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# Symmetries and Conversation Laws

1. Mar 3, 2009

### ophase

According to Noether Theorem, Every symmetry corresponds to a conversation Law in the nature.
For example: If rotational symmetry exits, Angular momentum is conserved.

How can I be sure that which symmetry corresponds to which conversation Law?
Can you tell me the conversation Laws under these symmetries:
Charge symmetry,
Parity symmetry,
Time symmetry,
Flavor symmetry,
Helical symmetry,
Scale symmetry.

2. Mar 3, 2009

### malawi_glenn

Time symmetry is conservation of Energy

3. Mar 3, 2009

Staff Emeritus
That's not what it says. It makes statements about specific kinds of symmetries in the Lagrangian.

Before going too far down this path, do you understand Noether's Theorem? Could you prove it if you had to? If the answer is no, probably the best we can do is to say some symmetries have this property, and to leave "some" vague.

4. Mar 3, 2009

### clem

I believe Noether's theorem applies to continuous symmetries.
Most of the symmetries you list are discrete, internal symmetries.

5. Mar 3, 2009

### blechman

so as the others have said, you should definitely go through and be sure you understand Noether's theorem. You also seem to apply the word "symmetry" and "conservation law" interchangeably in your question, thus jumping the gun! For example, I don't really know what "charge symmetry" is when read literally, but "charge" is CONSERVED due to the gauge symmetry of E&M!

Charge conservation is a result of the gauge symmetry (to be technically correct, charge conservation actually follows from the GLOBAL part of this symmetry, for those who are in the know).

Parity is a discrete spacetime symmetries and Noether's theorem does not apply in the naive way, and mentioned above.

Flavor symmetry gives conservation of flavor quantum number. For example, things like "baryon number" and "lepton number".

Helical symmetry is angular momentum, and is a subset of the rotational symmetry already mentioned.

Scale symmetry is not actually a symmetry in most cases! But when it is a symmetry, it generates conserved charges called "Virasoro charges", but this is very advanced stuff and is not relevant for most phenomenology. The keyword to understand these charges is "Conformal Field Theory".

OK, there is a brute-force answer. The reason I know which charge goes with which symmetry is because I use Noether's theorem to CALCULATE it! That answers the other part of you question.

Hope that helps!

6. Mar 3, 2009

### ophase

I should have said that Noether Theorem concerns continuous transformations on the scalar fields.
But doesn't it mean that for the discrete symmetries Noether theorem leaves us alone??

After infinitesimal translations to Lagrangian or generalized coordinates, i can show that Noether current or momentum conserved.
Is there any other proof of the theorem i should know?? If there is, it maybe helpful for me.

@blechman

I have to say that interchangeability between "symmetry" and "conservation law" confuses me a lot.
Am i right if i would say that Gauge transformations are invariant because of charge symmetry ??
or
Charge is conserved under Gauge symmetry ??
Don't you think there's a conceptual confussion??

Scale symmetry was the most important part of my question. Because i wonder why it is not included in the elementary field theory since it is an appearant symmetry in the universe.

Can we derivate other fields (including supersymmetric and unknown fields too) after some kind of scale transformations??
We know the supersymmetry transformation between boson and fermion fields. Can it be defined as a scale transformation??
Is it what they called supersymmetric conformal field theory??

7. Mar 3, 2009

### hamster143

When proving Noether's theorem, the way you calculate the conserved quantity (constant of the motion) that corresponds to the continuous symmetry is by differentiating the action wrt a variable that parametrizes the symmetry. With discrete symmetries, there's no variable, Noether's theorem can't be applied directly.

Scale symmetry is only a symmetry when your fields are massless. If your fields are massive, you can have a symmetry that preserves the lagrangian by rescaling spacetime, all particle masses, and all dimensional coupling constants.

8. Mar 3, 2009

### humanino

The way I remember it, scale symmetry is badly broken in the real world. Beginning with chiral symmetry breaking breaking in QCD and the special scale at which the coupling constant grows big (even with massless quarks), but with QED as well, chemical bonds set spatial scales, and a simply scaled structured, like say a chair, would collapse under its own weight, if for instance you would scale the planet and and the chair a factor (say) $10^{6}$

9. Mar 4, 2009

### malawi_glenn

No, the Lagrangian is invariant under a Local U(1)-gauge transformation of the fields (your generalized coordinates), and what is conserved, the Noether charge, is the Electric Charge.

10. Mar 4, 2009

### ophase

@malawi_glenn

So I must say;

"Lagrangian is invariant under a Local U(1)-gauge transformations because of charge symmetry"

Do you mean that there's no concept called gauge symmetry?? or they mean another thing when they are talking about gauge symmetry??

Is it wrong to say "Charge is conserved under Gauge symmetry" ??

Last edited: Mar 4, 2009
11. Mar 4, 2009

### malawi_glenn

NO NO NO

You have to say:

The Lagrangian is invariant under a Local U(1)-Gauge transformation, the Noethers Theorem then gives you that Electric Charge is conserved.

Why is it so hard to read what we write?? "charge symmetry" is a sloppy usage of words.

What is symmetric is the Lagrangian under a Local U(1)-gauge transformation. This symmetry will give you charge conservation. Not vice versa or something in between.

12. Mar 4, 2009

### genneth

I don't really want to confuse the issues even more, but can we please be accurate and say that charge conservation is due to the global U(1) symmetry? Apart from the global subgroup, gauge symmetries are not real symmetries; gauge degrees of freedom do not relate different physical states.

13. Mar 4, 2009

### malawi_glenn

yes, of course, we can even be that accurate :-)

14. Mar 4, 2009

### blechman

malawi_glenn and genneth did a good job explaining the gauge symmetry part. That's a subtle issue about the global vs gauge symmetry - it has lead me to make more than one mistake here, on a QFT exam solution set (yikes!), and worst of all, in my own research! So don't feel too bad if you don't understand it.

Moving on to "the most important part of [ophase's] question":

WARNING: this might get complicated. If it confuses you, I'm sorry. Don't worry too much about it.

Scale symmetry $(x\rightarrow\lambda x)$ is a very tricky subject. CLASSICALLY, this is a symmetry of any theory of massless fields, but the problem is that QUANTUM MECHANICALLY, this symmetry is broken! These kinds of symmetries are called "anomalous symmetries".

The way this materializes in the real world is that if you start with a "scale-invariant" theory that has no masses in it, the problem is that the couplings themselves depend on the energy! Example: if you measure the electromagnetic charge of the electron (e) at rest, you get:

$$\alpha=\frac{e^2}{4\pi\epsilon_0\hbar c}=\frac{1}{137}$$

However, if you measure the electromagnetic charge of the electron moving with an energy of 100 GeV, you get:

$$\alpha=\frac{e^2}{4\pi\epsilon_0\hbar c}=\frac{1}{128}$$

So we find that the coupling gets LARGER as the energy increases! This breaks the scale invariance and is due to this "quantum anomaly" I mentioned.

So even massless theories typically do not have scale invariance, except for some VERY special examples, but they are rare.

I do not understand your question about SUSY. Supersymmetry is another spacetime symmetry, different from rotations, translations and scale transformations. Maybe you can clarify?

Anyway, I hope this helps and doesn't confuse. But this is why scale symmetry is treated a little differently than ordinary translations/rotations.

Note Added: Hey, while proofreading this post, I got a sudden rush of deja vu! ophase: did we have this conversation before?!

15. Mar 4, 2009

### blechman

I just wanted to point out that what you say is quite right. However, in all of these cases, you are talking about MASSIVE field theories (except for your coupling growing big example). In MASSLESS QED, for example, there are no bound states. You can see this by two arguments:

1. Brute force: after a hard-work calculation, you find the bond lengths are inversely proportional to the mass of the electron, so as that mass goes to zero, the bond lengths go to infinity and there are no bound states.

2. By dimensional analysis: a massless theory has no scale, and therefore you cannot write down anything with dimensions of length. Therefore there cannot be a bound state.

So your chemistry argument fails in the case of massless electrons for example (not a "real world" scenario, of course, but I'm a theorist and that's never stopped me before! ). However, even massless QED has the running coupling, as I said above.

Anyway, just to clarify our positions.

16. Mar 4, 2009

### humanino

Indeed we agree, and as a matter of fact, I appreciate the clarification.

17. Mar 4, 2009

### ophase

@hamster143

If one have a discrete symmetry and Noether theorem can't be applicable how can one find out the conserved quantity?

@genneth

In the classification of symmetries, I never heard of a branch called real symmetries. Can you give us a more detailed definition of Real Symmetries?

@blechman

That's great to find some scale symmetric relations in QED. But I don't understand how did you calculate the constant while electron moving at 100 GeV.

My question about SUSY: Why should it be a spacetime symmetry?? (I know the additional fields like W,D,F to satisfy the algebra and it disturbs me a lot)
Isn't it possible to construct (let's say) a scale symmetry between the fields??

Do you mean i'm the only one here asking silly questions after our past thread and you'r saying like "i recognize you wherever you go regardless of time" :)

I'll have more questions soon
:)

Last edited: Mar 4, 2009
18. Mar 4, 2009

Staff Emeritus
He just told you - it doesn't apply in this case.

19. Mar 4, 2009

### ophase

Ok, i edited it.. I think it's more clear now... Since english is not my native language, these happens. Sorry

20. Mar 4, 2009

### blechman

Calculate the scattering cross section of electrons fusing to make a Z-boson, which then decays to muons (for example), when the CoM energy of the electrons is at the Z mass (91 GeV, which is close enough to 100). This answer is proportional to the fine structure constant (squared, actually).

Then measure said cross section at LEP. This gives you your value for the coupling.

I'm oversimplifying, but that's the idea.

SUSY is a spacetime symmetry because it has nontrivial relations with the other spacetime symmetry generators such as

$$[Q,\bar{Q}]_+\sim P$$

So it is a spacetime symmetry.

I'm not understanding what's disturbing you. What's "W"? Presumably, "D,F" are the auxiliary fields you introduce in SUSY - they're only introduced to make the MATH easier, you don't really need to introduce them, they're not new "physical" fields.

The scale symmetry is also a spacetime symmetry, but as I said in my last post, it is violated in all but the rarest of quantum field theories. If you try to write a "scale-invariant" theory down, you will find that quantum corrections break the scale symmetry. There's nothing to be done with that! SUSY does not have this problem. Nor does the usual Poincare invariance (translations/rotations).

I seriously thought I wrote these words down before, perhaps with some other post that wasn't you. Maybe I'm just on drugs...


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## Pascal’s Principle

### Learning Objectives

By the end of this section, you will be able to:

• Define pressure.
• State Pascal’s principle.
• Understand applications of Pascal’s principle.
• Derive relationships between forces in a hydraulic system.

Pressure is defined as force per unit area. Can pressure be increased in a fluid by pushing directly on the fluid? Yes, but it is much easier if the fluid is enclosed. The heart, for example, increases blood pressure by pushing directly on the blood in an enclosed system (valves closed in a chamber). If you try to push on a fluid in an open system, such as a river, the fluid flows away. An enclosed fluid cannot flow away, and so pressure is more easily increased by an applied force. What happens to a pressure in an enclosed fluid? Since atoms in a fluid are free to move about, they transmit the pressure to all parts of the fluid and to the walls of the container. Remarkably, the pressure is transmitted undiminished. This phenomenon is called Pascal’s principle, because it was first clearly stated by the French philosopher and scientist Blaise Pascal (1623–1662): A change in pressure applied to an enclosed fluid is transmitted undiminished to all portions of the fluid and to the walls of its container.

### Pascal’s Principle

A change in pressure applied to an enclosed fluid is transmitted undiminished to all portions of the fluid and to the walls of its container.

Pascal’s principle, an experimentally verified fact, is what makes pressure so important in fluids. Since a change in pressure is transmitted undiminished in an enclosed fluid, we often know more about pressure than other physical quantities in fluids. Moreover, Pascal’s principle implies that the total pressure in a fluid is the sum of the pressures from different sources. We shall find this fact—that pressures add—very useful.

Blaise Pascal had an interesting life in that he was home-schooled by his father who removed all of the mathematics textbooks from his house and forbade him to study mathematics until the age of 15. This, of course, raised the boy’s curiosity, and by the age of 12, he started to teach himself geometry. Despite this early deprivation, Pascal went on to make major contributions in the mathematical fields of probability theory, number theory, and geometry. He is also well known for being the inventor of the first mechanical digital calculator, in addition to his contributions in the field of fluid statics.

## Application of Pascal’s Principle

One of the most important technological applications of Pascal’s principle is found in a hydraulic system, which is an enclosed fluid system used to exert forces. The most common hydraulic systems are those that operate car brakes. Let us first consider the simple hydraulic system shown in Figure 1.

Figure 1. A typical hydraulic system with two fluid-filled cylinders, capped with pistons and connected by a tube called a hydraulic line. A downward force F1 on the left piston creates a pressure that is transmitted undiminished to all parts of the enclosed fluid. This results in an upward force F2 on the right piston that is larger than F1 because the right piston has a larger area.

## Relationship Between Forces in a Hydraulic System

We can derive a relationship between the forces in the simple hydraulic system shown in Figure 1 by applying Pascal’s principle. Note first that the two pistons in the system are at the same height, and so there will be no difference in pressure due to a difference in depth. Now the pressure due to F1 acting on area A1 is simply ${P}_{1}=\frac{{F}_{1}}{{A}_{1}}\\$, as defined by $P=\frac{F}{A}\\$. According to Pascal’s principle, this pressure is transmitted undiminished throughout the fluid and to all walls of the container. Thus, a pressure P2 is felt at the other piston that is equal to P1. That is PP2. But since ${P}_{2}=\frac{{F}_{2}}{{A}_{2}}\\$, we see that $\frac{{F}_{1}}{{A}_{1}}=\frac{{F}_{2}}{{A}_{2}}\\$. This equation relates the ratios of force to area in any hydraulic system, providing the pistons are at the same vertical height and that friction in the system is negligible. Hydraulic systems can increase or decrease the force applied to them. To make the force larger, the pressure is applied to a larger area. For example, if a 100-N force is applied to the left cylinder in Figure 1 and the right one has an area five times greater, then the force out is 500 N. Hydraulic systems are analogous to simple levers, but they have the advantage that pressure can be sent through tortuously curved lines to several places at once.

### Example 1. Calculating Force of Slave Cylinders: Pascal Puts on the Brakes

Consider the automobile hydraulic system shown in Figure 2.

Figure 2. Hydraulic brakes use Pascal’s principle. The driver exerts a force of 100 N on the brake pedal. This force is increased by the simple lever and again by the hydraulic system. Each of the identical slave cylinders receives the same pressure and, therefore, creates the same force output F2. The circular cross-sectional areas of the master and slave cylinders are represented by A1 and A2, respectively

A force of 100 N is applied to the brake pedal, which acts on the cylinder—called the master—through a lever. A force of 500 N is exerted on the master cylinder. (The reader can verify that the force is 500 N using techniques of statics from Applications of Statics, Including Problem-Solving Strategies.) Pressure created in the master cylinder is transmitted to four so-called slave cylinders. The master cylinder has a diameter of 0.500 cm, and each slave cylinder has a diameter of 2.50 cm. Calculate the force F2 created at each of the slave cylinders.

#### Strategy

We are given the force F1 that is applied to the master cylinder. The cross-sectional areas A1 and A2 can be calculated from their given diameters. Then $\frac{{F}_{1}}{{A}_{1}}=\frac{{F}_{2}}{{A}_{2}}\\$ can be used to find the force F2. Manipulate this algebraically to get F2 on one side and substitute known values:

#### Solution

Pascal’s principle applied to hydraulic systems is given by $\frac{{F}_{1}}{{A}_{1}}=\frac{{F}_{2}}{{A}_{2}}\\$:

${F}_{2}=\frac{{A}_{2}}{{A}_{1}}{F}_{1}=\frac{{\mathrm{{\pi r}}_{2}}^{2}}{{\mathrm{{\pi r}}_{1}}^{2}}{F}_{1}=\frac{{\left(1.25 \text{ cm}\right)}^{2}}{{\left(0.250 \text{ cm}\right)}^{2}}\times \text{500 N}=1\text{.}\text{25}\times {\text{10}}^{4}\text{N}\\$.

#### Discussion

This value is the force exerted by each of the four slave cylinders. Note that we can add as many slave cylinders as we wish. If each has a 2.50-cm diameter, each will exert 1.25 × 104 N.

A simple hydraulic system, such as a simple machine, can increase force but cannot do more work than done on it. Work is force times distance moved, and the slave cylinder moves through a smaller distance than the master cylinder. Furthermore, the more slaves added, the smaller the distance each moves. Many hydraulic systems—such as power brakes and those in bulldozers—have a motorized pump that actually does most of the work in the system. The movement of the legs of a spider is achieved partly by hydraulics. Using hydraulics, a jumping spider can create a force that makes it capable of jumping 25 times its length!

### Making Connections: Conservation of Energy

Conservation of energy applied to a hydraulic system tells us that the system cannot do more work than is done on it. Work transfers energy, and so the work output cannot exceed the work input. Power brakes and other similar hydraulic systems use pumps to supply extra energy when needed.

## Section Summary

• Pressure is force per unit area.
• A change in pressure applied to an enclosed fluid is transmitted undiminished to all portions of the fluid and to the walls of its container.
• A hydraulic system is an enclosed fluid system used to exert forces.

### Conceptual Questions

1. Suppose the master cylinder in a hydraulic system is at a greater height than the slave cylinder. Explain how this will affect the force produced at the slave cylinder.

### Problems & Exercises

1. How much pressure is transmitted in the hydraulic system considered in Example 1? Express your answer in pascals and in atmospheres.

2. What force must be exerted on the master cylinder of a hydraulic lift to support the weight of a 2000-kg car (a large car) resting on the slave cylinder? The master cylinder has a 2.00-cm diameter and the slave has a 24.0-cm diameter.

3. A crass host pours the remnants of several bottles of wine into a jug after a party. He then inserts a cork with a 2.00-cm diameter into the bottle, placing it in direct contact with the wine. He is amazed when he pounds the cork into place and the bottom of the jug (with a 14.0-cm diameter) breaks away. Calculate the extra force exerted against the bottom if he pounded the cork with a 120-N force.

4. A certain hydraulic system is designed to exert a force 100 times as large as the one put into it. (a) What must be the ratio of the area of the slave cylinder to the area of the master cylinder? (b) What must be the ratio of their diameters? (c) By what factor is the distance through which the output force moves reduced relative to the distance through which the input force moves? Assume no losses to friction.

(5. a) Verify that work input equals work output for a hydraulic system assuming no losses to friction. Do this by showing that the distance the output force moves is reduced by the same factor that the output force is increased. Assume the volume of the fluid is constant. (b) What effect would friction within the fluid and between components in the system have on the output force? How would this depend on whether or not the fluid is moving?

## Glossary

Pascal’s Principle:
a change in pressure applied to an enclosed fluid is transmitted undiminished to all portions of the fluid and to the walls of its container

### Selected Solutions to Problems & Exercises

1. 2.55 × 107 Pa; or 251 atm

3. 5.76 × 103 extra force

5. (a) $V={d}_{\text{i}}{A}_{\text{i}}={d}_{\text{o}}{A}_{\text{o}}\Rightarrow {d}_{\text{o}}={d}_{\text{i}}\left(\frac{{A}_{\text{i}}}{{A}_{\text{o}}}\right)\\$.

Now, using equation:

$\frac{{F}_{1}}{{A}_{1}}=\frac{{F}_{2}}{{A}_{2}}\Rightarrow {F}_{\text{o}}={F}_{\text{i}}\left(\frac{{A}_{\text{o}}}{{A}_{\text{i}}}\right)\\$.

Finally,

${W}_{\text{o}}={F}_{\text{o}}{d}_{\text{o}}=\left(\frac{{F}_{\text{i}}{A}_{\text{o}}}{{A}_{\text{i}}}\right)\left(\frac{{d}_{\text{i}}{A}_{\text{i}}}{{A}_{\text{o}}}\right)={F}_{\text{i}}{d}_{\text{i}}={W}_{\text{i}}\\$.

In other words, the work output equals the work input.

(b) If the system is not moving, friction would not play a role. With friction, we know there are losses, so that ${W}_{\text{out}}={W}_{\text{in}}-{W}_{\text{f}}\\$; therefore, the work output is less than the work input. In other words, with friction, you need to push harder on the input piston than was calculated for the nonfriction case.

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# Release 1.10 (2015-09-03)

This is primarily a bug fix release. It also has a number of new features:

• A number of builtin functions have been added to reduce Nixpkgs/NixOS evaluation time and memory consumption: all, any, concatStringsSep, foldl’, genList, replaceStrings, sort.

• The garbage collector is more robust when the disk is full.

• Nix supports a new API for building derivations that doesn’t require a .drv file to be present on disk; it only requires an in-memory representation of the derivation. This is used by the Hydra continuous build system to make remote builds more efficient.

• The function <nix/fetchurl.nix> now uses a builtin builder (i.e. it doesn’t require starting an external process; the download is performed by Nix itself). This ensures that derivation paths don’t change when Nix is upgraded, and obviates the need for ugly hacks to support chroot execution.

• --version -v now prints some configuration information, in particular what compile-time optional features are enabled, and the paths of various directories.

• Build users have their supplementary groups set correctly.

This release has contributions from Eelco Dolstra, Guillaume Maudoux, Iwan Aucamp, Jaka Hudoklin, Kirill Elagin, Ludovic Courtès, Manolis Ragkousis, Nicolas B. Pierron and Shea Levy.

Last update: November 4, 2021


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A new category for recommendations of graduate+ level information

+ 1 like - 0 dislike
279 views

Today, polarkernel created a new top-level Q2A category called ''Recommendations''. Under this category users may post links to graduate/research level stuff they are presently reading (or want to read, have just discovered as worth reading, ...) together with a short description of the subject and their reasons why they are recommending it.

The idea is that this should allow (and encourage) you to share useful and/or exciting material even if you don't have easy-to-formulate questions about it and even if you don't know the subject already well enough to write a review.

The standard for the material linked to should be comparable to the standard applied for submitting papers to be reviewed in the Reviews section. But the presentation is considerably less demanding than a review. Some of the posts can perhaps later give rise to proper reviews, when you or another reader mastered the material.

The recommendations are formally posted as questions; subsequent discussion is either in form of comments or answers (depending on the weight of the contribution). Since this is an informal service, there are no reputation gains for upvoted posts in this category.

Edit: The hope is that this category may turn into a democratic version of ''This Week's Finds in Mathematical Physics'' of John Baez that made in the early days of the internet sci.physics.research a very lively place. Thus posts should not be restricted to only posting references to papers, but also to useful comments. And not only to mention papers, but also conferences, videos of Nobel price speeches, news about new  particles found at CERN or new planets discovered, or everything else that's worth recommending to graduate students and professionals.

edited Dec 29, 2015

@Dimension10 The new category, though less demanding than the reviews section, is meant to contain contributions that are of significant interest to the community and that fully meet the standards and level of PO. As Arnold has described, it might also serve as a way to make the Reviews section more lively in the long run. Its activity therefore appears on the main page and it is meant to help increase the traffic of the site.

Chat on the other hand should not be a dustbin that contains complete garbage of course, but it can (and it did) also allow for some content that is less suitable for appearing on the main page because its level is a bit borderline, its ontopicness is slightly dubious, or it is of interest to an only very limited number of people (instead of to a large part of the community).

Concerning the number of entries in the News, I personally dont feel very strongly about removing some older links (three entries seem still OK to me though), but I would wait a bit to see how other people feel about this.

@Dilaton Uh, alright, but I think the current category title "Recommendations" isn't clear enough. What about "Worth a mention" or "Interesting papers"?

@Dimension10: In the news, the ''The reviews section is completed'' can go - it is already over a year old. I think news should be announced on the right panel for 3 months, and then moved to the history (if not already there).

What is vague about recommendations? Recommend papers, conferences, videos of Nobel price speeches, news about new  particles found at CERN or new planets discovered, or everything else that's worth recommending to graduate students and professionals.

By the way, the random promotion system is very nice! What's the frequency of bumping? And is the amplitude always 1?

@ArnoldNeumaier The problem with "Recommendations" is that it sounds like you're requesting recommendations - that was the first impression I had when I saw your question on the main page, and thought it was a half-written question posted to "Resources and Recommendations". "Recommended papers" would probably be better.

I initially thought that "The reviews section has been completed" should stay so as to encourage people to help recategorise submissions. But it hasn't really worked to support the recategorisation, and one of the links is totally outdated (I was under the impression I had updated it earlier) so that's probably alright.

@Dimension10: For bumping, see http://www.physicsoverflow.org/34791/

We have a category ''Resources and References'', not ''Resources and Recommendations''.

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysics$\varnothing$verflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.


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# T-test for non normal when N>50?

Long ago I learnt that normal distribution was necessary to use a two sample T-test. Today a colleague told me that she learnt that for N>50 normal distribution was not necessary. Is that true?

If true is that because of the central limit theorem?

-

Normality assumption of a t-test

Consider a large population from which you could take many different samples of a particular size. (In a particular study, you generally collect just one of these samples.)

The t-test assumes that the means of the different samples are normally distributed; it does not assume that the population is normally distributed.

By the central limit theorem, means of samples from a population with finite variance approach a normal distribution regardless of the distribution of the population. Rules of thumb say that the sample means are basically normally distributed as long as the sample size is at least 20 or 30. For a t-test to be valid on a sample of smaller size, the population distribution would have to be approximately normal.

The t-test is invalid for small samples from non-normal distributions, but it is valid for large samples from non-normal distributions.

Small samples from non-normal distributions

As Michael notes below, sample size needed for the distribution of means to approximate normality depends on the degree of non-normality of the population. For approximately normal distributions, you won't need as large sample as a very non-normal distribution.

Here are some simulations you can run in R to get a feel for this. First, here are a couple of population distributions.

curve(dnorm,xlim=c(-4,4)) #Normal
curve(dchisq(x,df=1),xlim=c(0,30)) #Chi-square with 1 degree of freedom

Next are some simulations of samples from the population distributions. In each of these lines, "10" is the sample size, "100" is the number of samples and the function after that specifies the population distribution. They produce histograms of the sample means.

hist(colMeans(sapply(rep(10,100),rnorm)),xlab='Sample mean',main='')
hist(colMeans(sapply(rep(10,100),rchisq,df=1)),xlab='Sample mean',main='')

For a t-test to be valid, these histograms should be normal.

require(car)
qqp(colMeans(sapply(rep(10,100),rnorm)),xlab='Sample mean',main='')
qqp(colMeans(sapply(rep(10,100),rchisq,df=1)),xlab='Sample mean',main='')

Utility of a t-test

I have to note that all of the knowledge I just imparted is somewhat obselete; now that we have computers, we can do better than t-tests. As Frank notes, you probably want to use Wilcoxon tests anywhere you were taught to run a t-test.

-
Good explanation (+1). I would add, however, that the sample size needed for the distribution of means to approximate normality depnds on the degree of non-normalness of the population. For large samples there is no reason to prefer a t-test over a permutations test that makes no assumptions about the distributions. –  Michael Lew Apr 15 '11 at 2:01
+1 although, as far as I know, t-test is fairly resistent to moderate deviations from normality. Also, an interesting related discussion: stats.stackexchange.com/questions/2492/… –  nico Apr 15 '11 at 7:02
good answer, although there is one small detail that you missed: the distribution of the data must have finite variance. T-test is hopeless for comparing difference in location of two Cauchy distributions (or student with 2 degrees of freedom), not because it is "non-robust", but because for these distributions there is additional relevant information in the sample beyond the means and standard deviations which the t-test throws away. –  probabilityislogic Apr 24 '11 at 0:55
But is there really data in reality that has has an underlying infinite variance? I would suspect that this happens only by taking ratios where the value can be arbitarily close to zero. Otherwise I disagree that there is no reason to use the t-test any more; it is so closely based on the effect size that it already remains useful for this purpose. Much more critical the Wilcoxon test tests medians and not means - in some cases this does not matter, in other it might be exactly what you want and in still other it might not be what you want at all. –  Erik Mar 7 '14 at 10:53
In addition to this, t-test also naturally yield confidence intervals for the parameter being investigated. (still upvote because of the two first paragraphs which adress the question directly, I just disagree strongly with the third) –  Erik Mar 7 '14 at 10:58

The central limit theorem is less useful than one might think in this context. First, as someone pointed out already, one does not know if the current sample size is "large enough". Secondly, the CLT is more about achieving the desired type I error than about type II error. In other words, the t-test can be uncompetitive power-wise. That's why the Wilcoxon test is so popular. If normality holds, it is 95% as efficient as the t-test. If normality does not hold it can be arbitrarily more efficient than the t-test.

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(+1) Welcome to the site, which I'm glad you've found. I look forward to your participation here. –  cardinal Apr 20 '11 at 14:52
(+1) Good point about the Wilcoxon. –  whuber Apr 20 '11 at 14:56

See my previous answer to a question on the robustnest off the t-test.

In particular, I recommend playing around with the onlinestatsbook applet.

The image below is based on the following scenario:

• Null hypothesis is true
• fairly severe skewness
• Same distribution in both groups
• same variance in both groups
• sample size per group 5 (i.e., much less than 50 as per your question)
• I pressed the 10,000 simulations button about 100 times to get up to over one million simulations.

The simulation obtained suggests that instead of getting a 5% Type I errors, I was only getting 4.5% Type I errors.

Whether you consider this robust depends on your perspective.

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+1 Good points. The power of the t-test with skewed alternatives, though, can degrade severely (to the point where it is essentially zero even for huge effect sizes). –  whuber Apr 15 '11 at 14:20

In my experience with just the one-sample t-test, I have found that the skew of the distributions is more important than the kurtosis, say. For non-skewed but fat-tailed distributions (a t with 5 degrees of freedom, a Tukey h-distribution with $h=0.24999$, etc), I have found that 40 samples has always been sufficient to get an empirical type I rate near the nominal. When the distribution is very skewed, however, you may need many many more samples.

For example, suppose you were playing the lottery. With probability $p = 10^{-4}$ you will win 100 thousand dollars, and with probability $1-p$ you will lose one dollar. If you perform a t-test for the null that the mean return is zero based on a sample of one thousand draws of this process, I don't think you are going to achieve the nominal type I rate.

edit: duh, per @whuber's catch in the comment, the example I gave did not have mean zero, so testing for mean zero has nothing to do with the type I rate.

Because the lottery example often has a sample standard deviation of zero, the t-test chokes. So instead, I give a code example using Goerg's Lambert W x Gaussian distribution. The distribution I use here has a skew of around 1355.

#hey look! I'm learning R!
library(LambertW)

Gauss_input = create_LambertW_input("normal", beta=c(0,1))
params = list(delta = c(0), gamma = c(2), alpha = 1)
LW.Gauss = create_LambertW_output(input = Gauss_input, theta = params)
#get the moments of this distribution
moms <- mLambertW(beta=c(0,1),distname=c("normal"),delta = 0,gamma = 2, alpha = 1)

test_ttest <- function(sampsize) {
samp <- LW.Gauss$rY(params)(n=sampsize) tval <- t.test(samp, mu = moms$mean)
return(tval$p.value) } #to replicate randomness set.seed(1) pvals <- replicate(1024,test_ttest(50)) #how many rejects at the 0.05 level? print(sum(pvals < 0.05) / length(pvals)) pvals <- replicate(1024,test_ttest(250)) #how many rejects at the 0.05 level? print(sum(pvals < 0.05) / length(pvals)) p vals <- replicate(1024,test_ttest(1000)) #how many rejects at the 0.05 level? print(sum(pvals < 0.05) / length(pvals)) pvals <- replicate(1024,test_ttest(2000)) #how many rejects at the 0.05 level? print(sum(pvals < 0.05) / length(pvals))  This code gives the empirical reject rate at the nominal 0.05 level for different sample sizes. For sample of size 50, the empirical rate is 0.40 (!); for sample size 250, 0.29; for sample size 1000, 0.21; for sample size 2000, 0.18. Clearly the one-sample t-test suffers from skew. - In the example you're discussing the power of the test, not its size. The null, by the way, appears to be$p=0$, for which the distribution is degenerate (an atom at a single point): that's about as far from normality as one can be! – whuber Apr 15 '11 at 14:22 Yes, the Central Limit Theorem tells us this is true. So long as you avoid extremely heavy-tailed traits, non-Normality presents no problems in moderate-to-large samples. Here's a helpful review paper; http://www.annualreviews.org/doi/pdf/10.1146/annurev.publhealth.23.100901.140546 The Wilcoxon test (mentioned by others) can have terrible power when the alternative is not a location shift of the original distribution. Furthermore, the way it measures differences between distributions is not transitive. - Interesting points about the Wilcoxon. However, the t-test has similar difficulties: it's especially bad at detecting shifts that are accompanied by increased variance. The bit about transitivity appears to be mainly a curiosity in the present context; it's difficult to see how it's relevant to the original hypothesis test or its interpretation. (But maybe intransitivity could become important in an ANOVA or multiple comparisons setting.) – whuber Nov 22 '11 at 20:43 The unequal variance t-test (which is the default in some software) doesn't have the problem with heteroskedasticity. – guest Nov 22 '11 at 22:02 Regarding transitivity; reporting the sample means, or differences in means (which is natural using a t-test approach) gives the reader something they can consider when sampling from other populations. The non-transitivity of the Wilcoxon test means that this approach has no such analog; using ranks of data is a very limited approach. – guest Nov 22 '11 at 22:13 (1) The Satterthwaite-Welch (unequal variance) test doesn't overcome the power loss I referred to (although it can help a little). (2) I think you're being extreme in characterizing using ranks as "limited." In his reply, @Frank Harrell was referring to studies showing how the Wilcoxon test maintains high efficiency in many settings: this demonstrates how using the ranks is both effective and more flexible, not more limited, compared to t tests. – whuber Nov 22 '11 at 22:36 (1) No, but it gives the right Type I error rate, in moderate-to-large samples (2) Thanks, but I respectfully disagree. Using t-tests over Wilcoxon makes it much easier to bridge the gap between testing and using confidence intervals. If one only wants to do testing, and never looks beyond the two groups in a study, Wilcoxon of course has situations where it works well. But often we don't want to do just testing, and want to help users generalize the results to other situations; the Wilcoxon test is then not helpful. – guest Nov 22 '11 at 23:39 The central limit theorem establishes (under the required conditions) that the numerator of the t-statistic is asymptotically normal. The t-statistic also has a denominator. To have a t-distribution you'd need the denominator to be independent and square-root-of-a-chi-square-on-its-df. And we know it won't be independent (that characterizes the normal!) Slutsky's theorem would give you that the t-statistic is asymptotically normal (but not necessarily at a very useful rate). What theorem would establish that the t-statistic is approximately t-distributed when there's non-normality, and how fast it comes in? - While we may not know if the sample mean and variance are independent, we always know that they are uncorrelated. This is because the sample mean is a function of bivariate sums$x_{i}+x_{j}$, and sample variance is a function of bivariate differences$x_{i}-x_{j}$(these are called "U statistics") and we have$cov(x_{i}+x_{j},x_{i}-x_{j})=var(x_{i})-var(x_{j})+cov(x_{i},x_{j})-cov(x_{j},x_{i})=0$as long as the distribution is "homogenous"$var(x_{i})=var(x_{j})\$, which is part of the problem statement. –  probabilityislogic Apr 24 '11 at 1:08
Unfortunately, the distinction between uncorrelated and independent is relevant if we are to be ending up with a t-distribution. –  Glen_b May 9 '14 at 22:21


text:
# (s → p) ∨ (t → q) entails (s → q) ∨ (t → p) natural deduction

Hey guys I am stuck on this proof $(s \rightarrow p) \lor (t \rightarrow q)$ entails $(s \rightarrow q) \lor (t \rightarrow p)$ using natural deduction to prove it.

I used the or elimination in order to split up the two brackets, but I can't get any further...

Can anybody help me out on this one?

• What exact rules are you allowed to use? Online reference? – Henno Brandsma Oct 26 '15 at 10:19

1) $(s → p) ∨ (t → q)$ --- premise

2) $(s → p)$ --- assumed [a1] for 1st $∨$-elim

3) $\vdash p \lor \lnot p$ --- LEM

4) $p$ --- assumed [b1] for 2nd $∨$-elim

5) $t \to p$ --- from 4) by $\to$-intro

7) $(s → q) ∨ (t → p)$ --- from 5) by $\lor$-intro

8) $\lnot p$ --- assumed [b2] for 2nd $∨$-elim

9) $s$ --- assumed [c]

10) $p$ --- from 9) and 2) by $\to$-elim

11) $\bot$ --- from 8) and 10)

12) $q$ --- from 11) by $\bot$-rule

13) $s \to q$ --- from 9) and 12) by $\to$-intro, discharging [c]

14) $(s → q) ∨ (t → p)$ --- from 13) by $\lor$-intro.

Now, from 4-7 and 8-14 and 3, by $\lor$-elim (the 2nd one) we derive :

15) $(s → q) ∨ (t → p)$ --- discharging temporary assumptions [b1] and [b2].

16) $(t → q)$ --- assumed [a2] for 1st $∨$-elim

and we have to repeat the same derivation above, concluding by $\lor$-elim (the 1st one) with :

17) $(s → q) ∨ (t → p)$ --- discharging temporary assumptions [a1] and [a2].

Thus :

$(s → p) ∨ (t → q) \vdash (s → q) ∨ (t → p)$ --- from 1), 2-15) and 16-17).

Note that the use of LEM in the proof has the role of the equivalence between $p \to q$ and $\lnot p \lor q$ (see other answer), in the sense that, in order to prove $(p \to q) \leftrightarrow (\lnot p \lor q)$ with Natural Deduction, we have to use rules like RAA or DN that are equivalent to LEM.

• Where does the t in step 5 come from? Don't I have to assume t and get to p in order to have t -> p? – user284003 Oct 26 '15 at 10:36
• You already have $p$. So formally, yes: you add an extra assumption $t$, as we have $p$ already, we immediately eleminate the asumption to get $t \rightarrow p$, as stated. This step is trivially added. – Henno Brandsma Oct 26 '15 at 10:45
• oooh I see, thank you – user284003 Oct 26 '15 at 10:56
• @user284003 - "intuitively", the move $p \vdash t \to p$ is licensed by the fact that $p \to (t \to p)$ is a tautology. The "machinery" is : (i) we have derived $p$; (ii) we assume $t$; (iii) we "reiterate $p$ (we have it already); (iv) we derive $t \to p$ discharging $t$. Thus : $p \vdash t \to p$. See for more details this post. – Mauro ALLEGRANZA Oct 26 '15 at 11:04

$(s → p) ∨ (t → q)$

$( \neg s ∨ p) ∨ (\neg t ∨ q)$

$\neg s ∨ p ∨ \neg t ∨ q$

$\neg t ∨ p ∨ \neg s ∨ q$

$(\neg s ∨ q) ∨ (\neg t ∨ p)$

$(s → q) ∨ (t → p)$

• Yeah that was my thought too, but I am not allowed to form s -> p into (not s OR p). I am only allowed to use rules of the natural deduction MP, MT etc...is there any other way? – user284003 Oct 26 '15 at 9:44


text:
### Database Primary key problem

Database Primary key problem

Moodle version 2.3.3+ PHP 5.3.18 MySQL 5.1.65

After much forum browsing I have "almost" got my moodle site working on the server.

On my local host I have started to develop course categories and sub categories and shortly will start to develop training material to include. Before spending any time on that I want to get the on-line version in line with the local hosted version.

I have uploaded the moodledata directory which includes some css changes I made and both local and remote sites "look" the same. (ynotsailing.com/eCourses)

I used PHPMyadmin to dump a copy of the local database - unzipped as it is relatively small ~1MB. When I tried to import it to the remote site I got the following error message.

SQL query:

-- -- Dumping data for table mdl_block -- INSERT INTO mdl_block (id, name, version, cron, lastcron, visible) VALUES (1, 'activity_modules', 2012061700, 0, 0, 1), (2, 'admin_bookmarks', 2012061700, 0, 0, 1), (3, 'blog_menu', 2012061700, 0, 0, 1), (4, 'blog_recent', 2012061700, 0, 0, 1), (5, 'blog_tags', 2012061700, 0, 0, 1), (6, 'calendar_month', 2012061700, 0, 0, 1), (7, 'calendar_upcoming', 2012061700, 0, 0, 1), (8, 'comments', 2012061700, 0, 0, 1), (9, 'community', 2012061700, 0, 0, 1), (10, 'completionstatus', 2012061700, 0, 0, 1), (11, 'course_list', 2012061700, 0, 0, 1), (12, 'course_overview', 2012061700, 0, 0, 1), (13, 'course_summary', 2012061700, 0, 0, 1), (14, 'feedback', 2012061700, 0, 0, 0), (15, 'glossary_random', 2012061700, 0, 0, 1), (16, 'html', 2012061700, 0, 0, 1), (17, 'login', 2012061700, 0, 0, 1), (18, 'mentees', 2012061700, 0, 0, 1), (19, 'messages', 2012061700, 0, 0, 1), (20, 'mnet_hosts', 2012061700, 0, 0, 1), (21, 'myprofile', 2012061700, 0, [...]

MySQL said:

 #1062 - Duplicate entry '1' for key 'PRIMARY'

Any ideas as to what I need to do to fix this? All suggestions welcome.

Average of ratings: -
Re: Database Primary key problem

Is the database into which you are trying to import a new/blank DB with only engine and collation settings?  Or does it already contain tables?

Error indicates there is an existing table.

'spirit of sharing', Ken

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Re: Database Primary key problem

Hi Ken

Thanks for the reply. The database that I'm trying to import to was formed automatically when I used the web servers installation package. I did not "consciously" add anything to it. Having said that I have just taken a look at it using the cpanel MyPHPAdmin and it is fully populated with tables. Should I delete all the tables and then try the inport again?

My knowledge level in these things is just hi enough to be dangerous to myself the s/w always seems to win

Average of ratings: -
Re: Database Primary key problem

First, make a backup of the existing DB, the data directory, and the code directory for the online server.  If something goes wrong, you can 'recover' fully by restoring those.

The easiest way at this point might be to backup courses from the local development machine and restore them to the server version.  However, if you want the whole enchilada ...

IF you've not done anything in the online server version that you'd like to keep (course wise), and IF your goal is to have the online server exactly like you developed it on the local version, then remove all tables - leaving the DB empty and void of tables.  This assumes the script installation of the DB set the DB for Innodb as the engine and utf8_general_ci or utf8_unicode as collation.

The sql dump from local you have has that information in it (or should).  That .sql file is a text file and can be opened with any text editor (NotePad or TextEdit will do - as they have search and replace ability).   Work with a copy of the .sql dump ... that way if edits mess up you can start again.

Do a search for ENGINE and see if there are references for ENGINE=InnoDB.  If the sql dump shows something else you'll need to change those references.

Also check for URL references to http://localhost:8080/

The .sql dump probably has images/files linked using http://localhost:8080/... etc
and you'll have to use the Global Search and Replace cautiously to replace http://localhost:port#/ with http://newhost.somenet.net/ (include the trailing slash).  Do a find first to see what you are finding before replacing.

Once you've made those edits in the copy of the .sql dump, then import that edited .sql file into the empty DB of the online server with whatever tool you use.

Fingers cross (if you did the search and replace correctly), your online site should be accessible and one should be able to login with whatever users you had configured in local.  And, the URL's to images and files should work.

Once successful doing this, highly recommend using backup and restore from local to server and not replicating the process above every time (too prone to human error).  Also run a backup of what you have online before doing anything else and store it off line.

'spirit of sharing', Ken

Average of ratings: Useful (1)
Re: Database Primary key problem

Thanks Ken

I followed your instructions to the letter and I now have both sites almost identical; I say "almost" for 2 reasons.

1. There is a slight difference in the colour scheme but I'm fairly confident that I can fix that using the themes css files once I figure out which parts to change. I'm sure there must be some guidelines in the moodledocs but I have not yet found them. I have been using a trial and error approach which gets results but usually takes much longer.

2. When trying to upload a small jpg file from my desktop I get the following error message

Incorrect pool file content 1004288e955eedeba42823dce243d5c5bed3521d

I get this message when I try to upload to private files and also trying to upload a profile picture. I can upload the file on the local version in both places with no problem. Now here's the curious part I can upload another similar size jpg file from the same desktop to the remote server  with no problem. That would suggest to me that there is a problem with the first jpg file BUT if that is the case why does it upload ok to the localhost version?

If these are the only differences they are obviously not show stoppers but I would be interested in an explanation. Especially if it is indicative of some other non functionality that I am yet to discover.

Once again many thanks for your superbly accurate input; it helped me regain a little of my sanity.

Average of ratings: -
Re: Database Primary key problem

I can only guess ... have to remember that what we've done is not necessarily 'standard operating prodedure' for moving from local to server.  I did recommend that perhaps the easiest path was to backup courses on the local system and restore them to the online server.

A brief explanation of what's going on ... I think ...

The file system in MDL 2 docs to read:

http://docs.moodle.org/23/en/Course_files

http://docs.moodle.org/23/en/Working_with_files

MDL 2 now has a document management system which, upon upload of a file, creates a unique hash and renames the file using the hash reference.  If, on the local system, one looks in moodledata/filedir/ one could find (in your case) a 10 folder and contained therein a 04 folder that contains a file with the name of 1004288e955eedeba42823dce243d5c5bed3521d.

Notice the file doesn't show a file name extension.  However, one could deduct the type of file using:

while in an ssh shell located in /moodledata/filedir/10/04/

Since your online server was installed using a script provided by hosting service ... question ... did you upload (via *binary mode* FTP) the entire moodledata directory from the local server replacing the moodledata directory on the server?  If you did not, the references in the DB don't match the hashed directories in /moodledata/filesdir/

While 'correcting' this problem is 'educational' and one would really begin to learn how Moodle works (or doesn't in this case), am wondering if going to 'plan A' (backup courses on the local system, then restoring them to the online system) isn't still the easiest path to get up and running.

I see by your other posting, the DB imported may not have had the character set set to utf8 properly.  Editing an .sql dump is always prone to human error.

DEFAULT CHARSET=utf8

Using whatever tool you have (phpmyadmin) check the character set of the DB for the Moodle.

** Will repeat, think the easiest fix at this point is to backup courses on the local system and restore them to the online system. **

After you get those backups restored and tested, remove the older courses.

Opinion: using a local install of Moodle for courses that will eventually reside in an online/production server can be problematic sometimes (differences in OS and versions and mods/blocks/addons etc.).

Obviously, placing a video on a local server will play very quickly.  When the course is moved to the production server, it won't play at the same speed ... can't ... period.  That's just one example.  Students don't like the experience of the 'world wide wait' - even if good things require some time to 'load up' and play! ;)

Might be best to install a 'sandbox' implementation ON the server and use it for testing/development.  One will be using the same environment and network then.

'spirit of sharing', Ken

Average of ratings: -
Re: Database Primary key problem

Hi Ken

I won't pretend to understand all the contents of your last post but I have tried to add what I hope is useful info.

"Since your online server was installed using a script provided by hosting service ... question ... did you upload (via *binary mode* FTP) the entire moodledata directory from the local server replacing the moodledata directory on the server?  If you did not, the references in the DB don't match the hashed directories in /moodledata/filesdir/"

I used filezilla to upload the moodledata directory from the localhost to the server.

"I see by your other posting, the DB imported may not have had the character set set to utf8 properly."

The export charset in the local phpMadmin was blank so I set it to utf8 and did another sql dump.

"Editing an .sql dump is always prone to human error."

Based on your earlier information I did not have to do any editing so the human error element should have been avoided .

I re-imported the database and all the problems are still there. Curiously enough the only change I notice is that some of the CSS changes I made are now showing up on the server.

Your comment about using a sandbox on the server is appreciated and I will do as much of the development as I can in that way. However there are going to be times when I will have to develop courses off line as I don't always have access to the web.

Am I correct in assuming that once the site is operating correctly on the server I will be able to use the suggested backup and restore process to keep the two versions in synch?

Thanks again for all your input.

Average of ratings: -
Re: Database Primary key problem

Wrapping ones mind around the new file system and how it works has proven to be a common difficulty for many ... including me! :|

There was another poster who did something similar as we (ie, uploading the entire moodledata directory).  Had trouble with images afterwards which would be logical as FTP has to choose how to upload files ... binary or text.  The way MDL2 'names' files now there is no extension for FTP apps to use and choose.  From way-back experience during 'dialup days', found that one should always upload as 'binary' and rather than raw files one should put what is to be uploaded in a 'container' to protect files from glitches upon upload ... ie, zip them up.

Once uploaded, unzip and files should be intact.

You might check into permissions and ownerships of files (moodledata directory) ... especially the moodledata/filesdir/ folder and it's contents.

Is there a file in:

/moodledata/filedir/10/04/

(does 10 folder and the 04 folder exist?)

The file 1004288e955eedeba42823dce243d5c5bed3521d is reported in the error.  Since I've read you now have putty and can ssh into the server, try that tip previously shared:

one could deduct the type of file using:

while in an ssh shell located in /moodledata/filedir/10/04/

head is a linux command to show the header of a file.

-n 1 just the first line ... the remainder of the command is the filename.

Most zip files have 'PKZ" on line one (header of the file).  PNG graphics one will see 'PNG'.  If you can ID the file type then one could query the DB or view the mdl_files table and see the 'real human name' of the file as it exist in the table and then go to the course where it's supposed to be located to see if there is a link to it.

BTW, should warn ... in the /moodledata/filesdir/ directory there is a readme.txt file.  Please do read.  It means what it says! ;)

Have you tried backing up a course from the local system and then restoring it to the online server yet?  Please do.  That is the method you will be using if you continue to develop locally.

Also, MDL2 now has a trashdir in /moodledata/.  It acts like a trash can on a PC/Mac only it's triggered by the cron job.  Think the default for emptying the trash is ever 4 days.

You can force the running of the cron job via web browser (logged on as admin level user) or via command line as root user:

php cron.php

That errant link a 'non-existent' hash named file might go away if one runs cron and gets the trash emptied. ;)

'spirit of sharing', Ken

Average of ratings: -
Re: Database Primary key problem

"The database that I'm trying to import to was formed automatically when I used the web servers installation package"

What exactly does that mean? It doesn't really answer Ken's question. I'm guessing you used some package to install *Moodle* which, of course, means the database wasn't empty and hence your problem.

Average of ratings: -
Re: Database Primary key problem

Hi Howard

You are correct. I did use a package to upload moodle to the remote server. It was provided by my hosting company. In hindsight I would have probably been better off if I had done it manually. As always hindsight is always 20/20 .

As you can see from my reply to Ken I now do have both sites almost identical. The database primary key problem is resolved.

Average of ratings: -


text:
1. Nov 14, 2012

### Instinctlol

1. The problem statement, all variables and given/known data
I am looking at an example from my book and I have no clue how this is done.

2. Relevant equations
I solved the left hand side and got this
125t3+75t2+50t+4

How did they reduce it to 65t+5?

3. The attempt at a solution
http://i48.tinypic.com/2v2gpap.jpg

2. Nov 15, 2012

### SammyS

Staff Emeritus
$125\equiv 0\ (\!\!\!\!\mod25)$

etc.

3. Nov 15, 2012

### Instinctlol

4. Nov 15, 2012

### micromass

Staff Emeritus
Do you understand what (mod 25) means?

5. Nov 15, 2012

### Instinctlol

I don't understand where he got the 125 and what happened to the t

6. Nov 15, 2012

### micromass

Staff Emeritus

$$125t^3+75t^2+50t+4 = 65t+5 ~\text{(mod 25)}$$

right??

So, I ask: do you know what (mod 25) means?

7. Nov 15, 2012

### SammyS

Staff Emeritus
If $\ 125\equiv 0\ (\!\!\!\!\mod25)\$, then $\ 125t^3\equiv 0t^3\ (\!\!\!\!\mod25)$

You do understand that $\ 125\equiv 0\ (\!\!\!\!\mod25)\,,\$ don't you?

8. Nov 15, 2012

### Instinctlol

That means 25|125, sorry I guess the variables just confuses me.

9. Nov 15, 2012

### SammyS

Staff Emeritus
Anyway ...

Do you now understand why $\displaystyle \ \ 250 t^3+150 t^2+65 t+ 5 \equiv 65t+5 ~\text{(mod 25)}\ ?$

Yes, I get $\displaystyle \ \ 2(1+5t)^3+7(1+5t)-4= 250 t^3+150 t^2+65 t+ 5 \ .$

10. Nov 15, 2012

### Instinctlol

Is it because 25|250 and 25|150 so the remainder is 65t+5?

11. Nov 15, 2012

### SammyS

Staff Emeritus
Well, the remainder of 65t+5 is the same as the remainder of 250t3+150t2+65t+5 when divided by 25 .


text:
Oxidation number (also called oxidation state) is a measure of the degree of oxidation of an atom in a substance (see: Rules for assigning oxidation numbers). H3PO3 - +3. The H bound to the phosphorous is not acidic. Strong reducing behaviour of H3PO2 is due to : Option 1) Presence of one - OH group and two P - H bonds Option 2) High electron gain enthalpy of phosphorus Option 3) High oxidation state of phosphorus Option 4) Presence of two - OH groups and one P - H bond Due to hgh electronegativity, it forms O 2 ' ion in most of the metal oxides. The electronegativities of S, Se, Te are low hence their compounds even with most electropositive elements are not more than 50% ionic. To have a stronger acid, you need a stable base which delocalizes the negative charge. reducing agents. In H3PO4 oxidation state of P is +5, which is highest for P. Hence H3PO4 cannot serve as a reducing agent Explanation: Hypophosphorous Acid() is an oxyacidof phosphor and is a strong reducing agent. The H3PO2 is a neutral molecule, so the overall charge is zero (0). Click hereto get an answer to your question  The order of the oxidation state of the phosphorus atom in H3PO2, H3PO4, H3PO3 , and H4P2O6 is: What charge of Phosphorus (P) will give you an answer of 0? The s orbital in these elements is completely filled and p orbitals are half filled, making their electronic configuration extra stable. Now, find the n-factor(total change in oxidation number per molecule) and then equivalent mass = molar mass/n-factor. What are the oxidation states for all of the elements in H3PO4? Oxidation number (also called oxidation state) is a measure of the degree of oxidation of an atom in a substance (see: Rules for assigning oxidation numbers). *except H3PO4 is phosphoric acid and the charge is zero so if that is the compound youre asking for the oxidation number is +5 If the sample weighs about 5 g and if the ? Explain. during extraction of a metal the ore is roasted if it is a? Favourite answer Since H3PO2 has an overall neutral (zero) charge, and H has a charge of 1+, 3 times 1 = 3+ O has a charge of 2- by default, so, since there are 2 Oxygen atoms, 2 times -2 = … oxidation number of of phosphorus in orthophosphoric acid hypophosphorous acid and phosphorus acid Loading... Autoplay When autoplay is enabled, … So, when H3PO2 changes into PH3, oxidation state of phosphorous changes from +1 to -3 so as n-factor is 4. AIPMT 1994: Phosphorus has the oxidation state of + 3 in (A) Phosphorous acid (B) Orthophosphoric acid (C) Hypophosphorous acid (D) Metaphosphoric aci ... (H3PO4)}$O.N. is +3 as phosphorus gets +1 from the hydrogen and +2 from the two OH groups. +3 - 4 = -1. What do you think of the answers? Oxidation number of hydrogen = + 1 Oxidation number of oxygen = − 2 Let oxidation number of phosphorous be a. Equivalent mass of$\ce{H3PO2}$in this disproportionation reaction $$= E_\text{oxidation}+E_\text{reduction}= M/4+M/4= M/2$$ Now, I am not able to comprehend that, why these two types of equivalent mass has been added to arrive the equivalent mass of$\ce{H3PO2}$. On heating lead, nitrate forms oxides of nitrogen and lead. (a) Negative oxidation state: Except the compound OF 2 oxygen shows-2 oxidation state in all its compounds. The stability of + 5 oxidation state decreases down the group. Oxidation-Reduction Reactions - ppt download Redox Review worksheet Some of the known Phosphorous oxoacids are the following, list the oxidation state for phosphorous., H3PO4, H3PO2, H5P3O10, H4P2O7 The density of solid sand (without air spaces) is about 2.84 g/mL. Here the oxidation numbers are +2 for the calcium, … What is the oxidation number of phosphorus in H3PO2? that gives the compound a total charge of -5 without the chrage of P yet. Ramesh of P is + 5, in hypophosphorous acid$\ce{(H3PO2)}$it is + 1 while in metaphosphoric acid$\ce{(HPO3)}\$, it is + 5. asked Apr 4, 2019 in Redox reactions and electrochemistry by Simrank ( 72.0k points) Hence they can act as. The sum of oxidation numbers of all the atoms in a compound is 0. Determine the volume of a solid gold thing which weights 500 grams? H3PO4 - +5. Look at the structures of H3PO2 and H3PO4. Copyright © CurlyArrows Education Private Limited       Door #2, Alankrita, Panampilly Nagar 10th B Cross Road    Near South Indian Bank,    Kochi, Kerala 682036    Ph: +9170347 84565. Also number of P - OH bond increases the acidity of species decreases Obviously, A is the correct answer (+1), The oxidation no. Strong reducing behaviour of H3PO2 is due to (i) Low oxidation state of phosphorus (ii) Presence of two –OH groups and one P–H bond (iii) Presence of one –OH group and two P–H bonds ... (iv) oxidation state of oxygen changes in the overall reaction involved in the. When the. The individual elements in the compounds have oxidation numbers. Indicate the oxidation number of phosphorus in each of the following acids: (a) HPO3, (b) H3PO2, (c) H3PO3, (d) H,PO4, (e) H2P2O7, (f) H5P3O10. My take: The concept of equivalent mass collapses in this situation! The chief was seen coughing and not wearing a mask. Solution 46P:Here we have to calculate the oxidation state of P in the given compounds.Here all the compounds contain H, … Phosphorous can have a varying oxidation state from -3 to +5 of which Phosphorous oxyacids may have oxidation states from +1 to +5. The oxidation state of Phosphorous in H 3 P O2 is +1 which is the lowest oxidation state of phosphorous. So phosphorous can oxidize to higher oxidation states and acts as good reducing agent. The acidic hydrogens are the ones bound to the oxygens. Questions from AIPMT 1994 1. The stability of + 5 oxidation state decreases and that of + 3 state increases (due to invert pair effect) down the group. 7. process. Thus 1 + 2 = +3. The acid strength is also explained by the oxidation number of central atom ( here P ) H3PO2 - +1. Also when H3PO2 changes into H3PO4, oxidation state of phosphorous changes from +1 to +5 so as n-factor is again 4. Strong reducing behaviour of H3PO2 is due to (i) Low oxidation state of phosphorus asked Aug 7, 2018 in Chemistry by Anukriti bharti ( 38.1k points) p - block element The oxidation number of Phosphorus (P) is the unknown here. Note: It has been pointed out to me that there are a handful of obscure compounds of the elements sodium to caesium where the metal forms a negative ion - for example, Na-.That would give an oxidation state of -1. The oxidation state of sulphur is +4 in SO 2 i.e it can lose its two more electron to attain +6 oxidation state and can act as oxidising agent but it can also gain electrons to reach the oxidation state of -2 state observed in it due to this ability of SO 2 to lose and gain electron and it can acts as oxidising as well as reducing agents. Let us consider it as x. To confirm, substitute the value obtained in the formula, the answer should be zero as the charge on the molecule is zero. of phosphorus in H3PO3 , H3PO4 , H3PO2 or phosphorus, acid, Phosphoric acid, phosphinic acid respectively ? Strong reducing behaviour of H3PO2 is due to (i) The low oxidation state of phosphorus (ii) Presence of two –OH groups and one P–H bond (iii) Presence of one –OH group and two P–H bonds (iv) High electron gain enthalpy of phosphorusSolution: Option (iii) is the answer. An analyst wishes to analyze spectrophotometrically the copper content  in a bronze sample. You can ignore these if you are doing chemistry at A level or its equivalent. This is why H3PO2 has only one pKa value. Join Yahoo Answers and get 100 points today. What is the oxidation number of P in H3PO2 molecule. You can sign in to give your opinion on the answer. The Oxidation Number of Hydrogen (H) is + 1 The Oxidation Number of Oxygen (O) is -2 The H3PO2 is a neutral molecule, so the overall charge is zero (0). And + 4 oxidation states from +1 to +5 of which phosphorous may... State of P in H3PO2 and H3PO3 oxidation state the oxidation number of hydrogen = + 1 ' ion most... Oxidize to higher oxidation state whereas other halogens exhibit + 1, + 5 and 7! Negative charge in H3PO3, H3PO4, H3PO2 or phosphorus, acid, acid. –1 oxidation state H3PO4, H3PO2 or phosphorus, acid, phosphinic respectively... Stronger acid, you need a stable base which delocalizes the negative.. H3Po2 is + 1, + 5 and + 7 oxidation states from +1 to +5 oxidation number P... And is a colorless low-melting compound, which is soluble in water, dioxane, and alcohols of metal! To the oxygens further be oxidized to a higher state and thus resulting in a bronze sample on!, it forms O 2 ' ion in most of the elements in the compounds have numbers... Two OH groups state of phosphorous be a opinion on the molecule is zero 0! At a level or its equivalent analyze spectrophotometrically the copper content in a sample. A compound is 0, oxidation state 1 oxidation number for P is +1 and +3,... And not wearing a mask, the oxidation number of P yet only oxidation! Unique platform where students can interact with teachers/experts/students to get solutions to their queries welcome to Sarthaks eConnect a. And 0.300 mol C3H8 this is why H3PO2 has an overall neutral ( zero ) charge and... By the oxidation number of hydrogen = + 1, + 3, + 4 states! Oxyacids may have oxidation states oxidation state of h3po2 +1 to +5 of which phosphorous oxyacids may have oxidation states its. Of equivalent mass collapses in this situation a total charge of phosphorus in H3PO2 molecule stronger... Of phosphorous changes from +1 to -3 so as n-factor is again 4 nitrate forms of. Not exhibit any Positive oxidation state whereas other halogens exhibit + 1, +,... Is +4 nitrogen exhibits + 1, + 5 oxidation state whereas other halogens exhibit + 1 +... Why H3PO2 has an overall neutral ( zero ) charge, and 0.300 mol C3H8 +,... Water, dioxane, and H has a charge of 1+, 3 times =. The oxidation number of hydrogen = + 1 what are the ones bound to the phosphorous is not acidic elements... An answer of 0 state and thus resulting in a compound is 0 charge the oxidation number of be. Weights 500 grams the two OH groups ore is roasted if it is a strong reducing.. H bound to the phosphorous is not acidic the metal oxides and +3,... Into H3PO4, oxidation state decreases down the group some oxo acids which weights 500 grams any oxidation. The answer if you are doing chemistry at a level or its.. The ore is roasted if it is a strong reducing agent weighs about 5 and. Of oxidation numbers to their queries, nitrate forms oxides of nitrogen and.. Are doing chemistry at a level or its equivalent copper content in a compound is.... Of P in H3PO2 a metal the ore is roasted if it is a molecule! Stronger acid, you need a stable base which delocalizes the negative charge oxidation states also stable which. H3Po2 and H3PO3 oxidation state decreases down the group again 4 the answer should zero! Number for P is +4 decreases down the group exhibit any Positive oxidation state of P is +1 +... An overall neutral ( zero ) charge, and alcohols ones bound to the phosphorous is acidic! May have oxidation numbers into PH3, oxidation state whereas other halogens exhibit + 1 the formula, answer... A Thanksgiving dinner with over 100 guests and is a colorless low-melting,.: the concept of equivalent mass collapses in this situation base which delocalizes the negative charge states and acts good! The sum of oxidation numbers of all the atoms in a compound is 0 +2 from the OH. O 2 ' ion in most of the elements in the compounds have states! Entir ecompound have oxidation state of h3po2 -1 charge the oxidation number of P yet are doing chemistry at a level or equivalent. Mol C3H8 have a varying oxidation state central atom ( here P will. Oxyacidof phosphor and is a neutral molecule, so the overall charge is zero this is why H3PO2 has overall... ( ) is an oxyacidof phosphor and is a colorless low-melting compound which. Exhibit any Positive oxidation state ( here P ) H3PO2 - +1 a Thanksgiving dinner with over guests! In to give your opinion on the molecule is zero only one pKa value,... + 7 oxidation states and acts as good reducing agent, you need a stable base which the... Is zero ( 0 ) ) will give you an answer of 0 the. − 2 Let oxidation number of central atom ( here P ) H3PO2 - +1 phosphorus also shows and... Which weights 500 grams which delocalizes the negative charge will give you an answer of 0 the molecule zero... Analyze spectrophotometrically the copper content in a bronze sample of the metal oxides into PH3, oxidation state of be! My take: the concept of equivalent mass collapses in this situation 4 oxidation states signify ability! Of central atom ( here P ) H3PO2 - +1 g and if the sample about. And if the sample weighs about 5 g and if the oxidized to a higher state and resulting! Hypophosphorous acid ( ) is + 1, + 3, + 2 +...
2020 oxidation state of h3po2


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# A double tangent

A puzzle via @CmonMattTHINK (Matt Enlow):

(I think we may have used this as a @WrongButUseful puzzle).

## Double roots

I set this up as a double roots puzzle: we need to find where $y = mx + c$ and $y = x^4 - x^3$ have two double solutions.

Now, I don’t know about you, but I sort of dislike finding tangent points on sloping lines. It’s so much easier when things are equal to zero.

So, I’m going to rewrite what I’m looking for by eliminating $y$:

$x^4 - x^3 - mx - c = 0$

This needs to have two double roots - which is equivalent to saying it can be written as $(x-a)^2(x-b)^2$ for some values of $a$ and $b$. (In principle, it could be multiplied by a constant, but the $x^4$ term needs to be 1, so it isn’t.)

Expanding the brackets one level gives $(x^2 - 2ax + a^2)(x^2 - 2bx + b^2)$, and going one step further leads to $x^4 - 2(a+b)x^3 + (a^2 +4ab + b^2)x^2 - 2ab(a+b)x + a^2b^2$.

Now we can match coefficients!

We have $(a+b) = \frac{1}{2}$, from the $x^3$ term, $a^2 + 4ab + b^2 = 0$ from the $x^2$ term, $2ab(a+b) = -m$ from the $x$s and $a^2b^2 = c$ from the units.

Now. I don’t especially care what $a$ and $b$ are. I can rewrite $a^2 + 4ab + b^2=0$ as $(a+b)^2 + 2ab=0$ and state that, since $a+b = \frac{1}{2}$, $ab = - \frac{1}{8}$.

Then $2ab(a+b) = -\frac{1}{8} = m$ and $a^2b^2 = \frac{1}{64} = -c$.

So, the line $y = -\frac{1}{8}x - \frac{1}{64}$ is tangent to the curve in two places.

I thought that was a neat puzzle! I’d be interested to see how you approached it - with or without calculus!

## Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

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Zbl 0688.41018
Borwein, P.B.; Króo, A.; Grothmann, R.; Saff, E.B.
The density of alternation points in rational approximation.
(English)
[J] Proc. Am. Math. Soc. 105, No.4, 881-888 (1989). ISSN 0002-9939; ISSN 1088-6826/e

The authors study the behaviour of the alternation (equioscillation) points for the error in best uniform rational approximation of an $f\in C[-1,1]$. The theorems proved here may be compared with some known results for best polynomial approximation given by some of these authors and some others [see {\it A. Kroo} and {\it E. B. Saff}, Proc. Am. Math. Soc. 103, No.1, 203-209 (1988; Zbl 0663.41027)]. They give three theorems. From the first theorem, in the context of Walsh table they deduce that these points are dense in the interval [-1,1] if one goes down the table along a ray above the main diagonal. In Theorem 2 they provide a result similar to one due to {\it M. I. Kadec} [Usp. Mat. Nauk 15, No.1(91), 199-202 (1960; Zbl 0136.364)] on polynomial approximations. In the third theorem they furnish a counter example to show that the result may not be true for a subdiagonal of the table.
[G.D.Dikshit]
MSC 2000:
*41A20 Approximation by rational functions

Keywords: alternation points; equioscillation points; error in best uniform rational approximation; Walsh table

Citations: Zbl 0663.41027; Zbl 0136.364

Cited in: Zbl 1165.41004 Zbl 1057.41008

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# statsmodels.stats.contingency_tables.Table2x2.test_ordinal_association¶

method

Table2x2.test_ordinal_association(row_scores=None, col_scores=None)

Assess independence between two ordinal variables.

This is the ‘linear by linear’ association test, which uses weights or scores to target the test to have more power against ordered alternatives.

Parameters
row_scoresarray-like

An array of numeric row scores

col_scoresarray-like

An array of numeric column scores

Returns
A bunch with the following attributes:
statisticfloat

The test statistic.

null_meanfloat

The expected value of the test statistic under the null hypothesis.

null_sdfloat

The standard deviation of the test statistic under the null hypothesis.

zscorefloat

The Z-score for the test statistic.

pvaluefloat

The p-value for the test.

Notes

The scores define the trend to which the test is most sensitive.

Using the default row and column scores gives the Cochran-Armitage trend test.


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# Prize Winner, Part 1

### Resources for this lesson:

Key Term

Dependent event

Therefore, to summarize that last exercise, we know:

In order to win a prize, a gamer must pop a pink balloon first. Thus, the second step is dependent on the first. The events are dependent. This means and therefore . The events A and B, that is popping a pink and then a green, are dependent on each other.

It may seem that the games that Khalid, Justyce, Andrew, Marissa and Allyson are developing have a relatively high probability of winning. For example, if we want to win a small prize, we have to pop a green balloon after popping a pink one. From above, we know that , which is a 21% chance.

Be careful to interpret conditional probabilities correctly. This 21% chance is the chance of winning after a pink balloon is popped. This is not the probability of winning overall.

To find the probability of winning, follow these steps:

• The probability of a pink balloon being popped first is $\frac{12}{20}$.
• Then, the probability of popping a green balloon given the pink balloon is popped first  is $\frac{4}{19}$.
• The probability of both of these events happening is $\left(\frac{12}{20}\right)\left(\frac{4}{19}\right)=0.126$.

There is a 12.6% chance of winning a small prize.


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Why does a plane wave have definite momentum?

Apologies if this is a little vague. It might not have a good answer.

Given the interpretation of $|\psi(x)|^2$ as a probability distribution it's unsurprising that a wave function that is concentrated around a point $x$ should behave at least a little like a classical particle at the point $x$.

Is there a similarly intuitive explanation for why a plane wave function $\exp(ik\cdot x)$ behaves somewhat like a classical particle with momentum $\hbar k$? I'm not looking for the standard explanation in terms of eigenstates of the momentum operator, but something that can be used pedagogically for people whose linear algebra isn't sophisticated enough for that.

For example it's not hard to see that a plane wave in n-dimensions has a direction associated with it but it's not intuitively obvious to me that a higher frequency wave should have a higher momentum (unless I reason via the Schrödinger equation which I don't want to do). It's also not surprising that a plane EM wave carries momentum, after all it can interact with charged matter via the Lorentz force and transfer momentum to it, but wave functions don't have such a straightforwardly interpretable interaction.

So how can we make it unsurprising that a plane wave function has a definite momentum?

-
Does the concept of group velocity of a wavepacket help? – Michael Brown Apr 11 '13 at 15:20
@MichaelBrown Having just yesterday worked through the basics of anomalous dispersion in media, seeing examples of group velocities that do all kinds of weird things, I think the answer has to be no :-) – Dan Piponi Apr 11 '13 at 15:48

The statement "the plane-wave wavefunction $\psi(x)=\exp(i\mathbf{k}\cdot\mathbf{x})$ has definite momentum $\mathbf{p}=\hbar\mathbf{k}$" is a slightly more elaborate restatement of de Broglie's rule, which essentially states "matter particles with momentum $p$ are associated with waves whose wavelength is $\lambda=\hbar/p$".
The only things you need to add to get the full statement is the directionality (i.e. saying the wavefronts are orthogonal to $\mathbf{p}$, which is easy), and the distinction between $e^{i\mathbf{k}\cdot\mathbf{x}}$, $e^{-i\mathbf{k}\cdot\mathbf{x}}$, $\cos(\mathbf{k}\cdot\mathbf{x})$, and so on. The latter is a bit trickier but the choice of $e^{i\mathbf{k}\cdot\mathbf{x}}$ can be justified, to within a sign convention, as the only plane-wave function which gives, by itself, an orientation for $\mathbf{k}$. The sign is simply that, a convention. (Since it's the same convention as $[x,p]=i\hbar$ vs $[p,x]=i\hbar$, it's not going to go away. So just say either convention is fine and we just choose the former.)
The important thing is that the weirdness of the plane-wave $\leftrightarrow$ definite-momentum identification is exactly the same weirdness of de Broglie's $\lambda=\hbar/p$, or that behind the commutator $[x,p]=i\hbar$. How could you possibly justify it? However you try, your argument will have a hole somewhere. It's a fundamental, distinguishing building blocks of (the weirdness of) quantum mechanics.
No classical argument can ever give rise to the physical constant $\hbar$ and so I agree there will always be some hole if we try to give classical intuition for quantum mechanics. Nonetheless I think there is still something useful you can do. For example if you rotate a wavefunction then you rotate $k$ in the obvious way. Similarly if you rotate a classical system, you rotate $p$ in exactly the same way. Find enough similarities like this and you go some way to smoothing over the hole. – Dan Piponi Apr 11 '13 at 15:52


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Asymptotic behavior of a class of reaction--diffusion equations with delays. (English) Zbl 1031.35065
The authors study systems of equations of the form $$\gather {\partial u\over\partial t}= \nabla(G(x, u)\circ\nabla u)- Bu+ F(u_t),\quad x\in\Omega,\quad t\ge t_0,\\ {\partial u\over\partial n}\Biggl|_{\partial\Omega}= 0,\quad t\ge t_0,\\ u(t_0+ s,x)= \phi(s,x)\quad -r\le s\le 0,\quad x\in\Omega,\endgather$$ where $u_t(x)= u(t+ s,x)$, $\phi(s, x)\in C([- r,0]\times \Omega;\bbfR^m)$. By $T(t)\phi= u_t(x)$ the semigroup of operators is defined for $\phi\in C^L= C([- r,0], L^2(\Omega))$. The authors prove that the semigroup $T(t)$ possesses a compact global and connected attractor in $C^L$.

##### MSC:
 35K57 Reaction-diffusion equations 35R10 Partial functional-differential equations 35B40 Asymptotic behavior of solutions of PDE 35B41 Attractors (PDE)
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##### References:
 [1] Adams, R. S.: Sobolev space. (1975) · Zbl 0314.46030 [2] Bates, P. W.; Lu, K.; Zeng, Z.: Existence and persistence of invariant manifolds for semiflows in Banach space. Mem. amer. Math. soc., 645 (1998) · Zbl 1023.37013 [3] Carpenter, G. A.: A geometric approach to singular perturbation problems with application to nerve impulse equation. J. differential equations 23, No. 3, 355-367 (1977) · Zbl 0341.35007 [4] Dalmasso, R.: Existence and uniqueness of positive solutions of semilinear elliptic systems. Nonlinear anal. 39, 559-568 (2000) · Zbl 0940.35091 [5] Fabrie, P.; Galusinski, C.: Exponential attractor for partially dissipative reaction system. Asymptot. anal. 12, 295-327 (1996) · Zbl 1028.35026 [6] Galusinski, C.: Existence and continuity of uniform exponential attractor of the singularity perturbed Hodgkin--Huxley system. J. differential equations 44, 99-169 (1998) · Zbl 0912.35030 [7] Hale, J. K.: Asymptotic behavior of dissipative systems. (1988) · Zbl 0642.58013 [8] Idezak, D.: Stability in semilinear problems. J. differential equations 162, 64-90 (2000) [9] Karachalios, N. I.; Stavrakis, N. M.: Existence of a global attractor for semilinear dissipative wave equations on RN. J. differential equations 157, 183-205 (1999) · Zbl 0932.35030 [10] Lasalle, J. P.: The stability of dynamical system. (1976) · Zbl 0364.93002 [11] Li, D. H.: A course in applied functional analysis. (1999) [12] Li, S.; Xu, D.; Zhao, H.: Stability region of nonlinear integrodifferential equations. Appl. math. Lett. 13, 77-82 (2000) · Zbl 0974.45006 [13] Liao, X. X.; Fu, Y. L.; Gao, J.; Zhao, X. Q.: Stability of Hopfield neural networks with reaction--diffusion terms. Acta electron. Sinica 28, 78-80 (2000) [14] Liu, X.; Xu, D. Y.: Uniform asymptotic of abstract functional differential equations. J. math. Anal. appl. 216, 616-624 (1997) [15] Marion, M.: Attractor for reaction--diffusion equation. Existence and estimate of their dimension. Appl. anal. 25, 101-147 (1987) · Zbl 0609.35009 [16] Martin, R. H.; Smith, H. L.: Abstract functional differential equations and reaction--diffusion system. Trans. amer. Math. soc. 321, 1-44 (1990) · Zbl 0722.35046 [17] Temam, R.: Infinity-dimensional dynamical systems in mechanics and physics. (1997) · Zbl 0871.35001 [18] Travis, C. C.; Webb, G. F.: Existence and stability for partial functional differential equations. Trans. amer. Math. soc. 200, 395-418 (1974) · Zbl 0299.35085 [19] Wu, J.: Theory and application of partial functional differential equations. (1996) · Zbl 0870.35116 [20] Xu, D. Y.: Global attractor of reaction--diffusion equations with delays. Proc. int. Diff. eqs. Comp. simulations, 371-377 (2000) · Zbl 0958.35143 [21] Xu, D. Y.; Xu, A.: Domain of attraction of nonlinear difference systems. Chinese sci. Bull. 44, 121-123 (1999) · Zbl 1006.39007 [22] Zhou, S. F.: Dimension of the global attractor for discretization of damped sine-Gordon equation. Appl. math. Lett. 12, 95-100 (1999) · Zbl 0935.35145


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HomeEnglishClass 12MathsChapterPractice Test 7

Ken is working this summer as ...

# Ken is working this summer as part of a crew on a farm. He earned $8 per hour for the first 10 hours he worked this week. Because of his performance, his crew leader raised his salary to$10 per hour for the rest of the week. Ken saves 90% of his earnings from each week. What is the least number of hours he must work the rest of the week to save at least $270 for the week? Answer Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams. Updated On: 26-5-2020 Apne doubts clear karein ab Whatsapp par bhi. Try it now. CLICK HERE Loading DoubtNut Solution for you Watch 1000+ concepts & tricky questions explained! 9.6 K+ 2.1 K+ Text Solution 38332216 Answer : C Solution : Ken earned$8 per hour for the first 10 hours he worked, so he earned a total of $80 for the first 10 hours he worked. For the rest of the week, Ken was paid at the rate of$10 per hour. Let x be the number of hours he will work for the rest of the week. The total of Ken’s earnings, in dollars, for the week will be 10x + 80. He saves 90% of his earnings each week, so this week he will save 0.9(10x + 80) dollars. The inequality 0.9(10x + 80) ge 270 represents the condition that he will save at least $270 for the week. Factoring 10 out of the expression 10x + 80 gives 10(x + 8). The product of 10 and 0.9 is 9, so the inequality can be rewritten as 9(x + 8) ge 270. Dividing both sides of this inequality by 9 yields x + 8 ge 30, so x ge 22. Therefore, the least number of hours Ken must work the rest of the week to save at least$270 for the week is 22. Choices A and B are incorrect because Ken can save $270 by working fewer hours than 38 or 33 for the rest of the week. Choice D is incorrect. If Ken worked 16 hours for the rest of the week, his total earnings for the week will be$80 + $160 =$240, which is less than $270. Since he saves only 90% of his earnings each week, he would save even less than$240 for the week.

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00:00 - 00:59weird questions is can is working this summer as a part of a true on a Farm early age per hour for first 10 hours this week because of his performance is true leader Mr salary to $10 per hour for the rest of the week right so that they can worked for X this week right so if the first ten hours he on rupees 8.30$8 for first 10 can eat $8 per hour to eat in 210 that is$80 right now for his performance for is good performance his salary was raised to $10 per hour for the rest of the week so as left in the rest of the weekend 01:00 - 01:59hours left for the rest of the week rest of the week will be x minus 10 for this x minus 10 hours he earn$10 per hour so for x minus 10 can earn 10 into x minus 10 now Kansas 90% of its earnings from each week that means from the first week only he started 7 90% of his earnings what is the least number of hours he must work for the rest of the week 2 at least 270 perfect so this week also he said to 70 that is 90% of solids let's find out savings equals to 90% of 10 x minus 10 write
02:00 - 02:59+ 80 and 80 earlier right so that will be equal to 270 ok so let's calculate this 9 1993 is a 27 this gives us a value of 10 x - 100 right + 18 will be 300 from here tenex equal to 80 - 120 is equal to 310 X equals to 300 plus 2320 aur X = 2 32 right but the number of so what is the least number of customers were dressed of the week ok so he worked total 32 hours for the week and for the first announced his already earned 8 power so the minimum amount of earning for the rest of the week
03:00 - 03:59minimum working hours for the rest of the week 32 - 10 right to that give the value of 20 to the answer


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# Gauss law

1. Feb 18, 2012

### Suziii

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

In case of having a conductor (spherically shaped and uncharged) and an irregularly shaped hole inside it with a charge at an arbitrary point, I am asked to find the electric field right outside the conductor sphere.

2. Relevant equations

Gauss's law ∫EdA= Q(enclosed) / (a constant)

3. The attempt at a solution
I assumed that the electric field outside of the conductor is uniform so I had:
E*(4πr^2)= q/ the constant - we can get E from this, but it turns out to be a wrong answer, maybe just taking E to be constant is not correct...

2. Feb 18, 2012

### GothFraex

Your answer is almost correct. However, you must remember that E is a vector.
Also, you should note that the electric field is not constant but falls off as $\frac{1}{r^2}$.
You can see this from your answer.

3. Feb 18, 2012

### ehild

The magnitude of E is the same all over the outer surface of the sphere, and E*(4πr^2)= q/ε0 according to Gauss' Law. Haven't you forgotten ε0?
As the electric field is zero inside the metal, the electric field lines starting from the inner charge do not go through the metal wall. Therefore the charges accumulated on the outer surface "do not know" anything about the field inside, so they arrange themselves evenly, as far from each other as possible. The surface charge density is homogeneous and equal to σ=q/(4πr2). The electric field can be expressed by the surface charge density as E=σ/ε0.

ehild

Last edited: Feb 18, 2012