text:
# R/K selection theory: Wikis

Note: Many of our articles have direct quotes from sources you can cite, within the Wikipedia article! This article doesn't yet, but we're working on it! See more info or our list of citable articles.

# Encyclopedia

(Redirected to r/K selection theory article)

In ecology, r/K selection theory relates to the selection of combinations of traits that trade off the quantity and quality of offspring to promote success in particular environments. The terminology of r/K-selection was coined by the ecologists Robert MacArthur and E. O. Wilson[1] based on their work on island biogeography.[2]

## Overview

In r/K selection theory, selective pressures are hypothesised to drive evolution in one of two generalized directions: r- or K-selection.[1] These terms, r and K, are derived from standard ecological algebra, as illustrated in the simple Verhulst equation of population dynamics:[3]

$\frac{dN}{dt} = rN \left(1 - \frac{N}{K}\right)$

where r is the growth rate of the population (N), and K is the carrying capacity of its local environmental setting. Typically, r-selected species exploit less-crowded ecological niches, and produce many offspring, each of which has a relatively low probability of surviving to adulthood. In contrast, K-selected species are strong competitors in crowded niches, and invest more heavily in fewer offspring, each of which has a relatively high probability of surviving to adulthood. In the scientific literature, r-selected species are occasionally referred to as "opportunistic", while K-selected species are described as "equilibrium".[4]

### r-selection (unstable environments)

In unstable or unpredictable environments, r-selection predominates as the ability to reproduce quickly is crucial. There is little advantage in adaptations that permit successful competition with other organisms, because the environment is likely to change again. Traits that are thought to be characteristic of r-selection include: high fecundity, small body size, early maturity onset, short generation time, and the ability to disperse offspring widely. Organisms whose life history is subject to r-selection are often referred to as r-strategists or r-selected. Organisms with r-selected traits range from bacteria and diatoms, through insects and weeds, to various semelparous cephalopods and mammals, especially small rodents.

### K-selection (stable environments)

In stable or predictable environments, K-selection predominates as the ability to compete successfully for limited resources is crucial and populations of K-selected organisms typically are very constant and close to the maximum that the environment can bear (unlike r-selected populations, where population sizes can change much more rapidly). Traits that are thought to be characteristic of K-selection include: large body size, long life expectancy, and the production of fewer offspring that require extensive parental care until they mature. Organisms whose life history is subject to K-selection are often referred to as K-strategists or K-selected. Organisms with K-selected traits include large organisms such as elephants, trees, humans and whales, but also smaller, long-lived organisms such as Arctic Terns. The climate that demonstrate predictably variable characteristics of a stable climate is the oceanic climate. The most stable climate in this sense would be found in Northwest Europe. The warm Gulf Stream and the cool temperature of the northern latitude form a natural front in which predictably variable characteristics are found to the greatest extent. The very location would be southern Lofoten which is also known for the greatest positive temperature anomaly in the world.

### As a continuous spectrum

Although some organisms are identified as primarily r- or K-strategists, the majority of organisms do not follow this pattern. For instance, trees have traits such as longevity and strong competitiveness that characterise them as K-strategists. In reproduction, however, trees typically produce thousands of offspring and disperse them widely, traits characteristic of r-strategists. Similarly, reptiles such as sea turtles display both r- and K-traits: although large organisms with long lifespans (should they reach adulthood), they produce large numbers of unnurtured offspring.

## In ecological succession

In areas of major ecological disruption or sterilisation (such as after a major volcanic eruption, as at Krakatoa or Mount Saint Helens), r- and K-strategists play distinct roles in the ecological succession that regenerates the ecosystem. Because of their higher reproductive rates and ecological opportunism, primary colonisers typically are r-strategists and they are followed by a succession of increasingly competitive flora and fauna. The ability of an environment to increase energetic content, through photosynthetic capture of solar energy, increases with the increase in complex biodiversity as r species proliferate to reach a peak possible with K strategies.[5] Eventually a new equilibrium is approached (sometimes referred to as a climax community), with r-strategists gradually being replaced by K-strategists which are more competitive and better adapted to the emerging micro-environmental characteristics of the landscape. Typically, biodiversity is maximised at this stage, with introductions of new species resulting in the replacement and local extinction of endemic species.[6]

## Status of the theory

Although r/K selection theory became widely used during the 1970s,[7][8][9][10] it also began to attract more critical attention.[11][12][13][14] In particular, an influential review by the ecologist Stephen Stearns drew attention to gaps in the theory, and to ambiguities in the interpretation of empirical data for testing it.[15] Nonetheless, the concept of r/K selection is still used in biological and ecological research.[16][17][18][19]

## In Humans

In the 1995 book Race, Evolution, and Behavior, J. Philippe Rushton, a psychologist and head of the Pioneer Fund, argues that different human races exhibit reproductive strategies characteristic of r/K selection theory. Rushton asserts that Asians, who have slightly longer lifespans and lower rates of reproduction relative to other races, are at the K end of the spectrum, while Africans, for whom the opposite is true, are more r-strategists[20].

Rushton's research has been heavily criticized,[21] and other studies have contradicted many of his claims.[22]

## References

1. ^ a b Pianka, E.R. (1970). On r and K selection. American Naturalist 104, 592-597.
2. ^ MacArthur, R. and Wilson, E.O. (1967). The Theory of Island Biogeography, Princeton University Press (2001 reprint), ISBN 0-691-08836-5M.
3. ^ Verhulst, P.F. (1838). Notice sur la loi que la population pursuit dans son accroissement. Corresp. Math. Phys. 10, 113-121.
4. ^ For example: Weinbauer, M.G.; Höfle, M.G. (1 October 1998). "Distribution and Life Strategies of Two Bacterial Populations in a Eutrophic Lake". Appl. Environ. Microbiol. 64 (10): 3776–3783. PMID 9758799.
5. ^ Gunderson, L. & Holling, C.S. (Eds) (2001), "Panarchy: Understanding Transformations in Human and Natural Systems" (Island Press)
6. ^ McNeely, J. A. (1994) Lessons of the past: Forests and Biodiversity. Biodiversity and Conservation 3, 3-20.
7. ^ Gadgil, M. and Solbrig, O.T. (1972). Concept of r-selection and K-selection - evidence from wild flowers and some theoretical consideration. Am. Nat. 106, 14.
8. ^ Long, T. and Long, G. (1974). Effects of r-selection and K-selection on components of variance for 2 quantitative traits. Genetics 76, 567-573.
9. ^ Grahame, J. (1977). Reproductive effort and r-selection and K-selection in 2 species of Lacuna (Gastropoda-Prosobranchia). Mar. Biol. 40, 217-224.
10. ^ Luckinbill, L.S. (1978). r and K selection in experimental populations of Escherichia coli. Science 202, 1201-1203.
11. ^ Wilbur, H.M. (1974). "Environmental certainty, trophic level, and resource availability in life history evolution". American Naturalist 108: 805–816. doi:10.1086/282956.
12. ^ Barbault, R. (1987). Are still r-selection and K-selection operative concepts? Acta Oecologica-Oecologia Generalis 8, 63-70.
13. ^ Kuno, E. (1991). Some strange properties of the logistic equation defined with r and K - inherent defects or artifacts. Researches on Population Ecology 33, 33-39.
14. ^ Getz, W.M. (1993). Metaphysiological and evolutionary dynamics of populations exploiting constant and interactive resources - r-K selection revisited. Evolutionary Ecology 7, 287-305.
15. ^ Stearns, S.C. (1977). "Evolution of life-history traits - critique of theory and a review of data" (PDF). Ann. Rev. Of Ecology and Systematics 8: 145–171. doi:10.1146/annurev.es.08.110177.001045.
16. ^ Caroli, L.; Capizzi, D. and Luiselli, L. (2000). "Reproductive Strategies and Life-history Traits of the Savi's Pine Vole, Microtus savii". Zoological Science 17: 209–216. doi:10.2108/zsj.17.209.
17. ^ Chikatsu, N.; Nakamura, Y., Sato, H., Fujita, T., Asano, S. and Motokura, T. (2002). "p53 mutations and tetraploids under r- and K-selection". Oncogene 21: 3043–3049. doi:10.1038/sj/onc/1205413.
18. ^ Okada, H.; Harada, H., Tsukiboshi, T. and Araki, M. (2005). "Characteristics of Tylencholaimus parvus (Nematoda: Dorylaimida) as a fungivorus nematode". Nematology 7: 843–849. doi:10.1163/156854105776186424.
19. ^ Hamer, A.J.; Mahony, M.J. (2007). "Life history of an endangered amphibian challenges the declining species paradigm". Australian Journal of Zoology 55: 79–88. doi:10.1071/ZO06093.
20. ^ Rushton, J. P. (1997). Race, Evolution, and Behavior: A Life History Perspective, Charles Darwin Research Institute, ISBN 0-9656836-2-1.
21. ^ Barash, D.P. (1995). "Book review: Race, Evolution, and Behavior". Animal Behaviour 49: 1131–1133.
22. ^ Cernovsky, Z.Z. (1995). "On the similarities of American blacks and whites: A reply to J.P. Rushton". Journal of Black Studies 25: 672–679.


text:
# What are the difficulties involved in proving that the Kunen inconsistency holds in $NGB$

or (contrariwise) that $$NGB$$ + "There exists a Reinhardt cardinal" is consistent?

The question is partially in the title. $$NGB$$ is used for the reasons stated in the Hamkins, Kirmayer, and Perlmutter paper, "Generalizations of the Kunen inconsistency", Annals of Pure and Applied Logic, 163 (2012) pp. 1873-1876 (Section 1, "A few metamathematical preliminaries").

In order that this question should not be deemed overbroad, answers must be limited in scope to specific mathematical difficulties (eg. the Erdos-Hajnal theorem used in in proving the Kunen inconsistency in $$NGBC$$ or $$NBC$$ + $$Choice$$ has no known proof in $$NGB$$ or is known not to be provable without choice--however, those who would say that the Erdos-Hajnal theorem cannot be proven without choice must show how this fact allows for a non-trivial elementary embedding $$j$$: $$V$$ $$\rightarrow$$ $$V$$ and the existence of the requisite critical point $$\kappa$$ in $$NGB$$). Those who are so inclined might make mention of Rupert McCallum's recent attempt in showing that "the existence of a non-trivial elementary embedding $$j$$: $$V_{\lambda + 2}$$ $$\rightarrow$$ $$V_{\lambda + 2}$$ for some ordinal $$\lambda$$...is not consistent with $$ZF$$" (quote from McCallum's withdrawn short preprint, The Choiceless Cardinals are Inconsistent, arXiv:1712.09768, which is discussed on Joel David Hamlin's blog — I would be especially interested in getting a nice explanation (i.e. analysis) of the gap in this argument and the relavance of the Laver paper which Gabriel Goldberg directed him to).

Though at first glance the answers to these questions might seem "primarily opinion-based", the mathoverflow community should realize the value of informed opinion and (yes) informed speculation regarding this topic, and use this question as an opportunity to provide these 'clues' (think of the proof of the 4-color theorem) to both the mathoverflow community and the mathematical community as a whole.

• Although the title question is clear enough (and answered by the dependence of the proof on the Erdös-Hajnal theorem), the parenthetical addendum in the question itself, "must show how this fact allows for a non-trivial elementary embedding $j:V\to V$," seems to require a proof of consistency (relative to large cardinals) of NBG + existence of such $j$. Oct 13 '18 at 17:02
• To elaborate on Andreas' comment: since at present we don't know a definitive answer, we can't possibly show that a given point is a necessary stumbling block. The best that can be done is to point out the key step which "no known argument can get around," and observe that this step doesn't go through without choice. So I don't understand what you're looking for in an answer, short of a new major result in set theory, if that doesn't satisfy you. Oct 13 '18 at 23:26
• @AndreasBlass: Well, at least a proof of relative consistency.... Oct 15 '18 at 13:49
• @NoahSchweber: While one may not have a definitive answer, one might have relevant facts that could lead to a definitive answer. An easy example, there might be, say, in some obscure journal, a proof of the relevant Erdos-Hajnal theorem that doesn't use choice, or a choice principle weaker than $AC$ (say, the Axiom of Dependent Choices). The relevant Erdos-Hajnal theorem has been around a long time so such a theorem might be out there. A proof that the relevant Erdos-Hajnal theorem cannot do without $AC$ may suggest that one might be able to construct a model of $NGB$ that has a Oct 15 '18 at 14:06
• (cont.) Reinhardt cardinal, but then again there might be some structural peculiarities (mimicking the 'weirdness' of $ZF$ relative to $ZFC$ Asaf Karagila speaks of in his blog or his answers and comments in mathoverflow and mathstackexchange) of models $NGB$ relative to models of$NGBC$ or $NGB$ + $Choice$ (but if one could localize $Choice$ in $NGB$ + $Choice$ to somwhere above the critical point of an $I0$ embedding, and above that, $Choice$ would fail-- where would the point of demarcation be?), how would those peculiarities relate to the success or failure of the Kunen inconsistency Oct 15 '18 at 14:29

Here is the strategy behind McCallum's attempted proof that there is no nontrivial elementary embedding $$j :V\to V$$.

Step 1 is to cite the theorem that if a $$j :V\to V$$ is consistent, then a $$j : V\to V$$ is consistent with $$V_\lambda\vDash \text{ZFC}$$ where $$\lambda = \sup_{n < \omega} j^n(\text{crt}(j))$$. This is proved by a minor modification of Woodin's technique for forcing choice assuming a supercompact. So it suffices to show this stronger assumption is inconsistent.

Step 2 is to try to reflect the failure of AC given by Kunen's theorem into $$V_\lambda$$. There is a technique for reflecting certain properties of tame substructures of $$V_{\lambda+2}$$ to cardinals below $$V_\lambda$$. This is called inverse limit reflection, originally invented by Laver and recently vastly extended by Scott Cramer. For example, a theorem of Cramer implies that in this situation, and assuming DC, there is some $$\bar \lambda < \lambda$$ such that there is an elementary $$\pi : L(V_{\bar \lambda+1})\to L(V_{\lambda+1})$$ and even a $$\pi : L(V_{\bar \lambda + 1},V_{\bar \lambda + 1}^\#)\to L(V_{\lambda + 1},V_{\lambda + 1}^\#)$$.

[Here is a successful application of this proof strategy. Note that $$I_0$$ holds at $$\bar \lambda$$ since $$I_0$$ at $$\lambda$$ is a first-order fact in $$(V_{\lambda+1},V_{\lambda+1}^\#)$$. Thus $$V_\lambda$$ is a model of ZFC + $$I_0$$. This is how I proved NBG + DC + $$j :V\to V$$ implies Con(ZFC + $$I_0$$).]

McCallum had rediscovered the inverse limit reflection technique, and he thought he could reflect the existence of an $$\omega$$-Jonsson cardinal into $$V_\lambda$$ using it. The issue, however, was that the statement that $$\lambda$$ is $$\omega$$-Jonsson is a fact about $$V_{\lambda+2}$$, and inverse limit reflection does not seem to be able to reflect statements about the whole structure $$V_{\lambda+2}$$ into $$V_\lambda$$. One can only reflect properties of canonical inner models over $$V_{\lambda+1}$$, like $$L(V_{\lambda+1})$$ and $$L(V_{\lambda+1},V_{\lambda+1}^\#)$$.

So maybe the obstruction to carrying out this proof idea is, very vaguely, that one must first extend inverse limit reflection / inner model theory to "reach" $$V_{\lambda+2}$$.

• Cramer showed that if $I^\#_0$ holds (i.e., there is a $j :L(V_{\lambda+1}^\#)\to L(V_{\lambda+1}^\#)$ with sup of critical sequence $\lambda$) then $I_0$ holds in $V_\lambda$. If you do the AC forcing and carry out Cramer's proof in DC then you get that $\text{ZF + DC + }I_0^\#$ implies $\text{Con(ZFC + }I_0)$. The fact that a Reinhardt implies $I_0^\#$ requires proving $V_{\lambda+1}^\#$ exists from a Reinhardt. In fact a Reinhardt implies every set has a sharp, another result of mine from that paper (which is still unpublished). Oct 23 '18 at 23:08
• It is in "Inverse limit reflection and the structure of $L(V_{\lambda+1})$." The result is Theorem 3.9. Oct 24 '18 at 15:04
• Yes $V_{\lambda+1}^\#$ can be viewed as a subset of $V_{\lambda+1}$. It is the theory of $\omega$-many Silver indiscernibles of $L(V_{\lambda+1})$ with parameters from $V_{\lambda+1}$. You should look at Caicedo's "A review of sharps." Oct 30 '18 at 12:29
• andrescaicedo.files.wordpress.com/2008/04/sharps.pdf Oct 30 '18 at 12:48
• $I_0^\#$ at $\lambda$ is precisely the statement that $V^\#_{\lambda+1}$ exists and is an Icarus set. But if $\lambda$ is least such that $I_0$ holds at $\lambda$ and $V_{\lambda+1}^\#$ exists, then $I_0^\#$ fails at $\lambda$ by Cramer's results. Oct 30 '18 at 14:41

format: plain


text:
# Infinite Dimensional Vector Spaces and Linear Operators

• March 8th 2011, 10:36 PM
I-Think
Infinite Dimensional Vector Spaces and Linear Operators
Give a example where V is an infinite dimensional vector space , $T_1, T_2$are linear operators on V and $T_1T_2$ (a composition) is a bijection but T_1 and T_2 are not bijections
• March 8th 2011, 11:07 PM
Drexel28
Quote:

Originally Posted by I-Think
Give a example where V is an infinite dimensional vector space , $T_1, T_2$are linear operators on V and $T_1T_2$ (a composition) is a bijection but T_1 and T_2 are not bijections

Think of the space $\mathbb{R}[x]$ and don't think too hard about the mappings...do what's natural.
• March 8th 2011, 11:30 PM
I-Think
What is the space R[x]?
• March 9th 2011, 12:10 AM
FernandoRevilla
Quote:

Originally Posted by I-Think
What is the space R[x]?

The real vector space whose elements are the polynomials with real coefficients on the unknown $x$ considering the standard operation sum an product by a real number.
• March 9th 2011, 02:28 AM
tonio
Quote:

Originally Posted by I-Think
Give a example where V is an infinite dimensional vector space , $T_1, T_2$are linear operators on V and $T_1T_2$ (a composition) is a bijection but T_1 and T_2 are not bijections

Take $V:=\{\,\{x_n\}_{n=1}^\infty\;;\;x_n\in\mathbb{R}\, \,\forall n\in\mathbb{N}\} =$ the vector space of all

real sequences with coordinatewise addition and multiplication by scalar, and take

$T_2(\{x_n\}):=\{0,x_1,x_2,\ldots\}\,,\,\,T_1(\{x_n \}):=\{x_2,x_3,\ldots\}$ , and prove these two do the trick.

Tonio


text:
# EPS-HEP2021 conference

26-30 July 2021
Zoom
Europe/Berlin timezone

## Flavorful leptoquarks at the LHC and beyond: Spin 1

29 Jul 2021, 10:30
15m
Zoom

#### Zoom

Parallel session talk Searches for New Physics

### Speaker

Ivan Nišandžić (Ruđer Bošković Institute)

### Description

Evidence for electron-muon universality violation that has been revealed in $b\to s \ell\ell$ transitions in the observables $R_{K,K^*}$ by the LHCb Collaboration can be explained with spin-1 leptoquarks in $SU(2)_L$ singlet $V_1$ or triplet $V_3$ representations in the ${\cal{O}}(1-10)$ TeV range. We explore the sensitivity of the high luminosity LHC (HL-LHC) and future proton-proton colliders to $V_1$ and $V_3$ in the parameter space connected to $R_{K,K^*}$-data. Future sensitivity projections based on extrapolations of existing ATLAS and CMS searches are worked out. We find that for $\kappa=1$ the mass reach for pair (single) production of $V_1$ can be up to 3 TeV (2.1 TeV) at the HL-LHC and up to 15 TeV (19.9 TeV) at the FCC-hh with $\sqrt{s}=100$ TeV and $20 \,\mbox{ab}^{-1}$. The mass limits and reach for the triplet $V_3$ are similar or higher, depending on flavor. While there is the exciting possibility that leptoquarks addressing the $R_{K,K^*}$-anomalies are observed at the LHC, to fully cover the parameter space $pp$-collisions beyond the LHC-energies are needed.

Collaboration / Activity Ruđer Bošković Institute

### Primary authors

Gudrun Hiller (Technische Universität Dortmund) Dennis Loose (Technische Universität Dortmund) Ivan Nišandžić (Ruđer Bošković Institute)


text:
### Home > CC3MN > Chapter 6 Unit 10 > Lesson INT1: 6.3.3 > Problem6-110

6-110.

Simplify each of the following expressions.

1. $(5x^3)^2$

Notice that when you put exponents outside of a pair of parentheses, the exponent applies to all of the factors within the parentheses.

$5^2{x}^{3(2)}$

$5^2{\it x}^6$

1. $\frac { 14 a ^ { 3 } b ^ { 2 } } { 21 a ^ { 4 } b ^ { 3 } }$

Look for Giant Ones.

$\frac{2}{3ab}$

1. $2m^{3}n^{2 }· 3mn^{4}$

When multiplying the same bases together, add the exponents.


text:
stratifyByPsAndCovariates uses the provided propensity scores and covariatesto stratify persons.

stratifyByPsAndCovariates(population, numberOfStrata = 5,
baseSelection = "all", cohortMethodData, covariateIds)

## Arguments

population A data frame with the three columns described below Into how many strata should the propensity score be divided? The boundaries of the strata are automatically defined to contain equal numbers of target persons. What is the base selection of subjects where the strata bounds are to be determined? Strata are defined as equally-sized strata inside this selection. Possible values are "all", "target", and "comparator". An object of type cohortMethodData as generated using getDbCohortMethodData. One or more covariate IDs in the cohortMethodData object on which subjects should also be stratified.

## Value

Returns a date frame with the same columns as the input population plus one extra column: stratumId.

## Details

The data frame should have the following three columns:

 rowId (integer) A unique identifier for each row (e.g. the person ID) treatment (integer) Column indicating whether the person is in the target (1) or comparator (0) group propensityScore (numeric) Propensity score

## Examples

# todo


text:
# MRI Graph Analysis

GUIMRIGraphAnalysis is a graphical user interface that allows the user to define the parameters to create the connectivity matrices to analyze MRI data, while simultaneously visualizing the resulting weighted or binary matrices. Binary connectivity matrices can be visualized as a function of density or threshold. The user can also define a community structure and restrict the analysis to a subset of brain regions. A list of the measures available for calculation is shown at the bottom of the interface. The MRI graph analysis can be saved in a file *.mga for future use within BRAPH; it also can be exported in xml format for use within other programs.

Figure 1: Screenshot of GUIMRIGraphAnalysis. On the top, there are the menu and the toolbar; in the middle there are the cohort panel and the graph analysis panel (on the left), and the main panel (on the right); on the bottom, there is the measures panel.

The layout of GUIMRICohort is shown in figure 1. It is composed of six main work areas:

• Menu permits one to access the basic functionalities of GUIMRIGraphAnalysis, including loading and saving an MRI graph analysis.
• Toolbar gives direct access to some of the most commonly employed functionalities, in particular loading and saving an MRI graph analysis as well as manipulating the graphic representations of the connectivity matrices.
• Cohort panel permits one to select an MRI cohort for the graph analysis or, if already selected, to view the cohort properties in GUIMRICohort.
• Graph analysis panel permits one to choose the properties of the graph analysis, to set a community structure, and to choose whether to perform the analysis on a subgraph.
• Main panel visualizes the connectivity matrix that will be used for the analysis.
• Measures panel shows the available measures for each type of graph (binary or weighted).

Example data and tutorial videos can be found on http://braph.org/videos/mri/mri-graph-analysis/.

Getting Started

As a first example of the use of GUIMRIGraphAnalysis, we will proceed to import the MRI cohort stored in the my_cohort.mc file. Then, we will define a binary undirected graph analysis with fixed density and positive Pearson correlation coefficients. We will further specify a dynamic community structure calculated with the Louvain algorithm by using the subject group gr1_MRI. Finally, we will choose to perform the analysis on the full connectivity matrix and save it as a *.mga file.

1. Push Select Cohort to select an MRI cohort as shown in figure 2. The cohort must be in *.mc format. After you select the file, the cohort panel is updated to show the properties of the cohort. The select button state is changed to View Cohort; pushing this button opens the uploaded cohort in GUIMRICohort with restricted access (i.e. no further changes to the cohort are allowed).

Figure 2: Importing an MRI cohort from a *.mc file into GUIMRIGraphAnalysis.

2. Select BUD from the ‘Graph’ popup menu in the graph analysis panel, Pearson from the ‘Correlation’ popup menu, and zero from the ‘Negative corrs.’ popup menu. With these settings, the connectivity matrix will be calculated using Pearson correlation coefficients where all negative coefficient are set to zero; this matrix will then be binarized at a fixed density.A number of settings are available to create different types of graph analyses.
More details about how these options affect the graphs can be found in the chapter ‘Brain Graphs’.
These options are accessible through the popup menus in the graph analysis panel:

• Graph sets the type of graph to be analyzed:
WU analyzes weighted undirected graphs.
BUT analyzes binary undirected graphs, i.e. graphs whose connectivity matrices are binarized by specifying the threshold.
BUD analyzes binary undirected graphs, i.e. graphs whose connectivity matrices are binarized by specifying the density.
• Correlation sets the correlation used to calculate connectivity
matrix coefficients:
Pearson is the Pearson product-moment correlation coefficient.
Spearman is the Spearman rank correlation coefficient.
Kendall is the Kendall rank correlation coefficient.
partial Pearson is the partial Pearson correlation coefficient.
partial Spearman is the partial Spearman correlation coefficient.
• Negative corrs. sets how to deal with the negative correlation
coefficients:
zero sets all negative correlation coefficients to zero.
none leaves all negative correlation coefficients as as they are.[1]
abs replaces all negative correlation coefficients with their absolute values.
3. Push Edit in the panel ‘Community structure’ to define a community structure. This opens a new interface where the parameters for the calculation of the community structure can be set. Check the Dynamic structure checkbox to define a dynamic structure, choose the Louvain algorithm and select the group group1 from the popup menu, as shown in figure 3.

Figure 3: Interface to define a community structure for the graph analysis.

The community structure interface consists of five main working areas as shown in figure 3:

• Menu permits one to generate the brain view of the community structure. This can be customized using the standard MatLab plotting tools. The figure can then be exported in several standard graphic formats.
• Toolbar gives direct access to various brain views and allows the user to choose how to represent the brain regions (spheres, symbols, or labels).
• Brain view panel visualizes the community structure on a brain surface. Different modules are represented with different colors.
• Parameter panel allows one to choose the parameters for the calculation of the community structure. The following parameters can be specified:

Fixed structure fixes the community structure. The same structure will be used throughout the analysis.

Dynamic structure creates a dynamic structure with the specified parameters. The structure will be recalculated with the selected parameters whenever needed throughout the analysis.

Louvain algorithm calculates the structure using the Louvain algorithm.

– Newman algorithm calculates the structure using the Newman algorithm.

Gamma sets the parameter $\gamma>0$ determining the resolution of the algorithm. The default setting is $\gamma = 1$. Larger values ($\gamma>1$) lead to more modules and smaller values ($0<\gamma<1$) to less modules.

Group permits one to choose the group whose data serve as basis for the community structure calculation.

The structure is calculated by pushing Calculate and reset by pushing Reset. Once all parameters are chosen, the structure can be set by pushing Set.

Alternatively, one can choose to perform the graph analysis with the default community structure by pushing Default. The default structure is a dynamic structure calculated with the Louvain algorithm with $\gamma=1$.

4. To start the graph analysis on the full connectivity matrix, push Start analysis. This opens a new interface, GUIMRIGraphAnalysisBUD, which allows one to calculate and visualize the graph measures. The details of this interface are discussed in chapter ‘MRI Graph Analysis BUD’. After this, the parameters of the analysis become fixed and, if any change is needed, a new graph analysis with different parameters should be created.

Figure 4: Interface to determine a subgraph on which to perform the graph analysis.

5. The analysis can be performed only on a subset of brain regions. To do this, push Subgraph analysis.This opens a new interface, shown in figure 4, with five main working areas:
• Menu permits one to generate the subgraph brain view, which can be customized using the standard MatLab plotting tools. The figure can then be exported in several standard graphic formats.
• Toolbar gives direct access to various brain views and allows the user to choose how to represent the brain regions (spheres, symbols, or labels).
• Table panel shows all brain regions. If the checkbox next to a brain region is checked, the region is included into the subgraph.
• Brain view panel permits one to visualize the subgraph on a brain surface. The included brain regions are highlighted with an orange color.
• Control panel allows one to choose the parameters for the calculation of the subgraph analysis as follows: (1) One of the modules from a previously calculated community structure can be selected as subgraph from the popup menu on the right. (2) Set sets the subgraph that will be used in the analysis. (3) Start analysis starts the analysis by opening a new interface, GUIMRIGraphAnalysisBUD. The analysis can be performed only after the subgraph has been set.
6. Select File $\rightarrow$ Save to save the MRI graph analysis as a *.mga file; alternatively you can also use the shortcut Ctrl+S or the Save icon on the toolbar.
7. Select File $\rightarrow$ Open to open an MRI graph analysis previously saved with GUIMRIGraphAnalysis; alternatively you can also use the shortcut Ctrl+O or the Open icon on the toolbar. Opening a *.mga file with with GUIMRIGraphAnalysis interface opens a new interface (i.e. GUIMRIGraphAnalysisWU, GUIMRIGraphAnalysisBUT, or GUIMRIGraphAnalysisBUD) depending on the type of analysis specified (weighted undirected, binary undirected fixed threshold, or binary undirected fixed density, respectively).

Main panel

The main panel allows one to visualize the connectivity matrix based on the parameters specified by the set of checkboxes on the right (figure 1). The available options are:

• Group selects the group whose connectivity matrix to show.
• Weighted correlation matrix, if checked, visualizes the correlation coefficients between any two brain regions: warmer colors denote higher coefficients.
• Histogram shows the distribution of the correlation coefficients.
• Binary correlation matrix (set density) shows the binarized connectivity matrix at the set density (text field and slider below).
• Binary correlation matrix (set threshold) shows the binarized connectivity matrix at the set threshold (text field and slider below).
• Rearrange to reflect community structure rearranges the rows and columns of the connectivity matrix to reflect the community structures (i.e. keeping together regions belonging to the same module).
• Divide communities draws lines (squares around each module) to emphasize the division of the brain into different modules. This option is available only if the matrix has been previously rearranged to reflect the community structure by selecting the option Rearrange to reflect community structure.

Measures panel
The measures panel lists the measures (together with a short description)
that are available for calculation given the kind of graph that
has been selected (WU, BUT, BUD).

File provides various options for importing and saving an MRI graph analysis:

• File $\to$ Open (Ctrl+O) opens a popup window to load an MRI graph analysis saved in *.mga format.
• File $\to$ Close (Ctrl+C) closes the GUIMRIGraphAnalysis.
• File $\to$ Save (Ctrl+S) saves the MRI current graph analysis in *.mga format.
• File $\to$ Save as opens a popup window to save the current MRI graph analysis in *.mga format possibly in a different file.
• File $\to$ Import (xml) imports an MRI graph analysis from an xml file.
• File $\to$ Export (xml) exports the current MRI graph analysis to an xml file.

Brain View $\to$ Generate figure (Ctrl+F) generates a figure that can be customized using the standard MatLab plotting tools. The figure can then be exported in several standard graphic formats.
About $\to$ About provides information about the current version of GUIMRIGraphAnalysis and BRAPH.

Toolbar

The toolbar provides different options to open and save the MRI graph analysis as well as to visualize the connectivity matrix. It is shown in figure 5.

Figure 5: GUIMRIGraphAnalysis toolbar.

Open and save commands

These commands allow the user to open and save an MRI graph analysis in the *.mga format. These are equivalent to the open and save menu options in the File menu.

opens a popup window to load an MRI graph analysis saved in *.mga format.
saves the current MRI graph analysis in *.mga format.

Visualization commands

These commands allow the user to control the visualization of the graphical representations of the connectivity matrix.

zooms in image.
zooms out image.
drags image.
shows/hides data cursor.
shows the colorbar.

Watch also:

MRI Graph Analysis

Footnotes and references

1. ^ Not all measures can be calculated in the presence of negative correlation coefficients.


text:
# 2011 AMC 12A Problems/Problem 14

## Problem

Suppose $a$ and $b$ are single-digit positive integers chosen independently and at random. What is the probability that the point $(a,b)$ lies above the parabola $y=ax^2-bx$?

$\textbf{(A)}\ \frac{11}{81} \qquad \textbf{(B)}\ \frac{13}{81} \qquad \textbf{(C)}\ \frac{5}{27} \qquad \textbf{(D)}\ \frac{17}{81} \qquad \textbf{(E)}\ \frac{19}{81}$

## Solution

If $(a,b)$ lies above the parabola, then $b$ must be greater than $y(a)$. We thus get the inequality $b>a^3-ba$. Solving this for $b$ gives us $b>\frac{a^3}{a+1}$. Now note that $\frac{a^3}{a+1}$ constantly increases when $a$ is positive. Then since this expression is greater than $9$ when $a=4$, we can deduce that $a$ must be less than $4$ in order for the inequality to hold, since otherwise $b$ would be greater than $9$ and not a single-digit integer. The only possibilities for $a$ are thus $1$, $2$, and $3$.

For $a=1$, we get $b>\frac{1}{2}$ for our inequality, and thus $b$ can be any integer from $1$ to $9$.

For $a=2$, we get $b>\frac{8}{3}$ for our inequality, and thus $b$ can be any integer from $3$ to $9$.

For $a=3$, we get $b>\frac{27}{4}$ for our inequality, and thus $b$ can be any integer from $7$ to $9$.

Finally, if we total up all the possibilities we see there are $19$ points that satisfy the condition, out of $9 \times 9 = 81$ total points. The probability of picking a point that lies above the parabola is thus $\frac{19}{81} \rightarrow \boxed{\textbf{E}}$


text:
# Background Check

11 September 2013

You are likely familiar with the cosmic microwave background. This background is a thermal remnant of the big bang. Because of the expansion of the universe, this remnant energy has a temperature of about 2.7 K, which means it exists primarily in the microwave wavelengths. We see this cosmic background as a diffuse, low-energy glow of microwave radiation.

But there is another background that exists, known as the cosmic x-ray background. Just as the cosmic microwave background is a diffuse microwave glow, the cosmic x-ray background is a diffuse x-ray glow. You can see an image of this x-ray background in the image above. It is a false-color image, where red, green and blue represent low, medium and high x-ray energies.

Unlike the microwave background, the x-ray background is not a remnant of the big bang. Instead it is generated through several processes. Most of the background is produced by localized sources such as active galactic nuclei, but other sources are the the local bubble of interstellar media that surrounds the Sun and other stars in our local spiral arm of Orion. But there is a small portion of the background that remains unexplained.

One of the difficulties in understanding the x-ray background is the sheer challenge of observing x-rays at high resolution. X-rays tend to penetrate materials, so you can’t simply make a mirror to focus x-rays the way we do visible light or radio waves. X-ray telescopes must have special materials to reflect x-rays, and they need to have a very long focal length.

But I’ll talk about that next time.


text:
# Energy time uncertainty principle

1. Nov 11, 2012

### bgq

Hi,

Consider a laser source of frequency f. This source is projected to a target at a distance D, so that the light needs time T = D/C to reach the target. I will consider the particle behavior of light in this situation. I will study the motion of one of the photons.

At the exact time T/2, the exact energy of the photon is hf.

How can we interpret this according to the uncertainty principle?

2. Nov 11, 2012

### cosmic dust

You have to clear on what exactly the energy-time uncertainty principle refers to. ΔE is the uncertainty of the energy distribution of the system, while Δt is the time approximately needed for the mean value of the energy to vary by ΔE. In your example, this principle cannot stand, because the photon has definite energy (i.e. ΔE = 0) and it is kept constant (Δt = infite), so you have the non-definite term ΔE Δt = zero x infinity.

3. Nov 11, 2012

### bgq

1) Why Δt is infinite?

2) I have read that uncertainty principle says that its impossible do determine the energy and the time accurately. In the example I provided, it seems that both time and energy are determined accurately.

4. Nov 11, 2012

### cosmic dust

1) Actually, Δt its self in not definite: rate of change = zero , because the energy is constant, therefore, Δt = (uncertainty)/(rate of change) = 0/0 -> not definite (if you considered that ΔE is small but finite, THEN Δt would be infinite)

2) This is a naive statement of the principle but, unfortunately, a famous one. First of all, there is no such observable as “time” . Therefore, Δt cannot stand for the “uncertainty of measuring time”, this is a non-sense expression. I recommend you to prove the principle (it’s not very difficult) in order to full understand what it means.

5. Nov 11, 2012

### bgq

Can you please give me a link to a simple clear proof? All what I found contain much of ambiguity that I can't understand.

6. Nov 11, 2012

### FedEx

I dont understand what you mean by rate of change.

A better way to put it : T is not a dynamical variable.

One must remember what the \delta T and the \delta E mean.

It is usually stated "The time-energy uncertainty principle a statement about how the statistical uncertainty in the energy controls the time scale for a change in the system."

When you started the laser, look at the first say 5 nanoseconds, (\delta T = 5 ns) \delta E is clearly non zero.

7. Nov 12, 2012

### cosmic dust

There was an inaccuracy in my previous post, I' m sorry for that. As FedEx correctly said, Δt is a characteristic tame scale of the system (how exactly we define that scale, it will be shown below)
I don’t have in mind a link, so I will show you a simple proof: consider an observable A and the corresponding operator $$\hat{A}$$ The expectation value of A will be: $$\left\langle A \right\rangle =\left( \psi ,\hat{A}\psi \right)$$ Assuming that the operator is time-independent, the rate of change of the mean value will be:
$$\frac{d}{dt}\left\langle A \right\rangle =\frac{d}{dt}\left( \psi ,\hat{A}\psi \right)=\left( \frac{\partial \psi }{\partial t},\hat{A}\psi \right)+\left( \psi ,\hat{A}\frac{\partial \psi }{\partial t} \right)$$
(FedEx, that’s what I mean by rate of change)and if you use Schrodinger’s equation $$\frac{\partial \psi }{\partial t}=\frac{1}{i\hbar }\hat{H}\psi$$ in the above expression, you get:
$$\frac{d}{dt}\left\langle A \right\rangle =\left( \frac{1}{i\hbar }\hat{H}\psi ,\hat{A}\psi \right)+\left( \psi ,\hat{A}\frac{1}{i\hbar }\hat{H}\psi \right)=-\frac{1}{i\hbar }\left( \psi ,\hat{H}\hat{A}\psi \right)+\frac{1}{i\hbar }\left( \psi ,\hat{A}\hat{H}\psi \right)=\frac{1}{i\hbar }\left( \psi ,\left[ \hat{A},\hat{H} \right]\psi \right)=\frac{1}{i\hbar }\left\langle \left[ \hat{A},\hat{H} \right] \right\rangle$$
Now, you can define a characteristic time scale by which “the mean value of the observable A varies by ΔA, assuming that the rate of change is constant” . I will call this characteristic time scale τ and not Δt, in order to avoid the naïve interpretation of Δt. So:
$$\tau \equiv \frac{\Delta A}{\left| d\left\langle A \right\rangle /dt\ \right|}=\hbar \frac{\Delta A}{\left| \left\langle \left[ \hat{A},\hat{H} \right] \right\rangle \right|}$$
Now, let’s take as granted the general uncertainty principle:
$$\Delta A\Delta B\ge \frac{1}{2}\left| \left\langle \left[ \hat{A},\hat{B} \right] \right\rangle \right|$$
Set in this expression $$\hat{B}=\hat{H}$$ in order to get:
$$\Delta A\Delta E\ge \frac{1}{2}\left| \left\langle \left[ \hat{A},\hat{H} \right] \right\rangle \right|$$
and according to the definition of τ, from the last expression we get:
$$\Delta A\Delta E\ge \frac{\hbar }{2}\frac{\Delta A}{\tau }\Rightarrow \tau \,\Delta E\ge \frac{\hbar }{2}$$
That is the time-energy uncertainty relation. I hope from the derivation of it, you clear it’s meaning.

8. Nov 14, 2012

### bgq

Thank you very much. Although I admit that I am not qualified enough to understand every step in the derivation, but I can see that this principle is based on the probabilistic point of view of quantum mechanics, that is there is uncertainty arose from this point of view regardless the measuring techniques. I also can see that the time is not an instant with a certain uncertainty, but it is a duration.

I still have two questions regardless the initial situation I have introduced:

1) Is it true that the calculated energy E = hf is nothing more than the most probable value, and so there is uncertainty associated with it?
2) How should we take the time? I assume it is the time needed by the light to travel from the source to the target. Is that true?

9. Nov 14, 2012

### cosmic dust

1) If the photon has some definite energy, it will also have some definite momentum. But a wavefunction with definite momentum is no localized in space (it is a plane wave), so the photon can be anywhere. In order to localize as much as possible the wavefunction, the distribution of energy has to be dispersed. So there will be some uncertainty…

2) Time is just a parameter that dictates the evolution of the system. Since we are dealing with quantum mechanics (i.e. with probabilities) you should ask the question "if photon's wavefunction at t=ti (the time of emission) gives probability = 1 to be at some place (the source), then what is the time tf for which the wavefunction gives probability = 1 at some other place (the target). The difference between tf and ti is the time needed for the photon to travel from the source to the target.


text:
# Question #ff1ed

There is no $N {a}^{+} H P {O}_{4}^{2 -}$ salt, and thus the question was not well-proposed......
Of course there is the $N {a}_{2} H P {O}_{4}$ salt, and the $N a {H}_{2} P {O}_{4}$ salt......we need concentration quoted.


text:
## Actions

Export to: EndNote  |  Zotero  |  Mendeley

## Collections

This file is in the following collections:

# Data from "An ϵ-pseudoclassical model for quantum resonances in a cold dilute atomic gas periodically driven by finite-duration standing-wave laser pulses" Open Access

Atom interferometers are a useful tool for precision measurements of fundamental physical phenomena, ranging from local gravitational field strength to the atomic fine structure constant. In such experiments, it is desirable to implement a high momentum transfer beam-splitter,'' which may be achieved by inducing quantum resonance in a finite-temperature laser-driven atomic gas. We use Monte Carlo simulations to investigate these quantum resonances in the regime where the gas receives laser pulses of finite duration, and demonstrate that an $\epsilon$-classical model for the dynamics of the gas atoms is capable of reproducing quantum resonant behavior for both zero-temperature and finite-temperature non-interacting gases. We show that this model agrees well with the fully quantum treatment of the system over a time-scale set by the choice of experimental parameters. We also show that this model is capable of correctly treating the time-reversal mechanism necessary for implementing an interferometer with this physical configuration.

## Descriptions

Resource type
Dataset
Contributors
Creator: Beswick, Benjamin T. 1
Contact person: Beswick, Benjamin T. 1
Creator: Astier, Hippolyte P. A. G. 2
Creator: Gardiner, Simon A. 1
Creator: Hughes, Ifan G. 1
Creator: Andersen, Mikkel F. 3
Creator: Daszuta, Boris 3
1 Durham University, UK
2 University of Cambridge, UK
3 University of Otago, New Zealand
Funder
Leverhulme Trust
Research methods
Other description
Data were produced using the Python programming language and plotted using the Matplotlib.pyplot module. Figure 1 was produced using Inkscape.
Keyword
quantum resonance
cold atomic gas
laser pulses
atomic physics
Subject

Location
Language
Cited in
Identifier
ark:/32150/rj4304528
Rights
Publisher

Date Created

## File Details

Depositor
B.T. Beswick
Date Modified
29 April 2016, 15:04:37
Audit Status
Audits have not yet been run on this file.
Characterization
File format: x-gzip (GZIP Format, GZIP)
Mime type: application/x-gzip
File size: 59193321


text:
Water in a paper conical filter drips into a cup as shown in the attachment. Let x denote the height of the water in the filter. Let y denote the height of the water in the cup. If 10 in^3 of water is poured into the filter, find the relationship between dy/dt and dx/dt

2. First, we need to find an equation which connects all the variables in this problem. From your description and the figure, it is easy to see that at any given time the volume of the water in the conical filter plus the volume of water in the cup is 10 in^3 (assume that it takes no time for the water to drip from the filter to the cup). Mathematically, we have the equation:

$\frac{1}{3}\pi r^2 x + \pi (4)^2 y = 10$

Also we are interested in the variables $x$ and $y$, so it is better to rewrite the third variable $r$ in terms of $x$ or $y$. By the properties of similar figures, it is easy to see that $\frac{r}{2}=\frac{x}{4} \rightarrow r=\frac{x}{2}$. Plug it into the above equation, we have:

$\frac{1}{3}\pi (\frac{x}{2})^2 x + \pi (4)^2 y = 10$ or $\frac{\pi}{12} x^3 + 16\pi y = 10$

Now differentiate both sides with respect to time $t$, we have:

$\frac{\pi}{4}x^2 \frac{dx}{dt}+16\pi\frac{dy}{dt}=0 \Rightarrow x^2\frac{dx}{dt}+64\frac{dy}{dt}=0$

We have the relations between $\frac{dx}{dt}$ and $\frac{dy}{dt}$


text:
# How to solve double limit $\lim_{m\to\infty}[\lim_{n\to\infty}(\cos(m!\cdot \pi\cdot x))^{2n}]$ [duplicate]

Please provide some hints as to how to solve questions with double limits such as this:

$$\lim_{m\to\infty}\left[\lim_{n\to\infty}(\cos(m!\cdot \pi\cdot x))^{2n}\right]$$

One of the things I did was convert the original function to: $$e^{n\ln(\cos(m!\cdot\pi\cdot x)^2)}$$ and then change cosine into sine and take $t=1/m$, and try to use $$\lim \frac{\sin(m!\cdot\pi\cdot x)}{m!\cdot\pi\cdot x}$$ but that only messed it up even further. I obviously can't use L'Hopital as not both the numerator and denominator go to zero.

Another thing was to try to use the power series expansion, but that seemed even more complicated as there is still the power of 2n to deal with.

-
Just to clarify, the power 2n is on the cosine function and not the part inside it. So it goes like (cos(m!.pi.x))^(2n). –  user130113 Feb 21 at 2:37
Hint: Look separately at $x$ rational, $x$ irrational. –  André Nicolas Feb 21 at 2:37
But is my approach right though? –  user130113 Feb 21 at 2:41
I do not see a direct way to use your approach. –  André Nicolas Feb 21 at 2:43
–  Martin Sleziak Feb 21 at 7:45

## marked as duplicate by Martin Sleziak, Sami Ben Romdhane, mau, heropup, Yiorgos S. SmyrlisFeb 21 at 9:50

Hint: Look separately at $x$ irrational, $x$ rational.

If $x$ is irrational, then $\cos(m!\pi x)$ has absolute value less than $1$.

If $x$ is rational, say $\frac{a}{b}$ where $a$ and $b$ are integers, what is the value of $\cos(m!\pi x)$ for large enough $m$?

Detail: Suppose that $x$ is irrational. Fix $m$. Then $m!\pi x$ is not an integer multiple of $\pi$. It follows that $|\cos(m!\pi x)|\lt 1$. Let $c_m=\cos(m\pi x)$. Since $c_m$ has absolute value less than $1$, we have $\lim_{n\to\infty} c_m^{2n}=0$. Thus our double limit is $\lim_{m\to\infty} 0$, which is $0$.

Now suppose that $x$ is rational. Then we can assume that $x=\frac{a}{b}$, where $a$ and $b$ are integers. Without loss of generality we may suppose that $b$ is positive.

Let $m\ge b$. Then $m!x=(b-1)!a$. Thus $m!\pi x$ is an integer multiple of $\pi$. It follows that $\cos(m!\pi x)=\pm 1$, and therefore $(\cos(m!\pi x))^{2n}=1$. Thus for any $m\ge b$, we have $$\lim_{n\to\infty} (\cos(m!\pi x))^{2n}=1.$$ We conclude that $$\lim_{m\to\infty}\left[ \lim_{n\to\infty} (\cos(m!\pi x))^{2n} \right]=1.$$

-
Would it still not be anything between 1 and -1? –  user130113 Feb 21 at 4:31
What is "it"? For $x$ rational, we can without loss of generality assume $b\ge 1$. If $m\ge b+2$, the cosine is $1$. –  André Nicolas Feb 21 at 4:35
This is what I have come to conclude: If m=1, x=1, cos(m!πx)^2n= 1 and for any other x=/ an integer, cos(m!πx)^2n=0. Since m=integer always, it comes down to the value of x. Is that correct at all? –  user130113 Feb 21 at 4:47
We are ultimately not interested in $m=1$, but in large $m$. (For $x$ not rational, the size of $m$ does not matter.) You do not mean that $(\cos(m!\pi x))^{2n}=0$, though the limit is $0$ if $x$ is not rational. At a certain point, if you wish, I can flesh out the hints. –  André Nicolas Feb 21 at 4:51
By the way, important, do check the parentheses, do you mean $\cos((m!\pi x)^{2n})$ or $(\cos(m!\pi x))^{2n}$? –  André Nicolas Feb 21 at 4:53


text:
Search results

Search: All articles in the CJM digital archive with keyword Lie algebra

 Expand all        Collapse all Results 1 - 6 of 6

1. CJM 2012 (vol 65 pp. 82)

Félix, Yves; Halperin, Steve; Thomas, Jean-Claude
 The Ranks of the Homotopy Groups of a Finite Dimensional Complex Let $X$ be an $n$-dimensional, finite, simply connected CW complex and set $\alpha_X =\limsup_i \frac{\log\mbox{ rank}\, \pi_i(X)}{i}$. When $0\lt \alpha_X\lt \infty$, we give upper and lower bound for $\sum_{i=k+2}^{k+n} \textrm{rank}\, \pi_i(X)$ for $k$ sufficiently large. We show also for any $r$ that $\alpha_X$ can be estimated from the integers rk$\,\pi_i(X)$, $i\leq nr$ with an error bound depending explicitly on $r$. Keywords:homotopy groups, graded Lie algebra, exponential growth, LS categoryCategories:55P35, 55P62, , , , 17B70

2. CJM 2009 (vol 62 pp. 52)

Deng, Shaoqiang
 An Algebraic Approach to Weakly Symmetric Finsler Spaces In this paper, we introduce a new algebraic notion, weakly symmetric Lie algebras, to give an algebraic description of an interesting class of homogeneous Riemann--Finsler spaces, weakly symmetric Finsler spaces. Using this new definition, we are able to give a classification of weakly symmetric Finsler spaces with dimensions $2$ and $3$. Finally, we show that all the non-Riemannian reversible weakly symmetric Finsler spaces we find are non-Berwaldian and with vanishing S-curvature. This means that reversible non-Berwaldian Finsler spaces with vanishing S-curvature may exist at large. Hence the generalized volume comparison theorems due to Z. Shen are valid for a rather large class of Finsler spaces. Keywords:weakly symmetric Finsler spaces, weakly symmetric Lie algebras, Berwald spaces, S-curvatureCategories:53C60, 58B20, 22E46, 22E60

3. CJM 2008 (vol 60 pp. 892)

Neeb, Karl-Hermann; Wagemann, Friedrich
 The Second Cohomology of Current Algebras of General Lie Algebras Let $A$ be a unital commutative associative algebra over a field of characteristic zero, $\k$ a Lie algebra, and $\zf$ a vector space, considered as a trivial module of the Lie algebra $\gf := A \otimes \kf$. In this paper, we give a description of the cohomology space $H^2(\gf,\zf)$ in terms of easily accessible data associated with $A$ and $\kf$. We also discuss the topological situation, where $A$ and $\kf$ are locally convex algebras. Keywords:current algebra, Lie algebra cohomology, Lie algebra homology, invariant bilinear form, central extensionCategories:17B56, 17B65

4. CJM 2006 (vol 58 pp. 1291)

Weimar-Woods, Evelyn
 The General Structure of $G$-Graded Contractions of Lie Algebras I. The Classification We give the general structure of complex (resp., real) $G$-graded contractions of Lie algebras where $G$ is an arbitrary finite Abelian group. For this purpose, we introduce a number of concepts, such as pseudobasis, higher-order identities, and sign invariants. We characterize the equivalence classes of $G$-graded contractions by showing that our set of invariants (support, higher-order identities, and sign invariants) is complete, which yields a classification. Keywords:Lie algebras, graded contractionsCategories:17B05, 17B70

5. CJM 1998 (vol 50 pp. 210)

Zhao, Kaiming
 Isomorphisms between generalized Cartan type $W$ Lie algebras in characteristic $0$ In this paper, we determine when two simple generalized Cartan type $W$ Lie algebras $W_d (A, T, \varphi)$ are isomorphic, and discuss the relationship between the Jacobian conjecture and the generalized Cartan type $W$ Lie algebras. Keywords:Simple Lie algebras, the general Lie algebra, generalized Cartan type $W$ Lie algebras, isomorphism, Jacobian conjectureCategories:17B40, 17B65, 17B56, 17B68

6. CJM 1997 (vol 49 pp. 133)

Reeder, Mark
 Exterior powers of the adjoint representation Exterior powers of the adjoint representation of a complex semisimple Lie algebra are decomposed into irreducible representations, to varying degrees of satisfaction. Keywords:Lie algebras, adjoint representation, exterior algebraCategories:20G05, 20C30, 22E10, 22E60


text:
2 Added

Assume we have a linear operator P_0: D->H $P_0: D->H$ where D $D$ is the domain and H $H$ some Hilbert space it is acting on. Assume moreover that the spectrum of P_0 $P_0$ is purely essential. If z $z$ is a complex number not in the spectrum of P_0 $P_0$ then the resolvent (P_0 $(P_0 - z)^{-1} z)^{-1}$ exists as a bounded operator H->D, $H->D$, meaning there is a constant (depending on z) $z$) such that |u|\leq C|(P_0-z)u| $|u|\leq C|(P_0-z)u|$ where the norm on then the left hand side is that of the domain D.

Now perturb P_0 $P_0$ by a "potential" V:R^n->C $V:R^n->C$ (complex-valued function on Euclidean n-space) and denote P=P_0+V $P=P_0+V$ where we assume V $V$ has compact support. Assume also that V $V$ generates an eigenvalue z_0 $z_0$ away from the essential spectrum of P_0 $P_0$ (which we may assume is also the essential spectrum of P). $P$). Thus there is a non-zero u $u$ in D $D$ with (P-z_0)u=0. $(P-z_0)u=0$. Let X $X$ be the characteristic function of some set (call it X $X$ as well) which is disjoint from the support of V. $V$. Then |Xu|\leq C|(P_0-z_0)Xu|=C|(P-z_0)Xu|=0$|Xu|\leq C|(P_0-z_0)Xu|=C|(P-z_0)Xu|=0$, i.e. the eigenfunction must be 0 wherever the potential is zero.

Am I forgetting something?

1

# Operators and eigenfunctions - is this correct?

Assume we have a linear operator P_0: D->H where D is the domain and H some Hilbert space it is acting on. Assume moreover that the spectrum of P_0 is purely essential. If z is a complex number not in the spectrum of P_0 then the resolvent (P_0 - z)^{-1} exists as a bounded operator H->D, meaning there is a constant (depending on z) such that |u|\leq C|(P_0-z)u| where the norm on then left hand side is that of the domain D.

Now perturb P_0 by a "potential" V:R^n->C (complex-valued function on Euclidean n-space) and denote P=P_0+V where we assume V has compact support. Assume also that V generates an eigenvalue z_0 away from the essential spectrum of P_0 (which we may assume is also the essential spectrum of P). Thus there is a non-zero u in D with (P-z_0)u=0. Let X be the characteristic function of some set (call it X as well) which is disjoint from the support of V. Then |Xu|\leq C|(P_0-z_0)Xu|=C|(P-z_0)Xu|=0, i.e. the eigenfunction must be 0 wherever the potential is zero.

Am I forgetting something?


text:
# How do you find the scalar and vector projections of b onto a given a = <3, -6, 2>, b = <1, 1, 1>?

Nov 14, 2016

The scalar projection is $= - \frac{1}{7}$
The vector projection is =-1/49〈3,-6,2〉

#### Explanation:

The scalar projection of $\vec{b}$ onto $\vec{a}$ is

=(veca.vecb)/(∥veca∥)

veca=〈3,-6,2〉

vecb=〈1,1,1〉

veca.vecb=〈3,-6,2〉.〈1,1,1〉=3-6+2=-1

∥veca∥=∥〈3,-6,2〉∥=sqrt(9+36+4)=sqrt49=7

The scalar projection is $= - \frac{1}{7}$

The vector projection is (veca.vecb)/(∥veca∥)^2veca

=-1/49〈3,-6,2〉


text:
# Multiplexing ? Newbie help

Status
Not open for further replies.

#### Lighty

##### New Member
Hi all

I've recently bought a PICKit 2 programmer and I am still learning the basics.

I have managed to program a couple of flashing patterns to the 18f690 on the dev board using GCBasicIDE. I'm a novice so please take it easy on me as I don't a grate deal on knowledge, I do how ever understand electronics to a degree.

So this brings me to my questions.

Multiplexing? This is where you would use a "group" of "share" output with individual Commons. Like when using 7 segment LED, you can use several segments all saring the same inputs but using individual grounds, the grounds are triggered in sync with the relivent segment signals....?

Ok so heres my problem, I busy with a project that has over 30 LEDs, 20x 2 way switches (with off in the middle), 7x 5 position rotory switches and 7x 7segment LEDs.......

My idea thought

Basically, the rotary switches are for controlling the speed of some motors via the PWM outputs, there is one 7seg per rotary switch, showing its position, Eg 1,2,3,4,5,0. I though of using a analog input per switch using the method I've seen on a 4x4 botton keypad, using resistors across the switch outputs to create a different voltage per position, changing the code to change the PWM output.... can this be done?

The next big user of I/O pins will be the 20 switches (on-off-on) for the "auto" (where the MCU will do the controlling), off, and "manual" where the output will be activated (override). Normally you would just push your signal from the MCU through the one pole of the switch to the divice and use the other pole to drive directly -thereby choosing the Auto or Man input - hope that make sence? - but there are 2 status LED for each switch, 1 shows Auto mode and other Man. mode.

So the only way I can think of doing this is multiplexing the LEDs so to use only 14 (10x4) pins and not 40.....?

I have not yet decided on the PIC chip, any ideas?, but would have to be at least a 40 pin PIC? could I not use 2 DIP PICs apposed to a TQFP mount as this is a home made DIY project................

#### Nigel Goodwin

##### Super Moderator
Check my tutorials - one of which multiplexes two 7 segment displays on a single port, and one 64 LED's on two 8 bit ports.

For 40 LED's, 4x10 (14 pins) would be a poor choice, 5x8 would only use 13 pins, and is more convenient for programming, as it only requires a port width, and not a port width plus 2.

There's also one for multiplexing a hex keypad - 16 keys (4x4).

Last edited:

#### Magen

##### New Member
Hi Lighty,
You could also look at using shift registers to increase the number of inputs/outputs. Nigel's tutorials will also help you with sending/receiving data serially.

Last edited:

#### Lighty

##### New Member
thanks guys, I'll check the tutorial and see if I can make heads or tails....

#### nickelflippr

##### Member
Can't help but think that the seven 5 pole switches are going to be expensive. Why not just look at it from a digital solution? It would simplify the project a lot, and get rid of unnecessary indicator leds.

If I have this right you are trying to control 7 motors? The 16f690 could do the job. So at a minimum there would need to be:

10 outputs:
7 for software PWM's
3 for the seven chained 74HCT595's (i.e. one for each 7 seg display)

7 inputs:
7 one each for the motor control (resistor divider for each fixed speed/button)

Arguably buying the seven 595's would put you in the range of a 40 pin PIC, so another interesting solution would be two 40 pin 18f4331's. You would then have 8 total hardware PWM's, and your 40 leds to boot (as long as you don't go over the max package current).

Here is a GCBasic example of using two 595's to light up some leds. To use for 7 segment displays, a select case scenario would be required for each digit.

Code:
'A program to flash 16 LEDs with two HC595's
'pins - OE tied low, Reset tied high

'Chip model
#chip 16f88, 20

#define Data PortB.0
#define Latch PortB.1
#define Clock PortB.2

dir PortB.0 out
dir PortB.1 out
dir PortB.2 out

dim loop as word

Start:
For loop = 1 to 65535
Shiftout Loop
wait 1 ms
Next
goto Start

'--------------------------------------------------
Sub Shiftout (Dataout as word) #NR

Latch = 0				'Start output data latch
For Clocks = 1 to 16
Clock = 0			'Shift register clocked
Data = Dataout.0
Clock = 1
Rotate Dataout Right Simple
Next Clocks
Latch = 1

End Sub
'--------------------------------------------------

#### Lighty

##### New Member
Can't help but think that the seven 5 pole switches are going to be expensive. Why not just look at it from a digital solution? It would simplify the project a lot, and get rid of unnecessary indicator leds.

I got the "control Panel" to fitted with most hardware already I'm not 100% but think they are Break Before Make, if not would this be a problem?

If I have this right you are trying to control 7 motors?

No, Only 2 motors

The 16f690 could do the job. So at a minimum there would need to be:

10 outputs:
7 for software PWM's
3 for the seven chained 74HCT595's (i.e. one for each 7 seg display)

7 inputs:
7 one each for the motor control (resistor divider for each fixed speed/button)

Arguably buying the seven 595's would put you in the range of a 40 pin PIC, so another interesting solution would be two 40 pin 18f4331's. You would then have 8 total hardware PWM's, and your 40 leds to boot (as long as you don't go over the max package current).

Here is a GCBasic example of using two 595's to light up some leds. To use for 7 segment displays, a select case scenario would be required for each digit.

Code:
'A program to flash 16 LEDs with two HC595's
'pins - OE tied low, Reset tied high

'Chip model
#chip 16f88, 20

#define Data PortB.0
#define Latch PortB.1
#define Clock PortB.2

dir PortB.0 out
dir PortB.1 out
dir PortB.2 out

dim loop as word

Start:
For loop = 1 to 65535
Shiftout Loop
wait 1 ms
Next
goto Start

'--------------------------------------------------
Sub Shiftout (Dataout as word) #NR

Latch = 0				'Start output data latch
For Clocks = 1 to 16
Clock = 0			'Shift register clocked
Data = Dataout.0
Clock = 1
Rotate Dataout Right Simple
Next Clocks
Latch = 1

End Sub
'--------------------------------------------------
Thx Nickelflippr but I'm

#### nickelflippr

##### Member
Thx Nickelflippr but I'm
Well, the example given shows how to light 16 leds (or two seven segment displays with modification) with just three outputs and two 74HC595's. There are tons of examples on the net, take a look, and read the data sheet. Is there confusion about that?

Using the 595's give you full led brightness, and uses less microcontroller pins. If you multiplex the leds, or the seven segment displays, the brightness will be diminished. GCBasic has a seven segment library that will multiplex 4 displays in a common cathode configuration. This would require one port plus 4 address pins (and transistors) to enable/disable each display. If you need code for common anode configuration let me know.

Since you have most of the parts, here is experiment you can try right now.
The GCBasic code for two multiplexed seven segment displays, can expand up to four displays:
Code:
#chip 18f4620,20
#config MCLRE = On
'This program will show "Hello" on a LED display
'The display should be connected to PORTB

#define DisplayPortA PORTB
#define DisplayPortB PORTB
#define DispSelectA nop
#define DispSelectB nop
#define Display1 PortA.0
#define Display2 PortA.1
#define waitDisplay wait 12 ms
#define CommonAnode  'extra code required in library!!!!
dir PortB out
dir PortA.0 out
dir PortA.1 out

main:
DIM Message(10)
Message() = "HelloYou"
For counter = 1 to 8
For seeDisplay = 1 to 25
Set Display2 Off:Set Display1 On
DisplayChar 1, Message(counter)
waitDisplay
Set Display1 Off:Set Display2 On
DisplayChar 2, Message(counter)
waitDisplay
next
next

for Counter = 0 to 99
Seg2 = Counter/10
Seg1 = Counter % 10
For seedisp = 1 to 7
Set Display2 Off:Set Display1 On
DisplayValue 1, SEG1
waitDisplay
Set Display1 Off:Set Display2 On
DisplayValue 2, SEG2
waitDisplay
next
next
goto main

#### Lighty

##### New Member
Thanks Nickelflippr,

If I understand correct, when you multiplex using the "ports" all the anodes for each segment is connected in parallel, and use another bit per segment to trigger the cathode, in other words, the bits will change in sync with the triggering of each segments common cathode....(sorry, I'm bad in trying to explain) but basically segment 1 will light then turn of, the seg 2 will light then turn off and so one, so quickly it looks like it stays on? but if too many segments are added, the time delay between "refreshing" is too long and would cause the segment to look dim?

And using the 74h595, it would stay light until its refreshed with something different? how many segments can you "chain up" together?

#### nickelflippr

##### Member
Yes, I think you have it.

Don't know how many cmos chips you can chain together. My guess is plenty, not in the device data sheet, maybe some analog person knows that one?

#### Lighty

##### New Member
but if they all linked, what makes them chain one after the other and not all the same time? would they have different ID or something?

#### Lighty

##### New Member
Ok, so I need to control plenty single LEDs and a couple 7 Segments, could this be done off the same "String" of 74HC595's? or would this cause a problem?

Could I run to seperate "strings" of 74HC595" or can only 1 set of pins be setup as the Latch, Data and Clock?

Status
Not open for further replies.


text:
1. ## Triangle hydrostatic force!

A triangular plate with height 4 ft and a base of 5 ft is submerged vertically in water so that the top is 2 ft below the surface. Express the hydrostatic force against one side of the plate as an integral and evaluate it. (Give your answer correct to the nearest whole number.)

Thanks! I don't know what to do at all, by the way, all I know is that I need the area and pressure. There's a picture of it of a triangle suspended in a square, and the hypotenuse side is facing the lower right.

2. Is the diagram I have what you mean?.

If so, we can use similar triangles:

$\frac{w(x)}{5}=\frac{6-x}{4}$

$w(x)=\frac{5}{4}(6-x)$

You did not state your weight density of the fluid. You can put that in.

$\int_{2}^{6}x\left(\frac{5}{4}(6-x)\right)dx$

3. Oh thanks!! They didn't give the density, but I think since we are in feet, I need to use 62.5 as the density, Im not too sure on the units. Also, where should I put that? I can pull it out in the front, right? Thank you!

4. Yeah, you just multiply by that. It's a constant.

5. Okay, thanks!!!


text:
# What else do I need to produce two dots above o?

In tex file:

  Toks\"{o}z

It produces

-
Welcome to TeX.SX! Please help us to help you and add a minimal working example (MWE) that illustrates your problem. It will be much easier for us to reproduce your situation and find out what the issue is when we see compilable code, starting with \documentclass{...} and ending with \end{document}. In this case, I can't reproduce the problem with the code you've provided; it works as expected and produces a proper ö. – Sean Allred Jul 2 '14 at 23:49
Make sure you are using " but not strange unicode madness in case you have copied it from a website etc. such as '' – percusse Jul 2 '14 at 23:52
It is very obvious I am using " not ' '. At least in my tex file it looks like " to me. I have used "shift" key to input ". – yanfyon Jul 3 '14 at 2:37

It is difficult to see in some editors, but in the first case you have \" acting on o, in the second: \' acting on ', hence the strange result.

\documentclass{article}

\begin{document}

Toks\"{o}z

Toks\''{o}z

\end{document}

BTW: your question gives an interesting example, why " and '' should never be mixed. Improper usage of " as a quote does not give so visible result.

-
I use TeXworks as editor. I agree that two separate ' without space between them looks same as ". But I use "shift" key to input ". The result is still same to me. – yanfyon Jul 3 '14 at 3:27
@yanfyon TeXworks, by default, auto-converts " to '' (Format -> Smart Quotes). I always turn this off from the Preferences, but some people clearly find it useful. (This happens even if you type in " using the standard input method, so you may not be aware of it.) – Joseph Wright Jul 3 '14 at 5:54

Maybe that's very naive, but why not simply input direcly ö in the editor?

\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{lmodern}

\begin{document}

Toksöz

\textsc{Toksöz}

TOKSÖZ

\end{document}

-

I just found out why this happened. In the TeXworks editor I use, the original setup is: "Format"--> "Smart Quotes"-->"Tex ligatures", after I changed the setup to: "Format"--> "Smart Quotes"-->"None", then it works as expected !!!!

-
I am really confused. When I changed the setup of "Smart Quotes" from "None" back to "Tex ligatures" again, it still works normal. But I still believe the abnormal behavior has something to do with the "Smart Quotes" setup. – yanfyon Jul 3 '14 at 3:55
I make one more test. If I set the "Smart Quotes" as "Tex ligatures" and re-input Toks\"{o}z, it still comes out as abnormal. So now I am sure the abnormal out is indeed related to the "Smart Quotes" setup. – yanfyon Jul 3 '14 at 4:16
The Smart Quotes setting has an effect only at the time you press the " key. Changing Smart Quotes on or off doesn't go through your document and replace existing " characters. – cjm Jul 3 '14 at 7:38
I petition to have this "feature" renamed "dumb quotes". (Especially since all the editor has to do is to check if the previous character is \  and, if so, leave the " alone.) – David Richerby Jul 3 '14 at 10:47


text:
# Thread: Pythagorean theorem and triangles Question

1. ## Pythagorean theorem and triangles Question

I need help with c) and d) and also e).

For c) they are asking for 2θ however I don't know what they did to get 1 - 2y^2.
d) Can you please explain to me what we do with the (pi - θ)?
e) Cos is adjacent over hypotenuse. However what do we do with the inverse of cos and y?

2. Originally Posted by florx

I need help with c) and d) and also e).

For c) they are asking for 2θ however I don't know what they did to get 1 - 2y^2.
d) Can you please explain to me what we do with the (pi - θ)?
e) Cos is adjacent over hypotenuse. However what do we do with the inverse of cos and y?

c) $\cos 2\theta = cos^2 \theta - sin^2 \theta$
d) $\sin (\pi - \theta) = \sin \theta$

"Why?" you ask. Imagine the unit circle if you will. Sine is positive in the first and second quadrant (between 0 and $\pi$.) Now $\theta < 90^{\cdot}$ so $\sin (\pi - \theta)$ lies in the second quadrant. By symmetry, it has the value of $\sin \theta$. Draw it out to see for yourself.

e) This one is trickier. Split it up into two parts. Let $sin^2 (arccos y) = sin^2 u$ where $u = arccos y$

Now $y = \cos u$. Looking at your triangle the only angle that satisfies this equation is $u = \frac{\pi}{2} - \theta$

So we now have $sin^2 (\frac{\pi}{2} - \theta)$. Look at your unit circle again and you'll realise that $\sin (\frac{\pi}{2} - \theta)$ is the same thing as $\cos \theta$

3. Thank you so much for your input. I understand what you about the sin(pi-θ). However i still don't quite understand how to get the answer.

Taken from the back of the book,

c) 1 - 2y^2
d) y
e) 1 - y^2

Thanks again for trying to help me and I hope you can help me through this and solve this problem.

4. Originally Posted by florx
Thank you so much for your input. I understand what you about the sin(pi-θ). However i still don't quite understand how to get the answer.

Taken from the back of the book,

c) 1 - 2y^2
d) y
e) 1 - y^2

Thanks again for trying to help me and I hope you can help me through this and solve this problem.
You know that $sin(\pi - \theta) = sin\theta$, right?

so, if you look at the triangle,

$sin\theta = \frac{opposite}{hypotenuse}= \frac{y}{1} = y$

Go through this link if you still have confusions:

Trigonometric functions - Wikipedia, the free encyclopedia


text:
# Field theory

A field theory is a mathematical description of physical phenomena which describes the behavior of a particle, not only by the physical qualities of the particle itself (e.g., mass, charge, etc.) but also the position of that particle in space relative to other objects. Such descriptions are called "field theories" because the effects of a particle's position exist regardless of the presence of an actual particle, and so describe physical phenomenon at every point in space - such a description is a "field."

## History

A "field theory" describes a force on a particle in terms of its position, or location in a "field". An example of a field theory is James Maxwell's theory of electromagnetism. Subsequent field theories are the theory of relativity, and modern quantum field theory.

Isaac Newton's (classical) theory of gravity is also considered a field theory, but it is based on "action at a distance" and does not have a force field. To the extent that a field theory has come to imply the existence of a force field, Newtonian gravity would not be one.

Newton described the force on a particle as related to the product of the particles mass $m_1 \$ and its distance from another mass:

$\frac{m_2}{r^2} \$

This is traditionally called a field theory because regardless of the first particle, the second gives rise to a "field" everywhere in space which would affect any particle at any position.


text: A single postage stamp costs $0.44. How much would a roll of 1,000 stamps cost? And there's really a couple of ways to do it. And I'll do it both ways just to show you they both work. One is kind of a faster way, but I want to make sure you understand why it works. And then we'll verify that it actually gives us the right answer using maybe the more traditional way of multiplying decimals. So we're starting at$0.44. I'll just write it as 0.44. Well, that's one stamp. So this is 1 stamp. I'll write it like this, 1 stamp. How much would 10 stamps cost? Well, if 1 stamp is $0.44, then 10 stamps, we could move the decimal to the right one place. And so it would be-- and now this leading 0 is not that useful. So it would now be$4.4. Or if you want to make it clear, it would be $4.40. Now, what happens if you want to have 100 stamps? Well, the same idea is going to happen. We're now taking 10 times more, so we're going to move the decimal to the right once. So 100 stamps are going to cost$44. And this should make sense for you. If 1 stamp is 44/100 of $1, then 100 stamps are going to be 44/100 of$100, or $44. Or you could view it as we've just moved the decimal over one place. So if we want 1,000 stamps, we'd move the decimal to the right one more time. Moving the decimal to the right is equivalent to multiplying by 10. So then it would be$440. And we could add another trailing zero just to make clear that there's no cents over here. So if you wanted to do it really quickly, you could have started with $0.44. And you say look, I'm not multiplying by 10. I'm not multiplying by 100. I'm multiplying by 1,000. So you're going to have to put another trailing zero over here. And you would move the decimal from over here to over here. You've essentially multiplied this times 10 times 10 times 10, which is 1,000. So then this would become$440. And let's verify that this works exactly the same if we multiply it the traditional way, the way we multiply decimals. So if you have 1,000 times $0.44. So you start over here. 4 times 0 is 0. 4 times 0 is 0. 4 times 0 is 0. 4 times 1 is 4. Or, you could just say, hey, this was 4 times 1,000. Then we're going to go one place over. So we're going to add a 0. And once again, we're going to have 4 times 0 is 0. 4 times 0 is 0. 4 times 0 is 0. 4 times 1 is 4. Or we just did 4 times 1,000. So that is 4,000 if you don't include this 0 that we added here ahead of time because we're going one place to the left. And then we have nothing left. I haven't at all thought about the decimals right now. So far I really just viewed it as 1,000 times 44. I've been ignoring the decimal. So if it was 1,000 times 44, we would get 0 plus 0 is 0. 0 plus 0 is 0. 0 plus 0 is 0. 4 plus 0 is 4. 4 plus nothing is 4. And if you ignore the decimal, that makes a lot of sense because 1,000 times 4 is 4,000. And 1,000 times 40 would be 40,000. So you would get 44,000. But this, of course, is not a 44. This is a 44/100. We have, between the two numbers, two numbers behind the decimal point. So we need to have two numbers behind or to the right of the decimal point in our answer, so one, two right over there. So once again, we get$440 for the 1,000 stamps.