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There is a formal definition for the Big O notation in . Up to now I have come across Big O in Numerical Analysis, Calculus and Algorithms which all are pretty distinct fields. What I am wondering is if that definition is global and is the only one that is used in every field where the Big O is involved or there are other ways that the Big O is defined in other areas  of Mathematics. I have to teach the Bayes formula to some undergraduates, in the form:  $$ P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{P(B|A)P(A)}{P(B)} $$  I was wondering if anyone had any really thoughtful or alternative ways of teaching it instead of the standard approach. I need to use a different version of circuitikz. How can I do that? I've made some high resolution pixel art in PS by drawing it down at the pixel level and then increasing the image size with the nearest neighbor setting to keep it crisp. In the web browser, however, I need to display these images as icons with a dynamic size, so I can't just size the images correctly to begin with, and the broswer seems to blur the images when rendering them smaller, while vector images seem to stay more crisp as they shrink.   So I'm wondering if there's a way to convert the pixel art neatly into vector. I'm aware of the trace function in illustrator, but it seems to do a rough job iirc. Defining coordinate reference system with rotation in GeoServer? I am using GeoServer and have a layer in EPSG:900913 ("Google Mercator").  I need to "rotate" the map around certain point (say, 1500000, 7000000) by certain degree (say, 30 degrees clockwise). How could I define such a coordinate system based on EPSG:900913?  GeoServer's  does not work for my purposes as I need to tile the map later on.  As far as I understand this, my only option is to define an own coordinate system. For GeoServer I'd need to . The configuration seems to be straightforward, but I have a difficulty defining my rotated CRS in .  I am wondering how to apply a rotation around certain point onto a CRS like Google Mercator:  PROJCS["WGS84 / Google Mercator",    GEOGCS["WGS 84",      DATUM["World Geodetic System 1984",        SPHEROID["WGS 84", 6378137.0, 298.257223563, AUTHORITY["EPSG","7030"]],        AUTHORITY["EPSG","6326"]],      PRIMEM["Greenwich", 0.0, AUTHORITY["EPSG","8901"]],      UNIT["degree", 0.017453292519943295],      AXIS["Longitude", EAST],      AXIS["Latitude", NORTH],      AUTHORITY["EPSG","4326"]],    PROJECTION["Mercator_1SP"],    PARAMETER["semi_minor", 6378137.0],    PARAMETER["latitude_of_origin", 0.0],    PARAMETER["central_meridian", 0.0],    PARAMETER["scale_factor", 1.0],    PARAMETER["false_easting", 0.0],    PARAMETER["false_northing", 0.0],    UNIT["m", 1.0],    AXIS["x", EAST],    AXIS["y", NORTH],    AUTHORITY["EPSG","900913"]]   My questions, specifically:   How to write a WKT which transform an existing CRS? My guess would be that I need a new PROJCS wrapping an existing one and adding a PROJECTION clause. How would I found out the projection id (like Mercator_1SP above) and the required parameters (the PARAMETER clauses)? Can I "reference" EPSG:900913 in CRS WKT instead of copy-pasting the whole PROJCS clause? How to gracefully handle files that exceed PHP's `post_max_size`? Is $\int_a^b f(x) dx = \int_{f(a)}^{f(b)} f^{-1}(x) dx$? Does a bad grade in one course (with otherwise OK record) affect graduate admissions? I'm currently an M. Phil student in Math and I failed a course last semester because I was not able to hand in a home work assignment (which was worth 40% of my grade) on time (I was 30 minutes late).   Apart from that I'd say I'm a fairly decent student, I got an upper second class honours bachelors degree, I have a B+ and A's for all my other courses in my current programme and I'm almost finished with my first paper to be published.   I've however been deeply worried about this failure as it would reflect poorly on my transcript if I were to apply to a Ph.D or another Master's programme.   Will my failing grade affect my chances of getting into a competitive programme or greatly lower my chances of getting a scholarship? Order of a homomorphism of groups Let $f: G \to H$ be a homomorphism of groups. Assume that $a\in G$ and $\operatorname{ord}(a)=n$. Prove that the order of $f(a)$ is a divisor of $n$. I know that if $H$ is a subgroup of $G$, then $$|G|=|H|\cdot|\text{distinct cosets of $H$}|$$ so $|H|$ must be divisor of $|G|$, but I don't understand how to apply that logic to homomorphisms (or if that will even work). Any help is appreciated! If $a$ and $b$ commute and $\text{gcd}\left(\text{ord}(a),\text{ord}(b)\right)=1$, then $\text{ord}(ab)=\text{ord}(a)\text{ord}(b)$. Prove if $\operatorname{ord}(a)=m$, $\operatorname{ord}(b)=n$, and $\operatorname{gcd}(m,n)=1$, then $\operatorname{ord}(ab)=mn$.   I was reading this and was thinking how this proof would look like. I tried to do it and am not sure if this is correct. Here is what I did:  If $a$ and $b$ commute then $(ab)^{mn} = a^{mn} * b^{mn} = (a^m)^n * (b^n)^m = 1 * 1 = 1.$ So $ord(ab) | mn$.  Now, take $k = \operatorname{ord}(ab) = m^\prime * n^\prime$ where m' is relatively prime with $n$ and $n'$ is relatively prime with $m$. By the result above $m' | m$ and $n' | n$. Now we have $((ab)^k)^{m/m'} = 1$ since $(ab)^k=1$. But on the other hand, by the commutativity:  $$((ab)^k)^(m/m') =$$  $$((ab)^(m' * n'))^(m/m') =$$  $$ a^{n'm} * b^{n'm} = $$  $$(a^m)^{n'} * b^{n'm'}=$$  $$b^{n'm'} = 1$$  This implies that $n' * m'$ is divisible by $n$. But $m'$ is relatively prime with $n$, so we must have $n' = n.$ By symmetry, $m' = m$. So $ord(ab) = mn$.  Just to say this again, I want to prove if $a$ and $b$ commute and $m$ and $n$ are relatively prime then   $$ord(ab)=mn.$$   The comments are vague. I guess this must be the answer then and there probably is not another way to do this. I dragged a file to the recycle bin a few days ago. I have Windows 7 premium on a 245 GB partition, a 240 GB SSD, and a 250 SSD. And somewhere in the past few days I incidentally installed Delta Search. I know I should remove it right away, but I just switched to Windows on the partition of a 3.5 1 TB drive and would like to restore that file and deal with the malware after Christmas is over.  I'm pretty sure I know what drive that deleted files should be on, but what is the real name of the recycle directory and exactly where is it located? In other words, if I know the name of the drive what would the rest of the path look like so I can just paste it in my Explorer window and change the drive letter. I want to load the new entries without reloading the page. I found a great solution for: The ajax script:  &lt;script&gt;     $(document).ready(function() {     var refreshId = setInterval(function()     {         $('#content').fadeOut("fast").load('http://neocsatblog.mblx.hu/new.php').fadeIn("fast");     }, 10000);      }); &lt;/script&gt;   And the php:  &lt;?php require_once("wp-blog-header.php"); ?&gt; &lt;div id="content" &lt;?php cyberchimps_filter_content_class(); ?&gt;&gt;          &lt;?php do_action( 'cyberchimps_before_content'); ?&gt;          &lt;?php if ( have_posts() ) : ?&gt;              &lt;?php while ( have_posts() ) : the_post(); ?&gt;                  &lt;?php get_template_part( 'content', get_post_format() ); ?&gt;              &lt;?php endwhile; ?&gt;          &lt;?php elseif ( current_user_can( 'edit_posts' ) ) : ?&gt;              &lt;?php get_template_part( 'no-results', 'index' ); ?&gt;          &lt;?php endif; ?&gt;          &lt;?php do_action( 'cyberchimps_after_content'); ?&gt;      &lt;/div&gt;   However, this is a special feature that works perfectly here:  However, the main page refuses to load new entries, I do not understand the reason for this, as we use everything the same. This isn't a particularly major issue, but I've had this thing nagging me for a while. When looking at QGIS 3.4's global variables, user_full_name is assigned my given and surname, which includes a diaeresis (¨) somewhere in there. I can't seem to find any way to get QGIS to switch/adapt encoding and display the name correctly. The accented character gets switched to that ubiquitous black rhombus.  I looked for info on this behavior and found moderately similar issues like  and  but it's not exactly the same problem as it's not a geodata encoding issue but rather how QGIS deals with values pulled from the OS.  I have tried unchecking the Override system locale tickbox in Options|General as I prefer to use American English even though my OS is set to Canadian French. No effect to observe.  There's somewhat of a workaround right now which is to fix the name in the Project variable project_author as it is editable (it's filled with the offending global variable by default).  I'm curious to know, though, if someone has encountered this before and if there's a solution in QGIS. A more succinct expression for "The day before yesterday" Is there a more succinct expression for "the day before yesterday"?   In German for example, gestern = 'yesterday.' The prefix vor roughly means before, so logically, vorgestern means 'the day before yesterday.'   Similarly, morgen = 'tomorrow', the prefix über roughly means over, so again, übermorgen means 'the day after tomorrow.'  (In Mandarin Chinese also you have respectively 前天 &amp; 後天.)  Presumably, there are also similarly logical ways to say "the page after the next" or "the paragraph before the last", etc.   Are there no similarly succinct, and graceful, expressions in English? Under what conditions is the exponential map on a Lie algebra injective? Let $G$ be a Lie group with Lie algebra $\mathfrak{g}$ and let $\exp :\mathfrak{g}\rightarrow G$ be the exponential map.  In his blog,  that if a Lie group is not simply-connected, then $\exp$ will not be injective.  Conversely, is it true that if a Lie group is simply-connected, then $\exp$ is injective?  If not, what is a counter-example? How do I get the next element in HTML using JavaScript?  Suppose I have three &lt;div&gt;s and I get a reference to one in JavaScript code, I want to get which is the next &lt;div&gt; and which is the previous. Is right hand rule arbitrary? In definition of vector multiplication, the direction of the resulting vector is given by the right hand rule. However I don't know any mathematical requirement to pick right hand instead of left for this purpose. Is there any? How To Program A Bash Script That Presses A Certain Combination Of Keys? Please, i need someone to help me in making a very simple bash script. The job of this script is just to press a certain combination of keys on the keyboard.  My objective is to create a script that would turn off the Kubuntu desktop effects completely. The key combination for that is: "Left-ALT + Left-Shift + F12". I am aware that there is a widget that might do this job, but i am trying to learn how to accomplish it with a bash script.  I have tried to do an online search but i guess i was choosing the wrong set of keywords. I did not manage to reach a tutorial or a reference that would mention this task.  Thanks in advance for your help!  P.S: Any suggestions, references, notes, comments or tutorials would also be greatly be appreciated. The wavefunction $\Psi(x,t)$ for a free particle is given by  $$\Psi(x,t) = A e^{i(kx-\frac{\hbar k}{2m}t)}$$  This wavefunction is non-normalisable. Does this mean that free particles do not exist in nature?  Why then do we use free particles $\psi(\vec{x}) = e^{ikz}$, for example, in scattering theory? Let $R$ be a commutative ring with unit, $I$ an ideal of $R$ and consider the following three constructions.   The  $R_I$ of $R$ at $I$ (i.e. the localization of $R$ at the multiplicative system $R\setminus I$) gives a morphism $$ f_1:\operatorname{Spec}(R_I)\to \operatorname{Spec}(R)=:X $$ The  $\widehat{R}_I$ of $R$ at $I$ gives a morphism $$ f_2:\operatorname{Spec}(\widehat{R}_I)\to X $$ For the special case $I=(a)$, the localization $R_a$ (i.e. the localization of $R$ at the multiplicative system $\{1,a,a^2,\ldots\}$) gives a morphism $$ f_3:\operatorname{Spec}(R_a)\to X $$   My question is:     What is the geometric meaning of all three constructions and how are they related?   This is what I ''know'' already or describes at least the style of answer I would appreciate. As remarked in the comments, $I$ has to be a prime ideal.   $R\to R_I$ is injective iff $R\setminus I$ contains no zero divisors which is the case if $I$ is a prime ideal. The scheme $\operatorname{Spec}(R_I)$ is the intersection of all the neightbourhoods of $I$ in $X$. This is a little contra-intuitive for me since the last statement sounds like ''$f_1$ is injective'' but the $\operatorname{Spec}$-operator should turn 'injective'' and ''surjective'' around somehow (I know this is literally not true but I only want to get a feeling like ''is contained'', ''is bigger'', ''is smaller'', etc.). $\operatorname{Spec}(R_a)$ is somehow the opposite of $\operatorname{Spec}(R_{(a)})$ (= the intersection of all the neightbourhoods of $(a)$ in $X$) because $\operatorname{Spec}(R_a)$ seems to be something like the union of all the open sets of $X$ not containing the point $(a)$. $R\to\widehat{R}_I$ is injective iff $\cap I^n=(0)$ and this holds very often (e.g. for $R$ noetherian and either an integral domain or a local ring). Hence (this is probably false intuition as remarked before), $f_2$ should be something like a ''projection'' (from something ''big'' into something ''small''). But what is geometrically the difference between localization and completion. I don't have a geometric idea of completion at all.   As remarked by Qiaochu Yuan in the comments below, one should not think of $\operatorname{Spec}$ as injective-surjective switching.