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## Algebra & Number Theory

### Vanishing of trace forms in low characteristics

Skip Garibaldi

#### Abstract

Every finite-dimensional representation of an algebraic group $G$ gives a trace symmetric bilinear form on the Lie algebra of $G$. We give criteria in terms of root system data for the existence of a representation such that this form is nonzero or nondegenerate. As a corollary, we show that a Lie algebra of type $E8$ over a field of characteristic 5 does not have a “quotient trace form”, answering a question posed in the 1960s.

#### Article information

Source
Algebra Number Theory, Volume 3, Number 5 (2009), 543-566.

Dates
Revised: 5 March 2009
Accepted: 6 April 2009
First available in Project Euclid: 20 December 2017

https://projecteuclid.org/euclid.ant/1513797448

Digital Object Identifier
doi:10.2140/ant.2009.3.543

Mathematical Reviews number (MathSciNet)
MR2578888

Zentralblatt MATH identifier
1282.20052

Subjects
Primary: 20G05: Representation theory
Secondary: 17B50: Modular Lie (super)algebras 17B25: Exceptional (super)algebras

#### Citation

Garibaldi, Skip. Vanishing of trace forms in low characteristics. Algebra Number Theory 3 (2009), no. 5, 543--566. doi:10.2140/ant.2009.3.543. https://projecteuclid.org/euclid.ant/1513797448

#### References

• <NAME> and <NAME>, “Étale slices for algebraic transformation groups in characteristic $p$”, Proc. London Math. Soc. $(3)$ 51:2 (1985), 295–317.
• <NAME>, “Trace forms on Lie algebras”, Canad. J. Math. 14 (1962), 553–564.
• <NAME> and <NAME>, “The Lie algebras with a nondegenerate trace form”, Illinois J. Math. 8 (1964), 543–549.
• <NAME>, Lie groups and Lie algebras. Chapters 4–6, Springer, Berlin, 2002.
• <NAME>, Finite groups of Lie type. Conjugacy classes and complex characters, Wiley, New York, 1985.
• <NAME>, “Representations of Lie algebras of classical type with applications to linear groups”, J. Math. Mech. 9 (1960), 307–326.
• <NAME> and <NAME>, Groupes algébriques. Tome I: Géométrie algébrique, généralités, groupes commutatifs, Masson & Cie, Éditeur, Paris, 1970. Avec un appendice Corps de classes local par Mich<NAME>.
• <NAME>, “Semisimple subalgebras of semisimple Lie algebras”, Mat. Sbornik N.S. 30(72) (1952), 349–462 (3 plates).
• <NAME>, “Orthogonal representations of twisted forms of $\rm SL\sb 2$”, Represent. Theory 12 (2008), 435–446.
• <NAME> and <NAME>, “Globally maximal arithmetic groups”, J. Algebra 272:2 (2004), 625–642.
• <NAME>, “Die adjungierten Darstellungen der Chevalley-Gruppen”, Arch. Math. $($Basel$)$ 42:5 (1984), 408–416.
• <NAME>, “Almost-classical Lie algebras. I, II”, Nederl. Akad. Wetensch. Indag. Math. 44:4 (1982), 441–452, 453–460.
• <NAME>, Conjugacy classes in semisimple algebraic groups, Mathematical Surveys and Monographs 43, American Mathematical Society, Providence, RI, 1995.
• <NAME>, Representations of algebraic groups, 2nd ed., Mathematical Surveys and Monographs 107, American Mathematical Society, Providence, RI, 2003.
• <NAME>, “Nilpotent orbits in representation theory”, pp. 1–211 in Lie theory, edited by <NAME> and <NAME>, Progr. Math. 228, Birkhäuser, Boston, 2004.
• <NAME> and <NAME>, “The line bundles on the moduli of parabolic $G$-bundles over curves and their sections”, Ann. Sci. École Norm. Sup. $(4)$ 30:4 (1997), 499–525.
• <NAME>, “Classification des algèbres de Lie simples”, pp. 245–286 in Séminaire Bourbaki, 1998/99, Astérisque 266, Soc. Math. de France, Paris, 2000.
• <NAME>, <NAME>, and <NAME>, Tables of representations of simple Lie algebras, I: Exceptional simple Lie algebras, Université de Montréal Centre de Recherches Mathématiques, Montreal, 1990.
• <NAME>, “Rost invariants of simply connected algebraic groups”, pp. 101–158 in Cohomological invariants in Galois cohomology, Univ. Lecture Ser. 28, Amer. Math. Soc., Providence, RI, 2003.
• <NAME>, “Irreducible representations of Lie algebras of reductive groups and the Kac–Weisfeiler conjecture”, Invent. Math. 121:1 (1995), 79–117.
• <NAME>, Jr., “Conjugacy classes in Lie algebras and algebraic groups”, Ann. of Math. $(2)$ 86 (1967), 1–15.
• <NAME>, Modular Lie algebras, Ergebnisse der Math. 40, Springer, 1967.
• <NAME> and <NAME>, “Conjugacy classes”, pp. 167–266 in Seminar on algebraic groups and related finite groups (Princeton, NJ, 1968/69), Lecture Notes in Mathematics 131, Springer, Berlin, 1970.
• <NAME>, “Automorphisms of classical Lie algebras”, Pacific J. Math. 11 (1961), 1119–1129. Reprinted as pp. 101–111 in his Collected Papers, American Math. Soc., Providence, 1997.
• <NAME>, “Générateurs, relations et revêtements de groupes algébriques”, pp. 113–127 in Colloque sur la Théorie des Groupes Algébriques (Bruxelles), Librairie Universitaire, Louvain, 1962. Reprinted as pp. 113–127 in his Collected Papers, American Math. Soc., Providence, 1997.
• <NAME>, “Representations of algebraic groups”, Nagoya Math. J. 22 (1963), 33–56. Reprinted as pp. 149–172 in his Collected Papers, American Math. Soc., Providence, 1997.
• <NAME>, Lectures on Chevalley groups, Yale University, New Haven, 1968.
• <NAME>, “Torsion in reductive groups”, Advances in Math. 15 (1975), 63–92. Reprinted as pp. 415-444 in his Collected Papers, American Math. Soc., Providence, 1997.
• <NAME>, Simple Lie algebras over fields of positive characteristic, I: Structure theory, Expositions in Mathematics 38, de Gruyter, Berlin, 2004.
• <NAME>, “Classification of algebraic semisimple groups”, pp. 33–62 in Algebraic groups and discontinuous subgroups (Boulder, CO, 1965), Amer. Math. Soc., Providence, R.I., 1966, 1966.


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Address 2026 <NAME> Dr, Beckley, WV 25801 (304) 253-1700

# latex error environment description undefined Princewick, West Virginia

Emilio. 14. text 2 c. What does Differential Geometry lack in order to "become Relativity" - References more hot questions question feed lang-tex about us tour help blog chat data legal privacy policy work here advertising flying-sheep commented Mar 29, 2016 sorry to be nitpicky, but i think what you really had to do is unload and reload repr in the kernel (probably easiest done by killing

You can find the package documentation here. When I try to reference an embedded enumeration, rather than getting just "I" I get "Ia". September 2012 at 10:10 Hey! Thanks again.

Thx in advance. October 2011 at 6:25 Hi Tom, How would I begin enumerating a later number than 1? l.423 \begin{description*} ! ==> Fatal error occurred, no output PDF file produced! Here is an example I got from this google group: \documentclass{article} \newcounter{description} \renewcommand{\descriptionlabel}[1]{% \refstepcounter{description}% \hspace\labelsep \normalfont\bfseries \thedescription. #1:} \begin{document} \begin{description} \item[Animals] Emus. \item[Cars] Ferraris. \end{description} \end{document} Cheers, Tom.

Reply asd 26. flying-sheep commented Mar 29, 2016 can you try with the newest IRkernel/repr from github? January 2016 at 13:09 Hi there, My suggestion would be to write a macro. Best, Tom.

However, it gives me error message of: 500 : Internal Server Error nbconvert failed: PDF creating failed However, the function of download as pdf would work for a iPython notebook with flying-sheep commented Jul 10, 2015 oh sorry, i meant package?repr Please don't forget about the issue with % characters not being escaped as well. share|improve this answer answered Jul 4 '12 at 17:06 <NAME> 28.8k7106178 5 Related Question: Why not use the minimal class. –<NAME> Jul 4 '12 at 17:13 add a text 3 Some more text.

The GUI download as feature still fails (500 internal server error nbconvert failed: PDF creating failed) gidden commented Mar 29, 2016 Ok, finally figured it out. See the LaTeX manual or LaTeX Companion for explanation. Best, Tom. Interestingly, the description* block matches the output of the IRKernel. \begin{description*} \item[X56403463.1727.4c23.a709.de5f168c4073] 50.0006258799053 \item[ca8afb2c.be3d.41e8.8dbc.19fe7ab6154b] 30.3254246898821 \item[CHC433] 21.169034458756 \item[X65e430e1.0c9a.4045.83a9.653b7cff811d] 14.2808853888554 \item[X6dcfc418.ab28.4365.95ea.c1ae254f2341] 11.8185201121792 \end{description*} json from notebook "text/latex": [ "\\begin{description*}\n", "\\item[X56403463.1727.4c23.a709.de5f168c4073] 50.0006258799053\n", "\\item[ca8afb2c.be3d.41e8.8dbc.19fe7ab6154b]

September 2012 at 10:14 Works exactly the same way. \begin{enumerate}\itemsep2pt \item \item \end{enumerate} Cheers, Tom. I tried converting to just latex and the table output is not the same as what repr::repr_latex provides (i.e., jupyter nbconvert does not appear to be using the latest IRkernel installed Reply Andreas 7. September 2015 at 8:37 Hi Sufipeace, Interesting questions, thanks! \item takes an optional argument, see description environment.

February 2012 at 2:27 Hi Tom, I have used \renewcommand{\labelenumi}{(\roman{enumi})} \begin{enumerate} \item First line \item Second line \end{enumerate} but my problem is the format. Mark 26. I don't know of any package which does that. You signed out in another tab or window.

I'm strugling with creating a list of symbols by using the description environment: \begin{description} \item[A]\hspace{0.8in} first letter \item[B]\hspace{0.8in} second letter \item[C]\hspace{0.8in} third \end{description} I want something like: A first letter B Underfull \hbox (badness 10000) in paragraph at lines 16--18 ! Reply tom 6. When your description has a text length longer then \textwidth, the text wont be cut of at \textwidth but at \textwidth + 1cm Do you maybe have a solution which prevents

October 2011 at 3:06 Hi Alison, Set the counter to 2, so your first item will have index 3: \begin{enumerate} \setcounter{enumi}{2} ... \end{enumerate} Cheers, Tom. Cheers, Tom. \documentclass[11pt]{article} \usepackage{etaremune} \begin{document} \renewcommand{\labelenumi}{[\theenumi]} \begin{etaremune} \item Item 3 \item Item 2 \item Item 1 \end{etaremune} \end{document} molom 8. I would like to know, if we can change the space between the nested lists, as there is too much space left between the first level items and second level items. Linked 68 Why should the minimal class be avoided?

I'm curious which solution you'll go with! December 2011 at 9:04 Hi there, Try this: \documentclass[11pt]{article} \usepackage{color} \begin{document} \renewcommand{\labelitemi}{${\color{red}\bullet}$} \begin{itemize} \item first \item second \end{itemize} \end{document} Cheers, Tom. Reply lailame 12. LaTeX Error: Environment description* undefined.

Personal Open source Business Explore Sign up Sign in Pricing Blog Support Search GitHub This repository Watch 21 Star 117 Fork 109 jupyter/nbconvert Code Issues 88 Pull requests 5 Projects Ooops, the webpage doesn't like my formatting. September 2010 at 14:42 You should try \usepackage .site-inner 0 and, depending on which computer language you use, to configure a "style" for it (including colorized output) with \lstset{}. Reload to refresh your session.

It seems so basic, but I can't figure it out. Thanks for pointing this out. Nevertheless, I think you will have problems creating a frame similar to the html above, as figure is a floating environment, which can't be put into a box. On several lines. [2] This is my second item.

Reply tom 10. How can I align them up? The determinant of the matrix Equation which has to be solved with logarithms How do you curtail too much customer input on website design? May 2012 at 15:17 Hi Rado, You can use \textnormal{} in any math environment form normal text.

format: plain


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## R code for fitting a multitype NHPP model to grouped temporal data

The following R code demonstrates how to fit a multitype NonHomogeneous Poisson Process (NHPP) model to grouped temporal data. Multitype Poisson processes belong to the family of marked point processes.
Note that grouped temporal data occurs when only the number of recurrences within a given time interval are known.

The R code may be used to fit 1) a multitype NHPP model with a loglinear intensity function, with the intensity at time t defined by $e^{\gamma_{0} + \gamma_{1}t}$, or 2) a multitype NHPP model with a power law intensity function, with the intensity at time t defined by $\gamma_{0}*t^{\gamma_{1}}$.

Furthermore, the R code also shows how to fit a multitype Homogeneous Poisson Process (HPP) model to grouped temporal data.

## R code for fitting a multitype NHPP model to temporal data

The following R code demonstrates how to fit a multitype NonHomogeneous Poisson Process (NHPP) model to temporal data. Multitype Poisson processes belong to the family of marked point processes.

The R code may be used to fit 1) a multitype NHPP model with a loglinear intensity function, with the intensity at time t defined by $e^{\gamma_{0} + \gamma_{1}t}$, or 2) a multitype NHPP model with a power law intensity function, with the intensity at time t defined by $\gamma_{0}*t^{\gamma_{1}}$.

Furthermore, the R code also shows how to fit a multitype Homogeneous Poisson Process (HPP) model to temporal data.

## R code for fitting a nonhomogeneous temporal Poisson process model using the spatstat package

In my previous blog post I showed how to fit power and loglinear nonhomogeneous Poisson process (NHPP) models to temporal data.

The following R code fits these same two models, but this time using the spatstat package.

## R code for fitting a nonhomogeneous temporal Poisson process model

The following R code demonstrates how to fit a nonhomogeneous Poisson process (NHPP) model to temporal data.

The R code may be used to fit 1) a NHPP model with a loglinear intensity function, with the intensity at time t defined by $e^{\gamma_{0} + \gamma_{1}t}$ , or 2) a NHPP model with a power law intensity function, with the intensity at time t defined by $\frac{\beta}{\eta}(\frac{t}{\eta})^{\beta-1}$. The latter NHPP model is sometimes referred to as the Crow-AMSAA model.

## R code for fitting the Crow-AMSAA model to grouped data

The Crow-AMSAA model is frequently used for analyzing and assessing reliability growth. In case of grouped data the Crow-AMSAA model parameters have to be estimated by numerical optimization. Grouped data occurs when only the number of failures within a given time period are known.

The Crow-AMSAA model in the R code fits a NonHomogenous Poisson Process (NHPP) having a power law intensity function. See this post for fitting a NHPP model having a loglinear intensity function. Continue reading R code for fitting the Crow-AMSAA model to grouped data


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Documentation

# relu

Apply rectified linear unit activation

## Description

The rectified linear unit (ReLU) activation operation performs a nonlinear threshold operation, where any input value less than zero is set to zero.

This operation is equivalent to

$f\left(x\right)=\left\{\begin{array}{cc}x,& x>0\\ 0,& x\le 0.\end{array}$

### Note

This function applies the ReLU operation to dlarray data. If you want to apply ReLU activation within a layerGraph object or Layer array, use the following layer:

example

dlY = relu(dlX) computes the ReLU activation of the input dlX by applying a threshold operation. All values in dlX that are less than zero are set to zero.

## Examples

collapse all

Use the relu function to set negative values in the input data to zero.

Create the input data as a single observation of random values with a height and width of 12 and 32 channels.

height = 12;
width = 12;
channels = 32;
observations = 1;

X = randn(height,width,channels,observations);
dlX = dlarray(X,'SSCB');

Compute the leaky ReLU activation.

dlY = relu(dlX);

All negative values in dlX are now set to 0.

## Input Arguments

collapse all

Input data, specified as a dlarray with or without dimension labels.

Data Types: single | double

## Output Arguments

collapse all

ReLU activations, returned as a dlarray. The output dlY has the same underlying data type as the input dlX.

If the input data dlX is a formatted dlarray, dlY has the same dimension labels as dlX. If the input data is not a formatted dlarray, dlY is an unformatted dlarray with the same dimension order as the input data.


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# Frequency problem

## Homework Statement

In a quartz oscillator, used as a stable clock in electronic devices, a transverse (shear) standing sound wave is excited across the thickness d of a quartz disk and its frequency f is detected electronically. The parallel faces of the disk are unsupported and so behave as "free ends" when the sound wave reflects from them as shown in the figure.

If the oscillator is designed to operate with the first harmonic, determine the required disk thickness if 88.0 MHz. The density and shear modulus of quartz are rho = 2650 kg/m^3 and G = 2.95 *10^10 N/m^2.

## Homework Equations

Possibly:
$$\Delta$$$$l=(1/G)(F/A)l_{0}$$
$$f=v/2l$$
$$l=(1/2)\lambda$$

## The Attempt at a Solution

I'm not even sure where to start with this. I tried looking in the section the question came from in the book, but I couldn't find anything about mediums other than air, let alone using a shear modulus. In the table provided, the speed of sound in quartz wasn't listed, so I assume I don't need to bother figuring that out. I'm not really certain where the shear modulus and density come into play either.

(edit:) I tried $$\lambda$$$$=[(3.7km/s)(1000)]/88000000Hz$$ (3.7km/s is the speed of sound in quartz according to google) and divided that by 2 to get $$l$$, but that's apparently not the correct answer...

Last edited:


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# Chapter 14 Trigonometric Graphs, Identities, and Equations - 14.7 Apply Double-Angle and Half-Angle Formulas - 14.7 Exercises - Skill Practice - Page 960: 23

$\dfrac{ 2 \tan \theta}{(1+\tan \theta) }$

#### Work Step by Step

Use the Identity $\tan 2 \theta =\dfrac{ 2 \tan \theta}{1-\tan^2 \theta}$ Thus, we have $(1-\tan \theta) \tan 2 \theta=(1-\tan \theta) \times \dfrac{ 2 \tan \theta}{1-\tan^2 \theta}$ or, $=(1-\tan \theta) \times \dfrac{ 2 \tan \theta}{(1-\tan \theta) (1+\tan \theta) }$ or, $=\dfrac{ 2 \tan \theta}{(1+\tan \theta) }$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.


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# Turbine Efficiency:Complete Insights and FAQs

The efficiency of any device is a measure of its performance

The turbine efficiency is expressed as the ratio of useful work energy output to the input energy.

There are mainly two types of a steam turbines. One is an impulse turbine, and another is a reaction turbine. The efficiency of both turbines is different, and it depends upon its action of work.

There are many factors on which the turbine efficiency depends. It can be blade angle, guide vanes angle at the inlet to the turbine, velocity of steam, etc.

## turbine efficiency formula

We can know and compare the performance of turbine with turbine efficiency

Here,

h3 – enthalpy on point 3

h4 – the enthalpy point 4 (actual turbine)

h4s – the enthalpy point 4s (isentropic turbine)

The isentropic efficiency of the turbine can give us an ideal performance of the turbine. The heat rejected to the surrounding is to be assumed as zero. The isentropic efficiency can be calculated by the ratio of actual work to ideal work.

The ideal turbine is a device that works with 100 % efficiency. It means that all the input energy is converted into useful work. This type of turbine is impossible in the real world. There are always some losses present in the turbine in the form of heat and friction.

## turbine efficiency curve

the turbine efficiency curve for impulse and reaction turbine is shown below

## What is stage efficiency of a turbine

The stage efficiency or gross efficiency of the turbine is related to the stages of blades.

The stage efficiency is the ratio work supplied on the blade by per kg of steam to the energy given per stage (per kg of steam)

Stage efficiency can also be expressed as the multiplication of nozzle efficiency and blade efficiency.

In an impulse turbine, there are stages of blades and nozzle set. The stage efficiency is the providing performance of stage by work done in the stages and enthalpy dropping in nozzles.

## how to improve turbine efficiency

The turbine efficiency can be improved by some factors discussed below

It can be enhanced by the superior blade design, Coatings of exhaust blade and geometry

## Why turbine has more efficiency than pump

The pump and turbine efficiency can be compared based on fluid handling

The pressure losses in the pumps are higher than in the turbine. This loss can be reduced by making the boundary layer thin in the turbine.

The working fluid in the turbine is expanding from the inlet. At the inlet, the static pressure and the velocity rise continuously. The rising in the kinetic energy of the fluid leads to a decrease in losses. The thin boundary layer can decrease the losses of power.

In the pump, the boundary layer thickness of fluid is more due to deceleration to raise static pressure. This increase in thickness leads to an increase in the losses in the pump.

## average turbine efficiency

The average efficiency of the turbine varies between some range depending on size and stage.

The average efficiency of a single-stage turbine is 40 % whereas, the efficiency for the multi-stage turbine is 65 – 90%.

The efficiency in the multistage turbine is high due to the moderate to high pressure ratio. The steam will get expanding through various stages, which is beneficial to efficiency.

## What is the ideal efficiency range for practical turbines

The ideal efficiency range of a practical turbine is dependent on various losses.

The overall plant thermal efficiency for the steam turbine is 42- 45%.

The power plant running on the superheated boiler is working with fossil fuels like coal. There are many heat losses present in a practical turbine. Due to thermal losses, the efficiency of the practical turbine is low.

## turbine vs piston efficiency

The turbine and piston engine can be compared for aircraft

The piston engine is more efficient compared to the turbine due to less specific fuel consumption.

The turbine is light in weight compared to piston engines. The weight of piston engines is more due to bore, turbocharger, etc. To reduce the unbalancing in an airplane, a long nose is provided.

## What is the maximum efficiency of a generator

The generator is a device that provides electrical energy at the expense of mechanical energy.

If the losses are proportional to the square of load current, then the generator’s efficiency is maximum.

The generator efficiency can be stated as the ratio of output electrical power to input mechanical power.

The efficiency of the dc generator can be maximum if the variable and constant losses are the same. The variable losses are armature losses, and the constant losses are field circuit losses and rotational losses.

<NAME>

I am <NAME>, Pursuing PhD in Mechanical- Renewable energy. I have five years of teaching and two-year research experience. My subject area of interest are thermal engineering, automobile engineering, Mechanical measurement, Engineering Drawing, Fluid mechanics etc. I have filed a patent on "Hybridization of green energy for power production". I have published 17 research papers and two books. I am glad to be part of Lambdageeks and would like to present some of my expertise in a simplistic way with the readers. Apart from academics and research, I like wandering in nature, capturing nature and creating awareness about nature among people. Let's connect through LinkedIn - https://www.linkedin.com/in/jani-deepak-b0558748/. Also refer my You-tube Channel regarding “Invitation from Nature”


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# Calculations with Legendre's Function

## The Prompt

How would Legendre's function (shown below) change with the coordinate transformation $z = cos(\theta)$

$$\left[\sin{\theta}\frac{d}{d\theta}\left(\sin{\theta}\frac{d}{d\theta}\right)-A\sin{\theta}^{2}\right]\Theta(\theta)=0$$ Powerpoint slide
PDF slide

How is $\frac{d}{d\theta}$ related to $\frac{d}{dz}$ by our previous coordinate transformation?

Do the product rule and write things in terms of first and second derivatives of z with our newly transformed Legendre function. $$\left[\left(1-z^{2}\right)\frac{d}{dz}\left(1-z^{2}\right)\frac{d}{dz}-A\left(1-z^{2}\right)\right]\Theta(z)=0$$ Powerpoint slide
PDF slide

How do we find the coefficients of the expression? $$f(z)=\sum_{l=0}^{\infty}c_{l}P_{l}(z)$$

This SWBQ

## Wrap Up

- Fix organizational structure here.

##### Views

New Users

Curriculum

Pedagogy

Institutional Change

Publications


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transfer function and error gain function

simulate this circuit – Schematic created using CircuitLab

I know the difference between those two functions , but i want to see the relation between them , i think that the error gain function , is given by the transfer function , and the best transfer function which is

1. stable.
2. has an error gain function equal to 1.

if this is true , how can i derive the error gain function , from transfer function ?

• The stable output condition for an input of 1.0 is 0.5. Does that help at all? Regarding the rest of your question, I'm not entirely sure what you mean. – Andy aka Apr 10 '14 at 18:50
• @Andyaka since E(s) = Y(s) - X(s)G(S) , where X : is the desired output(input to system) , and G is the error gain function , then X(s)G(S) = X(s) . which is equal to Y(s) , then E(s) = 0. – hbak Apr 10 '14 at 19:02

Let me label your circuit nodes for better understanding of your question

simulate this circuit – Schematic created using CircuitLab

• Error gain function: a gain k that multiplies the error e = r-y
• Transfer function: the behavior of the closed-loop system y/r

Now,

$$y=k\cdot e$$ $$e=r-y$$ $$y=k\cdot (r-y)$$

So, the error gain function follows as

$$k=\frac{y}{r-y}$$

So, you don't even need the transfer function to establish the value of k. You just need y and r.

Anyway the transfer function is

$$TF=\frac{y}{r}=\frac{k}{k+1}$$

So, if you're given the transfer function (note that this is just a constant for this theoretical system) you can find k as

$$k=\frac{TF}{1-TF}$$

Note also that, with negative feedback, in stable system conditions, and with finite k:

$$0\leq TF<1$$

• the transfer function that i meant , that the function that you entered on it an input and give you an output , but the transfer function that you define , is the function that give it desired output r , and give you an output y , in another woed , you define the system above as transfer function , and i want the transfer function that described by this error gain function .... – hbak Apr 19 '14 at 9:33
• Can you be clearer? r is an input, not an output. Anyway the transfer function described by the error gain function (assuming you agree that this error gain function is just k, otherwise I don't understand your question) is what I defined as TF. – raggot Apr 28 '14 at 13:11


text:
How do you solve this?
There is this kind of question in our test and I don't know how will I do it.

You're working in a company. Your starting income is 5000. Every year, the income will increase by 5%. What is your total income on your 25th year in the company?

Related Precalculus Mathematics Homework Help News on Phys.org
HallsofIvy
Homework Helper
This is a question, not a tutorial so I am moving out of "Learning Materials" to "Precalculus Homework and School Work".

You startwith income at 5000 and it increases by 5% each year.

Okay, the first year your increases by "5% of 5000"= .05(5000)= 250 so your income the second year is 5250= 5000+ (.05)5000= (1.05)(5000). At the end of that year it increases by "5% of 5250"= .05(5250)= 262.50 and the third year your income is 5512.50= 5250+ (.05)5250= (1.05)(5250)= 1.05(1.05(5000)). The reason I wrote it out like that is because neither you nor I want to do that 24 times! (During your 25th year, your income will have increased 24 times.) You should be able to see what is happening: each year your income is multiplied by 1.05. After 24 years, that initial 5000 is multiplied by 1.05 24 times: $(1.05)^{24}(5000)$.

$$a_1=5000$$

$$a_2=a_1+a_1*\frac{5}{100}=a_1*1.05$$

$$a_3=a_1*1.05 + a_1*1.05*0.05=a_1*1.05(1 + 0.05)=a_1*1.05*1.05$$

$$a_4=a_1*1.05*1.05*1.05$$

$$...................................$$

$$a_{n+1}=a_1*(1.05)^{n}$$

So a25=5000*(1.05)24

Regards.

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Is 1.05 constant?

Is 1.05 constant?
Yep, its a constant.

HallsofIvy
Homework Helper
It's certainly not going to change!

ok. where is that 1.05 come from?

ok. where is that 1.05 come from?
If you see my way of solving the problem, you'll spot that

$$a_2=a_1+a_1*\frac{5}{100}=a_1(1+\frac{5}{100})=a_1\frac{100+5}{100}=a_1\frac{105}{100}=a_1*1.05$$

See now where it comes from?

HallsofIvy


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# Vector p is 6m in the x-direction .vector q is 10m in the y-direction a) use the parallelogram method to work out p +q

Arun
25758 Points
4 years ago
Parallelogram method
R² = P² + Q² + 2PQ cos theta
Because x axis and y axis are perpendicular
Hence theta = 90 degree
R² = P² + Q² +0
R = $\large \sqrt$(P²+Q²)
R = $\large \sqrt$136
birtukan simachew
26 Points
8 months ago
vector p is 6m in the x direction and vector q is 10m in the y direcion use the phythagores therorem and trigonometry to work out p-q
<NAME>
26 Points
8 months ago
if vectoe p is 6m in the x-direction& vector q is 8m in the y-direction.Then what is the value of the vector p&q?


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# Find $\lim_{x\to0}\frac{\sin5x}{\sin4x}$ using $\lim_{\theta\to0}\frac{\sin\theta}{\theta}=1$.

I am trying to find $$\lim_{x\to0}\frac{\sin5x}{\sin4x}$$

My approach is to break up the numerator into $4x+x$. So,

$$\begin{equation*} \lim_{x\to0}\frac{\sin(4x+x)}{\sin4x}=\lim_{x\to0}\frac{\sin4x\cos x+\cos4x\sin x}{\sin4x}\\ =\lim_{x\to0}(\cos x +\cos4x\cdot\frac{\sin x}{\sin4x})\end{equation*}$$

Now the problem is with $\frac{\sin x}{\sin4x}$. If I use the double angle formula twice, it is going to complicate the problem.

The hint says that you can use $\lim_{\theta\to0}\frac{\sin\theta}{\theta}=1$.

I have little clue how can I make use of the hint.

Any helps are greatly appreciated. Thanks!

• Your approach will not lead you to use the given limit because whatever substitution you make, you will remain stuck with fractions involving trigonometric functions only (and not $x$ alone). – <NAME> Jul 30 '14 at 9:17
• – <NAME> Sep 24 '16 at 7:59

Hint: Try to insert some useful parts in the given fraction properly. I mean $$\frac{5x}{4x}$$ Note that we can do this cause while $x\to 0$ then $x\neq0$.

Hint:

$$\lim_{x\to 0} \frac{\sin 5x}{\sin 4x} = \frac{5}{4}\lim_{x\to 0} \frac{\sin 5x}{5x}·\frac{4x}{\sin 4x}$$

Actually it won't complicate the problem. You can get the same result by using formula

$$\sin(2x)= 2\sin(x)\cos(x)$$

i.e. in the denominator of second term expand $\sin(4x)$ as $$2\sin(2x)\cos(2x) = 4\sin(x)\cos(x)\cos(2x).$$ By cancelling sin(x) in numerator and denominator you get a whole function in cosine terms. Just substitute zero and you get the limit.

• I must say this is a very smart solution and it shows that the limit of $(\sin 4x)/(\sin 5x)$ can be calculated without using the result that $(\sin x)/x \to 1$ as $x \to 0$. +1 – <NAME> Jul 30 '14 at 10:54

$$\lim_{x\to0}\frac{\sin5x}{\sin4x}=\lim_{x\to0}\frac{5x\frac{\sin5x}{5x}}{4x\frac{\sin4x}{4x}}=\frac{5}{4}\frac{\lim_{x\to0}\frac{\sin5x}{5x}}{\lim_{x\to0}\frac{\sin4x}{4x}}=\frac{5}{4}$$

Applying the hint will need some trigonometric manipulations. Instead, the shortest possible way to find is to apply L'Hospital's Rule. So by that you have $\lim_{x\to 0} \frac{5 \cos 5x}{4 \cos 4x}$, in which case the limit $\frac{\cos 5x}{\cos 4x}$ goes to $1$ and are you are left with $\frac{5}{4}$.

$$\lim_{x\to0}{\sin5x\over\sin4x}=\lim_{x\to0}{\sin5x\over5x}{4x\over\sin4x}\frac54.$$ $$\lim_{x\to0}{\sin5x\over5x}=\lim_{x\to0}{4x\over\sin4x}=1,$$ thus we are done.

we have to find $\lim_{x\to0}\frac{\sin5x}{\sin4x}$. hence$$\lim_{x\to0}\frac{\sin5x}{\sin4x} = \frac{5}{4}\lim_{x\to0}\frac{\sin 5x}{5x}\cdot\frac{4x}{\sin 4x} =\frac{5}{4}\cdot\lim_{x\to0}\frac{\sin 5x}{5x}\cdot\lim_{x\to0}\frac{4x}{\sin 4x}$$ hope you got the idea.....

Firstly, there is a limit law that allows you to do "substitution" of limits:

Suppose you know that $$\lim_{x\to a} f(x) = L$$ Also suppose that $g$ is some function such that $g(x) \to a$ when $x \to b$.

Then: $$\lim_{x \to b} f(g(x)) = L$$

You can think of this as performing the substitution $t = g(x)$: $$t = g(x) \quad \text{so } x \to b \implies t \to a\\ \lim_{x \to b} f(g(x)) = \lim_{t \to a}f(t) = L$$

The upshot of this is that all of these limits are equal to 1: $$\lim_{x\to 0} \frac{\sin{6x}}{6x} = \lim_{x \to 0} \frac{\sin(3x^2)}{3x^2} = \lim_{x \to 1} \frac{\sin(x-1)}{x-1} = \lim_{x\to 0} \frac{\sin(\sin(x))}{\sin(x)} =\lim_{x\to 0} \frac{\sin(x)}{x} = 1$$ And, of course, innumerably many like them.

Knowing that all of these versions of the fundamental trig limit give the same result means that if we can arrange for our limit to include any of them, then that part of the limit will be done for us. This advice should help for other limits you might see in the future.

So we can do the following: \begin{align} \lim_{x\to 0} \frac{\sin(5x)}{\sin(4x)} &= \lim_{x\to 0} \frac{\sin(5x)}{1}\frac{1}{\sin(4x)}\\ &= \lim_{x\to 0} \frac{5\sin(5x)}{5x} \frac{4x}{4\sin(4x)}\\ &= \lim_{x\to 0} \frac{5}{4}\frac{\sin(5x)}{5x}\frac{4x}{\sin(4x)}\\ &= \frac{5}{4} \times 1 \times 1\\ &=\frac54 \end{align}

What I wanted to point out is that if $x\to0$, then $4x$ and $5x\to0$ as well.

Other than that earth shattering revelation, the rest of my answer is pretty much the same as most of the other posts.

$$\begin{array}{lll} \lim_{x \to 0}\frac{\sin5x}{\sin4x}&=&\lim_{x \to 0}\frac{\frac{\sin5x}{1}}{\frac{\sin4x}{1}}\\ &=&\lim_{x \to 0}\frac{\frac{\sin5x}{1}}{\frac{\sin4x}{1}}\cdot\frac{\frac{5x}{5x}}{\frac{4x}{4x}}\\ &=&\lim_{x \to 0}\frac{\frac{\sin5x}{5x}}{\frac{\sin4x}{4x}}\cdot\frac{\frac{5x}{1}}{\frac{4x}{1}}\\ &=&\lim_{x \to 0}\frac{\frac{\sin5x}{5x}}{\frac{\sin4x}{4x}}\cdot\frac{5x}{4x}\\ &=&\frac{\displaystyle\lim_{x \to 0}\frac{\sin5x}{5x}}{\displaystyle\lim_{x \to 0}\frac{\sin4x}{4x}}\cdot\lim_{x \to 0}\frac{5x}{4x}\\ &=&\frac{1}{1}\cdot\frac{5}{4}\\ &=&\frac{5}{4}\\ \end{array}$$ Notice how on the second line I essentially multiplied by 1 (e.g. $1=5x/5x=4x/4x$).


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# SOMANET EtherCat Drive Special Tuning Mode Application¶

This application is demonstrating the use of the tuning mode of SOMANET EtherCat Drive. It allows finding the commutation offset of the motor, tuning of the PID parameters and test the Position, Velocity and Torque controllers.

## Hardware setup¶

A minimal requirement for this application to run is having the complete SOMANET stack assembled consisting of a SOMANET Core-C22, SOMANET COM-EtherCAT, and a SOMANET IFM-Drive modules. The stack should be powered via the SOMANET IFM board. An example of a stack consisting of the SOMANET COM-EtherCAT, Core, and IFM-Drive-DC100 boards is shown below. In this case the IFM-DC100 board can be supplied with 12 - 24 V DC power source. Please refer to the corresponding to your IFM board hardware documentation and also motor specifications to select the proper supply voltage. For the best experience please make sure that your stabilized DC power supply is capable of delivering more that 2 Amperes of power for low power motors (up to 100W) and more than 10A for high power motors (up to 1kW). Please mind that at high motor accelerations starting current may be as high as 10 times the nominal.

Hardware Setup for SOMANET Cyclic Positioning Control with EtherCAT Demo

To setup the system:

1. If you don’t have the stack assembled, assemble it as shown in the image above. Make sure to connect the IFM side of the SOMANET Core module to the IFM-DC100 board and COM side to the Core Debug Adapter (see markings on the Core module)
3. Connect the xTAG to host PC.
4. Connect the motor supplied with the kit as shown in the image bellow.
5. Connect the IFM board to a 24 V DC power supply
6. Connect one side of the Ethernet cable to the node and plug the RS-45 connector to your PC.
7. Switch on the power supply. If everything is connected properly, drained current should not exceed 150mA.

Connecting the motor and cables to your kit

## Build the application¶

1. You need the EtherCAT master IgH EtherLab to be installed on your system to build the application.
2. You need the ncurses library installed on your system to build the application. There is probably a package existing for you linux distribution. It’s the °libncurses-dev° package on Ubuntu.
3. The app also needs libreadsdoconfig which is in sc_sncn_ethercat_drive/examples/libreadsdoconfig
4. Navigate with the terminal to the application directory sc_sncn_ethercat_drive/examples/app_demo_master_ethercat_tuning an run make to build the app.

## Commands¶

The application provides the following command line arguments

• -h print this help and exit
• -v print version and exit
• -o enable sdo upload
• -n <slave number> starting from 0. default is slave 0.
• -s <profile velocity> velocity in rpm used for position profiler
• -c <SDO config filename> filename of the sdo config csv file. Default is sdo_config/sdo_config.csv
• -F <record filename> filename where to record the position/velocity/torque in csv format.

The application is a ncurses graphical console application. It uses simple text commands to control execute some actions or set parameters. The commands consist of 1,2 or 3 characters with an optional number. The commands are case sensitive.

The following one character commands are executed directly without pressing enter after:

• q: stop the node and quit the app
• [enter]: just pressing enter will disable the motorcontrol block the brake.
• 0: pressing 0 will switch to torque control mode with a 0 torque command. Which stops the motor slower than by disabling the motorcontrol with [enter]. Be careful if the axis is loaded it could fall.
• r: reverse the current torque/velocity command
• .: start/stop record. this record the position/velocity/torque to a csv formated file.
• [backspace]: discard current command

The rest of the commands can be up to 3 characters with an optional number. The command is then executed by pressing [enter]. The number can be negative. Spaces are ignored. The default number value is 0.

• a: start the auto offset tuning. It automatically update the offset field display. If the offset detection fails it will print WRONG SENSOR POLARITY!.
Some possible causes of failure of the offset detection are:
• The sensor polarity is wrong. This can be fixed by changing the sensor polarity with the s command.
• The torque applied during tuning is too low. This setting is set in the SDO config file. 20% should be enough for an open motor but a motor with gears or a load can need more.
• The motor is blocked. If the motor does not freely turn during the detection the offset will be wrong.
• The sensor is disconnected or not working properly. If the angle feedback is not working the offset will be wrong (but the motor will probably still turn during the tuning). Check the position feedback and the eventual sensor errors.
• The motor phases are not connected properly. This will maybe prevent the motor to turn correctly and give a wrong offset.
• The pole pairs setting is wrong. The motor will probably still turn during the tuning but the offset will be wrong.

After the offset is found you need to make sure that a positive torque command result in a positive velocity/position increment (by testing). Otherwise the position and velocity controller will not work. If this is not the case use the m command to change the phase inverted parameter.

• ap2: start the automatic tuning of cascaded position controller. while cascaded position controller is being tuned, the dynamic values of PID controllers are shown on the terminal.
• ap3: start the automatic tuning of limited-torque position controller. During the automatic tuning procedure, the dynamic change of PID constants are updated on the terminal.
• av: start the automatic tuning of velocity controller. During the automatic tuning procedure, the dynamic change of PID constants are updated on the terminal.
• b: toggle the brake state between blocking and released.
• bs: set the brake release strategy parameter. 0 is to disable the brake. 1 to enable normal release. and 2 to 100 (percentage of maximum torque) for shaking release.
• p[number]: set a target position. the position control need to be enabled first.
• pp[number]: set a target position with profiler. The profile is executed on the master.
• ps[number]: start a position step command. For example ps1000 will move to 1000 then to -1000 then to 0. The motor need to be in 0 position before sending this command.
• psp[number]: start a position step command with profile. same as ps but with profiler.
• v[number]: set a velocity command. the velocity controller needs to be enabled first.
• ep[number]: enable position control. The value is to set the position control mode. 1 is for simple PID, 2 for velocity cascaded, and 3 for limited torque position controller. The modes don’t use the same parameters so check if the parameters are for the right mode before enabling.
• ev1: enable velocity control.
• et1: enable torque control.
• e: and any command starting with e like ep, ev, et will disable the motorcontrol. It’s the same as the command [enter].
• z: reset the multiturn position to 0 (the number of turn). This doesn’t change the offset. This command only works with the REM 16MT position sensor.
• zz: reset the multiturn and singleturn position to 0. The offset need to be found again. This command only works with the REM 16MT position sensor.
• o[number]: set the commutation offset. The range is [0 - 4095].
• s: toggle the sensor polarity. It will make the sensor to return the position in reverse. You can try this if the offset detection keep failing.
• d: toggle the motion polarity. It reverse the position/velocity/torque commands and feedback in the motion controller. Which will make you motor turn the other direction.
• m: toggle the phase inverted parameter. Use this if after finding the offset you have a positive torque resulting in a negative velocity.
• P[number]: set the pole pairs. If when using torque control and the motor moves a little bit then “hold” a position it can be because the pole pairs are incorrect. (it can also be caused by the position sensor).
• f: reset the motorcontrol fault. If the motor stops because of over/under current. Try adjusting you power supply settings and maybe set a lower maximum torque.
• tss: set the torque safe mode. in this mode all the phases are disconnected and the motor is free to move. Use this if you want to manually move the axis.
• kpp [number]: set the P coefficient of the Position controller.
• kpi [number]: set the I coefficient of the Position controller.
• kpd [number]: set the D coefficient of the Position controller.
• kpl [number]: set the Integral part limit the Position controller.
• kpj [number]: set the Moment of inertia of the Position controller.
• kpP [number]: set the Profiler mode the Position controller. 0 is disabled. 1 is enabled.
• kvp [number]: set the P coefficient of the Velocity controller.
• kvi [number]: set the I coefficient of the Velocity controller.
• kvd [number]: set the D coefficient of the Velocity controller.
• kpl [number]: set the Integral part limit the Velocity controller.
• ktr [number]: set the rated torque.
• kf [number]: set the cut-off frequency of the filter in motion control service in Hz (0 to disable)
• Lp [number]: set both the maximum and minimum position limit to [number] and -[number]. The motorcontrol will be automatically disable when the position limit is reached. You can use this feature if your axis has a limited movement. If you are past the limits move the axis manually (use b and tss to unlock the motor) or restart position/velocity/torque controller in the right direction (the position limiter has a threshold to allow to restart if the motor is right after the limit).
• Lpu [number]: set the maximum position limit.
• Lpl [number]: set the minimum position limit.
• Lt [number]: set the torque limit. The unit in in 1/1000 of rated torque. This command stops the motorcontrol.
• Lv [number]: set the velocity limit. Used in velocity control and in cascaded and limited-torque position control modes.
• [number]: just entering a number will switch to torque control mode and set a target torque.

Some of the command descriptions are also printed is the app.

## Run the application¶

When the application has been compiled, the next step is to run it on the Linux PC. Before doing that, make sure that the SOMANET EtherCAT stack is running a proper motor control software for the EtherCAT slave side, i.e. app_demo_slave_ethercat_motorcontrol.

1. Make sure your EtherCAT Master is up and running. To start the Master on a Linux machine, execute the following command:

sudo /etc/init.d/ethercat start

2. Make sure your SOMANET node is accessible by the EtherCAT master by typing:

ethercat slave

The output should indicate a presence of the SOMANET node and pre-operational state if the slave side software is running:

0  0:0  PREOP  +  CiA402 Drive

3. Set all the parameters for you motor in the sc_sncn_ethercat_drive/examples/app_demo_master_ethercat_tuning/sdo_config/sdo_config.csv file This is a Comma Separated Values formatted files. The parameters are in the format:

index, subindex,      axis 1,      axis 2,      axis 3,      axis 4,      axis 5,      axis 6

4. Navigate with the terminal to the application directory on the hard disk. The compiled binaray is in the bin folder. Then execute the application. Use the -n parameter to select the node number. Use the -o flag if you want to enable sdo parameters upload from the sdo_config.csv file:

bin/app_demo_master_ethercat_tuning -o -n 0

5. The application will display the actual position, velocity and torque of the selected slave. It also displays some other parameters or status such as the commutation offset, the brake and motorcontrol status, the PID parameters, etc. If there is an error with the motorcontrol, motion control or position sensor it will be displayed on the last line:

** Operation mode: off **
Position         122776 | Velocity               0
Offset             1200 | Pole pairs            10
Motion polarity normal  | Sensor polarity normal
Integrated Profiler off | Phases connection normal
Brake blocking
Speed  limit       5000 | Position min/max -2147483647 /  2147483647
Torque rated    270 mNm | Torque max    100 /    27 mNm
Position P        16000 | Velocity P        700000
Position I          280 | Velocity I         20000
Position D        41000 | Velocity D             0
Position I lim     1000 | Velocity I lim       900
Autotune Period    2000 | Amplitude          20000
Filter                0
* Motor Fault Under Voltage *

Commands:
b:          Release/Block Brake
a:          Find offset (also release the brake)
number:     Set torque command
r:          Reverse torque command
ep3:        Enable position control
p + number: Set position command
P + number: Set pole pairs
.:          Start/stop recording
L s/t/p + number: set speed/torque/position limit
** single press Enter for emergency stop **

>

6. Use the commands previously described to find the commutation offset then tune and test the position/velocity/torque controllers. After you found the optimal parameters please note them (don’t quit the app!) and update your sdo_config.csv file. You can also test the CSP,CSV,CST CiA 402 operation modes with the app_master_cyclic.

## Examine the code¶

Initialization:
• The master is initialized with ecw_master_init.
• If enabled we uploads the sdo parameters with write_sdo_config using the parameter parsed from the sdo_config.csv file.
• The master is started with ecw_master_start
• The rest is initialisation of various data structures used by the app. The profiler settings are initialized using values from the command line arguments.
Main loop:
• In the main loop the communication with the slave is done with ecw_master_cyclic_function.
• The pdo values are read and write with pdo_handler.
• First the app will try to switch the slave to the OPMODE_TUNING. It use a simple function go_to_state to control the CiA402 state machine on the slave to change the opmode.
• The tuning_input parse the received input pdo and unmux all the parameters and status sent by the slave.
• All this is then displayed with the tuning_display function.
• tuning_command is managing the console commands. It parses the text command and converts them to the corresponding numeric tuning commands.
• tuning_position generates a new position target. It uses the profiler when the pp or psp commands are active.
1.2


text:
SMS scnews item created by <NAME> at Wed 30 Apr 2014 1047
Type: Seminar
Distribution: World
Expiry: 2 May 2014
Calendar1: 1 May 2014 1300-1400
CalLoc1: New Law 024
Auth: seangard@cpe-124-188-249-104.ezsb1.cht.bigpond.net.au (sgar9702) in SMS-WASM

# SUMS: Tang -- Shapes of Triangles and Musical Chords

This week the SUMS talk is being given by graduate of USyd and recent postgraduate of
Warwick, Dr <NAME>.  (Please note the change from the usual location.)

Abstract: Suppose you wanted to write down a list of all possible triangles.
Eventually, you’ll realise that this isn’t a very good idea.  Instead of making a list,
one could try using a space which naturally captures how the triangles organise
themselves.  I will describe how to do this, and discuss its connections with the space
of 3-note musical chords.  Expect paper models and lots of pictures.

If you are registered you may mark the scnews item as read.
School members may try to .


text:
Display color management in darktable

The general picture on the modern Linux desktop

Modern Linux distros featuring either GNOME, Unity or KDE offer fairly easy configuration of color management, this system level configuration mostly pertains to the handling of an ICC display profile.

If you have set a display profile via your system configuration tool (The Color applet in System Settings for GNOME or Unity), there are a few things to keep in mind.

An ICC display profile consists of two main parts. First the so-called "vcgt", which corrects for whitepoint (this is most noticeable on laptops which shift from being very blueish to a bit more yellowish) and gamma. The "vcgt" is loaded into X11 and applied to your whole screen, so all applications automatically benefit. On a GNOME or Unity desktop this is done by GNOME Settings Daemon during login.

The second is the rest of the ICC profile, and this has to be processed in color management enabled application (typically via liblcms2). So we need a mechanism to pass the actual ICC profile to our applications without having to configure them all individually.

The oldest mechanism is the _ICC_PROFILE atom, which allows a single display profile to be defined for your system (this obviously fails for dual head configurations). You can check if a profile is setup like so:

3 thoughts on “Display color management in darktable”

1. Hi Pasacal,

This post and the link to your(?) pcode post were very helpful. However, I still have a question that I wonder if you could help me with. Suppose you have two monitors, one is standard gamut the other is wide gamut. Also suppose you have some raws that you have output to 8-bit sRGB and aRGB JPEGs and want to display. Does the pseudocode below capture sort of thing that you need to do?

sRGB = cmsCreate_sRGBProfile();
aRGB = // Load profile from image (assuming it is present)
sgMon = // Load profile for standard gamut monitor (using EDID?)
wgMon = // Load profile for wide gamut monitor (using EDID?)
// generate the transforms
// IN is one of sRGB and aRGB
// OUT is one of sgMon and wgMon
s2s = cmsCreateTransform(IN, TYPE_RGB_8, OUT,
TYPE_RGB_8, 0, 0);
// apply the transforms to the original image pixels (i.e. the JPG) then render the transformed image on the relevent screen

Now stepping back to how I got those JPEGs in the first place, what transform is applied to go from the raw to the sRGB (or aRGB) JPEG? Is there some camera profile that needs to be used at least as a starting point?

Thanks!

2. Hi dm

It’s a lot simpler than that isn’t it?

If you have profiled your screens and then view your images using a colour managed application they should render ‘correctly’ from either srgb or argb jpegs on either screen. Any transformation necessary should be transparent and handled by colord or whatever colour manager you are using.

Or are you trying to do something different?

3. Rob, please do read the article, dm isn’t far off at all… As I explicitly mention colord is just an infrastructure daemon which ties stuff together. It doesn’t do any color transforms by itself (and it shouldn’t).

As far as the pseudocode goes, you’d need to check Darktable code for the gritty details.

For a high level view, Darktable’s pipeline is completely transparant in the user interface, if you turn off plugin grouping (by clicking the active group once again), you can see all steps which are applied from bottom to the top.

Keep in mind that RAW files technically aren’t normal images to begin with, they inherently have their own sensor specific color space and are practically linear with regard to gamma. So a LOT of processing has to happen to get even a basic image.

The typical minimum steps are:

1. read sensor data from RAW
2. Whitebalance (RGB multipliers)
3. Demosaic (includes median filters/greeneq too)
4. Basecurve (corrects for gamma/contrast/tonality)
5. (CMS) Color In (Camera RGB -> color (XYZ) matrix -> LAB)
7. more plugins (for example denoise)
6. (CMS) Color Out (LAB -> output ICC -> sRGB or AdobeRGB)
8. more plugins (for example sharpen)

Some of this is discussed in the following video:


text:
# If $85_b = 58_c$, what is the smallest possible value of $b$?

I have searched and watched videos online and can't find a method to solve this problem:

If $$85_b=58_c$$ for some positive integer bases $$b$$ and $$c$$, what is the smallest possible value for $$b$$?

$$85_b = 58_c$$ means $$8b + 5 = 5 c + 8$$. Thus, $$b = (3+5c)/8$$. This is an integer whenever $$c = 8n + 1$$ for some integer $$n$$. In this case, $$b = 5n+1$$. Now, since $$8$$ appears as a digit, we must have $$b > 8$$. The smallest integer of the form $$5n+1$$ that is larger than $$8$$ is $$11$$.

So the smallest solution is $$b = 11$$, $$c = 17$$, as $$85_{11} = 58_{17} = 93_{10}$$

$$8b+5=5c+8\implies 5c =8b-3\implies 5\mid 8b-3$$

So $$5\mid 3b-3\implies 5\mid b-1\implies b=5k+1$$

Since $$b\geq 9$$ we have $$b_{\min}=11$$

• Although it doesn't affect the solution, $b > 8$, as base $8$ can't have $8$ appear as a digit. – eyeballfrog Mar 13 at 17:32

well think it out:

$$8b + 5 = 5c + 8$$.

So $$8b - 5c = 3$$.

$$b = 1; c= 1$$ is a solution but it is too small as we must have $$b,c > 8$$. But there are an infinite number of solutions:

If $$8b - 5c = 3$$ then $$8(b+ 5k)-(c+8k)= (8b-5c) + 40k - 40k$$ will also be a solution.

$$8$$ and $$5$$ are relatively prime so all solutions are of the form:

$$8(b + 5k) - (c + 8k) = 3$$. And as $$b=c = 1$$ is one solution all solutions are of form:

$$8(1 + 5k) - (1 + 8k)$$.

So we just need to find the smallest $$k$$ where $$1 + 5k, 1+8k > 8$$. That number is clearly $$k = 2$$ and $$b = 11> 8$$ and $$c = 17> 8$$.

And $$85_{11} = 58_{17}=93$$.


text:
# ATLAS Conference Slides

უკანასკნელი დამატებები:
2018-11-10
09:24
 Continuous signal modelling in a multidimensional space of coupling parameters / Brenner, Lydia (DESY, Hamburg and Zeuthen) /ATLAS Collaboration The plans for the second Run of the LHC changes the focus in the Higgs sector from searches to precision measurements. Effective Lagrangians can be used for parameterisation. [...] ATL-PHYS>-SLIDE-2018-1000.- Geneva : CERN, 2018 - 26 p. Fulltext: PDF; External link: Original Communication (restricted to ATLAS)

2018-11-10
09:11
 LCWS-2018: ttH measurements at the LHC / Narayan, Rohin (University of Texas at Austin) /ATLAS Collaboration ATLAS + CMS ttH Results ATL-PHYS-SLIDE-2018-999.- Geneva : CERN, 2018 Fulltext: PDF; External link: Original Communication (restricted to ATLAS)

2018-11-09
23:34
 2HDM-based and SUSY-inspired models / Cortes-Gonzale<NAME> (European Laboratory for Particle Physics, CERN) /ATLAS Collaboration Models with an extended Higgs sector, such as two-Higgs-doublet models (2HDMs), have traditionally played an important role in DM searches with supersymmetry as a prime example of a complete theory that addresses the nature of particle DM. In addition, UV-complete simplified models based on 2HDMs, such as the 2HDM with a pseudoscalar mediator investigated at the LHC, gain increasing attention. [...] ATL-PHYS-SLIDE-2018-998.- Geneva : CERN, 2018 Fulltext: PDF; External link: Original Communication (restricted to ATLAS)

2018-11-09
14:45
 Introduction to collider searches / <NAME> (DESY, Hamburg and Zeuthen) /ATLAS Collaboration "The goal of this talk is to introduce the scope of DM searches at colliders, focussing on the LHC programme (ATLAS, CMS, LHCb) and KEK (Belle2). This should include a discussion of then LHC and KEK timelines, including upgrades, energy ranges, etc [...] ATL-PHYS-SLIDE-2018-997.- Geneva : CERN, 2018 Fulltext: PDF; External link: Original Communication (restricted to ATLAS)

2018-11-09
14:45
 Measurements of the Higgs boson by ATLAS and CMS / <NAME> (Institute for Theoretical and Experimental Physics) /ATLAS Collaboration Recent results on Higgs boson production and decays in the ATLAS and CMS experiments at the LHC are reviewed. They are mostly based on the analyses of 13 TeV LHC proton-proton collision data accumulated during 2015--2016 or 2015--2017 year exposures. [...] ATL-PHYS-SLIDE-2018-996.- Geneva : CERN, 2018 Fulltext: PDF; External link: Original Communication (restricted to ATLAS) In : The 4th International Conference on Particle Physics and Astrophysics, Moscow, Russian Federation, 22 - 26 Oct 2018

2018-11-09
14:44
 Higgs Results at the LHC / Bell, <NAME> (Department of Physics and Astronomy, University College London) /ATLAS Collaboration Higgs results at the LHC ATL-PHYS-SLIDE-2018-995.- Geneva : CERN, 2018 Fulltext: PDF; External link: Original Communication (restricted to ATLAS)

2018-11-09
09:34
 Development of the ATLAS Liquid Argon Calorimeter Readout Electronics for the HL-LHC / <NAME> (Brookhaven National Laboratory (BNL)) /ATLAS Collaboration The ATLAS Liquid Argon Calorimeter electronics will be upgraded to readout the 182,500 calorimeter cells at 40 MHz with 16 bit dynamic range in the high pileup conditions of the high-luminosity LHC. Low-noise, low-power, radiation-tolerant and high-bandwidth electronics components are being developed in 65 and 130 nm CMOS technologies. [...] ATL-LARG-SLIDE-2018-994.- Geneva : CERN, 2018 - 1 p. Fulltext: PDF; External link: Original Communication (restricted to ATLAS) In : 2018 IEEE Nuclear Science Symposium and Medical Imaging Conference (NSS/MIC 2018), Sydney, Australia, 10 - 17 Nov 2018

2018-11-08
22:01
 Rich phenomenology of simplified models/EFTs in MET+X final states / Sevova, Stanislava (Stanford University) /ATLAS Collaboration Simplified models have played a major role in dark matter searches since the early days of LHC Run 2, providing a richer collider phenomenology that can be targeted by searches compared to the Effective Field Theory (EFT) approach. This talk will provide a short overview of the most common simplified models used at the LHC experiments and Belle/BelleII , with an emphasis on the complementarity of collider with direct (DD) and indirect detection (ID) results across the parameter space of the simplified models. [...] ATL-PHYS-SLIDE-2018-993.- Geneva : CERN, 2018 - 20 p. Fulltext: PDF; External link: Original Communication (restricted to ATLAS)

2018-11-08
19:02
 Beyond Standard Model searches by ATLAS and CMS / <NAME> (Fakultaet Physik, Technische Universitaet Dortmund, Dortmund) /ATLAS Collaboration An overview of recent searches for BSM physics with the ATLAS and CMS experiments is given. A focus is placed on searches with top quarks, in particular searches for top-quark partners. [...] ATL-PHYS-SLIDE-2018-992.- Geneva : CERN, 2018 Fulltext: PDF; External link: Original Communication (restricted to ATLAS) In : The 4th International Conference on Particle Physics and Astrophysics, Moscow, Russian Federation, 22 - 26 Oct 2018

2018-11-08
16:47
 Top-antitop charge asymmetry measurements in the lepton+jets channel with the ATLAS detector / Melo, Matej (Comenius University) /ATLAS Collaboration We report a measurement of the charge asymmetry AC in top quark pair production with the ATLAS experiment. The measurement focuses on the lepton+jets topology. [...] ATL-PHYS-SLIDE-2018-991.- Geneva : CERN, 2018 - 1 p. Fulltext: PDF; External link: Original Communication (restricted to ATLAS) In : 11th International Workshop on Top Quark Physics, Bad Neuenahr, Germany, 16 - 21 Sep 2018


text:
$$\frac{[(\omega_0^2-\omega^2)-2i\omega\gamma]^2}{[(\omega_0^2-\omega^2)^2+4\gamma^2\omega^2]^2}=\frac{1}{[(\omega_0^2-\omega^2)^2+4\gamma^2\omega^2]}$$

I don't understand how can I get to that solution. Any hint will be very thankful. The denominator for that fraction it can be write like adjoint but after that I will remain with a quadric equation and it won't be the same as the solution.

• Assuming that you want to solve the equation above. Let $z\in\mathbb{C}$. Suppose that $z^2$ is a positive real number. Show that $z\in\mathbb{R}$. If you want to prove that two sides of the equation are equal, then good luck. They are NOT equal except at very few values of $\gamma$ and $\omega$. – Batominovski Jul 29 '15 at 18:11

Note that on the numerator of the LHS of the equation, the [...]$^2$ operation is actually taking the modulus squared. That is, $|a+bi|^2 = a^2+b^2$.


text:
# Sigma algebra generated by a sigma algebra and another set

I'm really stuck.

If $\mathcal{A}$ is a sigma-algebra in $X$ and $C\subset X$ does not belong to $\mathcal{A}$, how do I need to show that the sigma-algebra $\sigma(\mathcal{A}\cup \{C\})$ (so generated by $\mathcal{A}$ and $C$) is equal to the collection of sets of the form $K=(A \cap C) \cup (B \cap C^c)$, with $A,B \in \mathcal{A}$.

Hope somebody can help me how to solve this problem.

• You probably mean $$\sigma(\mathcal{A}\cup\{C\})$$ instead of $$\sigma(\mathcal{A}\cup C)$$ What did you try "to solve this problem"? – Did Jan 8 '18 at 18:30
• Yes, that's what I mean. I will edit the original post. I already tried to write down all possibilities of the sigma-algebra generated by the set, but that are to much possibilities. After that I tried to show that $\sigma(\mathcal{A} \cup \{C\})$ is a subset of K and the other inclusion. But that also didn't work. I think drhab gives the most direct way to prove it. – MathI Jan 8 '18 at 19:06

$$[(A\cap C)\cup(B\cap C^{\complement})]^{\complement}=(A^{\complement}\cap C)\cup(B^{\complement}\cap C^{\complement})$$ so the collection of sets of that form is closed under complements.

$$\bigcup_{n=1}^{\infty}[(A_n\cap C)\cup(B_n\cap C^{\complement})]=((\bigcup_{n=1}^{\infty}A_n)\cap C)\cup ((\bigcup_{n=1}^{\infty}B_n)\cap C^{\complement})$$

so the collection of sets of that form is closed under countable unions.

Then it must be a $\sigma$-algebra that evidently contains $\mathcal A$ as a subcollection and has $C$ as one of its elements.

Further it is evident that any $\sigma$-algebra with these properties will contain every element of the form $(A\cap C)\cup(B\cap C^{\complement})$ where $A,B\in\mathcal A$.

edit:

\begin{aligned}\left[\left(A\cap C\right)\cup\left(B\cap C^{\complement}\right)\right]^{\complement} & =\left(A\cap C\right)^{\complement}\cap\left(B\cap C^{\complement}\right)^{\complement}\\ & =\left(A^{\complement}\cup C^{\complement}\right)\cap\left(B^{\complement}\cup C\right)\\ & =\left(A^{\complement}\cap B^{\complement}\right)\cup\left(A^{\complement}\cap C\right)\cup\left(B^{\complement}\cap C^{\complement}\right)\\ & =\left(A^{\complement}\cap B^{\complement}\cap C\right)\cup\left(A^{\complement}\cap C\right)\cup\left(A^{\complement}\cap B^{\complement}\cap C^{\complement}\right)\cup\left(B^{\complement}\cap C^{\complement}\right)\\ & =\left(A^{\complement}\cap C\right)\cup\left(B^{\complement}\cap C^{\complement}\right) \end{aligned}

• Thanks for your reaction. I just have problems to see why $[(A\cap C)\cup(B\cap C^{\complement})]^{\complement}=(A^{\complement}\cap C)\cup(B^{\complement}\cap C^{\complement})$. Even when I paint the diagrams it's not the same. Did you change intersection and union or is it me who see this wrong? – MathI Jan 8 '18 at 17:35
• Do you mean (one of) the 2 equalities that I mention, or is it something else? – drhab Jan 8 '18 at 17:37
• I now complete my reaction. I cannot use Enter here, because then I add a comment directly. I mean the $[(A\cap C)\cup(B\cap C^{\complement})]^{\complement}=(A^{\complement}\cap C)\cup(B^{\complement}\cap C^{\complement})$. – MathI Jan 8 '18 at 17:40
• I have edited to make things more clear. – drhab Jan 8 '18 at 18:24
• Thank you, now it's clear! I just needed a few more steps. – MathI Jan 8 '18 at 19:03


text:
# Forest Plot (with Horizontal Bands)

Forest plots are often used in clinical trial reports to show differences in the estimated treatment effect(s) across various patient subgroups. See, for example a review. The page on Clinical Trials Safety Graphics includes a SAS code for a forest plot that depicts the hazard ratios for various patient subgroups (this web page has links to other interesting clinical trial graphics, as well as links to notes on best practices for graphs, and is worth checking out).

The purpose of this blog post is to create the same forest plot using R.

It should be possible to create such a graphic from first principles, using either base R graphics or using the ggplot2 package such as posted here. However, there is a contributed package forestplot that makes it very easy to make forest plots interspersed with tables – we just need to supply the right arguments to the forestplot function in the package. Also, one question that arose was how easy it would be to get the horizontal grey bands for alternate rows in the forest plot. This too was not very difficult. Following is a short explanation of the entire process, as well as the relevant R code:

First, we store the data for the plot, ForestPlotData in any convenient file format and then read it into a dataframe in R:

workdir <- ""C:\\Path\\To\\Relevant\\Directory""
datafile <- file.path(workdir,"ForestPlotData.csv")

Then format the data a bit so that the column labels and columns match the required graphical output:

## Labels defining subgroups are a little indented!
subgps <- c(4,5,8,9,12,13,16,17,20,21,24,25,28,29,32,33)
data$Variable[subgps] <- paste(" ",data$Variable[subgps])

## Combine the count and percent column
np <- ifelse(!is.na(data$Count), paste(data$Count," (",data$Percent,")",sep=""), NA) ## The rest of the columns in the table. tabletext <- cbind(c("Subgroup","\n",data$Variable),
c("No. of Patients (%)","\n",np),
c("4-Yr Cum. Event Rate\n PCI","\n",data$PCI.Group), c("4-Yr Cum. Event Rate\n Medical Therapy","\n",data$Medical.Therapy.Group),
c("P Value","\n",data$P.Value))  Finally, include the forestplot R package and call the forestplot function with appropriate arguments. The way I got around to creating the horizontal band at every alternate row was by using settings for a very thick transparent line in the hrzl_lines argument! See below. The col=”#99999922″ option gives the light grey color to the line as well as sets it to be transparent. A graphics device (here, a png file) with appropriate dimensions is first opened and the forest plot is saved to the device. library(forestplot) png(file.path(workdir,"Figures\\Forestplot.png"),width=960, height=640) forestplot(labeltext=tabletext, graph.pos=3, mean=c(NA,NA,data$Point.Estimate),
lower=c(NA,NA,data$Low), upper=c(NA,NA,data$High),
title="Hazard Ratio",
xlab="     <---PCI Better---    ---Medical Therapy Better--->",
hrzl_lines=list("3" = gpar(lwd=1, col="#99999922"),
"7" = gpar(lwd=60, lineend="butt", columns=c(2:6), col="#99999922"),
"15" = gpar(lwd=60, lineend="butt", columns=c(2:6), col="#99999922"),
"23" = gpar(lwd=60, lineend="butt", columns=c(2:6), col="#99999922"),
"31" = gpar(lwd=60, lineend="butt", columns=c(2:6), col="#99999922")),
txt_gp=fpTxtGp(label=gpar(cex=1.25),
ticks=gpar(cex=1.1),
xlab=gpar(cex = 1.2),
title=gpar(cex = 1.2)),
col=fpColors(box="black", lines="black", zero = "gray50"),
zero=1, cex=0.9, lineheight = "auto", boxsize=0.5, colgap=unit(6,"mm"),
lwd.ci=2, ci.vertices=TRUE, ci.vertices.height = 0.4)
dev.off()

Here is the resulting forest plot:

## 18 thoughts on “Forest Plot (with Horizontal Bands)”

1. Hi Jyothi! Thanks for such a great R package. I use it a lot at my company actually. I have a naïve question because I am a novice R user – how do you add a footnote to the plot? I just want to add the date and time as well as the path indicating the location of the source documents. I guess it would take up multiple lines (3?) Thank you!

Like

• Hi! Thanks!, But I’m not the author of the forestplot package 🙂 I just used the package.
With respect to your question, it does look like the forestplot function does not have an argument to support footnotes. However, since it is based on grid graphics, without making too much of a change in the rest of the plot alignment, we can probably add two lines of text by including the following code just before the dev.off() statement:

grid.text(paste(“Date & time:”,date()), 0.01, 0.03, just=0, gp=gpar(cex=1))
grid.text(paste(“Source:”,”C:\\Path\\to\\source_file_location\\”), 0.01, 0.01, just=0, gp=gpar(cex=1))

The above should put the time stamp and file location below the plot. Hope this helps.

Like

2. Max says:

Hello I cannot obtain your same graphs: I received a message after running png(file.path(workdir,”Figures\\Forestplot.png”),width=960, height=640)

of Error in png(file.path(workdir, “Figures\\Forestplot.png”), width = 960, : unable to start png() device.

When I add type=”cairo” line is recognized however later:

Error in grid.Call(L_convert, x, as.integer(whatfrom), as.integer(whatto), : could not open file

Could you please provide an explanation?

Thanks

Like

• Hi! Thank you for trying the code.
The plot actually gets saved to a sub-directory named “Figures” within “workdir”. So if this directory is not present, the error is thrown. Unfortunately, I missed allowing the program to create the sub-directory if it did not exist. My mistake! Can you first try creating a “Figures” sub-directory within your “workdir” and then try again?

Like

3. Ramesh says:

I was trying to create similar plot. I still could not understand your data structure. How you had arranged your data. Could you please share your data. Thanks

Like

• Hi! The link to the dataset is given in the post itself as an MS Excel file. You may however have to save the file locally in .csv format before trying the read.csv() option.

Like

4. YS Lin says:

Hi

I have a quick question that I tried to follow the instruction:

data$Variable[subgps] <- paste(" ",data$Variable[subgps])

———————————————————————————–
Error in $<-.data.frame(*tmp*, "Variable", value = c("Overall", NA, : replacement has 33 rows, data has 1 YS Like 5. Amanda Ly says: Hi Jyothi, Many thanks for your post. I have a question, how do you italicise or made bold the column names and the subheadings for the subgroups (age, sex, race etc)? BW, Amanda Like 6. Tom says: Hey I like to reproduce your plot. Unfortunately I have some errors …. can you please help me ? Warning message: In is.na(data$Count) :
is.na() applied to non-(list or vector) of type ‘NULL’
Warning message:
In cbind(c(“Subgroup”, “\n”, data\$Variable), c(“No. of Patients (%)”, :
number of rows of result is not a multiple of vector length (arg 2)
Error in forestplot.default(labeltext = tabletext, graph.pos = 3, mean = c(NA, :
You have provided 2 rows in your mean arguement while the labels have 35 rows

Like

7. Bart says:

Thanks much Jyothi,

Your post was of great help.

Cheers,

Bart

Liked by 1 person

8. Tushar says:

Thank You Jyothi,

Only I have one concern, hope you should solve my issue.
My concern is, On my graph the point estimate point plotted on interval line and that is very big.
so can you help me please how would I decrease their size?

Like

• Hi! Try reducing the value of the boxsize argument and see if that resolves the issue.

Like

9. Allan says:

Only one suggestion. Your forest plot is amazing, and it’s exactly what I was looking for.
However, to be PERFECT, the size of the squares should be proportional to the number of patients in that analysis. Actually, in a meta-analysis, they are related to the inverse of variance, which in turn, is related to the sample size.
To be MORE THAN PERFECT, and now I know that I’m asking too much, it should be able to create the forest plot using the raw data. The same data you used to perform the Cox regression, instead of reorganizing the data in another sheet.
For what it’s worth, people have added a ggplot2-based function to create a forest plot for the cox model in the survminer package (https://www.r-bloggers.com/forest-plot-with-horizontal-bands/).
However, their forest plot is still wrong because, I believe, they were not able to perform the analyses by each subgroup.

Congrats!

Like

The default value of boxsize parameter should make the size of the square inversely proportional to the precision of the estimate. Getting the plot from raw data might be a little more difficult and I can try to do that in a future post!

Like

10. Kelly says:

This has been very helpful, thanks for the great example!
Do you know if there is a way to code multiple colors for the error bars in the forest plot?
I am trying to conditionally color the error bar lines of the plot based on statistical significance (using the p-value and CI in my data table (green, red, and grey). Do you have any suggestions on how to do this within fpcolor? I’ve tried a few things and it hasn’t worked…

thanks!
Kk

Like

11. Christa says:

Hi there,

Thanks for this! It has been super helpful.

I’m working on putting together a forest plot that has multiple confidence bands (like shown in the vignette https://cran.r-project.org/web/packages/forestplot/vignettes/forestplot.html), but would like to introduce horizontal bands as you’ve shown to set each row apart. However, when I do this, the bands always end up in between each of my row labels. Do you have any suggestions on how to rectify this? Thanks!

Like

• Hi.. You may try varying the lwd (line width) and the row numbers in the parameter list of the hrzl_lines argument of the forestplot function. Best luck!

Like

12. RTE says:

Hi- I’m getting a beautiful shell but no data shows up. And I get this error.
Error in grid.Call.graphics(C_setviewport, vp, TRUE) :
invalid ‘layout.pos.col’
I’m very new to this so sorry if this is stupid….

Like


text:
Proving a statement false by reverse induction

On a recent graph theory assignment, I proved a statement false by "reverse" induction, showing that $P(2)$ was false, and that $P(k) \implies P(k-1)$ for $k>2$, thus, by the contrapositive, $\neg P(k-1) \implies \neg P(k)$, so $P(n)$ should be false for all $n\in \mathbb{N}$. This seems natural and like it should work in general, but after a quick Google search into "reverse induction", nothing like this proof technique showed, at least in the first few results. For completeness, here's the statement and my proof.

Does there exist a graph with vertex set $\{v_1,\dots,v_n\}$ such that $\deg(v_i)= i-1$ for $i=1,\dots,n$?

Proof: It does not work for $n=2$, since the first vertex would be a singleton, and the second vertex must have degree one, so it must be connected to the first, but this is impossible since the first vertex would then have degree one, a contradiction. Now for $n \geq 2$, Consider a labeling $\{v_1,v_2,\dots,v_{n+1}\}$, and suppose that there is a set of edges such that $\deg v_i = i-1$. Then by removing $v_{n+1}$ an all edges connected to it, we would have a graph with vertex set $\{v_1,v_2,\dots,v_n\}$, such that $\deg v_i = i-1$. The contrapositive of this however, is that if there is no vertex set on $\{v_1,v_2,\dots,v_n\}$, such that $\deg v_i = i-1$, then there is no graph with vertex set $\{v_1,v_2,\dots,v_{n+1}\}$ such that $\deg v_i = i-1$. As we have shown this for $n=2$, and the inductive step is proved, we must have that there are no graphs with vertex set $\{v_1,v_2,\dots,v_n\}$, such that $\deg v_i = i-1$ for $n \geq 2$.

• Didn't follow the details of the proof, but the technique is a variant of proof by infinite descent. – dxiv Apr 11 '18 at 3:03
• In your opening paragraph how do you know it’s false for n = 1 – Prince M Apr 11 '18 at 3:12
• You said false for n in naturals but it looks like just n geq 2 – Prince M Apr 11 '18 at 3:14

Let me offer my five cents here.

(1) First of all, it appears to me that your proof is just the usual induction, rather than some kind of a "reverse" induction. You just need to understand what you proved: you applied the usual induction method to prove the statement $Q(n)=\neg P(n)$ for all $n\ge2$. Your basis was proving $Q(2)$, which is the same as demonstrating $\neg P(2)$, i.e. that $P(2)$ is false. And your induction step was proving $Q(k-1)\Rightarrow Q(k)$, which is the same as $\neg P(k-1)\Rightarrow \neg P(k)$, and for this particular proof you used a contrapositive method — you showed that $\neg Q(k)\Rightarrow \neg Q(k-1)$, i.e. that $\neg(\neg P(k))\Rightarrow \neg(\neg P(k-1))$, i.e. that $P(k)\Rightarrow P(k-1)$.

(2) Unrelated to your main question about interpreting such a "reverse" induction method, I still can't help but notice that you didn't have to use induction here at all. There's a pretty simple direct proof for this statement that works for any $n$. And you pretty much have it — just rephrase what you did for $n=2$ to apply to arbitrary $n$.

(3) And besides, I see a serious issue with your proposed proof, specifically with the following statement in it:

Then by removing $v_{n+1}$ an all edges connected to it, we would have a graph with vertex set $\{v_1,v_2,\ldots,v_n\}$, such that $\deg v_i=i−1$.

Here's my problem with this claim: if you removed a bunch of edges (the ones adjacent to $v_{n+1}$, which was given to have a nonzero degree), how come the degrees of the remaining vertices didn't change?

• Good answer, I was thinking about proving exactly that statement $Q(k)$, and it seems this was the right approach. – Nico Apr 11 '18 at 18:10

This resembles a proof by infinite descent, but is a bit difficult to read. You do not need to use the contrapositive here. Instead, the "standard" approach is to structure the proof like this:

1. Prove that $n > 2$ (i.e. prove $\neg P(2)$, as you did). You would also have to rule out $P(1)$, but in this example $P(1)$ is actually true, so you should probably exclude that in the problem statement instead of the proof.
2. Assume that $\exists n > 2: P(n)$, and further assume that $n$ is the smallest such value. This second assumption is only allowed because we are working with positive integers; in the rationals or reals, there might not be any smallest value of $n$.
3. Prove $P(n - 1)$ from $P(n)$, also as you did.
4. Derive a contradiction because we previously assumed that $n$ was minimal.

It is induction :

Instead of $P(2)$ take $Q(2) = \neg P(2)$ and more generally $Q(k) = \neg P(k)$

You have shown that $Q(2)$ is true, and that $Q(k) \implies Q(k+1)$ so by induction $Q(n)$ is false for all $n \in \mathbb{N}$

I think your confusion may come from you didn't specify the predicate statement correctly. You said:

$$P(k) \implies P(k-1).$$

So my question is: What's the domain of discourse of it? If you didn't make this clear, No one can prove/show that it is true. And you can't apply your "first domino", i.e. $P(2)$, to make it work like that of mathematical induction. So if you can show that for any $k\in \large\mathbb{N}$, the predicate is true down to $k-1=2,$ then yes your reverse induction technique should work, because you "connect" it to your "first domino".


text:
Search

### Journal Articles

#### The Effects of Natural and Human Disturbances on Soil Nitrogen Dynamics and Trace Gas Fluxes in a Puerto Rican Wet Forest

Author
<NAME>, <NAME>, <NAME>, <NAME>, <NAME>
Journal
• Biotropica SCI(E) SCOPUS
• view options
356p ~ 363p ISSN 0006-3606 E-ISSN 1744-7429
Publisher
in 1991
Cited Count
40
Partners
JSTOR Crossref
DOI
10.2307/2388252
Category
Natural Sciences > Biology , Natural Sciences > Ecology , Engineering > Environmental Engineering
##### Abstract

We examined the effects of two disturbances (Hurricane Hugo and forest clearcutting) on soil nitrogen dynamics and on the exchanges of N$_2$O, CO$_2$, and CH$_4$ between soils and the atmosphere of a subtropical wet forest in Puerto Rico. The disturbances resulted in prolonged increases in ammonium pools and short-term increases in rates of net N-mineralization and net nitrification. Nitrous oxide emissions increased following both disturbances. The most dramatic increase was observed 4 mo after clearcutting; N$_2$O emissions (109 49 $\mu g$ N/m$^2$-hr) from the cut plot were about two orders of magnitude higher than emissions from the reference plot (1.71 $\mu g$ N/m$^2$-hr). Carbon dioxide emissions from both disturbed plots (mean 102.47 mg C/m$^2$-hr) were about 30 percent lower than the reference (mean 15 1.28 mg C/m$^2$-hr). Soils at all sites were generally sinks for CH$_4$. Methane uptake, however, was suppressed by both disturbances. This suppression may be related to disturbance-induced changes in the nitrogen cycle, as we have previously observed in temperate zone forests.

##### References

No relevant information is available
If you register references through the customer center, the reference information will be registered as soon as possible.


text:
## CSDN博客

### 再转几个骗人的MAIL：借机学学英语。

DEAR FRIEND,
I am Mr.<NAME> Executive Director of the Hang Seng Bank Ltd, Hong Kong.
An Iraqi named <NAME>,a business man made a numbered fixed deposit of (346,736,899.68 TWD) for 18 calendar months, this is valued to Fourty Four million Five Hundred Thousand United State Dollars only in my branch. Upon maturity several notice was sent to him, even during the war, four years ago (2003). Again after the war another notification was sent and still no response came from him. We later found out that <NAME> and his family had been killed during the war in Gunfire that hit their home at Mukaradeeb where his personal oil well was.
After further investigation it was also discovered that <NAME> did not declare any next of kin in his official papers including the paper work of his bank deposit. And he also confided in me the last time we was at my office that no one except me knew of his deposit in my bank. So, Fourty Four million Five Hundred Thousand United State Dollars is still lying in my bank and no one will ever come forward to claim it. What bothers me is that according to the laws of my country at the expiration five years six months the funds will revert to the ownership of the Hong Kong Government if nobody applies to claim the funds.
My proposal, I am prepared to place you as the next of kin in a position to instruct HANG SENG BANK to release the deposit to you as the closest surviving relation. Upon receipt of the deposit, I am prepared to share the money with you in half. That is: I will simply nominate you as the next of kin and have them release the deposit to you. We share the procentange 70/30. I would have gone ahead to ask the funds be released to me, but that would have drawn a straight line in my involvement in claiming the deposit.I assure you that I could have the deposit released to you within a few days. I will simply inform the bank of the final closing of the file relating to Haider Hanoon I will then officially communicate with my Bank and instruct them to release the deposit to you. With these: all is done. I am aware of the consequences of this proposal.
I ask that if you find no interest in this project that you should discard this mail. I ask that you do not be vindictive and destructive. If my offer is of no appeal to you,delete this message and forget I ever contacted you. Do not destroy my career because you do not approve of my proposal. You may not know this but people like myself who have made tidy sums out of comparable situations run the whole private banking sector. I am not a criminal and what I do, I do not find against good conscience, this may be hard for you to understand, but the dynamics of my industry dictates that I make this move. Such opportunities only come ones way once in a lifetime. I cannot let this chance pass me by, for once I find myself in total control of my destiny. These chances wont pass me by. I ask that you do not destroy my chance, if you will not work with me let me know and let me move on with my life but do not destroy me. I am a family man and this is an opportunity to pro
vide them with new opportunities.
If you give me positive signals, I will initiate this process towards a conclusion. I wish to inform you that should you contact me via official channels; I will deny knowing you and about this project. I repeat, I do not want you contacting me through my official phone lines nor do I want you contacting me through my official email account. Contact me only through the numbers I will provide for you and also through the email account that i will provide therein in this email. Further more be informed that the fact that you are a foreigner give you the privilege to stand in as my deceased client beneficiary as my deceased client had no relation all his family died with him during the war and I can not use my relation because its not accepted here in Hong Kong. Also I am very confident that we will be able to establish the trust that is needed to complete this deal and all that I need from the time been is your willingness and commitment so that we can end this in
the next one weeks.
What you need to understand about this transaction is that I will make sure that it passes through all international banking laws regards to this I will take care of all the expenses and the cost of retaining the service of my Attorney to give the transaction the proper documentation that is required to perfect the finishing. Your only obligation in this transaction will be to set up an offshore account that can accommodate these funds and I will give you more light later once you agree to partner with me. Like I said before there is no risk involve as it will pass through international banking laws and all documents related to this transaction will be sent to you by post for your perusal and trust.
I do not want any direct link between you and me. My official lines are not secure lines as they are periodically monitored to assess our level of customer care in line with our Total Quality Management Policy. Please observe this instruction religiously. Please, again, note I am a family man, I have wife and children. I send you this mail not without a measure of fear as to what the consequences, but I know within me that nothing ventured is nothing gained and that success and riches never come easy or on a platter of gold. This is the one truth I have learned from my private banking clients. Do not betray my confidence. If we can be of one accord, we should plan a meeting, soon. Should you be interested I will prefer you to send me your FULL NAMES,ADDRESS,OCCUPATION AND PHONE NUMBER.on the email address below.
E-mail:<EMAIL>

Your earliest response to this letter will be appreciated.
Mr.<NAME>
－－－－－－－－－－－－－

 The Uk National  Lottery P O Box 1010 Dublin, 11 G Lower Dorset Street Ref: UK/9420X2/68     ATTN:   Pengtingjin, This is to formerly notify you that after the verification process conducted , I wish to inform you that the Administrative department and promotions department has just approved your payment of the 1,000,000.00 (ONE MILLION POUND STERLING) and the said sum is now deposited in our suspense account in your favour with  ANZ BANK AND SECURITIES.   Here is your Winning Contact Code (AQ11WWRZZA1) You Have to Contact the Bank with the Winning Numbers and also with your full details so that they Can Attend to you fast.    So in the light of the above explanation, I wish to inform you that you are to immediately contact Bank,   Here is the Bank Informations.   CONTACT BANK: ANZ BANK AND SECURITIES.   Managing Director: Mr. <NAME>  Tel: +44-7045-712-787  Fax:+44-7006-007-748  Email: <EMAIL>   Contact them and get back to us if you need More Details. And any Assistance from our side.   Congratulations once more. Yours faithfully Mr. James Hill.    U.K NATIONAL LOTTERY Sweepstakes International Program. Copyright @1994-2007 The UK National Lottery Inc. All rights reserved. Terms of Service - Guideline.

－－－－－－－－－－－－－－－－－－－－－－－－－－－－－－－－－－－－－－－－－－

OTTO NL

INTERNATIONAL PRIZE AWARD DEPT.

REF No:   FAQ/xxx-xxxxxxxx/xx
BATCH No:    EAPA/1xx/xxx/FAQ
TICKET No:    xxxxxxxx-xxxx
SERIAL No:   xxxx-xxx-xx
LUCKY No:    x-xx-xx-xx-xx-xx-x

We are pleased to inform you of the result of the Lottery Winners
International programs held on the 29th of December 2005. Your E-mail
address attached to Ticket number xxxxx with Serial number
xxxxx drew the lucky numbers xxxxxx, which consequently won
in the 1st category; you have therefore been approved for a lump sum payout
of US$2,000,000.00 (Two Million United States Dollars). This is from a total cash prize of USD$20,000,000.00 Million shared amongst Ten
International Lucky Winners in the Category A + (Plus).

CONGRATULATIONS!!!

This lottery is a promotional program by LOTTO, NL (Biggest lottery
Organization in the Netherlands) to advertise to the world its existence.
All participants were selected through a computer ballot system drawn from
over 50,000 companies and 2,000,000 individual email addresses from all over
the world, as part of our international promotions program, which we intend
to conduct several times a year.

Be informed that your documents have to pass our authority to obtain a
clearance, which shall be attached to your document in readiness of the
subsequent onward transfer into your nominated bank account with 24 hours of
completion of the authentication.
Due to the possibility of unscrupulous individuals filing a double claim, we
suggest that you keep this award strictly confidential until your claim has
been processed and notarized and your certificate of award obtained. This is
in conformity with the lottery claim regulations by our Organization & High
Court of Justice.

This is part of our security protocol measures to avoid double claiming and
unwarranted abuse of this program by some participants/non participants.
All winnings must be claimed not later than 12th of January 2005. After this
date all unclaimed funds will be included in the next stake. Anybody under
the age of 18 is automatically disqualified.

To file for your claim, please contact our /your processing agent
Mr. <NAME>
Email: <EMAIL>

NOTE: All winnings must be notarized and a certificate of award must be
obtained from the Gaming Board to complete the claims process, this
certificate can only be obtained through legal representation so winners are
to cover the legal charges for the notarization of the claims form and the
acquisition of the certificate of award not LOTTO.NL
In order to avoid unnecessary delays and complications, provide the

Name:
Nationality:
Country of residence:
Tel number:

Delays will compel us to disperse your funds, in other words, we shall
divert your funds as stated above. Kindly remember to quote your reference
number in all correspondence.
Finally, you make sure that you save a copy of this mail and note every
letter clearly as stated for we will not be held responsible should there be
any complications in this transaction due to laxity on your part.

Congratulations once more from our members of staff and thank you for being
part of our promotional program.

Yours truly,
<NAME>
Lotto.nl

NOTICE:  You have received this message from Lotto NL international prize
dept. at your e-mail address because you have visited one of our sponsored
sites and have voluntarily given your email address to receive mails from
their sponsors. If you wish to be taken out of this list do not reply to
this mail, reply to the agent with the words remove. If you are not the
intended recipient, you must not, directly or indirectly, use, disclose,
distribute, print or copy any part of t

0 0


text:
## Summary of Substitution

### Essential Concepts

• Substitution is a technique that simplifies the integration of functions that are the result of a chain-rule derivative. The term ‘substitution’ refers to changing variables or substituting the variable $u$ and du for appropriate expressions in the integrand.
• When using substitution for a definite integral, we also have to change the limits of integration.

## Key Equations

• Substitution with Indefinite Integrals
$\displaystyle\int f\left[g(x)\right]{g}^{\prime }(x)dx=\displaystyle\int f(u)du=F(u)+C=F(g(x))+C$
• Substitution with Definite Integrals
${\displaystyle\int }_{a}^{b}f(g(x))g\text{‘}(x)dx={\displaystyle\int }_{g(a)}^{g(b)}f(u)du$

## Glossary

change of variables
the substitution of a variable, such as $u$, for an expression in the integrand
integration by substitution
a technique for integration that allows integration of functions that are the result of a chain-rule derivative