text:
Home > Standard Error > The Standard Error Of The Mean Definition

# The Standard Error Of The Mean Definition

## Contents

So let me get my calculator back. Skip to main contentSubjectsMath by subjectEarly mathArithmeticAlgebraGeometryTrigonometryStatistics & probabilityCalculusDifferential equationsLinear algebraMath for fun and gloryMath by gradeK–2nd3rd4th5th6th7th8thHigh schoolScience & engineeringPhysicsChemistryOrganic chemistryBiologyHealth & medicineElectrical engineeringCosmology & astronomyComputingComputer programmingComputer scienceHour of CodeComputer animationArts And then when n is equal to 25, we got the standard error of the mean being equal to 1.87. Statistical Notes. Check This Out

It is useful to compare the standard error of the mean for the age of the runners versus the age at first marriage, as in the graph. It takes into account both the value of the SD and the sample size.•Both SD and SEM are in the same units -- the units of the data.• The SEM, by Because the age of the runners have a larger standard deviation (9.27 years) than does the age at first marriage (4.72 years), the standard error of the mean is larger for ISBN 0-7167-1254-7 , p 53 ^ Barde, M. (2012). "What to use to express the variability of data: Standard deviation or standard error of mean?". https://en.wikipedia.org/wiki/Standard_error

## Standard Error Example

It would be perfect only if n was infinity. Correlation Coefficient Formula 6. Let's see if it conforms to our formulas. And it actually turns out it's about as simple as possible.

Finding the sample mean is no different from finding the average of a set of numbers. The standard error can include the variation between the calculated mean of the population and once which is considered known, or accepted as accurate. and Keeping, E.S. (1963) Mathematics of Statistics, van Nostrand, p. 187 ^ Zwillinger D. (1995), Standard Mathematical Tables and Formulae, Chapman&Hall/CRC. Difference Between Standard Error And Standard Deviation The standard error of the mean (SEM) (i.e., of using the sample mean as a method of estimating the population mean) is the standard deviation of those sample means over all

Correction for correlation in the sample Expected error in the mean of A for a sample of n data points with sample bias coefficient ρ. Standard Error Vs Standard Deviation So here, when n is 20, the standard deviation of the sampling distribution of the sample mean is going to be 1. ISBN 0-8493-2479-3 p. 626 ^ a b Dietz, David; Barr, Christopher; Çetinkaya-Rundel, Mine (2012), OpenIntro Statistics (Second ed.), openintro.org ^ T.P. https://en.wikipedia.org/wiki/Standard_error So it turns out that the variance of your sampling distribution of your sample mean is equal to the variance of your original distribution-- that guy right there-- divided by n.

In each of these scenarios, a sample of observations is drawn from a large population. Standard Error Of Proportion And so this guy will have to be a little bit under one half the standard deviation, while this guy had a standard deviation of 1. Back to Top How to Find the Sample Mean Watch the video or read the steps below: How to Find the Sample Mean: Overview Dividing the sum by the number of To understand this, first we need to understand why a sampling distribution is required.

## Standard Error Vs Standard Deviation

The graphs below show the sampling distribution of the mean for samples of size 4, 9, and 25. https://explorable.com/standard-error-of-the-mean Here, n is 6. Standard Error Example Standard error is a statistical term that measures the accuracy with which a sample represents a population. Standard Error Formula Excel The true standard error of the mean, using σ = 9.27, is σ x ¯   = σ n = 9.27 16 = 2.32 {\displaystyle \sigma _{\bar {x}}\ ={\frac {\sigma }{\sqrt

The standard error of a proportion and the standard error of the mean describe the possible variability of the estimated value based on the sample around the true proportion or true his comment is here Well, let's see if we can prove it to ourselves using the simulation. If you know the variance, you can figure out the standard deviation because one is just the square root of the other. Step 2: Calculate the deviation from the mean by subtracting each value from the mean you found in Step 1. 170.5 - 162.4 = -8.1 161 - 162.4 = 1.4 160 Standard Error Regression

The mean age was 33.88 years. Let's see. The researchers report that candidate A is expected to receive 52% of the final vote, with a margin of error of 2%. http://learningux.com/standard-error/the-standard-error-of-the-mean.html What is the Sample Mean?

So maybe it'll look like that. Standard Error Symbol Let's say the mean here is 5. Here, when n is 100, our variance-- so our variance of the sampling mean of the sample distribution or our variance of the mean, of the sample mean, we could say,

## These assumptions may be approximately met when the population from which samples are taken is normally distributed, or when the sample size is sufficiently large to rely on the Central Limit

All Rights Reserved. Scenario 2. But let's say we eventually-- all of our samples, we get a lot of averages that are there. How To Interpret Standard Error It is rare that the true population standard deviation is known.

Let's say you had 1,000 people, and you sampled 5 people at a time and calculated their average height. And of course, the mean-- so this has a mean. ISBN 0-8493-2479-3 p. 626 ^ a b Dietz, David; Barr, Christopher; Çetinkaya-Rundel, Mine (2012), OpenIntro Statistics (Second ed.), openintro.org ^ T.P. http://learningux.com/standard-error/the-standard-error-is-the.html If values of the measured quantity A are not statistically independent but have been obtained from known locations in parameter space x, an unbiased estimate of the true standard error of

For example, when we take random samples of women's heights, while any individual height will vary by as much as 12 inches (a woman who is 5'10 and one who is Follow us! The sample standard deviation s = 10.23 is greater than the true population standard deviation σ = 9.27 years. This information should not be considered complete, up to date, and is not intended to be used in place of a visit, consultation, or advice of a legal, medical, or any

Then the mean here is also going to be 5. As an example of the use of the relative standard error, consider two surveys of household income that both result in a sample mean of \$50,000.


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# Is this case fit for the iPhone X ?

## 1 Answer from the Community

• Yes, but not perfectly. The camera bump on the XS is slightly larger so this case on the X has a bit of extra spacing around the camera. The XS also has a smaller microphone area (bottom left holes) so the cutout only fits half of the X’s microphone. This does not effect call quality.

2 of 2 people found this useful

format: plain


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112 mL of NO_2 at NTP was liquefied, the density of the liquid being 1.15 g mL^(-1) . The volume of the liquid

### Question Asked by a Student from EXXAMM.com Team

Q 2124123051.     112 mL of NO_2 at NTP was liquefied, the density of the liquid being 1.15 g mL^(-1) . The volume of the liquid is:
A

0.1 mL

B

0.2 mL

C

128.8 mL

D

1.15 mL

#### HINT

(Provided By a Student and Checked/Corrected by EXXAMM.com Team)

#### Access free resources including

• 100% free video lectures with detailed notes and examples
• Previous Year Papers
• Mock Tests
• Practices question categorized in topics and 4 levels with detailed solutions
• Syllabus & Pattern Analysis


text:
# Constructing a local nested base at a point in a first-countable space

I am trying to prove the following:

Let $X$ be a first countable space and $x$ a member of $X$. Prove that there is a local nested basis $\{S_n\}_{n=1}^\infty$ at $x$.

Since $X$ is first countable there is a countable local base $\mathcal{B}_x$ at $x$. Constructing a nested sequence of subsets of $\mathcal{B}_x$ is easy. Let $B_1 \in \mathcal{B}_x$. Then $B_1$ is an open set containing $x$, and so contains a member $B_2$ of $\mathcal{B}_x$ by the definition of a local base. Then $B_2$ is an open set containing $x$ and so contains a member $B_3$ of $\mathcal{B}_x$. Continuing in this fashion we obtain a nested sequence $\{B_n\}_{n=1}^\infty$ of members of $\mathcal{B}_x$ containing $x$.

I'm having trouble showing that this is a local base, in that I don't see how to prove that every open set in X contains some member of this sequence. Of course it's possible that this isn't true, and there's a different way to construct the nested sequence so that this can be done, but I'm having trouble with this as well.

Which way do I need to proceed, and how do we complete the proof? Thanks.

-

Note that although you are correct that there is some $B_2 \in \mathcal{B}_x$ such that $B_2 \subseteq B_1$, since $B_1 \in \mathcal{B}_x$ it might be that $B_2 = B_1$, and this could continue ad infinitum. If this were to happen (at any point along the inductive construction) it would almost certainly not generate a basis at $x$ (unless $x$ has a smallest open neighbourhood and we picked it at some point).
To avoid this problem we begin by first fixing any countable local base $\{ B_i \}_{i \in \mathbb{N}}$ at $x$. Next, for each $i \in \mathbb{N}$ we define $$S_i := B_1 \cap \cdots \cap B_i.$$ Note for each $i \in \mathbb{N}$ that
• as the set $S_i$ is a finite intersection of open neighbourhoods of $x$, it is itself an open neighbourhood of $x$; and
• $S_{i+1} \subseteq S_{i}.$
Therefore $\{ S_i \}_{i \in \mathbb{N}}$ is a nested family of open neighbourhoods of $x$, so we need only show that it is a local base at $x$. But this follows immediately from the fact that $S_i \subseteq B_i$ for each $i$, and we chose $\{ B_i \}_{i \in \mathbb{N}}$ to be a local base at $x$.


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# [Play Raw]Little Egret

(pphoto) #21

The start was a very nice photograph. The bird sharp and separated from the background, just two tiny overexposed areas on it.
Summary of the edits I made:

• I made the white balance much warmer, the original from the camera was too cold for my taste
• Some denoising (profile), but with focus on the bigger areas to denoise the background.
• Increased contrast on the bird’s neck and tail to bring out some details there
• Darkened the background a bit with filmic
• Highlights a little bit reduced
• Cutted out the borders to bring the bird more into the image’s focus
• The piece of wood in the background a bit darkened
• Highlighted the eye a bit

#22

(Matteo Costa) #23

My take with RawTherapee.
Many of you went for darker renditions, but I like it better colorful.
Usually I am a lurker … still I enjoy posting time to time

2018-07-20_10-23-51.10_DSC2161.jpg.out.pp3 (11,8 KB)

(senpai) #24

darktable 2.60

2018-07-20_10-23-51.10_DSC2161.NEF.xmp (6,8 Ko)

#25

Minimal processing using PhotoFlow by itself with its filmic (OCIO) and color correction modules. Remember to change the path in the PFI: 2018-07-20_10-23-51.10_DSC2161.pfi (23.0 KB).

<property name="file_name" value="path\to\NEF">

PS I am using the 20190210_2203 build; get it here: https://github.com/aferrero2707/PhotoFlow/releases/tag/continuous.

Zoom 100% and enjoy!

(Martin Scharnke) #26

What a nice result with only “minimal processing”. Looks like I’ll have to become PhotoFlow proficient.

Thanks!

#27

@martin.scharnke I neglected to mention the build no. of PhotoFlow that I was using, as OCIO is a recent addition. Updated the post to reflect that.


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Overview of the Student Precalculus Package - Maple Help

Home : Support : Online Help : Applications and Example Worksheets : Calculus : Overview of the Student Precalculus Package

The Student[Precalculus] Package

The Student[Precalculus] package contains routines that can be used to learn and understand precalculus concepts.  This worksheet demonstrates how you can launch the Precalculus tutors.

Getting Started

While any command in the package can be referred to using the long form, for example, Student[Precalculus][LineTutor], it is often easier to load the package and then use the short form command names.

 > $\mathrm{restart}$
 > $\mathrm{with}\left(\mathrm{Student}\left[\mathrm{Precalculus}\right]\right):$

The Student[Precalculus] Tutors

The following sections explain the functionality of each of the tutor routines and demonstrate how to launch the interactive tutors.


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## Bertrand's Conjecture

Theorem (Bertrand's Conjecture)
Fir
$n \ge 1$
there is at least one prime number between
$n$
and
$2n$
inclusive.
Proof
Proof. Suppose to the contrary that there is some
$n$
with no prime between
$n$
and
$2n$
. Consider the prime factors of
$C_n = \begin{pmatrix}2n\\n\end{pmatrix}= \frac{(2n)!}{n!n!}$
. If
$p$
is such a factor with
$\frac{2n}{3} \lt p \lt n$
then
$\left[ \frac{2n}{p} \right] =2 \left[ \frac{n}{p} \right] =2$
, so
$p$
is not a factor of
$2n$
, so
$p \le \frac{2n}{3}$
/
We need the following lemma.
Lemma
For any integer

, none of the prime powers in the prime factorisation of
$2n$
.
For example, if
$n=4$
,
$C_n=70=2^1 \times 5^1 \times 7^1$
. None of
$2^1, \; 5^1, \; 7^1$
exceeds
$2n=8$
/
Proof
Let
$v_p(n)$
be the number of times a prime
$p$
occurs in the prime factorisation of
$n!$
is
$\left[ \frac{n}{p} \right] + \left[ \frac{n}{p^2} \right] + \left[ \frac{n}{p^3} \right] +...$
.
$v_p(C_n)=v_p((2n)!)-2 v_p(n!)= \left( \left[ \frac{2n}{p} -2 \frac{n}{p} \right] \right) + \left( \left[ \frac{2n}{p^2} -2 \frac{n}{p^2} \right] \right) + \left( \left[ \frac{2n}{p^3} -2 \frac{n}{p^3} \right] \right) +...$

$p^k \gt 2n \rightarrow \left( \left[ \frac{2n}{p^k} -2 \frac{n}{p^k} \right] \right)=0$
else this term is at most 1. Hence
$v_p(C_n)$
is at less than or equal to the largest
$k$
satisfying
$p^k \le 2n \rightarrow p^{v_p(C_n)} \le 2n$
.
Let
$n \gt 4$
(
$n=1, \; 2, \; 3, \; 4$
are not counterexamples), then
$\sqrt{2n} \lt \frac{2n}{3}$
. We can divide factors of
$C_n$
into those less than
$\sqrt{2n}$
and those between
$\sqrt{2n}$
and
$\frac{2n}{3}$
.
$C_n=\underbrace{p_1^{a_1}p_2^{a_2}...p_r^{a_r}}_{p_i \le \sqrt{2n}} \underbrace{p_{r+1}^{a_{r+1}}...p_k^{a_k}}_{\sqrt{2n} \lt p_i \lt \frac{2n}{3}}$
.
By the above lemma none of the primes in the first factor on the right hand side exceeds
$2n$
so
$\underbrace{p_1^{a_1}p_2^{a_2}...p_r^{a_r}}_{p_i \le \sqrt{2n}} \le (2n)^{\sqrt{2n}}$
.
The primes in the second factor all have power equal to 1, since
$p \gt \sqrt{2n} \rightarrow p^2 \gt 2n$
and the power could not be 2 or greater again by the above lemma. It then follows that
$\underbrace{p_{r+1}^{a_{r+1}}...p_k^{a_k}}_{ \sqrt{2n} \lt p_i \lt \frac{2n}{3}} \le (\frac{2n}{3})! \le 4^{2n/3}$
.
We need another lemma.
Lemma 2
$\frac{4^n}{2n} \le C_n$

Proof
$4^n \le (1+1)^{2n} = \sum_{k=0}^{2n} \begin{pmatrix}2n\\k\end{pmatrix}=2+ \sum_{k=1}^{2n-1} \begin{pmatrix}2n\\k\end{pmatrix} \le 2+(2n-1) \begin{pmatrix}2n\\n\end{pmatrix} \le 2n \begin{pmatrix}2n\\n\end{pmatrix}$

Hence
$\frac{4^n}{2n} \le C_n$

With this lemma we have
$\frac{4^n}{2n} \le C_n= \underbrace{p_1^{a_1}p_2^{a_2}...p_r^{a_r}}_{p_i \le \sqrt{2n}} \underbrace{p_{r+1}^{a_{r+1}}...p_K^{r_k}}_{ \sqrt{2n} \lt p_i \lt \frac{2n}{3}} \le (2n)^{\sqrt{2n}} 4^{2n/3}$
.
This can be shown to be true for
$n=1, \; 2,...,467$
but false for
$n \ge 468$
.To show that there are no coun terexamples for
$n le 467$
, we find a sequence of primes beginning
$2 \le p \le 467$
such that each prime is no more than twice the previous one. Here is such a sequence:
2, 3, 5, 7, 13, 23, 43, 83, 163, 317, 631


text:
# Read the paragraph. Be wary of any individual who tries to sell you a device or pill that is promised to help you easily lose weight or cure some illness. Chances are good that this person is a charlatan who knows no more about fitness or disease than you do and is out to exploit the desire for quick fixes. Remember that there are few, if any, quick fixes in matters of health. What is a "charlatan"? A. an amateur B. an ignoramus C. a fraud D. a felon

###### Question:

Be wary of any individual who tries to sell you a device or pill that is promised to help you easily lose weight or cure some illness. Chances are good that this person is a charlatan who knows no more about fitness or disease than you do and is out to exploit the desire for quick fixes. Remember that there are few, if any, quick fixes in matters of health.
What is a "charlatan"?
A. an amateur
B. an ignoramus
C. a fraud
D. a felon

### Chandragupta maurya claimed the throne of the kingdom of magadha in about 321 b.c. which began the gupta empire

Chandragupta maurya claimed the throne of the kingdom of magadha in about 321 b.c. which began the gupta empire...

### The sum of 3 consecutive even integers is 78. What is the largest of the integers?

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### The ratio table below shows the number of eggs per omelette how many omelettes can be made with 18 eggs? 4 6 9 12

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### What causes the mechanical weathering of rocks and soil?

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### What is the next number in the following series: 15, 12, 17, 14, 19, ... a. 14 b. 13 c. 16 d. 15 (Explanation would be appreciated)

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### 1.Which phenomenon is responsible for sending electrically charged particles out into space?prominencessunspotssolar windO solar eclipsesw

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### A farmer watered 3/8 of a field what percentage is equivalent to the fraction Of the farmer watered

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### Can someone make this answer shorter like sum it up plz brainlyestWhite southerners would pass laws that controlled the movement of African-Americans because African-Americans were a large part of their workforce. This was especially true during ther era of Reconstruction. If a large amount of African-Americans moved away from the South, it would decrease the amount of workers available to work on farms on plantations. If there were less workers, that means farmers/plantation owners would make l

can someone make this answer shorter like sum it up plz brainlyestWhite southerners would pass laws that controlled the movement of African-Americans because African-Americans were a large part of their workforce. This was especially true during ther era of Reconstruction. If a large amount of Afric...

### Why is the Big Bang a theory and not a fact?a) Because no one is able to witness itb) Because there are no theories that could be possiblec) Because theories never turn into factsd) Because there are no such things as facts

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### During the jazz age, very famous and racially mixed night club in new york which was the

During the jazz age, very famous and racially mixed night club in new york which was the...

### Read the question and materials for the document-based essay on the beliefs of the Populist party. One step of writing document-based essays is analyzing the documents provided, many of which are primary sources. This activity will help you with that step. Using your knowledge and the documents provided for the question—political cartons, charts, and other documents—make two lists. One list should contain at least four reasons that farmers were discontent with the situations they faced at this t

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### Find the factors of 9q²-45pq+8q-40p

Find the factors of 9q²-45pq+8q-40p...

### 1. Solve the system using substitution (+20 pts) y = 2x – 2 y = –3x + 13

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### Explica que fue la resolución. francesa

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### Mixtures that would be considered buffers include which of the following? I. 0.10 M HCl + 0.10 M NaCl II. 0.10 M HF + 0.10 M NaF III. 0.10 M HBr + 0.10 M NaBr

Mixtures that would be considered buffers include which of the following? I. 0.10 M HCl + 0.10 M NaCl II. 0.10 M HF + 0.10 M NaF III. 0.10 M HBr + 0.10 M NaBr...


text:
# Simple proof that a $3\times 3$-matrix with entries $s$ or $s+1$ cannot have determinant $\pm 1$, if $s>1$.

Let $s>1$ and $A$ be a $3\times 3$ matrix with entries $s$ or $s+1$. Then $\det(A)\ne \pm 1$. The determinant has the form $as+b$ with integers $a$,$b$ and it has to be proven that $a>0$ if and only if $b>0$ and $a<0$ if and only if $b<0$. (The case $b=0$ is clear because the determinant is a multiple of $s$). I checked this with brute force, but I would like to have a simple proof.

• How do I get a plus sign over a minus-sign ? – Peter Jul 26 '14 at 19:26
• You do that with \pm – naslundx Jul 26 '14 at 19:27
• For square matrices with a size greater than $3\ x\ 3$, the above claim is not true. – Peter Jul 26 '14 at 19:57
• Your statement that ${\rm det}A$ has the form $as+b$ for integers $a$ and $b$ seems to assume that $s$ is an integer, which you have not stated in the problem. – Geoff Robinson Jul 26 '14 at 20:51
• s need not be an integer, the determinant has always the form as+b with a,b integers. But for non-integral s, the determinant might be non-integral. – Peter Jul 26 '14 at 20:55

$A$ is in the form $A=sU+V$ where the $u_{i,j}$ are $1$ and the $v_{i,j}\in\{0,1\}$. Then $\det(A)=as+b$ where $a=trace(Uadj(V))$ and $b=\det(V)$. If $\det(V)=0$, then we are done ; thus we assume that $V$ is invertible. Note that $ab$ has the same signum as $trace(UV^{-1})$ and it remains to show that $trace(UV^{-1})>0$ (the sum of the entries of $V^{-1}$). That can be done by brute-force with a PC in 0.2 second.


text:
ICorProfilerCallback::AppDomainShutdownStarted Method

Notifies the profiler that an application domain is being unloaded from a process.

Syntax

HRESULT AppDomainShutdownStarted(
[in] AppDomainID appDomainId);

Parameters

appDomainId
[in] Identifies the domain in which the application's assemblies are stored.

Remarks

The value of appDomainId is not valid for any information request after the AppDomainShutdownStarted method returns — this is the profiler's last chance to get information about this application domain.

Requirements

Platforms: See System Requirements.


text:
# A question on lowering the total spin

Is there a way to lower the total spin of the state and fixing the $S_z$ rather than lowering the $S_z$ by spin ladder operator? Or in other words, how to connect the $S=1$ state with $S=2$ or $S=0$ state? Is there such an operation though it might be unphysical?

-
For a spin-1 particle, the S=1 state has dimension 3, and the S=0 state has dimension 1. You can't connect these with a nice operator. –  Peter Shor Jun 13 '14 at 1:44
I mean multiple spins. Actually, from the tensor product, one sees that s=0 and s=1 states are in block form for S^2 and Sz. So I would guess this questions does not make any sense. However, I do not know whether there is a mathematical way to connect the two like ladder operators. –  user45765 Jun 13 '14 at 2:48


text:
## ln vs log [ENDORSED]

$E_{cell} = E_{cell}^{\circ}-\frac{RT}{nF}\ln Q$

Anne 2L
Posts: 35
Joined: Fri Sep 29, 2017 7:05 am

### ln vs log

I've noticed that sometimes we use log instead of ln in the Nernst equation. I was wondering if there's a reason we would prefer to use log over ln or if there is no difference.

Hazem Nasef 1I
Posts: 51
Joined: Tue Oct 10, 2017 7:13 am

### Re: ln vs log

There is no difference in using log versus using ln. The main reason log is preferred is because it is used in measuring pH, so it is easier to keep the logarithm the same base across chemistry, rather than switching from ln to log.

Julia Cheng 2J
Posts: 31
Joined: Thu Jul 13, 2017 3:00 am

### Re: ln vs log

They yield the same results though!

Meredith Steinberg 2E
Posts: 50
Joined: Thu Jul 13, 2017 3:00 am

### Re: ln vs log

They are both just different ways to write the Nernst Equation, but they should yield the same answer

Clara Hu 1G
Posts: 50
Joined: Sat Jul 22, 2017 3:00 am
Been upvoted: 1 time

### Re: ln vs log  [ENDORSED]

Using log makes it easier to find the pH because pH= - log [H+].

Alexandria Weinberger
Posts: 51
Joined: Fri Sep 29, 2017 7:03 am

### Re: ln vs log

Using ln versus log will yield the same answer. The only reasons one might want to use log over ln is to stay consistent within chemistry (usually use log for concentrations) and pH is found by using log.

Return to “Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)”

### Who is online

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text:
I believe that the confusion is coming from the fact that the /MONTH_CAL_ID is present in the record itself - and hence considered to be part of normal column names and then at the same time used as a partition column as well.
I would recommend to "move" the /MONTH_CAL_ID from record to its header using expression evaluator and then drop the /MONTH_CAL_ID from the record itself (using field remover). And finally changing the partition value expression to ${record:attribute('partition')} (or similar).  2 No.2 Revision LC 188 ●2 ●8 ●15 I believe that the confusion is coming from the fact that the /MONTH_CAL_ID is present in the record itself - and hence considered to be part of normal column names and then at the same time used as a partition column as well. I would recommend to "move" the /MONTH_CAL_ID from record to its header using expression evaluator the Expression Evaluator and then drop the /MONTH_CAL_ID from the record itself (using field remover). the Field Remover). And finally changing the partition value expression to ${record:attribute('partition')} (or similar).
I believe that the confusion is coming from the fact that the /MONTH_CAL_ID is present in the record itself - and hence considered to be part of normal column names and then at the same time used as a partition column as well.
I would recommend to "move" the /MONTH_CAL_ID from record to its header using the Expression Evaluator Evaluator and then drop the /MONTH_CAL_ID from the record itself (using the Field Remover). Remover). And finally changing the partition value expression to \${record:attribute('partition')} (or similar).


text:
## Miscellaneous Subjects

### Literature references and annotations by Dick Grune, dick@dickgrune.com. Last update: Tue Nov 29 11:13:41 2022.

##### These references and annotations were originally intended for personal use and are presented here only in the hope that they may be useful to others. There is no claim to completeness or even correctness. Each annotation represents my understanding of the text at the moment I wrote the annotation. No guarantees given; comments and content criticism welcome.

Jan Willen Klop, A Wonderful Stream, in Liber Amicorum Jaco de Bakker, ed. by Frank de Boer, et al., Amsterdam, CWI, 2002, pp. 125-142.
Entertaining but serious work on the Thue-Morse sequence. The sequence is related to square-free sequences, process algebra, the Toeplitz stream, various algebras of streams, the Mephistowals, and to various tilings (tessellations), which are shown in full color.

Jan Friso Groote, Bits, trits, qits, pits of hits, (in Dutch: Bits, Trits, Qits, Pits or Hits), in Liber Amicorum Jaco de Bakker, ed. by Frank de Boer, et al., Amsterdam, CWI, 2002, pp. 89-92.
In addition to the two-state bit, introduces the three-state trit, the four-state qit, the five-state pit, and the six-state hit. Then demonstrates that, for a given number of states, trits allow more numbers to be represented than any of the other N-its. For example, given 12 states, one could set up 6 (12/2) bits with them, allowing $2 sup 6 = 64$ numbers to be represented. But one can also acquire 4 trits (12/3) with them, allowing $3 sup 4 = 81$ numbers to be represented! Equipping 3 qits with the 12 states gives us $4 sup 3 = 64$ numbers again. No proof given.

Jan van Eijck, De Thue Morse reeks (Afscheid van Jaco), (in Dutch: The Thue Morse Sequence (Farewell to Jaco)), in Liber Amicorum Jaco de Bakker, ed. by Frank de Boer, et al., Amsterdam, CWI, 2002, pp. 61-77.
A playful treatment of the Thue-Morse sequence, the sequence S in which $S_2n ... S_2(n+1)-1$ is the inverse (and the reverse) of $S_0 ... S_2n-1$. With examples programmed in Haskell.

Henry G. Baker, March Möbius madness, with a polynomial postscript March 32, 1998, ACM SIGPLAN Notices, vol. 33, #3, March 1998, pp. 24-35.
Amusing account of finding roots of polynomials, with awful puns.

Helmut Kopka, Patrick W. Daly, A Guide to LaTeX 2e, (2nd Edn.), Addison-Wesley, Wokingham, UK, 1995, pp. 553.
Accessible guide, "for beginners and advanced users", as is says --correctly-- in the subtitle. Covers everything, from lists of all funny characters to advice on how to maintain a bibtex file. The last 125 pages are command and book indexes, very useful.

Günter Born, The File Formats Handbook, Int. Thomson Computer Press, London, 1995, pp. 1274.
Extensive descriptions of some 90 formats, mainly from the PC world: data bases, spreadsheets, word processors, graphics, Windows and OS/2, sounds, and page descriptions. Most of them are binary formats, but PCL, SGML, RTF and PostScript are included. No mention of uuencode or MIME formats. Some of the formats have been constructed by reverse engineering (for example the Windows GRP file format).
The book consists mainly of tables and the interspersed text is terse. Occasionally the author just uses a wrong word, which given the low redundancy of the text can be confusing. The JPEG format is described but its meaning and the process is not explained (perhaps wisely so).

Simon L. Peyton Jones, John Hughes, John Launchbury, How to give a good research talk, ACM SIGPLAN Notices, vol. 28, #11, Nov. 1993, pp. 9-12.
Tons of good advice, for example, ask yourself "If someone remembers only one thing from my talk, what would I like it to be?".

Henry G. Baker, Complex Gaussian integers for 'Gaussian graphics', ACM SIGPLAN Notices, vol. 28, #11, Nov. 1993, pp. 22-27.
Fascinating account of the properties of complex Gaussian integers, and their use in the design of graphic screens. Full of footnotes, puns and oblique references.

Yiya Yang, Anatomy of the design of an undo support facility, Int. J. Man-Machine Studies, vol. 36, pp. 81-95. 1992,
Describes the design decisions that went into an extended form of the Emacs undo facility, and is mainly concerned with the user aspects and ease of use; the implementation of the Emacs undo is taken for granted. Based on the author's PhD thesis.

Benjamin C. Pierce, Basic Category Theory for Computer Scientists, Foundations of Computing, MIT Press, Cambrigde, Mass., 1991, pp. 100.
Incomprehensible right from page 1, I'm sorry to say.

Igor Aleksander, Helen Morton, An Introduction to Neural Computing, Chapman & Hall, London, 1990, pp. 240.
A neuron is a specialized cell with about a dozen cord-like protuberances, called dendrites. One dendrite, the axon, is much longer than the others and serves as the output cord. The signal is passed down the axon along a number of relay stations, "nodes", to sequentialize things. The sides of the other dendrites serve as input connection places for the axons of other neurons; these connections are called "synapses". The synapses do not constitute electric connections. When the signal arrives at the synapse, it releases neurotransmitters, which cross the synaptic cleft to excite the post-synaptic membrane of the receiving dendrite. Depending on the nature of the post-synaptic membrane at that synapse, it will admit more or fewer positive or negative ions into the dendrite and thus into the neuron. If enough positive ions are amassed, the neuron fires.
An artificial neuron (according to McCulloch & Pitts) consists of a set of weights, one for each input, and a function. The weighted sum of the inputs is calculated at time t and fed into the function to obtain the output at time t+1. It is not unusual for one or more of the weights to be so negative as to inhibit any output if the corresponding input is non-zero.
There are two main questions here: can a given configuration of artificial neurons exhibit a given input-output behaviour; and if so, how do we find the weights.
A single neuron with N inputs can only implement a flat linear cut through N-space; so it certainly cannot detect parity in its inputs.
A perceptron is a neuron that can detect relations between its inputs. To this end, groups of its inputs are first fed into so-called association units. Association units are like neurons, except that their inputs are not weighted: they are just added and the sum is compared to a threshold. If the threshold is exceeded, the association unit fires. The outputs of the association units are the inputs to a traditional McCulloch & Pitts neuron. All this hardware together is called the perceptron.
A perceptron is more powerful than a single neuron: it is much better at recognizing shapes. Still, it cannot detect parity, connectedness, etc.
Full Boolean expressions can be handled by two-layer networks: one layer does the And-s and the second (one-neuron) layer does the Or-s, or vice versa. Single neurons are easy to train, perceptrons are more difficult and two-layer networks are very difficult.
An example is the multi-discriminator system WISARD, the input to which is supposed to consist of many bits, e.g. N x N. The WISARD contains a number of discriminators, one for each shape to be recognized (e.g. an A, a B, etc.) and a comparison unit. Each discriminators contains a number of decoders and a summing device. Each decoder is connected to a subset of the input, say k bits, uses the k bits to address an array of 2^k 1-bit integers and yields the integer found. A summing device sums the integers from the decoders, and yields the result of the the discriminator. The comparison unit picks the highest discriminator result and declares the corresponding shape recognized. When taught properly (to teach = to set the arrays) a WISARD can distinguish between a picture of a room with a person in it and one without one (intruder detection).
A second simplification to the arbitrarily connected network with arbitrary weights is the Hopfield net. In a Hopfield net, all connections are bidirectional and symmetric (which they certainly aren't in practice!). All neurons are supposed to fire at arbitrary but frequent intervals, until an equilibrium is reached.
For a Hopfield net, an energy function can be defined: each state change of a neuron will decrease the energy. Thus the system will settle in a local minimum-energy state. A network with N neurons has 2^N states, only a few of which are minima.
Sample application: a network to distinguish between predominantly horizontal and predominantly vertical patterns. Suppose 16 input values are derived from the pattern (in a 4x4 matrix). Then the net must have 16 neurons, and the input values are their initial states. The network is then trained to have two minima, horizontal and vertical, using a number of training patterns. Then, given an arbitrary pattern as initial state, the net will settle in one of these two minima.
During training false minima develop. Also, as with the McCulloch & Pitts model, the Hopfield net cannot recognize parity.
False minima cannot be avoided, but the system can be prevented from settling in one, by shaking ("heating") the system. Initially, a percentage of transitions in which the energy increases is allowed, depending on the virtual temperature. As this virtual temperature is reduced ("simulated annealing"), this percentage decreases. This increases the likelihood of the system settling in a deep minimum. A Hopfield net with simulated annealing is called a "Boltzmann machine".
The parity problem can again be solved by adding hidden neurons, i.e. neurons not directly initialized by the input. As usual, this makes teaching the net a lot harder. Techniques for teaching Hopfield and Boltzmann nets are: "clamping", in which a number of input values are clamped (fixed) to the correct output value; effectively this teaches the net to handle a subproblem, which is then extended by freeing more and more output values. And "error propagation", in which sets of differential equations are used to determine the direction in which errors must be propagated to achieve the desired effect; this is effectively curve fitting on the entire net.

Ben Shneiderman, Greg Kearsley, Hypertext Hands-On!: An Introduction to a New Way of Organizing and Accessing Information, Addison-Wesley, Reading, Mass., 1989, pp. 165.
Hypertext intends to be the screen version of book text. Information in Hypertext is stored in (small) records, each fitting on one or two screens; terms (words) in the text may be connected through hyperties to other records. Such words are highlighted and the trails laid out by them can be followed forward and backward. There are various other facilities.

David Barron, Mike Rees, Text Processing and Typesetting with Unix, Addison-Wesley, Reading, Mass., 1987,

Sandra L. Emerson, Karen Paulsell, troff Typesetting for UNIX Systems, Prentice-Hall, Englewood Cliffs, NJ, 1987, pp. 359.
Entertaining and knowledgeable. Treats troff primitives, macro writing, escape sequences, -ms, -mm, -man, tbl, eqn, the typesetting pipeline, and ditroff.

David Harel, Algorithms, The Spirit of Computing, Addison-Wesley, Wokingham, England, 1987, pp. 425.
Excellent introductory book to CS and algorithms.

Jeffrey D. Ullman, Computational Aspects of VLSI, Computer Science Press, Rockville, Md, 1984, pp. 495.
General treatment of same: models, lower bounds, layout algorithms, processor organization, systolic algorithms, high-wire, existing design systems, silicon compilation, optimization, and general algorithms.

Herman Koningsveld, Het verschijnsel wetenschap -- Een inleiding tot de wetenschapsfilosofie, (in Dutch: Science as a phenomenon -- Introduction to science philosophy), Boom, Meppel, 1978, pp. 211.
From Hume to Feyerabend.

Kim Gostelow, Vincent G. Cerf, Gerald Estrin, Saul Volansky, Proper termination of flow-of-control in programs involving concurrent processes, (ACM Annual Conference, Vol II, pp. 742-754), ACM SIGPLAN Notices, vol. 7, #11, Nov. 1972, pp. 15-27.
Communicating parallel programs are modeled using a token machine operating on a complex bigraph, which is an extension of a Petri net. A complex directed graph is a directed graph in which the arcs are complex arcs. A complex arc is a (named) arc that runs from a set of nodes to a set of nodes.
Resources (processors, data, open files, tape units, etc.) are modeled as tokens that are absorbed by and produced by nodes. For each node N there is a (restricted) Boolean expression, depending on the presence of tokens on the incoming complex arcs of N, and telling if N may fire; if N fires the required tokens are consumed. There is also a formula which tells how many tokens are produces on which outgoing complex arcs of N when N terminates.
The token machine starts with exactly one token on the start arc of the complex bigraph [why bi-??] of the program, and terminates properly when there is exactly one token on the outgoing arc of the bigraph, and no other tokens are left behind. A program is properly terminating if there is no run of the token machine that can terminate improperly.
To compute proper termination, the complex bigraph is transformed into a computation flow graph, using rules given in the paper. The program will terminate properly iff the computation flow graph is finite and all states have a path to the final state. This transformation is computation-intensive, and a faster algorithm is given, which reduces the computation flow graph. No cost analysis of the algorithm is given.

Friedrich L. Bauer, Andrei und das Untier, (in German: Andrei and the Monster), Bayrischer Schulbuch-Verlag, München, 1972, pp. 80.
Six amusing non-trivial lessons in computer science, for children and adults: 1. the nature of commands; 2. modification of commands; parameters; repetition; 3. conditions; recursion; 4. variables; 5. nesting; structs/records; 6. linked lists; graphs; transition diagrams. Programs are in a simple Algol68 with German keywords. Great pains are taken in explaining the concept of a variable, culminating in the Algol 68 definition 'ref' 'int' i = 'loc' 'int'. Parameter packs are recognized as an instance of structs! The chess example in Chapter 6 is very complicated. An appendix on the history of computer science in a much more academic style closes the book.
The story is set in Russia/Georgia, using Russian names and occasionally using Russian and Georgian letters for identifiers; the author explains this from the book being conceived during his stay in Novosibirsk.

Leo Geurts, Lambert Meertens, Reind van de Riet, Aad van Wijngaarden, Physionomie, psyche en chironomie, Mathematisch Centrum, Amsterdam, Oct. 4, 1969, pp. 32.
Very early example of morphing. The first 3 pages give a derivation of a morphing algorithm, presented jokingly as the "diffusion of physionomy and psyche", controlling the development using a variable $\chi$. It is the applied in the next 25 pages to the morphing of the portrait of the parting curator and geometrical algebra expert Prof. J.A. Schouten into the symbol for the Lie derivative $\math L \over v$. At the end $\chi$ becomes negative...

J. Riguet, Programmation et th\'eories de catégories, (in French: Programming and category theory), in Symposium on Symbolic Languages in Data Processing, Gordon and Breach, New York, 1962, pp. 83-98.
After a very accessible introduction to category theory, the latter is used to construct a model of computation. Function composition in category theory is compared to transition composition in a special kind of FS automaton. This FS automaton has two kinds of transitions: choice transitions, which are marked with strings from the language and which can be taken when a segment of the data matches the string; and action transitions, which are marked with an action to be performed on the data. Each state has either a set of outgoing choice transitions or a single outgoing action transition. An algorithm for multiplication in the unary system is given for this model; it is Turing-machine-like. Remarkably, the explanation of the programming model and the example can be understood without reference to the category theory part.

Grace Murray Hopper, The education of a computer, in Proc. ACM National Conference, Pittsburgh PA, ACM, 1952,
reprinted in Annals of the History of Computing, Vol. 9, No. 3/4, 1988, pp. 271-281.


text:
Caltech Home > Caltech HPC Homepage > Documentation > FAQ > How can I handle system signals in my job?
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# How can I handle system signals in my job?

In UNIX or unix-like systems (like Linux, which runs on Caltech's supercomputers), processes can be sent signals, and they can optionally handle those signals and change their behavior. Some of the signals cannot be caught by a signal handler, but many can. What signals are sent depends on the situation.

# What are some common signals that can be sent to a process?

The following signals are relatively commonly sent to a process:

• INT
• This signal is the most common way to interrupt a process. For example, if you are running a script, and you realize you made a mistake, and you use CTRL-C to cancel it, this is the signal that gets sent to your script.
• TERM
• This signal is sent when you use the kill command to kill a process that you own. It is also the first signal that is sent by the scheduling system to jobs that are being removed because one of the following has occurred:
• The job has gone beyond its walltime, and will be killed
• The job is running, and the user has decided to remove the job using a command like qdel
• KILL
• This is an unconditional termination signal that can be sent to a process. It cannot be caught and handled, so it causes processes to end immediately. If you have a job that has already been sent a TERM signal, and hasn't exited within a few seconds, the job will be sent a KILLsignal instead.

# What can I do to cleanup a process when it gets sent a signal?

A few signals cannot be caught, but for the most part, you can set up a job script or another process, to catch and handle common signals, and do something different. For example, you might want to copy back some in-processing files when a job is killed via a signal, that you wouldn't care about normally.

Signals are caught by signal handlers, which are usually functions that are associated with a signal. The function is not called until the signal is received. How this is done will depend on the programming language, but a few examples are shown below for common job scripting languages.

# Examples

## Bash

#!/bin/bash
#SBATCH --time=03:00:00  # walltime
#SBATCH --ntasks=8  # number of processor cores (i.e. tasks)
#SBATCH --nodes=1  # number of nodes
#SBATCH --mem-per-cpu=1024M  # memory per CPU core
#SBATCH --qos=test

# define the handler function
# note that this is not executed here, but rather
# when the associated signal is sent
term_handler()
{
echo "function term_handler called.  Exiting"
# do whatever cleanup you want here
exit -1
}

# associate the function "term_handler" with the TERM signal
trap 'term_handler' TERM

#Do your normal work here

## Perl

#!/usr/bin/perl -w
#SBATCH --time=03:00:00  # walltime
#SBATCH --ntasks=8  # number of processor cores (i.e. tasks)
#SBATCH --nodes=1  # number of nodes
#SBATCH --mem-per-cpu=1024M  # memory per CPU core
#SBATCH --qos=test
use strict;

# define the handler function
# note that this is not executed here, but rather
# when the associated signal is sent
sub TERM_handler {
print "TERM_handler called.  Exiting";
# do other stuff to cleanup here
exit(-1);
}

# associate the function "TERM_handler" with the TERM signal
\$SIG{'TERM'} = 'TERM_handler';

# Do your normal work here

## Python

#!/usr/bin/python
#SBATCH --time=03:00:00  # walltime
#SBATCH --ntasks=8  # number of processor cores (i.e. tasks)
#SBATCH --nodes=1  # number of nodes
#SBATCH --mem-per-cpu=1024M  # memory per CPU core
#SBATCH --qos=test
import signal
import sys

# define the handler function
# note that this is not executed here, but rather
# when the associated signal is sent
def term_handler(signum, frame):
print "TERM signal handler called.  Exiting."
# do other stuff to cleanup here
sys.exit(-1)

# associate the function "term_handler" with the TERM signal
signal.signal(signal.SIGTERM, term_handler)

# Do your normal work here


text:
Rotating position in a square grid

I was asked a question about the following problem (maybe puzzle would be a better word), which originally came from programming. The person, who asked me this, was programming something involving 3-by-3 square. He decided to index the positions on the square like this:

-------------
| 1 | 2 | 3 |
-------------
| 4 | 5 | 6 |
-------------
| 7 | 8 | 9 |
-------------

Later he found out that he wants to rotate the grid clockwise. So he needed map working like this: $1\mapsto 3$, $2\mapsto 6$, $3\mapsto9$, $4\mapsto2$, $5\mapsto5$, $6\mapsto8$ etc. And he noticed that he can get this map by using $n\mapsto 3n\bmod{10}$. His first question was:

Why does this work?

I'll add the way I was able to explain this, but feel free to include better explanation, if you have one.

I have added a coordinate system to the table:

    -1   0   1
-------------
-1 | 1 | 2 | 3 |
-------------
0 | 4 | 5 | 6 |
-------------
1 | 7 | 8 | 9 |
-------------

In these coordinates, the number in the square can be calculated as $x+3y+5$, where $x$ is the number of the of column and $y$ is the number of the row.

The clockwise rotation is given in this coordinate system by $(x,y)\mapsto(-y,x)$. Hence we want to have $f(x+3y+5)=-y+3x+5$. And the difference $$3(x+3y+5)-(-y+3x+5)=10y+10$$ is indeed a multiple of 10.

A very natural question was:

Can this be generalized?

The natural generalization I was able to think of was taking $n\times n$-square for $n=2k+1$. Again, I can index rows and columns by numbers $-k,\dots,-1,0,1,\dots,k$. Now the number in each row will be $x+ny+(n+1)k+1$. And multiplying by $n$ and taking reminder modulo $n^2+1$ will work: $$n(x+ny+(n+1)k+1)-(-y+nx+(n+1)k+1)=(n^2+1)y+(n^2-1)k+(n-1)=(n^2+1)(y+k)$$

I would be interested to see other generalizations (if there are some):

Another natural question is:

Can something similar be done, if I do not start indexing by 1, but by some other number (for example 0). If starting with 1 is the "best" or "most natural choice", what are the reasons behind it?

Can I get similar simple formula for rotating this table? (Since the numbers used here are 0,1,...,8; I would naturally expect something using modulus 9.)

-------------
| 0 | 1 | 2 |
-------------
| 3 | 4 | 5 |
-------------
| 6 | 7 | 8 |
-------------

-
One thing that could be helpful for your last question probably is to look at the numbers in base 3. Then they are all 2-digit numbers, the lower digit being the x coordinate and the higher the y coordinate. – mlk Jun 13 '14 at 14:01

For the $3 \times 3$ case, you wrote the number for given coordinates $(x,y)$ as $x + 3y + 5$. This is only the condensed form of the general formula.

For a $(2k+1) \times (2k+1)$ grid centered at the origin, the $0$-based number in one cell for given $(x,y)$ coordinates is $(x+k) + (2k+1) (y+k)$. You can just add $1$ to obtain the original, $1$-based numbers

With this description, the problem turns out to be a purely geometrical one. You can rotate grids of arbitrary sizes (i.e. with arbitrary values for $k$). You can even mirror the grid by letting $(x,y)\mapsto(-x,y)$. Basically, you can use any affine transform, but few of them will produce results that intuitively make sense.

An interesting task could be the generalization of this for even grid sizes, e.g. for a $4 \times 4$ grid (I don't have the time to do this right now, but maybe will have another look at this later).

-

In the ring $\mathbb{Z}/10\mathbb{Z}$, we may multiply by any number coprime to $10$ and permute the elements. Multiplying by $3$ partitions into two cycles of length 1 (namely $0$ and $5$), and two cycles of length $4$ (namely $2,6,8,4$ and $1,3,9,7$). Using a different modulus is possible, provided that some of the cycles are completely unused. Such a solution seems to break the spirit of the problem, though, unless the missing numbers are at the boundaries of the interval, like here (0 is at the boundary of $[0,9]$).

For this to work in a $4\times 4$ grid, you would need to work modulo $16$ or $17$, and find an automorphism that partitions into four cycles of length $4$. It turns out that $13$ works, with modulus $17$. They won't be arranged in the grid in the usual way; it was a great coincidence for the $3\times 3$.

  +----+----+----+----+
| 1  | 2  | 3  | 13 |
| 12 | 6  | 10 | 9  |
| 8  | 7  | 11 | 5  |
| 4  | 14 | 15 | 16 |
+----+----+----+----+

-


text:
# In $\mathbb R^p$:Every open subset is the union of a countable collection of closed sets & every open set is the countable union of disjoint open sets

Prove/Disprove that :

$(i)$ Every open Set in $\mathbb R^p$ can be written as the union of countable number of disjoint open Sets.

$(ii)$ Every open subset of $\mathbb R^p$ is the union of a countable collection of closed sets.

I was able to look at some similar posts asking this problem; but one seemed to be using the other and vice versa and seem convoluted.

Unfortunately, I have no idea on how to move forward. Can anyone please help me in preparing a proof for both of these problems?

Thank you for your help.

• An open set can be written as the union of itself. Maybe you require these disjoint open sets to satisfy a certain property. – Stefan Hamcke Sep 1 '14 at 20:47
• But, it's not disjoint from itself? – MathMan Sep 1 '14 at 20:49
• Maybe: (Pairwise) disjoint connected open sets? – Henno Brandsma Sep 1 '14 at 20:51
• Only several sets $(U_i)_J$ can be disjoint, meaning that $U_i\cap U_j=\emptyset$ whenever $i\ne j$. – Stefan Hamcke Sep 1 '14 at 20:51
• Consider for open $O \neq X$ and natural $n$: $F_n(O) = \{ x \in O: d(x, X \setminus O) \ge \frac{1}{n} \}$ – Henno Brandsma Sep 1 '14 at 20:53

## 1 Answer

Show:

1. Every open set is union of balls with rational radius and rational center.
2. Every open ball is a countable union of closed balls.

This gives (ii). For (i), given two points in your open set, say that they are equivalent iff there is a continuous path between them, completely contained in the open set. Argue that this is indeed an equivalence relation, and that its components (equivalence classes) are open. Now use that $\mathbb Q^n$ is dense in $\mathbb R^n$, so there can be no more than countably many equivalence classes.

• Thank you but I am afraid that the text I am reading has not even introduced dense sets .. – MathMan Sep 1 '14 at 20:59
• Surely you can show that every open set contains an element of $\mathbb Q^n$. – Andrés E. Caicedo Sep 1 '14 at 21:00
• Uhm, so, to prove that every open set in $\mathbb R^p$ contains an element of $\mathbb Q^n$, we need to prove that every open set in $\mathbb R$ contains an element of $\mathbb Q$ ? This should be true since, any open interval contains infinite rational numbers. Am I correct? – MathMan Sep 1 '14 at 21:15
• Sure, that's one way. And the stronger property you indicate is indeed true. – Andrés E. Caicedo Sep 1 '14 at 21:16
• If you have a path from $x$ to $y$ in your open set, and $z$ is in a ball centered at $y$ and contained in the set, then there is a path from $x$ to $z$ in the set: First go to $y$, and then go from $y$ to $z$ in a straight line. This shows the component that contains $x$ is open. – Andrés E. Caicedo Sep 1 '14 at 22:59


text:
# I would like to know if "imaginary velocity" has serious problem

I am writing a SF story, in this story, a life-form(You can just call them alien) from another universe comes to earth. They used "imaginary velocity" to make their time elapse faster than all beings on the earth. When the alien is using "imaginary velocity", they eat humans.

From SR, the formula for time dilation is: $$t^{'} = \frac{t}{\sqrt{1-\frac{v^{2}}{c^{2}}}}=\gamma t$$

if I plug in a imaginary number for $v$, for example $2ic$($i$ is the imaginary unit), then I get: $$t^{'} = \frac{t}{\sqrt{5}}$$ $\gamma$ is less than 1.

this means the time for the objects with imaginary velocity elapse faster than the objects in inertial frame.

In the story i will make it $10^{16}$ faster than inertial frame.(human's $10^{-16} s$ = alien's $1 s$, when the alien spent one day in their reference frame, only $86400*10^{-16} = 8.64*10^{-12} s$ passed from human's reference frame) So no people can see what's happening, also no machine can detect the actions of aliens.(in the latter part of the story, the situation will change)

Is their any contradiction with the theory of relativity or other physics?

ps. There is a known problem that all observable must be real numbers. However, I decided to ignore it. Since tachyons have imaginary mass and some SF story also used this in their story.

--recently added

$v=2ic$ means the object move $2ic$ m in 1 s, the distance is also imaginary. So, it is not moving in the real spacetime. You will see that object as "not moving" even if it has $2ic$ speed.

• I don't understand the physics either, but if the upshot of it is that time passes 10^16 times faster for the aliens, then i don't see how that prevents them from being detected by any machine or human. For example if they open a door to go through it, are you saying that they will open and close the door so quickly that (for example) an alarm wouldn't be triggered? Have you thought about the stress that would be placed on a door to be opened and closed so quickly, at billions of times the speed of sound? The door might just explode with the force acting upon it. Jun 26 '15 at 12:13
• In fact, just the action of moving within our atmosphere would generate massive shock waves. I think that they basically can't interact with earth at all without just smashing it into bits by accident. Jun 26 '15 at 12:16
• First point: Without further qualification imaginary velocities are fairly meaningless. Second point: At the magnitudes you are considering the aliens will not need any time dilation effects. If they can move an operate at 0.5c humans will not have time to register them - Other than in the way they interact with their environement ie. what-if.xkcd.com/1. Jun 26 '15 at 12:56
• @fairytale plus, all of Physics is a mathematical "trick". GR is a huge mathematical trick that makes no sense in the real world at all....unless we count the transit of venus, time dilation, gravitational lensing etc...
– Aron
Jun 26 '15 at 18:32
• @Taemyr I would say that "classically" TDSE on a "moving" particle, velocity of the particle is the group velocity of the particle....If we apply the same logic to a 4+0 particle and we rotate the particle on the time axis....OMG....
– Aron
Jun 26 '15 at 18:37

## 6 Answers

It is difficult to answer this question without you telling us what a velocity of $2ic$ means.

However my 5c is that this will not work.

Examine an alien moving from Human A to Human B, with a distance of $d$, at a constant speed of $2ic$. Lets ignore for now the fact that this statement is meaningless.

From the perspective of the humans the alien will spend time $t=\frac{d}{2ic}$ to cross this distance. Also, from the perspective of the human the alien will age $t'=\frac{d}{2\sqrt{5}ic}$.

From the perspective of the alien the humans are moving at a velocity of $2ic$. He will need to cross a distance $d'=d\sqrt{5}$ and will do so in time $\frac{d\sqrt{5}}{2ic}$. Meaning the formulae does not provide consistent results in the presence of imaginary velocities.

TLDR: If the alien is moving at $2ic$ with respect to the humans the humans are moving at $2ic$ with respect to the alien

• Right. Ask yourself at what time Oxford stops at the train.
– user
Jun 26 '15 at 12:58
• @fairytale i is a constant, you can't choose its value to suit your needs (make γ less than 1). Also, i times anything becomes a complex number, and the <, > operations are not defined for complex numbers (although you can compare their magnitudes, or define a lexicographic ordering) Jun 26 '15 at 13:36
• @fairytale "less than" and "greater than". Jun 26 '15 at 13:45
• @fairytale By "<, > operations" I meant the comparison operators (functions with two real numbers as input and a boolean output). They are canonically not defined for complex numbers, which aren't scalar (one dimensional) values like real numbers, but rather 2D vectors, consisting of a real and an imaginary component. Jun 26 '15 at 13:47
• @zovits complex numbers are scalar. Jun 26 '15 at 13:50
1. Instead of making time a complex dimension, you could have another (real) time dimension. That allows the state to evolve without moving forward in our time.

2. Are the aliens made of matter? Unless it's made of stuff other than our familiar protons and electrons, it could not operate differently than the normal rules of the universe.

3. If they are made of stuff from another brane or somesuch, how could they eat meat made from normal matter?

Perhaps the time freeze is a technological ability, using an enveloping field to allow time to (in effect) pass faster inside. That is a staple of SF, but generally unexplained: #1 above could be used .

• Imaginary time? Science Fiction with Imaginary TIME?! Quick someone email Prof Hawkings, he's been waiting for this day!
– Aron
Jun 26 '15 at 17:52
• The first point is a good suggestion. About point 2, I haven't decided yet. For 3, there are some settings that I didn't stated in this post. "Eat" is just an analogue. To be more detail, they absorb some kind of energy from human, then that human will vanish. Jun 27 '15 at 10:27

How can we get imaginary velocity? This is pretty obscure Physics I'm afraid.

Now lets first go to basics.

What is velocity?

$v = x / t$

Now as you already noted. If v is imaginary, then either x is imaginary or t is imaginary.

Let's assume we are in a 3,1 spacetime shall we? I know its pretty hard to imagine, but lets pretend ;).

If my understanding of the Wick rotation of 3, 1 spacetime over the time axis is correct, it would imply our friends with imaginary velocity inhabit a 4, 0 spacetime or 4 euclidean space.

Things get weird really quickly.

I can't seem to figure out how to type out my workings wrt to TDSE, but the jist of it is that, when you plug in imaginary time into TDSE, you get a particle that is confined to a single point in real time (or at least drops of exponentially from a single point in time), but is periodic on real space, covering the whole of space...

• what is the meaning of "covering the whole of space", do you mean the probability to find the particle at every point in the universe is equal? Jun 26 '15 at 18:57
• @fairytale I think it is periodic. There are those who believe that Santa is in fact made of particles that obeys such physics. It is why the effects of Santa are felt everywhere all at the same time, but when you try to observe him (stay up late), you collapse the wave form and he doesn't appear. It also explains how the Santa tunnels into people's homes.
– Aron
Jun 26 '15 at 19:00

This is pretty way over my head as far as the maths goes, but I do just want to point out the difficulties of moving extremely fast. Namely, wind resistance, relative force and all sorts of other nasty physics effects explored in various forms of fiction to date.

An excellent example of this is in The Thief of Time by Terry Pratchett. In the book there are history monks, who can bend time as they wish. Various examples of the difficulties and advantages of this can be found in the book, including the simple difficulty of walking, talking and breathing in an athmosphere which does not want to behave at your speed.

Another good example is in the recent film Xmen Days of Future Past. Quicksilver has speed as his "mutation" and as you may remember, can cause people to summersault over by simply touching their face.

You'd therefore have to make your aliens superhumanly strong, able to operate in near vacuum, and incredibly resilient so they can simply survive in the high speed environments of imaginary velocity.

Complex arithmetic is not a geometric construct, as much as it is a 'cheater' logic used to provide snobby academic types with means to impress and daze laypersons with negative answers to squared functions. All instances of complex arithmetic are trivial (in the philosophical sense, not the mathematical), and as such I find your attempt to actualize it in a work of SF refreshingly humorous!

I'm going to go ahead and assume we're using complex arithmetic to describe pan-dimensional motion. Let's insist the 'observable universe' exists in three dimensions and we give locations as vectors (v = [x,y,z]) and velocity as velocity over time (v/t). I want to make this an exercise in 3-manifolds rather than familiar euclidean space, but that makes me dizzy. It shouldn't matter over a small enough portion of space. Anyway, a human could be the classic example of Sagonian flatlander when viewed by our aliens. Or the aliens could be just as flat a we, yet able to project itself where our dimensions intersect. Many principles of flatland relativity might not 'scale' along the other dimensions, or 'proportions' of objects and energies change with respect to motion perpendicular to familiar references. Objects of this 'imaginary' universe might therefore react differently to gravitation, dilation, etc.

many questions arise, such as: if we accept motion with an 'imaginary' element to also satisfy complex arithmetic, doesn't that mean a force could decelerate what it might otherwise accelerate?

also: do we allow imaginary components to other aspects of reality, such as 'imaginary' mass, 'imaginary' charge, 'imaginary' gravity? Or are such imaginary components already provided by imaginary space?

Are our aliens pandi or transi? that is, are they Pan-dimensional bodies/intelligences, built of materials from both the real and 'imaginary' dimensions; or are they Trans-dimensional, able to, through some exploit, transition between our dimension and their home dimension. The former allow aliens to "eat" puny humans from every direction simultaneously, or even from within! Yikes!

Before we attempt to validate any sort of equation we must consider: TIME is a perception of and an attempt to measure and count the CHANGE in state of a three dimensional, mechanical device (clock, brain, planetary motion, etc). As a strictly human invention it might not have meaning to the aliens at all. SPEED is a distance traversed across three dimensional space by such a given time. It's hard to consider an alien as an being with capable intelligence while operating without time. For this reason, assume we cannot allow an 'imaginary' time component. There is another reason I think we should disallow (t+it) from our equations. If we are measuring time by 'seconds' than are assumed to have a clock. If we move trans dimensionally, are we carrying this clock with us? or are we leaving this clock to return to it later? This also affects the way in which we're comparing the motion and the metabolic rates of the extra-planers. This is similar in principle to the reason time appears to dilate near the speed of light. Remember, the speed of light is invariant at any reference. The motion of mechanics within the clock are not changing, rather the energy of a system increases as it accelerates, thus the underlying reference for motion is changing.

We are implying a sort of 'imaginary' time my simply allowing an imaginary component for each each spacial dimension.

We describe velocity thusly:

v = [(x+ix), (y+iy), (z+iz)] / t

this is 'time invariant'; the alien motion is trans-dimensional, yet occurs within the same reference of time as any human observer. For example, a human could move a mile in the 'real' universe in a minute. Or an alien could move at the same speed, over the same amount of time, along the 'imaginary' universe and appear to NOT MOVE AT ALL. Also, it is possible for aliens to exist as 'purely imaginary' constructs, meaning they can phase in and out of human perceptive existence. They do this as a function over topography, rather than an increase/decrease of relative time scales.

[x,y,z]/(t+it) is a 'positionally invariant' case, meaning aliens occupy the same space as us, and are at all times fully accessible, yet can modulate the rate and scale of physical forces acting upon them.

The later case is more restrictive, in that a change upon t affects x,y,z uniformly. The former, I like because thinking about moving along the -ix and the +iy at the same time makes me want to stand on my head! This concept is illustratable by considering the 3-sphere model of the universe:

In the case where our 'observable' universe is a 3-fold we allow 4 dimensions and say the universe is the group where any point is given by |∀p| = SQRT(w^2 + x^2 + y^2 + z^2) = 1. 'imaginary' motion would not simply be a scalar, otherwise it might be possible our aliens are able to attain the 'Singularity'; where |∀p| = 0 although I don't know why they'd bother to go there or how they'd return. Or are they already and always there?

In either case, adding complex geometry is not simply scaling the existing dimensions (thus increasing and decreasing speed). it is the proposition of doubling the number of dimensions. If you observe, moving from one, two, to three dimensions, the origin, or singularity, doesn't change. It is always 0; 0,0; 0,0,0; respectively. It follows more dimensions would share our singularity at the same point. For every dimension element of our frame, we add an imaginary element. Alien 'imaginary' dimensions confer the same constraint in that |∀p+ip| = 1. Velocity in any direction is given as a Magnitude. |v|. Because we're in a complex plane, we are essentially allowing velocity to be negative. If so, motion along a supposed 'imaginary' dimension is different from motion along the traditional plane, in that the classical Newtonian 'equal and opposite force' becomes 'equal and potentially perpendicular force'

This could mean the aliens, when they exert themselves upon the 'real' universe, would appear in a vastly decelerated frame as a static, immutable thing, such as a tree or rock, while absorb incoming energy (by projecting it along the perpendicular reference of it's 'imaginary' aspect), or it might be super fast like a plasma and amplify every inbound energy (causing air to heat, voltage to multiply, etc).

'imaginary' motion could be interpreted as a change in SCALE. Increased/decreased gravitation: Our aliens would become LIGHTER in the presence of a GREATER gravity, and crushed by the empty expanse between distant galaxies. They might even phase from our detection altogether.

Your provided equation is okay, but I don't like the (c^2) bit because the speed of light is SUPPOSE to be a constant and invariant. I think what you're attempting to do is give velocity as a fraction of the speed of light. Don't do that! let c = 1 and you simplify that part of your equation to 1-v^2. Even a human where moving at 0.999c faster than an alien, it would still be 1c less than the speed of light!

I'd argue that the point at which our aliens intersect with with an earthling, IE, when they eat one, speed is not an issue, rather, the alien's path of motion is equivocally perpendicular to the human. His intersection upon our plane would take the form of an expanding/contracting sphere which would, as it became more distant along the imaginary plane, simply become insubstantial to the point of phasing out completely. It would be like a blackhole emerging from nowhere, to suck you in. Don't even bother to ask Alice what is happening to that entangled-particle friendship bracelet you're share/wearing, but I assume the effect would be similar to ripping her arm out of her socket in a direction perpendicular to everywhere.

Even if there was nothing wrong with the imaginary numbers, there is another issue: moving $10^{16}$ times faster would cause very nasty side-effects as the air molecules cannot get out of the way.

There is an interesting explanation of this problem on xkcd:

http://what-if.xkcd.com/1/


text:
# clarification for disjoint union from any union

In the book "MEASURE THEORY AND FINE PROPERTIES OF FUNCTIONS", the authors claim: where

Can someone explain to me how can we use the observation to construct the disjoint union. I used $$F_1=A_1\times B_2$$, $$F_2=A_2 \times B_2 -A_1\times B_1$$, $$F_3=A_3 \times B_3 -(A_1\times B_1\cup A_2 \times B_2)$$ and so one.

I conclude that $$(F_i)$$ is a disjoint sequence of element of $$(A_i\times B_i)$$ which equals its union. But I don't know how to use there observation.

• I guess, $F_1=A_1\times B_1$. How do you write $F_2$ as a disjoint union of sets from $P_0$? Jan 28, 2020 at 0:03
• I dont understand your remark. I think i did F2 Jan 28, 2020 at 18:26
• You wrote $F_2=A_2\times B_2\,-\,A_1\times B_1$, but this is a set difference and not disjoint union of sets of the form $A\times B$. That's where you should use the framed identity. Jan 28, 2020 at 19:16
• But I considerd $F_3=A_3 \times B_3 -(A_1\times B_1\cup A_2 \times B_2)$ and $F_4$ and so one the all $(F_i)$ have a disjoint union. Still don't understand your remark Jan 28, 2020 at 22:10
• The problem is that $F_2\notin \mathcal P_0$, so this splitting is, though a disjoint union, but not of basic rectangles. But, $F_2, F_3,\dots$ can be split further, using the framed observation, this time to elements of $\mathcal P_0$. Jan 28, 2020 at 23:13

Your splitting to disjoint union in general does not consist of pure rectangles (elements of $$\mathcal P_0$$).
But, (repeatedly) using the given splitting of a difference of rectangles to smaller rectangles, you can further split $$F_2,F_3,\dots$$ so that you end up with a countable disjoint union of rectangles.


text:
# Improper integral =\

1. Oct 15, 2004

### FancyNut

I stopped at the last step while calculating this improper integral:

integral of x^3\ ( x^4 - 3)^1\2 with limits from 1 to infinity...

that's x cubed over the square root of x raised to 4 minus 3....

after replacing infinity with b and taking the limits it seems that I have to take limits of two expressions one that goes to infinity which is ( (b^4 - 3)^1\2) / 8 but the other is undefined since I have a negative 2 inside a square root. The second expression is the square root of negative two all over 8.

Thanks for any help XD

2. Oct 15, 2004

### NateTG

Just to make sure, that's:
$$\int_1^{\infty} \frac{x^3}{\sqrt{x^4-3}} dx$$
Which is divergent.

Perhaps there was an algebra error leading up to this?

3. Oct 15, 2004

### arildno

Not only is it divergent, but it makes no sense outside the field of complex analysis..

4. Oct 15, 2004

### FancyNut

Yeah that's it.

I end up with two expressions and when taking the limit of the first it's infinity but the second has a negative inside the sqaure root. I guessed it's divergent since whatever mistake I did doesnt change (I think..) that the second expression is a constant anyway... I guess. :p

Thanks dudes. ^_^