I'm using ubuntu 13.10 but i want to install windows so, i downloaded a windows iso  but how to Make bootable disk/usb in ubuntu 13.10 of windows iso
I'd like to create a Windows bootable USB stick, but I don't have a Windows machine with me at the moment to do so. How could I achieve the same using Ubuntu?
Why does the classical path give the dominant contribution in the path integral? Integral solutions of hyperboloid $x^2+y^2-z^2=1$ Are there integral solutions to the equation $x^2+y^2-z^2=1$? Create Trigger that updates the same table before (or after ??) the Insert I have just started using triggers and one thing that I would need to do is to create a trigger that updates the same table that's inserting the row.  TABLE:  CREATE TABLE `city` (   `city_id` smallint(5) unsigned NOT NULL AUTO_INCREMENT,   `city` varchar(50) NOT NULL,   `country_id` smallint(5) unsigned DEFAULT '0',   `country` varchar(150) DEFAULT NULL,   `last_update` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP,   PRIMARY KEY (`city_id`),   KEY `idx_fk_country_id` (`country_id`),   CONSTRAINT `city_ibfk_1` FOREIGN KEY (`country_id`) REFERENCES `country` (`country_id`) ON UPDATE CASCADE ) ENGINE=InnoDB AUTO_INCREMENT=6009 DEFAULT CHARSET=utf8;   Trigger:  DELIMITER $$ CREATE TRIGGER `populate_country_id` BEFORE INSERT ON `city` FOR EACH ROW BEGIN  IF NEW.country_id = 0  THEN     UPDATE city     INNER JOIN country B ON city.country = B.country     SET NEW.country_id = B.country_id;  END IF;  END $$ DELIMITER ;   MySQL is just returning: Can't update table 'city' in stored function/trigger because it is already used by statement which invoked this stored function/trigger.  Is there any way I can make this work?  Thanks everyone! How to get ongoing notifications of unread texts? One of my family members texts me at night when they are having nightmares. I want to know when they text me at 4 am instead of sleeping through it.  Is there a way to make text notifications keep going off until read? It seems that $\{e\}, \{e,s\}, \{e,rs\}, \{e,r^2s\},...,\{e,r^{n-1}s\}, \{e,r,r^2,...,r^{n-1}\}, D_n$ are always subgroups of $D_n$. Especially when $n$ is odd, these seem to be the only subgroups. But when n is even, say $n=4$, then there are also $\{e,s,r^2,r^2s\}$ and $\{e,rs,r^2,r^3s\}$. It makes me wonder, is there a general formula/algorithm for finding all subgroups of $D_n$ when $n$ is even? I switch back and forth between two computers constantly:   OSX Snow Leopard  Ubuntu 10.10    I'd like to be able to make Ubuntu use the same keyboard shortcuts as OSX System Wide, for the following keyboard sequences only (ie I don't want to make Ctrl act like Super)   Super+C -> Copy Super+V -> Paste Super+T -> Open a new tab in whatever browser I'm in.  Super+W -> Close a tab in whatever browser I'm in.    So in short, is there anyway for me to map just these keyboard sequences to the following, system wide ?   Super+C -> Ctrl+C Super+V -> Ctrl+V Super+T -> Ctrl+T  Super+W -> Ctrl+W   I know there are ways for me to do this for vim, and Firefox, and I'm sure specifically for most applications... but I would prefer to have to do this just once and have it work that way system wide!  I'm using a standard PC keyboard, that is "Generic 105 key (intl) PC" on Ubuntu. I'm also using the same keyboard on my Mac mini. Integrating $\int_0^1 \frac {\log(1-x)\log^2(1+x)}x \mathrm{d}x$ Question: Integrate$$\int\limits_0^1dx\,\frac {\log(1-x)\log^2(1+x)}x=-\frac {\pi^4}{240}$$   I'm curious as to if there is a way to integrate this. I've tried using integration by parts to get$$I=-\frac {\pi^2}6\log^22+2\int\limits_0^1dx\,\frac {\operatorname{Li}_2(x)\log(1+x)}{1+x}$$However, I'm not sure how to continue even further. The polylog in the second integrand seems a bit intimidating and I don't see how the first term even helps. Create a table with LaTeX I need to copy a table from a pdf file to my LaTeX document. I know I have to do it by hand but still don't understand how to create a table with LaTeX. Find an example where Markov's inequality is tight in the following sense: For each positive integer $a$, find a non-negative random variable $X$ such that $P(X\ge a)=E(X)/a$.   How to do this problem, I am really confused. Also, what is the definition of Markov's inequality? What is the Relationship between the Rank and the Number of Eigenvalues? for (a) it says it can't be rank 1 because there is only one non zero eigenvalue? I understand it can't be 0 or 3 but I don't understand how you can conclude we have 2 just by knowing how many eigenvalues we have. Is $\{0\}$ a field? Consider the set $F$ consisting of the single element $I$.  Define addition and multiplication such that $I+I=I$ and $I \times I=I$ .  This ring satisfies the field axioms:   Closure under addition.  If $x, y \in F$, then $x = y = I$, so $x + y = I + I = I \in F$. Closure under multiplication.  $x \times y = I \times I = I \in F$ Existence of additive identity.  $\forall x \in F$ (i.e., for $x=I$), $x + I = x$, so $I$ is the additive identity. Existence of mulitiplicative identity. $\forall x \in F, x \times I = x$, so $I$ is the multiplicative identity. Additive inverse. $\forall x \in F, \exists y = I \in F: x + y = I$ Multiplicative inverse. $\forall x \in F, \exists y = I \in F: x \times y = I$.  However, because the additive identity need not have a multiplicative inverse, this is a vacuous truth. Commutativity of addition.  $\forall x, y \in F, x + y = I = y + x$ Commutativity of multiplication.  $\forall x, y \in F, x \times y = I = y \times x$ Associativity of addition.  $\forall x, y, z \in F$, $(x + y) + z = I + I = I$ and $x + (y + z) = I + I = I$, so $(x + y) + z = x + (y + z)$ Associativity of multiplication.  $\forall x, y, z \in F$, $(x \times y) \times z = I \times I = I$ and $x \times (y \times z) = I \times I = I$, so $(x \times y) \times z = x \times (y \times z)$ Distributivity of multiplication over addition. $I \times (I + I) = I$ and $I \times I + I \times I = I$, so $\forall x,y,z \in F, x(y+z) = xy+xz$   Based on the above, $\{I\}$ seems to qualify as a field.  If $I$ is assumed to be a real number, then the unique solution of $I + I = I$ and $I \times I = I$ is, of course, $I = 0$.  So, is {0} a field, or is there generally considered to be an additional field axiom which would exclude it?  Specifically, is it required for the multiplicative identity to be distinct from the additive identity? Translation of words according to babel language I've got a template that is to be used in french or english, is there a name in babel defining the "of" of "page x of yy" than I want to translate automaticaly to "page x sur y"  If not, is there a way of setting a name according to language selected in babel. How to run "sudo apt-get update" through proxy in commandline? In my office i want to update and download through (wget) and here they are using proxy?  How can i do this by commandline? Accessing raster attribute table in QGIS? I have some Python code that makes Raster Attribute Tables for GeoTiffs using GDAL.    In ArcGIS I can see these attribute tables fine and they behave how I like (as classifications) but in QGIS the whole concept of an attribute table seems to be missing.    Is there any way to access a RAT in QGIS? Prevent VLC from starting a new instance every time a media file is opened Another point of view that $\mathbb{Z}/m\mathbb{Z} \times\mathbb{Z}/n\mathbb{Z}$ is cyclic. I was thinking that the product of groups $\mathbb{Z}/2\mathbb{Z} \times\mathbb{Z}/2\mathbb{Z}$ is not cyclic, but $\mathbb{Z}/2\mathbb{Z} \times\mathbb{Z}/p\mathbb{Z}$ is cyclic if p is an odd prime. Then I found logic that $\mathbb{Z}/m\mathbb{Z} \times\mathbb{Z}/n\mathbb{Z}$ is certainly cyclic if $m$ and $n$ are coprimes.  I searched on google a proof of that, to compare with mine and the answer is that $\mathbb{Z}/m\mathbb{Z} \times\mathbb{Z}/n\mathbb{Z}$ is isomorphic to $\mathbb{Z}/mn\mathbb{Z}$ (which is clearly cyclic) $\iff$ $m$ and $n$ are coprimes.  But I got another proof and I want to know what you think about it :     I claim that $([1]_m,[1]_n)$ generates $\mathbb{Z}/m\mathbb{Z}$   $\times\mathbb{Z}/n\mathbb{Z}$, where $[a]_d$ is the canonical   projection of $a\in \mathbb{Z}$ into $\mathbb{Z}/d\mathbb{Z}$.       Let's take multiples of $([1]_m,[1]_n)$ until we get $m \cdot$   $([1]_m,[1]_n) = ([0]_m,[m]_n)$ (where I suppose $m=\min(m,n)$...   otherwise, consider $\mathbb{Z}/n\mathbb{Z}$   $\times\mathbb{Z}/m\mathbb{Z}$ and $\min(m,n)=n$).       The next step is $([1]_m,[m+1]_n)$. This pair is not equal to   $([1]_m,[1]_n)$, since $m+1 \not\equiv1$ mod $n$ ($m$ and $n$ are   supposed to be coprimes). We can now continue adding $([1]_m,[1]_n)$   to the results.      The last step is precisely reached when you got   $([mn]_m,[mn]_n)=([0]_m,[0]_n)$. At this point, the fact that $m$ and $n$ are coprimes is absolutely necessary because if they are not coprimes, $lcm(m,n)\ne m\cdot n$ and   $([lcm(m,n)]_m,[lcm(m,n)]_n)=([0]_m,[0]_n)$ but all the $m \cdot n$ possible pairs are not reached, only the $lcm(m,n)$ are ! Maximum number of edges in a simple graph? I found that the maximum number of edges in a simple graph is equal to $$\sum\limits_{i=1}^{n-1} i$$  Where $n =$ number of vertices.  For example in a simple graph with $6$ vertices, there can be at most $15$ edges.  If there were any more edges then $2$ edges would connect the same pair of vertices and thus would not be a simple graph.  Is this a known theorem? How to let the user switch language in playframework 2 Where did the "Magic Blood" of the Harry Potter Universe Originate From? What is a wizard that makes them non-Muggle? Or, more directly, what ancestry does the witch/wizard gene follow? Is there a patriarchal wizard (or matriarchal witch) that Muggles have no part in? Elves perhaps? Stop reading here.  Purpose of the Question/History/Don't Read There seem to be solid consequences associated with being Muggle, or Muggle-related. A dominant theme in the novels is discrimination against the non-gifted race, but, when read into, the wizard bloodline doesn't appear all that well defined. JKR indicates that wizard-ness is genetic, or inherited:  &quot;Muggle-borns will have a witch or wizard somewhere on their family tree, in some cases many, many generations back. The gene resurfaces in some unexpected places.&quot;  By the above, it stands that a family that produces a squib (non-magical born of magical parents) must be a half-blooded family in some way, since half-blooded only requires one non-magical ancestor. Of course, we are dealing in magic genetics. Any other form of genetics would make this inherited, seemingly recessive trait (or squibs couldn't exist), never appear at all. Unless they were XMen. If it is so important, what does the wizard-bloodline follow? Other Mythos In other literature a wizard can be nearly anything, but magical powers are typically inherited as well as learned. Merlin of the Arthurian Legend (also mentioned in HP) is Cambian, meaning he is born of demonic intervention; this explains his source of enchantment. Gandalf of Middle Earth is a Maiar, as are all Wizards of Middle Earth. Wizards in Harry Potter seem to follow a bloodline, and it has been mentioned (in the question Gabe linked) that genetics have a heavy play in it, meaning that squibs and parents of Muggle-born wizards should have identical genetics. This bloodline must come from somewhere. Magical Creatures In Harry Potter it is common for magical creatures to &quot;mingle&quot; with humans, these beings are typically somewhat magical. Elves have been described to have very strong magic, but are for some unreasonable reason subservient to humans. I may be mistaken, but believe it was hinted that elves and the wizarding lineage had intersected at one time. Maybe I'm imagining this. Genocidal Allusions Voldemort hates muggles, and half-bloods, and whatever else is non-wizard. The presence of this theme, suggests that this wizardly bloodline must be at least defined if the main antagonist is so obsessed with keeping that mysterious bloodline pure. Since Slytherin himself was said to be a supremacist (circa a really long time ago) it suggests that this bloodline should have existed long before he was around, especially if he founded a school with people he was not related to, but were also magical. Primary ideals in Noetherian rings For an $R$-module $M$ I have the following definition for a submodule $N\subset M$ to be $\mathfrak{p}$-primary: this is the case when $\text{Ass}(M/N) = \{\mathfrak{p}\}$, that is, $M/N$ is coprimary or $(0)$ is a primary submodule.  Now for an ideal $\mathfrak{q} \subset R$ this definition gives that $\mathfrak{q}$ is $\mathfrak{p}$-primary iff $\text{Ass}(R/\mathfrak{q}) = \{\mathfrak{p}\}$. It can be shown that this is equivalent with $\mathfrak{p}^n \subset \mathfrak{q}$ for some $n\in \mathbb{N}$ and the condition \begin{align*} \forall r,s \in R: rs \in \mathfrak{q}, r\not\in \mathfrak{p}\Rightarrow s \in \mathfrak{q}. \end{align*}     Now I suspect that if $R$ is Noetherian, and if $\mathfrak{q}$ is $\mathfrak{p}$-primary, then we have $\sqrt{\mathfrak{q}} = \mathfrak{p}$. I have shown this for PID's, but does this generaly hold?    And if so, is every ideal $\mathfrak{q}\subset R$ then $\sqrt{\mathfrak{q}}$-primary?   Many thanks.