text:
# How do I run specific sudo commands without a password?

On one particular machine I often need to run sudo commands every now and then. I am fine with entering password on sudo in most of the cases.

However there are three sudo commands I want to run without entering password:

• sudo reboot
• sudo shutdown -r now
• sudo shutdown -P now

How can I exclude these commands from password protection to sudo?

• – fouric Aug 6 '12 at 0:37

## Use the NOPASSWD directive

You can use the NOPASSWD directive in your /etc/sudoers file.

If your user is called user and your host is called host you could add these lines to /etc/sudoers:

user host = (root) NOPASSWD: /sbin/shutdown
user host = (root) NOPASSWD: /sbin/reboot

This will allow the user user to run the desired commands on host without entering a password. All other sudoed commands will still require a password.

The commands specified in the sudoers file must be fully qualified (i.e. using the absolute path to the command to run) as described in the sudoers man page. Providing a relative path is considered a syntax error.

If the command ends with a trailing / character and points to a directory, the user will be able to run any command in that directory (but not in any sub-directories therein). In the following example, the user user can run any command in the directory /home/someuser/bin/:

user host = (root) NOPASSWD: /home/someuser/bin/

Note: Always use the command visudo to edit the sudoers file to make sure you do not lock yourself out of the system – just in case you accidentally write something incorrect to the sudoers file. visudo will save your modified file to a temporary location and will only overwrite the real sudoers file if the modified file can be parsed without errors.

## Using /etc/sudoers.d instead of modifying /etc/sudoers

As an alternative to editing the /etc/sudoers file, you could add the two lines to a new file in /etc/sudoers.d e.g. /etc/sudoers.d/shutdown. This is an elegant way of separating different changes to the sudo rights and also leaves the original sudoers file untouched for easier upgrades.

Note: Again, you should use the command visudo to edit the file to make sure you do not lock yourself out of the system:

sudo visudo -f /etc/sudoers.d/shutdown

This also automatically ensures that the owner and permissions of the new file is set correctly.

## If sudoers is messed up

If you did not use visudo to edit your files and then accidentally messed up /etc/sudoers or messed up a file in /etc/sudoers.d then you will be locked out of sudo.

The solution could be to fix the files using pkexec which is an alternative to sudo.

To fix /etc/sudoers:

pkexec visudo

To fix /etc/sudoers.d/shutdown:

pkexec visudo -f /etc/sudoers.d/shutdown

If the ownership and/or permissions are incorrect for any sudoers file, the file will be ignored by sudo so you might also find yourself locked out in this situation. Again, you can use pkexec to fix this.

The correct permissions should be like this:

\$ ls -l /etc/sudoers.d/shutdown
-r--r----- 1 root root 86 Jul 16 15:37 /etc/sudoers.d/shutdown

Use pkexec like this to fix ownership and permissions:

pkexec chown root:root /etc/sudoers.d/shutdown
pkexec chmod 0440 /etc/sudoers.d/shutdown

• The two lines refer specifically to these two commands. If user is not in other ways given sudo rights no other commands can be sudoed by this user. Have a look at man sudo and man sudoers. – mgd Jul 5 '12 at 15:27
• @StanKurdziel I tried with this line on Mac OS X Yosemite and it worked fine: mgd ALL=(root) NOPASSWD: /var/root/test. I am user mgd and ALL means "from all hosts". /var/root/test is a simple "Hello World" shell script with permissions: -rwx------ 1 root wheel 27 12 Jun 09:54 /var/root/test. I made no other changes to the system. – mgd Jun 12 '15 at 8:02
• What is host in the sudoers file? – user106563 Oct 20 '16 at 14:02
• It's not an answer, it's almost full guide about NOPASSWD directive. Thank you. – Eray Aug 14 '17 at 20:10
• @Michael please read the man page. It is all written there. To quote it: However, you may also specify command line arguments (including wildcards). Alternately, you can specify "" to indicate that the command may only be run without command line arguments. – mgd Aug 31 '17 at 18:35

Sorry but there's so much confusion over this and some really complicated answers, that i feel i must weigh in here before someone misunderstands and does something crazy.

Using visudo!!

Add the following lines to the config:

ALL ALL=NOPASSWD: /sbin/reboot,/sbin/shutdown

This allows the commands, reboot and shutdown with any parameters to be executed from any user.


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University_

# AJ Shearsby Prize

This scholarship supports undergraduate students enrolled in Geology units in their first year of studies.

## Highlights

Value Eligibility Open date Close date

## How to apply

No application is required, as this scholarship is awarded based on HSC results and University enrolments. Recipients will be informed by email.

## Who's eligible

You must:

• enrolled in any junior geology units in first year of studies
• have completed Earth and Environmental Science in the HSC in the previous year (equivalent is not allowed).

Note: This prize will be awarded to the student with the best result in Earth and Environmental Science.

## Background

This prize was founded in 1958 by a gift of £214 in honour of Mr A.J. Shearsby of Yass.

#### Terms and conditions

##### 1. Background

a. Established in 1958, this Prize has been generously donated in honour of Mr A.J. Shearsby of Yass from his geological friends and from the Rotary Club of Yass.

##### 2. Eligibility

a. To be eligible for this Prize, students must:

I. be enrolled in a Geology unit in at least one of their semesters during their first year of study at the University of Sydney, and

II. completed Earth and Environmental Science as part of the New South Wales Higher School Certificate in the previous year.

##### 3. Selection Criteria

a. The Prize will be awarded on the basis of academic merit.

b. The Prize will be awarded by the Dean of the Faculty of Science, or their nominee.

##### 4. Value

a. The Prize is valued at \$150 and is tenable for one year only.

b. The Prize annual value will be paid as a one-off payment.

c. Deferral of the Prize is not permitted without the prior permission of the Dean of the Faculty of Science or their nominee and it is not transferable to another University school, faculty or university.

d. The Prize will be offered subject to the availability of funds.

e. No other amount is payable.

##### 5. Ongoing eligibility

a. Not applicable.

##### 6. Termination

a. The Prize will be terminated if the recipient:

I. withdraws from an eligible course or fails to enrol,

II. does not maintain satisfactory academic performance,

III. is determined by the University to be guilty of serious misconduct, including, but not limited to, having provided false or misleading information,

IV. does not resume study at the end of a period of approved leave, or

V. any other provision as indicated in these terms that would lead to termination.

b. Once the Prize has been terminated, it will not be reinstated unless due to University error.

c. Where a student is found guilty of serious misconduct or withdraws from the relevant course, the University reserves the right to request that any Prize funds paid to the Prize recipient be reimbursed.

format: plain


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# Does imaginary part of complex number represents the meaning of down payment or stealing in real life??

I am new to complex numbers and trying to understand them and find real life conditions which could be interpreted by them "like positive one apple is gained apple and negative one apple is lost apple....so what is +1i apple and -1iapple" ....

I know that electromagnetism ,ac, wave analysis and quantum mechanics are explained by complex numbers but i can not hold them in my hand like apples...
So i had first thought that +1i may represent the situation when two persons A,B are making bets "gambling" over the result of match
And from point of view of person A : person A put -1i dollar with intermediate person ,person B put +1i dollar with the intermediate person .....
Here for person A, I consider the dollar he gives for the intermediary person -1i because if he loses, he transfers this -1i dollar to real -1 dollar ....
and the dollar given by B to the intermediary is +1i relative to A because if A wins he converts this +1i into real +1 dollar...
I do not know if my understanding is right but i continue to study complex numbers and i solve the following problem :
p(x)=-0.3x^2 +50 x-170 where p is the profit and x is the lamps produced per week.
"note here that i do not know the price of lamp or the cost of manufacturing and that the profit is nonlinear"
then I try to find number of lamps which make profit of 3000 dollars.
the solution is complex number with real part 83.3333 and imaginary part +-60.1849i.
Now i try to get profit for real part alone "1913.3333"and imaginary part alone"916.6667" then add them together but the result is 2830 not 3000.how this is possible????
Note that 2830 +170 =3000????!!!

During this thinking i tried to calculate the imaginary part after conversion to real 60.1849 ,then i thought of this positive imaginary part as taking full price in advance "down payment of 1752.5783" for future production "to get beyond the maximum profit per week of 1913.3333" and I thought of the negative imaginary part as stealing this 60 lamps from the deal and reselling them so we also get beyond barrier of weekly maximum profit...
but calculations for both cases were not equal to 3000 "1913.3333 + 1752.5783 = 3665.9117" does my guessings about meanings of imaginary part as price in advance or stealing true???

I know that my question may be vague ... this is because i am confused about meaning of complex numbers and i will accept all edits to make it more clear...

• There's no direct analog of complex numbers as physical entities. You shouldn't be surprised, though. There's also no good analog of negative numbers, or of the number $10^{10^{100}}$, or of a lot of other abstract mathematical concepts. Complex numbers, nevertheless, are useful, since they add lots of structure to concepts like exponentials or polynomials, which can then be used to prove more general results about them. Nov 9, 2019 at 4:23
• i see your point of view and i read about it in many sites....but there was this mathematician who described the area of land getting under water during sea tide as being negative area and thus if its shape is square then the side of this square will be square root of negative number....this is what encourages me to try to find real life examples of imaginary numbers till i gave up or find something which confirms that i really understand what complex numbers are for real Nov 9, 2019 at 4:30
• I think the term "imaginary" can be misleading. I've looked around and perhaps this article can help you understand things better: betterexplained.com/articles/…. Nov 9, 2019 at 4:30
• If you view a complex number as a positive magnitude (absolute vale) and an angular direction (a positive real number has angle of 0, and a negative real has angle of 180 and every non-zero complex number will have an angle between). Then multiplying multiples the magnitudes and adds the angle. I don't think any model will work unless a number represents a 2-d quantity with two different things being measured. And the best model will have one of the components is period and circular and the other is magnitudinal. Nov 9, 2019 at 4:56
• @URL Neither of those examples you gave are abstract. Negative numbers represent balance. If someone has $0$ dollars and they owe someone $5$ dollars, then they have $-5$ dollars. $\left({10^{10}}\right)^{100}$ is a natural number, so what's the deal there? Jun 23, 2020 at 3:12

Actually no number really exists. The natural numbers are just a convention for counting things. Like, an apple is represent by 1, and see an apple together with another apple is represent by 2, and so on. But for example, you can't say what does means having -1 apple, or $$\pi$$ apples.

Basically math is about making some "arbitrary" definitions that makes things good. You can define whatever you want (if it is not self contradictory, like let $$x=1$$ and $$x\neq 1)$$, but you try to define things in such way that you get good results.

For example, think about analytic geometry. Its about describing plane geometric things. The usually $$xy-$$plane are useful for, for example, describing straight lines, because the equations are simple, like $$y=2x$$. But describing rotations on the $$xy-$$plane, usually, leads to really complicated equations, involing $$\sin$$ and $$\cos$$. How can we avoid this?

Think you want to describe plane geometric things, but you are more interested in some rotations. Instead of thinking about the $$xy$$ plane, why don't you introduce a "vector component" $$w$$ that rotate things? For example, if you have some vector $$v$$ and multiply this component, you would have the same vector $$v$$, but rotated by $$90º$$ ($$\pi/2$$ radians) counter-clockwise. Now see that if you have the number 1 (which, in the $$xy-$$ plane would be the vector $$(1,0)$$) and apply this component $$w$$, since multiplying things by 1 intuitively keeps it fixed, you have $$1w=w.$$ Multiply $$w$$ again, and you have $$1ww=ww=w^{2}.$$ But applying $$w$$ twice means rotating $$180º$$ ($$\pi$$ radians). The vector $$(1,0)$$ in the $$xy-$$plane rotated $$180º$$ is the vector $$(-1,0),$$ namely $$-1$$. So, you have $$w^{2}=-1$$.

So, the component $$w$$ satisfies $$w^{2}=-1.$$ Well, the mathematicians use to denote this component $$w$$ by $$i$$, so $$i^{2}=-1.$$

That is a way to understand the complex number, and my favorite one.

• Your first sentence is very dubious. At best it is meaningless, or doesn't really exist, for what it's worth. Nov 9, 2019 at 18:06
• Which sentence? Nov 9, 2019 at 21:30
• The $1$st: Actually no number really exists... Nov 10, 2019 at 2:50

This is just an illustration of what Mateus Rocha has already said, but since it relates the usage of complex numbers to a real world problem, I thought it might help you in coming to grips with the use of complex numbers.

I work for an aircraft manufacturer, and we needed to figure out the location of the landing gear axles on an aircraft as it sits on the ground. For most of our aircraft models, the main landing gear use a trailing link configuration. This consists of a trunnion extending from the aircraft, with a bar (the "trailing link") connected to the trunnion by a pivot extending back to hold the wheel. A shock absorber (the "oleo") also connects to the trunnion to the trailing link. The location of the axle depends on how compressed the oleo is. Here is a simplified view of the geometry:

The location of the pivot and the anchor are known constants, as well as the distances from pivot to oleo attachment and axle. By taking a measurement of the oleo, we can know all sides of the triangle and can calculate the location of the attachment point, and therefore also the axle.

When I first did this calculation, I solved it using vectors, setting the origin at the Pivot. The Anchor was a known vector, and I rotated it down to the axle line by the amount needed to make the Oleo the right length.

The algebraic calculation was rather messy. But I reproduced the calculation in a number of variants of the tools we used, until at some point I noticed that a formula I was looking at was exactly like complex multiplication. Suddenly it occurred to me that I had been doing it the hard way.

If I view this as the complex plane, with $$0$$ at the pivot, and represent the Anchor, Attachment, and Axle as complex numbers, the rotation of Anchor to Attachment is just one complex multiplication, and converting the Attachment to the Axle is another (which in the actual calculation is not a real number, as the Pivot, Attachment, and Axle are not really in a straight line). The Oleo length is $$|\text{Anchor} - \text{Attachment}|$$, which made the calculation of their ratio more sensible as well.

The point is, any real-world calculation involving a complex number can be thought of instead as a calculation involving 2 real numbers. So it is never necessary to bring in complex numbers. But there are times when conceptually it is easier to work with complex numbers than pairs of real numbers (in the case of quantum mechanics, so much easier that it boggles the mind how hard it would be to express otherwise).

• The point is, any real-world calculation involving a complex number can be thought of instead as a calculation involving 2 real numbers. this is very helpful....that is what i am trying to do exactly....i am trying to convert the imag part of lamps to real and calculate their profit and then add them to the profit of real part ...but yesterday i was missing the 3rd term 2ab"a^2+b^2+2ab"... now i am recalculating and if real profit +imag profit +3rd term got me 3000 then my understanding will be true and imag part is just a definition of condition of "to be +real or to be -real for +i and -i" Nov 10, 2019 at 7:49
• I'm glad to help, but note that more often it is conceptually harder to do something in complex numbers than as a pair of real numbers. In my case, complex numbers were useful because they handily capture rotations and magnifications, as Mateus Rocha has said, and that is exactly what I needed. In your situation, I personally don't see where the properties of complex numbers would be helpful, so to me it seems like you are only adding complication. Nov 10, 2019 at 16:13

My take may be a little bit of a simplistic view, but every time that I have dealt with complex numbers, I have always been able to interpret them as some sort of "rotation". In some sense, this is what complex numbers describe.

To use your analogy of gambling, the way that I would interpret that situation would be something like 1+1i to mean the player is gaining a dollar per round but is also cycling a dollar. You would expect that the gambler will experience his wealth increasing by about a dollar per round, but will sometimes be ahead of that trendline by about a dollar, and sometimes will be below that trendline by about a dollar. You would expect an ordinary sinusoidal pattern, if you detrended the line.

If you understand the equation: $$e^{i\pi}+1=0$$ you will start to understand why complex numbers sort of imply rotations. This formula has deep ties to trigonometric functions, and thus circles.

Paul Sinclar's answer describes how he solved for the rotation of a shock absorbing system, and how he figured out that it was easier to work with complex numbers. The vector math was messy, and it was solved by rotating the system. But then he realized it was cleaned up the math to think about the system in the complex plane (because that is where rotations happen).

Quantum mechanics you are dealing with angular momentum which is just a fancy way of saying that things spin/rotate and that has to be conserved. So it is no wonder that complex numbers help to clean up some gnarly partial differential equations in that domain.

For your lamp example, the closest that you can get is by producing 83 lamps. Now, this answer isn't very helpful since the answer is not going to give you back \$3000. In some sense, though, the complex number tells you that you need to "spiral" out from 83 lamps (positive imaginary number) with something like an angular component of 60 lamps. Strange interpretation, but it is a static univariate problem and I'm doing my best.

The imaginary number in this case, is more likely telling you that it is not feasible to get your target profit with current production technologies. One interpretation that I like is a little sci-fi, it is how many lamps you would need to produce and rotate into our universe from a parallel universe to achieve your target. Silly interpretation, I suppose, but it gives you an idea that the target you are hoping to hit is not feasible.


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# Optimal Taxation

TODO

## Tagging

While Atkison-Stiglitz does promote labor income taxation, it leaves plenty of space for creativity within the regime. In particular, utility optimization should (all else being equal)

1. favor people who value the marginal dollar more than their income would suggest
2. favor people with low incomes due to immutable characteristics over people with low income due to their own choices

Some examples of (1):

• Suppose House A consists of an adult with an income of \$60k and House B consists of an adult and 3 kids with an income of \$60k. All else being equal, House B gains more from each additional dollar than House A.
• Suppose House A and B both make \\$60k but that someone in House B needs to buy cancer-treating drugs. Household B will (all else equal) value each additional dollar more than house B.

These examples aren't super controversial and our existing tax-system recognizes them by provide deductions for dependents and medical expenses. Our welfare system similarly takes these factors into account.

(2) is more controversial.

For instance, since tall people earn more than short people, (2) suggests they should pay more in taxes.

An important point here is that we don't care whether being tall causes higher incomes. The fact that we should tax tall people more is provable merely from the fact that height (1) correlates with higher incomes and (2) is essentially unchangeable.

Taxing height would be controversial enough, but there are lots of immutable characteristics that predict income that are very political charged such as race, sex, age - to name just the salient ones.

Needless to say, regardless of how optimal such factors make taxes in an ideal world. We won't be using them any time soon in this one.

Gini in the bottle. (2013, November 26). Retrieved April 24, 2020, from https://www.economist.com/democracy-in-america/2013/11/26/gini-in-the-bottle


text:
## CS700 and MAS583

#### Course summary

Deciding whether a graph on $$n$$ vertices has a clique of size $$k$$ is NP-hard. When $$k = 1000$$, on the other hand, there clearly is a polynomial-time algorithm: Just test all 1000-tuples of vertices. But this takes time $$\Theta(n^{1000})$$, and is clearly completely useless in practice!

Can the problem be solved in practice for small values of $$k$$? After all, it's not NP-hard for fixed $$k$$. Can we find an algorithm whose running time depends not on $$n^{k}$$, but is of the form $$f(k) n^{c}$$, for a fixed constant $$c$$ (that does not depend on $$n$$).

This question belongs to the area of parameterized algorithms. We consider NP-hard problems and look for algorithms that can solve them efficiently as long as some parameter of the problem instance is small.

We will study some techniques to develop parameterized algorithms:

• Kernelization,
• Bounded search trees
• Randomized methods

We will meet many problems that have efficient parameterized algorithms, but for the clique problem above, no such algorithm has been found—and in fact we have evidence that none exists.

What kind of evidence would this be? Ideally, we would like to prove a lower bound on the complexity of a problem. But this is generally very hard, and most students never meet any lower bound in their studies except for the $$\Omega(n \log n)$$ lower bound for comparison-based sorting.

We will generalize the sorting lower bound and prove $$\Omega(n \log n)$$ lower bounds for a few interesting problems. Then we look at some recent approaches for arguing about lower bounds when one cannot prove such a bound. In particular, we will study lower bounds based on several assumptions:

• 3SUM-hardness,
• the strong exponential time hypothesis (SETH),
• the unique games conjecture

Prerequisites: An algorithm course (CS300 or CS500 or equivalent), and some basic familiarity with NP-hardness. This course does not require any programming skills.

#### Course information

##### Lecturer:
Helmut Alt (Office: E3-1 1424, Phone: 3591) and Otfried Cheong (Office: E3-1 3434, Phone: 3542).
##### Lectures

The class meets Wednesday from 16:00 to 17:15 and Fridays from 10:30 to 11:45 in room 3445 in the EECS building (E3-1). Lectures are given in English.

The grade will be composed as follows (small changes reserved): Participation (20%), Homework (40%), Exam (40%).

##### Exams

We will have a final exam on Wednesday, June 14, from 16:00 to 19:00, in classroom #1 (room 1101) in building E3-1 (not our usual classroom).

#### Piazza

This term we will be using Piazza for class announcements, discussion, and asking questions.

Here is our Piazza class page.

You are responsible for checking the announcements on our Piazza class page regularly (if you make a Piazza account and enroll for the course, announcements will be mailed to you automatically.)

If you do not understand something, it is important that you ask questions. Piazza allows you to ask questions and get answers from the instructor, the teaching assistants, and your classmates. You can ask questions anonymously, so don't be shy and ask!

To ask questions, you need to register on Piazza and enroll as a student for CS700/MAS583. To do so, go to the enrollment page. Select "Join as student". You will then need to use your KAIST email address (ending with @kaist.ac.kr) to create an account.

#### Literature

Most of the material taught in this course stems from the last decade, so we will have to use some original papers as reference.

We will make use of the book Parameterized Algorithms, by Marek Cygan, Fedor V. Fomin, Łukasz Kowalik, Daniel Lokshtanov, Dániel Marx, Marcin Pilipczuk, Michał Pilipczuk, and Saket Saurabh. It is available on-line through the KAIST library.

We will also use lecture notes by Jeff Erickson. The originals are on his webpage (or on older versions of this page).

We will use slides made by Karl Bringmann and Sebastian Krinninger for their class.

The last topic of the course uses material from Introduction to the Theory of Computation, Michael Sipser, Second International Edition, Course Technology 2006. Check out its errata page, which contains a few substantive (as opposed to stylistic) errors.

The course will be taught on the blackboard, so do not expect slides or other such lecture materials. If you miss a class, you'll have to ask a friend for their notes.

#### Course progress

The material covered in the lectures so far:

 03-03 Introduction Chapter 1 03-08 Formal definitions of FPT and XP, Introduction to Kernelization Sections 1.1, 2.1, 2.2.1 03-10 Simple kernels Sections 2.2.2, 2.2.3 03-15 Crown decomposition Section 2.3 03-17 Kernels based on linear programming Section 2.5 03-22 Vertex cover search tree Section 3.1 03-24 Solving recursions, Feedback Vertex Set Section 3.2, 3.3 03-29 Vertex Cover above LP Section 3.4 03-31 VC above Matching, Variable-Deletion-Almost-2-SAT, Closest String Section 3.5 04-05 Lower bounds, decision trees, linear decision trees decision tree notes, Section 7.1 and 7.2 in notes 04-07 Algebraic decision trees Section 7.3 in the notes 04-12 Diameter, Maxgap, 3SUM-Hardness 3SUM notes 04-14 3SUM Midterm week (no class, no exam) 04-26 Parameterized reductions Section 13.1 04-28 Problems at least as hard as Clique Section 13.2 05-03 Holiday 05-05 Holiday 05-10 The W-hierarchy Sections 13.3, 13.5 05-12 ETH and SETH, SETH implies ETH Sections 14.1, 14.2 05-17 ETH bounds for FPT problems Section 14.3 05-19 ETH bound for CLIQUE, Orthogonal vectors hypothesis Section 14.4, slides (from here) 05-24 SETH implies OVH, lower bound for LCS slides (from here) 05-26 Cubic-time equivalence to APSP slides (from here) 05-31 3SUM revisited slides (from here) 06-02 Lower bounds for decidability of arithmetic Sipser 06-07 Lower bounds for decidability of arithmetic Sipser 06-09 Lower bounds for decidability of arithmetic Sipser Final exam on Wednesday, June 14, 16:00~19:00 in classroom 1 (1101)

#### Homework

There will be several homework assignments in this course. Homeworks must be turned in in English, and must be submitted on paper at the beginning of the lecture in the classroom.


text:
# Please solve this question of quadratic-: Q.1  A quadratic polynomial f (x)= x2 + ax +b is formed with one of its zeros being $\frac{4+3\sqrt{3}}{2+\sqrt{3}}$ where a and  b are integers. Also g (x) = x4 + 2x3 -10x2 + 4x- 10 is a biquadratic polynomial such that g $\left(\frac{4+3\sqrt{3}}{2+\sqrt{3}}\right)=\mathrm{c}\sqrt{3}+\mathrm{b}$ where c and d are also integers. Find the values of a. b. c and d.

• 1
What are you looking for?


text:
Solve time-dependent quantities/concentrations of different variables in PK model through the imported ode function in deSolve package. It can also be used to solve the function with analytical solution.

solve_fun(
x,
time = NULL,
initParmsfun = "initParms",
initState,
dllname = NULL,
func = "derivs",
initfunc = "initmod",
outnames,
method = "lsode",
rtol = 1e-08,
atol = 1e-12,
model = NULL,
lnparam = F,
vars = NULL,
tell = T,
...
)

## Arguments

x a list of storing information in the defined sensitivity function. a vector to define the given time sequence. a character for the given specific initial parameter function. a vector that define the initial values of state variables for the ODE system. a string giving the name of the shared library (without extension) that contains the compiled function. the name of the function in the dynamically loaded shared library. the name of the initialization function (which initialises values of parameters), as provided in dllname. the names of output variables calculated in the compiled function func. method used by integrator (deSolve). argument passed to integrator (deSolve). argument passed to integrator (deSolve). the defined analytical equation with functional output. a logical value that make the statement of the log-transformed parameter (default FALSE). a character for the selected output. a logical value to automatically combine the result y to decoupling simulation x. additional arguments for deSolve::ode method.

## References

Soetaert, K. E., Petzoldt, T., & Setzer, R. W. (2010). Solving differential equations in R: package deSolve. Journal of Statistical Software, 33(9), 1–25.

pksim

## Examples

  q <- "qunif"
q.arg <- list(list(min = 0.6, max = 1.0),
list(min = 0.5, max = 1.5),
list(min = 0.02, max = 0.3),
list(min = 20, max = 60))

params <- c("F","KA","KE","V")

set.seed(1234)
x <- rfast99(params = params, n = 200, q = q, q.arg = q.arg, rep = 20)

time <- seq(from = 0.25, to = 12.25, by = 0.5)
y <- solve_fun(x, model = FFPK, time = time, vars = "output")
#> Starting time: 2021-07-07 11:07:47#> Ending time: 2021-07-07 11:07:47
pksim(y) # Visualize uncertainty of model output


text:
# Understanding how to derive Feynman Rules for the 2-point correlation function ($\phi^3$ theory)

This is a follow-up question (first one here). I am aimed to work out $$V(\phi) = -\lambda \frac{\phi^3}{3!}$$ theory in detail to understand how Feynman rules can be derived from the functional integral

$$\begin{equation*} Z[J] := \int d[\phi] e^{iS[\phi] + i\int d^d x J(x) \phi(x)} \tag{1} \end{equation*}$$

I performed a completely analogous computation to that of Jeanbaptiste Roux, up to second order in perturbation theory, and obtained

\begin{align*} &\exp\left(\frac{i}{2}\int d^d x \int d^d y \frac{\delta}{\delta \phi(x)}\Delta_F(x-y) \frac{\delta}{\delta \phi(y)}\right)\times \exp\left(i \int d^d z \left( -\frac{\lambda}{3!}\phi^3+J\phi \right)\right)\Big|_{\phi=0} \\ &=1+\frac{i}{2}\times 0+\frac{i}{2}\int d^d x \int d^d y\,(i(J(x))\Delta_F(x-y)(iJ(y)) \\ &+ \frac{(i)^2}{4}\int d^d x \int d^d y \int d^d t \int d^d \xi\,(iJ(x))\Delta_F(x-y)(iJ(y)) (iJ(t))\Delta_F(t-\xi)(iJ(\xi)) + \mathcal{O}(\Delta_F^3) \end{align*}

So it seems to me that the perturbative expansion is determined by the number of propagators!

So, up to first order, we encounter one propagator and $$2$$ external legs $$iJ$$

Up to second order we encounter two propagators and $$4$$ external legs $$iJ$$

I've just learned that the source $$J$$ is represented by a vertex with one outgoing line i.e.

With this information, I was trying to understand why, for instance, the 2-point correlation function $$\langle \phi(x_1) \phi(x_2) \rangle$$ has the following contributions up to second order

Why should we draw a bubble like shown above? If I am not mistaken, we should have two external legs due to having two propagators and $$4$$ vertices due to having $$4$$ sources. However, I only see two vertices here... what am I missing?

PS: Please note this is not a homework question. I am studying Osborn notes, section 2.2. Interacting Scalar Field Theories, and I want to understand how he constructed the Feynman rules (page 23) via working out the simplest example I could find: $$\phi^3$$ theory

EDIT 0

Let me go slowly here. I will only focus on second order terms. As stated in the comments, we should get $$\propto \int \mathrm{d}^4 z \int \mathrm{d}^4 w J(x) \Delta_F(x-z) \Delta_F^2(z-w) \Delta_F(w-y) J(y)$$

This is what I have done so far

\begin{align*} &\exp\left(\frac{i}{2}\int d^d x \int d^d y \frac{\delta}{\delta \phi(x)}\Delta_F(x-y) \frac{\delta}{\delta \phi(y)}\right) \exp\left(i \int d^d t \left( -\frac{\lambda}{3!}\phi^3+J\phi \right)\right)\Big|_{\phi=0} \\ &=\left[\cdots +\frac{(i)^2}{4}\int d^{2d} z \int d^{2d} w \left(\frac{\delta}{\delta \phi(z)}\right)^2\Delta_F^2(z-w) \left(\frac{\delta}{\delta \phi(w)}\right)^2+\cdots \right]\left.e^{i \int d^d t \left( -\frac{\lambda}{3!}\phi^3+J\phi \right)}\right|_{\phi=0} \\ &=\left[1+\frac{(i)^2}{4}\int d^{2d} z\int d^{2d} w \left(\frac{\delta}{\delta\phi(z)}\right)^2\Delta_F^2(z-w)\left(\frac{\delta}{\delta\phi(w)}\right)\left( -\frac{\lambda}{2}i\phi^2(w)+iJ(w)\right) +\cdots\right] \\ &\times \left.e^{i \int d^d t \left( -\frac{\lambda}{3!}\phi^3+J\phi \right)}\right|_{\phi=0} \end{align*}

Is this OK so far?

• x, z, w, y are the 4 vertices... Mar 12 '21 at 15:01
• In the expansion you have written down, you have kept only the terms or order $\lambda^0$. This reproduces the free theory; $n$-point correlators are obtained from $Z[J]$ by differentiating $n$-times with respect to $J$ and then setting $J=0$. If you want the interaction effect you have to look for terms of order $\lambda$. The loops arise since in the interaction you have terms like $\phi(x)^3$. Mar 12 '21 at 16:12
• I've done the calculations. If I am right you should come with four tadpoles and a "double propagator" (two non-intersecting lines). Mar 12 '21 at 16:49
• Ah, I've done the calculations for the $\Delta_F (x-y) \Delta_F(z-\xi)$ term appearing in the expansion of the exponential of the functional differential operators, so not the right one... But yes given the form of the loop you seek, it should be something like you said. Mar 12 '21 at 17:17
• You can't do what you've done because the square of an integral is not the double integral of the square of the integrand: $\left[\int dx f(x) \right]^2=\int dx f(x) \int dy f(y) \neq \int d^{2x} f^2(x)$. Mar 12 '21 at 17:42

So, like it can be seen in the comments, I will perform the calculations for the second term of the expansion of the exponential containing the functional derivatives. I will denote $$\phi(x)$$ by $$\phi_x$$ and $$\frac{\delta}{\delta \phi(x)}$$ by $$\delta_{\phi_x}$$. \begin{align*} &-\frac{1}{4}\int d^d x \int d^d y \int d^d z \int d^d w \Delta_F(x-y) \Delta_F(z-w)\delta_{\phi_x}\delta_{\phi_y}\delta_{\phi_z}\delta_{\phi_w}\left.e^{i \int d^d \xi(-\frac{\lambda}{3!}\phi^3+J\phi)}\right|_{\phi=0} \\ &=-\frac{1}{4}\int d^d x \int d^d y \int d^d z \int d^d w \Delta_F(x-y) \Delta_F(z-w)\delta_{\phi_x}\delta_{\phi_y}\delta_{\phi_z}\left[i\left( -\frac{\lambda}{2}\phi^2_w+J_w \right) \right] \\ &\hphantom{==}\times\left.e^{i \int d^d \xi(-\frac{\lambda}{3!}\phi^3+J\phi)}\right|_{\phi=0} \\ &=-\frac{1}{4}\int d^d x \int d^d y \int d^d z \int d^d w \Delta_F(x-y) \Delta_F(z-w)\delta_{\phi_x}\delta_{\phi_y}\left[\vphantom{\frac{1}{2}}i\left( -\lambda \phi_w \delta_{w,z} \right) \right. \\ &\hphantom{==}\left. - \left(-\frac{\lambda}{2} \phi^2_w +J_w \right)\left( -\frac{\lambda}{2} \phi^2_z +J_z \right) \right]\left.e^{i \int d^d \xi(-\frac{\lambda}{3!}\phi^3+J\phi)}\right|_{\phi=0} \\ &=-\frac{1}{4}\int d^d x \int d^d y \int d^d z \int d^d w \Delta_F(x-y) \Delta_F(z-w)\delta_{\phi_x}\left[\vphantom{\frac{1}{2}} -i\lambda \delta_{w,y}\delta_{w,z}-\left( -\lambda \phi_w \delta_{w,y} \right) \right. \\ &\hphantom{==}\left.\times\left( -\frac{\lambda}{2}\phi^2_z+J_z \right)-\left( -\frac{\lambda}{2}\phi^2_w+J_w \right) \left( -\lambda \phi_z \delta_{zy} \right)+\left[-i\lambda \phi_w \delta_{w,z}-\left( -\frac{\lambda}{2}\phi^2_w+J_w\right)\right.\right. \\ &\hphantom{==}\left.\left.\times \left(-\frac{\lambda}{2}\phi^2_z+J_z\right)\right]\left( -i\frac{\lambda}{2}\phi^2_y+iJ_y\right)\right]\left.e^{i \int d^d \xi(-\frac{\lambda}{3!}\phi^3+J\phi)}\right|_{\phi=0} \end{align*} From here we just have to see that only the terms having zero $$\phi$$ in them after the ultimate derivation will survive to the limit $$\phi \rightarrow 0$$. So using one last time $$(uv)'=u'v+uv'$$ we arrive at the conclusion that the result ought to be: \begin{align*} &-\frac{1}{4}\int d^d x \int d^d y \int d^d z \int d^d w \Delta_F(x-y) \Delta_F(z-w)\left[\lambda \delta_{w,x}\delta_{w,y}J_z \right. \\ &\hphantom{==}+\lambda\delta_{z,y}\delta_{z,x}J_w+\lambda\delta_{w,x}\delta_{w,z}J_y+\lambda \delta_{w,y}\delta_{w,z}J_x+J_x J_y J_z J_w] \end{align*} We see that the integrand is composed of four tadpoles and one double propagator (two non-intersecting lines). Equating the terms that are equal to each other one finds: \begin{align*} &-\lambda \int d^d z \int d^d w \Delta_F (0) \Delta_F (z-w) \\ &-\frac{1}{4}\int d^d x \int d^d y \int d^d z \int d^d w \Delta_F(x-y) \Delta_F(z-w) J_x J_y J_z J_w \end{align*} I didn't check my calculations, hope there is no sign problem.


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# Tag Info

Accepted

### Intersection of two extended lines in TikZ without having to draw them

TikZ has an by now undocumented intersection coordinate system that can implicitly used as (intersection of <c11>--<c12> and <c21>--<c22>) which finds the intersections of a ...
• 90.7k

### How to automatize to draw a broken line with intersections with two straight lines

Besides a mathematical solution by Symbol 1 in the comments you can of course let PGF/TikZ do the math. If you don't need the zig-zag lines to be on one path you can just use one loop that finds the ...
• 90.7k

### Dimension too large for a simple 3D picture

You can use 3dtools \documentclass[tikz,border=3mm]{standalone} \usetikzlibrary{3dtools}%https://github.com/marmotghost/tikz-3dtools \begin{document} \begin{tikzpicture}[3d/install view={phi=140,...
• 8,997


text:
# 维基百科：Lipschitz continuity（英文）

For a Lipschitz continuous function, there exists a double cone (white) whose origin can be moved along the graph so that the whole graph always stays outside the double cone

In mathematical analysisLipschitz continuity, named after German mathematician Rudolf Lipschitz, is a strong form of uniform continuity for functions. Intuitively, a Lipschitz continuous function is limited in how fast it can change: there exists a real number such that, for every pair of points on the graph of this function, the absolute value of the slope of the line connecting them is not greater than this real number; the smallest such bound is called the Lipschitz constant of the function (or modulus of uniform continuity). For instance, every function that has bounded first derivatives is Lipschitz continuous.[1]

In the theory of differential equations, Lipschitz continuity is the central condition of the Picard–Lindelöf theorem which guarantees the existence and uniqueness of the solution to an initial value problem. A special type of Lipschitz continuity, called contraction, is used in the Banach fixed-point theorem.[2]

We have the following chain of strict inclusions for functions over a closed and bounded non-trivial interval of the real line:

Continuously differentiable ⊂ Lipschitz continuous ⊂ Hölder continuous,

where . We also have Lipschitz continuous ⊂ absolutely continuous ⊂ uniformly continuous.

# 维基百科：利普希茨连续（中文）

## 性质

• 符合利普希茨条件的函数连续，实际上一致连续
• 双李普希茨(bi-Lipschitz)函数是单射


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# Error G0155: Nested procedure definition

0

Hi,

I have a procedure within a procedure and an error comes up as:

Line 20 in C:\Users\youx\Code\Backward_05_20
Nested procedure definition G0155

So does GAUSS not allow nested procedure?

Thank you!
Laura

0

You cannot define a GAUSS procedure inside of another GAUSS procedure. You can, however, define a GAUSS fn inside of a procedure. An fn is essentially a one-line procedure. Here is a quick example:

```proc (1) = multiplyInputs(x, y);
local out;
fn multiply(arg1, arg2) = arg1*arg2;

out = multiply(x, y);
retp(out);
endp;
```

You an also, of course, define the sub-procedure above or below your main procedure.

0

You cannot define a GAUSS procedure inside of another GAUSS procedure. You can, however, define a GAUSS fn inside of a procedure. An fn is essentially a one-line procedure. Here is a quick example:

```proc (1) = multiplyInputs(x, y);
local out;
fn multiply(arg1, arg2) = arg1*arg2;

out = multiply(x, y);
retp(out);
endp;
```

You an also, of course, define the sub-procedure above or below your main procedure.

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# 微软的 C++ 编译器将自增自减炖成了一锅汤

《由一个自增引发的问题》的后续，用实验和文档证明微软挖大坑。

clang和llvm组合应该是这种结果

5.2.6 Increment and decrement [expr.post.incr]

The value of a postﬁx ++ expression is the value of its operand. [ Note: the value obtained is a copy of the original value — end note ] The operand shall be a modiﬁable lvalue. The type of the operand shall be an arithmetic type or a pointer to a complete object type. The value of the operand object is modiﬁed by adding 1 to it, unless the object is of type bool, in which case it is set to true. [ Note: this use is deprecated, see Annex D. — end note ] The value computation of the ++ expression is sequenced before the modiﬁcation of the operand object. With respect to an indeterminately-sequenced function call, the operation of postﬁx ++ is a single evaluation. [ Note: Therefore, a function call shall not intervene between the lvalue-to-rvalue conversion and the side effect associated with any single postﬁx ++ operator. — end note ] The result is a prvalue. The type of the result is the cv-unqualiﬁed version of the type of the operand. See also 5.7 and 5.17.

The operand of postﬁx — is decremented analogously to the postﬁx ++ operator, except that the operand shall not be of type bool. [ Note: For preﬁx increment and decrement, see 5.3.2. — end note ]

5.3.2 Increment and decrement [expr.pre.incr]

The operand of preﬁx ++ is modiﬁed by adding 1, or set to true if it is bool (this use is deprecated). The operand shall be a modiﬁable lvalue. The type of the operand shall be an arithmetic type or a pointer to a completely-deﬁned object type. The result is the updated operand; it is an lvalue, and it is a bit-ﬁeld if the operand is a bit-ﬁeld. If x is not of type bool, the expression ++x is equivalent to x+=1 [ Note: See the discussions of addition (5.7) and assignment operators (5.17) for information on conversions. — end note ]

The operand of preﬁx — is modiﬁed by subtracting 1. The operand shall not be of type bool. The requirements on the operand of preﬁx — and the properties of its result are otherwise the same as those of preﬁx ++. [ Note: For postﬁx increment and decrement, see 5.2.6. — end note ]

• 一堆的脚本加载失败
• 由此带来的不可上传附件和区域空缺

• MSDN 上的 C++ 版块充斥着语言应用的帖子，不是提交错误的地方。
• 这位老兄（年份很早了）想给微软提交一个崩溃报告，但是石沉大海。

• 这位老兄实在是幸运，在 Microsoft Connect 提交的编译器错误最终被修复了。
• 这位老兄的运气就很差，提交了一串bug都没有得到好的反馈，一位善良的客服陪着磨嘴皮子。


text:
# Article

Full entry | PDF   (0.4 MB)
References:
[1] JASEM M.: Weak isometries and direct decompositions of dually residuated lattice ordered semigroups. Math. Slovaca 43 (1993), 119-136. MR 1274597 | Zbl 0782.06012
[2] KOVÁŘ T.: Any $DR\ell$-semigroup is the direct product of a commutative $\ell$-group and a $DR\ell$-semigroup with the least element. Discuss. Math. Algebra Stochastic Methods 16 (1996), 99-105. MR 1429801
[3] KOVÁŘ T.: Note on axioms of a dually residuated lattice ordered semigroup. Discuss. Math. Algebra Stochastic Methods 17 (1997), 89-90. MR 1633244 | Zbl 0906.06012
[4] KOVÁŘ T.: Two remarks on dually residuated lattice ordered semigroups. Math. Slovaca 49 (1999), 17-18. MR 1804468 | Zbl 0943.06007
[5] SWAMY K. L. N.: Dually residuated lattice ordered semigroups. Math. Ann. 159 (1965), 105-114. MR 0183797 | Zbl 0138.02104
[6] SWAMY K. L. N.: Dually residuated lattice ordered semigroups II. Math. Ann. 160 (1965), 64-71. MR 0191851 | Zbl 0138.02104
[7] SWAMY K. L. N.: Dually residuated lattice ordered semigroups III. Math. Ann. 167 (1966), 71-74. MR 0200364 | Zbl 0158.02601

Partner of


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Home Molecular Weight 2-methyloct-3-en-2-ol

# Molecular Weight of 2-methyloct-3-en-2-ol (C9H18O)

## Identification of 2-methyloct-3-en-2-ol Chemical Compound

Chemical Formula C9H18O 142.23862 g/mol (3E)-2-methyloct-3-en-2-ol CCCCC=CC(C)(C)O InChI=1S/C9H18O/c1-4-5-6-7-8-9(2,3)10/h7-8,10H,4-6H2,1-3H3/b8-7+ ACFQKNIZLQUKQA-BQYQJAHWSA-N

## Molecular Weight Description

The 2-methyloct-3-en-2-ol molecule consists of 18 Hydrogen atom(s), 9 Carbon atom(s) and 1 Oxygen atom(s) - a total of 28 atom(s). The molecular weight of 2-methyloct-3-en-2-ol is determined by the sum of the atomic weights of each constituent element multiplied by the number of atoms, which is calculated to be:

$$142.23862 · \frac{g}{mol}$$

The exact term of the above molecular weight is “molar mass”, which is based on the atomic mass of each element. Molecular weight is actually an older term of “relative molar mass” or “molecular mass”, which is a dimensionless quantity equal to the molar mass divided by the molar mass constant defined by 1 g/mol.

Molecular masses are calculated from the standard atomic weights of each nuclide, while molar masses are calculated from the atomic mass of each element. The atomic mass takes into account the isotopic distribution of the element in a given sample.

## More Properties of 2-methyloct-3-en-2-ol

For physicochemical, thermodynamic, transport, spectra, and other property data & information, the followings are available from “Mol-Instincts”, a chemical database based on quantum mechanics:

## Additional Information for Identifying 2-methyloct-3-en-2-ol Molecule

• ### Structure Data File (SDF/MOL File) of 2-methyloct-3-en-2-ol

The structure data file (SDF/MOL File) of 2-methyloct-3-en-2-ol is available for download in the SDF page of 2-methyloct-3-en-2-ol, which provides the information on atoms, bonds, connectivity and coordinates of 2-methyloct-3-en-2-ol. The 2-methyloct-3-en-2-ol structure data file can be imported to most of the chemistry-related software, providing three-dimensional visualization and further analysis.

• ### Chemical structure of 2-methyloct-3-en-2-ol

The molecular structure image of 2-methyloct-3-en-2-ol is available in chemical structure page of 2-methyloct-3-en-2-ol, which provides the molecular geometry information, i.e., the spatial arrangement of atoms in 2-methyloct-3-en-2-ol and the chemical bonds that hold the atoms together.

• ### Chemical formula of 2-methyloct-3-en-2-ol

The molecular formula of 2-methyloct-3-en-2-ol is given in chemical formula page of 2-methyloct-3-en-2-ol, which identifies each constituent element by its chemical symbol and indicates the proportionate number of atoms of each element.

• ### InChI (IUPAC International Chemical Identifier) information of 2-methyloct-3-en-2-ol

In addition to the molecular weight information, the structural information of 2-methyloct-3-en-2-ol in a textual expression is available via InChi. The full standard InChI of 2-methyloct-3-en-2-ol is given below:

InChI=1S/C9H18O/c1-4-5-6-7-8-9(2,3)10/h7-8,10H,4-6H2,1-3H3/b8-7+

It can provide a way to encode the molecular information of 2-methyloct-3-en-2-ol to facilitate the search for the compound information in databases and on the web.

The condensed, 27 character standard InChIKey (hashed version of the full standard InChI) of 2-methyloct-3-en-2-ol is:

InChIKey=ACFQKNIZLQUKQA-BQYQJAHWSA-N

It may allow easier web searches for 2-methyloct-3-en-2-ol. The InChIKey, however, needs to be linked to the full InChI as well in order to get back to the original structure of the 2-methyloct-3-en-2-ol as the full standard InChI cannot be reconstructed from the InChIKey.

• ### Other names (synonyms) or registry numbers of 2-methyloct-3-en-2-ol

There may be different names of the 2-methyloct-3-en-2-ol compound depending on the various situations of industrial applications, which are given below including the registry numbers if available:

• 2-methyloct-3-en-2-ol
• 18521-06-7
• (E)-2-methyloct-3-en-2-ol

## 2-methyloct-3-en-2-ol Identification Summary Frequently Asked Questions (FAQs)

What’s the formula of 2-methyloct-3-en-2-ol?
C9H18O
How many atoms and what are the elements in the 2-methyloct-3-en-2-ol molecule?
28 atom(s) - 18 Hydrogen atom(s), 9 Carbon atom(s) and 1 Oxygen atom(s)
How many chemical bonds and what types of bonds are in the 2-methyloct-3-en-2-ol structure?
27 bond(s) - 9 non-H bond(s), 1 multiple bond(s), 4 rotatable bond(s), 1 double bond(s), 1 hydroxyl group(s) and 1 tertiary alcohol(s)
What’s the molar mass of 2-methyloct-3-en-2-ol?
142.23862 g/mol
What’s the SMILES format of 2-methyloct-3-en-2-ol?
CCCCC=CC(C)(C)O
What’s the InChI format of 2-methyloct-3-en-2-ol?
InChI=1S/C9H18O/c1-4-5-6-7-8-9(2,3)10/h7-8,10H,4-6H2,1-3H3/b8-7+
What’s the InChIKey string of 2-methyloct-3-en-2-ol?
ACFQKNIZLQUKQA-BQYQJAHWSA-N

8 The contents of this page can freely be shared if cited as follows:
Source: Mol-Instincts Chemical Database, Predicted on Quantum.


text:
# Writing a signed number for a real-world situation Online Quiz

Following quiz provides Multiple Choice Questions (MCQs) related to Writing a signed number for a real-world situation. You will have to read all the given answers and click over the correct answer. If you are not sure about the answer then you can check the answer using Show Answer button. You can use Next Quiz button to check new set of questions in the quiz.

Q 1 - Write a signed integer for following real-world situation.

16 degrees below zero temperature.

### Explanation

Step 1:

Below zero temperatures are represented by negative signed integers.

Step 2:

16 degrees below zero temperature is represented by the signed integer −16.

Q 2 - Write a signed integer for following real-world situation.

The opposite of -299.

### Explanation

Step 1:

The opposite of negative signed integer is represented by a positive signed integer.

Step 2:

The opposite of -299 is represented by the signed integer +299 or just 299.

Q 3 - Write a signed integer for following real-world situation.

A gain of eleven pounds in weight.

### Explanation

Step 1:

A gain is represented by a positive signed integer.

Step 2:

A gain of eleven pounds in weight is represented by the signed integer +11

Q 4 - Write a signed integer for following real-world situation.

The stock market went down 850 points today.

### Explanation

Step 1:

Stock market going down is represented by negative signed integer.

Step 2:

The given statement 'The stock market went down 850 points today' is represented by the signed integer −850

Q 5 - Write a signed integer for following real-world situation.

A pay cut of $490. ### Answer : B ### Explanation Step 1: A pay cut is represented by a negative signed integer. Step 2: A pay cut of$490 is represented by the signed integer −490

Q 6 - Write a signed integer for following real-world situation.

425 feet above sea level.

### Explanation

Step 1:

Position above sea level is represented by a positive signed integer.

Step 2:

425 feet above sea level is represented by the signed integer + 425

Q 7 - Write a signed integer for following real-world situation.

The opposite of 539.

### Explanation

Step 1:

The opposite of positive signed integer is represented by a negative signed integer.

Step 2:

The opposite of 539 is represented by the signed integer −539

Q 8 - Write a signed integer for following real-world situation.

84 feet below sea level.

### Explanation

Step 1:

Position below sea level is represented by a negative signed integer.

Step 2:

84 feet below sea level is represented by the signed integer −84

Q 9 - Write a signed integer for following real-world situation.

A deposit of $2,860 in bank account ### Answer : B ### Explanation Step 1: A deposit in a bank is represented by a positive signed integer. Step 2: A deposit of$2,860 in bank account is represented by the signed integer +2860

Q 10 - Write a signed integer for following real-world situation.

Stock market went up 3,690 points


text:
# What is the effect of measuring a node in a cluster state?

I'm still quite confused about the post-measurement state when a node of a cluster state is measured.

As far as I'm aware, for a cluster state, a given node $$j$$ will have a stabilizer

$$X_j\bigotimes_{k\in N_j}Z_k$$

for the set of nodes $$N_j$$ around $$j$$.

I don't understand the post-measurement state of one of these nodes. Say this $$j$$-th node is measured, the node will disappear from the cluster, but what will the remaining cluster state look like and what will the stabilizers be for each remaining node?

An explanation or a link to an explanation would be wonderful.

• Are you asking about measurement in a specific basis? Jun 7 at 6:19


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# Pig without Hadoop on Windows 7

I am trying to run PigUnit tests on a Windows 7 machine before running the actual pig script on a Ubuntu cluster and I start to think that my understanding of "withouthadoop" is not correct.

Do I need to install Hadoop to locally run a PigUnit test on a Windows 7 machine?

I installed:

1. eclipse Juno & ant
2. cygwin

I set up:

1. JAVA_HOME=C:\Program Files\Java\jdk1.6.0_39
2. PIG_HOME=C:\Users\john.doe\Java\eclipse\pig
3. PIG_CLASSPATH=%PIG_HOME%\bin

I created using eclipse's Ant builder jar-all and pigunit-jar:

1. pig.jar
2. pig-withouthadoop.jar
3. pigunit.jar

Still when I type pig -x local in cygwin I get:

$./pig -x local cygpath: can't convert empty path Exception in thread "main" java.io.IOException: Error opening job jar: /usr/lib/pig/pig-withouthadoop.jar at org.apache.hadoop.util.RunJar.main(RunJar.java:135) Caused by: java.io.FileNotFoundException: \usr\lib\pig\pig-withouthadoop.jar (the systen cannot find the given path) at java.util.zip.ZipFile.open(Native Method) at java.util.zip.ZipFile.<init>(ZipFile.java:127) at java.util.jar.JarFile.<init>(JarFile.java:136) at java.util.jar.JarFile.<init>(JarFile.java:73) at org.apache.hadoop.util.RunJar.main(RunJar.java:133)  When I try to run the test from http://pig.apache.org/docs/r0.10.0/test.html#pigunit from within eclipse using the option "Run as JUnit", I get: java.io.IOException at org.apache.pig.pigunit.pig.PigServer.registerScript(PigServer.java:62) at org.apache.pig.pigunit.PigTest.registerScript(PigTest.java:171) at org.apache.pig.pigunit.PigTest.assertOutput(PigTest.java:267) at org.apache.pig.pigunit.PigTest.assertOutput(PigTest.java:262) at da.utils.pigunit.PigUnitExample.testTop2Queries(PigUnitExample.java:72) at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method) at sun.reflect.NativeMethodAccessorImpl.invoke(Unknown Source) at sun.reflect.DelegatingMethodAccessorImpl.invoke(Unknown Source) at java.lang.reflect.Method.invoke(Unknown Source) at org.junit.runners.model.FrameworkMethod$1.runReflectiveCall(FrameworkMethod.java:47)
at org.junit.internal.runners.model.ReflectiveCallable.run(ReflectiveCallable.java:12)
at org.junit.runners.model.FrameworkMethod.invokeExplosively(FrameworkMethod.java:44)
at org.junit.internal.runners.statements.InvokeMethod.evaluate(InvokeMethod.java:17)
at org.junit.runners.ParentRunner.runLeaf(ParentRunner.java:271)
at org.junit.runners.BlockJUnit4ClassRunner.runChild(BlockJUnit4ClassRunner.java:70)
at org.junit.runners.BlockJUnit4ClassRunner.runChild(BlockJUnit4ClassRunner.java:50)
at org.junit.runners.ParentRunner$3.run(ParentRunner.java:238) at org.junit.runners.ParentRunner$1.schedule(ParentRunner.java:63)
at org.junit.runners.ParentRunner.runChildren(ParentRunner.java:236)
at org.junit.runners.ParentRunner.access$000(ParentRunner.java:53) at org.junit.runners.ParentRunner$2.evaluate(ParentRunner.java:229)
at org.junit.runners.ParentRunner.run(ParentRunner.java:309)
at org.eclipse.jdt.internal.junit4.runner.JUnit4TestReference.run(JUnit4TestReference.java:50)
at org.eclipse.jdt.internal.junit.runner.TestExecution.run(TestExecution.java:38)
at org.eclipse.jdt.internal.junit.runner.RemoteTestRunner.runTests(RemoteTestRunner.java:467)
at org.eclipse.jdt.internal.junit.runner.RemoteTestRunner.runTests(RemoteTestRunner.java:683)
at org.eclipse.jdt.internal.junit.runner.RemoteTestRunner.run(RemoteTestRunner.java:390)
at org.eclipse.jdt.internal.junit.runner.RemoteTestRunner.main(RemoteTestRunner.java:197)

I am starting to think that I missed some crucial basic information to run Pig on Windows, I have to say also that I am no experienced user with Windows 7 and cygwin, I come from the Unix world.

-

When you type pig -x local at cygwin's dollar prompt, the bash command script 'pig' is invoked and run. Have a look at it (from your PIG_HOME) with $cd bin$ cat pig. I've been right through it these last few days with vim (!) and near the end of the code is a little fork for cygwin users, in order to cast environment variables, that up until now have been in Unix format, into a form that the Windows version of java.exe will understand when 'exec java ...' is called right at the end of the script. Without this conversion, the Windows java.exe won't understand its parameters:

'#' cygwin path translation
if $cygwin; then CLASSPATH=cygpath -p -w "$CLASSPATH" PIG_HOME=cygpath -d "£PIG_HOME" PIG_LOG_DIR=cygpath -d "$PIG_LOG_DIR" fi Cygpath is a cygwin utility that converts UNIX-style file paths into Windows-style file paths, and vice versa. The error message: "cygpath: can't convert empty path" must come from here, I think. Check that CLASSPATH, PIG_HOME and PIG_LOG_DIR aren't empty, perhaps by placing your own echo commands in the script. On my machine and installation, there was an error generated here, but not the same as yours. I found replacing -w and -d with -m, which makes cygpath use the C:/Program Files/Java... syntax conversion, worked. But then other problems appear, which I shall leave for my own question. - Don't fight it. Install Hadoop HDInsight server on Windows from the Web Platform installer: http://www.microsoft.com/web/downloads/platform.aspx It doesn't take long or take up that much space, and the whole shebang is just set up and running for you. I can't get Pig scripts to take parameters, and there's no HBase, but you get HDFS, Pig, Hive. You can even get a whole local cluster going if just follow:: http://social.msdn.microsoft.com/Forums/en-US/hdinsight/thread/885efc22-fb67-4df8-8648-4ff38098dac6/ - I need HBase, that is not optional in my case. But this is a good thing to mention it, some reader may be in this case. – hadinbe Apr 24 '13 at 21:23 maybe you could have Hbase on a different machine, and still do your Pig thing on Win7 with HDInsight. Never wrestle with a pig, you both get dirty, and the pig likes it ... – TobyEvans Apr 25 '13 at 9:08 and HBase is now in HDP 1.3, works really well – TobyEvans Nov 13 '13 at 12:02 I have installed pig 0.12 in cygwin (I run windows 7 64-bit) without installing hadoop. As far as I can see, the steps I followed where: • Install Cygwin64 (with Perl package) • Download pig-0.12.1.tar.gz, copy to home folder • Extract to home folder in cygwin:$ tar xzf pig-0.12.1.tar.gz

• Export JAVA_HOME:

$export JAVA_HOME=/cygdrive/c/Program\ Files/Java/jre6/ • Add pig to path:$ export PATH=~/pig-0.12.1/bin/:$PATH • Rename pig-0.12.1.jar to pig.jar:$ cp pig-0.12.1.jar pig.jar

• Export PIG_CLASSPATH:

$export PIG_CLASSPATH=~/pig-0.12.1/pig.jar • Run pig in local mode (start Grunt):$ pig -x local

• There will be a warning: "cygpath: cannot create short name of C:\cygwin64\home\xxx\pig-0.12.1\logs" We can remove it simply by running:

$mkdir logs - Some follow up: I moved the exports to my .bash_profile and I also modified the JAVA_HOME definition to avoid issues with the space in "Program Files" (I made a junction in windows to map the java folder into c:/java_cygwin/). At that point I could run Pig in local mode and get the Grump shell. However, there seems to be an issue with Grump an Cygwin (you write commands but you basically get no response, this has been reported in other questions in Stack Overflow), so I tried to run pig scripts instead. – Alfonso Jul 16 '14 at 13:18 This did not work because the default hadoop version has a bug that make uncompatible with cygwin. What I did at this point was to install hadoop hadoop-0.20.2. This was useful: gist.github.com/tariqmislam/2159173 but the ssh part was a bit outdated, so I followed ebiquity.umbc.edu/Tutorials/Hadoop/05%20-%20Setup%20SSHD.html – Alfonso Jul 16 '14 at 13:20 Still I had another issue with data node not being alive. What I did in the end was to try to revert to the original configuration for the files in conf (I had modified some files according to the tutorial mentioned above). I think this is related to running hadoop in pseudo distributed mode instead on single node, or something like that. Anyway, in the end I managed to run the pig script shown in the gettings started guide in the official pig documentation. – Alfonso Jul 16 '14 at 13:23 The bottom line is that it is quite painful to get it working with Cywin, and I have just got working the script part, not Grunt. I think it is much simpler to download one virtual machine and do the setup with Linux. There are ready installed virtual machines where hadoop and pig are installed, so that that is what Im trying now... – Alfonso Jul 16 '14 at 13:25 According to this note [1], it is not possible to use the Hadoop native libraries on windows 7 directly using Cygwin: Hadoop native library is supported only on *nix platforms only. Unfortunately it is known not to work on Cygwin [...] I have traced down the error message cygpath: can't convert empty path to the line JAVA_LIBRARY_PATH=cygpath -w "$JAVA_LIBRARY_PATH" in hadoop-config.sh which I commented out following the advice from [2]

Comment some translation in hadoop-config.sh.

#if $cygwin; then #HADOOP_PREFIX=cygpath -w "$HADOOP_PREFIX"
#HADOOP_LOG_DIR=cygpath -w "$HADOOP_LOG_DIR" #JAVA_LIBRARY_PATH=cygpath -w "$JAVA_LIBRARY_PATH"
#fi

Now I get the following error:

Error before Pig is launched -- ERROR 2999: Unexpected internal error.
java.lang.UnsupportedOperationException: Not implemented by the DistributedFileSystem FileSystem implementation

So the conclusion I draw from this, is that Pig, even in local mode, requires the HDFS. And the HDFS requires the Hadoop native libraries. And the native libraries are known not to work on Cygwin. Hence: IMHO, Pig cannot run using Cygwin as it is.

-


text:
Next Meeting: October 20: Web Application Hacking: How to Make and Break Security on the Web Next Installfest: TBD Latest News: Oct. 10: LUGOD Installfests coming again soon Page last updated: 2001 Dec 30 16:52
Re: WD HDs (was Re: [vox] Hard Drives are cheap...)

# Re: WD HDs (was Re: [vox] Hard Drives are cheap...)

what zach is referring to, for those who haven't read this on tom's hardware,
is the "pin heighten adjustment" that people are using these days.

the theory goes like this:

IC chips need only a very small amount of current to work.  any current more
than required is simply wasted as thermal energy.

now heat conduction is proportional to the area of the surface in contact,
the surface area of an IC pin is 2 \pi r h, where h is the depth of the pin.

since only a very small amount of current is needed to keep the CPU active, a
common technique to keep wasted current from flowing is to clip off about 1
millimeter from each pin on your pentium or athlon chip.

this was dubbed, on tomshardware.com, as "shaving".

you need a *very* steady hand.  i would recommend practising on a socket PII
or something before attempting it on an expensive chip.

peter

begin: Zach Johnson <zjohnson@math.ucdavis.edu> quote
> Be sure to shave the cpu before applying the oil.
>
> On Fri, Oct 26, 2001 at 11:51:34AM -0700, Peter Jay Salzman wrote:
> > begin: John C. Alden <jcalden@ucdavis.edu> quote
> > > At 12:05 AM 10/26/01 -0700, you wrote:
> > > >begin: Mark K. Kim <markslist@cbreak.org> quote
> > > >> On Thu, 25 Oct 2001, Peter Jay Salzman wrote:
> > > >
> > > >it all goes into the equation of running a well-oiled system.
> > > >
> > > >pete
> > >
> > > You're supposed to OIL these things? Sheesh, no one told me!
> >
> > well, not all machines.  you're only supposed to oil the really hot CPU's.
> > there's a couple of benefits:
> >
> > 1. because of the heat generated by cpu's, the cpu pins have a tendancy to
> >    oxidize against the walls of the pin inserts.  oiling the pins will help
> >    prevent cpu's from getting stuck.  particular socket type cpu's.
> >
> > 2. oil helps transfer the heat away from the CPU.  this also helps the cpu
> >    run faster.
> >
> > 3. a light coating of oil also protects the CPU from dust and crud which gets
> >    trapped inside the case.
> >
> > for these reasons, it's highly recommended to apply a light coating of oil on
> > your cpu.  there are special oils made for the CPU.  if you have an AMD chip,
> > ONLY USE AMD APPROVED CPU OIL!!!
> >
> > the thing is, any light-weight cpu oil would work just as fine.  i personally
> > use normal cooking vegetable oil.  my guess is that something peanut oil
> > would probably be too heavy for this particular application.
> >
> > pete
> >
> > ps- oil also helps in the hot-swapping of IDE devices.
> >
> > --
> > "You may not use the Software in connection with any site that disparages
> > Microsoft, MSN, MSNBC, Expedia, or their products or services ..."
> >                     -- Clause from license for FrontPage 2002
>
> --
> -
> . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
>
>   Zachariah D. Johnson		Phone : 1-530-752-8801
>   Mathematics			Office: 657 Kerr Hall
>   UCDavis			WWW   : n/a

--
"You may not use the Software in connection with any site that disparages
Microsoft, MSN, MSNBC, Expedia, or their products or services ..."
-- Clause from license for FrontPage 2002


text:
# What question could be answered by “because the metabolites in the cell are in ionized form” in the topic of anaerobic metabolism?

I know it might sound a bit absurd, but my teacher gave us some "answers to question", that is, given one answer, write at least one question the answer to which is the given one.

I've solved most of the answers, but there's one I just can't figure out: "because the metabolites in the cell are in ionized form", and the topic is anaerobic respiration and fermentation. No additional details from her.

I'm in high school (a "scientific" school in Italy), if that can help for the knowledge level.

Any idea?

• Hmmm, scientific Jeopardy, interesting idea... – Martin - マーチン Apr 29 '20 at 10:38
• "Why does ethanol escape from cells but metabolites do not when anaerobic respiration and fermentation are happening?". BTW: Not all metabolites are in ionized form (e.g. glucose), but many are, including neutral molecules attached to charged carriers (phosphorylated glucose, fatty acids with CoA etc). – Karsten Theis Apr 29 '20 at 12:44
• That's a pretty valid question, thanks! Any other idea? – user256439 Apr 29 '20 at 13:35

The answer to the first one would be because the metabolite of lactose fermentation in the lactobacilli cells is lactic acid, that in ionized form releases $$\ce{H+}$$, lowering the medium pH until the isoelectric point of the milk proteines is reached, causing their agglutination. The answer to the second one is that because the venous blood carries away carbonic acid produced as a metabolite in cellular respiration, that has two ionized forms, $$\ce{HCO3-}$$(bicarbonate ion) and $$\ce{CO3^2-}$$(carbonate ion). The answer to the third one would be that, as in lactobacilli case, our body resorts to anaerobic lactic fermentation in activities that require a large power output, beyond the oxygen delivery capacity of the circulatory system. So lactic acid builds up in the muscles, lowering the pH and causing the fatigue sensation. This last explanation was broadly accepted, but seems to be controversial in light of recent studies, as we can see in the abstract for this article:


text:
# [NTG-context] Mutiple citations doesn't work with compress=no + interaction

Michael Murphy michael.murphy at uni-ulm.de
Wed Jun 9 11:21:29 CEST 2010

On 08/06/10 20:47, taco wrote:
>
> It also only happens in mkiv, something is broken there. As is
> usual in the mkiv version of the bib module, I can't figure
> it out :(
>
> However, here is a workaround:
>
> ...
>

... and indeed, this works for the minimal example, but in my actual
document, I get an error:

! Undefined control sequence.
\dododescriptioncomponent ...entdescriptioncoding
\s!tex \fi \ifx

and I have no idea why. I guess I'll play around some more.

--
Michael Murphy


text:
# LXX International conference "NUCLEUS – 2020. Nuclear physics and elementary particle physics. Nuclear physics technologies"

Oct 11 – 17, 2020
Online
Europe/Moscow timezone

## APPLICATION OF THE LIQUID SCINTILLATION ALPHA AND BETA SPECTROMETER QUANTULUS 1220 FOR DATING OF NATURAL OBJECTS

Oct 15, 2020, 6:15 PM
1h
Online

#### Online

Poster report Section 3. Modern nuclear physics methods and technologies.

### Speaker

Konstantin Gruzdov (FGBU «VSEGEI»)

### Description

The Quantulus 1220 is a liquid scintillation counting (LSC) system for the quantitative measurement of extremely low levels of alpha and beta activity. With both passive and active shielding, the Quantulus 1220 employs a universal background reduction system which is optimized according to type of analysis.
In the Centre of Isotopic Research (CIR) of FGBU «VSEGEI» Quantulus 1220 is used for radiocarbon dating of various organic objects (wood, peat, soil, bottom sediments, bones), dating young bottom sediments using ${}^{210}$Pb as well as determination the tritium content in water.
For radiocarbon dating the organic matter of a sample is chemically converted to benzene. The ${}^{14}$C activity is measured relative to the modern standard. Also it is necessary to measure the activity of the background sample (benzene without ${}^{14}$C).
When dating young bottom sediments by ${}^{210}$Pb, all the lead (99%) in the sample is chemically extracted. Then Optiphase HiSafe 3 liquid scintillator is added to the slightly acidic solution containing the lead.
When measuring the tritium content in water, a water sample is directly mixed with the Optiphase TriSafe 3 liquid scintillator. The minimum detectable concentration of tritium in water is approximately 1 Bq/L.
The obtained results are presented as the decay spectra of radioactive isotopes with age calculations.

### Primary author

Konstantin Gruzdov (FGBU «VSEGEI»)


text:
My Math Forum Differentiable or not

 Real Analysis Real Analysis Math Forum

March 26th, 2018, 10:13 AM   #11
Senior Member

Joined: May 2016
From: USA

Posts: 1,126
Thanks: 468

Quote:
 Originally Posted by Lalitha183 Which one do we take as RHD & LHD? Because to my question, answers are like is differentiable/ has L.H.D but not R.H.D/ has R.H.D but not L.H.D/ Neither L.H.D nor R.H.D.
If f(x) is not continuous when x = 0, is f(x) differentiable when x = 0?

March 27th, 2018, 03:53 AM   #12
Senior Member

Joined: Nov 2015

Posts: 232
Thanks: 2

Quote:
 Originally Posted by JeffM1 If f(x) is not continuous when x = 0, is f(x) differentiable when x = 0?
Every differentiable function is continuous. So, it won't.

March 27th, 2018, 04:39 AM   #13
Math Team

Joined: Jan 2015
From: Alabama

Posts: 3,261
Thanks: 894

Quote:
 Originally Posted by Lalitha183 Which one do we take as RHD & LHD? Because to my question, answers are like is differentiable/ has L.H.D but not R.H.D/ has R.H.D but not L.H.D/ Neither L.H.D nor R.H.D.
"RHD" means "Right hand derivative"- that is, coming toward x from that right (above) so is $\displaystyle \frac{f(x+h)- f(x)}{h}$ with h positive, x+ h> x. "LHD" means "Left hand derivative"- that is, coming toward x from the left (below) so is $\displaystyle \frac{f(x+ h)- f(x)}{h}$ with h negative, x+ h< x.

March 27th, 2018, 06:13 AM   #14
Senior Member

Joined: May 2016
From: USA

Posts: 1,126
Thanks: 468

Quote:
 Originally Posted by Lalitha183 Every differentiable function is continuous. So, it won't.
There you go.

Just remember that although

$\text {Not continuous } \implies \text { not differentiable}$,

$\text {Continuous } \not \implies \text { differentiable.}$

So you must address the existence and continuity of the Newton quotient after you find the function to be continuous. And like all limits, to be continuous, the limit of the Newton quotient must exist and be equal whether approached from left or right.

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text:
# What is an example of a ring in which the intersection of all maximal two-sided ideals is not equal to the Jacobson radical?

What is an example of a ring in which the intersection of all maximal two-sided ideals is not equal to the Jacobson radical? Wikipedia suggests that any simple ring with a nontrivial right ideal would work, but this is clearly false (take a matrix ring over a field, for instance).

Benson's Representations and Cohomology I, on the other hand, claims that the Jacobson radical is in fact the intersection of all maximal two sided ideals. He defines the Jacobson radical as the intersection of the annihilators of simple R-modules, which are precisely the maximal two-sided ideals. Since this is the same as the intersection of the annihilators of the individual elements of the simple modules, then this is the same as the intersections of the maximal left (or right) ideals.

I don't see the flaw in Benson's reasoning, but I seem to recall hearing somewhere else that the Jacobson radical is not always the intersection of the maximal two-sided ideals. Who is correct here?

-
Here's the flaw in Benson's reasoning: The annihilator of a simple right R-module is a <em>(right) primitive ideal</em>, which is not necessarily a maximal ideal as indicated by the answers to your question. The Jacobson radical is equal to the intersection of all maximal right ideals, and is equal to the intersection of all right primitive ideals (as well as the left-handed versions of these intersections) – Manny Reyes Oct 27 '09 at 2:16
Dave Benson has done a lot of interesting work, but he is most often thinking about finite groups and their (finite dimensional) group algebras. His books sometimes get too casual about the generalities, so should not be taken as a basic reference for this kind of ring theory. Jacobson himself wrote a pair of textbooks Basic Algebra which cover questions about arbitrary rings thoroughly. By now there are a number of such textbook treatments. – Jim Humphreys Dec 27 '10 at 15:47

A very important example is the quotient of $U(\mathfrak{g})$ (where $\mathfrak{g}$ is a simple complex Lie algebra) by the central elements killing a finite dimensional representation. This has a unique maximal ideal (the annihilator of the finite dimensional module), but its Jacobson radical is trivial, since the annihilator of the simple highest weight module in this block whose highest weight is in the anti-dominant Weyl chamber is actually faithful. There's actually an extremely interesting poset (by inclusion) of primitive ideals sitting in between.


text:
Source Finding¶

find_sources() is an iterative source-finding algorithm that uses peak detection on a TS map to find new source candidates. The procedure for adding new sources at each iteration is as follows:

• Generate a TS map for the test source model defined with the model argument.
• Identify peaks with sqrt(TS) > sqrt_ts_threshold and an angular distance of at least min_separation from a higher amplitude peak in the map.
• Order the peaks by TS and add a source at each peak starting from the highest TS peak. Set the source position by fitting a 2D parabola to the log-likelihood surface around the peak maximum. After adding each source, re-fit its spectral parameters.
• Add sources at the N highest peaks up to N = sources_per_iter.

Source finding is repeated up to max_iter iterations or until no peaks are found in a given iteration. Sources found by the method are added to the model and given designations PS JXXXX.X+XXXX according to their position in celestial coordinates.

Examples¶

model = {'Index' : 2.0, 'SpatialModel' : 'PointSource'}
srcs = gta.find_sources(model=model, sqrt_ts_threshold=5.0,
min_separation=0.5)

The method for generating the TS maps can be controlled with the tsmap_fitter option. TS maps can be generated with either tsmap() or tscube().

Reference/API¶

GTAnalysis.find_sources(prefix='', **kwargs)

An iterative source-finding algorithm that uses likelihood ratio (TS) maps of the region of interest to find new sources. After each iteration a new TS map is generated incorporating sources found in the previous iteration. The method stops when the number of iterations exceeds max_iter or no sources exceeding sqrt_ts_threshold are found.

Parameters: free_params (list) – (default : None) max_iter (int) – Maximum number of source finding iterations. The source finder will continue adding sources until no additional peaks are found or the number of iterations exceeds this number. (default : 5) min_separation (float) – Minimum separation in degrees between sources detected in each iteration. The source finder will look for the maximum peak in the TS map within a circular region of this radius. (default : 1.0) model (dict) – Dictionary defining the spatial/spectral properties of the test source. If model is None the test source will be a PointSource with an Index 2 power-law spectrum. (default : None) multithread (bool) – Split the calculation across all available cores. (default : False) sources_per_iter (int) – Maximum number of sources that will be added in each iteration. If the number of detected peaks in a given iteration is larger than this number, only the N peaks with the largest TS will be used as seeds for the current iteration. (default : 4) sqrt_ts_threshold (float) – Source threshold in sqrt(TS). Only peaks with sqrt(TS) exceeding this threshold will be used as seeds for new sources. (default : 5.0) tsmap_fitter (str) – Set the method for generating the TS map. Valid options are tsmap or tscube. (default : tsmap) tsmap (dict) – Keyword arguments dictionary for tsmap method. tscube (dict) – Keyword arguments dictionary for tscube method. peaks (list) – List of peak objects. sources (list) – List of source objects.


text:
# List of models

SSLR contains models created by developers and wrappers of different packages such as RSSL. From RSSL, we use S3VM methods.

The list of models is:

NOTE: In the Regression modelling section we can see more examples of use in regression tasks. In Decision Tree , Random Forest and coBC we only have examples for classification tasks.

NOTE: In the Clustering modelling section we can see how to plot clusters with factoextra package.


text:
# Question about a spanning set

Let $V$ be an $n$-dimensional vector space. Then any linearly independent set of vectors $\{v_1, v_2, \ldots, v_n\}$ is a basis for $V$

Proof:

Let $w \in V$. Then $\{v_1, v_2,\ldots, v_n, w \}$ is linearly dependent. Then exists at least one $c_i \ne 0$ s.t. $\displaystyle{c_0w +\sum_{i = 1}^nc_iv_i} = \vec 0.$ If $c_0 = 0,$ then there's at least one $c_i \ne 0$ in the sum $\displaystyle{\sum_{i = 1}^nc_iv_i} = \vec 0.$ Contradiction as $\{v_1, \ldots, v_n\}$ are linearly independent. Hence $c_0 \ne 0$ meaning we can solve for $w: \ w= \displaystyle{\sum_{i = 1}^n\left(-\frac{c_i}{c_0}\right)v_i}.$ Thus $v_i$ span $V$.

We know $\{v_1, v_2,\ldots, v_n, w \}$ is linearly dependent because $\dim(V) = n$ and $v_i \in V$ meaning $v_i$ span $V$. The proof above shows that $\text{span}(v_1,\ldots,v_n) = V.$ Are we proving something we already know from the definition of basis? Is there subtlety here that I am missing?

The definition of basis is a set of vectors that is linearly indepedent and spans the space. This theorem states that in a finite-dimensional vector space, as long as your set has $n$-many vectors and it's linearly indepedent, then it also spans the space. Therefore it is a basis.
Also, we know $B=\{v_1, v_2,\ldots, v_n, w \}$ is linearly depedent because the length of any linearly indepedent list must be less than or equal to the length of a spanning set. Since $\dim V=n$, the spanning sets of $V$ have length $n$, thus since $|B| \not \leq n$, it must be linearly depedent.


text:
# Thread: Find volume of solid using squares

1. ## Find volume of solid using squares

The base of a certain solid is a region in the first quadrant bounded by the curve and . Find the volume of the solid if each gross section perpendicular to the x-axis is a square with one edge in the base of the solid.

Attempt to solution:

$\displaystyle A= s^2= (x-x^2)^2$

$\displaystyle V=\int_0^1(x-x^2)^2$

How come the textbook has it the other way round

$\displaystyle V=\int_0^1(x^2-x)^2$

2. Does it matter? Not in this case.

It is traditional to write things in decreasing exponents.

3. I think what needs to be observed here is that $\displaystyle x-x^2=-\left(x^2-x\right)$ but $\displaystyle \left(x-x^2\right)^2=\left(-\left(x^2-x\right)\right)^2=(-1)^2\left(x^2-x\right)^2=\left(x^2-x\right)^2$