It shouldn't be possible to improve a suggested edit without changing anything
I improved an edit... but did I actually?
$f(x)=x/1$ is not injective? $C_c^{\infty}(\mathbb R^n)$ is dense in $W^{k,p}(\mathbb R^n)$ Let $G$ be a group and $\{H_i\}_{i\in I}$ a family of subgroups. I would like to find a condition that will imply that $\cup_{i\in I} H_i$ is a subgroup. I know that it's not true in general, I need help to find this condition.  Any help is welcome.  Thanks a lot. $$\dfrac{1}{\sqrt 1}+\dfrac{1}{\sqrt 2}+\dfrac{1}{\sqrt 3}+\cdots+\dfrac{1}{\sqrt n}\geq \sqrt n$$ I want to prove this by Induction  $$n=1 \checkmark\\ n=k \to \dfrac{1}{\sqrt 1}+\dfrac{1}{\sqrt 2}+\dfrac{1}{\sqrt 3}+\cdots+\dfrac{1}{\sqrt k}\geq \sqrt k\\ n=k+1 \to \dfrac{1}{\sqrt 1}+\dfrac{1}{\sqrt 2}+\dfrac{1}{\sqrt 3}+\cdots+\dfrac{1}{\sqrt {k+1}}\geq \sqrt {k+1}$$ so $$\dfrac{1}{\sqrt 1}+\dfrac{1}{\sqrt 2}+\dfrac{1}{\sqrt 3}+\cdots+\dfrac{1}{\sqrt k}+\dfrac{1}{\sqrt {k+1}}\geq \sqrt k+\dfrac{1}{\sqrt {k+1}}$$now we prove that $$\sqrt k+\dfrac{1}{\sqrt {k+1}} &gt;\sqrt{k+1} \\\sqrt{k(k+1)}+1 \geq k+1 \\ k(k+1) \geq k^2 \\k+1 \geq k \checkmark$$ and the second method like below ,   and I want to know is there more Idia to show this proof  ? forexample  combinatorics proofs  , or using integrals ,or fourier series ,....    Is there a close form for this summation ?  any help will be appreciated . It just struck me that Utility Companies world-wide supply domestic users at either 50/60Hz. It could have been selected to be 10 Hz, or 100 Hz ... whatever  Was 50/60 chosen at random? Is there some rationale involved (efficiency, easier to construct/control etc) in restricting domestic supply to either 50/60 Hz?  In other words, if I were designing a power system, why would I pick 50/60 Hz instead of some other frequency? Problem:Let $k$ be an odd positive integer. Prove that $(1+2+3...+n)|(1^k+2^k+3^k+...+n^k)$ for all positive integers $n$.  My Attempt: Proof (By Induction on $k$): Let $P(k)$ be the proposition that $(1+2+3...+n)|(1^k+2^k+3^k+...+n^k)$ for all positive integers $n$ when $k$ is odd. Then $P(1)$ and $P(3)$ are trivially true. So for $k\geq 5$, we assume that $P(k)$ holds. We would like to show that $P(k+2)$ also holds. In order to do that consider the following sums  $$1^{2}(1^k+2^k+3^k+...+n^k)=1^{k+2}+2^k+3^k+...+n^k$$ $$2^{2}(1^k+2^k+3^k+...+n^k)=2^2*1^k+2^{k+2}+2^2*3^k+...+2^2*n^k$$ $$..$$ $$n^2(1^k+2^k+3^k+...+n^k)=n^2*1^k+n^2*2^k+n^2*3^k+...+n^{k+2}$$ Note that $$1^2(1^k+2^k+3^k+...+n^k)+2^2(1^k+2^k+3^k+...+n^k)+...+n^2(1^k+2^k+3^k+...+n^k)\equiv 0\pmod {\frac{n(n+1)}{2}}$$  Therefore, if we denote $\sum_{i=1}^{n}i^2=s$, then    $(t)(1^k+2^k+3^k+...+n^k)-t+\sum_{i=1}^{n}i^{k+2}\equiv 0\pmod {\frac{n(n+1)}{2}}$ which implies that $P(k+2)$ holds. Hence by PMI we are done.  Now I am a bit hesitant in convincing myself whether this proof is correct or not. This is because I have never seen induction performed in this manner. Any reasons as to why this argument falls would be much appreciated. How do I place a bootable ISO on a USB drive? I am having a problem computing the pearson correlation coefficient of data sets with possibly zero standard deviation (i.e. all data has the same value).  Suppose that I have the following two data sets:  float x[] = {2, 2, 2, 3, 2}; float y[] = {2, 2, 2, 2, 2};   The correlation coefficient "r", would be computed using the following equation:  float r = covariance(x, y) / (std_dev(x) * std_dev(y));   However, because all data in data set "y" has the same value, the standard deviation std_dev(y) would be zero and "r" would be undefined.  Is there any solution for this problem? Or should I use other methods to measure data relationship in this case? Calling a function of a module by using its name (a string) I'm using python in QGIS. I have a layer of polygons (its a result of some earlier dissolving and has no overlapping polygons but some holes and islands in holes etc). I want to build a new layer which contains one polygon representing a bbox of the first layer with each polygon cut from it.  Essentially I want to "invert" the layer I have so that where the was polygon, now there is not, and where there was not polygon now there is. All inside a bounding box of the original layer.  I can build a bbox easily and I tried to "subtract" the polygon layer from my new bbox polygon layer but the process slows to a stop and then crashes my PC. Here's the algorythm  processing.runalg("qgis.difference","path/to/polygon.shp","path/to/bbox.shp","path/to/result.shp")   I dont care about attributes (obviously). I just need to cut one set of polygons from one larger polygon.  It seems so simple, I wish I could find a simple command or script to acheive this. As stated at the top I'm looking for a solution in python in QGIS. Suppose there is a message that is encrypted with AES-128-CBC.  The message is as follows, new lines are used to delimit the 16 byte boundary for each block:  Wire funds from: Alice to Bob in   the amount of $ 1   Because this message is encrypted using CBC mode,  any modification of the first block of cipher text would propagate throughout the message.   However, the amount of the fund transfer falls in the last block.  I am under the assumption that an attacker can use the property of  to alter the last block and produce a valid amount that is greater than \$1 in far fewer than $2^{128}$ operations.   Is this a correct assumption?  Is there a name for this attack or do you know of a real world example?    My instinct is that if you just need one byte of the 16 byte block to be of a specific value then it would only require $2^8$ operations because the other 120 bits of the plain text could be any value. For example lets say we wanted the last block of cipher text to decrypt to 9\0 (where \0 denotes a single null byte, and the message would be interpreted as \$9), then it would take only $2^{16}$ operations to find this cipher text block.  Is there something I am missing? Why is gksu no longer installed by default? I have always recommended pressing ALT+F2 and typing for example gksu nautilus or gksudo nautilus both are equivalent; to run graphical programs with root privileges.  However, I have just installed raring on a test system and this no longer works.  gksu and gksudo are not installed as part of a default raring install; though they are in the repositories. In 12.04 at least gksudo was just a symbolic link to gksu  The reasons I have always done this are because while it is often safe to use sudo under some circumstances it can cause problems with graphical applications because gksu has you running the program as root while with sudo you are running as yourself but with elevated privileges.  This can mean you inadvertently change the owner of critical system files.  This can cause problems.  This reasoning is explained better here:         Why was command removed from a fresh install and what should we be using instead? How can I print a variable with padded center alignment? The difference between two periodic functions converges to zero, is this two functions identical? Why can't a Federation Starship's transporter history be used to bring someone back to life? Correct spelling of Polish surname Dziubenofska? My Polish grandmothers last name was Dziubenofska, but I know that's not how it is spelt! I'm looking for the proper spelling.   Can you help find it for me?  She came over to Ellis Island and they changed it on her to dziuk. This is only a prefix to the full name. Nielsen & Chuang Exercise 2.2 - “Matrix representations: example” Reproduced from Exercise 2.2 of Nielsen &amp; Chuang's Quantum Computation and Quantum Information (10th Anniversary Edition):     Suppose $V$ is a vector space with basis vectors $|0\rangle$ and $|1\rangle$, and $A$ is a linear operator from $V$ to $V$ such that $A|0\rangle = |1\rangle$ and $A|1\rangle = |0\rangle$. Give a matrix representation for $A$, with respect to the input basis $|0\rangle, |1\rangle$, and the output basis $|0\rangle, |1\rangle$. Find input and output bases which give rise to a different matrix representation of $A$.   Note: This question is part of a series attempting to provide worked solutions to the exercises provided in the above book. Was this an oversight?  Or is it to do with the JVM? How do I get label of the selected item from a lightning:select drop down list? I would like to get the value of the label of the selected drop down item.  In this case, the label is the Name of the Contact.  &lt;aura:component&gt;  &lt;aura:attribute name="contacts" type="Contact[]"/&gt;  &lt;lightning:select label="Contact" aura:id="contactList" name="contactList"&gt;     &lt;aura:iteration items="{!v.contacts}" var="c" indexVar="index"&gt;        &lt;option value="{!c.Id}"&gt;{!c.Name}&lt;/option&gt;    &lt;/aura:iteration&gt;  &lt;/lightning:select&gt;  &lt;/aura:component&gt;   I know how to get the value:  component.find("contactList").get("v.value");   But how do I get the label? How does one write on the outside of a circle?  I am looking for sample files/packages which imitate a magazine lay-out. Specifically a page with two columns having a circular picture between the two columns.    Example of the desired layout, created using Word: