Is it a good idea to redirect http://www.example.com to http://example.com or keep both ?
After having read various articles on this site about WWW or Non-WWW website setup, ,  and , and after having setup a correct redirection rule in .Htaccess to at least eliminate duplicate contents (currently the non-www solution (e.g. http://example.com) I have a feeling that my site rank less since I changed this.   My confusion was strengthened by these two opponents:     &amp;     Could the change of www.example.com >> http://example.com have a factual impact on ranking?  say either is better or worse. Are they holding information back?  I am confused fellow folks...  As much as I would like to separate facts from fiction, I am aware that not all the good valid information is written. Not on google webmaster site and perhaps not elsewhere. If so it seems out of date!  Does or does not a WWW domain impact SEO/rank better than its NON-WWW version? (Given that you have setup a proper redirect to avoid duplicates and go with one)
One very convenient property of $\mathbb{R}$ as a ring is that there is an upper bound for the degree of irreducible polynomials in $\mathbb{R}[x]$, as      If $f\in\mathbb{R}[x]$ has degree larger than $2$, then $f$ is reducible.   However, the proof as I know depends highly on the fact that $\mathbb{C}$ is algebraically closed, and the very nice property: $z\in\mathbb{C}$ is in $\mathbb{R}$ if and only if $z=\bar{z}$.  This makes a generalization to other integral domains rather difficult. So the problem is      For what kind of integral domain $R$, we have a finite upper bound on the degree of irreducible elements in $R[x]$?   Some most familiar examples are ruled out: in $\mathbb{Z}[x]$, $\mathbb{Q}[x]$ and $\mathbb{F}_{p}[x]$, there is not such a bound. Unfortunately these exhaust all integral domains about which I have a working knowledge.  Another result might help is , which says if we can find a prime ideal $\mathfrak{p}$ in $R$ such that $\mathfrak{p}^2\neq\mathfrak{p}$, then by picking $a\in\mathfrak{p}^2\backslash\mathfrak{p}$ we have an irreducible $a+x^d$, where $d$ can be arbitrary, and hence the upper bound is not possible.  So we only need to focus on domains where $\mathfrak{p}^2=\mathfrak{p}$ for all prime ideals. This seems to be a quite strong restriction but I am not sure what to make of it.  Can someone give a hint?  Thanks! Interpreting the residuals vs. fitted values plot for verifying the assumptions of a linear model Consider the following figure from Faraway's Linear Models with R (2005, p. 59).    The first plot seems to indicate that the residuals and the fitted values are uncorrelated, as they should be in a homoscedastic linear model with normally distributed errors. Therefore, the second and third plots, which seem to indicate dependency between the residuals and the fitted values, suggest a different model.  But why does the second plot suggest, as Faraway notes, a heteroscedastic linear model, while the third plot suggest a non-linear model?  The second plot seems to indicate that the absolute value of the residuals is strongly positively correlated with the fitted values, whereas no such trend is evident in the third plot. So if it were the case that, theoretically speaking, in a heteroscedastic linear model with normally distributed errors  $$ \mbox{Cor}\left(\mathbf{e},\hat{\mathbf{y}}\right) = \left[\begin{array}{ccc}1 &amp; \cdots &amp; 1 \\ \vdots &amp; \ddots &amp; \vdots \\ 1 &amp; \cdots &amp; 1\end{array}\right] $$  (where the expression on the left is the variance-covariance matrix between the residuals and the fitted values) this would explain why the second and third plots agree with Faraway's interpretations.  But is this the case? If not, how else can Faraway's interpretations of the second and third plots be justified? Also, why does the third plot necessarily indicate non-linearity? Isn't it possible that it is linear, but that the errors are either not normally distributed, or else that they are normally distributed, but do not center around zero? Too much space between full paragraph and subequations env I have a document using subequations and the parskip package. If the paragraph before a subequations environment is completely filled up to the last line, then additional vertical whitespace is introduced between this paragraph and the following subequations. I guess this is a bug, but I'm not sure, nor do I have enough TeX knowledge to try and fix it. How can I fix this whitespace issue?  Below is a MWE which shows the behaviour, at least on LaTeX2e &lt;2003/12/01> with pdfeTeX 3.141592-1.21a-2.2 (unfortunately on this machine only an old tetex distribution is installed).  \documentclass[a4paper,11pt]{article}  \usepackage{parskip} \usepackage{amsmath}  \begin{document}  The subequations environment provides a convenient way to number a foos \begin{subequations}\label{foo}   \begin{align}     x^2 + y^2 &amp;= z^2\\     a &amp;= b + c   \end{align} \end{subequations}  \end{document} Please do not close this as a duplicate unless you can remove the “status-completed” tag from the  (because it’s obviously not been completed).  Steps to reproduce:   Have someone post an answer. Downvote the answer. Have the someone edit the answer within the initial 5 minutes of the answer’s existence. Try to undo the downvote more than 10 minutes later.   You cannot undo the downvote. Model Selection: Logistic Regression Can not add image inside SVG via jQuery:  tag becomes Is there a formula for $(1+i)^n+(1-i)^n$? I'm wondering if there is a formula for the value of $(1+i)^n+(1-i)^n$?  I calculated the first terms starting with $n=1$ to be, in order, $2$, $0$, $-4$, $-8$, $-8$, $0$, $16$, $\dots$  So it seems to be some sequence of positive and negative powers of $2$ with $0$s thrown in. Is there a more explicit formulation of what $(1+i)^n+(1-i)^n$ is, based on $n$?   With the binomial theorem, I get it equal to $$ \sum_{k=0}^n\binom{n}{k}i^{n-k}(1+(-1)^{n-k}). $$ Can this be made nicer? Thanks. Summarize the mathematical problem at the heart of breaking a Curve25519 public key It's pretty easy to generate a Curve25519 private key: generate 32 random bytes of data and then do:  e[0] &amp;= 248 e[31] &amp;= 127 e[31] |= 64   You can then generate a public key by doing  on the private key. For example, here's a random keypair I just generated:  public: e40ed268dff20db8497d0691421d00b086aad1efedcd5bd223d9eb6c0b85665f secret: e3c0e8b20ba7894882ddb37522dbe90867befcf9f31a5ce29c0d7997e0bd856d   The whole safety of public key encryption is that it is hard to take a public key and reverse the operation. Elliptic curve cryptography, as I understand it, relies on the fact that a  is hard to find. In particular, for a given large b and g, it is hard to find an integer k such that  $b^k = g$  How does that equation relate to Curve25519? Specifically, if I had a public key like the one above, how would I express, in math terms, the problem I need to solve to find the private key? How does it relate to the equation I see from time to time - $y^2 = x^3+486662x^2+x$, and the number $2^{255} - 19$?  I am guessing it is not as simple as "the private key is k and the public key is b and/or g" in the equation above. I'm also a little confused because the keys would appear to have roughly the same number of digits, which would seem to imply that there aren't many feasible values for k? Traditional search-as-you-type on newer Nautilus versions How do I change the Google Play country? Can someone tell me how to change the country preference for Google Play? I have searched high and low but in most of the forums people ask you to go to Gmail settings -> edit profile and change the country there. But there is no place to change the country in settings! There is a something called place (which can be marked as current) but that has no impact on Google Play.  I don't want to circumvent country restrictions, I want to permanently change the country. So it's NOT a duplicate of . Prove that $\binom{m+n}{m}=\sum\limits_{i=0}^m \binom{m}{i}\binom{n}{i}$ product of two uniformly continuous functions is uniformly continuous Suppose that $f$ and $g$ are uniformly continuous functions defined on $(a,b)$. Prove that $fg$ is also uniformly continuous on $(a,b)$.   My attempt: Since $f$ is uniformly continuous on $(a,b)$, for all $\epsilon&gt;0$, we have $\delta_f(\epsilon)&gt;0$ such that for all $x,y \in (a,b)$, $|x-y|&lt;\delta_f$, $|f(x)-f(y)|&lt;\epsilon$  Since $g$ is uniformly continuous on $(a,b)$, for all $\epsilon&gt;0$, we have $\delta_g(\epsilon)&gt;0$ such that for all $x,y \in (a,b)$, $|x-y|&lt;\delta_g$, $|g(x)-g(y)|&lt;\epsilon$  Notice that $$|f(x)g(x)-f(y)g(y)|=|f(x)g(x)-f(x)g(y)+f(x)g(y)-f(y)g(y)| \leq |f(x)||g(x)-g(y)| + |g(y)||f(x)-f(y)|$$  Here I don't know how to bound $|f(x)|$ and $|g(y)|$. I have proven that uniformly continuous functions preserve boundedness of an interval , i.e. $f$ is bounded on $(a,b)$. Can anyone help me? What color codes can I use in my PS1 prompt? I used several colors in my PS1 prompt such as   \033]01;31\] # pink \033]00m\]   # white \033]01;36\] # bold green \033]02;36\] # green \033]01;34\] # blue \033]01;33\] # bold yellow   Where can I find a list of the color codes I can use?  I looked at  but it didn't answer my question about a list of the actual codes.  It would be nice if there was a more readable form also.  See also C# Interfaces. Implicit implementation versus Explicit implementation What are the differences in implementing interfaces implicitly and explicitly in C#?  When should you use implicit and when should you use explicit?  Are there any pros and/or cons to one or the other?    Microsoft's official guidelines (from first edition ) states that using explicit implementations are not recommended, since it gives the code unexpected behaviour.  I think this guideline is very valid in a pre-IoC-time, when you don't pass things around as interfaces.  Could anyone touch on that aspect as well? Did Voldemort refer to Snape as one of the missing Death Eaters at his rebirth? In Goblet of Fire, Voldemort says of the missing six Death Eaters:     “And here we have six missing Death Eaters… three dead in my service. One,   too cowardly to return… he will pay. One, who I believe has left me forever… he   will be killed, of course… and one, who remains my most faithful servant, and   who has already reentered my service."   The answer to  speculates about who the three dead Death Eaters were.  We know who the last one is. I’m guessing one of the remaining two is Karkaroff. Is the other Snape? If so, which one? Perturbation property of Orthonormal basis for Hilbert space On my Toshiba Encore 8 Tablet I noticed that the CPU usage is very high. In Taskmgr / ProcessExplorer, I can see that OneDriveSetup.exe is hogging 1 CPU cores heavily for hours.    How an I solve this hang and finish the OneDrive Update? Why didn't Voldemort murder the Dursleys? Voldemort couldn't kill Harry as long as he called the Dursleys home. Why didn't he just murder the Dursleys? This would prevent Harry from being able to live there. Why can't helicopters reach mount everest? What, precisely, is a confidence interval?