As you see below I have two user account on my computer, and both of them have in  administrator group.     I want to run this below program :    When I log in with Abraham and double click on ELEnveloperGui it can't find my dongle that is connected to the computer :    But when I right-click on it and select Run as administrator it find my dongle !!!    Note: I have Windows-7 x64 professional  Q : What is the difference between Abraham and Administrator?     Update : I gave the program Full Permission, but problem still remain!
If my account is already a member of the administrator group why do some functions on Windows 7 require me to run them as administator? I kind of thought that was implied?
Does a family of a distribution have a different definition for statistics than in other disciplines?  In general, a  is a set of curves, each of which is given by a function or parametrization in which one or more of the parameters is varied. Such families are used, for example, to .  For statistics, a family according to  is the result of varying the shape parameter. How then can we understand that the gamma distribution has a shape and scale parameter and only the generalized gamma distribution has, in addition, a location parameter? Does that make the family the result of varying the location parameter? According to @whuber the meaning of a family is implicitly A "parameterization" of a family is a continuous map from a subset of ℝ$^n$, with its usual topology, into the space of distributions, whose image is that family.  What, in simple language, is a family for statistical distributions?  A question about relations among of the statistical properties of distributions from the same family has already generated considerable controversy for a  so it seems worthwhile to explore the meaning.  That this is not necessarily a simple question is born out by its use in the phrase , which has nothing to do with a family of curves, but is related to changing the form of the PDF of a distribution by reparameterization not only of parameters, but also substitution of functions of independent random variables. What's a word for a positive kind of "envy" without the sense of resentment? Imagine your college buddy is now running a multi-billion dollar company, while you're still stuck at the same white-collar day job 5 years after graduation. You chat with him and want to express the idea that you're happy for him and fancy having his lifestyle--something like "I wish I could buy a yacht and cruise around the world like you do; I'm glad you're doing really well!"  Basically, it's the feeling of envy when you take out the resentment from it. Is there a word for it?   (I thought about "admire", but found it too strong and broad) Where should the period go when using parentheses? For example:     In sentence one, I use this example (which has a parenthesis at the end.)   Should the period be inside, or outside of the parentheses? What about if the entire sentence is a parenthetical (as below)?     (Where does the terminal punctuation go here?)   Is there a hard and fast rule? Is there a way to achieve the equivalent of a  in javascript regular expressions?  I need to match a string that does not start with a specific set of characters.  It seems I am unable to find a regex that does this without failing if the matched part is found at the beginning of the string.  Negative lookbehinds seem to be the only answer, but javascript doesn't have one.  EDIT: This is the regex that I would like to work, but it doesn't:  (?&lt;!([abcdefg]))m  So it would match the 'm' in 'jim' or 'm', but not 'jam' I have the following code:  Using cmd As SqlCommand = Connection.CreateCommand     cmd.CommandText = "UPDATE someTable SET Value = @Value"     cmd.CommandText &amp;= " WHERE Id = @Id"     cmd.Parameters.AddWithValue("@Id", 1234)     cmd.Parameters.AddWithValue("@Value", "myValue")     cmd.ExecuteNonQuery End Using   I wonder if there is any way to get the final SQL statment as a String, which should look like this:  UPDATE someTable SET Value = "myValue" WHERE Id = 1234   If anyone wonders why I would do this:   for logging (failed) statements for having the possibility to copy &amp; paste it to the Enterprise Manager for testing purposes Changing units in QGIS: degrees to meters I have a global map with CRS WGS84-EPSG:4326 and I want to calculate the area of my polygons. From what I read, to change the unit from degrees to meters I need to change my CRS. However, there is no global system for this. Is there any solution to change the unit from degrees to meters for a global map? Can the real and imaginary parts of $\dfrac{\sin z}z$ be simplified? I have calculated the real and imaginary parts of $\dfrac{\sin z}z.$ I've obtained  $$\begin{eqnarray} \frac{\sin z}z&amp;=&amp;\frac{\sin(x+iy)}{(x+iy)}\\ &amp;=&amp; \frac{\sin(x)\cosh(y)+i\cos(x)\sinh(y)}{x+iy}\\ &amp;=&amp; \frac{x\sin(x)\cosh(y)+y\cos(x)\sinh(y)}{x^2+y^2}+i\frac{x\cos(x)\sinh(y)-y\sin(x)\cosh(y)}{x^2+y^2}. \end{eqnarray}$$  Before calculating it, I hoped I could use this in a solution to a certain problem, but I would need to differentiate the real and imaginary parts of $\dfrac{\sin z}z$ with regard to $x$ and $y$. But what I'm getting is very complicated. I understand that it could be that that's just the way it is. But maybe there are some simplifications to be done either in the functions or in their derivatives? I've never really used hyperbolic functions before so I'm not sure I'm not missing something.  Added. The problem I'm working on is the following.     Prove that the solutions of the equation $$\tan z=z$$ are all real.   I will sketch the solution I know here.  It can be calculated that  $$\tan(x+iy)=\frac{\sin(2x)}{\cos(2x)+\cosh(2y)}+i\frac{\sinh(2y)}{\cos(2x)+\cosh(2y)}.$$  Plugging this to the equation, we get  $$\begin{eqnarray} x&amp;=&amp;\frac{\sin(2x)}{\cos(2x)+\cosh(2y)}\\ y&amp;=&amp;\frac{\sinh(2y)}{\cos(2x)+\cosh(2y)}. \end{eqnarray}$$  From this we can get   $$\sin(2x)\cdot2y=\sinh(2y)\cdot2x.$$  By plotting the functions $\sin,$ $\mathrm{id}$ and $\sinh$ (and proving the right inequalities we can see from the plots), we can see that this is only possible when $xy=0$. But when we plug $x=0$ to the second equation, we can show that the only solution is $y=0$, which ends the proof.  What strikes me in this solution is the fact that $\dfrac{\sin z}z$ is used. (Not explicitly in my statement.) This function has another connection to the problem:     The non-zero solutions of the equation $\tan z=z$ are exactly the zeros of the derivative of $\dfrac{\sin z}z.$   This is easy to prove. I wondered if there shouldn't be a solution to the problem which uses this fact, given that $\dfrac{\sin z}z$ appears in the solution I know. What is the format for e-mail headers that display a name rather than the e-mail? If $f \in \mathcal{L}^1(\mathbb{R})$ then $\lim_{h\rightarrow 0}\int_{-\infty}^\infty |f(x+h)-f(x)|dx\rightarrow 0$ I am working on some past prelim problems and one of them is the following: If $f \in \mathcal{L}^1(\mathbb{R})$ then $\lim_{h\rightarrow 0}\int_{-\infty}^\infty |f(x+h)-f(x)|dx\rightarrow 0$   Update: Following Jackie Chong's advice, I am attempting this problem with the dominated convergence theorem. Since $f\in L^1(\mathbb{R})$, $f$ can be approximated from below by a sequence of continuous functions $f_n$. Then, I believe I can say that, for all $n$ sufficiently large, $|f_n(x+\frac{1}{n})-f_n(x)|\le 2|f(x)|$. (I'm not sure of this part thought). Then $2|f(x)|$ is integrable and hence a dominating function for $|f_n(x+\frac{1}{n})-f_n(x)|$. So by the dominated convergence theorem we can take the limit inside $\lim_{n\rightarrow \infty}\int|f_n(x+\frac{1}{n})-f(x)|dx=\int \lim_{n\rightarrow \infty}|f_n(x+\frac{1}{n})-f(x)|dx=0$ Is this on the right track?  Another approach that I thought may work, but which I'm less certain of is the following: If $f$ can be approximated by $C^1$ functions (I have no real justification for this, so please correct me if its wrong), then for each $n$ and each $x$ there is some $y(x,n)\in B_{1/n}(x)$ such that $|f(x+\frac{1}{n}-f(x)|=\frac{1}{n}|f'(y(x,n)|$ but from here I don't see how to proceed, because just because $|f|$ is integrable, I can't justify $|f'|$ being integrable, but if there is some way to fix this approach, hints would be appreciated.   Another thing I was thinking was to first approximate $f$ from below by continuous functions $f_n$ then to define $g_{m,n}=|f_n(x+\frac{1}{m})-f_n(x)| \chi_{[-m,m]}$. Then, for each $n$, the function $g_{m,n}$ is continuous on its support, and since it support is compact, it is uniformly continuous there. Can this be used to justify bringing the limit $\lim_{n\rightarrow \infty}$ inside the integral? Then apply the monotone convergence theorem? I know this idea is probably even more sketchy than my previous one, but if it can be fixed please let me know :)  Thanks! Question:  Is it possible to get multiple correct results when evaluating an indefinite integral?  If I use two different techniques to evaluate an integral, and I get two different answers, have I necessarily done something wrong?    Often, an indefinite integral can be evaluated using different techniques.  For example, an integrand might be simplified via partial fractions or other algebraic techniques before integration, or it might be amenable to a clever substitution.  These techniques give different results.  For example, looking over a few other questions on MSE:      From : evaluate $$ \int x(x^2+2)^4\,\mathrm{d}x. $$    Via the substitution $u = x^2+2$, this becomes $$ \int x(x^2+2)^4\,\mathrm{d}x       = \frac{1}{10}x^{10} + x^8 + 4x^6 + 8x^4 + 8x^2 + \frac{32}{5} + C.     $$ However, multiplying out the polynomial and integrating using the power rule gives $$ \int x(x^2+2)^4\,\mathrm{d}x       = \frac{1}{10}x^{10} + x^8 + 4x^6 + 8x^4 + 8x^2 + C    $$     From :  evaluate   $$ \int \frac{1-x}{(x+1)^2} \,\mathrm{d}x. $$    Simplifying the integrand using partial fractions then integrating gives $$ \int \frac{1-x}{(x+1)^2} \,\mathrm{d}x       = \frac{-2}{(x+1)} - \ln|x+1| + C.    $$ Via integration by parts, we get $$ \int \frac{1-x}{(x+1)^2} \,\mathrm{d}x       = \frac{x-1}{(x+1)} - \ln|x+1| + C.    $$     From : evaluate   $$ \int \frac {\tan(\pi x)\sec^2(\pi x)}2\,\mathrm{d}x. $$    Using the substitution $u = \sec(\pi x)$, this becomes $$ \int \frac {\tan(\pi x)\sec^2(\pi x)}2\,\mathrm{d}x       = \frac {\sec^2(\pi x)}{4\pi} + C.    $$ Using the substitution $u = \tan(\pi x)$, this becomes $$ \int \frac {\tan(\pi x)\sec^2(\pi x)}2\,\mathrm{d}x       = \frac {\tan^2(\pi x)}{4\pi} + C.    $$ Lightsabers are not made of light, nor have they anything to do with lasers (even if there are people who mistakenly refer to them as "laser swords").   Blaster bolts are also not lasers. They travel significantly slower than light, they travel even slower than ordinary bullets fired by gunpowder. (called slug throwers on the Star Wars universe)  Was there any use of lasers ever presented in the Star Wars universe, either as we use them today, or as weapons?  By laser I mean something which behaves like laser does, even if it's not called by that name in a movie or book. If something is called a laser (in the script or by fans) but clearly doesn't work like a laser (for example, it travels slowly, and in visible blocks) it doesn't count. Configuring dpkg stops processing when calling an HTTP request to Adobe Flash Player How do comment @replies work? How do the comment replies (aka &quot;pinging users&quot;) work?  Who can I send an @reply to? How do I respond to a specific user when entering my comment? Will they be notified? What do people mean when they talk about &quot;pinging&quot; another user?  Related: Can you make a custom mouse cursor for the game engine? I would like the mouse to have a custom look like most games and not just the default white.  Is there a way I can make an object as the mouse? Dynamic date range query I'm trying to create two cells that will update my query: A >= date"A1" and A &lt;= date "B1".  My goal is to create a search page for the data, where the end user could insert a date or the =today() function and it would update the query. I am currently getting a formula parse error from:  =QUERY(Data!6:1397,"SELECT A,B,C,D,E,F,G,H,I,J,K,L WHERE A &gt;= date"A1"and  A &lt;= date"B1"") How could I retrieve the exact duplicate value from conditionally formatted list that highlight duplicates? I have imported a webpage table in to Google Sheets. It frequently being updated. A series of calculations has been conducted on the sheet named Shawn Of the link spreadsheet.  I have filtered the my desired results into the Calculation sheet of that linked document... which is in columns A5:A. I then applied =ARRAYFORMULA(if(A5:A200=&quot;&quot;,,A5:A200*{12,13,37,101,111})) in columns B5. The Results of the formula extends from B5:F. This results from a5:F was subsequently conditionally formatted to highlighted duplicates in the aforementioned, i.e. a5:f using the lists of values in K5:K and I5:5. In Columns A5:F, is it possible to generate/place values in the range A5:f with matching name of states and aligned values in J5:K and H5:I? Suppose $f$ is entire and    $$\iint_\mathbb{C}|f(z)|^2dxdy &lt; \infty$$Prove that $f\equiv 0.$   So far I have:  Suppose $f$ is bounded.  Then $f$ is constant by virtue of Liouville and so the conclusion is obvious.  Thus, assume $f$ non-bounded.  Then  $$\Big|\iint_\mathbb{C}f^2(z)dA\Big| \leq \iint_\mathbb{C}|f(z)|^2dxdy &lt; \infty$$ We can parameterize \begin{align*} \iint_\mathbb{C}f^2(z)dA &amp;= \int_0^{\infty}\int_0^{2\pi}f^2(re^{i\theta})\;d\theta\;rdr \\ &amp;= \int_0^{\infty}\int_0^{2\pi}f^2(re^{i\theta})\;ire^{i\theta}[-i\frac{1}{r}e^{-i\theta}]d\theta\;rdr \\ &amp;= \int_0^{\infty}\oint_{C_R}f^2(z)\;[-i\frac{1}{z}]dz\;rdr \\ &amp;= \int_0^{\infty}2\pi r dr\frac{1}{2\pi i}\oint_{C_R}\frac{f^2(z)}{z}dz \\ \\ &amp;= 2\pi f^2(0)\int_0^\infty rdr  \end{align*}  Whence f(0) = 0.  And I have no idea where to go from there.  The "entire" bit seems to suggest Liouville but I have already dealt with the bounded case.  Please advise ... I'm looking for specific, real cases in which a causal relationship was inappropriately inferred from evidence of a correlation.   Specifically, I'm interested in examples that meet the following criteria:   Existence of the causal relationship was accepted as fact widely enough to have notable effects (on public policy, discourse, individual decisions, etc.). The link was inferred solely on the basis of correlative evidence (perhaps along with the existence of a coherent but unproven causal mechanism).  Causality has been objectively falsified or at least called into serious doubt.   The two examples that came to mind for me aren't quite ideal:   Sodium intake and blood pressure: As I understand it, it has since been determined that salt intake only increases blood pressure in sodium-sensitive individuals. The existence of a valid causal relationship (although not quite the one that was originally accepted) make this example less compelling. Vaccines and autism: I may have the background wrong, but I believe this link was surmised on the basis of both correlations and (fraudulent) experimental evidence. This example is weakened by the fact that (fake) direct evidence existed.   Note: I've seen this similar question:    My question differs primarily in that it focuses on notable, real-world examples and not on examples in which a causal link is clearly absent (e.g., weight and musical skill). Usually I don't mind about LaTeX's compilation speed, because a paper of less than 20 pages is compiled rather fast.  However, I am working on a large document - A report with more than 80 pages.  Compilation times are starting to slow me down.  I saw the couple of answers here on this topic, but most of my graphics are external and I don't use Tikz.  As a matter of fact, I'm not sure what's slowing the compilation down, other than the amount of text and required layout.   What's your best speeding-tex-up trick, other than externalizing graphics and pre-compiling the preamble? How can I know what's slowing TeX down? How to evaluate the integral $\int_0^{\pi}\frac{a^n\sin^2x+b^n\cos^2x}{a^{2n}\sin^2x+b^{2n}\cos^2x}dx$? Evaluate the integral $$\int_0^{\pi}\frac{a^n\sin^2x+b^n\cos^2x}{a^{2n}\sin^2x+b^{2n}\cos^2x}dx.$$   I have no idea.   $$\int_0^{\pi}\dfrac{a^n\sin^2x+b^n\cos^2x}{a^{2n}\sin^2x+b^{2n}\cos^2x}dx=\int_0^{\pi/2}\dfrac{a^n\sin^2x+b^n\cos^2x}{a^{2n}\sin^2x+b^{2n}\cos^2x}dx=\int_0^{\pi/2}\dfrac{a^n\tan^2x+b^n}{a^{2n}\tan^2x+b^{2n}}dx$$  I try the substitution $\tan{x}=t$, but it doesn’t work.