APA vs. tort comparison

Question: Any suggestions on where to find a detailed side-by-side comparison of the Administrative Procedures Act use versus the use of tort law for federal agency cases? Thanks for suggestions.
Searching on Scholar and other free law search engines. Expecting to, but not finding any review specific to my request.
Comment: "Tort" is not an acronym.  It is s word borrowed from French that means "wrong."
Answer: There isn't much to compare.
The Administrative Procedures Act creates a variety of statutory rights and remedies, primarily in connection with the issuance of regulations and to a lesser extent with the quasi-judicial adjudication of disputes with a governmental agency arising from agency action or a request for agency action.
Federal government agencies, except in a narrow class of lawsuits for money damages cases adjudicated by the U.S. Court of Claims (e.g. for automobile accidents involving a federal government agency employee) where sovereign immunity is expressly waived, reverse condemnation actions under the 5th Amendment to the U.S. Constitution, and select constitutional violations in Bivens actions (mostly against federal law enforcement officers), are immune from lawsuits arising on tort theories under the doctrine of sovereign immunity. Only one of these remedies - a reverse condemnation action - is allowed as a remedy for agency action also governed by the Administrative Procedures Act, and both are present only in very rare cases where a new regulation or agency action deprives someone of all economic use of their property without just compensation.
I suspect that what you are really looking for a different comparison about which there is a moderate sized academic literature that is discussed by economists and in legal theory (especially by tort law scholars).
This is the literature comparing two different ways other than criminal law for the government to regulate unlawful conduct by private actors.
One way to do that is with "private law" by authorizing someone who is harmed by the unlawful conduct of a private actor to sue the person who engaged in the unlawful conduct in a private tort law lawsuit (a tort is a "civil wrong" that may be remedied in a private lawsuit for money damages).
Another way to do that is to create a government agency that has the authority to have a government official enforce statutes or regulations enforcing or implementing statutes by bringing civil lawsuits, either in an administrative tribunal (like the National Labor Relations Board, or the Securities and Exchange Commission or the hybrid case of Equal Employment Opportunity Commission proceedings) subject to eventual Article III federal court review at least on appeal, or in a court. In these enforcement actions, remedies including fines, restitution to victims of the unlawful conduct, and/or injunctive relief.
There are many academic discussions comparing the features of these two ways to address wrongful, harmful conduct in violation of the law outside the criminal justice system.
But, generally speaking, an administrative agency enforcement mechanism arises from substantive federal regulatory statutes and not from the Administrative Procedures Act which governs primarily the process by which administrative agencies adopt regulations and to a much lesser extent to how administrative tribunals are constituted.
Comment: I appreciate your reply ohwilleke very much. I am traveling and will loose opportunity for a few days to provide a more meaningful reply. What I am seeking is a greater understanding of  recent suits against a federal agency by local and state governments and individuals using both the federal TORT and Administrative Procedures Act. Some litigation focuses more on one or the other.  Proving the federal agency violated their "discretionary function" or caused an "abuse of discretion" by the federal agency appears to be the biggest challenge for the plaintiffs under both laws. More later. TU
Comment: Note: There is significant documentation of harm caused by the federal agency. Specific non-discretionary regulation or "mandatory" policy directives also provide some level of proof the agency is in violation of their discretionary function. Yet, the courts often default to agency deference, seemingly without considering the evidence/facts against the fed agency that would put a private citizen in prison.
Comment: @Eliza You cannot (successfully) sue a federal agency in tort for money damages for abusing its discretion with regard to some discretionary function in a manner that causes serious harm (unless it is a 5th Amendment taking). It is immune from liability of this kind. So there is really nothing to compare.

APA vs. tort comparison


Can flags be used for "good" things?

Question: I know flags are usually to bring bad things to a moderator attention, but can they be used to bring a good thing to a moderator's attention?  For example, I came across a very helpful question: What "hacking" competitions/challenges exist? and it was closed for being too broad.  However, the information it contains is very helpful, so I flagged it, with a request to re-open it.  Was this the proper use of a flag here, and should I do this in the future if I come across something like that?
Comment: Oh yes. If you see a question repeatedly hit by trolls/spammers, you can ping a mod to protect the question.
Answer: "Good things" is a very broad term and doesn't really apply at all to what flags are for. Flags are for bringing important things to the attention of moderators. Most of the time that ends up being "bad things" but not always.
As for your specific example, your flag will probably get declined. Simply saying "this was helpful it should be reopened" is not a good flag. That question was closed for good reason, because it's not a good question for the site. It's usefulness is not at all relevant to its topicness on the site. Usefulness is determined by upvotes and downvotes, not the state of being open or closed.
Comment: @James It's apparently a [word I invented](http://meta.stackexchange.com/search?q=%22topicness%22).
Comment: [Ah, I see :D](http://meta.stackexchange.com/search?q=%22topicnessly%22)
Comment: "*topicness*"..? :P
Answer: Flags should be used whenever you feel that a moderator should look into something, just realize that they create work for someone. Asking a mod to review a closing is fine, if you feel that the question was closed inappropriately, but consider doing the following instead if you really think it belongs:

Edit the question to address any concerns folks raised in comments, or the reasons given in the close reason
Raise a discussion on the per-site meta to see if others feel that the question can now be reopened. If folks think that it can, then it will be.

Alerting the mods that something was closed without taking any other action is very likely going to result in a mod looking, and not taking action. If the concerns raised in comments or the close reason can't be addressed, then it's probably not a good fit for the site, and probably not the best use of time to have someone review it.
Put simply, flags to reopen something tend to carry quite a bit more weight if you've also put some work into the post yourself - just take care to spend your time wisely. If a really broad question managed to get an answer that you think is really useful, then make the question less broad, pair it more with the answer by narrowing it to the answer, and then we've got something of lasting value.
But yeah, flags are okay when you really feel "something isn't right here and someone needs to know about it" - just make sure there's something actionable there, at least to the best of your ability.
Answer: Moderator flags (that is, the ones where you type a custom message, like you did here) should be used for cases where a moderator should act.  Asking that a question be reopened, or closed, doesn't fall into that category unless it is stunningly obvious that there's a problem.  (That's more likely for "close" than "reopen".)  The reason close and reopen don't generally fall into this category is that these are things that should be done by the community.
If you cast a reopen vote, or if a question is edited after being put on hold, it goes into the reopen queue to be considered by other users.  That's the right outcome.  Similarly, if you cast a close vote it goes into the review queue for closing questions.  If you don't think a question is getting enough attention, you can also bring it up on your site's meta.
Here are some "good" cases where a moderator flag would be appropriate:

To point out that a bunch of comments on a post are obsolete, rude, or not constructive.  (If there are a bunch it's better to flag the post than to flag each individual comment.)
To point out an edit that changed the post in a way you think the author wouldn't approve of.
To point out an edit/rollback war.
To request that a deleted post be undeleted, if it was deleted by a moderator and then edited to address whatever problems got it deleted in the first place.
Answer: Given that particular situation, you might be better off making a meta.security post.  Posting in meta can draw additional attention to the question and kick off a discussion of whether or not it should be closed.  A flag may only get the attention of a single moderator, which is much less useful.
It's also worth noting that closing a question doesn't necessarily mean that a question will be deleted.  The deletion vote is separate from the closure vote unless the question has a non-positive score and no answers.  Closure itself just means that no new answers can be posted.  Maybe the thing to do is to post a comment arguing for its retention, since those will be visible in the Delete queue.
You can read more in How does deleting work?
Note:  I'm not arguing that you can't flag for "good" -- just that you might not want to do so in this particular case.

Can flags be used for "good" things?


Character creation for more experienced mages

Question: What is the rule to create more experienced mages in Mage: The Ascension? How many points do you have to create an elder mage?
Answer: In Mage, the "elder" equivalent is "archmage."
And you can find the rules for making those in Masters of the Art. Character creation rules in that book start on page 81 — like elder vampires, they get a lot of additional dots to start with and aren't limited to five dots in their traits, including Spheres.

Attributes: 8/6/4
Abilities: 19/13/7
Spheres: 12, with at least one at six (••••••)
Willpower: Starts at 6
Arete: Starts at 6
Backgrounds: 10
Freebie Points: 30
Merits and Flaws: Up to 7, as normal
Answer: There are no rules in the rulebook for creating more experienced characters, nor is "elder mage" a defined status for mages (though it is used by vampires).
From my experience, the easiest way to create experienced characters is to create a starting character to make sure you have a solid base, and then reward experience to spend after the initial creation.
To create an "elder mage" of sorts, the player should take the Status background to confer that importance onto them.  Points in Chantry, Node, or Cult could also establish the character as an elder of sorts.
If you mean elder as in just an old mage, no special background would be required.  Some life spheres to extend their longevity may be all that they need.
Comment: This is explicitly how *Vampire: the Masquerade* approaches creating elder-level vampires as PCs: you choose how old you want them to be upfront and then assign extra experience points (drawn from a table) based on that. I'm less familiar with *M:tA*, but if the systems are as similar as they were in Revised the *V:tM* approach might be directly applicable for *M:tA*.


Common proof for $(1+x)(1+x^2)(1+x^4)...(1+x^{2^n})=\dfrac{1-x^{2^{n+1}}}{1-x} $

Question: I'm asking for an alternative (more common?) proof of the following equality, more specifically an alternative proof for the inductive step:
$$(1+x)(1+x^2)(1+x^4)...(1+x^{2^n})=\dfrac{1-x^{2^{n+1}}}{1-x} (x\neq 1)$$
This is how I proved it:
Basecase: substitute $1$ for $n$, everything works out.
Inductive step: assume that $$\prod _{i=1}^{n}(1+x^i)=\dfrac{1-x^{2^{n+1}}}{1-x}$$
then $$\prod _{i=1}^{n+1}(1+x^i)=\dfrac{1-x^{2^{n+1}}}{1-x} (1+x^{n+1})$$
Let $a$, $b$ and $c$ be positive real numbers, then $\dfrac {c}{a}=b\Leftrightarrow ab=c$, thus
$$\dfrac {\left( \dfrac{1-x^{2^{n+2}}}{1-x}\right)}{\left( \dfrac{1-x^{2^{n+1}}}{1-x}\right)}=1-x^{2^{n+1}}  \Leftrightarrow \dfrac{1-x^{2^{n+1}}}{1-x} (1+x^{n+1}) = \dfrac{1-x^{2^{n+2}}}{1-x}$$
$$\dfrac {\left( \dfrac{1-x^{2^{n+2}}}{1-x}\right)}{\left( \dfrac{1-x^{2^{n+1}}}{1-x}\right)}=\dfrac {1-x^{2^{n+2}}}{1-x^{2^{n+1}}}$$
Applying polynomial division, we see that indeed $\dfrac {1-x^{2^{n+2}}}{1-x^{2^{n+1}}} = 1+x^{2^{n+1}}$.
Thus $\dfrac{1-x^{2^{n+1}}}{1-x} (1+x^{n+1}) = \dfrac{1-x^{2^{n+2}}}{1-x}$.
However, the exercise was in a chapter on binomial coefficients and pascal's triangle, furthermore we didn't mention polynomial division in class. Which makes me think that there was another solution that I was "supposed" to see. How was I supposed to prove it?
Answer: In the inductive step, you assume that
$$(1+x)(1+x^2)(1+x^4)...(1+x^{2^n}) = \dfrac{1-x^{2^{n+1}}}{1-x}$$
Which means that
$$(1-x)(1+x)(1+x^2)(1+x^4)...(1+x^{2^n}) = 1-x^{2^{n+1}}$$
Multiply by $(1+x^{2^{n+1}})$ and you get
$$(1-x)(1+x)(1+x^2)(1+x^4)...(1+x^{2^n})(1+x^{2^{n+1}}) = (1-x^{2^{n+1}})(1+x^{2^{n+1}})$$
Using $(a-b)(a+b)=a^2 - b^2$, the left side simplifies to
$$(1-x^{2^{n+1}})(1+x^{2^{n+1}}) = 1 - \left(x^{2^{n+1}}\right)^2 = 1-x^{2^{n+2}}.$$
This means that the previous equation becomes
$$(1-x)(1+x)(1+x^2)(1+x^4)...(1+x^{2^n})(1+x^{2^{n+1}}) = (1-x^{2^{n+2}})$$
and, after dividing by $(1-x)$, you are done. No polynomial division required.
Comment: nice solution +1
Answer: Expanding $(1+x)(1+x^2)(1+x^4)\cdots(1+x^{2^n})$ gives $1+x+x^2+\cdots +x^{2^{n+1}-1}$ because every integer $k$ with $0 \le k \le 2^{n+1}-1$ has a unique binary representation (as a sum of powers of $2$).
Answer: On the left-hand side, the ratio of $\Pi_n$ over $\Pi_{n-1}$ is
$$1+x^{2^n},$$
and on the right-hand side,
$$\frac{\dfrac{1-x^{2^{n+1}}}{1-x}}{\dfrac{1-x^{2^n}}{1-x}}=\frac{1-(x^{2^n})^2}{1-x^{2^n}}.$$

Alternatively,
$$\frac{1-x^{2^{n+1}}}{1-x}=\sum_{k=0}^{2^n}x^k.$$
Then
$$(1+x^{2^n})\sum_{k=0}^{2^n}x^k=\sum_{k=0}^{2^n}x^k+\sum_{k=0}^{2^n}x^{k+2^n}=\sum_{k=0}^{2^{n+1}}x^k.$$
Answer: A Proof with a Bit of Field Theory
Over a field of characteristic $2$, $1+x^{2^r}=(1+x)^{2^r}$ and $1-x=1+x$, so the LHS is $$
\begin{align}
\prod_{i=0}^n\,\left(1+x^{2^i}\right)
&=\prod_{i=0}^n\,(1+x)^{2^i}=(1+x)^{\sum\limits_{i=0}^n\,2^i}=(1+x)^{2^{n+1}-1}
\\
&=\frac{(1+x)^{2^{n+1}}}{1+x}=\frac{1+x^{2^{n+1}}}{1+x}=\frac{1-x^{2^{n+1}}}{1-x}\,.
\end{align}$$
Over a field of characteristic not equal to $2$, we may assume to be algebraically closed, hence every nonconstant polynomial has roots in the field. The polynomial $1-x^{2^{n+1}}$ has the derivative $-2^{n+1}\,x^{2^{n+1}-1}$, which is nonzero at any root of $1-x^{2^{n+1}}$.  Thus, $1-x^{2^{n+1}}$ has only simple roots, one of which is $1$.  Therefore, $\frac{1-x^{2^{n+1}}}{1-x}$ is the product of $x-\omega$, where $\omega\neq 1$ are other roots of $1-x^{2^{n+1}}$.  However, if $r$ is the smallest positive integer such that $\omega^{2^r}=1$, then $\omega^{2^{r-1}}=-1$, or $\omega$ is a root of $1+x^{2^{r-1}}$.  Furthermore, all the roots of $1+x^{2^{r-1}}$, for $r=1,2,\ldots,n+1$, are clearly roots of $1-x^{2^{n+1}}$ and they are not equal to $1$.  Finally, if $r$ and $s$ are distinct positive integers, then we can easily show that $\gcd\left(1+x^{2^{r-1}},1+x^{2^{s-1}}\right)=1$ (well, the roots of $1+x^{2^{r-1}}$ are all elements $\omega$ in the field such that $\omega^k=1$ if and only if $2^{r}$ divides $k$).  Hence, $\prod\limits_{r=1}^{n+1}\,\left(1+x^{2^{r-1}}\right)=\prod\limits_{i=0}^n\,\left(1+x^{2^i}\right)$ is a factor of $\frac{1-x^{2^{n+1}}}{1-x}$.  As both of these polynomials have the same degree and are monic, we conclude that $$\prod\limits_{i=0}^n\,\left(1+x^{2^i}\right)=\frac{1-x^{2^{n+1}}}{1-x}\,.$$


Does Axton's Sabre Turret allow for legendary drops?

Question: When I'm farming for legendaries as Axton, is it a bad idea to let my sabre turrets kill the enemy? Do I need to kill them myself to allow a legendary to drop?
Answer: No.
The idea that killing things with your action skill would reduce the legendary drop rate seemed so absurd initially that I almost downvoted the question. I mean, depending on your build, that could easily be the main way you kill everything. It would be extremely odd to penalize a large segment of players playing the game the way it was intended.
That being said, whether you or your follower kills something in Diablo 2 can have an impact on the magic find rate in certain rare circumstances, so I decided it wasn't completely ludicrous.
After about 15 minutes searching message boards, wikis, and guides, I see no reason to think that you should avoid killing things with your turret.
Comment: Thanks for the answer... and for not down voting the question :)
Comment: Well, it's also not "_completely_ ludicrous" or "absurd" because the game says that the turret is a separate entity, but I like the point you made.
Comment: +1 For all the reasons above. It just doesn't make sense for the turret to have this effect. Or any of the characters action skills.


How do I change my account name?

Question: How do I change my account name?
Comment: Given that it changed in February it's not likely to be an April Fool's joke.
Answer: Go to your profile on the main site and click on the "edit" link.
The first field is the display name. Change that from "trystero" to what ever you like. I don't know whether just deleting what's there will work - that is you don't have to set up a name but once you have you might have to always have a name.
This will get propagated to the meta site.
You can then copy that information to all other sites by going to you "accounts" tab and clicking the "Copy Code Review Profile to Related Accounts".
This will update everything apart from the OpenIDs you use on each site.

How do I change my account name?


You 'looks' ugly or you 'look' ugly?

Question: Which one is grammatically correct? Look or looks in a sentence "You ____ ugly". And difference between look & looks
Answer: "You look ugly" is correct.
"You" is second person.  the "-s" is used for singular third person nouns and pronouns.

"I look ugly."
"You look ugly."
"He/She/It looks ugly."
"They look ugly."
Comment: As FumbleFingers has said, '… we should seek to discourage people from providing trivial answers to trivial questions.... We all like to be helpful, but the place for that level of help is English Language Learners'
Comment: @EdwinAshworth Fair enough

You 'looks' ugly or you 'look' ugly?


references for abelian schemes

Question: Hi,
I have a very basic question.
I am looking for references explaining how to construct explicitily a Jacobian starting from a curve or examples of projective equations for an abelian scheme. I know a little bit the theory in general so I need examples to fix it, at least in the cases which are not too complicate( or when it is possible).
Thank you
Comment: Look at the book of Birkenhake-Lange "Complex abelian varieties", Chapters 10 and 11.
Comment: Volume 2 of Mumford's tata lectures on theta, chapter IIIa, is entitled "an elementary construction of hyperelliptic jacobians".
Comment: Also, take a look at Milne's article on Jacobians in the book "Arithmetc geometry".
Comment: and this question:
http://mathoverflow.net/questions/51204/is-there-a-crash-course-book-on-abelian-varieties-e-g-an-introduction-for-ph/51232#51232
Answer: The equations defining the Jacobian of a curve as a projective variety become very complicated as soon as the genus of the curve is bigger than 1.  In the case of genus 2 curves, say $\mathcal{C}:y^2 = f(x)$, Grant [1] gives an explicit embedding in $\mathbb{P}^8$ and the defining equations when $\deg(f) = 5$ and Flynn [2] gives an explicit embedding in $\mathbb{P}^{15}$, the 72 (!) defining equations of the projective variety, and the biquadratic forms defining the addition law for when $\deg(f) = 6$ (see also Cassels and Flynn's book [3] for an "updated" version of Flynn's work in the early 90s among other things).  For this reason, most computations with Jacobians use the Mumford representation of points in $\operatorname{Sym}^2(\mathcal{C})$ together with Cantor's algorithm for the addition law.
[1] Grant, D. Formal groups in genus two. J. Reine Angew. Math. 411 (1990), 96–121.
[2] Flynn, E. V. The Jacobian and formal group of a curve of genus 2 over an arbitrary ground field. Math. Proc. Cambridge Philos. Soc. 107 (1990), no. 3, 425–441.
[3] Cassels, J. W. S.; Flynn, E. V. Prolegomena to a middlebrow arithmetic of curves of genus 2. London Mathematical Society Lecture Note Series, 230. Cambridge University Press, Cambridge, 1996. xiv+219 pp. ISBN: 0-521-48370-0

references for abelian schemes


Read Pre defined Style in MapinfoTab File in Geotools

Question: We have open a MapInfoTab File using OGRFeatureSource the file open on map successfully. But one question we have there is a function named GetStyleString in OGRLibrary of Ogr2Ogr but in our OGRSource the same method is not present. How we can override the same method.
And one more thing if we have a Style feature wise means each feature has a different style is there any option to open style feature wise without creating a Rule.
Answer: Even if you could get the MapInfo style string out of the file it has no real value to GeoTools (at present).
If you want to style each feature differently the only way to do it is to construct an SLD object based on properties (attributes) of the features to control the styling, possibly with additional filtering (or judicious use of the if-then-else function).
Comment: thanks for this , but one more thing there is an explanation for all the functions in `BridgeOgr` in class `OgrLibrary.java` . What is the equivalent of all those methods present in that Class  and how they load style on map?


Using and accessing Webresource

Question: I know this has been asked before. Yes, i did my research but it doesn't seem to be working for me so i hope you experts can help me out :)
Here's what my Project looks like
https://i.stack.imgur.com/nnPZJ.png
Yes, the Build Action is Embedded Resource. I also added this in the AssemblyInfo
Assembly: WebResource("WFL.WebResource.EXT.XXX.png", "image/png")

So now, in the default.aspx i say
Protected Sub Page_Load(ByVal sender As Object, ByVal e As System.EventArgs) Handles Me.Load

        Dim cs As ClientScriptManager = Page.ClientScript
        Dim strReturn As String = cs.GetWebResourceUrl(Me.GetType(), "WFL.WebResource.EXT.XXX.png")
        Dim strReturn2 As String = cs.GetWebResourceUrl(GetType(_Default), "WFL.WebResource.EXT.XXX.png")

        Image1.ImageUrl = "http://localhost" + strReturn
        Image2.ImageUrl = "http://localhost" + strReturn2

        Response.Write("http://localhost" + strReturn)
        Response.Write("http://localhost" + strReturn2)

    End Sub

But when accessing the returned URL, i get The resource cannot be found.
What am i doing wrong?
Thanks in advance.
Answer: This being VB and not C#, you don't need the .EXT portion in WFL.WebResource.Ext.XXX.png. That is something C# needs (to specify folder paths), but VB doesn't. You just want your namespace then resource. Try WFL.WebResource.XXX.png and see if that works.

Using and accessing Webresource


Solution for Airprint with a router that is not Bonjour compatible

Question: I'm trying to troubleshoot remotely for my mother and her iPad. I sent her a Canon Airprint-compatible printer which is visible/usable on her wireless network by a desktop PC and a netbook, but not visible by the iPad. After trying software updates and power cycling without success, we finally determined that her Actiontec router is not Bonjour compatible. We're about to replace her router, but I thought I'd check to see if there might be a cheaper solution to be found here.
Comment: Have you tried turning off the IGMP proxy?http://cooper7.wordpress.com/2012/03/02/multicast-bonjour-on-actiontec-wireless/
Comment: I should add that Mom's not willing to leave her PC on to use a Printopia-type solution. She's 90, I need to let her have this, and give her credit for learning the iPad!
Comment: Funny that you should mention that. I recently disabled IGMP proxy on my Actiontec Verizon FIOS router to fix Bonjour problems I was having. As for Mom's network, I replaced her old router almost 2 years ago with one that was Bonjour capable.
Comment: I found for a HP Laser Printer with Wifi and Airprint that the configuration software from HP gave it (at least in my case) a static IP-number.  This gave trouble.  Worked much better when I changed it to get an IP-number from DHCP.
Comment: @AlanShutko I have been fighting this issue with an AirPort Time Capsule for WEEKS. Enabling "IGMP Snooping" on the Time Capsule made it start working instantly. God bless you.
Answer: I don't know if it will work, mostly because the product hasn't been released yet but xPrintServer might work. It is a bit pricey for this usage though.
Comment: Thanks for the link - I saw that too in the CES coverage.
Comment: Just got my xPrintServer by FedEx. OMG, works great.
Answer: Just thought I'd check- I assume she only has one wireless network? (at 90, I'd be surprised if not, but you never know!)
I have problems with both AirPrint and AirPlay on my network if the devices that are attempting to connect are connected through different routers.
Comment: Good question - she only has one wireless network. But I suppose I should have her check to make sure that the iPad is on her network, and not on a neighbor's open network. I only set her up for her network when we bought the iPad a year ago.
Comment: I hadn't thought of that possibility- but yes, thats worth checking for.

(in my case both routers are physically wired together, and all IP addresses allocated through a single DHCP server, but its still an issue for Bonjour)
Comment: Just to add though that we already know that her router isn't Bonjour compatible, and isn't "broadcasting" the printer. So getting her on her own network isn't going to solve our problem.
Answer: Bonjour is a special form of DNS broadcasting - perhaps if you reset the router to default DNS settings. If you're running multiple subnets for any reason (I doubt so for a 90 year old, but you never know) this will also cause the printer to only be available on one subnet. You'd likely have multiple subnets if you have more than one router.
Comment: I've seen on other forums that other users have been waiting for many, many months for a software update for her particular Actiontec router. I've checked the dsl-provider's site recently - she's on the most recent version. I don't believe she's running multiple subnets, but I'll have to check that out too.


Combinations n sided polygon

Question: Three vertices of a convex n sided polygon are selected. If the number of triangles that can be constructed such that none of the sides of the triangle is also the side of the polygon is 30 then 'n' is
Answer: We proceed to solve this problem in finding the good cases by subtracting the total cases from the "bad" ones:

Total possiblities:
Any combination of $3$ vertices can serve to form a triangle.
Number of combinations of $3$ that can be formed from n choices = $\binom{n}{3}$.

Bad case $1$: triangles with exactly $2$ sides in common with the polygon:
Any set of three CONSECUTIVE vertices can serve to form a triangle with exactly $2$ sides in common with the polygon. Thus, given a polygon with $n$ sides, the total number of ways = $n$.

Bad case $2$: triangles with exactly $1$ side in common with the polygon:
Each side can be combined with any NON-ADJACENT vertex to form a triangle with exactly 1 side in common with the polygon. Since there are $n$ sides, each with $n-4$ non-adjacent vertices, the total number of ways = $n(n-4)$.

Thus, the required total is $\binom{n}{3}-n-n(n-4)$. This can be simplified to $$\frac{n(n-4)(n-5)}{6} = 30$$ The answer is thus $\boxed{n = 9}$. Hope it helps.
Answer: Assuming $n\ge 5$:
Total number of triangles:
$$\binom n3$$
Triangles such that exactly one of its sides is a side of the polygon:
$$n(n-4)$$
Triangles such that exactly two of its sides are sides of the polygon:
$$n$$
Then
$$\binom n3-n(n-4)-n=30$$
$$n^3-9n^2+20n-180=0$$
$$(n-9)(n^2+20)=0$$

Combinations n sided polygon


Open file with any application using xdg-open

Question: How to open files without default application using xdg-open file_name? xdg-open file_name opens the file in default application. To play video files, it always opens with vlc which is set to default. If I want to open with media player, what will be the xdg-open code?
Comment: Just a side note, since it's already been answered that xdg-open will only open the default application for the file type. Check out http://manpages.ubuntu.com/manpages/cosmic/man1/xdg-open.1.html especially the links to xdg-mime and xdg-settings for info on how to list the default types and change them from the command line.
Answer: If you want to open a file with a specific application, use that application's command directly instead of xdg-open. xdg-open's purpose is to open the file with the associated application.
For example, with the default media player, Totem, the command is totem, so you'd do:
totem path/to/file.avi

If you want to use mplayer, do:
mplayer path/to/file.avi

and so on.


Is 2008 Server Core installation the most secure deployment?

Question: In a Windows environment where security is more important than flexibility is Windows 2008 R2 Core Server installation the best option? How much more difficult is it to setup an environment with Server Core?
Comment: Welcome to the site.  Please make sure you have read the FAQ, at:  http://security.stackexchange.com/faq  The question you have asked, as it is, is fairly subjective and open to interpretation.  Can you please refine it so that it is more definite and objectively answerable?  Otherwise, it may be closed.
Comment: How is this any more subijective than the other product-specific questions asked?
Comment: @SteveSyfuhs - I'm not sure which questions in particular you're referring to, but anything including "best" in the question is by definition subjective.  Some can be modified to still be essentially the same question while not being out-of-bounds, but that is not always the case.  Essentially, any question for which there is cannot be an objective, factual, and definite answer, is a subjective one.
Answer: Theoretically Core is more secure than a full installation because of a lower attack surface.
Setting up a core installation requires everything be done via command line, but there are tons of resources out there to help you in the process.  With that being said, it is harder to set up than a full installation because it is all manual.
I would recommend you move this over to Sever Fault though.
Comment: Thanks. That's helpful. Also question raised on server fault.
Comment: yfuhs I don't think "is it more secure" should go over to SF, though the "is it more difficult" probably should. It's 2 different questions... Your "lower attack surface" bit would probably get missed at SF (+1 for that, btw)


IP history of a No-IP host

Question: I use No-IP so that hello.ddns.net points to a server that has a dynamic IP.
Is it possible to see the history of IPs a No-IP host took?
Answer: This answer may be overly simplistic. I apologize if that is the case. I am think of all of those who may stumble across this page.
NoIP is a DDNS (dynamic domain name service). Every domain name must have an IP address in order for the network to know how to address packets and route them through the Internet and indeed any TCP/IP based network. How a DDNS system works is that a computer must report it's current IP address to the DDNS using software along with the domain name. From there, the DDNS hosts the domain name and links it to an IP address dynamically meaning that a user does not have to specifically identify an IP address with a host name in the traditional way but does so in an automatic and convenient way. This is an important feature for some who need to participate in the Internet in a larger way than a casual user but does not have a static IP address assigned to them. Most users of the Internet are subscribers to a service which uses dynamically assigned IP addresses taken from a pool of addresses. These can change every time a user accesses the Internet or in the case of DSL where the access is for a longer period of time, when the user connects and the address lease expires.
That said, it is possible to resolve any domain name be tied to an IP address at any given point, however, this requires resolving the domain name which is a specific action normally taken in the background as the user accesses the network. There are services that keep track of domain name and IP address assignments, however, because of the scale of the Internet, this is not done an a granular way. This means that a user can connect several times a day and have several different IP addresses assigned to a domain name, but because the services that track these things cannot realistically track all of these accesses, many will go unnoticed. One of the reasons for this is because DDNS is not the large market it was years ago with broadband now so prevalent and the need to track these domain names is simply not there.
Part of the problem is that most DDNS services do not set PTR (reverse) records which means that a domain name can resolve to an IP address, but the same IP address will not likely resolve back to the domain name. This means that the domain name must be known in order to track it. With a PTR record, this is not necessary. It is far simpler to reverse look-up IP addresses since they are well defined and known. More specifically, it is possible to know what IP addresses are subscriber IP addresses and in particular which pools of IP addresses belong to a DDNS service.
So to answer your question, it is possible to get some of the IP addresses of a domain name that uses DDNS, but not likely all of them. Most whois websites can give you this information if they are aware of the domain name. But again, it is not very granular.
Answer: If you have a "free" account it is not possible to see a history of IP address changes for a particular hostname. You can only see when the current IP address was last updated.
However, if you have subscribed to the "Plus Managed DNS Service" ($29.95/yr) then you have access to the last "90 days of historical IP address information".
Reference:

https://www.noip.com/blog/2006/06/16/what-was-my-ip-address-last-week/

IP history of a No-IP host


Trace of density matrix

Question: Hi friends of Mathematica, i´m a physics student and try to calculate Tr[rho^4] for a general two qubit system analytically. Since this is a very long calculation, the probability of making mistakes is almost 1. Therefore i wanted to double check my calculation with the help of Mathematica, but unfortunately it gives me no helpful result. Here is my code:
Id = IdentityMatrix[2];
sig[0] = SparseArray[PauliMatrix[0]];
sig[1] = SparseArray[PauliMatrix[1]];
sig[2] = SparseArray[PauliMatrix[2]];
sig[3] = SparseArray[PauliMatrix[3]];
sigv = Table[PauliMatrix[i], {i, 1, 3}];

v1 = {vx, vy, vz};
w1 = {wx, wy, wz};
cM = {{c11, c12, c13}, {c21, c22, c23}, {c31, c32, c33}};

rho = 1/4[
    Id\[CircleTimes]Id + Id\[CircleTimes]v1.sigv +
     w1.sigv\[CircleTimes]Id +
     Sum[cM[[i]][[j]].(sig[i]\[CircleTimes]sig[j]), {i, 1, 3}, {j, 1,
       3}]];

Tr[rho.rho.rho.rho]

I want my result to depend only on the matrix cM and the vectors v1 and w1. But it yields some sparse array notifications. It seems as if the calculation is to complicated for Mathematica, although i doubt this highly, since it is my task to do it with pen and paper :) Has someone any suggestions how to obtain a useful result?
Comment: Oh its a typo, but only here in the forum. In my original code its a dot :)
Answer: Maybe you want this?
rho = 1/4 (KroneckerProduct[Id, Id] + KroneckerProduct[Id, v1.sigv] +
     KroneckerProduct[w1.sigv, Id] +
     Sum[cM[[i]][[j]] KroneckerProduct[sig[i], sig[j]], {i, 1, 3}, {j,
        1, 3}]);
Tr[rho.rho.rho.rho] // Simplify

Most importantly, you must not use [ ]  to group mathematical expression, use ( ) instead.
Comment: Nice. Here's how I'd do the sum: `Total[Normal[Outer[KroneckerProduct, Array[sig, 3], Array[sig, 3], 1] cM], 2]`. Additionally, `Tr[MatrixPower[rho, 4]]` is much more compact.
Comment: Thanks,a lot, this seems to work :D

Trace of density matrix


Dictionary of email server settings for different companies and sites?

Question: Is there a public database of common email settings for different domains for email? Getting Yahoo!, Gmail and others is pretty easy but what if I wanted the IMAP settings from some random company. Is there a good database out there that has a lot of IMAP and POP settings and non-autodiscovery Exchange URLs?
Answer: Mozilla maintains such a database, see https://wiki.mozilla.org/MailServerList


Collectivity in nuclei

Question: Why do inelastic scattering of nuclei as well as their electric multipole radiative transition (via say Coulomb excitation) show similar collective enhancement, even though electromagnetic field interacts only with protons while inelastic scattering affects all nucleons?
Comment: This question is too vague, you have to give more details preferably in links of what you are talking about and the terminology you are using, the energy range .
Answer: It's due to the strong force binding of the nucleus; early experiments which noted these very results you describe imply that there is something extra going on in the nucleus to hold it together.
Of course, as the energy scale goes up, and enters the nuclear regime, things begin to break up.
Comment: So in the region of Coulomb-nuclear interference, the collective enhancement due to inelastic scattering would not be similar to that due to Coulomb excitation? The collectivity parameters would no longer be the same?

Collectivity in nuclei


What's the best retention policy for past client information?

Question: Do you have a policy for how long you retain client data after they no longer work with you? I have old files, emails, contracts, etc., for clients that I haven't spoken with in several years. I also have records from client companies that no longer exist (I work primarily B2B). Is there an ideal amount of time after which you just delete old records? If the company shuts down, do you delete everything sooner? I would keep items that can be used for my portfolio, but I don't know the value or liability in keeping anything else. I have had clients leave and then return a year or so later, so I do want to keep records for at least a year.
Answer: As regards myself, I kept all data/messages/documentation/source code (software is my business) since I have been freelancing, say 20 years ago.
Every now and then, the client may return after years, and this is useful. In some cases, an NDA was concluded with the client; but regardless, in all cases I keep that material strictly confidential.
For safety, or if you are bound to the rules of the General Data Protection Regulation, you can mention this in your general sales terms, your offers/contracts or your GDPR declaration.
Note that "the company shuts down" is usually an event that you are not informed about.
Comment: In this case, I am aware of when my clients closed their businesses. In the case of one, the owner died. In other cases, they shut down due to the pandemic. I design, build, and host websites, so I have a lot of code, graphics, backups, photos, etc. They take up a lot of space.
Comment: @LMKWeb: terabyte disks are cheap.
Comment: True! I actually hadn't considered a hard copy backup since most everything is currently in the cloud, but that's a great idea to get them out of my line of sight without deleting entirely.


Using .ai (adobe illustrator) variables with Drupal variables

Question: As we know that we can assign the variables to Adobe illustrator file.  Is it possible to access these variables by using Drupal 7 variables feature?
Comment: I think the word variables is getting abused here. Content in an Illustrator file is likely inaccessible from Drupal. Can you be any more specific in what you're asking?
Comment: Adobe illustrator have the feature of adding the text to an image and it can be assigned as object & variable (for eg: object=name & variable=aby) and in Drupal there is a feature of Global Variables.  So i was thinking that is it possible to create a template with adobe illustrator file (.ai file) in drupal.  So that we can assign dynamic text to the Template depending on the user input.
Answer: Short answer: no.
Longer answer: This is theoretically possible, but.
What you're referring to really amounts to rewriting some of the fundamental underpinnings of the internet. This would require, at minimum, extremely innovative development and entirely recreating several major software components. For example:
1) Users' browsers read and render hypertext. They would need to be rewritten to understand AI equivalents of links, pages, and other internet standards.
2) Javacript, Jquery, and other client-side components would need to be rebuilt from scratch. You would also need to invent a new CSS and DOM that their replacements can understand.
3) Apache .... would be mostly okay with some minor tweaks. One or two new extensions at most.
4) PHP (which stands for "PHP hypertext preprocessor" and not "PHP advanced graphical tool") would need to be entirely redone, along with all of it's extensions, integrations, and fundamental concepts.
5) Drupal and all its modules (which are build on the assumption that the output will be hypertext) would need to be substantially retooled. In particular, you would need a replacement for PHPTemplate that accesses AI objects.
So: There's a lot to do. I would say "let's get started," except that 6) AI is a proprietary product and we don't have licenses to develop and extend it.
Answer: I think it depends on what you want to do with the resulting file. Are you thinking of a variable data Illustrator document that would be generated from the values in Drupal? If so, I think that is very possible. The Illustrator file format specification is somewhat available, you would just need to process the file without the help of Illustrator (which may present some challenges).
If you want to generate something that would be viewable in a browser you're better off using SVG with some sort of placeholder that you could replace with regular expressions or maybe some XPath queries to get the nodes you want to manipulate and then adding your values there.
As Dominic said, the end use isn't very clear here so the answers are really dependent on what you need to accomplish.


Shifting Between Gears To Achieve Higher Acceleration

Question: If the car's acceleration is higher when the wheel's angular acceleration is higher(tangential acceleration=alpha x radius). Why  are higher gears not used to accomplish the acceleration of the car instead of lower gears?
Comment: Low gear provides more available torque so accelerates better so long as you can still spin the wheels faster than the speed you are currently spinning at. Is that what you're asking? You might want to try and clean up your question. I find the detailed description rather incomprehensible. But from what I can interpret, the line "no longer remain constant" seems to follow from nothing.
Comment: basically I tried so  hard to clarify my question in simplest way . editing it further will lead me to a long essay.
Answer: The Rotational Inertia of individual gears doesn't matter.
You are way overestimating the effect of rotating gears on overall acceleration.  The mass of gears is inconsequential compared to the weight of the vehicle (and the rest of the rotating mass.  So a larger diameter first gear does have a slightly larger rotational inertia.  But gear ratios overpower this small effect and renders it inconsequential.
In the end, acceleration is related to both the rotational and non-rotational mass.  The small change in rotational inertia between first and second physical gears is immaterial.
Answer: It is down to power against drag ie rolling & aerodynamic. After about 40mph, the aerodynamic drag is larger and increases with the square of the speed.
Basically the vehicle has to push more and more air out of the way the faster it goes. Once the power equals the sum of the drag, acceleration will be zero.
And it won't matter if you have 4, 14 or 40 gears or an automatic or cvt.
Comment: @JesseFlynn won't the wheels be in contact withe the ground so any wheel acceleration is linked to vehicle acceleration? And you asked for "good reasons"...
Comment: Thank you for your answer > But that did not answer my question . I actually know this information , but I am asking about something else if you carefully read my question.
Comment: YES INDEED . I've just bold the keywords in my question . I'm talking about wheel ANGULAR acceleration not the vehicle's acceleration which is measured in m/s .
Answer: You are mixed UP. The angular acceleration of the wheel does not effect the entire car's acceleration. What will, is the force applied to the ground from the tires.
Comment: That was my point.
Answer: Every engine will start to choke when demand torque from the transmission, or better said from the tires, gets close to the engine's output torque at that state, RPM, breathing, engine capacity to shed the heat, and many other factors.
All engines need a burst of torque to accelerate their angular speed and that torque comes from the gap between available and demand torque.
A good driver or a good automatic transmission will keep the demand torque under available torque to let the engine breathe and rev up and deliver the best performance till the RPM nears the maximum engine's design RPM and available torque is so much more than the demand torque that will allow up-shifting for speed without choking.
And then and only then what you mean to say (if I understand correctly) is applied, higher gears are engaged or even overdrive.


Android webView Load in Background

Question: I have an app which i want to open a webview
right now i am loading the webview the most simple way
i would like to load the webpage in the background so the user will not have a waiting time
maybe in a service running in the background or in a prevoius activity
i looked around the forums and could not find a refrence to the subject
help will be appriciated.
Thx
Answer: You can set the webview's visibility to GONE and load the page and when onPageFinished is fired, you can then show the webview.
Hope this helps.
Comment: An onPageFinished event is fired when a HTML-code is loaded into a WebView. If later some JavaScript-code will run, a webpage will continue to create, it takes time. Visibility = GONE does not affect a time of loading (have just tested).
Answer: I hade the same problem a while ago, I know what you want. Asynctask! here is a good link I read on to get it to work!
http://www.brighthub.com/mobile/google-android/articles/82805.aspx
How you get it to work properly.
Load your webview in the doInBackground and perhaps a ProgressDialog that shows that your app i working?
GL mate
Answer: you can use asynctaskLoader take a look it's better than the asynctask because it can manage the orientation changed and some other cool things...

Android webView Load in Background


How bad is a leak-through?

Question: I replaced a Fluidmaster in a toilet, and I suspect the fill valve was slightly loose because, after about 12 hours there was a small puddle in the bathroom and the ceiling below was damp. I turned off the water and dried everything, but I'm not sure where to go from here. Can I just spackle and repaint the affected portion of ceiling, or does the drywall need to be removed? Will a one-time incident like this likely cause significant water damage?
Answer: If the damage is fairly small and the paper on the sheetrock has not bubbled or separated, just let it dry out well and prime the water stained area with Bins pigmented shellac, then paint the area. Hopefully you will get a good color match as a touch up.
Comment: Will using a sealing primer like KILZ do the same thing?
Comment: If the stains are not too bad, a couple coats of Kilz should work. Don't be surprised if you see some bleed through of the stain after one coat. Any latex product is susceptible to bleed through. A shellac primer is really the best solution.

How bad is a leak-through?


Does having an afro hairstyle have an advantage in doing headers?

Question: Would the hair have a cushion effect that makes them have more grip and control of the ball in the air?
Comment: Having played football as a defender with both short and long hair at various times, when a header is mistimed and you strike it with the top of your head, having long hair actually does make it hurt less. However, the ball should not be headed that high and is normally struck with the forehead, so the hair has little difference.
Answer: Generally, the last thing you want when heading the ball is more cushioning as that equals less power - as a defensive header, you want it going as far downfield as possible, and as an attacking header you again want it going as fast as possible.
Answer: In a perfect header, the ball is played to the forhead. As far as I know, there is no player with an afro on his forhead (or hair at least). Most players, with long hair, use auxiliaries, such as hair bands, so that the vision is not impaired by the face falling hair.


CVIVI CRM setup on Wordpress

Question: Please can an expert assist me in setting up Civi CRM on wordpress.
Cronjobs and mail
I have done basic configurations, but not sure about cron jobs.
Need also to add in new forms, and how to add them.
Need someone with full setup from start to finish please
Comment: Agree with Aidan - unclear what 'need someone with full set up from start to finish pls' really means. this is a voluntary support system. paid support is available via other means
Answer: Welcome to SE!
If you want to pay someone to do it for you then see https://civicrm.org/partners-contributors
If you want free community help to do it yourself then you'll need to describe in more detail what you have already done and the problems you have encountered.  Have you read the documentation on cron jobs and checked the other cron questions here? https://docs.civicrm.org/sysadmin/en/latest/setup/jobs/

CVIVI CRM setup on Wordpress


blog.example.com(example.com/blog) or another-domain.example

Question: I have a website which is going to show my shareware, another site for my blog. I will post all my thinkings/development tips/design tips/business experience on my blog, but most of them are not related to my shareware. Software users may not interested in my blog. In this case, is it still better to host my blog to another domain?
Comment: i'd say "think branding" in this case. you will brand you url/domain with shareware and blog. is that good? if so, go ahead!
Comment: I don't think this has strictly to do with programming. Anyway, I like it when a company has blog.example.com or example.com/blog. It's probably better also for SEO purposes.
Answer: Depending on just how disjoint the blog posts are from the shareware I would create a new domain for the blog. Then I would use specific tags on posts about the shareware and use a tag RSS feed to put those posts into a news section on the shareware site. Doing this would give you the flexibility to have multiple sites pulling from your blog on different topics. If someone really wants to see all your posts they can click through to your blog from the other site.
Answer: Simple answer - both, but the blog.domain.com should only list posts related to domain.com using domain.com design/layout (RSS feed solution sounds ok).
Answer: I like Carson's approach. Both my product site and my blog site are created using wordpress, and the feeds are managed by feedburner. I searched the web and found feedburner may have limited support for category, but wordpress can support feed for category very well. So The solution can be:

Create 2 feedburner feeds from my blog, one for all the posts of blog (feed1), and the other for one category specified for products (feed2).
Feed1 will be set as my feed for blog
Feed2 will be set as my feed for product site

Looks not too bad, the only concern is that subscribers may getting confused by 2 domains.
Edit: However, when I start to think more about branding. Host 2 domains have no contribution to brand, just confusion. So the answer should be clear, www.domain.com/blog is better. Better for SEO, rank, and less confusion to visitors. But we do can make 2 different feeds for different types of visitors.


degree of an etale cover of the affine line

Question: Let $X\subset \mathbb{A}^N_k$ be an irreducible smooth variety over an algebraically closed field $k$. Suppose we have an etale map $\pi:X\to \mathbb{A}^1_k$. Are there any bounds on the degree of $\pi$? Here etale means flat with smooth, finite fibers. I'm interested mainly in the characteristic zero case.
Comment: If $X \to \mathbb{A}^1_k$ is a finite étale cover and $X$ is connected then it is an isomorphism. No? For $k = \mathbb{C}$ this corresponds to the fact that $\mathbb{C}$ is simply connected.
Comment: @ZhenLin: Thanks! you are right. I think I confused myself by thinking that degree 2 plane curves have genus zero and must give an etale degree 2 map to the affine line.
Answer: Here your $X$ is just a covering space of an open subset of $\mathbb{A}^1_k$ (this is because etale maps are open). If $k$ is characteristic 0, then this is basically reduces to the case of $k = \mathbb{C}$. In this case the Riemann-existence theorem says that everything can be done topologically.
In particular you may be interested in the theory of covering spaces, which among many things says that isomorphism classes of degree $n$ covering spaces of a "nice" topological space $X$ are in bijection with equivalence classes of permutation representations $\pi_1(X)\rightarrow S_n$, where the equivalence relation is "modulo composition by inner automorphisms of $S_n$". Here connected covers correspond to transitive permutation representations (ie, where the image of $\pi_1(X)$ is a transitive subgroup of $S_n$).
You're basically asking for a bound on the degrees of finite covers of subsets of $\mathbb{C}$, for which there is none.
Suppose for example you want covers of $\mathbb{C}\setminus\{0\}$. The fundamental group of the punctured plane is $\mathbb{Z}$, and there are transitive representations $\mathbb{Z}\rightarrow S_n$ for every $n$. In fact the universal cover here is connected and has infinite degree.
If you're only interested in surjective etale maps $X\rightarrow\mathbb{A}^1_k$, then since $\mathbb{C}$ is simply connected, there are no nontrivial covers of $\mathbb{A}^1_k$.
Even if you allow $k$ to be a number field you can use the etale fundamental group to compute that every cover of a subset of $\mathbb{A}^1$ realizable over $\mathbb{C}$ is realizable over $\mathbb{Q}$ (and hence over any number field).
Though, note that if $k$ is not algebraically closed then there exist plenty of covers of $\mathbb{A}^1_k$ coming from taking separable extensions of $k$.
If $k$ is characteristic $p$ then things get quite interesting. For example see https://mathoverflow.net/questions/868/etale-covers-of-the-affine-line


After upgrading my nexus 4 to 4.3, adb in not recognizing it

Question: My device driver for nexus 4 is not installing on my windows, after trying several times.
My sdk tools and platform tools are updated to the latest version.
Answer: I had the exact same thing, i'm using windows 7 btw.
On your phone:
1) Connect your Nexus 4 to your pc with usb
2) Revoke USB debugging authorizations
On Windows
1) In windows go to your device manager (called MTP for my case)
2) Uninstall the current driver
3) Re-install the driver from disk (\extras\google\usb_driver)
In my case the device was visible again in Eclipse, and after authorizing it on my phone I was able to use it like before. Hope this helps
Comment: In my case, I just updated the driver searching the folder you've mentioned.


What is the electric field produced by a conductor "plate"?

Question: I know that the electric field is zero inside a conductor. However, a plate does not have some kind of "interior space." So, if there is a infinitely large conductor plate with a uniform charge density, what is the electric field made by the plate? Is it just the same as a plate with uniform charge distribution? It is so much confusing...
Comment: An infinitely large plate with a finite net change has a charge density of zero and no electric field. Maybe change it to a charge density of $\sigma$ instead of a net charge of $Q$.
Comment: Ok I will edit it. What is the answer then? Could you tell me?
Comment: The field only cares where the charges are, not whether they're in a conductor or not. So it's the same. But in general, you're not allowed to specify the distribution of charge on a conductor: it will rearrange itself to eliminate the field inside and make the field just outside normal to the surface.
Answer: The electric field from an infinite single plane of charge is given by $$\vec{E}=\frac{\sigma}{2\epsilon_0}\hat{n},$$
where $\sigma$ is the area charge density and $\hat{n}$ is the unit vector normal to the plane and away from the plane on both sides.
A conducting plate necessarily has 2 planes of charge because to be a conductor it must be material object with 2 different sides. The charge will rearrange until there is a plane of charge on each side of the plate. At some distance away from the surface, the net electric field will be the sum of the fields which will be
$$\vec{E}=\frac{\sigma}{\epsilon_0}\hat{n}.$$
While there is no truly infinite plate, if the distance from the plate (or the planes) is small compared to the size of the plate, these relationships are very good approximations, especially if one is not close to the edge of the plates.
Answer: If you refer to an ideal plate, with zero width, it's right that there is no "internal space", and so there is no point on asking what is the electric field in that space.
Said so, if with an "infinitely large conductor" you mean a conductor which cover the entire universe, the electric field inside it will be zero, and nothing will exist outside it.
In a more real-world example, let's pick a square plate with it's width: inside it the electric field will be zero, and outside it will be perpendicular to the surface only in the real proximity of the plate


Is it possible to set a page layout when creating a site OOTB (using UI)?

Question: I have created a custom page layout for a publishing site (SharePoint 2010). When i create a subsite using SharePoint UI, the default.aspx page is created with a another page layout. Is it possible to set my page layout as the default one for each new site creation ?
Answer: The changes can be made in site definition files. If the site created by you is using custom site template(You have created template) you can set it in the onet.xml file of the Site definition.
In the onet.xml, refer the module in the configuration and define the module int he following way
<code><Modules>
    <Module Name="Default" Url="$Resources:osrvcore,List_Pages_UrlName;" Path="">
        <File Url="default.aspx" Type="GhostableInLibrary" >
            <Property Name="PublishingPageLayout" Value="~SiteCollection/_catalogs/masterpage/tabviewpagelayout.aspx, $Resources:spscore,TabViewPageLayoutTitle;" />
            <Property Name="Title" Value="$Resources:spscore,SiteDirectoryPageTitle;" />
</code>
........
If the site collection is not created by created by you, the onet.xml will be of sharepoint provided template onet.xml and any changes made to this file will impact the server farm.


Why is an electrodes standard reduction potential does not depend on amount of chemicals?

Question: The standard reduction potential is fixed for the type of chemical reaction. Why is it so?
Why is an emf of a cell or an electrode's Standard reduction potential does not depend on say, quantity of chemical, concentration of chemical present in cell etc.
If there are more chemicals, then more ions. Then why does the potential increase with increase in amount of chemicals?
Answer: Suppose we have some cell with a reaction:
$$ A + B \rightleftharpoons C + D $$
the the cell emf does depend on the concentrations of all the reagents. Specifically we can write it as:
$$ E = E^\circ - \frac{RT}{zF}\ln\left(\frac{a_Ca_D}{a_Aa_B}\right) $$
where the $a_A$ etc are the activities of the chemicals in the cell and $E^\circ$ is a constant. This constant $E^\circ$ is the standard cell emf, and it's just the cell emf when everything is as unit activity. It's a constant because it's defined that way.
We use the standard emf because it's a nice way to separate out concentration effects from the fundamental properties of the cell. We use the same idea for electrode potentials i.e. we write them as a constant plus a term that depends on concentration, and the constant part is then the standard electrode potential.
Comment: Thank you. Is there a way to calculate Eo  from a chemical reaction's fundamental properties? I am just looking at  a way to understand what gives a cell it's emf
Comment: @KarthickS yes there is because it is related to the Gibbs free energy change $\Delta G$ for the reaction. The Gibbs free energy is basically a form of potential energy, and the cell EMF is also a potential energy, and the two are proportional. $\Delta G = -|z|FE$ where $E$ is the cell potential and $F$ is a constant of proportionality called [Faraday's constant](https://en.wikipedia.org/wiki/Faraday_constant). $z$ is the number of electrons transferred in the reaction.
Comment: @Karthick S basically it is really intrinsic to the species involved.  Microscopically you can see it as potential between particles. All the rest brings quantities to a easy to handle and practical amounts.


The process cannot access the file because it is being used by another process

Question: I am trying to do the following:
var path = Server.MapPath("File.js"));

// Create the file if it doesn't exist or if the application has been restarted
// and the file was created before the application restarted
if (!File.Exists(path) || ApplicationStartTime > File.GetLastWriteTimeUtc(path)) {
    var script = "...";

    using (var sw = File.CreateText(path)) {
        sw.Write(script);
    }
}

However occasionally the following error is sometimes thrown:

The process cannot access the file '...\File.js' because it is being
  used by another process

I have looked on here for similar questions however mine seems slightly different from the others. Also I cannot replicate it until the server is under heavy load and therefore I wish to make sure it is correct before I upload the fix.
I'd appreciate it if someone could show me how to fix this.
Thanks
Comment: If two requests arrive at roughly the same time, the two IIS threads can try to write the file at the same time and step on each other's toes. You could implement some kind of locking, but I would advise against that strategy -- maybe you can write a handler to generate the script on the fly instead of writing it to disk every once in a while?
Comment: Is your application run on demand / multiple times? You may need a mutex to protect file access.
Comment: @helb it's a web application and this code is trigger on every request.
Comment: Have you tried sw.Close(); ?

I know that the using statement is closing the file auto. but just to be sure.
Comment: You must guard mutual access since your application may run multiple times in parallel.
Answer: It sounds like two requests are running on your server at the same time, and they're both trying to write to that file at the same time.
You'll want to add in some sort of locking behavior, or else write a more robust architecture. Without knowing more about what specifically you're actually trying to accomplish with this file-writing procedure, the best I can suggest is locking. I'm generally not a fan of locking like this on web servers, since it makes requests depend on each other, but this would solve the problem.

Edit: Dirk pointed out below that this may or may not actually work. Depending on your web server configuration, static instances may not be shared, and the same result could occur. I've offered this as a proof of concept, but you should most definitely address the underlying problem.

private static object lockObj = new object();

private void YourMethod()
{
    var path = Server.MapPath("File.js"));

    lock (lockObj)
    {
        // Create the file if it doesn't exist or if the application has been restarted
        // and the file was created before the application restarted
        if (!File.Exists(path) || ApplicationStartTime > File.GetLastWriteTimeUtc(path))
        {
            var script = "...";

            using (var sw = File.CreateText(path))
            {
                sw.Write(script);
            }
        }
    }
}

But, again, I'd be tempted to reconsider what you're actually trying to accomplish with this. Perhaps you could build this file in the Application_Start method, or even just a static constructor. Doing it for every request is a messy approach that will be likely to cause issues. Particularly under heavy load, where every request will be forced to run synchronously.
Comment: I'm not an expert on web servers, but if the server uses multiple app domains then this won't work as each app domain has its own static instances.
Comment: You could use a named mutex instead of the standard lock. That even works across process boundaries.
Comment: Thanks I took your advice and moved the code to the Application_Start event.
Comment: @Dirk Ah, interesting. I hadn't thought of that, that's a good point. As I said, I'd never actually do this, it's more of just a proof of concept. But yes, if that's true (and I believe it), it's definitely worth bringing this code out into something that only runs once.


Silverlight Datagrid binding with Object with a List Property

Question: I'm having trouble with some Silverlight functionality.  My goal is to get some data from the database and display it in my grid.  Sounds simple, however, there are 7 columns that are always going to be there.
The rest of the columns depends on the account.  There could be 2 (called Actions) associated with a certain account, compared to another account that could have 5, and the action names can be completely different.
I have successfully been able to return a list of all the possible actions for the given account, and then adding the columns to the grid in the code bihind(VB), and I can bind my data to the grid for all the columns that
are already known.  The way I store the item's actions, is a Property of type List.  My problem is finding a way to iterate through the actions list of the Item object in order to bind those actions to the grid.
Any suggestions?
Answer: You can create extension that will create additional columns and bindings that will bind each value from list to column. Also you should create an additional Dependency Property which you'll bind
to your List.


Can't post file data when request is sent over SSL

Question: I have problems trying to use custom basic authentication module similar to this. The client uses HttpWebRequest class.
The client runs the following code:
void uploadFile( string serverUrl, string filePath )
{
    HttpWebRequest request = (HttpWebRequest)HttpWebRequest.
        Create( serverUrl );
    CredentialCache cache = new CredentialCache();
    cache.Add( new Uri( serverUrl ), "Basic", new NetworkCredential( "User", "pass" ) );
    request.Credentials = cache;
    request.Method = "POST";
    request.ContentType = "application/octet-stream";
    request.Timeout = 60000;
    request.KeepAlive = true;

    using( BinaryReader reader = new BinaryReader(
        File.OpenRead( filePath ) ) ) {

        request.ContentLength = reader.BaseStream.Length;
        using( Stream stream = request.GetRequestStream() ) {
            byte[] buffer = new byte[1024];
            while( true ) {
                int bytesRead = reader.Read( buffer, 0, buffer.Length );
                if( bytesRead == 0 ) {
                    break;
                }
                stream.Write( buffer, 0, bytesRead );
            }
        }
    }

     HttpWebResponse result = (HttpWebResponse)request.GetResponse();
     //handle result - not relevant
 }

If the request is created for a URI starting with http:// it works okay - a request reaches the server, the authentication module is passed the request, it replies with WWW-Authenticate, the request is repeated now with authentication parameters, the module validates it and it passes further.
If the request is created for a URI starting with https:// it doesn't work. The initial request gets to the module and the module replies with WWW-Authenticate
void ReplyWithAuthHeader()
{
    HttpContext currentContext = HttpContext.Current;
    context.Response.StatusCode = 401;
    context.Response.AddHeader( "WWW-Authenticate",
       String.Format("Basic realm=\"{0}\"", "myname.mycompany.com"));
}

an an exception is thrown at the client with "Unable to write data to the transport connection: An established connection was aborted by the software in your host machine." text.
I tried System.Net tracing and discovered that after sending the initial request the client gets back the following header:
Date: Fri, 04 Feb 2011 12:15:04 GMT
Server: Microsoft-IIS/5.1
X-Powered-By: ASP.NET

while when the URI started with http:// the client received the following:
Content-Length: 1894
Cache-Control: private
Content-Type: text/html; charset=utf-8
Date: Fri, 04 Feb 2011 12:12:11 GMT
Server: Microsoft-IIS/5.1
WWW-Authenticate: Basic realm="myname.mycompany.com"
X-AspNet-Version: 2.0.50727
X-Powered-By: ASP.NET

so clearly the WWW-Authenticate response is swallowed somewhere and doesn't reach the client.
Also if I exclude the code that writes the file data into the request it also authenticates okay.
How do I fix this? How do I make the WWW-Authenticate response get to the client?
Answer: What are the http status codes returned from the server?
Is the certificate of the server valid?
Can you post the stack trace?
Try to disable your firewall and other networking tools like fiddler etc on your client.
Also try with the KeepAlive property to false.


Why am I seeing "Your network settings prevent content from loading privately" in the mail apps after updating Pi-hole?

Question: After a recent update to my Pi-hole server, I'm now seeing the warning "Your network settings prevent content from loading privately" above email messages in the Mail apps in iOS/iPadOS 15.x and macOS v12 (Monterey) when I'm connected to my home Wi-Fi, preventing me from using the "Protect Mail Activity" feature in iOS 15.
Answer: Recent versions of Pi-hole automatically sink hole requests to two key Apple domains (mask.icloud.com and mask-h2.icloud.com) to prevent the Apple Private Relay feature from bypassing Pi-hole. Due to this, the "Protect Mail Activity" feature in iOS 15 may not work when connected to a network using a Pi-hole server.
In iOS, the "Protect Mail Activity" feature is found under Settings → Mail → Privacy Protection. In macOS, it's found under Mail → Preferences → Privacy.
To resolve this, add the BLOCK_ICLOUD_PR=false setting to your Pi-hole server's pihole-FTL.conf file using the following steps:

SSH to your Pi-hole server
Run the following command: sudo nano /etc/pihole/pihole-FTL.conf
Append the following setting to this file and save the file: BLOCK_ICLOUD_PR=false
Either restart your Pi-hole server, or just restart the DNS Resolver service with pihole restartdns

Pi-hole documentation: BLOCK_ICLOUD_PR=true|false (PR #1171)
Comment: This is a fine, fine Q&A.  Much obliged
Comment: It should be noted that this may cause traffic to bypass the Pi-hole, i.e., instead of more privacy you may actually get less. Tracking may be obscured though the privacy relay, however, telemetry sent by the Apps may contain sensitive information and will not be blocked anymore.
Comment: @Christoph I'm not sure which resources could be useful here but all your conclusions are already spot on. Tracking pixels are typically individualized, so the argument by Apple is some like "the sender cannot know if it was read because all tracking pixels will always be loaded" - which makes them less reliable as metric on one hand but giving wrong results upstream, on the other hand (like a collection agency thinks you have read the mail even when you didn't). I'm still convinced that bypassing what I wanted to get blocked isn't the right solution here.
Comment: @MrD This is a very good point which I'd like to understand better. So what Apple's privacy relay does is anonymise/obscure loading of objects in emails (though I guess that doesn't help if the tracking pixels themselves are individualised?) but it does not prevent anything from being loaded (which is the aim of pihole). Is that correct? Can you point us to any resources related to this?
Comment: ©MrD Thanks for clarifying/confirming. But I think saying that Apple is bypassing "what I wanted to get blocked" is slightly misleading because it's not simply bypassing your pi-hole block list. What is bypassed or not depends on the content of the email/website, right? In that case, maybe it would be more accurate to say that Apple is opening loop holes for emails and websites to bypass your pihole. You may say, that ultimately that amounts to the same thing, but I think it's good to understand the difference.


Can a node on a terminated end of a CAN bus self-identify that it is the final node?

Question: Using the example below, is there any way for Node1 or Node3 to be able to recognize that they are the final nodes on the CAN bus? I want to allow my device to automatically enable a 120Ohm termination in its circuit if it knows that it is the final node on the bus.
Comment: The easiest (and cheapest) way to deal with this is usually just to provide a jumper or connector on each side of the bus.
Comment: You should add something to the cable- maybe a pulled up input that is connected to GND of the next node. Orhewise the method is TDR (time domain radar) which is an overkill.
Comment: I like the reasoning! If there are 120ohm (termination) at the both end of the bus (terminals), none of the nodes would need to worry about the termination.
Answer: Not really. The bus won't work properly (at least not to spec) if it is not terminated. Likely there is no way in the CAN network to communicate to each node to organize how they would separately detect correct (or not) termination.
In principle, each node could (randomly ?) send signals on the bus, and watch for reflections. If there was only 1 reflection, it could presume it is on the end of the bus (and the reflection is from the far end); if there were 2 reflections (one from each end; these would be distinguishable only if it was not in the precise middle of the bus), then it could presume it is in the middle.
However ensuring that all nodes work independently (before communication is established) would not be possible over the (unterminated) CAN network.
Alternatively, a node could measure the effective impedance of the bus. All nodes not at the ends would see a transmission line impedance of Z/2, while the ones at the end would see Z. in practice this would be difficult because the (low cost) CAN transceivers couldn't measure this TDR (time-domain reflectometry) signal accurately enough. In addition, when there might be a cluster of nodes at the end of the bus, they couldn't easily distinguish which is the dominant one.
Answer: Do you have a CAN_IN and CAN_OUT connector on each device? Can you customize the connector? If so, maybe you could add a pair of pins to each connector. On the cable, the two pins would be shorted together. On the device, one pin would have a pullup and the other would be grounded. Any time a cable is plugged in, you can detect it because the loopback on the cable will force the pulled-up node to GND.
If a device has only one cable plugged in, it must be the last device, and should enable termination.
Answer:
Can a node on a terminated end of a CAN bus self-identify that it is
the final node?

Yes, if you have control over what connector you use to connect to the bus, and the design of the bus cable itself.
The cable that forms the backbone of the CAN bus will have connectors on it.  If you are going to use one of the CAN nodes themselves to terminate the bus then that node must be on a specific connector that is physically at the end of the bus.
On all the bus connectors you can include GND, CAN_H, and CAN_L signals plus a fourth pin.  On the nodes in the middle of the bus the 4th pin is blank.  On the nodes at the end of the bus its not.  On the end-nodes you can either jumper the 4th pin to ground to signal to the node that it should switch in termination.  Or even easier, just jumper the 4th wire directly to the termination as shown below.

simulate this circuit – Schematic created using CircuitLab


How to setup a back test step by step?

Question: I would like to setup a back test for Indian equities, Kindly help out with step by step procedure, No need to go into details, outlining of procedure are enough in bullet point, I will research further, In case of further queries, I'll raise my question in the same forum.
Answer: There are many ways of doing this, but what most traders do is program their strategies into their trading platform and use the backtest feature to make the program give you the set of indicator parameters that will give you the best returns.
You ideally perform the backtest optimizations on only a portion of your historical prices. Once you are content with the results, run backtests on out of sample data to see if they come close to the optimized results. Also, perform walk-forward tests to make sure the returns (or losses) are distributed throughout history and not in isolated instances.
If you cannot program your strategy due to it being discretionary in nature, you will have to go through your charts and log the buys and sells manually in a spreadsheet.
Try out the Ninjatrader platform. It is free and really good for simple individual tests. You might even be able to make your strategies with a visual wizard without having to program much if your strategy is simple enough.
Answer: The most difficult aspect of a backtest is making sure you are working with all stocks that were available at the time each decision is made and using only information that would have been available at the time each decision is made.
This means (1) the database must include  stocks that were once in the index but now are not ("dead stocks" like Enron or Northern Rock Bank), (2) the database must include "point in time" figures that capture the date that earnings were released as well as the amount announced, if the earnings are restated there should be a second date and figure with the restated earnings and restatement date, and so on. Many databases include only the currently accepted earnings figure and are not clear on the date they were released.
Also, you have to be careful about overfitting, which is the testing of a large number of hypotheses at random in the attempt to generate a statistically significant one.

How to setup a back test step by step?


Cauchy-Schwarz inequality on double-summation term

Question: I have the following, where $v$ is a vector
$$
v\cdot (v\cdot \nabla)v
$$
which in index notation becomes $v_jv_id_iv_j$. I want to apply the Cauchy-Schwarz inequality on this, which is given by
$$
|\sum_nA_nB_n|^2 \leq \sum_nA_n^2\sum_mB_m^2
$$
How do I use this (which only has a single summation index on the LHS) on $v_jv_id_iv_j$, which has two indices?
Answer: Assuming $\nabla$ is not just some vector, but a differential operator, and $v$ is a function, and for simplicity $n=3$, there are two vectors involved: $$v=\begin{pmatrix} v_1 \\v_2\\v_3\end{pmatrix} \text{ and } w := \begin{pmatrix} (v_1 \partial_x + v_2 \partial_y + v_3 \partial_z )v_1   \\ (v_1 \partial_x + v_2 \partial_y + v_3 \partial_z )v_2\\ (v_1 \partial_x + v_2 \partial_y + v_3 \partial_z )v_3\end{pmatrix}.$$
Applying Cauchy-Schwarz to their scalar product yields
$$ \vert v \cdot (v \cdot \nabla)v \vert = \vert v \cdot w \vert\leq \Vert v \Vert \Vert w \Vert = \left(\sum_j v_j^2\right)^{1/2}\left(\sum_j ((v_1 \partial_x + v_2 \partial_y + v_3 \partial_z )v_j)^2\right)^{1/2}.$$
I'm deliberately not using a double summation here, but if you wanted to, you could reintroduce it to shorthand the rows of $w$, i.e. the second sum above.
Edit: Loosely speaking, keep the summation over $i$ attached to the second $v_j$, and do Cauchy-Schwarz for $j$.
Comment: @BillyJean: It depends on what 'it' is. I'll edit my answer to give some more insight; we can play with $w$.
Comment: @BillyJean: No edit, just a comment due to time constraints: If we can show that $\vert w \vert \leq \vert v \vert \vert \nabla v \vert$, then we are done. It's not clear what's meant by $\nabla v$ since both are vectors. Is it a scalar? Is it a matrix? For a matrix $A$, we have $\vert Av \vert \leq \Vert A \Vert \vert v \vert$, where $\Vert A \Vert$ denotes the induced matrix norm of $A$.
Comment: Thanks for your answer (which I agree with). According to my book, it should equal $|v|^2 |\nabla v^|$, but that can't be right?
Comment: thanks, I will give it a shot and report back


How to separate in object mode?

Question: I have successfully separated two rings out of a UV sphere using "separate" (Y key) option in Edit mode. But as soon as I switch to Object mode, there is no separation and the objected is treated as one. What can I do to separate it in Object mode for real?
Answer: You have made two different meshes but they are still parts of the same object. To make a new object, go back in Edit mode, select the separate mesh with L, then press P > Selection, this part will be a new object.
By the way, to create a separate object you don't have to separate the part of the mesh with Y or V, you can just select the part of the mesh you want to separate, then P (separate) > Selection.
Comment: Gotcha. I first selected my mesh separated it using Y and then did the Separate by Selection. It was Compulsory to first separate the object or else the object got embedded/fix/stuck into the mesh and after that, I even couldn't Select that part of the mesh.
Comment: you don't have to separate the part of the mesh with Y or V, you can just select it, then P (separate) > Selection

How to separate in object mode?


Action hook for editing post

Question: I'm looking for the add_action() hook for when a current post is edited and saved. I know there's edit_post and save_post, but I'm looking for something that ONLY fires when a current post is edited, not created or altered in any other way.
Answer: Not a real answer.
Take a look the files that use save_post and edit_post
It will lead you to the hook post_updated
Take a look and see if you can find a hook you need. Keep in mind that you might need to do some checks in the hook.
You'll figure it out.
Comment: Yeah, I have an idea of how I could do it. Thanks, I'll just except this as the answer.
Answer: The codex doesn't really give much information but it does look like edit_post is the hook to use.

edit_post
  Runs when a post or page is updated/edited, including when a comment is added or updated (which causes the comment count for the post to update). Action function arguments: post or page ID.

Action hook for editing post


Particle in infinite well but not centred at origin

Question: When the box is centered at origin, we get the wavefunction
\begin{equation}
\psi = \sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})
\end{equation}
But what will be the wavefunction of the particle is not centered at origin? If the box extends from $x = x_0$ to $x = x_0+l$. To solve this, I used the general solution which is
\begin{equation}
\psi = A\sin(\frac{\sqrt{2mE}x}{\hbar})+B\cos(\frac{\sqrt{2mE}x}{\hbar})
\end{equation}
then applied boundary conditions which gave me
\begin{equation}
\frac{A}{B} = \tan(\frac{\sqrt{2mE}(l+2x_0)}{\hbar})
\end{equation}
Now I'm stuck here. I don't know how to solve further. Any help will be appreciated
Answer: You can trivially obtain the solutions to the shifted problem by replacing $x$ with $x-x_0$ in the original solutions.  From there, it's a matter of elementary trigonometry (i.e. the angle addition formula) to express the result in terms of $\sin([\ldots]x)$ and $\cos([\ldots]x)$.
Answer: Another general solution for the wave function is:
$$\psi = A\sin([...]x+a)$$
where $a$ is a constant. You can easily put the boundary conditions you like.


Why does electric field intensity $E$ can be uniquely determined by its divergence and curl?

Question: My question is, the number of following equations
$$\nabla\cdot E=\frac{\rho}{\varepsilon}$$
$$\nabla\times E=-\frac{\partial B}{\partial t}$$
is 4 while the number of unknown variables $E=(E_1,E_2,E_3)$ is 3. Intuitively, the equation is overdetermined and the solution may not exist unless four equations are correlated. Is my intuition right?
Comment: Comment to the question (v2): If OP is viewing $B$ as an _arbitrary_ source vector field, then the eqs. system (Gauss's law+Faraday's law) indeed generically overdetermines the $E$ field. However, this is because one needs to impose the remaining Maxwell's equations as well, which turns the overdetermined part of OP's question into a duplicate of [this](http://physics.stackexchange.com/q/20071/2451) Phys.SE post.
Comment: related:  http://math.stackexchange.com/q/41844/
Comment: @ArtBrown I'm not asking the uniqueness of decomposition but the converse that given the divergence and curl, if it is unique
Answer: The divergence and curl do not uniquely determine a vector field. For example, if all the derivatives are zero, any constant field is a solution. In order to determine the field, we must include boundary conditions.
As far as your question, you are not looking at equations for the three components of $E$. You are looking at equations for the partial derivatives of those components, of which there are nine ($\frac{\partial E_x}{\partial x}, \frac{\partial E_x}{\partial y}$), etc.
Comment: Yes, once time-dependent boundary conditions are established, Maxwell's equations completely determine the electromagnetic field.
Comment: I don't know how to make that more clear. You know that the divergence and curl are derivatives right? You know how to write them in component form? If so, I don't see how to explain any more that they are equations for the derivatives.
Comment: Thank you very much. And could you explain more? Why am I looking to all 9 partial derivatives rather than 3 components?
Comment: Indeed. Then 4 underdetermined equations for 9 unknowns? Can boundary conditions give the rest 5?
Comment: Could you explain all boundary conditions? How many of them
Comment: different boundary conditions are used in different circumstances http://en.wikipedia.org/wiki/Maxwell's_equations#Solutions
Answer: The electric field $E$ is uniquely determined by its curl (Faraday's law) and its divergence (Poisson's) if it is assumed that the field must fall to zero at infinity.  The two operators are indeed correlated so that they do not overdetermine the field.
To see what's going on, suppose $E_2$ is a solution of the second (curl) equation (Faraday's law).  Then so is $E_1+E_2$, where $E_1=-\nabla  \phi$ is a gradient of a scalar function.  Any scalar function (within reason). (In math parlance, gradients are in the kernel of the curl operator.)
So, you need 1 more equation to determine the scalar function $\phi$.  And, voila, there it is, the first equation (Poisson's).
To be tidy about it, move all the gradients (stuff with non-zero divergences) into the $E_1$ field, so that $E$ is the sum of two components with one ($E_1$, known as the longitudinal component) curl-free and the second ($E_2$, aka the transverse component) divergence-less.  Qmechanic provides explicit formulas for these components in his answer to this question.
Now, given such a field $E$ has been constructed, are there other fields which also satisfy the two equations (i.e. have the same divergence and curl)?  By the Helmholtz decomposition theorem, this solution is unique, if it is required that the field falls off to zero at infinity "sufficiently quickly".  (You can see the precise definition of "sufficiently quickly" in the reference, as well as the required reasonableness of the functions involved.)  For example, you can add a constant field and the equations are still satisfied.
All this assumes that:
1) $B$ is known and is divergence-less
2) the currents $j$ are such that the last Maxwell equation is satisfied.

Update.  More on uniqueness:  If $E_a$ and $E_b$ have identical divergences and curls, their difference $E_d$ has zero curl (and is hence the gradient of a scalar function $\psi$ in a simply connected domain) and zero divergence, so $E_d$ is the gradient of a harmonic function (solution of Laplace's equation):
$$E_d=-\nabla \psi \quad , \quad  \nabla^2 \psi = 0$$
Assuming $E$ goes to zero at infinity forces $E_d=0$ (and proves uniqueness), via a Green's Theorem application.  See for example here (pdf).
Comment: It would be good to mention the conditions whereunder your version of Helmholtz decomposition holds (swiftness of decay at infinity), so that your answer is consistent with Mark Eichenlaub's, but +1
Comment: @WetSavannaAnimalakaRodVance: Thanks.  I've added in the appropriate caveats.
Answer: Let's look at this from a numerical perspective.  Take Maxwell's equations and solve them for the time derivatives.  (Here, I use $\epsilon_0 = \mu_0 = c = 1$ units.)
$$\begin{align}
\nabla \cdot E &= \rho \\
\frac{\partial E}{\partial t} &= \nabla \times B - j \\
\nabla \cdot B &= 0 \\
\frac{\partial B}{\partial t} &= - \nabla \times E
\end{align}$$
From initial data on some spatial hypersurface, the time evolution of the EM field is determined by the 6 time derivative equations (2) and (4) above.  The other two equations involve only spatial derivatives and form constraints on the initial data.
For general differential equations of this form, I think the takeaway here is that boundary conditions are constrained.  Let me restrict myself to the case $\nabla \cdot E = \rho$ and $\nabla \times E = 0$.  We know that the electric field at any point can be found using a convolution with the Green's function--that is, the field of a point charge--along with a surface integral:
$$E(r) = \int_M \frac{\rho(r') (r-r')}{4\pi|r-r'|^3} \, dV' + \text{nasty surface integral over $\partial M$}$$
Usually the surface integral is ignored because we take the boundary condition at infinity, but it is definitely there, and it involves integrating $E$ directly on that surface (along with the Green's function).
But once $\rho$ is chosen, $E$ is constrained on the surface to respect the choice of $\rho$.  It must because we have, through Gauss's law,
$$\int_M \rho dV = \oint_{\partial M} E \cdot dS = Q$$
In other words, you cannot simply put any choice of $E$ on the surface as a boundary condition.  You can put some $E$ that respects the total charge within the volume, and then perhaps add any extra $E'$ that corresponds to charge from outside the volume (and thus is divergence- and curl-free), but that's all the freedom you have.
In short: boundary condition data is constrained.  The additional constraint equations eliminate degrees of freedom and should make everything match up.
Answer: Maxwell's equations do not uniquely determine $\mathbf{E}$ and $\mathbf{B}$.
Instead, they place constraints on how the fields may coexist given different setups. Usually, one must provide boundary conditions to get the actual answer; in physical situations there are almost always boundary conditions which bring  back the deterministic nature of the system.
This is rather evident when you attempt to solve the equations for a moving rod in a time varying (but space-constant) magnetic field. Quoting from my answer to the relevant question

Your fundamental issue is that Maxwell's equations (of which Faraday's
  law is one) are not "cause and effect". You cannot "plug in" a value
  of magnetic field and get a corresponding value of $\bf E$ field
  induced by the $\bf B$ field. All Maxwell's equations tell you is
  "which kinds of $\bf E$ and $\bf B$ fields can coexist given so-and-so
  conditions".
Trying to solve the situation via Maxwell's equations
I remember solving a similar situation via Maxwell's equations and
  being surprised by the answer.
The "initial conditions" were $\mathbf {B}=\beta t\hat k$, $\rho=0$
  (no charge), $\mathbf{J}=0$ (no current).
Solving{*} for $\mathbf{E}$, using the differential+microscopic form
  of Maxwell's equations(since the integral form can only get you the
  value of $\bf E$ at certain positions at many times), I got:
$$\mathbf{E}=\hat i (lx + \frac{\beta}{2}y+az+c_1)+\hat
> j(-\frac\beta{2}x+my+bz+c_2)+\hat k(ax+by+nz+c_3)$$
where $a,b,l,m,n,c_1,c_2,c_3$ are arbitrary constants subject to
  $l+m+n=0$
Note that this is a family of electric fields (Setting certain
  constants to zero, you get concentric ellipses IIRC). All this means
  is that any $\bf E$ field of this type can coexist with a $\bf B$
  field.
Implication for your problem
This means that your initial conditions are insufficient/inconsistent.
  Along with such a magnetic field, any type of electric field
  satisfying the above equations can exist--and must exist.
So, in addition to knowing how your magnetic field is changing with
  time, you need to know:

Which one of these bajillion electric fields is present
Where is the rod in relation to this electric field?
Comment: Thank you very much. I benefit much from your answer. Would you mind specifying all boundary conditions (and also how many) for me?
Comment: @ShuchangZhang It really depends. I'd hazard a guess that the value of E on every point on a given loop ought to be enough.


How to make a batch create and write on a text file

Question: I am trying to make a program with batch that will allow me to add and see customers for a business I work for. I have all the input stuff correct, but I can't get the program to write all the info to the file. This is what I have for that:
echo %first%, %last%, %address%, %phone%, %compmodel%, %date%, %problem%, %email% >> Customers\%date%/%last%.txt

When it gets to this part of the script, it says that it cant find the path specified. I've tried to tell it exactly where, but that doesn't work either.
Any help would be great!!!
Answer: there is a / character in the filename you are trying to write to. That is not allowed.
echo %first%, %last%, %address%, %phone%, %compmodel%, %date%, %problem%, %email% >> Customers\%date%/%last%.txt