for (actuarial) ratemaking. On that topic, I usually claim that the family is not the most important parameter in the regression model. Consider the following dataset
> db <- data.frame(x=c(1,2,3,4,5),y=c(1,2,4,2,6))
> plot(db,xlim=c(0,6),ylim=c(-1,8),pch=19)
To visualize a regression model, use the following code
> nd=data.frame(x=seq(0,6,by=.1))
> add_predict = function(reg){
+ prd1=predict(reg,newdata=nd,se.fit = TRUE,type="response")
+ y1=prd1$fit + y1_upp=prd1$fit+prd1$residual.scale*1.96* prd1$se.fit
+ y1_low=prd1$fit-prd1$residual.scale*1.96*
prd1$se.fit + polygon(c(nd$x,rev(nd$x)),c(y1_upp, rev(y1_low)),col="light green",angle=90, density=40,border=NA) + lines(nd$x,y1,col="red",lwd=2)
+ }
For instance, with a Poisson regression (with a log link function) we get
> plot(db)
> reg1=glm(y~x,family=poisson(link="log"),
+ data=db)
> add_predict(reg1)
while, with a Gaussian regresion (but still with a log link function), we get
> plot(db)
> reg2=glm(y~x,family=gaussian(link="log"),
+ data=db)
> add_predict(reg2)
If we just care about the expected value of our prediction, the output is more or less the same
> plot(db)
> lines(nd$x,predict(reg1,newdata=nd, + type="response"),col="red",lwd=1.5) > lines(nd$x,predict(reg2,newdata=nd,
+ type="response"),col="blue",lwd=1.5)
So, indeed, forget about the (distribution) law when running a GLM. Not convinced? Consider – on the same dataset – a Poisson regression (with an identity link function this time)
> plot(db)
> reg1=glm(y~x,family=poisson(link="identity"),
+ data=db)
> add_predict(reg1)
while, with a Gaussian regresion (but still with an identity link function), we get
> plot(db)
> reg2=glm(y~x,family=gaussian(link="identity"),
+ data=db)
> add_predict(reg2)
Again, if we just plot the expected value of our prediction, the output is more or less the same
> plot(db)
> lines(nd$x,predict(reg1,newdata=nd, + type="response"),col="red",lwd=1.5) > lines(nd$x,predict(reg2,newdata=nd,
+ type="response"),col="blue",lwd=1.5)
So clearly, the simplistic message you should not care too much about the (distribution) law seems to be valid…
# GLM, non-linearity and heteroscedasticity
Last week in the non-life insurance course, we’ve seen the theory of the Generalized Linear Models, emphasizing the two important components
• the link function (which is actually the key component in predictive modeling)
• the distribution, or the variance function
Just to illustrate, consider my favorite dataset
lin.mod = lm(dist~speed,data=cars)
A linear model means here
$Y_i=\beta_0+\beta_1 X_i +\varepsilon_i$
where the residuals are assumed to be centered, independent, and with identical variance. If we visualize that linear regression, we usually see something like that
The idea here (in GLMs) is to assume
$Y\vert X=x\sim\mathcal{N}( \beta_0+\beta_1 x,\sigma^2)$
which will produce the same model as the one describe previously, based on some error term. That model can be visualized below,
attach(cars)
n=2
X= cars$speed Y=cars$dist
df=data.frame(X,Y)
vX=seq(min(X)-2,max(X)+2,length=n)
vY=seq(min(Y)-15,max(Y)+15,length=n)
mat=persp(vX,vY,matrix(0,n,n),zlim=c(0,.1),theta=-30,ticktype ="detailed", box = FALSE)
reggig=glm(Y~X,data=df,family=gaussian(link="identity"))
x=seq(min(X),max(X),length=501)
C=trans3d(x,predict(reggig,newdata=data.frame(X=x),type="response"),rep(0,length(x)),mat)
lines(C,lwd=2)
sdgig=sqrt(summary(reggig)$dispersion) x=seq(min(X),max(X),length=501) y1=qnorm(.95,predict(reggig,newdata=data.frame(X=x),type="response"), sdgig) C=trans3d(x,y1,rep(0,length(x)),mat) lines(C,lty=2) y2=qnorm(.05,predict(reggig,newdata=data.frame(X=x),type="response"), sdgig) C=trans3d(x,y2,rep(0,length(x)),mat) lines(C,lty=2) C=trans3d(c(x,rev(x)),c(y1,rev(y2)),rep(0,2*length(x)),mat) polygon(C,border=NA,col="yellow") C=trans3d(X,Y,rep(0,length(X)),mat) points(C,pch=19,col="red") n=8 vX=seq(min(X),max(X),length=n) mgig=predict(reggig,newdata=data.frame(X=vX)) sdgig=sqrt(summary(reggig)$dispersion)
for(j in n:1){
stp=251
x=rep(vX[j],stp)
y=seq(min(min(Y)-15,qnorm(.05,predict(reggig,newdata=data.frame(X=vX[j]),type="response"), sdgig)),max(Y)+15,length=stp)
z0=rep(0,stp)
z=dnorm(y, mgig[j], sdgig)
C=trans3d(c(x,x),c(y,rev(y)),c(z,z0),mat)
polygon(C,border=NA,col="light blue",density=40)
C=trans3d(x,y,z0,mat)
lines(C,lty=2)
C=trans3d(x,y,z,mat)
lines(C,col="blue")}
We do have two parts here: the linear increase of the average, $\mathbb{E}(Y\vert X=x)=\beta_0+\beta_1 x$ and the constant variance of the normal distribution $\text{Var}(Y\vert X=x)=\sigma^2$.
On the other hand, if we assume a Poisson regression,
poisson.reg = glm(dist~speed,data=cars,family=poisson(link="log"))
we have something like
This time, two things have changed simultaneously: our model is no longer linear, it is an exponential one $\mathbb{E}(Y\vert X=x)=e^{\beta_0+\beta_1 x}$, and the variance is also increasing with the explanatory variable $\text{Var}(Y\vert X=x)=e^{\beta_0+\beta_1 x}$, since with a Poisson regression,
$Y\vert X=x\sim\mathcal{P}(e^{ \beta_0+\beta_1 x})$
If we adapt the previous code, we get
The problem is that we changed two things when we introduced the Poisson regression from the linear model. So let us look at what happens when we change the two components independently. First, we can change the link function, with a Gaussian model but this time a multiplicative model (with a logarithm link function)
gaussian.reg = glm(dist~speed,data=cars,family=gaussian(link="log"))
which is still, here, an homoscedasctic model, but this time non-linear. Or we can change the link function in the Poisson regression, to get a linear model, but heteroscedastic
poisson.lin = glm(dist~speed,data=cars,family=poisson(link="identity"))
So this is basically what GLMs are about….
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2017-08-22 07:15:36
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https://people.bath.ac.uk/mbr20/felipe/manual.htm
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# Introduction
FELIPE (Finite Element Learning Package) is a software package whose primary objective is to help students understand the finite element method in mathematics and engineering, and develop their own f.e. programs. Its advantage over the f.e. textbooks which provide their example programs printed or on ftp or cd-rom, is that it combines full, commented and documented source code (in standard Fortran77) for the f.e. main engines', with powerful interactive graphics pre- and post-processors capable of generating complex, detailed meshes. Because of this, it is also very suitable for practising engineers and researchers as a low-cost alternative to the many commercial black box'' packages on the market (which do not provide source code).
The principal components of FELIPE are:
• PREFEL, a pre-processor, used to create the input data file which defines the finite element mesh, boundary conditions, material properties, loading, etc. The resulting data file is given a .dat filename extension. The pre-processor is provided as the executable file PREFEL.EXE;
• Three Basic-level Fortran77 main engines' (for the 2D Poisson's equation, plane elasticity, and beam/frame analyses), the theory for which is summarized in this Manual. Each main engine' reads an input file such as prob1.dat (created by the PREFEL pre-processor) and produces an output file prob1.out (for use with the post-processor) and a results file prob1.prt in printable format;
• Six further Advanced-level Fortran77 main engines' for analysing a range of mathematics and engineering applications (e.g. viscoplasticity, thermoelasticity), which illustrate the main practical aspects of finite element programming. In particular, a wide range of 1D and 2D finite and infinite element types are used, and coding is provided for all the most important algorithms for equation solution (from Gaussian elimination to conjugate gradients with Incomplete Choleski preconditioning). The individual main engines' are summarized below. There is also a file of input/output subroutines common to all the main engines'. Each engine' is provided in source code .for and executable .exe form;
• linking files ( .inf) for each of the main engines' if they are to be compiled, linked and run using the Salford FTN77 compiler;
• FELVUE, a post-processor which reads in a .out file and displays the results graphically. Displays include contouring, stress crosses, displacement vectors, deformed mesh, etc. This processor is also provided in executable form, namely in the file FELVUE.EXE. It is possible to produce PostScript files (with a .ps extension) of the graphical displays, for later printing;
• SALFLIBC.DLL. The Salford Fortran library needed to allow the programs to run on any PC;
• Sample .dat and .out files, for one or more example problems for each main engine', also documented in this Manual and shown on the FELIPE website.
By this means, users of FELIPE have the interactive graphics processing power of a commercial finite element package, combined with fully-documented source code which they can modify and extend for their own purposes. The package is completely self-standing; the only supporting software needed is a Fortran compiler if the user wishes to modify the source code and re-compile it. Since the pre- and post-processors communicate with the main f.e. programs through formatted ASCII data files, they can also be interfaced with other finite element programs, whether in Fortran or another language.
The pre- and post-processors are compiled under FTN77 (as are the executable versions of the main engines'), which includes a memory extender; thus, large meshes can thus be created, and problems of real mathematical and engineering significance solved. (The processors are dimensioned to handle a maximum of 900 elements, and 3,000 nodes.) They make extensive use of the graphics and mouse routines available with the FTN77 compiler, and can also produce PostScript graphics files for subsequent processing by, for example, GhostScript software (available from www.hensa.ac.uk by ftp) and printing on LaserJet or DeskJet printers. The main engines' can also be compiled and run using FTN77 (which is available for personal use free from the Salford software website), but if this is not available any other Fortran compiler can be used, as they are written in standard Fortran77.
# The nine main engines'
The three Basic-level main engines', and their principal features, are now listed:
1. POISS.FOR
Application: solves Poisson's equation -a_x u_{xx} - a_y u_{yy} = f(x,y) on an arbitrary 2D domain.
Material properties: diffusion coefficients a_x, a_y
Primary nodal unknown: potentials U
Secondary unknowns: flow rates U_x, U_y
Elements used: 3-noded linear triangles.
Boundary conditions: reflecting, radiating and Dirichlet boundaries
File size: 25.8KB
Analysis type: linear
Matrix storage: symmetric band
Solution algorithm: Choleski (L.L^T) decomposition
2. ELAST.FOR
Application: solves plane strain, plane stress or axisymmetric linear elasticity problems
Material properties: Young's modulus E, Poisson's ratio \nu, thickness t, tensile strength \sigma_{\mbox{ten}}
Primary nodal unknown: displacements u,v
Secondary unknowns: stresses \sigma_x, \sigma_y, \tau_{xy}
Boundary conditions: fixities in x or y planes
File size: 34.8KB
Analysis type: linear
Matrix storage: symmetric band
Solution algorithm: Choleski (L.L^T) decomposition
3. FRAME.FOR
Application: analyses plane frames comprising elastic beams.
Material properties: Young's modulus E, Moment of Inertia I, cross-sectional area A
Primary nodal unknown: displacements x,y and rotations \theta
Elements used: 2-noded cubic beam elements.
Boundary conditions: displacement and rotation fixities
File size: 24.4KB
Analysis type: linear
Matrix storage: element-by-element matrices on scratch file
Solution algorithm: preconditioned conjugate gradients, with diagonal preconditioning
The six Advanced-level main engines' are:
Application: Large-scale 2D elasticity analyses
Material properties: Young's modulus E, Poisson's ratio \nu, thickness t, tensile strength \sigma_{\mbox{ten}}
Primary nodal unknown: displacements u,v
Secondary unknowns: stresses \sigma_x, \sigma_y, \tau_{xy}
Elements used: 3- and 6-noded triangles, 4- and 8-noded quadrilaterals, mapped infinite elements, 2- and 3- noded (cubic and quartic) beam elements
Boundary conditions: fixities in x or y planes
File size: 91.9KB
Analysis type: linear
Solution algorithm: symmetric frontal algorithm
2. PLAST.FOR
Application: plane strain associated-flow Mohr-Coulomb elasto-plasticity analyses
Material properties: Young's modulus E, Poisson's ratio \nu, triaxial stress factor k, strength \sigma_c
Primary nodal unknown: displacements u,v
Secondary unknowns: stresses \sigma_x, \sigma_y, \tau_{xy}
Boundary conditions: fixities in x or y planes
File size: 50.3KB
Analysis type: nonlinear, iterative, incremental
Matrix storage: symmetric band
Solution algorithm: Choleski (L.L^T) decomposition
3. VPLAS.FOR
Application: plane strain Mohr-Coulomb elasto-viscoplasticity analyses, with non-associated flow
Material properties: Young's modulus E, Poisson's ratio \nu, triaxial stress factor k, strength \sigma_c, fluidity \gamma, dilation factor l
Primary nodal unknown: displacements u,v
Secondary unknowns: stresses \sigma_x, \sigma_y, \tau_{xy}
Elements used: 8-noded serendipity' quadrilaterals
Boundary conditions: fixities in x or y planes
File size: 75.5KB
Analysis type: nonlinear, incremental, time-dependent
Solution algorithm: Frontal algorithm, for non-symmetric matrices
Application: as PLAST, but with a range of solution algorithms
Material properties: Young's modulus E, Poisson's ratio \nu, triaxial stress factor k, strength \sigma_c
Primary nodal unknown: displacements u,v
Secondary unknowns: stresses \sigma_x, \sigma_y, \tau_{xy}
Boundary conditions: fixities in x or y planes
File size: 67.7KB
Analysis type: nonlinear, iterative, incremental
Matrix storage: symmetric skyline, element-by-element
Solution algorithms: Choleski (L.L^T) and L.D.L^T decomposition, conjugate gradients with diagonal or Incomplete Choleski preconditioning
5. THERM.FOR
Application: plane stress/strain thermoelasticity
Material properties: Young's modulus E, Poisson's ratio \nu, thickness t, conductivity coefficient k, coefficient of thermal expansion \alpha
Primary nodal unknown: displacements u,v, temperatures T
Secondary unknowns: stresses \sigma_x, \sigma_y, \tau_{xy}
Elements used: 4-noded (linear) and 8-noded (serendipity) quadrilaterals with (u,v,T) degrees of freedom at all nodes
Boundary conditions: fixities in x or y planes, reflecting and Dirichlet temperature boundaries
File size: 44.2KB
Analysis type: linear, coupled
Matrix storage: nonsymmetric band
Solution algorithm: Gauss elimination for non-symmetric matrices
6. CONSL.FOR
Application: plane strain soil consolidation (poroelasticity)
Material properties: Young's modulus E, Poisson's ratio \nu, effective permeabilities \frac{k_x}{\gamma_w}, \frac{k_y}{\gamma_w}, effective porosity \frac{\eta}{K_f}
Primary nodal unknown: displacements u,v, pore-pressures p
Secondary unknowns: effective stresses \sigma_x, \sigma_y, \tau_{xy}
Elements used: 8-noded serendipity' quadrilaterals with pore-pressure d.o.f.s at corner nodes only
Boundary conditions: fixities in x or y planes, impermeable or permeable boundaries
File size: 53.4KB
Analysis type: linear, coupled, time-dependent
Matrix storage: symmetric band
Solution algorithm: L.D.L^T decomposition
# Installation
To install the package from the floppy disk, run the self-extracting zip file FELIPE.EXE which is on the disk. You can do this by locating the file on your disk drive (normally the A: drive) using My Computer or Windows Explorer, and double-clicking on it. Alternatively, type a:\felipe.exe into the command line copy using the Run... utility from the Start menu. You will be prompted to nominate a directory into which the FELIPE files should be unzipped; the default is C:\FELIPE.
If you have already downloaded the evaluation version of FELIPE from the website into the C:\FELIPE directory, you can still use the same directory; the demonstration versions of the files will be overwritten by the full versions, and new files added.
The installation process does not alter any Windows settings on your PC. To uninstall FELIPE, simply delete the directory containing the files.
\vspace{1cm}
In this manual, Chapter 2 describes how to use the PREFEL pre-processor.
The next three chapters describe the three Basic-level main engines': Chapter 3 covers the theory and programming of the Poisson solver, while Chapter 4 describes the solver for elasticity problems, and Chapter 5 deals with beam theory.
Chapter 6 describes the use of the FELVUE post-processor. Then
Chapter 7 summarizes the operation and use of the other six, Advanced-level main engines'. Chapter 8 covers the various algorithms used for equation solution.
Chapter 9 documents the sample datafiles provided in the FELIPE package.
The final Chapter suggests ways in which the main engines' may be modified, and new `main engines' written, and gives recommendations for textbooks for further reading about the finite element method.
Acknowledgements: I am very grateful to Prof. J.R. Whiteman and Dr. M.K. Warby for permission to use in FELIPE some of the basic graphics and PostScript subroutines developed by Dr. Warby, and to Dr. T.-Y. Chao for working with me on the programming and documentation of the elasticity module. I also acknowledge the support of the Enterprise in Higher Education Unit at Brunel University, in enabling me to work on this project. The Fortran coding of the elasticity 'main engines' is based on the FINEPACK program developed at the Dept. of Civil Engineering, University College Swansea, and I am grateful to Dr. D.J. Naylor for permission to use this.
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2021-04-17 01:40:38
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|
https://www.physicsforums.com/threads/raising-and-lowering-operators.319870/
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# Raising and lowering operators
1. Jun 14, 2009
### mkhurana
1. The problem statement, all variables and given/known data
Can anyone please explain why any term which is a product of 4 raising and lowering operators with a lowering operator on the extreme right (eg. A-A+A+A-) has zero expectation value in the ground state of a harmonic oscillator?
2. Relevant equations
3. The attempt at a solution
2. Jun 14, 2009
### Cyosis
What happens if you let the lowering operator work on $\psi_1$? What do you get when you let the lowering operator work on $\psi_0$?
Calculate $a \psi_1$ and $a \psi_0$. Does $a \psi_0$ exist?
3. Jun 14, 2009
### mkhurana
I get it now... thank you. When the lowering operator operates on v1, I get v0. When it operates on v0, I get zero. A- v0 doesn't exist so the rest of the operators in the product have nothing to operate on.
Cheers.
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2017-11-19 11:15:01
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http://dsp.stackexchange.com/unanswered
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206 views
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I'm trying to wrap my head around the proper use of a Wiener or error-prediction filter for filtering data. It seems to me that it is only a whitening filter, so how is it used when the data you want ...
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So I'm trying to write a frequency-domain interpolator that zero-pads the frequency response of a signal and inverse transforms. There's two cases I have to deal with: Even-length response - have ...
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I want to implement an adaptive thresholding algorithm in MATLAB for filtering data received by an underwater sonar receiver. The data received has an interactive noise component resulting from ...
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While working on this problem, I started to have doubts that my initial definition of $$SINAD = 10 \log_{10} \left( \frac{p_f} {\sum_i{(p_i)} - p_0 - p_f} \right)dB$$ is correct. In this equation, ...
219 views
### Kalman Filter - best way to handle “derived” measurements?
Ie, if you have as state variables position (p) and velocity (v), and I make low-frequency measurements of p, this also indirectly gives me information about v (since it's the derivative of p). What ...
379 views
### How to deal with a negative pole (unstable) in the pre-filter of a control system?
So while answering how to design a PI controller for a first order time delayed system (Question Here ) Here is the closed loop equation to a control system: G_C(s) = ...
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### Calculating a homography based on detected lines
I know you can calculate homographies from image to camera plane using correspondence points between a "perfect model" and the image points. I'm doing it for a football pitch/field, and have used ...
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### Estimating the input to a system from a system state using EKF
[ Cross-posted from: http://math.stackexchange.com/questions/164169/estimating-the-input-to-a-system-from-a-system-state ] I have a system for which I have obtained a non-linear time-varying ...
421 views
### Ensemble learning, multiple classifier system
I am trying to use a MCS (Multi classifier system) to do some better work on limited data i.e become more accurate. I am using K-means clustering at the moment but may choose to go with FCM (Fuzzy ...
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### How to find the lower eyelid in an image?
I have an image of 480x360 px that contains eye portion only i.e., sclera iris, pupil, eyebrows etc. Now I have to find the lower eyelid in the image. What I've done till now is that cut the image ...
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### What is the normalization factor for the quincunx lattice wavelet transform, and how do you find it?
On pages 57-60 (preview was available last I checked, images here in case), there is a quincunx lattice transform described. Lattice: ...
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### How do you choose the optimal number of taps in a windowed image scaling filter?
I'm attempting to use a filter to scale an image and for the most part, I have something that appears to work. But now I want to be able to generate a filter with an optimal number of taps. I know ...
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### Implementing a Matlab filter to cancel cross talk between two sensors
I have two sensors that have a cross talk between them. I would like to cancel the cross talk. For this I recorded two tracks, where sensor no.1 (called x1) has some input, and sensor no.2 (called x2) ...
I want to know how to solve those types of problems.. is it by inspection ? Consider the linear system below. When the inputs to the system $x_1[n]$, $x_2[n]$ and $x_3[n]$, the responses of the ...
I am tracking a car using particle filter by making a rectangle around it, and I am using the state vector $[x, y, u, v, a, h]$, where: $x$, $y$ is the position of the body $u$ and $v$ are velocity ...
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2014-09-22 14:13:22
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http://mathcentral.uregina.ca/QQ/database/QQ.09.13/h/david1.html
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SEARCH HOME
Math Central Quandaries & Queries
Question from David: I have small equipment hut that has a small attic space that I want to blow insulation into it. The attic size is shaped like a triangle that is about 8 feet by 10 feet by 7 inches at the highest point. I need to estimate the volume in square feet or cubic feet to know how many bags of insulation to blow into the space. Thanks
Hi David,
I think the attic looks something like the diagram below.
If so the volume is $\large \frac12 \normalsize 10 \times 8 \times \large \frac{7}{12} \normalsize = 23 \large \frac13 \normalsize \mbox{ cubic feet.}$
If I have the incorrect shape write back, preferably with a diagram.
Penny
Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.
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2017-11-21 17:36:42
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https://dev.socrata.com/docs/transforms/to_multiline.html
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Socrata was acquired by Tyler Technologies in 2018 and is now the Data and Insights division of Tyler. The platform is still powered by the same software formerly known as Socrata but you will see references to Data & Insights going forward.
# to_multiline
##### Function: to_multiline
convert a line into a multiline
parse a WKT (text) representation of a multiline into a multiline value
Examples:
to_multiline(to_line('LINESTRING (30 10, 10 30, 40 40)'))
-- Result: {"type":"MultiLineString","coordinates":[[[30,10],[10,30],[40,40]]]}
to_multiline(to_line('LINESTRING (30 10)'))
-- Result: {"type":"invalid_geometry","english":"Geometry invariant violation: LineString must have at least 2 coordinates","data":{"value":{"type":"LineString","coordinates":[[30,10]]},"reason":"LineString must have at least 2 coordinates"}}
to_multiline('MULTILINESTRING ((10 10, 20 20, 10 40),(40 40, 30 30, 40 20, 30 10))')
-- Result: {"type":"MultiLineString","coordinates":[[[10,10],[20,20],[10,40]],[[40,40],[30,30],[40,20],[30,10]]]}
to_multiline(a_wkt_multiline)
-- Result: {"type":"MultiLineString","coordinates":[[[10,10],[20,20],[10,40]],[[40,40],[30,30],[40,20],[30,10]]]}
to_multiline('MULTILINESTRING ((10 10),(40 40, 30 30, 40 20, 30 10))')
-- Result: {"type":"invalid_geometry","english":"Geometry invariant violation: LineString must have at least 2 coordinates","data":{"value":{"type":"MultiLineString","coordinates":[[[10,10]],[[40,40],[30,30],[40,20],[30,10]]]},"reason":"LineString must have at least 2 coordinates"}}
###### Signatures
line -> multiline
text -> multiline
multiline -> multiline
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2023-03-25 22:36:33
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http://forum.math.toronto.edu/index.php?topic=850.0
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### Author Topic: Typos? (Read 358 times)
#### Luyu CEN
• Jr. Member
• Posts: 12
• Karma: 0
##### Typos?
« on: November 23, 2016, 01:36:19 PM »
http://www.math.toronto.edu/courses/apm346h1/20169/PDE-textbook/Chapter9/S9.1.html#mjx-eqn-eq-9.1.8
The coefficient should be $\frac{1}{4\pi c^2 t}$ instead of $\frac{1}{4\pi c t}$
http://www.math.toronto.edu/courses/apm346h1/20169/PDE-textbook/Chapter9/S9.1.html#mjx-eqn-eq-9.1.18
$v^\pm_\omega$ should be$-\frac{1}{4\pi}\iiint |x-y|^{-1}e^{\mp i\omega c^{-1}|x-y|}f(y)\, dy$ instead of
$v^\pm_\omega= -\frac{1}{4\pi}\iiint |x-y|^{-1}e^{\mp i\omega c^{-1}|x-y|}\, dy.$
http://www.math.toronto.edu/courses/apm346h1/20169/PDE-textbook/Chapter9/S9.2.html#mjx-eqn-eq-9.2.3
This one I am not sure but I think the coefficient $\frac{1}{2}$ is missing if we follow equation (2)
Otherwise we could not use the quadratic form in (4) which leads to theorem 1 in this section.
Therefore, it should be
\iint_{\Sigma} \Bigl(
\frac{1}{2}\bigl(u_{t}^2+|\nabla u|^2\bigr)\nu_t -c^2 u_t \nabla u \cdot \nu_x\Bigr) \,d\sigma=0
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2018-06-18 00:11:35
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http://mathhelpforum.com/differential-geometry/88519-measure-theory.html
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# Math Help - Measure theory
1. ## Measure theory
Show that (a,b) = $\bigcup_{n\in N}$( $a_{n},b_{n}$) , where $a,b\in\mathbb{R}$ , { $a_{n}$} is a decreasing sequence of rational numbers with $a_{n}\rightarrow a$ , { $b_{n}$} is an increasing sequence of rational numbers with $b_{n}\rightarrow b$ .
2. Suppose x is in (a, b). That is, that a< x< b. Show that, for some n, $a_n< x< b_n$. That should be easy. Take $\epsilon= x-a$. Since $a_n \to a$, there exist n such that $|a- a_n|< \delta$ so $a. Since $b_n\to b$, there exist m such that $|b-b_n|< \delta$ so $b< b_n< b$.
It should be clear that is x< a, then x is in NONE of $(a_n, b_n)$ and so not in their union. Similarly for x> b. The important part is x= a and x= b. You need to show that $a< a_n$ and $b_n< b$ for all n.
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2015-04-25 11:10:20
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|
https://howlingpixel.com/i-en/Surface_brightness
|
# Surface brightness
In astronomy, surface brightness quantifies the apparent brightness or flux density per unit angular area of a spatially extended object such as a galaxy or nebula, or of the night sky background. An object's surface brightness depends on its surface luminosity density, i.e., its luminosity emitted per unit surface area. In visible and infrared astronomy, surface brightness is often quoted on a magnitude scale, in magnitudes per square arcsecond in a particular filter band or photometric system.
Measurement of the surface brightnesses of celestial objects is called surface photometry.
## General description
The total magnitude is a measure of the brightness of an extended object such as a nebula, cluster, galaxy or comet. It can be obtained by summing up the luminosity over the area of the object. Alternatively, a photometer can be used by applying apertures or slits of different sizes of diameter.[1] The background light is then subtracted from the measurement to obtain the total brightness.[2] The resulting magnitude value is the same as a point-like source that is emitting the same amount of energy.[3]
The apparent magnitude of an astronomical object is generally given as an integrated value—if a galaxy is quoted as having a magnitude of 12.5, it means we see the same total amount of light from the galaxy as we would from a star with magnitude 12.5. However, a star is so small it is effectively a point source in most observations (the largest angular diameter, that of R Doradus, is 0.057 ± 0.005 arcsec), whereas a galaxy may extend over several arcseconds or arcminutes. Therefore, the galaxy will be harder to see than the star against the airglow background light. Apparent magnitude is a good indication of visibility if the object is point-like or small, whereas surface brightness is a better indicator if the object is large. What counts as small or large depends on the specific viewing conditions and follows from Ricco's law.[4] In general, in order to adequately assess an object's visibility one needs to know both parameters.
## Calculating surface brightness
Surface brightnesses are usually quoted in magnitudes per square arcsecond. Because the magnitude is logarithmic, calculating surface brightness cannot be done by simple division of magnitude by area. Instead, for a source with a total or integrated magnitude m extending over a visual area of A square arcseconds, the surface brightness S is given by
${\displaystyle S=m+2.5\cdot \log _{10}A.}$
For astronomical objects, surface brightness is analogous to photometric luminance and is therefore constant with distance: as an object becomes fainter with distance, it also becomes correspondingly smaller in visual area. In geometrical terms, for a nearby object emitting a given amount of light, radiative flux decreases with the square of the distance to the object, but the physical area corresponding to a given solid angle or visual area (e.g. 1 square arcsecond) decreases by the same proportion, resulting in the same surface brightness.[5] For extended objects such as nebulae or galaxies, this allows the estimation of spatial distance from surface brightness by means of the distance modulus or luminosity distance.
## Relationship to physical units
The surface brightness in magnitude units is related to the surface brightness in physical units of solar luminosity per square parsec by
${\displaystyle S({\rm {mag}}/{\rm {arcsec}}^{2})=M_{\odot }+21.572-2.5\log _{10}S(L_{\odot }/{\rm {pc}}^{2}),}$
where ${\displaystyle M_{\odot }}$ and ${\displaystyle L_{\odot }}$ are the absolute magnitude and the luminosity of the Sun in chosen color-band[6] respectively.
Surface brightness can also be expressed in candela per square metre using the formula [value in cd/m2] = 10.8×104 × 10(-0.4*[value in mag/arcsec2]).
There is an online calculator available here http://unihedron.com/projects/darksky/magconv.php?ACTION=SOLVE&txtMAGSQA=21.83
## Examples
A truly dark sky has a surface brightness of 2×10−4 cd m−2 or 21.8 mag arcsec−2.[7]
The peak surface brightness of the central region of the Orion Nebula is about 17 Mag/arcsec2 (about 14 millinits) and the outer bluish glow has a peak surface brightness of 21.3 Mag/arcsec2 (about 0.27 millinits).[8]
## References
1. ^ Daintith, John; Gould, William (2006). The Facts on File dictionary of astronomy. Facts on File science library (5th ed.). Infobase Publishing. p. 489. ISBN 0-8160-5998-5.
2. ^ Palei, A. B. (August 1968). "Integrating Photometers". Soviet Astronomy. 12: 164. Bibcode:1968SvA....12..164P.
3. ^ Sherrod, P. Clay; Koed, Thomas L. (2003). A Complete Manual of Amateur Astronomy: Tools and Techniques for Astronomical Observations. Astronomy Series. Courier Dover Publications. p. 266. ISBN 0-486-42820-6.
4. ^ Crumey, Andrew (2014). "Human contrast threshold and astronomical visibility". Monthly Notices of the Royal Astronomical Society. 442: 2600. arXiv:1405.4209. Bibcode:2014MNRAS.442.2600C. doi:10.1093/mnras/stu992.
5. ^ Sparke & Gallagher (2000, § 5.1.2)
6. ^ Absolute magnitudes of the Sun in different color-bands can be obtained from Binney & Merrifield (1998) or Absolute Magnitude of the Sun in Several Bands Archived 2007-07-18 at the Wayback Machine
7. ^ Based on the equivalence 21.83 mag arcsec−2 = 2×10−4 cd m−2, from description of a "truly dark sky", Section 1.3 of Crumey, A. (2014). Human contrast threshold and astronomical visibility. MNRAS 442, 2600–2619.
8. ^ Clark, Roger (2004-03-28). "Surface Brightness of Deep Sky Objects". Retrieved 2013-06-29.. The conversion to nits is based on 0 magnitude being 2.08 microlux.
### General references
Andromeda XXI
Andromeda XXI (And 21, And XXI) is a moderately bright dwarf spheroidal galaxy about 859 ± 51 kiloparsecs (2.80 ± 0.17 Mly) away from the Sun in the constellation Andromeda. It is the fourth largest Local Group dwarf spheroidal galaxy.
This large satellite of the Andromeda Galaxy (M31) has a half-light radius of nearly 1 kpc.The discovery arose from the first year data of a photometric survey of the M31/M33 subgroupings of the Local Group by the Pan-Andromeda Archaeological Survey (PAndAS). This survey was conducted with the Megaprime/MegaCam wide-field camera mounted on the Canada-France-Hawaii Telescope.
Andromeda XXI appears as a spatial overdensity of stars. It has red giant branches at the distance of M31/M33, and follows metal-poor, [Fe/H]=-1.8 when plotted in a color-magnitude diagram.
Although moderately bright (MV=-9.9 ± 0.6), it has low surface brightness. This indicates that numerous relatively luminous M31 satellites remain undiscovered.
Antlia 2
Antlia 2 (Ant 2) is a low-surface-brightness dwarf satellite galaxy of the Milky Way at a galactic latitude of 11.2°. It spans 1.26° in the sky just southeast of Epsilon Antliae. The galaxy is similar in size to the Large Magellanic Cloud, despite being 10,000 times fainter. Antlia 2 has the lowest surface brightness of any galaxy discovered and is ~ 100 times more diffuse than any known ultra diffuse galaxy. It was discovered by the European Space Agency's Gaia spacecraft in November 2018.
Crater 2 Dwarf
Crater 2 is a low-surface-brightness dwarf satellite galaxy of the Milky Way, located approximately 380,000 ly from Earth. Crater 2 was identified in imaging data from the VST ATLAS survey.The galaxy has a half-light radius of ∼1100 pc, making it the fourth largest satellite of the Milky Way. It has an angular size about double of that of the moon.
Dwarf spiral galaxy
A dwarf spiral galaxy is the dwarf version of a spiral galaxy. Dwarf galaxies are characterized as having low luminosities, small diameters (less than 5 kpc), low surface brightnesses, and low hydrogen masses. The galaxies may be considered a subclass of low-surface-brightness galaxies.
Dwarf spiral galaxies, particularly the dwarf counterparts of Sa-Sc type spiral galaxies, are quite rare. In contrast, dwarf elliptical galaxies, dwarf irregular galaxies, and the dwarf versions of Magellanic type galaxies (which may be considered transitory between spiral and irregular in terms of morphology) are very common.It is suggested that dwarf spiral galaxies can transform into dwarf elliptical galaxies, especially in dense cluster environments.
The effective radius (${\displaystyle R_{e}}$) of a galaxy is the radius at which half of the total light of the system is emitted. This assumes the galaxy has either intrinsic spherical symmetry or is at least circularly symmetric as viewed in the plane of the sky. Alternatively, a half-light contour, or isophote, may be used for spherically and circularly asymmetric objects.
${\displaystyle R_{e}}$ is an important length scale in de Vaucouleurs ${\displaystyle {\sqrt[{4}]{R}}}$ law, which characterizes a specific rate at which surface brightness decreases as a function of radius:
${\displaystyle I(R)=I_{e}\cdot e^{-7.67\left({\sqrt[{4}]{\frac {R}{R_{e}}}}-1\right)}}$
where ${\displaystyle I_{e}}$ is the surface brightness at ${\displaystyle R=R_{e}}$. At ${\displaystyle R=0}$,
${\displaystyle I(R=0)=I_{e}\cdot e^{7.67}\approx 2000\cdot I_{e}}$
Thus, the central surface brightness is approximately ${\displaystyle 2000\cdot I_{e}}$.
Galactic disc
A galactic disc is a component of disc galaxies, such as spiral galaxies and lenticular galaxies. Galactic discs consist of a stellar component ( composed of most of the galaxy's stars) and a gaseous component (mostly composed of cool gas and dust). The stellar population of galactic discs tend to exhibit very little random motion with most of its stars undergoing nearly circular orbits about the galactic center. Discs can be fairly thin because the disc material's motion lies predominantly on the plane of the disc (very little vertical motion). The Milky Way's disc, for example is approximately 1 kpc thick but thickness can vary for discs in other galaxies.
Low Surface Brightness galaxy
A low-surface-brightness galaxy, or LSB galaxy, is a diffuse galaxy with a surface brightness that, when viewed from Earth, is at least one magnitude lower than the ambient night sky.
Most LSBs are dwarf galaxies, and most of their baryonic matter is in the form of neutral gaseous hydrogen, rather than stars. They appear to have over 95% of their mass as non-baryonic dark matter. There appears to be no supernova activity in these galaxies.Rotation curve measurements indicate an extremely high mass-to-light ratio, meaning that stars and luminous gas contribute only very little to the overall mass balance of an LSB. The centers of LSBs show no large overdensities in stars, unlike e.g. the bulges of normal spiral galaxies. Therefore, they seem to be dark-matter-dominated even in their centers, which makes them excellent laboratories for the study of dark matter.
In comparison to the high-surface-brightness galaxies, LSBs are mainly isolated field galaxies, found in regions devoid of other galaxies. In their past, they had fewer tidal interactions or mergers with other galaxies, which could have triggered enhanced star formation. This is an explanation for the small stellar content.
LSB galaxies were theorized to exist in 1976 by Mike Disney.
Malin 1
Malin 1 is a giant low surface brightness (LSB) spiral galaxy. It is located 1.19 billion light-years (366 Mpc) away in the constellation Coma Berenices, near the North Galactic Pole. As of February 2015, it is arguably the largest known spiral galaxy, with an approximate diameter of 650,000 light-years (200,000 pc), six and a half times the diameter of our Milky Way. It was discovered by astronomer David Malin in 1986 and is the first LSB galaxy verified to exist. Its high surface brightness central spiral is 30,000 light-years (9,200 pc) across, with a bulge of 10,000 light-years (3,100 pc). The central spiral is a SB0a type barred-spiral.Malin 1 is peculiar in several ways: its diameter alone would make it the largest barred spiral galaxy ever to have been observed.Malin 1 was found later to be interacting with two other galaxies, Malin 1B and SDSS J123708.91+142253.2. Malin 1B is located 46,000 light-years (14,000 pc) away from the high surface brightness central spiral of Malin 1, which may be responsible for the formation of the galaxy's central bar. Meanwhile, SDSS J123708.91+142253.2 is located within the huge, faint halo of Malin 1 and might have caused the formation of the extended low surface brightness disc through tidal stripping.
Observations by Galaz et al. in April 2014 revealed a detailed view of the spiral structure of Malin 1 in optical bands. The galaxy exhibits giant and very faint spiral arms, with a thickness of up to one-third the diameter of the Milky Way. Other details, such as possible stellar streams and formation regions, are revealed as well.
NGC 1090
NGC 1090 is a barred spiral galaxy located in the constellation Cetus.
NGC 1090 has a pseudo inner ring. The disc has a very low surface brightness.
This galaxy has been the site of two known supernovae (in 1962 and 1971).
NGC 1090 is not part of a galaxy group, even though it appears close to NGC 1087, M-77 (NGC 1068), NGC 1055, NGC 1073, and five other small irregular galaxies.
The distance to NGC 1090 is approximately 124 million light years and its diameter is about 144,000 light years.
NGC 1264
NGC 1264 is a low-surface-brightness barred spiral galaxy located about 145 million light-years away in the constellation Perseus. The galaxy was discovered by astronomer Guillaume Bigourdan on October 19, 1884. NGC 1264 is a member of the Perseus Cluster.
NGC 296
NGC 296 is a low surface brightness unbarred spiral galaxy in the constellation of Pisces. The designation NGC 295 is sometimes mistakenly used for NGC 296.
NGC 3794
NGC 3794, also cataloged in the New General Catalogue as NGC 3804, is a low-surface-brightness galaxy in the constellation Ursa Major. It is very far from Earth, with a distance of about 68,470,000 light-years (20,990,000 pc).
NGC 3821
NGC 3821 is a low surface brightness spiral galaxy and a ring galaxy about 270 million light-years away in the constellation Leo. The galaxy was discovered by astronomer William Herschel on April 26, 1785 and is a member of the Leo Cluster.
NGC 3883
NGC 3883 is a large low surface brightness spiral galaxy located about 330 million light-years away in the constellation Leo. NGC 3883 has a prominent bulge but does host an AGN. The galaxy also has flocculent spiral arms in its disk. It was discovered by astronomer William Herschel on April 13, 1785 and is a member of the Leo Cluster.
NGC 45
NGC 45 is a low surface brightness spiral galaxy in the constellation of Cetus. It was discovered on 11 November 1835 by the English astronomer John Herschel.
Unlike the Milky Way, NGC 45 has no clear defined spiral arms, and its center bar nucleus is also very small and distorted. NGC 45 thus does not have a galactic habitable zone. For the Milky Way, the galactic habitable zone is commonly believed to be an annulus with an outer radius of about 10 kiloparsecs and an inner radius close to the Galactic Center, both of which lack hard boundaries.
NGC 5774
NGC 5774 is an intermediate spiral galaxy approximately 71 million light-years away from Earth in the constellation of Virgo. It was discovered by Irish engineer Bindon Stoney on April 26, 1851.NGC 5774 belongs to a small group of galaxies, together with nearby NGC 5775 and IC 1070.
It has been classified as a "low surface brightness" (LSB) galaxy, but its central surface brightness is 5 times brighter than the brightest LSB galaxies.
It has a multiple spiral pattern with bright blue knotty structure all along the arms.It is an extremely low star forming galaxy with five X-ray sources plus three ultraluminous X-ray source candidates.
NGC 765
NGC 765 is an intermediate spiral galaxy located in the constellation Aries. It is located at a distance of circa 220 million light years from Earth, which, given its apparent dimensions, means that NGC 765 is about 195,000 light years across. It was discovered by Albert Marth on October 8, 1864. The galaxy has an extensive hydrogen (HI) disk with low surface brightness, whose diameter is estimated to be 240 kpc (780,000 light years).
NGC 779
NGC 779 is a spiral galaxy seen edge-on, located in the constellation Cetus. It is located at a distance of circa 60 million light years from Earth, which, given its apparent dimensions, means that NGC 779 is about 70,000 light years across. It was discovered by William Herschel on September 10, 1785.NGC 779 features a bright nucleus and an elliptical or boxy bulge. It is seen with high inclination. The inner arms are tightly wound and form an inner pseudoring with high surface brightness. A break is seen at the northwest side of the pseudoring and may be due to dust extinction. The disk has lower surface brightness and is smooth, with no pronounced star-forming knots. The spiral pattern of the galaxy gas been described either as multiple-armed or grand-design two-armed spiral.NGC 779 forms a small galaxy group with UGCA 024, known as the NGC 779 group. NGC 779 is considered to be part of the Cetus II cloud, which also includes NGC 584, NGC 681, NGC 720, and their groups, although it could also lie in the foreground.The galaxy is included in the Herschel 400 Catalogue. It lies about five degrees northeast from Zeta Ceti. It can be seen with a small telescope at moderate magnification, with its core being more easily detected.
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http://www.besjournal.com/en/article/doi/10.3967/bes2020.011
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Volume 33 Issue 1
Feb. 2020
Turn off MathJax
Article Contents
SONG Xiu Ping, ZHANG Hai Bin, LIU Qi Yong, SUN Ji Min, XU Lei, GU Shao Hua, SUN Wan Wan, YUE Yu Juan, REN Dong Sheng, WANG Jun, LI Dong Mei. Seroprevalence of Bartonella henselae and Identification of Risk Factors in China[J]. Biomedical and Environmental Sciences, 2020, 33(1): 72-75. doi: 10.3967/bes2020.011
Citation: SONG Xiu Ping, ZHANG Hai Bin, LIU Qi Yong, SUN Ji Min, XU Lei, GU Shao Hua, SUN Wan Wan, YUE Yu Juan, REN Dong Sheng, WANG Jun, LI Dong Mei. Seroprevalence of Bartonella henselae and Identification of Risk Factors in China[J]. Biomedical and Environmental Sciences, 2020, 33(1): 72-75.
# Seroprevalence of Bartonella henselae and Identification of Risk Factors in China
##### doi: 10.3967/bes2020.011
Funds: This study was supported by the National Science and Technology Major Projects [No.2018ZX10712001 and 2017ZX10303404]; Major Infectious Diseases Such as AIDS and Viral Hepatitis Prevention and Control Technology Major Projects [No.2018ZX10712-001]
• Author Bio:
SONG Xiu Ping, female, born in 1972, Associate Senior Technologist, majoring in infectious disease
ZHAND Hai Bin, female, born in 1973, Associate Senior Technologist, majoring in public health research
• Corresponding author: LI Dong Mei, E-mail: lidongmei@icdc.cn
• Accepted Date: 2019-12-03
• Serum samples were tested for Bartonella henselae IgG antibodies using indirect immunofluorescence assays. We then analyzed associated risk factors. Serum samples were considered positive when reactive at a dilution of more than 1:320. Differences between groups and risk factors associated with Bartonella exposure were statistically analyzed using Chi-square tests and the generalized linear model. 122 of 1,260 samples (9.68%) were positive for B. henselae infection. The infection rate ranged from 0% to 30.43% and differed significantly among age groups (P < 0.01); infection rate in the 50–59 years group was significantly higher than that in other age groups. The seroprevalence of Bartonella varied significantly among sites within the four provinces, and the infection rate of field workers was significantly higher than that of urban workers.
• [1] Boulouis HJ, Chang CC, Henn JB, et al. Factors associated with the rapid emergence of zoonotic Bartonella infections. Vet Res, 2005; 36, 383−410. [2] Raoult D, Tissot Dupont H, Enea-Mutillod M. Positive predictive value of Rochalimaea henselae antibodies in the diagnosis of cat-scratch disease. Clin Infect Dis, 1994; 19, 335−40. [3] Rothova A, Kerkhoff F, Hooft HJ, et al. Bartonella serology for patients with intraocular infl -ammatory disease. Retina, 1998; 18, 348−55. [4] R Core Team. R. A language and environment for statistical computing. R Foundation for Statistical Computing, Vienna, Austria. 2018. https://www.R-project.org/. [5] Sun JM, Fu GM, Lin JF, et al. Seroprevalence of Bartonella in Eastern China and analysis of risk factors. BMC Infect Dis, 2010; 10, 121. [6] Yang H, Bai HM, Yang FL, et al. Serological survey on Bartonella infection in Yunnan. Chin J Nat Med, 2007; 1, 4. [7] Tuneoka H, Fujii R, Kurashige H, et al. Prevalence of serum IgG antibody against Bartonella henselae in an asymptomaatic Japanese population. Kansenshogaku Zasshi, 1999; 73, 90−1. [8] Zupan S, Poljak M, Avsic-Zupanc T. Prevalence of Bartonella infections in Slovenian intravenous drugs users. Ann N Y AcadSci, 2003; 990, 414−8. [9] TG HARR ISON, N DOSH I. Serological evidence of Bartonella spp. infection in the UK. Epidemiol Infect, 1999; 123, 233−40. [10] Klein JD. Cat scratch disease. Pediatr Rev, 1994; 15, 348−53.
###### 通讯作者: 陈斌, bchen63@163.com
• 1.
沈阳化工大学材料科学与工程学院 沈阳 110142
Tables(1)
## Seroprevalence of Bartonella henselae and Identification of Risk Factors in China
##### doi: 10.3967/bes2020.011
###### 1. State Key Laboratory of Infectious Disease Prevention and Control, Collaborative Innovation Center for Diagnosis and Treatment of Infectious Diseases, National Institute for Communicable Disease Control and Prevention, Chinese Center for Disease Control and Prevention, Beijing 102206, China2. The Qinghuangdao Municipal Center for Disease Prevention and Control, Qinhuangdao 066000, Hebei, China3. Zhejiang Provincial Center for Disease Control and Prevention, Hangzhou 310051, Zhejiang, China4. Ningbo Center for Disease Control and Prevention, Ningbo 315010, Zhejiang, China
Funds: This study was supported by the National Science and Technology Major Projects [No.2018ZX10712001 and 2017ZX10303404]; Major Infectious Diseases Such as AIDS and Viral Hepatitis Prevention and Control Technology Major Projects [No.2018ZX10712-001]
• Author Bio:
• ###### Corresponding author:LI Dong Mei, E-mail: lidongmei@icdc.cn
Abstract: Serum samples were tested for Bartonella henselae IgG antibodies using indirect immunofluorescence assays. We then analyzed associated risk factors. Serum samples were considered positive when reactive at a dilution of more than 1:320. Differences between groups and risk factors associated with Bartonella exposure were statistically analyzed using Chi-square tests and the generalized linear model. 122 of 1,260 samples (9.68%) were positive for B. henselae infection. The infection rate ranged from 0% to 30.43% and differed significantly among age groups (P < 0.01); infection rate in the 50–59 years group was significantly higher than that in other age groups. The seroprevalence of Bartonella varied significantly among sites within the four provinces, and the infection rate of field workers was significantly higher than that of urban workers.
SONG Xiu Ping, ZHANG Hai Bin, LIU Qi Yong, SUN Ji Min, XU Lei, GU Shao Hua, SUN Wan Wan, YUE Yu Juan, REN Dong Sheng, WANG Jun, LI Dong Mei. Seroprevalence of Bartonella henselae and Identification of Risk Factors in China[J]. Biomedical and Environmental Sciences, 2020, 33(1): 72-75. doi: 10.3967/bes2020.011
Citation: SONG Xiu Ping, ZHANG Hai Bin, LIU Qi Yong, SUN Ji Min, XU Lei, GU Shao Hua, SUN Wan Wan, YUE Yu Juan, REN Dong Sheng, WANG Jun, LI Dong Mei. Seroprevalence of Bartonella henselae and Identification of Risk Factors in China[J]. Biomedical and Environmental Sciences, 2020, 33(1): 72-75.
• Bartonella henselae (B. henselae) is a fastidious, slow-growing, gram-negative, hemotropic bacterium that infects blood erythrocytes and endothelial cells in its host[1]. B. henselae is responsible for cat scratch disease (CSD), bacillary angiomatosis, bacteremia, endocarditis, and peliosis hepatitis in patients with acquired immunodeficiency syndrome. Importantly, B. henselae is transmitted to humans via trauma caused by cats (cat scratches or bites) or via bites from infected fleas. Laboratory methods to diagnose Bartonella infections include culture-based isolation, serological assays, histopathological examinations, and molecular detection of Bartonella DNA in affected tissues. However, Bartonella spp. are not easily detected by routine bacterial culture protocols. Therefore, in order to determine the incidence of Bartonella infection, it is necessary to investigate the seroprevalence of this bacterium in the general population and determine the primary reservoirs and vectors involved in transmission of infection. In recent years, serological tests for detection of antibodies for different Bartonella species have been developed, and serologic testing for antibodies targeting B. henselae has been proposed as an alternative to skin testing[2].
The seroprevalence of B. henselae varies among different countries and geographic regions within countries The seropositivity rates among the healthy adults are 30% in Germany, 19.8% in Greece. Furthermore, similar or lower seroprevalence rates have been found in healthy populations from the Netherlands (3.3%)[3]. However, the seroprevalence of B. henselae in China and the risk factors for infection have not been clarified.
Accordingly, in this study, we evaluated 1,260 blood samples from healthy populations in China in order to detect anti-B. henselae antibodies using indirect immunofluorescence assays (IFAs) and then analysed the risk factors associated with infection.
Serum samples (n = 1,260) were collected from healthy populations from four regions in China (Table 1). All participants provided clinical and demographic information, including age, sex, pet contact, bite/scratch history, and living and working environments. Of the donors, 51.03% were male, and 48.97% were female (age range: 1–83 years). Serum samples were stored at −20 ℃ before testing.
Group Number of participants Positive cases Positive rate (%) Age (years) 1–9 213 14 6.57 10–19 180 10 5.56 20–29 200 33 16.50 30–39 522 39 7.47 40–49 116 19 16.38 50–59 23 7 30.43 > 60 6 0 0 Sampling sites Qinhuangdao city, Hebei PRVN 327 54 16.51 Xingshan county, Hubei PRVN 388 24 6.19 Weifang city, Shandong PRVN 95 18 18.95 Dongying city, Shandong PRVN 450 26 5.78 Gender Male 643 54 8.40 Female 617 68 11.02 Occupation Urban workers 210 2 0.95 Field workers 240 25 10.41 Contact with animal Yes 51 11 21.57 No 371 61 16.44
Table 1. Characteristics of serosurvey participants, in China, 2008–2013
The experimental research reported in this study was approved by the ethics committee of National Institute for Communicable Disease Control and Prevention, Chinese Center for Disease Control and Prevention. Human research was carried out in compliance with the Helsinki Declaration. All participants provided informed consent for participation in this study.
A commercially available B. henselae-based IFA test kit (Euroimmun, Order No. FI219b-1005G, Hangzhou) was used to assess the presence of IgG antibodies against B. henselae in serum samples. Sera were diluted to 1:100, 1:320, or 1:1,000 in phosphate-buffered saline (PBS-Tween buffer; provided in the test kit), and IFAs were then conducted according to the manufacturer’s protocol. Positive and negative sera (provided in the kit) were used for quality control in each test. Immunofluorescence was observed using an epifluorescence microscope at a magnification of 400×. According to the manufacturer’s guidelines, a titer of 1:320 was considered indicative of an infection.
Data were analyzed using R3.4.3[4]. Univariate analyses (Chi-square tests) were performed to determine the effects of different independent risk factors. We conducted generalized linear model (GLM) analysis with a binomial family using human prevalence data, with the following models:
Model 1. ${P_i} = \text{a} + {b_1} \times {\text{G}_i} + {b_2} \times {\text{A}_i} + {b_3} \times {\text{L}_i} + \text{ε}$
Model 2. ${P_i} = \text{a} + {b_1} \times {\text{G}_i} + {b_2} \times {\text{A}_i} + {b_3} \times {\text{L}_i} + {b_4} \times {\text{J}_i} + \text{ε}$
Model 3. ${P_i} = \text{a} + {b_1} \times {\text{G}_i} + {b_2} \times {\text{A}_i} + {b_3} \times {\text{L}_i} + {b_4} \times {\text{C}_i} + \text{ε}$
where Pi indicates the prevalence of B. henselae at location I; Gi, Ai, and Ji indicate the sex, age, and occupation of the patients, respectively; Li indicates the four different locations; Ci indicates whether the patient had contact with a potentially infected animal host; a indicates the overall intercept; and bn indicates the partial regression coefficient of the model.
B. henselae antibodies were evaluated in 1,260 serum samples, and the positive rate was 9.68% (Table 1). According to the results of Chi-square tests, the seroprevalence of B. henselae varied significantly among sites in the four provinces (5.78%–18.95%; χ2 = 40.05, P < 0.001). The seropositivity rates in Dongying city and Xingshan county were significantly lower than those in other cities (Table 1).
Among seropositive individuals, males (8.40%) had a lower seroprevalence than females (11.02%); however, this difference was not statistically significant (χ2 = 0.755, P = 0.385). Similar results were observed for the GLM (Table 1).
In our study, the highest seropositivity was observed in patients 50–59 years old (χ2 = 41.22, P < 0.001), and similar results were observed for the GLM. The seropositivity rates among individuals with a history of raising and being scratched by cats/dogs was not significantly different (χ2 = 0.51, P = 0.48), similar to the results in the GLM (P > 0.05). The seroprevaleces of urban workers and field workers were significantly different (χ2 = 16.15, P < 0.001, OR = 12.09) in the GLM (P < 0.01, Table 1).
Modern diagnostic approaches, including serological tests and polymerase chain reaction, provide precise information concerning B. henselae infection. Serological testing is the reference test for the diagnosis of CSD and other Bartonella-based infections. Although determination of the seroprevalence of B. henselae in the general population is important for controlling and preventing CSD, few large-scale epidemiological surveys of B. henselae have been reported among such populations.
To the best of our knowledge, this study report the prevalence of anti-Bartonella IgG antibodies in a large-scale population in China advancedly. We found that 9.68% of healthy populations had IgG antibodies against B. henselae; this rate was slightly lower than those reported by other investigators in other countries[3]. Some previous studies have indicated that B. henselae may be endemic in China and that its seroprevalence may be 5.78%–19.60% in Zhejiang[5] and 14.28% in Yunnan[6]. Indeed, the seroprevalence of Bartonella infection varies according to geographical region. Warm and humid environments are associated with higher seroprevalence rates for B. henselae because these areas have the highest number of potential vectors[7]. Zhejiang and Yunnan are located in the central and southwestern part of China, and the mild, wet climate is suitable for propagation of arthropod vectors. These B. henselae seropositivity rates were higher than that in our study, indicating that the climate may just be one of several factors influencing the prevalence of Bartonellosis. Therefore, new surveys should be conducted with larger numbers of participants and covering broader regions, particularly southern China.
Because of the inherent serological crossreactivity among Bartonella species in IFAs[5], the observed seropositivity for B. henselae antigen may represent a prior or current infection with another Bartonella species. According to our analysis, crossreactivity with other Bartonella species cannot be ruled out; however, the kit used in this study has more than 84% specificity and 88% sensitivity for B. henselae IgG. Thus, our data suggested that the antibodies detected in this study were representative of all Bartonella species, although most were against B. henselae. Despite this assumption, the observed 9.95% seroprevalence suggested that the rate of Bartonella infection in the general population may be underestimated in China. The epidemiology of and risk factors for Bartonella infections have not been fully elucidated. In this study, there were no significant differences in sex or animal contact, and only residential area was significantly between urban workers and field workers. Working conditions with a high probability of close contact with rodents, bacteremic cats, fleas, lice, and other potential vectors; repeated parenteral exposures; and insufficient medical care may lead to increased infection rates[5, 8]. Accordingly, it is recommended that field workers take appropriate protective measures to reduce bites by hosts and blood-sucking vectors. Previously published studies have indicated that B. henselae infections are more common in children than adults[9]. However, we did not observe a significantly higher seropositivity among children than adults in our study. In contrast, in our study, the highest seropositivity was observed in patients 50–59 years of age. Similar results have been reported in previous literatures[5]. With the growth of age, people has been exposed to risk factors such as cats and ticks for a longer time, and his/her immunity begins to decline, so the positive rate of infection is higher. There are few people over the age of 60 years old in this experiment, so the results are biased. In the future, we will make further investigation on the people over the age of 50 years old. Moreover, sex has been shown to not affect the prevalence of B. henselae infection[5]. Similarly, in our current study, we found no significant relationship between seroprevalence of B. henselae and sex. However, in one previous study, males were found to be more likely to acquire CSD than females[10].
B. henselae infection is a zoonotic, vector-borne disease, and animals and their parasitic vectors play important roles in the transmission of this bacterium. Cats and cat fleas are the most important hosts and vectors, respectively, although other animals and insects, including dogs, rodents, lice, and flies, play important roles as well. In our study, we found that the seroprevalence rate of healthy individuals with a pet contact history was 21.57%, which was higher than that of donors without pet contact history (16.44%); however, this difference was not statistically significant. In future studies, we will separate people based on the contact history of cats and dogs then analyze, and the results could be more accurate.
In this study, we found that Bartonella infections were widespread among healthy individuals, indicating that a substantial portion of healthy individuals may be asymptomatic carriers of B. henselae or have been infected previously with B. henselae. Because Bartonella spp. are emerging and re-emerging pathogens that can cause various symptoms, it is important to identify certain potential risk factors and dissect the degree of seropositivity to various Bartonella agents. Our current findings therefore provide an epidemiological and serological framework for future Bartonella studies.
Reference (10)
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2021-04-15 10:44:59
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https://dsp.stackexchange.com/questions/28667/can-i-create-a-transformation-matrix-from-rotation-translation-vectors
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# Can I create a transformation matrix from rotation/translation vectors?
I'm trying to deskew an image that contains an element of known size. Here's a test image:
I can use aruco::estimatePoseBoard which returns rotation and translation vectors. Is there a way to use that information to deskew everything that's in the same plane as the marker board? (Presumably by using the rotation and translation vectors to create a transformation matrix to pass to warpPerspective.)
I know how to deskew the marker board itself. What I want to be able to do is deskew the other things (in this case, the cloud-shaped object) that are in the same plane as the marker board. I'm trying to determine whether or not that's possible and, if so, how to do it.
I can already put four individual markers around the object I want to deskew and use the detected corners along with the known distance between them as input to getPerspectiveTransform. That works very well. But for our real-world application it may be difficult for the user to place markers exactly. It would be much easier if they could place a single marker board in the frame and have the software deskew the other objects.
• From the principles of computer graphics (3D transformations) there are methods of creating transformation matrices from such vectors.Btw, skew is not the best way to describe what happen in this image, which seems like a perspective projecton of a rotated plane, to me. – Fat32 Feb 11 '16 at 2:02
• @Fat32 No doubt I'm using incorrect terminology and that you're correct about this being a perspective issue, not skewing. So far, I haven't come up with the right search to find a source to teach me how to create the transformation matrix from the vectors. If you can point me to something I'd appreciate it. – SSteve Feb 11 '16 at 2:10
• Computer Graphics_Principles and Practice_2ed_Foley, chs 5 & 6 provide all the necessary mathematics of both 3D transformations and perspective projection issues. Give it a try. Don't expect a solution in 2 days though. It can take more than a week (depending on how many hours you spend) – Fat32 Feb 11 '16 at 3:26
• @Fat32 Thanks. I ordered the book. In the meantime I've been reading chapter 10 (Transformations in Two Dimensions) in the 3rd ed which is available free on the book's website and it's already shedding some light. – SSteve Feb 12 '16 at 12:37
• I was actually referring to 2nd edition (from 1990s), but 3rd is also quite fine, at least a modern version. So chapter 10 in 3rd edition refers to 2D transforms. It's one of the most reliable sources. – Fat32 Feb 12 '16 at 22:45
I was stuck on the assumption that the destination points in the call to getPerspectiveTransform had to be the corners of the output image. Once it dawned on me that the destination points could be somewhere within the output image I had my answer.
float boardX = 1240;
float boardY = 1570;
float boardWidth = 1730;
float boardHeight = 1400;
vector<Point2f> destinationCorners;
destinationCorners(Point2f(boardX+boardWidth, boardY));
destinationCorners(Point2f(boardX+boardWidth, boardY+boardHeight));
destinationCorners(Point2f(boardX, boardY+boardHeight));
destinationCorners(Point2f(boardX, boardY));
Mat h = getPerspectiveTransform(detectedCorners, destinationCorners);
Mat bigImage(image.size() * 3, image.type(), Scalar(0, 50, 50));
warpPerspective(image, bigImage, h, bigImage.size());
This fixed the perspective of the board and everything in its plane. (The waviness of the board is due to the fact that the paper wasn't lying flat in the original photo.)
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2019-11-17 12:08:06
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https://ftp.mcs.anl.gov/pub/fathom/moab-docs/classmoab_1_1WriteCCMIO_1_1MeshInfo.html
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MOAB: Mesh Oriented datABase (version 5.3.1)
moab::WriteCCMIO::MeshInfo Class Reference
contains the general information about a mesh More...
#include <WriteCCMIO.hpp>
Collaboration diagram for moab::WriteCCMIO::MeshInfo:
MeshInfo ()
## Public Attributes
unsigned int num_dim
unsigned int num_nodes
unsigned int num_elements
unsigned int num_matsets
unsigned int num_dirsets
unsigned int num_neusets
Range nodes
## Detailed Description
contains the general information about a mesh
Definition at line 77 of file WriteCCMIO.hpp.
## Constructor & Destructor Documentation
moab::WriteCCMIO::MeshInfo::MeshInfo ( ) [inline]
Definition at line 88 of file WriteCCMIO.hpp.
: num_dim( 0 ), num_nodes( 0 ), num_elements( 0 ), num_matsets( 0 ), num_dirsets( 0 ), num_neusets( 0 )
{
}
## Member Data Documentation
Definition at line 86 of file WriteCCMIO.hpp.
unsigned int moab::WriteCCMIO::MeshInfo::num_dim
Definition at line 80 of file WriteCCMIO.hpp.
unsigned int moab::WriteCCMIO::MeshInfo::num_dirsets
Definition at line 84 of file WriteCCMIO.hpp.
unsigned int moab::WriteCCMIO::MeshInfo::num_elements
Definition at line 82 of file WriteCCMIO.hpp.
unsigned int moab::WriteCCMIO::MeshInfo::num_matsets
Definition at line 83 of file WriteCCMIO.hpp.
unsigned int moab::WriteCCMIO::MeshInfo::num_neusets
Definition at line 85 of file WriteCCMIO.hpp.
unsigned int moab::WriteCCMIO::MeshInfo::num_nodes
Definition at line 81 of file WriteCCMIO.hpp.
List of all members.
The documentation for this class was generated from the following file:
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2021-12-09 00:35:59
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https://pablo.tools/posts/computers/custom-kali-box/
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# Custom Kali with Ansible and Vagrant
Having spend way too much time perfecting my personal development setup, I’ve reached a point, where it annoys me to not be able to use the muscle memory, keyboard shortcuts and tools I have gotten used to. Like probably most Linux users I started out with a ready-made Linux distribution, I think it was Debian in my case. Over the years I have tried countless distributions, window managers, editors, themes, shells and tools. Some of them stuck with me, others proved not to be what I was looking for.
This “natural selection” has stabilized in some instances, but I still try to optimize workflows where I can and look into new stuff on a regular basis. While I would like to think differently, critics might say I could have spent all the time I tried to fine-tune vim plugins to save a few keystrokes on writing actual code and would have been more productive. This might be true, but let’s face it: I had and still have a lot of fun on the way, have learned about things I would never have discovered.
This endless rabbit hole even led me to build my own keyboard and program it with a modified firmware along with learning the colemak keyboard layout, just to name a few.
## The plan
While Kali Linux is a great tool for security research and pentesting, I have trying to avoid using it for the reasons above. It’s great, but it seemed easier in most cases to just install an AUR package on arch Linux, than to have to work in an uncustomized VM.
This approach is starting to reach its limits. My OS is starting to get full of one-use tools I installed for a particular use-case and forgot to get rid of afterwards. Also, quite a lot of them need root privileges to work properly and I don’t feel like running everything on my main installation.
My dotfiles have been managed with Ansible for some time now. In conjunction with Vagrant this seemed like a good solution to create Kali VMs with one command, use them and dispose them to create them fresh if something goes sideways.
I adapted them and created a Vagrantfile with an Ansible playbook that allows me to generate and start a fresh Kali VM with one command, including all my personal customizations. And here is how I did it.
## Ingredients
Ansible: Wikipedia says: Ansible is an open-source software provisioning, configuration management, and application-deployment tool. I will be using it to automate all the setup steps I would normally have to do manually on a fresh install to get it in the state I want it.
Vagrant: Wikipedia (again): Vagrant is an open-source software product for building and maintaining portable virtual software development environments.
The setup will be based around the i3 tiling window manager and my dotfiles.
## Vagrantfile
Kali has announced officially maintained Images on Vagrant Cloud.
The so called Vagrantfile defines a VM based on some Image. If you wanted to just use Kali in its default configuration, you could just start the Vagrant file provided and be done.
To get it, just create an empty directory and initialize it with vagrant init
vagrant init kalilinux/rolling
This will create a Vagrantfile that can be used to spawn new VMs with ease. Vagrant itself doesn’t do the virtualization, it only manages it. While multiple virtualization backends are supported, I opted for the default: VirtualBox.
There are a few basic settings I changed to add a shared folder to the VM, set the amount of RAM it can use and enable the GUI.
# Share an additional folder to the guest VM. The first argument is
# the path on the host to the actual folder. The second argument is
# the path on the guest to mount the folder.
config.vm.synced_folder "~/CTF", "/vagrant_data"
# VirtualBox settings
config.vm.provider "virtualbox" do |vb|
# Display the VirtualBox GUI when booting the machine
vb.gui = true
# Customize the amount of memory on the VM:
vb.memory = "4096"
end
Later on, I encountered an error that occurs from time to time and prevents Ansible from finding the playbook we will add. There is an issue on their GitHub page about it and I’m still not sure if it’s a proper bug. The solution was to add this line just before the final end in the Vagrantfile.
# Workaround to make Ansible find the playbook
vagrant_synced_folder_default_type = ""
Now comes the most important customization of the file, adding Ansible provisioning. The Vagrantfile is a plain ruby file, but even without knowing the language it should be mostly self-explanatory.
We use the :ansible_local build-in provisioning. This allows for Ansible to be executed inside the VM. That way, you don’t have to even have Ansible installed on your system. The ansible.install = true takes care of magically installing it when needed in the VM.
config.vm.provision :ansible_local do |ansible|
ansible.playbook = "playbook.yml"
ansible.verbose = true
ansible.install = true
ansible.limit = "all"
ansible.galaxy_role_file = "requirements.yml"
ansible.galaxy_roles_path = "roles"
ansible.become = true
end
If you have worked with Ansible before, you will know what the playbook.yml and requirements.yml files are. Here we just instruct Ansible to look for them in the current directory. Internally Vagrant creates a shared folder and copies those files to the VM so Ansible can run on it.
And that’s it for the Vagrantfile. We now only have to create our Ansible project to set everything up. One note about Ansible’s inventory: Vagrant will generate that for us dynamically. The ansible.limit = "all" directive takes care of everything else.
## Ansible
Let’s start with what I already had. To manage my dotfiles on my main OS I have created already roles that take care of setting up different applications. This can be added to the roles list in the playbook and will be executed. If you want to look into what each of them does, look at my dotfiles repository.
roles:
- ansible-xresources
- ansible-i3
- ansible-vim
- ansible-xfce4-terminal
- ansible-tmux
- ansible-zsh
- ansible-rofi
These roles only set up the configuration files for my personal user and do not actually install the tools. I kept them that way to make them compatible with different distributions and being able to deploy them on systems where I don’t have root privileges to install stuff. For this reason, we need to install the packages that are not in Kali per default.
I added a task with a list of packages to be installed via apt. It’s easy to add or remove entries here. The list is not that long so putting them in a separate file didn’t make sense at this point.
tasks:
- name: Install missing packages
apt:
pkg:
- i3-wm
- i3lock
- i3status
- zsh
- neovim
- apt-file
- htop
- rofi
- nodejs
- nitrogen
state: latest
update_cache: yes
Since my font of choice, Roboto Mono is not available in the repositories I also added a straight forward task to pull it from google and place it in the correct location. It includes a creates directive that checks if the fonts are already present and skips downloading if true.
- name: Install fonts and update font cache
script: install_fonts.sh
args:
creates: /usr/share/fonts/truetype/robotomono
The script just downloads the font files to the correct folder and updates the font cache. The --content-disposition enables the use of “Content-Disposition” headers to describe what the name of a downloaded file should be.
#!/usr/bin/env bash
wget --content-disposition -P \
/usr/share/fonts/truetype/robotomono \
# Update font cache
fc-cache -f -v
I also set the shell of root to zsh in a separate task.
- name: Set shell of user root to zsh
user:
name: root
shell: /bin/zsh
The last two tasks are not absolutely necessary but make the experience a bit more enjoyable. I guess you can guess what they do.
- name: Copy wallpaper
copy:
src: wallpaper.png
dest: /usr/share/backgrounds/kali/kali-i3.png
- name: Set wallpaper
copy:
src: nitrogen.cfg
dest: "{{ansible_env.HOME}}/.config/nitrogen/bg-saved.cfg"
While talking about eye-candy I should add, that all my dotfiles are based on the great base16 color schemes by chriskemptson. You can find all the supported applications in this list. Each color scheme is defined is made up of 16 colors and has it’s own repository. More specifically, the color schemes definitions are defined in yaml syntax, which is great because Ansible can use it in its templates directly.
To use a different color scheme just pick one that you like and copy it to the group_vars/all file in the Ansible project. I chose onedark, you can see the result in the screenshot above.
# group_vars/all
---
scheme: "OneDark"
scheme: "base16-onedark"
author: "Lalit Magant (http://github.com/tilal6991)"
base00: "282c34"
base01: "353b45"
...
base0F: "be5046"
Finally there are a few differences between the configuration files I use on my main installation and the ones inside the VM. For this purpose I added a few control variables to the vars: section of the playbook. Setting these values will change the outcome of the rendered templates.
As an example, I use the i3-gaps fork of i3 on my system, which has some patches applied to it. While there is a Linux Arch package for it, there is none in the repositories of Kali. The configuration is mostly compatible, except for the gaps directives, which are provided by one of the patches.
In this case the i3_use_gaps variable is set to false in the playbook (default value being true) explicitly like this:
vars:
- i3_use_gaps: false
- i3_terminal: xfce4-terminal
- i3_use_polybar: false
- i3_optional: false
The relevant parts of my i3 configuration template are fenced with special jinja2 directives, which evaluate those and only include the output between them if the value is true.
{% if i3_use_gaps %}
gaps inner 6
gaps outer 0
smart_gaps on
smart_borders no_gaps
{% endif %}
### A note about Ansible Galaxy
Ansible Galaxy provides prepackaged units of work known to Ansible as roles. Think of it like a sharing platform for Ansible roles, directly integrated with GitHub. This allows us to specify a requirements.yml file with all the roles needed for the playbook. Vagrant also passes this file to Ansible which in turn automatically downloads them, so they can be used.
The syntax is pretty straightforward and just defines where to get them from, the name and the protocol to use. Here is an example for the i3 role:
- src: https://github.com/pinpox/ansible-i3.git
name: ansible-i3
scm: git
## Give me the VM already!
And that’s it. If all that sounded complicated to just create a virtual machine, be relived to know that you now just need one single command to generate the VM:
vagrant up
In case you make changes to the playbook after the VM has already been created, you can run the modified playbook again without the need to create a new machine with vagrant provision
|
2021-01-25 03:11:09
|
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|
https://tel.archives-ouvertes.fr/tel-01927948
|
# Projection-based in-situ 4D mechanical testing
Abstract : The quantitative analysis of 3D volumes obtained from tomography allows models to be identified and validated. It consists of a sequence of three successive inverse problems: (i) volume reconstruction (ii) kinematic measurement from Digital Volume Correlation (DVC) and (iii) identification. The required very long acquisition times prevent fast phenomena from being captured.A measurement method, called Projection-based DVC (P-DVC), shortens the previous sequence and identifies the kinematics directly from the projections. The number of radiographs needed for tracking the time evolution of the test is thereby reduced from 500 to 1000 down to 2.This thesis extends this projection-based approach to further reduce the required data, letting faster phenomena be captured and pushing the limits of time resolution. Two main axes were developed:- On the one hand, the use of different spatial and temporal regularizations of the 4D fields (space/time) generalizes the P-DVC approach (with a known reference volume) to the exploitation of a single radiograph per loading step. Thus, the test can be carried out with no interruptions, in a few minutes instead of several days.- On the other hand, the measured motion can be used to correct the reconstructed volume itself. This observation leads to the proposition of a novel procedure for the joint determination of the volume and its kinematics (without prior knowledge) opening up new perspectives for material and medical imaging where sometimes motion cannot be interrupted.end{itemize}The development of these two axes opens up new ways of performing tests, faster and driven to the identification of key quantities of interest. These methods are compatible with the recent hardware" developments of fast tomography, both at synchrotron beamlines or laboratory and save several orders of magnitude in acquisition time and radiation dose.
Keywords :
Document type :
Theses
Cited literature [325 references]
https://tel.archives-ouvertes.fr/tel-01927948
Contributor : Abes Star <>
Submitted on : Tuesday, November 20, 2018 - 11:24:07 AM
Last modification on : Thursday, September 12, 2019 - 3:37:23 AM
### File
72543_JAILIN_2018_archivage.pd...
Version validated by the jury (STAR)
### Identifiers
• HAL Id : tel-01927948, version 1
### Citation
Clément Jailin. Projection-based in-situ 4D mechanical testing. Mechanics of the solides [physics.class-ph]. Université Paris-Saclay, 2018. English. ⟨NNT : 2018SACLN034⟩. ⟨tel-01927948⟩
Record views
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2019-09-15 22:15:07
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http://mathematica.stackexchange.com/questions?page=9&sort=newest
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I stumbled upon this problem when I was trying to make some nice smooth ListDensityPlots of my numerical data. In short, my problem is that I let Mathematica plot two copies of basically the exact ...
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### Finding the max points of a surface, maximize
So, this should not be hard at all, in fact no one seems so be having my issue so it should be me missing out on a point of information or just experience. So my mission is to with Mathematica find ...
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dS/dt = a - dS - λ S I + β R dI/dt = λ S I - (d + m)I dR/dt = m I- (d + β)R Solve the above equation and plot the curve.
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### MatchQ[x y, (x | y) (x | y)] returns False
MatchQ[x y, (x | y) (x | y)] It returns false. Why? I want to eliminate terms like x^2, ...
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### How can I combine NDSolve[] with NSolve[]
Consider the two below ODEs with following dependent B.C. In the boundary conditions $k$ is an unknown. $$y_1''(x)=x$$ $$y_2''(y)=0$$ $$y_1(1)=y_2(1)$$ $$y_1(-k)=y_2(-k)$$ \frac{\text{dy}_1(1)}{\...
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### How to graph a 3D function of two variables? [closed]
How to graph this: \begin{alignat*}{3} x(s, t) &= a\cos(mt) \cos^{k}(ns) &&\cos(t) &&\cos(s), \\ y(s, t) &= a\cos(mt) \cos^{k}(ns) &&\sin(t) &&\cos(s), \\ z(s, ...
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I'm trying to compile what I think is a pretty simple function that does matrix multilplications in a loop (see below). But Mathematica issues the message: Could not complete external evaluation ...
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### Adding labels to close points in a ListPlot [duplicate]
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### Can evaluation show what was evaluated along with results? [duplicate]
After I write and evauate the following lines in Mathematica: a = 1 b = 2 x = a + b It will show only the result : 3. I want mathematica to show the following ...
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2016-06-25 03:36:44
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https://www.lmfdb.org/knowledge/show/nf.index
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show · nf.index all knowls · up · search:
Let $\mathcal{O}$ be an order in a number field $K$. The index of $\mathcal{O}$ is its index as a sublattice of the ring of integers of $K$.
Authors:
Knowl status:
• Review status: reviewed
• Last edited by John Jones on 2018-08-08 17:43:27
Referred to by:
History:
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2020-12-04 05:49:27
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https://www.typhoon-hil.com/documentation/typhoon-hil-software-manual/References/zeta_converter.html
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# Zeta Converter
Description of the Zeta Converter component in Schematic Editor
A block diagram and input parameters for the Zeta Converter switching block are given in Table 1
Table 1. The Zeta Converter component in the HIL Toolbox
component component dialog window component parameters
• Property tabs
• General
• Electrical
Weight = 1 or 2
## Component dialogue box and parameters
The Zeta Converter component dialogue box consists of two tabs:
1. The General tab contains control-related properties. A detailed description can be seen in Control options.
2. Electrical tab parameters are shown in Table 2.
Table 2. Electrical tab properties
Property Code name Description
Inductance type ind_type Chooses between the inductor and the switched inductor
Double switch/inductor dbl_ind Enables the use of two equal inductances, each with a corresponding controllable switch. More details in Converter configurations section.
Inductance inductance Value of inductance of a single inductor. [H]
Initial current init_current Initial current of the inductors. [A]
Capacitance capacitance Value of capacitance of a single capacitor. [F]
Hybrid-switched capacitor hybrid_cap Enables the use of two equal capacitors, along with two diodes instead of one. More details in Converter configurations section.
Capacitor's ESR esr Equivalent series resistance of the capacitors. [Ω]
Initial voltage init_voltage Initial voltage of the capacitors. [V]
## Control options
Selecting Digital inputs as the Control parameter enables assigning gate drive inputs to any of the digital input pins (from 1 to 32(64)). For example, if S1 is assigned to 1, the digital input pin 1 will be routed to the S1 switch gate drive. In addition, the gate_logic parameter selects either active high (High-level input voltage VIH turns on the switch), or active low (Low-level input voltage VIL turns on the switch) gate drive logic, depending on the external controller design.
Selecting Internal modulator as the Control parameter, enables use of the internal PWM modulator for driving S1 switch instead of the digital input pins. In this configuration, two additional component inputs will be present. En input is used to enable/disable the internal PWM modulator, while In is used as a reference signal input.
Selecting Model, as the Control parameter, enables setting the IGBTs' gate drive signals directly from the signal processing model. The input pin gates appears on the component and requires a scalar input of gate drive signal for S1. When controlled from the model, logic is always active high.
## Converter configurations
The Zeta Converter component supports multiple modifications to its topology. For further details, see Schematic block diagrams of converter configurations. Modifications that can be done are based on the following three principles:
1. Inductor modification: The idea is to use a so-called switched inductor configuration instead of a single inductor. In steady state operation, when the controllable switch is in 'ON' state, diodes Dp1 and Dp2 are conducting, meaning that the inductors L1 and L2 are charged while connected in parallel. When the controllable switch is in 'OFF' state, only the Ds diode is conducting, thus connecting these inductors in series during discharge. For this reason, it is necessary to use the inductors of same value. This means that it is possible to obtain a larger voltage gain, since the inductance is 'easier to charge and harder to discharge'.
2. Capacitor modification: Similarly to the switched inductor, it is possible to use two capacitors, here refered to as a 'hybrid-switched capacitor', which further increases the voltage step-up ability of the converter.
3. It is possible to use two inductances with corresponding switches. This way, two inductances are charged while in a parallel connection and discharged while in a series connection. The two switches are controlled simultaneously.
Available converter configurations are given in Table 3. Their schematic diagrams can be seen in Schematic block diagrams of converter configurations. The input to output voltage ratio formula is derived under the assumptions of continuous current conduction mode and constant capacitor voltage, with D representing the duty cycle of controllable switch(es). The output LC filter should be connected externally for each of the configurations in order for the converter to be functional.
Table 3. Converter configurations
Property value Vout/Vin Configuration name
Inductance type Double switch/inductor Use hybrid switched capacitor Input to output voltage ratio of the converter Name of the configuration variant that is configured with each set of property values
Inductor False False
$\frac{D}{1-D}$
Single inductor
Inductor False True
$\frac{2D}{1-D}$
Single inductor with hybrid-switched capacitor
Inductor True False
$\frac{2D}{1-D}$
Double inductor
Inductor True True
$\frac{3D+1}{1-D}$
Double inductor with hybrid-switched capacitor
Switched inductor False False $\frac{D\left(D+1\right)}{1-D}$ Switched inductor
Switched inductor False True $\frac{D\left(D+3\right)}{1-D}$ Switched inductor with hybrid-switched capacitor
Switched inductor True False $\frac{2D\left(D+1\right)}{1-D}$ Double switched inductor
Switched inductor True True $\frac{{2D}^{2}+5D+1}{1-D}$ Double switched inductor with hybrid-switched capacitor
## Schematic block diagrams of converter configurations
Configurations that utilize a double inductor contain two switches. Since these switches are controlled simultaneously, only one switching signal is used. Therefore, the switch designated as 'S1' in the component parameters can represent one or two controllable switches, depending on how many are included.
## Digital Alias
If a converter is controlled by digital inputs, an alias for every digital input used by the converter will be created. Digital input aliases will be available under the Digital inputs list alongside existing Digital input signals. The alias will be shown as Converter_name.Switch_name, where Converter_name is name of the converter component and Switch_name is name of the controllable switch in the converter.
## References
1. B. Axelrod, Y. Berkovich and A. Ioinovici, 'Hybrid Switched - Capacitor Ćuk/Zeta/SEPIC Converters in Step-up Mode', 2005 IEEE International Symposium on Circuits and Systems, May 2005.
2. M. S. Bhaskar et al., "A New Hybrid Zeta-Boost Converter With Active Quad Switched Inductor for High Voltage Gain," in IEEE Access, vol. 9, pp. 20022-20034, 2021, doi: 10.1109/ACCESS.2021.3054393.
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2023-01-26 22:26:07
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|
http://mathoverflow.net/questions/112223/probability-measures-with-entropy-equal-to-nonnegative-number
|
# probability measures with entropy equal to nonnegative number
Is it true that for a given nonnegative number, there exists a measure-theoretical entropy value (supremum of entropies of all partitions under a measure-preserving transformation) that equals this number?
-
Take a shift on $G^{\mathbb Z}$ where $G$ is a finite group. The entropy will be $\log |G|$ (let's say logarithm with base $2$). So if you choose $G$ such that $|G|=2^n$ (for $n$ your favourite positive integer) you have the example you are looking for. – Simone Virili Nov 12 '12 at 21:14
In general, using Cantor sets constructions, one can find sub-systems of the shift system of any given entropy between 0 and $\log(n)$. I believe this appears in Furstenberg's disjointness paper for example, although the construction is obviously much older than the paper. This is not a real research question. – Asaf Nov 12 '12 at 22:11
Use the intermediate value theorem for Bernoulli measures on $\lbrace 1,\ldots,n\rbrace^{\mathbb Z}$. – Anthony Quas Nov 12 '12 at 23:58
1. Let $S_2$ be a surface of genus 2 with constant curvature $-1$. Then the geodesic flow $\phi_t$ on the unit tangent bundle preserves the Liouville measure $\mu$ and has positive metric entropy: $h_\mu(\phi_t)=|t|\cdot h_\mu(\phi_1)$ with $h_\mu(\phi_1)>0$. So for any given number $r$, just pick $t=r/h_\mu(\phi_1)$. In fact any flow with positive entropy works.
2. We use the affineness of measure-theoretical entropy $h_{\bullet}(f):\mathcal{M}(f)\to \mathbb{R},\mu\mapsto h_\mu(f)$. Pick a large integer $n\ge e^r$ and consider the full shift $\sigma$ on $\lbrace 1,\cdots,n\rbrace^{\mathbb{Z}}$. Then $\mu=(1/n,\cdots,1/n)^{\oplus\mathbb{Z}}$ has entropy $h_\mu(\sigma)=\log n>r$. Let ${\bf 1}=(1111\cdots)$. Then the enropy of $\mu_p=p\mu+(1-p){\delta_{\mathbf{1}}}$ has entropy $p\log n$. So we set $p=r/\log n$.
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2016-06-25 16:01:58
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https://encyclopediaofmath.org/wiki/Arakelov_geometry
|
# Arakelov geometry
Arakelov theory
A combination of the Grothendieck algebraic geometry of schemes over $\mathbf{Z}$ with Hermitian complex geometry on their set of complex points. The goal is to provide a geometric framework for the study of Diophantine problems in higher dimension (cf. also Diophantine equations, solvability problem of; Diophantine problems of additive type).
The construction relies upon the analogy between number fields and function fields: the ring $\mathbf{Z}$ has Krull dimension (cf. Dimension) one, and "adding a point" $\infty$ to the corresponding scheme $\operatorname {Spec}( \mathbf{Z})$ makes it look like a complete curve. For instance, if $f \in \mathbf{Q} ^ { * }$ is a rational number, the identity
\begin{equation*} \sum _ { p } v _ { p } ( f ) \operatorname { log } ( p ) + v _ { \infty } ( f ) = 0, \end{equation*}
where $v _ { p } ( f )$ is the valuation of $f$ at the prime $p$ and where $v _ { \infty } ( f ) = - \operatorname { log } | f |$, is similar to the Cauchy residue formula
\begin{equation*} \sum _ { x \in C } v _ { x } ( f ) = 0 \end{equation*}
for the differential $d f / f$, when $f$ is a non-zero rational function on a smooth complex projective curve $C$.
In higher dimension, given a regular projective flat scheme $X$ over $\mathbf{Z}$, one considers pairs $( Z , g )$ consisting of an algebraic cycle $Z$ of codimension $p$ over $X$, together with a Green current $g$ for $Z$ on the complex manifold $X ( \mathbf{C} )$: $g$ is real current of type $( p - 1 , p - 1 )$ such that, if $\delta _{\text{Z}}$ denotes the current given by integration on $Z ( \mathbf{C} )$, the following equality of currents holds:
\begin{equation*} d d ^ { c } g + \delta _ { Z } = \omega, \end{equation*}
where $\omega$ is a smooth form of type $( p , p )$. Equivalence classes of such pairs $( Z , g )$ form the arithmetic Chow group $\widehat { \operatorname {CH} } ^ { p } ( X )$, which has good functoriality properties and is equipped with a graded intersection product, at least after tensoring it by $\mathbf{Q}$.
These notions were first introduced for arithmetic surfaces, i.e. models of curves over number fields [a1], [a2] (for a restricted class of currents $g$). For the general theory, see [a7], [a9] and references therein.
Given a pair $( E , h )$ consisting of an algebraic vector bundle $E$ on $X$ and a $C ^ { \infty }$ Hermitian metric $h$ on the corresponding holomorphic vector bundle on the complex-analytic manifold $X ( \mathbf{C} )$, one can define characteristic classes of $( E , h )$ with values in the arithmetic Chow groups of $X$. For instance, when $E$ has rank one, if $s$ is a non-zero rational section of $E$ and $\operatorname { div } ( s )$ its divisor, the first Chern class of $( E , h )$ is the class of the pair $( Z , g ) = ( \operatorname { div } ( s ) , - \operatorname { log } ( h ( s , s ) ) )$. The main result of the theory is the arithmetic Riemann–Roch theorem, which computes the behaviour of the Chern character under direct image [a8], [a6]. Its strongest version involves regularized determinants of Laplace operators and the proof requires hard analytic work, due to J.-M. Bismut and others.
Since $\widehat { CH } ^ { 1 } ( \operatorname { Spec } ( \mathbf{Z} ) ) = \mathbf{R}$, the pairings $$\widehat { CH }^p (X) \otimes \widehat { CH } ^ {n-p}(X) \rightarrow \widehat { CH } ^ { 1 } ( \operatorname { Spec } ( \mathbf{Z} ) ) \,,$$ $p \geq 0$, give rise to arithmetic intersection numbers, which are real numbers when their geometric counterparts are integers. Examples of such real numbers are the heights of points and subvarieties, for which Arakelov geometry provides a useful framework [a3].
When $X$ is a semi-stable arithmetic surface, an important invariant of $X$ is the self-intersection $\omega ^ { 2 }$ of the relative dualizing sheaf equipped with the Arakelov metric [a1]. L. Szpiro and A.N. Parshin have shown that a good upper bound for $\omega ^ { 2 }$ would lead to an effective version of the Mordell conjecture and to a solution of the ABC conjecture [a10]. G. Faltings and E. Ullmo proved that $\omega ^ { 2 }$ is strictly positive [a4], [a11]; this implies that the set of algebraic points of $X$ is discrete in its Jacobian for the topology given by the Néron–Tate height.
P. Vojta used Arakelov geometry to give a new proof of the Mordell conjecture [a12], by adapting the method of Diophantine approximation. More generally, Faltings obtained by Vojta's method a proof of a conjecture of S. Lang on Abelian varieties [a5]: Assume $A$ is an Abelian variety over a number field and let $X \subset A$ be a proper closed subvariety in $A$; then the set of rational points of $X$ is contained in the union of finitely many translates of Abelian proper subvarieties of $A$.
#### References
[a1] S.J. Arakelov, "Intersection theory of divisors on an arithmetic surface" Math. USSR Izv. , 8 (1974) pp. 1167–1180 MR472815 Zbl 0355.14002 [a2] S.J. Arakelov, "Theory of intersections on an arithmetic surface" , Proc. Internat. Congr. Mathematicians Vancouver , 1 , Amer. Math. Soc. (1975) pp. 405–408 MR466150 [a3] J.-B. Bost, H. Gillet, C. Soulé, "Heights of projective varieties and positive Green forms" J. Amer. Math. Soc. , 7 (1994) pp. 903–1027 MR1260106 Zbl 0973.14013 [a4] G. Faltings, "Calculus on arithmetic surfaces" Ann. of Math. , 119 (1984) pp. 387–424 MR0740897 Zbl 0559.14005 [a5] G. Faltings, "Diophantine approximation on Abelian varieties" Ann. of Math. , 133 (1991) pp. 549–576 MR1109353 Zbl 0734.14007 [a6] G. Faltings, "Lectures on the arithmetic Riemann–Roch theorem" Ann. Math. Study , 127 (1992) (Notes by S. Zhang) MR1158661 Zbl 0744.14016 [a7] H. Gillet, C. Soulé, "Arithmetic intersection theory" Publ. Math. IHES , 72 (1990) pp. 94–174 MR1087394 Zbl 0741.14012 [a8] H. Gillet, C. Soulé, "An arithmetic Riemann–Roch Theorem" Invent. Math. , 110 (1992) pp. 473–543 MR1189489 Zbl 0777.14008 [a9] C. Soulé, D. Abramovich, J.-F. Burnol, J. Kramer, "Lectures on Arakelov geometry" , Studies Adv. Math. , 33 , Cambridge Univ. Press (1992) MR1208731 Zbl 0812.14015 [a10] L. Szpiro, "Séminaire sur les pinceaux de courbes elliptiques (à la recherche de Mordell effectif)" Astérisque , 183 (1990) [a11] E. Ullmo, "Positivité et discrétion des points algébriques des courbes" Ann. of Math. , 147 : 1 (1998) pp. 167–179 MR1609514 Zbl 0934.14013 [a12] P. Vojta, "Siegel's theorem in the compact case" Ann. of Math. , 133 (1991) pp. 509–548 MR1109352 Zbl 0774.14019
How to Cite This Entry:
Arakelov geometry. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Arakelov_geometry&oldid=50891
This article was adapted from an original article by Christophe Soulé (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article
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2023-03-31 19:07:14
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https://math.stackexchange.com/questions/3052569/finding-structures-where-the-singleton-axiom-the-axiom-of-extensionality-and-th
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finding structures where the singleton axiom, the axiom of extensionality and the axiom of empty set do hold
Hey friends of maths,
I am trying to find an example that fulfills the requirements mentioned above, more precisely:
I'd appreciate your help with finding two (or better more than 200) non-isomorphic finite structures where the axiom $$\forall x \exists P \forall z (z \in P \leftrightarrow z = x)$$ does hold, and also, if possible, where the axioms $$\forall A \forall B ([\forall x: x \in A \Leftrightarrow x \in B] \Rightarrow A = B)$$ and $$\exists N:(\forall x: x \in N \Leftrightarrow x \neq x)$$ - do hold.
This will help: it is impossible to satisfy both the axiom of the empty set and the singleton axiom using a finite sized structure:
First, by axiom of the empty set, you need an object, call it $$d_0$$, and for whatever other objects you have in the domain, those objects cannot stand in the relation as denoted by $$\in$$ (let's call this relarion $$R$$) with it.
Second, by the axiom of the singleton, there needs to be an object such that $$d_0$$ is the one and only object standing in the relation $$R$$ with it. For the reason just stated earlier, this cannot be $$d_0$$ itself, and so we need a new object $$d_1$$. So, we have $$(d_0,d_1)\in R$$ ... and there cannot be any other objects standing in relation $$R$$ to $$d_1$$.
OK, but then by the axiom of the singleton again, there needs to be some objects such that $$d_1$$ is the one and only object standing in relation $$R$$ with it ... and this cannot be $$d_0$$ (since nothing stands in relarion $$R$$ to $$d_0$$), not can it be $$d_1$$, since there cannot be any object other than $$d_0$$ standing in relation $$R$$ to $$d_1$$. So, we need to have anew object, $$d_2$$, and we need that $$(d_1,d_2)\in R$$, and we need to make sure that nothing but $$d_1$$ stands in the relation $$R$$ to $$d_2$$.
... ok, I think you see where this is going ... you indefinitely need to keep adding one more object $$d_{i+1}$$ to the already existing objects $$d_0$$ through $$d_i$$, because in order to satisfy the singleton axiom for $$d_i$$ we need some object that $$d_i$$ stands in the relation $$R$$ to, and by the axiom of the empty set that object cannot be $$d_0$$ and since we have $$(d_j,d_{j+1})\in R$$ for all $$0\le j \le i-1$$, all objects $$d_1$$ through $$d_i$$ already have some object other than $$d_i$$ standing in the relation $$R$$ to them and are not allowed to have any other object stand in relation $$R$$ to them.
• This answers the title question with room to spare, since extensionality wasn't needed. But the text of the question demands only singletons, with the empty set thrown in "if possible". Since it's impossible in a finite universe, it might be worth noting that the singleton axiom and extensionality have lots of finite models: Arrange any finite number of entities in a cyclic order and let the "membership" relation hold just between each element and the next in the cyclic ordering. – Andreas Blass Dec 26 '18 at 2:22
• @AndreasBlass Ah, I missed the 'if possible' in the body of the question. Thanks! – Bram28 Dec 26 '18 at 2:30
• @Studentu I meant it just the other way around: $(d_0,d_1)$ would stand for $d_0 \in d_1$ – Bram28 Dec 26 '18 at 21:55
• @Studentu Also, to explain what AndreasBlass was saying: suppose you have a domain with exactly one object: $d_1$. Also assume that $(d_1,d_1)\in R$, i.e. we basically have $d_1 \in d_1$. Then notice that for all objects (which is just $d_1$ of course) there is something $again$d_1$!) such that that second something has exactly one 'element' ... so, the singleton axiom is satisfied. And, the axiom of extensionality is satisfied ... there is only one object that has the same 'elements' 'inside' it. OK, so this is one model – Bram28 Dec 26 '18 at 21:59 • @Studentu OK, now consider the following: we have two obejcts,$d_1$and$d_2$. we have$d_1 \in d_2$, and$d_2 \in d_1$... a '2-cycle' of the kind Andres was talking about. Do you see how this satisfies the axioms as well? And then of course you can do 3 objects, and have$d_1 \in d_2$,$d_2 \in d_3$, and$d_3 \in d_1\$ .... – Bram28 Dec 26 '18 at 22:01
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2019-04-25 17:43:45
|
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https://socratic.org/questions/how-do-you-condense-2-ln-4-ln-2
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# How do you condense 2 ln 4 - ln 2?
May 7, 2016
$2 \ln 4 - \ln 2 = \ln 8$
#### Explanation:
As $a \ln b = \ln {b}^{a}$ and $\ln p - \ln q = \ln \left(\frac{p}{q}\right)$
$2 \ln 4 - \ln 2$
= $\ln {4}^{2} - \ln 2$
= $\ln 16 - \ln 2$
= $\ln \left(\frac{16}{2}\right) = \ln 8$
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2019-10-19 01:48:55
|
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|
https://www.mrmath.com/lessons/pre-calculus/real-number-exponents/
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# Real Number Exponents
Lesson Features »
Lesson Priority: Normal
Objectives
• Quickly recap all previously learned exponent rules
• Make connections between rational exponents, which we've seen, and terminating decimal exponents, which we haven't yet seen
• Understand how irrational exponents work and what they mean, even though they are not as intuitive
Lesson Description
Building up to the next several lessons on exponential functions and relationships, we need to understand that we can let any real number be an exponent, from integer and fraction exponents that we have already seen, to irrational numbers like $\sqrt{2}$ and $\pi$.
Practice Problems
Practice problems and worksheet coming soon!
## Exponents on Steroids
As you progress through Algebra, exponents become much more complicated than they seemed when you were first introduced to them. The first milestone complication happens when we need to apply exponent rules to numbers and variables, such as what happens when you multiply $x^{4}$ by $x^{12}$. Later on, we combined some knowledge of roots and radicals with that of exponents, and introduced the idea of fraction exponents, such as $x^{4/3}$.This lesson will open up Pandora's Box all the way and consider exponents of any type of number - including terminating decimals, repeating decimals, and even irrational numbers like $\pi$.But first, let's make sure we're up to speed on everything exponents up to this point.
## Exponent Recap
Let's quickly look at what we know about exponents.First, we were introduced to exponents » as repeated multiplication. That is indeed what exponents are when they are integers - for example, $3^5$ equals $3 \cdot 3 \cdot 3 \cdot 3 \cdot 3$. This will always be a valid statement and will always be a useful perspective, since we commonly work with integer exponents.Later on, we learned some exponent shortcut rules » for simplifying and working with expressions that contain multiple appearances of exponents, as well as rules for zero and negative exponents ». Specifically, make sure all of the following are familiar:$$x^{a} \cdot x^{b} = x^{a+b}$$$$\frac{x^a}{x^b} = x^{a-b}$$$$\left(x^a\right)^b = x^{ab}$$$$x^{-a} = \frac{1}{x^a}$$$$x^{0} = 1, \; \mathrm{for} \; x \ne 0$$Finally, in Algebra Two, we learned that exponents could be fractions » and gained an understanding of what fraction exponents actually mean.Recall that $x^{a/b}$ tells us to raise $x$ to the $a$ power and then take the $b$-th root (or vice versa - we also learned that the order doesn't matter).For example, $64^{2/3}$ is equal to $16$, because the cube root of $64^2$ is $16$ (or once again, the square of $\sqrt[3]{64}$ is also $16$).Now that we're caught up on exponents up to this point, let's do more.
## Decimal Exponents
Decimal exponents are actually not a far stretch from fraction exponents in most cases. As we've learned over the course of our Algebra journey, decimals come in three major flavors: terminating, repeating, and non-terminating non-repeating.Terminating Decimal ExponentsDue to the structure and definition of decimals, terminating decimals can always be written as fractions with a denominator of a power of $10$ (and possibly reduced to a different denominator).Examples of this include$$0.47 = \frac{47}{100}$$$$0.6 = \frac{6}{10} = \frac{3}{5}$$$$0.837629 = \frac{837,629}{1,000,000}$$Simply stated, if a decimal can be written as an equivalent fraction, then we can use what we already know about fraction exponents to deal with decimal exponents.For example, since $0.6$ equals $3/5$, we can say$$x^{0.6} = x^{3/5} = \sqrt[5]{x^3}$$Repeating DecimalsRepeating decimals require more work but ultimately give us the same end result as terminating decimals. This is because both terminating and repeating decimals represent the same type of real number - a rational number, which by definition can always be represented as a simple fraction.If $0.\bar{3}$ is really $1/3$, then$$x^{0.\bar{3}} = x^{1/3} = \sqrt[3]{x}$$Again, this is no different from what we stated a minute ago about terminating decimals. The biggest difference is that repeating decimals can be a pain to convert to rational form.These days we're often allowed to rely on technology for this conversion - especially in advanced Algebra and Pre-Calculus courses. On the TI-84, you can have it convert a repeating decimal to a fraction in two super quick steps.
• First, enter twelve or more characters of repeating pattern into the calculator.
• Then, press Math $\longrightarrow$ Enter $\longrightarrow$ Enter. This works because the [ $\triangleright$ Frac ] option is the first choice in the Math menu.
Just make sure you enter enough digits. If you only enter $0.333$, your calculator will convert it to the fraction $333/1000$. If you enter twelve or more $3's it will decide that you're actually trying to decode a repeating decimal.It's beyond uncommon, but if you have to do any work to convert repeating decimals to fractions without the aide of a calculator, make sure to check out the lesson on converting repeating decimals algebraically ».Irrational DecimalsDecimals that never end and don't repeat a pattern represent irrational numbers. Unlike rational numbers, irrational ones cannot be expressed as a simple fraction. Some examples of irrational numbers that we've encountered along the way are$\pi$or$\sqrt{2}$. Your calculator will give you decimal representations of these numbers to as many decimals as you need when asked, but the true decimal representations of these numbers never end.Without a fraction to convert to, irrational number exponents cannot be dealt with using knowledge we currently have. ## Irrational Exponents We need to allow for the possibility that an exponent could be any real number so that we can begin to allow exponents to be variables. In order to do that, we cannot settle for allowing only rational numbers to be exponents, as the set of real numbers consists of the union of the sets of both rational and irrational numbers.We can intuitively understand why the results we get are what they are, but it's important to know right away that there is no perfect conceptual translation of irrational exponents to roots and powers in the same way that exists for fraction exponents.Let's start with an example. Example 1Find two expressions that use rational exponents such that$2^{\pi}$is in between the two expressions.$\blacktriangleright$This question is intentionally somewhat open-ended to illustrate a few new ideas.First, if you grab your calculator and ask it for the decimal approximation of$2^{\pi}$, it gives us$$2^{\pi} \approx 8.824977827$$Nine decimal places is a strong approximation for most practical purposes, so we'll loosely refer to this result as the "true" value.Recall that$\pi$is approximately$3.1415927$. Recall also that as we raise$2$to bigger and bigger powers, the result grows. Specifically we know that$2^2 = 4$,$2^3 = 8$,$2^4 = 16$, etc. Putting this all together, it is reasonable to make two proposals:1)$2^{3.1} \lt 2^{3.2}$and furthermore,2)$2^{3.1} \lt 2^{\pi} \lt 2^{3.2}$What we should understand here is that while$2^{3.1}$can be written as$\sqrt[10]{2^{31}}$and$2^{\pi}$cannot, both results follow a logical progression in that$2^{\pi}$is in between$2^{3.1}$and$2^{3.2}$due to the fact that$\pi$is in between$3.1$and$3.2$, even though$2^{\pi}$doesn't have a clean interpretation in terms of roots and powers.Therefore, to answer the question, we can say that$2^{3.1}$and$2^{3.2}$are two expressions that use rational exponents and also have$2^{\pi}$in between them. You Should Know We could have found "tighter" boundaries to contain$2^{\pi}$between simply by using more decimal places. For example, the same steps and logic above could lead us to state that$2^{\pi}$is between$2^{3.1415}$and$2^{3.1416}$, which is a narrower corridor. ## Working with Functions As an introduction to the idea of having variables in the exponent rather than the base, consider the following function:$$f(x) = 4^{x}$$One reason we want to understand why both rational and irrational exponents are allowable is so that we can examine functions like this and correctly identify their domain ». As we have been discussing, both rational and irrational numbers are allowed to be exponents, so the domain of this particular function is all real numbers ($x \in \mathbb{R}$).Additionally, we should be prepared to take a function like this and evaluate it at various specific values of$x$. Example 2For the function$f(x) = 3^x$, evaluate the following:$f(1)$$f(0)$$f(\sqrt{3})$$f(-2)$$\blacktriangleright$Part of what we learn here is the intuition about what we could and could not do without a calculator. We can actually be expected to calculate each of these quantities calculator-free, except for the third one.$$f(1) = 3^{1} = 3$$$$f(0) = 3^{0} = 1$$$$f(-2) = 3^{-2} = \frac{1}{9}$$Now, the third quantity we were asked for literally can't be simplified once plugged in.$$f(\sqrt{3})=3^{\sqrt{3}}$$However, we can get a decimal approximation with a calculator, if needed.$$3^{\sqrt{3}} \approx 6.705$$ The key takeaway is that exponents can be any real number, not just integers and fractions. We can't interpret irrational exponents the same way that we can for rational ones, but that's not unusual since irrational numbers have a never-ending and never-repeating decimal expansion. But we can get decimal approximations to things like$5^{\sqrt{3}}$just like we can get decimal approximations to things like$\sqrt{3}$. And when we prefer exact answers we just leave it be. ## Put It To The Test As you become an expert on exponential relationships over the next few lessons, these basic practice problems will quickly become "too easy" to be included on unit tests. However, at this point as you are starting out with this unit, it's important that everything we are saying above is well-understood. Make sure you don't have any hang-ups with the following extra practice problems. Example 3Find two rational exponent expressions that bound the value of$4^{\sqrt{10}}$. Show solution$\blacktriangleright$Since$\sqrt{10}$is between$3$and$4$(since$\sqrt{9}$is$3$and$\sqrt{16}$is$4$) then we know that$4^{\sqrt{10}}$must be between$4^3$and$4^4$. These are the two expressions we can turn in as our answer.If we had a calculator, which can tell us that$\sqrt{10} \approx 3.162$, we could be even more precise and pick something like$4^{3.16}$and$4^{3.17}$as our answers. Harder questions will require this kind of precision by asking for two numbers such that the error from the true value is very small, say less than$1/100$-th. Example 4With the given function, evaluate the exact result at each specified$x$values, and then use a calculator to evaluate a decimal approximation to those function values.$$f(x) = 9^{x}$$for$x \in {0.5,\; 1,\; 2,\; \pi,\; 4}$. Show solution$\blacktriangleright$Except for the$x$value of$\pi$, these are all non-calculator computations.$$f(0.5) = 9^{1/2} = 3$$$$f(1) = 9^1 = 9$$$$f(2) = 9^2 = 81$$$$f(4) = 9^4 = 6561$$Now, when$x=\pi\$, we should turn in the exact answer as instructed, in addition to the decimal approximation (the previous four answers do not need any approximating since they are exact integers):$$f(\pi) = 9^{\pi} \approx 995.04$$
Lesson Takeaways
• Review and re-master all exponent rules from Algebra
• Understand the connection between decimal exponents and fraction exponents
• See how irrational exponents fit into the grand scheme of numbers similar to how irrational numbers fit in between rational ones
• Begin to work with functions that have variable exponents and domains of all real numbers
• Popular Content
• Get Taught
• Other Stuff
Lesson Metrics
At the top of the lesson, you can see the lesson title, as well as the DNA of Math path to see how it fits into the big picture. You can also see the description of what will be covered in the lesson, the specific objectives that the lesson will cover, and links to the section's practice problems (if available).
Key Lesson Sections
Headlines - Every lesson is subdivided into mini sections that help you go from "no clue" to "pro, dude". If you keep getting lost skimming the lesson, start from scratch, read through, and you'll be set straight super fast.
Examples - there is no better way to learn than by doing. Some examples are instructional, while others are elective (such examples have their solutions hidden).
Perils and Pitfalls - common mistakes to avoid.
Mister Math Makes it Mean - Here's where I show you how teachers like to pour salt in your exam wounds, when applicable. Certain concepts have common ways in which teachers seek to sink your ship, and I've seen it all!
Put It To The Test - Shows you examples of the most common ways in which the concept is tested. Just knowing the concept is a great first step, but understanding the variation in how a concept can be tested will help you maximize your grades!
Lesson Takeaways - A laundry list of things you should be able to do at lesson end. Make sure you have a lock on the whole list!
Special Notes
Definitions and Theorems: These are important rules that govern how a particular topic works. Some of the more important ones will be used again in future lessons, implicitly or explicitly.
Pro-Tip: Knowing these will make your life easier.
Remember! - Remember notes need to be in your head at the peril of losing points on tests.
You Should Know - Somewhat elective information that may give you a broader understanding.
Warning! - Something you should be careful about.
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2021-06-22 16:30:13
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https://puzzling.stackexchange.com/questions/57628/landing-4-on-a-broken-die
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# Landing 4 on a Broken Die [closed]
This is not hard, but fun.
Suppose you have a die such that its 1,5,6 and 4 face are broken. Whenever 1 face lands, it gets rolled over to an adjacent face. If 5 face lands, it is equally likely that it may stick or get rolled over to the adjacent face. If 4 lands, it has 1/4 probability of rolling over to the adjacent face. While if 6 shows up it always sticks. (EDIT: Not to confuse 6 is no different from 2,3. They behave like regular die faces.That 6 statement is just to provide element to the puzzle.)
(By land I mean on top)
What is the probability of getting 4 on this die?
• will the roll over probability for 5, 1 and 4 only occur during the first time the die lands, or does it always happen whenever that side moves to the top? i.e. if it lands on 5 which rolls over to 1, will it continue rolling over to another adjacent face? – votbear Dec 5 '17 at 3:43
• What about 2 and 3? What makes them different from 6? – Deusovi Dec 5 '17 at 3:43
• (it's actually kind of amusing to imagine a die repeatedly flopping back and forth between landing on 1 and 5) – votbear Dec 5 '17 at 3:45
• Any face on a die has 4 adjacent faces. If 4 lands, I assume the 1/4 probability of it rolling over is to any of its 4 neighbours (with equal chance)? – Xenocacia Dec 5 '17 at 3:48
• If it lands on say 5, and rolls over to 1, is that then the end? Or will it roll over again since it's on 1? – Lolgast Dec 5 '17 at 6:54
A slightly differently explained answer from Jesse's, assuming the dice can only roll over once:
base Roll over from 1 Roll over from 4 Roll over from 5 Total
P(1) = 1/6*0 +0 +1/4*1/4*1/6 +1/4*2/4*1/6 =( 0+0+1+2)/96
P(2) = 1/6 +1/4*1/6 +1/4*1/4*1/6 +0 =(16+4+1+0)/96
P(3) = 1/6 +1/4*1/6 +0 +1/4*2/4*1/6 =(16+4+0+2)/96
P(4) = 1/6*3/4 +1/4*1/6 +0 +1/4*2/4*1/6 =(12+4+0+2)/96
P(5) = 1/6*1/2 +1/4*1/6 +1/4*1/4*1/6 +0 =( 8+4+1+0)/96
P(6) = 1/6 +0 +1/4*1/4*1/6 +1/4*2/4*1/6 =(16+0+1+2)/96
Giving total probabilities:
- 1: 3/96
- 2: 21/96
- 3: 22/96
- 4: 18/96
- 5: 13/96
- 6: 19/96
• I think you miss something, when 4 lands up (1/6) we got : 1/6 * 1/4 for the 4 stays on top, and 1/6 * 3/4 that another number roll, so if 4 land's up 1,2,5 and 6 can land's up after the last roll, so we got 1/6 * 3/4 *1/4 for the others numbers – Skyvask Dec 5 '17 at 9:11
• @Skyvask From OP: "If 4 lands, it has 1/4 probability of rolling over to the adjacent face" – Lolgast Dec 5 '17 at 10:08
• Ok i'm confused, just need to read more carrefuly next time – Skyvask Dec 5 '17 at 10:09
I used a probability tree to figure this out:
First you roll the die and it lands with equal probability on 1 of 6 sides. For each side it lands on it has varying probabilities to end up on one of 5 numbers (1-6 excluding whichever is on the opposite side ie. not adjacent. note: Opposite = 7 - Landed) so you have the given stick probability for the same number again and then an equal probability for all the 4 adjacent sides (*1/4) the given amount.
Once I did this for every number, go through and add all the last branches for each number to get the total probability that the die will end up on each number, I got:
• 1: 3/96
• 2: 21/96
• 3: 22/96
• 4: 18/96
• 5: 13/96 (edit: thanks for the find)
• 6: 19/96
Edit: i added them all to verify and it equals 100%
• You miss +2 on the 1 probability, it's 5/96 ( since 3/96 when 4 lands up, and 2/96 when 5 lands up). – Skyvask Dec 5 '17 at 8:09
• It's actually 1/96 from 4, so that one is correct. The probability for 5 is incorrect, it should be 13/96 – Lolgast Dec 5 '17 at 9:00
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2021-06-18 03:31:18
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https://math.stackexchange.com/questions/3162999/combinatorial-proof-for-sum-i-kn-2i-k-binomi-1k-12-k-binomnk
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# Combinatorial proof for $\sum_{i=k}^n (2i-k) \binom{i-1}{k-1}^2 = k \binom{n}{k}^2$
Give combinatorial proof for:
$$\sum_{i=k}^n (2i-k) \binom{i-1}{k-1}^2 = k \binom{n}{k}^2$$
RHS: We want to make sequence A and B with their element only 0 and 1, and the number of element 1 is k in A and B. From sequence A, pick one element 1, and that number colored by red.
LHS: Try to counting that sequence A and B with another way. I'm trying to consider element 1 k-th position. So this element can be in k,k+1,...,n-1,n in sequence
$$k\binom{n}k^2$$ counts the number of ways to select two subsets $$A$$ and $$B$$ of $$\{1,2,\dots,n\}$$, and then select a special element of $$A$$.
How many ways are there to do this so the maximum element of $$A\cup B$$ is $$i$$? There are three cases:
• If $$\max A=\max B$$, then the remaining $$k-1$$ elements of $$A$$ and $$B$$ can be freely chosen in $$\binom{i-1}{k-1}^2$$ ways.
• If $$\max A>\max B$$, then $$A$$'s other elements can be chosen in $$\binom{i-1}{k-1}$$ ways, and all of $$B$$'s elements can be chosen in $$\binom{i-1}k$$ ways, for a total of $$\binom{i-1}{k-1}\binom{i-1}k$$.
• If $$\max A<\max B$$, the number of ways to choose $$A$$ and $$B$$ is still $$\binom{i-1}{k-1}\binom{i-1}k$$.
Finally, we must multiply by $$k$$ to account for choosing the special element of $$A$$. The result is
$$k\left(\binom{i-1}{k-1}^2+2\binom{i-1}{k-1}\binom{i-1}k\right)$$ This simplifies to $$(2i-k)\binom{i-1}{k-1}^2$$.
• Nice answer (+1). – Markus Scheuer Mar 26 at 20:04
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2019-12-16 11:15:33
|
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https://www.x-mol.com/paper/1221575107905867776
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Machine Learning ( IF 2.809 ) Pub Date : 2020-01-24 , DOI: 10.1007/s10994-020-05868-6
Dimitris Bertsimas, Bart Van Parys
We present a novel method for sparse polynomial regression. We are interested in that degree r polynomial which depends on at most k inputs, counting at most $$\ell$$ monomial terms, and minimizes the sum of the squares of its prediction errors. Such highly structured sparse regression was denoted by Bach (Advances in neural information processing systems, pp 105–112, 2009) as sparse hierarchical regression in the context of kernel learning. Hierarchical sparse specification aligns well with modern big data settings where many inputs are not relevant for prediction purposes and the functional complexity of the regressor needs to be controlled as to avoid overfitting. We propose an efficient two-step approach to this hierarchical sparse regression problem. First, we discard irrelevant inputs using an extremely fast input ranking heuristic. Secondly, we take advantage of modern cutting plane methods for integer optimization to solve the remaining reduced hierarchical $$(k, \ell )$$-sparse problem exactly. The ability of our method to identify all k relevant inputs and all $$\ell$$ monomial terms is shown empirically to experience a phase transition. Crucially, the same transition also presents itself in our ability to reject all irrelevant features and monomials as well. In the regime where our method is statistically powerful, its computational complexity is interestingly on par with Lasso based heuristics. Hierarchical sparsity can retain the flexibility of general nonparametric methods such as nearest neighbors or regression trees (CART), without sacrificing much statistical power. The presented work hence fills a void in terms of a lack of powerful disciplined nonlinear sparse regression methods in high-dimensional settings. Our method is shown empirically to scale to regression problems with $$n\approx 10{,}000$$ observations for input dimension $$p\approx 1000$$.
down
wechat
bug
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2020-05-27 06:30:01
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5527060627937317, "perplexity": 774.5086260524777}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590347392141.7/warc/CC-MAIN-20200527044512-20200527074512-00477.warc.gz"}
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https://www.ncatlab.org/nlab/show/generalized+elliptic+operator
|
# Contents
## Idea
In the context of noncommutative geometry/KK-theory, a generalized elliptic operator is a linear operator of a Hilbert bimodule over two C*-algebras $A, B$ in C*Alg, such that in the special case that $A = C_0(Y)$ and $B = C_0(Y)$ for $X,Y$ smooth manifolds, this reduces to a $Y$-parameterized collection of ordinary elliptic operators on $X$.
Created on April 10, 2013 at 14:05:44. See the history of this page for a list of all contributions to it.
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2021-09-27 07:12:34
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 6, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6959089040756226, "perplexity": 334.3997353738024}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780058373.45/warc/CC-MAIN-20210927060117-20210927090117-00141.warc.gz"}
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http://stackoverflow.com/questions/2135339/using-latex-beamer-to-display-code
|
# Using LaTeX Beamer to display code
I'm using the following LaTeX code in a Beamer presentation:
\begin{frame}
\begin{figure}
\centering
\tiny
\lstset{language=python}
\lstinputlisting{code/get_extent.py}
\end{figure}
\end{frame}
Is it possible to select specific lines from my get_extent.py file rather than displaying it all?
-
This has nothing to do with beamer; it's about a listings feature. Its excellent manual has more. For example:
\lstinputlisting[firstline=2,lastline=5]{code/get_extent.py}
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2014-03-17 05:27:12
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6316019296646118, "perplexity": 5295.976932341087}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-10/segments/1394678704872/warc/CC-MAIN-20140313024504-00001-ip-10-183-142-35.ec2.internal.warc.gz"}
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https://eprint.iacr.org/2021/020
|
Catching the Fastest Boomerangs - Application to SKINNY
Stéphanie Delaune, Patrick Derbez, and Mathieu Vavrille
Abstract
In this paper we describe a new tool to search for boomerang distinguishers. One limitation of the MILP model of Liu et al. is that it handles only one round for the middle part while Song et al. have shown that dependencies could affect much more rounds, for instance up to 6 rounds for SKINNY. Thus we describe a new approach to turn an MILP model to search for truncated characteristics into an MILP model to search for truncated boomerang characteristics automatically handling the middle rounds. We then show a new CP model to search for the best possible instantiations to identify good boomerang distinguishers. Finally we systematized the method initiated by Song et al. to precisely compute the probability of a boomerang. As a result, we found many new boomerang distinguishers up to 24 rounds in the TK3 model. In particular, we improved by a factor $2^{30}$ the probability of the best known distinguisher against 18-round SKINNY-128/256.
Metadata
Available format(s)
Category
Secret-key cryptography
Publication info
Published elsewhere. IACR-TOSC ISSUE 4-2020
DOI
10.46586/tosc.v2020.i4.104-129
Keywords
BoomerangMILP modelSKINNY
Contact author(s)
patrick derbez @ irisa fr
History
2021-01-06: received
Short URL
https://ia.cr/2021/020
License
CC BY
BibTeX
@misc{cryptoeprint:2021/020,
author = {Stéphanie Delaune and Patrick Derbez and Mathieu Vavrille},
title = {Catching the Fastest Boomerangs - Application to SKINNY},
howpublished = {Cryptology ePrint Archive, Paper 2021/020},
year = {2021},
doi = {10.46586/tosc.v2020.i4.104-129},
note = {\url{https://eprint.iacr.org/2021/020}},
url = {https://eprint.iacr.org/2021/020}
}
Note: In order to protect the privacy of readers, eprint.iacr.org does not use cookies or embedded third party content.
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2023-03-29 00:41:31
|
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|
http://stackoverflow.com/questions/18763849/return-java-system-exit-value-to-bash-script?answertab=votes
|
# Return Java system exit value to bash script
I am trying to get the return value from a java program ( System.exit(1);) into a shell script, but it seems like its returning the jvm exit code, which is always 0, if it doesnt crash. For testing purposes, this is the very first line in my main(). Anyone know how to do this?
My bash code:
....................
java bsc/cdisc/ImportData $p$e $t #----------------------------------------- # CATCH THE VALUE OF${?} IN VARIABLE 'STATUS'
# STATUS="${?}" # --------------------------------------- STATUS="${?}"
cd ../scripts
echo "${STATUS}" Thanks - Interesting, since that should just work. Any more details? Also, why the curly braces? – Anders R. Bystrup Sep 12 '13 at 12:20 ## 2 Answers If your script has only the two lines then you are not checking for the correct exit code. I am guessing you are doing something like: $ java YourJavaBinary
$./script where script contains only: STATUS="${?}"
echo "${STATUS}" Here, the script is executed in a subshell. So when you execute the script, $? is the value of last command in that shell which is nothing in the subshell. Hence, it always returns 0.
What you probably wanted to do is to call the java binary in your script itself.
java YourJavaBinary
STATUS="${?}" echo "${STATUS}"
Or simply check the exit code directly without using the script:
$java YourJavaBinary ; echo$?
-
Sorry, I am calling the java binary within my code, and no thats not all that's in the script. I just put up the error reporting part. – mike628 Sep 12 '13 at 12:39
So you have java YourJavaBinary ; STATUS="${?}" ; echo "${STATUS}" in the script and still it doesn't give the correct exit code in your script? If that's case, then it's strange. May be, can you show the whole part relating to the call to java and exit code checking and Can you produce the same issue with a simple java program that does only System.exit(1); ? – Blue Moon Sep 12 '13 at 12:40
No, I get a "0" – mike628 Sep 12 '13 at 12:43
Just to reiterate it, you don't have any statements (excluding comments) between java ... and STATUS="${?}" right? :) Can you reproduce the problem I described in my previous comment? – Blue Moon Sep 12 '13 at 13:06 You should do like this: Test.java: public class Test{ public static void main (String[] args){ System.exit(2); } } test.sh #!/bin/bash java Test STATUS=$?
echo \$STATUS
-
Yes! You're right. I had the solution in my pocket and I used it, but I don't know why a understood wrong. Should I vote to delete my answer? – Danilo Muñoz Sep 12 '13 at 12:22
Now it's corrected :-) – Danilo Muñoz Sep 12 '13 at 12:26
Danilo, Thats exactly what I have.( and had) . I didnt post the entire script, because most is irrelevent. – mike628 Sep 12 '13 at 12:40
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2015-01-27 14:42:51
|
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|
http://tex.stackexchange.com/questions/165384/to-make-one-environment-flow-into-another-without-empty-space
|
# To make one environment flow into another without empty space
I have this design at the moment.
However, when you have many of those, they distract much.
I propose to have Answer just on the same line with Question like
Question 41 Lorem ipsun. Answer 38 Lorem again.
My tex
\documentclass{article}
\usepackage[framemethod=tikz]{mdframed}
\newtheorem{question}{Question}
\let\masiquestion\question
\AtBeginDocument{%
\let\umquestion\question
\let\question\masiquestion
}
\mdfdefinestyle{ans}{
linecolor=cyan,
backgroundcolor=yellow!20
}
\usepackage{environ}
\usepackage{ifthen}
{%
{%
\noindent
\begin{minipage}[t]{\linewidth}
\BODY
\end{minipage}% here put what the command has to do when outside
}%
}%
\begin{document}
\begin{question}
Why is the pressure same in Arteries and aorta?
\end{question}
Because they do not coil (only arch of aorta), so the Frank-Starling equation does not hold but Laplace law instead.
Laplace law is the reason why you can have the same pressure in aorta (big) and arteries (small), because the pressure depends inversely on the pressure.
\end{document}
I am thinking what would be the right way of changing these Question and Answer environments. I do not want to remove them because they offer me flexibility. I just want to remove the empty line and put the answer flowing just flowing after the question.
How can you remove the space between the two environments?
-
question being in the color box too? – percusse Mar 13 at 20:38
The \tcolorbox package might be worth to look at. – Christian Hupfer Mar 13 at 20:40
@percusses color box is not necessary. It can be easily disabled. The easy way to read is most important without distracting the reader much. – Masi Mar 13 at 20:45
I would recomend using the frametitle of the mdframed environment to show the question and the body to show the answer:
## Notes:
• You can probably select better colors.
• Also need to add the numbering if that is required.
## Code:
\documentclass{article}
\usepackage[framemethod=tikz]{mdframed}
\mdfdefinestyle{ans}{
linecolor=cyan,
backgroundcolor=yellow!20,
frametitlebackgroundcolor=green!40,
frametitlerule=true,
}
\newenvironment{question}[1]{%
\begin{mdframed}[style=ans,frametitle={Question: #1}]
}{%
\end{mdframed}%
}%
\begin{document}
\begin{question}{Why is the pressure same in Arteries and aorta?}
Because they do not coil (only arch of aorta), so the Frank-Starling equation does not hold but Laplace law instead.
Laplace law is the reason why you can have the same pressure in aorta (big) and arteries (small), because the pressure depends inversely on the pressure.
\end{question}
\begin{question}{Why is the pressure same in Arteries and aorta?}
Because they do not coil (only arch of aorta), so the Frank-Starling equation does not hold but Laplace law instead.
Laplace law is the reason why you can have the same pressure in aorta (big) and arteries (small), because the pressure depends inversely on the pressure.
\end{question}
\end{document}
-
I had conflicting packages - problem solved. Thank you for your answer! – Masi Mar 14 at 6:35
With tcolorbox:
\documentclass{article}
\usepackage[most]{tcolorbox}
\newtcolorbox[auto counter,number within=section]{question}[2][]{%
colback=magenta!15!white,colframe=blue!40!green,fonttitle=\bfseries,
title=Question.~\thetcbcounter: #2,#1}
\begin{document}
\section{Some section}
\begin{question}{Why is the pressure same in Arteries and aorta?}
Because they do not coil (only arch of aorta), so the Frank-Starling equation does not hold but Laplace law instead.
Laplace law is the reason why you can have the same pressure in aorta (big) and arteries (small), because the pressure depends inversely on the pressure.
\end{question}
\begin{question}{Why is the pressure same in Arteries and aorta?}
Because they do not coil (only arch of aorta), so the Frank-Starling equation does not hold but Laplace law instead.
Laplace law is the reason why you can have the same pressure in aorta (big) and arteries (small), because the pressure depends inversely on the pressure.
\end{question}
\end{document}
-
\documentclass{article}
\usepackage[framemethod=tikz]{mdframed}
\newtheorem{question}{Question}
\mdfdefinestyle{ans}{
linecolor=cyan,
backgroundcolor=yellow!20
}
\makeatletter
\def\endquestion{\ifhmode\unskip\fi\begingroup\let\par\relax}
\makeatother
|
2014-10-24 07:05:33
|
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|
https://www.physicsforums.com/threads/product-proof.226051/
|
# Product Proof
1. Apr 2, 2008
### gop
1. The problem statement, all variables and given/known data
Proof that for n>2 and n is a natural number it holds that
$$\prod_{k=1}^{n}\frac{k^{2}+2}{k^{2}+1}<3$$
and
$$\prod_{k=1}^{n}\frac{k^{2}+2}{k^{2}+1}<\frac{3n}{n+1}$$
2. Relevant equations
3. The attempt at a solution
My best approach was to split the product over the fraction and then to arrive at a statement that looks like
$$\prod_{k=2}^{n}k^{2}+2<\prod_{k=1}^{n}k^{2}+1$$
I then tried to prove by induction that this statement holds but that doesn't really work. The best result I got (for n+1) is then
$$(\prod_{k=2}^{n}k^{2}+2)<(\prod_{k=1}^{n}k^{2}+1)\cdot\frac{n^{2}+2n+2}{n^{2}+2n+3}$$
But I can't do anything usefuel with that...
2. Apr 2, 2008
### Avodyne
You could try writing the product as the exponential of a sum, and then bounding the sum by an integral.
|
2017-10-19 01:58:00
|
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|
https://www.transtutors.com/questions/a-manager-is-deciding-whether-or-not-to-build-a-small-facility-demand-is-uncertain-a-3180201.htm
|
# A manager is deciding whether or not to build a small facility. Demand is uncertain and can be... 1 answer below »
1. A manager is deciding whether or not to build a small facility. Demand is uncertain and can be either at a high or low level. If the manager chooses a small facility and demand is low, the payoff is $300. If the manager chooses a small facility and demand is high, the payoff is$100. On the other hand, if the manager chooses a large facility and demand is low, the payoff is–$200, but if demand is high, the payoff is$800.
Shalini
The pay off matrix can be shown as below:
States of Nature (Demand) Strategy 1 Large facility Strategy 2 Small Facility High $800$100 Low -$200$300
Ans a
Maxi max criterion involves taking the maximum value for each column (strategy) and then taking the maximum of these column maximums. The corresponding strategy is chosen as the best.
## Plagiarism Checker
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|
2020-06-01 23:03:53
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http://poincare101.herokuapp.com/post/12
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Here it is:
If $x, y, z \text{ and } n$ are all natural numbers with $n>1$ and $x^n + y^n = z^n$, prove that $x, y, z > n$.
So, how do we go about this? $x^n + y^n = z^n$ seems like a pretty difficult equation to mess about with (Fermat's Last Theorem, anyone?), but, we might gain some insight if we move some stuff around, plug in a few numbers, etc.
But, we should try to find out some more information by just observing the problem; it might be useful.
Obviously, $z > x,y$ (from the equation given). And, we can also say, without losing generality, $y \ge x$, since the equation is symmetric about these two terms (i.e. we can interchange $x$ and $y$, and it won't make any difference whatsoever).
Let's try to separate out one of the variables, and see what we can do.
$x^n = z^n-y^n$
If you remember some factorization tools:
$x^n = (z-y)(z^{n-1} + z^{n-2}\cdot y + z^{n-3} \cdot y^2 + ... + z^{1}\cdot y^{n-2} + y ^ {n-1})$
Since $z > y$ is the same as $z > y+1$ since $z, y$ are natural numbers:
$x^n > ((y+1)-y)(z^{n-1} + z^{n-2}\cdot y + z^{n-3} \cdot y^2 + ... + z^{1}\cdot y^{n-2} + y ^ {n-1})$
Therefore,
$x^n > ((y+1)-y)(nx^{n-1}) = nx^{n-1}$
So, finally,
$x > n$
And, since the equation is symmetric for $x$ and $y$, we get $y > n$, and we can use very similar methods to prove $z > n$.
So, what are some important moves here?
First of all, noticing the symmetric nature of the equation helped us get down some initial observations and assumptions, which made our lives much easier.
Factoring the $z^n - y^n$ allowed us to tackle it much more effectively.
Then, going from an equation to an inequality by replacing $z$ with $y+1$ and simplifying the expansion to having terms of $x$ allowed us to complete the proof.
But, the biggest move was just separating out the $x$ on the left side so that we would be able to use methods on the right side.
This problem would probably be considered as part of number theory, which has a host of very interesting and easy to understand problems that don't require much knowledge to solve, but, require quite a measure of creativity.
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2015-08-01 18:07:44
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https://www.taylorfrancis.com/books/e/9780429136177/chapters/10.1201/9781420028249-13
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chapter 8
8 Pages
## Weakly Krull Inside Factorial Domains
ByDaniel D. Anderson, Muhammed Zafrullah, Gyu Whan Chang
Chapman, Halter-Koch, and Krause [8] introduced the notion of an inside factorial monoid and integral domain. Throughout we will confine ourselves to the integral domain case, but the interested reader may easily supply definitions and proofs for the monoid case. An integral domain D is inside factorial if there exists a divisor homomorphism ϕ: F → D∗ = D − {0} where F is a factorial monoid and for each x ∈ D∗, there exists an n ≥ 1 with xn ∈ ϕ(F ). They showed [8, Proposition 4] that an inside factorial domain may be defined in terms of a Cale basis and it will be this characterization that we use for the definition of an inside factorial domain. A subset Q ⊆ D∗ is a Cale basis for D if 〈Q〉 = {uqα1 · · · qαn | u ∈ U(D), qαi ∈ Q} is a factorial monoid with primes Q and for each d ∈ D∗ there exists an n ≥ 1 with dn ∈ 〈Q〉. Here U(D) is the group of units of D. A domain D is inside factorial if and only if D has a Cale basis. They showed [8, Theorem 4] that D is inside factorial if and only if D, the integral closure of D, is a generalized Krull domain with torsion t-class group Clt(D), for each P ∈ X(1)(D), the valuation domain DP has value group order-isomorphic to a subgroup of (Q,+) (we say that D is rational), and D ⊆ D is a root extension (i.e., for each x ∈ D, there exists an n ≥ 1 with xn ∈ D).
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2020-02-26 01:58:47
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https://www.physicsforums.com/threads/schwarzschild-radius-explanation.712371/
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Hello,
First off, I'm not a physics student by any means, but it does intrigue me. I've been wondering about the Schwarzschild radius, and how the density of an object dictates the escape speed. Why is it, that the more you compress an object, the more gravity it "owns?" If you were able to theoretically compress the earth down to 9.00mm, you'd have reached it's Schwarzschild radius, and essentially would have created a black hole... but why? I understand why that happens when stars collapse upon themselves due to mass, but I don't understand how taking a fixed amount of mass and compressing it, increases the amount of gravity to that object.
A "simple" explanation would be much appreciated. Again, I am not a physicist, and I only understand the basics.
Thanks
-Sean
phinds
Gold Member
2021 Award
It doesn't increase the amount of gravity as felt by an object outside the Schwarzschild radius, but it warps space-time so much that objects inside that radius cannot escape. Think of it somewhat not as the AMOUNT of gravity but the DENSITY of gravity. Does that help?
mathman
The simplest explanation (not using General Relativity) is that the force of gravity is proportional to m/r2, where m is the mass of the earth (the attracting body) and r is its radius. Shrinking the radius down to 9 mm makes the force that much larger in comparison to that of the earth in its present size.
The gravitational field strength drops like 1/r^2 as you get further away, right? Imagine going closer to the mass, then the strength will keep increasing: get twice as close and the field is four times as strong. So what happens when you get too close?
In most cases, this trend is halted when you hit the surface of the object, because once you start digging to get even closer, some of the mass starts being above you and the effective mass is reduced. Once you pass the surface of the object (assuming homogeneous density), the 1/r^2 relationship no longer applies, and it becomes now a linear relationship with r. This is the key:
The gravitational field strength attains a maximum at this surface of the object.
But if you concentrate the same amount of mass into a smaller volume, you can keep going closer to the centre without hitting the surface, and the 1/r^2 increase in gravitational field strength continues. At the Schwarzschild radius (distance from the centre of mass), if you still haven't hit the surface, then the field strength at this point is so high you have the event horizon, and the result is a black hole.
It's essentially all down to this 1/r^2 relationship, which comes from the surface area of a sphere: the "flux" of gravity through a spherical surface centred on a mass is the same for all radii greater than the radius of the mass, and the surface area of a sphere is proportional to the square of the radius. So if you halve the radius of a sphere, you quarter its surface area, and to conserve gravitational flux, the gravitational field strength must be four times bigger. So that's where the 1/r^2 term comes from.
Ahh ok, so using Earth as an example, does this mean that gravity's force is stronger as I get closer to the core? As in, if an object weighs 100lbs on the earth's crust, it weighs more in the mantle, etc?
-Sean
Nugatory
Mentor
Ahh ok, so using Earth as an example, does this mean that gravity's force is stronger as I get closer to the core? As in, if an object weighs 100lbs on the earth's crust, it weighs more in the mantle, etc?
-Sean
No - that's in the second paragraph of mikeph's post #4 above.
"The gravitational field strength attains a maximum at this surface of the object. "
Assuming you can compress the object, you can go further, without touching the surface of the object, correct?
Hello,
First off, I'm not a physics student by any means, but it does intrigue me. I've been wondering about the Schwarzschild radius, and how the density of an object dictates the escape speed. Why is it, that the more you compress an object, the more gravity it "owns?" If you were able to theoretically compress the earth down to 9.00mm, you'd have reached it's Schwarzschild radius, and essentially would have created a black hole... but why? I understand why that happens when stars collapse upon themselves due to mass, but I don't understand how taking a fixed amount of mass and compressing it, increases the amount of gravity to that object.
A "simple" explanation would be much appreciated. Again, I am not a physicist, and I only understand the basics.
Thanks
-Sean
A very crude example. I think this is how it works. The weight of the entire Earth occupying only 9mm of space would cause space to warp so much that it would cause a blackhole. Just the same as say a marble compressed to the size of an atom would cause a blackhole (example)
Also to note that the gravity remains proportional, as in the gravity at the centre of the Earth relative to the surface would stay proportional even if the Earths radius was 9mm
I'm pretty sure that anything with mass can cause a blackhole should you make it dense enough. Earth with a radius of 9mm would warp space the same as the Sun if it were the size of Earth. Both would create a blackhole (for example). The key to creating blackholes is if you take any object and increase is density, it will eventually become too dense and warp space so much that nothing can get out of the "dip"
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stevendaryl
Staff Emeritus
Hello,
First off, I'm not a physics student by any means, but it does intrigue me. I've been wondering about the Schwarzschild radius, and how the density of an object dictates the escape speed. Why is it, that the more you compress an object, the more gravity it "owns?" If you were able to theoretically compress the earth down to 9.00mm, you'd have reached it's Schwarzschild radius, and essentially would have created a black hole... but why? I understand why that happens when stars collapse upon themselves due to mass, but I don't understand how taking a fixed amount of mass and compressing it, increases the amount of gravity to that object.
A "simple" explanation would be much appreciated. Again, I am not a physicist, and I only understand the basics.
Thanks
-Sean
Actually, before Einstein developed his theory of General Relativity there was an argument that something weird would happen near a very massive star.
If you shoot a rocket straight up in the air, it will return to the Earth unless it's going fast enough to escape the Earth's gravity. To escape, the rocket must have a velocity greater than or equal to what's called the "escape velocity" v given by:
$v_{escape} = \sqrt{(2GM/r)}$
where M is the mass of the Earth, G is Newton's gravitational constant, and r is the radius of the Earth. The bigger M is, the bigger the escape velocity must be. But also, the smaller r is, the bigger the escape velocity must be. So based on this formula, there would seem to be a special radius $R$ such that even light can't escape:
If $R < 2GM/c^2$, where $c$ is the speed of light, then $v_{escape}> c$ and so not even light could escape.
This analysis (and I don't remember who first made it, but it was a long time ago) is based on an old theory of gravity, Newton's theory of gravity. Newton's theory is known to be wrong, only an approximation to the current best guess as to how gravity works, which is Einstein's General Relativity.
However, even though the argument uses Newton's theory, by one of those strange quirks, it turns out that if you use Einstein's theory instead, you get exactly the same answer: If the mass of a star or planet is squeezed down to a radius $R$ where $R < 2GM/c^2$, then it becomes a "black hole" where nothing, not even light, can escape.
This analysis (and I don't remember who first made it, but it was a long time ago) is based on an old theory of gravity, Newton's theory of gravity. Newton's theory is known to be wrong, only an approximation to the current best guess as to how gravity works, which is Einstein's General Relativity.
Michell-Laplace dark star. Note the crucial difference though: for dark star light can leave its surface much like a ball can leave earth's surface -- but it will fall back down due to the strong gravity. In GR, light cannot ever leave the horizon.
Michell-Laplace dark star. Note the crucial difference though: for dark star light can leave its surface much like a ball can leave earth's surface -- but it will fall back down due to the strong gravity.
This is an important difference. Light traveling radially does not have 'escape velocity' to reach infinity (i.e. never to return) given the dark star's mass, but can be 'observed' by someone outside the event horizon, and even provided additional momentum to attain escape velocity.
In GR, light cannot ever leave the horizon.
Interesting to note that gravity itself (a form of energy with a finite velocity of c) is able to escape a black hole (leave the event horizon) and be observed/measured from outside.
A.T.
Interesting to note that gravity itself (a form of energy with a finite velocity of c) is able to escape a black hole (leave the event horizon) and be observed/measured from outside.
Gravitational waves / disturbances have a finite velocity of c, and they cannot leave the event horizon.
Gravity itself having a velocity is a tricky concept, because gravity itself is not propagating, just its disturbances are. For example, an inertially moving mass pulls you towards its current position, not towards a retarded position due to finite signal speed.
Nugatory
Mentor
Interesting to note that gravity itself (a form of energy with a finite velocity of c) is able to escape a black hole (leave the event horizon) and be observed/measured from outside.
It doesn't have to escape because it wasn't inside to begin with. The body that collapsed to form the black hole had a gravitational field above its surface even before that surface fell through the Scwarzchild radius and the horizon appeared.
jambaugh
Gold Member
You can only go so far with "stretching space" analogies because space-time is not Euclidean but has a "hyperbolic" signature. Picturing fingers poking rubber sheets is just the wrong analog.
But think of it this way... even when an object is sitting still, it is moving through time so when we combine space and time considering 4-dim space-time it turns out that all objects are moving at the same "rate" of 1 second per second and their velocity is just the direction of their motion which ends up being what they see as their time direction.
You see an object starting to fall towards a mass and what is happening is that its time axis is getting turned toward the mass. You and I being farther away and unaffected by gravity see the objects time motion turn into spatial motion... it was sitting there ticking away and now it is falling toward the planet.
All futures for the object are a bit closer to the mass than if that mass wasn't there. That is true even if those futures involve motion away from the mass as for an object moving away from the mass. By this I mean if your moving away from the mass your future will be farther away but not so far as if the mass hadn't been there.
So at the Schwarzschild radius of a Black Hole what has happened is that the gravity has turned time so severely that at that point all futures lead explicitly closer to the mass. To get back out one would have to build the equivalent of a time machine/ FTL spacecraft (each equates to the other in relativity).
If you understand what a light cone is, the future light cones of all events near a mass get turned toward the mass until at the Sch.Rad. all future light-cones point inward.
The analog of a stretched sheet is not sufficient to understand gravitational forces one should rather imagine a twisting effect due to mass on the space vs time directions of space-time. This is harder (gross understatement) to visualize as it involves non-Euclidean geometry. You have to parse it logically and ultimately you have to "do the math" to get a deep understanding.
A.T.
You can only go so far with "stretching space" analogies because space-time is not Euclidean but has a "hyperbolic" signature. Picturing fingers poking rubber sheets is just the wrong analog.
Problem of the rubber sheet is lack of time dimension. As for the Euclidean/pseudo-Euclidean part, you can use space-propertime diagrams where coordinate time is the Euclidean path distance. But still you have to include the propertime as one coordinate of the sheet, to show gravity:
You see an object starting to fall towards a mass and what is happening is that its time axis is getting turned toward the mass. You and I being farther away and unaffected by gravity see the objects time motion turn into spatial motion... it was sitting there ticking away and now it is falling toward the planet.
You can also interpret it as distances between propertime coordiantes getting larger towards the mass (gravitational time dilation):
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Gravitational waves / disturbances have a finite velocity of c, and they cannot leave the event horizon.
Gravity itself having a velocity is a tricky concept, because gravity itself is not propagating, just its disturbances are. For example, an inertially moving mass pulls you towards its current position, not towards a retarded position due to finite signal speed.
I find this distinction hard to understand the way you word it. Since the parallelism with light is being considered (dark stars vs. black holes), if you were to say "disturbances of the EM field have a finite velocity but light itself is not propagating, just the disturbances are, it would sound odd, wouldn't it?
Also the "current versus retarded position" example obviates the difficulties that these terms have to be defined for spacelike separated points in the presence of the relativity of simultaneity, and if a gravitational field pulls you it is hard to maintain that this field is not propagating.
It doesn't have to escape because it wasn't inside to begin with. The body that collapsed to form the black hole had a gravitational field above its surface even before that surface fell through the Scwarzchild radius and the horizon appeared.
Again, I don't know how this is relevant to the analogy with light, how is this different from the EM fields and the light of the body that collapsed?
stevendaryl
Staff Emeritus
I find this distinction hard to understand the way you word it. Since the parallelism with light is being considered (dark stars vs. black holes), if you were to say "disturbances of the EM field have a finite velocity but light itself is not propagating, just the disturbances are, it would sound odd, wouldn't it?
The closer analogy with electromagnetism would be: The electric field is not propagating, only disturbances in it. Electromagnetic waves are, as the name indicates, waves in the electromagnetic field. But there are such things as static electric fields. For example, the electric field of a stationary charged particle. The electric field in that case isn't "propagating" from the particle to wherever it is measured. The electric field is everywhere in space, and it's static.
In the same way that a black hole can have a gravitational field outside the event horizon, even though it is impossible for gravity waves to propagate from inside the black hole to outside, a charged black hole can have an electric field, even though it is impossible for electromagnetic waves to propagate from inside to outside.
Fields are properties of space (or spacetime). It is only disturbances to fields that propagate, not the fields themselves.
A.T.
if you were to say "disturbances of the EM field have a finite velocity but light itself is not propagating, just the disturbances are
Light is a disturbance of the EM field, so light is propagating at a finite speed. But is a static E-field propagating?
Again, I don't know how this is relevant to the analogy with light,
You mean the retarded vs current position thing? If you have a charged object moving inertially, and you carry the opposite charge, then the Coulomb force will pull you towards its current position, while you still see it at the retarded position.
how is this different from the EM fields and the light of the body that collapsed?
If the body was charged, the E-field outside the horizon remains, just like the G-field. That doesn't require anything propagating from below the horizon. Neither EM-waves nor G-waves can leave the horizon.
.... but is a static E-field propagating?
Part of my questions lead to this question, yes. Is it safe to assume E-fields or gravitational fields don't propagate? Is that assumption just valid as a useful approximation, given the fact we know the universe is not static?
You mean the retarded vs current position thing?.......
No, that was meant for Nugatory's quoted comment.
stevendaryl
Staff Emeritus
Part of my questions lead to this question, yes. Is it safe to assume E-fields or gravitational fields don't propagate? Is that assumption just valid as a useful approximation, given the fact we know the universe is not static?
I don't think that changes anything: A charged black hole has an electric field. The field far from the black hole does not involve anything propagating from the black hole to the distant observer. The same is true for gravity.
A.T.
Is it safe to assume E-fields or gravitational fields don't propagate?
I guess it depends on what "propagate" means. My point in the previous posts was that you have to distinguish between the field and disturbances of the field, when it comes to propagation.
The closer analogy with electromagnetism would be: The electric field is not propagating, only disturbances in it. Electromagnetic waves are, as the name indicates, waves in the electromagnetic field. But there are such things as static electric fields. For example, the electric field of a stationary charged particle. The electric field in that case isn't "propagating" from the particle to wherever it is measured. The electric field is everywhere in space, and it's static.
In the same way that a black hole can have a gravitational field outside the event horizon, even though it is impossible for gravity waves to propagate from inside the black hole to outside, a charged black hole can have an electric field, even though it is impossible for electromagnetic waves to propagate from inside to outside.
All this is fine for sure, but see my questions in the post above.
Fields are properties of space (or spacetime). It is only disturbances to fields that propagate, not the fields themselves.
This is clear and understood as a general definition for undergrad physics, but I fear that it gives a somewhat simplistic account of the physics in GR and QFT.
Once you work with quantum fields the distinction wave-field is not so clear cut, and the same thing happens in GR, originating some confusion and debate even amongst quantum gravity experts regarding fields that include time(spacetime fields), background independence, the hole argument and all that.
Anyway, the term disturbances would need a clear definition here, if it is simply a way to refer to changes of the field in time, spacetime fields carry them within themselves.
WannabeNewton
Fields definitely propagate on space-time.
stevendaryl
Staff Emeritus
Fields definitely propagate on space-time.
I think maybe we're arguing about semantics. I would say that CHANGES to fields propagate. I don't know what it would mean for the fields themselves to propagate.
A toy model of a field might be a string stretched between two trees. It doesn't really make sense to say that the string propagates from one tree to the other. But if you tap the string near one tree, the tap will cause a vibration that will propagate to the other tree.
WannabeNewton
The term "propagation of such and such field" is a standard one. One says, for example, that the massive Klein Gordon field ##\varphi## propagates in a background Minkowski space-time ##(\mathbb{R}^{4},\eta_{ab})## according to ##\partial^a \partial_a \varphi - m^2 \varphi = 0## i.e. ##\varphi## evolves in ##(\mathbb{R}^{4},\eta_{ab})## according to the aforementioned evolution equations.
stevendaryl
Staff Emeritus
The term "propagation of such and such field" is a standard one. One says, for example, that the massive Klein Gordon field ##\varphi## propagates in a background Minkowski space-time ##(\mathbb{R}^{4},\eta_{ab})## according to ##\partial^a \partial_a \varphi - m^2 \varphi = 0## i.e. ##\varphi## evolves in ##(\mathbb{R}^{4},\eta_{ab})## according to the aforementioned evolution equations.
"Evolution" and "propagation" are both about change. Static fields don't propagate.
WannabeNewton
Clearly but you were saying this for any field whatsoever as per the quote in post #22 and the statement in that quote is incorrect.
stevendaryl
Staff Emeritus
Clearly but you were saying this for any field whatsoever as per the quote in post #22.
You're arguing about semantics, without really paying attention to the context. The original question was the old conundrum: If nothing can propagate from inside a black hole to outside, then how does gravity itself escape? (Same question for the electric field, in the case of a charged black hole.)
The notion of "propagate" in this conundrum is not "obeys a differential equation", but spreading out from one point to other points. The gravitational field DOESN'T propagate from inside the black hole to outside the black hole, in this sense of the word "propagate". Neither does the electric field.
The notion of "propagate" that is appropriate for talking about whether an effect inside a black hole can propagate outside is in terms of disturbances. If I make a tiny change to the distribution of mass inside a black hole (or distribution of charge), is that change visible from outside? The answer is "no".
PeterDonis
Mentor
The gravitational field DOESN'T propagate from inside the black hole to outside the black hole, in this sense of the word "propagate". Neither does the electric field.
No, but there is a sense in which those fields "propagate" from the collapsing object that originally formed the hole. The law that governs the propagation is the vacuum Einstein Field Equation (coupled with the source-free Maxwell Equations if charge is present). For the idealized case of spherically symmetric collapse, the field in the entire vacuum region to the future of the collapsing object is completely determined by these laws of propagation from the source.
stevendaryl
Staff Emeritus
No, but there is a sense in which those fields "propagate" from the collapsing object that originally formed the hole. The law that governs the propagation is the vacuum Einstein Field Equation (coupled with the source-free Maxwell Equations if charge is present). For the idealized case of spherically symmetric collapse, the field in the entire vacuum region to the future of the collapsing object is completely determined by these laws of propagation from the source.
Sure. But I think that the false intuition that is behind the question about how gravity (or the electric field) gets out of a black hole is something like this: We feel gravity because something (gravitons, or whatever) is emitted by the gravitational source and travels across space to hit us. We measure an electric field because something (photons) are emitted by charged particles and travel across space to hit us. So if we could only block those carrier particles, we could eliminate the gravitational effects of the mass, or we could eliminate the electric field due to a charge particle. But that's an incorrect intuition. Nothing has to travel from the source to us in order for us to detect gravity or an electric field.
PeterDonis
Mentor
I think that the false intuition that is behind the question about how gravity (or the electric field) gets out of a black hole is something like this: We feel gravity because something (gravitons, or whatever) is emitted by the gravitational source and travels across space to hit us. We measure an electric field because something (photons) are emitted by charged particles and travel across space to hit us.
Well, on one view (the quantum field theory view of forces being mediated by virtual particles), this intuition is correct as far as it goes. We measure an electric field because of an exchange of virtual photons between the source of the field and our detector. We measure a gravitational field because of the exchange of virtual gravitons between the gravitating mass and our detector.
Where the intuition goes wrong, on this view, is thinking that the virtual particle exchange is limited to the speed of light. It isn't, which is why, on this view, it's perfectly possible for virtual gravitons to get out of a black hole and make us feel gravity from it.
The virtual particle picture certainly has limitations, one of which is that it's based on perturbation theory, and in the case of gravity, perturbation theory implies that the metric of spacetime can't be too different from a flat metric. I'm not sure how well that applies to spacetime near a black hole's horizon (much less inside it). But I think pop science discussions of the virtual particle picture are often the source (sometimes vague and indirect) of people's intuition that particles have to get out of the black hole to make us feel gravity from it, so I think it's important to recognize that there is a sense in which that picture is valid, if limited.
Nothing has to travel from the source to us in order for us to detect gravity or an electric field.
I don't agree, because this statement, as it stands, is even stronger than the one I quoted above; it rules out, not just the virtual particle view, but also the view I have advocated in a number of threads on PF, that the source of the black hole's gravity is the object that originally collapsed to form the hole. The metric in the vacuum region to the future of that collapsing object *is* caused by the object, and one way to describe how that causation occurs, as I said in my previous post, is that the spacetime metric propagates from the source to where you feel the gravity, governed by the vacuum EFE as the law of propagation.
stevendaryl
Staff Emeritus
Well, on one view (the quantum field theory view of forces being mediated by virtual particles), this intuition is correct as far as it goes. We measure an electric field because of an exchange of virtual photons between the source of the field and our detector. We measure a gravitational field because of the exchange of virtual gravitons between the gravitating mass and our detector.
I think that's taking terms in Feynmann diagrams a little too literally. Virtual photons are calculational aids, they are terms in a Taylor series expansion. I don't think that they necessarily correspond to any real particles. Anyway, but that view, that forces are all due to the exchange of virtual photons or gravitons, is what is behind the intuition that it should be impossible for gravity to get out of a black hole.
Yeah, I know, you can say that virtual particles can escape in a way that real particles, but I think that's a little hoky. Virtual particles were developed for perturbation theory, which tries to understand interacting physics as small perturbations on free field theory. It's been a while since I've studied stuff like that, but I think that there are nonperturbative effects that can't be understood in terms of summing Feynmann diagrams. Bound states of particles I think is an example, and I would believe that black holes would be another example.
Where the intuition goes wrong, on this view, is thinking that the virtual particle exchange is limited to the speed of light. It isn't, which is why, on this view, it's perfectly possible for virtual gravitons to get out of a black hole and make us feel gravity from it.
I suppose that that's a way of viewing things, but I don't like it. As I said, I think that it's viewing terms of a Taylor series a little too literally.
The other reason I don't like that resolution in the case of black holes is that I feel like it misses out on the topological reasons that a charged star collapsing into a black hole can't possibly cause the electric field or gravitational field to disappear. Whatever is going on inside a collapsing star, we know that before collapse, the electric field has a certain value for the surface integral over a surface surrounding the star. According to Maxwell's equations, you can't change that value unless charge crosses the surface. So what goes on inside the surface is just irrelevant.
Now, that is a classical picture, which is certainly changed by quantum field theory, but the macroscopic predictions of quantum field theory have to agree with classical field theory.
The virtual particle picture certainly has limitations, one of which is that it's based on perturbation theory, and in the case of gravity, perturbation theory implies that the metric of spacetime can't be too different from a flat metric. I'm not sure how well that applies to spacetime near a black hole's horizon (much less inside it). But I think pop science discussions of the virtual particle picture are often the source (sometimes vague and indirect) of people's intuition that particles have to get out of the black hole to make us feel gravity from it, so I think it's important to recognize that there is a sense in which that picture is valid, if limited.
I don't agree, because this statement, as it stands, is even stronger than the one I quoted above; it rules out, not just the virtual particle view, but also the view I have advocated in a number of threads on PF, that the source of the black hole's gravity is the object that originally collapsed to form the hole. The metric in the vacuum region to the future of that collapsing object *is* caused by the object, and one way to describe how that causation occurs, as I said in my previous post, is that the spacetime metric propagates from the source to where you feel the gravity, governed by the vacuum EFE as the law of propagation.
Hmm. An eternal black hole that has always existed (rather than collapsing from a star) is certainly a solution to the Einstein Field Equations. So insisting that gravity has to have a "source" goes beyond GR. But it's probably more realistic. So maybe that point of view has something to be said for it.
PeterDonis
Mentor
I think that's taking terms in Feynmann diagrams a little too literally.
I did say this viewpoint had limitations.
you can say that virtual particles can escape in a way that real particles, but I think that's a little hoky.
"Limited" might be a better word.
I think that there are nonperturbative effects that can't be understood in terms of summing Feynmann diagrams. Bound states of particles I think is an example, and I would believe that black holes would be another example.
Yes, those are among the limitations I was referring to.
The other reason I don't like that resolution in the case of black holes is that I feel like it misses out on the topological reasons that a charged star collapsing into a black hole can't possibly cause the electric field or gravitational field to disappear. Whatever is going on inside a collapsing star, we know that before collapse, the electric field has a certain value for the surface integral over a surface surrounding the star. According to Maxwell's equations, you can't change that value unless charge crosses the surface. So what goes on inside the surface is just irrelevant.
This is a good point, but note that it assumes that the spacetime is simply connected. A wormhole, for example, blurs the distinction between what is "inside" the surface and what is "outside"; charge could migrate through the wormhole from the region "inside" the surface to some point way "outside" it, without passing through the surface. (Also, on "charge without charge" views such as, IIRC, John Wheeler proposed, you could have a nonzero surface integral without having any charge at all; the field lines would pass through the wormhole and wrap back around without ever having to end on a source.) But that's certainly not a vindication of the virtual particle view either.
An eternal black hole that has always existed (rather than collapsing from a star) is certainly a solution to the Einstein Field Equations. So insisting that gravity has to have a "source" goes beyond GR.
I would say it goes beyond a "naive" application of the EFE, but is still within the purview of "GR"; GR is not just the EFE, it also includes discussion of which solutions of the EFE apply to which physical scenarios, and which solutions are not physically reasonable at all.
Gravitational waves / disturbances have a finite velocity of c, and they cannot leave the event horizon.
Gravity itself having a velocity is a tricky concept, because gravity itself is not propagating, just its disturbances are. For example, an inertially moving mass pulls you towards its current position, not towards a retarded position due to finite signal speed.
I understand that is so from GR theory. Also I think one explanation from Feynmann is that this is a matter of virtual particles whose speeds are not restricted to c (hope I got this correct!).
However, that still does mean that "some information" is passing from the event horizon to the outside - otherwise, without gravitational waves or anything leaving the event horizon, how does the 'disturbance' (presumably outside the horizon) know that it has to propagate? Isn't some sort of 'information' continuity from the event horizon to the outside necessary for this? How is this reconciled?
A.T.
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2022-07-03 14:38:33
|
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https://magnusson.io/
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## Lyttle Lytton 2020
My Lyttle Lytton entry for 2020:
"Actually, you do like this," maverick CEO Eric Davies, Ph.D., insisted as he pulled my foreskin back and cunnilingussed my pee hole.
I’m not exactly proud of it, but I’m glad it’s no longer in my head.
## Not made for this world
It’s hot tonight, and this is going to be mostly the whiskey talking while I wait for it to get cool enough to sleep. I’ve killed the mosquitos I’ve seen, however, so until then I have only Haskell to keep me company.
This morning, tef tweeted about monads, which sent the Haskell pack his way with barks of not getting it. Just now, pinboard was reminded of some guy’s rage against Esperanto, from back in the 90’s when the web was fun and mostly devoted to things like explaining how "The Downward Spiral" is a concept album or destroying Unix instead of each other’s mental health.
For the Haskell pack, I did a PhD in the type of math that necessitates a lot of category theory, and I have looked at your use of category theory, and judged it to be unnecessary and pretentious and mainly focused on making you look smart while being entirely trivial. But this is not that kind of blog post, one that gets too tangled up in whether category theory is useful to get to the point. (If nothing else, Pijul proves that category theory is useful.) We’re here to discuss how Haskell as a whole is nonsense if you’re not an academic. Our claim is that Haskell is a useless language for writing software that has users.
Our point is simple, and focused on IO. We propose that you can measure how user-facing a program or language is by measuring how much of its time it spends or worries about doing IO. That is, after all, the medium through which anyone who is not a program’s author (of which there may be many) will interact with the program. The time spent doing IO can be on the command line, via a GUI, over a network, or wherever; but to be a serious contender for user-facing programs, a language has to make IO be easy.
C is a terrible language for most new things today. Anyone writing new software in C, that they expect to be used by other than thouroughly vetted people, needs to be able to explain why they’ve chosen C. At the same time, a lot of us are still exposed to C through the BSD or Linux kernels and syscalls, the undying popularity of K&R, random software on the internet, or other vectors. The culture around C invented the modern language textbook, K&R, and the modern user-facing program, "Hello world", both of which spend most or all of their time dealing with IO to talk to you or other users.
I claim making IO as simple as possible, which C does for all its faults, to do is analogous to trying to making it as simple as possible for other people to talk to you in designed languages as you can.[1]. Esperanto shows you can fail at that goal, if you even had it, for it favors sounds native to European languages above others. Likewise, Haskell shows you can fail at the goal of making IO easy, if you even had it, for it does not.
Haskell is a purely functional, lazily evaluated language, with a type system. Like tef explains, that is great, until you run into IO. Up until that point, you could rearrange computations in any order you liked, if they needed to be done at all. As soon as you need to do IO, though, you need something to happen before another thing, which makes you very unhappy if you’re Haskell. It in fact makes you so unhappy that you’ll drag the entire lost-at-sea community of category theorists into the orbit of your language just so you can have an abstraction for doing IO that fits into your model of the world. This abstraction, monoids, then comes with the added benefit of being abstract enough that all of your programmers can spend their time explaining it to each other instead of writing programs that use the abstraction to do IO, and therefore deal with any actual users.
Haskell is where programmers go to not have users.
1. One could say that what I mean is something more like making FFI as easy as possible, but that’s missing the point and would just move this discussion to some other, less inflammatory, level that we’re keen to avoid.
## Curvature as a bilinear form
Here’s something I thought of when I couldn’t sleep last night.
The curvature tensor of a Kähler metric can be viewed as a Hermitian form on $$\bigwedge^{1,1} T_X^*$$ by mapping $$\operatorname{End} T_X \to \bigwedge^{1,1} T_X^*$$ via the metric.
If we’re on a compact Kähler manifold with zero first Chern class, then for each Kähler class $$\omega$$ and $$(1,1)$$-classes $$u, v$$, we can pick the Ricci-flat metric in $$\omega$$ and the harmonic representatives of $$u, v$$. If $$R$$ is the curvature tensor of the metric, viewed as a Hermitian form, we can then set $b(u,v)(\omega) := \int_X R(u, v) \, dV_{\omega}.$ This defines a smooth bilinear form $$b$$ on the tangent space of the Kähler cone of $$X$$.
Besides being fun times, can we say anything interesting about $$b$$? For example, what is its norm with respect to the Riemannian metric on the Kähler cone, or its trace with respect to that metric? Can we integrate it over some subset of the cone?
## Choices
I only have time to work on the arXiv project so often, so I’m taking a lot of time between sessions to think about what I’m doing. When I look at my system design notes, I feel like all the decisions I’m making are the obvious choices, but they also were not at all obvious when I started thinking about them. It’s a good feeling.
## Queuesim
I haven’t made a lot of progress on my projects. I did create a Scaleway VM and shoved an OAI harvester on there that’s happily downloading the arXiv’s backlog of metadata. I can also parse the XML it fetches, and have some ideas about how I’m going to store it.
This lack of progress mostly comes from me being nerd-sniped into thinking about bounded queues under load. My old work project wanted to use a FIFO queue to hold its requests. That is a bad idea, because FIFO queues perform very poorly under load, as I went a little overboard in demonstrating.
Funnily enough, that very simple thing is one of my most popular Github projects ever.
## Ignore Mastodon
Counterpoint: I truly, madly, deeply want to ignore all new and existing social networks.
## Shame
To motivate me to actually finish some of the projects I start, I’m going to try announcing them to the world. Shame me into finishing these:
• A nicer arXiv frontend of daily new additions.
• A modern typesetting of Beauville’s Surfaces algebriques complexes.
Shame. Shame. Shame.
## Request-level configuration invariance in Go
Suppose we have an HTTP service. The behaviour of service depends on some configuration that may change at runtime; it may reload a static configuration file on SIGHUP, need to react to changes in its service discovery mechanism, or have A/B test state or features toggled on and off.
In Go, a naive way of handling this is by writing our configuration state to a struct and updating it in background goroutines:
type State struct {
frobinate bool
}
var state State{}
func handler(w http.ResponseWriter, r *http.Request) {
if s.frobinate {
fmt.Fprintln(w, "Great success")
} else {
fmt.Fprintln(w, "Great non-success")
}
}
This naive approach has at least two problems:
1. The state may change while we’re processing a request, causing us to process part of the request with one state, and another part with another. This isn’t a big deal in our example, but becomes more of a problem as the time needed to handle a request increases.
2. There are no synchronization primitives in play, so updating the state has data race conditions.
Check out the working example in this commit to see the first problem in action.
One way to resolve these problems is to add a mutex and to pass copies of the state to the request handlers:
type State struct {
frobinate bool
*sync.Mutex
}
// Copy may be arbitrarily complicated if State contains slices, maps,
// pointers, or other structs.
func (s *State) Copy() State {
s.Lock()
defer s.Unlock()
return *s
}
var state &State{}
func handler(w http.ResponseWriter, r *http.Request) {
s := state.Copy()
if s.frobinate {
fmt.Fprintln(w, "Great success")
} else {
fmt.Fprintln(w, "Great non-success")
}
}
The background goroutine that updates the state then either does so through a dedicated method that locks/unlocks the mutex, or does the locking itself.
While this works fine, it relies on global state and uses none of Go’s built-in concurrency features. We were promised a brave new world, and encouraged to "share memory by communicating". This points the way to another solution to our two problems that leverages Go’s primitives better:
type State struct {
frobinate bool
}
// Copy may be arbitrarily complicated if State contains slices, maps,
// pointers, or other structs.
func (s State) Copy() State {
return s
}
var stateCh chan State
var toggle chan struct{}
func stateManager() {
state := State{}
for {
select {
case stateCh <- state.Copy():
case <-toggle:
state.frobinate = !state.frobinate
}
}
}
func handler(w http.ResponseWriter, r *http.Request) {
s := <-stateCh
if s.frobinate {
fmt.Fprintln(w, "Great success")
} else {
fmt.Fprintln(w, "Great non-success")
}
}
A complete working example is here. Note that the working example doesn’t rely on any global state to pass state to the handler functions, using closures instead. Modifying the state is also only possible within the stateManager function. The working example could also be extended to have middleware copy the state to each request processing function instead of the ad-hoc way done there.
It is still up to the developers to ensure the Copy method doesn’t pass mutable state around, but they no longer need to deal with mutexes and locks themselves. This also means that any future additions to the program that use the same pattern won’t need to worry about those locks either.
There are no silver bullets in heavily concurrent systems, but in Go we can choose to not deal with some of the footguns we would need to handle in C, Java or other similar languages.
|
2019-10-14 18:53:58
|
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|
http://gurneyjourney.blogspot.com/2011/05/pencil-sketch-vignette.html?showComment=1306774405385
|
Friday, May 27, 2011
Pencil Sketch Vignette
Ernest Watson shows a nice trick for vignetting a pencil sketch.
He allows the outer edges of the drawing to gradually dissolve into simple construction lines. The windows are nicely varied, too. No two are alike.
This 1916 drawing is from Arthur Guptill’s book “Sketching and Rendering in Pencil" (republished as "Drawing and Sketching in Pencil,” available here in book form and here for free download.
Christopher Thornock said...
Great find! Thanks for the link.
Tony W. said...
This is a good principle as well: http://www.archive.org/stream/sketchingrenderi00guptuoft#page/64/mode/2up
James Gurney said...
Tony, yes, thanks. Guptill illustrates this idea in one of his other books, too (Color in Sketching and Rendering)--the idea that you can control the eye by adding or reducing contrast.
|
2014-08-22 09:52:14
|
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|
http://blog.sigfpe.com/2008_04_01_archive.html
|
A Neighborhood of Infinity
Sunday, April 27, 2008
Infinitesimal rotations and Lie algebras
A little while back I tried to give a rough idea of what Lie algebras were about. What I want to do now is show how with a bit of Haskell code we can directly get our hands on one. I'm going to assume you know how to make matrices for rotations, and multiply them, and some knowledge from previous posts, for which I'll give links.
> {-# OPTIONS -fno-warn-missing-methods #-}
Now we need some quick and dirty matrix code:
> data Matrix a = M [[a]] deriving (Eq,Show)> instance Functor Matrix where> fmap f (M a) = M $map (map f) a> instance Num a => Num (Matrix a) where> M a * M b = M$ mult a b
A Lie Group
What I'm going to do is start by constructing elements of the group of 3D rotations, otherwise known as SO(3), and show how there's another algebraic structure hidden inside it. So let's make some rotation matrices:
> rx theta = M $[[1,0,0],> [0,cos theta,-sin theta],> [0,sin theta,cos theta]]> ry theta = M$ [[cos theta,0,sin theta],> [0,1,0],> [-sin theta,0,cos theta]]> rz theta = M $[[cos theta,-sin theta,0],> [sin theta,cos theta,0],> [0,0,1]] These are the three rotations around the x-, y- and z-axes. It's traditional to build arbitrary rotations through the use of Euler angles: > euler [a,b,c] = rx a*ry b*rz c The 3D rotations form an example of a Lie group. (A Lie group is essentially a group where the operations like multiplication are differentiable.) Any 3D rotation can be constructed from a single application of euler. But notice how there's a bit of ugliness in this function. I've made an arbitrary choice of which order to apply the rotations in. I could have defined: > euler' [a,b,c] = rz c*ry b*rx a And it's easy to show that eulereuler'. This is because rotations don't commute. In other words, for rotations, a*b≠b*a. We can measure the non-commutativity. Remember that any rotation has an inverse. For example rx theta*rx (-theta) gives us the identity matrix because one of these two rotations 'undoes' the other. Given any two rotations we can construct what is known as their commutator: > commutator a b = inverse a*inverse b*a*b The idea is that we first perform b, then a, then undo b and then undo a. If a and b commute then this expression can be rearranged to inverse a*inverse b*b*a and then the inverses cancel leaving us with the identity matrix. If they don't commute then we end up with a non-identity matrix. So the comuutator measures the extent to which matrices don't commute. As I'm feeling lazy, I don't feel like writing inverse. Instead, as I'm only going to work with rotations, I'll use the fact that the inverse of a rotation matrix is the transpose and define: > inverse = transpose Try playing with expressions like commutator (rx 1) (ry 2). Note how the numbers quickly get messy. Try to write down closed form expressions for applications of euler and you'll see how complex things can get. A Lie Algebra But there's another way to approach the rotation group - through 'infinitesimal' rotations. In my earlier article I just talked about infinitesimal group operations in a hand-wavey way. Now I'm going to make the notion more rigorous. We just need to introduce an infinitesimal number, d, whose square is zero. I've talked about this a lot before so I'm borrowing my earlier code and defining: > d :: Num a => Dual a> d = D 0 1 If you try it you'll see that d*d is zero. Now we can try making infinitesimal rotations: > rot1 = euler [d,2*d,3*d] Note how when we evaluate this we get 'nice' numbers. No need to worry about those trig functions any more. And if you look closely at rot1 you'll see that it's essentially the identity matrix plus an infinitesimal part. We can pull the infinitesimal part out using fmap im rot1. You may be able to guess how to build it from the arguments to euler. But first, try evaluating rot2: > rot2 = euler' [d,2*d,3*d] It's the same! Try other infiitesimal rotations. When dealing with infinitesimal rotations it doesn't matter what order you apply them in, you get the same result. Working with infinitesimal rotations is looking much easier than working with full=size rotations. In fact, it gets better. Try defining > rot3 = euler [5*d,-d,2*d] Now look at fmap im (rot1*rot3) and compare with fmap im rot1 and fmap im rot3. We can multiply infinitesimal rotations simply by adding their infinitesimal parts. In fact, we can define > star [a,b,c] = M$ [[0, -c, b],> [c,0,-a],> [-b,a,0]]
So we have:
fmap im (euler [a*d,b*d,c*d]) == star [a,b,c]
and
fmap im (euler [a*d,b*d,c*d]*euler [u*d,v*d,w*d]) == fmap im (euler [(a+u)*d,(b+v)*d,(c+w)*d])
Not a single trig expression anywhere!
So we have a simplified way of viewing rotations by looking at infinitesimal rotations. A triple [a,b,c] can be thought of as representing an infinitesimal rotation through star and instead of multiplying matrices we just add the triples elementwise. These triples, together with the binary operation of addition form an example of a Lie algebra. But there's a piece missing. We have lost all information about the non-commutativity of the rotations. It's one thing to simplify, but it's another to lose an important feature of what you're looking at.
The problem is that d is 'too small'. We need an infinitesimal that doesn't go to zero the moment you square it, but is still, in some sense, infinitesimally small. We could rewrite the Dual type. But there's a trick. Define:
> e :: Num a => Dual (Dual a)> e = D (D 0 1) (D 1 0)
(If you think of Dual as forming a tensor product as I described here then e=1⊗d+d⊗1.)
You can check that e^2 is non-zero but e^3 is zero.
Now when we compute a commutator we get something a little different:
> comm1 = commutator (euler [e,0,0]) (euler [0,e,0])
fmap re comm1 is essentially the identity as before. But if we look at fmap im comm2 there's a non-zero infinitesimal piece which is different from what we had when we worked with d. This infinitesimal piece is in fact proportional to e^2. As (im . im) (e^2) is a half, we can extract the coefficient of e^2 from comm1 using
> im2 x = im (im x)/2
In fact, we have fmap im2 comm1 == star [0,0,1]. So by choosing infinitesimals that aren't too small, we haven't lost information about non-commutativity. In fact, a bit of experimentation may convince you that with:
> shrink u = fmap (e*) u
we have:
fmap im2 (commutator (euler (shrink u)) (euler (shrink v))) == star (u cross v)
So let's step back a bit and recapitulate all this in something approaching English:
Given a tiny rotation we can represent it as three Euler angles a, b, c, all of which are tiny. We can think of a, b and c as forming a vector [a,b,c]. When we do this, apart from an even smaller error, multiplication of rotations becomes ordinary addition of vectors and the order of rotations isn't significant. But if we choose not to ignore this small error we see that a rotation represented by u and a rotation represented by v don't quite commute and the order does matter. The size of this error is measured by the cross product of u and v. This is intuitively plausible, we'd expect that rotations defined by vectors in a similar direction would be closer to commuting, and this is reflected in the fact that the cross product is zero for parallel vectors.
So now I can fill in the missing piece from the description of the Lie algebra I gave above. The Lie algebra so(3) consists of the 3d vectors (representing infinitesimal rotations), addition of vectors (representing multiplcation of infinitesimal rotations) and the cross product (measuring the amount by which small rotations fail to commute). This picture doesn't just apply to the rotations, a similar one applies for any Lie group. We get the same pattern of a simplified form of multiplication and a binary operation that measures non-commutativity.
But you might still ask if there's something missing. What if we defined f so that f^4 is zero but f^3 isn't? Would we extract more information about the non-commutativity of rotations? Well the interesting fact, which I won't prove here, is that it doesn't. Almost everything you need to know about a Lie group can be extracted from its Lie algebra. In fact, from a group's Lie algebra you can almost recover the original group. (In fact, what you recover is its universal cover.) But notice how there are no trig formulae involved when talking about Lie algebras. Lie algebras give a really nice way to study Lie groups without getting your hands too dirty. But this isn't the only reason to study Lie algebras, many physical properties arising from symmetries are more naturally studied through the Lie algebra because Lie algebras arise from Lie groups as soon as you start differentiating things. In particular, Lie algebras play a major role in Noether's theorem, one of the cornerstones of modern theoretical physics.
In summary then, I hope I've given some flavour of Lie algebras. Using infinitesimals you can get your hands on them directly without the use of a large amount of mathematical machinery. There has been some difficult stuff here, but I'm hoping that the freedom you now have at the Haskell prompt to play with the things I've been talking about will make up for the inadequacies of my explanations.
One last word: in principle we could do the same with E8. But we'd need 248 by 248 matrices, and the equivalent of euler would need 248 parameters. (That's a coincidence, in general the number of parameters needed to define an element of a Lie group isn't equal to the dimension of the matrix, but it is for SO(3) and E8).
Appendix (the bits of code I left out above)
Defining the infinitesimals:
> data Dual a = D a a deriving (Show,Eq)
Extract the 'full-size' and 'infinitesimal' parts of a number:
> re (D a _) = a> im (D _ b) = b> instance Num a => Num (Dual a) where> fromInteger i = D (fromInteger i) 0> (D a a')+(D b b') = D (a+b) (a'+b')> (D a a')-(D b b') = D (a-b) (a'-b')> (D a a')*(D b b') = D (a*b) (a*b'+a'*b) > instance Floating a => Floating (Dual a) where> cos (D a a') = D (cos a) (-sin a*a')> sin (D a a') = D (sin a) (cos a*a')> instance Fractional a => Fractional (Dual a)
Some useful matrix and vector operations:
> mult a ([]:_) = map (const []) a> mult a b = zipWith (:) (map (dot (map head b)) a) (mult a (map tail b))> transpose (M a) = M $transpose' a where> transpose' [] = repeat [] > transpose' (xs : xss) = zipWith (:) xs (transpose' xss) Some simple vector operations > dot a b = foldr (+) 0$ zipWith (*) a b> cross [a,b,c] [d,e,f] = [b*f-c*e,c*d-a*f,a*e-b*d]
Saturday, April 12, 2008
Negative Probabilities
I'm always interested in new ways to present the ideas of quantum mechanics to a wider audience. Unfortunately the conventional description of quantum mechanics requires knowledge of complex numbers and vector spaces making it difficult to avoid falling back on misleading analogies and metaphors. But there's another approach to quantum mechanics, due to Feynman, that I have never seen mentioned in the popular science literature. It assumes only basic knowledge of probability theory and negative numbers to get a foot on the ladder.
We start with tweaking probability theory a bit. One of the axioms of probability theory says that all probabilities must lie in the range zero to one. However, we could imagine relaxing this rule even though on the face of it it seems meaningless. For example, suppose we have a coin that has a 3/2 chance of landing heads and a -1/2 chance of landing tails. We can still reason that the chance of getting two heads in a row is 3/2×3/2=9/4 by the usual multiplication rule. But obviously no situation like this could ever arise in the real world because after tossing such a coin 10 times we'd expect to see -5 tails on average.
But what if we could contrive a system with some kind of internal state governed by negative probabilities even though we couldn't observe it directly? So consider this case: a machine produces boxes with (ordererd) pairs of bits in them, each bit viewable through its own door. Let's suppose the probability of each possible combination of two bits is given by the following table:
First bit Second bit Probability 0 0 -1/2 0 1 1/2 1 0 1/2 1 1 1/2
Obviously if we were able to look through both doors we'd end up with the meaningless prediction that we'd expect to see 00 a negative number of times. But suppose that things are arranged so that we can only look through one door. Maybe the boxes self-destruct if one or other door is opened but you still get enough time to see what was behind the door. Now what happens?
If you look through the first door the probability of seeing 1 is P(10)+P(11)=1. We get the same result if we look through the second door. We only get probabilities in the range zero to one. As long as we're restricted to one door we get meaningful results.
If we were to perform this experiment repeatedly with different runs of the machine, each time picking a random door to look through, we'd eventually become very confident that every box contained 11. After all, if we freely choose which door to look through, and we always see 1, there's no place 0's could be 'hiding'.
But now suppose a new feature is added to the box that allows us to compare the two bits to see if they are equal. It reveals nothing about what the bits are, just their state of equality. And of course, after telling us, it self-destructs. We now find that the probability of the two bits being different is P(01)+P(10)=1. So if we randomly chose one of the three possible observations each time the machine produced the box we'd quickly run into the strange situation that the two bits both appear to be 1, and yet are different. But note that although the situation is weird, it's not meaningless. As long as we never get to see both bits at the same time we never directly observe a paradox. If we met such boxes in the real world we'd be forced to conclude that maybe the boxes knew which bit you were going to look at and changed value as a result, or that maybe you didn't have the free will to choose door that you thought you had, or maybe, even more radically, you'd conclude that the bits generated by the machine were described by negative probabilities.
That's all very well, but obviously the world doesn't really work like this and we never see boxes like this. Except that actually it does! The EPR experiment has many similarities to the scenario I described above. The numbers aren't quite the same, and we're not talking about bits in boxes, but we do end up with a scenario involving observations of bits that simply don't add up. If we do try to explain what's going on using probability theory, we either conclude there's something weird about our assumptions of locality or causality or we end up assigning negative probabilities to the internal states of our systems. In fact, you can read the details in an article by David Schneider. Being forced to conclude that we have negative probabilities in a physical system is usually taken as a sign that we have a contradiction. In the case of Bell's theorem it shows that we can't interpret what we see in terms of probability theory and hence that the weirdness of quantum mechanics can't be explained in terms of some hidden random variable that we can't see. QM simply doesn't obey the rules you'd expect of hidden variables we can't see.
But in a paper called Negative Probability, Feynman tried taking the idea of negative probabilities seriously. He showed that you could reformulate quantum mechanics completely in terms of them so that you no longer needed to think in terms of the complex number valued 'amplitudes' that physicists normally use. This means the above isn't just an analogy, it's actually a formal system within which you can do QM, although I haven't touched on the bit that refers to the dynamics of quantum systems. So if you can get your head around the ideas I've talked about above you're well on your way to understanding some reasons why quantum mechanics seems so paradoxical.
At this point you may be wondering how nature contrives to hide these negative probabilities from direct observation. Her trick is that making one kind of observation disturbs up the state of what you've observed so that you can't make the other kind of observation on a pristine state. You have to pick one kind of observation or the other. Electrons and photons really are a lot like the boxes I just described.
So why don't physicists use this formulation? Despite the fact that negative numbers seem simpler to most people than imaginary numbers, the negative number formulation of QM is much more complicated. What's more, because it makes exactly the same predictions as regular QM there's no compelling reason to switch to it. And anyway, it's not as if directly observing negative probabilities is any more intuitive or meaningful than imaginary ones. Once you've introduced negative ones, you might as well go all the way!
This all ties in with what I said a while back. The important thing about QM is that having two ways to do something can make it less likely to happen, not more.
For a different perspective this is an interesting comment.
Footnote: We can embed QM in negative probability theory. But can we do the converse? Can every negative probability distribution be physically realised in a quantum system? I've a hunch the answer is obvious but I'm too stupid to see it.
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2016-10-21 00:28:04
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http://openstudy.com/updates/55b6a60de4b04559507b8098
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anonymous one year ago Absolute zero is A. –273.15°C. B. the lowest possible temperature. C. the temperature at which the average kinetic energy of particles would theoretically be zero. D. all of the above
C. Temperature at which the average kinetic energy of the particles would be theoretically zero is correct. because temperature is defined as the average kinetic energy $KE = \frac{ 1 }{ 2 }mV ^{2}$ B. this is theoretically the lowest possible temperature. this is also correct. A this is true because Kelvin = C+273 so the absolute zero is 0K or -273C
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2016-10-24 14:36:19
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https://mathhelpboards.com/threads/quasi-local-commutative-ring-with-maximal-ideal-m.8033/
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# Quasi-local commutative ring with maximal ideal M
#### Peter
##### Well-known member
MHB Site Helper
I am reading R.Y. Sharp: Steps in Commutative Algebra.
In Chapter 3: Prime Ideals and Maximal Ideals, Exercise 3.19 reads as follows:
Let R be a quasi-local commutative ring with maximal ideal M.
Show that the ring $$\displaystyle R[[X_1, ... \ ... , X_n]]$$ of formal power series in indeterminates $$\displaystyle X_1, ... \ ... , X_n$$ with coefficients in R is again a quasi-local ring, and that its maximal ideal is generated by $$\displaystyle M \cup \{ X_1, ... \ ... , X_n \}$$
Can someone please help me get started on this problem?
Peter
Note: On page 41, Chapter 3, Sharp defines a quasi-local ring as follows:
3.12 Definition. A commutative ring R which has exactly one maximal ideal, M say, is said to be quasi-local.
#### Turgul
##### Member
It is worth noting that these days, what you are calling a quasi-local ring is often referred to as a simply a local ring. Classically one required the ring to also be noetherian to be called local (as opposed to quasi-local), but over the last 50 years, it has become standard to omit this demand.
There is one very important observation to be made about a (quasi-)local ring: any element not contained in the (unique) maximal ideal must be a unit (why?). Hence all things in the maximal ideal are the non-units of the ring and everything else is a unit. So it is good enough to understand when something is a unit if you want to understand the maximal ideal.
With that observation in mind, try to simplify the problem and consider the ring $R[[x_1]]$. What are the units of a power series ring in one variable?
#### Peter
##### Well-known member
MHB Site Helper
It is worth noting that these days, what you are calling a quasi-local ring is often referred to as a simply a local ring. Classically one required the ring to also be noetherian to be called local (as opposed to quasi-local), but over the last 50 years, it has become standard to omit this demand.
There is one very important observation to be made about a (quasi-)local ring: any element not contained in the (unique) maximal ideal must be a unit (why?). Hence all things in the maximal ideal are the non-units of the ring and everything else is a unit. So it is good enough to understand when something is a unit if you want to understand the maximal ideal.
With that observation in mind, try to simplify the problem and consider the ring $R[[x_1]]$. What are the units of a power series ring in one variable?
Thank you for the hint Turgul.
I suspect that the units of R[X] are simply the units of R, but I am not sure how to use this to show that R[X] is (quasi-)local.
Peter
#### Turgul
##### Member
The units of $R[x]$ are indeed the units of $R$, but we're not looking for that. We are looking for the units of $R[[x]]$. Note for example that $(1-x)(1+x+x^2+x^3+ \cdots) = (1+x+x^2+x^3+ \cdots) - x(1+x+x^2+x^3+ \cdots) = 1$.
In fact, as per my comment on the maximal ideal of local rings, if we are to believe the problem, the units of $R[[x]]$ are precisely those things not in the ideal generated by $M$ and $x$. What does that mean? It means that any power series which has a nonzero constant term which is also a unit in $R$ (ie the constant term is not in $M$) should be a unit.
Note that the observation about local rings goes the other way as well; if all elements of a ring outside of a given ideal happen to be units, then that ideal must be maximal and the ring local with that as the unique maximal ideal (why?).
But this means that if we were to show that any power series with constant term not in $M$ were in fact a unit of $R[[x]]$, then $R[[x]]$ would have to be a local ring with maximal ideal exactly those power series which happen to have constant term in $M$. But this is exactly the ideal generated by $M \cup \{x\}$.
#### Deveno
##### Well-known member
MHB Math Scholar
Thank you for the hint Turgul.
I suspect that the units of R[X] are simply the units of R, but I am not sure how to use this to show that R[X] is (quasi-)local.
Peter
Well, no, that's not right. We certainly can write any power series in $R[[x_1]]$ as:
$a - x_1f$, where $a \in R$, and $f$ is some other power series in $R[[x_1]]$.
So if:
$(a - x_1f)(b - x_1g) = ab - x_1(f+g) + x_1^2(fg) = ab - x_1(f+g - x_1(fg)) = 1$
We must have:
$ab = 1$ and $f+g - x_1(fg) = 0$.
The first equation tells us the constant term of a unit must be a unit of $R$.
On the other hand, if $a$ is a unit, so that there is $b \in R$ with $ab = 1$, then taking:
$g = -b(f + bx_1f^2 + bx_1^2f^3 + \cdots)$, we have:
$(a - x_1f)(b - x_1g) = (a - x_1f)(b(1 + bx_1f + (bx_1f)^2 + (bx_1f)^3 + \cdots))$
$= ab + (ab^2x_1f - bx_1f) + (ab^3x_1^2f^2 - b^2x_1^2f^2) + (ab^4x_1^3f^3 - b^3x_1^3f^3) + \cdots$
$= 1 + 0 + 0 + 0 + \cdots = 1$
So the units of $R[[x_1]]$ are precisely the power series with constant term in $U(R)$.
#### Peter
##### Well-known member
MHB Site Helper
Well, no, that's not right. We certainly can write any power series in $R[[x_1]]$ as:
$a - x_1f$, where $a \in R$, and $f$ is some other power series in $R[[x_1]]$.
So if:
$(a - x_1f)(b - x_1g) = ab - x_1(f+g) + x_1^2(fg) = ab - x_1(f+g - x_1(fg)) = 1$
We must have:
$ab = 1$ and $f+g - x_1(fg) = 0$.
The first equation tells us the constant term of a unit must be a unit of $R$.
On the other hand, if $a$ is a unit, so that there is $b \in R$ with $ab = 1$, then taking:
$g = -b(f + bx_1f^2 + bx_1^2f^3 + \cdots)$, we have:
$(a - x_1f)(b - x_1g) = (a - x_1f)(b(1 + bx_1f + (bx_1f)^2 + (bx_1f)^3 + \cdots))$
$= ab + (ab^2x_1f - bx_1f) + (ab^3x_1^2f^2 - b^2x_1^2f^2) + (ab^4x_1^3f^3 - b^3x_1^3f^3) + \cdots$
$= 1 + 0 + 0 + 0 + \cdots = 1$
So the units of $R[[x_1]]$ are precisely the power series with constant term in $U(R)$.
Thanks Deveno
Thanks to you and Turgul I am getting a much better idea of the ring of formal power series.
Just a couple of questions regarding your analysis above:
Question 1
You write:
"So if:
$(a - x_1f)(b - x_1g) = ab - x_1(f+g) + x_1^2(fg) = ab - x_1(f+g - x_1(fg)) = 1$
We must have:
$ab = 1$ and $f+g - x_1(fg) = 0$."
How can we be sure ab = 1 as a and b may be zero divisors since we are not given that R is an integral domain.
Question 2
In showing that
$$\displaystyle (a - x_1f)(b - x_1g) = 1$$ you assume $$\displaystyle g = -b(f + bx_1f^2 + bx_1^2f^3 + \cdots)$$
How on earth did you determine the expression for g? What were your considerations and figuring?
By the way, do you know of a text that treats the manipulation of power series in the way you and Turgul were doing in your posts? Most analysis books that I know just treat convergence issues and most algebra books assume that one is already familiar with such things and just treat abstract aspects of the ring R[[X]]. ... ...
... ... now must get back to thinking about a solution to the exercise ...
Peter
Last edited:
#### Peter
##### Well-known member
MHB Site Helper
Thanks Deveno
Thanks to you and Turgul I am getting a much better idea of the ring of formal power series.
Just a couple of questions regarding your analysis above:
Question 1
You write:
"So if:
$(a - x_1f)(b - x_1g) = ab - x_1(f+g) + x_1^2(fg) = ab - x_1(f+g - x_1(fg)) = 1$
We must have:
$ab = 1$ and $f+g - x_1(fg) = 0$."
How can we be sure ab = 1 as a and b may be zero divisors since we are not given that R is an integral domain.
Question 2
In showing that
$$\displaystyle (a - x_1f)(b - x_1g) = 1$$ you assume $$\displaystyle g = -b(f + bx_1f^2 + bx_1^2f^3 + \cdots)$$
How on earth did you determine the expression for g? What were your considerations and figuring?
By the way, do you know of a text that treats the manipulation of power series in the way you and Turgul were doing in your posts? Most analysis books that I know just treat convergence issues and most algebra books assume that one is already familiar with such things and just treat abstract aspects of the ring R[[X]]. ... ...
... ... now must get back to thinking about a solution to the exercise ...
Peter
Just summarizing where I am with this interesting exercise ...
From Deveno's post plus some hints from Turgul, we have the following:
... the units of R[[X]] are precisely those power series with a with a constant term, say K, where k is a unit of R.
These elements of R[[X]] cannot possibly belong to the (only) maximal ideal, M' of R[[X]] because if a unit belonged to M' then M' = R[[X]].
Now working on Turgul's hints we have the following ideas ...
If we can show that all the elements outside M are units, then the maximal ideal M' would consist of precisely those power series with a constant term in M. (Any power series with a constant term outside of M would be a unit ...)
... ...
BUT
(1) how do we show that all elements of R outside M are units?
(2) how do we show that M' is the only maximal ideal in R[[X]]?
I would appreciate some further help.
Peter
#### Deveno
##### Well-known member
MHB Math Scholar
If it is *given* that $R$ is quasi-local, then if $d$ is any non-unit of $R$, we have that:
$(d)$ is a proper ideal of $R$ (why?), so $(d)$ is contained in a maximal ideal, so (because there is only one maximal ideal) $(d) \subseteq M$, so $d \in M$.
Thus $M$ contains every non-unit, and none of the units.
On the other hand, if an ideal contains every non-unit, and is not the entire ring, it must be maximal, and no other maximal ideal can possibly exist (since any proper ideal only has non-units in it, and is thus contained in our ideal of every non-unit).
The discussion up to now indicates we should look at the constant terms of elements of $R[[x_1]]$. So this is what you need to do:
A) Prove that the set of power series whose constant term is not a unit forms an ideal. By our discussion above, this will be a unique maximal ideal, showing that $R[[x_1]]$ is quasi-local. If you are perceptive enough, you will see that this boils down to a similar question about $R$, where the matter is already settled.
B) Next, consider the case $R[[x_1,x_2]]$. Once you have this settled, you can use an induction argument on the number of indeterminates.
#### Peter
##### Well-known member
MHB Site Helper
If it is *given* that $R$ is quasi-local, then if $d$ is any non-unit of $R$, we have that:
$(d)$ is a proper ideal of $R$ (why?), so $(d)$ is contained in a maximal ideal, so (because there is only one maximal ideal) $(d) \subseteq M$, so $d \in M$.
Thus $M$ contains every non-unit, and none of the units.
On the other hand, if an ideal contains every non-unit, and is not the entire ring, it must be maximal, and no other maximal ideal can possibly exist (since any proper ideal only has non-units in it, and is thus contained in our ideal of every non-unit).
The discussion up to now indicates we should look at the constant terms of elements of $R[[x_1]]$. So this is what you need to do:
A) Prove that the set of power series whose constant term is not a unit forms an ideal. By our discussion above, this will be a unique maximal ideal, showing that $R[[x_1]]$ is quasi-local. If you are perceptive enough, you will see that this boils down to a similar question about $R$, where the matter is already settled.
B) Next, consider the case $R[[x_1,x_2]]$. Once you have this settled, you can use an induction argument on the number of indeterminates.
Thanks for the help and guidance, Deveno
I wish to try to deal with your point "$(d)$ is a proper ideal of $R$ (why?)". Can you please confirm the argument given is OK.
---------------------------------------------------------------------------
we have that (d) is not proper, that is (d) = R if and only if (d) contains a unit.
Assume (d0 is not proper. Then (d) contains a unit.
Elements of (d) have the form rd = dr for $$\displaystyle r \in R$$ so then if an element of (d) is a unit then there is some $$\displaystyle r_1 \in R$$ such that:
$$\displaystyle (dr)r_1 = 1$$
$$\displaystyle \Longrightarrow d(rr_1) = 1$$
$$\displaystyle \Longrightarrow dr_2 = 1$$ for some $$\displaystyle r_2 \in R$$
BUT d is not a unit in R so this equation cannot hold
Therefore d does not contain a unit ...
Therefore (d) does not contain the unit 1 and so $$\displaystyle (d) \ne R$$, that is (d) is a proper ideal.'
Can you confirm this is a sound argument?
Peter
#### Deveno
##### Well-known member
MHB Math Scholar
Yes, an ideal is proper iff it does not contain a unit (or equivalently is the entire ring iff it does contain a unit). You have a couple of typos but your reasoning is sound.
So the group of units of a ring is important, because it helps us classify the ideals.
A word about motivation:
Rings can be very complicated structures, and the presence of a lot of non-units can make solving equations from them daunting. Zero divisors also pose similar challenges to isolating epxressions we want to evaluate. So the hope is we can find an ideal containing all the "bad" elements, and mod it out, giving a "nicer" ring in which calculation is more straight-forward.
And...if we are fortunate, this nicer ring will allow us to "lift" information gleaned from it, to our original ring. For example, equations involving integers are often solved by passing to the integers mod n (if n = p, a prime, this is a field and we can divide...something we CAN'T do in the original integers), and then going back to the integers. Any argument involving even and odd integer cases is actually using this "trick" (although many people employing this do not even realize this).
Maximal ideals are important, because for commutative rings with unity, we get a field as our quotient ring. Fields are the very pinnacle of "niceness" for rings, algebra works just like our intuition tells us it should. Conversely, the size of a maximal ideal, gives us a rough idea of how much of the ring is going to be hard to work with.
In answer to your earlier question about "how" i inverted the power series:
$a - xf$ it really is just a variation of inverting:
$1 - x$ by noting that:
$1(1 + x + x^2 + x^3 + \cdots) - x(1 + x + x^2 + x^3 + \cdots)$
$= 1 + (x - x) + (x^2 - x^2) + (x^3 - x^3) + \cdots$
$= 1 + 0 + 0 + 0 + \cdots = 1$
Now if $a$ is a unit, so that we have some $b$ with $ab = 1$, we have:
$1 = 1 + 0 + 0 + 0 + \cdots$
$= ab + 0 + 0 + 0 + \cdots$
$= ab + (bxf - bxf) + ((bxf)^2 - (bxf)^2) + ((bxf)^3 - (bxf)^3) + \cdots$
$= ab + (ab)(bxf - bxf) + (ab)((bxf)^2 - (bxf)^2) + (ab)((bxf)^3 - (bxf)^3) + \cdots$
$= ab + ab^2xf - ab^2xf + ab^3x^2f^2 - ab^3x^2f^2 + ab^4x^3f^3 - ab^4x^3f^3 + \cdots$
$= a(b + b^2xf + b^3x^2f^2 + b^4x^3f^3 + \cdots) - xf(b + b^2xf + b^3x^2f^2 + b^4x^3f^3 + \cdots)$
$= (a - xf)(b + (b^2xf + b^3x^2f^2 + b^4x^3f^3 + \cdots))$
$= (a - xf)(b - x(-b)(bf + b^2xf^2 + b^3x^2f^3 + \cdots))$
so that we may take:
$g = (-b^2)(f + bxf^2 + b^2x^2f^3 + \cdots)$
(it appears that I have a typo in my original post. Oh dear. *Sigh*. I try to check these things, but my mind wanders...)
Anyway, that's how you derive the formula for $g$.
#### Peter
##### Well-known member
MHB Site Helper
If it is *given* that $R$ is quasi-local, then if $d$ is any non-unit of $R$, we have that:
$(d)$ is a proper ideal of $R$ (why?), so $(d)$ is contained in a maximal ideal, so (because there is only one maximal ideal) $(d) \subseteq M$, so $d \in M$.
Thus $M$ contains every non-unit, and none of the units.
On the other hand, if an ideal contains every non-unit, and is not the entire ring, it must be maximal, and no other maximal ideal can possibly exist (since any proper ideal only has non-units in it, and is thus contained in our ideal of every non-unit).
The discussion up to now indicates we should look at the constant terms of elements of $R[[x_1]]$. So this is what you need to do:
A) Prove that the set of power series whose constant term is not a unit forms an ideal. By our discussion above, this will be a unique maximal ideal, showing that $R[[x_1]]$ is quasi-local. If you are perceptive enough, you will see that this boils down to a similar question about $R$, where the matter is already settled.
B) Next, consider the case $R[[x_1,x_2]]$. Once you have this settled, you can use an induction argument on the number of indeterminates.
Thanks so much for the help Deveno.
You write that the first thing we must do is as follows:
"Prove that the set of power series whose constant term is not a unit forms an ideal."
Consider f to be a power series, that is of the form:
$$\displaystyle f = a_0 + a_1x + a_2x^2 + ... \ ...$$
then define the set M* defined as follows:
M* = $$\displaystyle \{ f \ | \ f \in R[[x_1]]$$ and $$\displaystyle a_0$$ is a non-unit $$\displaystyle \}$$
We wish to show that M* is an ideal under the operations of addition and multiplication of power series. That is, specifically, we wish to show:
(i) $$\displaystyle M^* \ne \emptyset$$
(ii) whenever $$\displaystyle f,g \in M^*$$ , then $$\displaystyle f+g \in M^*$$ and
(iii) whenever $$\displaystyle f \in M^*$$ and $$\displaystyle h \in R[[x_1]]$$ then $$\displaystyle hf \in M^*$$
To show $$\displaystyle M^* \ne \emptyset$$
The zero element of $$\displaystyle R[[x_1]]$$ (viz. $$\displaystyle 0 = 0 + 0.x_1 + 0.x^2 + ... \ ...$$ ) belongs to M* since 0 is a non-unit, Therefore M* is not the empty set.
To show that whenever $$\displaystyle f,g \in M^*$$, then $$\displaystyle f+g \in M^*$$
Consider $$\displaystyle f, g \in M^*$$
where $$\displaystyle f = a_0 + a_1x + a_2x^2 + ... \ ...$$
and $$\displaystyle g = b_0 + b_1x + b_2x^2 + ... \ ...$$
Then $$\displaystyle f + g = (a_0 + b_0) + ... \ ...$$
Now $$\displaystyle a_0, b_0 \in M$$ where M is the (unique) maximal ideal in R containing all non-units of R
We have $$\displaystyle a_0, b_0 \in M \Longrightarrow a_0 + b_0 \in M$$
$$\displaystyle \Longrightarrow a_0 + b_0$$ is a non-unit in R
$$\displaystyle \Longrightarrow f + g \in M^*$$
Finally ... ...
To show that whenever $$\displaystyle f \in M^*$$ and $$\displaystyle h \in R[[x_1]]$$ then $$\displaystyle hf \in M^*$$
Consider $$\displaystyle f \in M^*$$ and $$\displaystyle h \in R[[x_1]]$$
where $$\displaystyle f = a_0 + a_1x + a_2x^2 + ... \ ...$$
and $$\displaystyle h = h_0 + h_1x + h_2x^2 + ... \ ...$$
Then $$\displaystyle hf = h_0a_0 + ... \ ...$$
But $$\displaystyle h_0a_0 \in M$$
$$\displaystyle \Longrightarrow h_0a_0$$ is a non-unit
$$\displaystyle \Longrightarrow h_0a_0 \in M^*$$
Thus M* is an ideal
Can you confirm that the above proof is sound?
Will now move on to thinking about Deveno's suggested second step ... namely
"Next, consider the case $R[[x_1,x_2]]$. Once you have this settled, you can use an induction argument on the number of indeterminates"
Peter
Last edited:
#### Deveno
##### Well-known member
MHB Math Scholar
That's the idea exactly: reduce the problem of $R[[x_1]]$ where we are a bit uncertain of how things work, to the same problem of $R$, where we already know the answer.
Now, for the next step:
Is there an isomorphism between $R[[x_1,x_2]]$ and $(R[[x_1]])[[x_2]]$? If it existed, what would it have to look like?
(I'll give you a hint: use the analogous isomorphism that takes $R[x][y] \to R[x,y]$ that is, map the terms "polynomially").
#### Peter
##### Well-known member
MHB Site Helper
That's the idea exactly: reduce the problem of $R[[x_1]]$ where we are a bit uncertain of how things work, to the same problem of $R$, where we already know the answer.
Now, for the next step:
Is there an isomorphism between $R[[x_1,x_2]]$ and $(R[[x_1]])[[x_2]]$? If it existed, what would it have to look like?
(I'll give you a hint: use the analogous isomorphism that takes $R[x][y] \to R[x,y]$ that is, map the terms "polynomially").
Again, thank you for the guidance and help ...
You write:
"Now, for the next step:
Is there an isomorphism between [FONT=MathJax_Math]R[/FONT][FONT=MathJax_Main][[/FONT][FONT=MathJax_Main][[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]][/FONT][FONT=MathJax_Main]][/FONT] and [FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]R[/FONT][FONT=MathJax_Main][[/FONT][FONT=MathJax_Main][[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]][/FONT][FONT=MathJax_Main]][/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main][[/FONT][FONT=MathJax_Main][[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]][/FONT][FONT=MathJax_Main]][/FONT]? If it existed, what would it have to look like?
(I'll give you a hint: use the analogous isomorphism that takes [FONT=MathJax_Math]R[/FONT][FONT=MathJax_Main][[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]][/FONT][FONT=MathJax_Main][[/FONT][FONT=MathJax_Math]y[/FONT][FONT=MathJax_Main]][/FONT][FONT=MathJax_Main]→[/FONT][FONT=MathJax_Math]R[/FONT][FONT=MathJax_Main][[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Math]y[/FONT][FONT=MathJax_Main]][/FONT] that is, map the terms "polynomially"). "
I checked a number of textbooks on the construction of the two rings $$\displaystyle R[x][y]$$ and $$\displaystyle R[x,y]$$ including the isomorphism between them, and found Papantonopoulou: Algebra: Pure and Applied most helpful.
To understand the isomorphism between $$\displaystyle R[x][y]$$ and $$\displaystyle R[x,y]$$ we must first understand the nature of the two rings.
Drawing on Papantonopoulou we have the following:
The elements of $$\displaystyle R[x][y]$$ are polynomials in y whose coefficients are taken from the ring of polynomials in x, namely R[x].
Thus for example, one element would be
$$\displaystyle f_0 + f_1y + f_2y^2$$
$$\displaystyle = (x^2 +1) + (x^3 + 2x)y + (x+2)y^2$$ ... ... ... ... (1)
However, we can also define a ring R[x,y] whose elements are finite sums $$\displaystyle \sum a_{ij}x^iy^j$$
An element of R[x,y] for example would be
$$\displaystyle x^2 + 1 + x^3y + 2xy + xy^2 + 2y^2$$ ... ... ... ... (2)
where (2) for illustrative purposes is (1) multiplied out
---------------------------------------------------------------------------
Now to focus on the isomorphism between R[x][y] and R[x,y] ... ...
Let $$\displaystyle \phi$$ map an element of R[x][y] to an element of R[x,y]
obtained by "multiplying out" in the way (1) is turned in (1) above is turned into (2).
So, consider an element h of R[x][y]:
$$\displaystyle h = f_n(x)y^n + ... \ ... f_1(x)y + f_0$$
where $$\displaystyle f_i(x) = {\sum}_{j=1}^{m_i} a_{ij}x^j$$
Then we have
$$\displaystyle \phi (h) = {\sum}_{i=0}^{n} {\sum}_{j=0}^{m_i} a_{ij} x^j y^i$$
Papantonopoulou claims that $$\displaystyle \phi$$ is an isomorphism ...
If we accept this, then where to now ... can you help?
Peter
#### Deveno
##### Well-known member
MHB Math Scholar
Use the same mapping just change the $m$ and $n$ to $\infty$....
#### Peter
##### Well-known member
MHB Site Helper
Use the same mapping just change the $m$ and $n$ to $\infty$....
Sorry Deveno ... I can see that this will give us an isomorphism between power series rings (as distinct from polynomials) but I am not understanding how this can be used establishing what we need to establish in the exercise viz. in the following:
"Let R be a quasi-local commutative ring with maximal ideal M.
Show that the ring $$\displaystyle R[[X_1, ... \ ... , X_n]]$$ of formal power series in indeterminates $$\displaystyle X_1, ... \ ... , X_n$$ with coefficients in R is again a quasi-local ring, and that its maximal ideal is generated by $$\displaystyle M \cup \{ X_1, ... \ ... , X_n \}$$"
Can you help me to see the connection and progress the problem a little ... would be grateful for some help ...
Peter
#### Deveno
##### Well-known member
MHB Math Scholar
Outline of the entire string of beads strung together:
1) Prove that if an ideal of a ring is such that it contains every non-unit, and no units, it is the unique maximal ideal of the ring, and therefore that the ring is quasi-local.
2) Given a quasi-local ring $R$, it possesses a unique maximal ideal $M$.
3) The units of $R[[x_1]]$ are precisely the power series with unit constant terms, that is to say, the constant term is in $R - M$ (the set difference). Hence the non-units of $R[[x_1]]$ are all power series with constant terms in $M$. These non-units form an ideal, which is equal to the ideal generated by $M$ and the set $\{x_1\}$. By (1), this is a unique maximal ideal, so $R[[x_1]]$ is quasi-local.
4) Since $R[[x_1,x_2]] \cong (R[[x_1]])[[x_2]]$, letting $S = R[[x_1]]$, we have from (1) through (3), that $S$ is quasi-local, so $S[[x_2]]$ is quasi-local, with unique maximal ideal generated by:
$(M \cup \{x_1\}) \cup \{x_2\} = M \cup \{x_1,x_2\}$
By induction on $k$, we see that:
$R[[x_1,\dots,x_k]]$ is quasi-local for any positive integer $k$ (mimicking the steps taken going from (3) to (4)).
Taking $k = n$, we are finished.
Credit where credit is due: Turgul's observations are really the key to understanding this...when you understand the essence of what he was saying, my proof is just "formal machinery".
#### Peter
##### Well-known member
MHB Site Helper
Outline of the entire string of beads strung together:
1) Prove that if an ideal of a ring is such that it contains every non-unit, and no units, it is the unique maximal ideal of the ring, and therefore that the ring is quasi-local.
2) Given a quasi-local ring $R$, it possesses a unique maximal ideal $M$.
3) The units of $R[[x_1]]$ are precisely the power series with unit constant terms, that is to say, the constant term is in $R - M$ (the set difference). Hence the non-units of $R[[x_1]]$ are all power series with constant terms in $M$. These non-units form an ideal, which is equal to the ideal generated by $M$ and the set $\{x_1\}$. By (1), this is a unique maximal ideal, so $R[[x_1]]$ is quasi-local.
4) Since $R[[x_1,x_2]] \cong (R[[x_1]])[[x_2]]$, letting $S = R[[x_1]]$, we have from (1) through (3), that $S$ is quasi-local, so $S[[x_2]]$ is quasi-local, with unique maximal ideal generated by:
$(M \cup \{x_1\}) \cup \{x_2\} = M \cup \{x_1,x_2\}$
By induction on $k$, we see that:
$R[[x_1,\dots,x_k]]$ is quasi-local for any positive integer $k$ (mimicking the steps taken going from (3) to (4)).
Taking $k = n$, we are finished.
Credit where credit is due: Turgul's observations are really the key to understanding this...when you understand the essence of what he was saying, my proof is just "formal machinery".
Hi Deveno,
You and Turgul have made this exercise a great learning experience for me
Thanks to you both!
Peter
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2021-09-26 22:40:31
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# A milkman mixes 20 litres of water with 80 litres of milk.
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A milkman mixes 20 litres of water with 80 litres of milk. [#permalink]
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01 Nov 2018, 05:44
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A milkman mixes 20 litres of water with 80 litres of milk. After selling one-fourth of this mixture, he adds water to replenish the quantity that he had sold. What is the current proportion of water to milk?
a. 2:3
b. 1:2
c. 1:3
d. 3:4
e. 1:1
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Re: A milkman mixes 20 litres of water with 80 litres of milk. [#permalink]
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01 Nov 2018, 06:03
Total quantity = 100 litres. the ratio of water to milk = 1:4
25% is sold.
remaining = 75 litres.
the quantity of water remaining = 75/5 = 15 litres.
quantity of milk remaining = 75*4/5 = 60 litres.
Now 25 liters of water is added.
so 40 liters of water and 60 liters of milk.
The ratio becomes $$2:3$$
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Re: A milkman mixes 20 litres of water with 80 litres of milk. [#permalink]
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01 Nov 2018, 06:44
RhythmGMAT wrote:
A milkman mixes 20 litres of water with 80 litres of milk. After selling one-fourth of this mixture, he adds water to replenish the quantity that he had sold. What is the current proportion of water to milk?
a. 2:3
b. 1:2
c. 1:3
d. 3:4
e. 1:1
Total Quantity =100 litres
Water =20 litres, Milk =80 litres
W/M=1/4
W/(W+M)= 1/5
He sold 1/4th of the mixture means 25 litres of the mixture.
here W+M = 25 Litres
W= 5 litres , M = 20 Litres
so the remaining quantity in the mixture
W= 15 litres and M= 60 Litres
As he adds 25 litres of water in the mixture
W= 15+25= 40 Litres
M= 60 Litres
W/M=40/60
W/M=2/3
A:)
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Joined: 21 Jun 2018
Posts: 32
A milkman mixes 20 litres of water with 80 litres of milk. [#permalink]
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02 Nov 2018, 03:11
RhythmGMAT wrote:
A milkman mixes 20 litres of water with 80 litres of milk. After selling one-fourth of this mixture, he adds water to replenish the quantity that he had sold. What is the current proportion of water to milk?
a. 2:3
b. 1:2
c. 1:3
d. 3:4
e. 1:1
Initial ratio = 2:8
After 1/4th is sold, quantity left is 75litres
Water in mixture = 2/10 * 75 = 15litres
Milk in mixture = 8/10 *75 = 60 litres
new ratio = [15 litres + 25 litres (qty sold) ] : 60 litres
After simplification , answer is 2:3
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A milkman mixes 20 litres of water with 80 litres of milk. &nbs [#permalink] 02 Nov 2018, 03:11
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2018-11-21 20:14:41
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https://www.jiskha.com/questions/1590636/a-ferris-wheel-is-elevated-1-meter-above-ground-when-car-reaches-the-highest-point-of-the
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# Pre-Calculus
a ferris wheel is elevated 1 meter above ground. When car reaches the highest point of the Ferris wheel, it's altitude from ground level is 31 metres. How far away from the center, horizontally, is the car when it is at an altitude of 25 meters?
1. 👍 2
2. 👎 0
3. 👁 5,076
1. so, the radius of the wheel is 15 m
...(31 - 1) / 2
when the altitude is 25 m, the car is 9 m above the center
... 25 - 15 - 1
using Pythagoras,
... hypotenuse (radius) is 15 m
... vertical side is 9 m
... horizontal side is ?
HINT (a big one)
... it's a Pythagorean triple, 3-4-5
1. 👍 4
2. 👎 0
2. x2 + (y - 16)2 = 225
x2 + (25 + 16) = 225
x2 + 81 = 225
x2 = 225 - 81
x2 = 144 -> squared 144 to illiminate the square in
x2
so.. x = 12
1. 👍 8
2. 👎 1
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2020-11-25 08:15:19
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Conformal Dimension: Theory and Application
John M. Mackay University of Illinois at Urbana-Champaign, Urbana, IL
Jeremy T. Tyson University of Illinois at Urbana-Champaign, Urbana, IL
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AMS Member Price: $56.40 Click above image for expanded view Conformal Dimension: Theory and Application John M. Mackay University of Illinois at Urbana-Champaign, Urbana, IL Jeremy T. Tyson University of Illinois at Urbana-Champaign, Urbana, IL Available Formats: Softcover ISBN: 978-0-8218-5229-3 Product Code: ULECT/54 List Price:$47.00 MAA Member Price: $42.30 AMS Member Price:$37.60
Electronic ISBN: 978-1-4704-1649-2 Product Code: ULECT/54.E
List Price: $44.00 MAA Member Price:$39.60 AMS Member Price: $35.20 Bundle Print and Electronic Formats and Save! This product is available for purchase as a bundle. Purchasing as a bundle enables you to save on the electronic version. List Price:$70.50 MAA Member Price: $63.45 AMS Member Price:$56.40
• Book Details
University Lecture Series
Volume: 542010; 143 pp
MSC: Primary 30; 28;
Conformal dimension measures the extent to which the Hausdorff dimension of a metric space can be lowered by quasisymmetric deformations. Introduced by Pansu in 1989, this concept has proved extremely fruitful in a diverse range of areas, including geometric function theory, conformal dynamics, and geometric group theory.
This survey leads the reader from the definitions and basic theory through to active research applications in geometric function theory, Gromov hyperbolic geometry, and the dynamics of rational maps, amongst other areas. It reviews the theory of dimension in metric spaces and of deformations of metric spaces. It summarizes the basic tools for estimating conformal dimension and illustrates their application to concrete problems of independent interest. Numerous examples and proofs are provided.
Working from basic definitions through to current research areas, this book can be used as a guide for graduate students interested in this field, or as a helpful survey for experts. Background needed for a potential reader of the book consists of a working knowledge of real and complex analysis on the level of first- and second-year graduate courses.
Graduate students and research mathematicians interested in geometric function theory.
• Chapters
• Chapter 1. Background material
• Chapter 2. Conformal gauges and conformal dimension
• Chapter 3. Gromov hyperbolic groups and spaces and their boundaries
• Chapter 4. Lower bounds for conformal dimension
• Chapter 5. Sets and spaces of conformal dimension zero
• Chapter 6. Gromov–Hausdorff tangent spaces and conformal dimension
• Chapter 7. Ahlfors regular conformal dimension
• Chapter 8. Global quasiconformal dimension
• Requests
Review Copy – for reviewers who would like to review an AMS book
Permission – for use of book, eBook, or Journal content
Accessibility – to request an alternate format of an AMS title
Volume: 542010; 143 pp
MSC: Primary 30; 28;
Conformal dimension measures the extent to which the Hausdorff dimension of a metric space can be lowered by quasisymmetric deformations. Introduced by Pansu in 1989, this concept has proved extremely fruitful in a diverse range of areas, including geometric function theory, conformal dynamics, and geometric group theory.
This survey leads the reader from the definitions and basic theory through to active research applications in geometric function theory, Gromov hyperbolic geometry, and the dynamics of rational maps, amongst other areas. It reviews the theory of dimension in metric spaces and of deformations of metric spaces. It summarizes the basic tools for estimating conformal dimension and illustrates their application to concrete problems of independent interest. Numerous examples and proofs are provided.
Working from basic definitions through to current research areas, this book can be used as a guide for graduate students interested in this field, or as a helpful survey for experts. Background needed for a potential reader of the book consists of a working knowledge of real and complex analysis on the level of first- and second-year graduate courses.
Graduate students and research mathematicians interested in geometric function theory.
• Chapters
• Chapter 1. Background material
• Chapter 2. Conformal gauges and conformal dimension
• Chapter 3. Gromov hyperbolic groups and spaces and their boundaries
• Chapter 4. Lower bounds for conformal dimension
• Chapter 5. Sets and spaces of conformal dimension zero
• Chapter 6. Gromov–Hausdorff tangent spaces and conformal dimension
• Chapter 7. Ahlfors regular conformal dimension
• Chapter 8. Global quasiconformal dimension
Review Copy – for reviewers who would like to review an AMS book
Permission – for use of book, eBook, or Journal content
Accessibility – to request an alternate format of an AMS title
Please select which format for which you are requesting permissions.
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2023-03-30 07:36:56
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https://www.ncatlab.org/nlab/show/deduction
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Contents
Idea
In logic, a deduction or derivation is a part of a proof, drawing a conclusion from certain premises. See:
Properties
Last revised on January 28, 2015 at 17:41:06. See the history of this page for a list of all contributions to it.
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2019-10-15 11:19:16
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http://www.oalib.com/relative/2208374
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Home OALib Journal OALib PrePrints Submit Ranking News My Lib FAQ About Us Follow Us+
Title Keywords Abstract Author All
Search Results: 1 - 10 of 100 matches for " "
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Electronic Journal of Differential Equations , 2006, Abstract: In this paper we study the existence of global weak solutions for a hyperbolic differential inclusion with a discontinuous and nonlinear multi-valued term. Also we investigate the asymptotic behavior of solutions.
Guo Xingming International Journal of Mathematics and Mathematical Sciences , 2001, DOI: 10.1155/s0161171201003994 Abstract: A mixed-typed differential inclusion with a weakly continuous nonlinear term and a nonmonotone discontinuous nonlinear multi-valued term is studied, and the existence and decay of solutions are established.
Mathematics , 2014, Abstract: In this work we construct reliable a posteriori estimates for some discontinuous Galerkin schemes applied to nonlinear systems of hyperbolic conservation laws. We make use of appropriate reconstructions of the discrete solution together with the relative entropy stability framework. The methodology we use is quite general and allows for a posteriori control of discontinuous Galerkin schemes with standard flux choices which appear in the approximation of conservation laws. In addition to the analysis, we conduct some numerical benchmarking to test the robustness of the resultant estimator.
Electronic Journal of Differential Equations , 2003, Abstract: In this paper we study the existence of generalized solutions for a hyperbolic system with a discontinuous multi-valued term and nonlinear second-order damping terms on the boundary.
Mathematics , 2015, Abstract: The paper proposes a scheme by combining the Runge-Kutta discontinuous Galerkin method with a {\delta}-mapping algorithm for solving hyperbolic conservation laws with discontinuous fluxes. This hybrid scheme is particularly applied to nonlinear elasticity in heterogeneous media and multi-class traffic flow with inhomogeneous road conditions. Numerical examples indicate the scheme's efficiency in resolving complex waves of the two systems. Moreover, the discussion implies that the so-called {\delta}-mapping algorithm can also be combined with any other classical methods for solving similar problems in general.
Piotr Kalita Mathematics , 2014, Abstract: In this paper we show the convergence of a semidiscrete time stepping \theta-scheme on a time grid of variable length to the solution of parabolic operator di?erential inclusion in the framework of evolution triple. The multifunction is assumed to be strong-weak upper-semicontinuous and to have nonempty, closed and convex values, while the quasilinear operator present in the problem is required to be pseudomonotone, coercive and satisfy the appropriate growth condition. The convergence of piecewise constant and piecewise linear interpolants constructed on the solutions of time discrete problems is shown. Under an additional assumption on the sequence of time grids and regularity of quasilinear operator strong convergence results are obtained.
Mathematics , 2011, Abstract: Inclusion relations of metric balls defined by the hyperbolic, the quasihyperbolic, the $j$-metric and the chordal metric will be studied. The hyperbolic metric, the quasihyperbolic metric and the $j$-metric are considered in the unit ball.
Mathematics , 2013, Abstract: In this paper, we study superconvergence properties of the discontinuous Galerkin (DG) method for one-dimensional linear hyperbolic equation when upwind fluxes are used. We prove, for any polynomial degree $k$, the $2k+1$th (or $2k+1/2$th) superconvergence rate of the DG approximation at the downwind points and for the domain average under quasi-uniform meshes and some suitable initial discretization. Moreover, we prove that the derivative approximation of the DG solution is superconvergent with a rate $k+1$ at all interior left Radau points. All theoretical finding are confirmed by numerical experiments.
Mathematics , 2007, DOI: 10.1007/s002050050145 Abstract: This paper is concerned with the initial-boundary value problem for a nonlinear hyperbolic system of conservation laws. We study the boundary layers that may arise in approximations of entropy discontinuous solutions. We consider both the vanishing viscosity method and finite difference schemes (Lax-Friedrichs type schemes, Godunov scheme). We demonstrate that different regularization methods generate different boundary layers. Hence, the boundary condition can be formulated only if an approximation scheme is selected first. Assuming solely uniform L\infty bounds on the approximate solutions and so dealing with L\infty solutions, we derive several entropy inequalities satisfied by the boundary layer in each case under consideration. A Young measure is introduced to describe the boundary trace. When a uniform bound on the total variation is available, the boundary Young measure reduces to a Dirac mass. Form the above analysis, we deduce several formulations for the boundary condition which apply whether the boundary is characteristic or not. Each formulation is based a set of admissible boundary values, following Dubois and LeFloch's terminology in Boundary conditions for nonlinear hyperbolic systems of conservation laws'', J. Diff. Equa. 71 (1988), 93--122. The local structure of those sets and the well-posedness of the corresponding initial-boundary value problem are investigated. The results are illustrated with convex and nonconvex conservation laws and examples from continuum mechanics.
Electronic Journal of Differential Equations , 2007, Abstract: This paper considers a hyperbolic system with discontinuous coefficients in a bounded, open, connected set with smooth boundary and controlled through the Robin boundary condition. Uniform stabilization of the solutions are established. Exact boundary controllability is obtained through the Russell's "Controllability via Stabilizability" principle.
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2019-10-15 07:01:48
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https://socratic.org/questions/how-to-do-i-solve-this-problem-3x-2-12-0-x
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# How to do I solve this problem? 3x^2-12=0
## I can't remember how to do these and just need step by step so I can finish the problem. This is only a portion of my calculus problem but I forget how to do this part. The whole problem is f(x) = x^3-12x+17 find the relative extreme points of the function if they exist. I did the first part and found the derivatives of the problem but now I am stuck at the next step which is what I posted for my question. It's been years since I took algebra so I forget. Thanks!
Feb 3, 2017
$x = - 2 , x = 2$
#### Explanation:
You need to factor $3 {x}^{2} - 12$, which ends up being
$3 \left(x + 2\right) \left(x - 2\right)$
and then set each of these (but ignore the $3$) equal to zero and solve.
$x + 2 = 0$
$x + 2 - 2 = 0 - 2$
$x = - 2$
and
$x - 2 = 0$
$x - 2 + 2 = 0 + 2$
$x = 2$
So $x = 2 , - 2$
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2021-06-16 08:15:35
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http://nrich.maths.org/274/note
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### Baby Circle
A small circle fits between two touching circles so that all three circles touch each other and have a common tangent? What is the exact radius of the smallest circle?
### Absurdity Again
What is the value of the integers a and b where sqrt(8-4sqrt3) = sqrt a - sqrt b?
### Strange Rectangle 2
Find the exact values of some trig. ratios from this rectangle in which a cyclic quadrilateral cuts off four right angled triangles.
# Ab Surd Ity
##### Stage: 5 Challenge Level:
Why do this problem?
The problem provides good practice in the manipuation of surds and in algebra (involving the expansions of $(p+q)^2$ and $(p+q)^3$, the difference of two squares and the use of the Remainder Theorem to factorise a cubic equation). If care is taken with the algebra the result comes out in a satisfyingly neat way. The question looks complicated but it turns out to be simple.
Possible approach
Although this is a longer, two part, question, the Hint gives sufficient guidance for this to be set to a class to work on independently.
Key questions
If 'extra' solutions are introduced by squaring or cubing, how do you decide which are the correct solutions?
This is how the problem was used by Peter Thomas, a Sixth Form College teacher:
I was absent at a meeting and set the class work to consolidate topics taught the previous lesson. The work was routine exercises from a textbook (Emanuel and Wood) which I encouraged them to approach selectively (what I called 'bread and butter' with some specific questions as a 'doggy bag' for homework).
Alongside this I set them the four nrich problems as 'cake' with the instruction to tackle at least one.
Ab Surd Ity (this problem)
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2016-10-25 12:11:17
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http://aimsciences.org/article/doi/10.3934/dcdsb.2018067?viewType=html
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# American Institue of Mathematical Sciences
## The modified Camassa-Holm equation in Lagrangian coordinates
1 Department of Mathematics of Harbin Institute of Technology, Harbin, 150001, China 2 Department of Physics and Department of Mathematics, Duke University, Durham, NC 27708, USA
* Corresponding author: Yu Gao
Received June 2017 Revised September 2017 Published January 2018
Fund Project: The second author is supported by KI-Net NSF RNMS (Grant No. 1107444) and NSF DMS (Grant No. 1514826)
In this paper, we study the modified Camassa-Holm (mCH) equation in Lagrangian coordinates. For some initial data $m_0$, we show that classical solutions to this equation blow up in finite time $T_{max}$. Before $T_{max}$, existence and uniqueness of classical solutions are established. Lifespan for classical solutions is obtained: $T_{max}≥ \frac{1}{||m_0||_{L^∞}||m_0||_{L^1}}.$ And there is a unique solution $X(ξ, t)$ to the Lagrange dynamics which is a strictly monotonic function of $ξ$ for any $t∈[0, T_{max})$: $X_ξ(·, t)>0$. As $t$ approaching $T_{max}$, we prove that the classical solution $m(·, t)$ in Eulerian coordinates has a unique limit $m(·, T_{max})$ in Radon measure space and there is a point $ξ_0$ such that $X_ξ(ξ_0, T_{max}) = 0$ which means $T_{max}$ is an onset time of collisions of characteristics. We also show that in some cases peakons are formed at $T_{max}$. After $T_{max}$, we regularize the Lagrange dynamics to prove global existence of weak solutions $m$ in Radon measure space.
Citation: Yu Gao, Jian-Guo Liu. The modified Camassa-Holm equation in Lagrangian coordinates. Discrete & Continuous Dynamical Systems - B, doi: 10.3934/dcdsb.2018067
##### References:
[1] J. T. Beale, T. Kato, A. Majda, Remarks on the breakdown of smooth solutions for the 3-d euler equations, Comm. Math. Phys., 94 (1984), 61-66. doi: 10.1007/BF01212349. [2] L. Brandolese, Local-in-space criteria for blowup in shallow water and dispersive rod equations, Comm. Math. Phys., 330 (2014), 401-414. doi: 10.1007/s00220-014-1958-4. [3] A. Bressan, Hyperbolic Systems of Conservation Laws: The One-Dimensional Cauchy Problem, volume 20. Oxford University Press on Demand, 2000. [4] A. Bressan, A. Constantin, Global conservative solutions of the Camassa-Holm equation, Arch. Ration. Mech. Anal., 183 (2007), 215-239. doi: 10.1007/s00205-006-0010-z. [5] A. Bressan, A. Constantin, Global dissipative solutions of the Camassa-Holm equation, Anal. Appl., 5 (2007), 1-27. doi: 10.1142/S0219530507000857. [6] R. Camassa, D. D. Holm, An integrable shallow water equation with peaked solitons, Phys. Rev. Lett., 71 (1993), 1661-1664. doi: 10.1103/PhysRevLett.71.1661. [7] R. Camassa, J. Huang, L. Lee, Integral and integrable algorithms for a nonlinear shallow-water wave equation, J. Comput. Phys., 216 (2006), 547-572. doi: 10.1016/j.jcp.2005.12.013. [8] R. M. Chen, Y. Liu, C. Qu, S. Zhang, Oscillation-induced blow-up to the modified Camassa-Holm equation with linear dispersion, Adv. Math., 272 (2015), 225-251. doi: 10.1016/j.aim.2014.12.003. [9] A. Chertock, J.-G. Liu, T. Pendleton, Convergence of a particle method and global weak solutions of a family of evolutionary PDEs, SIAM J. Numer. Anal., 50 (2012), 1-21. doi: 10.1137/110831386. [10] A. Constantin, Existence of permanent and breaking waves for a shallow water equation: A geometric approach, In Ann. Inst. Fourier, 50 (2000), 321-362. doi: 10.5802/aif.1757. [11] A. Constantin, J. Escher, Wave breaking for nonlinear nonlocal shallow water equations, Acta. Math., 181 (1998), 229-243. doi: 10.1007/BF02392586. [12] A. Constantin, L. Molinet, Global weak solutions for a shallow water equation, Comm. Math. Phys., 211 (2000), 45-61. doi: 10.1007/s002200050801. [13] R. Danchin, A few remarks on the Camassa-Holm equation, Diff. Int. Eq., 14 (2001), 953-988. [14] A. S. Fokas, The Korteweg-de Vries equation and beyond, Acta Appl. Math., 39 (1995), 295-305. doi: 10.1007/BF00994638. [15] Y. Fu, G. Gui, Y. Liu, C. Qu, On the cauchy problem for the integrable modified Camassa-Holm equation with cubic nonlinearity, J. Differential Equations, 255 (2013), 1905-1938. doi: 10.1016/j.jde.2013.05.024. [16] B. Fuchssteiner, Some tricks from the symmetry-toolbox for nonlinear equations: generalizations of the Camassa-Holm equation, Phys. D, 95 (1996), 229-243. doi: 10.1016/0167-2789(96)00048-6. [17] Y. Gao, J.-G. Liu, Global convergence of a sticky particle method for the modified Camassa-Holm equation, SIAM J. Math. Anal., 49 (2017), 1267-1294. doi: 10.1137/16M1102069. [18] G. Gui, Y. Liu, P. J. Olver, C. Qu, Wave-breaking and peakons for a modified Camassa-Holm equation, Comm. Math. Phys., 319 (2013), 731-759. doi: 10.1007/s00220-012-1566-0. [19] H. Holden, X. Raynaud, Convergence of a finite difference scheme for the Camassa-Holm equation, SIAM J. Numer. Anal., 44 (2006), 1655-1680. doi: 10.1137/040611975. [20] H. Holden, X. Raynaud, Global conservative solutions of the Camassa-Holm equation-a lagrangian point of view, Comm. Partial Differential Equations, 32 (2007), 1511-1549. doi: 10.1080/03605300601088674. [21] A.A. Himonas, G. Misiolek, G. Ponce, Y. Zhou, Persistence properties and unique continuation of solutions of the Camassa-Holm equation, Comm. Math. Phys., 271 (2007), 511-522. doi: 10.1007/s00220-006-0172-4. [22] Z. Jiang, L. Ni, Y. Zhou, Wave breaking of the Camassa-Holm equation, J. Nonlinear Sci., 22 (2012), 235-245. doi: 10.1007/s00332-011-9115-0. [23] G. Leoni, A First Course in Sobolev Spaces, volume 105, American Mathematical Society Providence, RI, 2009. [24] Y. Liu, P. J. Olver, C. Qu, S. Zhang, On the blow-up of solutions to the integrable modified Camassa-Holm equation, Anal. Appl., 12 (2014), 355-368. doi: 10.1142/S0219530514500274. [25] H. P. McKean, Breakdown of a shallow water equation, Asian J. Math., 2 (1998), 867-874. doi: 10.4310/AJM.1998.v2.n4.a10. [26] L. Molinet, On well-posedness results for Camassa-Holm equation on the line: A survey, J. Nonlinear Math. Phy., 11 (2013), 521-533. doi: 10.2991/jnmp.2004.11.4.8. [27] L. Ni, Y. Zhou, Well-posedness and persistence properties for the Novikov equation, J. Differential Equations, 250 (2011), 3002-3021. doi: 10.1016/j.jde.2011.01.030. [28] P. J. Olver, P. Rosenau, Tri-hamiltonian duality between solitons and solitary-wave solutions having compact support, Phys. Rev. E, 53 (1996), 1900-1906. doi: 10.1103/PhysRevE.53.1900. [29] Z. Qiao, A new integrable equation with cuspons and w/m-shape-peaks solitons J. Math. Phys. , 47 (2006), 112701, 9 pp. doi: 10.1063/1.2365758. [30] C. Villani, Topics in Optimal Transportation, Number 58. American Mathematical Soc., 2003. [31] Z. Xin, P. Zhang, On the weak solutions to a shallow water equation, Comm. Pure. Appl. Math., 53 (2000), 1411-1433. doi: 10.1002/1097-0312(200011)53:11<1411::AID-CPA4>3.0.CO;2-5. [32] Z. Xin, P. Zhang, On the uniqueness and large time behavior of the weak solutions to a shallow water equation, Comm. Partial Differential Equations, 27 (2002), 1815-1844. doi: 10.1081/PDE-120016129. [33] Q. Zhang, Global wellposedness of cubic Camassa-Holm equations, Nonlinear Anal., 133 (2016), 61-73. doi: 10.1016/j.na.2015.11.020. [34] Y. Zhou, Blow-up of solutions to the DGH equation, J. Funct. Anal., 250 (2007), 227-248. doi: 10.1016/j.jfa.2007.04.019. [35] Y. Zhou, Z. Guo, Blow up and propagation speed of solutions to the DGH equation, Discrete Contin. Dyn. Syst. Ser. B, 12 (2009), 657-670. doi: 10.3934/dcdsb.2009.12.657.
show all references
##### References:
[1] J. T. Beale, T. Kato, A. Majda, Remarks on the breakdown of smooth solutions for the 3-d euler equations, Comm. Math. Phys., 94 (1984), 61-66. doi: 10.1007/BF01212349. [2] L. Brandolese, Local-in-space criteria for blowup in shallow water and dispersive rod equations, Comm. Math. Phys., 330 (2014), 401-414. doi: 10.1007/s00220-014-1958-4. [3] A. Bressan, Hyperbolic Systems of Conservation Laws: The One-Dimensional Cauchy Problem, volume 20. Oxford University Press on Demand, 2000. [4] A. Bressan, A. Constantin, Global conservative solutions of the Camassa-Holm equation, Arch. Ration. Mech. Anal., 183 (2007), 215-239. doi: 10.1007/s00205-006-0010-z. [5] A. Bressan, A. Constantin, Global dissipative solutions of the Camassa-Holm equation, Anal. Appl., 5 (2007), 1-27. doi: 10.1142/S0219530507000857. [6] R. Camassa, D. D. Holm, An integrable shallow water equation with peaked solitons, Phys. Rev. Lett., 71 (1993), 1661-1664. doi: 10.1103/PhysRevLett.71.1661. [7] R. Camassa, J. Huang, L. Lee, Integral and integrable algorithms for a nonlinear shallow-water wave equation, J. Comput. Phys., 216 (2006), 547-572. doi: 10.1016/j.jcp.2005.12.013. [8] R. M. Chen, Y. Liu, C. Qu, S. Zhang, Oscillation-induced blow-up to the modified Camassa-Holm equation with linear dispersion, Adv. Math., 272 (2015), 225-251. doi: 10.1016/j.aim.2014.12.003. [9] A. Chertock, J.-G. Liu, T. Pendleton, Convergence of a particle method and global weak solutions of a family of evolutionary PDEs, SIAM J. Numer. Anal., 50 (2012), 1-21. doi: 10.1137/110831386. [10] A. Constantin, Existence of permanent and breaking waves for a shallow water equation: A geometric approach, In Ann. Inst. Fourier, 50 (2000), 321-362. doi: 10.5802/aif.1757. [11] A. Constantin, J. Escher, Wave breaking for nonlinear nonlocal shallow water equations, Acta. Math., 181 (1998), 229-243. doi: 10.1007/BF02392586. [12] A. Constantin, L. Molinet, Global weak solutions for a shallow water equation, Comm. Math. Phys., 211 (2000), 45-61. doi: 10.1007/s002200050801. [13] R. Danchin, A few remarks on the Camassa-Holm equation, Diff. Int. Eq., 14 (2001), 953-988. [14] A. S. Fokas, The Korteweg-de Vries equation and beyond, Acta Appl. Math., 39 (1995), 295-305. doi: 10.1007/BF00994638. [15] Y. Fu, G. Gui, Y. Liu, C. Qu, On the cauchy problem for the integrable modified Camassa-Holm equation with cubic nonlinearity, J. Differential Equations, 255 (2013), 1905-1938. doi: 10.1016/j.jde.2013.05.024. [16] B. Fuchssteiner, Some tricks from the symmetry-toolbox for nonlinear equations: generalizations of the Camassa-Holm equation, Phys. D, 95 (1996), 229-243. doi: 10.1016/0167-2789(96)00048-6. [17] Y. Gao, J.-G. Liu, Global convergence of a sticky particle method for the modified Camassa-Holm equation, SIAM J. Math. Anal., 49 (2017), 1267-1294. doi: 10.1137/16M1102069. [18] G. Gui, Y. Liu, P. J. Olver, C. Qu, Wave-breaking and peakons for a modified Camassa-Holm equation, Comm. Math. Phys., 319 (2013), 731-759. doi: 10.1007/s00220-012-1566-0. [19] H. Holden, X. Raynaud, Convergence of a finite difference scheme for the Camassa-Holm equation, SIAM J. Numer. Anal., 44 (2006), 1655-1680. doi: 10.1137/040611975. [20] H. Holden, X. Raynaud, Global conservative solutions of the Camassa-Holm equation-a lagrangian point of view, Comm. Partial Differential Equations, 32 (2007), 1511-1549. doi: 10.1080/03605300601088674. [21] A.A. Himonas, G. Misiolek, G. Ponce, Y. Zhou, Persistence properties and unique continuation of solutions of the Camassa-Holm equation, Comm. Math. Phys., 271 (2007), 511-522. doi: 10.1007/s00220-006-0172-4. [22] Z. Jiang, L. Ni, Y. Zhou, Wave breaking of the Camassa-Holm equation, J. Nonlinear Sci., 22 (2012), 235-245. doi: 10.1007/s00332-011-9115-0. [23] G. Leoni, A First Course in Sobolev Spaces, volume 105, American Mathematical Society Providence, RI, 2009. [24] Y. Liu, P. J. Olver, C. Qu, S. Zhang, On the blow-up of solutions to the integrable modified Camassa-Holm equation, Anal. Appl., 12 (2014), 355-368. doi: 10.1142/S0219530514500274. [25] H. P. McKean, Breakdown of a shallow water equation, Asian J. Math., 2 (1998), 867-874. doi: 10.4310/AJM.1998.v2.n4.a10. [26] L. Molinet, On well-posedness results for Camassa-Holm equation on the line: A survey, J. Nonlinear Math. Phy., 11 (2013), 521-533. doi: 10.2991/jnmp.2004.11.4.8. [27] L. Ni, Y. Zhou, Well-posedness and persistence properties for the Novikov equation, J. Differential Equations, 250 (2011), 3002-3021. doi: 10.1016/j.jde.2011.01.030. [28] P. J. Olver, P. Rosenau, Tri-hamiltonian duality between solitons and solitary-wave solutions having compact support, Phys. Rev. E, 53 (1996), 1900-1906. doi: 10.1103/PhysRevE.53.1900. [29] Z. Qiao, A new integrable equation with cuspons and w/m-shape-peaks solitons J. Math. Phys. , 47 (2006), 112701, 9 pp. doi: 10.1063/1.2365758. [30] C. Villani, Topics in Optimal Transportation, Number 58. American Mathematical Soc., 2003. [31] Z. Xin, P. Zhang, On the weak solutions to a shallow water equation, Comm. Pure. Appl. Math., 53 (2000), 1411-1433. doi: 10.1002/1097-0312(200011)53:11<1411::AID-CPA4>3.0.CO;2-5. [32] Z. Xin, P. Zhang, On the uniqueness and large time behavior of the weak solutions to a shallow water equation, Comm. Partial Differential Equations, 27 (2002), 1815-1844. doi: 10.1081/PDE-120016129. [33] Q. Zhang, Global wellposedness of cubic Camassa-Holm equations, Nonlinear Anal., 133 (2016), 61-73. doi: 10.1016/j.na.2015.11.020. [34] Y. Zhou, Blow-up of solutions to the DGH equation, J. Funct. Anal., 250 (2007), 227-248. doi: 10.1016/j.jfa.2007.04.019. [35] Y. Zhou, Z. Guo, Blow up and propagation speed of solutions to the DGH equation, Discrete Contin. Dyn. Syst. Ser. B, 12 (2009), 657-670. doi: 10.3934/dcdsb.2009.12.657.
At $T_{max}$, $X_\xi(\cdot,T_{max})\geq0$ and $X_\xi(\xi,T_{max}) = 0$ for $\xi\in\{\xi_1,\xi_4\}\cup[\xi_{21},\xi_{22}]\cup[\xi_{31},\xi_{32}]$. $F_{T_{max}} = \{x_1,x_2,x_3,x_4\}$ and $\widehat{F}_{T_{max}} = \{x_2,x_3\}$
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2016 Impact Factor: 0.994
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2018-01-22 22:27:18
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http://www.conservapedia.com/Speed
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# Speed
In physics, speed refers to the magnitude of the velocity of an object.
Speed is the total distance traveled divided by the time it takes to travel it[1].
$Speed=\frac{d D}{d t}$
Note that speed is a scalar, therefore it has no direction, and because it is a measure of magnitude, it is never negative.
Like velocity, units for its measurement are meters/second (m/s), and more commonly in informal circumstances kilometers/hour (kph) or (km/h) and miles/hour (mph) or (MPH).
## References
1. Serway and Beichner, Physics for Scientists and Engineers, Fifth Edition
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2014-09-01 21:42:09
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http://www.computer.org/csdl/trans/tc/2008/04/ttc2008040472-abs.html
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Subscribe
Issue No.04 - April (2008 vol.57)
pp: 472-480
ABSTRACT
In this contribution we introduce a low-complexity bit-parallel algorithm for computing square roots over binary extension fields. Our proposed method can be applied for any type of irreducible polynomials. We derive explicit formulae for the space and time complexities associated to the square root operator when working with binary extension fields generated using irreducible trinomials. We show that for those finite fields, it is possible to compute the square root of an arbitrary field element with equal or better hardware efficiency than the one associated to the field squaring operation. Furthermore, a practical application of the square root operator in the domain of field exponentiation computation is presented. It is shown that by using as building blocks squarers, multipliers and square root blocks, a parallel version of the classical square-and-multiply exponentiation algorithm can be obtained. A hardware implementation of that parallel version may provide a speedup of up to 50% percent when compared with the traditional version.
INDEX TERMS
Computations in finite fields, Computer arithmetic, Algorithms
CITATION
Guillermo Morales-Luna, Julio López, "Low-Complexity Bit-Parallel Square Root Computation over GF(2^{m}) for All Trinomials", IEEE Transactions on Computers, vol.57, no. 4, pp. 472-480, April 2008, doi:10.1109/TC.2007.70822
REFERENCES
[1] “IEEE P1363: Standard Specifications for Public Key Cryptography” IEEE Standards documents, Draft Version D18. IEEE, http://grouper.ieee.org/groups1363/, Nov. 2004. [2] J. Daemen and V. Rijmen, “The Design of Rijndael,” AES: The Advance Encryption Standard. Springer-Verlag, 2002. [3] D. Hankerson, A. Menezes, and S. Vanstone, Guide to Elliptic Cryptography. Springer-Verlag, 2004. [4] R. Schroeppel, C. Beaver, R. Gonzales, R. Miller, and T. Draelos, “A Low-Power Design for an Elliptic Curve Digital Signature Chip,” Proc. Fourth Int'l Workshop Cryptographic Hardware and Embedded Systems, B. Kaliski, Ç. Koç, and C. Paar, eds., pp. 366-380, Aug. 2002. [5] G. Orlando and C. Paar, “An Efficient Squaring Architecture for ${\rm GF}(2^{m})$ and Its Applications in Cryptographic Systems,” IEE Electronic Letters, vol. 36, no. 13, pp. 1116-1117, June 2000. [6] D.E. Knuth, The Art of Computer Programming, third ed. Addison-Wesley, 1997. [7] R. Avanzi, “Another Look at Square Roots and Traces (and Quadratic Equations) in Fields of Even Characteristic,” Cryptology ePrint Archive Report 2007/103, http:/eprint.iacr.org/, 2007. [8] M. Scott, “Optimal Irreducible Polynomials for ${\rm GF}(2^{m})$ Arithmetic,” Cryptology ePrint Archive Report 2007/192, http:/eprint.iacr.org/, 2007. [9] D. Hankerson and F. Rodríguez-Henríquez, “Parallel Formulation of Scalar Multiplication on Koblitz Curves,” Report CACR 2007-17, Center for Applied Cryptographic Research, http:/www.cacr.math.uwaterloo.ca/, 2007. [10] F. Rodríguez-Henríquez, G. Morales-Luna, N. Saqib, and N. Cruz-Cortés, “Parallel Itoh-Tsujii Multiplicative Inversion Algorithm for a Special Class of Trinomials,” Cryptology ePrint Archive Report 2006/035, http:/eprint.iacr.org/, 2006. [11] K. Fong, D. Hankerson, J. López, and A. Menezes, “Field Inversion and Point Halving Revisited,” IEEE Trans. Computers, vol. 53, no. 8, pp. 1047-1059, Aug. 2004. [12] R. Dahab, D. Hankerson, F. Hu, M. Long, J. Lopez, and A. Menezes, “Software Multiplication Using Normal Bases,” Technical Report CACR 2004-12, Dept. of Combinatorics and Optimization, Univ. of Waterloo, p. 21, 2004. [13] F. Rodríguez-Henríquez, G. Morales-Luna, and J. López-Hernández, “Low-Complexity Bit-Parallel Square Root Computation over ${\rm GF}(2^{m})$ for All Trinomials,” Cryptology ePrint Archive Report 2006/133, http://eprint.iacr.org/, 2006. [14] A. Menezes, P. van Oorschot, and S. Vanstone, Handbook of Applied Cryptography. CRC Press, Oct. 1996. [15] H. Wu, “Low Complexity Bit-Parallel Finite Field Arithmetic Using Polynomial Basis,” Proc. First Int'l Workshop Cryptographic Hardware and Embedded Systems, Ç. Koç and C. Paar, eds., pp. 280-291, Aug. 1999. [16] H. Wu, “On Complexity of Squaring Using Polynomial Basis in ${\rm GF}(2^{m})$ ,” Proc. Seventh Ann. Int'l Workshop Selected Areas in Cryptography, pp. 118-129, Sept. 2000. [17] Recommended Elliptic Curves for Federal Government Use, special publication, Nat'l Inst. Standards and Tech nology, http://csrc.nist.gov/csrcfedstandards.html , July 1999. [18] R. Schroeppel, “Elliptic Curve Point Ambiguity Resolution Apparatus and Method,” Int'l Application Number PCT/US00/31014, 9 Nov. 2000. [19] E.W. Knudsen, “Elliptic Scalar Multiplication Using Point Halving,” Advances in Cryptology—Proc. Fifth Int'l Conf. Theory and Applications of Cryptology and Information Security, pp. 135-149, 1999.
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2014-09-18 18:04:01
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http://uncyclopedia.wikia.com/wiki/Urza?diff=prev&oldid=5265246
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# Urza
(Difference between revisions)
For those without comedic tastes, the so-called experts at Wikipedia have an article about Urza.
Urza (934 PR - 4232? BS) was a near mythological figure and assumed dictator, until he lost his sanity, when he became famous. He is now buried in a monolithic tomb in Urborg and presumed dead.
## editEarly Life (the first sixty-three years or so)
Urza was born in a city called Penregon, which can be located on a map in Silicon Valley somewhere. He had a younger, far cooler brother named Mishra, who had a talent for doing, while Urza had a reputation for thinking. Unfortunately, Urza's father had AIDS and thus the two brothers needed to be sent into the desert because their psychotic mother Hillary Clinton couldn't handle them.
It was in the desert that both brothers developed a talent for digging and consuming, a somewhat widespread obscure practice in which the participants dig into the desert and eat whatever they can find. Consequently, the sport has caught on in Vermont, but however all they can find are rotten leaves and penguins.
### editDiscovery of the stones
Both Urza and Mishra grew exceptional at this sport (Mishra would later be the creator of the DCSL, which would overtake the NBA in popularity in the 1920s). Mishra got so good that he dug a tunnel that led all the way to the first airplane. Urza immediately got into a heated argument on how an airplane could possibly be buried, but Mishra decided to fly it out anyway. Unfortunately, when engaging the jet engines, the flames encountered the fart of a Montana moose, which caused an implosion that propelled Urza and Mishra all the way to the barren wasteland caves of southern France. Inside the caves, they found a large door which was wedged shut with a giant shiny thing (which was later confirmed to be solidified alcohol). Unfortunately, Urza tripped over his own feet while breathing air and fell onto the shiny thing, breaking it in two with his pointy nose. Mishra immediately grabbed a piece of the thing. This sparked another argument, which ended by both brothers braining each other with their things. This led to two hallucinations caused by alcohol seepage (although Mishra has always accused Urza of being on pot, and Urza has always accused Mishra of being on Skittles).
Later, after hallucinations ended and the two brothers struggled back to the primitive campsite (later known as Phoenix, Arizona), Urza discovered how the stones worked. Seemingly releasing infinite alcohol-related energy (this has been furiously debated by six thousands scholars and eight drunks in the Council of Nicea), Urza's shiny thing made things stronger and Mishra's made things weaker. Urza named his shiny thing the Mightstone (though the official name was the Kick-Supreme-Ass Thing) and named Mishra's the Weakstone (the Not-So-Kick-Supreme-Ass Thing). Mishra was then furious that he wasn't invited to his stone's naming ceremony (he missed the bus the day after) that he attacked Urza. The resulting explosion of alcohol related energy spawned insanity, craziness, the moons of Saturn and Uranus, and Donald Trump, who immediately fired Mishra, exiled him to Mexico, and sent Urza back to Silicon Valley for his safety (nobody liked Urza, but unlike Raistlin Majere, Urza believed the world loved him, whereas Oscar Wilde really didn't care less for an extended period of time starting later).
### editUrza's Marriage
So then Urza spent the next few years wandering about California, Nevada, and Afghanistan, somehow never getting laid or high in the process (thus breaking the world record). Finally, he wound up in a city named Kroog, found in Kentucky (not be confused with Kitchener, which was found in Ontario, but their resemblance is uncanny). There, Urza became a modest clockmaker and builder of IPods, later masterminding the creation of the Cuckoo Clock of Doom. However, one day he met some beautiful girl in his shop, who was the princess of the city. Urza promptly ignored her and continued his public debate with Einstein on how time would not stop if you rode on a light beam, but you would just have a burn streak on your pants and that clock would keep on ticking.
But later, Urza decided to get married to the beautiful girl, who turned out to be Eowyn from Rohan. Her father had declared that she would be wed to the strongest person in Kentucky. After repeated groin pulls by contesters like Godzilla, St. Peter, and Rocky, Urza decided to give the test a try. It was one where one had to move the Statue of Liberty all the way to Kentucky without getting shot by the FBI. Considering Urza had little muscle to speak of, he designed a gigantic, mechanized monkey that carried the statue all the way to Kentucky. Her father (later confirmed to be Dr. Phil) was infuriated, until his adviser Johnny Depp told him Eowyn could marry the statue instead. Instead of discussing the prospect of female-to-stone copulation (which is a current topic before the White House), Dr. Phil communicated with his feelings and decided to let her marry Urza after all. Urza was just happy he didn't have to go on Dr. Phil's new diet.
### editThe Brother's War
But Mishra wasn't finished with Urza just yet. Harnessing the power of his shiny thing, he summoned a mak fawa from the sands (a dragon engine channelling the spirit of Chuck Norris). Driving his fearsome beast through the sands of Alaska, he also discovered the metropolis of Anchorage and his new apprentice, Ashnod (who may or may not be human, have a gender, or drink the blood of hamsters). Urza also had an apprentice who was a buff guy who went under the alias of Tawnos, but was really Ashton Kutcher in disguise. Urza also revitalized the creation of airplanes, creating the company which became WestJet. Unfortunately, Mishra had harnessed the evil powers of Martha Stewart, who proceeded to sue Urza for insider trading, run his business into the ground, and get him fired by Donald Trump (the master of firing things, people, and local landscapes around North Dakota).
And so the war finally started. It was later called, by Urza's wife, the "Brothers War," and it remains perhaps the bloodiest war ever to take place in North America. Several scholars have argued against the existence of the war, but it was proven when Urza showed up and built a factory right over the council building.
The war also created the largest industrial development since the creation of McDonalds. Factories, mines, power plants, and towers were built, and resources were stolen from the earth's bones, flesh, and muscles (thus paralleling the Industrial Revolution). The war got so bad, people didn't even have time to get high or laid. This resulted in the suicide of Hitler, the explosion south-east of Oregon, and the spontaneous implosion of several irregularly spaced objects.
A key part of this war was the gathering of resources. Both Mishra and Urza needed vast amounts of lumber, minerals, chicken, and oil to fuel their artifact creations (the alcohol was supplied by their shiny thing). Considering how Urza's shiny thing could provide alcohol of slightly more potency than Mishra's, he began bribing world leaders with increased supplies of whisky. Unfortunately, Mishra also had access to pot fields, and thus he could blackmail more leaders than Urza (this also explains George Bush's support of Mishra and his willing diversion of all the Texas oil to Mishra's Factories).
#### editArgoth and Phyrexia
Sometime later, Urza's son (whose name was unavailable and forgotten last week) crash-landed on an mysterious, forested island (later identified to be Ukraine), which was called "Argoth" by its inhabitants. Urza was so overjoyed when he found out, he immediately began chartering strip miners (not to be confused with strippers) to level the area, driving out the native inhabitants. This great decision, later ingrained in the US Constitution, proved to be the beginning of the end of the world (as a mysterious author quoted when writing Urza's history, which was promptly set on fire).
Mishra, meanwhile, had discovered a dark new land he called "Phyrexia". This particular area is a hellish wasteland where flesh and metal are amalgamated in grotesque proportion (in short, it was a combination of California and Florida stored in nine little spheres located slightly west of Hell). Mishra was fascinated by Phyrexia, and even eventually became a Phyrexian (although he did not, however, get a sex change in the process; that's just a urban myth). He met up with a mysterious creature named Gix (who also goes by the aliases Peter Jackson, Donald Rumsfeld, and Jennifer Lopez). Gix told Mishra of the mysterious island, and Mishra sent his own armies there for "the mid-section of the end of the world". Unfortunately, the big boss of Phyrexia thought Gix was being a showoff, so he excoriated him to the negative-seventh floor, where he was nearly consumed by grinding machines (see blender) and he was tortured by being forced to watch endless episodes of Trading Spaces.
Finally, Urza and Mishra met in epic combat (only described as epic because there were no CNN reporters there at the time). The concentration of alcohol related energies oozing out of the shiny things, combined with drunken hallucination memories, an ornamental bowl called the Sylex, and Urza's capitalist instincts, caused a great disaster, which literally leveled the mysterious island, causing Chernobyl.
But Urza transcended, while Mishra ascended into Phyrexia, and Ukraine descended, thus causing "the end of the end of the world." But something far worse happened. The two shiny things replaced Urza's eyes, and the resulting infusion of the purest of pure energy caused a miraculous transformation.
In short, Urza became a planeswalker, a person able to leap from jet to Concorde to space shuttle in the amount of time it takes an ordinary man to get high.
But Urza's sanity was not completely gone yet, and California would regret that later.
## editPlaneswalking the Planes
### editWar on Evil
Planeswalkers happen to be exceptionally powerful, but are only created through explosions, implosions, atomic detonations, and viewing a whole season of Survivor. Urza was no exception to this rule. Planeswalkers also had the unfortunate benefit of being insane, and Urza was likewise. However, his insanity was compounded by the fact that the two shiny things in his head were stimulating his brain with vast amounts of alcohol-related energy. This allowed him to maintain the purest insight and clarity of vision, although it did cause him to pass out every now and then.
Another unfortunate characteristic with planeswalkers is that they all had a specific goal. These goals could include anything from becoming President, all-powerful, or consuming three tons of lasagna within ten seconds. Urza's goal, once he purchased it on EBay, was to take revenge for his brother Mishra (which he had killed) and destroy the evil realm slightly west of Hell known as Phyrexia.
Unfortunately, even though Urza was soooo close to being all powerful, he could not find the door to Phyrexia, even after using a Spot check. So he enlisted a renegade Phyrexian (not to be confused with a Las Vegan, although the resemblance, with so much metal and plastic amalgamated with flesh, is uncanny) named Xantcha. Xantcha was a rather odd fellow, who had zero concept of clothing, silverware, string theory, or gender, although she was able to show Urza how to get into Phyrexia. Luckily for Urza, the Phyrexians were so happy to have Urza come and visit, they threw a massive party with dragon engines, slayers, and demonic bombs. Unfortunately, Urza, rather lacking in social graces, immediately tried to become the center of attention by blowing holes in the furnaces, thus waking up the big boss of Phyrexia named Yawgmoth, who was napping five floors down.
### editSerra's Realm
Xantcha, who had been fired from Phyrexia for sleeping with the big boss's wife (although the gender on both sides has been debated, it has been confirmed that this is at least partially lesbian), was so scared she dragged Urza all the way to the local Alcoholics Anonymous Center, perhaps to remove the alcohol-producing stones in his head. This particular center, run by a pretty chick named Serra, allowed Urza to recuperate, recover, rebuild, and refill the local toilets with clockwork mechanisms and other, fouler things (like carrots and cable bills). Unfortunately, Urza's head was so infused with alcohol that he eventually was kicked out, with Xantcha in tow. He then spent the next couple thousand years wandering around the local universes, trying to find his home, a walk-in diner that served onion rings, and Texas. Meanwhile, Serra's Center was invaded by Phyrexians after she told them that only biological beings infused with alcohol were allowed, not mechanical (this caused the Pope to draw the Line of Demarcation, separating what was mechanical and biological; in other words, the Portuguese from the Spanish).
Eventually, something happened called a temporal flux. Whatever the hell that is has not yet been decided since yesterday, but during this mysterious flux, Urza and Xantcha managed to find their way back to Dominaria. Unfortunately, both were taking a shower at the time, and when they arrived in the middle of the main city {see London), they were both naked and were laughed all the way to Pluto. Urza, who was somewhat incensed at this, took the name Rainz and assassinated Lord British, thus getting banned from all MMORPGs for the rest of his life (although Urza was so smashed at that time, he didn't really care)
## editUrza Goes Crazy (more than normal)
During this following period, after Urza returned to Dominaria, he fell into a sedentary life-style, lured on by pornography, communism, and Mr. Clean commercials. The planeswalker abandoned his habit from leaping from Cessnas to Concordes in favour of playing with little action figures (not Barbie dolls, this occurred later in Urza's life) trying to attract 'motes of time' to them. As Xantcha, fuming from the nose at this inane development tried to find some way to occupy her (living out hie/her female part) self, she met a strange fellow named Ratepe (pronounced Rat-A-PEE, thus a prophetic name, since all nearby rodents wet themselves when he was around). Freeing Ratepe from his enslavement to Waldo and the Iranian government, Xantcha convinced the poor boy that he was in fact Mishra. Urza, being rather confused at the time as well, agreed with this.
### editDefeat of Gix
Unfortunately, Phyrexians invaded Urza's quaint homestead in Kansas, led by the demon Gix, who had somehow escaped the negative-seventh floor (which was also a sphere). The final battle came down by the same doorway that Urza and Mishra first opened millennia earlier. Unfortunately, it ended in a similar fashion. Urza tripped over his own feet and Gix caught him in his arms, trying to pull the planeswalker's eyes out. This descended into a rather gory scene that is not suited for R-rated audiences, but suffice to say that Ratepe and Xantcha were furious they weren't part of the experience, so they kicked Gix in the ass and pushed him all the way through the doorway back into Phyrexia. In some sort of mystical ritual, the two made out at the same time as well. Urza, disgruntled that he couldn't join them, popped the alcohol-energy-emitting shiny things that were his eyes back and in and imploded Xantcha, Gix, and Ratepe for the fun of it. Unfortunately, the planeswalker forgot that implosion tends not to be survivable for the majority of people, except for God, Michael Jackson, most strippers and people who work for Revenue Canada. Only one thing survived of the three imploded corpses - a battery that had been embedded in Xantcha's thigh that she called her heart (this has baffled scholars for decades, as the heart is found either the anus or behind the jugular vein, not in the thigh). Somehow, this battery was still running, and thus Urza, claiming this a marvelous discovery, created the 'Duracell' company of batteries, which, while being somewhat obscure, did not damage his fortunes that badly. However, this particular battery had also been infused with Xantcha's life force, which confounded Urza to the highest degree. He resolved his feelings by nuking a large, rather vacant landscape in Nevada.
Somehow, through all of this, Urza had created one of his plans. He decided that he would build a time machine, go back in time nine thousand or so years, and stop the Phyrexians. Thus, when he would return to the present, none would ever exist and his brother, who he thought was destroyed by Phyrexia, would still be alive (one fails to mention that Urza killed Mishra, not the Phyrexians; although one pointing this out to Urza would be subject to spontaneous implosions).
To harness his powers, Urza decided to build a place where only smart people could live (a novelty on Earth). Transmuting a huge chunk of Saskatchewan land into the middle of the ocean, Urza created the very definition of "the middle of nowhere." Unfortunately, since Urza didn't bother to take any bedrock, just topsoil, the land sunk straight to the bottom of the sea. Undaunted, Urza triggered several volcanic eruptions to create more land by cracking the crust of the world open. He was fairly certain that the planetary ecosystem wouldn't be harmed.
He later said he cast this spell based on an equation that went something like:
$Tolaria(x) = m \cdot c^2 \cdot (rock\;-\;gravel\;-\;|water|\;\cdot\; (-1)\;\cdot\; 10^5)\;+\;\frac{\frac{mana^5}{cheese}}{\log(alcohol)}\;+\;\infty\;\cdot\;energy\;\cdot\;heat$
He later commented that the simplicity of the math is simply remarkable, and that everyone, presuming they have a gram of intellect, could solve for x.
And so Urza began to build an academy of various wizards, sorcerers, and other various beings, all the aspects of studying time and magic. At this point, he met his new second-in-command, Barrin, who served as Urza's connection to reality and sanity. Unfortunately, Barrin was a wizard himself and also ended up committing suicide a thousand years later while burninating the entire island of Tolaria (that spell did not require an equation). Barrin also was a master of all five colors of magic, which proved to be rather useful when Urza kept leaving him to die in various locations. This certainly explains Urza's great compassion for his friends.
### editStudents of the Tolarian Academy
Urza also had some famous students, most of whom he abducted from their homes to work at his academy (thus proving everything said about how school is morally, ethically, painfully, and politically wrong). Many of his students and their chancellors felt as imprisoned as the next fellow, although the next fellow was currently holidaying in Cuba. Many of them died when Urza's time machine imploded, reloaded, and exploded later, thus proving that living on a tiny island with no boats to escape might not be the most intelligent idea.
Urza had several famous students (actually, since Urza did more accidental killing than teaching, they were more like pseudo-friends/enemies/opponents/sandwiches/students). One of them was a young black boy named Teferi. This boy grew up in the tropical hood called Jamuraa, which, of course, is a land infested with alleys, smokers, booze, and crack smugglers, not to mention griffons, phoenixes, rabbits, and other magical monsters. Teferi's later exploits include overthrowing governments, screwing with time and gravity, and spending forty years on fire so he could become a planeswalker like Urza (thus proving all planeswalkers are perfectly sane, in a way).
Another one of his students was a girl named Jhoira, who specialized in building mechanical monsters out of steel, wood, lemons, and sporks. Perhaps the most sane person ever to live on Tolaria, Jhoira was also driven to find a way off the island. She also specialized in normal pursuits like firebreathing, implosions, and warping the laws of physics. She did however view Tolaria as somewhat of a blessing, since the cable bills were so high, she never was exposed to the terror that was Oprah.
And so Urza spent his time trying to build his time machine. He spent the remainder of his time eating, sleeping, debating, building, destroying, shitting, teaching, and criticizing others for wasting time. Barrin promptly called him a hypocrite. Urza just as promptly got angry and immolated Barrin, who promptly teleported into a bathtub and left the scene.
Perhaps Urza's greatest friend was a silver golem named Karn, who Urza built with Xantcha's heart that he found in her thigh. Karn was reportedly going to be the one who went back in time for Urza (which really doesn't make a whole lot of sense). Unfortunately for the planeswalker, Karn was sentient, brilliant, cogitant, patient, and thereby completely not fun to be around. Teferi capitalized on this by promptly naming the silver golem (not be confused with the sliver golem, which the big boss of Phyrexia made) Arty Shovelhead, a truly prophetic and impressive sounding name. Karn found the name mildly annoying, but he was such a bore to be around, Teferi eventually left him alone, leaving only Urza who cared about Karn left (unfortunately, Urza's cares were like his plans: over-complicated, implosive, rather west of Chicago, and covered in lemon juice).
### editThe Time Spiral
But Jhoira, when walking on the beach contemplating how to break the elementary theory of peaches, discovered a castaway named Kerrick (not be confused with Ben Affleck, although the fact both spent extensive time in California is suspicious). Jhoira, tired of dealing with imbeciles like Teferi, fell in love with Kerrick and they reportedly spent the next year making out in a cave. Unfortunately for her, Kerrick was also a Phyrexian sleeper agent, a badass who spent the majority of his time sleeping, the minority of his time wreaking havoc, and the rest making out. Note, since the publication of this article, thousands of people have applied to become Phyrexian sleepers, including assumed asskickers such as Chuck Norris, Darth Vader, Jesus, and the Backstreet Boys.
Unfortunately for Jhoira, her lovemaking was interrupted by Teferi, who was spying on them. Worse still, Teferi decided to be a bit of a jackass and tell Urza, who was so thrilled about the development that he sent Karn to go kill Kerrick. The sneaky bastard (Kerrick, not Teferi, and definitely not Urza) summoned a group of Phyrexian negators to the academy. These monsters, a cross between dragons, ducks, and robots were quite the terrifying beasts and definitely needed to be fought off by Chuck Norris, who was holidaying in Nebraska at the time, and they managed to kill everyone except Urza and Karn. Now a bit desperate, Karn was sent back in time to kill Kerrick before the disaster ever happened. Unfortunately, Urza forgot to send himself back in time, and in the hurry to pack his toothbrush into his suitcase, he accidentally reverse-inverse-perverse-triggered his time machine.
The resulting explosion created several things (explosions do create things besides ash and dust and ugly naked people, you know). The first creation were bubbles of fast time. In these areas, time went faster. Jhoira learned to capitalize on these areas by getting laid for years on end in fast-time, then getting into regular time and discovering that only two hours had passed (that is sooo totally kick-ass).
Another odd phenomenon was slow-time, areas where time was slower inside the bubbles than outside. These areas proved to be a bit annoying for Jhoira, as she would step inside a slow time bubble to relax for a few hours, except then she would leave and discover six weeks had passed. Fortunately the academy had already been destroyed, otherwise the slow-time would have put a serious problem in the school system (thwarting the diabolical process of [homework]).
Teferi, however, was rather close to the time-machine when it exploded. Luckily for him, he got caught in the blast wave, which only set his robes on fire. Seconds later, Teferi ran straight into a bubble of extreme-slow time, which for him only three seconds he was on fire, but instead he was on fire for forty years.
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2014-11-25 00:45:27
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https://myaptitude.in/cat/quant/the-number-of-common-terms-in-the-two-sequences-17-21-25
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The number of common terms in the two sequences 17, 21, 25, ..., 417 and 16, 21, 26, ..., 466 is
1. 19
2. 20
3. 77
4. 78
Total number of terms in the sequence 17, 21, 25 …417 is 101
Total number of terms in the sequence 16, 21, 26 …466 is 91
nth term of the first sequence = 4n + 13
mth term of the second sequence = 5m + 11
4n + 13 = 5m + 11
5m - 4n = 2
Possible integral values of n that satisfy 5m = 2 + 4n are (2, 7, 12 …97)
The total number of terms common in both the sequences is 20.
The correct option is B.
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2020-02-24 17:37:27
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http://the-property-lab.com/topic/extreme-value-distribution-calculator-afe004
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The distributive property of 3(10n+9) is 30n+27. This method is generally referred to as the "Peak Over Threshold" [1] method (POT) and can lead to several or no values being extracted in any given year. The inverse of the Gumbel distribution is of probability distributions. Required fields are marked *. To document the efficacy of that correction and the package as … = 36j + 4 + 24j As you will see below, the differences in the distribution for maximums and the distribution for … = 60j + 4. 0th. generalized extreme value (GEV) distribution The web apps in this project use only the extreme value distributions for maximums. The distributive property is the one which allows us to multiply the number by a group of numbers, which are added together. the same data. What is the distributive property of 3(10n+9). The GEV combines three distributions into a single framework. a(b+c) = ab+ ac The first method relies on deriving block maxima (minima) series as a preliminary step. In each graph the ξ-labeled horizontal axis is for the true value of that parameter while the μ, lnσ, and ξ–labeled vertical axes are for the various measures of estimator accuracy and stability (mean, SD, RMSE). Decimal to Fraction Fraction to Decimal Hexadecimal Scientific Notation Distance Weight Time. These distributions are based on the extreme types theorem, and they are widely used in risk management, finance, economics, material science and other industries. Checking some double-double precision (about 32 decimals) fast code for bugs, sometimes in extreme areas for ratio of cdf to pdf (Mills Ratio). Normal distribution is important in statistics and is often used in the natural and social sciences to represent real-valued random variables whose distributions are not known. BYJU’S online distributive property calculator tool makes the calculations faster and it displays the simplification of numbers in a fraction of seconds. (as required by the Frechet distribution). of long (finite) sequences of random variables. For each distribution and value of ξ, I ran 5000 Monte Carlo simulations, applying four variants of the estimator: The graphs below show, for each distribution, the average bias, standard deviation, and root-mean-square error of the estimators for all parameters. The calculator will find the critical points, local and absolute (global) maxima and minima of the single variable function. We know that the distributive property is given by: One is based on the smallest extreme and the other is based on the largest extreme. m(v-v)=, What is the distributive property of 4(2+m)=. Show Instructions. Therefore, 4 (2+8) is 40. [4] Like the extreme value distribution, the generalized extreme value distribution is often used to model the smallest or largest value among a large set of independent, identically distributed random values representing measurements or observations. Given expression: 4(2+ 8) extval_gev function. Example 6 (below) offers a sample block maxima approach based on extracting the maximum rate of river flow ineach year. NCL has a small number of basic extreme value (EV) and recurrence statistical functions. \(\normalsize Normal\ distribution\ N(x,\mu,\sigma)\\. Conversions. Step 3: Finally, the simplification of the given expression will be displayed in a new window. maximum-liklihood estimates of these parameters for the GEV and Gamma distributions, respectively. Here the maximum river flow rate The only variation between these two is that they have different shapes. In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. can readily be used to derive parameter estimates for other extreme value distributions.
.
Vegetable Garden Layout App, Mall Tycoon Emulator, Logitech 4k Webcam, 9 Year Cycle Calculator, Patriots Color Rush Jersey 2020, New England Colonies Relationship With Natives,
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2022-05-22 22:37:09
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http://www2.warwick.ac.uk/fac/sci/maths/research/events/seminars/areas/combinatorics/
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Combinatorics Seminar
Organisers:
Term 1 2016/17 - Room MS.04
Date Name Title
7 Oct Felix Reidl (Aachen) Characterising structural sparseness by neighbourhood complexity
Or: If your neighbourhood is boring you probably live in a sparse region.
14 Oct David Ellis (QMUL) Some applications of the $p$-biased measure to Erd\H{o}s-Ko-Rado type problems
21 Oct Maryam Sharifzadeh (Warwick) Ramsey-Turan-type of extremal problems
28 Oct Ping Hu (Warwick) Tilings in graphons
4 Nov Katherine Staden (Warwick) The number of triangles in a graph with a given number of vertices and edges
11 Nov Alexey Pokrovskiy (ETH) Rainbow cycles and rainbow expanders
18 Nov Agnes Backhausz (Eötvös Loránd University) On the graph limit approach to eigenvectors of random regular graphs
25 Nov Richard Montgomery (Cambridge) Subdivisions in C_4-free graphs
2 Dec Tim Netzer (Innsbruck) Checking Inclusion of Polytopes and Spectrahedra
9 Dec Will Perkins (Birmingham)
An occupancy approach to bounding graph polynomials
Term 2 2016/17 - Room MS.04
Date Name Title
13 Jan Christoph Koch (Warwick)
Jigsaw percolation on random hypergraphs
20 Jan Mihyun Kang (TU-Graz)
27 Jan Angelika Steger (ETH Zurich)
3 Feb
10 Feb Jakub Sosnovec (Warwick)
17 Feb Amin Coja-Oghlan (Frankfurt)
24 Feb Jakub Konieczny (Oxford)
3 Mar
10 Mar
17 Mar
Term 3 2016/17 - Room MS.04
Date Name Title
28 April
5 May
12 May
19 May
26 May
2 June
9 June
16 June
23 June tbc
30 June tbc
7 Oct Felix Reidl (Aachen) Characterising structural sparseness by neighbourhood complexity
Or: If your neighbourhood is boring you probably live in a sparse region.
Structurally sparse graphs always had a special place in algorithm theory:
almost any problem becomes much more tractable if we restrict ourselves to
nice graph classes like trees, planar graphs, or graphs excluding a minor.
On the other hand, algorithms on such classes are only of limited use
in real-world settings---hardly any real-world network exhibits these stricter
notions of structural sparseness.
In this talk we will explore a notion of sparseness (so-called bounded expansion
classes introduced by Nesetril and Ossona de Mendez) that is general enough
to hold in real-world settings while still providing use with a rich algorithmic
toolkit. In particular, we will see a novel characterisation of bounded expansion
of structurally sparse graphs.
14 Oct: David Ellis (QMUL) Some applications of the $p$-biased measure to Erd\H{o}s-Ko-Rado type problems
Erd\H{o}s-Ko-Rado type problems' are well-studied in extremal combinatorics; they concern the sizes of families of objects in which any two (or any $r$) of the objects in the family 'agree', or 'intersect', in some way.
If $X$ is a finite set, the '$p$-biased measure' on the power-set of $X$ is defined as follows: choose a subset $S$ of $X$ at random by including each element of $X$ independently with probability $p$. If $\mathcal{F}$ is a family of subsets of $X$, one can consider the {\em $p$-biased measure} of $\mathcal{F}$, denoted by $\mu_p(\mathcal{F})$, as a function of $p$. If $\mathcal{F}$ is closed under taking supersets, then this function is an increasing function of $p$. Seminal results of Friedgut and Friedgut-Kalai give criteria under which this function has a 'sharp threshold'. Perhaps surprisingly, a careful analysis of the behaviour of this function also yields some rather strong results in extremal combinatorics which do not explicitly mention the $p$-biased measure - in particular, in the field of Erd\H{o}s-Ko-Rado type problems. We will discuss some of these, including a recent proof of an old conjecture of Frankl that a symmetric three-wise intersecting family of subsets of $\{1,2,\ldots,n\}$ has size $o(2^n)$, and some 'stability' results characterizing the structure of 'large' $t$-intersecting families of $k$–element subsets of $\{1,2,\ldots,n\}$. Based on joint work with (subsets of) Nathan Keller, Noam Lifschitz and Bhargav Narayanan.
21 Oct: Maryam Sharifzadeh (Warwick) Ramsey-Turan-type of extremal problems
Motivated by the fact that the extremal example for Tur\'an theorem has linear-sized independent sets, Erd\H os and S\'os initiated the so-called Ramsey-Tur\'an theory, where they studied the maximum size of an $H$-free graph $G$ with the additional condition that $\alpha(G)=o(|G|)$. During this talk, I will discuss the Ramsey-Tur\'an variation of some classical results, whose extremal graphs are close to the Tur\'an graph, including Corr\'adi-Hajnal theorem on triangle factors in graphs and Erd\H os-Rothschild problem on the number of edge-colorings with no monochromatic clique.
Joint work partly with J\'ozsef Balogh and Hong Liu, and partly with J\'ozsef Balogh and Theodore Molla.
28 Oct: Ping Hu (Warwick) Tilings in graphons
We introduce a counterpart to the notion of vertex disjoint tilings by copy of a fixed graph F to the setting of graphons. The case F be an edge gives the notion of matchings in graphons. We give a transference statement that allows us to switch between the finite and limit notion, and derive several favorable properties, including the LP-duality counterpart to the classical relation between the fractional vertex covers and fractional matchings/tilings. As an application of our theory, we determine the asymptotically almost surely F-tiling number of inhomogeneous random graphs G(n,W). As another application, we give a strengthening of a theorem of Komlos, which gives minimum degree threshold that guarantees a prescribed lower bound on the fractional F-cover number of a graphon W.
Joint work with Jan Hladky and Diana Piguet.
4 Nov Katherine Staden (Warwick) The number of triangles in a graph with a given number of vertices and edges
The famous theorem of Tur\'an from 1940 states that every $n$-vertex graph with at least $n^2/4$ edges contains at least one triangle. Erd\H{o}s asked for a quantitative version of this statement: for every n and m, how \emph{many} triangles an must an n-vertex m-edge graph contain? This question has received a great deal of attention, and a long series of partial results culminated in an asymptotic solution by Razborov, extended to larger cliques by Nikiforov and Reiher. Currently, an exact solution is only known for a small range of edge densities, due to Lov\'asz and Simonovits. In this talk, I will discuss the history of the problem and recent work which gives an exact solution for almost the entire range of edge densities. This is joint work with Hong Liu and Oleg Pikhurko.
4 Nov: Richard Montgomery (Cambridge) Subdivisions in C_4-free graphs
Bollob\'as and Thomason, and independently Koml\'os and Szemer\'edi, showed in 1994 that any graph $G$ with average degree $d(G)$ contains a subdivision of a clique with at least $c\sqrt{d(G)}$ vertices, for some universal constant $c>0$. In 1999, Mader conjectured that $C_4$-free graphs $G$ in fact contain a subdivision of a larger clique, one with at least $c d(G)$ vertices, for some universal constant $c>0$. I will discuss a proof of this conjecture as well as a generalisation concerning $K_{s,t}$-free graphs.
This is joint work with Hong Liu.
11 Nov: Alexey Pokrovskiy (ETH) Rainbow cycles and rainbow expanders
A subgraph of an edge-coloured complete graph is called rainbow if all its edges have different colours. Andersen conjectured that every properly n-edge-coloured complete graph Kn has a rainbow Hamiltonian path. This seminar will be about a proof of an approximate version of this conjecture - that every properly edge-coloured Kn has a rainbow cycle of length n - O(n^{3/4}). One of the main ingredients of our proof, which is of independent interest, shows that a random subgraph of a properly edge-coloured Kn formed by the edges of a random set of colours has a similar edge distribution as a truly random graph with the same edge density. In particular it has very good expansion properties.
This is joint work with Noga Alon and Benjamin Sudakov.
18 Nov: Agnes Backhausz (Eötvös Loránd University) On the graph limit approach to eigenvectors of random regular graphs
The goal of the talk is to show how the graph limit approach
can be used to understand spectral properties of large random
regular graphs. In our work, we consider eigenvectors of
random regular graphs of fixed degree. As the number of
vertices tends to infinity, as a "limit", we investigate
invariant random processes on the infinite tree that
satisfy the eigenvalue equation and that can be "modelled" on
random regular graphs. In this infinite setting, based on
certain entropy inequalities, we could prove that (under
appropriate conditions) all these processes are Gaussian.
As a consequence, we could prove that the distribution of
delocalized eigenvectors of finite random regular graphs
is close to the Gaussian distribution. Joint work with Balazs
Szegedy.
2 Dec: Tim Netzer (Innsbruck) Checking Inclusion of Polytopes and Spectrahedra
The question whether one polytope is contained in another arises in interesting applications. Its computational complexity depends on the type of the input, and reaches from polynomial time to co-NP-hard. Spectrahedra are generalizations of polyhedra, and appear as feasible sets in semidefinite programming. In many applications, one or both of the given sets are spectrahedra, and inclusion testing becomes even more complicated, even at a conceptual level. There are certain relaxations of the problem, that work well in practice. After introducing the necessary background, we show that these relaxations are only reliable for simplices, and we derive some quantitative error bounds in the general case. The results use methods from operator theory, and some nice elementary constructions.
9 Dec: Will Perkins (Birmingham) An occupancy approach to bounding graph polynomials
I will present a new method for proving tight bounds on graph polynomials and on the number of independent sets, matchings, and colorings in regular graphs. The method is based on optimizing various observables in relevant probabilistic models from statistical physics (e.g., the hard-core, monomer-dimer, and Potts models) via linear programming relaxations. No previous knowledge of statistical physics is needed for the talk, and I will present many related open problems. Based on joint work with E. Davies, M. Jenssen, and B. Roberts.
13 Jan: Christoph Koch (Warwick) Jigsaw percolation on random hypergraphs
Jigsaw percolation was introduced by Brummit, Chatterjee, Dey, and Sivakoff as a model for interactions within a social network. It was inspired by the idea of collectively solving a puzzle. The premise is that each individual of a group of people has a piece of a puzzle all of which must be combined in a certain way to solve the puzzle. The compatibility of different puzzle pieces and the information which pairs of people meet are stored in two graphs (on a common vertex set), the puzzle graph and the people graph. Bollobás, Riordan, Slivken, and Smith studied the process when both graphs are given by independent binomial random graphs. More abstractly the process can be seen as a notion of simultaneous connectedness of a pair of (random) graphs. We transfer the process to hypergraphs in the context of high-order connectedness. We provide the asymptotic order of the critical threshold probability for percolation when both hypergraphs are chosen binomially at random, extending the result of Bollobás, Riordan, Slivken, and Smith. The evolution of the process is closely related to the presence of traversable triples in the pair of random hypergraphs.
This is joint work with Béla Bollobás, Oliver Cooley, and Mihyun Kang.
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2016-12-11 11:54:36
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https://datascience.stackexchange.com/tags/statistics/hot
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# Tag Info
20
In linear regression overfitting occurs when the model is "too complex". This usually happens when there are a large number of parameters compared to the number of observations. Such a model will not generalise well to new data. That is, it will perform well on training data, but poorly on test data. A simple simulation can show this. Here I use R: ...
4
Overfitting happens when the model performs well on the train data but doesn't do well on the test data. This is because the best fit line by your linear regression model is not a generalized one. This might be due to various factors. Some of the common factors are Outliers in the train data. Train and Test data are from different distributions. So before ...
2
Large number of parameters compared to data points In general, one aspect of overfitting is trying to "invent information out of knowthing" when you want to determine a comparably large number of parameters from a limited amount of actual evidence data points. For a simple linear regression y = ax + b there are two parameters, so for most sets of ...
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I'll try and provide some intuition for you here, instead of focusing on the mechanics of the math behind the methods. Imagine you are evaluating whether a coin is fair or not, so you collect a sequence of heads and tails as your data set. In MLE, we simply look at the data we collected and find the maximum likelihood... this works well when we have no prior ...
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To start with, you could use a simple thresholding. If you have the dataset $S$ where an element has the form $(x,y,c) \in S$, $x$ denotes the year, $y$ is a binary value (exam passed or not), and $c$ is the student id. you can obtain a classifier by using $\{(x,y,c) \in S \mid x \leq \theta\}$ and $\{(x,y,c) \in S \mid x > \theta\}$. Now you can check ...
1
You can use a Sigmoid function on the Force values(Scaled to [0-10] based on a max value) The Threshold should become 5 after scaling. def predict_proba(y_pred): y_pred = y_pred*10/100000 # scaled to [0-10] thresold = 5 proba = np.exp(y_pred - threshold)/(1 + np.exp(y_pred - threshold)) return proba predict_proba(100000), predict_proba(...
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If you are using pandas, all you need to do is: import pandas as pd import seaborn as sns import matplotlib.pyplot as plt corrMatrix = df.corr() Then you can print the correlation matrix and also plot it using seaborn or any other plotting method. sns.heatmap(corrMatrix, annot=True) plt.show() Hope this helps.
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Under/overfitting depends on two things: the amount of data in your dataset and the complexity of your model. To identify when each of these is happening, you will have to split the data you have into two parts: training data and test data. You then train your model only on the training data, and then evaluate its performance (e.g. calculate its accuracy or ...
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It's a good idea to remember what you're trying to predict and that is: $\mathbb{E}[Y | X=x]$. The simplest estimate is a single tree: $$\hat{\mu}(x) = \sum_{i \leq n} w_i(x, \theta) Y_i$$ with: $w_i(x, \theta)$ the single tree node weights (equation 4 in the paper) $Y_i$ your observations As we all know this is not a great estimate (high variance among ...
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This would be done with a controlled experiment, similar to clinical tests: Randomly assign customers into to groups 1 and 2. Group 1 is given the attractive offer, group 2 is not (control group). It's important to avoid any external bias, so for example ideally nobody in the company should know whether a customer is in group 1 or group 2. The goal is to ...
1
You could create a second label for your usernames according to whether they contain london or not (pseudocode below): for idx, username in df['Usernames']: if 'London' in username: df['London'].iloc[idx] = 1 else: df['London'].iloc[idx] = 0 Consequently given you want to go with correlation and that you are comparing binary ...
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Rationale Some of the terms are a little vague, particularly what you refer to as eligible students and returned students. I'll set some variables for clarity, but tell me if I defined them incorrectly. I assume them to mean: eligible students $= A$ being the set of all students in the after-school program 2019-2020 returning students $= A\cap S$ where ...
1
No, that isn't what it means. For one thing it is not clear what parameter the confidence interval that you calculated is for. In any case, some care is needed in the interpretation of (frequentist) confidence intervals. In frequentist statistics, a confidence interval is random, and the parameter that the interval is for is fixed. In the case of a 99% ...
Only top voted, non community-wiki answers of a minimum length are eligible
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2020-09-25 06:36:17
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https://raisingthebar.nl/tag/complex-number/
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14
Aug 18
## Complex numbers: time to turn on the beacon
Math is motivated by analogy and the desire to solve new problems, among other things. We'll see that this is the pattern that leads from real numbers to complex numbers and further to matrix algebra.
### Complex numbers are not that complex
Problem. For any real number $x,$ the square $x^2$ is non-negative. Therefore the equation $x^2=-1$ doesn't have solutions in the set of real numbers.
Idea. Expand the set of real numbers in such a way that a) the square root $\sqrt{-1}$ exists in the new set of numbers and b) the properties of real numbers involving addition/subtraction and multiplication/division are preserved.
Solution to the problem. Step 1. Formally introduce the imaginary unit $i=\sqrt{-1}.$ This implies, in particular, that $i^{2}=-1$ and that everywhere $i^{2}$ is encountered it should be replaced by $-1.$
Step 2. Formally introduce complex numbers as linear combinations $c=a+ib$ of the real unit $1$ and imaginary unit $i$ with real $a,b.$ Manipulate them using the properties of real numbers involving addition/subtraction and multiplication/division. For example,
(1) $(a-ib)(a+ib)=a^2-abi+abi-b^2i^2=a^2+b^2.$
Step 3. A complex number $c=a+ib$ can be interpreted as a vector $(a,b)$ on the plane because summation of complex numbers
$c_1+c_2=a_1+ib_1+a_2+ib_2=(a_1+a_2)+i(b_1+b_2)$
corresponds to summation of vectors. In particular, use the norm of the vector $c$ as the absolute value of $c:$
(2) $|c|\overset{def}{=}(a^2+b^2)^{1/2}=\|c\|.$
These formalities plus the geometric interpretation is all one needs to know about the set of complex numbers $C.$
Definition 1. The number $\bar{c}=a-ib$ is called a conjugate of the number $c=a+ib.$
With this definition, we have from (1) and (2)
(3) $\bar{c}c=a^2+b^2,$ $|c|=(\bar{c}c)^{1/2}.$
Besides, it's easy to see that $c=\bar{c}$ if and only if $b=0.$ This is the way to identify real numbers in the set of complex numbers:
(4) a number $c$ is real if and only if $c=\bar{c}.$
Exercise 1. Express the ratio $u/v$ of two complex numbers in the form $a+ib.$
The set of complex numbers $C$ corresponds to the whole plane and the set of real numbers $R$ corresponds to the $x$ axis. Similarly to $R^n$ we can define $C^n$ as the set of all vectors $x=(x_1,...,x_n)$ with components $x_i\in C.$
Definition 2. The scalar product in $C^n$ is defined by $x\cdot y=\sum{x_i}\bar{y}_i.$
Exercise 2. Check that it has all properties of the scalar product in $R^n$ except that instead of $x\cdot (ay+bz)=ax\cdot y+bx\cdot z$ one has $x\cdot(ay+bz)=\bar{a}x\cdot y+\bar{b}x\cdot z$ where $a,b\in C.$
### Polar representation
Figure 1. Polar form
Let us write
$c=a+ib=|c|\left(\frac{a}{|c|}+i\frac{b}{|c|}\right).$ Here the numbers $a_1=\frac{a}{|c|},$ $b_1=\frac{b}{|c|}$ satisfy $a_1^2+b_1^2=1.$ Therefore there exists an angle $\theta$ such that $a_1=\cos\theta,$ $b_1=\sin\theta$ (see Figure 1, sorry about the notation discrepancy). This implies
(5) $c=|c|\left(\cos\theta+i\sin\theta\right).$
Euler established a wonderful formula
(6) $e^{i\theta}=\cos\theta+i\sin\theta.$
For example, with $\theta=\pi$ we have an interesting relationship between three most important numbers in mathematics: $e^{i\pi }=-1.$ Combining (5) and (6) we get the polar representation:
(7) $c=|c|e^{i\theta}.$
The additive form $c=a+ib$ is better for addition/subtraction and the polar (multiplicative) form is better for multiplication/division. In particular, by (7) $c^n=|c|^ne^{in\theta}.$
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2022-09-27 02:39:34
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https://flyingcoloursmaths.co.uk/theres-one-way-direct-inverse-proportion/
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$y$ is directly proportional to $x^3$, you say? And when $x = 4$, $y = 72$? Well, then.
The traditional method is to say:
$y = kx^3$ and substitute in what you know.
$72 = 64k$ $k = \frac{72}{64} = \frac{9}{8}$
That gives $y = \frac98 x^3$. Easy enough.
### But you can do it without the $k$
Rather than introduce an arbitrary constant, you can do this directly by writing it as:
$\frac {y}{y_0} = \frac {x^3}{x_0^3}$, where $x_0$ and $y_0$ are the numbers you’re given:
$\frac y {72} = \frac{x^3}{64}$
$y = \frac{72}{64}x^3$, which cancels down to the same thing as before.
If you know $B$ is inversely proportional to $r^2$, and when $r = 3,$ $B = 1$, you can do this:
$\frac {B}{B_0} = \frac{r_0^2}{r^2}$ $\frac{B}{1} = \frac{9}{r^2}$ $B = \frac{9}{r^2}$
I find that a bit easier to get my head around than setting up an arbitrary $k$ and finding its value - but I’d be interested to hear what you think!
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2022-07-06 04:00:30
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