text:
# QED electron radius

1. Mar 19, 2009

### Orion1

Penning trap upper limit for the electron particle radius:
$$r_e \leq 10^{-22} \; \text{m}$$

QED Compton wavelength electron radius:
$$\boxed{r_e = \alpha_e \overline{\lambda}_c}$$

QED Compton electron radius:
$$\boxed{r_e = \frac{\hbar \alpha_e}{m_e c}}$$

QED electron fine structure constant:
$$\alpha_e = \frac{r_e}{\overline{\lambda}}_c = \frac{m_e c r_e}{\hbar}$$

$$\boxed{\alpha_e = \frac{m_e c r_e}{\hbar}}$$

Weak nuclear fine structure constant:
$$\alpha_w = \sqrt{ \frac{T_{\Delta}}{T_{\Sigma}}} = 2.738 \cdot 10^{-7}$$
$$T$$ - Delta and Sigma particle lifetimes

$$\boxed{\alpha_e = 2.589 \cdot 10^{-10}}$$
$$\boxed{\alpha_w = 2.738 \cdot 10^{-7}}$$

The QED electron fine structure constant magnitude corresponds with the weak nuclear fine structure constant.

How is $$\alpha_w$$ measured via weak $$\beta$$ decay?

Are these equations correct?

Reference:
http://en.wikipedia.org/wiki/Electron#Fundamental_properties"
http://en.wikipedia.org/wiki/Weak_nuclear_force" [Broken]
http://hyperphysics.phy-astr.gsu.edu/HBASE/forces/couple.html#c4"
http://hyperphysics.phy-astr.gsu.edu/HBASE/particles/weastr.html#c1"

Last edited by a moderator: May 4, 2017
2. Mar 19, 2009

### malawi_glenn

what do you want?

3. Mar 19, 2009

Staff Emeritus
No, it's not. It's $\alpha/\sin^2\theta_w$.

No, it's not. That doesn't even make any sense, as QED doesn't know anything about the weak nuclear force.

4. Mar 19, 2009

### Orion1

Vanadium, that is an element, not an equation.

Is this the equation that you attempted to define?

Weak nuclear fine structure constant:
$$\alpha_w = \frac{\alpha}{\sin^2 \theta_w}$$
$$\theta_w$$ - Weinberg angle (weak mixing angle)

Please state the numerical value for that solution and cite a reference.

Have you examined my references in post #1?

Why is the university level hyperphysics weak nuclear fine structure constant equation incorrect?

Reference:
http://en.wikipedia.org/wiki/Weinberg_angle" [Broken]

Last edited by a moderator: May 4, 2017
5. Mar 19, 2009

### malawi_glenn

Look in mandl's textbook in Quantum Field theory for instance. Or any book that contains the gauge theory of the electroweak unification.

Also, what is the purpose of your original post? Are you asking something?

6. Mar 19, 2009

### Orion1

Have you examined my references in post #1?

Are my equations listed in post #1 correct?

Anyone have mandl's textbook in Quantum Field Theory?

Last edited: Mar 19, 2009
7. Mar 19, 2009

### malawi_glenn

as Vanadium pointed out, no.

The QED fine structure constant is 1/137

why are you doing all this?

A good, easy, reference on this, I consider particle physics by Martin

And yes, Manls textbook on QFT is quite common to have... your Library should definitely have one copy. Otherwise, there should be millions of tutorials on www, try search for "Antonio Pich" on ArXiV

Your references in post 1 is wikipedia and hyperphysics, what should I say?... they are not rigour

8. Mar 19, 2009

### Orion1

Vanadium only stated one equation as incorrect. You are now stating all the equations are incorrect? Including the university level hyperphysics equations?

Negative, that value for $$\alpha$$ is not 'rigorous'.

malawi_glenn, you are unable to state a numerical value for Mandl and Martin's $$\alpha_w$$.

rig·or·ous:
. precise
. severely accurate

malawi_glenn, based upon the 'incorrect' equations in post #1, what is the numerical value for the electron radius if the QED fine structure constant for electromagnetism is used?

And why exactly do all these equations fail dimensional examination?

Last edited: Mar 19, 2009
9. Mar 19, 2009

### malawi_glenn

why would anyone use "electron radius" since it is not a 'real' concept, just a unit?

ok, the fine structure constant is 1/137.03599907 which is a numerical value of 2.39934 e-4

If you want to define a constant which has the units of length by introducing $$r_e = \alpha \lambda_r$$ you are free to do it, it is just a unit, nothing 'physical'

The relation between 'the classical electron radius' and the electron comptonwavelngth is: $$r_e{\text{ }}_{class} = {\alpha \lambda_e \over 2\pi} = \alpha^2 a_0$$

where a_0 is the bohr radius.

The weak 'fine structure' constant is $\alpha _W \approx 0.58\alpha$ (martin)

in QED, the electron is a point-particle, so don't call it "QED electron radius"..

Last edited: Mar 19, 2009
10. Mar 19, 2009

### Orion1

And why exactly do all these equations fail dimensional examination?

The Penning trap experiment that determined the electron radius upper limit considers the electron as a QED point-particle.

malawi_glenn, then by your own supposition:
$$\alpha_w = \frac{\alpha}{\sin^2 \theta_w} = 0.58 \cdot \alpha = 4.232 \cdot 10^{-3}$$

A rigorously determined value: (Martin)
$$\alpha_w = 4.232 \cdot 10^{-3}$$

Is this equation correct?

malawi_glenn, based upon the 'incorrect' equations in post #1 and post #9, what is the numerical value for the electron radius if the QED fine structure constant for electromagnetism is used?

Reference:
http://en.wikipedia.org/wiki/Penning_trap" [Broken]

Last edited by a moderator: May 4, 2017
11. Mar 19, 2009

### malawi_glenn

because physics is more advanced then this...

e.g. the bohr radius is just a unit, it is not an orbital radius of the electron in hydrogen (as we know, the radial distance in hydrogen is a probabilistic quantity)

the number 1/137 is more rigour than your $$\boxed{\alpha_e = 2.589 \cdot 10^{-10}}$$ ... rigour in that sense that it is close to the experimental measured value.

I mean, what is the physical (first principle reason) for someone to suggest that the electron radius is alpha time electron Compton wavelength? What is the underlying idea?

12. Mar 19, 2009

### Orion1

malawi_glenn, based upon the 'incorrect' equations in post #1 and post #9, what is the numerical value for the electron radius if the QED fine structure constant for electromagnetism is used?

And why exactly do all these equations fail dimensional examination?

Last edited: Mar 19, 2009
13. Mar 19, 2009

### malawi_glenn

you don't have a calculator or what?

and "what" electron radius? There are many kinds of electron radii units, you mean your suggestion $$\boxed{r_e = \alpha_e \overline{\lambda}_c}$$? Which you call "QED compton wavelenght radius"?

14. Mar 19, 2009

### Orion1

$$\boxed{r_e = \alpha_e \overline{\lambda}_c}$$

Affirmative.

malawi_glenn, you still have not stated why exactly do all these equations fail dimensional examination?

Negative, that value for $$\alpha$$ is numerology and not 'rigorous' at all.

Negative, that numerical value is incorrect, and certainly not even close to rigorous.

malawi_glenn, what are the physical (first principle reasons) for these equations?

$$r_e =\frac{\alpha \lambda_e}{2 \pi}$$

Rigorous classical electron radius:
$$r_e = \frac{1}{4 \pi \epsilon_0} \frac{e^2}{m_e c^2} = 2.8179402894(58) \cdot 10^{-15} \; \text{m}$$

Negative, the numerical value for the classical electron radius from Wikipedia is excellent, they are both extremely rigorous.

Negative, I do not believe that I am the one that requires a calculator at the moment.

Reference:

Last edited by a moderator: May 4, 2017
15. Mar 19, 2009

### malawi_glenn

roger copral

16. Mar 19, 2009

Staff Emeritus
What in blazes are you talking about?

17. Mar 19, 2009

### malawi_glenn

Only God knows... he usally does this kind of approach =/

1/137 is not rigour, it is just "numerology" LOL

18. Mar 20, 2009

### Orion1

malawi_glenn's numerical value for the electromagnetic fine structure constant is incorrect.

malawi_glenn, you still have not stated a correct numerical value for the electromagnetic fine structure constant.

Vanadium_50, why is the university level hyperphysics weak nuclear fine structure constant equation incorrect?

Vanadium_50, and why exactly do all these equations fail dimensional examination?

Vanadium_50, is the equation and numerical values correct in post #10?

Reference:
https://www.physicsforums.com/showpost.php?p=2123562&postcount=10"

Last edited by a moderator: Apr 24, 2017
19. Mar 20, 2009

### malawi_glenn

orion, in post #9 I gave you a better (closer to experimental) value of alpha (1/137 is a VERY good number for alpha, orion you are a pain in our *** with your style..

the figure 2.39934 e-4 is of course wrong, i made that on purpose LOL, why can't you just take 1/137.03599907 in your own calculator??

This is an open forum, I can answer to anything I want to.

Why do you want to calculate all this crap anyway?

The value hyperphysics gives you for the weak finestructure is just an approximation, the hadrons are composite objects etc, you have to go to elementary particle level to examine the couplings of interactions. If you want the latest values, go to particle data group, not wikipedia or hyperphysics.

You are the one who proposed that a unit called "QED electron radius" should be invented, you have to motivate "why". I am not the guy who should say on what first principles it relies on.

The classical electron radius is something which you evaluate from classical electrodynamics, when you evaluate Thomson scattering etc. It should be done in great detail in e.g. Jackson.

I now invent a thing called "Glenn electron radius"

$$r_e^{\text{Glenn}} = \alpha_e^3 \frac{1}{4\pi\epsilon_0}\frac{e^2}{m_e c^2}$$

And what does it mean? I have no idea, it is just a unit.

You see my point why all your 'effort' is pointless?

20. Mar 20, 2009

### Orion1

Negative, that equation is dimensionally incorrect.

An accurately measured approximation is not a failure in dimensional examination.

malawi_glenn, instead of deliberately introducing known unbalanced and non-formulated dimensionally incorrect equations on this forum without any referencing at all, why not state exactly why do all these equations fail dimensional examination?

malawi_glenn, you have not stated what 'first principle' the classical electron radius is based upon.

$$r_e = \frac{1}{4 \pi \epsilon_0} \frac{e^2}{m_e c^2}$$

$$\boxed{\frac{e^2}{4 \pi \epsilon_0} = \hbar c \alpha}$$

$$r_e = \frac{1}{4 \pi \epsilon_0} \frac{e^2}{m_e c^2} = \frac{\hbar \alpha}{m_e c} = \alpha \overline{\lambda}_c$$

Electromagnetic fine structure constant:
$$\boxed{\alpha = \frac{e^2}{4 \pi \epsilon_0 \hbar c}} = 7.2973525705 \cdot 10^{-3}$$

$$\boxed{\alpha = 7.2973525705 \cdot 10^{-3}}$$

QED Compton wavelength electron radius:
$$\boxed{r_e = \alpha \overline{\lambda}_c}$$

Reference:
http://en.wikipedia.org/wiki/Fine-structure_constant" [Broken]


text:
# Fitting Incidence Data with Multi-Stage Clonal Expansion Models using msce

## Motivation

The Two-Stage Clonal Expansion Model was invented in the late 70's to explain cancer incidence curves by viewing carcinogenesis as a sequence of stochastic processes like mutations and proliferation of cells (Moolgavkar and Venzon 1979). Later on the model was successively extended to Multi-Stage Clonal Expansion (MSCE) Models (Little, Vineis, and Li 2008) and is regularly used to describe incidence data after time-dependend exposure to risk factors, especially in the field of radiation epidemiology (Rühm, Eidemüller, and Kaiser 2017).

Effects of risk factors on incidence (or mortality) data are fitted in MSCE models based on their presumed effect on disease devlopment. The model depends on so called biological parameters including e.g. mutation or proliferation rates. A risk factor is assumed to affect one or several of these biological parameters and the model relates this (time dependent) change in biological parameters to a change in the incidence risk. The resulting time-dependence of incidence risk may be perceived to be more realistic compared to fitting simple mathematical functions, and some insight between age- or time-dependence of risk and underlying processes may be gained. On the other hand, applying Multi-Stage Clonal Expansion Models for fitting of incidence data can be very intense regarding computing power, especially if the data set contains many strata (or individual person data) and biological parameters depend on age or time. To improve on this issue, the package msce was developed and takes advantage of the package RcppParallel to speed up calculations.

## A glimpse on Multi-Stage Clonal Expansion Models

MSCE models are a sort of Markov process with countably infinite state space. Here, we do not present any mathematical details (Tan 1991) but give only one possible though simplified interpretation of its use in cancer epidemiology. Let's assume there is some mass $$N$$ of potentially cancerogenous cells (e.g. stem cells of a certain organ). Each of these cells has the same probability to acquire a certain transformation (mutation). This happens with rate $$\nu_0$$. Other transformations may follow with rates $$\nu_1$$ ... $$\nu_{s}$$. In the model, each transformation corresponds to the transition into a higher stage and the last stage corresponds to malignant cells. Here, we assume transformation to occur during cell division, i.e. the number of cells in the lower stage is not affected by a transition into a higher stage. With each stage the cell may have gained a proliferative advantage compared to the neighboring tissue thus expanding clonally. Cell division rate in stage $$i$$ is denoted $$\alpha_i$$ and (stem) cell extinction rate (by e.g. differentiation) is denoted $$\beta_i$$. See the figure below for a schematic presentation of MSCE models.

In the msce package, the logarithm of the survival function and the hazard are calculated. The survival function is the age-dependent probability for no malignant cell. As usual, the hazard is the derivative of the negative logarithm of the survival function. Therefore, the survival function is interpreted as the probability of not having (a specific) cancer, the hazard is the modeled incidence rate. In many applications a fixed lag-time is introduced to approximately take into account the time difference between occurrence of the first malignant cell and observation of (or even death from) cancer.

## General cautions on fitting incidence data with Multi-Stage Clonal Expansion Models

In this package, there are essentially three different types of biological parameters: transition rates $$\nu_i$$, $$i=0,1,...,s$$, proliferation rates $$\alpha_i$$ and clonal expansion rates $$\gamma_i$$, $$i=1,...,s$$. Dynamics of the first transition is governed only by the product of the number of potentially cancerogenous cells $$N$$ with $$\nu_0$$. Therefore, the absolute rate of first transitions $$N\nu_0$$ is expected as function argument and not $$N$$ and $$\nu_0$$ separately. Moreover, clonal expansion rates $$\gamma_i$$ are defined by $\gamma_i = \alpha_i-\beta_i$ for $$\beta_i$$ being a death (or differentiation) rate. One might argue which of $$\alpha_i$$, $$\beta_i$$ and $$\gamma_i$$ most naturally describe the disease process. Homeostasis is regulated in the body and not an accidental result of the opposing effects of proliferation and death/differentiation. Deviation from homeostasis between the clone and surrounding tissue is described by $$\gamma_i$$. Therefore, $$\gamma_i$$ are expected as function arguments, together with $$\alpha_i$$ mainly for ease of internal calculations. But in any case, this choice of function arguments does not hinder the user to apply any combination of parameters as fit parameters.

Finally, it has to be noted that not all parameters are identifiable from incidence data. This is true already for the Two-Stage Clonal Expansion Model (Heidenreich, Luebeck, and Moolgavkar 1997) because the incidence curve is insensitive to a certain parameter combination. For models with more stages, best estimates can typically not be calculated for several parameters. Therefore, at least some parameters or combinations thereof need to be fixed in the fitting procedure. It is therefore necessary to have an idea about reasonable values of the parameters. Moreover, parameters may be correlated, and fits with different models may yield similar goodness-of-fit. Therefore knowledge on the modeled processes should be taken into account and results of fitting need to be interpreted cautiously.

## A fitting approach

For this example, a hypothetical lung cancer data set is provided.

library(msce)
data("lungCancerSmoking")

The data set is organized in strata. Each stratum is defined by a range in risk factors and presents one row in the data set. To keep our example simple, we assume only age and smoking cigarettes to be relevant risk factors for lung cancer incidence. Therefore, next to the variables age, there are ageStart and ageQuit for the ages of smoking start and cessation, cigsPerDay for smoking intensity, and pyr and cases for the corresponding number of person years followed up and the number of lung cancer cases that occurred during these person years. For lung cancer, we keep with the simplest, i.e. the Two-Stage Clonal Expansion model. It corresponds to $$s=1$$ in the above figure. A specific function tsce is provided for the Two-Stage Clonal Expansion model based on exact analytical formulas.

But before, there is some preparatory work to do. For lung cancer we may assume a lag-time of 5 years. This is equivalent to the assumption that it takes 5 years for a malignant cell to grow to an observed cancer and means that observation of cancer is not affected by the smoking history within the last five years.

lungCancerSmoking$laggedAge <- lungCancerSmoking$age-5

Define binary variables to classify each stratum to belong to non-smokers, former, or current smokers.

#consider smoking for less than one year as non-smoker
lungCancerSmoking$nonSmoke <- (lungCancerSmoking$ageStart==lungCancerSmoking$ageQuit) | (lungCancerSmoking$ageStart >= lungCancerSmoking$laggedAge) lungCancerSmoking$exSmoke <- (lungCancerSmoking$ageQuit < lungCancerSmoking$laggedAge) &
!lungCancerSmoking$nonSmoke lungCancerSmoking$smoke <- !lungCancerSmoking$nonSmoke & !lungCancerSmoking$exSmoke

Next define a matrix of time intervals during which biological parameters are assumed to be constant and an indicator function for the time intervals smoked. (Only cigarette smoking is assumed to alter biological parameters.)

# t: nonSmoke: 0        0        laggedAge
#    exSmoke:  ageStart ageQuit  laggedAge
#    smoke:    0        ageStart laggedAge
t<-matrix(0,nrow=NROW(lungCancerSmoking),ncol=3)
t[,3] <- lungCancerSmoking$laggedAge t[lungCancerSmoking$smoke,2] <-lungCancerSmoking$ageStart[lungCancerSmoking$smoke]
t[lungCancerSmoking$exSmoke,2] <- lungCancerSmoking$ageQuit[lungCancerSmoking$exSmoke] t[lungCancerSmoking$exSmoke,1] <- lungCancerSmoking$ageStart[lungCancerSmoking$exSmoke]

# smInd: nonSmoke: 0 0 0
#        exSmoke:  0 1 0
#        smoke:    0 0 1
smInd <- matrix(0,nrow=NROW(lungCancerSmoking),ncol=3)
smInd[lungCancerSmoking$smoke,3] <- 1 smInd[lungCancerSmoking$exSmoke,2] <- 1

It is convenient to write a wrapper function to relate fit parameters to the function arguments of tsce.

wrap <-function(pars)
{
# assume alpha to be small and constant
alpha <- matrix(1,nrow=1,ncol=3)

Nnu0 <- matrix(exp(pars[1]),nrow=1,ncol=3)
nu1 <- matrix(exp(pars[2]),nrow=1,ncol=3)

# assume only gamma to depend on smoking
gamma <- matrix(pars[3],nrow=NROW(lungCancerSmoking),ncol=3) +
pars[4]*smInd +
pars[5]*smInd * (lungCancerSmoking$cigsPerDay>5) parList <- list(Nnu0=Nnu0, alpha=alpha,gamma=gamma,nu1=nu1) result <- tsce(t,parList) return (result$hazard)
}

Moreover, we need the objective function to minimize, in this case the normal Poisson log-likelihood

loglik <- function(pars)
{
return (-sum(dpois(lungCancerSmoking$cases, lungCancerSmoking$pyr*wrap(pars), log = TRUE)))
}

Finally everything is set up to fit the model against the data. Here we use nlminb to find the minimum of the log-likelihood

# use upper bounds to ensure gamma < alpha
minResult <- nlminb(start = c(-3,-14,0.1,0.05,0.0), objective = loglik, upper=c(0,0,0.3,0.3,0.3))
bestEstimates <- minResult\$par
bestEstimates
#> [1]  -2.62085682 -12.98733657   0.08941883   0.02084876   0.03663581

Assuming validity of the data and appropriateness of the model and assumptions, this result could be interpreted such that smoking increases the growth advantage of initiated cells from 0.09 to 0.11 per year for smoking with less than 5 cigarettes per day and to 0.15 for more intense smoking. As a side remark it may be noted that similar best estimates would have been obtained with the more general msce_numerical instead of the exact tsce function, a bestEstimate of -2.62 -13.0 0.0898 0.0210 0.0368.

To conclude, let's illustrate the fit result with a plot on the incidence risk for life-long non-smokers, and on the effect of starting intense smoking at age 20, and quitting at age 50.

## References

Heidenreich, W F, E G Luebeck, and S H Moolgavkar. 1997. “Some properties of the hazard function of the two-mutation clonal expansion model.” Risk Anal. 17 (3): 391–99.

Little, Mark P, Paolo Vineis, and Guangquan Li. 2008. “A stochastic carcinogenesis model incorporating multiple types of genomic instability fitted to colon cancer data.” J Theor Biol 254 (2). Elsevier: 229–38.

Moolgavkar, S H, and D J Venzon. 1979. “Two-Event Models for Carcinogenesis: Incidence Curves for Childhood and Adult Tumors.” Math Biosci 47: 55–77.

Rühm, Werner, Markus Eidemüller, and Jan Christian Kaiser. 2017. “Biologically-Based Mechanistic Models of Radiation-Related Carcinogenesis Applied to Epidemiological Data.” International Journal of Radiation Biology 93 (10). Taylor & Francis: 1093–1117. doi:10.1080/09553002.2017.1310405.

Tan, W Y. 1991. Stochastic models of carcinogenesis. New York: Marcel Dekker.

format: plain


text:
UMA001 MATHEMATICS-I

Successive Differentiation: Higher order derivatives, nth derivatives of standard functions, nth derivatives of   rational functions, Leibnitz theorem.

Applications of Derivatives: Mean value theorems and their geometrical interpretation, Cartesian graphing using first and second order derivatives, Asymptotes and dominant terms, Graphing of polar curves, Polar equations for conic sections.

Sequences and Series: Introduction to sequences and Infinite series, Tests for convergence/divergence: Limit comparison test, Ratio test, Root test, Cauchy integral test, Cauchy condensation test. Alternating series, Absolute convergence and conditional convergence.

Series Expansions: Power series, Taylor series, Convergence of Taylor series, Error estimates, Term by term differentiation and integration, Multiplication and division process in power series.

Partial Differentiation: Functions of several variables, Limits and continuity, Chain rule, Change of variables, Partial differentiation of implicit functions, Taylor series of two variables, Directional derivatives and its properties, Maxima and minima by using second order derivatives.

Multiple Integrals: Change of order of integration, Change of variables, Applications of multiple integrals to areas and volumes.

Vector Calculus
: Differentiation and integration of vector valued functions, velocity, acceleration, tangent, principle normal and binormal vectors, Curvature, Torsion and TNB frame. Scalar and vector fields, Gradient, Divergence and Curl. Line integrals, Work, Circulation and Flux. Green’s theorem in Plane, Gauss-divergence and Stoke’s theorem (without proof).

Text Books

1.      Thomas, G.B. and Finney, R.L., Calculus and Analytic Geometry, Pearson Education (2007) 9th ed.

2.      Stewart James, Essential Calculus; Thomson Publishers (2007) 6th ed.

Reference Books

1.      Wider David V, Advanced Calculus: Early Trancedentals, Cengage Learning (2007).

2.      Apostol Tom M, Calculus, Vol I and II John Wiley (2003).


text:
## Algebra & Number Theory

### The smallest prime that does not split completely in a number field

Xiannan Li

#### Abstract

We study the problem of bounding the least prime that does not split completely in a number field. This is a generalization of the classic problem of bounding the least quadratic nonresidue. Here, we present two distinct approaches to this problem. The first is by studying the behavior of the Dedekind zeta function of the number field near $1$, and the second by relating the problem to questions involving multiplicative functions. We derive the best known bounds for this problem for all number fields with degree greater than $2$. We also derive the best known upper bound for the residue of the Dedekind zeta function in the case where the degree is small compared to the discriminant.

#### Article information

Source
Algebra Number Theory, Volume 6, Number 6 (2012), 1061-1096.

Dates
Revised: 8 September 2011
Accepted: 26 September 2011
First available in Project Euclid: 20 December 2017

https://projecteuclid.org/euclid.ant/1513729861

Digital Object Identifier
doi:10.2140/ant.2012.6.1061

Mathematical Reviews number (MathSciNet)
MR2968634

Zentralblatt MATH identifier
1321.11105

#### Citation

Li, Xiannan. The smallest prime that does not split completely in a number field. Algebra Number Theory 6 (2012), no. 6, 1061--1096. doi:10.2140/ant.2012.6.1061. https://projecteuclid.org/euclid.ant/1513729861

#### References

• S. Bessassi, “Bounds for the degrees of CM-fields of class number one”, Acta Arith. 106:3 (2003), 213–245.
• D. A. Burgess, “The distribution of quadratic residues and non-residues”, Mathematika 4 (1957), 106–112.
• H. Davenport, Multiplicative number theory, 3rd ed., Graduate Texts in Mathematics 74, Springer, New York, 2000.
• H. G. Diamond, H. L. Montgomery, and U. Vorhauer, “Beurling primes with large oscillation”, Math. Ann. 334:1 (2006), 1–36.
• M. Einsiedler, E. Lindenstrauss, P. Michel, and A. Venkatesh, “The distribution of periodic torus orbits and Duke's theorem for cubic fields”, Ann. of Math. $(2)$ 173:2 (2011), 815–885.
• A. Granville and K. Soundararajan, “The spectrum of multiplicative functions”, Ann. of Math. $(2)$ 153:2 (2001), 407–470.
• A. Granville and K. Soundararajan, “Notes on Burgess's theorem”, In preparation.
• H. Halberstam and H.-E. Richert, “On a result of R. R. Hall”, J. Number Theory 11:1 (1979), 76–89.
• D. R. Heath-Brown, “Zero-free regions for Dirichlet $L$-functions, and the least prime in an arithmetic progression”, Proc. London Math. Soc. $(3)$ 64:2 (1992), 265–338.
• J. Hoffstein, “Some analytic bounds for zeta functions and class numbers”, Invent. Math. 55:1 (1979), 37–47.
• J. C. Lagarias, H. L. Montgomery, and A. M. Odlyzko, “A bound for the least prime ideal in the Chebotarev density theorem”, Invent. Math. 54:3 (1979), 271–296.
• S. Louboutin, “Explicit bounds for residues of Dedekind zeta functions, values of $L$-functions at $s=1$, and relative class numbers”, J. Number Theory 85:2 (2000), 263–282.
• S. Louboutin, “Explicit upper bounds for residues of Dedekind zeta functions and values of $L$-functions at $s=1$, and explicit lower bounds for relative class numbers of CM-fields”, Canad. J. Math. 53:6 (2001), 1194–1222.
• V. K. Murty, “The least prime which does not split completely”, Forum Math. 6:5 (1994), 555–565.
• J. B. Rosser and L. Schoenfeld, “Approximate formulas for some functions of prime numbers”, Illinois J. Math. 6 (1962), 64–94.
• H. M. Stark, “The analytic theory of algebraic numbers”, Bull. Amer. Math. Soc. 81:6 (1975), 961–972.
• S. B. Stechkin, “The zeros of the Riemann zeta-function”, Mat. Zametki 8 (1970), 419–429. In Russian; translated in Math. Notes 8: 4 (1971), 706–711.
• G. Tenenbaum, Introduction to analytic and probabilistic number theory, Cambridge Studies in Advanced Mathematics 46, Cambridge University Press, 1995.
• J. D. Vaaler and J. F. Voloch, “The least nonsplit prime in Galois extensions of ${\bf Q}$”, J. Number Theory 85:2 (2000), 320–335.
• J. M. Vinogradov, “On the bound of the least non-residue of $n$th powers”, Trans. Amer. Math. Soc. 29:1 (1927), 218–226.


text:
# check (method)¶

check(self)[source]

assert that the arc is correct (begin != end; radius !=0)

Parameters

self (Arc1) – An Arc1 object

Returns

Return type

None

Raises
• PointArc1Error – The beginning point and the ending point of an Arc1 can’t be the same


text:
# What are the polarization states of the photons in a polarized and unpolarized light?

The photons are completely polarized, i.e their polarization states can be expressed as $a|R\rangle+b|L\rangle$, where $|R\rangle$ and $|L\rangle$ are two helicity eigenstates of the photon. For example, the $|R\rangle$ photon is right circularly polarized and the $|H\rangle=\frac{1}{\sqrt{2}}(|R\rangle+|L\rangle)$ is horizontally linearly polarized photon.

Is the polarized light a pure state and the unpolarized light a statistical mixture of photons with different polarizations?

-

Yes, a photon in a polarized light is found in a pure state such as $|H\rangle$, $|V\rangle$, $|L\rangle$, $|R\rangle$, or any complex linear combination of them. A photon in (completely) unpolarized light is described by the density matrix $$\rho = \frac{1}{2} \left( |L\rangle \langle L| + |R\rangle \langle R| \right) = \frac{1}{2} \left( |H\rangle \langle H| + |V\rangle \langle V| \right)$$ Note that you omitted the relationship for the vertically polarized state, $|V\rangle = i(|R\rangle - |L\rangle)/\sqrt{2}$, up to an overall sign which is a convention (well, the whole phase including $i$ is physically inconsequential, so it doesn't matter at all but one must be self-consistent with the conventions).
Your density matrix corresponds to a completely unpolarized light. Partially unpolarized light density matrix is expressed via projection operators with different "weights" $w_n$ (probabilities), not 1/2. –  Vladimir Kalitvianski Nov 7 '11 at 10:37
Dear @ANKU, to discuss frequency (and/or direction), you need to extend the Hilbert space by tensor-multiplying it with the space of different $\vec k$. Again, one may have pure states and mixed states with respect to frequencies, too. The totally non-monochromatic, unpolarized light is given by the density matrix like mine, but it would also have extra labels $\vec k$ in all the bra,ket states and one would integrated over some interval of $\vec k$. However, one may also have pure states relatively to the position/frequencies. There are many states, pure and mixed; what's your exact question? –  Luboš Motl Nov 7 '11 at 22:24


text:
### Calculate each of the following: (a) Mass fracti…

View
University of Pittsburgh - Main Campus

This question is in the process of being solved. The video shown is an answer to a question that covers similar topics.

Problem 21

Calculate each of the following:
(a) Mass $\%$ of I in strontium periodate
(b) Mass $\%$ of $\mathrm{Mn}$ in potassium permanganate

Check back soon!
Chapter 3
Stoichiometry of Formulas and Equations
CHEMISTRY: The Molecular Nature of Matter and Change 2016

## Discussion

You must be signed in to discuss.

## Video Transcript

No transcript available


text:
# Show that the system has no limit cycle.

Show that the system with $\dot{x} = y$ and $\dot{y} = -2x-y-3x^4+y^2$ has no limit cycle in $\mathbb{R}^2$. (HINT: a function in the form $\mathrm{e}^{\beta x}$ might be useful)

I know that I need to use Dulac's Criteria here, but I can't find the correct $\beta$ to go along with the hint. Any advice is welcome!

Thanks!

For any planar system of the form

$\dot x = X(x, y), \tag 1$

$\dot y = Y(x, y), \tag 2$

the Bendixson-Dulac criterion states that there are no periodic orbits in a region $\Omega$ where

$\nabla \cdot (\phi(X, Y)) = \nabla \cdot (\phi X, \phi Y) \ne 0, \tag 3$

where $\phi(x, y)$ is some differentiable scalar function on $\Omega$. For the present system, we may take $\Omega = \Bbb R^2$ and

$\dot x = y = X(x, y), \tag 4$

$\dot y = -2x - y - 3x^4 + y^2 = Y(x, y); \tag 5$

if we set

$\phi(x, y) = e^{\beta x}, \tag 6$

then we calculate:

$\nabla \cdot (\phi(X,Y)) = \dfrac{\partial (\phi X)}{\partial x} + \dfrac{\partial (\phi Y)}{\partial y} = \dfrac{\partial (e^{\beta x}y)}{\partial x} + \dfrac{\partial (e^{\beta x}(-2x - y - 3x^4 + y^2))}{\partial y}$ $= \beta e^{\beta x}y + (e^{\beta x}(-1 + 2y)) = e^{\beta x}(\beta y - 1 + 2y) = e^{\beta x}((\beta + 2)y - 1); \tag 7$

we observe that the function $e^{\beta x}((\beta + 2)y - 1)$ occurring on the right-hand side of (7) takes on the value zero precisely at that $y_0$ for which

$(\beta + 2)y_0 - 1 = 0, \tag 8$

that is, along the line

$y = y_0 = \dfrac{1}{\beta + 2}, \tag 9$

in the $xy$-plane. If we choose

$\beta = \epsilon - 2, \tag{10}$

for some real $\epsilon$, then since

$\epsilon = \beta + 2, \tag{11}$

we see that

$y_0(\epsilon) = \dfrac{1}{\beta + 2} = \dfrac{1}{\epsilon}, \tag{12}$

and we further observe that by choosing $\epsilon > 0$ sufficiently small, not only may we ensure that $y_0(\epsilon)$ is arbitraritly large, but also that the function

$(\beta + 2)y - 1 = \epsilon y - 1 \tag{13}$

increases with increasing $y$; thus it is positive for $y > y_0$ and negative for $y < y_0$.

Now if we had a periodic orbit, it would form the boundary of a compact subset $K \subset \Bbb R^2$ upon which the function $y$ is continuous, hence bounded above by some

$y_M = \sup \{y \mid y \in K \}; \tag{14}$

thus for suffiiently small $\beta + 2 = \epsilon > 0$ we have

$y_M < y_0; \tag{15}$

that is, on $K$,

$(\beta + 2)y - 1 < 0 \Longrightarrow e^{\beta x}((\beta + 2)y - 1) < 0; \tag{16}$

but now by Bendixson-Dulac, via (7), this contradicts the existence of a periodic trajectory of the system (1)-(2); thus in particular this system cannot have a limit cycle in $\Bbb R^2$.

• This is a much more rigorous answer than I was expecting. I agree in it's correctness, however, is there a simpler way to do this? Can we find a $\phi$ such that $\nabla (\phi(X,Y)) \neq 0$ as you mention in line 3 and just use that? Or is that not an option for this problem? – obewanjacobi Jul 9 '18 at 2:24
• @obewanjacobi: Well, I figured that no matter what $\phi$ you choose you have to show it works. I went with the suggestion and tried to find the right range of $\beta$, one thing led to another, and voila! It's really not that complicated but you have to get $\beta$ in the right range. By the way, love your screen name--very clever. Cheers! – Robert Lewis Jul 9 '18 at 2:34
• @obewanjacobi: obviously the whole trick in my method is to realize that the line where $(2 + \beta)y - 1 = 0$ can be moved out of the region surrounded by the periodic orbit! – Robert Lewis Jul 9 '18 at 2:37
• @RobertLewis I am still trying to understand how the line can be moved out of the region surrounding the periodic orbit? – BAYMAX Jul 9 '18 at 5:00
• @BAYMAX Well, the position of the line depends on $\beta$, a free parameter. I just chose $\beta$ so that the zero set of $\nabla \cdot e^{\beta x} (X, Y)$ does not intersect the closed orbit or its interior, which then implies this divergence is of fixed sign on that region. Of course, it is a "happy accident" that this can be so simply done. Obviously if $X, Y$ or $\phi$ was more complicated, things could get messy fast. – Robert Lewis Jul 9 '18 at 5:08

Let us take $\beta=-2$. Then: \begin{aligned} &\frac{\partial}{\partial x}\Big(\ e^{-2x}\cdot y\ \Big) + \frac{\partial}{\partial y}\Big(\ e^{-2x}\cdot (-2x-y-3x^4+y^2)\ \Big) \\ &\qquad = -2e^{-2x}\cdot y+ e^{-2x}\cdot (-1+2y)=-e^{-2x} \\ &\qquad<0\ . \end{aligned} This is enough to apply Dulac's criteria.


text:
## Intermediate Algebra (12th Edition)

$F=86^{\mathrm{o}}$
We got the following equation in Exercise 100: $F=\displaystyle \frac{9}{5}C+32$ We plug in $C=30$ and solve for $F$: $F=\displaystyle \frac{9}{5}(30)+32$ $=54+32$ $=86^{\mathrm{o}}$


text:
# lsst-texmf: The LSST LaTeX classes¶

The LSST lsst-texmf package contains LaTeX document classes, style files, and bibliographies that can be used to help you write LSST documentation in LaTeX to match the project style defined in [2] and [1], as documented in [3].

To get started, install lsst-texmf on your computer and check out the example projects. Next, learn more about writing documents with the lsstdoc class and making Beamer presentations with the LSST-beamer style file.

lsst-texmf is on GitHub at https://github.com/lsst/lsst-texmf.


text:
# Find value of serie with complex number

I need to find the value of the following serie:

$\sum_{n=0}^\infty \frac{1}{2^n}+\frac{(\pi i)^n}{n!}$

Our professor didn't show us how to do that with complex numbers.
$\frac{1}{2^n}$ gets smaller and smaller as n increases. When I expand $\frac{(\pi i)^n}{n!}$, I get $1+\pi i -\pi^2/2- i \pi^3/6 + \pi^4/24+ i\pi^5/120 - \pi^6/720 - ...$

As expected, the sign changes every two times because of the imaginary number, but I don't really see what to do next.

• $e^z=\sum \frac {z^n}{n!}$ even for complex $z$. – lulu Feb 21 '18 at 19:22


text:
# What is the equation of the line that is normal to f(x)= x^3 ln(x^2+1)-2x at  x= 1 ?

Dec 23, 2017

Solve for the derivative, plug in $x = 1$ to get the slope of the tangent line, then take the negative reciprocal to get the slope of the normal line, use the point-slope form and finally simplify to get

$y = \frac{x - 1}{1 - 3 \ln \left(2\right)} + \ln \left(2\right) - 2$

#### Explanation:

First, find the derivative of this function,

$f \left(x\right) = {x}^{3} \ln \left({x}^{2} + 1\right) - 2 x$.

Using the sum rule, we could split this into two parts,

$g \left(x\right) = {x}^{3} \ln \left({x}^{2} + 1\right)$

and

$h \left(x\right) = - 2 x$,

so that

$f \left(x\right) = g \left(x\right) + h \left(x\right)$

and

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{\mathrm{dg}}{\mathrm{dx}} + \frac{\mathrm{dh}}{\mathrm{dx}}$.

We could then proceed to find the derivative of each function, starting with $g \left(x\right)$:

$\frac{\mathrm{dg}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left({x}^{3} \ln \left({x}^{2} + 1\right)\right)$

Using the product rule:

$\mathrm{dg} = {x}^{3} \cdot d \left(\ln \left({x}^{2} + 1\right)\right) + \ln \left({x}^{2} + 1\right) \cdot d \left({x}^{3}\right)$

Let's solve for each differential, from derivatives, starting from $d \left({x}^{3}\right)$:

$\frac{d \left({x}^{3}\right)}{\mathrm{dx}} = 3 {x}^{2} \rightarrow d \left({x}^{3}\right) = 3 {x}^{2} \mathrm{dx}$

Putting that back in:

$\mathrm{dg} = {x}^{3} \cdot d \left(\ln \left({x}^{2} + 1\right)\right) + \ln \left({x}^{2} + 1\right) \cdot 3 {x}^{2} \mathrm{dx}$

Now, how about the other differential? I think solving for that derivative requires breaking it up into smaller functions. Let's set

$p \left(x\right) = \ln \left({x}^{2} + 1\right) = {p}_{3} \left({p}_{2} \left({p}_{1} \left(x\right)\right)\right)$

where

${p}_{1} \left(x\right) = {x}^{2}$

${p}_{2} \left(x\right) = x + 1$

${p}_{3} \left(x\right) = \ln \left(x\right)$

then use the chain rule:

$\frac{\mathrm{dp}}{\mathrm{dx}} = \frac{{\mathrm{dp}}_{1}}{\mathrm{dx}} \cdot \frac{{\mathrm{dp}}_{2}}{{\mathrm{dp}}_{1}} \cdot \frac{{\mathrm{dp}}_{3}}{{\mathrm{dp}}_{2}}$

Starting from the derivative of ${p}_{1}$ to $x$:

$\frac{{\mathrm{dp}}_{1}}{\mathrm{dx}} = \frac{d \left({x}^{2}\right)}{\mathrm{dx}} = 2 x$

Then solving for the differential, by "multiplying" by $\mathrm{dx}$:

$\frac{{\mathrm{dp}}_{1}}{\mathrm{dx}} = 2 x \rightarrow {\mathrm{dp}}_{1} = 2 x \mathrm{dx}$

Next, ${p}_{2}$ to ${p}_{1}$:

$\frac{{\mathrm{dp}}_{2}}{{\mathrm{dp}}_{1}} = \frac{d \left({p}_{1} + 1\right)}{{\mathrm{dp}}_{1}} = 1$

$\frac{{\mathrm{dp}}_{2}}{{\mathrm{dp}}_{1}} = 1 \rightarrow {\mathrm{dp}}_{2} = {\mathrm{dp}}_{1}$

Awesome! Evaluate ${\mathrm{dp}}_{1}$:

${\mathrm{dp}}_{2} = 2 x \mathrm{dx}$

Then ${\mathrm{dp}}_{3}$ to ${\mathrm{dp}}_{2}$:

$\frac{{\mathrm{dp}}_{3}}{{\mathrm{dp}}_{2}} = \frac{d \left(\ln \left({p}_{2}\right)\right)}{{\mathrm{dp}}_{2}} = \frac{1}{{p}_{2}}$

$\frac{{\mathrm{dp}}_{3}}{{\mathrm{dp}}_{2}} = \frac{1}{{p}_{2}} \rightarrow {\mathrm{dp}}_{3} = \frac{1}{{p}_{2}} {\mathrm{dp}}_{2}$

${\mathrm{dp}}_{3} = \frac{1}{{p}_{1} + 1} 2 x \mathrm{dx} = \frac{2 x}{{x}^{2} + 1} \mathrm{dx}$

So:

$\frac{{\mathrm{dp}}_{3}}{\mathrm{dx}} = \frac{\mathrm{dp}}{\mathrm{dx}} = \frac{2 x}{{x}^{2} + 1}$

$\frac{\mathrm{dp}}{\mathrm{dx}} = \frac{2 x}{{x}^{2} + 1} \rightarrow \mathrm{dp} = \frac{2 x}{{x}^{2} + 1} \mathrm{dx}$

Substituting $\mathrm{dp}$ back in $\mathrm{dg}$:

$\mathrm{dg} = {x}^{3} \cdot \mathrm{dp} + \ln \left({x}^{2} + 1\right) \cdot 3 {x}^{2} \mathrm{dx}$

$\mathrm{dg} = {x}^{3} \cdot \frac{2 x}{{x}^{2} + 1} \mathrm{dx} + \ln \left({x}^{2} + 1\right) \cdot 3 {x}^{2} \mathrm{dx}$

Simplifying:

$\mathrm{dg} = \frac{2 {x}^{4}}{{x}^{2} + 1} \mathrm{dx} + 3 {x}^{2} \ln \left({x}^{2} + 1\right) \mathrm{dx}$

Dividing by $\mathrm{dx}$ to solve for the derivative:

$\frac{\mathrm{dg}}{\mathrm{dx}} = \frac{2 {x}^{4}}{{x}^{2} + 1} + 3 {x}^{2} \ln \left({x}^{2} + 1\right)$

And substituting this back into $\frac{\mathrm{df}}{\mathrm{dx}}$:

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{2 {x}^{4}}{{x}^{2} + 1} + 3 {x}^{2} \ln \left({x}^{2} + 1\right) + \frac{\mathrm{dh}}{\mathrm{dx}}$

Now, on to $h \left(x\right) = - 2 x$:

$\frac{\mathrm{dh}}{\mathrm{dx}} = - 2$

So:

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{2 {x}^{4}}{{x}^{2} + 1} + 3 {x}^{2} \ln \left({x}^{2} + 1\right) - 2$

Let's solve for $x = 1$:

$\rightarrow \frac{2 {\left(1\right)}^{4}}{{\left(1\right)}^{2} + 1} + 3 {\left(1\right)}^{2} \ln \left({\left(1\right)}^{2} + 1\right) - 2$

$= 1 + 3 \ln \left(2\right) - 2$

${m}_{t} = 3 \ln \left(2\right) - 1$

That gives us the slope of the tangent line. Taking the negative reciprocal:

${m}_{n} = - {\left({m}_{t}\right)}^{- 1} = - {\left(3 \ln \left(2\right) - 1\right)}^{- 1} = \frac{1}{1 - 3 \ln \left(2\right)}$

Now that we have the slope of the normal line, we just need the $y$-intercept. We can start by using the point-slope form:

$y - {y}_{1} = m \left(x - {x}_{1}\right)$

The slope, $m$, in this case is ${m}_{n} = \frac{1}{1 - 3 \ln \left(2\right)}$ :

$y - {y}_{1} = \frac{x - {x}_{1}}{1 - 3 \ln \left(2\right)}$

We need a point on the graph. Well, there's our input, $x = 1$, and the output, $f \left(1\right) = {\left(1\right)}^{3} \ln \left({\left(1\right)}^{2} + 1\right) - 2 \left(1\right) = \ln \left(2\right) - 2$, so substitute that for ${x}_{1}$ and ${y}_{1}$ respectively:

$y - \left(\ln \left(2\right) - 2\right) = \frac{x - 1}{1 - 3 \ln \left(2\right)}$

Finally, isolate $y$:

$y = \frac{x - 1}{1 - 3 \ln \left(2\right)} + \ln \left(2\right) - 2$

That's the equation for the line normal to $f \left(x\right) = {x}^{3} \ln \left({x}^{2} + 1\right) - 2 x$ at $x = 1$.


text:
# Skew heap

A skew heap (or self-adjusting heap) is a heap data structure implemented as a binary tree. Skew heaps are advantageous because of their ability to merge more quickly than binary heaps. In contrast with binary heaps, there are no structural constraints, so there is no guarantee that the height of the tree is logarithmic. Only two conditions must be satisfied:

• The general heap order must be enforced
• Every operation (add, remove_min, merge) on two skew heaps must be done using a special skew heap merge.

A skew heap is a self-adjusting form of a leftist heap which attempts to maintain balance by unconditionally swapping all nodes in the merge path when merging two heaps. (The merge operation is also used when adding and removing values.)

With no structural constraints, it may seem that a skew heap would be horribly inefficient. However, amortized complexity analysis can be used to demonstrate that all operations on a skew heap can be done in O(log n).[1] In fact, with φ=(1+√5)/2 denoting the golden ratio, the exact amortized complexity is known to be logφ n (approximately 1.44 log2 n).[2][3]

## Definition

Skew heaps may be described with the following recursive definition:

• A heap with only one element is a skew heap.
• The result of skew merging two skew heaps ${\displaystyle sh_{1}}$ and ${\displaystyle sh_{2}}$ is also a skew heap.

## Operations

### Merging two heaps

When two skew heaps are to be merged, we can use a similar process as the merge of two leftist heaps:

• Compare roots of two heaps; let p be the heap with the smaller root, and q be the other heap. Let r be the name of the resulting new heap.
• Let the root of r be the root of p (the smaller root), and let r's right subtree be p's left subtree.
• Now, compute r's left subtree by recursively merging p's right subtree with q.

```template<class T, class CompareFunction>
SkewNode<T>* CSkewHeap<T, CompareFunction>::_Merge(SkewNode<T>* root_1, SkewNode<T>* root_2)
{
SkewNode<T>* firstRoot = root_1;
SkewNode<T>* secondRoot = root_2;

if(firstRoot == NULL)
return secondRoot;

else if(secondRoot == NULL)
return firstRoot;

if(sh_compare->Less(firstRoot->key, secondRoot->key))
{
SkewNode<T>* tempHeap = firstRoot->rightNode;
firstRoot->rightNode = firstRoot->leftNode;
firstRoot->leftNode = _Merge(secondRoot, tempHeap);
return firstRoot;
}
else
return _Merge(secondRoot, firstRoot);
}
```

Before:

after

### Non-recursive merging

Alternatively, there is a non-recursive approach which is more wordy, and does require some sorting at the outset.

• Split each heap into subtrees by cutting every path. (From the root node, sever the right node and make the right child its own subtree.) This will result in a set of trees in which the root either only has a left child or no children at all.
• Sort the subtrees in ascending order based on the value of the root node of each subtree.
• While there are still multiple subtrees, iteratively recombine the last two (from right to left).
• If the root of the second-to-last subtree has a left child, swap it to be the right child.
• Link the root of the last subtree as the left child of the second-to-last subtree.

Adding a value to a skew heap is like merging a tree with one node together with the original tree.

### Removing values

Removing the first value in a heap can be accomplished by removing the root and merging its child subtrees.

### Implementation

In many functional languages, skew heaps become extremely simple to implement. Here is a complete sample implementation in Haskell.

```data SkewHeap a = Empty
| Node a (SkewHeap a) (SkewHeap a)

singleton :: Ord a => a -> SkewHeap a
singleton x = Node x Empty Empty

union :: Ord a => SkewHeap a -> SkewHeap a -> SkewHeap a
Empty              `union` t2                 = t2
t1                 `union` Empty              = t1
t1@(Node x1 l1 r1) `union` t2@(Node x2 l2 r2)
| x1 <= x2                                 = Node x1 (t2 `union` r1) l1
| otherwise                                = Node x2 (t1 `union` r2) l2

insert :: Ord a => a -> SkewHeap a -> SkewHeap a
insert x heap = singleton x `union` heap

extractMin :: Ord a => SkewHeap a -> Maybe (a, SkewHeap a)
extractMin Empty        = Nothing
extractMin (Node x l r) = Just (x, l `union` r)
```


text:
# A praise for JSR223 with Jython (highly recommended)

Hi all,

about two month ago I started creating rules with the JSR223-Engine and Jython.
I was instantly amazed by how simple I could do stuff and by the endless possibilities it offers.
No more need for lambdas, just straight function calls.
These Z-Wave remotes with multiple buttons - just create a base class with all the logic and just override the function for the button press.
Functions I want to use in multiple files - no problem!
And don’t get me started on performance. I was looking into an ODROID C2 for faster response times.
JSR223 runs at least 10 times faster than the stock rule ( at least for me - search for another thread from me there I do the comparison).
I am now still keeping my Raspberry PI2 because it is fast enough and noticeably running smoother.

For me my conclusion is easy:
With Jython Openhab finally became what I always wanted it to be!
Easy, well documented (just search for python code samples) and fast.

If I would had it in the beginning I would have way less problems learning Openhab.
For all the users who are thinking about giving it a try - I highly encourage you to do so!

I just wanted to say thanks for implementing all this!

2 Likes

Would you be interested in sharing your rules/config? I would be very keen to see move across if I could see some examples.

I have been pondering a switch to JSR223 myself for while, but haven’t gotten around to it yet.

Is it possible to mix JSR232 rules with “traditional” rules, as a migration step? Obviously I do not mean to mix JSR233 and “traditional” in the same rule file, but starting out with converting some of my rule to JSR233 to see how it works out…

@KjetilA: Yes - this is exactly what I have done. I started with one rule and left all the other ones as they are. Then I migrated more and more to jython. Now I only have few traditional rule files left.

@ben_jones12: Sure - I can post some examples.

1 Like

I’m not going to argue about your results. I’m more than willing to accept jython will be 10x to 20x faster. However, I do want to caveat that that in that other thread the test performed was not really an apples to apples comparison of the two’s performance.

@rlkoshak: I changed my top post. It sais now that for me it is over 10 times faster.

@ben_jones12:
Here is an example even if it is a little bit more sophisticated.

I have a couple of contacts (door/windows). Every time I open one, I want the amount of movements in the last half hour (for presence detection) and the last time stamps (for information). Additionally, I count all the movements just for fun!

First there is a function which immediately increases and after half an hour decreases the value:

def IncreaseDecrease( item, time):
BusEvent.postUpdate( item, int(str(item.state)) + 1)
oh.createTimer(DateTime.now().plusMinutes(time), lambda : BusEvent.postUpdate( item, max(int(str(item.state)) - 1,0) ))

Then there is my contact class, it basicly handles all the code. Note that it is very easy to get an item just by its name! My timestamps are all named like the item, but with a “ts_” prefix.

class cContact(Rule):
def execute(self, event):

logger.debug( "event: '" + str(event) + "'")

item = event.item
if not item:
return None

name = item.getName()

#Zeitpunkte
BusEvent.postUpdate( "ts3_" + name, str(ItemRegistry.getItem( "ts2_" + name).state) )
BusEvent.postUpdate( "ts2_" + name, str(ItemRegistry.getItem( "ts1_" + name).state) )
BusEvent.postUpdate( "ts1_" + name, str(ItemRegistry.getItem( "ts0_" + name).state) )
BusEvent.postUpdate( "ts0_" + name, str(DateTimeType()))

#30 min counter
IncreaseDecrease(ItemRegistry.getItem( "ctr_" + name + "_30min"), 30)

#normaler counter
counter_item = ItemRegistry.getItem( "ctr_" + name)
BusEvent.postUpdate( counter_item, int(str(counter_item.state)) + 1)

Now for all my contacts I just have to derive from the class:

class AlarmContact(cContact):
def getEventTrigger(self):
return [
ChangedEventTrigger("Flur_Schlafzimmertuere_Tuer",      None, OpenClosedType.CLOSED),
ChangedEventTrigger("Flur_Wohnungstuere_Tuer",          None, OpenClosedType.CLOSED),
ChangedEventTrigger("Wohnzimmer_Balkontuere_Tuer",      None, OpenClosedType.CLOSED),
ChangedEventTrigger("Schlafzimmer_Fenster_Fenster",     None, OpenClosedType.CLOSED)
]

Note, that it is also possible to know which item triggered the rule execution. This allows very elegant code!

1 Like

That is looking very powerful - how do you get an openHAB contact item to derive from your custom class tho?

Sorry, probably a stupid question but I obviously have a lot to learn about this Jython stuff!

Jython is basicly Python for Java. So you can search the web for python tutorials.
The real magic is the JSR223-Engine.

I’ll post a simpler example:

class UpdateSensorManuellKueche(Rule):
def getEventTrigger(self):
return [
UpdatedEventTrigger("Kueche_Arbeitsplatte_Manuell")
]

def execute(self, event):

item = event.item
if not item:
return None

logger.debug("{:15s} | manuell: {}".format( "Arbeitsplatte", event.newState))

manuell = str( event.newState)

if( manuell == "ON"):
#sleep nur, wenn es an ist
if int(str(ItemRegistry.getItem("Kueche_Arbeitsplatte_Out_R").state)) > 0:
BusEvent.sendCommand( "Kueche_Arbeitsplatte_Out_R",  "0")
time.sleep( 1.5)

BusEvent.sendCommand( "Kueche_Arbeitsplatte_Out_R",  "100")
else:
BusEvent.sendCommand( "Kueche_Arbeitsplatte_Out_R",  "0")

Each Rule ist a class.
With getEventTrigger you can return different types of triggers.
Based on these triggers the execute-function will be called.

Thanks Sebastian - I know a little python - I was more interested to see how rules got attached to items etc. From what you are saying you create a python class for each rule, and in the getEventTrigger method return one or more triggers for that rule - do you reference the item names by string?

Then you can access other items via the ItemRegistry and BusEvent.

Thanks for posting those, I think it is high time I started having a play and migrating some of my rules across.

I’m also a JSR223/Jython fan. I wrote a tutorial about using the JSR223 rules that might help.

In addition to rules, I’ve also written full OH1 bindings in JSR223 Jython.

1 Like

I would love to move to JSR223 but the documentation only shows a setup procedure for Linux. Applying the procedure on MacOS which has great similarity to LInux was not straight forward and a big barrier for me.
Unfortunately for some reasons which I cannot change I have to be on windows.

If someone could create a documentation on how to setup jython on windows that would be really appreciated.

http://www.jython.org/archive/21/MacOS_Install.html

I think he was asking for directions about how to configure openHAB to use Jython. Installying Jython itself is relatively straightforward but to use it openHAB requires updating startup scripts with correct Java classpaths and system properties. It can be a bit challenging for some who isn’t a Java programmer.

@steve1. correct.
When I tried to get it running under MacOS, I indeed strugggled for sometime to change the start.sh because the Linux start.sh did not work.
Patching the start.bat in Windows exceeds my capabilities.

@martin_klimke: here is a working startup.bat. Just change the JYTHON_HOME accordingly.
It also references “configurations/scripts/lib” as a import/library folder where you can write library functions.

@echo off

:: set path to eclipse folder. If local folder, use '.'; otherwise, use c:\path\to\eclipse
set ECLIPSEHOME=server

:: set ports for HTTP(S) server
set HTTP_PORT=8080
set HTTPS_PORT=8443
set JYTHON_HOME=c:\Progs\jython2.7.0

:: get path to equinox jar inside ECLIPSEHOME folder
for /f "delims= tokens=1" %%c in ('dir /B /S /OD %ECLIPSEHOME%\plugins\org.eclipse.equinox.launcher_*.jar') do set EQUINOXJAR=%%c

:: start Eclipse w/ java
echo Launching the openHAB runtime...
set ARGS=-Dorg.osgi.framework.bundle.parent=ext -Dpython.home="%JYTHON_HOME%" -Dpython.path="configurations/scripts/lib" -Dosgi.clean=true -Declipse.ignoreApp=true -Dosgi.noShutdown=true -Djetty.port=%HTTP_PORT% -Djetty.port.ssl=%HTTPS_PORT% -Djetty.home=. -Dlogback.configurationFile=configurations/logback.xml -Dfelix.fileinstall.dir=addons -Dfelix.fileinstall.filter=.*\\.jar -Djava.library.path=lib -Djava.security.auth.login.config=./etc/login.conf -Dorg.quartz.properties=./etc/quartz.properties -Dequinox.ds.block_timeout=240000 -Dequinox.scr.waitTimeOnBlock=60000 -Djava.awt.headless=true -Dfelix.fileinstall.active.level=4 -cp %JYTHON_HOME%\jython.jar;%EQUINOXJAR% org.eclipse.equinox.launcher.Main %* -console
::echo %ARGS%
java %ARGS%
1 Like

I have spent a frustrating hour trying to install this on a standard Ubuntu package installation. I have installed jython and added

#For Jython
JYTHON_HOME="/opt/jython"

OPENHAB_ARGS="-Dpython.home=${JYTHON_HOME} -Dpython.path=${OPENHAB_CONFIGURATIONS_DIR}/scripts/lib -classpath \${JYTHON_HOME}/jython.jar"

to openhab.in.sh but it does not seem to be recognised

12:14:21.306 [DEBUG] [.r.internal.RuleModelActivator:42   ] - Registered 'rules' configuration parser
12:14:21.376 [DEBUG] [m.r.internal.engine.RuleEngine:77   ] - Started rule engine
12:14:24.775 [DEBUG] [j.internal.Jsr223CoreActivator:33   ] - Registered 'jsr223' scripting-engine
12:14:24.781 [DEBUG] [j.internal.engine.Jsr223Engine:96   ] - itemRegistry set
12:14:24.790 [DEBUG] [j.internal.engine.Jsr223Engine:68   ] - activate()
12:14:24.795 [INFO ] [o.o.c.j.i.e.s.ScriptManager   :58   ] - Available engines:
12:14:24.827 [INFO ] [o.o.c.j.i.e.s.ScriptManager   :60   ] - Oracle Nashorn
12:14:24.839 [WARN ] [o.o.c.j.i.e.s.ScriptManager   :92   ] - No Engine found for File: test.py
12:14:24.849 [DEBUG] [j.internal.engine.Jsr223Engine:78   ] - Started jsr223 engine

What am I missing?

Edit:
I found the problem. The packages run a jar file and from man java

           When you use the -jar option, the specified JAR file is the source of all user classes, and
other class path settings are ignored.

You learn something every day.

Edit2:
I learned something today as well. You can use the -Xbootclasspath in this situation. I changed /etc/default/openhab to include this line and now I can load python scripts

JAVA_ARGS="-Xbootclasspath/a:/opt/jython2.7.0/jython.jar -Dpython.home=/opt/jython2.7.0 -Dpython.path=configurations/scripts/lib"

@Spaceman_Spiff
worked like charm.
So easy to setup.

Thank you so much.

Just wondering if it is all that sensible to start migrating all my rules over to Jython, when it isn’t currently supported in OH v2.0. Do we know what the rules engine will be in v2.0?

Don’t want to have to do this migration twice!

A JSR223 extension is being developed for OH2…

1 Like

@steve1
Can you please comment the stability on the existing JSR233 solution based on your practical experience ?


text:
Documentation

# zero

Zeros and gain of SISO dynamic system

## Description

example

Z = zero(sys) returns the zeros of the single-input, single-output (SISO) dynamic system model, sys. The output is expressed as the reciprocal of the time units specified in sys.TimeUnit.

example

[Z,gain] = zero(sys) also returns the zero-pole-gain of sys.

example

[Z,gain] = zero(sys,J1,...,JN) returns the zeros and gain of the entries in the model array sys with subscripts J1,...,JN.

## Examples

collapse all

Compute the zeros of the following transfer function:

$sys\left(s\right)=\frac{4.2{s}^{2}+0.25s-0.004}{{s}^{2}+9.6s+17}$

sys = tf([4.2,0.25,-0.004],[1,9.6,17]);
Z = zero(sys)
Z = 2×1

-0.0726
0.0131

Calculate the zero locations and zero-pole gain of the following transfer function:

$sys\left(s\right)=\frac{4.2{s}^{2}+0.25s-0.004}{{s}^{2}+9.6s+17}$

sys = tf([4.2,0.25,-0.004],[1,9.6,17]);
[z,gain] = zero(sys)
z = 2×1

-0.0726
0.0131

gain = 4.2000

The zero locations are expressed in ${\mathrm{second}}^{-1},$ because the time unit of the transfer function (H.TimeUnit) is seconds.

For this example, load a 3-by-1 array of transfer function models.

size(sys)
3x1 array of transfer functions.
Each model has 1 outputs and 1 inputs.

Find the zeros and gain values of the models in the array.

[Z, gain] = zero(sys);
Z(:,:,1,1)
ans = 2×1

-0.0726
0.0131

gain(:,:,1,1)
ans = 4.2000

zero returns an array each for the zeros and the gain values respectively. Here, Z(:,:,1,1) and gain(:,:,1,1) corresponds to the zero and the gain value of the first model in the array, that is, sys(:,:,1,1).

## Input Arguments

collapse all

Dynamic system, specified as a SISO dynamic system model, or an array of SISO dynamic system models. Dynamic systems that you can use include continuous-time or discrete-time numeric LTI models such as tf, zpk, or ss models.

If sys is a generalized state-space model genss or an uncertain state-space model uss, zero returns the zeros of the current or nominal value of sys. If sys is an array of models, zero returns the zeros of the model corresponding to its subscript J1,...,JN in sys. For more information on model arrays, see Model Arrays (Control System Toolbox).

Indices of models in array whose zeros you want to extract, specified as a positive integer. You can provide as many indices as there are array dimensions in sys. For example, if sys is a 4-by-5 array of dynamic system models, the following command extracts the zeros for entry (2,3) in the array.

Z = zero(sys,2,3);

## Output Arguments

collapse all

Zeros of the dynamic system, returned as a column vector or an array. If sys is:

• A single model, then Z is a column vector of zeros of the dynamic system model sys

• A model array, then Z is an array containing the zeros of each model in sys

Z is expressed as the reciprocal of the time units specified in sys.TimeUnit. For example, zero is expressed in 1/minute if sys.TimeUnit = 'minutes'.

Zero-pole-gain of the dynamic system, returned as a scalar. In other words, gain is the value of K when the model is written in zpk form.

## Tips

• If sys has internal delays, zero sets all internal delays to zero, creating a zero-order Padé approximation. This approximation ensures that the system has a finite number of zeros. zero returns an error if setting internal delays to zero creates singular algebraic loops. To assess the stability of models with internal delays, use step or impulse.

• To calculate the transmission zeros of a multi-input, multi-output (MIMO) system, use tzero.


text:
# American Institute of Mathematical Sciences

September  2010, 9(5): 1311-1318. doi: 10.3934/cpaa.2010.9.1311

## On the regularity of minimizers to degenerate functionals

 1 Dipartimento di Matematica e Informatica Università, degli Studi di Salerno Via Ponte don Melillo, 84084 Fisciano (SA), Italy 2 Dipartimento di Statistica e Matematica per la Ricerca Economica Università, “Parthenope ”Via Medina 40, 80131 Napoli, Italy

Received  October 2009 Revised  January 2010 Published  May 2010

In this paper, we prove a higher integrability result for the gradient of a minimizer of a functional of the type

$I(\Omega , u)=\int_{\Omega}\sum_{i,j} a_{i,j} D_i u D_jv dx$

whose coefficient matrix $A(x)= ^tA(x)$ satisfies the anisotropic bounds

$\frac{|\xi |^2}{K(x)}\leq < A(x) \xi, \xi > \leq K(x) |\xi |^2\quad \forall \xi \in R^n,$ for a.e. $x\in \Omega,$

where $K:\Omega \subset R^n \rightarrow [1,+\infty),$ a locally integrable function in $\Omega$, belongs to $A_2 \cap G_n$ and has a majorant $Q(x)\geq K(x)$ of finite mean,

limsup$_{R \rightarrow 0} \int_{B_R(x)} Q(y)dy < \infty$ at every point $x \in \Omega.$

Citation: P. Di Gironimo, L. D’Onofrio. On the regularity of minimizers to degenerate functionals. Communications on Pure & Applied Analysis, 2010, 9 (5) : 1311-1318. doi: 10.3934/cpaa.2010.9.1311
 [1] Herbert Koch. Partial differential equations with non-Euclidean geometries. Discrete & Continuous Dynamical Systems - S, 2008, 1 (3) : 481-504. doi: 10.3934/dcdss.2008.1.481 [2] Wilhelm Schlag. Spectral theory and nonlinear partial differential equations: A survey. Discrete & Continuous Dynamical Systems - A, 2006, 15 (3) : 703-723. doi: 10.3934/dcds.2006.15.703 [3] Eugenia N. Petropoulou, Panayiotis D. Siafarikas. Polynomial solutions of linear partial differential equations. Communications on Pure & Applied Analysis, 2009, 8 (3) : 1053-1065. doi: 10.3934/cpaa.2009.8.1053 [4] Arnulf Jentzen. Taylor expansions of solutions of stochastic partial differential equations. Discrete & Continuous Dynamical Systems - B, 2010, 14 (2) : 515-557. doi: 10.3934/dcdsb.2010.14.515 [5] Nguyen Thieu Huy, Ngo Quy Dang. Dichotomy and periodic solutions to partial functional differential equations. Discrete & Continuous Dynamical Systems - B, 2017, 22 (8) : 3127-3144. doi: 10.3934/dcdsb.2017167 [6] Barbara Abraham-Shrauner. Exact solutions of nonlinear partial differential equations. Discrete & Continuous Dynamical Systems - S, 2018, 11 (4) : 577-582. doi: 10.3934/dcdss.2018032 [7] Paul Bracken. Exterior differential systems and prolongations for three important nonlinear partial differential equations. Communications on Pure & Applied Analysis, 2011, 10 (5) : 1345-1360. doi: 10.3934/cpaa.2011.10.1345 [8] Frédéric Mazenc, Christophe Prieur. Strict Lyapunov functions for semilinear parabolic partial differential equations. Mathematical Control & Related Fields, 2011, 1 (2) : 231-250. doi: 10.3934/mcrf.2011.1.231 [9] Enrique Zuazua. Controllability of partial differential equations and its semi-discrete approximations. Discrete & Continuous Dynamical Systems - A, 2002, 8 (2) : 469-513. doi: 10.3934/dcds.2002.8.469 [10] Paul Bracken. Connections of zero curvature and applications to nonlinear partial differential equations. Discrete & Continuous Dynamical Systems - S, 2014, 7 (6) : 1165-1179. doi: 10.3934/dcdss.2014.7.1165 [11] Michela Eleuteri, Pavel Krejčí. An asymptotic convergence result for a system of partial differential equations with hysteresis. Communications on Pure & Applied Analysis, 2007, 6 (4) : 1131-1143. doi: 10.3934/cpaa.2007.6.1131 [12] Tomás Caraballo, José Real, T. Taniguchi. The exponential stability of neutral stochastic delay partial differential equations. Discrete & Continuous Dynamical Systems - A, 2007, 18 (2&3) : 295-313. doi: 10.3934/dcds.2007.18.295 [13] Yuhki Hosoya. First-order partial differential equations and consumer theory. Discrete & Continuous Dynamical Systems - S, 2018, 11 (6) : 1143-1167. doi: 10.3934/dcdss.2018065 [14] Hernán R. Henríquez, Claudio Cuevas, Alejandro Caicedo. Asymptotically periodic solutions of neutral partial differential equations with infinite delay. Communications on Pure & Applied Analysis, 2013, 12 (5) : 2031-2068. doi: 10.3934/cpaa.2013.12.2031 [15] Frédéric Gibou, Doron Levy, Carlos Cárdenas, Pingyu Liu, Arthur Boyer. Partial Differential Equations-Based Segmentation for Radiotherapy Treatment Planning. Mathematical Biosciences & Engineering, 2005, 2 (2) : 209-226. doi: 10.3934/mbe.2005.2.209 [16] Rongmei Cao, Jiangong You. The existence of integrable invariant manifolds of Hamiltonian partial differential equations. Discrete & Continuous Dynamical Systems - A, 2006, 16 (1) : 227-234. doi: 10.3934/dcds.2006.16.227 [17] Seyedeh Marzieh Ghavidel, Wolfgang M. Ruess. Flow invariance for nonautonomous nonlinear partial differential delay equations. Communications on Pure & Applied Analysis, 2012, 11 (6) : 2351-2369. doi: 10.3934/cpaa.2012.11.2351 [18] Antonio Cañada, Salvador Villegas. Lyapunov inequalities for partial differential equations at radial higher eigenvalues. Discrete & Continuous Dynamical Systems - A, 2013, 33 (1) : 111-122. doi: 10.3934/dcds.2013.33.111 [19] Zhongkai Guo. Invariant foliations for stochastic partial differential equations with dynamic boundary conditions. Discrete & Continuous Dynamical Systems - A, 2015, 35 (11) : 5203-5219. doi: 10.3934/dcds.2015.35.5203 [20] Shigui Ruan, Junjie Wei, Jianhong Wu. Bifurcation from a homoclinic orbit in partial functional differential equations. Discrete & Continuous Dynamical Systems - A, 2003, 9 (5) : 1293-1322. doi: 10.3934/dcds.2003.9.1293

2018 Impact Factor: 0.925


text:
# UY1: Phasors & Alternating Currents

Any device that supplies a sinusoidally varying voltage (potential difference)

$$v = V \cos{\omega t}$$

, where V is the voltage amplitude,

or current

$$i = I \cos{\omega t}$$

, where I is the current amplitude,

is an alternating current (ac) source with angular frequency:

\begin{aligned} \omega &= 2 \pi f \\ &= \frac{2 \pi}{T} \end{aligned}

You can represent the information in a Phasor diagram, as shown below.

\begin{aligned} i^{2} &= I^{2} \cos^{2}{\omega t} \\ i^{2} &= \frac{1}{2} I^{2} \left( 1 + \cos{2 \omega t} \right) \\ \frac{1}{T} \int\limits_{0}^{T} i^{2} \, dt &= \frac{1}{2} I^{2} \\ I_{\text{rms}} &= \frac{1}{\sqrt{2}} I \end{aligned}

A diode (rectifier) is a device that conducts better in one direction than in the other – an ideal diode has zero resistance for one direction of current and infinite resistance for the other.

For a full-wave rectifier circuit, the rectified average current is given by:

\begin{aligned} I_{rav} &= \frac{2}{T} \int\limits_{0}^{\frac{T}{2}} I \sin{\omega t} \, dt \\ &= \frac{4I}{\omega T} \\ &= \frac{2}{\pi} I \end{aligned}

Next: Resistors, Inductors & Capacitors In A.C. Circuits

Previous: R-L Circuit

Back To Electromagnetism (UY1)


text:
# Math Help - Simplifying Equations

1. ## Simplifying Equations

Hi,

I know te following equation simplifies as shown, but how?? I also know I should know this but I don't!

$\frac{ (\ln{x})(4x^3)-(x^4)(1/x)} {(ln{x})^2}$

= $\frac{x^3(4\ln{x}-1)}{(\ln{x})^2}$

Hi,

I know te following equation simplifies as shown, but how?? I also know I should know this but I don't!

$\frac{ (\ln{x})(4x^3)-(x^4)(1/x)} {(ln{x})^2}$

= $\frac{x^3(4\ln{x}-1)}{(\ln{x})^2}$

in the numerator, note that $x^4 \cdot \frac{1}{x} = x^3$
therefore ... $(\ln{x})(4x^3) - x^3 = x^3(4\ln{x} - 1)$


text:
# Exercise with Linear Transformation

1. Jan 27, 2016

### Kernul

1. The problem statement, all variables and given/known data
Being $f : \mathbb R^4\rightarrow\mathbb R^4$ the endomorphism defined by:
$$f((x,y,z,t)) = (13x + y - 2z + 3t, 10y, 9z + 6t, 6z + 4t)$$

1) Determine the basis and dimension of $Ker(f)$ and $Im(f)$. Complete the base chosen in $Ker(f)$ into a base of $\mathbb R^4$;
2) Determine the basis and dimension of the subspaces of $\mathbb R^4 Ker(f) \cap Im(f)$ and $Ker(f) + Im(f)$;
3) Determine the preimage of the vector $\vec v = (0, 2h, 1, h - 4)$, with the variation of the real parameter $h$;
4) Find, if they exist, the values of the $h$ parameter with $[2\vec e_1 + \vec e_2, 3\vec e_3, \vec v, -\vec e_2 + \vec e_4]$ being a base of $\mathbb R^4$. $ε = (\vec e_1, \vec e_2, \vec e_3, \vec e_4)$ is the canonic base of $\mathbb R^4$.
5) Having with $A$ the matrix $M_{εε}(f)$ associated to $f$ in respect to the canonic base, find out if $A$ is diagonalizable and, if so, determine a diagonalizing (?) matrix $P$ and the corresponding diagonal matrix $D$ to which $A$ is similar.
6) Find out if the matrix $B = \begin{pmatrix} 3 & 0 & 0\\ 1 & 3 & 0\\ 1 & 0 & 1\end{pmatrix}$ is diagonalizable and show a maximum independent system of eigenvectors of $B$

2. Relevant equations
Rank-Nullity Theorem
Operation on the vector subspaces (intersection and sum)
Canonic Base
Diagonalization
Eigenvectors and eigenvalues

3. The attempt at a solution
So, I know how to do the first point of the exercise but I found myself stuck.
In order to find the dimension and basis of both $Ker(f)$ and $Im(f)$ I first have to find the rank of the matrix $$A = \begin{pmatrix} 13 & 1 & -2 & 3\\ 0 & 10 & 0 & 0\\ 0 & 0 & 9 & 6\\ 0 & 0 & 6 & 4 \end{pmatrix}$$
From what I see, the matrix' rank is 4. Knowing that $ρ(A) = dim(Im(f))$, the base of $Im(f)$ is made of all the columns of $A$. So: $$B_{Im(f)} = [A^1, A^2, A^3, A^4]$$
Now, knowing that $dim(Ker(f)) = dim(A) - dim(Im(f))$, I have $dim(Ker(f)) = 4 - 4 = 0$. This means that $Ker(f)$ has no basis, right? Then how do I complete the basis of $Ker(f)$(as the exercise asks) if there is none?
For now I would like to concentrate on this first point and then ask other questions for the following points, if needed.

2. Jan 27, 2016

### Staff: Mentor

Item 6 below seems unrelated to this problem. Please start a new thread with that problem.
No.
It's fairly obvious (at least to me) that the rank of A is 3. To see this for yourself, row-reduce the matrix.

3. Jan 27, 2016

### Kernul

Yeah, I know it wasn't part of the exercise but it was put there. I don't know why.

Oh yeah, I'm sorry. I tried that but it seems I didn't quite try enough...
Anyway, doing the calculations, I find myself with: $B_{Ker(f)} = [\begin{pmatrix} -1/13\\ 0\\ -1\\ 1 \end{pmatrix}]$
In order to complete the base of $Ker(f)$ I'll put it together with the base of $Im(f)$, so:$$B_{Ker(f)} = [\begin{pmatrix} -1/13\\ 0\\ -1\\ 1 \end{pmatrix}, \begin{pmatrix} 13\\ 0\\ 0\\ 0 \end{pmatrix}, \begin{pmatrix} 1\\ 10\\ 0\\ 0 \end{pmatrix}, \begin{pmatrix} -2\\ 0\\ 9\\ 6 \end{pmatrix}]$$
I hope I'm right until now.

Now the second point.
I have to find the base and dimension of $Ker(f) \cap Im(f)$ and $Ker(f)+Im(f)$, knowing that $dim(Im(f)) = 3$ and $dim(Ker(f)) = 1$. At this point, do I have to find a vector that is in both of them? If yes, should I do something like this:$$\vec v = αA^1 + βA^2 +γA^3 + ωKer(f)$$

4. Jan 27, 2016

### Staff: Mentor

No, this isn't right. If you multiply A times the vector above, you don't get the zero vector. Check your work in row reducing A.
???
I don't know what you're doing here. As it turns out, dim(Ker(A)) = 1, so a basis for the kernel would consist of one vector.
For the first part of item 2, $Ker(f) \cap Im(f)$, yes you need to find the vectors that are in both subspaces.
For the second part, $Ker(f)+Im(f)$, how is '+' defined for two sets?
I'm not sure you have the right vectors for your basis of Im(f).

5. Jan 27, 2016

### Kernul

I made some miscalculations. It's $-1/3$ instead of $-1/13$ and $-2/3$ instead of $-1$.

so I know that I have to put together both $Ker(f)$ and $Im(f)$ in order to have a basis of $\mathbb R^4$. Or am I doing something wrong? Because it's something I learned recently.

I know that the dimension of $Ker(f) + Im(f) = dim(Ker(f)) + dim(Im(f)) - dim(Ker(f) \cap Im(f))$ and that the sum is the subspace $$Ker(f) + Im(f) = {\vec v + \vec w : \vec v \in Ker(f), \vec w \in Im(f)}$$. I have to first find the dimension of the intersection in order to find the dimension of the sum.

Why? Now that I know that the dimension of $Im(f)$ is 3, the basis would be $$B_{Im(f)} = [A^1, A^2, A^3]$$ So these are the vectors I will have to use in order to find the intersection.

6. Jan 27, 2016

### Staff: Mentor

OK, that's better. You could just as well have chosen $\begin{bmatrix} 1 \\ 0 \\ 2 \\ -3 \end{bmatrix}$ for your basis vector.
For the four vectors, I would use the kernel basis vector you found, plus three linearly independent vectors that form a basis for the image of f.
Can any vector be in both the kernel and the image?
Writing them as $A^1, A^2, A^3$ is confusing, as it suggests you are raising them to powers. It's better to write them as $A_1, A_2, A_3$, with subscripts, not exponents. Again, these three vectors should be a basis for Im(f).

7. Jan 28, 2016

### Kernul

Yes, that's exactly what I wrote.
with $\begin{pmatrix} -1/13\\ 0\\ -1\\ 1 \end{pmatrix}$ being the kernel basis vector and $\begin{pmatrix} 13\\ 0\\ 0\\ 0 \end{pmatrix}, \begin{pmatrix} 1\\ 10\\ 0\\ 0 \end{pmatrix}, \begin{pmatrix} -2\\ 0\\ 9\\ 6 \end{pmatrix}$ being the image basis vectors which are $A^1, A^2, A^3$ respectively.

I guess no? I don't know how to check that. Do I have to calculate something?

I'm sorry but my lecturer taught us that when it is a column of a matrix we write $A^x$, if it is a row of a matrix we write $A_x$.
And yes, that's what I wrote:

8. Jan 28, 2016

### Staff: Mentor

How are kernel and image defined?

9. Jan 29, 2016

### HallsofIvy

Staff Emeritus
I would NOT write the first problem in terms of matrices. To find the kernel simply means finding x, y, z, and t such that
13x+y−2z+3t= 0,10y= 0, 9z+6t= 0, and 6z+4t= 0. From the second equation, y= 0 and then the other equations are easy.

10. Jan 29, 2016

### Kernul

$Ker(f) = \{\vec v \in V : f(\vec v) = \vec 0\}$
$Im(f) = \{\vec w \in W : \exists \vec v \in V, f(\vec v) = \vec w\} = \{f(\vec v) : \vec v \in V\}$
So it is a vector $\vec v = \vec 0$? The null vector?

And yeah, I know I wasn't bound to use the matrix. I actually put the first three rows in a system and assigned $t = a$.

11. Jan 29, 2016

### HallsofIvy

Staff Emeritus
For any linear transformation, T, T(0)= 0 so the 0 vectors is always in the kernel- in fact the kernel is a subspace. But it is not necessarily the only vector in the kernel. As I said before, solve the equations 13x+y−2z+3t= 0,10y= 0, 9z+6t= 0, and 6z+4t= 0 for x, y, z, and t. Since you have only three equations in three unknowns you will have at least some of the variables in terms of others.

12. Jan 29, 2016

### Kernul

I have something like this:
\begin{array}{l}
13x + y - 2z + 3t = 0\\
10y = 0\\
9z + 6t = 0\\
6z + 4t = 0
\end{array}
\begin{array}{l}
x = + \frac{2z}{13} - \frac{3t}{13}\\
y = 0\\
z = - \frac{6t}{9} = - \frac{2t}{3}\\
t = - \frac{6z}{4} = - \frac{3z}{2}
\end{array}

13. Jan 29, 2016

### HallsofIvy

Staff Emeritus
So finish it! If x= 2z/13- 3t/13 and z= -2t/3, what is x in terms or 6 only?

14. Jan 29, 2016

### Kernul

I get this, which doesn't give me a number:
$\begin{array}{l} x = - \frac{1z}{3}\\ y = 0\\ z = - \frac{2t}{3}\\ t = - \frac{3z}{2} \end{array}$
And if I replace $z$ in $t$ as $- \frac{2t}{3}$ I get $t = t$.


text:
Verify (and optionally write) checksums. Checksums are computed using .digest, which is simply a wrapper around digest::digest.

Checksums(
path,
write,
quickCheck = FALSE,
checksumFile = file.path(path, "CHECKSUMS.txt"),
files = NULL,
verbose = getOption("reproducible.verbose", 1),
...
)

# S4 method for character,logical
Checksums(
path,
write,
quickCheck = FALSE,
checksumFile = file.path(path, "CHECKSUMS.txt"),
files = NULL,
verbose = getOption("reproducible.verbose", 1),
...
)

# S4 method for character,missing
Checksums(
path,
write,
quickCheck = FALSE,
checksumFile = file.path(path, "CHECKSUMS.txt"),
files = NULL,
verbose = getOption("reproducible.verbose", 1),
...
)

## Arguments

path Character string giving the directory path containing CHECKSUMS.txt file, or where it will be written if checksumFile = TRUE. Logical indicating whether to overwrite CHECKSUMS.txt. Default is FALSE, as users should not change this file. Module developers should write this file prior to distributing their module code, and update accordingly when the data change. Logical. If TRUE, then this will only use file sizes, rather than a digest::digest hash. This is generally faster, but will be much less robust. The filename of the checksums file to read or write to. The default is CHECKSUMS.txt located at file.path(path, module, "data", checksumFile). It is likely not a good idea to change this, and should only be used in cases such as Cache, which can evaluate if the checksumFile has changed. An optional character string or vector of specific files to checksum. This may be very important if there are many files listed in a CHECKSUMS.txt file, but only a few are to be checksummed. Numeric, -1 silent (where possible), 0 being very quiet, 1 showing more messaging, 2 being more messaging, etc. Default is 1. Above 3 will output much more information about the internals of Caching, which may help diagnose Caching challenges. Can set globally with an option, e.g., options('reproducible.verbose' = 0) to reduce to minimal Passed to digest and write.table. For digest, the notable argument is algo. For write.table, the notable argument is append.

## Value

A data.table with columns: result, expectedFile, actualFile, checksum.x, checksum.y, algorithm.x, algorithm.y, filesize.x, filesize.y indicating the result of comparison between local file (x) and expectation based on the CHECKSUMS.txt file.

## Note

In version 1.2.0 and earlier, two checksums per file were required because of differences in the checksum hash values on Windows and Unix-like platforms. Recent versions use a different (faster) algorithm and only require one checksum value per file. To update your CHECKSUMS.txt files using the new algorithm, see https://github.com/PredictiveEcology/SpaDES/issues/295#issuecomment-246513405.

## Examples

if (FALSE) {
moduleName <- "my_module"
modulePath <- file.path("path", "to", "modules")

## verify checksums of all data files
Checksums(moduleName, modulePath)

## write new CHECKSUMS.txt file

# 1. verify that all data files are present (and no extra files are present)
list.files(file.path(modulePath, moduleName, "data"))

# 2. calculate file checksums and write to file (this will overwrite CHECKSUMS.txt)
Checksums(moduleName, modulePath, write = TRUE)
}


text:
×
INTELLIGENT WORK FORUMS
FOR ENGINEERING PROFESSIONALS

Are you an
Engineering professional?
Join Eng-Tips Forums!
• Talk With Other Members
• Be Notified Of Responses
• Keyword Search
Favorite Forums
• Automated Signatures
• Best Of All, It's Free!

*Eng-Tips's functionality depends on members receiving e-mail. By joining you are opting in to receive e-mail.

#### Posting Guidelines

Promoting, selling, recruiting, coursework and thesis posting is forbidden.

# CRN ferrules are 7 times more expensive! would like to avoid that2

## CRN ferrules are 7 times more expensive! would like to avoid that

(OP)
thread794-423427: Welded Clamp Ferrules on Code stamped vessels and CRN

designer here, trying to obtain a CRN for a vessel with standard Tri-Clamp ferrules (gasket and clamp not included). I have provided calculations for the ferrules as standard nozzles (weld, thickness and reinforcement check). The reviewer requests appendix 24 calcs. I say it's not applicable as I'm not including the gasket and clamp in the CRN.

one solution (in the reference thread) calls for ferrules with CRN. OK, but they're 7 times more expensive than regular ones that I want to buy (I still provide mill tests). do you have experience like that ?

regards,

### RE: CRN ferrules are 7 times more expensive! would like to avoid that

Standard Designation: BPV Section VIII Div 1
Para./Fig./Table No:
Subject Description: Section VIII, Division 1 (1986 Edition, 1988 Addenda), U-1(e)
Date Issued: 07/07/1989
Record Number: BC89-186
Interpretation Number : VIII-1-89-112
Question: Is a clamp which fastens a piping component to a pressure vessel nozzle included in the Scope of Section VIII, Division 1?

### RE: CRN ferrules are 7 times more expensive! would like to avoid that

(OP)
david, I understand the clamp is not in the scope. But what about the ferrule that is welded to my vessel ? that's the issue.
my calcs treat the welded ferrules as standard nozzle (machined from bar stock SA-479 material)

this interpretation does not solve my issue.

regards,

### RE: CRN ferrules are 7 times more expensive! would like to avoid that

I don't agree... App. 24 applies to the design of clamp connections....and the design of a clamp connection involves the selection of the "gasket, bolting, hub and clamp geometry". If you don't know these you can only really do one equation...hub hoop stress, equation 9 of 24-6..which is not per UG-27.

### RE: CRN ferrules are 7 times more expensive! would like to avoid that

Tri clamp ferrules don't comply with Appendix 24. The scope of Appendix 24 states that the hub faces shall have metal to metal contact outside the gasket seal diameter. Sanitary type ferrules have a gasket across the whole hub face. Typically these need to be validated with a burst test.

Of course the CRN registered ones are more expensive. Someone has to pay for that certification.

### RE: CRN ferrules are 7 times more expensive! would like to avoid that

(OP)
the reviewer also asks for either CRN or burst test. help me understand the logic behind this requirement.
On the same vessel, I have a custom machined flange for a future agitator. I have provided calculations for this flange and they are accepted. So why is it acceptable for a machined flange but not accepted for a machined ferrule. I have mill tests, approved material and all that. why a burst test ?

### RE: CRN ferrules are 7 times more expensive! would like to avoid that

Flange calculations are well established in ASME VIII-1 Appendix 2 for flanges. I am not aware of any calculations that cover these ferrules. Mill test reports only give material properties, they don't prove that the design of the ferrule is sufficient. How do you know the dimensions are acceptable for the pressure and temperature rating? What analysis have you provided the reviewer to show that these are acceptable for ASME VIII-1 use?

When no rules are given for a particular component, the component shall be verified with a burst test. See UG-19(c). U-2(g) would also permit alternative analysis procedures that are outside the code (FEA, textbook formulas, calculations from another code, etc.)

### RE: CRN ferrules are 7 times more expensive! would like to avoid that

2

#### Quote (wydim)

CRN ferrules are 7 times more expensive!

"Everyone is entitled to their own opinions, but they are not entitled to their own facts."

### RE: CRN ferrules are 7 times more expensive! would like to avoid that

(OP)
cbPVme, rules are given for a standard nozzle wall thickness, welds and opening reinforcement. That's what I request to use. To my understanding, burst test are usefull for a complicated and complex geometrical shape like valve bodies, and very specific equipment.

ironic metallurgist: it literraly takes 5 min. with Compress software to add a custom nozzle size to a vessel. The $is real. I have for (2100$ of CRN ferrules) vs (300\$ normal ferrules + 15 min. of calculations in Compress). that's for only 1 project. We have between 4 and 12 ASME rated project per year.

### RE: CRN ferrules are 7 times more expensive! would like to avoid that

(OP)
Another thing that poped into my mind while thinking about this : how can I do a burst test with an open component. If I put a tri-clamp cap and clamp the ferrule, then i'm not evaluating the ferrule alone, but the whole assembly. Nothing proves the ferrule by itself is strong enough! my CRN design request stops at the ferrule sealing face, not a mm further.

another thought: an existing CRN for ferrules covers the exact geometrical shape and the specific material (what else is there?!). If that's the case, then every ferrule that has this geometrical shape and material is covered ! by default, every ferrule has a CRN, only they are not marked ! I can prove the geometrical shape by drawing and the mill test takes care of the material.

### RE: CRN ferrules are 7 times more expensive! would like to avoid that

wydim, making your case in this forum is all well and good, but gets you nowhere with the jurisdiction. Going to need to take it up there...

Regards,

Mike

The problem with sloppy work is that the supply FAR EXCEEDS the demand

### RE: CRN ferrules are 7 times more expensive! would like to avoid that

You are saying that your calculations only take into account the nozzle/shell junction (can be done in Compress in 15mins).

What calculations have you provided to justify the design of the sealing surface of the ferrule?

### RE: CRN ferrules are 7 times more expensive! would like to avoid that

(OP)
SnTMan : I guess I was trying to see if anyone had previous experience with the non-CRN ferrule on a code-vessel.

Marty007: none. And the discussion in this forum has showed that the sealing surface of ferrule/gasket is excluded from the scope of the code (and from my CRN request)

### RE: CRN ferrules are 7 times more expensive! would like to avoid that

It's not just the design of a component, but also the manufacturer and their quality program, that is covered with a CRN.

Bottom line is that you need a CRN for every pressure component in a Canadian system. Full stop. Either you get it yourself or you buy one from a supplier that has them. I recommend the latter.

### RE: CRN ferrules are 7 times more expensive! would like to avoid that

(OP)
"Bottom line is that you need a CRN for every pressure component in a Canadian system"

TGS4, no. I don't need a CRN for my machined flange that I provided calcs for.
UG-14 permit the use of rod and bars (provided with accepted material).

I'll let you know how it ends.

### RE: CRN ferrules are 7 times more expensive! would like to avoid that

For the machined flange that you provided calcs, you get a CRN for either that component or the piping system. But it does have a CRN.

### RE: CRN ferrules are 7 times more expensive! would like to avoid that

(OP)
TGS4, I'm sorry, I don't understand what you mean.
my machined flange is welded to my pressure vessel. I will eventually get a CRN for the vessel (as a whole), but the machined flange doesn't have a CRN on it's own. I'm looking to apply the same principle to the ferrules.

#### Red Flag This Post

Please let us know here why this post is inappropriate. Reasons such as off-topic, duplicates, flames, illegal, vulgar, or students posting their homework.

#### Red Flag Submitted

Thank you for helping keep Eng-Tips Forums free from inappropriate posts.
The Eng-Tips staff will check this out and take appropriate action.

#### Resources

White Paper - Addressing Tooling and Casting Requirements at the Design Stage
Several of the tooling and casting requirements of a part can be addressed at the design stage. If these requirements are not addressed at the design stage, lot of time is spent in design iteration when the design reaches the die caster. These design issues lead to increase in time and cost of production leading to delay in time to market and reduced profits for the organization. Download Now
White Paper - The Other Side of Design for Assembly
Assembly level constraints need to be satisfied before the design can move downstream. This white paper will go through the various assembly level issues, which need to be tackled by various organizations on a regular basis. Know more about DFMPro, a design for assembly software. Download Now

Close Box

# Join Eng-Tips® Today!

Join your peers on the Internet's largest technical engineering professional community.
It's easy to join and it's free.

Here's Why Members Love Eng-Tips Forums:

• Talk To Other Members
• Notification Of Responses To Questions
• Favorite Forums One Click Access
• Keyword Search Of All Posts, And More...

Register now while it's still free!


text:
# What fractions are equivalent to 4/8

Decimal and Fraction Conversion Chart
Fraction Equivalent Fractions
4/7 8/14 48/84
5/7 10/14 60/84
6/7 12/14 72/84
1/8 2/16 12/96

23 more rows

## Are 2 4 and 4/8 are equivalent?

The fraction 1/2 and 2/4 and 4/8 show the same amount of shaded or colored parts. 1/2 and 2/4 and 4/8 are equivalent fractions. Equivalent fractions are fractions that have different forms but the same value.

## What fraction is equivalent to?

Equivalent Fractions ChartUnit FractionEquivalent Fractions1/32/6, 3/9, 4/12..1/42/8, 3/12, 4/16..1/52/10, 3/15, 4/20,..1/62/12, 3/18, 4/24,..4 more rows

## What is 4/8 as a whole number?

4 / 8 = 0.5 If you insist on converting 4/8 to a whole number, the best we can do is round the decimal number up or down to the nearest whole number.

## What is 4 9 equal to as a fraction?

8/18Decimal and Fraction Conversion ChartFractionEquivalent Fractions4/98/1816/365/910/1820/367/914/1828/368/916/1832/3623 more rows

## What is 1/3 the same as?

Answer: The fractions equivalent to 1/3 are 2/6, 3/9, 4/12, etc. Equivalent fractions have the same value in the reduced form. Explanation: Equivalent fractions can be written by multiplying or dividing both the numerator and the denominator by the same number.

## What is 4/8 in the simplest form?

Let’s take a look at the same example, 4/8 :Find the GCF of 4 and 8 , which is 4 .4 divided by 4 is 1 .8 divided by 4 is 2 ./8 in simplest form is 1/2 .You can always convert a fraction to decimal, in this case 0.5 .

## Which of the following is equivalent to 4 8?

Equivalent fractions are defined as those fractions which are equal to the same value irrespective of their numerators and denominators. For example, both 6/12 and 4/8 are equal to 1/2, when simplified, which means they are equivalent in nature.

## Which of the following is equal to 4 8?

Decimal and Fraction Conversion ChartFractionEquivalent Fractions4/78/1436/635/710/1445/636/712/1454/631/82/169/7223 more rows

## What is the fraction 3/4 equivalent to?

Equivalent fractions of 3/4 : 6/8 , 9/12 , 12/16 , 15/

## What fraction is 2/3 equivalent to?

An equivalent fraction of two-thirds (2/3) is sixteen twenty-fourths (16/24).

## What is 3/6 equivalent to as a fraction?

3/6 is 3 between 6, which is 0.5.

## What is 2/8 equivalent to as a fraction?

1/4Fractions equivalent to 1/4: 2/8, 3/12, 4/16, 5/20 and so on … Fractions equivalent to 2/4: 4/8, 6/12, 8/16, 10/20 and so on … Fractions equivalent to 3/4: 6/8, 9/12, 12/16, 15/20 and so on … Fractions equivalent to 1/5: 2/10, 3/15, 4/20, 5/25 and so on …

## How to find equivalent fractions?

To find equivalent fractions, you just need to multiply the numerator and denominator of that reduced fraction ( 1 2) by the same natural number, ie , multiply by 2, 3, 4, 5 , 6 …

## Is 4 8 a fraction?

Important: 4 8 looks like a fraction, but it is actually an improper fraction.

## Can you convert fractions to decimals?

This Equivalent Fractions Table/Chart contains common practical fractions. You can easily convert from fraction to decimal, as well as, from fractions of inches to millimeters.

## What are Equivalent Fractions?

Equivalent fractions are fractions with different numbers representing the same part of a whole. They have different numerators and denominators, but their fractional values are the same.

## How to make a fraction equivalent?

Multiply both the numerator and denominator of a fraction by the same whole number. As long as you multiply both top and bottom of the fraction by the same number, you won’t change the value of the fraction , and you’ll create an equivalent fraction.

## What is half of a fraction?

For example, think about the fraction 1/2. It means half of something. You can also say that 6/12 is half, and that 50/100 is half. They represent the same part of the whole. These equivalent fractions contain different numbers but they mean the same thing: 1/2 = 6/12 = 50/100

## How to find equivalent fractions?

To find equivalent fractions, you just need to multiply the numerator and denominator of that reduced fraction ( 38) by the same natural number, ie , multiply by 2 , 3, 4, 5, 6 …

## Can you convert fractions to decimals?

This Equivalent Fractions Table/Chart contains common practical fractions. You can easily convert from fraction to decimal, as well as, from fractions of inches to millimeters.

## How to find equivalent fractions?

To find equivalent fractions, you just need to multiply the numerator and denominator of that reduced fraction ( 14) by the same natural number, ie, multiply by 2, 3, 4, 5, 6

## Is 2 8 a fraction?

Important: 2 8 looks like a fraction, but it is actually an improper fraction.

## Can you convert fractions to decimals?

This Equivalent Fractions Table/Chart contains common practical fractions. You can easily convert from fraction to decimal, as well as, from fractions of inches to millimeters.


text:
# Extension of pseudometrics to Hausdorff completion

Let $(X,\mathcal{U})$ be a uniform space with Hausdorff completion $(X',\mathcal{U}')$ (made by the minimal Cauchy filters). Since $X$ is uniform, $\mathcal{U}$ is generated by pseudometrics $(d_i)_{i \in I}$, which are uniformly continuous for $\mathcal{U}$. Consequently, there are for every $i \in I$ extensions $d_i': X'\times X'\rightarrow \mathbb{R}_{\geq 0}$, which can be proven to be pseudometrics using the fact that the image of $X$ in $X'$ is dense. Is it true that $U'$ is generated by the $(d_i')_{i \in I}$ and if so, how do you prove it ? Necessarily, the uniformity generated by $(d_i')_{i \in I}$ is part of $\mathcal{U}'$, but how do you prove the converse ?

Any help will be appreciated.

-


text:
If we introduce the notion of a massless string to denote the fact that net force on a massless string will always be $0$, since it is massless . How can these massless strings ever accelerate when force on them is $0$ .

As an example consider two blocks joined by a massless string , you push a block , the other block accelerates and so does the massless string . But massless string has no net force .

How to resolve this paradox ?

-

$F=ma$. If $F=0$, and $m=0$, $a$ can be anything. Most physical laws are not "A causes B". They usually say that "A and B can coexist in these conditions". So, it is not necessarily "Force causes acceleration". It is "an accelerating body can coexist with a force if $F=ma$"

The net force on a massless string is always 0 -- it has to be (otherwise it will have infinite acceleration). Whenever we draw free body diagrams of systems that contain massless strings, we always take a tension force $T$ that represents the string "pulling" the body. Take the reaction force of $T$ on the string and you'll notice that the string is always in equilibrium.

For example, take this system, where someone is pulling a set of two boxes interconnected by a string:

Note that the reaction force of $T$ (in red) on the string balances itself.

For a more complicated system, take the following:

(I've taken a massless smooth pulley here. If the pulley wasn't smooth, then the tensions in the two portions of string would be different. If it wasn't massless, the force from the ceiling would be different)

In this system $T$ balances out on the string as well. This is precisely why we say that a section of string exerts $T$ on both ends — to maintain equilibrium.

-

Non-relativistic mechanics can't. Massless objects travel at the speed of light. The only reason to introduce a massless string is so that you can get some effect from the string without having to worry about the string in calculations. As soon as you start worrying about forces on the string causing it to accelerate you've violated the whole reason for making it massless in the first place.

-
Yes , so what is a massless object in non-relativistic mechanics ?. I tagged the question with newtonian mechanics to make it clear . – user23503 May 16 '13 at 7:06
Get some effect ? – user23503 May 16 '13 at 7:08
A massless object in non-relativistic mechanics is no object at all. It doesn't exist. It's only there to enforce some rule and make calculations simple. – Brandon Enright May 16 '13 at 7:08
But we are always considering constraints due to that and implying force on it is 0 – user23503 May 16 '13 at 7:09
For your massless string connecting two blocks together example, the massless string is just a conceptual short-hand. You don't worry about what happens to the massless string, you just use it as a guide to figure out what force is being applied through it to a second block. Put another way, the "mass" of the massless string connecting the two blocks together is the sum of the masses of the two blocks. You've attached them all together as though they are one system. – Brandon Enright May 16 '13 at 7:15