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# Documents  Charles, François | enregistrements trouvés : 6

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## Bertini theorems in arithmetic geometry Charles, François | CIRM H

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Research talks;Algebraic and Complex Geometry;Number Theory

The classical Bertini irreducibility theorem states that if $X$ is an irreducible projective variety of dimension at least 2 over an infinite field, then $X$ has an irreducible hyperplane section. The proof does not apply in arithmetic situations, where one wants to work over the integers or a finite fields. I will discuss how to amend the theorem in these cases (joint with Bjorn Poonen over finite fields).

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## Relative integral $p$-adic Hodge theory Morrow, Matthew | CIRM H

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Research talks

Given a smooth scheme $X$ over the ring of integers of a $p$-adic field, we introduce the notion of a relative Breuil-Kisin-Fargues module $M$ on $X$. Each such $M$ simultaneously encodes the data of a lisse étale sheaf, a module with flat connection, and a crystal, whose cohomologies are then intertwined by a relative form of the $A_{inf}$ cohomology introduced in "Integral $p$-adic Hodge theory" by Bhatt-M-Scholze. They are moreover closely related to other work in relative $p$-adic Hodge theory, notably Faltings small generalised representations and his relative Fontaine Lafaille theory. Joint with Takeshi Tsuji. Given a smooth scheme $X$ over the ring of integers of a $p$-adic field, we introduce the notion of a relative Breuil-Kisin-Fargues module $M$ on $X$. Each such $M$ simultaneously encodes the data of a lisse étale sheaf, a module with flat connection, and a crystal, whose cohomologies are then intertwined by a relative form of the $A_{inf}$ cohomology introduced in "Integral $p$-adic Hodge theory" by Bhatt-M-Scholze. They are moreover closely ...

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## Symmetric power moments of Kloosterman sums Fresán, Javier | CIRM H

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Research talks

We construct motives over the rational numbers associated with symmetric power moments of Kloosterman sums, and prove that their $L$-functions extend meromorphically to the complex plane and satisfy a functional equation conjectured by Broadhurst and Roberts. Although the motives in question turn out to be classical, the strategy consists in first realizing them as exponential motives and computing their Hodge numbers by means of the irregular Hodge filtration. We show that all Hodge numbers are either zero or one, which implies potential automorphicity thanks to recent results of Patrikis and Taylor. The first talk will be concerned with the arithmetic aspects and in the second one we will present the Hodge theoretic computations. Joint work with Claude Sabbah and Jeng-Daw Yu. We construct motives over the rational numbers associated with symmetric power moments of Kloosterman sums, and prove that their $L$-functions extend meromorphically to the complex plane and satisfy a functional equation conjectured by Broadhurst and Roberts. Although the motives in question turn out to be classical, the strategy consists in first realizing them as exponential motives and computing their Hodge numbers by means of the irregular ...

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## Periods of $1$-motives Huber-Klawitter, Annette | CIRM H

Multi angle

Research talks

(joint work with G. Wüstholz) Roughly, $1$-dimensional periods are the complex numbers obtained by integrating a differential form on an algebraic curve over $\bar{\mathbf{Q}}$ over a suitable domain of integration. One of the alternative characterisations is as periods of Deligne $1$-motives.
We clear up the linear relations between these numbers, proving Kontsevich's version of the period conjecture for $1$-dimensional periods. In particular, a $1$-dimensional period is shown to be algebraic if and only if it is of the form $\int_\gamma (\phi+df)$ with $\int_\gamma\phi=0$. We also get formulas for the spaces of periods of a given $1$-motive, generalising Baker's theorem on logarithms of algebraic numbers.
The proof is based on a version of Wüstholz's analytic subgroup theorem for $1$-motives.
(joint work with G. Wüstholz) Roughly, $1$-dimensional periods are the complex numbers obtained by integrating a differential form on an algebraic curve over $\bar{\mathbf{Q}}$ over a suitable domain of integration. One of the alternative characterisations is as periods of Deligne $1$-motives.
We clear up the linear relations between these numbers, proving Kontsevich's version of the period conjecture for $1$-dimensional periods. In particular, ...

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## Hilbert schemes of K3 surfaces Negut, Andrei | CIRM H

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Research talks

We give a geometric representation theory proof of a mild version of the Beauville-Voisin Conjecture for Hilbert schemes of K3 surfaces, namely the injectivity of the cycle map restricted to the subring of Chow generated by tautological classes. Although other geometric proofs of this result are known, our approach involves lifting formulas of Lehn and Li-Qin-Wang from cohomology to Chow, and using them to quickly solve the problem by invoking the irreducibility criteria of Virasoro algebra modules, due to Feigin-Fuchs. Joint work with Davesh Maulik. We give a geometric representation theory proof of a mild version of the Beauville-Voisin Conjecture for Hilbert schemes of K3 surfaces, namely the injectivity of the cycle map restricted to the subring of Chow generated by tautological classes. Although other geometric proofs of this result are known, our approach involves lifting formulas of Lehn and Li-Qin-Wang from cohomology to Chow, and using them to quickly solve the problem by invoking ...

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## Vanishing theorems for Shimura varieties at infinite level Caraiani, Ana | CIRM H

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Research talks

#### Filtrer

##### Audience

Ressources Electroniques (Depuis le CIRM)

Books & Print journals

Recherche avancée

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# Simple analysis proof: x^n -> a^n as x -> a

1. Sep 29, 2009

### applegatecz

1. The problem statement, all variables and given/known data
Prove that x^n approaches a^n as x approaches a.

2. Relevant equations

3. The attempt at a solution
I understand the concept here ... need to find a delta>0 for epsilon>0 s.t. |x-a|<delta implies |x^n-a^n|<epsilon. For some reason I can't solve this one. Thanks.

2. Sep 29, 2009

### Dick

You know you can factor x^n-a^n=(x-a)*(x^(n-1)+x^(n-2)*a+ ... + a^(n-1)), right?

3. Sep 29, 2009

### applegatecz

Oh! Somehow I missed that. Thank you very much for your help.

4. Sep 29, 2009

### g_edgar

Dick assumes n is a positive integer, which you did not tell us under "all variables and given/known data" ... so are we to assume n is NOT know to be a positive integer?

5. Oct 2, 2009

### economist1985

still, how do you show that (x^(n-1)+x^(n-2)*a+ ... + a^(n-1)) is < epsilon?

6. Oct 2, 2009

### HallsofIvy

You don't, and no one has claimed it is. You want $|x^n- a^n|= |x- a||x^{n-1}+ ax^{n-2}+ ...+ a^{n-1}|< \epsilon$ for |x- a| small enough. That is the same as choosing $|x- a|< \epsilon/|x^{n-1}+ ax^{n-2}+ ...+ a^{n-1}|$ and you can do that by finding any positive lower bound on $|x^{n-1}+ ax^{n-2}+ ...+ a^{n-1}|$ for x close to a.

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# George B. Moody PhysioNet Challenge

Quick links for this year's Challenge:

# Detecting and Quantifying T-Wave Alternans: The PhysioNet/Computing in Cardiology Challenge 2008

### George Moody

Community forum for the 2008 PhysioNet/CinC Challenge (Sept. 23, 2008, 2:46 p.m.)

If you have any questions or comments regarding this challenge, please post it directly in our Community Discussion Forum. This will increase transparency (benefiting all the competitors) and ensure that all the challenge organizers see your question.

Results from the PhysioNet/CinC Challenge 2008 (Sept. 22, 2008, midnight)

Congratulations to the winners of the PhysioNet/CinC Challenge 2008 on the subject of T-wave alternans. Final results are now available: scores, reference rankings of the Challenge 2008 data set, and the open source software contributed by participants in the open source division of the Challenge.

PhysioNet/CinC Challenge 2008 (March 17, 2008, midnight)

The PhysioNet/Computers in Cardiology Challenge 2008 is underway. The topic of this, the ninth open challenge in our annual series, is detection and quantification of T-wave alternans.

## Citations

When using this resource, please cite the following publications:

### Introduction

The ninth annual PhysioNet/Computers in Cardiology Challenge aims to improve understanding of methods for identification and analysis of T-wave alternans in the ECG.

### Background

One hundred years after the first observation of T-wave alternans was reported by HE Hering [1], the phenomenon is widely understood to be an important indicator of risk of sudden cardiac death. Yet for most of that time TWA was believed to be rare. In 1981, at the eighth annual meeting of Computers in Cardiology, Dan Adam, Solange Akselrod, and Richard Cohen first reported the existence of microvolt-level T-wave alternans [2], too small in amplitude to be detected visually at standard display scales.

T-wave alternans (TWA) is a pattern in the ECG characterized by two (rarely more) distinct forms of T-waves appearing in alternation, at or above a patient-specific heart rate generally in the range of 90 to 120 beats per minute. Although the mechanisms have not been fully elucidated, a large amount of empirical evidence collected during the past 25 years has demonstrated an association between the amount of TWA, the heart rate at which it appears, and the risk of sudden cardiac death (SCD). In particular, the absence of significant TWA in a patient with congestive heart failure, low ejection fraction, or a recent myocardial infarction is strongly predictive of a low risk of SCD. A positive finding in such a patient, though less specific, may indicate that an implantable cardiac defibrillator would be appropriate, an indication that can be confirmed using invasive testing. Since TWA analysis is performed on the surface ECG, it is an inexpensive and non-invasive test. In clinical applications, TWA analysis can be done as part of an exercise stress test, but there is interest in the research community in using conventional long-term (Holter) ECG recordings to observe TWA in the context of activities of daily living. A review by Armoundas, Tomaselli, and Esperer discusses mechanisms that may account for the associations among TWA and other risk factors for SCD, as well as clinical applications of TWA [3].

A variety of algorithms for detecting and quantifying TWA have been proposed, employing techniques from linear and nonlinear signal processing such as spectral analysis, complex demodulation, counting zero-crossings in a series of correlation coefficients, periodogram and complex demodulation analysis of T-wave principal components, Capon filtering, Poincaré maps, periodicity transforms, statistical tests, moving averages, maximum likelihood estimators and generalized likelihood ratio tests, and more. For a comprehensive and systematic discussion of methods for TWA detection and analysis, see the review by Martínez and Olmos [4].

### The Challenge Data

PhysioNet has a wide variety of data that might be appropriate for this challenge. Visitors were invited to nominate some of these data for inclusion in the challenge data set, from among those in these databases:

• The Long-Term ST Database, which consists of 86 records, each 21 to 24 hours in duration. The records are digitized 2- or 3-channel long-term (Holter) ECG tape recordings, mostly from subjects who had transient myocardial ischemia. Each record is annotated beat-by-beat, and with respect to ST and rhythm changes, and is accompanied by a detailed clincal summary.
• The PTB Diagnostic ECG Database, consisting of 549 records from 294 subjects, most of which are about 2 minutes long (some are shorter). These are of very high technical quality (12 standard leads and 3 Frank leads, sampled at 1 KHz per signal with 16-bit resolution), mostly post-MI patients but with representation of other types of subjects including healthy controls. Each record is accompanied by a detailed clinical summary. This database was previously used in the 2006 QT challenge, and it might be interesting to relate QT intervals and TWA in the same subjects.
• The MIT-BIH ST Change Database, which includes 23 high-quality records of exercise stress tests, each typically 20 to 40 minutes in length. The recordings each contain two ECG signals sampled at 360 Hz; there are no annotations or clinical correlates available.
• The Sudden Cardiac Death Holter Database, a collection that currently includes 23 complete 2-channel Holter recordings (7 to 25 hours each) of patients who experienced sustained ventricular tachyarrhythmias and cardiac arrest during the recordings. The records have been digitized at 250 Hz per signal. Twelve records have been meticulously annotated beat-by-beat; algorithm-generated beat annotations are available for the others. No clinical correlates are available.
• The BIDMC Congestive Heart Failure Database, 15 2-channel Holter recordings each about 20 hours in duration, from patients with severe (NYHA class 3-4) congestive heart failure. The records have been digitized at 250 Hz per signal. Age, gender, and NYHA class are available for each patient.

A good challenge data set should include not only clear-cut cases, but also a sampling of cases that might be expected to pose problems (for example, because of changes in conduction patterns, ectopy, or noise).

Participants were asked send a list of the records they wished to nominate for inclusion in the official challenge data set by email to webmaster@physionet.org with the subject “Challenge 2008 data nomination”, no later than noon GMT on Thursday, 17 April 2008. Those who wished to nominate records other than those in the databases listed above were asked to state briefly why they believed their choices are particularly appropriate for this challenge. Participants also were invited to contribute data to PhysioNet to be considered for this challenge, and several did so. Thanks to all who participated in this process.

We have compiled all nominations and selected the most promising cases for inclusion in the official data set, together with new cases that have not been available previously. The official PhysioNet/Computers in Cardiology Challenge 2008 data set is now available (as individual files) and, for convenience, it is also available as a tarball.

### Reporting Results and Scoring

2. For each record in the challenge data set, participants must detect or estimate the peak magnitude of T-wave alternans using a fully-automated method, and enter that estimate on the entry form. If your method detects but does not quantify TWA, enter 1 as the estimate for each record in which TWA was detected, and 0 as the estimate for the remaining records. Valid entries must include a TWA estimate for each record in the challenge data set.
3. Use your favorite text editor to fill in the entry form. Please note that the entry form is a plain text file, not a word processor document; keep it in plain text format. Be sure that your entry form includes the email address where you wish to have your scores sent.
4. Send the filled-in entry by email to webmaster@physionet.org, with the subject line “Challenge 2008 entry”. Entries will be accepted until noon GMT on 1 September 2008.

Scores will be determined by the following procedure, once a minimum number of entries have been received:

1. For each entry, the records will be ranked by the magnitudes of the associated TWA estimates. Thus the record with the lowest TWA estimate in a given entry receives the rank of 1 for that entry, the record with the second-lowest TWA estimate gets a rank of 2, etc.
2. Each record receives a median rank, which is the median of the ranks assigned it by the entries, and a reference ranking is made by sorting the median ranks (i.e., the record with the lowest median rank gets a reference rank of 1, etc.)
3. The score for each entry is the Kendall (tau) rank correlation coefficient between the entry ranking and the reference ranking, where 1 is perfect agreement and -1 is perfect disagreement [5].

You will receive your score by return email. The first scores will be sent on or about 27 April 2008; scores for entries received after that date will usually be sent within a day of receipt.

The reference rankings are initially determined by the entries received. Final reference rankings will be compiled in August. Scores reported before that time are preliminary and are subject to change. Final scores used to determine the winners of the challenge will be sent to all participants shortly after the deadline for entries (1 September 2008).

### Entering the Open Source Division

As in previous years, the Challenge will include an open source division. You may enter the open source division by sending your source code by email, before noon GMT on Monday, 1 September 2008, to PhysioNet. Use the subject line “Challenge 2008 entry source”, and be sure to include:

• All sources needed to produce a working version of your software (except for readily available standard libraries and header files)
• A note describing how to produce a working version of your software (a commented Makefile is ideal), and how to run your software

Each source file submitted should begin with a comment block containing the names of its authors and a reference to the open source license you have chosen for it, if any; for example:

     /* twarng.c - Detect and measure T-wave alternans using a random number generator
Copyright (C) 2008  Herman Foobar <hbar@uncertain.org>
This software is released under the terms of the GNU General
*/

Source files in C, C++, Fortran, or Matlab m-code are preferred; other languages may be acceptable, but please ask first. Do not submit any code that cannot be freely redistributed. Following the conclusion of the Challenge, selected entries will be posted, with full credit to their authors, on PhysioNet.

### Events and Awards

If you submit an abstract describing your work on the challenge no later than 1 May 2008 to Computers in Cardiology, and a challenge entry no later than 1 September 2008 at noon GMT, you may be eligible for one or more awards that will be presented during the final plenary session of the conference on Wednesday, 17 September 2008.

The Board of Directors of Computers in Cardiology provides an annual grant of US$1000 for challenge awards. This year, a matching contribution from Electrogram, Inc. has allowed us to double the prize fund. We are deeply grateful for the generosity and enthusiastic support of Electrogram and Computers in Cardiology. An award of US$1000 will be received by the eligible participant who achieves the best final score overall in this year’s Challenge.

An additional award of US\$1000 will be given to the winner of the open source division. Participants in the open source division are eligible for both awards.

Regarding the estimate of the peak T-wave alternans magnitude, in what units it should be expressed? Does the estimate include the T-wave magnitude or only the added alternans?

You may use whatever units you wish. Estimate the magnitude of the alternans, not the magnitude of the T-waves. (Some methods involve expressing alternans magnitude or power as a fraction of total T-wave magnitude or power; it’s fine to do this, or not to do this, as you wish — but don’t add a T-wave measurement to your alternans estimate.)

The scoring is based on how your estimates rank the records (from least to most TWA) compared with a reference ranking of the records. Thus the units you choose don’t matter, so long as you use the same units for all of your measurements.

This is a hard problem, and I won’t be able to get final results in time for the abstract deadline. What needs to be submitted in the abstract?

Briefly, you will need to submit a short (~300 words) abstract, which may not include illustrations, by 1 May. Abstracts must be submitted on-line.

It would be ideal if you have submitted rough results for the Challenge problem no later than 27 April, so they can be scored and so that you can report your preliminary score in your abstract. If this is not possible, please discuss in your abstract relevant studies you have performed to date, touching on any novel aspects of your methods and quoting whatever results you have. Generally abstracts containing only “we will show ...” without any results are not accepted, so it is very important that you demonstrate in the abstract that you have actually accomplished something already. Given the complexity of the challenge and the short time available for working on it, the abstract reviewers will certainly make some allowance for those who are unable to obtain results on the challenge problem by the abstract deadline, if it appears they will be able to do so before the conference in September.

May I revise my entry?

You may submit entries at any time until the final deadline of noon GMT on 1 September. Each entry will be scored, and you may attempt to improve your score by submitting up to four revised entries (five in all).

You may choose any of your (up to five) entries as the basis for ranking, by sending an email specifying your choice to webmaster@physionet.org, on or before noon GMT on 1 September. Please remember that it may take up to 24 hours after submitting an entry to receive scores, so try to submit your last entry at least a day before the deadline if you think you may want to exercise your choice. Your last entry will be used for ranking unless you specify otherwise.

If my abstract is accepted, what else must I do?

In September, if your abstract is accepted, you will be expected to submit a four-page paper, which may be illustrated, for publication in Computers in Cardiology on-line and in print. You will also be expected to attend the conference (14-17 September 2008 in Bologna, Italy) and to present your work in one of the scientific sessions of the conference, either as a poster or as a 10-minute oral presentation. Your paper and presentation should include your final results.

Why don’t you have a challenge about ...?

Each year, we receive many suggestions for challenge topics. We encourage you to contact us with further suggestions.

### References

1. Hering HE. Das Wesen des Herzalternans. Münchener med Wochenschr 1908;4:1417-21.
2. Adam DR, Akselrod S, Cohen RJ. Estimation of ventricular vulnerability to fibrillation through T-wave time series analysis. Computers in Cardiology 1981;8:307-310.
3. Armoundas AA, Tomaselli GF, Esperer HD. Pathophysiological basis and clinical application of T-wave alternans. JACC 2002;40:207-217.
4. Martínez JP, Olmos S. Methodological Principles of T-wave alternans: a unified framework. IEEE Trans Biomed Eng 2005;52(4):599-613. [requires subscription]
5. Abdi H.“The Kendall Rank Correlation Coefficient”, in Salkind N (Ed.) Encyclopedia of Measurement and Statistics. Thousand Oaks (CA): Sage (2007).

### Challenge Results

Final results are now available: papers presented by participants at CinC 2008, scores, reference rankings and additional details for the Challenge 2008 data set, and the open source software contributed by participants in the open source division of the Challenge.

### Papers

The papers below were presented at Computers in Cardiology 2008. Please cite this publication when referencing any of these papers. These papers have been made available by their authors under the terms of the Creative Commons Attribution License 2.5 (CCAL). We wish to thank all of the authors for their contributions.

The first of these papers is an introduction to the challenge topic, with a summary of the challenge results and a discussion of their implications.

The remaining papers were presented by participants in the Challenge, who describe their approaches to the challenge problem.

An Open-Source Standard T-Wave Alternans Detector for Benchmarking
A Khaustov, S Nemati, GD Clifford

## Access Policy

Anyone can access the files, as long as they conform to the terms of the specified license.

## Files

##### Access the files

• Access the files using the Google Cloud Storage Browser here. Login with a Google account is required.

• Access the data using the Google Cloud command line tools (please refer to the gsutil documentation for guidance):

gsutil -m -u YOUR_PROJECT_ID cp -r gs://challenge-2008-1.0.0.physionet.org DESTINATION

wget -r -N -c -np https://physionet.org/files/challenge-2008/1.0.0/


text:
International
Tables for
Crystallography
Volume C
Mathematical, physical and chemical tables
Edited by E. Prince

International Tables for Crystallography (2006). Vol. C, ch. 4.2, pp. 192-193

## Section 4.2.1.2. The continuous spectrum

U. W. Arndta

#### 4.2.1.2. The continuous spectrum

| top | pdf |

The shape of the continuous spectrum from a thick target is very simple: , the energy per unit frequency band in the spectrum, is given by the expression derived by Kramers (1923): where Z is the atomic number of the target and A and B are constants independent of the applied voltage . B/A is of the order of 0.0025 so that the term in can usually be neglected (Fig. 4.2.1.3 ). is the maximum frequency in the spectrum, i.e. the Duane–Hunt limit at which the entire energy of the bombarding electrons is converted into the quantum energy of the emitted photon, where Using the latest adjusted values of the fundamental constants (Cohen & Taylor, 1987): Equation (4.2.1.5) can be rewritten in a number of forms. If is the number of photons of energy E per incident electron, where  photons eV−1  electron−1, and is known as Kramer's constant.

 Figure 4.2.1.3 | top | pdf |Intensity per unit frequency interval versus frequency in the continuous spectrum from a thick target at different accelerating voltages. From Kuhlenkampff & Schmidt (1943).

From (4.2.1.7), it follows that the total energy in the continuous spectrum per electron is Since the energy of the bombarding electron is , the efficiency of production of the continuous radiation is Crystallographers are more accustomed to thinking of the spectrum in terms of wavelength. Equation (4.2.1.7) can be transformed into which has a maximum at . In practice, the emerging spectrum is modified by target absorption, which is greatest for the longer wavelengths and moves the maximum more nearly to .

It is of interest to compare the X-ray flux in a narrow wavelength band selected by an appropriate monochromator with the flux in a characteristic spectral line, in order to examine the practicability of XAFS (X-ray absorption fine-structure spectroscopy) or optimized anomalous-dispersion diffractometry experiments. For these purposes, the maximum permissible wavelength band is about 10−3 Å. From equation (4.2.1.10), we see that, for a tungsten-target X-ray tube operated at 80 kV, is about 1.1 × 10−5 photons with the Kα energy electron−1 steradian−1 (10−3 δλ/λ)−1 for an X-ray wavelength in the neighbourhood of 1.5 Å. By comparison, from equation (4.2.1.2), a copper-target tube operated at 40 kV produces about 5 × 10−4 Kα photons electron−1 steradian−1. In spite of this shortcoming by a factor of about 45, laboratory XAFS experiments are sufficiently common to have merited at least one specialized conference (Stern, 1980; see also Tohji, Udagawa, Kawasak & Masuda, 1983; Sakurai, 1993; Sakurai & Sakurai, 1994).

The use of continuous radiation for diffraction experiments is complicated by the fact that the radiation is polarized. The degree of polarization may be defined as where and are the intensities of radiation with the electric vector parallel and perpendicular to the plane containing the incident electrons and the direction of the emitted photons. For an angle of π/2 between the electrons and the emitted beam, p varies smoothly through the spectrum; it is negative for the softest radiation, approximately zero at and reaches values between +0.7 and +0.9 near the Duane–Hunt limit (Kirkpatrick & Wiedmann, 1945). Since practical use of white radiation is likely to be in the vicinity of , the effect is not a large one.

It should also be noted that the spatial distribution of the white spectrum, even after correction for absorption in the target, is not isotropic. The intensity has a maximum at about 50° to the electron beam and non-zero minima at 0 and 180° to that beam (Stephenson, 1957).

### References

Cohen, E. R. & Taylor, B. N. (1987). The 1986 adjustment of the fundamental physical constants. Rev. Mod. Phys. 89, 1121–1148.Google Scholar
Kirkpatrick, P. & Wiedmann, L. (1945). Theoretical continuous X-ray energy and polarization. Phys. Rev. 67, 321–339.Google Scholar
Kramers, H. A. (1923). On the theory of X-ray absorption and of the continuous X-ray spectrum. Philos. Mag. 46, 836–871.Google Scholar
Kulenkampff, H. & Schmidt, L. (1943). Die Energieverteilung im Spektrum der Röntgen Bremsstrahlung. Ann. Phys. (Leipzig), 43, 494–512.Google Scholar
Sakurai, K. (1993). High-intensity X-ray line-focal spot for laboratory EXAFS measurements. Rev. Sci. Instrum. 64, 267–268.Google Scholar
Sakurai, K. & Sakurai, H. (1994). Comment on high intensity low tube-voltage X-ray source for laboratory EXAFS measurements. Rev. Sci. Instrum. 65, 2417–2418.Google Scholar
Stephenson, S. T. (1957). The continuous X-ray spectrum. Handbuch der Physik. XXX: X-rays, edited by S. Flügge, pp. 337–370. Berlin: Springer.Google Scholar
Stern, E. A. (1980). Editor. Laboratory EXAFS facilities 1980. AIP Conf. Proc. No. 64.Google Scholar
Tohji, K., Udagawa, Y., Kawasaki, T. & Masuda, K. (1983). Laboratory EXAFS spectrometer with a bent-crystal, a solid-state detector, and a fast detection system. Rev. Sci. Instrum. 54, 1482–1487.Google Scholar


text:
# Equilateral triangle

912pages on
this wiki

Equilateral triangle

Edges and vertices3
Schläfli symbols{3}
Coxeter–Dynkin diagrams
Symmetry groupDihedral (D3)
Internal angle
(degrees)
60°

In geometry, an equilateral triangle is a triangle in which all three sides are equal. In traditional or Euclidean geometry, equilateral triangles are also equiangular; that is, all three internal angles are also congruent to each other and are each 60°. They are regular polygons, and can therefore also be referred to as regular triangles.

## PropertiesEdit

Assuming the lengths of the sides of the equilateral triangle are $a\,\!$, we can determine that:

• The area is $A=a^2\frac{\sqrt{3}}{4}$
• The perimeter is $p=3a\,\!$

These formulas can be derived using the Pythagorean theorem.

An equilateral triangle is the most symmetrical triangle, having 3 lines of reflection and rotational symmetry of order 3 about its center. Its symmetry group is the dihedral group of order 6 D3.

Equilateral triangles are found in many other geometric constructs. The intersection of circles whose centers are a radius width apart is a pair of equilateral arches, each of which can be inscribed with an equilateral triangle. They form faces of regular and uniform polyhedra. Three of the five Platonic solids are composed of equilateral triangles. In particular, the regular tetrahedron has four equilateral triangles for faces and can be considered the three dimensional analogue of the shape. The plane can be tiled using equilateral triangles giving the triangular tiling.

A result finding an equilateral triangle associated to any triangle is Morley's trisector theorem.

## Geometric constructionEdit

An equilateral triangle is easily constructed using a compass. Draw a straight line, and place the point of the compass on one end of the line, and swing an arc from that point past halfway of the line segment. Repeat with the other side of the line. Finally, connect the point where the two arcs intersect with each end of the line segment

Alternate method: Draw a circle with radius r, place the point of the compass on the circle and draw another circle with the same radius. The two circles will intersect in two points. An equilateral triangle can be constructed by taking the two centers of the circles and either of the points of intersection.

## Almost-equilateral Heronian trianglesEdit

A Heronian triangle is a triangle with rational sides, area and inradius. Since the area of an equilateral triangle with rational sides is an irrational number, no equilateral triangle is Heronian. However, there is a unique sequence of Heronian triangles that are "almost equilateral" because the three sides, expressed as integers, are of the form n − 1, n, n + 1. The first few examples of these almost-equilateral triangles are set forth in the following table.

Side length Area Inradius
n − 1 n n + 1
3 4 5 6 1
13 14 15 84 4
51 52 53 1170 15
193 194 195 16296 56
723 724 725 226974 209

Subsequent values of n can be found by multiplying the last known value by 4, then subtracting the next to the last one (52 = 4 × 14 − 4, 194 = 4 × 52 − 14, etc), as expressed in

$q_n = 4q_{n-1} - q_{n-2}.\,\!$

This sequence A003500 can also be generated from the solutions to the Pell equation x² − 3y² = 1, which can in turn be derived from the regular continued fraction expansion for √3. [1]

## In culture and societyEdit

Equilateral triangles have frequently appeared in man made constructions:

## ConstructionEdit

• Use a compass to draw a circle
• Draw a pair of perpendicular lines that meet at the center of the circle. One horizontal and one verticle.
• Make a circle cetered at the spot where the bottom of the verticle line intersects the original circle's curve. Make sure the curve of the new circle goes through the center of the original circle.
• Make a line AB where A and B are where the curve of the new circle meets the curve of the original circle.
• Make lines AC and BC where  C is the spot where the top of the verticle line meets the original circle's curve.
• Erase everything but triangle ABC.

## ReferencesEdit

1. Takeaki Murasaki (2004), On the Heronian Triple (n+1, n, n−1), Sci. Rep. Fac. Educ., Gunma Univ. 52, 9-15.


text:
How to draw tile edges when you don't know where they're going.

This is more of an art question than a programing one but still game development.

I have a tile engine that makes a map randomly from tiles, each tile is a square 3x3 grid. The problem is that while the elements on each tile work well together such as having forests along the top three squares and grass on the other 6, the engine could put the forrests against anything such as rivers, grass, mountains or more forest.

how can i draw the edges of the tiles so the look good no matter what they are places against ?

You are should look into auto tiling. There are two prevailing methods as to how to implement this.

The first assigns a bit value to each tile type, tile_bit, and a bit value to each tile surrounding the location, loc_bit. Multiplying tile_bit by loc_bit will give a unique value for every combination of surrounding locations. This very simple to do with only two or three types of tiles. More details here : http://www.saltgames.com/2010/a-bitwise-method-for-applying-tilemaps/

The second method instead divides each tile up into for sections. These four sections each get a bit value based on their neighbor tile's value. Then you can draw the appropriate tile that matches up with the four bit values. More details here : http://www.codeproject.com/Articles/106884/Implementing-Auto-tiling-Functionality-in-a-Tile-M

Both of these articles assume that you have every combination of transition drawn. This gives you greater control over the look of your map. However, this can get extremely tedious if you have a many tile types. Instead you can dynamics generate edges.

These articles talk about just having generic predrawn edge tiles and assembling them based on your map: http://www.xnaresources.com/default.asp?page=Tutorial:TileEngineSeries:6

http://www.gamedev.net/page/resources/_/technical/game-programming/tilemap-based-game-techniques-handling-terrai-r934

This other topic also talks about even generating edges instead of drawing them. However I think this will just lead to a very crude edge most of the time: Map tile terrain transitions with 3-4 different types

It depends I suppose on how the different regions are rendered? (I.e, Do you generate a forest, then a river, then grass and glue them together? If that is the case you could manually provide the tiles that should go in between? A nice tile that indicates a tree next to a river and so on?

You could even provide a small region in between that interpolates between region A & B and makes the transition smooth. Let the forest end and provide some grass until the river region starts.


text:
# $C$ is an uncountable set. Show that $B(x,\epsilon)\cap C$ is uncountable.

Suppose $X$ is a separable metric space, and $C$ is an uncountable subset of $X$. Prove that there is a point $x \in C$ such that for each $\epsilon>0$, $B(x;\epsilon)\cap C$ is uncountable.

I missed first couple of classes so I don't really know what is the general strategy to prove something is uncountable (or countable)? The only thing that comes to my mind is that, since $X$ is separable, then for any open cover, it has a countable sub cover (may not be finite though), and I was hoping I could get a contradiction out of here?

I tried contradiction, but that would means for every $x\in C$, there is a corresponding $\epsilon(x)$ such that $B(x;\epsilon(x))\cap C$ is countable. But I cannot deduce that $C$ is then countable because that intersection may be empty.

Any idea?

Even more is true, and not much harder to prove.

HINT: Let $\mathscr{B}$ be a countable base for $X$. Let $$\mathscr{B}_0=\{B\in\mathscr{B}:B\cap C\text{ is countable}\}\;.$$

Let $C_0=C\setminus\bigcup\mathscr{B}_0$. Show that $C_0$ is actually an uncountable set such that $B\cap C_0$ is uncountable for every open nbhd $B$ of every $x\in C_0$.

• OK, so I'll have to look up what a countable base is, too... – 3x89g2 Apr 21 '15 at 5:14
• @Misakov: As long as you know what a base for a topology is, you're okay: it's just a base (or, as some prefer, a basis) that contains only countably many sets. For example, the family of open intervals with rational endpoints is a countable base for the usual topology on $\Bbb R$. – Brian M. Scott Apr 21 '15 at 5:23
• I don't think we ever talked about what a base is, but I'll look it up. Thanks! – 3x89g2 Apr 21 '15 at 6:40
• @Misakov: Short definition: if $\tau$ is a topology on a set $X$ (i.e., the collection of all open sets), then a family $\mathscr{B}$ of subsets of $X$ is a base for $\tau$ if $\mathscr{B}\subseteq\tau$, and for each $U\in\tau$ there is a $\mathscr{B}_U\subseteq\mathscr{B}$ such that $U=\bigcup\mathscr{B}_U$. One of the basic(!) theorems of metric spaces is that every separable metric space has a countable base: if $D$ is a countable dense subset of $X$, $\{B(x,r):x\in D\text{ and }r\in\Bbb Q^+\}$ is a countable base for the the topology on $X$. – Brian M. Scott Apr 21 '15 at 6:55

Initially I was proving existence of $x\in X$ since I misread the problem this way. Now I made the small changes to prove existence of such $x\in C$.

Let $\{x_n\}\subset C$ be dense in $C$. Assume no such $x\in C$ exists. Then there are $r_n>0$, largest, such that $B(x_n,r_n)\cap C$ is countable because, in particular, none of the $x_n$ is a candidate for $x$.

But $\bigcup_n B(x_n,r_n)\supset C$. This implies that $C$ is countable.

Assume $y\in C\setminus\bigcup_n B(x_n,r_n)$ (in particular $y\neq x_n$ for all $n$). Let $r>0$ be such that $B(y,r)\cap C$ is countable, which by the assumption exists. Let $x_n$ be inside $B(y,r/4)$. Then $r_n\leq d(y,x_n)<r/4$. But $B(x_n,3r/4)\subset B(y,r)$ and therefore $B(x_n,3r/4)\cap C$ is countable contradicting the choice of $r_n$ (since $3r/4>r/4>r_n$).

Therefore $\bigcup_n B(x_n,r_n)\supset C$ as claimed.

Existence of the largest $r$:

Under the assumptions that $C$ is uncountable, and that $x$ is such that there is $\epsilon>0$ such that $B(x,\epsilon)\cap C$ is countable. Let $r:=\sup\{s\in\mathbb{R}^+:\ B(x,s)\cap C\text{ is countable}\}$. Since $B(x,r)=\bigcup_{n\in\mathbb{N}}B(x,r-1/n)$ we have that $B(x,r)\cap C=\bigcup_{n\in\mathbb{N}}\left(B(x,r-1/n)\cap C\right)$ is countable.

• Okay, you're now taking advantage of the fact that separability is hereditary in metric spaces (which is not true for arbitrary topological spaces). – user642796 Apr 21 '15 at 6:09
• @ArthurFischer ... and which is a pretty trivial fact to prove. – Alamos Apr 21 '15 at 6:11


text:
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
You are viewing an older version of this Concept. Go to the latest version.

# Graphs of Rational Functions

## Graphs of functions with x in the denominator of a fraction

0%
Progress
Practice Graphs of Rational Functions
Progress
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Graphs of Rational Functions

Can you use your graphing calculator to help you sketch the graph of the rational function? Does this function have any zeros or asymptotes?

y=4x2+1\begin{align*}y=\frac{4}{x^2+1}\end{align*}

### Guidance

A rational function is the quotient of two polynomial functions. In general,

f(x)=A(x)B(x)\begin{align*}f(x)=\frac{A(x)}{B(x)}\end{align*}

where A\begin{align*}A\end{align*} and B\begin{align*}B\end{align*} are polynomials and B0\begin{align*}B \ne 0\end{align*}. You can use your graphing calculator to graph a rational function and look for important features. Consider the function y=xx3\begin{align*}y=\frac{x}{x-3}\end{align*}.

Notice that the function has two pieces. In between those two pieces are the asymptotes.

• A vertical asymptote occurs at the x\begin{align*}x\end{align*} value(s) that cause the denominator of the function to be equal to zero (which is undefined). This function has a vertical asymptote at x=3\begin{align*}x=3\end{align*}.
• A horizontal asymptote occurs at the y\begin{align*}y\end{align*} value(s) that cause the denominator of the function to be zero if the function is rewritten and solved for x\begin{align*}x\end{align*} instead of y\begin{align*}y\end{align*}. This is what it looks like to solve for x\begin{align*}x\end{align*}:
Thus, \begin{align*}y \ne 1\end{align*} and this function has a horizontal asymptote at \begin{align*}y = 1\end{align*}.

The image below shows the graph with the asymptotes drawn in and labelled. For rational functions, the asymptotes represent the lines that the function will approach but never touch.

It is also important to note the x-intercepts (zeros) of the function. The zeros of the function will be the values for \begin{align*}x\end{align*} that cause the numerator, but not also the denominator, to be equal to zero.

#### Example A

Use technology to sketch the graph of the rational function. Find all zeros and asymptotes and label those on your sketch.

\begin{align*}y=\frac{1}{x^2-9}=\frac{1}{(x+3)(x-3)}\end{align*}

Solution: Here is the sketch from the calculator:

It can sometimes be hard to interpret what you see on the graphing calculator screen. Use algebra to find the asymptotes and zeros and sketch those first.

• There are no zeros for the numerator. Therefore, there are no x-intercepts for the function.
• The zeros of the denominator are 3 and –3. This means that there are two vertical asymptotes. One vertical asymptote is the line \begin{align*}x = 3\end{align*} and the other is the line \begin{align*}x = -3\end{align*}.
• Another way to determine a horizontal asymptote besides solving the equation for \begin{align*}x\end{align*} is to look at the degrees of the numerators and denominators. The degree is the highest exponent. The degree of the numerator is 0 and the degree of the denominator is 2. In general, if the degree of the numerator is less than the degree of the denominator, there will be a horizontal asymptote at \begin{align*}y=0\end{align*}.

Once you have sketched the asymptotes, use the table and/or graph from the graphing calculator to decide what the rest of the graph looks like. Here is the graph with the asymptotes labelled.

#### Example B

Use technology to sketch the graph of the rational function. Find all zeros and asymptotes and label those on your sketch.

\begin{align*}y=-\frac{1}{x}\end{align*}

Solution: Here is the sketch from the calculator:

Use algebra to find the asymptotes and zeros and sketch those first.

• There are no zeros for the numerator. Therefore, there are no x-intercepts for the function.
• The zero of the denominator is 0. This means that there is one vertical asymptote, the line \begin{align*}x = 0\end{align*}.
• The horizontal asymptote is \begin{align*}y = 0\end{align*}. You can use algebra to solve the equation for \begin{align*}x\end{align*} and look for the values of \begin{align*}y\end{align*} that will cause the denominator to be equal to zero:

Once you have sketched the asymptotes, use the table and/or graph from the graphing calculator to decide what the rest of the graph looks like. Here is the graph with the asymptotes labelled.

#### Example C

Use technology to sketch the graph of the rational function. Find all zeros and asymptotes and label those on your sketch.

\begin{align*}y=\frac{x+2}{x-3}\end{align*}

Solution: Here is the sketch from the calculator:

Use algebra to find the asymptotes and zeros and sketch those first.

• The zero of the numerator is –2 so the zero (x-intercept) of the function is (–2, 0).
• The zero of the denominator is 3. This means that there is one vertical asymptote, the line \begin{align*}x = 3\end{align*}.
• The horizontal asymptote is \begin{align*}y = 1\end{align*}. You can use algebra to solve the equation for \begin{align*}x\end{align*} and look for the values of \begin{align*}y\end{align*} that will cause the denominator to be equal to zero:

Once you have sketched the asymptotes, use the table and/or graph from the graphing calculator to decide what the rest of the graph looks like. Here is the graph with the asymptotes labelled.

#### Concept Problem Revisited

\begin{align*}y=\frac{4}{x^2+1}\end{align*}

Here is a sketch of the function:

• There are no zeros for this function since there are no zeros for the numerator. The graph does not cross the \begin{align*}x\end{align*}-axis.
• There are no zeros for the denominator. Therefore, there are no vertical asymptotes.
• Because the degree of the numerator is less than the degree of the denominator, there is a horizontal asymptote at \begin{align*}y=0\end{align*}.

### Vocabulary

Rational Function
A rational function is the quotient of two polynomial functions. In general,

\begin{align*}f(x)=\frac{A(x)}{B(x)}\end{align*}

where \begin{align*}A\end{align*} and \begin{align*}B\end{align*} are polynomials and \begin{align*}B \ne 0\end{align*}.

### Guided Practice

1. Sketch the graph of the rational function: \begin{align*}y=\frac{x+1}{x}\end{align*}

2. Sketch the graph of the rational function: \begin{align*}y=\frac{x^2}{x-3}\end{align*}

3. Find the asymptotes of the function: \begin{align*}y=\frac{1}{x^2-16}\end{align*}

1. The zero of the denominator is 0. This means that there will be a vertical asymptote at \begin{align*}x=0\end{align*} (the y-axis). Because the degree of the denominator and numerator are the same, you can solve the equation for \begin{align*}x\end{align*} and get \begin{align*}x=\frac{1}{y-1}\end{align*}. The zero of the denominator is now 1. This means there is a horizontal asymptote at \begin{align*}y=1\end{align*}. The numerator has a zero at -1, so there is an x-intercept at -1.

2. The zero of the denominator is 3. This means that there will be a vertical asymptote at \begin{align*}x=3\end{align*}. Because the degree of the numerator is greater than the degree of the denominator, there are no horizontal asymptotes. The zero of the numerator is 0, so there is an x-intercept at 0.

3. The zeros of the denominator are 4 and –4. This means that there will be two vertical asymptotes. One vertical asymptote will be the line \begin{align*}x = 4\end{align*} and the other will be the line \begin{align*}x = -4\end{align*}. Because the degree of the numerator is less than the degree of the denominator, there is a horizontal asymptote at \begin{align*}y=0\end{align*}. Though you did not need to sketch the graph, here is the graph of the function:

### Practice

Sketch the graph of each of the following rational functions.

1. \begin{align*}y=\frac{2}{x+3}\end{align*}
2. \begin{align*}y=\frac{x}{x-1}\end{align*}
3. \begin{align*}y=\frac{1}{x^2-4}\end{align*}
4. \begin{align*}y=\frac{x+2}{x}\end{align*}
5. \begin{align*}y=\frac{1}{x^2+2}\end{align*}
6. \begin{align*}y=\frac{x}{x+2}\end{align*}
7. \begin{align*}y=\frac{1}{x^2-x-12}\end{align*}
8. \begin{align*}y=\frac{x-1}{x+3}\end{align*}
9. \begin{align*}y=\frac{x-1}{x+4}\end{align*}
10. \begin{align*}y=\frac{5}{x^2+1}\end{align*}

Without graphing the following rational functions, state what you know about their asymptotes and zeros.

1. \begin{align*}y=\frac{1}{x^2-x-2}\end{align*}
2. \begin{align*}y=-\frac{2}{x-4}\end{align*}
3. \begin{align*}y=-\frac{2}{x^2+1}\end{align*}
4. \begin{align*}y=\frac{6}{x^2+1}\end{align*}
5. \begin{align*}y=\frac{x-1}{x+3}\end{align*}

### Vocabulary Language: English

compression

compression

A stretch or compression is a function transformation that makes a graph narrower or wider, without translating it horizontally or vertically.
Function

Function

A function is a relation where there is only one output for every input. In other words, for every value of $x$, there is only one value for $y$.
Horizontal Asymptote

Horizontal Asymptote

A horizontal asymptote is a horizontal line that indicates where a function flattens out as the independent variable gets very large or very small. A function may touch or pass through a horizontal asymptote.
Oblique Asymptote

Oblique Asymptote

An oblique asymptote is a diagonal line marking a specific range of values toward which the graph of a function may approach, but generally never reach. An oblique asymptote exists when the numerator of the function is exactly one degree greater than the denominator. An oblique asymptote may be found through long division.
Polynomial Function

Polynomial Function

A polynomial function is a function defined by an expression with at least one algebraic term.
Reflection

Reflection

Reflections are transformations that result in a "mirror image" of a parent function. They are caused by differing signs between parent and child functions.
shift

shift

A shift, also known as a translation or a slide, is a transformation applied to the graph of a function that does not change the shape or orientation of the graph, only the location of the graph.
shifts

shifts

A shift, also known as a translation or a slide, is a transformation applied to the graph of a function that does not change the shape or orientation of the graph, only the location of the graph.
Slant Asymptote

Slant Asymptote

A slant asymptote is a diagonal line marking a specific range of values toward which the graph of a function may approach, but will never reach. A slant asymptote exists when the numerator of the function is exactly one degree greater than the denominator. A slant asymptote may be found through long division.
stretch

stretch

A stretch or compression is a function transformation that makes a graph narrower or wider.
Transformations

Transformations

Transformations are used to change the graph of a parent function into the graph of a more complex function.
Vertical Asymptote

Vertical Asymptote

A vertical asymptote is a vertical line marking a specific value toward which the graph of a function may approach, but will never reach.


text:
mersenneforum.org > Data processed dc and tc posts
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 2019-03-06, 21:40 #1 Mark Rose     "/X\(‘-‘)/X\" Jan 2013 B8716 Posts processed dc and tc posts You may wish to not delete claims immediately lest they not be completed. It happens from time to time.
 2019-03-07, 14:56 #2 Uncwilly 6809 > 6502     """"""""""""""""""" Aug 2003 101×103 Posts 10,739 Posts processed dc and tc posts a collection thread for the posted cleared from the DC and TC thread.
2019-05-07, 10:18   #3
LaurV
Romulan Interpreter

"name field"
Jun 2011
Thailand

2·5,059 Posts

Quote:
 Originally Posted by Uncwilly You are a supermod. Next time you can put it on the list yourself. GP2 and ATH have been editing the lists too.
Which is a bit frown about, in this part of the lake... I would not recommend that all of us who, due to benevolence of the admin, appeared like the mushrooms after the rain, we start now modifying the assignment lists and so on. For many projects we have no much idea what the guyz and galz are doing there... Just my two pence... Job of the mod is to be arbiter, intervene if some sheep happens, but let the projects managers do their job... they know better what to do and what is requested, etc...

Last fiddled with by LaurV on 2019-05-07 at 10:19

2019-05-07, 16:10   #4
Uncwilly
6809 > 6502

"""""""""""""""""""
Aug 2003
101×103 Posts

10,739 Posts

Quote:
 Originally Posted by LaurV Which is a bit frown about, in this part of the lake... I would not recommend that all of us who, due to benevolence of the admin, appeared like the mushrooms after the rain, we start now modifying the assignment lists and so on.
Sure, but I think that I am ok with Luigi doing it. George, Alex, and some others I would have no problem with.

You, on other hand, must beg for assignments and pay a steep fee.

2019-05-09, 13:45   #5
LaurV
Romulan Interpreter

"name field"
Jun 2011
Thailand

278616 Posts

Quote:
 Originally Posted by Uncwilly You, on other hand, must beg for assignments and pay a steep fee.
Haha, ok, ok, you won't see me here, at least, no until this ends (ETA ~20 days, over 90% done, LL+DC in parallel).

2019-05-17, 15:04   #6
kriesel

"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest

1ABB16 Posts

Quote:
 Originally Posted by ATH You should register them in Primenet as well, everyone might not read this thread.
I would if I could. I checked they are not assigned to anyone else, via exponent details, individually, then tried to manually reserve them, individually, and got error messages.

Edit: these are being run in CUDALucas on two fast gpus. Should be done in a few days. The method I described above usually works with assignments for gpus. The error message that occurs for manual checkout of these is:
Code:
Error code: 40

Error text: No assignment available meeting CPU, program code and work preference requirements
Attached Thumbnails

Last fiddled with by kriesel on 2019-05-17 at 15:56

2019-05-17, 15:23   #7
Uncwilly
6809 > 6502

"""""""""""""""""""
Aug 2003
101×103 Posts

10,739 Posts

Quote:
 Originally Posted by ATH You should register them in Primenet as well, everyone might not read this thread.
Quote:
 Originally Posted by kriesel I would if I could. I checked they are not assigned to anyone else, via exponent details, individually, then tried to manually reserve them, individually, and got error messages.
From the first post in this thread.
Quote:
 Originally Posted by Uncwilly The easy way to get some assignments (with Prime95):Grab the lines of the exponents that you will do. Stop Prime95. Edit your worktodo.txt, inserting the new lines, save it, and close the editor. Restart Prime95 Go to the menu Advanced-> Manual Communication Make sure "Contact PrimeNet server now" and "Send new expected completion dates to server" are checked. Hit OK After Prime95 is done communicating, look and see if any of the new exponents don't have an Assignment ID (because they were already assigned.) If they don't, stop Prime95, reopen the worktodo.txt, remove those lines, save, then restart Prime95
That should do it. mprime probably has a similar technique.

2019-05-17, 16:41   #8
kriesel

"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest

152738 Posts

Quote:
 Originally Posted by Uncwilly From the first post in this thread. That should do it.
It does indeed. Although for gpu usage, and my crappy slow ISP, there are additional steps:
repeat the manual communication in prime95 as needed until communication completes successfully instead of giving CURL library errors;
stop prime95, cut the doublecheck assignments with new AIDs from the worktodo.txt of the prime95 instance used to reserve them, restart prime95;
stop the CUDALucas session, paste the doublecheck assignments with newly issued AIDs into the gpu worktodo.txt, and restart the CUDALucas session;
update local assignment tracking log.

Effective, though a bit tedious compared to my usual manual checkout method when it works.
I wonder if primenet will get confused by these 4 assignments completing and being reported manually in a few days, after being reserved on an old i3 laptop that could not finish one of the assignments in 3 months, and having been removed from the worktodo.txt of the regularly reporting i3 system.

2019-05-23, 15:14   #9
kriesel

"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest

3×2,281 Posts
methods for gpus

Quote:
 Originally Posted by Uncwilly The easy way to get some assignments (with Prime95): Stop Prime95. Edit your worktodo.txt, inserting the new lines, save it, and close the editor. Restart Prime95 Go to the menu Advanced-> Manual Communication Make sure "Contact PrimeNet server now" and "Send new expected completion dates to server" are checked. Hit OK After Prime95 is done communicating, look and see if any of the new exponents don't have an Assignment ID (because they were already assigned.) If they don't, stop Prime95, reopen the worktodo.txt, remove those lines, save, then restart Prime95 If someone can post tips for the GPU's that would help.
The following assumes gpu application(s) CUDALucas etc. are already running and are not near completion of the current work item(s). Add gpu application stops and starts as appropriate if that assumption is not valid.
• Post which DC or TC candidates you're taking.
• Grab the lines of the exponents that you will do.
• Edit your gpu(s)' worktodo.txt, inserting the new lines, save it, and close the editor.
continue, with either Method 1:
• Go to https://www.mersenne.org/manual_assignment/
• Set preferred work type to Double check ll tests
• set Optional exponent range fields to one less, and one more than the exponent in the first worktodo entry added that does not yet have an AID
• click Get Assignments
• If it succeeds, copy the assignment with AID, and overwrite the corresponding gpu worktodo.txt entry that had no AID, with it, and save.
• If it fails, likely it is with error 40, no assignment available. Use a prime95 session as uncwilly described, to get the assignment, then copy that to the gpu worktodo.txt, again, overwriting the no-AID version entry.
• Repeat previous steps of method 1 for each DC or TC. Depending on how many you're doing, it can get tedious.
or continue with Method 2:
• reserve as described by uncwilly in a prime95 session.
• Copy the new assignments with AIDs from the prime95 worktodo file to the gpu(s) worktodo file(s), and save, then close the gpu worktodo file(s).
• Move the new assignments in the prime95 worktodo file, to the end of the prime95 worktodo.txt file behind some ongoing lengthy work intended for the cpu, save the prime95 worktdo file, stop and restart prime95, and check with Test, Status, that the assignments order in effect on the cpu is as intended.
For either method, wrapup:
• If you're unable to get a reservation for any exponent you posted you were taking, by either method, edit your initial post to exclude it. If the edit period already expired, post which you were unable to reserve in a new message.
Optionally: if any gpu assignment is about to expire, because manual extension does not work, try
• Open the prime95 worktodo file, temporarily put its worktodo entry at the front of the prime95 work list, save prime95 worktodo file, stop and restart prime95 computation to run at least 0.1% of completion.
• Make sure "Contact PrimeNet server now" and "Send new expected completion dates to server" are checked. Click OK, and wait for prime95 to finish talking with the primenet server
• Stop prime95, move the extended assignment to the end of the prime95 worktodo file, save and close prime95 worktodo file, restart prime95, check with Test, Status.
• After the gpu assignment completes, remove the corresponding assignment entry from the prime95 worktodo file, save and close the prime95 worktodo file, stop computation and resume in prime95, delete the prime95 save file and backup files for the few dozen iterations performed on the cpu for the completed gpu assignment.
A variant of this extension method seemed to work, in which I commented out the prime95 assignment copy and deleted the exponent's save file and bu files, on an exponent I had that was going to expire yesterday. It gained 10 weeks in the assignments report. But I see today that it is marked expired for me and assigned to someone else. Its chance of working to maintain the extension may be better if it remains in the prime95 instance's worktodo as an active entry that will be included in periodic automatic reporting of progress. By putting it at the end of the work list, behind a lengthy DC I'm running on the prime95 instance, it should report status periodically from there, and yet not use up cpu cycles partially duplicating gpu work.

A side effect of using this method for extension is it changes the assignment from manual to the prime95 instance used to extend it. At least that is what the assignments page shows.

Because this assignment or extension method might confuse Primenet about what's happening on the prime95 instance, and any expirations may count against that instance/system in getting new assignments, consider using an old slow cpu for making these reservations or extensions, where it will make little difference regarding assignment of future work via primenet.

Last fiddled with by Uncwilly on 2019-06-10 at 13:48 Reason: done at the request of the op

2019-05-30, 08:54   #10
GP2

Sep 2003

13×199 Posts

Quote:
 Originally Posted by Uncwilly GP2 can you handle the honours of posting the worktodo lines? I have no clue about them.
Lines suitable for doing a double-check with mprime would be:
Code:
PRP=1,2,87667499,-1,76,0,3,4
PRP=1,2,87595463,-1,76,0,3,4
I assume gpuOwL can understand the same format.

2019-06-10, 12:44   #11
kriesel

"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest

3·2,281 Posts
methods for gpus (amended)

Quote:
Substitute "at least 0.1% of completion" in place of the phrase highlighted in red.

Last fiddled with by kriesel on 2019-06-10 at 12:47

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text:
Publication
Title
Magnetization measurements and Ginzburg-Landau simulations of micron-size $\beta$-tin samples : evidence for an unusual critical behavior of mesoscopic type-I superconductors
Author
Abstract
 We describe investigations of the largely unexplored field of mesoscopic type-I superconductors. Micromagnetometry and 3D Ginzburg-Landau simulations of our single crystal β-tin samples in this regime reveal size- and temperature-dependent supercritical fields whose behavior is radically different from the bulk critical field HcB. We find that complete suppression of the intermediate state in medium-size samples can result in a surprising reduction of the critical field significantly below HcB. We also reveal an evolution of the superconducting-to-normal phase transition from the expected irreversible first order at low temperatures through the previously unobserved reversible first-order to a second-order transition close to Tc, where the critical field can be many times larger than HcB. Finally, we have identified striking correlations between the mesoscopic Hc3 for nucleation of surface superconductivity and the thermodynamic Hc near Tc. All these observations are entirely unexpected in the conventional type-I picture.
Language
English
Source (journal)
Physical review letters. - New York, N.Y.
Publication
New York, N.Y. : 2012
ISSN
0031-9007
Volume/pages
109:19(2012), 5 p.
Article Reference
197003
ISI
000310853100017
Medium
E-only publicatie
Full text (Publisher's DOI)
Full text (open access)
UAntwerpen
 Faculty/Department Research group Publication type Subject Affiliation Publications with a UAntwerp address


text:
ISSN 1816-9791 (Print)
ISSN 2541-9005 (Online)

# short sums of characters

## Distribution of Values of Dirichlet Characters in the Sequence of Shifted Primes

The new estimate for the sum of the values of a primitive Dirichlet character modulo an integer q has been obtained over the sequence of shifted primes p − l, (l, q) = 1, p ≤ x. This estimate is nontrivial for x ≥ q 5 6 +ε and refines the estimate obtained by J. B. Friedlander, K. Gong, I. E. Shparlinskii. Their estimate holds provided that x ≥ q 8/9+ε


text:
# Faulty Combinatorial Reasoning?

I have 10 books, 4 of which are biographies while the remaining 6 are novels. Suppose I have to choose 4 total books with AT LEAST 2 of the 4 books being biographies. How many different combinations of choosing 4 books in such a way are there?

The following line of reasoning is faulty, but I can't figure out why:

First we figure out how many ways there are of choosing 2 biographies from 4. Then we multiply this by the number of ways there are of choosing 2 of any of the remaining books from 8. This way we will ensure that we get at least two biographies (perhaps more) when we enumerate the choices. Then we have:

1. BIOGRAPHIES: There are (4*3)/2! choices for the two biographies (we divide by 2! since the order in which the two biographies are chosen doesn't matter).
2. REMAINING BOOKS: There are now 8 books left (6 novels, 2 biographies), which can be chosen in any order. This leaves us with (8*7)/2! choices.
3. Overall we have [(4*3)/2!]*[(8*7)/2!] = 168 total choices.

Where did I go wrong?

-
How could I adjust for the over-counting I did here? (Rather than constructing the answer of 115 by adding together the discrete cases of choosing 2 bios, 3 bios, and 4 bios)? –  George Sep 4 '12 at 22:15

In your reasoning, you are counting some cases several times. For example, if you take the biographies $B_1$ and $B_2$ as your mandatory biographies and take $B_3$ and $B_4$ as the two other ones, or if you take $B_£$ and $B_4$ as the mandatory ones and $B_1$ and $B_2$ as the other books, it is the same choice of $4$ books, but it will be counted twice.

To solve the problem:

1. the number of ways of choosing $4$ books is $A_4 = \frac{10!}{4!\times6!} = 210$
2. the number of ways of choosing $4$ books with no biographies is $B_0 = \frac{6!}{4!\times2!} = 15$
3. the number of ways of choosing $4$ books with exactly $1$ biographies is $B_1 = 4\times\frac{6!}{3!\times3!} = 80$ (you pick $1$ biography amongst $4$ and then choose $3$ novels).
4. the number of ways of choosing at least two biographies is $B_2^+ = A_4 - (B_1+B_2) = 115$.
-

Your enumeration would count the following possibilities (and many other similar examples) separately:

1. First choose Bio1 and Bio2, then choose Bio3 and Novel1.
2. First choose Bio1 and Bio3, then choose Bio2 and Novel1.

Notice the only difference is that I switched when Bio2 and Bio3 would be chosen.

In general, it is best to split "at least"-type questions into several instances of "exactly". Here, you should try to count the ways to get exactly two, exactly three, and exactly 4 biographies as separate problems, then add up all the cases.

-

When you do this you sum two times the cases when you have more than two biographies books. Then need to do this

2 BIOGRAPHIES books: $\dfrac{4\cdot3}{2!} \cdot \dfrac{6\cdot5}{2!}=90$

3 BIOGRAPHIES books: $\dfrac{4\cdot 3 \cdot 2}{32!} \cdot 6 =24$

4 BIOGRAPHIES books: 1

then the result is 115.

-

Suppose the biographies are of $A$, $B$, $C$, and $D$. Among the ways you counted when initially you chose two biographies, there were the biographies of $A$ and of $B$. Among the choices you counted when you chose two more books was the biography of $C$ and novel $N$. So among the choices counted in your product was choosing $A$ and $B$, then choosing $C$ and $N$. That made a contribution of $1$ to your $168$.

But among the choices you counted when you chose two biographies, there were the biographies of $A$ and $C$. And among your "two more" choices, there was the biography of $B$ and novel $N$. So among the choices counted in your product, there was the choice of $A$ and $C$, and then of $B$ and $N$. That made another contribution of $1$ to your $168$.

Both of these ways of choosing end us up with $A$, $B$, $C$, and $N$. So does choosing $B$ and $C$ on the initial choice, and $A$ and $N$ on the next. Still another contribution of $1$ to your $168$.

So your product counts the set $\{A, B, C, N\}$ three times. This it does for every combination of three biographies and one novel. It also overcounts the set $\{A, B, C, D\}$.

One could adjust for the overcount. In some problems that is a useful strategy. Here it takes some care.

But a simple way to solve the problem is to count separately the ways to choose two bios, two novels; three bios, one novel; four bios, no novels and add up.

-


text:
# Statistics- Normal Distribution Question

#### mattaj

##### New member
If there is no change in interest rate, the profit on a stock after 1 year has normal distribution with mean 500 and standard deviation 100. However, if interest rates are raised in this period, then the profit from investment is normally distributed with a mean 300 and a standard deviation of 80.
If there is 30% chance of interest rate to be raised and a 70% chance of remaining unchanged, find the probability, that the profit exceeds 400.

Hi - im really stuck with this question- i understand how to solve normal distributions but im unsure of how to go about finding the answer to this question.

#### mattaj

##### New member
is this definately the way to do it?


text:
# zbMATH — the first resource for mathematics

The essential norms and spectra of composition operators on $$H^ \infty$$. (English) Zbl 1053.47022
Composition operators, acting on various analytic function spaces on the unit disk of the complex plane, have been extensively studied over the last three decades. Most of such studies have focused on well-known function spaces such as the Hardy and Bergman spaces. However, composition operators acting on $$H^\infty$$ have been barely studied so far.
In this paper, the author proves two results for composition operators on $$H^\infty$$. One is a result exhibiting a phenomenon quite different from the $$H^2$$ case and the other one is a result demonstrating a phenomenon quite similar to the $$H^2$$ case. More explicitly, the first result asserts that every noncompact composition operator on $$H^\infty$$ has essential norm 1. This result appears to be quite interesting, because it reveals a very different phenomenon from Shapiro’s essential norm formula on $$H^2$$.
The second result is a characterization of spectra of composition operators on $$H^\infty$$ in the case that symbols have interior fixed points. This result appears to be an $$H^\infty$$ analogue of the $$H^2$$ results obtained by H. Kamowitz [J. Funct. Anal. 18, 132–150 (1975; Zbl 0295.47003)] and C. C. Cowen B. D. MacCluer [J. Funct. Anal. 125, 223–251 (1994; Zbl 0814.47040)].

##### MSC:
 47B33 Linear composition operators 46E15 Banach spaces of continuous, differentiable or analytic functions 47B38 Linear operators on function spaces (general) 47A10 Spectrum, resolvent
Full Text:


text:
# Max and min value, multi variable (open sets)

#### Petrus

##### Well-known member
Calculate max and min value of the function $$\displaystyle f(x,y)=x^2+y^2-2x-4y+8$$
in the range defined by the $$\displaystyle x^2+y^2≤9$$
Progress:
$$\displaystyle f_x(x,y)=2x-2$$
$$\displaystyle f_y(x.y)=2y-4$$
So I get $$\displaystyle x=1$$ and $$\displaystyle y=2$$ We got one end point that I don't know what to do with $$\displaystyle x^2+y^2≤9$$
If I got this right it should be a elips that x can max be 3,-3 and y 3,-3

Last edited:

#### MarkFL

Staff member
You are correct to first find any critical points in the interior of the region, which as you found is (1,2).

In order to examine $f$ on the boundary of the region, we may represent the circle $x^2+y^2=3^2$ by means of the parametric equations $x=3\cos(t),\,y=3\sin(t),\,0\le t\le 2\pi$. Thus, on the boundary we can write $f$ as a function of a single variable $t$:

$$\displaystyle f(\cos(t),\sin(t))=3^2\cos^2(t)+3^2\sin^2(t)-2\cos(t)-4\sin(t)+8=17-2\cos(t)-4\sin(t)$$

Can you proceed now, differentiating $f(t)$ and equating to zero to find the critical point(s)? And don't neglect the endpoints...

#### Petrus

##### Well-known member
You are correct to first find any critical points in the interior of the region, which as you found is (1,2).

In order to examine $f$ on the boundary of the region, we may represent the circle $x^2+y^2=3^2$ by means of the parametric equations $x=3\cos(t),\,y=3\sin(t),\,0\le t\le 2\pi$. Thus, on the boundary we can write $f$ as a function of a single variable $t$:

$$\displaystyle f(\cos(t),\sin(t))=3^2\cos^2(t)+3^2\sin^2(t)-2\cos(t)-4\sin(t)+8=17-2\cos(t)-4\sin(t)$$

Can you proceed now, differentiating $f(t)$ and equating to zero to find the critical point(s)? And don't neglect the endpoints...
I got to this $$\displaystyle 2(\sin(t)-2\cos(t))=0$$ how do I solve that
Can't I also solve this two way? The way I try to use early?

#### MarkFL

Staff member
You cannot use the extrema of the variables, you want the extrema of the function on the boundary.

Now, to solve the equation you found, first divide through by 2 and rearrange to get:

$$\displaystyle \sin(t)=2\cos(t)$$

What do you think you should do next?

Also, you may find the critical points through use of Lagrange multipliers rather than parametrization, using the boundary of the region as the constraint. I would highly recommend doing both just for practice and insight.

#### Petrus

##### Well-known member
You cannot use the extrema of the variables, you want the extrema of the function on the boundary.

Now, to solve the equation you found, first divide through by 2 and rearrange to get:

$$\displaystyle \sin(t)=2\cos(t)$$

What do you think you should do next?

Also, you may find the critical points through use of Lagrange multipliers rather than parametrization, using the boundary of the region as the constraint. I would highly recommend doing both just for practice and insight.
I would love to solve this as lagrange multipliers as well
rewrite sin(t) as 2cos(t) in our orginal function before we derivate?

#### MarkFL

Staff member
We'll deal with the parametrization method first.

You want to find the value(s) of $t$ which satisfy the equation that resulted from equating the derivative to zero. Now, we could observe that $\sin(t)=2\cos(t)$ implies $y=2x$ (which is what you will find with Lagrange) and plug this into the equation of the circle, but let's solve for $t$ first. What happens if we divide through by $\cos(t)$?

#### Petrus

##### Well-known member
We'll deal with the parametrization method first.

You want to find the value(s) of $t$ which satisfy the equation that resulted from equating the derivative to zero. Now, we could observe that $\sin(t)=2\cos(t)$ implies $y=2x$ (which is what you will find with Lagrange) and plug this into the equation of the circle, but let's solve for $t$ first. What happens if we divide through by $\cos(t)$?
$$\displaystyle \tan(t)=2$$

#### MarkFL

Staff member
Correct. We should be aware that we needn't be worried about division by zero as there is no real value of $t$ such that $\sin(t)=\cos(t)=0$.

Now, in which quadrants will we find solutions for $\tan(t)=2$?

#### Petrus

##### Well-known member
Correct. We should be aware that we needn't be worried about division by zero as there is no real value of $t$ such that $\sin(t)=\cos(t)=0$.

Now, in which quadrants will we find solutions for $\tan(t)=2$?

#### MarkFL

Staff member
Yes, there is a first quadrant solution, but there is another...think of the periodicity of the tangent function.

#### Petrus

##### Well-known member
and second, so first and second ?

#### MarkFL

Staff member
No, the period of the tangent function is $\pi$ units, i.e., $\tan(\theta+k\pi)=\tan(\theta)$. To stay within the given domain of $t$, we let $k=1$, so we will add $\pi$ radians to the first quadrant solution to find we are in which quadrant?

#### Petrus

##### Well-known member
No, the period of the tangent function is $\pi$ units, i.e., $\tan(\theta+k\pi)=\tan(\theta)$. To stay within the given domain of $t$, we let $k=1$, so we will add $\pi$ radians to the first quadrant solution to find we are in which quadrant?
pi is half way (180 degree) and if se are in first quadant and add pi Then we get to third quadant

#### MarkFL

Staff member
Correct!

This means there are first and third quadrant solutions. Can you state the first quadrant solution and how the third quadrant solution relates to this in terms of the $x$ and $y$ coordinates?

#### Petrus

##### Well-known member
Correct!

This means there are first and third quadrant solutions. Can you state the first quadrant solution and how the third quadrant solution relates to this in terms of the $x$ and $y$ coordinates?
What you mean:O? There is a normal tangent function and Then i draw à line in y=2 and it Will hit tan function in y=2 when x is something? I dont know what third quadant got with this problem but i guess y=-2? Hmm is this correct?

#### MarkFL

Staff member
You have the right idea, but if $\tan(t)=2$, then you essentially want to find the points of intersection of the circle and the line $y=2x$.

There are many ways to look at this though, so first let's just work with the first quadrant solution. Can you state this solution?

T=Arctan(2)

#### MarkFL

Staff member
Good, yes $t=\tan^{-1}(2)$. Now, when you plug this value of $t$ into the parametric equations for $x$ and $y$, what do you find?

#### Petrus

##### Well-known member
Good, yes $t=\tan^{-1}(2)$. Now, when you plug this value of $t$ into the parametric equations for $x$ and $y$, what do you find?
$$\displaystyle x=3\cos(\arctan(2))$$
$$\displaystyle y=3\sin(\arctan(2))$$
I am correct?

#### MarkFL

Staff member
Yes, now can you evaluate them, using a right triangle if necessary? Recall you want the side of the triangle opposite the angle $t$ to be 2 and the side adjacent to be 1, then use the Pythagorean theorem to compute the hypotenuse, and finally use the definitions of sine and cosine to complete the evaluations.

While there are simpler ways to go about this, which we will explore next, I think this is good practice.

#### Petrus

##### Well-known member
Yes, now can you evaluate them, using a right triangle if necessary? Recall you want the side of the triangle opposite the angle $t$ to be 2 and the side adjacent to be 1, then use the Pythagorean theorem to compute the hypotenuse, and finally use the definitions of sine and cosine to complete the evaluations.

While there are simpler ways to go about this, which we will explore next, I think this is good practice.
I get hypotenusan to $$\displaystyle \sqrt{5}$$ is that correct?

#### MarkFL

Staff member
Yes, as $$\displaystyle \sqrt{2^2+1^1}=\sqrt{5}$$. Good work!

So, what do you find as the critical coordinates in the first quadrant?

#### Petrus

##### Well-known member
Yes, as $$\displaystyle \sqrt{2^2+1^1}=\sqrt{5}$$. Good work!

So, what do you find as the critical coordinates in the first quadrant?
$$\displaystyle x=\frac{3}{\sqrt{5}}$$
$$\displaystyle y=\frac{6}{\sqrt{5}}$$

#### MarkFL

Staff member
Excellent! Now, what do you find when you let $t=\pi+\tan^{-1}(2)$ to get the 3rd quadrant critical point? And remember to consider the endpoints of the domain for $t$.

Hint: what are $\sin(\theta+\pi)$ and $\cos(\theta+\pi)$ using the angle-sum identities?

Note: I must once again run off for about 4.5 hours, so if anyone else wants to jump in, that's fine. I will check back here first thing when I return, and we can look at some other methods for obtaining these critical points so you can see how they relate.

#### Petrus

##### Well-known member
Excellent! Now, what do you find when you let $t=\pi+\tan^{-1}(2)$ to get the 3rd quadrant critical point? And remember to consider the endpoints of the domain for $t$.

Hint: what are $\sin(\theta+\pi)$ and $\cos(\theta+\pi)$ using the angle-sum identities?

Note: I must once again run off for about 4.5 hours, so if anyone else wants to jump in, that's fine. I will check back here first thing when I return, and we can look at some other methods for obtaining these critical points so you can see how they relate.
Well in third quadrant x and y is negative so $$\displaystyle x=-\frac{3}{\sqrt{5}}$$
$$\displaystyle y=-\frac{6}{\sqrt{5}}$$? I am correct?


text:
# Current density

This page is about the electric current density in electromagnetism. For the probability current density in quantum mechanics, see Probability current.

In electromagnetism, current density is the electric current per unit area of cross section. It is defined as a vector whose magnitude is the electric current per cross-sectional area at a given point in space (i.e. it is a vector field). In SI units. The electric current density is measured in amperes per square metre.[1] It also has applications in related fields in solid state physics, condensed matter physics, etc.

## Definition

Electric current density J is simply the electric current I (SI unit: A) per unit area A (SI unit: m2). Its magnitude is given by the limit:[2]

$J = \lim\limits_{A \rightarrow 0}\frac{I(A)}{A}$

For current density as a vector J, the surface integral over a surface S, followed by an integral over the time duration t1 to t2, gives the total amount of charge flowing through the surface in that time (t2t1):

$q=\int_{t_1}^{t_2}\iint_S \bold{J}\cdot\bold{\hat{n}}{\rm d}A{\rm d}t$

The area required to calculate the flux is real or imaginary, flat or curved, either as a cross-sectional area or a surface. For example, for charge carriers passing through an electrical conductor, the area is the cross-section of the conductor, at the section considered.

The vector area is a combination of the magnitude of the area through which the mass passes through, A, and a unit vector normal to the area, $\bold{\hat{n}}$. The relation is $\bold{A} = A \bold{\hat{n}}$.

If the current density J passes through the area at an angle θ to the area normal $\bold{\hat{n}}$, then

$\bold{J}\cdot\bold{\hat{n}}= J\cos\theta$

where · is the dot product of the unit vectors. This is, the component of current density passing through the surface (i.e. normal to it) is J cos θ, while the component of current density passing tangential to the area is J sin θ, but there is no current density actually passing through the area in the tangential direction. The only component of current density passing normal to the area is the cosine component.

## Importance

Current density is important to the design of electrical and electronic systems.

Circuit performance depends strongly upon the designed current level, and the current density then is determined by the dimensions of the conducting elements. For example, as integrated circuits are reduced in size, despite the lower current demanded by smaller devices, there is trend toward higher current densities to achieve higher device numbers in ever smaller chip areas. See Moore's law.

At high frequencies, current density can increase because the conducting region in a wire becomes confined near its surface, the so-called skin effect.

High current densities have undesirable consequences. Most electrical conductors have a finite, positive resistance, making them dissipate power in the form of heat. The current density must be kept sufficiently low to prevent the conductor from melting or burning up, the insulating material failing, or the desired electrical properties changing. At high current densities the material forming the interconnections actually moves, a phenomenon called electromigration. In superconductors excessive current density may generate a strong enough magnetic field to cause spontaneous loss of the superconductive property.

The analysis and observation of current density also is used to probe the physics underlying the nature of solids, including not only metals, but also semiconductors and insulators. An elaborate theoretical formalism has developed to explain many fundamental observations.[3][4]

The current density is an important parameter in Ampère's circuital law (one of Maxwell's equations), which relates current density to magnetic field.

In special relativity theory, charge and current are combined into a 4-vector.

## Calculation of current densities in matter

### Free currents

Charge carriers which are free to move constitute a free current density, which are given by expressions such as those in this section.

Electric current is a coarse, average quantity that tells what is happening in an entire wire. At position r at time t, the distribution of charge flowing is described by the current density:[5]

$\mathbf{J}(\mathbf{r}, t) = \rho(\mathbf{r},t) \; \mathbf{v}_\text{d} (\mathbf{r},t) \,$

where J(r, t) is the current density vector, vd(r, t) is the particles' average drift velocity (SI unit: ms−1), and

$\rho(\mathbf{r}, t)= qn(\mathbf{r},t)$

is the charge density (SI unit: coulombs per cubic metre), in which n(r, t) is the number of particles per unit volume ("number density") (SI unit: m−3), q is the charge of the individual particles with density n (SI unit: coulombs).

A common approximation to the current density assumes the current simply is proportional to the electric field, as expressed by:

$\mathbf{J} = \sigma \mathbf{E} \,$

where E is the electric field and σ is the electrical conductivity.

Conductivity σ is the reciprocal (inverse) of electrical resistivity and has the SI units of siemens per metre (S m−1), and E has the SI units of newtons per coulomb (N C−1) or, equivalently, volts per metre (V m−1).

A more fundamental approach to calculation of current density is based upon:

$\mathbf{J} (\mathbf{r}, t) = \int_{-\infty}^t \mathrm{d}t' \int \mathrm{d}^3\mathbf{r}' \; \sigma(\mathbf{r}-\mathbf{r}', t-t') \; \mathbf{E}(\mathbf{r}',\ t') \,$

indicating the lag in response by the time dependence of σ, and the non-local nature of response to the field by the spatial dependence of σ, both calculated in principle from an underlying microscopic analysis, for example, in the case of small enough fields, the linear response function for the conductive behaviour in the material. See, for example, Giuliani or Rammer.[6][7] The integral extends over the entire past history up to the present time.

The above conductivity and its associated current density reflect the fundamental mechanisms underlying charge transport in the medium, both in time and over distance.

A Fourier transform in space and time then results in:

$\mathbf{J} (\mathbf{k}, \omega) = \sigma(\mathbf{k}, \omega) \; \mathbf{E}(\mathbf{k}, \omega) \,$

where σ(k, ω) is now a complex function.

In many materials, for example, in crystalline materials, the conductivity is a tensor, and the current is not necessarily in the same direction as the applied field. Aside from the material properties themselves, the application of magnetic fields can alter conductive behaviour.

### Polarization and magnetization currents

Currents arise in materials when there is a non-uniform distribution of charge.[8]

In dielectric materials, there is a current density corresponding to the net movement of electric dipole moments per unit volume, i.e. the polarization P:

$\mathbf{J}_\mathrm{P}=\frac{\partial \mathbf{P}}{\partial t}$

Similarly with magnetic materials, circulations of the magnetic dipole moments per unit volume, i.e. the magnetization M lead to volume magnetization currents:

$\mathbf{J}_\mathrm{M}=\nabla\times\mathbf{M}$

Together, these terms form add up to the bound current density in the material (resultant current due to movements of electric and magnetic dipole moments per unit volume):

$\mathbf{J}_\mathrm{b}=\mathbf{J}_\mathrm{P}+\mathbf{J}_\mathrm{M}$

### Total current in materials

The total current is simply the sum of the free and bound currents:

$\mathbf{J} = \mathbf{J}_\mathrm{f}+\mathbf{J}_\mathrm{b}$

### Displacement current

There is also a displacement current corresponding to the time-varying electric displacement field D:[9][10]

$\mathbf{J}_\mathrm{D}=\frac{\partial \mathbf{D}}{\partial t}$

which is an important term in Ampere's circuital law, one of Maxwell's equations, since absence of this term would not predict electromagnetic waves to propagate, or the time evolution of electric fields in general.

## Continuity equation

Main article: Continuity equation

Since charge is conserved, current density must satisfy a continuity equation. Here is a derivation from first principles.[11]

The net flow out of some volume V (which can have an arbitrary shape but fixed for the calculation) must equal the net change in charge held inside the volume:

$\int_S{ \mathbf{J} \cdot \mathrm{d}\mathbf{A}} = -\frac{\mathrm{d}}{\mathrm{d}t} \int_V{\rho \; \mathrm{d}V} = - \int_V{ \frac{\partial \rho}{\partial t}\;\mathrm{d}V}$

where ρ is the charge density, and dA is a surface element of the surface S enclosing the volume V. The surface integral on the left expresses the current outflow from the volume, and the negatively signed volume integral on the right expresses the decrease in the total charge inside the volume. From the divergence theorem:

$\int_S{ \mathbf{J} \cdot \mathrm{d}\mathbf{A}} = \int_V{\mathbf{\nabla} \cdot \mathbf{J}\; \mathrm{d}V}$

Hence:

$\int_V{\mathbf{\nabla} \cdot \mathbf{J}\; \mathrm{d}V}\ = - \int_V{ \frac{\partial \rho}{\partial t} \;\mathrm{d}V}$

This relation is valid for any volume, independent of size or location, which implies that:

$\nabla \cdot \mathbf{J} = - \frac{\partial \rho}{\partial t}$

and this relation is called the continuity equation.[12][13]

## In practice

In electrical wiring, the maximum current density can vary from 4A∙mm−2 for a wire with no air circulation around it, to 6A∙mm−2 for a wire in free air. Regulations for building wiring list the maximum allowed current of each size of cable in differing conditions. For compact designs, such as windings of SMPS transformers, the value might be as low as 2 A∙mm−2.[14] If the wire is carrying high frequency currents, the skin effect may affect the distribution of the current across the section by concentrating the current on the surface of the conductor. In transformers designed for high frequencies, loss is reduced if Litz wire is used for the windings. This is made of multiple isolated wires in parallel with a diameter twice the skin depth. The isolated strands are twisted together to increase the total skin area and to reduce the resistance due to skin effects.

For the top and bottom layers of printed circuit boards, the maximum current density can be as high as 35 A∙mm−2 with a copper thickness of 35 µm. Inner layers cannot dissipate as much heat as outer layers; designers of circuit boards avoid putting high-current traces on inner layers.

In semiconductors, the maximum current density is given by the manufacturer. A common average is 1 mA∙µm−2 at 25 °C for 180 nm technology. Above the maximum current density, apart from the joule effect, some other effects like electromigration appear in the micrometer scale.

In biological organisms, ion channels regulate the flow of ions (for example, sodium, calcium, potassium) across the membrane in all cells. Current density is measured in pA∙pF−1 (picoamperes per picofarad), that is, current divided by capacitance, a de facto measure of membrane area.[clarification needed]

In gas discharge lamps, such as flashlamps, current density plays an important role in the output spectrum produced. Low current densities produce spectral line emission and tend to favour longer wavelengths. High current densities produce continuum emission and tend to favour shorter wavelengths.[15] Low current densities for flash lamps are generally around 1000 A∙cm−2. High current densities can be more than 4000 A∙cm−2.

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