so when I try to convert my end direction (ex. N, S, E, W) into a positive or negative 1.0 to multiply to the eventual degrees/minutes'seconds&quot; calculation, my double sign  always returns as 0.0. I've tried testing each letter in separate if statements and have tried putting the calculations into the if statements. I've been testing with the following coordinates: 40°35'6.9288&quot;N public static double convertDegreesToDecimals(String strValue) {    double decimal = 0;    double sign = 1.0;     String degree = strValue.substring(0, strValue.indexOf(DEGREE));    double deg = Double.parseDouble(degree);    String minute = strValue.substring(strValue.indexOf(DEGREE)+1,strValue.indexOf(MINUTE));    double min = Double.parseDouble(minute);    String second = strValue.substring(strValue.indexOf(MINUTE)+1,strValue.indexOf(SECOND));    double sec = Double.parseDouble(second);    String direction = strValue.substring(strValue.indexOf(SECOND)+1,strValue.length());    System.out.println(direction);   //prints N, S, or whatever the direction is correctly       if(direction==&quot;S&quot; || direction==&quot;E&quot;){       double sign = -1.0;    }else if(direction==&quot;N&quot; || direction==&quot;W&quot;){       double sign = 1.0;    }       decimal = ( (deg + (min / 60.0) + (sec / 3600.0) ) * sign);    return decimal; }  I expect for the double sign to equal -1.0 when the direction is S or W, and to equal 1.0 when the direction is N or E. My calculations for degrees/minutes'seconds&quot; are correct except for the sign, and String direction always returns the expected result(N,S,E,W). When I put println statements in the if statements, they did not execute which led me to think that something must be wrong with the if statements.
I've been using the == operator in my program to compare all my strings so far. However, I ran into a bug, changed one of them into .equals() instead, and it fixed the bug.  Is == bad? When should it and should it not be used? What's the difference?
"Firstly" or "first"? Why should I use a private variable in a property accessor? How can I send same email multiple times in iOS? I wanted to send a single email to different users. I save an email to drafts and send the email to a user and it goes to the sent box instead of staying in the drafts box. Is there a way I can send an email repeatedly from the drafts box? How do I get the logged in user's window's credentials in a WPF application I am calculating the class group of $\mathbb Q(\sqrt 6)$. My working is as follows:  The Minkowski bound is $\lambda(6)=\sqrt 6&lt;3$ so we only need to look at prime ideals of norm $2$. $2$ divides the discriminant, $24$, of $\mathbb Q(\sqrt 6)$, so $2$ ramifies, and so for any prime ideal $\mathfrak p$ of norm $2$, $\mathfrak p^2$ is principal. But $(2, \sqrt 6)^2=(2)$, so $\mathfrak p=(2, \sqrt 6)$ has norm $2$. If I can show $\mathfrak p$ is not principal, then I can conclude that $Cl(\mathbb Q(\sqrt 6))\cong C_2.$ But if there was an element of $\mathcal O_{\mathbb Q(\sqrt 6)}$ of norm 2, this would correspond to a solution of the diophantine equation $x^2-6y^2=2$. Clearly $x$ is even, so writing $x=2n$ we get $2n^2-3y^2=1$. Now reducing modulo $3$, we get $2n^2\equiv 1 \pmod 3$ which has no soultions. Hence $\mathfrak p$ is not principal and $Cl (\mathbb Q(\sqrt 6)) \cong C_2.$  However,  says that the class group is trivial. Where have I gone wrong? What am I misunderstanding? Is $(2, \sqrt 6)$ principal? What is the probability that all three tosses result in the same outcome? A coin is tossed three times: The probability of zero heads is 1/8 and the probability of zero tails is 1/8.  And my question is: What is the probability that all three tosses result in the same outcome?   So, if P(zero heads)= 1/8 , then that should be the same of p(all tails)?  If so, we could use the Addition Rule which is $P(A \cup B) = P(A) + P(B)$  where A and B are disjoint events, i.e. $A \cap B = \emptyset$, A is the event of tossing all heads and B is the event of tossing all tails.  I'm not sure how to continue after that... would the complement be used? Could I use a 1/2TB USB 2/3 external HDD with an XBox 360? How do I resolve "WILL_NOT_PERFORM" MS AD reply when trying to change password in scala w/ the unboundid LDAP SDK? Help with completing the proof: Any unbounded sequence is either infinitely large or has a convergent subsequential limit Given $\{x_n\}$ is an unbounded sequence, prove:   $$ \lim_{n\to\infty}x_n = \infty\ \  \text{or}\ \ \lim_{n_k \to \infty} = A $$   In other words any unbounded sequence either infinitely large or has a convergent subsequence.     Everything before udpate section is wrong.  Part 1. Proving $|x_n| &gt; M \implies \lim_{n\to\infty} x_n = \infty$  $\Box$ Start with the definition of an unbounded sequence: $$ \forall M &gt; 0 \in \Bbb R \ \exists N \in \Bbb N: \forall n &gt; N\implies |x_n| &gt; M $$  Suppose that: $$ \lim_{n\to\infty}x_n \ne \infty $$ That means: $$ \forall n\in\Bbb N\ \exists C &gt;0 \in\Bbb R : |x_n| &lt; C $$  Now taking $M^\prime = \max\{M, C\}$ we have that: $$ \begin{cases} \begin{align} \exists N \in \Bbb N : &amp;\forall n&gt;N \implies &amp;|x_n| &gt; M^\prime \\ &amp;\forall n \in \Bbb N: &amp;|x_n| &lt; M^\prime \end{align} \end{cases} $$  So we have arrived to a contradiction which means the assumption is wrong and: $$ \lim_{n\to\infty}x_n = \infty $$ $\Box$     Part 2. It's not clear to me how to prove that if unboundedness does not imply infinity limit then it must imply the existence of a convergent subsequence.   Two questions is my mind:   Is part 1 correct?  How do I proceed with the second part?     I'm using the following definition. $$ \lim_{n\to\infty}x_n = +\infty \stackrel{\text{def}}{\iff} \forall \epsilon &gt; 0 \exists N \in \Bbb N :\forall n &gt; N \implies x_n &gt; \epsilon \\ \lim_{n\to\infty}x_n = -\infty \stackrel{\text{def}}{\iff} \forall \epsilon &gt; 0 \exists N \in \Bbb N :\forall n &gt; N \implies x_n &lt; -\epsilon \\ $$ A sequence is considered infinitely large when: $$ \lim_{n\to\infty}x_n = \infty\ \ \text{when}\ \ \lim_{n\to\infty}x_n = + \infty \ \ \text{or} \ \ \lim_{n\to\infty}x_n = - \infty $$    Update.   Looks like i've messed up a lot of things. I will try once again. As shown in comments and answers the statement holds only in case: $$ \lim_{n\to\infty} |x_n| = \infty \iff \forall \epsilon &gt; 0 \exists N \in \Bbb N: \forall n &gt; N \implies |x_n| &gt; \epsilon $$  Consider the following: $$ \lnot P = \lnot\left(\lim_{n\to\infty}|x_n| = \infty \right) \iff \exists \epsilon &gt; 0\ \forall N_1 \in \Bbb N : \exists n &gt; N_1 \land |x_n| &lt; \epsilon $$  But on the other hand it is given that $x_n$ is unbounded: $$ Q = \forall M &gt; 0\ \exists N_2 \in \Bbb N : |x_{N_2}| &gt; M $$  Construct a negative expression for boundedness: $$ \lnot P = \exists M &gt; 0\ \forall N_2 \in \Bbb N : |x_{N_2}| &lt; M $$  If $S = \lnot P \implies \lnot Q$ then $S = P \lor \lnot Q$  Let $\epsilon = M$, choose $N = \max\{N_1, N_2\}$ then both statements are true and: $$ \exists \epsilon &gt; 0\ \forall N = \max\{N_1, N_2\}: \exists n &gt; N \land |x_n| &lt; \epsilon $$  Now either $P$ is true, which would mean $\lim |x_n| = \infty$, or $\lnot Q$ is true which would mean the sequence is bounded and hence contains a convergent subsequence. Why is gradient the direction of steepest ascent? Ex-dividend date: How long do I have to hold a stock in order to get the next dividend? I want to be a stockholder on record for a dividend. So lets assume I bought stock before the ex dividend date. Can I immediately sell the stock after the ex-dividend date to collect the dividend, or do I need to hold for a few weeks to collect the dividend?  Lets say the ex-dividend date is the 15th of the month. I buy stock on the 14th (I think this is soon enough to be a shareholder on record) and then I sell on the 15th. Do I get the dividend in my account when it is distributed?  Don't worry &ndash; I have a proprietary way of hedging the stock's post-dividend price adjustment, and it costs less than the actual dividend so all is well.  I just don't know what the shortest time I can hold a position is. I'd like to know the shortest theoretical time possible in order to get the dividend.  Insight appreciated. Is there any way to erase personal information from my phone that has already been stolen? My phone has been stolen. I do not want the thief to have access to any of my personal information such as calls, texts, and information saved in the phones contents. Is there any possible way I can get this information erased from my phone? What are the visa (or any other crucial) requirements of Bhutan for tourists? I heard that access to Bhutan is quite restricted for all people of outside South-Asian countries?  Can I have the details on that? Wilson and Fermat's Theorem Evaluating $\sum_{n=1}^{\infty}\frac{(n-1)!}{\prod_{r=1}^{n}(x+r)}$ for $x\in\mathbb{R}^{+}$ I am trying to evaluate the following sum. Here $x\in\mathbb{R^{+}}$. $$f(x)=\sum_{n=1}^{\infty}\frac{(n-1)!}{\prod_{r=1}^{n}(x+r)}$$ At present I do not know how to proceed. Any hints are appreciated. Thanks. I found Minecraft's marvelous large worlds extremely slow to navigate, even with a quad core and meaty graphics card.  I assume Minecraft's slowness comes from:   Java, as spatial partitioning and memory management are faster in native C++. Weak world partitioning.   I could be wrong on both assumptions. However, this got me thinking about the best way to manage large voxel worlds. As it is a true 3D world, where a block can exist in any part of the world, it is basically a big 3D array [x][y][z], where each block in the world has a type (i.e BlockType.Empty = 0, BlockType.Dirt = 1, etc.)  I assume that to make this sort of world perform well, you would need to:   Use a tree of some variety (//) to split all the cubes out; it seems like an oct/kd would be the better option, as you can just partition on a per cube level not a per triangle level. Use some algorithm to work out which blocks can currently be seen, as blocks closer to the user could obfuscate the blocks behind, making it pointless to render them. Keep the block object themselves lightweight, so it is quick to add and remove them from the trees.   I guess there is no right answer to this, but I would be interested to see peoples' opinions on the subject.  How would you improve performance in a large voxel-based world? Every convergent sequence in a metric space has a unique limit. Bounding box is larger than expected when drawing a curved path I have been working on logo for my department and I have the following code. The problem is that whenever I compile the diagram seems to be larger than the actual logo. I do not know what is the problem. I know the numbers I have selected are not the best but any insights into the matter will be highly appreciated.   \documentclass[letterpaper]{article}  \usepackage[top=2cm,bottom=2cm,left=2cm,right=2cm]{geometry}  %\usepackage{amsmath,amssymb,units}  %\usepackage{enumitem,multicol}  \usepackage{tikz}  %\usetikzlibrary{arrows}  \usepackage{lipsum}   \begin{document}   \lipsum[1-2]  \begin{tikzpicture}[y=0.80pt, x=0.8pt,yscale=-1]  \path[draw=black,fill=white]    (258.9527,290.5199) .. controls (173.9885,538.4766) and (568.5860,261.2969) ..    (306.5098,257.1141) .. controls (44.4337,252.9312) and (429.9845,542.5624) ..    (352.9767,292.0206) .. controls (275.9689,41.4788) and (119.6549,497.6604) ..    (334.1376,346.9999) .. controls (548.6203,196.3394) and (66.4622,188.6439) ..    (276.0276,346.0724) .. controls (485.5930,503.5010) and (343.9169,42.5633) ..    (258.9527,290.5199) -- cycle;  \end{tikzpicture}  \lipsum[1-2]  \end{document}    See image below: Turning vector grid into individual shapefiles using QGIS? I am using QGIS and I created a vector grid on a crop field in order to cut it into plots. I want to save each grid cell into individual polygon shapefiles automatically.  Here is an example of a vector grid I would like to split into 117 shapefiles:    Eventually, I did it manually with the selection tool and "Save as" but it is really time-consuming when there are more than 100 plots to save.    Is there any way to do it faster and more automatically?  Each plot as an attribute named "Plot" that can discriminate against them (basically being id + 1). It's the attribute used here to label them. Shortest sequence containing all permutations Given an integer $n$, define $s(n)$ to be the length of the shortest sequence $S = (a_1, \cdots a_{s(n)})$ such that every permutation of $\{1,\cdots,n\}$ is a subsequence of $S$.  If $n=1$, then $S = (1)$ is the shortest sequence containing all permutations of $\{1\}$, so s(1) = 1.  If $n=2$, then $S = (1, 2, 1)$ contains all permutations of $\{1,2\}$ as a subsequence, so $s(2)=3$.  Is there a general formula for $s(n)$?