Why do electrons revolve around the atom's nucleus?  Where does it get the energy for the revolution?  Do the electrons stop revolving at absolute zero temperatures?
I'm having trouble understanding the simple "planetary" model of the atom that I'm being taught in my basic chemistry course.   In particular,    I can't see how a negatively charged electron can stay in "orbit" around a positively charged nucleus. Even if the electron actually orbits the nucleus, wouldn't that orbit eventually decay?  I can't reconcile the rapidly moving electrons required by the planetary model with the way atoms are described as forming bonds. If electrons are zooming around in orbits, how do they suddenly "stop" to form bonds.     I understand that certain aspects of quantum mechanics  to address these problems, and that there are  of atoms. My question here is whether the planetary model itself addresses these concerns in some way (that I'm missing) and whether I'm right to be uncomfortable with it.
Currently I am using Outlook 2010, and sometimes customers claim that they will received an attachment in Windows.dat format.  I have searched online for possible solutions, but mostly the suggestion is to change the sending format to plain text. However, by using plain text format, there is some limitation.  So I just wonder, besides the plain text format, is there a ways to solve the winmail.dat issue?  Besides, does plain text format really 100% avoid the issue? Unable to upgrade pip Compute $e^{e^z}$, where z is a complex number I am self-studying the MAT 67 linear algebra course and I got stuck at a problem in Chapter 2:     Compute the real and imaginary parts of $e^{e^z}$ for $z$ $\in$ $\Bbb C$.   I tried to expand it using Euler's formula but it seemed endless. Any help? I am stuck on the following question:     Find the volume of the torus obtained by rotating the circle $(x-a)^2+y^2=b^2$, where $a&gt;b$, around the y-axis. How to access a SQL Server database from other computer connected to the same workgroup? I have a NOPASSWD line in /etc/sudoers (edited with visudo)  gatoatigrado    ALL=(ALL) NOPASSWD: /bin/set-slow-cpufreq   However, the output is,  gatoatigrado@coral:~&gt; sudo -n /bin/set-slow-cpufreq sudo: sorry, a password is required to run sudo   This kind of command works on an OpenSuSE machine, but not on Ubuntu 11.10. What am I doing wrong?  Note: I cannot find any relevant system log messages, e.g. via tail -f /var/log/syslog.  edit  Here is /etc/sudoers.  Defaults    env_reset  # things I've tried copying from an opensuse machine Defaults always_set_home Defaults env_keep = "LANG LC_ADDRESS LC_CTYPE LC_COLLATE LC_IDENTIFICATION LC_MEASUREMENT LC_MESSAGES LC_MONETARY LC_NAME LC_NUMERIC LC_PAPER LC_TELEPHONE LC_TIME LC_ALL LANGUAGE LINGUAS XDG_SESSION_COOKIE"  root    ALL=(ALL:ALL) ALL gatoatigrado ALL=NOPASSWD: /bin/set-slow-cpufreq %admin ALL=(ALL) ALL %sudo   ALL=(ALL:ALL) ALL I am working on vector auto-regression (VARs) and impulse response function (IRFs) estimation based on panel data with 33 individuals over 77 quarters.  How should this type of situation be analyzed?  What algorithm's exist for this purpose?  I would prefer to conduct these analyses in R, so if anyone is familiar with R code or a package designed for this purpose that they could suggest, that would be especially helpful. Find $cot\space x_n$ if $sin(x_{n+1}-x_n)+2^{-(n+1)}sin\space x_nsin \space x_{n+1}=0$ for $n\ge1$ Also Given     Suppose $x_1=tan^{-1}2&gt;x_2&gt;x_3&gt;......$ are positive real numbers   I cannot understand as to how I should approach but I have a hunch which is a bit hand-wavy way to prove it but tried to show that  $lim_{n \to \infty}x_n =\dfrac{\pi}{4}$ (For the proof see my edits in @hypernova's answer)  But can anybody prove it and then use it to find $cot \space x_n$ please? Methods to compute $\sum_{k=1}^nk^p$ without Faulhaber's formula I remember that a couple of years ago a friend showed me and some other people the following expression:  $$\lim_{n\to \infty}\frac{1^n+2^n+\cdots+(n-1)^n}{n^n}.$$  As shown below, I can prove that this limit exists by the monotone convergence theorem. I also remember that my friend gave a very dubious "proof" that the value of the limit is $\frac{1}{e-1}$. I cannot remember the details of the proof, but I am fairly certain that it made the common error of treating $n$ as a variable in some places at some times and as a constant in other places at other times. Nevertheless, numerical analysis suggests that the value my friend gave was correct, even if his methods were flawed. My question is then:  What is the value of this limit and how do we prove it rigorously?  (Also, for bonus points, What might my friend's original proof have been and what exactly was his error, if any?)  I give my convergence proof below in two parts. In both parts, I define the sequence $a_n$ by $a_n=\frac{1^n+2^n+\cdots+(n-1)^n}{n^n}$ for all integers $n\ge 2$. First, I prove that $a_n$ is bounded above by $1$. Second, I prove that $a_n$ is increasing.   (1) The sequence $a_n$ satisfies $a_n&lt;1$ for all $n\ge 2$.  Note that $a_n&lt;1$ is equivalent to $1^n+2^n+\cdots+(n-1)^n&lt;n^n$. I prove this second statement by induction. Observe that $1^2=1&lt;4=2^2$. Now suppose that $1^n+2^n+\cdots+(n-1)^n&lt;n^n$ for some integer $n\ge 2$. Then  $$1^{n+1}+2^{n+1}+\cdots+(n-1)^{n+1}+n^{n+1}\le(n-1)(1^n+2^n+\cdots+(n-1)^n)+n^{n+1}&lt;(n-1)n^n+n^{n+1}&lt;(n+1)n^n+n^{n+1}\le n^{n+1}+(n+1)n^n+\binom{n+1}{2}n^{n-1}+\cdots+1=(n+1)^{n+1}.$$  (2) The sequence $a_n$ is increasing for all $n\ge 2$.  We must first prove the following preliminary proposition. (I'm not sure if "lemma" is appropriate for this.)  (2a) For all integers $n\ge 2$ and $2\le k\le n$, $\left(\frac{k-1}{k}\right)^n\le\left(\frac{k}{k+1}\right)^{n+1}$.  We observe that $k^2-1\le kn$, so upon division by $k(k^2-1)$, we get $\frac{1}{k}\le\frac{n}{k^2-1}$. By , we may find:  $$\frac{k+1}{k}\le 1+\frac{n}{k^2-1}\le\left(1+\frac{1}{k^2-1}\right)^n=\left(\frac{k^2}{k^2-1}\right)^n.$$  A little multiplication and we arrive at $\left(\frac{k-1}{k}\right)^n\le\left(\frac{k}{k+1}\right)^{n+1}$.  We may now first apply this to see that $\left(\frac{n-1}{n}\right)^n\le\left(\frac{n}{n+1}\right)^{n+1}$. Then we suppose that for some integer $2\le k\le n$, we have $\left(\frac{k}{n}\right)^n\le\left(\frac{k+1}{n+1}\right)^{n+1}$. Then:  $$\left(\frac{k-1}{n}\right)^n=\left(\frac{k}{n}\right)^n\left(\frac{k-1}{k}\right)^n\le\left(\frac{k+1}{n+1}\right)^{n+1}\left(\frac{k}{k+1}\right)^{n+1}=\left(\frac{k}{n+1}\right)^{n+1}.$$  By backwards (finite) induction from $n$, we have that $\left(\frac{k}{n}\right)^n\le\left(\frac{k+1}{n+1}\right)^{n+1}$ for all integers $1\le k\le n$, so:  $$a_n=\left(\frac{1}{n}\right)^n+\left(\frac{2}{n}\right)^n+\cdots+\left(\frac{n-1}{n}\right)^n\le\left(\frac{2}{n+1}\right)^{n+1}+\left(\frac{3}{n+1}\right)^{n+1}+\cdots+\left(\frac{n}{n+1}\right)^{n+1}&lt;\left(\frac{1}{n+1}\right)^{n+1}+\left(\frac{2}{n+1}\right)^{n+1}+\left(\frac{3}{n+1}\right)^{n+1}+\cdots+\left(\frac{n}{n+1}\right)^{n+1}=a_{n+1}.$$  (In fact, this proves that $a_n$ is strictly increasing.) By the monotone convergence theorem, $a_n$ converges.  I should note that I am not especially well-practiced in proving these sorts of inequalities, so I may have given a significantly more complicated proof than necessary. If this is the case, feel free to explain in a comment or in your answer. I'd love to get a better grip on these inequalities in addition to finding out what the limit is.  Thanks! Why does stdafx.h work the way it does? As usual, when my brain's messing with something I can't figure out myself, I come to you guys for help :)  This time I've been wondering why stdafx.h works the way it does? To my understanding it does 2 things:   Includes standard headers which we might (?) use and which are rarely changed Work as a compiler-bookmark for when code is no longer precompiled.   Now, these 2 things seems like two very different tasks to me, and I wonder why they didn't do two seperate steps to take care of them? To me it seems reasonable to have a #pragma-command do the bookmarking stuff and to optionally have a header-file a long the lines of windows.h to do the including of often-used headers... Which brings me to my next point: Why are we forced to include often-used headers through stdafx.h? Personally, I'm not aware of any often used headers I use that I'm not already doing my own includes for - but maybe these headers are necessary for .dll generation?  Thx in advance Obfuscation is one way, but it can't protect from breaking the piracy protection security of the application. How do I make sure that the application is not tampered with, and how do I make sure that the registration mechanism can't be reverse engineered?  Also it is possible to convert a C# application to native code, and  is too costly.  C# provides lot of features, and is the ideal language for my code, so writing the whole codebase again in C++ is out of the question.  Secure certificates can be easily removed from the signed assemblies in .NET. In Spivak's Calculus 3rd Edition, there is an exercise to prove the following:  $$x^n - y^n = (x-y)(x^{n-1} + x^{n-2} y + ... + x y^{n-2} + y^{n-1})$$  I can't seem to get the answer. Either I've gone wrong somewhere, I'm overlooking something, or both. Here's my (non) proof:  $$\begin{align*} x^n - y^n &amp;= (x - y)(x^{n-1} + x^{n-2}y +\cdots+ xy^{n-2} + y^{n-1}) \\           &amp;= x \cdot x^{n-1} + x \cdot x^{n-2} \cdot y + \cdots + x \cdot x \cdot y^{n-2} + x \cdot y^{n-1}\\ &amp;\qquad + (-y) \cdot x^{n-1} + (-y) \cdot x^{n-2} \cdot y + \cdots + (-y) \cdot x \cdot y^{n-2} + (-y) \cdot y^{n-1}\\           &amp;= x^n + x^{n-1} y + \cdots + x^2 y^{n-2} + x y^{n-1} - x^{n-1}y - y^2 x^{n-2}  - \cdots- x y^{n-1} - y^n \\           &amp;= x^n + x^2 y^{n-2} - x^{n-2} y^2 - y^n  \\           &amp;\neq x^n - y^n \end{align*}$$  Is there something I can do with $x^n + x^2 y^{n-2} - x^{n-2} y^2 - y^n$ that I'm not seeing, or did I make a mistake early on?  EDIT:  I should have pointed out that this exercise is meant to be done using nine of the twelve basic properties of numbers that Spivak outlines in his book:     Associate law for addition   Existence of an additive identity   Existence of additive inverses   Commutative law for additions   Associative law for multiplication   Existence of a multiplicative identity   Existence of multiplicative inverses   Commutative law for multiplication   Distibutive law It seems that panel shadow is somehow hardcoded in KDE 4 and devs do not have any plans to change it and/or make it customizable. So, what I'm actually asking is some kind of hack (if there is any) which will remove the shadow of the panel in KDE 4.x completely (but without need to disable all desktop effects in the process).  Possible application: When a user wants a completely transparent panel, no top/bottom borders at all, the shadow is completely out of place and destroys the experience. I'm working under 12.04 and suppose I have a Firefox windows active (or in foreground) with English as input language and I need to type a document in other language using some text editor.   With the text editor in foreground (or active) and the input language set to a non-English one, when I bring Firefox in foreground (or making it active) the input language remains set to the non-English and the language flag does not switch to English (as it would be expected, since I do not alter the language during the whole Firefox session).  Because of this, I have to make extra moves and change the input language manually every time I switch from the text editor to Firefox and back to text editor. This was not happening with 10.04, and each application windows had the corresponding input language set to its default or previous session every time I was bringing it to the foreground!   How will I make 12.04 to behave the same way? Fastest way to convert string to integer in PHP Stopping a Thread in Java? I'm in the process of writing a piece of code that connects to a server spawns a bunch of threads using that connection and does a bunch of "stuff".  There are certain instances where the connection fails and I need to stop everything and start from scratch with a new object.  I wanted to clean up after the object but calling thread.stop on the threads, but this method is seemingly deprecated.  What is the recommended alternative to doing this? Should I write my own cleanup and exit method for each of the threads? Set the thread to null? or something else? Selecting features that do NOT intersect in ArcGIS Desktop? Suppose $f$ and $g$ are entire functions, and $|f(z)| \leq |g(z)|$ for all $z \in \mathbb{C}$, Prove that $f(z)=cg(z)$. Suppose $f$ and $g$ are entire functions, and $|f(z)| \leq |g(z)|$ for all $z \in \mathbb{C}$, Prove that $f(z)=cg(z)$.   My try :  I consider $h(z)=\frac{f(z)}{g(z)}$. If I prove that $h(z)$ is entire, using the fact that $|h(z)|\leq 1$, the result follows immediately from Liouville's theorem.  To prove that $h(z)$ is entire, I have to prove that $h(z)$ has removable singularities at possible zeros of $g(z)$.  Suppose $g(z_0)=0 \Rightarrow |g(z_0)|=0 \Rightarrow |f(z_0)|=0$, BUT what if $z_0$ be a root of of order $k$ of $g(z)$ (i.e. $f^{(n)}(z_0) = 0 $ for all $ 0 \leq n \leq k-1$) and be a zero of order $m$ of $f(z)$ and $k&gt;m$ ?  And if that's not the case, how to rigorously show that $z_0$ is a removable singularity ?  Thank you all in advance. In the Wikipedia article , it says     After Newtonian gravitation was found to be inconsistent with special relativity, . . .   I don't see how Newtonian gravitation itself is inconsistent with special relativity.  After all, Newton's Universal Law of Gravitation is completley analogous to Coulomb's Law, so it would seem that, if there were an analogous "gravitational magnetic field", one could formulate a theory of gravitation in exact analogy with Maxwell's Theory of Electromagnetism, and of course, this would automatically be consistent with Special Relativity.  What about this approach to gravitation does not work?  The only problem I could see with this is the lack of evidence for a "gravitational magnetic field".  That being said, gravity is "weak" as it is, and my guess is that it would be extremely difficult to set up an experiment in which a massive body were moving fast enough for this "magnetic field effect" to be obsevable.  EDIT:  As has been pointed out to me in the answers, Newton itself is inconsistent with SR.  Similarly, however, so is Coulomb's Law, yet, electromagnetism is still consistent with SR.  Thus, how do we know that it is not the case that Newton's Law is the special "static" case of a more general gravitomagnetic theory exactly analogous to Maxwell's Theory:  Let $\mathbf{G}$ be the gravitation field, let $\rho$ be the mass density, and $\mathbf{J}$ be the mass current density, let $\gamma _0$ be defined so that $\frac{1}{4\pi \gamma _0}=G$, the Universal Gravitation Constant, let $\nu _0$ be defined so that $\frac{1}{\sqrt{\gamma _0\nu _0}}=c$, and suppose there exists a field $\mathbf{M}$, the gravitomagnetic field, so that the following equations hold: $$ \vec{\nabla}\cdot \mathbf{G}=-\frac{\rho}{\gamma _0} $$ $$ \vec{\nabla}\cdot \mathbf{M}=0 $$ $$ \vec{\nabla}\times \mathbf{G}=-\frac{\partial \mathbf{M}}{\partial t} $$ $$ \vec{\nabla}\times \mathbf{M}=\nu _0\mathbf{J}+\gamma _0\nu _0\frac{\partial \mathbf{G}}{\partial t} $$ where theses fields would produce on a mass $m$ moving with velocity $\mathbf{v}$ in our inertial frame the Lorentz force $\mathbf{F}=m\left( \mathbf{G}+\mathbf{v}\times \mathbf{M}\right)$.  This theory would automatically be consistent with SR and would reduce to Newton's Law of Gravitation for the case of gravitostatics:  $\frac{\partial \mathbf{G}}{\partial t}=\frac{\partial \mathbf{M}}{\partial t}=\mathbf{J}=0$.  (To be clear, you can't just set the time derivatives equal to $\mathbf{0}$ as it seems I have done.  Upon doing so, you obtain the corresponding static theory, which is technically incorrect (as you can easily see because it won't be relativistically invariant), but is nevertheless often a useful approximation.)  My question can thus be phrased as:  without appealing directly to GR, what is wrong with this theory?