text:
Need Help?

Need help finding the right class? Have a question about how classes work?

Alcumus

Alcumus is our online adaptive learning system, and it is the main source of practice you'll have in your course. We include Alcumus assignments in all of our introductory-level subject courses – Prealgebra 1, Prealgebra 2, Introduction to Algebra A, Introduction to Counting, Introduction to Number Theory, Introduction to Algebra B, and Introduction to Geometry.

Your Alcumus problems are tailored to you! Alcumus monitors your progress and carefully selects problems, just for you, based on your performance on earlier problems. It will check your answer, provide you with solutions, and pick more problems.

There isn't a fixed number of problems to complete within a topic, and you'll likely see different problems than most of your classmates. Alcumus chooses your problems based on your level in that topic. You'll be able to more quickly pass topics you understand well, and you'll spend more time on topics you need more practice with.

You should work through Alcumus for as short or as long a time as needed to gain an understanding of the topic. Alcumus will tell you when you've passed a topic. After passing all of the week's assigned Alcumus topics, you should then move on to the week's Challenge Problems on the Homework tab of the course homepage.

Alcumus can be used by all AoPS users. This means that you'll still have access to it after your course ends, and all of the progress you made during the course will be saved! You can always access Alcumus from this page.

How To Use Alcumus

Each week, a list of 1-3 Alcumus topics will be listed on your Homework tab. Your goal is to "pass" those topics by answering questions in Alcumus. The Alcumus bar for the topic that you are working in will turn green when it's decided that you've demonstrated enough of an understanding of the topic to "pass." It will turn blue if you go above and beyond passing to achieve "mastery" of a topic.

Alcumus is on the Course Homepage

If your class has Alcumus assignments, there will be a Tab for Alcumus on the Course Homepage. You can work on your Alcumus assignments in the Alcumus tab there.

You can also solve problems on the Alcumus page, if you prefer. When using the Alcumus page, be sure to choose the Alcumus subject associated with your class and the topics within that are assigned for the week.

You have up to two tries at each Alcumus problem. After solving correctly, giving up, or two incorrect answers, you'll be shown the correct answer and a solution. Whether or not you've answered a problem correctly, be sure to read the solutions! For questions you didn't answer correctly, read about how you could have solved the problem. For questions that you did answer correctly, read over the solution to ensure that they understand it —you might learn something, such as a new method to solve the problem. Also, the solutions provided are great examples of well-written solutions, which you can refer to when writing up your own solutions for Writing Problems.

Even More Help

If you still need help after reading this page, you can visit the Alcumus Help Page. If you have any technical questions, you can find support on the Alcumus Forum.

A Little More About How Alcumus Works

Alcumus has seven subjects: Prealgebra, Algebra, Counting & Probability, Number Theory, Geometry, Intermediate Algebra, and Precalculus. The problems in these subjects are independent, except a small number of problems are shared between Prealgebra and other subjects and between Algebra and Intermediate Algebra.

Alcumus is a complicated system. The main idea is that it considers the difficulty level of the problems it has given you and how you've answered those problems. Alcumus will primarily give you problems it thinks are just the right level for you, while mixing in some problems that are a little easier and a little harder. It then uses all of this information to determine about how well you understand the material in each topic. This level of understanding is reflected in your "rating" within the topic.

Review Problems

Alcumus will also occasionally throw in some Review Problems from topics you've already passed. Topics in mathematics typically build off of things you've studied in the past, so Alcumus wants to make sure you're holding on to all of that mathematical knowledge! You won't be able to skip these problems. Answering a review problem incorrectly could cause you to "un-pass" a topic you previously passed. If that happens, be sure to revisit that topic when you have a little time, to refresh your memory.

Hard Problems

It's OK if you find some of the problems in an Alcumus topic to be too hard. There's a Give Up button in Alcumus for a reason! If you've given a problem your very best effort, it can sometimes be a better choice to give up on the problem and carefully read over the solution than to spend hours agonizing over it. Occasionally giving up or answering a problem incorrectly will not prevent you from passing the topic. However, make sure you aren't skipping over all the hard problems.

XP, Levels, Quests, and Achievements

Alcumus also tracks a lot of other data for you. You'll find several types of "XP" (Experience Points) that you can earn in Alcumus, as well as a number of entertaining Quests and Achievements. These have no impact on progressing through Alcumus, but they sure can be fun! Learn more about these extra features on the Alcumus Help Page.

If You're Having Trouble Passing A Topic

First of all, remember that your Alcumus rating in a topic is what Alcumus believes about your ability to solve problems in that topic. When you solve a problem correctly, your rating will go up. When you solve a problem incorrectly, your rating will go down. Once you've tried enough problems, you should expect your rating to be an honest measure of your ability to solve problems in that topic.

Your Alcumus ratings may fluctuate a lot when you start a new topic because Alcumus only has a little data from you. The more Alcumus learns about you in a topic, the less your rating in that topic will fluctuate.

If your rating stops moving and you haven't passed the topic, that means that you need to try something different! Look over the suggestions below. You can also check out our When You Get Stuck guide and the Alcumus Help page for more tips.

• Read the solutions. The problems are organized into topics for a reason. You should always be reading the solutions to problems to see if your ideas match up with ours, but you should especially be reading the solutions to problems you miss! There's likely to be something in the solution that is going to generalize to the next problem or the problem after that.
• Go back to the book. If you keep trying at a topic for a while but haven't been able to pass, then it's probably time to take a break. The solution to every Alcumus problem also lists related textbook sections. Go back to those textbook sections and see what it is that you might be missing.
• Don't skip all the hard problems. It will be nearly impossible to pass a topic if you are skipping a lot of problems, especially a lot of hard problems. When you encounter a tough problem, stop and try to figure it out. Even if you don't answer correctly or choose to ultimately give up, be sure to read the solution! Try to understand the important steps of the given solution —was there a step you missed? Should you go back to the textbook or ask a question on the message board to clarify something about the topic?
• Check your difficulty level. You can find this on the Alcumus settings page. If your difficulty rating is set to "hard" or "insanely hard," consider dialing it back one step while you're working through this topic.
• Ask questions. You can always ask your classmates and instructors on the class Message Board when you get stuck.
• Revisit solved problems. You can look over Alcumus problems you have answered on your Alcumus Report page. Click the time-and-date stamp above any problem statement to look over the answers you gave and view the official solution.

Formatting Tips

Entering answers that aren't numbers can sometimes be tricky. Computers aren't always as good as people are at reading answers written in a format they do not expect. Sometimes, the problem statement will include instructions for how to enter your answer. ("Enter all possible values of x, separated by commas.") Sometimes, you'll be given a link to formatting tips on how to enter a particular answer. Most of those tips come from the list given below. The formatting tip you're given for a problem may or may not be relevant to the answer that you're trying to enter—we don't want to spoil that your answer should be a fraction or square root!

• Do not include units of measurement or dollar signs in your answers.
• Express probabilities, decimal values, and ratios as simplified common fractions unless told otherwise.
• Enter variable values without including the variable and an equals sign. For example, enter
4
and not
x=4
.
• Enter fractions with x as the numerator and y as the denominator either as
x/y
or with the LaTeX code
\frac{x}{y}
. For example, enter the fraction "three fourths" as
3/4
or
\frac{3}{4}
.
• Enter mixed numbers only when the problem says so. Mixed numbers with z as the whole number, x as the numerator, and y as the denominator should be entered as
z x/y
. For example, enter "one and two thirds" as
1 2/3
.
• Enter "the square root of x" as
\sqrt{x}
.
• Enter powers with x as the base and y as the exponent as
x^y
. For example, enter "n to the 7th power" as
n^7
.
Make sure to include any necessary grouping symbols. For example,
(3x+2)^2
is different from
3x+2^2
.
• Enter coordinate pairs as
(x,y)
. For example, enter the pair "x=2 and y=8" as
(2,8)
.
• Write coefficients of variables and square roots without the multiplication sign. For example, enter
3x
3*x
and
4\sqrt{y}
4*\sqrt{y}
. The only exception is when you are asked to entering scientific notation, such as
4.01*10^21
.
• Enter polynomials in decreasing degree order. For example, enter
x^2+3x+2
3x+2+x^2
.
• Enter expressions in alphabetical order when using multiple variables. For example, enter
x^3yz^2
z^2yx^3
pi


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# Scanning Electron Microscopy: 6 SEM Sample Preparation Pointers for Successful Imaging

An image is worth a thousand words.

Scanning Electron Microscopy (SEM) is one of the more popular imaging techniques in life science [1] because of the pretty and informative micrographs you get from it.

Like most experiments, however, your data (or in this case your image) is only as good as your sample.

In this article, we’ll cover 6 essential SEM sample prep points to maximize your chances of successful SEM imaging.

Need to take a step back? Read our ‘Applications of Electron Microscopy‘ article to learn more about how electron microscopy works and what it can be used for.

## The Working Principles of SEM

Ever wondered how SEM works? Here are the very basics:

An electron gun at the top of the microscope directs a stream of electrons vertically down a set of electromagnetic lenses inside a vacuum chamber. Usually, one of the following electron guns is present in an SEM microscope:

1. Thermionic guns: these use filament heating to to produce a stream of electrons.
2. Field emission guns: these produce a strong electric field in order to eject electrons from their atoms.

The lenses are placed in a vacuum chamber in order to avoid obstruction and contamination by other particles. The lenses help direct the electrons towards the sample. Then, the electron imprint is converted to a three-dimensional image of the sample which is digitally constructed.

## A Note on Conductivity and Sputter Coating

Samples for SEM need to be conductive. Or at least their surface does.

This is because non-conductive materials “collect” electrons on the surface. This build-up of charge leads to image artifacts.

So, metal samples require no additional preparation due to their inherent ability to conduct electricity.

Non-metal samples, however, need to be coated with a conductive material during SEM sample prep to make them compatible with SEM. Usually, a thin layer of gold works fine. Deposition of conductive material onto the sample is performed using an instrument called a sputter coater. Check out Figure 1 below for an example.

With all that out of the way, let’s get on with the SEM sample preparation.

## SEM Sample Prep: 6 Pointers

For scanning electron microscopy, you need to consider the sample’s size, shape, state, and conductive properties prior to imaging.

Ideally, the smallest representative sample size is the one to use. The microscope’s detection capacity is as much as 1 $\mu$m from the sample surface. So, consider all the following:

### 1. Sample Cleaning

A clean sample is essential for image clarity. For biological samples, use appropriate buffers or distilled water for cleaning the samples. Use a surfactant if the sample requires more vigorous cleaning.

If the biological property of the sample is known, you might be able to use proteolytic enzyme cleaning. To remove oils on the sample surface, wash with appropriate solvents. Additionally, you can use ultrasonic baths for cleaning the sample.

Be careful as ultrasonic baths are quite aggressive and may physically harm your sample.

### 2. Sample Fixation and Dehydration

Use a fixative like glutaraldehyde or osmium vapor to maintain the structural details of the sample.

Note: if a fixative uses a phosphate-based buffer for its preparation, salt deposits may interfere with the sample’s image quality.

For dehydration, use a graded series of alcohol and finish off the final dehydration step with 100% alcohol or acetone.

### 3. Drying

Prior to placing the sample in a high vacuum environment, it must be totally dry. Otherwise, water vaporization will obstruct the electron beam and reduce the clarity of the micrograph.

When using biological samples, be careful with your critical point drying (CPD), so as to not compromise the structural integrity of the sample. A suitable CPD instrument can help achieve this.

Alternatively, you can try using freeze-drying. Freeze-drying causes the least amount of sample shrinkage in comparison to air drying or CPD. However, freeze-drying carries the risk of ice crystal formation on the sample.

### 4. Sample Preparation of Tissue Sections

To observe details from tissue sections, remove the epoxy resin using organic solvents, ion beam etching, or plasma etching. You can also break the sample in the appropriate direction to reveal its internal details.

### 5. Sample Stubs, Adhesives, and Mounting Approach

Sample stubs or supports are available in different diameters, making them convenient for imaging different kinds of samples. Be sure the stubs are clean and handle the sample with clean forceps.

Also, use gloves during the entire sample preparation stage. The material you use as an adhesive to glue the sample to the stub should be non-toxic and should not tumble into the sides of the sample.

Conductive, double-coated carbon tape is the most common sample mounting adhesive. You need to ensure that a conducting path exists as you mount the sample.

This is because, to generate a micrograph, the electrons need to pass through your sample and mounting apparatus (the whole setup is grounded/earthed).

So, if the sample requires a conductive coating, be sure to mount it before coating so that both the sample and plug receive the coating.

### 6. Sample Storage

Store the sample and stubs in a dry, clean environment. Use clean forceps and gloves whenever you handle the stubs.

## Wrapping Up

I hope that, after reading this article, you have a feel for what’s important in SEM sample prep to get successful images like the one below.

Happy imaging!

Got any great SEM sample prep tips of your own? Leave them in the comments below if so. We’ve already updated the article based on them, so they are useful.

Originally published April 2017. Reviewed and updated December 2021.

## Reference

1. Golding C, Lamboo L, and Beniac D et al. (2016) The scanning electron microscope in microbiology and diagnosis of infectious diseaseSci Rep 6:26516

1. Oyesolape on January 2, 2019 at 4:28 am

What of if my sample is a Nanoparticles synthesized with a plant extract, what cleaning procedure would you suggest

2. Aaron Leong on December 8, 2017 at 4:49 am

For Sample Fixation and Dehydration, if my sample used is metal can I just ignore this step?

3. Bob Kyeyune on November 12, 2017 at 10:45 pm

Thanks for the article, its been helpful.

4. Sandeep on November 8, 2017 at 7:23 pm

Thanks for having this precious knowledge
I am very thankful to you for sharing this knowledge
Thank you

5. ujek on April 14, 2017 at 8:10 am

nice article. thanks for the info

what is the purpose of conductivity for the non metal sample? thanks in advance

• Blanca Carbajal Gonzalez on June 1, 2017 at 3:39 pm

If your sample is not conductive, electrons will build up on your sample. This results in bright white spots or lines across your image and is referred to as “charging”. Coating non-conductive samples minimizes this effect.

• Indrani on January 20, 2019 at 5:47 pm

Non metals do not bear conductance which is the essence of any sample to be visualized under SEM. Samples need to have properly absorbed electron beam. So covering non metal surface with any conducting material will be able to absorb electron beam.

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# Relating Radiancy to Energy Density?

1. Nov 9, 2005

### eck

I'm looking through some of my physics notes where we derived radiancy equations for black body radiation (Rayleigh-Jeans and Planck), and I have a quick question. In class, we went through the derivation for the energy density as a function of frequency and temperature. However, I can't find anything in my notes about how radiancy is related to energy density. I tried to look online, and didn't find anything helpful, other than the equation that relates them: $$R(\nu ,T) = c/4 u(\nu ,T)$$, where u is the energy density and R is the radiancy. From the units it looks like radiancy is the power per area per time, but I am totally mystified by the factor c/4. Can anyone quickly go over where this comes from?

2. Nov 9, 2005

### eck

Sorry, the TeX came out wrong. It should be one-fourth times c times u (nu, T).


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# Output of a LTI system [closed]

I'm trying to resolve this exercise but I don't even know where to start. Do I have to apply convolution?

**It appears that this question is classified as homework - this an exame problem which I was/am studying **, not homework per se

The "simple" doubt was I had no idea where to start -The answer by Fat32 helped me a lot: "express the input x1[n] in terms of shifted and scaled copies of x0[n]"

I can upload the resolution with the graphics if this help the question.

Thank you, Eddy

• No, you must apply fundamental LTI property: Express the input $x_1[n]$ in terms of shifted and scaled copies of $x_0[n]$, and then the output will be obtained in terms of the same shifted and scaled copies of the response, $y_0[n]$ , of the system to the basic input $x_0[n]$... May 9 at 17:44
• Eddy, this is not a place to post your homework. May 9 at 17:44
• I simply don't know how to do this question, but thank you for your help. If you wish I will gladly send my homework May 9 at 18:24
• @Eddygrinder: Why don't you try to follow Fat32's advice? May 9 at 18:25
• @MattL. sure I will May 9 at 18:26

$$x_1[n] = 2x_0[n-4] + x_0[n-2] + x_0[n-6]$$
$$-\delta[n]-2\delta[n-1]-2\delta[n-2]-2\delta[n-3]+2\delta[n-5]+2\delta[n-6]+2\delta[n-7]+\delta[n-8]$$


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If you like us, please share us on social media.
The latest UCD Hyperlibrary newsletter is now complete, check it out.

MindTouch
Copyright (c) 2006-2014 MindTouch Inc.
http://mindtouch.com

This file and accompanying files are licensed under the MindTouch Master Subscription Agreement (MSA).

At any time, you shall not, directly or indirectly: (i) sublicense, resell, rent, lease, distribute, market, commercialize or otherwise transfer rights or usage to: (a) the Software, (b) any modified version or derivative work of the Software created by you or for you, or (c) MindTouch Open Source (which includes all non-supported versions of MindTouch-developed software), for any purpose including timesharing or service bureau purposes; (ii) remove or alter any copyright, trademark or proprietary notice in the Software; (iii) transfer, use or export the Software in violation of any applicable laws or regulations of any government or governmental agency; (iv) use or run on any of your hardware, or have deployed for use, any production version of MindTouch Open Source; (v) use any of the Support Services, Error corrections, Updates or Upgrades, for the MindTouch Open Source software or for any Server for which Support Services are not then purchased as provided hereunder; or (vi) reverse engineer, decompile or modify any encrypted or encoded portion of the Software.

A complete copy of the MSA is available at http://www.mindtouch.com/msa

# Gibbs Free Energy

Gibbs free energy, denoted G, combines enthalpy and entropy into a single value. The change in free energy, ΔG, is equal to the sum of the enthalpy plus the product of the temperature and entropy of the system. ΔG can predict the direction of the chemical reaction under two conditions: (1) constant temperature and (2) constant pressure. If ΔG is positive, then the reaction is nonspontaneous (it requires  the input of external energy to occur) and if it is negative, then it is spontaneous (occurs without external energy input).

### Introduction

In 1875, Josiah Gibbs introduced a thermodynamic quantity combining enthalpy and entropy into a single value called Gibbs free energy. This quantity is the energy associated with a chemical reaction that can be used to do work, and is the sum of its enthalpy (H) and the product of the temperature and the entropy (S) of the system. This quantity is defined as follows:

$G= H-TS$

or more completely as

$G= U+PV-TS$

where

• U = internal energy (SI unit: joule)
• P = pressure (SI unit: pascal)
• V = volume (SI unit: $$m^3$$)
• T = temperature (SI unit: kelvin)
• S = entropy (SI unit: joule/kelvin)
• H = enthalpy (SI unit: joule)

### Free energy of reaction (ΔG)

In chemical reactions involving the changes in thermodynamic quantities, a variation on this equation is often encountered:

$\Delta G \qquad \qquad = \qquad \qquad \Delta H \qquad \qquad - \qquad \qquad T \Delta S$

$\text {change in free energy} \qquad \text {change in enthalpy} \qquad \text {(temperature) change in entropy}$

The sign of ΔG indicates the direction of a chemical reaction:

Terminology

• If $$\Delta H^o$$ < 0 and  $$\Delta S^o$$ > 0,  then the reaction is spontaneous ($$\Delta G^o$$ < 0 ) at any temperature.
• If $$\Delta H^o$$ > 0  and entropy $$\Delta S^o$$ < 0,  then the reaction is nonspontaneous ($$\Delta G^o$$ > 0 ) at any temperature.

Standard-state free energy of reaction ( $$\Delta G^o$$) is defined as the free energy of reaction at standard state conditions:

$\Delta G^o = \Delta H^o - T \Delta S^o$

### Standard-State Free Energy of Formation (ΔGfo)

• The partial pressure of any gas involved in the reaction is 0.1 MPa.
• The concentrations of all aqueous solutions are 1 M.
• Measurements are generally taken at a temperature of 25° C (298 K).

The standard-state free energy of formation is the change in free energy that occurs when a compound is formed from its elements in their most thermodynamically stable states at standard-state conditions. In other words, it is the difference between the free energy of a substance and the free energies of its constituent elements at standard-state conditions.

The standard-state free energy of reaction can be calculated from the standard-state free energies of formation.  It is the sum of the free energies of formation of the products minus the sum of the free energies of formation of the reactants:

$\Delta G^o = \sum \Delta G^o_{f_{products}} - \sum \Delta G^o_{f_{reactants}}$

### Free energy and equilibrium constants

The following equation relates the standard-state free energy of reaction with the free energy at any point in a given reaction (not necessarily at standard-state conditions):

$\Delta G = \Delta G^o + RT \ln Q$

• $$\Delta G$$ = free energy at any moment
• $$\Delta G^o$$ = standard-state free energy
• R is the ideal gas constant = 8.314 J/mol-K
• T is the absolute temperature (Kelvin)
• $$\ln Q$$ is natural logarithm of the reaction quotient

At equilibrium, ΔG = 0 and Q=K. Thus the equation can be arranged into:

$\Delta G^o = - RT \ln K$

This equation is particularly interesting as it relates the free energy difference under standard conditions to the properties of a system at equilibrium (which is very rarely at standard conditions).

### Free energy and cell potentials

The Nernst equation relates the standard-state cell potential with the cell potential of the cell at any moment in time:

$E = E^o - \frac {RT}{nF} \ln Q$

• $$E$$ = cell potential in volts (joules per coulomb)
• $$n$$ = moles of electrons
• $$F$$ = Faraday's constant: 96,485 coulombs per mole of electrons

By rearranging this equation we obtain:

$E = E^o - \frac {RT}{nF} \ln Q \text {multiply the entire equation by nF}$

$nFE = nFE^o - RT \ln Q$

which is similar to:

$\Delta G = \Delta G^o + RT \ln Q$

By juxtaposing these two equations:

$nFE = nFE^o - RT \ln Q$

$\Delta G = \Delta G^o + RT \ln Q$

it can be concluded that:

$\Delta G = -nFE$

Therefore,

$\Delta G^o = -nFE^o$

Example 1

These values can be substituted into the free energy equation

### References

1. General Chemistry: Principles of Modern Applications 9th Edition (pages 792-795)
2. SAT Subject Test: Chemistry 6th Edition (page 66)
3. http://en.wikipedia.org/wiki/Gibbs_free_energy

13:08, 13 Jan 2015

## Classifications

(not set)
(not set)

### Textbook Maps

An NSF funded Project


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# Grouped Bar Chart with logarithmic scale

I am creating a grouped bar chart with logarithmic scale but I encountered some problems. In the below picture you can see that the ticks on the x-scale are not positioned right. The beginning is correct, but "St.8" for example is too far left.

\documentclass[12pt,a4paper,onecolumn, openright]{report}
\usepackage{xcolor}
\usepackage{pgfplots}
\usepackage{tikz}

\definecolor{bblue}{HTML}{4F81BD}
\definecolor{rred}{HTML}{C0504D}
\definecolor{ggreen}{HTML}{9BBB59}

\begin{document}

\begin{tikzpicture}
\begin{axis}[
width  = 0.9\textwidth,
height = 8cm,
bar width=4pt,
major x tick style = transparent,
symbolic x coords={St.~1,St.~2,St.~3,St.~4,St.~5,St.~6,St.~7,St.~8},
enlarge x limits=0.1,
xlabel={Stufe des Algorithmus},
x label style={at={(axis description cs:0.5,-0.1)},anchor=north},
ymin = 0,
ybar = \pgflinewidth,
ymajorgrids = true,
ylabel = {Anzahl},
ymode=log,
log basis y={10},
legend cell align=left,
legend style={
at={(1,1.05)},
anchor=south east,
column sep=1ex
}
]

\addplot[style={bblue,fill=bblue,mark=none}] coordinates {(St.~1, 1) (St.~2,10006) (St.~3,99895) (St.~4, 99867)};
\addplot[style={rred,fill=rred,mark=none}] coordinates {(St.~1,1) (St.~2,10006) (St.~3,99448) (St.~4, 99487)};
\addplot[style={ggreen,fill=ggreen,mark=none}] coordinates {(St.~1,1) (St.~2,10) (St.~3,10) (St.~4, 10)};

\addplot[style={bblue!50,fill=bblue!50,mark=none}] coordinates {(St.~5,99810) (St.~6,99857) (St.~7, 80000) (St.~8, 40000)};
\addplot[style={rred!50,fill=rred!50,mark=none}] coordinates {(St.~5,99416) (St.~6,99463) (St.~7, 79685) (St.~8, 39842)};
\addplot[style={ggreen!50,fill=ggreen!50,mark=none}] coordinates {(St.~5,10) (St.~6,10) (St.~7, 10) (St.~8, 0)};

\legend{Muster,Subgruppen,Gekürzt}
\end{axis}
\end{tikzpicture}

\end{document}

I tried rearranging the code for the plots and got a better result, but the bars are not grouped anymore. What can I do to keep the grouping and correct the x-axis?

% blue
\addplot[style={bblue,fill=bblue,mark=none}] coordinates {(St.~1, 1) (St.~2,10006) (St.~3,99895) (St.~4, 99867)};
\addplot[style={bblue!50,fill=bblue!50,mark=none}] coordinates {(St.~5,99810) (St.~6,99857) (St.~7, 80000) (St.~8, 40000)};

% red
\addplot[style={rred,fill=rred,mark=none}] coordinates {(St.~1,1) (St.~2,10006) (St.~3,99448) (St.~4, 99487)};
\addplot[style={rred!50,fill=rred!50,mark=none}] coordinates {(St.~5,99416) (St.~6,99463) (St.~7, 79685) (St.~8, 39842)};

% green
\addplot[style={ggreen,fill=ggreen,mark=none}] coordinates {(St.~1,1) (St.~2,10) (St.~3,10) (St.~4, 10)};
\addplot[style={ggreen!50,fill=ggreen!50,mark=none}] coordinates {(St.~5,10) (St.~6,10) (St.~7, 10) (St.~8, 0)};

So you would like to have something like the following?

% used PGFPlots v1.14
\documentclass[border=5pt]{standalone}
\usepackage{pgfplots}
\pgfplotsset{compat=1.3}
\definecolor{bblue}{HTML}{4F81BD}
\definecolor{rred}{HTML}{C0504D}
\definecolor{ggreen}{HTML}{9BBB59}
\begin{document}
\begin{tikzpicture}
\begin{axis}[
width=0.9\textwidth,
height=8cm,
bar width=4pt,
% commented to see, if the result is correct
%        major x tick style=transparent,
symbolic x coords={St.~1,St.~2,St.~3,St.~4,St.~5,St.~6,St.~7,St.~8},
enlarge x limits=0.1,
xlabel={Stufe des Algorithmus},
x label style={at={(axis description cs:0.5,-0.1)},anchor=north},
% changed from \pgflinewidth' to 1 easier see that the result is correct
ymin=1,
% changed to 0 to make the computation of the shift easier
ybar=0pt,
ymajorgrids=true,
ylabel={Anzahl},
ymode=log,
legend cell align=left,
legend style={
at={(1,1.05)},
anchor=south east,
column sep=1ex
},
]
% add a scope around the to shift bars
\begin{scope}[
% shift the bars accordingly
xshift={0.75*\pgfplotbarwidth},
% and just "fill" the bars
% (to avoid overlapping of the more right bars to the more left ones)
draw=none,
]
% your style could heavily be simplified to just providing the +'
% sign and the color as option
{(St.~1, 1) (St.~2,10006) (St.~3,99895) (St.~4, 99867)};
{(St.~1,1) (St.~2,10006) (St.~3,99448) (St.~4, 99487)};
{(St.~1,1) (St.~2,10) (St.~3,10) (St.~4, 10)};
\end{scope}

\begin{scope}[
xshift={-0.75*\pgfplotbarwidth},
draw=none,
]
{(St.~5,99810) (St.~6,99857) (St.~7, 80000) (St.~8, 40000)};
{(St.~5,99416) (St.~6,99463) (St.~7, 79685) (St.~8, 39842)};
{(St.~5,10) (St.~6,10) (St.~7, 10) (St.~8, 0)};
\end{scope}

\legend{Muster,Subgruppen,Gekürzt}
\end{axis}
\end{tikzpicture}
\end{document}

• That looks good. The smaller bars are nearly invisible because they are so small. How can I conveniently write the total count above each bar (vertically)? Maybe that would make it more understandable. – robinki Feb 12 '17 at 12:30
• I would just state another value for ymin, e.g. 1e-1. Does that fit your needs as well? (Because when using nodes near coords this will overlap the values and you have another problem.) – Stefan Pinnow Feb 12 '17 at 12:54
• That should do the trick for now. Thank you. I would upvote, if I had enough reputation. – robinki Feb 12 '17 at 13:09
• Not a problem. What a little bit longer and just remember to to it when you have ;) – Stefan Pinnow Feb 12 '17 at 14:00
• I finally upvoted now. I used ymin = 0 in the final version, btw. You can slightly see the bar of the first stage. I will see how it looks printed. – robinki Feb 15 '17 at 8:51


text:
The complex conjugate $$z^*$$ of a complex number $$z=x+iy$$ is found by replacing every $$i$$ by $$-i\text{.}$$ Therefore $$z^*=x-iy\text{.}$$ (A common alternate notation for $$z^*$$ is $$\bar{z}\text{.}$$) Geometrically, you should be able to see that the complex conjugate of ANY complex number is found by reflecting in the real axis, as shown in Figure 2.2.1.
Notice that this product is ALWAYS a positive, real number. The (positive) square root of this number is the distance of the point $$z$$ from the origin in the complex plane. We call the square root the norm or magnitude of $$z$$ and we use the same notation as "absolute value," i.e. $$\vert z\vert\text{.}$$ In this way, we see that the definition of absolute value, as in $$\vert -2\vert=2\text{,}$$ was never "strip off the minus sign," but really "how far is $$-2$$ from the origin."


text:
# Equivalent Fractions

Equivalent fractions are the fractions having the same value. Same fraction can be represented in many ways. Let us take the following example.

In picture (i) the shaded part is represented by fraction $$\frac{1}{2}$$.

The shaded part in picture (ii) is represented by fraction $$\frac{2}{4}$$. In picture (iii) the same part is represented by fraction $$\frac{4}{8}$$. SO, the fraction represented by these shaded portions are equal. Such fractions are called equivalent fractions.

We say that $$\frac{1}{2}$$ = $$\frac{2}{4}$$ = $$\frac{4}{8}$$

Hence, for a given fraction there can be many equivalent fractions.

Making Equivalent Fractions:

We have seen in the above example that $$\frac{1}{2}$$, $$\frac{2}{4}$$ and $$\frac{4}{8}$$ are equivalent fractions.

Therefore, $$\frac{1}{2}$$ can be written as $$\frac{1}{2}$$ = $$\frac{1 × 2}{2 × 2}$$ = $$\frac{1 × 3}{2 × 3}$$ = $$\frac{1 × 4}{2 × 4}$$ and so on.

Hence, an equivalent fraction of any given fraction can be obtained by multiplying its numerator and denominator by the same number.

Same way, when the numerator and denominator of a fraction are divided by the same number, we get its equivalent fractions.

$$\frac{1}{2}$$ = $$\frac{1 ÷ 1}{2 ÷ 1}$$ = $$\frac{2}{4}$$ = $$\frac{2 ÷ 2}{4 ÷ 2}$$ = $$\frac{3}{6}$$ = $$\frac{3 ÷ 3}{6 ÷ 3}$$

We have,

2/4 = (1 × 2)/(2 × 2)

3/6 = (1 × 3)/(2 × 3)

4/8 = (1 × 4)/(2 × 4)
We observe that 2/4, 3/6 and 4/8 are obtained by multiplying the numerator and denominator of 1/2 by 2, 3 and 4 respectively.

Thus, an equivalent fraction of a given fraction can be obtained by multiplying its numerator and denominator by the same number (other than zero).

2/4 = (2÷ 2)/(4 ÷ 2) = 1/2

3/6 = (3÷ 3)/(6 ÷ 3) = 1/2

4/8 = (4 ÷ 4)/(8 ÷ 4) = 1/2

We observe that if we divide the numerators and denominators of 2/4, 3/6 and 4/8 each by their common factor 2, we get an equivalent fraction 1/2.

Thus, an equivalent fraction of a given fraction can be obtained by dividing its numerator and denominator by their common factor (other than 1), if ant.

Note:

(i) Multiplying its numerator (top) and denominator (bottom) by the same number (other than 0).

(ii) Dividing its numerator (top) and denominator (bottom) by their common factor (other than 1).

For Example:

1. Write three equivalent fraction of 3/5.

Equivalent fractions of 3/5 are:

(3 × 2)/(5× 2) = 6/10,

(3 × 3)/(5 × 3) = 9/15,

(3 × 4)/(5 × 4) = 12/20

Therefore, equivalent fractions of 3/5 are 6/10, 9/15 and 12/20.

2. Write next three equivalent fraction of $$\frac{2}{3}$$.

We multiply the numerator and the denominator by 2.

We get, $$\frac{2 × 2}{3 × 2}$$ = $$\frac{4}{6}$$

Next, we multiply the numerator and the denominator by 3. We get

$$\frac{2 × 3}{3 × 3}$$ = $$\frac{6}{9}$$.

Next, we multiply the numerator and the denominator by 4. We get

$$\frac{2 × 4}{3 × 4}$$ = $$\frac{8}{12}$$.

Therefore, equivalent fractions of $$\frac{2}{3}$$ are $$\frac{4}{6}$$, $$\frac{6}{9}$$ and $$\frac{8}{12}$$.

3. Write three equivalent fraction of 1/4.

Equivalent fractions of 1/4 are:

(1× 2)/(4× 2) = 2/8,

(1 × 3)/(4 × 3) = 3/12,

(1× 4)/(4× 4) = 4/16

Therefore, equivalent fractions of 1/4 are 2/8, 3/12 and 4/16.

4. Write three equivalent fraction of 2/15.

Equivalent fractions of 2/15 are:

(2× 2)/(15 × 2) = 4/30,

(2 × 3)/(15 × 3) = 6/45,

(2× 4)/(15 × 4) = 8/60

Therefore, equivalent fractions of 2/15 are 4/30, 6/45 and 8/60.

5. Write three equivalent fraction of 3/10.

Equivalent fractions of 3/10 are:

(3× 2)/(10× 2) = 6/20,

(3 × 3)/(10 × 3) = 9/30,

(3× 4)/(10× 4) = 12/40

Therefore, equivalent fractions of 3/10 are 6/20, 9/30 and 12/40.

## You might like these

To add two or more like fractions we simplify add their numerators. The denominator remains same.

• ### Worksheet on Addition of Like Fractions | Addition of Fractions

In worksheet on addition of fractions having the same denominator, all grade students can practice the questions on adding fractions. This exercise sheet on fractions can be practiced by the students to get more ideas how to add fractions with the same denominators.

• ### Worksheet on Subtraction of Like Fractions |Subtracting Like Fractions

In worksheet on subtraction of fractions having the same denominator, all grade students can practice the questions on subtracting fractions. This exercise sheet on fractions can be practiced by the students to get more ideas how to subtract fractions with the same

• ### Addition and Subtraction of Like Fractions |Addition of Like Fractions

Addition and subtraction of like fractions. Addition of Like Fractions: To add two or more like fractions we simplify add their numerators. The denominator remains same. To subtract two or more like fractions we simply subtract their numerators and keep the same denominator.

• ### Worksheet on Add and Subtract Fractions | Word Problems | Fractions

Recall the topic carefully and practice the questions given in the math worksheet on add and subtract fractions. The question mainly covers addition with the help of a fraction number line, subtraction with the help of a fraction number line, add the fractions with the same

• ### 4th Grade Fractions Worksheet | Fractions Worksheets | Math Only Math

In 4th grade fractions worksheet we will circle the like fractions, circle the greatest fraction, arrange the fractions in descending order, arrange the fractions in ascending order, addition of like fractions and subtraction of like fractions.

• ### Fractions in Ascending Order | Arranging Fractions an Ascending Order

We will discuss here how to arrange the fractions in ascending order. Solved examples for arranging in ascending order: 1. Arrange the following fractions 5/6, 8/9, 2/3 in ascending order. First we find the L.C.M. of the denominators of the fractions to make the denominators

• ### Comparison of Unlike Fractions | Compare Unlike Fractions | Comparing

In comparison of unlike fractions, we change the unlike fractions to like fractions and then compare. To compare two fractions with different numerators and different denominators, we multiply by a number to convert them to like fractions. Let us consider some of the

• ### Comparison of Like Fractions | Comparing Fractions | Like Fractions

Any two like fractions can be compared by comparing their numerators. The fraction with larger numerator is greater than the fraction with smaller numerator, for example $$\frac{7}{13}$$ > $$\frac{2}{13}$$ because 7 > 2. In comparison of like fractions here are some

• ### Like and Unlike Fractions | Like Fractions |Unlike Fractions |Examples

Like and unlike fractions are the two groups of fractions: (i) 1/5, 3/5, 2/5, 4/5, 6/5 (ii) 3/4, 5/6, 1/3, 4/7, 9/9 In group (i) the denominator of each fraction is 5, i.e., the denominators of the fractions are equal. The fractions with the same denominators are called

• ### Worksheet on Equivalent Fractions | Questions on Equivalent Fractions

In worksheet on equivalent fractions, all grade students can practice the questions on equivalent fractions. This exercise sheet on equivalent fractions can be practiced by the students to get more ideas to change the fractions into equivalent fractions.

• ### Verification of Equivalent Fractions | Exploring Equivalent Fractions

We will discuss here about verification of equivalent fractions. To verify that two fractions are equivalent or not, we multiply the numerator of one fraction by the denominator of the other fraction. Similarly, we multiply the denominator of one fraction by the numerator

• ### 5th Grade Fractions Worksheets | Comparison of Fractions | Addition

In 5th Grade Fractions Worksheets we will solve how to compare two fractions, comparing mixed fractions, addition of like fractions, addition of unlike fractions, addition of mixed fractions, word problems on addition of fractions, subtraction of like fractions

• ### Reciprocal of a Fraction | Multiply the Reciprocal of the Divisor

Here we will learn Reciprocal of a fraction. What is 1/4 of 4? We know that 1/4 of 4 means 1/4 × 4, let us use the rule of repeated addition to find 1/4× 4. We can say that $$\frac{1}{4}$$ is the reciprocal of 4 or 4 is the reciprocal or multiplicative inverse of 1/4

• ### Division in Terms of Reciprocal | Division of Fractions | Reciprocal

To divide a fraction or a whole number by a fraction or a whole number, we multiply the reciprocal of the divisor. We know that the reciprocal or the multiplicative inverse of 2 is $$\frac{1}{2}$$.

Representations of Fractions on a Number Line

Fraction as Division

Types of Fractions

Conversion of Mixed Fractions into Improper Fractions

Conversion of Improper Fractions into Mixed Fractions

Equivalent Fractions

Fractions in Lowest Terms

Like and Unlike Fractions

Comparing Like Fractions

Comparing Unlike Fractions

Addition and Subtraction of Like Fractions

Addition and Subtraction of Unlike Fractions

Inserting a Fraction between Two Given Fractions


text:
+0

# Since I got 100 points (I created my account several days ago), I want to make some problems for it! Thanks all for helping me!

+1
156
4
+161

The easiest problem:

Nathan and Nick went to the Tiger Carnival. They earned 12 stickers, 23 pencils, and 89 markers. They wanted to put them in groups. What is the largest group they can put it in?

The intermediate problem:

$$10^{3}+100^{90}$$

What is the answer? Put it in powers of ten. For example: $$10^{1}$$

What is the area, in square units, of the interior region formed by the lines y=2x-4,y=-3x+16 and they y axis? (this problem, I didn't make it. It is a problem I like from AoPS alcumus)

The master problem: (precalc, I don't expect an answer from this though)

Find $$\begin{pmatrix} -5 \\ 1 \\ -4 \end{pmatrix} + \begin{pmatrix} 0 \\ 8 \\ -4 \end{pmatrix}$$

The bonus question:

Jan has a 20 by 20 by 20 gift box that that needs to be placed carefully into a 2 by 2 by 2 shipping carton, surronded by packing peanuts. How many half-cubic-foot bags of peanuts does Jan need to buy?

Hope you can solve the problems! If you can answer it, put a solution next to it so others can understand. Thanks so much for 100 points!

-tigernathan

Jul 15, 2020
edited by tigernathan  Jul 15, 2020

#1
+2141
0

Yay! More of tigernathan's questions!

Number 1: Twelve groups, since the smallest number of anything is 12.

Jul 15, 2020
#2
+161
+1

You got it! A lot of people think this as a trick question, since it never said equal.

tigernathan  Jul 15, 2020
#3
+2141
0

10^3 plus 10^180

Simplify that!

Jul 15, 2020
#4
+161
+1

Remember the rule: you add the exponents. So, what is the answer?

tigernathan  Jul 15, 2020


text:
시간 제한 메모리 제한 제출 정답 맞은 사람 정답 비율
2 초 512 MB 0 0 0 0.000%

## 문제

Type checking is the process of verifying that a computer program satisfies the typing rules of a programming language. In some languages this is very difficult, and just checking if two types are equal can be tricky!

For this problem, a type will be represented as a tree made up of nodes. Each node has zero or more child nodes, corresponding to component types. Here is an example of a tree:

Figure 1: A simple finite tree. The root is node 0 and its children are nodes 1 and 2. Node 1 does not have any children, and node 2 has one child (node 3).

Types are allowed to be recursive. This means that a node can have any node as a child, including its parent and even itself. The result is an infinite tree:

Figure 2: An infinite tree. Node 0 has node 2 as a child, which has node 0 as a child, which has node 2 as a child, and so on. The tree keeps going forever, and has been truncated to fit on this page.

Given two possibly infinite trees, your task is to determine whether or not they have the same structure. Two nodes A and B have the same structure if

• They have the same number of children, and
• The i'th child of node A has the same structure as the i'th child of node B, for each child index i.

Two trees have the same structure if their root nodes have the same structure.

## 입력

Input will consist of a sequence of problems. The first line for each problem will have two space-separated integers, N and M, the number of nodes in the two trees (1 ≤ N, M ≤ 100,000).

The following N lines will describe the first tree. The i'th of these lines (counting from 0) will consist of a number of space-separated integers describing node i. The first integer will be c[i], the number of children that node i has (0 ≤ c[i]; sum of c[i]'s ≤ 100,000). The rest of the line will contain c[i] space-separated integers, the children of node i. The following M lines will describe the second tree similarly. The root of each tree is always node 0.

Each problem will be followed by a blank line. End of input will be denoted by a line with two zeroes (which should not be processed).

## 출력

For each problem, output "YES" if the two trees have the same structure, or "NO" if they don't.

## 예제 입력 1

4 4
2 1 2
0
1 3
0
2 2 1
0
1 3
0

3 4
2 1 2
0
1 0
2 1 2
0
1 3
2 1 2

3 1
2 1 1
2 2 2
2 0 0
2 0 0

0 0

## 예제 출력 1

NO
YES
YES

## 힌트

For the first problem, Figure 1 shows the first tree and the other tree is its mirror image. They have different structures, because the order of the children is significant.

Figure 2 shows the first tree in the second problem. The other tree is represented using more nodes but has the same structure.


text:
Thread: Quadratic Substitution.

1. Quadratic Substitution.

Solve Using substitution.

$x^{3/2}+8x^{1/2}+16x^{-1/2}=0$

$Let A= x^{1/2}$

$A^2+8A+\frac{16}{A}=0$

$A^3+8A^2+16=0$

$A^2(A+8)=-16$

I have no clue if i even did this correctly...Help me please.

2. Re: Quadratic Substitution.

Originally Posted by theloser
Solve Using substitution.

$x^{3/2}+8x^{1/2}+16x^{-1/2}=0$

$Let A= x^{1/2}$

$A^2+8A+\frac{16}{A}=0$
This should be

$A^3+8A+\frac{16}{A}=0$

Now multiply through by A.

3. Re: Quadratic Substitution.

So even if the the first factor is not ^2 of the second term, A substitution is still possible?

$A^4+8A^2+16=0$

$A^2(A^2+8)=-16$

$A=\pm\sqrt{-16}$
$Or$
$A=\pm\sqrt{-24}$

4. Re: Quadratic Substitution.

Originally Posted by theloser
So even if the the first factor is not ^2 of the second term, A substitution is still possible?

$A^4+8A^2+16=0$

$A^2(A^2+8)=-16$

$A=\pm\sqrt{-16}$
$Or$
$A=\pm\sqrt{-24}$
$A^4+8A^2+16=0$

Now $A^2 = B$ thus

$B^2+8B+16=0$

$(B+4)^2=0$ , back substitute twice.

5. Re: Quadratic Substitution.

$A=\pm\sqrt{-4}$

6. Re: Quadratic Substitution.

Originally Posted by theloser
$A=\pm\sqrt{-4}$
Yes, now you know that $\displaystyle x^{\frac{1}{2}} = \pm \sqrt{-4}$, what is $\displaystyle x$?

7. Re: Quadratic Substitution.

Not sure since there is already a $\pm$ on one side, how do I square both sides?

8. Re: Quadratic Substitution.

Originally Posted by theloser
Not sure since there is already a $\pm$ on one side, how do I square both sides?
What happens when you square a positive number? What happens when you square a negative number?

9. Re: Quadratic Substitution.

$x=\pm4$

$x=\pm-4$

10. Re: Quadratic Substitution.

Originally Posted by theloser
$x=\pm4$

$x=\pm-4$
Not even close, sorry.

$\displaystyle \left(\sqrt{-4}\right)^2 = \sqrt{-4} \times \sqrt{-4} = -4$.

$\displaystyle \left(-\sqrt{-4}\right)^2 = \left(-\sqrt{-4}\right)\left(-\sqrt{-4}\right) = \sqrt{-4}$.

Therefore $\displaystyle \left(\pm \sqrt{-4}\right)^2 = -4$.


text:
# Definition of a limit

1. Apr 20, 2015

### Yoni V

1. The problem statement, all variables and given/known data
It is not exactly a homework question, but why does the definition of a limit use strict inequalities as follows:
if 0 < |x - a| < δ, then |f(x) - l| < ε
rather than weak inequalities, for example
if 0 < |x - a| < δ, then |f(x) - l| ≤ ε

Could the addition of the equality option make a difference?

2. Relevant equations

3. The attempt at a solution
I tried thinking of functions that would yield different limits to the limit produced by the formal definition, but couldn't find any.
I also tried to rule it out somehow with formal deduction, but couldn't.
Any hints or ideas?

Thanks

2. Apr 20, 2015

### Yoni V

Ok, I realized that the fact I couldn't disprove it is because it indeed holds.
It might not be as nice to be the definition, but given the definition,
if 0 < |x - a| < δ, then |f(x) - l| ≤ ε
implies that the limit of f is l.

3. Apr 20, 2015

### nuuskur

We proved early in our analysis course as a "recreational" activity that the two are, in fact, equivalent statements, but we just agreed to use the strict inequality.

4. Apr 20, 2015

### Fredrik

Staff Emeritus
It's obvious that if $\lim_{x\to a}f(x)=l$ in the sense of the standard definition, then $\lim_{x\to a}f(x)=l$ in the sense of the alternative definition.

Suppose that $\lim_{x\to a}f(x)=l$ in the sense of the alternative definition. Let $\varepsilon>0$. Let $\delta>0$ be such that the following implication holds for all $x\in\mathbb R$,
$$0<|x-a|<\delta~\Rightarrow~|f(x)-l| <\frac\varepsilon 2.$$ (Our assumption ensures that such a $\delta$ exists). For all $x\in\mathbb R$ such that $0<|x-a|<\delta$, we have $|f(x)-l|<\frac\varepsilon 2<\varepsilon$. This implies that $\lim_{x\to a}f(x)=l$ in the sense of the standard definition.

5. Apr 23, 2015

### Yoni V

Thanks for your responses.
Could you refer me to the analysis course you mentioned?

Edit: Oh, I guess you meant a course you took elsewhere, not some section here in PF. Nvm...


text:
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abstract: |
This work concerns the problem of reconstructing a collection of manifolds $\{M_1, \dots, M_n \}$, by using the data $d(f_i,f_j)$ of their distance functions from each other. The class of examples we consider is that of arithmetic manifolds ${\mathbb{C}}^n/L$ with Hermitian lattices $L$. We prove that if $n \leq 3$ and $L$ is of sufficiently large volume then for any finite collection of probability densities $\{f_i\}$, with $f_i \in L^2(M_i)$, the data $\{d(f_i,f_j)\}$ uniquely determines the
6a5afdab4c

## SuperEasy Backup Pro Crack +

Keep in mind that you can get apps at free of cost, but we only put some as the download link in the download section. We never claim that these apps are entirely free. We only link them to the original sites. Some apps require paid subscription to use or get in-app purchases. It all depends on the developers of each app.Q:

redirect after successful user registration using jsp

RegisterUser

On successful registration, I want to redirect the user to DisplayRegistrationPage.jsp
But its giving a null pointer exception on response.redirect..
I think I am doing something wrong.

A:

Well you just need to remove error variable and simply put
response.sendRedirect(“DisplayRegistrationPage.jsp”);

That’s all, nothing else.
What you’re doing there is calling a method, and no method definition is given.
And you could even try without the error var, since it’s the response’s job to redirect a user.
String error = request.getParameter(“errormsg”);
response.sendRedirect(“DisplayRegistrationPage.jsp”);
// ^^

A:

You do not need to pass request parameter with “errormsg”.
response.sendRedirect(“DisplayRegistrationPage.jsp”);

## What’s New In SuperEasy Backup Pro?

• Create a local or cloud-based backup
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What’s new in version 2017 build 100:
• Added “Folders only” backup mode
• Improved compatibility with Firefox.
• Included setup instructions in the readme file.Q:

How to insert a function in the middle of a bash script using source?

I have a bash script which reads the user input and executes the appropriate commands.
I need to make the script throw an error if the user enters a command that the script doesn’t support.
Is there a way that I can insert a function before the script is executed and get the result of the function so that I can examine it?
For example,
If the user enters the command :

The script should throw an error because ls is not valid command.
So, I want to insert a function before the script is executed.
If user enters the command :
help|

the script should throw an error
How to do this?

A:

One of those options might work for you (though it doesn’t store the results so you could run into problems down the road with out-of-sync files).
It does assume that the script knows it’s being run interactively (i.e. it isn’t wrapped in a function). If it is, it will cause an error.
Here we create a function call_ls_with_help that runs the ls command and tells you what it does:
f() {
# Run the command and capture its stdout:
command -v ls > /dev/null 2>&1
if [[ $?!= 0 ]]; then echo ls is not a valid command. else # Run it and print its exit code: echo “Exits with code$?”
echo “stdout:”
cat >/dev/null
echo “stderr:”
cat >/dev/null
fi
}

f

Here is how you might call it:
set -u
set -o pipe

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RAM: 1 GB
Processor: Intel® Core™ 2 Duo 1.5 GHz or better
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# Using the Tensorboard Callback¶

In this note we will cover the use of the TensorBoard callback. This is one of three callbacks in torchbearer which use the TensorboardX library. The PyPi version of tensorboardX (1.4) is somewhat outdated at the time of writing so it may be worth installing from source if some of the examples don’t run correctly:

pip install git+https://github.com/lanpa/tensorboardX

The TensorBoard callback is simply used to log metric values (and optionally a model graph) to tensorboard. Let’s have a look at some examples.

## Setup¶

We’ll begin with the data and simple model from our quickstart example.

BATCH_SIZE = 128

normalize = transforms.Normalize(mean=[0.485, 0.456, 0.406],
std=[0.229, 0.224, 0.225])

transform=transforms.Compose([transforms.ToTensor(), normalize]))
splitter = DatasetValidationSplitter(len(dataset), 0.1)
trainset = splitter.get_train_dataset(dataset)
valset = splitter.get_val_dataset(dataset)

traingen = torch.utils.data.DataLoader(trainset, pin_memory=True, batch_size=BATCH_SIZE, shuffle=True, num_workers=10)
valgen = torch.utils.data.DataLoader(valset, pin_memory=True, batch_size=BATCH_SIZE, shuffle=True, num_workers=10)

transform=transforms.Compose([transforms.ToTensor(), normalize]))
testgen = torch.utils.data.DataLoader(testset, pin_memory=True, batch_size=BATCH_SIZE, shuffle=False, num_workers=10)

class SimpleModel(nn.Module):
def __init__(self):
super(SimpleModel, self).__init__()
self.convs = nn.Sequential(
nn.Conv2d(3, 16, stride=2, kernel_size=3),
nn.BatchNorm2d(16),
nn.ReLU(),
nn.Conv2d(16, 32, stride=2, kernel_size=3),
nn.BatchNorm2d(32),
nn.ReLU(),
nn.Conv2d(32, 64, stride=2, kernel_size=3),
nn.BatchNorm2d(64),
nn.ReLU()
)

self.classifier = nn.Linear(576, 10)

def forward(self, x):
x = self.convs(x)
x = x.view(-1, 576)
return self.classifier(x)

model = SimpleModel()

loss = nn.CrossEntropyLoss()

The callback has three capabilities that we will demonstrate in this guide:

1. It can log a graph of the model
2. It can log the batch metrics
3. It can log the epoch metrics

## Logging the Model Graph¶

One of the advantages of PyTorch is that it doesn’t construct a model graph internally like other frameworks such as TensorFlow. This means that determining the model structure requires a forward pass through the model with some dummy data and parsing the subsequent graph built by autograd. Thankfully, TensorboardX can do this for us. The TensorBoard callback makes things a little easier by creating the dummy data for us and handling the interaction with TensorboardX. The size of the dummy data is chosen to match the size of the data in the dataset / data loader, this means that we need at least one batch of training data for the graph to be written. Let’s train for one epoch just to see a model graph:

from torchbearer import Trial
from torchbearer.callbacks import TensorBoard

torchbearer_trial = Trial(model, optimizer, loss, metrics=['acc', 'loss'], callbacks=[TensorBoard(write_graph=True, write_batch_metrics=False, write_epoch_metrics=False)]).to('cuda')
torchbearer_trial.with_generators(train_generator=traingen, val_generator=valgen)
torchbearer_trial.run(epochs=1)

To see the result, navigate to the project directory and execute the command tensorboard --logdir logs, then open a web browser and navigate to localhost:6006. After a bit of clicking around you should be able to see and download something like the following:

The dynamic graph construction does introduce some weirdness, however, this is about as good as model graphs typically get.

## Logging Batch Metrics¶

If we have some metrics that output every batch, we might want to log them to tensorboard. This is useful particularly if epochs are long and we want to watch them progress. For this we can set write_batch_metrics=True in the TensorBoard callback constructor. Setting this flag will cause the batch metrics to be written as graphs to tensorboard. We are also able to change the frequency of updates by choosing the batch_step_size. This is the number of batches to wait between updates and can help with reducing the size of the logs, 10 seems reasonable. We run this for 10 epochs with the following:

torchbearer_trial = Trial(model, optimizer, loss, metrics=['acc', 'loss'], callbacks=[TensorBoard(write_graph=False, write_batch_metrics=True, batch_step_size=10, write_epoch_metrics=False)]).to('cuda')
torchbearer_trial.with_generators(train_generator=traingen, val_generator=valgen)
torchbearer_trial.run(epochs=10)

Runnng tensorboard again with tensorboard --logdir logs, navigating to localhost:6006 and selecting ‘WALL’ for the horizontal axis we can see the following:

## Logging Epoch Metrics¶

Logging epoch metrics is perhaps the most typical use case of TensorBoard and the TensorBoard callback. Using the same model as before, but setting write_epoch_metrics=True we can log epoch metrics with the following:

torchbearer_trial = Trial(model, optimizer, loss, metrics=['acc', 'loss'],  callbacks=[TensorBoard(write_graph=False, write_batch_metrics=False, write_epoch_metrics=True)]).to('cuda')
torchbearer_trial.with_generators(train_generator=traingen, val_generator=valgen)
torchbearer_trial.run(epochs=10)

Again, runnng tensorboard with tensorboard --logdir logs and navigating to localhost:6006 we see the following:

Note that we also get the batch metrics here. In fact this is the terminal value of the batch metric, which means that by default it is an average over the last 50 batches. This can be useful when looking at over-fitting as it gives a more accurate depiction of the model performance on the training data (the other training metrics are an average over the whole epoch despite model performance changing throughout).

## Source Code¶

The source code for these examples is given below:


text:
Home > Research > Publications & Outputs > On the nature of H$α$ emitters at $z \sim 2$ fr...

### Electronic data

• 1506.02670v1

Rights statement: This is a pre-copy-editing, author-produced PDF of an article accepted for publication in Monthly Notices of the Royal Astronomical Society following peer review. The definitive publisher-authenticated version I. Oteo, D. Sobral, R. J. Ivison, I. Smail, P. N. Best, J. Cepa, and A. M. Pérez-García On the nature of Hα emitters at z~ 2 from the HiZELS survey: physical properties, Lyα escape fraction and main sequence MNRAS 2015 452: 2018-2033. is available online at: http://mnras.oxfordjournals.org/content/452/2/2018

Accepted author manuscript, 2.51 MB, PDF document

• 1506.02670v1

Accepted author manuscript, 2.51 MB, PDF document

## On the nature of H$α$ emitters at $z \sim 2$ from the HiZELS survey: physical properties, Ly$α$ escape fraction, and main sequence

Research output: Contribution to journalJournal articlepeer-review

Published
• I. Oteo
• D. Sobral
• R. J. Ivison
• I. Smail
• P. N. Best
• J. Cepa
• A. M. Perez-Garcia
Close
Journal publication date 11/09/2015 Monthly Notices of the Royal Astronomical Society 2 452 16 2018-2033 Published 21/07/15 English

### Abstract

We present a detailed multi-wavelength study (from rest-frame UV to far-IR) of narrow-band (NB) selected, star-forming (SF) H$\alpha$ emitters (HAEs) at $z \sim 2.23$ taken from the High Redshift(Z) Emission Line Survey (HiZELS). We find that HAEs have similar SED-derived properties and colors to $sBzK$ galaxies and probe a well-defined portion of the SF population at $z \sim 2$. This is not true for Ly$\alpha$ emitters (LAEs), which are strongly biased towards blue, less massive galaxies (missing a significant percentage of the SF population). Combining our H$\alpha$ observations with matched, existing Ly$\alpha$ data we determine that the Ly$\alpha$ escape fraction ($f_{\rm esc}$) is low (only $\sim$ 4.5\% of HAEs show Ly$\alpha$ emission) and decreases with increasing dust attenuation, UV continuum slope, stellar mass, and star formation rate (SFR). This suggests that Ly$\alpha$ preferentially escapes from blue galaxies with low dust attenuation. However, a small population of red and massive LAEs is also present in agreement with previous works. This indicates that dust and Ly$\alpha$ are not mutually exclusive. Using different and completely independent measures of the total SFR we show that the H$\alpha$ emission is an excellent tracer of star formation at $z \sim 2$ with deviations typically lower than 0.3 dex for individual galaxies. We find that the slope and zero-point of the HAE main-sequence (MS) at $z \sim 2$ strongly depend on the dust correction method used to recover SFR, although they are consistent with previous works when similar assumptions are made.

### Bibliographic note

This is a pre-copy-editing, author-produced PDF of an article accepted for publication in Monthly Notices of the Royal Astronomical Society following peer review. The definitive publisher-authenticated version I. Oteo, D. Sobral, R. J. Ivison, I. Smail, P. N. Best, J. Cepa, and A. M. Pérez-García On the nature of Hα emitters at z~ 2 from the HiZELS survey: physical properties, Lyα escape fraction and main sequence MNRAS 2015 452: 2018-2033. is available online at: http://mnras.oxfordjournals.org/content/452/2/2018


text:
MathSciNet bibliographic data MR1765981 (2001i:11032) 11D45 (11B68 11D41) Brindza, B.; Pintér, Á. On the number of solutions of the equation \$1\sp k+2\sp k+\cdots+(x-1)\sp k=y\sp z\$$1\sp k+2\sp k+\cdots+(x-1)\sp k=y\sp z$. Dedicated to Professor Kálmán Győry on the occasion of his 60th birthday. Publ. Math. Debrecen 56 (2000), no. 3-4, 271–277. Journal

For users without a MathSciNet license , Relay Station allows linking from MR numbers in online mathematical literature directly to electronic journals and original articles. Subscribers receive the added value of full MathSciNet reviews.


text:
# Submitted Papers

1. Complete classification of global solutions to the obstacle problem (with S. Eberle and G.S. Weiss)
2. A quantitative stability result for the Prékopa-Leindler inequality for arbitrary measurable functions (with K. J. Böröczky and J. P. G. Ramos)
3. The singular set in the Stefan problem (with X. Ros-Oton and J. Serra)


text:
# What is the intuitive meaning of autocorrelation - when is $r_xx[0] = 1$

Consider a linear model $y_n = \beta x_n + v_n$ where $n$ represents the time index, $x$ is the input data source $\hat{\beta}_{OLS} = (X^TX)^{-1}X^Ty$ is the ordinary least square estimate of the parameter $\beta$. This parameter is obtained assuming the distribution of the observation $y$ is $i.i.d$ Gaussian.

Consider, the covariance of the estimate $\hat{\beta} = \sigma^2{(X^TX)}^{-1}$

The answers to the following questions are extremely hard for me to obtain from books, therefore I have posted the question. Shall appreciate help. Also, please correct me if any information is wrong.

1) What is the meaning of autocorrelation and how it will change in the following scenario -- If the measurement noise $v$ is White Gaussian then it means that the noise samples are $i.i.d$. If the measurement noise is colored Gaussian, then the noise samples are not $i.id$ but correlated which means that the property of independence and uncorrelated samples cannot be applied. When the noise is white Gaussian then expectation, $E[vv^T] = \sigma^2$ I think. What does the autocorrelation function look like in white and colored Gaussian case and what would expectation of colored noise samples give?

2) UPDATE: Consider, a new data source denoted by $x_{new}$ which has some different properties than $x$.

If the covariance $\sigma^2{(X^TX)}^{-1}$ is higher than another estimator of $\hat{\beta}$ given as $\hat{\beta_{new}} = \sigma^2{(X_{new}^TX_{new})}^{-1}$ obtained using the new data input $x_{new}$, then what does having a lower variance say about the nature of the data $x_{new}$? Can I say that $x_{new}$ input transfers more energy, has higher power than the data $x$?

3) Does $X^TX$ term have a name?

• Saying that $y$ is i.i.d and then wondering about autocorrelation, doesn't make sense - they are mutually exclusive. I am further confused because the model you propose is not indexed over time, so where would the autocorrelation come from? Finally, the class of linear of estimators is very very broad, and again, for the proposed model there is no point in switching to a non linear estimator. Maybe you can give us some to context? What are you trying to do? What lead you to wonder about these questions? – Repmat Dec 16 '17 at 11:32
• @Repmat: I have updated the question, thank you for your suggestions. – Ria George Dec 16 '17 at 17:23

Disclaimer: I'm not entirely clear on everything myself, this is my best-effort, open for improvements.

1. If $$y_{n + 1} = \beta\cdot y_n + \epsilon_n$$ and $$\epsilon_n \sim \text{norm}(0, 1)$$, then autocorrelation can be simplified by linearity and because the errors are independent, $$\text{cov}(y_n, y_{n - 1}) = \beta\cdot\text{cov}(y_{n-1}, y_{n-1}) = \beta\cdot\text{var}(y_{n-1})$$. If $\beta$ is zero then this is "white noise" and the autocorrelation is zero. For larger gaps the same simplifications can be used to show $\text{cov}(y_n, y_{n-k}) = \beta^k\cdot\text{var}(y_k)$.

2. OLS is the maximum likelihood estimate of $\beta$ if the error is Gaussian. But if noise is non-Gaussian, then this may not work that well. Let's say the noise is skewed "upwards", so errors are a bit bigger upwards, let's say lognormal, then the maximum likelihood estimate for the mean is different. So if you use it, it's not optimal.

3. If you are estimating the mean of a lognormal distribution, see the link above for the nonlinear estimator of that.

EDIT, some topics below

• I think the formula in the original question changed, now it's $y_n = \beta\cdot x_n + \epsilon_n$. Unless there is information on how $x_n$ or $y_n$ depend on variables at time $n-1$, there is not much to say about autocorrelation.
• Autocorrelation is correlation with previous values, which is covariance with previous values divided by standard deviations. To simplify, I've only shown a derivation with covariance.
• In the comments below, I think you mean variance of $\beta$, not it's covariance (otherwise, covariance with what?).
• The assumptions of the question regarding $X_{\text{new}}$ are unclear to me, and I suspect it's a question I don't necessarily know the answer to, or maybe there is not a single answer. Basically, if you assume Gaussian error, then the OLS estimate is the maximum likelihood estimate and also the unbiased estimator of $\beta$ with the lowest possible variance. If you change the distribution of the noise, then the bets are off. Maybe you can start with assuming different noise model and the see what happens, before trying to get the general answer. To be precise, the OLS estimate is the best unbiased estimator under broader assumptions, see https://en.wikipedia.org/wiki/Gauss%E2%80%93Markov_theorem.
• Thank you for your answer. You are correct by indexing the variables with respect to time. My autocorrelation for the noise in the model in the question is $v_n v_n^T$. I don't follow why you have covariance when explaining autocorrelaton. For Point 2) assume I am estimating a parameter using the same dynamical model by assuming different distribution of the noise. I want to select the best estimate of the parameter. – Ria George Dec 16 '17 at 15:27
• The variance of the estimate should be minimum in order to make this choice. Can variance of the estimate be less when the distribution is non-Gaussian and what happens to autocorrelation for non-Gaussian case? – Ria George Dec 16 '17 at 15:27