How to prove that it exists an integer $n$ such that the decimal expansion of $2^n$ starts with $123456789$?
Prove that for any finite sequence of decimal digits, there exists an $n$ such that the decimal expansion of $2^n$ begins with these digits.
Replace one Ubuntu with another from USB pen drive Grant group traversal rights on a folder Redefining \listoffigures and \listoftables as tables Rendered image looks pixelated The help pages in R assume I know what those numbers mean, but I don't. I'm trying to really intuitively understand every number here. I will just post the output and comment on what I found out. There might (will) be mistakes, as I'll just write what I assume. Mainly I'd like to know what the t-value in the coefficients mean, and why they print the residual standard error.   Call: lm(formula = iris$Sepal.Width ~ iris$Petal.Width)  Residuals:      Min       1Q   Median       3Q      Max  -1.09907 -0.23626 -0.01064  0.23345  1.17532    This is a 5-point-summary of the residuals (their mean is always 0, right?). The numbers can be used (I'm guessing here) to quickly see if there are any big outliers. Also you can already see it here if the residuals are far from normally distributed (they should be normally distributed).  Coefficients:                  Estimate Std. Error t value Pr(&gt;|t|)     (Intercept)       3.30843    0.06210  53.278  &lt; 2e-16 *** iris$Petal.Width -0.20936    0.04374  -4.786 4.07e-06 *** --- Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1    Estimates $\hat{\beta_i}$, computed by least squares regression. Also, the standard error is $\sigma_{\beta_i}$. I'd like to know how this is calculated.  I have no idea where the t-value and the corresponding p-value come from. I know $\hat{\beta}$ should be normal distributed, but how is the t-value calculated?  Residual standard error: 0.407 on 148 degrees of freedom   $\sqrt{ \frac{1}{n-p} \epsilon^T\epsilon }$, I guess. But why do we calculate that, and what does it tell us?  Multiple R-squared: 0.134,  Adjusted R-squared: 0.1282    $ R^2 = \frac{s_\hat{y}^2}{s_y^2} $, which is $ \frac{\sum_{i=1}^n (\hat{y_i}-\bar{y})^2}{\sum_{i=1}^n (y_i-\bar{y})^2} $. The ratio is close to 1 if the points lie on a straight line, and 0 if they are random.  What is the adjusted R-squared?  F-statistic: 22.91 on 1 and 148 DF,  p-value: 4.073e-06    F and p for the whole model, not only for single $\beta_i$s as previous. The F value is $ \frac{s^2_{\hat{y}}}{\sum\epsilon_i} $. The bigger it grows, the more unlikely it is that the $\beta$'s do not have any effect at all. Why do we have wrinkly fingers after swimming? I found an explanation for my question here , which suggests that water washes out sebum (special oil, which covers our skin) and water gets into outer layer of now unprotected skin by osmosis. This article also suggests that fingers are not really shriveled - they are water-logged. But is this right? I think I observed many times my wrinkly fingers after swimming, but I never felt that they are water-logged. Any thought on this matter would be highly appreciated. rotatebox symbol in caption generates error Following minimal example doesn't work and generates an error  \documentclass{article} \usepackage{graphicx} \usepackage{amssymb} \begin{document} \begin{figure}   FIGURE \caption{$\blacksquare$ result of; \rotatebox[origin=c]{45}{$\blacksquare$}}% \end{figure} \rotatebox[origin=c]{45}{$\blacksquare$} \end{document}   The error message says:  rotatebox-in-caption.tex|7| Argument of \@caption has an extra }. || &lt;inserted text&gt;  ||                 \par  || l.7 ...; \rotatebox[origin=c]{45}{$\blacksquare$}}   Do I have to wrap the \rotatebox in a certain box to make it work? How do I escape a single quote in SQL Server? How does "proof of effort" make a question better? It seems like many questions by new users are downvoted immediately.  Sometimes there is a comment to the tune of "lacks research", "we're not here to do your homework", or "show us some effort".  Now I wonder, what does it matter how much work someone has done before posting?  If someone can ask a good question in 5 seconds, more power to them.  To illustrate this point, look at how you could "improve" a question to include proof of effort.  You could put in 5 failed code snippets and their in- and output.  This would make for very long questions with a high noise ratio.  Another suggestion is to add "research".  By "research", I suppose most programmers mean "Google".  So this is basically a suggestion to add a list of fruitless Google queries.  That doesn't sound interesting either.  How can "proof of effort" turn a bad question into a good one? If $\sum_{n=1}^{\infty} a_n^{2}$ converges, then so does $\sum_{n=1}^{\infty} \frac {a_n}{n}$ SEO (or other) problems with a 404? The official Checkpoint out command line tool from CheckPoint, for setting up a SSL Network Extender VPN is not longer working from the Linux command line. It is also no longer actively supported by CheckPoint.  However, there is a promising  project, that tries to replicate the Java applet for authentication, that talks with the snx command line utility, called snxconnect.   I was trying to put snxconnect text utility to work in Debian Buster, doing:  sudo pip install snxvpn   and  export PYTHONHTTPSVERIFY=0 snxconnect -H checkpoint.hostname -U USER    However, it was mostly dying either with an HTTP error of:   HTTP/1.1 301 Moved Permanently:   or:  Got HTTP response: HTTP/1.1 302 Found   or:  Unexpected response, try again.   What to do about it?  PS. The EndPoint Security VPN official client is working well both in a Mac High Sierra and Windows 10 Pro. My internet connection used to be a direct LAN connection to my provider. Back then, everything would load fine on both Windows and Ubuntu (dual boot). However, a while ago they started needing me to dial (PPPoE) using a username and password. Gateway, subnet mask, IP, DNS servers all stayed the same. But since then, I haven't been able to browse certain websites on Ubuntu, even though there have been no such issues on Windows. Some example websites are -  (although share.ovi.com loads fine, and nokia.com loads fine),  (works on Chrome(ium) and Opera but not on Firefox (both 3.6 and 4)) Mozilla  and other random websites.  Some of the websites that don't load show timeout messages and for some websites (like the moz addons one), the browser will keep trying to load without an end (I've left it like that even for hours but not noticed anything different happen).  I have tried changing the DNS servers to public ones. I have even tried booting from a Fedora LiveCD and then changing the DNS to those (and even to the ones of OpenDNS), but the exact same thing happens. What could be inherently wrong with some config within Linux itself that is causing this problem?  Does anyone know why this is happening and how it can be fixed?  Note: This question has been  on SU, but not gotten any responses.  Update: Just saw  that someone else was having similar problem and solved it by putting a NetworkManager.conf file in /etc/NetworkManager. What needs to be in that file? Prove that $\phi(n) \geq \sqrt{n}/2$ So I'm trying to prove the following two inequalities: $$\frac{\sqrt{n}}{2} \leq \phi(n) \leq n.$$ The upper bound we get from simply noting that $\phi(n) = n \prod_{p | n}\left( 1 - \frac{1}{p}\right)$ and the fact that $(1 - \frac{1}{p}) \leq 1$. But how can we get the lower bound? I tried using the same expression for $\phi(n)$ but it seems to not really give me the inequality. Can you help? A single row background colour I saw there are various manga with Akame ga KILL! in their names.  The ones I have found are:       Which one should I read if I want to continue the storyline of the anime? How can I use wget to get files recursively if the username contains an @? Firstly my main goal is to try and get FTP files recursively in Centos.  However the actual FTP login contains an @ which ruins the syntax.  Is there some way to place the username in "double quotes" or some such equivalent such that I can use @in the username? Zero to the zero power – is $0^0=1$? Could someone provide me with a good explanation of why $0^0=1$?  My train of thought:  $x&gt;0$  $0^x=0^{x-0}=0^x/0^0$, so  $0^0=0^x/0^x=\,?$  Possible answers:   $0^0\cdot0^x=1\cdot0^0$, so $0^0=1$ $0^0=0^x/0^x=0/0$, which is undefined   PS. I've read the  on mathforum.org, but it isn't clear to me. How to remove a filter that is an anonymous object? Slit screen and wave-particle duality