text:
## 8 Factors that Affect the Workability of Fresh Concrete

Lubricating nature of workable concrete is the most important nature. Workability is dependent on the proportion of the constituent material as well as on their individual characteristics.

The Factors that Affect the Workability of Concrete are Given Below:

### 01. Water Content of the Concrete Mix:

Water content will have important influences on the workability in given volume of concrete. The higher the water content per cubic meter of concrete, the higher will be the fluidity of concrete, which affect the workability.

Water requirement is mainly associated with absorption by aggregates surface& filling up the voids between aggregates.

However, Adding Extra Water to the Concrete Mix can be Disadvantageous, as Given Below:

• The strength of the concrete may get reduced.
• More quantity of water comes out from the surface of concrete resulting into bleeding.
• Cement slurry also escapes through the joints of formwork resulting into the loss of cement from concrete.

### 02. The Size of Aggregates:

Workability is mainly governed by the maximum size of aggregates.Water and paste require, will be not less if a chosen size of aggregates for concrete is bigger. Consequently, for a given quantity of water content & paste, bigger size aggregate will give higher workability.

Note: On the site, the maximum size of aggregate to be used will depend upon the many factors such as the handling, mixing and placing equipment, the thickness of section and quantity of reinforcement. Later two are very important.

##### Also Read: Why Aggregates have an Important Role in Concrete
Courtesy - 123rf

### 03. The shape of Aggregates:

The Shape of Aggregates Seriously Influences the Workability of Concrete:

• Angular, flaky & elongated aggregate reduces the workability of concrete.
• Rounded or subrounded aggregates increase the workability due to the reduction of surface area for a given volume or weight.Therefore, an excess paste is available to give better lubricating effect.
• Rounded shape aggregate has less frictional resistance and gives a high workability as compared to angular, flaky or elongated aggregates.

Note: River sand & gravel provide greater workability to concrete than crushed sand.

### 04. Surface Texture of Aggregates:

Courtesy - 123rf

The roughly textured aggregates have more surface area than Smoothly rounded aggregates of the same volume. Smooth rounded or glassy aggregates will give better workability than roughly textured aggregates. A reduction of interparticle frictional resistance offered by smooth aggregates also contributes to higher workability.

### 05. The Porosity of Aggregates:

Porous and non-saturated aggregate will require more water than non-absorbent aggregates. For the same degree of workability, latter will require less water. Overall, this factor is only of secondary importance.

### 06. Grading of Aggregates:

Grading of aggregates has the greatest influence on workability. The better the grading of aggregates, the less is the amount of void in concrete so well-graded aggregates should be used. When total voids are less in concrete, the excess paste is available to give better lubricating effect.

With excess amount of concrete paste present in the mixture, it becomes cohesive&fatty that prevents segregation of particles & least amount of compacting efforts is required to compact the concrete.

For a given workability, there is one value of coarse aggregate / Fine aggregate ratio, which needs the lower water content.

### 07. Uses of Concrete Admixtures:

This is one of the commonly used methods to enhance the workability of concrete. Concrete admixtures such as plasticizer and superplasticizers greatly improve the workability.

Air entraining agents are also used to increase the workability. Air entraining agents creates a large number of very tiny air bubbles. These bubbles get distributed throughout the mass of concrete and act as rollers and increase the workability.

Mineral admixtures like Pozzolanic materials are also used to improve the workability of concrete.

### 08. Ambient Temperature:

In hot weather, if temperature increases, the evaporation rate of mixing water also increases and hence fluid viscosity increases, too. This phenomenon affects the flowability of concrete and due to fast hydration of concrete, it will gain strength earlier which decreases the workability of fresh concrete.

Also Read:

## Material Exhibition

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Exhibit your Brands/Products.

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text:
# Linear transformations

This applet allows you to experiment with 2x2-matrices and linear transformations of the plane. You can move the vector x (the blue arrow) and see how the vector y = Mx (the black arrow) moves. The red lattice illustrates how the entire plane is effected by the transformation represented by M. You can redefine the matrix .
Try out different matrices. Try for example: a=d=1 and b=c=0, a=b=c=d=0, a=2, b=c=0 and d=3, a=0, b=1, c=-1 and d=0, a=b=c=d=1/2. For each matrix consider the following: What happens to x = (1,0) and x = (0,1)? Notice the connection between these and the matrix M. Can you make a matrix that reflects all vectors through the x-axis? The y-axis? rotates every vector through an angle of 45 degrees? Play on!

format: plain


text:
# [OS X TeX] creating a "smart macro"

Alex Hamann mailinglist at alex-hamann.de
Thu Apr 23 17:57:30 CEST 2009

Am 22.04.2009 um 15:08 schrieb Alan Munn:

> At 10:58 AM +0200 4/22/09, Alex Hamann wrote:
>> Hi,
>>
>> I need to create a number of name tags for a class. The idea is to
>> create a list of the students' names and have a macro that will
>> fetch the first name, then reexecute using the second name and so on.
>> This is the layout I created. Can anybody give me a hint how to
>> make a smart macro out of this that will work its way through a
>> list of 109 students and create a pdf in which every single page
>> looks like the output of the code below while replacing \1stname
>> and \lastname with the names in the list?
>>
>> TIA,
>> Alex
>>
>
> Use the datatool package.  Store your names in an Excel or
> OpenOffice csv file and then read the data in.
>
> and is called students.csv the following will do the trick. (You
> might want to make the first name a bit bigger, though...)  -Alan
>
> \documentclass[landscape,a5paper,pagesize,DiV=14]{scrartcl}
> \usepackage
> {geometry}
> \usepackage{xltxtra}
> \setmainfont[Mapping=tex-text]{Hoefler Text}
> \pagestyle{empty}
> \usepackage{fix-cm}
> \usepackage{datatool}
> \begin{document}
> \DTLforeach{students}{%
> \firstname=FirstName,\lastname=LastName}{%
> \vspace*{7,5cm}
> \center
> \huge{\firstname}\\
> \fontsize{60}{70}\selectfont
>  \textbf{\lastname}
> \vspace*{-1cm}
> \pagebreak
> }
> \end{document}
> --
> Alan Munn

Alan,
thank you very much. Just tested and it works like a charm.
Cheers,
Alex


text:
# multiple decoupling capacitor circuits (integrating multiple circuits)

I'm integrating two circuits into one. One being an Arduino Pro Mini (ATMega328P) and the other being a CAN controller+transceiver.

For the Aduino Pro Mini, I'm using a TLF80511TF50 (5V LDO). I've not yet calculated the values here, but they don't matter for the question, I think. The Pro Mini uses two tantalum 10uF caps here (and a third ceramic 0.1uF behind a jumper), but that's for a MIC52505 regulator.

The CAN board:

Of course, I can just copy the two boards onto a single PCB, but I'm wondering, do I have to duplicate the decoupling circuits?

The TLF80511TFV50 will power both the MCU, and the two CAN chips. It will also power three I2C I/O extenders (PCF8575), but these barely require any power.

The CAN board uses a 8MHz crystal, the Arduino an 16MHz crystal.. In case that matters (I'm thinking noise...)

Q1: Is it advisable to use a set of decoupling capacitors near the VCC input of each chip/pcb section? Or what are the factors to consider here?

Q2: If this is not required, should the number of logic chips on the board influence the values/number of the decoupling capacitors?

.

UPDATE

Based on reply by @peufeu which I feel makes sense, I removed the basically double decoupling and left with decoupling at the regulator, following recommended setup from the datasheet. See schematic:

What I am left unanswered with is how to deal with decoupling at the various logic chips. I've added a cap at the VDD rail to the three expanders, as peufeu pointed out will have a switching behaviour. I remain unsure if I should add caps to the Arduino chip, and CAN chip(s). -- It will all probably work well without them, on the other hand adding caps is easy and cheap.. What is the rule of the tumb here??

If this is a personal project,

• Put 100nF (or better 1µF) SMD ceramic cap (X7R 16V) on each VCC pin.

The caps are worth cents each, if you build a million boards yes you can spend the time to accurately determine if each and every cap is actually necessary, but for a DIY project, invest 50 cents in caps and don't spend hours thinking about it...

Using the same philosophy, you can pay $10 for a 2-layer board, or$35 for a 4-layer, if this saves you 4 hours of layout time, then it's worth it...

The TLF80511TFV50 will power both the MCU, and the two CAN chips. It will also power three I2C I/O extenders (PCF8575), but these barely require any power.

IO expanders will draw current pulses when switching their outputs, so they need caps too.

Now check LDO datasheet for caps, it says "minimum capacitance 1µF, maximum ESR 5 ohms".

So, there is no maximum capacitance, and there is no minimum ESR.

So if you put a 1µF MLCC on each chip, you're good to go.

Add another cap at the LDO input. Don't bother with values, just have a roll of 1µF MLCCs... the roll of 100 will be cheaper than 10x100nF, 10x10nF, etc...

Now if you're feeling extra paranoid, you can add aluminium electrolytic capacitor at the regulator input, and even at the output... Probably won't be necessary, according to the LDO datasheet. If you have the room for it on your board, a capacitor footprint costs you nothing if there's no cap in it.

Don't bother with tantalums unless you have strict height requirements.

--

EDIT

The job of the cap at the chip's pin is both to supply current quickly when needed (which the inductance of VCC traces hampers) and to keep the current loop antenna small (prevents emission of noise from the chip's supply currents).

For a 0805 cap, inductance would be a little bit below 1nH. Now, you are using mostly low-frequency chips (by modern standards) and also keep in mind that the inductance of the bondwires which connect the chip to the pins will be >5nH, so you don't need to put several caps in parallel on each pin with staggered values (like 10nF 100nF in parallel) to shave off the caps'inductance down to 0.5nH... that wouldn't matter at all.

Also (unless you use fancy multilayer) vias & traces will have more inductance than the cap. So it's all about placing the cap close to the pin and making sure the GND return loop to the same chip's GND pin is tight.

On 2 layer, the irony is that what you're doing is no different than, say, laying out the board for a Commodore 64... Notice 1 decoupling cap per chip, and all the traces on toplayer going horizontal, all the traces on the other side would go vertical. Thus you can make a sort of a grid for power and GND. Check this link.

If you use 4 layer with power/gnd planes then the supply will be very low impedance, so placement of caps is less critical.

• Thanks. I'm not sure I have more or less questions now tho ;-). First off, yes, it is personal, but I'll be making more for my friends plus it is somewhat mission critical, so I do want to make it as good as I can. That said, I am all for not going overboard on details. I'm a few months into EE, so a lot of this is new and I'm still learning, so I want to also understand the choices I'm making. --- Also, good point in adding the footprints now, and have options later. --- Back to my original question, I'm not sure your're saying to add caps at all the chips, or just at the regulator.. – svenema Aug 28 '17 at 8:29
• 1 cap per IC pin, and placement/routing is more important than value, see edit – peufeu Aug 28 '17 at 11:49


text:
# zbMATH — the first resource for mathematics

##### Examples
 Geometry Search for the term Geometry in any field. Queries are case-independent. Funct* Wildcard queries are specified by * (e.g. functions, functorial, etc.). Otherwise the search is exact. "Topological group" Phrases (multi-words) should be set in "straight quotation marks". au: Bourbaki & ti: Algebra Search for author and title. The and-operator & is default and can be omitted. Chebyshev | Tschebyscheff The or-operator | allows to search for Chebyshev or Tschebyscheff. "Quasi* map*" py: 1989 The resulting documents have publication year 1989. so: Eur* J* Mat* Soc* cc: 14 Search for publications in a particular source with a Mathematics Subject Classification code (cc) in 14. "Partial diff* eq*" ! elliptic The not-operator ! eliminates all results containing the word elliptic. dt: b & au: Hilbert The document type is set to books; alternatively: j for journal articles, a for book articles. py: 2000-2015 cc: (94A | 11T) Number ranges are accepted. Terms can be grouped within (parentheses). la: chinese Find documents in a given language. ISO 639-1 language codes can also be used.

##### Operators
 a & b logic and a | b logic or !ab logic not abc* right wildcard "ab c" phrase (ab c) parentheses
##### Fields
 any anywhere an internal document identifier au author, editor ai internal author identifier ti title la language so source ab review, abstract py publication year rv reviewer cc MSC code ut uncontrolled term dt document type (j: journal article; b: book; a: book article)
Geometric stability switch criteria in delay differential systems with delay dependent parameters. (English) Zbl 1013.92034

Summary: In most applications of delay differential equations in population dynamics, the need of incorporation of time delays is often the result of the existence of some stage structure. Since the through-stage survival rate is often a function of time delays, it is easy to conceive that these models may involve some delay dependent parameters. The presence of such parameters often greatly complicates the task of an analytical study of such models. The main objective of this paper is to provide practical guidelines that combine graphical information with analytical work to effectively study the local stability of some models involving delay dependent parameters.

Specifically, we show that the stability of a given steady state is simply determined by the graphs of some functions of $\tau$ which can be expressed explicitly and thus can be easily depicted by Maple and other popular software. In fact, for most application problems, we need only look at one such function and locate its zeros. This function often has only two zeros, providing thresholds for stability switches.

The common scenario is that as time delay increases, stability changes from stable to unstable to stable, implying that a large delay can be stabilizing. This scenario often contradicts the one provided by similar models with only delay independent parameters.

##### MSC:
 92D25 Population dynamics (general) 34K20 Stability theory of functional-differential equations 34K18 Bifurcation theory of functional differential equations
Maple


text:
# How do you find the inverse of A=((0, 7), (0, -3))?

$\left\mid \begin{matrix}0 & 7 \\ 0 & - 3\end{matrix} \right\mid = \left(0 \cdot - 3\right) - \left(7 \cdot 0\right) = 0$ so $A$ has no inverse.


text:
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# Factoring out a common term?

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Factoring out a common term?  [#permalink]

### Show Tags

23 Oct 2017, 22:52
11³+11⁴-->11³(11⁰+11¹)-->11³(1+11)-->11³(12)

Sorry guys, but I don't really understand that concept. Do you know more about it?

thank you
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Factoring out a common term?  [#permalink]

### Show Tags

24 Oct 2017, 05:43
Hi thegame12 ,

Welcome to GMATClub

$$11^3 + 11^4 = 1*11^3 + 11*11^3$$(One multiplied with any number gives the same number)

This can be further simplified by taking the common term outside.
We need to follow the distributive law a·b+a·c = a·(b+c)

In this case the common term is 11^3 which if taken outside gives
$$1*11^3 + 11*11^3 = 11^3(1+11) = 11^3(12)$$

Hope that helps!
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Re: Factoring out a common term?  [#permalink]

### Show Tags

24 Oct 2017, 06:56
thegame12 wrote:
11³+11⁴-->11³(11⁰+11¹)-->11³(1+11)-->11³(12)

Sorry guys, but I don't really understand that concept. Do you know more about it?

thank you
pushpitkc has explained it, but are there any specific things here that you're not comfortable with? For example, are you okay "seeing" 11⁰ as 1 (any non-zero number raised to the power of 0 is 1)?
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Re: Factoring out a common term? &nbs [#permalink] 24 Oct 2017, 06:56
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# Factoring out a common term?

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text:
# Math Help - [SOLVED] Binomial Probability

1. ## [SOLVED] Binomial Probability

Hi,

I am having trouble with the following question:

During winter it rains on average 18 out of 30 days. Five winter days are selected at random. Find (4 d.p) the probability that:

(a) the first two days chosen will be fine and the remainder wet
My approach is this is an ordered selection:

I have defined p = it will rain = $\frac{18}{30}$ or 0.6
q = 1 - 0.6 = 0.4
= $0.4^2 \times 0.6^3$
= 0.03456

The answers I have are yielding 0.0124.

(b) more rainy days than fine days have been chosen
P(3 rainy days) = $_5C_3 \times 0.6^3 \times 0.4^2$
P(4 rainy days) = $_5C_4 \times 0.6^4 \times 0.4$
P(5 rainy days) = $_5C_5 \times 0.6^5$
= 0.3456 + 0.2592 + 0.007776
= 0.612576

I wonder where I am going wrong
TIA for any assistance!

2. Originally Posted by lindah
Hi,

I am having trouble with the following question:

During winter it rains on average 18 out of 30 days. Five winter days are selected at random. Find (4 d.p) the probability that:

(a) the first two days chosen will be fine and the remainder wet
My approach is this is an ordered selection:

I have defined p = it will rain = $\frac{18}{30}$ or 0.6
q = 1 - 0.6 = 0.4
= $0.4^2 \times 0.6^3$
= 0.03456 .... Looks right to me

The answers I have are yielding 0.0124.

(b) more rainy days than fine days have been chosen
P(3 rainy days) = $_5C_3 \times 0.6^3 \times 0.4^2$
P(4 rainy days) = $_5C_4 \times 0.6^4 \times 0.4$
P(5 rainy days) = $_5C_5 \times 0.6^5$
= 0.3456 + 0.2592 + 0.007776 ..... 0.07776
= 0.612576 ...... 0.68256


text:
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />

# 8.6: Percents, Decimals and Fractions

Difficulty Level: At Grade Created by: CK-12

## Introduction

Sweeping and Mopping

Sam’s least favorite part of her afterschool job at the supermarket is closing. When closing the store, the break room has to be swept and mopped. Each of the students who work at the store part-time take turns closing, and every Friday night is Sam’s turn.

On Friday, Sam got her mop and broom and headed up to the break room. It seemed to be even messier than usual.

“Oh no, I will never get done,” Sam sighed, but she picked up the broom and began to sweep.

In just fifteen minutes, Sam had swept four-fifths of the room. She was amazed at how quickly the task was getting done with a little focus and effort.

What fraction of the room does Sam still have to sweep?

What percent of the room has she finished? What percent of the room is still left?

This lesson will teach you all about converting fractions and percents. By the end of the lesson, you will know how to answer these questions.

What You Will Learn

By the end of this lesson, you will be able to demonstrate the following skills.

• Write fractions as percents.
• Write decimals as percents.
• Compare and order fractions, decimals and percents.
• Write fractional or decimal real-world data as percents.

Teaching Time

In our last lesson, you learned how to understand what a percent is. Remember that a percent means a part of a whole, and the whole is 100. You also learned how to take a percent and write it as a decimal and as a fraction. In this lesson, you will learn to work the other way around. Let’s begin by learning how to write fractions as a percent.

I. Write Fractions as a Percent

When writing a percent as a fraction, we can drop the % sign and make that the denominator of 100. Then we take the quantity and write it in the numerator above the 100. This is our fraction.

Example

Write 14% as a fraction.

First, we drop the % sign. % means out of 100, so 100 becomes our new denominator.

14% becomes 14100\begin{align*}\frac{14}{100}\end{align*}.

Yes! We can change the denominator of 100 to a % sign and add that to the quantity in the numerator. Here is what that looks like.

Example

Write 47100\begin{align*}\frac{47}{100}\end{align*} as a percent.

First, drop the 100. Then write 47 with a % sign.

47100\begin{align*}\frac{47}{100}\end{align*} becomes 47 %

Write each fraction as a percent.

1. 48100\begin{align*}\frac{48}{100}\end{align*}
2. 82100\begin{align*}\frac{82}{100}\end{align*}
3. 91100\begin{align*}\frac{91}{100}\end{align*}

Not all fractions have a denominator of 100. How do we write a fraction as a percent when the denominator is not 100?

This is the next thing that we need to learn. If a fraction does not have a denominator of 100, to write it as a percent we need to rewrite it as an equal fraction with a denominator of 100.

Here is an example.

Example

Write 25\begin{align*}\frac{2}{5}\end{align*} as a percent.

To write 25\begin{align*}\frac{2}{5}\end{align*} as a percent, we must first rewrite it as a fraction with a denominator of 100. Write a proportion to show this comparison.

25=100\begin{align*}\frac{2}{5}=\frac{}{100}\end{align*}

What number was multiplied by 5 to get 100 as a product?

20! So we multiply the numerator by 20 and we will have an equivalent fraction with a denominator of 100.

2×2025=40100=40=40%\begin{align*}2 \times 20 &= 40\\ \frac{2}{5}=\frac{40}{100}&=40\%\end{align*}

One special fraction to work with is one-third. To convert one-third to a percent is a little tricky because 3 does not divide evenly into 100. Take a look.

13=100\begin{align*}\frac{1}{3}=\frac{}{100}\end{align*}

When completing this problem, we end up with a repeating decimal; the 3’s just continue on and on and on.

.33333333 etc.

To work with this fraction, we call it 33 13\begin{align*}\frac{1}{3}\end{align*} %.

Write each fraction below as a percent. Remember you will have to convert each to a fraction with a denominator of 100 first.

1. 14\begin{align*}\frac{1}{4}\end{align*}
2. 15\begin{align*}\frac{1}{5}\end{align*}
3. 34\begin{align*}\frac{3}{4}\end{align*}

Check your work and take a few notes on how to convert a fraction to a percent before moving on.

II. Write Decimals as Percents

In our last lesson you learned to write percents as a decimal. Here is an example.

Example

Write 31% as a decimal.

You will recall that to do this, we drop the % sign and move the decimal point two places in to the left. These two places represent the hundredths place of the decimal.

31% becomes .31

We can also write a decimal as a percent. To do this, we are going to move the decimal point two places to the right and add a percent sign.

Example

Write .14 as a percent.

To do this, we move the decimal point two places to the right because two decimal places represent hundredths and percents are out of 100. Then we add in the % sign.

.14.=14%\begin{align*}\underrightarrow{.14.} = 14\%\end{align*}

Sometimes, you will have a decimal that is written with a zero for the tenths place. We do the same thing to convert to a percent. Move the decimal two places to the right and add a percent sign.

Example

Write .03 as a percent.

.03 becomes 3%

What about a decimal that does not have two decimal places represented?

Example

Write .2 as a percent.

To do this, we move the decimal point two places to the right, which will require adding a zero. Then we can see that two tenths becomes 20 percent.

.20=20%\begin{align*}\underrightarrow{.20} = 20\%\end{align*}

What about if you have a decimal with more than two places?

When this happens, it is an interesting case, because you have to move the decimal point two places to the right, but you will have a percent that is also a decimal.

Example

.345

.345 becomes 34.5%.

Now it is time for you to practice. Write each decimal as a percent.

1. .85
2. .09
3. .5

Take a few minutes to check your work with a partner. Take a few notes on how to write decimals as percents before moving on to the next section.

III. Compare and Order Fractions, Decimals and Percents

We have already established that fractions, decimals and percents are all related to one another. Because they are related and we can establish equivalents of each, we can also compare each using greater than, less than or equal to. We can also write them in order.

To compare fractions, decimals and percents, we should have them in the same form. If we are comparing a fraction and a percent, we have to write both of them either as fractions or percents so we can figure out which is greater.

Example

Compare 45% and 45\begin{align*}\frac{4}{5}\end{align*}

To compare these two quantities, first write them in the same form. Let’s change four-fifths to a percent. We do that by writing it as a fraction out of 100, which we can then change to a percent.

45=80100=80%\begin{align*}\frac{4}{5}=\frac{80}{100}=80\%\end{align*}

45% is less than 80%.

Our answer is 45%<45\begin{align*}45\% < \frac{4}{5}\end{align*}.

We can do the same thing when working with decimals and percents.

Example

Compare 18% and .9

To complete this, we have to convert both of these to either percents or decimals. Let’s change .9 to a percent. To do this, we move the decimal point two places to the right.

.9 = 90%

18% is less than 90%.

Our answer is 18% < .9.

Try a few of these on your own. Use greater than >, less than < or = .

1. .19 and 19%
2. 25\begin{align*}\frac{2}{5}\end{align*} and 45%
3. 56% and 21%

Check your work with a peer.

What about ordering fractions, decimals and percents? When we order a set of numbers or quantities, we write them from least to greatest or from greatest to least. Fractions, decimals and percents are no exception, but it is to order them if they are in the same form.

Example

Write .56, 34%, 910\begin{align*}\frac{9}{10}\end{align*} and 12\begin{align*}\frac{1}{2}\end{align*} in order from least to greatest

To do this, we need to write them all in the same form. Let’s convert all of them to percents.

.56=56%\begin{align*}.56 = 56\%\end{align*}

34% stays the same

91012=90100=90%=50100=50%\begin{align*}\frac{9}{10} &= \frac{90}{100}=90\%\\ \frac{1}{2} &= \frac{50}{100}=50\%\end{align*}

So we have 56%, 34%, 90% and 50%, now it becomes easy to write them in order.

34%, 50%, 56%, and 90%

Our answer is 34%, 12\begin{align*}\frac{1}{2}\end{align*}, .56, 910\begin{align*}\frac{9}{10}\end{align*}.

Take a few notes on ordering fractions, decimals and percents.

## Real Life Example Completed

Sweeping and Mopping

Sam is working away and you have been too. Here is the original problem once again. Reread the problem and then use what you have learned to figure out how much of the floor Sam has completed and how much she has left. Be sure to start by underlining the important information.

Sam’s least favorite part of her after school job at the supermarket is closing. When closing the store, the break room has to be swept and mopped. Each of the students who work at the store part-time take turns working on this and every Friday night is Sam’s turn.

On Friday, Sam got her mop and broom and headed up to the break room. It seemed to be even messier than usual.

“Oh no, I will never get done,” Sam sighed, but she picked up the broom and began to sweep.

In just fifteen minutes, Sam had swept four-fifths of the room. She was amazed at how quickly the task was getting done with a little focus and effort.

What fraction of the room does Sam still have to sweep?

What percent of the room has she finished? What percent of the room is still left?

Let’s work through this solution.

If Sam has completed four-fifths of the room, then she has one-fifth left to complete.

What percent of the room has she completed?

To figure this out, we have to figure out what four – fifths is as a percent. To do this, we can figure it out by using an equal ratio out of 100.

45=80100=80%\begin{align*}\frac{4}{5}=\frac{80}{100}=80\%\end{align*}

Sam has completed 80% of the room.

What percent of the room does she have left?

You can figure this out two different ways. The first way is to simply subtract 80% from 100%. 100% - 80% (what Sam has completed) =20%\begin{align*}= 20\%\end{align*}

The other way is to convert one-fifth (the amount left) to a percent. We can do this by creating an equal ratio out of 100.

15=20100=20%\begin{align*}\frac{1}{5}=\frac{20}{100}=20\%\end{align*}

Since it only took Sam 15 minutes to complete 80% of the room, if she continues with her great effort she will be finished in no time at all.

## Vocabulary

Here are the vocabulary words that are found in this lesson.

Percent
is "per-cent" or "per-hundred", it is a quantity written with a % sign, a part of a whole (100)
Fraction
a part of a whole, related to decimals and percents.
Decimal
a part of a whole shown by a decimal point, hundredths means two decimal places.
Equivalent
means equal
Compare
to determine whether a quantity is greater than, less than, or equal to another quantity.
Order
to write in order from least to greatest or from greatest to least.

## Technology Integration

Other Videos:

http://www.teachertube.com/members/viewVideo.php?video_id=141903 – This video compares fractions, decimals and percents. You'll need to register at the site to view the video.

http://www.mathplayground.com/howto_perfracdec.html – This video shows how to convert fractions and decimals to percents.

## Time to Practice

Directions: Write each fraction as a percent.

1. 4100\begin{align*}\frac{4}{100}\end{align*}

2. 24100\begin{align*}\frac{24}{100}\end{align*}

3. 20100\begin{align*}\frac{20}{100}\end{align*}

4. 76100\begin{align*}\frac{76}{100}\end{align*}

5. 61100\begin{align*}\frac{61}{100}\end{align*}

6. 14\begin{align*}\frac{1}{4}\end{align*}

7. 34\begin{align*}\frac{3}{4}\end{align*}

8. 36\begin{align*}\frac{3}{6}\end{align*}

9. 25\begin{align*}\frac{2}{5}\end{align*}

10. 45\begin{align*}\frac{4}{5}\end{align*}

11. 810\begin{align*}\frac{8}{10}\end{align*}

12. 610\begin{align*}\frac{6}{10}\end{align*}

Directions: Write each decimal as a percent.

13. .31

14. .56

15. .43

16. .08

17. .01

18. .4

19. .6

20. .65

21. .33

22. .19

23. .3

24. .9

25. .11

26. .18

27. .34

28. .99

29. .21

30. .88

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# Synopsis: A “Magic Frequency” for Atomic Spectroscopy

When light is chosen with a special frequency, its absorption by a cloud of atoms is independent of their internal orientation.

Physicists use lasers to track the quantum evolution of atomic states, but interpreting the measurements can be tricky when the light absorption depends on the orientation of the atom’s internal angular momentum. Menachem Givon at Ben-Gurion University of the Negev, Israel, and his colleagues propose and demonstrate a method to monitor how many atoms are in a particular state, no matter how they are oriented.

Atoms can be in the same electronic state, yet differ in the alignment between the angular momenta of their nucleus and their electrons. And even atoms in the same such “hyperfine” state have different energies depending on how they align with a magnetic field, an effect called Zeeman splitting. In addition to having different peak absorption frequencies, when atoms in different Zeeman levels are illuminated with linearly polarized light at a particular angle, some levels absorb more light and some absorb less light than the average absorption, which is what is measured using unpolarized light. But in a cloud of moving atoms, the absorption peaks are broadened by the Doppler effect. At frequencies between these peaks, light is absorbed both by atoms with a higher transition frequency moving in one direction and by other atoms in the lower-frequency state moving in the opposite direction. If the frequency of light is tuned away from a level that absorbs more than the average absorption and toward one that absorbs less, at some intermediate frequency, the level will absorb just the average amount.

Givon and his colleagues realized that, for any particular motional broadening, this intermediate—or “magic”—frequency is the same for every Zeeman level, so the absorption just measures the total number of atoms in the hyperfine state. The researchers confirmed experimentally that when they illuminated a vapor of rubidium-$87$ atoms with light at the magic frequency, a well-known sloshing of atoms between different levels became invisible. – Don Monroe

### Announcements

More Announcements »

## Subject Areas

Atomic and Molecular Physics

## Previous Synopsis

Atomic and Molecular Physics

Nanophysics

## Related Articles

Atomic and Molecular Physics

### Synopsis: Taking Pictures with Single Ions

A new ion microscope with nanometer-scale resolution builds up images using single ions emitted one at a time from an ion trap. Read More »

Atomic and Molecular Physics

### Viewpoint: Squeezed Light Reengineers Resonance Fluorescence

By bathing a superconducting qubit in squeezed light, researchers have been able to confirm a decades-old prediction for the resulting phase-dependent spectrum of resonance fluorescence. Read More »

Gravitation

### Synopsis: Skydiving Spins

Atom interferometry shows that the free-fall acceleration of rubidium atoms of opposite spin orientation is the same to within 1 part in 10 million. Read More »


text:
# Find a homomorphism $\phi$ from $U(30)$ to $U(30)$ such that $\ker(\phi) = \{1,11\}$ and $\phi(7) = 7$

Find a homomorphism $$\phi$$ from $$U(30)$$ to $$U(30)$$ such that $$\ker(\phi) = \{1,11\}$$ and $$\phi(7) = 7$$.

Note: I have seen this Homomorphism from U(30) to U(30) with a given kernel but I have more general questions about the First Isomorphism Theorem that I would like to address. This answer also feels case specific in that it uses the generators, and I am more interested in the general process.

Note 2: We have not covered rings, only groups.

I am having some confusion with the First Isomorphism Theorem, and I want to confirm/deny how I am supposed to interpret this. The statement given in the book is:

Let $$\phi$$ be a homomorphism from $$G$$ to $$\bar{G}$$. Then the mapping from $$\frac{G}{\ker(\phi)} \to \phi(G)$$ given by $$g\ker(\phi) \to \phi(g)$$ is an isomorphism. In symbols, $$\frac{G}{\ker(\phi)} \approx \phi(G)$$

Now, so far in the problems I've done, and the examples we did in class, $$\phi(G) = \bar{G}$$ has been used. As soon as I got to this problem though, this claim no longer seems to make sense. The kernel is non-trivial, so having $$\frac{U(30)}{\{1,11\}} \approx U(30)$$ makes no sense because the orders would not match up.

What I am hoping to confirm is how to find $$\phi(G)$$ because no examples were done in class nor in the book. I would also like to know when it is appropriate to take $$\phi(G) = \bar{G}$$ so that I avoid this confusion again.

It says the map is given by $$g \ker(\phi) \to \phi(g)$$. I suppose what I would do is find this map $$\forall g \in G$$. So, in $$U(30) = \{1,7,11,13,17,19,23,29\}$$, I would have

$$\phi(1) = 1 * \{1,11\} = \{1,11\}$$

$$\phi(7) = 7 * \{1,11\} = \{7,77\} = \{7,17\}$$

$$\phi(11) = 11 * \{1,11\} = \{11,121\} = \{11,1\} = \phi(1)$$

$$\phi(13) = 13 * \{1,11\} = \{13,143\} = \{13,23\}$$

$$\phi(17) = 17 * \{1,11\} = \{17,187\} = \{17,7\} = \phi(7)$$

$$\phi(19) = 19 * \{1,11\} = \{19,209\} = \{19,29\}$$

$$\phi(23) = 23 * \{1,11\} = \{23,253\} = \{23,13\} = \phi(13)$$

$$\phi(29) = 29 * \{1,11\} = \{29,319\} = \{29,19\} = \phi(19)$$

Then, $$\phi(U(30)) = \{1,7,13,19\}$$

Is this the correct process? If so, is there not an issue with the fact that the result is isomorphic to $$\mathbb{Z}_4$$ (i.e. isomorphic to a cyclic group)?

From here though, I could claim that $$\phi$$ is the following:

$$\{1,11\} \to 1$$

$$\{7,17\} \to 7$$

$$\{13,23\} \to 13$$

$$\{19,29\} \to 19$$

I am having a hard time believing this is indeed a homomorphism though. I feel like this maps to $$\phi(U(30))$$, not $$U(30)$$ itself. Can someone convince me that this is an homomorphism?

• Which book are you referring to? Oct 10, 2021 at 17:51
• @Shaun this is from Gallian, Contemporary Abstract Algebra Oct 10, 2021 at 18:03
• @GEdgar We have not covered rings, sorry about that. Let me add that to clarify Oct 10, 2021 at 18:10
• Can you use (or have you learned) the fact that every finite abelian group is isomorphic to some direct product of finite cyclic groups? (In particular, $U(30)$ has to be isomorphic to $\mathbb{Z}_8$, $\mathbb{Z}_4 \times \mathbb{Z}_2$, or $\mathbb{Z}_2^3$, and it's not hard to figure out which.) Oct 10, 2021 at 18:34
• @ConnorHarris yes I can, but I was told I don't even need the abelian assumption. I can just say that $U(n * m) \approx U(n) \oplus U(m)$ where $gcd(n,m) = 1$. In this case, I can write $U(30) \approx U(2) \oplus U(5) \oplus U(3) \approx \mathbb{Z}_1 \oplus \mathbb{Z}_4 \oplus \mathbb{Z}_{2} \approx \mathbb{Z}_4 \oplus \mathbb{Z}_2$. This is actually what I tried first believe it or not, but it didn't get me anywhere. Oct 10, 2021 at 18:47

Well we see that $$\phi(7)=7\implies\phi(7^{2})=7^{2}$$ So $$\phi(19)=19$$

Similarly $$\phi(7^{3})=\phi(13)=13$$.

So we know the images of $$1,11,7,13,19$$.

$$\phi(11.7)=\phi(11).\phi(7)=7$$

So $$\phi(17)=7$$. (We are using the fact that $$\phi(11)=1$$ and $$11.7=17)$$

again $$\phi(11.19)=\phi(11)\phi(19)=19$$.

So $$\phi(29)=19$$. (We are using the fact that $$\phi(11)=1$$ and $$11.19=29)$$

Again $$\phi(11.13)=\phi(11)\phi(13)$$

So $$\phi(23)=13$$. (We are using the fact that $$\phi(11)=1$$ and $$11.13=23)$$

Namely $$\phi$$ is the homomorphism which takes

$$1\to 1\,,7\to 7\,,11\to 1\,,13\to 13\,,17\to 7\,, 19\to 19\,, 23\to 13\,,29\to 19$$

Also since all the information have been generated by the kernel and the known image of the element $$7$$ this is unique . Now we have to just basically check that this is indeed a homomorphism( it is obvious that it is....but for a complete proof we do have to just do the laborous task and verify it ). You can try and device a computer program for this. I tried to work on something like that once....but I could not finish it.

you will see that for finite sets of same cardinality a surjection is a injection. So if $$\phi(G)=\bar{G}$$. You will have that $$\phi$$ is an isomorphism. In that case the kernel will be trivial. However when you are given that the kernel is non-trivial. Then you have to have that the inamge will only have as many elements as $$|\frac{G}{\ker(\phi)}|$$. Ofcourse I am only talking about finite groups. So if we were to only use these concepts of First isomorphism Theorem I will argue like this:-

We are given that $$\phi(7)=7$$ and $$\ker(\phi)=\{1,11\}$$. So as $$\frac{G}{\ker(\phi)}\cong \phi(G)$$.

$$\phi(G)$$ can only have $$4$$ elements.

But since $$\phi(7)=7$$,$$\phi(7^{2})=7^{2}$$ , $$\phi(7^{3})=7^{3}$$ and $$\phi(7^{4})=7^{4}$$. We already have $$4$$ distinct elements.

So the images of the remaining elements must be one of the above $$4$$. We are given $$\phi(11)=\phi(1)=1$$. So we are left with $$3$$ other elements. Now the results have to be such that they agree with the facts $$\phi(7)=7$$ and $$\phi(11)=1$$. But as I showed above, the only way to have that is by the way I did . Hence we have our argument.

• +1 For a clear answer assuming no more theory than necessary. One might want to verify that the map you have determined is indeed a homomorphism; this answer shows that if there is such a homomorphism then it must be this map. But a priori there may be no such homomorphism at all. Oct 10, 2021 at 19:29
• Yeah you're right. I verified uniqueness...but for existence part we do have to actually sit and verify that this is a homomorphism. I'll edit it. Oct 10, 2021 at 19:34
• @Mr.GandalfSauron I agree that this answers the part of my question that this is indeed a homomorphism. I am satisfied with that. I also found that the comments from earlier verified that I found it correctly. Are you able to answer the remaining part? When is it sufficient to take $\phi(G) = \bar{G}$? Oct 10, 2021 at 23:14
• You don't have to have $\phi(G)=\bar{G}$ in order to use first isomorphism theorem. You can say that $\frac{G}{ker(\phi)}\cong \phi(G)$ where $\phi(G)$ is viewed as a subgroup of $\bar{G}$. You will see that this notion will be used several times in gallian. For example in Normalizer-Centralizer theorem. Or Cayley's theorem or even Generalized Cayley. Oct 11, 2021 at 6:42
• Also you will see that for finite sets a surjection is a injection. So if $\phi(G)=\bar{G}$. You will have that $\phi$ is an isomorphism. In that case the kernel will be trivial. However when you are given that the kernel is non-trivial. Then you have to have that the inamge will only have as many elements as $|\frac{G}{ker(\phi)}|$. Ofcourse I am only talking about finite groups. Wait I'll edit in these parts. Oct 11, 2021 at 6:48


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# Math Help - factorize

1. ## factorize

1. Factorise:
a. $8x^2 - 2x - 15$
b. $2y^3 + y^2 - 2y - 1$

2. By inspection,

$8x^2 - 2x - 15 = (2x - 3)(4x + 5)$

You could have used the quadratic formula here to find the 2 roots and thus factorise the equation.

$\text{Let} \,\,f(y) = 2y^3 + y^2 - 2y - 1$

Using the remainder theorem, we can see $f(1) = 2 + 1 - 2 - 1 = 0$
And therefore $y = 1$ is a root, therefore $(y - 1)$ is a factor of the polynomial.

Factorising: $2y^3 + y^2 - 2y - 1 = (y - 1)(2y^2 + 3y + 1) = (y - 1)(2y + 1)(y + 1)$

3. Originally Posted by nzmathman
By inspection,

You could have used the quadratic formula here to find the 2 roots and thus factorise the equation.
thanks A LOT! for the solution of the first equation....but for the second one...can you please show the long division..cuz even i understood HOW to do it....but i go wrong in the procedure....ALWAYS!!!! so i kinda needed help with that....

4. Hmmm I know how to do the long division but cannot present it here nicely for you to see...I am not that advanced in the math coding system. I'm just able to factorise such expressions straight away...

5. hmmm...sure .. no probz..ill try the long division again and if i still have a doubt..i'll tell u...

6. Originally Posted by sudha_14

1. Factorise:
a. $8x^2 - 2x - 15$
b. $2y^3 + y^2 - 2y - 1$
There is no need for using polynomial division

\begin{aligned}2y^3+y^3-2y-1&=(2y^3+y^2)-(2y+1)\\
&=y^2(2y+1)-(2y+1)\\
&=(y^2-1)(2y+1)\\
&=(y-1)(y+1)(2y+1)\end{aligned}

7. b) Another approach, without the long division:
$2y^3+y^2-2y-1=y^2(2y+1)-(2y+1)=(2y+1)(y^2-1)=(2y+1)(y-1)(y+1)$


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# Why is there an emphasis on tensor equations in GR?

In my understanding the purpose of using tensor equations in GR is to ensure that they are true in all coordinate systems. I understand that writing equations tensorially ensures this will be the case; however, are there not non-tensor equations that would also be true in all coordinate systems?

For example, one can define a tensor by its components and how they transform from one coordinate system to another (the tensor transformation law). It seems to me that you could define some other quantity that transforms according to another transformation law, and that equations written in this quantity would also be valid in all coordinate systems.

I've also seen tensors defined as geometric objects on the manifold that act as linear forms on the tangent and cotangent spaces on the manifold. This geometric definition immediately guarantees coordinate independence. Again, I don't see why we can't define a more general geometric object (i.e., not a tensor) and make that the basis of our coordinate independent equations.

To summarise, why is there an emphasis on tensor equations in GR when it seems to me that there should be plenty of non-tensor equations that are valid in all coordinate systems as well?

EDIT: As an example, consider some arbitrary mapping from the tangent space to the reals that is not linear in the tangent vectors. This is a coordinate independent definition. The only difference between these objects and tensors is that for tensors, the mapping is linear. I suppose non-linearity means that these objects won't have straightforward, easily interpretable 'components' in each coordinate system, but I don't see why we still couldn't make important statements about the geometry of spacetime using them.

• What do you want to do with these "other quantities"?
– Ihle
Jul 31, 2017 at 6:47
• I think, rather than hypothesising these objects, you should try to give examples of them.
– user107153
Jul 31, 2017 at 6:50
• I think the question is not just limited to GR, but to any field that uses tensors, for e.g. fluid mechanics.
– Deep
Jul 31, 2017 at 7:34
• Is the nonlinear function a single function that is set for once and for all, applying to every object within a certain class of objects? Or is the nonlinear function different for every function in a particular class?
– user4552
Jul 31, 2017 at 15:06

There are coordinate-independent objects that aren't tensors.

Connections, densities, spinors, sections of fiber bundles in general etc.

Tensors however are

• related to manifold geometry (contrast this with sections of an arbitrary vector bundle)

• have linear dependence on directions.

I'm going to illustrate this with the same example Wald uses in his GR book. Imagine a magnetic field $B$ permeating space. You have a detector that measures the magnetic field in the direction the detector probe is pointing. How do you measure the magnetic field at point $x$?

You choose and record three linearly independent probe orientations. Since the probe probably uses the same units in all directions, and has the same sensitivity, all three orientations can be taken to be unit vectors.

Let the three directions be $e_1,e_2$ and $e_3$. You take the three measurements, these return the values $$B_1=B(e_1),\ B_2=B(e_2),\ B_3=B(e_3).$$

As you can see, the magnetic field plays the role of a covector here, instead of a vector. From this, you can assemble the magnetic field as $$B=B_1 e^1+B_2 e^2+B_3e^3,$$ where $e^i$ is the dual basis element to $e_i$.

The metric tensor is $$g_{ij}=\left(\begin{matrix} 1 && \cos\alpha_{12} && \cos\alpha_{13} \\ \cos{\alpha_{12}} && 1 && \cos{\alpha_{23}} \\ \cos{\alpha_{13}} && \cos{\alpha_{23}} && 1\end{matrix}\right)$$ where $\alpha_{ij}$ is the angle between $e_i$ and $e_j$.

The magnetic field vector is given by $\sharp B=g^{1i}B_ie_1+g^{2i}B_ie_2+g^{3i}B_ie_3$, the magnitude is given by $||B||=g^{ij}B_iB_j$ etc.

Now, if instead of having a linear dependence on directions, $B$ was some arbitrary smooth function $B:T_xM\rightarrow\mathbb{R}$, then you'd need an infinite amount of measurements (in an infinite amount of directions) to reconstruct it at a point.

Clearly, these "direction-dependent" quantities in physics behave in such a way that you don't need an infinite amount of measurements to measure them at a point. If they did, physics as we know it would not exist! So the reason we use tensors is that physics is measurable.

• I am pretty sure that, given linear functions $\left\{\omega_n: T_xM \to \mathbb{R}\right\}$ such that, for $\vec{v}\in T_xM$, $\lim_{n\to\infty}\sum_{n=0}^m \left(\omega_n(\vec{v})\right)^n$ converges (note no ESC here!), then, well, that's a Taylor series, and so you can express quite general nonlinear functions using a set of tensors. But you need an infinite set in general, which is the point you were making.
– user107153
Jul 31, 2017 at 12:01
• I'll give you +20 if you can find me a probe that uses different units when oriented in different directions.
– Jim
Jul 31, 2017 at 13:33
• In addition to the examples you give like Christoffel symbols and tensor densities, I guess one can simply build inhomogeneous n-tuples, e.g., an object $(f,\omega)$ that is an ordered pair built out of a scalar and a covector.
– user4552
Jul 31, 2017 at 15:09
• Thanks for the answer. One thought: the Einstein tensor is usually motivated by being the only tensor formed of the metric and its first and second derivatives. We seem to be really lucky that the way matter curves spacetime is perfectly encapsulated by a tensor equation! If it were a bit more complicated (non-tensorial), we would seem to have no hope of measuring this relationship easily. (Granted, SR teaches us that energy-momentum is a tensor quantity, so maybe that would motivate the tensorial nature of the curvature measure on the LHS of Einstein's Field Equations) Aug 1, 2017 at 8:37

Good question, and one that isn't commonly gone into in the physics literature when they introduce the tensor transformation law.

Try visualising it geometrically: take a simple example, deforming the surface of a sphere to an ellipsoid; if we take a small (ie infinitesimal) patch of the ball and see how it transforms we see that there is a multilinear dependence; this example can be generalised to arbitrary manifolds in any cartesian space, and we can also, with some thought, drop the embedding.

A multilinear transformation is characterised universally by tensors, and then by taking bases we get the usual coordinate transformation property that characterises tensors common in the physics literature.

The best reference I've seen for this is Lees book on Differential Geometry, and Dodsons Tensor Geometry, though he tends to use some idiosyncratic terminology.

Tensors by themselves don't ensure that a formula is correct in all coordinate systems. Navier-Stokes equations for example can be written in tensor form bu are not coordinate independent. What you need in fact is the covariant derivative property. The tensor form on the other hand is needed to describe stresses on a surface. To describe the shear stress on a surface for example, you need one vector lying in the surface for describing the force on that surface with magnitude and direction. But then you need another vector to describe the position and orientation of the surface itself- hence the two indices of the second rank tensor.


text:
# Math Help - Vending machine problem

1. ## Vending machine problem

Hey guys here's my problem:
A vending machine takes 10, 20 and 50 cent coins only.
In how many ways can we put n cents in the machine provided that n is a multiple of 10(i.e. n =10*k, where $k\geq0$)

2. Sorry for the bump here, but I was also wondering how i could relate this problem to the Coin Problem:
Coin Problem -- from Wolfram MathWorld


text:
# Topology of the Hamel basis in a TVS

Let $V$ be a complex topological vector space, and let $I$ be a Hamel basis of it. Then as a subset $I\subset V$ acquires an induced topology, becoming a topological space. For a topological space $X$ consider the space $C_\#(X,\mathbb{C})$ of finitely supported complex functions on $X$. Then $C_\#(I,\mathbb{C})$ is algebraically isomorphic to $V$ by definition of a Hamel basis, $$\forall v\in V,\quad v=\sum_{e\in I}\lambda_e^v e,\quad I\ni e\mapsto\lambda_e^v\in\mathbb{C}.$$

Question: Does there exist a topology on $C_\#(X,\mathbb{C})$, such that $C_\#(I,\mathbb{C})$ is isomorphic to $V$ as topological vector spaces?

In finite dimensions $I$ is finite and hence discrete, so that all reasonable topologies on $C_\#(I,\mathbb{C})=\mathbb{C}^I$ coincide and give a positive answer. The question therefore pertains to infinite dimensions.

I am mostly interested in separable Hausdorff $V$. Thank you.

• So if I understand correctly, you are looking for a general construction which, given a topological space $X$, outputs a TVS topology $\tau_X$ on $C_\#(X, \mathbb{C})$. And it should have the property that whenever $V$ is a (separable, Hausdorff) TVS and $I \subset V$ is a Hamel basis with the subspace topology, then $V \cong (C_\#(I, \mathbb{C}), \tau_I)$ as TVSes (i.e. there is a linear homeomorphism between them). Have I got that right? – Nate Eldredge Jun 2 '17 at 14:16
• So your claim would be disproved if we could find a pair of TVSes $V_1, V_2$, not isomorphic as TVSes, with respective Hamel bases $I_1, I_2$ which are homeomorphic in their subspace topologies. Right? – Nate Eldredge Jun 2 '17 at 14:17
• Exactly. Both comments are correct. – Bedovlat Jun 2 '17 at 16:18
• Then just take a linearly independent Cantor set $C$ in $\ell_2$ whose linear hull $lin(C)$ is dense in $\ell_2$ and observe that $lin(C)$ with the subspace topology inherited from $\ell_2$ is not topologically isomorphic to $lin(C)$ endowed with the topology inherited from the Banach space $c_0$. If these two TVS would be isomorphic, then their completions (i.e., $\ell_2$ and $c_0$) would be isomorphic too, but this is not the case. – Taras Banakh Jun 6 '17 at 15:11
• But are the two induced topologies on $C$ equivalent? – Bedovlat Jun 6 '17 at 16:57


text:
# hydrateZoomA-Z

## Subject - Organic Chemistry

In organic chemistry, hydrates are defined as geminal diols generated by addition of water to the carbonyl group of or .

## Subject - Inorganic Chemistry

In inorganic chemistry, hydrates are defined as solid or liquid molecular compounds in which water is one of the components and in which the two $OH$ bonds in each molecule stay intact. Crystalline hydrates contain the crystallization water in stoichiometric amounts, for example in gypsum (calcium sulfate dihydrate, CaSO4·2H2O).


text:
## shielding constant, $$\sigma$$

https://doi.org/10.1351/goldbook.S05645
In NMR the difference between the external @M03686@ and the local @M03686@ at a resonating @N04256@ affected by the neighbouring electrons divided by the external flux density, $$\sigma =\frac{B_{0}\,-\,B}{B_{0}}$$.
Source:
Green Book, 2nd ed., p. 25 [Terms] [Book]


text:
Blogging, LaTeX and millionaires

February 20, 2007 at 8:07 pm | Posted in comics, facts and figures, tools | 2 Comments

wooo, wordpress now supports latex! Like I’d need to use it for inserting equations here ^^”. Just for the hell of it:

$E=mc^2$

hmmm…something we all have seen sometime in our lives

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

this is one very powerful typesetting tool. Hardly any of my professors use word to type their papers. LateX is what they use. And it is freeware. Sometimes they make us write our reports in LaTeX just to see us die.

There was a report released a while back, on Hong Kong’s millionaires:

#####

Hong Kong Millionaires

Of the 276,000 Hong Kong people who are millionaires,

51% are women
88% are married
82% invest in stocks
75% own properties
44% have a car

#####

Just…food for thought. This is what real life financially successful people do. What about you?

Blogging 1

I kid you not. The other one on the list for today was “cute.pdf” which is logical, because a previous post was about cute pdf, the interesting pdf exporting tool.

1 Chapter of quantum physics down! Only 13 more to go! aah…lovely stree-free revision//


text:
# Elements of Euclidean Domain have Greatest Common Divisor

## Theorem

Let $\struct {D, +, \times}$ be a Euclidean domain.

Then any two elements $a, b \in D$ have a greatest common divisor $d$ such that:

$d \divides a \land d \divides b$
$x \divides a \land x \divides b \implies x \divides d$

and $d$ is written $\gcd \set {a, b}$.

For any $a, b \in D$:

$\exists s, t \in D: s a + t b = d$

Any two greatest common divisors of any $a, b$ are associates.

## Proof

Let $a, b \in D$.

Let $U \subseteq D$ be the set of all elements $h a + k b$ of $D$ where $h, k \in D$.

Then $U$ is an ideal of $D$.

Note that $U = \ideal a + \ideal b$ where $\ideal a$ and $\ideal b$ are Principal Ideal.

By Sum of Ideals is Ideal, $U$ is an ideal.

By Euclidean Domain is Principal Ideal Domain, $U$ is a principal ideal, $\ideal d$ say.

As $a, b \in U$ it follows that $d$ is a divisor of $a$ and $b$, that is:

$d \divides a \land d \divides b$

Since $d$ itself is in $U$, we have:

$\exists s, t \in D: s a + t b = d$
$x \divides a \land x \divides b \implies x \divides d$

So $d$ is a greatest common divisor of $a$ and $b$.

If $d$ and $d'$ are both greatest common divisors of $a$ and $b$, then $d \divides a \land d \divides b$ and so $d \divides d'$.

Similarly $d' \divides d$.

So $d$ and $d'$ are associates.

$\blacksquare$


text:
# Can not find Windows platform

• I am trying to distribute the application I developed. I know I have to include a sub folder named platforms with the proper Windows DLLs. I also have added lines in my main function as follows:

However, I still get the error "This application failed to start because it could not find or load the Qt platform plugin 'Windows'."

"Available platform plugins are: minimal, offscreen, windows."

The DLLs are in the sub folder, however for some reason the application does not seem to find them.

Thanks,
Bear35645

• Did you try to use the windeployqt tool (http://doc.qt.io/qt-5/windows-deployment.html) to copy the right dependencies into your applications folder?

• @sneubert Thanks. That was the help I needed.


text:
# Find the Exact Value tan((8pi)/3)

Remove full rotations of until the angle is between and .
Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because tangent is negative in the second quadrant.
The exact value of is .
The result can be shown in multiple forms.
Exact Form:
Decimal Form:
Find the Exact Value tan((8pi)/3)


text:
# Picard's Method

1. Jun 21, 2011

### trojansc82

1. The problem statement, all variables and given/known data

y' = -y , y(0) = 1

2. Relevant equations

Picard's method

3. The attempt at a solution

I found y1 as 1 + 1/2x2, however y1 is really 1 + x

Last edited: Jun 21, 2011
2. Jun 21, 2011

### LCKurtz

Neither of those is correct for the equation you wrote. Show us what you did so we can find your mistake(s).

3. Jun 21, 2011

### lanedance

you should show you method, so we can see where you're going wrong - what was your first approximation? starting with a constant should integrate to a constant times x and you should have the solution after one iteration

4. Jun 21, 2011

### lanedance

also as LCKurtz points out neither is correct, you should alway check by substituting into the original DE

5. Jun 22, 2011

### trojansc82

Sorry I meant y1 is really 1 -x. It was a typo.

6. Jun 22, 2011

### HallsofIvy

Staff Emeritus
I think LCKurtz and lanedance are misunderstanding your notation. Neither $y= 1+ x^2/2$ nor $y= 1+ x$ is a solution to the equation but you are not claiming it is.

Picard's method of solving the differential equation y' = f(x,y), with initial condition $y(0)= y_0$ is an iterative method. Taking $y_0$ to be the initial value, $y_1= \int f(x, y_0)dx$ is the first iteration, then $y_2= \int f(x, y_1(x))dx$, $y_3= \int f(x,y_2(x))dx$, etc.

For the problem as given, the first iteration of Picard's method is, indeed, $y_1(x)= x+ 1$. But no one can show you where you went wrong until you show us what you did.

7. Jun 22, 2011

### LCKurtz

Nor are we claiming that.

No, it isn't.

8. Jun 22, 2011

### HallsofIvy

Staff Emeritus
Oh, I completely misread the equation! It is y'= -y, not y'= y as I was seeing!

My apologies to both LCKurtz and lanedance.

trojansc82, I still don't see how you got "1+ x^2/2". Please show us what you did.

9. Jun 22, 2011

### trojansc82

Again, I apologize, I mistyped the answer.

The answer for y1 = 1 - x

10. Jun 22, 2011

### LCKurtz

Is that the answer you got? In your original post you said you got something else. Have you figured it out? You never did show us what you did...

11. Jun 22, 2011

### trojansc82

Within the integral I multiplied -1 (since y was -y) by t. I ended up integrating -t, which came to -1/2 x2.

I have trouble within the integral.

12. Jun 22, 2011

### LCKurtz

What integral? We aren't mind readers. My guess is you have the integral wrong in the first place. Show us your work.


text:
# Problem with high-side current measurement

I am having strange failures in a high-side current measurement circuit that I have not yet understood. The U3 measurement chip is getting damaged and shorts out. Perhaps someone can help identify potential failure points in this type of measurement circuit?

+VSA is a "safe zone" discrete supply (below) designed to hopefully minimize the chances of this measurement circuit creating a failure point (sigh). It's typically between 3-4V. M3 lets a very small (microseconds) pulse of voltage (up to ~7-12V maybe) through before the gate turns it off... then M3 is turned on in a slow-start (but still fairly quick) manner. Because of this, +VSA is potentially lower than S+ for that very short period (That circuit's input is fed from the same point as the top of M3). I don't see any reason in the datasheet for that to be a problem, though ("Independent supply and input common-mode voltage").

I added D10 and D11 thinking the differential voltage spec of 0.8V was being violated with inrush current... but I've had another failure with even this new design.

It only seems to fail during power cycle events.

The failures seem to be related to parts of the downstream circuitry getting shorted out during/before power on... but in testing only this circuit I can't seem to kill it no matter what I do downstream. It only seems to fail in combination with the rest of the unit. So, I'm wondering if anyone can identify additional failure modes I can look into.

Here's the datasheet for U3: ZXCT1086 datasheet

+VSA circuit:

EDIT: Ok, so the +VSA circuit above is perhaps confusing. The +24V node refers to the system voltage input at the top of M3 (well, system input after the fuse, switch, GDTs, and reverse voltage protection). The node after R26 is +24_AUX, which is not the same.

However, I did notice the diodes D10 and D11 are perhaps not clamping to ~400mV like I expected. They're rated for 410mV forward current, but at 1mA. At even 15mA the forward voltage goes up to 1V. I did not expect that, and I can't believe that this far into my circuit designing "career" I haven't heard of how much they vary by current before. I knew they varied a little, but that's way more than a little. I'm adding the datasheet for the diodes as well. It appears I need to re-think the protection the diodes are trying to accomplish.

Another note: In previous testing, I was unable to force the voltage across R26 to go any higher than 100mV.

Diode datasheet: BAS70H,115 datasheet

• It only seems to fail during power cycle events. When I measure voltage regulation during power on, I typically get my diff probe between S+ and S- to see if there's any potential noise that occurs. This datasheet indicates the differential voltage should be between 0 to 0.5V. If you find one your oscope that it's very sinusoidal, it could indicate a large amount of noise. I don't think I've ever seen any designs at work that have diodes between your sense lines. I have seen capacitors but not diodes.
– user103380
Dec 14, 2019 at 17:44
• I didn't realize the diodes probably don't help at all for reasonably high inrush currents. I thought ~400mV Vf diodes would be okay for the 800mV maximum differential... wow, nope. I haven't measured over 100mV between them yet, though. Dec 14, 2019 at 19:35


text:
Sympy and EMT – Comparison

In Euler Math Toolbox, symbolic expressions are seamlessly integrated by the &-syntax. The expressions and functions are handled externally via the CAS Maxima, but mirrors exists in the numerical side of EMT as simple strings or numerical functions.

Let me give an example.

>function fx(t) &= t*cos(2*pi*t)

t cos(2 pi t)

>function fy(t) &= t*sin(2*pi*t)

t sin(2 pi t)

>plot2d("fx","fy",xmin=0,xmax=1,r=1):

The functions „fx“ and „fy“ are symbolic functions. I.e., Maxima defines the functions in its system, and they can be used in Maxima.

>&fx(z^2+1)

2                 2
(z  + 1) cos(2 pi (z  + 1))

But the functions are also defined in EMT as numerical functions. Thus, they can be used like any other numerical function in EMT.

>fx(linspace(0,1,10))
[0,  0.0809017,  0.0618034,  -0.0927051,  -0.323607,  -0.5,  -0.48541,
-0.216312,  0.247214,  0.728115,  1]

Because the functions contain only elementary operations, they can work on vectors too. This is called a „vectorized“ or „mapped“ function. Consequently, they can be called in other functions of EMT. E.g., the „plot2d“ function can plot curves, if two functions (or two expressions) are given.

Continuing our example, we compute the length of the curve using

$$l(f) = \int\limits_0^1 \|f'(t)\| \,dt.$$

>function f(t) &= [fx(t),fy(t)]

[t cos(2 pi t), t sin(2 pi t)]

>function df(t) &= diff(f(t),t)

[cos(2 pi t) - 2 pi t sin(2 pi t),
sin(2 pi t) + 2 pi t cos(2 pi t)]

>integrate(&norm(df(x)),0,1)
3.3830442855
>&integrate(norm(df(t)),t,0,1), %()

2
asinh(2 pi) + 2 pi sqrt(4 pi  + 1)
----------------------------------
4 pi

3.3830442855

To compute the length of the curve, we define a vector valued symbolic function „f“. That function does work in Maxima symbolically, or in EMT numerically.

A side remark: If „f“ is applied to a vector instead of one point, it will compute „fx“ and „fy“ and concatenate these vectors, which is not useful most of the time. In future versions, EMT should be able to generate a matrix of values if a vector valued function is applied to a vector of values. In our application, we only need „f“ as a symbolic function.

In the same way, we can then define a function „df(t)“ for the derivative vector simply by symbolic differentiation. „df“ will also work symbolically and numerically. The function „norm“ is a pre-defined symbolic function. You can plug in a symbolic vector and get an expression which represents a function of „t“.

>&norm(v)

sqrt(v . transpose(v))

>&norm(df(t))

2
sqrt((cos(2 pi t) - 2 pi t sin(2 pi t))
2
+ (sin(2 pi t) + 2 pi t cos(2 pi t)) )

We then use the adaptive integral of EMT to integrate the expression „norm(df(x))“ between 0 and 1.

Finally, Maxima can compute the „exact“ integral in symbolic form, which we evaluate numerically to verify our previous result.

How does this all translate to Python?

The first problem is that there are no symbolic functions in Sympy. There are only expressions, which can be handled by the symbolic features of Sympy. And, of course, there is no automatic parallel definition of a numerical function for Numpy as in EMT. To evaluate a symbolic expression at a point, „subs“ can be used, or it can be made a numerical function by „lambdify“.

Here is an example.

import sympy as sp
import numpy as np

t = sp.symbols('t')
fe = t*sp.cos(2*np.pi*t)
print(fe.subs(t,0.2))
f = sp.lambdify(t,fe)
print(f(0.2))

-->
0.0618033988749895
0.061803398874989326

As you see, „subs“ just substitutes a value for „t“. This is not going to work of „sp.pi“ is in the expression instead of „np.pi“.

t = sp.symbols('t')
fe = t*sp.cos(2*sp.pi*t)
print(fe.subs(t,0.2))
f = sp.lambdify(t,fe)
print(f(0.2))

-->
0.2*cos(0.4*pi)
0.06180339887498949

For Sympy, „pi“ is a mathematical variable with some properties and a value. Because „cos(2pi) „is known, „f(1.0)“ does work as expected. But „f(0.2)“ leaves the expression and is not evaluated numerically.

„lambdify“ translates a symbolic expression so that it can be used by Numpy, effectively translating „sp.pi“ to „np.pi“. Thus, „f“ works in both cases.

Back to our example. We use what we learned to trick Python into doing what we want.

import sympy as sp
import numpy as np

t = sp.symbols('t')
fxe = t*sp.cos(2*sp.pi*t)
fye = t*sp.sin(2*sp.pi*t)
fx = sp.lambdify(t,fxe)
fy = sp.lambdify(t,fye)

vt = np.linspace(0,1,400)
x = fx(vt)
y = fy(vt)
fig,ax = plt.subplots(figsize=(8,8))
ax.set_xlim(-1,1)
ax.set_ylim(-1,1)
ax.grid(True)
ax.plot(x,y);

It might be a good idea to define a function „plot_curve“ to produce the plot, like the following.

def plot_curve (fx,fy,xmin,xmax,n=100,figsize=(8,8),r=1):
vt = np.linspace(0,1,400)
x=fx(vt); y=fy(vt);
fig,ax = plt.subplots(figsize=figsize)
ax.set_xlim(-r,r); ax.set_ylim(-r,r)
ax.grid(True)
return ax.plot(x,y)

plot_curve(fx,fy,0,1,400);

Such functions can go into a Python module. I already started a module „renestools.py“ with functions I need to make my life in Python easier.

Now, let us compute the length. The integration is done using an adaptive Gauss quadrature in my package. I explained in a previous posting, how this is working. Note that we need to pass a vectorized function to the integration. „fnorm“ is not vectorized (in contrast to the expression „norm(f(t))“ which we used in EMT). „np.vectorize“ transforms any function for single real values into vectorized functions which will work on Numpy matrices or vectors.

fe = sp.Matrix([fxe,fye])
fde = sp.diff(fe,t)

def norm(v):
return np.sqrt(np.sum(v*v))

fd = sp.lambdify(t,fde)
def fnorm(t):
return norm(fd(t))

import renestools as rt
rt.integrate(np.vectorize(fnorm),0,1)

--> 3.3830442855028062

To compute the integral in symbolic form, we need to help Sympy with „trigsimp“. We evaluate the result to 20 digits. (We could have done that in Maxima too, by the way.)

res = sp.integrate(sp.trigsimp(sp.sqrt(fde[0]**2+fde[1]**2)),(t,0,1))
print(res)
print(res.evalf(20))

-->
asinh(2*pi)/(4*pi) + sqrt(1 + 4*pi**2)/2
3.3830442855028069597

As you see, Maxima and EMT are much more intuitive than Sympy and Numpy. I say this in the hope that many of you find their way to Euler Math Toolbox and help the development. Currently, Python has a lot more potentially useful toolboxes than EMT.

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text:
f

#### How do I get rid of the folder "\My Documents\IBM\Client Access"?

Client Access seems to create the folder "\My Documents\IBM\Client
Access" automatically.

If I move / delete the folder, it gets recreated following a Windows
restart.

If I change / delete the registry entry
"HKEY_CURRENT_USER\Software\IBM\Client Access
Express\CurrentVersion\(Default)", the entry gets reset and the folder
recreated following a Windows restart.

I am running:
IBM iSeries Access for Windows
Version 5
Release 3
Modification level 0
Service level SI23492

Anyone know how to get rid of it for good?
Even being able to move it to somewhere else would be good.

Thanks.

 0
8/2/2006 8:46:24 AM
comp.sys.ibm.as400.misc 9219 articles. 4 followers.

2 Replies
948 Views

Similar Articles

[PageSpeed] 19

andyrobertsonuk@gmail.com wrote:
> Client Access seems to create the folder "\My Documents\IBM\Client
> Access" automatically.
>
> If I move / delete the folder, it gets recreated following a Windows
> restart.
>
> If I change / delete the registry entry
> "HKEY_CURRENT_USER\Software\IBM\Client Access
> Express\CurrentVersion\(Default)", the entry gets reset and the folder
> recreated following a Windows restart.
>
> I am running:
>   IBM iSeries Access for Windows
>   Version 5
>   Release 3
>   Modification level 0
>   Service level SI23492
>
> Anyone know how to get rid of it for good?
> Even being able to move it to somewhere else would be good.
>
> Thanks.
>
Maybe I would try "Control Panel/iSeries Access for Windows" properties
in the PC5250 tab, selecting something different from "Documents" in
"PC5250 file path" section.

--
Dr.Ugo Gagliardelli,Modena,ItalyCertifiedUindoscrasherA�ejoAlcoolInside
Spaccamaroni andate a cagare/Spammers not welcome/Spammers v�o � merda
Spamers iros a la mierda/Spamers allez vous faire foutre/Spammers loop
schijten/Spammers macht Euch vom Acker/Spamerzy wypierdalac'

 0
info1229 (187)
8/2/2006 11:59:11 AM
Dr.UgoGagliardelli wrote:
> Maybe I would try "Control Panel/iSeries Access for Windows" properties
> in the PC5250 tab, selecting something different from "Documents" in
> "PC5250 file path" section.

Thanks for the tip, but that didn't seem to do the trick. I'm still
stuck with the folder.
Andy

 0
8/14/2006 7:18:48 PM

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I just bought an X31 off ebay am having a few config issues. It didn't come with a disk. So, I put a clean copy of XP on a 2.5 drive and installed it and got a Blue Screen error: "A problem has been detected and Windows has been shut down to prevent damage to your computer .... ***Stop: 0x0000007B (0xF8973640,0xC0000034,0x0000000,0x0000000) " I put the bios into debug mode and subsequent reboots it does the memory, device checkout and then goes to the boot which gives the above message. It says at the bottom, to interrupt normal startup, press the blue Access IBM button. However, when I press the button, it does not bring me back to the BIOS setup screen. No matter what I have tried (FN key with f1, f2, f3 etc), I have not been able to get back. At this point, I have 3 questions : 1. How do I get back to BIO setup? 2. Does the X31 expect something other than what Windows places in the MBR (to get past the blue screen message) 3. Is there any web references on the X31 I appreciate any help, thanks Lou Lou Lipnickey wrote: > I just bought an X31 off ebay am having a few config issues. It didn't > come with a disk. So, I put a clean copy of XP on a 2.5 drive and > installed it and got a Blue Screen error: > > "A problem has been detected and Windows has been shut down to prevent > damage to your computer .... > > ***Stop: 0x0000007B (0xF8973640,0xC0000034,0x0000000,0x0000000) " > > I put the bios into ...

Cannot get IBM 3275 HDD to go "Available"
Having a problem trying to get an IBM 3275 HDD go from defined to available. - Have updated firmware for adapter and system - Updated to 5.2.0.6 OS --------------------------------------------------------------------------------------- hdisk2 U787A.001.DPM1DTH-P1-T10-L3-L0 16 Bit LVD SCSI Disk Drive (146800 MB) Manufacturer................IBM H0 Machine Type and Model......HUS103014FL3800 FRU Number..................03N5267 ROS Level and ID............52505152 Serial Number...............00B32AC5 EC Level....................H17923D...

IBM announced newest release of product including "IBM COBOL for Windows"
This week, IBM announced the latest release of the *ONLY* product that they now sell that include a currently supported version of "IBM COBOL for Windows". See: http://www-01.ibm.com/common/ssi/rep_ca/0/897/ENUS209-380/ENUS209-380.PDF It appears (but I haven't seen the details) that there are new installation options that will allow a user to better "tailor" which features of RDz they want installed. NOTE WELL: This does NOT mean that there are options for only BUYING those features that you want to use. If you are a shop that *only* wants COBOL (not PL/I...

"My Documents" Folder.
I want to write a FileLocator that is relative to "My Documents" folder. How do you locate "My Documents"? umur wrote: > I want to write a FileLocator that is relative to "My Documents" folder. > How do you locate "My Documents"? You could try the appended small package, which provides access to some of the pre-defined locations. Once loaded, use: ShellLibrary default getSpecialFolderPath: CSIDL_PERSONAL. -- chris =========== CU Windows Shell Extensions.pac =========== | package | package := Package name: 'CU W...

Cannot get BIOS setting change to "take" on IBM NetVista
Hello All: In the BIOS of an IBM NetVista running XP I am setting the floppy drive to 3.5-inch/1.44Mb, but the save doesn't "take". It keeps returning to 5.25-inch. Is there something/somewhere that can keep a BIOS setting change from taking? Thanks in advance, -Richard Vaughn "Richard Vaughn" <rvaughn9@verizon.net> wrote in message news:543tc.22226\$yc4.19203@nwrdny02.gnilink.net... > Hello All: > > In the BIOS of an IBM NetVista running XP I am setting > the floppy drive to 3.5-inch/1.44Mb, but the save doesn't "take&q...

Urgent JAVA Requirement in """"""NEW YORK"""""""""
Hello Partners, How are you ? Please find the requirement below. Location : NY Duration : 8 mnths Rate :Open Job description: Java/J2EE Web Service Developer =B7 (4+ years of application development experience in Java/J2EE and Web service technologies. =B7 Experience with spring & Hibernate. =B7 Experience with J2EE Application Server (preferably Web logic). =B7 Preferable Aqua logic DSP Experience =B7 Preferable Sonic ESB Composite Service experience Experience working in UNIX environment. Good working knowledge of Oracle RDBMS and PL/SQL) Note : Please send your updated resume along with contact no to Karthik@bhaninfo.com. Thanks & Regards Karthik BhanInfo karthik@bhaninfo.com # 1572, 7 th Floor Suites, 1 Blue Hill Plaza Pearl River,NY-10965 www.bhaninfo.com ...

how to get rid of "today" "yesterday" etc bars in kmail 1.12.4?
Hi: I looked around in the setup of kmail, I googled but for the life of me I can't find any way to get rid of the superfluous and molesting bars which devide my message list into "today", "yesterday", "last week", "NAME OF MONTH". etc [rant] I am able to read the dates and sort against the dates column - yessir! I don't want such "Outlook-Mania for dummies". Oh wait! Maybe it's not for dummies - 'cause dummies aren't able to get rid of it. [rant] And yes, I would like to stay with kmail. Kind regards, Eike -- "T...

How can Client get "Client IP" ?
Hi, All : I wrote a simple server in Python, and I can get the client IP by following script: s = socket.socket( socket.AF_INET, socket.SOCK_STREAM ) s.bind( (HOST, PORT) ) s.listen(1) conn, addr = s.accept() clientIP = addr s.close() That's wonderful. However, I hope that my client can get his own IP "by himself". ( Not by server :Q ) So what can I do ? Is there a CONST I can use ? Thanks. --pWyiEgJYm5f9v55/ Content-Type: text/plain; charset=us-ascii Content-Disposition: inline For a socket object s, s.getsockname() "Return the socket's own address. This is usefu...

puts "\\".gsub("\\", "\\\\")
Hello, I have a mini-ruby quiz. Guess what this line of code writes to the console, then try it for yourself: puts "\\".gsub("\\", "\\\\") Why is that so? Martin From: martinus [mailto:martin.ankerl@gmail.com]=20 # Hello, I have a mini-ruby quiz. Guess what this line of code writes to # the console, then try it for yourself: # puts "\\".gsub("\\", "\\\\") puts "\\".gsub("\\", "\\\\") \ #=3D> nil # Why is that so? faq. escaping the escape in sub/gsub. search the archives. maybe you want somethin...

Question about "sprintf" "@" "do for"
Hello, this works: A1=3D1 A2=3D2 A3=3D3 i=3D1 vari=3Dsprintf("A%.f",i) print vari,"=3D",@vari i=3Di+1 vari=3Dsprintf("A%.f",i) print vari,"=3D",@vari i=3Di+1 vari=3Dsprintf("A%.f",i) print vari,"=3D",@vari do for [i=3D1:3]{ vari=3Dsprintf("A%.f",i) print vari } But I want to have "print vari,"=3D",@vari" in the loop. But it dosen't=20 work. Why can't I use "print vari,"=3D",@vari" in the loop? Is there a=20 solution for? J=C3=B6rg Jörg ...

Gary Sokolich """"""
"""""""""" http://www.manta.com/c/mmlq5dm/w-gary-sokolich W Gary Sokolich 801 Kings Road Newport Beach, CA 92663-5715 (949) 650-5379 http://www.tbpe.state.tx.us/da/da022808.htm TEXAS BOARD OF PROFESSIONAL ENGINEERS February 28, 2008 Board Meeting Disciplinary Actions W. Gary Sokolich , Newport Beach, California �V File B-29812 - It was alleged that Dr. Sokolich unlawfully offered or attempted to practice engineering in Texas (...) Dr. Sokolich chose to end the proceedings by signing a Consent Order that was accepted by ...

WHERE did Joseph Smith get the terms "Mormon" and "Moroni" and "Cumorah" from???
Nick thinks he is a historian, and has his own kooky Victor von Frankenstein interpretation of how The Mormons holy "book of mormon" came into being. He is the only one in the world who thinks that Joseph smith actually had gold plates that were smuggled to the americas where he then translated them giving a coded history of Europe. Mormons including their founding prophet think Joe's supposed gold plates, which he said an angel gave to him, detailed a history of american indians. No one other than mormons believe that nonsense, but nick has a theory that only he can be stupid...

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