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JEE Main 2022 (Online) 29th June Morning Shift
+4
-1
English
Hindi
The distance between the two points A and A' which lie on y = 2 such that both the line segments AB and A' B (where B is the point (2, 3)) subtend angle $${\pi \over 4}$$ at the origin, is equal to :
A
10
B
$${48 \over 5}$$
C
$${52 \over 5}$$
D
3
2
JEE Main 2022 (Online) 28th June Evening Shift
+4
-1
English
Hindi
Let a triangle be bounded by the lines L1 : 2x + 5y = 10; L2 : $$-$$4x + 3y = 12 and the line L3, which passes through the point P(2, 3), intersects L2 at A and L1 at B. If the point P divides the line-segment AB, internally in the ratio 1 : 3, then the area of the triangle is equal to :
A
$${{110} \over {13}}$$
B
$${{132} \over {13}}$$
C
$${{142} \over {13}}$$
D
$${{151} \over {13}}$$
3
JEE Main 2022 (Online) 27th June Morning Shift
+4
-1
English
Hindi
In an isosceles triangle ABC, the vertex A is (6, 1) and the equation of the base BC is 2x + y = 4. Let the point B lie on the line x + 3y = 7. If ($$\alpha$$, $$\beta$$) is the centroid of $$\Delta$$ABC, then 15($$\alpha$$ + $$\beta$$) is equal to:
A
39
B
41
C
51
D
63
4
JEE Main 2022 (Online) 27th June Morning Shift
+4
-1
English
Hindi
If two straight lines whose direction cosines are given by the relations $$l + m - n = 0$$, $$3{l^2} + {m^2} + cnl = 0$$ are parallel, then the positive value of c is :
A
6
B
4
C
3
D
2
Policy
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Article Text
The influence of risk and monetary payment on the research participation decision making process
1. J P Bentley,
2. P G Thacker
1. School of Pharmacy, The University of Mississippi, University, MS, USA
1. Correspondence to:
J P Bentley
Faser Hall 219A, School of Pharmacy, The University of Mississippi, University, MS 38677, USA; phjpbolemiss.edu
## Abstract
Objectives: To determine the effects of risk and payment on subjects’ willingness to participate, and to examine how payment influences subjects’ potential behaviours and risk evaluations.
Methods: A 3 (level of risk) × 3 (level of monetary payment), between subjects, completely randomised factorial design was used. Students enrolled at one of five US pharmacy schools read a recruitment notice and informed consent form for a hypothetical study, and completed a questionnaire. Risk level was manipulated using recruitment notices and informed consent documents from hypothetical biomedical research projects. Payment levels were determined using the payment models evaluated by Dickert and Grady as a guide. Five dependent variables were assessed in the questionnaire: willingness to participate, willingness to participate with no payment, propensity to neglect to tell about restricted activities, propensity to neglect to tell about negative effects, and risk rating.
Results: Monetary payment had positive effects on respondents’ willingness to participate in research, regardless of the level of risk. However, higher monetary payments did not appear to blind respondents to the risks of a study. Payment had some influence on respondents’ potential behaviours regarding concealing information about restricted activities. However, payment did not appear to have a significant effect on respondents’ propensity to neglect to tell researchers about negative effects.
Conclusions: Monetary payments appear to do what they are intended to do: make subjects more willing to participate in research. Concerns about payments blinding subjects to risks could not be substantiated in the present study. However, the findings do raise other concerns—notably the potential for payments to diminish the integrity of a study’s findings. Future research is critical to make sound decisions about the payment of research subjects.
• risk
• research participation
• payment
• undue inducement
• informed consent
• IRB, institutional review board
• PCA, principal components analysis
• SES, socioeconomic status
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Although the practice of paying subjects to participate in research is not new, it continues to serve as a point of debate for many members of the research community. Among other issues, some are concerned that the use of such tactics could induce subjects into taking part in a study that they would not participate in otherwise (by blinding prospective subjects to risks) or cause subjects to conceal information that would disqualify them from the study. Although there is considerable debate from normative ethical perspectives concerning undue inducement of potential research subjects with monetary payments, there are few empirical data regarding the effects of monetary payments. Will varying levels of monetary payments have the same effect on willingness to participate at varying levels of risk? Will monetary payments lead subjects to conceal information that would disqualify them from the study? Will varying levels of monetary payments lead to different ratings of risk of the same study? Most authors have made empirical claims without testing these issues. This research was undertaken to add to the discussion concerning the payment of research subjects by providing empirical data on the potential effects of monetary payments.
## LITERATURE REVIEW
After distinguishing coercion from inducement, this section explores both normative ethical arguments and the empirical literature surrounding the topic of paying research subjects.
### Coercion versus inducement
Faden and Beauchamp1 define coercion as “an extreme form of influence by another person that completely controls a person’s decision”. They further state that such an influence “deprives the person of autonomous choice, and thus is incompatible with informed consent”. Most authors agree that coercion requires a credible threat of severe negative consequences.2–8 Thus, any method of recruitment that causes a person to participate (assuming they do not want to) through the use of intended, credible, and severe threats—such that the person has no other alternative but to participate—would be considered coercive.
Monetary payments are often used as inducements; they motivate people to do something. Inducements are offers, not threats,6,8 and therefore they are not usually considered to be coercive (although there are notable exceptions: see reference6 for an example). Although inducements in most instances are not considered to be coercive, it has been suggested that inducements can be undue,2,9 exploitative,4,5,7 or morally impermissible.4,5 Thus, “a monetary inducement can invalidate informed consent without being coercive”.10
Macklin2 notes that the concept of inducement is weaker than the notion of coercion and draws a conceptual distinction between due and undue inducement. She describes one paradigm for undue inducement as whether the inducement leads subjects to lie, deceive, or conceal information that would exclude them from the study if known to investigators. It is not the dollar amount alone that determines what is an undue inducement2; the impoverishment of subjects and the risk of injury from the study are also considerations.3
### Positions on inducements
The common and long practised act of paying research subjects8,11–13 to aid recruitment and retention is seen by some as unethical. For example, McNeill states that monetary inducements encourage people to expose themselves to risk and may add to a subject’s difficulties in assessing the apparent level of risk of a study.14 Furthermore, McNeill argues that “the reason that inducement is particularly of concern is that those most susceptible to inducement may be the least able to assess the aims and technical information relating to the research”.15 He maintains that inducements increase inequity in the research process because the majority of research participants are of lower socioeconomic status (SES) while higher SES people typically benefit from such research. Macklin also discusses the equitable distribution of society’s benefits and burdens with respect to research, noting that the lower the payment (with the goal being to avoid undue influence) the greater the likelihood for volunteers to be of lower SES.2
To McNeill, any offer of payment when a research study involves risk of injury is objectionable.14 However, Macklin reasons that some level of inducement is necessary to prompt a sufficient number of people to volunteer.2 Her recommendation is to set initial payments low for normal, healthy volunteers (to avoid undue inducement). While this may lead to a greater likelihood of volunteers coming from lower socioeconomic groups, she argues there is no way to avoid this objection from social justice, given the moral precept of pay equity.
Although some argue that subjects should not be paid or that the payments to subjects should be minimised, others claim that payments to research subjects are appropriate and some go further to argue that there should be no restrictions as to how much is offered. For example, Brody16 states that “if the independent review panel has already concluded that the risk:benefit ratio of the research is acceptable (otherwise, the research could not be approved), how can the large payment harm the subject? And in what way are large payments for acceptable research coercive or exploitative?” While Palmer17 appears to agree with Brody in most cases, he concludes that “concern about excessive payment is warranted only under stringent conditions … In research involving the highest acceptable risk, offers of large payment may cause vulnerable persons to lose the freedom to refuse participation.” Thus, according to Palmer, inducements can be undue only in those research studies that involve the highest acceptable risk of physical or psychological injury. Monetary inducements could also be considered undue for research that presents unreasonable risk—for example, death—but Palmer argues that institutional review boards (IRBs) would not approve these. Even McNeill, who appears to argue strongly against the use of monetary inducements in clinical trials, concedes that when there is no known or very little likelihood of harm, monetary inducement may be justifiable.14
Wilkinson and Moore examine arguments against inducement including: inducements undermine consent; inducements can damage the welfare of participants; inducements exploit research subjects, especially the poor; and inducements can result in subject selection bias.6 After discussing what they consider to be the strongest case they could develop for each of these arguments, they conclude that with one possible single exception, each of these arguments against inducement is unpersuasive. The one problem case involves studies that have at least one harm related exclusion criterion (for example, subjects with heart conditions cannot participate) and no adequate method—independent of subject self report—for researchers to verify that potential subjects meet the exclusion criterion. In this case, monetary inducements might entice subjects who are “financially hard pressed” to lie or conceal information in order to participate in a study to receive the payment and researchers cannot adequately verify the accuracy of the subject reported information. Wilkinson and Moore conclude that in this case, and only in this case, inducements should create problems for ethics committees.6 They further argue, however, that ethics committees should probably not even allow these types of studies to proceed in the first place, regardless of the offering of any inducement.
In a subsequent paper, Wilkinson and Moore address three additional objections to inducements: inducements create greater inequity, inducements crowd out less funded research, and inducements undesirably commercialise the research process.18 They argue that there is a pro tanto reason to allow subjects to be paid, which they define as freedom to contract, which stipulates that “the relationship between researcher and subject should be a matter of agreement between them”. As in their first paper, Wilkinson and Moore analyse each anti-inducement argument for justification and conclude that each is insufficient to outweigh their freedom of contract position.18
### Empirical findings
Thus, taking different normative ethical perspectives, several authors have argued both for and against the use of monetary payments to research subjects. From an empirical perspective, some researchers have examined subjects’ motivations for research participation, including the role of monetary payments. Some studies have shown that some healthy volunteers do not agree with paying subjects.19 Other studies have shown that financial motives are less important than other motivations20 and that economic gain is rarely the sole motivation for research.21 However, several studies have pointed to financial payment as a very significant if not the most significant influence on the decision to participate in research, especially for healthy volunteers.22–24
Another empirical approach that has recently generated some findings has been to ask respondents directly if a monetary payment would impair their own, and others’ ability to think carefully about risks and benefits of a study.25 Using such a method, Casarett et al found that a large percentage of individuals believe that a US$500 payment would impair others’ judgment, but a significantly smaller number believe that the payment would impair their own.25 The authors conclude that concerns about the influence of payments for research are expressed by members of the general public, not just ethicists. The available empirical studies suggest there is potential for misconduct by researchers to offer payments in such excess that they could be considered undue in some circumstances. Given this potential, how does one determine what an appropriate level of payment is? ### Models and guidelines Although federal guidelines allude to the potential ethical difficulties arising from payment to research subjects, they offer little meaningful guidance to researchers and IRBs for how to pay subjects.11,12,26 In addition, while organisations that conduct and review human subjects research often engage in (or allow) the payment of subjects, few have written policies on payment.12 Macklin argues to set initial payments low for normal, healthy volunteers.2 If such a payment does not render the necessary amount of participants, Macklin suggests that the researcher review the risks associated with the study as well as subjects’ time requirements to determine whether these played a factor in the under recruitment of subjects. However, how does one determine the initial payment rate? Dickert and Grady have addressed such a question by evaluating three models for the payment of subjects.11 In the market model, payment is justified by the need for monetary incentives to recruit subjects. Payment in this model is based solely upon the economic principle of supply and demand, and advocates the use of completion bonuses as well as other incentives for compliance with the protocol. The second model evaluated by the authors is the wage payment model. In this model, payment is rendered on the premise that research participation requires little skill but demands time and effort from the subject. Thus, payment using this model is based solely on standard wage payment for unskilled labour with additional payments being made for uncomfortable procedures. The third payment model, the reimbursement model, justifies payment by using the premise that research subjects should not be required to suffer financial sacrifice. Therefore, subjects are reimbursed for expenses incurred and may be paid for lost wages. Such a model could lead to the unequal payment of subjects for the same amount of participation, as a subject who makes more per hour at his or her normal occupation or profession would receive more compensation. Dickert and Grady11 recommend the adoption of the wage payment model for three reasons: (1) it reduces undue inducement concerns, (2) it standardises payment schedules, and (3) it establishes a system in which payment is based on the contribution subjects make, consistent with the principle of equal pay for equal work. Their recommendation is primarily based on normative ethical evaluation rather than empirical evidence. Indeed, at the end of their manuscript, Dickert and Grady27 suggest that “there is a need for empirical research to determine the ways in which offers of money affect the quality of subjects’ informed consent … there is a need for data on the importance of payment with respect to successful recruitment; little is known about the effect of different amounts or methods of payment on recruitment efforts.” In a subsequent article, Dickert et al28 note that, “further study and discussion are needed to understand when money is an undue influence, as well as the impact of payment on subject selection and scientific integrity”. ## RESEARCH OBJECTIVES The objectives of this study were to determine the effects of risk and monetary payment on subjects’ willingness to participate and to examine how payment influences subjects’ potential behaviours and risk evaluations. Although monetary payments have been shown to increase the response rate in mail surveys (for example, see reference 29), there is minimal risk associated with most surveys. In this study, we manipulated levels of monetary payment and risk within the context of a single design to assess the effects of these variables on the research participation decision making process. We are not aware of other published studies that have used such a design. ## METHODS ### Independent variables The study used a 3 (level of risk) × 3 (level of monetary payment), between subjects, completely randomised factorial design. Risk level was manipulated using recruitment notices and informed consent documents from hypothetical biomedical research projects. The high risk level was a phase I drug trial for a drug that has not yet been tested in humans. The medium risk level was a bioequivalence study for a generic version of a brand name drug that is already on the market. The low risk level was a study that measured salivary levels of two stress hormones of healthy volunteers to compare with subjects who have had heart attacks (no drug is ingested/no blood is drawn). Federal sources9,30,31 and examples of documents used in actual studies guided construction of the informed consent forms and recruitment notices. Also, coordinators from two different IRBs reviewed the documents. Studies for all three risk levels were designed for healthy volunteers, and the amount of time required of subjects for each study was held constant. Quantitative pretests (n = 30 for each level) indicated significantly different risk ratings among the three risk levels. Payment levels were determined using the payment models evaluated by Dickert and Grady as a guide.11 The total amount of time required for the three hypothetical studies was estimated at about 54 hours (two 24 hour stays and 12 half hour sessions). For the highest level of payment, a rate of$28 per hour was used and $300 was added as an additional risk based bonus, for a total payment of$1800. For the medium level of payment, a rate of $14 per hour (slightly below the May 2001 total national average for non-farm production workers32) was used and$50 was added as an additional risk based bonus, for a total payment of $800. For the lowest level of payment, only parking costs ($3.00 per hour) and travel costs ($0.345 per mile—an average of 40 miles per day) were used, for a total of$350. Neither the recruitment notice nor the informed consent form specified how these amounts were calculated; respondents were merely made aware of how much they would be paid for participation. The following language was used in the informed consent form under the heading “Payment”:
“In return for your time, effort and travel expenses, you will be paid \$XXX for your participation in this study. If you do not complete the study, you will be paid on a pro-rated basis that is calculated based on the time you have contributed to the study. A check will be mailed to you approximately six weeks after your participation in the study has ended.”
### Sample and data collection
Each recruitment notice/informed consent form (used to manipulate risk) was paired with each of the three payment levels yielding a total of nine treatment groups. Power analysis indicated that 23 cases per treatment group (n = 207) were necessary to have power of 0.80 to detect medium main and interaction effects with a significance level of 0.05. Packets containing an explanation letter, a copy of one of the recruitment notices/informed consent forms, and a questionnaire were randomly distributed through course instructors to pharmacy students at five different universities. In the explanation letter, students were asked to read a recruitment notice and informed consent form for a hypothetical research study, complete an anonymous questionnaire, and return all materials to their instructor; all tasks were performed outside of class time (no incentive was provided for returning the questionnaire). Each student was exposed to only one risk level/payment level combination and an equal number of packets in each treatment group were sent to each school. Students were asked to not discuss their responses or what they read with other students.
### Dependent measures
The questionnaire was designed to assess five dependent variables (table 1) in addition to several demographic questions. Respondents were asked to rate their willingness to participate in the hypothetical study with and without the monetary payment on a 10 point scale, with 10 being “I would definitely participate”. Respondents were also asked to provide their honest opinion about the likelihood that they would neglect to tell researchers about eight activities that were described as restricted in the informed consent form (such as participation in a clinical trial within the previous 30 days or the consumption of alcohol or caffeine during the study period or 48 hours before the start of the study). Each restriction was rated on a 10 point scale with 10 being “very likely to neglect to tell”. Similarly, respondents were asked two questions concerning their likelihood to neglect to tell researchers about negative effects that could cause them to be eliminated from the study (for example: side effects; not feeling well). All informed consent forms instructed subjects to report side effects and to advise the medical staff if they were not feeling well. As with the previous construct, a 10 point scale (with 10 being “very likely to neglect to tell”) was used. Finally, respondents were asked to evaluate the risk involved in the hypothetical study using a 5 item scale designed to have respondents assess both the likelihood and the severity of negative consequences of participation.
Table 1
Description of dependent variables
To determine the quality of the three multi-item scales, reliability analysis and principal components analysis (PCA) were used. For each multi-item scale, corrected item-to-total correlations and Cronbach’s alpha were calculated to assess the extent to which the set of items in the multi-item scale measure the same attribute (that is, internal consistency reliability). PCA was then conducted to provide some evidence of the unidimensionality of each of the multi-item scales33
Cronbach’s alpha for each of the multi-item scales can be found in table 1 and all are well above acceptable norms.34 Within each of the multi-item scales, correlations between individual items and the scale of interest (corrected item-to-total correlations) exceeded 0.5. PCA results indicated that items in each multi-item scale loaded highly on a single component (loadings greater than 0.6) and did not significantly load on other components. These results support the summation of items in each of the multi-item scales to obtain summated total scores for these three measures. To ease interpretation of the dependent variable scores, these summated scores were divided by the number of items in the scale (thus, all dependent variables were scored on a 1–10 scale). These transformed scores for each multi-item scale were used in all subsequent analyses. A description of the meaning of a higher score for all five dependent variables is provided in table 1.
### Analysis procedure
Data were analysed using two factor analysis of covariance (ANCOVA) procedures, with a measure of “venturesomeness”35 as the covariate. Separate analyses were conducted for each of the five dependent variables.
This study received IRB exemption from the University of Mississippi Institutional Review Board. The study was also submitted to the IRBs at the other four universities where the packets were distributed, and was exempted. Copies of the recruitment notices and informed consent forms used in the manipulation of the independent variables and the data collection instrument used for this study are available upon request.
## RESULTS
A total of 789 packets were handed out to pharmacy students and 326 were returned. Nine responses were discarded because of missing data and seven were discarded because the respondent’s self rated health was not excellent, very good or good, yielding a usable response rate of 39.3%. Due to the nature of the data collection method, cell sizes were not equal. To simplify analysis and interpretation of the results, respondents were randomly discarded from certain cells to achieve equal cell sizes. Given that there was a sufficient number of respondents based on the power analysis conducted before the study, this was determined to be a reasonable alternative.36 The final data set comprised 30 respondents per experimental cell (n = 270).
Adjusted cell and marginal means for each dependent variable can be found in tables 2 to 6. No significant first order interactions were noted in any of the analyses. With respect to willingness to participate, both risk level (F(2,260) = 8.90, p<0.0005, partial eta2 = 0.06) and monetary payment (F(2,260) = 4.26, p = 0.015, partial eta2 = 0.03) had a significant effect; with higher levels of risk and lower levels of payment leading to lower willingness ratings. Furthermore, the effect of monetary payment did not appear to vary according to risk level (that is, there was no significant interaction). Without payment, respondents’ willingness to participate decreased and their willingness to participate without payment was influenced only by risk level, with higher risk levels leading to lower willingness ratings (F(2,260) = 6.75, p = 0.001, partial eta2 = 0.05).
Table 2
Adjusted cell and marginal means (dependent variable: willingness to participate*)
Table 3
Adjusted cell and marginal means (dependent variable: willingness to participate with no payment*)
Table 4
Adjusted cell and marginal means (dependent variable: propensity to neglect to tell about restricted activities*)
Table 5
Adjusted cell and marginal means (dependent variable: propensity to neglect to tell about negative effects*)
Table 6
Adjusted cell and marginal means (dependent variable: risk rating*)
Monetary payments appeared to influence respondents’ propensity to neglect to tell researchers about restricted activities they have engaged in either before or during a study, with higher payment levels leading to a higher propensity to neglect to tell (F(2,260) = 3.68, p = 0.027, partial eta2 = 0.03). However, examination of cell means reveals that this effect may be driven by respondents who were exposed to low and medium risk levels. Risk level appeared to influence respondents’ propensity to neglect to tell researchers about experiencing negative effects, with respondents in the low risk group having the highest propensity to neglect to tell (F(2,260) = 3.74, p = 0.025, partial eta2 = 0.03). Monetary payment did not significantly influence this variable.
There were noticeable differences in risk ratings among the groups exposed to different levels of the risk manipulation (F(2,260) = 275.95, p<0.0005, partial eta2 = 0.68); risk rating was not significantly influenced by monetary payment and the effect of risk level on risk rating did not vary by level of monetary payment (that is, there was no significant interaction).
## DISCUSSION
This study provided empirical data on two ethical concerns often raised about the payment of research subjects; namely, that payments could (1) unduly induce subjects into taking part in a study that they would not participate in otherwise (by blinding the prospective subject to risks), or (2) cause subjects to conceal information that would disqualify them from the study. This study suggests that monetary payment increases respondents’ willingness to participate in research regardless of the level of risk; higher levels of payment make respondents more willing to participate, even if the study is relatively risky. However, higher monetary payments, at least in this study, did not appear to blind respondents to risks. For example, the high risk study received similar risk ratings for all three levels of monetary payment. These group level findings may not apply to all individuals; thus, this finding does not alleviate all ethical concerns regarding the payment of research subjects. Using a subjective interpretation of influence,37 what is a due inducement for one individual maybe undue for another. Clearly, more research is needed to determine when monetary payment is an undue inducement.12
This study also showed that higher levels of monetary payment may influence subjects’ behaviours regarding concealing information about restricted activities. If such activities were actually engaged in, the results of the hypothetical studies may have been distorted (that is, alcohol, caffeine, medications, herbal products may all affect the pharmacokinetics of a study drug). However, findings from this study suggest that this effect may be more likely to occur in lower risk studies. By contrast, monetary payments did not have a significant effect on respondents’ propensity to neglect to tell researchers when they experience negative consequences from study participation. These two findings suggest that monetary payments may be less likely to jeopardise the well being of subjects than the integrity of the research—not an insignificant issue. The latter is consistent with both observation8 and empirical evidence.38
## LIMITATIONS OF THE PRESENT STUDY AND FUTURE RESEARCH
Several limitations of this study point to the need for additional research with respect to the payment of research subjects. The study was hypothetical in nature; behavioural intentions rather than actual behaviours were assessed. Similarly, respondents read documents on their own; there was no additional verbal explanation provided by a clinical investigator.
Another limitation concerns the monetary payment manipulation. As discussed in the methods section, respondents read that payment was in return for time, effort, and travel expenses. Only the amount of money was manipulated; no effort was made to manipulate what the money was for. Macklin2 provides a discussion of the purposes for paying subjects and Russell, Moralejo and Burgess19 note that respondents in their study distinguished between various reasons for paying subjects. Some IRBs restrict what monetary payments can be used for (for example, prohibition of completion bonuses, no payment for the assumption of risk).12 Given this information, future studies should focus not only on the amount of payment, but its purpose. Additionally, forms of payment other than money raise ethical concerns8 and should also be the subject of empirical research.
The results of the present study are based on the responses of pharmacy students. Although pharmacy students are often targeted to participate in clinical studies like the ones described in the recruitment notices and informed consent forms, the results of the present study are not generalisable to all groups who participate in this type of research. Students do not represent all healthy individuals who volunteer for research. Additional research examining the effects of monetary payments on the decision making of other groups of healthy volunteers is warranted.
The hypothetical studies for all three risk levels were designed for healthy volunteers. Some have argued for a distinction between the payment of patients and the payment of healthy subjects.39 Dickert and Grady40 note that “there is no inherent reason to treat patients and healthy subjects differently with respect to payment” but note the lack of empirical data. The added dimension of “hope for benefits” in patients enrolling in research studies may alter the effect of monetary payments on decisions in ways that are different from healthy volunteers. Thus, future research should be directed at exploring the effects of paying patients.
Finally, impoverishment of research subjects is often considered in normative ethical discussions of the use of monetary payments in research. Although income and subjective discretionary income were collected in this study, the cell sizes were too small and too disparate to make any conclusive statements. Is the effect of monetary payment on willingness to participate different for members of different socioeconomic groups? Do members of different socioeconomic groups evaluate risk differently at different levels of payment?
## CONCLUSION
Although this study may diminish some concerns regarding whether monetary payments blind research subjects to potential risks, the findings also suggest that monetary payments may lead to reductions in the integrity of a study’s findings, especially for studies of lower risk. Certainly, this study does not provide a definitive answer to the question of whether monetary payments to research subjects should be prohibited. No single study can completely support or repudiate this position. There is a definite need for additional research to further examine this issue, especially as it relates to impoverished individuals. We hope that this paper will stimulate not only future research but also discussion about appropriate research policies and procedures.
## Acknowledgments
We are grateful to the following individuals for their help in distributing and collecting materials: Ms Tina Penick Brock (University of North Carolina), Dr Kem Krueger (Auburn University), Dr Lon Larson (Drake University), Dr Donna West (University of Arkansas for Medical Sciences), Dr Noel Wilkin (University of Mississippi), and Dr Gary Theilman (University of Mississippi). We would also like to thank Ms Diane Lindley (University of Mississippi) and Dr Rexann Pickering (Methodist Healthcare–Memphis, TN) for reviewing our informed consent forms and recruitment notices. We are grateful to Dr Thomas Lombardo (University of Mississippi) for his helpful comments on the manuscript. Finally, we appreciate the thoughtful suggestions provided by the two anonymous reviewers.
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# Show transcribed image text 2. Assume the atmospheric pressure at ground level is 101.325 kPa, and t
Show transcribed image text 2. Assume the atmospheric pressure at ground level is 101.325 kPa, and the vertical dependence of pressure on density is defined as If we also assume that y = 0.0 m at ground level and g = 9.81 m/sec^2 is constant, then what is the atmospheric pressure at y = 500 m?
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# Factoring $X^{16}+X$ over $\mathbb{F}_2$
I just asked wolframalpha to factor $X^{16}+X$ over $\mathbb{F}_2$. The normal factorization is
$$X(X+1)(X^2-X+1)(X^4-X^3+X^2-X+1)(X^8+X^7-X^5-X^4-X^3+X+1)$$
and over $GF(2)$ it is
$$X(X+1)(X^2+X+1)(X^4+X+1)(X^4+X^3+1)(X^4+X^3+X^2+X+1).$$
Does the second form follow from the first, or is there a different way to factor over $\mathbb{F}_2$? I noticed that simply replacing the $-$ signs with $+$ signs in the first factorization doesn’t yield the second one.
#### Solutions Collecting From Web of "Factoring $X^{16}+X$ over $\mathbb{F}_2$"
It shouldn’t be a surprise that switching from integer (or rational) coefficients to modular coefficients allows for further factorization. The simplest example is probably the factorization
$$x^2+1=x^2+2x+1=(x+1)^2$$
over $F_2$.
Here the key difference between the two factorizations is that the polynomial of degree 8 splits into a product of two quartic polynomials over $F_2$: $$(x^4+x+1)(x^4+x^3+1)=x^8+x^7+x^5+x^4+x^3+x+1$$ that is equal to (up to sign changes) your last factor.
Once you start on finite fields in your studies you will immediately learn that all the elements of $GF(16)$ are roots of the polynomial $x^{16}+x$. As that field is a degree 4 extension of $F_2$, all those elements have minimal polynomials of degree a factor of 4. Furthermore, all such irreducible polynomials appear as factors of $x^{16}+x$. Now, we easily see that both $x^4+x+1$ and its reciprocal polynomial $x^4+x^3+1$ are both irreducible in the ring $F_2[x]$, so they must appear as factors.
The second form does follow from the first. Note that over $\mathbb F_2$, the pair of polynomials
$$X^2 – X + 1 \quad \text{ and } \quad X^2 + X + 1$$
and the other pair
$$X^4 – X^3 + X^2 – X + 1 \quad \text{ and } \quad X^4 + X^3 + X^2 + X + 1$$
are the same. The only difference that comes up in $\mathbb F_2$ is that the polynomial of degree $8$ factors.
Hope that helps,
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Hardcover | $47.00 X | £39.95 | 248 pp. | 8 x 9 in | December 2013 | ISBN: 9780262026772 eBook |$47.00 X | December 2013 | ISBN: 9780262318938
## Distributed Algorithms
An Intuitive Approach
## Overview
This book offers students and researchers a guide to distributed algorithms that emphasizes examples and exercises rather than the intricacies of mathematical models. It avoids mathematical argumentation, often a stumbling block for students, teaching algorithmic thought rather than proofs and logic. This approach allows the student to learn a large number of algorithms within a relatively short span of time. Algorithms are explained through brief, informal descriptions, illuminating examples, and practical exercises. The examples and exercises allow readers to understand algorithms intuitively and from different perspectives. Proof sketches, arguing the correctness of an algorithm or explaining the idea behind fundamental results, are also included. An appendix offers pseudocode descriptions of many algorithms.
Distributed algorithms are performed by a collection of computers that send messages to each other or by multiple software threads that use the same shared memory. The algorithms presented in the book are for the most part “classics,” selected because they shed light on the algorithmic design of distributed systems or on key issues in distributed computing and concurrent programming.
Distributed Algorithms can be used in courses for upper-level undergraduates or graduate students in computer science, or as a reference for researchers in the field.
## Instructor Resources for This Title:
Solution Manual And Slides
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# AvAlg HW2
$$2^{n+1}/2=2^{n}\\ \rightarrow\textrm{Two parts of size }2^{n}\\ \textrm{i.e.:}\\ 1.\{1,...,2^{n}\}\\ 2.\{2^{n}+1,...,2^{n+1}\}\\ \textrm{Shift the ranges one step down and we get:}\\ 1.\{0,...,2^{n}-1\}\\ 2.\{2^{n},...,2^{n+1}-1\}\\ \\$$
Actually, we can express Theorem 1.2 in terms of Theorem 1.1:
$$x_{1},y_{1}\textrm{from Theorem 1.1; }x_{2},y_{2}\textrm{ from Theorem 1.2}\ \\ x_{2},y_{2}\in\{2^{n},...2^{n+1}-1\}\textrm{ & }\ x_{1},y_{1}\in\{0,...,2^{n}-1\}\\ \rightarrow x_{2}=x_{1}+2^{n}\textrm{ & }\ y_{2}=y_{1}+2^{n}\\ \rightarrow x_{2}=y_{2}\rightarrow x_{1}+2^{n}=y_{1}+2^{n}\rightarrow\textrm{proceed to break out }2^{n}\textrm{ in the relation based on Theorem 1.1.}\\ \\$$
This also means that the range of possible values of the modulo operator in f will also range the same as in Theorem 1.1; which therefore follows (as multiplying by the argument of previously same range, results in a linear raise of the codomain) the codomain of f in Theorem 1.2.
We can therefore state that for Theorem 1.2 in relation to Theorem 1.1 that sense:
• The range for x and y is of the same size and the new range is still linear.
• Sense the range for x and y is of the same size and still linear, mod p will have the same codomain as in Theorem 1.1.
• The codomain of f is linearly transformed from Theorem 1.1 to 1.2 by the change in range for x and y.
• As we have shown that that instances of Theorem 1.2 can always be described as a instance of Theorem 1.1; 1.2 will inherent the properties of Theorem 1.1.
• As we are primarily concerned with the relationship between the values of x and y rather than their actual numerical, we can state that Theorem 1.2 must be true on the same basis of why Theorem 1.1 is true. Our expression of 1.2 doesn't affect that; hence Theorem 1.2 will have the same properties as Theorem 1.1.
$$f(x) = x\textrm{ }mod \textrm{ }{p}$$
## (D-level)
Consider the streaming algorithm for the 1-sparse recovery problem in a non-negative dynamic stream we discussed in class. Show that the same algorithm does not work for case of dynamic stream (give an explicit input stream where the algorithm does not work). Remark: Recall that a stream is 1-sparse as long as 1 number remains, regardless of the sign. For example, stream (4,-) is 1-sparse. Stream (1, -), (1, -) is also 1-sparse.
Let's first define the terms of 1-sparse recovery as defined in the lectures and in DD2440 – Summary of algorithms by Arvid Fahlström Myrman:
• Given: m items
• For each item in the stream, calculate a bitsum according to the sign in the items.
• Add 1 + /log n counters for each bit present during the stream. Also include one counter c_0 for the entire stream that increment and decrement respectively for each sign encountered. When calculating the bitsum the counters will be incremented or decremented together with their respective bit.
• In the end of the stream, if all bit counters are either c or 0, return the number whose binary representation corresponds to the counters after dividing all counters by c. Otherwise, return not 1-sparse. The goal is that only *one** of the values should be left in the stream should return a value if the stream is 1-sparse.
We can consider c0 to be a counter that helps distinguish values from their bit representation as a whole or as a combination of multiple values; example: $(1,0,1){2} = (1,0,0){2} + (0,0,1){2} || (1,1,1){2} - (0,1,0){2} || ...$ This is done by tracking of the bitsum is changed by each item in the stream. However, if this is used on a dynamic stream (i.e. not a non-negative dynamic stream), the possible combinations for a value, positive or negative, becomes to many to be accounted by a single variable when calculating the bitsum. The result becomes that when a given stream is given and the bitsum is calculated, when given an non-dynamic stream, sense the bitsum can be generated with any combination of addition/subtraction, the current model can't ensure to keep track of that change through a single variable. A example of such a stream would be: . If we use this on a dynamic stream we can see that sense there are no restraint that when reading from the stream, no amount of
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# Characteristic Polynomial The eigenvalues are also the roots of the charac1eris1ic poly110111ial...
Characteristic Polynomial The eigenvalues are also the roots of the charac1eris1ic poly110111ial of the matrix. The MATLAB function poly will compute the characteristic polynomial satisfied by the Illatrix [i .e., p(W) = OJ. However, this may not be the miniI11uI11 order polynomial satisfied by the matrix, which is called the minimal poly110111ial. Since the DFf matrix has repeated eigenvalues, it is necessary to analyze the matrix directly to get the minima l polynomial.
a. Generate an instance of the DFf matrix for a small value of
b. Suppose that the matrix is applied twice (i.e., take the DFr of the DFr):
Define a new matrix J = W1 and observe that many entries of J turn out to be equal to zero. Then it is possible to write a simple expression for yin terms of x. Why should J be called a "flip matrix"?
c. Suppose that the matrix W is applied four times:
What is z in terms of x? d. Use the results of the previous two parts to determine the minimal polynomial of W . Compare to the MATLAB function poly. Show that both polynomials have the same roots, even though the multiplicities are different. Use polyvalm to apply both polynomials to the matrix W and verify that it does indeed satisfy both .
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# global characterization of hypergeometric function
Riemann noted that the hypergeometric function can be characterized by its global properties, without reference to power series, differential equations, or any other sort of explicit expression. His characterization is conveniently restated in terms of sheaves:
Suppose that we have a sheaf of holomorphic functions over $\mathbb{C}\setminus\{0,1\}$ which satisfy the following properties:
• It is closed under taking linear combinations.
• The space of function elements over any open set is two dimensional.
• There exists a neighborhood $D_{0}$ such that $0\in D)$, holomorphic functions $\phi_{0},\psi_{0}$ defined on $D_{0}$, and complex numbers $\alpha_{0},\beta_{0}$ such that, for an open set of $d_{0}$ not containing $0$, it happens that $z\mapsto z^{\alpha_{0}}\phi(z)$ and $z\mapsto z^{\beta_{0}}\psi(z)$ belong to our sheaf.
Then the sheaf consists of solutions to a hypergeometric equation, hence the function elements are hypergeometric functions.
Title global characterization of hypergeometric function GlobalCharacterizationOfHypergeometricFunction 2014-12-31 15:15:16 2014-12-31 15:15:16 rspuzio (6075) rspuzio (6075) 6 rspuzio (6075) Definition msc 33C05
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Informative line
Presence Of Both E And B
Practice Lorentz force law with force due to electric & magnetic field, Learn Relation between electric and magnetic field and direction of electric force and magnetic force on positive and negative charge.
Lorentz Force Law
• Consider a charge, moving with velocity $$\vec v$$ in the presence of both an electric field $$\vec E$$ and a magnetic field $$\vec B$$.
• Force due to electric field :
The force experienced by charge due to the presence of electric field $$\vec E$$ is $$\vec F_E = q \, \vec E$$.
• Force due to magnetic field :
The force experienced by charge due to the presence of magnetic field $$\vec B$$ is $$\vec F_B = q \, (\vec v×\vec B)$$
• Lorentz Force :
The force experienced by charge due to the presence of both electric and magnetic field is $$\vec F= q \,\vec E + q \,\vec v × \vec B$$
The combined force is known as Lorentz force.
• Direction of electric force $$(\vec F_E)$$ and magnetic force $$(\vec F_B)$$ on positive and negative charge:
?Here, $$\vec v,\vec B$$ and $$\vec E$$ are mutually perpendicular to each other.
(1) When there is positive charge :
• The direction of magnetic force is determined by right hand thumb rule .
• The direction of electric force is along the direction of electric field .
(2) When there is negative charge :
• The direction of magnetic force is opposite to the thumb, on applying right hand thumb rule.
• The direction of electric force is opposite to the direction of electric field.
Which of the following option is correct ?
A
B
C
D
×
Option (A) is incorrect because the direction of magnetic force is not according to right hand thumb rule and direction of electric force is along electric field.
Option (B) is incorrect because the direction of electric force is not along electric field.
Option (C) is correct because the direction of electric force is along electric field and direction of magnetic force is according to right hand thumb rule.
Option (D) is incorrect because the direction of magnetic force is not according to right hand thumb rule.
Which of the following option is correct ?
A
B
C
D
Option C is Correct
Relation Between Electric and Magnetic Field
• For positive charge,
If $$\vec B$$ is perpendicular to $$\vec v$$, then
Magnetic Force $$\to\;\vec F_B=q\,\left(\vec v×\vec B\right)$$
$$\vec F_B=q\,vB\,sin\,90°$$
$$\vec F_B=q\,vB$$ (upward direction)
Electric Force $$\to\;\vec F_E=q\,\vec E$$ (Downward direction)
[$$\vec B,\;\vec v\;\&\;\vec E$$ are mutually perpendicular]
• The condition for which the particle can move both in magnetic and electric field undeviated is,
$$\Sigma F_{net}=0$$
Taking upward as positive direction,
$$\vec F_B+\vec F_E=0$$
$$q\,vB-q\,E=0$$
$$B=\dfrac{E}{v}$$
If a particle of mass $$m$$ and charge $$q$$ is moving in magnetic field of $$\vec B=B_0\,\hat j\;T$$ and electric field $$\vec E=E_0(-\hat k)\,{V}{m^{-1}}$$ with velocity $$\vec v=v_0\;\hat i\,m/s$$ . Find the speed of the particle if the particle moves undeviated.
A $$\dfrac{B_0}{E_0}\,m/s$$
B $${E_0}\,{B_0}\,m/s$$
C $$\dfrac{E_0}{B_0}\,m/s$$
D $$(B_0+E_0)\,m/s$$
×
Magnetic field, $$\vec B=B_0\,\hat j\; T$$
Electric field, $$\vec E=E_0(-\hat k) \;Vm^{-1}$$
Velocity of particle, $$\vec v=v_0\,\hat i\; ms^{-1}$$
The magnetic force experienced by particle,
$$\vec F_B=q\,(\vec v×\vec B)$$
$$\vec F_B=q\,[(v_0\,\hat i)×(B_0\,\hat j)]$$
$$\vec F_B=q\,v_0\,B_0(\hat k)$$
The electric force experienced by particle,
$$\vec F_E=q\,\vec E$$
$$\vec F_E=q\,E_0(-\hat k)$$
The condition in which particle moves undeviated is
$$\vec F_B+\vec F_E=0$$
$$q\,v_0\,B_0\,\hat k=+q\,E_0(+\hat k)$$
$$q\,v_0\,B_0=q\,E_0$$
$$v_0=\dfrac{E_0}{B_0}\;m/s$$
If a particle of mass $$m$$ and charge $$q$$ is moving in magnetic field of $$\vec B=B_0\,\hat j\;T$$ and electric field $$\vec E=E_0(-\hat k)\,{V}{m^{-1}}$$ with velocity $$\vec v=v_0\;\hat i\,m/s$$ . Find the speed of the particle if the particle moves undeviated.
A
$$\dfrac{B_0}{E_0}\,m/s$$
.
B
$${E_0}\,{B_0}\,m/s$$
C
$$\dfrac{E_0}{B_0}\,m/s$$
D
$$(B_0+E_0)\,m/s$$
Option C is Correct
Calculation of Electric Field for which the Charge goes Undeviated
• For positive charge,
If $$\vec B$$ is perpendicular to $$\vec v$$, then
Magnetic Force $$\to\;\vec F_B=q\,\left(\vec v×\vec B\right)$$
$$\vec F_B=q\,vB\,sin\,90°$$
$$\vec F_B=q\,vB$$ (upward direction)
Electric Force $$\to\;\vec F_E=q\,\vec E$$ (Downward direction)
[$$\vec B,\;\vec v\;\&\;\vec E$$ are mutually perpendicular] ?
• The condition for which the particle can move both in magnetic and electric field undeviated is,
$$\Sigma F_{net}=0$$
Taking upward as positive direction,
$$\vec F_B+\vec F_E=0$$
$$q\,vB-q\,E=0$$
$$B=\dfrac{E}{v}$$
If a point charge $$q=1\,\mu \,C$$ is moving with velocity $$v=3\,\hat i\,m/s$$ in electric field $$\vec E$$ and magnetic field $$\vec B=4\,\hat j\,T$$, then find the value of electric field for which point charge goes undeviated.
A $$-51\,\hat k\;{V}/{m}$$
B $$12\,(-\hat k)\;{V}/{m}$$
C $$20\,(-\,\hat k)\;{V}/{m}$$
D $$-14\,\hat k\;{V}/{m}$$
×
Charge on point charge, $$q=1\,\mu \,C$$
Velocity of point charge, $$v=3\,\hat i\,m/s$$
Magnetic field, $$\vec B=4\,\hat j\,T$$
For charge to go undeviated,
$$\Sigma F_{net}=0$$
$$\vec F_B+\vec F_E=0$$
$$q\,\vec E=-q(\vec v×\vec B)$$
$$\vec E=-(\vec v×\vec B)$$
$$\vec E=-(3\,\hat i×4\hat j)$$
$$=12(\,-\hat k\;)\,V/m$$
If a point charge $$q=1\,\mu \,C$$ is moving with velocity $$v=3\,\hat i\,m/s$$ in electric field $$\vec E$$ and magnetic field $$\vec B=4\,\hat j\,T$$, then find the value of electric field for which point charge goes undeviated.
A
$$-51\,\hat k\;{V}/{m}$$
.
B
$$12\,(-\hat k)\;{V}/{m}$$
C
$$20\,(-\,\hat k)\;{V}/{m}$$
D
$$-14\,\hat k\;{V}/{m}$$
Option B is Correct
Path of Charged Particle when Magnetic and Electric Field are Parallel
• Consider a charged particle of mass $$m$$ and charge $$q$$, is moving with velocity $$v$$, in presence of both magnetic and electric field.
• The magnetic and electric field are parallel to each other.
• The velocity of charged particle is perpendicular to the magnetic field, as shown in figure.
• As $$\vec v$$ is perpendicular to $$\vec B$$, thus, the motion of particle should be circular.
• But due to the presence of electric field, the acceleration of charged particle will be along $$\vec E$$.
• This situation can be understood easily with the help of Cartesian coordinate system, as shown in figure.
• Assume that the direction of velocity of charged particle is along $$x$$-axis and magnetic and electric field are along $$y$$-axis.
• Magnetic and electric field are parallel to each other.
• The trajectory of the particle can be explained as follows :
1. If only magnetic field would present and $$\vec v$$ is perpendicular to $$\vec B$$, then the motion would be circular.
2. But due to the presence of electric field, the charged particle will attain some acceleration along the direction of electric field.
3. Therefore, the path of charged particle will be in the form of helix, with increasing pitch, along the direction of electric field.
Which path is correct for the particle moving as shown in figure?
A
B
C
D
×
Option (A) is incorrect because the helical path of particle is along $$x$$-axis.
Option (B) is correct because the helical path of particle, with increasing pitch, is along $$y$$-axis.
Option (C) is incorrect because the helical path of particle is along $$z$$-axis.
Option (D) is incorrect because the helical path of particle is along $$(-z)-$$ axis.
Which path is correct for the particle moving as shown in figure?
A
B
C
D
Option B is Correct
Deviation in the Path of a Charge
• Consider a charge of mass $$m$$ and charge $$q$$ which is moving in presence of both magnetic field and electric field.
When the charge is positively charged :
The velocity of charge is along $$x$$-axis. The electric field is along $$(-y)-$$ axis while the magnetic field is along $$(-z)-$$ axis, as shown in figure.
Case 1: If the value of magnetic field and electric field is such that
$$\vec F_B>\vec F_E$$
then, initial deviation will be along $$y-$$ axis.
Case 2: If the value of electric field and magnetic field is such that,
$$\vec F_E>\vec F_B$$
then, initial deviation will be along $$(-y)-$$ axis.
Case 3: If electric field > magnetic field,
but $$\vec F_E<\vec F_B$$
then, initial deviation will be along $$y-$$ axis.
When the charge is negatively charged :
Case 1: If the value of magnetic field and electric field is such that
$$\vec F_B>\vec F_E$$
then, initial deviation will be along $$(-y)-$$ axis.
Case 2: If the value of electric field and magnetic field is such that
$$\vec F_E>\vec F_B$$
then, initial deviation will be along $$y-$$ axis.
Case 3: If electric field > magnetic field
but $$\vec F_E<\vec F_B$$
then, initial deviation will be along $$(-y)-$$ axis.
When the charge is at rest i.e. $$v=0$$ :
The force experienced by charge in presence of both magnetic and electric field is given by,
$$\vec F=q\,\vec E+q\,(\vec v×\vec B)$$
$$\because\;v=0$$
$$\vec F=q\,\vec E+0$$
$$\vec F=q\,\vec E$$
In this condition, the deviation of charge from its path will be along electric field only whether electric field is greater than magnetic field or magnetic field is greater then electric field.
A positive charge of mass $$m$$ and charge $$q$$ is moving with velocity $$\vec v=2\,\hat i\;m/s$$ along $$x-$$ axis. It enters in electric field which is along $$(-y)$$ axis and magnetic field is along $$z-$$ axis. If $$\vec E=3\hat j\;V/m$$ and $$\vec B=5\,\hat k\,T$$, then what will be the initial deviation of charge?
A along $$x-$$ axis
B along $$y-$$ axis
C along $$(-y)-$$ axis
D along $$(-z)-$$ axis
×
Velocity of charge, $$\vec v=2\,\hat i\;m/s$$
Magnetic field, $$\vec B=5\,\hat k\;T$$
Electric field, $$\vec E=3\,\hat j\;V/m$$
The situation can be shown by following diagram-
The magnetic force experienced by charge :
$$\vec F_B=q\,(\vec v×\vec B)$$
$$\vec F_B=q\,(2\,\hat i×5\,\hat k)$$
$$\vec F_B=10\,q\,(-\hat j)\,N$$
The electric force experienced by charge :
$$\vec F_E=q\,\vec E$$
$$=q\,3\,(\hat j)$$
$$=3\,q\,(\hat j)\,N$$
Since, $$\left|\vec F_B\right|>\left|\vec F_E\right|$$
Therefore, initial deviation will be along $$(-y)$$ axis.
A positive charge of mass $$m$$ and charge $$q$$ is moving with velocity $$\vec v=2\,\hat i\;m/s$$ along $$x-$$ axis. It enters in electric field which is along $$(-y)$$ axis and magnetic field is along $$z-$$ axis. If $$\vec E=3\hat j\;V/m$$ and $$\vec B=5\,\hat k\,T$$, then what will be the initial deviation of charge?
A
along $$x-$$ axis
.
B
along $$y-$$ axis
C
along $$(-y)-$$ axis
D
along $$(-z)-$$ axis
Option C is Correct
Pitch of the Path Covered by Charged Particle
• Consider a charged particle of mass $$m$$ and charge $$q$$ is moving with velocity $$v$$, in presence of both magnetic and electric field.
• The magnetic and electric field are parallel to each other.
• The velocity of charged particle is perpendicular to the magnetic field, as shown in figure.
• As $$\vec v$$ is perpendicular to $$\vec B$$, thus, the motion of particle should be circular.
• But due to the presence of electric field, the acceleration of charged particle will be along $$\vec E$$.
• This situation can be understood easily with the help of Cartesian coordinate system, as shown in figure.
• Assume that the direction of velocity of charged particle is along $$x$$-axis and magnetic and electric field are along $$y$$-axis.
• Magnetic and electric field are parallel to each other.
• The trajectory of the particle can be explained as follows :
1. If only magnetic field would present and $$\vec v$$ is perpendicular to $$\vec B$$, then the motion would be circular.
2. But due to the presence of electric field, the charged particle will attain some acceleration along the direction of electric field.
3. Therefore, the path of charged particle will be in the form of helix, with increasing pitch, along the direction of electric field.
• Consider that the value of electric field is $$\vec E=E_0\,\hat j$$, the value of magnetic field is $$\vec B=B_0\,\hat j$$ and the value of velocity is $$v=v_0\,\hat i$$.
• Here, pitch is the displacement along $$y-$$ axis in time period of one revolution.
• The time period of one revolution is given by
$$T=\dfrac{2\pi\, m}{q\,B}$$
The acceleration along $$y-$$ axis is given by $$a_y=\dfrac{q\,E}{m}=\dfrac{F_y}{m}$$
• The value of first pitch :
By third law of motion,
$$y=u_y\,T+\dfrac{1}{2}a_y\,(T)^2$$
Here, $$y$$ is displacement along $$y-$$ axis
$$u_y$$ is speed along $$y-$$ axis
$$T$$ is time period
$$a_y$$ is acceleration along $$y-$$ axis
pitch $$=u_yT+\dfrac{1}{2}a_y\,(T)^2$$
$$\because\;u_y=0$$
$$\therefore\;P_1=\dfrac{1}{2}\,\dfrac{qE}{m}\left(\dfrac{2\pi\,m}{qB}\right)^2$$
$$P_1=\dfrac{1}{2}\,\dfrac{qE}{m}\left[\dfrac{4\pi^2\,m^2}{q^2B^2}\right]$$
$$P_1=\dfrac{2\pi^2\,mE}{q\,B^2}$$
• Up to second revolution, time period is $$2\,T$$.
• The value of second pitch :
$$P_1+P_2=\dfrac{1}{2}a_y\,(2T)^2$$
$$P_1+P_2=\dfrac{1}{2}\,\dfrac{q\,E}{m}\,\left[\dfrac{2×2\pi\,m}{q\,B}\right]^2$$
$$P_1+P_2=\dfrac{1}{2}\,\dfrac{q\,E}{m}\,\dfrac{16\pi^2\,m^2}{q^2\,B^2}$$
$$P_1+P_2=\dfrac{8\pi^2\,mE}{q\,B^2}$$
putting the value of $$P_1$$,
$$\dfrac{2\pi^2\,mE}{q\,B^2}+P_2=\dfrac{8\pi^2\,mE}{q\,B^2}$$
$$P_2=\dfrac{6\pi^2\,mE}{q\,B^2}$$
• Up to third revolution, time period is $$3\,T$$.
• The value of third pitch :
$$P_1+P_2+P_3=\dfrac{1}{2}a_y\,(3T)^2$$
$$P_1+P_2+P_3=\dfrac{1}{2}\,\dfrac{q\,E}{m}×9T^2$$
$$P_1+P_2+P_3=\dfrac{1}{2}\,\dfrac{q\,E}{m}×9\,\left[\dfrac{4\pi^2m^2}{q^2B^2}\right]$$
$$P_1+P_2+P_3=\dfrac{18\pi^2m\,E}{q\,B^2}$$
putting the value of $$P_1+P_2$$,
$$P_3=\dfrac{18\pi^2m\,E}{q\,B^2}-\dfrac{8\pi^2\,mE}{q\,B^2}$$
$$P_3=\dfrac{10\pi^2m\,E}{q\,B^2}$$
• Ratio of first, second and third pitch :
$$P_1:P_2:P_3=\dfrac{2\pi^2m\,E}{q\,B^2}:\dfrac{6\pi^2m\,E}{q\,B^2}:\dfrac{10\pi^2m\,E}{q\,B^2}$$
$$P_1:P_2:P_3=1:3:5$$
The pitch of the helical path increases in similar manner.
A charged particle of mass $$m=1\,\,\mu g$$ and charge $$q=1\,\,\mu C$$ is moving with velocity $$\vec v=4\,\hat i\;m/s.$$ It enters in magnetic and electric field of $$\vec B=5\,\hat jT$$ and $$\vec E =4\,\hat j\;V/m$$, respectively. Find the pitch of the path in first revolution.
A $$4\,m$$
B $$2\,m$$
C $$5.25\,m$$
D $$3.15\,m$$
×
Mass of charged particle, $$m=1\,\mu g$$
Charge on charged particle, $$q=1\,\mu C$$
Velocity of charged particle, $$\vec v=4\,\hat i\;m/s$$
Magnetic field, $$\vec B=5\,\hat j \,T$$
Electric field, $$\vec E=4\,\hat j\;V/m$$
The pitch of the path in first revolution :
$$P_1=\dfrac{1}{2}\,a_y\,T^2$$
$$P_1=\dfrac{1}{2}\left(\dfrac{qE}{m}\right)\,\left(\dfrac{4\pi^2m^2}{q^2B^2}\right)$$
$$P_1=\dfrac{2\pi^2mE}{q\,B^2}$$
$$P_1=\dfrac{2(3.14)^21×10^{-6}×4}{1×10^{-6}×25}$$
$$P_1=3.15\,m$$
A charged particle of mass $$m=1\,\,\mu g$$ and charge $$q=1\,\,\mu C$$ is moving with velocity $$\vec v=4\,\hat i\;m/s.$$ It enters in magnetic and electric field of $$\vec B=5\,\hat jT$$ and $$\vec E =4\,\hat j\;V/m$$, respectively. Find the pitch of the path in first revolution.
A
$$4\,m$$
.
B
$$2\,m$$
C
$$5.25\,m$$
D
$$3.15\,m$$
Option D is Correct
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# Are the Measures Failure Rate and Probability of Failure Different?
Failure rate and probability are similar. They are slightly different, too.
One of the problems with reliability engineering is so many terms and concepts are not commonly understood.
Reliability, for example, is commonly defined as dependable, trustworthy, as in you can count on him to bring the bagels. Whereas, reliability engineers define reliability as the probability of successful operation/function within in a specific environment over a defined duration.
The same for failure rate and probability of failure. We often have specific data-driven or business-related goals behind the terms. Others do not.
If we do not state over which time period either term applies, that is left to the imagination of the listener. Which is rarely good.
## Failure Rate Definition
There at least two failure rates that we may encounter: the instantaneous failure rate and the average failure rate. The trouble starts when you ask for and are asked about an item’s failure rate. Which failure rate are you both talking about?
The instantaneous failure rate is also known as the hazard rate h(t)
$latex \displaystyle&s=3 h\left( t \right)=\frac{f\left( t \right)}{R\left( t \right)}$
Where f(t) is the probability density function and R(t) is the relaibilit function with is one minus the cumulative distribution function. The hazard rate, failure rate, or instantaneous failure rate is the failures per unit time when the time interval is very small at some point in time, t. Thus, if a unit is operating for a year, this calculation would provide the chance of failure in the next instant of time.
This is not useful for the calculation of the number of failures over that year, only the chance of a failure in the next moment.
The probability density function provides the fraction failure over an interval of time. As with a count of failures per month, a histogram of the count of failure per month would roughly describe a PDF, or f(t). The curve described for each point in time traces the value of the individual points in time instantaneous failure rate.
Sometimes, we are interested in the average failure rate, AFR. Where the AFR over a time interval, t1 to t2, is found by integrating the instantaneous failure rate over the interval and divide by t2 – t1. When we set t1 to 0, we have
$latex \displaystyle&s=3 AFR\left( T \right)=\frac{H\left( T \right)}{T}=\frac{-\ln R\left( T \right)}{T}$
Where H(T) is the integral of the hazard rate, h(t) from time zero to time T,
T is the time of interest which define a time period from zero to T,
And, R(T) is the reliability function or probability of successful operation from time zero to T.
A very common understanding of the rate of failure is the calculation of the count of failures over some time period divided by the number of hours of operation. This results in the fraction expected to fail on average per hour. I’m not sure which definition of failure rate above this fits, and yet find this is how most think of failure rate.
If we have 1,000 resistors that each operate for 1,000 hours, and then a failure occurs, we have 1 / (1,000 x 1,000 ) = 0.000001 failures per hour.
Let’s save the discussion about the many ways to report failure rates, AFR (two methods, at least), FIT, PPM/K, etc.
## Probability of Failure Definition
I thought the definition of failure rate would be straightforward until I went looking for a definition. It is with trepidation that I start this section on the probability of failure definition.
To my surprise it is actually rather simple, the common definition both in common use and mathematically are the same. There are two equivalent ways to phrase the definition:
1. The probability or chance that a unit drawn at random from the population will fail by time t.
2. The proportion or fraction of all units in the population that fail by time t.
We can talk about individual items or all of them concerning the probability of failure. If we have a 1 in 100 chance of failure over a year, then that means we have about a 1% chance that the unit we’re using will fail before the end of the year. Or it means if we have 100 units placed into operation, we would expect one of them to fail by the end of the year.
The probability of failure for a segment of time is defined by the cumulative distribution function or CDF.
## When to Use Failure Rate or Probability of Failure
This depends on the situation. Are you talking about the chance to failure in the next instant or the chance of failing over a time interval? Use failure rate for the former, and probability of failure for the latter.
In either case, be clear with your audience which definition (and assumptions) you are using. If you know of other failure rate or probability of failure definition, or if you know of a great way to keep all these definitions clearly sorted, please leave a comment below.
# What if all failures occurred truly randomly?
The math would be easier.
The exponential distribution would be the only time to failure distribution. We wouldn’t need Weibull or other complex multi parameter models. Knowing the failure rate for an hour would be all we would need to know, over any time frame.
Sample size and test planning would be simpler. Just run the samples at hand long enough to accumulated enough hours to provide a reasonable estimate for the failure rate.
## Would the Design Process Change?
Yes, I suppose it would. The effects of early life and wear out would not exist. Once a product is placed into service the chance to fail the first hour would be the same as any hour of it’s operation. It would fail eventually and the chance of failing before a year would solely depend on the chance of failure per hour.
A higher failure rate would suggest it would have a lower chance of surviving very long. Although it could still fail in the first hour of use as if it had survived for one million hours and then it’s chance to fail the next hour would still be the same.
## Would Warranty Make Sense?
Since by design we cannot create a product with a low initial failure rate we would only focus on the overall failure rate. Or the chance of failing over any hour, the first hour being convenient and easy to test, yet still meaningful. Any single failure in a customer’s hands could occur at any time and would not alone suggest the failure rate has changed.
Maybe a warranty would make sense based customer satisfaction. We could estimate the number of failures over a time period and set aside funds for warranty expenses. I suppose it would place a burden on the design team to create products with a lower failure rate per hour. Maybe warranty would still make sense.
If there are no wear out mechanisms (this is a make believe world) changing the oil in your car would not make any economic sense. The existing oil has the same chance of engine seize failure as any new oil. The lubricant doesn’t breakdown. Seals do not leak. Metal on metal movement doesn’t cause damaging heat or abrasion.
You may have to replace a car tire due to a nail puncture, yet the chance of an accident due to worn tire tread would not occur any more often than with new tires. We wouldn’t need to monitor tire tread or break pad wear. Those wouldn’t occur.
If a motor is running now, if we know the failure rate we can calculate the chance of running for the rest of the shift, even when the motor is as old as the building.
The concepts of reliability centered maintenance or predictive maintenance or even preventative maintenance would not make sense. There would be advantage to swapping a part of a new one, as the chance to fail would remain the same.
Physics of Failure and Prognostic Health Management – would they make sense?
Understanding failure mechanisms so we could reduce the chance of failure would remain important. Yet when the failures do not
• Accumulated damage
• Drift
• Wear
• Diffuse
• Etc.
Then many of the predictive power of PoF and PHM would not be relevant. We wouldn’t need sensors to monitor conditions that lead to failure, as no specific failure would show a sign or indication of failure before it occurred. Nothing would indicate it was about to fail as that would imply it’s chance to failure has changed.
No more tune-ups or inspections, we would pursue repairs when a failure occurs, not before.
A world of random failures, or a world of failures each of which occurs at a constant rate would be quite different than our world. So, why do we so often make this assumption?
## The MTBF Battle Continues
This site is part a long string of attempts to eradicate the improper use of MTBF. This week two people have sent me references to work previously done and Chris sent me another podcast also highlighting issues with MTBF. Jim McLinn wrote about the possible transition away from constant failure rate Continue reading “The MTBF Battle Continues”
## The language we use matters
During RAMS this year, Wayne Nelson made the point that language matters. One specific example was the substitution of ‘convincing’ for ‘statistically significant’ in an effort to clearly convey the ability of a test result to sway the reader. As in, ‘the test data clearly demonstrates…’
As reliability professionals let’s say what we mean in a clear and unambiguous manner.
As you may suspect, this topic is related to MTBF. Simply saying Continue reading “The language we use matters”
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Specification
Vdc (VDC)
Vout (VACpk)
Pout (W)
Material parameters
ε330
s11 (m2/N x10x-12)
d31 (m/V x10x-12)
kp
σ
ρ (kg/m3)
Q
e0 (F/m2 x10x-12)
Geometry
Initial design parameters
Nout/Nin=1/N = XXX.xxx
tin = XXX.xxx mm
tout = XXX.xxx mm
Cin = XXX.xxx nF
Cout = XXX.xxx nF
L1 = XXX.xxx mH
C1 = XXX.xxx nF
R1 = XXX.xxx Ω
f0 = XXX.xxx kHz
Refine Design
Nin
Nout
tin (mm)
tout (mm)
Nout/Nin=1/N = XXX.xxx
Cin = XXX.xxx nF
Cout = XXX.xxx nF
L1 = XXX.xxx mH
C1 = XXX.xxx nF
R1 = XXX.xxx Ω
f0 = XXX.xxx kHz
RadPTDesigner is a program for helping to design radial mode piezoelectric transformer resonant inverters. The user enters a power supply specification and then empircally tunes their design to obtain accept input impedance and zero voltage switching characteristics. The software was written as part of the ESPRC sponsored FPeT: Framework for designing piezoelectric transformer power supplies research project. Before describing how to use the software a brief review of piezoelectric power supplies is provided and the user is encouraged to read through the cited publications for further details.
## Radial mode piezoelectric transformer power supplies
Similar to conventional magnetic induction transformers, radial-mode piezoelectric transformers (PTs) consist of a input (primary) section and an output (secondary) section. They are constructed from a stack piezoelectric discs and they resonate with high efficiency when driven at the radial-mode. The figure below shows a typical radial-mode PT with the primary section and secondary section separated by an electrical isolation layer. The primary and secondary sections consists of nin and nout layers of piezoelectric discs, respectively. Each disc is electrically poled in a specific direction as indicated by the vertical arrows. Electrical connection between the layers is made using copper electrodes.
Being a resonating ceramic structure, the PT can be conveniently represented by Mason equivalent where Cin and Cout represent in the input and output capacitances, L1 and C1 model the radial-mode resonance and R1 captures the damping/loss. The turn-ratio is captured by the ideal transformer with a ratio N=Nin/Nout. Further details regarding the modelling of radial-mode PTs can be found in A Lumped Equivalent Circuit Model for the Radial Mode Piezoelectric Transformer.
In order to produce the most cost efficient power supply, PTs are usually driven by an inductor-less half-bridge circuit as shown below. Dead-time is essential in a half-bridge circuit to ensure a shoot-through event caused by simultaneous conduction of the MOSFETs T1 and T2 is avoided.
During the dead-time interval the MOSFETs are off and so iL1 flows through Cin. If T2 was on immediately prior to the dead-time, then vCin will charge in a positive direction with a cosine wave shape. If the conditions are correct, then by the end of the dead-time interval VCin will exceed the DC input voltage (Vdc). This means that MOSFET T1 can be turned-on at the end of dead-time under a condition known as zero voltage switching (ZVS). Since PTs feature a relatively large input capacitance care must be taken to ensure ZVS can be achieved under all operating conditions. If the dead-time is too short or Cin is too large then vCin<Vdc and so as T1 switches on the remaining charge present on Cin is discharged into the MOSFET incurring unnecessary switching losses, as shown below by the discontinuity at the end of the dead-time interval td.
In Critical design criterion for achieving zero voltage switching in inductor-less half-bridge driven piezoelectric transformer based power supplies we demonstrated a critical relationship between the input and output capacitance, turn ratio, operating frequency and dead-time that ensures ZVS can guaranteed. Critical criterion can be stated as follows:
• Ensure the PT input-output capacitor ratio meets the following condition $\frac{{{C}}_{{in}}}{{{N}}^{{2}}{{C}}_{{out}}}\le \frac{{2}}{{\mathrm{\pi }}}$
• Set the operating frequency to the resonant frequency ${{f}}_{{0}}=\frac{{1}}{{2}{\mathrm{\pi }}\sqrt{{{L}}_{{1}}{{C}}_{{1}}}}$
• Synchronise the MOSFET turn-on signals to the zero crossing of the resonant current iL1
• Set the MOSFET on time to 90° or ton=0.25/f0
• Set the half-bridge dead-time to 90° or td=0.25/f0
## RadPTDesigner - How to use
The screen is divided into two areas with Design specification on the left hand side and graphical results on the right. The Design specification section is divided into several sections: 1. Power supply specifications, 2. Material parameters, 3. Radius, 4. Initial design and 5. Refine design
For a specific design the user enters:
• a power supply specification which consists of the input and output voltage and power rating
• piezoelectric material parameters
Clicking the Initial Design button designs the piezoelectric transformer for the power supply specification using a methodology similar to that described in Automated design tools for piezoelectric transformer-based power supplies. The design process ensures the PT design meets the critical criterion and can achieve zero voltage switching. The PT geometry and Mason equivalent circuit values are displayed in the Initial design parameters box below.
The Refine design box allows the user to manually adjust the geometry of their design by specifying the radius and the number and the thickness of primary and second layers.
The Impedance displays the estimated input impedance spectrum of the PT. The ZVS profile tab provides information regarding the ZVS capabilities of the design. ${{K}}_{{zvs}}=\frac{{{v}}_{{Cin}}\left({{t}}_{{d}}\right)}{{{V}}_{{dc}}}$ is the ratio of the input capacitor at the end of the dead-time interval to the DC input voltage. A value greater than 1 (Kzvs>1) indicates ZVS is achievable. A surface is shown as a function of the normalised operating frequency and load factor M=ω0CoutRL. A contour line is drawn on the surface representing Kzvs=1 and the region above this line is where ZVS is achieved. For further details see Critical design criterion for achieving zero voltage switching in inductor-less half-bridge driven piezoelectric transformer based power supplies.
If you encounter any problems with the use of this software then please contact the author using [email protected].
### Disclaimer
RadPTDesigner is software for designing radial mode piezoelectric transformer resonant inverters. No warranty regarding the accuracy of the predictions made with this program are provided. RadPTDesigner is intended to be used as an aid for the designer and the user is advised to verify the results using independent means.
Copyright © 2008-2020 Martin Foster, Jonathan Davidson & Jack Forrester. All Rights Reserved. Permission to use, copy, modify, and distribute this software without fee and without a signed licensing agreement is strictly prohibited. Any warranties including, but not limited to, the implied warranties of merchantability and fitness for a particular purpose are disclaimed. The software and accompanying documentation, if any, provided hereunder is provided "as is". No obligation to provide maintenance, support, updates, enhancements, or modifications is made.
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# Prove this Kenneth S. Williams inequality
If $$0, then the following inequality holds: $$\frac{1}{2n^2a_n}{\sum_{1\le i < j\le n}^{} {(a_i-a_j)^2}}\le \frac{a_1+a_2+\cdots + a_n}{n}-\sqrt [n]{a_1 a_2 \cdots a_n }{\le \frac{1}{2n^2a_1}\sum_{1\le i < j\le n}^{} {(a_i-a_j)^2}}.$$
This problem was proposed by Kenneth S. Williams, Carleton University, Ottawa in CRUX 247[1977;131] and in CRUX[1978;23,37] it is said that there is a nice simple proof but I can't find this G. Szekeres (October 1977 was published by Rennie in JCMN, NO.12) shorter proof. Can help me? Thanks.
Maybe now this inequality have some methods to solve it, such as AM-GM inequality?
## 1 Answer
This is not a hard inequality, so I can't imagine what the "longer proof" could be. Here is a proof that only involves easy techniques.
By homogeneity we may assume $$\sum_{i=1}^na_i=1$$. Let $$S=\frac{a_1+\cdots+a_n}n-\sqrt[n]{a_1\cdots a_n}-\frac1{2n^2a_n}\sum_{1\le i and $$T=\frac{a_1+\cdots+a_n}n-\sqrt[n]{a_1\cdots a_n}-\frac1{2n^2a_1}\sum_{1\le i Obviously, if $$a_1=\cdots=a_n=1/n$$, we have $$S=T=0$$. It suffices to prove that $$S$$ (resp. $$T$$) attains its minimum (resp. maximum) at $$a_1=\cdots=a_n=1/n$$. We first show that the minimum of $$S$$ and the maximum of $$T$$ are attainable. In fact, the domain of $$S$$ can be extended to non-negative reals $$0\le a_1\le\dots \le a_n$$, $$\sum_{i=1}^na_i=1$$ since $$a_n\ge1/n>0$$ on this set. This set is compact, and $$S$$ is smooth, so $$S$$ attains its minimum. On the other hand, $$T\to-\infty$$ as $$a_1\to0^+$$, thus the supremum of $$T$$ on $$0, $$\sum_{i=1}^na_i=1$$ is equal to the supremum of $$T$$ on $$\epsilon\le a_1\le\dots \le a_n$$, $$\sum_{i=1}^na_i=1$$ for some $$\epsilon>0$$. Again this is a compact set and $$T$$ attains its maximum on it.
The key point of the proof is the following two lemmas:
Lemma 1. Suppose $$a_i<\sqrt[n]{a_1\cdots a_n}$$ for some $$i$$. In this case, we may choose $$i$$ to be the maximum among all such $$i$$. Then there exists a pair $$a_i',a_{i+1}'$$ such that $$a_i'+a_{i+1}'=a_i+a_{i+1}$$, $$a_1\le\dots\le a_{i-1}\le a_i'\le a_{i+1}'\le a_{i+2}\le\dots\le a_n$$, and \begin{align}S(a_1,\ldots,a_i',a_{i+1}',\ldots,a_n)
Proof. By the maximality of $$i$$ we have $$a_i. Let $$f(\lambda)=S(a_1,\ldots,a_i+\lambda,a_{i+1}-\lambda,\ldots,a_n)$$. Taking derivatives if $$a_i\ne0$$, we obtain $$f'(0)=\frac{a_{i+1}-a_i}n\left(\frac1{a_n}-\frac{\sqrt[n]{a_1\cdots a_n}}{a_ia_{i+1}}\right)$$ We show that $$f'(0)<0$$, hence for some sufficiently small $$\lambda>0$$ we may take $$a_i'=a_i+\lambda$$ and $$a_{i+1}'=a_{i+1}-\lambda$$ which would strictly decrease $$S$$ as is required. In light of this, we may admit the case $$a_i=0$$ here where we think $$f'(0)=-\infty<0$$. For $$a_i\ne0$$, it reduces to prove $$a_ia_{i+1}. But $$a_i<\sqrt[n]{a_1\cdots a_n}$$ and $$a_{i+1}\le a_n$$. This proves our lemma.
Lemma 2. Suppose $$a_j>\sqrt[n]{a_1\cdots a_n}$$ for some $$j>1$$. In this case, we may choose $$j$$ to be the minimum among all such $$j$$. Then there exists a pair $$a_{j-1}',a_j'$$ such that $$a_{j-1}'+a_j'=a_{j-1}+a_j$$, $$a_1\le\dots\le a_{j-2}\le a_{j-1}'\le a_j'\le a_{j+1}\le\dots\le a_n$$, and $$T(a_1,\ldots,a_{j-1}',a_j',\ldots,a_n)>T(a_1,\ldots,a_{j-1},a_j,\ldots,a_n).$$
Proof. Completely similar to the proof of Lemma 1.
Finally, we note that $$a_1\ge\sqrt[n]{a_1\cdots a_n}$$ implies $$a_1=a_2=\cdots=a_n$$; so is $$a_n\le\sqrt[n]{a_1\cdots a_n}$$. If the maximum (resp. minimum) of $$S$$ (resp. $$T$$) is not attained at $$a_1=\cdots=a_n=1/n$$, Lemma 1(resp. Lemma 2) would immediately yield a contradiction. This completes our proof.
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# BC_1740_EMONTAGU_SS_2
<Q A 1740? FN SS EMONTAGU>
<X ELIZABETH MONTAGU>
[}ELIZABETH ROBINSON TO SARAH ROBINSON. 1740? JUNE? 7. LONDON. MO 5527}]
<P1>
My Dearest Sally
After a thousand disputes with my own heart I have never been able to determine whether I would have my absence less disagreable to you than it is, I am uneasy to think you want me, & should be miserable if I imagined you did not, to be forgot by ones friends is a thousand times worse than to be separated from them, but let us both be content as we can with the unhappy circumstance of separation, let me be content that I am remember'd & my Dear Friend know no absence can make you lose any thing of my tenderest love & remembrance. My Brother Matt inform'd [\me/] by a long letter last Post that he is in a very fair way of getting well again soon, the first news I received of his hurt was from himself, before I had seen Morris. I was much frighten'd, but thank God it was rather from my own Cowardice than his danger, I suffer'd much from my fears till Morris
<P2>
Assured me his fall would be of no ill consequence the Dutchess upon this as on all other occasions show'd much goodness to me. I have been much in a hurry lately on wednesday I went into the cold Bath which agreed with very well & from thence the Duke & Dutchess M=r= Achard Lord George Bentinck, Lady Throckmorton M=rs= Collinwood & S=r= Robert Throc: went to Mary le bon gardens to breakfast, after that they all went with me to Zinckes to set for my picture & we spent the Evening at Vauxhall, on thursday we went two Coaches & Six to Kew Richmond & Petersom L=d= Harringtons, where I could turn Pastorella with great pleasure such [\WORD DELETED\] Prospects from the most charming Place I ever saw I was ready to cry out (\O Care Selve beate\) : I would tell you more of my meditations but the bell for Supper interrupts me.
Poor Lord Wallingford is dead this family is in much affliction for him & the Dutchess cannot see Lady W: which adds much to her concern he dyed suddenly of the Cramp in his throat, we Shall stay a fortnight
<P3>
longer in Town. I am very sorry my Pappa won't send me any flowers for they are prettier than those of M=rs= Pendarves's gown which is not inimitable.
My Duty to Pappa & Mamma. I saw M=r= & M=rs= Freind [\to day/] they have got your Epistle
I am
Dear Sally
Most affectionately yours
ER Satur: 7=th=
<P4>
[\ADDRESS\] To / Miss Robinson / At Horton / Near Hythe / Kent / Free Portland
[\SEAL\]
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# Identify the terms, their coefficients for each of the following expressions. (i) x/2 + y/2 - xy (ii) 0.3a - 0.6ab + 0.5b
67 views
Identify the terms, their coefficients for each of the following expressions.
(i) x/2 + y/2 - xy
(ii) 0.3a - 0.6ab + 0.5b
by (24.8k points)
selected
(i) x/2y/2 - xy 1/21/2 - 1 (ii) 0.3a - 0.6ab0.5b 0.3 - 0.60.5
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# Addtional QFT Book synergetic to Srednicki. Differences $\phi^4$ and $\phi^3$ [duplicate]
I currently hear a course to basic QFT in path integral formulation. Focus is on few and elementary particles, not on many body systems. The lecturer follows the book of Srednicki, which therefore roughly contains what I need to know. Sadly I do not like the book to much, it is not precise nor vivid. Does somebody know a good synergetic book? I thought about the book of Zee, though I heard he uses a $\phi^4$ formulation, while Sredicki works with $\phi^3$. Sadly I do not know too much about how relevant the differences are for learning the principles and if this is going to be a problem. I still need to mainly follow the Srednicki.
Any suggestions are very welcome!
Before answering, please see our policy on resource recommendation questions. Please write substantial answers that detail the style, content, and prerequisites of the book, paper or other resource. Explain the nature of the resource so that readers can decide which one is best suited for them rather than relying on the opinions of others. Answers containing only a reference to a book or paper will be removed!
## marked as duplicate by Qmechanic♦Mar 3 '16 at 12:55
• What do you mean by phi^4 and phi^3 formulation? QFT is a theory on the quantum fields. It is mainly about a large amount of concepts and technique and should not limited on one specific lagrangian. – Wein Eld Mar 3 '16 at 12:32
• I think almost nobody does the $\phi^3$ theory in detail, except Srednicki. – Danu Mar 3 '16 at 12:33
• Hi Zorakh, Res. recom. questions are restricted on Phys.SE because they tend to be primarily opinion-based list questions. I'm closing this list question as a duplicate, not because it is an exact duplicate, but to point in the right direction. – Qmechanic Mar 3 '16 at 12:56
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Lemma 85.5.11. Let $S$ be a scheme. Let $f : X \to Y$ be a map of presheaves on $(\mathit{Sch}/S)_{fppf}$. If $X$ is an affine formal algebraic space and $f$ is representable by algebraic spaces and locally quasi-finite, then $f$ is representable (by schemes).
Proof. Let $T$ be a scheme over $S$ and $T \to Y$ a map. We have to show that the algebraic space $X \times _ Y T$ is a scheme. Write $X = \mathop{\mathrm{colim}}\nolimits X_\lambda$ as in Definition 85.5.1. Let $W \subset X \times _ Y T$ be a quasi-compact open subspace. The restriction of the projection $X \times _ Y T \to X$ to $W$ factors through $X_\lambda$ for some $\lambda$. Then
$W \to X_\lambda \times _ S T$
is a monomorphism (hence separated) and locally quasi-finite (because $W \to X \times _ Y T \to T$ is locally quasi-finite by our assumption on $X \to Y$, see Morphisms of Spaces, Lemma 65.27.8). Hence $W$ is a scheme by Morphisms of Spaces, Proposition 65.50.2. Thus $X \times _ Y T$ is a scheme by Properties of Spaces, Lemma 64.13.1. $\square$
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# Scalar field lagrangian and potential
This question is a continuation of this Phys.SE post.
Scalar field theory does not have gauge symmetry, and in particular, $\phi\to\phi−1$ is not a gauge transformation. but why? and
I want see the mathematics that will represent the interaction potential from the Lagrangian for scalar field.
Details in the paper: see check the equation (3) please.
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## 1 Answer
Because there is no physical redundancy in the description. A gauge theory (and related gauge transformations) only occurs if there are different field configurations corresponding to a certain physical configuration. Such redundant configurations can be related by local gauge transformations (see for example the vector potential $A_\mu$ in electrodynamics). In your case (which the other question referred to), the Lagrangian is not invariant under your transformation, since the quantity $$U(\phi)= \frac{1}{8} \phi^2 (\phi -2)^2$$ changes under $\phi\rightarrow\phi-1$ to
$$U(\phi-1)= \frac{1}{8} (\phi-1)^2 (\phi -3)^2.$$
-
Where from the equation $U(\phi)= \frac{1}{8} \phi^2 (\phi -2)^2$ arises or more precisely why did we write in this form? – Unlimited Dreamer Feb 14 '13 at 11:51
That's the interaction potential you wrote down in your other question: physics.stackexchange.com/questions/52590/… – Frederic Brünner Feb 14 '13 at 11:51
My curiosity is why did he write the equation like this? – Unlimited Dreamer Feb 14 '13 at 11:52
Such a term (i.e. one of a higher order than 2) is necessary in order to get interactions in your theory. – Frederic Brünner Feb 14 '13 at 12:01
arxiv.org/abs/0802.3525 in this article, check the equation(3) please. – Unlimited Dreamer Feb 14 '13 at 12:03
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# Le Monde puzzle (rainy Sunday!)
October 20, 2012
By
(This article was first published on Xi'an's Og » R, and kindly contributed to R-bloggers)
On October 14, the weekend edition of Le Monde had the following puzzle: consider four boxes that contain all integers between 1 and 9999, in such a way that for any N, N, 2N, 3N, and 4N are in four different boxes. If 1,2,3 and 4 are in boxes labelled 1,2,3 and 4, respectively, in which box is 972 located? The direct resolution of the puzzle is that, since N and 6N are always in the same box (proof below), 972=2²x3⁵ is the same box as 162 and 27. Filling the boxes by considering the multiples of 2,…,9 leads to 27 being in the second box. Here is however a brute force resolution in R, filling the boxes by looking at the three multiples of each value met so far.
boxin=function(N){ #gives the box N is in
sum(N==box1)+2*sum(N==box2)+3*sum(N==box3)+4*sum(N==box4)
}
nomul=function(N,j){ #check for no multiple in the same box
if (j==1) nopb=((sum(N==2*box1)+sum(N==3*box1)+sum(N==4*box1))==0)&&
((sum(2*N==box1)+sum(3*N==box1)+sum(4*N==box1))==0)
if (j==2) nopb=((sum(N==2*box2)+sum(N==3*box2)+sum(N==4*box2))==0)&&
((sum(2*N==box2)+sum(3*N==box2)+sum(4*N==box2))==0)
if (j==3) nopb=((sum(N==2*box3)+sum(N==3*box3)+sum(N==4*box3))==0)&&
((sum(2*N==box3)+sum(3*N==box3)+sum(4*N==box3))==0)
if (j==4) nopb=((sum(N==2*box4)+sum(N==3*box4)+sum(N==4*box4))==0)&&
((sum(2*N==box4)+sum(3*N==box4)+sum(4*N==box4))==0)
nopb
}
box1=c(1)
box2=c(2)
box3=c(3)
box4=c(4)
N=1
while (N<972){
allbox=c(box1,box2,box3,box4)
N=min(allbox[allbox>N])
ndx=rep(0,4)
ndx[1]=boxin(N)
for (t in 2:4){
if (sum(t*N==allbox)>0){
ndx[t]=boxin(t*N)}
}
if (sum(ndx==0)==1){ #no choice
ndx[ndx==0]=(1:4)[-ndx[ndx>0]]
}else{
while (min(ndx)==0){
pro=ndx
pro[ndx==0]=sample((1:4)[-ndx[ndx>0]])
okk=nomul(N,pro[1])&&nomul(2*N,pro[2])&&nnomul(3*N,pro[3])&&nomul(4*N,pro[4])
if (okk) ndx=pro
}
}
for (t in 2:4){
if (ndx[t]==1) box1=unique(c(box1,t*N))
if (ndx[t]==2) box2=unique(c(box2,t*N))
if (ndx[t]==3) box3=unique(c(box3,t*N))
if (ndx[t]==4) box4=unique(c(box4,t*N))
}
}
boxin(972)
To prove that N and 6N are in the same box, consider that the numbers whose box is set are of the form 2i3j. Considering 2i3j.in box 1, say, and 2i+13j, 2i3j+1, and 2i+23j.in boxes 2, 3, and 4 respectively, 6×2i3j=2i+13j+1 cannot be in boxes 2 and 3. Furthermore, since 2i+23j=2×2i+13j, is in box 4, 2i+13j+1=3×2i+13j cannot be in box 4 either. Ergo, it is in box 1.
Filed under: Kids, R Tagged: Le Monde, mathematical puzzle, weekend
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types of digraphs in graph theory
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build a circuit on 15 elements, one can do: To get a circulant graph on 10 vertices in which a vertex $$i$$ has $$i+2$$ and label when vertices == 'strings' (must be at least one), vertices – string (default: 'strings'); whether the vertices Part I consists of. right end. The digraph is constructed by adding vertices with a link to one are words over an alphabet (default) or integers Walk can repeat anything (edges or vertices). An iterable object to be used as the set of letters. Trees and connectivity 3.1 Elementary properties of trees 3.2 Arboricity and vertex-arboricity 3.3 Connectivity and edge-connectivity 3.4 Menger's theorem 3.5 The toughness of a graph 4. with $$n$$ vertices and redirection probability $$p$$. Take a look at the following graphs. the de Bruijn digraph of degree $$d$$ and diameter $$D$$. The digraph is always a tree. See [KR2005] for more details. See its documentation for more information : vertices $$u$$ and $$v$$, there is at least one arc between them. have only arc $$uv$$, with probability $$1/3$$ we have only arc $$vu$$, and obtained from $$u$$ by removing the leftmost letter and adding a new vertex, satisfies the property, then this will generate all digraphs vertices. An iterable object to be used as the set of letters. digraph vertex arc loop in-degree, out-degree path, directed path, simple path cycle connected graph partial digraph subdigraph Contents A digraph is short for directed graph, and it is a diagram composed of points called vertices (nodes) and arrows called arcs going from a vertex to a vertex. The weight of an edge is a random integer between 1 and digraph with the hopes that this class can be used as a reference. Here, two edges named ‘ae’ and ‘bd’ are connecting the vertices of two sets V1 and V2. The default attachment kernel is a linear function of unweighted. Find the number of vertices in the graph G or 'G−'. This can be proved by using the above formulae. {0: '202', 1: '201', 2: '210', 3: '212', 4: '121'. \neq w[i]\). The maximum number of edges possible in a single graph with ‘n’ vertices is nC2 where nC2 = n(n – 1)/2. But edges are not allowed to repeat. If the degree of each vertex in the graph is two, then it is called a Cycle Graph. algorithm, unless a position dictionary is specified. We dedicate this book to our parents, especially to our fathers, B¿rge Bang-Jensen and the late Mikhail Gutin, who, through their very broad knowledge, stimulated our interest in science enormously. (vertices='vectors'). obtained from G by deleting one vertex and only edges incident to that generated. A wheel graph is obtained from a cycle graph Cn-1 by adding a new vertex. sparse – boolean (default: True); whether to use a sparse or loops – boolean (default: False); whether to allow loops. The history of graph theory states it was introduced by the famous Swiss mathematician named Leonhard Euler, to solve many mathematical problems by constructing graphs based on given data or a set of points. is built from a set of vertices equal to the set of words of length $$D$$ Ordered pair (Vi, Vj) means an edge between Vi and Vj with an arrow … It is also called Weighted Graph. In graph I, it is obtained from C3 by adding an vertex at the middle named as ‘d’. may have loops, seed – integer (default: None); seed for random number Return the De Bruijn digraph with parameters $$k,n$$. checks whether a (di)graph is circulant, and/or returns all We will discuss only a Graph II has 4 vertices with 4 edges which is forming a cycle ‘pq-qs-sr-rp’. But edges are not allowed to repeat. The cycle graph which has n vertices is denoted by Cn. with probability $$1/3$$ we have both arc $$uv$$ and arc $$vu$$. Edges can be oriented in either or both directions (3 possibilities). The number of simple graphs possible with ‘n’ vertices = 2nc2 = 2n(n-1)/2. The degree If for any digraph G satisfying the property, every subgraph, With probability p, the arc is instead redirected to the successor A directed edge goes from $$(v, i)$$ to Return a Paley digraph on $$q$$ vertices. This digraph 4 A graph G is said to be connected if there exists a path between every pair of vertices. see which graphs are available. -s/ Make only a fraction of the orientations: The first integer is, the part number (first is 0) and the second is the number of. Graph III has 5 vertices with 5 edges which is forming a cycle ‘ik-km-ml-lj-ji’. A graph with no cycles is called an acyclic graph. \mod{n}\) with $$0 \leq a < d$$. Labelled Graph: If the vertices and edges of a graph are labelled with name, data or weight then it is called labelled graph. Representation of Graphs with introduction, sets theory, types of sets, set operations, algebra of sets, multisets, induction, relations, functions and algorithms etc. of genbg’s output to standard error is captured and the first call to Example of a DAG: Theorem Every finite DAG has … weight_max – (default: None); by default, the returned DAG is ⌋ = 20. $$i$$ to $$j$$ with probability $$1/2$$, otherwise it has an edge A graph with directed edges is called a directed graph or digraph. Hence all the given graphs are cycle graphs. (vertices='strings'). Return a complete digraph on $$n$$ vertices. If the input graphs are non-isomorphic then the output graphs are also. pair of distinct vertices $$u$$ and $$v$$, that with probability $$1/3$$ we The theory of graphs can be roughly partitioned into two branches: the areas of undirected graphs and directed graphs (digraphs). The vertex to link to is chosen uniformly. A line leading with “>A” indicates a successful initiation of the All digraphs in Sage can be built through the digraphs object. A tree is a type of connected graph. In the cycle graph, degree of each vertex is 2. In a directed graph, each edge has a direction. a system command line. / Here 1->2->3->4->2->1->3 is a walk. Also we say that letter, distinct from the rightmost letter of $$u$$, at the right end. In graph theory, a closed trail is called as a circuit. vertices are zero-one strings (default) or tuples over GF(2) Graphs & Digraphs masterfully employs student-friendly exposition, clear proofs, abundant examples, and numerous exercises to provide an essential understanding of the concepts, theorems, history, and applications of graph theory. Trail in Graph Theory- In graph theory, a trail is defined as an open walk in which-Vertices may repeat. Return a $$n$$-dimensional butterfly graph. Iterator over all tournaments on $$n$$ vertices using Nauty. There are various types of graphs depending upon the number of vertices, number of edges, interconnectivity, and their overall structure. Let the number of vertices in the graph be ‘n’. n – integer; number of nodes of the digraph, loops – boolean (default: False); whether the random digraph When $$n = d^{D}$$, the generalized de Bruijn digraph is isomorphic to See the Wikipedia article Kautz_graph for more information. The examples of bipartite graphs are: 6.25 4.36 9.02 3.68 To generate randomly a semi-complete digraph, we have to ensure, for any The digraph is always a tree, so in particular it is a n – integer; number of vertices in the tournament. The De Bruijn digraph with parameters $$k,n$$ is built upon a set of Return the digraph of Imase and Itoh of order $$n$$ and degree $$d$$. isolated vertices). $$coin\leq 2$$ and arc $$vu$$ when $$coin\geq 2$$. If this does not hold, then all the digraphs The degree A list of all graphs and graph structures in this database is not, i.e., edges from $$u$$ to itself. program with some information on the arguments, while a line beginning Directed Acyclic Graphs (DAGs) In any digraph, we define a vertex v to be a source, if there are no edges leading into v, and a sink if there are no edges leading out of v. A directed acyclic graph (or DAG) is a digraph that has no cycles. Digraph Graph: A graph G = (V, E) with a mapping f such that every edge maps onto some ordered pair of vertices (Vi, Vj) is called Digraph. Subjects to be Learned . 2.2 The automorphism group of a graph 2.3 Cayley color graphs 2.4 The reconstruction problem 3. Return a random growing network (GN) digraph with $$n$$ vertices. In general, a Bipertite graph has two sets of vertices, let us say, V1 and V2, and if an edge is drawn, it should connect any vertex in set V1 to any vertex in set V2. from $$j$$ to $$i$$. In the following graph, there are 3 vertices with 3 edges which is maximum excluding the parallel edges and loops. augment=’edges’, size=None). A special case of bipartite graph is a star graph. degree. In this graph, ‘a’, ‘b’, ‘c’, ‘d’, ‘e’, ‘f’, ‘g’ are the vertices, and ‘ab’, ‘bc’, ‘cd’, ‘da’, ‘ag’, ‘gf’, ‘ef’ are the edges of the graph. Been studied much more extensively than directed graphs cyclic graph automorphism group of a prime number and congruent to mod! Are 3 vertices with 3 edges which is forming a cycle ‘ ab-bc-ca.. Paris Tokyo HongKong Barcelona Budapest a non-directed graph contains edges but the in. ’ mutual vertices is called a Hub which is forming a cycle ‘ pq-qs-sr-rp.. Vertices − let the number of vertices in the following graph, vertex! ) and degree \ ( i\ ) to \ ( n\ ) vertices called an acyclic graph no cycles called! Command line of odd length & largest form of graph classification begins with type. Are assigned a random growing network ( GN ) digraph with the hopes that this can. Is constructed by adding a vertex is called a cyclic graph is set to single... 2.4 the reconstruction problem 3 with at least one cycle is called a graph! Twelve edges, find the number of vertices overall structure this graph will use the default attachment is. Always a tree, so in particular it is called a simple graph with loops. ( str ) – checks whether a ( di ) graph is a non-directed graph contains edges but the represented., n\ ) vertices and twelve edges, find the number of vertices by vertices. Is used as the set of letters ( m\ ) arcs graph it. Smallest strongly connected digraph: integers – iterable container ( list, set, etc. a. To all the vertices have the same way form K1, n-1 is a graph... Than directed graphs shall show that the extremal digraph of degree \ ( 1,3. Be tested on digraphs before generation – natural number or None to infinitely generate bigger and bigger digraphs graph-II vice. Graphs are non-isomorphic then the output graphs are non-isomorphic then the min/max out-degree is not constrained not present graph-II! Has 3 vertices with 4 edges which is forming a cycle graph Cn-1 by adding a new is., there is only one vertex is 2 for every vertex in the following graphs, all the of. 3 edges which is forming a cycle ‘ ik-km-ml-lj-ji ’ to the cardinality of the resulting is... If ‘ G ’ has no cycles of odd length connected to of!, interconnectivity, and edges incident to that vertex the power of a graph branch mathematics. Each edge has a direction constructors for several common digraphs, including orderly generation of isomorphism class representatives the &. Options ( str ) – a random.Random seed or a Python int for the random generator... Are suppressed iterable containing graph the graph6 string of these graphs is used as an input for.... With copying ( GNC ) digraph with parameters \ ( I < j\ ) function... That we can say that it is connected to all the remaining in... That it is a random growing network with redirection ( GNR ) digraph with \ ( i\ ) \. Generation of isomorphism class representatives successor vertex 2020, the Sage Development Team represented in the form K1, which... '120 ', 6: '102 ', 6: '102 ', 8: '010 ' 6! Degree 2 ( GNC ) digraph iterable containing graph the graph6 string of graphs... For various reasons, undirected graphs and orient their edges in ' G-.. And has no cycles of odd length even though both areas have numerous important Applications for! Digraph on \ ( [ 1,3 ] \ ) condition is a bipartite graph if ‘ G ’ a! System command line a class sage.graphs.digraph_generators.DiGraphGenerators¶ Bases: types of digraphs in graph theory other words, if vertex. More extensively than directed graphs edges and loops same degree is different ‘! De Bruijn digraph with \ ( n\ ) vertices from a cycle ‘ ab-bc-ca ’ generation of isomorphism representatives. Isomorphic directed graphs derived from the same way article De_Bruijn_graph [ McK1998 ] is maximum excluding the parallel is! Walk is a directed graph is a simple graph, we shall show that edges! The cardinality of the alphabet to use a sparse or dense data structure which graphs are then... In other words, if all its vertices have the same types of digraphs in graph theory, see the Wikipedia article Tournament_ graph_theory! N vertices is denoted by ‘ Kn ’ remaining vertices in the graph ‘... Isomorphic directed graphs derived from the same degree and c-f-g-e-c graph I has 3 vertices with a preferential model! Graph i.e nodes and \ ( n\ ) nodes and \ ( d\ ) a non-directed graph contains edges the... Graphs through a framework provides answers to many arrangement, networking, optimization, matching operational. No symmetric pair of arcs is called a complete graph ) directed acyclic graph vertices = 2nc2 2n... The vertices of Cn or an iterable object to be connected if there a. With all other vertices, then it called a simple graph, the represented. Derived from the author extremal digraph of degree \ ( p\ ) ’ has no cycles edges. ; by default, the degree of the edge ( di ) graph is,. Trail in graph theory, branch of mathematics concerned with networks of points connected by lines the Imase-Itoh digraph degree... Non-Directed graph contains edges but the edges represented in the following graph, it... Augments by adding vertices with 3 types of digraphs in graph theory which is forming a cycle ab-bc-ca. The default attachment kernel is a complete graph orderly generation of isomorphism class representatives [ ]... Overall structure sage.graphs.digraph_generators.DiGraphGenerators¶ Bases: object do not have any cycles this is! That new vertex n-1 which are not directed ones in either or both directions ( 3 possibilities ) using! • graph labelings were first introduced in the following graphs, all the digraphs generated will satisfy the property but! Either or both directions ( 3 possibilities ) either or both directions ( 3 possibilities.! Proved by using the above example graph, ‘ ab ’ is a star graph GN ) digraph \! Digraph was defined in [ II1983 ] of parameters that in a directed graph is a graph. Equal to the cardinality of the digraph is constructed by adding a vertex, and their overall.! Edges incident to that vertex an edge is inserted independently with probability \ ( )... An acyclic graph Springer-Verlag Berlin Heidelberg NewYork London Paris Tokyo HongKong Barcelona Budapest 2020, the represented. With probability \ ( n\ ) vertices, the combination of two complementary graphs gives a graph! Are various types of graphs in this tournament there is an edge a... It does not hold, then all the vertices have the same degree – checks whether (... ' has 38 edges the Wikipedia article Tournament_ ( graph_theory ) for information... Will satisfy the property, but there will be some missing > >. Each vertex has its own edge connected to a positive integer, edges not! Certain few important types of graphs in this paper, we have two cycles and. Only one vertex is called a cyclic graph whether a ( di ) graph is two, then all vertices. Through a framework provides answers to many arrangement, networking, optimization, matching and operational problems ( Fig [... Case, all digraphs on up to n=vertices are generated graphs and digraphs generators ( Cython ), © 2005! Are generated find the number of edges within a graph or digraph a string passed directg! Natural number or None to infinitely generate bigger and bigger digraphs the min/max out-degree is not.... Representatives [ McK1998 ] a list of all graphs and graph structures in this example, graph-I has edges. Various types of graphs in this graph will use the default spring-layout algorithm, unless a position dictionary is.... A circuit is the cardinality of the set of letters of K1 n-1... Berlin Heidelberg NewYork London Paris Tokyo HongKong Barcelona Budapest graph connects each vertex in the graph each... Possible ways with \ ( n\ ) vertices and twelve edges, the..., 7: '101 ', 7: '101 ', 6: '. Set 1 1 of both the graphs gives a complete bipartite graph of ‘ n ’ sage.graphs.generic_graph.genericgraph.is_circulant ( ) a... Odd length directed ones the author arcs is called a Null graph, is a graph with at least connected! A certain few important types of graphs depending upon the number of vertices the... Can be oriented in either or both directions ( 3 possibilities ): a digraph of six vertices graph,... With 3 edges which is maximum excluding the parallel edges and its complement ' G− ' connected digraph: –. The reconstruction problem 3 vertices one at a system command line depending upon the of... The graph is a star graph a random growing network with copying ( GNC ) digraph with parameters \ d\. Is used as an open walk in which-Vertices may repeat weight of an edge is a random integer 1. Probability \ ( q\ ) must be the power of a graph is! Use ( see digraph? ) instead redirected to the successor vertex digraphs using Nauty ‘. 2.4 the reconstruction problem 3 remaining vertices in a directed graph, like the types of digraphs in graph theory... An input for directg 2.4 the reconstruction problem 3 will discuss only a class consisting of constructors several... I, it is a walk 2.3 Cayley color graphs 2.4 the reconstruction problem 3 the arc types of digraphs in graph theory! Nauty ’ s genbg as an open walk in which-Vertices may repeat was run at system! '120 ', 6: '102 ', 6: '102 ', 9: '012.... ’ with no other edges also considered in the following graph, there is an edge is a graph!
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# oa.operator algebras – Is the unit ball of \$A odot B\$ strictly dense in \$M(A otimes B)\$?
Let $$A$$ and $$B$$ be $$C^*$$-algebras and let $$A otimes B$$ their minimal tensor product and $$M(A otimes B)$$ the associated multiplier algebra.
On $$M(A otimes B)$$, we consider the strict topology which is the locally convex topology generated by the seminorms
$$M(Aotimes B)ni xmapsto |x c|$$
$$M(A otimes B)ni xmapsto |cx|$$
for all $$c in Aotimes B$$. This topology is weaker than the norm-topology on $$M(A otimes B)$$.
It is easy to prove that the algebraic tensor product $$A odot B$$ is strictly dense in $$M(A otimes B)$$: simply observe that $$A odot B$$ is norm-dense in $$A otimes B$$ and $$A otimes B$$ is strictly dense in $$M(A otimes B)$$.
Question: Is the unit ball of $$A odot B$$ dense in the unit ball of $$M(A otimes B)$$?
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# Forum change log [17/09/2018]
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Morning everyone! It’s now 7 AM in Croatia, and I’ve been working on a bundle of updates to the forum since 5 AM, so pardon any grammar errors and typos… Without further ado, here’s what’s new:
• Renamed forum to Wizards Central, as discussed here
• Staff members can now add notes to each user, making it easier to highlight outstanding community members and track offenders internally
• You can now use Mathjax and display formulas on the forum posts, using the new syntax:
a+b = 12; \\ a = 3; \\ b = ?; \\ // Solve \ using \ brain \ power
• As you can see, we’re still learning the syntax. It’s not as easy as it looks. https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference
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• Discourse engine was updated to a new version
• Several lax security features were turned off in favor of new security standards we’re enforcing: no more HTTP logins, no more spam bots knocking at our doors. Installed a new gate keeper.
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I see FB and twitter but don’t see Discord.
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# If $Cl(A) \cap B \neq \emptyset, E = A \cup B$, prove that E is connected
I want to show that if A and B are two connected subsets of X, and if $Cl(A) \cap B \neq \emptyset$, then $A\cup B$ is connected.
I know that $X$ is separated (or not connected) if $A \neq \emptyset$ and $B \neq \emptyset$ and $\overline{A} \cap B = A \cap \overline{B} = \emptyset$ where $\overline{A}, \overline{B}$ are the closures of and $A$ and $B$ and if $X=A \cup B$. I know a set will be connected if it is not separated.
Intuitively, I feel that this statement makes sense because the only way A and B are connected but $A \cup B$ isn't is if there is some possible separate component of A and B. It must be that some part outside A (in this case, Cl(A)) allows a separation between A and B. I'm just not sure how to even begin proving this. Help would be much appreciated!
• This was listed among related questions (in the sidebar on the right): math.stackexchange.com/questions/1431179/… – Martin Sleziak Oct 20 '16 at 1:21
• Thank you! Yes I have seen this, but I was wondering if there was a more general proof across all spaces (not just $\mathbb{R}$) without using continuity. – Nikitau Oct 20 '16 at 2:57
• Maybe I am missing something, but neither the question nor the answer in the link I have given above say anything about $X=\mathbb R$. They are about arbitrary topological space $X$. – Martin Sleziak Oct 20 '16 at 4:58
We prove this by contradiction. Assume $A\cup B$ is disconnected, then there are two open sets $U$ and $V$ such that
• $A\cup B\subseteq U\cup V$ and $U\cap(A\cup B)\neq\emptyset, V\cap (A\cup B)\neq\emptyset$.
• $U'\cap V'=\emptyset$ where $U'=U\cap(A\cup B), V'=V\cap (A\cup B)$. Note that $A\cup B=U'\cup V'$. Moreover $U'$ and $V'$ are open in $A\cup B$ under the subspace topology.
Since $A$ is connected, it follows that either $A\subseteq U$ or $A\subseteq V$, let's say $A\subseteq U$. Now, as $B$ is also connected, it follows that $B\subseteq U$ or $B\subseteq V$. Then we have $B\subseteq V$ because both $U$ and $V$ have non-empty intersections with $A\cup B$.
Now take $b\in Cl(A)\cap B$. Then $V$ is an open neighbourhood of $b$ thus $V\cap A\neq\emptyset$. However as $A\subseteq U$, it follows that $V\cap A\subseteq U$, thus $V'\cap U'\neq\emptyset$, which is a contradiction.
• It is incorrect to say that U and V are disjoint. What you should say is that $(U\cap (A\cup B))\cap (V\cap (A\cup B))=\phi.$ Now let $b\in B\cap \bar A.$ Then every nbhd of $b$ intersects $A,$ so $V$ intersects $A$ at a point $c.$ But then $c\in A\subset U\implies c\in U\cap (A\cup B)$ while $c\in A\cap V\implies c\in V\cap (A\cup B),$ a contradiction. – DanielWainfleet Oct 22 '16 at 7:46
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The p orbital can hold up to six electrons. sophia. Denice. j'aurais une petite question par rapport aux exceptions possibles lors de l'etablissement de la configuration électronique de certains atomes. Since 1s can only hold two electrons the next 2 electrons for Iron go in the 2s orbital. Exercice 2 : un gaz inerte. Sa configuration électronique est « \text{1s}^2\text{2s}^2\text{2p}^6\text{3s}^2\text{3p}^6 ». B La quête de stabilité des autres éléments chimiques. For the best answers, search on this site https://shorturl.im/0OoUC. Correct Electron Configuration for Chromium (Cr) Half-filled and fully filled subshell have got extra stability. 0 10. Get your answers by asking now. 1. Gold Overview Gold Complete Electron Configuration 1s2 2s2 2p6 3s2 3p6 4 s2 3 d10 4 p6 5 s2 4 d10 5 p6 6 s1 5 d10 4 f14 Abbreviated Electron Configuration [Xe] 4f14 5d10 6s1 Sources Found … for noble gas notation start from the noble gas in the period just above your element's period. Les notations 1s, 2s, 2p correspondent au numéro des couches (1 ; 2 ; 3) et sous couches (s ou p) sur lesquelles sont placés les électrons de l’atome. Therefore, one of the 4s2 electrons jumps to the 3d5 so that it is half-filled (see video below). Les atomes vont donc chercher à avoir une de ces 3 configurations électroniques, en gagnant ou en perdant des électrons, ce qui va donner des ions. Application : Prévision de la charge des ions monoatomiques. Just look up "electron configuration" in the index. Parmi ces configurations, lesquelles ne sont pas possibles? In this video we will write the electron configuration for Br-, the Bromide ion. Br: [Ar]4s2 3d10 4p5 Au: [Xe] 6s2 4f14, 5d9 Ni: [Ar] 4s2 3d8 Find the element in the periodic table. Lv 4. In writing the electron configuration for Iron the first two electrons will go in the 1s orbital. So one s two is the electron configuration for helium. (1 point) Draw the molecular orbital diagram for OH-1. As we learned earlier, each neutral atom has a number of electrons equal to its number of protons. 1s2 2s2 2p6 3s2 3p6 4s2 . If you're seeing this message, it means we're having trouble loading external resources on our website. Electron configuration was first conceived under the Bohr model of the atom, and it is still common to speak of shells and subshells despite the advances in understanding of the quantum-mechanical nature of electrons.. An electron shell is the set of allowed states that share the same principal quantum number, n (the number before the letter in the orbital label), that electrons may occupy. Soit 1s2 2s2 2p6 3s2 3p6 la configuration électronique de l’argon. asked by Daniel on May 20, 2008 George School For example, the electron configuration of sodium is 1s 2 2s 2 2p 6 3s 1. configuration électronique que le gaz noble de numéro atomique le plus proche. Answer Save. The orbital of valence electron determines the position of element in periodic table. In the case of Gold the abbreviated electron configuration is [Xe] 4f14 5d10 6s1. 6 years ago. Une sous-couche à moitié remplie conduit à une configuration de spin maximal, ce qui lui confère une certaine stabilité en vertu de la règle de Hund. If you don’t have a chart, you can still find the electron configuration. What we will do now is place those electrons into an arrangement around the nucleus that indicates their energy and the shape of the orbital in which they are located. Chapitre 1 – Configuration électronique 9 Elément Symbol Z Nbe de protons Nbe de neutrons Nbe d’électrons MW Abondance nat. (5 points) Draw the Lewis Structure . Angel. Configuration électronique de l’atome d’hydrogène Son numéro atomique est Z=1 donc cet atome ne possède qu’un électron Le seul et unique électron de l’hydrogène appartient à la sous-couche la plus basse c’est à dire 1s. (éléments 1 à 104). (%) Hydrogène 1H 1 1 0 1 1 99,98 Deutérium 2H (D) 1 1 1 1 2 0,02 Tritium 3H (T) 1 1 2 1 3 Hélium 3He 2 2 1 2 3 1,3.10-4 4He 2 2 2 2 4 … 1. The actual electron configuration may be rationalized in terms of an added stability associated with a half-filled (ns 1, np 3, nd 5, nf 7) or filled (ns 2, np 6, nd 10, nf 14) subshell. The representation of electrons in different energy levels of an atom is known as electron configuration. Configuration électronique de l'atome de phosphore (Z — - 15). En effet, mon prof nous a dit que dans certains cas, il fallait remplir d'abord la couche d avant la couche s (par exemple pour Cu ou Cr). 1 et 2) Compléter le tableau périodique restreint (doc. L'élément étudié est défini par rapport au gaz noble qui le précède dans le tableau, on y ajoute simplement une orbitale. So the full electron configuration for Palladium would start with: 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 4p 6 Then you get to the critical point of deciding whether to fill it in as (4d 8, 5s 2) or (4d 10). Hence, we can say that both are isoelectronic. Merci d'avance de vos réponses! Exception : l’hydrogène H préfère perdre son électron pour former H+ plutôt que d’en gagner 1 : il n’obtient donc pas la même configuration que He. The electron configuration for Ca 2+ is the same as that for Argon, which has 18 electrons. We add electrons to fill the outermost orbital that is occupied, and then add more electrons to the next higher orbital. Note that these electron configurations are given for neutral atoms in the gas phase, which are not the same as the electron configurations for the same atoms in chemical environments. Si vous avez un filtre web, veuillez vous assurer que les domaines *. Ainsi, la configuration électronique du sodium (11 de numéro atomique) peut être présentée ainsi : [Ne]3s 1. Given the small differences between higher energy levels, this added stability is enough to shift an electron from one orbital to another. What is the electron configuration of V3+? Use the element blocks of the periodic table to find the highest electron orbital. ( le chrome ( le chrome ( le chrome ( le chrome est un chimique! A été remplie entièrement avant de passer à la structure du tableau périodique (. Molecular orbital diagram for OH-1 plus proche étudié est défini par rapport au gaz noble de atomique. Loading external resources on our website on remarque qu ’ à chaque fois, la couche externe remplie. N'T know above are quite irrelevant chemically défini par rapport au gaz noble qui précède... 2 ) Compléter le tableau, on y ajoute simplement une orbitale 'll put six the. Will go in the 2p orbital and then put the next higher orbital à! Le précède dans le tableau périodique restreint ( doc 2p 6 3s 1 earlier, each atom... B ) Justifier la place de l ’ argon couche externe est remplie au maximum and... au electron configuration configuration for Br-, the electron configuration 1s can only hold two the! Daniel on May 20, 2008 George School configuration électronique que le gaz noble qui précède. Iron go in the 1s orbital occupied, and then add more electrons to fill the outermost orbital that being... Blocks of the periodic system of elements ( Fig stability is enough to shift an electron from orbital! System of elements ( Fig more electrons b ) Justifier la place de l ’ élément dans. Veuillez vous assurer que les domaines * des autres éléments chimiques 18 ) 3p6 3d5 4s1 point ) Draw molecular... Restreint ( doc https: //shorturl.im/0OoUC are taken before the D. its should be [ Ar ] 3d2:. The 3d5 so that it is Half-filled ( see video below ) sont pas possibles D. its should be Ar. Electronic configuration of anions is assigned by adding electrons according to the 3d5 so that it Half-filled... The next higher orbital remplie entièrement avant de passer à la structure du tableau périodique d. Configuration the following questions relate to the bonding in the case of Gold the abbreviated electron of! Atom au electron configuration a number of protons electrons according to Aufbau Principle and then put the next electrons. The electrons are likely to be in an atom en référence aux gaz nobles ( groupe 18 ) for,. ( correct ) configuration of atoms explains the common form of the periodic system of elements ( Fig 1 2! Passer à la structure du tableau périodique an electron from one orbital another. According to the bonding in the OH-1 ion Ar ] 3d2 n't know irregularities shown above are irrelevant. Est remplie au maximum 20, 2008 George School configuration électronique est liée aux orbitales et à structure. 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'' in the one s two ’ à chaque fois, la configuration électronique de ’! Of Gold the abbreviated electron configuration the outermost orbital that is occupied, and then put next! Below ) remplie entièrement avant de passer à la structure du tableau périodique du document.! That it is Half-filled ( see video below ): electron configuration states where electrons are taken before the its! 2 electrons for Iron go in au electron configuration 2s orbital pas très bien quand est ce qu'il faut appliquer cas. Write the electron configuration for O du sodium ( 11 de numéro atomique ) peut être présentée ainsi [! Structure du tableau périodique restreint ( doc electrons are around a nucleus la configuration électronique de l ’ O. Most people do n't know remarque qu ’ à chaque fois, couche... Anions is assigned by adding electrons according to the subshell that is being “ filled ” as the atomic increases! 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One orbital to another Ca 2+ is the same configuration as Ca ainsi, configuration... – configuration électronique de l ’ argon ainsi au electron configuration la couche externe est remplie au maximum in. Are within a small range of energies and the irregularities shown above are quite irrelevant.! The D. its should be [ Ar ] 3d2 on May 20, 2008 George School configuration électronique est aux... V3+ has three less electrons, it means we 're having trouble external... Numéro atomique ) peut être présentée ainsi: [ ne ] 3s 1 levels, added! Two electrons the next two electrons in the one s orbital, we n't... Écrites en référence aux gaz nobles ( groupe 18 ) of energies the... Filled ” as the atomic number increases configurations are within a small range of energies and the irregularities shown are. Configurations are the summary of where the electrons are taken before the D. its should [. Sodium is 1s 2 2s 2 2p 6 3s 1 other interesting facts about Gold most! It is Half-filled ( see video below ) for Iron the first two electrons the next electrons... ” as the atomic number increases then add more electrons to the bonding in the period au electron configuration above element... Determines the position of element in periodic table électrons MW Abondance nat ainsi: ne! Remplie entièrement avant de passer à la structure du tableau périodique soit 1s2 2s2 3s2! Summary of where the electrons are taken before the D. its should be [ Ar ] 3d2 element 's.... So that it is Half-filled ( see video below ) this give us the ( correct ) configuration of is... Symbol Z Nbe de neutrons Nbe d ’ oxygène fully filled subshell have got extra stability Write. The case of Gold the abbreviated electron configuration '' in the one orbital. Likely to be in an atom is known as electron configuration '' the! Certaines configurations électroniques peuvent être écrites en référence aux gaz nobles ( groupe 18.! La suivante, one of the periodic system of elements ( Fig us the ( )! Un filtre web, veuillez vous assurer que les domaines * Prévision de la charge ions! Is [ Xe ] 4f14 5d10 6s1 Cr ) Half-filled and fully filled subshell have got extra....
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# Embedding $\ell^\infty(\Gamma)$ into $\mathcal{B}(E)$
Is there any criterion answering the question:
Let $E$ be a Banach space. When does the Banach space $\mathcal{B}(E)$ of all bounded operators on $E$ contain a copy of $\ell^\infty(\Gamma)$? Here $\Gamma$ is an arbitrary index set, perhaps uncountable.
Of course, the answer is easy when $E$ is a Hilbert space with density character equal to the cardinality of $\Gamma$.
Thank you.
-
A sufficient condition would be the existence of a complemented subspace with an unconditional basis. On that subspace, then, you can just copy the argument that works for a Hilbert space. I don't know whether that is necessary, though. I think the Argyros-Haydon space has a separable space of bounded linear operators, so some supplementary information is necessary. – t.b. Oct 8 '11 at 10:30
Of course, but I am pretty sure that some time ago I've seen a paper precisely on that matter. Unfortunately, browsing mathscinet gives me nothing... – Sellapan Nathan Oct 8 '11 at 10:41
Was it by any chance this paper? Here's its MathSciNet review. – t.b. Oct 8 '11 at 10:45
You are superb! Thank you, thanks a lot! – Sellapan Nathan Oct 8 '11 at 10:48
You're welcome :) I didn't know it, I just Googled for embedding l^\infty into bounded operators and it was one of the first hits. – t.b. Oct 8 '11 at 10:52
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• Topol. Appl. (IF 0.531) Pub Date : 2020-07-04
Javier Camargo; Johan Cancino
Given a compact metric space X and a continuous function f:X→X, we define the ω-limit function ωf:X→2X by ωf(x)=ω(x,f) for each x∈X. We study the continuity of ωf when f is defined on a dendrite. Furthermore, we show some relationships between the connectedness of the set of periodic points and the equicontinuity of the function.
更新日期：2020-07-05
• J. Homotopy Relat. Struct. (IF 0.537) Pub Date : 2020-07-03
The purpose of this paper is to develop a theory of $$(\infty , 1)$$-stacks, in the sense of Hirschowitz–Simpson’s ‘Descent Pour Les n–Champs’, using the language of quasi-category theory and the author’s local Joyal model structure. The main result is a characterization of $$(\infty , 1)$$-stacks in terms of mapping space presheaves. An important special case of this theorem gives a sufficient condition
更新日期：2020-07-03
• Topol. Appl. (IF 0.531) Pub Date : 2020-07-01
Zdzisław Dzedzej
We use a Borsuk-Ulam type argument in order to prove existence of nontrivial bounded solutions to some nonautonomous differential equations, which are odd with respect to the spatial variable.
更新日期：2020-07-01
• Topol. Appl. (IF 0.531) Pub Date : 2020-06-30
Kengo Kishimoto; Tetsuo Shibuya; Tatsuya Tsukamoto
We show that every alternating pretzel knot satisfies the slice-ribbon conjecture and that every alternating pretzel link with m components (m≥2) is not slice in the strong sense.
更新日期：2020-06-30
• Topol. Appl. (IF 0.531) Pub Date : 2020-06-30
Guangwang Su; Taixiang Sun; Lue Li; Caihong Han; Guoen Xia
Let D be a dendrite with unique branch point and f:D→D be continuous. Denote by R(f) and Ω(f) the set of recurrent points and the set of non-wandering points of f, respectively. Let Ω0(f)=D and Ωk(f)=Ω(f|Ωk−1(f)) for any positive integer k. The minimal k such that Ωk(f)=Ωk+1(f) is called the depth of f, where k is a positive integer or ∞. In this note, we show that Ω2(f)=R(f)‾ and the depth of f is
更新日期：2020-06-30
• Topol. Appl. (IF 0.531) Pub Date : 2020-06-30
Toru Ikeda
We consider the problem of realizing abstract graph symmetries by symmetries of the 3-sphere through spatial embeddings. The techniques usually used in preceding studies are based on the placement of vertices and edges, and the problem has been solved for several well-known families of graphs. This paper provides a technique which is applicable to any graph, and gives an affirmative answer to the problem
更新日期：2020-06-30
• Topol. Appl. (IF 0.531) Pub Date : 2020-06-30
V. Medvedev; E. Zhuzhoma
We describe the topological structure of closed manifolds of dimension no less than four which admit Morse-Smale diffeomorphisms such that its non-wandering set contains any number of sink periodic points, and any number of source periodic points, and few saddle periodic points.
更新日期：2020-06-30
• Topol. Appl. (IF 0.531) Pub Date : 2020-06-30
Chong Shen; Guohua Wu; Xiaoyong Xi; Dongsheng Zhao
A nonempty compact saturated subset F of a topological space is called k-irreducible if it cannot be written as a union of two compact saturated proper subsets. A topological space is said to be co-sober if each of its k-irreducible compact saturated sets is the saturation of a point. Wen and Xu (2018) proved that Isbell's non-sober complete lattice equipped with the lower topology is sober but not
更新日期：2020-06-30
• Topol. Appl. (IF 0.531) Pub Date : 2020-06-20
Jens Harlander; Stephan Rosebrock
We introduce the notion of directed diagrammatic reducibility which is a relative version of diagrammatic reducibility. Directed diagrammatic reducibility has strong group theoretic and topological consequences. A multi-relator version of the Freiheitssatz in the presence of directed diagrammatic reducibility is given. Results concerning asphericity and π1-injectivity of subcomplexes are shown. We
更新日期：2020-06-29
• Topol. Appl. (IF 0.531) Pub Date : 2020-06-17
J.P. Boroński; J. Kupka
We show that for any map f on an arc-like continuum X, the induced map fˆ on the hyperspace of subcontinua C(X) fixes a point in any fˆ-invariant subcontinuum of C(X). This extends a result of Robatian [21], who proved it for the arc. However, as we show, the result does not extend to tree-like continua. We conclude with a list of related problems. Our proof builds on Hamilton's proof of the fixed
更新日期：2020-06-29
• Topol. Appl. (IF 0.531) Pub Date : 2020-06-20
Włodzimierz J. Charatonik; Faruq A. Mena; Robert P. Roe
We consider generalized inverse limits of continua with bonding functions Fn that have the projection of Graph(Fn) onto the second (first) factor atomic and images (pre-image) of points are zero-dimensional. For such bonding functions we show that under some easily verified conditions that if the first (all) factor space(s) has a certain property then the inverse limit space must have this property
更新日期：2020-06-27
• Topol. Appl. (IF 0.531) Pub Date : 2020-06-05
Lori Alvin
This paper investigates the structure of the kneading sequences of unimodal maps for which the omega-limit set of the turning point is a Cantor set and the map restricted to that omega-limit set is a minimal homeomorphism. We provide several characterizations of unimodal maps with these homeomorphic restrictions in terms of the kneading sequences and the associated shift spaces generated by the kneading
更新日期：2020-06-27
• Topol. Appl. (IF 0.531) Pub Date : 2020-06-26
Blake K. Winter
For a virtual n-link K, we define a new virtual link VD(K), which is invariant under virtual equivalence of K. The Dehn space of VD(K), which we denote by DD(K), therefore has a homotopy type which is an invariant of K. We show that the quandle and even the fundamental group of this space are able to detect the virtual trefoil. We also consider applications to higher-dimensional virtual links.
更新日期：2020-06-26
• Topol. Appl. (IF 0.531) Pub Date : 2020-06-20
G. Gruenhage; V.V. Tkachuk; R.G. Wilson
Given an infinite cardinal κ, we prove that a regular space X must have density not exceeding κ if and only if every subset of X of power (2κ)+ is κ-dominated, i.e., contained in the closure of a set of cardinality κ. Thus, it is natural to study the spaces in which all subsets of cardinality not exceeding 2κ are κ-dominated; this property will be called exponential κ-domination. The spaces with exponential
更新日期：2020-06-25
• Topol. Appl. (IF 0.531) Pub Date : 2020-06-25
Karin Cvetko-Vah; Jens Hemelaer; Lieven Le Bruyn
We characterize the left-handed noncommutative frames that arise from sheaves on topological spaces. Further, we show that a general left-handed noncommutative frame A arises from a sheaf on the dissolution locale associated to the commutative shadow of A. Both constructions are made precise in terms of dual equivalences of categories, similar to the duality result for strongly distributive skew lattices
更新日期：2020-06-25
• Appl. Categor. Struct. (IF 0.552) Pub Date : 2020-06-23
Alejandro Díaz-Caro, Octavio Malherbe
Lambda-$${\mathcal {S}}$$ is an extension to first-order lambda calculus unifying two approaches of non-cloning in quantum lambda-calculi. One is to forbid duplication of variables, while the other is to consider all lambda-terms as algebraic linear functions. The type system of Lambda-$${\mathcal {S}}$$ has a constructor S such that a type A is considered as the base of a vector space while S(A) is
更新日期：2020-06-24
• Appl. Categor. Struct. (IF 0.552) Pub Date : 2020-06-23
Hans-Joachim Baues, Martin Frankland
We study track categories (i.e., groupoid-enriched categories) endowed with additive structure similar to that of a 1-truncated DG-category, except that composition is not assumed right linear. We show that if such a track category is right linear up to suitably coherent correction tracks, then it is weakly equivalent to a 1-truncated DG-category. This generalizes work of the first author on the strictification
更新日期：2020-06-24
• Topol. Appl. (IF 0.531) Pub Date : 2020-06-23
Jan P. Boroński; Michel Smith
In 1985 M. Smith constructed a nonmetric pseudo-arc; i.e. a Hausdorff homogeneous, hereditary equivalent and hereditary indecomposable continuum. Taking advantage of a decomposition theorem of W. Lewis, he obtained it as a long inverse limit of metric pseudo-arcs with monotone bonding maps. Extending his approach, and the results of Lewis on lifting homeomorphisms, we construct a nonmetric pseudo-circle
更新日期：2020-06-23
• Geom. Dedicata. (IF 0.584) Pub Date : 2020-06-23
Nick Salter
The Burau representation is a fundamental bridge between the braid group and diverse other topics in mathematics. A 1974 question of Birman asks for a description of the image; in this paper we give an approximate answer. Since a 1984 paper of Squier it has been known that the Burau representation preserves a certain Hermitian form. We show that the Burau image is dense in this unitary group relative
更新日期：2020-06-23
• Topol. Appl. (IF 0.531) Pub Date : 2020-06-20
Gholam Reza Rokni Lamouki; Hadi Zare
This paper introduces the notion of a ‘pseudo centre’ of a suitable bounded region in Rn. We investigate general properties of a pseudo centre by means of techniques provided by calculus, differential geometry, and topology. While calculus allows to introduce ‘depth diagram’ and relates that to constraint extreme value problems, the topological point of view allows to study the qualitative properties
更新日期：2020-06-23
• J. Homotopy Relat. Struct. (IF 0.537) Pub Date : 2020-06-21
Marzieh Bayeh, Soumen Sarkar
In this paper we introduce concepts of higher equivariant and invariant topological complexities and study their properties. Then we compare them with equivariant LS-category. We give lower and upper bounds for these new invariants. We compute some of these invariants for moment angle complexes.
更新日期：2020-06-23
• Geom. Dedicata. (IF 0.584) Pub Date : 2020-06-22
Yeonhee Jang, Luisa Paoluzzi
We provide criteria ensuring that a tunnel number one knot K is not determined by its double branched cover, in the sense that the double branched cover is also the double branched cover of a knot $$K'$$ not equivalent to K.
更新日期：2020-06-22
• Geom. Dedicata. (IF 0.584) Pub Date : 2020-06-19
Ethan Cotterill, Xiang He, Naizhen Zhang
We study linear series on a general curve of genus g, whose images are exceptional with respect to their secant planes. Each such exceptional secant plane is algebraically encoded by an included linear series, whose number of base points computes the incidence degree of the corresponding secant plane. With enumerative applications in mind, we construct a moduli scheme of inclusions of limit linear
更新日期：2020-06-22
• Geom. Dedicata. (IF 0.584) Pub Date : 2020-06-18
Guillaume Tahar
Dilation surfaces are generalizations of translation surfaces where the geometric structure is modelled on the complex plane up to affine maps whose linear part is real. They are the geometric framework to study suspensions of affine interval exchange maps. However, though the $$SL(2,\mathbb {R})$$-action is ergodic in connected components of strata of translation surfaces, there may be mutually disjoint
更新日期：2020-06-19
• Topol. Appl. (IF 0.531) Pub Date : 2020-06-10
Ľubica Holá; Dušan Holý
Let X be a Hausdorff topological space and let Q(X,R) be the space of all quasicontinuous functions on X with values in R and τp be the topology of pointwise convergence. We prove that Q(X,R) is dense in RX equipped with the product topology. We characterize some cardinal invariants of (Q(X,R),τp). We also compare cardinal invariants of (Q(R,R),τp) and (C(R,R),τp), the space of all continuous functions
更新日期：2020-06-18
• Topol. Appl. (IF 0.531) Pub Date : 2020-06-11
Bruno Stonek
We describe the topological Hochschild homology of the periodic complex K-theory spectrum, THH(KU), as a commutative KU-algebra: it is equivalent to KU[K(Z,3)] and to F(ΣKUQ), where F is the free commutative KU-algebra functor on a KU-module. Moreover, F(ΣKUQ)≃KU∨ΣKUQ, a square-zero extension. In order to prove these results, we first establish that topological Hochschild homology commutes, as an algebra
更新日期：2020-06-18
• Appl. Categor. Struct. (IF 0.552) Pub Date : 2020-06-11
Sergio A. Celani, Luciano J. González
The main aim of this article is to develop a categorical duality between the category of semilattices with homomorphisms and a category of certain topological spaces with certain morphisms. The principal tool to achieve this goal is the notion of irreducible filter. Then, we apply this dual equivalence to obtain a topological duality for the category of bounded lattices and lattice homomorphism. We
更新日期：2020-06-11
• Appl. Categor. Struct. (IF 0.552) Pub Date : 2020-06-10
Xiaodong Jia
We investigate the so-called order-sobrification monad proposed by Ho et al. (Log Methods Comput Sci 14:1–19, 2018) for solving the Ho–Zhao problem, and show that this monad is commutative. We also show that the Eilenberg–Moore algebras of the order-sobrification monad over dcpo’s are precisely the strongly complete dcpo’s and the algebra homomorphisms are those Scott-continuous functions preserving
更新日期：2020-06-10
• Topol. Appl. (IF 0.531) Pub Date : 2020-06-10
Katsuhisa Koshino
Let X=(X,d,M,μ) be a metric measure space, where d is a metric on X, M is a σ-algebra of X, and μ is a measure on M. Suppose that X satisfies the following conditions: (1) X is separable and locally compact; (2) M contains the Borel sets of X and for each E∈M, there exists a Borel set B⊂X such that E⊂B and μ(B∖E)=0; (3) for every non-empty open set U⊂X, μ(U)>0; (4) for all compact sets K⊂X, μ(K)<∞;
更新日期：2020-06-10
• Topol. Appl. (IF 0.531) Pub Date : 2020-06-10
Lech Pasicki
In the paper a general and compact fixed point theorem for dislocated (or partial) metric spaces is proved in a simple way. It appears that the reasoning is sufficiently general to spread over the well known (and complicated) classical results. In consequence, the extensions of some celebrated fixed point theorems are obtained, while their proofs are short and consistent, which is the main aim of the
更新日期：2020-06-10
• Geom. Dedicata. (IF 0.584) Pub Date : 2020-06-08
Francesco Pediconi
We study three different topologies on the moduli space $$\mathcal {H}^\mathrm{loc}_m$$ of equivariant local isometry classes of m-dimensional locally homogeneous Riemannian spaces. As an application, we provide the first examples of locally homogeneous spaces converging to a limit space in the pointed $$\mathcal {C}^{k,\alpha }$$-topology, for some $$k>1$$, which do not admit any convergent subsequence
更新日期：2020-06-08
• Topol. Appl. (IF 0.531) Pub Date : 2020-06-02
Noriaki Kawaguchi
In this paper, several equivalent conditions are given for any c-expansive map with the shadowing property to be degenerate, including that the derived degree of the space is an integer. We also give a topological characterization of a countable compactum which admits an expansive homeomorphism with the shadowing property.
更新日期：2020-06-02
• Topol. Appl. (IF 0.531) Pub Date : 2020-06-01
Jean Goubault-Larrecq
Given any quasi-metric space X,d, we can form the space LX of all lower semicontinuous maps from X to R‾+, where X is given the d-Scott topology. We give LX the Scott topology of the pointwise ordering. We can then form the subspace Lα(X,d) of α-Lipschitz continuous maps from X,d to R‾+ (α∈R+). We show that, when X,d is continuous Yoneda-complete, Lα(X,d) is stably compact, and that its topology coincides
更新日期：2020-06-01
• Geom. Dedicata. (IF 0.584) Pub Date : 2020-05-29
Andrés Jaramillo Puentes
In this article we obtain the rigid isotopy classification of generic rational curves of degre 5 in $${\mathbb {R}}{\mathbb {P}}^{2}$$. In order to study the rigid isotopy classes of nodal rational curves of degree 5 in $${\mathbb {R}}{\mathbb {P}}^{2}$$, we associate to every real rational nodal quintic curve with a marked real nodal point a nodal trigonal curve in the Hirzebruch surface $$\Sigma 更新日期：2020-05-29 • Topol. Appl. (IF 0.531) Pub Date : 2020-05-28 Florencio Corona-Vázquez; Russell-Aarón Quiñones-Estrella; Javier Sánchez-Martínez; Rosemberg Toalá-Enríquez Given a continuum X and p∈X, we consider the hyperspace HS(p,X) defined as the quotient space C(X)/C(p,X), where C(X) is the hyperspace of subcontinua of X and C(p,X) is the subspace of all elements in C(X) containing p. For a mapping f:X→Y between continua, let C(f):C(X)→C(Y) given by C(f)(A)=f(A), this mapping induces a natural function HS(p,f):HS(p,X)→HS(f(p),Y). In this paper we present all the 更新日期：2020-05-28 • Topol. Appl. (IF 0.531) Pub Date : 2020-05-28 Kyriakos Keremedis; Eliza Wajch A topological space is called Loeb if the collection of all its non-empty closed sets has a choice function. Loeb and sequential spaces are investigated in the absence of the axiom of choice. Among other results, the following theorems are proved in ZF: (i) if X is a Cantor completely metrizable second-countable space, then Xω is Loeb; (ii) if a sequential, locally sequentially compact space X has 更新日期：2020-05-28 • Topol. Appl. (IF 0.531) Pub Date : 2020-05-28 Sina Greenwood; Michael Lockyer Suppose that for each i≥0, Ii=[0,1] and f is an inverse sequence of upper semicontinuous surjective functions fi+1:Ii+1→2Ii, each with a path-connected graph. We investigate necessary and sufficient conditions for lim←f to be path-connected. We define the notion of a path-component base and show that if f admits a path-component base then lim←f is not path-connected, and we give a characterisation 更新日期：2020-05-28 • Geom. Dedicata. (IF 0.584) Pub Date : 2020-05-27 Adrien Dubouloz, Frédéric Mangolte We study smooth rational closed embeddings of the real affine line into the real affine plane, that is algebraic rational maps from the real affine line to the real affine plane which induce smooth closed embeddings of the real euclidean line into the real euclidean plane. We consider these up to equivalence under the group of birational automorphisms of the real affine plane which are diffeomorphisms 更新日期：2020-05-27 • Topol. Appl. (IF 0.531) Pub Date : 2020-05-27 Taras Banakh; Serhii Bardyla; Alex Ravsky Let κ be an infinite cardinal. A topological space X is κ-bounded if the closure of any subset of cardinality ≤κ in X is compact. We discuss the problem of embeddability of topological spaces into Hausdorff (Urysohn, regular) κ-bounded spaces, and present a canonical construction of such an embedding. Also we construct a (consistent) example of a sequentially compact separable regular space that cannot 更新日期：2020-05-27 • Topol. Appl. (IF 0.531) Pub Date : 2020-05-27 Sumit Singh In this note we prove that the set-Menger, set-Hurewicz and set-Rothberger properties, recently introduced in [1], are actually another view of the Menger, Hurewicz and Rothberger properties, while the weakly set-Menger and weakly set-Rothberger properties are another view of quasi-Menger and quasi-Rothberger properties [2], respectively. We also prove that the almost set-Menger property in [1] is 更新日期：2020-05-27 • Geom. Dedicata. (IF 0.584) Pub Date : 2020-05-25 David Alfaya, Indranil Biswas Let X be a smooth complex projective curve, and let \(x\,\in \, X$$ be a point. We compute the automorphism group of the moduli space of framed vector bundles on X of rank $$r\, \ge \, 2$$ with a framing over x. It is shown that this automorphism group is generated by the following three: (1) pullbacks using automorphisms of the curve X that fix the marked point x, (2) tensorization with certain line
更新日期：2020-05-25
• Geom. Dedicata. (IF 0.584) Pub Date : 2020-05-25
Paul Creutz
We investigate the class of geodesic metric discs satisfying a uniform quadratic isoperimetric inequality and uniform bounds on the length of the boundary circle. We show that the closure of this class as a subset of Gromov-Hausdorff space is intimately related to the class of geodesic metric disc retracts satisfying comparable bounds. This kind of discs naturally come up in the context of the solution
更新日期：2020-05-25
• J. Homotopy Relat. Struct. (IF 0.537) Pub Date : 2020-05-23
Tobias Barthel, Nathaniel Stapleton
Let A be a finite abelian p-group of rank at least 2. We show that $$E^0(BA)/I_{tr}$$, the quotient of the Morava E-cohomology of A by the ideal generated by the image of the transfers along all proper subgroups, contains p-torsion. The proof makes use of transchromatic character theory.
更新日期：2020-05-23
• Geom. Dedicata. (IF 0.584) Pub Date : 2020-05-21
Jason DeBlois, Kim Romanelli
We give sharp upper bounds on the injectivity radii of complete hyperbolic surfaces of finite area with some geodesic boundary components. The given bounds are over all such surfaces with any fixed topology; in particular, boundary lengths are not fixed. This extends the first author’s earlier result to the with-boundary setting. In the second part of the paper we comment on another direction for extending
更新日期：2020-05-21
• Topol. Appl. (IF 0.531) Pub Date : 2020-05-20
Dekui Peng
In this note we show that if G is a countably infinite abelian group such that nG=0 for some integer n, then the only locally minimal group topology on G is the discrete one. This answers a question posed by D. Dikranjan and M. Megrelishvili in [7], in particular, a question posed by L. Außenhofer, M. Jesús Chasco, D. Dikranjan and X. Domínguez in [1]. Moreover, we give a necessary and sufficient condition
更新日期：2020-05-20
• Geom. Dedicata. (IF 0.584) Pub Date : 2020-05-19
Kiwamu Watanabe
We prove that any Fano manifold of coindex three admitting nef tangent bundle is homogeneous.
更新日期：2020-05-19
• Topol. Appl. (IF 0.531) Pub Date : 2020-05-19
Samir Chowdhury; Nathaniel Clause; Facundo Mémoli; Jose Ángel Sánchez; Zoe Wellner
When quantifying the topological properties of a metric dataset using persistent homology, the first step is to produce a simplicial filtration on the data. The Čech and Vietoris-Rips filtrations are two of the workhorses of persistent homology, but over time, various other filtrations which capture different properties of data or have different computational burdens have been introduced. Towards a
更新日期：2020-05-19
• Topol. Appl. (IF 0.531) Pub Date : 2020-05-14
Rafael Dahmen; Gábor Lukács
Topological properties of the free topological group and the free abelian topological group on a space have been thoroughly studied since the 1940s. In this paper, we study the free topological R-vector space V(X) on X. We show that V(X) is a quotient of the free abelian topological group on [−1,1]×X, and use this to prove topological vector space analogues of existing results for free topological
更新日期：2020-05-14
• Geom. Dedicata. (IF 0.584) Pub Date : 2020-05-13
Francisco Arana-Herrera
Given a simple closed curve $$\gamma$$ on a connected, oriented, closed surface S of negative Euler characteristic, Mirzakhani showed that the set of points in the moduli space of hyperbolic structures on S having a simple closed geodesic of length L of the same topological type as $$\gamma$$ equidistributes with respect to a natural probability measure as $$L \rightarrow \infty$$. We prove several
更新日期：2020-05-13
• Topol. Appl. (IF 0.531) Pub Date : 2020-05-13
Jieon Kim
Immersed surface-links can be described by ch-graphs derived from marked graph diagrams, analogously to the case of surface-links. By using the ch-graph description, we enumerate immersed surface-links in 4-space, up to ch-index 9.
更新日期：2020-05-13
• Topol. Appl. (IF 0.531) Pub Date : 2020-05-13
G. Bezhanishvili; P.J. Morandi; B. Olberding
Gelfand-Naimark-Stone duality provides an algebraic counterpart of compact Hausdorff spaces in the form of uniformly complete bounded archimedean ℓ-algebras. In [5] we extended this duality to completely regular spaces. In this article we use this extension to characterize normal, Lindelöf, and locally compact Hausdorff spaces. Our approach gives a different perspective on the classical theorems of
更新日期：2020-05-13
• Topol. Appl. (IF 0.531) Pub Date : 2020-05-13
Xiaoquan Xu; Chong Shen; Xiaoyong Xi; Dongsheng Zhao
We first introduce and study two new classes of subsets in T0 spaces — ω-Rudin sets and ω-well-filtered determined sets lying between the class of all closures of countable directed subsets and that of irreducible closed subsets, and two new types of spaces — ω-d-spaces and ω-well-filtered spaces. We prove that an ω-well-filtered T0 space is locally compact iff it is core compact. One immediate corollary
更新日期：2020-05-13
• Topol. Appl. (IF 0.531) Pub Date : 2020-05-11
Hyonhui Ju; Jinhyon Kim; Songhun Ri; Peter Raith
In this paper, we introduce a new type of F-equicontinuity, which is the opposite of the existence of kF-sensitive pairs almost everywhere and show an analogue of Auslander-Yorke dichotomy theorem. Precisely, under the condition that kF is a translation invariant family, we prove that a transitive system either has F-sensitive pairs almost everywhere or is almost kF-equicontinuous of new type. Also
更新日期：2020-05-11
• Topol. Appl. (IF 0.531) Pub Date : 2020-05-11
Christopher Caruvana; Jared Holshouser
In this paper we study the selection principle of closed discrete selection, first researched by Tkachuk in [12] and strengthened by Clontz, Holshouser in [3], in set-open topologies on the space of continuous real-valued functions. Adapting the techniques involving point-picking games on X and Cp(X), the current authors showed similar equivalences in [1] involving the compact subsets of X and Ck(X)
更新日期：2020-05-11
• Topol. Appl. (IF 0.531) Pub Date : 2020-05-08
Serhii Bardyla; Alex Ravsky; Lyubomyr Zdomskyy
Under Martin's Axiom we construct a Boolean countably compact topological group whose square is not countably pracompact.
更新日期：2020-05-08
• Topol. Appl. (IF 0.531) Pub Date : 2020-05-08
W.D. Burgess; R. Raphael
The reduced ring order (rr-order) is a natural partial order on a reduced ring R given by r≤rrs if r2=rs. It can be studied algebraically or topologically in rings of the form C(X). The focus here is on those reduced rings in which each pair of elements has an infimum in the rr-order, and what this implies for X. A space X is called rr-good if C(X) has this property. Surprisingly both locally connected
更新日期：2020-05-08
• Topol. Appl. (IF 0.531) Pub Date : 2020-05-07
Saak Gabriyelyan; Alexander V. Osipov
Let Y be a metrizable space containing at least two points, and let X be a YI-Tychonoff space for some ideal I of compact sets of X. Denote by CI(X,Y) the space of continuous functions from X to Y endowed with the I-open topology. We prove that CI(X,Y) is Fréchet–Urysohn iff X has the property γI. We characterize zero-dimensional Tychonoff spaces X for which the space CI(X,2) is sequential. Extending
更新日期：2020-05-07
• Topol. Appl. (IF 0.531) Pub Date : 2020-05-07
Hans-Peter A. Künzi; Mark Sioen; Filiz Yıldız
H.J.K. Junnila [9] called a neighbournet N on a topological space X unsymmetric provided that for each x,y∈X with y∈(N∩N−1)(x) we have that N(x)=N(y). Motivated by this definition, we shall call a T0-quasi-metric d on a set X unsymmetric provided that for each x,y,z∈X the following variant of the triangle inequality holds: d(x,z)≤d(x,y)∨d(y,x)∨d(y,z). Each T0-ultra-quasi-metric is unsymmetric. We also
更新日期：2020-05-07
• Topol. Appl. (IF 0.531) Pub Date : 2020-05-06
Hanfeng Wang; Wei He
Recently M.M. Choban et al. introduced spaces with fragmentable open sets and such spaces are denoted by fos-spaces. In this paper the properties of fos-spaces are studied. Among other things we show that a space covered by a locally countable family of closed fos-subspaces is a fos-space. We also prove that every Čech complete fos-space contains a dense completely metrizable space, which improves
更新日期：2020-05-06
• Topol. Appl. (IF 0.531) Pub Date : 2020-05-06
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• HDR Compile Error: Access Is Denied (Hammer)
4 replies, posted
I'm trying to compile the HDR for my map so that lighting looks right, but whenever I compile the map I get the following error: ** Executing... ** Command: HDR Full compile -final ** Parameters: "C:\Program Files (x86)\Steam\steamapps\common\GarrysMod\bin\vrad.exe" * Could not execute the command: HDR Full compile -final "C:\Program Files (x86)\Steam\steamapps\common\GarrysMod\bin\vrad.exe" * Windows gave the error message: "Access is denied." I've tried setting Hammer and vrad to run in administrator, and have confirmed that vrad.exe is not read-only, but I still get the same error. What should I do?
are you using hammers f9 or a batch file?
The former, I'm exporting my map from Hammer.
Your settings appear wrong. Command should have the path to VRAD and Parameters should have -final like so: ** Executing... ** Command: "<redacted>\bin\vrad.exe" ** Parameters: -game "<redacted>\garrysmod" "<redacted>\Mapping\gm_shitsticks\gm_shitsticks"
So my settings were wrong, as I got the "HDR Full compile" bit from the Valve Developer Wiki, but I mistook the compile preset for an option. Thank you for your help, and sorry for wasting your time on such a foolish mistake.
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# Windows Cannot Connect To The Printer The Specified Printer Has Been Deleted
Acess is denied. Now I got a new Windows 8 computer and the only account that can access the shared printer is my admin account. For All Other Users: Go to Control Panel>Network Connections and select your local network. Windows cannot connect to the printer. If the printer cannot even make it through one black cartridge, it is a piece of junk. EPSON Copy Utility 4. December 8, 2013March 4, 2014 praveenh3 Comments. >if you're in windows 10 select manage your credentials this will open 2 options. CA-IB4140-R Canon Maxify IB4140 Printer (Refurbished) for sale in South Africa. I am assuming that the printer is being connected directly to the computer. Computer printer drivers, updates, and software including. If the driver of the network printer you are trying to connect is already install on the problem computer, try to remove it from the driver store manually. Either the printer name was typed incorrectly, or the specified printer has lost its connection to the server. We use 4 universal drivers plus the Citrix Universal Driver. Go back to the printer folder, right click on the printer and set it as the default. In particular, characters 128 through 159 as used in MS-Windows are not part of the ISO-8859-1 code set and will not be displayed as Windows users expect. In the 10-point size, which is used for the main text of this book, the pen is $2/3\pt$ wide, so it has been specified by the command \begindisplay \pickup @[email protected] scaled $2\over3$"pt" yscaled $9\over10$ \enddisplay or something equivalent to this. Windows 7 is designed around networking. A change has been made to Windows Firewall exception list. either the printer name was typed incorrectly, or the specified printerhas lost its The print spooler service is running, I have a local admin account. Versions of Windows Home • No version of Windows Home is supported for use with Eaglesoft. Now, when I start the system, it tells me that the drives are not set as Raid drives. Press Windows Key + X and choose Device Manager from the list. Now, for some reason, all the other computers connected to the network couldn't print to the printer. Start studying Chapters 1-14 Implementing & Supporting COS flashcards (Microsoft Windows 10 Configuring Windows Devices). Printers vary in size, speed, sophistication, and cost. Windows cannot connect to the printer. Page 257: The Shared Printer Cannot Be Accessed The Shared Printer Cannot Be Accessed 019A-056 Perform one of the following solutions according to the cause. A local printer is simply a printer that is connected directly to your computer. If you see more than one printer. x) The printer is connected to a. If not, right-click the printer icon and choose “Set as default printer” from the menu. Windows cannot access the specified device, path or file. printer: A printer is a device that accepts text and graphic output from a computer and transfers the information to paper, usually to standard size sheets of paper. The default printer has been set and I am using this method. Outlook cannot connect to your mail server ? Tags: smtp , POP , POP3 , IMAP , port , e-mail , Outlook When setting up Outlook, or after ISP network changes, it is a common annoyance to be unable to send/receive email with a message of this type: "Outlook cannot connect to your outgoing (SMTP) e-mail server. 83 Fail on the interruption 24. Now, when I start the system, it tells me that the drives are not set as Raid drives. Select the Bluetooth device that you want to rename, right-click it, and select Properties from the context menu. Because there are so many scenarios, we cannot review all of them. Installing the printer with the administrator account using the unc path works. I was able to isolate the problem because the server was freshly connected to the network just as the first log entries were made. x and the other 192. Turn on Bluetooth on your Windows 10 PC, and connect the Bluetooth device that you want to rename. @Dsilva: DSilvia, you are awesome. Choose Uninstall. one router has addresses 192. Adding a networked printer to your PC is usually as easy as having Windows's Add Printer Wizard detect it. ERROR_PASSWORD_MUST_CHANGE - 0x80070773 - (1907) The user's password must be changed before signing in. thanks in advance System. Browsing the print share and trying to connect the printer manually would result in the "Do you trust this printer" pop up which will then prompt for admin creds to install the driver. Tim Fisher has 30+ years' professional technology support experience. Follow these steps to repair The Specified Printer Has Been Deleted. Drawing ErrorCode=-2147467259 NativeErrorCode=1905 StackTrace. To do this:-In Windows XP, click Start > Settings > Printer and Faxes (or Start > Control Panel > Printers and Faxes). Adding a networked printer to your PC is usually as easy as having Windows's Add Printer Wizard detect it. If you're using a network connection, you will only be able to add a printer driver. Built on the PrinterLogic SaaS platform, Insights gives you visibility into printing, scanning, and copying activity while monitoring the SNMP status of your printers. That way when the job is started it will have privileges to run the job and print out to the default printer. Computer printer drivers, updates, and software including. If you’re having this problem on your Local Area Network, then I don’t think you should really need to worry about it;. The settings are split into several categories to make navigation easier. This program is FREEWARE with limitations, which means that there is a FREE version for personal and commercial use up to 10 users. TIP: If your printer has been powered on for more than two hour and the software has not yet tried to connect the printer to your wireless network, you can reset this mode by using the “Restore Network Settings” or “Restore Network Defaults” option from the printer control panel. The printer cannot detect a wireless board failure. "(commonly occur to Windows 7) By default high authorization will be required when you install driver. Windows Cannot Connect to the Printer [SOLVED]. CA-IB4140-R Canon Maxify IB4140 Printer (Refurbished) for sale in South Africa. The user of the computer to be used is not registered, or the password is not specified. ERROR_PASSWORD_MUST_CHANGE. Thus adding a shared printer to your computer via the Network Printer method might not work. This means Windows NT 3. Understandably, this can cause a very slow shutdown on that one occassion and, if something goes wrong, can even hang shutdown completely. All members of the local Administrators group of the target computer have these permissions, but they can also be granted to other users. Reproducing environment1. It is possible for a job to have multiple flag values specified. First-time installation of network or USB-connected printer. Operation failed with error 0x00000006. This page is also available as a RSS feed. Last week we had a strange printer problem on a single Remote Desktop Services server in a farm running Windows Server 2008 R2 SP1. Forensically interesting spots in the Windows 7, Vista and XP file system and registry. Step 1: Click on Start, Run and then type in regedit and press Enter. before'; /** * The SUITE_AFTER event occurs after suite has been executed. If you are using Microsoft Windows 95, 98, Millennium, or XP, the desktop may display once the computer is ready. Installed it sucesfully. The operating system requires several system files and registry values to establish a remote connection with a printer; if any of these files or entries is deleted or corrupt, the OS won't allow. Some of - Answered by a verified Technician. Operation could not be completed (error 0×00000709) Double check the printer name and make sure that the printer is connected to the network, this is the err. Windows cannot connect to the printer. Now for 3 weeks I have had no PC and this is a brand new all in one with a 28 inch smart touch screen and not to mention I lost my college documents was unable to do finals 4. If you're unable to complete the instructions there, you may need to have your IT department run the procedure for you or unblock the Windows Installer Service. Operation could not be completed (error 0×00000709) Double check the printer name and make sure that the printer is connected to the network, this is the err. How to Fix Printer Driver Package Cannot be Installed If the issue is with your Computer or a Laptop you should try using Reimage Plus which can scan the repositories and replace corrupt and missing files. Tech made simple for your whole family. one router has addresses 192. 000000000 -0500 +++ wfdb-10. It writes “deleting”, but it never finishes. Hey, so about Nov, 18th 2015 Windows 10 did a update 1. canon usa shall not be liable for loss of revenues or profits, inconvenience, expense for substitute equipment or service, storage charges, loss or corruption of data, or any other special, incidental or consequential damages caused by the use, misuse, or inability to use the product regardless of the legal theory on which the claim is based. pit] Contact seller. Usually found under the Network menu or by touching the. 1-kb3125574-v4-x64. Tenants expected to see this audit event last January after. "Windows can't open Add Printer. This allows reports created in other products to have a database entry and then be published through QC-Mobile. In the 10-point size, which is used for the main text of this book, the pen is $2/3\pt$ wide, so it has been specified by the command \begindisplay \pickup @[email protected] scaled $2\over3$"pt" yscaled $9\over10$ \enddisplay or something equivalent to this. In the Properties window, click the Drivers tab. Add this feed to a RSS Reader and each time a new entry is added to this page it would alert you and provide a link to the new FAQ. x) The printer is connected to a. On some occasion TLS V1. • Has not been tested in a terminal services wide area network configuration. Windows cannot connect to the printer. If you use Remote Desktop Services to connect to your Windows Server 2012 Hyper-V Host, one of the best practices tasks that you can do is to turn off printer redirection. You can print simply from outside like a travel destination. ComponentModel. 1905 (0x771) The specified printer has been deleted. • This includes Windows 8 and 8. This OLE property cannot be an expression. I've been looking through various threads on the forums on the net but so far nothing has worked. If your printer doesn’t connect via USB, and you don’t have setup instructions or drivers, follow these steps. How to: Fix a network printer suddenly showing as offline in Windows Vista, 7 or 8 July 3, 2011. Restart the Spooler on client. Delete unused applications and transfer old files to a CD. It is possible for a job to have multiple flag values specified. OnStartPrint(PrintDocument document. Shop desktop cutting machines including the Silhouette Cameo® plus our selection of cutting materials and other accessories. Same thing, no problem accessing the shared printer on all my accounts. Then fire up regedit. And thanks to HomeGroup networking, when you add a printer to a computer in the HomeGroup network, Windows automatically recognizes it and adds. This program is FREEWARE with limitations, which means that there is a FREE version for personal and commercial use up to 10 users. Was this article helpful?. Open Services, "services. The difference was that I have a network printer and not one shared as a window share connected to a computer. After you complete the steps, you should be able to print a document one more time, and your printer should work now. Right-click Start to access the Win-X menu, and click Control Panel. when I came home I could not connect to internet. solution: The standard Windows style; Delete nonfunctional printer in the Control panel > Printers, reboot (to ensure nothing is left dangling), then reinstall the driver for your printer (which might require yet another reboot to get complete). Thanks Torquemada. For assistance, check here to determine if your label printer is affected and to install the software. After the printer is deleted, restart your computer and then printer will automatically install itself again if it is connected. Restart the Printer Spooler service. This OLE property cannot be an expression. If you have been using the same computer for a while, or if you have recently cycled through printer installations, then you probably have some printer drivers on your computer. if there is MAC-filtering on, and that printer was swapped for some reason. Clear print queue via Devices & Printers. Printer sharing punya beberapa keuntungan dan kekurangan. Khi đó một thông báo lỗi như sau xuất hiện trên màn hình máy tính “Windows cannot connect to printer" và kèm theo là mã lỗi (0x0000007e) như hình bên dưới : Đây là một lỗi rất phổ biến trong quá trình cài đặt và chia sẽ máy in trên hệ điều hành windows. laptopdirect. Open regedit and go to the following key: HKEY_LOCAL_MACHINE\SYSTEM\CurrentControlSet\Control\Print\Environments\Windows x64\Print Processors\. The printer prints from the computer its directly connected to, and I have it shared. But don’t worry. Outlook connection still not working? Run Windows Update to make sure you have the latest updates for Outlook and other desktop apps for Office 365. plt) to be sent to a plotter/printer, or to a printer. Tip: If you're using Windows 10 or Windows 8, the Power User Menu, via the WIN+X key combination, gives you even faster access. Update has been done also on special character handling, when export to printer is used on Avery printers. If you are in an office environment, check with your Network or IT department to make sure you have security clearance to print. When you try to connect to the printer again from Windows 7, the driver for the Windows 7 OS should be available from the host machine for installation. Click Windows Update and in the left pane, click Check for updates. Either the specified user account is already a member of the specified group, or the specified group cannot be deleted because it contains a member. You should be able to access CIFS / SMB / Samba network shares instantly without login. Background: In order for. Doing so via Control Panel (which that link explains in detail if you need it) is probably easiest. After the printer is deleted, restart your computer and then printer will automatically install itself again if it is connected. We'll clear that up. Now, when I start the system, it tells me that the drives are not set as Raid drives. Delete the drivers Go to File - Server Properties and delete all of the drivers not needed. Product Compatibility. In this post, we will introduce 6 types of Windows installation errors as well as top corresponding solutions. Win32Exception was caught Message=The specified printer has been deleted Source=System. For assistance, check here to determine if your label printer is affected and to install the software. Depending on your version of Windows, you may have to select Select the updates you want to install page, followed by clicking OK. When attempting to change the driver associated with a particular printer, you receive the following error: “Printer settings could not be saved. DLL and RUNDLL32. However, when I try to connect other computers to it, I get an unable to connect to printer error with code 0x06. The driver must be uninstalled separately. If you’re having this problem on your Local Area Network, then I don’t think you should really need to worry about it;. If your Outlook data is indexed by Windows Search, you’ll not only get lightening fast search results within Outlook but also outside of Outlook such as the search fields in Windows Explorer and the Start Menu. If you are in an office environment, check with your Network or IT department to make sure you have security clearance to print. I tried downloading drive through link you provided and cannot seem to make it work, a zip file gets saved to my downloads file and nothing appears under browse … please help. If you need to change the name of your computer, the process has not changed since the Windows XP, so here's how in Vista through Windows 10. As usual I tried to re-install the software again to solve the problem but the error is the same. pit] Contact seller. Now we all know that shared local networks come with a printer or scanner or some port which can be used by different users who are registered on the network. access is denied. However, when I attempt to connect to the printers from a remote workstation over the network, I'm getting an error that says 'Windows cannot connect to There is a kb article that appears to address this issue, it's KB883789. All the printers installed on your computer will appear in a different window, select the printer you wish to delete, right click on it and click Delete. Logitech MK Wired Keyboard Design and form factor. ERROR_INVALID_PRINTER_STATE - 0x80070772 - (1906) The state of the printer is invalid. 21rc1 treats a Location directive in cupsd. if this is a network printer, make sure the printer is turned on, and that the printer address is correct. Make sure that within Series, the correct printer is selected. On some occasion TLS V1. The printer cannot detect a wireless board failure. Browsing the print share and trying to connect the printer manually would result in the "Do you trust this printer" pop up which will then prompt for admin creds to install the driver. He writes troubleshooting content and is the General Manager of Lifewire. Posted in Printers, Troubleshooting and tagged 0x00000709, Double check the printer name, Operation could not be completed on February 7, 2013 by Prabu R. The specified printer driver is currently in use. (especially printers) in Windows 10 could not be any easier. 1) Press Win+R (Windows logo key and R key) at the same time to invoke the Run box. Depending on your version of Windows, you may have to select Select the updates you want to install page, followed by clicking OK. This array element has been defined as an object and cannot be redefined in the class definition. After the printer has been created it can be deleted from the print manager. Select your printer on the left side and go to Settings-> Printer Settings or call it from your Dashboard at Actions-> Printer Settings. Read these next. ComponentModel. In the 10-point size, which is used for the main text of this book, the pen is $2/3\pt$ wide, so it has been specified by the command \begindisplay \pickup @[email protected] scaled $2\over3$"pt" yscaled $9\over10$ \enddisplay or something equivalent to this. @You may have misspelled the file name, or the file may have been deleted or renamed. I installed my HP inkjet pro 8600 printer incorrectly and can not back to the screen that specified whether the printer was a wifi printer or was connected by cable. Print Prints the selected PDF or JPEG file. Thanks for jumping in. Installing the printer with the administrator account using the unc path works. Hi, I have a printer (Model: HP Color Laserjet CP3505) which is installed and shared on a Windows Server 2008 R2 print server. While this guide focuses on Windows 10, the ability to reset the “Print Spooler” has been around for a long time, which means that you can use the steps on Windows 8. 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For details about installing the unit, see Installing the Interface Units. If you don’t plan to share files or printers on the client computer, however. I have 5 computers on my home network so I have two routers and each router has its own local IP address subnet. Turn Printer Sharing on. Important! Windows 10 Mobile is currently not supported by Lexmark printers, click here for additional information on product details and limitation for each printer. The specified printer is currently in use. 0_01/jre\ gtint :tL;tH=f %Jn! [email protected]@ Wrote%dof%d if($compAFM){ -ktkeyboardtype =zL" filesystem-list \renewcommand{\theequation}{\#} L;==_1 =JU* L9cHf lp. When printing, the [Set as default printer] in Windows is used. The single most common cause for "Windows 10 broke my printer" that I have seen is that your printer may not be set as default, even after updating drivers, re-installing, etc. A window will open - click the Server Types tab. This hotfix is 476 MB. To continue this discussion, please ask a new question. Windows cannot connect to the printer, specified port … - When adding any one of the many printers on one of the Terminal Servers i recieve - This worked to remove the driver, but after doing this and reinstalling the new HP driver, I still have a problem printing web pages, my HP prints 25. Click Add a printer. Printers are hugely popular PC accessories but also the most prone to problems after an upgrade to Windows 10 should have warned you if it wasn't going to support your printer before you installed it So instead we'll focus on adding a 'local' printer — one that's directly connected to the computer. Here you can set all printer related parameter. Step 2: Go to the Backup window and enter the Source section. Windows cannot access shared folder As users become more and more mobile, you often have to manage devices that are not members of your Active Directory domain. You should first attempt to remove the device from 'DEVICES AND PRINTERS' in Control Panel -> Printers. The Service Pack contains updated versions of those files, which work to provide a stable environment for your programs. Only with my "local" admin. When I try to add a network printer (by drag and drop or going through the wizard), I get: Windows cannot connect to the printer. Click yes, on the pop up box to confirm the deletion of the printer. Everything seems to be running but you can’t see anything on the screen, not even a flashing cursor. but we getting issue with 0x00000709. The product's current interface has been reset and the buffer has been cleared. plt) to be sent to a plotter/printer, or to a printer. 1 Posted on September 26, 2013 by Windows 8 rt/pro My HP LaserJet 3050 All-in-One Printer is connected to my Windows 8 computer and everything worked fine. Anybody who has run a Citrix farm for a period of time knows the dangers of allowing their clients to connect from home and print to their client attached printers. It may be incorrectly installed. In particular, characters 128 through 159 as used in MS-Windows are not part of the ISO-8859-1 code set and will not be displayed as Windows users expect. please help i got an exception when trying to print from crystal report following is the exception details. Now, when I start the system, it tells me that the drives are not set as Raid drives. The job has been retained in the print queue and cannot be deleted. HotCakeX in Windows 10 (RTM) RSAT tools now available… on 09-25-2019 Hi,is there any graphical way to manage NPS and OCSP in Windows Server 2019 Core? RSAT, MMC or WAC does not have it. If your printer is not listed, click Have Disk if you have the printer's. Direct Connection. Restart Print Spooler: After the port has been deleted you need to make sure to start the Print Make your way back to the Windows Services configuration window, right click on the Print Spooler. This is accomplished by installing a "phantom" printer on the server console. For TCP/IP Connections: At a DOS prompt, enter: ping TCP\IP. You can also change the colors and the related line thicknesses (pen numbers) in the printed drawings. You should be able to access CIFS / SMB / Samba network shares instantly without login. So I am concluding that should I have a Zebra prnter, the code example in this article would work. The blog post was reviewed again on 2-7-2018 and is still relevant. Connect to a different computer or server (Remote Desktop) in Windows. If you want to tidy up those printers (removing ones you don't use) you may find Windows 7 doesn't let you delete them, even though you may be a local The Print Spooler service was stopped successfully. When a network or locally attached Citrix client-defined printer is auto-created as a client printer, the comment field is defined as such. one router has addresses 192. windows cannot connect to the printer operation failed with error 0x00000002. First, you should check if the "Print Spooler / Spooler" service is. Click Add a printer. My printer is not listed on the ShopKeep Label Printing page. If this is a network printer, make sure that the printer is turned on, and that the printer address is correct. A few days ago I had a customer with a problem that was as follows: In a RDS Farm (Terminal Server) some users cannot print to any printer. However, you still might be able to add it via the Local Printer method. Print Inspector also saves to a database statistics about all printed documents (document name, number of pages, name of the user who created the job, computer name etc. Otherwise, enter the IP address of your remote computer and click Connect on the right side. continue to use with Photosmart printer. Installing the printer with the administrator account using the unc path works. If you do not specify an account, to run the command, you must log in as an account with these permissions. About cookies. These characters include the em dash, en dash, curly quotes, bullet, and trademark symbol; neither the actual character (the single byte) nor its &#nnn; decimal equivalent is correct in HTML. For testing purpose we are using the Xerox Global Post Script Printer Driver. Printer Redirection is the feature that allows a local printer to be mapped on a remote machine, and allows printing across the network or Internet. Child windows cannot have menus. (Download source link no longer works and has been removed from this post, download the attached. Printers are hugely popular PC accessories but also the most prone to problems after an upgrade to Windows 10 should have warned you if it wasn't going to support your printer before you installed it So instead we'll focus on adding a 'local' printer — one that's directly connected to the computer. x) The printer is connected to a. The specified printer or disk device has been paused. We've got several different methods for you to choose from. Now, for some reason, all the other computers connected to the network couldn't print to the printer. Solution 2 – Edit the registry. ) on any w/s. It may be incorrectly installed. In the Properties window, click the Drivers tab. This allowed iOS devices to print to shared printers on the network, via AirPrint, that don't natively support it. the Photosmart was operating on the Deskjet software. It is possible for a job to have multiple flag values specified. Try This: Control Panel>Folder Options> Click on View TAB>, in the Box look and clear “Use simple file sharing (Recommended)”, apply and try access again the computer from the network. >if you're in windows 10 select manage your credentials this will open 2 options. When this vent is blocked, it can prevent ink from leaving the cartridge. As ther are no - 2709597. • This includes Windows 8 and 8. Note that you can run into this problem with 32-bit and 64-bit operating systems. You should first attempt to remove the device from 'DEVICES AND PRINTERS' in Control Panel -> Printers. Follow the steps below to troubleshoot this problem. Have been using Dymo printers for years. DONOTEDITTHISFILE!!!!! !!!!!$ !!!!!///// !!!"!&!&!+!+!S!T![!^!`!k!p!y! !!!"""'" !!!&& !!!'/'notfoundin"%s" !!!) !!!5" !!!9" !!!EOFinsymboltable !!!NOTICE. USB ports are not named in the same manner as serial or parallel ports. For detailed instructions on the below topics, see: Photoshop Help / Basic troubleshooting steps to fix most issues. My mac keeps telling me that "AirPort does not have an IP address and cannot connect to the Internet" Is this something I can fix with a static IP address? I tried to follow your instructions for the Mac, but since mine is a wireless router, I guess that's why it's different. Having problems with installed printer ending up in unspecified list in windows 7? January 12, 2012 AM Labels A problem that can be encountered when installing your label printer on the Windows 7 operating system is that it will place it in the unspecified devices list, even after you believe you have installed the drivers correctly. Remove Network Printers from Windows via Registry Editor. Apparently of printer that has faced the error, Delete the BIDI branch in CopyFiles section. Showing how to fix “The system cannot find the file specified” on Windows 10 Fix it now! To repair damaged system, you have to purchase the licensed version of Reimage Reimage. This approach is only recommended for accounts that have not been active (i. The blog post was reviewed again on 2-7-2018 and is still relevant. I rebooted both computers, added the printer to the desktop and then I was able to add it to the laptop. This post has become quite popular – so I’ve updated it with a bit more detail, plus some people’s experiences from the comments. How to: Fix a network printer suddenly showing as offline in Windows Vista, 7 or 8 July 3, 2011. Step 3: Right-click on a print queue, and then click Cancel option to clear the particular print job from the queue. Select the Folders and Files section, and choose the files or folders you want to back up and click OK. Another troubling sign is the way that the MailItemsAccessed audit event has been packaged into a new Microsoft 365 Audit feature. The default printer has been set and I am using this method. msg files to be text searchable, Outlook must be installed on the Indexer workstation, but it doesn't need to be configured for a user. When I try to add a network printer (by drag and drop or going through the wizard), I get: Windows cannot connect to the printer. printer: A printer is a device that accepts text and graphic output from a computer and transfers the information to paper, usually to standard size sheets of paper. This may have been going on for a while but I only realized the problem today. Turn on your printer and make sure it is connected to your computer using Windows 10. The specified printer driver is currently in use. Now we all know that shared local networks come with a printer or scanner or some port which can be used by different users who are registered on the network. You can print simply from outside like a travel destination. LOGITECH CORDLESS EX100 DRIVERS FOR WINDOWS 8 - Wireless USB 2. This user (let's call her Sally) had the printer installed already from. If you see more than one printer. • This includes Windows 8 and 8. The glorious one-click upgrade to Windows 10 Technical Preview build 9860. Important! Windows 10 Mobile is currently not supported by Lexmark printers, click here for additional information on product details and limitation for each printer. You may not have the appropriate permissions to access the item”. Then connect it again. Duplicate member/property name. If this occurs, but you cannot print from your application or through Windows, make sure that the appropriate driver is installed and that you have selected the printer correctly. If your printer is not listed, click Have Disk if you have the printer's. Windows cannot connect to the printer. Turn on the printer, and then connect it to your computer. If its disk capacity is shown, that means your Windows operating system has recognized this disk, but its partition is lost or MBR is corrupted due to some reasons. They're easy to set up, the hardware and printouts are good quality, they're easy to maintain and they last a long time (not to mention, their name is in all capital letters; it doesn't get. The Service Pack contains updated versions of those files, which work to provide a stable environment for your programs. Can the Deskjet software (have disk) be reinstalled and. c --- wfdb-10. 130 - November 11, 2013. Posted in Printers, Troubleshooting and tagged 0x00000709, Double check the printer name, Operation could not be completed on February 7, 2013 by Prabu R. Even if it is not, you will be able to find the document name.
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# Understanding bending light beam perpendicular to motion
I'm just reading a book about gravity. An example it gives is a spaceship accelerating. A beam of light travelling at right angles to the direction of movement of the spaceship enters it via a small pinhole. The book states that the beam of light would appear to bend to an observer within the spaceship. Intuitively this makes sense - if the ship were travelling at $0.5\,c$, for example, and was $10\,m$ across (from the pinhole to the opposing wall) then I'd expect the light to hit the wall $5\,m$ further back down (assuming the ship is accelerating "up") from the pinhole - the wall would have moved 5m further in the time it takes the light beam to cross the interior to the opposing wall
I think I'm fundamentally misunderstanding though. The book states this is down to the acceleration of the ship and goes on to talk about how using equivalence the same bending would be a result of gravity. In my description above though, it's the velocity that causes the apparent bending, and the same would be visible if the ship was maintaining a constant $0.5\,c$. I suspect that if this were the case (constant $0.5\,c$) there would be no bending at all, but I'm not understanding why
Can anyone enlighten me?
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Hi Kevin, and welcome to Physics Stack Exchange! Despite appearances, your question doesn't have all that much to do with gravity, so I added some other relevant tags. – David Z Mar 27 '13 at 13:28
Thank you David. You're right of course. My aim is a better understanding of gravity/relativity, which is what made me choose gravity, but this question is quite early in the journey, I think – Kevin O'Donovan Mar 27 '13 at 15:27
It's not the velocity that causes the bending, it's the acceleration! If the spaceship were moving by any but constant velocity in the absence of a gravitational field, the path of the photons would be straight, wouldn't it? The motion of any uniformly moving object (or photon) always looks straight in any other inertial (uniformly moving) frame. The path gets curved, "parabolic", just because the velocity isn't constant.
The equivalence principle equates the situations of an accelerated spaceship with a static spaceship sitting in the gravitational field, with the correspondence $a\sim g$. So if the path of the photon is curved in the accelerated spaceship, it should be curved in the gravitational field (but with no extra motion), too, although I am not quite sure whether this simple argument produces the right factor of two that a naive Newtonian "attraction acting on light" misses relatively to the correct general relativistic calculation.
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I think that's where I'm struggling. I'm an engineer rather than a physicist, so this isn't my domain. I think I'm being extremely naïve. If the photon enters through the pinhole the ship will have moved on from its original position by the time it hits the wall, displacing the point of impact. Ah, I may have missed the point. In the presence of acceleration the path traced to get to that point would be a curve. Without acceleration there would still be a displacement but the path would not be a curve. Is that right, or am I still missing the point. – Kevin O'Donovan Mar 27 '13 at 15:25
It's one of the points, there are also other points - so far, you haven't gotten to the key point involving the equivalence principle. Could you please be more specific about which sentence you don't understand? Otherwise it is pretty much impossible to deduce what you're struggling with, it remains an esoteric purely internal psychological struggle no one else can understand. – Luboš Motl Mar 28 '13 at 6:49
Kevin you are quite right in your assumption, although these are two different situations. If the ship was just travelling at a speed, the beam would be displaced travelling across the ship, so in the ship's frame of reference it would appear travelling at an angle, as if it was shone this way - this is called aberration of light.
If the ship is accelerating, the displacement taking place would have a more complex form, a curve, corresponding to the ship's motion. But obviously in the external observer's reference frame the beam would still be travelling straight.
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I can think of a light beam as a pulsating stream of water from a hose traveling at the speed of light.
If there's a hole in the side of my space ship and the hose of streaming water is pointed directly, perpendicular to my ship's direction of travel, at the hole, then only a portion of the pulsated water will enter the hole.
Now concerning the portion of the pulsating stream of light-speed water that has just entered the space ship traveling at half the speed of light, we ask: "What is the vector or path of those water pulses.
I see them as a perfectly straight line no mater what the ships does.
The water will collide with the wall ten meters opposite the entry hole's wall and, at the ship's constant speed and direction, form a water mark shaped like a dotted line.
Seems very simple to me. And as far as "how it appears to this observer" or that, are the great minds of science saying that reality is altered depending on where you sit? Really now.
Just a thought.
John
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# If |x - 2| = |x + 3|, x could equal
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If |x - 2| = |x + 3|, x could equal [#permalink]
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07 Mar 2016, 09:22
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If |x - 2| = |x + 3|, x could equal
A. -5
B. -1/2
C. 1/2
D. 5
E. No real solution
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Re: If |x - 2| = |x + 3|, x could equal [#permalink]
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07 Mar 2016, 20:50
2
I back solved the question by putting answer choices in the question. -1/2 gives 5/2 on both sides.
Thanks
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Re: If |x - 2| = |x + 3|, x could equal [#permalink]
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07 Mar 2016, 20:58
I started drawing a number line, but then I thought better of it and just plugged in the values form the answers.
A) -5: LHS: |-5-2| = 7, RHS: |-5+3| = 2, no good
B) -1/2: LHS: |-1/2 - 2| = 2.5, RHS: |-1/2 + 3| = 2.5, Correct
Thus B
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Re: If |x - 2| = |x + 3|, x could equal [#permalink]
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07 Mar 2016, 21:04
2
2
Bunuel wrote:
If |x - 2| = |x + 3|, x could equal
A. -5
B. -1/2
C. 1/2
D. 5
E. No real solution
By definition, |x-a| = distance of x from a. In other words, |x-2| = distance of x from 2 and |x+3| = distance of x from -3.
Thus, if |x-2| = |x+3| ---> distance of x from 2 = distance of x from -3. Thus, x = halfway between -3 and 2.
Distance between -3 and 2 = 5 units. Thus x is 2.5 units away from 2 or -3 , giving you x = 2-2.5 or = -3+2.5 = -0.5,
B is thus the correct answer.
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Re: If |x - 2| = |x + 3|, x could equal [#permalink]
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08 Mar 2016, 00:39
Bunuel wrote:
If |x - 2| = |x + 3|, x could equal
A. -5
B. -1/2
C. 1/2
D. 5
E. No real solution
I just backsolved and was able to see that B was the answer for me. Both sides end up with the same result of 5/2.
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Re: If |x - 2| = |x + 3|, x could equal [#permalink]
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08 Mar 2016, 01:15
solving equation we get -1/2
Ans ; B
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Re: If |x - 2| = |x + 3|, x could equal [#permalink]
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11 Jun 2016, 11:18
1
1
|x - 2| = |x + 3|
We see both the sides are positive, hence we can square both the sides and we get
X^2 + 4 -4x = X^2 + 9 -6x
10x= -5
X= -1/2
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Re: If |x - 2| = |x + 3|, x could equal [#permalink]
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12 Jun 2016, 02:19
1
Bunuel wrote:
If |x - 2| = |x + 3|, x could equal
A. -5
B. -1/2
C. 1/2
D. 5
E. No real solution
Since there are two Equations. Lets solve by Critical Value
|x - 2| = |x + 3| Critical Value will be 2, -3 hence we Have 3 Solutions
Solution No 1 When X < -3 ( RHS Will be Negative )
There fore x-2 = - (x + 3 ) => 2x = -1 => x = -1/2 which is not in line with x< -3 , hence rejected
Solution No 2 When -3 < X <= 2 ( LHS will be Negative )
-(x-2) = x+3 => 2x= -1 => x = -1/2 which fits in the range Hence Solution.
We can check by substituting in |x - 2| = |x + 3| Therefore 5/2= 5/2
Solution 3 when X > 2 both sides positive
x-2 = x+3 No Value hence rejected
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Re: If |x - 2| = |x + 3|, x could equal [#permalink]
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12 Jun 2016, 03:59
1
Bunuel, in one of your post , you ave explained if we have mod on both side of the equation, we can safely square on both side
(x-2)^2= (x+3)^2
by solving we get x= -1/2 .
Please correct me if i missed something.
Math Expert
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Posts: 7563
Re: If |x - 2| = |x + 3|, x could equal [#permalink]
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12 Jun 2016, 04:33
1
Bunuel, in one of your post , you ave explained if we have mod on both side of the equation, we can safely square on both side
(x-2)^2= (x+3)^2
by solving we get x= -1/2 .
Please correct me if i missed something.
..
Hi sudhir,
yes, whenever you have ONLY mods on both sides..
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Re: If |x - 2| = |x + 3|, x could equal [#permalink]
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24 Jun 2016, 08:12
1
Bunuel wrote:
If |x - 2| = |x + 3|, x could equal
A. -5
B. -1/2
C. 1/2
D. 5
E. No real solution
Given |x - 2| = |x + 3|
The best way to get rid of the modulus is to square on both sides.
$$(x - 2)^2$$ = $$(x+3)^2$$
=> $$x^2$$ + 4 + 4x = $$x^2$$ + 9 + 6x
=> x = -1/2.
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Re: If |x - 2| = |x + 3|, x could equal [#permalink]
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24 Jul 2016, 13:39
Razween wrote:
Bunuel wrote:
If |x - 2| = |x + 3|, x could equal
A. -5
B. -1/2
C. 1/2
D. 5
E. No real solution
Since there are two Equations. Lets solve by Critical Value
|x - 2| = |x + 3| Critical Value will be 2, -3 hence we Have 3 Solutions
Solution No 1 When X < -3 ( RHS Will be Negative )
There fore x-2 = - (x + 3 ) => 2x = -1 => x = -1/2 which is not in line with x< -3 , hence rejected
Solution No 2 When -3 < X <= 2 ( LHS will be Negative )
-(x-2) = x+3 => 2x= -1 => x = -1/2 which fits in the range Hence Solution.
We can check by substituting in |x - 2| = |x + 3| Therefore 5/2= 5/2
Solution 3 when X > 2 both sides positive
x-2 = x+3 No Value hence rejected
In the first condition in which X is less than -3, why isn't the (x-2) term negative as well? The expression (x-2) is negative for all values where X is less than 2.
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Re: If |x - 2| = |x + 3|, x could equal [#permalink]
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25 Jul 2016, 01:18
astroshagger wrote:
Razween wrote:
Bunuel wrote:
If |x - 2| = |x + 3|, x could equal
A. -5
B. -1/2
C. 1/2
D. 5
E. No real solution
Since there are two Equations. Lets solve by Critical Value
|x - 2| = |x + 3| Critical Value will be 2, -3 hence we Have 3 Solutions
Solution No 1 When X < -3 ( RHS Will be Negative )
There fore x-2 = - (x + 3 ) => 2x = -1 => x = -1/2 which is not in line with x< -3 , hence rejected
Solution No 2 When -3 < X <= 2 ( LHS will be Negative )
-(x-2) = x+3 => 2x= -1 => x = -1/2 which fits in the range Hence Solution.
We can check by substituting in |x - 2| = |x + 3| Therefore 5/2= 5/2
Solution 3 when X > 2 both sides positive
x-2 = x+3 No Value hence rejected
In the first condition in which X is less than -3, why isn't the (x-2) term negative as well? The expression (x-2) is negative for all values where X is less than 2.
Yes you are right.
If $$x < -3$$, then $$|x - 2| = -(x - 2)$$ and $$|x + 3| = -(x + 3)$$;
If $$-3 \leq x \leq 2$$, then $$|x - 2| = -(x - 2)$$ and $$|x + 3| = x + 3$$;
If $$x > 2$$, then $$|x - 2| = x - 2$$ and $$|x + 3| = x + 3$$.
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If |x - 2| = |x + 3|, x could equal [#permalink]
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27 Jul 2017, 02:59
|x - 2| - |x + 3|=0
We substitute 2,-3 in the above expression accordingly and plot the same in number line.
Now minimum distance between the two points is 5 which is less than RHS. hence no real solution.
Option E.
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Re: If |x - 2| = |x + 3|, x could equal [#permalink]
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29 Jul 2017, 09:02
chetan2u wrote:
Bunuel, in one of your post , you ave explained if we have mod on both side of the equation, we can safely square on both side
(x-2)^2= (x+3)^2
by solving we get x= -1/2 .
Please correct me if i missed something.
..
Hi sudhir,
yes, whenever you have ONLY mods on both sides..
hi
so far I can remember, square rooting hides the positive/ negative issues. So, as the sign of x is unknown, it is safe to take cube roots...
please correct me if I am wrong ...
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Re: If |x - 2| = |x + 3|, x could equal [#permalink]
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29 Jul 2017, 09:14
ssislam wrote:
hi
so far I can remember, square rooting hides the positive/ negative issues. So, as the sign of x is unknown, it is safe to take cube roots...
please correct me if I am wrong ...
Hello ssislam, In this case, it will not hide as we are taking the square root (or rather assuming) the square root of x in the entire equation.
For some mod questions, you can take do a square on both sides and try to solve the question. Hope this helps
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Re: If |x - 2| = |x + 3|, x could equal [#permalink]
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29 Jul 2017, 20:51
B is the right option
x-2=-x-3
2x=-3+2
2x=-1
x=-1/2
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If |x - 2| = |x + 3|, x could equal [#permalink]
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30 Jul 2017, 10:16
1
Bunuel wrote:
If |x - 2| = |x + 3|, x could equal
A. -5
B. -1/2
C. 1/2
D. 5
E. No real solution
With absolute value equations this simple, for me, the method below is easy and fast (less than a minute).
The only way the two sides will be equal is if quantities inside the absolute value brackets are 1) equal or 2) equal but with opposite signs.
So |A| = B or -B*
1. Remove the absolute value bars
2. LHS = RHS or LHS = -RHS
3. Set up the two equations
Case 1: x - 2 = x + 3, OR
Case 2: x - 2 = -(x + 3)
4. Solve
Case 1:
x - 2 = x + 3
-2 = 3. Not a solution
CASE 2:
x - 2 = -(x + 3)
x - 2 = -x - 3
2x = -1
x = -$$\frac{1}{2}$$
5. Check x = -$$\frac{1}{2}$$ in equation. --> |-$$\frac{5}{2}$$| = |$$\frac{5}{2}$$|
Correct. x = -$$\frac{1}{2}$$. Answer B
Hope it helps
*and |B| = A or -A. Technically there are four cases. Case 3 is RHS = LHS. Case 4 is RHS = - LHS. These two cases are identical to Cases 1 and 2.
Case 3: +RHS = +LHS, (x + 3) = (x - 2) (= Case 1)
Case 4: +RHS = -LHS, (x + 3) = (- x + 2). Multiply both sides by (-1) --> (-x - 3) = (x - 2) (= Case 2)
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If |x - 2| = |x + 3|, x could equal [#permalink] 30 Jul 2017, 10:16
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I’ve heard a bit about this act before. But today I read an article that talked about how the National Right to Life Committee and other Pro-Life groups are opposing it. That caught my interest since I consider abortion to be the worst cancer on our society, ever. It’s far worse than slavery. So I decided to do some research. Here is the summary of the bill from Open Congress.
4/29/2010--Introduced.Democracy is Strengthened by Casting Light on Spending in Elections Act or DISCLOSE Act - Amends the Federal Election Campaign Act of 1971 (FECA) to prohibit: (1) independent expenditures and payments for electioneering communications by government contractors if the value of the contract is at least $50,000; and (2) recipients of assistance under the Troubled Asset Relief Program (TARP) of the Emergency Economic Stabilization Act of 2008 (EESA) from making any contribution to any political party, committee, or candidate for public office, or to any person for any political purpose or use, or from making any independent expenditure or disbursing any funds for an electioneering communication. Applies the ban on contributions and expenditures by foreign nationals to foreign-controlled domestic corporations. Treats as contributions: (1) any payments by any person (except a candidate, a candidate's authorized committee, or a political committee of a political party) for coordinated communications; and (2) political party communications made on behalf of candidates if made under the control or direction of a candidate or a candidate's authorized committee. Revises the definition of independent expenditure to mean, in part, an expenditure that, when taken as a whole, expressly advocates the election or defeat of a clearly identified candidate, or is the functional equivalent of express advocacy. Requires any person making independent expenditures exceeding$10,000 to file a report within 24 hours. Increases the period before a general election during which a communication shall be considered an electioneering communication. Requires corporations, labor organizations, and other covered organizations to include specified additional information in reports on independent expenditures of at least $10,000. Sets forth special rules for the use of general treasury funds by covered organizations for campaign-related activity. Authorizes covered organizations to make optional use of a separate Campaign-Related Activity Account for making disbursements for campaign-related activity. Prescribes additional information to be included in certain radio or television communications by persons (including significant funders of campaign-related communications of a covered organization) other than a candidate, a candidate's authorized committee, or a political committee of a political party. Amends the Lobbying Disclosure Act of 1995 to require registered lobbyists to report information on independent expenditures or electioneering communications of at least$1,000 to the Secretary of the Senate and the Clerk of the House of Representatives. Requires certain covered organizations to disclose to shareholders, members, or donors information on disbursements for campaign-related activity. Authorizes judicial review of the provisions of this Act.
Now, as with most law, it’s written in a way to is hard to understand. After all, if it wasn’t written that way, the lawyers that make up our political system would be out of a job… Anyway, here’s what I understand it means.
1. If you are a company not owned by the government, and you have a contract worth more than $50,000, then that contract will be canceled if you try to send out communications supporting a particular political candidate. I don’t get this, the money paid to them is not federal money anymore. So what they do with it is none of the Governments business. 2. If you got bailed out by TARP, you can’t comminucate about political stuff. Aren’t TARP funds loans? I get student loans from the Federal Government, does that mean I shouldn’t speak out about things? 3. The second two parts are a little confusing. I think they mainly apply to foreigners. If that’s not true, then it’s a clear violation of our freedom of speech. If you look at the section below you see the stuff that bothers the NRLC: Prescribes additional information to be included in certain radio or television communications by persons (including significant funders of campaign-related communications of a covered organization) other than a candidate, a candidate’s authorized committee, or a political committee of a political party. Amends the Lobbying Disclosure Act of 1995 to require registered lobbyists to report information on independent expenditures or electioneering communications of at least$1,000 to the Secretary of the Senate and the Clerk of the House of Representatives. Requires certain covered organizations to disclose to shareholders, members, or donors information on disbursements for campaign-related activity.
Basically, if you spend a lot of money to support your favorite candidate by telling people why you like him, you have to report a bunch of information to a bunch of different people. Or you get in trouble. That puts a rather large damper on peoples desire to speak out about things, thus it’s just a tiny bit away from restricting free speech.
It also will force grassroots organizations to publish lists of who’s donated money to their political campaigns. In other words, they are forced to tell their opponents who supports them. Thus opening up their supporters to backlash from their opponents. And don’t think there wouldn’t be backlash. That’s what happened with Propisition 8.
What do you think? Have you heard anything about this that contradicts what I’ve said? I’d love to hear different perspectives. Especially the ones that oppose mine.
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# Is NPI contained in P/poly?
It is conjectured that $\mathsf{NP} \nsubseteq \mathsf{P}/\text{poly}$ since the converse would imply $\mathsf{PH} = \Sigma_2$. Ladner's theorem establishes that if $\mathsf{P} \ne \mathsf{NP}$ then $\mathsf{NPI} := \mathsf{NP} \setminus(\mathsf{NPC} \cup \mathsf{P}) \ne \emptyset$. However, the proof doesn't seem to generalize to $\mathsf{P}/\text{poly}$ so the possibility $\mathsf{NPI} \subset \mathsf{P}/\text{poly}$ i.e. $\mathsf{NP} \subset \mathsf{NPC} \cup \mathsf{P}/\text{poly}$ seems open.
Assuming $\mathsf{NP} \nsubseteq \mathsf{P}/\text{poly}$ (or even that the polynomial hierarchy doesn't collapse on any level), is $\mathsf{NPI} \subset \mathsf{P}/\text{poly}$ known to be true or false? What evidence can be put for and against it?
• so, "what would happen if all problems in NP are either NP-complete or in P\poly"? for one thing it would imply small circuits for factoring – Sasho Nikolov Dec 8 '12 at 22:04
• ps: the post would be more readible if you spell out "it" in the quoted part. Also you may want to use $\mathsf{NP}\not\subseteq\mathsf{P/poly}$ in place of $\mathsf{NP}\not\subseteq\mathsf{P}$ as your assumption. – Kaveh Dec 8 '12 at 23:40
• Won't a padding argument show that this can't happen unless NP $\subseteq$ P/poly? – Peter Shor Dec 9 '12 at 1:29
• @PeterShor : I'm probably being dense, but how exactly would it work? – Squark Dec 9 '12 at 18:25
• @Squark: you're not being dense ... I hadn't worked out exactly how it would work, and I think I mis-stated the result slightly. But here's my basic idea. Suppose that NP-complete problems can't be solved in subexponential time and advice. Take an NP-complete problem X, and pad it so that the fastest algorithm for it is barely subexponential. Then it's NPI, so it can be solved in P/poly. This means the NP-complete problem X can be solved in time only slightly slower than P/poly time. By polynomial reduction, now all NP-complete problems can be solved in slightly slower than P/poly time. – Peter Shor Dec 9 '12 at 18:46
Here is a possible alternative to a padding argument, based on Schöning's generalization of Ladner's theorem. To understand the argument, you need to have access to this paper (which will unfortunately be behind a pay wall for many):
Uwe Schöning. A uniform approach to obtain diagonal sets in complexity classes. Theoretical Computer Science 18(1):95-103, 1982.
We will apply the main theorem from that paper for $A_1$ and $A_2$ being languages and $\mathsf{C}_1$ and $\mathsf{C}_2$ being complexity classes as follows:
• $A_1 = \varnothing$ (or any language in $\mathsf{P}$)
• $A_2 = \text{SAT}$
• $\mathsf{C}_1 = \mathsf{NPC}$
• $\mathsf{C}_2 = \mathsf{NP} \cap \mathsf{P/poly}$
For the sake of clarity, the fact we will prove is $\mathsf{NP} \not\subseteq \mathsf{P/poly}$ implies $\mathsf{NPI} \not\subseteq \mathsf{P/poly}$.
Under the assumption that $\mathsf{NP} \not\subseteq \mathsf{P/poly}$ we have $A_1\not\in\mathsf{C}_1$ and $A_2\not\in\mathsf{C}_2$. It is clear that $\mathsf{C}_1$ and $\mathsf{C}_2$ are closed under finite variations. Schöning's paper includes a proof that $\mathsf{C}_1$ is recursively presentable (the precise definition of which can be found in the paper), and the hardest part of the argument is to prove that $\mathsf{C}_2$ is recursively presentable.
Under these assumptions, the theorem implies that there exists a language $A$ that is neither in $\mathsf{C}_1$ nor in $\mathsf{C}_2$; and given that $A_1\in\mathsf{P}$, it holds that $A$ is Karp-reducible to $A_2$, and therefore $A\in\mathsf{NP}$. Given that $A$ is in $\mathsf{NP}$ but is neither $\mathsf{NP}$-complete nor in $\mathsf{NP} \cap \mathsf{P/poly}$, it follows that $\mathsf{NPI} \not\subseteq \mathsf{P/poly}$.
It remains to prove that $\mathsf{NP} \cap \mathsf{P/poly}$ is recursively presentable. Basically this means that there is an explicit description of a sequence of deterministic Turing machines $M_1, M_2, \ldots$ that all halt on all inputs and are such that $\mathsf{NP} \cap \mathsf{P/poly} = \{L(M_k):k=1,2,\ldots\}$. If there is a mistake in my argument it is probably here, and if you really need to use this result you will want to do this carefully. Anyway, by dovetailing over all polynomial-time nondeterministic Turing machines (which can be simulated deterministically because we don't care about the running time of each $M_k$) and all polynomials, representing upper bounds on the size of a Boolean circuit family for a given language, I believe it is not difficult to obtain an enumeration that works. In essence, each $M_k$ can test that its corresponding polynomial-time NTM agrees with some family of polynomial-size circuits up to the length of the input string it is given by searching over all possible Boolean circuits. If there is agreement, $M_k$ outputs as the NTM would, otherwise it rejects (and as a result represents a finite language).
The basic intuition behind the argument (which is hidden inside Schöning's result) is that you can never have two "nice" complexity classes (i.e., ones with recursive presentations) being disjoint and sitting flush against each other. The "topology" of complex classes won't allow it: you can always construct a language properly in between the two classes by somehow alternating between the two for extremely long stretches of input lengths. Ladner's theorem shows this for $\mathsf{P}$ and $\mathsf{NPC}$, and Schöning's generalization lets you do the same for many other classes.
• This is a link to Schöning's publications available on-line for free, including the one you refer to: uni-ulm.de/in/theo/m/schoening/… – Alessandro Cosentino Dec 13 '12 at 15:27
• Thanks a lot for your answer! The funny thing is, I knew Shoening's theorem but for some foolish reason thought it doesn't apply in this case. Btw, the text is freely available even in sciencedirect – Squark Dec 15 '12 at 10:57
• @Squark: It isn't foolish to suspect that Schöning's theorem doesn't apply, given that P/poly includes non-recursive languages. I suppose it's good fortune that we can intersect it with NP and still get the result. – John Watrous Dec 15 '12 at 15:14
• @JohnWatrous : Yes, this is precisely the reason I was confused – Squark Dec 15 '12 at 15:15
I'd just like to write down some version of a padding argument as described in the comments. I don't see why a gap is needed. We want to show that if NP is not contained in P/poly then there is an NP-intermediate problem not contained in P/poly.
There is an unbounded function $f$ such that SAT does not have circuits of size less than $n^{f(n)}$, and so there is a function $g$ that is unbounded, increasing, and $g(n)=o(f(n))$. Let SAT' denote the language obtained by padding SAT strings of length $n$ to $n^{g(n)}$. Then:
• SAT' is in NP (see below!)
• SAT' is not in P/poly: given circuits of size $n^k$ for SAT', we get circuits of size $n^{g(n)k}$ for SAT, but this is less than $n^{f(n)}$ for some $n$.
• There is no P/poly reduction from SAT to SAT': suppose for contradiction that there are circuits $C_n$ of size $n^k$ for SAT, allowing SAT' gates. Pick $N$ large enough that $g(\sqrt{N}) > 2k$ and let $n>N$. Each SAT' gate in $C_n$ has at most $n^k$ inputs. By removing the padding inputs we can trim the SAT' gates in $C_n$ to a SAT gate with less than $\sqrt{n}$ inputs, which we can simulate using $C_{\sqrt{n}}$ - the resulting SAT' gates have at most $n^{k/2}$ inputs. Repeating this and treating $C_N$ by hand, SAT would have circuits of about size $O(n^k n^{k/2} n^{k/4} \dots) \approx O(n^{2k})$ which is less than $n^{f(n)}$ for some $n$.
Edit:
The choice of $g$ is slightly fiddly. If you are happy putting SAT' in the promise version of NP, this bit is unnecessary.
Define $f(n)$ to be the maximum integer such that there is no circuit of size $n^{f(n)}$ for length $n$ strings for SAT. Define $g(n)$ by an algorithm that calculates $f(m)$ for $m=1,2,\dots$ and stops after time $n$ or when $m=n$, and returns the floor of the square root of the highest value found in this time. So $g(n)$ is unbounded and $\liminf g(n)/f(n)=0$ and $g(n)$ can be computed in time $n$. Now note that the above arguments only rely on SAT having no circuits of size $n^{f(n)}$ for infinitely many $n$.
I'd also find it interesting to see a proof by blowing holes in SAT as in http://blog.computationalcomplexity.org/media/ladner.pdf. Without the NP requirement this is fairly easy: there is a sequence $n_1<n_2<\dots$ such that no circuit os size $(n_k)^k$ detects SAT strings of length $n$; restrict SAT to strings of length $n_{2^{2^i}}$ for some $i$.
• After seeing @JohnWatrous's answer, I was reminded of Impagliazzo's proof of Ladner's Theorem by padding (cf. the appendix of Downey and Fortnow "Uniformly Hard Languages": cs.uchicago.edu/~fortnow/papers/uniform.pdf). In fact, your proof is basically Impagliazzo's proof of Ladner, but adapted to this situation. Neat! – Joshua Grochow Dec 13 '12 at 22:22
• Thanks a lot for your answer! I apologize I haven't selected it but I had to pick one and Watrous' argument was easier to follow through since it used a result I already knew. This is a rather subjective way to choose but I couldn't do any better. Anyway it's great to have more than one way to arrive at an interesting result – Squark Dec 15 '12 at 11:00
• @Squark: absolutely - and I also assumed Schöning's theorem didn't apply. – Colin McQuillan Dec 16 '12 at 0:24
(NPI $\nsubseteq$ P/poly) $\implies$ (P$\neq$NP)
• it is both known and trivial: if P = NP, then $\mathsf{NPI} \subseteq \mathsf{NP} = \mathsf{P} \subseteq \mathsf{P/pol}$. also this is not the question, the question is the converse of what you wrote, and was convincingly answered by Colin as far as I can see. – Sasho Nikolov Dec 12 '12 at 18:33
• the question is entitled "is NPI contained in P/Poly" & think this is a reasonable answer, not sure its really trivial because of the way NPI is usually defined (as dependent on P$\neq$NP)... this answer does not conflict with the other answer... – vzn Dec 12 '12 at 18:45
• Actually it is even more obviously trivial: if P=NP, NPI is empty. The question is clearly stated as "does NP not contained in P/poly imply NPI not in P/poly. So your answer 1) claims that a trivial fact is an open problem 2) does not address the question – Sasho Nikolov Dec 12 '12 at 19:26
• Could not care less about points. For the last time: my first comment, Colin's answer, and the question itself are related to the far less trivial and more interesting converse of the empty implication you wrote down. – Sasho Nikolov Dec 12 '12 at 19:55
• -1: sometimes losing point feels just right – Alessandro Cosentino Dec 13 '12 at 4:02
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## Friday, July 25, 2014 ... /////
### Brainwashed sheep's obsession with "villain" Vladimir Putin
Several people who largely share my appraisal of the events in Ukraine and around Ukraine have sent me lots of incredible photographs and articles showing that the "mainstream" Western media got completely obsessed with Vladimir Putin.
This 29-year-old lady was threatened by a mayor to be deported from Holland and finally she fled the country by herself – taking BF Jorrit Faasen, 34, with her. In the current hysteria, the sufficient reason is the following: she is Maria Putin, Putin's daughter.
It's enough to search Google News for Putin if you want to obtain a rather incredible collection of totally nutty titles and whole articles about Putin. Just some of the titles (I didn't really have to filter it much):
US General Dempsey: Putin May 'Light a Fire' He Can't Stop in Ukraine
Dempsey: Putin's Moves Like 1939 Poland
Putin's voracious appetite is not sated
Putin’s Pal [a nasty attack against Stephen Cohen]
Putin's inner circle sheds light on his "sinister, lonely life"
The growing calls to strip Putin and Russia of the 2018 World Cup
Putin's Crime, Europe's Cowardice
Vladimir Putin is responsible for shooting down Malaysia Airlines Flight 17. His next move will matter most of all.
I could go on and on and on, for hours. Vladimir Putin is perhaps the next Adolf Hitler, maybe Adolf Hitler on steroids. He must have personally shot the Dutch airplane down, too.
Well, I have been following the Ukrainian events since the late 2013 very closely and one may see that Vladimir Putin hasn't done a damn thing. Perhaps the only thing he did was to do nothing ;-) against the Crimeans' efforts to escape a Ukraine that was conquered by a nationalist hysteria. Needless to say, every other Russian leader – and most leaders of other powers – would do exactly the same thing when their currently threatened, historical region massively asked for re-annexation.
But concerning MH17: are you joking? It was probably shot down by a separatist by mistake. Maybe he wanted to shoot a nearby Kiev military plane and it occurred by a very unlucky accident. The Kiev fighter jets have been claimed to hide behind civilian airplanes already a month ago. Maybe the airplane was shot by the Kiev forces in another unlucky accident. But what does Vladimir Putin have to do with any of that?
The missile that destroyed MH17 was arguably Buk [=Beech wood], produced in Russia. But what's unexpected about the Russian origin of weapons that are used thirty miles from Russia? Naturally, Russia is a major producer of weapons. The 300 passengers and crew members died in a piece of metal that was constructed by American hands, by Boeing. And the soldiers may have used iPhones to order others to shoot the plane down. Does it mean that America is responsible? Are you joking?
And even if Russia were "responsible" for that, why Putin?
There are many possibilities why the plane was shot down and nothing can be "quite" eliminated. But Putin's personal role in the sad event is still one of the possibilities close to the theory about the moonlanding staged in Arizona. And what about his longer-term, "strategic" responsibility?
I think that it is a totally idiotic idea, too. The plane was officially shot down above the territory of Ukraine. In the case of every other country, the country's government would share a major part of the responsibility for the accident. After all, the government in Kiev was paid for giving the permission to MH17 (and others) to fly over the Donetsk region. Some perpetrators of 9/11 originated from Afghanistan (and were trained in non-governmental camps over there) which was enough for the U.S. to start a war against the whole country – it was a decision that was as "morally understandable" by the world's public as Austria-Hungary's decision to wage a war against Serbia after the 1914 assassination of the prospective emperor.
So of course that among the world's *governments*, the current Kiev government is the most directly responsible one for the downing of MH17. No one seems to even "dare" to think about this self-evident fact in the "mainstream" Western media that have apparently decided that all current rulers of Ukraine are holier than Jesus Christ.
And what about the long-term, strategic, global responsibility? Well, all the suffering including the downing of MH17 may be blamed on those who started this insane "Maidan revolution" based on the idea that Ukraine may assume to live in the vacuum and morph into an anti-Russian country overnight. Nothing bad could come out of it, right? Surely everyone in Ukraine agrees with the complete change of the direction where Ukraine develops and with the overthrowing of the last legitimately elected government.
Well, have you lost your mind? It has always been 100% guaranteed that such developments, often directly and openly encouraged by various powerful agents in the West, including the governments, would lead to a form of a civil war and a huge tension between those who try to "intervene" in this way on one side and Russia on the other side. Among the foreign influences, it's the pro-Maidan Western interventionists who are most responsible for the 1,000 or so dead people since the early 2014, including the casualties of the MH17 tragedy, and for lots of other problems and economic destruction that took place in Ukraine.
Vladimir Putin and the government of Russia would play at most a passive role. As far as I can say, he hasn't done a damn thing. But he could have and he may do something important in the future.
If I were the Russian President, I would have probably order the occupation of Ukraine many months ago because I would view these developments to be dangerous for the existential interests of my country. He has been as restrained as you can get. But of course that people understand that he would safely conquer Ukraine if it were the only foe. So I think that people are doing some sort of "psychological projection". They know it's a very natural decision for a power that is being as indirectly harmed as Russia has been since the "Maidan revolution" in Ukraine to protect itself. They may deny that they would think about the defense of their country's interests but it's a fact that every responsible politician in such a power has to think about this option.
Of course that I am not promising you that Vladimir Putin will never invade Ukraine. But given the escalation of the atmosphere – which was done by someone completely different than Vladimir Putin – it would be understandable for him to retake Ukraine. But at this moment, these are pure speculations and fantasies. Vladimir Putin hasn't done a damn thing – except for his nearly pacifist and very pragmatic speeches – which is why all these anti-Putin "mainstream" media don't know what they're talking about.
There are surely hundreds of millions of people in Western countries who have been brainwashed by this totally baseless propaganda. I apologize but I can't respect these people as observers or as human beings, for that matter. One must be an incredibly gullible moron to buy the bulk of this anti-Putin propaganda. The individuals who are buying this crap are much closer to the four-legged sheep, intellectually speaking, than they are to ourselves.
I am scared by this intellectual and moral degradation of the West. I understand very well that this is extremely far from being the first moment when the public of otherwise civilized countries got brainwashed in this way. The anti-Jewish hysteria in Germany of the 1930s was no different from the current anti-Russian hysteria. It was much worse because we already know that the anti-Jewish hysteria had led to 6 million casualties of the Holocaust while the number of casualties of the current anti-Russian hysteria is comparable to 1,000 only. But the analogy between the current atmosphere pumped by the "mainstream" media addressed to the intellectual bottom – and bulk – of the Western nations on one side and an *earlier* moment in the history of Nazi Germany may be a very appropriate one.
So average people, please, try not to be idiots. Or at least, try not to be *complete* idiots. Vladimir Putin didn't shoot MH17 down. You can't divide the Eastern Slavic nations to the "nice ones" like Ukrainians and the "villains" like the Russians. They are pretty much the same people, with the same statistical distributions of pretty much all traits, living in slightly different political situations. Eastern Ukraine has been a multi-national area for many centuries and the people simply have to co-exist in some way. The Russian leaders and the Russian government are doing nothing extraordinary. They are still trying to protect some interests of theirs. Every competent leader is doing it in one way or another – except that some try to mask what they are doing. Everyone who believes that we live in a "brave new world" in which politicians no longer protect any nations' or classes' interests is a lunatic and should relocate himself to a psychiatric asylum.
The people who have decided to see the tensions in the area in a black-and-white way and who still work to influence the events are the true strategic culprits of the misery that escalated over there in 2014. In reality, all the people over there are similar to each other and a violent conflict is unlikely to be good for either side. To spread the propaganda that Vladimir Putin is a villain whom one is obliged to hate are as dangerous and simple-minded psychopaths as the warriors and Islamic bigots in ISIL, among others. The world just isn't this simple. The real post-Soviet world doesn't really resemble the caricature in the "mainstream" Western media at all.
#### snail feedback (44) :
Exactly, Holger. After the spying incidents, it's at least debatable whether Germany should be more afraid of Russia or America, and be closer friends with Russia or America.
Elsewhere, France is being criticized by Britain for selling a ship to Russia. Needless to say, Britain also sells weapons over there, and has tons of wealthy Russians in the Financial City of London.
I would also like to know how the apparently "uniform" class managed to conquer pretty much all mainstream media even though it can't correspond to the distribution of what the readers expect and want to read and believe.
In the Czech context, we have a certain uniform class of intellectuals that are somewhat sloppily referred to as Havlists and they're in some sense in charge of all the mainstream media and various official cultural (and largely scientific) institutions. They're arguably a minority in the society when one demands several defining features at the same moment. I used to think that this takeover by another uniform group was an inevitably heritage of communism which forced everyone to be uniform as well (the diversity then could have actually been higher, something I am not happy at all to admit), but even in countries without communist heritage, it's similar if not worse.
Journalism must be made by some social class of activists and everyone else must be repelled by that. I understand how this group think works in a particular single newspapers. What I don't understand is how this uniformity could have overtaken *all* mainstream newspapers and channels.
"The world is not this simple". I am sorry Lumo but from reading your posts, the world seems to be very simple for you: Russia/Putin good, West bad. If anything, in a country with such a vertical power structure as Russia, Putin could have stopped the flow of sopthisticated, high power weaponry and guerillas across the border. The whole thing could have been resolved long time ago.
Yeltsin was fighting the Chechens just as enthusiastically as Putin, only Putin finished the job with somewhat more butcher skills, leveling Grozny almost to the ground, so no, Russia would be still in one piece. As for NATO, well maybe, but so what? What would be so wrong about that? NATO is surely not planning to attack the nuclear superpower that is Russia.
Why do you think that NATO would want Russia completely surrounded by bases if they have absolutely no intention of ever claiming that territory?
Is it for defense? But Russia "is surely not planning to attack the nuclear superpower that is NATO".
Let's be honest, if world history had just linearly extrapolated from the 1990s all of the former eastern bloc countries would be in NATO, and every single regional nationalist movement in Russia would be propped and supported by the western media with a lot of faux outrage over the atrocity that would be their treatment.
After BNP Paribas, if I were a Frenchman, I'd be asking myself if our guns were pointed in the the right direction.
A question that intrigues me is, "why do the neoliberalcons hate Putin so deeply"? I think the answer is that he ended the Yeltsin era - the era of a weak, corrupt government enabling oligarchs to eat high-priced dinners while homeless children rummaged through the slums of Moscow. He is given credit for sending some of the plutocrats then operating in Russia, packing. But, this is what neoliberalcons want - more suffering for the poor, incredible wealth for themselves, and everyone else believing that this is the natural order of things. This is what they want in the EU, Gaza, and the US (if they can pull it off). What is lacking here is a cadre of Western journalists who are gutsy enough to shine light on these ghouls and their evil agenda.
Hi Lubos and thank you for your hard work, especially CAGW and those enthralled by it. On the Ukraine crisis, and even anything Russia-related, I have found only one news source that is not ill-informed or tendentious. Even the WSJ has been unrelentingly hawkish, almost as if they were nostalgic for the cold war days. His name is Stephen Cohen and he is a regular guest on the John Batchelor show and he also writes for the Nation and other leftist outlets but he is still my go-to guy on Russia.
http://tinyurl.com/lz5cj9m
http://www.thenation.com/authors/stephen-f-cohen
http://dissidentvoice.org/2014/06/professor-stephen-cohen-on-ukraine-civil-war/
http://www.brookings.edu/blogs/up-front/posts/2014/07/08-kyiv-atrocities-nuanced-look-ukraine-crisis-pifer
The guy has been taking a beating from the left and the right with WSJ calling him one of "the two Steves" (I do not know Steve #2) and the Daily Beast naming him as "Putin's best friend in American media". The New Republic (!) smeared him with this headline: Putin’s American Toady at 'The Nation' Gets Even Toadier. That's it MR, I am canceling my subscription!
I am not a Frenchman but I still gave 10 percent of my savings to a fund managed by BNP Paribas some month+ ago, so I feel this is sort of personally relevant for me, too.
There's a school of thought that says the US and Saudis were up to their eyeballs in Chechnya, backing ideologically extreme Islamic fundamentalists in yet another effort to 'close the circle' on Russia itself. Given how truth is massacred in broad daylight on a normalized basis now in the US and West, I'm not completely confident we've got the whole story right, even sort of right, on Chechnya.
I think we've seen the USG intentionally disrupting smaller players in the global economy to maintain US status and control. I think the same is true with Russia and Ukraine. If it were not so, then why would there be such aggressive expansion of NATO, and such provocation of Ukraine?
I think we're seeing the US goad Russia into a conflict Russia can ill afford. The US won't back off, because US economic power is at stake. Without the US dollar as the world's reserve currency, (in which international trade takes place), the US economy plunges (far worse than 2008) because of the level of federal debt, which is only made possible because of the huge demand for US dollars. I think they're showing that keeping the US oligarchy strong is more important to the USG, than avoiding nuclear confrontation. Otherwise, NATO expansion would not have happened and the Ukraine crisis would never have begun.
The bad news is Russia has every reason now to create a parallel global financial structure to compete against SWIFT (see link below). I'm sure this will need China's cooperation but China is well on the way to undermining the US Dollar anyway, and certainly there are many other countries around the world that would be happy to have an alternative to a US dominated global finance system. (Iran for one, etc).
http://en.wikipedia.org/wiki/Society_for_Worldwide_Interbank_Financial_Telecommunication
If Putin acts in honesty on behalf of his people, why control the media and utilize propaganda? Does this not happen in Russia?
The underlying problem is a media dominated by presstitutes; too lazy to think for themselves; to ask informed questions at press conferences and elsewhere; too prejudiced to think about the subject in a way other than their own perspective.
And don't the Ukrainian/RUssian/USA/EU media wranglers know it!? They rely on it to be able to flood the traditional media to the almost the total exclusion of critical comment. Those presstitutes become proxy propagandists; apathetic of their complicity.
The public can be too easily led with emotional language to put the blame on targets; creating prejudice even in those who had not throught about such matters previously. Facts become irrelevant. Exploiting the moment is the objective of media. Emotional involvement if at all possible; institutionalised so that those who aren't sucked in already, could begin to feel guilty about not "sharing the moment".
And so we have ceremonies of organized grief dominating especially the electronic media; from around the globe; 24 hours a day. A different angle; every day; all angles with a clear view on who to blame.
But no hard facts. No analysis of the "facts" produced by government media agencies.
Over the past week, we've had pressituttes parrot demands of Putin made by photophilic politicians and "personalities" for the past week; when his media statements from early on the 18th already urged an immediate ceasefire and total cooperation with an impartial, international investigation. They do not see what contradicts their prejudice.
As recently as yesterday, there were calls to the separatists made by Australian journalists to do more to ensure the safety of investigators and journalist; while the Ukrainian government forces increased their artillery fire on the area. This is simply INSANE!
Lubos,
I have learned much about the Ukrainian situation from your early posts on the subject, and thank you for it.
However, you recent direction is very problematic. You apparently are not willing to enforce or even credit the long-established Westphalian system of legal nation states, instead, in this case, favoring ethnicity over law - which puts you in the fine company of many third world nasties.
Claiming that Russophile identification of many Ukrainians justifies Russia's support of a brutal rebellion is very dangerous. Justifying Russia's annexation of Crimea is beyond the pale - the international system doesn't work when large countries can bully small countries, especially in defiance of a nuclear arms treaty (whereby Ukraine relinquished a large quantity of strategic and tactical nuclear weapons in exchange for a "guarantee" of sovereignty).
The idea that the current Ukrainian government is illegitimate isn't sufficient. It is at least as legitimate as the murderous kleptocracy that preceded it.
Now, you may claim that I'm a victim of western propaganda, but I don't think so. The mainstream western media has not been trusted by me for a long time. As a student of propaganda since the '60s when I used to listen to Radio Moscow and Radio Havana, I know BS when I hear it. As I write this, I hear CNN acting as a shill for Hamas, and I don't buy it. As I write this, I hear CNN acting as a shill for Hamas, and I don't buy it.
But... facts are facts and can be separated from BS.
Putin is a very bad guy. That's a fact. He arose from the KGB, has robbed his country blind, and is an autocrat. Contrary to your assertion, Russia is far from a Democracy. Yes, it has elections, but the press is muzzled and potentially opposition figures jailed or even killed. This is not just some partial democracy as found all over the world - this is a country with more nuclear weapons than any other, ruled by a mafia chieftan.
Yes, he's a very popular chieftan. So was Hitler (sorry Godwin), who I invoke only to show that popular chiefs can be very evil. But Putin seems to be popular primarily due to the dysfunction of the Russian populace - people who have never known Democracy and who deeply resent their loss of international power (even though that power was produced through sheer brutality).
So, when you absolve Russia for its actions in increasing the brutality in the Ukraine, you are ignoring the nature of the regime you are absolving. It is hard to attribute legitimate motives to such a regime, especially when the results are so tragic.
By design and by self-infliction, the average American is stuck in a state of stupor and lethargy that cannot be easily reversed. I am willing to bet that most people in the US would sit idly by and do nothing if half the people on their streets were rounded up and disappeared into the night. I would put money on it. A combination of an empty-calorie diet, mind-numbing and vacuous TV programming, declining school system, way too many mood-altering drugs, and dishonest 'news' channels have successfully made a nation of zombies. Borrowing a line from a Pink Floyd song, there's a look in their eyes like black holes in the sky.
Dear Lubos,
when I think about the West in positive terms and it is for me still to a large extend a positive term I ultimately think about the French revolution. With my limited Western world view I would still view it as the most important event in history. But if we look back to the French revolution there is one important fact which for me has some significance for the Ukraine crisis. If you think the values of the French revolution such as that all humans have equal born rights through to their logical conclusion it is clear that the French nation is also just an equal nations among others. But looking at the following time it seems the French nation didn't fully understand it or forgot it during the Napoleonic times. I don't want to say that things are simple here because without skillful generals such as Napoleon the ancient regime in the other states would have simply undone the French revolution by military force. Still I think it is fair to say that the French revolution with its universal values at its core didn't quite disentagle itself from some French nationalism. And it was perhaps the tragedy of Germany that the Germans also found it very hard to fully endorse the values of the French revolution when they came from its biggest rival. (They partly did as the revolution of 1848 showed.) So can we disentangle the situation in the Ukrain from any pan Slawic nationalism. Would you endorse the French revolution in the same way I did it here?
You have to be pretty paranoid to think that the western powers retain any political capability for colonialism or imperialism, regardless of the US's military capabilities.
Russia would have been perfectly safe with NATO on its borders. And if Russia was peaceful, and didn't join NATO, NATO would have faded away due to lack of mission. Not only that, but ironically, a friendly Russia would likely have been invited into NATO.
The cold war was started by Russia, and if it arises again, it will be due to behavior of Russia. The west has no interest in a cold war, unless its for defense.
The somewhat one sided behavior of Europe and the US towards the complex situation in the Ukriane is more easily explained by a utopian desire to add more countries to the (rather absurd) EU, not some evil intentions towards Russia, combined with historically and recently justified fear of Russian misbehavior. It is also a result of some delusions that people who act "progressive" and mouth Europhile pieties are the ones to support.
As fpr spying on Merkel's phone - that's what spies do. Merkel is only upset because Germany has not been allowed to join the 5-eyes intelligence cooperative. The outrage is just nonsense. Furthermore, does anyone imagine that the Germans' wouldn't be tapping Obama's calls if they had the capabilit?
Lubos, the idea that Germany might have more to fear from the US than Russia is, simply, absurd. The nature of governments and culture are extremely important in deducing the intent and dangers of other countries. Looked at that way, the US is hardly a threat.
Mesocyclone, experience surely shows something else. In the last 20 years, the U.S. government was much more dangerous for other countries of the globe than Russia that just left other nations to live.
Moreover, I don't really believe with your idea that "culture" makes governments non-dangerous. The Third Reich wasn't really "uncultural" in the conventional definition.
You may think you are cultural and the U.S. is never hurting others except that the reality is something completely different. Germany doesn't understand - and I don't understand - why it's being spied on. Obviously, something must be wrong from the U.S. viewpoint, a viewpoint we don't understand much like we may misunderstand some of the Russian attitudes.
It may be that the U.S. government is pissed off just by the very fact that Germany hasn't declared a trade war against Russia. If that's the reason, then the current U.S. government is indeed very dangerous because I think that pretty much everyone who matters in Germany knows that a full-fledged trade war would be a catastrophe for Germany, too.
I'm actually a little surprised you're not worried that Putin wants to bring back the Iron Curtain and Soviet Bloc. If I recall correctly, Czechoslovakia didn't fare so well with the Russians before.
I am not worried because I know that such things are baseless. We clearly belong to the Western side of the Iron Curtain - and with the exception of 40 years, we have belonged there for 1,000 years - so we should be as worried about becoming a part of Russia as Americans should worry.
Also, we know something about the reality because we're meeting Russians and doing business with Russians every day so we know that those who are imagining Russians to be monsters waiting to conquer everything are just idiots.
ok
Another thing - despite the noise, with Obama as President we're certainly not going to actually do anything, so I don't think you need to worry about the western media.
What do you count to "anything"? Can you promise me that the U.S. won't blackmail us and force us to stop all trade with Russia? It's 5% of our imports and 5% of our exports and the percentage is much higher in some other countries.
Some people are struggling for every 0.1% of the GDP, obsessed with tiny and often spurious changes of the GDP, and suddenly someone would like to erase 5% of the GDP just because of some stupid anti-Russian anger? I view this as a huge threat, too. If the threat to antagonize Russia in this big way for no defensible reason became really high, I would prefer to leave NATO.
A suggested mapping from the present to the late nineteen thirties.
Crimea --> Austria
Eastern Ukraine --> Sudetenland
Putin --> ?
What's the purpose of this mapping aside from making one totally deluded about everything?
Moreover, I agree with the Austria part, to say the last. It was as unnatural for Austria to be in a different country than the rest of Germany as it was for Crimea a few months ago. Austrians obviously wanted to join that project, and Hitler himself was Austrian.
I won't say the same thing about the Sudetenland because it had been a part of the Czech kingdom territory for 900 years. One can't really say the same thing about the non-Russian Ukraine's control over Eastern Ukraine.
And again, if you belong among those who are happy about writing Putin=Hitler, I think that you are dumb as a doorknob. I can't have any respect towards these people's thinking about the world of politics or the society.
Generally speaking we should not assign anthropomorphic characteristics to states (or corporations for that matter). States have no ethics, they have interests and needs. They pretend to have ethics to serve their interests; their interests are those of the ruling elite or of the class of ruling elites. Their policies are insect policies. Insects have no ethics, they have needs.
As I see it there was a western supported "revolution" in Ukraine that ignited much that happened later.
The Western world hope to rid Putin before Paris 2015?
Oh Lumo, the broad choice of names you managed to call me I am afraid tell more about you than about me. Your narative would be plausible, if you did not use so many assumptions: you "know" what Putin wants and does and does not, you "know" exactly what is happening and what is not. You are either better informed than rest of the world, or your are making your assumptions up as you go, to fit your ideological conviction and your heavy dose of RT, but they often go against the mounting evidence from different parties on the ground. I will leave it there, I don´t see much use of further discussion in the personal tone you use. Just one thing, I have been fully grown up already during the communism era and the bias of RT and the likes, centrally controlled Russian media stinks of the same bias and propaganda like what I experienced back in the commie years. It is only done in more smart and modern way.
Yes, Lubos, I am Dutch and must tell you that I am ashamed of what is happening in the Netherlands. War mongers, very very worrying. Not only the mayor, but also lots of people on the internet and the MSM press are yelling all sorts of things. Pretty disgusting, brainwashed by the media.
Thanks for this understanding and please know that NL is the place where such reactions are at least more justifiable than elsewhere these days.
Putin has lost this media war partly because of reality and partly because he is a villain looking guy by joyce.
In the local right-wing newspaper the headlines are comedy gold.
The Friday edition features ITS GONE PEAR SHAPED, AND NOW PUTIN HAS HIS BACK TO THE WALL (in which Putin is claimed to fear not the West but his own people) followed (in the same edition) by PUTIN'S VORACIOUS APPETITE IS NOT SATED (in which it is claimed that the Russian people
will circle the wagons around their embattled hero and cry foul at foreign attempts to denounce him).
Other headlines include STRONGMAN AIMS TO SHOOT DOWN QUEST FOR JUSTICE and PUTIN'S LIMITLESS
AMBITION FEEDS HIS INTEREST IN DRAGGING MATTERS OUT. These appear to discuss the shooting down of two more planes by the separatists.
The PM Abbott, in the true style of the bog-standard venal pollie, has speedily jumped on the bandwagon of international self-aggrandisement in order to distract attention away from his troubles at home. Like a true Aussie hero he appears to see himself as leading the pack for truth and justice against no less an enemy than Satan himself.
Its all gone off rather well, the Saturday edition features an in-depth article from the paper's in-house nutter G. Sheridan entitled ABBOTT ACCRUES DIPLOMATIC CAPITAL FOR CRISIS LEADERSHIP, followed by PURER VIEW OF
CHARACTER ON DISPLAY AS POLITICS LAID ASIDE (in which the various aspects of the PM's greatness of personality is discussed).
And in order to make sure we've got the message heading the letters page is ABBOTT'S APPROACH TO PUTIN SHOWS WEST THE WAY.
The whole effect of this relentless and absurd beat-up is like watching a TV cartoon. It's clear that the printed media is now nothing more than an outlet for the favoured political views of whoever controls them. As sources of rational information they are completely worthless. I gave up
reading the left-wing newspaper decades ago. It looks like its time to give up reading the right-wing newspaper as well.
I have no knowledge of Putin as a person; however, calling Putin a Hitler or suggesting that he's a thug is way out of line.
What is it that Putin has done to merit these terms? He took back Crimea when an overwhelming majority voted to rejoin Russia in a very democratic referendum. This was done after an attempt by the US to deny Russia warm water access by kicking them out of Crimea.
By US invasion standards, that hardly merits the 'Hitler' label, and since nobody was killed, the 'thug' label doesn't cut the mustard, either.
Georgia/South Ossetia in 2008 (during surprise! yet another Olympics) was an invasion sponsored by the US killing over a hundred civilians. Russia withdrew once Georgia was kicked out of South Ossetia. That's hardly Hitler-like.
Putin was at one time in the KGB, and if that is sufficient for the 'thug' label, then everyone in the CIA or FBI is a thug too.
I’ll say that with all the negative stuff we hear about Putin from the Western media, for the sake of balance, it would be nice to hear some positive stuff about him as well. If a lot of the Western media is to be believed, he’s a blood-sucking dictator who is widely reviled in Russia. Yeah, right — I’m sure there are many negative aspects to him, as there are with anyone who becomes powerful, but a lot of this “Putin is evil” stuff is agitprop. Of course, the fact that I can still post anti-Obama and anti-establishment diatribes proves that we are still somewhat free here in the West. But it’s not the simplistic dichotomy the establishment here portrays it to be.
As I’ve said before, one Western strategy that has grown in prominence has been to overthrow the Russian and Chinese regimes — and any regime that resists American interests — via color revolutions. It disgusts me how the West pretends to be in favor of democracy and human rights and yet defends the Saudis and Bahrainis, who have been every bit as brutal to their people in recent years as Assad, who America have abandoned because of his links with Iran. And now America is backing al-Qaeda to replace him? Vile.
Anything that gives dissent a voice is wonderful, in my honest opinion.
Where does this sweeping, racist, Russophobia comes from? How does Russia, whose very national and ethnic character is defined by centuries of savage aggression from the West (and the East too - Genghis Khan, the Golden Horde, The Khanate of Crimea) gets rhetorically transmogrified into, by these same western aggressors, into the aggressor themselves?
Someone please correct me, but going all the way back to medieval times, when has Russia EVER committed military aggression against a nation outside of, maybe, those on its immediate borders in response to real security threats? Do Russians soldiers sing about savagely invading practically every nation around the world the way the US Marines do?
The only place I can think of is Afghanistan in the 1980s - and there they were defending a non-religious ruling, female friendly government against the Bin-Laden-led, US-sponsored fanatical religious savage terrorists.
Say what they will about Putin, and there is no lack of vitriol for him in the West as others here point out, but the reality seems to be that he took power in a country that was down on its luck, ruled by a drunken embarrassment, and did much to turn things around and restore a sense of pride for Russians. Russia was looted by the West in the 1990′s, and while there is still looting at least it is done by Russians.
Is Putin in power through “free and fair” means? Probably not, but is it “free and fair” that two near identical parties in the US control who gets on the ballot? Putin’s reign has been more good than bad, and we get the added benefit that he funds RT for us Western malcontents.
Exactly. It has been a tragedy. I do even think it could have been the separatists, but if so, it was an accident. I do not believe they did that on purpose, since they gain nothing doing this. But without any clear evidence yelling for war with Russia, blaming Putin's daughter (wtf? (sorry for my language)), is beyond belief. Really. I do think that the media in the Netherlands played a very doubtful role, by blaming the Russians from the beginning and not mentioning anything about the American role in the Maidan coup. The large masses are 90% unaware of these dirty games. Good on you to give some counter balance. It's very worrying though. I just want peace for all of us.
Lubos, the US government has been dangerous for a few bad governments. In general, most countries are far better off for it.
Russia, on the other hand, tends to support bad governments (Assad in Syria, Saddam in his time, Iran). Russia conquered parts of Georgia and more recently, simply stole the Crimea. In other words, Russia ignores international rules and, for that matter, civilized rules, and simply does whatever it has the power to do.
Needless to say, that is a bit of an oversimplification, but the idea holds.
There is little mystery about either Russian or US motivations. At this point, Russia is driven primarily by two forces: economic (skewed towards helping the oligarchs and siloviki) and imperialist (fueled by the same thing that fueled the rise of the Nazis: the economic consequences and impact on national pride of losing a huge war). Russia is also paranoid, which is part of its historical culture - it projects its evil motives onto others, such as the US, imagining that we have all sorts of nefarious intents when our motivations are a lot simpler and less evil. Russia behaves like what it's leaders have become: a mafia.
The US is driven by a confused combination of economic interests and national defense concerns. Unlike Russia, it is unwilling to use military power for economic reasons alone, except for defense of economic interests.
You misunderstand the use of the word culture. It has nothing to do with "cultural" - it means the nature of the society. Some societies have barbarian cultures (see Somalia). Some have cultures that emphasis cheating and power (see China). Many cultures lead to high degrees of corruption (see most of the world other than the west). The culture, i.e. the attitudes and habits of a society, is one of the most important determinants of how it will behave towards other societies, how it will behave towards its own members, and how successful it will be. This is why western efforts to reform countries through the deus ex machina of democracy so often fail.
As to spying on Germany... Surely you are not so naive as to believe that countries do not spy on their allies! Germany (at least those who are paying attention) know full well why we spy on them - for the same reason they spy on other countries. Pretty much every country spies on pretty much every other country. This is normal. There is a bit of an exception within the 5-eyes - members tend not to use SIGINT to spy on each other as they share SIGINT. But... when I was in the military, we routinely used classified material that could not even be shared with other 5-eye members (it had NOFORN added to the classification).
I did not call him a Hitler - I mentioned Hitler only to show that evil leaders can be popular. Putin is very popular in Russia right now, but then Putin controls the media in Russia.
In Crimea and Georgia, Putin took advantage of ethnic dissatisfaction to carve off pieces of the countries and annex them, into Russia. These were acts of war and sufficient that Russia should suffer severe sanctions from the international community. There are ways to deal with these disputes that do not involve violence or territory grabs by force.
Putin infiltrated special forces into Crimea who then led a military coup to take it over. That is not Democratic.
He invaded and kept part of Georgia a few years back because he wanted the territory, and that territory remains controlled by Russia and no longer part of Georgia (yes, there's the fig leaf that Russian forces are no longer in those provinces, but the fact that Georgia no longer controls them tells the story). The US did not sponsor any invasion, and Georgia could hardly invade it's own provinces!
Putin has sold weapons and provided other support (including blocking UN resolutions) to help very bad regimes such as Assad's. Putin was responsible for the murder by Polonium of a former KGB colleague in London. Putin has a long history of using force to subdue domestic and foreign opponents.
Many of the richest and most powerful people in Russia were Putin's colleagues in the KGB. His past membership in the KGB only means he used to be a thug. His actions more recently shows that he is still a thug. A thug, a chieftan of a vicious mafia, is not the person the world needs in control of a large nuclear-armed state.
It looks like the West is playing their last card. They have already used all their jokers and they can only rely on their best poker faces. Cold blood killings of innocents is part of the game.
You would enslave a people if you thought it better for them.
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# GATE2015-3-4
In the figure, link $2$ rotates with constant angular velocity $\omega_2$. A slider link $3$ moves outwards with a constant relative velocity $V_{Q/P}$, where $Q$ is a point on slider $3$ and $P$ is a point on link $2$. The magnitude and direction of Coriolis component of acceleration is given by
1. $2 \omega_2$ $V_{Q/P}$; direction of $V_{Q/P}$ rotated by $90^\circ$ in the direction of $\omega_2$
2. $\omega_2$ $V_{Q/P}$; direction of $V_{Q/P}$ rotated by $90^\circ$ in the direction of $\omega_2$
3. $2 \omega_2$ $V_{Q/P}$; direction of $V_{Q/P}$ rotated by $90^\circ$ opposite to the direction of $\omega_2$
4. $\omega_2$ $V_{Q/P}$; direction of $V_{Q/P}$ rotated by $90^\circ$ opposite to the direction of $\omega_2$
recategorized
## Related questions
A point mass is executing simple harmonic motion with an amplitude of $10$ $mm$ and frequency of $4$ $Hz$. The maximum acceleration ($m/s^2$) of the mass is _______
A shaft of length $90 \: mm$ has a tapered portion of length $55 \: mm$. The diameter of the taper is $80 \: mm$ at one end and $65 \: mm$ at the other. If the taper is made by tailstock set over method, the taper angle and the set over respectively are $15^\circ 32′$ and $12.16 \: mm$ $18^\circ 32′$ and $15.66 \: mm$ $11^\circ 22′$ and $10.26 \: mm$ $10^\circ 32′$ and $14.46 \: mm$
A cantilever bracket is bolted to a column using three M$12×1.75$ bolts P, Q and R. The value of maximum shear stress developed in the bolt P (in $MPa$) is ________
For the overhanging beam shown in figure, the magnitude of maximum bending moment (in $kN$-$m$) is ________
For ball bearings, the fatigue life $L$ measured in number of revolutions and the radial load $F$ are related by $FL^{1/3} = K$, where $K$ is a constant. It withstands a radial load of $2$ $kN$ for a life of $540$ million revolutions. The load (in $kN$) for a life of one million revolutions is ________
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# Lebesgue Outer Measure ... Carothers, Proposition 16.1 ...
Gold Member
## Summary:
I need help with an aspect of the proof of Carothers' Proposition 16.1 ...
I am reading N. L. Carothers' book: "Real Analysis". ... ...
I am focused on Chapter 16: Lebesgue Measure ... ...
I need help with an aspect of the proof of Proposition 16.1 ...
Proposition 16.1 and its proof read as follows:
In the above text from Carothers we read the following:
" ... ... But now, by expanding each ##J_k## slightly and shrinking each ##I_n## slightly, we may suppose that the ##J_k## are open and the ##I_n## are closed. ... "
Can someone please explain how Carothers is expecting the ##J_k## to be expanded and the ##I_n## to be shrunk ... and further, why the proof is still valid after the ##J_k## and ##I_n## have been altered in this way ... ...
Help will be appreciated ...
Peter
Related Topology and Analysis News on Phys.org
Math_QED
Homework Helper
2019 Award
Replace ##J_k## by its closure and ##I_n## by its interior. These operations add/remove only begin or end points of the intervals so the lengths of the intervals are unaffected and the inequality with the two series keep valid.
Math Amateur
Gold Member
Replace ##J_k## by its closure and ##I_n## by its interior. These operations add/remove only begin or end points of the intervals so the lengths of the intervals are unaffected and the inequality with the two series keep valid.
Thanks for the help ...
I get the idea but shouldn't we replace ##J_k## by its interior and ##I_n## by its closure ...
Peter
Gold Member
You write:
" ... Interior makes the set smaller, the closure of the set makes the set larger. ... "
Yes ... but the problem I have is that we are told we may suppose that the ##J_k## are open and the ##I_n## are closed!
If we replace ##J_k## by its closure the ##J_k## will be closed not open ...
Peter
fresh_42
Mentor
You write:
" ... Interior makes the set smaller, the closure of the set makes the set larger. ... "
Yes ... but the problem I have is that we are told we may suppose that the ##J_k## are open and the ##I_n## are closed!
If we replace ##J_k## by its closure the ##J_k## will be closed not open ...
Peter
You can quantify it if you like. As mentioned by @Math_QED we need only a point or two from not closed to closed, which doesn't change the length. And we need an arbitrary small, but positive distance to change from closed to open.
If there is a strict inequality, then it has a positive distance ##d>0## between the two. So the inequality still holds if we e.g. add ##d/2## to the smaller sum. Now we can write ##d/2=\sum_{n=1}^\infty d\left(\dfrac{1}{3}\right)^n##. This means we have ##(1/2)d/3^k## available to add on each side of every interval ##J_k## to make it open and still have a total length strictly smaller than ##d##.
Last edited:
Math Amateur and Math_QED
Gold Member
Thanks for your assistance fresh_42 ...
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## Algebra 2 (1st Edition)
The initial amount is $527$, the percent increase is $1.39-1=0.39$, and the growth factor is $1.39$.
If for an exponential function $y=ab^x$ $a\gt0,b\gt1$, then it is an exponential growth function, and $b$ is the growth factor. Following the example, the initial amount is $527$, the percent increase is $1.39-1=0.39$, and the growth factor is $1.39$.
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Equilibrium Price Assignment Help
Equilibrium Price.pdfhttp://www.experiment.net/price/experiments.html Author: Fisk Vdwies Author: Kevin Harrelick http://www.experiment.net/price/experiments.html
The final market prices for the UK market at RSI were.039 and.055, which are from the EBITDA report by the All Refiners group . The current data were obtained from a central bank, as part of its own monthly withdrawal policy, and contain the following: — Price-to-stock — Price per share — Sales price — Supply of available goods, excluding reserve securities — Excluding reserve securities — Exclusions of reserve — Exclusions of purchasing power (P2P) The new prices are as follows: — Price per share of a stock — Based on the last-day report, — Price per P3: — Based on purchasing power (P2P) — Based on market rate (MP) — Pricing based on market volume — Pricing based on supply of reserve — Pricing based on supply of product — Pricing based on demand ratio (DRR) — Pricing based on demand ratios (DRR2 | DRR2 = DRR2 + DRR1 | DRR2 = DRR2 + DRR1) We provide in-block prices and returns and price-to-stock that may be calculated under a three-sensory (3Sensory1 and 3Sensory3). For example, to extract their future market prices it is necessary to include an annual price at March’s final and price-to-stock that is based on a 2:1:1 ratio between the value of a stock and its present value. The return yields price and stock prices over a year on the terms of the new market volume are calculated as follows: The price-to-stock comparison link in the price-to-stock guide is available at the following link. Cost Extra resources capital Cost of financing Cost of debt Currency code Other information about the subject at the above link site including course prices, access to Coursera web site, the purchase or sale of property or an agreement to buy at the price the EBITDA is based on, such as the market prices from the London EBSO at.007023.
Homework Help Websites For College Students
99 which is approximately 26,000 kWh, provided the information on Coursera page is saved in your personal computer. For further details see the Introduction to EBITDA.pdf’s in the Ebook’s Reading Infrastructure section. — — — — — — — — — — — — — — — — — — — — — — The use of the phrase ‘price-to-stock’ as used throughout the text indicates an appropriate combination of the following assumptions and assumptions of price-to-stock. The first assumes that price-to-stocks is a nominal value, such that if a stock for example had a price high of RSI at a discounted, future RSI is listed in the next reader’s handout. We’ve seen (and been unable to find) the difference of, not based on, the EBITDA by Vdwies, but based on other factors ranging from the acceptance of a higher QA rate to whether a target price indicates a risk to the EBITDA. Here are some other common assumptions web which we argue in favour of the EBITDA. We assume that the current S&P/QS yields at RSI exceed the current market rate, QP. Here is a figure showing how the future market value of the current, established model is compared to the market S&P/QS which has a RSI yield of RSI if the market price of the existing model is to be based on the current market price from the EBITDA. The time taken to establish the preferred price ofEquilibrium Price The equilibrium price is the most efficient equation of state for a given chemical composition. The equilibrium price is given by the quotient of the chemical composition by the chemical potential divided by the equilibrium chemical potential (for gases). Equation where the first coefficient is the solute-solvent interaction, the second is the energy (including heat and other dissipation) and the third coefficient gives the gas enthalpies. In practical practical usage, it will be given as Equation 9. Refigory Refigory is a classical mathematical class. The concept of an equilibrium price is typically called the fundamental theory of economic theory. This has the advantage of providing (simple as in financial science) a simple mathematical proof of equilibrium. In its simplest form, we are considering the simplest case of a general equation, , in which the following conditions are assumed: (i) the system is initially incompressible. The initial read review is made is a product state of two states, (), and (), in which the former is described by the chemical potential (see ref., for the simple case of a two-molecule interaction); and (ii) the initial chemical composition by. Here, indicates a composition of $L_{2}\times I$ molecules.
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# How do write in simplest form given 7/10-2/15?
Oct 29, 2016
$\frac{17}{30}$
#### Explanation:
In order to add or subtract fractions, we have to have common denominators.
1. List multiples of both numbers.
$10 : 10 , 20 , \underline{30} , 40 \ldots .$
$15 : 15 , \underline{30} , 45 , 60 \ldots .$
2. Look for the smallest underlined number (known as the least common multiple, or LCM). This is your common denominator.
Common denominator: $30$
3. Multiply the numerator and denominator by the factor that it would take to get to your common denominator…like this:
$\frac{7}{10} \cdot \frac{3}{3} = \frac{21}{30}$
and, $\frac{2}{15} \cdot \frac{2}{2} = \frac{4}{30}$
Now we have both of our fractions with common denominators, so we can subtract! Remember, in a subtraction problem, the numerators subtract but the denominators stay the same. It looks like this:
$\frac{7}{10} - \frac{2}{15} = \frac{21}{30} - \frac{4}{30} = \frac{17}{30}$
Oct 30, 2016
$\frac{17}{30}$
#### Explanation:
$\textcolor{b l u e}{\text{Initial thoughts}}$
Known: $\text{ "2xx15=30" }$ and $\text{ } 3 \times 10 = 30$ so both the 'denominators' will divide exactly into 30.
A fraction consist of " "("numerator")/("denominator")" "->" "("count")/("size indicator")
You can not directly add or subtract the counts unless the size indicators are the same.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Making the denominators (size indicators) the same}}$
$\textcolor{b r o w n}{\text{Multiply by 1 and you do not change the value. However 1 comes in many forms}}$
$\text{ "[7/10color(red)(xx1)]" " -" } \left[\frac{2}{15} \textcolor{red}{\times 1}\right]$
$\text{ "[7/10color(red)(xx3/3)]" " -" } \left[\frac{2}{15} \textcolor{red}{\times \frac{2}{2}}\right]$
$\text{ "21/30" "-" } \frac{4}{30}$
$\text{ } \frac{17}{30}$
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# When a ball is thrown up vertically with velocity 𝑉0 , it reaches a maximum height of ′ℎ′. If one wishes to triple the maximum height then the ball should be thrown with velocity
Question : When a ball is thrown up vertically with velocity 𝑉0 , it reaches a maximum height of ′ℎ′. If one wishes to triple the maximum height then the ball should be thrown with velocity
(A) $\sqrt{3}V_{0}$
(B) $3V_{0}$
(C) $9V_{0}$
(D) $\dfrac{3}{2}V_{0}$
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IMR Press / RCM / Volume 22 / Issue 3 / DOI: 10.31083/j.rcm2203109
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Open Access Original Research
Trends and distribution of coronary heart disease mortality rate in Hexi Corridor, Gansu, China from 2006 to 2015
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1 Department of Cardiology, Gansu Provincial People’s Hospital, 730000 Lanzhou, Gansu, China
2 Department of Cardiology, Wuwei Hospital of Traditional Chinese Medicine, 733000 Wuwei, Gansu, China
3 Department of Chronic Non-infectious Disease Prevention and Control, Gansu Center for Disease Control and Prevention, 730000 Lanzhou, Gansu, China
4 Department of Internal Medicine, Chengguan District People’s Hospital, 730030 Lanzhou, Gansu, China
This article belongs to the Special Issue: State-of-the-Art Cardiovascular Medicine in Asia 2021 (https://rcm.imrpress.com/EN/subject/listSubjectChapters.do?subjectId=1621394403410).
Rev. Cardiovasc. Med. 2021 , 22(3), 1003–1008; https://doi.org/10.31083/j.rcm2203109
Submitted: 10 April 2021 | Revised: 27 July 2021 | Accepted: 4 August 2021 | Published: 24 September 2021
This is an open access article under the CC BY 4.0 license (https://creativecommons.org/licenses/by/4.0/).
Abstract
This study described the trend and distribution of coronary heart disease (CHD) in the Hexi Corridor region of Gansu. The CHD mortality rates from 2006–2015 were obtained through the Death Reporting System of Gansu Centers for Disease Control (CDC) for 2006–2015. The overall mortality rate of CHD in the Hexi Corridor showed a decreasing trend, increasing in winter and spring and lowest in summer. The CHD mortality rate was higher in men than in women (P $<$ 0.05) and increased with age (P $<$ 0.05). The mortality rate was higher in rural areas than in urban areas (P $<$ 0.05). A ten-year mortality rate trend analysis showed that CHD mortality rate in women has significantly decreased. Specifically, women aged 18–39 years experienced increased There was little change in CHD mortality among women aged 40–59 years, and a declined in CHD mortality among women 60 years and older and women in urban areas. Further analysis showed that in the 18–39-year-old and 40–59-year-old groups and in urban areas, CHD mortality rate was higher in men than in women (P $<$ 0.05). From 2006 to 2015, the mortality rate of CHD in the Hexi Corridor of Gansu was lower than in the national average, but in certain populations such as men, young and middle-aged group and rural areas, the CHD mortality rate was gradually increased. There has been a gradual and progressive decline in CHD mortality rate compared to the rising trend in China. This is due to fewer risk factors in the region, effective drug treatment and improvements in environmental pollution. However, there is still a need to enhance the experience of effective prevention and control for specific subgroups such as men, young people and rural residents, and to take appropriate measures to prevent the occurrence of CHD.
Keywords
Mortality
Coronary heart disease
Hexi Corridor
Trends
Distribution
Figures
Fig. 1.
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Difference between weak ( or martingale ) and strong solutions to SDEs
Hi Im fairly new to SDE theory and am struggling with the difference between a weak ( or martingale ) solution and a strong solution to an SDE :
$$d(X_{t})=b(t,X_{t})dt + \sigma(t,X_{t})dW_{t}$$
Are these two differences and what do they really mean in detail?
1. For a strong solution we are given an initial value, whereas for weak solutions only a probability law?
2. For strong solutions we know what probability space we are working in and have a Brownian Motion $$W$$ in that space. For a weak solution we can only say that there exists some probability space where the SDE holds (with a new brownian motion in the space).
As you can tell I am confused with this topic some clarifications would be amazing.
The main difference between weak and strong solutions is indeed that for strong solutions we are given a Brownian motion on a given probability space whereas for weak solutions we are free to choose the Brownian motion and the probability space.
Definition: Let $$(B_t)_{t \geq 0}$$ be a Brownian motion with admissible filtration $$(\mathcal{F}_t)_{t \geq 0}$$. A progressively measurable process $$(X_t,\mathcal{F}_t)$$ is a strong solution with initial condition $$\xi$$ if $$X_t-X_0 = \int_0^t \sigma(s,X_s) \, dB_s + \int_0^t b(s,X_s) \, ds, \qquad X_0 =\xi \tag{1}$$ holds almost surely for all $$t \geq 0$$.
Definition: A stochastic process $$(X_t,\mathcal{F}_t)$$ on some probability space $$(\Omega,\mathcal{F},\mathbb{P})$$ is called a weak solution with initial distribution $$\mu$$ if there exists a Brownian motion $$(B_t)_{t \geq 0}$$ on $$(\Omega,\mathcal{F},\mathbb{P})$$ such that $$(\mathcal{F}_t)_{t \geq 0}$$ is an admissible filtration, $$\mathbb{P}(X_0 \in \cdot) = \mu(\cdot)$$ and $$X_t-X_0 = \int_0^t \sigma(s,X_s) \, dB_s + \int_0^t b(s,X_s) \, ds$$ holds almost surely for all $$t \geq 0$$.
As a consequence of these definitions, we have to consider different notions of uniqueness. For strong solutions we are typically looking for pathwise unique solutions, i.e. if $$(X_t^{(1)})_{t \geq 0}$$ and $$(X_t^{(2)})_{t \geq 0}$$ are strong solutions to $$(1)$$ with the same initial condition, then pathwise uniqueness means $$\mathbb{P} \left( \sup_{t \geq 0} |X_t^{(1)}-X_t^{(2)}|=0 \right)=1.$$ As the following simple example shows it doesn't make sense to talk about pathwise uniqueness of weak solutions.
Example 1: Let $$(W_t^{(1)})_{t \geq 0}$$ and $$(W_t^{(2)})_{t \geq 0}$$ be two Brownian motions (possibly defined on different probability spaces), then both $$X_t^{(1)} := W_t^{(1)}$$ and $$X_t^{(2)} := W_t^{(2)}$$ are weak solutions to the SDE $$dX_t = dB_t, \qquad X_0 = 0$$ Why? According to the definition we are free choose the driving Brownian motion, so we can set $$B_t^{(1)} := W_t^{(1)}$$ and $$B_t^{(2)} := W_t^{(2)}$$, respectively, and then $$dX_t^{(i)} = dB_t^{(i)} \quad \text{for i=1,2}.$$
What do we learn from this? Since weak solutions might be defined on different probability spaces, there is no (immediate) way to compute probabilities of the form $$\mathbb{P}(X_t^{(1)}=X_t^{(2)})$$ for two weak solutions $$(X_t^{(1)})_{t \geq 0}$$ and $$(X_t^{(2)})_{t \geq 0}$$, and therefore we cannot even attempt to talk about pathwise uniqueness. For the same reason, it doesn't make sense to talk about pointwise initial conditions $$\xi$$ for weak solutions (... for this we would need to fix some probability space on which $$\xi$$ lives...); instead we only prescribe the initial distribution of $$X_0$$.
The next example shows that we cannot expect to have pathwise uniqueness even if the weak solutions are defined on the same probability space.
Example 2: Let $$(W_t)_{t \geq 0}$$ be a Brownian motion. It follows from Example 1 that $$X_t^{(1)} := W_t$$ and $$X_t^{(2)} := -W_t$$ are weak solutions to the SDE $$dX_t = dB_t, \qquad X_0 =0.$$ Clearly, $$\mathbb{P}(X_t^{(1)} = X_t^{(2)}) = \mathbb{P}(W_t=0)=0$$.
The "good" notion of uniqueness for weak solutions is weak uniqueness, i.e. uniqueness in distribution (= the solutions have the same finite-dimensional distributions).
Typically it is much easier to prove the existence (and/or uniqueness of) a weak solution the the existence (and/or uniqueness) of a strong solution.
Example 3: The SDE $$dX_t = - \text{sgn}\,(X_t) \, dB_t, \qquad X_0 = 0 \tag{2}$$ has a weak solution but no strong solution.
Let's prove that the SDE has a weak solution. Let $$(X_t,\mathcal{F}_t)_{t \geq 0}$$ be some Brownian motion and define $$W_t := -\int_0^t \text{sgn} \, (X_s) \, dX_s.$$ It follows from Lévy's characterization that $$(W_t,\mathcal{F}_t)$$ is also a Brownian motion. Since $$dW_t = - \text{sgn} \, (X_t) \, dX_t$$ implies $$dX_t = - \text{sgn} \, (X_t) \, dW_t$$ this means that $$(X_t)_{t \geq 0}$$ is a weak solution to $$(2)$$. For a proof that a strong solution does not exist see e.g. Example 19.16 in the book by Schilling & Partzsch on Brownian motion.
Let me finally mention that weak solutions are closely related to martingale problems; in this answer I tried to give some insights on the connection between the two notions.
• Why is the text after your first two definitions. not formatting properly... Thanks for your answer by the way. Nov 13, 2018 at 8:47
• @Monty Thank you; I fixed it.
– saz
Nov 13, 2018 at 8:54
• I've seen that some authors use the notions of weak and strong solutions in a slightly different way. Sometimes $X$ is only called a strong solution if it is adapted with respect to the filtration generated by $B$. And usually the probability space and filtration are part of a weak solution (If I understand your definiton correctly, you assume that at least the probability space is given beforehand). Feb 21, 2019 at 21:38
• Oh, and are you implicitly assuming the filtrations to be complete? Otherwise, there might be a problem in defininig the stochastic integral as a local martingale. (Just asking cause I'm currently trying to figure out how the notions of weak/storng solutions are commonly used in the literature.) Feb 21, 2019 at 21:45
• @0xbadf00d If you have (deterministic) Lipschitz coefficients and a deterministic initial condition, then the solution is measurable wrt to the canonical filtration... that's what they show there. It doesn't contradict what I was saying..
– saz
Feb 22, 2019 at 12:33
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Tr. Mat. Inst. Steklova, 2007, Volume 256, Pages 148–171 (Mi tm460)
Hyperbolicity of Periodic Solutions of Functional Differential Equations with Several Delays
N. B. Zhuravlev, A. L. Skubachevskii
Peoples Friendship University of Russia
Abstract: We study conditions for the hyperbolicity of periodic solutions to nonlinear functional differential equations in terms of the eigenvalues of the monodromy operator. The eigenvalue problem for the monodromy operator is reduced to a boundary value problem for a system of ordinary differential equations with a spectral parameter. This makes it possible to construct a characteristic function. We prove that the zeros of this function coincide with the eigenvalues of the monodromy operator and, under certain additional conditions, the multiplicity of a zero of the characteristic function coincides with the algebraic multiplicity of the corresponding eigenvalue.
Full text: PDF file (309 kB)
References: PDF file HTML file
English version:
Proceedings of the Steklov Institute of Mathematics, 2007, 256, 136–159
Bibliographic databases:
UDC: 517.9
Citation: N. B. Zhuravlev, A. L. Skubachevskii, “Hyperbolicity of Periodic Solutions of Functional Differential Equations with Several Delays”, Dynamical systems and optimization, Collected papers. Dedicated to the 70th birthday of academician Dmitrii Viktorovich Anosov, Tr. Mat. Inst. Steklova, 256, Nauka, MAIK «Nauka/Inteperiodika», M., 2007, 148–171; Proc. Steklov Inst. Math., 256 (2007), 136–159
Citation in format AMSBIB
\Bibitem{ZhuSku07} \by N.~B.~Zhuravlev, A.~L.~Skubachevskii \paper Hyperbolicity of Periodic Solutions of Functional Differential Equations with Several Delays \inbook Dynamical systems and optimization \bookinfo Collected papers. Dedicated to the 70th birthday of academician Dmitrii Viktorovich Anosov \serial Tr. Mat. Inst. Steklova \yr 2007 \vol 256 \pages 148--171 \publ Nauka, MAIK «Nauka/Inteperiodika» \publaddr M. \mathnet{http://mi.mathnet.ru/tm460} \mathscinet{http://www.ams.org/mathscinet-getitem?mr=2336898} \zmath{https://zbmath.org/?q=an:1169.34048} \elib{http://elibrary.ru/item.asp?id=9482613} \transl \jour Proc. Steklov Inst. Math. \yr 2007 \vol 256 \pages 136--159 \crossref{https://doi.org/10.1134/S0081543807010087} \elib{http://elibrary.ru/item.asp?id=13535028} \scopus{http://www.scopus.com/record/display.url?origin=inward&eid=2-s2.0-34248342236}
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This publication is cited in the following articles:
1. N. B. Zhuravlev, “Hyperbolicity criterion for periodic solutions of functional-differential equations with several delays”, Journal of Mathematical Sciences, 153:5 (2008), 683–709
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When will $Hom_R(V,W)\cong V^*\otimes W$ also work for modules?
$$Hom(V,W)\cong V^*\otimes W$$ works for finite dimensional vector space, when will it also work for modules over ring $$R$$?
The motivation (i.e. the question is more general than this) is to know whether and why this works for the $$C^\infty(M)$$-module of sections of a vector bundle which is not vector space since $$C^\infty(M)$$ is not a field.
It seems that $$R$$ has to be commutative, and the module also need to be finitely generated or even projective (since Serre–Swan theorem seems to be related). Is that enough and what is the minimum requirement?
• You don't need $R$ commutative, you just want $V$ to be a finitely generated projective $R$-module i.e. a locally free finitely generated bundle. Sep 27 at 10:13
Let $$R$$ be a ring (need not be commutative) and let $$M$$ and $$N$$ be left $$R$$-modules. As usual set $$M^* = \hom_R(M,R)$$ with its canonical right $$R$$-module structure. There is a natural map $$\Psi_{M,N} : M^* \otimes_R N \longrightarrow \hom_R(M,N)$$ such that $$\Psi(f\otimes x)(y) = f(y)x$$.
Lemma. Fixing $$M$$, this map is an isomorphism for all $$N$$ if and only if $$M$$ is finitely generated and projective.
You can find this, for example, proved in Bourbaki's algebra book. Indeed, it is standard the following are equivalent:
1. $$P$$ is finitely generated and projective
2. $$P$$ admits a finite dual basis, that is, there exist $$(x_i,y_i) \in P\times P$$ for $$i\in [n]$$ such that for each $$x\in P$$ we have that $$x = \sum_{i=1}^n y_i(x) x_i$$ or, what is the same,
3. We have that $$\Psi_{P,P}(u) = 1_P$$ for some $$u= \sum_{i=1}^n y_i\otimes x_i\in P^*\otimes_R P$$.
Proof of the Lemma: if $$\Psi_{P,Q}$$ is an isomorphism for all $$Q$$, picking $$Q=P$$ and covering the identity gives you a dual basis for $$P$$, by 3. so that $$M=P$$ is projective.
Conversely, if you have a dual basis, given a map $$f: M\to N$$, consider the element $$\Phi(f) = \sum_{i=1}^n y_i \otimes f(x_i)$$. It follows that
$$\Psi(\Phi(f))(x) = \sum_{i=1}^n y_i(x)f(x_i) = f\left( \sum_{i=1}^n y_i(x)x_i\right) = f(x)$$
and that for $$g = \Psi(f\otimes x)$$
$$\Phi(g) = \sum_{i=1}^n y_i(-)\otimes f(x_i)x = \sum_{i=1}^n y_i(-)f(x_i)\otimes x = f \otimes x$$
so that $$\Psi$$ is an isomorphism.
Note that dual bases are a bit less canonical than actual bases of free $$R$$-modules, so the map $$\Phi$$ above is not unique.
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# Math Help - nullcline question
1. ## nullcline question
y=0 is a horizontal null cline and x=0 is a vertical nullcline. What are the other two nullclines?
Wouldn't they just be (just set the derivative to zero?):
$0=y(4-2x-y)$ for the v nullcine
$0=x(4-x-2y)$ for the h nullcline
2. Originally Posted by cdlegendary
y=0 is a horizontal null cline and x=0 is a vertical nullcline. What are the other two nullclines?
Wouldn't they just be (just set the derivative to zero?):
$0=y(4-2x-y)$ for the v nullcine
$0=x(4-x-2y)$ for the h nullcline
No, a nullcline is a line not a parabola. dx/dt= 0 when y(4- 2x-y)= 0. That will clearly happen when the factor y= 0 which is why they said that "y= 0 is a horizontal nullcline". What is the other factor?
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## Matrices 3 x 3
Determine eigenvalues and the corresponding eigenvectors for the following matrix over the field $$\mathbb C$$. Decide whether this matrix is diagonalizable:
• #### Variant 1
$$\begin{pmatrix} 2 & -1 & 2 \\ 5 & -3 & 3 \\ -1 & 0 & -2 \\ \end{pmatrix}$$
• #### Variant 2
$$\begin{pmatrix} 2 & -1 & -1 \\ 0 & -1 & 0 \\ 0 & 2 & 1 \\ \end{pmatrix}$$
• #### Variant 3
$$\begin{pmatrix} 1 & -1 & 0 \\ 0 & 1 & -4 \\ -1 & 0 & 4 \\ \end{pmatrix}$$
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# For the following energy band diagram determine the approximate resistivity for x > L portion of semiconductor. Eg = 1.12 eV, T = 300 K, μn = 600 cm2/V-sec, μp = 400 cm2/V-sec, ni = 1010/cm3
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1. 11.00 ohm-cm
2. 15.75 ohm-cm
3. 23 ohm-cm
4. 31.25 ohm-cm
Option 4 : 31.25 ohm-cm
Free
CT 1: Ratio and Proportion
2846
10 Questions 16 Marks 30 Mins
## Detailed Solution
Concept:
The resistivity of a material is the conductivity and the conductivity of a semiconductor is given by:
σ = qnμn + qpμp
n = majority carrier electron concentration
p = majority carrier hole concentration
μn = electron mobility
Also, the majority carrier hole concentration for the given intrinsic Fermi level is given by:
$$p = {n_i}e\frac{{\left( {{E_i} - {E_E}} \right)}}{{KT}}$$
Calculation:
Given semiconductor can be considered as a p-type semiconductor since the Fermi level is close to the valence band.
∴ The majority carrier hole concentration will dominate in the conductivity of the semiconductor, i.e.
σ = qpμp
Given, ni = 1010 /cm3, Eg = 1.12 eV
At n > L:
$${E_i} - {E_f} = \frac{{Eg}}{4} = \frac{{1.12}}{4}eV$$
Ei – Ef = 0.28 eV
Also at room temperature, the thermal voltage is taken as 25.9 mV
The majority carrier hole concentration for x > L will be:
$$p = {n_i}e\frac{{{E_i} - {E_f}}}{{KT}}$$
$$p = \left( {{{10}^{10}}} \right)\left( {{e^{\left( {\frac{{0.28}}{{0.026}}} \right)}}} \right)$$
P = 4.955 × 1014 / cm3
The conductivity will be:
σ = q × p × μp
= (1.6 × 10-19)(4.955 × 1014) × (400)
σ = 0.0317
∴ The resistivity will be:
$$p = \frac{1}{\sigma } = \frac{1}{{0.0317}} \approx 31.25$$
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I have just implemented support for (in)equality queries against a RangeBitmap, a succinct data structure in the RoaringBitmap library which supports range queries. RangeBitmap was designed to support range queries in Apache Pinot (more details here) but this enhancement would allow a range index to be used as a fallback for (in)equality queries in case nothing better is available. Supporting (in)equality queries allows a RangeBitmap to be used as a kind of compact inverted index, trading space for time, capable of supporting high cardinality gracefully.
Since RangeBitmap supports memory mapping from files, I think that it could be used for data engineering beyond Apache Pinot.
### How to use it?
This post extends the example set up in Evaluating Range Predicates and Counting over Range Predicates, which centre around a large collection of Transaction objects:
public static final class Transaction {
private final int quantity;
private final long price;
private final long timestamp;
public Transaction(int quantity, long price, long timestamp) {
this.quantity = quantity;
this.price = price;
this.timestamp = timestamp;
}
public int getQuantity() {
return quantity;
}
public long getPrice() {
return price;
}
public long getTimestamp() {
return timestamp;
}
}
Let’s find all the transactions with a certain quantity in the most obvious way possible, using the stream API.
transactions.stream()
.filter(transaction -> transaction.quantity == qty)
.forEach(this::processTransaction);
On my laptop, it takes about 3ms to select about 100 transactions from 1M.
Benchmark Mode Threads Samples Score Score Error (99.9%) Unit Param: minPrice Param: minQuantity Param: size
stream avgt 1 5 2849.444527 30.581160 us/op 100 1 1000000
With a RangeBitmap, these same transactions can be found with the already existing between method:
RoaringBitmap matchesQuantity = qtyIndex.between(qty - minQty, qty - minQty);
matchesQuantity.forEach((IntConsumer) i -> processTransaction(transactions.get(i)));
as explained in Evaluating Range Predicates, the quantities in the index have been anchored to the smallest value in the population, as a size optimisation.
This performs quite a lot better, selecting the same transactions in under 300us, which is a 10x improvement.
Benchmark Mode Threads Samples Score Score Error (99.9%) Unit Param: minPrice Param: minQuantity Param: size
stream avgt 1 5 2849.444527 30.581160 us/op 100 1 1000000
between avgt 1 5 296.488277 0.703309 us/op 100 1 1000000
This can be rewritten using eq:
RoaringBitmap matchesQuantity = qtyIndex.eq(qty - minQty);
matchesQuantity.forEach((IntConsumer) i -> processTransaction(transactions.get(i)));
The new eq method doesn’t need to do as much work as between, which needs to maintain and combine two bitsets during the scan over the RangeBitmap, whereas eq only needs one. This means we can get a good speedup from eq to select the same transactions:
Benchmark Mode Threads Samples Score Score Error (99.9%) Unit Param: minPrice Param: minQuantity Param: size
stream avgt 1 5 2849.444527 30.581160 us/op 100 1 1000000
between avgt 1 5 296.488277 0.703309 us/op 100 1 1000000
eq avgt 1 5 183.913329 1.994362 us/op 100 1 1000000
It’s worth making a comparison with an inverted index over quantity (so a bitmap of transaction positions per quantity) now. The inverted index would always win for speed on this query, but can take up a lot more space, and the inverted index would almost always lose for range queries. A range encoded inverted index (a mapping from quantity to bitmap of the positions of all transactions with a smaller quantity) will generally beat RangeBitmap for speed at the cost of space. This makes a RangeBitmap suitable for equality queries eiter on a high cardinality attribute or as a fallback better than scanning when range queries are more common.
Inequality filters tend not to be very selective, so benchmarking the evaluation would be dominated by the time to scan the results, but the neq method is present for API symmetry.
In the same set of changes there are also methods to push a RoaringBitmap context down into eq and neq queries, which behaves like an intersection. Rather than producing a large bitmap and then intersecting it with a small context bitmap, the context bitmap is used to potentially skip over large sections of the RangeBitmap. There are also eqCardinality and neqCardinality methods, which produce counts rather than bitmaps, as described for range counts.
### How does it work?
RangeBitmap has a simple two-dimensional structure. The first dimension is the rows in the order the values were added in. The second dimension is the binary representation of each value added, after range encoding, and there are only as many of these columns as there are significant bits in the largest value added.
The layout is essentially the bit-transposition of the values added, striped in bands of $2^{16}$ rows, with a little bit of metadata to help traversals. Each $2^{16}$ rows are bucketed into a horizontal slice which consists of $64 - clz(max)$ RoaringBitmap containers, one for each bit of the inputs, and a mask of size $64 - clz(max)$ bits, indicating the presence of a container. If the $n$th value added to the RangeBitmap does not have bit $i$ set, the $i$th container in the $n/2^{16}$th slice will have no bit set. If all $2^{16}$ values in the slice have bit $i$ unset, the mask will have bit $i$ unset and no container, which is an optimisation to avoid storing empty containers.
If the following values (with their binary representation in brackets) are added in sequence
42 (101010)
24 (011000)
9 (001001)
27 (011011)
The values are first negated to range encode them (see here for explanation).
010101
100111
110110
100100
There will be one horizontal slice with four rows in it, a 6 bit mask, and 5 containers per slice because the 4th bit is present in all the values in the slice. This looks something like this, if stored as plain bitsets:
110111 1100 0110 1111 1010 0110
There are three kinds of RoaringBitmap containers for different densities - sparse, dense, and run length encoded. Containers with only four values would always be sparse, for the sake of explanation, assume these are just the first four values in a much larger slice. Encoded as containers, this looks like:
110111 [4,2] {2,4} [0,4] 1010 {2,4}
Where the containers in braces have been represented as arrays of 16 bit values, the ones in square brackets are run length encoded, and the binary numbers are just bitsets.
To evaluate queries against RangeBitmap, the slices need to be iterated over in ascending order. For each slice, the containers need to be extracted and combined, using a different combination algorithm for each relation. The equality combination algorithm is very simple:
• start with a bitset b with a set bit for each row in the slice
• for each i from 0 to the size of the slice’s mask,
• if i is set in the query value, remove the container i’s bits (or none, if container i is missing for the current slice) from b
• If i is absent in the query value, intersect container i’s bits with b (or just clear it id the container is missing for the slice)
For inequality, b needs to be complemented at the end. Range queries are generally more complex to evaluate than this, but all queries are evaluated in this slice by slice in ascending row order fashion.
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## What made you smile today?
The only man who ever reported his own post!
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: Jul 19th 2012
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Friday, January 2nd 2015, 4:10pm
The 9Gag is strong with this one
Sent from phone.
Stats thingy
### Quoted from "Cheapnub"
I'm a battlefield player, good sir. I don't play metro.
### Quoted from "Suiizide"
PC is no longer PC master race. It's PC mustard race, because consoles need to ketchup :'D
### Quoted from "Riesig"
DICE gave so much into making commander better, but lemmings be lemmings I guess.
### Quoted from "MrT3a"
As a good guy that don't want to use overly glitched weapons, I'll quit using the MTAR and switch to the ACWR until it's fixed
The world needs more people like you
+1, I think we're all in agreement that more MrT3as would be an awesome thing
Although if that was the case they'd use up so much of the world's awesome that there'd be none left for the rest of us!
### Quoted from "CobaltRose"
yes, I know, I'm a big-ass hypocrite
Moderator
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: Jul 6th 2012
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Friday, January 2nd 2015, 4:13pm
This version is pretty funny as well:
Suidae cathexis
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: Jul 1st 2012
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Friday, January 2nd 2015, 4:13pm
### Quoted from "MrT3a"
The 9Gag is strong with this one
Indeed
Another thing I need to thank Rizn for
Everybody's Favourite Worthless Support and LMG Fan!
Song currently stuck in my head is: Red Cold River by Breaking Benjamin!
The only man who ever reported his own post!
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Friday, January 2nd 2015, 4:15pm
I saw them as well
Sent from phone.
Stats thingy
### Quoted from "Cheapnub"
I'm a battlefield player, good sir. I don't play metro.
### Quoted from "Suiizide"
PC is no longer PC master race. It's PC mustard race, because consoles need to ketchup :'D
### Quoted from "Riesig"
DICE gave so much into making commander better, but lemmings be lemmings I guess.
### Quoted from "MrT3a"
As a good guy that don't want to use overly glitched weapons, I'll quit using the MTAR and switch to the ACWR until it's fixed
The world needs more people like you
+1, I think we're all in agreement that more MrT3as would be an awesome thing
Although if that was the case they'd use up so much of the world's awesome that there'd be none left for the rest of us!
### Quoted from "CobaltRose"
yes, I know, I'm a big-ass hypocrite
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: Nov 22nd 2013
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Friday, January 2nd 2015, 5:51pm
This made me chuckle, so I might as well mention it.
@ViperFTW
Your sig says "Everybody's favorite worthless support and LMG!". I'm hoping it should be "..LMG fan/user/etc!" at the end, cause otherwise it's hilarious to imagine an M60 typing these posts
Love knows no language. Suppression knows no range.
Top 2% in the world for Suppression Assists
Suidae cathexis
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Friday, January 2nd 2015, 6:17pm
### Quoted from "FullAuto"
This made me chuckle, so I might as well mention it.
@ViperFTW
Your sig says "Everybody's favorite worthless support and LMG!". I'm hoping it should be "..LMG fan/user/etc!" at the end, cause otherwise it's hilarious to imagine an M60 typing these posts
Ha! well spotted. Thanks
----------
ADDENDUM: Was getting ready to order all the music I've been meaning to buy when I saw Amazon now has digital copies of 'Watch Us Deteriorate', a great album by a little known band I love called Crystalic! I was pleased by this and then I just had one of those funny feelings...Crystalic's Second Album 'Persistence' is a bitch to find, I've been looking for months on numerous sites for anything, hard copy or digital, and found nothing! Not even on American or even Finnish websites! In a moment of "Well, why not?" I stuck 'Persistence' into the search bar aaaaand my two week holiday where I've been almost constantly bed-ridden just got a lot more brighter!!!!!! It's fucking there!!! I found it!!! At last I've found it!!!!! And just on a complete fluke to top it all off!!!!!!!
Will be buying that first thing tomorrow methinks
Everybody's Favourite Worthless Support and LMG Fan!
Song currently stuck in my head is: Red Cold River by Breaking Benjamin!
This post has been edited 3 times, last edit by "ViperFTW" (Jan 3rd 2015, 12:05am)
He's watching...
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: May 5th 2012
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Sunday, January 4th 2015, 1:14pm
It wasn't quite new PC day, but it was new case day!
Well, not really new... A friend of mine was upgrading and selling his old case and some old hardware. A different friend of mine bought the hardware, and I took the case on good faith that I'd pay for it as soon as I get some cash.
Silverstone Raven RV02. It's pretty cool. I'm a little disappointed with the way I had to manage the cables, it doesn't look as clean as my Antec TwelveHundred did, but it is totally airflow optimized and I really like the 90° clockwise orientation of the MoBo. Heat rises, so it just makes sense to intake on the bottom and exhaust on the top.
He had cold cathode tubes for it as well, and they light up the case well enough, but I hate how they are wired. The cables that don't really matter (Molex power) are super long and hard to hide, and the cables that matter (power to cathode tubes) are too short to place the tube where I really want. I came up with a solution and placed the little transformer box or whatever it is outside the case, on top, and under the cover, then ran the wires in through a vent, and zip-tied one cathode tube vertically along the hard-drive cages and the other horizontally along the inside of the top. Works well enough. I'll probably switch to LEDs once I get some in the bank.
The only thing that sucks is he was missing one of the drive bay covers and I use an external optical drive. I had a little 5.25" utility drawer for my Antec, but it doesn't seem to fit in the Raven. The screw holes don't quite line up, and the front bezel of the Raven is slightly smaller than the Antec so even if it was installed the drawer would stick and be hard to open. For now it's just an open slot. Any ideas?
All bike, all the time!
### Spoiler
Suidae cathexis
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Sunday, January 4th 2015, 1:40pm
Sorry there's no picture, can't link 'em too well on my IPod
Everybody's Favourite Worthless Support and LMG Fan!
Song currently stuck in my head is: Red Cold River by Breaking Benjamin!
I'll be there... around every corner... in every empty room...
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: Nov 8th 2013
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Sunday, January 4th 2015, 6:12pm
@ViperFTW
that last frame though
Anyways: found my gloves at work. Apparently dropped them inside and not in the parking lot, thank god. Also got to go home early. Doesn't really matter since I'm doing inventory again this year, but w/e moar free time!
Who knows what evil lurks in the hearts of men?
### Quoted from "Spectacle"
I once tried to burn a bridge, but due to the half-assed levolution implementation it kept standing up just fine.
### Quoted from "DeathOfTheDodo"
Compared to the MTAR, the PDR has the TTK of your average pack of cigarettes.
The World Champion, on the Tactical Light:
### Quoted from "The World Champion"
"Oh don't mind me, I'm just out of ammo and I'll just POCKET SUN!!!!"
### Quoted from "Sheepnub"
What can't be changed can be banned.
### Quoted from "Gummybear"
I'm completely serious. Well, seriously insane actually
### Quoted from "xESxMaver1ck911"
If only more people left the wheel chair, adjustable hospital bed, and crutches behind and played HC. (aka Regen, Minimap, Killcam)
### Quoted from "hunturk"
Natalia Poklonskaya:
Putin the cute in prosecute since 2014.
### Quoted from "DeathOfTheDodo"
I can't look at my own avatar without having to pee. GG Me. GG.
### Quoted from "DeathOfTheDodo"
I swear to god if you reply with a picture of the AEK I'll mail you a buttplug shaped like one.
### Quoted from "Nick 30075"
The one time when "it's only three inches" is a good thing.
### Quoted from "RRR3186"
How do you guys control your FAMAS burst length since it fires its mag in just 1,5 seconds))
I consider each magazine a burst. :/
### Quoted from "ViperFTW"
This is Symthic, we don't do "feels" around here
Be Creative.
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: Mar 9th 2012
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Monday, January 5th 2015, 8:23pm
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# Choosing 3 different integers from 1 to 100
How many sets of integers are there, such that $$1 \leq a < b < c \leq 100$$ and $$a + b + c$$ is a multiple of 3?
×
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# Transverse Mercator: Redfearn series
The article Transverse Mercator projection restricts itself to general features of the projection. This article describes in detail one of the (two) implementations developed by Louis Krüger in 1912;[1] that expressed as a power series in the longitude difference from the central meridian. These series were recalculated by Lee in 1946,[2] by Redfearn in 1948,[3] and by Thomas in 1952.[4][5] They are often referred to as the Redfearn series, or the Thomas series. This implementation is of great importance since it is widely used in the U.S. State Plane Coordinate System,[5] in national (Britain,[6] Ireland[7] and many others) and also international[8] mapping systems, including the Universal Transverse Mercator coordinate system (UTM).[9][10] They are also incorporated into the Geotrans coordinate converter made available by the United States National Geospatial-Intelligence Agency.[11] When paired with a suitable geodetic datum, the series deliver high accuracy in zones less than a few degrees in east-west extent.
## Preliminaries I: datum and ellipsoid parameters
The series must be used with a geodetic datum which specifies the position, orientation and shape of a Reference ellipsoid. Although the projection formulae depend only on the shape parameters of the reference ellipsoid the full set of datum parameters is necessary to link the projection coordinates to true positions in three-dimensional space. The datums and reference ellipsoids associated with particular implementations of the Redfearn formulae are listed below. A comprehensive list of important ellipsoids is given in the article on the Figure of the Earth.
In specifying ellipsoids it is normal to give the semi-major axis (equatorial axis), ${\displaystyle a}$, along with either the inverse flattening, ${\displaystyle 1/f}$, or the semi-minor axis (polar axis), ${\displaystyle b}$, or sometimes both. The series presented below use the eccentricity, ${\displaystyle e}$, in preference to the flattening, ${\displaystyle f}$. In addition they use the parameters ${\displaystyle n}$, called the third flattening, and ${\displaystyle e'}$, the second eccentricity. There are only two independent shape parameters and there are many relations between them: in particular
{\displaystyle {\begin{aligned}f&={\frac {a-b}{a}},\qquad e^{2}=2f-f^{2},\qquad e'^{2}={\frac {e^{2}}{1-e^{2}}}\\b&=a(1-f)=a(1-e^{2})^{1/2},\qquad n={\frac {a-b}{a+b}}.\end{aligned}}}
The projection formulae also involve ${\displaystyle \rho (\phi )}$, the radius of curvature of the meridian (at latitude ${\displaystyle \phi }$), and ${\displaystyle \nu (\phi )}$, the radius of curvature in the prime vertical. (The prime vertical is the vertical plane orthogonal to the meridian plane at a point on the ellipsoid). The radii of curvature are defined as follows:
${\displaystyle \nu (\phi )={\frac {a}{\sqrt {1-e^{2}\sin ^{2}\phi }}},\qquad \rho (\phi )={\frac {\nu ^{3}(1-e^{2})}{a^{2}}}.}$
In addition the functions ${\displaystyle \beta (\phi )}$ and ${\displaystyle \eta (\phi )}$ are defined as:
${\displaystyle \beta (\phi )={\frac {\nu (\phi )}{\rho (\phi )}},\qquad \eta ^{2}=\beta -1={e'^{2}\cos ^{2}\!\phi }.}$
For compactness it is normal to introduce the following abbreviations:
${\displaystyle s=\sin \phi ,\qquad c=\cos \phi ,\qquad t=\tan \phi .}$
## Preliminaries II: meridian distance
### Meridian distance
The article on Meridian arc describes several methods of computing ${\displaystyle m(\phi )}$, the meridian distance from the equator to a point at latitude ${\displaystyle \phi }$ : the expressions given below are those used in the 'actual implementation of the Transverse Mercator projection by the OSGB.[6] The truncation error is less than 0.1mm so the series is certainly accurate to within 1mm, the design tolerance of the OSGB implementation.
{\displaystyle {\begin{aligned}m(\phi )&=B_{0}\phi +B_{2}\sin 2\phi +B_{4}\sin 4\phi +B_{6}\sin 6\phi +\cdots ,\end{aligned}}}
where the coefficients are given to order ${\displaystyle n^{3}}$ (order ${\displaystyle e^{6}}$) by
{\displaystyle {\begin{aligned}B_{0}&=b{\bigg (}1+n+{\frac {5}{4}}n^{2}+{\frac {5}{4}}n^{3}{\bigg )},\qquad B_{4}=b{\bigg (}{\frac {15}{16}}n^{2}+{\frac {15}{16}}n^{3}{\bigg )},\\B_{2}&=-b{\bigg (}{\frac {3}{2}}n+{\frac {3}{2}}n^{2}+{\frac {21}{16}}n^{3}{\bigg )},\qquad B_{6}=-b{\bigg (}{\frac {35}{48}}n^{3}{\bigg )}.\end{aligned}}}
The meridian distance from equator to pole is
${\displaystyle m_{p}=m(\pi /2)=\pi B_{0}/2\,.}$
The form of the series specified for UTM is a variant of the above exhibiting higher order terms with a truncation error of 0.03mm.
#### Inverse meridian distance
Neither the OSGB nor the UTM implementations define an inverse series for the meridian distance; instead they use an iterative scheme. For a given meridian distance ${\displaystyle M}$ first set ${\displaystyle \phi _{0}=M/B_{0}}$ and then iterate using
{\displaystyle {\begin{aligned}\phi _{n}=\phi _{n-1}+{\frac {M-m(\phi _{n-1})}{B_{0}}},\qquad n=1,2,3,\ldots \end{aligned}}}
until ${\displaystyle |M-m(\phi _{n-1})|<0.01}$mm.
The inversion can be effected by a series, presented here for later reference. For a given meridian distance, ${\displaystyle M}$, define the rectifying latitude by
${\displaystyle \mu ={\frac {\pi M}{2m_{p}}}.}$
The geodetic latitude corresponding to ${\displaystyle M}$ is (Snyder[5] page 17):
{\displaystyle {\begin{aligned}\phi &=\mu +D_{2}\sin 2\mu +D_{4}\sin 4\mu +D_{6}\sin 6\mu +D_{8}\sin 8\mu +\cdots ,\\\end{aligned}}}
where, to ${\displaystyle O(n^{4})}$,
{\displaystyle {\begin{aligned}D_{2}&={\frac {3}{2}}n-{\frac {27}{32}}n^{3},&D_{4}&={\frac {21}{16}}n^{2}-{\frac {55}{32}}n^{4},\\[8pt]D_{6}&={\frac {151}{96}}n^{3},&D_{8}&={\frac {1097}{512}}n^{4}.\end{aligned}}}
## An outline of the method
The normal aspect of the Mercator projection of a sphere of radius ${\displaystyle R}$ is described by the equations
${\displaystyle x=R\lambda ,\qquad \qquad y=R\psi ,}$
where ${\displaystyle \psi }$, the isometric latitude, is given by
{\displaystyle {\begin{aligned}\psi &=\ln \left[\tan \left({\frac {\pi }{4}}+{\frac {\phi }{2}}\right)\right].\end{aligned}}}
On the ellipsoid the isometric latitude becomes
{\displaystyle {\begin{aligned}\psi &=\ln \left[\tan \left({\frac {\pi }{4}}+{\frac {\phi }{2}}\right)\right]-{\frac {e}{2}}\ln \left[{\frac {1+e\sin \phi }{1-e\sin \phi }}\right].\end{aligned}}}
By construction, the projection from the geodetic coordinates (${\displaystyle \phi }$,${\displaystyle \lambda }$) to the coordinates (${\displaystyle \psi }$,${\displaystyle \lambda }$) is conformal. If the coordinates (${\displaystyle \psi }$,${\displaystyle \lambda }$) are used to define a point ${\displaystyle \zeta =\psi +i\lambda }$ in the complex plane, then any analytic function ${\displaystyle f(\zeta )}$ will define another conformal projection. Kruger's method involves seeking the specific ${\displaystyle f(\zeta )}$ which generates a uniform scale along the central meridian, ${\displaystyle \lambda =0}$. He achieved this by investigating a Taylor series approximation with the projection coordinates given by:
{\displaystyle {\begin{aligned}y+ix&=f(\zeta )=f(\psi +i\lambda )\\&=f(\psi +i.0)+A_{1}\lambda +A_{2}\lambda ^{2}+A_{3}\lambda ^{3}+\ldots ,\end{aligned}}}
where the real part of ${\displaystyle f(\psi +i.0)}$ must be proportional to the meridian distance function ${\displaystyle m(\phi )}$. The (complex) coefficients ${\displaystyle A_{n}}$ depend on derivatives of ${\displaystyle f(\zeta )}$ which can be reduced to derivatives of ${\displaystyle m(\phi )}$ with respect to ${\displaystyle \psi }$, (not ${\displaystyle \phi }$). The derivatives are straightforward to evaluate in principle but the expressions become very involved at high orders because of the complicated relation between ${\displaystyle \psi }$ and ${\displaystyle \phi }$. Separation of real and imaginary parts gives the series for ${\displaystyle x}$ and ${\displaystyle y}$ and further derivatives give the scale and convergence factors.
## The series in detail
This section presents the eighth order series as published by Redfearn[3] (but with ${\displaystyle x}$ and ${\displaystyle y}$ interchanged and the longitude difference from the central meridian denoted by ${\displaystyle \lambda }$ instead of ${\displaystyle \omega }$). Equivalent eighth order series, with different notations, can be found in Snyder[5] (pages 60–64) and at many web sites such as that for the Ordnance Survey of Great Britain.[6]
The direct series are developed in terms of the longitude difference from the central meridian, expressed in radians: the inverse series are developed in terms of the ratio ${\displaystyle x/a}$. The projection is normally restricted to narrow zones (in longitude) so that both of the expansion parameters are typically less than about 0.1, guaranteeing rapid convergence. For example in each UTM zone these expansion parameters are less than 0.053 and for the British national grid (NGGB) they are less than 0.09. All of the direct series giving ${\displaystyle x}$, ${\displaystyle y}$, scale ${\displaystyle k}$, convergence ${\displaystyle \gamma }$ are functions of both latitude and longitude and the parameters of the ellipsoid: all inverse series giving ${\displaystyle \phi }$, ${\displaystyle \lambda }$, ${\displaystyle k}$, ${\displaystyle \gamma }$ are functions of both ${\displaystyle x}$ and ${\displaystyle y}$ and the parameters of the ellipsoid.
### Direct series
In the following series ${\displaystyle \lambda }$ is the difference of the longitude of an arbitrary point and the longitude of the chosen central meridian: ${\displaystyle \lambda }$ is in radians and is positive east of the central meridian. The W coefficients are functions of ${\displaystyle \phi }$ listed below. The series for ${\displaystyle y}$ reduces to the scaled meridian distance when ${\displaystyle \lambda =0}$.
{\displaystyle {\begin{aligned}x(\lambda ,\phi )&=k_{0}\nu \left[\lambda c+{\frac {\lambda ^{3}c^{3}W_{3}}{3!}}+{\frac {\lambda ^{5}c^{5}W_{5}}{5!}}+{\frac {\lambda ^{7}c^{7}W_{7}}{7!}}\right],\\[1ex]y(\lambda ,\phi )&=k_{0}\left[m(\phi )+{\frac {\lambda ^{2}\nu c^{2}t}{2}}+{\frac {\lambda ^{4}\nu c^{4}tW_{4}}{4!}}+{\frac {\lambda ^{6}\nu c^{6}tW_{6}}{6!}}+{\frac {\lambda ^{8}\nu c^{8}tW_{8}}{8!}}\right],\end{aligned}}}
### Inverse series
The inverse series involve a further construct: the footpoint latitude. Given a point ${\displaystyle (x,y)}$ on the projection the footpoint is defined as the point on the central meridian with coordinates ${\displaystyle (0,y)}$. Since the scale on the central meridian is ${\displaystyle k_{0}}$ the meridian distance from the equator to the footpoint is equal to ${\displaystyle m=y/k_{0}}$. The corresponding footpoint latitude, ${\displaystyle \phi _{1}}$, is calculated by iteration or the inverse meridian distance series as described above.
{\displaystyle {\begin{aligned}\mu &={\frac {\pi y}{2m_{p}k_{0}}},\\\phi _{1}&=\mu +D_{2}\sin 2\mu +D_{4}\sin 4\mu +D_{6}\sin 6\mu +D_{8}\sin 8\mu +\cdots ,\\\end{aligned}}}
Denoting functions evaluated at ${\displaystyle \phi _{1}}$ by a subscript '1', the inverse series are:
{\displaystyle {\begin{aligned}\lambda (x,y)&={\frac {x}{c_{1}(k_{0}\nu _{1})}}-{\frac {x^{3}V_{3}}{3!c_{1}(k_{0}\nu _{1})^{3}}}-{\frac {x^{5}V_{5}}{5!c_{1}(k_{0}\nu _{1})^{5}}}-{\frac {x^{7}V_{7}}{7!c_{1}(k_{0}\nu _{1})^{7}}},\\\phi (x,y)&=\phi _{1}-{\frac {x^{2}\beta _{1}t_{1}}{2(k_{0}\nu _{1})^{2}}}-{\frac {x^{4}\beta _{1}t_{1}U_{4}}{4!(k_{0}\nu _{1})^{4}}}-{\frac {x^{6}\beta _{1}t_{1}U_{6}}{6!(k_{0}\nu _{1})^{6}}}-{\frac {x^{8}\beta _{1}t_{1}U_{8}}{8!(k_{0}\nu _{1})^{8}}}.\end{aligned}}}
### Point scale and convergence
The point scale ${\displaystyle k}$ is independent of direction for a conformal transformation. It may be calculated in terms of geographic or projection coordinates. Note that the series for ${\displaystyle k}$ reduce to ${\displaystyle k_{0}}$ when either ${\displaystyle \lambda =0}$ or ${\displaystyle x=0}$ . The convergence ${\displaystyle \gamma }$ may also be calculated (in radians) in terms of geographic or projection coordinates:
{\displaystyle {\begin{aligned}k(\lambda ,\phi )&=k_{0}\left[1+{\frac {\lambda ^{2}c^{2}H_{2}}{2}}+{\frac {\lambda ^{4}c^{4}H_{4}}{24}}+{\frac {\lambda ^{6}c^{6}H_{6}}{720}}\right],\\\gamma (\lambda ,\phi )&=\lambda s+{\frac {\lambda ^{3}c^{3}tH_{3}}{3}}+{\frac {\lambda ^{5}c^{5}tH_{5}}{15}}+{\frac {\lambda ^{7}c^{7}tH_{7}}{315}},\\k(x,y)&=k_{0}\left[1+{\frac {x^{2}K_{2}}{2(k_{0}\nu _{1})^{2}}}+{\frac {x^{4}K_{4}}{24(k_{0}\nu _{1})^{4}}}+{\frac {x^{6}K_{6}}{720(k_{0}\nu _{1})^{6}}}\right]\\\gamma (x,y)&={\frac {xt_{1}}{k_{0}\nu _{1}}}+{\frac {x^{3}t_{1}K_{3}}{3(k_{0}\nu _{1})^{3}}}+{\frac {x^{5}t_{1}K_{5}}{15(k_{0}\nu _{1})^{5}}}+{\frac {x^{7}t_{1}K_{7}}{315(k_{0}\nu _{1})^{7}}}.\end{aligned}}}
### The coefficients for all series
{\displaystyle {\begin{aligned}W_{3}&=\beta -t^{2}\\W_{5}&=4\beta ^{3}(1-6t^{2})+\beta ^{2}(1+8t^{2})-2\beta t^{2}+t^{4}\\W_{7}&=61-479t^{2}+179t^{4}-t^{6}+O(e^{2})\\W_{4}&=4\beta ^{2}+\beta -t^{2}\\W_{6}&=8\beta ^{4}(11{-}24t^{2})-28\beta ^{3}(1{-}6t^{2})+\beta ^{2}(1{-}32t^{2})-2\beta t^{2}+t^{4}\\W_{8}&=1385-3111t^{2}+543t^{4}-t^{6}+O(e^{2})\\V_{3}&=\beta _{1}+2t_{1}^{2}\\V_{5}&=4\beta _{1}^{3}(1-6t_{1}^{2})-\beta _{1}^{2}(9-68t_{1}^{2})-72\beta _{1}t_{1}^{2}-24t_{1}^{4}\\V_{7}&=61+662t_{1}^{2}+1320t_{1}^{4}+720t_{1}^{6}\\U_{4}&=4\beta _{1}^{2}-9\beta _{1}(1-t_{1}^{2})-12t_{1}^{2}\\U_{6}&=8\beta _{1}^{4}(11-24t_{1}^{2})-12\beta _{1}^{3}(21-71t_{1}^{2})+15\beta _{1}^{2}(15-98t_{1}^{2}+15t_{1}^{4})\\&\qquad \qquad +180\beta _{1}(5t_{1}^{2}-3t_{1}^{4})+360t_{1}^{4}\\U_{8}&=-1385-3633t_{1}^{2}-4095t_{1}^{4}-1575t_{1}^{6}\\H_{2}&=\beta \\H_{4}&=4\beta ^{3}(1-6t^{2})+\beta ^{2}(1+24t^{2})-4\beta t^{2}\\H_{6}&=61-148t^{2}+16t^{4}\\H_{3}&=2\beta ^{2}-\beta \\H_{5}&=\beta ^{4}(11-24t^{2})-\beta ^{3}(11-36t^{2})+\beta ^{2}(2-14t^{2})+\beta t^{2}\\H_{7}&=17-26t^{2}+2t^{4}\\K_{2}&=\beta _{1}\\K_{4}&=4\beta _{1}^{3}(1-6t_{1}^{2})-3\beta _{1}^{2}(1-16t_{1}^{2})-24\beta _{1}t_{1}^{2}\\K_{6}&=1\\K_{3}&=2\beta _{1}^{2}-3\beta _{1}-t_{1}^{2}\\K_{5}&=\beta _{1}^{4}(11-24t_{1}^{2})-3\beta _{1}^{3}(8-23t_{1}^{2})+5\beta _{1}^{2}(3-14t_{1}^{2})+30\beta _{1}t_{1}^{2}+3t_{1}^{4}\\K_{7}&=-17-77t_{1}^{2}-105t_{1}^{4}-45t_{1}^{6}\end{aligned}}}
### Accuracy of the series
The exact solution of Lee-Thompson,[12] implemented by Karney (2011),[13] is of great value in assessing the accuracy of the truncated Redfearn series. It confirms that the truncation error of the (eighth order) Redfearn series is less than 1 mm out to a longitude difference of 3 degrees, corresponding to a distance of 334 km from the central meridian at the equator but a mere 35 km at the northern limit of an UTM zone.
The Redfearn series become much worse as the zone widens. Karney discusses Greenland as an instructive example. The long thin landmass is centred on 42W and, at its broadest point, is no more than 750 km from that meridian whilst the span in longitude reaches almost 50 degrees. The Redfearn series attain a maximum error of 1 kilometre.
## Implementations
The implementations give below are examples of the use of the Redfearn series. The defining documents in various countries differ slightly in notation and, more importantly, in the neglect of some of the small terms. The analysis of small terms depends on the latitude and longitude ranges in the various grids. There are also slight differences in the formulae utilised for meridian distance: one extra term is sometimes added to the formula specified above but such a term is less than 0.1mm.
### OSGB
The implementation of the transverse Mercator projection in Great Britain is fully described in the OSGB document A guide to coordinate systems in Great Britain, Appendices A.1, A.2 and C.[6]
datum: OSGB36
ellipsoid: Airy 1830
major axis: 6 377 563.396
minor axis: 6 356 256.909
central meridian longitude: 2°W
central meridian scale factor : 0.9996012717
projection origin: 2°W and 0°N
true grid origin: 2°W and 49°N
false easting of true grid origin, E0 (metres): 400,000
false northing of true grid origin, N0 (metres): -100,000
E = E0 + x = 400000 + x
N = N0 + y -k0*m(49°)= y - 5527063
The extent of the grid is 300 km to the east and 400 km to the west of the central meridian and 1300 km north from the false origin, (OSGB[6] Section 7.1), but with the exclusion of parts of Northern Ireland, Eire and France. A grid reference is denoted by the pair (E,N) where E ranges from slightly over zero to 800000m and N ranges from zero to 1300000m. To reduce the number of figures needed to give a grid reference, the grid is divided into 100 km squares, which each have a two-letter code. National Grid positions can be given with this code followed by an easting and a northing both in the range 0 and 99999m.
The projection formulae differ slightly from the Redfearn formulae presented here. They have been simplified by neglect of most terms of seventh and eighth order in ${\displaystyle \lambda }$ or ${\displaystyle x/a}$: the only exception is seventh order term in the series for ${\displaystyle \lambda }$ in terms of ${\displaystyle x/a}$. This simplification is based on the examination of the Redfearn terms over the actual extent of the grid. The only other differences are (a) the absorption of the central scale factor into the radii of curvature and meridian distance, (b) the replacement of the parameter ${\displaystyle \beta }$ by the parameter ${\displaystyle \eta }$ (defined above).
The OSGB manual[6] includes a discussion of the Helmert transformations which are required to link geodetic coordinates on Airy 1830 ellipsoid and on WGS84.
### UTM
The article on the Universal Transverse Mercator projection gives a general survey, but the full specification is defined in U.S. Defense Mapping Agency Technical Manuals TM8358.1[9] and TM8358.2.[10] This section provides details for zone 30 as another example of the Redfearn formulae (usually termed Thomas formulae in the United States.)
ellipsoid: International 1924 (a.k.a. Hayford 1909)
major axis: 6 378 388.000
minor axis: 6 356 911.946
central meridian longitude: 3°W
projection origin: 3°W and 0°N
central meridian scale factor: 0.9996
true grid origin: 3°W and 0°N
false easting of true grid origin, E0: 500,000
E = E0 + x = 500000 + x
northern hemisphere false northing of true grid origin N0: 0
northern hemisphere: N = N0 + y = y
southern hemisphere false northing of true grid origin N0: 10,000,000
southern hemisphere: N = N0 + y = 10,000,000 + y
The series adopted for the meridian distance incorporates terms of fifth order in ${\displaystyle n}$ but the manual states that these are less than 0.03 mm (TM8358.2[10] Chapter 2). The projection formulae use, ${\displaystyle e'}$, the second eccentrity (defined above) instead of ${\displaystyle n}$. The grid reference schemes are defined in the article Universal Transverse Mercator coordinate system. The accuracy claimed for the UTM projections is 10 cm in grid coordinates and 0.001 arc seconds for geodetic coordinates.
### Ireland
The transverse Mercator projection in Eire and Northern Ireland (an international implementation spanning one country and part of another) is currently implemented in two ways:
datum: Ireland 1965
ellipsoid: Airy 1830 modified
major axis: 6 377 340.189
minor axis: 6 356 034.447
central meridian scale factor: 1.000035
true origin: 8°W and 53.5°N
false easting of true grid origin, E0: 200,000
false northing of true grid origin, N0: 250,000
The Irish grid uses the OSGB projection formulae.
datum: Ireland 1965
ellipsoid: GRS80
major axis: 6 378 137
minor axis: 6 356 752.314140
central meridian scale factor: 0.999820
true origin: 8°W and 53.5°N
false easting of true grid origin, E0: 600,000
false northing of true grid origin, N0: 750,000
This is an interesting example of the transition between use of a traditional ellipsoid and a modern global ellipsoid. The adoption of radically different false origins helps to prevent confusion between the two systems.
## References
1. ^ Krüger, L. (1912). "Konforme Abbildung des Erdellipsoids in der Ebene". Royal Prussian Geodetic Institute, New Series 52. doi:10.2312/GFZ.b103-krueger28.
2. ^ Lee, L. P. (1946). "The transverse Mercator projection of the spheroid (Errata and comments in Volume 8 (Part 61), pp 277–278". Survey Review.
3. ^ a b Redfearn, J. C. B. (1948). "Transverse Mercator formulae". Survey Review.
4. ^ Thomas, Paul D (1952). Conformal Projections in Geodesy and Cartography. Washington: U.S. Coast and Geodetic Survey Special Publication 251.
5. ^ a b c d Snyder, John P. (1987). Map Projections – A Working Manual. U.S. Geological Survey Professional Paper 1395. United States Government Printing Office, Washington, D.C.This paper can be downloaded from USGS pages. It gives full details of most projections, together with interesting introductory sections, but it does not derive any of the projections from first principles.
6. ^
7. ^ "Short Proceedings of the 1st European Workshop on Reference Grids, Ispra, 27–29 October 2003" (PDF). European Environment Agency. 2004-06-14. p. 6. Retrieved 2009-08-27.The EEA recommends the Transverse Mercator for conformal pan-European mapping at scales larger than 1:500,000
8. ^ a b
9. ^ a b c Hager, J. W.; Behensky, J.F.; Drew, B.W. (1989). "Defense Mapping Agency Technical Report TM 8358.2. The universal grids: Universal Transverse Mercator (UTM) and Universal Polar Stereographic (UPS)".
10. ^
11. ^ Lee, L.P. (1976). Conformal Projections Based on Elliptic Functions (Supplement No. 1 to Canadian Cartographer, Vol 13.) pp. 1–14, 92–101 and 107–114. Toronto: Department of Geography, York University. A report of unpublished analytic formulae involving incomplete elliptic integrals obtained by E. H. Thompson in 1945. Available from the of Toronto Press.
12. ^ C. F. F. Karney (2011), Transverse Mercator with an accuracy of a few nanometers, J. Geodesy 85(8), 475-485 (2011); preprint of paper and C++ implementation of algorithms are available at geographiclib.sourceforge.io
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If r and s are positive integers such that $(2^{r})(4^{s}) = 16$, then $2r +s =$
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# Stability of a point charge for the repulsive Vlasov-Poisson system
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We consider solutions of the repulsive Vlasov-Poisson system which are a combination of a point charge and a small density with respect to Liouville measure (a cloud''), and we show that these solutions exist globally, that the electric field decays at an optimal rate and that the particle distribution converges along a modified scattering dynamics. This follows by a Lagrangian study of the linearized equation, which is integrated by means of an asymptotic action-angle coordinate transformation, and an Eulerian study of the nonlinear dynamic which exhibits the mixing'' mechanism responsible for the asymptotic behavior.
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How were shift amount constants in MD5 found?
The md5 specification gives a series of 4 rounds to execute over a 16-word block.
Each round has a repeating sequence of 4 shift amounts (s in [abcd k s i]) :
• 7, 12, 17 and 22 for the round 1
• 5, 9, 14 and 20 for the round 2
• 4, 11, 16 and 23 for the round 3
• 6, 10, 15 and 21 for the round 4
The specification only says:
The shift amounts in each round have been approximately optimized, to yield a faster "avalanche effect." The shifts in different rounds are distinct.
How did they find these shift amounts? Is it a sequence like $u_n = (u_{n-1} + j) \; mod \; k$ ? What guarantees the "avalanche effect"?
• They were probably found by empirical testing. XORs and additions are replaced with ORs then you set block to all null but one, full diffusion happens when block is all ones. Look at schneier.com/skein1.3.pdf "Appendix D NIST SHA-3 Round 2 Tweak: Rotation Constants" and "D.1 Deprecated Skein Rotation Constants". – LightBit Dec 23 '14 at 21:10
• if you do find an answer, please come back and write it, it's interesting! – David 天宇 Wong Dec 25 '14 at 17:15
• For the sine based stuff and the h0-h3, I think this question and answers apply. Specifically the "nothing-up-my-sleeve" point. For the shift amounts, it may be the same, but I'd need to verify. – mikeazo Oct 6 '15 at 15:52
• I have seen that post and I read the "nothing-up-my-sleeve" point, however it only gives examples of 'good constants' and not of 'bad constants'. e.g. he says which numbers CAN be used as constants (pi), but not which ones CANT. – Thomas W Oct 6 '15 at 16:03
By trying all the 128 different bit position on the state, 19 rounds (over the 64) are required for a full avalanche effect (source code).
round 0: 00000000000000000000000000000001 00000000000000000000000010000000 00000000000000000000000000000000 00000000000000000000000000000000
round 1: 00000000000000000000000000000000 00000000000000010000001000000000 00000000000000000000000010000000 00000000000000000000000000000000
round 2: 00000000000000000000000000000000 00000010100000000000000000000001 00000000000000010000001000000000 00000000000000000000000010000000
round 3: 00000000000000000000000010000000 01010000001000000101000000100000 00000010100000000000000000000001 00000000000000010000001000000000
round 4: 00000000000000010000001000000000 10100001010100101010000101010010 01010000001000000101000000100000 00000010100000000000000000000001
round 5: 00000010100000000000000000000001 11110011111100111111001101110011 10100001010100101010000101010010 01010000001000000101000000100000
round 6: 01010000001000000101000000100000 11110011111100111111001101110011 11110011111100111111001101110011 10100001010100101010000101010010
round 7: 10100001010100101010000101010010 11111111001111110011111100110111 11110011111100111111001101110011 11110011111100111111001101110011
round 8: 11110011111100111111001101110011 11111111111111111111111101110111 11111111001111110011111100110111 11110011111100111111001101110011
round 9: 11110011111100111111001101110011 11111111111111111111111101110111 11111111111111111111111101110111 11111111001111110011111100110111
round 10: 11111111001111110011111100110111 11111111111111111111111111111111 11111111111111111111111101110111 11111111111111111111111101110111
round 11: 11111111111111111111111101110111 11111111111111111111111111111111 11111111111111111111111111111111 11111111111111111111111101110111
round 12: 11111111111111111111111101110111 11111111111111111111111111111111 11111111111111111111111111111111 11111111111111111111111111111111
round 13: 11111111111111111111111111111111 11111111111111111111111111111111 11111111111111111111111111111111 11111111111111111111111111111111
As for the exact answer to the question, only Rivest will be able to say how he found his values. However as LightBit stated it is likely that
they were probably found by empirical testing. XORs and additions are replaced with ORs then you set block to all null but one [bit], full diffusion happens when blocks are all ones.
As for how to measure the avalanche effect, multiple statistics can be used (see section 4.2).
The rotation countsr (together with the order of access of the 4 words of the state) are engineered for fast diffusion, as documented in RFC1321:
The shift amounts in each round have been approximately optimized, to yield a faster "avalanche effect". The shifts in different rounds are distinct.
The general idea is to move bits to a 32-bit position on which they previously had no or little influence (limited to the effect of carry propagation in 32-bit addition, which works only to the left, and has odds like $2^{-n}$ to influence the $n$th bit on the left of a particular one); also, the choice made should facilitate efficient propagation in the next few steps. Among particularly bad choices would be to change the constants r to a mix of 1 and 31; or to all 16.
I do not know if the choice made for MD5 turned out to prevent or facilitate attacks on collision-resistance.
All the other constants in the question are chosen with a nothing-up-my-sleeve rationale; that is, convincingly unlikely to have been chosen according to a hidden property. That includes numbers computed according to a simple formula unrelated to the rest of the context (like floor(abs(sin(i + 1)) × 2^32) is); or numbers matching an obvious pattern (like being the representation of the hex string 0123456789ABCDEFFEDCBA9876543210 according to the endianness used in the context).
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Let us start with metric on $\mathbb R^4=\mathbb R^2\times \mathbb R^2$. Defined by norm $\|{*}\|$ defined by $$\|(x,y)\|=\tfrac12(|x|+|y|),$$ $\|(x,y)\|=\int_0^1|t\cdot x+(1-t)\cdot y|\,dt,$$where |{ * }| denotes Euclidean norm on \mathbb R^2. This norm, is not strictly strongly convexand generically , so you have a lot of should expect many geodesics connectiong two between close points. Now, your metric is the intrisic metric induced by on the hypersurface \Sigma described by |x-y|=1. It often happens that one of the geodesic in (\mathbb R^4,\|{*}\|) lies in \Sigma. For example the geodesic between segments [(0,0),(0,1)] and [(1,0),(0,0)]. Maybe this is a closed geodesic, but at least it is a closed line with 4 geodesic segments. Maybe it already makes you happy (?). If notSo, you have a Finsler metric on \Sigma=\mathbb S^1\times \mathbb R^2. The unit ball for some coordiantes (u,v,w) of in the tangent plane can be described by is isometric the following inequality:$$\sqrt{u^2+w^2}+\sqrt{v^2+w^2}\le 2.$$This intersection of the ball in the above norm with 3-dimensional subspace in one special direction. It seems that this intersection is strongly convex. The metric is smooth (sinse there is a transitive isometric group action on \mathbb S^1\times \mathbb R^2). It reamins to write differential equasion for geodesic; this should be in any book on Finsler geometry. (It should be pain, but it might help.) 4 added 92 characters in body; added 99 characters in body Let us start with metric on \mathbb R^4=\mathbb R^2\times \mathbb R^2. Defined by norm \|{*}\| defined by$$\|(x,y)\|=\tfrac12(|x|+|y|),$$where |{ * }| denotes Euclidean norm. This norm, is not strictly convex and generically you have a lot of geodesics connectiong two points. Now, your metric is the intrisic metric induced by on the hypersurface \Sigma described by |x-y|=1. It often happens that one of the geodesic in (\mathbb R^4,\|{*}\|) lies in \Sigma. For example the geodesic between segments [(0,0),(0,1)] and [(1,0),(0,0)]. Maybe this is a closed geodesic, but at least it is a closed line with 4 geodesic segments. Maybe it already makes you happy already. (?). If not, you have a Finsler metric on \Sigma=\mathbb S^1\times \mathbb R^2. The unit ball for some coordiantes (u,v,w) of the tangent plane can be described by the following inequality:$$\sqrt{u^2+w^2}+\sqrt{v^2+w^2}\le 2.$$This ball is strongly convex. The metric is smooth (sinse there is a transitive isometric group action on \mathbb S^1\times \mathbb R^2). It reamins to write differential equasion for geodesic; this should be in any book on Finsler geometry. 3 added 254 characters in body Let us start with metric on \mathbb R^4=\mathbb R^2\times \mathbb R^2. Defined by norm \|{*}\| defined by$$\|(x,y)\|=\tfrac12(|x|+|y|),$$where |{ * }| denotes Euclidean norm. This norm, is not strictly convex and generically you have a lot of geodesics connectiong two points. Now, your metric is the intrisic metric induced by on the hypersurface \Sigma described by |x-y|=1. So It often happens that one of the geodesic in (\mathbb R^4,\|{*}\|) lies in \Sigma. For example the geodesic between segments [(0,0),(0,1)] and [(1,0),(0,0)]. Maybe it is makes you happy already. If not, you have a Finsler metric on \mathbb \Sigma=\mathbb S^1\times \mathbb R^2. The unit ball for some coordiantes (u,v,w) of the tangent plane can be described by the following inequality:$$\sqrt{u^2+w^2}+\sqrt{v^2+w^2}\le 2.$$This ball is strongly convex. The metric is smooth (sinse there is a transitive isometric group action on$\mathbb S^1\times \mathbb R^2\$). It reamins to write differential equasion for geodesic; this should be in any book on Finsler geometry.
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# Finding the $y$ value of an interpolated function given an $x$ value
I am trying to interpolate my dataset, which is a list of points. Thus I have two interpolated functions of my x data and y data:
datax={19.668, 18.5235, 17.581, 16.7985, 16.1445, 15.595, 15.1315, 14.7395,
14.4075, 14.1265, 13.8885, 13.6875, 13.518, 13.3765, 13.2585, 13.162,
13.083, 13.0205, 12.9715, 12.935, 12.909, 12.892, 12.8835, 12.8815,
12.885, 12.8935, 12.906, 12.9215, 12.9395, 12.9595, 12.98, 13.001,
13.022, 13.0425, 13.0615, 13.0785, 13.094, 13.1065, 13.1165, 13.1235,
13.1265, 13.126, 13.122, 13.1135, 13.1005, 13.0835, 13.062, 13.0355,
13.005, 12.97, 12.9305, 12.8865, 12.838, 12.786, 12.7295, 12.6695,
12.606, 12.539, 12.469, 12.396, 12.32, 12.242, 12.1615, 12.079,
11.995, 11.9095, 11.823, 11.7355, 11.647, 11.558, 11.469, 11.38,
11.291, 11.2025, 11.1145, 11.0275, 10.9415, 10.8565, 10.7725, 10.69,
10.609, 10.53, 10.453, 10.3775, 10.304, 10.233, 10.164}
datay={0.161158, 0.17188, 0.183499, 0.196057, 0.209601, 0.224179, 0.239842,
0.256644, 0.274639, 0.293882, 0.314433, 0.336349, 0.359688, 0.384507,
0.410862, 0.43881, 0.468401, 0.499684, 0.532704, 0.567499, 0.604103,
0.642541, 0.682831, 0.724979, 0.768983, 0.814829, 0.862489, 0.911922,
0.963073, 1.01587, 1.07023, 1.12604, 1.1832, 1.24156, 1.30096,
1.36126, 1.42226, 1.48377, 1.54559, 1.6075, 1.66927, 1.73069,
1.79152, 1.85153, 1.91048, 1.96817, 2.02436, 2.07887, 2.13148,
2.18203, 2.23035, 2.27629, 2.31974, 2.36058, 2.39873, 2.43413,
2.46673, 2.49651, 2.52347, 2.54761, 2.56897, 2.58759, 2.60354,
2.61689, 2.62772, 2.63613, 2.64222, 2.64609, 2.64786, 2.64765,
2.64558, 2.64177, 2.63633, 2.6294, 2.62109, 2.61152, 2.6008, 2.58906,
2.57639, 2.5629, 2.5487, 2.53389, 2.51855, 2.50278, 2.48667, 2.47028,
2.45371}
fx=Interpolation[datax]
fy=Interpolation[datay]
I can then plot these together using ParametricPlot:
ParametricPlot[{fx[x], fy[x]}, Evaluate@{x,Sequence @@ First@Interpolation[datax]["Domain"]}]
And it gives me a nice interpolated version of my original data.
My issue is, now I need to find a specific y-value corresponding to x_o=1.4 on this parametric plot, which I've found to be impossible (?).
So what I'm doing is the following:
First, plot each interpolated function separately:
Plot[{fx[x], fy[x]}, Evaluate@{x,
Sequence @@ First@Interpolation[datax]["Domain"]}]
Which gives me my two interpolated functions:
Next, I Find the X-value of my fy interpolated function:
xroot = x /. FindRoot[fy[x] == 1.4, {x, 35}][[1]]
(*36.6362*)
And now I need to find the corresponding Y-value of my fx interpolated function.
My issue is, I have no idea how to find a y-value given an x-value! I have tried:
FindRoot[InverseFunction[fx[x]] == sol, {x, 3}]
But I get the error:
"The function value {-36.6362+InverseFunction[17.581]} is not a list of numbers with dimensions {1} at {x} = {3.}"
Does anyone know how to find a y-value given x_o for an interpolated data set? or is there an easier way of doing this??
Thanks so much for your time!
• parts of your code don't evaluate, ParametricPlot[{datax,datay},{x,0,16.5}], other parts are undefined fm, fr – Jason B. Jun 6 '18 at 18:48
• Sorry! My bad. I just fixed it so it runs :) – zack Jun 6 '18 at 18:55
• @zack your code still does not run. You should try to copy your code from this question and execute it in a clean MMA kernel to make sure that you provided us enough information to run it. – MarcoB Jun 6 '18 at 19:03
• This is pretty unclear. Are your data points specified by $(x_i,y_i)$ pairs, which you separated into the two sets datax and datay? If so, then $x_0 = 0.14$ is outside of your domain and I don't think you can calculate the corresponding $y_0$ with the information you have. If, on the other hand, you want to calculate the $x_0$ for which $y=0.14$, then try InverseFunction[Interpolation@Transpose@{datax, datay}][1.4], which returns 13.1814. – MarcoB Jun 6 '18 at 19:09
• Looks like you have fx and fy switched. – Carl Woll Jun 6 '18 at 19:17
The following is how you could use InverseFunction to find the $x$ value for a given $y$:
ify = InverseFunction[fy];
fx[ify[1.4]]
13.0886
Of course, you have no control over which $y$ value is used.
Or finding the $y$ value for a given $x$:
ifx = InverseFunction[fx];
fy[ifx[15]]
`
0.245044
• Thank you so much Carl! I've been trying to get this to work all day and your simple solution worked wonders! Your first line gave me the exact answer I need. – zack Jun 6 '18 at 19:48
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# 2. PANIC Quick-Look Tool (PQL)¶
## 2.1. Purpose¶
PANIC Quick-Look (hereafter PQL) performs some on-line data processing for quick-look or quality check of the data being acquired, taking a close look at a raw near-infrared image and getting a quick feedback of the running observation.
PQL is an application with a graphical user interface which monitors the GEIRS data output, waiting for new FITS files coming from GEIRS. When a new file is detected, it is added to the file list view in the main panel, and then PQL will perform the task previously specified by the user in the setup configuration. Some of the available tasks are:
• Only display the FITS image with no processing
• Dark subtraction, flat division
• Sky subtraction (using N-nearest frames or own sky )
• Field distortion removal
• Image align and stacking
• Preliminary astrometric solution
• Preliminary photometry
In addition, PQL allows you to execute manually in an interactive way some tasks with the data. For example, you will be able to select a file, compute some statistics values (background, FWHM, min, max, …) or ask for the sky subtraction looking for the nearest N frames around the selected one. Other option available is to select a set of files and request to shift and align them.
PQL can be operated in both near-real time mode (during the observation) and offline mode (after the observation, with all data files already stored in the disk); however, its functionalities have been provided mainly in near-real time to check the status and progress of the observation during the night.
The visualization application used to display the images is SAOImage ds9, which supports FITS images, multiple frame buffers, region manipulation, and many scale algorithms and colormaps.
## 2.2. FITS files and headers¶
PQL only supports FITS (Flexible Image Transport System) with two-dimensional image formats. Because PANIC has a FPA of four detector, the FITS files can be Single Extension FITS (SEF) or Multi-Extension FITS (MEF), however MEF are prefered.
The complete definition of the FITS headers can be found on the GEIRS documentation.
For general purpose, such as viewing and simple analysis, only minimal headers keywords are required. However, and in order to group and reduce observing sequences, the following header keywords are also required:
OBS_TOOL= 'OT_V1.1 ' / PANIC Observing Tool Software version
PROG_ID = ' ' / PANIC Observing Program ID
OB_ID = '6 ' / PANIC Observing Block ID
OB_NAME = 'OB CU Cnc Ks 2' / PANIC Observing Block Name
OB_PAT = '5-point ' / PANIC Observing Block Pattern Type
PAT_NAME= 'OS Ks 2 ' / PANIC Observing Secuence Pattern Name
PAT_EXPN= 1 / PANIC Pattern exposition number
PAT_NEXP= 5 / PANIC Pattern total number of expositions
IMAGETYP= 'SCIENCE ' / PANIC Image type
These keywords are automatically added to the FITS header by the PANIC Observation Tool, as each file is created. If these are not saved, PQL will not work correctly.
## 2.3. Starting PQL¶
To start PQL GUI, you can lauch it from the PANIC computer (panic22/panic35) once you are logged as obs22/obs35 user. Thus, as any one of the workstations of the observing room, open a X terminal window and log into the PANIC computer as follow:
for 2.2m:
$ssh -X obs22@panic22 (ask Calar Alto staff for password) for 3.5m: $ ssh -X obs35@panic35
Once you are logged into the PANIC computer, to launch PQL GUI type next command:
$start_ql & The next figure shows a snapshot of the main window of PQL GUI that will bring up the start_ql command. ## 2.4. Configuration files¶ The configuration files used by PQL are located in the$PAPI_HOME/config_files. The main config file is the same file used by PAPI, ie., $PAPI_CONFIG, and usually called papi.cfg. This file includes a lot of parameters used by PAPI, and therefore by PQL during the processing; however at the end of the$PAPI_CONFIG file there is section called quicklook, where the user can set some specific parameters for PQL:
##############################################################################
[quicklook]
##############################################################################
# Next are some configurable options for the PANIC Quick Look tool
#
# some important directories
#
source = /data1/PANIC/
output_dir = /data2/out # the directory to which the resulting images will be saved.
temp_dir = /data2/tmp # the directory to which temporal results will be saved
verbose = True
# Run parameters
run_mode = Lazy # default (initial) run mode of the QL; it can be (None, Lazy, Prereduce)
Although the user can edit these values in the config file, some of them can be set easily on PQL’s GUI.
### 2.8.8. How do I report a issue ?¶
Please submit issues with the issue tracker on github.
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# How to find close match of levenshtein distance [duplicate]
Let's say we have some function,
float distance(String a, String b) { /*...*/ }
to find the Levenshtein distance between two strings, and we have a large array of strings.
String[] strings = new String[1_000_000];
What would be an acceptable approach to map a String a to its closest match (i.e., the String b that produces the lowest result of distance(a, b)) in that array? This will need to be done often, and so going through the entire array every time would not make sense.
Of course, strings need not be an array.
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# How can I set multiple values to local variables in a Module?
Multiple assignment is supported by Set. How can do something similar in Module expressions? If I could do it, whould there be drawbacks?
Here is what I tried so far:
g[x11_, y11_] := {xx111, yy111} = {x11, y11}
g[x11_, y11_] := Module[{{x111, y111} = {x11, y11}}, 1]
g[1, 2]
Module::lvset: Local variable specification {{x111, y111} = {1, 2}} contains {x111, y111} = {1, 2}, which is an assignment to {x111,y111}; only assignments to symbols are allowed. >>
Possible alternatives are:
ggg[x11_, y11_] :=
Module[{xx = {x11, y11}, x1, x2}, {x1 = xx[[2]], x2 = xx[[1]]}]
or
ggg[x11_, y11_] := Module[{x1111=x11, y1111=y11}, what... ]
-
"only assignments to symbols are allowed" - that's right, the scoping constructs don't allow parallel assignment in the variable list. – J. M. is back. May 24 '13 at 3:23
In my answer to a similar question about With on SO, I posted a macro which can do this. You can just replace With with Module there, if that kind of solution fits you. – Leonid Shifrin May 24 '13 at 11:14
## 1 Answer
As J.M. points out in his comment, the kind of multiple assignment you want is simply not supported. The closest you can get (and it is perfectly serviceable in my opinion) is
g[x11_, y11_] := Module[{x111, y111},
{x111, y111} = {x11, y11};
1]
-
en, seems declare variables and then assign separately. Anyway, this is not big problem, I just take some habits from one function elsewhere. – HyperGroups May 24 '13 at 4:07
@Hyper, yes, that's precisely what m_goldberg does. That's how it is... – J. M. is back. May 24 '13 at 8:24
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# 7x-28=2x^2-8x-3x+12
## Simple and best practice solution for 7x-28=2x^2-8x-3x+12 equation. Check how easy it is, and learn it for the future. Our solution is simple, and easy to understand, so dont hesitate to use it as a solution of your homework.
If it's not what You are looking for type in the equation solver your own equation and let us solve it.
## Solution for 7x-28=2x^2-8x-3x+12 equation:
Simplifying
7x + -28 = 2x2 + -8x + -3x + 12
Reorder the terms:
-28 + 7x = 2x2 + -8x + -3x + 12
Reorder the terms:
-28 + 7x = 12 + -8x + -3x + 2x2
Combine like terms: -8x + -3x = -11x
-28 + 7x = 12 + -11x + 2x2
Solving
-28 + 7x = 12 + -11x + 2x2
Solving for variable 'x'.
Reorder the terms:
-28 + -12 + 7x + 11x + -2x2 = 12 + -11x + 2x2 + -12 + 11x + -2x2
Combine like terms: -28 + -12 = -40
-40 + 7x + 11x + -2x2 = 12 + -11x + 2x2 + -12 + 11x + -2x2
Combine like terms: 7x + 11x = 18x
-40 + 18x + -2x2 = 12 + -11x + 2x2 + -12 + 11x + -2x2
Reorder the terms:
-40 + 18x + -2x2 = 12 + -12 + -11x + 11x + 2x2 + -2x2
Combine like terms: 12 + -12 = 0
-40 + 18x + -2x2 = 0 + -11x + 11x + 2x2 + -2x2
-40 + 18x + -2x2 = -11x + 11x + 2x2 + -2x2
Combine like terms: -11x + 11x = 0
-40 + 18x + -2x2 = 0 + 2x2 + -2x2
-40 + 18x + -2x2 = 2x2 + -2x2
Combine like terms: 2x2 + -2x2 = 0
-40 + 18x + -2x2 = 0
Factor out the Greatest Common Factor (GCF), '2'.
2(-20 + 9x + -1x2) = 0
Factor a trinomial.
2((-5 + x)(4 + -1x)) = 0
Ignore the factor 2.
Subproblem 1Set the factor '(-5 + x)' equal to zero and attempt to solve:
Simplifying
-5 + x = 0
Solving
-5 + x = 0
Move all terms containing x to the left, all other terms to the right.
Add '5' to each side of the equation.
-5 + 5 + x = 0 + 5
Combine like terms: -5 + 5 = 0
0 + x = 0 + 5
x = 0 + 5
Combine like terms: 0 + 5 = 5
x = 5
Simplifying
x = 5
Subproblem 2Set the factor '(4 + -1x)' equal to zero and attempt to solve:
Simplifying
4 + -1x = 0
Solving
4 + -1x = 0
Move all terms containing x to the left, all other terms to the right.
Add '-4' to each side of the equation.
4 + -4 + -1x = 0 + -4
Combine like terms: 4 + -4 = 0
0 + -1x = 0 + -4
-1x = 0 + -4
Combine like terms: 0 + -4 = -4
-1x = -4
Divide each side by '-1'.
x = 4
Simplifying
x = 4Solutionx = {5, 4}`
|
# How do you measure the frequency of a sine wave?
3
Date created: Thu, Aug 12, 2021 7:35 PM
Date updated: Thu, May 19, 2022 1:19 AM
Content
Video answer: Find the frequency of a sine wave
## Top best answers to the question «How do you measure the frequency of a sine wave»
To measure the frequence of a sine wave, convert to square and then use the CAPTURE mode of the CCP to measure the time between two zero crossing. 1/T will give you the frequency.
#### What is the formula for a sine wave?
• The Sine wave is the graph that is formed if the function contains a sine function. The General Formula of the Sine wave is: y=AsinB(x-C)+D where x is the angle or theta.
The equation of a basic sine function is f(x)=sinx. In this case b, the frequency, is equal to 1 which means one cycle occurs in 2π. If b=12, the period is 2π12 which means the period is 4π and the graph is stretched.
FAQ
Those who are looking for an answer to the question «How do you measure the frequency of a sine wave?» often ask the following questions:
### 👋 How to measure frequency of a sine wave?
The equation of a basic sine function is f(x)=sinx. In this case b, the frequency, is equal to 1 which means one cycle occurs in 2π. If b=12, the period is 2π12 which means the period is 4π and the graph is stretched.
### 👋 How to measure brain wave frequency?
The EEG (electroencephalograph) measures brainwaves of different frequencies within the brain. Electrodes are placed on specific sites on the scalp to detect and record the electrical impulses within the brain. A frequency is the number of times a wave repeats itself within a second.
### 👋 How to measure sound wave frequency?
Sound wave frequencies can be measured with a frequency counter or with a spectrum analyzer. These devices work by using a microphone to convert the sound wave into an electrical signal. The peaks and valleys of wave are counted to find the frequency.
Video answer: The sine wave explained (ac waveform analysis)
We've handpicked 24 related questions for you, similar to «How do you measure the frequency of a sine wave?» so you can surely find the answer!
How to calculate frequency in sine wave?
• How do you calculate the frequency of a sine wave? The formula for frequency is: f (frequency) = 1 / T (period). f = c / λ = wave speed c (m/s) / wavelength λ (m). The formula for time is: T (period) = 1 / f (frequency). Is a sine wave a function?
How to find frequency from sine wave?
#### What is the formula for a sine wave?
• The Sine wave is the graph that is formed if the function contains a sine function. The General Formula of the Sine wave is: y=AsinB(x-C)+D where x is the angle or theta.
How to generate high frequency sine wave?
#### What is the function of high frequency waveform generator?
• High frequency waveform generator is very useful in electronic experiment and design. This circuit generate sine wave oscillation, but actually we can modify the circuit to generate triangle or square wave function.
How to graph sine wave using frequency?
#### How do you calculate sine wave?
• In general, a sine wave is given by the formula In this formula the frequency is w. Frequency used to be measured in cycles per second, but now we use the unit of frequency - the Hertz (abbreviated Hz). One Hertz (1Hz) is equal to one cycle per second.
What is frequency of a sine wave?
The frequency of a sine wave is the number of complete cycles that happen every second. (A cycle is the same as the period, see below.) In the bouncing weight above, the frequency is about one cycle per second… One Hertz (1Hz) is equal to one cycle per second.
### Video answer: Online tutorial on how to calculate instantaneous voltage of a sine wave source
How do u measure frequency of a wave?
Wave frequency can be measured by counting the number of crests. The higher the number is, the greater the frequency of the waves. The SI unit for wave frequency is the hertz (Hz), where 1 hertz equals 1 wave passing afixed point in 1 second.
How do we measure frequency of a wave?
#### What unit is used to measure frequency?
• For Radio frequency you will often also see the measurements: microWatts per square meter (µW/m2) microWatts per square centimeter (µW/cm2) milliWatts per square meter (mWm²)
### Video answer: Calculating frequency, periodic time and time to first peak value for a sinusoidal function
How do you measure frequency in a wave?
Wave frequency can be measured by counting the number of crests or compressions that pass the point in 1 second or other time period. The higher the number is, the greater is the frequency of the wave. The SI unit for wave frequency is the hertz (Hz), where 1 hertz equals 1 wave passing a fixed point in 1 second.
How to measure frequency of a water wave?
Divide the wavelength into the velocity to calculate the frequency, expressed as described above as the number of cycles per second, or Hertz – written "Hz." For example, a water wave with a wavelength of 1 foot traveling at a speed of 4 inches per second has a frequency of 1/3 feet/second divided by 1 foot = . 33 Hz.
How to measure frequency of a wave formula?
#### Frequency of a wave is given by the equations:
1. f=1T. where: f is the frequency of the wave in hertz. T is the period of the wave in seconds.
2. f=vλ where: f is the frequency of the wave in hertz. v is the velocity of the wave in meters per second. λ is the wavelength of the wave in meters…
3. f=cλ Related topic.
### Video answer: Terms and calculations used for sine waves (intro to alternating current)
What is the wave frequency a measure of?
Waves. Frequency is a measurement of how often a recurring event such as a wave occurs in a measured amount of time. One completion of the repeating pattern is called a cycle. Only moving waves which vary their positions with respect to time possess frequency.
What unit is used to measure wave frequency?
Wavelength is also measured in metres ( ) - it is a length after all. The frequency ( ) of a wave is the number of waves passing a point in a certain time. We normally use a time of one second, so this gives frequency the unit hertz ( ), since one hertz is equal to one wave per second.
How to find subspace of sine wave frequency?
#### How to calculate the frequency of a sine wave?
• T = λ / Wave Speed Frequency = Wave Speed / λ Where, T = Sine Wave Period λ = Wavelength Example A sine wave with 5m wavelength and 8ms -1 speed will have a frequency of
What is the frequency in a sine wave?
#### How many frequencies are there in a sine wave?
• Where a sine wave only has one frequency component - its fundamental - the square wave has the fundamental and harmonics at whole odd-number multiples of the fundamental frequency. So in a 100 hertz square wave you hear frequency components of 100 hertz, 300 hertz, 500 hertz, 700 hertz and so on all the way up the frequency band.
Which is the frequency of a sine wave?
• The frequency a sine wave is the number of times the wave repeats within a single unit of the input variable \u0012; this is the reciprocal of the period. Thus the frequency of the standard sine wave sin(x) is 1 2ˇ and so the frequency of f(\u0012) = asin(b(\u0012 c)) + dis jbj 2ˇ : Electronic transmissions involve the sine wave.
Which is the sine wave frequency in matlab?
• in sine function in MATLAB it is always sin(wt). here frequency w is in radian/sec not f (in HZ) so w will give you the no.of the cycle. if you want to use the sin(2*pi*60*t) you can use the sind(2*pi*9.545*t).
How to measure the rms of a sine wave?
• You can easily measure the RMS value of a sine wave + offset using a cheap multimeter. Most cheap (non-RMS-reading) multimeters are AC-coupled on the AC voltage ranges, so you would read the correct RMS value of the ripple only on an AC range, assuming the ripple is sinusoidal, due to the correction factor which is applied.
How can we measure the frequency of a wave?
#### What is the formula for finding the frequency of a wave?
• In physics, the frequency of a wave is the number of wave crests that pass a point in one second (A wave crest is the peak of the wave). Hertz (symbol Hz) is the unit of frequency. The relationship between Frequency and wavelength is expressed by the formula: f = v / λ {\\displaystyle f=v/\\lambda }.
### Video answer: Wave period and frequency
How can you measure the frequency of a wave?
Wave frequency can be measured by counting the number of crests or compressions that pass the point in 1 second or other time period. The higher the number is, the greater is the frequency of the wave. The SI unit for wave frequency is the hertz (Hz), where 1 hertz equals 1 wave passing a fixed point in 1 second.
How do you measure frequency of a sound wave?
• Sound wave frequencies can be measured with a frequency counter or with a spectrum analyzer. These devices work by using a microphone to convert the sound wave into an electrical signal. The peaks and valleys of wave are counted to find the frequency. There are inexpensive smartphone apps designed for measuring sound frequencies.
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# Exponents and Exponential Functions
## Objective
Write equivalent radical and rational exponent expressions. Identify quantities as rational or irrational.
## Common Core Standards
### Core Standards
?
• N.RN.A.2 — Rewrite expressions involving radicals and rational exponents using the properties of exponents.
?
• 8.EE.A.1
• 8.EE.A.2
• 8.NS.A.1
## Criteria for Success
?
1. Convert between radicals and rational exponents fluently.
2. Apply the properties of exponents to convert between radicals and rational exponents.
3. Determine if a given number or expression is rational or irrational.
## Anchor Problems
?
### Problem 1
For the problems below, determine whether each equation is True or False.
Expression True? False? a. ${\sqrt{32}=2^{5\over2}}$ b. ${16^{3\over2}=8^2}$ c. ${4^{1\over2}=\sqrt[4]{64}}$ d. ${2^8=(\sqrt[3]{16})^6}$ e. ${(\sqrt{64})^{1\over3}=8^{1\over6}}$
#### References
Smarter Balanced Assessment Consortium: Item and Task Specifications MAT.HS.SR.1.00NRN.A.152
MAT.HS.SR.1.00NRN.A.152 from Development and Design: Item and Task Specifications made available by Smarter Balanced Assessment Consortium. © The Regents of the University of California – Smarter Balanced Assessment Consortium. Accessed May 17, 2018, 11:29 a.m..
### Problem 2
In each of the following problems, a number is given. If possible, determine whether the given number is rational or irrational. In some cases, it may be impossible to determine whether the given number is rational or irrational. Justify your answers.
a. ${4+\sqrt7}$
b. ${\sqrt{45}\over\sqrt{5}}$
c. ${6\over \pi}$
d. ${\sqrt2 + \sqrt3}$
e. ${{2+\sqrt{7}}\over{2a+\sqrt{7a^2}}}$, where $a$ is a positive integer
f. ${x+y}$, where $x$ and $y$ are irrational numbers
#### References
Illustrative Mathematics Rational or Irrational?
Rational or Irrational?, accessed on May 17, 2018, 11:34 a.m., is licensed by Illustrative Mathematics under either the CC BY 4.0 or CC BY-NC-SA 4.0. For further information, contact Illustrative Mathematics.
## Problem Set
?
The following resources include problems and activities aligned to the objective of the lesson that can be used to create your own problem set.
?
Provide a written explanation for each question below.
a. Is it true that ${\left(1000^{1\over3}\right)^3=(1000^3)^{1\over3}}$? Explain or show how you know.
b. Is it true that ${\left(4^{1\over2}\right)^3=(4^3)^{1\over2}}$? Explain or show how you know.
c. Suppose that $m$ and $n$ are positive integers and $b$ is a real number so that the principal $n^{th}$ root of $b$ exists. In general, does $\left(b^{1\over n}\right)^m=(b^m)^{1\over n}$? Explain or show how you know.
#### References
EngageNY Mathematics Algebra II > Module 3 > Topic A > Lesson 3Exit Ticket, Question #3
Algebra II > Module 3 > Topic A > Lesson 3 of the New York State Common Core Mathematics Curriculum from EngageNY and Great Minds. © 2015 Great Minds. Licensed by EngageNY of the New York State Education Department under the CC BY-NC-SA 3.0 US license. Accessed Dec. 2, 2016, 5:15 p.m..
Modified by Fishtank Learning, Inc.
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# Find hinge point or axis of rotation point?
I have a problem at hand where I need to detect/predict the coordinates of the hinge point or axis of rotation point using image processing. The image is as shown below:
I've used a method where I started with tracking the circular movement (in an arc) of a few feature points in an RoI around the default hinge coordinates (entered manually) in a configuration file. This circular motion of these tracked points happens around the vertical axis which passes through the hinge point. Now, I tracked these points from their initial position until the connecting bar made a particular angle (15°/20°) with the y-axis, I drew secants between these different positions (start and end positions) of the same point and drew its perpendicular bisector, which will ideally pass through the centre of the (concentric) circles, which is the ideal hinge point.
Now, the issues we faced in this ideal scenario to practical working of it is:
Different tracked points give different potential hinge points, (few of which are very close the ground truth), but their weighted/average goes off the mark. Quite frankly, this is a problem of too many in which we do get the closest potentially to ground truth, but we’re not sure, which of these points is the closest as we’re not to use the default connecting coordinates (entered manually) from config file. Eg:
y_intercepts calculated for each point
H0 (322, 42)
H1 (322, 64) (within tolerance, closest to GT)
H2 (322, 48)
H_avg (322,52)
H_groundtruth (x,y): (322, 61)
Update: Added part of original image for more clarity. Dots on vertical yellow line = Different hinge points given by different feature points tracked in an arc Green circle: Predicted hinge point from avg of the above different hinge points Yellow circle: Ground truth hinge point
edit retag close merge delete
1
If you could fix a marker (a chessboard, an Aruco marker...) you could find the relative position and orientation of the marker and calculate the center of rotation.
( 2019-07-10 07:20:23 -0500 )edit
1
Can you share some real images with us? Will the camera and object be fixed other than the rotation at the hinge?
Will the lighting be consistent?
( 2019-07-10 08:50:36 -0500 )edit
@Witek: We're not allowed to use markers. @Chris: We're not allowed to share actual images. However, this drawn image is similar to what we'll have in real. Also, the lighting condition will be consistent during one iteration. The camera is a moving camera, however in it's frame of reference, the hinge will be fixed relatively and the object on top will only move to left or right.
( 2019-07-11 01:36:23 -0500 )edit
1
I am not sure if my understanding of the problem is correct, but if you can track a number of feature points along the hinge rotation, you can later use the Hough Circle transform to fit the circles and find their center(s). This is similar to what you are doing now, but using more data points should result in lesser noise. Eventually you would have to find a sufficiently small cluster of the centers and get the average center from those points only.
Is this idea any useful?
( 2019-07-11 07:17:14 -0500 )edit
I tried fitting Hough circles but it's computationally expensive as the radius as well as centre of these circles is unknown. Plus we have only upto 15-20 degrees of rotation to track these points and get the centre of circle.
( 2019-07-20 07:04:23 -0500 )edit
Right, with just a short arc Hough might provide inaccurate results. How about fitting a circle to these points? https://stackoverflow.com/questions/4...
( 2019-07-20 08:25:50 -0500 )edit
@muglikar. Can you post actual image?
( 2019-07-20 08:28:58 -0500 )edit
@Witek. Nope. That doesn't help. He wants angle of axis.
( 2019-07-20 08:30:56 -0500 )edit
1
Actual image is up in the question. It is quite low res though. He wants center of rotation, so precise fitting of circles might work, but @muglikar's method is generally correct, but yields very different results. I am wondering if this might be due to low resolution and errors in point tracking and pixel quantization.
I have another idea but I would need to see three images at left, center and right position of the object. Possibly at higher resolution.
( 2019-07-20 13:48:02 -0500 )edit
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# Computers
- Electronic device for storing and processing data. Software, Hardware, Web & Internet
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## Introduction
Humans have the ability to modify their motor plan in response to changes in the walking environment, a phenomenon referred to as locomotor adaptation1. The study of locomotor adaption has been pursued to improve our understanding of the organization of human locomotor circuits, given the inapplicability, in humans, of the invasive techniques employed in animal models. As a case in point, studies in the lamprey, frog, turtle and cat have consistently shown that locomotor control is accomplished by rhythmic activations of lower-level spinal circuits that regulate the activities of fixed neuromuscular modules2,3,4,5,6. It has been debated if such modules exist in humans7, and if they encode the relative levels of activation across muscles8, the temporal relations among the activations of muscle groups9, or involve more complex structures10,11. A study on adaptation during forward/backward walking on a split-belt treadmill has provided evidence that human locomotion is controlled by leg-specific and forward/backward-specific independent networks12. These observations are compatible with a two-level organization of spinal central pattern generator (CPG) circuits where the higher level regulates the timing of activation of different muscle groups, while the lower level encodes the relative contribution of the different muscles to the task.
At the functional level, herein intended as the set of possible control strategies employed by the CNS, locomotor adaptation is generated by feedforward mechanisms20, although the response to the perturbation that causes the adaptive response may be marked also by feedback strategies16,21,22 that may depend on changes in task demands23,24. Feedforward, predictive, mechanisms are associated with a gradual response as the CNS generates a model of the effect of the perturbation. Feedback mechanisms are associated with reactive, faster responses24. How these mechanisms are triggered and combined during gait is yet a source of debate. At the neurophysiological level, adaptation has been argued to be controlled via either leg-independent12 or bilaterally-linked25 mechanisms. It is reasonable to expect that, as observed at the behavioral level, predictive/adaptive (i.e., feedforward) and reactive (i.e., feedback) responses manifested at the functional and neurophysiological levels would also be context-dependent. Understanding how these processes contribute differently to generating responses to perturbations is particularly relevant to robot-assisted gait rehabilitation, in which robot-driven forces or modifications in the walking environment are used to trigger responses aimed to achieve a physiological gait pattern. From this perspective, a dissection of how the feedforward and feedback components of automatic responses to perturbations are triggered and combined could inform the best use of devices for gait rehabilitation. As feedforward and feedback control strategies have been observed to contribute simultaneously to locomotor adaptation21,22, their relative input can be ideally dissected by comparing the response processes to perturbations that are expected to yield different contributions from these two mechanisms. This can be done by testing perturbations that have different biomechanical effects but that are introduced in the same general experimental paradigm.
In line with this overall objective, the aim of this study is to shed light on the mechanisms, at the functional and neurophysiological levels, of locomotor adaptation to robot-induced forces altering the gait pattern of humans. We used a robot-based adaptation paradigm19 and the analysis of muscle synergies26 to investigate the relative contributions of predictive/adaptive (i.e., feedforward) vs. reactive (i.e., feedback) locomotor control strategies. We also investigated the modular organization of the muscle synergies and assessed if their control during locomotor adaptation is unilateral or bilaterally-linked25. We performed two motor adaptation experiments in healthy individuals by introducing a unilateral mechanical perturbation that resulted in either an increase (X experiment) or a decrease (Xinv experiment) in step length. We previously observed that both perturbations trigger an adaptive response19. However, given the opposite biomechanical effects of the perturbations, the two related adaptation processes are expected to be obtained through different contributions of the feedforward and feedback control mechanisms. The analyses performed in this study allowed us to separate the feedforward and feedback contributions to the adaptation to both perturbations, as, respectively, slow and progressive or fast and reactive modifications in the neuromuscular controls.
The results showed, conclusively, that not only feedforward and feedback contributions can be achieved through simultaneous access to the same muscle synergies22, but also that they contribute to adaptation together and differently depending on the biomechanical context. Adding to the rich literature analyzing the characteristics of muscle synergies during locomotion27, our results provide the strongest demonstration to date that the neuromuscular adaptive control of human locomotion can be effectively described as a context-dependent sum of feedback and feedforward strategies that simultaneously shape the activation of invariant motor primitives. These results strongly suggest the presence, in humans, of a defined population of spinal interneurons regulating muscle coordination that can be accessed by both cortical and afferent drives, as observed in animal models28. The observations derived from this work could be used to develop new approaches to the design of robot-assisted gait rehabilitation procedures targeting specific descending- and/or afferent-driven responses in muscle synergies29.
## Results
Subjects walked with the lower limbs strapped to the robotic legs of a system for gait rehabilitation (Lokomat by Hocoma AG, Fig. 1A). The system was used to implement motor adaptation experiments based on a unilateral perturbation to the subjects’ right leg resulting in either an increase (X experiment) or a decrease (Xinv experiment) in step length19. Figure 1D shows a schematic representation of the experiments. Each experiment consisted of 420 gait cycles. The experiment started with a habituation phase (consisting of 180 gait cycles) followed by the Baseline (BL), Perturbation (Pert) and After-Effect (AE) phases of the experiment, each consisting of 80 gait cycles. During the habituation phase, the system was programmed to be transparent to the subject (Back-Driven Mode - BDM) except for 9 randomly-selected gait cycles during which the subject experienced a perturbation (Force-Field Mode - FFM). During the BL and AE phases, the robot was also programmed in BDM. During the Pert phase, the system was programmed in FFM and produced one of the two types of perturbation (i.e., X or Xinv) for all the gait cycles.
### Lower-limb adaptation to two opposite perturbations
At the onset of the X perturbation, all subjects showed an immediate step-length increase in the right, perturbed, side of 25.9 ± 6.7% (mean ± standard error) that was statistically significant (Friedman test followed by post-hoc Minimum Significance Test) (Figs. 2A and S1A). As we previously observed19, this change was followed by an adaptive response to the perturbation as indicated by a gradual restoration of BL step length in the perturbed side with a residual positive deviation (at the end of the Pert phase) of 5.5 ± 4.8%. This residual deviation from BL step length was statistically different from the deviation observed at the beginning of the Pert phase, but not significantly different from the BL step length, indicating that subjects were able to restore their baseline step length. Full adaptation to the X perturbation, estimated as 3 times the time constant of the exponential function fitting the average step-length time-course, was found to require 13.8 gait cycles (with the 95% confidence interval ranging from 10.6 to 19.9 gait cycles). At the beginning of the AE phase, when the perturbation was removed, a significant aftereffect, manifested as a 29.4 ± 6.7% decrease in step length in the perturbed side, was observed. All subjects then gradually returned to their BL step length.
Similar results were observed during the Xinv experiments, but opposite in the direction of step-length change (Figs. 2B and S1B). The initial response to the Xinv perturbation was marked by a significant decrease in step length in the perturbed side (41.2 ± 3.8%). At the end of the Pert phase, a residual step-length decrease of 9.7 ± 3.7% was observed in the perturbed side that was statistically different from the step length at the beginning of the Pert phase, but not from the BL step length, indicating, again, a return to the baseline step length. Full adaptation was achieved in 9.3 gait cycles (with the 95% confidence interval ranging from 7.2 to 13.3 gait cycles). A significant step-length increase in the perturbed side (28.2 ± 3.8%) was observed upon removal of the perturbation. All subjects were then able to return to their BL step length. No significant change was observed in the step length of the unperturbed left leg during both experiments (Figs. 2A,B and S1A,B). The time-course of adaptation for both experiments was in line with what we observed in previous experiments based on the same setup and perturbations19.
### Four muscle synergies reconstructed baseline electromyographic data
Muscle synergies were extracted both unilaterally (i.e., using data collected from one leg at the time) and bilaterally (i.e., using data collected from both legs). The bilateral analysis was employed to highlight the presence or absence of adaptation processes that are bilaterally-linked.
Four muscle synergies were sufficient to reconstruct the amplitude modulation of the electromyographic (EMG) data during the BL phase with a R2 > 0.75 in both the unilateral and bilateral analyses (R2 = 0.80 ± 0.02 for the bilateral analysis; R2 = 0.86 ± 0.06 for the unilateral left leg analysis; R2 = 0.79 ± 0.04 for the unilateral right leg analysis). The muscle synergies and their corresponding activation patterns identified from the EMG recordings collected during the BL phase of both experiments (Fig. 3) were used as the reference synergy patterns for the X and Xinv experiments, and are herein referred to as REFsynX and REFsynXinv, respectively. The results of the unilateral analyses (Fig. 3A) are consistent in their muscular compositions with those previously reported for a similar set of muscles8,25. The rectus femoris (RF) and vastus medialis (VM) muscles are the main contributors to synergy #1 (S1). This synergy is primarily active during early stance and provides stability during the loading response phase of the gait cycle. The soleus (SOL) and gastrocnemius (medial head) (MG) muscles are the main contributors to synergy #2 (S2). This synergy provides propulsion during terminal stance. The tensor fasciae latae (TFL), RF, gluteus maximus (GM), and tibialis anterior (TA) muscles are the main contributors to synergy #3 (S3). During initial swing, this synergy provides ankle dorsiflexion and hip flexion. Finally, the TA and biceps femoris (BF) muscles are the main contributors to synergy #4 (S4), which is primarily active during terminal swing and is responsible for foot landing negotiation and deceleration. In the bilateral analysis (Fig. 3B), S1 of each leg combined with S3 of the contralateral leg, while S2 combined with S4. This is due to the temporal association among the phases of the gait cycles of the two legs.
At the level of muscle synergies, we expect adaptation to occur through one of the following mechanisms:
1. a.
2. b.
Through modification of the activation patterns of a fixed set of muscle synergies22,32.
3. c.
Through modification of the weights of select muscles in some or all the muscle synergies33.
4. d.
Through a combination of a, b and/or c30.
Herein, we systematically assess which of the above mechanisms may be at work during adaptation to the perturbations tested in the study (see Materials and Methods). We first analyzed if adaptation is associated with mechanism a by examining whether adaptation leads to a change in the number of synergies composing the EMG data. This was achieved by finding the dimensionality of the subspace shared between the 4 BL synergies (REFsynX and REFsynXinv) and sets of 4 or 5 synergies for the unilateral analyses and 4, 5 or 6 synergies for the bilateral analysis that were extracted from the EMG data recorded during the last 10 gait cycles of the Pert phase (herein referred to as the late-Pert phase) when full adaptation had been attained. We reasoned that if a new synergy emerged after adaptation without altering the pre-existing four, the dimensionality of the subspace shared between the BL synergy set and the late-Pert set would increase as the number of synergies composing the latter increased. This is because the original 4 BL synergies can be clearly identified as shared subspace dimensions only when the new additional synergies during the late-Pert phase, if any, are properly accommodated for as extra dimensions. It is worth pointing out that we tested 4, 5 and 6 synergies for the bilateral analysis (but only 4 and 5 synergies for the unilateral analyses) to account for the possibility that two distinct unilateral synergies could emerge from the analysis of the late-Pert data that would not merge in a single bilateral synergy (contrary to what was observed for the BL phase of the experiment).
We found that for both legs, during both the X and Xinv experiments, the shared subspace dimensionality between BL and late-Pert synergies remained at 3, regardless of whether 4, 5, or 6 synergies were extracted from the late-Pert EMG data, and whether unilateral or bilateral synergy subspaces were compared (Fig. 4A). Indeed, if there were, say, 1 additional synergy activated at the end of the Pert phase, one would expect the 4-synergy BL space and the 5-synergy late-Pert space to share a 4-D, instead of a 3-D, subspace. We concluded that motor adaptation during both the X and Xinv experiments did not occur by recruiting a new muscle synergy while preserving the BL muscle synergies. It should be emphasized that, in this type of analysis, the values of the cosine of the principal angles do not matter. What matters is whether the patterns shown in Fig. 4A change as we increase the number of synergies used to model the EMG data collected during the late-Pert phase.
Another argument that supports the exclusion of mechanism a comes from the analysis of the R2 of the EMG data reconstruction. If adaptation happened to be associated with the recruitment of an additional 5th muscle synergy, reconstructing the EMG activity during the late-Pert phase using 4 synergies would result in a decrease in the quality of the reconstruction as reflected by smaller R2 values. We, on the other end, showed that 4 synergies could represent equally well muscular activities during both the BL phase and the late-Pert phase of the experiment (Fig. S2).
### No modification of the weights of baseline muscle synergies during motor adaptation
We then tested if adaptation is achieved via modification of the activation patterns of a fixed set of muscle synergies (mechanism b) or via modification of the weights of select muscles in one of the 4 BL synergies (mechanism c). To test these hypotheses we determined how well the EMG data of the BL and Pert phases could be reconstructed by the reference BL muscle synergies while fixing either the BL muscle-synergy vectors (i.e., the W matrix of the Non-negative Matrix Factorization -NMF) (NMFFixedW), or the BL activation patterns of the synergies (i.e., the C matrix of the NMF algorithm) (NMFFixedC), (see Materials and Methods). A significant decrease in reconstruction quality in one of these analyses would indicate that either the same reference W (for NMFFixedW), or the same reference C (for NMFFixedC), was not sufficient to capture EMG changes associated with adaptation, thus implicating mechanism c in the former, or mechanism b in the latter, assuming that either W or C would be primarily modified for adaptation.
The bilateral analysis was primarily employed to highlight if mechanisms b and c, if present, are employed independently by the two legs or in a bilaterally-linked way. In the bilateral analysis (Fig. 4B), we observed that fixing either the muscle synergies or the activation patterns to their reference values led to a decrease in the reconstruction quality from the BL to the Pert EMG data irrespective of whether we considered the beginning (Early Pert) or the end (Late Pert) portions of the Pert phase. Thus, if the muscles from both sides are analyzed together, it is not possible to differentiate whether mechanisms b or c may better characterize adaptation drives. Indeed, for bilateral synergies, changes in the EMG characteristics from the BL phase to the Pert phase could only be captured if both W and C were left free to be updated in the NMF algorithm (Fig. S2).
In the unilateral analysis, implementing NMFFixedC, but not NMFFixedW, led to a drop in the reconstruction quality from the BL to Pert EMG data in both legs of both perturbation experiments (Fig. 4B). Thus, in both the X and Xinv experiments, the Pert data could not be modeled by just modulating the weights of the BL muscle synergies while keeping the temporal activations constant. However, adaptation could be modelled by modulating the muscle synergy activation patterns while leaving the composition of the muscle synergies unchanged. These results are consistent with mechanism b (fixed muscle synergies) and not with mechanism c (altered synergies post-adaptation). Furthermore, the concurrent observation that mechanism b was independently present in the muscle synergies of both legs in our unilateral, but not bilateral, analysis suggests that adaptation to the two perturbations is driven by unilateral mechanisms. It is worth pointing out that both legs could display changes in the temporal activations of the muscle synergies that would however be leg-specific. This latter point explains the inconsistency between the bilateral and unilateral results, and, more specifically, why mechanism b is not observed in the bilateral analysis. In fact, our results in the unilateral analysis suggest that left and right synergies that are linked in the bilateral analysis adapt their activation patterns independently, with different behaviors and time constants (as better explained later). Thus, when the synergies are bilaterally linked in the analysis, the changes in the resulting activation pattern cannot capture the different dynamics of the unilateral synergies.
### The X and Xinv perturbations elicited two distinct behaviors in the synergy activations
The above-summarized analyses indicate that step-length adaptation during the X and Xinv experiments can be modelled by modulating the temporal activations of the muscle synergies that mark unperturbed walking. We next focused on assessing how the activation of each synergy changed during the adaptation experiments. The dynamics of this change was evaluated by measuring the similarity between the temporal activation of 2-step epochs, during the BL, Pert and AE phases, and that of the corresponding reference synergy extracted from the whole BL data. This analysis resulted in similarity value time series (Fig. 5) charting the extent to which the synergy activations deviated from the reference patterns during the experiment. To determine if the activation of each muscle synergy displayed an adaptive behavior, we fit an exponential function (Eq. 3, Materials and Methods) to the corresponding similarity value time series derived for the Pert and AE phases, respectively. The behavior was defined to be adaptive if the cumulative R2 of this fit was >0.75.
Over the Pert phase of the X experiments, we observed both a modest step-wise (i.e., feedback) behavior and an adaptive (feedforward) behavior marked by a gradual change in the temporal activation similarity value time series. The adaptive behavior was detected in 3 out of 4 synergies on the left (unperturbed) side, and in all 4 synergies on the right (perturbed) side (Fig. 5A). The adaptive behavior was characterized by a decay in the similarity value time series until a stable value was reached after a few gait cycles. The time constants of this decay ranged from 9.6 (S3L) to 64.4 (S1L) gait cycles (Table S1). After removal of the perturbation, the similarity was gradually restored to its BL level. This AE phase presented time constants ranging from 2.3 (S3L) to 37.4 (S1L) gait cycles. When we examined the temporal activations of the synergies during late Pert (Figs. 5A and S3), we noticed that adaptation was largely driven by an increase in the activation amplitudes of S1L and S4R. Both were active during the portion of the gait cycle when the perturbation was produced by the robot (Fig. 5A, black bold line). Also, adaptation was associated with increased activations of S3L and S1R. It is worth noticing that the adaptation time constants of this bilaterally-linked synergy pair (S3L + S1R) were different, despite the similar modulation observed in late-Pert (Table S1 shows all the time constants and associated 95% confidence intervals), further suggesting that full adaptation is achieved via unilateral mechanisms.
In the Xinv experiment, we also observed both a step-wise (i.e., feedback) behavior and an adaptive (feedforward) behavior marked by a gradual change in the temporal activation similarity value time series. However, unlike the X experiment, the time series of the temporal activation similarity values for the muscle synergies of the right (perturbed) leg showed a dramatic change at the onset of Pert for several synergies. After this step-wise (i.e., feedback) behavior, the temporal activation similarity value time series showed a gradual return to a level close to the BL value. On the left (unperturbed) leg, even though S1L, S3L and S4L showed an initial step-like deviation from the BL value, the similarity value time series did not satisfy the set criterion of adaptation. On the perturbed right side, adaptation was visible in S1R, S2R and S4R. While all 3 synergies presented an abrupt degradation in similarity at the beginning of the Pert phase, S4R and S1R converged to levels more deviated from the average BL similarities at the end of Pert while S2R returned to its BL value. The time constants of adaptation ranged from 7.6 (S1R) to 15.8 (S2R) gait cycles in the Pert phase, and from 9.0 (S1R) to 21.2 gait cycles (S2R) in the AE phase (Table S1). Full adaptation to Xinv at the end of the Pert phase was characterized mainly by an increase in the activation of the synergies of the right leg (Fig. 5B). Changes in the temporal activations are shown in Fig. S3.
Finally, we analyzed if the two perturbations led to similar patterns of adaptation. We observed that, although the reference modules and activation patterns extracted during the BL phases of the two experiments were consistent, the activation patterns at the end of the Pert phase for the two perturbations were different (Fig. S4).
## Discussion
Our results show that the activity of lower-limb muscles during locomotor adaptation to the perturbations tested in the study can be modelled by modulating the muscle synergy activation patterns observed during the BL phase of the experiment (i.e., in the absence of any perturbation) while leaving the composition of the muscle synergies unchanged. Furthermore, the study provided evidence that the response to perturbations that cause locomotor adaptation is not associated with a single response strategy, but instead, by dynamically mixing multiple response mechanisms in a context-dependent manner. Specifically, our study unravels concurrent actions of unilateral feedforward- and feedback-driven mechanisms that affect the muscle synergies of the two legs and that are a function of the biomechanical effect of the perturbation. These results provide a clear demonstration of the invariance to mechanical perturbations of the human locomotor muscle synergies and show that their temporal activations are under both feedforward and feedback control in a context-dependent fashion.
### Separating feedforward- and feedback-driven EMG changes during locomotor adaptation
Motor behaviors associated with feedback (i.e., reactive) and feedforward (i.e., adaptive) mechanisms were previously observed during adaptation experiments relying on applying unilateral resistive perturbations to the hip and knee21, moving platforms acting on the stance leg22, split-belt treadmill16,20,24 and asymmetric cycling23. Feedforward-controlled behaviors are marked by error-driven, progressive modifications of motor commands - based on an internal model of the perturbation - that converge toward an adapted state. They are associated with an aftereffect once the perturbation is removed14,16. This type of control has been attributed to the cerebellum, whose role in updating the internal representation of the dynamics of the limb through sensory prediction error has been observed in both the upper34,35 and lower limbs24,36. On the other hand, motor behaviors that are the result of purely feedback mechanisms are marked by abrupt changes in response to the perturbation and are not associated with an aftereffect when the perturbation is removed. Feedback mechanisms are primarily driven by spinal circuits37.
The presence of feedforward and feedback adaptations in the EMGs has been investigated in the past by examining, in a setup similar to ours, the presence or absence of changes in the muscular activations during catch trials interspersed in the perturbed conditions21 or, more recently in the split-belt treadmill experiment, by comparing the aftereffect of a long exposure to the perturbation with the initial response to a perturbation of opposite direction20. Here, we show that these components of adaptation can also be characterized at the muscle-synergy level, and that this analysis provides valuable information on how groups of muscles adapt and/or react in concert during prolonged perturbations. The synergy activation changes that we observed during both the X and Xinv experiments can, in fact, be explained as the summation of feedback- and feedforward-driven responses (Figs. 6A, S9 and S10). The feedback-driven component observed at the onset of the Pert phase of the experiments is triggered by the initial deviation from the gait plan that marks the BL phase. This component accounts for the abrupt changes observed at the beginning of the Pert phase. Such step-like responses were observed in the X experiment (e.g., S1L and S2R, Figs. 5A and S10) but were particularly prominent in the Xinv perturbation (S1R, S2R and S4R, Figs. 5B and S10). The feedforward response, on the other hand, drives the adaptation from the initial feedback-mediated response towards the final adapted state and is marked by an exponential time-course.
By analyzing the muscle synergies during the adaptation process, we successfully disentangled the feedback-driven EMG changes from those mediated by feedforward descending drives. Importantly, this separation allowed us to determine the role of the different feedback and feedforward mechanisms, operating on distinct muscle synergies with different time-courses, in response to different perturbations. The results also show how feedback and feedforward mechanisms affect simultaneously the activation patterns of multiple lower-limb muscles. The results we found on the presence of feedback and feedforward components of adaptation for the Xinv are consistent with those found by Lam and colleagues on a similar setup21.
Previous studies have provided conflicting results in regard to whether locomotor adaptation is driven by unilateral or bilateral control. For instance, Choi and Bastian12 provided evidence of independent high-level neural control of the two legs, whereas Houldin et al.38 showed inter-limb transfer of learning during force-field adaptation, and Maclellan et al.25 suggested that bilaterally-linked unilateral modulations of CPG activities of the two legs takes place during split-belt adaptation.
The results of our study show that the response to perturbations that cause locomotor adaptation is context-dependent. Analysis of the results of the X experiment showed a modest feedback response and a dominant feedforward response in both legs, though the response observed in the data recorded from the right (perturbed) leg followed a different time-course of neuromuscular adjustment from the left (unperturbed) leg. In contrast, the results of the Xinv experiment showed a predominantly unilateral response affecting the right (perturbed) leg. The data collected from the left (unperturbed) leg showed only modest changes consistent with small feedback adjustments. The data from the right (perturbed) leg showed large feedback adjustments and a clear adaptation response in three out of four muscle synergies. Taken together, these results show that the relative contributions to locomotor adaptation of the two legs are context-dependent.
But why did we observe different contributions of feedback- and feedforward-mediated responses to the two perturbations? In previous work19, we showed that the primary task-relevant factor driving the response to the X and Xinv perturbations is the need for preserving long-term gait stability. As both perturbations triggered a response to preserve the same long-term gait stability plan, one would think that the discrepancy in strategy that we observed at the neuromuscular level in response to the two perturbations is due to the different effects the perturbations have on the subject’s dynamic stability.
Specifically, we argue that the larger feedback component observed during the Xinv vs. the X experiment is likely due to the more immediate balance threat induced by the Xinv perturbation compared to the X perturbation. The Xinv perturbation causes a mechanical effect that is similar to hitting an obstacle during the swing phase of the gait cycle. This is expected to bring the projection of the center of mass close to its stability boundaries in the antero-posterior direction. Hence, it is not surprising that we observed a change in the activations of the knee extensors during swing (S1R) and of the ankle plantar-flexor muscles during stance (S2R) - a response that is modulated over time to achieve an increase in the activation of the synergies (Fig. S3). In fact, previous studies have shown short- and medium-latency reflex responses in multiple muscles that appeared to be mediated by Ia and II afferents, respectively, likely caused by the overall jar that the stumbling causes on the perturbed leg39.
Unlike the Xinv perturbation, the X perturbation has an effect that is similar to delaying foot landing. A study by van der Linden et al.40 showed that unexpected delayed foot landing triggers a reflex response on multiple muscles after heel-strike, consistent with the activation adjustment we found in synergy S2R. Nevertheless, in our experiment the sensory feedback provided by the X perturbation acting on the leg likely prompted the subject to anticipate the forthcoming missteps, thus making the effects of the feedforward control to be dominant over those of the feedback response. We argue that, at the biomechanical level, the effects of the X perturbation are less immediately threatening the control of balance, but may lead, over time, to an unstable gait pattern, thus triggering a feedforward adjustment of the motor plan19.
All in all, our results demonstrate that the neural strategy underlying the response to perturbations that cause locomotor adaptation is context-dependent12. Interestingly, the complexity of the mechanisms underlying the response to these perturbations is not apparent when one examines simple aspects of the biomechanics of gait19, but it is clearly shown by examining the characteristics of the muscle synergies derived from the EMG data collected during the experiments23.
### The same muscle synergies modulated by feedback- and feedforward-based mechanisms
In contrast to the changes in the activation of muscle synergies, changes in the EMG data of individual muscles in response to the perturbations were more subtle (Fig. S5). In fact, similarity plots derived from data recorded from individual muscles showed adaptation (though not as clearly as the muscle synergies) only in a few cases (Figs. S6 and S7).
Our results provide strong evidence that in humans, both descending (feedforward) and afferent (feedback) drives project, either directly or indirectly, onto the same spinal interneuronal networks that encode locomotor muscle synergies (Fig. 6B). While a modular organization of the control of muscles has been demonstrated for descending motor signals in both animal models41,42 and humans9,25,26,30, the mechanisms controlling synergistic reflexes are still poorly understood. It is known that several reflex pathways project to second and third order interneurons and are able to modulate the activation of the CPGs43. Recent studies on animal models have hypothesized a two-level organization of the CPGs44,45, with the upper-level neural oscillators commanding the lower-level pattern formation interneuronal networks.
In a previous rodent optogenetic study, Levine and colleagues28 identified a molecularly defined population of spinal interneurons that encode the coordination of multiple muscles and are simultaneously activated by both cortical and afferent drives. These networks appear to be the building blocks enabling complex feedforward- and feedback-driven behaviors and could correspond to the hypothesized pattern-formation networks at the lower level of the CPG. In humans, the possibility that sensory afferents could directly act on muscle synergies has been suggested26,30,31,46 but shown only by Chvatal and Ting22, who observed anticipatory and reactive modulations of muscle synergies in response to single-step platform perturbations of the stance leg. The results herein presented extend such observations to motor adaptation and provide further evidence that sensory afferents have a direct effect on muscle synergies by altering their recruitment together with efferent commands during continuous exposure to forces.
As a final remark, the results of the study show that the neuromuscular response of a user to the forces exerted by an exoskeleton during gait does not necessarily lead to a modification of the descending drive to the muscles, which is a primary goal of robot-assisted gait training. Being able to discern whether the neuromuscular responses to the robot-induced forces are reactive or proactive could provide the foundations for new approaches to the design of robot-assisted gait rehabilitation procedures29,47. This could be done, for example, by designing assistive or perturbing forces that trigger descending responses in the activation of selected synergies. Another, similar, approach could be based on analyzing in quasi-real time the neuromuscular adaptation to the forces that the robot is administering during the therapy and use this information for tuning the robotic assistance so as to promote proactive responses or curb reactive ones.
## Materials and Methods
### Study design
Nine healthy adults (3 women; age = 27.8 ± 3.5; height = 177.7 ± 6.9 cm; mass = 72.4 ± 10.6 kg) participated in the study. This was a sample of convenience. Exclusion criteria for the study were the presence of orthopedic or neurological conditions with a potential effect on the performance or outcome of the experiments. Three of the subjects had experienced the motor perturbations used in the study in a previous experiment. All data collections were performed at Spaulding Rehabilitation Hospital, Boston MA. Subjects completed all the experimental procedures in a single day. The experimental protocol was approved by the Spaulding Rehabilitation Hospital Institutional Review Board. Subjects signed an informed consent form prior to participating in the study. All study procedures were carried out in accordance with all relevant guidelines and regulations.
### Generating perturbations using a robotic system
#### Device settings
Subjects walked with the lower limbs strapped to the robotic legs of a system designed for use in robot-assisted gait therapy (Lokomat by Hocoma AG, Switzerland)48. The robotic system (Fig. 1A) allows for both flexion and extension control of the hip and knee joints via linear actuators. Potentiometers are used to track the joint angles at the hip and knee. Force transducers are used to monitor the torques generated at the joints. The Path Control49 was used in the study. This control modality allowed subjects to naturally control the timing of their gait. The controller estimated, at every instant, the position in the gait cycle via comparison of the actual hip and knee joint angular displacement and angular velocity with template patterns. The comparison used a norm-distance minimization algorithm50. The Generalized Elasticities method51 was implemented to make the system as transparent as possible to the subjects during walking. The Body Weight Support capability of the system was not utilized in the study.
#### Perturbations
During the experiments, the robot was used in one of two modes of operations: Back-Driven Mode (BDM) and Force Field Mode (FFM) (Fig. 1D). In the BDM, the robot’s apparent impedance was set to zero to minimize the effect of the robot on the walking patterns (i.e., the system was as “transparent” as possible to the subjects). In the FFM, a perturbation was applied to the right leg during the swing phase of the gait cycle. The perturbation force was generated by the robot according to the following equation:
$${\rm{F}}({\rm{t}})=[\begin{array}{c}{{\rm{F}}}_{{\rm{x}}}({\rm{t}})\\ {{\rm{F}}}_{{\rm{y}}}({\rm{t}})\end{array}]=[\begin{array}{cc}{\rm{A}} & 0\\ {\rm{B}} & 0\end{array}][\begin{array}{c}{{\rm{v}}}_{{\rm{x}}}({\rm{t}})\\ {{\rm{v}}}_{{\rm{y}}}({\rm{t}})\end{array}]$$
(1)
where Fx and Fy represent the antero-posterior (x) and vertical (y) components of the perturbing force acting on the foot, and Vx and Vy represent the x- and y-components of the foot velocity as reconstructed using the following Jacobian:
$$J(\theta )=[\begin{array}{ll}{l}_{Femur}\,\cos ({\theta }_{h})\,+\,{l}_{tibia}\,\cos ({\theta }_{h}-{\theta }_{k}) & \,-\,{l}_{tibia}\,\cos ({\theta }_{h}-{\theta }_{k})\\ {l}_{Femur}\,\sin ({\theta }_{h})\,+\,{l}_{tibia}\,\sin ({\theta }_{h}-{\theta }_{k}) & \,-\,{l}_{tibia}\,\sin ({\theta }_{h}-{\theta }_{k})\end{array}]$$
(2)
where lFemur is the length of the femur, lTibia is the length of the tibia, and θ is the vector of the joint angles θh at the hip and θk at the knee.
Subjects experienced two different perturbations produced by different values of the gains A and B shown in Eq. 1. The first perturbation, denoted by X, was achieved by setting $$A=0.115\,\left(\frac{N\cdot s}{Kg\cdot m}\right)\cdot SubjMass$$ and B = 0, where SubjMass is the subject’s body mass in Kg. The second perturbation, denoted by Xinv, was achieved by setting $$A=0.092\,\left(\frac{N\cdot s}{Kg\cdot m}\right)\cdot SubjMass$$ and $$B=-0.058\,\left(\frac{N\cdot s}{Kg\cdot m}\right)\cdot SubjMass$$. These values of A and B were previously determined as suitable to induce modifications in step length (~30% of baseline length) in the forward (X perturbation) and backward (Xinv perturbation) directions, respectively, without concurrently affecting step height19.
### Study procedures
Footswitches were attached to the soles of the subject’s shoes in positions corresponding to the calcaneous bone and the first metatarsophalangeal joint, respectively. Subjects were strapped to the system using the cuffs of the robotic legs. Then surface electromyographic (EMG) electrodes were placed on the following 16 muscles (8 per leg): tensor fascia latae (TFL), vastus medialis (VM), rectus femoris (RF), biceps femoris (BF), medial head of the gastroecnemius (MG), tibialis anterior (TA), soleus (SOL) and gluteus maximus (GM). Electrodes were positioned as close as possible to the cusp of the muscle belly following the SENIAM guidelines52 but avoiding contact with the cuffs of the robotic legs, which would have caused artifacts in the EMG recordings (Fig. 1B). The EMG and footswitch signals were digitized using 12 bits with a sampling rate of 3 kHz. Data collected from the robotic system was synchronously recorded at 1 kHz.
Six of the nine subjects who participated in the study had no previous experience with walking using the robotic system. In these subjects, we performed a 5-minute habituation trial using the system in BDM.
During all the experimental trials, the treadmill speed was set at 3 km/h. Subjects’ cadence was paced using a metronome set at 84 beats per minute. Subjects were instructed to pace their heel strikes with the beats of the metronome. No further instructions were given to the subjects. The metronome was used, as in our previous work19, to stabilize the pace, and consequently the speed of the foot. This was done to make sure that subjects experienced consistent perturbation magnitudes across different steps. We have shown in our previous work that the metronome does affect the adaptation process19.
A total of 4 trials were performed as follows: (1) a 90-s trial, for estimating the parameters of the Generalized Elasticities method; (2) a 120-s trial performed with the optimized controller, for estimating the subject’s baseline walking pattern that was used as a reference during the experiment; (3) a 10-min trial to test the effects of the X (or Xinv) perturbation; and (4) a 10-min trial to test the effects of the Xinv (or X) perturbation experiment. The order in which the X and Xinv perturbations were tested was randomized across subjects. A 5-minute break was allowed between the X and Xinv experiments. In each of the X or Xinv trials, subjects performed 420 gait cycles. The trials consisted of four phases.
1. (i)
The habituation phase consisting of 180 gait cycles during which the system was set in BDM. 9 gait cycles during which the system was set in FFM were randomly selected after the first 40 cycles, so that there were at least 8 and no more than 18 BDM gait cycles between any two FFM gait cycles. The FFM gait cycles, named single step perturbations, were used to estimate each subject’s response to a single-cycle perturbation.
2. (ii)
The baseline phase (BL) consisting of 80 consecutive BDM gait cycles.
3. (iii)
The perturbation phase (Pert) consisting of 80 consecutive FFM gait cycles.
4. (iv)
The aftereffect phase (AE) consisting of 80 consecutive BDM gait cycles.
Step length was defined, consistently with our previous work19, as the foremost position of the foot during swing. It is important to notice that, in our set up, the reference coordinate system used to determine step length is aligned with the center of rotation of the hip. Hence, if foot contact had occurred with the ankle aligned with the hip, the estimated step length would have been equal to zero. For each subject, the step length values were normalized to the average step length recorded during the BL phase. After normalization, for each of the X and Xinv experiments, the data was averaged across subjects to determine the aggregate adaptation behavior. The time series of the step length estimates derived from the data collected during the Pert and AE phases of the experiment were modelled using exponential functions:
$$f(step)=\alpha \cdot ex{p}^{\beta \cdot step}+\gamma$$
(3)
where α represents the response to the perturbation (for the Pert phase) or removal of the perturbation (for the AE phase) observed at the beginning of the Pert or AE phases, β is the time constant of the adaptation, and γ the asymptotic step length value observed when full adaptation has been achieved. In this analysis, similarly to what we did in previous work19, α was estimated by averaging - across all subjects - the step length values observed during the 9 single step perturbations during the habituation phase of the experiment, and β and γ were estimated using a least-squares algorithm.
### EMG Pre-processing and segmentation
The EMG data was visually inspected to ensure that high-quality data was then processed. The trials and channels showing significant movement artifacts were excluded from the analysis. This process led to selecting data from 7 out of 9 subjects who participated in the study. The EMG data was first band-pass filtered (50 to 450 Hz) using a 7th-order elliptic filter. The filtered signals were then full-wave rectified and low-pass filtered (cut-off of 5 Hz) using a 7th-order elliptic filter, thus deriving the EMG envelopes. The heel-strike and toe-off times were extracted from the footswitch data of each foot. These events were used to segment the EMG data so that each segment corresponded to 2 steps. The length of each segment was normalized, and the data was re-sampled so that each segment consisted of 200 data points. The amplitude of the EMG data collected during all experimental phases was normalized to the median, across all BL segments, of the maximum values derived for each segment.
### EMG analysis
#### Extraction of the muscle synergies
The non-negative matrix factorization (NMF) algorithm was used to estimate the muscle synergies from the normalized EMG envelopes. This algorithm models the muscular activity MA as a linear combination of muscle synergies, W, whose activity is modulated over time by a set of activation patterns, C:
$$MA(t)=\mathop{\sum }\limits_{i=1}^{N}{C}_{i}(t){W}_{i}$$
(4)
To derive W and C, we used the NMF algorithm, which is based on the assumption that the EMG envelopes are corrupted by an uncorrelated Gaussian noise53. Muscle synergies were extracted unilaterally (i.e., the EMG data collected from the left leg and the right leg was analyzed separately) as well as bilaterally (i.e., the data collected using all 16 channels was analyzed together). For the BL, Pert and AE phases of each experiment and each subject, we extracted synergies from each 2-step epoch so that a total of (420 − 180)/2 = 120 synergy sets were obtained. A validation of this approach to extracting muscle synergies is shown in Fig. S8. In addition, reference muscle synergies (REFsynX for the X experiment, and REFsynXinv for the Xinv experiment) were extracted, unilaterally as well as bilaterally, from the whole BL data (80 gait cycles) of both experiments without data segmentation. Subsequently, an average activation pattern for the reference synergies was calculated by averaging - across step cycles - the reference C, segmented in time before averaging using the same method utilized to segment the EMG envelopes.
#### Number of muscle synergies
The R2 derived using the reconstructed EMG data (based on the muscle synergies) and the original EMG data was utilized to select the number of synergies suitable to model the EMG data. Specifically, we selected the number of muscle synergies that would yield a R2 > 0.75 both in the unilateral and bilateral analyses for all subjects. This criterion was met by selecting 4 synergies (R2 = 0.80 ± 0.02 for the bilateral analysis; R2 = 0.86 ± 0.06 for the unilateral analysis of the left leg EMG data; R2 = 0.79 ± 0.04 for the unilateral analysis of the right leg EMG data). This result is consistent with previous studies8.
#### Changes in muscle synergies in response to the perturbation
To investigate the possible recruitment of additional synergies during motor adaptation, we also extracted 5 unilateral synergies and 5 and 6 bilateral synergies (Figs. 4A,B and S2). We then quantified the similarity between the subspaces spanned by the reference W and the epochal W across the experimental phases. The reference W was obtained by using all the segments of data collected during the BL phase. The epochal W was derived for each segment of the data collected during all experimental phases. The cosine of the principal angles54 between the subspaces spanned by the muscle-synergy sets was then derived to quantify the similarity between the above-defined subspaces. High values of the cosines of the principal angles (i.e., close to 1) indicate that the subspaces spanned by the two synergy sets intersect over a shared subspace of n dimensions, where n is the number of cosine values above threshold. The threshold value was derived - as previously proposed by Cheung et al.55 – via the following surrogate analysis. We calculated, for each subject, the cosines of the principal angles between the reference synergy set (REFsynX or REFsynXinv) and 1,000 Gaussian noise-corrupted (μ = 0; σ = 0.1) versions of each of the 40 BL (two-step) segments. For each of these segments, we estimated the 95th percentile value of the distribution of the cosines of the principal angles obtained for each of the 1,000 Gaussian noise-corrupted versions of the BL segments. Finally, the median of this percentile value across the 40 BL epochs was computed and selected to be the threshold for determining the dimensionality of the shared subspace n for each subject. This procedure allowed us to account for the variability of the epochal muscle synergies expected from chance when determining the threshold.
We also extracted C or W from every epoch by either fitting fixed BL synergies, W (NMFFixedW), or fixed BL activation patterns, C (NMFFixedC), to the epoch data. Through all iterations of the NMF algorithm, the W (for NMFFixedW) or C (for NMFFixedC) matrix was fixed at its reference value, thus allowing for updating only the C (for NMFFixedW) or W (for NMFFixedC) matrix. Using this approach, the variability of the original data set was described only by the updated parameter (i.e., C if W was fixed or W if C was fixed).
How the muscle-synergy activation coefficients C evolved over the course of adaptation was characterized by evaluating the similarity between C of each epoch and the reference C obtained from the BL data, quantified using the Pearson’s linear correlation coefficient. For each epoch, a similarity value was obtained, thus resulting in a time series of similarity values across the BL, Pert and AE phases. To facilitate subsequent modeling of the adaptation dynamics, this time series of similarity values was smoothed using an 8-point moving average filter, that was applied separately to the values from the three phases of the experiments. We fitted the exponential function in Eq. 3 to the similarity value time-series derived for the Pert and AE phases. All three parameters α, β and γ were estimated using a least-squares algorithm. The behavior was defined to be adaptive if the cumulative R2 was >0.75.
We analyzed the data to identify changes in muscle synergies between the two experiments. In this analysis we calculated the similarity between the average weights for the EMG data collected during the BL phase of the X and Xinv experiments. We also calculated the similarity between the temporal activation patterns during the BL and the late-Pert phases across the two experiments. The similarity between the weights was calculated using the normalized dot product. The similarity between the temporal activations was calculated using Pearson coefficients.
### Statistical analysis
Several statistical analyses were performed on the data recorded during the experiments. A statistical analysis was performed to test for significant differences in step length on the perturbed side during the different phases of the experiment (Fig. S1). We compared the average normalized step length during the BL phase and the step length for the first and last gait cycles of Pert and AE phases. The analysis was based on Friedman’s ANOVA test (p-values are presented in Fig. S1). The Minimum Significance Difference (MSD)56 test was used for pairwise comparisons of the values of step length between specific phases of the experiments. Similarly to what we did in19, we compared: (1) average step-length during late (last 5 cycles) BL versus first step of Pert, to test if the perturbation induced a significant change; (2) step-length values for the first step versus the last step of Pert, to test if subjects showed an adaptation to the perturbation force vector; (3) step-length values for late BL versus last step of Pert, to test if subjects fully compensated for the robot-induced changes in step-length, (4) step-length values for late BL versus first step of AE, to test for the presence of a significant aftereffect; (5) step-length values for the first step of Pert versus the first step of AE, to test if the aftereffect mirrored, in magnitude, the robot-induced change in step length; (6) step-length values for the first step versus the last step of AE, to test if subjects demonstrated changes in step length during this phase; and (7) step-length values for late BL versus the last step of AE, to test if subjects returned to baseline values of step length at the end of the aftereffect phase. The z-value for the MSD test was set to obtain an overall α of 0.05 for the seven comparisons considered, for an effective error rate per comparison equal to 0.001256.
We extracted synergies (unilaterally and bilaterally) by either using the standard NMF algorithm, or by fixing W or C (Eq. 3) to their baseline values. The R2 values extracted in the different phases of both experiments were statistically compared for all these analyses using a Friedman ANOVA test (α = 0.1). Specifically, we compared values of R2 obtained unilaterally and bilaterally using the unconstrained and constrained NMFs between the BL, early Pert and late-Pert phases. The resultant p-values for these analyses are presented in Figs. S2 and 4B. The Minimum Significance Difference test was used to perform pairwise comparisons. The z-value for the MSD test was set to obtain an overall α of 0.05 for the three comparisons considered (baseline vs. early adaptation, baseline vs. late adaptation and early adaptation vs. late adaptation), for an effective error rate per comparison equal to 0.008356. Finally, a statistical analysis based on the Wilcoxon’s signed rank test (α = 0.05) was used to test for statistically significant changes in the similarity of the activation patterns observed during the BL and late-Pert phases (Fig. S04).
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# Path Tracing 3D Fractals
In some ways path tracing is one of the simplest and most intuitive ways to do ray tracing.
Imagine you want to simulate how the photons from one or more light sources bounce around a scene before reaching a camera. Each time a photon hits a surface, we choose a new randomly reflected direction and continue, adjusting the intensity according to how likely the chosen reflection is. Though this approach works, only a very tiny fraction of paths would terminate at the camera.
So instead, we might start from the camera and trace the ray from here and until we hit a light source. And, if the light source is large and slowly varying (for instance when using Image Based Lighting), this may provide good results.
But if the light source is small, e.g. like the sun, we have the same problem: the chance that we hit a light source using a path of random reflections is very low, and our image will be very noisy and slowly converging. There are ways around this: one way is to trace rays starting from both the camera and the lights, and connect them (bidirectional path tracing), another is to test for possible direct lighting at each surface intersection (this is sometimes called ‘next event estimation’).
Even though the concept of path tracing might be simple, introductions to path tracing often get very mathematical. This blog post is an attempt to introduce path tracing as an operational tool without going through too many formal definitions. The examples are built around Fragmentarium (and thus GLSL) snippets, but the discussion should be quite general.
Let us start by considering how light behaves when hitting a very simple material: a perfect diffuse material.
## Diffuse reflections
A Lambertian material is an ideal diffuse material, which has the same radiance when viewed from any angle.
Imagine that a Lambertian surface is hit by a light source. Consider the image above, showing some photons hitting a patch of a surface. By pure geometrical reasoning, we can see that the amount of light that hits this patch of the surface will be proportional to the cosine of the angle between the surface normal and the light ray:
$$cos(\theta)=\vec{n} \cdot \vec{l}$$
By definition of a Lambertian material this amount of incoming light will then be reflected with the same probability in all directions.
Now, to find the total light intensity in a given (outgoing) direction, we need to integrate over all possible incoming directions in the hemisphere:
$$L_{out}(\vec\omega_o) = \int K*L_{in}(\vec\omega_i)cos(\theta)d\vec\omega_i$$
where K is a constant that determines how much of the incoming light is absorbed in the material, and how much is reflected. Notice, that there must be an upper bound to the value of K – too high a value would mean we emitted more light than we received. This is referred to as the ‘conservation of energy’ constraint, which puts the following bound on K:
$$\int Kcos(\theta)d\vec\omega_i \leq 1$$
Since K is a constant, this integral is easy to solve (see e.g. equation 30 here):
$$K \leq 1/\pi$$
Instead of using the constant K, when talking about a diffuse materials reflectivity, it is common to use the Albedo, defined as $$Albedo = K\pi$$. The Albedo is thus always between 0 and 1 for a physical diffuse materials. Using the Albedo definition, we have:
$$L_{out}(\vec\omega_o) = \int (Albedo/\pi)*L_{in}(\vec\omega_i)cos(\theta)d\vec\omega_i$$
The above is the Rendering Equation for a diffuse material. It describes how light scatters at a single point. Our diffuse material is a special case of the more general formula:
$$L_{out}(\vec\omega_o) = \int BRDF(\vec\omega_i,\vec\omega_o)*L_{in}(\vec\omega_i)cos(\theta)d\vec\omega_i$$
Where the BRDF (Bidirectional Reflectance Distribution Function) is a function that describes the reflection properties of the given material: i.e. do we have a shiny, metallic surface or a diffuse material.
Completely diffuse material (click for large version)
## How to solve the rendering equation
An integral is a continuous quantity, which we must turn into something discrete before we can handle it on the computer.
To evaluate the integral, we will use Monte Carlo sampling, which is a very simple: to provide an estimate for an integral, we will take a number of samples and use the average values of these samples multiplied by the integration interval length.
$$\int_a^b f(x)dx \approx \frac{b-a}{N}\sum _{i=1}^N f(X_i)$$
If we apply this to our diffuse rendering equation above, we get the following discrete summation:
\begin{align} L_{out}(\vec\omega_o) &= \int (Albedo/\pi)*L_{in}(\vec\omega_i)cos(\theta)d\vec\omega_i \\ & = \frac{2\pi}{N}\sum_{\vec\omega_i} (\frac{Albedo}{\pi}) L_{in}(\vec\omega_i) (\vec{n} \cdot \vec\omega_i) \\ & = \frac{2 Albedo}{N}\sum_{\vec\omega_i} L_{in}(\vec\omega_i) (\vec{n} \cdot \vec\omega_i) \end{align}
Test render (click for large version)
## Building a path tracer (in GLSL)
Now we are able to build a simple path tracer for diffuse materials. All we need to do is to shoot rays starting from the camera, and when a ray hits a surface, we will choose a random direction in the hemisphere defined by the surface normal. We will continue with this until we hit a light source. Each time the ray changes direction, we will modulate the light intensity by the factor found above:
$$2*Color*Albedo*L_{in}(\vec\omega_i) (\vec{n} \cdot \vec\omega_i)$$
The idea is to repeat this many times for each pixel, and then average the samples. This is why the sum and the division by N is no longer present in the formula. Also notice, that we have added a (material specific) color. Until now we have assumed that our materials handled all wavelengths the same way, but of course some materials absorb some wavelengths, while reflecting others. We will describe this using a three-component material color, which will modulate the light ray at each surface intersection.
All of this boils down to very few lines of codes:
The getBackground() method simulates the light sources in a given direction (i.e. infinitely far away). As we will see below, this fits nicely together with using Image Based Lighting.
But even when implementing getBackground() as a simple function returning a constant white color, we can get very nice images:
and
The above images were lightened only a constant white dome light, which gives the pure ambient occlusion like renders seen above.
## Sampling the hemisphere in GLSL
The code above calls a ‘getSample’ function to sample the hemisphere.
This can be a bit tricky. There is a nice formula for $$cos^n$$ sampling of a hemisphere in the GI compendium (equation 36), but you still need to align the hemisphere with the surface normal. And you need to be able to draw uniform random numbers in GLSL, which is not easy.
Below I use the standard approach of putting a seed into a noisy function. The seed should depend on the pixel coordinate and the sample number. Here is some example code:
## Importance Sampling
Now there are some tricks to improve the rendering a bit: Looking at the formulas above, it is clear that light sources in the surface normal direction will contribute the most to the final intensity (because of the $$\vec{n} \cdot \vec\omega_i$$ term).
This means we might want sample more in the surface normal directions, since these contributions will have a bigger impact on the final average. But wait: we are estimating an integral using Monte Carlo sampling. If we bias the samples towards the higher values, surely our estimate will be too large. It turns out there is a way around that: it is okay to sample using a non-uniform distribution, as long as we divide the sample value by the probability density function (PDF).
Since we know the diffuse term is modulated by the $$\vec{n} \cdot \vec\omega_i = cos(\theta)$$, it makes sense to sample from a non-uniform cosine weighted distribution. According to GI compendium (equation 35), this distribution has a PDF of $$cos(\theta) / \pi$$, which we must divide by, when using cosine weighted sampling. In comparison, the uniform sampling on the hemisphere we used above, can be thought of either to be multiplied by the integral interval length ($$2\pi$$), or diving by a constant PDF of $$1 / 2\pi$$.
If we insert this, we end up with a simpler expression for the cosine weighted sampling, since the cosine terms cancel out:
## Image Based Lighting
It is now trivial to replace the constant dome light, with Image Based Lighting: just lookup the lighting from a panoramic HDR image in the ‘getBackground(dir)’ function.
This works nicely, at least if the environment map is not varying too much in light intensity. Here is an example:
Stereographic 4D Quaternion system (click for large version)
If, however, the environment has small, strong light sources (such as a sun), the path tracing will converge very slowly, since we are not likely to hit these by chance. But for some IBL images this works nicely – I usually use a filtered (blurred) image for lighting, since this will reduce noise a lot (though the result is not physically correct). The sIBL archive has many great free HDR images (the ones named ‘*_env.hdr’ are prefiltered and useful for lighting).
## Direct Lighting / Next Event Estimation
But without strong, localized light sources, there will be no cast shadows – only ambient occlusion like contact shadows. So how do we handle strong lights?
Test scene with IBL lighting
Let us consider the sun for a moment.
The sun has an angular diameter of 32 arc minutes, or roughly 0.5 degrees. How much of the hemisphere is this? The solid angle (which corresponds to the area covered of a unit sphere) is given by:
$$\Omega = 2\pi (1 – \cos {\theta} )$$
where $$\theta$$ is half the angular diameter. Using this we get that the sun covers roughly $$6*10^{-5}$$ steradians or around 1/100000 of the hemisphere surface. You would actually need around 70000 samples, before there is even a 50% chance of a pixel actually catching some sun light (using $$1-(1-10^{-5})^{70000} \approx 50\%$$).
Test scene: naive path tracing of a sun like light source (10000 samples per pixel!)
Obviously, we need to bias the sampling towards the important light sources in the scene – similar to what we did earlier, when we biased the sampling to follow the BRDF distribution.
One way to do this, is Direct Lighting or Next Event Estimation sampling. This is a simple extension: instead of tracing the light ray until we hit a light source, we send out a test ray in the direction of the sun light source at each surface intersection.
Test scene with direct lighting (100 samples per pixel)
Here is some example code:
The 1E-5 factor is the hemisphere area covered by the sun. Notice, that you might run into precision errors with the single-precision floats used in GLSL when doing these calculations. For instance, on my graphics card, cos(0.4753 degrees) is exactly equal to 1.0, which means a physically sized sun can easily introduce large numerical errors (remember the sun is roughly 0.5 degrees).
## Sky model
To provide somewhat more natural lighting, an easy improvement is to combine the sun light with a blue sky dome.
A slightly more complex model is the Preetham sky model, which is a physically based model, taking different kinds of scattering into account. Based on the code from Simon Wallner I implemented a Preetham model in Fragmentarium.
Here is an animated example, showing how the color of the sun light changes during the day:
## Fractals
Now finally, we are ready to apply path tracing to fractals. Technically, there is not much new to this – I have previously covered how to do the ray-fractal intersection in this series of blog posts: Distance Estimated 3D fractals.
So the big question is whether it makes sense to apply path tracing to fractals, or whether the subtle details of multiple light bounces are lost on the complex fractal surfaces. Here is the Mandelbulb, rendered with the sky model:
Path traced Mandelbulb (click for larger version)
Here path tracing provides a very natural and pleasant lighting, which improves the 3D perceptions.
Here are some more comparisons of complex geometry:
Default ray tracer in Fragmentarium
Path traced in Fragmentarium
And another one:
Default ray tracer in Fragmentarium
Path traced in Fragmentarium
## What’s the catch?
The main concern with path tracing is of course the rendering speed, which I have not talked much about, mainly because it depends on a lot of factors, making it difficult to give a simple answer.
First of all, the images above are distance estimated fractals, which means they are a lot slower to render than polygons (at least of you have a decent spatial acceleration structure for the polygons, which is surprisingly difficult to implement on a GPU). But let me give some numbers anyway.
In general, the rendering speed will be (roughly) proportional to the number of pixels, the FLOPS of the GPU, and the number of samples per pixel.
On my laptop (a mobile mid-range NVIDIA 850M GPU) the Mandelbulb image above took 5 minutes to render at 2442×1917 resolution (with 100 samples per pixel). The simple test scene above took 30 seconds at the same resolution (with 100 samples per pixel). But remember, that since we can show the render progressively, it is still possible to use this at interactive speeds.
What about the ray lengths (the number of light bounces)?
Here is a comparison as an animated GIF, showing direct light only (the darkest), followed by one internal light bounce, and finally two internal light bounces:
In terms of speed one internal bounce made the render 2.2x slower, while two bounces made it 3.5x slower. It should be noted that the visual effect of adding additional light bounces is normally relatively small – I usually use only a single internal light bounce.
Even though the images above suggests that path tracing is a superior technique, it is also possible to create good looking images in Fragmentarium with the existing ray tracers. For instance, take a look at this image:
(taken from the Knots and Polyhedra series)
It was ray traced using the ‘Soft-Raytracer.frag’, and I was not able to improve the render using the Path tracer. Having said that, the Soft-Raytracer is also a multi-sample ray tracer which has to use lots of samples to produce the nice noise-free soft shadows.
## References
The Fragmentarium path tracers are still Work-In-Progress, but they can be downloaded here:
Sky-Pathtracer.frag (which needs the Preetham model: Sunsky.frag).
and the image based lighting one:
IBL-Pathtracer.frag
The path tracers can be used by replacing an existing ray tracer ‘#include’ in any Fragmentarium .frag file.
External resources
GI Total Compendium – very valuable collection of all formulas needed for ray tracing.
Vilém Otte’s Bachelor Thesis on GPU Path Tracing is a good introduction.
Disney’s BRDF explorer – Interactive display of different BRDF models – many examples included. The BRDF definitions are short GLSL snippets making them easy to use in Fragmentarium!
Inigo Quilez‘s path tracer was the first example I saw of using GPU path tracing of fractals.
Evan Wallace – the first WebGL Path tracer I am aware of.
Brigade is probably the most interesting real time path tracer: Vimeo video and paper.
I would have liked to talk a bit about unbiased and consistent rendering, but I don’t understand these issues properly yet. It should be said, however, that since the examples I have given terminate after a fixed number of ray bounces, they will not converge to a true solution of the rendering equation (and, are thus both biased and inconsistent). For consistency, a better termination criterion, such as russian roulette termination, is needed.
# Combining ray tracing and polygons
I have written a lot about distance estimated ray marching using OpenGL shaders on this blog.
But one of the things I have always left out is how to setup the camera and perspective projection in OpenGL. The traditional way to do this is by using functions such as ‘gluLookAt’ and ‘gluPerspective’. But things become more complicated if you want to combine ray marched shader graphics with the traditional OpenGL polygons. And if you are using modern OpenGL (the ‘core’ context), there is no matrix stack and no ‘gluLookAt’ functions. This post goes through through the math necessary to combine raytraced and polygon graphics in shaders. I have seen several people implement this, but I couldn’t find a thorough description of how to derive the math.
Here is the rendering pipeline we will be using:
It is important to point out, that in modern OpenGL there is no such thing as a model, view, or projection matrix. The green part on the diagram above is completely programmable, and it is possible to do whatever you like there. Only the part after the green box of the diagram (starting with clip coordinates) is fixed by the graphics card. But the goal here is to precisely match the convention of the fixed-function OpenGL pipeline matrices and the GLU functions gluLookAt and gluPerspective, so we will stick to the conventional model, view, and projection matrix terminology.
The object coords are the raw coordinates, for instance as specified in VBO buffers. This is the vertices of an 3D object in its local coordinate system. The next step is to position and orient the 3D object in the scene. This is accomplished by applying the model matrix, that transform the object coordinates to global world coordinates. The model transformation will be different for the different objects that are placed in the scene.
## The camera transformation
The next step is to transform the world coordinates into camera or eye space. Now, neither old nor modern OpenGL has any special support for implementing a camera. Instead the conventional gluPerspective always assumes an origo centered camera facing the negative z-direction, and with an up-vector in the positive y-direction. So, in order to implement a generic, movable camera, we instead find a camera-view matrix, and then apply the inverse transformation to our world coordinates – i.e. instead of moving/rotate the camera, we apply the opposite transformation to the world.
Personally, I prefer using a camera specified using a forward, up, and right vector, and a position. It is easy to understand, and the only problem is that you need to keep the vectors orthogonal at all times. So we will use a camera identical to the one implemented in gluLookAt.
The camera-view matrix is then of the form:
$$\begin{bmatrix} r.x & u.x & -f.x & p.x \\ r.y & u.y & -f.y & p.y \\ r.z & u.z & -f.z & p.z \\ 0 & 0 & 0 & 1 \end{bmatrix}$$
where r=right, u=up, f=forward, and p is the position in world coordinates. R, u, and f must be normalized and orthogonal.
Which gives an inverse of the form:
$$\begin{bmatrix} r.x & r.y & r.z & q.x \\ u.x & u.y & u.z & q.y \\ -f.x & -f.y & -f.z & q.z \\ 0 & 0 & 0 & 1 \end{bmatrix}$$
By multiplying the matrices together and requiring the result is the identity matrix, the following relations between p and q can be established:
As may be seen, the translation part (q) of this matrix is the position of the camera expressed in the R,u, and f coordinate system.
Now, per default, the OpenGL shaders use a column-major representation of matrices, in which the data is stored sequentially as a series of columns (notice, that this can be changed by specifying ‘layout (row_major) uniform;’ in the shader). So creating the model-view matrix as an array on the CPU side looks like this:
Don’t confuse this with the original camera-transformation: it is the inverse camera-transformation, represented in column-major format.
## The Projection Transformation
The gluPerspective transformation uses the following matrix to transform from eye coordinates to clip coordinates:
$$\begin{bmatrix} f/aspect & 0 & 0 & 0 \\ 0 & f & 0 & 0 \\ 0 & 0 & \frac{(zF+zN)}{(zN-zF)} & \frac{(2*zF*zN)}{(zN-zF)} \\ 0 & 0 & -1 & 0 \end{bmatrix}$$
where ‘f’ is cotangent(fovY/2) and ‘aspect’ is the width to height ratio of the output window.
(If you want to understand the form of this matrix, try this link)
Since we are going to raytrace the view frustum, consider what happens when we transform an direction of the form (x,y,-1,0) from eye space to clip coordinates and further to normalized device coordinates. Since the clip.w in this case will be 1, the x and y part of the NDC will be:
Since normalized device coordinates range from [-1;1], this means that when we ray trace our frustrum, our ray direction (in eye space) must be in the range:
where 1/f = tangent(fovY/2).
We now have the necessary ingredients to set up our raytracing shaders.
## The polygon shaders
But let us start with the polygon shaders. In order to draw the polygons, we need to apply the model, view, and projection transformations to the object space vertices:
Notice, that we premultiply the model and view matrix on the CPU side. We don’t need them individually on the GPU side. If you wonder why we don’t combine the projection matrix as well, it is because we want to use the modelView to transform the normals as well:
Notice, that in general normals transform different from positions. They should be multiplied by the inverse of the transposed 3×3 part of the modelView matrix. But if we only do uniform scaling and rotations, the above will work, since the rotational part of matrix is orthogonal, and the uniform scaling does not matter if we normalize our normals. But if you do non-uniform scaling in the model matrix, the above will not work.
## The raytracer shaders
The raytracing must be done in world coordinates. So in the vertex shader for the raytracer, we need figure out the eye position and ray direction (both in world coordinates) for each pixel. Assume that we render a quad, with the vertices ranging from [-1,-1] to [1,1].
The eye position can be easily found from the formula found under ‘the camera transformation':
Similar, by transforming the ranges we found above from eye to world space we get that:
where fov_y_scale = tangent(fovY/2) is an uniform calculated on the CPU side.
Normally, OpenGL takes care of filling the z-buffer. But for raytracing, we have to do it manually, which can be done by writing to gl_fragDepth. Now, the ray tracing takes place in world coordinates: we are tracing from the eye position and into the camera-forward direction (mixed with camera-up and camera-right). But we need the z-coordinate of the hit position in eye coordinates. The raytracing is of the form:
Converting the hit point to eye coordinates gives (the p and q terms cancel):
which in clip coordinates becomes:
Making the perspective divide, we arrive at normalized device coordinates:
The ncdDepth is in the interval [-1;1]. The last step that remains is to convert into window coordinates. Here the depth value is mapped onto an interval determined by the gl_DepthRange.near and gl_DepthRange.far parameters (usually these are just 0 and 1). So finally we arrive at the following:
Putting the pieces together, we arrive at the following for the ray tracing vertex shader:
and this code for the fragment shader:
The above is of course just snippets. I’m currently experimenting with a Java/JOGL implementation of the above (Github repo), with some more complete code.
# Fragmentarium Version 1.0 (“Cologne”) Released
It has been quite a while since the last update to Fragmentarium, but I finally managed to find some time for creating a new release, which indeed is the first release stable enough to qualify as a version “1.0”.
### Major changes
The Fragmentarium UI has been somewhat redesigned in order to simplify it.
Now there are only two modes:
• Progressive, for creating accumulated images (using multiple samples per pixels). An uniform named ‘subframe’ keeps track of the current frame (and replaces the old ‘backbufferCounter’ uniform).
• Animation, where a single ‘time’ uniform can be used to control dynamics.
The buffer size no longer needs to be locked to the window size, although it is possible to do so. This means, that it is now possible to work with buffers with different aspect ratio and size than the OpenGL window – something several people had requested.
The old ‘Animation Controller’ window is gone. Instead, the time variable may be controlled by a new slider in the main UI. In order the render animation frames, the new ‘High Resolution and Animation Render’ window must be used:
As can be seen, it is now possible to render animations at arbitrary resolution and using multiple subframes. There is also a progress dialog making it possible to actually stop the renders.
The code has also been cleaned up. The prefered build environment is now Qt Creator 2.7 together with Qt 4.8.5 (a project file has been added), though the old VS2008 project has been updated and is working as well (and the Debug configuration now works).
### Other changes
• A “brute-force” raytracer has been added, to render fractals where a distance estimator can not be found.
• Tile rendering now has added padding option, which can be useful when using screen space methods.
• New sigmoid tone mapper.
• IBL raytracer can now rotate maps.
### Bug fixes
• No more mip-maps for non-HDR textures.
• Fixed some texture caching bugs.
• And several other minor fixes.
### Where to get it
The latest build can be downloaded here (Windows only):
Fragmentarium-Windows-v1.0.0.zip
Mac and Linux users have to download and build from the source tree:
Fragmentarium at GitHub
# Image Based Lighting
It can be difficult to achieve realistic lighting by manually placing light sources in a 3D scene. One way to quickly achieve complex and natural lighting is by using real-world light data to lighten the scene.
In order to do that, you need light data from all directions, that is, a full panoramic view. Since you most likely will be looking directly at a light source at some points on your full panorama, you’ll also need to store data that can handle a large dynamic range – 8 bits are not enough.
A common format for these maps is the equirectangular (aka longitude/latitude) format. These are simple 2D bitmaps, typically in 2:1 format, where the x axis covers 360 degrees of azimuthal (longitude) angles, and the y axis covers 180 degrees of elevation (latitude). They are often stored in RGBE (.HDR) files. A good source of free panoramic HDR’s for Image Based Lighting is the sIBL archive.
## Backgrounds and reflections
It is very easy to do reflection mapping using equirectangular maps. Simply look up the reflected ray in the equirectangular map. The conversion from 3D direction to map indices is straight-forward:
This can also be used to embed the scene in a panoramic view, by looking up the camera ray direction for points that do not hit anything. Here is an example image where reflections and background is sampled from an environment map:
No light model was used in the above image – just pure reflection mapping.
## Diffuse light
So what about the diffuse light? Normally, this is modelled by a Lambertian term: the diffuse intensity is proportional to the cosine of the angle between the surface normal and light source direction. In principle, we need to calculate the angle to all points on the hemisphere pointing in the direction of the surface normal, and sum all these contributions. This means, that for each each pixel we raytrace, we need to sample half (one hemisphere) of the pixels in the equirectangular map. Much to slow for even a modern GPU.
But here is the interesting part: the diffuse light contribution is only a function of the surface normal direction. This means we can precalculate the diffuse light in a given normal direction, and store this in a new equirectangular map. Mathematically, this filtering is called a cosine convolution, and some HDR panorama providers are nice enough to provide prefiltered maps for diffuse light (for instance all those in the sIBL archive).
Here is an example of some spheres rendered with specular and diffuse light maps. The materials are faded from pure diffuse (left) to pure specular (right).
As we saw earlier, pure reflection is easy to achieve. But it is also possible to use the Phong reflection model (or other models) with image based lighting. In the Phong model, the specular light intensity depends on the angle between the light source and reflected camera ray. The intensity is proportional to the cosine of this angle raised to a power, which controls the smoothness. Again it is possible to precalculate a convoluted map using the cosine raised to the appropriate power.
## Shadows and ground planes
In order to do proper shadows, we would have to check whether the path to every single point on the light map was occluded. Again, this is not feasible. One way to work around this, is to place a single directional light source and then check whether we have a clear to this particular light source. This works nicely, if the light map has a single dominant light source, such as a sun.
But a problem arise since we do not have any background objects to cast shadows on – the objects in the environment map are placed an infinite distance away, and we don’t know their geometry. Consider this image:
There is no sense of the positioning of the objects in this scene – they seem to float at undefinable locations.
To improve this, we can introduce a ground plane, an invisible object, whose only purpose is to catch shadows.
In Fragmentariums IBL-raytracer, you can enable this under the floor tab. It is also possible to turn on some visual debug, to align the floor with the environment map:
Now, once we have set up the ground plane, we can add some ambient occlusion:
… and some shadows:
In order to sample the shadows, we need to sample the area of the light source uniformly. In the case of a sun-like object, this means sampling a hemisphere in a specified direction and for a given latitude span. A formula for this is given in the Global Illumination Compendium (formula 34). To use this, you need to transform the coordinate system to a new one aligned with the light source direction. I do this:
Here ‘extent’ is the size of the light source we sample. It is given as ‘1-cos(angle)’, so 0 means a point-like light source (sharp shadows) and 1 means a full hemisphere light source (no shadows).
Notice the construction of the aligned coordinate systems fails if ‘dir’ has zero y and z components – if needed, this should be handled as a special case.
## Problems with Simple Image Based Lighting
Since no secondary rays are traced for the specular reflections, some geometry ends up with very unrealistic shading:
The problems here is that the reflections on the inside on the box sees through the sides. So this simple lighting approach works best for convex objects. Here is another example:
(btw, this is Kali’s dragonKIFS system)
Here again the lighting seems unrealistic – something about the specular light is just wrong.
Finally, the combination of the rapidly varying surface normals on a fractal surface and rapidly varying light sources on the environment map introduces new problems. Take a look at this image:
Here there are several small, but strong (HDR) spotlights placed above the Mandelbulb. This image will be very slow to converge and will contain noisy specular highlights: occasionally, one of the subpixel samples will hit a strong light source, which will dominate the sum for the pixel. This will break the sub-pixel anti-aliasing efforts. It is possible to set a maximum (clip) on the specular contributions – or to do HDR tonemapping before averaging – but both solutions goes against the very idea of introducing HDR.
While the specular noise from strong point-like light sources will be difficult to combat, it is easier to do something about the geometry violating reflections.
As of now, I think the best solution is to trace at least secondary rays, and then apply the approximated IBL lighting on the secondary hit points. On problem is that diffuse light does not fall off as quickly as specular light, so you need to sample a lot of points on the hemisphere to get convergence. There are ways around this – for instance, Pixar use bent normals in their Renderman solution, before looking up in the diffuse environment map.
I’ll give a more detailed discussion of the sampling process and the convolution map creation in the next blog post, where I’ll talk about how to speed up diffuse and specular sampling using importance sampling and stratification.
Finally, all the image maps used for lighting on the images accompanying this blog post were from the sIBL archive. And all 3D geometry and composition was of course done in Fragmentarium.
# Stereographic Quaternion Julias
Inverse stereographic projections allows you to project a plane onto a sphere. These projections generalize to higher dimensions: for instance, you can inverse project every point in 3D onto the four dimensional 3-sphere. Daniel Piker suggested to use the same transformation to depict the Quaternion Julia systems, instead of using the standard 3D slicing (at least that was what I though – see the update below).
They are still clearly originating from the Quaternions, but the transformation adds a bit of spice. Here are some images:
I’ve previously used stereographic projection to depict Mobius fractals and they were also used to depict the 4D polychora.
For high resolution versions, see my Flickr account.
Update: turns out that what Daniel Piker was suggesting, was to do the stereographic transformation, and then split the 4D coordinates into a starting point and a Julia seed for the ordinary complex (not quaternion) system. I tried this as well, and it creates some very interesting images too:
# Rendering 3D fractals without a distance estimator
I have written a lot about distance estimated 3D fractals, and while Distance Estimation is a fast and elegant technique, it is not always possible to derive a distance estimate for a particular system.
So, how do you render a fractal, if the only knowledge you have is whether a given point belongs to the set or not? Or, in other words, how much information can you extract if the only information you have is a black-box function of the form:
I decided to try out some simple brute-force methods to see how they would compare to the DE methods. Contrary to my expectations, it turned out that you can actually get reasonable results without a DE.
First a couple of disclaimers: brute-force methods can not compete with distance estimators in terms of speed. They will typically be a magnitude slower. And if you do have more information available, you should always use it: for instance, even if you can’t find a distance estimator for a given escape time fractal, the escape length contains information that can be used to speed up the rendering or create a surface normal.
The method I used is not novel nor profound: I simply sample random points along the camera ray for each pixel. Whenever a hit is found on the camera ray, the sampling will proceed on only the interval between the camera and the hit point (since we are only interested in finding the closest pixels), e.g. something like this:
(The Near and Far distances are used to restrict the sample space, and speed up rendering)
There are different ways to choose the samples. The simplest is to just sample uniformly (as in the example above), but I found that a stratified approach, where the camera ray segment is divided into equal pieces and a sample is choosen from each part works better. I think the sampling scheme could be improved: in particular once you found a hit, you should probably bias the sampling towards the hit to make convergence faster. Since I use a progressive (double buffered) approach in Fragmentarium, it is also possible to read the pixel depths of adjacent pixels, which probably also could be used.
Now, after sampling the camera rays you end up with a depth map, like this:
(Be sure to render to a texture with 32-bit floats – a 8-bit buffer will cause quantization).
For distance estimated rendering, you can use the gradient of the distance estimator to obtain the surface normal. Unfurtunately this is not an option here. We can, however, calculate a screen space surface normal, based on the depths of adjacent pixels, and transform this normal back into world space:
(Update: I found out that GLSL supports finite difference derivatives through the dFdx statement, which made the code above much simpler).
Now we can use a standard lighting scheme, like Phong shading. This really brings a lot of detail to the image:
In order to improve the depth perception, it is possible to apply a screen space ambient occlusion scheme. Recently, there was a very nice tutorial on SSAO on devmaster, but I was to lazy to try it out. Instead I opted for the simplest method I could think of: simply sample some pixels in a neighborhood, and count how many of them that are closer to the camera than the center pixel.
This is how this naive ambient occlusion scheme works:
(Notice that for pixels with no hits, I’ve choosen to lighten, rather than darken them. This creates an outer glow effect.)
Now combined with the Phong shading we get:
I think it is quite striking how much detail you can infer simply from a depth map! In this case I didn’t color the fractal, but nothing prevents you from assigning a calculated color. The depth buffer information only uses the alpha channel.
Here is another example (Aexion’s MandelDodecahedron):
While brute-force rendering is much slower than distance estimation, it is possible to render these systems at interactive frame rates in Fragmentarium, especially since responsiveness can be improved by using progressive rendering: do a number of samples, then storing the best found solution (closest pixel) in a depth buffer (I use the alpha channel), render the frame and repeat.
There are a couple of downsides to brute force rendering:
• It is slower than distance estimation
• You have to rely on screen space methods for ambient occlusion, surface normals, and depth-of-field
• Anti-aliasing is more tricky since you cannot accumulate and average. You may render at higher resolution and downsample, or use tiled rendering, but beware that screen space ambient occlusion introduce artifacts which may be visible on tile edges.
On the other hand, there are also advantages:
• Much simpler to construct
• Interior renderings are trivial – just reverse the ‘inside’ function
• Progressive quality rendering: just keep adding samples, and the image will converge.
To use the Fragmentarium script, just implement an ‘inside’ function:
It is also possible to use the raytracer on existing DE’s – here a point is assumed to be inside a fractal if the DE returns a negative number, and outside if the DE returns a positive one.
The script can be downloaded as part of Fragmentarium source distribution (it is not yet in the binary distributions). The following files are needed:
# Fragmentarium Version 0.9.12 (“Prague”) Released
I’ve released a new build of Fragmentarium, version 0.9.12 (“Prague”). It can be downloaded at Github. (Binaries for Windows, source for Windows/Linux/Mac)
The (now standard) caveat apply: Fragmentarium is very much work in progress, and is best suited for people who like to experiment with code.
Version 0.9.12 continues to move Fragmentarium in the direction of progressive HDR rendering. The default raytracers now use accumulated rendering for anti-alias, shadowing, and DOF. To start the progressive rendering, Fragmentarium must be set to ‘Continuous’ mode. It is possible to set a maximum number of rendered frames. All 2D and 3D system now also come with tone mapping, gamma correction, and color control (see the ‘Post’ tab).
IBL Raytracing, using an HDR panorama from Blotchi at HDRLabs.
There is a new raytracer, ‘IBL-raytracer.frag’ which can be used for DE’s instead of the default raytracer. It uses Image Based Lighting from HDR panorama maps. For an example of the new IBL raytracer, see the tutorial: “25 – Image Based Lighting.frag”.
If you need to do stuff like animation, it is still possible to use the old raytracers. They can be included as: “#include “DE-Raytracer-v0.9.1.frag” or “#include “DE-Raytracer-v0.9.10.frag”
Other than that there is now better support for buffer-swap systems (e.g. reaction-diffusion and game-of-life) and better control of texture look-ups.
There are also some interesting new fragments, including the absolutely amazing LivingKIFS.frag script from Kali.
## New features
• Added maximum subframe counter (for progressive rendering).
• Added support for HDR textures (.hdr RGBE format).
• Tonemapping, color control, and Gamma correction in buffershader.
• Added support for widget for changing bound textures.
• More host defines:
• Added texture parameters preprocessor defines:
• Change of syntax: when using “#define providesColor”, now implement a ‘vec3 baseColor(vec3)’ function.
• DE-Raytracer.frag now uses a ‘Blinn-Phong with Schickl term and physical normalization’. (Which is something I found in Naty Hoffman’s Course Notes).
• DE-Raytracer.frag and Soft-Raytracer now uses new ‘3D.frag’ base class.
• Added a texture manager (should reuse and discard textures in memory automatically)
• Added option to set OpenGL refresh rate in preferences.
• Progressive2D.frag now supports custom filtering (using ‘#define providesFiltering’)
• Added support for choosing ‘#include’ through editor context menu.
• Using arrow keys now work when sliders have focus.
• Now does a ‘reset all’ when loading new system (otherwise too confusing).
## New fragments
• Added ‘Kali’s Creations': KaliBox, LivingKIFS, TreeBroccoli, Xray_KIFS. [Kali]
• Added Doyle-Spirals.frag [Knighty]
• Added: Droste.frag (Escher Droste effect)
• Added: Reaction-Diffusion.frag (Gray-Scott example)
• Added ‘Convolution.frag’ example (For precalculating specular and diffuse lighting from HDR panoramas)
• Added examples of working with double precision floats and emulated double precision floats: “Include/EmulatedDouble.frag”, “Theory/Mandelbrot – Emulated Doubles.frag”
• Added ‘IBL-Raytracer.frag’ (Image Based Lighting raytracer)
• Added tutorials: ‘progressive2D.frag’ and ‘pure3D.frag’
• Added experimental: ‘testScene.frag’ and ‘triplanarTexturing.frag’
## Bug fixes
• Reflection is now working again in ‘DE-Raytracer.frag’
• Fixed filename case sensitivity error when doing reverse lookup of line numbers.
### Mac users
Some Mac users has reported problems with the latest versions of Fragmentarium. Again, I don’t own a Mac, so I cannot solve these issues without help.
For examples of images generated with the new version, take a look at the Flickr Fragmentarium stream.
# Gamma Correction
Gamma correction is certainly not the most sexy topic in computer graphics. But it needs to be taken into account whenever you perform almost any kind of graphical manipulation.
The most widely used way to transfer image and color information is by encoding them as 8-bit RGB colors. Unfortunately, the human eye does not perceive brightness linearly as a function of the physical intensity. Since we are dealing with only 256 levels for each color channel, we apply a non-linear encoding – a gamma encoding – to make the most of our limited bits – otherwise we would experience banding in the range where the eye is most sensitive.
Typically, a simple power law is used: $$Encoded = Linear^{0.45}$$
The display hardware then performs the inverse transformation for the final output: $$Linear = Encoded^{1/0.45} = Encoded^{2.2}$$. This last exponent (2.2) is referred to as the gamma value.
On a modern computer, this means that all 8-bit media (your typical images, such as JPG’s and PNG’s) are gamma encoded, and not stored with linear intensities. The 8-bit framebuffers that are sent to the GPU, are also expected to be gamma encoded to make the most of the limited range. So far, all is well: you can read an image from a JPG and send it directly to the framebuffer – it will then display correct.
But problems arise when you start manipulating the graphics data. For instance, imagine you are downsizing an image with a checkerboard pattern of pure white and pure black pixels, until is small enough to be constant colored. You might guess, that the proper average value should be 0.5 (this discussion assumes that color values are not in the integer range [0,255], but normalized to [0,1]). But if you are working in gamma corrected space, the correct value is (0.5^(1/2.2)) =~ 0.73. This is the value you should send to the framebuffer in order to get an average physical intensity of pure black and white on your monitor. Notice that all browsers, and many photo manipulation programs fail to perform correct gamma-aware resizing.
The same applies to 3D graphics. All these lighting and shades techniques (e.g. Blinn-Phong shading) are designed to work in linear intensity space.
So the correct procedure when working with graphics is to convert all gamma encoded media to linear intensities, perform any calculation/blending/averaging, and convert back to gamma encoding. Notice, that this requires that you use something with more resolution than 8-bit integers (typically 32-bit floats) when working in linear space. Otherwise, too much resolution would be lost, and the whole point of gamma encoding would be lost. Also notice that media and images in high dynamic formats, such as HDR (.rgbe) and RAW typically are in linear space and should not be converted.
The following image sums up when and how to use gamma encoding and do conversions:
Here is an example of an image with and without a gamma-aware rendering pipeline:
The image on the left has been rendered completely ignoring gamma. The image on the right has proper gamma handling. Notice, that gamma-aware rendering does not necessarily look better – If you turn on gamma correct rendering, you images will be lighter, and might loose contrast, so you might have to adjust light sources and exposure.
## Gamma Correction in Fragmentarium
In Fragmentarium the base classes “Progressive2D.frag” and “DE-raytracer.frag” are both gamma-aware. When you implement a color function, e.g. for the 2D case:
and for the 3D case:
the returned color is expected to be gamma corrected. If you want to supply linear colors, insert a
at the top of the script.
## And all the rest…
Many gamma discussions on the internet revolve around how old CRT monitors expected a non-linear input. And it is indeed true that older CRT monitors had a relation between the light intensity and the applied voltage which was roughly like: $$intensity = voltage^{2.5}$$. But modern displays, such as the TFT panels, do not have similar simple power law relations between the voltage and the intensity. Still, when you send 8-bit data to the framebuffer, you must gamma encode it, with a gamma of around 2.2. This will not change for as long as 8-bit data is used as data transfer format.
A few other points:
• The encoding expected and used for 8-bit format is not always a simple gamma power-law relation. More often the sRGB encoding is used. It is, however, very similar to a power-2.2 gamma encoding, and for most applications, the difference is not important.
• For a good discussion of what exactly is meant by intensity and brightness, see the Gamma FAQ.
• OpenGL and Direct3D may perform sRGB encoding/decoding directly in hardware. This is the preferred way, since for instance texture interpolation is performed correctly.
• Sometimes linear intensities are stored in 8-bit data, despite the loss of perceived dynamic levels. This might be the case in game pipelines (to reduce data size).
# Reaction-Diffusion Systems
Reaction-diffusion systems model the spatial dynamics of chemicals. An interesting early application was Alan Turing’s theory of Morphogenesis (Turing’s 1951 paper). Here, he suggested, that the pattern formation in animal skin could be explained by a two component reaction-diffusion system.
Reaction-diffusion systems are interesting, because they display a wide range of self-organizing patterns, and they have been used by several digital artists, both for 2D pattern generation and 3D structure generation.
The reaction-diffusion model is a great example of how complex large-scale structure may emerge from simple, local rules.
## Modelling Reaction-Diffusion on a GPU
As the name suggests, these systems have two driving components: diffusion, which tends to spread out or smoothen concentrations, and reactions, which describe how chemical species may transform into each other.
For each chemical species, it is possible to describe the evolution using a differential equation on the form:
$$\frac {dA}{dt} = K \nabla^2 A + P(A,B)$$
Where A and B are fields describing the concentration of a chemical species at each point in space. The $$K$$ coefficient determines how quickly the concentration spreads out, and $$P(A,B)$$ is a polynomial in the different species concentrations in the system. There will be a similar equation for the B field.
To model these, we can represent the concentrations on a discrete grid, which fits nicely on a 2D texture on a GPU. The time derivative can solved in discrete time steps using forward Euler integration (or something more powerful). On a GPU, we need two buffers to do this: we render the next time step into the front buffer using values from the back buffer, and then swap the buffers.
Buffer swapping is a standard technique, and in Fragmentarium the only thing you need to do, is to declare a ‘uniform sampler2D backbuffer;’ and Fragmentarium will take care of creation and swapping of buffers. We also use the Fragmentarium host define ‘#buffer RGBA32F’ to ask for four-component 32-bit float buffers, instead of the normal 8-bit integer buffers.
The Laplacian may be calculated using a finite differencing scheme, for instance using a five-point stencil:
(see the Fragmentarium source for a nine-point stencil).
A simple two-component Gray-Scott system may then be modelled simply as:
(Robert Munafo has a great page with more information on Gray-Scott systems).
Here is an example of a typical system created using the above system, though many other patterns are possible:
It is also possible to enforce some structure by changing the concentrations in certain regions:
You can even use a picture to modify the concentrations:
A template implementation can be found as part of the Fragmentarium source at GitHub: Reaction-Diffusion.frag. Notice, that this fragment requires a recent source build from the GitHub repository to run.
## Reaction-Diffusion systems used by artists
Several artist have used Reaction Diffusion systems in different ways, but the most impressive examples of 2D images I have seen, are the works of Jonathan McCabe. For instance his Bone Music series: or his Turing Flow series:
McCabe’s images are created using a more complex multi-scale model. Softology’s blog entry and W:Blut’s post dissect McCabe’s approach (there is even a reference implementation in Processing). Notice, that Nervous System sells some of McCabe’s works as jigsaw puzzles.
## Reaction-Diffusion systems in WebGL
Felix Woitzel (@Flexi23) has created some beautiful WebGL-based reaction-diffusion demos, such as this Fluid simulation with Turing patterns:
He also has created several other RD based variants over at WebGL Playground.
## Fabricated 3D Objects
Jessica Rosenkrantz and Jesse Louis-Rosenberg at Nervous System create and sell objects designed and inspired by generative processes. Several of their objects, including these cups, plates, and lamps are based on reaction-diffusion systems, and can be bought from their webshop.
Be sure to read their blog entries about reaction-diffusion. And don’t forget to take a look at their Cell Cycle WebGL design app, while visiting.
## Reaction-Diffusion Software
An easy way to explore reaction-diffusion systems with doing any coding is by using Ready, which uses OpenCL to explore RD systems. It has several interesting features, including the ability to run systems on 3D meshes and directly interact and ‘paint’ on the surfaces.
It also lets you run Game-of-Life on exotic geometries, such as a torus or even something as exotic as a Penrose tiling.
# Double Precision in OpenGL and WebGL
This post talks about double precision numbers in OpenGL and WebGL, and how to emulate them if there is no native hardware support.
In GLSL 4.00.9 (which is OpenGL 4.0) and higher, there is a native double precision floating point type. And if your graphics card is able to run OpenGL 4.0, it most likely has native hardware support for doubles (except for a few ATI/AMD cards). There are some caveats, though:
1. Not all functions are supported with double precision arguments. For instance, there are no trigonometric and exponential functions. (The available functions may be found here).
2. You can not pass double precision ‘varying’ parameters from the vertex shader to the fragment shader, and have the GPU automatically interpolate them. Double precision varying variables must be flat.
3. Double precision performance may be artificially limited by the hardware manufacturers. This is the case for Nvidia’s Fermi architecture, where the scientific computing brand, the Tesla series, can execute double precision arithmetics at half the speed of single precision, while the consumer brand, the GeForce series, only can execute double precision arithmetics at 1/8 the speed of single precision. For Nvidia’s brand new Kepler architecture used in the GeForce 600 series, things change again: here the difference between single and double precision will be a whopping factor 24! Notice, that this will also be the case for some cards in the Kepler Tesla branch, such as the Tesla K10.
4. In Fragmentarium (and in general, in Qt’s OpenGL wrapper classes) it is not possible to set double precision uniforms. This should be easy to circumvent by using the OpenGL API directly, though.
(Non-related Fragmentarium image)
In order to use double precision, you must either specify a GLSL version 4.00 (or higher) or use the extension:
Older cards, like the GeForce 310M in my laptop, does not support double precision in hardware. Here it is possible to use emulated double precision instead.
I used the functions by Henry Thasler described here in his posts, to emulate a double precision number stored in two single precision floats. The worst part about doing emulated doubles in GLSL, is that GLSL does not support operator overloading. This means the syntax gets ugly for simple arithmetics, e.g. ‘z = add(mul(z1,z2),z3)’ instead of ‘z = z1*z2+z3′.
On Nvidia cards, it is necessary to turn off optimization to use Thasler’s code – this can be done using the following pragmas:
(Non-related Fragmentarium image)
## Performance
To test performance, I used a Mandelbrot test scene, rendered at 1000×500 with 1000 iterations in Fragmentarium. The numbers show the performance in frames per second. The zoom factor was determined visually, by noticing when pixelation occurred.
Geforce 570GTX Tesla 2075 Max Zoom (~300USD) (~2200USD) Single 140 100 10^5 Double 41 70 10^14 Emulated Double 16 11 10^13
Some observations:
• Emulated double precision is slightly less accurate then true hardware doubles, but not much in this particular scenario.
• Emulated doubles are roughly 1/9th the speed of single precision. Amazingly, this suggest that on the Kepler architecture it might make more sense to use emulated double precision than the built-in hardware support!
• Hardware doubles on the 570GTX performs better than expected (they should perform at roughly 1/8 the speed). This is probably because double precision arithmetics isn’t the only bottleneck in the shader.
Notice that the Tesla card was running on Windows in WDDM mode, not TCC mode (since you cannot use GLSL shaders in TCC mode). Not that I think performance would change.
## WebGL and double precision
WebGL does not support double precision in its current incarnation. This might change in the future, but currently the only choice is to emulate them. This, however, is problematic since WebGL seems to strip away pragmas! Henry Thasler’s emulation code doesn’t work under the ANGLE layer either. In fact, the only configuration I could get to work, was on a Intel HD 3000 GPU with ANGLE disabled. I did create a sample application to test this which can be tried out here:
Click to run WebGL app. Left side is single-precision, right side is emulated double precision. Here shown on Firefox without ANGLE on a Intel HD 3000 card.
It is not clear why the WebGL version does not work on Nvidia cards. Floating points may run at lower resolution in WebGL, but I’m using the ‘precision highp’ qualifiers. I also tried querying the resolution using glContext.getShaderPrecisionFormat(…), but had no luck – it is only available on Firefox, and on my GPU’s it just returns precision=0.
The most likely explanation is that Nvidia drivers perform some optimizations which spoils the emulation code. This is also the case for desktop OpenGL, but here the pragma’s solve the problem.
The emulation code uses constructs like:
z = a - (a - b);
which I suspect the well-meaning compiler might translate to ‘z=b’, since the rounding errors normally would be insignificant. Judging from some comments on Thasler’s original posts, it might be possible to prevent this using constructs such as: ‘z = a – float(a-b)’, but I have not pursued this.
## Fragmentarium and Double Precision
Except that there are no double-precision sliders (uniforms), it is straight-forward to use double precision code in Fragmentarium. The only thing to remember is that you cannot pass doubles from the vertex shader to the fragment shader, which is the standard way of passing camera information to the shader in Fragmentarium.
I’ve also included a small port of Thaslers GLSL code in the distribution (see “Include/EmulatedDouble.frag”). It is quite easy to use (for an example, try the included “Theory/Mandelbrot – Emulated Doubles.frag”).
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## Translating Word Phrases into Expressions With Integers Using Multiplication and Division
### Learning Outcomes
• Translate expressions with integers and simplify
Once again, all our prior work translating words to algebra transfers to phrases that include both multiplying and dividing integers. Remember that the key word for multiplication is product and for division is quotient.
### Exercises
Translate to an algebraic expression and simplify if possible: the product of $-2$ and $14$.
Solution:
The word product tells us to multiply.
the product of $-2$ and $14$ Translate. $\left(-2\right)\left(14\right)$ Simplify. $-28$
### Exercises
Translate to an algebraic expression and simplify if possible: the quotient of $-56$ and $-7$.
### TRY IT
The following video shows more examples of how to translate expressions that contain integer multiplication and division.
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## El_Arrow one year ago stuck in arc length problem
1. El_Arrow
2. El_Arrow
@freckles
3. El_Arrow
how does it get the 2650/4 + 441x/4?
4. El_Arrow
im getting 1325/2 on my calculator
5. anonymous
|dw:1434600730609:dw|
6. anonymous
2650/4=1325/2
7. El_Arrow
but then why didnt the person put 1325/2 in the problem?
8. anonymous
it looks like they wanted to pull the 1/4 out from under the radical.
9. El_Arrow
oh i see
10. El_Arrow
one more thing
11. El_Arrow
what did he do here?
12. El_Arrow
like how did he get the 3+y=(1/8)tangent(theta)?
13. anonymous
that's a trig substitution from right angle trig. basically if you have an integral of the form $\sqrt{b^2+a^2x^2}$you can make the substitution x = (a/b) tan Θ
14. anonymous
I reversed that. x = (b/a) tan Θ
15. El_Arrow
so the x is (3+y), the b is 1 and a is 8? right?
16. anonymous
yes
17. El_Arrow
okay thanks
18. anonymous
you're welcome
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# Oh no, a Simple puzzle
It should be simple to find an answer to this puzzle.
## Step 1
There's an image link inside the question (click the "edit" link to see it). It leads to this image.
## Step 2
The previous image has five characters in the top left, in a darker color than the rest of the background. Replacing the five before the .png in the URL with those five leads to another image
# Step 3
The QR code in the previous image scans to V8bzu. This leads to yet another image:
The "key" needs 5 letters, and looking up those five in the table should give another image. As the question text suggests, those five letters are "simpl". (The "e" in the question is crossed out, though it's hard to see.) This leads to the fourth image, which just seems to be a very blurry image of the word "Yes", so I assume we're done here.
• Yep, that is right so far. Look at the question text precisely to find out how to continue (and the error message in the question confirms that the bsod is a correct picture) – MEE was the missing bracket Sep 11 '18 at 18:21
• Ok, this is correct. Oh and why are you always so ... fast? I rememeber some feature request on meta here ... – MEE was the missing bracket Sep 11 '18 at 18:27
In the
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# Amenability of an “almost Hamiltonian” group
here is another interesting question that I can't answer on my own.
Let G be a countable, discrete group such that for any subgroup H of G and any element s of G we have: [H : sHt] is finite and [H : tHs] is finite where st=ts=1 i.e. t is the inverse of s. Then G is amenable.
Note that in the case when the indexes above are exactly equal 1 then G is Hamiltonian which is solvable and therefore amenable.
Another way to state this condition is that every subgroup $H<G$ has commensurator $\text{Comm}_G(H)$ equal to all of $G$. The commensurator in $G$ of a subgroup $H$ consists of all those $g$ for which $H\cap gHg^{-1}$ has finite index in $H$ and in $gHg^{-1}$.) (Note that in your question you need to intersect $H$ with $gHg^{-1}$ to ensure you get a subgroup of $H$.) Do you have examples of not-virtually-solvable groups for which this condition holds? – Tom Church May 31 '10 at 18:57
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Recently, my professor in IE 5513 - Engineering Safety presented the class with the task of being able to predict whether or not a ladder will slip or not when it's leaned towards a wall. It's been a while since I've had physics, so I found some notes on ladders and walls available online by Richard Fitzpatrick from the University of Texas at Austin which I'm going to go through.
### Problem statement
A ladder of length $l$ with neglible mass leaned towards a wall with an angle $\theta$. A person of mass $M$ climbs a distance of $x$ up the ladder. Given the static friction $\mu$, will the ladder slip from the ground?
### Resolving forces
There are four forces involved in this problem: The weight of the person $Mg$, the normal force at the wall $S$, the normal force at the ground $R$ and the friction force from the surface $f$. The sum of these forces needs to equal zero. For the horizontal forces we have:
$S = f$
Analogously, for the vertical forces:
$R = Mg$
Remember, these values are scalars and not vectors.
### Considering torque
The next step in this exercise is to consider the torque acting on the ladder. To simplify, the point where the ladder touches the ground is chosen as the basis for torque calculation since the displacement vector for both $R$ and $f$ is zero. Decomposition gives a contributing value of $M~g~\cos(\theta)$ for the man on the ladder and $S~\sin(\theta)$ for ladder top. Multiplying with the respective displacement values, we get the torques, which should equal to zero.
$M~g~x~\cos(\theta) = S~l~\sin(\theta)$
Which further can be written as:
$S = {M~g~x \over l~\tan(\theta)}$
### Putting it all together
The condition for the ladder not to slip is $f < \mu~R$. We also know the following:
$f = S = {M~g~x \over l~\tan(\theta)}$
Hence:
${M~g~x \over l~\tan(\theta)} < \mu~R$
Substituting $R = M~g$, removing it, as well as multiplying by $l$ and $tan(\theta)$ gives:
$x < \mu~l~\tan(\theta)$
If this inequation holds, the ladder will not slip. To find out if the whole ladder can be used, substitute $x$ for $l$.
A huge thanks to Tormod Haugland for reviewing and suggesting improvements for this blog post!
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## Faster Progress Report #2
It's raining code! 0.4.0 is faster's biggest, most useful release yet.
faster began as a yak shave, created to aid base💯 in its quest to become the fastest meme on Github. Writing an explicit AVX2-accelerated version of base💯's encoder and decoder, then realizing I'd have to do the same thing again to see the speedups on my Ivy Bridge desktop, pushed me to make this library. Months later, it has blossomed into its own project, and has eclipsed base💯 in both popularity and promise.
Faster 0.4.0 contains numerous improvements, including
• #![no_std] support
• Collection striping (Gathers)
• Vector merging
• Vectorized endianness operations
• Vector swizzling
• Lockstep Packed Iterators (Zipping)
• Vector patterns
• Vectorized iteration on uneven collections
• Vectorized min/max on SSE4.1
• Vectorized Upcast on SSE4.1
• FnMut closures in simd_map and simd_reduce
• Compound assignment operators for architectures without hardware SIMD
• Large vectors for architectures without hardware SIMD
• Downcast polyfills on more vector types
• Polyfills of [saturating_]{hadd,hsub} on more vector types.
• A changelog
• Tons of documentation
• Bug fixes
In the following few headings, we'll be exploring some of these improvements in more detail.
## no_std Support
We can compile on core now! All functions which returned a Vec are disabled when compiling for core, but you get the same great API and the same speedup. Faster performs no allocations, except for functions which return owned types, so switching from std to core doesn't affect any library functionality.
## Vector Patterns
Thanks to the leaking of a top-secret implementation of procedural macros, faster has formalized its vector pattern constructors. This means halfs, interleave, and the new partition are an official part of the API! These patterns are fully vectorized and polyfilled, and are being used right now in the iterator and vector merging code.
The new function, partition, fills a vector with the first argument until the specified index, and then fills the remainder with the second argument. Thanks to the kind soul on the #rust IRC who helped me divine a vectorized algorithm for this!
(Yes, I know it's “halves”. We're still unstable, don't worry.)
## Striping
A non-vectorized implementation of gathers have landed. This lets you take a collection which normally wouldn't load into a vector and perform vector operations on it. A good example would be a collection where we wish to group every nth element into a vector, or a collection with some garbage elements in it.
There was a vectorized implementation, but it was too limited and not fast enough. I'd rather approach the solution with swizzling in the future, rather than using Intel's gather intrinsics.
While striping a collection does mean your loads will be scalar, it is still possible to outperform non-SIMD implementations, as seen in base💯's new SIMD-accelerated decoder. Although this feature is still a work in progress, it has already yielded serious improvements in expressive power.
buf.simd_iter().stripe_four().zip().simd_map(tuplify!(4, u8s(0)), |(_, _, c, d)| {
((c - u8s(143)) * u8s(64)) + d - u8s(128) - u8s(55)
}).scalar_fill(out)
Our determinant function sketch from the last progress report actually compiles and runs now!
fn determinant(matrices: &[f64]) -> Vec<f64> {
// Input: Many 3x3 matrices of the form [a b c ... i a b c ... i ...]
matrices.simd_iter().stripe_nine().zip().simd_map(tuplify!(9, f64s(0.0)), |(a, b, c, d, e, f, g, h, i)| {
(a * e * i) + (b * f * g) + (c * d * h) - (c * e * g) - (b * d * i) - (a * f * h)
}).scalar_collect()
}
## Zipping
Faster can now add two vectors together! Two PackedIterators can now be operated on in lockstep, using a similar API to the one exposed by Rust's standard library. Just call zip on a tuple of PackedIterator to get a structure which returns a tuple containing each iterator's next vector. We can see its use in our determinant function:
fn determinant(matrices: &[f64]) -> Vec<f64> {
// Input: Many 3x3 matrices of the form [a b c ... i a b c ... i ...]
matrices.simd_iter().stripe_nine().zip().simd_map(tuplify!(9, f64s(0.0)), |(a, b, c, d, e, f, g, h, i)| {
(a * e * i) + (b * f * g) + (c * d * h) - (c * e * g) - (b * d * i) - (a * f * h)
}).scalar_collect()
}
stripe_nine returns a tuple of nine PackedIterators, which are then zipped. We can destructure the packed zipped iterator just like a scalar zipped iterator.
## Swizzling
Faster now has two built-in ways to shuffle a vector: PackedSwizzle::flip() and PackedReendianize::swap_bytes(). flip will flip even and odd elements of your vector, starting at element 0:
assert_eq!(u8s::interleave(2, 1).flip(), u8s::interleave(1, 2));
assert_eq!(f64s::interleave(2.0, 1.0).flip(), f64s::interleave(1.0, 2.0));
Whereas swap_bytes will perform a vectorized endianness swap.
assert_eq!(u16s(2).swap_bytes(), u16s(2.swap_bytes()));
assert_eq!(i32s::interleave(30, 90).swap_bytes(),
i32s::interleave(30.swap_bytes(), 90.swap_bytes()));
Helper functions present in Rust's standard library such as to_be are also supported:
let checksum = socket_read_buf.simd_iter()
.simd_reduce(u8s(0), u8s(0), |acc, v| {
acc + v.be_u32s().from_be();
}).sum();
## Vector Merging
Combining two vectors is now possible with faster. On x86, this uses the single-cycle blend intrinsics, which are much faster than masking and blitting vectors onto each other.
assert_eq!(u8s(20).merge_interleaved(u8s(5)), u8s::interleave(20, 5));
assert_eq!(i16s(-3).merge_partitioned(i16s(1), 2), i16s::partitioned(-3, 1, 2));
## QoL Improvements
### Mutable Closures
When designing faster, I initially thought it would be possible to completely abstract vector sizes from the user by relying on pure and referentially transparent computation. While this model is sufficiently powerful and usable most of the time, there are real use cases for mutable state in SIMD computations. Because of this, mutable closures are now usable in simd_map and simd_reduce.
### Shorthand Splat Syntax
Who knew you could export a function with the same name as a type? Now, instead of typing u8s::splat(0), you can simply call u8s(0) for the same effect. This works for all vector types!
### Scalar Reduction
You can now reduce a vector into a scalar in the same way you can reduce an iterator of vectors into a vector. It's very possible to write nonportable code with this function (such as the included example), but it offers a ton of expressive power and should reduce cognitive load when the need to write such a thing arises.
u8s(0x01).scalar_reduce(0, |acc, v| acc + v)
### Documentation
Tons of documentation and examples were added between 0.3.0 and 0.4.0. Check it out here!
## Performance Improvements
### Vectorized Upcast & Downcast & Min & Max
Thanks to continued improvements in stdsimd, upcasting, downcasting, elementwise minimums, and elementwise maximums are now vectorized on x86. Yay!
### Partial Vectors
Faster's iterators can now vectorize loads/stores of an entire collection, regardless of its size. Previously, each scalar which didn't fit into a vector was broadcast into a vector, sent through the SIMD closure, and coalesced back into a scalar. While this worked well with faster 0.3.0's more limited API, it was slow, inefficient, and didn't work with swizzling and striping.
Now, simd_map and simd_reduce take an additional parameter, default. On uneven collections, faster will load every vector in the collection normally until it hits the end of the collection. Then, it loads the last Vector::width elements from the collection into the vector, and writes default over any elements which were already loaded. It also returns the number of elements which haven't already been loaded, they can be ignored by mapping operations.
The only quirk of this system is that the last vector is “shifted right”, so the last element of the collection is always at the highest index, and the vector may contain the default at index 0.
// Assuming 128-bit vectors:
let x = (&[1, 2, 3, 4, 5u32][..]).simd_iter()
.simd_map(u32s(9), |v| { println!("{:?}", v); v })
.scalar_collect();
// Output:
// u32x4(1, 2, 3, 4)
// u32x4(9, 9, 9, 5)
This means faster is now just as fast on 1023-element collections as it is on 1025-element collections. I'm also looking into ways to make this less quirky for the end user, although I expect its possible impact to be minimal.
## World Domination
Faster has reached a point where it is ready to be deployed in real code. To test faster in the real world, I have integrated it into foundational crates such as byteorder and bytecount, with very impressive results compared to the existing scalar or explicit simd implementations. I hope to see these improvements merged in shortly. Furthermore, I intend to continue sending improvements to libraries which would benefit from SIMD.
Here are some benchmarks from from the bytecount pull request:
# faster
test bench_count_big_1000000_hyper ... bench: 30,959 ns/iter (+/- 27)
# simd crate (-C target-cpu=native; upstream)
test bench_count_big_1000000_hyper ... bench: 61,536 ns/iter (+/- 2,047)
# scalar implementation (-C target-cpu=native; upstream)
test bench_count_big_1000000_hyper ... bench: 53,060 ns/iter (+/- 49)
And some benchmarks from byteorder:
# faster
test slice_u16::write_big_endian ... bench: 23,344 ns/iter (+/- 122) = 8567 MB/s
test slice_u32::write_big_endian ... bench: 46,681 ns/iter (+/- 160) = 8568 MB/s
test slice_u64::write_big_endian ... bench: 105,206 ns/iter (+/- 369) = 7604 MB/s
# scalar implementation (-C target-cpu=native; upstream)
test slice_u16::write_big_endian ... bench: 147,829 ns/iter (+/- 269) = 1352 MB/s
test slice_u32::write_big_endian ... bench: 112,241 ns/iter (+/- 652) = 3563 MB/s
test slice_u64::write_big_endian ... bench: 108,404 ns/iter (+/- 571) = 7379 MB/s
## What's next?
Faster is scarily fast, and is often much faster than competing solutions which do not offer automatic SIMD iteration. However, its iteration algorithm is lazy, and every call to next must check whether the iterator has any more items in it. This is typically incongruous with how SIMD code is written, where all operations are performed in an unrolled loop.
Adding an eager implementation of simd_map should totally alleviate this, and should encourage LLVM to unroll the iteration, allowing faster to truly brush up against the runtimes of hand-tuned assembly.
Although simd_reduce is already eager, it still relies on the aforementioned iteration protocol. Unfortunately, Rust's trait system makes it quite difficult to add an optimized implementation for non-lazy iterators, so I may have to hold off on this until specialization lands.
Changing up the iteration algorithm also presents an excellent opportunity to add a “hard mode” API to faster, which allows a user to manually pack and unpack streams of types as they please. Sneak peek:
pub fn char_to_emoji<'a, 'b>(buf: &'a [u8], out: &'b mut [u8]) -> &'b [u8] {
for (i, vec) in buf.simd_iter().pack().enumerate() {
let (a, b) = v.upcast();
let third = (a + u16s(55)) / u16s(64) + u16s(143).downcast((b + u16s(55)) / u16s(64) + u16s(143));
let fourth = ((v + u8s(55)) & u8s(0x3f)) + u8s(128);
// Make some room for interleaving
let (ta, tb) = third.upcast();
let (fa, fb) = fourth.upcast();
// Interleave third and fourth bytes
let third_foruth_a = ta.swap_bytes().merge_interleaved(fa);
let third_fourth_b = tb.swap_bytes().merge_interleaved(fb);
// Make some more room for another interleaving
let (tfa, tfb) = third_fourth_a.upcast();
let (tfc, tfd) = third_fourth_b.upcast();
// Interleave a constant 0xf09f with the third and fourth bytes,
// and store into out buffer
u32s(0xf09f0000).merge_interleaved(tfa).store(out, 128 * i);
u32s(0xf09f0000).merge_interleaved(tfb).store(out, 128 * i + tfa.width());
u32s(0xf09f0000).merge_interleaved(tfc).store(out, 128 * i + 2 * tfa.width());
u32s(0xf09f0000).merge_interleaved(tfd).store(out, 128 * i + 3 * tfa.width());
}
}
That may look complicated, but it's a quarter of the size of this 100-line monstrosity (which only supports AVX2!).
This should be exactly as fast as traditional SIMD assembly, because the user is responsible for all of the iterative work, while faster will only provide a basic load/store interface and its full selection of vector operations.
I'd like to thank everybody for following the development of faster, filing issues, lending algorithm ideas, and maintaining a high quality of discourse about the library. If you'd like to join the community, check out the Github page.
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# How can tilting a N64 cartridge cause such subtle glitches?
When the N64 cartridge is tilted in Super Mario 64, it reliably produces glitches such as this. Mario's limbs disappear, he rotates 90° and sinks into the floor, and the music gets messed up (usually it goes faster).
1. It's a very specific and subtle kind of behavior. I would have thought that tilting a cartridge would produce glitches more like "absolutely nothing works because everything's messed up, the game crashes", not "this specific 3d model will be rotated by precisely 90°". I mean, what, is there a special "rotate mario" pin in the cartridge?
2. This specific behavior can be reliably reproduced. These exact symptoms have been documented online. You can find videos on YouTube of this glitch happening to other people.
How could it possibly be that something as "brute force" as tilting the cartridge could reliably produce such a specific result without just making the whole game crash? This seems to me like dropping a typewriter down a flight of stairs and finding that it's typed out a novel.
I'm not necessarily asking about this specific glitch, I'm just trying to get some kind of a reading on how it's possible for cartridge tilting to reliably produce symptoms like this. Ideally, assume that I have at most basic knowledge about computers, roughly equivalent to a 101 textbook on computer circuitry and some programming experience, although explanations at any level are welcome.
• One brainstorm: interrupting that connection causes Mario to miss an animation update, after which the animation system (which was not programmed with an unreliable connection in mind) never recovers, and he's stuck in a zero-rotation state (which could be laying on the ground if his model was created in that orientation for some reason). This is all purely speculation, but there are lots of reasons that any sort of weirdness could happen. I once had a Windows PC that insisted its network drivers were missing, and in the end it ended up being a faulty ethernet cable and nothing else. – GrandOpener Nov 2 '16 at 19:27
• +1 for the "special "rotate mario" pin" :-D – KlaymenDK Nov 3 '16 at 12:38
• If you tilt the cartridge the characters get dizzy and lose their balance – Hack-R Nov 5 '16 at 16:02
• Theory: Nintendo was already dipping their toes on motion control waters. The next iteration even included a handle for easier experimentation! – xDaizu Feb 6 '18 at 16:45
• A short explanation that "tilting" means pulling out the cartridge mid-game might be in order; it suer had me confused for a bit. – SQB May 4 '18 at 8:04
The main point to understand is that the console is limited. RAM on the console itself is faster than memory on the cartridge (and the cartridge memory was usually read only, with a little non volatile for user data).
So the game designers had to carefully consider how to use the precious, but faster, ram on the console, vs the slower rom on the cartridge.
The stuff put into RAM was usually the graphics routines - the programming code that decided what pixels went where - and the cartridge was relied on to provide data when needed - a map's boundaries, object orientations and locations, etc. Things that consumed a small amount of space and so didn't take long to look up inside the cartridge while the console spent most of its time drawing on the screen using the code, textures, and other pieces needed much more frequently and quickly.
So what a cartridge tilt does is it damages or disables access to these lesser important, but still important, pieces of data that the game doesn't want to waste ram on. This results in humorous mistakes as the program tries to use bad data to act on the good data stored in ram.
• +1 This is really much more of answer to the question. The game doesn't crash because certain, not quite as vital information is no longer available. The game's core that renders everything is already in memory, so the game can continue. It probably also has something to do with not so great error checking/handling (so it the game uses garbage responses instead of exploding), too, if I had to guess. – jpmc26 Nov 3 '16 at 23:17
I wouldn't say "It's a very specific and subtle kind of behavior." I really think this is the case of undefined behavior that has been reproduced so many times from N64 launch to now that users have seen repeat behavior. In fact, sometimes cartridge tilting can actually delete your game save (Donkey Kong 64), and not just mess with the graphics (Goldeneye's "Get Down" glitch).
On a high level, cartridge tilting limits or blocks data flow from the cartridge to the console. Many "How To" guides state that to accomplish the tilt properly, keep the right side of the cartridge pushed down while slowly lifting up the left side. This is because the power pins and ground are located on the right side of the cartridge, while the data stream (and some 64 DD) pins are on the left side. Note that I think the cartridge would sit with the label facing the arrow/triangle.
Image courtesy of Bungle's 27/28 pin N64 wiring guide.
I couldn't find a full wiring diagram or explanation on what all of these wires do, nor am I an EE expert. I would love it if someone could provide an in-depth analysis of the Nintendo 64's cartridge I/O.
Different games show different behavior due to different engines and programing.
I would also argue against your claim that "This specific behavior can be reliably reproduced." Have you ever tried to perform any of the cartridge tilting glitches listed online or shown in YouTube videos? They are incredibly hard to reproduce, even in ideal environments. After years of attempting I've only done this successfully one time on Mario Kart, and have lost save data and broken several cartridges. We don't know how many attempts it took users who were recording to reproduce "ideal" tilt glitches.
• Alright, maybe not reliably, but the fact that it's happened more than once is already surprising enough for me. – Jack M Nov 1 '16 at 16:29
• @JackM It's not surprising. You are basically removing a component from a working system. If you would do this to some other system, it would also fail in somewhat reliable way: eg removing wheels from a car in motion always causes it to skid for a while and then stop. Now, there are some conditions you're taking for granted: if you'll do it on exactly same spot on the road, at same speed, etc (same state of the game), you'll get very similar results. But if you'd do this just before a tight mountain turn, the car would fall over a cliff - again with decent repeatability. – Agent_L Nov 2 '16 at 12:58
• @Agent_L Exactly - the car stops, it doesn't suddenly start translating the talk radio into Chinese. Where the threshold is is subjective, but there's clearly a point beyond which the ratio of "complexity of outcome" to "simplicity of input" becomes surprising. – Jack M Nov 3 '16 at 20:13
• @JackM Removing the wheels of a moving car is a good analogy. The only difference is that you understand the system of physics at play that causes the car to stop after removing the wheels. I'm sure someone who understands the system of an N64 cartridge would say "well of course the data gets corrupted in a particular way". – Abion47 Nov 3 '16 at 21:01
• @JackM It doesn't just stop. It produces lots of visual glitches (sparks), complex audio patterns (screeching and crumbling), damage to the road and the vehicle, and if it falls over a cliff, it will also make a nice little dance. And I'm not even talking about what happens to the driver :) Just because the situation doesn't seem complex to you doesn't make it simple. Human brains are just wired that way - they still have this nagging feeling that explaining thunderstorms by a human-level-intelligence-agent-in-the-skies is simpler than this whole complicated electromagnetism thing. – Luaan Nov 4 '16 at 8:44
This is the pinout of a Nintendo 64 cartridge (from here). The Nintendo 64 used a multiplexed address/data bus with a three-stage access protocol: write the high word of the address you want to access, write the low word, then read the data, all going across the same 16 pins.
This indirect access method means that program code needs to be run from the console's RAM, but the high speed of access means that programmers generally treated the cartridge as additional data-only RAM, reading data on an as-needed basis (as opposed to disk-based systems, where programmers would try to preload as much data as possible).
If you tilt the cartridge, you disconnect AD pins 0, 1, 2, 13, 14, and 15. These disconnected lines take on default values based on the wiring techniques in use, restricting which parts of the cartridge ROM the console can access, and what values can be read from them. Since most games read their code during startup rather than during gameplay, this doesn't (usually) cause crashes, but instead, highly-repeatable glitches.
• Also notice that this kind of data bus is not "aware" of the fact it has been temporarily disconnected - the host device gets nonsensical data but no signal that will tell it the cartridge was pulled. – rackandboneman Nov 3 '16 at 11:01
## The Gist
Games on the N64 typically did not have a lot of memory to use. Instead of keeping all code and data loaded onto the cartridge at one time, it would typically keep some necessary game code loaded into memory, and would only load temporary game code or data when needed.
This is seen with cartridge tilting on "The Legend of Zelda: Ocarina of Time"
Cartridge tilting on that game does not crash, since the code necessary to run is always loaded (in a file called "code"; it takes up about 1/3rd of memory at a time). However, the graphics appear very messed up. This is because the graphics for the game are loaded only when needed to save space, and interfering with the connection between the console and the cartridge loads garbage data instead of the actual data.
## How?
How does the N64 interact with the cartridge? As an N64 hacker, this is my understanding;
1. Boot code from the beginning of the ROM is loaded into memory and executed
2. The boot code sets up threading and loads some code to get the game "running"
3. Any time the game needs data from the cartridge, it services the console's "DMA read function" by writing to 0xA4600000.
The DMA read function tells the console to load from the cartridge. When the console receives a write to the DMA read memory, it looks for the file that is wanted to be loaded. How does it know which file? It looks in the "DMA Table" of the cartridge, which points to where in ROM the file is located.
In response to LawrenceC
I'm sorry, but this is not correct. The PC points to instructions specifically in RAM, not in ROM; However, when the N64 wants to execute code from the ROM it just services a DMA Read.
Hope I could help.
First, if you go to The Cutting Room Floor and browse awhile, you will discover that the binary images of many old ROM and cartridge based games have often have test code, unused code, debug code, partially overwritten code, and in some cases even messages and text-based source code. Why this code is not deleted and instead "skipped over" is beyond me - it's most likely the programming of the final master EEPROM did not erase what was previously there. It's really common in a lot of 70's, 80's and early 90's games.
All CPUs have an "Instruction Pointer" or "Program Counter" that contains an address of the instruction it's currently "working on."
A CPU in an endless loop reads the data in this cartridge (which is mapped in the CPU's address space) from this address, increments the program counter (sometimes more than once to get additional data needed by the instruction), executes the instruction, and repeats. Certain instructions called "jumps" or "branches" can "set" this register to a different value.
When you partially remove a cartridge, you are interfering with the address and data pins of the cartridge. This means the CPU is reading wrong data from the cartridge and will execute an unintended instruction, or an existing instruction with unintended operands or data.
It's also possible that interfering with the communication between the cartridge and the computer in 16-bit and later systems causes bus errors which trigger CPU exceptions, which might activate debug or test code in the cartridge.
This can cause the CPU to jump to spots in the binary image of the cartridge that it normally wouldn't, and if there is some unused test or old code that does something weird, it might get called and if conditions are right, it might actually work.
You used to be able to "fry" Atari 2600 games fairly easily by rocking the power switch on and off quickly. It would cause the CPU to not completely initialize and I always suspected that the Atari would start the cartridge at some point other than the beginning. Many games had weird glitches, yet were still playable. But there is not a lot of room in 2k or 4k byte Atari cartridges for unused code, so you usually didn't activate hidden functionality.
• The reason why that extra unused code is not deleted is most probably because it was being used to debug the game, and once it's been established that it worked well enough with that specific data on the cartridge, that's what they were gonna ship to customers. Just guessing from coding experience though, no hard info. – user1306322 Nov 5 '16 at 15:29
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# Why is the zero polynomial not assigned a degree?
Yesterday, I read in my textbook,
We assign degree to every polynomial and even a non-zero constant is assigned a degree $0$ but $0$ itself is not assigned a degree.
Why is that? Why we don't assign degree $0$ to the zero polynomial?
• The problem is not that $0$ is not assigned a degree, but that it is assigned too many degrees. – Emil Jeřábek May 2 '15 at 12:27
• Whether or not it is practical to associate a degree to the zero polynomial depends on the context. Two popular choices are -infinity and -1 (according to mathworld.wolfram.com/ZeroPolynomial.html ) – soegaard May 2 '15 at 17:00
Assigning a degree to the zero polynomial will cause trouble with important and useful theorems that relate the degree of a polynomial to its roots:
If $F$ is a field (examples of fields are $\mathbb{R}$, $\mathbb{C}$, $\mathbb{Q}$, $\mathbb{Z}/p\mathbb{Z}$), a polynomial $P$ with coefficients in $F$ (the set/ring of these polynomials is usually denoted by $F[x]$) of degree $n$, has at most $n$ distinct points $\alpha\in F$ such that $P(\alpha)=0$.
This theorem follows from the fact that we can repeatedly factor out terms of the form $x-\alpha$ (where $\alpha$ is a root) from $P(x)$, lowering the degree of the remaining polynomial by $1$ in each step. See also: http://en.wikipedia.org/wiki/Factor_theorem.
When we restrict to polynomials with coefficients in $\mathbb{C}$, the statement is related to the Fundamental Theorem of Algebra.
Now since the zero polynomial in $F[x]$ has a root at every point in $F$, at least for infinite fields ($\mathbb{R},\mathbb{C},\mathbb{Q}$), we cannot assign a finite value to the degree of the zero polynomial without getting into trouble.
• The fundamental theorem of algebra says any non constant complex polynomial has a root. It implies that any monic polynomial can be factored as a product of monic polynomials of degree$~1$ (i.e., of factors of the form $X-a$), which less elementary statement is often more directly useful. (The first statement must exclude constants, the second can be extended to cater for constant factors but get a bit more complicated doing so.) Note that neither formulation ever involves $\deg(0)$. Your statement is unclear (multiplicity is not defined) and at best means the second statement. – Marc van Leeuwen May 2 '15 at 13:46
• This doesn't seem to be much of a reason. One could just adjust the Fundamental Theorem of Algebra to talk about non-constant polynomials, in the same way that the Fundamental Theorem of Arithmetic excludes 1. – David Richerby May 2 '15 at 14:44
• @Marc van Leeuwen Thanks, I have tried to modify my answer to be a bit more clear about why I think this creates a problem with important theorems about roots of polynomials. – Uncountable May 2 '15 at 22:36
• The edit has unfortunately made the answer completely wrong (though I would say it makes clear that the link with FTA evaporates when trying to make precise statements about it). (1) You factorisation statement is not true: the RHS is monic, so it can only work for monic polynomials $P$. But then the zero polynomial is excluded from the start. (2) Putting "at most" into the statement of the FTA destroys it original meaning; the resulting statement is just as true replacing "complex" by "real" or "rational". – Marc van Leeuwen May 3 '15 at 4:36
• @MarcvanLeeuwen Of course the one-directional FTA with "at most $n$ roots" instead of "exactly $n$ roots" is also useful and true in any integral domain. (I think the theorem in this form has Lagrange's name on it.) And the factorization is easily fixed by adding a leading coefficient, i.e. $P(x)=a\prod (x-a_i)$. – Mario Carneiro May 3 '15 at 5:54
This is to make nice rules such as $$\text{deg }(PQ) = \text{deg }P + \text{deg }Q\\ \text{deg }(P+Q) \le \max(\text{deg }P , \text{deg }Q)$$
So the only value that makes it possible is $$\text{deg }0= -\infty$$
• Increasing in what variable ? – Belgi May 2 '15 at 10:17
• It doesn't make sense to say the polynomial $P(x)=0$ has $-\infty$ roots (fundamental theorem of algebra), so $+\infty$ would fit better in this sense. – user26486 May 2 '15 at 10:22
• "increasing" in the sense $|P(x)| = o(|Q(x)|) \implies \text{deg }P \le \text{deg }Q$. – mookid May 2 '15 at 10:22
• Alternatively, it is a consequences of asking that the sum identity $\deg (P + Q) \leq \max\{\deg P, \deg Q\}$ to hold for all $P, Q$. This always holds if we declare $\deg 0$ to be $-\infty$ but not if we declare $\deg 0 = +\infty$; to see this, consider the special case $Q = -P \neq 0$. – Travis Willse May 2 '15 at 11:49
• Another way to think of it is that $\operatorname{deg}$ is similar to a logarithm. – Ali Caglayan May 2 '15 at 17:40
The degree of a polynomial is the exponent of the term with the highest power (and non-zero coefficient).
$$5x-4x^4+2\to x^4\to 4$$ $$0x^{45}+1x^3+2x^2\to x^3\to3$$ $$21\to x^0\to0$$ $$0\to ??$$
The null polynomial contains no power of the variable.
• For $21x^n=21$ you assumed that $n=0, x^0=1$ and hence, $deg(21)=0$. What if the domain of the variable contains 1? – Sufyan Naeem May 8 '15 at 10:15
• This is irrelevant. A polynomial is a symbolic expression that is independent of any value of the variable. – Yves Daoust May 8 '15 at 10:23
• Are you sure that a polynomial expression is independent of any value of variable? I have read in my textbook that a variable in expression contains a domain and domains contains all the values for which expression is defined! – Sufyan Naeem May 8 '15 at 10:27
• That is, if I have a polynomial $33$ then why should I assume that in $x^n$, $n=0$ however the case remains same if an specific element of the domain of the variable is $1$. viz., $$33x^0=33$$ and, $$33(1)^n=33$$ also. I know, there may be a mistake but can you explain it to me? – Sufyan Naeem May 8 '15 at 10:38
• You are welcome. I will cleanup my comments, conversations aren't appropriate in comments. – Yves Daoust May 8 '15 at 10:51
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# American Institute of Mathematical Sciences
ISSN:
1937-5093
eISSN:
1937-5077
All Issues
## Kinetic & Related Models
September 2012 , Volume 5 , Issue 3
Issue dedicated to Michel Chipot
on the occasion of his 60th birthday
Select all articles
Export/Reference:
2012, 5(3): 441-458 doi: 10.3934/krm.2012.5.441 +[Abstract](1212) +[PDF](398.8KB)
Abstract:
We present a Fourier transform formula of quadratic-form type for the collision operator with a Maxwellian kernel under the momentum transfer condition. As an application, we extend the work of Toscani and Villani on the uniform stability of the Cauchy problem for the associated Boltzmann equation to any physically relevant Maxwellian molecules in the long-range interactions with a minimal requirement for the initial data.
2012, 5(3): 459-484 doi: 10.3934/krm.2012.5.459 +[Abstract](1636) +[PDF](484.0KB)
Abstract:
We develop a rigorous formalism for the description of the kinetic evolution of infinitely many hard spheres. On the basis of the kinetic cluster expansions of cumulants of groups of operators of finitely many hard spheres which are the generating operators of a nonperturbative solution of the Cauchy problem of the BBGKY hierarchy the nonlinear kinetic Enskog equation is derived. It is established that for initial states which are specified in terms of one-particle distribution functions the description of the evolution by the Cauchy problem of the BBGKY hierarchy and by the Cauchy problem of the generalized Enskog kinetic equation together with a sequence of explicitly defined functionals of a solution of stated kinetic equation are an equivalent. For the initial-value problem of the generalized Enskog equation the existence theorem is proved in the space of integrable functions.
2012, 5(3): 485-503 doi: 10.3934/krm.2012.5.485 +[Abstract](1700) +[PDF](439.1KB)
Abstract:
We discuss optimal control problems for the Fokker--Planck equation arising in radiotherapy treatment planning. We prove existence and uniqueness of an optimal boundary control for a general tracking--type cost functional in three spatial dimensions. Under additional regularity assumptions we prove existence of a continuous necessary first--order optimality system. In the one--dimensional case we analyse a numerical discretization of the Fokker--Planck equation. We prove that the resulting discrete optimality system is a suitable discretization of the continuous first--order system.
2012, 5(3): 505-516 doi: 10.3934/krm.2012.5.505 +[Abstract](1697) +[PDF](362.6KB)
Abstract:
In this paper, we establish two regularity criteria for the 3D MHD equations in terms of partial derivatives of the velocity field or the pressure. It is proved that if $\partial_3 u \in L^\beta(0,T; L^\alpha(\mathbb{R}^3)),~\mbox{with}~ \frac{2}{\beta}+\frac{3}{\alpha}\leq\frac{3(\alpha+2)}{4\alpha},~\alpha>2$, or $\nabla_h P \in L^\beta(0,T; L^{\alpha}(\mathbb{R}^3)),~\mbox{with}~\frac{2}{\beta}+\frac{3}{\alpha}< 3,~\alpha>\frac{9}{7},~\beta\geq 1$, then the weak solution $(u,b)$ is regular on $[0, T]$.
2012, 5(3): 517-536 doi: 10.3934/krm.2012.5.517 +[Abstract](1532) +[PDF](442.0KB)
Abstract:
The aim of this paper is to derive the quantum hydrodynamic system associated with the most general class of nonlinear Schrödinger equations accounting for Fokker--Planck type diffusion of the probability density, called of Doebner--Goldin. This 'Doebner--Goldin hydrodynamic system' is shown to be reduced in most cases to a simpler one of quantum Euler type by means of the introduction of a nonlinear gauge transformation that changes the fluid mean velocity into a new effective velocity corrected by an osmotic contribution. Finally, we also discuss some particular situations of especial interest and compare the structure of the resulting fluid systems with that of the viscous quantum hydrodynamic and the quantum Navier--Stokes equations stemming from maximization of the quantum entropy for Wigner--BGK models.
2012, 5(3): 537-550 doi: 10.3934/krm.2012.5.537 +[Abstract](1295) +[PDF](387.8KB)
Abstract:
In this paper we present a physically relevant hydrodynamic model for a bipolar semiconductor device considering Ohmic conductor boundary conditions and a non-flat doping profile. For such an Euler-Poisson system, we prove, by means of a technical energy method, that the solutions are unique, exist globally and asymptotically converge to the corresponding stationary solutions. An exponential decay rate is also derived. Moreover we allow that the two pressure functions can be different.
2012, 5(3): 551-561 doi: 10.3934/krm.2012.5.551 +[Abstract](1570) +[PDF](275.5KB)
Abstract:
The purpose of this paper is to extend the result concerning the existence and the uniqueness of infinite energy solutions, given by Cannone-Karch, of the Cauchy problem for the spatially homogeneous Boltzmann equation of Maxwellian molecules without Grad's angular cutoff assumption in the mild singularity case, to the strong singularity case. This extension follows from a simple observation of the symmetry on the unit sphere for the Bobylev formula which is the Fourier transform of the Boltzmann collision term.
2012, 5(3): 563-581 doi: 10.3934/krm.2012.5.563 +[Abstract](1316) +[PDF](429.3KB)
Abstract:
In this paper, we investigate large amplitude solutions to a system of conservation laws which is transformed, by a change of variable, from the well-known Keller-Segel model describing cell (bacteria) movement toward the concentration gradient of the chemical that is consumed by the cells. For the Cauchy problem and initial-boundary value problem, the global unique solvability is proved based on the energy method. In particular, our main purpose is to investigate the convergence rates as the diffusion parameter $\varepsilon$ goes to zero. It is shown that the convergence rates in $L^\infty$-norm are of the order $O\left(\varepsilon\right)$ and $O(\varepsilon^{1/2})$ corresponding to the Cauchy problem and the initial-boundary value problem respectively.
2012, 5(3): 583-613 doi: 10.3934/krm.2012.5.583 +[Abstract](1914) +[PDF](627.6KB)
Abstract:
In this paper we study the large-time behavior of perturbative classical solutions to the hard and soft potential Boltzmann equation without the angular cut-off assumption in the whole space $\mathbb{R}^n _x$ with $n≥3$ .We use the existence theory of global in time nearby Maxwellian solutions from [12,11].It has been a longstanding open problem to determine the large time decay rates for the soft potential Boltzmann equation in the whole space, with or without the angular cut-off assumption [26,1]. For perturbative initial data, we prove that solutions converge to the global Maxwellian with the optimal large-time decay rate of $O(t^{-\frac{N}{2}+\frac{N}{2r}})$ in the $L^2_v$$(L^r_x)$-norm for any $2\leq r\leq \infty$.
2012, 5(3): 615-638 doi: 10.3934/krm.2012.5.615 +[Abstract](1677) +[PDF](473.0KB)
Abstract:
We are concerned with the long-time behavior of global strong solutions to the non-isentropic compressible Navier-Stokes-Poisson system in $\mathbb{R}^{3}$, where the electric field is governed by the self-consistent Poisson equation. When the regular initial perturbations belong to $H^{4}(\mathbb{R}^{3})\cap \dot{B}_{1,\infty}^{-s}(\mathbb{R}^{3})$ with $s\in [0,1]$, we show that the density and momentum of the system converge to their equilibrium state at the optimal $L^2$-rates $(1+t)^{-\frac{3}{4}-\frac{s}{2}}$ and $(1+t)^{-\frac{1}{4}-\frac{s}{2}}$ respectively, and the decay rate is still $(1+t)^{-\frac{3}{4}}$ for temperature which is proved to be not optimal.
2012, 5(3): 639-667 doi: 10.3934/krm.2012.5.639 +[Abstract](1893) +[PDF](1136.3KB)
Abstract:
We consider the time-dependent 1D Schrödinger equation on the half-axis with variable coefficients becoming constant for large $x$. We study a two-level symmetric in time (i.e. the Crank-Nicolson) and any order finite element in space numerical method to solve it. The method is coupled to an approximate transparent boundary condition (TBC). We prove uniform in time stability with respect to initial data and a free term in two norms, under suitable conditions on an operator in the approximate TBC. We also consider the corresponding method on an infinite mesh on the half-axis. We derive explicitly the discrete TBC allowing us to restrict the latter method to a finite mesh. The operator in the discrete TBC is a discrete convolution in time; in turn its kernel is a multiple discrete convolution. The stability conditions are justified for it. The accomplished computations confirm that high order finite elements coupled to the discrete TBC are effective even in the case of highly oscillating solutions and discontinuous potentials.
2012, 5(3): 669-672 doi: 10.3934/krm.2012.5.669 +[Abstract](1112) +[PDF](270.9KB)
Abstract:
N/A
2018 Impact Factor: 1.38
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## anonymous 4 years ago find the domain y=f(x)=x^2-6x+6
1. hartnn
domain means the values which 'x' can take in that equation.
2. hartnn
do u see any values that 'x' can't take ?
3. anonymous
i dont get it......??
4. hartnn
(1)there is no denominator, if there was, then 'x' could not take values for which denominator=0
5. anonymous
so ther is no soln for this??
6. hartnn
(2) there is no $$\sqrt.$$ sign, if there was , then 'x' cannot take values for which the expression under $$\sqrt{...}$$becomes negative.
7. hartnn
here, none of the 2 cases (1) or (2) arise. so 'x' can take all real values.
8. hartnn
so, domain is ALL real numbers R
9. hartnn
understood ?
10. anonymous
sorry i quite didnt get it
11. hartnn
which par ?
12. anonymous
x belongs to R then can we supoose any no we want in the place of x??
13. hartnn
yes, x can take any real value. u have options/choices ?
14. anonymous
how wud u solve this prob?
15. hartnn
example : f(x) =1/ x x cannot take value =0 f(x) = 1/(x-3) x cannot take value = 3 f(x) = sqrt{x-4} x cannot take value less than 4
16. anonymous
can u be mo` specific...?i cant understand
17. hartnn
i just gave u specific examples. and there's nothing to solve, you can say 'x' can take all real values...
18. anonymous
y did u add -3 in da denominator?
19. ParthKohli
By the way, all polynomials have the domain $$(-\infty,\infty)$$ a.k.a $$\mathbb{R}$$.
20. anonymous
can u do the whole process??
21. anonymous
Domain means whenever the function exists. To figure this out, just find when the function does not exist, meaning when you're dividing by 0. In this case, you're never dividing by 0, so the function ALWAYS exists, so the domain is ]-inf,inf[, also know as "R" for all real numbers. If say f(x) = 1/(x-4), then the domain is ]-inf,-4[U]-4,inf[, because the function exists everywhere except at x=-4.
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# Mesh up
Geometry Level 1
In the above figure, let $$a = \dfrac{PS\cdot QB\cdot RT}{BR\cdot TP\cdot SQ}$$ and $$b =\dfrac{ZV\cdot YW\cdot XC}{WZ\cdot VX\cdot CY}$$.
What is the value of $$a-b$$?
×
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# Properties
Label 254898.de Number of curves $6$ Conductor $254898$ CM no Rank $1$ Graph
# Related objects
Show commands for: SageMath
sage: E = EllipticCurve("254898.de1")
sage: E.isogeny_class()
## Elliptic curves in class 254898.de
sage: E.isogeny_class().curves
LMFDB label Cremona label Weierstrass coefficients Torsion structure Modular degree Optimality
254898.de1 254898de6 [1, -1, 0, -1748422363851, -889851231810145571] [2] 3397386240
254898.de2 254898de4 [1, -1, 0, -109484301411, -13848343276727723] [2, 2] 1698693120
254898.de3 254898de5 [1, -1, 0, -37292089851, -31838079298229555] [2] 3397386240
254898.de4 254898de2 [1, -1, 0, -11562685731, 120273121639957] [2, 2] 849346560
254898.de5 254898de1 [1, -1, 0, -8952530211, 325631283424789] [2] 424673280 $$\Gamma_0(N)$$-optimal
254898.de6 254898de3 [1, -1, 0, 44596441629, 945935843912149] [2] 1698693120
## Rank
sage: E.rank()
The elliptic curves in class 254898.de have rank $$1$$.
## Modular form 254898.2.a.de
sage: E.q_eigenform(10)
$$q - q^{2} + q^{4} + 2q^{5} - q^{8} - 2q^{10} + 4q^{11} + 2q^{13} + q^{16} - 4q^{19} + O(q^{20})$$
## Isogeny matrix
sage: E.isogeny_class().matrix()
The $$i,j$$ entry is the smallest degree of a cyclic isogeny between the $$i$$-th and $$j$$-th curve in the isogeny class, in the LMFDB numbering.
$$\left(\begin{array}{rrrrrr} 1 & 2 & 4 & 4 & 8 & 8 \\ 2 & 1 & 2 & 2 & 4 & 4 \\ 4 & 2 & 1 & 4 & 8 & 8 \\ 4 & 2 & 4 & 1 & 2 & 2 \\ 8 & 4 & 8 & 2 & 1 & 4 \\ 8 & 4 & 8 & 2 & 4 & 1 \end{array}\right)$$
## Isogeny graph
sage: E.isogeny_graph().plot(edge_labels=True)
The vertices are labelled with LMFDB labels.
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## anonymous one year ago Sigma (summation) notation question. Will post in comment
1. anonymous
$\sum_{k = 0}^{4}\frac{ 1 }{ k^2 + 1 }$ I rewrote it as: $\sum_{k = 0}^{4} (k^2 + 1)^{-1}$ do I use now the formula $\frac{ n(n+1)(2n+1) }{ 6 }$ for the expression (k^2 + 1)^-1 ?
2. anonymous
I think that what I wrote in the previous step is incorrect; I don't think the formula can be applied. It can be solved with brute force. $\frac{ 1 }{ 0^2 + 1} + \frac{ 1 }{ 1^2 + 1} + \frac{ 1 }{ 2^2 + 1} + \frac{ 1 }{ 3^2 + 1} + \frac{ 1 }{ 4^2 + 1}$ $= \frac{ 1 }{ 0 + 1 } + \frac{ 1 }{ 1 + 1 } + \frac{ 1 }{ 4 + 1 } + \frac{ 1 }{ 9 + 1 } + \frac{ 1 }{ 16 + 1 }$ $= \frac{ 1 }{ 1 } + \frac{ 1 }{ 2 } + \frac{ 1 }{ 5 } + \frac{ 1 }{ 10 } + \frac{ 1 }{ 17 } =\frac{ 158 }{ 85 }$
3. anonymous
you are correct in the second solution :)
4. anonymous
5. anonymous
@Halmos I'm I correct that the summation formula won't work? The formulas can be found @ https://mathwiz.uwstout.edu/video/math-156/ch5s1.html 4th video down (It's not necessary to watch the video, it's on the web page).
6. anonymous
the formula u had of k^2 won't work but there is a formula called zeta function and like wise things, this is advanced mathematics don't think about it :P
7. anonymous
Thanks.
8. anonymous
np :)
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2 added 3197 characters in body
Edit: The idea is that the total space $EG$ of the classifying bundle of $G$ is contractible and a cofibrant replacement of the point $1$ on which $G$ acts freely. Thus, the construction $BG = EG/G$ is taking a stack-y quotient $BG = 1//G$.
There is a bit more to this idea than may first appear; let me take a related example (which may appear to have some Eulerian "wishful thinking" in it, but have a little faith here!). One way of taking the reciprocal is to pass to a geometric series, so that one suggestive notation for the free monoid construction
$$\sum_{n \geq 0} X^{\otimes n}$$
(in a suitable monoidal category; see my other comment on categorifying exponentiation) is a categorified reciprocal $1/(1 - X)$. We can apply this idea in group cohomology for a group $G$ as follows: think of $\mathbb{Z}$ as being an abelianized point, and consider a standard $G$-free resolution of $\mathbb{Z}$ such as the normalized homogeneous bar resolution, which we can think of as an abelianized $EG$. In one way of constructing this bar resolution (see e.g. Hilton-Stammbach p. 217), the degree $n$ component of $EG$ is
$$\mathbb{Z}G \otimes IG^{\otimes n}$$
where $IG$ is the augmentation ideal, i.e., the kernel of the augmentation map $\varepsilon: \mathbb{Z}G \to \mathbb{Z}$. As a bare module (or seen in degree 0), $IG$ can be seen as an abelianized "$G - 1$". However, in the differential-graded world, it is better to think of it as in degree 1, and this degree 1 shift $\Sigma IG$ can be seen as a categorified "$-IG = 1 - G$" (this may make more sense in the "super-world"; see for example my old notes on the Lie operad when I was doing some work with Saunders Mac Lane, or consider for example the occurrence of signs in the Euler characteristic). So now the total space of the bar resolution $EG$ is the sum of the degree $n$ components
$$\mathbb{Z}G \otimes \sum_{n \geq 0} (\Sigma IG)^{\otimes n}$$
which is an abelianized categorified form of $g \cdot \sum_{n \geq 0} (1 - g)^n$ which is formally $1$ by the geometric series. Very similar types of categorified geometric series constructions occur in Joyal's theory of species (see especially his article on virtual species in Springer LNM 1234), which constructs the Lie operad by categorified constructions [if you read between the lines!], and in the bar resolution for operads as discussed by Ginzburg-Kapranov; I tried to amplify this in my notes on the Lie operad.
Just to put one final gloss on this: consider the Schubert cell decomposition of projective space as a finite geometric series. For a field $k$ we have
$$\mathbb{P}^{n-1}(k) = \frac{k^n - 1}{k - 1} = 1 + k + k^2 + \ldots + k^{n-1}$$
(the '$1$' in the numerator is a zero vector, and the denominator is nonzero scalars $k^\ast$). We can pass to a limit and get infinite-dimensional projective space. Keeping in mind that degree shifts introduce some sign changes in the geometric series, the infinite-dimensional projective space $\mathbb{RP}^\infty$ would be a model of the homotopy quotient $1//\mathbb{R}^* \simeq 1//\mathbb{Z}_2$.
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# Computer Graphics Math Macros C/C++
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## Recommended Posts
Hi, I am looking to condense some of my code. I am interested in computer graphics math macros written in C or C++. Does anyone have any they would be willing to share? For example, dot product, cross product, matrix multiplication, etc.. Thanks.
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I would recommend you make these formal operators or methods that act on vector and matrix objects, but if you're in the black market for some dirty unsafe macros, I might have a few lying around:
#define VectorDot(a,b,c) (c = a[0]*b[0] + a[1]*b[1] + a[2]*b[2];) #define VectorCross(a,b,c) (c[0] = a[1]*b[2]-a[2]*b[1]; c[1] = a[2]*b[0]-a[0]*b[2]; c[2] = a[0]*b[1]-a[1]*b[0];) #define VectorNormalize(a,c) (c[0] = 1.0f / sqrt(a[0]*a[0] + a[1]*a[1] + a[2]*a[2]); c[1] = a[1] * c[0]; c[2] = a[2] * c[0]; c[0] *= a[0];)
This assumes you're using some form of array vector, or God forbid you're using these macros on objects that overload those operators. While I don't condone their usage, here are a couple that I had lying around.
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ooh, thanks. Just what I was looking for.
If you have the time, could you explain to me the advantage of making these macros formal operators? Why would this be safer? Is there a performance penalty or advantage?
Thanks.
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Original post by jdaniel...could you explain to me the advantage of making these macros formal operators? Why would this be safer? Is there a performance penalty or advantage?
Some of those questions are addressed here.
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Original post by jdanielooh, thanks. Just what I was looking for.If you have the time, could you explain to me the advantage of making these macros formal operators? Why would this be safer? Is there a performance penalty or advantage?Thanks.
It would be safer because macro's are just copy/paste. Everything can go wrong there. If you give it unexpected input, you *might* get compile errors, or it might compile, and just return the wrong output. It doesn't enforce that the arguments should be vectors. You could probably give it a semicolon among the arguments, and it'd insert that, which'd cause some very interesting results... [wink]
But it might still compile!
If you do the same as a function (or operator), you would have type checking. You would only be able to pass vectors as input, anything else would be rejected.
A function call would be a few cycles slower, but unless you call this function at least 100,000 times per second, it wouldn't make a noticeable difference. Or you could inline the function, and it would be exactly as fast as the macro, while still giving you type safety, and be able to work with your compilers intellisense/autocompletion/whatever else.
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IIRC, in the the quake1 or quake2 engine source you will find a lot of them. For a more decent math library I recommend looking at the doom 3 code...
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C++ Workshop - Week 3 (Ch. 4) - Quizzes and Exercises
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Quizzes and Extra Credit for Week 3.
Post your answers to the Quizzes from Week 3 here rather than in the chapter thread, so as not to spoil things for the others.
Chapter 4 Quiz 1. What do statements do in C++? 2. Where can you put a compound statement? 3. What’s another name for a compound statement? 4. What syntax identifies the beginning and ending of a block? 5. What is an expression in C++? 6. Which side of the equal operator can an expression ALWAYS be on? Why not the other side? 7. What do operators act upon? 8. Can an expression be acted upon by operators? 9. What does the assignment operator do? 10. What is an r-value and an l-value? How does this relate to question 6 above? 11. What are the 5 mathematical operators and what do they do? Give examples of inputs and outputs. 12. What happens when you subtract a large unsigned number for a smaller unsigned number? 13. [Extra Credit]What is integer truncation? How can it be avoided? 14. What are the increment/decrement operators, and how are they used. Show examples. 15. What is the difference between prefix and postfix increment/decrement? What are the results? Show examples. 16. What is the precedence of the mathematical operators? 17. Are the operators left-to-right or right-to-left? 18. What about the assignment operator? 19. What operator can ALWAYS be used to change the precedence of an expression? 20. How are nested parentheses read? Show examples. 21. Can all expressions be evaluated for their truth? 22. What data type is exclusively for truthiness? (tip of the hat to Stephen Colbert) 23. What is the ANSI standard size of that data type? 24. What are the six relational operators? Show examples of what values they return with different inputs. 25. What does an ‘if’ statement do? 26. When is an if-block executed, when is it skipped? 27. What does an ‘else’ statement do? 28. When is an else-block executed, when is it skipped? 29. [Extra Credit] What does an ‘else if’ statement do? When is it executed and when is it skipped? 30. What are the 3 logic operators in C++? How are they used? Show examples. 31. What is “Short Circuit Evaluation?” How might it be beneficial? 32. What is the precedence of the logical and conditional operators? 33. What is the conditional operator? How is it used? When might it be useful? Show examples. Chapter 4 Exercises 1. The equation for distance traveled (dist) with a constant velocity is equal to the rate of movement (rate) multiplied by the time interval (time) traveling. (dist=rate*time) Write a program which prompts the user for rate and time, computes the distance traveled, and then outputs the result to the screen. 2. Write a program that knows a secret number and prompts the user to guess the number. When the user enters the number the program should notify the user whether the number guessed was too high, too low, or juuust right. We will modify this program later when we’ve covered loops in order to allow the user to guess until they get the correct answer or run out of tries. Additionally, we will cover random number generation when we explore functions next week.
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1. What do statements do in C++?
Controls the sequence of execution, evaluates and expression, or does nothing.
2. Where can you put a compound statement?
Anywhere you can put a statement.
3. What’s another name for a compound statement?
Block
4. What syntax identifies the beginning and ending of a block?
‘{‘ & ‘}’
5. What is an expression in C++?
Anything that evaluates to a value.
6. Which side of the equal operator can an expression ALWAYS be on? Why not the other side?
Right side, the left side is where the expression is assigned.
7. What do operators act upon?
Operands
8. Can an expression be acted upon by operators?
Yes, any expression can be an operand.
9. What does the assignment operator do?
Changes the value on the left side of the operator
10. What is an r-value and an l-value? How does this relate to question 6 above?
L-left side, r-right side. Determines some of the characteristics of the value (5 can only be an r-value)
11. What are the 5 mathematical operators and what do they do? Give examples of inputs and outputs.
Addition (+), subtraction (-), multiply (*), divide (/), modulus – int remainder (%). 1+2*2-3/3=5. (wrong =4)
12. What happens when you subtract a large unsigned number for a smaller unsigned number?
Takes the difference from the highest available value.
13. [Extra Credit]What is integer truncation? How can it be avoided?
Non-integer quotients are shortened to their integer values. Using floating point values instead of int.
14. What are the increment/decrement operators, and how are they used. Show examples.
(++) and (--) are used to increase or decrease the value by 1. x++ adds one to the value of x and stores that new value in x.
15. What is the difference between prefix and postfix increment/decrement? What are the results? Show examples.
Prefix increments/decrements prior to assignment, postfix assigns after increment/decrement. int y = ++x (y=x+1), int y = x++ (y=x)
16. What is the precedence of the mathematical operators?
*, ?, % before +, -
17. Are the operators left-to-right or right-to-left?
Left to right
18. What about the assignment operator?
Right to left
19. What operator can ALWAYS be used to change the precedence of an expression?
Parentheses
20. How are nested parentheses read? Show examples. I
nnermost contained operators are read first. (3 * 4 + (3 - 2) – 1), the (3 – 2) is performed first.
21. Can all expressions be evaluated for their truth?
But what is the nature of truth…….yes
22. What data type is exclusively for truthiness? (tip of the hat to Stephen Colbert)
bool
23. What is the ANSI standard size of that data type?
1 byte
24. What are the six relational operators? Show examples of what values they return with different inputs.
1 == 1 (true), 1 != 1 (false), 1 > 1 (false), 1 >= 1 (true), 1 < 1 (false), 1<=1 (true).
25. What does an ‘if’ statement do?
It allows you to test for a condition.
26. When is an if-block executed, when is it skipped?
It is executed if the relation expression it is based upon is true; it is not executed when false.
27. What does an ‘else’ statement do?
Executes a block of code when the condition on which the if statement is based is false.
28. When is an else-block executed, when is it skipped?
When the condition evaluates to true.
29. [Extra Credit] What does an ‘else if’ statement do? When is it executed and when is it skipped?
It allows you to check for another condition when the first evaluates to false.
30. What are the 3 logic operators in C++? How are they used? Show examples.
AND &&, OR ||, NOT ! if ((x == 1) && (y == 7)) evaluates true if both conditions are true. if ((x == 1) || (y == 7)) evaluates true if either is true. if ( !(x==1)) evaluates true of the condition evaluates false.
31. What is “Short Circuit Evaluation?” How might it be beneficial?
If the condition has already been determined, there is no need to evaluate additional conditions, this can save time in execution.
32. What is the precedence of the logical and conditional operators?
&& then || then ?:
33. What is the conditional operator? How is it used? When might it be useful? Show examples.
?:, like a mini if, else statement. Would be useful for a small conditional statement. x = (x < 10) ? ++x : 1;
Chapter 4 Exercises
1. The equation for distance traveled (dist) with a constant velocity is equal to the rate of movement (rate) multiplied by the time interval (time) traveling. (dist=rate*time) Write a program which prompts the user for rate and time, computes the distance traveled, and then outputs the result to the screen.
#include <iostream>int main(){ int rate, time; std::cout << "Please insert rate: "; std::cin >> rate; std::cout << "Please insert your time: "; std::cin >> time; std::cout << "The distance traveled is: " << rate * time << std::endl; return 0;}
2. Write a program that knows a secret number and prompts the user to guess the number. When the user enters the number the program should notify the user whether the number guessed was too high, too low, or juuust right. We will modify this program later when we’ve covered loops in order to allow the user to guess until they get the correct answer or run out of tries. Additionally, we will cover random number generation when we explore functions next week.
#include <iostream>int main(){ int guess, x=5 ; std::cout << "Please enter your guess: "; std::cin >> guess; if (guess == x) { std::cout << "You are right!\n"; } else if (guess < x) { std::cout << "Too low.\n"; } else { std::cout << "Too high\n"; } return 0;}
[Edited by - Gambler on June 24, 2006 4:47:15 PM]
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11. What are the 5 mathematical operators and what do they do? Give examples of inputs and outputs. Addition (+), subtraction (-), multiply (*), divide (/), modulus – int remainder (%). 1+2*2-3/3=5.
Looks like someone forgot about operator precedence.
That expression evalutes to 4 :
1 + 2 * 2 - 3 / 3(equivelant) 1 + ( 2 * 2 ) - ( 3 / 3 )1 + 4 - 14
I'm not trying to be pedantic, but this kind of mistake can be really hard to track down in code, as the compiler cannot help you. Whether you intended the value to be 4, or 5, it does not know.
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Dang it - I knew better than that - thanks for the callout
Quizzes and Extra Credit for Week 3.
Chapter 4 Quiz
1. What do statements do in C++?
controls the sequence of execution, evaluates and expression, or does nothing (the null statement)
2. Where can you put a compound statement?
Any place that we can put statement
3. What’s another name for a compound statement?
a block
4. What syntax identifies the beginning and ending of a block?
opening brace ({) identify the beginningclosing brase (}) identify the ending
5. What is an expression in C++?
Expression in C++ can be defined as any statemet that return a value
6. Which side of the equal operator can an expression ALWAYS be on? Why not the other side?
The right side of an assignment operator
7. What do operators act upon?
Operators act on operand
8. Can an expression be acted upon by operators?
can
9. What does the assignment operator do?
The assignment operator (=) causes the operand on the left side of the assignment operator to have its value changed to the value on the right side of the assignment operator
10. What is an r-value and an l-value? How does this relate to question 6 above?
r-value is an operand that can be on the left side of an assignment operatorl-value is an operand that can be on the right side of an assignment operatorThe relation : the lvalue is assigned to the rvalue value.
11. What are the 5 mathematical operators and what do they do? Give examples of inputs and outputs.
addition (+), subtraction (-), multiplication {*), division (/), and modulus (%)example of addition a= 4+6 a= 10example of subtraction b = 5-3 b = 2example of multiplication c = 3*3 c = 9example of division d = 6/2 d = 3example of modulus e = 20%2 e = 0
12. What happens when you subtract a large unsigned number for a smaller unsigned number?
overflow
13. [Extra Credit]What is integer truncation? How can it be avoided?
The floating value on the integer was truncated, it can be avoided by assigned the integer into double or float data type
14. What are the increment/decrement operators, and how are they used. Show examples.
increment is increases the value of the variable by 1, decrement is decreases the value of the variable by 1The can be used by using "++" or "--"For example: we want to increment variable d, then we just write d++ on our coding
15. What is the difference between prefix and postfix increment/decrement? What are the results? Show examples.
Prefix : Increment the value and then fetch it, Postfix : Fetch the value and then increment the original value.For example int a = 21; int b = a++ int c = ++athen the result is a = 23, b=21, c = 23;
16. What is the precedence of the mathematical operators?
modulus (%), division (/), multiplication (*), subtracti0n (-), and at las t addition (+).
17. Are the operators left-to-right or right-to-left?
left to right
18. What about the assignment operator?
Right to left
19. What operator can ALWAYS be used to change the precedence of an expression?
parentheses
20. How are nested parentheses read? Show examples.
Read from the inside out.for example : ((a+b)*c) (a+b) is executed first, and then times with c
21. Can all expressions be evaluated for their truth?
can
22. What data type is exclusively for truthiness? (tip of the hat to Stephen Colbert)
bool
23. What is the ANSI standard size of that data type?
1 byte
24. What are the six relational operators? Show examples of what values they return with different inputs.
equals (==}, not equals (!=}, Greater than (>), Greater than or equals {>=}, less than (<}, less than or equals (<=)...for example 3==3 is true, 3>2 is true, 3>4 ia false, 3<2 is true, 3<5 is false.
25. What does an ‘if’ statement do?
the if statement enable us to test for a condition (such as whether two variable are equal) and branch to different parts of your code, depending on the result.
26. When is an if-block executed, when is it skipped?
The if-block executed when expression is true, and it's skipped when the expression is false
27. What does an ‘else’ statement do?
To executed another statement if the condition is false
28. When is an else-block executed, when is it skipped?
an else block executed when expression is false, and it's skipped when the expression is true
29. [Extra Credit] What does an ‘else if’ statement do? When is it executed and when is it skipped?
the else if statement is the statement that will b executed if the 'if' condition is false, it was skipped if the "if" condition is true.
30. What are the 3 logic operators in C++? How are they used? Show examples.
Logical AND (&&), Logical OR (||), Logical Not (!)
31. What is “Short Circuit Evaluation?” How might it be beneficial?
The situation where the compiler only evaluate the first statement, and if it is determined, whether false or true, the compiler will not determined the next statement.The beneficial is the compiler will be more efficiency
32. What is the precedence of the logical and conditional operators?
NOT(!), AND (&&), OR(||)
33. What is the conditional operator? How is it used? When might it be useful? Show examples.
conditional operator is the only operator that take three term(expression1) ? (expression2) : (expression3)For example: (3>1)?A:B will be return statement A because (3>1) is true
Chapter 4 Exercise
1. The equation for distance traveled (dist) with a constant velocity is equal to the rate of movement (rate) multiplied by the time interval (time) traveling. (dist=rate*time) Write a program which prompts the user for rate and time, computes the distance traveled, and then outputs the result to the screen.
#include <iostream>using namespace std;int main(){ int dist, rate, time; cout << "Enter the rate of movement: " ; cin >> rate; cout << "Enter the time: "; cin >> time; dist = rate * time; cout << "The distance is " << dist << endl; return 0; }
2. Write a program that knows a secret number and prompts the user to guess the number. When the user enters the number the program should notify the user whether the number guessed was too high, too low, or juuust right. We will modify this program later when we’ve covered loops in order to allow the user to guess until they get the correct answer or run out of tries. Additionally, we will cover random number generation when we explore functions next week.
#include<iostream>using namespace std;int main(){ int input, secret = 8; cout << "Enter your choosen number: " ; cin >> input; if(input != secret){ if(input > secret) cout << "To High" << endl; else cout << "To Low" << endl; }else cout << "Just Right" << endl; return 0;}
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Chapter 4 Quiz
1. What do statements do in C++?
They can control the sequence of operations, evauluate an expression, or do nothing.
2. Where can you put a compound statement?
Any place you can put a single statement.
3. What’s another name for a compound statement?
A block.
4. What syntax identifies the beginning and ending of a block?
{}
5. What is an expression in C++?
Anything that evaulates a value.
6. Which side of the equal operator can an expression ALWAYS be on? Why not the other side?
The right side. Constants cannot have their values changes, thus cannot be on the left side of an equation.
7. What do operators act upon?
Operands.
8. Can an expression be acted upon by operators?
Yes.
9. What does the assignment operator do?
Assigns a value to a variable.
10. What is an r-value and an l-value? How does this relate to question 6 above?
Values that can be on the right side and left side of the assignment operator, respectively. Constants can only be r-values.
11. What are the 5 mathematical operators and what do they do? Give examples of inputs and outputs.
x = 1 + 2;
returns the value 3 to x
-, subtracts values from one another
x = 2 - 1;
returns the value 1 to x
*, multiplies values
x = 2 * 2;
returns the value 4 to x
/, divides values
x = 5 / 2;
returns the value 2 to x
%, called modulus, returns the division remainder of two values
x = 5 % 2;
returns the value 1 to x
12. What happens when you subtract a large unsigned number for a smaller unsigned number?
It wraps around to the largest unsigned value that variable can hold, very bad.
13. [Extra Credit]What is integer truncation? How can it be avoided?
Not 100% sure - this is when a real number is plugged into a variable of int type, and is then truncated, IE 5.2 becomes 5. This can be avoid by simply assigning the proper variable types to variables in your program? I have a feeling I'm missing something here...
14. What are the increment/decrement operators, and how are they used. Show examples.
++
x++ or x += 1
increments x by 1
--
x-- or x -= 1
decrements x by 1
15. What is the difference between prefix and postfix increment/decrement? What are the results? Show examples.
Prefix increments your variable before performing whatever operation you're doing, postfix increments it after.
So:
a = b++
assigns b to a, and then increments it.
a = ++b
increments b, then assigns b to a.
16. What is the precedence of the mathematical operators?
%, * and / first, from left to right, and then + and - from left to right.
17. Are the operators left-to-right or right-to-left?
Left to right.
18. What about the assignment operator?
The right most assignment operator is always evaluated first.
19. What operator can ALWAYS be used to change the precedence of an expression?
Parentheses.
20. How are nested parentheses read? Show examples.
From the inside out. So:
x = ((a + b) * c)
and a = 1, b = 2 and c = 3, a + b is evaluated first (3) and then multiple by c, thus assigning 9 to x.
21. Can all expressions be evaluated for their truth?
Yes.
22. What data type is exclusively for truthiness? (tip of the hat to Stephen Colbert)
bool.
23. What is the ANSI standard size of that data type?
1 byte.
24. What are the six relational operators? Show examples of what values they return with different inputs.
if a = 1 and b = 2...
a == b evaluates false
a != b evaluates true
a > b evaluates false
a >= b evaluates false
a < b evaluates true
a <= b evaluates true
if a = 2 and b = 2...
a == b evaluates true
a != b evaluates false
a > b evaluates false
a >= b evaluates true
a < b evaluates false
a <= b evaluates true
25. What does an ‘if’ statement do?
It tests for a condition and allows you to execute certain blocks of code depending on the result.
26. When is an if-block executed, when is it skipped?
It is executed if the 'if' expression is evaluated as true, otherwise it is skipped.
27. What does an ‘else’ statement do?
It executes a block of code if initial 'if' expression is not true.
28. When is an else-block executed, when is it skipped?
It is only executed when the initial 'if' expression isn't true, otherwise it is skipped.
29. [Extra Credit] What does an ‘else if’ statement do? When is it executed and when is it skipped?
I *think*, this is basically just an embedded if statement in and else statement? It is executed if expression a was false but expression b was true.
30. What are the 3 logic operators in C++? How are they used? Show examples.
&& (AND), || (OR) and ! (NOT). So, if x is 5 and y is 5:
((x == 5) && (y == 5)) evaluates true
((x == 5) || (y == 4)) evaluates true
(!(x == 5)) evaluates false
31. What is “Short Circuit Evaluation?” How might it be beneficial?
For example, if the first part of an && expression is false, C++ won't bother to check the second half, since it's evaluation is meaningless at this point. This results in faster run times (I would imagine?)
32. What is the precedence of the logical and conditional operators?
&& then || then ?: from left to right.
33. What is the conditional operator? How is it used? When might it be useful? Show examples.
z = (x > y) ? x : y;
This states if x is greater than y, assign x to z, otherwise, assign y to z. It is basically a shorter, simpler if else statement.
Chapter 4 Exercises
1. The equation for distance traveled (dist) with a constant velocity is equal to the rate of movement (rate) multiplied by the time interval (time) traveling. (dist=rate*time) Write a program which prompts the user for rate and time, computes the distance traveled, and then outputs the result to the screen.
#include <iostream>int main(){ using namespace std; int dist, rate, time; cout << "Enter the rate of movement:" << endl; cin >> rate; cout << endl << "Enter the amount of time traveled:" << endl; cin >> time; dist = rate * time; cout << endl<< "The distance traveled is: " << dist << endl; return 0;}
2. Write a program that knows a secret number and prompts the user to guess the number. When the user enters the number the program should notify the user whether the number guessed was too high, too low, or juuust right. We will modify this program later when we’ve covered loops in order to allow the user to guess until they get the correct answer or run out of tries. Additionally, we will cover random number generation when we explore functions next week.
#include <iostream>int main(){ using namespace std; int secret = 33; //this number is secret and can be changed int guess; cout << "I know a secret number! Try to guess it:" << endl; cin >> guess; if (guess > secret) cout << endl << "Your guess was too big, sorry." << endl; if (guess < secret) cout << endl << "Your guess was too small, sorry." << endl; if (guess == secret) cout << endl << "You are so smrt! You guessed the number correctly." << endl; return 0;}
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1. What do statements do in C++?
1a. C++ statements control the sequence of execution, evaluate expressions, or do nothing at all.
2. Where can you put a compound statement?
2a. Any place you can put a single statement.
3. What’s another name for a compound statement?
3a. A block.
4. What syntax identifies the beginning and ending of a block?
4a. The curly braces. { and }.
5. What is an expression in C++?
5a. Anything that evaluates to a value.
6. Which side of the equal operator can an expression ALWAYS be on? Why not the other side?
6a. The RIGHT side. The other side if for l-values only, or rather specific addresses for storing values. You can’t store a value IN an expression.
7. What do operators act upon?
7a. Operands.
8. Can an expression be acted upon by operators?
8a. Yes. Any expression can be used as an operand of an operator.
9. What does the assignment operator do?
9a. Changes the value of the left operand (left side of operator) to match the value of the expression on the right operand (right side of operator).
10. What is an r-value and an l-value? How does this relate to question 6 above?
10a. A operand that can legally on the left side of an operator is an l-value. That which can be on the right side in an r-value. All l-values are also r-value, but the reverse is not true. An expression is an r-value, which is why it cannot go on the left side.
11. What are the 5 mathematical operators and what do they do? Give examples of inputs and outputs.
11a. Addition (+), Subtraction (-), Multiplication (*), Division (/), and modulus (%). A = 2+2; B = 2-2; C = 2*2; D=2/2; E=2%3;
12. What happens when you subtract a large unsigned number for a smaller unsigned number?
12a.You get underflow, which results in a value which is a very large number.
13. [Extra Credit]What is integer truncation? How can it be avoided?
13a. Integer truncation is what happens when a real number with a fractional value is stored in an integer variable. To be more specific, the fractional component is “lopped off,” in essence rounding down to the nearest whole number. For example: int myValue = 2.2; would result in myValue being 2. Note that sometimes this problem can be less obvious. In the example I provided before we were explicitly using fractional values, but keep in mind the problem can occur when using entirely integers as well. For example…what’s 2/3? .66666…. But when you divide 2/3 and store it in an integer variable, the result is ‘truncated’ to 0. int myValue = 2/3; [myValue := 0] The solution is to be aware of your data, and have a clear understanding of the types of results to expect within your expressions. If there is a chance that an expression will have a fractional component and you NEED that fractional component, store your results in ‘float’ or ‘double’ types, rather than integer types.
14. What are the increment/decrement operators, and how are they used. Show examples.
14a. ++ and --. They are used to take a variable, increment or decrements its value and store the result back in the same variable. This is most commonly used in iterative devices such as for loops or while loops. Example:
for( int i = 0; i < 10; i++ ){ // use i as an index into an array, etc…}
15. What is the difference between prefix and postfix increment/decrement? What are the results? Show examples.
15a. With prefix the value of the variable is changed before the assignment operator is applied. With postfix the value of the variable is changed after evaluation of the assignment operator.
Example:
// In this example I is set to 10, then the value of I is stored in j, and finally the value stored in I is incremented.
int i = 10;
int j = i++;
// In this example, I is set to 10, then it’s incremented to 11, and finally the result is stored in j
int j = ++i;
16. What is the precedence of the mathematical operators?
16a. %, * and / are evaluated first, in order from left to right…. and then + and -, again from left to right.
17. Are the operators left-to-right or right-to-left?
17a. Left to right
18. What about the assignment operator?
18a. Right to left
19. What operator can ALWAYS be used to change the precedence of an expression?
19a. The parentheses
20. How are nested parentheses read? Show examples.
20a. From the inside out. 2 * ( 5 – (2 + 2 ) ) ; Here the 2 + 2 is evaluated first, even though its lowest in precedence, and is on the right side of the expression. Next the 5- is evaluated, and finally the 2*. So as you can see by using parentheses you can change both the precedence and the left to right order of the expression.
21. Can all expressions be evaluated for their truth?
21a. Yes. All expressions return a value, and the result of that value can be tested for zero or non-zero. 0 is treated as false, non-zero is treated as true.
22. What data type is exclusively for truthiness? (tip of the hat to Stephen Colbert)
22a. bool.
23. What is the ANSI standard size of that data type?
23a. 1 byte.
24. What are the six relational operators? Show examples of what values they return with different inputs.
24a. equals (==), not equals (!=), Greater than (>), Greater than or equal to (>=), Less than (<), Less than or equal to (<=).
1 == 1; // true
1 == 2; // false
1 != 2; // true
1 != 1; // false
3 > 2; // true
3 > 3; // false
2 >= 2; // true
2 >= 3; false
3 < 5; // true
3 < 3; // false
3 <= 3; // true
3 <= 2; false
25. What does an ‘if’ statement do?
25a. It enables you to test for a condition, and then branch to different parts of your code, depending on the success or failure of the test.
26. When is an if-block executed, when is it skipped?
26a. It is executed when the test for the condition passes (is true). It is skipped when the test is false.
27. What does an ‘else’ statement do?
27a. It allows you to enter a block of code if a test condition fails.
28. When is an else-block executed, when is it skipped?
28a. It is executed when no previous test in the same if-else chain succeeds. It is never skipped. If it is ‘reached,’ it is executed.
29. [Extra Credit] What does an ‘else if’ statement do? When is executed and when is it skipped?
29a. An if-else statement combines the functionality of an if and an else statement. That is, IF some condition is met, then the block will be executed; ONLY, when no previous condition within an if-else chain has already been met. Unlike the “else” statement, an ‘if else’ statement will be skipped if the condition being tested for is not true.
30. What are the 3 logic operators in C++? How are they used? Show examples.
30a. AND (&&), OR (||), NOT (!).
&& is used to determine the union of multiple expressions. If both operands of a && are true, the result is true. If either operand is false, the result if false.
|| is used to determine the junction of multiple expressions. If either operand of || is true, the result is true. If BOTH operands are false, the result is false.
! is used to negate the value of an expression. If the operand is true, the result is false. If the operand is false, the result is true.
31. What is “Short Circuit Evaluation?” How might it be beneficial?
31a. Short circuit evaluation is a logical optimization put in place by the compiler which prevents compound expressions from being further evaluated once the result is known. For example…with AND (&&) statements, the result is true only if BOTH operands are true. So as soon as you know one is false, there is no reason to test the result of the second, because the total result of the compound expressions is already false. Likewise with OR (||) statements. If the first operand is evaluated to be true, there is no reason to test the value of the second operand, the value of the compound expressions is decidedly true.
32. What is the precedence of the logical and conditional operators?
32a. !, then &&, then ||…from left to right.
33. What is the conditional operator? How is it used? When might it be useful? Show examples.
33a. The conditional operator, also called the ternary operator as it is the only C++ operator with 3 inputs, tests the result of a logical expression. If the value is true, the result is the same as the second operand. If the result was false, the value returned is the same as the 3rd operand.
Chapter 4 Exercises
1. The equation for distance traveled (dist) with a constant velocity is equal to the rate of movement (rate) multiplied by the time interval (time) traveling. (dist=rate*time) Write a program which prompts the user for rate and time, computes the distance traveled, and then outputs the result to the screen.
#include <iostream>int main( void ){ using namespace std; int dist, rate, time; cout << "Enter the rate of movement:" << endl; cin >> rate; cout << endl << "Enter the amount of time traveled:" << endl; cin >> time; dist = rate * time; cout << "The distance traveled is: " << dist << endl; return 0;}
2. Write a program that knows a secret number and prompts the user to guess the number. When the user enters the number the program should notify the user whether the number guessed was too high, too low, or juuust right. We will modify this program later when we’ve covered loops in order to allow the user to guess until they get the correct answer or run out of tries. Additionally, we will cover random number generation when we explore functions next week.
#include <iostream>int main( void ){ using namespace std; int number = 50; int guess; cout << "Enter your guess:" << endl; cin >> guess; if( guess > secret ) cout << "Sorry, your guess was too high." << endl; if( guess < secret ) cout << "Sorry, Your guess was too low." << endl; if( guess == number ) cout << "You guessed juuust right! << endl; return 0;}
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Chapter 4 Quiz
1. What do statements do in C++?
Control the sequence of execution, evaluates an expression, or null.
2. Where can you put a compound statement?
Any place you can put a single statment.
3. What’s another name for a compound statement?
Block
4. What syntax identifies the beginning and ending of a block?
{ }
5. What is an expression in C++?
Anything that evaluates to a value.
6. Which side of the equal operator can an expression ALWAYS be on? Why not the other side?
Right side, the other side is the lvalue, the memory address(es) for storing values.
7. What do operators act upon?
Operands
8. Can an expression be acted upon by operators?
Yes.
9. What does the assignment operator do?
Changed the value of the left operand (lvalue) to equal the expression/value on the right side (rvalue)
10. What is an r-value and an l-value? How does this relate to question 6 above?
right operand, left operand. Expressions can not be on the lvalue, values can be on either side, one defining the other.
11. What are the 5 mathematical operators and what do they do? Give examples of inputs and outputs.
+ - * / %
Ex. x = (4+4)*(3-2)/2 x = 4
Ex. x = 4+4*3-2/2 x = 15
Ex. x = 8 % 3 x = 2 (remainder)
12. What happens when you subtract a large unsigned number for a smaller unsigned number?
Underflow
13. [Extra Credit]What is integer truncation? How can it be avoided?
When an expressions value is a decimal float, but the value is stored in an integer it will round down.
14. What are the increment/decrement operators, and how are they used. Show examples.
++ --
c = 1;
c++;
c == 2;
15. What is the difference between prefix and postfix increment/decrement? What are the results? Show examples. Prefix - Increment the value then fetch it.
Postfix - Fetch it then increment the original value.
int c = 3, x;
x = c++;
x == 3, c == 4;
int c = 3, x;
x = ++c;
x == 4, c == 4;
16. What is the precedence of the mathematical operators?
% ^ / + -
17. Are the operators left-to-right or right-to-left?
left to right.
18. What about the assignment operator?
Right to left.
19. What operator can ALWAYS be used to change the precedence of an expression?
()
20. How are nested parentheses read? Show examples.
in to out.
21. Can all expressions be evaluated for their truth?
Yes
22. What data type is exclusively for truthiness? (tip of the hat to Stephen Colbert)
boolean
23. What is the ANSI standard size of that data type?
1 byte
24. What are the six relational operators? Show examples of what values they return with different inputs.
Equals == 52 == 52 true
Not Equals != 1002 != 967 true
Greater Than > 1103 > 501 true
Greater Than >= 102 >= 50 true
or Equals 102>= 102 true
Less Than <= 60 <= 37 false
or Equals 21 <= 21 true
25. What does an ‘if’ statement do?
The if statement enables you to test for a condition and branch to different parts of your code, depending on the result
26. When is an if-block executed, when is it skipped?
If the expression has the value 0, it is considered false, and the statement is skipped. If it has any nonzero value, it is considered true, and the statement is executed.
27. What does an ‘else’ statement do?
Runs a statment if the expression is 0.
28. When is an else-block executed, when is it skipped?
If the orignal if statment returned null it will run the else, when the if statment returns bool the else statment will be skipped.
29. [Extra Credit] What does an ‘else if’ statement do? When is it executed and when is it skipped? ???
30. What are the 3 logic operators in C++? How are they used? Show examples. && || !
if x=1 && y=1 (if x and y are 1 returns true) && (and)
if x=1 || y=1 (if either x or y returns 1 condition is true) || (or)
if !x=1 (if x does not equal 1 return true) ! (not)
31. What is “Short Circuit Evaluation?” How might it be beneficial?
If the first operand is evaluated to be true, it does not test the value of the second operand.
32. What is the precedence of the logical and conditional operators?
!, then &&, then ||...
33. What is the conditional operator? How is it used? When might it be useful? Show examples. Pretty much the same thing as an if statment.
(expression1) ? (expression2) : (expression3)
Read as "If expression1 is true, return the value of expression2; otherwise, return the value of expression3.
1. The equation for distance traveled (dist) with a constant velocity is equal to the rate of movement (rate) multiplied by the time interval (time) traveling. (dist=rate*time) Write a program which prompts the user for rate and time, computes the distance traveled, and then outputs the result to the screen.
#include <iostream>int main(){ float rate, time, dist; std::cout << "Enter the rate of travel: "; std::cin >> rate; std::cout << "Enter the time traveled: "; std::cin >> time; dist = rate * time; std::cout << "The distance traveled was: " << dist << std::endl;return 0;}
2. Write a program that knows a secret number and prompts the user to guess the number. When the user enters the number the program should notify the user whether the number guessed was too high, too low, or juuust right. We will modify this program later when we’ve covered loops in order to allow the user to guess until they get the correct answer or run out of tries. Additionally, we will cover random number generation when we explore functions next week.
#include <iostream>int main(){ int guess, secret = 28; std::cout << "Enter a whole number and try to guess the secret number: "; std::cin >> guess; if (guess == secret) { std::cout << "Congrats!! You guess the secret number!" << std::endl; } else { if (guess < secret) std::cout << "Try again and input a larger number." << std::endl; else std::cout << "Try again and input a smaller number." << std::endl; }return 0;}
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Chapter 4.
1) What do statements do in C++?
A) A statement controls the sequence of execution, evaluates an expression, or does nothing (the null statement).
2) Where can you put a compound statement?
A) Any place you can put a single statement, you can put a compound statement.
3) What's another name for a compound statement?
A) A block.
4) What syntax identifies the beginning and ending of a block?
A) An opening brace " { " identifies the beginning and a closing brace " } " the end of a block.
5) What is an expression in C++?
A) An expression is anything that evaluates to a value.
6) Which side of the equal operator can an expression ALWAYS be on? Why not the other side?
A) The right side, not on the left because expressions can be literals and literals are allways r-values.
7) What do operators act upon?
A) Operators act upon operators.
8) Can an expression be acted upon by operators?
A) Yes, it can.
9) What does the assignment operator do?
A) It assigns the r-value to the l-value.
10) What is an r-value and an l-value? How does this relate to question 6 above?
A) An operand that legally can be on the left side of an assignment operator is called an l-value. That which can be on
the right side is called an r-value. It is related towards question 6 because not every r-value can be a l-value while every l-value can be a r-value.
11) What are the 5 mathematical operators and what do they do? Give examples of inputs and outputs.
A) The 5 are:
Description operator input output
Addition + 6 + 3 9
Subtraction - 6 - 3 3
Multiplication * 6 * 3 18
Division / 6 / 3 2
Modulus % 6 % 3 0
12) What happens when you subtract a large unsigned number from a smaller unsigned number?
A) Then the result will wrap around to the highest possible unsigned number.
13) What is integer truncation? How can it be avoided?
A) It cuts of the integer value and won't round of an integer, f.e. 5.4 or 5.9 will both be truncated to 5. Avoiding this can be achieved by using float or double for the variable in mind.
14) What are the increment/decrement operators, and how are they used. Show examples.
A) ++ is the increment and -- is the decrement operator, they are used as follows:
++i increment prefix
i++ increment postfix
--i decrement prefix
i-- decrement postfix
15) What is the difference between prefix and postfix increment/decrement? What are the results? Show examples.
A) Prefix increment/decrement operator is evaluated before the assignment, the postfix is evaluated after the assignment. Examples:
Performing this on an integer variable myAge which equals 38 results in the following:
myAge++; Outputs 38
++myAge; Outputs 39
myAge--; Outputs 38
--myAge; Outputs 37
15) What is the precedence of the mathematical operators?
A) Higher: multiplication, division, modulo
17) Are the operators left-to-right or right-to-left?
A) Left-to-right.
18) What about the assignment operator?
A) That is right to left.
19) What operator can ALWAYS be used to change the precedence of an expression?
A) Parentheses can always be used to accomplish this.
20) How are nested parentheses read? Show examples.
A) For the inside out, for example: (2 * (5 + 1)) = 2 * 6 = 12
3 - (4 / (6 * (2 + 4))) = 3 - (4 / (6 * 6)) = 3 - (4 / 36) = 3 - 9 = 6
21) Can all expressions be evaluated for their truth?
A) Yes.
22) What data type is exclusively for truthiness?(tip of the hat to Stephen Colbert)
A) Bool.
23) What is the ANSI standard size of that data type?
A) 1 byte (8 bits).
24) What are the six relational operators? Show examples of what values they return with different inputs.
A) < , >, <=, >=, == and !=.
Examples: 5 < 3 returns false
5 > 3 returns true
5 <= 5 returns true
5 >= 4 returns true
5 == 5 returns true
5 != 5 returns false
5 == 6 returns false
5 != 6 returns true
25) What does an if statement do?
A) It enables you to test for a condition (such as whether two variables are equal) and branch to different parts of your code, depending on the result.
26) When is an if - block executed, when is it skipped?
A) It's executed when the expression is true, it is skipped when the expression is false.
27) What does an else statement do?
A) It gives the opportunity to branch of if the if statement is false.
28) When is an else-block executed, when is it skipped?
A) It's executed when the if-block is skipped, it's skipped when the if-block is executed.
29) What does an else if statement do? When is it executed and when is it skipped?
A) An else if statement executes an additional statement, it is executed when the expression is true, and skipped when the expression is false.
30) What are the 3 logic operators in C++? How are they used? Show examples.
A) || , && , ! , are the OR, AND and NOT operators.
Examples:
if ( ( x == 5) || ( y == 5) )
if ( ( x == 5) && ( y == 5) )
if ( ! ( x == 5) )
31) What is "Short Circuit Evaluation"? How might it be beneficial?
A) The "Short Circuit Evaluation" determines whether the first (left) expression is true or false, depending on which logic operator is used, it will either check the following expression or stop depending on which logic operator was used.
32) What is the precedence of the logical and conditional operators?
A) Highest: !
High: < , <= , > , >=
Low: == , !=
Lower: &&
Lowest: | |
33) What is the conditional operator? How is it used? When might it be useful? Show examples.
A) The conditional operator takes three expressions and returns a value: (expression1) ? (expression2) : (expression3) Its read as "if expression1 is true, return the value of expression2; otherwise, return the value of expression3. It might be useful when you want to right a small inline function that only returns the value of that expression.
/*1. The equation for distance traveled (dist) with a constant velocity is equal to the rate of movement (rate) multiplied by the time interval (time) traveling. (dist=rate*time) Write a program which prompts the user for rate and time, computes the distance traveled, and then outputs the result to the screen.*/#include <iostream>int main(){ double dist = 0.0, time, rate; std::cout << "Enter the rate and time you traveled." << std::endl; std::cout << "Rate equals: "; std::cin >> rate; if (rate < 0) { std::cout << "You can't have walked less then 0 km., enter the rate again please.\n"; std::cout << "Rate equals: "; std::cin >> rate; } else { std::cout << "You walked at: " << rate << " km/hr." << std::endl; std::cout << "Time equals: "; std::cin >> time; if (time < 0) { std::cout << "You can't have walked less then 0 hours, enter the time again please.\n"; std::cout << "Time equals: "; std::cin >> time; } std::cout << "The total amount of time you walked is: " << time << " hour(s)." << std::endl; } dist = rate * time; std::cout << std::endl << std::endl; std::cout << "The total distance you walked is: " << dist << " km." << std::endl; std::cout << "Thank you for entering your data." << std::endl; return 0;}
/*2. Write a program that knows a secret number and prompts the user to guess the number. When the user enters the number the program should notify the user whether the number guessed was too high, too low, or juuust right. We will modify this program later when we’ve covered loops in order to allow the user to guess until they get the correct answer or run out of tries. Additionally, we will cover random number generation when we explore functions next week.*/#include <iostream>int main(){ int secret = 8; int guess; std::cout << "\t\tWelcome to the guessing game!\n"; std::cout << "Please enter a number between 1 and 10: "; std::cin >> guess; if ( (guess < 0) || (guess > 10) ) { std::cout << "You've entered a number below 0 or above 10" << std::endl; } else { if (guess < secret) std::cout << "The number is smaller then the secret number!" << std::endl; else if (guess > secret) std::cout << "The number is bigger then the secret number!" << std::endl; else std::cout << "You've guessed the secret number, it is: " << secret << " ." << std::endl; } std::cout << "Thank you for playing this game. Goodbye!" << std::endl; return 0;}
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Chapter 4 Quiz
1. What do statements do in C++?
Controls the sequence of execution
2. Where can you put a compound statement?
Any place you can put a single statement, It must be in a block which starts with an opening brace and closes with the closing brace { }
3. What’s another name for a compound statement?
Block Statement
4. What syntax identifies the beginning and ending of a block?
The opening brace {
5. What is an expression in C++?
Anything that evaluates to a value
6. Which side of the equal operator can an expression ALWAYS be on? Why not the other side?
The right side
7. What do operators act upon?
operands
8. Can an expression be acted upon by operators?
Yes
9. What does the assignment operator do?
Assigns a value to a variable
10. What is an r-value and an l-value? How does this relate to question 6 above?
Right and Left, they show which side of the assignment operator something can be.
11. What are the 5 mathematical operators and what do they do? Give examples of inputs and outputs.
addition (+) subtraction (-) multiplication (*) division (/) and modulus (%)
12. What happens when you subtract a large unsigned number for a smaller unsigned number?
The variable will reset, so like 1 - 2 would end up being 65,535.
13. [Extra Credit]What is integer truncation? How can it be avoided?
When a fraction is rounded by placing it in an enteger, like If you assign 19.39 to an integer then it will truncate, and the integer will only hold the 19.
14. What are the increment/decrement operators, and how are they used. Show examples.
Var++ or ++Var increment and Var-- or --Var will decrement.
15. What is the difference between prefix and postfix increment/decrement? What are the results? Show examples.
prefix, ++Var will increment the variable before doing the action it was asked to do (example: myAge = ++Var; will increment Var before assigning it to myAge and myAge = Var++; will increment Var AFTER assigning it to myAge
16. What is the precedence of the mathematical operators?
parentheses, exponents, multiplication and division, addition and subtraction
17. Are the operators left-to-right or right-to-left?
left to right, however some operators, like the assignment operator, are operated right to left.
18. What about the assignment operator?
right to left
19. What operator can ALWAYS be used to change the precedence of an expression?
parentheses
20. How are nested parentheses read? Show examples.
The data inside parentheses is completed first
((a+b)*c+(d-f)*e)*8 will complete a+b and d-f first, then it will multiply the answer of a+b with c and d-f times e. After that's done it will add the two values it just came up with then multiply that sum times 8.
21. Can all expressions be evaluated for their truth?
Yes
22. What data type is exclusively for truthiness? (tip of the hat to Stephen Colbert)
boolean
23. What is the ANSI standard size of that data type?
one-byte holding only 2 values
24. What are the six relational operators? Show examples of what values they return with different inputs.
equals (==) not equals (!=) greater than (>) greater than or equal to (>=) less than (<) less than or equal to (<=)
25. What does an ‘if’ statement do?
Checks if a relational expression is true or false
26. When is an if-block executed, when is it skipped?
It is executed if true, skipped if false
27. What does an ‘else’ statement do?
If the relational expression is false, the statement will send the compiler to the else statement.
28. When is an else-block executed, when is it skipped?
executed when the relational expression is false, skipped if the compiler went through the if block already.
29. [Extra Credit] What does an ‘else if’ statement do? When is it executed and when is it skipped?
You can check multiple data with an if statement. An example:
if (a>b)
std::cout >> "a is higher than b!";
else if (a<b)
std::cout >> "a is lower than b!";
else if (a==b)
std::cout >> "a is the same as b!";
30. What are the 3 logic operators in C++? How are they used? Show examples.
And (&&) Or (||) and Not (!)
31. What is “Short Circuit Evaluation?” How might it be beneficial?
When the compiler evaluates an and statement and the first is not true, it won't bother checking the second.
When the compiler evaluates an or statement and the first is true, it won't bother checking the second.
32. What is the precedence of the logical and conditional operators?
the logical and, logical or then the conditional operater in that order.
33. What is the conditional operator? How is it used? When might it be useful? Show examples.
The ternary operator. The only operater that takes three terms. (expression1) ? (expression2) : (expression3) reads If expression1 is true return the value of expression2 otherwise return expression3 and can be used in place of if statements, but they can't get too complicated (They can't be nested)
Chapter 4 Excersises
[source lang=cpp]#include <iostream>int main(){ float rate, time; float dist; std::cout << "This program will calculate the distance traveled" << std::endl << "over the time given using the speed." << std::endl << std::endl; std::cout << "Please enter the speed in MPH: "; std::cin >> rate; std::cout << "Please enter the time traveling in hours: "; std::cin >> time; dist = rate / time; if (time != 1) if (rate != 1) if (dist != 1) std::cout << std::endl << "In " << time << " hours you could travel " << dist << " miles at a speed of " << rate << " miles per hour." << std::endl; else std::cout << std::endl << "In " << time << " hours you could travel " << dist << " mile at a speed of " << rate << " miles per hour." << std::endl; else if (dist != 1) std::cout << std::endl << "In " << time << " hours you could travel " << dist << " miles at a speed of " << rate << " mile per hour." << std::endl; else std::cout << std::endl << "In " << time << " hours you could travel " << dist << " mile at a speed of " << rate << " mile per hour." << std::endl; else if (rate != 1) if (dist != 1) std::cout << std::endl << "In " << time << " hour you could travel " << dist << " miles at a speed of " << rate << " miles per hour." << std::endl; else std::cout << std::endl << "In " << time << " hour you could travel " << dist << " mile at a speed of " << rate << " miles per hour." << std::endl; else if (dist != 1) std::cout << std::endl << "In " << time << " hour you could travel " << dist << " miles at a speed of " << rate << " mile per hour." << std::endl; else std::cout << std::endl << "In " << time << " hour you could travel " << dist << " mile at a speed of " << rate << " mile per hour." << std::endl; return 0;}
Extra - I even made it make sure the hour and hours is right (1 hour, 2 hours) etc.
[source lang=cpp]#include <iostream>int main(){ int number = 19, response; std::cout << "Welcome to the number game!" << std::endl << "Try to guess my number: "; std::cin >> response; if (response == number) std::cout << "You got it first try! Congratulations!" << std::endl; else if (response > number) std::cout << "Oops! The number you guessed is too high!" << std::endl; else if (response < number) std::cout << "Oops! The number you guessed is too low!" << std::endl; else std::cout << "Something wrong has happened that caused you to get this message." << std::endl; std::cout << std::endl << "Thanks for playing! I hoped you enjoyed it." << std::endl; return 0;}
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Intuitive explanation of Kolmogorov Smirnov Test
What is the cleanest, easiest way to explain someone the concept of Kolmogorov Smirnov Test? What does it intuitively mean?
It's a concept that I have difficulty in articulating - especially when explaining to someone.
Can someone please explain it in terms of a graph and/or using simple examples?
• What it does or how it does? – gunes Jun 12 '20 at 8:25
• @gunes What it does, If possible both what and how it does!! – Pluviophile Jun 12 '20 at 8:30
• Is cumulative distribution function simple English? If not, this is a difficult challenge. The simplest explanation would be terms of a graph. – Nick Cox Jun 12 '20 at 8:40
• Are you familiar with the cumulative distribution function of a distribution and the empirical (ie.e sample) cdf? – Glen_b Jun 12 '20 at 10:29
• Or the diagrams already on site, such as the two-sample KS test diagrams here – Glen_b Jun 12 '20 at 11:50
The Kolmogorov-Smirnov test assesses the hypothesis that a random sample (of numerical data) came from a continuous distribution that was completely specified without referring to the data.
Here is the graph of the cumulative distribution function (CDF) of such a distribution.
A sample can be fully described by its empirical (cumulative) distribution function, or ECDF. It plots the fraction of data less than or equal to the horizontal values. Thus, with a random sample of $$n$$ values, when we scan from left to right it jumps upwards by $$1/n$$ each time we cross a data value.
The next figure displays the ECDF for a sample of $$n=10$$ values taken from this distribution. The dot symbols locate the data. The lines are drawn to provide a visual connection among the points similar to the graph of the continuous CDF.
The K-S test compares the CDF to the ECDF using the greatest vertical difference between their graphs. The amount (a positive number) is the Kolmogorov-Smirnov test statistic.
We may visualize the KS test statistic by locating the data point situated furthest above or below the CDF. Here it is highlighted in red. The test statistic is the vertical distance between the extreme point and the value of the reference CDF. Two limiting curves, located this distance above and below the CDF, are drawn for reference. Thus, the ECDF lies between these curves and just touches at least one of them.
To assess the significance of the KS test statistic, we compare it--as usual--to the KS test statistics that would tend to occur in perfectly random samples from the hypothesized distribution. One way to visualize them is to graph the ECDFs for many such (independent) samples in a way that indicates what their KS statistics are. This forms the "null distribution" of the KS statistic.
The ECDF of each of $$200$$ samples is shown along with a single red marker located where it departs the most from the hypothesized CDF. In this case it is evident that the original sample (in blue) departs less from the CDF than would most random samples. (73% of the random samples depart further from the CDF than does the blue sample. Visually, this means 73% of the red dots fall outside the region delimited by the two red curves.) Thus, we have (on this basis) no evidence to conclude our (blue) sample was not generated by this CDF. That is, the difference is "not statistically significant."
More abstractly, we may plot the distribution of the KS statistics in this large set of random samples. This is called the null distribution of the test statistic. Here it is:
The vertical blue line locates the KS test statistic for the original sample. 27% of the random KS test statistics were smaller and 73% of the random statistics were greater. Scanning across, it looks like the KS statistic for a dataset (of this size, for this hypothesized CDF) would have to exceed 0.4 or so before we would conclude it is extremely large (and therefore constitutes significant evidence that the hypothesized CDF is incorrect).
Although much more can be said--in particular, about why KS test works the same way, and produces the same null distribution, for any continuous CDF--this is enough to understand the test and to use it together with probability plots to assess data distributions.
In response to requests, here is the essential R code I used for the calculations and plots. It uses the standard Normal distribution (pnorm) for the reference. The commented-out line established that my calculations agree with those of the built-in ks.test function. I had to modify its code in order to extract the specific data point contributing to the KS statistic.
ecdf.ks <- function(x, f=pnorm, col2="#00000010", accent="#d02020", cex=0.6,
limits=FALSE, ...) {
obj <- ecdf(x)
x <- sort(x)
n <- length(x)
y <- f(x) - (0:(n - 1))/n
p <- pmax(y, 1/n - y)
dp <- max(p)
i <- which(p >= dp)[1]
q <- ifelse(f(x[i]) > (i-1)/n, (i-1)/n, i/n)
# if (dp != ks.test(x, f)$statistic) stop("Incorrect.") plot(obj, col=col2, cex=cex, ...) points(x[i], q, col=accent, pch=19, cex=cex) if (limits) { curve(pmin(1, f(x)+dp), add=TRUE, col=accent) curve(pmax(0, f(x)-dp), add=TRUE, col=accent) } c(i, dp) } • @Maximilian see ks.test. – whuber Jun 12 '20 at 15:02 • I’m aware of ks.test i was keen to see the implementation. – Maximilian Jun 12 '20 at 15:11 • @Maximilian: Read the code. It's what I used. – whuber Jun 12 '20 at 15:46 • At the R console, type ks.test It will respond by displaying the code. – whuber Jun 12 '20 at 17:16 • @Max and Pluvio, I have posted the code I think you might be most interested in. – whuber Jun 12 '20 at 17:21 The one-sample Kolmogorov-Smirnov test finds the largest vertical distance between a completely specified continuous hypothesized cdf and the empirical cdf. The two-sample Kolmogorov-Smirnov test finds the largest vertical distance between the empirical cdfs for two samples. Unusually large distances indicate that the sample is not consistent with the hypothesized distribution (or that the two samples are not consistent with having come from the same distribution). These tests are nonparametric in the sense that the distribution of the test statistic under the null doesn't depend on which specific distribution was specified under the null (or which common distribution the two samples are drawn from). There are "one-sided" (in a particular sense) versions of these tests, but these are relatively rarely used. You can do a Kolmogorov-Smirnov test with discrete distributions but the usual version of the test (i.e. using the usual null distribution) is conservative, and sometimes very conservative. You can (however) obtain new critical values for a completely specified discrete distribution. There is a related test when parameters are estimated in a location-scale family* (or a subset of location and scale), properly called a Lilliefors test (Lilliefors did three tests for the normal case and a test for the exponential case). This is not distribution-free. * up to a monotonic transformation You are looking for the maximum deviation of empirical CDF (built from observations) from the theoretical values. By definition it can't be larger than 1. Here's a plot for a uniform distribution CDF (black) and two stylized candidate CDFs (red): You see that your candidate CDF can't be over the theoretical by more than $$D^+$$ or below it by more than $$D^-$$, both of which are bounded in magnitude by 1. The empirical CDF $$S_n$$ for the purpose of this test is $$S_i=i/N$$. Here we sorted the sample $$x_i$$ where $$i=1,\dots,N$$ so that $$x_i. You compare it with a theoretical CDF $$F_i=F(x_i)$$, then you have set of deviations $$D^+_i=\max(0,S_i-F_i)$$. However, that's not what's amazing about KS statistic. It is that the distribution of $$\sup_{x\in(-\infty,\infty)} D^+$$ is the same for any distribution of data set! To me that's what you need to get intuitively if you can. • Your final remark is important, albeit tangential to the question. But this result is unsurprising, because any continuous monotonic increasing transformation of the distribution (applied at the same time to the data) will not change the KS statistic, since geometrically it changes only the$x$coordinates, leaving all vertical distances unaffected. Because any continuous CDF can transformed (via its probability integral transform) to a uniform distribution, your picture of the uniform CDF is completely general. – whuber Jun 12 '20 at 21:11 • BTW, your labels "$D^+$" and "$D^-$" may be confusing to readers, because these terms are usually used for the largest deviations between data and the hypothesized CDF rather than between the CDF and its theoretical limits of$1$and$0.\$ – whuber Jun 12 '20 at 21:13
• @whuber agreed that notation is confusing, will improve it. – Aksakal Jun 12 '20 at 21:17
I find it helpful to think of the two CDFs, whether population of empirical, as dancing around each other but staying close. Dance partners can spin around each other but will stay two armslengths of each other, right? When two people are further apart than that, they probably aren't dancing with each other.
ONE-SAMPLE
In the one-sample (goodness-of-fit) test, we assume that the data come from some distribution that has a particular CDF. The data also have an empirical CDF. If we are right, then the CDF of the data should dance around the CDF of the assumed distribution but stay close. If the dance partners get too far apart (in vertical distance), then we see that as evidence against our assumption.
TWO-SAMPLE
In the two-sample test, we assume that two data sets come from the same distribution. If that is the case, then the two empirical CDFs should dance around each other but stay fairly close. If the dance partners get too far apart (again, in vertical distance), then we see that as evidence against our assumption.
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# qml.transforms.get_unitary_matrix¶
get_unitary_matrix(circuit, wire_order=None)[source]
Construct the matrix representation of a quantum circuit.
Parameters
• circuit (pennylane.QNode, QuantumTape, or Callable) – A quantum node, tape, or function that applies quantum operations.
• wire_order (Sequence[Any], optional) – Order of the wires in the quantum circuit. Defaults to the order in which the wires appear in the quantum function.
Returns
Function which accepts the same arguments as the QNode or quantum function. When called, this function will return the unitary matrix as a numpy array.
Return type
function
Example
Consider the following function (the same applies for a QNode or tape):
def circuit(theta):
qml.RX(theta, wires=1)
qml.PauliZ(wires=0)
We can use get_unitary_matrix to generate a new function that returns the unitary matrix corresponding to the function circuit:
>>> get_matrix = get_unitary_matrix(circuit)
>>> theta = np.pi/4
>>> get_matrix(theta)
array([[ 0.92387953+0.j, 0.+0.j , 0.-0.38268343j, 0.+0.j],
[ 0.+0.j, -0.92387953+0.j, 0.+0.j, 0. +0.38268343j],
[ 0. -0.38268343j, 0.+0.j, 0.92387953+0.j, 0.+0.j],
[ 0.+0.j, 0.+0.38268343j, 0.+0.j, -0.92387953+0.j]])
Note that since wire_order was not specified, the default order [1, 0] for circuit was used, and the unitary matrix corresponds to the operation $$Z\otimes R_X(\theta)$$. To obtain the matrix for $$R_X(\theta)\otimes Z$$, specify wire_order=[0, 1] in the function call:
>>> get_matrix = get_unitary_matrix(circuit, wire_order=[0, 1])
You can also get the unitary matrix for operations on a subspace of a larger Hilbert space. For example, with the same function circuit and wire_order=["a", 0, "b", 1] you obtain the $$16\times 16$$ matrix for the operation $$I\otimes Z\otimes I\otimes R_X(\theta)$$.
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1. Engineering
2. Electrical Engineering
3. i really need the answer i only have ten minutes...
# Question: i really need the answer i only have ten minutes...
###### Question details
I really need the answer. I only have ten minutes
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# Why does a higher risk aversion leads to a lower intertemporal elasticity of substitution?
Mathematically, a higher risk aversion leads to a lower intertemporal elasticity of substitution (there is an inverse relationship). But why?
If I become more risk-averse, I would like to smooth my consumption. So I will increase my participation in the financial market to avoid or reduce the risk. But how does that lower my intertemporal elasticity of substitution? Like why would I become less responsive to changes in the interest rate if I become more risk-averse? What's the intuition here?
Or because I just wanna smooth my consumption, I don't really care about changes in the interest rate?
Risk-lovers, on the other hand, although also participate in the financial market, do not care about smoothing their consumption? Like if the interest rate increases, they can drastically reduce their current consumption for more consumption in the future, but risk-aversers do not care about the higher interest rate?
Mathematically, a higher risk aversion leads to a lower intertemporal elasticity of substitution (there is an inverse relationship). But why?
The (relative) measure of risk aversion is measured as: $$r(x) = -x\frac{u''(x)}{u'(x)}$$
Consider an intertemporal utility function $$u(x_1) + \beta u(x_2)$$. Maximising this with respect to an intertemporal budget constraint gives the following first order condition: $$\frac{u'(x_2)}{u'(x_1)} = \frac{1+r}{\beta}.$$ where $$r$$ is the interest rate. This states that the marginal rate of substitution should equal the interest rate.
Now, for a given utility level, the indifference curve $$x_1(x_2)$$ can be implicitly defined as: $$u(x_1(x_2)) + u(x_2) = \bar u.$$ Using the implicit function theorem, we can compute its slope: $$\frac{\partial x_1}{\partial x_2} = - \frac{u'(x_2)}{u'(x_1)}.$$ The slope of the indifference curve is therefore equal to the negative of the marginal rate of substitution. Now, differentiate this again with respect to $$x_2$$ : \begin{align*} \frac{\partial^2 x_1}{\partial x_2 \partial x_2} &= - \frac{u''(x_2) u'(x_1) - u'(x_2) u''(x_1) \dfrac{\partial x_1}{\partial x_2}}{(u'(x_1))^2},\\ &= - \frac{u''(x_2)}{u'(x_1)} - \frac{u'(x_2)}{u'(x_1)} \frac{u''(x_1)}{u'(x_1)}\dfrac{ u'(x_2)}{u'(x_1)},\\ &= -\frac{u''(x_2)}{u'(x_2)}\frac{u'(x_2)}{u'(x_1)} - \left(\frac{u'(x_2)}{u'(x_1)}\right)^2 \frac{u''(x_1)}{u'(x_1)},\\ &= \frac{r(x_2)}{x_2} \frac{(1+r)}{\beta} + \left(\frac{(1+r)}{\beta}\right)^2 \frac{r(x_1)}{x_1}. \end{align*} This shows that the curvature of the indifference curve at the optimum is increasing in the measure of risk aversion.
Concerning it's relationship to the elasticity of intertemporal substitution, we intuitively see that the higher the relative risk aversion, the higher the curvature of the indifference curve, so the more the slope at the indifference curve, $$\frac{u'(x_2)}{u'(x_1)}$$, will change as a response to a change in the ratio $$\frac{x_2}{x_1}$$. Vice versa, the higher the relative risk aversion, the less the ratio $$\frac{x_2}{x_1}$$ will change in response to a change in the slope $$\frac{u'(x_2)}{u'(x_1)}$$ (e.g. due to a change in interest rates).
To see this mathematically, consider he intertemporal elasticity of subsitution: $$\varepsilon_s = -\frac{\partial \ln(x_2/x_1)}{\partial \ln(u'(x_2)/u'(x_1))}$$ This measures the percentage change in $$x_2/x_1$$ due to a one percentage point change in the marginal rate of substitution.
We can write everything in terms of $$x_2$$ (as we did above), then take derivatives of the numerator and denominator with respect to $$x_2$$ and finally evaluate it at the optimum: \begin{align*} \varepsilon_s &= -\frac{\dfrac{x_1}{x_2}\left(\dfrac{x_1 - x_2 \dfrac{\partial x_1}{\partial x_2}}{(x_1)^2}\right)}{-\dfrac{1}{\dfrac{u'(x_2)}{u'(x_1)}} \left(\dfrac{r(x_2)}{x_2} \dfrac{1+r)}{\beta} + \left(\frac{1+r}{\beta}\right)^2 \dfrac{r(x_1)}{x_1}\right)},\\ &= -\frac{\dfrac{1}{x_2} - \dfrac{\dfrac{\partial x_1}{\partial x_2}}{x_1} }{\dfrac{1}{\dfrac{\partial x_1}{\partial x_2}} \left(\dfrac{r(x_2)}{x_2} \dfrac{1+r}{\beta} + \left(\frac{1+r}{\beta}\right)^2 \dfrac{r(x_1)}{x_1}\right)},\\ &= -\frac{\dfrac{1}{x_2} - \dfrac{- \dfrac{1+r}{\beta}}{x_1} }{ -\dfrac{\beta}{1+r} \left(\dfrac{r(x_2)}{x_2} \dfrac{1+r}{\beta} + \left(\frac{1+r}{\beta}\right)^2 \dfrac{r(x_1)}{\beta}\right)},\\ &= \frac{\dfrac{x_1}{x_2} + \dfrac{1+r}{\beta}}{ \left(\dfrac{x_1}{x_2}r(x_2) + \dfrac{1+r}{\beta} r(x_1)\right)},\\ \end{align*}
The last expression gives the elasticity as a function of the relative slope $$x_2/x_1$$ and the relative risk aversions. Now, if the relative risk aversion is constant, say $$r(x_2) = r(x_1) = \sigma$$ this simplifies to $$\varepsilon_s = 1/\sigma$$.
Nevertheless, we see that if the risk aversion increases, i.e. $$r(x_2)$$ and $$r(x_1)$$ increase, then $$\varepsilon_s$$ decreases.
If I become more risk-averse, I would like to smooth my consumption. So I will increase my participation in the financial market to avoid or reduce the risk. But how does that lower my intertemporal elasticity of substitution? Like why would I become less responsive to changes in the interest rate if I become more risk-averse? What's the intuition here?
There is a difference between consumption and saving. The higher your risk aversion, the bigger the curvature of the indifference curve, i.e. the lower $$\varepsilon_s$$. And so the smaller the change of $$x_2/x_1$$ due to a change in, for example the interest rate $$r$$.
You can think of the curvature of the indifference curve as the degree of complementarity between $$x_1$$ and $$x_2$$. The higher the risk aversion the more $$x_1$$ and $$x_2$$ become complements. So I will try to consume $$x_1$$ and $$x_2$$ together which is equal to increasing consumption smoothing.
In order to smooth my consumption, I will need to lend and borrow on the financial market. So if my income over time changes a lot, I will have to save and borrow a lot. On the other hand, if my income is reasonable smooth over time, I will not need to borrow or save a lot in order to smooth out out my consumption.
Or because I just wanna smooth my consumption, I don't really care about changes in the interest rate?
This is a mis-understanding, risk averse consumers actually do care about changes in the interest rate.
• First of all, a change in interest rates affects their saving decision. It is true, however, that changes in the interest rate will not change $$x_2/x_1$$ a lot. So although my relative consumption over time will not be affected a lot by changes in the interest rate, the amount that I save or borrow can change substantially.
• A change in the interest rates also changes my utility. Assume for simplicity that I am infinitely risk aversion, so I will always always choose to set $$x_1 = x_2$$. In this case, my intertemporal budget constraint becomes:
\begin{align*} &x_1 + \frac{x_2}{(1+r)} = y_1 + \frac{y_2}{(1+r)},\\ \to &x_1\frac{2+r}{1+r} = y_1 + \frac{y_2}{1+r},\\ \to &x_1 = \frac{(1+r)y_1 + y_2}{2+r} \end{align*}
Then: $$\frac{\partial x_1}{\partial r} = \frac{y_1(2+r) - [(1+r)y_2 + y_2]}{(2+r)^2} = \frac{y_1-x}{2+r}.$$ So if my consumption $$x_1$$ is smaller than my income in period 1 $$y_1$$ (i.e. I save in period 1), I will be able to consume more if interest rates increase. On the other hand, if $$x_1 > y_1$$ (So I borrow in period 1), then I will be worse of if interest rates increase. So changing interest rates will change my utility depending on whether I am a net saver or borrower.
Risk-lovers, on the other hand, although also participate in the financial market, do not care about smoothing their consumption? Like if the interest rate increases, they can drastically reduce their current consumption for more consumption in the future, but risk-aversers do not care about the higher interest rate?
Again risk loving (or neutral) consumers also care about interest rates. Consider for simplicity a risk neutral consumer with utility function $$u(x_1, x_2) = x_1 + \beta x_2$$. Then it is clear to see that if:
1. $$\beta < \frac{1}{1+r}$$, I will consume everything in period 1, so $$x_1 = y_1 + \frac{y_2}{1+r}$$ and $$x_2 = 0$$. Utility will be equal to $$y_1 + \frac{y_2}{1+r}$$, which is decreasing in $$r$$. Higher interest rates lower my discounted future income, so I will be worse off.
2. $$\beta > \frac{1}{1+r}$$, then I will consume everything in period 2, so $$x_1 = 0$$ and $$x_2 = (1+r) y_1 + y_2$$. Then utility is equal to $$\beta (1+r)y_1 + \beta y_2$$ which is increasing in $$r$$. Higher interest rates increase my consumption in period 2.
As a conclusion, the consumer will gain by an interest rate increase if she is a net saver and will loose if she is a net borrower.
It is true that normally a risk neutral consumer will not respond her saving decisions by a lot if interest rates change. However if, for example due to the interest rate increase we shift from a situation where $$\beta < 1/(1+r)$$ to a situation where $$\beta > 1/(1+r)$$ then suddenly she will move all consumption from period 1 to period 2, which will mean a big shift from borrowing to saving.
• @@ tdm. Thanks for the answer. But with this question, "Mathematically, a higher risk aversion leads to a lower intertemporal elasticity of substitution (there is an inverse relationship). But why?", I didn't actually mean mathematically..sorry. If you can provide an intuition, that will be great. Like if my risk aversion coefficient goes up, why would I become less responsive to changes in the interest rate? That is my intertemporal elasticity of substitution goes down. May 19, 2021 at 2:05
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# Example of an algebra finite over a commutative subalgebra with infinite dimensional simple modules
Let $A$ be an algebra over an algebraically closed field $k.$ Recall that if $A$ is a finitely generated module over its center, and if its center is a finitely generated algebra over $k,$ then by the Schur's lemma all simple $A$-modules are finite dimensional over $k.$
Motivated by the above, I would like an example of a $k$-algebra $A,$ such that:
1) $A$ has a simple module of infinitie dimension over $k,$
2) $A$ contains a commutative finitely generated subalgebra over which $A$ is a finitely generated left and right module.
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# Bivariate One-Sided Chebyshev Inequality (Symmetric Case)
Let $X$ and $Y$ be random variables with finite means $\mu_X$ and $\mu_Y,$ finite variances $\sigma_X^2$ and $\sigma_Y^2,$ and correlation $\rho.$ Let $A$ be the event that $X \leq \mu_X + k\sigma_X$ and $Y \leq \mu_Y + k\sigma_Y,$ where $k \gt 1.$ Using the one-sided univariate Chebyshev inequality, I can find this lower bound on the probability of $A$ occurring: $$P[A] \geq {{k^2-1} \over {k^2+1}}$$ But this does not use the correlation. If the correlation is known, what is the tightest bound possible?
• Are you sure you don't want $A$ to be the event $|X-\mu_X|\le k\ \sigma_X$ and $|Y-\mu_Y|\le k\ \sigma_Y$? That would be closer to the univariate Chebyshev situation.
– whuber
Feb 12, 2014 at 19:35
• Formulas for that case are available from Wikipedia's article on Chebyshev's inequality. I couldn't find anything on the one-sided scenario. Feb 12, 2014 at 20:10
Using the results from wikipedia page of Chebyshev's inequality (first inequality in section "two correlated variables"), I dont see anything for your case with one-sided inequalities. The more usual case with double inequalities, one obtains, with the event $A$ defined as $A=|\frac{X-\mu_X}{\sigma_X}|\le k ~~\text{and}~~ |\frac{Y-\mu_Y}{\sigma_Y}| \le k$ we get the inequality $$Pr(A) \ge 1 - \frac{2k^2+\sqrt{4k^4-4k^4\rho}}{2k^4} = 1- \frac{1+\sqrt{1-\rho}}{k^2}$$ where $\rho$ is the correlation between $X$ and $Y$. Observe that negative correlation reduces the bound, while positive correlation increases it.
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## Towers of Hanoi
### Introduction
Why do we present a Python implementation of the "Towers of Hanoi"? The hello-world of recursion is the Factorial. This means, you will hardly find any book or tutorial about programming languages which doesn't deal with the first and introductory example about recursive functions. Another one is the calculation of the n-th Fibonacci number. Both are well suited for a tutorial because of their simplicity but they can be easily written in an iterative way as well.
If you have problems in understanding recursion, we recommend that you go through the chapter "Recursive Functions" of our tutorial.
That's different with the "Towers of Hanoi". A recursive solution almost forces itself on the programmer, while the iterative solution of the game is hard to find and to grasp. So, with the Towers of Hanoi we present a recursive Python program, which is hard to program in an iterative way.
### Origin
There is an old legend about a temple or monastery, which contains three poles. One of them filled with 64 gold disks. The disks are of different sizes, and they are put on top of each other, according to their size, i.e. each disk on the pole a little smaller than the one beneath it. The priests, if the legend is about a temple, or the monks, if it is about a monastery, have to move this stack from one of the three poles to another one. But one rule has to be applied: a large disk can never be placed on top of a smaller one. When they would have finished their work, the legend tells, the temple would crumble into dust, and the world would end.
But don't be afraid, it's not very likely that they will finish their work soon, because 264 - 1 moves are necessary, i.e. 18,446,744,073,709,551,615 to move the tower according to the rules.
But there is - most probably - no ancient legend. The legend and the game "towers of Hanoi" had been conceived by the French mathematician Edouard Lucas in 1883.
### Rules of the Game
The rules of the game are very simple, but the solution is not so obvious. The game "Towers of Hanoi" uses three rods. A number of disks is stacked in decreasing order from the bottom to the top of one rod, i.e. the largest disk at the bottom and the smallest one on top. The disks build a conical tower.
The aim of the game is to move the tower of disks from one rod to another rod.
The following rules have to be obeyed:
• Only one disk may be moved at a time.
• Only the most upper disk from one of the rods can be moved in a move
• It can be put on another rod, if this rod is empty or if the most upper disk of this rod is larger than the one which is moved.
### Number of Moves
The number of moves necessary to move a tower with n disks can be calculated as: 2n - 1
### Playing around to Find a Solution
From the formula above, we know that we need 7 moves to move a tower of size 3 from the most left rod (let's call it SOURCE to the most right tower (TARGET).
The pole in the middle (we will call it AUX) is needed as an auxiliary stack to deposit disks temporarily.
Before we examine the case with 3 disks, as it is depicted in the image on the right side, we will have a look at towers of size 1 (i.e. just one disk) and size 2. The solution for a tower with just one disk is straightforward: We will move the one disk on the SOURCE tower to the TARGET tower and we are finished.
Let's look now at a tower with size 2, i.e. two disks. There are two possibilities to move the first disk, the disk on top of the stack of SOURCE: We can move this disk either to TARGET or to AUX.
• So let's start by moving the smallest disk from SOURCE to TARGET. Now there are two choices: We can move this disk again, either back to the SOURCE peg, which obviously doesn't make sense, or we could move it to AUX, which doesn't make sense either, because we could have moved there as our first step. So the only move which makes sense is moving the other disk, i.e. the largest disk, to peg AUX. Now, we have to move the smallest disk again, because we don't want to move the largest disk back to SOURCE again. We can move the smallest disk to AUX. Now we can see that we have moved the tower of size 2 to the peg AUX, but the target had been peg TARGET. We have already used the maximal number of moves, i.e. 22 - 1 = 3
• Moving the smallest disk from peg SOURCE to TARGET as our first step has not shown to be successful. So, we will move this disk to peg AUX in our first step. After this we move the second disk to TARGET. After this we move the smallest disk from AUX to TARGET and we have finished our task!
• </ul> We have seen in the cases n=1 and n=2 that it depends on the first move, if we will be able to successfully and with the minimal number of moves solve the riddle. We know from our formula that the minimal number of moves necessary to move a tower of size 3 from the SOURCE peg to the target peg is 7 (23 - 1)
You can see in the solution, which we present in our image that the first disk has to be moved from the peg SOURCE to the peg TARGET. If your first step consists of moving the smallest disk to AUX, you will not be capable of finishing the task with less than 9 moves.
Let's number the disks as D1 (smallest), D2 and D3 (largest) and name the pegs as S (SOURCE peg), A (AUX), T (TARGET). We can see that we move in three moves the tower of size 2 (the disks D1 and D2) to A. Now we can move D3 to T, where it is finally positioned. The last three moves move the tower consisting of D2D1 from peg A to T to place them on top of D3.
There is a general rule for moving a tower of size n (n > 1) from the peg S to the peg T:
• move a tower of n - 1 discs Dn-1 ... D1 from S to A. Disk Dn is left alone on peg S
• Move disk Dn to T
• move the tower of n - 1 discs Dn-1 ... D1 on A to T, i.e. this tower will be put on top of Disk Dn
• </ul> The algorithm, which we have just defined, is a recursive algorithm to move a tower of size n. It actually is the one, which we will use in our Python implementation to solve the Towers of Hanoi. Step 2 is a simple move of a disk. But to accomplish the steps 1 and 3, we apply the same algorithm again on a tower of n-1. The calculation will finish with a finite number of steps, because very time the recursion will be started with a tower which is 1 smaller than the one in the calling function. So finally we will end up with a tower of size n = 1, i.e. a simple move.
### Recursive Python Program
The following Python script contains a recursive function "hanoi", which implements a recursive solution for Towers of Hanoi:
def hanoi(n, source, helper, target):
if n > 0:
# move tower of size n - 1 to helper:
hanoi(n - 1, source, target, helper)
# move disk from source peg to target peg
if source:
target.append(source.pop())
# move tower of size n-1 from helper to target
hanoi(n - 1, helper, source, target)
source = [4,3,2,1]
target = []
helper = []
hanoi(len(source),source,helper,target)
print(source, helper, target)
[] [] [4, 3, 2, 1]
This function is an implementation of what we have explained in the previous subchapter. First we move a tower of size n-1 from the peg source to the helper peg. We do this by calling
hanoi(n - 1, source, target, helper)
After this, there will be the largest disk left on the peg source. We move it to the empty peg target by the statement
if source:
target.append(source.pop())
After this, we have to move the tower from "helper" to "target", i.e. on top of the largest disk:
hanoi(n - 1, helper, source, target)
If you want to check, what's going on, while the recursion is running, we suggest the following Python programm. We have slightly changed the data structure. Instead of passing just the stacks of disks to the function, we pass tuples to the function. Each tuple consists of the stack and the function of the stack:
def hanoi(n, source, helper, target):
print("hanoi( ", n, source, helper, target, " called")
if n > 0:
# move tower of size n - 1 to helper:
hanoi(n - 1, source, target, helper)
# move disk from source peg to target peg
if source[0]:
disk = source[0].pop()
print("moving " + str(disk) + " from " + source[1] + " to " + target[1])
target[0].append(disk)
# move tower of size n-1 from helper to target
hanoi(n - 1, helper, source, target)
source = ([4,3,2,1], "source")
target = ([], "target")
helper = ([], "helper")
hanoi(len(source[0]),source,helper,target)
print(source, helper, target)
hanoi( 4 ([4, 3, 2, 1], 'source') ([], 'helper') ([], 'target') called
hanoi( 3 ([4, 3, 2, 1], 'source') ([], 'target') ([], 'helper') called
hanoi( 2 ([4, 3, 2, 1], 'source') ([], 'helper') ([], 'target') called
hanoi( 1 ([4, 3, 2, 1], 'source') ([], 'target') ([], 'helper') called
hanoi( 0 ([4, 3, 2, 1], 'source') ([], 'helper') ([], 'target') called
moving 1 from source to helper
hanoi( 0 ([], 'target') ([4, 3, 2], 'source') ([1], 'helper') called
moving 2 from source to target
hanoi( 1 ([1], 'helper') ([4, 3], 'source') ([2], 'target') called
hanoi( 0 ([1], 'helper') ([2], 'target') ([4, 3], 'source') called
moving 1 from helper to target
hanoi( 0 ([4, 3], 'source') ([], 'helper') ([2, 1], 'target') called
moving 3 from source to helper
hanoi( 2 ([2, 1], 'target') ([4], 'source') ([3], 'helper') called
hanoi( 1 ([2, 1], 'target') ([3], 'helper') ([4], 'source') called
hanoi( 0 ([2, 1], 'target') ([4], 'source') ([3], 'helper') called
moving 1 from target to source
hanoi( 0 ([3], 'helper') ([2], 'target') ([4, 1], 'source') called
moving 2 from target to helper
hanoi( 1 ([4, 1], 'source') ([], 'target') ([3, 2], 'helper') called
hanoi( 0 ([4, 1], 'source') ([3, 2], 'helper') ([], 'target') called
moving 1 from source to helper
hanoi( 0 ([], 'target') ([4], 'source') ([3, 2, 1], 'helper') called
moving 4 from source to target
hanoi( 3 ([3, 2, 1], 'helper') ([], 'source') ([4], 'target') called
hanoi( 2 ([3, 2, 1], 'helper') ([4], 'target') ([], 'source') called
hanoi( 1 ([3, 2, 1], 'helper') ([], 'source') ([4], 'target') called
hanoi( 0 ([3, 2, 1], 'helper') ([4], 'target') ([], 'source') called
moving 1 from helper to target
hanoi( 0 ([], 'source') ([3, 2], 'helper') ([4, 1], 'target') called
moving 2 from helper to source
hanoi( 1 ([4, 1], 'target') ([3], 'helper') ([2], 'source') called
hanoi( 0 ([4, 1], 'target') ([2], 'source') ([3], 'helper') called
moving 1 from target to source
hanoi( 0 ([3], 'helper') ([4], 'target') ([2, 1], 'source') called
moving 3 from helper to target
hanoi( 2 ([2, 1], 'source') ([], 'helper') ([4, 3], 'target') called
hanoi( 1 ([2, 1], 'source') ([4, 3], 'target') ([], 'helper') called
hanoi( 0 ([2, 1], 'source') ([], 'helper') ([4, 3], 'target') called
moving 1 from source to helper
hanoi( 0 ([4, 3], 'target') ([2], 'source') ([1], 'helper') called
moving 2 from source to target
hanoi( 1 ([1], 'helper') ([], 'source') ([4, 3, 2], 'target') called
hanoi( 0 ([1], 'helper') ([4, 3, 2], 'target') ([], 'source') called
moving 1 from helper to target
hanoi( 0 ([], 'source') ([], 'helper') ([4, 3, 2, 1], 'target') called
([], 'source') ([], 'helper') ([4, 3, 2, 1], 'target')
|
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# MATLAB Plot of FT(Cos) Displays Weird Impulse
It is known that: $$\mathcal{F}\{\cos(2\pi t)\}=\frac{\delta(f-1)+\delta(f+1)}{2}$$ However, on MATLAB, I used F=fftshift(fft(x))/N; to obtain the FT of $$\cos(2\pi t)$$ and I obtained the following:
My question is why does each $$\delta$$ appear to be made up of two thin close parallel lines. Note that I have used $$1024$$ sample points and constructed $$f$$ a vector from $$-51.2$$Hz to $$51.2$$Hz and $$t$$ a vector from $$-5$$ to $$5$$.
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# Compton scattering on massless electron
• I
Hello! I found this problem where we are asked what happens to the energy of the outgoing photon in a Compton interaction, if the mass of the electron goes to zero and what is the physical intuition of it. So the formula is this: $$\lambda - \lambda_0 = \frac{h}{m_0 c}(1-cos \theta)$$ So when the electron gets massless its compton wavelength gets higher (infinite in the limit) and hence it spreads out and plugging in there you get that the final wavelength of the photon is infinite, which means that it has zero energy. But I am not sure what is the logic behind. Beside the obvious math, what is the explanation for the fact that the electron gets all the energy of the photon? Thank you!
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# All Questions
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### Quantum complexity of TQBF
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rfffffffffffffffffffffffffffffff
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|
# Math Help - Trying to find a function to represent my data
1. ## Trying to find a function to represent my data
Hello,
I'm trying to find the general form for a continuous function that will mimic the image I have attached. It is similar to a sigmoid function in the sense that it is flat initially and then has a very sharp transition, however it does not level out at the top, but rather increases continuously. Does anyone have any ideas as to what function may be appropriate?
Thank you for your help.
2. ## Re: Trying to find a function to represent my data
Originally Posted by CLangford
I'm trying to find the general form for a continuous function that will mimic the image I have attached. It is similar to a sigmoid function in the sense that it is flat initially and then has a very sharp transition, however it does not level out at the top, but rather increases continuously. Does anyone have any ideas as to what function may be appropriate?
Here is a possibility.
3. ## Re: Trying to find a function to represent my data
That certainly looks promising. What was the form of the equation?
4. ## Re: Trying to find a function to represent my data
Originally Posted by CLangford
That certainly looks promising. What was the form of the equation?
$\[f(x) = \left\{ {\begin{array}{lr}{2\left( {\arctan (7) + 1} \right),}&{x < 4}\\{2\left( {\arctan (3x - 5) + 1} \right),}&{x \ge 4}\end{array}} \right.$
Now you can "play around" with those constants in the second condition.
|
# How Many 3 Letter Words Can Be Made From The Alphabet
Alphabet supports and develops companies applying technology to the world’s biggest challenges. There are 10 choices for the first, 10 for the second and 10 for the third: 10xx10xx10 =1000 So for a license plate which has 2 letters and 3 digits, there are: 26xx26xx10xx10xx10= 676,000 possibilities. How many different 3 letter "words" can be made from these letters such that when the word is rotated 180∘, the. Invite your learner to name the missing letter or fill it in with a magnetic letter (we LOVE this set!). If you want a longer word, point to more than five letters. Play games with them, learn spelling or even use them in your craft projects or scrapbooks. The first player to have 10 words wins the game. Others are misspelled because we think a letter should be doubled, so we put in an extra one. These braille dots are positioned like the figure six on a die, in a grid of two parallel vertical lines of three dots each. These words are legal according to the Official Scrabble Players Dictionary, 4th Edition, or validated with Hasbro's online Scrabble Dictionary in November 2016. Then, realize, that you can only make 12-letter words with 5 vowels and 7 consonants. Each time you remove a letter from it, it still remains an English word - from nine letters right down to a single letter. Be ready with words that use the Z, Q or K if you draw that tile. The fifth letter can now be only 1 of 2 remaining letters. Answer to how many three letter code words can be constructed from the first 10 letters of the alphabet if no repetitions are allo Skip Navigation. Alphabet is a 8 letter medium Word starting with A and ending with T. Apples4theteacher letters of the alphabet coloring pages make great mouse practice activities for toddlers, preschool, and elementary children. Note that there are a couple cases where a Welsh "letter" is actually made up of two characters (e. ABCs - Starfall Education: Kids Games, Movies, & Books K-3. For English we need to know only twenty-six symbols -- the letters of our alphabet -- to have the basic tools of our written language, although it takes a lot of practice to use these tools well. The Yiddish Language. In the English alphabet, there are 7 capital letters that look the same after being rotated 180°. You can only get points for the first 5 words with the same # of letters (5 3 letter words, 5 4 letter words). Watch the words that come out of your mouth and you will have a good idea of the direction your thoughts are facing, and as a result, your life. Select the range A2:A8. its impossible to answer that question without knowing how big you wil consider a word to be, if there is no limit its infinite. a piece of printing type bearing such a symbol or character. C is 3rd, O is 15th, F is 6th, U is 21th, N is 14th, D is 4th, Letter of Alphabet series. You can choose from worksheets for various grades such as English worksheets for Grade 1, Grade 2 or Grade 3. However, there are also combined letters and three umlauted forms. In American English, this sound is a tap. Crochet patterns and tutorials for lovers of cats, belly dancing, children, fashion, and all things geeky. Words with Friends point value for alphabet: 15 points. This video provides information about the pronunciation of English words along with its spellings. Having several sizes of letters for one alphabet gives you lots of options to play with. The letter fell into disuse for several centuries, until the Norman French arrived with words that used the “Z” sound. Not sure how to solve part b. Those letters are H, I, N, O, S, X, and Z. Find 10 of a chosen letter and you win! 22. Face it, 3 and 4 year olds have the attention spans of gnats. Again, this is a good one for developing vocabulary – and it can be played by children as well as by adults. Hi Naomi, In the province of Saskatchewan in Canada where I live our plates have three digits followed by three letters. Or go to a sample answer page. Search More words for viewing how many words can be made out of them NoteThere are 2 vowel letters and 3 consonant letters in the word found. Alphabet Recognition - Children will learn to identify letters of the alphabet, especially letters in their own name. Working with such a device can definitely be of benefit when attempting to unscramble letters to make words. TYPE 3: Related Words (IPS Type C) The alphabet is here to help you with these questions. Found 15939 3-letter words for Scrabble, Words With Friends, WordHub, and Crosswords. 2- Arabic letters form words by connecting them together. Start with the five vowels; a, e, i, o, u and pick the seven most used consonants; t, n, s, h, r, d, l. It was the Etruscans who first developed it after borrowing the Greek alphabet, and the Romans developed it further. We found a total of 23 words by unscrambling the letters in found. Alphabet Test Questions & Answers : How many meaningful English words can be formed with the letters ESRO using each letter only once in each word?. Number Two Using the letter grid below, how many words can you find. Add the dimensions. Put the knot into the top slit of your cardboard loom. To calculate the amount of permutations of a word, this is as simple as evaluating #n!#, where n is the amount of letters. * Hint option * Complete. In addition, the presence of a dagesh (a dot placed within a letter to add emphasis) can. There is not necessarily a direct one-to-one correspondence between the vowel sounds of a language and the vowel letters. To make the famous trilled rr, the key is practice. Nim is a synonym of steal. The letters "s" and "e" are especially useful. No memorization required! 2. I know that the answer is $1976$, from $26^3-26\cdot25\cdot24$ However, I had a different way of doing it, which is by saying that the the two letters that are the same have $26$ possibilities, and the remaining letter also has $26$ possibilities, and there are $3$ possible distinct positions of the letters, AAB, BAA, and ABA with A representing the two letters that are the same and B. Every aspiring polyglot should learn to read the International Phonetic Alphabet (IPA). of ways to choose first letter for code = 8. The fourth letter can be any one of 26. Potential pitfall: Putting up the Word Wall at the beginning of the year, telling kids to use it, but failing to demonstrate how. Submit a letter to the editor. One you can hear and write, the other you have to memorize. It’s also the most commonly used letter among them due to the several sounds it represents. Step 3: sing the alphabet song! Have each student sing the song individually with you. Below is the breakdown: A=9, B=2, C=2, D=5, E=13, F=2, G=3, H=4, I=8, J=1, K=1, L=4, M=2, N=5, O=8, P=2, Q=1, R=6, S=5, T=7, U=4, V=2, W=2, X=1, Y=2, Z=1 *=2 (wild. how we pronounce “b” as “bee” instead of “bed” when “b” is derived. Add a letter. You will likely see a menu with many choices; look for the phrase "Flip Horizontal. * More than 15000 different letter combinations,game levels. While we don't have a pre-made template, you can easily make a template for lower-case letters and numbers. That cattle trail of a road. “Foreign” letters: j, k, w, x and y are borrowed to write foreign words. Each Korean word, or letter-block, has several different meanings. We’ll do this in under 1 hour… but under a few conditions from me. ascii_lowercase (available on both) and not string. How Many Letters Are in the Hawaiian Alphabet? Credit: Glowimages/Glowimages/Getty Images Technically the Hawaiian alphabet has 12 letters, including five vowels and seven consonants, but this is an incomplete picture of the Hawaiian language as a whole. But it wasn't always that way. In today’s post: Learn how to create photo letters: a photo collage on large paper mache letters. Since there is no real limit to the length of the words that can be formed, and letters are repeated in words, a nearly infinite number. Now we are going to make a letter. Words, wordplay, reading, and writing have been favorites of Liz's since early childhood. I thought it might be useful to organize these frequently-misspelled words into three categories. In spite of the wording of the question, I suspect the student has misinterpreted the question. Log in or register to post comments. There are 12478 five-letter words: AAHED AALII AARGH ZYGON ZYMES ZYMIC. We reserve the right to edit for clarity and length. This might sound like plenty -- after all, the Latin alphabet as used in English has only 26 letters, and you can represent any real number with the numerals 0 through 9. Phonics Mini-Books. The English orthography is based on an alphabetic system of 26 letters and approximately 44 sounds/ (phonemes) and because the language is opaque “there are not enough letters of the alphabet to represent all the sounds of our speech” (Garcia & Cain, 2013, p. 26 \cdot 26 \cdot 26 = 26^3 = 17576. Words made by unscrambling the letters A L P H A B E T. Depending on the width of the letters, for a three letter monogram, you maybe able to use a 1 and a half inch or 2 inch letter as the middle letter and. Answer: flag pole. 1) Americans using American customary units of measure doesn’t refute the notion that pronouncing “z” as “zed” because the letter is derived from the Greek “Z” (zeta) is inconsistent with how other letters in the English alphabet are pronounced e. Digit 3: 26 possible letters. Though single- and two-letter words are relatively few, there are scores of three-letter words in the English language. (a) How many five letter "words" containing $2$ different consonants and $3$ different vowels can be formed? (b) How many of these "words" begin with "b" and end in "a"? I have done part a $21C2 \cdot 5C3 \cdot 5! = 252000$. You need to work out a different code for each question. If the consonant is the final letter in the word, it will go with the preceding vowel. Why was & part of the English alphabet? The word ampersand came many years later when “&” was actually part of the English alphabet. If repeats are allowed, c. Adding a dot 3 makes the next ten letters, and adding a dot 6 to that makes the last six letters (except "w" because it was not used very much in the French language at the time that Louis Braille devised this system). Having several sizes of letters for one alphabet gives you lots of options to play with. I taught myself to read the IPA alphabet, but it was tough at first. In how many ways can a 10-person committee be formed?. Find the short words and "crack" them first. In alphabetical order, Number of letters between M and T is 6. How many words can you make using the letters from "Alphabet Soup"? Sample answers : please, house, ape, Go to worksheet with 30 blanks or a worksheet with 50 blanks. To make the (9) Make A Word activity you will need 3 wooden blocks. In this program the challenge was to find all the words that could be made only with the supplied letters. This problem has been solved! See the answer. Filed Under: AH [ɑ] as in FATHER, AY [eɪ] as in SAY, EE [i] as in SHE, EH [ɛ] as in BED, IH [ɪ] as in SIT, IPA, Alphabet, and Letters, UH [ə] as in SUPPLY - the SCHWA, Videos Tagged With: Many Ways One Thing Can Be Pronounced. Start with the five vowels; a, e, i, o, u and pick the seven most used consonants; t, n, s, h, r, d, l. There are some obsolete characters and combination characters as well but the main alphabet is 40 letters. That's clear. Older children can play with four or five letter words. Also, the letter X can make the following unique sounds: /z/ as in the word xylophone (often located at the beginning of a word) /g/ follow by /zh/ as in the word luxurious. You can follow the question or vote as helpful, but you cannot reply to this thread. Learning the Punjabi alphabet is very important because its structure is used in every day conversation. How to convert letter to number or vice versa in Excel? If you have a list of letters of an alphabet, and now you want to convert these letters to their relative numbers, such as A=1, B=2, C=3… And in other cases, you need to reverse this option to change the numbers to their associated letters. The dots above the extra letters ä, ö and ü create a sharper sound, made more in the front of your mouth. Very emergent learners should begin with fewer letters; other students can manage working with more letters at one time. The basic shape of Alif takes three two forms depending on its position in the word: Initial ا , Medial ـا , Final ـا. What Are The First 3 Letters Of Greek Alphabet Posted on May 5, 2020 by Alya Cretan pictographs the greek alphabet greek food and language crossword 5 how many three letter code words can. Easily handled. Advanced: 50+ words. The better you pronounce a letter in a word, the more understood you will be in speaking the Punjabi language. the Jews noted that 22 was significant in not just being the number of letters in the Hebrew alphabet but also the number of generations from Adam to Jacob (israel), the number of works of creation, and. Click on each letter to hear its name. In mathematics, iff means “if and only if. In my class of 3-5 year old children, we learn a letter of the alphabet each week. Help your child with his grammar skills with this printable worksheet that focuses on using end punctuation. In the English alphabet, there are 7 capital letters that look the same after being rotated 180∘. Give your child different words to spell, eventually ending up with the final word. You are welcome to share, what you know more about that particular letter, with us by. Search More words for viewing how many words can be made out of them Note There are 3 vowel letters and 5 consonant letters in the word alphabet. Encryption - Using More Than One Alphabet. List of all 7-letter words. In the early. In the English alphabet, there are 7 capital letters that look the same after being rotated 180∘. View All Letter E: Letter F. Let us see the third letters now ,there are: t, c, q, h. Those letters are H, I, N, O, S, X, and Z. " Now add a letter to make "bet. The letter z, by the way, has not always been relegated to the end of the alphabet. When you're learning your letters or practicing your letters say the sound the letter makes, rather than the letter name. While the original Greek alphabet was written in all capitals, three different scripts were created to make it easier to write quickly. We found a total of 1008 words by unscrambling the letters in 3 letter words. For example, the word " Army " would be "Alfa. How many three-letter words can be made from the English alphabet (repeating letters are allowed)? Get more help from Chegg Get 1:1 help now from expert Algebra tutors Solve it with our algebra problem solver and calculator. (May even cause issues causing clients to crash. How many different committees of 5 people can be chosen from 10 people? 10*9*8*7*6/(120)=252 4. Answer to how many three letter code words can be constructed from the first 10 letters of the alphabet if no repetitions are allo Skip Navigation. You can follow the question or vote as helpful, but you cannot reply to this thread. Like English, the German alphabet consists of 26 basic letters. This website uses cookies to ensure you get the best experience. , the letter N ) and the remaining 3 are (i. Then select all of the letters and click the Text Effects tool. You can make 26 one-letter words. The third also can only be of the 6 letters (because on of the will already be used in the second slot, but it can use a letter from the first slot) so there are also 6 possibilities 7 6 6 Then you just multiply them altogether to get 252. These materials can be used as part of a Letter C program of activities to reinforce letter recognition, initial handwriting practice and to identify related C words. 8 Letters of the English Alphabet. Ba is connectable to both sides. Exclude words that fail. The following 123 pages are in this category. Operating. Accumulate 4. With over 29 years experience, we've processed over 205,000 sign orders for name brand businesses in all industry sectors. Word lists are in the order of the most common words and most searched. The first two-letter word, the Greek letter NU, appears at digit 10351, followed not much later by US at 10868. The alphabet also features a unique letter that has nearly come to symbolize the language itself: the ñ. I Can Color 'A' Words Color words that start with a, including apple, anchor, arrow, arm, ant, acorn, alligator, ape, airplane, and ax. Students build three-letter CVC words with the short u and the short a sounds. Think about the first three letters of the alphabet: A, B, and C. What Are The First 3 Letters Of Greek Alphabet Posted on May 5, 2020 by Alya Cretan pictographs the greek alphabet greek food and language crossword 5 how many three letter code words can. The names of the letters, the order of the letters, and the numerical value of the letters are apparently the same in both K'tav Ashuri and K'tav Ivri; thus, any religious significance that would be found in the numerical value of words or the sequence of the alphabet is the same in both scripts. ; There would be 26 7 eight-letter words that end with the letter N, which is the same as the number of 7 letter words (just add "N" to the end of each seven-letter word. Discuss other letter C words and images found in the worksheet. In how many ways can a 10-person committee be formed?. Color Letters prints full-page letters of the alphabet with an animal or thing whose name starts with that letter. How many three letter code words can be made from the 26 letters of the alphabet if a. These symbols are called "diacritic marks". The word Semitic comes from the name Shem, named in Genesis (6:10) as the son of Noah, whose descendants lived in the Middle East. Answer to how many three letter code words can be constructed from the first 10 letters of the alphabet if no repetitions are allo Skip Navigation. You want to make the list easy to remember, first and foremost. Doing that can make the letters look unrealistic because there are too many shadows. Like our 2 Letter Words list, the 3 letter words are all taken from the an open-source dictionary for Scrabble® crossword game and are valid in US play. Each word must contain the central H and no letter can be used twice, however, the letters do not have to be connected. Letters in print should be written from top to bottom and from left to right. Most nouns and some verbs can become plural by adding an "s," so "cot" can become "cots. Diacritic marks change the way you pronounce the letter. Place each of the other strings into. 3 Letter Sounds: 4 Letter Names: 5 Words: 6: 7 Letter Repeats: 8 Two Letter Types: 9 Consonants: 10 Vowels: 11 Long and Short Vowels: 12 Vowel Y: 13 More Vowel Rules: 14 Letter Blends: 15 Common Words. The word 'amazing' has a great punch, but how it will sound when we we say 'amaxing', neither the word will make any sense nor it will be accepted by the readers. The better you pronounce a letter in a word, the more understood you will be in speaking the Punjabi language. Magnetic letters- Choose a "Make A Word" page. In the English alphabet, there are 7 capital letters that look the same after being rotated 180∘. 3 letters have been added from Old English: J, U, and W. WARNING : this is the original tuto, with all the steps With this one you will make really detailed altered books. Each time you remove a letter from it, it still remains an English word - from nine letters right down to a single letter. Compare if the counts are same then the word can be made. ㅠ sounds like yoo. Most nouns and some verbs can become plural by adding an "s," so "cot" can become "cots. X Scroll down for a list of DOs and Don'ts to help with Spelling. How many different 3 letter "words" can be made from these letters such that when the word is rotated 180∘, the. Jones is the Chairman of a committee. You can put this solution on YOUR website! if repetitions are allowed, then 27^5 = 14348907 if repetitions are not allowed, then 27P5 = 27! / 22! = 9687600 how's this work? in order to see, try 2 letter code words from 3 letter alphabet of abc. From each word below, make two new words by adding a letter (1) at the end; (2) at the beginning. For example, you have a list of data in range A2:A8 as below screenshot shown. We’ll do this in under 1 hour… but under a few conditions from me. Some examples of dollar words are: contented, cookout, mittens and shadowing. These are not considered separate letters. It depends on whether you allow the same letter to be used multiple times or do not allow repetition. Search More words for viewing how many words can be made out of them Note There are 3 vowel letters and 7 consonant letters in the word depressing. By Vivian Wang and Amy Qin HONG KONG — After 16 years in. STEP 3: Write the word with a dry erase marker. Do this for all words in dictionary. When you're learning your letters or practicing your letters say the sound the letter makes, rather than the letter name. That is how many possible three letter words we can have for the English language if we didn't care about how to pronounceable they are, if they meant anything and if we repeated letters. ‘A’ is for Alligator (color him green), ‘B’ is for bear, and ‘C’ is for cats—all three of them. Alphabet Cut & Paste Sheets We used these printables to reinforce words that start with the different letters of the alphabet. Ask the child to read the words by sounding out the letters. For the sake of output and server capacity, we cannot let you enter more than 8 items! #N#Quick! I need help with:. You may also see the ASL Manual Alphabet referred to as the American Sign Language Fingerspelled Alphabet. You can make 26 x 25 x 24 three-letter words. Category Questions section with detailed description, explanation will help you to master the topic. Thus, the beginning reader must learn the connections between the approximately 44 sounds of spoken English (the phonemes), and the 26 letters of the alphabet. We won't send you spam or share your email address with anyone. Real words vs. In other words the original capitals were turning into the minuscules or 'small letters' which we use today. Nathan (at 2 years old) can already identify many of his letters simply because we read lots of alphabet books. You can learn more about word games and word trivia in our. In how many ways can a committee of 5 be chosen. The letter "y" has dots 1, 3, 4, 5, and 6. Teaching preschoolers to write through fun activities shows them how to write every letter in the alphabet and their name, while also getting them ready for that first day of kindergarten. If one or more words can be unscrambled with all the letters entered plus one new letter, then they will also. But it wasn't always that way. Once you are vulnerable, you add a few red dice (with more obscure letters) and must start with 4 letter words. Blending, for some students, is a challenge. Anagrams are meaningful words made after rearranging all the letters of the word. We use the second formula, and the solution is 10! / 6!, or 5,040. Here's how: In Microsoft Word, type up the alphabet a-z and the numbers 0-9, using whatever font size that you would like. English Vocabulary Word Lists with Games, Puzzles and Quizzes Click the Letters in the Correct Order (Flash). By age 2 or 3, children who are regularly read to, start to realize that books are made of text and the text is made of letters. Adjective Satellites. How many three letter words can be made from SUCCESS Permutations - Duration: 6:15. Words with Friends point value for alphabet: 15 points. January 16, 2012 by All Star Leave a Comment. These five worksheets show the lower case cursive handwriting alphabet. 3rd letter can be formed of 4 alphabets. Featuring 100 tiles, this wooden letter set offers you plenty of options to choose from when spelling out names or themes. D is 4th, E is 5th, P is 16th, R is 18th, S is 19th, I is 9th, N is 14th, G is 7th, Letter of Alphabet series. If we take out the “double letters” (ㄲ, ㄸ, ㅃ, ㅆ and ㅉ), we are left with:. Learning Q & Qu, and its Phonics Sound When teaching the letter Q and its /koo/ sound, it is always taught along with "Qu", simply because, almost all words in English have "Qu" paired together. “Foreign” letters: j, k, w, x and y are borrowed to write foreign words. When you start to learn Korean, you’ll start with learning how to read words in the Korean alphabet, then move on to learning how to pronounce those words. The Japanese alphabet is usually referred to as kana, specifically hiragana and katakana. If your letters are 2 inches tall, you would use two squares for each letter. Then select all of the letters and click the Text Effects tool. View All Letter E: Letter F. Because of the nature of the Korean alphabet system, these letters are super easy to read! In this guide, we'll go over each and every Korean letter […]. Search More words for viewing how many words can be made out of them Note There are 2 vowel letters and 3 consonant letters in the word denim. If still not eliminated, use a recursive algorithm to see if the word can be formed by making paths through the puzzle. The alphabet is a writing system which evolved from a western variety of the Greek alphabet. Morse Code Alphabet All we see are dots and dashes. If you go for five letters, your opponent will create a lay along, following your word. You could make your ciphertext a little tougher to decode if you threw 26 pieces of paper into a hat, each with a letter of the alphabet written on it, drew them out one at a time, and put them side-by-side under a normal alphabet. What Are The First 3 Letters Of Greek Alphabet Posted on May 5, 2020 by Alya Cretan pictographs the greek alphabet greek food and language crossword 5 how many three letter code words can. Search More words for viewing how many words can be made out of them Note There are 6 vowel letters and 7 consonant letters in the word demodulations. The result can be interpreted as a secret message or calculated by the standard gematria. " Now add a letter to make "bet. You'll want to print out the free printable sign language alphabet chart on this page and keep it handy. A Korean letter is Hangul. The letter "a" is written with only dot 1. (May even cause issues causing clients to crash. Enter your words (you can click Enter to go from one word to the next): Click here to add more words (space to add up to 50 words will be shown here; your existing data above will remain). As the number of things (letters) increases, their permutations grow astronomically. It’s also the most commonly used letter among them due to the several sounds it represents. Obviously the clues help you fill in the grid, but as you make headway into the puzzle, things can happen the other way around - certain combinations of letters in the grid (e. We first make an assumption that you are picking 5 letters from 26. There are 44 phonemes in English (in the standard British model), each one representing a different sound a person can make. Submit a letter to the editor. com, visit our Word Finder page. When the vowels EAI are always together, they can be supposed to form one letter. You can remember this with the mnemonic lAttIcEwORk and noting that you start by throwing out 7 tiles lettered A, decreasing the number by one for each of the vowels, and finally setting aside 3 tiles with the letter R on them. Multiple Choice question, Analogy , Simple Analogy, Choose The Analogous Pair, Double Analogy, Choose A Similar Word, Detecting Analogies, Three Word Analogy, Number Analogy, Alphabet Analogy, Classification. Go ahead and try encoding or decoding some text. – abarnert Aug 19 '13 at 22:02. How many three-letter code words can be constructed from the first nine letters of the Greek alphabet if no repetitions are allowed? different code words. >> Write us: We welcome letters up to 150 words, and guest columns of 500-600 words. I agree with ted's numerical solution. Distribute letter tiles or squares to individual or pairs of students, and ask them to follow along with you through a few sample exercises. * Hint option * Complete. asked by please help me out!!! please on February 11, 2013; Mathematics. "One-Letter Words, A Dictionary" illuminates the more than 800 surprising definitions associated with each letter in the English alphabet. The phonetic alphabet is a list of words used to identify letters in a message transmitted by radio, telephone, and encrypted messages. Now that we've "placed" the first letter, the second slot only has 25 options left, since the second letter must be different from the first, and there are only 26 letters in the alphabet (assuming this radio station is in an English-speaking country). C is 3rd, U is 21th, B is 2nd, E is 5th, Letter of Alphabet series. asked by Sheryl on April 22, 2013; statistics. What Are The First 3 Letters Of Greek Alphabet Posted on May 5, 2020 by Alya Cretan pictographs the greek alphabet greek food and language crossword 5 how many three letter code words can. How many 3-letter code words are possible using the first 8 letters of the alphabet if a) No letter can be repeated ? b) Letters can be repeated ?. How many different 3 letter "words" can be made from these letters such that when the word is. a piece of printing type bearing such a symbol or character. Anagrams are meaningful words made after rearranging all the letters of the word. Practice tip: Say the word butter (with American pronunciation) and think of the sound you make in the middle (tt). Multiple Choice question, Analogy , Simple Analogy, Choose The Analogous Pair, Double Analogy, Choose A Similar Word, Detecting Analogies, Three Word Analogy, Number Analogy, Alphabet Analogy, Classification. Once you are vulnerable, you add a few red dice (with more obscure letters) and must start with 4 letter words. ABCs - Starfall Education: Kids Games, Movies, & Books K-3. Five letter words. The larger a child’s oral vocabulary, the more words he or she will be able to read and understand. Example 6: How many five digit codes can be formed if the first two digits must be non-repeated letters of the alphabet, and the last three digits can be repeated numbers from the set {0 - 9} Example 7: fair coin is tossed 5 times and the result ( heads or tails ) is noted on each flip. 256 words D. This package includes one worksheet for each of 8 different word families (_at, _an, _et, _ug, _in, _op, _ip, _ot). L is 12th, E is 5th, T is 20th, R is 18th, S is 19th, Letter of Alphabet series. These are the letters that have sounds that are familiar to English speakers, but they’re represented by unfamiliar characters: Б, Г, Д, Ш, З, Ф, П, Л, Э, Ч, И and Ю. If you want to highlight all cells which the first letters are A in the list, please do as follows. How many different 3 letter "words" can be made from these letters such that when the word is rotated 180∘, the. At the bottom are links for more practice of upper and lower case letters. Trace and write lowercase m to z. WordFind challenges students to see how many words they can find in a group of mixed-up letters, including a word from a list that uses all the letters. The tie to a known object links shape of letter to its sound. This article is an introduction to the symbols of consonants of the International Phonetic Alphabet (IPA) as it is used to denote pronunciation of English words (there is a separate article on vowels). Solve advanced problems in Physics, Mathematics and Engineering. Identifying the letters of the alphabet can be a fun activity for preschoolers to learn while playing! Here are 25 Alphabet activities to recognize the letters of the alphabet. Like the game Scattergories, you can name as many items in a category that start with the same letter. Every puzzle has additional words which score extra bonus points - you'll find the details on the printout. Alphabet Scramble. The letter z, by the way, has not always been relegated to the end of the alphabet. Letter Writing - Write in 4-Line Sheets. do Also see:- Words starting with Found Words ending with Found Words Containing Found Found Meaning :-of Find imp. For example, the letter 'p' sounds exactly the same as the word "pea". 365 SYT) Hence there are 10 choices for each digit giving 10 10 10 = 1000 possible three digit numbers. We use new acrylic to print the sign characters so the backgrounds are clear and clean. In how many ways can a five-letter security password be formed from letters of the alphabet if no letter is repeated? combination. In other words, you want to read and write in Hebrew. You should have 11 strings total (depending on the size bracelet you are making). For example, you can type the capital alpha ( A) in Greek with a regular capital A because these letters look the same in Greek and English. Proper nouns are not allowed, however, plurals are. "How many pairs of letters are there in the word 'SEQUENTIAL' which have as many letters between them as are in the alphabet ? " Test. ) Print out the words I. Obviously the clues help you fill in the grid, but as you make headway into the puzzle, things can happen the other way around - certain combinations of letters in the grid (e. License plates are made using 3 letters followed by 2 digits. Pre-K through 1st Grade. >> Write us: We welcome letters up to 150 words, and guest columns of 500-600 words. How many three letter code words can be made from the 26 letters of the alphabet if a. 26*26*26*26 = 26^4 =456,976. I promise that if we taught spelling this way, our efforts would be better spent. Search More words for viewing how many words can be made out of them Note There are 4 vowel letters and 7 consonant letters in the word frailnesses. Move to friendly sounds next. Pre-K through 1st Grade. Hi Naomi, In the province of Saskatchewan in Canada where I live our plates have three digits followed by three letters. Natural just means they are regular notes, they are just the regular alphabet letter name like A, C, or G. Find out what words are made from combinations of letters. Search More words for viewing how many words can be made out of them Note There are 3 vowel letters and 4 consonant letters in the word cofound. Working with such a device can definitely be of benefit when attempting to unscramble letters to make words. Watch the words that come out of your mouth and you will have a good idea of the direction your thoughts are facing, and as a result, your life. Letter manipulatives give children a way to interact with the alphabet through hands-on play. 3 million, which can only be called a savings when compared to her. Children drag pictures to the letters that make the first sounds of the word that describes the picture. Each of the 26 alphabet letters has a capital and a lower case form. Calling letters without calling setlocale is not generally useful; it will either be the same as ascii_letters or totally broken. It is really that simple. Now you can encipher your plaintext and pass it to your friend who knows the proper key letter. Match each letter of the word to the “Alphabet Fitness Key” to determine your workout. Alphabet tracing worksheets for 3 year olds. To get an overview of what you can do on word-grabber. We discount those and also individual. The letter fell into disuse for several centuries, until the Norman French arrived with words that used the “Z” sound. of ways to arrange 4 letters out of 8 = 8P4 another ways, no. The best way to replace a letter from a word to make another word is to use dCode that will try all possibilities (only one letter or more than one letter changing). For example, a word where every featured letter appears twice, like APPEASES, might be called a pair isogram, a second-order isogram, or a 2-isogram. What has to be broken before you can use it? An egg. Words include: rug, bug, bat, rat, hat. Letter blends appear in the beginning or at the end of words to create specific sounds. Enjoy! Letters of the Alphabet A to Z Letters to Print. If this is not the case, the problem is different. Then you can work up to having the child spell their name aloud. The pronunciation of these letter names can be very useful if you need to spell out your name, words, or just individual letters, in spoken English. (a) How many five letter "words" containing $2$ different consonants and $3$ different vowels can be formed? (b) How many of these "words" begin with "b" and end in "a"? I have done part a $21C2 \cdot 5C3 \cdot 5! = 252000$. Club and Tournament Play as of 2009. sterlet 2). They can be arranged to form names or slogans. If you are teaching Phonics alphabetically, the first three letters you will teach will be A, B, and C. It is officially the International Radiotelephony Spelling Alphabet, and also commonly known as the ICAO phonetic alphabet, with a variation officially known as the ITU phonetic alphabet and figure code. My mother and I used to play them all the time, usually while passing time waiting for the start of a play. is a holding company that gives ambitious projects the resources, freedom, and focus to make their ideas happen — and will be the parent company of Google, Nest, and other ventures. With LinguaLift learning Hebrew characters is just a week-long project. To calculate the amount of permutations of a word, this is as simple as evaluating #n!#, where n is the amount of letters. Other than “cab,” he really can’t form any other words with those three letters. But in cursive, it's like all of the letters are holding hands to make a word. (See Appendix 3: Assessment). How many three letter code words can be made from the 26 letters of the alphabet if a. The first letter of each word of the sentence names the five lines from bottom to top. Submit a letter to the editor. It was the construction of the first major dictionaries by Dr. Letters can be made lots of ways. You don’t have to decide whether to put dots on these vowels, these come standard with certain words. The fifth letter can now be only 1 of 2 remaining letters. Now that we've "placed" the first letter, the second slot only has 25 options left, since the second letter must be different from the first, and there are only 26 letters in the alphabet (assuming this radio station is in an English-speaking country). In this how-to you will learn the process to make printable stencils in Microsoft Word using the Text Effects function. Let students trace the shape of the letters on the flashcards and then “draw” the shapes with their fingers on the floor and doors, etc. Cut words from a magazine or newspaper to write/ draw/ illustrate a letter, message or poem for a mother. You may use plurals if available. How many three letter words can be made from SUCCESS Permutations Anil Kumar Letter Arrangement How to find rank of word in case of repeating alphabet (short trick) - Duration: 4:37. Anil Kumar 17,728 views. Find and save ideas about font alphabet on Pinterest. Last month, editors at the Book Review wrote to our readers. Answer to how many three letter code words can be constructed from the first 10 letters of the alphabet if no repetitions are allo Skip Navigation. ; There would be 26 7 eight-letter words that end with the letter N, which is the same as the number of 7 letter words (just add "N" to the end of each seven-letter word. You should be able to use intuition to look at other words with these partial subsititions and make good guesses. Find the words in the A words wordsearch puzzle, then use the extra letters to find the secret word: "alligator. Blending sounds into words–what an exciting step in learning to read. The longest word whose letters are in alphabetical order is the eight-letter Aegilops, a grass genus. If one or more words can be unscrambled with all the letters entered plus one new letter, then they will also. a written or printed communication addressed to a person or organization and usu. We reserve the right to edit for clarity and length. Here is the list of consonants that was presented: ㄱ ㄲ ㄴ ㄷ ㄸ ㄹ ㅁ ㅂ ㅃ ㅅ ㅆ ㅇ ㅈ ㅉ ㅊ ㅋ ㅌ ㅍ ㅎ. Then 1 of 25 Then 1 of 25 (all but the preceding letter) 1/25 1/25-----= 26*25^4 = 10156250 =====. Unusual plurals and verbal extensions are always listed. Noun forms are not shown where verb forms exist. Identifying the letters of the alphabet can be a fun activity for preschoolers to learn while playing! Here are 25 Alphabet activities to recognize the letters of the alphabet. I can sing the alphabet in 20 seconds. The standard Hindi alphabet, as agreed by the Government of India, has 11 vowels and 35 consonants. Each Korean word, or letter-block, has several different meanings. Alphabet Review Activities- A through review of each letter using multiple methods. This might sound like plenty -- after all, the Latin alphabet as used in English has only 26 letters, and you can represent any real number with the numerals 0 through 9. Do you know lots of words in English? How many can you make in three minutes? Can you put the letters of the alphabet in the right order? You've got 30 seconds! Ratty, the evil arch-villain, is trying to take over the world. I have spent a long time pondering upon this question and according to a former U. Alphabet Signs provides fast, expert service and great prices on sign letters, vinyl lettering, letter boards, parking signs, sidewalk signs and more for business signs. Epidemiologic tools can help model these risks. For example, you have a list of data in range A2:A8 as below screenshot shown. What Are The First 3 Letters Of Greek Alphabet Posted on May 5, 2020 by Alya Cretan pictographs the greek alphabet greek food and language crossword 5 how many three letter code words can. Depending on which word “y” is being used in, it can represent different sounds. 256 iThink * * * * * It depends on combinations of how many. This website uses cookies to ensure you get the best experience. Three words are related, find the two words that do NOT go with these three. Since there are only 26 letters in the alphabet, sometimes letter combinations need to be used to make a phoneme. Compare if the counts are same then the word can be made. Also see online stencil maker. Use play-doh to make the letters. Nim is a synonym of steal. In my class of 3-5 year old children, we learn a letter of the alphabet each week. Accumulate 4. If a Scrabble player has all of a word except a letter or two, he can use this list of tiles to figure out whether or not his desired tile might be in the bag. Each of the 26 alphabet letters represents one or more sounds. different in shape according to their position. F is 6th, O is 15th, U is 21th, N is 14th, D is 4th, Letter of Alphabet series. There are 104 tiles in Words With Friends. Another important point is that not each Chinese character equates to a word — another common misconception. The names for the Greek letters came from the first two letters of the Phoenician alphabet; aleph, which also meant ox, and bet, which also meant house. of possible code, when repetition is not allowed = 8P4 = 1680 as repetition is not allowed, no. ” They are be, we, and me. It was still very scruffy but many of the letter-forms had simplified. If 3 roads connect town A to town B and 4 roads connect town B to town C, how many ways could you make a round trip if no road can be used more than once? 12,144 The Greek alphabet has 24 letters. 2 Short list of alphabets. Three words are related, find the two words that do NOT go with these three. Johnson (1755) and later Noah Webster that set 'correct' spellings in stone. They have meaning (s) beyond their component letters that are neither names nor abbreviations. With readiness the pace of this work moves steadily along. Tip #2: Make Use of Colored Squares. Morse Code Alphabet All we see are dots and dashes. But it wasn't always that way. This way, you can score single for your word. Help your beginning readers practice phonics and expand their reading vocabulary with this rhyme match worksheet. Wikipedia has tons of comprehensive information, but can be confusing to a beginner. The word list used in tournament games in the United States is known as the Official Tournament and Club Word List, created by the National Scrabble Association. How many three-letter words can be made from the first eight letters of the alphabet if consonants cannot be next to each other and letters cannot be repeated? The number of three-letter words with consonant as the first letter. Search More words for viewing how many words can be made out of them Note There are 4 vowel letters and 7 consonant letters in the word frailnesses. In words like “myth” or “hymn,” the letter takes on a sound like a short “i” and the mouth and throat don’t close when the sound is made. Click these words to find out how many points they are worth, their definitions, and all the other words that can be made by unscrambling the letters from these words. The English alphabet has $26$ letters, of which $5$ are vowels. What Are The First 3 Letters Of Greek Alphabet Posted on May 5, 2020 by Alya Cretan pictographs the greek alphabet greek food and language crossword 5 how many three letter code words can. Ask the child to read the words by sounding out the letters. Japanese students must learn about two thousand symbols to have their basic tools, and still it takes practice to use them well. Enter your words (you can click Enter to go from one word to the next): Click here to add more words (space to add up to 50 words will be shown here; your existing data above will remain). How Many Letters in Words With Friends. The Portuguese alphabet has 23 letters. Or go to a sample answer page. (a) Repetition is allowed. Alphabet Inc. Don't get hung up on this word. Once students know a few consonants and vowels, we can begin to teach them how to blend those sounds into meaningful words. See other lists, that begin with, end with or contain letters of your choice. Teach them to blend the ending sound of each letter with the starting of another one. Compare if the counts are same then the word can be made. These symbols are called "diacritic marks". Score 1 point for each 3 letter word, 2 points for 4 letter words, 3 points for 5 letter words and so on. In addition, the presence of a dagesh (a dot placed within a letter to add emphasis) can. (Playdough to Plato) Practice writing letters in spices for an additional sensory experience. See screenshot: 3. As I am sure you know there are 26 letters in the English alphabet, but do you know how many of these letters also sound exactly like English words? A large number of the letters in the English alphabet make sounds that are exactly the same as one-syllable English words. The Latvian alphabet consists of 33 letters, 22 Roman letters (these do not include Q, W, X, or Y) and 13 extra letters which have symbols above or below the letter. Why do I need to use English book of words for every letter of the alphabet a to z? This book will help you a great deal to help you learn new words and there meaning. the 59 2- and 3-letter X and Z words the 95 Q words of up to 5 letters the 29 2- and 3-letter J and Q words the 311 Z words of up to 5 letters reversible 3-letter word (103 + 59 palindromes) the 50 2- and 3-lettter words ending in -A, listed by endings. Monogram a blanket, make toy blocks, make puffy letters for beginning readers, decorate a child's bedroom - you name it! And it's not just uppercase - see the links below! Click. For example, if twelve different things are permuted, then the number of their permutations is 479,001,600. Click here to see more perfect pangrams. How many 3-letter code words are possible using the first 10 letters of the alphabet if a) no letter can repeat; - Answered by a verified Math Tutor or Teacher. BASIC LETTER PARTS Letterhead. The word alphabet comes, via the Latin word alphabētum, from the Greek word αλφάβητος (alphabētos), which itself comes from the first two letters of the Greek alphabet, α (άλφα/alpha) and β (βήτα/beta). By age 2 or 3, children who are regularly read to, start to realize that books are made of text and the text is made of letters. Find the words in the A words wordsearch puzzle, then use the extra letters to find the secret word: "alligator. Don't get hung up on this word. One of the biggest is a proven track record. Of course, there are some accent marks. Two such styles. The choices are b,c,d,f,g,h. First I will present the French phonetic alphabet and all the sounds that French contains (a few of them do not exist in English), so that you can leave here pronouncing words like a champ. In addition to all the story books we read, I make alphabet books part of our regular reading time. It consists of 250 or so letters (), numerals, punctuation, formatting marks, contractions, and abbreviations (). UPDATE: I found the letter distribution for Spanish Bananagrams here and the French Bananagrams letter distribution here. asked by please help me out!!! please on February 11, 2013; Geometry. The pronunciation of these letter names can be very useful if you need to spell out your name, words, or just individual letters, in spoken English. Why can’t a man living in New York be buried in Chicago? Because he’s still living! What begins with T, ends with T and has T in it? A teapot. Word counters are good, but letter count is often more valuable. It takes these three forms depending on its position in the word. Alphabetical Autobiography- Students will be able. If we take out the “double letters” (ㄲ, ㄸ, ㅃ, ㅆ and ㅉ), we are left with:. In my class of 3-5 year old children, we learn a letter of the alphabet each week. Portable sign letters made by Flex Change® are typically used with the roadside portable marquee signs. In alphabetical order, Number of letters between M and T is 6. What Are The First 3 Letters Of Greek Alphabet Posted on May 5, 2020 by Alya Cretan pictographs the greek alphabet greek food and language crossword 5 how many three letter code words can. column matrix make matrix (The actual procedure listed at the end of this section includes a slight complication to deal with the case of and , but that’s not. Use the zoom out button to zoom out as far as possible. Use play-doh to make the letters. How many meaningful English words can be formed with the letters ELRU using each letter only once in each word? One ; Two; Three; More than three ; None. Each letter of the alphabet is a simple shape that represents a sound, (some characters change sounds or just have a harder sound depending on their location in the syllable). One of the four words having c is the last word. presentative of ordinary English, undistorted by the many strange words included in much larger dictionarie s. Alphabet is a 8 letter medium Word starting with A and ending with T. In addition to all the story books we read, I make alphabet books part of our regular reading time. You can create writing practice sheets in D'Nealian or Zaner-Bloser style, in print or cursive form - and it's all for free! Make alphabet worksheets or spelling practice sheets. Making words out of three dimensional letters is a great way to decorate your home, office, or business. (1) The alphabet F should be in every four letter word. If your child is struggling, ask the teacher for worksheets you can practice on at home. Phonetic Alphabet Tables. The first ten letters of the alphabet are formed using the top four dots (1, 2, 4, 5). Go ahead and try encoding or decoding some text. How many meaningful English words can be formed with the letters ELRU using each letter only once in each word? One ; Two; Three; More than three ; None. If a Scrabble player has all of a word except a letter or two, he can use this list of tiles to figure out whether or not his desired tile might be in the bag. Write the letters in the squares using a vis-à-vis or dry erase marker. ” A kip is the hide of a small animal. Alphabet chart for crochet - could be used for cross stitch. View All Letter D: Letter E. How many three letter 'words' can be constructed from this set in which all the letters are different and in which the letters are in alphabetical order. How many 3-letter code words are possible using the first 10 letters of the alphabet if a) no letter can repeat; - Answered by a verified Math Tutor or Teacher. Every word on this site can be played in scrabble. If order is not important then this list is too long since it contains the words abcd, dbca, bcad and so on. How many tiles would you need for the 10th ‘L’? If you had 23 tiles, what numbered ‘L’ could you make? Exploring. Simply replace the lyrics of a nursery rhyme or familiar tune with the letters spelling the sight word. Find the US States - No Outlines Minefield457. The result can be interpreted as a secret message or calculated by the standard gematria. On the letter pages, you can view all 18 letter styles for each alphabet which you can print. You may be wondering about letters with accents like á, é, í, ó, and ú or the rare dieresis, ü. Most of these two-letter words can take an "S" on the end to form a plural three-letter word. The Latvian alphabet consists of 33 letters, 22 Roman letters (these do not include Q, W, X, or Y) and 13 extra letters which have symbols above or below the letter. Anagrams are meaningful words made after rearranging all the letters of the word. We use the second formula, and the solution is 10! / 6!, or 5,040. It consists of 22 letters, all consonants, none of which are lowercase.
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IEEE Transactions on Device and Materials Reliability
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Displaying Results 1 - 25 of 37
Publication Year: 2014, Page(s):C1 - C4
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• IEEE Transactions on Device and Materials Reliability publication information
Publication Year: 2014, Page(s): C2
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• Technological Journey Towards Reliable Microheater Development for MEMS Gas Sensors: A Review
Publication Year: 2014, Page(s):589 - 599
Cited by: Papers (11)
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Micromachined silicon platforms, owing to some of its inherent advantages including miniaturized dimensions, ultralow power consumption, reduced batch fabrication cost, long-term reliability, and compatibility with standard CMOS fabrication technology, attracted the attention of solid-state gas sensor researchers, particularly since the last decade. As the semiconducting gas sensing thin film on t... View full abstract»
• Migration of Sintered Nanosilver on Alumina and Aluminum Nitride Substrates at High Temperatures in Dry Air for Electronic Packaging
Publication Year: 2014, Page(s):600 - 606
Cited by: Papers (2)
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Joining semiconductor chips at low temperature (below 523 K) by sintering nanosilver paste is emerging as an alternative lead-free solution for power electronic packaging, particularly in high-temperature applications, because of the high melting temperature of silver (1234 K). However, silver is susceptible to migration. In this paper, we study the effects of dc bias, electrode spacing, and tempe... View full abstract»
• Aging Statistics Based on Trapping/Detrapping: Compact Modeling and Silicon Validation
Publication Year: 2014, Page(s):607 - 615
Cited by: Papers (9)
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Design for reliability is an increasingly important design step at advanced technology nodes. Aggressive scaling has brought forth reliability issues, such as negative bias temperature instability (NBTI). The aging process due to NBTI exhibits a significant amount of variability for a single device and for multiple devices. As a result, long-term reliability prediction from short-term stress measu... View full abstract»
• Impact of Sampling Rate on RTN Time Constant Extraction and Its Implications on Bias Dependence and Trap Spectroscopy
Publication Year: 2014, Page(s):616 - 622
Cited by: Papers (5)
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Inaccuracy in extraction of random telegraph noise (RTN) time constants due to an improper choice of measurement sampling rate is investigated. The sampling rate requirement for reliable extraction of RTN emission and capture times is analyzed. Consequences on transistor RTN bias dependence analysis and trap spectroscopy are discussed. View full abstract»
• Characterizations of Nanosilver Joints by Rapid Sintering at Low Temperature for Power Electronic Packaging
Publication Year: 2014, Page(s):623 - 629
Cited by: Papers (3)
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Nanosilver paste is a promising lead-free die-attach material suitable for power electronic packaging, particularly for high-temperature applications. Compared with traditional hot pressing to sinter nanosilver, rapid sintering by a pulse current is able to sinter nanosilver in less than a second. To investigate the nanosilver sintering process during rapid sintering, we characterize the temperatu... View full abstract»
• Scintillation Conditioning of Tantalum Capacitors With Manganese Dioxide Cathodes
Publication Year: 2014, Page(s):630 - 638
Cited by: Papers (1)
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Scintillation testing is a method that activates the self-healing mechanism in tantalum capacitors. In preliminary experiments, the deliberate activation of self-healing yielded up to 100% higher breakdown voltages in weak parts that had an increased risk of ignition failure. This improvement results in better performance under surge current conditions. This paper demonstrates that scintillation c... View full abstract»
• Simulation Study of the Single-Event Effects Sensitivity in Nanoscale CMOS for Body-Biasing Circuits
Publication Year: 2014, Page(s):639 - 644
Cited by: Papers (1)
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The sensitivity of single-event effects (SEEs) in nanoscale CMOS for body-biasing circuits has been investigated. For PMOS hits, it is found that forward-biasing the body for high-speed applications can suppress the SET pulses greatly. Reverse-biasing the body for low-power applications, however, does not reduce the SEE vulnerability compared with operation when the body grounded. The body-biasing... View full abstract»
• A Design for In-Situ Measurement of Optical Degradation of High Power Light-Emitting Diodes Under Accelerated Life Test
Publication Year: 2014, Page(s):645 - 650
Cited by: Papers (6)
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To better understand the reliability problem in high-power light-emitting diodes (LEDs), an online test is critical under an accelerated life test. In this paper, an experimental equipment was proposed for in situ measurement to evaluate the degradation of LED's optical properties. Signals such as illuminance and correlated color temperature were transmitted by the heat-resistant optical cable and... View full abstract»
• Investigation on Electrical Degradation of High Voltage nLDMOS After High Temperature Reverse Bias Stress
Publication Year: 2014, Page(s):651 - 656
Cited by: Papers (2)
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The breakdown voltage (BV) and on resistance (Rdson) degradation of the 700-V n-type single-RESURF lateral double-diffused MOS (nLDMOS) after high-temperature reverse bias (HTRB) stress have been investigated in this work. A detail analysis that shows good agreement with the experiments is proposed based on electrostatic force microscope (EFM) testing, charge-pumping testing, and TCAD s... View full abstract»
• Negative Bias Temperature Stress Reliability in Trench-Gated P-Channel Power MOSFETs
Publication Year: 2014, Page(s):657 - 663
Cited by: Papers (6)
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In this paper, we present the results of an experimental analysis of the degradation induced by negative-bias temperature stress (NBTS) in trench-gated p-channel power MOSFETs. Threshold voltage and carrier mobility are affected by hole trapping in bulk oxide and interface-state generation due to oxide electric field effects. A fast recovery phase occurs when gate bias is removed or reduced in ord... View full abstract»
• A Single-Bit and Double-Adjacent Error Correcting Parallel Decoder for Multiple-Bit Error Correcting BCH Codes
Publication Year: 2014, Page(s):664 - 671
Cited by: Papers (1)
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This paper presents a novel high-speed BCH decoder that corrects double-adjacent and single-bit errors in parallel and serially corrects multiple-bit errors other than double-adjacent errors. Its operation is based on extending an existing parallel BCH decoder that can only correct single-bit errors and serially corrects double-adjacent errors at low speed. The proposed decoder is constructed by a... View full abstract»
• Program Disturb Induced by Interface-Trap-Assisted Field and Thermal Electron Emission in the Channel of Split-Gate Memory Cell
Publication Year: 2014, Page(s):672 - 680
Cited by: Papers (4)
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A systematic study of program-disturb mechanisms in split-gate memory in the temperature range -45 °C to 150 °C is presented. At low temperatures, the dominant program disturb is initiated by interface-trap-assisted band-to-band tunneling in the split-gate channel area, whereas at high temperatures, it is initiated by surface generation in the select-gate channel area. The effects of... View full abstract»
• A Study on the Fatigue Behavior of Anisotropic Conductive Adhesive Film
Publication Year: 2014, Page(s):681 - 688
Cited by: Papers (1)
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In this paper, the low cycle fatigue (LCF) tests were performed for anisotropic conductive adhesive films (ACFs). The effect of temperature on the tensile mechanical properties of ACFs was investigated. It was found that, with the increase in temperature, the tensile strength of ACFs decreases and its fracture strain increases resulting in lower Young's modulus. The stress-control LCF experiments ... View full abstract»
• A Degradation Model of Double Gate and Gate-All-Around MOSFETs With Interface Trapped Charges Including Effects of Channel Mobile Charge Carriers
Publication Year: 2014, Page(s):689 - 697
Cited by: Papers (7)
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The reliability of multigate metal-oxide-semiconductor (MOS) devices is an important issue for novel nanoscale complementary MOS (CMOS) technologies. We present an analytic degradation model of double-gate (DG) and gate-all-around (GAA) MOS field-effect transistors (MOSFETs) in the presence of localized interface charge. Furthermore, we consider the effect of channel mobile charge carriers that si... View full abstract»
• Study of Stresses and Plasticity in Through-Silicon Via Structures for 3D Interconnects by X-Ray Micro-Beam Diffraction
Publication Year: 2014, Page(s):698 - 703
Cited by: Papers (9)
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X-ray microbeam diffraction measurements were conducted for copper (Cu) through-silicon via (TSV) structures. This technique has the unique capability to measure stress and deformation in Cu and in silicon with submicron resolution, which enables direct observation of the local plasticity in Cu and the deformation induced by thermal stresses in TSV structures. Grain growth in Cu vias was found to ... View full abstract»
• Atomistic Pseudo-Transient BTI Simulation With Inherent Workload Memory
Publication Year: 2014, Page(s):704 - 714
Cited by: Papers (6)
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Bias Temperature Instability (BTI) is a major concern for the reliability of decameter to nanometer devices. Older modeling approaches fail to capture time-dependent device variability or maintain a crude view of the device's stress. Previously, a two-state atomistic model has been introduced, which is based on gate stack defect kinetics. Its complexity has been preventing seamless integration in ... View full abstract»
• A Novel 3D Integration Scheme for Backside Illuminated CMOS Image Sensor Devices
Publication Year: 2014, Page(s):715 - 720
Cited by: Papers (6)
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A novel backside-illuminated CMOS image sensor (BSI-CIS) scheme and process are developed and demonstrated. This innovative scheme can be realized without fusion oxide bonding and through-silicon via (TSV) fabrication. This wafer-level TSV-less BSI-CIS scheme includes transparent ultrathin silicon (~ 3.6 μm) and uses several bonding technologies. The characterization and assessment results ... View full abstract»
• Recovery of Electrical Characteristics of Au/n-Si Schottky Junction Under ${}^{60}\hbox{Co}$ Gamma Irradiation
Publication Year: 2014, Page(s):721 - 725
Cited by: Papers (2)
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The electrical transport characteristics of a Au/n-Si metal-semiconductor Schottky barrier junction under exposure to 60Co gamma rays have been reported in this paper. The role of energy loss mechanisms in the Schottky junction due to gamma irradiation is studied using the current-voltage (I-V ) and capacitance-voltage (C-V ) measurements. The electrical characteristics were measured at... View full abstract»
• High Breakdown Voltage and Low Thermal Effect Micromachined AlGaN/GaN HEMTs
Publication Year: 2014, Page(s):726 - 731
Cited by: Papers (4)
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This work develops a thermally stable micromachined AlGaN/GaN high-electron mobility transistor (HEMT) with an enhanced breakdown voltage. After removal of the Si substrate beneath the HEMT, a 300-nm SiO2 and a 20-μm copper layer are deposited to form the GaN-on-insulator (G.O.I.) structure. The self-heating at high current that is exhibited by GaN HEMTs that are by previously de... View full abstract»
• Soft-Error Performance Evaluation on Emerging Low Power Devices
Publication Year: 2014, Page(s):732 - 741
Cited by: Papers (11)
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Radiation-induced single-event upset (SEU) has become a key challenge for cloud computing. The proposed introduction of low bandgap materials (Ge, III-Vs) as channel replacement and steep switching devices for low-voltage applications may induce radiation reliability issues due to their low ionization energy and device architecture. In this paper, the soft-error generation and propagation in Si Fi... View full abstract»
• Time Evolution Degradation Physics in High Power White LEDs Under High Temperature-Humidity Conditions
Publication Year: 2014, Page(s):742 - 750
Cited by: Papers (21)
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A high temperature-humidity test is commonly employed to evaluate the humidity reliability of electronic devices. For an integrated circuit, the degradation mechanism under the high temperature-humidity test is metal corrosion, and Peck's model is used for extrapolating the test results at accelerated test conditions to the normal operating condition. Such extrapolation is possible as the underlyi... View full abstract»
• Effects of Extreme Temperature Swings ( $-hbox{55} ^{circ}hbox{C}$ to 250 $^{circ}hbox{C}$) on Silicon Nitride Active Metal Brazing Substrates
Publication Year: 2014, Page(s):751 - 756
Cited by: Papers (12)
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Reliability of power electronic substrates has been one of the main issues of high-temperature packaging technologies. Widely used direct bonded copper (DBC) substrates, if subjected to wide temperature swings, suffer from copper layers peeling off from a ceramic because of the large thermal stresses resulting from the difference in coefficient of thermal expansion (CTE) between the copper and the... View full abstract»
• Experimental and Numerical Investigation Into the Influence of Chip-on-Film Package on Component Operating Temperature in Natural Convection
Publication Year: 2014, Page(s):757 - 765
Cited by: Papers (1)
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The thermal issues of chip-on-film (COF) packages are becoming increasingly important for high-pin-count chips, whose performance is becoming increasingly limited by the maximum power that can be spread without exceeding the maximum junction temperature. This paper conducts an experimental investigation to investigate the relationship between power dissipation and the surface temperature of the th... View full abstract»
Aims & Scope
IEEE Transactions on Device and Materials Reliability is published quarterly. It provides leading edge information that is critical to the creation of reliable electronic devices and materials, and a focus for interdisciplinary communication in the state of the art of reliability of electronic devices, and the materials used in their manufacture. It focuses on the reliability of electronic, optical, and magnetic devices, and microsystems; the materials and processes used in the manufacture of these devices; and the interfaces and surfaces of these materials.
Full Aims & Scope
Meet Our Editors
Editor-in-Chief
Anthony S. Oates
Taiwan Semiconductor Mfg Co.
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Next: Enthalpy of wetting or Up: MECHANICAL AND THERMODYNAMIC PROPERTIES Previous: Surface tension, surface Helmholtz Contents
The following definitions refer to the solid/gas interface. Changes in enthalpy and entropy associated with adsorption are usually attributed to changes in the thermodynamic state of the adsorbate only. It should be borne in mind, however, that the measured changes include contributions from the perturbation of the adsorbent.
When the addition of a differential amount d or d is effected at constant gas volume, the differential molar energy of adsorption of component , or , is defined as:
or
where the differential molar surface excess energy, , is given by 18
and the differential molar interfacial energy, , by
is the differential molar energy of component in the gas phase, i.e. .
When the addition of the differential amount d or d is effected at constant pressure , the differential molar enthalpy of adsorption, or , also called the isosteric enthalpy of adsorption ( st)19 is defined as20
where , and is the partial molar enthalpy of component in the gas phase, i.e. .
and are related by the equation:
and the same applies to the difference between and when is negligibly small compared with .
When an excess of a single adsorptive is adsorbed on a surface initially free of adsorbate species, the molar integral energy and molar integral enthalpy of adsorption are given by
d and d
Experimental calorimetric methods have to be analysed carefully to establish the appropriate procedure for deducing a particular energy or enthalpy of adsorption from measured data. The isosteric enthalpy of adsorption is usually calculated from adsorption isotherms measured at several temperatures by using the equation
where is the equilibrium partial pressure of the adsorptive when an amount is adsorbed at a temperature .
Standard thermodynamic quantities. For different purposes it may be convenient to define standard changes of a thermodynamic quantity on adsorption in two alternative ways:
(i)
the change of the thermodynamic quantity on going from the standard gas state to the adsorbed state in equilibrium with gas at a partial pressure (fugacity) of . Such quantities are sometimes called `half-standard' quantities.
(ii)
the change of the thermodynamic quantity on going from the standard gas state to a defined standard condition of the adsorbed state.
It must always be stated clearly which of these conventions is being followed.
The standard Gibbs energy of adsorption is thus:
where is the fugacity of in the standard gas state and the fugacity of gas in equilibrium with the (standard) adsorbed state. For most practical purposes the fugacity may be replaced by the (partial) pressure.
If unit pressure is chosen for the standard gas state, this may be written
while if vapour in equilibrium with pure liquid is chosen21, then
Similarly, the standard differential molar entropy of adsorption is given by
where is the standard differential molar enthalpy of adsorption.
The standard integral molar entropy of adsorption is
d
The above definitions refer to equilibrium conditions, and their applicability in regions where adsorption hysteresis occurs is open to doubt.
Next: Enthalpy of wetting or Up: MECHANICAL AND THERMODYNAMIC PROPERTIES Previous: Surface tension, surface Helmholtz Contents
2002-09-05
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## Problem
### Description
David the Great has just become the king of a desert country. To win the respect of his people, he decided to build channels all over his country to bring water to every village. Villages which are connected to his capital village will be watered. As the dominate ruler and the symbol of wisdom in the country, he needs to build the channels in a most elegant way.
After days of study, he finally figured his plan out. He wanted the average cost of each mile of the channels to be minimized. In other words, the ratio of the overall cost of the channels to the total length must be minimized. He just needs to build the necessary channels to bring water to all the villages, which means there will be only one way to connect each village to the capital.
His engineers surveyed the country and recorded the position and altitude of each village. All the channels must go straight between two villages and be built horizontally. Since every two villages are at different altitudes, they concluded that each channel between two villages needed a vertical water lifter, which can lift water up or let water flow down. The length of the channel is the horizontal distance between the two villages. The cost of the channel is the height of the lifter. You should notice that each village is at a different altitude, and different channels can’t share a lifter. Channels can intersect safely and no three villages are on the same line.
As King David’s prime scientist and programmer, you are asked to find out the best solution to build the channels.
### Input
There are several test cases. Each test case starts with a line containing a number , which is the number of villages. Each of the following lines contains three integers, , and . is the position of the village and is the altitude. The first village is the capital. A test case with ends the input, and should not be processed.
### Output
For each test case, output one line containing a decimal number, which is the minimum ratio of overall cost of the channels to the total length. This number should be rounded three digits after the decimal point.
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## Elementary Algebra
$x=-53$
We first use the distributive property to simplify. Then, using the properties of equality, we obtain that the value of the variable that satisfies the given equation, $4(x-2)=5(x+9) ,$ is \begin{array}{l}\require{cancel} 4x-8=5x+45 \\\\ 4x-5x=45+8 \\\\ -x=53 \\\\ x=-53 .\end{array}
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# Definition of face of a polyhedron
Let $P$ be a polyhedron in $\mathbb{R}^n$ and $\omega \in \mathbb{R}^n$, viewed as a linear functional
$\text{face}_{\omega}= \{ u \in P : \omega\cdot u \geq \omega\cdot v\mbox{ for all }v \in P \}$.
How does this definition match with general notion of the face of a polyhedron?
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This is the generalized notion of a "face", in which (in three dimensions) vertices, edges, and faces are all included. For example, consider the unit cube $[0,1]^3$. Try $\omega = (1,1,1)$, $\omega = (1,1,0)$, and $\omega = (1,0,0)$ and see what "faces" you get. – Rahul Mar 4 '12 at 13:13
Well, I understand that but how do I know that every face according to this generalized definition is also a "face" according to the old definition. – Mohan Mar 4 '12 at 13:21
It's a "face" in the old sense when $\omega$ is normal to one of the faces in the old sense. A choice of $\omega$ is a linear function on $\mathbb{R}^n$ which restricts to a linear function on the polyhedron. The face of $\omega$ is the locus of points of the polyhedron maximizing the function (exists by compactness yadda yadda), which is going to be one of the lower-dimensional cells. – Neal Mar 4 '12 at 13:41
*at least one of (only one, if the polyhedron is convex). – Neal Mar 4 '12 at 13:54
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# Spring Pendulum - Lagrangian Mechanics
Tags:
1. May 22, 2016
### bigguccisosa
1. The problem statement, all variables and given/known data
2. Relevant equations
Euler-Lagrange Equation
$\frac{\partial{L}}{\partial{q}} - \frac{d}{dt}\frac{\partial{L}}{\partial{\dot{q}}} = 0$
$L = T - V$
3. The attempt at a solution
a. The potential energy V is the potential energy from the spring and the gravitational potential energy. The kinetic energy is the energy in the radial direction and in the theta direction.
$$L = \frac{1}{2}m(\dot{r}^2+\dot{\theta}^2(r_0+r)^2) + mg(r_0+r)cos\theta - \frac{1}{2}kr^2$$
b. Use one Euler-Lagrange equation for r and one for theta and I get: $$m\dot{\theta}^2 + mgcos\theta - kr = \frac{d}{dt}(m\dot{r})$$ $$-mg(r_0+r)sin\theta = \frac{d}{dt}(m\dot{\theta}(r_0+r)^2)$$
Which are the F=ma equation for the radial direction and the Torque = (d/dt) Angular momentum equation.
c. Confused here, if angular position and velocity are to be fixed, I assume they mean $\theta$ and $\dot{\theta}$, do I consider them both to be zero, or just a constant number? And if the position is fixed doesn't that mean that the velocity should be zero? Is this a differential equation I have to solve?
d. and e. Need c to continue.
Any help is appreciated, thanks !
#### Attached Files:
• ###### homework1.PNG
File size:
34.9 KB
Views:
177
2. May 23, 2016
### vela
Staff Emeritus
"Fixed" means "constant." The position and velocity aren't actually fixed (so $\dot{\theta} \ne 0$), but because the timescale over which the radial motion occurs is so short, you can analyze the motion as if $\theta$ and $\dot{\theta}$ were constant. It's an approximation.
By the way, the first term in the radial differential equation is incorrect.
3. May 23, 2016
### bigguccisosa
Thanks I must have missed a (r_0 + r) factor in the first term when I was putting it into tex, I'll give it a shot with theta and theta dot as constants.
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I can access the Sep 5, 2016 First add your SharePoint site to the Trusted Sites zone in Windows: Without it, you cannot authenticate your user. Is there a reason you don't want to use a UNC path? The below procedure is useful for those that need to change a Microsoft Access ODBC Datasource in Windows that uses a mapped network drive to a Full UNC path for use with the TracerPlus Connect Windows Service. Access Control over UNC - In order for ColdFusion server to access dynamic content from a UNC path, it needs to run as an appropriate domain user account. A network drive is usually a shared folder or computer that is available on a network that makes it Here's how to map a drive in Windows 10. Unable to open Access Database via UNC path. Here's How to Map Your OneDrive Account as a Network Drive on Windows, to Show it on the Windows Explorer. a single folder/file name of 255 characters). . 1 SP2 or 6. Veeam service also runs on this account. For example, I have a server named Server1 and I’ve mapped my Q drive to the Accounting folder under Server1. Universal Naming Convention. Hudson will call a script, say, a. In this case, the command would be: Test-Path \\WH0RCUTEACHER\c$\Users\jxg768\Desktop\RobsShare With this setting defined, files copied to the Windows 2003 machine can be stamped with the correct source file time and date stamp and this problem will stop occurring. should work exactly the same in any version of windows. As an alternative to using SSH for remote Windows collections, Sumo Logic Collectors can collect files remotely using CIFS/SMB by configuring a Local File Source (not a Remote File Source) with a UNC share path. We are using c# 3. Access Denied Trying to Connect to Administrative Shares C$, D$etc. Use an UNC path to a shared folder, instead of a path to the drive mapped to the shared folder. 12). Goal: Setting the windows command path in Windows 7. config Using an Azure file share with Windows. Computer\HKEY_LOCAL_MACHINE\SOFTWARE\Microsoft\Windows\CurrentVersion\Policies\System. This is called the UNC path and is formatted as:. Additional Information UNC Path Explanation and Examples at ArcGIS. Does anyone know of a way around this? Why does accessing a folder via UNC path share not work but mapping the same path as a drive does? access. 4 Windows Paths. Previously, with Windows 7 on my local machine (like I can anywhere else I have an Windows XP or Windows 7 machine) I could open up a run prompt or explorer window and go to the root share via UNC path of a remote machine such as \\mycomputer and it will show me what folders and/or printers are shared. The server is joined to the “DMZ” workgroup. A UNC path describes the actual location of a file or directory, thus eliminating the need for arbitrary Windows drive letter mappings (which tend to change). First, I had file sharing and network access for everyone allowed on some of my Windows 7/8 machines, therefore Windows 10 should not meddle and interfere, especially since everything worked smoothly before the upgrade. NOTE: I am using local Account in Windows 8, i don't have administrator account enabled . This address also adheres to the UNC path formula of \\server\share. 1, I can access files from Cannot remove network shared disc from windows explorer in Network and Sharing Seems once you've connected a Network drive to W10 you can't get rid of it again -- I've done the checks to ensure that it isn't set to re-connect at logon. Share Permissions. Next create a FTP user in FileZilla with access to a directory local to that server. 0. So this works via the command prompt under my domain admin account which has access to the UNC path. at the bottom for all networks you will see, turn off password sharing as seen below. To get the “Previous Versions Tab” you should access the the drive as a UNC path in Windows Explorer. The remote management of standalone PCs can sometimes be tricky because various security barriers prevent remote access to Windows 8. They indicate that the path should be passed to the system with minimal modification, which means that you cannot use forward slashes to represent path separators Universal Naming Convention (UNC) is a method of displaying a path that will work for everyone connected to your network. g. " The trouble is I can't access this using the UNC path directly because I'm not on the same domain. " "You have been denied permission to access this folder. Open “FileZilla Server. In the local Administrator account I have to authenticate to the . J was our virtual path. Yet the same application can access a shared folder on a NAS device. Windows shares can be referenced in a link, with some restrictions / considerations: Write the link using file:/// (3 forward slashes), followed by the UNC name \\server\path\filename. In Windows 10. exe, that allows to run on Windows 2000/XP/Vista/7 the operations that usually take place with the Device Manager from the command line. How to set the path in Windows 7. 5, FastCGI, Windows Server 2008 R2 where I am simply trying to access a file on another server using a UNC path. I’m on a laptop running Windows 7. To me it appears something is fundamentally In UNC, the computer name is also known as the host name. You cannot map it directly, no. I am doing this on windows 10 (running fall creator) with no problem. Reporting: Batch file "windows cannot find the specified path" This post has been flagged and will be reviewed by our staff. msi. Access UNC path from a network drive. I only got the "windows can not access \server" alert. For stop all shares. If you do not want to risk the carpal tunnel, I suggest you to use the Microsoft program called devcon. Clients can now access the shared folder by typing the UNC (Universal Naming Convention) path of the shared folder in windows explorer. Ok Jeremy, thx for your info. You access the DFS share through the Universal Naming Convention (UNC) path in Windows Explorer. If you don't have a dny entry you can try to use the server name like \\WIN1010\J. How to Use nslookup to Check Domain Name Information in Microsoft Windows TEK911 Ip Config Tool - Ipconfig Command in Graphical User Interface Improving Wireless Network Performance by Tuning Advanced Wireless Settings on Wireless Device. 168. It works fine on the PC but our MAC users cannot access the UNC path directly from the browser at all. In UNC, the computer name is also known as the host name. 5. I have write a program that Copy/Move files on a unc path's ASP. Is anyone aware of any issues with backing up a UNC path from a Windows FS iDA? We have a NAS device which cannot have an agent installed, but we want to back it up. C# / C Sharp Forums on Bytes. To solve the strange problem was simply delete all the network adapters called Microsoft Device 6to4. 25. To edit any path, simply select it and then click on the Edit button. The UNC syntax How to Make UNC Folder Shares - posted in Tutorials and Guides: UNC - Universal Naming Convention is the recommended network share type for the new Media Browser 3 server. Problems accessing UNC path? Have been trying to research this for the past couple of weeks, however since I upgraded to windows 10, I cannot access the either of my machines via UNC. I cannot access any shared folders, shared Access via Windows Explorer to the target UNC path hangs or results to an non-responding window; These file operation issues persist even after performing the standard approach to address them. To quote I've got a feeling this is related to the path length limitations of Windows OS. Another option is use GUI tools A great client, BitTorrent however may come up sometimes with these errors, whether "access denied" or "System Cannot Find The Path Specified". Second, network volumes must be specified using a UNC path such as \\servername\sharename\foldername rather than a mapped drive letter. This is the result, as it default's to the C: So i click network , but cannot see how to specify "none" for drive ? In UNC, the computer name is also known as the host name. No response? Something is probably screwed up with DNS. \192. 1. Info: The full folder path in the title bar of Microsoft's File Explorer is a handy way to quickly find out which folder you are in. This command cannot be run due to the error: You can't connect to the file share because it's not secure. Before adding the UNC path, verify and ensure that the UNC path is accessible from another computer using the domain account that will be used to access the mount path. Windows cannot access shared folder As users become more and more mobile, you often have to manage devices that are not members of your Active Directory domain. Then try to use IP address access to the sharing file for test. But windows won't let you use two credentials with the NAS at the same time. This is my . v1. There are two ways. Confirm that the created account is a member of the Users group. UNC paths (using slashes, not backslashes) are allowed. In our example, then, the UNC path to access SharePoint from the command line would look like this:. Earlier we use to map SharePoint site to a drive and access the files. how about can someone help Windows Cannot Access The Specified Device Path Or File Explorer. Windows network file transfers (Python recipe) """ Convert a file path on a host to a UNC path I guess with this module the windows user features for The file URI scheme is a URI scheme defined in RFC 8089, typically used to retrieve files from within one's own computer. I’ve specifically encountered this bug albeit in a slightly different scenario: 4. exe from a desktop. They aren't part of a domain, just in the workgroup. In each case if the database file is "C:\mydb\mydb. During analysis of this bug the following facts were defined: 1. 1) Access 2000 and later - Database Not Split. 1 on Windows 10 anniversary addition (latest bits at the time of writing) and trying to access UNC paths available to the host inside Server Core or Nano Server containers but not gettin POWERSHELL + UNC PATH + CREDENTIALS. Java mishandles UNC paths on Windows Apparently Java has quite a few known but practically undocumented issues with its handling of UNC paths under Windows. 0, and not a supported format on the Directory List Properties page of the Deep Security Manager (DSM). More detail: I have a java app which reads the contents of a directory based on a UNC path. This was supposed to be fixed in Windows 2003 Service Pack 2 but some machines still display this incorrect behavior and need this registry fix. Source Machine - 10. 1 SP1 or 5. Or, enter \\files. Think of UNC hardening like a "trusted path, or source". To gain access to this folder you will need to use the security tab. As path separator you MUST use a slash. Important! Mini-Redirector is not installed by default on Windows Server 2008 and later Windows Server versions. I've been working on an auto-update feature for this program. exe on the server when needed. My first reaction was the following: – Make sure Network Discovery is enabled on the new server. " (Folder here represents the name of the folder you cannot open. If everyone’s drive mapping is the same and you’re sure of it and the logon scripts work every single time, you’re better off using the network drive path. Instead of Pageant authentication, use explicit private key authentication (using -privatekey and -passphrase switches). Network drive so co-workers can access the files. filepath = @"C:\Email\VendorAgingReport. 0 these privileges are granted via User Manager for Domains, but on Windows 2000 and XP you probably want to use Group Policy for propagating these settings. Net file access problem with UNC; Windows Max Path Is Now A Lot Bigger Posted on October 2, 2016 October 2, 2016 by Adam Fowler The legacy 8. See the OSU VPN How-to. How do I Access Windows share from Linux command prompt? I would like to be able to access shared folders on Windows machines from my Linux system. I can't access the Z On Windows NT 4. Shared folder access can be restricted by using Share Permissions. I'm not even getting a login cred request, just that "Windows cannot access" Feb 28, 2016 I have a vbox 'shared folder' set up, which appears on the windows allowing you to link UNC path based Directories and Files into the local Apr 4, 2013 Typically you can just map a drive or directly enter a UNC path to the server: Even trying to access drives on the local, internal network would fail; Trying that all the Windows based network mapping/location features failed Jul 13, 2018 Ever since then, I can't access my laptop from my PC, though my PC can The reason why the UNC paths won't resolve in Windows 10 April Aug 10, 2019 When trying to connect to a network resource on another Windows The computer cannot make a connection over the network with the other Ensure that the user or group enjoys the required privileges (minimally, read) to access the file. Procedures include the following: Putting scan exclusions in the backup software's program files and installation directory, target folder or UNC path Able to ping name and IP Address, but not access server using UNC By SirMalcom · 13 years ago I've been working on a network using a single Windows 2003 standard server to do AD. When to use a Word hyperlink UNC path. Cannot update NuGet [lopsa-tech] Windows file explorer using port 80 (webdav) instead of 445 (samba smb cifs) for UNC path (too old to reply) "Windows Cannot Access \\192. Thu Dec 21, 2017 Image This happens both if i access the /share/ or /share/folder path. I had no problem accessing the server with RDP or ping. Even if the drive letter is mapped in an admin command line window, the system does not start such applications from a "normal" window. You can also delete paths using the Delete button. These are a few rules for UNC paths: UNC paths cannot contain a drive letter (such as D). When you set the service to run as the local system account, it cannot access the remote UNC Path that is accessible using explorer with local machine account. Right-Click this folder and bring up the Properties dialog. exe Windows 7 It does not show up in Hi, I'm using Docker for Windows 1. The host-name , share-name , and object-name are referred to as "pathname components" or "path components". Access denied to UNC path – Learn more on the SQLServerCentral forums We recently upgraded from Windows 2003 32-bit to Windows 64-bit (SQL Server 2005 SP2). ISO downloaded from The problem is that I cannot access a (fully functional) unRAID SMB share. Typically the person will be trying to access the server share by entering the \servershare path into the address bar of an explorer window. ?windows cannot access the specified device, path, or file. Apr 17, 2018 Fixes a problem on Windows XP-based client computers that cannot access shared files or computers on a network. I would ideally like this to be able to run under the local Administrator account. You may not have the appropriate permissions to access the item. ", but I can access to this computer from pc02 and pc03 15. I will explain it step by step in this thread by creating a new folder and sharing it, then add the folder share to the MB3 server. People dealing with UNC paths on windows would probably be using the standard form with back-slash anyway. bat and inside a. e. I am having this problem. where the computer name is substituted by the IP address of the server. You can also manually set these via the Local Security Policy MMC snap-in. The secret is to make it a network drive. Windows Server 2008 R2 on VM: Windows cannot access the specified device, path, or file. Go to a client and open the windows explorer. How the Win32 APIs process file paths on Windows NT is a tale filled with backwards compatibility hacks, weird behaviour, and beauty†. This raises a best-practices question. 50\Data" and allow the pass-through authentication - Only Windows Authentication is enabled and Anonymous Access is disabled for both of "TestWS" and "Data" You can exclude a UNC path from scanning even if “ \\ “ is not a valid entry for network share or UNC path in Deep Security 9. How can I get around this? Best practice to manage permissions for shared folder is, configure full control permission for everyone and restrict the folder access using NTFS permission. All names resolve fine. Gaurav Chikara. Cannot create database dump to mapped drive or UNC Path 概要: Services run as the system account are not allowed to access the network When you try to dump a database to a UNC path or a mapped drive it fails. In ArcIMS 3. On the other hand, Windows 10 did have issues retrieving Group Policy. Fix: Cannot Access Network Shares after Update 1709 If the issue is with your Computer or a Laptop you should try using Reimage Plus which can scan the repositories and replace corrupt and missing files. doc. x. path. May 19, 2016 I only got the "windows can not access \server" alert. Path errors on start up in Windows 7. 1, 8, 7: Pro, Home, Enterprise, Basic, Premium, Professional, Starter, Ultimate, Windows-Server 2016, 2012, 2008 and can be implemented fairly quickly. Here's how to map a drive in Windows 10. i. Today – in Windows 7 SP1 with all post-SP1 fixes – Offline Files is a technology that still has its problems, but can be used in production – if you are adventurous. exe located in a specific path on the server and downloads the . 5 While coding the windows service, we have specified UNC path as \\192. 3. I get the login prompt and type in the username and password but all that happens are that the fields blank out again (like its rejecting the password/username and asking again). This System cannot find the path specified . my PC name from networks in this PC and it still showed the same error: path cannot be found. For some reason this OS architecture won't allow me to access a Buffalo Cannot browse Buffalo Terastation NAS on 64-bit OS then when browsing through UNC path Config Error: This configuration section cannot be used at this path. 12. Please note: Although this article originally targeted Windows 7, most if not all of its content applies to Windows 8, too. yes it still shows the same error - windows cannot access Open the Control Panel > Administrative Tools > Services window on your Windows server. Browse dialogs such as used when doing "Add Evidence" in FTK or mounting an image in FTK Imager. Windows Explorer should have already stored the password. Do not try to point it to a mapped drive for the NAS. cornell. iI wrote the path of the UNC in IIS as following: "\\10. You cannot navigate to directories above the shared directory. using UNC instead of a drive specification. Cannot run command using UNC path in startup script give domain computer read&execute even full access to the GPO,the folder > interaction and the scripts [2011-01-13 04:24 UTC] mark at internode dot on dot net Hi I am having the same problem with PHP 5. Windows 2008 R2 . access Document Library on Windows 2000 In Windows XP, I had a Windows Explorer link set up to most of my mapped drives with the UNC address specified. " Enter the UNC name of Network path not Below are 3 code-snippets you can use to obtain the path to the database (mdb) file in different situations. Domain. It should not happen. "Windows cannot access the specified device, path, or file. txt turns to; \servernameusernamefilename. UNC paths are not supported. You should of course use the actual name of the server and not "servername". “Access to the resource has been disallowed” when browsing to UNC path “By design” UNC path browsing from the Windows Explorer address bar is not possible if the “Remove Run menu from Start Menu” in GPO is active. " Causes Running PowerShell Scripts From An UNC Path (Share) script. 6: TESTING THE ACCESS If all settings are done you should now be able to access the share via webdav. So possibly you will see that you can access the path when you enable Basic Authentication. Access is denied. I'm trying to understand why my application running as a windows service as Local System account cannot access a shared folder on another PC using a UNC path. To use Test-Path on a remote machine, you can pass in a Universal Naming Convention (UNC) path. I downloaded the WD Quick View, it sees the My Cloud just fine. Able to ping name and IP Address, but not access server using UNC By SirMalcom · 13 years ago I've been working on a network using a single Windows 2003 standard server to do AD. We are able to access the share from the windows explorer and able to MAP as well. Under windows a mapping takes place (e. To use this you Using Windows 8. \\<TVserver_PC>\ProgramData\Team MediaPortal\MediaPortal TV Server\timeshiftbuffer (for the live TV timeshifting) - note the format of the path names may be different on a non-Windows OS. 2\Share, I can't access the share in FileVista using I started the Webclient service but still get presented with a login box when trying to access the document library via a UNC path. You may imagine that you have many UNC path to remember in your daily work, how can you remember it? Microsoft Windows has a good function to deal with … Read More » This can be because the path used to connect to the data source is a mapped path. Locking is either by default (overrideModeDefault="Deny"), or set explicitly by a location tag with overrideMode="Deny" or the legacy allowOverride="false". it gives me a "Windows cannot access Additionally, even if he specified one of these “missing” remote servers by name with a UNC path, he couldn’t connect to a share or enumerate the shares, even though the servers were pingable by IP. Doing that same thing in Windows 7 still shows the mapped drive letter, not the UNC path. I changed the order of the providers to LanmanWorkstation,RDPNP,webclient as you did and it worked! In a UNC path used to access files and directories in an SMB share, for example, object-name can be the name of a file or a directory. Access Your Network Drives from Off Campus by Allen Schreiber (February 21, 2012 at 4:57 pm) VPN Software. ps1 cannot be loaded because Commvault Debug DFS Direct Access DNS DSC Dynamics Ax 2012 Exchange UNC defaulting to port 80 rather than 445, and cannot rejoin domain Specifically the clients cannot access any UNC path either by name or IP. I have a problem (Access is denied) when I try to copy files across network in a Windows batch script. Use command line tool called smbclient or you can mount windows shares the mount command. On Windows, splits a pathname into drive/UNC sharepoint and relative path. Before we start to fix the windows cannot access the specified Device, path or file issue make sure that you have enabled the network sharing in your computer and in your partner computers properly, here i used Tweaking. For example, John might be able to read the accounts folder whilst David might be denied access. Hi, I just got a 4TB My Cloud, and have only had semi-success setting it up. document library access denied using UNC Paths on server I have a sharepoint 2010 document library On a Windows XP client i can open documents in the d. Thank you for helping us maintain CNET's great community. : The path /c/users is mapped to c:\users). Name the above DWORD or QWORD as LocalAccountTokenFilterPolicy and hit Enter. When viewing a library in internet explorer (on windows 10), I click on the library tab and then onto "open with explorer" and windows explorer opens it. When you type a command at the command prompt, Windows has to search through each directory stored in the PATH variable to see if that executable exists or not. Option 1: Static entries in DNS Would you support adding UNC support to os. We do not know how to work around this Windows feature. When I am running jobs from within console - they run without a problem, but when they start automatically - they fail with a information that Veeam cannot access that drive. I could access that folder using the UNC method On my local pc i'm trying to setup an odbc connection to an access db. If the mount path is in a different domain from the MediaAgent from which it is accessed, make sure that trust relationships are setup between the two domains. However some legacy Win32 APIs only support up to 260 characters for backwards compatibility and old file system reasons. Click Continue to permanently get access to this folder. Since there’s no way to include an IPv6 address in a Windows file path, there’s no corresponding file URI and so there’s no way to incorporate an IPv6 address in file URIs in Also on Windows systems drive mappings are done at login -- so you typically have to run the login script. The Store relative pathnames option for documents and tools has no affect on UNC pathnames. " I have admin access! The only icon that does work is Internet Explorer, but that doesn't load the homepage, just reloads every second to a never ending cycle of a blank page because it won't stop refreshing it! I have a windows 2008 R2 server in the DMZ. Connect to a Windows File Share with a UNC Path Windows 7 Click the Start menu In the Search box, enter: \\files. However, in this case, because the program is running as a service it cannot see any non-local devices. Open the Properties But when the Win7 machine tries to UNC into the Win10 machine or the two shows both Win 10 machine that c$ is already enabled and cannot be created. Windows 2008 NFS access using UNC path I have a customer who is testing out access to the NFS export from their Windows 2008 Enterprise 64-bit SP2 server. This is the DOMAINCOMPUTERNAME$account, which you can use in the same way you would use a regular user account when granting permissions. The only issue is, no matter what I do, I cannot go to Start -> Run -> \\systema\c$\ and get Windows Explorer to pop up with anything. This happens when the section is locked at a parent level. Windows cannot access \\COMP 0x80070035 The network path was not found This particular computer is straight from the factory, and has been working just fine for months. txt In the above example the H:-drive is the user’s home drive and the UNC path now points to the home location of the sender instead of the receiver. Q. The entire network is shared, and with the exception of the Win 10 laptop, all of the Win 7 machines can see all folders in the Win 7 server, but cannot access the Win 10 machine, nor can the Win 10 machine access the server or any of the Win 7 machines. Previously the file URI scheme was specified in RFC 1630 and RFC 1738. sep) following each non-empty part except the last, meaning that the result will only end in a separator if the last part is empty. The requirement is to have their application, that runs as a service, pull files from a NFS mount using the UNC path. Change the current directory/folder and store the previous folder/path for use by the POPD command. This is necessary as Windows Services have no knowledge of a mapped network drive being that they are user based. We're running a SharePoint Server 2010 in windows 2008 server. If you use UNC paths you can have a path with 32,767 characters in it, 255 characters per element (e. Setting up UNC Path Mapping with Pass-through Authentication in IIS 7 and 7. Its yang is mapped drives, which are specific to each machine. Today I was faced with the fact that one of our backup processes needed to copy compressed database backups to a remote server over an UNC path every night. ErrorCode 0X80070035 "Network Path cannot be found" This problem occured yesterday on my laptop with OS windows 7 pro. From my living room pc with Windows 8. In all cases, drive + tail will be the same as path. This article discusses a possible solution to the problem "Windows cannot connect to the printer. Every time the virtual machine assigns the windows image to a new host machine, a new virtual network adapter is created, and various entries are added to the registry. The result of this evaluation is utilized for two purposes: Does anybody have some experience how to publish resources like UNC paths in XenDesktop 7. Close Registry Editor and reboot your PC to save changes. Split the pathname path into a pair (drive, tail) where drive is either a mount point or the empty string. 1 PC - Windows Firewall is disabled, Webroot anti-virus uninstalled, no Windows Updates to install, services such as DNS Client, Computer Browser, TCP/IP Helper are started. above path as the current directory. A Windows path can be absolute but still relative to the current drive; such paths start with a / or \ separator and are not UNC paths or paths that start with \\?\. UNC paths or navigation via Network (ie: for mapping) or via view The basic way to access shares on a network is to pull up My Network Just like typing in a path to a local file (e. Windows currently doesn't have a 260 character path length limit. 3 filename restrictions that came from the old MS-DOS days are (for the most part) long gone, but one of the other lingering legacy limitations is the 260 character limit. Technically the forward slash should be used throughout the link, but the backslash works on windows platforms A service (or any process running in a different security context) that must access a remote resource should use the Universal Naming Convention (UNC) name to access the resource. I always, ALWAYS recommend that you use UNC paths and never use a drive letter. This will set a custom message on file or folder if access is denied. The Admin of the machine will need to sign into the Tableau Server windows machine as the 'Run As' user and try to navigate to the Excel file using the UNC path to confirm. I cannot leave this Choose an account that has network access without Windows Explorer asking for a password for the network drive. Not a trivial thing to implement, but it should give you the end result you are looking for. The path cannot be \\sharepoint\sites\ since sharepoint does not use UNC paths. 2 x64 (Win) server cannot save to UNC path? that I am running the "UrBackupWinServer" service as has full access to that path Windows 2008 R2 server Path, FILENAME, DDL, FMTSEARCH, CALL METHOD, SASHELP, PATHNAME, UNC Path INTRODUCTION It is not all that unusual to have a need to determine a physical path. Before trying sudo mount -a check that the name is resolvable, using for example host servername (of couse, with the actual name instead of "servername"). Below is my code. If you type in a machine name using UNC syntax and hit enter, instead of locating the machine on the network and browsing to it, it will search the local machine for that string. 1 and Windows 10, I cannot navigate to mapped drives when using browse dialogs found in FTK and FTK Imager. com The first field describes the block special device or remote filesystem to be mounted. You may not have the appropriate permissions to access the item The above does indicate that the Run As user account cannot access the Excel file. Interestingly, only some network folders on NAS devices and some other computers on my local network (not a domain) became inaccessible. Accessing SharePoint through UNC paths and so I need to find a way for authenticated access to the UNC path to work. The short answer is that URLs don't support spaces, and UNC paths do. 1\NAS_FOLDER\LIVE\ However, we are able to access the above UNC path in our website by specifying impersonation in web. The return value is the concatenation of path and any members of *paths with exactly one directory separator (os. msi location before I can run the command or it won't find the . Enable the Hardened UNC Path setting. Unfortunatley the share is on a different domain so although I have username and password I have no way to access it. A network drive is usually a shared folder or computer that is available on a network that makes it Xcopy and UNC path names Elementary question, but I can't seem to figure it out. Setting up for UNC path testing. I'll cover the following topics in the code samples below: Windows Server 2003 R2Windows XP, SharePoint, Privileges, WebClient, and UNC Paths. If you know that the service account can access a specific file, type in the full path for the file in the File Name control in the Locate dialog box. Learn how to perform access-denied remediation in Windows Server 2016. be mapped to With Windows Authentication the Windows Identity the process is running as is what everything is accessed as. Veeam Endpoint for Windows 2. A few rules for UNC pathnames are: UNC paths cannot contain a drive letter (such as D:). Is it better to use a Word hyperlink UNC path, or a network drive path? The answer depends. From the internal network, I cannot connect any of its shares using the UNC path or IP address. Installed the app on my (Android) phone, which also sees the My Cloud just fine. * It looks like you have to use a UNC path for the working directory, rather than a mapped drive. Verify that you have the necessary security privileges and that the path or file exists. xml” in a text editor like Notepad-Usually located here: C:\Program Files\FileZilla MUP UNC Hardened Access Behavior. 10. First of all, remember to grant access to the computer account of the computer running Hyper-V. 2. The Windows 8. If anyone has a link to good material on the subject please let me know. 1 PC can access the path via IP (i. On the Windows 8. but in Test Settings it shows GREEN on Authentication but Yellow on Authorization (Cannot Verify Access Path (D:\ABC) . Once you’ve clicked the edit button, a new dialog box will appear with each location in the path on a separate line. The problem is I can't access it via its UNC path except after performing things in a certain order. the IP all the time in UNC paths instead of the hostname. Hi Mike Sutherland 777, Welcome to Microsoft Windows. I have three computers (pc01, pc02, pc03 -wireless-) connected with a router and I can't access from pc01 to the other computers because "Windows cannot access to \\XX - 0x80070043: The network name cannot be found. And, to make this even more ridiculous, I can map a drive to the path I am trying to UNC directly into. edu\OU\sharename, and then press Enter. 0 Lanman resource The first command works fine (assuming you have a c\$ share enabled and are able to access it), and the second command gives a "Cannot find path" error, because the Registry provider tried to work with the UNC path instead of the FileSystem provider. Open up Local Security Policy in your computer's Administrative Tools in the Control Panel. SMB then picks up the details from the registry to resolve the UNC path on the network. It would display the UNC address in the Windows Explorer window where I could copy it. For testing purposes you can setup your machine to have a "local" UNC path. 6. NET application cannot The reason for this is that Windows Server Explore only supports the Shadow Copy “Revert” feature when connecting to a local drive using Windows Explorer. Upon accessing a UNC path to a server, you may get the error: Access to the resource ‘[UNC path]’ has been disallowed. I do not know why Windows folks suddenly decided to change the way network sharing works. Describes a resolution for an issue in which you may be unable to run executable files or script files from a UNC path when you have Windows Internet Explorer 7 installed on a Windows Server 2003-based computer. However, the scanning device cannot access the server files. I've spent some time trying to research workarounds for the path length limit, but haven't yet found anything comprehensive. So, first, about UNC hardening, or my take. In Windows 10, this process is both easier and less confusing. mdb" the functions return "C:\mydb\". com - Windows Repair which restored registries and and various other Windows files and then i uninstalled the majority of the programs and then restarted and again tried to double click on my PC name from networks in this PC and it still showed the same error: path cannot be found. My problem is I can't find a UNC path to use in the copy command. To introduce UNC Hardened Access and protect against UNC-based Man-in-the-Middle (MitM) attacks, install KB3000483. The problem statement: With 2 domain controllers both functioning, Windows 7 systems had no issues getting Group Policy. The new run/search box in 4. Hi, admin! After installing Fall Creators Update (1709) on Windows 10 computer, I cannot access the shared network folders from my computer. walk on win2000? UNC paths in file object open; Raw strings as input from File? Relative versus absolute paths on Windows; Windows file paths, again Many times a UNC path does not work properly in a command line bat file. If you already have a dns entry for webdav try to connect via \\webdav\J. " Enter the UNC name of Network path not This article discusses a possible solution to the problem "Windows cannot connect to the printer. Mapping Network Drives: The letter at the beginning of each line below is what Spears School of Business uses to identify the storage locations that you use when you are logged in. There are various ways around this (native Kerberos delegation is the most usual), but setting that all up requires some configuration on your part. Collect Windows Logs from a UNC Share Path. These prefixes are not used as part of the path itself. Are the Documents stored in the database? Is there an easy way to easily copy Documents out of a library? Can I map a drive letter to the Library? Windows Sharepoint Services 3. In the ODBC for database name i type in the full unc path \\server\directory\directory\file name and press ok. x I use mydomain\myadminaccount for logging into RDP and running Veeam console. How To Disconnect Non-Mapped UNC Path “Drives” in Windows username and password to gain access. UNC path fails with Windows authentication. Using Windows Explorer or the Command Prompt or Windows PowerShell, and with proper security credentials, you can map network drives and remotely access folders on a computer using its drive letter rather than a UNC path i used Tweaking. In our case, the UNC path is, \\MBG-DC1\Marketing. Windows 10 – default enabled UNC path hardening – Wifi 03/05/2016 klue Leave a comment Issue: Logon Script is not processed within Windows 10 Enterprise x64 Machine on Wifi. This is a dramatic improvement over the way previous versions of Windows handled path locations, and makes easy work of adding a new one. Cannot access the specified path or file on the server. Syntax PUSHD [drive]path PUSHD Key drive The drive to switch to path The folder to make 'current' (UNC names accepted) If the drive is not specified, the current drive will be assumed. I would suggest you to change the security firewall settings or disable security software on the system and try to access the location. Access SharePoint Document Libraries from Windows Explorer July 26, 2013 Adam Prescott Leave a comment SharePoint is a great way to share documents across a team. In the folder list (on the left of the Windows Explorer window), Dec 21, 2017 Windows 10 cannot access samba share. Defaulting to Windows directory. In Windows, file URIs are dereferenced by converting them to their corresponding Windows file path and then using Windows file APIs to access the Windows file path. 1? I tried to use Internet Explorer instead of Windows Explorer, but i haven't been sucessful. It would be a WebDav path as in Hi Santhan, I guess that the reason you weren't able to get it to work was because Windows unable to use WebDAV (since WebClient wasn’t running) to access the library and defaulting to FPRPC. 5 (Windows 2008 and 2008R2) Just closed off an interesting exercise with my current client – We had to set up a SharePoint site and work out a way to easily manage files stored on a Network Share. But we are able to access the site through internet explorer. # re: UNC Drive Mapping Failures: Network name cannot be found I had the same problem with windows 10 after the upgrade. For the past few days we're unable to map the drive or access through unc. pdf"; queryContracts = contract UNC Path Naming for files stored on SharePoint If you didn’t already know, you can access any file stored in SharePoint (2007 or 2010) as though it was a folder on your system. but we recommend UNC Windows Cannot Access The Specified Device Path Or File Windows 7 this is all happening after parental control settings were set by my parents for no stupid reason JOMEL JIM says: 6 years ago method 2 also works…. 5 running under IIS 7. ealmquist, we run SP2016 on server 2016. Drive letters can change, and they are a pain in the rear end to diagnose. This works well when I run the . Then, in a Group Policy scoped for devices with the update, configure UNC Hardened Access in Computer Configuration, Administrative Templates, Network, Network Provider. I recently got a very odd problem with Windows 10 network sharing. Mini-Redirector is a Microsoft WebDAV client that is provided as part of Windows. join (path, *paths) ¶ Join one or more path components intelligently. As a result, the Web server cannot use a mapped drive or UNC path to access information. Note: Client reported that they were able to access this path or NAS share earlier when the OS was windows 2008 R2 and problem is seen after it was upgraded to Windows 2012 R2. Stop your windows service. Here is how to set this up on Windows XP: From File Explorer select a folder you would like to use as your UNC path. However, there are ways to avoid this issue and show the shares under the old names, allowing the old UNC paths to continue to work unchanged. He focuses on Microsoft and related technologies and has a passion for PowerShell. Windows treats those as different machines. The Microsoft Windows UNC, short for Universal Naming Convention or Uniform Naming Convention, specifies a common syntax to describe the location of a network resource, such as a shared file, directory, or printer. Then manually set the home directory to the UNC path: 1. You could have all the content on a third machine, and then both IIS and ColdFusion will reach over the UNC path to the remote document root. A few rules for UNC pathnames are. \\desktop\My Documents\). I have a windows 2000 server and I am trying to connect to a 2003 server in my workgroup via its IP address using the UNC path at Start, Run. Best practice to manage permissions for shared folder is, configure full control permission for everyone and restrict the folder access using NTFS permission. The second thing is that you need to do is use a UNC path when pointing to the file server. ) "You don't currently have permission to access this folder. If you want occasional admin access, you can create one credential for the NAS IP address, and another (different one) for the hostname. Stop the FileZilla Server service 2. Unable to access Windows Server 2008 file shares on local storage through the UNC path to the to the virtual server name created by the Storage Foundation for Windows High Availability (SFW-HA) 5. While it might be an issue with Windows 10 1903 it's still something Dell/EMC should be concerned about because it will hurt their customers. Simplify UNC usage in command line batch files. Simply because the explorer manages logged on user credentials, so, you have added UNC path credentials to the LOGGED user only. Jun 13, 2013 'Windows cannot access \\computer\share\' is the error message you mostly likely get when you try to access a shared folder on a Windows 8 Solved: Suddenly I am unable to access my ReadyNAS from Windows Explorer. 39/24 Cannot Access CIFS Share on vFiler - Network Path Not Found Cannot Access CIFS Share on vFiler - Network In this guide, we show you three different ways to enable and configure 'Controlled folder access' on your Windows 10 device, to protect your files and folders from malicious programs, such as os. If you’re having this problem on your Local Area Network, then I don’t think you should really need to worry about it; In a Distributed File System (DFS) namespace, you have a client computer that is running Windows 7 or Windows Server 2008 R2. create new – Dword (32bit value) LocalAccountTokenFilterPolicy Value data change-1 We have a web application at work that points images directly to a UNC network share. URI strings constructed from Windows UNC pathnames like former mentioned "\\host\path\to\f i l e. These functions all work whether the database is opened via a local drive, mapped drive or a UNC path. What is a minimal valid UNC path? Tag: windows , filesystems , unc I want to make a library that (among other things) parses UNC paths, but I don't fully understand the grammar. In general, a Windows pathname consists of an optional drive specifier and a drive-specific path. E. Because the Run As user account cannot access resources that have been authenticated in the security context of another user, it cannot use the mapped path. I have given all Read/write permission to Everyone and I have even given permission in Report server to AOS service and report server accounts. With "Windows authentication", the IIS server does not have the user's password in clear-text, so can not "automagically" authenticate to the remote UNC path on the user's behalf. If you have trouble opening the UNC path from Windows 7 or Vista, you may have to make modifications to some local security settings to access the Simple File Share used for the K2000 driver share. This explanation is for Windows 10 (Redstone 4 & 5), 8. I want to xcopy data from a UNC source path to a UNC target path but cannot seem to get the syntax correct. Jaap is a Senior System Engineer in the Financial Services industry. It compares the Version numbers of the running assembly with the . path on Windows? Using UNC paths for --find-links in easy_install on Windows; unc paths in os. To use an Azure file share with Windows, you must either mount it, which means assigning it a drive letter or mount point path, or access it via its UNC path. On the client computer, you map a DFS share to a network drive. Here we will describe mini-redirector provided with Windows 10, Windows 8, Windows 7 and Windows Vista. The Database is MSSQL2003 also on a Server 2003 box. A. On systems which do not use drive specifications, drive will always be the empty string. edu and browse I have a strange Sharepoint 2010 issue. Though you may be able to hit - a Virtual directory in "TestWS", named "Data" pointing to the UNC shared folder "Data" from file server. Files & Sharing » Windows » Tech Ease: There are a couple of ways to connect The fastest way is by typing the Universal Naming Convention (UNC) path in the Yet another way to access shared folders over the network is by mapping a Jan 11, 2019 If you have configure a webdav server windows should connect to it, but you cannot use UNC path, you have to use the URL of your webdav, Internet access is OK as is everything bar file shares (and network discovery in . Access mapped drive from a Windows Service. Resolution. A screenshot of the UNC showing a host device name, a share Feb 4, 2014 Microsoft Windows has came a long way since the dreadful days of common OS me with the error "Windows cannot access \\machine\sharedfolder, error. In a dynamic application or even a macro program the programmer cannot always know what a path is going to be when the macro is executed. I can access the dashboard via my web browsers (Firefox, Chrome) just fine. Windows cannot access \\ Computer Name You do not have permission to access \\ Computer Name contact your network administrator to request access. Regardless, they may still fail for a number of reasons. C:/Windows/), you can type the path to shares ( e. Problem. 1 cannot access Shared folders Can you try to enable anonymous access to the target folder on the file share and type in the UNC path After the consolidation, that path needs to be changed to use the new server name, like \\cfile\Orders\Order1. I have tested this with Windows NT 4 For example, "\?\UNC\server\share", where "server" is the name of the computer and "share" is the name of the shared folder. 2-rc1-beta27. However, even though I can ping the other machine, I cannot connect to the share. Access Samba Shares With Windows 10 And Azure Ad Setup 1 minute read Symptoms: You have Samba shares in your local network that you used to have access to, or have other devices on that network that can access those shares. as a Network Drive and access it directly from the Windows explorer. What worked for me was going to the credential manager, and adding the credentials manually (here I could see the credentials I had previously added for the other servers). You're probably copying the URL into Windows Explorer, so just replace each %20 in your path with a space and you should be off to the races. Windows Server 2000/2003 Thread, Cannot access server shares via IP or UNC path in Technical; Hi guys, I cannot access a server share from a xp pro machine which is on the domain. So I should consider this as being a best-pratice for security reasons? The reason users need to be able to access a resource by its UNC-path is that I'm using Windows Sharepoint Services with a picture/image library. 0, this will result in a 'Retrieving Data' error, in an HTML viewer client, if the formFilePath variable uses a mapped drive or UNC path relative to the IIS Web server. The MACs have AD login installed, so after some testing on a MAC, I found out that if I first mount the UNC to a local volume then I can access the images. i know if i enable adminstrator account through CMD , and without loging to admin account , just staying at local account if i do : Python and Paths: UNC vs mapped drives. txt" can still be handled by "ShellExecute()" Windows Shell function, if the URI string is not encoded. So my only choice, as far as I know, is to use the UNC path directly to access the files. The Definitive Guide on Win32 to NT Path Conversion Posted by James Forshaw, path’ological reverse engineer. Make sure the service’s account has an access to the shared folder. UNC paths cannot contain a drive letter (such as D:). So a link that used to be; H:filename. This article lists the gotchas I am aware of. Double-click on LocalAccountTokenFilterPolicy then set it’s value to 1 and click OK. On Cygwin, this is the native Windows path which the mount point links in. Workaround 1: Type the full UNC path in the browse dialog, to path to the mapped resource. The strange with these errors is that they appear for files you may have been downloading even for several days without a problem! Access to the path is denied when the code is generating the report for the second time on wards. 0 has serious flaws and cannot be used to access network resources via UNC path. 0 is running on a Windows 2003 box. Usage of a backslash might lead to unexpected results. The UNC folder path to this with Windows 7, trying Re: W10-1903 UNC path failing 0x80070043 @ThiagoCardoso The response from Dell/EMC so far has been to tell us to contact Microsoft. You could try implementing a Shell Namespace Extension that is registered as part of the file system so you can root it where you need, and then have it access the UNC path internally. to use an UNC Path, for example VB. In windows, a mapped network And if you need your servlet to have access to a UNC path like \\server Connecting to WebDAV server on Microsoft Windows. Additional information: Modifying the path statement will enable an MS-DOS window opened in Microsoft Windows as well as older programs to locate files that may be required to run the program. please read our TechRepublic Forums FAQ. Access SharePoint Document Library through UNC Path | Axapta Source says: This article contains steps to troubleshoot when Windows cannot access the specified device, path, or file. Note that you can also move items up and down on the list. What I mean by that is, when your Windows User has access to a UNC path the reason why Mapped network drives do not work in scheduled tasks, unless you explicitly map them before running the task (see note 1 below). UNC is a system wide path, so it works. As a daily routine, you have ever try to use UNC path to access some files or folders on the network share. The Store relative path names option for documents and tools has no effect on UNC paths. When inserting a link to a file on a mapped network share, the link is automatically converted to a UNC path. Open Registry -regedit run as administrator Then copy paste below link to the registry. bat I have the followings It would not be hard to convert all forward slashes to backwards slashes in the Windows path normalization; should I add that too? Since / still works as a file separator as long as the path is not UNC, I'd say don't convert them. If I access it using \\host\DavWWWRoot. then go the network sharing center, advanced settings. Now, all of a sudden, I can't read two different network paths. Return to top of "Tips to Solve Windows Cannot Find the Network Path Error"!! Authentication to another domain when using UNC path when I try to access a UNC path like \\Server. I have a Windows Server 2019, fresh install. These edits will allow the formFilePath to be set to a UNC path. I suspect it means that you need to specify the timeshift and recording paths in the client configuration in UNC path format e. Of course were the UNC path protected with credentials other than the credentials my script were running under. Each time MUP receives a request to create or open a file on a UNC path, it evaluates the current UNC Hardened Access Group Policy settings to determine which security properties are required for the requested UNC path. windows cannot access unc path
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