I have an old file named ˇ¡␀␀ˇ¡␀␀␀"9 ␀"9 which is not deleted when I empty the trash. I have tried renaming it but it does not accept any changes. Using the Terminal didn't work either so far:  Mac-Pro:deletethisfolder me$ find . -inum 425884608 -delete find: -delete: rmdir(./ˇ¡␀␀ˇ¡␀␀␀"9\006␀"9): Invalid argument   what can I do? Thanks!
When I try to move, rename, or delete a file named ␀in Finder, I get this error message:     The operation can't be completed, because an unexpected error occured (error code -50).   With mv, rm or find -delete in Terminal   rm: /Library/␀: Invalid argument  mv: rename /Library/␀ to /tmp/␀: Invalid argument  find: -delete: unlink(/Library/␀): Invalid argument   Can I move or delete this file in Mac OS X or do I have to use another operating system?  I even tried creating a file with this name in /tmp and it has not gone away since 8 reboots.
Get top results for each group (in Oracle) What is the difference between linear equation and linear regression? I am using linear equation to find the correlation of values of x, y and z  3x + 4y + 2z = 6 4x + 1y + 4z = 17 8x + 2y + 5z = 9   I would like to know the difference between linear equation and linear regression. Is it the same in approach? I have noticed that the confidence interval for predicted values in an linear regression tends to be narrow around the mean of the predictor and fat around the minimum and maximum values of the predictor. This can be seen in plots of these 4 linear regressions:    I initially thought this was because most values of the predictors were concentrated around the mean of the predictor. However, I then noticed that the narrow middle of the confidence interval would occur even if many values of were concentrated around the extremes of the predictor, as in the bottom left linear regression, which lots of values of the predictor are concentrated around the minimum of the predictor.  is anyone able to explain why confidence intervals for the predicted values in an linear regression tend to be narrow in the middle and fat at the extremes? What is the bug about: Keyword gets duplicated even when the query is not modified.  Impact: As far as i can tell - cosmetic.  How to replicate: Go to:  and keep pressing Search button. With each search there will be a duplicate python keyword appended to search box.  Best guess for what is causing it: I think you have somewhere a piece of code that appends tag as a keyword without checking if it is already present. My reasoning comes from this url:  not having the same issue - the tag is not present in it. Bolzano-Weierstrass Theorem Proof Abbott (Ax. of Choice) In Abbott's Proof of the Bolzano-Weierstrass Theorem, does Abbott use any form of the Axiom of Choice ?   I think he is since he chooses an $a_{n_k} \in I_k$ where there are multiple such $a_{n_k}$. I think I remember reading somewhere that the Baire Category Theorem is supposedly quite powerful. Whether that is true or not, it's my favourite theorem (so far) and I'd love to see some applications that confirm its neatness and/or power.  Here's the theorem (with proof) and two applications:    (Baire) A non-empty complete metric space $X$ is not a countable union of nowhere dense sets.  Proof: Let $X = \bigcup U_i$ where $\mathring{\overline{U_i}} = \varnothing$. We construct a Cauchy sequence as follows: Let $x_1$ be any point in $(\overline{U_1})^c$. We can find such a point because $(\overline{U_1})^c \subset X$ and $X$ contains at least one non-empty open set (if nothing else, itself) but $\mathring{\overline{U_1}} = \varnothing$ which is the same as saying that $\overline{U_1}$ does not contain any open sets hence the open set contained in $X$ is contained in $\overline{U_1}^c$. Hence we can pick $x_1$ and $\varepsilon_1 &gt; 0$ such that $B(x_1, \varepsilon_1) \subset (\overline{U_1})^c \subset U_1^c$.   Next we make a similar observation about $U_2$ so that we can find $x_2$ and $\varepsilon_2 &gt; 0$ such that $B(x_2, \varepsilon_2) \subset \overline{U_2}^c \cap B(x_1, \frac{\varepsilon_1}{2})$. We repeat this process to get a sequence of balls such that $B_{k+1} \subset B_k$ and a sequence $(x_k)$ that is Cauchy. By completeness of $X$, $\lim x_k =: x$ is in $X$. But $x$ is in $B_k$ for every $k$ hence not in any of the $U_i$ and hence not in $\bigcup U_i = X$. Contradiction. $\Box$    Here is one application (taken from ):  Claim: $[0,1]$ contains uncountably many elements.  Proof: Assume that it contains countably many. Then $[0,1] = \bigcup_{x \in (0,1)} \{x\}$ and since $\{x\}$ are nowhere dense sets, $X$ is a countable union of nowhere dense sets. But $[0,1]$ is complete, so we have a contradiction. Hence $X$ has to be uncountable.     And here is another one (taken from ):  Claim: The linear space of all polynomials in one variable is not a Banach space in any norm.  Proof: "The subspace of polynomials of degree $\leq n$ is closed in any norm because it is finite-dimensional. Hence the space of all polynomials can be written as countable union of closed nowhere dense sets. If there were a complete norm this would contradict the Baire Category Theorem." Correct usage of "persons" (vs. "people") I had a little fight about persons vs. people.  Could you advise if both of the following are correct, if possible with reference to a dictionary?        A table for two people please.   A table for two persons please.      Besides that, how would you ask for a table in a restaurant?  I read  and , but they didn't really solve the question if both are correct.  I also found the last paragraph of , which suggests that (1) is "more correct", but doesn't say (2) is wrong. An example of a group such that $G \cong G \times G$ I was trying to find an example such that $G \cong G \times G$, but I am not getting anywhere. Obviously no finite group satisfies it. What is such group? Latest earned tag badge doesn't update when I earn a new one When you earn one tag badge, and then don't click "Track the next one" in the tag badge progress tracking box for a while, and then earn a second tag badge, that box continues to show "Congratulations, you earned [the same badge as before]", rather than updating to show that you've actually earned another tag badge since then.  For instance, right now  looks like this:    Even though I  a few weeks ago and then  yesterday, my profile hasn't updated to show the new achievement.  I also didn't get the "cipher" tag badge until after the 3am watershed, even though tag badges usually appear as soon as I visit my profile after earning them (for instance, I earned the "tags" tag badge on meta.SFF.SE today the very instant I clicked on my own profile there after having become eligible for the badge). Even when I went to look at the list of my progress towards different tag badges on Puzzling, which I thought would jog the system into realising I'd actually earned the badge, "cipher" still showed up as 99/100 score even after I'd gone well past 100, until last night when the badge was finally awarded and "cipher" disappeared from my list of badges to track.  TL;DR: if I don't click "Track the next one", my tag badges update slower than they should.  It's a very unimportant bug, but there must be more people than just me who like to see the "Congratulations" thing there for a while and rest on their laurels before moving on to track their progress towards the next milestone. From this paper:     Shouldn't $du$ be $dt$? And also how do you get from that line to the final result if $du$ is not $dt$? The search parameter isanswered ignores acceptances Why is it “a defeated Napoleon, not “the defeated Napoleon” who rode off the battlefield and into exile? I noticed an infinite article; ‘a’ was used before ‘defeated Napoleon’ in the following sentence of Jeffery Archer’s novel, "Fales Impression" :  “General Sir Harry Wentworth sat at the right hand of the Iron Duke that night, and was commanding his left flank when a defeated Napoleon rode off the battlefield and into exile."  I’m curious to know why it should be “a defeated Napoleon,” not “the defeated Napoleon” or simply “defeated Napoleon.” Isn’t it only one Napoleon who was defeated the battle and exiled (to St. Helena). It looks like as though there were many (defeated) Napoleons who were exiled.  It’s always headache for me to handle articles as we Japanese (I think Chinese too) don’t have article as the basic part of speeches in our language system. I’ve studied it on English grammar text books at school, but have never reached full understanding of the usage of articles, because we don’t have that practice. Is there any special knack to master how to use articles? Real Analysis: Continuity of a Composition Function Suppose $f$ and $g$ are functions such that $g$ is continuous at $a$, and $f$ is continuous at $g(a)$. Show the composition $f(g(x))$ is continuous at $a$.  My idea: Can I go straight from definition and take $\delta=\min\{\delta_1,\delta_2\}$, where $\delta_1$ is used for the continuity of $g$ at $a$ and $\delta_2$ is used for f being continuous at $g(a)$. In my proof I just treat $g(a)$ as a point when referring to the composition. So it goes like this:  Proof: Given $\epsilon&gt;0$, take $\delta=\min\{\delta_1,\delta_2\}$. Then $0&lt;|x-g(a)|&lt;\delta$ which implies $|f(g(x))-f(g(a))|&lt;\epsilon$. Needing help picturing the group idea. I have a question;   X and Y are subgroups of a group G.  If $|X|, |Y| &lt; \infty$, show that   $|XY| = \frac{|X||Y|}{|X \cap Y|}$  but I can't really picture what it is talking about to even get started.  Is it to do with looking for different ways so combine the elements {x, y, xy} ?  Thanks Formal basis for variable substitution in limits So we do a lot of variable substitution in limits at school. Stuff like $\lim\limits_{x\to5}\ (x+1)^2\ =\ \lim\limits_{y\to6}\ y^2$, where we define the substitution $y = x + 1$.  But I've never been clear on what exactly the theoretical basis for this is. What is the formula that you're actually applying when you do variable substitution? What are the formal conditions under which it is possible?  My conjecture would be the following:  For all continuous $f$, and all real $a$: $\lim\limits_{x\to a}\ (f\circ g)(x)\ =\lim\limits_{x\to g(a)}\ f(x)$, where $g$ is a continuous function  So to take my first example, $f$ would be $x^2$, $g$ would be $x + 1$, and $a$ would be $5$.  Am I in the right area? If this is correct, can it be proven using $\epsilon$-$\delta$? I had a half-hearted shot at it the other night and didn't get anywhere. Why does the browser renders a newline as space? Input field for decimal property My VF tags:  &lt;apex:column &gt;     &lt;apex:facet name="header"&gt;Sales Price&lt;/apex:facet&gt;     &lt;apex:inputText value="{!prod.SalesPrice}"/&gt;     &lt;/apex:column&gt;   Controller code:  public string SalesPrice{get; set;} oli.UnitPrice =decimal.valueOf(pi.SalesPrice);   It works when a user insert 15 and 15.00 but it doesn´t with 15,00(comma).  I m using apex:inputText linked to a controller apex property because apex:inputfield is not allowed.  I´ve tried with apex:input with type=number but it doesn´t look good, (small button to increase/decrease inserted value). Is there another Visualforce tag that i can use with a property? Is the only solution to replace comma with dot in the controller? $\varepsilon$-number countability without choice Let $\alpha\mapsto\varepsilon_\alpha$ be the enumeration of the $\varepsilon$-numbers--that is, those $\alpha$ such that $\omega^\alpha=\alpha$--by the ordinals.  If we know that countable unions of countable sets are countable, (slightly weaker than countable choice), then we can show fairly easily that $\varepsilon_\alpha$ is countable iff $\alpha$ is countable.  I think I saw a result at some point that $\varepsilon_0$ is countable without relying on any choice principle (though I can't recall where I saw that). Is this true? More generally, for which $\alpha$ can we conclude that $\varepsilon_\alpha$ is countable in ZF?  Update: I have been able to prove (in ZF) that if $\varepsilon_\alpha$ is countable, then $|\alpha|&lt;\aleph_1$, so that's one direction. I'm still a bit uncertain about the answers I've got so far. I have a sword that currently rotates around a pivot point thats off of the actual object. Im new to blender and im not sure how i would go about changing it so that the sword rotates around the handle. Any help would be appreciated! How to find out what's accessing a particular hard drive on Windows? Possible Duplicate:          My HDD on a dock just started making noises like it's being accessed but I don't see any program accessing it.  Is there a program that shows which HDD is accessed by which program in percentages similar to how the task manager shows which programs use the most CPU?