text:
# RC Circuit Discharge

1. Jul 1, 2008

### purduegirl

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

A circuit consists of 3 components connected between 2 terminals: A 6 V source, a 60 ohms resistor and a 0.02 F capacitor. The capacitor is initially uncharged. The voltage source is turned on. What is the current flowing through the above capacitor after a very long time?

2. Relevant equations

I(t) = E/R *e^(-t/time constant)
time constant = (R)(C)

3. The attempt at a solution
time constant = 1.2 s

Because the time is a long time, I thought that I wouldn't have to enter anything in for time. I then used the I= E/R*e^(-1/time constant) to solve for the current. I got I = (6 V/60 ohms)*e^(-1/1.2s) = 4.35E-1. I'm pretty sure I'm wrong here so any help to tell me what I am doing wrong would be greatly appreciated.

2. Jul 1, 2008

### alphysicist

Hi purduegirl,

But you did enter something here for time--you entered 1, so you found the current at 1 second.

When they say the time is a very long time, what does that mean? Or alternatively, what is $e^{-t/(\mbox{time constant})}$ for very large t (if you plot it, for example)?

3. Jul 1, 2008

### purduegirl

They didn't give a specified time for " a long time". That's why I'm not sure what they want here.

4. Jul 1, 2008

### alphysicist

No, there's not a specified number; that's not what they want here.

But try answering my second question; the problem is understanding the behavior of the exponential function, which is important for understanding RC circuits. Make a plot of $\exp(-t/1.2)$. What happens to the value of that as t gets larger and larger?

Once you know what happens to the exponential function, you'll know what happens to the current, since it's just a constant multiplied by the exponential function.

5. Jul 1, 2008

### Gear300

If this is a series circuit, remember that capacitors are technically breaks in the circuit. After a long time, the circuit is in steady state, and the current across branches involving capacitors can set to 0A. When referring to a long time, take the limit as t approaches infinite.

6. Jul 2, 2008

### purduegirl

So from what I just read in my textbook, long after the switch is closed, the potential difference across the capacitor is nearly equal to the emf and the current is small. Looking at a graph of current as a function of time, I can see that the longer the circuit is flowing, the smaller the current gets.

7. Jul 2, 2008

### purduegirl

Unforunatly, the only exponential examples my book gives are for finding the voltage across the capacitor (0.632*Voltage of Battery) and for finding the current when the time equals the time constant (0.368*(Voltage of Battery/Resisitivity). I tried the last equation to no avail.

8. Jul 2, 2008

### alphysicist

That's right; and as t gets larger and larger (goes to infinity), what would you say the current is? That is the answer they are looking for here.

9. Jul 2, 2008

### Dick

What could they mean by 'after a very long time'? There's no number to put in corresponding to a 'very long time' is there? I think you've just stated the answer, metaphorically.

10. Jul 2, 2008

### purduegirl

Thanks to all! My prof gave us homework on RC circuits, but never lectured over it. Thanks so much for the help!


text:
CommitLineData
c7dab9ff 1% Generated by roxygen2: do not edit by hand
2% Please edit documentation in R/modelSelection.R
3\name{modelSelection}
4\alias{modelSelection}
5\title{Among a collection of models, this function constructs a subcollection of models with
6models having strictly different dimensions, keeping the model which minimizes
7the likelihood if there were several with the same dimension}
8\usage{
9modelSelection(LLF)
10}
11\arguments{
12\item{LLF}{a matrix, the first column corresponds to likelihoods for several models
13the second column corresponds to the dimensions of the corresponding models.}
14}
15\value{
16a list with indices, a vector of indices selected models,
17 and D1, a vector of corresponding dimensions
18}
19\description{
20Among a collection of models, this function constructs a subcollection of models with
21models having strictly different dimensions, keeping the model which minimizes
22the likelihood if there were several with the same dimension
23}
24

format: plain


text:
# Combination of dielectrics in capacitor

For figure 2 in the above diagram, I derived the following equation:

$$\\E_{o}= E_{1} + E_{2}$$ Where $\\E_1$ and $\\E_2$ are the net electric fields or dielectric 1 and 2, respectively (i.e., $\\E_{1/2}= E - E_p$) And since the net electric field is positive in both cases, the above equation should be correct.

Taking $\\E_1 = \sigma/(\epsilon*k_1)$ and $\\E_2=\sigma/(\epsilon*k_2)$, I'm getting the final equation $$\\C_o*k_1 k_2/ (k_1+k_2)$$

But the answer key says it's $$\\C_o*k_1 k_2/ (k_1-k_2)$$ instead. I don't understand how this is happening. Can someone help, please?

• Well,what's $E_p$ and $E_{1/2}$...? – Nehal Samee Feb 24 '18 at 16:40
• I think your answer is the correct one. Particularly, if you consider the case where $k_1=k_2$, the formula of the answer key blows up; which should not happen. – Sahand Tabatabaei Feb 24 '18 at 16:42
• @NehalSamee $\\E_p$ is the internal E of the dielectric due to polarization and $\\E_1/_2$ is $\\E_1$ or $\\E_2$, respectively – Antara Kulkarni Feb 24 '18 at 16:57
• How do you transit from “E” to “$C_0$”...? – Nehal Samee Feb 24 '18 at 17:00
• I hope you are correct $V=E_1 d + E_2 d$. $$V=\frac {E_0}{k_1} d+ \frac {E_0}{k_2}d$$ ... This leads to your answer ... – Nehal Samee Feb 24 '18 at 17:02


text:
Multiple Conditioning on Event Probabilities

I am trying to understand what's wrong with the following logic related to "multiple conditioning." Why is the probability of [(A given B) given C] not the same as the probability of [A given (B and C)] ? I know it's not true, but only because numbers disagree. I am having a hard time parsing what's wrong with the logic.

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Both are the same thing. What numbers disagree? –  leonbloy Dec 18 '11 at 0:32
There is no such thing as [A given B]. There is such a thing as the probability given B, of A. –  Michael Hardy Dec 19 '11 at 22:53

4 Answers

Peter Milne's paper "Bruno de Finetti and the Logic of Conditional Events" is an account of attempts to base probability theory upon conditional events, with special regard to de Finetti's subjectivistic "coherent betting behavior" approach using a three-valued logic. The latter approach results in a conditional event algebra with the property that both $$(A|B)|C \ \equiv \ A|(B \And C)$$ and $$A|(B|C) \equiv A|(B \And C)$$ where $A, B, C$ are ordinary propositions, implying that the respective probabilities must be equal. (This is contrary to your expectation that they should differ. See p. 218 of the cited paper, where what you've called "multiple conditioning" is called "iterated conditioning".)

NB: In de Finetti's approach ...

To introduce the notion of conditional probability means extending the definition of $P(X)$ from the field of ordinary events $X$ to the field of conditional events.

This, of course, differs from standard presentations of probability theory, in which there are no such objects as "conditional events".

De Finetti's ideas has been rediscovered and developed in the late 80' (Goodman I. R., Nguyen, H. T. and E. A, Walker Conditional Inference and Logic for Intelligent systems, North Holland 1991). More recently, A. Mura presented a modified version of de Finetti's logic, equipped with a semantics that fits and generalizes Adam's probabilistic logic (see Adams E. W., The Logic of Conditionals, Reidel, 1975; Mura A. 'Probability and the Logic of de Finetti's Trievents', in Galavotti M. C. (ed.) De Finetti Radical Probabilist, College Publications, 2008, pp. 201-42; Mura A. 'Towards a New Logic of Indicative Conditionals', L&PS – Logic and Philosophy of Science, 9, 2011, pp. 17-31).

-

There's no such thing as [A given B].

It is NOT "the probabilty of {A given B}".

Rather, it is "{the probability, given B} of A".

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What is wrong with saying "Given $C$, what is the probability of $A$ given $B$?" which becomes "Given $B$ and $C$, what is the the probability of $A$?" –  Henry Dec 18 '11 at 14:10
+1 A lot of confusion is caused by the common notation $P(A|B)$, which to a newcomer may wrongly suggest (in standard presentations) that there is some object $A|B$ (i.e. a conditional event). In this sense, $P_B(A)$ is a much better notation for what is intended. –  r.e.s. Dec 18 '11 at 15:25
@Henry : It seems as if you're responding to my comment under your own answer rather than to what I wrote in my answer. –  Michael Hardy Dec 19 '11 at 6:49

You cannot apply standard probability rules to conditional events of the form (A | B) except when A and B are ordinary two-valued events. Conditional events may be true, false or null (so the logic of conditional events is a three-valued logic). The calculus of probability for conditional events requires generalised formulas. The matter becomes very simple if two unary truth-functional connectives are introduced: T(p) and H(p). The first returns true if p is true and false if p is either false or null. The second returns true if p is either true or false and false if p is null. Every conditional event p may be written as (T(p) | H(p)) and its probability is given by the following general equation: Pr(p) = Pr(T(p))/Pr(H(p)). If p = B|C where B and C are ordinary events, the formula yields Pr(T(B|C))/Pr(H(B|C)). But, in this special case T(B|C) = (B & C) and H(B|C) = C, so that if B and C are ordinary events it holds: Pr(B | C) = Pr(B & C)/Pr(C).

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$\Pr(X | Y) = \dfrac{\Pr(X \cap Y)}{\Pr(Y)}$ so $\Pr((A|B)|C) = \dfrac{\Pr(A \cap B|C)}{\Pr(B|C)} =\dfrac{\frac{\Pr(A \cap B\cap C)}{\Pr(C)}}{\frac{\Pr(B\cap C)}{\Pr(C)}} = \dfrac{\Pr(A \cap B\cap C)}{\Pr(B\cap C)}=\Pr(A | B \cap C)$ so they are the same.

Here is a way to check your numbers. If you have

• $\Pr(A \cap B \cap C) = d/s$
• $\Pr(A \cap B \cap C^c) = e/s$
• $\Pr(A \cap B^c \cap C) = f/s$
• $\Pr(A \cap B^c \cap C^c) = g/s$
• $\Pr(A^c \cap B \cap C) = h/s$
• $\Pr(A^c \cap B \cap C^c) = i/s$
• $\Pr(A^c \cap B^c \cap C) = j/s$
• $\Pr(A^c \cap B^c \cap C^c) = k/s$

where $s=d+e+f+g+h+i+j+k$, then $\Pr(B \cap C) = \dfrac{d+h}{s}$ and $\Pr(A | B \cap C) = \dfrac{d}{d+h}$.

Given $C$, we just need to look at

• $\Pr(A \cap B | C) = d/t$
• $\Pr(A \cap B^c | C) = f/t$
• $\Pr(A^c \cap B | C) = h/t$
• $\Pr(A^c \cap B^c | C) = j/t$

where $t=d+f+h+j$ (and $\Pr(C)=t/s$), then $\Pr(B|C) = \dfrac{d+h}{t}$ and $\Pr((A|B)|C) = \dfrac{d}{d+h}$, the same as before.

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How do you get $\Pr((A|B)|C) = \dfrac{\Pr(A \cap B|C)}{\Pr(B|C)}$? The expression before "${}={}$" makes no sense. It's almost as if you're taking $A|B$ to mean $A\cap B$. –  Michael Hardy Dec 18 '11 at 2:09
@Michael: I applied the initial expression $\Pr(X | Y) = \dfrac{\Pr(X \cap Y)}{\Pr(Y)}$. $\Pr((A|B)|C)$ is in the question "probability of [(A given B) given C]". I realise you think it makes no sense, but I attempted to give it an interpretation, namely that in the universe in which $C$ is true what is the probability of $A$ given $B$. I also gave the questioner a method of checking whether "numbers disagree". If I had intended $A|B$ to mean $A \cap B$ then I would have had $\Pr((A|B)|C) = {\Pr(A \cap B|C)}$ but clearly I did not. –  Henry Dec 18 '11 at 9:12
Whatever you meant, it's not a standard notation. –  Michael Hardy Dec 19 '11 at 1:34


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Succeeding in First-Year Calculus: My guides to MAT135 and MAT137

Last year I wrote a guide for students taking the University of Toronto’s big (2000 student) first year calculus class MAT135. It was so successful that I wrote another guide to MAT137, the more specialized first year calculus class. Let me share them with you:

They are both links to Reddit, but you don’t need an account there to read them.

Delta-Epsilon Magic

2022 update: I’ve added a couple of sentences about the context of this event to emphasize how positive this experience was.

In my second year of undergrad I had a formative experience with Delta-Epsilon proofs that stuck with me for a long time. Last week I was able to provide a similar experience for some first year calculus students.

Van der Waerden’s Theorem is false for $latex omega_1$

One of my main research problems involves something I think is related to arithmetic progressions in $mathbb{Z}$. After learning a nice proof of Van der Waerden’s Theorem, and sharing it with colleagues, Daniel Soukup asked a nice question about Van der Waerden’s Theorem on $omega_1$. We answered it, and the example was sufficiently nice that I would like to share it.

Van der Waerden’s Theorem is false for $latex \omega_1$

One of my main research problems involves something I think is related to arithmetic progressions in $\mathbb{Z}$. After learning a nice proof of Van der Waerden’s Theorem, and sharing it with colleagues, Daniel Soukup asked a nice question about Van der Waerden’s Theorem on $\omega_1$. We answered it, and the example was sufficiently nice that I would like to share it.

Secret Santa 4: The Surprise

After a long hiatus (7 months!) I am finally back to writing. This week I revisited an old problem that Sam Coskey told me a couple of years ago. Some of you will remember this “Secret Santa problem”, which I wrote about before. The problem is:

Sam’s Problem. Is it possible for two people to each choose a natural number so that both numbers are exactly 1 apart and neither person knows who has the larger of the two numbers?

Math Anxiety

A study was done at the University of Chicago looking at whether math can cause physical pain. (The rather laughable title of the article is “Doing Math Really Does Make Your Head Hurt, Says Science”. Here‘s the actual published study.)

The idea is that for people with high levels of math anxiety, just knowing that they are about to do a math problem is enough to make the pain part of their brain go off. The actual solving of the problem wasn’t such a big deal. This suggests that presenting math is an important part of teaching it and discussing it.

Some other anecdotal examples are that people are notoriously good at calculating retail discounts – “It will be 12.50 for that 25 dollar shirt that is half off”. Compare this to the more difficult “What is 25 divided by 2?” or even worse “What is 25 x 0.5?”.

Similarly compare these two questions:

1. After the half-hour 6 o’clock news you watch an episode of Jeopardy, then Survivor, then two episodes of Wheel of Fortune. You read until it’s a quarter to 10 and then go to bed. How long did you read before you went to bed?
2. Here’s an easy question about fractions: Solve 6 + 1/2 + 1/2 + 1 + 2 (1/2) + R = 10 – 1/4.

I got a little bit anxious looking at that second question.

What experiences do you have about teaching people with math-anxiety?

“When someone asks you what you do as a mathematician, they take the most advanced training they have in mathematics and assume that you do a much harder version of that.”

This is a paraphrase of something I will attribute to Leo Goldmakher.

The idea is that if someone has only ever taken high-school math, they might think that mathematicians spend their time multiplying and dividing very large numbers; (“Are you even allowed to use calculators in your work?“). If someone has been lucky enough to take a first-year calculus course they might imagine that mathematicians solve harder and harder related rates questions. Going a step further, telling a mathematician that I study set theory often invokes brow-furrowing and questions like: “Don’t we already know everything about ordinals and cardinal arithmetic?“.

Is Algebra Necessary? – Andrew Hacker

Today on The Sunday Edition, on CBC Radio 1, host Michael Enright interviewed Andrew Hacker, about his controversial views on mathematics education. Basically, he thinks that demanding that all high school students master Algebra is “an onerous stumbling block for all kinds of students”. Listening to the interview this morning made me angry. I don’t get riled up easily, but my blood was boiling this morning.

Here is the link to the radio interview. (It starts around the 6:30 mark and is about 24 minutes long.) Here is the article Andrew Hacker wrote for the New York Times.

I intend to write a detailed response to the claims of Andrew Hacker later this week, but first let me share some reckless assertions that pissed me off. (These are paraphrases.)

“Should we be learning mathematics for learning’s sake? Don’t get me wrong, I’m a liberal-arts professor; I teach multi-disciplinary courses. However, I think that we should be focusing on learning about mathematics: the history of mathematics, the philosophy of mathematics. Not the dry stuff.”

“Hacker: 1.7 million Americans enter American universities every year, all required to have some high school mathematics. How many do you think choose to major in mathematics?

Micheal Enright: I don’t even want to guess.

Hacker: 15 Thousand. This tells me that even though teachers have had the chance to show the students the ‘beauty of mathematics’ they just aren’t getting it. [Goes on to imply that it’s because this is too hard to do.] Why are we training all of these students in high school when so few are going into STEM [science, technology engineering and mechanics] careers?”

“Do people need to learn about Fermat’s Theorem [sic] or Rye-man’s Hypothesis [sic]?”

I agree that something is broken with the way we learn, teach and perceive mathematics, especially in middle years and high school. This is a conversation we must have. However, it is troubling when the social scientists try to use their irrelevant authority to persuade the masses, leveraging the common man’s bias against mathematics.

I’m mad. I don’t like when people piss on what I love, and I especially dislike when they do it while tricking the public. As people we need emotions: love, anger, joy, contempt; but as a people we need thought, analysis, reason and mathematics to universally prevail.

Again, I plan on writing a cool-headed, but strong reply, and this is just me venting. Until then, let me leave you with this, taken from JFK’s famous speech:

“But why, some say, the moon? Why choose this as our goal? And they may well ask why climb the highest mountain? Why, 35 years ago, fly the Atlantic? Why does Rice play Texas? We choose to go to the moon. We choose to go to the moon in this decade and do the other things, not because they are easy, but because they are hard, because that goal will serve to organize and measure the best of our energies and skills, because that challenge is one that we are willing to accept, one we are unwilling to postpone, and one which we intend to win, and the others, too.”

Every Day I’m Simulating

Have you ever tried to explain/inflict forcing on your non-set theory friends? Let me tell you it is hard. In my ongoing effort to try to explain everything to everyone here is my attempt at explaining the idea of forcing, without explaining forcing.

First, I direct you to this post which explains the simulation argument.

tl; dr – Imagine that our civilization could simulate (artificially) intelligent civilizations. They wouldn’t know that they are in a simulation and could also run their own simulations.

Reading this post reminded me about forcing. Here is my reply to the OP (who happens to be my brother-in-law), and I would appreciate feedback on this so that I can refine my analogy:

Cool. This illustrates a key observation that underlies a lot of the mathematics (set theory) that I do. It goes like this for the interested parties:

We run a simulation as you’ve described, but we make sure that the simulation only has “a small number of things in it”. Perhaps we have some sort of minimal simulation, like we don’t include the letter Z in their languages or something (Call this SIM1). Now we check that the simulation can come up with its own simulations. It can? Great! So now we know that “having the letter Z in your language” is not a requirement for coming up with simulations. Or we could add a whole bunch of new crazy letters (in say SIM2) and see if they can still run simulations. Lets say they can still come up with simulations. Continue reading Every Day I’m Simulating


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# [OS X TeX] OT: Film wheel font

Simon Spiegel simon at simifilm.ch
Wed Aug 24 06:43:56 EDT 2005

On 24.08.2005, at 12:10, Bruno Voisin wrote:

> Le 24 août 05 à 11:47, Simon Spiegel a écrit :
>
>
>> Thanks, the Webdings font is already quite close. A related
>> question: What's the easiest way to get a single glyph of a font
>> into a LaTeX document? Of course with XeLaTeX I could easily
>> include the font, but I'm not sure yet if my final output will be
>> XeLaTeX. Now if possible, I want to avoid all the hassle of font
>> converting etc. just to get this one glyph.
>>
>
> Herb Schulz did similar work for including Mac OS symbols (such as
> the Apple key symbol) in TeX documents, so there should be a way.
> See his applekeys package <http://homepage.mac.com/herbs2/.Public/
> applekeys.zip>.
>
>
>> What I want to do: I write my PhD in film studies. There are
>> screenshots, and I plan to create a CD that comes with the printed
>> text which includes short clip. The film wheel (or projector)
>> symbol is to indicate that a whole clip exist on the CD for the
>> screenshot printed. So I need this glyph in the caption.
>>
>> I thought I could avoid setting up another font by creating a
>> small PDF which just includes this one glyph. This is a bit of a
>> hack, and unfortunately, LaTeX doesn't seem happy about
>> \includegraphics inside \caption I get:
>> ! Argument of \Hy at tempa has an extra }.
>> <inserted text>
>>                 \par
>>
>> The PDF gets created the way I want, but I have the errors, and
>> scaling the PDF doesn't work at all.
>>
>
> It looks like you're using the hyperref package, and the error
> message relates to a version of \caption modified by this package.
> The hyperref package is very useful, but I've found extremely
> difficult to understand/debug/customize the code of its versions of
> standard LaTeX commands.
>
> That said, the above message might as well be caused by a lack of
> \protect'ion. Have you attempted to insert a \protect before the
> \includegraphics? I remember a time when many LaTeX commands (such
> as \ref) needed to be \protect'ed when put inside a \caption,
> though things have got much better.
>

Bruno, you're the best! \protect does indeed help and the applekeys
package was very illuminating. I now can easily include the font. Now
I just need a font that actually looks the way I want, the projector
font is bit too delicate for small sizes.

Thank you very much!

simon
--
Simon Spiegel
Mutschellenstr. 97
8038 Zürich

Telephon: ++41 43 535 81 71
Mobophon: ++41 76 459 60 39

http://www.simifilm.ch

"I'm not that easy to impress. – O, a blue car." Homer Simpson

--------------------- Info ---------------------
Mac-TeX Website: http://www.esm.psu.edu/mac-tex/
& FAQ: http://latex.yauh.de/faq/
TeX FAQ: http://www.tex.ac.uk/faq
List Post: <mailto:MacOSX-TeX at email.esm.psu.edu>


text:
# How do you write the electron configuration for Ni^(+2)?

Oct 11, 2016

$1 {s}^{2} 2 {s}^{22} {p}^{6} 3 {s}^{2} 3 {p}^{6} 4 {s}^{2} 3 {d}^{6}$ in the ground state
More correctly the electrons lost are the $4 {s}^{2}$leaving
$1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6} 3 {d}^{8}$

#### Explanation:

Nickel $N {i}_{28}$ has 28 electrons. 18 electrons fill up the third electron shell leaving 10 valance electrons. 2 electrons in the 4s and 8 elections in the 3d.

When Nickel becomes $N {i}^{+} 2$ Nickel has lost 2 electrons leaving the atom with only 8 valance electrons.

The 4s electrons a lower energy level that the 3d electrons because of the simpler electron path of the S orbital. So in the ground state the electrons being lost should be the 3d electrons.

However the 4s electrons are further from the nucleus so losing the the 2 4s electrons leaves only the third shell electrons making the atoms more stable than losing the 3d electrons.

So the somewhat stable electron configuration of $N {i}^{+} 2$ is
$1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6} 3 {d}^{8}$

Oct 11, 2016

$1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6} 3 {d}^{8}$ or you can shorten it to $\left[A r\right] 3 {d}^{8}$

#### Explanation:

Electron configuration for Ni is $1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 4 {s}^{2} 3 {d}^{8}$.

$N {i}^{2 +}$ has two electrons less than Ni ( that is why $N {i}^{2 +}$ is positively charged).

So when writing electron configuration for $N {i}^{2 +}$ we exclude last two electrones from the last shell of Ni electron confgiruation.

I suggest to watch Tyler DeWitt's YouTube video on electron configuration. It helped me a lot.


text:
Integer type:  int32  int64  nag_int  show int32  show int32  show int64  show int64  show nag_int  show nag_int

Chapter Contents
Chapter Introduction
NAG Toolbox

# NAG Toolbox: nag_lapack_dopmtr (f08gg)

## Purpose

nag_lapack_dopmtr (f08gg) multiplies an arbitrary real matrix C$C$ by the real orthogonal matrix Q$Q$ which was determined by nag_lapack_dsptrd (f08ge) when reducing a real symmetric matrix to tridiagonal form.

## Syntax

[ap, c, info] = f08gg(side, uplo, trans, ap, tau, c, 'm', m, 'n', n)
[ap, c, info] = nag_lapack_dopmtr(side, uplo, trans, ap, tau, c, 'm', m, 'n', n)

## Description

nag_lapack_dopmtr (f08gg) is intended to be used after a call to nag_lapack_dsptrd (f08ge), which reduces a real symmetric matrix A$A$ to symmetric tridiagonal form T$T$ by an orthogonal similarity transformation: A = QTQT$A=QT{Q}^{\mathrm{T}}$. nag_lapack_dsptrd (f08ge) represents the orthogonal matrix Q$Q$ as a product of elementary reflectors.
This function may be used to form one of the matrix products
 QC , QTC , CQ  or  CQT , $QC , QTC , CQ  or  CQT ,$
overwriting the result on C$C$ (which may be any real rectangular matrix).
A common application of this function is to transform a matrix Z$Z$ of eigenvectors of T$T$ to the matrix QZ$QZ$ of eigenvectors of A$A$.

## References

Golub G H and Van Loan C F (1996) Matrix Computations (3rd Edition) Johns Hopkins University Press, Baltimore

## Parameters

### Compulsory Input Parameters

1:     side – string (length ≥ 1)
Indicates how Q$Q$ or QT${Q}^{\mathrm{T}}$ is to be applied to C$C$.
side = 'L'${\mathbf{side}}=\text{'L'}$
Q$Q$ or QT${Q}^{\mathrm{T}}$ is applied to C$C$ from the left.
side = 'R'${\mathbf{side}}=\text{'R'}$
Q$Q$ or QT${Q}^{\mathrm{T}}$ is applied to C$C$ from the right.
Constraint: side = 'L'${\mathbf{side}}=\text{'L'}$ or 'R'$\text{'R'}$.
2:     uplo – string (length ≥ 1)
This must be the same parameter uplo as supplied to nag_lapack_dsptrd (f08ge).
Constraint: uplo = 'U'${\mathbf{uplo}}=\text{'U'}$ or 'L'$\text{'L'}$.
3:     trans – string (length ≥ 1)
Indicates whether Q$Q$ or QT${Q}^{\mathrm{T}}$ is to be applied to C$C$.
trans = 'N'${\mathbf{trans}}=\text{'N'}$
Q$Q$ is applied to C$C$.
trans = 'T'${\mathbf{trans}}=\text{'T'}$
QT${Q}^{\mathrm{T}}$ is applied to C$C$.
Constraint: trans = 'N'${\mathbf{trans}}=\text{'N'}$ or 'T'$\text{'T'}$.
4:     ap( : $:$) – double array
Note: the dimension of the array ap must be at least max (1, m × (m + 1) / 2 ) $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{m}}×\left({\mathbf{m}}+1\right)/2\right)$ if side = 'L'${\mathbf{side}}=\text{'L'}$ and at least max (1, n × (n + 1) / 2 ) $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}×\left({\mathbf{n}}+1\right)/2\right)$ if side = 'R'${\mathbf{side}}=\text{'R'}$.
Details of the vectors which define the elementary reflectors, as returned by nag_lapack_dsptrd (f08ge).
5:     tau( : $:$) – double array
Note: the dimension of the array tau must be at least max (1,m1)$\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{m}}-1\right)$ if side = 'L'${\mathbf{side}}=\text{'L'}$ and at least max (1,n1)$\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}-1\right)$ if side = 'R'${\mathbf{side}}=\text{'R'}$.
Further details of the elementary reflectors, as returned by nag_lapack_dsptrd (f08ge).
6:     c(ldc, : $:$) – double array
The first dimension of the array c must be at least max (1,m)$\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{m}}\right)$
The second dimension of the array must be at least max (1,n)$\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$
The m$m$ by n$n$ matrix C$C$.

### Optional Input Parameters

1:     m – int64int32nag_int scalar
Default: The first dimension of the array c.
m$m$, the number of rows of the matrix C$C$; m$m$ is also the order of Q$Q$ if side = 'L'${\mathbf{side}}=\text{'L'}$.
Constraint: m0${\mathbf{m}}\ge 0$.
2:     n – int64int32nag_int scalar
Default: The second dimension of the array c.
n$n$, the number of columns of the matrix C$C$; n$n$ is also the order of Q$Q$ if side = 'R'${\mathbf{side}}=\text{'R'}$.
Constraint: n0${\mathbf{n}}\ge 0$.

ldc work

### Output Parameters

1:     ap( : $:$) – double array
Note: the dimension of the array ap must be at least max (1, m × (m + 1) / 2 ) $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{m}}×\left({\mathbf{m}}+1\right)/2\right)$ if side = 'L'${\mathbf{side}}=\text{'L'}$ and at least max (1, n × (n + 1) / 2 ) $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}×\left({\mathbf{n}}+1\right)/2\right)$ if side = 'R'${\mathbf{side}}=\text{'R'}$.
Is used as internal workspace prior to being restored and hence is unchanged.
2:     c(ldc, : $:$) – double array
The first dimension of the array c will be max (1,m)$\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{m}}\right)$
The second dimension of the array will be max (1,n)$\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$
ldcmax (1,m)$\mathit{ldc}\ge \mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{m}}\right)$.
c stores QC$QC$ or QTC${Q}^{\mathrm{T}}C$ or CQ$CQ$ or CQT$C{Q}^{\mathrm{T}}$ as specified by side and trans.
3:     info – int64int32nag_int scalar
info = 0${\mathbf{info}}=0$ unless the function detects an error (see Section [Error Indicators and Warnings]).

## Error Indicators and Warnings

info = i${\mathbf{info}}=-i$
If info = i${\mathbf{info}}=-i$, parameter i$i$ had an illegal value on entry. The parameters are numbered as follows:
1: side, 2: uplo, 3: trans, 4: m, 5: n, 6: ap, 7: tau, 8: c, 9: ldc, 10: work, 11: info.
It is possible that info refers to a parameter that is omitted from the MATLAB interface. This usually indicates that an error in one of the other input parameters has caused an incorrect value to be inferred.

## Accuracy

The computed result differs from the exact result by a matrix E$E$ such that
 ‖E‖2 = O(ε) ‖C‖2 , $‖E‖2 = O(ε) ‖C‖2 ,$
where ε$\epsilon$ is the machine precision.

The total number of floating point operations is approximately 2m2n$2{m}^{2}n$ if side = 'L'${\mathbf{side}}=\text{'L'}$ and 2mn2$2m{n}^{2}$ if side = 'R'${\mathbf{side}}=\text{'R'}$.
The complex analogue of this function is nag_lapack_zupmtr (f08gu).

## Example

```function nag_lapack_dopmtr_example
side = 'Left';
uplo = 'L';
trans = 'No transpose';
ap = [2.07;
-5.825753170191817;
0.4331793442217867;
-0.1186086299654892;
1.474093708197552;
2.624045178795586;
0.8062881532775791;
-0.6491595075457843;
0.9162727563219193;
-1.694934200651768];
tau = [1.664291789738249;
1.212047324162142;
0];
c = [0.5657591788223874, -0.2328424308031574;
0.6869179572505918, -0.1626170961491636;
-0.4395889372131648, -0.3017273343882724;
0.1217449705930083, 0.9101102670229791];
[apOut, cOut, info] = nag_lapack_dopmtr(side, uplo, trans, ap, tau, c)
```
```

apOut =

2.0700
-5.8258
0.4332
-0.1186
1.4741
2.6240
0.8063
-0.6492
0.9163
-1.6949

cOut =

0.5658   -0.2328
-0.3478    0.7994
-0.4740   -0.4087
0.5781    0.3737

info =

0

```
```function f08gg_example
side = 'Left';
uplo = 'L';
trans = 'No transpose';
ap = [2.07;
-5.825753170191817;
0.4331793442217867;
-0.1186086299654892;
1.474093708197552;
2.624045178795586;
0.8062881532775791;
-0.6491595075457843;
0.9162727563219193;
-1.694934200651768];
tau = [1.664291789738249;
1.212047324162142;
0];
c = [0.5657591788223874, -0.2328424308031574;
0.6869179572505918, -0.1626170961491636;
-0.4395889372131648, -0.3017273343882724;
0.1217449705930083, 0.9101102670229791];
[apOut, cOut, info] = f08gg(side, uplo, trans, ap, tau, c)
```
```

apOut =

2.0700
-5.8258
0.4332
-0.1186
1.4741
2.6240
0.8063
-0.6492
0.9163
-1.6949

cOut =

0.5658   -0.2328
-0.3478    0.7994
-0.4740   -0.4087
0.5781    0.3737

info =

0

```


text:
# Enneper surface

Figure 1. A portion of the Enneper surface

In mathematics, in the fields of differential geometry and algebraic geometry, the Enneper surface is a self-intersecting surface that can be described parametrically by:

${\displaystyle x=u(1-u^{2}/3+v^{2})/3,\ }$
${\displaystyle y=-v(1-v^{2}/3+u^{2})/3,\ }$
${\displaystyle z=(u^{2}-v^{2})/3.\ }$

It was introduced by Alfred Enneper 1864 in connection with minimal surface theory.[1][2][3][4]

The Weierstrass–Enneper parameterization is very simple, ${\displaystyle f(z)=1,g(z)=z}$, and the real parametric form can easily be calculated from it. The surface is conjugate to itself.

Implicitization methods of algebraic geometry can be used to find out that the points in the Enneper surface given above satisfy the degree-9 polynomial equation

${\displaystyle 64z^{9}-128z^{7}+64z^{5}-702x^{2}y^{2}z^{3}-18x^{2}y^{2}z+144(y^{2}z^{6}-x^{2}z^{6})\ }$
${\displaystyle {}+162(y^{4}z^{2}-x^{4}z^{2})+27(y^{6}-x^{6})+9(x^{4}z+y^{4}z)+48(x^{2}z^{3}+y^{2}z^{3})\ }$
${\displaystyle {}-432(x^{2}z^{5}+y^{2}z^{5})+81(x^{4}y^{2}-x^{2}y^{4})+240(y^{2}z^{4}-x^{2}z^{4})-135(x^{4}z^{3}+y^{4}z^{3})=0.\ }$

Dually, the tangent plane at the point with given parameters is ${\displaystyle a+bx+cy+dz=0,\ }$ where

${\displaystyle a=-(u^{2}-v^{2})(1+u^{2}/3+v^{2}/3),\ }$
${\displaystyle b=6u,\ }$
${\displaystyle c=6v,\ }$
${\displaystyle d=-3(1-u^{2}-v^{2}).\ }$

Its coefficients satisfy the implicit degree-6 polynomial equation

${\displaystyle 162a^{2}b^{2}c^{2}+6b^{2}c^{2}d^{2}-4(b^{6}+c^{6})+54(ab^{4}d-ac^{4}d)+81(a^{2}b^{4}+a^{2}c^{4})\ }$
${\displaystyle {}+4(b^{4}c^{2}+b^{2}c^{4})-3(b^{4}d^{2}+c^{4}d^{2})+36(ab^{2}d^{3}-ac^{2}d^{3})=0.\ }$

The Jacobian, Gaussian curvature and mean curvature are

${\displaystyle J=(1+u^{2}+v^{2})^{4}/81,\ }$
${\displaystyle K=-(4/9)/J,\ }$
${\displaystyle H=0.\ }$

The total curvature is ${\displaystyle -4\pi }$. Osserman proved that a complete minimal surface in ${\displaystyle \mathbb {R} ^{3}}$ with total curvature ${\displaystyle -4\pi }$ is either the catenoid or the Enneper surface.[5]

Another property is that all bicubical minimal Bézier surfaces are, up to an affine transformation, pieces of the surface.[6]

It can be generalized to higher order rotational symmetries by using the Weierstrass–Enneper parameterization ${\displaystyle f(z)=1,g(z)=z^{k}}$ for integer k>1.[3] It can also be generalized to higher dimensions; Enneper-like surfaces are known to exist in ${\displaystyle \mathbb {R} ^{n}}$ for n up to 7.[7]

## References

1. ^ J.C.C. Nitsche, "Vorlesungen über Minimalflächen" , Springer (1975)
2. ^ Francisco J. López, Francisco Martín, Complete minimal surfaces in R3
3. ^ a b Ulrich Dierkes, Stefan Hildebrandt, Friedrich Sauvigny (2010). Minimal Surfaces. Berlin Heidelberg: Springer. ISBN 978-3-642-11697-1.
4. ^
5. ^ R. Osserman, A survey of Minimal Surfaces. Vol. 1, Cambridge Univ. Press, New York (1989).
6. ^ Cosín, C., Monterde, Bézier surfaces of minimal area. In Computational Science — ICCS 2002, eds. J., Sloot, Peter, Hoekstra, Alfons, Tan, C., Dongarra, Jack. Lecture Notes in Computer Science 2330, Springer Berlin / Heidelberg, 2002. pp. 72-81 ISBN 978-3-540-43593-8
7. ^ Jaigyoung Choe, On the existence of higher dimensional Enneper's surface, Commentarii Mathematici Helvetici 1996, Volume 71, Issue 1, pp 556-569


text:
Enter the total number of sales missed due to understocking and the average cost per sale ($) into the Cost of Understocking Calculator. The calculator will evaluate the Cost of Understocking. ## Cost of Understocking Formula The following two example problems outline the steps and information needed to calculate the Cost of Understocking. CUS = SM * ACPS Variables: • CUS is the Cost of Understocking ($)
• #SM is the total number of sales missed due to understocking
• ACPS is the average cost per sale ($) To calculate the cost of understocking, multiply the number of sales missed due to understocking by the average cost per sale. ## How to Calculate Cost of Understocking? The following steps outline how to calculate the Cost of Understocking. 1. First, determine the total number of sales missed due to understocking. 2. Next, determine the average cost per sale ($).
3. Next, gather the formula from above = CUS = #SM * ACPS.
4. Finally, calculate the Cost of Understocking.
5. After inserting the variables and calculating the result, check your answer with the calculator above.

Example Problem :

Use the following variables as an example problem to test your knowledge.

total number of sales missed due to understocking = 500

average cost per sale (\$) = 30

CUS = #SM * ACPS = ?


text:
# Function that returns another function, for a programming task

This is a task I was given for an online course. This is my solution to the problem.

Task: Write a function called general_poly, that meets the specifications below.

For example, general_poly([1, 2, 3, 4])(10) should evaluate to 1234 because

1*10^3 + 2*10^2 + 3*10^1 + 4*10^0

So in the example the function only takes one argument with general_poly([1, 2, 3, 4]) and it returns a function that you can apply to a value, in this case x = 10 with general_poly([1, 2, 3, 4])(10).

def general_poly (L):
""" L, a list of numbers (n0, n1, n2, ... nk)
Returns a function, which when applied to a value x, returns the value
n0 * x^k + n1 * x^(k-1) + ... nk * x^0 """

My Code:

def general_poly (L):
""" L, a list of numbers (n0, n1, n2, ... nk)
Returns a function, which when applied to a value x, returns the value
n0 * x^k + n1 * x^(k-1) + ... nk * x^0 """

def function_generator(L, x):
k = len(L) - 1
sum = 0
for number in L:
sum += number * x ** k
k -= 1
return sum

def function(x, l=L):
return function_generator(l, x)

return function

• Why do you use L[i] instead of number? – Arnial Oct 30 '16 at 12:24
• wow! Facepalm haha i was stressed when i wrote that, thank you very much lol! – ChrisIkeokwu Oct 30 '16 at 12:26

1. The name function_generator could be improved. What this function does is to evaluate the polynomial at the given value x. So a name like evaluate would be better.

2. The body of function_generator can be simplified. First, it's simpler to iterate over L in reversed order, because then you can start with k = 0 rather than k = len(L) - 1:

def evaluate(L, x):
k = 0
sum = 0
for number in reversed(L):
sum += number * x ** k
k += 1
return sum

Now that k goes upwards, you can use enumerate to generate the values for k:

def evaluate(L, x):
sum = 0
for k, a in enumerate(reversed(L)):
sum += a * x ** k
return sum

And now that the body of the loop is a single expression, you can use the built-in sum function to do the addition:

def evaluate(L, x):
return sum(a * x ** k for k, a in enumerate(reversed(L)))

Another approach to polynomial evaluation is to generate the power series in $x$ separately:

def power_series(x):
"Generate power series 1, x, x**2, ..."
y = 1
while True:
yield y
y *= x

and then combine it with the coefficients using zip:

def evaluate(L, x):
"Evaluate the polynomial with coefficients in L at the point x."
return sum(a * y for a, y in zip(reversed(L), power_series(x)))

This is more efficient for large polynomials because each step in power_series is a multiplication by x, which is cheaper to compute than the exponentiation x ** k.

Finally, you could take advantage of this transformation: $$a_k x^k + a_{k-1} x^{k-1} + \dots + a_1x + a_0 = (\cdots ((0 + a_k)x + a_{k-1})x + \cdots + a_1)x + a_0$$ and write:

def evaluate(L, x):
"Evaluate the polynomial with coefficients in L at the point x."
r = 0
for a in L:
r = r * x + a
return r

(This is Horner's method.)

3. There's no need for function, because it's fine for the body of function_generator to refer to nonlocal variables like L. So you can write:

def general_poly(L):
"Return function evaluating the polynomial with coefficients in L."
def evaluate(x):
r = 0
for a in L:
r = r * x + a
return r
return evaluate

4. An alternative approach is to use functools.partial:

from functools import partial

def evaluate(L, x):
"Evaluate the polynomial with coefficients in L at the point x."
r = 0
for a in L:
r = r * x + a
return r

def general_poly(L):
"Return function evaluating the polynomial with coefficients in L."
return partial(evaluate, L)

This would be useful if you sometimes wanted to call evaluate directly (instead of via general_poly).


text:
Binary cartesian set product

Let $X$ and $Y$ each be a D11: Set.
The binary cartesian product of $X$ with $Y$ is the D11: Set $$X \times Y : = \{ (x, y) : x \in X \text{ and } y \in Y \}$$
Child definitions
Results


text:
# Paragraph Part 1: Interest and Inventory Costs Instructions: Calculate the following HNL finance scenarios. 1. On...

###### Question:

Paragraph Part 1: Interest and Inventory Costs Instructions: Calculate the following HNL finance scenarios. 1. On January 3, 2016. HNL negotiated a short-term loan of $685,000. The loan was due October 1 2016, and carried a 6.86% interest rate. a What is the total amount HNL paid on the maturity date? Use ordinary interest to calculate the interest. Show your work in the space provided. b. In a few complete sentences, explain how you calculated the answer. 2. On January 1, HNL had an inventory costing$95.000. During January, HNL had net purchases of 5118,900. Over recent years. HNL's gross profit in January has averaged 45% on sales. The company's net sales in January were \$210,800. a. What is the estimated cost of ending inventory using the gross profit method? Show your work in the space provided. b. In a few complete sentences, explain how you calculated the answer

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text:
Reward book (Combat Academy) can no longer be obtained in RuneScape after an update but still exists in-game for those who had done so.

The reward book is a reward given to players who tested the Combat Academy. After finishing a section in the combat academy, the player can open the course interface and right-click on the rank and select Claim-Reward to claim the book. If all 21 sections were completed, 21 books will be given. They can also be claimed from Lady Deathknell.

The books can give experience in one of any of the following skills: Attack, Strength, Defence, Constitution, Ranged, Magic, Prayer, Herblore, Summoning, Slayer

The experience ${\displaystyle x}$ obtained at level ${\displaystyle l}$ is:

${\displaystyle x=l^{2}-2l+100}$
``` template  = Template:Combat Academy calc
form      = ccr
result    = reward
param     = level|Skill level|1|int|1-99
param     = books|Amount of books|1|int|1-21
```


text:
# Is this a correct application of Rice-Shapiro theorem?

Let $$\langle M\rangle$$ be the encoding of a Turing machine as a string over $$\Sigma=\{0,1\}$$, and consider the language $$L=\{\langle M\rangle| \text{ M is a Turing machine that accepts a string of length 2014} \}$$

I want to use The Rice-McNaughton-Myhill-Shapiro Theorem (page 14 in this) to prove that $$L$$ is acceptable:

Say $$\mathcal{L}$$ is the set of all acceptable languages that contain strings of length 2014.

• This set is monotone: every superset of a $$L \in \mathcal{L}$$ also contains strings of length 2014 so belongs to $$\mathcal{L}$$

• This set is compact: for every language in $$\mathcal{L}$$ I can choose the finite subset with strings of length 2014.

• This set is finitely acceptable: this is the part I'm confused about. Given every finite language of $$\mathcal{L}$$ is there a TM that can accept this set of finite languages?

• You misread the last condition. It does not say what you wrote. – Andrej Bauer Jan 14 at 19:54
• the set of all finite languages of $\mathcal{L}$ should be acceptable. Is that correct? – sprajagopal Jan 14 at 20:10
• I guess the $\langle L \rangle$ is an encoding of the whole language itself. If each finite language of $\mathcal{L}$ is encoded like this, then set of encodings as a language should be acceptable – sprajagopal Jan 14 at 20:12
• @sprajagopal You're right. Also, if I'm not mistaken, I think the problem is with the first condition. I've added an answer. In general, I think the theorem is more appealing for proving undecidability results (or showing that a language is not acceptable). – Bader Abu Radi Jan 15 at 12:31

The application of the theorem is not correct. Note that $$\mathcal{L}$$ is a set of acceptable languages, and consider your first point. If you take an acceptable language $$L$$ that contains a string of length 2014, then it is not necessarily that every superset of $$L$$ is acceptable.
Regarding the last point, the answer is yes. Given an encoding $$\langle L\rangle$$ of a finite language $$L \in \mathcal{L}$$, it is easy to check whether $$L$$ contains a string of length $$2014$$. Indeed, we're assuming that the encoding encodes all the words of $$L$$ in some order, so you can simply traverse the encoding to look for a word of length 2014. Specifically, quoting from the link you attached:
If $$L$$ is a finite language over $$\Sigma$$, then you can assume that $$\langle L\rangle$$ is the unique string over the alphabet $$\Sigma \cup \{ \{,•,\}, \epsilon\}$$ that contains the strings in $$L$$ in lexicographic order, separated by dots • and surrounded by braces {}, with $$\epsilon$$ representing the empty string. For example, if $$\Sigma = \{ 0, 1\}$$, and $$L = \{\epsilon, 0,01,0110,01101001\}$$, then $$\langle L\rangle = \{ \epsilon•0•01•0110•01101001\}$$.
Finally, you can simply show that $$L$$ is acceptable directly by defining a TM that accepts it. Indeed, you can simply simulate the run of $$M$$ on all strings of length 2014 in parallel, and accepts iff $$M$$ accepts at least one of them.
• That last paragraph makes the original problem clear. Still wondering why every superset of $L$ is not in $\mathcal{L}$. Every superset of $L$, say $S$, will contain the strings of 2014 length. I thought: since $S$ contains length 2014 strings, it must be in $\mathcal{L}$ but I think by definition $\mathcal{L}$ only contains acceptable strings so I cannot assume $S$ is acceptable. But is there a concrete way to prove that $S$ is not acceptable? Only then can we conclude this $\mathcal{L}$ is not monotone. The whole thing seems circular and going round and round... – sprajagopal Jan 15 at 15:05
• You're right, in order to conclude that $S\in \mathcal{L}$, we need also to prove that $S$ is acceptable, for every suprset $S$. Unfortunately, this is far from being true. Consider for example the language $L = \{ 1^{2014}\}$, and the superset $L\cup A$, where $A$ is some non-acceptable problem (e.g., $A = \overline{Halt_{TM}} = \{ \langle M, w\rangle: \text{$M$does not halt on$w$}\}$) – Bader Abu Radi Jan 15 at 17:39


text:
# Radical expressions or absolute value?

Algebra Level 3

$$\sqrt{10-4 \sqrt{6}}$$ can also be represented as $$\text{__________}$$.

×


text:
# Measurement of the b-Hadron Production Cross Section Using Decays to mu^- D^0 X Final States in ppbar Collisions at sqrt s = 1.96 TeV

Abstract : We report a measurement of the production cross section for b hadrons in p-pbar collisions at sqrt{s}=1.96 TeV. Using a data sample derived from an integrated luminosity of 83 pb^-1 collected with the upgraded Collider Detector (CDF II) at the Fermilab Tevatron, we analyze b hadrons, H_b, partially reconstructed in the semileptonic decay mode H_b -> mu^- D^0 X. Our measurement of the inclusive production cross section for b hadrons with transverse momentum p_T > 9 GeV/c and rapidity |y|<0.6 is sigma = 1.30 mub +/- 0.05 mub(stat) +/- 0.14 mub (syst) +/- 0.07 mub (B), where the uncertainties are statistical, systematic, and from branching fractions respectively. The differential cross sections dsigma/dp_T are found to be in good agreement with recent measurements of the H_b cross section and well described by fixed-order next-to-leading logarithm predictions.
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Journal articles

http://hal.in2p3.fr/in2p3-00368655
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Submitted on : Tuesday, March 17, 2009 - 10:46:37 AM
Last modification on : Wednesday, October 27, 2021 - 2:00:14 PM

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### Citation

T. Aaltonen, J. Adelman, T. Akimoto, B. Alvarez Gonzalez, S. Amerioy, et al.. Measurement of the b-Hadron Production Cross Section Using Decays to mu^- D^0 X Final States in ppbar Collisions at sqrt s = 1.96 TeV. Physical Review D, American Physical Society, 2009, 79, pp.092003. ⟨10.1103/PhysRevD.79.092003⟩. ⟨in2p3-00368655⟩

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# Probability question involving two independent poisson random variables?

A restaurant has two waiters. suppose that the number of customers serviced daily by each waiter can be viewed as independent Poisson random variables with parameters $\lambda$ and $\mu$. whats the probability that the two waiters together handle less than $6$ customers during the day?

I was thinking along the lines of... $$P(X\leq 6,Y\leq 6) = P(X\leq6)\cdot P(Y\leq 6),$$ but for the probability mass function of Poisson, what would $\lambda$ be?? We are not given any data regarding the average number of customers served.

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$X \sim P(\lambda),\ Y \sim P(\mu)$. If X and Y are independent, then $X+Y \sim P(\lambda+\mu)$ –  Coiacy Feb 21 '13 at 11:12

Let's let $X$ be the number of customers served by the first waiter, and $Y$ the number served by the second waiter.
Note that $P(X \le 6, Y \le 6)$ is not what you want: it's the probability that each waiter serves at most 6 customers. But it includes the event that, say, the first waiter serves 5 and the second waiter serves 4, in which case the total is more than 6.
What you want is $P(X + Y \le 6)$. It's a fact, which you should be able to find in your textbook, that the sum of independent Poisson random variables is another Poisson random variable, whose mean is the sum of the means (here, $\lambda + \mu$).
So you are looking for the probability for the Poisson random variable $X+Y$, whose mean is known to you, to be less than or equal to 6. I expect you can take it from here.


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# Taylor–Green vortex

In fluid dynamics, the Taylor–Green vortex is an unsteady flow of a decaying vortex, which has an exact closed form solution of the incompressible Navier–Stokes equations in Cartesian coordinates. It is named after the British physicist and mathematician Geoffrey Ingram Taylor and his collaborator A. E. Green.[1]

Vector plot of the Taylor-Green Vortex

## Original work

In the original work of Taylor and Green,[1] a particular flow is analyzed in three spatial dimensions, with the three velocity components ${\displaystyle \mathbf {v} =(u,v,w)}$ at time ${\displaystyle t=0}$ specified by

${\displaystyle u=A\cos ax\sin by\sin cz,}$
${\displaystyle v=B\sin ax\cos by\sin cz,}$
${\displaystyle w=C\sin ax\sin by\cos cz.}$

The continuity equation ${\displaystyle \nabla \cdot \mathbf {v} =0}$ determines that ${\displaystyle Aa+Bb+Cc=0}$. The small time behavior of the flow is then found through simplification of the incompressible Navier–Stokes equations using the initial flow to give a step-by-step solution as time progresses.

An exact solution in two spatial dimensions is known, and is presented below.

## Incompressible Navier–Stokes equations

The incompressible Navier–Stokes equations in the absence of body force, and in two spatial dimensions, are given by

${\displaystyle {\frac {\partial u}{\partial x}}+{\frac {\partial v}{\partial y}}=0,}$
${\displaystyle {\frac {\partial u}{\partial t}}+u{\frac {\partial u}{\partial x}}+v{\frac {\partial u}{\partial y}}=-{\frac {1}{\rho }}{\frac {\partial p}{\partial x}}+\nu \left({\frac {\partial ^{2}u}{\partial x^{2}}}+{\frac {\partial ^{2}u}{\partial y^{2}}}\right),}$
${\displaystyle {\frac {\partial v}{\partial t}}+u{\frac {\partial v}{\partial x}}+v{\frac {\partial v}{\partial y}}=-{\frac {1}{\rho }}{\frac {\partial p}{\partial y}}+\nu \left({\frac {\partial ^{2}v}{\partial x^{2}}}+{\frac {\partial ^{2}v}{\partial y^{2}}}\right).}$

The first of the above equation represents the continuity equation and the other two represent the momentum equations.

## Taylor–Green vortex solution

In the domain ${\displaystyle 0\leq x,y\leq 2\pi }$, the solution is given by

${\displaystyle u=\cos x\sin y\,F(t),\qquad \qquad v=-\sin x\cos y\,F(t),}$

where ${\displaystyle F(t)=e^{-2\nu t}}$, ${\displaystyle \nu }$ being the kinematic viscosity of the fluid. Following the analysis of Taylor and Green[1] for the two-dimensional situation, and for ${\displaystyle A=a=b=1}$, gives agreement with this exact solution, if the exponential is expanded as a Taylor series, i.e. ${\displaystyle F(t)=1-2\nu t+O(t^{2})}$.

The pressure field ${\displaystyle p}$ can be obtained by substituting the velocity solution in the momentum equations and is given by

${\displaystyle p=-{\frac {\rho }{4}}\left(\cos 2x+\cos 2y\right)F^{2}(t).}$

The stream function of the Taylor–Green vortex solution, i.e. which satisfies ${\displaystyle \mathbf {v} =\nabla \times {\boldsymbol {\psi }}}$ for flow velocity ${\displaystyle \mathbf {v} }$, is

${\displaystyle {\boldsymbol {\psi }}=-\cos x\cos yF(t)\,{\hat {\mathbf {z} }}.}$

Similarly, the vorticity, which satisfies ${\displaystyle {\boldsymbol {\mathbf {\omega } }}=\nabla \times \mathbf {v} }$, is given by

${\displaystyle {\boldsymbol {\mathbf {\omega } }}=-2\cos x\cos y\,F(t){\hat {\mathbf {z} }}.}$

The Taylor–Green vortex solution may be used for testing and validation of temporal accuracy of Navier–Stokes algorithms.[2][3]

## References

1. ^ a b c Taylor, G. I. and Green, A. E., Mechanism of the Production of Small Eddies from Large Ones, Proc. R. Soc. Lond. A, 158, 499–521 (1937).
2. ^ Chorin, A. J., Numerical solution of the Navier–Stokes equations, Math. Comp., 22, 745–762 (1968).
3. ^ Kim, J. and Moin, P., Application of a fractional-step method to incompressible Navier–Stokes equations, J. Comput. Phys., 59, 308–323 (1985).


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## Tuesday, June 20, 2017

### About grants: What are "indirect costs"?

Before blogging further about science, I wanted to explain something about the way research grants work in the US.  Consider this part of my series of posts intended to educate students (and perhaps the public) about careers in academic research.

When you write a proposal to a would-be source of research funding, you have to include a budget.  As anyone would expect, that budget will list direct costs - these are items that are clear research expenses.  Examples would include, say, $30K/yr for a graduate student's stipend, and$7K for a piece of laboratory electronics essential to the work, and $2K/yr to support travel of the student and the principal investigator (PI) to conferences. However, budgets also include indirect costs, sometimes called overhead. The idea is that research involves certain costs that aren't easy to account for directly, like the electricity to run the lights and air conditioning in the lab, or the costs to keep the laboratory building maintained so that the research can get done, or the (meta)costs for the university to administer the grant. So, how does the university to figure out how much to tack on for indirect costs? For US federal grants, the magic (ahem) is all hidden away in OMB Circular A21 (wiki about it, pdf of the actual doc). Universities periodically go through an elaborate negotiation process with the federal government (see here for a description of this regarding MIT), and determine an indirect cost rate for that university. The idea is you take the a version of the direct costs ("modified total direct costs" - for example, a piece of equipment that costs more than$5K is considered a capital expense and not subject to indirect costs) and multiply by a negotiated factor (in the case of Rice right now, 56.5%) to arrive at the indirect costs.  The cost rates are lower for research done off campus (like at CERN), with the argument that this should be cheaper for the university.  (Effective indirect cost rates at US national labs tend to be much higher.)

Foundations and industry negotiate different rates with universities.  Foundations usually limit their indirect cost payments, arguing that they just can't afford to pay at the federal level.  The Bill and Melinda Gates Foundation, for example, only allows (pdf) 10% for indirect costs.   The effective indirect rate for a university, averaged over the whole research portfolio, is always quite a bit lower than the nominal A21 negotiated rate.  Vice provosts/presidents/chancellors for research at major US universities would be happy to explain at length that indirect cost recovery doesn't come close to covering the actual costs associated with doing university-based research.

Indirect cost rates in the US are fraught with controversy, particularly now.  The current system is definitely complicated, and reasonable people can ask whether it makes sense (and adds administrative costs) to have every university negotiate its own rate with the feds.   It remains to be seen whether there are changes in the offing.

## Saturday, June 17, 2017

Summer travel and other activities have slowed blogging, but I'll pick back up again soon.  In the meantime, here are a couple of interesting things to read:

• Ignition!  An Informal History of Liquid Rocket Propellants (pdf) is fascinating, if rather chemistry-heavy.  Come for discussions of subtle side reactions involved in red fuming nitric acid slowly eating its storage containers and suggested (then rejected) propellants like dimethyl mercury (!!), and stay for writing like, "Miraculously, nobody was killed, but there was one casualty — the man who had been steadying the cylinder when it split. He was found some five hundred feet away, where he had reached Mach 2 and was still picking up speed when he was stopped by a heart attack."  This is basically the story from start to finish (in practical terms) of the development of liquid propellants for rockets.   That book also led me to stumbling onto this library of works, most of which are waaaaay too chemistry-oriented for me.  Update:  for a directly relevant short story, see here.
• Optogenetics is the idea of using light to control and trigger the activation/inactivation of genes.  More recently, there is a big upswing in the idea of magnetogenetics, using magnetic fields to somehow do similar things.  One question at play is, what is the physical mechanism whereby magnetic fields can really do much at room temperature, since magnetic effects tend to be weak.  (Crudely speaking, the energy scale of visible photons is eV, much larger than the thermal energy scale of $k_{\mathrm{B}}T \sim ~$26 meV, and readily able to excite vibrations or drive electronic excitations.  However, one electron spin in a reasonably accessible magnetic field of 1 Tesla is $g \mu_{\mathrm{B}}B \sim ~$ 0.1 meV.)  Here is a nice survey article about the constraints on how magnetogenetics could operate.
• For a tutorial in how not to handle academic promotion cases, see here.

## Tuesday, June 06, 2017

### Follow-up: More 5nm/7nm/10nm transistors

Two years ago I wrote this post about IBM's announcement of "7 nm transistors", and it still gets many pageviews every week.   So, what's been going on since then?

I encourage you to read this article from Semiconductor Engineering - it's a nice, in-depth look at what the major manufacturers are doing to get very small transistors actually to market.  It also explains what those numerical designations of size really mean, and changes in lithography technology.  (This is one part of my book that I really will have to revise at some point.)

Likewise, here is a nice article about IBM's latest announcement of "5 nm" transistors.  The writing at Ars Technica on this topic is usually of high quality.  The magic words that IBM is now using are "gate all around", which means designing the device geometry so that the gate electrode, the one that actually controls the conduction, affects the channel (where the current flows) from all sides.  In old-school planar transistors, the gate only couples to the channel from one direction.

Later I will write a long, hopefully accessible article about transistors, as this year is the 70th anniversary of the Bell Labs invention that literally reshaped global technology.

## Thursday, June 01, 2017

### What should everyone know about physics?

A couple of weeks ago, Sean Carroll made an offhand remark (the best kind) on twitter about what every physics major should know.  That prompted a more thoughtful look at our expectations for physics majors by Chad Orzel, with which I broadly agree, as does ZapperZ, who points out that most physics majors don't actually go on to be physics PhDs (and that's fine, btw.)

Entertaining and thought-provoking as this was, it seems like it's worth having a discussion among practicing physicist popularizers about what we'd like everyone to know about physics.  (For the pedants in the audience, by "everyone" I'm not including very young children, and remember, this is aspirational.  It's what we'd like people to know, not what we actually expect people to know.)  I'm still thinking about this, but here are some basic ingredients that make the list, including some framing topics about science overall.
• Science is a reason- and logic-based way to look at the natural world; part of science is figuring out "models" (ways of describing the natural world and how it works) that have explanatory (retrodictive) and predictive power.
• Basic science is about figuring out how the world works - figuring out the "rules of the game".  Engineering is about using that knowledge to achieve some practical goal.  The line is very blurry; lots of scientists are motivated by and think about eventual applications.
• Different branches of science deal with different levels of complexity and have their own vocabularies.  When trying to answer a scientific question, it's important to use the appropriate level of complexity and the right vocabulary.  You wouldn't try to describe how a bicycle works by starting with the molecular composition of the grease on the chain....
• Physics in particular uses mathematics as its language and a tool.  You can develop good intuition for how physics works, but to make quantitative predictions, you need math.
• Ultimately, observation and experiment are the arbiters of whether a scientific model/theory is right and wrong, scientifically.  "Right" means "agrees with observation/experiment whenever checked", "wrong" means "predicts results at odds with reality".  Usually this means the model/theory needs an additional correction, or has only a limited range of applicability.  (Our commonplace understanding of how bicycles work doesn't do so well at speeds of thousands of miles an hour.  That doesn't mean we don't understand how bikes work at low speeds; it means that additional effects have to be considered at very high speeds.)
• There are many branches of physics - it's not all particle physics and astrophysics, despite the impression you might get from TV or movies.
• Physics explains light in all its forms (the microwaves that heat your food; the radio waves that carry your wifi and cell phone traffic; the light you see with your eye and which carries your internet data over fibers; the x-rays that can go through your skin; and the really high energy gamma rays that do not, in fact, turn you into an enormous green ragemonster).
• Physics includes not just looking at the tiniest building blocks of matter, but also understanding what happens when those building blocks come together in very large numbers - it can explain that diamond is hard and transparent, how/why water freezes and boils, and how little pieces of silicon in your computer can be used to switch electric current.  Physics provides a foundation for chemistry and biology, but in those fields often it makes much more sense to use chemistry and biology vocabulary and models.
• Quantum mechanics can be unintuitive and weird, but that doesn't mean it's magic, and it doesn't mean that everything we don't understand (e.g., consciousness) is deeply connected to quantum physics.  Quantum is often most important at small scales, like at the level of individual atoms and molecules.  That's one reason it can seem weird - your everyday world is much larger.
• Relativity can also be unintuitive and weird - that's because it's most important at speeds near the speed of light, and again those are far from your everyday experience.
• We actually understand a heck of a lot of physics, and that's directly responsible for our enormous technological progress in the last hundred and fifty years.
• Physicists enjoy being creative and speculative, but good and honest ones are careful to point out when they're hand waving or being fanciful.
I'll add more to this list over time, but that's a start....

## Wednesday, May 24, 2017

### Hot electrons and a connection to thermoelectricity

The two recent posts about the Seebeck effect and hot electrons give some context so that I can talk about a paper we published last month.

We started out playing around with metal nanowires, and measuring the open-circuit voltage (that is, hook up a volt meter across the device, which nominally doesn't allow current to flow) across those wires as a function of where we illuminated them with a near-IR laser.  Because the metal absorbs some of the light, that laser spot acts like a local heat source (though figuring out the temperature profile requires some modeling of the heat transfer processes).   As mentioned here, particles tend to diffuse from hot locations to cold locations; in an open circuit, a voltage builds up to balance out this tendency, because in the steady state no net current flows in an open circuit; and in a metal, the way electron motion and scattering depend on the energy of the electrons gives you the magnitude and sign of this process.   If the metal is sufficiently nanoscale that boundary scattering matters, you end up with a thermoelectric response that depends on the metal geometry.  The end result is shown in the left portion of the figure.  If you illuminate the center of the metal wire, you measure no net voltage - you shouldn't, because the whole system is symmetric.  The junction where the wire fans out to a bigger pad acts like a thermocouple because of that boundary scattering, and if you illuminate it you get a net thermoelectric voltage (sign depends on how you pick ground and which end you're illuminating).   Bottom line:  Illumination heats the electrons a bit (say a few Kelvin), and you get a thermoelectric voltage because of that, to offset the tendency of the electrons to diffuse due to the temperature gradient.  In this system, the size of the effect is small - microvolts at our illumination conditions.

Now we can take that same nanowire, and break it to make a tunnel junction somewhere in there - a gap between the two electrodes where the electrons are able to "tunnel" across from one side to the other.  When we illuminate the tunnel junction, we now see open-circuit photovoltages that are much larger, and very localized to the gap region.  So, what is going on here?  The physics is related, but not true thermoelectricity (which assumes that it always makes sense to define temperature everywhere).   What we believe is happening is something that was discussed theoretically here, and was reported in molecule-containing junctions here.   As I said when talking about hot electrons, when light gets absorbed, it is possible to kick electrons way up in energy.  Usually that energy gets dissipated by being spread among other electrons very quickly.  However, if hot electrons encounter the tunnel junction before they've lost most of that energy, they have a higher likelihood of getting across the tunnel junction, because quantum tunneling is energy-dependent.  Producing more hot electrons on one side of the junction than the other will drive a tunneling current.  We still have an open circuit, though, so some voltage has to build up so that the net current in the steady state adds up to zero.  Bottom line:  Illumination here can drive a "hot" electron tunneling current, and you get a photovoltage to offset that process.  This isn't strictly a thermoelectric effect because the electrons aren't thermally distributed - it's the short-lived high energy tail that matters most.

It's fun to think about ways to try to better understand and maximize such effects, perhaps for applications in photodetection or other technologies....

## Friday, May 19, 2017

### What are "hot" electrons?

In basic chemistry or introductory quantum mechanics, you learn about the idea of energy levels for electrons.  If you throw a bunch of electrons into some system, you also learn about the ground state, the lowest energy state of the whole system, where the electrons fill up* the levels from the bottom up, in accord with the Pauli principle.   In statistical physics, there are often a whole lot of energy levels and a whole lot of electrons (like $10^{22}$ per cc), so we have to talk about distribution functions, and how many electrons are in the levels with energies between $E$ and $E + dE$.   In thermal equilibrium (meaning our system of interest is free to exchange energy in the form of heat with some large reservoir described by a well-defined temperature $T$), the distribution of electrons as a function of energy is given by the Fermi-Dirac distribution.

So, what are "hot" electrons?  If we have a system driven out of equilibrium, it's possible to have the electrons arranged in a non-thermal (non-FD distribution!) way.  Two examples are of particular interest at the nanoscale.  In a transistor, say, or other nanoelectronic device, it is possible to apply a voltage across the system so that $eV >> k_{\mathrm{B}}T$ and inject charge carriers at energies well above the thermally distributed population.  Often electron-electron scattering on the 10-100 fs timescale redistributes the energy across the electrons, restoring a thermal distribution at some higher effective temperature (and on longer timescales, that energy cascades down into the vibrations of the lattice).  Electrons in a metal like Au at the top of the distribution are typically moving at speeds of $\sim 10^{6}$ m/s (!!), so that means that near where the current is injected, on distance scales like 10-100 nm, there can be "hot" electrons well above the FD distribution.

The other key way to generate "hot" electrons is by optical absorption.  A visible photon (perhaps a green one with an energy $\hbar \omega$ of 2 eV) can be absorbed by a metal or a semiconductor, and this can excite an electron at an energy $\hbar \omega$ above the top of the FD distribution.  Often, on the 10-100 fs timescale, as above, that energy gets redistributed among many electrons, and then later into the lattice.  That's heating by optical absorption.  In recent years, there has been an enormous amount of interest in trying to capture and use those hot electrons or their energy before there is a chance for that energy go become converted to heat.  See here, for instance, for thoughts about solar energy harvesting, or here for a discussion of hot electron photochemistry.  Nanoscale systems are of great interest in this field for several reasons, including the essential fact that hot electrons generated in them can access the system surface or boundary in the crucial timespan before energy relaxation.

*Really, the whole many-body electron wavefunction has to be antisymmetric under the exchange of any two electrons, so it's wrong to talk as if one particular electron is sitting in one particular state, but let's ignore that for now.  Also, in general, the energy levels of the many-electron system actually depend on the number and arrangement of the electrons in the system (correlation effects!), but let's ignore that, too.

## Tuesday, May 16, 2017

### More coming, soon.

I will be posting more soon.  I'm in the midst of finally shifting my group webpage to a more modern design.  In the meantime, if there are requests for particular topics, please put them in the comments and I'll see what I can do.

Update:  Victory.  After a battle with weird permissions issues associated with the way Rice does webhosting, it's up here:  natelson.web.rice.edu/group.html

Still a few things that should be updated and cleaned up (including my personal homepage), but the major work is done.


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## Found 174 Documents (Results 1–100)

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### $$\alpha$$-limit sets and Lyapunov function for maps with one topological attractor. (English)Zbl 07560338

MSC:  37B25 37B35 54H20
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### Global attracting set and exponential decay of coupled neutral SPDEs driven by fractional Brownian motion. (English)Zbl 1492.60191

MSC:  60H15 60G22 60H05
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### Impulsive-integral inequalities for attracting and quasi-invariant sets of neutral stochastic integrodifferential equations with impulsive effects. (English)Zbl 1472.45008

MSC:  45J05 45R05 34K45
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### Forward omega limit sets of nonautonomous dynamical systems. (English)Zbl 1439.35056

MSC:  35B40 35B41 37L30
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### Well-posedness and dynamics of impulsive fractional stochastic evolution equations with unbounded delay. (English)Zbl 07264427

MSC:  34Gxx 43Axx 26Axx
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### Existence of global attractors for a class of reaction-diffusion system. (Chinese. English summary)Zbl 1449.35093

MSC:  35B41 35K57
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### The numerical computation of unstable manifolds for infinite dimensional dynamical systems by embedding techniques. (English)Zbl 1435.37103

MSC:  37M21 35B42 37L25
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### Attracting graphs of skew products with non-contracting fiber maps. (English)Zbl 1414.37014

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### On a one-parameter discrete-time $$Z_4$$-equivariant cubic dynamical system. (English)Zbl 1411.37027

MSC:  37C80 39A12 37D45 37C25 37D10
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### Forward attracting sets of reaction-diffusion equations on variable domains. (English)Zbl 1409.37077

MSC:  37L30 35K57
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### The existence and asymptotic behavior of solutions to fractional stochastic evolution equations with infinite delay. (English)Zbl 1406.35469

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### Hyberbolic saddle measures and laminarity for holomorphic endomorphisms of $$\mathbb{P}^2{\mathbb{C}}$$. (English)Zbl 1414.32019

Reviewer: Xu Zhang (Weihai)
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### Codimension-one attracting sets in $$\mathbb{P}^k(\mathbb{C})$$. (English)Zbl 1401.37006

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### Prescribed cycles of König’s method for polynomials. (English)Zbl 1382.37045

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### Is there switching without suspended horseshoes? (Portuguese)Zbl 1458.37028

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### Global attracting set, exponential decay and stability in distribution of neutral SPDEs driven by additive $$\alpha$$-stable processes. (English)Zbl 1354.60072

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### Lattice structures for attractors. I. (English)Zbl 1346.37013

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### Asymptotic behavior of impulsive stochastic functional differential equations. (English)Zbl 1294.34076

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### Difference inequality for attracting and quasi-invariant sets for a class of impulsive stochastic difference equations with continuous time. (English)Zbl 1291.30171

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### Attracting and invariant sets of nonlinear stochastic neutral differential equations with delays. (English)Zbl 1268.39021

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### Asymptotic behavior of impulsive non-autonomous delay difference equations with continuous variables. (English)Zbl 1388.39001

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### Global attracting set for a class of nonautonomous neutral functional differential equations. (English)Zbl 1277.26030

MSC:  26D10 26D15 34K40
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### Global attracting sets for a class of functional differential equations. (Chinese. English summary)Zbl 1265.34275

MSC:  34K25 34K12 34K20
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### Global attracting set and stability of stochastic neutral partial functional differential equations with impulses. (English)Zbl 1250.93124

MSC:  93E15 35R60 60H15
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### Attracting and quasi-invariant sets for a class of impulsive stochastic difference equations. (English)Zbl 1267.39014

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### The gentlest ascent dynamics. (English)Zbl 1237.37052

MSC:  37L25 37J10 37C25 37M99
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### Attracting cycles for the relaxed Newton’s method. (English)Zbl 1215.65089

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### Julia sets of Newton’s method for a class of complex-exponential function $$F(z)=P(z)e^{Q(z)}$$. (English)Zbl 1215.37034

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### On the dynamics of the singularly perturbed rational maps. (English)Zbl 1205.30024

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### Instability of closed invariant sets of semidynamical systems: method of sign-constant Lyapunov functions. (English. Russian original)Zbl 1185.34070

Math. Notes 85, No. 3, 374-384 (2009); translation from Mat. Zametki 85, No. 3, 382-394 (2009).
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### $$p$$-attracting and $$p$$-invariant sets for a class of impulsive stochastic functional differential equations. (English)Zbl 1165.60329

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# Find the Y component of Electric field at a point

## Homework Statement

The figure below shows a thin, vertical rod of length L with total charge Q. The indicated point P is a horizontal distance x from the one end of the rod. What is the electric field at point P. Express your answer in component notation in the two blanks below.

L = 5.0 cm, Q = 3.0 nC, and x = 3.0 cm.

kq/r^2

## The Attempt at a Solution

I found the x component to be 15000N/C

theta = arctan(5/3) = 59 deg
total charge = [(8.99x10^9)(3x10^-9)]/((0.03)^2) = 29966
x component = 29966cos59=15417 with sig figs is 15000

so now the y component should just be 29966sin59 = 25685 rounded to 26000 N/C but that doesn't make sense because the y component should be facing down

#### Attachments

• physs2.jpg
3.6 KB · Views: 315

even when I did theta = -59 degrees the answer was incorect

haruspex
Homework Helper
Gold Member
total charge = [(8.99x10^9)(3x10^-9)]/((0.03)^2) = 29966
Not all the charge is at distance 3cm from P.

isukatphysics69
Not all the charge is at distance 3cm from P.
So did I just get lucky with the first answer??

im confused

Ok I think I see what youre saying. I got the distance from the top of the rod as
[(8.99x10^9)(3x10^-9)]/((0.058^2))=7932
7932sin59 = 6800

incorrect

@haruspex

I tried generalizing from the middle

[(8.99x10^9)(3x10^-9)]/((0.039^2))=17685
(new angle arctan(2.5/3) = 39 deg
17685sin(39) = 11321

incorrect

haruspex
Homework Helper
Gold Member
Ok I think I see what youre saying. I got the distance from the top of the rod as
[(8.99x10^9)(3x10^-9)]/((0.058^2))=7932
7932sin59 = 6800

incorrect

@haruspex
Consider an element length dy at distance y up the rod. What is the charge on it? What field does it generate at P? What is the x component of that?

Consider an element length dy at distance y up the rod. What is the charge on it? What field does it generate at P? What is the x component of that?
so the charge on dy is kq/r^2 at point P the x component is (kq/r^2)cos(theta)

my x component is correct tho, I thought the y should be simple just sin of the angle @haruspex

haruspex
Homework Helper
Gold Member
so the charge on dy is kq/r^2
That s not a charge.
What is the charge on a section of length dy?
If it is distance y up the rod, how far is it from point P?

Last edited:
isukatphysics69
That s not a charge.
What is the charge on a section of length dy?
Of it is distance y up the rod, how far is it from point P?
would it be dy/3? im really not understanding. dy is sqrt(y^2+9) away

haruspex
Homework Helper
Gold Member
would it be dy/3?
No.
What is the charge density along the rod? If the whole rod length L carries charge Q, how much charge is there on section length dy?
dy is sqrt(y^2+9) away
Units! But better still, don't plug in numbers yet, just call it x2, not 9.

isukatphysics69
No.
What is the charge density along the rod? If the whole rod length L carries charge Q, how much charge is there on section length dy?
Is it Q*dy????

Help?

GOT IT!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! >=[

haruspex
Homework Helper
Gold Member
Is it Q*dy????
The charge Q is spread out along a rod length L, what is the charge density (charge per unit length)?

isukatphysics69
The charge Q is spread out along a rod length L, what is the charge density (charge per unit length)?
Then I think it is dQ*dL

I got the answer partial correct 1.8/2 points with the answer -7900N/C

Im confused why im getting partial credit

[(8.99x10^9)(3x10^-9)]/((sqrt(.05^2+.03^2)^2) = 7932 taking the negative because +/- sqrt = -7932

1.8/2 points

SammyS
Staff Emeritus
Homework Helper
Gold Member
Is it Q*dy????
You seem to be guessing, rather than answering the specific question haruspex asked you.

To ask basically the same thing:
If the length of the rod is 5.0 cm, and the charge on the rod is 3.0 nC, how much charge is on each 1 cm of the rod?
How would you write that using symbols: L for length and Q for the charge?

SammyS
Staff Emeritus
Homework Helper
Gold Member
Im confused why im getting partial credit

[(8.99x10^9)(3x10^-9)]/((sqrt(.05^2+.03^2)^2) = 7932 taking the negative because +/- sqrt = -7932

1.8/2 points
Slow down. You're nowhere near ready to solve this. The solution will require that you use Calculus to do integration.

Slow down. You're nowhere near ready to solve this. The solution will require that you use Calculus to do integration.
Oh god.. I have to go to physics class right now. im getting frustrated awith this anyway I have to step away

SammyS
Staff Emeritus
Homework Helper
Gold Member
So did I just get lucky with the first answer??
You mean to ask if you got lucky with getting the correct numerical answer for the x component of the leectric field (which is in your "other" thread.)

The answer is absolutely yes, you got lucky.

You used cosine, when what was required there was to use sine.

Both components require the use of integral calculus.

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hey @SammyS do I have to use trig substitution?

SammyS
Staff Emeritus
Homework Helper
Gold Member
hey @SammyS do I have to use trig substitution?
Most likely you will.

You will first have to come up with a linear charge density and will need to set up an integral.

Most likely you will.

You will first have to come up with a linear charge density and will need to set up an integral.
omfg I hate doing trig sub thought I was done with that after calc 2 im really pissed right now

im just going to sleep

linear algebra test tomorrow and 2 programming assignments due theres no way ill get this frikkin degree I don't even have a job right now and I cant even keep up with the work

Im thinking I can use a double integral so that I don't have to do the frikking trig substitution.. plus I took calc 3 in the summer so I should be using double integrals

does this look good?

#### Attachments

• double integral.JPG
27.4 KB · Views: 255
SammyS
Staff Emeritus
Homework Helper
Gold Member
Im thinking I can use a double integral so that I don't have to do the frikking trig substitution.. plus I took calc 3 in the summer so I should be using double integrals
[ ATTACH=full]230927[/ATTACH]

does this look good?
I can't imagine how a double integral can even be used here. Don't even be concerned about Trig substitution at this point in solving the problem. That integral doesn't correspond to this problem at all.

Let me remind you of the advice haruspex gave you in post #8
Consider an element of charge with length dy at distance y up the rod. What is the charge on it? What field does it generate at P? What is the x component of that?
To use this idea, you will need to use the linear charge density, usually represented by the variable λ. You can get a value for λ from the given information, but there's no need bother with that yet.

( By the way: It looks like you will not need to use Trig. substitution for finding the y-component. )

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I just did the integration and got -140677.0588 and that was incorrect.
I had my integral limits from 0 to 0.05 and ∫(kqy(dy)/(a(x^2+y^2)3/2))

used U substitution and was a pretty straight forward integral. not sure what I did wrong here

SammyS
Staff Emeritus
Homework Helper
Gold Member
I just did the integration and got -140677.0588 and that was incorrect.
I had my integral limits from 0 to 0.05 and ∫(kqy(dy)/(a(x^2+y^2)3/2))

used U substitution and was a pretty straight forward integral. not sure what I did wrong here
It's much more helpful to do the integration using symbols. Plug in the numerical values later. There is virtually no way for anyone to look at the result, −140677.0588 and tell what might be wrong with it. It doesn't even have units.

That is pretty close to being the correct integral.
Having ##\ q\ ## there isn't correct. That should be ##\ \lambda\ ##, the linear charge density.
What is ' a ' ? (It's in the denominator.)
For integration limits, 0 to L .

What do you get for the integral, prior to substituting the integration limits?.

isukatphysics69
It's much more helpful to do the integration using symbols. Plug in the numerical values later. There is virtually no way for anyone to look at the result, −140677.0588 and tell what might be wrong with it. It doesn't even have units.

That is pretty close to being the correct integral.
Having ##\ q\ ## there isn't correct. That should be ##\ \lambda\ ##, the linear charge density.
What is ' a ' ? (It's in the denominator.)
For integration limits, 0 to L .

What do you get for the integral, prior to substituting the integration limits?.
I had 'a' as the length of the entire rod, I will try again

This is really frustrating but im staying calm like a monk

SammyS
Staff Emeritus
Homework Helper
Gold Member
I had 'a' as the length of the entire rod, I will try again
Well in that case, that does give you a factor of q/L, which is correct and it's needed because, q/L is λ .

Making that replacement, makes your integral:
∫(kqy(dy)/(L(x2+y2)3/2)) with limits from 0 to L .

In more readable form, using LaTeX that can be written:
##\displaystyle \frac{kq}{L} \int_0^L \frac {y\,dy}{(x^2+y^2)^{3/2}} \,,\ ## where I factored out the constant factors.
.


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# R

## Estimating variance: should I use n or n - 1? The answer is not what you think

Estimates of population parameters based on samples are not exact: there is always some error involved. In principle, one can estimate a population parameter with any estimator, but some will be better than others. There is one particular case which was always very confusing to me (because of the multiple alternatives) and that is the estimation of the variance of a Normal population from a sample. In this article I describe the properties and tradeoffs among the different alternatives and discuss how important these differences are.

## Maximum likelihood estimation from scratch

Maximum likelihood estimates of a distribution Maximum likelihood estimation (MLE) is a method to estimate the parameters of a random population given a sample. I described what this population means and its relationship to the sample in a previous post. Before we can look into MLE, we first need to understand the difference between probability and probability density for continuous variables. Probability density can be seen as a measure of relative probability, that is, values located in areas with higher probability will get have higher probability density.

## Beyond normality: the bootstrap method for hypothesis testing

tl;dr: Parametric bootstrap methods can be used to test hypothesis and calculate p values while assuming any particular population distribution we may want. Non-parametric bootstrapping methods can be used to test hypotheses and calculate p values without having to assume any particular population as long as the sample can be assumed to be representative of the population and one can transform the data adequately to take into account the null hypothesis.

## Where do p-values come from? Fundamental concepts and simulation approach

tl;dr: P-values are tail probabilities calculated from the sampling distribution of a sample-based statistic. This sampling distribution will depend on the size of the sample, the statistic being calculated and assumptions about the random population from which the data could have been sampled. For a few cases, analytical p-values are available and, for the rest of cases, approximations based on Monte Carlo simulation can be computed by generating the sampling distribution from the population.


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# Definition:Nonzero Ring Element

## Definition

Let $R$ be a ring.

A nonzero element of $R$ is an element that is not its ring zero.


text:
# How do you solve the following system: 3x+4y=-10, x-2y=4 ?

Jan 8, 2016

The solution for the system of equations is
color(blue)(x=-2/5, y=-11/5

#### Explanation:

$\textcolor{b l u e}{3 x} + 4 y = - 10$......equation $\left(1\right)$

$x - 2 y = 4$, multiplying by $3$
$\textcolor{b l u e}{3 x} - 6 y = 12$.............equation $\left(2\right)$

Solving by elimination

Subtracting equation $2$ from $1$

$\cancel{\textcolor{b l u e}{3 x}} + 4 y = - 10$
$- \cancel{\textcolor{b l u e}{3 x}} + 6 y = - 12$

$10 y = - 22$

$y = - \frac{22}{10}$

color(blue)(y =-11/5

Finding $x$ from equation $2$
$x - 2 y = 4$

$x = 4 + 2 y$

$x = 4 + 2 \times \left(- \frac{11}{5}\right)$

$x = 4 - \frac{22}{5}$

$x = \frac{20}{5} - \frac{22}{5}$

color(blue)(x= -2/5