Coordinates of an n-sphere I'm a little embarrassed to ask this question because it should be easy but it's stumped me for over a week now. The answer will determine how I write some code, so it matters.  According to  the coordinates of an n-sphere in $n$ coordinates is given as  \begin{align} x_1 &amp;= r \cos(\phi_1) \\ x_2 &amp;= r \sin(\phi_1) \cos(\phi_2) \\ x_3 &amp;= r \sin(\phi_1) \sin(\phi_2) \cos(\phi_3) \\     &amp;\vdots\\ x_{n-1} &amp;= r \sin(\phi_1) \cdots \sin(\phi_{n-2}) \cos(\phi_{n-1}) \\ x_n &amp;= r \sin(\phi_1) \cdots \sin(\phi_{n-2}) \sin(\phi_{n-1}) \,. \end{align}  By this formula, the coordinates of an n-sphere in 4 coordinates should be  \begin{align} x_1 &amp;= r \cos(\phi_1) \\ x_2 &amp;= r \sin(\phi_1) \cos(\phi_2) \\ x_3 &amp;= r \sin(\phi_1) \sin(\phi_2) \cos(\phi_3) \\ x_4 &amp;= r \sin(\phi_1) \sin(\phi_2) \sin(\phi_3) \end{align}  Whereas  gives coordinates as (allowing for a change of notation and ordering)  \begin{align} x_1 &amp;= r \cos(\phi_1) \\ x_2 &amp;= r \sin(\phi_1) \cos(\phi_2) \\ x_3 &amp;= r \sin(\phi_1) \sin(\phi_2) \sin(\phi_3) \\ x_4 &amp;= r \sin(\phi_1) \sin(\phi_2) \cos(\phi_3) \end{align}  Which is correct? Note that $x_3$ and $x_4$ are swapped in my version and wolfram's.
Transformations from n-sphere coordinates to cartesian coordinates. I was wondering how one would proceed to convert between coordinate systems in $ \mathbb R^n $.  For $ \mathbb R^2 $ the conversion is easy and just basic trigonometry. Given $(r, \theta)$ we can convert these polar coordinates to cartesian coordinates to $(x=r\cos\theta, y=r\sin\theta)$  The same follows for $ \mathbb R^3 $. Given $(r,\theta_1,\theta_2)$ in spherical coordinates we can convert the spherical coordinates to cartesian coordinates with the following transformation $(x=r\cos\theta_2\cos\theta_1,y=r\cos\theta_2\sin\theta_1, z=r\sin\theta_2)$, just so my point of reference is clear:$\theta_1$is the angle lying in the x-y plane, and  $\theta_2$ is the angle the vector makes with the x-y plane.  So what about for four dimensions and up? For the sake of simplicity we can define the n-sphere vector as $(r,\theta_1,\theta_2,...,\theta_n)$ and the cartesian vectors $(x_1,x_2,...,x_n)$ and maintain the same reference of $\theta_1,\theta_2$ as before.  How do we define a transformation in $ \mathbb R^4 $?  A look into this wikipedia page under the  gave me something that I don't believe is quite right.  As according to the section  $(x_1=r\cos\theta_1,x_2=r\sin\theta_1 \cos\theta_2,x_3=\sin\theta_1 \sin\theta_2\cos\theta_3,...)$ which is already in contradiction to my statement about $ \mathbb R^3$ transformations.   What is the general transformation rule? Am I missing something that I should know from linear algebra that I might be forgetting because my linear algebra is rusty? If so, please let me know what I should review or point me in the right direction.
Often (via Google Now, or when searching for a solution to some problem), I'll open a question from a non-favourite site in the app. Sometimes, I'd like to see what else is going on in the site where the question was posted. Currently, I have one of two ways:   Via tag icon:   Click one of the tag icons beneath the title Delete the [tag] text (at which point, thankfully, the sorting changes to Active from Votes).  Via manually searching for the site, which is more cumbersome. I type in hard and Hardware Recommendations is the fourth entry (below Ask Different, Raspberry Pi and Arduino)! This more annoying for meta sites, where I have to pick Meta from the dropdown menu.   Can I pretty please have a way to head to the site directly from the question? Maybe:   the site name in the top bar could become clickable and lead to the site, or an option in the More menu?     This would be especially useful for Meta sites, where questions posted around the same time could be related. Interestingly enough, most questions that Google Now are for metas of sites I visit on occasion, but haven't pinned to the list. Title of a book telling about people teleported to a hostile planet while in deep meditation I read it sometime in the 90's but it seemed at least 15 years old at that time. The title might be containing something with 'wolves' or 'wilderness'.  Alert - Spoilers below      The book tells of some (I think) earth cult that dictates the practice of meditation. This cult is somewhat of a scam as When reaching the top level of meditation the meditator is either teleported or his consciousness abducted (I can't remember..) to be used as a calculation machine by the abductors.   Any references would be greatly appreciated. Who is the Community user? I would like to revise the default new user registration form in Drupal 7. Rather than to requst "username", I want to request "full name" with helper instructions to enter the text "username is your full name as you would normally use it."  There may be some other minor formatting changes I would like to make as well.  I cannot find an admin menu item to access anything relating to this question. And I am unable to locate a file within any director that has any text that would indicate it is the source for the new user registration form.  I realize this is a bit unconventional, however it is useful for our needs. I am aware of a module that allows the input of First Name / Last Name, but in processing it changes the input to first initial/lastname, ie Tom Thumb = username of "tthumb". The module "Realname Registration" would be just what I am looking for if it would only translate the username the same as the input.   That wouldn't necessarily solve the other formatting edits, but I would be very happy just to be able to manually edit the registration form help text under the username field, instructing the new user to register using thier real name.  Thanks for any help that you can provide.  J.M. For example:  me$ FOO="BAR * BAR" me$ echo $FOO BAR file1 file2 file3 file4 BAR   and using the \ escape character:  me$ FOO="BAR \* BAR" me$ echo $FOO BAR \* BAR   I'm obviously doing something stupid.  How do I get the output BAR * BAR? Prove that $$1+\frac{1}{3}+\frac{1\cdot 3}{3\cdot 6}+\frac{1\cdot 3\cdot 5}{3\cdot 6 \cdot 9}+.........=\sqrt{3}$$   $\bf{My\; Try::}$ Using Binomial expansion of  $$(1-x)^{-n} = 1+nx+\frac{n(n+1)}{2}x^2+\frac{n(n+1)(n+2)}{6}x^3+.......$$  So we get $$nx=\frac{1}{3}$$ and $$\frac{nx(nx+x)}{2}=\frac{1}{3}\cdot \frac{3}{6}$$  We get $$\frac{1}{3}\left(\frac{1}{3}+x\right)=\frac{1}{3}\Rightarrow x=\frac{2}{3}$$  So we get $$n=\frac{1}{2}$$  So our series sum is $$(1-x)^{-n} = \left(1-\frac{2}{3}\right)^{-\frac{1}{2}} = \sqrt{3}$$  Although I know that this is the simplest proof, can we solve it any other way someting like defining $a_{n}$ and then use Telescopic sum.  Thanks. Wielding two weapons without "Two-Weapon Fighting" or "Dual Wielder"? We have a Barbarian on our team that very much likes to switch up his weapons; he goes from a Dwarven Hammer, to a pair of Hand-Axes, to occasionally just going bare-fisted. No issues there, he's fairly adept with all 3.  However, I have only just recently picked up on the way that he rolls his attacks for the two Hand-Axes: He rolls the first attack, adds his modifiers etc. Then he rolls the second attack, this time without modifiers, which is all well and good. But I'm not sure if this action uses his bonus action or not, or is considered a separate action (though, being a level 4 Barbarian, he only has one action per turn), or is merely part of the same action.  If we have a look at the Two-Weapon Fighting feature from the Fighter's fighting styles:     Two-Weapon Fighting   When you engage in two-weapon fighting, you can add your ability modifier to the damage of the second attack. (PHB5e p.72)   Similarly, the Dual Wielder feat also uses the Two-Weapon Fighting feat:     Dual Wielder   You master fighting with two weapons, gaining the following benefits:      ...         You can use two-weapon fighting even when the one-handed melee weapons you are wielding aren’t light. (PHB5e p.164, emphasis mine)      So; yes, this gives a bonus to the second weapon attack, but having asked a  to clarify which action this secondary weapon attack belongs to, it was determined that this uses your bonus action;     [...] the second attack requires a bonus action   So after that long winded explanation, here is the question: Without any additional feats, can you wield and use two weapons at once, and if so, how many actions does it require to make a full attack with both weapons in a turn? extract lines that match a list of words in another file I have file 1 which have those lines:  ATM 1434.972183 BMPR2 10762.78192 BMPR2 10762.78192 BMPR2 1469.14535 BMPR2 1469.14535 BMPR2 1738.479639 BMS1 4907.841667 BMS1 4907.841667 BMS1 880.4532628 BMS1 880.4532628 BMS1P17 1249.75 BMS1P17 1249.75 BMS1P17 1606.821429 BMS1P17 1606.821429 BMS1P17 1666.333333 BMS1P17 1666.333333 BMS1P17 2108.460317 BMS1P17 2108   And file 2 have a list of words:  ATM BMS1   So, the output will be like this:  ATM 1434.972183 BMS1 4907.841667 BMS1 4907.841667 BMS1 880.4532628 BMS1 880.4532628   I know it's really a duplicate question, but I tried all types of grep and sed and awk, maybe it will works with you guys with this tiny example  but I have a very huge file &gt; 1M lines and all previous way doesn't help  it return part of the lines that containing those words although there are other words in file 2 that matches the lines from file 1 How does one actually show from associativity that one can drop parentheses? How can I find posts related to lost reputation due to a "user was removed" message? Today I noticed that my reputation got truncated by 64 points, and when I looked at my reputation history, I saw  notifying me of points removal along to a "user was removed" message. Each has a slightly different timestamp, ranging from 12:39:04 to 12:39:43.    I can't find the questions/answers concerned by this (and I don't remember having won as much as 65 points once, making this even stranger). I would like to find out how I could have lost 65 points while I never won as much in a single question/answer.  Is there a way I can check which of my posts were impacted? What characters/symbols are allowed within the CSS class selectors? I know that the following characters are invalid, but what characters are valid?  ~ ! @ $ % ^ &amp; * ( ) + = , . / ' ; : " ? &gt; &lt; [ ] \ { } | ` # Do I need a transit visa for a layover in the United Arab Emirates? I am really having trouble working a problem in my course. I will post it and then add what I have thought etc.  Let p be an odd prime, then using the Euclidian Algorithm, calculated the GCD of of $x^{p-1}-1$ and $x^2+1$ in $\mathbb{Z} / p\mathbb{Z}[x]$  Also, how does $p=1 (mod4)$ tell us that  $-1$ is a square in $\mathbb{Z} / p\mathbb{Z}$, infact that $-1$ is a square here if and only if $p=1(mod4)$.  My thoughts:   I know that $$x^{p}-x=(x)(x-1)(x-2)…(x-(p-1))$$ and so that in this sense our roots will be the congruence classes of $1,2,…p-1$  and also that $$x^{p-1}-1=(x-1)(x-2)…(x-(p-1))$$  But I get lost when trying to apply the EA to these polynomials,  like how can I do it when p is arbitrary?  Preferably id like to be able to understand/see how to do the first part first i.e. using division with residue and then move on to the second part. Any insight/hint or solution to this would be great,  Thank you all  Update:  When I try to do division with residue, I just get  $$x^{p-1}-1=(x^2+1)(x^{p-3}-x^{p-5}+x^{p-7}-x^{p-9}...)$$  for example,  $p=7$  $$(x^{6}-1)=(x^2+1)(x^4-x^2+1)$$  for $p=11$  $$(x^{10}-1)=(x^2+1)(x^{8}-x^{6}+x^{4}-x^{2}+1)$$  which for specific cases tells me that 1 is the gcd, but I dont know how to generalize that for the case with arbitrary prime p. Bevel is not symmetrical What's the fastest way to find mineshafts? , and one place I'm fairly sure I can find them is a mineshaft.  Which leaves me back at square one, how do I find a mine shaft?  It seems they were made substantially harder to find between 1.8 and 1.0.  What's the fastest way to find a mineshaft? How do you match only valid roman numerals with a regular expression? Predict() - Maybe I'm not understanding it Prove that $J^+AJ$ has positive eigenvalues Let $J \in \mathbb{R}$ be a $m \times n$ matrix of rank $n$ and a $A \in \mathbb{R}$ be a $m \times m$ diagonal positive definite matrix. Denote by $J^+$ a  of $J$.   As was shown  $J^+AJ$ is not positive definite in general.   However, I was wondering whether one can prove or disprove that $J^+AJ$ has real and positive eigenvalues. How do I correctly tag my questions? Here are some tips for tagging questions. These guidelines will help you more accurately tag your questions, which in turn will help them get more attention and lead to better answers faster.     For more information, see "" in the . QR code scanner