Let $X$ be a vector space, equipped with two norms $\|\cdot\|_1$ and $\|\cdot\|_2$ Which are equivalent.   What is the easiest way to prove that these two equivalent norms induce same topology?
Note that (1) all norms on finite dimensional vector space are equivalent.  And here (2) norms on vector space are equivalent iff they have same topology.   Question : How can we prove (2) ? Please recommend reference  Thank you in anticipation
Due to the implementation of Java generics, you can't have code like this:   public class GenSet&lt;E&gt; {     private E a[];      public GenSet() {         a = new E[INITIAL_ARRAY_LENGTH]; // error: generic array creation     } }   How can I implement this while maintaining type safety?  I saw a solution on the Java forums that goes like this:  import java.lang.reflect.Array;  class Stack&lt;T&gt; {     public Stack(Class&lt;T&gt; clazz, int capacity) {         array = (T[])Array.newInstance(clazz, capacity);     }      private final T[] array; }   But I really don't get what's going on. Recovering from a move of /home when it's encrypted (ecryptfs) I had a 20GB / partition at work that dual-booted with Windows and it quickly ran out of space. I usually do 20GB / and 160GB (or whatever the rest of the hard drive is) as /home, so I intended on recovering the space Windows was taking up and turning that into /home.  That went fine and I was able to mount the partition and rsync all my files over.  However, I forgot that the partition was encrypted, so now all my files are sitting in /home/.ecrypt and /home/$USER is empty.  How can I recover my files from /home/.ecrypt and move them to /home/$USER?  I followed this tutorial: What are the basis for co-countable and co-finite topologies I've recently updated to Ubuntu 15.04 and after installing the latest nvidia drivers (346.59) which worked on ubuntu 14.10 seems to have caused some trouble on boot.  All I get is the following when booting:   A purple screen for about a second and then  [0.514409] ACPI PPC Probe failed. Starting version 219  A black screen with a message of my monitor saying : No Signal detected.   I am using a GTX 970. Showing that $\frac{\sqrt[n]{n!}}{n}$ $\rightarrow \frac{1}{e}$ Show:$$\lim_{n\to\infty}\frac{\sqrt[n]{n!}}{n}= \frac{1}{e}$$  So I can expand the numerator by geometric mean. Letting $C_{n}=\left(\ln(a_{1})+...+\ln(a_{n})\right)/n$. Let the numerator be called $a_{n}$ and the denominator be $b_{n}$ Is there a way to use this statement so that I could force the original sequence into the form of $1/\left(1+\frac{1}{n}\right)^n$ Are Goodberries affected by the Life domain class features? Using Georeference information stored in tfwx (from ArcGIS) in QGIS? I have georeferenced some images (ArcGIS -> Update Georeferencing) which creates a *.tfwx file with the new parameters for georeferencing.   I renamed it to *.tfw but QGIS is not able to read the georeferencing information.   Does anybody has any useful tips how to fix this? I've got some lenses of brand X, but I've just got a new camera of brand Y. How can I tell if it's possible to use my lenses on my camera? I accept that I may lose a lot of automatic functionality like autofocus, aperture control and metering when using an adapter to connect the two. Is there a way to set up a default folder for Windows Explorer? I often use Win + E to pop up Windows Explorer, and then I usually have to navigate down into a few network folders to find my primary directory. Is there a way to set this folder as the default so that Windows Explorer always opens to this folder? I can't not be pregnant This is a quote from a  where she talks about her miscarriage:     "It was awful," she told the Daily Mail's You magazine. "It happened really late, and I wasn't expecting it at all. "For a long time I just couldn't understand that it had happened and why. I was like 'But I can't not be pregnant. I'm already past the first trimester, and I have three car seats in my head.' To this day I feel like I'm missing that kid. But it wasn't meant to be and you have to trust the universe. "   What does she mean by saying "I can't not be pregnant?" Is this "can't not" construction a grammatical and common usage? When would you use the Builder Pattern? What are some common, real world examples of using the Builder Pattern?  What does it buy you?  Why not just use a Factory Pattern? What is the fastest algorithm for finding the natural logarithm of a big number? Let $a_n$ be the number of those permutation $\sigma $ on $\{1,2,...,n\}$  such that $\sigma $ is a product of exactly two disjoint cycles. Then find $a_4$ and $a_5$.   Calculating $a_4$: Possible cases which can happen are $(12)(34),(13)(24),(14)(23)$, any cycle of the form $(123)$ or $(12)$ i.e. two-cycles and three cycles thus we have in total $3+\frac{1}{3}4P_3+\frac{1}{2}4P_2=3+8+6=17$ but the correct answer is given to be either $11$ or $14$.  Where am I wrong? Please help. Inverse of a Positive Definite Let K be nonsingular symmetric matrix, prove that if K is a positive   definite so is $K^{-1}$ .   My attempt:  I have that $K = K^T$ so $x^TKx = x^TK^Tx = (xK)^Tx = (xIK)^Tx$ and then I don't know what to do next. I have been/am being taught that end punctuation should always go inside quotes. For example, you are supposed to write:     Marvin thought it was "awful."   The problem is I do not see how does this make sense. Intuitively, I always wrote:     Marvin thought it was "awful".   as that makes more logical sense — you want a quote to be an exact replication of what somebody else said, so why should you add punctuation inside?  I always thought it made more sense to not touch the quote and add anything after or before if it must be added.  So, why should I put end punctuation inside quotes? Deriving even odd function expressions What is the logic/thinking process behind deriving an expression for even and odd functions in terms of $f(x)$ and $f(-x)$?  I've been pondering about it for a few hours now, and I'm still not sure how one proceeds from the properties of even and odd functions to derive: $$\begin{align*} E(x) &amp;= \frac{f(x) + f(-x)}{2}\\ O(x) &amp;= \frac{f(x) - f(-x)}{2} \end{align*}$$ What is the logic and thought process from using the respective even and odd properties, $$\begin{align*} f(-x) &amp;= f(x)\\    f(-x) &amp;= -f(x) \end{align*}$$   to derive $E(x)$ and $O(x)$?    The best I get to is: For even: $f(x)-f(-x)=0$ and for odd: $f(x)+f(-x)=0$  Given the definition of $E(x)$ and $O(x)$, it makes a lot of sense (hindsight usually is) but starting from just the properties. Wow, I feel I'm missing something crucial. Do I need to activate the firewall? I only use Ubuntu for a home desktop use? In my Computer science class I was given a problem where I have to simulate a bouncing ball using "real physics". I have been trying to find a equation that will simulate the height of the bounce given a gravity and an arbitrary mass. And I will need to calculate the next bounce and it's height. A lot of the equations I've found require a time. But I don't really care about a time, all I want is to get the next height after the previous bounce until it finally hits height of 0 or close to it. I haven't taken a physics class since high school( 5 years ago) and that was basic physics. My non-native English speaking friend just asked me: "Data is..." or "Data are..."?  I said both but that's because I've been desensitized from reading/writing both (especially from writing code and adding quick comments).  My question: Is it acceptable to utilize either for a university paper? Or is one safer than the other (when confronted with stickler professors)?    Related questions: Find all continuous functions satisfying $\int_0^xf=(f(x))^2+C$ for some constant $C \neq 0$. Find all continuous functions $f$ satisfying   $$\int_0^xf=(f(x))^2+C$$ for some constant $C \neq 0$, assuming that $f$ has at most one $0$.  I have a question about the solution to this problem. It says that   Clearly $f^2$ is differentiable everywhere, its derivative at $x$ is $f(x)$ (I'm assuming that this is because the left hand side is differentiable, the right hand side must be differentiable as well). So $f$ is differentiable at $x$ whenever $f(x) \neq 0$, and  $$f(x)=2f(x)f'(x),$$ so $f'(x) = \frac{1}{2}$ at such points. Thus, $f= \frac{1}{2}x+b$ for some $b$ on any interval where it is non-zero; if $f$ has a zero, this gives two possible solutions, with two possible values of $b$, but since $f$ is assumed continuous, they must be the same. So we need $$\int_0^x(\frac{1}{2}t+b)dt=(\frac{1}{2}x+b)^2+C$$  so we must have $b=\sqrt{-C}$ or $-\sqrt{-C}$, leading to two solutions for $C \lt 0$.  I don't understand the bolded statements. First, how does the differentiability of $f^2$ guarantee that $f$ is differentiable, and why only at $x$ where $f(x) \neq 0$?  Next, why are there two possible solutions if $f$ has a zero, but since $f$ is continuous they must be the same?  Finally, does the last statement mean that if $C \gt 0$, then there is no solution, since we're restricted to real numbers here?  I would appreciate it if anyone answers my questions.