This particular question was asked in quiz yesterday and I was unable to solve it.  Suppose f and g are entire functions and g(z) $\neq$0 for all z $\epsilon \mathbb{C} $ . If |f(z) |$\leq$ |g(z) | , the which one of them is true.  1.f is a constant function. 2.f(0) =0 . 3.for some C $\epsilon \mathbb{C} $ , f(z) = C g(z) . 4.f(z) $\neq$ 0 for all z $\epsilon \mathbb{C} $ . 4 th option can be removed by taking f(z) =0 and g(z) =2 . But I am unable to find any way to remove rest of options.
Let $f$ and $g$ be entire functions and $g(z)\neq 0$ for all $z\in \mathbb{C}$. If $|f(z)|\le |g(z)|$, can we say $f$ is constant?  Liouville theorem says that an entire bounded function is constant, but g is not given bounded here. So I think it should be false. What else can we say about $f$?
How to reinstall Windows 10 on new hard disk How did Sauron remain in control of the Nazgûl? The One Ring gave Sauron control of the Nine, yet when it was taken from him he still possessed the power to control them (or, at least he regained it towards the end of the third age).  Why is this?   mentions that he drew the Nine to him, but not how. How do I delete a board in Trello? Correlation between dichotomous and continuous variable I am trying to find the correlation between a dichotomous and a continuous variable.  From my ground work on this I found that I have to use independent t-test and the precondition for it is that the distribution of the variable has to be normal.  I performed Kolmogorov-Smirnov test for testing the normality and found that the continuous variable is non-normal and is skewed (for about 4,000 data points).   I did the Kolmogorov-Smirnov test for the entire range of variables. Should I split them into groups and do the test? I.e., say if I have risk level (0 = Not risky, 1 = Risky) and cholesterol levels, then should I:    Divide them into two groups, like  Risk level =0 (Cholestrol level) -&gt; Apply KS Risk level =1 (Cholestrol level) -&gt; Apply KS  Take them together and apply the test? (I performed it on the whole dataset only.)   After that, what test should I do if it is still non-normal?    EDIT: The above scenario was just a description I tried to provide for my problem. I have a dataset which contains more than 1000 variables and about 4000 samples. They are either continuous or categorical in nature. My task is to predict a dichotomous variable based on these variables (maybe come up with a logistic regression model). So I thought the initial investigation would involve finding the correlation between dichotomous and a continuous variable.  I was trying to see how the distribution of the variables are and hence tried to go to t-test. Here I found the normality as an issue. The  Kolmogorov-Smirnov test gave a significance value of 0.00 in most of these variables.  Should I assume normality here? The skewness and kurtosis of these variables also show that the data is skewed (>0) in almost all cases.  As per the note given below I will investigate the point-biserial correlation further. But about the distribution of variables I am still unsure. Let $A$, $B$ be two reduced $k$-algebras. Then if an element of the form $$\sum a_{i}\otimes b_{j}$$ is nilpotent, we can compose it with any $k$-homomorphism $f$ from $A$ to $k$ to get a homomorphism from $A\otimes B$ to $B$. This map's image is $$\sum f(a_{i})\otimes b_{j}$$must be nilpotent as well. So if we assume $b_{j}$ is linearly independent in the first place, we may reason that $f(a_{i})=0$. This showed that $a_{i}$ are in the Jacobson radical of $A$. However, we only know $\mathrm{Nil}(A)=0$, and it is not clear to me how to pass from the Jacobson radical to the nilradical without using some commutative algebra machinery.   Here is what I think that might work. If $A$ is of finite type, then by Nullstellensatz we can assert the above is true. But if $A$ is not of finite type, then I can write $A=k[x_{1},...,x_{n}][X]$, where $n$ is the transcendence degree of $k(A)$ and $k[X]$ is finite over $k$. Then I should be able to pass from $k[X]$ to $A$ by inductive argument on the transcendental degree of $k(A)$, since prime ideals are maximal in a UFD (I am working with $k[x]$ by introducing $[x]$).   The proof I have for the case of finite type is not satisfying. Given $I\subset k[X]$, using Nullstellensatz one can show that $$I_{X}V_{X}[I]=\sqrt{I}$$ where $V_{X}[I]$ is the set of all points in $X$ on which $I$ vanishes, and $I_{X}(V_{X}[I])$ be the ideal in $k[X]$ which vanishes on $V_{X}[I]$. Granting this, let $I=(a_{i})$, then we want to show $I=0$, which is $V_{X}[I]=X$. This follows since if $U\subset W$, then $V_{X}[U]\supset V_{X}[W]$. Since $V_{X}[I]\supset V_{X}[m_{x}]$ for all maximal ideal corresponding to points, it must be $X$. This proof is unnecessarily winding by using the ideal-variety correspondence. Ideally I should be able to construct a pure algebraic proof.   The thing not clear to me is whether the above strategy really works. Namely I am not sure if I can really induct on the transcendental degree of $k(A)$. To make matters simple assume we showed the statement holds for $A$ with transdental degree $&lt;n$. For $A$ with transdental degree $n$ we can thus write it as $k[A'][x]$. Then any maximal ideal $m'$ of $k[A'][x]$ must correspond to a maximal ideal $m=m'\cap k[A']$ in $k[A']$. But it seems not possible to show that any prime ideal is maximal. Indeed considering the simple example of $k[x,y]$, where $(x)$ is a prime ideal but not maximal since $k[x,y]/(x)=k[y]$ is not a field. The problem is in the inducting process the base ring is not neccessarily a field, and so this strategy cannot work.   I am thinking maybe I can show $m'=m[f(x)]$ with $f(x)$ some irreducible polynomial in $k[A'][x]$. Because we know $k[A'][x]$ is noetherian by Hilbert's basis theorem, so in particular $m'$ must have a finite generating set. And we can reduce the number of irreducible polynomial to 1 since $m'$ is a maximal ideal. But I still do not know how to use this to finish the proof. Second order partial of $f(x,y)=\frac{xy(x^2-y^2)}{x^2+y^2}$ Consider the function $f(x,y)=\dfrac{xy(x^2-y^2)}{x^2+y^2}$ for $(x,y) \neq (0,0)$, $f=0$ otherwise.  I have to compute $\dfrac{d^2f}{dydx}(0,0)$.  I know that I have to calculate $\frac{df}{dx}$ first. But that is ,  $\frac{df}{dx} = \frac{\partial f}{\partial y}\frac{dy}{dx}+\frac{\partial f}{\partial x}$. When I put it in wolframalpha, it gives me this calculation, and also just $\frac{\partial f}{\partial x}$ as an alternative form.   Why do I just ignore the y'? I don't know what y(x) is, as a function of x.  y and x are independent functions, no? How does the /proc//exe symlink differ from ordinary symlinks? If I start a process and then delete the binary of it, I can still recover it from /proc/&lt;pid&gt;/exe:  $ cp `which sleep` . $ ./sleep 10m &amp; [1] 13728 $ rm sleep $ readlink /proc/13728/exe                            /tmp/sleep (deleted) $ cp /proc/13728/exe ./sleep-copy $ diff sleep-copy `which sleep` &amp;&amp; echo not different not different $ stat /proc/13728/exe    File: ‘/proc/13728/exe’ -&gt; ‘/tmp/sleep (deleted)’   Size: 0           Blocks: 0          IO Block: 1024   symbolic link   On the other hand, if I make a symbolic link myself, delete the target and attempt to copy:  cp: cannot stat ‘sleep’: No such file or directory   /proc is an interface to the kernel. So does this symbolic link actually point to the copy loaded in memory, but with a more useful name? How does the exe link work, exactly? Imagine you're trapped inside a free falling elevator. Would you decrease your impact impulse by jumping during the fall? When? canonical-livepatch, Doesn't seem to patch anything Profile tab hyperlink should match name change I applied for a B1 visa to attend a conference in the USA in 2013. I had another nationality (Non- VWP) at the time and my B1 application rejected under section 306 of border security. I now can apply for ESTA under my new nationality. Can my ESTA application get approved? If not and I need to apply for visa is it every time I need to travel to stated or is it just the first time? Why did Sirius Black's family live among Muggles? How do I create my own URL protocol? (e.g. so://...) I have seen:   http://www... ftp://blah.blah... file://blah.blah... unreal://blah.blah... mailto://blah.blah...   What is that first section where you see http and the like called?  Can I register my own? Can "mustn't" be used for conclusions? I heard this sentence in an American film a while ago as I was watching it on DVD (the part after but is verbatim):  "I'm doing my best but I mustn't be doing it right."  This is something I occasionally hear in American films: phrases like "he mustn't have done it" or "she mustn't be studying now", where a logical conclusion is expressed. So far I've thought that the normal thing to say is "he can't have done it" or "she can't be studying now".  My question is: Can mustn't be used to express a logical conclusion when the speaker is certain that something didn't happen or isn't happening, at least in informal speech? Is this an American usage? (I've never heard this usage in British English, but this doesn't mean it doesn't exist.) Is there a change going on in the usage of the particular modal verb?  Note: All the references I've checked don't even mention this use of mustn't. Google books aren't of help either.   EDIT: I should clarify that I'm asking this question because if I wanted to express that I'm certain I'm not doing something right (as in the sentence quoted from the film) I'd say "I'm doing my best but I can't be doing it right". I would think that the use of mustn't/must not wouldn't be standard usage (although the meaning is perfectly clear to me; I never mistook it for an injuction). Is there a specific reason for having a very short window for passengers to get to the boarding gate before closing it? I am not a frequent traveler but it's been like this with all the flights I have been on. Here is the most recent example:  Below is my boarding pass    As you can see Gate Closes at 20:20  Here is the departures board at 18:32, clearly it says that they will announce the boarding gate at 20:10  ?!    I was hoping they will do that earlier but at 19:41 it's still the same    In fact they announced it at 20:11 but since I was freaking out and had to leg it to the gate I didn't have the time to take a photo.  You might think 9 minutes is enough, well below are the times required to reach the gates from the closest place to them     And that's for a healthy young person, no way my grandmother could be able to be at the gates within 9 minutes.  Is there a specific reason? it's not like they allocate the gate dynamically! The staff must have known about the gate hours ago as I don't expect the pilot to drive the airplane from the garage to the gate within 9 minutes? Is there any difference between like best and like the best? Is there a difference between these two sentences? Are they the same?   What color do you like the best? What color do you like best?   Is there a difference in their meaning? Story about a deadly alien creature on another planet that has some chemical in its body that makes it faster I know that a similar question has been asked in:        although I didn't really understand the answer and couldn't really make it work.  However, I have tried something different. I decided to read the manual page for ls and it mentions different environment variables that can be set when dealing with the ls command. If one goes to the -G option it says to look at the CLICOLOR environment variables. I did and that one links you to documentation for the LSCOLORS environment variable. I went to it and it mentions:     "The value of this variable describes what color to use for which attribute when colors are enabled with CLICOLOR. This string is a concatenation of pairs of the format fb, where f is the foreground color and b is the background color. "   It also mentions that the default value of LSCOLORS is "exfxcxdxbxegedabagacad" and that each part specifies the color of some specific thing in the ls command. For example, the first pair ex in example specifies the color of directories. the second fx the color of symbolic links etc... e means blue and x means default "foreground".  1) First what does this mean: "when colors are enabled with CLICOLOR"?  2) What does "foreground" mean?  What must environment variables be for the ls command to color the output however I desire?    I also tried:  ls --color   But it threw me the following error:  ls: illegal option -- - usage: ls [-ABCFGHLOPRSTUWabcdefghiklmnopqrstuwx1] [file ...]   The same error happened when I did ls --color=auto: it alone also throws an error. I have no idea why, and am not sure if it was due to iTerm2 or OS X or why that happened. It seems it works on other systems... Make two vector points move together (Sketch or Illustrator) $M + N$ and $M \cap N$ are finitely generated $A$-modules implies $M$ and $N$ finitely generated using exact sequences Let $0 \to M_1 \to M \to M_2 \to 0$ be an exact sequence of $A$-modules.  i) Prove: If $M_1$ and $M_2$ are finitely generated, then $M$ is too.  ii) Let $M$ and $N$ be sub-modules of an $A$-module $L$. Assume that $M+N$ and $M\cap N$  are finitely generated $A$-modules. Prove that $M$ and $N$ are finitely generated $A$-modules.  If have proved the first part, but i don't see how to do the second part. I tried to use the first part, so to find an exact sequence with $M$ in the middle and finitely generated modules in place of $M_1$ and $M_2$, but I can't find a surjective map $M \to P$ (for some f.g. $A$-module) in the sequence:  $$0 \to M\cap N \to M \to P \to 0$$  any hints/help would be appreciated.