'get off of me' or 'get off me'. what is the difference?
Difference between get "off of" and "off"
This error message is being presented, any suggestions?     Allowed memory size of 33554432 bytes exhausted (tried to allocate   43148176 bytes) in php How do I connect to a MySQL Database in Python? How do I connect to a MySQL database using a python program? I designed a poster in CorelDraw X5 for printing purposes. I was told by the Printshop guy to use colors only in CMYK. For the most part I did, I don't know why I missed it but I used one color in RGB. I liked this color so much that I've used it at many places. So, current situation is that some colors are in RGB and some are in CMYK. So, I want to convert all RGB to CMYK and thought the problem would be solved. But now I'm actually facing a lot of trouble. The colors are not at all same. I hate the colors that it is showing me.  This is the color I used in RGB:    When I changed the "Model" to CMYK, the color it shows is really different. As you can see in the picture below, it's too dull:    The RGB values of both colors are not the same. What can I do to get the same color in CMYK? I did a little bit of reading here and there. They say, that you can't get same. In that case, what is the closest color that I can get? In the book &quot;Quantum Mechanics and Path Integrals&quot; Feynman &amp; Hibbs state that  the probability $P(b,a)$ to go from point $x_a$ at time $t_a$ to the point $x_b$ at the time $t_b$ is $P(b,a) = \|K(b,a)\|^2$ of an amplitude $K(b,a)$ to go from $a$ to $b$. This amplitude is the sum of contributions $\phi[x(t)]$ from each path. $$ K(b,a) = \sum_{\text{paths from $a$ to $b$}} \phi[x(t)]$$ The contributions of a path has a phase proportional to the action $S$: $$ \phi[x(t)] = \text{const}\ e^{(i/\hbar)S[x(t)]}$$  Why must the contribution of a path be $\sim e^{(i/\hbar)S[x(t)]}$? Can this be somehow derived or explained? Why can't the contribution of a path be something else e.g. $\sim \frac{S}{\hbar}$, $\sim \cos(S/\hbar)$, $\log(S/\hbar)$ or $e^{- (S[x(t)]/\hbar)^2}$ ? Edit: I have to admit that in the first version of this question, I didn't exclude the possibility to derive the contribution of a path directly from Schrödinger's equation. So answers along this line are valid although not so interesting. I think when Feynman developed his formalism his goal was to find a way to quantize systems, which cannot be treated by Schrödinger's equation, because they cannot be described in terms of a Hamiltonian (e.g. the Wheeler-Feynman absorber theory). So I think a good answer would explain Feynman's Ansatz without referring to Schrödinger's equation, because I think Schrödinger's equation can only handle a specific subset of all the systems that can be treated by Feynman's more general principle. lognormal distribution, standard-deviation and (physical) units I have a distribution of microparticles that follows a lognormal distribution. The cumulative distribution function thus is given by:  $$ F_X(x;\mu,\sigma) = \frac12 \operatorname{erfc}\!\left(-\frac{\ln x - \mu}{\sigma\sqrt{2}}\right) $$ $$ \mu = ln(M) + \sigma^2 $$  Now, the plot of the distribution function should be exactly the same no matter if the particle diameter $x$ is given in micrometers or meters (as long as I adapt the x-axis accordingly of course). However, this only works if I only convert $x$ and $M$, while not touching the numerical value of $\sigma$, and I don't understand why. $F_X$ has to be unitless, so $x$, $\mu$ and $\sigma$ should all have the same unit, right? Find $\lim\limits_{n \to \infty} \frac{\sqrt 1 + \sqrt 2 + \dots + \sqrt{n}}{n\sqrt{n}}$. $$\lim_{n \to \infty} \dfrac{\sqrt 1 + \sqrt 2 + \dots + \sqrt{n}}{n\sqrt{n}}$$   $$\lim_{n \to \infty} \dfrac{\sqrt 1 + \sqrt 2 + \dots + \sqrt{n}}{n\sqrt{n}} =\lim_{n\to \infty} \dfrac1{n}\sum^{n}_{k = 1} \sqrt{\dfrac k n} $$  While searching this question I found, .   Like in the accepted answer I first compared my series to LRAM,    $$\int_a^b f(x)\,dx=\lim_{n\to\infty}\frac 1n\sum_{i=1}^n f\left(a+\frac{b-a}n i \right)$$  I got $a = 0$, $b = 1$ and $f(x) =\sqrt{x}$ so,  $$\int_0^1 \sqrt{x}\ dx = \dfrac2 3$$ should be the answer.  Is there any simpler method to do this sum ? I have not learnt this method to do infinite sums so I can't use it. I'm trying to prove that the converse of Lagrange's theorem is true for finite abelian groups (i.e. "given an abelian group $G$ of order $m$, for all positive divisors $n$ of $m$, $G$ has a subgroup of order $n$"). This is an exercise from a book, and it is in the section on finite abelian groups, so I know I have to use the fundamental theorem of finite abelian groups. I have come up with a proof, but it seems a bit messy, and I'm not entirely sure if it's correct. It is given below.     Let the order of $G$ be $m$ = $p_1^{\alpha_1} \ldots p_k^{\alpha_k}$. It is known that $G$ is a direct product of $p$-groups, say:      $$G = G_1 \times \ldots \times G_k$$      where each $G_i$ is a $p_i$-group. By the fundamental theorem of finite abelian groups, each $G_i$ is isomorphic to a direct product of cyclic groups of the form      $$\mathbb{Z}_{{p_i}^{\beta_1}} \times \ldots \times \mathbb{Z}_{{p_i}^{\beta_l}},$$      where $\beta_1, \ldots, \beta_l$ are positive integers such that $\sum_{j=1}^l \beta_j = \alpha_i$. Now if $n$ divides $m$, then we must have      $$n = p_1^{\gamma_1} \ldots p_k^{\gamma_k}$$      for some $\gamma_1, \ldots, \gamma_k$ with $0 \leq \gamma_i \leq \alpha_i$.      Claim: Each $G_i$ has a subgroup of order $p_i^{\gamma_i}$      Proof: As above, we have that       $$ G_i \cong \mathbb{Z}_{{p_i}^{\beta_1}} \times \ldots \times \mathbb{Z}_{{p_i}^{\beta_l}} $$      where $\beta_1, \ldots, \beta_l$ are positive integers such that $\sum_{j=1}^l \beta_j = \alpha_i$.      Now since $0 \leq \gamma_i \leq \alpha_i$, we can find   $l$ numbers $\delta_1, \ldots , \delta_l$ such that $\gamma_i = \sum_{j=1}^l \delta_j$, and $0 \leq \delta_j \leq \beta_j$. (This choice of numbers is not necessarily unique).      Then $p_i^{\delta_j} | p_i^{\beta_j}$ for each $j = 1, \ldots , l$. Hence, for each factor $\mathbb{Z}_{{p_i}^{\beta_j}}$, there exists a subgroup of order $p_i^{\delta_j}$, namely $\mathbb{Z}_{{p_i}^{\delta_j}}$ (using the fact that the converse of Lagrange's theorem is true for finite cyclic groups). Taking the direct product of each of these subgroups, we get a new subgroup $G_i'$ of $G_i$:      $$G_i' \cong \mathbb{Z}_{{p_i}^{\delta_1}} \times \ldots \times \mathbb{Z}_{{p_i}^{\delta_l}}$$      The order of this subgroup is $p_i^{\delta_1} \times \ldots \times p_i^{\delta_l} = p_i^{\delta_1 + \ldots + \delta_l} = p_i^{\gamma_i} $. So we have found a subgroup of $G_i$ of order $p_i^{\gamma_i}$, as required.      So each factor $G_i$ in the product $G = G_1 \times \ldots \times G_k$ has a subgroup $G_i'$ of order $p_i^{\gamma_i}$.       Therefore, $G$ has a subgroup   $$G_1' \times G_2' \times \ldots \times G_k'$$       of order $p_1^{\gamma_i}...p_k^{\gamma_k} = n$, which completes the proof.   I have two questions about this: firstly, does this proof seem to work? Secondly, is there a way to make the proof more concise (e.g. a way to prove the statement without using all these indices)? where does lspci gather its information from? In  I found "[The   latest   version    of   this   document   ". There I says: "… lspci is a synonym for cat /proc/pci".  Which isn't the case here, on a Crunchbang 10 system (Debian based). No such directory. I do get the basic idea and, as far as I  know, the content of proc is (mainly) created during runtime. (Wrong?). This made me curious:  Question: Where does lspci gather it's information from? And where is this documented? (Where did I miss something?)  &nbsp; Another difference I found: In kernel.org's documentation, under "Table 1-5: Kernel info in /proc"      pci &nbsp;&nbsp;&nbsp; Deprecated info of PCI bus (new way -> /proc/bus/pci/, decoupled by `lspci) What is Snell's law's main formula? My book states it as $(\sin i)/(\sin r)=n$. However, University Physics has a different say.  It says, $n_{a} \sin a =n_{b} \sin b$. Which of the following is correct? Positive operator has a positive spectrum? Can an integer of the form $4n+3$ written as a sum of two squares? Files greater than 1 GB and older than 6 months I want to find files which are greater than 1 GB and older than 6 months in entire server. How to write a command for this? What are the best practices for battery longevity? How do you maximize the cycle count of your battery?  There seems to be a debate between:   Leave plugged in 24/7 but do a full cycle once a month Charge to ~100%, drain to ~10%, repeat.   NOTE: Taking your   Which method is better and why? How much of a difference does it really make?  Here are some of the better sources I've found about battery behavior so far. Even still I feel they don't conclusively answer the question above. See my elaborations below:           Elaborating on Option 1  Now IF Apple engineers were smart and optimized for AC draw when plugged in to not use up your charge cycles, then it seem logical to assume that you are using effectively none of your 1000 charge cycles while plugged in. If that's the case, it would seem that the LiPo drop from holding at 100% would be vastly outweighed by the fact you're not using any charge cycles.  However, this is a purely speculative assumption and I could find no evidence to either confirm or deny.  If this is NOT true, and the AC adapter only goes to charging the battery, then there would be no difference. Your battery would drain to 99%, then back up to 100%, then back down to 99%, etc. Those micro-charge cycles would add up at the same rate as 100% -> 0% -> 100%, and you would get no gain. In this case the negative effect of holding LiPo batteries at 100% would outweigh everything else.  Elaborating on Option 2  There are lots of good reasons why Option 2 is the way it is:   Because of known LiPo chemistry issues, holding a 100% charge for a long time causes the battery to degrade Apple specifically recommends not to leave it plugged in all the time  Draining a battery all the way to zero all the time is bad for the cell (Which is why ~10% is used) Heat is a killer, and when plugged in AND charging, you get extra heat that causes damage.   There are a couple reasons why I'm challenging Option 2:   It seems silly that the most intuitive use case for thousands of people that use their macs as primary computers is wrong. If the hypothesis in option 1 is true, you're just burning unnecessarily through your finite charge cycles. It's a pain to have to remember to keep plugging in and plugging out and be worried about whether or not your device is plugged in. Calculating extent of custom tiles? Problem  The picture is just an example of my case, but I am left with blank space on the left when I load all the tiles and I get error message in the console:   Failed to load resource: the server responded with a status of 404 (Not Found) http://path/to/the/tile//base/8/6/53.png  etc.     I could guess and limit the extent of the layer, but I would really like to know is there any way to calculate it and also be able to get rid of the error message.  What I know about the tiles?   Originally they are served as TMS They represent png images 256*256 They are in projection: EPSG:3857 In OpenLayers2 they were called like TMS layer In OpenLayers3 (v.3.8.2) they are called like XYZ layer (that includes -y):     var baseLayer = new ol.layer.Tile({   source: new ol.source.XYZ({     url: 'http://path/to/the/tiles/base/{z}/{x}/{-y}.png'   }) });   Thoughts and questions   How can I calculate extent of the tiles? Or in another words how can I find out zmin zmax xmin ymin xmax ymax or whole range of x/y?  What does x/y really represent? Are they coordinates of the left bottom corner or center of the tile? In which direction does x/y increase? How can I figure out x/y of at least one of the tile, for example left bottom one?    (I have modified the question for better understanding of the problem) Is there a reason why ls does not have a --zero or -0 option Ain't and negatives To prove that if $ab\equiv ac\bmod n$ and $(a,n)=1$ then $b\equiv c\bmod n$ I'm trying to create a unit test to test the case for when the timezone changes on a machine because it has been incorrectly set and then corrected.  In the test I need to be able to create DateTime objects in a none local time zone to ensure that people running the test can do so successfully irrespective of where they are located.  From what I can see from the DateTime constructor I can set the TimeZone to be either the local timezone, the UTC timezone or not specified.  How do I create a DateTime with a specific timezone like PST? What type of lens will auto-focus on Nikon bodies without a focus motor? I know my D40 doesn't have a motor in it to auto-focus lenses.  What type of lens has the motor in it? How can I know which lenses will and will not work? Smooth structure of embedded submanifolds Here is a rather long quotation from "Introduction to smooth manifolds" by J. Lee.     Proposition 5.2 (Images of Embeddings as Submanifolds). Suppose $M$ is a   smooth manifold with or without boundary, $N$ is a smooth manifold, and   $F: N \to M$ is a smooth embedding. Let $S = F(N)$. With the subspace   topology, $S$ is a topological manifold, and it has a unique smooth   structure making it into an embedded submanifold of $M$ with the   property that $F$ is a diffeomorphism onto its image.      Proof. If we give $S$ the subspace topology that it inherits from $M$,   then the assumption that $F$ is an embedding means that $F$ can be   considered as a homeomorphism from $N$ onto $S$, and thus $S$ is a   topological manifold. We give S a smooth structure by taking the   smooth charts to be those of the form $(F(U), \phi \circ F^{-1})$, where $(U, \phi)$   is any smooth chart for $N$; smooth compatibility of these charts   follows immediately from the smooth compatibility of the   corresponding charts for $N$. With this smooth structure on $S$, the map   $F$ is a diffeomorphism onto its image (essentially by definition), and   this is obviously the only smooth structure with this property. ...   My question is about the bold sentence. He says that it is obvious, but it is not obvious to me. Why is the constructed smooth structure the only smooth structure that makes $F$ a diffeomorphism between $N$ and $S$?