If log 1 to the base 1 can be any number. So we can invent a new mathematical idea called the &quot;any&quot;. If log 1(1) = &quot;that idea&quot;, so is that wrong?
What is $\log(1)$ to the base of $1$? My teacher says it is $1$. I beg to differ, I think it can be all real numbers! i.e., $1^x = 1$, where $x\in \mathbb{R}$.  So I was wondering where I have gone wrong.
Problem statement  Prove that if $A$ is an uncountable set and $B$ is a countable set, then $A\setminus B$ must be uncountable.  What I think  The statement does not mention $A$ and $B$ relationship. I think there are two possibilities:   If $A \cap B = \emptyset $, then $A\setminus B$ is trivially uncountable If $A \cap B = B$, then $B \subset A$ and as a bijection can not be made between $A\setminus B$ and $\mathbb{N}$, $A\setminus B$ is uncountable.   And there is where I'm stuck. How can I prove that a bijecton can't be made?  TIA. Given an InputStream called in which contains audio data in a compressed format (such as MP3 or OGG), I wish to create a byte array containing a WAV conversion of the input data. Unfortunately, if you try to do this, JavaSound hands you the following error:  java.io.IOException: stream length not specified   I managed to get it to work by writing the wav to a temporary file, then reading it back in, as shown below:  AudioInputStream source = AudioSystem.getAudioInputStream(new BufferedInputStream(in, 1024)); AudioInputStream pcm = AudioSystem.getAudioInputStream(AudioFormat.Encoding.PCM_SIGNED, source); AudioInputStream ulaw = AudioSystem.getAudioInputStream(AudioFormat.Encoding.ULAW, pcm); File tempFile = File.createTempFile("wav", "tmp"); AudioSystem.write(ulaw, AudioFileFormat.Type.WAVE, tempFile); // The fileToByteArray() method reads the file // into a byte array; omitted for brevity byte[] bytes = fileToByteArray(tempFile); tempFile.delete(); return bytes;   This is obviously less desirable. Is there a better way? Custom unique constraint, only enforced if one column has a specific value A question about $f(x)\equiv 0$ What are the minimum root filesystem applications that are required to fully boot linux? where instance variables get initialized Is there more to an interface than having the correct methods So lets say I have this interface:  public interface IBox {    public void setSize(int size);    public int getSize();    public int getArea();   //...and so on }   And I have a class that implements it:  public class Rectangle implements IBox {    private int size;    //Methods here }   If I wanted to use the interface IBox, i can't actually create an instance of it, in the way:  public static void main(String args[]) {     Ibox myBox=new Ibox(); }   right? So I'd actually have to do this:  public static void main(String args[]) {     Rectangle myBox=new Rectangle(); }   If that's true, then the only purpose of interfaces is to make sure that the class which implements an interface has got the correct methods in it as described by an interface? Or is there any other use of interfaces? How would you evaluate $\liminf\limits_{n\to\infty} \ n \,|\mathopen{}\sin n|$ How would you evaluate the limit inferior of the sequence $n\,|\mathopen{}\sin n|$? That is, $$\liminf\limits_{n\to\infty} \ n \,|\mathopen{}\sin n|$$    Edit. Let $\mu$ be the irrationality measure of $\pi$. Since $\mu$ is not known, I will split the question:   Assuming $\mu &gt; 2$, what is $\liminf\limits_{n\to\infty} \ n\,|\mathopen{}\sin n|$? Assuming $\mu = 2$, what is $\liminf\limits_{n\to\infty} \ n\,|\mathopen{}\sin n|$?     I kind of like its graph... I have a question related to modeling short time-series. It is not a question , but how. What method would you recommend for modeling (very) short time-series (say of length $T \leq 20$)? By "best" I mean here the most robust one, that is the least prone to errors due the fact of limited numbers of observations. With short series single observations could influence the forecast, so the method should provide a cautious estimate of errors and possible variability connected to the forecast. I am generally interested in univariate time-series but it would be also interesting to know about other methods. Continuity of absolute value of a function Let $f(x)$ be a continuous function. Prove that $\left|f(x)\right|$ is also continuous.  Is it correct to say that, by the reverse triangle inequality, $\left|f(x)-f(c)\right| \geq \left|f(x)\right|-\left|f(c)\right|$ in all cases, so we will always have that for all $\left|x-c\right|&lt;\delta$ implies $\left|f(x)-f(c)\right|\leq \epsilon$?  I am not sure if my solution adequately uses the functional form, since I just used the equation and not my actual function. Please help with the proper solution! We are in the early stages of development of the official Stack Exchange Android app and we need your help with testing. To set a few expectations: there will be bugs, crashes, and missing features, all of which will be fixed in .    We are initially targeting Android devices running  (Android 4.1 or higher) starting with phones (up to ~5 inch) and eventually moving our way to tablets.    To register for alpha testing .  After you register we will send you an invite to a private Google community where you can find a link to download our app directly from the  store. The invitations will be sent out in waves, so if you don’t get an invite immediately, be patient.  Crash reports will be automatically reported back to us using . Bugs or feature requests should be reported here on meta and tagged . Please do not use the Google community to report bugs.  As always, we welcome your feedback and are committed to building the best mobile app possible for browsing the Stack Exchange sites. Are virtual particles only a fictive tool in equations? There is no "action at a distance" in nature. Attraction of a piece of iron by a magnet, attraction between distant electric charges of opposite sign, have to be mediated by something. The  are proposed as an explanation.  If they have an observable effect, it seems like they must exist. If so, why do we call them virtual? It can't be that they have short lifetimes, as there are short-lived particles (resonances) that we don't call "virtual."  On the other hand, some people say that the so-called virtual particles only exist on paper, or that they are just a technique to help us handle our equations.  So what is the meaning of the word "virtual" here? Do these particles really exist, or not? Java: Cannot find symbol? Simplest Example of a Poset that is not a Lattice A partially ordered set $(X, \leq)$ is called a lattice if for every pair of elements $x,y \in X$ both the infimum and suprememum of the set $\{x,y\}$ exists. I'm trying to get an intuition for how a partially ordered set can fail to be a lattice. In $\mathbb{R}$, for example, once two elements are selected the completeness of the real numbers guarantees the existence of both the infimum and supremum. Now, if we restrict our attention to a nondegenerate interval $(a,b)$ it is clear that no two points in $(a,b)$ have either a suprememum or infimum in $(a,b)$.   Is this the right way to think of a poset that is not a lattice? Is there perhaps a more fundamental example that would yield further clarity? Why commutator subgroup is normal to G? If $G$ is a group and $G'$ is generated by $\{xyx^{-1}y^{-1}|x,y\in G\}$, then $G'\trianglelefteq G$ and  $G/G'$ is Abelian.  At first, I thought this is easy because I thought $g[x,y]g^{-1}=[gxg^{-1},gyg^{-1}]$. But it was a mistake. I could not prove that it is also subgroup of G and $G/G'$ is abelian. I need help. Time dependent Schrödinger equation with time independent potential and separation of variables suppose we have a potential that's independent of time $V(x,t) = V(x)$ so in Schrödinger equation we get: $$i\hbar \frac{\partial \Psi (x,t)}{\partial t}=-\frac{\hbar^2}{2m}\frac{\partial^2 \Psi (x,t)}{\partial x^2}+V(x)\Psi(x,t)$$ since the LHS involves a variation of $\Psi$ with $x$ and the LHS involves a variation of $\Psi$ with $t$ can we say that a basis for the solutions will be with the form $\Psi(x,t)=\psi(x)T(t)$? Is the reason why the time evolution operator is unitary based on purely physical arguments, i.e. that the physical processes that an isolated system undergoes shouldn't depend on any particular instant in time (homogeneity of time); thus two experimenters who conduct the same experiment starting from the same initial state, but at different times, should have the same probability amplitude for that state?! Or is there some mathematical argument as well?  Also, is the reason why the time evolution operator is linear implied by the superposition principle (as an arbitrary state can be expressed as a linear combination of basis states the operator should act linearly as otherwise the state as a whole would evolve differently to the superposition of states that it was initially represented by)?! I try to create a common extent for rasters in QGIS 2.99. Here is my script:  import processing  processing.runalg("script:unifyextentandresolution",                   "/qgis_data/rasters/Image2.tif;                   /qgis_data/rasters/Image1.tif",                  -9999,"/qgis_data/rasters",True)   However, processing.runalg is being used for QGIS 2.99. I need it for QGIS 2.99 Looking for intro to Conformal Bootstrap Just wanted some feedback to ensure I did not make any mistakes with this proof. Thanks!  Since $G$ is abelian, every subgroup is normal. Since $G$ is simple, the only subgroups of $G$ are $1$ and $G$, and $|G| &gt; 1$, so for some $x\in G$ we have $x\neq 1$ and $\langle x\rangle\leq G$, so $\langle x\rangle = G$. Suppose $x$ has either infinite order or even order. Then $\langle x^2\rangle \leq G$ but $\langle x^2\rangle \neq \langle x \rangle$, a contradiction. So $x$, and therefore $G$, has odd order, and by Feit-Thompson, $G\cong Z_p$ for some prime $p$.  Edit: Thanks, I see that Feit-Thompson is too much.  Since $G$ is abelian, every subgroup is normal. Since $G$ is simple, the only subgroups of $G$ are $1$ and $G$, and $|G| &gt; 1$, so for some $x\in G$ we have $x\neq 1$ and $\langle x\rangle\leq G$, so $\langle x\rangle = G$. Suppose $x$ has infinite order. Then $\langle x^2\rangle \leq G$ but $\langle x^2\rangle \neq \langle x \rangle$, a contradiction. So $x$, and therefore $G$, has finite order. Suppose $x$ has composite order $n$ so for some $p &gt; 1$ that divides $n$, $\langle x^p \rangle$ is a proper non-trivial subgroup of $G$, so $G$ is not simple. So $G$ is a cyclic group of prime order.