I want to know the function name that tells the location of where the session is stored.
Are $_SESSION variables stored on the client or the server?
Did spacetime start with the Big bang? I've discovered a new pattern.  Is this pattern well known or what is the opinion about it?  Basically, I have a hard time scrubbing up and down source files to figure out what module imports are available and so forth, so now, instead of  import foo from bar.baz import quux  def myFunction():     foo.this.that(quux)   I move all my imports into the function where they're actually used., like this:  def myFunction():     import foo     from bar.baz import quux      foo.this.that(quux)   This does a few things.  First, I rarely accidentally pollute my modules with the contents of other modules.  I could set the __all__ variable for the module, but then I'd have to update it as the module evolves, and that doesn't help the namespace pollution for code that actually lives in the module.  Second, I rarely end up with a litany of imports at the top of my modules, half or more of which I no longer need because I've refactored it.  Finally, I find this pattern MUCH easier to read, since every referenced name is right there in the function body. Why is $\mathbb{R}^2$ not a subspace of $\mathbb{R}^3$? I cannot understand why $\mathbb{R}^2$ is not a subspace of $\mathbb{R}^3$.   My reasoning is as follows: Choose any elements $v_1$ and $v_2$ from $\mathbb{R}^2$, add them together you get an element of $\mathbb{R}^2$ and same for scalar multiplication and $0$ vector is an element of $\mathbb{R}^2$. So where did I go wrong? What is an intuitive way of understanding Ostrogradsky instability? Is there a way of explaining why no differential equations in physics exceed order two without delving into Lagrangian and Hamiltonian mechanics - i.e. from Newtonian mechanics? Moreover, is there such an alternative proof, which focuses more on physical reasoning than abstract mathematics? Where should I put software I compile myself? For the following pretty straightforward task: query a list of products from a DB and present it on a webpage, consider 2 setups:  Setup1: PHP script queries. All content is built on the server and the entire page is served back to the client.  Setup2: Static HTML "page skeleton" requesting content using AJAX. Received content is parsed on the client side using Javascript and rendered using innerHTML or similar.  Of course the second setup only makes sense when you have pages, categories and tags for the client user to choose from.  I need to compare these two, at least by means of:   time it will take content to be served user experience (setup1 is delivered as a whole, setup2 is delivered in "two parts") scalability - how do the setups compare when I have 100,000 queries daily   Any thoughts on the issue will be much appreciated. 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'm trying to decide on a higher-end digital SLR, and I'm down to choosing between an APS-C model and a full-frame model.  I understand that the sensors are of different sizes, and as such have an effect on the perceived magnification of the lens, with the smaller APS-C sensor having an effective focal length greater than what it otherwise would be with a full frame sensor. But why does this matter?     What things should drive my choice between one or the other?   In which situations is one better than the other, and why? Below is an attempt at a proof of the following problem. Any feedback would be greatly appreciated. Thx!  Let $E$ be a set and $O_n = \{x: d(x, E) &lt; \frac{1}{n}\}$. Show  If $E$ is compact then $m(E) = lim_{n \rightarrow \infty} m(O_n)$.  This is false for $E$ closed and unbounded or $E$ open and bounded.   Note that all sets are real and measurable refers to Lebesgue measurable.  Suppose $E$ is compact. Then $m(E) &lt; \infty$. Also, since each $O_n$ is an open set of $\mathbb{R}^d$, $m(O_n) &lt; \infty$. If either of the following is true, we have that $lim_{n \rightarrow \infty}m(O_n)=m(E)$: $O_n \nearrow E$ $O_n \searrow E$ Now  $\bigcap_{n=1}^{\infty} O_n = \{ x:d(x,E) = 0\}$. That is, $x \in \bigcap_{n=1}^{\infty}O_n \iff \exists n&gt; \frac{1}{\epsilon},\ n \in \mathbb{N} \iff x\in \{x \mid d(x,E) =0\}$. Then we have that $\bigcap_{n=1}^{\infty}O_n = E$ and $O_n \searrow E$. Thus $ \ m(E) = lim_{n \rightarrow \infty} m(O_n)$. Suppose $E$ is closed and unbounded. Let $E = \{(x,y): y = 2 \}$ and $O_n = \{(x,y): d(x,E) &lt; \frac{1}{n}\}$. Since $E$ is a line in $\mathbb{R}^2$, $m(E) = 0$. Also, since the measure of a rectangle in $\mathbb{R}^d$ is its volume, $m(O_n) = \left| O_n \right| = \infty$, since the rectangle $O_n$ is unbounded. It is therefore apparent that $m(E) \not = lim_{n \rightarrow \infty} m(O_n)$. Suppose $E$ is open and bounded. Let $E=(0,1)$ and $O_n = \{x \mid d(x,E) &lt; \frac{1}{n}\}$. $\mathbb{R}-E$ is closed and $O_n \in \mathbb{R}-E$. Since $\mathbb{R}-E$ contains all its limit points $lim_{n \rightarrow \infty}O_n = O \in \mathbb{R}-E$. Thus $lim_{n \rightarrow \infty} m(O_n) \neq m(E)$. Whenever my cheese gets too old and has a bit of blue/white mold on the side, I'm not sure whether I should throw it completely out or not. Is it safe to eat if I cut off the edge? Should I cut a bit off of all the sides just to be safe?  I'm particularly interested in the answer for cheddar cheese, but answers about other cheese are welcome. How to prove that the only subgroup of the symmetric group $S_n$ of order $n!/2$ is $A_n$?  Why isn't there other possibility?  Thanks :) How can I make Windows Paint save by default as JPEG format? Every time I want to save an image I've pasted into the  application, it chooses a BMP/DIB format (which is the worst for saving things from the clipboard).  How can I hack Paint into defaulting to the JPEG format when saving images?  I am using Windows XP SP3 and Paint 5.1 at the moment. I hope though that any hacks might be generic and I'd be able to use them across all my Windows machines. I am having an access problem to some files and directories that I have tracked down to the group assigned to access them.    When I manually change the name of the group in the properties/permissions menu for a directory from the default setting (MyGroup) to www-data used by my Drupal Website, the directory error messages I get are gone.    There are a large number of files and directories that require this change.  If believe using chmod will change the owner which is set correctly and being new to Ubuntu, I am reluctant to experiment without a better understanding of how owner and groups work in the permissions settings.   What command changes the Group setting for a directory? I'm trying to identify a book I read in school, though I can't recall at what grade level; let's say it had to have been published before I graduated high school in 2000. Despite being required reading, it didn't feel much like a classic.  The story took place on Earth, after some catastrophe had caused the air to become poisonous. However, one teenage girl was able to survive because her house was in a valley which was somehow protected from the poisonous winds and was able to sustain its own ecosystem. One day, a man wearing a protective full-body suit and gas mask appears in the valley, and both are rather surprised to find someone else alive. She feeds and shelters the man and shows him how she had come to survive in the valley. Rather abruptly…      the man attempts to rape her, but she overpowers him somehow. She steals his protective suit and starts to leave the valley. He tells her that he had seen birds in a certain direction away from the valley, indicating there might be life there as well, and she leaves.   I believe there was an author's note either at the beginning or end of the book stating that the author had actually died while writing the manuscript, and that their friends or family had finished the book by going off the author's original notes, which I suppose may explain the abruptness of the ending.  In case it wasn't clear from the above, I didn't really like the book. But I'm curious about it regardless. If $E/F$ is algebraic and every $f\in F[X]$ has a root in $E$, why is $E$ algebraically closed? Create ColorRamp How can I create a ColorRamp for a Cycles material through Python?  I am able to create a new shader output node with nodes.new(type='ShaderNodeOutputMaterial'). Searching online, I see that it is only possible through C code (however, these pages dates back several years).  Is there a way to fake a ColorRamp (maybe with multiple Math nodes in greater/less than mode) or create a ColorRamp for a Cycles material through Python? How to calculate relatedness in haplodiploid organisms (mainly full sisters and full brothers)? I have tried to calculate the relatedness for haplodiploid organisms, but cannot understand the calculations behind full sister and full brother.  (taken from  I have managed to solve the other cases. Since female will have ZZ:  female will share 1/2 of her genes with daughter female will share 1/2 of her genes with son female will share 1/2 of her genes with mother female will share 1/2 of her genes with father  Since male will have only Z:  male will share 1 (100%) of his genes with his daughter male won't share anything with his son (unfertilized eggs leads to male in haplodiploidy system) male will share 1(100%) of his genes with his mother male won't share anything with his father (unfertilized eggs leads to male in haplodiploidy system)  Even though I figured out how others work, I wasn't able to calculate r for fullsister fullbrother? How can it be 3/4,1/4,1/2,1/2? Volume generated by revolving the region in the first quadrant bounded by the parabolas $y^2=x$, $y^2=8x$, $x^2=y$, $x^2=8y$ about the $x$-axis. I came across this question from Schaum's Outlines of Advanced Calculus:     Show that the volume generated by revolving the region in the first quadrant bounded by the parabolas $y^2=x$, $y^2=8x$, $x^2=y$, $x^2=8y$ about the $x$-axis is $\dfrac{279\pi}{2}$.   My solution was to divide the region into three parts:    $$\pi\int_{1}^{2}(x^2)^2 - (\sqrt{x})^2\;dx = 4.7\pi$$  $$\pi\int_{4}^{8}(\sqrt{8x})^2 - (\sqrt{x})^2\;dx = 42\pi$$  $$\pi\int_{2}^{4} (x^2)^2 - \left(\frac{1}{8}x^2 \right)^2\;dx = 44.9\pi.$$  Hence the total volume will be  $$\pi(4.7 + 42 + 44.9) = 91.6\pi$$  which is wrong.  Can someone point out my mistake? Prove that "injective function $f:X\to Y$ exists" and "surjective function $g:Y\to X$ exists" is logically equivalent. Prove that   injective function $f:X\to Y$ exists surjective function $g:Y\to X$ exists   is equivalent.$(x,y\neq \emptyset)$   My approach  First,prove left to right.  If injective function $f:X\to Y$ exists, a set A can be defined like this. $$A=\{f(x):x\in X\}$$ then it is sure that $A\subset Y$, and $|A|=|X|$  And, if we define function $g$ like $$g(y)=\begin{cases}x &amp; \text{if $y\in A$ and } f(x)=y,\\x_0&amp; \text{if }y\notin A.\end{cases}$$ ($x_0$ is a fixed element of $X$)  Then, $g$ is a surjective function, so surjective function $g:Y\to X$ exists.  Now,prove right to left.  If surjective function $g:Y\to X$ exists, $\forall x \in X$, we can select a element $y\in Y$ that $g(y)=x$, which is clear that we can pick a different $y\in Y$ for $\forall x \in X$  if we define function $f$ as $f(x)=\text{one element in Y that g(that element)=$x$}$, f is a injective function, so injective function $f:X\to Y$ exists.    But, I am not sure that my approach is right, especially on proving right to left. Please check whether my approach is right, ways to improve it,  and some other ways of proving this question. How to change the shadow color in Blender Cycles? I want to change the color of the shadow casting by a cube to green. While it's very easy to do it in Blender Internal (you select the lamp, go to the Lamp data header and change the shadow's color in the Shadow panel) in Cycles there're no such options. How to achieve it in Cycles?