I don't mean 1080p, but actual unscaled image size so when I set the wallpaper to an image with that size, it won't ask me to crop. It will fit without scaling.
On an Android phone the wallpaper spans across all of the home screens.  So what size should pictures be to fit properly as custom wallpapers?
I got this question in homework:     Let $\{0,1\}^A$ the set of all functions from A (not necessarily a finite set)       to $\{0,1\}$. Find a correspondence (function) between $\{0,1\}^A$ and       $\mathcal P(A)$ (The power set of $A$).       Prove that this correspondence is one-to-one and onto.   I don't know where to start, so I need a hint. What does it mean to find a correspondence? I'm not really supposed to define a function, right?  I guess once I have the correspondence defined somehow, the proof will be easier.  Any ideas? Thanks! Clearing Flag Bits By Writing 1 Mathematical induction for inequalities: $\frac1{n+1} + \frac1{n+2} + \cdots +\frac1{3n+1} > 1$ Why didn't the Alliance send for Yoda instead of Obi-Wan? Continuously Differentiable Curves in $\mathbb{R}^{d}$ and their Lebesgue Measure Show that the image of the curve $\Gamma\in\mathscr{C}^{1}\left([a,b]\to\mathbb{R}^{d}\right)$ has d-dimensional Lebesgue measure zero (of course, $d\geq2$).  This can be proved using the absolute continuity of $\Gamma'$ (since $[a,b]$ is compact and $\Gamma'$ is assumed continuous, hence in $L^{1}([a,b])$) together with the fundamental theorem of calculus to obtain an $\epsilon$-small cover of $\Gamma$ by balls.  But I am trying to prove this using more elementary means (i.e. without integration).  Intuitively, since $\Gamma$ is smooth, we ought to (for fine enough partitions) be able to cover $\Gamma$ by boxes which arise from its tangent line.  And by taking the partition of $[a,b]$ to be finer and finer, the "tangent box" cover ought to also get smaller and smaller.  More rigorously, the vector-version of the mean value theorem can be applied: $$|\Gamma(t_{i-1})-\Gamma(t_{i})|\leq(t_{i}-t_{i-1})|\Gamma'(t_{i}^{\star})|\leq M_{i}\Delta t$$ where $t_{i}^{\star}\in(t_{i-1},t_{i})$ and $M_{i}=\sup_{t\in[t_{i-1},t_{i}]}|\Gamma'(t)|$ which exists and is finite since $\Gamma'$ is continuous.  But to me, it's not quite clear how to rigorously construct a cover by boxes from here.  NOTE:  In the proof I mentioned (using integration), essentially you sum the left hand side over all partition intervals of uniform length $\delta$ (which depends on the $\Gamma'$), and "integrate" the right hand side.  Actually, to be more specific, for each $\epsilon&gt;0$ there exists a $\delta&gt;0$ such that $||P||&lt;\delta$ implies $\int_{t_{i-1}}^{t_{i}}|\Gamma'(t)|dt&lt;\epsilon$ (e.g. absolute continuity).  This allows you to define numbers $\epsilon_{i}=\sup_{t,\bar{t}\in[t_{i-1},t_{i}]}|\Gamma(t)-\Gamma(\bar{t})|\leq\epsilon$ so that $\sum_{i=1}^{\#P}\epsilon_{i}\leq||\Gamma'||_{L^{1}([a,b])}$.  Then you can use these $\epsilon_{i}$ to put balls at each point $\Gamma(t_{i})$ of radius (say) $2\epsilon_{i}$, thus giving you an $\epsilon$-small cover.  Again though, this is harder to establish without integration theory. Image of Jacobson Radical is the Jacobson Radical I'm stuck on the following problem:      Let $R$ be a commutative ring with $1$, and suppose $R$ is semilocal with maximal ideals $\mathfrak m_1, ... , \mathfrak m_s$.  Let $\mathfrak a$ be an ideal of $R$. The problem is to show that under the usual surjection $\pi:R \rightarrow R/\mathfrak a$, the Jacobson radical of $R$ is sent to the Jacobson radical of $R/\mathfrak a$.   My attempt: suppose that $\mathfrak m_1, ... , \mathfrak m_t$ contain $\mathfrak a$, but $\mathfrak m_{t+1}, ... , \mathfrak m_s$ do not.  Then $Jac(R/\mathfrak a) = (\mathfrak m_1/ \mathfrak a) \cap \cdots \cap (\mathfrak m_t/ \mathfrak a)$.  Obviously $\pi Jac(R) \subseteq Jac(R/ \mathfrak a)$, and the converse inclusion is equivalent to the following claim: if $x \in \mathfrak m_1 \cap \cdots \cap \mathfrak m_t$, then there exists a $y \in \mathfrak m_1 \cap \cdots \cap \mathfrak m_s$ such that $x - y \in \mathfrak a$.    This should be some application of the Chinese remainder theorem, but I'm not seeing it.  All I have to go on is that there exist $m_i \in \mathfrak m_i$ ($t+1 \leq i \leq s$) with $m_i \not\in I$.  I also haven't found out a way to use the fact that $\mathfrak m_1 ,... , \mathfrak m_s$ are all the maximal ideals of $R$.  I was also thinking about looking at some explicit maps.  If we let $\Gamma$ be the composition $$\frac{(R/\mathfrak a)}{Jac(R/ \mathfrak a)} \xrightarrow{\cong} \frac{(R/\mathfrak a)}{(\mathfrak m_1/\mathfrak a)} \oplus \cdots \oplus \frac{(R/\mathfrak a)}{(\mathfrak m_t/\mathfrak a)} \xrightarrow{\cong} R/\mathfrak m_1 \oplus \cdots \oplus R/\mathfrak m_t$$ then we have a commutative diagram $$\begin{matrix} R/Jac(R) &amp;  \rightarrow &amp; \frac{(R/\mathfrak a)}{Jac(R/\mathfrak a)} \\ \downarrow &amp; &amp; \downarrow \Gamma \\ R/\mathfrak m_1 \oplus \cdots R/\mathfrak m_s &amp; \rightarrow &amp; R/\mathfrak m_1 \oplus \cdots \oplus R/\mathfrak m_t \end{matrix}$$ where the vertical arrows are isomorphisms, and the horizontal arrows are surjections. We all know that in an idealised world all objects accelerate at the same rate when dropped regardless of their mass.  We also know that in reality (or more accurately, in air) a lead feather falls much faster than a duck's feather with exactly the same dimensions/structure etc.  A loose explanation is that the increased mass of the lead feather somehow defeats the air resistance more effectively than the duck's feather.  Is there a more formal mathematical explanation for why one falls faster than the other? Using Squeeze Theorem to show the following limits Consider a function f (x) defined on R satisfying:   for all x distinct 0.  Calculate: lim f(x) with x→∞ and limf(x) with x→−∞   Issue information:   if x → ∞ then we can consider | x | = x  if x → -∞ then we can consider | x | = -x  I can use x² = | x |? Is there a noun that doesn't have a plural in any of its forms? While attempting to answer  @ProgramFox has been trying to come up with an example of a noun with no plural.   The best I've come up with is egotism and Judaism.  ()  Some of the contenders I've already shot down are air, water, and money.  (Airs, waters, and monies are all legitimate plurals.)  My question is, a) Do my two examples have actual plurals?  b) Are there any other non-ism nouns that don't have a plural? How to keep rice in a biryani from burning and sticking together? The bottom part of my  Biryani always  gets burnt, rice always gets sticky but I want each grain separated. How can I prevent burning &amp; stickiness of rice? How to avoid the wrinkle at the poles of the UV sphere? Can I use Named Credential merge fields as URL parameters on the endpoint? I have a named credential as follows:  Name   Test_Credential URL    https://www.somesite.com Identity Type    Named Principal Authentication Protocol    Password Authentication Username   foobar Password   mySomeSiteDeveloperKey Generate Authorization Header   FALSE Allow Merge Fields in HTTP Header    TRUE Allow Merge Fields in HTTP Body     TRUE   The service that I am trying to authenticate against uses URL parameters to pass in the developer key and username. Is there any way to use  to pass those in to the URL?   The Named Credential attributes seem not to get decoded until the request is actually sent, as debugging the endpoint, or headers if I set them per the linked doc, show the un-merged data. I can get the callout to go to the right place, and return the expected error because the auth information is not provided. Anybody know of a "show me my HTTP request" site that I could point to to get a better idea of how this works?  I've tried both of these:  req.setEndpoint('callout:Test_Credential/api/action?id={!$Credential.Password}&amp;account={!$Credential.UserName}');  req.setEndpoint('callout:Test_Credential/api/action' + '?id=' + '{!$Credential.Password}' + '&amp;account=' + '{!$Credential.UserName}');   (Edited to remove the part about $Credential vs. $Test_Credential as I was able to determine that it is the former, and see the merged creds by calling out to .) Redirect based on country Let's say I have a site that has multiple versions for different countries, e.g.:  www.example.com/us www.example.com/uk www.example.com/ca   and I want to use IP geolocation to redirect users who go to www.example.com to the appropriate country version of the site.  What is the best way to do that without confusing crawlers such as Googlebot or running into other SEO type trouble? Let $\pi : \widetilde{X} \longrightarrow X$ be the blow up morphism of $X$ a long of $Y \subset X$, with exceptional divisor $E$ and $\text{dim}Y &gt; 0$,   where $X$ and $Y$ are smooth projectives schemes.   For an answer to a question asked here in the forum (), Blowing up and direct image of line bundle), we have: $$\pi_{*}\mathcal{O}_{\widetilde{X}}(-E) = I_{Y/X} \tag{$**$}$$ here $I_{_{Y/X}}$ denotes the ideal sheaf of $Y$.  $(**)$ was obtained through the following exact sequence: $$0 \longrightarrow \mathcal{O}_{\widetilde{X}}(-E) \longrightarrow \mathcal{O}_{\widetilde{X}} \longrightarrow \mathcal{O}_{E} \longrightarrow  0$$  My doubt is: How to find $\pi_{*}\mathcal{O}_{\widetilde{X}}(-nE)$ for $n \geq 0$? What if $n &lt; 0 $?  Thank you. My ubuntu partition was deleted and I can't boot from either a DVD or USB "Screaming" noise while steaming milk? I have question about ArcGIS and python script. There are following sample code, So I would like to change Another Field , not OIDFieldName. Please let me know how to change Another Field.  import arcpy import itertools  arcpy.env.workspace = r"C:\path\to\workspace"  base_shapefile = "some_shapefile.shp"  oid_field = arcpy.Describe(base_shapefile).OIDFieldName feature_count = int(arcpy.GetCount_management(base_shapefile).getOutput(0))  for (i, j) in itertools.combinations(range(feature_count), 2):     arcpy.Select_analysis(base_shapefile , "combination_%i_%i.shp" % (i, j), '"%s" in (%i, %i)' % (oid_field, i, j))   These sample code is combination for 5C2. I want to create all combiantions 5C3 , 5C4 , 5C5. $ which echo echo: shell built-in command. $ which ls /bin/ls $ which cat /bin/cat   Why is echo not an independent utility like ls, ps, cat etc? Why is it shell specific? Any good reasons? How can I control a servo motor with the PWM output of a microcontroller? I wanted to control the direction of servo motor with a switch. I am using an Atmega32.  Outputs are PIND4 and PIND5 and the input is PINA3, but the two servo motors turn only in 1 direction.  #ifndef F_CPU  #define F_CPU 1000000UL   #endif  #include &lt;avr/io.h&gt; #include &lt;util/delay.h&gt; #include &lt;avr/interrupt.h&gt;   int main(void) { DDRD= 0xFF ; DDRA=0x00; PORTA =0x00; // FAST PWM TCCR1A |= 1&lt;&lt;COM1A0 | 1&lt;&lt;COM1A1 | 1&lt;&lt;WGM11 |1&lt;&lt;COM1B0|1&lt;&lt;COM1B1 ; TCCR1B= 1&lt;&lt;WGM12|1&lt;&lt;WGM13 ; ICR1 = 19999 ;  TCCR1B |=1&lt;&lt;CS10 ; while (1) {    if(PORTA==0x08) //if pina3 is 1    {     OCR1A = ICR1 -2000 ;   //turn servo 1     _delay_ms(1000);          OCR1B=ICR1-2000;    //turn servo 2     _delay_ms(1000);     }     if(PORTA==0x00)// if pina3 is 0     {       OCR1A = ICR1 -900 ; //turn servo 1 in the other direction       _delay_ms(1000);               OCR1B=ICR1-900;     //turn servo 2 in the other direction       _delay_ms(1000);      } }       return 0 ;  } "It is to be discussed", what is the infinitive doing in this sentence? It is to be discussed.   Is be + infinitive forming the future tense here?      You are to be dressed and ready by 8:00.   I was thinking it's almost commanding (or speaking of a command) but this doesn't seem to be the case, as sentences like this are commanding without this structure:     You will be dressed and ready by 8:00.   or      You are going to be dressed and ready by 8:00.   And there's sentences like this that don't imply commandment in any way:     I am to go home tomorrow!   or     I am to be home tomorrow!   (Slightly different meaning, same general sentiment)  So what's going on with the be + infinitive structure?