If $\triangle$ is the area of triangle with side lengths $a,b,c$, then show that $\triangle \le\dfrac{1}{4}\cdot\sqrt{(a+b+c)abc}$
proving the inequality $\triangle\leq \frac{1}{4}\sqrt{(a+b+c)\cdot abc}$
Regular Expression to match valid dates I'm trying to write a regular expression that validates a date.  The regex needs to match the following   M/D/YYYY MM/DD/YYYY Single digit months can start with a leading zero (eg: 03/12/2008) Single digit days can start with a leading zero (eg: 3/02/2008) CANNOT include February 30 or February 31 (eg: 2/31/2008)   So far I have   ^(([1-9]|1[012])[-/.]([1-9]|[12][0-9]|3[01])[-/.](19|20)\d\d)|((1[012]|0[1-9])(3[01]|2\d|1\d|0[1-9])(19|20)\d\d)|((1[012]|0[1-9])[-/.](3[01]|2\d|1\d|0[1-9])[-/.](19|20)\d\d)$   This matches properly EXCEPT it still includes 2/30/2008 &amp; 2/31/2008.  Does anyone have a better suggestion?  Edit: I found  on RegExLib  ^((((0[13578])|([13578])|(1[02]))[\/](([1-9])|([0-2][0-9])|(3[01])))|(((0[469])|([469])|(11))[\/](([1-9])|([0-2][0-9])|(30)))|((2|02)[\/](([1-9])|([0-2][0-9]))))[\/]\d{4}$|^\d{4}$   It matches all valid months that follow the MM/DD/YYYY format.  Thanks everyone for the help. Which groups have precisely two automorphisms Which groups $G$ have precisely two automorphisms, i.e., precisely one non-trivial automorphism?  Examples: $G= C_3, \mathbf{Z},\ldots$.  I think $G$ has to be abelian. In fact, we have $ \vert G\vert \geq 3$. Therefore, if $G$ is not abelian, we have at least two non-trivial inner automorphisms.  If we can show that $G$ is cyclic the above examples are all of them. When I choose the Connection Type in VPN Connections, I have only PPTP and I can't find OpenVPN, but I already installed it via terminal with the command sudo apt-get install openvpn easy-rsa  Why I can't find it? Proof by induction of summation inequality: $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots+\frac1{2^n}\ge 1+\frac{n}2$ Prove by induction the summation of $\frac1{2^n}$ is greater than or equal to $1+\frac{n}2$.  We start with $$1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots+\frac1{2^n}\ge 1+\frac{n}2$$ for all positive integers.   I have resolved that the following attempt to prove this inequality is false, but I will leave it here to show you my progress. In my proof, I need to define P(n), work out the base case for n=1, and then follow through with the induction step. Strong mathematical induction may be used.  This is equivalent to $$\sum_{k=0}^n\frac1{2^k}\ge 1+\frac{n}2\;.$$  Let $P(n)$ be summation shown above.   Base case for $n=1$, the first positive integer,   $$\sum_{k=0}^1\frac1{2^k}=\frac1{2^0}+\frac1{2^1}=1+\frac12=\frac32\ge 1+\frac12=\frac32\;,$$  so base case is true.  Induction step: Assume $P(n)$ is true and implies $P(n+1)$. Thus   $$\sum_{k=0}^{n+1}\frac1{2^k}\ge\frac1{2^{n+1}}+\sum_{k=0}^n\frac1{2^k}\ge 1+\frac{n+1}2\;.$$  This can be written as   $$\sum_{k=0}^{n+1}\frac1{2^k}\ge \frac1{2^{n+1}}+1+\frac{n}2\ge 1+\frac{n+1}2\;.$$  I work the math out but I get stuck contradicting my statement. Please show your steps hereafter so I can correct my mistakes. Huge paged pool probably caused by fltMgr.sys , what to do? I have been trying to solve this problem for quite a while. I am still unsure of whether any of the avenues I have pursued have been of any use. Any advice will be much appreciated. Question: Let $V$ be a finite-dimensional inner product space, and let $E$ be an idempotent linear operator on $V$. Prove that if $EE^* = E^*E$, then $E$ is self-adjoint. (This is essentially exercise 5(a) in sec. 80 on p.162 of Paul R. Halmos, Finite-Dimensional Vector Spaces, but Halmos didn't assume that the dimension of $V$ is finite.) Antonym of "phobia" If a phobia is to have an irrational fear of something, what is the word for having an irrational affinity for something?  For example a numerologist may fear the number 13, but be attracted to (or even have a love for) the number 8. (Some would say this is irrational.) Disable or prevent multitouch in Activity Getting the starting shortcut in c# Calculation of $\lim_{n\to \infty}\prod_{r=1}^{n} \left(1+\frac{r}{n}\right)^{\frac{1}{n}}$ Calculation of  $\displaystyle \lim_{n\to \infty}\prod_{r=1}^{n} \left(1+\frac{r}{n}\right)^{\frac{1}{n}}$  $\bf{My\; Solution::}$  Let $y=\displaystyle \lim_{n\to \infty}\prod_{r=1}^{n} \left(1+\frac{r}{n}\right)^{\frac{1}{n}}$, Now taking $\log_{e}$ on both side  $\displaystyle \log_{e}(y) = \lim_{n\to \infty}\log_{e}\left(\prod_{r=1}^{n} \left(1+\frac{r}{n}\right)^{\frac{1}{n}}\right)=\lim_{n\to \infty}\frac{1}{n}\cdot \sum_{r=1}^{n}\log\left(1+\frac{r}{n}\right)$  Now Using Reinman Sum of Integral ...  $\displaystyle \log_{e}(y) = \lim_{n\to\infty}\frac{1}{n}\cdot \sum_{r=1}^{n}\log\left(1+\frac{r}{n}\right) = \int_{0}^{1}\ln(1+x)dx = \log_{e}(4)-\log_{e}(e) = \log_{e}\left(\frac{4}{e}\right)$  So $\displaystyle y =\lim_{n\to \infty}\prod_{r=1}^{n} \left(1+\frac{r}{n}\right)^{\frac{1}{n}} =\frac{4}{e}$  But how can i solve above question using Stirling Approximation,  May I ask for help?  Thank you in advance! I want to write $ k = \{k1,k2,k3,...,kn\} $. How could I do this? Call blocking for iPhone? How do I block unwanted callers without jailbreaking? I want to get user input in one page, store that in a php variable and use it in another php page. I have tried using 'sessions' but it doesn't seem to be working. Is there another safe alternative? This information is likely to be usernames and passwords. Which meta am I supposed to post on? I'm mostly active on  and we have a separate . Most of my questions on IT Security meta have answers with links here. So my question is, where am I supposed to post what i.e. feature request, bug report, behavior questions? Characterization of weak convergence in $\ell_\infty$ Is there some simple characterization of weak convergence of sequences in the space $\ell_\infty$? If yes, is there some similar claim for nets?   I was only able to come up with a characterization of sequential weak convergence using limit along ultrafilters, which I will describe below. I wonder whether there is some insight into this characterization. (E.g. whether there is some simple reformulations which does not use ultralimits.) Moreover, I do not know whether at least this characterization works for nets, too.  We have the following result describing weak convergence in $C(K)$, see e.g. Corollary 3.138,  in Banach Space Theory by Fabian, Habala et al. (It is a consequence of Rainwater theorem and characterization of extreme points of unit ball in $C(K)^*$.)     Let $K$ be a compact topological space. Let $\{f_n\}$ be a bounded sequence in $C(K)$ and $f\in C(K)$. Then, if $f_n\to f$ pointwise, we have $f_n\overset{w}\to f$.   Moreover we have isometric isomorphism between $\ell_\infty$ and $C(\beta\mathbb N)$, which is described e.g. in the Wikipedia article on  or in  of Carothers' book A short course on Banach space theory. This isomorphism assigns to each bounded sequence $(x_n)$ the continuous function $\overline x$ on $\beta\mathbb N$ defined by $$\overline x(\mathscr U) = \operatorname{\mathscr U-lim} x_n,$$ where $\operatorname{\mathscr U-lim} x_n$ denotes the  of $x_n$ w.r.t the ultrafilter $\mathscr U$.  Combining the above results we get the following characterization:      Let $f^{(n)},f\in\ell_\infty$. The sequence $f^{(n)}$ converges to $f$ weakly if and only if for every ultrafilter $\mathscr U$    $$\lim_{n\to\infty} \operatorname{\mathscr U-lim} f^{(n)}= \operatorname{\mathscr U-lim} f.$$   (The above claim for principal ultrafilters is just a pointwise convergence. But in the above claim the equality is required for free ultrafilters, too.) Regression - How do I know if my residuals are normally distributed? Performing a regression and need to find out if my residuals are normally distributed. I'm looking for a word that means     someone (especially a boss) who is not pleased (with your work, attitude etc.) no matter how hard you try. How FluentWait is different from WebDriverWait? How is the FluentWait is different from WebDriverWait? I am using WebDriverWait having the polling interval set to : 500 miliseconds. Still, WebDriverWait returns after some delay (i.e. 2 - 3 seconds after the element load is completed)? Any opinion here? How to round a number to n decimal places in Java What I would like is a method to convert a double to a string which rounds using the half-up method - i.e. if the decimal to be rounded is 5, it always rounds up to the next number. This is the standard method of rounding most people expect in most situations.  I also would like only significant digits to be displayed - i.e. there should not be any trailing zeroes.  I know one method of doing this is to use the String.format method:  String.format("%.5g%n", 0.912385);   returns:  0.91239   which is great, however it always displays numbers with 5 decimal places even if they are not significant:   String.format("%.5g%n", 0.912300);   returns:  0.91230   Another method is to use the DecimalFormatter:  DecimalFormat df = new DecimalFormat("#.#####"); df.format(0.912385);   returns:  0.91238   However as you can see this uses half-even rounding. That is it will round down if the previous digit is even. What I'd like is this:  0.912385 -&gt; 0.91239 0.912300 -&gt; 0.9123   What is the best way to achieve this in Java? Windows INSERT key anti-functionality accidentally triggers; how to stop it permanently?