What is the difference between `cuTexObjectCreate` and `cudaCreateTextureObject `?
CUDA Driver API vs. CUDA runtime
I know there are many ways to prevent image caching (such as via META tags), as well as a few nice tricks to ensure that the current version of an image is shown with every page load (such as image.jpg?x=timestamp), but is there any way to actually clear or replace an image in the browsers cache so that neither of the methods above are necessary?  As an example, lets say there are 100 images on a page and that these images are named "01.jpg", "02.jpg", "03.jpg", etc. If image "42.jpg" is replaced, is there any way to replace it in the cache so that "42.jpg" will automatically display the new image on successive page loads? I can't use the META tag method, because I need everuthing that ISN"T replaced to remain cached, and I can't use the timestamp method, because I don't want ALL of the images to be reloaded every time the page loads.  I've racked my brain and scoured the Internet for a way to do this (preferrably via javascript), but no luck. Any suggestions? How do I create a 50x3 cell array from a 50x3 matrix. Keep just getting a 1x1 cell array with a 50x3 submatrix? How can I distribute a native executable for a Python program? Lately all of my work has been in Python (and Ruby), and I'm in love with those two languages. There are many libraries for game development on Python but the problem I see is that many of those frameworks don't cover the "distribution" aspect of game creation. I'd like to code my game on Python (and any other pure Python extra libraries I need) and be able to distribute an executable for the main PC platforms (Windows, Mac and Linux).  Any options out there or what are the best combinations for this? (maybe pyglet + cx_freeze) Surjectivity of Composition of Surjective Functions Suppose we have two functions, $f:X\rightarrow Y$ and $g:Y\rightarrow Z$. If both of these functions are onto, how can we show that $g\circ f:X\rightarrow Z$ is also onto? How to prevent alt-f1 from selecting Dash error C2143: syntax error : missing ';' before 'type' Intertext vertical spacing How do I reduce the vertical spacing of \intertext? How can a "proper" function have a vertical slope? Plotting the function $f(x)=x^{1/3}$ defined for any real number $x$ gives us:   Since $f$ is a function, for any given $x$ value it maps to a single y value (and not more than one $y$ value, because that would mean it's not a function as it fails the vertical line test). This function also has a vertical tangent at $x=0$.   My question is: how can we have a function that also has a vertical tangent? To get a vertical tangent we need 2 vertical points, which means that we are not working with a "proper" function as it has multiple y values mapping to a single $x$. How is it possible for a "proper" function to have a vertical tangent?  As I understand, in the graph I pasted we cannot take the derivative of x=0 because the slope is vertical, hence we cannot see the instantaneous rate of change of x to y as the y value is not a value (or many values, which ever way you want to look at it). How is it possible to have a perfectly vertical slope on a function? In this case I can imagine a very steep curve at 0.... but vertical?!? I can't wrap my mind around it. How can we get a vertical slope on a non vertical function? Where to locate "sockets" I am new to this, so sorry if its obvious. I am running a Debian server and installing the likes of UWSGI, NGinx etc on there. The configurations keep talking about pointing to "sockets".  In the build options I seem to be able to specify where the sockets for each program go. By default it looks like most of them go in /tmp (not all of them).  Is this a good place for them to go? I'm trying to keep things as organized as possible but just bunging them in /tmp doesn't seem like the best option. iPhone asks for Mac password (or Mac asks for iOS password) During iPhone update to iOS 10.3.1, I was asked for my password using the correct machine name for my Mac mini. To be clear, the request was not for my iCloud password, my actual local user account password for macOS. This can also happen in the other direction - setting up a Mac may ask for verification for you to enter your iPhone / iPad passcode (and not your iCloud password)  I could not find any information about this kind of access. I was thinking it was security for iCloud and possibly related to two step authorization or two factor authorization. This is the first time I can think of that I entered a desktop password outside of that actual desktop.  I don't have a screen shot but I recall an implication that some files in my Mac are encrypted and need my password. '... need to enter the password for .'   Are there settings I can I control this assuming it's a legitimate part of iCloud setup process or iOS setup process? Can the Force be used to fly? Simplifying and evaluating $\cot 70^\circ+4\cos 70^\circ$ I hava Lenovo g580 laptop. I want to change disk to SSD drive, but I'm not sure if my laptop support SATA 3 (6Gbps).  I write following output:  dmesg | grep SATA [    3.190743] ahci 0000:00:1f.2: AHCI 0001.0300 32 slots 6 ports 6 Gbps 0x5 impl SATA mode [    3.199450] ata1: SATA max UDMA/133 abar m2048@0xd3617000 port 0xd3617100 irq 43 [    3.199455] ata3: SATA max UDMA/133 abar m2048@0xd3617000 port 0xd3617200 irq 43 [    3.518365] ata1: SATA link up 3.0 Gbps (SStatus 123 SControl 300) [    3.857921] ata3: SATA link up 1.5 Gbps (SStatus 113 SControl 300)   Is that means I Can use SSD disk in SATAa 6Gbps mode? Meaning of symbols $\vdash$ and $ \models$ I'm confused about the use of symbols $\vdash$ and  $ \models$. Reading the answers to  and  I see that:  the turnstile symbol $ \vdash $ denotes syntactic implication. Then $ S \vdash \psi$ means that $\psi$ can be derived from the formulae in $S$  the double turnstile $ \models $, denotes semantic implication. Then the notation $A \models B$ means simply that B is true in every model in which A is true.  So I understand that, e.g., $(ZFC+ A) \vdash B$ means that $A \Rightarrow B$ is a theorem in ZFC and, in this sense $\vdash$ is a mathematical symbol for a theorem. While $(ZFC+ A) \models B$ means that $ B$ is true in every model of ZFC in which $A$ is true, and, in this sense $\models$ is not a mathematical but a metamathematical symbol.  I understand correctly? And those are the only uses of the two symbols? Proving $\sqrt2$ is irrational I used the method of contradiction by assuming that $\sqrt 2$ is a rational number. Then, by the definition of rational number, there exist two integers $p$ and $q$ whose ratio equals $\sqrt 2$. Thus, $$\frac pq = \sqrt2\tag{x}$$  Squaring both sides, $$p^2/q^2 = 2\tag{a}$$  or $$p^2 = 2q^2\tag{b}$$ This means that $p^2$ is an even number, implying that $p$ is even.  Now, any even integer can be written as $2^kf$ where $f$ is any odd integer and $k$ is some positive integer (the minimum value of $k$ and $f$ is $1$ since even numbers start from $2 = 2^1\cdot1$). For odd numbers, $k=0$.  For example, $8 = 2^3 \cdot 1$, $4= 2^2 \cdot 1$, $18= 2^1 \cdot 9$, $24 = 2^3 \cdot 3$, $-12= 2^2 \cdot(-3)$, etc.  Now, from $(b)$, $q^2$ can be even or odd.  Case 1: $q^2$ is even (thus meaning $q$ is even). Then $p= 2^{k_1} \cdot f_1$ (say) and $q = 2^{k_2} * f_2$ (say). Note here in this case, both conditions $k_1=k_2$ and $f_1=f_2$ can't hold simultaneously since that would mean $p/q =1$ and here $p/q =\sqrt2$ (from $(x)$)). Let's consider the condition when $k_1=k_2=k$ but $f_1\ne f_2$. Then $$\frac{p^2}{q^2}=\frac{(2^kf_1)^2}{(2^kf_2)^2} = \frac{f_1^2}{f_2^2} =2$$ (from $(a)$), i.e.: odd/odd ($f_1$ and $f_2$ are odd) can never equal $2$.  Now lets consider $f_1=f_2=f$ but $k_1\ne k_2$, thus $(2^{k_1}f)^2/(2^{k_2}f)^2= 2^{2(k_1-k_2)}= 2$ meaning $k_1 - k_2 = 0.5$ but $k_1,k_2$ are integers, so their difference can't be $0.5$. (Also here, $k_1-k_2$ must be greater or equal to $1$ since $2^{k_1-k_2}=\sqrt 2$, $k_1-k_2$ can't be negative since $\sqrt 2&gt;1$, but $k_1-k_2\ge1$ can't satisfy $(a)$ since the minimum value of $p/q$ in this case will be $2$ which is surely greater than $\sqrt2$.)  Now lets consider $f_1\ne f_2$ and $k_1\ne k_2$, thus $\frac{(2^{k_1} f_1)^2}{(2^{k_2} f_2)^2}= 2^{2(k_1-k_2)}\frac{f_1^2}{f_2^2}$ can never equal $2$ since $\frac{f_1^2}{f_2^2}$ is either odd or "odd/odd" ie: it doesn't contain 2 as a factor and $2^{2(k_1-k_2)}$ is a power of $2$. So in $2^{2(k_1-k_2)}\frac{f_1^2}{f_2^2}$, there's no chance of cancellation of the odd factor $\frac{f_1^2}{f_2^2}$.  Case 2: $q^2$ is odd (thus meaning $q$ is odd). $q=f'$ (say, here $k=0$ for $q$ but not for $p$), thus from $(a)$ $2^{2k}\frac{f^2}{f'^2} = 2$ but this isn't possible since a power of $2$ multiplied with odd factor can't equal $2$.  Thus, both case 1 and 2 suggest that for any possible combination of $k_1,k_2,f_1$ and $f_2$, $p/q\ne\sqrt2$, i.e.: for no value of integers $p$ and $q$, $p/q = \sqrt2$. Thus, this contradicts our assumption that $\sqrt2$ is rational. Therefore it must be irrational.  PLS NOTE that i would like to clarify that $f_1^2$/$f_2^2$ is either an odd integer or a fraction of odd/odd or 1/odd form, thus not containing  2 as a factor,for any $2^n$, if it has to be reduced to 2 , must be multiplied with $1/ 2^(n-1)$ but $f_1^2$/$f_2^2$ can't be of $1/2^[n-1]$ form since for odd nos, in $2^k$*f notation, k is 0 , so 2 vanishes as $2^k$ becomes 1 in this case thus f1/f2 can't be represented as  1/ $2^(n-1)$ in which n has to be existent.  therefore $2^{2(k_1-k_2)}\frac{f_1^2}{f_2^2}$ can never equal 2 as to further explain the 3rd condition of case 1  Difficulty: is my approach correct? This proof which I thought is different from proofs found on the internet or books since I have used $p$ and $q$ as any integers, which may have a common factor. So I am not sure if I am on right track. Will someone please check this out and make kind comments? Recently I created a website with hundreds of thousands of unique pages. A week ago, Google had indexed about 250,000 pages. It has now decreased to 90k URLs (using a site:domain.com query).   What should I do to have Google index all the pages again? Why did all these pages get de-indexed from Google?. Compact way to assign values by slicing list in Python Prove that for any positive integer $n$, $$2\sqrt{n + 1} − 2 \le 1 +\frac 1 {\sqrt 2}+\frac 1 {\sqrt 3}+ \dots +\frac 1 {\sqrt n} \le 2\sqrt n − 1$$ Progress I think Riemann sum should be used for the middle term. I got the limit as $n\to \infty$ of the function inside using Riemann sum but what I cannot get are the lower and upper bound. After Hoth, why did Vader pursue the Millennium Falcon instead of Luke's X-Wing? In The Empire Strikes Back film, the opening crawl says:     The evil lord Darth Vader, obsessed with finding young Skywalker, has dispatched thousands of remote probes into the far reaches of space....   It has also been established in Return of the Jedi that Force users can sense each other:     Chewie whines nervously. Luke stares at the Huge Super Star Destroyer    that looms ever larger before them.      LUKE: Vader's on that ship.   If Darth Vader is obsessed with finding Luke and he can sense his presence, the obvious thing to do would be to follow Luke's X-Wing after the battle of Hoth rather than the Millennium Falcon. Why didn't he? Last minute flight bookings - better to book online or at the airport? We had , which covered ground on what happens to flight prices over the period.  What I'm interested to know is if I had to fly out at short notice, say within 1-2 days, then am I better off booking within that 1-2 day gap online, or am I better off trying my luck directly at the airport (as long as I'm willing to wait at the airport) for a cheaper price? Is there any truth to rumours that airlines sell tickets at cheaper prices at the airport right before flights to fill empty seats or is that a myth?