if i browse localhost/visiondms it shows, Fatal error: require_once(): Failed opening required 'C:\xampp\htdocs\visiondms/wp-admin.php' (include_path='C:\xampp\php\PEAR') in C:\xampp\htdocs\visiondms\index.php on line 10 Please tell my what can i do?
I'm trying to install , an open-source Microblogging application on my Windows development server using XAMPP as per the . The website cannot find PEAR, and throws the below errors:  Warning: require_once(PEAR.php) [function.require-once]: failed to open stream: No such file or directory in C:\xampplite\htdocs\laconica\lib\common.php on line 31 Fatal error: require_once() [function.require]: Failed opening required 'PEAR.php' (include_path='.;\xampplite\php\pear\PEAR') in C:\xampplite\htdocs\laconica\lib\common.php on line 31   PEAR is located in C:\xampplite\php\pear phpinfo() shows me that the include path is .;\xampplite\php\pear  What am I doing wrong? Why isn't the PEAR folder being included?
Is it better to have an ErrorDocument 404 redirect back to the homepage or a standard 404 error page? I want to block my server talking to an external IP address continously.  The IP address in question is I think is being hijacked or spoofed. When Used who.is it resolves to Netherlands. But if you use any other tools it resolves to Jordan.  Any step by step configuration is highly appreciated.  Thank you Kind Regards N How to blacklist kernel modules? How do I disable loading of unnecessary kernel modules. Kernel 3.2.4 Disproving the statement "If $f(z)$ is not an entire function, then $g(z) = f^2(z)$ is not entire.” If $f(z)$ is not an entire function, then $g(z) = (f(z))^2$ cannot be an entire function.”  From this statement why would saying &quot;Let $f(z) = \sqrt z$, assuming that it is a branch that takes $−1$ to $i$. Then it is not analytic on the branch cut, but $g(z) = (f(z))^2 = z$ is obviously an entire function.&quot; Not be a proper example to disprove this statement/ be a incorrect counterexample? I understand that for $\sqrt z$ to be a incorrect counterexample means it is not analytic on the branch -1 to I, therefore not entire. However I thought $\sqrt z$ would be analytic throughout the whole branch cut from -1 to i. Since on the complex plane, $\lim_{z \to -1} \sqrt z$ exists as it would approach $i$ in that case, and in the other case $\lim_{z \to i} \sqrt z$ it approaches $\sqrt i$ meaning that at least on that branch cut in the complex plane it is continuous everywhere, therefore the partial derivatives exist everywhere on the branch cut meaning it is analytic, but this assumption seems to be incorrect. What am I missing or not understanding here. Precisely, I don't understand why $\sqrt z$ would not be a proper counterexample to the statement. How to enable top bar background opacity? There is a way to change top bar background opacity in ubuntu 20.04 lts.  I installed the hide top bar extension but is not a option to change it. Updating list definition using features I have a feature that deploys content types, a list definition and template that uses these content types and finally a list instance. This feature is already in production.  Now we have a change request to add new views to this existing list. So I make the changes to schema.xml and upgrade the solution. But alas, the new views don't show up. If I create a new site, the list instance is created with the new views. I tried deactivating/reactivating, iisreset but still no success. Is there anything I'm missing? How do I change the extension of multiple files? In XFCE4 desktop environments, alt+scroll wheel zooms my entire desktop. I'm finding this very annoying because sometimes I flick the scroll wheel to go to the top of a document then press the alt key for a some unrelated keyboard shortcut and find myself zoomed in so far I can't see anything at all. Is there any way to disable this feature? What damage will powering down instead of shutting down do? Sorry if this is very basic but here's a question.   Let $\mathbf{v}=(v_1,\ldots, v_n)\in k^n$ where $k=\bar{k}$.      Why do we have   $$ \sqrt{\sum_{i=1}^n |v_i|^2} \leq \sum_{i=1}^n |v_i|,  $$   where the left-hand side can be thought of as the $2$-norm $\|\mathbf{v}\|_2$ on $L^2(k^n)$?   $\mathbf{General \; case}$: If this is true for $p=2$-norm, I am guessing that this is true for all $p\geq 1$:  $$ \left( \sum_{i=1}^n |v_i|^p\right)^{1/p}\leq \sum_{i=1}^n |v_i|.  $$     In fact, does this inequality hold when $p$ is a rational number? Oversized \underbrace's label causes unwanted spacing Consider this:  \left( x-1 \right) \underbrace{ \left( ? \right) }_{ \text{what is this?} }   The label is longer than the braced portion, which causes unwanted space between the parentheses.  Is there a way around that, so that the parentheses "touch" as they would normally?  Please note that I just started on LaTeX, and I'm using TexMaths inside LibreOffice Writer. LaTeX on Stack Overflow? in questions, answers, and comments. Can we get something like this going on Stack Overflow?  I think it’d be appropriate as I at least pretty often want to write something like n^2 and would benefit greatly from prettier markup. Intuition behind the definition of a measurable set This week I saw the definition of a measurable set for an outer measure.     Let $\mu^*$ be an outer measure on a set $X$. We call $A \subseteq X$ measurable if $$\mu^*(E) = \mu^*(A\cap E) + \mu^*(A^c\cap E)$$ for every $E \subseteq X$.   This is not the first time I've seen this definition. Unlike most other things in mathematics, over time I have gained absolutely no intuition as to why this is the definition.   The only explanation I've ever seen is that a set is measurable if it 'breaks up' other sets in the way you'd want. I don't really see why this is the motivation though. One reason I am not comfortable with it is that you require a measurable set to break up sets which, according to this definition, are non-measurable; why would you require that? Of course, you can't say what a non-measurable set is without first defining what it means to be measurable so I suppose no matter what your condition is, it will have to apply to all subsets of $X$.     Is there an intuitive way to think about the definition of measurable sets? Is there a good reason why we should use this definition, aside from "it works"? Why does Windows only show about 3.5 GB of my 4 GB of RAM? How do I run a file on localhost? Why is the electric field zero inside a hollow conducting sphere? Folland, "Real Analysis", Chapter 5.3, Exercise 36. Folland, "Real Analysis", Chapter 5.3, Exercise 36:     Let $\mathcal{X}$ be a separable Banach space and let $\mu$ be   counting measure on $\mathbf{N}$.  Suppose that   $\left\{x_n\right\}_1^\infty$ is a countable dense subset of the unit   ball of $\mathcal{X}$, and define $T \colon L^1(\mu) \to \mathcal{X}$   by $Tf = \sum_1^\infty f(n) x_n$.  (a)  $T$ is bounded.  (b)  $T$ is   surjective.   I have proved (a).  I would like help on (b).  Here are my ideas so far.    Say we want to show $y \in \mathcal{X}$ is in the image of $T$.  Reduce to the case $||y|| = 1$ and let $x_{n_m} \to y$.  My thought is to produce for each $N$ a function $f_N \in L^1(\mu)$ such that $||y - \sum_n^\infty f(n)x_n|| &lt; 1/N$ and also such that the $f_N$ converge in $L^1(\mu)$ to some $f \in L^1(\mu)$.  Then use the continuity of $T$ to conclude $T(f) = y$.    The way I have been setting up my inequalities is as follows. Pick $M_N$ such that for all $m &gt; M_N$, $||y - x_{n_m}|| &lt; 1/N$.  Then I want to define $f_N$ so that $$\sum_n^\infty f_N(n) = \sum_{m &gt; M_N}^\infty f_N(m) = 1, $$ hence $$||y - T(f_N)|| = ||\sum_{m &gt; M_N}^\infty f_N(m) y + \sum_n^\infty f_N(n)x_n|| \leq \sum_{m &gt; M_N}^\infty f_N(m) || y - x_{n_m} || &lt; \frac{1}{N}. $$  Presumably, if the $f_N$ are appropriately chosen, then I can find a dominating function and apply the Dominated Convergence Theorem to finish up.    However, getting all these hypotheses to hold simultaneously has been difficult.  In essence, the difficulty seems to be in requiring that the $f_N$ be "normalized" as well as possess a dominating function.  If this method is feasible, at least it seems to require a bit of analysis to show the desired series all converge.  Perhaps I am missing something cleaner.  If so, a simpler solution would be much appreciated.    -Thanks. How to calculate polygon areas and perimeters using QGIS? What's a good phrase for "refining a process which is hopelessly broken"? I'm looking for a turn of phrase to describe a situation where the powers that be wish to continue making small improvements to a process which, due to deep-rooted flaws, will never be close to accurate.  I'm working on a process for estimating the states of business objects, and there's interest in fixing some edge cases which have been noticed.  This could be worthwhile, but because of missing data we're already making heroic guesses.  So if implemented, we might get from 80.0 to 80.1% accuracy.  If we were at 99.5%, I'd be happy to get it to 99.6%, but in the current situation it doesn't seem productive.  We have some CS people on E.SE; what would you call it when someone wants to optimize a two-second process to load data, though once loaded it will be chewed on by an inefficient algorithm for hours?  Something like "bailing the ocean" or "fixing a leaky faucet in a burning building" is close to what I have in mind.  There's also "rearranging deck chairs on the Titanic," but that carries an implication that the changes are wholly useless, and that events are coming to a head. Create default home directory for existing user in terminal How do I do this? I created a user without a home directory and now I want to create one. Not just a folder called new_user, I want the default home directory with all it's folders and hidden files, etc.