Prove the estimate $$\binom{n}{k} \le \left(\frac{en}{k}\right)^k$$ directly from $$e\left(\frac{n}{e}\right)^n \le n! \le en\left(\frac{n}{e}\right)^n.$$
This is from page 3 of  ().  Copying the relevant segment:     Stirling’s approximation tells us $\sqrt{2\pi n} (n/e)^n \leq n! \leq e^{1/12n} \sqrt{2\pi n} (n/e)^n$. In particular we can use this to say that   $$ {n \choose k} \leq \left(\frac{en}{ k}\right)^k$$   I tried the tactic of combining bounds from $n!$, $k!$ and $(n-k)!$ and it didn't work. How does this bound follow from stirling's approximation?
The standard font packages make global changes to the fonts in my document. What if I want to use a particular font for just a portion of my text?   How do I find the right name of the font? How do I select the font? How do I restrict the scope of the font change?     There are separate answers for different flavors of TeX (quick links): A composition of a periodic function that is somehow not periodic. Given a matrix $A \in R^{n \times n}$ which is normal ($AA^H=A^HA$ where $A^H$ is hermitian of $A$) and nilpotent ($A^k=0$ for some $k$). Now we need to show that $A=0$. (This is essentially exercise 5(b) in sec. 80 on p.162 of Paul R. Halmos' Finite-Dimensional Vector Spaces.) I tried to show in the following way, we know that, $AA^H=A^HA$ pre-multiply by $A^{k-1}\implies A^kA^H=A^{k-1}A^HA$ Now, we have $0 = A^{k-1}A^HA$, since $A$ nilpotent. I am not sure how to proceed from here to show $A=0$. Can someone help me in this problem? I am playing a bit around with Qgis 2.14.6 to understand how it deals with distance measurement.  I took a line-shapefile that I saved in 3 different coordinate systems (4326, 32642 and a national one that doesn't exist in Qgis yet).  Using 32642 and the national one, the length is quite similar (6442m vs. 6428m)  But using 4326 (unprojected WGS84), I get 8057m  In all cases, I just used the Field Calculator and introduced $length.  I was already positively surprised that the result using 4326 was in meters and not in degrees.  To calculate the distance on the dataset in 4326, I assume that Qgis does some kind of re-projection in the background. Does anyone know which one?   Shapefiles here if anyone wants to play around:   EDIT - adding more info after more tests  As you can see, now both my shapefile and the project are in EPSG:4326, so unprojected:    Here the rough measurement of my shapefile, with the Qgis measure tool:   When I use the field calculator, the preview gives me a value in degrees, while the result seems to be in meters:      This would mean that the function $length, like the measurement tool, always gives a result in meters, but using a very coarse approximation of the degrees => meters transformation:    Is this a bug? or should it really behave like this? I find it at best counter-intuitive (but maybe it's just me) and, at worse, completely misleading (it makes very little sense to use, everywhere on earth, the length of 1 degree at the Equator). Any input or opinion here? Should I open a bug report (at least for the inconsistency between the preview and the actual result)? General Information about Eigenvalues for an 3x3 symmetric matrix This is my circuit. I need to find the power dissipated by Vlamp. So the lamp turns on if the Vlamp is more than 2V. When the lamp is on it has an internal resistance of 10 Ohms, and when it is off it acts like an open circuit.  How would I find the power dissipated? I know how to do it for Voltage sources and Resistors, but not elements like these.  One more thing, I already figured out that: Battery has open circuit voltage of 4V R1 = 5 Ohms R2 = 5 Ohms k = 0.2A/V When lamp on and Vlamp = 2V, Rbat = 10 Ohms   . Calculate $I=\int_0^1\frac{\ln(1+x)}{1+x^2}\,dx$ How can I calculate the integral $I=\int_0^1\frac{\ln(1+x)}{1+x^2}\,dx$ by substituting $t=\frac{1-x}{1+x}$. Number of Homomorphisms from $\Bbb{Z}_m$ to $\Bbb{Z}_n$ This question coutesy of Allan Clark's "Elements of Abstract Algebra" (60$\zeta$).   Find the number of homomorphisms from $\Bbb{Z}_m\to \Bbb{Z}_n$ as a function of $m$ and $n$.  This is stumping me, can anyone help? Can I use Wubi for manual installation? Is it possible to use Wubi for Ubuntu 13.04? I read that it will not be supported in the official release due to some problems. Is this going to change in the future, and can we expect to see a future release with support for 13.04? Rooting Stock Firmware - GSII with NFC Word to describe someone who goes to all the events in town! I'm looking for a short word that could describe people who are always going to every event in town. It doesn't need to be an existing word, feel free to create one of your own. It can also be made of words related to events and parties. (ex: club, event, guest list, etc.)  Edit  The word will serve as a new term to describ these people in a favorable way. It will probably be used for an app name so it need to be catchy and fun. ( Preferably ) What is the reason behind the multiple Spider-Man reboots? Everybody has to obey their own parents I am new here and I don't really know how to post, so please forgive my mistakes. My question is this: Is a sentence like "everybody has to obey their own parents" nowadays completely acceptable under a grammatical point of view or it is still preferable, in written form, to use "everybody has to obey his/her own parents"? Thank you very much and sorry if this has been asked before. Sum of Permutations from 0 to n In a question I asked on algorithms with time complexity of $f(x) = n^n$ I was told that enumerating the number of strings that can be formed from a string of length $n$ qualifies.   I.e the sum of all permutations of $n$ from $n$ to $0$ is $n^n$  $\sum ^n_{i=0} nPi$ $= n^n$  Can I please see an easy to understand derivation of that formula.   EDIT The above identity is wrong. I just tested it. Can I get a derivation of the formula for the sum of permutations. If we disbelieve $H_0$, why quote a p value calculated assuming $H_0$ was true? Smaug first appeared in TA 2770 as a mature Fire-drake. I'm not sure if there are canonical sources which say his age or maturity, but he must have been quite large and quite old at the time in order to destroy the city of Dale and capture Erebor.  Since it is safe to assume that Smaug would have been attacking towns and stealing all of the gold and jewels he could get his claws on for most of his life, then he must have amassed a huge fortune prior to taking residence at Erebor. How did he transport his existing horde to Erebor? It seems unlikely that he would have simply abandoned all of his treasure that he had previously accumulated.  EDIT: It has been established that Smaug was not a mature drake at the time of his appearance. However, he may have already had a small horde (compared to Erebor). Why would a slug thrower be more effective against a lightsaber/Jedi than a blaster? From  question about gunpowder in the Star Wars universe it is  that the shot from a  slug thrower is not completely deflectable by a lightsaber. This seems to be listed as an advantage (over blasters) against lightsabers and Jedi.     For example, during the days of the Galactic Republic, mercenaries fearing an intervention by Jedi Knights used rapid-fire slugthrowers that were impossible to completely deflect, unlike blaster bolts.   This doesn't make much sense though. A slug, while maybe not as deflectable should be just as ineffective against a lightsaber as a blaster shot is, due to the nature of a . That is, a blade made of plasma should simply vaporize the slug, should the slug manage to penetrate the containment field.     The weapon consisted of a blade of pure plasma emitted from the hilt and suspended in a force containment field. The field contained the immense heat of the plasma, protecting the wielder, and allowed the blade to keep its shape.   While a slug would seem to be more accurate than a blaster bolt (at least when used by a skilled marksman), I can't imagine that the slug would be faster than a bolt. Or if it is faster I wouldn't imagine it would be so fast that a Jedi would be unable to react with enough speed to block it, either with their saber or with the force. So what is the implied advantage of a slug thrower, versus a blaster, when used against a Jedi? Show: $\sum_{k=1}^{\infty}\mu(\left\{f\geq k\right\})\leq\int f\, d\mu\leq\sum_{k=0}^{\infty}\mu(\left\{f>k\right\})$ Show that für $f\colon (\Omega,\mathcal{A})\to\mathbb{R}_{\geq 0}$ measurable, it is     $$ \sum_{k=1}^{\infty}\mu(\left\{f\geq k\right\})\leq\int f\, d\mu\leq\sum_{k=0}^{\infty}\mu(\left\{f&gt;k\right\}). $$      I do not know if one needs it here, but we defined the integral of a non-negative measurable function $f\colon (\Omega,\mathcal{A})\to (\overline{\mathbb{R}},\overline{\mathcal{B}})$ by $$ \int_{\Omega}f\, d\mu=\sup\left\{\sum_{k=1}^{n}(\inf\left\{f(x): x\in A_k\right\})\mu(A_k): n\in\mathbb{N}, A_1,\ldots,A_n\in\mathcal{A}\mbox{ disjoint },\Omega=\bigcup_{k=1}^{n}A_k\right\}. $$    My first idea was to use this definition, but I did not come along with it...  Maybe you can help me to show that?  (I think, maybe it has something to do with monotone convergence theorem or majorated convergence?) How to pass data from Javascript to PHP and vice versa? I’m currently working on a solar powered boat with a modest sized team of people. For our competition we are only allowed to have 3 12v batteries on board but we are allowed to have up to 52v at any point in our setup. We are hoping to kick up the speed of our outboard motor for the sprint portion of the completion and are looking into using a boost converter for this.   After doing a little research I’m a bit confused about how they fundamentally work. From what I can tell they take in a voltage and current and put out a higher voltage with a lower current with some level of efficiency. But it seems to be independent from the load( in this case the motor) which doesn’t make much sense to me because I thought the current would be determined by whatever my load is. How do loads actually interact with boost converters?( The simpler the better.)  For reference we are using the Saietta 119r motor system and I’m considering using the MPS MPQ3425 boost converter.