text:
## Declaring and Initializing Variables

In Python, you can declare and initialise variables in the following formats:

age = 5height = 1.75name = "john"

Python is smart enough to know that the data type for age is intheight is float and name is str.

In Java, you declare and initialise variables in the following formats:

int age = 5;double height = 1.75;String name = "john";

You have to state the data type of the variable as part of variable declaration. (Take note of the semi-colon at the end of each statement).

Point to note:

For python, you can change the data type of a variable. For example,the same variable a can be used to store the value 1 (an integer) and the value "john" (a string).

This is not allowed in Java. If you have declared the variable a as an integer.

int a = 1;

You will get an error if you attempt to do the following:

a = "john";

format: plain


text:
# prinsenvlag¶

The prinsenvlag colormap is a visual representation of the flag with the same name, which one can read about right here. It covers the special $$[50, 100]$$ lightness range and uses the colors orange and blue. Because the entire colormap is fairly bright, prinsenvlag is great for usage in plots where all values are important and should be distinguishable.


text:
Article Text

Smoking in correctional facilities: a survey of employees
1. Matthew J Carpentera,
2. John R Hughesa,
3. Laura J Solomona,
4. Thomas A Powellb
1. aDepartment of Psychiatry, University of Vermont, Burlington, Vermont, USA, bThe Vermont Department of Corrections
1. Matthew Carpenter, University of Vermont, Department of Psychiatry, 38 Fletcher Place, Burlington, VT 05401, USAmatthew.carpenter{at}uvm.edu

## Abstract

OBJECTIVE To assess level of endorsement and expected consequences of worksite smoking restriction policies among correctional employees.

DESIGN Mailed survey to Vermont state correctional employees.

MAIN OUTCOME MEASURES Support for various policy alternatives for both staff and inmate smoking; expected consequences of restrictive smoking policies and smoking behaviour.

PARTICIPANTS 321 of 640 (50%) state correctional employees responded.

RESULTS Employees were somewhat receptive to smoking restrictions for inmates, but less supportive of staff smoking restrictions. A complete ban on inmate smoking both indoors and outdoors was supported by 56% and 49% of never and ex-smokers, respectively, but only 15% of current smokers (p < 0.01). A similar ban on employee smoking was supported by 38% of never and ex-smokers, but only 3% of current smokers (p < 0.01). Overall, employees were most supportive (52%) of a policy for themselves that banned indoor smoking and restricted it to certain areas outdoors. Current smokers were more likely to expect negative consequences as a result of further restrictions than were never or ex-smokers.

CONCLUSIONS Although our findings are limited by a low response rate, most employees support an indoor ban, but not a total ban on smoking. Employees generally favoured a policy that was slightly more restrictive than the current policy, but were less supportive of tighter smoking restrictions for themselves. However, a more restrictive smoking policy is likely to result in some degree of resistance among current smoking employees, who may require specific attention to address their opposition.

• smoking restrictions
• policy
• environmental tobacco smoke
• correctional facilities

## Statistics from Altmetric.com

Exposure to environmental tobacco smoke (ETS) has long been identified as a health risk, both within the home1 and the worksite.2 3 As a result, there has been a proliferation of restrictive indoor smoking policies in the USA in the past decade.4 5 Currently, about 85% of US worksites have some form of a smoking restriction policy (SK Hammond, KM Emmons, unpublished data). Policies that restrict smoking indoors not only decrease exposure to ETS,6 but also reduce cigarette consumption7 and increase desire to quit8 as well as actual smoking cessation.9 10 In most hospital and worksite settings, the majority of employees, including smokers, approve of some smoking restrictions,11 12 although current smokers are less supportive even in these settings.13

However, not all worksites are keen to impose smoking restrictions. Long term residential care facilities, especially psychiatric and correctional institutions, have been more resistant to restrictions on smoking.14 While the Joint Commission on Accreditation of Healthcare Organizations (JCAHO) mandates smoking policies for psychiatric and residential hospitals,15 comparable standards are not required by either the American Correctional Association (ACA) or the National Commission on Correctional Healthcare (NCCHC). These organisations offer recommendations for non-smoking policies but do not enforce these guidelines.16 In a national survey of over 900 correctional facilities, 45% of institutions allowed some form of smoking, either by staff or inmates. In fact, unlike hospitals and most worksites, the rate at which prisons are adopting new smoking bans has declined in recent years.17 Furthermore, exactly what restrictions are being implemented is unclear. Unfortunately, the previously published research on smoking restrictions in correctional facilities does not state whether different policies exist for inmates versus employees, or indoor versus outdoor policies.17-20

The threat of ETS is especially potent in prison settings for several reasons. First, estimates of smoking prevalence among prisoners in the USA range from 60–80%.16 21 Secondly, poor ventilation systems along with inadequate enforcement of existing smoking restrictions may lead to very high levels of ETS. For example, ambient air nicotine levels taken before implementation of a smoking ban in Vermont prisons averaged 6 μg/m3, with even higher levels found in common rooms and cell areas (SK Hammond, KM Emmons, unpublished data). These levels were 10 times above the 0.75 μg/m3 level that has been suggested as an acceptable level of ETS.22 Finally, the prison environment creates almost constant involuntary exposure to ETS because inmates spend the vast majority of time indoors.

ETS creates not only public health but also legal problems for jails as well. Inmates and staff complain of hazardous exposure to ETS, while at the same time prison administrators are reluctant to ban smoking completely because of fears of increased tension among inmates and staff. In a landmark case, the US supreme court held that inmates have a constitutional right to be free from unreasonable risks associated with ETS.23

In 1992, the state of Vermont completely banned smoking indoors and outdoors for both staff and inmates of all correctional facilities, and ETS levels dropped by as much as 80% following implementation of this policy (SK Hammond, KM Emmons, unpublished data). Because of the creation of a black market for cigarettes and the perception of increased staff–inmate tension, the policy was amended to prohibit indoor smoking but to allow smoking outdoors by staff and inmates. This policy has not been fully enforced. Smoking in individual cells and common areas remains a problem and prisoners still retain smoking paraphernalia inside the facilities. In 1996, an inmate with emphysema sued the state of Vermont claiming he was wrongfully exposed to ETS.24 A superior court subsequently found in favour of the inmate and noted that the state must protect the health needs of the inmates. The state of Vermont then asked us to examine and refine the current smoking policy for its jails and prisons.

While there is general information on tobacco restrictions in correctional facilities,16 17 there is little information on what specific smoking restriction policies will work and how they can be adequately enforced. It has been suggested that when there is resistance to smoking policy changes, it typically comes from prison employees more than from inmates.16 25 Smoking restrictions in non-prison worksites are typically accepted, even among smokers (50–60%).26 27 However, the acceptability of smoking restrictions in correctional settings is largely unknown. In the only study of employees in a prison setting, only 44% of Australian correctional employees favoured a complete ban on smoking and smokers advocated for less restrictive options than non-smokers.19 20 Given that employee support is thought to be crucial for implementation of smoking restrictions,28 we sought to first determine correctional employees' support for various policy alternatives and to investigate their expectations regarding the consequences of further smoking restrictions in the worksite.

## Methods

At the time of this survey, the Vermont Department of Corrections (DOC) employed approximately 800 staff. The present study surveyed the 640 employees who worked in one of the three prisons, the five jails, or the administrative office. Prisons are generally for those inmates who have committed more serious crimes and have extended sentences. Jails are for those held in pre-adjudication detention or those with shorter terms. We created a brief questionnaire based on previous research on smoking restrictions in hospitals and prisons.16 17 19 20

Several policy alternatives were offered for both staff and inmate smoking, and respondents indicated their degree of support for each using a five point Likert scale (1 = definitely would not support, 5 = definitely would support). These policies ranged from the least restrictive (smoking permitted indoors and outdoors) to most restrictive (complete ban on smoking on the grounds). The policies were listed such that an employee could support more than one; therefore, low support for a policy could reflect a wish for tighter or looser restrictions. Seventeen questions assessed employee beliefs about positive and negative consequences of smoking restrictions (for example, health outcomes, rule violations). These items were also presented in a five point Likert format (1 = strongly disagree, 5 = strongly agree). Additionally, three questions assessed perceptions of the difficulty of implementing smoking restrictions. We also asked employees to indicate their smoking status, and for current smokers, cigarettes smoked per day, age of first use, and intention to quit in the next six months. We did not include demographic questions because some of the employees were concerned that they could be identified by such items.

We assigned all employees a confidential ID number, and sent the questionnaires, with self addressed stamped envelopes, to employee homes. We offered a lottery of 16 cash incentives ranging from $10 to$100 to those who completed the survey, and conducted a second mailing for those who did not initially respond. This study was approved by the University of Vermont Committee on Human Research.

Following the survey, we conducted five focus groups (three prisons, two jails) to discuss smoking policy in further detail. Specific officers (smokers and non-smokers alike) were invited to participate based upon their superintendent's belief that they would provide meaningful input to the discussion of smoking restrictions. We included only officers in these discussions because they are on the front line of policy enforcement. A total of 23 officers agreed to participate in the focus groups. Each group met once for approximately 75 minutes. The structure of each group was loosely framed around issues of policy enforcement and implementation, although the moderator of the groups (MJC) allowed the discussion to turn to other issues that the officers believed to be relevant.

## Results

A total of 321 questionnaires were completed for a 50% response rate. Seventy eight (24%) of the respondents were current smokers, 120 (38%) were ex-smokers, and 121 (38%) were never smokers. Two employees did not indicate their smoking status. The mean (SD) age of first daily cigarette smoking among current smokers was 17.7 (4.8) years. Smokers smoked an average of less than a pack of cigarettes per day (17.9 (8.3)); 41% of daily cigarettes were smoked while at work. Most smokers (70%) had at least some desire to stop smoking, and the majority (52%) reported they “possibly” intended to quit in the next six months.

Responses were initially compared by type of facility (jailv prison) and employee position (uniformed employees v non-uniformed employees). For clarity, responses to Likert items were collapsed to indicate support (“probably would support” and “definitely would support”) and χ2 analyses were conducted. Support for inmate and employee smoking policies generally did not differ as a function of type of facility. Uniformed employees, who included correctional officers and shift supervisors, supported less restrictions than did non-uniformed employees (table 1).

Table 1

Support for inmate and employee smoking policies by type of employment (alternatives are independent of each other)

Because responses differed by position, further analysis of smoking policies controlled for employee position via analysis of covariances (ANCOVAs) (table 2). In general, employees favoured some restrictions for inmates, but were less supportive of such policies for themselves. In terms of inmate smoking, 56% of never smokers and 49% of ex-smokers supported a complete ban indoors and outdoors, while only 15% of current smokers supported this option (p < 0.001). There was, however, some common ground for an inmate smoking policy. Employees indicated a moderate amount of support (over 60%) for a complete ban indoors and limited smoking outdoors for inmates.

Table 2

Support for inmate and employee smoking policies by employee smoking status, after controlling for employee position (alternatives are independent of each other)

In terms of employee smoking, never and ex-smokers were concordant, but current smokers favoured decidedly less restrictive options. The most comprehensive option, a total ban on employee smoking indoors and outdoors, received support from approximately 38% of never and ex-smokers and only 3% of current smokers (p < 0.001). The next most restrictive option for employee smoking, a complete ban indoors and limited outdoor smoking, received support from approximately 57% of never and ex-smokers, but only 36% of current smokers (p < 0.02).

A similar pattern occurred with general beliefs about the consequences of further smoking restrictions in the worksite (table 3). More current smokers anticipated negative consequences of further smoking restrictions than did never or ex-smokers. With two exceptions, over 45% of current smokers anticipated negative outcomes in morale, quality of work, tension between staff and inmates, other rule violations, and unrest by inmates. In contrast, with one exception, less than half of never and ex-smokers expected these negative consequences. On a positive note, approximately two thirds of all respondents agreed that further restrictions would be acceptable if help were offered to those who wanted to quit smoking.

Table 3

Rather than conduct multiple comparisons for each of the beliefs regarding consequences of restrictions, a factor analysis was conducted for all items. This resulted in a two factor solution; one factor reflected the negative consequences of restrictions (Cronbach α = 0.87), while the other factor reflected the positive benefits (Cronbach α = 0.71) (table 3). After controlling for employee position, only endorsement of the negative consequence factor significantly differed across smoking groups (F 2 319 = 43.8, p < 0.001). Post hoc (Bonferroni) comparisons indicated that negative beliefs about smoking restrictions were endorsed significantly more by current smokers than by never or ex-smokers. There was also a main effect for type of facility on negative beliefs (F 2 302 = 3.9, p = 0.02), such that prison employees believed more negative consequences would result than did jail employees. Finally, we observed a significant interaction between correctional facility and smoking on beliefs about negative outcomes of further smoking restrictions (F 16 284 = 1.7, p < 0.05). At all facilities, current smokers expected more negative consequences than did never or ex-smokers, but whether ex-smokers expected more negative consequences than did never smokers varied across sites.

Current smokers differed from never and ex-smokers in their beliefs about the difficulty of implementing more restrictive smoking policies. Among current smokers, 46% believed that further restrictions would be very difficult to institute, and 59% believed that problems would persist. In contrast, only 23% of both never and ex-smokers believed that restrictions would be very difficult to implement and few (30%) believed that problems would be long lasting.

Qualitative information derived from the focus groups indicated that officers generally believed that smoking restrictions are possible given appropriate and consistent enforcement. Officers noted the necessity of both standardised disciplinary procedures and a strict penalty system as critical to the success of any smoking restriction. In addition, officers also believed that restrictions are feasible when smoking assistance is provided, both to smoking inmates and officers. Finally, the focus group participants acknowledged that smoking restrictions would result in some degree of unrest initially, but that successful implementation requires a long term commitment.

## Discussion

Our findings indicate that correctional employees are generally favourable towards smoking restrictions for inmates, but are less supportive of further restrictions for themselves. The current state wide policy in Vermont correctional facilities bans indoor smoking for both inmates and staff, but does allow for outdoor smoking. Most employees agree with this policy; however, most never and ex-smokers would like to see slightly tighter controls, including restricted areas for outdoor smoking. In contrast, most current smokers generally favoured policies that were less restrictive than the existing policy, including limited indoor smoking areas for both inmates and employees, with unlimited smoking outdoors.

Beliefs regarding the potential consequences of further smoking restrictions also differed across smoking groups. Current smokers were more likely to anticipate negative outcomes for tighter smoking restrictions than were either never or ex-smokers. Examples of these negative outcomes included beliefs that further restrictions would result in increased tension between staff and inmates and an increase in other rule violations. Current smokers believed that these negative outcomes would be more severe and long lasting than did never and ex-smokers.

A limitation of this study was the 50% response rate. Twenty four per cent of respondents were current smokers. We suspect the actual prevalence of smoking in correctional employees is higher. We could not find data on the prevalence of smoking either in state of Vermont correctional employees or in prior published reports of other correctional systems. To ensure confidentiality and thereby increase response rates, we purposely omitted demographic questions. Unfortunately, this prevents us from estimating any response bias. If, as we suspect, we undersampled smokers, then our results may overestimate support for restrictions.

The only other study to assess support of smoking restrictions among correctional employees was conducted in Australian correctional facilities.19 20 Unfortunately, these researchers did not distinguish whether their results applied to inmate smoking, employee smoking, or both. In that study, 15% of 225 employees favoured a complete smoking ban, and 44% of the workers favoured a ban in all indoor areas. Non-smokers favoured tighter restrictions than did smokers.

Our results suggest that specific interventions may be necessary to build acceptance among employees who currently smoke. Such interventions may include providing a clear rationale for restrictive smoking policies (for example, to prevent ETS), dispelling paranoia (for example, the policy is not meant to force employees to stop smoking), and citing legal precedents (for example, there is no “right to smoke”). In light of health and legal concerns, smoking policies in correctional facilities need to be evaluated. Although anecdotal reports suggest that there are few problems among prisons and jails that do impose smoking restrictions,25 empirical research on this is lacking. The current study, using both survey and focus group data, suggests a moderate degree of smoking restrictions is not only supported, but feasible as well.

## Acknowledgments

This study was funded by NIDA training grant DA-07242 (MJC), a Career Development Award, DA-00109 (JRH), and grant DA11557 from the National Institute on Drug Abuse. The authors would like to thank Jonathan Hirsch and Ashley Bryant for their assistance.

## References

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# Help with particular integral on 2nd order differential equation

Discussion in 'Math' started by u-will-neva-no, Sep 29, 2011.

1. ### u-will-neva-no Thread Starter Member

Mar 22, 2011
230
2
Hello everyone, my second order to solve is:

y'' - 2y' = 12e^2x -8e^-2.

I have found the complementary function, which is Ae^2x + B but im not sure what my trial solution should be to find the particular solution.

I tried y = axe^2x - be^-2x, for the trial solution and had a nice answer. I don't think this is correct however...please let me know! Thankyou.

2. ### steveb Senior Member

Jul 3, 2008
2,433
469
Can you verify your equation? Is it the following?

$y'' - 2y' = 12{\rm e}^{2x} -8 {\rm e}^{-2}$

Or is it this?

$y'' - 2y' = 12{\rm e}^2 x -8 {\rm e}^{-2}$

Or, is it this?

$y'' - 2y' = 12{\rm e}^{2x} -8 {\rm e}^{-2x}$

3. ### u-will-neva-no Thread Starter Member

Mar 22, 2011
230
2
Sorry, i messed up. Its the last one you typed.

4. ### steveb Senior Member

Jul 3, 2008
2,433
469
The way I would solve this is to make a couple of substitutions.

First, define an input function u=12exp(2x) - 8exp(-2x).

Then, define a new state variable z=dy/dx.

With these substitutions, the state space form is.

dy/dx=z
dz/dx=2z+u

which in matrix notation is

dW/dx=AW+Bu

where W=[y; z] as a column vector, A=[0 1; 0 2] as a square matrix and B=[0; 1] as a column vector

From there you can solve using standard methods such as state transition matrix approach, or Laplace transform method and the response to your input function will give you a particular solution matched to given initial conditions.

Last edited: Sep 30, 2011
u-will-neva-no likes this.
5. ### steveb Senior Member

Jul 3, 2008
2,433
469
Maybe you haven't had a chance to look at this, but I had a few minutes at coffee break and solved this via state transition matrix. The solution made it clear that a valid particular solution is.

$y=6x{\rm e}^{2x}-{\rm e}^{-2x}$

So, the form of your suggested solution is correct, but you needed to solve for the "a" and "b" coefficients.

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• ###### page2.pdf
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Last edited: Sep 30, 2011
u-will-neva-no likes this.
6. ### u-will-neva-no Thread Starter Member

Mar 22, 2011
230
2
Wow, I have never come across this method to solve it although I think I will be taught this sometime soon at university. I will keep hold of this page and refer to it once I get my head around this method. Thank you for helping and putting all that effort in!

7. ### steveb Senior Member

Jul 3, 2008
2,433
469
You are welcome, and yes you should be learning this method a little later. It is one of the corner-stones of signals and systems theory and used extensively in controls system design. It probably looks complex to you now, but you will eventually see that it is straightforward, and wasn't much effort at all. It was just a nice little problem to solve on coffee break while my coworkers were chatting about nonsense.

Since you say you want to save the pages, I should point out that there is a little error in there that I realized later. The solution for z(x) is a little off. It should be the following.

$z(x)=y'(0){\rm e}^{2x}+12x{\rm e}^{2x}+2{\rm e}^{-2x}-2{\rm e}^{2x}$

and NOT the following

$z(x)=y'(0){\rm e}^{2x}+24x{\rm e}^{2x}+4{\rm e}^{-2x}-2{\rm e}^{2x}$

Last edited: Oct 1, 2011


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# How do you solve x^2-x= -1 using the quadratic formula?

Jul 26, 2016

$x = \frac{1 \pm i \sqrt{3}}{2}$

#### Explanation:

The quadratic formula for a general quadratic equation $a {x}^{2} + b x + c = 0$ is given by:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

For the equation:

${x}^{2} - x = - 1$

or ${x}^{2} - x + 1 = 0$

you get

a=1; b=-1 and c = 1

by substituting these values in the quadratic formula:

$x = \frac{- \left(- 1\right) \pm \sqrt{{\left(- 1\right)}^{2} - 4 \cdot 1 \cdot 1}}{2 \cdot 1}$

$x = \frac{1 \pm \sqrt{1 - 4}}{2}$

or $x = \frac{1 \pm \sqrt{- 3}}{2}$

or $x = \frac{1 \pm i \sqrt{3}}{2}$


text:
Finite square well

I know how and why we use this form of stationary Schrödinger equation for finding $\psi$ outside the finite square potential well:

$$\frac{d^2 \psi}{dx^2}=\kappa^2 \psi$$

I Also know that the general solution to this equation is:

$$\psi = Ae^{\kappa\, x} + Be^{-\kappa\, x}.$$

But why do we use only left part $\psi = Ae^{\kappa \, x}$ for the left outer side $x<0$, and the right part $\psi = B e^{-\kappa\, x}$ on the right outer side part $x>0$?

-
If you're using the usual conventions, don't you mean $Ae^{\kappa x}$ on the left, not right? –  Dan Piponi Mar 29 '13 at 20:25

I think you are asking for a finite well of width L that is from $-L/2< x< L/2$. Why do we only use

$\psi(x) = A e^{+\kappa x}$ for $x<-L/2$

and

$\psi(x) = B e^{-\kappa x}$ for $x>+L/2$.

The reason is we want to be able to interpret the wavefunction as the probability density for finding the particle at x. For this to make sense the probability of finding the particle anywhere should be 1 or

$\int_{-\infty}^{-\infty}dx \ |\psi(x)|^2 = 1$.

If you have solutions of the form

$\psi(x) = A e^{+\kappa x} + B e^{-\kappa x}$

this is only possible if $A = 0$ in the $x > +L/2$ region and $B = 0$ in the $x < +L/2$ region.

-
I think i get it but please correct me if i am wrong. On the interval $x > +\tfrac{L}{2}$ i can only increase $x$ towards $+\infty$. Soo i have to make sure that i have a negative exponent or else i ll get $\psi = \infty$! This is why on this interval i need only $Ae^{-\kappa x}$. It is vice versa for the interval $x<-\frac{L}{2}$ where i can only decrease $x$ towards $-\infty$. This means i have to take a part with a positive exponent so i won't get $\psi = \infty$. It is all about the signs of the exponents if i am not mistaken. –  71GA Mar 29 '13 at 20:39
Pretty much. You can go a step further and actually try integrating $\int_{-L/2}^\infty \ dx \ e^{\kappa x}$ and you will see that you will get $\infty$, so there isn't any hope of normalizing this to anything finite. –  DJBunk Mar 29 '13 at 20:50
You need to ensure that the wavefunction is normalisable: $$\int\limits_{-\infty}^{\infty} \mathrm{d}x\, |\psi(x)|^2 = 1.$$ This ensures that the wave function yields a valid probability distribution upon application of the Born rule. If you use the normal convention of taking positive $x$ values to the right and negative $x$ values to the left, this implies that you can only keep $e^{-\kappa x}$ for the right outer side, etc. Otherwise the value of the wavefunction grows without bound as $x\to \pm \infty$
@71GA You should be able to convince yourself that in general, the normalisation condition is violated for any function that does not vanish at $x = \pm \infty$. This means that the solution $e^{\kappa x}$ is not a valid solution on the right side of the box, and a similar argument applies to the left side. Have you actually tried evaluating the normalisation integral for the various possibilities to see what comes out? –  Mark Mitchison Mar 29 '13 at 20:36


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## Building a polynomial from its roots

Suppose we are given a set of $n \geq 2$ distinct real numbers $\mathcal{R} = \{r_1, r_2, \ldots, r_n\}$, and we build a monic univariate polynomial (over field $\mathbb{R}$) of degree $n$ whose $n$ distinct roots are the elements of set $\mathcal{R}$

$p_n(x) = \displaystyle\prod_{i=1}^n (x - r_i) = \displaystyle\sum_{m=0}^n a_{m,n} x^m$,

where $a_{m,n} \in \mathbb{R}$ is the coefficient of monomial $x^m$. The second subscript in $a_{m,n}$ tells us that the coefficient corresponds to a polynomial whose degree is $n$.

Problem: for $n \geq 2$, what are the values of the coefficients $\{a_{m,n}\}_{m=0}^{n}$ in terms of the roots $\{r_i\}_{i=1}^n$?

At first glance, this problem seems elementary. However, I had never thought of it and I found it quite interesting. I prefer to think of examples before attempting to see the “big picture”, so let us consider the simplest cases.

_____

Case $n=1$

The monic polynomial of degree $1$ whose root is $r_1$ is

$p_1(x) = x - r_1$

and the coefficients are

$\left[\begin{array}{c}a_{0,1}\\a_{1,1}\end{array}\right] = \left[\begin{array}{c} -r_1\\1\end{array}\right]$.

_____

Case $n=2$

The monic polynomial of degree $2$ whose roots are $r_1, r_2$ is

$\begin{array}{rl} p_2(x) &= (x - r_1) (x - r_2)\\ &= x^2 - (r_1 + r_2) x + r_1 r_2\end{array}$

and the coefficients are

$\left[\begin{array}{c}a_{0,2}\\a_{1,2}\\a_{2,2}\end{array}\right] = \left[\begin{array}{c}r_1 r_2\\-(r_1+r_2)\\1\end{array}\right]$.

_____

Case $n=3$

The monic polynomial of degree $3$ whose roots are $r_1, r_2, r_3$ is

$\begin{array}{rl} p_3(x) &= (x - r_1) (x - r_2) (x - r_3)\\ &= x^3 - (r_1+r_2 + r_3) x^2 + (r_1 r_2 + r_1 r_3 + r_2 r_3) x - r_1 r_2 r_3\\\end{array}$

and the coefficients are

$\left[\begin{array}{c}a_{0,3}\\a_{1,3}\\a_{2,3}\\a_{3,3}\end{array}\right] = \left[\begin{array}{c} - r_1 r_2 r_3\\ r_1 r_2 + r_1 r_3 + r_2 r_3\\ -(r_1+r_2 + r_3)\\1\end{array}\right]$.

_____

The coefficients $\{a_{m,n}\}_{m=0}^{n}$ can be computed in a recursive fashion. Consider the monic polynomials

$p_{k}(x) = \displaystyle\prod_{i=1}^{k} (x - r_i)$

and

$p_{k+1}(x) = \displaystyle\prod_{i=1}^{k+1} (x - r_i) = (x - r_{k+1}) \displaystyle\prod_{i=1}^k (x - r_i)$,

then $p_{k+1} = (x - r_{k+1}) p_k(x)$, and thus

$\begin{array}{rl} p_{k+1}(x) &= (x - r_{k+1})\displaystyle\sum_{m=0}^k a_{m,k} x^m\\ &= \displaystyle\sum_{m=0}^k a_{m,k} x^{m+1} - \displaystyle\sum_{m=0}^k r_{k+1} a_{m,k} x^m.\\\end{array}$

Note that

$\begin{array}{rl} \displaystyle\sum_{m=0}^k a_{m,k} x^{m+1} &= \displaystyle\sum_{m=1}^k a_{m-1,k} x^m + a_{k,k} x^{k+1}\\ \displaystyle\sum_{m=0}^{k} r_{k+1} a_{m,k} x^m &= r_{k+1} a_{0,k} + \displaystyle\sum_{m=1}^k r_{k+1} a_{m,k} x^m\end{array}$

and therefore

$p_{k+1}(x) = a_{k,k} x^{k+1} + \displaystyle\sum_{m=1}^k \left(a_{m-1,k} - r_{k+1} a_{m,k}\right) x^m - r_{k+1} a_{0,k}$.

We can write $p_{k+1}(x)$ in the expanded form in terms of the coefficients $\{a_{m,k+1}\}_{m=0}^{k+1}$

$p_{k+1}(x) = a_{k+1,k+1} x^{k+1} + \displaystyle\sum_{m=1}^k a_{m,k+1} x^m + a_{0,k+1}$

and therefore we can write the coefficients $\{a_{m,k+1}\}_{m=0}^{k+1}$ in terms of the coefficients $\{a_{m,k}\}_{m=0}^{k}$, as follows

$a_{m,k+1} = \displaystyle\left\{\begin{array}{rl} - r_{k+1} a_{0,k} & \text{if} \quad{} m = 0\\ a_{m-1,k} - r_{k+1} a_{m,k} & \text{if} \quad{} 1 \leq m \leq k\\ a_{k,k} & \text{if} \quad{} m=k+1\\ \end{array}\right.$

Hence, starting with the zero-degree monic polynomial $p_0(x)$, whose only coefficient is $a_{0,0} = 1$, we can build $p_1(x)$, the monic polynomial of degree $1$ whose root is $r_1$. From $p_1(x)$, and given a second root $r_2$, we can build $p_2(x)$, the monic polynomial of degree $2$ whose roots are $r_1, r_2$. Iterating successively, we can build $p_n(x)$, the monic polynomial of degree $n$ whose roots are $r_1,r_2, \ldots, r_n$.


text:
MMN-2601

# A note on radicals of associative rings and alternative rings

• S. Tumurbat, Department of Mathematics, National University of Mongolia, Ikh Surguuliin Gudamj-1, Ulaanbaatar, Mongolia and, School of Applied Science, Mongolian University of Science and Technology, P.O.Box 75, Ulaanbaatar, Mongolia, stumurbat@hotmail.com
• T. Khulan, National University of Mongolia, Department of Mathematics, Ikh Surguuliin Gudamj-1, Ulaanbaatar, Mongolia, P.O.Box-46A/523, 210646, hulangaaa@yahoo.com
• D. Dayantsolmon, National University of Mongolia, Department of Mathematics, Ikh Surguuliin Gudamj-1, Ulaanbaatar, Mongolia, P.O.Box-46A/523, 210646, dayantsolmon@num.edu.mn

## Abstract

Let $U$ be a universal subclass of a universal class $V$ of rings. We investigate connections of radicals in $U$ and $V$. In \cite{T} we defined $T$ and $T_s$. Let $\gamma \in \{ T, T_s \}$ and let $A$ be a commutative ring with minimum condition on ideals. We give a sufficient and necessary condition when $A$ is $\gamma$-semisimple.

Vol. 23 (2022), No. 1, pp. 175-182
DOI: 10.18514/MMN.2022.2601


text:
for Journals by Title or ISSN for Articles by Keywords help
 Subjects -> STATISTICS (Total: 130 journals)
 The end of the list has been reached or no journals were found for your choice.
 Metrika   [SJR: 0.605]   [H-I: 30]   [2 followers]  Follow         Hybrid journal (It can contain Open Access articles)    ISSN (Print) 1435-926X - ISSN (Online) 0026-1335    Published by Springer-Verlag  [2335 journals]
• Conditional empirical likelihood for quantile regression models
• Authors: Wu Wang; Zhongyi Zhu
Pages: 1 - 16
Abstract: In this paper, we propose a new Bayesian quantile regression estimator using conditional empirical likelihood as the working likelihood function. We show that the proposed estimator is asymptotically efficient and the confidence interval constructed is asymptotically valid. Our estimator has low computation cost since the posterior distribution function has explicit form. The finite sample performance of the proposed estimator is evaluated through Monte Carlo studies.
PubDate: 2017-01-01
DOI: 10.1007/s00184-016-0588-6
Issue No: Vol. 80, No. 1 (2017)

• Robust feature screening for varying coefficient models via quantile
partial correlation
• Authors: Xiang-Jie Li; Xue-Jun Ma; Jing-Xiao Zhang
Pages: 17 - 49
Abstract: This article is concerned with feature screening for varying coefficient models with ultrahigh-dimensional predictors. We propose a new sure independence screening method based on quantile partial correlation (QPC-SIS), which is quite robust against outliers and heavy-tailed distributions. Then we establish the sure screening property for the QPC-SIS, and conduct simulations to examine its finite sample performance. The results of simulation study indicate that the QPC-SIS performs better than other methods like sure independent screening (SIS), sure independent ranking and screening, distance correlation-sure independent screening, conditional correlation sure independence screening and nonparametric independent screening, which shows the validity and rationality of QPC-SIS.
PubDate: 2017-01-01
DOI: 10.1007/s00184-016-0589-5
Issue No: Vol. 80, No. 1 (2017)

• Shrinkage estimation of the linear model with spatial interaction
• Authors: Yueqin Wu; Yan Sun
Pages: 51 - 68
Abstract: The linear model with spatial interaction has attracted huge attention in the past several decades. Different from most existing research which focuses on its estimation, we study its variable selection problem using the adaptive lasso. Our results show that the method can identify the true model consistently, and the resulting estimator can be efficient as the oracle estimator which is obtained when the zero coefficients in the model are known. Simulation studies show that the proposed methods perform very well.
PubDate: 2017-01-01
DOI: 10.1007/s00184-016-0590-z
Issue No: Vol. 80, No. 1 (2017)

• On the residual lifetime of coherent systems with heterogeneous components
• Authors: P. Samadi; M. Rezaei; M. Chahkandi
Pages: 69 - 82
Abstract: The residual lifetime is of significant interest in reliability and survival analysis. In this article, we obtain a mixture representation for the reliability function of the residual lifetime of a coherent system with heterogeneous components in terms of the reliability functions of residual lifetimes of order statistics. Some stochastic comparisons are made on the residual lifetimes of the systems. Some examples are also given to illustrate the main results.
PubDate: 2017-01-01
DOI: 10.1007/s00184-016-0591-y
Issue No: Vol. 80, No. 1 (2017)

• Asymptotics of self-weighted M-estimators for autoregressive models
• Authors: Xinghui Wang; Shuhe Hu
Pages: 83 - 92
Abstract: In this paper, we consider a stationary autoregressive AR(p) time series $$y_t=\phi _0+\phi _1y_{t-1}+\cdots +\phi _{p}y_{t-p}+u_t$$ . A self-weighted M-estimator for the AR(p) model is proposed. The asymptotic normality of this estimator is established, which includes the asymptotic properties under the innovations with finite or infinite variance. The result generalizes and improves the known one in the literature.
PubDate: 2017-01-01
DOI: 10.1007/s00184-016-0592-x
Issue No: Vol. 80, No. 1 (2017)

• Fiducial inference in the classical errors-in-variables model
• Authors: Liang Yan; Rui Wang; Xingzhong Xu
Pages: 93 - 114
Abstract: For the slope parameter of the classical errors-in-variables model, existing interval estimations with finite length will have confidence level equal to zero because of the Gleser–Hwang effect. Especially when the reliability ratio is low and the sample size is small, the Gleser–Hwang effect is so serious that it leads to the very liberal coverages and the unacceptable lengths of the existing confidence intervals. In this paper, we obtain two new fiducial intervals for the slope. One is based on a fiducial generalized pivotal quantity and we prove that this interval has the correct asymptotic coverage. The other fiducial interval is based on the method of the generalized fiducial distribution. We also construct these two fiducial intervals for the other parameters of interest of the classical errors-in-variables model and introduce these intervals to a hybrid model. Then, we compare these two fiducial intervals with the existing intervals in terms of empirical coverage and average length. Simulation results show that the two proposed fiducial intervals have better frequency performance. Finally, we provide a real data example to illustrate our approaches.
PubDate: 2017-01-01
DOI: 10.1007/s00184-016-0593-9
Issue No: Vol. 80, No. 1 (2017)

• Robust Dickey–Fuller tests based on ranks for time series with
• Authors: V. A. Reisen; C. Lévy-Leduc; M. Bourguignon; H. Boistard
Pages: 115 - 131
Abstract: In this paper the unit root tests proposed by Dickey and Fuller (DF) and their rank counterpart suggested by Breitung and Gouriéroux (J Econom 81(1): 7–27, 1997) (BG) are analytically investigated under the presence of additive outlier (AO) contaminations. The results show that the limiting distribution of the former test is outlier dependent, while the latter one is outlier free. The finite sample size properties of these tests are also investigated under different scenarios of testing contaminated unit root processes. In the empirical study, the alternative DF rank test suggested in Granger and Hallman (J Time Ser Anal 12(3): 207–224, 1991) (GH) is also considered. In Fotopoulos and Ahn (J Time Ser Anal 24(6): 647–662, 2003), these unit root rank tests were analytically and empirically investigated and compared to the DF test, but with outlier-free processes. Thus, the results provided in this paper complement the studies of the previous works, but in the context of time series with additive outliers. Equivalently to DF and Granger and Hallman (J Time Ser Anal 12(3): 207–224, 1991) unit root tests, the BG test shows to be sensitive to AO contaminations, but with less severity. In practical situations where there would be a suspicion of additive outlier, the general conclusion is that the DF and Granger and Hallman (J Time Ser Anal 12(3): 207–224, 1991) unit root tests should be avoided, however, the BG approach can still be used.
PubDate: 2017-01-01
DOI: 10.1007/s00184-016-0594-8
Issue No: Vol. 80, No. 1 (2017)

• Bayesian estimation based on ranked set sample from Morgenstern type
bivariate exponential distribution when ranking is imperfect
• Authors: Manoj Chacko
Abstract: In this paper we consider Bayes estimation based on ranked set sample when ranking is imperfect, in which units are ranked based on measurements made on an easily and exactly measurable auxiliary variable X which is correlated with the study variable Y. Bayes estimators under squared error loss function and LINEX loss function for the mean of the study variate Y, when (X, Y) follows a Morgenstern type bivariate exponential distribution, are obtained based on both usual ranked set sample and extreme ranked set sample. Estimation procedures developed in this paper are illustrated using simulation studies and a real data.
PubDate: 2016-12-08
DOI: 10.1007/s00184-016-0607-7

• An ergodic theorem for proportions of observations that fall into random
sets determined by sample quantiles
• Authors: Anna Dembińska
Abstract: Assume that a sequence of observations $$(X_n; n\ge 1)$$ forms a strictly stationary process with an arbitrary univariate cumulative distribution function. We investigate almost sure asymptotic behavior of proportions of observations in the sample that fall into a random region determined by a given Borel set and a sample quantile. We provide sufficient conditions under which these proportions converge almost surly and describe the law of the limiting random variable.
PubDate: 2016-12-07
DOI: 10.1007/s00184-016-0606-8

• On the shape of the cross-ratio function in bivariate survival models
induced by truncated and folded normal frailty distributions
• Authors: Steffen Unkel
Abstract: In shared frailty models for bivariate survival data the frailty is identifiable through the cross-ratio function (CRF), which provides a convenient measure of association for correlated survival variables. The CRF may be used to compare patterns of dependence across models and data sets. We explore the shape of the CRF for the families of one-sided truncated normal and folded normal frailty distributions.
PubDate: 2016-12-07
DOI: 10.1007/s00184-016-0608-6

• Acceleration of the stochastic search variable selection via componentwise
Gibbs sampling
• Authors: Hengzhen Huang; Shuangshuang Zhou; Min-Qian Liu; Zong-Feng Qi
Abstract: The stochastic search variable selection proposed by George and McCulloch (J Am Stat Assoc 88:881–889, 1993) is one of the most popular variable selection methods for linear regression models. Many efforts have been proposed in the literature to improve its computational efficiency. However, most of these efforts change its original Bayesian formulation, thus the comparisons are not fair. This work focuses on how to improve the computational efficiency of the stochastic search variable selection, but remains its original Bayesian formulation unchanged. The improvement is achieved by developing a new Gibbs sampling scheme different from that of George and McCulloch (J Am Stat Assoc 88:881–889, 1993). A remarkable feature of the proposed Gibbs sampling scheme is that, it samples the regression coefficients from their posterior distributions in a componentwise manner, so that the expensive computation of the inverse of the information matrix, which is involved in the algorithm of George and McCulloch (J Am Stat Assoc 88:881–889, 1993), can be avoided. Moreover, since the original Bayesian formulation remains unchanged, the stochastic search variable selection using the proposed Gibbs sampling scheme shall be as efficient as that of George and McCulloch (J Am Stat Assoc 88:881–889, 1993) in terms of assigning large probabilities to those promising models. Some numerical results support these findings.
PubDate: 2016-11-22
DOI: 10.1007/s00184-016-0604-x

• Efficient paired choice designs with fewer choice pairs
• Authors: Aloke Dey; Rakhi Singh; Ashish Das
Abstract: For paired choice experiments, two new construction methods of designs are proposed for the estimation of the main effects. In many cases, these designs require about 30–50% fewer choice pairs than the existing designs and at the same time have reasonably high D-efficiencies for the estimation of the main effects. Furthermore, as against the existing efficient designs, our designs have higher D-efficiencies for the same number of choice pairs.
PubDate: 2016-11-17
DOI: 10.1007/s00184-016-0605-9

• Imputation based statistical inference for partially linear quantile
regression models with missing responses
• Authors: Peixin Zhao; Xinrong Tang
Abstract: In this paper, we consider the confidence interval construction for partially linear quantile regression models with missing response at random. We propose an imputation based empirical likelihood method to construct confidence intervals for the parametric components and the nonparametric components, and show that the proposed empirical log-likelihood ratios are both asymptotically Chi-squared in theory. Then, the confidence region for the parametric component and the pointwise confidence interval for the nonparametric component are constructed. Some simulation studies and a real data application are carried out to assess the performance of the proposed estimation method, and simulation results indicate that the proposed method is workable.
PubDate: 2016-11-01
DOI: 10.1007/s00184-016-0586-8

• Qualitative robustness of estimators on stochastic processes
• Authors: Katharina Strohriegl; Robert Hable
Abstract: A lot of statistical methods originally designed for independent and identically distributed (i.i.d.) data are also successfully used for dependent observations. Still most theoretical investigations on robustness assume i.i.d. pairs of random variables. We examine an important property of statistical estimators—the qualitative robustness in the case of observations which do not fulfill the i.i.d. assumption. In the i.i.d. case qualitative robustness of a sequence of estimators is, according to Hampel (Ann Math Stat 42:1887–1896, 1971), ensured by continuity of the corresponding statistical functional. A similar result for the non-i.i.d. case is shown in this article. Continuity of the corresponding statistical functional still ensures qualitative robustness of the estimator as long as the data generating process satisfies a certain convergence condition on its empirical measure. Examples for processes providing such a convergence condition, including certain Markov chains or mixing processes, are given as well as examples for qualitatively robust estimators in the non-i.i.d. case.
PubDate: 2016-11-01
DOI: 10.1007/s00184-016-0582-z

• Tree based diagnostic procedures following a smooth test of
goodness-of-fit
• Authors: Gilles R. Ducharme; Walid Al Akhras
Abstract: This paper introduces a statistical procedure, to be applied after a goodness-of-fit test has rejected a null model, that provides diagnostic information to help the user decide on a better model. The procedure goes through a list of departures, each being tested by a local smooth test. The list is organized into a hierarchy by seeking answers to the questions “Where is the problem?” and “What is the problem there?”. This hierarchy allows to focus on finer departures as the data becomes more abundant. The procedure controls the family-wise Type 1 error rate. Simulations show that the procedure can succeed in providing useful diagnostic information.
PubDate: 2016-11-01
DOI: 10.1007/s00184-016-0585-9

• A test of linearity in partial functional linear regression
• Authors: Ping Yu; Zhongzhan Zhang; Jiang Du
Abstract: This paper investigates the hypothesis test of the parametric component in partial functional linear regression. We propose a test procedure based on the residual sums of squares under the null and alternative hypothesis, and establish the asymptotic properties of the resulting test. A simulation study shows that the proposed test procedure has good size and power with finite sample sizes. Finally, we present an illustration through fitting the Berkeley growth data with a partial functional linear regression model and testing the effect of gender on the height of kids.
PubDate: 2016-11-01
DOI: 10.1007/s00184-016-0584-x

• AR(1) model with skew-normal innovations
• Authors: M. Sharafi; A. R. Nematollahi
Abstract: In this paper, we consider an autoregressive model of order one with skew-normal innovations. We propose several methods for estimating the parameters of the model and derive the limiting distributions of the estimators. Then, we study some statistical properties and the regression behavior of the proposed model. Finally, we provide a Monte Carlo simulation study for comparing performance of estimators and consider a real time series to illustrate the applicability of the proposed model.
PubDate: 2016-11-01
DOI: 10.1007/s00184-016-0587-7

• Nonparametric estimation in a mixed-effect Ornstein–Uhlenbeck model
• Authors: Charlotte Dion
Abstract: Two adaptive nonparametric procedures are proposed to estimate the density of the random effects in a mixed-effect Ornstein–Uhlenbeck model. First a kernel estimator is introduced with a new bandwidth selection method developed recently by Goldenshluger and Lepski (Ann Stat 39:1608–1632, 2011). Then, we adapt an estimator from Comte et al. (Stoch Process Appl 7:2522–2551, 2013) in the framework of small time interval of observation. More precisely, we propose an estimator that uses deconvolution tools and depends on two tuning parameters to be chosen in a data-driven way. The selection of these two parameters is achieved through a two-dimensional penalized criterion. For both adaptive estimators, risk bounds are provided in terms of integrated $$\mathbb {L}^2$$ -error. The estimators are evaluated on simulations and show good results. Finally, these nonparametric estimators are applied to neuronal data and are compared with previous parametric estimations.
PubDate: 2016-11-01
DOI: 10.1007/s00184-016-0583-y

• A study on the conditional inactivity time of coherent systems
• Authors: S. Goli; M. Asadi
Abstract: The study on the inactivity times is useful in evaluating the aging and reliability properties of coherent systems in reliability engineering. In the present paper, we investigate the inactivity time of a coherent system consisting of n i.i.d. components. We drive some mixture representations for the reliability function of conditional inactivity times of coherent systems under two specific conditions on the status of the system components. Some ageing and stochastic properties of the proposed conditional inactivity times are also explored.
PubDate: 2016-10-25
DOI: 10.1007/s00184-016-0600-1

• Reliability parameters estimation for parallel systems under imperfect
repair
• Authors: Soumaya Ghnimi; Soufiane Gasmi; Arwa Nasr
Abstract: We consider in this paper a parallel system consisting of $$\eta$$ identical components. Each component works independently of the others and has a Weibull distributed inter-failure time. When the system fails, we assume that the repair maintenance is imperfect according to the Arithmetic Reduction of Age models ( $$ARA_{m}$$ ) proposed by Doyen and Gaudoin. The purpose of this paper is to generate a simulated failure data of the whole system in order to forecast the behavior of the failure process. Besides, we estimate the maintenance efficiency and the reliability parameters of an imperfect repair following $$ARA_{m}$$ models using maximum likelihood estimation method. Our method is tested with several data sets available from related sources. The real data set corresponds to the time between failures of a compressor which is tested by Likelihood Ratio Test (LR). An analysis of the importance and the effect of the memory order of imperfect repair classes ( $$ARA_{m}$$ ) will be discussed using LR test.
PubDate: 2016-10-22
DOI: 10.1007/s00184-016-0603-y

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text:
## Illuminance vs. Luminance: Know Differences!

Architect / Content Writer

In the lighting world, many of the lighting terms often sound very  similar but have vast different meanings which can lead you to the confusion and a misunderstanding of lighting terminology. Illuminance and luminance both the terms are the most confusing term and are more or less used incorrectly by people, sometimes even by the people in the same industry!

### Luminance

Simply the illuminance determines the amount of light that covers a surface. Illuminance is invisible, generally it is light passing through space and not seen unless you look at the source (e.g. a light bulb) or a surface it reflects off or we can only see luminance.

OR

The illuminance or light level is the amount of light energy reaching a given point on a defined surface area, which is the luminous flux (i.e. lumens) per square meter.

OR

In other words, At a point on a surface, the ratio of the luminous flux incident on an infinitesimal element of the surface containing the point under consideration to the area of the element.

(NOTE — The unit of illuminance (the measurement of illumination) is Iux which is 1 lumen per square metre.)

Luminance is an amount of visible light that comes to the eye from surface or it can also be said as a state or quality of radiating or reflecting light.

Luminance is often used to describe emission or reflection from flat or diffuse surfaces. The luminance indicates how much luminous power will be detected by an eye looking at the surface from a particular angle of view.

Therefore luminance is thus an indicator of how bright the surface will appear.

Note: Luminance = Illuminance + Reflection

OR

In other words, it is the quotient of the luminous intensity in the given direction of an infinitesimal element of the surface containing the point under consideration by the orthogonally projected area of the element on a plane perpendicular to the given direction. The unit is candela per square metre (cd/m2).

##### Also Read: Components of Daylight Factor
Courtesy - 123rf
 Standard unit for illuminance is Lux (lx) which is lumens per square meter (lm/m2). The SI unit for luminance is candela per square metre (cd/m2).

Typical illuminance values/ranges for our understanding are as follows:

 1 lx Full moon 10 lx Street lighting 100-1’000 lx Workspace lighting 10’000 lx Surgery lighting 100’000 lx Plain sunshine

## Material Exhibition

Explore the world of materials.


text:
# How do you solve the inequality abs(3-x)<=3?

Mar 29, 2017

$0 \le x \le 6$

#### Explanation:

$\text{Given the inequality } | x | \le a$

$\text{the solution is always of the form } - a \le x \le a$

$\Rightarrow - 3 \le \textcolor{red}{3 - x} \le 3$

Isolate x in the centre interval while obtaining numeric values in the 2 end intervals.

subtract 3 from ALL intervals.

$- 3 - 3 \le \cancel{3} \cancel{- 3} - x \le 3 - 3$

$\Rightarrow - 6 \le - x \le 0$

multiply by - 1 to obtain x

$\textcolor{b l u e}{\text{Note}}$ when multiplying/dividing an inequality by a $\textcolor{b l u e}{\text{negative}}$ value we must $\textcolor{red}{\text{ reverse the inequality symbol}}$

$\Rightarrow 6 \ge x \ge 0 \leftarrow \textcolor{red}{\text{ reverse symbol}}$

$\Rightarrow x \le 6 \textcolor{red}{\text{ and }} x \ge 0$

$\Rightarrow 0 \le x \le 6 \text{ is the solution}$

$x \in \left[0 , 6\right] \leftarrow \textcolor{red}{\text{ in interval notation}}$

Mar 29, 2017

0≤x≤6, $x \setminus \in \left[0 , 6\right]$
We can rewrite the equation as -3≤3-x≤3. Subtract both sides by $3$ to get -6≤-x≤0. Now, we multiply both sides by $- 1$ and flip the inequality signs: 6≥x≥0. This can be rewritten to 0≤x≤6.
Other notations for this answer: $x \setminus \in \left[0 , 6\right]$


text:
# Even Odd or Neither

If f(x) and g(x) are both even functions, is f + g even? If f(x) and g(x) are both odd functions, is f + g odd? What if f(x) is even and g(x) is odd?

Now intuitively I get that for the first one, the resulting answer will be EVEN

For second one it will be ODD

And for the last one it will be Neither.

But I don't know how to justify that... it just came to me intuitively, Can someone show me a way through al-gebra to prove my claim.

-
use the fact (f+g)(x)=f(x)+g(x) and the definition of odd/even function :-) –  Ale Jan 15 '14 at 20:35
The easiest way to prove the 3rd one is to use an example. $f(x)=1$ and $g(x)=x$. You can prove that $h(x)=x+1$ is neither odd nor even by comparing with $h(-x)$. –  ex0du5 Jan 15 '14 at 20:38

For the first two, just use the definition of even and odd to draw the desired conclusion. You will not be able to justify your conclusion on the last one. The sum of an even function and an odd function maybe even, odd, both, or neither. Bear in mind that the constant 0 function is both even and odd, as that should help you construct explicit examples for each of the four possibilities. For example, consider $f(x)=x^2$ and $g(x)=x^3.$ Then $f$ is even and $g$ is odd, but $$(f+g)(-1)=0\neq2=(f+g)(1)$$ and $$(f+g)(-1)=0\neq-2=-(f+g)(1),$$ so $f+g$ is neither even nor odd. I leave the other three cases to you.

-
oh okay I see.. so if f(x) was the zero function the last problem will be an odd function.... but if g(x) was the zero function it would be an even? –  Naqvi Jan 15 '14 at 20:43
Precisely correct! –  Cameron Buie Jan 15 '14 at 20:45
Actually,we can say a bit more. Assuming that $f$ and $g$ are functions $E\to\Bbb R$ for some $E\subseteq\Bbb R,$ where $f$ is even and $g$ is odd, we have that $f+g$ is even if and only if $g$ is the constant $0$ function on $E$, and that $f+g$ is odd if and only if $f$ is the constant $0$ function on $E.$ That's a bit more complicated to prove, though. –  Cameron Buie Jan 15 '14 at 20:51

Let $f(x)$ and $g(x)$ be even functions. Then we have $$(f+g)(-x) = f(-x) + g(-x) = f(x)+g(x) = (f+g)(x)$$

Let $f(x)$ and $g(x)$ be odd functions. Then we have $$(f+g)(-x) = f(-x)+g(-x) = -f(x) -g(x) = -(f+g)(x)$$

Let $f(x)$ be an even function and $g(x)$ be an odd function. Then we have $$(f+g)(-x) = f(-x)+g(-x) = f(x) - g(x) = (f-g)(x)$$ We can't say whether $f-g$ is even or odd.

-
This only transforms the 3rd problem to an equivalent intuition. –  ex0du5 Jan 15 '14 at 20:36
Oh okay that makes sence... as we are assuming here... okay what you did. Alright thanks ! –  Naqvi Jan 15 '14 at 20:38
@ex0du5.. what do you mean? –  Naqvi Jan 15 '14 at 20:38
I suspect that ex0du5 means that nothing has actually been shown by your manipulations in the third case. –  Cameron Buie Jan 15 '14 at 20:40
@CameronBuie.. really? but it shows that (f-g)(x) is nether -(f+g)(x) nor is it (f+g)(x).... hence neither even nor odd ? –  Naqvi Jan 15 '14 at 20:42


text:
# Randomized quicksort expected running time analysis

I am following the quicksort analysis in CLRS (pp. 181-184, 3rd edition). Let me summarize the setting of the analysis.

### Setting in CLRS

First let $Z = \{z_1, ..., z_n\}$ be the set of elements of the input array $A$ in sorted order. $X$ is defined as the number of comparison we make (i.e., between pivot and an element). Because any two elements of the array $A$ is compared at most once, we have $X = \sum_{i=1}^{n-1} \sum_{j = i+1}^{n} X_{ij}$ where $X_{ij} = 1$ if $z_i$ and $z_j$ are compared. So we just need to calculate the expectation, $\mathbb{E} \left[ X \right] = \sum_{i=1}^{n-1} \sum_{j=i+1}^{n} \mathbb{P} \left(X_{ij} = 1\right)$.

So we just need to calculate $\mathbb{P} \left(X_{ij} = 1\right)$. According to the book, $\mathbb{P} \left(X_{ij} = 2\right) = \frac{1}{j - i + 1}$ and I don't follow this result.

### My analysis

Let $k$ be the index of the randomly selected pivot. Conditioning on the event that $k \in Z_{ij}$ where $Z_{ij} = \{z_i, z_{i+1}, ..., z_j\}$,

\begin{aligned} \mathbb{P} \left(X_{ij} = 1\right) &= \mathbb{P} \left( \left.X_{ij} = 1 \right| k \in Z_{ij}\right) \mathbb{P} \left(k \in Z_{ij} \right) + \mathbb{P} \left( \left.X_{ij} = 1 \right| k \not\in Z_{ij}\right) \mathbb{P} \left(k \not\in Z_{ij} \right) \\ &= \frac{2}{j - i + n} \frac{j - i + 1}{n} + \mathbb{P} \left( \left.X_{ij} = 1 \right| k \not\in Z_{ij}\right) \left(1 - \frac{j - i + 1}{n} \right). \end{aligned}

From this, we can see that if $\mathbb{P} \left( \left.X_{ij} = 1 \right| k \not\in Z_{ij}\right) = \mathbb{P} \left(X_{ij} = 1 \right)$, then indeed $\mathbb{P} \left(X_{ij} = 1 \right) = \frac{2}{j-i+n}$ agreeing the book's result. But I don't think that equality holds.

Suppose we choose pivot $k > j$ which is not in $Z_{ij}$. Then, we reduce the problem size to $\{1, ..., k-1\}$. Because $z_i$ and $z_j$ still belong to the same partition, we are given another chance to pick another pivot such that it is either $i$ or $j$. So I would expect that $\mathbb{P} \left( \left.X_{ij} = 1 \right| k \not\in Z_{ij}\right) = \mathbb{P} \left(X_{ij} = 1 \right)$ does not hold.

### Question

Is my analysis correct, i.e., $\mathbb{P} \left( \left.X_{ij} = 1 \right| k \not\in Z_{ij}\right) \not= \mathbb{P} \left(X_{ij} = 1 \right)$? If not, how to derive $\mathbb{P} \left(X_{ij} = 1 \right)$ considering the fact that if the chose pivot is not in $Z_{ij}$ we are given another chance?

### Update

Yuval Filmus' answer is very good and complete in itself. But let me just add a little more step just for completeness. In particular steps leading to $\frac{2}{n} + \frac{2}{n} \cdot \frac{i-1}{j-i+1} + \frac{2}{n} \cdot \frac{n-j}{j-i+1}$ were not trivial to me. So hopefully this may help others.

As Yuval Filmus said we can prove that $p(n,i,j) = 2/(j-i+1)$ using (strong) induction (over $n$). First check the base case which is $n = 2$ because we need at least two elements in order to even think about comparing $z_i$ and $z_j$. With $n = 2$, $j = 2$, $i = 1$ and $p(2,1,2) = \frac{2}{2 - 1 + 1} = 1$. If we have 2 elements, then pivot has to be one of them, and $\mathbb{P} \left(X_{ij} =1; 2\right) = 1$. This proves the base case.

Now consider the inductive step. Assume that $p(m,i,j) = \frac{2}{j-i+1}$ for all $m < n$. In particular, note that all the values of the first argument of $p(\cdot, \cdot, \cdot)$ inside both summations are less than $n$. Hence, we can plug in $p(m,i,j) = \frac{2}{j-i+1}$.

\begin{aligned} p(n,i,j) &= \frac{2}{j - i + 1} \frac{j - i + 1}{n} + \frac{1}{n} \sum_{k=1}^{i-1} \frac{2}{(j-k)-(i-k)+1} + \frac{1}{n} \sum_{k=j+1}^{n} \frac{2}{j-i+1} \\ &= \frac{2}{j - i + 1} \frac{j - i + 1}{n} + \frac{2}{n} \frac{i-1}{j-i+1} + \frac{2}{n} \frac{n-j}{j-i+1}, \end{aligned}

which is the expression given by Yuval Filmus. The rest of the step is in his answer.

Let us denote by $p(n,i,j)$ the probability that $X_{ij} = 1$ given that there are $n$ elements.
Let $k$ be the first chosen pivot. If $k \in \{i,j\}$ then $X_{ij} = 1$. If $i < k < j$, then $X_{ij} = 0$. Otherwise, if $k < i$ then $X_{ij} = 1$ with probability $p(n-k, i-k, j-k)$, and if $k > j$ then $X_{ij} = 1$ with probability $p(k-1, i, j)$. Altogether, we get $$p(n,i,j) = \frac{2}{n} + \frac{1}{n} \sum_{k=1}^{i-1} p(n-k, i-k, j-k) + \frac{1}{n} \sum_{k=j+1}^n p(k-1,i,j).$$ We can now prove that $p(n,i,j) = 2/(j-i+1)$ by induction, by noticing that $$\frac{2}{n} + \frac{2}{n} \cdot \frac{i-1}{j-i+1} + \frac{2}{n} \cdot \frac{n-j}{j-i+1} = \\ \frac{2}{n} + \frac{2}{n} \cdot \frac{n-(j-i+1)}{j-i+1} = \\ \frac{2}{n} + \frac{2}{j-i+1} - \frac{2}{n} = \\ \frac{2}{j-i+1}.$$ Another way to get this result is to follow the algorithm until the first time when a pivot (at any level of recursion) in $\{i,\ldots,j\}$ gets selected. Elements $i$ and $j$ are compared if and only if this pivot happens to be $i$ or $j$, which happens with probability $2/(j-i+1)$.


text:
What was the result of the Korean War?

Question:

What was the result of the Korean War?

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# Plots Made for the RoPP Review Article on Dark Energy

This page contains original plots made for our Reports on Progress in Physics review paper on dark energy.

D. Huterer and D.L. Shafer, "Dark energy two decades after: Observables, probes, consistency tests" (PDF ), Rep. Prog. Phys., 2017

## Hubble diagram with current (binned) SNIa and BAO data:

[Caption:] Evidence for the transition from deceleration in the past to acceleration today. The blue line indicates a model that fits the data well; it features acceleration at relatively late epochs in the history of the universe, beginning a few billion years ago but still billions of years after the big bang. For comparison, we also show a range of matter-only models in green, corresponding to $0.3 \leq \Omega_m \leq 1.5$ and thus spanning the open, flat, and closed geometries without dark energy. Finally, the red curve indicates a model that \emph{always} exhibits acceleration and that also does not fit the data. The black data points are binned distance moduli from the Supercal compilation \cite{Supercal} of 870~SNe, while the three red data points represent the distances inferred from the most recent BAO measurements (BOSS DR12 \cite{Alam:2016hwk}).

## Energy densities vs. redshift:

[Caption:] Energy density of species in the universe as a function of $(1 + z)$, where $z$ is the redshift. The dashed vertical line indicates the present time ($z = 0$), with the past to the left and future to the right. Note that matter ($\propto (1 + z)^3$) and radiation ($\propto (1 + z)^4$) energy densities scale much faster with the expanding universe than the dark energy density, which is exactly constant for a cosmological constant $\Lambda$. The shaded region for dark energy indicates the energy densities allowed at 1$\sigma$ (68.3\% confidence) by combined constraints from current data assuming the equation of state is allowed to vary as $w(z) = w_0 + w_a \, z/(1 +z)$.

## $\Omega_M$-$w$ plane, current constraints:

[Caption:] Constraints on cosmological parameters from our analysis of current data from three principal probes: SN Ia (JLA \cite{Betoule:2014frx}; blue), BAO (BOSS DR12 \cite{Alam:2016hwk}; green), and CMB (\textit{Planck} 2015 \cite{Ade:2015xua}; red). We show constraints on $\Omega_m$ and constant $w$. The contours contain 68.3\%, 95.4\%, and 99.7\% of the likelihood, and we assume a flat universe.

## $\Omega_M$-$w$ plane, historical constraints (1999-2009-2015):

[Caption:] History of constraints on key dark energy parameters $\Omega_m$ and a constant equation of state $w$, assuming a flat universe such that $\Omega_\text{de} = 1 - \Omega_m$. The three sets of contours show the status of measurements around the time of dark energy discovery (circa 1998; green), roughly a decade later following precise measurements of CMB anisotropies and the detection of the BAO feature (circa 2008; red), and in the present day, nearly two decades after discovery (circa 2016; blue).

## $w_0$-$w_a$ plane, current constraints:

[Caption:] Constraints on cosmological parameters from our analysis of current data from three principal probes: SN Ia (JLA \cite{Betoule:2014frx}; blue), BAO (BOSS DR12 \cite{Alam:2016hwk}; green), and CMB (Planck 2015 \cite{Ade:2015xua}; red). We show constraints on $w_0$ and $w_a$ in the parametrization from Eq. (17), marginalized over $\Omega_m$. The contours contain 68.3\%, 95.4\%, and 99.7\% of the likelihood, and we assume a flat universe.

## Illustration of the quantities in the $(w_0, w_a)$ parametrization:

[Caption:] Illustration of the main features of the popular parametrization of the equation of state \cite{Linder_wa,Chevallier_Polarski} given by $w(z) = w_0 + w_a \, z/(1 + z)$. We indicate the pivot redshift $z_p$, the corresponding value of the equation of state $w_p$, the intercept $w_0$, the slope (proportional to $w_a$), and a visual interpretation of the approximate uncertainties in $w_0$ and $w_p$.

## Effect of DE on cosmological observables:

[Caption:] Dependence of key cosmological observables on dark energy. The top left and right panels show, respectively, the comoving distance and growth suppression (relative to the matter-only case). The bottom left and right panels show, respectively, the CMB angular power spectrum $\mathcal{C}_\ell$ as a function of multipole $\ell$ and the matter power spectrum $P(k)$ as a function of wavenumber $k$. For each observable, we indicate the prediction for a fiducial $\Lambda$CDM model ($\Omega_m = 0.3$, $w = -1$) and then illustrate the effect of varying the indicated parameter. In each case, we assume a flat universe and hold the combination $\Omega_m h^2$ fixed.

## Current constraints in the $f$-$\sigma_8$ plane:

[Caption:] Constraints on the quantity $f\sigma_8$ at different redshifts from RSD and peculiar velocity surveys. At the lowest redshifts $z \approx 0$, peculiar velocities from galaxies and SNe Ia (leftmost \cite{Johnson:2014kaa} and rightmost \cite{Huterer:2016uyq} red points) and SNe Ia alone (purple data point; \cite{Turnbull:2011ty}) constrain the velocity power spectrum and effectively the quantity $f\sigma_8$. At higher redshifts, constraints on $f\sigma_8$ come from the RSD analyses from 6dFGS (maroon at $z = 0.067$; \cite{Beutler:2012px}), GAMA (pink points; \cite{Blake:2013nif}), WiggleZ (dark green; \cite{Blake:2011rj}), BOSS (dark blue; \cite{Beutler:2016arn}), and VIPERS (orange; \cite{delaTorre:2013rpa}). The solid line shows the prediction corresponding to the currently favored flat $\Lambda$CDM cosmology.

## Cluster counts:

[Caption:] Predicted cluster counts for a survey covering 5,000~deg$^2$ that is sensitive to halos more massive than $10^{14} \, h^{-1} M_\odot$, shown for a fiducial $\Lambda$CDM model as well as three variations as in Figure ( "Effect of DE on observables" above ).


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ACTA issues

## A divergence criterion and an elementary proof of the divergence of ergodic averages along special subsequences

Minh Dzung Ha

Acta Sci. Math. (Szeged) 68:3-4(2002), 697-703
2863/2009

 Abstract. Consider ${\bf T}=\{z \in {\bf C}:|z|=1\}$, the unit circle with the usual normalized arc-length measure ${\cal L}$. We give a simple sufficient condition (a Divergence Criterion), with a completely self-contained and elementary proof, for the divergence of ergodic averages along subsequences in ${\bf N}$. As an application, we give a very elementary argument of the following result. Let $(n_k)_1^\infty$ be any increasing sequence in ${\bf N}$ with strictly increasing gaps, i.e., $n_{k+1}-n_{k}>n_{k}-n_{k-1}, k\geq 2$. Let $0<\rho<1$ be given. Then there exists an ergodic rotation $\tau \colon {\bf T}\to {\bf T}$ such that for any given $\epsilon >0$, there are infinitely many $f \in L^\infty({\bf T})$ satisfying $${\cal L}\Big( \big\{z \in {\bf T}: \overline{\lim}{1 \over l}\sum_{k=1}^{l}f \circ \tau^{10^{n_k}}(z)- \underline{\lim}{1 \over l}\sum_{k=1}^lf \circ\tau^{10^{n_k}}(z) \geq \rho \big\}\Big)\geq 1 -\epsilon.$$ AMS Subject Classification (1991): 28D99, 60F99 Received March 27, 2001. (Registered under 2863/2009.)


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#### Vol. 5, No. 2, 2012

 Recent Issues
 The Journal About the Journal Editorial Board Editors’ Interests Subscriptions Submission Guidelines Submission Form Policies for Authors Ethics Statement ISSN: 1948-206X (e-only) ISSN: 2157-5045 (print) Author Index To Appear Other MSP Journals
Dispersion and controllability for the Schrödinger equation on negatively curved manifolds

### Nalini Anantharaman and Gabriel Rivière

Vol. 5 (2012), No. 2, 313–338
##### Abstract

We study the time-dependent Schrödinger equation $ı\frac{\partial u}{\partial t}=-\frac{1}{2}\Delta u$, on a compact Riemannian manifold on which the geodesic flow has the Anosov property. Using the notion of semiclassical measures, we prove various results related to the dispersive properties of the Schrödinger propagator, and to controllability.

##### Keywords
Schrödinger equation, semiclassical analysis, control theory
Primary: 35B37


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# Equations with rational degrees

#### mathmaniac

##### Active member
Is there a way to make these a polynomial?
1)$$\displaystyle \sqrt{A}+\sqrt{B}+\sqrt{C}=D$$
2)A1/3 + B1/3 = C
Or how else can it be solved?

#### MarkFL

Staff member
One can always appropriately arrange, then raise both sides to an exponent that changes exponents to integral values. This may have to be repeated to complete the process. If course, then solutions must be checked, since extraneous solutions may be introduced.

#### mathmaniac

##### Active member
Edit

1)$$\displaystyle \sqrt{A}+\sqrt{B}+\sqrt{C}+\sqrt{D}=E$$

Please show how to solve modified 1) and 2),i.e, make them polynomial

#### MarkFL

Staff member
I'll demonstrate the process for 2), and this will give you an idea how to do 1).

2.) $$\displaystyle \sqrt[3]{A}+\sqrt[3]{B}=C$$

Cube both sides to get:

$$\displaystyle A+3\sqrt[3]{A^2B}+3\sqrt[3]{AB^2}+B=C^3$$

$$\displaystyle 3\sqrt[3]{AB}\left(\sqrt[3]{A}+\sqrt[3]{B} \right)=C^3-(A+B)$$

The original equation allows us to simplify as:

$$\displaystyle 3C\sqrt[3]{AB}=C^3-(A+B)$$

Cube again:

$$\displaystyle 27ABC^3=C^9-3C^6(A+B)+3C^3(A+B)^2+(A+B)^3$$

#### mathmaniac

##### Active member
Nice
Please give an answer for 1) too
I tried and failed

General Question:Can this be done for any eq with rational powers?

#### HallsofIvy

##### Well-known member
MHB Math Helper
Nice
Please give an answer for 1) too
I tried and failed

General Question:Can this be done for any eq with rational powers?
It should be obvious that the answer is yes.

As for your previous question please show what you tried.

#### mathmaniac

##### Active member
Squaring,we get
$$\displaystyle 2\sqrt{AB}+2\sqrt{AC}+2\sqrt{AD}+2\sqrt{BC}+2\sqrt{BD}+2\sqrt{CD}....$$

I can take common $$\displaystyle 2\sqrt{A}$$and sub for $$\displaystyle \sqrt{B}+\sqrt{C}+\sqrt{D}$$

Whatever I tried to do then didn't help....

Whoever's going to reply only needs to show how to reduce the number of rooted terms,make it less than 4...
I think there's some trick or maybe just my problem.

#### Opalg

##### MHB Oldtimer
Staff member
Squaring,we get
$$\displaystyle 2\sqrt{AB}+2\sqrt{AC}+2\sqrt{AD}+2\sqrt{BC}+2\sqrt{BD}+2\sqrt{CD}....$$

I can take common $$\displaystyle 2\sqrt{A}$$and sub for $$\displaystyle \sqrt{B}+\sqrt{C}+\sqrt{D}$$

Whatever I tried to do then didn't help....

Whoever's going to reply only needs to show how to reduce the number of rooted terms,make it less than 4...
I think there's some trick or maybe just my problem.

Take one term across to the other side before squaring: $\sqrt A + \sqrt B = D - \sqrt C$.

#### mathmaniac

##### Active member
Take one term across to the other side before squaring: $\sqrt A + \sqrt B = D - \sqrt C$.
I meant the modified 1
See my edit.

#### Opalg

##### MHB Oldtimer
Staff member
I meant the modified 1
See my edit.
[I should have read the whole thread – I didn't see the modification.]

Try an inductive method. Start with the equation $\sqrt{A}+\sqrt{B}+\sqrt{C}=F$. That only has three roots, so you (presumably) know how to remove them to get a polynomial equation in powers of $A$, $B$, $C$ and $F$. Now substitute $F = E - \sqrt D$ into that equation. You will then have an equation in which the only radical is $\sqrt D$. So collect all the terms with $\sqrt D$ onto one side of that equation and square again.

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Is there a way to make these a polynomial?
1)$$\displaystyle \sqrt{A}+\sqrt{B}+\sqrt{C}=D$$
2)A1/3 + B1/3 = C
Or how else can it be solved?
Alternatively, substitute $A=x^6, B=y^6, C=z^6$ with $x,y,z \ge 0$.
Then you get:
1) $$\displaystyle x^3+y^3+z^3=D$$
2) $$\displaystyle x^2+ y^2= z^6$$

#### mathmaniac

##### Active member
Alternatively, substitute $A=x^6, B=y^6, C=z^6$ with $x,y,z \ge 0$.
Then you get:
1) $$\displaystyle x^3+y^3+z^3=D$$
2) $$\displaystyle x^2+ y^2= z^6$$
No,I was thinking about A,B,C...representing polynomial functions on a variable (say x) but RHS a constant.There you can't do these kind of techniques when your purpose is to solve for x.

But nice (funny?) answer,for my less informative post.

#### mathmaniac

##### Active member
[I should have read the whole thread – I didn't see the modification.]

Try an inductive method. Start with the equation $\sqrt{A}+\sqrt{B}+\sqrt{C}=F$. That only has three roots, so you (presumably) know how to remove them to get a polynomial equation in powers of $A$, $B$, $C$ and $F$. Now substitute $F = E - \sqrt D$ into that equation. You will then have an equation in which the only radical is $\sqrt D$. So collect all the terms with $\sqrt D$ onto one side of that equation and square again.
Very very nice trick,impressive!!!

using this,i can make any eq with only degree as 1/2 a polynomial no matter,how many terms there are...

I'll be soon bringing another question about rational degrees(maybe) and I think the general question can also be solved by induction.Right???
Let me see.

Last edited:


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# What Is Xavier Initialization?

The 365 Team 1 Oct 2021 8 min read

As humans, we exercise regularly in order to stay in shape. In technical terms, we’re optimizing our bodies by training and gaining muscle mass. Well, the same can be said for machine learning and deep learning models. While you won’t see a deep neural network hitting the gym, it still requires an equal amount of exercise as any regular Joe looking to get into shape.

Training ensures that our models will be optimized to their maximum performance. But, just like you wouldn’t bench 100kg on your first trip to the gym, you wouldn’t expect your model to pull the same weight on its first go either – not without going through the proper procedures first. That is why it’s important not to skip certain steps, such as initialization.

In this tutorial, we’ll introduce you to the concept of initialization, show you exactly why this process is necessary, and explain what the state-of-the-art Xavier initialization is all about.

## What Is Initialization?

Without beating around the bush, initialization is a crucial part of machine learning.

As we already mentioned, training is highly important. We have to use the appropriate methods on our models. Using clumsy or inappropriate methods results in poor performance – in such cases, even the fastest computer in the world won’t be able to help. They say that the devil is in the details, and this saying is as true as it gets when talking about initialization.

Initialization is the process of setting the initial values of weights for our models, neural networks or other deep learning architecture. You’re probably wondering why that matters – surely these initial weights don’t matter, right? Well, it’s quite the opposite actually.

## Why Is Initialization Important?

For starters, an inappropriate initialization would lead to an unoptimizable model.

Choosing the right initializer is important to our model’s performance and training. In addition, we’d be also shortening the convergence time, as well as minimizing the loss function by setting our initial weights. As you can see, the benefits are plenty.

But in case you’re still not convinced, let us illustrate the importance of initialization with an example of a model with a single hidden layer:

Let’s initialize our weights and biases in such a way they are equal to a constant – it doesn’t matter which one:

As you can see, the three hidden units are completely symmetrical to the inputs.

Each hidden unit is a function of one weight coming from x1 and one from x2. If all of these weights are equal, there is no reason for the algorithm to learn that h1, h2, and h3 are different. Forward propagating, there is no reason for the algorithm to think that even our outputs are different:

Based on this symmetry, when we’re backpropagating, all the weights are bound to be updated without distinguishing between the different nodes in the net. Some optimization would still take place, so it won’t be the initial value. Still, the weights would remain useless.

So, how are we supposed to initialize the weights then? We’ll look at a few types of initialization next.

## Types of Simple Initializations

Now that we know initialization matters, let’s see how we can deal with it.

### Random Initialization

A simple approach would be to initialize weights randomly within a small range. We’ll use the NumPy method: random uniform with a range between minus 0.1 and 0.1.

Even though the initialization with random weights picks the values indiscriminately, they are still chosen in a uniform manner. That means each one has the exact same probability of being chosen – it might sound intuitive, but it is important to stress it.

### Normal (Naïve) Initialization

Another way we could go about this is by choosing a normal initializer. The idea is basically the same. This time, though, we pick the numbers from a zero-mean normal distribution. The chosen variance is arbitrary, but should be small. As you can guess, since it follows the normal distribution, values closer to 0 are much more likely to be chosen than other values.

An example of such initialization is to draw from a normal distribution with a mean 0 and a standard deviation 0.1:

Our initial weights and biases will be picked randomly from the interval [-0.1, 0.1] in a random normal manner, where the mean is 0 and the standard deviation is 0.1 (variance 0.01).

Although they were the norm until 2010, both methods are somewhat problematic, as they use the sigmoid function. It was only recently that academics came up with a solution.

Let’s explore what the problem with the sigmoid activation function is.

## What Are the Disadvantages of the Sigmoid Function?

Here is an example that illustrates why sigmoid could be “bad”.

Weights are used in linear combinations which we then activate:

In this case, we use the sigmoid activator. This function, like other commonly used non-linearities, is peculiar around its mean and its extremes:

Activation functions take as inputs the linear combination of the units from the previous layer, right? Well, if the weights are too small, this will cause values that fall around this range:

As you can see, unfortunately, the sigmoid is almost linear. If all our inputs are in this range (which will happen if we use small weights), our chosen function won’t apply a non-linearity to the linear combination as we want it to. Instead, it’ll apply a linearity:

Such an outcome is not ideal as non-linearities are essential for deep networks.

If values are too large or too small, however, the sigmoid is almost flat, meaning the output will be only 1s or only 0s, respectively:

A static output of the activations will minimize the gradient while we still haven’t fully trained the algorithm.

So, what we want here is a wide range of inputs for the sigmoid. These inputs depend on the weights, which will have to be initialized in a reasonable range so we can have a nice variance along the linear combinations:

We can achieve this by using a more advanced strategy – the Xavier initialization.

## What Is Xavier Initialization?

This method is also known as Glorot initialization. But first, who are Xavier and Glorot?

The truth is, this is one person actually - Xavier Glorot. He is the only academic (that we know of) whose work is named after his first name, rather than his last. Quite intriguing, isn’t it?

Mr. Glorot – or simply Xavier – proposed this method in 2010 and it was quickly adopted on a large scale. This type of initialization is a state-of-the-art technique that we believe anyone interested in neural networks should get properly acquainted with.

So, let’s view an example of how Xavier initialization works.

## Example of Xavier Initialization in TensorFlow

We arbitrarily chose the range for the first two cases, right? Well, the Xavier initialization addresses this issue itself.

The main idea is that the method used for randomization isn’t so important. It is the number of outputs in the following layer that matters. With the passing of each layer, the Xavier initialization maintains the variance in some bounds so that we can take full advantage of the activation functions.

There are two formulas for this strategy.

Uniform Xavier initialization: draw each weight, w, from a random uniform distribution in in [-x,x] for $x = \sqrt {\frac {6}{inputs\,+\,outputs}}$

Normal Xavier initialization: draw each weight, w, from a normal distribution with mean of 0, and a standard deviation $\sigma = \sqrt {\frac {2}{inputs\,+\,outputs}}$

### Uniform Xavier Initialization

The Uniform Xavier initialization states we should draw each weight w from a random uniform distribution in the range from minus x to x, where x is equal to square root of 6, divided by the number of inputs, plus the number of outputs for the transformation.

### Normal Xavier Initialization

For the normal Xavier initialization, we draw each weight w from a normal distribution with a mean of 0, and a standard deviation equal to 2, divided by the number of inputs, plus the number of outputs for the transformation.

The numerator values 2 and 6 vary across sources, but the main idea is the same.

### Why Are Inputs and Outputs in Xavier Initialization Important?

Another detail we should highlight here is that the number of inputs and outputs matters.

Outputs are clear – that’s where the activation function goes. So, the higher the number of outputs, the higher the need to spread weights:

What about inputs? Well, since we achieve optimization through backpropagation, we would obviously have the same problem, but in the opposite direction:

Finally, this is the default initializer in TensorFlow:

So, if you initialize the variables without specifying how, the model will automatically adopt the Xavier initializer.

## Xavier Initialization: Next Steps

Deep learning is an evolving practice. It’s important to stay on top of the trends and follow the adopted best practices that will ensure you’re on the right track, moving forward.

As we’ve seen, initialization is a crucial step in training models or neural networks. Moreover, the Xavier initialization is an innovative method that will not only save you time, but also expertly initialize your model’s weights by taking on the brunt of the work. So, it’s more than a good idea to explore it as you go along your machine learning and deep learning journey.


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# [Haskell-cafe] OutsideIn(X) question

Alejandro Serrano Mena trupill at gmail.com
Wed Feb 26 15:07:56 UTC 2014

Dear Haskell-café,
I don't know if this is the best forum to ask questions about the
OutsideIn(X) paper that lies below type inference in GHC. But given that
this is the place where I know many GHC devs look at, I'll just try :)

My question related about the proof of soundness and principality,
specifically Lemma 7.2 (to be found in page 67). In that lemma, it's stated
that QQ and \phi' Q_q ||- \phi Q_w <-> \phi' Q_w'. I'm trying to recover
the proof (which is omitted in the text), but I stumble upon a wall when
trying to work out what happens in the case an axiom is applied.

In particular, I'm playing with an example where

QQ (the set of axioms) = { forall. C a => D a } (where C and D are
one-parameter type classes)
Q_q = { }
Q_w = { D Int }

Thus, if I apply the rule DINSTW (to be found in page 65), I get a new

Q_w' = { C Int }

Now, if the lemma 7.2 is true, it should be the case that

(1)  QQ ||- C Int <-> D Int

which in particular means that I have the two implications

(2)  { forall. C a => D a, C Int } ||- D Int
(3)  { forall. C a => D a, D Int } ||- C Int

(2) follows easily by applying the AXIOM rule of ||- (as shown in page 54).
However, I don't see how to make (3) work :(
I think that understanding this example will be key for my understanding of
the whole system.

Anybody could point to the error in my reasoning or to places where I could


text:
# How to get line-breaking / multirow to work in my beginner's table?

I can't seem to get the multirow or pbox to give me a line break (of "automatic" width). The following won't compile successfully: note the two variations in the "TOTALLY" line. How can I fix it?

\documentclass{article}
\usepackage{tabulary}
\usepackage{threeparttable}
\usepackage{array}
\usepackage{siunitx}
\usepackage{booktabs}
\usepackage{multirow}
\usepackage{pbox}
\usepackage{lipsum}

\renewcommand{\arraystretch}{1.2}

\begin{document}

\lipsum[1]

\begin{table}
\begin{threeparttable}
\caption{this is my first table's captive caption}
\begin{tabulary}{\textwidth}{@{}*{2}{L}*{6}{S[table-format=3.2]}@{}} \toprule
× & TOTALLY bla percentage of bla & {1000} & {2000} & {3000} & \pbox[t]{1.5cm}{Bonjour monde!} & {5000} & {6000\tnote{1}}\\ \midrule
% × & TOTALLY bla percentage of bla & {1000} & {2000} & {3000} & \multirow{2}*{4000 apples and pears \\ and whatnot} & {5000} & {6000\tnote{1}}\\ \midrule
DDD 1 & 47.6 & 29.1 & 1.0 & 0.2 & 1.9 & 15.2 & 0.0\\
UUU & 24.8 & 10.8 & 6.4 & 0.0 & 3.2 & 4.5 & 0.0\\
× & × & × & × & × & × & × & ×\\
Unweighted average: & × & × & × & × & × & × & ×\\
BBB & 33.8 & 11.3 & 9.1 & 0.4 & 1.8 & 11.0 & 0.2\\
GGG & 32.9904 & 8.60325 & 9.3845 & 0.0495 & 1.43225 & 10.79525 & 0.119\\
DDD & 39.4545 & 9.8695 & 15.3365 & 0.6915 & 2.246 & 10.6705 & 0.5105\\ \bottomrule
\end{tabulary}

\begin{tablenotes}
\item [1] the first note ...
\end{tablenotes}

\end{threeparttable}
\end{table}

\lipsum[2]

\end{document}

related (but didn't help me, yet):

so this question is probably just yet-another-beginner's-desperate-duplicate... plz forgive!

Edit

just to supply a working example (for a quick comparison, note I got rid of the \\ in the uncommented "TOTALLY" line):

\documentclass{article}
\usepackage{tabulary}
\usepackage{threeparttable}
\usepackage{array}
\usepackage{siunitx}
\usepackage{booktabs}
\usepackage{multirow}
\usepackage{pbox}
\usepackage{lipsum}

\renewcommand{\arraystretch}{1.2}

\begin{document}

\lipsum[1]

\begin{table}
\begin{threeparttable}
\caption{this is my first table's captive caption}
\begin{tabulary}{\textwidth}{@{}*{2}{L}*{6}{S[table-format=3.2]}@{}} \toprule
× & TOTALLY bla percentage of bla & {1000} & {2000} & {3000} & {4000 apples and pears and whatnot} & {5000} & {6000\tnote{1}}\\ \midrule
%× & TOTALLY bla percentage of bla & {1000} & {2000} & {3000} & {4000 apples and pears \\ and whatnot} & {5000} & {6000\tnote{1}}\\ \midrule
% × & TOTALLY bla percentage of bla & {1000} & {2000} & {3000} & \pbox[t]{1.5cm}{Bonjour monde!} & {5000} & {6000\tnote{1}}\\ \midrule
% × & TOTALLY bla percentage of bla & {1000} & {2000} & {3000} & {\multirow{2}*{\pbox[t]{1.5cm}{Bonjour monde!}}} & {5000} & {6000\tnote{1}}\\ \midrule
DDD 1 & 47.6 & 29.1 & 1.0 & 0.2 & 1.9 & 15.2 & 0.0\\
UUU & 24.8 & 10.8 & 6.4 & 0.0 & 3.2 & 4.5 & 0.0\\
× & × & × & × & × & × & × & ×\\
Unweighted average: & × & × & × & × & × & × & ×\\
BBB & 33.8 & 11.3 & 9.1 & 0.4 & 1.8 & 11.0 & 0.2\\
GGG & 32.9904 & 8.60325 & 9.3845 & 0.0495 & 1.43225 & 10.79525 & 0.119\\
DDD & 39.4545 & 9.8695 & 15.3365 & 0.6915 & 2.246 & 10.6705 & 0.5105\\ \bottomrule
\end{tabulary}

\begin{tablenotes}
\item [1] the first note ...
\end{tablenotes}

\end{threeparttable}
\end{table}

\lipsum[2]

\end{document}

-
from the multirow manual: If the width was set explicitly, the text will be set in a \parbox of that width; you can use \\ to force linebreaks where you like. Is there a conflict to the other packages, e.g. siunitx ? –  nutty about natty Apr 12 '13 at 13:55
this works: × & TOTALLY bla percentage of bla & {1000} & {2000} & {3000} & \multirow{2}{*}{\parbox{2cm}{Principal actor in rule adoption process}} & {5000} & {6000\tnote{1}}\\ \midrule; so maybe its a problem specific to pbox vs. ??? –  nutty about natty Apr 12 '13 at 14:08

You have a utf08 times in the first row but had not declared any encoding so this was being silently dropped, I added

\usepackage[utf8]{inputenc}\usepackage{textcomp}

so it worked, then easiest way to get a multi-line heading is to use \multicolumn{1} to get out of S and then nest a tabular (I probably should have used \begin{tabular}[t] but the image shows the default vertically centred arrangement.

\documentclass{article}
\usepackage{tabulary}
\usepackage{threeparttable}
\usepackage{array}
\usepackage{siunitx}
\usepackage{booktabs}
\usepackage{multirow}
\usepackage{pbox}
\usepackage{lipsum}
\usepackage[utf8]{inputenc}\usepackage{textcomp}
\renewcommand{\arraystretch}{1.2}

\begin{document}

\lipsum[1]

\begin{table}
\begin{threeparttable}
\caption{this is my first table's captive caption}
\begin{tabulary}{\textwidth}{@{}*{2}{L}*{6}{S[table-format=3.2]}@{}} \toprule
× & TOTALLY bla percentage of bla & {1000} & {2000} & {3000} &
\multicolumn{1}{c}{\begin{tabular}{@{}c@{}}Bonjour\tabularnewline monde!\end{tabular}} &
{5000} & {6000\tnote{1}}\\ \midrule
% × & TOTALLY bla percentage of bla & {1000} & {2000} & {3000} & \multirow{2}*{4000 apples and pears \\ and whatnot} & {5000} & {6000\tnote{1}}\\ \midrule
DDD 1 & 47.6 & 29.1 & 1.0 & 0.2 & 1.9 & 15.2 & 0.0\\
UUU & 24.8 & 10.8 & 6.4 & 0.0 & 3.2 & 4.5 & 0.0\\
× & × & × & × & × & × & × & ×\\
Unweighted average: & × & × & × & × & × & × & ×\\
BBB & 33.8 & 11.3 & 9.1 & 0.4 & 1.8 & 11.0 & 0.2\\
GGG & 32.9904 & 8.60325 & 9.3845 & 0.0495 & 1.43225 & 10.79525 & 0.119\\
DDD & 39.4545 & 9.8695 & 15.3365 & 0.6915 & 2.246 & 10.6705 & 0.5105\\ \bottomrule
\end{tabulary}

\begin{tablenotes}
\item [1] the first note ...
\end{tablenotes}

\end{threeparttable}
\end{table}

\lipsum[2]

\end{document}

-
regarding \\ vs \tabularnewline: tex.stackexchange.com/questions/78796/… –  nutty about natty Apr 12 '13 at 14:20
Regarding pbox vs parbox: Why does × & TOTALLY bla percentage of bla & {1000} & {2000} & {3000} & \multirow{2}{*}{\parbox{2cm}{Principal actor in rule adoption process}} & {5000} & {6000\tnote{1}}\\ \midrule  work, while × & TOTALLY bla percentage of bla & {1000} & {2000} & {3000} & \multirow{2}{*}{\pbox{2cm}{Principal actor in rule adoption process}} & {5000} & {6000\tnote{1}}\\ \midrule fails? –  nutty about natty Apr 12 '13 at 14:27
@nuttyaboutnatty I don't know. pbox is rather fragile I suspect (I've never used it) –  David Carlisle Apr 12 '13 at 14:30
How could I make the second line smaller - or even better: narrower (as in "horizontally squashed" text)? That's "monde!" in our example. Would I need to use this from ctan.org/pkg/array ? If so, how? Else, what other (sensible) options would there be? –  nutty about natty Apr 12 '13 at 15:13
@nuttyaboutnatty \small monde! (also it's best not to ask followup questions in the comments, best to ask new questions) –  David Carlisle Apr 12 '13 at 15:32


text:
Note that Michael Brickenstein's "slimgb" (which is often used in Singular and thus also in Sage) is sometimes referred to as a version of F4. If you create a multivariate ideal J and want to compute its Gröbner basis with a particular algorithm, you can use something like
J.groebner_basis(algorithm='slimgb')