I need to prove that circles and lines on the complex plane correspond to circles on Riemann sphere. This is intuitively easy to understand, as lines on the complex plane can be viewed as a circle going through the point $\infty$. But the thing is, how to describe a circle on Riemann sphere? How to make sure that circles and lines map to circles instead of ellipses or other shapes? So I need some features to determine that the images under this projection are indeed circles. Any help would be appreciated!
Suppose $ T \subset \mathbb{C} $. Show that the corresponding set $ S \subset \Sigma $ is  a. a circle if $ T $ is a circle. b. a circle minus (0, 0, 1) if $ T $ is a line.  Here we are defining $ \Sigma $ to be the Riemann sphere, given by the set: $$ \Sigma = \left \{(\xi, \eta, \zeta) : \xi^{2} + \eta^{2} + (\zeta - \frac{1}{2})^{2} = \frac{1}{4} \right \} $$  To take a point from $ \mathbb{C} $ to $ \Sigma $ we can use the following:  $$ \xi = \frac{x}{x^{2} + y^{2} + 1}; \eta = \frac{y}{x^{2} + y^{2} + 1}; \zeta = \frac{x^{2} + y^{2}}{x^{2} + y^{2} + 1} $$  We define a circle on $ \Sigma $ to be the intersection of a plane of the form $ A\xi + B\eta + C\zeta = D $ with $ \Sigma $. We also know the converse of this problem is true, that the intersection above yeilds a set in $ \mathbb{C} $ with the following property:  $ (C - D)(x^{2} + y^{2}) + Ax + By = D $. As you can see, when C = D, then an equation for a line is yeilded, otherwise it is a circle.   I really am at a  loss about how to solve this problem. The only thing I can think to do is to pick 3 points on a circle or radius $ r $ with center $ z_{0} $, use these points to find two vectors in $ \Sigma $, take their cross product to get a normal vector, use this normal vector to get a plane. Once I have the plane in form $ A\xi + B\eta + C\zeta = D $ then I could prove that the circle I had chosen corresponds exactly with $ (C - D)(x^{2} + y^{2}) + Ax + By = D $. Is there not an easier, less computation way to do this?
Find the image of (1,1,1) by T How to remove all files in a directory except hidden ones? First, I'd like to apologize for the awkwardly-phrased title.   Instead of trying to explain the structure I'm thinking of, I'd like to give some context and then an example or two. I've used and seen this construction used conversationally, and if my memory and intuition serve, conversationally only. I'm curious to know if there is a formal term for it.  Ex 1  "If you don't like what I'm offering, the door is that way."  A purely logical interpretation would make little sense with this example. In this context, the inference is that the speaker is indicating to whomever he is speaking that they should leave, which suggests that there might be an elision, e.g.  "If you don't like what I'm offering, ... [then you may leave. T]...he door is that way."  Ex 2  "If you're hungry, I have food."  This is a similar context. The implication is that the speaker will offer food to whom she is speaking if they are hungry. A (strictly) logical interpretation would, again, make little sense (if you are hungry, then I have food doesn't convey any intent to provide it, nor does it make sense; the speaker always has food, she just will offer it if her guest is hungry).  I recognize that this might be too specific to have a term associated with it. I'm just curious to know what it is if it does exist. The height of a string in a gravitational field in 2-dimensions is bounded by $h(x_0)=h(x_l)=0$ (nails in the wall)  and also $\int_0^l ds= l$. ($h(0)=h(l)=0$, if you take $h$ as a function of arc length) .  What shape does it take?  My try so far: minimise potential energy of the whole string,  $$J(x,h, \dot{h})=\int_0^l gh(x) \rho \frac{ds}{l}=\frac{g \rho }{l}\int_0^l h(x) \sqrt{1+\dot{h}^2} dx$$  With the constraint  $$\int_0^l \sqrt{1+\dot{h}^2} dx- l=0$$ If it helps, it's evident that $\dot{h}(\frac{l}{2})=0$.  Generally, this kind of equation is a case of a constrained variational problem, meaning that the integrand in  $$\int_0^l \frac{g \rho }{l}h(x) \sqrt{1+\dot{h}^2} +\lambda(\int_0^l \sqrt{1+\dot{h}^2} dx- l)dx$$  Must satisfy the Euler Lagrange equation. The constraint must also be satisfied.  But, in truth, by this point I am clueless. $\lambda$ is worked through $\nabla J  = \lambda \nabla(\int_0^l \sqrt{1+\dot{h}^2} dx- l)$. I have tried this , but get nonsensical answers.  Is this method the best? If so, in what ways am I going about it wrongly thusfar? Unicode filenames on Windows with Python & subprocess.Popen() SSH Server stops working after reboot, caused by missing /var/run/sshd $H_0(X,A) = 0 \iff A \cap X_i \neq \emptyset \forall $ path-components $X_i$ I've been doing some more exercises in , in particular the following:  Show that $H_0(X,A) = 0$ iff $A$ meets each path-component of $X$.  "$\Leftarrow$": Let $x_i \in A \cap X_i \neq \emptyset \forall $ path-components $X_i$. Then for any $x \in X_i$ there is a path $\gamma$ from $x_i$ to $x$. They therefore differ by a boundary: $\partial \gamma = x - x_i$ and therefore $\{&#160;x \} = \{ x_i \}$ in $H_0(X,A) $. Therefore $H_0(X,A) = 0$.  "$\Rightarrow$" Now for this direction I'm not so sure and I'd be glad if you could give me a hint. Let $H_0(X,A) = 0$. By definition, the relative homology group is calculated from the following sequence of chain groups:  $$ 0 \rightarrow C_0(A) \rightarrow C_0(X) \xrightarrow{\partial_1} C_0(X) / C_0(A) \xrightarrow{\partial_0} 0$$  Then $H_0(X, A) = H_0(C_0(X) / C_0(A)) = ker \partial_0 / im \partial_0 = (C_0(X) / C_0(A) ) / im \partial_1 = 0$.  There are two cases where this equality holds:  (i) $C_0(X) / C_0(A) = 0$  (ii) $im \partial_1 = C_0(X) / C_0(A)$  In case (i), $A \cap X_i \neq \emptyset$ for all $X_i$.  In case (ii) I'm not sure how to proceed. Can you give me a hint? Many thanks for your help!  Edit Or maybe I could do the second direction like this:  $$0 = H_0(X,A) = \oplus_i H_0(X_i, A) \implies  H_0(X_i, A) = 0 \forall i$$  $$ \implies A \cap X_i \neq \emptyset \forall i$$? When does a Square Matrix have an LU Decomposition? See dashboard as non logged user in community Is it possible, as a non logged user, to see report and dashboard in community when it is open to non logged user? And  for analytic dashboards? "Print" and "ui.Label" return different strings in console and in the UI in Google Earth Engine I am working on a variable label in Google Earth Engine and noticed that the label as printed in the console is not the same as the label shown on the map. Why is that? Code below:  var aoi = ee.Geometry.Polygon([[[-106.75701284923326, 42.31332966408458],           [-106.75701284923326, 42.003856624775146],           [-106.31755972423326, 42.06913460511273]]]);  var chirpsSelect = ee.ImageCollection("UCSB-CHG/CHIRPS/DAILY")             .filterDate('1985-01-01','2018-01-01')             .filter(ee.Filter.dayOfYear(3,6))             .filterBounds(aoi)  var chirpsStartImage = ee.String(chirpsSelect.first().get('system:index')) var dateStartString = ((((chirpsStartImage.slice(6,8)).cat('/')).cat(chirpsStartImage.slice(4,6))).cat('/')).cat(chirpsStartImage.slice(2,4))   var sortChirpsSelect = chirpsSelect.sort('system:index',false)  var chirpsEndImage = ee.String(sortChirpsSelect.first().get('system:index')) var dateEndString = ((((chirpsEndImage.slice(6,8)).cat('/')).cat(chirpsEndImage.slice(4,6))).cat('/')).cat(chirpsEndImage.slice(2,4)) var dateTitle = ee.String((dateStartString.cat(' to ')).cat(dateEndString))  print(dateTitle)  var dateLabel = ui.Label({   value: dateTitle })  var panel = ui.Panel() panel.add(dateLabel) Map.add(panel) Can any one tell me the advantage of synchronized method over synchronized block with an example? I believe it's a common practice in some communities to name children after their grandparents or parents or relatives and sometimes even friends and popular people. Places, roads, streets etc., get named after famous people, too. Many inventions and discoveries have been named after people who invented or discovered them.  But I am not yet aware of a term that refers to the person whose name is given to people, places or objects this way. Is there a suitable word to fill in the blank below?     They named their son after the famous football player, David Beckham.      David Beckham is .......... of their son. How do I enable hardware virtualization technology (VT-x) for use in Virtualbox? I have a geotiff raster image that has a coordinate system with longitudes from 0 to 360. The horizontal center of the image is 180 longitude. See image below:    I want to transform it to EPSG:4326 SRS with -180 180 longtitude range.  And I want the center of the image to be at Greenwich meridian (0). I guess this srs is very widely used. I expect the result look like this:    So I use a gdalwarp command to reproject:  gdalwarp -s_srs '+proj=latlong +datum=WGS84 +pm=180dW' -t_srs EPSG:4326 test_col.tif test_4326.tif   But I only get a tiff with bigger dimensions (more pixels) and EPSG:4326 metadata. The image itself looks the same, as the initial one. But I expect it to swap the hemispheres.  How do I gdalwarp an image to be strictly -180 180 EPSG:4326 with the center in 0 longitude?  This is gdalinfo of my initial file:  Origin = (-0.102272598067084,89.946211604095552) Pixel Size = (0.204545196134167,-0.204423208191126) Metadata:   AREA_OR_POINT=Area Image Structure Metadata:   INTERLEAVE=BAND Corner Coordinates: Upper Left  (  -0.1022726,  89.9462116) (  0d 6' 8.18"W, 89d56'46.36"N) Lower Left  (  -0.1022726, -89.9462116) (  0d 6' 8.18"W, 89d56'46.36"S) Upper Right (     359.897,      89.946) (359d53'50.18"E, 89d56'46.36"N) Lower Right (     359.897,     -89.946) (359d53'50.18"E, 89d56'46.36"S) Center      ( 179.8975000,  -0.0000000) (179d53'51.00"E,  0d 0' 0.00"S)   This is gdalinfo after gdalwarp     Origin = (-180.102727401932952,89.946211604095552) Pixel Size = (0.091397622896436,-0.091420837939082) Metadata:   AREA_OR_POINT=Area Image Structure Metadata:   INTERLEAVE=PIXEL Corner Coordinates: Upper Left  (-180.1027274,  89.9462116) (180d 6' 9.82"W, 89d56'46.36"N) Lower Left  (-180.1027274, -89.9699975) (180d 6' 9.82"W, 89d58'11.99"S) Upper Right ( 179.8211116,  89.9462116) (179d49'16.00"E, 89d56'46.36"N) Lower Right ( 179.8211116, -89.9699975) (179d49'16.00"E, 89d58'11.99"S) Center      (  -0.1408079,  -0.0118929) (  0d 8'26.91"W,  0d 0'42.81"S) Removing elements that have consecutive duplicates does not appear to actually define what a rigid shape is. Rather it defines 'same shape' and 'rigid transformations' without giving any definitions of what is necessary and sufficient for a shape to be considered rigid.  For instance, I've seen the following image:    I understand intuitively why the triangle is rigid and quadrilateral is non-rigid. It is also my understanding that inserting a single diagonal connection into the quadrilateral would make the shape rigid? However, given the following image (labeling the shapes $S_1$, $S_2$ and $S_3$ in sequence left to right):    If $S_1$ had a single diagonal connection AB, it would be possible to flip ADB over axis q to obtain a shape similar to $S_2$ with the addition of connection AB (my explanation may not be the greatest). This shape is clearly not the same as $S_1$, so how can we say $S_1$ with a single diagonal connection CD is rigid? I feel I am unclear on what the definition of rigidity actually is. Background: I have the opportunity to pursue a Computer Science graduate degree at a well respected program (around top 20). I spoke with the professors at the university and I feel like I am a good fit with their research program. My ultimate pursuits are to complete a PhD in CS. I've done research as an undergraduate and it has only confirmed my notion that I want to pursue a research track career.  More background: I applied to jobs (in case I was rejected to all the programs I applied for) prior to hearing back from my respective graduate programs. I accepted a position for an industry job (this was to secure I wouldn't be both unemployed &amp; not in school).  Dilemma: Turns out my top graduate school is very interested in me, and I'm very interested in them. I would rather go to graduate school than work in industry forever. The industry job pays very well. I am split between A) working for a maximum of 1 year(It would only be 1 year, seriously I do not care about the money enough to work past a year) and B) going straight to graduate school.   My concerns are as follows, if I choose A):    Could I potentially defer my admissions? If I am not allowed to defer my admissions, would I have a good chance of re-applying and being accepted a year later? Would my potential advisers look down on me for deciding to work a year?   I already made my mind up it would be a 1 year gig if I decide A). I know some people say once you make money, you may not be able to readjust to the graduate salary pay. But I don't think that will apply to because I'm going to live very frugally with or without industry pay. Recently the cranks on my bike started to creak. I checked with my usual "bike-bible" . They recommended dis-assemble, lubricate, re-assemble.  So I did that.  Remove cranks. Remove spindle. Clean. Lubricate. Reassemble.  No, the creak is still there.  (To find out the source-area of the sound I had a friend riding the bike slowly while I ran alongside listening. And the sound where coming from the area of the bottom bracket/cranks. I am happy no neighbor saw me...)   For each revelation of the cranks there are two creak-noises.  If I, when on the bike, push down with both feet I hear a noise. Turn 180 degrees without pressure, push down again and there is the noise again. If I am "too soft", low pressure there are no sound. Since I have clips I have tried to only work with one foot to keep a constant load during a revelation, there where no detectable sound then.  Anyone have any idea what is wrong on the bike? And how I should fix it?  /Leif How to get the ASCII value of a character How do I get the  value of a character as an int in ? Is it possible to convert a Physical Ubuntu Install (Intel based PC with Ubuntu 10.10) to a virtual box disk, so that I can use that same environment on a separate pc or can keep it virtual for testing?  Something similar to "Disk2VHD" which converts a physical Windows 2K3 and above box into a VHD or VMware image.  Will be grateful for your assistance ! Kind Regards