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https://labs.tib.eu/arxiv/?author=James%20Geach
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• ### Evolution of Dust-Obscured Star Formation and Gas to z=2.2 from HiZELS(1702.06182)
Feb. 20, 2017 astro-ph.GA
We investigate the far-infrared properties of galaxies selected via deep, narrow-band imaging of the H$\alpha$ emission line in four redshift slices from $z=0.40$--$2.23$ over $\sim 1$deg$^2$ as part of the High-redshift Emission Line Survey (HiZELS). We use a stacking approach in the Herschel PACS/SPIRE bands, along with $850\,\mu$m imaging from SCUBA-2 to study the evolution of the dust properties of H$\alpha$-emitters selected above an evolving characteristic luminosity threshold, $0.2L^\star_{{\rm H}\alpha}(z)$. We investigate the relationship between the dust temperatures and the far-infrared luminosities of our stacked samples, finding that H$\alpha$-selection identifies cold, low-$L_{\rm IR}$ galaxies ($T_{\rm dust}\sim 14$k; $\log[L_{\rm IR}/{\rm L}_\odot]\sim 9.9$) at $z=0.40$, and more luminous, warmer systems ($T_{\rm dust}\sim 34$k; $\log[L_{\rm IR}/{\rm L}_\odot]\sim 11.5$) at $z=2.23$. Using a modified greybody model, we estimate "characteristic sizes" for the dust-emitting regions of HiZELS galaxies of $\sim 0.5$kpc, nearly an order of magnitude smaller than their stellar continuum sizes, which may provide indirect evidence of clumpy ISM structure. Lastly, we measure the dust masses from our far-IR SEDs along with metallicity-dependent gas-to-dust ratios ($\delta_{\rm GDR}$) to measure typical molecular gas masses of $\sim 10^{10}$M$_\odot$ for these bright H$\alpha$-emitters. The gas depletion timescales are shorter than the Hubble time at each redshift, suggesting probable replenishment of their gas reservoirs from the intergalactic medium. Based on the number density of H$\alpha$-selected galaxies, we find that typical star-forming galaxies brighter than $0.2L^{\star}_{{\rm H}\alpha}(z)$ host a significant fraction ($35\pm10$%) of the total gas content of the Universe, consistent with the predictions of the latest cosmological simulations.
• ### A unifying evolutionary framework for infrared-selected obscured and unobscured quasar host haloes(1610.03493)
Oct. 11, 2016 astro-ph.GA
Recent measurements of the dark matter halo masses of infrared-selected obscured quasars are in tension --- some indicate that obscured quasars have higher halo mass compared to their unobscured counterparts, while others find no difference. The former result is inconsistent with the simplest models of quasar unification that rely solely on viewing angle, while the latter may support such models. Here, using empirical relationships between dark matter halo and supermassive black hole masses, we provide a simple evolutionary picture that naturally explains these findings and is motivated by more sophisticated merger-driven quasar fueling models. The model tracks the growth rate of haloes, with the black hole growing in spurts of quasar activity in order to "catch-up" with the $M_{\textrm{BH}}$ - $M_{\textrm{star}}$ - $M_{\textrm{halo}}$ relationship. The first part of the quasar phase is obscured and is followed by an unobscured phase. Depending on the luminosity limit of the sample, driven by observational selection effects, a difference in halo masses may or may not be significant. For high luminosity samples, the difference can be large (a few to 10 times higher masses in obscured quasars), while for lower luminosity samples the halo mass difference is very small, much smaller than current observational constraints. Such a simple model provides a qualitative explanation for the higher mass haloes of obscured quasars, as well as rough quantitative agreement with seemingly disparate results.
• ### The SCUBA-2 Cosmology Legacy Survey: The clustering of submillimetre galaxies in the UKIDSS UDS field(1604.00018)
Sept. 22, 2016 astro-ph.CO, astro-ph.GA
Submillimetre galaxies (SMGs) are among the most luminous dusty galaxies in the Universe, but their true nature remains unclear; are SMGs the progenitors of the massive elliptical galaxies we see in the local Universe, or are they just a short-lived phase among more typical star-forming galaxies? To explore this problem further, we investigate the clustering of SMGs identified in the SCUBA-2 Cosmology Legacy Survey. We use a catalogue of submillimetre ($850\mu$m) source identifications derived using a combination of radio counterparts and colour/IR selection to analyse a sample of 914 SMGs in the UKIDSS Ultra Deep Survey (UDS), making this the largest high redshift sample of these galaxies to date. Using angular cross-correlation techniques, we estimate the halo masses for this large sample of SMGs and compare them with passive and star-forming galaxies selected in the same field. We find that SMGs, on average, occupy high-mass dark matter halos (M$_{\text{halo}} >10^{13}$M$_{\odot}$) at redshifts $z > 2.5$, consistent with being the progenitors of massive quiescent galaxies in present-day galaxy clusters. We also find evidence of downsizing, in which SMG activity shifts to lower mass halos at lower redshifts. In terms of their clustering and halo masses, SMGs appear to be consistent with other star-forming galaxies at a given redshift.
• ### The Star-Formation History of BCGs to z = 1.8 from the SpARCS/SWIRE Survey: Evidence for significant in-situ star formation at high-redshift(1508.07302)
Aug. 28, 2015 astro-ph.CO, astro-ph.GA
We present the results of a MIPS-24um study of the Brightest Cluster Galaxies (BCGs) of 535 high-redshift galaxy clusters. The clusters are drawn from the Spitzer Adaptation of the Red-Sequence Cluster Survey (SpARCS), which effectively provides a sample selected on total stellar mass, over 0.2 < z < 1.8 within the Spitzer Wide-Area Infrared Extragalactic (SWIRE) Survey fields. 20%, or 106 clusters have spectroscopically confirmed redshifts, and the rest have redshifts estimated from the color of their red sequence. A comparison with the public SWIRE images detects 125 individual BCGs at 24um > 100uJy, or 23%. The luminosity-limited detection rate of BCGs in similar richness clusters (Ngal> 12) increases rapidly with redshift. Above z ~ 1, an average of ~20\% of the sample have 24um-inferred infrared luminosities of LIR > 10^12 Lsun, while the fraction below z ~ 1 exhibiting such luminosities is < 1 \%. The Spitzer-IRAC colors indicate the bulk of the 24um-detected population is predominantly powered by star formation, with only 7/125 galaxies lying within the color region inhabited by Active Galactic Nuclei (AGN). Simple arguments limit the star-formation activity to several hundred million years and this may therefore be indicative of the timescale for AGN feedback to halt the star formation. Below redshift z ~ 1 there is not enough star formation to significantly contribute to the overall stellar mass of the BCG population, and therefore BCG growth is likely dominated by dry-mergers. Above z~ 1, however, the inferred star formation would double the stellar mass of the BCGs and is comparable to the mass assembly predicted by simulations through dry mergers. We cannot yet constrain the process driving the star formation for the overall sample, though a single object studied in detail is consistent with a gas-rich merger.
• ### An Extreme Starburst in Close Proximity to the Central Galaxy of a Rich Galaxy Cluster at z=1.7(1508.04982)
Aug. 20, 2015 astro-ph.CO, astro-ph.GA
We have discovered an optically rich galaxy cluster at z=1.7089 with star formation occurring in close proximity to the central galaxy. The system, SpARCS104922.6+564032.5, was detected within the Spitzer Adaptation of the red-sequence Cluster Survey, (SpARCS), and confirmed through Keck-MOSFIRE spectroscopy. The rest-frame optical richness of Ngal(500kpc) = 30+/-8 implies a total halo mass, within 500kpc, of ~3.8+/-1.2 x 10^14 Msun, comparable to other clusters at or above this redshift. There is a wealth of ancillary data available, including Canada-France-Hawaii Telescope optical, UKIRT-K, Spitzer-IRAC/MIPS, and Herschel-SPIRE. This work adds submillimeter imaging with the SCUBA2 camera on the James Clerk Maxwell Telescope and near-infrared imaging with the Hubble Space Telescope (HST). The mid/far-infrared (M/FIR) data detect an Ultra-luminous Infrared Galaxy spatially coincident with the central galaxy, with LIR = 6.2+/-0.9 x 10^12 Lsun. The detection of polycyclic aromatic hydrocarbons (PAHs) at z=1.7 in a Spitzer-IRS spectrum of the source implies the FIR luminosity is dominated by star formation (an Active Galactic Nucleus contribution of 20%) with a rate of ~860+/-30 Msun/yr. The optical source corresponding to the IR emission is likely a chain of of > 10 individual clumps arranged as "beads on a string" over a linear scale of 66 kpc. Its morphology and proximity to the Brightest Cluster Galaxy imply a gas-rich interaction at the center of the cluster triggered the star formation. This system indicates that wet mergers may be an important process in forming the stellar mass of BCGs at early times.
• ### GLACE survey: OSIRIS/GTC Tuneable Filter H$\alpha$ imaging of the rich galaxy cluster ZwCl 0024.0+1652 at z = 0.395. Part I -- Survey presentation, TF data reduction techniques and catalogue(1502.03020)
Feb. 10, 2015 astro-ph.CO, astro-ph.GA
The cores of clusters at 0 $\lesssim$ z $\lesssim$ 1 are dominated by quiescent early-type galaxies, whereas the field is dominated by star-forming late-type ones. Galaxy properties, notably the star formation (SF) ability, are altered as they fall into overdense regions. The critical issues to understand this evolution are how the truncation of SF is connected to the morphological transformation and the responsible physical mechanism. The GaLAxy Cluster Evolution Survey (GLACE) is conducting a study on the variation of galaxy properties (SF, AGN, morphology) as a function of environment in a representative sample of clusters. A deep survey of emission line galaxies (ELG) is being performed, mapping a set of optical lines ([OII], [OIII], H$\beta$ and H$\alpha$/[NII]) in several clusters at z $\sim$ 0.40, 0.63 and 0.86. Using the Tunable Filters (TF) of OSIRIS/GTC, GLACE applies the technique of TF tomography: for each line, a set of images at different wavelengths are taken through the TF, to cover a rest frame velocity range of several thousands km/s. The first GLACE results target the H$\alpha$/[NII] lines in the cluster ZwCl 0024.0+1652 at z = 0.395 covering $\sim$ 2 $\times$ r$_{vir}$. We discuss the techniques devised to process the TF tomography observations to generate the catalogue of H$\alpha$ emitters of 174 unique cluster sources down to a SFR below 1 M$_{\odot}$/yr. The AGN population is discriminated using different diagnostics and found to be $\sim$ 37% of the ELG population. The median SFR is 1.4 M$_{\odot}$/yr. We have studied the spatial distribution of ELG, confirming the existence of two components in the redshift space. Finally, we have exploited the outstanding spectral resolution of the TF to estimate the cluster mass from ELG dynamics, finding M$_{200}$ = 4.1 $\times$ 10$^{14}$ M$_{\odot} h^{-1}$, in agreement with previous weak-lensing estimates.
• ### The dependence of star formation activity on environment and stellar mass at z~1 from the HiZELS H-alpha survey(1007.2642)
Sept. 14, 2010 astro-ph.CO, astro-ph.GA
(Abridged) This paper presents an environment and stellar mass study of a large sample of star-forming (SF) galaxies at z=0.84 from the HiZELS survey, over 1.3 deg^2 in the COSMOS and UKIDSS UDS fields. By taking advantage of a truly panoramic coverage, from the field to a rich cluster, it is shown that both mass and environment play crucial roles in determining the properties of SF galaxies. The median specific SFR declines with mass in all environments, and the fraction of galaxies forming stars declines from ~40%, for M~10^10M_sun to effectively zero at M>10^11.5M_sun, confirming that mass-downsizing is generally in place by z~1. The fraction of SF galaxies also falls as a function of local environmental density from ~40% in the field to approaching zero at rich group/cluster densities. When SF does occur in high density regions, it is merger-dominated and, if only non-merging SF galaxies are considered, then the environment and mass trends are even stronger and largely independent, as in the local Universe. The median SFR of SF galaxies is found to increase with density up to intermediate (group or cluster outskirts) densities; this is clearly seen as a change in the faint-end slope of the H-alpha LF from steep (-1.9), in poor fields, to shallow (-1.1) in groups and clusters. Interestingly, the relation between median SFR and environment is only found for low to moderate-mass galaxies (below ~10^10.6M_sun), and is not seen for massive SF galaxies. Overall, these observations provide a detailed view over a sufficiently large range of mass and environment to reconcile previous observational claims: mass is the primary predictor of SF activity at z~1, but the environment, while enhancing the median SFR of (lower-mass) SF galaxies, is ultimately responsible for suppressing SF activity in all galaxies above surface densities of 10-30 Mpc^-2 (groups and clusters).
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2021-04-14 22:12:10
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https://edx.readthedocs.io/projects/open-edx-building-and-running-a-course/en/open-release-lilac.master/course_components/create_problem.html
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# 8.4. Working with Problem Components¶
This section introduces the core set of problem types that course teams can add to any course by using the problem component. It also describes editing options and settings for problem components.
For information about specific problem types, and the other exercises and tools that you can add to your course, see Problems, Exercises, and Tools.
To add interactive problems to a course in Studio, in the course outline, at the unit level, you select Problem. You then choose the type of problem that you want to add from the Common Problem Types list or the Advanced list.
The common problem types include relatively straightforward CAPA problems such as multiple choice and text or numeric input. The advanced problem types can be more complex to set up, such as math expression input, open response assessment, or custom JavaScript problems.
The common and advanced problem types that the problem component lists are the core set of problems that every course team can include in a course. You can also enable more exercises and tools for use in your course. For more information, see Enabling Additional Exercises and Tools.
When you establish the grading policy for your course, you define the assignment types that count toward learners’ grades: for example, homework, labs, midterm, final, and participation. You specify one of these assignment types for each of the subsections in your course.
As you develop your course, you can add problem components to a unit in any subsection. The problem components that you add automatically inherit the assignment type that is defined at the subsection level. For example, all of the problem components that you add to a unit in the midterm subsection are graded.
For more information, see Set the Assignment Type and Due Date for a Subsection.
## 8.4.2. The Learner View of a Problem¶
All problems on the edX platform have these component parts, some of which can be configured. For configurable options, you can specify whether and when an option is available in problems.
1. Problem text. The problem text can contain any standard HTML formatting.
Within the problem text for each problem component, you must identify a question or prompt, which is, specifically, the question that learners need to answer. This question or prompt also serves as the required accessible label, and is used by screen readers, reports, and Insights. For more information about identifying the question text in your problem, see The Simple Editor.
2. Response field. Learners enter answers in response fields. The appearance of the response field depends on the type of the problem.
3. Rendered answer. For some problem types, the LMS uses MathJax to render plain text as “beautiful math.”
4. Submit. When a learner selects Submit to submit a response for a problem, the LMS saves the grade and current state of the problem. The LMS immediately provides feedback about whether the response is correct or incorrect, as well as the problem score. The Submit option remains available if the learner has unused attempts remaining, so that she can try to answer the problem again.
Note
If you want to temporarily or permanently hide learners’ results for the problem, see Set Problem Results Visibility.
5. Attempts. You can set a specific number of attempts or allow unlimited attempts for a problem. By default, the course-wide Maximum Attempts advanced setting is null, meaning that the maximum number of attempts for problems is unlimited.
In courses where a specific number has been specified for Maximum Attempts in Advanced Settings, if you do not specify a value for Maximum Attempts for an individual problem, the number of attempts for that problem defaults to the number of attempts defined in Advanced Settings.
6. Save. The learner can select Save to save his current response without submitting it for grading. This allows the learner to stop working on a problem and come back to it later.
7. Reset. You can specify whether the Reset option is available for a problem. This setting at the problem level overrides the default setting for the course in Advanced Settings.
If the Reset option is available, learners can select Reset to clear any input that has not yet been submitted, and try again to answer the question.
• If the learner has already submitted an answer, selecting Reset clears the submission and, if the problem includes a Python script to randomize variables and the randomization setting is On Reset, changes the values the learner sees in the problem.
• If the problem has already been answered correctly, Reset is not available.
• If the number of Maximum Attempts that was set for this problem has been reached, Reset is not available.
8. Show Answer. You can specify whether this option is available for a problem. If a learner selects Show Answer, the learner sees both the correct answer and the explanation, if any.
If you specify a number in Show Answer: Number of Attempts, the learner must submit at least that number of attempted answers before the Show Answer option is available for the problem.
9. Feedback. After a learner selects Submit, an icon appears beside each response field or selection within a problem. A green check mark indicates that the response was correct, a green asterisk (*) indicates that the response was partially correct, and a red X indicates that the response was incorrect. Underneath the problem, feedback text indicates whether the problem was answered correctly, incorrectly, or partially correctly, and shows the problem score.
Note
If you want to temporarily or permanently hide learners’ results for the problem, see Set Problem Results Visibility.
In addition to the items above, which are shown in the example, problems also have the following elements.
• Correct answer. Most problems require that you specify a single correct answer.
Note
If you want to temporarily or permanently hide learners’ results for the problem, see Set Problem Results Visibility.
• Explanation. You can include an explanation that appears when a learner selects Show Answer.
• Grading. You can specify whether a group of problems is graded.
• Due date. The date that the problem is due. Learners cannot submit answers for problems whose due dates have passed, although they can select Show Answer to show the correct answer and the explanation, if any.
Note
Problems can be open or closed. Closed problems, such as problems whose due dates are in the past, do not accept further responses and cannot be reset. Learners can still see questions, solutions, and revealed explanations, but they cannot submit responses or reset problems.
There are also some attributes of problems that are not immediately visible. You can set these attributes in Studio.
• Accessible Label. Within the problem text, you can identify the text that is, specifically, the question that learners need to answer. The text that is labeled as the question is used by screen readers, reports, and Insights. For more information, see The Simple Editor.
• Randomization. In certain types of problems, you can include a Python script to randomize the values that are presented to learners. You use this setting to define when values are randomized. For more information, see Randomization.
• Weight. Different problems in a particular problem set can be given different weights. For more information, see Problem Weight.
## 8.4.3. Editing a Problem in Studio¶
When you select Problem and choose one of the problem types, Studio adds an example problem of that type to the unit. To replace the example with your own problem, you select Edit to open the example problem in an editor.
The editing interface that opens depends on the type of problem you choose.
• For common problem types, the simple editor opens. In this editor, you use Markdown-style formatting indicators to identify the elements of the problem, such as the prompt and the correct and incorrect answer options.
• For advanced problem types (with the exception of open response assessment), the advanced editor opens. In this editor you use open learning XML (OLX) elements and attributes to identify the elements of the problem.
For open response assessment problem types, you define the problem elements and options by using a graphical user interface. For more information, see Create an Open Response Assessment Assignment.
You can switch from the simple editor to the advanced editor at any time by selecting Advanced Editor from the simple editor’s toolbar.
Note
After you save a problem in the advanced editor, you cannot open it again in the simple editor.
### 8.4.3.1. The Simple Editor¶
When you edit one of the common problem types, the simple editor opens with a template that you can use as a guideline for adding Markdown formatting. The following templates are available.
Blank common problems also open in the simple editor but they do not provide a template.
The following image shows the multiple choice template in the simple editor.
The simple editor includes a toolbar with options that provide the required Markdown formatting for different types of problems. When you select an option from the toolbar, formatted sample text appears in the simple editor. Alternatively, you can apply formatting to your own text by selecting the text and then one of the toolbar options.
Descriptions of the Markdown formatting that you use in the simple editor follow.
1. Heading: Identifies a title or heading by adding a series of equals signs (=) below it on the next line.
2. Multiple Choice: Identifies an answer option for a multiple choice problem by adding a pair of parentheses (( )) before it. To identify the correct answer option, you insert an x within the parentheses: ((x)).
3. Checkboxes: Identifies an answer option for a checkboxes problem by adding a pair of brackets ([ ]) before it. To identify the correct answer option or options, you insert an x within the brackets: ([x]).
4. Text Input: Identifies the correct answer for a text input problem by adding an equals sign (=) before the answer value on the same line.
5. Numerical Input: Identifies the correct answer for a numerical input problem by adding an equals sign (=) before the answer value on the same line.
6. Dropdown: Identifies a comma-separated list of values as the set of answer options for a dropdown problem by adding two pairs of brackets ([[ ]]) around the list. To identify the correct answer option, you add parentheses (( )) around that option.
7. Explanation: Identifies the explanation for the correct answer by adding an [explanation] tag to the lines before and after the text. The explanation appears only after learners select Show Answer. You define when the Show Answer option is available to learners by using the Show Answer setting.
8. Advanced Editor link: Opens the problem in the advanced editor, which shows the OLX markup for the problem.
9. Toggle Cheatsheet: Opens a list of formatting hints.
10. Question or Prompt: Identifies the question that learners need to answer. The toolbar does not have an option that provides this formatting, so you add two pairs of inward-pointing angle brackets (>> <<) around the question text. For example, >>Is this the question?<<.
• You must identify a question or prompt in every problem component. In problems that include multiple questions, you must identify each one.
• The Student Answer Distribution report uses the text with this formatting to identify each problem.
• Insights also uses the text with this formatting to identify each problem. For more information, see Using edX Insights.
11. Description: Identifies optional guidance that helps learners answer the question. For example, when you add a checkbox problem that is only correct when learners select three of the answer options, you might include the description, “Be sure to select all that apply.” The toolbar does not have an option that provides this formatting, so you add it after the question within the angle brackets, and then you separate the question and the description by inserting a pair of pipe symbols (||) between them. For example, >>Which of the following choices is correct? ||Be sure to select all that apply.<<.
#### 8.4.3.1.2. Adding Text, Symbols, and Mathematics¶
You can also add text, without formatting, to a problem. Note that screen readers read all of the text that you supply for the problem, and then repeat the text that is identified as the question or prompt immediately before reading the answer choices for the problem. For problems that require descriptions or other text, you might consider adding an HTML component for the text immediately before the problem component.
When you enter unformatted text, note that the simple editor cannot interpret certain symbol characters correctly. These symbols are reserved HTML characters: greater than (>), less than (<), and ampersand (&). If you enter text that includes these characters, the simple editor cannot save your edits. To resolve this problem, replace these characters in your problem text with the HTML entities that represent them.
• To enter >, type >.
• To enter <, type <.
• To enter &, type &.
To add mathematics, you can use LaTeX, MathML, or AsciiMath notation. Studio uses MathJax to render equations. For more information, see Using MathJax for Mathematics.
When you edit one of the advanced problem types, the advanced editor opens with an example problem. The advanced editor is an XML editor that shows the OLX markup for a problem. You edit the following advanced problem types in the advanced editor.
For the Open Response Assessment advanced problem type, a dialog box opens for problem setup.
Blank advanced problems do not provide an example problem, but they also open in the advanced editor by default.
The following image shows the OLX markup in the advanced editor for the same example multiple choice problem that is shown in the simple editor above.
For more information about the OLX markup to use for a problem, see the topic that describes that problem type.
## 8.4.4. Defining Settings for Problem Components¶
In addition to the text of the problem and its Markdown formatting or OLX markup, you define the following settings for problem components. To access these settings, you edit the problem and then select Settings.
If you do not edit these settings, default values are supplied for your problems.
Note
If you want to temporarily or permanently hide problem results from learners, you use the subsection-level Results Visibility setting. You cannot change the visibility of individual problems. For more information, see Set Problem Results Visibility.
### 8.4.4.1. Display Name¶
This required setting provides an identifying name for the problem. The display name appears as a heading above the problem in the LMS, and it identifies the problem for you in Insights. Be sure to add unique, descriptive display names so that you, and your learners, can identify specific problems quickly and accurately.
The following illustration shows the display name of a problem in Studio, in the LMS, and in Insights.
### 8.4.4.2. Maximum Attempts¶
This setting specifies the number of times that a learner is allowed to try to answer this problem correctly. You can define a different Maximum Attempts value for each problem.
A course-wide Maximum Attempts setting defines the default value for this problem-specific setting. Initially, the value for the course-wide setting is null, meaning that learners can try to answer problems an unlimited number of times. You can change the course-wide default by selecting Settings and then Advanced Settings. Note that if you change the course-wide default from null to a specific number, you can no longer change the problem-specific Maximum Attempts value to unlimited.
Only problems that have a Maximum Attempts setting of 1 or higher are included in the answer distribution computations used in edX Insights and the Student Answer Distribution report.
Note
EdX recommends setting Maximum Attempts to unlimited or a large number when possible. Problems that allow unlimited attempts encourage risk taking and experimentation, both of which lead to improved learning outcomes. However, allowing for unlimited attempts might not be feasible in some courses, such as those that use primarily multiple choice or dropdown problems in graded subsections.
### 8.4.4.3. Problem Weight¶
Note
The LMS scores all problems. However, only scores for problem components that are in graded subsections count toward a learner’s final grade.
This setting specifies the total number of points possible for the problem. In the LMS, the problem weight appears near the problem’s display name.
By default, each response field, or answer space, in a problem component is worth one point. You increase or decrease the number of points for a problem component by setting its Problem Weight.
In the example shown above, a single problem component includes three separate questions. To respond to these questions, learners select answer options from three separate dropdown lists, the response fields for this problem. By default, learners receive one point for each question that they answer correctly.
For information about how to define a problem that includes more than one question, see Including Multiple Questions in One Component.
#### 8.4.4.3.1. Computing Scores¶
The score that a learner earns for a problem is the result of the following formula.
Score = Weight × (Correct answers / Response fields)
• Score is the point score that the learner receives.
• Weight is the problem’s maximum possible point score.
• Correct answers is the number of response fields that contain correct answers.
• Response fields is the total number of response fields in the problem.
Examples
The following are some examples of computing scores.
Example 1
A problem’s Problem Weight setting is left blank. The problem has two response fields. Because the problem has two response fields, the maximum score is 2.0 points.
If one response field contains a correct answer and the other response field contains an incorrect answer, the learner’s score is 1.0 out of 2 points.
Example 2
A problem’s weight is set to 12. The problem has three response fields.
If a learner’s response includes two correct answers and one incorrect answer, the learner’s score is 8.0 out of 12 points.
Example 3
A problem’s weight is set to 2. The problem has four response fields.
If a learner’s response contains one correct answer and three incorrect answers, the learner’s score is 0.5 out of 2 points.
### 8.4.4.4. Randomization¶
Note
This Randomization setting serves a different purpose from “problem randomization”. This Randomization setting affects how numeric values are randomized within a single problem and requires the inclusion of a Python script. Problem randomization presents different problems or problem versions to different learners. For more information, see Problem Randomization.
For problems that include a Python script to generate numbers randomly, this setting specifies how frequently the values in the problem change: each time a different learner accesses the problem, each time a single learner tries to answer the problem, both, or never.
Note
This setting should only be set to an option other than Never for problems that are configured to do random number generation.
For example, in this problem, the highlighted values change each time a learner submits an answer to the problem.
If you want to randomize numeric values in a problem, you complete both of these steps.
• Make sure that you edit your problem to include a Python script that randomly generates numbers.
• Select an option other than Never for the Randomization setting.
The edX Platform has a 20-seed maximum for randomization. This means that learners see up to 20 different problem variants for every problem that has Randomization set to an option other than Never. It also means that every answer for the 20 different variants is reported by the Answer Distribution report and edX Insights. Limiting the number of variants to a maximum of 20 allows for better analysis of learner submissions by allowing you to detect common incorrect answers and usage patterns for such answers.
Important
Whenever you choose an option other than Never for a problem, the computations for the Answer Distribution report and edX Insights include up to 20 variants for the problem, even if the problem was not actually configured to include randomly generated values. This can make data collected for problems that cannot include randomly generated values, (including, but not limited to, all multiple choice, checkboxes, dropdown, and text input problems), extremely difficult to interpret.
You can choose the following options for the Randomization setting.
Option
Description
Always
Learners see a different version of the problem each time they select Submit.
On Reset
Learners see a different version of the problem each time they select Reset.
Never
All learners see the same version of the problem. For most courses, this option is supplied by default. Select this option for every problem in your course that does not include a Python script to generate random numbers.
Per Student
Individual learners see the same version of the problem each time they look at it, but that version is different from the version that other learners see.
This setting adds a Show Answer option to the problem. The following options define when the answer is shown to learners.
### 8.4.4.6. Show Answer: Number of Attempts¶
This setting limits when learners can select the Show Answer option for a problem. Learners must submit at least the specified number of attempted answers for the problem before the Show Answer option is available to them.
### 8.4.4.7. Show Reset Button¶
This setting defines whether a Reset option is available for the problem.
Learners can select Reset to clear any input that has not yet been submitted, and try again to answer the problem.
If the learner has already submitted an answer, selecting Reset clears the submission and, if the problem contains randomized variables and randomization is set to On Reset, changes the values in the problem.
If the number of Maximum Attempts that was set for this problem has been reached, the Reset option is not visible.
This problem-level setting overrides the course-level Show Reset Button for Problems advanced setting.
### 8.4.4.8. Timer Between Attempts¶
This setting specifies the number of seconds that a learner must wait between submissions for a problem that allows multiple attempts. If the value is 0, the learner can attempt the problem again immediately after an incorrect attempt.
Adding required wait time between attempts can help to prevent learners from simply guessing when multiple attempts are allowed.
If a learner attempts a problem again before the required time has elapsed, she sees a message below the problem indicating the remaining wait time. The format of the message is, “You must wait at least {n} seconds between submissions. {n} seconds remaining.”
## 8.4.5. Including Multiple Questions in One Component¶
In some cases, you might want to design an assessment that combines multiple questions in a single problem component. For example, you might want learners to demonstrate mastery of a concept by providing the correct responses to several questions, and only giving them credit for problem if all of the answers are correct.
Another example involves learners who have slow or intermittent internet connections. When every problem appears on a separately loaded web page, these learners can find the amount of time it takes to complete an assignment or exam discouraging. For these learners, grouping several questions together can promote increased engagement with course assignments.
When you add multiple questions to a single problem component, the settings that you define, including the display name and whether to show the Reset button, apply to all of the questions in that component. The answers to all of the questions are submitted when learners select Submit, and the correct answers for all of the questions appear when learners select Show Answer. By default, learners receive one point for each question they answer correctly. For more information about changing the default problem weight and other settings, see Defining Settings for Problem Components.
Important
To assure that the data collected for learner interactions with your problem components is complete and accurate, include a maximum of 10 questions in a single problem component.
### 8.4.5.1. Adding Multiple Questions to a Problem Component¶
To design an assignment that includes several questions, you add one problem component and then edit it to add every question and its answer options, one after the other, in that component. Be sure to identify the text of every question or prompt with the appropriate Markdown formatting (>> <<) or OLX <label> element, and include all of the other required elements for each question.
• In the simple editor, you use three hyphen characters (---) on a new line to separate one question and its answer options from the next.
• In the advanced editor, each question and its answer options are enclosed by the element that identifies the type of problem, such as <multiplechoiceresponse> for a multiple choice question or <formularesponse> for a math expression input question.
• You can provide a different explanation for each question with the appropriate Markdown formatting ([explanation]) or OLX <solution> element.
As a best practice, edX recommends that you avoid including unformatted paragraph text between the questions. Screen readers can skip over text that is inserted among multiple questions.
The questions that you include can all be of the same problem type, such as a series of text input questions, or you can include questions that use different problem types, such as both numerical input and math expression input.
Note
You cannot use a Custom JavaScript Display and Grading Problem in a problem component that contains more than one question. Each custom JavaScript problem must be in its own component.
An example of a problem component that includes a text input question and a numerical input question follows. In the simple editor, the problem has the following Markdown formatting.
>>Who invented the Caesar salad?||Be sure to check your spelling.<<
= Caesar Cardini
[explanation]
[explanation]
---
>>In what year?<<
= 1924
[explanation]
Cardini invented the dish at his restaurant on 4 July 1924 after the rush of holiday business left the kitchen with fewer supplies than usual.
[explanation]
That is, you include three hyphen characters (---) on a new line to separate the problems.
In the advanced editor, the problem has the following OLX markup.
<problem>
<description>Be sure to check your spelling.</description>
<textline size="20"/>
<solution>
<div class="detailed-solution">
<p>Explanation</p>
</div>
</solution>
</stringresponse>
<label>In what year?</label>
<formulaequationinput/>
<solution>
<div class="detailed-solution">
<p>Explanation</p>
<p>Cardini invented the dish at his restaurant on 4 July 1924 after
the rush of holiday business left the kitchen with fewer supplies
than usual.</p>
</div>
</solution>
</numericalresponse>
</problem>
## 8.4.6. Adding Feedback and Hints to a Problem¶
You can add feedback, hints, or both to the following core problem types.
By using hints and feedback, you can provide learners with guidance and help as they work on problems.
### 8.4.6.1. Feedback in Response to Attempted Answers¶
You can add feedback that displays to learners after they submit an answer.
For example, the following multiple choice problem provides feedback in response to the selected option when the learner selects Submit. In this case, feedback is given for an incorrect answer.
### 8.4.6.2. Best Practices for Providing Feedback¶
The immediacy of the feedback available to learners is a key advantage of online instruction and difficult to do in a traditional classroom environment.
You can target feedback for common incorrect answers to the misconceptions that are common for the level of the learner (for example, elementary, middle, high school, college).
In addition, you can create feedback that provides some guidance to the learner about how to arrive at the correct answer. This is especially important in text input and numeric input problems, because without such guidance, learners might not be able to proceed.
You should also include feedback for the correct answer to reinforce why the answer is correct. Especially in questions where learners are able to guess, such as multiple choice and dropdown problems, the feedback should provide a reason why the selection is correct.
### 8.4.6.3. Providing Hints for Problems¶
You can add one or more hints that are displayed to learners. When you add hints, the Hint button is automatically displayed to learners. Learners can access the hints by selecting Hint beneath the problem. A learner can view multiple hints by selecting Hint multiple times.
For example, in the following multiple choice problem, the learner selects Hint after having made one incorrect attempt.
The hint text indicates that it is the first of two hints. After the learner selects Next Hint, both of the available hints appear. When all hints have been used, the Hint or Next Hint option is no longer available.
### 8.4.6.4. Best Practices for Providing Hints¶
To ensure that your hints can assist learners with varying backgrounds and levels of understanding, you should provide multiple hints with different levels of detail.
For example, the first hint can orient the learner to the problem and help those struggling to better understand what is being asked.
The second hint can then take the learner further towards the answer.
In problems that are not graded, the third and final hint can explain the solution for learners who are still confused.
### 8.4.6.5. Create Problems with Feedback and Hints¶
You create problems with feedback and hints in Studio. Templates with feedback and hints configured are available to make creating your own problems easier.
While editing a unit, in the Add New Component panel, select Problem. In the list that opens, select Common Problem Types. Templates for problems with feedback and hints are listed.
Add the problem type you need to the unit, then edit the component. The exact syntax you use to configure hints and feedback depends on the problem type. See the topic for the problem type for more information.
## 8.4.7. Awarding Partial Credit for a Problem¶
You can configure the following problem types so that learners can receive partial credit for a problem if they submit an answer that is partly correct.
By awarding partial credit for problems, you can motivate learners who have mastered some of the course content and provide a score that accurately demonstrates their progress.
### 8.4.7.1. How Learners Receive Partial Credit¶
Learners receive partial credit when they submit an answer in the LMS.
In the following example, the course team configured a multiple choice problem to award 25% of the possible points (instead of 0 points) for one of the incorrect answer options. The learner selected this incorrect option, and received 25% of the possible points.
### 8.4.7.2. Partial Credit and Reporting on Learner Performance¶
When a learner receives partial credit for a problem, the LMS only adds the points that the learner earned to the grade. However, the LMS reports any problem for which a learner receives credit, in full or in part, as correct in the following ways.
• Events that the system generates when learners receive partial credit for a problem indicate that the answer was correct. Specifically, the correctness field has a value of correct.
• The AnswerValue in the Student Answer Distribution report is 1, for correct.
• The edX Insights insights:student performance reports count the answer as correct.
Course teams can see that a learner received partial credit for a problem in the learner’s submission history. The submission history shows the score that the learner received out of the total available score, and the value in the correctness field is partially-correct. For more information, see Learner Answer Submissions.
### 8.4.7.3. Adding Tooltips to a Problem¶
To help learners understand terminology or other aspects of a problem, you can add inline tooltips. Tooltips show text to learners when they move their cursors over a tooltip icon.
The following example problem includes two tooltips. The tooltip that provides a definition for “ROI” is being shown.
Note
For learners using a screen reader, the tooltip expands to make its associated text accessible when the screen reader focuses on the tooltip icon.
To add the tooltip, you wrap the text that you want to appear as the tooltip in the clarification element. For example, the following problem contains two tooltips.
<problem>
<text>
<p>Given the data in Table 7 <clarification>Table 7: "Example PV
Installation Costs", Page 171 of Roberts textbook</clarification>,
compute the ROI <clarification><strong>ROI</strong>: Return on
Investment</clarification> over 20 years.
</p>
. . .
## 8.4.8. Problem Randomization¶
Presenting different learners with different problems or with different versions of the same problem is referred to as “problem randomization”.
You can provide different learners with different problems by using randomized content blocks, which randomly draw problems from pools of problems stored in content libraries. For more information, see Randomized Content Blocks.
Note
Problem randomization is different from the Randomization setting that you define in Studio. Problem randomization presents different problems or problem versions to different learners, while the Randomization setting controls when a Python script randomizes the variables within a single problem. For more information about the Randomization setting, see Randomization.
Creating randomized problems by exporting your course and editing some of your course’s XML files is no longer supported.
## 8.4.9. Modifying a Released Problem¶
Warning
Be careful when you modify problems after they have been released. Changes that you make to published problems can affect the learner experience in the course and analysis of course data.
After a learner submits a response to a problem, the LMS stores that response, the score that the learner received, and the maximum score for the problem. For problems with a Maximum Attempts setting greater than 1, the LMS updates these values each time the learner submits a new response to a problem. However, if you change a problem or its attributes, existing learner information for that problem is not automatically updated.
For example, you release a problem and specify that its answer is 3. After some learner have submitted responses, you notice that the answer should be 2 instead of 3. When you update the problem with the correct answer, the LMS does not update scores for learners who originally answered 2 for the problem and received the wrong score.
For another example, you change the number of response fields to three. Learners who submitted answers before the change have a score of 0, 1, or 2 out of 2.0 for that problem. Learners who submitted answers after the change have scores of 0, 1, 2, or 3 out of 3.0 for the same problem.
If you change the weight setting for the problem in Studio, however, existing scores update when the learner’s Progress page is refreshed. In a live section, learners will see the effect of these changes.
### 8.4.9.1. Workarounds¶
If you have to modify a released problem in a way that affects grading, you have two options to ensure that every learner has the opportunity to submit a new response and be regraded. Note that both options require you to ask your learners to go back and resubmit answers to a problem.
• In the problem component that you changed, increase the number of attempts for the problem, and then ask all of your learners to redo the problem.
• Delete the entire problem component in Studio and replace it with a new problem component that has the content and settings that you want. Then ask all of your learners to complete the new problem. (If the revisions you must make are minor, you might want to duplicate the problem component before you delete it, and then revise the copy.)
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2021-09-28 02:32:53
|
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https://www.hrwhisper.me/deep-learning-word2vec/
|
0%
• Onehot vs 分布式表示
• CBOW和skip gram
• Huffman改进
• 负采样
## 为什么要分布式表示
word2vec是一种词的embedding方法，在介绍之前，首先了解一下啥叫embedding呢？
Embedding在数学上表示一个mapping, f: X -> Y， 也就是一个function，其中该函数是injective（就是我们所说的单射函数，每个Y只有唯一的X对应，反之亦然）和structure-preserving (结构保存，比如在X所属的空间上X1 < X2,那么映射后在Y所属空间上同理 Y1 < Y2)。那么对于word embedding，就是将单词word映射到另外一个空间，其中这个映射具有injective和structure-preserving的特点。
word embedding，就是找到一个映射或者函数，生成在一个新的空间上的表达，该表达就是word representation。
## Skip-Gram
### 训练
$\mathcal{L} = -\sum_{-m \leq j \leq m,\ j \neq 0} \text{log}\, P(w^{(i+j)} \mid w^{(i)})\tag{3-5}$
3-5的式子其实也可以看成是交叉熵损失，标签y为one-hot的结果，则 $L({\bf W}) = - \sum_{-m \leq j \leq m,\ j \neq 0} \sum_{a=1}^V y_a^{(i+j)} \log P(w^{(i+j)} \mid w^{(i)})\tag{3-9}$ 但每个$$\bf y^{(i+j)}$$中只有一个为1，其余的为0，因此最后就和3-5等价了。
## CBOW 模型
CBOW(continuous bag-of-word)即连续词袋模型，即用一个中心词前后距离d内的背景词来预测该中心词出现的概率
CBOW模型可以用如下的一层神经网络表示：
### 训练
4-3取对数得： $\log P(w_j|w_1,\dots,w_C) = {\bf {\bar v}^T u_j } - \log{\sum_{k = 1}^{V}\exp({\bf{\bar v}^T u_k}})\tag{4-3}$ 4-3对中心词向量$$\bf u_j$$求导得： \begin{aligned} \frac{\partial \text{log}\, P}{\partial \bf{u}_j} &= {\bf {\bar v}} - \frac{\exp({\bf {\bar v}^T u_j}){\bf {\bar v}}}{\sum_{a=1}^{V} \exp({\bf {\bar v}^T u_a})}\\ &= {\bf {\bar v}} - P(w_j|w_1,\dots,w_C) {\bf {\bar v}} \end{aligned}\tag{4-4} 对于任意的背景词向量$$\bf v_i$$，4-3对其求导有： \begin{aligned} \frac{\partial \text{log}\, P}{\partial \bf{v}_i} &= \frac{1}{C} {\bf u_j} - \frac{\sum_{b=1}^{V}\exp({\bf {\bar v}^T u_b})\frac{1}{C} {\bf u_b} }{\sum_{a=1}^{V} \exp({\bf {\bar v}^T u_a})}\\ &= \frac{1}{C} \left({\bf u_j} - \sum_{b=1}^VP(w_b|w_1,\dots,w_C) {\bf u_b} \right) \end{aligned}\tag{4-5}
## 加速计算
### 负采样 negative sampling
• 正例：即中心词$$w_c$$和窗口范围内每个背景词$$w_i$$分别组成的pair就是正例。（和之前的一样）
• 负例：用相同的中心词$$w_c$$，然后在词库中随机抽取k个词，中心词和这k个词分别组成pair，得到的k个pair都标记为负例。这里的k一般取5~20。这里有个小细节，比如我的中心词是fox，窗口大小为2，但是我从词库中抽取的词是brown，在该句子中其实是在fox前面的，但是仍然算作负例
• 如果用均等的概率选择的话，其实对于英文的文本是没有代表性的。
• 如果用出现的频率来取词，像"the"这些stop words的采样得到的概率会很大，但是("fox", "the") 并不能告诉我们很多的信息，因为the这个单词在很多上下文都出现过。
### 欠采样 subsample
• 负采样的方式通过考虑同时含有正例样本和负例样本的相互独立事件来构造损失函数,其训练中每一步的梯度计算开销与采样的负例个数线性相关.
• 层次Softmax使用了Huffman树,并根据根节点到叶子结点的路径来构造损失函数,其训练中每一步的梯度计算开销与词典大小对数相关.
## CBOW 和 Skip-gram对比
In CBOW the vectors from the context words are averaged before predicting the center word. In skip-gram there is no averaging of embedding vectors. It seems like the model can learn better representations for the rare words when their vectors are not averaged with the other context words in the process of making the predictions.
As we know, CBOW is learning to predict the word by the context. Or maximize the probability of the target word by looking at the context. And this happens to be a problem for rare words. For example, given the context yesterday was really [...] day CBOW model will tell you that most probably the word is beautiful or nice. Words like delightful will get much less attention of the model, because it is designed to predict the most probable word. Rare words will be smoothed over a lot of examples with more frequent words.
On the other hand, the skip-gram is designed to predict the context. Given the word delightful it must understand it and tell us, that there is huge probability, the context is yesterday was really [...] day, or some other relevant context. With skip-gram the word delightful will not try to compete with word beautiful but instead, delightful+context pairs will be treated as new observations. Because of this, skip-gram will need more data so it will learn to understand even rare words.
from https://stats.stackexchange.com/a/261440
## 参考资料
1. Word2Vec Tutorial - The Skip-Gram Model
2. 词嵌入（word2vec）
3. Mikolov T, Sutskever I, Chen K, et al. Distributed representations of words and phrases and their compositionality[C]//Advances in neural information processing systems. 2013: 3111-3119.
4. 漫谈Word2vec之skip-gram模型
5. Meyer D. How exactly does word2vec work?[J]. 2016.
6. DeepLearning.ai 吴恩达 第五课第二周， 自然语言处理与词嵌入
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2021-04-16 07:15:29
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https://math.stackexchange.com/questions/2548942/no-l-hospital-lim-x-rightarrow-0-left-1-frac-1-arctan-x-right/2548965
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# No L' Hospital $\lim_{x \rightarrow 0} \left (1+\frac {1} {\arctan x} \right)^{\sin x}$ and $\lim_{x \rightarrow 0} \frac {\tan ^7 x} {\ln (7x+1)}$
What would be the best approach to calculate the following limits
$$\lim_{x \rightarrow 0} \left (1+\frac {1} {\arctan x} \right)^{\sin x}, \qquad \lim_{x \rightarrow 0} \frac {\tan ^7 x} {\ln (7x+1)}$$ in a basic way, using some special limits, without L'Hospital's rule?
• One question per post. – Simply Beautiful Art Dec 3 '17 at 14:18
• For the second limit, note that $$\tan x\sim\ln(1+x)$$as $x\to0$. – Simply Beautiful Art Dec 3 '17 at 14:19
Use the standard limits $$\lim\limits_{x\to 0}\frac{\ln(1+x)}{x}=1$$ and $$\lim\limits_{x\to 0}\frac{\sin x}{x}=1$$
In the first example take a log first.
Note that the later implies that
$$\lim\limits_{x\to 0}\frac{\tan x}{x}=1$$
and thus that $$\lim\limits_{x\to 0}\frac{\arctan x}{x}=1$$
The first limit can be written as
$$\ln \left(1+\frac{1}{\arctan x}\right)^{\sin x}=\frac{\sin x}{x}\frac{\ln(1+\frac{1}{\arctan x})}{\frac{1}{\arctan x}}\frac{x}{\arctan x}$$
• Does the first standard limit you suggested apply here? Limit of $\frac{1}{\arctan x}$ as $x$ goes to $0$ actually doesn't exist. Could you tell me how I should evaluate the expression in the middle? – Theta Dec 3 '17 at 15:54
• @Theta: you are correct. The middle expression tends to $0$ as $x\to 0^{+}$ via the standard limit $\lim\limits_{t\to\infty} \dfrac{\log t} {t} =0$. Thus the first limit in question is $1$ but you must consider $x\to 0^{+}$. – Paramanand Singh Dec 4 '17 at 7:24
The second is: $$\lim_{x\to0}\frac{\sin^7x}{\cos^7x}\frac{1}{\log(7x+1)}= \lim_{x\to0}\frac{\sin^7x}{\cos^7x}\frac{1}{\log(7x+1)}\frac{x^7}{x^7}\frac{7x}{7x}=0$$
For the first use the substitution method.
For the second:
$$\frac {\tan ^7 x} {\ln (7x+1)}=\frac {\tan ^7 x} {x^7}\ \frac {x^7} {7x} \ \frac {7 x} {\ln (7x+1)}=1\cdot0\cdot1=0$$
A solution for the first by Taylor series:
we can write the limit as follow: $$\left (1+\frac {1} {\arctan x} \right)^{\sin x}=e^{sinx \ \log{\left (1+\frac {1} {\arctan x} \right)}}$$
Calculate Taylor series expansion for each term at the first order: $$\sin x = x+o(x)$$
$$\log{\left (1+\frac {1}{\arctan x} \right)} =\log{\left (\frac {1+ \arctan x}{\arctan x} \right)} =-\log{\left (\frac {\arctan x}{1+\arctan x} \right)}\\ =-\log{\left (\frac {x+o(x)}{1+x+o(x)} \right)} =-\log{\left [(x+o(x))\cdot(1-x+o(x)) \right]} =-\log{(x+o(x))}$$
Thus: $$\sin x \ \log{\left (1+\frac {1}{\arctan x} \right)}=(x+o(x))\cdot [-\log{(x+o(x))}]=-x \log x + o(x)\to 0$$
Finally:
$$\left (1+\frac {1} {\arctan x} \right)^{\sin x}\to e^0 =1$$
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2019-10-23 07:39:09
|
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https://math.stackexchange.com/questions/110457/closed-form-for-int-0-infty-fracxn1-xmdx/110467
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# Closed form for $\int_0^\infty {\frac{{{x^n}}}{{1 + {x^m}}}dx }$
I've been looking at
$$\int\limits_0^\infty {\frac{{{x^n}}}{{1 + {x^m}}}dx }$$
It seems that it always evaluates in terms of $\sin X$ and $\pi$, where $X$ is to be determined. For example:
$$\displaystyle \int\limits_0^\infty {\frac{{{x^1}}}{{1 + {x^3}}}dx = } \frac{\pi }{3}\frac{1}{{\sin \frac{\pi }{3}}} = \frac{{2\pi }}{{3\sqrt 3 }}$$
$$\int\limits_0^\infty {\frac{{{x^1}}}{{1 + {x^4}}}dx = } \frac{\pi }{4}$$
$$\int\limits_0^\infty {\frac{{{x^2}}}{{1 + {x^5}}}dx = } \frac{\pi }{5}\frac{1}{{\sin \frac{{2\pi }}{5}}}$$
So I guess there must be a closed form - the use of $\Gamma(x)\Gamma(1-x)$ first comess to my mind because of the $\dfrac{{\pi x}}{{\sin \pi x}}$ appearing. Note that the arguments are always the ratio of the exponents, like $\dfrac{1}{4}$, $\dfrac{1}{3}$ and $\dfrac{2}{5}$. Is there any way of finding it? I'll work on it and update with any ideas.
UPDATE:
The integral reduces to finding
$$\int\limits_{ - \infty }^\infty {\frac{{{e^{a t}}}}{{{e^t} + 1}}dt}$$
With $a =\dfrac{n+1}{m}$ which converges only if
$$0 < a < 1$$
Using series I find the solution is
$$\sum\limits_{k = - \infty }^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{{a + k}}}$$
• Digamma function argument will work, though unnatural. Feb 17 '12 at 20:57
• possible duplicate of Interesting integral formula Feb 17 '12 at 21:46
• @Listing Please, do not close this thread. I know nothing about complex or countour integration and thus I can't use the solutions given in the thread. I'm sticking to series and Digammas. Feb 17 '12 at 22:19
• @sos440 What do you mean by "unnatural"? I usually use that adjective but I guess it is rather a personal use. See here where I use the Digamma to solve this. Feb 21 '12 at 3:36
• The solution to this problem is also given here: math.stackexchange.com/questions/48740/int-0-infty-fracdx1xn/… Feb 29 '12 at 17:09
## 10 Answers
I would like to make a supplementary calculation on BR's answer.
Let us first assume that $0 < \mu < \nu$ so that the integral $$\int_{0}^{\infty} \frac{x^{\mu-1}}{1+x^{\nu}} \; dx$$ converges absolutely. By the substitution $x = \tan^{2/\nu} \theta$, we have $$\frac{dx}{1+x^{\nu}} = \frac{2}{\nu} \tan^{(2/\nu)-1} \theta \; d\theta.$$ Thus \begin{align*} \int_{0}^{\infty} \frac{x^{\mu-1}}{1+x^{\nu}} \; dx & = \int_{0}^{\frac{\pi}{2}} \frac{2}{\nu} \tan^{\frac{2\mu}{\nu}-1} \theta \; d\theta \\ & = \frac{1}{\nu} \beta \left( \frac{\mu}{\nu}, 1 - \frac{\mu}{\nu} \right) \\ & = \frac{1}{\nu} \Gamma \left( \frac{\mu}{\nu} \right) \Gamma \left( 1 - \frac{\mu}{\nu} \right) \\ & = \frac{\pi}{\nu} \csc \left( \frac{\pi \mu}{\nu} \right), \end{align*} where the last equality follows from Euler reflexion formula.
• That's better. Thanks. Feb 17 '12 at 20:48
• Any info on the series representation? Feb 17 '12 at 20:55
• Check my other question, I have found a way to prove it with the Digamma function Feb 18 '12 at 23:47
• @sos440: nice way.(+1) Aug 11 '12 at 12:07
• Why is $\mu<0$ not allowed? Feb 15 '20 at 18:36
The general formula (for $m > n+1$ and $n \ge 0$) is $\frac{\pi}{m} \csc\left(\frac{\pi (n+1)}{m}\right)$. IIRC the usual method involves a wedge-shaped contour of angle $2 \pi/m$.
EDIT: Consider $\oint_\Gamma f(z)\ dz$ where $f(z) = \frac{z^n}{1+z^m}$ (using the principal branch if $m$ or $n$ is a non-integer) and $\Gamma$ is the closed contour below:
$\Gamma_1$ goes to the right along the real axis from $\epsilon$ to $R$, so $\int_{\Gamma_1} f(z)\ dz = \int_\epsilon^R \frac{x^n\ dx}{1+x^m}$. $\Gamma_3$ comes in along the ray at angle $2 \pi/m$. Since $e^{(2 \pi i/m) m} = 1$, $\int_{\Gamma_3} f(z)\ dz = - e^{2 \pi i (n+1)/m} \int_{\Gamma_1} f(z)\ dz$. $\Gamma_2$ is a circular arc at distance $R$ from the origin. Since $m > n+1$, the integral over it goes to $0$ as $R \to \infty$. Similarly, the integral over the small circular arc at distance $\epsilon$ goes to $0$ as $\epsilon \to 0$. So we get
$$\lim_{R \to \infty, \epsilon \to 0} \int_\Gamma f(z)\ dz = (1 - e^{2 \pi i (n+1)/m}) \int_0^\infty \frac{x^n\ dx}{1+x^m}$$
The meromorphic function $f(z)$ has one singularity inside $\Gamma$, a pole at $z = e^{\pi i/m}$ where the residue is $- e^{\pi i (n+1)/m}/m$. So the residue theorem gives you
$$\int_0^\infty \frac{x^n\ dx}{1+x^m} = \frac{- 2 \pi i e^{\pi i (n+1)/m}}{ m (1 - e^{2 \pi i (n+1)/m})} = \frac{\pi}{m} \csc\left(\frac{\pi(n+1)}{m}\right)$$
• "$\Gamma_3$ comes in along the ray at angle 2π/m. Since $e^{(2πi/m)m}=1, \int_{\Gamma_3}(z) \,dz=−e^{2πi(n+1)/m}∫_{\Gamma_1}f(z)\, dz$". How is it that the integral over $\Gamma_3$ becomes a multiple of the integral over $\Gamma_1$? Nov 21 '13 at 18:30
• Hi @RobertIsrael, I like the method of using a wedge contour - it obviously minimizes computation of residues, and the conversion from complex number (lots of Euler formulas) to real number as a final answer to the original, real integral is a much easier process. However, say there are 3 roots, and a wedge going from 0 argument to pi/2 is enough to enclose 1 of the poles, why do we insist on using a wedge of angle 2pi/3 = 120 degrees? If we use a wedge of angle pi/2, does the positive imaginary axis pose a problem? Thanks Professor Israel, Oct 23 '15 at 5:52
• The wedge contour works if the integrals over the radial parts are related by symmetry: multiplying $z$ by $e^{i\theta}$, where $\theta$ is the angle of the wedge, multiplies the function by a constant. The choice of angle is dictated by the symmetry of the function. Oct 23 '15 at 6:13
• why you used the small arc at the origin? Our function doesnt have a singularity at zero, it is not easier just use two straight lines and the big arc? Sep 20 '18 at 10:37
• @Masacroso Habit, I guess. But it might be useful if $n$ is not an integer. Sep 20 '18 at 16:24
Contour Integration Approach
Assuming only that $$m>0$$ and $$-1, let $$z=x^m$$ with the exaggerated contour $$\gamma$$:
$$\hspace{4.5cm}$$
$$\gamma$$ is actually tight above and below the positive real axis and around the origin and circles back at an arbitrarily large distance from the origin.
The part just above the positive real axis captures the integral. The part just below the positive real axis gets $$-e^{2\pi i\frac{n-m+1}{m}}$$ times the integral. The residue at $$z=-1$$ of $$\frac{z^{\frac{n-m+1}{m}}}{1+z}$$ is $$e^{\pi i\frac{n-m+1}{m}}$$. Putting this all together yields
$$\frac1m\int_\gamma\frac{z^{\frac{n-m+1}{m}}}{1+z}\,\mathrm{d}z=\left(1-e^{2\pi i\frac{n-m+1}{m}}\right)\int_0^\infty\frac{x^n}{1+x^m}\,\mathrm{d}x$$ $$\frac{2\pi i}{m}e^{\pi i\frac{n-m+1}{m}}=\left(1-e^{2\pi i\frac{n-m+1}{m}}\right)\int_0^\infty\frac{x^n}{1+x^m}\,\mathrm{d}x$$ $$\frac{\pi}{m}=\sin\left(\pi\frac{n+1}{m}\right)\int_0^\infty\frac{x^n}{1+x^m}\,\mathrm{d}x$$ $$\frac{\pi}{m}\csc\left(\pi\frac{n+1}{m}\right)=\int_0^\infty\frac{x^n}{1+x^m}\,\mathrm{d}x\tag1$$
Relation to $$\bf{\Gamma(\alpha)\Gamma(1-\alpha)}$$
Setting $$m=1$$, $$n=\alpha-1$$, and using the substitution $$s=tu$$ yields Euler's Reflection Formula: \begin{align} \Gamma(\alpha)\Gamma(1-\alpha) &=\int_0^\infty s^{\alpha-1}e^{-s}\,\mathrm{d}s\int_0^\infty t^{-\alpha}e^{-t}\,\mathrm{d}t\\ &=\int_0^\infty\int_0^\infty u^{\alpha-1}e^{-(tu+t)}\,\mathrm{d}u\,\mathrm{d}t\\ &=\int_0^\infty\frac{u^{\alpha-1}}{1+u}\,\mathrm{d}u\\[6pt] &=\pi\csc(\pi\alpha)\tag2 \end{align}
Another Proof of $$\bf{(1)}$$ \begin{align} \int_0^\infty\frac{x^{\alpha-1}}{1+x}\,\mathrm{d}x &=\int_0^1\frac{x^{-\alpha}+x^{\alpha-1}}{1+x}\,\mathrm{d}x\\ &=\sum_{k=0}^\infty(-1)^k\int_0^1\left(x^{k-\alpha}+x^{k+\alpha-1}\right)\mathrm{d}x\\ &=\sum_{k=0}^\infty(-1)^k\left(\frac1{k-\alpha+1}+\frac1{k+\alpha}\right)\\ &=\sum_{k\in\mathbb{Z}}\frac{(-1)^k}{k+\alpha}\\[9pt] &=\pi\csc(\pi\alpha)\tag3 \end{align} where the last step of $$(3)$$ is proven in $$(3)$$ of this answer. Then apply $$(3)$$ to get \begin{align} \int_0^\infty\frac{x^n}{1+x^m}\,\mathrm{d}x &=\frac1m\int_0^\infty\frac{x^{\frac{n-m+1}m}}{1+x}\,\mathrm{d}x\\ &=\frac\pi{m}\csc\left(\pi\frac{n+1}m\right)\tag4 \end{align}
• Hi @robjohn, when faced with p.v. integrals, I always try to use this keyhole contour as my first method -- but if I remember correctly, this method has only ever worked nicely and correctly...once. Every other time, I've had to use a wedge contour, similar to what Robert Israel shows above. I wanted to ask you: are there times when the keyhole contour is the clearly better choice of contour - perhaps even necessary? If not, then using the wedge contour minimizes computation of residues and seems like the safest method to go with, especially on an exam, with the clock ticking. Oct 23 '15 at 5:43
• What do you think? Thanks @robjohn, Oct 23 '15 at 5:43
• Hi @robjohn, just a heads up, I've posted this comment as a new question on MSE. Thanks, Oct 23 '15 at 6:13
• @robjohn , Hi ! May you detailled, ( I BEG YOU ) how can you compute the integral beneath the real positive axis ? I understand that the above part gives the original integral, but it appears to me that this is the same for the part under... where is my mistake and how to do you fix this ? THANK YOU. Jan 1 '19 at 17:14
• @MarineGalantin: There is a branch cut along the real axis. $z^{\frac{n-m+1}m}$ limits to the usual real value from above the real axis, and it limits to $e^{2\pi i\,\frac{n-m+1}m}$ times the usual real value from below the real axis.
– robjohn
Jan 1 '19 at 23:41
Use Ramanujan's Master theorem, with $u=x^b\Rightarrow dx=\frac{du}{bu^{1-\frac{1}{b}}}$:
$$\int_0^{\infty}\frac{x^{a-1}}{1+x^b}dx=\frac{1}{b}\int_0^{\infty}\frac{u^{\frac{a-1}{b}-1+\frac{1}{b}}}{1+u}du$$ $$=\frac{1}{b}\int_0^{\infty}u^{\frac{a}{b}-1}\sum_{k \ge 0}\frac{(-u)^k}{\Gamma(k+1)}\Gamma(k+1)du$$ $$=\frac{1}{b}\Gamma\left(\frac{a}{b}\right)\Gamma\left(1-\frac{a}{b}\right)=\frac{\pi}{b \sin \left(\frac{a\pi}{b}\right)}.$$
\begin{align} &\sum_{k = - \infty }^\infty{\pars{-1}^{k} \over a + k} = {1 \over a} + \sum_{k = 1}^{\infty}\bracks{ {\pars{-1}^{-k} \over a - k} + {\pars{-1}^{k} \over a + k}} = {1 \over a} + \sum_{k = 1}^{\infty}\pars{-1}^{k + 1} {2a \over \pars{k - a}\pars{k + a}} \\[5mm]&= {1 \over a} + \sum_{k = 0}^{\infty}{2a \over \pars{2k + 1 - a}\pars{2k + 1 + a}} - \sum_{k = 0}^{\infty}{2a \over \pars{2k + 2 - a}\pars{2k + 2 + a}} \\[5mm] = &\ {1 \over a} + {a \over 2} \sum_{k = 0}^{\infty}{1 \over \pars{k + \pars{1 - a}/2}\pars{k + \pars{1 + a}/2}} \\[2mm] &\ - {a \over 2}\sum_{k = 0}^{\infty} {1 \over \pars{k + \pars{1 - a/2}}\pars{k + \pars{1 + a/2}}} \\[5mm] = &\ {1 \over a} + {a \over 2}\,{\Psi\pars{\bracks{1 - a}/2} - \Psi\pars{\bracks{1 + a}/2} \over \pars{1 - a}/2 - \pars{1 + a}/2} - {a \over 2}\,{\Psi\pars{1 - a/2} - \Psi\pars{1 + a/2} \over \pars{1 - a/2} - \pars{1 + a/2}} \\[5mm] = &\ {1 \over a} + \half\bracks{ -\Psi\pars{1 - a \over 2} + \Psi\pars{1 + a \over 2} + \Psi\pars{1 -{a \over 2}} -\Psi\pars{1 + {a \over 2}}} \end{align}
With the identities: $$\Psi\pars{1 - z} = \Psi\pars{z} + \pi\cot\pars{\pi z}\,,\quad \Psi\pars{\half + z} = \Psi\pars{\half - z} + \pi\tan\pars{\pi z}\,,\quad \Psi\pars{1 + z} = \Psi\pars{z} + {1 \over z}$$ we'll get
\begin{align} &\color{#0000ff}{\large \sum_{k = - \infty }^\infty{\pars{-1}^{k} \over a + k}} = {1 \over a} + \half\bracks{ \pi\tan\pars{\pi a \over 2} + \Psi\pars{a \over 2} + \pi\cot\pars{\pi a \over 2} - \Psi\pars{a \over 2} - {2 \over a}} \\[3mm]&= {\pi \over 2}\,{\sec^{2}\pars{\pi a/2} \over \tan\pars{\pi a/2}} = {\pi \over \sin\pars{\pi a}} = \color{#0000ff}{\large\pi\,\csc\pars{\pi a}} \end{align}
• You seem to be forgetting the $k=0$ term. The result should be $\pi\csc \pi a$ as seen here. Dec 2 '13 at 1:38
• @PedroTamaroff Yes. I already checked. I forgot to add the $k = 0$ term and the $1/2$ factor. Thanks. Dec 2 '13 at 2:09
• Nicely done! +1 Nov 5 '16 at 16:30
• @Dr.MV Thanks a lot. Nov 5 '16 at 19:32
This is in Gradshteyn/Ryzhik, formula 3.241.2: $$\int_0^\infty {x^{\mu-1}\over 1+x^\nu}\ dx={\pi\over\nu}\csc\bigg({\mu\pi\over\nu}\bigg)={1\over\nu}B\bigg({\mu\over\nu},{\nu-\mu\over\nu}\bigg)$$ assuming, of course, $Re(\nu)>Re(\mu)>0$, and where $B(x,y)$ denotes the Beta function $B(x,y)={\Gamma(x)\Gamma(y)\over\Gamma(x+y)}$.
To see the Beta function part of the equality, use the integral representation $$B(x,y)=\int_0^\infty {t^{x-1}\over (1+t)^{x+y}}\ dt$$ Then ${1\over\nu}B\big({\mu\over\nu},{\nu-\mu\over\nu}\big)$ is $${1\over\nu}\int_0^\infty {t^{{\mu\over\nu}-1}\over 1+t}\ dt$$ Send $t$ to $t^\nu$, and you are done!
• The Beta function has an integral representation $$B(x,y)=\int_0^\infty {t^{x-1}\over (1+t)^{x+y}}\ dx$$ Since we have the answer, using this formula, it is not hard to show it is correct :)
– B R
Feb 17 '12 at 20:16
• But my integral has $1+x^v$, not $(1+x)^v$. How did you transform it? Feb 17 '12 at 20:25
• @B R: nice way (+1) Aug 11 '12 at 12:06
The answers on both of these previous threads
Simpler way to compute a definite integral without resorting to partial fractions?
$\int_{0}^{\infty}\frac{dx}{1+x^n}$
provide a solution to exactly this problem.
Also this problem arise again in this newer thread:
Is there an elementary method for evaluating $\int_0^\infty \frac{dx}{x^s (x+1)}$?
The meta thread discussing the similarities between these 4 questions can be found here: http://meta.math.stackexchange.com/questions/3746/abstract-duplicates-what-to-do-in-this-scenario
• Note that the previous qustions don't ask for the general solution but rather a particular case, and it is the answerers that decide to provide a solution to the general integral I present here. Also, I wanted another approach in the solution, see here. Feb 29 '12 at 23:48
$$t=\frac1{1+x^m}\qquad\iff\qquad x=\left(\frac1t-1\right)^\frac1m=\left(\frac{1-t}t\right)^\frac1m=t^{^{-\frac1m}}\cdot(1-t)^{^\frac1m}\qquad\iff$$
Can you already see where this is going ? ;-)
$$\iff\frac{dx}{dt}=\frac{d}{dt}\left[\left(\frac1t-1\right)^\frac1m\ \right]=\frac1m\left(\frac{1-t}t\right)^{\frac1m-1}\left(-\frac1{t^2}\right)=-\frac{t^{^{-\frac1m-1}}\cdot(1-t)^{^{\frac1m-1}}}m\iff$$
$$I=\int_0^\infty\frac{x^n}{1+x^m}dx=\frac1m\int_0^1t^{^{-\frac{n+1}m}}\cdot(1-t)^{^{\frac{n+1}m-1}}dt=\frac{B\left(1-\frac{n+1}m,\frac{n+1}m\right)}m$$
Now all we have to do is use the fact that $B(m,n)=\displaystyle\frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}$ , along with Euler's reflection formula, $\displaystyle\Gamma(z)\cdot\Gamma(1-z) = \frac{\pi}{\sin{(\pi z)}}$ , and we're done ! :-)
Letting $u =x^m$ entails $dx =\frac{1}{m}t^{1/m-1}$ and hence,
$$\int\limits_0^\infty {\frac{{{x^n}}}{{1 + {x^m}}}dx } =\int\limits_0^\infty {\frac{t^{\frac{n+1}{m}-1}}{(1 + t)^{1}}dt } =\frac{1}{m}B\left(\frac{n+1}{m}, 1-\frac{n+1}{m}\right)\\=\color{red}{\frac{1}{m}\Gamma\left(\frac{n+1}{m}\right)\Gamma\left( 1-\frac{n+1}{m}\right)=\frac{\pi}{m\sin\left(\frac{\pi(n+1)}{m}\right)}}$$
where we made use of the beta and Gamma function and the Schwartz duplication formula
Integrate over a closed contour that is in the shape of a wedge in the 1st quadrant, having an angle of $\pi/5$. That is, consider
$$\oint_C dz \frac{z^2}{z^{10}+1}$$
where $C$ is that wedge contour. As mentioned above, the contour splits into 3 pieces:
$$\oint_C dz \frac{z^2}{z^{10}+1} = \int_0^R dx \frac{x^2}{x^{10}+1} + i R \int_0^{\pi/5}d\phi \, e^{i \phi} \frac{R^2 e^{i 2 \phi}}{R^{10} e^{i 10 \phi}+1}+ e^{i 3 \pi/5} \int_R^0 dx \frac{x^2}{x^{10}+1}$$
As $R \to \infty$, the second integral vanishes as $1/R^7$ (Note that the denominator never vanishes). The rest is then just equal to $i 2 \pi$ times the residue at the pole $z=e^{i \pi/10}$. Thus, combining the integrals, we have
$$\left (1-e^{i 3 \pi/5}\right) \int_0^{\infty} dx \frac{x^2}{x^{10}+1} = i 2 \pi \frac{e^{i 2 \pi/10}}{10 e^{i 9 \pi/10}}$$
Doing the algebra, I get
$$\int_0^{\infty} dx \frac{x^2}{x^{10}+1} = \frac{\pi}{10 \cos{(\pi/5)}}$$
• This seems similar to Robert Israel's completed answer.
– robjohn
May 15 '13 at 11:24
• Were you answering this question?
– robjohn
May 15 '13 at 11:44
• @robjohn: yes. As I had invested the time to do so, I placed the answer in the non-dup section. May 15 '13 at 11:59
• I don't understand why you have answered with a particular case. I also don't know any complex integration, so I cannot make use of this, yet. May 15 '13 at 13:18
• @PeterTamaroff: The original question was closed. due to being a duplicate of this; it was closed just as I was finishing typing this solution. May 15 '13 at 14:35
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2021-12-03 19:06:37
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https://www.gamedev.net/forums/topic/640974-single-big-shader-buffer-for-hardware-instancing/
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# DX11 Single big shader buffer for hardware instancing
## Recommended Posts
I saw this part from the battlefield3 presentation (http://dice.se/wp-content/uploads/GDC11_DX11inBF3_Public.pdf) on page 30 where they show how they transfer transformations of each instance to the shader using a single big buffer and a const buffer with indices to index this buffer at the right position using the SV_InstanceID semantic. Can someone tell me what exactly this buffer is ?
This part here:
Buffer<float4> instanceVectorBuffer : register(t0);
I'm not too familiar with buffers in directx11 beyond constant buffers...
Isn't register(t0) for texture slots ? Also they seem to use 3x4 floats for their transformations, so does that mean they construct their transformation matrix inside the shader using translation, rotation and scale vectors ?
And last how would I set such a buffer in the DX11 API ?
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The MSDN documentation is very lacking in explaining these concepts...
IIRC:
The t registers are similar to the c registers, but they're designed for random memory access patterns (like texture lookups) rather than constant memory access patterns (e.g. it's assumed that every pixel will read the same constants, but may read different texels). You can bind both textures and DX11 buffer objects to the t registers and then load values from them in the shader.
You bind values to the t registers using *SSetShaderResources, which takes "shader resource views". You can create a view of a texture, which is the common case, but you can also create a view of a buffer.
As mentioned above, you should prefer this method of binding buffers to t registers when you're going to be performing random lookups into the buffer. If every pixel/vertex/etc is going to read the same values from the buffer, then you should bind it to a c register using the usual method.
Also they seem to use 3x4 floats for their transformations, so does that mean they construct their transformation matrix inside the shader using translation, rotation and scale vectors ?
No, in a regular transformation matrix that's been constructed from a traslation + rotation + scale, the 4th row/column (depending on your conventions) will always be [0,0,0,1] so you can hard-code that value in the shader to save some space in the buffer.
I'm still not as experienced with DX11 as I am with DX9, so I hope that's correct
Edited by Hodgman
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Yes I meant 3 x float4 (sorry for the confusing syntax )
Alright thanks for the info.
EDIT: By the way this buffer then should be dynamic, correct? Since I need to put in or remove elements during rendering.
Edited by lipsryme
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Yeah by the looks of it, they'd potentially be updating the entire buffer every frame.
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EDIT: Nevermind got it fixed...my shader wasn't up to date when I ran it
Edited by lipsryme
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The advantage of a buffer over cbuffer is the size (128mb vs 64kb, although in typical scenario 2-4mb should be enough) which allows you to store a frame worth of data to the buffer, also it allows you to fill the buffer once so you'll minimize buffer updates to minimum. The minor inconvenience is that you'll be able to store only float data (probably you can get around this with some bit manipulation method).
The buffer is pratically a texture and you'd bind it as vertex shader resource.
[source]
float4x4 GetInstanceMatrix(uint InstID,uint Offset)
{
uint BufferOffset = InstID * ElementsPerInstance + StartIndex + Offset;
// InstID = InstanceID
// ElementsPerInstance, how many float4's there are per drawn instance (typically 3 for a single mesh if 4x3 matrices are used)
// StartIndex = location of the first float4 inside the buffer for the first instance
// Offset (used to retrieve bone matrices)
float4 r0 = InstanceData.Load(BufferOffset + 0);
float4 r1 = InstanceData.Load(BufferOffset + 1);
float4 r2 = InstanceData.Load(BufferOffset + 2);
float4 r3 = float4(0.0f,0.0f,0.0f,1.0f);
return float4x4(r0,r1,r2,r3);
}
[/source]
Inside the vertex shader you may use the above code to retrieve the transform matrix for each vertex.
Cheers!
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The advantage of a buffer over cbuffer is the size (128mb vs 64kb, although in typical scenario 2-4mb should be enough)
Another important advantage from a tbuffer over a cbuffer is that cbuffer suffer from constant waterfalling, which make them a horrible fit for scalable hw accelerated vertex skinning (indexing the constant buffer with a different index in each vertex).
When you're not suffering constant waterfalling, cbuffers can be faster though, for well.... constant data.
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Ok this may sound stupid, but what about just using regular instancing?Has this method been benchmarked against regular instancing with a second vertex buffer?
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Ok this may sound stupid, but what about just using regular instancing?Has this method been benchmarked against regular instancing with a second vertex buffer?
If you mean that regular instancing means using a second vertex buffer, I remember reading some years ago that one of the big companies wrote that shaders using a second vertex stream will take a performance hit. It had something to do with the fact that the vertex declaration structure gets pretty big.
I think that the problem with classic instancing is also, that you'll need to define a vertex declaration for each different vertex shader type where the parameters are different. Normally of course you would just have a transform matrix as instance data for the vertex shader, but what if you need also some other per instance parameters, such as color, inverse world matrix or some other parameter which makes your instance different from the others. When using second vertex stream you'll need a different vertex declaration for each case.
In the case of the generic buffer, you'll need just one vertex declaration since the vertex data won't change. After it is the job of the vertex shader to extract the desired data from the generic buffer. Of course you'll need to make sure that the buffer contains the expected data. Also, since the buffer can be pretty big you'll be updating data much less often to the graphic card, which reduces API calls. Also, the program side logic get simpler when you don't need to worry so much about "will the data fit to the buffer or not, do I have to make several draw calls instead of one".
Of course, under D3D11 the instancing just another parameter that you can access in the vertex shader (InstanceID) and after that you may read or generate the instancing data in the best way you can imagine, though vertex streams can't be randomly accessed.
Cheers!
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So..I've created the buffer like this:
D3D11_BUFFER_DESC instanceBufferDesc = {};
instanceBufferDesc.Usage = D3D11_USAGE_DYNAMIC;
instanceBufferDesc.ByteWidth = sizeof(XMFLOAT4);
instanceBufferDesc.CPUAccessFlags = D3D11_CPU_ACCESS_WRITE;
instanceBufferDesc.MiscFlags = 0;
instanceBufferDesc.StructureByteStride = 0;
hr = this->device->CreateBuffer(&instanceBufferDesc, NULL, &this->instanceTransformBuffer);
is that correct so far ?
Now I'm a little clueless on how to create the shader resource view from this.
Any ideas ? Do I set the shader resource view description to NULL ?
This seems to work without errors:
D3D11_SHADER_RESOURCE_VIEW_DESC shaderResourceDesc;
Now if the above is correct, how can I basically push float4's onto the buffer ?
Edited by lipsryme
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Use DeviceContext->Map(...) with D3D11_MAP_WRITE_DISCARD to get a pointer to an empty the buffer.
Fill the data (with memcpy for example) and the call Unmap. Don't write outside of the buffer! It will cause memory corruption and may even hang your computer.
It doesn't really make sense to use the buffer for one float, but I assume it to be a test. I used DXGI_FORMAT_R32G32B32A32_FLOAT as the buffer format since handling things as float4 makes maybe a bit more sense. You'll need at least 3x4 floats to store a typical matrix, but of course probably you'll want to allocate some megabytes of storage (within reason of course).
Cheers!
Edited by kauna
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Got it working perfectly thanks !
So the only way is to define a constant max size for this buffer inside the code...?
Worst case would be 3 x numInstances which could add up to quite a lot.
By the way I'm currently assigning each instance the 3 float4's inside a loop. I assume it would be more efficient to add them inside a structure which I then copy once using memcopy or is that neglectable?
I'm only doing map / unmap once, but I'm filling the data like this:
// Update buffer
for(int u = 0; u < 3; u++)
{
XMStoreFloat4(&pInstanceData[(numInstances * 3) + u], worldTransform.r[u]);
}
basically in the loop that goes through every instance and counts them to do the instanced draw call at the end.
Also am I correct in thinking that I am restricted to one specific material (texture and properties from const buffer) for every unique instance group ?
update: Just read about putting them inside a Texture2DArray...so for specific material properties would it make sense to also store them inside a buffer like I did with the transforms ? Might be overkill for e.g. booleans but still...
Edited by lipsryme
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Got it working perfectly thanks !
So the only way is to define a constant max size for this buffer inside the code...?
Worst case would be 3 x numInstances which could add up to quite a lot.
By the way I'm currently assigning each instance the 3 float4's inside a loop. I assume it would be more efficient to add them inside a structure which I then copy once using memcopy or is that neglectable?
I'm only doing map / unmap once, but I'm filling the data like this:
// Update buffer
for(int u = 0; u < 3; u++)
{
XMStoreFloat4(&pInstanceData[(numInstances * 3) + u], worldTransform.r[u]);
}
basically in the loop that goes through every instance and counts them to do the instanced draw call at the end.
Also am I correct in thinking that I am restricted to one specific material (texture and properties from const buffer) for every unique instance group ?
update: Just read about putting them inside a Texture2DArray...so for specific material properties would it make sense to also store them inside a buffer like I did with the transforms ? Might be overkill for e.g. booleans but still...
Wait, can I ask something - how does your draw call look like?I mean, do you call DrawIndexedInstanced with only 1 buffer?Or with 1 vertex buffer and 1 empty instance buffer/or having the ID inside the instance buffer data?
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I only use a vertex buffer yes, not using instance buffer at all but giving it per instance transformations via a float4 buffer that holds these.
They're then accessed inside the shader using the instanceID.
Edited by lipsryme
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I only use a vertex buffer yes, not using instance buffer at all but giving it per instance transformations via a float4 buffer that holds these.
They're then accessed inside the shader using the instanceID.
Hm I tried to do the same as you to see how it compares to normal instancing, however nothing leaves the vertex shader.The Shader Debugger detects that the sphere goes in:
However nothing gets rasterized.It can't be the view matrix, since it works on normal rendering, so it has to be something to do with the large buffer object.Did you ever experience such an issue?
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Hmm I do remember having some issues that no pixel shader was being executed....can't remember exactly what it was though...
Should look something like this:
D3D11_INPUT_ELEMENT_DESC lo[] =
{
{"POSITION", 0, DXGI_FORMAT_R32G32B32_FLOAT, 0, 0, D3D11_INPUT_PER_VERTEX_DATA, 0 },
{"TEXCOORD", 0, DXGI_FORMAT_R32G32_FLOAT, 1, D3D11_APPEND_ALIGNED_ELEMENT, D3D11_INPUT_PER_VERTEX_DATA, 0},
{"NORMAL", 0, DXGI_FORMAT_R32G32B32_FLOAT, 2, D3D11_APPEND_ALIGNED_ELEMENT, D3D11_INPUT_PER_VERTEX_DATA, 0},
{"TANGENT", 0, DXGI_FORMAT_R32G32B32_FLOAT, 3, D3D11_APPEND_ALIGNED_ELEMENT, D3D11_INPUT_PER_VERTEX_DATA, 0},
};
fourth parameter is important here as this is the vertex buffer slot...
Also try to make sure the float4s are combined in the correct order.
I've had a lot of problems with this. In the end I had to transpose the matrix first and then pass them to the cbuffer because otherwise you'd loose the 4th row because
if you pass your 3 rows something like:
cbufferData.data[i] = worldMatrix.r[i];
you're passing only 3 vectors of this matrix and you will loose important data if you pass the rows instead of the columns.
That's why I changed it over to columns and it worked:
float4x4 GetInstanceTransform(uint instID, uint offset)
{
uint BufferOffset = instID * elementsPerInstance + startIndex + offset;
float4 c0 = InstanceDataBuffer.Load(BufferOffset + 0);
float4 c1 = InstanceDataBuffer.Load(BufferOffset + 1);
float4 c2 = InstanceDataBuffer.Load(BufferOffset + 2);
float4 c3 = float4(0.0f, 0.0f, 0.0f, 1.0f);
float4x4 _World = { c0.x, c1.x, c2.x, c3.x,
c0.y, c1.y, c2.y, c3.y,
c0.z, c1.z, c2.z, c3.z,
c0.w, c1.w, c2.w, c3.w };
return _World;
}
I don't see a way around that...but maybe someone does ?
Edited by lipsryme
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That's why I changed it over to columns and it worked
oh I see, cause in your other post you specify #pragma pack_matrix( row_major )
I'll try to get it to work tonight when i can to see what happens
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Yes but that only applies to float4x4 values coming from the cbuffer. It doesn't apply to what you use inside the vertex shader and also not for float4's.
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• 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):
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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;
|
2017-11-18 23:35:15
|
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|
https://codereview.stackexchange.com/questions/159810/root-finding-and-integration
|
# Root finding and integration [closed]
I am trying to implement the modified Next Reaction Method with time varying propensities as mentioned in sections IV and V of this paper. At one step in the process this expression must be evaluated: $\int_t^{t+\Delta t\mu}a(X(t,s), s) \textrm{d}s - (P_k - T_k) = 0$. In the end this evaluation should tell us $t + \Delta t\mu$.
The code turns out to be quite slow, and since I need to run a huge system (many $a(X(t,s))$'s), this slow code is not an option. Below is the code, and comments on what it does, and what I've tried to optimize so far.
Maybe I should try Cython, but I do not know how, and for which parts it would be most beneficial; tips are welcome. Maybe other packages for integration, and root finding would help. But again, I am not aware of any such packages, or whether it would help me greatly, and tips are very welcome.
In the real project, this code is in a subclass of some more general class of simulation methods. This means that much less arguments are passed around, due to making objects attributes of this class.
import numpy as np
from scipy import integrate, optimize
from functools import partial, update_wrapper
# In my original code, the propensity functions (propensity_1, propensity_2)
# are generated automatically from a matrix and a bunch of parameter vectors,
# that is why setup might seem redundant in this code snippet.
# In one futile effort I tried @jit as shown below, but this made the code
# drastically slower.
def wrapped_partial(func, *args, **kwargs):
partial_func = partial(func, *args, **kwargs)
update_wrapper(partial_func, func)
return partial_func
#@jit
def sinputt(t, amplitude=5.0, frequency=0.2):
return amplitude * (1 + np.sin(frequency * t)) + 1
#@jit
def propensity_1(state, t, k):
return k * sinputt(t=t)
#@jit
def propensity_2(state, t, k):
return k * state
# Function to be integrated
def functionieren(t, state, func):
return func(t=t, state=state)
# Returns integral
def integrieren(t, dt, state, func):
return np.abs(
integrate.quad(func=functionieren, a=t, b=dt, args=(state, func), epsabs=1.e-4, epsrel=1.e-4, limit=50)[0])
# Function for which root is to be found
def optimizieren(dt, t, state, func, P_T):
return P_T - integrieren(t, dt, state, func)
# Function that finds root, thus t+dt
def solve(t, dt, state, func, P_T):
return optimize.newton(func=optimizieren,x0=dt,args=(t,state,func,P_T))
# One step in the algorithm
def run_step(P, T, state, t, propensities, funcs):
dtk = np.divide((P - T), propensities)
r_i = np.argmin(dtk)
dt = dtk[r_i, 0]
t += dt
if r_i == 0:
state += 1
elif r_i == 1:
state -= 1
T += propensities * dt
P[r_i, 0] -= np.log(np.random.random(1))
prop_func1,prop_func2=funcs
#HERE IS THE BUGGER
dtp=solve(t=t,dt=t+1.0,state=state,func=prop_func1,P_T=P[0,0]-T[0,0])
propensity1=prop_func1(state=state, t=dtp)
propensity2=prop_func2(state=state,t=t)
propensities = np.array([[propensity1],
[propensity2]])
return P, T, state, t, propensities
#The algorithm, final time=100
def run_sim(P, T, state, t, propensities, funcs):
state_track = [state0]
t_track = [t0]
while t < 100:
P, T, state, t, propensities = run_step(P, T, state, t, propensities,funcs)
state_track.append(np.copy(state))
t_track.append(np.copy(t))
return np.array(state_track), np.array(t_track)
# Parameters
k1 = 2.0
k2 = 2.0
# Initial conditions
t0 = 0
state0 = 5
# Initial conditions internal clocks
T0 = np.zeros((2, 1))
P0 = -np.log(np.random.random((2, 1)))
# Generation of propensity functions
prop_func1 = wrapped_partial(propensity_1, k=k1)
prop_func2 = wrapped_partial(propensity_2, k=k2)
# Initial propensities
propensity1=solve(t=t0,dt=t+1.0,state=state0,func=prop_func1,P_T=P0[0,0]-T0[0,0])
propensity2=prop_func2(state=state0,t=t0)
propensities0=np.array([[propensity1],
[propensity2]])
# If dividing by a 0 propensity inf is returned
np.seterr(divide="ignore")
state_track,t_track=run_sim(P0,T0,state0,t0,propensities0,(prop_func1,prop_func2))
# Plot of the results as they are intended
import matplotlib.pyplot as plt
t_sin = np.arange(0, 100, 0.1)
input_sin = sinputt(t=t_sin)
plt.plot(t_sin, input_sin,label='input')
plt.step(t_track, state_track,label='state')
plt.show()
# Uncomment the %timeit below to see which steps are slow
# %timeit run_step(P, T, state, t, propensities,(prop_func1,prop_func2))
# %timeit solve(t=t,dt=t+1.0,state=state,func=prop_func1,P_T=P[0,0]-T[0,0])
# %timeit integrieren(t, t+1.0, state, prop_func1)
# %timeit functionieren(t, state, prop_func1)
# Output timeit
# 1000 loops, best of 3: 294 µs per loop
# 1000 loops, best of 3: 265 µs per loop
# 10000 loops, best of 3: 42.7 µs per loop
# The slowest run took 7.98 times longer than the fastest. This could mean
# that an intermediate result is being cached.
# 1000000 loops, best of 3: 1.74 µs per loop
• I tried running the code, and it failed with a NameError. – Peter Taylor Aug 2 '19 at 10:05
• Spacing Between variables and operators, there should be spaces to improve readability. Any code like prop_func1,prop_func2=funcs should instead be prop_func1, prop_func2 = funcs.
• Lots of Parameters: When you have tons of parameters you are passing/receiving to/in a method, having them all on one line can decrease readability significantly. You should put each parameter on its own line, to make it clear what is being passed/received.
• Docstrings: You should have a docstring at the beginning of every method, class, and module you write. This helps documentation determine what your code is supposed to accomplish. You had the right idea with comments above the methods explaining what they do, now just move them inside the function between """ docstring here """.
• Import Statements: All import statements should be at the top of the program, regardless of when you use them in your code. My personal preference is to have them ordered alphabetically, but it's up to you. The convention is to have them at the top of the program. (PEP-8 import statement info here)
• Main Guard: Let's say you want to reuse this system for another program. You import the file, write your code, and WHAT??? Everything is messed up. Having a main guard is a way to prevent this from happening. Just include everything outside of methods in a if __name__ == '__main__' statement, and once you import the file, since it's not the main file being run, that code won't run. An answer from StackOverflow better explains this.
• Constants: Any and all constants you have in your program should be UPPERCASE, to make it clear they are constants.
Updated Code
"""
Program implements the modified Next Reaction Method with time varying propensities
"""
import matplotlib.pyplot as plt
import numpy as np
from functools import partial, update_wrapper
from scipy import integrate, optimize
# In my original code, the propensity functions (propensity_1, propensity_2)
# are generated automatically from a matrix and a bunch of parameter vectors,
# that is why setup might seem redundant in this code snippet.
# In one futile effort I tried @jit as shown below, but this made the code
# drastically slower.
def wrapped_partial(func, *args, **kwargs):
""" Calculation Method """
partial_func = partial(func, *args, **kwargs)
update_wrapper(partial_func, func)
return partial_func
#@jit
def sinputt(t, amplitude=5.0, frequency=0.2):
""" Calculation Method """
return amplitude * (1 + np.sin(frequency * t)) + 1
#@jit
def propensity_1(state, t, k):
""" Calculation Method """
return k * sinputt(t=t)
#@jit
def propensity_2(state, t, k):
""" Calculation Method """
return k * state
def functionieren(t, state, func):
""" Returns the function to be integrated """
return func(t=t, state=state)
def integrieren(t, dt, state, func):
""" Returns the integral """
return np.abs(
func=functionieren,
a=t,
b=dt,
args=(state, func),
epsabs=1.e-4,
epsrel=1.e-4,
limit=50)[0]
)
def optimizieren(dt, t, state, func, P_T):
""" Function for which root is to be found """
return P_T - integrieren(t, dt, state, func)
def solve(t, dt, state, func, P_T):
""" Function that finds root, thus t+dt """
return optimize.newton(
func=optimizieren,
x0=dt,
args=(t, state, func, P_T)
)
def run_step(P, T, state, t, propensities, funcs):
""" One step in the algorithm """
dtk = np.divide((P - T), propensities)
r_i = np.argmin(dtk)
dt = dtk[r_i, 0]
t += dt
if r_i == 0:
state += 1
elif r_i == 1:
state -= 1
T += propensities * dt
P[r_i, 0] -= np.log(np.random.random(1))
prop_func1, prop_func2 = funcs
#HERE IS THE BUGGER
dtp = solve(t=t, dt=t+1.0, state=state, func=prop_func1, P_T=P[0, 0]-T[0, 0])
propensity1 = prop_func1(state=state, t=dtp)
propensity2 = prop_func2(state=state, t=t)
propensities = np.array([[propensity1], [propensity2]])
return P, T, state, t, propensities
def run_sim(P, T, state, t, propensities, funcs):
""" The algorithm, final time=100 """
state_track = [STATE0]
t_track = [t0]
while t < 100:
P, T, state, t, propensities = run_step(P, T, state, t, propensities, funcs)
state_track.append(np.copy(state))
t_track.append(np.copy(t))
return np.array(state_track), np.array(t_track)
if __name__ == '__main__':
# Parameters
k1 = 2.0
k2 = 2.0
# Initial conditions
t0 = 0
STATE0 = 5
# Initial conditions internal clocks
T0 = np.zeros((2, 1))
P0 = -np.log(np.random.random((2, 1)))
# Generation of propensity functions
prop_func1 = wrapped_partial(propensity_1, k=k1)
prop_func2 = wrapped_partial(propensity_2, k=k2)
# Initial propensities
PROPENSITY1 = solve(t=t0, dt=t0+1.0, state=STATE0, func=prop_func1, P_T=P0[0, 0]-T0[0, 0])
PROPENSITY2 = prop_func2(state=STATE0, t=t0)
PROPENSITIES0 = np.array([[PROPENSITY1], [PROPENSITY2]])
# If dividing by a 0 propensity inf is returned
np.seterr(divide="ignore")
STATE_TRACK, T_TRACK = run_sim(P0, T0, STATE0, t0, PROPENSITIES0, (prop_func1, prop_func2))
# Plot of the results as they are intended
T_SIN = np.arange(0, 100, 0.1)
INPUT_SIN = sinputt(t=T_SIN)
plt.plot(T_SIN, INPUT_SIN, label='input')
plt.step(T_TRACK, STATE_TRACK, label='state')
plt.show()
# Uncomment the %timeit below to see which steps are slow
# %timeit run_step(P, T, state, t, propensities,(prop_func1,prop_func2))
# %timeit solve(t=t,dt=t+1.0,state=state,func=prop_func1,P_T=P[0,0]-T[0,0])
# %timeit integrieren(t, t+1.0, state, prop_func1)
# %timeit functionieren(t, state, prop_func1)
# Output timeit
# 1000 loops, best of 3: 294 µs per loop
# 1000 loops, best of 3: 265 µs per loop
# 10000 loops, best of 3: 42.7 µs per loop
# The slowest run took 7.98 times longer than the fastest. This could mean
# that an intermediate result is being cached.
# 1000000 loops, best of 3: 1.74 µs per loop
|
2020-10-22 09:22:40
|
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|
https://share.cocalc.com/share/6ca61bdb47a9fd152e59906a5016c2b2a36e3155/Last%20Bracket%20in%20comment.sagews?viewer=share
|
CoCalc Public FilesLast Bracket in comment.sagews
Author: Dominique Laurain
Views : 54
Description: Post at google groups sage-support
Compute Environment: Ubuntu 18.04 (Deprecated)
# Strangely enough : after ] at end of line comment at end of cell, we cannot open new line after it. Try it below
#res = [
#x == -(w*(4*z + 7) + 7*z + 11)/(2*w*(z + 2) + 4*z + 7),
#x == -(w*(2*z + 5) + 5*z + 10)/(2*w*(z + 2) + 4*z + 7)
#]
# that is new cel
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2020-10-31 02:30:12
|
{"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": 0.2928714156150818, "perplexity": 12444.590531088143}, "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-2020-45/segments/1603107912593.62/warc/CC-MAIN-20201031002758-20201031032758-00600.warc.gz"}
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https://yutsumura.com/a-positive-definite-matrix-has-a-unique-positive-definite-square-root/
|
# A Positive Definite Matrix Has a Unique Positive Definite Square Root
## Problem 514
Prove that a positive definite matrix has a unique positive definite square root.
In this post, we review several definitions (a square root of a matrix, a positive definite matrix) and solve the above problem.
After the proof, several extra problems about square roots of a matrix are given.
## Definitions (Square Roots of a Matrix)
Let $A$ be a square matrix. If there exists a matrix $B$ such that $B^2=A$, then we call $B$ a square root of the matrix $A$.
### Examples
For example, if $A=\begin{bmatrix} 2 & 2\\ 2& 2 \end{bmatrix}$, then it is straightforward to see that
$\begin{bmatrix} 1 & 1\\ 1& 1 \end{bmatrix} \text{ and } \begin{bmatrix} -1 & -1\\ -1& -1 \end{bmatrix}$ are square roots of $A$.
(The less trivial question is that these are the only square roots of $A$.
See the post “Find All the Square Roots of a Given 2 by 2 Matrix” .)
Some matrices do not have a square root at all.
For example, the matrix $A=\begin{bmatrix} 0 & 1\\ 0& 0 \end{bmatrix}$ does not have a square root matrix.
(See the post “No/Infinitely Many Square Roots of 2 by 2 Matrices” Part (a).)
On the other hand, some matrices have infinitely many square roots.
For example, the $2\times 2$ identity matrix has infinitely many distinct square roots.
(See the post “No/Infinitely Many Square Roots of 2 by 2 Matrices” Part (b).)
## Analogy with Positive Real Number
For a positive real number $a$, there are two square roots $\pm \sqrt{a}$.
Here $\sqrt{a}$ is the unique positive number whose square is $a$.
The corresponding notion of positive number in matrices is positive definite.
### Definition (Positive Definite Matrix)
An $n\times n$ real symmetric matrix $A$ is said to be positive definite if $\mathbf{v}^{\trans}A\mathbf{v}$ is positive for all nonzero vector $\mathbf{v}\in \R^n$.
We will use the following fact in the proof.
Fact.
A real symmetric matrix $A$ is positive definite if and only if the eigenvalues of $A$ are all positive.
For a proof of this fact, see the post “Positive definite Real Symmetric Matrix and its Eigenvalues”.
### Problem
Just like a positive real number $a$ has a unique positive square root $\sqrt{a}$, we can prove the following (which is the problem of this post).
Problem.
A positive definite matrix has a unique positive definite square root.
## Proof.
Let $A$ be a positive definite $n\times n$ matrix.
### Existence of a Square Root
We first show that there exists a positive definite matrix $B$ such that $B^2=A$.
Let $\lambda_1, \dots, \lambda_n$ be eigenvalues of $A$.
Since $A$ is a real symmetric matrix, it is diagonalizable by an orthogonal matrix $S$.
That is, we have
$S^{\trans}AS=D,$ where $D$ is the diagonal matrix
$D=\begin{bmatrix} \lambda_1 & 0 & 0 & 0 \\ 0 &\lambda_2 & 0 & 0 \\ \vdots & \cdots & \ddots & \vdots \\ 0 & 0 & \cdots & \lambda_n \end{bmatrix}.$ (Note that $S^{-1}=S^{\trans}$ since $S$ is orthogonal.)
Recall that the eigenvalues of a positive definite matrix are positive real numbers by Fact.
So eigenvalues $\lambda_i$ of $A$ are positive real numbers.
Hence $\sqrt{\lambda_i}$ is a positive real number for $i=1, \dots, n$.
Define the matrix
$B=SD’S^{\trans},$ where $D’$ is the diagonal matrix
$D’=\begin{bmatrix} \sqrt{\lambda_1} & 0 & 0 & 0 \\ 0 &\sqrt{\lambda_2} & 0 & 0 \\ \vdots & \cdots & \ddots & \vdots \\ 0 & 0 & \cdots & \sqrt{\lambda_n} \end{bmatrix}.$
Then $B$ is a symmetric matrix because
$B^{\trans}=(SD’S^{\trans})^{\trans}=(S^{\trans})^{\trans}D’^{\trans}S^{\trans}=SD’S^{\trans}=B.$
It follows from the definition of $B=SD’S^{\trans}$ that the eigenvalues of $B$ are positive numbers $\sqrt{\lambda_1}, \dots, \sqrt{\lambda_n}$.
Thus by Fact the matrix $B$ is positive-definite.
The matrix $B$ is a square root of $A$ since we have
\begin{align*}
B^2=(SD’S^{\trans})(SD’S^{\trans})=SD’^2S^{\trans}=SDS^{\trans}=A.
\end{align*}
### Uniqueness of a Square Root
Now we prove the uniqueness of the square root.
Suppose that $C$ is another positive definite square roots of the matrix $A$.
Since $C$ is a real symmetric matrix, there is an orthogonal matrix $P$ such that
$P^{\trans}CP=T.$ Here $T$ is the diagonal matrix
$T=\begin{bmatrix} \mu_1 & 0 & 0 & 0 \\ 0 &\mu_2 & 0 & 0 \\ \vdots & \cdots & \ddots & \vdots \\ 0 & 0 & \cdots & \mu_n \end{bmatrix},$ where $\mu_1, \dots, \mu_n$ are eigenvalues of $C$.
Since $C^2=A$, we have
\begin{align*}
P^{\trans}AP&=P^{\trans}C^2P=(P^{\trans}CP)^2=T^2\6pt] &=\begin{bmatrix} \mu_1^2& 0 & 0 & 0 \\ 0 &\mu_2^2 & 0 & 0 \\ \vdots & \cdots & \ddots & \vdots \\ 0 & 0 & \cdots & \mu_n^2 \end{bmatrix}. \end{align*} Thus, the matrix P diagonalizes A, and it follows that up to permutation \mu_1^2, \dots, \mu_n^2 are equal to \lambda_1, \dots, \lambda_n. Hence we can modify P (by interchanging columns vectors), and without loss of generality we may assume that \mu_i^2=\lambda_i for i=1, \dots, n. Thus P^{\trans}CP=D’ or equivalently, \[ C=PD’P^{\trans}.
Since $B^2=A=C^2$, we have
\begin{align*}
&(SD’S^{\trans})^2=(PD’P^{\trans})^2\\
&\Leftrightarrow SD’^2S^{\trans}=PD’^2P^{\trans}\\
&\Leftrightarrow SDS^{\trans}=PDP^{\trans}\\
&\Leftrightarrow (P^{\trans}S) D=D (P^{\trans}S).
\end{align*}
Let $Q:=P^\trans S$. Then the last equality is $QD=D Q$, and hence $Q$ commutes with $D$.
Without loss of generality, we may assume that the matrix $D$ can be expressed as a block matrix
$D= \left[\begin{array}{c|c|c|c} \lambda_1 I_1 & 0 &\dots &0\\ \hline 0 & \lambda_2 I_2 & \dots & 0\\ \hline \vdots & \dots & \ddots& \vdots\\ \hline 0 & \dots & \dots & \lambda_k I_k \end{array} \right] ,$ where $\lambda_1, \dots, \lambda_k$ are distinct eigenvalues of $A$ and $I_k$ are some identity matrix. (These $\lambda_i$ and the previous $\lambda_i$ are different. The size of the identity matrix $I_j$ is the algebraic multiplicity of $\lambda_j$.)
Express the matrix $Q$ as the block matrix with the same partition as $D$, and write it as
$Q=\left[\begin{array}{c|c|c|c} Q_{11} & Q_{12} &\dots &Q_{1 k}\\ \hline Q_{21} & Q_{22}& \dots & Q_{2k}\\ \hline \vdots & \dots & \ddots& \vdots\\ \hline Q_{k1} & \dots & \dots & Q_{k k} \end{array} \right].$
Comparing the $(i,j)$-block of both sides of $QD=DQ$ yields that
$\lambda_j Q_{ij}=\lambda_i Q_{ij}.$
It follows that if $i\neq j$ then $Q_{ij}$ is the zero matrix.
Thus, $Q$ is a block diagonal matrix
$Q=\left[\begin{array}{c|c|c|c} Q_{11} & 0 &\dots &0\\ \hline 0 & Q_{22}& \dots & 0\\ \hline \vdots & \dots & \ddots& \vdots\\ \hline 0 & \dots & \dots & Q_{k k} \end{array} \right].$ Note that $D’$ is also a block diagonal matrix with the same partition.
It follows that
\begin{align*}
QD’&=\left[\begin{array}{c|c|c|c}
Q_{11} & 0 &\dots &0\\
\hline
0 & Q_{22}& \dots & 0\\
\hline
\vdots & \dots & \ddots& \vdots\\
\hline
0 & \dots & \dots & Q_{k k}
\end{array}
\right] \left[\begin{array}{c|c|c|c}
\lambda_1 I_1 & 0 &\dots &0\\
\hline
0 & \lambda_2 I_2 & \dots & 0\\
\hline
\vdots & \dots & \ddots& \vdots\\
\hline
0 & \dots & \dots & \lambda_k I_k
\end{array}
\right]\6pt] &=\left[\begin{array}{c|c|c|c} \lambda_1 Q_{11} & 0 &\dots &0\\ \hline 0 & \lambda_2 Q_{22} & \dots & 0\\ \hline \vdots & \dots & \ddots& \vdots\\ \hline 0 & \dots & \dots & \lambda_k Q_{kk} \end{array} \right]\\[6pt] &=\left[\begin{array}{c|c|c|c} \lambda_1 I_1 & 0 &\dots &0\\ \hline 0 & \lambda_2 I_2 & \dots & 0\\ \hline \vdots & \dots & \ddots& \vdots\\ \hline 0 & \dots & \dots & \lambda_k I_k \end{array} \right] \left[\begin{array}{c|c|c|c} Q_{11} & 0 &\dots &0\\ \hline 0 & Q_{22}& \dots & 0\\ \hline \vdots & \dots & \ddots& \vdots\\ \hline 0 & \dots & \dots & Q_{k k} \end{array} \right] =D’Q. \end{align*} It yields that \[D’=Q^{\trans}D’Q since $Q=P^{\trans}S$ is an orthogonal matrix.
Then we obtain
\begin{align*}
B&=SD’S^{\trans}=S(Q^{\trans}D’Q)S^{\trans}\\
&=SS^{\trans} PD’ P^{\trans} S S^{\trans}\\
&=PD’P^{\trans}=C.
\end{align*}
Therefore, any square root of $A$ must be equal to the matrix $B$.
This completes the proof of the uniqueness, hence the proof of the problem.
## Remark.
The above problem is still true if “positive definite” is replaced by “positive semi-definite”.
## Related Questions About Square Roots of a Matrix
If you want to solve more problems about square roots of a matrix, then try the following problems.
Problem. Does there exist a $3 \times 3$ real matrix $B$ such that $B^2=A$ where
$A=\begin{bmatrix} 1 & -1 & 0 \\ -1 &2 &-1 \\ 0 & -1 & 1 \end{bmatrix}\,\,\,\,?$
This is a part of Princeton University’s Linear Algebra exam problems.
See the post ↴
A Square Root Matrix of a Symmetric Matrix
for a solution.
Problem. Find a square root of the matrix
$A=\begin{bmatrix} 1 & 3 & -3 \\ 0 &4 &5 \\ 0 & 0 & 9 \end{bmatrix}.$ How many square roots does this matrix have?
This is one of Berkeley’s qualifying exam problems.
See the post ↴
Square Root of an Upper Triangular Matrix. How Many Square Roots Exist?
for a solution.
### 6 Responses
1. 07/18/2017
[…] For a solution of this problem, see the post A Positive Definite Matrix Has a Unique Positive Definite Square Root […]
2. 07/18/2017
[…] For a solution of this problem, see the post A Positive Definite Matrix Has a Unique Positive Definite Square Root […]
3. 07/18/2017
[…] For a solution of this problem, see the post A Positive Definite Matrix Has a Unique Positive Definite Square Root […]
4. 07/18/2017
[…] For a solution of this problem, see the post A Positive Definite Matrix Has a Unique Positive Definite Square Root […]
5. 07/18/2017
[…] For a solution of this problem, see the post A Positive Definite Matrix Has a Unique Positive Definite Square Root […]
6. 07/18/2017
[…] For a solution of this problem, see the post A Positive Definite Matrix Has a Unique Positive Definite Square Root […]
##### Find All the Square Roots of a Given 2 by 2 Matrix
Let $A$ be a square matrix. A matrix $B$ satisfying $B^2=A$ is call a square root of $A$. Find all...
Close
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2018-03-22 13:38:19
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https://testbook.com/question-answer/when-the-lower-temperature-of-the-refrigerating-ma--624543790318f7c3e925577b
|
# When the lower temperature of the refrigerating machine is fixed then the coefficient of performance .can be improved by
This question was previously asked in
RPSC Lecturer Tech Edu Previous Year Paper 2020 (Mechanical Engineering Paper - 2)
View all RPSC Lecturer Tech Edu Papers >
1. Operating the machine at high speed
2. Operating the machine at low speed
3. Raising the higher temperature
4. Lowering the higher temperature
Option 4 : Lowering the higher temperature
Free
ST 1: Engineering Materials (Crystal Geometry)
0.8 K Users
16 Questions 8 Marks 20 Mins
## Detailed Solution
Explanation:
Refrigeration:
• Refrigeration is the process of removing heat from a confined space so that its temperature is first lowered and then maintained at a low temperature compared to that of the surroundings, i.e. it is a phenomenon by the virtue of which one reduces the temperature of a confined space compared with that of surroundings.
• The device used to produce the refrigeration effect is called a refrigeration system or refrigerator.
• The basic components of a refrigeration system are evaporator, compressor, condenser, and expansion valve. In addition to these, there may be a refrigerant accumulator, temperature controller, etc.
Refrigeration effect, RE = Q1
Work Input = Q2 – Q1
Coefficient of Performance (COP) = $$\frac{{Refrigeration\;effect}}{{Work\;input}} = \;\frac{{{Q_1}}}{W}$$
For reversible engines, $$\frac{{{Q_2}}}{{{Q_1}}} = \;\frac{{{T_2}}}{{{T_1}}}$$
COPrefrigerator = $$\frac{{{Q_1}}}{{{Q_2} - \;{Q_1}}}$$ = $$\frac{{{T_1}}}{{{T_2} - \;{T_1}}}$$
If lower temperature T1 is fixed then the COP depends on higher temperature T2. We can see that if we decrease the higher temperature, the denominator decreases, and the COP increases.
So, when the lower temperature of the refrigerating machine is fixed then the coefficient of performance can be improved by lowering the higher temperature.
Latest RPSC Lecturer Tech Edu Updates
Last updated on Sep 22, 2022
The Rajasthan Public Service Commission (RPSC) has declared the Interview Result and Cut Off for RPSC Lecturer Tech Edu (Mathematics Lecturer) recruitment exam. The Rajasthan Public Service Commission had released 39 vacancies in 7 subjects for the post of Lecturer for the Technical Education Department. The RPSC Lecturer selection process consists of a written examination (objective type) and an interview. The candidates can check their RPSC Lecturer Tech Edu Result from here.
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2023-02-04 09:13:08
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http://physics.stackexchange.com/questions/63974/why-is-this-thought-experiment-flawed-a-vast-lever-rotating-faster-than-the-spe/63975
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# Why is this thought experiment flawed: A vast lever rotating faster than the speed of light [duplicate]
If there were a vast lever floating in free space, a rigid body with length greater than the width of a galaxy, made of a hypothetical material that could endure unlimited internal stress, and this lever began to rotate about its middle like a propeller so that a person looking at the universe would simply see it spinning at a gentle pace like a windmill, would not its ends be moving many, many times the speed of light?
Or am I making so many errors in my thought process that the whole question is absurd? I'm trying to establish why, in essence, a sufficiently large mechanical device (large beyond reason) could not exceed the speed of light.
-
## marked as duplicate by John Rennie, Qmechanic♦May 9 '13 at 19:01
This question was marked as an exact duplicate of an existing question.
To imagine this, take your bicycle and try to rotate one of its wheels. As you put your finger closer to he centre, you need stronger force (but same energy of course) to get the wheel moving. So the distance from the centre ($r$) matters somehow obviously. And when you try to give the mass at the perimeter speed over $c$ (lightspeed), you are in the same situation as here.
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2016-05-01 17:42:19
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https://www.zbmath.org/serials/?q=se%3A00001217
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## Journal of Hebei Normal University. Natural Science Edition
Short Title: J. Hebei Norm. Univ., Nat. Sci. Ed. Parallel Title: Hebei Shifandaxue Xuebao. Ziran Bexue Ban Publisher: Hebei Normal University, Shijiazhuang ISSN: 1000-5854 Online: http://www.hebtu.edu.cn/http://scholar.ilib.cn/P-hbsfdxxb.html Comments: No longer indexed
Documents Indexed: 717 Publications (1987–2021)
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### Latest Issues
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### Authors
27 Gao, Gailiang 16 Zhou, Haiyun 15 Chen, Dongqing 14 Li, Cuixiang 14 Li, Qiaoluan 11 Liu, Lihong 11 Wu, Chenyu 11 Xue, Zhiqun 10 Liu, Lixia 9 Yang, Jianfa 8 An, Xinlei 8 Li, Lixia 8 Qiao, Yuying 8 Wang, Zhijun 8 Xie, Yonghong 8 Zhang, Jiangang 7 Li, Fachao 7 Liu, Qiming 7 Liu, Yujun 7 Song, Zhanjie 7 Su, Zhanjun 7 Xu, Rui 6 Begehr, Heinrich 6 Chen, Yingwei 6 Dong, Wenlei 6 Gao, Hui 6 Gao, Yinzhi 6 Guo, Jinti 6 Kang, Qingde 6 Liu, Zhaoshuang 6 Qi, Qiulan 6 Tian, Zihong 6 Yu, Jianning 6 Zhang, Gengsheng 6 Zhang, Jinlian 5 Li, Yuanfei 5 Liu, Qingsong 5 Liu, Xiqiang 5 Wang, Rongxin 5 Yang, Heju 5 Yu, Xiuping 5 Yuan, Landang 5 Zhang, Caishun 5 Zhang, Lina 5 Zhou, Xiangquan 4 Bi, Ping 4 Cai, Bingling 4 Chen, Jinmei 4 Ding, Yanhong 4 Dong, Mei 4 Du, Tingsong 4 Du, Wenju 4 Feng, Mei 4 Feng, Yingjie 4 Guo, Jingfang 4 Guo, Shunsheng 4 Guo, Zhifen 4 Hao, Jianhong 4 Jin, Tingliang 4 Kang, Na 4 Li, Fumei 4 Li, Gang 4 Li, Gang 4 Li, Gaoming 4 Li, Jinshan 4 Li, Wei 4 Li, Yanling 4 Li, Zengti 4 Liang, Zhihe 4 Liu, Guofen 4 Liu, Shuxia 4 Lu, Guiwen 4 Mi, Yuzhen 4 Rong, Xiaojian 4 Wang, Liping 4 Wang, Zhanjing 4 Wang, Zhiming 4 Wei, Li 4 Xiao, Zhuofeng 4 Yin, Guoju 4 Zhang, Qingnian 4 Zhang, Yuqin 4 Zhao, Guanhua 4 Zhao, Hongtao 4 Zhao, Suqian 4 Zhao, Xiaoqing 4 Zhou, Heyue 4 Zuo, Dawei 3 Cao, Nanbin 3 Duan, Liling 3 Duan, Shenggui 3 Feng, Junqing 3 Gao, Haixia 3 Gu, Liyan 3 Guo, Jun 3 Guo, Zengxiao 3 Hao, Yanpeng 3 Hou, Bo 3 Hou, Wenyu 3 Jia, Huixian ...and 793 more Authors
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### Fields
102 Operator theory (47-XX) 89 Combinatorics (05-XX) 78 Ordinary differential equations (34-XX) 77 Partial differential equations (35-XX) 44 Dynamical systems and ergodic theory (37-XX) 39 Approximations and expansions (41-XX) 37 Biology and other natural sciences (92-XX) 32 Probability theory and stochastic processes (60-XX) 31 Associative rings and algebras (16-XX) 29 Difference and functional equations (39-XX) 28 Functions of a complex variable (30-XX) 25 Numerical analysis (65-XX) 25 Information and communication theory, circuits (94-XX) 22 Operations research, mathematical programming (90-XX) 22 Game theory, economics, finance, and other social and behavioral sciences (91-XX) 19 Convex and discrete geometry (52-XX) 19 Statistics (62-XX) 19 Systems theory; control (93-XX) 17 Mathematical logic and foundations (03-XX) 16 Functional analysis (46-XX) 16 Geometry (51-XX) 15 Real functions (26-XX) 14 Harmonic analysis on Euclidean spaces (42-XX) 14 Computer science (68-XX) 13 Group theory and generalizations (20-XX) 13 General topology (54-XX) 12 Manifolds and cell complexes (57-XX) 11 Measure and integration (28-XX) 11 Quantum theory (81-XX) 9 Linear and multilinear algebra; matrix theory (15-XX) 9 Nonassociative rings and algebras (17-XX) 9 Integral equations (45-XX) 7 Calculus of variations and optimal control; optimization (49-XX) 6 Order, lattices, ordered algebraic structures (06-XX) 6 Fluid mechanics (76-XX) 4 Abstract harmonic analysis (43-XX) 3 General algebraic systems (08-XX) 3 Differential geometry (53-XX) 3 Algebraic topology (55-XX) 3 Global analysis, analysis on manifolds (58-XX) 3 Statistical mechanics, structure of matter (82-XX) 2 Potential theory (31-XX) 2 Integral transforms, operational calculus (44-XX) 1 Number theory (11-XX) 1 Commutative algebra (13-XX) 1 Topological groups, Lie groups (22-XX) 1 Several complex variables and analytic spaces (32-XX) 1 Special functions (33-XX) 1 Mechanics of particles and systems (70-XX) 1 Mechanics of deformable solids (74-XX) 1 Classical thermodynamics, heat transfer (80-XX) 1 Relativity and gravitational theory (83-XX) 1 Astronomy and astrophysics (85-XX)
### Citations contained in zbMATH Open
21 Publications have been cited 24 times in 21 Documents Cited by Year
The existence of solution of nonlinear Neumann boundary value problem involving the generalized $$p$$-Laplacian operator. Zbl 1093.35509
Li, Wei; Hou, Wenyu
2004
The distribution of zeros of solutions to delay differential equations. Zbl 0962.34051
Du, Ying
1999
Approximation of Bernstein-Sikkema operators and their derivatives. Zbl 1207.41016
Li, Cuixiang; Liu, Yana
2006
Asymptotic average and Lipschitz shadowing property of the product map under group action. Zbl 1449.37015
Ji, Zhanjiang; Zhang, Gengrong; Tu, Jingxian
2019
$$(K_{2,3}+e)$$-design of $$\lambda K_\nu$$. Zbl 1013.05016
Tian, Zihong; Kang, Qingde
2002
Stechkin-Marchaud-type inequalities for Bernstein polynomials. Zbl 1023.41009
Guo, Shunsheng; Yu, Jinhua
2002
The overlarge sets of Mendelsohn triple systems. Zbl 1014.05013
Zhang, Jie; Tian, Zihong
2002
A primal-dual infeasible interior point algorithm for separable convex quadratic programming with box constraints. Zbl 1127.90413
Wang, Junling; Zhang, Mingwang; Du, Tingsong
2002
A new second-order accurate linearized difference scheme for the Rosenau-Burgers equation. Zbl 1289.65186
Ma, Weiyuan; Tang, Yurong; Hua, Chuanning
2012
On triangle-decomposition of $$\lambda K_v\cup\lambda DK_v$$. Zbl 1151.05310
Tian, Zihong; Zhao, Hongtao; Ma, Chunping
2004
The pointwise results of weighted simultaneous approximation by Baskakov operator. Zbl 1152.41305
Chen, Yingwei; Lv, Guiwen
2004
Decomposition of $$\lambda K_v$$ into a bipartite graph. Zbl 1084.05053
Liu, Shuxia
2004
An Anzahl formula for the number of subspaces in finite vector spaces and its application. Zbl 1084.05003
Guo, Jun; Huo, Yuanji; Zhao, Dongming
2004
Periodic solutions of a type of second order functional differential equation with complicated deviating argument. I. Zbl 1095.34544
Wang, Haiqing; Suo, Xiaohui
2004
Probabilistic teleportation of two-particle state of general formation based on four-particle entanglement. Zbl 1107.81305
Huo, Hairui; Yan, Fengli
2006
Graph designs of a graph with six vertices and eight edges. Zbl 1174.05388
Yuan, Landang; Kang, Qingde
2008
On chaotic analysis and control of a new chaotic system. Zbl 1174.37325
Peng, Jiankui; Yu, Jianning; Zhang, Li; Zhang, Jiangang
2008
Graph designs of two graphs with six vertices and eight edges. Zbl 1199.05031
Yuan, Landang
2008
Research of the Erdős-Ko-Rado theorem based on symplectic spaces over finite fields. Zbl 1424.05281
Liu, Xuemei; Fan, Qianyu; Sun, Qingfeng
2018
Explicit solutions, conservation laws of the extended $$(2+1)$$-dimensional Jaulent-Miodek equation. Zbl 1374.35352
Li, Yu; Liu, Xiqiang; Xin, Xiangpeng
2016
Hopf bifurcation control of a van der Pol-Duffing system. Zbl 1389.34224
Qin, Shuang; Zhang, Jiangang; Du, Wenju; Yu, Jianning
2017
Asymptotic average and Lipschitz shadowing property of the product map under group action. Zbl 1449.37015
Ji, Zhanjiang; Zhang, Gengrong; Tu, Jingxian
2019
Research of the Erdős-Ko-Rado theorem based on symplectic spaces over finite fields. Zbl 1424.05281
Liu, Xuemei; Fan, Qianyu; Sun, Qingfeng
2018
Hopf bifurcation control of a van der Pol-Duffing system. Zbl 1389.34224
Qin, Shuang; Zhang, Jiangang; Du, Wenju; Yu, Jianning
2017
Explicit solutions, conservation laws of the extended $$(2+1)$$-dimensional Jaulent-Miodek equation. Zbl 1374.35352
Li, Yu; Liu, Xiqiang; Xin, Xiangpeng
2016
A new second-order accurate linearized difference scheme for the Rosenau-Burgers equation. Zbl 1289.65186
Ma, Weiyuan; Tang, Yurong; Hua, Chuanning
2012
Graph designs of a graph with six vertices and eight edges. Zbl 1174.05388
Yuan, Landang; Kang, Qingde
2008
On chaotic analysis and control of a new chaotic system. Zbl 1174.37325
Peng, Jiankui; Yu, Jianning; Zhang, Li; Zhang, Jiangang
2008
Graph designs of two graphs with six vertices and eight edges. Zbl 1199.05031
Yuan, Landang
2008
Approximation of Bernstein-Sikkema operators and their derivatives. Zbl 1207.41016
Li, Cuixiang; Liu, Yana
2006
Probabilistic teleportation of two-particle state of general formation based on four-particle entanglement. Zbl 1107.81305
Huo, Hairui; Yan, Fengli
2006
The existence of solution of nonlinear Neumann boundary value problem involving the generalized $$p$$-Laplacian operator. Zbl 1093.35509
Li, Wei; Hou, Wenyu
2004
On triangle-decomposition of $$\lambda K_v\cup\lambda DK_v$$. Zbl 1151.05310
Tian, Zihong; Zhao, Hongtao; Ma, Chunping
2004
The pointwise results of weighted simultaneous approximation by Baskakov operator. Zbl 1152.41305
Chen, Yingwei; Lv, Guiwen
2004
Decomposition of $$\lambda K_v$$ into a bipartite graph. Zbl 1084.05053
Liu, Shuxia
2004
An Anzahl formula for the number of subspaces in finite vector spaces and its application. Zbl 1084.05003
Guo, Jun; Huo, Yuanji; Zhao, Dongming
2004
Periodic solutions of a type of second order functional differential equation with complicated deviating argument. I. Zbl 1095.34544
Wang, Haiqing; Suo, Xiaohui
2004
$$(K_{2,3}+e)$$-design of $$\lambda K_\nu$$. Zbl 1013.05016
Tian, Zihong; Kang, Qingde
2002
Stechkin-Marchaud-type inequalities for Bernstein polynomials. Zbl 1023.41009
Guo, Shunsheng; Yu, Jinhua
2002
The overlarge sets of Mendelsohn triple systems. Zbl 1014.05013
Zhang, Jie; Tian, Zihong
2002
A primal-dual infeasible interior point algorithm for separable convex quadratic programming with box constraints. Zbl 1127.90413
Wang, Junling; Zhang, Mingwang; Du, Tingsong
2002
The distribution of zeros of solutions to delay differential equations. Zbl 0962.34051
Du, Ying
1999
all top 5
### Cited by 43 Authors
3 Kang, Qingde 3 Wei, Li 2 Agarwal, Ravi P. 2 Tian, Zihong 2 Wu, Hongwu 1 Chen, Shangdi 1 Cheng, Sui Sun 1 Cheng, Xiujun 1 Dahiya, Kalpana 1 Du, Yanke 1 Duan, Liling 1 Erbe, Lynn Harry 1 Hu, Jingsen 1 Ji, Zhanjiang 1 Jian, Jigui 1 Jiang, Xuejiao 1 Li, Fengjun 1 Liu, Chunxia 1 Lv, Benjian 1 Qi, Jianming 1 Suneja, Surjeet Kaur 1 Tang, Xiaojun 1 Tao, Yuanhong 1 Verma, Vanita 1 Wang, Haiqing 1 Wang, Jianjun 1 Wang, Kaishun 1 Wang, Wenquan 1 Wang, Xinchang 1 Wong, Patricia J. Y. 1 Xie, Linsen 1 Xu, Ling-Shan 1 Xue, Yinchuan 1 Yan, Yan 1 Yang, Jiaqin 1 Yao, Tian 1 Yi, Hua 1 Yong, Xinlei 1 Yuan, Landang 1 Zhang, Xiaolian 1 Zhao, Pan-ru 1 Zhao, ZhiHua 1 Zhou, Haiyun
all top 5
### Cited in 16 Journals
3 Applied Mathematics and Computation 2 Computers & Mathematics with Applications 2 Acta Mathematicae Applicatae Sinica. English Series 2 Abstract and Applied Analysis 1 International Journal of Theoretical Physics 1 Chaos, Solitons and Fractals 1 Journal of Statistical Planning and Inference 1 Computational Optimization and Applications 1 Finite Fields and their Applications 1 Complexity 1 Journal of Difference Equations and Applications 1 Journal of Applied Mathematics and Computing 1 Analysis in Theory and Applications 1 Nonlinear Analysis. Hybrid Systems 1 Advances in Mathematical Physics 1 Open Mathematics
all top 5
### Cited in 13 Fields
5 Ordinary differential equations (34-XX) 5 Partial differential equations (35-XX) 4 Combinatorics (05-XX) 3 Approximations and expansions (41-XX) 2 Operator theory (47-XX) 2 Systems theory; control (93-XX) 2 Information and communication theory, circuits (94-XX) 1 Dynamical systems and ergodic theory (37-XX) 1 Geometry (51-XX) 1 Numerical analysis (65-XX) 1 Mechanics of deformable solids (74-XX) 1 Quantum theory (81-XX) 1 Operations research, mathematical programming (90-XX)
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2022-05-23 15:30:47
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https://www.physicsforums.com/threads/question-about-finding-limits-with-lhopitals-rule.642177/
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# Question About Finding Limits with L'Hopital's Rule
## Homework Statement
A question we had for homework was: Which of the following is equivalent to limh->0 [arcsin((3(x+h))/4) – arcsin (3x/4)]/h ?
## The Attempt at a Solution
We've already walked through the problem in class so I know the steps to solve it, but there's one step that I didn't understand and he couldn't explain quite clearly to me. Why are we allowed to just calculate the derivative of arcsin(3x/4)? I know that according to L'Hopital's Rule we can calculate the limit by finding the limit of the derivatives, but why can we skip the other parts of the function altogether?
It is not an application of L'Hopital's rule. The limit you are given is the definition of the derivative of f(u) = arcsin(3u/4) at the point x. Recall that the derivative of f(x) at the point a is equal to the limit of the difference quotient:
$$f'(a) = \lim_{h\rightarrow 0} \frac{f(a + h) - f(a)}{h}$$
If you see a limit that is of the form on the right side of this equation, and you know the derivative f'(a) by another method, then there is no need to go through the rigmarole of algebraic limit deconstruction: just use your knowledge that it is equal to f'(a).
So you would just be plugging 0 in for h and solving the limit normally?
We cannot replace h by 0 directly, as division by 0 is undefined, and in any case would only be equal to the limit if the expression is continuous at h = 0. Not all functions are continuous or have continuous derivatives!
While it may be possible to simplify the limit by going through extended analytical means, the fact that the limit is equal to the derivative, and we already know the derivative by simpler methods, means we can go ahead and replace the entire limit by its equivalent: the derivative. You should review the definition of the derivative (not just the shortcuts for its calculation) and practice recognizing this type of limit.
I'll definitely do that. Thank you!
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2021-09-28 10:15:44
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https://codereview.stackexchange.com/questions/6488/graph-generator-for-undirected-graphs
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Graph generator for undirected graphs
I am still pretty new to Haskell and am working on a graph generator for undirected unlabeled graphs containing loops and I have a bottleneck in the following functions. So far, I have not given any thought to performance and just looked for correctness.
I know this kind of question is not really popular, but I'd be interested in general guidelines for improving performance given some naive but correct implementation like the following.
For instance:
1. Where to enforce strictness?
2. Should I use vector/array instead of lists? When, where and why?
3. Should I improve single functions by replacing recursion using folds/maps/etc. or similar?
1. I'm using -O2 with ghc
2. profiling shows (when normalizing runtime of connections to 100%)
• arqSeq 40% (with almost all time spend in boundsequences)
• connectionCombinations 60% with a quarter of the time spend in occurences
I don't think you need to understand all the details. I'm looking more for micro-improvements (the following determines arcs from a given node to equivalence classes of vertices grouped by their degree).
import Data.List
type Arcs = Int
type Count = Int
type Vertex = Int
type MSequence = [Int]
data EquivClass = EC { order :: Int, verts :: [Vertex] } deriving (Eq, Show)
type ECPartition = [EquivClass]
type NodeSelection = [(Arcs,Count)]
type ECNodeSelection = (EquivClass, NodeSelection)
-- number of occurences of unique elements in a list
occurences :: (Ord a) => [a] -> [(a, Int)]
occurences = map (\xs@(x:_) -> (x, length xs)) . group . reverse . sort
-- number of vertices in an equivalence class
ecLength :: EquivClass -> Int
ecLength = length . verts
-- for a given y = (y_1,...,y_n) and a bound m, find all vectors
-- x = (x_1,...,x_n) such that |x| = m and x_i <= y_i
boundSequences :: (Num a, Ord a, Enum a) => a -> [a] -> [[a]]
boundSequences m x | m <= sum x = (fByM . sequence . ranges) x
| otherwise = [[]]
where fByM = filter (\x -> sum x == m)
ranges = map (\x -> [0..x])
-- return m-sequences (combinations of number of arcs) for the given
-- order to an ECPartition
arcSeq :: Int -> ECPartition -> [MSequence]
arcSeq m x = boundSequences m (map ecProd x)
where ecProd e = ecLength e * order e
-- return all the possible arc combinations to an equivalence
-- class for a given number of arcs
connectionCombinations :: Int -> EquivClass -> [NodeSelection]
connectionCombinations arcs = map groupOcc . prune arcs . sequence . orderRep
where orderRep (EC o v) = replicate (length v) [0..o]
prune a = nub . map (reverse . sort) . filter ((== a) . sum)
groupOcc = filter ((/= 0) . fst) . occurences
-- return all the possible lists of equivalence class node selections
-- from a Partition and a givne number of arcs
connections :: ECPartition -> Int -> [[ECNodeSelection]]
connections p order = concatMap (con p) $arcSeq order p where con p arcs = map (zip p)$ zipWithM connectionCombinations arcs p
main = do
-- small example. In the complete app this is called 10,000's of
-- times in case of high degrees or vertex counts
let cons = connections [EC 4 [1], EC 1 [2,3,4]] 5
mapM_ (print) cons
migrated from stackoverflow.comDec 2 '11 at 23:51
This question came from our site for professional and enthusiast programmers.
• Possibly the most important thing is to figure out how to effectively measure performance, and determine based on that data where the bottlenecks are. Haskell's performance can be very hard to accurately predict without quantitative measurements. – Matt Fenwick Dec 1 '11 at 18:03
• Performance can be very hard to predict accurately without measurements in any language. Laziness just makes it harder to analyze those results (also in any language, but pervasive in Haskell). – C. A. McCann Dec 1 '11 at 18:09
• @MattFenwick Please see my update, I have chosen this bunch of functions after profiling. Sorry I forgot to mention this. – bbtrb Dec 1 '11 at 18:14
• Could you put up an example that compiles? – Nathan Howell Dec 1 '11 at 18:20
• @NathanHowell: Code updated, should compile and run – bbtrb Dec 1 '11 at 18:26
-- for a given y = (y_1,...,y_n) and a bound m, find all vectors
-- x = (x_1,...,x_n) such that |x| = m and x_i <= y_i
boundSequences :: (Num a, Ord a, Enum a) => a -> [a] -> [[a]]
boundSequences m x | m <= sum x = (fByM . sequence . ranges) x
| otherwise = [[]]
where fByM = filter (\x -> sum x == m)
ranges = map (\x -> [0..x])
Excepting some fortunate cases, you are wasting a lot of work here. sequence . ranges produces (y_1+1) * ... * (y_n+1) lists you have to check. If you write a function to produce only the successful lists, you can gain a lot of performance for larger input (it won't do too much for short lists with small elements, but it will help for those too).
Shouldn't the case sum x < m return [] rather than [[]]?
-- return all the possible arc combinations to an equivalence
-- class for a given number of arcs
connectionCombinations :: Int -> EquivClass -> [NodeSelection]
connectionCombinations arcs = map groupOcc . prune arcs . sequence . orderRep
where orderRep (EC o v) = replicate (length v) [0..o]
prune a = nub . map (reverse . sort) . filter ((== a) . sum)
groupOcc = filter ((/= 0) . fst) . occurences
Instead of reverse . sort, use sortBy (flip compare). That won't make much difference for short lists, but it's cleaner, IMO. prune arcs . sequence contains another occurrence of the boundSequences problem, creating a lot of lists and filtering out most of them immediately. nub is not good, it is quadratic (worst case, O(total*distinct) in general). The things you're nubbing have an Ord instance, if the order doesn't matter, map head . group . sort is a much faster nub than nub (still better Data.Set.toList . Data.Set.fromList), if order matters, keep a Set of seen elements and for each new, output that and add it to the set of seen.
I've not yet profiled, but it may be that the profile is misleading because it attributes costs to the functions where results are forced rather than where they are calculated, that needs some investigation, check back later.
Okay, the given data doesn't run long enough to produce a meaningful profile, so I changed it to
let cons = connections [EC 4 [1 .. 6], EC 6 [2 .. 9]] 5
That runs long enough to collect a handful of samples. Unfortunately, the profile wasn't very informative:
total time = 7.90 secs (395 ticks @ 20 ms)
total alloc = 8,098,386,128 bytes (excludes profiling overheads)
COST CENTRE MODULE %time %alloc
connectionCombinations Main 100.0 100.0
So I inserted a couple of {-# SCC #-} pragmas. The biggo in that programme is the sequence . orderRep, unsurprisingly. A rewrite along the lines mentioned above,
boundSequences :: (Num a, Ord a, Enum a) => a -> [a] -> [[a]]
boundSequences m xs
| sm < m = []
| sm == m = [xs]
| otherwise = go sm m xs
where
sm = sum xs
go _ r []
| r == 0 = [[]]
| otherwise = []
go _ r [y]
| y < r = []
| otherwise = [[r]]
go s r (y:ys) = do
let mny | s < r+y = r+y-s
| otherwise = 0
mxy = min y r
c <- [mny .. mxy]
map (c:) (go (s-y) (r-c) ys)
and using that in connectionCombinations instead of prune arcs . sequence (requires a small change in orderRep),
connectionCombinations :: Int -> EquivClass -> [NodeSelection]
connectionCombinations arcs = map groupOcc . nub . map (sortBy (flip compare)) .
boundSequences arcs . orderRep
where orderRep (EC o v) = replicate (length v) o
groupOcc = filter ((/= 0) . fst) . occurences
brought the running time (ghc -O2, no profiling) down significantly:
dafis@schwartz:~/Cairo> time ./orArcs > /dev/null
real 0m5.836s
user 0m5.593s
sys 0m0.231s
dafis@schwartz:~/Cairo> time ./arcs > /dev/null
real 0m0.008s
user 0m0.005s
sys 0m0.003s
Whether any further optimisations (e.g. the abovementioned nub) are necessary, and where most of the time in the actual programme is spent, so that one could identify the most important points, I cannot tell without more realistic data.
• Wow, thank you for the elaborate answer. It wasn't obvious to me that the algorithm (using sequence) is so flawed, I've tried the most obvious approach first and hoped for haskell to figure the rest out for me. – bbtrb Dec 1 '11 at 22:35
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2019-07-17 18:22:11
|
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|
http://www.onemathematicalcat.org/algebra_book/online_problems/frac_exps.htm
|
BASIC EXPONENT PRACTICE WITH FRACTIONS
• PRACTICE (online exercises and printable worksheets)
DEFINITIONS: properties of exponents base; exponent; power Let $\,x\in\Bbb{R}\,$. In the expression $\,x^n\,$, $\,x\,$ is called the base and $\,n\,$ is called the exponent or the power. positive integers If $\,n\in\{1,2,3,\ldots\}\,$, then $\,x^n = x\cdot x\cdot x \cdot \ldots \cdot x\,$, where there are $\,n\,$ factors in the product. In this case, $\,x^n\,$ is just a shorthand for repeated multiplication. Note that $\,x^1 = x\,$ for all real numbers $\,x\,$. zero If $\,x\ne 0\,$, then $\,x^0 = 1\,$. The expression $\,0^0\,$ is not defined. negative integers If $\,n\in\{1,2,3,\ldots\}\,$ and $\,x\ne 0\,$, then $\displaystyle\,x^{-n} = \frac{1}{x^n} = \frac{1}{x\cdot x\cdot x\cdot \ldots \cdot x}$, where there are $\,n\,$ factors in the product. In particular, $\,\displaystyle x^{-1} = \frac{1}{x}\,$ for all nonzero real numbers $\,x\,$. That is, $\,x^{-1}\,$ is the reciprocal of $\,x\,$.
With fractions, it looks like this:
$\displaystyle (\frac{a}{b})^{-1} = \frac{1}{\frac{a}{b}} = 1 \div \frac{a}{b} = 1\cdot\frac{b}{a} = \frac{b}{a}$
That is, the reciprocal of $\displaystyle\,\frac{a}{b}\,$ is $\displaystyle\,\frac{b}{a}\,$.
Now that you've mired through this calculation once,
you'll never have to do it this long way again!
When a fraction is raised to the $\,-1\,$ power, it just flips.
The numerator becomes the new denominator, and the denominator becomes the new numerator.
EXAMPLES:
$\displaystyle \left(\frac23\right)^{-1} = \frac32$
$\displaystyle \left(\frac23\right)^{0} = 1$
$\displaystyle \left(\frac23\right)^{1} = \frac23$
$\displaystyle \left(-\frac23\right)^{-1} = -\frac32$
$\displaystyle 5^{-1} = \frac15$
$\displaystyle (-5)^{-1} = -\frac15$
$\displaystyle \left(\frac13\right)^{-1} = 3$
$\displaystyle \left(-\frac13\right)^{-1} = -3$
Master the ideas from this section
Practice with$\,x^mx^n = x^{m+n}$
|
2016-08-26 01:22:45
|
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|
https://www.oercommons.org/browse?f.alignment=CCSS.Math.Content.8.EE.A.4
|
# 20 Results
View
Selected filters:
• CCSS.Math.Content.8.EE.A.4
Rating
This lesson unit is intended to help teachers assess how well students are able to: translate between decimal and fraction notation, particularly when the decimals are repeating; create and solve simple linear equations to find the fractional equivalent of a repeating decimal; and understand the effect of multiplying a decimal by a power of 10.
Subject:
Algebra
Ratios and Proportions
Material Type:
Assessment
Lesson Plan
Provider:
Shell Center for Mathematical Education
U.C. Berkeley
Provider Set:
Mathematics Assessment Project (MAP)
04/26/2013
Rating
This lesson unit is intended to help teachers assess how well students are able to: solve linear equations in one variable with rational number coefficients; collect like terms; expand expressions using the distributive property; and categorize linear equations in one variable as having one, none, or infinitely many solutions. It also aims to encourage discussion on some common misconceptions about algebra.
Subject:
Algebra
Material Type:
Assessment
Lesson Plan
Provider:
Shell Center for Mathematical Education
U.C. Berkeley
Provider Set:
Mathematics Assessment Project (MAP)
04/26/2013
Rating
This lesson unit is intended to help teachers assess how well students are able to classify solutions to a pair of linear equations by considering their graphical representations. In particular, this unit aims to help teachers identify and assist students who have difficulties in: using substitution to complete a table of values for a linear equation; identifying a linear equation from a given table of values; and graphing and solving linear equations.
Subject:
Algebra
Material Type:
Assessment
Lesson Plan
Provider:
Shell Center for Mathematical Education
U.C. Berkeley
Provider Set:
Mathematics Assessment Project (MAP)
04/26/2013
Rating
This lesson unit is intended to help teahcers assess how well students are able to interpret speed as the slope of a linear graph and translate between the equation of a line and its graphical representation.
Subject:
Algebra
Material Type:
Assessment
Lesson Plan
Provider:
Shell Center for Mathematical Education
U.C. Berkeley
Provider Set:
Mathematics Assessment Project (MAP)
04/26/2013
Conditions of Use:
No Strings Attached
Rating
This task requires students to work with very large and small values expressed both in scientific notation and in decimal notation (standard form). In addition, students need to convert units of mass.
Subject:
Mathematics
Material Type:
Activity/Lab
Provider:
Illustrative Mathematics
Provider Set:
Illustrative Mathematics
Author:
Illustrative Mathematics
08/21/2012
Rating
This lesson unit is intended to help teachers assess how well students are able to: interpret a situation and represent the variables mathematically; select appropriate mathematical methods to use; explore the effects of systematically varying the constraints; interpret and evaluate the data generated and identify the break-even point, checking it for confirmation; and communicate their reasoning clearly.
Subject:
Mathematics
Material Type:
Assessment
Lesson Plan
Provider:
Shell Center for Mathematical Education
U.C. Berkeley
Provider Set:
Mathematics Assessment Project (MAP)
04/26/2013
Rating
This lesson unit is intended to help you assess how well students working with square numbers are able to: choose an appropriate, systematic way to collect and organize data, examining the data for patterns; describe and explain findings clearly and effectively; generalize using numerical, geometrical, graphical and/or algebraic structure; and explain why certain results are possible/impossible, moving towards a proof.
Subject:
Algebra
Geometry
Measurement and Data
Material Type:
Assessment
Lesson Plan
Provider:
Shell Center for Mathematical Education
U.C. Berkeley
Provider Set:
Mathematics Assessment Project (MAP)
04/26/2013
Rating
This lesson unit is intended to help teachers assess how well students are able to create and solve linear equations. In particular, the lesson will help you identify and help students who have the following difficulties: solving equations with one variable and solving linear equations in more than one way.
Subject:
Algebra
Material Type:
Assessment
Lesson Plan
Provider:
Shell Center for Mathematical Education
U.C. Berkeley
Provider Set:
Mathematics Assessment Project (MAP)
04/26/2013
Conditions of Use:
Remix and Share
Rating
This course will place a emphasis on the continued study of integers, order of operations, variables, expressions, equations and polynomials. You will solve equations, write and solve proportions, explore polynomials and build an understanding of important mathematical properties.
Subject:
Mathematics
Material Type:
Full Course
Author:
Deanna Mayers
06/10/2018
Conditions of Use:
No Strings Attached
Rating
This is a task from the Illustrative Mathematics website that is one part of a complete illustration of the standard to which it is aligned. Each task has at least one solution and some commentary that addresses important asects of the task and its potential use. Here are the first few lines of the commentary for this task: A penny is about $\frac{1}{16}$ of an inch thick. In 2011 there were approximately 5 billion pennies minted. If all of these pennies were placed in a s...
Subject:
Mathematics
Material Type:
Activity/Lab
Provider:
Illustrative Mathematics
Provider Set:
Illustrative Mathematics
Author:
Illustrative Mathematics
03/17/2013
Rating
This lesson unit is intended to help teachers assess how well students are able to: estimate lengths of everyday objects; convert between decimal and scientific notation; and make comparisons of the size of numbers expressed in both decimal and scientific notation.
Subject:
Mathematics
Material Type:
Assessment
Lesson Plan
Provider:
Shell Center for Mathematical Education
U.C. Berkeley
Provider Set:
Mathematics Assessment Project (MAP)
04/26/2013
Conditions of Use:
Rating
In a hands-on way, students explore light's properties of absorption, reflection, transmission and refraction through various experimental stations within the classroom. To understand absorption, reflection and transmission, they shine flashlights on a number of preselected objects. To understand refraction, students create indoor rainbows. An understanding of the fundamental properties of light is essential to designing an invisible laser security system.
Subject:
Engineering
Physics
Material Type:
Activity/Lab
Provider:
TeachEngineering
Provider Set:
TeachEngineering
Author:
Meghan Murphy
09/18/2014
Conditions of Use:
No Strings Attached
Rating
This is a task from the Illustrative Mathematics website that is one part of a complete illustration of the standard to which it is aligned. Each task has at least one solution and some commentary that addresses important aspects of the task and its potential use.
Subject:
Numbers and Operations
Material Type:
Activity/Lab
Provider:
Illustrative Mathematics
Provider Set:
Illustrative Mathematics
Author:
Illustrative Mathematics
08/06/2015
Conditions of Use:
Remix and Share
Rating
Students will create an interactive Powerpoint game where they will create real world problems that are used as clues to move on to the next level. Problems include all 8th grade math standards.
Subject:
Mathematics
Material Type:
Interactive
Author:
Linda Ridings
01/28/2016
Conditions of Use:
No Strings Attached
Rating
This is an instructional task meant to generate a conversation around the meaning of negative integer exponents. While it may be unfamiliar to some students, it is good for them to learn the convention that negative time is simply any time before t=0.
Subject:
Mathematics
Material Type:
Activity/Lab
Provider:
Illustrative Mathematics
Provider Set:
Illustrative Mathematics
Author:
Illustrative Mathematics
05/01/2012
Conditions of Use:
No Strings Attached
Rating
My goal is to merge New York State standards with Common Core Standards and Integrated Algebra Regent Standards for our 8th grade curriculum.
Subject:
Mathematics
Material Type:
Full Course
Lesson Plan
Syllabus
Author:
Shaun Errichiello
01/28/2016
Conditions of Use:
Rating
Students learn how engineers navigate satellites in orbit around the Earth and on their way to other planets in the solar system. In accompanying activities, they explore how ground-based tracking and onboard measurements are performed. Also provided is an overview of orbits and spacecraft trajectories from Earth to other planets, and how spacecraft are tracked from the ground using the Deep Space Network (DSN). DSN measurements are the primary means for navigating unmanned vehicles in space. Onboard spacecraft instruments might include optical sensors and an inertial measurement unit (IMU).
Subject:
Engineering
Astronomy
Material Type:
Activity/Lab
Lesson Plan
Provider:
TeachEngineering
Provider Set:
TeachEngineering
Author:
Janet Yowell
Malinda Schaefer Zarske
09/18/2014
Conditions of Use:
Remix and Share
Rating
In Grade 8 Module 1, students expand their basic knowledge of positive integer exponents and prove the Laws of Exponents for any integer exponent. Next, students work with numbers in the form of an integer multiplied by a power of 10 to express how many times as much one is than the other. This leads into an explanation of scientific notation and continued work performing operations on numbers written in this form.
Subject:
Numbers and Operations
Material Type:
Module
Provider:
New York State Education Department
Provider Set:
EngageNY
05/14/2013
Conditions of Use:
Remix and Share
Rating
In this unit of study, students will research topics in nanotechnology and attempt to identify a mystery mixture of different powders using a scanning-electron microscope. This unit integrates nine STEM attributes and was developed as part of the South Metro-Salem STEM Partnership's Teacher Leadership Team. Any instructional materials are included within this unit of study.
Subject:
Physical Science
Material Type:
Activity/Lab
Assessment
Homework/Assignment
Lecture Notes
Lesson Plan
|
2020-03-29 11:19:42
|
{"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": 0.3138180673122406, "perplexity": 3135.1309262175914}, "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-2020-16/segments/1585370494331.42/warc/CC-MAIN-20200329105248-20200329135248-00445.warc.gz"}
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http://www.chegg.com/homework-help/questions-and-answers/-q3602023
|
## Question 1.4 Part D) Find Probability of Independent X's
• given g(a)= 0 for a<0 a2 for 0=<a<1 1 for a=>1
a)given to find the probability for Pr(X=>0.5 tolal out puts are three probability of getting this output outof one output is 0.5 so this probability will be 0.5/3=1/6
b) Pr(0.5=<X=<0.75 probability of getting this from one output is 1/4=0.25 so the probability will be 0.25/3
=1/12
c)given to find Pr(x1+X2=<1 will be theprobality getting this from all 3 out puts is 2/3
|
2013-05-21 03:22:18
|
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|
https://tex.stackexchange.com/questions/349870/can-you-use-beamer-minted-tikz-to-single-step-code
|
# Can you use beamer + minted + tikz to single-step code?
I have some C pointer code that I would like to "single-step" to show what happens with e.g. dangling pointers. I use minted for syntax highlighting, and that seems to be causing the problem.
The code below compiles (with -shell-escape), but the code that should be escaped isn't. The reason seem to be a bug in pygments (see https://github.com/gpoore/minted/issues/70). I have also tried to solution of marvin2k that re-defined !, but failed to get that code to work.I have also tried various combinations of \begin{minted}[highlightlines=\only<1>{3}\only<2>{5}]{c} to no avail.
Does anyone have a solution? Edit: My platform is ubuntu 14.04, pygmentize 1.6, minted 2.4.1.
\documentclass{beamer}
\usepackage{minted,tikz}
\begin{document}
\begin{frame}[fragile]{Pointers}
\begin{minted}[linenos,escapeinside=||]{c}
int main(void) {
char *p;
p=(char *)malloc(5); |\only<1>{$\Leftarrow$}|
/* do stuff */
p=(char *)malloc(7); |\only<2>{$\Leftarrow$}|
free(p);
return 0;
}
\end{minted}
\begin{tikzpicture}
\node<1->[rectangle,draw] (p) {p};
\node<1->[rectangle,draw,right of=p] (q1) {q1};
\draw<1>[->] (p) -- (q1);
\uncover<2->{\node[rectangle,draw,below right of=p] (q2) {q2};};
\draw<2->[->] (p) -- (q2);
\end{tikzpicture}
\end{frame}
\end{document}
• Completely unrelated but you shouldn’t cast the return value of malloc in C, it’s unnecessary and error-prone (since you could accidentally cast to the wrong pointer type and invoke inadvertent undefined behaviour): stackoverflow.com/q/605845/1968 – Konrad Rudolph Jan 23 '17 at 10:42
• @KonradRudolph, unrelated...and you are right. It was a long time since I last did C and I haven't kept track of 'best practices'. I tried to modify my question but I don't know how to modify the image. – Niclas Börlin Jan 24 '17 at 13:03
As < and > are problematic with minted, you may define the \myonly command:
\newcommand\myonly[2]{\only<#1>{#2}}
MWE:
\documentclass{beamer}
\usepackage{minted,tikz}
\newcommand\myonly[2]{\only<#1>{#2}}
\begin{document}
\begin{frame}[fragile]{Pointers}
\begin{minted}[linenos,escapeinside=||]{c}
int main(void) {
char *p;
p=(char *)malloc(5); |\myonly{1}{$\Leftarrow$}|
/* do stuff */
p=(char *)malloc(7); |\myonly{2}{$\Leftarrow$}|
free(p);
return 0;
}
\end{minted}
\begin{tikzpicture}
\node<1->[rectangle,draw] (p) {p};
\node<1->[rectangle,draw,right of=p] (q1) {q1};
\draw<1>[->] (p) -- (q1);
\uncover<2->{\node[rectangle,draw,below right of=p] (q2) {q2};};
\draw<2->[->] (p) -- (q2);
\end{tikzpicture}
\end{frame}
\end{document}
• That's odd...when I compile your solution (ubuntu 14.04, pygmentize v1.6, minted v2.4.1), my problem remains. What setup did you use? – Niclas Börlin Jan 22 '17 at 15:49
• @NiclasBörlin I use TeXLive 2016, pygmentize 2.1.3 and minted 2.4.1. – Paul Gaborit Jan 22 '17 at 17:24
• It looks like I need to upgrade pygmentize then... Do you run Linux or something else? – Niclas Börlin Jan 22 '17 at 17:29
• @NiclasBörlin I use Ubuntu 16.10. – Paul Gaborit Jan 22 '17 at 17:29
The best solution I have so far is based on @VZ.'s overprint solution (https://tex.stackexchange.com/a/51618/35602). However, it would be nice with a solution that did not require the C code to be repeated...
\documentclass{beamer}
\usepackage{minted}
\usepackage{tikz}
\begin{document}
\begin{frame}[fragile]{Foo}
\begin{overprint}
\onslide<1>
\begin{minted}[linenos,highlightlines={3}]{c}
int main(void) {
char *p;
p=(char *)malloc(5);
/* do stuff */
p=(char *)malloc(7);
free(p);
return 0;
}
\end{minted}
\onslide<2>
\begin{minted}[linenos,highlightlines={5}]{c}
int main(void) {
char *p;
p=(char *)malloc(5);
/* do stuff */
p=(char *)malloc(7);
free(p);
return 0;
}
\end{minted}
\end{overprint}
\begin{tikzpicture}
\node<1->[rectangle,draw] (p) {p};
\node<1->[rectangle,draw,right of=p] (q1) {q1};
\draw<1>[->] (p) -- (q1);
\uncover<2->{\node[rectangle,draw,below right of=p] (q2) {q2};};
\draw<2->[->] (p) -- (q2);
\end{tikzpicture}
\end{frame}
\end{document}
|
2019-06-18 04:40:51
|
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|
https://phys.libretexts.org/Courses/College_of_the_Canyons/Physci_101_Lab%3A_Physical_Science_Laboratory_Investigations_(Ciardi)/08%3A_Investigation_7_-_Traveling_in_a_Circle/8.6%3A_General_Questions
|
# 8.6: General Questions
1. Assume that you have the object on the rope circling overhead. At which point in the orbit should you release the object if you want to hit a target (X marks the spot) directly in front of you? Draw a diagram. Do not attempt to confirm this experimentally!
1. If you maintain a constant rate of speed for the orbiting object, is there acceleration? Explain.
2. Draw the sketch below. Draw arrows to indicate the direction of the force keeping the spaceship in orbit.
1. Assume the spaceship has a deflector screen which makes it immune to the force from Earth. If the ship is traveling along in its orbit, and the ship turns off its engines at the same time it turns on its deflector shield, what will happen? Choose one answer and defend it.
1. The ship will continue traveling in a circle.
2. The ship will travel in a straight line tangent to the circular orbit.
3. The ship will spiral down to the Earth.
4. The ship will come to a full stop, hovering above the Earth.
|
2022-12-09 23:28:44
|
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|
https://www.physicsforums.com/threads/iterating-through-an-array.526109/
|
Iterating through an array
I'm trying to understand a sorting algorithm (selection sort, to be exact). I started out with this:
public static void selectionSort1(int[] x) {
for (int i=0; i<x.length-1; i++) {
for (int j=1; j<x.length; j++) {
if (x > x[j]) {
//... Exchange elements
int temp = x;
x = x[j];
x[j] = temp;
}
}
}
}
but that didn't work so i made one slight modification by changing j to i + 1
public static void selectionSort1(int[] x) {
for (int i=0; i<x.length-1; i++) {
for (int j=i+1; j<x.length; j++) {
if (x > x[j]) {
//... Exchange elements
int temp = x;
x = x[j];
x[j] = temp;
}
}
}
}
i'm having trouble figuring out what's the difference between the codes?
The first one will not work because you check:
first iteration:
x[0] > x[1]
x[0] > x[2]
x[0] > x[3]
and so on. This will work for first iteration.
some iterations later:
x[3] > x[1]
x[3] > x[2]
x[3] > x[3]
and so on. This will not work because now you will switch places with numbers already sorted.
For example x[3] will for sure be larger than x[1] because x[1] is sorted to be the second smallest number. And when they switch a larger number is placed in x[1].
Second code works because
first iteration:
x[0] > x[1]
x[0] > x[2]
x[0] > x[3]
and so on
some iterations later:
x[3] > x[4]
x[3] > x[5]
x[3] > x[6]
this will work because you discard the already checked values which is not the case in the first code
Last edited:
Mark44
Mentor
When you include code in your post, please put (code) and (/code) tags around it (with the tags in brackets [] though). Doing this preserves your formatting and makes your code easier to read.
I have edited your post to do this.
I'm trying to understand a sorting algorithm (selection sort, to be exact). I started out with this:
Code:
public static void selectionSort1(int[] x) {
for (int i=0; i<x.length-1; i++) {
for (int j=1; j<x.length; j++) {
if (x[i] > x[j]) {
//... Exchange elements
int temp = x[i];
x[i] = x[j];
x[j] = temp;
}
}
}
}
but that didn't work so i made one slight modification by changing j to i + 1
Code:
public static void selectionSort1(int[] x) {
for (int i=0; i<x.length-1; i++) {
for (int j=i+1; j<x.length; j++) {
if (x[i] > x[j]) {
//... Exchange elements
int temp = x[i];
x[i] = x[j];
x[j] = temp;
}
}
}
}
i'm having trouble figuring out what's the difference between the codes?
|
2021-10-20 14:36:34
|
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|
https://thatsmaths.com/2022/03/
|
## Archive for March, 2022
### Infinitesimals: vanishingly small but not quite zero
Abraham Robinson (1918-1974) and his book, first published in 1966.
A few weeks ago, I wrote about Hyperreals and Nonstandard Analysis , promising to revisit the topic. Here comes round two.
### The Chromatic Number of the Plane
To introduce the problem in the title, we begin with a quotation from the Foreword, written by Branko Grünbaum, to the book by Alexander Soifer (2009): The Mathematical Coloring Book: Mathematics of Coloring and the Colorful Life of its Creators:
If each point of the plane is to be given a color, how many colors do we need if every two points at unit distance are to receive distinct colors?
About 70 years ago it was shown that the least number of colours needed for such a colouring is one of 4, 5, 6 and 7. But which of these is the correct number? Despite efforts by many very clever people, some of whom had solved problems that appeared to be much harder, no advance has been made to narrow the gap
${4\le\chi\le 7}$.
### The Improbability Principle and the Seanad Election
A by-election for the Seanad Éireann Dublin University constituency, arising from the election of Ivana Bacik to Dáil Éireann, is in progress. There are seventeen candidates, eight men and nine women. Examining the ballot paper, I immediately noticed an imbalance: the top three candidates, and seven of the top ten, are men. The last six candidates listed are all women. Is there a conspiracy, or could such a lopsided distribution be a matter of pure chance?
To avoid bias, the names on the ballot paper are always listed in alphabetical order. We may assume that the name of a randomly chosen candidate is equally likely to appear at any of the positions on the list; with 17 candidates, there about 6% chance for each of the 17 positions; the distribution for a single candidate is uniform. However, when several candidates are grouped, the distribution is more complicated [TM231 or search for “thatsmaths” at irishtimes.com].
Following the invention of calculus, serious concerns persisted about the mathematical integrity of the method of infinitesimals. Leibniz made liberal use of infinitesimals, with great effect, but his reasoning was felt to lack rigour. The Irish bishop George Berkeley criticised the assumptions underlying calculus, and his objections were not properly addressed for several centuries. In the 1800s, Bolzano, Cauchy and Weierstrass developed the ${\varepsilon}$${\delta}$ definition of limits and continuity, which allowed derivatives and integrals to be defined without recourse to infinitesimal quantities. Continue reading ‘Hyperreals and Nonstandard Analysis’
|
2022-10-06 04:26:09
|
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|
https://math.berkeley.edu/wp/apde/dmitry-jakobson-mcgill/
|
# Dmitry Jakobson (January 26)
Speaker: Dmitry Jakobson (McGill)
Title: Probability measures on manifolds of Riemannian metrics
Abstract: This is joint work with Y. Canzani, B. Clarke, N. Kamran, L. Silberman and J. Taylor. We construct Gaussian measures on the manifold of Riemannian metrics with the fixed volume form. We show that diameter, Laplace eigenvalue and volume entropy functionals are all integrable with respect to our measures. We also compute the characteristic function for the L^2(Ebin) distance from a random metric to the reference metric.
|
2019-06-18 03:09:56
|
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|
http://math.stackexchange.com/questions/153798/question-about-continuity-of-path-integrals?answertab=votes
|
Question about Continuity of Path Integrals
I have this continuous function $f:\mathbb{C}\rightarrow\mathbb{C}$ defined on an open set $\Omega$.
I also have a family of identical smooth curves up to translation $z_{t}:[a,b]\rightarrow\mathbb{C}$ in $\Omega$ and $t \in [0, T]$, where the parameter $t$ specifies some linear translation, i.e. $z_{t+\delta} = z_{t} + \delta K$ where $K$ is some constant. (You can imagine that the curve $z_{t}$ is "shifting" with the passage of time, $t$.)
So my question is, is the integral $\int_{z_{t}}f(x) dx$ continuous as a function of $t$? In other words, as my curve $z_{t}$ moves a little bit in $\Omega$, does the value of the integral also move a little bit?
More specifically, in my case I know that the value of the integral is $0$ for all $t \in (\alpha, \beta]$. Must the value of the integral also be zero for $t=\alpha$?
-
Write down the definitions: $$\int_{z_t} f(x)dx = \int_a^b f(z_t(s))\frac{d}{ ds} z_t(s) ds$$ With your 'translation' property of the $z_t$ the derivative wrt $s$ does not depend on $t$, so only the $f(z_t(s))$ plays a role. Now write down the difference for $t = \alpha$ and $t$ close to $\alpha$ and estimate the difference: $$\int_a^b \left(f(z_\alpha(s)) - f(z_t(s))\right)\frac{d}{ ds} z_t(s) ds$$ Since the image of $[0,T]\times [a,b]$ under $z$ is compact und $f$ continuous, $f$ is uniformly continuous on that image, hence the difference can be made uniformly arbitrarily small as $t\rightarrow \alpha$, given your definition of $z_t$.
(This reasoning, of course, relies on the assumption that the images of all $z_t$ are contained in $\Omega$ and do not, e.g. tend to the boundary of $\Omega$ as $t\rightarrow \alpha$. But since you've chosen closed intervals as domain of definition, I assume this to be true.)
Let $F(t,s) = f(z_{t}(s))$. Then $F:[0,T]\times[a,b]\rightarrow\mathbb{C}$ It seems reasonable that $F$ is continuous on its domain and therefore F is uniformly continuous. Now choose an $\epsilon$ and we get $|F(\alpha, s) - F(0, s)| < \epsilon$ when we choose $\alpha < \delta$. Then $\int_{a}^{b}|F(\alpha,s)-F(0,s)|z'(s)ds < \epsilon\int_{a}^{b}z'(s)ds$. Since $\epsilon$ is arbitrary, the integral is $0$. Is this correct? – Mark Jun 4 '12 at 17:05
@Mark Kind of. I think you mixed up $\alpha, t$ and $0$ when compared with your posting. $\alpha$ already has a special meaning in your original question. Then, first of all, yes, your $F$ is continuous as composition of continuous maps. Then choose $\epsilon$ such that $|F(\alpha,s) - F(t,s)|\le \epsilon$ for $|t-\alpha|< \epsilon$. Then replace $0$ in the integral by $t$, and in order to estimate the (norm of the) integrals you want to take the norm of the derivative of $z_t$ wrt to $s$ otherwise you might get in trouble with the signs. Or replace $\alpha, \beta$ in your OP by $0,T$. – user20266 Jun 4 '12 at 17:28
|
2015-10-09 03:19:55
|
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|
https://gis.stackexchange.com/questions/339068/why-do-arcpy-getcount-management-and-arcpy-arcsdesqlexecute-yield-different-row
|
# Why do arcpy.GetCount_management and arcpy.ArcSDESQLExecute yield different row counts?
I was benchmarking which method was faster for counting the number of rows in a table, when something curious happened:
import arcpy
import os
import time
test_tuple = ("path\to\connection", "feature_class_name")
fc = os.path.join(*test_tuple)
i = 0
while i < 3:
# Method 1
time.sleep(5)
start_time = time.clock()
count = int(arcpy.GetCount_management(fc).getOutput(0))
end_time = time.clock()
print("Method 1 ({count}) finished in {time} seconds".format(count=count, time=(end_time - start_time)))
print("The count was: {count}".format(count=count))
# Method 2
time.sleep(5) # Let the cpu/ram calm before proceeding!
start_time = time.clock()
query = """SELECT COUNT(*) FROM {}""".format(test_tuple[-1].upper())
execute_object = arcpy.ArcSDESQLExecute(test_tuple[0])
result = execute_object.execute(query)
count = int(result)
end_time = time.clock()
print("Method 2 ({count}) finished in {time} seconds".format(count=count, time=(end_time - start_time)))
print("The count was: {count}".format(count=count))
i += 1
This gave the following results:
Method 1 finished in 5.018752999999999 seconds
The count was: 22645
Method 2 finished in 0.24905560000000015 seconds
The count was: 21473
Method 1 finished in 0.7440046999999996 seconds
The count was: 22645
Method 2 finished in 0.7023353000000014 seconds
The count was: 21473
Method 1 finished in 0.7402944999999974 seconds
The count was: 22645
Method 2 finished in 0.684015500000001 seconds
The count was: 21473
Why are the counts different between the two methods?
• I've not personally used the arcpy.ArcSDESQLExecute method, so I don't know if this is it or not, but my gut reaction is to assume it likely has something to do with a versioned dataset and resulting discrepancies. If so, using Versioned Views rather than the dataset directly may assist. – John Oct 17 '19 at 19:51
• Is the feature class versioned? If so, that's your answer. – Vince Oct 17 '19 at 20:06
• It is versioned, but I don't understand why versioning would change counts if both methods are looking at DEFAULT. – jesnes Oct 17 '19 at 20:09
• Because there are additional rows in the Adds and Deletes tables which haven't yet manifested changes in the business table. Versioning 101 is the place to start. – Vince Oct 17 '19 at 20:13
• Does that mean that the discrepancy would go away if the database was compressed? – jesnes Oct 17 '19 at 20:16
When querying the table of a versioned (or archive-enabled) feature class directly using SQL, your results would not be expected to reflect the current state of the feature class. You should use the versioned view instead. The table could contain records that have been deleted, or records that have been changed, or may be missing records that have been added.
Try changing the line:
query = """SELECT COUNT(*) FROM {}""".format(test_tuple[-1].upper())
to:
query = """SELECT COUNT(*) FROM {}_EVW""".format(test_tuple[-1].upper())
Ie, run it against the versioned view, not the table. The versioned view usually has the same name as the table but with "_evw" appended and it is what ArcGIS/arcpy would be actually using in the background and not the table directly.
If you are using versioning or archiving, you should NEVER interact directly with the database table outside of ArcGIS, but always with the versioned view. Otherwise you will get invalid data such as superceded data, deleted records, missing records, etc (and if you make edits directly to the table, you could corrupt your database, as far as ArcGIS is concerned).
ESRI have a lot of documentation on this sort of thing. Eg:
and
• I'm glad I asked this question, because the answer is more informative than I imagined. Thanks! Also, there's a small error in where you say to add "_evw": it should be added to the line query = """SELECT COUNT(*) FROM {}""".format(test_tuple[-1].upper() + "_EVW") because that's where the view is being queried. As you have it, you are adding "_EVW" to the workspace's path. – jesnes Oct 18 '19 at 15:42
• Oops. Don’t know what I was thinking. I’ve fixed the post now - slightly different to how you wrote it to be a little more efficient I think. – Son of a Beach Oct 18 '19 at 21:07
• I wholeheartedly agree, thanks again for your informative answer. – jesnes Oct 19 '19 at 0:39
In addition to using the versioned view, consideration should be made for the length of the table name. In this case, I've discovered that if adding "_EVW" to the end of the table name exceeds 30 characters, an error will be thrown. I got around this with the following:
test_tuple = ("path\to\connection", "OWNER.table_name")
own, name = test_tuple[-1].split(".")
fc = os.path.join(*test_tuple)
desc = arcpy.Describe(fc)
query_name = test_tuple[-1]
char_limit = 30 # Oracle has a 30 character limit on table names
if desc.isVersioned:
if len(name) + 4 > char_limit: # +4 for the addition of "_EVW"
query_name = ".".join([own, name[:char_limit - 4]])
query = """SELECT COUNT(*) FROM {}_EVW""".format(query_name.upper())
else:
if len(self.name) > char_limit:
query_name = ".".join([self.owner, self.name[:char_limit]])
query = """SELECT COUNT(*) FROM {}""".format(query_name.upper())
execute_object = ArcSDESQLExecute(self.connection)
result = execute_object.execute(query)
count = int(result)
• Good find. I've just checked for MS SQL Server, and the limit there is 128 characters. Less likely to be an issue, but still good to be aware of! Thanks for posting this update. – Son of a Beach Oct 22 '19 at 21:31
|
2020-10-29 22:28:50
|
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|
http://www.ask.com/web?qsrc=6&o=102140&oo=102140&l=dir&gc=1&q=Local+Extrema
|
Web Results
## Maxima and minima
en.wikipedia.org/wiki/Maxima_and_minima
In mathematical analysis, the maxima and minima of a function, known collectively as extrema (the plural of extremum), are the largest and smallest value of the function, either within a given range...
## How to Find Local Extrema with the First Derivative ... - Dummies.com
www.dummies.com/how-to/content/how-to-find-local-extrema-with-the-first-derivativ.html
## Local Extrema
www.math.utah.edu/lectures/math1210/18PostNotes.pdf
18B Local Extrema. 2. Definition. Let S be the domain of f such that c is an element of S. Then,. 1) f(c) is a local maximum value of f if there exists an interval (a,b) ...
## Local Extremum -- from Wolfram MathWorld
mathworld.wolfram.com/LocalExtremum.html
A local extremum, also called a relative extremum, is a local minimum or local maximum. SEE ALSO: Extremum, Global Extremum, Local Maximum, Local ...
## Critical Points - SOS Math
www.sosmath.com/calculus/diff/der13/der13.html
Analogously, f(x) is said to have a local minimum at c iff there exists an interval I around c such that. \begin{displaymath}f(c) \leq f(x) \. A local extremum is a local ...
## Local Extrema Finder - Desmos
www.desmos.com/calculator/4bcktfbguj
Local Extrema Finder.
Aug 26, 2009 ... WEBSITE: http://www.teachertube.com First derivative test to determine local minimum and maximum.
## First Derivative Test for Local Extrema - CliffsNotes
www.cliffsnotes.com/study-guides/calculus/calculus/applications-of-the-derivative/first-derivative-test-for-local-extrema
If the derivative of a function changes sign around a critical point, the function is said to have a local (relative) extremum at that point. If the derivative.
## 1. Find all local extrema of the function f(x) = x x2 + 4 The derivative ...
www.math.uiuc.edu/~muncast/math220/Quizzes/quiz5Solutions.pdf
numbers and hence the local extrema occur where f (x) = 0. From the deriva- ... To determine which of these are local maxs or local mins, we need to do either.
## Graphing Polynomials - Cool math Algebra Help Lessons - Relative ...
www.coolmath.com/algebra/22-graphing-polynomials/10-relative-extrema-maximums-minimums-01
Check out this graph: a graph of a polynomial with relative extrema at points ( -3 , 2 ) ... (Relative extrema (maxes and mins) are sometimes called local extrema.).
### Introduction to local extrema of functions of two variables - Math Insight
mathinsight.org
Introduction to the idea of critical points and local extrema of two variable functions.
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2016-09-25 17:08:08
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https://tutorme.com/tutors/51959/interview/
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TutorMe homepage
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Faisal U.
Teaching assistant at UofT for three years now Software Engineer at Achievers
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Python Programming
TutorMe
Question:
Write a $$\textbf{recursive}$$ function $$perm(s)$$ that takes a string and returns a list of all the permutations of the given string. For example $$>>>perm('cat')$$ $$>>> ['cat', 'act', 'atc', 'cta', 'tca', 'tac']$$
Faisal U.
Before writing any code we shall first come up with a recursive algorithm. To do that, it helps thinking about how can we solve this problem if we had the solution to the simpler problem. In other words, if the string is '$$cat$$' and we had the list of permutations of the smaller string, lets say '$$at$$' ('$$at$$' is substring of '$$cat$$') which is $$['at', 'ta']$$ how can we construct the final answer. With a little observation, you can see that we can obtain the final answer by taking the remaining letter '$$c$$' and inserting it into all the possible places in the provided string. With '$$at$$' we can construct three new permutation by inserting the letter '$$c$$' before, middle and end of '$$at$$' to get '$$cat$$', '$$act$$', and '$$atc$$'. Doing the same to '$$ta$$' to get '$$cta$$', '$$tca$$', and '$$tac$$' which is all the permutation of the string. Now that we have an algorithm we will go ahead with writing the code. $( \text{def} \ perm(s): \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \text{if (len(}s\text{) == 0):} \\ \ \text{return [s]} \\ \text{else:} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ res = [] \ \ \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{for } term \text{ in }perm(s[1:]): \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{for } i \text{ in range(len(} s \text{)):} \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \hspace{4.5 cm} res\text{.append(}term[:i] + s[0] + term[i:]\text{)} \\ \text{return }res$) $$\textbf{Note:}$$ this can be done in less line of code using list comprehension. $( \text{def} \ perm(s): \hspace{12 cm} \\ \text{if (len(}s\text{) == 0):} \hspace{11 cm} \\ \ \text{return [s]} \hspace{12 cm} \\ \text{else:} \hspace{11 cm} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \ \ \ \ \ \ \ \ \text{return} \text{ [} term[:i] + s[0] + term[i:] \text{ for } term \text{ in }perm(s[1:]) \text{ for } i \text{ in range(len(} s \text{))]} \\$)
Physics (Electricity and Magnetism)
TutorMe
Question:
Suppose there is a metal sphere of radius $$R$$ and total charge of $$Q$$. Prove that the electric field outside of the sphere is the same as a point charge $$Q$$ centered at the center of the sphere.
Faisal U.
First, we notice that the charge $$Q$$ is distributed evenly so there is a spherical symmetry of the electric field that points radially outward (positive charge Q). To take advantage of the symmetry we shall consider a Gaussian surface to be a spherical surface of radius $$r > R$$ concentric with the charged sphere in question. Gauss's law is $(\Phi_e = \oint \vec{E} \cdot d\vec{A} = \frac{Q_{in}}{\epsilon_0}$) where $$\vec{A}$$ area vector and $$Q_{in}$$ is the total enclosed charge inside the Gaussian surface. Because the Gaussian surface encloses the entire sphere of charge, $$Q_{in} = Q$$ To calculate the flux, notice that the electric field is perpendicular to the surface everywhere. And due to spherical symmetry, the electric field magnitude $$E$$ must have the same value along the surface of the Gaussian surface. This means that $$E$$ can be pulled out of the integral, so $(E\oint dA = \frac{Q}{\epsilon_0}$) The integral now is just the area of the Gaussian surface, which is just the surface is a sphere with radius $$r$$, so $$A_{sphere} = 4\pi r^2$$. $(EA_{sphere} = E4\pi r^2 = \frac{Q}{\epsilon_0}$) and solving for the electrical filed $$E$$ gives, $(E = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}$) Which is exactly the field of a point charge Q?
Linear Algebra
TutorMe
Question:
Assume vector space of polynomials of degree 2 with coefficients of module 7. Let $$W = span(1+6x+2x^2, 3+x, 5+6x+4x^2, 5+5x+2x^2)$$. Find the basis for W.
Faisal U.
We know that basis must be linearly independent of one another. In this case, $$W$$ is a subspace that is a span of 4 vectors. Since space has polynomials of degree 2 then any basis must have three vectors. Therefore, we must a subset of $$1+6x+2x^2, 3+x, 5+6x+4x^2, 5+5x+2x^2$$ that spans $$W$$. To achieve this will turn the four vectors as columns for a matrix then row reduce the matrix and find which columns have pivots once the matrix is in row echelon form. The question is how do we turn a polynomial into a vector. To do that we write each polynomial as a coefficient vector of the natural basis $$(1, x, x^2)$$. The matrix will look like $(\begin{bmatrix} 1 & 3 & 5 & 5\\ 6 & 1 & 6 & 5 \\ 2 & 0 & 4 & 2 \end{bmatrix}$) and the row echelon form $(\begin{bmatrix} 1 & 3 & 5 & 5\\ 0 & 1 & \frac{24}{17} & \frac{25}{17} \\ 0 & 0 & 1 & \frac{1}{3} \end{bmatrix}$) So from this we can deduce that $$(1+6x+2x^2, 3+x, 5+6x+4x^2)$$ can be a basis for $$W$$.
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2018-12-14 16:58:02
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https://zbmath.org/?q=an%3A1106.42026
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# zbMATH — the first resource for mathematics
Semi-orthogonal parseval frame wavelets and generalized multiresolution analyses. (English) Zbl 1106.42026
Summary: We study Parseval frame wavelets in $$L^2(\mathbb{R}^d)$$ with matrix dilations of the form $$(Df)(x)=\sqrt 2f(Ax)$$, where $$A$$ is an arbitrary expanding $$n\times n$$-matrix with integer coefficients, such that $$|\det A|=2$$. In our study we use generalized multiresolution analyses (GMRA) $$(V_j)$$ in $$L^2(\mathbb{R}^d)$$ with dilations $$D$$. We describe, in terms of the underlying multiresolution structure, all GMRA Parseval frame wavelets and, a posteriori, all semi-orthogonal Parseval frame wavelets in $$L^2(\mathbb{R}^d)$$. As an application, we include an explicit construction of an orthonormal wavelet on the real line whose dimension function is essentially unbounded.
##### MSC:
42C40 Nontrigonometric harmonic analysis involving wavelets and other special systems 42C30 Completeness of sets of functions in nontrigonometric harmonic analysis
##### Keywords:
parseval frame; wavelet; multiresolution analysis
Full Text:
##### References:
[1] Baggett, L.; Medina, H.; Merrill, K., Generalized multiresolution analyses, and a construction procedure for all wavelet sets in $$\mathbb{R}^n$$, J. Fourier anal. appl., 5, 6, 563-573, (1999) · Zbl 0972.42021 [2] D. Bakić, I. Krishtal, E.N. Wilson, Parseval frame wavelets with $$E_n^{(2)}$$-dilations, Appl. Comput. Harmon. Anal. (special issue), in press · Zbl 1090.42020 [3] B. Behera, S. Madan, On a class of band-limited wavelets not associated with an MRA, Preprint · Zbl 1096.42024 [4] Benedetto, J.J.; Li, S., The theory of multiresolution analysis frames and applications to filter banks, Appl. comput. harmon. anal., 5, 389-427, (1998) · Zbl 0915.42029 [5] Bownik, M., The structure of shift-invariant subspaces of $$L^2(\mathbb{R}^n)$$, J. funct. anal., 112, 282-309, (2000) · Zbl 0986.46018 [6] M. Bownik, Z. Rzeszotnik, On the existence of multiresolution analysis for framelets, Preprint, 2004 · Zbl 1081.42026 [7] Bownik, M.; Rzeszotnik, Z.; Speegle, D., A characterization of dimension function of orthonormal wavelets, Appl. comput. harmon. anal., 10, 1, 71-92, (2001) · Zbl 0979.42018 [8] Hernández, E.; Weiss, G., A first course on wavelets, (1996), CRC Press Boca Raton, FL · Zbl 0885.42018 [9] Kim, H.O.; Lim, J.O., On frame wavelets associated with multiresolution analyses, Appl. comput. harmon. anal., 10, 61-70, (2001) · Zbl 1022.94001 [10] Paluszyński, M.; Šikić, H.; Weiss, G.; Xiao, S., Tight frame wavelets, their dimension functions, MRA tight frame wavelets and connectivity properties, Adv. comput. math., 18, 297-327, (2003) · Zbl 1018.42020 [11] Papadakis, M., On the dimension function of orthonormal wavelets, Proc. amer. math. soc., 128, 7, 2043-2049, (2000) · Zbl 0956.42022
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.
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2021-09-21 18:33:07
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https://www.gradesaver.com/textbooks/math/algebra/elementary-and-intermediate-algebra-concepts-and-applications-6th-edition/chapter-6-rational-expressions-and-equations-6-4-addition-and-subtraction-with-unlike-denominators-6-4-exercise-set-page-402/91
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## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)
$$\frac{10x-14}{\left(x+4\right)\left(x-5\right)}$$
The perimeter of a rectangle is equal to the sum of twice the length of each of the adjacent sides. Doing this, we find: $$\frac{6}{x+4}+\frac{4}{x-5}\\ \frac{6\left(x-5\right)}{\left(x+4\right)\left(x-5\right)}+\frac{4\left(x+4\right)}{\left(x-5\right)\left(x+4\right)}\\ \frac{6\left(x-5\right)+4\left(x+4\right)}{\left(x+4\right)\left(x-5\right)}\\ \frac{10x-14}{\left(x+4\right)\left(x-5\right)}$$
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2019-12-08 07:43:38
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https://xslates.com/post/chebyshev-inequality-for-standard-deviation/
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The Chebyshev inequality goes like this. Suppose that $$x$$ is an $$n$$-vector, and that $$k$$ of its entries satisfy $$|x_i| \geq a$$, where $$a > 0$$. Then $$k$$ of its entries satisfy $$x_i^2 \geq a^2$$. It follows that
$||x||^2 = x_1^2 + \cdots + x_n^2 \geq ka^2$
since $$k$$ of the numbers in the sum are at least $$a^2$$, and the other $$n - k$$ numbers are nonnegative. We conclude that $$k \leq ||x||^2 / a^2$$, which is the Chebyshev inequality.
When $$||x||^2 / a^2 \geq n$$, the inequality tells us nothing, since we always have $$k \leq n$$. In other cases, it limits the number of entries in a vector that can be large. In general, the inequality states that no entry of a vector can be larger in magnitude than the norm of the vector. Another interpretation, using the RMS value, looks like this
$\frac{k}{n} \leq \bigg( \frac{\mathbf{rms}(x)}{a} \bigg)^2$
where $$k$$ is the number of entries of $$x$$ with absolute value at least $$a$$. The left-hand side is the fraction of entries of the vector that are at least $$a$$ in absolute value. The right hand side is the inverse square of the ratio of $$a$$ to $$\mathbf{rms}(x)$$. It says, for example, that no more than 1/25 = 4% of the entries of a vector can exceed its RMS value by more than a factor of 5. Some statements that follow from the inequality are that:
• not too many of the entries of a vector can be much bigger (in absolute value) than its RMS value
• at least one entry of a vector has absolute value as large as the RMS value of the vector
import numpy as np
a = 6
x = np.array([3, 4, 1, 2, 4, 8, 5])
n = x.size
k = np.count_nonzero(x >= a)
rms = np.sqrt(np.mean(x ** 2))
k / n <= (rms / a) ** 2
## True
Relation To Standard Deviation
The Chebyshev inequality can be transcribed to an inequality expressed in terms of the mean and standard deviation: If $$k$$ is the number of entries of $$x$$ that satisfy $$|x_i - \mathbf{avg}(x)| \geq a$$, then $$k/n \leq (\mathbf{std}(x)/a)^2$$.
In probability theory the Chebyshev inequality is used to state something along the lines of “at most, approximately 11.11% of a distribution will lie at least three standard deviations away from the mean.”.
Another way to state this is: The fraction of entries of $$x$$ within $$\alpha$$ standard deviations of $$\mathbf{avg}(x)$$ is at least $$1 - 1/\alpha^2$$ (for $$a > 1$$).
As an example, if we consider a time series of return on an investment, with a mean return of 8%, and a risk (standard deviation) of 3%. By the Chebyshev inequality, the fraction of periods with a loss ($$x_i \leq 0$$) is no more than $$(3/8)^2$$, or 14.1%.
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2020-07-03 09:48:34
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http://www.chegg.com/homework-help/questions-and-answers/suppose-potential-seen-neutron-nucleus-berepresented-dimension-square-width10-8722-12-cm-h-q136271
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## Quantum Physics
(a) Suppose the potential seen by a neutron in a nucleus can berepresented (in one dimension) by a square well of width10−12 cm with very high walls. What is the minimumkinetic energy of the neutron, in MeV ?
(b) Can the electron be confined in the nucleus ? Answer thisquestion using the following outline or some other method:
1. Using the same assumptions as in (a), calculate the minimumkinetic energy, in
MeV, of an electron bound inside the nucleus.
2. Calculate the approximate Coulomb potential energy, in MeV, ofan electron at the
surface of the nucleus, compared to its potential energy atinfinity. Take the nuclear
charge to be +50e.
3. Is the potential energy calculated in (2) sufficient to bind theelectron with kinetic
energy calculated in (1) ?
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2013-05-24 09:12:15
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https://www.nature.com/articles/s41598-020-79074-8?error=cookies_not_supported&code=9cb05a2b-55b7-4838-9a00-fff507b2b92c
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## Introduction
The prevailing spatial unevenness in the distribution of biodiversity globally1,2 imposes pressing challenges to the understanding and conservation of ecosystems given that the distribution of threats is also spatially asymmetric3,4,5,6. Therefore, in an era of alarming worldwide biodiversity declines7,8,9,10, where the majority of species still remain to be discovered11, it is crucial to accurately identify the geographic regions of primary conservation concern12,13. Hotspots of biodiversity—areas characterised by exceptional relative species-richness, or by unusually high numbers of endemic and endangered species14,15—have largely been considered primary targets for such conservation actions16,17,18,19. Ecological criteria such as vulnerability20,21, irreplaceability22 and representativeness23 have been used to identify biodiversity areas of priority concern17,24,25. However, a major challenge intrinsic to studies of biodiversity distributions at large spatial or taxonomic scales is the difficulty that assembling comprehensive datasets that cover high proportions of the diversity of entire regions or clades involves26,27 (often referred to as ‘the Wallacean shortfall’28). This challenge is aggravated in some countries that tend to concentrate less comprehensive biodiversity datasets29 (i.e., data-poor regions), whilst often hosting the biodiversity-richest regions and undergoing the toughest pressures on biodiversity28,30,31.
Over the last 2 decades, increased efforts have been made to expand the spatial and/or taxonomic coverage of biodiversity datasets, with the aim to advance our ability to quantify biodiversity patterns and to make evidence-based decisions about conservation actions18,32,33,34,35. As comprehensive biodiversity datasets become available, evidence has revealed, among other core findings, that different measures of biodiversity hotspots are not consistently congruent in space36,37. Additionally, conservation actions are not effective across or even within the same clades, given the biological idiosyncrasies of each group32—for example, given that ectothermic tetrapods (amphibians, reptiles) do not experience the same environmental pressures that endothermic organisms (mammals, birds) do, their biological responses to the same environments can differ importantly38. The spatial and time scales considered are also expected to impact on the patterns being observed, and thus, finer-scale data tend to offer better resources for more effective inferences about the spatial organisation of biodiversity39,40. Therefore, efforts devoted to create truly comprehensive primary biodiversity datasets are expected to hold a vital key to understand and manage biodiversity41,42.
Within South America—one of the world’s most biodiverse continents14,16–, Uruguay, which spans one of the world’s most extensive natural grasslands43,44, stands-out given the convergence of tropical and non-tropical eco-regions that combined bring together the geographic boundaries of an important ‘mix’ of species representative of those major biological domains45,46,47,48,49,50. Yet, Uruguay’s network of protected areas spans < 1.5% of the country’s territory51—the lowest for any country in South America52,53. A dominant reason for such discrepancies between the country’s diversity and conservation actions can be attributed to the limited knowledge on the distribution of its biodiversity. While previous attempts to identify the geographic location of Uruguay’s biodiversity hotspots have been made54,55,56,57, the quantification of multiple forms of biodiversity hotspots remains fundamentally neglected given the severe gaps in geographic data and the lack of comprehensive open-access databases on species distributions. Thus, despite its unique ecosystem and biodiversity attributes, the availability of scientific tools for evidence-based management and longer-term planning of Uruguay’s environments lags behind most countries in the continent. To address this historical issue, Biodiversidata—Uruguay’s first Consortium of Biodiversity Data (https://biodiversidata.org/) has recently been created58,59. This initiative gathers a collaborative scientific community of the country’s biodiversity experts with the aim of assembling a constantly growing database for Uruguay’s biodiversity. The first version of Biodiversidata presented a comprehensive dataset for all tetrapod species58.
In this study, we investigate for the first time the distribution, diversity and congruence of Uruguay’s hotspots of tetrapods, using > 69,000 geographic records for 664 species of amphibians, reptiles, birds and mammals (Table 1, Supplementary Fig. S1) compiled as part of the emerging Biodiversidata initiative58. We aim to (1) elucidate the spatial patterns of tetrapod species-richness, endemism and threatened species (both measured as absolute and relative numbers of species facing extinction risk), and quantitatively assess their spatial congruence, (2) measure the sampling bias and evaluate estimated diversities, (3) assess whether the current protected area network in Uruguay is capturing the different hotspots, and (4) identify areas where future sampling to foster robust environmental management and planning should be prioritised.
## Results
### The spatial distribution of biodiversity
The distributional patterns of all three types of hotspot (species-richness, endemism and threatened species number/proportion) varied across taxa (Fig. 1 and Supplementary Fig. S2,S3). Species-richness for all groups of tetrapods was aggregated in the south and south-east coast of the country (Fig. 1, top row), with the highest numbers of species detected in the surroundings of Montevideo (capital city) and other coastal cities. High numbers of species were also detected in the mid-east border with Brazil and in the north-west side of the Uruguay River. Each group also exhibited high concentrations of species in exclusive grid-cells—not shared between groups (Fig. 1, top row). Endemism patterns were more homogeneously distributed for all tetrapods, with moderate values all across the territory and few high endemism areas towards the south and south-east coast, north-west and the mid-east borders of the country (Fig. 1, second row from the top). For each separate tetrapod class, we detected a similar pattern. Amphibians presented exclusive areas of endemism in the northwest, northeast and mid-east of the country, birds in the south coast, and mammals in the south-east Atlantic coast. We did not observe any reptile-exclusive areas of endemism. The hotspots of threatened species for all tetrapods combined and separated exhibited a few grid-cells of high value, for both species proportions (Fig. 1, third and fourth rows) and numbers (Fig. 1, fifth and bottom rows), irrespective of the IUCN categories used to assess threatened species. However, when using national IUCN assessments, additional peak values were revealed. Each tetrapod class showed unique areas with a high proportion of threatened species which, contrary to species-richness and endemism, were principally located distant to the coast and towards the centre of the country. The scarce hotspots of threatened species identified in all cases are mainly due to the low number of species that are assessed as threatened (i.e., critically endangered, endangered and vulnerable), either globally with 4 species of amphibians (8% of the total species), 9 reptiles (13.2%), 15 birds (3.5%) and 8 mammals (6.8%), or at the national level, with 12 species of amphibians (24% of the total species), 8 reptiles (11.8%) and 40 birds (9.3%).
### The spatial congruence of biodiversity hotspots
We found a tendency for low spatial congruence across and within different measures of biodiversity hotspots when taking tetrapod classes combined or separately. The extent of spatial congruence across types of hotspots varied according to grid-cell size (12.5, 25 and 50 km), if threatened species proportion or number was considered, and the percentage of the hotspot area definition (% of area/number of cells occupied by hotspots; from 0 to 100%) (Fig. 2). However, we observed a tendency towards greater levels of congruence when relaxing the hotspot definition to 10% of the richest grid-cells, using the major grid-cell size and numbers of threatened species instead of proportions (Table 2a). Exceptionally for amphibians, birds and mammals, using the 50 × 50 km unit and considering the smallest definition of the area occupied by hotspots (2–2.5%), we observed complete congruence as the three types of hotspots were localised in the same unique grid-cell (Fig. 2a,c,d: red line). Amphibians and birds showed higher congruence when the national IUCN assessment was used (Fig. 2a,c). Yet, this was not the case for reptiles, which had a prevailing tendency for incongruence between diversity hotspots (Table 2a). The extent of spatial congruence of hotspots for tetrapods combined was generally higher than for each separate group (Fig. 2e).
Similarly, we observed very few overlapping grid-cells for each separate hotspot type when combining all tetrapods (Supplementary Fig. S4). Congruence increased by varying the hotspot definition criteria to the richest 10% of grid-cells, though it always remained less than 18% for all types of hotspots (Table 2b). The spatial overlap of hotspots of species-richness combining groups decreased with the increase in the grid-cell size, while the congruence of hotspots of endemism was less sensitive to it (Supplementary Fig. S4a, b). The congruence of hotspots of threatened species was very low regardless of grid’s size, for both species proportion and numbers (Supplementary Fig. S4c, d).
### Spatial correlation of sampling effort and biodiversity hotspots
The reconstruction of biodiversity hotspots critically depends on the historical distribution of sampling efforts. The spatial patterns of species-richness and endemism were significantly highly correlated with those of sampling effort for all tetrapod species and each single group (r ≥ 0.8, p < 0.001) (Supplementary Table S1). The spatial correlations between sampling effort and threatened species numbers were moderate to high for all groups (0.54 < r < 0.73, p < 0.001), while there were no significant correlations between threatened species proportions and sampling effort, or any metric of hotspot diversity.
Despite the lack of congruence across types of hotspots for each tetrapod group (Fig. 2), we saw moderate to high levels of spatial correlations between each pair of diversity measure (Supplementary Table S1), ranging from r = 0.706 to r = 0.802 (p < 0.001) for species-richness versus endemism, r = 0.521 to r = 0.793 (p < 0.001) for species-richness versus threatened species number, and r = 0.613 to r = 0.828 (p < 0.001) for endemism versus threatened species number, while there were no significant correlations between threatened species proportions and any other metric. For those groups with national red lists assessments (amphibians, reptiles and birds), we found higher correlations when using national IUCN threat categories rather than global, except in the case of reptiles’ spatial association between endemism and threat patterns (Supplementary Table S1).
When assessing how similarly distributed hotspot types between groups were, we found low to moderate correlation values for each type of diversity hotspot, all significant (p < 0.001) except when measuring threatened species proportions for any pair of tetrapod group (Supplementary Table S2). Species-richness patterns were moderately correlated, with amphibians versus reptiles showing the highest levels (r = 0.748), while the lowest were between amphibians and birds (r = 0.514, p < 0.001). Endemism revealed weaker associations between groups, ranging from r = 0.360 (p < 0.001) for amphibians versus birds, to a maximum of r = 0.697 (p < 0.001) between birds and mammals. The association between groups regarding the threatened species numbers showed similar low correlation values, below r = 0.595 (p < 0.001) for all pairs of classes analysed using global threat categories, and under r = 0.467 (p < 0.001) when using the national ones.
### Comparing true diversities among groups
The number of grid-cells that were sufficiently sampled to be considered for the analyses varied across scales and taxa (see Supplementary Table S3 for specific numbers at each spatial resolution). Yet, for each taxonomic group, only between 7–21% and 34–62% of the grid-cells were included, for the 25 × 25 and 50 × 50 km grid-cell size, respectively. Amphibians were the group with the highest coverage levels (at Cmax and C5%) regardless of the scale, while reptiles were the group with the lowest values of sampling coverage (Supplementary Table S3). The distribution patterns of the observed species-richness levels, across groups and scales, were usually not congruent with those of the species-richness standardised for sampling coverage at Cmax and C5% values (Fig. 3, Supplementary Fig. S5), though, the patterns of richness at Cmax and at C5% were generally consistent. For reptiles, birds and mammals, estimated species-richness was higher in northern areas, opposite to the peaks of observed richness seen in the coast (Fig. 3). For amphibians, maximum values of observed or estimated values of species-richness were more similarly distributed, occupying both northern and southern coastal regions (Fig. 3).
### Protected area network and hotspots of tetrapods
The network of protected areas (Supplementary Fig. S6) only partially encompasses hotspots in Uruguay (Supplementary Table S4). On average 56.7% of the richest grid-cells of observed tetrapod species-richness overlapped in part with a protected area (i.e., some extent of the hotspot was within a declared area), whilst overlapping decreased to 43.4% when considering richness standardised for sampling coverage at Cmax and to 48.6% for richness estimated at C5%. Mammals species-richness peaks were better covered (50–66.7%) while reptiles’ hotspots were the poorest integrated (16.7–50%). In the case of hotspots of endemism, we saw that on average for all tetrapods 54.6% of the peaks were located within protected areas. For threatened species number we found that an average of 58.8% and 65.6.3% of the peaks were covered, considering global and national IUCN assessments respectively. Hotspots of threatened species proportion, however, did not overlap with protected areas, except for the amphibian’ group for which we found a 25% of overlapping.
### Areas of ‘ignorance’
The spatial evenness in the distribution of the > 69,000 geographic records was low, and the levels of incompleteness per-area were considerably high (Fig. 4), with most of the territory (> 95.5%) identified as under-sampled (see SAC slope > 0.05; Supplementary Table S5; Supplementary Fig. S7). For amphibians, the data covered 61% of the territory (Table 1: grid-cell size 25 × 25 km), and yet, only two grid-cells can be considered as well sampled (SAC slope ≤ 0.5), covering only 0.3% of Uruguay’s area (Supplementary Fig. S7a). Birds, the group with the highest sampling effort, also had low levels of geographic coverage with 29.5% of the area unsampled (Table 1, Supplementary Fig. S7c), and with 20 grid-cells (4.5% of the national territory) considered as well-sampled. Reptiles and mammals showed the highest spatial coverage, with 77.2% and 79.5% of the total area covered, respectively (Table 1, Supplementary Fig. S7b,d). However, mammals did not present any well-sampled grid-cell and reptiles only 1, covering 0.2% of Uruguay’s area.
A major part of the Uruguayan territory was considered under the high and very-high sampling priority categories and in average for all tetrapods 67.5% of the area has been completely neglected (Fig. 4, Supplementary Table S5). These areas were mostly concentrated in the centre lowlands of the country.
## Discussion
Our study provides a detailed empirical case revealing the severe consequences that the lack of open-access biodiversity databases can have for the implementation of effective conservation actions directed to a country’s biodiversity management. By focusing on the tetrapod biodiversity of Uruguay—one of America’s most neglected countries in terms of availability of scientific data on its biodiversity58—our results show how the non-systematic (i.e., lacking a structured strategic approach for symmetric coverage of areas) and geographically concentrated sampling in only a few areas (at the expense of the majority of the country’s surface) have prevented the opportunities to identify areas of potential conservation and management priority. To address this issue, we have created the first open-access biodiversity initiative (Biodiversidata) in Uruguay, which reveals the distribution of different hotspots of biodiversity, and shows low levels of congruence among these measures regardless of the spatial scale or the IUCN assessment level used to calculate threatened species hotspots (i.e., national or global). Additionally, we identified well-surveyed sites, spatial gaps, and priority areas for future sampling efforts of amphibians, reptiles, birds and mammals in Uruguay. Thus, we believe that the novel evidence presented in our study will provide a critical scientific tool to effectively allocate resources for the exploration and monitoring of Uruguay’s biodiversity, and to ultimately enhance the efficiency in the process of evidence-based decision making towards conservation.
### Biodiversity hotspots: real or fabricated?
We found that number of species and endemism tend to concentrate around southern coastal cities. Studies performed at regional scales (e.g., sampling units > 2500 km2), have reported positive correlations between human population density and species-richness60,61. However, we cannot distinguish if the distribution of hotspots surrounding Uruguay’s major Atlantic-coast cities are a true pattern or an artefact product of sampling effort62, given that similar diversity levels could be found at other locations if sampling was as intense. By assessing true diversities (i.e., species-richness for a standardised sampling coverage), we found that the number of comparable areas is highly limited, yet those analysed tended not to exhibit the same distribution of species-richness peaks across scales and taxa. Thus, the question whether richness levels in the centre of the country are biologically (rather than artifactually) low, remains open, given the predominant knowledge gaps at these locations. Likewise, the high spatial correlation of species-richness for reptiles and amphibians, and of endemism for birds and mammals, cannot be disentangled from the effect of what could be essentially coordinated efforts of data collection linking these groups (e.g., herpetologist collectors). Importantly, incompleteness in the inventories may also correspond to existing knowledge that is not digital or accessible. In this sense, our analyses consider all the existing information that is possible to analyse, data that we rescued63 and made available.
### Spatial incongruence among hotspots of biodiversity
Discrepancies in the congruence of hotspots of biodiversity have previously been reported for other regions37,64,65,66,67. As the establishment of protected areas usually relies on the use of species-richness as a proxy of biodiversity39, the lack of congruence between different cross-taxon metrics debilitates the premise that a subset of taxa or features can be representative of biodiversity for conservation planning. Our findings suggest that the patterns we observe in Uruguay are strongly influenced by historical biases in sampling efforts that have dominated scientific practice. In this context, it is critical for future conservation assessments that we are able to quantify the spatial distribution of the different hotspots types, across taxa and at different scales of analysis40, to accurately recognise the limitations of the selection of reserve areas for the whole biodiversity protection and to monitor their effective representativeness.
### Towards effective conservation of biodiversity
The conservation prioritisation and planning strategies adopted by nations around the world are myriad. However, they all depend on comprehensive, high-resolution, up-to-date spatial information about species, ecosystems, and ecosystem services. The National System of Protected Areas of Uruguay was created in 2000, inaugurating the newest protected area system of Latin America. To prioritise which areas to include, the government collated a large biodiversity database (in grids of 660 km2—a very crude spatial resolution) that led to the decision of including specific areas despite the incompleteness of this nation-wide resource68—the Biodiversidata initiative aims to overcome this limitation58. In recent years, studies on the distribution of biodiversity have been performed using more complex quantitative methods54, yet, the data limitations have remained mostly the same. The bias that results from uneven sampling effort highly affects the estimation of richness69,70 and may lead to ineffective conservation prioritisation71,72, particularly in developing countries73. The efficiency of biodiversity conservation of the protected area system in Uruguay has not been tested. Precisely, in our study we observe that some areas need additional conservation attention to reach the most complete representation of the different tetrapod groups in the current network.
Uncertainty in the selection of suitable environments for conservation may lead to inadequate reserve selection and inappropriate habitat protection to higher extinction vulnerabilities74. Consequently, the allocation of investment for the study of neglected areas (particularly in countries such as Uruguay, where the dominant proportion of the country can be classed within this category) is likely to impact considerably on the efficiency of decisions and ultimately, on the expected outcomes75. Conservationists are often required to make decisions with incomplete and biased data, however, in order to improve and project better decision-making, there is an urgent need to focus in the knowledge gaps76. For instance, in the past 15 years, new species to science have been described and others have been recorded for the first time in the country expanding their distribution ranges, most of which are not considered on conservation prioritisation schemes. A range of new tetrapod species, including reptiles (Contomastix charrua77 and Liolaemus gardeli78), amphibians (e.g., Rhinella achavali79, Melanophryniscus langonei80 and Odontophrynus maisuma81), as well as first time species records including charismatic mammals (e.g., Puma yagouaroundi82 and Alouatta caraya83), amphibians (e.g., Leptodactylus furnarius84, Boana albopunctata85 and Physalaemus cuvieri86) and birds (e.g., Piculus aurulentus, Myiarchus tyrannulus and Anthus nattereri87, Ramphastos toco88 and Tyrannus tyrannus89).
### Future directions targeting knowledge gaps
Gaps in digital accessible information about the geographical distribution of species are a well-known and global issue28,90 that precludes from informing or monitoring the accomplishment of conservation targets across continents76. Open-access standardised datasets91 on species taxonomy, distribution, abundance, and evolutionary patterns remain largely unavailable in Uruguay, for all groups across the tree of life—this makes Uruguay one of America’s most neglected countries in this sense. Remarkably, our results reveal that for tetrapods, > 95.5% of the country’s land area remains insufficiently sampled. Thus, in the near future, biodiversity data mobilisation92 is amongst the greatest challenges the country will face93,94. Currently, the major scientific collections (i.e., Universidad de la República and the Museo Nacional de Historia Natural de Uruguay) are digitally inaccessible and, therefore, at latent risk of being lost95. Key efforts need to be made to support research institutions, researchers, policy makers and other stakeholders to digitise and store biodiversity data, and to guarantee its availability for evidence-based environmental planning and management76. Importantly, field research and data-sharing practices need to be encouraged. In this regard, our work provides a detailed roadmap of areas where to increase efforts for each tetrapod group. Lastly, as it is to many other non-western countries96, citizen science data (e.g., eBird, iNaturalist) has proven to bear a remarkable potential in documenting and monitoring biodiversity97,98, and therefore, the promotion of public engagement and knowledge democratization processes in countries like Uruguay can play an important role in channelling the needed scientific-culture change.
## Materials and methods
### Data
Geographic occurrence data of the tetrapods of Uruguay were collated from original sources collected by Biodiversidata expert members, from online databases and from the scientific literature (see details and protocols in Grattarola, et al.58). To avoid over-inflation of the data, all duplicates species per locality/year (i.e., same geographic coordinates) were removed. After this process of data filtering, the total number of records was 69,364 (Table 1), covering 664 tetrapod native species. This is the most geographically and taxonomically comprehensive database of Uruguay’s biodiversity that has been collected to date. The complete database is available at Grattarola, et al.99 and Grattarola, et al.100.
### Mapping biodiversity metrics and hotspots
We considered two diversity metrics to define biodiversity hotspots19 (number of species per area, or species-richness, and the proportion of species restricted to a particular area, or endemism); and two measures of species vulnerability (the proportion of threatened species relative to total species numbers, and the number of threatened species). We used the range-size-weighted species-richness (rswSR) as our measure of endemism, a parameter that considers the rarity/prevalence of the species over the study area. This enabled us to account for the predominance of species with restricted geographic distribution in the country, for simplicity we refer to it as ‘endemism’. The rswSR was calculated following Roll, et al.32 Eq. (1),
$${\text{rswr}}_{{\text{i}}} = \sum_{{\text{j}}} {\text{q}}_{{{\text{ij}}}} ,$$
(1)
where qij is the fraction of the distribution of the species j in the cell i. Threatened species number was calculated counting the number of species listed as threatened and threatened species proportion as the fraction of species listed as threatened per grid-cell, following Böhm, et al.18 Eq. (2),
$${\text{Prop}}_{{{\text{Threat}}}} = \left( {{\text{CR}} + {\text{EN}} + {\text{VU}}} \right){\text{/N}},$$
(2)
including critically endangered (CR), endangered (EN) and vulnerable (VU) categories by the total number of species (N). For both measures of threatened species we used the IUCN Red List of Threatened Species global assessment101 and the IUCN Red List national assessments for amphibians and reptiles102, and birds103. We only used the global assessment for mammals given there is no IUCN assessment for this group at the national level. Thus, threatened species analyses combining all tetrapods were done considering only global categories.
We defined hotspots as a measure of the spatial distribution of diversity/vulnerability metrics (i.e., species-richness, endemism or threatened species proportion and number), as a function of grid-cells rather than an arbitrary cut-off point (e.g., 2.5% of the richest areas). Therefore, we assumed that hotspots were the highest extremes of a gradient of continuous variation.
Analysing spatial patterns using different scales of observation can be useful when the size of the unit (grid-cell in this case) at which the spatial structure can be characterised is unknown104. Thus, we performed all analyses using three different sizes: 50 × 50, 25 × 25 and 12.5 × 12.5 km. Although all the analyses were made for the three different grid-cells sizes, we report results from the analyses with the 25 km grid-cell size (see Supplementary Fig. S2,S3 for analyses with the other two grid-cell sizes). All maps for each group separately (amphibians, birds, reptiles and mammals) and for all tetrapods combined were created using ArcGis 10.6. Sampling effort was evaluated as the number of records in each cell (after filtering for pseudo-replication) and species-richness as the number of species corresponding to those records.
### Hotspots congruence
We assessed the extent of congruence for each hotspot type (species-richness, endemism and threatened species proportion and number), within each group and across the tetrapod group, by calculating the number of overlapping grid-cells37. In cases where data are not randomly distributed, measuring metrics with small sampling units will increase the variance while using large sampling units will reduce the variability104, which therefore limits our capacity to determine the congruence among hotspots. For this reason, to analyse the extent of congruence between the biodiversity hotspots we varied both the size of the sampling unit (12.5 × 12.5, 25 × 25 and 50 × 50 km) and the criterion to define a hotspots (% of area/number of cells occupied by hotspots). First, grid-cells without records were removed. Then we sorted all the cells from high to low values of the corresponding metrics. Finally, at each definition criterion (from 0 to 100% overlapping over the total area by 0,5%), we computed the percentage of congruence as the number of matching grid-cells over the total number of unique cells. See the script with a working example in Grattarola105.
### Spatial correlations
We assessed the spatial association between: (1) sampling effort versus each hotspot type, (2) pair of hotspot types within each tetrapod group (e.g., amphibians’ species-richness versus amphibians’ endemism), and (3) pair of tetrapod classes within each hotspot type (e.g., reptiles’ endemism vs. birds’ endemism). Ecological data often have some degree of spatial structure104,—grid-cells can show a tendency to have similar values for a given variable with closely distributed grid-cells (i.e., spatial autocorrelation). Therefore, it is important to control whether spatial autocorrelation exists among grid-cells for each hotspot metric, and if so, estimate the ‘effective sample size’ given the dependency among the values104. Thus, to measure the association between the number of records and the biodiversity metrics per grid-cell we used a corrected Pearson’s correlation for spatial autocorrelation104,106 of the ‘SpatialPack’ R package107. Cells without records were eliminated from all correlations to remove double zeros.
### Comparisons of observed and estimated diversities
Although the question of whether patterns of species-richness are real (the recovered pattern represents the true distribution of biodiversity) or fabricated (the recovered pattern is an artefact of the distribution of sampling efforts) cannot be accurately assessed for severely under-sampled assemblages, we can still infer species-richness for a standardised coverage and make comparisons between observed and estimated diversity values108,109. Fair comparisons across multiple groups can be performed using coverage-based rarefaction and extrapolation sampling curves up to a maximum value of Cmax (i.e., the level of coverage reached by the sample that attains the lowest coverage when all samples are extrapolated to double the reference sample size)109.
We defined the frequency of species incidence at two spatial resolutions: 25 × 25 and 50 × 50 km grid-cells (we did not use the 12.5 × 12.5 km resolution due to low numbers of available grid-cells with data for analyses). Each grid-cell was further divided into sub-grids of 1 × 1 km to create a species incidence dataset at the grid-cell level by counting the number of sub-gridded cells that contained occurrence records for individual species (see the script with a working example in Grattarola105). For interpolation (rarefaction) and extrapolation of species-richness (Hill's number of order q = 0) we used the R package ‘iNext’110. To obtain reliable estimates, grid-cells with few occurrence records were excluded from analyses111: whenever the number of species observed in the grid was less than six, the number of sub-gridded cells with at least one incidence was less than six, and the total number of species incidences was equal to the number of unique species (species that are each detected in only one sub grid-cell). To compare among grid-cells in terms of sampling coverage, we estimated species-richness for each tetrapod group at Cmax109 and at 5% percentile (C5%) of sampling coverage at doubled sample sizes111. To determine Cmax, each sample within the grid unit was first extrapolated to double the reference sample size, then Cmax was calculated as the minimum among the coverage values obtained from those extrapolated samples. See Supplementary Table 3 for specific values at each group and spatial resolution.
### Congruence between protected areas and hotspots of biodiversity
To determine whether the location of existing protected areas in Uruguay (as established by the current National System of Protected Areas of Uruguay) covers hotspots of species-richness, endemism and threatened species, all maps were overlapped with the 16 currently operating protected areas (see Supplementary Figure S4 for a map showing: the network of protected areas, areas under assessment for potential consideration as protected areas, and areas for which a proposal for consideration has been prepared). Congruence for each hotspot type was then calculated as the proportion of the 2.5% of the richest grid-cells, removing empty grid-cells, that were at least partially covered by a protected area (i.e., some amount of the area was within a hotspots).
### Identification of ‘areas of ignorance’
As a critical first step to interrogate historically-established public policies and efforts about the sampling of Uruguay’s biodiversity, we quantified the levels of inventory incompleteness for each group per area by using curvilinearity of smoothed species accumulation curves (SACs)112,113. This method assumes that SACs of poorly sampled grid-cells tend towards a straight line, while those of better sampled ones have a higher degree of curvature114. Smoothed SACs were calculated with the method ‘exact’ of the function ‘specaccum’ in the vegan R package115. As a proxy for inventory incompleteness we calculated the degree of curvilinearity as the mean slope of the last 10% of SACs112. Steep slopes (values close to one) reflected high levels of incompleteness, whereas shallow slopes (values close to zero) indicated saturation in the sampling and thus low levels of incompleteness. We considered grids with slope values > 0.05 to be under-sampled and those with slope values ≤ 0.05 to be well sampled. The R scripts used for these analyses can be found at Grattarola105.
Finally, with the aim to outline a plan to suggest where future sampling efforts should be allocated across the territory of Uruguay (e.g., for funding considerations), we generated a map of ‘priority areas of sampling’ for each tetrapod group. Priority levels were established considering the levels of inventory incompleteness. The scale ranged from: Null (i.e., grid-cells where mean slope of the last 10% of SACs was lower or equal to 0.05), ‘Low’ (between 0.05 and 0.25), ‘Medium’ (between 0.25 and 1), ‘High’ (grid-cells where the sampling effort was so low that it was not possible to calculate SACs), and ‘Very High’ (i.e., grid-cells where no records were found).
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https://math.stackexchange.com/questions/1710385/time-shifted-trigonometric-fourier-series-coefficients
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# Time-Shifted Trigonometric Fourier Series Coefficients
I'm trying to find the Trigonometric Fourier series coefficients for a particular periodic function.
Given
$$f(t) = 2 - \frac9\pi \sum_{n=1}^\infty\frac1{2n-1}\sin\left(\frac{2n-1}2 \pi t\right)$$
The $a_0$, $a_n$, and $b_n$ coefficients would be
$$a_0=2$$
$$a_n=0 \ for \ all \ positive \ n$$
$$b_n=-\frac9\pi\frac1{2n-1} \ for \ all \ positive \ n$$
But in the case that I need to find the coefficients for $f(t - 1)$, how would I go about doing it?
My first thought was to substitute the value of $t - 1$ into the equation, followed by trigonometric identity to expand the function, giving:
$$f(t-1) = 2 - \frac9\pi \sum_{n=1}^\infty\frac1{2n-1}\sin\left(\frac{2n-1}2 \pi (t-1)\right)$$
$$= 2 - \frac9\pi \sum_{n=1}^\infty\frac1{2n-1}\sin\left(\frac{2n-1}2 \pi t - \frac{2n-1}2 \pi \right)$$
$$= 2 - \frac9\pi \sum_{n=1}^\infty\frac1{2n-1}\left[\sin\left(\frac{2n-1}2 \pi t\right)\cos\left(\frac{2n-1}2 \pi\right) - \cos\left(\frac{2n-1}2 \pi t\right)\sin\left(\frac{2n-1}2 \pi\right)\right]$$
At which point I get stuck. I can't find a way to factorize the cosine and sine terms out to form a "proper" Fourier Series representation in order to determine the coefficients.
Is there something that I'm doing very wrong here? Any help you can provide would be great! Thanks in advance!
• What do you mean by "proper"? It sure is a Fourier series (the general form is $$a_0/2+\sum_{n=1}^{\infty}\Big(a_n\sin\frac{n\pi x}{L}+b_n\cos\frac{n\pi x}{L}\Big) ~).$$ – Nikolaos Skout Mar 23 '16 at 16:00
• Wait, so we can count ${\frac9 \pi} {\frac 1 {2n - 1}} {\sin\left(\frac{2n - 1} 2 \pi \right)}$ as $a_n$, and $-{\frac9 \pi} {\frac 1 {2n - 1}} {\cos\left(\frac{2n - 1} 2 \pi \right)}$ as $b_n$? Huh...did not know that... Always thought the coefficients weren't supposed to contain functions... – Arjuna Mar 23 '16 at 16:09
• basically, note that the expressions $\sin(\ldots),~ \cos(\ldots)$ you noted above are not functions but ... numbers (there is no $x$)! – Nikolaos Skout Mar 23 '16 at 16:10
• Right. That just kicked me in the head just now. Thanks a lot for your help! – Arjuna Mar 23 '16 at 16:13
• you're welcome ;-) ! – Nikolaos Skout Mar 23 '16 at 16:14
You are actually very close to solving the problem
$$f(t-1) = {2 - \frac9\pi \sum_{n=1}^\infty\frac1{2n-1}\left[\sin\left(\frac{2n-1}2 \pi t\right)\cos\left(\frac{2n-1}2 \pi\right) - \cos\left(\frac{2n-1}2 \pi t\right)\sin\left(\frac{2n-1}2 \pi\right)\right] }$$
$$\space$$
The new coefficients are as follows
$$a_0 = 2$$
$$a_n = \frac{9}{\pi\left(2n-1\right)}\sin\left(\frac{2n-1}2 \pi\right)$$
$$b_n = \frac{-9}{\pi\left(2n-1\right)}\cos\left(\frac{2n-1}2 \pi\right)$$
You might think that it is wrong to have $$cosine$$ and $$sine$$ factors in $$a_n$$ and $$b_n$$, but it is perfectly acceptable as these are not functions of the variable $$t$$, so relative to $$t$$ they are constants.
$$\space$$
Note that the two trigonometric factors that do not contain the variable $$t$$ can be simplified as follows
$$\cos\left(\frac{2n-1}2 \pi\right) = 0 \space, \forall \space n \in \mathbb{Z^+}$$
$$\sin\left(\frac{2n-1}2 \pi\right) = (-1)^{n-1} \space, \forall \space n \in \mathbb{Z^+}$$
$$\space$$
This allows us to simplify $$a_n$$ and $$b_n$$ as follows
$$a_n = \frac{9(-1)^{n-1}}{\pi\left(2n-1\right)}$$
$$b_n = 0$$
|
2019-09-16 21:00:09
|
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https://harryrschwartz.com/2019/11/16/canadian-permanent-residency
|
# Harry R. Schwartz
Software engineer, nominal scientist, gentleman of the internet.
Member, ←Hotline Webring→.
· 1B41 8F2C 23DE DD9C 807E A74F 841B 3DAE 25AE 721B
ovo-lacto vegetarian
he
him
his
### Canadian Permanent Residency
Published .
Tags: personal, travel.
I recently applied for, and received, Canadian permanent residency. Several friends have expressed interest in doing the same, so this article describes the steps involved and my experience with the process.
#### Why do it?
Well, first, my fiancée Jenny is a dual citizen, and we’d love to spend (at least) a few years up north! British Columbia is beautiful, with tons of options for mountaineering and skiing and cold-water diving, and Vancouver has a reasonably good tech scene. I’ve never lived outside the US for more than a month or two before, so it would be a good opportunity to grow a bit. And, of course, who doesn’t love a civilized healthcare system?
Second, American politics have started feeling a little higher-stakes than usual over the past few years. Jenny’s a US citizen, but she’s also an immigrant, which isn’t a comfortable thing to be just now. Without sounding too paranoid, one of the lessons of history is that situations can deteriorate surprisingly quickly. Obtaining residency takes about a year, so I didn’t see a good reason to delay.
#### What’s the process?
For context, I’m a fairly young US citizen and a native English speaker. I’m also an experienced software engineer with a graduate degree. I’m absolutely playing the immigration game in Easy Mode. If you’re in a similar situation, this might be a useful guide for you. If not, well, I apologize and sincerely wish you the best—I hope this is still helpful, but I can only speak from my own experience.
I applied through Canada’s Express Entry program for skilled workers. This process has three stages:
1. First, I applied to join the pool of candidates.
2. After being accepted into the pool I was invited to submit a more exhaustive application for permanent residence.
3. Once that was approved, I needed to visit Canada to formally register as a permanent resident.
This process involved a lot of paperwork and coordination, but it was fundamentally easy. My application’s status and next steps were readily available online through a portal run by IRCC (that’s “Immigration, Refugees, and Citizenship Canada,” the Canadian counterpart to the USA’s INS).
Notably, I didn’t even need to visit Canada before I was granted my Confirmation of Permanent Residence, the penultimate step before the formality of a landing. Let those who claim US immigration is too lax take note!
#### Entering the Express Entry pool
First, you need to apply to enter the Express Entry pool as a skilled worker. To do that, you’ll first need to:
• Take an English test through IELTS and forward the results to IRCC, and
• Send your college transcripts to a credentialing agency, who will verify that your foreign degree is equivalent to a Canadian degree. Again, the agency should forward the results to IRCC. I used WES to handle my verification. They were adequate, but it took at least a month.
Once both of those reports have been sent to IRCC, apply as a federal skilled worker on their website. You’ll need to provide some verification numbers to link the reports to your application. Once you’ve submitted the application, you should hear back within an hour or so (it’s automated, I think).
If you’ve met the requirements for Express Entry you’ll be assigned a CRS number (“Comprehensive Ranking System”) which numerically represents your desirability as an immigrant. You’ll also enter the Express Entry pool.
Every two weeks or so, IRCC selects the top candidates in the pool (for the last year or two that’s been the 3,900 candidates with the highest CRS scores, which are usually above about 440–450) and invites them to apply for permanent residency (via the “invitation to apply,” or “ITA”).
As I said before, I’m fairly young, well-educated, a native English speaker, and working in a high-demand profession. While I’m currently engaged to a lovely Canadian citizen, that fact wasn’t taken into account in the application. But the other factors were more than sufficient, my CRS score was 465, and I was invited to apply in the first round after I entered the pool.
#### Applying for permanent residency
The full permanent residence application involves verifying all your claims, so it’s much more thorough than the Express Entry application. It’s also quite a bit more hectic, since invited candidates have only 60 days to submit a full application before being thrown back into the pool.
This full application involves:
• Getting an official letter from every employer over the last ten years verifying your salary, title, responsibilities, and dates of employment,
• providing details about every international trip you’ve taken in the last decade, with scans of your passport stamps and visas,
• proof of means of support, in the form of official letters from your banks and investment brokers,
• a medical exam performed by a certified doctor (of which there is exactly one in the state of Colorado and who charges accordingly),
• scans of your diplomas, and
• a police certificate from the FBI detailing your criminal record. This is utterly routine and just done through the mail, but involves the thrill1 of fingerprinting yourself.
A couple of months after sending in the application, IRCC requested my biometric information. That involved getting fingerprinted at a local USCIS office. There was one within an hour’s drive, and the process wasn’t too complicated. I didn’t even need an appointment.
If everything’s shipshape you’ll be asked to send a self-addressed envelope and some Canadian-sized passport-style photos to IRCC. They’ll use that envelope to return your Confirmation of Permanent Residence (“CoPR”) letter. The photos have fussy requirements and I’m useless with a camera, so I threw up my hands and had them done at a local camera shop which guaranteed that the requirements were met.
Getting a self-addressed envelope that can be sent back across international borders was similarly trickier than I expected, and I had to create an account on UPS’s Web site and call up their customer service to have them walk me through the process.
My CoPR arrived at my place in the States a few weeks after sending off the paperwork.
#### Landing in Canada
You’ll need to visit a Canadian port of entry to present your CoPR and actually become a permanent resident. That process is called a “landing.” And you’ll need to do your landing fairly promptly, since the CoPR expires one year after your medical exam and therefore probably about five months after receiving your CoPR.
The landing can be done at any Canadian border or airport.2 Go to immigration as you usually would, tell the officer that you’re there for your landing, and they’ll funnel you through the process.
Bear in mind that if you’ll immediately want health care coverage you should avoid landing in Ontario or British Columbia; both of those provinces have a three-month waiting period before new residents can get coverage. Other provinces have different rules—Alberta gives coverage immediately, I believe, for example—so you’ll want to look into that if that’s something you’ll need.
Nevertheless, we decided to do my landing in Vancouver. We were still living in the States, so I didn’t need Canadian insurance immediately, and we wanted to scope out the city as a potential place to live. The landing itself took an hour or so and was thoroughly painless.
Within 60 days of your landing you’ll be mailed your permanent resident card (colloquially, your “maple leaf card”). However, IRCC can only post to a Canadian address, so you’ll need one of those! Since we weren’t living in Canada, and therefore didn’t have a mailing address, I had the card sent to a family friend living in British Columbia who forwarded it to me in the States. IRCC formally discourages mailing PR cards across borders, since they could be lost, but I couldn’t find any specific prohibition against it and it seems to be fairly common practice.
Once you’ve got permanent residency, the next steps are to apply for a Social Insurance Number, which you’ll need to work in Canada, and a provincial healthcare card. But that’s outside the scope of this article!
#### How long did it take?
I started my application at the end of June 2018 and received my permanent resident card in October 2019. So, in my case, about 16 months. I made a few missteps that slowed the process down; notably, I started applying before I’d taken my English test, and later I didn’t realize I had to have my transcripts verified. Those errors probably added two or three months to the process, especially since I could have done them concurrently. I also could have done my landing immediately after receiving my CoPR, which would have saved another couple of months, but we had some other commitments scheduled.
An eligible and dedicated candidate could probably go through the whole process in a bit under a year, but I think it would be hard to do it much faster than that. If you think you might want residency a year from now, it’s definitely time to get started!
#### What did it cost?
There were some other miscellaneous expenses here and there (countless stamps, transportation to the IELTS testing site and USCIS office, and so on), but these were the main ones:
• IELTS English test: 203 USD
• Education credential verification through WES: 159 USD (plus a bit more to buy copies of my transcripts from schools)
• FBI report: 18 USD
• Medical exam: 500 USD
• Permanent residence application fee: 550 CAD (about 410 USD)
• “Right of Permanent Residence” fee (paying for the card, essentially): 490 CAD (about 366 USD)
• Flights and hotels in Vancouver: I paid for these with travel points (thanks, churning)!
The whole process cost about \$1,700 and maybe 40 hours of my time. Neither of those numbers includes the trip to Vancouver. That’s objectively a big chunk of change, but I think it’s totally worth it to give us the experience of indefinitely settling in a whole new country.
If you need more tangible value to convince you, well, I encourage you to consult your annual health insurance bill and compare.
#### What are your next steps?
We’re happily living in Colorado at the moment, but we’re considering giving up our apartment next summer and spending a few months traveling. We’re both working remote jobs, so it seems like we should take advantage of that opportunity.
After that, though, we’re discussing making the move north! One of the conditions of maintaining permanent residency is living in Canada for two years out of every five.
An interesting option is that it’s possible, after living in Canada for only three years, to apply for citizenship. That would be dual citizenship: Canada isn’t a jealous nation, so you don’t need to give up your previous passport to get a Canadian one. And citizenship doesn’t expire, of course. We’ll see!
1. Another thrill is that this verifiably-not-a-criminal certification arrives as a formidably official-looking document. It’s now framed and hanging above my desk. Rest easy, visitors: I am provably not a felon.
2. Though, as often seems to be the case, Quebec may have additional requirements. I’m not sure about that.
You might like these related articles:
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2020-01-18 08:19:18
|
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http://developer.openmono.com/en/latest/datasheets/datasheets.html
|
# Datasheets¶
If you need to dive a little deeper into the inner workings of Mono, we have collected the datasheets for the components in Mono
You might need to consult specific datasheets for the components in Mono, if you are debugging or just need some advanced features not provided by the API.
## Accelerometer¶
Mono’s accelerometer is a MMA8652FC chip by Freescale. It is connected to Mono’s I2C bus.
The accelerometer is handled by the MonoAccelerometer class in the software framework. If you need specific features, or just wish to play with the component directly, you should consult the datasheet.
## MCU¶
Mono’s Micro Controller Unit (MCU) is a Cypress PSoC5LP, that is an Arm Cortex-M3 CPU. You can use all its registers and functions for your application, the SDK includes headers for all pins and registers. (You must explicitly include the project.h file.)
The MCU model we use has 64 Kb SRAM, 256 Kb Flash RAM and runs at 66 Mhz.
The software framework encapsulates most MCU features in the mbed layer, such as GPIO, interrupts and timers. Power modes is also controlled by the registers in the MCU and utilized by the PowerManagement class.
## Display Chip¶
The display is driven by an ILITEK 9225G chip. On mono we have hardwired the interface to 16 bit 5-6-5 color space and the data transfer to be 9-bit dedicated SPI, where the 9th bit selects data/command registers. (This should make sense, when you study the datasheet.)
In the framework the display controller class ILI9225G utilizes the communication and pixel blitting to the display chip.
## Wireless¶
Mono uses the Redpine Wifi chip to achieve wireless communication. (The same chip includes Bluetooth for the Maker model, also.) The chip is connected via a dedicated SPI interface, and has a interrupt line connected as well.
The communication interface is quite advanced, including many data package layers. You can find our implementation of the communication in the ModuleSPICommunication class. This class utilizes the SPI communication from and to the module, it does not know anything about the semantics of the commands sent.
## Temperature Sensor¶
The temperature sensor is an Amtel AT30TS74 chip, connected via the internal I2C bus.
The temperature interface is used in the AT30TS74Temperature class.
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2019-04-22 04:31:39
|
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http://rsnapshot.org/faq.html
|
### How do I restore files with rsnapshot?
If you have super-user access on the rsnapshot server, you can just copy the files from the snapshot root (eg: /.snapshots/daily.0/server/directory/file). The daily backups will be more recent than the weeklys, and the weekly more recent than the monthlys, etc. Your system administrator may have set up a read-only copy of the snapshot root (eg: with read-only NFS or read-only Samba), as suggested in the HOWTO. If so, it is better (safer and possibly more convenient) to copy files from this read-only copy of the snapshot root.
### I have a snapshot root or backup point with a space (or other special character) in it and this is not working at all with rsnapshot 1.3.1. Why?
rsnapshot version 1.3.1 has an issue where the rsync command is interpreted by a shell rather than directly executed by rsnapshot. This bug is expected to be fixed in rsnapshot CVS on 24 March 2009, and so should appear an a forth-coming release (probably version 1.3.2). This bug was not present in rsnapshot 1.3.0.
rsnapshot versions 1.2.9 and 1.3.0 have a bug where –link-dest is omitted in these circumstances for any backup where the .sync directory exists when the backup is started. This bug was fixed in rsnapshot version 1.3.1. With older versions of rsnapshot, please take care to avoid this combination, for example by disabling link_dest or sync_first.
### Why aren’t sync_first and include_conf documented in the rsnapshot HOWTO?
That’s a good question. sync_first and include_conf were new features added between rsnapshot 1.2.3 and 1.2.9. Unfortunately the maintainers have been short of time and haven’t found time to update the HOWTO. If updating the HOWTO is something you can help with, please tell the rsnapshot-discuss list. At least you can find some documentation for these features in the rsnapshot man page.
### Where do I report a bug in rsnaspshot?
You can report bugs, request features and submit patches to the Github issue tracker.
### How do I backup from Windows machines to Linux?
You can run rsnapshot on a Linux machine (where the backups will be stored), and run cwRsync Server on the Windows machines so that you can connect to them with ssh protocol or rsyncd protocol. Many people have reported backups hanging in the middle of a backup if they use ssh protocol to Windows machines. You should be able to avoid these hangs by using rsync://user@host/dir (rsyncd protocol) in your backup line rather than user@host:/dir (rsync over ssh protocol). But bear in mind that rsyncd protocol is unencrypted, in case you are transferring over an untrusted network (like the Internet). There is a suggestion secure connections between linux rsync clients and cwRsync servers in the cwrsync faq, but it would need to be adapted for rsnapshot and customised for your environment.
### I run rsnapshot for the first time, but nothing happens. Why?
rsnapshot does two major things - actual backup (with rsync) and rotation (moving snapshots around). Before it can do rotations, it needs to have at least one actual backup. So you need to understand which rsnapshot invocation will make an actual backup.
For example, if you have sync_first enabled, then you need to run rsnapshot sync (which makes a backup) before you can do a rotation like rsnapshot hourly or rsnapshot daily. With sync_first enabled, all intervals (hourly, daily, etc) just do rotation.
If you do not have sync_first (it is disabled by default), then the backup is made by the lowest interval (that is, the first one that you listed in your rsnapshot.conf). The other (higher) intervals do rotation. For example, if you have intervals hourly, daily, weekly and monthly, then you need to run rsnapshot hourly to do a backup before the other intervals (daily, weekly and monthly which do rotations) will do anything.
In fact, you need a complete set of hourly backups before a rsnapshot daily will do anything. Similarly, you need a complete set of daily backups (usually 7) before rsnapshot weekly will do anything.
You should be able to fix the warnings by installing the perl Lchown module. Get it from the CPAN module-page. Or if you don’t care about symlinks having the wrong ownership in your snapshots, then you could ignore the warnings.
### How do I exclude files/directories with spaces in their names, like Documents and Settings?
You can make use of the wildcard matching and replace the space with a ?, for example: exclude=Documents?and?Settings/
### My rsnapshot setup seems to eat the processor on the machines I’m backing up from. How can I prevent this?
rsnapshot itself is a low-overhead program, but rsync can drive processor utilization uncomfortably high. To address this, tell your rsync to run with high nice and ionice values, like 10 and -c3 respectively.
Depending on how rsync was packaged for your system, your installation may have an /etc/default/rsync file. If it does, set RSYNC_NICE and RSYNC_IONICE as recommended. Then restart the rsync daemon. You should notice a difference immediately.
Thanks to Eric Raymond for writing this entry!
### I’m trying to backup up Windows machines onto a Linux box using rsnapshot (on Linux) and cwrsync as a daemon (on Windows). When I back things up, though, it doesn’t seem to delete files in the backup that were deleted in the source, so my backups keep getting bigger and bigger. How can I fix this?
The simple-but-dangerous approach is to add –ignore-errors to your rsync_long_opts line in your rsnapshot configuration file. The tedious-and-fragile-but-probably-safer approach is to find out where the I/O errors are occurring in the backup and exclude those files from the backup set. It seems that not all cwrsync and rsync versions handshake well on some Windows filenames, such as “Shortcut to 3½ Drive” (chokes on the ½ character), some URLs held in Internet Explorer’s cache (chokes on length?), some entries in the Recycle Bin (chokes on the curly braces?). If there’s any I/O error, it seems that rsync elects not to honor the –delete that rsnapshot passes in, unless you also say –ignore-errors, which could get you in trouble when real errors occur.
Thanks to Mark Murphy for writing this entry!
### Can I set the snapshot_root to a remote SSH path? I want to push my backups to a remote server, rather than pull them from a remote server.
Rsnapshot does not support a remote snapshot root via SSH. However you should be able to use a remote snapshot root that is NFS mounted on the machine that runs rsnapshot but hosted on another machine (NFS server).
If you are running rsnapshot as user root (which is the normal case), make sure that the NFS server allows root access for that NFS mount to the rsnapshot machine as an NFS client. For a solaris NFS server, see root= in share_nfs(1). For a Linux NFS server, see no_root_squash in exports(5). Otherwise you might get errors about chown, removing directories/files, etc if permissions on the NFS server are mapped from “root” to “nobody”.
For advanced users, Matt McCutchen suggested the following alternative. What you can do instead is put the rsnapshot configuration file on the destination server (with a local snapshot root), allocate a “staging” area on that server, and define it as the sole backup point. To make a snapshot, execute an ordinary rsync push to the staging area and then invoke rsnapshot on that server to incorporate the new data into a snapshot. To do this conveniently over SSH, create the following script rsync-and-kick-rsnapshot on the destination server:
#!/bin/bash
rsync "$@" && rsnapshot$interval
And then pass
--rsync-path=*/path/to/*rsync-and-kick-rsnapshot.
With an rsync daemon, create this script kick-rsnapshot:
#!/bin/bash
if [ "$RSYNC_EXIT_STATUS" == "0" ]; then rsnapshot$interval
fi
And specify it as the post-xfer exec command for the module containing the staging area.
If multiple machines need to be backed up, instead of trying to coordinate their pushes and rotations, the easiest thing to do is to make a completely separate rsnapshot configuration file and snapshot root for each. Some more hints, including two approaches to avoid the use of double the disk space on the destination server, may be found in this message and this message.
### I’m backing up to a firewire or USB external drive. When the drive isn’t mounted, rsnapshot writes all the backups to the /mnt/ directory on the local hard drive, filling up the disk. How can I prevent this?
Set the no_create_root option to 1 in the config-file.
### How do I set up public keys for ssh so that rsnapshot can run from cron?
Run commands like these on your rsnapshot server as root (or whichever user you run rsnapshot as from cron) :-
rsnapshotserver# ssh-keygen -N "" -f ~/.ssh/rsnapshot_dsa
rsnapshotserver# ssh-copy-id -i ~/.ssh/rsnapshot_dsa.pub root@<I>client1</I>
rsnapshotserver# ssh-copy-id -i ~/.ssh/rsnapshot_dsa.pub root@<I>client2</I>
To make rsnapshot use this DSA key, add
ssh_args -i /root/.ssh/rsnapshot_dsa
to your config-file (assuming you don’t have ssh_args currently).
### After I’ve taken a few snapshots, why do they all show up the same size in df? I thought rsnapshot was meant to only take one full snapshot and then a bunch of incrementals?
You thought right, and it does! It looks like you’ve got a bunch of full snapshots because any file which hasn’t changed between two consecutive snapshots will be a hard link, so potentially several directory entries in consecutive snapshots may actually point at the same data on disk, so the only space taken up by a snapshot is whatever is different between it and the previous one. Using the du utility to gauge your disk usage will normally produce confusing results because of this. The rsnapshot-diff utility solves this. You could also use the -c argument to du.
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2017-05-23 06:51:08
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https://www.math.ias.edu/seminars/abstract?event=44685
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On the Conjectures of Nonnegative $k$-Sum and Hypergraph Matching
Computer Science/Discrete Mathematics Seminar II Topic: On the Conjectures of Nonnegative $k$-Sum and Hypergraph Matching Speaker: Hao Huang Affiliation: University of California, Los Angeles; Member, School of Mathematics Date: Tuesday, October 9 Time/Room: 10:30am - 12:30pm/S-101 Video Link: https://video.ias.edu/csdm/huang
A twenty-year old conjecture by Manickam, Mikl\'os, and Singhi asked whether for any integers $n, k$ satisfying $n \ge 4k$, every set of $n$ real numbers with nonnegative sum always has at least $\binom{n-1}{k-1}$ $k$-element subsets whose sum is also nonnegative. In this talk we discuss the connection of this problem with an old question by Erd\H{o}s, who asks to determine the maximum possible number of edges in a $k$-uniform hypergraph on $n$ vertices with no matching of size $t$, and with the question of estimating the probability that the sum of nonnegative independent random variables exceeds its expectation by a given amount. Using these connections and probabilistic techniques, we verify the Manickam-Mikl\'os-Singhi conjecture for $n \ge 33k^2$. In the remaining time of the talk, we will give an overview of the progress on Erd\H os' conjecture, and discuss its application to a problem of existence of perfect matchings in uniform hypergraphs, and to a question about finding an optimal data allocation in a distributed storage system. This talk is based on joint works with Alon, Frankl, Loh, R\"odl, Ruci\'nski and Sudakov.
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2018-06-25 07:45:56
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https://www.physicsforums.com/threads/the-need-for-the-dirac-delta-function.765162/
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# The need for the Dirac delta function
1. Aug 8, 2014
### albega
So part of the idea presented in my book is that:
div(r/r3)=0 everywhere, but looking at this vector field it should not be expected. We would expect some divergence at the origin and zero divergence everywhere else.
However I don't understand why we would expect it to be zero everywhere but the centre, because if you draw it, the arrows get smaller as we move out radially. If you consider placing a little cube somewhere in the field not at the centre, the arrows entering that cube would be larger than those leaving. Surely that would give a negative divergence at these points. I obviously understand why it should be large at the centre.
This leads me to my other point. Consider the r/r2 field - it is similar to the above field, just falling off less rapidly. This field has 1/r2 divergence however. My first issue is, why does it not give negative divergences by my above argument of arrows into a little cube, and because of the similarities for the above field, why does it not give zero divergence apart from at the origin? Secondly, why does the maths behave so much more nicely for such a similar field?
Thanks for any help :)
2. Aug 8, 2014
### Ray Vickson
You say "However I don't understand why we would expect it to be zero..."; well, have you actually sat down and computed $\text{div}\: (\vec{r}/r^3)$ at $\vec{r} \neq \vec{0}$? If you keep track of things during the calculation you will see how and why some cancellations occur, leaving you with 0.
3. Aug 8, 2014
### vanhees71
This is an utmost important exercise. So you should do it very carefully. The correct equation to prove is
$$\vec{\nabla} \cdot \frac{\vec{r}}{r}=4 \pi \delta^{(3)}(\vec{r}).$$
The proof for $\vec{r}$ is almost trivial, just brute-force derivatives will do.
To check that the $\delta$-distribution is correct, use Gauss's Law for an arbitrary volume containing the origin in its interior with a ball of infinitesimal radius around the origin taken out to integrate the expression on the left-hand side together with an arbitrary test function.
4. Aug 9, 2014
### vela
Staff Emeritus
Because of the spherical symmetry of this scenario, a cube isn't the best choice to use to analyze the situation. Try using the infinitesimal volume element for the spherical coordinate system instead.
For an infinitesimal cube, I expect any differences in flux correspond to second-order terms and can therefore be neglected as the volume of the cube tends to 0. For a finite cube, I suspect the difference in angle at which the field lines intersect the surfaces is just enough to cancel out the effect of the changing field strength. It's left to the reader (you) to verify this is indeed the case.
In this case, the field doesn't fall off fast enough with increasing $r$ so that there's a net positive flux.
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2017-10-18 05:11:04
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http://diskopumkmbandunggoid.somee.com/cheap-term-paper/page-340-2020-12-03.html
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The Karatsuba algorithm is Multiplication algorithm - Wikipedia fast multiplication Multiplication algorithm - Wikipedia. It was discovered by Anatoly Karatsuba Multiplication algorithm - Wikipedia and published in The Karatsuba algorithm was the first multiplication algorithm Multiplication algorithm - Wikipedia faster than the quadratic "grade school" algorithm. Kolmogorov was very excited Springsummer2013vista by Villanova the discovery; he communicated it at the next meeting of the seminar, which was then terminated.
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2021-04-10 12:01:53
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https://mathematica.stackexchange.com/questions/195665/select-two-of-three-in-a-triplet-list-x1-y1-z1-x2-y2-z2
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# Select two of three in a triplet list {{x1,y1,z1}, {x2,y2,z2}} [closed]
I have a list of triplets {{x1,y1,z1}, {x2,y2,z2}.....} and would like to plot $$x$$ vs. $$y$$, then $$x$$ vs. $$z$$. ListLinePlot will plot these if they were doublets, but how do I select pairs to plot?
## closed as off-topic by C. E., happy fish, m_goldberg, MarcoB, Henrik SchumacherApr 21 at 16:23
This question appears to be off-topic. The users who voted to close gave this specific reason:
• "This question arises due to a simple mistake such as a trivial syntax error, incorrect capitalization, spelling mistake, or other typographical error and is unlikely to help any future visitors, or else it is easily found in the documentation." – C. E., happy fish, m_goldberg, MarcoB, Henrik Schumacher
If this question can be reworded to fit the rules in the help center, please edit the question.
• Look at the documentation page for the Part function. You'll want to use something like data[[All, {1,2}]] for {x,y} and data[[All, {1,3}]] for {x,z}. – user6014 Apr 20 at 18:09
• Thank you -- I did try Part, but I'm still learning the fundamentals of Mathematica (so far, I love it!). I'll use your note as a template and come back with the results, when I give it another shot. Much appreciated! - Andy OK, so I tried it right away -- Yes, it does work. I was close, but it's not second nature yet. Thank you! I will return the favor to someone else..... – alaz Apr 20 at 21:11
Contrived data producing a list of 10 triples of the form {{1, y1, z1}, {2, y2, z2}, ...}.
SeedRandom[1]; data = Table[{i, Sequence @@ RandomReal[1., 2]}, {i, 10}];
Now we plot x & y and x & z.
ListLinePlot[{data[[All, ;; 2]], data[[All, ;; 3 ;; 2]]}]
or
ListLinePlot[{data[[All, {1, 2}]], data[[All, {1, 3}]]}]
Both display
Note: Look up Part and Span in the docs,
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2019-07-17 07:27:35
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https://yorikuwa.com/m1507/
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# 相関係数
スポンサーリンク
スポンサーリンク
① ② ③ ④ ⑤ ⑥ $$x$$ $$9$$ $$6$$ $$2$$ $$5$$ $$8$$ $$6$$ $$y$$ $$5$$ $$2$$ $$7$$ $$4$$ $$4$$ $$8$$
スポンサーリンク
スポンサーリンク
## 共分散と相関係数
Point：共分散と相関係数相関係数を求めるときは、２種類のそれぞれのデータの $$x$$ の標準偏差 $$S_x$$ と、$$y$$ の標準偏差 $$S_y$$ と、$$x~,~y$$ の共分散 $$C_{xy}$$ を求めて以下の式より求めます。
$$r=\frac{C_{xy}}{S_x\cdot S_y}$$
となります。
また、$$x~,~y$$ の共分散 $$C_{xy}$$ の求め方は、
$$x$$ の偏差 $$y$$ の偏差 $$x$$ の偏差× $$y$$ の偏差 $$\vdots$$ $$\vdots$$ $$\vdots$$
この表より、$$x$$ の偏差× $$y$$ の偏差の和データの個数で割ったものが共分散 $$C_{xy}$$ となります。
## 問題解説：相関係数
① ② ③ ④ ⑤ ⑥ $$x$$ $$9$$ $$6$$ $$2$$ $$5$$ $$8$$ $$6$$ $$y$$ $$5$$ $$2$$ $$7$$ $$4$$ $$4$$ $$8$$
データ $$x$$ の $$6$$ 個の平均値は、$$~~~~~~\frac{9+6+2+5+8+6}{6}$$$$~=\frac{36}{6}$$$$~=6$$データ $$x$$ の偏差と偏差の２乗を表にまとめると、
$$x$$ 偏差 偏差の２乗 $$9$$ $$3$$ $$9$$ $$6$$ $$0$$ $$0$$ $$2$$ $$-4$$ $$16$$ $$5$$ $$-1$$ $$1$$ $$8$$ $$2$$ $$4$$ $$6$$ $$0$$ $$0$$
$$y$$ 偏差 偏差の２乗 $$5$$ $$0$$ $$0$$ $$2$$ $$-3$$ $$9$$ $$7$$ $$2$$ $$4$$ $$4$$ $$-1$$ $$1$$ $$4$$ $$-1$$ $$1$$ $$8$$ $$3$$ $$9$$
$$x$$ の偏差 $$y$$ の偏差 偏差のかけ算 $$3$$ $$0$$ $$0$$ $$0$$ $$-3$$ $$0$$ $$-4$$ $$2$$ $$-8$$ $$-1$$ $$-1$$ $$1$$ $$2$$ $$-1$$ $$-2$$ $$0$$ $$3$$ $$0$$
よって、答えは $$-0.34$$ となります。
## 今回のまとめ
【問題一覧】数学Ⅰ：データの分析
このページは「高校数学Ⅰ：データの分析」の問題一覧ページとなります。解説の見たい単元名がわからな...
スポンサーリンク
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2019-08-23 12:05:57
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https://socratic.org/questions/how-do-you-solve-x-2-12x-36-25-by-completing-the-square
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# How do you solve x^2 - 12x + 36 = 25 by completing the square?
${x}^{2} - 12 x + 36 = 25$
or, ${x}^{2} - 2 \cdot 6 \cdot x + {\left(6\right)}^{2} = 25$
or, ${\left(x - 6\right)}^{2} = {5}^{2}$
either, $x - 6 = 5$ or, $x - 6 = - 5$
so,$x = 11$ or, $x = 1$
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2019-08-22 13:11:37
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http://tex.stackexchange.com/questions/100181/a-question-on-feasible-region
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# A question on feasible region
I faced a difficulty to name the corner point of the feasible region determined by the below LaTeX/TikZ code, who can help me.
\begin{figure}[htpb]
\centering
\begin{tikzpicture}
\draw[gray!50, thin, step=0.5] (-2,-2) grid (14,14);
\draw[very thick,->] (-2,0) -- (14,0) node[right] {$x$};
\draw[very thick,->] (0,-2) -- (0,14) node[above] {$y$};
\foreach \x in {-2,...,14} \draw (\x,0.5) -- (\x,-0.5) node[below] {\tiny\x};
\foreach \y in {-2,...,14} \draw (-0.5,\y) -- (0.5,\y) node[right] {\tiny\y};
\fill[blue!50!cyan,opacity=0.3] (0,0)-- (0,5) -- (6,11) -- (8,10) --(13,0) --cycle;
\put(0,0){A}\put(0,5){B}\put(6,11){C}\put(8,10){D}\put(13,0){E}
\draw (-2,3) -- node[above,sloped] {\tiny$-x+y\leq5$} (9,14);
\draw (6,14) -- node[above,sloped] {\tiny$2x+y\leq26$} (14,-2);
\draw (0,14) -- node[above right,sloped] {\tiny$2x+4y\leq56$} (14,7);
\end{tikzpicture}
\caption{Feasible Region of the problem}
\label{fig:ch3:c2}
\end{figure}
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Are you trying to mix xypic or PSTricks or picture mode with TikZ syntax? \put commands should not be there. – percusse Feb 27 '13 at 16:16
I want to give name for each corner points. I tried to use for instance \put (0,0){A}\put (0,5){B} \put(6,11){C} \put(8,10){D} \put(13,0){E}, but it doesn't work. – ftolessa Feb 27 '13 at 16:21
I think the command you need is \node at (0,0) {A};. – JLDiaz Feb 27 '13 at 16:23
Or also \coordinate (A) at (0,0); – Claudio Fiandrino Feb 27 '13 at 16:28
Welcome to TeX.sx! No need to add thanks, simply upvote any good replies you may receive. It is our preferred way to show appraciation, it makes posts more concise. – Peter Jansson Feb 27 '13 at 16:30
As percusse said in his comment: \put has nothing to do in a TikZ picture.
You probably want to use \node (with text) or \coordinate (saves the coordinate under a name). I used a combination of both (and a loop to save TikZ macros and my mind).
First the actual points are saved under a name (A, B, …):
\coordinate (<name>) at (<x>, <y>);
and with the help of the label option a node is added so that a reader can identify the points by their name. Instead of the label option, I could have used a \node which adds the text.
\foreach \x/\y/\t/\p in {0/0/A/below left,
0/5/B/above left,
6/11/C/above,
8/10/D/below left,
13/0/E/above right}{% as two commands are used, those need to be
% enclosed in { } so that they both are looped.
\coordinate (\t) at (\x,\y);
\node[\p] at (\t) {\t};
}
In the \fill command I have used the stored coordinate names.
In the second example I have used the the intersections library to let TikZ find the right points. You can now change the lines and TikZ does the rest for you.
## Code
\documentclass[tikz]{standalone}
\usetikzlibrary{intersections}
\begin{document}
\begin{tikzpicture}
\draw[gray!50, thin, step=0.5] (-2,-2) grid (14,14);
\draw[very thick,->] (-2,0) -- (14,0) node[right] {$x$};
\draw[very thick,->] (0,-2) -- (0,14) node[above] {$y$};
\foreach \x in {-2,...,14} \draw (\x,0.5) -- (\x,-0.5) node[below] {\tiny\x};
\foreach \y in {-2,...,14} \draw (-0.5,\y) -- (0.5,\y) node[right] {\tiny\y};
\foreach \x/\y/\t/\p in {0/0/A/below left,
0/5/B/above left,
6/11/C/above,
8/10/D/below left,
13/0/E/above right}
\coordinate[label={\p:\t}] (\t) at (\x,\y);
\fill[blue!50!cyan,opacity=0.3] (A) -- (B) -- (C) -- (D) -- (E) -- cycle;
\draw (-2,3) -- node[above,sloped] {\tiny$-x+y\leq5$} (9,14);
\draw (6,14) -- node[above,sloped] {\tiny$2x+y\leq26$} (14,-2);
\draw (0,14) -- node[above right,sloped] {\tiny$2x+4y\leq56$} (14,7);
\end{tikzpicture}
\begin{tikzpicture}
\draw[gray!50, thin, step=0.5] (-2,-2) grid (14,14);
\draw[very thick,->, name path=x axis] (-2,0) -- (14,0) node[right] {$x$};
\draw[very thick,->, name path=y axis] (0,-2) -- (0,14) node[above] {$y$};
\foreach \x in {-2,...,14} \draw (\x,0.5) -- (\x,-0.5) node[below] {\tiny\x};
\foreach \y in {-2,...,14} \draw (-0.5,\y) -- (0.5,\y) node[right] {\tiny\y};
\draw[name path=line 1] (-2,3) -- node[above,sloped] {\tiny$-x+y\leq5$} (9,14);
\draw[name path=line 2] (6,14) -- node[above,sloped] {\tiny$2x+y\leq26$} (14,-2);
\draw[name path=line 3] (0,14) -- node[above right,sloped] {\tiny$2x+4y\leq56$} (14,7);
\tikzset{
name intersections={of=y axis and x axis, name=A},
name intersections={of=y axis and line 1, name=B},
name intersections={of=line 1 and line 3, name=C},
name intersections={of=line 3 and line 2, name=D},
name intersections={of=x axis and line 3, name=E},
}
\foreach \t/\p in {A/below left,
B/above left,
C/above,
D/below left,
E/above right}
\node[fill,circle,minimum size=6\pgflinewidth,inner sep=0pt,label={\p:\t}] at (\t) {};
\fill[blue!50!cyan,opacity=0.3] (A) -- (B) -- (C) -- (D) -- (E) -- cycle;
\end{tikzpicture}
\end{document}
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2014-08-30 06:26:50
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https://www.physicsforums.com/threads/absorbed-dose-of-radiation.203690/
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1. Dec 9, 2007
### Red88
1. The problem statement, all variables and given/known data
Suppose you ingest a microgram radioactive material with an initial activity of 10^5/sec and a physical lifetime of 4 hours. Suppose also that all of the energy of each emission of alpha particles with energy of 0.3MeV per particle is deposited throughout your tissue. If you eliminate the material, making a body lifetime of 12 hours, what will be the total dose given to your body, of mass 50kg, after a month?
2. Relevant equations
According to my professor,
Activity = (No/T)exp(-t/T) where T is the lifetime of the substance
With both a physical half-life, tp, and a body or metabolic half-life, tm,
N = No(1/2)^(t(1/tp + 1/tm))
Dose = J/kg
1 eV = 1.602 x 10^-19 J
3. The attempt at a solution
10^5 dis/s X 2419200s (in a month) X (4hrs/12hrs) X (4.806 x 10^-14J)/(dis) X (1/50kg)
=> Dose = 7.75 x 10^-5 J/kg
Please respond ASAP - the homework is due tomorrow!!! I'll appreciate any help........
Regards,
Red88
2. Dec 10, 2007
### dynamicsolo
I'm just finding this problem now, so I suppose it may be too late. It seems to me that an integration is required, since you need to know the total number of decays that are released within your body in the course of the month. The combination of radioactive decay and physical elimination gives a lifetime of
1/(LT) = 1/4 + 1/12 = 1/3 ,
so the effective lifetime is 3 hours. The total number of decays within your body would then be
(10^5)·[exp -(t/3)] per second, with t in hours, or
(3.6·10^8)·[exp -(t/3)] per hour. Integrating this over the 720 hours in a month would give
-(1/[1/3])·(3.6·10^8)·[exp -(t/3)] , evaluated from t = 0 to t = 720 hours, or
-(1.08·10^9)·[exp -(720/3)] - { -(1.08·10^9)·[exp -(0/3)] } , which is essentially
(1.08·10^9) · 1 , since exp -(720/3) is really tiny.
The isotope is pretty much "gone" within the first 36 hours, since exp(-12) is about 6·10^-6 , so the activity would be less than 1/sec by then.
With an exposure of 1.08·10^9 decays, the energy release is (1.08·10^9)·(4.806 x 10^-14 J) = 5.19·10^-5 J , or a dosage of 5.19·10^-5 J / 50 kg = 1.04·10^-6 J/kg .
We're in the same neighborhood, so I suspect you are given an approximate formula for making this calculation. What did they say the answer was?
3. Dec 12, 2007
### Red88
Alas we haven't been provided with a solution to the problem yet.........your procedure seems more concise than mine so I hope that this is indeed the solution to this problem. Thanks for your help!
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2017-02-20 07:11:24
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https://peadarcoyle.wordpress.com/tag/data/
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# One weird tip to improve the success of Data Science projects
Standard
I was recently speaking to some data science friends on Slack, and we were discussing projects and war stories. Something that came across was that ‘data science’ projects aren’t always successful.
Source: pixabay
Somewhere around this discussion a lightbulb went off in my head about some of the problems we have with embarking on data science projects. There’s a certain amount of Cargo cult Data Science and so collectively we as a community – of business people, technologists and executives don’t think deeply enough about the risks and opportunities of projects.
So I had my lightbulb moment and now I share it with everyone.
The one weird trick is to write down risks before embarking on a project.
Here’s some questions you should ask you start a project – preferably gather all data .
• What happens if we don’t do this project? What is the worse case scenario?
• What legal, ethical or reputational risks are there involved if we successfully deliver results with this project?
• What engineering risks are there in the project? Is it possible this could turn into a 2 year engineering project as opposed to a quick win?
• What data risks are there? What kinds of data do we have, and what are we not sure we have? What risks are there in terms of privacy and legal/ ethics?
I’ve found that gathering stakeholders around helps a lot with this, you hear different perspectives and it can help you figure out what the key risks in your project are. I’ve found for instance in the past that ‘lack of data’ killed certain projects. It’s good to clarify that before you spend 3 months on a project.
Try this out and let me know how it works for you! Share your stories with me at myfullname[at]google[dot]com.
# Interviews with Data Scientists: NLP for the win
Standard
Recently I decided to do some quick Data Analysis of my interviews with data scientists.
It seems natural when you collect a lot of data to explore it and do some data analysis on it.
You can access the code here.
The code isn’t in much depth but it is a simple example of how to use NLTK, and a few other libraries in Python to do some quick data analysis of ‘unstructured’ data.
First question:
What does a word cloud of the data look like?
Word cloud of my Corpus based on interviews published on Dataconomy
Here we can see above that science, PHD, science, big etc all pop up a lot – which is not surprising given the subject matter.
Then I leveraged NLTK to do some word frequency analysis. Firstly I removed stop words, and punctuation.
I got the following result – unsurprisingly the most common word was data followed by science, however the other words are of interest – since they indicate what professional data scientists talk about in regards their work.
Source: All interviews published on Dataconomy by me until the end of last week – which was the end of September 2015.
# An interview with a data artisan
Standard
J.D.Long is the current AVP Risk Management at RenaissanceRe and has a 15 year history of working as an analytics professional.
I sent him an interview recently to see what he would say.
1. What project have you worked on do you wish you could go back to, and do better?
Longer answer: Interestingly, what I find myself thinking about when asked this question is not analytics projects where I wish I could redo the analysis, but rather instances where I felt I did good analysis but did a bad job explaining the implications to those who needed the info. Which brings me to #2…
2. What advice do you have to younger analytics professionals?
2) Learn technical skills and enjoy learning new things, naturally. But, 1) always plot your data to visualize relationships and 2) remember at the end of the analysis you have to tell a story. Humans are hard wired to remember stories and not numbers. Throw away your slide deck pages with a table of p values and instead put a picture of someone’s face and tell their story. Or possible show a graph that illustrates the story. But don’t forget to tell the story.
3. What do you wish you knew earlier about being a data artisan?
3) Inside of a firm, cost savings of $1mm seems like it should be the same as generating income of$1mm. It’s not. As an analyst you can kick and whine and gripe about that reality, or you can live with it. One rational reason for the inequality is that income is often more reproducible than cost savings. However, the real reason is psychological. Once a cost savings happens it’s the new expectation. So there’s no ‘credit’ for future years. Income is a little different in that people who can produce \$1mm in income every year are valued every year. That’s one of the reasons I listed “be a profit center” in the post John referenced. There are many more reasons, but that alone is a good one.
4. How do you respond when you hear the phrase ‘big data’?
4) I immediately think, “buzz word alert”. The phrase is almost meaningless. I try to listen to what comes next to see if I’m interested.
5) Everybody loves a good “ah-ha!” moment. Analytics is full of those. I think most of us get a little endorphin drop when we learn or discover something. I’ve always been very open about what I like about my job. I like being surrounded by interesting people, working on interesting problems, and being well compensated. What’s not to love!
Cheers,
P.s. the post J.D.Long mentioned is http://www.johndcook.com/blog/2011/11/21/career-advice-regarding-tools/
# Information Retrieval
Standard
Attention conservation notice: 680 words about Information Retrieval, and highly unoriginal.
The following is very much inspired by a course by Cosma Shalizi but I felt it was worth rewriting to get to grips with the concepts. This is the first of what is hopefully a series of posts on ‘Information Retrieval’, and applications of Mathematics to ‘Data Mining’.
1. Textual features
I’d like to introduce the concepts of how features are extracted, and we shall consider these proxies of actual meanings. One classic representation we can use is bag-of-words (BoW). Let us define this representation, this means we list all of the distinct words in the document together with how often each one appears. This is easy to calculate from the text. Vectors One way we could try to code up the bag-of-words is to use vectors. Let each component of the vector correspond to a different word in the total lexicon of our document collection, in a fixed, standardised order. The value of the component would be the number of times the word appears, possibly including zero. We use this vector bag-of-words representation of documents for two big reasons:
• There is a huge pre-existing technology for vectors: people have worked out, in excruciating detail, how to compare them, compose them, simplify them, etc. Why not exploit that, rather than coming up with stuff from scratch?
• In practice, it’s proved to work pretty well.
We can store data from a corpus in the form of a matrix. Each row corresponds to a distinct case (or instance instance, unit, subject,…) – here, a document – and each column to a distinct feature. Conventionally, the number of cases is n and the number of features is p. It is no coincidence that this is the same format as the data matrix X in linear regression.
2. Measuring Similarity
Right now, we are interested in saying which documents are similar to each other because we want to do a search by content. But measuring similarity – or equivalently measuring dissimilarity or distance</b> – is fundamental to data mining. Most of what we will do will rely on having a sensible way of saying how similar to each other different objects are, or how close they are in some geometric setting. Getting the right measure of closeness will have a huge impact on our results. \paragraph{} This is where representing the data as vectors comes in so handy. We already know a nice way of saying how far apart two vectors are, the ordinary or Euclidean distance, which we can calculate with the Pythagorean formula:
$\displaystyle \|\bar{x} - \bar{y}\| = \sqrt{\sum_{i=1}^{p}(x_i - y_i)^2}$
where ${x_i}$, ${y_i}$ are the ${i^{th}}$ components of ${\bar{x}}$ and ${\bar{y}}$. Remember that for bag-of-words vectors each distinct word – each entry in the lexicon – is a component or a fecture. We can also use our Linear Algebra skills to calculate the Euclidean norm or Euclidean distance . Of any vector this is ${\|\bar{x}\| = \sqrt{\sum_{i=1}^{p}x^{2}_{i}}}$ so the distance between two vectors is the norm of their distance ${\bar{x} - \bar{y}}$. Equivalently, the norm of a vector is the distance from it to the origin, ${\bar{0}}$ Obviously, one can just look up a topology textbook and remind oneself of other metrics such as the taxicab metric.
2.1. Normalisation
Just looking at the Euclidean distances between document vectors doesn’t work, at least if the documents are at all different in size. Instead, we need to normalise by document size, so that we can fairly compare short texts with long ones. There are (at least) two ways of doing this.
Document length normalisation Divide the word counts by the total number of words in the document. In symbols,
$\displaystyle \bar{x} \mapsto \frac{\bar{x}}{\sum_{i=1}^{p} x_{i}}\ \ \ \ \ (1)$
Notice that all the entries in the normalised vector are non-negative fractions, which sum to 1. The i-th component is thus the probability that if we pick a word out of the bag at random, it’s the i-th entry in the lexicon.
Cosine ‘distance’ is actually a similarity measure, not a distance`:
$\displaystyle d_{cos} \bar{x}, \bar{y} = \frac{\sum_{i} x_{i}y_{i}}{\|\bar{x}\|\|\bar{y}\|} \ \ \ \ \ (2)$
It’s the cosine of the angle between the vectors ${\bar{x}}$ and ${\bar{y}}$.
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2017-10-22 15:40:05
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https://brilliant.org/discussions/thread/proofathon-spring-competition/
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×
Proofathon Spring Competition
Dear Proofathoners,
We are just days away from the Proofathon Spring Competition. This competition will cover the areas of Algebra, Number Theory, Geometry, and Combinatorics. The prizes are as follows, the upper places receiving all of the prizes of the lower places as well:
1st gets his/her name added to the Proofathon Super Hall of Fame. 3rd gets a copy of Mathematica, as well as a Proofathon t-shirt. 10th gets a commemorative score certificate. Everyone gets the experience in solving maths problems.
I wish you luck, I will see you at proofathon.org this Friday, the 28th at 0:00 EST. The contest is one week long, so that should give you ample time to compete and succeed.
Good luck, Proofathoners!
Note by Cody Johnson
2 years, 9 months ago
Sort by:
An example of a past Proofathon problem:
Prove that for all real $$x, y,$$ and $$z$$,
$|\cos (x)| + |\cos (y)| + |\cos (z)| + |\cos(y+z)| + |\cos(z+x)| + |\cos(x+y)| + 3|\cos(x+y+z)| \geq 3.$ Staff · 2 years, 9 months ago
Hey Cody , just saw your solutions on proofathon. Amazing work there. Could tell me which books you follow to study maths.? Also could you give me ur email id? mine's sbrdacm@gmail.com · 2 years, 9 months ago
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2017-01-19 00:21:39
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https://proxieslive.com/tag/ab/
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## With Σ = {a,b}, give a dfa for L= w1aw2 : |w1 |≥ 3, |w2 |≤ 5}
I racked my brain,I saw other people’s solutions and it don’t make sense. I think my biggest problem is I don’t know when one string ends,like for example w1 is >=3 and it can have how many ever b’s or a’s as long as it’s 3 and above,so when do we get to have the First a that’s after the W1?how do I make sure it’s not a part of the W1? that’s whats so confusing the W1 string can have as many a’s as it likes which means it should have an a loop,but you can’t have an a loop cause YOU NEED 1 A EXACTLY after the W1 string. this is a catch 22,damned if you don’t damned if you do..help please I spent an hour on this to no avail. I tried every possible thing and saw people’s solutions and I feel like when they solved it they are either dismissing what I mentioned above,or something is going COMPLETELY over my head.
## Computing automaton for $L(A) / L(B)$ gives ones for $A,B$
I’m trying to figure out whether infinite language change the answer.
Show that the following language is decidable: $$L=\{\langle A,B \rangle : \text{ A,B are DFAs, L(B) is finite, and L(A)/ L(B)=L(0^*1^*) }\}.$$
(I am talking about right division.)
We know how to check whether the language of a DFA is finite, and given two DFAs, we know how to check whether their languages are equal. The algorithms I know to the above problems uses the DFA’s, so it is necessary having the DFA’s in order to decide those problems.
I’m trying to figure out whether $$|L(B)|=\infty$$ changes the answer. To the best of my understanding, because $$|L(B)|<\infty$$, we can explicitly construct a DFA that accepts $$L(A)/ L(B)$$, whereas if $$L (B)=\infty$$ all we know is about the existence of $$DFA$$ that accepts $$L(A)/ L(B)$$.
However, even if $$L(B)$$ is an infinite language, since there is a finite number of DFAs, one of which accepts $$L(A) / L(B)$$, I can certainly know that there is a Turing machine that decides the language $$L$$. Right?
## If I can efficiently uniformly sample both $A$ and $B\subset A$, can I efficiently uniformly sample $A-B$?
As posed in the question; the statement naively seems like it should be self-evident but there are no algorithms that come immediately to mind. Suppose I have some domain $$A$$ (in my case a subset of $$\mathbb{Z}^n$$ for some $$n$$, but I would expect any answer to be independent of that structure), and a domain $$B\subset A$$. If I have an efficient algorithm for uniformly sampling from $$A$$, and an efficient algorithm for uniformly sampling from $$B$$, is there any way of ‘combining’ these to get an efficient algorithm for uniformly sampling $$A-B$$? I can certainly rejection-sample, but if $$|A-B|\ll|A|$$ then there’s no guarantee that that will be efficient.
## Is this language L = {w $\in$ {a,b}$^*$ : ($\exists n \in \mathbb{N}$)[$w|_b = 5^n$]} regular?
Let’s say we have the language L = {w $$\in$$ {a,b}$$^*$$ : ($$\exists n \in \mathbb{N}$$)[$$w|_b = 5^n$$]}. I want to know if this is a regular language or not. How do I go about doing this? I’m familiar with the Myhill-Nerode theorem but I don’t know how to apply it.
## Design a CFG that generates the language { x in {a,b}* | the length of x is odd and its middle symbol is a b }
I am trying to design a context-free grammar that generates the language { x in {a,b}* | the length of x is odd and its middle symbol is a b }.
This is really confusing me, I’m having trouble with making sure that b is always in the middle. Any help?
## Convert PDA to CFG – Language of Palindromes over $\{a,b\}^+$
I tried to convert the following PDA $$M$$
to a CFG and got the following results. However I’m not quite sure if I did the algorithm the right way. I feel like I maybe misunderstood some the algorithm when it comes to the $$A_{p,q} \to xA_{r,s}y$$ part. Could you help me to clear things up?
$$G = (V,\Sigma,P,S)$$ with \begin{align*} V = \{&A_{0,0},A_{0,1},A_{0,2},A_{0,3},A_{0,4},A_{1,0},A_{1,1},A_{1,2},A_{1,3},A_{1,4},A_{2,0},A_{2,1},A_{2,2},A_{2,3},A_{2,4},\ &A_{3,0},A_{3,1},A_{3,2},A_{3,3},A_{3,4},A_{4,0},A_{4,1},A_{4,2},A_{4,3},A_{4,4},S\}, \end{align*} $$\Sigma = \{a,b\}$$, $$S = A_{0,4}$$ and the following productions: \begin{align*} S &\to A_{0,4}\ A_{0,0} &\to \epsilon\ A_{1,1} &\to \epsilon\ A_{2,2} &\to \epsilon\ A_{3,3} &\to \epsilon\ A_{4,4} &\to \epsilon\ A_{0,0} &\to A_{0,0}A_{0,0} \vert A_{0,1} A_{1,0} \vert A_{0,2}A_{2,0} \vert A_{0,3}A_{3,0} \vert A_{0,4}A_{4,0}\ A_{0,1} &\to A_{0,0}A_{0,1} \vert A_{0,1} A_{1,1} \vert A_{0,2}A_{2,1} \vert A_{0,3}A_{3,1} \vert A_{0,4}A_{4,1}\ A_{0,2} &\to A_{0,0}A_{0,2} \vert A_{0,1} A_{1,2} \vert A_{0,2}A_{2,2} \vert A_{0,3}A_{3,2} \vert A_{0,4}A_{4,2}\ &\vdots\ A_{4,4} &\to A_{4,0}A_{0,4} \vert A_{4,1} A_{1,4} \vert A_{4,2}A_{2,4} \vert A_{4,3}A_{3,4} \vert A_{4,4}A_{4,4}\ A_{0,4} &\to A_{1,3}\ A_{1,3} &\to aA_{1,3}a\ A_{1,3} &\to bA_{1,3}b\ A_{1,3} &\to aA_{1,1}\ A_{1,3} &\to bA_{1,1} \end{align*}
## What does {a,b}* for DFA’s mean?
For instance when the question contains {a,b}* does this mean that the DFA must have at least 1 a and 1 b on top of whatever conditions it has? For example a DFA that accepts {w E {a,b}* : w contains bbb} should it reject the actual string bbb because it does not contain an a?
## Use the pumping lemma for context free languages to prove L = {w#w | w \in {a,b}*} is not context free
I know the basics of using the pumping lemma for CFG to prove a language L is not context-free, however, the # symbol seems to be throwing me off or my understanding is not complete.
## Prove that there isn’t a DFA characterized by (a*)+(b*) for which |Q| = 3
Let $$R=(a*)+(b*)$$ be a regular expression. Prove that there cannot exist any DFA $$M=(Q,\Sigma,\delta , q0, A)$$ such that $$|Q| = 3$$ and $$L(M)=L(R)$$.
The problem is, I think IT IS possible to construct such a dfa that has 3 states in total.
Am I missing something?
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2020-07-14 00:43:37
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http://www.physicsforums.com/showthread.php?t=581687
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# find f(x) for the tangent line of the graph
by lalahelp
Tags: graph, line, tangent
P: 75 1. The problem statement, all variables and given/known data Suppose line tangent to graph of y=f(x) at x =3 passes through (-3, 7) & (2,-1). Find f'(3), what is the equation of the tangent line to f at 3? 2. Relevant equations I found the slope of which equals -8/5 Im not sure how to find the equation... do I do y-3=-8/5(x-3)? or is that wrong?
P: 615 The general form of a straight line is y=m x + c Try fitting that to your points
P: 8 The tangent line passes through those two points. Since it is a LINE, you may use the equation to solve for the slope: m= (y2-y1) / (x2-x1) Once you have found the slope, use any of of the two coordinates to solve for 'c', the constant/y-intercept. Once you have the constant, simply write it in the form y=(#)x + (#). That is the equation of the tangent line at that point x=3.
P: 75
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2014-03-09 20:46:28
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https://mathshistory.st-andrews.ac.uk/Biographies/Malfatti/
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# Gian Francesco Malfatti
### Quick Info
Born
26 September 1731
Ala, Trento (now Italy)
Died
9 October 1807
Ferrara (now Italy)
Summary
Gianfrancesco Malfatti was an Italian mathematician who worked on geometry, probability and mechanics and made contributions to the problem of solving polynomial equations.
### Biography
Gianfrancesco Malfatti name is often written as Gian Francesco Malfatti. His parents were Giovanni Battista Malfatti and his wife Giuseppa. He began his education at the Jesuit College in Verona before continuing his studies at Bologna. He studied under Laura Bassi, Vincenzo Riccati, Francesco Maria Zanotti (1692-1777) and Gabriele Manfredi at the College of San Francesco Saverio in Bologna. Bassi, who taught physics, and Zanotti, who taught both philosophy and physics, were both strong supporters of Newtonian physics and gave lectures designed to introduce their students to that approach.
The University of Ferrara was erected by Alberto V who reigned over Ferrara from 1388 to 1393. The University was obtained by Pope Boniface IX as a concession in 1391 and this is usually taken as its founding date. Copernicus studied there in 1503 but the buildings which house the University today date from the end of the 16th-century. The University of Ferrara was re-established in its 16th-century buildings in 1771 and, the president of the University, Giovanni Maria Riminaldi (1718-1789), knowing that by this time Malfatti was gaining a reputation as a leading mathematician, wanted to appointed him to the chair of mathematics. This, however, presented a problem since Malfatti, although he had been educated by the Jesuits, was not himself a member of the Order and it was expected that professors would be Jesuits. Despite the difficulties, Malfatti's scholarship won the day and he was appointed to the chair.
The Jesuit Order was at this time at the centre of a major controversy due, in part, to their wish to accept pagan ideas into Christianity. The Order had been made illegal in France in 1764, then Spain and the Kingdom of the Two Sicilies acted against them in 1767 and finally pope Clement XIV dissolved the Jesuits in 1773. Although this did not affect Malfatti directly, it did have a major impact on Alessandro Zorzi (1747-1779) who was given employment by the Marquis Bevilacqua as a tutor for his grandchildren. In this way Malfatti got to know Zorzi and became interested in the project that Zorzi was planning to undertake. This was to produce an Italian encyclopaedia, Nuova Enciclopedia italiana, to be an Italian equivalent of Diderot and d'Alembert's Encyclopédie. Soon Malfatti agreed to write articles for the encyclopaedia and Zorzi enlisted the help of other scholars such as Gerolamo Tiraboschi (1731-1794), Larraro Spallanzani (1729-1799), Gregorio Fontana (1735-1803), Paolo Frisi, Antonio Mario Lorgna, Giuseppe Louis Lagrange (1736-1813), Giuseppe Toaldo (1719-1797), Giambattista Beccaria (1716-1781), Leopoldo Marco Antonio Caldani (1725-1813), Saverio Bettinelli (1718-1808) and several others. Tiraboschi was a historian of Italian literature and professor at the University of Milan before becoming a librarian to the duke of Modena. Spallanzani served as professor of logic, metaphysics, and Greek, then as professor of physics at the University of Modena before gaining worldwide recognition as a physiologist while holding a chair at the University of Pavia. Fontana was a mathematician who succeeded Boscovich in the chair of mathematics at the University of Pavia. Giuseppe Lagrange was a mathematician and astronomer who was director of mathematics at the Prussian Academy of Sciences in Berlin. Toaldo was a physicist who held the chair of astronomy at the University of Padua. Beccaria held the chair of experimental physics at Turin. Caldani was professor of anatomy in Padua. Bettinelli was a poet and literary critic who had been professor at Modena until the suppression of the Jesuits after which he retired to Mantua. This project had great promise but never came to anything since Zorzi died before it had progressed very far and nobody took over the reins.
Dirk Struik, describing the papers in [5], writes that Malfatti:-
... was a quiet, scholarly man who spent most of his life as a librarian and professor in Ferrara. He was one of the founders of the "Società Italiana delle Scienze" (1782) and was active in academic reform, especially in the Napoleonic period.
Let us give more details on the founding of the Società Italiana delle Scienze, the Society of the XL, which has become the National Academy of Sciences of Italy. In addition to Malfatti, four men played major roles in founding the Society. These were Antonio Mario Lorgna, Carlo Barletti (1735-1800) the professor of physics at Pavia University, the mathematician Ruggero Boscovich, and Lazzaro Spallanzani who we mentioned above. These five were friends and they discussed setting up a Society over many years. On 28 June 1766 Lorgna wrote to Malfatti saying that it was important for Italy to have a scientific journal. Over the following ten years discussions went on and by 1776 Lorgna and Malfatti were discussing Lorgna's "idea of forming an Academy for all Italian scholars". Malfatti thought that "the idea of forming an Academy for all the Italian Letterati is noble and glorious" but was worried that they would not get deep enough contributions from mathematicians and other scientists. Barletti was also worried that they might not get the highest quality research for its journal Memorie della Società italiana, but Lorgna, who was confident that the project would succeed, pushed their doubts aside. Another problem was over the decision to publish articles in Italian in the Memorie della Società italiana. Malfatti strongly supported writing papers in Italian rather than Latin and from 1779 he wrote all his papers in Italian. Several thought that this would limit the readership of the journal to Italy but they pressed ahead with the proposal and the Society, together with its publication, played a small part in the unification of Italy. Malfatti took his obligations to the Society very seriously, and membership spurred him on to undertake intensive research. He submitted fourteen articles to the journal of the Society between 1782 and 1807.
One of the topics that Malfatti worked on was the logarithmic curve. Florian Cajori writes [9]:-
In 1795 ... Malfatti ... discussed at length the question whether the logarithmic curve has one or two branches. The equation $ydx = dy$ has an infinite number of integrals even under the restriction that $x = 0$ when $y = 1$. For any constant $n, e^{nx} = y^{n}$ gives rise to the differential equation$ydx = dy$. Before integrating one must agree upon the number of values $y$ shall have. If it is to have one value, we get $e^{x} = y$; if two values, we get $e^{2x} = y^{2}$ or $(e^{x} - y)(e^{x} + y) = 0$. That $e^{x} = y$ has only one branch is argued from geometric and analytic considerations.
As we have seen above, Malfatti wrote an important work on equations of the fifth degree. In 1799 Paolo Ruffini published a paper on the theory of equations with his claim that quintics could not be solved by radicals as the title shows: General theory of equations in which it is shown that the algebraic solution of the general equation of degree greater than four is impossible. Melvin Kiernan writes [19]:-
As representative of the "old school" [Malfatti] had long believed that the general quintic equation could be solved algebraically, and had once published a supposed proof of this result. His objections were not so much to individual mathematical points in Ruffini's paper as they were an attack on the general concept and methodology. It is easy today to underestimate the enormity of the difficulty faced by these mathematicians. The mathematics of their day was largely computational, and they were unprepared to accept, and unable to understand, the new approach through theoretical and existential arguments. Thus the most frequent objection to the works of the new mathematicians was "vagueness." ... Ruffini's exchanges with Malfatti ... gave rise to six published versions of Ruffini's proof, the last appearing in 1813.
Malfatti replied to a 1802 proof by Ruffini by publishing Dubbi proposti al socio Paolo Ruffini sulla sua dimostrazione dell'impossibilità di risolvere le equazioni superiori al quarto grado (1804). However, Ruffini was not the only mathematician with whom Malfatti had a dispute. Another was Pietro Paoli, also a founding member of the Società Italiana delle Scienze, and they carried out their dispute in the Society's journal. The dispute concerned the problem of determining the pressures on the corners of a solid figure resulting from its weight.
In 1802 Malfatti considered the problem of describing in a triangle three circles that are mutually tangent, each of which touches two sides of the triangle, the so-called Malfatti problem. His solution, involving heavy calculations and complicated formulas, was published in a paper of 1803 on Un problema stereotomica. However, the situation was more complicated as Howard Eves explains in [11]:-
In 1803 Malfatti proposed the problem of cutting three circular cylindrical holes out of a solid right triangular prism in such a way that the cylinders and the prism have the same altitude and that the combined volumes of the cylinders be a maximum. Malfatti intuitively reduced this problem to another, now commonly referred to as the "Malfatti problem": To inscribe within a triangle three circles each of which be tangent to the other two and to two sides of the triangle. This reduced problem has enjoyed a long history and has been solved and generalised in many ways. Malfatti considered this reduced problem as equivalent to the original problem. That such is not the case was indicated in 1929, when Lob and Richmond pointed out that in the case of the equilateral triangle the inscribed circle and two of the smaller circles that can be fitted into the corners have a combined area which is greater than that given by the Malfatti arrangement ...
Michael Goldberg, in 1967, showed in [17] that the Malfatti circles are never the solution to the original Malfatti problem. Jacob Bernoulli had solved the Malfatti problem for an isosceles triangle while, after Malfatti, the problem was also solved by an elegant geometric solution by Jacob Steiner in 1826 and Clebsch solved it using elliptic functions. Let us give a generalisation of the Malfatti problem. A configuration of circles in a triangle is called a 'greedy configuration' if the circles are chosen one after the other in such a way that each circle is the largest that does not intersect any previous circle [24]:-
H Mellisen has conjectured that for any arrangement of $n$ non-overlapping circles in a triangle, the greedy arrangement has the largest total area. In 1994, V A Zalgaller and G A Los verified this conjecture for $n = 3$, thus settling Malfatti's original problem.
Malfatti's interests extended beyond the results we have mentioned above for his papers dealt with many subjects from probability to mechanics. In fact he published twenty-two memories on mathematical analysis, geometry, the theory of algebraic equations, mathematical physics, the theory of difference equations, combinatorics and probability. There are several papers in [5] which describe Malfatti's work. These include: Problems and methods of mathematical analysis in the work of Gianfrancesco Malfatti, Contributions of Gianfrancesco Malfatti to combinatorial analysis and to the theory of finite difference equations, The work of Malfatti in the realm of mechanics, The geometrical research of Gianfrancesco Malfatti, Gianfrancesco Malfatti and the theory of algebraic equations, and Gianfrancesco Malfatti and the support problem.
We should note one final interesting result which he obtained. In 1781 he showed in the paper Della curva cassiniana that the lemniscate had the property that [1]:-
... a mass point moving on it under gravity goes along any arc of the curve in the same time as it traverses the subtending arc.
Let us return to make final comments on Malfatti's career. He lived through the military and political events that shook Italy in the late eighteenth century and the beginning of the next. Like almost all scientists of his time, he participated in reforming ideas with enthusiasm and actively collaborated with the republican government. In December 1796 Napoleon Bonaparte created the Cispadane Republic by merging of the duchies of Reggio and Modena and the legate states of Bologna and Ferrara. Malfatti participated in the Cispadane Republic as a member of its Education Committee in 1796 and, in the following year, he served on the committee drafting proposals to reform schools in Ferrara. On 23 May 1799 combined Austrian and Russian forces entered Ferrara and Malfatti was removed from his chair of mathematics. The Austrians dismantled the educational reforms which had been introduced, returning to the ways before the Republic was formed. However, in 1801 French forces drove out the Austrians and retook the city. Malfatti was reinstated to his chair of mathematics but he retired shortly after this, having gained thirty years of service at the University. Since the early 1790s he had suffered from respiratory ailments and eye strain. Reduced to almost complete blindness, he regained his sight after cataract surgery and was able to continue his research up until shortly before his death. After his death in Ferrara he was buried in the chapel of St Maurelio in Ferrara cathedral.
### References (show)
1. A Natucci, Biography in Dictionary of Scientific Biography (New York 1970-1990). See THIS LINK.
2. M T Borgato, L Capra, A Fiocca and L Pepe (eds.), Opere matematiche della pubblica Biblioteca di Ferrara (1753-1815) (Comune di Ferrara, Assessorato alie Istituzioni Culturali, Université degli Studi di Ferrara, Ferrara, 1981).
3. L Franchini, La matematica e il gioco del lotto - Una biografia di Gianfrancesco Malfatti (Edizioni Stella, Rovereto, 2007).
4. K Gavroglu (ed.), The Sciences in the European Periphery During the Enlightenment (Springer Science & Business Media, 2013).
5. L Pepe, L Biasini, L Capra and M Fiorentini (eds.), Gianfrancesco Malfatti nella cultura del suo tempo, Proceedings of the Conference on Gianfrancesco Malfatti 23-24 October 1981, Ferrara (Bologna, 1982).
6. M Andreatta, A Bezdek and J P Boronski, The problem of Malfatti: two centuries of debate, Math. Intelligencer 33 (1) (2011), 72-76.
7. G Arrighi, Gian Francesco Malfatti matematico trentino del '700: Il 'Lotto' e la 'Cassiana' in due lettere inedite, Stud. Trent. Sci. Stor. 61 (1982), 397-401.
8. E Bortolotti, Sulla risolvente di Malfatti, Atti dell'Accademia di Modena 7 (1906).
9. F Cajori, History of the Exponential and Logarithmic Concepts, Amer. Math. Monthly 20 (4) (1913), 107-117
10. Y-C Chan and M-K Siu, Historical notes: the Malfatti problem in 19th-century China, Math. Today (Southend-on-Sea) 49 (5) (2013), 229-231.
11. H Eves, Malfatti problem, Amer. Math. Monthly 53 (5) (1946), 285-286.
12. A Fiocca, Malfatti, Gianfrancesco, Dizionario Biografico degli Italiani 68 (2007).
13. A Fiocca, Malfatti's problem in nineteenth-century mathematical literature (Italian), Ann. Univ. Ferrara Sez. VII (N.S.) 26 (1980), 173-202.
14. H Gabai and E Liban, On Goldberg's Inequality Associated with the Malfatti Problem, Mathematics Magazine 41 (5) (1968), 251-252.
15. Gianfrancesco Malfatti, Centro PRISTEM, Università Commerciale L Bocconi, Milan. http://matematica.unibocconi.it/autore/gianfrancesco-malfatti
16. E Giusti, Problemi e metodi di analisi matematica nell'opera di Gianfrancesco Malfatti, in L Pepe, L Biasini, L Capra and M Fiorentini (eds.), Gianfrancesco Malfatti nella cultura del suo tempo, Proceedings of the Conference on Gianfrancesco Malfatti 23-24 October 1981, Ferrara (Bologna, 1982), 37-56.
17. M Goldberg, On the Original Malfatti Problem, Mathematics Magazine 40 (5) (1967), 241-247.
18. A Hirayama, Malfatti's problem in wasan (Japanese), Sûgakushi Kenkyû No. 85 (1980), 16-18.
19. B M Kiernan, The Development of Galois Theory from Lagrange to Artin, Archive for History of Exact Sciences 8 (1/2) (1971), 40-154.
20. A Laguzzi, Carlo Barletti e le 'Encyclopédies', Studi Storici, Anno 33 (4) (1992), 833-862.
21. J Lorenat, Not set in stone: nineteenth-century geometrical constructions and the Malfatti problem, Bull. British Society for the History of Mathematics 27 (3) (2012), 169-180.
22. L Miani, I Ventura and S Giuntini, The Gianfrancesco Malfatti - Sebastiano Canterzani correspondence (Italian), Boll. Storia Sci. Mat. 3 (2) (1983), 1-198.
23. P Peak, Review: On the Original Malfatti Problem, Mathematics Magazine, by Michael Goldberg, The Mathematics Teacher 62 (2) (1969), 120.
24. P D S, Review: The Problem of Malfatti: Two Centuries of Debate, by Marco Andreatta, András Bezdek and Jan P Boronski, The College Mathematics Journal 42 (4) (2011), 340.
25. R Taton, Review: M T Borgato, L Capra, A Fiocca and L Pepe (eds.), Opere matematiche della pubblica Biblioteca di Ferrara (1753-1815), Revue d'histoire des sciences 37 (3/4) (1984), 360-361.
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2021-11-27 19:59:19
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https://stats.stackexchange.com/questions/422887/treating-missing-data-in-making-bayesian-inference/422986
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# Treating missing data in making Bayesian inference
Suppose we have two biased coins $$X_1,X_2$$ that are possibly correlated to each other.
In each round, when both the coins are tossed, there can be four possible outcomes: $$(HH,HT,TH,TT).$$ Let's denote the associated probabilities by $$(p_{HH},p_{HT},p_{TH},p_{TT})$$.
To make a Bayesian inference, let's assume $$(p_{HH},p_{HT},p_{TH},p_{TT})\sim Dirichlet(a_{HH},a_{HT},a_{TH},a_{TT}).$$
From this prior, we can update the probabilities of outcomes based on the observed coin-tossings.
My question is what happens if only one coin is tossed in some rounds? For example, outcomes are (H,H) in round 1, (H,no tossing) in round 2, (no tossing, T) in round 3, and so on. How can we deal with this missing data cases if we still want to make a Bayesian inference?
I initially thought I could construct the Bayesian inference by expanding the outcome space into $$(HH,HN,HT,NH,NN,NT,TH,TN,TT)$$, where the outcome $$N$$ means "no tossing" or missing data. However, this doesn't seem to be right because I'm allocating some probabilities to missing data. e.g., $$p_{NN}$$ is positive but "no tossing" is not even an outcome of the random variable.
Moreover, suppose that the prior is given by $$(p_{HH},p_{HN},p_{HT},p_{NH},p_{NN},p_{NT},p_{TH},p_{TN},p_{TT})\sim Dirichlet(1,1,1,1,1,1,1,1,1).$$
When we observe an outcome of $$(HN)$$, the posterior becomes $$(p_{HH},p_{HN},p_{HT},p_{NH},p_{NN},p_{NT},p_{TH},p_{TN},p_{TT})\sim Dirichlet(1,2,1,1,1,1,1,1,1).$$
Then, the expected probability of $$H$$ of the first coin is $$\frac{1+2+1}{1+2+1+1+1+1+1+1+1}=\frac{4}{10}$$, which does not seem to be correct.
How should I deal with this missing data case in Bayesian approach using Multinomial distribution?
• I believe your mistake is in expanding the outcome space to iimclude missing data/not tossed and incorporating that into the prior. Depending on the assumptions you want to make on the missing data process, you can simply use the observed data likelihood. Also, it might help to think about it this way, if you don't toss the coin, your posterior probability should be the same as your prior. – jsk Aug 20 at 16:45
• Why is it missing? Is whether or not the data is missing for coin 2 dependent on the results for coin 1? – Louis Cialdella Aug 20 at 16:49
In theory, your problem is easy:
• Compute what you need conditionnaly upon you missing data
p(what I need| My Data, The missing Data)
• Then marginalize the missing Data
p(what I need| My Data) =
Sum_The missing Data p(what I need, The missing Data| My Data) =
Sum_The missing Data p(what I need| My Data, The missing Data) p(The missing Data)
In practice, you might run into numerical trouble if the number N of missing data is too large because the sum has 2^N terms for binary data.
• Thanks @Fabrice Pautot. Could you be more specific about the posterior? For example, what would be the posterior after observing (Head,No tossing) when the prior is given by $Dir(a_{HH},a_{HL},a_{LH},a_{LL})$? – Andeanlll Sep 6 at 14:25
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2019-12-13 07:34:44
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https://quantumcomputing.stackexchange.com/tags/nonclassicality/new
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# Tag Info
$$I(A:B)=S(A)+S(B)-S(AB)$$ $$J(A_{\{\Pi_{i}\}}:B)=S(A_{\{\Pi_{i}\}})+S(B)-S(A_{\{\Pi_{i}\}}B)$$ $$I(A:B)-J(A_{\{\Pi_{i}\}}:B)=S(A)-S(AB)-S(A_{\{\Pi_{i}\}})+S(A_{\{\Pi_{i}\}}B)$$ Since $$\rho = \sum_j p_j \pi_j\otimes \rho_j,$$ it is invariant to the measurements performed on the subsystem A, after the results are forgotten, so $$S(A)=S(A_{\{\Pi_{i}\}})$$ and ...
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2021-09-18 01:36:06
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https://stats.stackexchange.com/questions/321135/sequence-to-sequence-models-why-is-the-start-sequence-e-g-s-required
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# Sequence to Sequence models: Why is the start sequence (e.g. “</s>”) required?
Many people add a symbol to the beginning of the decoder sentence for training, also known as the start sequence. For example,
Encoder input: "How are you today?"
Desired decoder input: "</s> I'm very good! </e>"
Why is this? Will the Seq2Seq model not converge if the symbol wasn't there?
a seq2seq is typically implemented using an rnn. Mathematically, an rnn looks something like:
$$h_{t} = f(h_{t-1}, i_t) \\ o_{t} = g(h_t)$$
... where:
• $h_t$ is the hidden state at timestep $t$
• $i_t$ is the input at timestep $t$
• $o_t$ is the output at timestep $t$
• $f( \cdot, \cdot)$ is some function, possibly linear, eg $\mathbf{W}\mathbf{x} + \mathbf{b}$, but could be more complicated than this
• $g(\cdot)$ is another function, again could be linear
Then, seq2seq happens in two phases:
• in phase 1, we take in the input sentence, one token at a one, and update the hidden state
• in phase 2, we initialize the hidden state with the final hidden state from state 1, then predict outputs
For phase 2, looking at the equation, it is not enough to have just the hidden state to generate an output: we also need an input at each timestep.
What should be the input to the rnn at each timestep in phase 2?
• for all but the first tokens, we can use the previous output
• but what about for the very first output token
• so we create a special 'start' token, which we feed in as the first token, when we are generating the output
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2019-12-06 20:32:36
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https://brilliant.org/problems/time-period-of-oscillations-of-a-particle/
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# Time period of oscillations of a particle revolving in a paraboloid
Consider a vertical paraboloid $$y = r^2$$. A small particle kept on its smooth inner surface at $$y_{0} = 1 m$$ is given an angular velocity of $$\omega = \omega_{0}(1+ \eta)$$, where $$\eta <<1$$, and $$\omega_{0} = \sqrt{19.62} \text{rad}/s^2$$. It has no vertical velocity initially.
Find the time period of its vertical oscillations in $$\text{ seconds }$$. Assume $$g = 9.81 m/s^2$$.
Hint : Use conservation of angular momentum and conservation of energy. Afterwards, use approximations in your integral ($$\eta<<1$$)
Sorry for not posting problems for a long time. Hope you enjoy!
×
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2018-06-25 06:31:54
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https://eccc.weizmann.ac.il/eccc-reports/2006/TR06-089/index.html
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Under the auspices of the Computational Complexity Foundation (CCF)
REPORTS > DETAIL:
### Paper:
TR06-089 | 16th July 2006 00:00
TR06-089
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2021-01-21 06:11:38
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http://math.ac.vn/en/ho%E1%BA%A1t-%C4%91%E1%BB%99ng-trong-tu%E1%BA%A7n/icalrepeat.detail/2019/04/11/3509/-/-.html
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# Weekly Activities
New results on stability and $L_ {\infty}$-gain analysis for positive linear differential-algebraic equations with unbounded time-varying delays
Speaker: Nguyen Huu SauTime: 9h, Friday, April 11, 2019 Location: Room 513, Building A6, Institute of MathematicsAbstract: This paper addresses the problems of stability and $L_ {infty}$ gain analysis for positive linear differential-algebraic equations with unbounded time-varying delays. First, we consider the stability problem of a class of positive linear differential-algebraic equations with unbounded time-varying delays. A new method, which is based on the upper bounding of the state vector by a decreasing function, is presented to analyze the stability of the system. Then, by investigating the monotonicity of state trajectory, the $L_ {infty}$ -gain for differential-algebraic systems with unbounded time-varying delay is characterized. It is shown that the $L_ {infty}$-gain for differential-algebraic systems with unbounded time-varying delay is also independent of the delays and fully determined by the system matrices. A numerical example is given to illustrate the obtained results
### Highlights
12/04/21, Conference:CIMPA School “Functional Equations: Theory, Practice and Interactions” 22/04/21, Conference:Hội thảo TỐI ƯU VÀ TÍNH TOÁN KHOA HỌC lần thứ 19 29/06/21, Conference:The 7th International Conference on Random Dynamical Systems
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2021-02-25 02:55:29
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https://128.84.21.199/list/hep-ph/recent
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# High Energy Physics - Phenomenology
## Authors and titles for recent submissions
[ total of 122 entries: 1-25 | 26-50 | 51-75 | 76-100 | 101-122 ]
[ showing 25 entries per page: fewer | more | all ]
### Mon, 25 Sep 2017 (showing first 25 of 26 entries)
[1]
Title: Spin Precession Experiments for Light Axionic Dark Matter
Subjects: High Energy Physics - Phenomenology (hep-ph); Cosmology and Nongalactic Astrophysics (astro-ph.CO); High Energy Physics - Experiment (hep-ex); Atomic Physics (physics.atom-ph)
[2]
Title: Freeze-out of baryon number in low-scale leptogenesis
Subjects: High Energy Physics - Phenomenology (hep-ph); Cosmology and Nongalactic Astrophysics (astro-ph.CO)
[3]
Title: Positivity bounds on gluon TMDs for hadrons of spin $\le$ 1
Subjects: High Energy Physics - Phenomenology (hep-ph)
[4]
Title: QED and QCD self-energy corrections through the loop-tree duality
Comments: 9 pages, 1 figure, To be published in Journal of Physics Conference Series (IOP). Joint Proceedings of the XXXI Annual Meeting of the Division of Particles and Fields of the Mexican Physical Society
Subjects: High Energy Physics - Phenomenology (hep-ph)
[5]
Title: Reconstruction of top-quark mass effects in Higgs pair production and other gluon-fusion processes
Subjects: High Energy Physics - Phenomenology (hep-ph)
[6]
Title: Effects of the temperature and magnetic-field dependent coupling on the properties of QCD matter
Authors: Li Yang, Xin-jian Wen
Comments: 7 pages, 7 figures, version accepted for publication in Phys. Rev. D
Subjects: High Energy Physics - Phenomenology (hep-ph)
[7]
Title: Two-Loop Renormalization Group Equations and Gauge Couplings Unification
Comments: 20 pages, 23 figures, 1 table
Subjects: High Energy Physics - Phenomenology (hep-ph); High Energy Physics - Theory (hep-th)
[8]
Title: Photon-jet correlations at RHIC and the LHC
Authors: M. Klasen
Subjects: High Energy Physics - Phenomenology (hep-ph)
[9]
Title: Some Phenomenologies of a Simple Scotogenic Inverse Seesaw Model
Authors: Yi-Lei Tang
Subjects: High Energy Physics - Phenomenology (hep-ph)
[10]
Title: The five-loop Beta function for a general gauge group and anomalous dimensions beyond Feynman gauge
Comments: 17 pages, 2 ancillary files files available with the source
Subjects: High Energy Physics - Phenomenology (hep-ph); High Energy Physics - Theory (hep-th)
[11]
Title: Calculation of the Decay Rate of Tachyonic Neutrinos against Charged-Lepton-Pair and Neutrino-Pair Cerenkov Radiation
Journal-ref: J.Phys.G 44 (2017) 105201
Subjects: High Energy Physics - Phenomenology (hep-ph)
[12]
Title: Phenomenology of Majorons
Authors: Julian Heeck
Comments: 4 pages. Contributed to the 13th Patras Workshop on Axions, WIMPs and WISPs, Thessaloniki, May 15 to 19, 2017
Subjects: High Energy Physics - Phenomenology (hep-ph)
[13]
Title: Spontaneous mass generation suggests small dimension of the SM group U(1)xSU(2)xSU(3)
Comments: Contribution to the EPS conference on High Energy Physics, Venice, Italy, 5th-12th July 2017. 3+1 pages, 2 figures
Subjects: High Energy Physics - Phenomenology (hep-ph)
[14]
Title: Possible roles of Peccei-Quinn symmetry in an effective low energy model
Authors: Daijiro Suematsu
Subjects: High Energy Physics - Phenomenology (hep-ph)
[15]
Title: Gravitational Waves from Hidden QCD Phase Transition
Subjects: High Energy Physics - Phenomenology (hep-ph)
[16]
Title: Massless on-shell box integral with arbitrary powers of propagators
Authors: O. V. Tarasov
Subjects: High Energy Physics - Phenomenology (hep-ph)
[17]
Title: Solving differential equations for Feynman integrals by expansions near singular points
Subjects: High Energy Physics - Phenomenology (hep-ph)
[18]
Title: On the observable spectrum of theories with a Brout-Englert-Higgs effect
Subjects: High Energy Physics - Phenomenology (hep-ph); High Energy Physics - Lattice (hep-lat); High Energy Physics - Theory (hep-th)
[19]
Title: How the axial anomaly controls flavor mixing among mesons
Subjects: High Energy Physics - Phenomenology (hep-ph); High Energy Physics - Lattice (hep-lat)
[20] arXiv:1709.07612 (cross-list from hep-lat) [pdf, ps, other]
Title: Thermal quarkonium physics in the pseudoscalar channel
Subjects: High Energy Physics - Lattice (hep-lat); High Energy Physics - Phenomenology (hep-ph)
[21] arXiv:1709.07513 (cross-list from hep-lat) [pdf, ps, other]
Title: Progress in computing parton distribution functions from the quasi-PDF approach
Comments: 8 pages, 3 figures; Proceedings of the 35th International Symposium on Lattice Field Theory, Granada, Spain
Subjects: High Energy Physics - Lattice (hep-lat); High Energy Physics - Phenomenology (hep-ph)
[22] arXiv:1709.07503 (cross-list from gr-qc) [pdf, other]
Title: Can the graviton have a large mass near black holes?
Subjects: General Relativity and Quantum Cosmology (gr-qc); High Energy Astrophysical Phenomena (astro-ph.HE); High Energy Physics - Phenomenology (hep-ph); High Energy Physics - Theory (hep-th)
[23] arXiv:1709.07479 (cross-list from astro-ph.CO) [pdf, other]
Title: Primordial non-Gaussianity and power asymmetry with quantum gravitational effects in loop quantum cosmology
Comments: revtex4, one figure and no tables
Subjects: Cosmology and Nongalactic Astrophysics (astro-ph.CO); General Relativity and Quantum Cosmology (gr-qc); High Energy Physics - Phenomenology (hep-ph); High Energy Physics - Theory (hep-th)
[24] arXiv:1709.07451 (cross-list from astro-ph.CO) [pdf, other]
Title: Vacuum dynamics in the Universe versus a rigid $Λ=$const
Comments: Published in Int. J. Mod. Phys. A32 (2017) 1730014. arXiv admin note: text overlap with arXiv:1705.06723
Subjects: Cosmology and Nongalactic Astrophysics (astro-ph.CO); General Relativity and Quantum Cosmology (gr-qc); High Energy Physics - Phenomenology (hep-ph); High Energy Physics - Theory (hep-th)
[25] arXiv:1709.07445 (cross-list from physics.ins-det) [pdf, other]
Title: Analysis of Ultra High Energy Muons at the INO-ICAL Using Pair-Meter Technique
Comments: Presented in Advanced Detectors for Nuclear, High Energy and Astroparticle Physics - Bose Institute Kolkata, India
Subjects: Instrumentation and Detectors (physics.ins-det); High Energy Physics - Phenomenology (hep-ph)
[ total of 122 entries: 1-25 | 26-50 | 51-75 | 76-100 | 101-122 ]
[ showing 25 entries per page: fewer | more | all ]
Disable MathJax (What is MathJax?)
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2017-09-25 09:38:37
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https://www.ideals.illinois.edu/handle/2142/79385
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Title: DUAL-COMB SPECTROSCOPY IN THE OPEN AIR Author(s): Coddington, Ian Contributor(s): Newbury, Nathan R.; Swann, William C; Ycas, Gabriel; Truong, Gar-Wing; Baumann, Esther; Sinclair, Laura; Giorgetta, Fabrizio; Diddams, Scott; Klose, Andrew; Rieker, Greg B Subject(s): Mini-symposium: High-Precision Spectroscopy Abstract: Dual-comb spectroscopy is arguably the natural successor to FTIR. Based on the interference between two frequency combs, this technique can record broadband spectra with a resolution better than 0.0003 wn. Like FTIR, dual-comb spectroscopy measures an entire spectrum simultaneously, allowing for suppression of systematic errors related to temporal dynamics of the sample. Unlike FTIR it records the entire spectrum with virtually no instrument lineshape or error in the frequency axis. The lack of moving parts in dual-comb spectroscopy means that spectra can be recorded in milliseconds to microseconds with the desired signal-to-noise being the only real constrain on the minimum recording time. Finally the high spacial beam quality of the frequency combs allows for increased sensitivity through long interaction paths either in free-space, multi-pass cells or enhancement cavities. This talk will explore the recent use of dual-comb spectroscopy in the near-infrared to measure atmospheric carbon dioxide, methane and water concentrations over a 2-km outdoor open-air path. Due to many of the strengths just mentioned, precisions of $<$1 ppm for CO$_2$ and $<$3 ppb for CH$_4$ in 5 min are achieved making this system very attractive for carbon monitoring at length scales relevant to carbon transport models. Additionally this presentation will address recent work on robust, compact, and portable dual-comb spectrometers as well as dual-comb spectroscopy further into the IR. Issue Date: 24-Jun-15 Publisher: International Symposium on Molecular Spectroscopy Citation Info: ACS Genre: CONFERENCE PAPER/PRESENTATION Type: Text Language: English URI: http://hdl.handle.net/2142/79385 Date Available in IDEALS: 2016-01-05
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2017-04-23 08:18:22
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https://math.stackexchange.com/questions/1733438/asymptotic-equivalence-between-harmonic-numbers-and-logarithm
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# asymptotic equivalence between harmonic numbers and logarithm
I have recently learned about asymptotical equivalence, defined as $$a(n) \sim b(n) \Leftrightarrow \lim\limits_{n \to \infty} \frac{a(n)}{b(n)} = 1$$
Now I would need to prove that $H_n \sim \ln(n)$ where $H_n = \sum_{i=1}^n \frac{1}{i}$ is the harmonic series and $\ln(n)$ is the natural logarithm, but I have no idea how to get started.
I wrote down the definition as \begin{align} \lim\limits_{n \to \infty} \frac{H_n}{\ln(n)} & = \lim\limits_{n \to \infty} \left(\frac{1}{\ln(n)} \cdot \sum_{i=1}^n \frac{1}{i}\right) \\ \end{align} and tried to apply l'Hospital's rule which leads to $\lim\limits_{n \to \infty} \left(\frac{-1}{n\ln^2(n)} \cdot \sum_{i=1}^n \frac{-1}{i^2}\right)$, but that does not seem to be helping me any further. I tried out some other things, but everything seemed to be incorrect or not helpful.
I am looking at it for quite some time now already and I also tried to find some hints online, but I just don't seem to find anything that uses this definition. I must admit I am not really the biggest hero in analysis or calculus, but if anyone could guide me in (one of) the right directions to get to a solution that would already be very helpful. Thanks in advance.
• L'Hopital's rule won't work here because a sum is a discrete sequence and therefore has no derivative. However, there is a "discrete L'Hopital's rule" which works here. That's what Jack uses in his answer below. – Antonio Vargas Apr 9 '16 at 15:54
By the Stolz-Cesàro theorem, $$\lim_{n\to +\infty}\frac{H_n}{\log n} = \lim_{n\to +\infty}\frac{H_{n+1}-H_{n}}{\log(n+1)-\log n} =\lim_{n\to +\infty}\frac{1}{(n+1)\log\left(1+\frac{1}{n}\right)}=\color{red}{1}.$$
As an alternative, $$0\leq H_n-\log(n+1) = \sum_{k=1}^{n}\left(\frac{1}{k}-\log\left(1+\frac{1}{k}\right)\right)\leq\sum_{k=1}^{n}\frac{1}{2k^2}<\frac{\zeta(2)}{2}=\frac{\pi^2}{12}$$ is enough. It comes from a telescoping trick and the fact that $x-\log(1+x)\leq\frac{x^2}{2}$ over $[0,1]$.
• That's really a useful theorem. I find it strange that I've never heard about it before... I wanted to try to understand the alternative, but I'm afraid I'm missing a bit too much in order to get it, but thanks anyway! If I get some spare time I have something new to think about ;) – Mr Tsjolder Apr 8 '16 at 19:07
It is
$H_n - 1 = \sum_{k=2}^{n} \frac{1}{k} \leq \sum_{k=2}^{n} \int_{k-1}^{k} \frac{1}{k} = \int_{1}^{n} \frac{1}{x} \mathrm{d}x = \ln (x)$
and
$H_n - \frac{1}{n} = \sum_{k=1}^{n-1} \frac{1}{k} \geq \sum_{k=1}^{n-1} \int_{k}^{k+1} \frac{1}{x} = \ln(x)$.
Hence $\ln (n) + \frac{1}{n} \leq H_n \leq \ln (n) + 1$.
Therefore $1 + \frac{1}{n \ln (n)} \leq \frac{H_n}{\ln (n)} \leq 1 + \frac{1}{\ln (n)}$.
You will get the desired result by $n \to \infty$.
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2019-06-20 05:07:23
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https://imcs.dvfu.ru/cats/problem_text?cid=790664;pid=1220137;sid=
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Problem H. Horticulture ≡
• problems
• en ru
Author: M. Sporyshev Time limit: 1 sec Input file: input.txt Memory limit: 256 Mb Output file: output.txt
Statement
Young programmer James decided to practice horticulture and brought home N plants.
Plant number i has starting height of ai centimeters and grows by vi centimeters per day. Vasya has a girlfriend Alice, who is currently traveling away from home. Alice does not like tall plants, so Vasya decided to cut his plants shorter from time to time.
Vasya can reduce height of any plant by exactly X centimeters (but not make it negative) in one cut. Vasya's endurance in not very good, so he can make no more than M cuts per day. Every day, plants grow after all cutting is finished.
Alice will return in T days. Vasya would like to cut plants in such a way as to make on the day of her arrival the height of the highest plant as small as possible.
Input file format
First line of input file contains integers N, T, M, X.
Second line contains N integers ai — initial plant heights.
Third line contains N integers vi — plant growth speeds.
Output file format
Output file must contain a single integer — minimal height of the tallest plant after T days.
Constraints
1 ≤ N ≤ 105
1 ≤ T ⋅ M ≤ 105
1 ≤ X, ai, vi ≤ 105
Sample tests
No. Input file (input.txt) Output file (output.txt)
1
4 1 2 4
8 9 10 11
8 8 3 2
13
2
1 100000 1 100000
100000
100000
100000
0.075s 0.010s 13
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2023-03-24 08:40:38
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http://lotsasplainin.blogspot.com/2009/07/wednesday-math-vol-80-implication.html
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# Lotsa 'Splainin' 2 Do
This blog is still alive, just in semi-hibernation.
When I want to write something longer than a tweet about something other than math or sci-fi, here is where I'll write it.
## Wednesday, July 22, 2009
### Wednesday Math, Vol. 80: Implication
Logic has been considered a part of philosophy for several thousand years. In the mid 1800's, a group of English mathematicians took it upon themselves to turn logic into a part of the mathematical field, foremost among them being George Boole, Augustus DeMorgan and Charles Dodgson, known better by his pen name Lewis Carroll. In mathematical logic, the variables like P and Q stand in for statements like "The moon is made of green cheese." or "It's Tuesday." or "The door is open." A statement can either be true (T) or false (F). There are modern extensions of logic where statements can be somewhere between true and false, and this field is called fuzzy logic. Regular logic is tricky enough, so let's stick to that for the time being.
Once we have statements, logic starts dealing with compound statements. The simplest logical operators are AND (denoted with a ^) and OR (denoted with a v). The compound statement "P AND Q" is only true is both P and Q are true, while "P OR Q" is true unless both P is false and Q is false.
Then there's implication, stated either as "P IMPLIES Q" or "IF P, THEN Q". Implication has been a very important concept in logic ever since the Greeks started talking about Socrates being a man and all men being mortal, but it's a little trickier than AND or OR.
The first statement in an implication is the premise. If the premise of an implication is false, the entire implication is considered to be true, the only way for an implication to be false is for the premise to be true while the conclusion is false.
This is an odd idea, but let's think of it in terms of a law like the legal age to buy alcohol. We can think of the law as "Buying alcohol IMPLIES you are 21 years or older." There are four possible situations, three of them legal and one illegal.
You buy alcohol AND you are 21 years or older (legal.)
You buy alcohol AND you are NOT 21 years or older (illegal.)
You do NOT buy alcohol AND you are 21 years or older (legal.)
You do NOT buy alcohol AND you are NOT 21 years or older (legal.)
If you don't buy alcohol, the premise of the implication is false and the entire implication is true. There is no way to break this law if you never buy alcohol. The only illegal situation is for the premise to be true, in this case, buying alcohol, and the conclusion to be false, being less than 21 years old.
While this can be explained to most people's satisfaction relatively easily, implication can make logic something of a hornet's nest. One of the important concepts in logic is the tautology, a compound statement that is always true. The simplest tautology is "P OR NOT P", which is to say, either the moon is made of green cheese or the moon is not made of green cheese. It has to be one way or the other, so we can't go wrong with this statement, it's always true. The next simplest tautology is reverse implication, "P IMPLIES Q OR Q IMPLIES P". Because implication works the same as "NOT P OR Q", reverse implication can be translated at "P OR NOT P OR Q OR NOT Q", and it's always true.
But it doesn't sound always true.
It's raining implies it's Tuesday or it's Tuesday implies it's raining.
I'm a man implies I'm a woman or I'm a woman implies I'm a man.
Those last two statements are 100% true, regardless of what day it is or if it's raining or not or in the second case, the gender of the speaker. They work because if one of the statements is false, it is the premise in an implication which must be true, and with an OR statement, one true part is enough. The other possibility is that both statements are true, and then the reverse implication is fine as well.
We are lead to believe that logic is the way the mind is supposed to work if operating properly. I teach my students that logic is a game with rules, and you have to learn the rules, even if sometimes it creates always true statements that don't sound like they make sense.
dguzman said...
This kind of logic talk makes my brain tired. I like your point, that logic is a game with rules; otherwise, it's too overwhelming to think about.
Matty Boy said...
Hi, dg. I just want you to know that I appreciate your comments on Wednesday Math posts, though I know they aren't your favorites.
47th Problem of Euclid said...
At UNC, I used to take the bus with a logic PhD student. He used to constantly remind me that the kind of logic that mathematicians use is hopelessly obsolete by logician standards: using the Law of the Excluded Middle, that contemporary logic has rejected along with bivalent truth schemes. The problem is that proofs fall apart without the Law of the Excluded Middle (try proving that $\sqrt{2}$ is irrational without it).
Matty Boy said...
I wasn't aware of the modern version of logic. I remember in my undergraduate career having people talk about doing math without the Axiom of Choice. Lots of proofs crumble to the ground without that one as well.
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2013-05-20 01:28:05
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https://cracku.in/ssc-questions/ssc-cgl-reasoning
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# SSC SSC CGL Reasoning Questions with Answers
#### SSC CGL Reasoning
An excellent collection of SSC SSC CGL Reasoning questions and answers with detailed explanations for competitive exams. Given below are some of the most repeated practice questions on SSC CGL Reasoning and General Intelligence for SSC CGL, CHSL, JE, GD constable, Stenographer, MTS, and CPO exams. Go through this very important online quiz based on model and previous year asked questions from SSC SSC CGL Reasoning with solutions to ace the exam.
### Take a free SSC CHSL Mock
Thousands of students have taken Cracku's Free SSC CHSL Mock.
Instructions
For the following questions answer them individually
Question 1
If 5$125 = 25, 12$48 = 4 then what is the value of 4$24 = ? Question 2 If 45 % 11 = 7, 59 % 34 = 7 then what is the value of 55 % 4 = ? Question 3 If 18 x 12 = 206 and 19 x 22 = 408, then 23 x 36 = ? Question 4 If 35 % 31 = 12, 92 % 30 = 14 then what is the value of 15 % 24 = ? Question 5 In India the reform policies were first introduced in which year? Question 6 In the following question, correct the equation by interchanging two signs. 6 + 8 ÷ 4 - 4 = 8 Question 7 In the following question, by using which mathematical operator will the expression become correct? 18 ? 6 ? 9 ? 27 Question 8 The weights of 4 boxes are 20, 40, 50 and 30 kilograms. Which of the following cannot be the total weight, in kilograms, of any combination of these boxes and in a combination a box can be used only once Question 9 If 19 (36) 13 and 37 (81) 28, then what is the value of 'A' in 43 (A) 38? Question 10 If 11$25 = 18, 12$20 = 16 then what is the value of 4$50 = ?
Question 11
The weights of 4 boxes are 20, 90, 40 and 60 kilograms. Which of the following cannot be the total weight,in kilograms, of any combination of these boxes and in a combination a box can be used only once?
Question 12
The weights of 4 boxes are 30, 20, 60 and 70 kilograms. Which of the following cannot be the total weight,in kilograms, of any combination of these boxes and in a combination a box can be used only once?
Question 13
The weights of 4 boxes are 100, 70, 50 and 90 kilograms. Which of the following cannot be the total weight, in kilograms, of any combination of these boxes and in a combination a box can be used only once?
Question 14
In the following figure, square represents Pharmacists, triangle represents Dancers, circle represents Gynecologists and rectangle represents Women. Which set of letters represents Gynecologists who are neither Women nor Dancers?
Question 15
The weights of 4 boxes are 30, 20, 50 and 90 kilograms. Which of the following cannot be the total weight, in kilograms, of any combination of these boxes and in a combination a box can be used only once?
Question 16
In the following figure, square represents Psychologists, triangle represents Chemists, circle represents Actors and rectangle represents Fathers. Which set of letters represents Psychologists and actors who are also fathers?
Question 17
Pearl Towers is taller than Sky Towers but shorter than Unity Towers. Unity Towers and Cyber Towers are of same height. Pearl Towers is shorter than Indus Towers. Amongst the buildings, who is the second shortest?
Question 18
In the following question, correct the given equation by interchanging two numbers.
8 x 3 ÷ 4 + 9 - 5 = 16
Question 19
If yesterday was Tuesday, then what day of the week will the tenth day from today be?
Question 20
In the following figure, square represents Optometrists, triangle represents Painters, circle represents Vegetarians and rectangle represents Men. Which set of letters represents painters who are either men or vegetarians?
Question 21
Which of the following interchanges in signs will make the given equation correct?
17- 15 x 5 = 250
Question 22
If 9@3=12; 15@4=22; 16@14=4; then what is the value of 6@2=?
Question 23
In a row of people Manu is 7th from bottom end of row. Shrey is 10 ranks above Manu. If Shrey is 8th from top end, then how many people are there in this row?
Question 24
If 26#14=80; 5#3=16; 6#5=22; then what is the value of 14#5=?
Question 25
If 4 x 9 x 3 = 4 and 5 x 3 x 1 = 3, then 9 x 9 x 7 = ?
Question 26
In a row of 74 girls, Shweta is 27th from left end. Palak is 7th to the right of Shweta. What is Palak's position from the right end of the row?
Question 27
The average temperature of a town in the first six days of a month was 410C and the sum of the temperatures of the first five days of the same month was 2010C. What was the temperature on the sixth day of the month?
Question 28
In the following figure, square represents Painters, triangle represents women, circle represents Accountants and rectangle represents Americans. Which set of letters represents Americans who are not accountants?
Question 29
In a company all Mondays and Sundays are offs. If a month starts with a Monday and has 31 days then how many offs will be there in that month?
Question 30
A, B, C,D and E are standing in a row. D is the immediate neighbour of A and E. B is at the right of E and C is in the extreme right. Who is fourth to the left of C?
Question 31
Chandan is 2 years older than Ankit but 1 year younger than Sumit. Ankit is twice as old as Khushboo. If Khushboo's age is 25 years, what is the age (in years) of Sumit?
Question 32
In the following figure, square represents Pharmacists, triangle represents singers, circle represents Surgeons and rectangle represents Mothers. Which set of letters represents surgeons who are either mothers or singers?
Question 33
If 17@1=8; 9@1=4; 6@4=1; then what is the value of 8@2=?
Question 34
If 5@1=60; 12@8=200; 16@2=180; then what is the value of 16@10=?
Question 35
If $$(3)^2 @ 1 * 7 = 98$$ and $$(4)^2 @ 2 * 16 = 178$$, then $$(5)^2 @ 3 * 9 = ?$$
Question 36
Nine years later, age of B will be equal to the present age of A. Sum of A's age 3 years later and B's age 4 years ago is 76. If C is half of the present age of B, then what will be C's age (in years) after 10 years?
Question 37
If $$9^{2}$$ A $$4^{2}$$ B $$3^{2}$$ = 56 and $$7^{2}$$ A $$2^{2}$$ B $$1^{2}$$ = 44, then
$$11^{2}$$ A $$5^{2}$$ B $$7^{2}$$ = ?
Question 38
In the following figure, square represents Therapists, triangle represents Geneticists, circle represents yoga practitioners and rectangle represents Fathers. Which set of letters represents yoga practitioners who are neither geneticists nor fathers?
Question 39
If 18 (9) 3 and 36 (30) 5, then what is the value of A in 19 (A) 18?
Question 40
There are 45 trees in a row. The lemon tree is 20th from right end. What is the rank of lemon tree from left end?
Question 41
Ankit scored 32, 45 and 49 marks in Physics, English and Philosophy respectively. The maximum marks in each subject is 50. Determine his overall percentage for the three subjects.
Question 42
The weights of 4 boxes are 40, 70, 80 and 50 kilograms.Which of the following cannot be the total weight,in kilograms, of any combination of these boxes and in a combination a box can be used only once
Question 43
Present age of A is 2 times the present age of B. After 8 years the B's age will be 4 times of C's present age. If C celebrated his fifth birthday 9 years
ago, then what is the present age (in years) of A?
Question 44
There are five students - P, Q, R, S and T having different heights in a class. P's height is more than only one student. Q's height is more than S and P but not more than R. S's height is more than P. R is not the smallest. Who is having the maximum height in the class?
Question 45
Which of the following interchanges in signs and numbers will make the given equation correct?
8 x 9 + 2 = 74
Question 46
Vishu,Pooja,Vishakha,Rani and Ram are sitting in a line. Pooja is third to the extreme right end. Vishu is second to the left of Pooja. Vishakha is to the right of Pooja. Rani is third to the right of Ram, who is the immediate neighbour of Vishu. Who is sitting in the middle?
Question 47
Hitesh, Sunny, Vicky, Nitin and Bharat are arranged in ascending order of the height from the top. Hitesh is at third place. Bharat is between Nitin and Hitesh while Nitin is not at the bottom. Who has the maximum height among them?
Question 48
The weights of 4 boxes are 70, 50, 30 and 90 kilograms. Which of the following cannot be the total weight,in kilograms, of any combination of these boxes and in a combination a box can be used only once
Question 49
The weights of 4 boxes are 20, 40, 80 and 90 kilograms. Which of the following cannot be the total weight, in kilograms, of any combination of these boxes and in a combination a box can be used only once?
Question 50
In the given figure , how many bottles are not brown
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2019-09-19 19:05:22
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https://codeforces.com/blog/entry/96337
|
### Skeef79's blog
By Skeef79, history, 5 weeks ago, translation,
Hello codeforces!
I am wondering is there a way to caculate prefix sums $k$ times modulo $p$ fast, especially when we can't use matrix multiplication?
By calculating prefix sums $k$ times I mean this piece of code:
for(int it = 0; it < k; it++) {
for(int i = 1; i < n; i++)
pref[i]+= pref[i-1];
}
So if we start with array: $1, 0, 0, 0$ we will get:
1. $1, 1, 1, 1$;
2. $1, 2, 3, 4$;
3. $1, 3, 6, 10$;
4. $1, 4, 10, 20$; ...
• +40
» 5 weeks ago, # | +17 you have written your blog in russian so the users with the english interface don't see it
• » » 5 weeks ago, # ^ | 0 Thank you for mentioning
» 5 weeks ago, # | 0 Auto comment: topic has been translated by Skeef79 (original revision, translated revision, compare)
» 5 weeks ago, # | 0 Fast meaning you want O(n) time independent of the value of k? or what exactly
• » » 5 weeks ago, # ^ | +1 Yes, $k$ can be upto $10^{18}$, but modulo I think should be around $10^5$
» 5 weeks ago, # | +110 If $b$ is the array after these operations then $b_i = \sum\binom{k+j}{k}a_{i-j}$, which is the $i$-th coefficient of some polynomial product, so this works in $O(n\log{n})$.
• » » 5 weeks ago, # ^ | +22 How did you get this formula? Is there any article?
• » » » 5 weeks ago, # ^ | +77 Rotate the second half of your post 45 degrees clockwise and it is Pascal's triangle.
• » » » » 5 weeks ago, # ^ | +17 Really...Thank you!
• » » » 5 weeks ago, # ^ | +16 Here is some informal handwaving. Each occurrence of $a_{i-j}$ into $b_i$ is basically a sequence $(d_1, \ldots, d_k)$ of nonnegative "steps" with sum being equal to $j$. It has the following meaning: when we expand the last a[i] := sum(a[0..i]), we look specifically at $a_{i-d_1}$ from the previous step. During the previous step, this value was assigned the sum of its prefix, but we look only at $a_{i-d_1-d_2}$, and so on. In the end we will look at $a_{i-j}$, and this is how it contributed into the final value along this path: through positions $(i-j)\to (i-j+d_k)\to\ldots \to i$.On the other hand, the number of such sequences can be computed using stars and bars technique and equals that binomial coefficient.
» 5 weeks ago, # | ← Rev. 2 → +21 we can get k-times prefix sum of an array by a transformation reference. $$f_{k} = (\sum_{i=0}^{n-1} a_{i}*x^i)(1/(1-x)^k)$$
» 5 weeks ago, # | ← Rev. 2 → +11 Btw similar problem exists on project euler 739Golovanov399 was 4th person to solve it. Its soln is similar to what Golovanov399 mentioned above. If you AC it you would be able to access its discussion forum were people have mentioned how did they solved that problem.
» 5 weeks ago, # | +24 Here's the same problem on Codeforces: 223C. Partial Sums
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2021-11-29 03:09:53
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|
https://www.nag.co.uk/numeric/dt/nagdotnet_dtw02/html/M_NagLibrary_S_s11ac.htm
|
s11ac returns the value of the inverse hyperbolic cosine, $\mathrm{arccosh} x$. The result is in the principal positive branch.
# Syntax
C#
```public static double s11ac(
double x,
out int ifail
)```
Visual Basic
```Public Shared Function s11ac ( _
x As Double, _
<OutAttribute> ByRef ifail As Integer _
) As Double```
Visual C++
```public:
static double s11ac(
double x,
[OutAttribute] int% ifail
)```
F#
```static member s11ac :
x : float *
ifail : int byref -> float
```
#### Parameters
x
Type: System..::..Double
On entry: the argument $x$ of the function.
Constraint: ${\mathbf{x}}\ge 1.0$.
ifail
Type: System..::..Int32%
On exit: ${\mathbf{ifail}}={0}$ unless the method detects an error or a warning has been flagged (see [Error Indicators and Warnings]).
#### Return Value
s11ac returns the value of the inverse hyperbolic cosine, $\mathrm{arccosh} x$. The result is in the principal positive branch.
# Description
s11ac calculates an approximate value for the inverse hyperbolic cosine, $\mathrm{arccosh} x$. It is based on the relation
$arccosh x=lnx+x2-1.$
This form is used directly for $1, where $k=n/2+1$, and the machine uses approximately $n$ decimal place arithmetic.
For $x\ge {10}^{k}$, $\sqrt{{x}^{2}-1}$ is equal to $\sqrt{x}$ to within the accuracy of the machine and hence we can guard against premature overflow and, without loss of accuracy, calculate
$arccosh x=ln 2+ln x.$
# References
Abramowitz M and Stegun I A (1972) Handbook of Mathematical Functions (3rd Edition) Dover Publications
# Error Indicators and Warnings
Errors or warnings detected by the method:
${\mathbf{ifail}}=1$
The method has been called with an argument less than $1.0$, for which $\mathrm{arccosh} x$ is not defined. The result returned is zero.
${\mathbf{ifail}}=-9000$
An error occured, see message report.
# Accuracy
If $\delta$ and $\epsilon$ are the relative errors in the argument and result respectively, then in principle
$ε≃xx2-1arccosh x×δ.$
That is the relative error in the argument is amplified by a factor at least $\frac{x}{\sqrt{{x}^{2}-1}\mathrm{arccosh} x}$ in the result. The equality should apply if $\delta$ is greater than the machine precision ($\delta$ due to data errors etc.) but if $\delta$ is simply a result of round-off in the machine representation it is possible that an extra figure may be lost in internal calculation and round-off. The behaviour of the amplification factor is shown in the following graph:
Figure 1
It should be noted that for $x>2$ the factor is always less than $1.0$. For large $x$ we have the absolute error $E$ in the result, in principle, given by
$E∼δ.$
This means that eventually accuracy is limited by machine precision. More significantly for $x$ close to $1$, $x-1\sim \delta$, the above analysis becomes inapplicable due to the fact that both function and argument are bounded, $x\ge 1$, $\mathrm{arccosh} x\ge 0$. In this region we have
$E∼δ.$
That is, there will be approximately half as many decimal places correct in the result as there were correct figures in the argument.
None.
None.
# Example
This example reads values of the argument $x$ from a file, evaluates the function at each value of $x$ and prints the results.
Example program (C#): s11ace.cs
Example program data: s11ace.d
Example program results: s11ace.r
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2018-12-10 15:05:56
|
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|
https://gmatclub.com/forum/is-a-b-a-b-105457.html?kudos=1
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# Is |a| + |b| > |a + b| ?
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28 Nov 2010, 07:05
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Is |a| + |b| > |a + b| ?
(1) a^2 > b^2
(2) |a| * b < 0
[Reveal] Spoiler: OA
Last edited by Bunuel on 04 Dec 2012, 01:57, edited 1 time in total.
Renamed the topic and edited the question.
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chiragatara wrote:
Hi Bunuel,
I am not very much comfortable with Absolute value Problems, e.g.
Is |a| + |b| > |a + b| ?
(1) a2 > b2
(2) |a| × b < 0
Is there any GYAN Material for improvement? Kindly help.
Hi, Don't worry. You will learn the things and GMAT absolute value tricks sooooon.
The be low is my approcah for any modulus qtn in GMAT.
Remember.
The meaning of |x-y| is "On the number line, the distance between X and +Y"
The meaning of |x+y| is "On the number line, the distance between X and -Y"
The meaning of |x| is "On the number line, the distance between X and 0".
On the # line, Left to 0 are all the -ve #s and right to 0 are all +ve #s
Original Qtn:
Is |a| + |b| > |a + b| ?
menas "is the SUM of the distance between a and 0, and the distance bewtween b and 0 > the distance bewtween a+b and 0"
let us take some cases to simplify the qtn.
case1: if a and b both are RIGHT to 0 on the # line then a+b will also be RIGHT to zero and even more far away from 0 than a or/and b is/are.
Then |a| + |b| > |a + b| is FLASE
and infact LHS is always = RHS
e.g: a = 1 and b = 3 then a+b = 4
a is 1 unit away right to 0
b is 3 units away right to
a+b is 4 units away right to 0
==>
the SUM of the distance between a and 0, and the distance between b and 0 = 1+3 = 4
and the distance b/w a+b and 0 is 4
-------------0----------b-----a---------a+b---
case2: if a and b both are LEFT to 0 on the # line then a+b will also be LEFT to zero and even more far away from 0 than a and/or b is.
Then |a| + |b| > |a + b| is FLASE
and infact LHS is always = RHS
e.g: a = -1 and b = -3 then a+b = -4
a is 1 unit away left to 0
b is 3 units away left to
a+b is 4 units away left to 0
==>
the SUM of the distance between a and 0, and the distance between b and 0 = 1+3 = 4
and the distance b/w a+b and 0 is 4
----a+b------a---------b-------0-------------
HENCE,
for |a| + |b| > |a + b| or
|a| + |b| < |a + b| or
|a| + |b| not= |a + b| to be true,
a and b, on the # line, should NOT be on the same side with respect to 0.
So BASICALLY the qtn is ASKING IF a AND b ARE on DEFFERENT SIDES, with respect to 0, on THE # LINE>
stmnt1: a^2 > b^2
from this we can say that the magnitude of a is > the magnitude of b. magnitude means with out sign.
==> it says that |a| > |b|
FROM THIS,
a and b can be on the same side on the # line with respect to 0
-------0------b----------------a , obviosly a is toofar ways from 0 than b is ==> |a| > |b|
OR,
a can be on a different side than b
-----a-------------0---b------ again, a is toofar ways from 0 than b is ==> |a| > |b|
so we have both the cases...hence stmnt 1 is NOT suff.
stmnt2: |a| × b < 0
|a| is always +ve hence b shud be -ve for the product to be -ve.
so all that stmnt 2 says is b is LEFT to 0 on the # line but it does not say whether a is on LEFT or RIGHT to 0. hence NOT suff.
stmnts 1&2 together
stmnt:1 says that a is too far from zero than b is.
stmnt2 says that b is LEFT to 0 on the # line
hence using both too, we can not say if a and b are on different sides to 0 on the #line or on the on the same sides
hence NOT suff.
Regards,
Murali.
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28 Nov 2010, 08:13
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chiragatara wrote:
Hi Bunuel,
I am not very much comfortable with Absolute value Problems, e.g.
Is |a| + |b| > |a + b| ?
(1) a2 > b2
(2) |a| × b < 0
Is there any GYAN Material for improvement? Kindly help.
You should notice that inequality $$|a|+|b|>|a+b|$$ holds true if and only $$a$$ and $$b$$ have opposite sign, as only in this case absolute value of positive+negative will be less than |positive|+|negative|. For example |-2|+|3|>|-2+3|. In all other cases $$|a|+|b|=|a+b|$$.
So, basically the question is whether $$a$$ and $$b$$ have opposite sign.
(1) a^2 > b^2 --> can not determine whether $$a$$ and $$b$$ have opposite sign. Not sufficient.
(2) |a|*b<0 --> just tells us that $$b<0$$, but we don't know the sign of $$a$$. Not sufficient.
(1)+(2) $$b<0$$ and $$a^2>b^2$$ --> still can not get the sign of $$a$$. Not sufficient.
For theory check About Value chapter of Math Book created by Walker: math-absolute-value-modulus-86462.html
Absolute value PS questions: search.php?search_id=tag&tag_id=58
Absolute value DS questions: search.php?search_id=tag&tag_id=37
Hard absolute value questions: inequality-and-absolute-value-questions-from-my-collection-86939.html
Hope it helps.
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16 Jun 2011, 00:32
clean E here.
b < 0 but a can be < > 0
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24 Aug 2011, 20:09
|a| + |b| > |a+b|?
if a and b are of same sign then LHS = RHS. Hence No.
if a and b are of opposite sign then LHS > RHS . Then Yes.
1. Not sufficient
|a|>|b|
we cannot say anything about the sign of a and b.
2. Not sufficient
b<0, but we dont know the sign of a.
together,
|a|>|b| and b<0 => a can be positive or negative
Hence not sufficient.
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30 Aug 2011, 02:08
|a| + |b| > |a + b|
the expression will be true if "a,b have opposite signs e.g. a<0, b>0 etc"
st1) insufficient. a or b could have same or different sign
st2) this proves that b<0. but it is also possible that a<0. so, insufficient.
E is the ans
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04 Dec 2012, 01:54
Is |a| + |b| > |a + b| ?
(1) Both positive |5| + |5| > |5 + 5| NO!
(2) Both negative |-5| + |-5| > |-10| NO!
(3) Oppositve signs |5| + |-5| > |5-5| YES!
(1) a2 > b2
|a| > |b|
This doesn't tell us anything about the sign. INSUFFICIENT.
(2) |a| × b < 0
This tells us b is negative. INSUFFICIENT.
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06 Jun 2013, 08:03
$$Is |a| + |b| > |a + b| ?$$
$$(1) a^2>b^2$$
($$2)|a| * b<0$$
got me confusing for a while and choose E.
used the std rule of | A + B | < |A| + | B | when XY > 0
Since it is difficult to find the sign of A in any case selected E.
Please let me know if the approach to this problem is correct ?
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Re: Ds Modules and inequalities [#permalink]
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06 Jun 2013, 09:43
Normally I solve it by plugging numbers.
1. a^2>b^2
lets take a=-3 and b=-2; so, |a|+|b|=5 & |a+b|=5
Lets take a=-3 and b=2; so |a|+|b| = 5 & |a+b|=1
two different values, hence insufficient.
2. |a|*b<0
This implies b<0.
Lets take a=-3 and b=-2; so |a| + |b| = 5 & |a+b|=5
Lets take a=3 and b=-2; so |a| + |b| = 5 & |a+b|=1
two different values, hence insufficient.
Taking both together:
we can use the same numbers in statement 2. Hence, insufficient.
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Re: Ds Modules and inequalities [#permalink]
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06 Jun 2013, 10:01
Manhnip wrote:
$$Is |a| + |b| > |a + b| ?$$
$$(1) a^2>b^2$$
($$2)|a| * b<0$$
got me confusing for a while and choose E.
used the std rule of | A + B | < |A| + | B | when XY > 0
Since it is difficult to find the sign of A in any case selected E.
Please let me know if the approach to this problem is correct ?
Merging similar topics.
Look at the original post here: is-a-b-a-b-105457.html#p824216 The name of the topics MUST be "Is |a| + |b| > |a + b| ?"
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Re: Is |a| + |b| > |a + b| ? [#permalink]
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30 Jun 2013, 12:46
Is |a| + |b| > |a + b| ?
Is a^2 + b^2 > (a+b)^2?
is a^2 + b^2 > a^2 + 2ab + b^2?
Is 0 > 2ab?
Is ab negative?
(1) a^2 > b^2
This tells us that a^2 is greater than b^2 but it tells us nothing about the signs of a and b. For example, (-3)^2 > (2)^2 but -3 is less than 2. In this case, ab would be negative. On the other hand, (3)^2 > (2)^2 in which case 3 > 2 and ab would be positive.
INSUFFICIENT
(2) |a| * b < 0
This tells us that b must be negative as |a| will always be positive. however, a could be negative in which case ab would be positive or a could be negative in which case ab would be negative.
INSUFFICIENT
1+2) a^2 > b^2 and b is negative: that means the absolute value of |a| > |b|.
|-4| > |-1| ===> 4>1 Valid ===> ab = (-4)*(-1) = 4 (positive)
|4| > |-1| ===> 4>1 Valid ===> ab = (4)*(-1) = -4 (negative)
In other words, ab could be either positive or negative.
INSUFFICIENT
(E)
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Re: Is |a| + |b| > |a + b| ? [#permalink]
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22 Jul 2013, 13:41
Is |a| + |b| > |a + b| ?
|4| + |2| > |4 + 2| 6>6 Invalid
|-4| + |-2| > |-4 + -2| 6>6 Invalid
|-4| + |2| > |-4 + 2| 6>2 Valid
|4| + |-2| > |4 + -2| 6> 2 Valid
(Take note that the only two valid cases are when a and b have opposite signs)
The question then becomes, do a and b have opposite signs?
(1) a^2 > b^2
|a| > |b| but we don't know the signs of either.
INSUFFICIENT
(2) |a| * b < 0
For this to hold true, regardless of what a is (any number but zero) b must be negative. Like #1, this tells us nothing about the sign of a.
INSUFFICIENT
1+2) a^2 > b^2, |a| * b < 0
we know that b is negative but a could be positive or negative and a^2 > b^2 could still hold true. For example:
-6^2 > -5^2
36>25
-a, -b
6^2 > -5^2
36 > 25
a, -b
INSUFFICIENT
(E)
There is a similar problem in which you plug in numbers to solve (is-xy-0-1-x-3-y-5-x-y-2-0-2-x-y-x-y-86780.html) but when I try to solve it, it becomes a confusing mess. Could someone explain to me where I went wrong?
So, valid when:
|-a| > |b|
|a| > |-b|
(1) a^2 > b^2
The absolute value of a must be greater than the absolute value of b. For #1:
|a| > |b|
|-a| > |b|
|-a| > |-b|
Therefore, according to #1, the values of a and b could make the inequality |4| + |2| > |4 + 2| invalid (for example, if |a|>|b| or valid if |-a| > |b|)
INSUFFICIENT
(2) |a| * b < 0
|a|*b < 0
This is saying [positive or zero] * [b] < 0. If a is positive, b is negative. a and b cannot = 0. b is always negative regardless of what a is as |a| will always be positive or zero.
|-a| * (-b) < 0
|a| * (-b) < 0
The problem is valid when |-a| > |-b|. We don't know if |-a| > |b| as is stated in the stem. For example, |-1| * (-4) < 0 but |-1| is not greater than |-4|
INSUFFICIENT
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Re: Is |a| + |b| > |a + b| ? [#permalink]
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07 Apr 2014, 05:55
From question we need only to know if ab<0.
Statement 1 tells us that abs A > abs B, which is clearly insufficient.
Statement 2 tells us that b<0.
From both statements we can't tell whether a is negative.
Therefore E is the correct answer choice
Hope this helps
Cheers
J
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Is |a| + |b| > |a + b| ? Manhattan Inquality [#permalink]
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14 Dec 2014, 05:15
Is |a| + |b| > |a + b| ?
(1) a2 > b2
(2) |a| × b < 0
I got this from one of the closed group, thought to share for the advantage of vast global aspirant.
N.B. a2 = a X a = a square
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Re: Is |a| + |b| > |a + b| ? Manhattan Inquality [#permalink]
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14 Dec 2014, 05:30
honchos wrote:
Is |a| + |b| > |a + b| ?
(1) a2 > b2
(2) |a| × b < 0
I got this from one of the closed group, thought to share for the advantage of vast global aspirant.
N.B. a2 = a X a = a square
st.1
a^2>b^2
if a=3, b=1, then answer to the question is no.
if a=-3,b=1, then answer to the question is yes
hence insufficient
st.2
|a| × b < 0
from here we can conclude that b<0 as |a| will always be positive. but here a can be both positive as well as negative.
if a=4, b=-2 then answer to the question is yes.
if a=-4 ,b=-1 then answer to the question is no.
st.1 and st.2
we know for sure than b<0, but a can still take both positive as well as negative values. hence both yes and no. answers are possible. therefore answer must be E.
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Re: Is |a| + |b| > |a + b| ? Manhattan Inquality [#permalink]
### Show Tags
14 Dec 2014, 17:46
The question 'Is |a| + |b| > |a + b| ?' is basically asking if a and b have opposite signs.
(1) Insufficient. Says nothing about the signs of a and b.
(2) Insufficient. Just tells that b is negative, but we still don't know about the sign of a.
(1)+(2) Insufficient. We still don't know about the sign of a.
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Re: Is |a| + |b| > |a + b| ? [#permalink]
### Show Tags
15 Dec 2014, 08:08
honchos wrote:
Is |a| + |b| > |a + b| ?
(1) a2 > b2
(2) |a| × b < 0
I got this from one of the closed group, thought to share for the advantage of vast global aspirant.
N.B. a2 = a X a = a square
Merging topics. Please refer to the discussion above.
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Re: Is |a| + |b| > |a + b| ? [#permalink]
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02 Jun 2016, 11:55
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Re: Is |a| + |b| > |a + b| ? [#permalink]
### Show Tags
02 Jun 2016, 13:07
chiragatara wrote:
Is |a| + |b| > |a + b| ?
(1) a^2 > b^2
(2) |a| * b < 0
|a| + |b| will have both a and b +ve
|a + b| will have positive sum of a and b
But we don't know signs of both a and b
If both the signs are +ve then |a| + |b| = |a + b|
If both the signs are -ve then |a| + |b| = |a + b|
If both have opposite signs then |a| + |b| > |a + b|
(1) a^2 > b^2
It doesn't tell us anything about signs of a or b
(2) |a| * b < 0
|a| will be +ve, hence b is -ve
but a can be +ve or -ve. We don't get the actual sign.
Combining both statements is not giving us answer
(-6)^2> -5^2
6^2> -5^2
Not sufficient
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Re: Is |a| + |b| > |a + b| ? [#permalink]
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12 Jul 2017, 04:09
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Re: Is |a| + |b| > |a + b| ? [#permalink] 12 Jul 2017, 04:09
Display posts from previous: Sort by
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2017-11-19 18:32:57
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https://stats.stackexchange.com/questions/183973/how-not-to-overfit-a-random-forest-in-r
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# How (not) to (over)fit a random forest in R
I'm reaching out to you because I am unsure whether my implementation of a group of random forests in R (using library randomForest) is valid or whether I have an error in reasoning.
I have a sales dataset with a binary outcome (1: Sale, 0: No Sale) and a set of possibly significant predictors x1-x14. My data is highly imbalanced, with ~124k '0' observations (No Sale) and ~18k '1' observations (Sale). I balance it by randomly cutting down the 124k observations to 18k, as suggested in http://bit.ly/1I7F0AC.
Cross-validation is not necessary due to the nature of random forests, however: In order to find a random forest with a good F-score, I loop through a set of possible predictors and a set of tree-numbers for the forest:
possiblyUsefulPredictors=
c("x1",..."x14") # Shortened to pseudo-code
treerange=c(1,2,3,4,5,6,7,8,9,10,15,20,25,30,35,40,45,50,60,70,80,90,100,
200,300,400,500,750,1000)
# Create a multitude of models by looping
# through different settings for parameters
for (i in 2:length(possiblyUsefulPredictors)){
for (j in treerange){
### Choose model here by setting data, outcome and predictors:
x=possiblyUsefulPredictors[1:i] # Set predictors
ntree=j # Set number of trees
# Tune mtry
bestMtry=tuneRF(x=x, y=y, ntreeTry=1,
stepFactor=1, improve=0.01, trace=FALSE,
plot=FALSE, doBest=FALSE)
# Run random forest
rf=randomForest(y=y,x=x,data=df,mtry=bestMtry,ntree=ntree,
type="classification",importance=T)
}
}
I then store model diagnostics precision, recall, and F-score in a table and choose the model that created the highest F-score (13 predictors, 90 trees, mtry=1, which leads to an F-score of 78%).
Specific questions:
1. Obviously, the way I subset and loop through the predictors is highly arbitrary. Could a more sophisticated approach (e.g. looping through all possible subsets) get me anywhere, or does a random forest inherently choose significant predictors, so that I wouldn't have to try to find a meaningful subset myself (like I do when using step-wise in linear regression)?
2. By building a set of 416 random forests, do I simply overfit the dataset? I am skeptical that the predictors are as good as my best model suggests.
Thank you and kind regards, Jan
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2020-01-24 23:51:26
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https://engtech.spbstu.ru/en/article/2013.84.6/
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# Investigation of hydrodynamic characteristics of laminar flow fluid with transverse viscosity in the convergent channel
Authors:
Abstract:
We study the hydrodynamics in the laminar flow of an aqueous solution of hydroxyethyl cellulose, is a non-linear viscous fluid with a viscosity shear, in a convergent channel. A nonlinear rheological model defining the Reiner — Rivlin fluid is used.
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2020-10-01 02:37:52
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http://roussev.net/sdhash/tutorial/03-quick.html
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Installation
# Quick start¶
sdhash is designed to work right out of the box with minimal number of command line parameters. At the same time, the tool provides considerable flexibility for advanced users that can be introduced incrementally, as needed.
Invoking the tool from the command line with no arguments shows a brief summary of available command line options (on Linux, the man sdhash commands provides more details):
sdhash 3.3 by Vassil Roussev, Candice Quates [sdhash.org] 07/2013
Usage: sdhash <options> <files>
Configuration:
-r [ --deep ] generate SDBFs from directories and files
-f [ --target-list ] generate SDBFs from list(s) of filenames
-c [ --compare ] compare SDBFs in file, or two SDBF files
-g [ --gen-compare ] compare all pairs in source data
-t [ --threshold ] arg (=1) only show results >=threshold
-b [ --block-size ] arg hashes input files in nKB blocks
-s [ --sample-size ] arg (=0) sample N filters for comparisons
-z [ --segment-size ] arg set file segment size, 128MB default
-o [ --output ] arg send output to files
--separator arg (=pipe) for comparison results: pipe csv tab
--hash-name arg set name of hash on stdin
--validate parse SDBF file to check if it is valid
--index generate indexes while hashing
--index-search arg search directory of reference indexes
--config-file arg (=sdhash.cfg) use config file
--verbose warnings, debug and progress output
--version show version info
-h [ --help ] produce help message
Note
Whenever we discuss command line options, we will quote both the short and long form as in -c [--compare].
## Digest generation¶
Invoked with one, or more, file names, sdhash produces the similarity digests of each one files and prints them to standard output. Each digest is completely self-contained and is exactly one (potentially very, very long) line of printable ASCII characters. It consists of several header fields separated by semicolons, followed by base64-encoding of the (binary) digest data. (The details of the format are explained in [?].)
Example
Generate the sdhash of file 014.html:
$sdhash 014.html sdbf:03:8:014.html:38053:sha1:256:5:7ff:160:3:75:HZvUjQGMcYEfsk8IQgIggA ... Lines from different files can be freely combined using standard text processing utilities. For all practical purposes, there are only two “user serviceable” fields–the file/object name (prefaced by its length) followed by the original object/file length. Note Minimum file size if 512 bytes; sdhash will skip over smaller files quietely. If you need warning about these, use the [--verbose] option. Standard pathname patterns expansion (globbing) works as expected, courtesy of the shell: Example Hash all html files in the current directory: $ sdhash *.html
sdbf:03:8:014.html:38053:sha1:256:5:7ff:160:3:75:HZvUjQGMcYEfsk8IQgIggA ...
sdbf:03:8:056.html:6664:sha1:256:5:7ff:160:1:99:AAAYTGCQsiAOBEkANREgAIh ...
sdbf:03:8:057.html:15893:sha1:256:5:7ff:160:2:102:KSxoLJfAKAAaIQiRoAMZw ...
sdbf:03:8:058.html:734:sha1:256:5:7ff:160:1:11:AAAAAAAAAAAAAAAAAAAAAAAQ ...
...
Note
There are practical limits to globbing pattern expansion in shell environments. In our experience, bash maxes somewhere between 40,000 and 50,00 files. To get around these limitations, use the -f [ --hash-list ] or -r [--deep]. See [?].
To save the result to a file you can either use output stream redirection:
$sdhash *.html > html.sdbf or the -o [--output] option (which could be anywhere in the command line): $ sdhash -o html.sdbf *.html
$sdhash -c html.sdbf 195.html|206.html|001 201.html|206.html|007 428.html|608.html|058 428.html|607.html|069 061.html|062.html|026 060.html|059.html|031 199.html|198.html|066 ... The output consists of three columns separated by the pipe (vertical bar) symbol. The first two columns are names of the files being compared, whereas the third one gives the similarity score, which is a number between -1 and 100. By default, only positive scores are shown. ### Two-set comparison¶ In this case, every hash in the first set compared against every hash in the second one. For example, using the html.sdbf from above, we can compare the set against itself (twice!) using the two-set format: $ sdhash -c html.sdbf html.sdbf | sort
014.html|014.html|100
056.html|056.html|100
057.html|057.html|100
059.html|059.html|100
059.html|060.html|031
060.html|059.html|031
060.html|060.html|100
061.html|061.html|100
061.html|062.html|026
062.html|061.html|026
...
The main use case for the two-set comparison (and sdhash in general) is the querying of a reference database with unknown data in an effort to find correlations:
$sdhash -o unknown.sdbf unknown/* sdhash -c unknown.sdbf reference.sdbf In our example set, we could compare html versus text files: $ sdhash *.html > html.sdbf
$sdhash *.txt > txt.sdbf$ sdhash -c txt.sdbf html.sdbf
In this case, we find no matches, which is common when comparing files of different encodings. The test set includes a set of text files that have been derived from the html files in the set using the html2text utility, which strips away the markup and leaves the plain text. Intuitively, we would expect html files that have large chunks of plain text inside to come out in the following experiment:
$sdhash *.html-txt > html-txt.sdbf$ sdhash -c html.sdbf html-txt.sdbf
740.html|740.html-txt|002
448.html|448.html-txt|024
418.html|418.html-txt|010
136.html|136.html-txt|016
597.html|597.html-txt|064
434.html|434.html-txt|095
As it turns out, the last file–434.html–is almost all text.
## Threshold (-t)¶
The threshold parameter instructs sdhash to display only results that are greater than or equal to the parameter; by default, it is set to 1, so all positive results are shown. The proper setting of the threshold is inherently case-specific but in the following sections we give a good starting point that works for most cases.
Setting the threshold to zero (-t 0) will display the results of all comparisons; setting it to negative one (-t -1) will also show comparisons that have not been performed due to insufficient data (see below).
## Result interpretation¶
As already mentioned, the result of the comparison is a number between 0 and 100; its interpretation depends on the scenario to which the tool is applied and the data encoding (file type) of the compared objects.
### Scenarios¶
There are two basic usage scenarios:
• Fragment identification–we search for (traces of) something “small” in something bigger. This can include comparing the content of a disk block/network packet to a file, a file to a RAM capture, or file to a disk image.
• Version correlation–we look for similarity between two comparably-sized objects, usually files.
Note that the distinction between the two is entirely in the eyes of the user–the tool works exactly the same way in all cases. The distinction is only important in understanding what the result means.
For example, we could compare an executable to a raw RAM snapshot, in search for clues that the executable had been loaded. We may get a result as high as 100, which obviously does not mean that the two objects are identical; it merely reflects that sdhash is highly confident that large pieces of the file are present.
Another example case would be to search for known embedded images (e.g., a corporate logo) inside a collection of documents.
Version correlation is very helpful in finding files that have non-trivial amount of commonality. For example, executables tend to change on a function-by-function basis and we have found that we can quite reliably identify new versions of the files from hashes of previous versions. For data, the commonality may come from common boilerplate (html, pdf, etc.), or as a result of normal editing operation.
### Significance¶
Despite its range (0-100), the sdhash comparison result should not be intepreted as a percentage of common content. Rather, it should be viewed as a confidence value that indicates how certain the tool is that the two data objects have non-trivial amounts of commonality.
The proper use of the sdhash score is to examine the results in descending order, until the false positives (as defined by the user) exceed the true positives by some margin.
The following is a basic guide to intepreting the results from sdhash comparisons. In [?], we will add some finer points and caveats:
• Strong (range: 21-100). These are reliable results with very few false positives. When used to evaluate resemblance of two comparable in size objects (files) the number is loosely related to the level of commonality but this is not a guarantee. When used as part of a containment query (find a small object inside a bigger one), the number can vary widely depending on the particular position of the embedding. In other words the larger object may contain the small one 100% but the score may be as small as 25.
• Marginal (11-20). The significance of resemblance comparisons in this range depends substantially on the underlying data. For many composite file types (PDF, MS Office) there tends to be some embedded commonality, which is a function of commonly used applications leaving their imprint on the file; we’ve observed this on occasion even with JPEG files that contain lots of (Adobe Photoshop) metadata. In that sense, the tool is not wrong but the discovered correlation is usually not of interest. Other embedded artifacts, such as fonts, are also among the discovered commonalities but are rarely significant. For simpler file types, results in this range are much more likely to be significant and should be examined in decreasing order until the false positives start to dominate.
• Weak (1-10). These are generally weak results and, typically, most would be false positives. However, when applied to simple file types, such as text, scores as low as 5 could be significant.
• Negative (0). The correlation between the targets is statistically comparable to that of two blobs of random data. Special care needs to be taken when comparing large targets to each other as discovered commonality could be avaraged out to zero. For example, if two 100GB have 1GB in common, the tool will discover that fact but when averaged with the results from the remaining 99GB, the final score will almost certainly be zero, and definitely no more than one.
• Unknown (-1). This is a rare occurance for files above 4KB unless they contain large regions of low-entropy data. Recall that the absolute minimum file size that sdhash will consider hashing is 512 bytes. (If a case requires the comparison of lots of tiny files, sdhash is likely the wrong tool.)
Warning
A result of 100 does not guarantee that two objects are identical. Use a crypto hash to establish identity. (We plan to incorporate a crypto hash to test for identity, starting with version 4.0.)
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2020-10-29 04:58:07
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https://www.physicsforums.com/threads/product-of-diagonal-matrices.601031/
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# Product of Diagonal Matrices
1. Apr 28, 2012
### jsgoodfella
If we take an nxn diagonal matrix, and multiply it by an nxn matrix C such that AC=CA, will C be diagonal? I know, for instance, if C is a matrix with ones in every entry, AC=CA holds. But is there a more general way to format such a counterexample, or have I already provided a sufficient "proof"?
Thanks in advance. This isn't a homework question.
2. Apr 28, 2012
### DonAntonio
$$\left(\begin{array}{cc}2&0\\0&2\end{array}\right) \left(\begin{array}{cc}1&1\\0&1\end{array}\right)=\left(\begin{array}{cc}2&2\\0&2\end{array}\right)=\left(\begin{array}{cc}1&1\\0&1\end{array}\right) \left(\begin{array}{cc}2&0\\0&2\end{array}\right)$$
DonAntonio
3. Apr 28, 2012
### phyzguy
When you disprove a statement by providing a counter-example, one counter-example is sufficient. There is no need to provide more.
4. Apr 28, 2012
### DonAntonio
Yes, of course. Whom are you addressing and why?
DonAntonio
5. Apr 28, 2012
### phyzguy
The OP asked whether he had already provided a sufficient proof. The answer is yes - since had already provided one counter-example, this is sufficient to disprove the original statement. That's all I'm saying.
6. Apr 28, 2012
### DonAntonio
Good. This time you provide a quote of whom you're addressing and thus we know. The last time we, or at least I, didn't know.
DonAntonio
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2017-08-20 19:56:13
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https://quantumcomputing.stackexchange.com/questions/21793/qiskits-paulitrotterevolution-yields-weird-gates
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# Qiskit's PauliTrotterEvolution yields weird gates
I am trying to work with Qiskit's PauliTrotterEvolution() module, but the resulting circuits contain weird gates that I know nothing about.
Here is a simple example: I want to implement the fermionic creation operator on one of two qubits
from qiskit_nature.mappers.second_quantization import BravyiKitaevMapper
from qiskit_nature.operators.second_quantization import FermionicOp
from qiskit.opflow.evolutions import PauliTrotterEvolution
fermi_op = FermionicOp("+I", display_format="dense")
The fermionic operator is mapped to a sum of Pauli gates using the Bravyi-Kitaev mapper:
mapper = BravyiKitaevMapper()
bosonic_op = mapper.map(fermi_op)
Now we apply the PauliTrotterEvolution to the exponential of the bosonic operator:
pauli_trotter = PauliTrotterEvolution("trotter", reps=1)
conv = pauli_trotter.convert(bosonic_op.exp_i())
circ = conv.to_circuit()
circ.draw("latex")
I have no clue what these "circuit" gates are. Has anyone seen something similar before?
These "circuit" gates are empty, if you transpile the circuit to a basis gate set they will disappear:
from qiskit import transpile
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2021-12-06 13:53:12
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https://www.shaalaa.com/question-bank-solutions/a-diet-two-foods-f1-f2-contains-nutrients-thiamine-phosphorous-iron-amount-each-nutrient-each-food-in-milligrams-25-gms-given-following-table-graphical-method-of-solving-linear-programming-problems_47671
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# A Diet of Two Foods F1 And F2 Contains Nutrients Thiamine, Phosphorous and Iron. the Amount of Each Nutrient in Each of the Food (In Milligrams per 25 Gms) is Given in the Following Table: - Mathematics
Sum
A diet of two foods F1 and F2 contains nutrients thiamine, phosphorous and iron. The amount of each nutrient in each of the food (in milligrams per 25 gms) is given in the following table:
Nutrients Food F1 F2 Thiamine 0.25 0.10 Phosphorous 0.75 1.50 Iron 1.60 0.80
The minimum requirement of the nutrients in the diet are 1.00 mg of thiamine, 7.50 mg of phosphorous and 10.00 mg of iron. The cost of F1 is 20 paise per 25 gms while the cost of F2 is 15 paise per 25 gms. Find the minimum cost of diet.
#### Solution
Let 25x grams of food F1 and 25y grams of food F2 be used to fulfil the minimum requirement of thiamine, phosphorus and iron.
As, we are given
Nutrients Food F1 F2 Thiamine 0.25 0.10 Phosphorous 0.75 1.50 Iron 1.60 0.80
And the minimum requirement of the nutrients in the diet are 1.00 mg of thiamine, 7.50 mg of phosphorous and 10.00 mg of iron.$\text{ Therefore, }$
$0 . 25x + 0 . 10y \geq 1$
$0 . 75x + 1 . 50y \geq 7 . 5$
$1 . 6x + 0 . 8y \geq 10$
$\text{ Since, the quantity cannot be negative }$
$\ ∴ x, y \geq 0$
The cost of F1 is 20 paise per 25 gms while the cost of F2 is 15 paise per 25 gms.Therefore, the cost of 25x grams of food F1 and 25y grams of food F2 is
Rs (0.20x + 0.15y).
Hence,
Minimize Z = $0 . 20x + 0 . 15y$
subject to
$0 . 25x + 0 . 10y \geq 1$
$0 . 75x + 1 . 50y \geq 7 . 5$
$1 . 6x + 0 . 8y \geq 10$
$x, y \geq 0$
First, we will convert the given inequations into equations, we obtain the following equations:
0.25x + 0.10y = 1, 0.75x + 1.50y = 7.51.6x + 0.8y = 10, x = 0 and y = 0.
The line 0.25x + 0.10y = 1 meets the coordinate axis at
A(4,0) and B(0, 10). Join these points to obtain the line 0.25x + 0.10y = 1.
Clearly, (0, 0) does not satisfies the inequation 0.25x + 0.10y ≥ 1. So, the region in xy-plane that does not contains the origin represents the solution set of the given equation.
The line 0.75x + 1.50y = 7.5. meets the coordinate axis at C(10, 0) and D(0, 5). Join these points to obtain the line 0.75x + 1.50y = 7.5.
Clearly, (0, 0) does not satisfies the inequation 0.75x + 1.50y ≥ 0.75. So, the region in xy-plane that does not contains the origin represents the solution set of the given equation.
The line 1.6x + 0.8y = 10 meets the coordinate axis at
$E\left( \frac{25}{4}, 0 \right)$ and $F\left( 0, \frac{25}{2} \right)$ Join these points to obtain the line 1.6x + 0.8y = 10.
Clearly, (0, 0) does not satisfies the inequation 1.6x + 0.8y ≥ 10. So, the region in xy-plane that does not contains the origin represents the solution set of the given equation.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations.
These lines are drawn using a suitable scale.
The corner points of the feasible region are F(0, 12.5), G(5, 2.5), C(10, 0)
The value of the objective function at these points are given by the following table
Points Value of Z F 0.20(0)+0.15(12.5) = 1.875 G 0.20(5)+0.15(2.5) = 1.375 C 0.20(10) + 0.15(0) = 200
Thus, the minimum cost is at G which is Rs 1.375
Concept: Graphical Method of Solving Linear Programming Problems
Is there an error in this question or solution?
#### APPEARS IN
RD Sharma Class 12 Maths
Chapter 30 Linear programming
Exercise 30.3 | Q 1 | Page 38
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2022-05-22 17:54:31
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http://blogs.msdn.com/b/oldnewthing/default.aspx?PageIndex=368
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• #### High-performance multithreading is very hard
Among other things, you need to understand weak memory models.
Hereby incorporating by reference Brad Abrams' discussion of volatile and MemoryBarrier(). In particular, Vance Morrison's discussion of memory models is important reading.
(Though I think Brad is being too pessimistic about volatile. Ensuring release semantics at the store of "singleton" is all you really need - you want to make sure the singleton is fully constructed before you let the world see it. volatile here is overkill.)
Vance's message also slyly introduces the concepts of "acquire" and "release" memory semantics. An interlocked operation with "acquire" semantics prevents future reads from being advanced to before the acquisition. An interlocked operation with "release" semantics prevents previous writes from being delayed until after the release.
In the absence of explicitly-named memory semantics, the Win32 Interlocked* functions by default provide full memory barrier semantics. However, some functions, like InterlockedIncrementAcquire, forego the full memory barrier semantics and provide only acquire or release semantics.
• #### What is the difference between Minimize All and Show Desktop?
The keyboard shortcut for "Minimize All" is ÿ+M and The keyboard shortcut for "Show Desktop" is ÿ+D. How are they different?
"Minimize All" is easier to describe. It minimizes all the windows that support the "Minimize" command. You can minimize a window by selecting "Minimize" from its System menu, or by clicking the 0 button in the title bar. So "Minimize All" is effectively the same as going to each window that is open and clicking the Minimize button. If there is a window that doesn't have a Minimize button, then it is left alone.
"Show Desktop" takes "Minimize All" one step further. After minimizing all the windows that can be minimized, it then takes the desktop and "raises" it to the top of the window stack so that no other windows cover it. (Well, okay, topmost windows continue to cover it.)
So "Show Desktop" manages to get a few more windows out of your way than "Minimize All".
Note, however, that when you return the desktop to its normal state (either by selecting "Show Open Windows" or just by switching to another window), all the un-minimizeable windows come back because the desktop has "lowered" itself back to the bottom of the window stack.
• #### Meet Anton Chekhov
(Not to be confused with Star Trek's Pavel Chekov, who spells his name with a plain "k" instead of a "kh". Of course, since this is all transliteration from Cyrillic, of which there are multiple systems, arguing about spelling is rather dubious anyway.)
The guerilla performance group Improv Everywhere stages events in public places and carefully records the reaction of their unwitting "audience". Perhaps one of their best-known stunts is the appearance of author Anton Chekhov at a New York City Barnes and Noble.
After the reading, they set up a table in Union Square Park across the street and sold autographed copies of The Cherry Orchard from a table.
The young man in the jean jacket above insisted that Chekov only sign his autograph, rather than making it out to anyone in particular. He proudly announced, "When you die, this is going to be worth lots of money!"
Though I think their best performance from an artistic performance point of view was The Moebius.
• #### How does the desktop choose the icon label color?
Many system colors come in pairs, one for background and one for foreground. For example, COLOR_MENU and COLOR_MENUTEXT; COLOR_WINDOW and COLOR_WINDOWTEXT; COLOR_HIGHLIGHT and COLOR_HIGHLIGHTTEXT. More on these colors in a future blog entry.
But there is COLOR_DESKTOP and no COLOR_DESKTOPTEXT.
How does the desktop choose the icon label color?
It's not very exciting.
The COLOR_DESKTOP color is studied to see if it is mostly dark or mostly light. If mostly dark, then the icon text is white. If mostly light, then the icon text is black.
This works great if you don't have a wallpaper. If you do, then the desktop color doesn't typically correspond to the dominant color in your wallpaper, which may lead to a poor choice of black vs. white.
To remedy this, you can manually set the desktop color by going to the Display control panel and going to the Desktop tab. Change the desktop color to something dark to get white icon text and to something light to get black icon text.
• #### Art too bad to be ignored
The Athlete
Crayon and pencil on canvas by Unknown
30"x40"
Acquired from trash in Boston by Scott Wilson
A startling work, and one of the largest crayon on canvas pieces that most people can ever hope to see. The bulging leg muscles, the black shoes, the white socks, the pink toga, all help to make this one of the most popular pieces in the MOBA collection.
The Athlete also happens to be one of the most controversial works in the MOBA collection. Read the scandalous details!
• #### When you change the insides, nobody notices
I find it ironic when people complain that Calc and Notepad haven't changed. In fact, both programs have changed. (Notepad gained some additional menu and status bar options. Calc got a severe workover.)
I wouldn't be surprised if these are the same people who complain, "Why does Microsoft spend all its effort on making Windows 'look cool'? They should spend all their efforts on making technical improvements and just stop making visual improvements."
And with Calc, that's exactly what happened: Massive technical improvements. No visual improvement. And nobody noticed. In fact, the complaints just keep coming. "Look at Calc, same as it always was."
The innards of Calc - the arithmetic engine - was completely thrown away and rewritten from scratch. The standard IEEE floating point library was replaced with an arbitrary-precision arithmetic library. This was done after people kept writing ha-ha articles about how Calc couldn't do decimal arithmetic correctly, that for example computing 10.21 - 10.2 resulted in 0.0100000000000016.
(These all came from people who didn't understand how computers handle floating point. I have a future entry planned to go into floating point representations in more detail.)
Today, Calc's internal computations are done with infinite precision for basic operations (addition, subtraction, multiplication, division) and 32 digits of precision for advanced operations (square root, transcendental operators).
Try it: 1 / 3 * 10000000000 - 3333333333 =. The result is one third exactly. Type 1/x - 3 = and you get zero back. (Of course, if you don't believe that, then repeat the sequence "* 10000000000 - 3333333333 =" until you're bored and notice that the answer always comes back as 0.33333333333333333333333333333333. If it were fixed-precision, then the 3's would eventually stop coming.)
Thirty-two positions of precision for inexact results not good enough? The Power Calculator PowerToy uses the same arithmetic engine as Calc and lets you crank the precision to an unimaginable 512 digits.
Anyway, my point is that - whether you like it or not - if you don't change the UI, nobody notices. That's so much effort is spent on new UI.
• #### The F*deral Bur*au of Inv*stigations
According to the Chicago Division,
Use of the NAME, INITIALS, or SEAL of the F*I
is restricted by law
and may be used only with written permission of the F*I.
Asterisks inserted to avoid being arrested by the F*I for using their name and initials without written permission.
(I hope they don't go after me for using their initials without permission late last year.)
• #### Extending the Internet Explorer context menu
In a comment, Darrell Norton asked for a "View in Mozilla" option for Internet Explorer.
Internet Explorer's context menu extension mechanism has been in MSDN for years. Let me show you how you can create this extension yourself.
First, create the following registry key:
[HKEY_CURRENT_USER\Software\Microsoft\Internet Explorer\MenuExt\View in Mozilla]
@=REG_SZ:"C:\some\path\to\ViewInMozilla.htm"
Contexts=REG_DWORD:1
Of course, you need to change C:\some\path\to to an actual path.
How did I know to do this? Because steps 1, 2 and 3 in the "Implementation Steps" section tell me (1) what key to create, (2) what to set the default value to, and (3) what to set Contexts to. I chose a Context value of 1, which means "Default".
Okay, now to write the script ViewInMozilla.htm. Well, the documentation says that I can access context from the menuArguments property of the external object. So let's start with that.
<SCRIPT>
</SCRIPT>
Okay, let's run this puppy. Launch IE, right-click on a blank space in the web page, select "View in Mozilla", and you get...
[object]
Woo-hoo! This is a major accomplishment: Something happened at all. Doing things in small steps makes it easy to identify where a problem is. If we had run full steam ahead to completion and then it didn't work, we wouldn't have known whether it was due to a bug in the script, a bad registration, a bad filename...
Now that I have the menu arguments, I can use that to suck out information about the item that the context menu applies to. Let's try this:
<SCRIPT>
</SCRIPT>
Woo-hoo, now it gives me the URL. Almost there. All that's left to do is to run a program with that URL as the command line parameter.
<SCRIPT>
var shell = new ActiveXObject("WScript.Shell");
shell.run("mozilla \"" + external.menuArguments.document.URL + "\"");
</SCRIPT>
Mission accomplished.
Now you too can create Internet Explorer context menu extensions.
In fact, go ahead and do it, since Darrell asked for it: Create an Internet Explorer context menu extension that operates on anchors and opens the linked-to page in Mozilla.
(Bonus: Tony Schreiner cooked up something similar for zooming.)
• #### Callback, the safety newsletter for the aviation community
Yes the "S" in NASA stands for "Space", but don't forget that the "A" stands for "Aeronautics". One of the programs that I (amateur aviation wannabe) find fascinating is the Aviation Safety Reporting System, where pilots can submit anonymous reports of "stupid things I have done" in order to teach other pilots "Don't do what I did."
• #### Do you know when your destructors run? Part 2.
Continuing from yesterday, here's another case where you have to watch your destructors. Yesterday's theme was destructors that run at the wrong time. Today, we're going to see destructors that don't run at all!
Assume there's an ObjectLock class which takes a lock in its constructor and releases it in its destructor.
DWORD ThreadProc(LPVOID p)
{
... do stuff ...
ObjectLock lock(p);
... do stuff ...
return 0;
}
Pretty standard stuff. The first batch of stuff is done without the lock, and the second batch is done inside the lock. When the function returns, the lock is automatically released.
But suppose somebody adds a little code to this function like this:
DWORD ThreadProc(LPVOID p)
{
... do stuff ...
ObjectLock lock(p);
...
...
return 0;
}
The code change was just to add an early exit if the object was cancelled.
But when does that ObjectLock destructor run?
It runs at the return statement, since that's when the lock goes out of scope. In particular, it is not run before you call ExitThread.
Result: You left an object locked permanently.
You can imagine how variations on this code could lead to resource leaks or other problems.
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2013-12-12 11:48:12
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http://chempaths.chemeddl.org/services/chempaths/?q=book/General%20Chemistry%20Textbook/Nuclear%20Chemistry/1552/nuclear-fusion
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# Nuclear Fusion
Submitted by ChemPRIME Staff on Thu, 12/16/2010 - 16:05
In addition to fission, a second possible method for obtaining energyA system's capacity to do work. from nuclear reactions lies in the fusing together of two light nuclei to form a heavier nucleus. As we see when discussing Figure 1 from Mass-Energy Relationships, such a process results in nucleons which are more firmly bonded to each other and hence lower in potential energy. This is particularly true if ${}_{\text{2}}^{\text{4}}\text{He}$ is formed, because this nucleus is very stable. Such a reaction occurs between the nuclei of the two heavy isotopes of hydrogen, deuteriumThe isotope of hydrogen having one neutron in its nucleus. and tritiumThe isotope of hydrogen that has two neutrons in its nucleus.:
${}_{\text{1}}^{\text{2}}\text{D + }{}_{\text{1}}^{\text{3}}\text{T }\to \text{ }{}_{\text{2}}^{\text{4}}\text{He + }{}_{\text{0}}^{\text{1}}n$ (1)
For this reaction Δm = – 0.018 88 g mol–1 so that ΔHm = – 1700 GJ mol–1. Although very large quantities of energy are released by a reaction like Eq. (1) such a reaction is very difficult to achieve in practice. This is because of the very high activation energyThe energy barrier over which a reaction must progress in order for reactants to form products; the minimum energy that reactants must have if they are to be converted to products., about 30 GJ mol–1, which must be overcome to bring the nuclei close enough to fuse together. This barrier is created by coulombic repulsion between the positively charged nuclei. The only place where scientists have succeeded in producing fusion reactions on a large scale is in a hydrogen bomb. Here the necessary activation energy is achieved by exploding a fission bomb to heatEnergy transferred as a result of a temperature difference; a form of energy stored in the movement of atomic-sized particles. the reactants to a temperatureA physical property that indicates whether one object can transfer thermal energy to another object. of about 108 K. Attempts to carry out fusion in a more controlled way have met with only limited success. At the very high temperatures required, all molecules dissociate and most atomsThe smallest particle of an element that can be involved in chemical combination with another element; an atom consists of protons and neutrons in a tiny, very dense nucleus, surrounded by electrons, which occupy most of its volume. ionize. A new state of matterAnything that occupies space and has mass; contrasted with energy. called a plasma is formed. It is neither solid, liquid, or gas and behaves much like the universal solventThe substance to which a solute is added to make a solution. of the alchemists by converting any solid material which it contacts into vaporThe gaseous state of a substance that typically exists as a liquid or solid; a gas at a temperature near or below the boiling point of the corresponding liquid..
Two techniques for producing a controlled fusion reaction are currently being explored. The first is to restrict the plasma by means of a strong magnetic field rather than the walls of a container. This has met with some success but has not yet been able to contain a plasma long enough for usable energy to be obtained. The second technique involves the sudden compression and heating of pellets of deuterium and tritium by means of a sharply focused laser beam. Again, only a limited success has been obtained.
Though these attempts at a controlled fusion reaction have so far been only partially successful, they are nevertheless worth pursuing. Because of the much readier availability of lighter isotopes necessary for fusion as opposed to the much rarer heavier isotopes required for fission, controlled nuclear fusion would offer the human race an essentially limitless supply of energy. There would still be some environmental difficulties with the production of isotopes such as tritium, but these would be nowhere near the seriousness of the problem caused by the production of the witches brew of radioactiveDescribes a substance that gives off radiation‐alpha particles, beta particles, or gamma rays‐by the disintegration of its nucleus. isotopes in a fission reactor. It must be confessed, though, that at the present rate of progress, the prospect of limitless clean energy from fusion seems unlikely in the next decade or two.
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2015-03-04 11:16:07
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https://zbmath.org/?q=an:05315117
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# zbMATH — the first resource for mathematics
Intersections of polynomial rings and modules with applications. (English) Zbl 1145.13300
Informatik. Aachen: Shaker Verlag; München: Tech. Univ. München, Fak. für Informatik (Diss.) (ISBN 978-3-8322-5349-3/pbk). 120 p. (2006).
Preface: We are concerned with two problems, namely with computing the intersection of (some) graded structures of commutative algebra and computing the stratification of linear actions of compact Lie groups. The connection between these seemingly unrelated problems is invariant theory, more precisely, the construction of fundamental invariants and, additionally, of fundamental equivariants of compact Lie groups. The principle of symmetry breaking allows to decompose in a unique way the representation and orbit space in finitely many disjoint basic open semi-algebraic sets, called strata. A set of fundamental invariants allows to construct this stratification of the representation- and the orbit space.
We provide algorithms for computing the intersection of finitely generated graded subalgebras, which can be specialized to compute invariants of algebraic groups and invariant rings of compact Lie groups, and for computing the intersection of graded submodules, which can be specialized to compute equivariants of algebraic groups. Moreover we propose a new approach for stratifying actions of compact Lie groups and we present algorithms for computing s stratification of the representation space and of all or selected strata (and their closures) of the orbit space. In addition, we show that the dimension of a stratum of the orbit space is an upper and lower bound for the number of inequalities needed for a description. Finally we describe the computer algebra package STRATIFY for computing stratifications of compact Lie groups.
##### MSC:
13-02 Research exposition (monographs, survey articles) pertaining to commutative algebra 22-02 Research exposition (monographs, survey articles) pertaining to topological groups 13P10 Gröbner bases; other bases for ideals and modules (e.g., Janet and border bases) 22E45 Representations of Lie and linear algebraic groups over real fields: analytic methods
STRATIFY
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2021-09-28 22:56:51
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http://eprint.iacr.org/2007/203
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Cryptology ePrint Archive: Report 2007/203
Kipnis-Shamir's Attack on HFE Revisited
Xin Jiang and Jintai Ding and Lei Hu
Abstract: In this paper, we show that the claims in the original Kipnis-Shamir's attack on the HFE cryptosystems and the improved attack by Courtois that the complexity of the attacks is polynomial in terms of the number of variables are invalid. We present computer experiments and a theoretical argument using basic algebraic geometry to explain why it is so. Furthermore we show that even with the help of the powerful new Gr\"{o}bner basis algorithm like $F_4$, the Kipnis-Shamir's attack still should be exponential not polynomial. This again is supported by our theoretical argument.
Category / Keywords: public-key cryptography /
Contact author: ding at math uc edu
Available format(s): PDF | BibTeX Citation
Short URL: ia.cr/2007/203
[ Cryptology ePrint archive ]
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2016-05-31 19:08:40
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http://forum.mackichan.com/node/1785
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## SWP 6.0.23 Import TeX Problem
I attach a .rap file created in Scientific Word 5.5. When I unwrap this and try to import it into SWP 6.0.23, I have the following problems:
1. Although the file opens, I cannot edit it
2. When I click the Source tab to inspect the source file, SWP 6 either crashes or else gives the error message “Command failure: Error executing command ‘sourcemode’" and, when I click on More Info, gives the message “Failure details: msiGetBodyElement editor is null”
If I click the PDF Preview tab to generate a PDF file, the following problems occur:
1. Some of the references do not appear, either in the text or in the list of references
2. The figures do not appear
3. The table captions and the notes to the tables do not appear
I am running SWP 6 on a Mac with OS X 10.12.3. But the document can be edited and compiled correctly in TeXShop using the same TeX installation as SWP 6.0.23, so I don’t think these problems have anything to do with the TeX installation.
I notice that, when using Import TeX on the file in SWP 6.0.23, I receive an Alert:
“The file resource: //app/res/shells/-Standard_LaTeX/Standard_LaTeX_Article.sci cannot be found. Please check the location and try again.”
The file Standard_LaTeX_Article.sci is present on the computer in the location
“Mackintosh HD/Applications/MacKichan/SWP.app/Contents/Resources/shells/-Standard_LaTeX”
but I do not see anywhere that I can instruct the Import TeX procedure how to locate that file.
Are these problems because of a bug in SWP 6 Import TeX process? If not, please tell me URGENTLY what I need to do to get a properly functioning .sci file that works with SWP 6.
AttachmentSize
MalcomsonMavroeidis18.rap267.9 KB
### Same thing occurs on Windows,
Same thing occurs on Windows, if that is any help in diagnosing the problem. Also, on Windows, pretex.exe takes a very long time (upwards of 6 mins) to run. Once imported, you can save the file as a sci file; but if you then try to open the sci file you get the same error message, and the file is still un-editable on screen.
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2017-06-24 15:34:06
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http://kazune-lab.net/contest/2020/01/10/abc150/
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AtCoder Beginner Contest 150
Updated:
AtCoder Beginner Contest 150
Solutions
A - 500 Yen Coins
Flush “Yes” if $500K \geq X$ or “No” otherwise.
B - Count ABC
Use substr to get the substrings of $S$.
C - Count Order
Use next_permutation to enumerate all possibilities of the permutations.
D - Semi Common Multiple
Note
We have a constraint that all $a _ i$ are even. But this is actually not guaranteed at the contest.
Solution
If there exists an odd $a _ i$, the answer is $0$. Suppose that all $a _ i$ are even. Let $a _ i’ = a _ i / 2$. We have $X = a _ i’ (2p + 1)$. Let $f(x)$ be the function that presents how many times we can divide $x$ by $2$. We have $f(X) = f(a _ i’)$; otherwise the answer is $0$. Suppose that all $f(a _ i’)$ are the same. We use $t$ to denote this number. Let $a _ i’’ = a _ i’ / 2 ^ t$ and $X’ = X / 2 ^ t$. Then $X’$, $a _ i’’$ and $2p + 1$ are all odd numbers with $X’ = a _ i’’ (2p + 1)$. Let $L = \mathrm{lcm}(a’’ _ i)$. Then, we have $X’ = L \times (\text{odd number})$ and this presents all possibilities. The answer is $(M’ + 2L - 1) / 2L$, where $M’ = M / 2 ^ t$.
E - Change a Little Bit
Solution
Suppose that we fix $S = 00 \dots 0$. Then we find that the answer will be $2 ^ N$ times. We assume without loss of generalities that $\{ C _ i \}$ is in ascending order. We can see that we always should change $1$ into $0$ ion ascending order.
How many times do we add $C _ i$? We can say that when we consider changing $1$ to $0$ on the $i$-th digit, we have solved the $j$-th digit for $0 \leq j < i$. The final count will be $2 ^ i$ times. Then, on the $i$-th digit, we always solve the conflict. Thus we count $2 ^ {N - 1 - i}$ times. For $j$-th digit for $j > i$, we count the half: $2 ^{N - 1 - i - 1}$. Thus we have $2 ^ i (2 ^ {N - 1 - i} + (N - 1 - i) 2 ^ {N - 1 - i - 1} ) = 2^{N - 2} (N - i + 1)$ counts. Therefore the answer is $2 ^ N \sum _ {i = 0} ^ {N - 1} 2^{N - 2} (N - i + 1) C _ i = 4 ^ {N - 1} \sum _ {i = 0} ^{N - 1} C _ i (N - i + 1).$
F - Xor Shift
Observation
We work on the condition $(\exists x \in \mathbb{N})(\forall i \in N)(a _ {i + k} \oplus x = b _ i). \tag{F.1}$ This is equivalent to the following condition. $(\forall i \in N)(a _ {i + k} \oplus a _ {i + k + 1} = b _ i \oplus b _ {i + 1}). \tag{F.2}$ We can see this equivalence as follows. Suppose we have (F.1). Then (F.2) follows immediately. Suppose we have (F.2). Then we set $x = a _ k \oplus b _ 0$ to have (F.1).
Let $a _ i’ = a _ i \oplus a _ {i + 1}$ and $b _ i’ = b _ i \oplus b _ {i + 1}$. Then, (F.2) is equivalent to $(\forall i \in N)(a _ {i + k}’ = b _ i’). \tag{F.3}$ Then, (F.3) means that we find $A$ in $BB$ starting by the $k$-th element. This can be done by some good string search algorithm, such as KMP algorithm.
Implement
We create $A$ and $BB$. We create $A$’s MP array. Then, search $BB$ for $A$. When we find $l$, it will mean that $k = N - l$. Note that the cases $k = 0$ and $k = N$ are the same. Letting $x = a _ k \oplus b _ 0$, we flush the answer.
Tags:
Categories:
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2020-05-29 20:25:59
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https://www.lmfdb.org/EllipticCurve/Q/458640na/
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# Properties
Label 458640na Number of curves $6$ Conductor $458640$ CM no Rank $1$ Graph
# Related objects
Show commands for: SageMath
sage: E = EllipticCurve("458640.na1")
sage: E.isogeny_class()
## Elliptic curves in class 458640na
sage: E.isogeny_class().curves
LMFDB label Cremona label Weierstrass coefficients Torsion structure Modular degree Optimality
458640.na6 458640na1 [0, 0, 0, 105693, -5296606] [2] 4718592 $$\Gamma_0(N)$$-optimal*
458640.na5 458640na2 [0, 0, 0, -458787, -44019934] [2, 2] 9437184 $$\Gamma_0(N)$$-optimal*
458640.na3 458640na3 [0, 0, 0, -3986787, 3033101666] [2, 2] 18874368 $$\Gamma_0(N)$$-optimal*
458640.na2 458640na4 [0, 0, 0, -5962467, -5599434526] [2] 18874368
458640.na1 458640na5 [0, 0, 0, -63609987, 195270223106] [2] 37748736 $$\Gamma_0(N)$$-optimal*
458640.na4 458640na6 [0, 0, 0, -811587, 7731762626] [2] 37748736
*optimality has not been proved rigorously for conductors over 400000. In this case the optimal curve is certainly one of the 4 curves highlighted, and conditionally curve 458640na1.
## Rank
sage: E.rank()
The elliptic curves in class 458640na have rank $$1$$.
## Modular form 458640.2.a.na
sage: E.q_eigenform(10)
$$q + q^{5} + 4q^{11} - q^{13} - 6q^{17} + 4q^{19} + O(q^{20})$$
## Isogeny matrix
sage: E.isogeny_class().matrix()
The $$i,j$$ entry is the smallest degree of a cyclic isogeny between the $$i$$-th and $$j$$-th curve in the isogeny class, in the Cremona numbering.
$$\left(\begin{array}{rrrrrr} 1 & 2 & 4 & 4 & 8 & 8 \\ 2 & 1 & 2 & 2 & 4 & 4 \\ 4 & 2 & 1 & 4 & 2 & 2 \\ 4 & 2 & 4 & 1 & 8 & 8 \\ 8 & 4 & 2 & 8 & 1 & 4 \\ 8 & 4 & 2 & 8 & 4 & 1 \end{array}\right)$$
## Isogeny graph
sage: E.isogeny_graph().plot(edge_labels=True)
The vertices are labelled with Cremona labels.
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2020-12-05 10:03:22
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|
https://hssliveguru.com/plus-one-physics-notes-chapter-7/
|
# Plus One Physics Notes Chapter 7 Systems of Particles and Rotational Motion
Students can Download Chapter 7 Systems of Particles and Rotational Motion Notes, Plus One Physics Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.
## Kerala Plus One Physics Notes Chapter 7 Systems of Particles and Rotational Motion
Summary
Introduction
In the earlier chapters we discussed the motion of particle. We applied the results of our study to the motion of bodies of finite size.
A large class of problems with extended bodies can be solved by considering them to be rigid bodies. Ideally a rigid body is a body with a perfectly definite and unchanging shape. The distances between different pairs of particles of such a body do not change.
(i) Basically a rigid body can have two types of motion.
1. translational
2. Rotational motion
1. Translational motion:
In pure translational motion, at any instant of time every particle of the body has the same velocity. Explanation
Consider a rectangular block moving down along an inclined plane as shown in the figure: This body is rigid. Hence all the particles have same velocity. Any points like P1 or P2 of the body moves with the same velocity at any instant of time.
2. Rotational motion:
In pure rotational motion at any instant of time every particle of the body have different velocities. Explanation
Consider a cylinder rolling down along inclined plane as shown in figure. Points P1, P2, P3 and P4 have different velocities (shown by arrows) at any instant of time.
Note: The above figure shows a combination motion of both translational and rotational motion.
Centre Of Mass
The centre of mass of a system of particles is the point where all the mass of the system may be assumed to be concentrated.
Position vector of two particle system:
Consider two particles of masses m1 and m2 with position vectors $$\vec{r}_{1}$$ and $$\vec{r}_{2}$$ respectively with respect to the origin O. Now the position coordinate of the center of mass C is defined as.
X, Y and Z coordinate of centre of mass of two particle system
Position vector of N particle system:
Consider a system of N particles of mass m1, m2…….,mN with position vectors $$\vec{r}_{1}, \vec{r}_{2}, \ldots \ldots \ldots \vec{r}_{N}$$ respectively. Then the centre of mass of the system of N particle is defined as
Position vector of CM of continuous distribution of mass:
If body is continuous distribution of mass, we divide the body into n small elements of mass: Dm1, Dm2,………..DmN. Let r1, r2,……….rN be the position vectors of those small elements respectively.
Then position vector of CM is,
when we take N as bigger ((Dmi) becomes smaller) the summation can be changed into integration.
Motion Of Centre Of Mass
Velocity of centre of mass. The position vector of the centre of mass of N particle system is given
Differentiating the position vector of C.M., we get velocity of CM. ie;
Acceleration of centre of mass:
Acceleration of centre of mass a = $$\frac{\mathrm{d}}{\mathrm{dt}} \overrightarrow{\mathrm{v}}$$
Force acting on the centre of mass:
Force acting on the centre of mass = Total mass at the centre of mass × acceleration of the centre of mass.
ie; $$\overrightarrow{\mathrm{F}}$$ = M$$\overrightarrow{\mathrm{a}}$$ ______(1)
But F = Finternal + Fextenal
By Newton’s third law, sum of the internal forces is zero.
∴ F = Fextenal
Therefore eq(1) becomes
Therefore the centre of mass moves as if it were a particle of mass equal to the total mass of the system and all the external forces are acting on it.
Linear Momentum Of A System Of Particles
Momentum of centre of mass
Velocity of centre of mass,
The equation means that, the total momentum of a system of particles is equal to the product of the total mass of the system and the velocity of its centre of mass.
Momentum conservation and centre of mass motion:
statement:
The total momentum of the centre of mass is conserved if no external force acts.
Proof:
According to Newton’s second law, rate of change momentum of centre of mass is directly proportional s to force acting on it.
If the total external force acting on the system is zero; the centre of mass moves with a constant velocity.
Application of the idea of centre of mass:
1. Explosion of a shell inflight
Consider a shell projected into air. If it is not exploded, its path will be a parabola. If it is exploded in air, the centre of mass follows the same parabolic path. Because the forces due to explosion are internal. Internal force can’t change the direction of centre of mass.
Note: External force can change the direction of centre of mass
2. Decay of radio active nuclei at rest:
Consider radioactive decay of radium nucleus moving along a straight line. A radium nucleus disintegrates into a nucleus of radon and an alpha particle. The two particles produced in the decay move in different directions. But the centre of mass (of radon and α particle) moves along a straight line, because the force leading to decay is internal. Internal force can’t change the direction of centre of mass.
Centre of mass frame:
If we observe this decay from the frame of reference in which the centre of mass (of a particle and radon) at rest, the product particle seems to be moving in opposite direction along a straight line.
In centre of mass frame, the motion of product particles become simple.
3. Binary stars:
Consider the trajectories of the two stars of equal mass as shown in figure. If there are no external forces, the centre of mass moves along a straight line as shown in figure.
Centre of mass frame of Binary stars:
If we look the trajectories of S1 and S2 from the centre of mass frame, we find that these two stars are moving in a circle as shown in figure.
In this frame of reference, the trajectories of the stars are a combination of
1. uniform motion in a straight line of the centre of mass and
2. circular orbits of the stars about the centre of mass.
Vector Product Of Two Vectors
The vector product of two vectors $$\overrightarrow{\mathrm{a}}$$ and $$\overrightarrow{\mathrm{b}}$$ is
Screw rule:
Rotate a right handed screw from $$\overrightarrow{\mathrm{a}}$$ to $$\overrightarrow{\mathrm{b}}$$. Then the direction of advance of the. tip of the screw gives the direction of ($$\overrightarrow{\mathrm{a}}$$ × $$\overrightarrow{\mathrm{b}}$$)
Right hand rule:
Curl the fingers of the right hand from $$\overrightarrow{\mathrm{a}}$$ to $$\overrightarrow{\mathrm{b}}$$ along the shorter angle. Then the direction in which the thumb points gives the direction of ($$\overrightarrow{\mathrm{a}}$$ × $$\overrightarrow{\mathrm{b}}$$).
Properties of cross product:
Note:
If $$\hat{i}, \hat{j}, \hat{\mathbf{k}}$$ occur cyclically in the above vector product relation, the vector product is positive. If $$\hat{i}, \hat{j}, \hat{\mathbf{k}}$$ do not occur in cyclic order, the vector product is negative.
Question 1.
Angular Velocity And Its Relation With Linear Velocity
In earlier chapter, we treated angular velocity as scalar. But angular velocity is a vector quantity. The relation between angular velocity and linear velocity can be written in vector notation as
Direction of angular velocity:
Consider a body moving along the circumference of circle of radius ‘r’with velocity v. Then the direction of angular velocity will be perpendicular to both $$\overrightarrow{\mathbf{V}}$$ and $$\overrightarrow{\mathbf{r}}$$ (along the axis of rotation).
The figure(1) shows the direction of w, if body rotates in clockwise direction.
The figure(2) shows the direction of w, if body rotates in anticlockwise direction.
1. Angular acceleration:
Rate of change of angular velocity is called angular acceleration.
angular acceleration $$\vec{\alpha}=\frac{\mathrm{d} \vec{\omega}}{\mathrm{dt}}$$
If the axis of rotation is fixed, the direction of ω and hence α is fixed. In this case the vector equation reduces to scalar equation.
Torque And Angular Momentum
Torque and angular momentum are important quantities to discuss the motion of rigid bodies.
1. Moment of force (Torque)
Torque (or the moment) of a force of about a point is the rotating effect of the force about that point.
Explanation
If f is the force acting at a point A, then torque about a point O is given by
Where $$\overrightarrow{\mathbf{r}}$$ is the position vector of the point A from the point O. Torque is a vector quantity. The direction of torque is (given by right hand screw rule). Perpendicular to both $$\overrightarrow{\mathbf{r}}$$ and $$\overrightarrow{\mathbf{f}}$$.
Unit:
The unit of torque is Nm.
Note:
• If the turning tendency of the force is anticlockwise, then the torque is positive and if it is clockwise, then torque is negative.
• Torque has dimensions ML2T-2. Its dimensions are the same as those of work or energy. Torque is. vector, while work is scalar.
• Torque plays the same role in rotational motion as force does in translational motion.
Couple:
Two equal and opposite forces, separated by a distance, constitute a couple.
Moment of couple (Torque):
The moment of couple is the product of either of the forces and the perpendicular distance between them.
Question 2.
The door is a rigid body which can rotate about a fixed axis passing through the hinges. What makes the door rotate?
A force applied to the hinge line can’t produce any rotation. But a force of given magnitude applied at right angles to the door at its outer edge is most effective in producing motion. The rotational analogue of force is moment of force. It is also referred to as torque.
Question 3.
When you fix a handle on a door where do you fix it?
You fix it in such a way that the torque is maximum. For this the lever arm must be maximum. So the handle is fixed as far away from the hinges as possible Note: A couple produces rotation without translation motion.
2. Angular momentum of a particle:
Angular momentum of a particle about a point is defined as the moment of its linear momentum about that point.
Explanation
Considers particle of mass ‘m’ and linear momentum $$\overrightarrow{\mathbf{p}}$$ at a position $$\overrightarrow{\mathbf{r}}$$ relative to the origin O. The angular momentum $$\overrightarrow{\mathbf{l}}$$ of the particle can be written as (with respect to the origin O).
Note:
• The quantity angular momentum is the rotational analogue of linear momentum.
• Direction of $$\overrightarrow{\mathbf{l}}$$ is given by right hand screw rule, (angular momentum is perpendicular to both $$\overrightarrow{\mathbf{r}}$$ and $$\overrightarrow{\mathbf{p}}$$).
Relation between angular momentum and torque:
Angular momentum of a particle,
When differentiate on both side, we get
ie The rate of change of angular momentum is the torque applied to it. This is similar to force equal to rate of change of linear momentum.
Torque and angular momentum for a system of particles:
Consider a system having n particles. The total angular momentum of system is the vector sum of angular momentum of individual particles.
∴ total angular momentum of system,
$$\frac{\mathrm{dL}}{\mathrm{dt}}=\tau$$ _____(1)
But Στi = τ is the total torque acting on the system of particle. Actually torque acting on a system is sum of external torque (τext) and internal torque (τint)
ie. τ = τext + τint
But τint = 0, because internal torque arises due to action – reaction pair. Action reaction pair for a system is zero.
∴ τ = τext ______(2)
substituting eq(2) is eq (1) we get
$$\frac{\mathrm{d} \overrightarrow{\mathrm{L}}}{\mathrm{dt}}=\tau_{\mathrm{ext}}$$
The above equation means that, the time rate of the total angular momentum of a system of particles about a point is equal to the sum of the external torques acting on the system taken about the same point.
Conservation of angular momentum of a system:
For a system of particles,
If the total external torque on a system of particles is zero, then the total angular momentum of the system is conserved.
Note: The statement that the total angular momentum is conserved means that each of these three components (Lx, Ly, Lz) is conserved.
Equilibrium Of A Rigid Body
A rigid body is said to be in a mechanical equilibrium, if both it’s linear momentum and angular momentum are not changing with time.
(or)
A body is said to be in mechanical equilibrium, if the body has neither linear acceleration nor angular acceleration. A body moving with constant momentum is said to . be in translational equilibrium. Similarly a body with constant angular momentum is said to be in rotational equilibrium.
Condition for translational equilibrium:
If the total force acting on a rigid body is zero, the body, moves with constant linear momentum (or zero linear acceleration). The body moving with constant linear momentum is said to be in translational equilibrium Undertranslational equilibrium, the total force acting on the rigid body is zero.
Condition for rotational equilibrium:
If the total torque acting on rigid body is zero, the total angular momentum of the body does not change. Then the body is said to be in rotational equilibrium. Mathematical condition for rotational equilibrium is
Principles of moments:
Consider a lever pivoted at some origin (say fulcrum) in mechanical equilibrium. Let $$\overrightarrow{\mathrm{f}}_{1}$$ and $$\overrightarrow{\mathrm{f}}_{2}$$ he the forces acting at A and B as shown in figure. Let $$\overrightarrow{\mathrm{R}}$$ be the reaction of the support at the fulcrum. For translational equilibrium
For rotational equilibrium, sum of moments must be zero.
ie d1f1 + -d2f2 = 0
d1f1 = d2f22 ______(2)
In the case of lever, f1 is used to lift some weight. Hence f1 is called load and its distance from the fulcrum d1 is called load arm. Force f2 is the effort applied to lift the load, distance d2 is called effort arm.
Hence d1f1 = d2f2 can be written as
The above equation expresses the principle of moments for a lever.
The ratio $$\frac{f_{1}}{f_{2}}$$ is called mechanical advantage (MA)
Note:
High value of mechanical advantage means that a small effort can be used to lift a large load.
Examples for level
• See – saw
• Beam balance
1. Centre of gravity:
Centre of gravity of a body is that point through which the net gravitational force acts. The centre of gravity of a body may not be the same as its centre of mass. For a body of very large dimensions, the value of acceleration due to gravity is different for its different parts. In this situation the centre of gravity does not coincide with the centre of mass.
If the size of the body is small, the value of acceleration due to gravity is same for its different parts. In this situation, the centre of gravity of the body coincides with the centre of mass. Torque due to gravity.
Moment Of Inertia
A body at rest cannot rotate by itself. A body in uniform rotation cannot stop by itself. This inability of a material body is called rotational inertia. It depends on the mass of the body and axis of rotation. In other words it depends on a quantity called moment of inertia.
a. Moment of inertia of a particle:
Consider a particle of mass ‘m’ capable of rotation about an axis AB.
Let ‘r’ be the perpendicular distance of particle from AB. The moment of inertia of the particle about AB is defined as the product of mass of the particle and square of the distance between the particle and the axis of rotation.
b. Moment of inertia of a rigid body:
Consider a rigid body capable of rotation about an axis AB. Let the body consisting of particles of , masses, m1, m2, m3……….mn at distances r1, r2, r3………..rn
Then by definition, moment of inertia of m, about AB = m1r12.
M.I of m2 about AB = m2r22
————————–
M.I. of mn about AB = mnrn2
Therefore total moment of inertia of the body about
I = m1r12 + m2r22 + ………………mnrn2
c. Moment of inertia of a ring about an axis through its centre and perpendicular to its plane:
Consider ring of mass M and radius R. AB the axis of rotation. Let ‘m’ be the mass of small section on the circumference of the ring.
M.I. of ‘m’about AB = mR2
∴ Total moment of inertia about AB,
I = ΣmR2
where Σm = M.
d. Moment of inertia of pair of small masses attached to the two ends of massless rod:
Consider a rigid massless rod of length / with a pair of small masses, rotating about an axis through the centre of mass perpendicular to the rod. Each mass M/2 is at a distance 1/2 from the axis of rotation.
∴ Total moment of inertia
Radius of gyration of a body is the square root of ratio of moment of inertia and total mass of the body
Question 10.
Why do the concept of radius of gyration introduce?
The moment of inertia of a body is given by
I = Σmr2
From the above equation it is clear that, we have to find the product of mass and the square of distances from the axis of rotation for all particles and then sum all these products. But the concept of radius of gyration simplifies the above problem.
In this method we find the centre of mass. Then we find a distance to any axis from this centre of mass, in such a way that the moment of inertia of this point (centre of mass) may be equal to that of I = Σmr2. This distance from the axis of rotation to centre of mass is called radius of gyration. Practical utility of moment of inertia.
Question 11.
In a fly wheel (or) wheels of vehicles, most of the mass is concentrated at the rim? Explain why?
(i) Flywheel:
The machines (such as steam engine, automobile engine etc) that produce rotational motion have a disc with large moment of inertia.
This disc is called fly wheel. Because of its large moment of inertia, the flywheel resists the sudden increase or decrease of the speed of the vehicle. It allows a gradual change in the speed and prevents the jerky motions. Thus fly wheel gives a smooth ride forthe passengers on the vehicle.
(ii) The wheels of vehicles:
The moment of inertia of wheels is increased by concentrating most of the mass at the rim of the wheel. If such a wheel gain or loses some K.E of rotation $$\left(\frac{1}{2} I \omega^{2}\right)$$ brings a relatively smaller change in its angular speed w(∵ I is large) Hence such a flywheel helps in maintaining uniform rotation.
1. K.E of rotating body:
Consider a body rotating about an axis passing through some point O with uniform angular velocity ω. The body can be considered to be made up of a number of particles of masses m1, m2, m3 etc.
At distances r1, r2, r3 etc. All the particles will have same angular velocity ω. But their linear velocities will be different say v1, v2, v3 etc.
Theorems Of Perpendicular And Parallelaxes
Theorem of perpendicular axes:
The moment of inertia of a planar body (lamina) about an axis perpendicular to its plane is equal to the sum of its moments of inertia about two perpendicular axes concurrent with perpendicular axis lying in the plane of the body.
(or)
Moment of inertia of a plane lamina about the z-axis is equal to sum of the moments of inertia about x axis and the y axis, if planer lamina lies in xy plane.
Explanation
Let Ix and Iy be the moments of inertia of the lamina about ox and oy. Let Iz be the moment of inertia of the lamina about an axis perpendicular to the lamina and passing through ‘0’ (about z axis) Then by perpendicular axis theorem.
Question 12.
M.l of disc about one of its diameters.
According perpendicular axis theorem, Iz = Ix + Iy
But in case Ix = Iy
Iz = 2 Ix _____(1)
But we know Iz = MR2/2 _____(2)
sub(2) in (1), we get $$\frac{\mathrm{MR}^{2}}{2}$$ = 2 Ix
Ix = $$\frac{M R^{2}}{4}$$
The moment of inertia about any diameter is $$\frac{M R^{2}}{4}$$.
1. Theorem of parallel axis:
The moment of inertia of a body about any axis is equal to the sum of the moment of inertia of the body about a parallel axis passing through its centre of mass and the product of its mass and the square of the distance between the two parallel axes.
Explanation
Let I be the moment of inertia of a-body about an axisAB. Let I0 be moment of inertia about the axis CD parallel to AB and passing through the centre of gravity (G of the body). Let M be the mass of the body and ‘a’ be the distance between the two axes. Then by parallel axes theorem.
I = Io + Ma2
Question 13.
What is the moment of inertia of rod of mass M, length l about an axis perpendicular to it through one end.
Consider a rod of mass M and length l, rotating about an axis passing through one end. Let dx be a small element at a distance x from the axis of rotation.
mass of the element dx, dm = $$\frac{M}{l}$$dx
∴ M.I of length small element, dl = $$\frac{M}{l}$$ dx x2
Moment of inertia of a ring about a tangent:
Consider a ring of mass M and radius R. If this ring is rotating about tangent, we can use parallel axis theorem.
According to parallel axis theorem,
Kinematics Of Rotational Motion About A Fixed Axis
The quantities ‘q’, w and ‘α’ in rotational motion has corresponding quantities in translational motion (x, ‘v’ and a respectively). For translational motion, we have
v = u + at
v2 = u2 + 2as
s = ut+ $$\frac{1}{2}$$ at2
putting u = w1, v = w2 and s = q, we get the equations of motion in rotational motion.
w2 = w1 + αt
ω22 = ω12 + 2αθ
θ = ω1t + $$\frac{1}{2}$$ αt2
Dynamics Of Rotational Motion About A Fixed Axis
Table 7.2 lists quantities associated with linear motion and their analogues in rotational motion.
a. Work done by a torque:
Consider a particle at P1. Let r1 be the position vector at time t = 0. This position vector makes an angle q with x – axis. Let particle be acted by a force $$\overrightarrow{\mathrm{F}}_{1}$$. Due to this force the particle subtends an angle dq and reaches at p11.
ds is the linear displacement due to the force $$\overrightarrow{\mathrm{F}}_{1}$$ This force makes angle with α1, the position vector r1.f1 is the angle made by $$\overrightarrow{\mathrm{F}}_{1}$$ with linear displacement.
Form the triangle the workdone for small displacement ds,
Substituting the eq(2) in (1) we get
dW1 = τ1
For rigid body, there are many particles. Hence total work done on it.
dW1 = (τ1 + τ2 +……….)dθ
Where τ is the total torque acting oh the body.
b. Rateofwork done by torque:
Instantaneous power due to torque.
We know dw = τ dθ
dividing both sides by dt, we have $$\frac{d w}{d t}=\tau \frac{d \theta}{d t}$$
P = τω
c. Angular acceleration:
Rate of change of angular velocity is called angular acceleration.
Angular acceleration α = $$\frac{\mathrm{d} \omega}{\mathrm{dt}}$$.
d. Relation between torque and angular acceleration:
The rate at which work is done on the body is equal to the rate at which kinetic energy increases.
τω = Iωα
τ = Iα
Note:
Just as force (F = ma) produces acceleration, torque produces angular acceleration in a body.
Newtons second law in rotation about a fixed axis
The angular acceleration is directly proportional to the applied torque and is inversely proportional to the moment of inertia of the body for rotation about a fixed axis.
Angular Momentum In Case Of Rotation About A Fixed Axis
Consider a rigid body rotating about a given axis with a uniform angular velocity w. Let the body consist of n particles of masses m1, m2, m3…………mn at perpendicular distances r1, r2, r3………..rn respectively from the axis of rotation.
If v1, v2, v3…………vn are the linear velocities of the
respective particles, then
v1 = r1w, v2 = r2w, v3 = r3w……….
The linear momentum of particle of mass m1 is,
P1 = m1v1
P1 = m1r1w.
The angular momentum of this particle about the given axis,
l1 = p1 x r1
= (m1 v1) x r1
v1 = r1w
= m1r12w
Similarly angular momentum of second particle
l2 = m2r22w
Angular momentum of the about the given axis,
1. Conservation of angular momentum:
Conservation of angular momentum and moment of inertia. When there is no external torque, the total angular momentum of a body or a system of bodies are a constant.
L = constant
But L = Iω
∴ Iω = a constant
Application
Question 14.
If the polar ice cap melts what will happen to the length of the day?
For earth, angular momentum is a constant (Lw = constant, ie no torque acts on the earth). When the polar ice cap melts, the water thus formed will flow down to the equtorial region. The accumulation of water in equatorial line will increase the moment of inertia I of earth. In order to keep the angular momentum as a constant, w will decrease. The decrease in ‘w’ will increase the length of day.
Question 15.
If the earth loses the atmosphere what will happen to the length of the day?
For earth, the angular momentum (L = Iw) is a constant, because there is no torque acting on it. When earth loses the atmosphere, I decreases and w increases to keep L as constant. Hence length of the day decreases.
Question 16.
A girl standing on a turn table. What happens to the rotation speed, if she stretches her hand?
If a girl rotating with a uniform speed on turn table, it’s angular momentum (L = Iw) will be a constant. When she suddenly stretches her hand, I increases and w decreases to keep L as constant.
Question 17.
How does a circus acrobat and a divertake advantage of conservation of angular momentum?
The diver while leaving the spring board, is throwing himself in a rotating, motion. When he brings his hands and legs close, I decrease and w increases. But before reaching water he will stretch his hands and legs. Hence I increases and w decreases. So that he gets a smooth entry into the water.
Rolling Motion
Question 18.
A wheel rolling uniformly along a level road is shown in the figure. The centre is moving with speed (VCM). Find resultant velocities at P1, P2 and P0.
We know that the translational velocity of body is equal to the velocity of centre of mass. The velocity of centre of mass VCM = Rw. Where R is the radius of wheel.
Velocity at P1:
The point at P1 has two velocities.
• Linear velocity (Vl)
• Translational velocity (Vt)
The linear velocity at P1, Vl = Rw.
Translational velocity at P1, Vt = Rw
[∵ translational velocities are same for all points on the wheel and its value equal to velocity of centre of mass].
The direction of Vl and Vt are same at P1.
Hence total velocity at
Velocity at P2:
Linear velocity at P2, Vl = rw
[where r is the distance of P2 from centre of mass]
translational velocity at P2, Vt = Rw.
∴ Total velocity at
Velocity at P0:
Linear velocity at P0, $$\overrightarrow{\mathrm{V}}_{l}$$ = Rω Translational at P0, Vt = Rw.
The direction of Vland Vt are opposite.
∴ Hence total velocity,
= -Rw + Rw = 0
which means that the point P0 is instantaneously at rest. Hence the friction at P0 is zero. As a result rolling friction becomes less than the kinetic friction.
Note: The condition that P0 is instantaneously at rest requires VCM = Rw. Thus for the disc the condition for rolling without slipping is,
VCM = RW.
1. Kinetic energy of rolling motion (without slipping):
In this case kinetic energy has two parts,
• due to the linear motion of centre of mass
• due to the rotational motion of the body.
K.E in terms of radius of gyration:
Moment of inertia I = mk2
where K is the radius of gyration of the body and v = Rw.
Substituting I, and V in eq(1), we get
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2023-03-25 07:53:00
|
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http://brahma.tcs.tifr.res.in/print/2534
|
Modem Design for MultiGigabit OFDM System
Speaker:
Tapan Shah School of Technology and Computer Science Tata Institute of Fundamental Research Homi Bhabha Road
Time:
Monday, 15 November 2010 (All day)
Venue:
• AG-80
In multiGigabit systems having digital signal processing (DSP) based transceiver architectures, the analog-to-digital converter (ADC) is a critical component due to the cost and power requirements of high precision ADC. Thus, we may be forced to reduce the precision to 1-3 bits. We study the performance degradation of a conventional OFDM receiver at low precision and investigate an alternate receiver structure, which we call analog OFDM receiver (AOR), where we incorporate some analog processing before sampling. In an ideal implementation, the AOR yields near full precision performance. We also discuss some challenges in implementing such a receiver.
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2019-09-18 18:03:25
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|
http://www.drumtom.com/q/proof-that-4-n-c-3-n
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# Proof that 4^n>c(3)^n?
• Proof that 4^n>c(3)^n?
Proof that 4^n>c(3)^n? Find answers now! No. 1 Questions & Answers Place. More questions about Science & Mathematics
Positive: 82 %
A visual proof that 1+2+3+...+n = n(n+1)/2 We can visualize the sum 1+2+3+ ... (n) = 1 + 2 + 3 + ... + n = n(n + 1)/2. The same proof using algebra!
Positive: 79 %
### More resources
Induction Problems Tom Davis ... Show that 3n 2n. Proof: If n= 0, 1 = 30 20 = 1, so it’s true in this case. Assume 3k 2k. Multiply both sides by 3:
Positive: 82 %
2 Direct Proof 1; 3 Direct Proof 2; 4 Proof by Induction. 4.1 Base case; 4.2 Induction Hypothesis; ... {n - 1}\right) + n + \left({n - 1}\right) + \cdots + 1\)
Positive: 77 %
Math 3210-3 HW 27 Solutions Applications of Uniform Convergence 1. Let fn(x) = nx 1+nx for x ∈ [0,1]. ... n=2 n2 2n. Proof: Since f(x) = x +x2
... 3,$.... First, if n happens to be zero,$f(x) = x^0 = 1,\$ a constant, ... Step 3: Proof of the Power Rule for Rational Exponents. For this step, ...
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2017-01-19 21:34:52
|
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https://src.rampantmonkey.com/dcphr/tree/cyoa/walkthrough/family_tree.tex?id=0eee44e78f23ed3dc9eec074449726b0a11fa84f
|
summaryrefslogtreecommitdiffstats log msg author committer range
blob: 9bda87289b356710e198f99c5bbd0258667eee3f (plain)
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 \section{Family Tree} \subsection*{Upon Arrival} \begin{itemize} \item Insert bookmark into book on page 98 \item Turn book to page 120 \item Give team a shoebox full of pictures \end{itemize} \subsection*{Upon Departure} \begin{itemize} \item Ensure that the correct answer is entered on page 98 \item Add the following sticker to page 98 \\ \begin{mdframed} I suppose this Rosalind person will be the next member of the family to inherit me… assuming we all make it out of this mess. Go back to page 24. \\ \end{mdframed} \item Return to page 98 \end{itemize}
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2020-04-10 03:13:50
|
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|
https://tex.stackexchange.com/questions/555169/refer-to-an-equation-with-multiple-labels
|
# Refer to an equation with multiple labels
I have the following equation:
$$\tag{E_{n}}\label{E} f_n(x)=a$$
When I write \eqref{E} I am referred to the equation and it is written E_n
My question is: how to be referred to the same equation and write for example E_{n-1} rather then E_n?
• As a writer but more importantly as reader of scientific papers I beg you: please, please don't do that ;-) – campa Jul 26 '20 at 19:46
• Welcome to TeX.SE! Instead of writing See Equation \eqref{E} you can also just write see Equation (E$_{n-1}$), so don't use \eqref at all if you don't want to use the tag. But I guess that would be very confusing for the reader? – Marijn Jul 26 '20 at 19:47
• The question/intent is ill formed, as the reference is just that, a reference to the equation, which you have given a specific (non-numeric) name to. If you wish to change the formula, then this produces a dangling reference, which does not refer to anything. – oliversm Jul 26 '20 at 19:49
• I want that when the reader click on E_{n-1}, he will be directed to the equation. Indeed, the equation is written as a function of n, and when we write E_{n-1} it will be clear for the reader how he can deduce it. – Ghizlane Jul 26 '20 at 19:53
You can do like this:
\documentclass{article}
\usepackage{amsmath}
\usepackage{hyperref}
\DeclareRobustCommand{\tagindex}{\ensuremath{_{\defaulttagindex}}}
\newcommand{\defaulttagindex}{n}
\DeclareRobustCommand{\indexedref}[2][n]{%
\begingroup\def\defaulttagindex{#1}\ref{#2}\endgroup
}
\DeclareRobustCommand{\indexedeqref}[2][n]{%
\begingroup\def\defaulttagindex{#1}\eqref{#2}\endgroup
}
\begin{document}
$$\tag{E\tagindex}\label{E} f_n(x)=a$$
We refer to~\indexedeqref{E}, but also to~\indexedeqref[n-1]{E}.
\end{document}
For this use case you can use the command \hyperref[label]{text} from the hyperref package. It creates a (clickable) link to the label with the text being shown in the document (and not the default label text).
MWE:
\documentclass{article}
$$\tag{E_{n}}\label{E} f_n(x)=a$$
See Equation \eqref{E}, also known as Equation (\hyperref[E]{E$_{n-1}$}).
|
2021-06-25 09:21:03
|
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|
http://www.citizendia.org/Geoid
|
Map of the undulations of the geoid, in meters (based on the EGM96 gravity model and the WGS84 reference ellipsoid). EGM96 ( Earth Gravitational Model 1996) is a Geopotential model of the Earth consisting of Spherical harmonic coefficients complete to degree and order 360 The World Geodetic System defines a reference frame for the earth for use in Geodesy and Navigation. [1]
The geoid is that equipotential surface which would coincide exactly with the mean ocean surface of the Earth, if the oceans were to be extended through the continents (such as with very narrow canals). Equipotential surfaces are Surfaces of constant Scalar potential. According to C.F. Gauss, who first described it, it is the "mathematical figure of the Earth," a smooth but highly irregular surface that corresponds not to the actual surface of the Earth's crust, but to a surface which can only be known through extensive gravitational measurements and calculations. Johann Carl Friedrich Gauss (ˈɡaʊs, Gauß Carolus Fridericus Gauss ( 30 April 1777 – 23 February 1855) was a German Gravitation is a natural Phenomenon by which objects with Mass attract one another Despite being an important concept for almost two hundred years in the history of geodesy and geophysics, it has only been defined to high precision in recent decades, for instance by works of P. Vaníček and others. Geodesy (dʒiːˈɒdɪsi also called geodetics, a branch of Earth sciences, is the scientific discipline that deals Geophysics, a major discipline of Earth sciences, is the study of the Earth by quantitative physical methods especially by seismic, electromagnetic Petr Vaníček (born 1935 in Sušice, Czechoslovakia, today in Czech Republic) is a Czech It is often described as the true physical figure of the Earth, in contrast to the idealized geometrical figure of a reference ellipsoid. The expression figure of the Earth has various meanings in Geodesy according to the way it is used and the precision with which the Earth's size and shape is to be defined In Geodesy, a reference ellipsoid is a mathematically-defined surface that approximates the Geoid, the truer Figure of the Earth, or other planetary body
## Description
1. Ocean
2. Ellipsoid
3. Local plumb
4. Continent
5. Geoid
The geoid surface is irregular, unlike the reference ellipsoids often used to approximate the shape of the physical Earth, but considerably smoother than Earth's physical surface. In Geodesy, a reference ellipsoid is a mathematically-defined surface that approximates the Geoid, the truer Figure of the Earth, or other planetary body While the latter has excursions of +8,000 m (Mount Everest) and −11,000 m (Mariana Trench), the total variation in the geoid is less than 200 m (-106 to +85 m[2](compared to a perfect mathematical ellipsoid). Mount Everest, also called Sagarmatha (सगरमाथा meaning Head of the Sky) or Chomolungma, Qomolangma or Zhumulangma (in The Mariana Trench (or Mariana's Trench) is the deepest part of the world's Oceans and the deepest location on the surface of the Earth 's
Sea level, if undisturbed by tides and weather, would assume a surface equal to the geoid. If the continental land masses were criss-crossed by a series of tunnels or narrow canals, the sea level in these canals would also coincide with the geoid. In reality the geoid does not have a physical meaning under the continents, but geodesists are able to derive the heights of continental points above this imaginary, yet physically defined, surface by a technique called spirit leveling. Geodesy (dʒiːˈɒdɪsi also called geodetics, a branch of Earth sciences, is the scientific discipline that deals Spirit leveling is a technique for determining differences in height between points on the Earth's surface
Being an equipotential surface, the geoid is by definition a surface to which the force of gravity is everywhere perpendicular. Equipotential surfaces are Surfaces of constant Scalar potential. This means that when travelling by ship, one does not notice the undulations of the geoid; the local vertical is always perpendicular to the geoid and the local horizon tangential component to it. In Mathematics, given a vector at a point on a Surface, that vector can be decomposed uniquely as a sum of two vectors one Tangent to the surface called Likewise, spirit levels will always be parallel to the geoid.
## Spherical harmonics representation
Three-dimensional visualization of geoid undulations, using units of gravity.
Spherical harmonics are often used to approximate the shape of the geoid. In Mathematics, the spherical harmonics are the angular portion of an Orthogonal set of solutions to Laplace's equation represented in a system of The current best such set of spherical harmonic coefficients is EGM96 (Earth Gravity Model 1996)[3], determined in an international collaborative project led by NIMA. EGM96 ( Earth Gravitational Model 1996) is a Geopotential model of the Earth consisting of Spherical harmonic coefficients complete to degree and order 360 The National Geospatial-Intelligence Agency ( NGA) is an agency of the United States Government with the primary mission of collection analysis and The mathematical description of the non-rotating part of the potential function in this model is
$V=\frac{GM}{r}\left(1+{\sum_{n=2}^{n_{max}}}\left(\frac{a}{r}\right)^n{\sum_{m=0}^n}\overline{P}_{nm}(\sin\phi)\left[\overline{C}_{nm}\cos m\lambda+\overline{S}_{nm}\sin m\lambda\right]\right),$
where $\phi\$ and $\lambda\$ are geocentric (spherical) latitude and longitude respectively, $\overline{P}_{nm}$ are the fully normalized Legendre functions of degree $n\$ and order $m\$, and $\overline{C}_{nm}$ and $\overline{S}_{nm}$ are the coefficients of the model. Note This article describes a very general class of functions Note that the above equation describes the Earth's gravitational potential $V\$, not the geoid itself, at location $\phi,\;\lambda,\;r,\$ the co-ordinate $r\$ being the geocentric radius, i. The Mathematical study of potentials is known as Potential theory; it is the study of Harmonic functions on Manifolds This mathematical e, distance from the Earth's centre. The geoid is a particular[4] equipotential surface, and is somewhat involved to compute. Equipotential or isopotential in Mathematics and Physics (especially Electronics) refers to a region in space where every point in it is at the The gradient of this potential also provides a model of the gravitational acceleration. EGM96 contains a full set of coefficients to degree and order 360, describing details in the global geoid as small as 55 km (or 110 km, depending on your definition of resolution). One can show there are
$\sum_{k=2}^n 2k+1 = n(n+1) + n - 3 = 130,317$
different coefficients (counting both $\overline{C}_{nm}$ and $\overline{S}_{nm}$, and using the EGM96 value of n = nmax = 360). For many applications the complete series is unnecessarily complex and is truncated after a few (perhaps several dozen) terms.
New even higher resolution models are currently under development. For example, many of the authors of EGM96 are working on an updated model[5] that should incorporate much of the new satellite gravity data (see, e. g. , GRACE), and should support up to degree and order 2160 (1/6 of a degree, requiring over 4 million coefficients). The goal of the Gravity Recovery And Climate Experiment ( GRACE) space mission is to obtain accurate global and high-resolution determination of both the static and the time-variable
## Precise geoid
The 1990s saw important discoveries in theory of geoid computation. The Precise Geoid Solution [6] by Vaníček and co-workers improved on the Stokesian approach to geoid computation. Petr Vaníček (born 1935 in Sušice, Czechoslovakia, today in Czech Republic) is a Czech Sir George Gabriel Stokes 1st Baronet FRS ( 13 August 1819 &ndash 1 February 1903) was a mathematician and physicist Their solution enables millimetre-to-centimetre accuracy in geoid computation, an order-of-magnitude improvement from previous classical solutions [7] [8] [9]. Computation is a general term for any type of Information processing. An order of magnitude is the class of scale or magnitude of any amount where each class contains values of a fixed ratio to the class preceding it
## References
1. ^ data from http://earth-info.nga.mil/GandG/wgs84/gravitymod/wgs84_180/wgs84_180.html
2. ^ http://www.csr.utexas.edu/grace/gravity/gravity_definition.html visited 2007-10-11
3. ^ NIMA Technical Report TR8350. 2, Department of Defense World Geodetic System 1984, Its Definition and Relationships With Local Geodetic Systems, Third Edition, 4 July 1997. [Note that confusingly, despite the title, versions after 1991 actually define EGM96, rather than the older WGS84 standard, and also that, despite the date on the cover page, this report was actually updated last in June 23 2004. Available electronically at: http://earth-info.nga.mil/GandG/publications/tr8350.2/tr8350_2.html]
4. ^ There is no such thing as "The" EGM96 geoid
5. ^ Pavlis, N. K. , S. A. Holmes. S. Kenyon, D. Schmit, R. Trimmer, "Gravitational potential expansion to degree 2160", IAG International Symposium, gravity, geoid and Space Mission GGSM2004, Porto, Portugal, 2004.
6. ^ UNB Precise Geoid Determination Package, page accessed 02 October 2007
7. ^ Vaníček, P. , Kleusberg, A. The Canadian geoid-Stokesian approach, Pages 86-98, Manuscripta Geodaetica, Volume 12, Number 2 (1987)
8. ^ Vaníček P., Martinec Z. Compilation of a precise regional geoid (pdf), Pages 119-128, Manuscripta Geodaetica, Volume 19 (1994)
9. ^ Vaníček et al. Compilation of a precise regional geoid (pdf), pp. 45, Report for Geodetic Survey Division - DSS Contract: #23244-1-4405/01-SS, Ottawa (1995)
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2013-06-19 17:44:57
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https://greprepclub.com/forum/for-each-of-the-following-functions-3058.html
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# For each of the following functions
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For each of the following functions [#permalink] 23 Dec 2016, 12:37
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For each of the following functions, give the domain and a description of the graph y=f(x) in the xy plane, including its shape, and x and y intercepts?
1) f(x) = -4
2) 100 - 900x
3) f(x) = 5 - (x+20)^2
4) f(x) = $$\sqrt{x+2}$$
5) f(x) = x + I x I
Does anyone know how to solve this question?
Thank you
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Re: For each of the following functions [#permalink] 23 Dec 2016, 15:14
Expert's post
They key is to understand the meaning of domain. For all values of x for which f(x) is a real number a set of such value is called the domain.
So try and find things like values where the function is imaginary or a point where the value of f(x) suddenly goes to infinity.
Case no 4 i your question is example of imaginary value of f(x).
Example of a function suddenly going to infinity is $$f(x)= \frac{1}{x}$$ Here domain of f(x) is all real number except x=0.
1) f(x) = -4
This is constant function, ie for any value of x you will have a constant value of y so f(x) is defined for all real values of x.
Thus domain of this is the set of all real numbers R.
2) f(x) = 100 - 900x
This is a sum of a constant and a linear function. Both functions are defined for all real numbers
Thus domain of this is the set of all real numbers R.
3) f(x) = 5 - (x+20)^2
This is a sum of a constant and quadratic function. Both functions are defined for all real numbers
Thus domain of this is the set of all real numbers R.
4) f(x) = $$\sqrt{x+2}$$
Here is x < -2 then f(x) is undefined. for any other value f(x) is defined.
Here the domain x all values of $$x \geq -2$$.
5) f(x) = x + |x|
This is a sum of a constant and modulous function. Both functions are defined for all real numbers
Thus domain of this is the set of all real numbers R.
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Re: For each of the following functions [#permalink] 23 Dec 2016, 15:14
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2018-12-09 22:41:46
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http://physics.stackexchange.com/tags/antimatter/new
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# Tag Info
4
There was a lot of hype $10$ to $15$ years ago over the hydrogen economy. It was of course rather odd that anyone could take this seriously. How much free hydrogen gas is available? Answer: virtually none. The problem is that you have to either put electrical energy into water to split it into $H_2$ and $O_2$, or if you chemically condition methane $CH_4$ ...
2
Positronium is an exotic species composed of positron and electron. Depending on the allignment of the spins of postron and electron, the total spin of the positron can be zero or 1 (for the ground state for which l=0) and are designated as: ^1S (para-positronium) and ^3S (orto-positronium)
0
We should start by saying that we don't understand the physics that produced a matter-antimatter asymmetry. It seems likely that the asymmetry was around one extra matter particle per billion or ten billion matter-antimatter particles, but this figure is based on general principles rather than any precise calculation. What this means is that if we take a ...
1
Not really. If we were made of antimatter, we would think of matter as antimatter. Antimatter is really the same as matter, just with an opposite charge and opposite lepton/baryon numbers. The laws of physics would be virtually the same. However, if we were made of matter in a universe that was mostly antimatter, we'd have to be very careful not to get ...
2
The electron and positron are two point charges with opposite sign, and classically , as the field lines are an iconal representation of the charge, when the charge becomes zero there will be no electric field lines from the spot where the two point particles overlap. BUT electrons and positrons are quantum mechanical particles and when close enough ...
35
To maintain lepton number as a conserved quantity. Consider, in detail, what's going on in a beta decay (well, I'm going to ignore the nuclear context). The reaction is then $$n \longrightarrow p^+ + e^- + \nu \,,$$ where you should take the symbol $\nu$ to mean some neutrino (without prejudice about matter-type or anti-matter-type for the moment). There ...
10
So there's this funny rule whose provenance I can't recall, but whose essence is: everything that is not forbidden, eventually happens. This rule is particularly fecund in quantum mechanics. If the process you describe is allowed, then every neutron already is a superposition of neutron and antineutron, and the question is just whether the oscillations ...
0
Whatsoever properties of matter or antimatter are related to "charge" at particle level.........are responsible for annihilation phenomena .Now an interesting question could be that if so then how is it possible that two particles neutron and its anti-particle can annihilate each other...? The reason for this is the , fact of composite nature of neutron and ...
1
A neutron has baryon number = 1, while the anti-neutron has baryon number = -1. Physics Guy has much the same here with quarks, that works as well. In the language of CPT the charge operator reverses the charge of the quarks, so the two up quarks with charge $-1/3$ is flipped to $1/3$ and the down quark from $2/3$ to $-2/3$. I gave this question a 1-vote, ...
4
Every particle has (or can have) an antiparticle. Sometimes, it is even his own antiparticle. An antiparticle $D'$ is (easily said) defined as a particle $D$ after a CPT-transformation. CPT-Symmetry is believed to be a fundamental concept of physical nature. A CPT-transformation is a complete changing of observables of a particle. The C stands for ...
0
Somehow, I don't believe the question of "quickly" needs to be applied here. The reason is simple. For a fission bomb, there is a need to maintain confinement in order for the nuclear reaction to continue - as density reduces, the chain reaction can extinguish. On the other hand, for an "antimatter bomb", as the antimatter heats up, it will start to expand; ...
0
What I mean is, suppose we could somehow get a kilogram of antimatter and contain it safely. "contain it safely" is science fiction. There is no way to keep antimatter in a matter environment safely except by the use of electric and magnetic fields, suspended in them . Even suppose one can create ( another science fiction scenario which I will not ...
1
The obvious answer, to my mind, would be to build an antimatter bomb like a fission bomb. You have a shell of matter (probably the containment unit) around the antimatter which is crushed by some kind of explosive, perhaps a nuclear explosive, creating a rapidly contracting dense shell that can partially overcome the centrifugal tendencies introduced by the ...
0
Likely the dominant process for the positron in the lead plate is simply Bhabha scattering. This is no surprise. We must consider the lab energy of the positron. At center-of-momentum energies less than twice the muon rest mass, the electron and positron annihilation is limited primarily to radiation of multiple photons. At energies above the eV scale, ...
1
As Jon said in the comments, the mean free path before the electron-positron annihilation is longer than 6 millimeters, or at least not much shorter, and it's simply hard for them to annihilate. Most of the space is empty, the electron and positron are pointlike particles, and they're unlikely to hit each other. The mean free path that indicates the "...
0
Books tell us that electricity is constituted only by electrons. But in reality protons also cause electricity. In some materials like conductors this is an exception as protons are fixed at their places and are too heavy to move. So basically electricity is a flow of charge which could be positive or negative.
Top 50 recent answers are included
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2016-06-28 03:54:35
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https://heart.bmj.com/content/100/11/862?ijkey=25efcde6fb27851bbad3dd6e052dc61fa8f14f32&keytype2=tf_ipsecsha
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Article Text
Original article
Systemic inflammation and brachial artery endothelial function in the Multi-Ethnic Study of Atherosclerosis (MESA)
1. Shepard D Weiner1,
2. Hanna N Ahmed1,
3. Zhezhen Jin2,
4. Mary Cushman3,
5. David M Herrington4,
6. Jennifer Clark Nelson5,
7. Marco R Di Tullio1,
8. Shunichi Homma1
1. 1Department of Medicine, Columbia University College of Physicians and Surgeons, New York, New York, USA
2. 2Department of Biostatistics, Joseph P. Mailman School of Public Health, Columbia University, New York, New York, USA
3. 3Departments of Medicine and Pathology, University of Vermont College of Medicine, Burlington, Vermont, USA
4. 4Department of Medicine, Wake Forest University School of Medicine, Winston-Salem, North Carolina, USA
5. 5Biostatistics Unit, Group Health Research Institute and Department of Biostatistics, University of Washington, Seattle, Washington, USA
1. Correspondence to Dr Shepard D Weiner, Division of Cardiology, Columbia University College of Physicians and Surgeons, Columbia University Medical Center, New York Presbyterian Hospital, Milstein Family Heart Center, 4th Floor, 173 Fort Washington Avenue, New York, NY 10032, USA; sw2150{at}columbia.edu
## Abstract
Background and objective Inflammation and endothelial dysfunction have been implicated in the pathogenesis of atherosclerotic vascular disease. Brachial artery flow-mediated dilation (FMD) is a reliable, non-invasive method of assessing endothelial function. We hypothesised that increased levels of systemic inflammatory markers are associated with impaired endothelial function as assessed by FMD in a multi-ethnic cohort.
Methods We assessed brachial artery FMD in 3501 participants (1739 men, 1762 women; median age 61 years) in the Multi-Ethnic Study of Atherosclerosis and measured serum concentrations of interleukin (IL)-6, C reactive protein (CRP) and tumour necrosis factor (TNF)-α receptor 1. Spearman correlation coefficients were used to evaluate the association of each inflammatory marker with FMD, adjusting for the effect of other variables associated with FMD.
Results There was a significant inverse correlation between IL-6 levels and FMD (−0.042; p=0.02) after adjustment for age, gender, race/ethnicity, education, income, low-density lipoprotein, diabetes, glucose, hypertension status and treatment, waist circumference, triglycerides, baseline brachial diameter, recent infection and use of medications that may alter inflammation. There was no significant correlation between CRP and FMD (0.008; p=0.64) or TNF-α receptor 1 and FMD (0.014; p=0.57). There was no evidence of effect modification by race/ethnicity.
Conclusions In this multi-ethnic cohort, increased levels of the pro-inflammatory cytokine IL-6 were associated with impaired endothelial function assessed by FMD. Elevated IL-6 levels may reflect a state that promotes vascular inflammation and development of subclinical atherosclerosis independent of traditional cardiovascular risk factors.
• Endothelium
View Full Text
## Introduction
Advancements in understanding of the pathobiology of atherothrombosis have implicated inflammation as a central contributor to atherosclerotic vascular disease. Inflammation plays a major role in all stages of atherogenesis and its systemic complications.1 In addition to its role in established cardiovascular disease, inflammation contributes to the earliest steps in atherogenesis, supporting the concept of using pro-inflammatory biomarkers to predict future cardiovascular risk.2
Ultrasound assessment of brachial artery flow-mediated dilation (FMD) during reactive hyperaemia is a reproducible method for the assessment of endothelial dysfunction.3 Endothelial dysfunction is a frequent precursor of early atherogenesis.4 Impaired endothelium-dependent dilation in the coronary circulation is associated with coronary risk factors and atherosclerosis and improves with risk reduction therapies.5 Impaired brachial artery FMD is predictive of endothelial dysfunction in the coronary circulation.6 Therefore, brachial artery FMD has emerged as an attractive non-invasive technique for the study of endothelial function as a marker of subclinical atherosclerosis.
There is prior evidence to suggest that IL-6 may play a role in the pathobiology of atherogenesis. IL-6 gene transcripts have been found to be expressed in human atherosclerotic lesions.7 The main biological effect of IL-6 is regulation of the acute-phase response. During acute inflammation, IL-6 promotes activation of T cells and differentiation of B cells, and regulates liver production of acute phase reactants such as CRP. Additionally, IL-6 stimulates platelet aggregation and expression of tissue factor.8 Elevated levels of interleukin-6 (IL-6) have been shown to predict the development of type II diabetes9 and are associated with increased risk of future cardiovascular events after adjustment for traditional cardiovascular risk factors in healthy individuals.10
Previous studies examining the relationship between inflammatory biomarkers and endothelial dysfunction have shown conflicting results.11–14 To date, however, no large multi-ethnic study has examined the relationship between inflammatory biomarkers and endothelial dysfunction as assessed by FMD. We thus sought to investigate whether IL-6 or CRP levels were associated with vascular endothelial function in a large multi-ethnic community-based sample while adjusting for known cardiovascular risk factors.
Tumour necrosis factor (TNF)-α is a pro-inflammatory cytokine that also induces secretion of CRP from the liver. While CRP has been well studied as an independent predictor of cardiovascular events, few studies have examined the relationship between TNF-α and endothelial function or cardiovascular events, and those that have show mixed results.15–17 To extend prior findings, we also studied circulating TNF-α receptor levels. TNF-α receptor levels are more stable and can be measured with greater sensitivity and reliability than TNF-α itself.18
## Methods
### Study cohort
The Multi-Ethnic Study of Atherosclerosis (MESA) was initiated in July 2000 to investigate the prevalence, correlates and progression of subclinical atherosclerosis in a population-based sample of individuals without known cardiovascular disease.19 This is a population-based study of 6814 women and men aged 45–84 years old recruited from six US communities (Baltimore, Maryland; Chicago, Illinois; Forsyth County, North Carolina; Los Angeles, California; New York, New York; and St. Paul, Minnesota). Cohort participants were 38% Caucasian (n=2624), 28% African American (n=1895), 22% Hispanic (n=1492) and 12% Chinese (n=803). Details of the MESA recruitment strategy have previously been reported.20 The MESA study was approved by institutional review boards at each site, and all participants gave written informed consent.
Medical history, anthropometric measurements and laboratory data were taken from the first examination of the MESA cohort (July 2000 to August 2002). FMD determination and blood collection for inflammatory marker assessment were performed on the same day. Information about age, gender and ethnicity were obtained by questionnaires. Education was classified into three groups: less than high school, complete high school or equivalent certification, and complete college or more. Annual household income was bracketed into three categories: <$20 000,$20 000–$49 999, and >$50 000. Current smoking was defined as having smoked a cigarette in the last 30 days. Diabetes was defined as having a fasting glucose of ≥126 mg/dL, use of hypoglycaemic medications or self-report of diabetes. Resting blood pressure was measured three times in the seated position, and the average of the second and third readings was recorded. Hypertension status was defined as a systolic blood pressure ≥140 mm Hg or diastolic blood pressure ≥90 mm Hg. Waist circumference was measured in centimetres. Triglycerides and high-density lipoprotein (HDL) cholesterol were measured from blood samples obtained after a 12 h fast. Low-density lipoprotein (LDL) cholesterol was estimated by the Friedewald equation. Participants were categorised as having a recent infection if they reported cold/flu, sinus, urinary or tooth infection, bronchitis or pneumonia in the preceding 2 weeks. Medication use that alters inflammation was defined as current use of lipid-lowering medications, non-steroidal anti-inflammatory drugs or aspirin.
Brachial artery FMD was documented with the use of ultrasound in the MESA study during the first examination. Of the 6814 MESA participants, 6489 (95.2%) had successful FMD testing. Participants were excluded from the FMD examination if they had uncontrolled hypertension (systolic blood pressure >180 mm Hg or diastolic blood pressure >110 mm Hg) or blood pressures in the left and right arms that differed by >15 mm Hg (n=158), a history of Raynaud phenomenon (n=55), a congenital abnormality of the arm or hand (n=12) or mastectomy (n=100). Brachial ultrasound videotapes were analysed at the Wake Forest University Cardiology Image Processing Laboratory. Images from 5731 participants were of sufficient quality for reading. Due to financial constraints, images were read for a subset of these participants (n=3501). These participants were a nested case-cohort that included all participants who had an adjudicated cardiovascular event (n=188), and a random sample of the remaining participants in the MESA study (n=3313). Of these 3501 participants, 1739 were men and 1762 were women.
### Measurement of inflammatory biomarkers
Inflammatory markers examined include IL-6, CRP and TNF-α receptor 1 (R1). Standardised methods were used for blood collection, processing and shipping, with fasting morning phlebotomy and use of a central laboratory (Laboratory for Clinical Biochemistry Research, University of Vermont, Burlington).21 IL-6 and TNF-α R1 were measured by ultrasensitive ELISA from R&D Systems (Minneapolis, Minnesota, USA). TNF-α R1 was successfully measured in 2871 participants of the Exam 1 MESA cohort. These participants correspond to 2880 ‘group 4’ participants who were selected for an early ancillary candidate gene effort, where participants were selected equally from the four race/ethnic groups (n=720 selected at random from each race/ethnic group). High-sensitivity CRP was measured by nephelometry using the BNII nephelometer (N High Sensitivity CRP, Dade Behring, Deerfield, Illinois, USA).
### Determination of flow-mediated dilation
Participants were examined in the supine position after 15 min of rest, after at least 6 h fasting and at least 6 h without smoking. An automated sphygmomanometer (Dinamap device) was used to monitor blood pressure and pulse in the left arm at 5 min intervals throughout the examination. A standard blood pressure cuff was positioned around the right arm, two inches below the antecubital fossa, and the artery was imaged 5–9 cm above the antecubital fossa. A linear-array multifrequency transducer operating at 9 MHz (GE Logiq 700 Device) was used to acquire images of the right brachial artery. After baseline images were obtained, the cuff was inflated to 50 mm Hg above the participant's systolic blood pressure for 5 min. Digitised images of the right brachial artery were captured continuously for 30 s before cuff inflation and for 2 min beginning immediately before cuff deflation to document the vasodilator response. Semiautomated readings (media-adventitial interfaces to media-adventitial interfaces) of these digitised images generated the baseline and maximum diameters of the brachial artery from which % FMD was determined as follows: % FMD=((maximum diameter − baseline diameter)/baseline diameter)×100.
To evaluate intra-reader reproducibility for baseline diameter, maximum diameter and % FMD, ultrasound examinations for 40 MESA participants (32 men and 8 women; 18 Caucasian, 2 Chinese, 10 African American and 10 Hispanic) were reanalysed. The intraclass correlation coefficients were 0.99, 0.99 and 0.93, for Caucasian, African American and Hispanic groups, respectively. The intraclass correlation coefficient was not calculated for the Chinese group due to small sample size. A more detailed description of the scanning and reading protocol can be found at the MESA website (http://www.mesa-nhlbi.org).
### Statistical analysis
Either medians and ranges or mean and SD were calculated for continuous variables and proportions for categorical variables. Either the two-sided Wilcoxon rank sum test or the two-sided t test was used for the comparison for continuous variables, and the χ2 test was used for the comparison for categorical variables. Since the distributions of FMD and inflammatory markers were skewed, we used Spearman correlation coefficients to evaluate the association of each inflammatory marker with FMD. We estimated raw Spearman correlations, Spearman partial correlations that accounted for age and gender, and Spearman partial correlations that accounted for age, gender, race/ethnicity, education, income, LDL, diabetes, glucose, hypertension status and treatment, waist circumference, triglycerides, baseline brachial diameter, recent infection and use of medications. These included clinical covariates previously reported to correlate with FMD (cigarette use, hypertension status, fasting glucose, diabetes mellitus, LDL cholesterol, HDL cholesterol, triglycerides and waist circumference). Recent infection and use of medications were also included as they were known to alter inflammation. Statistical significance was determined at an α-level of 0.05 with two-sided tests. Statistical analyses were conducted with SAS V.9.3.
## Results
### Participant characteristics, measures of vascular function and biomarkers
Clinical characteristics of the 3501 participants with FMD measurements are provided in table 1. IL-6 assays were available for 97% of these participants (n=3404), CRP for 99% (n=3481) and TNF-α R1 for 52% (n=1812).
Table 1
Clinical characteristics of the study sample
The median (range) baseline brachial artery diameter was 3.8 mm (2.2–6.0 mm) for women and 4.8 mm (2.5–7.1 mm) for men. The median (range) FMD for women was 0.16 mm (0–0.74 mm) or 4.2% (0–24.8%). The median (range) FMD for men was 0.17 mm (−0.07 to 0.63 mm) or 3.5% (0–14.8%). The summary statistics for these measures of vascular function and for inflammatory markers are displayed in table 2. In adjusted Spearman correlation analysis, the measured variables that had the strongest correlation with FMD were baseline brachial artery diameter (−0.379; p<0.0001), age (−0.242; p<0.0001) and waist circumference (0.083; p<0.0001). Unadjusted Spearman correlation coefficients among the three markers of inflammation were positively correlated (IL-6 and CRP 0.536, p<0.0001; IL-6 and TNF-α R1 0.336, p<0.0001; CRP and TNF-α R1 0.237, p<0.0001). For further descriptive information, see the online supplementary table.
Table 2
Markers of inflammation and measures of vascular function
### Inflammatory markers and vascular function
Table 3 shows the correlations between inflammatory markers and brachial artery FMD. In univariate analysis, there was a significant inverse correlation between IL-6 levels and FMD (−0.115; p<0.001) and between TNF-α R1 and FMD (−0.096; p<0.001), but no correlation between CRP and FMD (−0.010; p=0.58). In multivariate analysis, the inverse relationship between elevated IL-6 levels and brachial artery FMD remained significant (−0.042; p=0.02), but there was no correlation between TNF-α R1 and FMD (0.014; p=0.570) or CRP and FMD (0.008; p=0.64). The multivariable modelling did account for some of the relationship between IL-6 levels and FMD.
Table 3
Relationship between flow-mediated dilatation and markers of inflammation
### Inflammatory biomarkers and FMD by race/ethnic group
The correlation between IL-6 and FMD was significant among Caucasian, Chinese and Hispanic race/ethnic groups in univariate analyses. In complete multivariate analysis however, there was no evidence for a difference in the association of IL-6 with FMD by race/ethnicity; that is, there was no evidence for effect modification by race. Although it did not reach statistical significance, the association was larger among non-Caucasian groups, particularly for Hispanics (see table 4). For African Americans, the correlation between TNF-α R1 and FMD was significant in the complete multivariate model. There was no significant association of CRP with FMD in any race/ethnic group.
Table 4
Relationship between flow-mediated dilation and inflammatory markers by race/ethnic group
## Discussion
In this large multi-ethnic community-based sample of adults 45–84 years of age who were free of clinical cardiovascular disease, higher levels of the pro-inflammatory cytokine IL-6 were associated with endothelial dysfunction as assessed by impaired FMD on multivariate analyses that adjusted for traditional cardiovascular risk factors. The other inflammatory biomarkers, CRP and TNF-α R1, were not related to FMD on multivariable analysis. Although the association between elevated IL-6 levels and impaired FMD remained present on multivariable analysis, part of the association was mediated by traditional cardiovascular risk factors as well as baseline brachial artery diameter.
Although the association between elevated IL-6 levels and impaired FMD in our study was not individually significant in any of the race/ethnic groups studied, the association was weakest in Caucasians. This trend, along with other data that demonstrate ethnic variations in atherosclerosis and IL-6,22 seems to suggest that the significant association between elevated IL-6 levels and impaired FMD in this study might at least partially relate to the inclusion of non-Caucasian race/ethnic groups in the MESA cohort. While our study did not show a significant relationship between TNF-α R1 and FMD overall, in the subgroup of African Americans, TNF-α R1 was significantly associated with FMD in both the univariate and the complete multivariable models. Whether this occurred by chance or is a true association is not known. This is a novel finding which warrants further study.
Our study results are in line with recent publications on inflammatory biomarkers and endothelial function in community-based cohorts, which have demonstrated weak or no associations of circulating inflammatory biomarkers with FMD. Two previous community-based cohort studies involving the largely Caucasian Framingham Offspring Study have examined the relationship between inflammatory biomarkers and endothelial dysfunction. One study observed modest unadjusted correlations between FMD and the inflammatory markers IL-6 and CRP that were rendered non-significant after adjustment for cardiovascular disease risk factors in 2701 participants. In this study, CRP and IL-6 were inversely related to FMD and reactive hyperaemia in the unadjusted models. After multivariable adjustment, CRP and IL-6 were no longer correlated with FMD, but were still correlated with reactive hyperaemia though the association was weaker.12 Another study conducted in the same cohort a few years later confirmed the finding of no association of CRP with FMD.23 The Framingham investigators did not evaluate TNF-α R1. The lack of association between CRP and brachial artery FMD in the current study is consistent with the results of those studies. However, we found a significant association between IL-6 and FMD that was not observed in the first Framingham Offspring study. In another study of healthy Caucasian men, the investigators also found a significant negative independent association of IL-6 with endothelium-dependent vasodilation in their multivariable analysis. In this study, IL-6 was also found to be negatively associated with insulin sensitivity.11
If further studies confirm these findings, therapies that target IL-6, either indirectly or directly, may be of use. A significant reduction in IL-6 levels and improvement in endothelium-dependent vasodilation of the brachial artery has been demonstrated in patients treated with atorvastatin,24 supporting the anti-inflammatory role of statin pleiotropy. There is also evidence that the role IL-6 plays in promoting endothelial dysfunction is related to the renin–angiotensin system. Treatment with the angiotensin receptor blocker (ARB) irbesartan for 4 weeks resulted in statistically significant reductions in plasma levels of IL-6 and a 67% increase in FMD of the brachial artery compared to subjects who received placebo.25 Another study demonstrated that in post-myocardial infarction subjects with impaired left ventricular systolic function, baseline FMD was low and levels of TNF-α and CRP were high. After 8 and 12 weeks, in the group treated with quinapril, but not in the group treated with enalapril, FMD significantly improved and levels of TNF-α and CRP significantly decreased.26
In a cross-sectional analysis such as this, causal pathways between inflammatory biomarkers and FMD cannot be inferred. The present study identifies an association between an inflammatory marker IL-6 and impaired FMD which serves as a surrogate for endothelial dysfunction. There may be unmeasured confounders that have not been accounted for that could contribute to the association between increased IL-6 levels and impaired FMD. There is always the possibility of measurement errors in FMD, although this should have been minimised by the use of a standardised protocol for FMD measurement. While our study did not show a significant relationship between TNF-α R1 and FMD overall, the sample size of subjects with TNF-α R1 measurements available was small and thus we may not have been able to detect a significant association. However, despite the small sample size, in the subgroup of African Americans, TNF-α R1 was significantly associated with FMD in both the univariate and the complete multivariable models.
Strengths of the present study include the large sample size, use of a standardised protocol for FMD measurement, availability of clinical covariates and use of multivariate analyses in a multi-ethnic community-based sample. Due to the multi-ethnic nature of the cohort, there is increased generalisability of the study. In addition, both men and women were included in this analysis. Further analyses need to be conducted to explore the role of gender with inflammatory biomarkers and FMD.
## Conclusion
Increased level of the pro-inflammatory cytokine IL-6 was associated with impaired endothelial function as assessed by brachial artery FMD in a large multi-ethnic population cohort. The results of the present study suggest that increased circulating IL-6 levels may impair endothelial function, particularly in the brachial artery. Elevated IL-6 concentration may reflect a state that promotes vascular inflammation and development of subclinical atherosclerosis independent of traditional risk factors.
### Key messages
What is already known on this subject?
• Previous studies examining the relationship between inflammatory biomarkers and endothelial dysfunction have shown either weak or no associations of circulating inflammatory biomarkers with flow-mediated dilation.
What this study adds?
• Our study adds to the literature in that it includes data derived from a multi-ethnic community-based cohort that includes non-Caucasians, which to our knowledge has not been previously reported. In addition, we show data divided by race/ethnic group.
How might this impact on clinical practice?
• If further studies confirm these findings, therapies that target IL-6, either indirectly or directly, may be of clinical use.
## Acknowledgments
The authors thank the other investigators, the staff and the participants of the MESA study for their valuable contributions. A full list of participating MESA investigators and institutions can be found at http://www.mesa-nhlbi.org.
View Abstract
• ## Supplementary Data
This web only file has been produced by the BMJ Publishing Group from an electronic file supplied by the author(s) and has not been edited for content.
Files in this Data Supplement:
## Footnotes
• Collaborators Mohammed F Saad, MD.
• Contributors All authors have contributed to the planning, conduct and reporting of the work described in the article. SDW, HNA, ZJ, MRD and SH assume responsibility for the overall content as guarantors.
• Funding This research was supported by contracts N01-HC-95159 through N01-HC-95167 from the National Heart, Lung, and Blood Institute.
• Competing interests None.
• Ethics approval All IRBs at sites involved with the MESA study.
• Provenance and peer review Not commissioned; externally peer reviewed.
## Request Permissions
If you wish to reuse any or all of this article please use the link below which will take you to the Copyright Clearance Center’s RightsLink service. You will be able to get a quick price and instant permission to reuse the content in many different ways.
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2019-07-23 22:34:20
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https://financecitycenter.com/business-and-personal-finance-individual-retirement-accounts-iras/
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# Business and Personal Finance: Individual Retirement Accounts (IRAs)
## Retirement Plans – Individual Retirement Accounts (IRAs)
For decades, IRAs have been among the most popular ways to put away money for retirement while getting a current tax deduction. You also don’t have to pay any current income taxes on any money the account earns— until you start taking distributions, that is, and then you get taxed on every dollar you take out. A new twist was added with the arrival of the Roth IRA; this version gives you no tax deduction on contributions now in exchange for no taxes at all on the earnings (when you wait at least five years to take a qualified distribution). Both traditional and Roth IRAs continue to grow in popularity as more people get cracking on their retirement savings.
### FACT
Many of the most popular small-business retirement plans are based on the IRA. In fact, some of them (such as the SEP and some SIMPLEs) are really just a collection of IRAs brought together under the umbrella of the company’s general retirement plan.
As a business owner, you do have other options when it comes to put- ting away money for the future, but none is as simple as an IRA—regardless of their catchy acronyms (you’ll read more about these later in this chapter). If you want to save money for yourself without the hassle of a formal retirement plan and without having to make contributions for your employees, you can just open a personal IRA.
All you have to do to open an IRA is fill out a single form and send in a check. You can do it at your bank, do it with your broker, or even do it online with an electronic bank transfer. It can be a standard interest-bearing savings account, or you can fill it with stocks, bonds, and mutual funds. Multiple IRAs are also possible; for example, you might do that when you want to invest directly in mutual funds (as opposed to going through a broker and paying his fees) but want funds from different fund families.
### The Common Rules
Here are the basic rules for both kinds of IRAs. You can’t contribute more than your taxable income; for example, if your sole proprietorship earns $2,000 this year, that’s the most you can put into an IRA (unless you have other income, such as a salary). For 2006 and 2007, the maximum contribution is$4,000 per person (no matter how many IRAs you have), but if you’re 50 or older, you can bump that up by another $1,000. In 2008, that max jumps to$5,000 for everyone, and the $1,000 add-on can still be used for anyone who’s at least 50 years old. Contributions have to be made with money; you can do it by credit card, but you can’t do it with a diamond necklace. You have to make the contribution by the time you file your tax return in order for it to count in that tax year; for example, if you make a$4,000 contribution on April 15, 2007, when you file your 2006 tax return, the IRA deduction counts for 2006. Other than these rules, the two types of IRAs don’t have much in common.
### More Rules for Traditional IRAs
In addition to the tax differences mentioned earlier, the two IRA types veer onto different paths in other ways:
• You have to start taking distributions from a traditional IRA the year after you hit age 70 1/2 or the year you retire, whichever comes last; there are no mandatory distribution requirements with Roth IRAs.
• Once you hit age 70 1/2, you can’t make any more contributions, even during that year; there’s no age limit on contributions into a Roth IRA.
• There are many restrictions on when and how you can take money out of your traditional IRA, and restricted distributions can be subject to hefty penalties; with Roth IRAs, once the cash has been sitting there for five years and you hit age 59 1/2 you can do whatever you want with it.
For the specific rules that apply to your unique situation, talk to your accountant or IRA plan manager before you take out any money, especially if you have a traditional IRA. Mistakes can lead to IRS penalties, and those can be pretty big.
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2023-03-25 05:07:48
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https://handwiki.org/wiki/Correlate_summation_analysis
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# Correlate summation analysis
Correlate summation analysis is a data mining method. It is designed to find the variables that are most covariant with all of the other variables being studied, relative to clustering. Aggregate correlate summation is the product of the totaled negative logarithm of the p-values for all of the correlations to a given variable and its (normalized) standard deviation-to-mean quotient. Discrete correlate summation is the product of the totaled absolute value of the logarithm of the p-value ratios between two groups' correlations to a given variable and its absolute value of the logarithm of the group mean ratios.
## Correlate summation template
This zipped Excel template performs a correlate summation analysis for up to 100 variables for 4 groups of 15 subjects:
The paper [1] describing the method is embedded in the spreadsheet.
## Discrete correlate summation
Given two groups, a correlation matrix (m by m) was constructed for m variables for each group. Each column represents all of the correlations (r) between a given variable and each of the other variables. For variables with either heterogeneous or homogeneous numbers of data points (n), the n for each individual correlation was calculated by assigning each data point with a value of one and taking the sum of the products for each pair in that correlation.
The correlations were tested for linearity using Student's t-distribution to evaluate:
$\displaystyle{ t=\frac{|r|}{\sqrt{\frac{1-r^2}{n-2}}} }$
for (n − 2) degrees of freedom, returning two tails.[2]
The correlation matrices were thus transformed into linear probability matrices. For the two groups, the absolute value of the logarithm of the ratio of each comparison’s p-value gives a log correlation ratio that is larger as the ratio approaches zero or infinity. Each column was totaled to form the discrete correlate summation array. As in the log correlation ratio (logcr), the log mean ratio (logmr) for the two groups’ means was acquired for each variable. The correlate summation was then multiplied by the log mean ratio, to yield the discrete mean-correlate summation (DCΣx).[1]
## Aggregate correlate summation
As in the discrete correlate summation, a linear probability matrix was calculated for all of the data (no grouping). The negative logarithm was taken for all of the p-values; the columns were totaled to give the aggregate correlate summation (ACΣ) array. The standard deviation for each variable is divided by its mean to normalize the variances between variables. Data with a bimodal distribution will have a larger normalized standard deviation (nSD) than will data with a normal distribution. The nSD array multiplied by the ACΣ array yielded the aggregate mean-correlate summation (ACΣx).[1]
## Non-linear modeling
A linear correlation between variables for a given sample set is typically the initial step in the investigation of relationships, which may lead to an underlying mechanism. The variation (either inherent or in response to a challenge) in a given population gives rise to correlations of variables of which only a portion of the sigmoidal (control) relationship may be evident. Generally in the face of data that defies linear regression, data patterns indicate power relationship of the general type:
$\displaystyle{ y=mx^a }$
Type 1: a < 0 is a hyperbolic function
Type 2: a = 0 is a horizontal line
Type 3: 0 < a < 1 is a root function
Type 4: a = 1 is actually a linear function
Type 5: a > 1 is a power function
(In all five cases a log-log plot yields a linear curve.) [3]
On a positive sigmoidal/logistic curve, the initial, intermediate and late portions resemble power, linear and root functions, respectively. Also, the late portion of a negative control function is reminiscent of a hyperbolic curve.
In an analysis of variable correlation, the sigmoidal relationship of the entire (unsampled in some cases) data range should be considered. This type of analysis is accomplished by regression with either a logistic curve or simple linear regression with further investigation of the Type 1, 3 and 5 power relationships.[1]
## References
1. Westwood, B; Chappell, M. (2006). "Application of correlate summation to data clustering in the estrogen- and salt-sensitive female mRen2.Lewis rat". Proceedings of the 1st international workshop on Text mining in bioinformatics - TMBIO '06. TMBIO '06 (ACM). pp. 21–26. doi:10.1145/1183535.1183542. ISBN 1-59593-526-6.
2. Swinscow, T. (1997) Statistics at Square One. BMJ Publishing Group.
3. Mandel, J. (1984) The Statistical Analysis of Experimental Data. Dover Publications, Mineola, NY.
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2022-01-23 15:28:43
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https://math.stackexchange.com/questions/1874773/does-int-infty-0-frac-cos-x1x-mathrm-dx-converge
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# Does $\int^{\infty}_0 \frac{\cos x}{1+x}\,\mathrm dx$ converge?
The integral $$\int^{\infty}_0 \frac{\cos x}{1+x}\,\mathrm dx$$ is shown to be equal to $$\int^{\infty}_0 \frac{\sin x}{{(1+x)}^2}\,\mathrm dx$$ through integration by parts. The latter one converges absolutely (i.e the integral still exists if you take the absolute value of the function) as the function is dominated by the function $1/{(1+x)}^2$. But the first integral does not converge absolutely as stated in Rudin "Principles of Mathematical Analysis" ch 6 exercise 9.
Why is it so?
• Someone edit this question. – Tahir Imanov Jul 29 '16 at 10:43
• If $A$ and $B$ is a sum or an integral then just knowing that $A = B$ does not tell you anything about the internal properties of $A$ and $B$ (like if they both converge absolutely or if both have just positive terms or if both have infinitely many non-zero terms etc. etc. etc.). Here is a similar example with series: $\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n} = \frac{6\log(2)}{\pi^2}\sum_{n=1}^\infty \frac{1}{n^2}$. The series on the right hand side converges absolutely while the left hand side does not. – Winther Jul 29 '16 at 11:25
• @Winther Pretty example. – Olivier Oloa Jul 29 '16 at 14:06
By integrating by parts, $$\int_0^M \frac{\cos x}{x+1}\:dx=\frac{\sin M}{M+1}+\int_0^M \frac{\sin x}{(x+1)^2}\:dx, \quad M\ge0,$$ and letting $M \to \infty$, one has $$\left|\int_0^\infty \frac{\cos x}{x+1}\:dx\right|=\left|\int_0^\infty \frac{\sin x}{(x+1)^2}\:dx\right|<\int_0^\infty \frac1{(x+1)^2}\:dx<\infty$$ the given integral is then convergent, there is no contradiction with $$\int_0^M \left|\frac{\cos x}{x+1}\right|\:dx \to \infty$$ as $M \to \infty$, since we have $$\left|\int_0^M \frac{\cos x}{x+1}\:dx\right| <\int_0^M \left|\frac{\cos x}{x+1}\right|\:dx.$$
Remark. One may observe that $$\int_{(2n-1)\frac{\pi}2}^{(2n+1)\frac{\pi}2} \left|\cos x\right|\:dx=2,\qquad n=1,2,\cdots,$$ giving $$\int_{(2n-1)\large\frac{\pi}2}^{(2n+1)\large\frac{\pi}2} \left|\frac{\cos x}{x+1}\right|\:dx \ge \frac1{(2n+1)\large\frac{\pi}2+1}\int_{(2n-1)\large\frac{\pi}2}^{(2n+1)\frac{\pi}2} \left| \cos x \right| dx \ge \frac{2}{(n+1)\pi}$$ and by summing $$\int_{\large \frac{\pi}2}^{(2N+1)\large \frac{\pi}2} \left|\frac{\cos x}{x+1}\right|\:dx \geq \frac{2}{\pi}\sum_{n = 2}^{N+1} \frac{1}{n}\qquad N\ge1,$$ which shows that $$\int_0^\infty \left|\frac{\cos x}{x+1}\right|dx$$ is divergent.
• Thanks a lot, this answers my question (sorry for late feed back due to my vacation) – Manfred Aug 19 '16 at 11:56
The first one does converge, actually. The catch is that if an integral does not converge absolutely, it doesn't mean it is not convergent.
The integration by parts technique is one way to show that it actually is convergent! To understand intuitively why it is so, it is helpful to look at the graph of $f(x) = (\cos x){(1+x)}^{-1}$:
As you can see, the function is alternating between positive and negative values. As the Riemann integral takes into account the sign of the function, all these areas will cancel out each other. In this case, the result is approximately $0.343378$, and that's because the area between the function and the $x$ axis is progressively smaller and smaller.
Conversely, being $|f(x)|$ non-negative, if you integrate it you will be summing all those areas, which are all positive, and the sum does not converge because its general term is asymptotically equivalent to $1/x$, whose sum diverges.
In general, it can be proved that the Riemann integral has the property $$\left | \int_a^b f(x)\,\mathrm dx \right | \leq \int_a^b \left | f(x) \right |\,\mathrm dx$$
and it can be extended to improper integrals as well.
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2020-04-09 14:51:11
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https://casflo-app.com/learning-center/manuals/financial-analysis/project-valuation-through-fcfe/equity-value/
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# 5.1.4. Equity value
The ultimate objective of using the FCFE is to identify the equity value. Unlike the use of the FCFF, using FCFE does not require the deduction of debt value, but the non-operational assets were not accounted for in the value generated through the FCFE, so it is necessary to add them.
• $$Equity\ value=operations\ equity\ value+NOA$$
End of Manual: Back to Learning Center
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2020-10-26 14:58:34
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https://www.groundai.com/project/thickness-dependent-ferromagnetic-metal-to-paramagnetic-insulator-transition-in-la_06sr_04mno_3-thin-films-studied-by-x-ray-magnetic-circular-dichroism/
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Thickness-dependent ferromagnetic metal to paramagnetic insulator transition in La{}_{0.6}Sr{}_{0.4}MnO{}_{3} thin films studied by x-ray magnetic circular dichroism
# Thickness-dependent ferromagnetic metal to paramagnetic insulator transition in La0.6Sr0.4MnO3 thin films studied by x-ray magnetic circular dichroism
G. Shibata Department of Physics, University of Tokyo, Bunkyo-ku, Tokyo 113-0033, Japan K. Yoshimatsu Department of Physics, University of Tokyo, Bunkyo-ku, Tokyo 113-0033, Japan Department of Applied Chemistry, University of Tokyo, Bunkyo-ku, Tokyo 113-8656, Japan E. Sakai Department of Applied Chemistry, University of Tokyo, Bunkyo-ku, Tokyo 113-8656, Japan Photon Factory, Institute of Materials Structure Science, High Energy Accelerator Research Organization (KEK), Tsukuba, Ibaraki 305-0801, Japan V. R. Singh Department of Physics, University of Tokyo, Bunkyo-ku, Tokyo 113-0033, Japan V. K. Verma Department of Physics, University of Tokyo, Bunkyo-ku, Tokyo 113-0033, Japan K. Ishigami Department of Complexity Science and Engineering, University of Tokyo, Bunkyo-ku, Tokyo 113-0033, Japan T. Harano Department of Physics, University of Tokyo, Bunkyo-ku, Tokyo 113-0033, Japan T. Kadono Department of Physics, University of Tokyo, Bunkyo-ku, Tokyo 113-0033, Japan Y. Takeda Condensed Matter Science Division, Japan Atomic Energy Agency, Sayo-cho, Sayo-gun, Hyogo 679-5148, Japan T. Okane Condensed Matter Science Division, Japan Atomic Energy Agency, Sayo-cho, Sayo-gun, Hyogo 679-5148, Japan Y. Saitoh Condensed Matter Science Division, Japan Atomic Energy Agency, Sayo-cho, Sayo-gun, Hyogo 679-5148, Japan H. Yamagami Condensed Matter Science Division, Japan Atomic Energy Agency, Sayo-cho, Sayo-gun, Hyogo 679-5148, Japan Department of Physics, Kyoto Sangyo University, Kyoto 603-8555, Japan A. Sawa National Institute of Advanced Industrial Science and Technology (AIST), Tsukuba, Ibaraki 305-8562, Japan H. Kumigashira Department of Applied Chemistry, University of Tokyo, Bunkyo-ku, Tokyo 113-8656, Japan Photon Factory, Institute of Materials Structure Science, High Energy Accelerator Research Organization (KEK), Tsukuba, Ibaraki 305-0801, Japan M. Oshima Department of Applied Chemistry, University of Tokyo, Bunkyo-ku, Tokyo 113-8656, Japan T. Koide Photon Factory, Institute of Materials Structure Science, High Energy Accelerator Research Organization (KEK), Tsukuba, Ibaraki 305-0801, Japan A. Fujimori Department of Physics, University of Tokyo, Bunkyo-ku, Tokyo 113-0033, Japan Department of Complexity Science and Engineering, University of Tokyo, Bunkyo-ku, Tokyo 113-0033, Japan Condensed Matter Science Division, Japan Atomic Energy Agency, Sayo-cho, Sayo-gun, Hyogo 679-5148, Japan
July 26, 2019
###### Abstract
Metallic transition-metal oxides undergo a metal-to-insulator transition (MIT) as the film thickness decreases across a critical thickness of several monolayers (MLs), but its driving mechanism remains controversial. We have studied the thickness-dependent MIT of the ferromagnetic metal LaSrMnO by x-ray absorption spectroscopy and x-ray magnetic circular dichroism. As the film thickness was decreased across the critical thickness of the MIT (6-8 ML), a gradual decrease of the ferromagnetic signals and a concomitant increase of paramagnetic signals were observed, while the Mn valence abruptly decreased towards Mn. These observations suggest that the ferromagnetic phase gradually and most likely inhomogeneously turns into the paramagnetic phase and both phases abruptly become insulating at the critical thickness.
###### pacs:
75.47.Lx, 75.70.-i, 78.20.Ls, 78.70.Dm
## I Introduction
The physical properties of 3 transition-metal oxides (TMOs) are usually controlled by the bandwidth and/or band filling of the 3 bands Imada et al. (1998). Recently, attempts have also been made to control them by the thickness of thin film samples, namely, by dimensionality Hong et al. (2005); Huijben et al. (2008); Yoshimatsu et al. (2009); Boschker et al. (2012); Xia et al. (2009); Toyota et al. (2005); Yoshimatsu et al. (2010, 2011); Scherwitzl et al. (2011). In many metallic oxide thin films, including ferromagnetic ones such as LaSrMnO (LSMO) Hong et al. (2005); Huijben et al. (2008); Yoshimatsu et al. (2009); Boschker et al. (2012) and SrRuO (SRO) Xia et al. (2009); Toyota et al. (2005), and paramagnetic ones such as SrVO (SVO) Yoshimatsu et al. (2010, 2011) and LaNiO Scherwitzl et al. (2011), the resistivity increases when the film thickness is decreased, and metal-to-insulator transitions (MITs) occur at a critical thickness of several monolayers (MLs). Evidence for the MITs was found by transport measurements Hong et al. (2005); Huijben et al. (2008); Xia et al. (2009); Scherwitzl et al. (2011) and photoemission spectroscopy (PES) Yoshimatsu et al. (2009); Toyota et al. (2005); Yoshimatsu et al. (2010, 2011). According to the PES studies of TMO thin films, the density of states at the Fermi level () disappears below 4 ML (SRO Toyota et al. (2005), SVO Yoshimatsu et al. (2010)) to 8 ML (LSMO Yoshimatsu et al. (2009)), resulting in a large insulating gap (of order 1 eV) at . In order to explain such MITs, transport theories for conventional metal thin films Fuchs (1938); Sondheimer (1950), which take into account only surface roughness, predict extremely small critical thicknesses, and one has to invoke strong electron-electron scattering, namely, strong electron correlation. For example, the thickness-dependent MIT of SVO thin films has been considered as a bandwidth-controlled Mott transition caused by the decreased number of nearest-neighbor V atoms Yoshimatsu et al. (2010). Furthermore, the decrease of the film thickness leads to the lowering of spatial dimension and symmetry, and the increase of interfacial effects, resulting in the changes of the electric and magnetic properties. In LSMO Huijben et al. (2008); Yoshimatsu et al. (2009) as well as in SRO Xia et al. (2009), not only metallic conduction but also ferromagnetism disappears simultaneously. For LSMO thin films, where the number of electrons is not an integer, the simple bandwidth-controlled Mott transition mechanism alone is not sufficient to explain the MIT and one has also to take into account the effect of disorder or local lattice distortion to induce the localization of charge carriers. Therefore, it is important to consider mechanisms such as charge ordering and/or the splitting of the bands due to the lower structural symmetry at the surface and interface. Such mechanisms are usually accompanied by changes in the magnetic properties. Thus, it is strongly desired to probe both the electronic states and magnetic properties on the microscopic level at the same time as a function of film thickness.
For this purpose, we have investigated the electric and magnetic properties of LSMO () thin films as functions of film thickness using x-ray magnetic circular dichroism (XMCD). XMCD in core-level x-ray absorption spectroscopy (XAS) is a powerful tool to obtain information about the magnetism of specific elements, together with information about the valence states, and is especially suitable for the study of thin films and nanostructures, because it can probe the intrinsic magnetism without contribution from the substrate and other extrinsic effects.
## Ii Experiment
LSMO thin films with various thicknesses were fabricated on the TiO-terminated (001) surface of SrTiO (STO) substrates by the laser molecular beam epitaxy (laser-MBE) method. The LSMO thickness ranged from 2 ML to 15 ML. After the deposition, the films were capped with 1 ML of LaSrTiO (LSTO) and then 2 ML of STO [Fig. 1(a)]. The LSTO layer was inserted to keep the local environment of the topmost MnO layer the same as that of the deeper MnO layers Yoshimatsu et al. (2009), namely, each MnO layer is sandwiched by LaSrO (LSO) layers. All the fabrication conditions were identical to those of Ref. Yoshimatsu et al., 2009. The surface morphology of the multilayers was checked by atomic force microscopy, and atomically flat step-and-terrace structures were clearly observed in all the films studied. Four-circle x-ray diffraction measurements confirmed the coherent growth of the films. Prior to measurements, the samples were annealed in 1 atm of O at 400 for 45 minutes in order to eliminate oxygen vacancies. The XAS and XMCD measurements were performed at polarization-variable undulator beamlines BL-16A of the Photon Factory (PF) and at BL23SU of SPring-8 Saitoh et al. (2012). The maximum magnetic field was at PF and at SPring-8. At both beamlines, the magnetic field was applied perpendicular to the film surface. Photons were incident normal to the sample surface and their helicity was reversed to measure XMCD. The measurements were performed at K. All the spectra were taken in the total electron yield mode. Most of the spectra shown in the figures were taken at PF, while data taken at high magnetic fields () were measured at SPring-8.
## Iii Results and Discussion
Figures 1(b) and 1(b)(c) show the XAS and XMCD spectra of all the LSMO samples. The spectral line shapes of XMCD [Fig. 1(c)] are similar to those of bulk LSMO Koide et al. (2001). Reflecting the thickness-dependent magnetic properties of LSMO thin films Hong et al. (2005); Huijben et al. (2008); Yoshimatsu et al. (2009), the XMCD intensity decreased with decreasing thickness. Using the XMCD sum rules Thole et al. (1992); Carra et al. (1993), we have estimated the spin () and orbital () magnetic moments of the Mn ions [Sincethespinsumruletendstounderestimatethemagneticmomentwhenspin-orbitsplittingofthetransition-metal$2p$corelevelsisnotlargeenoughcomparedtothe$2p$-$3d$exchangesplittinginteraction; wehavedivided$M_\text{spin}$bya`correctionfactor'0.587givenin][]Teramura. The ratio was found to be less than for all the LSMO thicknesses; therefore mainly contributes to the magnetic moment of Mn. Figure 2 shows thus estimated magnetization curves () of the LSMO thin films with various thicknesses. In Fig. 2(a), ’s are plotted as functions of applied magnetic field . In Fig. 2(b) the same data are replotted as functions of magnetic field corrected for the demagnetizing field perpendicular to the film as . After the demagnetization-field correction, the saturation field is reduced to - , but is still an order of magnitude larger than the saturation field when the external magnetic field is applied parallel to the film and the demagnetizing field is absent () Ishii et al. (2005). Therefore, we conclude that the LSMO thin films have in-plane easy magnetization axes due to magnetocrystalline anisotropy, which probably originates from the tensile strain from the STO substrate Konishi et al. (1999); Pesquera et al. (2012) (note that for bulk LSMO while for STO).
Next, we decompose the magnetization curves into two components as shown in the inset of Fig. 2(b): the ferromagnetic component which saturates below and the paramagnetic component which increases linearly with up to the highest magnetic fields [Examplesofsuchadecompositionofthemagnetizationcurvescanbeseeninreferences; e.g.][]Takeda_GaMnAs. Thus the intercept of the magnetization curve gives the ferromagnetic moment and the slope of the magnetization curve gives paramagnetic susceptibility of the Mn ions. We confirmed the linear increase of as a function of up to higher magnetic fields () for several samples (not shown), which justifies the separation of the magnetization curves into the ferromagnetic and paramagnetic components. In Fig. 3(a), the and values thus obtained are plotted as functions of film thickness. With decreasing film thickness, decreases and increases, indicating a gradual transition from the ferromagnetic state to the paramagnetic state. The measured is, however, somewhat larger than the one predicted by the Curie law of the Mn-Mn mixed valence state in the entire thickness range 111 We have compared the magnitude of with that of non-interacting Mn/Mn local moments calculated for the Curie paramagnetic state, namely, where is the -factor, is the saturation magnetization of LSMO, and is the number ratio of the paramagnetic Mn atoms. . This indicates that the system exhibits a ferromagnetic-to-paramagnetic phase separation and that even in the paramagnetic state there are ferromagnetic correlations between the Mn local moments.
We note that while the thickness-dependent MIT occurs rather abruptly according to PES Yoshimatsu et al. (2009), Fig. 3(a) shows a gradual decrease and increase of the ferromagnetic and paramagnetic components, respectively, with decreasing film thickness. Indeed, the thickness dependence of the valence-band maximum (VBM) measured by PES Yoshimatsu et al. (2009), as shown in Fig. 3(b), indicates an abrupt opening of the energy gap below the critical thickness. This means that the paramagnetic component already exists slightly above the critical thickness of the MIT and the ferromagnetic component persists below it as an ferromagnetic insulating (FM-I) phase. Therefore, the present results suggest that a paramagnetic metallic (PM-M) or paramagnetic insulating (PM-I) phase starts to appear in the ferromagnetic metallic (FM-M) phase from slightly above the critical thickness, and that the FM-I and the PM-I phases coexist below the critical thickness of MIT.
In order to obtain the information about changes in the electronic structure across the MIT, we examine the thickness dependence of the line shapes and the energy positions of the XAS and XMCD spectra. Comparing the experimental XAS spectra with those of LaMnO (Mn) Burnus et al. (2008) and SrMnO (Mn) Sahu et al. (2002) [Fig. 1(b)], the intensities of structures a and c, which originate from Mn, become stronger when the LSMO thickness is reduced. In addition, as shown in Figs. 1(b) and 3(b), the peak positions of the Mn L and L edges are abruptly shifted to lower energies by between 8 ML and 6 ML, where the thickness-dependent MIT occurs Yoshimatsu et al. (2009). Similar peak shifts are also observed in the XMCD spectra, as shown in Fig. 1(c). In the reference XAS spectra in Fig. 1(b), both the L and L edges are located at lower photon energies for Mn than for Mn. From these spectral changes, we conclude that the effective hole concentration decreases as the LSMO thickness decreases, and that it suddenly drops at the critical thickness of MIT. Considering that bulk LSMO enters the FM-I phase in the low hole concentration region Urushibara et al. (1995), the observed valence shift towards Mn in the thin films is certainly related with the FM-M to FM-I transition with decreasing thickness.
The observed valence change towards Mn with decreasing film thickness may be partly explained by the presence of the LSTO layer in the cap. As mentioned in Sec. II a 1-ML-thick LSTO layer is inserted between the LSMO film and the STO cap layer in our samples [Fig. 1(a)]. The (LSO) layer in LSTO acts as an electron donor. Because the work function of STO is smaller than that of LSMO, the electrons supplied by the LSO layer are doped into the LSMO side rather than the STO side Kumigashira et al. (2006). Thus the average Mn valence is shifted towards the side from the nominal valence . This effect is more significant in the thinner films because the number of the doped electrons per monolayer is larger. However, the amount of electron charges is not enough to explain the observed valence change. There is also a possibility that some oxygen vacancies may exist in the LSMO films and/or STO substrates, which leads to additional electron doping. We note that such a valence change towards Mn with decreasing film thickness has also been reported in previous studies Lee et al. (2010); Sandiumenge et al. (2013).
The coincidence of the MIT and the abrupt valence change at 6-8 ML implies some connection between the MIT and the valence change. One possible scenario is that the MIT induces changes in the charge distribution in the LSMO film, leading to the apparent decrease of the hole concentration over the entire LSMO film. When the film is thick and metallic, the free electric charges (holes) will be distributed at the top and bottom interfaces of LSMO, so that there is no potential gradient inside the film. When the film becomes thinner and insulating, the holes are distributed over the entire LSMO layer. If the holes at the bottom interface are distributed in a more extended region, it may cause the abrupt Mn valence change observed by XMCD (the probing depth of which is 3-5 nm Nakajima et al. (1999) or 8-12 ML). In order to clarify the relationship between the observed valence change and the MIT, further experiments would be required, especially on the depth profiles of the electronic states and magnetism in the LSMO thin films.
## Iv Summary
We have performed XAS and XMCD studies of LSMO thin films with varying thickness in order to investigate the origin of the thickness-dependent MIT and the concomitant loss of ferromagnetism. With decreasing film thickness, a gradual decrease of the ferromagnetic component and an increase of the paramagnetic component were observed. The experimental paramagnetic susceptibility was larger than the Curie law, indicating that spin correlations between Mn atoms are ferromagnetic. The Mn valence was found to approach below the critical thickness of the MIT. The ferromagnetic-to-paramagnetic transition occurred gradually as a function of thickness, whereas the MIT and the valence change towards took place abruptly. These results can be understood within the picture of mixed phases: the films above the critical thickness as a mixture of the FM-M and PM-I or PM-M phases while the films below the critical thickness as a mixture of the FM-I and PM-I phases. The mechanism of the valence change has to be investigated as a future theoretical problem.
###### Acknowledgements.
This work was supported by a Grant-in-Aid for Scientific Research from the JSPS (No. S22224005) and the Quantum Beam Technology Development Program from the JST. The experiment was done under the approval of the Photon Factory Program Advisory Committee (Proposal No. 2010G187 and No. 2010S2-001) and under the Shared Use Program of JAEA Facilities (Proposal No. 2011A3840/BL23SU). G.S. acknowledges support from Advanced Leading Graduate Course for Photon Science (ALPS) at the University of Tokyo and the JSPS Research Fellowships for Young Scientists (Project No. 26.11615).
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# On a Class of Parametric (p,2)-equations
On a Class of Parametric (p,2)-equations We consider parametric equations driven by the sum of a p-Laplacian and a Laplace operator (the so-called (p, 2)-equations). We study the existence and multiplicity of solutions when the parameter $$\lambda >0$$ λ > 0 is near the principal eigenvalue $$\hat{\lambda }_1(p)>0$$ λ ^ 1 ( p ) > 0 of $$(-\Delta _p,W^{1,p}_{0}(\Omega ))$$ ( - Δ p , W 0 1 , p ( Ω ) ) . We prove multiplicity results with precise sign information when the near resonance occurs from above and from below of $$\hat{\lambda }_1(p)>0$$ λ ^ 1 ( p ) > 0 . http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Applied Mathematics and Optimization Springer Journals
# On a Class of Parametric (p,2)-equations
, Volume 75 (2) – Jan 22, 2016
36 pages
/lp/springer_journal/on-a-class-of-parametric-p-2-equations-7l6VWr3g0v
Publisher
Springer US
Subject
Mathematics; Calculus of Variations and Optimal Control; Optimization; Systems Theory, Control; Theoretical, Mathematical and Computational Physics; Mathematical Methods in Physics; Numerical and Computational Physics, Simulation
ISSN
0095-4616
eISSN
1432-0606
D.O.I.
10.1007/s00245-016-9330-z
Publisher site
See Article on Publisher Site
### Abstract
We consider parametric equations driven by the sum of a p-Laplacian and a Laplace operator (the so-called (p, 2)-equations). We study the existence and multiplicity of solutions when the parameter $$\lambda >0$$ λ > 0 is near the principal eigenvalue $$\hat{\lambda }_1(p)>0$$ λ ^ 1 ( p ) > 0 of $$(-\Delta _p,W^{1,p}_{0}(\Omega ))$$ ( - Δ p , W 0 1 , p ( Ω ) ) . We prove multiplicity results with precise sign information when the near resonance occurs from above and from below of $$\hat{\lambda }_1(p)>0$$ λ ^ 1 ( p ) > 0 .
### Journal
Applied Mathematics and OptimizationSpringer Journals
Published: Jan 22, 2016
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Computer Engineering and Applications ›› 2021, Vol. 57 ›› Issue (1): 213-218.
### Target Segmentation Algorithm Based on SLIC and Region Growing
HAN Jipu, DUAN Xianhua, CHANG Zhen
1. School of Computer Science, Jiangsu University of Science and Technology, Zhenjiang, Jiangsu 212000, China
• Online:2021-01-01 Published:2020-12-31
### 基于SLIC和区域生长的目标分割算法
1. 江苏科技大学 计算机学院，江苏 镇江 212000
Abstract:
The segmentation result of the traditional region growing algorithm depends on the selection of the seed point. The noise of the image and the uneven grayscale value are easy to form the segmentation cavity in the process of segmentation. Aiming at the above problems, an improved region growing algorithm based on superpixel is proposed. Frist of all, the Laplacian sharpening is used to enhance the boundary of the target to be segmented. According to the features of gray similarity, the SLIC（Simple Linear Iterative Clustering） superpixel segmentation method is used to segment the original image into several irregular regions. Then an undirected weighted graph based on irregular regions will be established. A region is selected as a seed, the region is grown in units of the segmented irregular regions according to the undirected weighting map. To clarify the edge area, the region growing algorithm in pixels runs at the edge of the segmentation target finally. Compared with the traditional region growing algorithm, the improved algorithm is less affected by the seed point selection in the segmentation result, and the improved algorithm can effectively solve the problem of segmentation holes. Compared with clustering segmentation, Otsu threshold segmentation method, the proposed algorithm has obvious advantages in segmentation accuracy.
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2023-02-05 21:24:55
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https://www.gradesaver.com/textbooks/math/algebra/intermediate-algebra-connecting-concepts-through-application/chapter-9-conic-sections-sequences-and-series-9-3-arithmetic-sequences-9-3-exercises-page-727/2
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Chapter 9 - Conic Sections, Sequences, and Series - 9.3 Arithmetic Sequences - 9.3 Exercises - Page 727: 2
$-7,-4,-1,2$
Work Step by Step
$a_n=-7+(n-1)3$ $a_1=-7$ $a_2=-7+3=-4$ $a_3=-7+2\times3=-1$ $a_4=-7+3\times3=2$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
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2018-12-16 11:16:05
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https://zbmath.org/?q=an%3A1024.76028
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## ADER: Arbitrary high-order Godunov approach.(English)Zbl 1024.76028
The ADER Toro approach is extended to the case of nonlinear scalar hyperbolic conservation laws. In this approach, which is related to the ENO/WENO methodology, a generalized Riemann problem is solved as a sequence of conventional Riemann problems for the $$k$$-th derivatives of the solution. The presented schemes enjoy very high order of accuracy in space and time and, while being explicit one-step methods, show optimal stability properties.
### MSC:
76M12 Finite volume methods applied to problems in fluid mechanics 76M20 Finite difference methods applied to problems in fluid mechanics
Full Text:
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2022-12-02 08:41:03
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https://laculinaria.de/monster-hunter-frontier-offline-pc-downloadinstmank-best/
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monster hunter world free download. Free games for windows. Create a costume and search for. theres a video of monster hunter universe on youtube and its. was given to butler in the new recruit training.Monster Hunter Frontier Online Download PC Games Monster Hunter Frontier Gwent PC Game Fan.Immunogenicity, pharmacokinetics and safety of a novel composite, chimeric HIV-1 subtype C Gag/Env gp140 protein vaccine: an extended follow-up study of a Phase I clinical trial.
This is an extended follow-up study of a Phase I clinical trial that tested the safety, immunogenicity and gp140 shedding of a novel construct designed to incorporate both plasma and vaginal MHC I and II epitopes from HIV-1 subtype C Gag and HIV-1 subtype B Env. A composite protein vaccine, ZP3P-CVF15, was administered intramuscularly to 10 HIV-1 seronegative men and women. The median time of the first repeat vaccination was 15 months. No safety signals were detected. A multivalent HIV-1 vaccine induced strong and sustained immune responses against Gag, Pol and Env. One participant in the gp140 group and two in the gp160 group developed HIV-1 antibody to Pol. No gp140 and gp160 viral antigen was detected in the serum. Three participants showed occasional transient CCR5-tropic HIV-1 shedding, with a peak of 44 and 45 HIV RNA copies/ml at Weeks 14 and 15. The extent of V2-loop and V4-stem epitope conservation within Gag and Env from each participant was 100% and 98%, respectively. These data suggest that a multivalent HIV-1 vaccine can be safely administered to seronegative individuals who already have an intact immune system and that this approach may be safe and immunogenic.Kinetics and spectroscopy of the interaction of methyl-3-hydroxy-2-oxindole with tryptophan in aqueous solution.
The reaction kinetics and the nature of the adducts formed when tryptophan (Trp) is reacted with methyl-3-hydroxy-2-oxindole (HOI-Mn) were investigated. HoI-Mn reaction with Trp was studied in aqueous solutions, monitored by visible spectroscopy at 785 nm and by HPLC-UV. UV-vis spectra revealed two linear components, one
Download Monster Hunter Frontier Offline Pc Downloadinstmank |Dansa.com / For the brave souls that want to play without needing an internet connection. can play offline.Q:
Why do we use the axiom of choice for finding a non-empty subset of a given set?
I’m having a hard time understanding why the axiom of choice is necessary in the following question,
Question: Is there a non-empty subset $X$ of the set of real numbers with the property that $\forall n \in \mathbb{N}, nx \in X$?
By non-empty I mean that it is $ot\subseteq\emptyset$, and by $n \in \mathbb{N}$ that the $n$ is a natural number.
I’m thinking that if we can have a one-to-one correspondence between $X$ and the natural numbers, then $\mathbb{N} \to X$ would be injective and hence we’d have a proof without the axiom of choice.
My question is: what is the precise reason why we use the axiom of choice for this question?
A:
The axiom of choice is needed because you want to choose an element from every set that you look into (what is usually denoted by the symbol $\bigcup_{n\in\Bbb N}A_n$), not just every element of every set.
In your proposed line of argumentation, the point is that the natural number $0$ corresponds to the empty subset of $\Bbb R$, but the point is that there are larger and larger subsets with the property that $n\cdot x\in X$. This is not possible in a naive fashion, since there is no largest non-empty subset of $\Bbb R$ with this property.
That is, the property is a very strong property that actually requires a different proof than for every $n\in\Bbb N$ there is an $x\in X$ with $n\cdot x\in X$ (which in turn cannot be shown without the axiom of choice).
In fact, you can already see this in the statement of the problem: the statement only looks for a non-empty
6d1f23a050
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2022-10-02 13:15:03
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https://math.stackexchange.com/questions/1095786/relation-between-lattice-theorem-in-groups-and-in-rings
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# Relation between lattice theorem in groups and in rings
I was studying my abstract algebra notes, and couldn't help but notice a striking similarity between the following two statements:
1. Let $G$ be a group, and $H\triangleleft G$. The canonical projection $$\pi:G\to G/H$$ induces a bijection between subgroups of $G$ that contain $H$, and subgroups of $G/H$. As a corollary, $G/H$ is simple iff $H$ is maximal among proper normal subgroups of $G$.
2. Let $R$ be a ring, and $\mathfrak m$ an ideal. The canonical projection $$\pi:R\to R/\mathfrak m$$ induces a bijection between ideals in $R$ that contain $\mathfrak m$, and ideals of $R/\mathfrak m$. As a corollary, $R/\mathfrak m$ is a field iff $m$ is a maximal ideal.
There has to be a relationship between the two, right? I might even go so far as to say that $1. \Leftrightarrow 2.$ (and obviously $\text{true}\Leftrightarrow\text{true}$ always holds but, well, you know what I mean). Can someone shed some light?
Here are a few ways that those facts are connected.
1. Rings are groups with extra structure. This fact is how you prove isomorphism theorems for rings, for the most part: use the corresponding group theorem for the additive group of the ring, and then check that the other properties that need to be true are true.
2. This lies under the umbrella of universal algebra: there is a way to generalize the idea of a group or a ring to other structures, and in these structures, generalized versions of the isomorphism theorems hold. (See the Wikipedia page on this subject.) You'll see another instance of the theorems when you learn about modules.
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2020-01-27 10:10:34
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https://labs.tib.eu/arxiv/?author=O.%20Rader
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• ### Band renormalization of blue phosphorus on Au(111)(1803.08862)
March 31, 2018 cond-mat.mes-hall
Most recently, theoretical calculations predicted the stability of a novel two-dimensional phosphorus honeycomb lattice named blue phosphorus. Here, we report on the growth of blue phosphorus on Au(111) and unravel its structural details using diffraction, microscopy and theoretical calculations. Most importantly, by utilizing angle-resolved photoemission spectroscopy we identify its momentum-resolved electronic structure. We find that Au(111) breaks the sublattice symmetry of blue phosphorus leading to an orbital-dependent band renormalization upon the formation of a (4x4) superstructure. Notably, the semiconducting two-dimensional phosphorus realizes its valence band maximum at 0.9 eV binding energy, however, shifted in momentum space due to the substrate-induced band renormalization.
• ### Quantum oscillations and Dirac dispersion in the BaZnBi2 semimetal guaranteed by local Zn vacancy order(1801.07474)
March 27, 2018 cond-mat.str-el
We have synthesized single crystals of Dirac semimetal candidates AZnBi2 with A=Ba and Sr. In contrast to A=Sr, the Ba material displays a novel local Zn vacancy ordering, which makes the observation of quantum oscillations in out-of-plane magnetic fields possible. As a new Dirac semimetal candidate, BaZnBi2 exhibits small cyclotron electron mass, high quantum mobility, and non-trivial Berry phases. Three Dirac dispersions are observed by ARPES and identified by first- principles band-structure calculations. Compared to AMn(Bi/Sb)2 systems which host Mn magnetic moments, BaZnBi2 acts as non-magnetic analogue to investigate the intrinsic properties of Dirac fermions in this structure family.
• ### Negative longitudinal magnetoresistance from anomalous N=0 Landau level in topological materials(1704.02021)
Aug. 11, 2017 cond-mat.mes-hall
Negative longitudinal magnetoresistance (NLMR) is shown to occur in topological materials in the extreme quantum limit, when a magnetic field is applied parallel to the excitation current. We perform pulsed and DC field measurements on Pb1-xSnxSe epilayers where the topological state can be chemically tuned. The NLMR is observed in the topological state, but is suppressed and becomes positive when the system becomes trivial. In a topological material, the lowest N=0 conduction Landau level disperses down in energy as a function of increasing magnetic field, while the N=0 valence Landau level disperses upwards. This anomalous behavior is shown to be responsible for the observed NLMR. Our work provides an explanation of the outstanding question of NLMR in topological insulators and establishes this effect as a possible hallmark of bulk conduction in topological matter.
• ### Suppression of electron scattering resonances in graphene by quantum dots(1707.05577)
Transmission of low-energetic electrons through two-dimensional materials leads to unique scattering resonances. These resonances contribute to photoemission from occupied bands where they appear as strongly dispersive features of suppressed photoelectron intensity. Using angle-resolved photoemission we have systematically studied scattering resonances in epitaxial graphene grown on the chemically differing substrates Ir(111), Bi/Ir, Ni(111) as well as in graphene/Ir(111) nanopatterned with a superlattice of uniform Ir quantum dots. While the strength of the chemical interaction with the substrate has almost no effect on the dispersion of the scattering resonances, their energy can be controlled by the magnitude of charge transfer from/to graphene. At the same time, a superlattice of small quantum dots deposited on graphene eliminates the resonances completely. We ascribe this effect to a nanodot-induced buckling of graphene and its local rehybridization from sp$^{2}$ to sp$^{3}$ towards a three-dimensional structure. Our results suggest nanopatterning as a prospective tool for tuning optoelectronic properties of two-dimensional materials with graphene-like structure.
• ### Laser-induced persistent photovoltage on the surface of a ternary topological insulator at room temperature(1705.07078)
May 19, 2017 cond-mat.mes-hall
Using time- and angle-resolved photoemission, we investigate the ultrafast response of excited electrons in the ternary topological insulator (Bi$_{1 x}$Sb$_{x}$)$_2$Te$_3$ to fs-infrared pulses. We demonstrate that at the critical concentration $x$=0.55, where the system becomes bulk insulating, a surface voltage can be driven at room temperature through the topological surface state solely by optical means. We further show that such a photovoltage persists over a time scale that exceeds $\sim$6 $\mu$s, i.e, much longer than the characteristic relaxation times of bulk states. We attribute the origin of the photovoltage to a laser-induced band-bending effect which emerges near the surface region on ultrafast time scales. The photovoltage is also accompanied by a remarkable increase in the relaxation times of excited states as compared to undoped topological insulators. Our findings are relevant in the context of applications of topological surface states in future optical devices.
• ### Surface Fermi arc connectivity in the type-II Weyl semimetal candidate WTe$_{2}$(1608.05633)
Aug. 19, 2016 cond-mat.mtrl-sci
We perform ultrahigh resolution angle-resolved photoemission experiments at a temperature T=0.8 K on the type-II Weyl semimetal candidate WTe$_{2}$. We find a surface Fermi arc connecting the bulk electron and hole pockets on the (001) surface. Our results show that the surface Fermi arc connectivity to the bulk bands is strongly mediated by distinct surface resonances dispersing near the border of the surface-projected bulk band gap. By comparing the experimental results to first-principles calculations we argue that the coupling to these surface resonances, which are topologically trivial, is compatible with the classification of WTe$_{2}$ as a type-II Weyl semimetal hosting topological Fermi arcs. We further support our conclusion by a systematic characterization of the bulk and surface character of the different bands and discuss the similarity of our findings to the case of topological insulators.
• ### Nonmagnetic band gap at the Dirac point of the magnetic topological insulator (Bi$_{1-x}$Mn$_x)_2$Se$_3$(1603.02881)
March 9, 2016 cond-mat.mes-hall
Magnetic doping is expected to open a band gap at the Dirac point of topological insulators by breaking time-reversal symmetry and to enable novel topological phases. Epitaxial (Bi$_{1-x}$Mn$_{x}$)$_{2}$Se$_{3}$ is a prototypical magnetic topological insulator with a pronounced surface band gap of $\sim100$ meV. We show that this gap is neither due to ferromagnetic order in the bulk or at the surface nor to the local magnetic moment of the Mn, making the system unsuitable for realizing the novel phases. We further show that Mn doping does not affect the inverted bulk band gap and the system remains topologically nontrivial. We suggest that strong resonant scattering processes cause the gap at the Dirac point and support this by the observation of in-gap states using resonant photoemission. Our findings establish a novel mechanism for gap opening in topological surface states which challenges the currently known conditions for topological protection.
• ### Spin Mapping of Surface and Bulk Rashba States in Ferroelectric a-GeTe(111) Films(1512.01363)
Dec. 4, 2015 cond-mat.mtrl-sci
A comprehensive mapping of the spin polarization of the electronic bands in ferroelectric a-GeTe(111) films has been performed using a time-of-flight momentum microscope equipped with an imaging spin filter that enables a simultaneous measurement of more than 10.000 data points (voxels). A Rashba type splitting of both surface and bulk bands with opposite spin helicity of the inner and outer Rashba bands is found revealing a complex spin texture at the Fermi energy. The switchable inner electric field of GeTe implies new functionalities for spintronic devices.
• ### Rashba splitting of 100 meV in Au-intercalated graphene on SiC(1510.02291)
Intercalation of Au can produce giant Rashba-type spin-orbit splittings in graphene but this has not yet been achieved on a semiconductor substrate. For graphene/SiC(0001), Au intercalation yields two phases with different doping. Here, we report the preparation of an almost pure p-type graphene phase after Au intercalation. We observe a 100 meV Rashba-type spin-orbit splitting at 0.9 eV binding energy. We show that this giant splitting is due to hybridization and much more limited in energy and momentum space than for Au-intercalated graphene on Ni.
• ### Ultrafast spin polarization control of Dirac fermions in topological insulators(1505.02742)
May 11, 2015 cond-mat.mes-hall
Three-dimensional topological insulators (TIs) are characterized by spin-polarized Dirac-cone surface states that are protected from backscattering by time-reversal symmetry. Control of the spin polarization of topological surface states (TSSs) using femtosecond light pulses opens novel perspectives for the generation and manipulation of dissipationless surface spin currents on ultrafast timescales. Using time-, spin-, and angle-resolved spectroscopy, we directly monitor for the first time the ultrafast response of the spin polarization of photoexcited TSSs to circularly-polarized femtosecond pulses of infrared light. We achieve all-optical switching of the transient out-of-plane spin polarization, which relaxes in about 1.2 ps. Our observations establish the feasibility of ultrafast optical control of spin-polarized Dirac fermions in TIs and pave the way for novel optospintronic applications at ultimate speeds.
• ### Anisotropic effect of warping on the lifetime broadening of topological surface states in angle-resolved photoemission from Bi$_2$Te$_3$(1504.01691)
April 7, 2015 cond-mat.mes-hall
We analyze the strong hexagonal warping of the Dirac cone of Bi$_2$Te$_3$ by angle-resolved photoemission. Along $\overline{\Gamma}$$\overline{\rm M}, the dispersion deviates from a linear behavior meaning that the Dirac cone is warped outwards and not inwards. We show that this introduces an anisotropy in the lifetime broadening of the topological surface state which is larger along \overline{\Gamma}$$\overline{\rm K}$. The result is not consistent with nesting. Based on the theoretically predicted behavior of the ground-state spin texture of a strongly warped Dirac cone, we propose spin-dependent scattering processes as explanation for the anisotropic scattering rates. These results could help paving the way for optimizing future spintronic devices using topological insulators and controlling surface-scattering processes via external gate voltages.
• ### Samarium hexaboride: A trivial surface conductor(1502.01542)
Feb. 5, 2015 cond-mat.str-el
Recent theoretical and experimental studies suggest that SmB$_6$ is the first topological Kondo insulator: A material in which the interaction between localized and itinerant electrons renders the bulk insulating at low temperature, while topological surface states leave the surface metallic. While this would elegantly explain the material's puzzling conductivity, we find the experimentally observed candidates for both predicted topological surface states to be of trivial character instead: The surface state at $\bar{\Gamma}$ is very heavy and shallow with a mere $\sim 2$ meV binding energy. It exhibits large Rashba splitting which excludes a topological nature. We further demonstrate that the other metallic surface state, located at $\bar{X}$, is not an independent in-gap state as supposed previously, but part of a massive band with much higher binding energy (1.7 eV). We show that it remains metallic down to 1 K due to reduced hybridization with the energy-shifted surface 4$f$ level.
• ### Photoemission of Bi$_2$Se$_3$ with Circularly Polarized Light: Probe of Spin Polarization or Means for Spin Manipulation?(1310.1160)
Oct. 4, 2013 cond-mat.mes-hall
Topological insulators are characterized by Dirac cone surface states with electron spins aligned in the surface plane and perpendicular to their momenta. Recent theoretical and experimental work implied that this specific spin texture should enable control of photoelectron spins by circularly polarized light. However, these reports questioned the so far accepted interpretation of spin-resolved photoelectron spectroscopy. We solve this puzzle and show that vacuum ultraviolet photons (50-70 eV) with linear or circular polarization probe indeed the initial state spin texture of Bi$_2$Se$_3$ while circularly polarized 6 eV low energy photons flip the electron spins out of plane and reverse their spin polarization. Our photoemission calculations, considering the interplay between the varying probing depth, dipole selection rules and spin-dependent scattering effects involving initial and final states explain these findings, and reveal proper conditions for light-induced spin manipulation. This paves the way for future applications of topological insulators in opto-spintronic devices.
• ### Tunneling Tuned Spin Modulations in Ultrathin Topological Insulator Films(1307.5485)
July 21, 2013 cond-mat.mes-hall
Quantitative understanding of the relationship between quantum tunneling and Fermi surface spin polarization is key to device design using topological insulator surface states. By using spin-resolved photoemission spectroscopy with p-polarized light in topological insulator Bi2Se3 thin films across the metal-to-insulator transition, we observe that for a given film thickness, the spin polarization is large for momenta far from the center of the surface Brillouin zone. In addition, the polarization decreases significantly with enhanced tunneling realized systematically in thin insulating films, whereas magnitude of the polarization saturates to the bulk limit faster at larger wavevectors in thicker metallic films. Our theoretical model calculations capture this delicate relationship between quantum tunneling and Fermi surface spin polarization. Our results suggest that the polarization current can be tuned to zero in thin insulating films forming the basis for a future spin-switch nano-device.
• ### Reversal of the circular dichroism in the angle-resolved photoemission from Bi2Te3(1303.6161)
The helical Dirac fermions at the surface of topological insulators show a strong circular dichroism which has been explained as being due to either the initial-state spin angular momentum, the initial-state orbital angular momentum, or the handedness of the experimental setup. All of these interpretations conflict with our data from Bi2Te3 which depend on the photon energy and show several sign changes. Our one-step photoemission calculations coupled to ab initio theory confirm the sign change and assign the dichroism to a final-state effect. The spin polarization of the photoelectrons, instead, remains a reliable probe for the spin in the initial state.
• ### Massless Dirac fermions in epitaxial graphene on Fe(110)(1212.6866)
Dec. 31, 2012 cond-mat.mtrl-sci
Graphene grown on Fe(110)by chemical vapor deposition using propylene is investigated by means of angle-resolved photoemission. The presence of massless Dirac fermions is clearly evidenced by the observation of a fully intact Dirac cone. Unlike Ni(111) and Co(0001), the Fe(110) imposes a strongly anisotropic quasi-one-dimensional structure on the graphene. Certain signatures of a superlattice effect appear in the dispersion of its \sigma-bands but the Dirac cone does not reveal any detectable superlattice or quantum-size effects although the graphene corrugation is twice as large as in the established two-dimensional graphene superlattice on Ir(111).
• ### Magnetically induced spin reorientation on the surface of a topological insulator (a surface magnetic topological insulator MBE film)(1206.2090)
Dec. 13, 2012 cond-mat.mes-hall
The surface of topological insulators is proposed as a promising platform for spintronics and quantum information applications. In particular, when time- reversal symmetry is broken, topological surface states are expected to exhibit a wide range of exotic spin phenomena for potential implementation in electronics. Such devices need to be fabricated using nanoscale artificial thin films. It is of critical importance to study the spin behavior of artificial topological MBE thin films associated with magnetic dopants, and with regards to quantum size effects related to surface-to-surface tunneling as well as experimentally isolate time-reversal breaking from non-intrinsic surface electronic gaps. Here we present observation of the first (and thorough) study of magnetically induced spin reorientation phenomena on the surface of a topological insulator. Our results reveal dramatic rearrangements of the spin configuration upon magnetic doping contrasted with chemically similar nonmagnetic doping as well as with quantum tunneling phenomena in ultra-thin high quality MBE films. While we observe that the spin rearrangement induced by quantum tunneling occurs in a time-reversal invariant fashion, we present critical and systematic observation of an out-of-plane spin texture evolution correlated with magnetic interactions, which breaks time-reversal symmetry, demonstrating microscopic TRB at a Kramers' point on the surface.
• ### Probing two topological surface bands of Sb2Te3 by spin-polarized photoemission spectroscopy(1201.4323)
Oct. 30, 2012 cond-mat.mes-hall
Using high resolution spin- and angle-resolved photoemission spectroscopy, we map the electronic structure and spin texture of the surface states of the topological insulator Sb2Te3. In combination with density functional calculations (DFT), we directly show that Sb2Te3 exhibits a partially occupied, single spin-Dirac cone around the Fermi energy, which is topologically protected. DFT obtains a spin polarization of the occupied Dirac cone states of 80-90%, which is in reasonable agreement with the experimental data after careful background subtraction. Furthermore, we observe a strongly spin-orbit split surface band at lower energy. This state is found at 0.8eV below the Fermi level at the gamma-point, disperses upwards, and disappears at about 0.4eV below the Fermi level into two different bulk bands. Along the gamma-K direction, the band is located within a spin-orbit gap. According to an argument given by Pendry and Gurman in 1975, such a gap must contain a surface state, if it is located away from the high symmetry points of the Brillouin zone. Thus, the novel spin-split state is protected by symmetry, too.
• ### Graphene for spintronics: giant Rashba splitting due to hybridization with Au(1208.4265)
Aug. 21, 2012 cond-mat.mtrl-sci
Graphene in spintronics has so far primarily meant spin current leads of high performance because the intrinsic spin-orbit coupling of its pi-electrons is very weak. If a large spin-orbit coupling could be created by a proximity effect, the material could also form active elements of a spintronic device such as the Das-Datta spin field-effect transistor, however, metal interfaces often compromise the band dispersion of massless Dirac fermions. Our measurements show that Au intercalation at the graphene-Ni interface creates a giant spin-orbit splitting (~100 meV) in the graphene Dirac cone up to the Fermi energy. Photoelectron spectroscopy reveals hybridization with Au-5d states as the source for the giant spin-orbit splitting. An ab initio model of the system shows a Rashba-split dispersion with the analytically predicted gapless band topology around the Dirac point of graphene and indicates that a sharp graphene-Au interface at equilibrium distance will account for only ~10 meV spin-orbit splitting. The ab initio calculations suggest an enhancement due to Au atoms that get closer to the graphene and do not violate the sublattice symmetry.
• ### High spin polarization and circular dichroism of topological surface states on Bi2Te3(1108.1053)
May 25, 2012 cond-mat.mes-hall
Topological insulators have been successfully identified by spin-resolved photoemission but the spin polarization remained low (~20%). We show for Bi2Te3 that the in-gap surface state is much closer to full spin polarization with measured values reaching 80% at the Fermi level. When hybridizing with the bulk it remains highly spin polarized which may explain recent unusual quantum interference results on Bi2Se3. The topological surface state shows a large circular dichroism in the photoelectron angle distribution with an asymmetry of ~20% the sign of which corresponds to that of the measured spin.
• ### Topological surface state under graphene for two-dimensional spintronics in air(1104.3308)
Spin currents which allow for a dissipationless transport of information can be generated by electric fields in semiconductor heterostructures in the presence of a Rashba-type spin-orbit coupling. The largest Rashba effects occur for electronic surface states of metals but these cannot exist but under ultrahigh vacuum conditions. Here, we reveal a giant Rashba effect ({\alpha}_R ~ 1.5E-10 eVm) on a surface state of Ir(111). We demonstrate that its spin splitting and spin polarization remain unaffected when Ir is covered with graphene. The graphene protection is, in turn, sufficient for the spin-split surface state to survive in ambient atmosphere. We discuss this result along with evidences for a topological protection of the surface state.
• ### Quantitative determination of spin-dependent quasiparticle lifetimes and electronic correlations in hcp cobalt(1008.3414)
We report on a quantitative investigation of the spin-dependent quasiparticle lifetimes and electron correlation effects in ferromagnetic hcp Co(0001) by means of spin and angle-resolved photoemission spectroscopy. The experimental spectra are compared in detail to state-of-the-art many-body calculations within the dynamical mean field theory and the three-body scattering approximation, including a full calculation of the one-step photoemission process. From this comparison we conclude that although strong local many-body Coulomb interactions are of major importance for the qualitative description of correlation effects in Co, more sophisticated many-body calculations are needed in order to improve the quantitative agreement between theory and experiment, in particular concerning the linewidths. The quality of the overall agreement obtained for Co indicates that the effect of non-local correlations becomes weaker with increasing atomic number.
• ### About the strength of correlation effects in the electronic structure of iron(0910.4360)
The strength of electronic correlation effects in the spin-dependent electronic structure of ferromagnetic bcc Fe(110) has been investigated by means of spin and angle-resolved photoemission spectroscopy. The experimental results are compared to theoretical calculations within the three-body scattering approximation and within the dynamical mean-field theory, together with one-step model calculations of the photoemission process. This comparison indicates that the present state of the art many-body calculations, although improving the description of correlation effects in Fe, give too small mass renormalizations and scattering rates thus demanding more refined many-body theories including non-local fluctuations.
• ### Correlated Electrons Step-by-Step: Itinerant-to-Localized Transition of Fe Impurities in Free-Electron Metal Hosts(0907.2911)
High-resolution photoemission spectroscopy and realistic ab-initio calculations have been employed to analyze the onset and progression of d-sp hybridization in Fe impurities deposited on alkali metal films. The interplay between delocalization, mediated by the free-electron environment, and Coulomb interaction among d-electrons gives rise to complex electronic configurations. The multiplet structure of a single Fe atom evolves and gradually dissolves into a quasiparticle peak near the Fermi level with increasing the host electron density. The effective multi-orbital impurity problem within the exact diagonalization scheme describes the whole range of hybridizations.
• ### Observation of Hubbard Bands in $\gamma$-Manganese(cond-mat/0112430)
Dec. 22, 2001 cond-mat.str-el
We present angle-resolved photoemission spectra of the $\gamma$-phase of manganese as well as a theoretical analysis using a recently developed approach that combines density functional and dynamical mean field methods (LDA+DMFT). The comparison of experimental data and theoretical predictions allows us to identify effects of the Coulomb correlations, namely the presence of broad and undispersive Hubbard bands in this system.
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2021-01-23 10:39:15
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https://www.khanacademy.org/math/ap-calculus-ab/limits-basics-ab/limits-from-graphs-ab/e/two-sided-limits-from-graphs
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# Limits from graphs
Practice finding two sided limits by looking at graphs.
### Problem
What appears to be the value of $\displaystyle\lim_{x\to 0}h(x)$?
Please choose from one of the following options.
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2016-08-24 23:22:14
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https://mathhelp.fandom.com/wiki/Integral_of_1/x
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## FANDOM
49 Pages
A student asks, I need to learn about the integral of 1/x.
## solutionEdit
Let's consider the indefinite integral,
$\int \! {\frac {1} {x}} \,dx \,$
The solution is $\log(x)+C$
Now, let's consider the definite integral,
$\int_a^b \! {\frac {1} {x}} \,dx \,$
Evaluating the indefinite integral at its limits, we get
$\log(x)+C \: \bigg{|}_a^b = \log(b) - \log(a)$
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2019-02-24 01:42:08
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http://nips.cc/Conferences/2009/Program/event.php?ID=1548
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IBM T.J. Watson Research; UC Berkeley; RPI & NSF
Unsupervised Feature Selection for the $k$-means Clustering Problem
7:00 – 11:59pm Monday, December 07, 2009
This is part of the Poster Session which begins at 19:00 on Monday December 7, 2009
M27
We present a novel feature selection algorithm for the $k$-means clustering problem. Our algorithm is randomized and, assuming an accuracy parameter $\epsilon \in (0,1)$, selects and appropriately rescales in an unsupervised manner $\Theta(k \log(k / \epsilon) / \epsilon^2)$ features from a dataset of arbitrary dimensions. We prove that, if we run any $\gamma$-approximate $k$-means algorithm ($\gamma \geq 1$) on the features selected using our method, we can find a $(1+(1+\epsilon)\gamma)$-approximate partition with high probability.
Download the PDF
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2014-11-01 07:12:05
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