Attempt:  I thought that by fixing one of the index and doing it conversely I can find accumulation points as follows:  For every fix $n_0\in \mathbb N$ $2^{-n_0}+5^{-m}$ converges to $2^{-n_0}$  and For every fix $m_0\in \mathbb N$ $2^{-n}+5^{-m_0}$ converges to $5^{-m_0}$  and letting free both indices we have a accumulation point $0$  So all accumulation points are $  {2^{-n},n\in\mathbb N^+ } \cup {5^{-n},m\in\mathbb N^+} \cup 0 $$ is it my reasoning right?
Determine all of the accumulation points of the following sets in $\mathbb{R}^1$ and decide whether the sets are open or closed or neither.  The set $X$ of all numbers of the form $2^{-n} + 5^{-m}$ where $m,n = 1,2,\ldots$.   Claim: The set of accumulation points is the following: $$\left\{\frac{1}{2^n}: n \in \mathbb{N} \right\} \cup \left\{\frac{1}{5^n}: n \in \mathbb{N}\right\} \cup \{0\}.$$ Let $n \in \Bbb{N}$ and $r&gt;0$, and consider the open neighbourhood $B\left(\frac{1}{2^n},r\right) = \left(\frac{1}{2^n} - r,\frac{1}{2^n} + r\right)$. For the given $r$ we can find $m$ large enough such that $5^{-m} &lt; r \implies \frac{1}{2^n} + \frac{1}{5^m} &lt; \frac{1}{2^n} + r$, and $\frac{1}{2^n} - \frac{1}{5^m} &gt; \frac{1}{2^n} - r$; therefore $5^{-m} + 2^{-n} \in B\left(\frac{1}{2^n},r\right)$. A similar argument shows that each element of $\{5^{-m}: m \in \mathbb{N}\} \cup \{0\}$ is an accumulation point of $X$. Now consider $y \in X$. $B(y,r)$ isn't a subset of $X$ since for all $r &gt; 0$, $B(y,r) \subset (\mathbb{R} - \mathbb{Q}) \not\subset \Bbb{Q}$, hence $B(y,r) \not\subset X$.
I'm putting some type on a path, a circle; but when I do it, the type is upside down. How can I flip it so it reads the right way up? See image below: How to load and parse SVG documents replacement wood threshold insert for metal door I have some late 70's doors and one is missing a threshold insert (the adjustable wood kind), anybody know of where one could by just the "wood strip" that goes in and screws down (there is a metal piece that sticks up into the wood, though you can't see it)?    They have about 4 screws that go down into them? How to divide two 64-bit numbers in Linux Kernel? Some code that rounds up the division to demonstrate (C-syntax):  #define SINT64 long long int #define SINT32 long int  SINT64 divRound(SINT64 dividend, SINT64 divisor) {   SINT32 quotient1 = dividend / divisor;    SINT32 modResult = dividend % divisor;   SINT32 multResult = modResult * 2;   SINT32 quotient2 = multResult / divisor;    SINT64 result = quotient1 + quotient2;    return ( result ); }   Now, if this were User-space we probably wouldn't even notice that our compiler is generating code for those operators (e.g. divdi3() for division). Chances are we link with libgcc without even knowing it. The problem is that Kernel-space is different (e.g. no libgcc). What to do?  Crawl Google for a while, notice that pretty much everyone addresses the unsigned variant:  #define UINT64 long long int #define UINT32 long int  UINT64 divRound(UINT64 dividend, UINT64 divisor) {   UINT32 quotient1 = dividend / divisor;    UINT32 modResult = dividend % divisor;   UINT32 multResult = modResult * 2;   UINT32 quotient2 = multResult / divisor;    UINT64 result = quotient1 + quotient2;    return ( result ); }   I know how to fix this one: Override udivdi3() and umoddi3() with do_div() from asm/div64.h. Done right? Wrong. Signed is not the same as unsigned, sdivdi3() does not simply call udivdi3(), they are separate functions for a reason.  Have you solved this problem? Do you know of a library that will help me do this? I'm really stuck so whatever you might see here that I just don't right now would be really helpful.  Thanks, Chad Please Join Us in Welcoming Our New Meta Stack Exchange Moderators! Sometimes after I fry something, I wash it immediately in water because it's a bit easier to wash (before the remains stick to the pan).  Does this damage the pan in any way? Process descendants literally as a hyperbole Possible Duplicate:          I have seen a lot of backlash in internet media against people using the word literally to mean something not literal. Something like "he was literally as big as a house" to mean someone was very big I think this comic sums up the anti literal movement well   Is this actually grammatically incorrect or can the word "literally" be a hyperbole strengthening a metaphor? Euclidean Algorithm for GCD of polynomials I am struggling to use the Euclidean algorithm for polynomials.  Given something like $$GCD(x^5+1, x^3+1)$$  I can easily use it as follows:   $$x^5+1 = x^2(x^3+1) -x^2 +1 \\ x^3+1 = -x(-x^2+1) + x +1 \\ -x^2+1 = (x+1)(-x+1)$$  GCD = x+1  But for something like $$GCD(2x^2+6x+3, 2x+1)$$   I cannot figure out how to do it using the same method. I start like this:  $$2x^2+6x+3 = x(2x+1) + 5x+3\\ 2x+1 = \frac{2}{5}(5x+3) - \frac{1}{5}\\$$  and I am not sure how to continue. Any help would be appreciated, sorry for my poor attempt at using LaTeX. Removal of statistically significant intercept term increases $R^2$ in linear model In a simple linear model with a single explanatory variable,  $\alpha_i = \beta_0 + \beta_1 \delta_i + \epsilon_i$  I find that removing the intercept term improves the fit greatly (value of $R^2$ goes from 0.3 to 0.9). However, the intercept term appears to be statistically significant.  With intercept:   Call: lm(formula = alpha ~ delta, data = cf)  Residuals:      Min       1Q   Median       3Q      Max  -0.72138 -0.15619 -0.03744  0.14189  0.70305   Coefficients:             Estimate Std. Error t value Pr(&gt;|t|)     (Intercept)  0.48408    0.05397    8.97   &lt;2e-16 *** delta        0.46112    0.04595   10.04   &lt;2e-16 *** --- Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1   Residual standard error: 0.2435 on 218 degrees of freedom Multiple R-squared: 0.316,    Adjusted R-squared: 0.3129  F-statistic: 100.7 on 1 and 218 DF,  p-value: &lt; 2.2e-16    Without intercept:   Call: lm(formula = alpha ~ 0 + delta, data = cf)  Residuals:      Min       1Q   Median       3Q      Max  -0.92474 -0.15021  0.05114  0.21078  0.85480   Coefficients:       Estimate Std. Error t value Pr(&gt;|t|)     delta  0.85374    0.01632   52.33   &lt;2e-16 *** --- Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1   Residual standard error: 0.2842 on 219 degrees of freedom Multiple R-squared: 0.9259,   Adjusted R-squared: 0.9256  F-statistic:  2738 on 1 and 219 DF,  p-value: &lt; 2.2e-16    How would you interpret these results? Should an intercept term be included in the model or not?  Edit  Here's the residual sums of squares:  RSS(with intercept) = 12.92305 RSS(without intercept) = 17.69277 What are the “spam” and “rude or abusive” (offensive) flags, and how do they work? Roots of unity filter, identity involving $\sum_{k \ge 0} \binom{n}{3k}$ How do I see that$$\sum_{k \ge 0} \binom{n}{3k} = (1 + 1)^n + (\omega + 1)^n + (\omega^2 + 1)^n,$$where $\omega = \text{exp}\left({2\over3}\pi i\right)$? What is the underlying intuition behind this equality? What Canon lens would one use for Sartorialist-like street photography? You probably know Scott Schuman and his blog (). What kind of lens (that is not expensive) would you recommend for Canon 550D to achieve similar effect (in terms of DoF etc.)? I'm a newbie so I'm not sure how to properly interpret those photos' EXIF info. My guess is, that a similar effect can be achieved using .   PS:   Update: Thank you guys for all the answers, it's much more clear for me right now. I think I'll try the Canon 50mm f/1.4 lens. What is the value of $\arctan(1/2)+\arctan(1/5)+\arctan(1/8)$? How does the tester screwdriver work? How does the tester screwdriver work? If I put the tester screwdriver inside the "hot wire" of an electrical socket, it lits up if I press my finger against the metal cap on top of the screwdriver. This happens also if I stand on a surface of isolating material, such as wood. I read elsewhere that this happens due to stray capacitance formed by the "hot wire", the human body, and the ground. One has  $$ Z = R + \dfrac{1}{j \omega C}  $$  for the impedance, so if C  is high enough, the impedance should be close to r , the "effective resistance" of the formed circuit. Here I get lost; why r  is small enough to cause a current in the range of mA  even if I stand on an isolating surface?  So really what I am asking is how one may represent the system hot wire - screwdriver - human body - wooden floor - building - ground as an electrical circuit, and which parts of the physical system contribute to the resistance, capacitance (and inductance?) and in which proportion, even very approximately. "I have" vs "I have got" I don't know what is the difference between using "I have got" and "I have" For example:      He's got broad shoulders.    Is it possible to say:      He has broad shoulders.    If yes, so what is the difference? I have posted a proof below, and would appreciate it if someone could review it for accuracy. Thanks!   Problem:   Let n $\in$ $\mathbb{Z}$ with $n$ $\ge$ 3. Prove the following:   (a) Z(D$_{2n}$) = 1 if $n$ is odd.   (b) Z(D$_{2n}$) = {1, r$^k$} if $n$ = $2k$.   Note that $r$ and $s$ generate D$_{2n}$ with the group presentation { $r$,$s$ | $r$$^n$ = $s$$^2$ = 1, $rs$ = $sr$$^{-1}$ }   Proof:  part (a)   Let $n$ $\ge$ 3 where $n$ $\in$ $\mathbb{Z}$ and $n$ is odd.   For any x $\in$ Z(D$_{2n}$), x must commute with both $s$ and $r$, since both are in $D_{2n}$. Note that if $x$ commutes with both $s$ an $r$, then $x$ commutes with any element of $D_{2n}$ since $r$ and $s$ generate D$_{2n}$.    Then for any x$\in$ Z(D$_{2n}$), we have $xr$ = $rx$, where x is of the form $x$ = s$^j$r$^w$ (any such element can be arranged into this form via the relation $rs$ = $sr$$^{-1}$ and the fact that $r$ and $s$ generate D$_{2n}$). Note that since r and s have finite order, the exponents are modulo n and modulo 2 for $r$ and $s$ respectively.   Then we have  s$^j$r$^w$$r$ = $r$s$^j$r$^w$.   Which implies s$^j$r$^{w+1}$ = s$^j$r$^{w +- 1}$  In the case where the power of r on the RHS is $w$ - 1, it is clear we cannot have equality unless |r| = 1, which it is not. Note that if $j$ is even then we have only the $w$ + 1 case. Hence $x$ may be of the form $1$$r$$^w$ = $r$$^w$.   Also we have that x must commute with $s$, hence $s$$x$ = $x$$s$.   Then $s$$r$$^w$ = $r$$^w$$s$.  Which implies $s$$r$$^w$ = $s$$r$$^{-w}$.   Applying s$^{-1}$ to both sides we arrive at the task of finding when r$^w$ = r$^{-w}$ i.e when $w$ = $-w$. But since r has finite order we have:   $$ \bar{w} = (n-1)*w $$   where $\bar{w}$ is the residue class of w modulo n. Hence for some a $\in$ $\mathbb{Z}$$_+$   $$ w + an = wn - w $$  $$ 2w = wn - an $$ $$ 2w = n(w-a) $$   Now since 0 $\lt$ w $\le$ n we have that the LHS is greater than 0. Also since n $\ge$ w then (w-a) $\le$ 2, since otherwise the RHS would be greater than the LHS. So since n is odd we must have that (w-a) is even and hence (w-a) = 2, implying that 2w = 2n and hence n = w.   Hence x = $r$$^w$ = $r$$^n$ = 1. So Z(D$_{2n}$) = 1.   part (b)   From the argument above any x $\in$ Z(D$_{2n}$) must be of the form $s$$^j$$r$$^w$ for j even and j $\in$ $\mathbb{Z}$. Then x is of the form $r$$^w$ for $0$ $\lt$ w $\le$ $n$. Since $n$ is even we have the equation below is satisfied when $2w$ = $n$ or $w$ = $n$, since (w-a) $\le$ 2 but also (w-a) $\ge$ 0 for (w-a) $\in$ $\mathbb{Z}$$_+$.   $$ 2w = n(w - a) $$   Hence $r$$^w$ $\in$ Z(D$_{2n}$) when $2w$ = $n$, as desired. Is it possible to prove a null hypothesis? How do I use the latest GCC on Ubuntu? I want to compile my program with the latest version of gcc.  Ubuntu 14.04 comes with gcc 4.8.2, however there's 4.9.0 available, moreover, I see that it is available as a package: . I tried to install it  sudo apt-get install gcc-4.9   but it says  Reading package lists... Done Building dependency tree        Reading state information... Done Note, selecting 'gcc-4.9-base' for regex 'gcc-4.9' gcc-4.9-base is already the newest version. 0 upgraded, 0 newly installed, 0 to remove and 0 not upgraded.   Looks like it is already installed, just not as the default one? How do I utilize it to build my program? Sum of First $n$ Squares Equals $\frac{n(n+1)(2n+1)}{6}$ I am just starting into calculus and I have a question about the following statement I encountered while learning about definite integrals:  $$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$$  I really have no idea why this statement is true. Can someone please explain why this is true and if possible show how to arrive at one given the other?