I was reading about how we can ascertain whether a cosmic object is receding from or approaching us by studying the doppler shift of its spectrum- I mean the redshift and blue-shift.  But we have no idea about the actual frequency of light emitted by the cosmic object. Then how can we say that the light is getting red-shifted or blue-shifted?  Some texts suggest that the frequency of the emitted radiation depends on the hydrogen, helium, lithium, carbon etc. content of the cosmic object. But we can at most make a guess about its percentage composition. Then how can we identify what the frequency of the emitted radiation is?  And, if the above mentioned way is not how we understand the doppler shift of the light emitted by cosmic objects, how do we confirm that the light is getting doppler shifted?  EDIT:  My question is almost similar to the duplicate question. It has solved my problem to a good extent but it doesn't totally resolve my problem.  Say we are receiving a pattern of lines from a particular star and the lines are characteristic of an element $X_1$. What happens if the red-shifted line pattern of $X_1$ resembles the pattern of lines for some other distinct element $X_2$?  How do we understand if the star is emitting light characteristic of the element $X_1$ or $X_2$?   If the question still looks unclear, you can refer to the comments section where I have tried to elaborate my query to John Rennie.
Doppler redshift of distant galaxies gave first hint that the universe is expanding. I am curious to know how this redshift is actually measured and interpreted from observation. Suppose I observe some  visible line say red with wavelength \lambda_1.  This is not enough to tell that this line is redshifted, i.e., the  original line emitted from that galaxy had a smaller wavelength. How do I know the value of that particular line when it was emitted from that galaxy? It appears that mere observation of a line is not enough. So how do we proceed?
Not able to install Win Xp on Lenovo Ideapad z570 I want to install Win Xp on my Ideapad Z 570 laptop but i always get blue screen with message with some drive related problem which asks me to run CHKDSK command. I tried with that but no luck. Is there any way i can install Win Xp with SP 2 ?  Edit: i searched across the sites and came to know there could be some Compatibility problems, and i could fix it by setting HDD mode to IDE; will it solve the problem? I installed Drush for the first time (so it may be a configuration error) and I'm trying to update to the latest core version of Drupal.  Drush is giving me this error - I'm on an OSX box.  dev5:clientnamedirectory my.name$ drush pm-update projects drupal-7.25 &lt;h1&gt;Additional uncaught exception thrown while handling exception.&lt;/h1&gt;  &lt;h2&gt;Original&lt;/h2&gt;&lt;p&gt;PDOException: SQLSTATE[HY000] [2002] No such file or directory in drupal_get_installed_schema_version() (line 155 of /Users/my.name/Sites/proface/includes/install.inc).&lt;/p&gt;  &lt;h2&gt;Additional&lt;/h2&gt;  &lt;p&gt;PDOException: SQLSTATE[HY000] [2002] No such file or directory in system_list() (line 165 of /Users/my.name/Sites/clientnamedirectory/includes/module.inc).&lt;/p&gt;&lt;hr /&gt;  Drush command terminated abnormally due to an unrecoverable error. I'm using Rules 7.x-2.6 and the way it works has changed a bit since I last saw.  I have a rule as follows:  EVENT A node has been flagged, under "Report content"  CONDITIONS None  ACTIONS Send mail to [site:mail]   However, I'd like to, instead of sending directly to [site:mail], send an email to all users flagged with an "admin" role.   Now I know that I'll need to provide a loop do so, but my sticking point is generating the list of users to pass to the loop. So far, I have attempted to add a new action:   "Fetch entity by property" Entity type: User Property: User roles Value: admin   But this doesn't seem to give me what I want in data selection, when passing it to the loop. Can someone recommend what they would set here? I read a lot about the classical twin paradox recently. What confuses me is that some authors claim that it can be resolved within , others say that you need . Now, what is true (and why)? Only squares instead of letters and numbers are displayed in my calculator and other apps in Ubuntu 18.04 I have downloaded Python 3.3 from the official site but no idea how to install it.   I'm using Ubuntu 12.04 Opposite of Alternatives What are the pros and cons of engineered hardwood? Are there any downsides to engineered hardwood? I'm looking to install hardwood in dining room and kitchen areas. What make is my BMX frame? I have a BMX forsale, but I would like to know what make the frame is so I can determine how much it is worth. Does anyone recognise this symbol on the frame?  Thanks for your help Jon Proof for two sequences producing the maximum value when sorted I am trying to improve my discrete Math skills by doing some proof exercises. But I am struggling to understand how to start proving the following hypothesis:  $$ S = {x}_1{y}_1 + {x}_2{y}_2 + \ldots + {x}_n{y}_n $$ Where ${x}_1,{x}_2\ldots{x}_n$ and ${y}_1,{y}_2\ldots{y}_n $ are ordering of two different sequences of positive real numbers, each containing n elements. S takes its maximum value when ${x}_1,{x}_2\ldots{x}_n$ and ${y}_1,{y}_2\ldots{y}_n $ are sorted in non decreasing order.  I was thinking that I should maybe use a proof by contradiction. Where the negation of the hypothesis is: S takes it minimum value over all ordering of the two sequences when both sequences are sorted into non increasing order.  However I am not sure my negation of the hypothesis is correct or how proceed from there. I do not want the complete proof because I am trying to learn how to do it myself but if some one can tell me how to make a start on this I would be grateful. Ubuntu/Kubuntu 16.04.1 installation in VirtualBox I'm trying to install Kubuntu 16.04.1 on a virtual machine (I'm using VirtualBox). When I boot from the disk image Ubuntu begins to load and then I get this screen:   It looks like it's loading but the display is not working properly (the pink blob is my mouse pointer, only very large and pink).  I also tried installing Ubuntu 16.04.1 and I get a very similar result:   Am I missing something in VirtualBox? Thanks! Show $\text{Hom}_{Ab}(A,\mathbb{Q}) = \{0\}$ for finite $A$ I have a finite abelian group $A$ and I have shown that $A\otimes\mathbb{Q} \cong \{0\}$, the trivial group.  Now I want to show that $\text{Hom}_{\textbf{Ab}}(A,\mathbb{Q}) = \{0\}$, i.e we only have the trivial zero group homomorphism from $A$ to $\mathbb{Q}$.  The question I'm trying to solve is phrased in such a way that $\text{Hom}_{\textbf{Ab}}(A,\mathbb{Q}) = \{0\}$ should fall out almost automagically once you have that the tensor product collapses but I'm really stuck on this even though some people I talked to said it was really easy.  Some thoughts:   I've tried to assume the existence of a nontrivial homorphism $A \to    \mathbb{Q}$ to generate a contradiction but with no success. We have the Tensor-hom adjunction theorem stating that for abelian groups $A,B,C$ we have a natural bijection  $$\text{Hom}_{\textbf{Ab}}(A\otimes B,C) \cong \text{Hom}_{\textbf{Ab}}(B ,\text{Hom}_{Ab}(A,C))$$ and by plugging in our particular groups we get that:   $$\text{Hom}_{\textbf{Ab}}(A\otimes \mathbb{Q},\mathbb{Q})\cong\text{Hom}_{\textbf{Ab}}(\mathbb{Q} ,\text{Hom}_{Ab}(A,\mathbb{Q}))$$  Since $A\otimes\mathbb{Q} \cong \{0\} \implies \text{Hom}_{\textbf{Ab}}(\mathbb{Q} ,\text{Hom}_{Ab}(A,\mathbb{Q})) = \{0\}$ so we only have the trivial morphism from $\mathbb{Q} \to \text{Hom}_{Ab}(A,\mathbb{Q})$ but I don't think this implies that $\text{Hom}_{Ab}(A,\mathbb{Q}) = \{0\}$ and I'm not sure if this reasoning helps me at all. Target filesystem doesn't have requested /sbin/init Although we didn't do any changes, this new error appeared in the GSC Structured Data section:    I don't really get what it means. There are many fields for Schema.org, is homeLocation suddenly a required field? Does that make any sense? How to sanity check a date in Java Giving edit reputation bonus for same user for same post multiple times On Stack Overflow, I saw some users editing the same question multiple times, and getting the 2 point reputation bonus for each time.  I saw it during review of a suggested edit (some months ago, I forgot the question that's why I'm unable to paste the link here). I rejected it and gave the reason in the "Other" reason field.  Most of the time I saw that users first formats the code, and then after approval they edit the title and fix the grammar.  Why is this allowed?  My point is, if an user gained 2 reputation points for a particular edit, then why give another 2 points again for the same post? Because the second edit on same question/answer means that there were some mistakes or misses in their previous edit. I’m looking for an alien movie I watched years ago, mid to late 2000’s about the Earth being cooked or heated up by aliens. The poster is a reflection of the Earth heating up with an orange glow on an aliens eye. It was a low budget movie and I think the one I watched was pirated from China. Again low budget, bad quality and the name could’ve been something worthy he word scorch or something like that. The alien in the movie had legs with knees that went inwards and it had red eyes. Prove $| \sum_{i \leq n} \frac{\mu(i)}{i} | \leq 1$ So I am trying to prove the above stated formula. I am completely new to number theory so I thank you in advance for baring with me. I start off by using the definition:  $\frac{\phi(n)}{n} = \sum_{d|n} \frac{\mu(d)}{d}$  Where $\phi$ is Eulers function, and $\mu$ is the Mobius mu function. From here, I use the Mobius inversion formula to give:  $\frac{\mu(n)}{n} = \sum_{d|n}d\frac{\phi(\frac{n}{d})}{n}\mu(d)$  Now summing $n$ up to $x$:  $\sum_{n \leq x}\frac{\mu(n)}{n} = \sum_{n \leq x}\sum_{d|n}d\frac{\phi(\frac{n}{d})}{n}\mu(d)$  Now I'm not fully sure on how I can rearrange things inside the two summations on the right hand side, so leaving things as they are; using that $\sum_{d|n} \mu(d) = 1$, if and only if $n =1$. Hence:  $\sum_{n \leq x}\frac{\mu(n)}{n} = \sum_{d|1}d\phi(\frac{1}{d})\mu(d)$  Thus $d=1$. Subbing this in:  $\sum_{n \leq x}\frac{\mu(n)}{n} = \phi(1) = 1$  I am not sure what I have missed (or done wrong) as to have an equality and not a triangle relation. Any suggestions or ideas would be greatly appreciated How do I pipe each command given to the shell? I'd like to edit my .bashrc so that every command executed on the shell is piped to something, for example:   $ sudo apt update   _________________ &lt; sudo apt update &gt;  -----------------     \   ^__^      \  (oo)\_______         (__)\       )\/\             ||----w |             ||     ||   I have managed something rather similar, but not entirely:  $ bash $ exec &gt; &gt;(cowsay) $ echo "Hello AU!" $ exit  _______ &lt; Hello AU! &gt;  -------     \   ^__^      \  (oo)\_______         (__)\       )\/\             ||----w |             ||     ||   It is not the desired result, as it only happens after exiting the current shell.  It's mainly for fun/learning purposes. How do I reset an old iPod? My daughter gave her little brother her iPod Touch 8g. I have been trying to update the software and would like to get it rest to a factory clean state so I can linked it to the iTunes that already syncs to his iPad.   How can I do a clean up and wipe of everything from an iPod?