A question about a solution to an ODE problem
Comparison theorem for ODE
If my account is already a member of the administrator group why do some functions on Windows 7 require me to run them as administator? I kind of thought that was implied? If you copy the contents of   httpd.conf   and then paste it into a cat command.. such as this one..  #!/bin/bash cat &gt; /test &lt;&lt; EOF pasted here.. EOF   you run into this error:  -bash: command substitution: line 1: unexpected EOF while looking for matching `'' -bash: command substitution: line 4: syntax error: unexpected end of file   perhaps the solution is to escape the dollar signs and maybe even quotes and so on..  but given that this is such a large file.. can dollar signs be automatically be escaped ?  is my only option to escape them via another program and then give it to the cat command ? How to know if Mac OS X machine is being remotely viewed? Compute the (multiplicative) inverse of $4x+3$ in the field $\frac {\Bbb F_{11}[x]}{\langle x^2+1 \rangle}$? So I am finding a polynomial $px+q$ ($p,q \in \Bbb F_{11}$) which is multiplicative inverse of $4x+3$ in $\frac {\Bbb F_{11}[x]}{\langle x^2+1 \rangle}$.  i.e. $[(4x+3)+\langle x^2+1 \rangle][(px+q)+\langle x^2+1 \rangle]=1+\langle x^2+1 \rangle$   $\Rightarrow$  $(4x+3)(px+q)+\langle x^2+1 \rangle=1+\langle x^2+1 \rangle$  $\Rightarrow$  $4px^2+(4q+3p)x+3q+\langle x^2+1 \rangle=1+\langle x^2+1 \rangle$.  We see that the remainder,when $(4x+3)(px+q)$ is divided by $x^2+1$, is $1$. So by Division algorithm,  $$ \require{enclose} \begin{array}{r} 4p \\[-3pt] x^2+1 \enclose{longdiv}{4px^2+(4q+3p)x+3q} \\[-3pt] \underline{4px^2+4p} \\[-3pt] (4q+3p)x+(3q-4p) \\[-3pt] \end{array} $$    So I equate $(4q+3p)x+(3q-4p)=1$ and solve the simultaneous linear equations  $4q+3p=0, 3q-4p=1$. I get $p=6,q=1$  Hence $6x+1$ is the required inverse.  I am pretty sure that the answer is correct but is the method to achieve it right? Can we exempt downvoted accepted answers from getting the top spot? Customize equation numbering for Equation environment? I want to manually label equation number ? For example I want to label this equation 3.1 instead of (1) which automatically pops up. Is it also possible in split environment as well or not ?      \begin{equation}     \log \mu_{ijk}= \lambda + \lambda_i ^I + \lambda_j ^J+\lambda_k ^K+\lambda_{ij} ^{IJ}+\lambda_{ik} ^{IK}+\lambda_{jk} ^{JK}+\lambda_{ijk} ^{IJK} \label{eq:3.1}     \end{equation} LED light fixture glows when not turned on This past week at Menards I bought 2 Smart Electrician® 3650 Lumens 46" LED Linkable Shop Light Model Number: 0053-0011  |  Menards® SKU: 3481400.  The LED shop light fixture has a glow to it. Looked like the LEDs were turned on, but not all the way. The power switch is turned off. Don’t know why I did this but I took one of those three prong adaptors converters to two prong and to my surprise the light fixture didn’t light up. I unplug the cord with the adaptor and turned it around and it still didn’t light up. The power supply electronics inside the light fixture must be using the earth ground wire somehow. Using my volt meter. I measured to earth ground to the green wire that comes out of the adaptor and to my surprise depending how I plug in the adaptor. I’m getting 112 volts AC. That's because the adaptor isn't keyed and I'm switching the HOT and NEUTRAL around depending how I  plug it in.  I connected the LED light fixture to “Safety Analyzer tester”. When I press the “Open Ground” button, the fixture stops glowing (lightly lite LED’s).  Looks like the power supply circuit inside the fixture has a connection to “Earth Ground”.  I then switched the “Normal” on the polarity switch to “Reverse” and the LED’s were off. Switching the Neutral to open position has no effect. I'm thinking of taking them apart and see inside what's going on or take them back.  I figured out the problem.    In this light fixture you can clearly see the white wire (neutral/common) is connected to the light switch, turning the power off the LED light strip. Commonly we switch off the “HOT” to your light socket so when you change your bulb you have a less of a chance of getting electrocuted (that’s another story for later). Sometimes in Power supply circuits they place capacitors in there to help filter electrical noise out. Also known as suppressers (electronics to reduce unwanted noise or oscillation in a circuit or unwanted frequencies in a signal). Depending on how or where they are placed in the circuit it could possible cause some issue if not installed correctly or if the circuit isn’t designed correctly.   Since the BLACK live wire is connected to the LED PCB all the time, anything connected to that line will potentially be LIVE voltage, that way when I disconnected the earth ground from the socket and measured 112 volts on it. My guess is they have capacitors connected to both LIVE and NEUTRAL to the earth ground wire.  The current is flowing through capacitor C2, just enough to turn the LED’s on slightly. If the switch was on the hot side, this problem wouldn’t be happening. I still may use them just not connect the “earth ground” wire. Because that’s providing the return path for lighting the LED’s. I would appreciate it if someone could refer me to a proof (or simply give one here) for the statement in the title. That is:  If $k&lt;n$, then every $k-$dimensional subspace of $\mathbb{R}^{n}$ has $n-$dimensional Lebesgue measure zero.  I've seen some proofs that use Sard's lemma but I'm not really familiar with that subject and I've never seen a proof of said lemma so I'd appreciate a proof that doesn't use it if possible.  Thanks in advance! recursion-consecutive numbers what is the number of subsets of the set {k∈N|1≤k≤n} with no two consecutive numbers? The answer says: $$a_n=a_{n-1}+a_{n-2}$$ with the starting conditions: $$a_0=1, a_1=2$$1. why does $a_1=2$? $$$$ 2. How does this recursion was created, I understand that if $a_n$ mean all the places with no consecutive numbers so, $a_{n-1}$ come from the fact that after the first place all those places satisfy this condition, but why there is need to "look" at what happen after the two places? Baryons have charges that are the result of a polynomial calculation of their building blocks (quarks)'s fractional charges. But what gives these quarks electric charges? What interactions do they have with photons? And what about leptons, like muons and electrons? How do they get their -1 charge? And then, how do positrons and anti-muons get +1 charge? Same goes for quarks: how does up-antiquarks get -2/3 charge? Or I may want to formulate my question this way:  How does any elementary particle interact with the EM-Force to get an electric charge? I am not a native English speaker, I usually hear “something caught my eye” but never “something caught (both) my eyes”. This seems pretty strange to me.  Cambridge Dictionaries Online provide these examples but without any explanation        (i) to get someone's ​attention:   A ​sudden ​movement ​caught my ​eye.        (ii) to get someone's ​attention, ​especially by ​looking at them:   I ​tried to catch the waiter's ​eye, so we could ​order.        (iii) to be ​attractive or different enough to be ​noticed by someone:   It was the ​unusual ​colour of his ​jacket that ​caught my ​eye.   Why is only one eye ‘caught”? Doesn't something catch our sight? We normally see with both eyes, not with one. If $f: D(0,1)\to \mathbb C$ is holomorphic, then there is a sequence $\{z_n\}$ in $D(0,1)$ such that $|z_n| \to 1$ and $\{f(z_n)\}$ is bounded Prove $\int_0^\infty \frac{\sin^4x}{x^4}dx = \frac{\pi}{3}$ Are delphi variables initialized with a value by default? In Shakespeare's Macbeth, the Third Witch says to Banquo: "Thou shalt get kings, though thou be none" (1.3.65); if I want to quote that line and write it in the third person singular, how should I conjugate the verbs "shall" and "be" correctly? Prove that if $n \in \mathbb{N}$, then $\sum_{d|n}{(d(n))^3}=(\sum_{d|n}{d(n)})^2$ where $d(n)$ is the divisor function. Sort by author in Alpha bibliography Question on groups of order $pq$ How to disable all Vim-LaTeX mappings? I like some of the features provided by the Vim-LaTeX suite, but not all of them.  In particular, I dislike all of the mappings like [[ = \left[ \right] and the automatic placeholder stuff with &lt;++&gt;. I am a very fast typist but with that comes more typing mistakes, which I correct almost instantly by pressing backspace. Unfortunately, if something has automatically added a bunch of place-holders or expanded some expression, I then have to move around the document and remove more characters than necessary. For example, if I accidentally press [ twice, instead of only having to delete one character, I have to delete 10.  Having already looked through the documentation, and some of the source, I have managed to disable some of these "features", but not all of them:  In particular:  let g:Tex_SmartKeyBS = 0 let g:Tex_SmartKeyQuote = 0 let g:Tex_SmartKeyDot = 0   (Turn off maps for backspace, double quote and triple dot.)  let g:Imap_UsePlaceHolders = 0   (Turn off &lt;++&gt; placeholder insertion.)  But I can't seem to find a way of turning off the other stuff. Anyone have any suggestions?  Edit: I was able to "disable" some more things by setting:  let g:Tex_Leader = '`tex' let g:Tex_Leader2 = ',tex'   (Which stops some expansions but still not all. It's not really disabling it but just making it less likely that you're going to type in `texa as opposed to the potentially more common `a, for example.)  PS: Personally, I think it is very poor practice to make an optional package that, when installed, greatly changes the default handling of your system. I don't mind some intelligent source highlighting etc, or some new commands (that don't clobber any old ones), but it seems unhelpful to enable this drastic kind of key remapping by default. How do I handle the window close event in Tkinter?