I've studied ellipses. I've studied physics. But when it comes down to the elliptical orbits of the planets is where I get confused. Ellipses contain two foci — and in the orbits of our solar system the sun is consistently stated as one of them. Okay. What about the second focus point? Why is it never brought up? And if there is no second focus — then how do we even move in an elliptical orbit?
1.In classical mechanics, using Newton's laws, the ellipticity of orbits is derived. It is also said that the center of mass is at one of the foci.   2.Each body will orbit the center of the mass of the system.  My question is : Are the assumptions in 1 and 2 correct?   Follow up question : Assuming the distance from the centre of the mass to each body remains the same, do we have two bodies orbiting the centre of the mass of the system in an elliptical or circular orbit?   Finally : With elliptical orbits, if the heavier mass is supposed to be in one of the foci, if there is any significance to second focus, what is it? Is it a Lagrange point by any chance or does it have some other mathematical property?
What is the difference between building from source and using an install package? I was wondering: when installing something, there's an easy way of double clicking an install executable file, and on the other hand, there is a way of building it from source.   The latter one, downloading a source bundle, is really cumbersome.  But what is the fundamental difference between these two methods? Which arena do free and crown chests use when determining the cards you will get? Silencing "Your disk is almost full" notification After upgrading to macOS Sierra, I'm getting the notification that "Your disk is almost full. Save space by optimizing storage.":    The options appear to be to store my files in iCloud, automatically delete files, or manually delete files:    My problem/irritation is that I have 80GB free of my 440GB volume.  The question: is there a way to silence this notification in a (semi-)permanent fashion, or to change the threshold it uses for the notification? I am studying homology groups and I am looking to try and develop, if possible, a little more intuition about what they actually mean. I've only been studying homology for a short while, so if possible I would prefer it if this could be kept relatively simple, but I imagine it is entirely possible there is no real answer to my query anyway.   As I said above, I want to gain a little deeper understanding of what the n-th homology group actually means: I can happily calculate away using Mayer-Vietoris but it doesn't really give me a great deal of intuition about what the n-th homology group actually means. For example, with homotopy groups, the fundamental group is in some sense a description of how loops behave on the object in question, and it is obvious to me why that is what it is for say, the torus or the circle. However, I have no idea what, if anything, I am actually saying about a triangulable object when I talk about it having 0-th homology group this or 1st homology group that.  The best I have been able to find online or in my limited book selection is the brief description "intuitively, the zeroth homology group counts how many disjoint pieces make up the shape and gives that many copies of $\Bbb Z$, while the other homology groups count different types of holes". What 'different types of holes' are there, roughly speaking? I appreciate that it may often be completely non-obvious what the low-order homology groups are for some complicated construction, but perhaps in simpler examples it might be more explicable. Are there (simple) cases where I could say, just from looking something like e.g. the torus, what its zero-th or first or second etc. homology group was based on the nature of the object? I guess in the zero-th case it is, as my source () above says, related to the number of disjoint pieces. Can we delve deeper than this for the other homology groups?   Any book/website suggestions would be welcomed (preferably websites as I am nowhere near a library!) - I have Hatcher but not a great deal else, and I haven't gleaned as much as I wish to from that alone. Of course I know that there is a great deal we don't know about homology groups even today, so I don't expect some magical all-encompassing answer, but any thoughts you could provide would be appreciated. I hope this question is appropriate for SE Mathematics, apologies if not! -M What process created the /var/lib/dpkg/lock-frontend file in Ubuntu? I have just booted my clean Ubuntu 18.04 LTS VM and tried to install some application with sudo apt install.  But got error message:     E: Could not get lock /var/lib/dpkg/lock-frontend - open (11: Resource temporarily unavailable)       E: Unable to acquire the dpkg frontend lock (/var/lib/dpkg/lock-frontend), is another process using it?   What process created the /var/lib/dpkg/lock-frontend file in Ubuntu? How to gently stop it and take manual control on APT? After Hoth, why did Vader pursue the Millennium Falcon instead of Luke's X-Wing? In The Empire Strikes Back film, the opening crawl says:     The evil lord Darth Vader, obsessed with finding young Skywalker, has dispatched thousands of remote probes into the far reaches of space....   It has also been established in Return of the Jedi that Force users can sense each other:     Chewie whines nervously. Luke stares at the Huge Super Star Destroyer    that looms ever larger before them.      LUKE: Vader's on that ship.   If Darth Vader is obsessed with finding Luke and he can sense his presence, the obvious thing to do would be to follow Luke's X-Wing after the battle of Hoth rather than the Millennium Falcon. Why didn't he? If a PIC MCU provides multiple Vdd/Vss should you provide power to them all? Take a look at this example schematic:  The chip in question is the  and as you can see power is provided on both sides (with a 100nF cap to smooth noise I guess). Is this strictly required or could you put power into one side and just ignore the other? I know that I have just put power in one side and everything still seems to work fine but I feel uneasy with doing that. I feel that they would not waste pins that could be doing something useful by having an extra pair of power in/out pins. For what purpose does this PIC MCU (and others like it) have multiple power pins? My question is from Apostol's Vol. 1 One-variable calculus with introduction to linear algebra textbook.  Page 57. Exercise 12. Show that the following are polynomials by converting them to the form $\sum_{k=0}^{m}a_kx^k$ for a suitable $m$. In each case $n$ is a positive integer. $a)$ $(1+x)^{2n}.$ $b)$ $\frac{1-x^{n+1}}{1-x}, x\not=1.$ $c)$ $\prod_{k=0}^{n}(1+x^{2^k}).$  The attempt at a solution: a) part of the problem is pretty easy I guess, it is example of binomial theorem, so the answer would be $(1+x)^{2n}=\sum_{k=0}^{2n}(^{2n}_{k})x^k.$ Answer to part b) would be the following: $$\frac{1-x^{n+1}}{1-x}=\frac{(1-x)(1+x+x^2+\cdots+x^n)}{1-x}=1+x+x^2+\cdots+x^n=‌​\sum_{k=0}^{n}x^k.$$thanks to @DiegoMath's hint. As for part c), we have $$\prod_{k=0}^{n}(1+x^{2^k})=(1+x)(1+x^2)(1+x^4)(1+x^8)\cdots(1+x^{2^n})$$ and I have trouble &quot;converting&quot; this to sum which would be of a form of polynomial. If the union of two sets is contained in the intersection, then one is contained in the other ($[A\cup B\subseteq A\cap B]\implies A \subseteq B$) Let $A$ and $B$ be sets. Let $A\cup B\subseteq A\cap B.$ Prove that $A\subseteq  B.$   My understanding of this question is that all the elements of set $A$ that intersects with set $B$ exists in the union of sets $A$ and $B$ and because in order for the elements to intersect, the elements must exist in both sets $A$ and $B$. Therefore set $A$ is a subset of $B$.  My question is how do I write the proof down so that it is an acceptable answer? Also does this mean set $B$ is a subset of $A$? Community isn't on ~15 sites! Using PyQGIS to transform layer of polygons into layer of holes inside bbox? I'm using python in QGIS. I have a layer of polygons (its a result of some earlier dissolving and has no overlapping polygons but some holes and islands in holes etc). I want to build a new layer which contains one polygon representing a bbox of the first layer with each polygon cut from it.  Essentially I want to "invert" the layer I have so that where the was polygon, now there is not, and where there was not polygon now there is. All inside a bounding box of the original layer.  I can build a bbox easily and I tried to "subtract" the polygon layer from my new bbox polygon layer but the process slows to a stop and then crashes my PC. Here's the algorythm  processing.runalg("qgis.difference","path/to/polygon.shp","path/to/bbox.shp","path/to/result.shp")   I dont care about attributes (obviously). I just need to cut one set of polygons from one larger polygon.  It seems so simple, I wish I could find a simple command or script to acheive this. As stated at the top I'm looking for a solution in python in QGIS. How do I get back my suspend option in Gnome Shell? With the new version of Gnome Shell in Ubuntu 12.10 they turned  around by just displaying the shutdown option which becomes suspend on pressing Alt.  How do I get back my beloved suspend option? Problem Statement  Let $f(x) = x^3 + ax^2 + bx + c$ and $g(x) = x^3 + bx^2 + cx + a$ where $a,b,c$ are integers with $c\not=0$   Suppose that the following conditions hold:   $f(1)=0$ the roots of $g(x)=0$ are the squares of the roots of $f(x)=0$      $$\text{Find the value of} \: \: a^{2013} + b^{2013} + c^{2013}$$   My attempt  Let $f(x) = (x-p)(x-q)(x-r) \: \:$ and $g(x) = (x - p^2)(x-q^2)(x-r^2)$  Expanding we get $$f(x) = x^3 - (p + q + r)x^2 + (pq + pr + qr)x - pqr$$  $$g(x) = x^3 - (p^2 + q^2 + r^2)x^2 + (p^2q^2 + p^2r^2 + r^2q^2)x - p^2q^2r^2$$  And so we have $$a = -(p+q+r) = -p^2q^2r^2$$ $$b = pq + pr + qr = p^2 + q^2 + r^2$$ $$c = -pqr = p^2q^2 + p^2r^2 + r^2q^2$$  Then we get the following relations  $$a = -c^2$$ $$b = a^2 - 2b \implies 3b = a^2 \implies b = \frac{c^4}{3}$$  From $f(1) = 0$ we have $a + b + c = -1 \implies \dfrac{c^4}{3} + c - c^2 = -1$  However, upon graphing this function, I saw that there are no integer solutions for $c$. Have I done something wrong?  Edit  For future readers, my approach can simplified alot if you replace $r$ with $1$ since $f(1) = g(1) = 0$ I hear so many terms involving the word "linear". Linear function, linear equation, linear system, linear operator, linear transformation, linear mapping, linear space, linear algebra, linear electrical circuits, linear filters, linear electrical elements, linear approximation, linear optimization.  I'm getting crazy trying to understand the application of the "linear concept" to all this aspects (function, equation, mapping, system, operator, transformation, algebra, etc.) and I wish to know the one essence that is to be linear. What something has to be to be linear? If I say something is linear, what do I know for sure about that something (no matter what the something is)?       I heard a definition of linearity is by homogeinity (scaling the input results in a scaled output) and addition (summing the inputs results in summing the outputs). Can I apply this simple definition to all the branches (operator, mapping, system, transformation, algebra, ...) I mentioned ? Do they all behave like a line ?  y = ax + b, for example, is a line but doesn't behave like a line because y is not linear. Java and manually executing finalize My book uses the argument that the multiplicities of a few macrostates in a macroscopic object take up an extraordinarily large share of all possible microstates, such that even over the entire lifetime of the universe, if each microstate had an equal chance of being accessed, fluctuations away from equilibrium would never occur.  My question to this statistical proof is this: In the real world, is there really an infinitesimal but nonzero chance that macroscopic systems could access some of the more unlikely macrostates (e.g. heat transfer from a cold object to a hot object)? Exists a dedicated non-O(n²) Algorithm for avoiding to have overlapping elements? I am generating positions in a random fashion.  In the game world, instantiated elements have a radius, and shouldn't overlap.  The problem is that they will overlap because I use a usual random number generator combined with other fanciness. I can't use other random number generation algorithms where I can guarantee that the elements will never overlap.  A way to solve this is to use a spatial data structure and search for the nearest neighbors in that tree. When I insert an element into that tree, I update it accordingly.  This solution would do the work and it would be fast but my question is:  Is there a dedicated algorithm without that Tree implementation complexity which is faster than O(n²)? centering column headers in siunitx table column Is it feasible to compile Python to machine code? Find a non-negative function on [0,1] such that $t\cdot m(\{x:f(x) \geq t\}) \to 0$ that is not Lebesgue Integrable Problem:  Find a non-negative function $f$ on $[0,1]$ such that $$\lim_{t\to\infty} t\cdot m(\{x : f(x) \geq t\}) = 0,$$ but $f$ is not integrable, where $m$ is Lebesgue measure.  My Attempt:  Let $f(x) = \frac{\chi_{(0,1]}}{\sqrt{x}}$. Then,  \begin{align*} \lim_{t\to\infty} t\cdot m(\{x \in [0,1]: f(x) \geq t\}) &amp;= \lim_{t\to\infty} t \cdot m(\{x\in (0,1]: 1/\sqrt{x} \geq t\})\\ &amp;=\lim_{t\to\infty} t \cdot m(\{x\in (0,1]: x \leq (1/t^2)\})\\ &amp;= \lim_{t\to\infty} t \cdot m((0, (1/t^2)))\\ &amp;= \lim_{t\to\infty} \frac{t}{t^2} = 0. \end{align*}  However, $f(x)$ is integrable over this interval. I have tried functions looking like $f(x) = 1/x^p$ but I cannot find any that will work here. There is a hint that says there is a monotonic function that fits this description.    Also, does anyone know of a list of non-Lebesgue integrable functions on $[0,1]$? I feel as though I could use this for many counterexamples if one were to exist. Thanks!