Prove that  $|e^z - 1| \le e^{|z|} - 1$ for all complex number $z$  How to prove this? I have tried a lot, like z=x+yi. But it doesn't give a way to correctly solve this problem
I am trying to prove this inequality  $$|e^z-1|\leq e^{|z|}-1\leq |z|e^{|z|}$$  I've tried calculating the difference in their power series   $$\sum_{k=0}^\infty\frac{|z|^k}{k!}-1-\left|\,\sum_{k=0}^\infty\frac{z^k}{k!}-1\,\right|$$   but did not get anywhere.
Are there any keyboard shortcuts for Windows 10 developer preview's virtual desktops ("Task View")? What is the best way to find out a researcher's current affiliation? Let's say I know a name and a field, but not much more.   A google search seems like the most obvious approach, but unfortunately university personnel or research group pages often seem to be quite poorly maintained, and it is not uncommon that someone is listed even if they left more than a year ago.  Early-career researchers are particularly difficult to track down since they often have less of a "presence" on institution webpages and are likely to be moving relatively frequently. So I just got this Nikon D3100 Camera.  And I am trying to take a picture using autofocus with the kit lens(18-55mm).    The system is in Auto mode, AF-A, single frame and the lens A-M switch is on "A",  VR is "ON".  When I press shutter half way down, focus confirmation light in lower left corner of viewfinder neither lit nor blinks(hear clicking). When I press the shutter all the way down, the shutter seems locked (hear clicking, but shutter not released).    The manual says the shutter will lock itself if it can't get focus. But the place is well lit and have distinct color and I can easily focus using my old Canon.    The camera works fine in manual focusing, but I can't get it to take picture under auto-focusing.  I wonder is there something wrong with the setting that I don't know or is that I got a lemon?  Thanks I am already a little bit familiar with Linux distros like Debian or Ubuntu (yeah, very similar) but I wanted to try Red Hat based - CentOS 6.2. I have installed it on my Windows 7 host in VirtualBox and tried to play with it a little. I have come across a small problem, namely: the default eth0 interface is down by default. I use the option with NAT (the virtual machine is 'behind' the host). Even if I bring the interface up with ifconfig eth0 up, it does not work right away. I get this after bringing the interface up: eth0      Link encap:Ethernet  HWaddr 08:00:27:0F:00:8A           inet6 addr: fe80::a00:27ff:fe0f:8a/64 Scope:Link           UP BROADCAST RUNNNING MULTICAST MTU:1500 Metric:1           RX packets:0 errors:0 dropped:0 overruns:0 frame:0           TX packets:6 errors:0 dropped:0 overruns:0 carriers:0           collisions:0 txqueuelen:1000           RX bytes:0 (0.0 b)  TX bytes:468 (468.0 b)           Interrupt:19 Base address:0xd020  lo        Link encap:Local Loopback           inet addr:127.0.0.1  Mask:255.0.0.0           inet6 addr: ::1/128 Scope:Host           UP LOOPBACK RUNNING  MTU:16436  Metric:1           RX packets:0 errors:0 dropped:0 overruns:0 frame:0           TX packets:0 errors:0 dropped:0 overruns:0 carrier:0           collisions:0 txqueuelen:0           RX bytes:0 (0.0 b)  TX bytes:0 (0.0 b)  virbr0    Link encap:Ethernet  HWaddr 52:54:00:75:C2:9B           inet addr:192.168.122.1  Bcast:192.168.122.255  Mask:255.255.255.0           UP BROADCAST RUNNING MULTICAST  MTU:1500  Metric:1           RX packets:0 errors:0 dropped:0 overruns:0 frame:0           TX packets:0 errors:0 dropped:0 overruns:0 carrier:0           collisions:0 txqueuelen:0 [root@centos ~]# _  What should be done more to configure the network on CentOS machine? Setting default values (conditional assignment) In Ruby you can easily set a default value for a variable   x ||= "default"   The above statement will set the value of x to "default" if x is nil or false  Is there a similar shortcut in PHP or do I have to use the longer form:  $x = (isset($x))? $x : "default";   Are there any easier ways to handle this in PHP? What exactly is a kilogram-meter? Was the Wookiee language fully designed or is it just random emotionally-relevant sounds? We know that some of the other Star Wars languages were fairly well developed by the movie's creators, in terms of vocabulary, grammar, etc.  Was the Wookiee language designed in this level of detail or did they just try to put together sounds that made sense emotionally? How to prove by induction that $\frac{a^n+b^n}{2}\geq\left(\frac{a+b}{2}\right)^n$? I'm about to prove that for any $a,b&gt;0$ and $n\in\mathbb{N},$ the inequality: $\frac{a^n+b^n}{2}\geq\left(\frac{a+b}{2}\right)^n$ holds.   By induction I get: $$\left(\frac{a+b}{2}\right)\cdot\frac{a^n+b^n}{2}$$ $$=\frac{1}{2}\cdot\frac{a^{n+1}+b^{n+1}+ab^n+ba^n}{2}$$ $$=\frac{1}{2}\left(\frac{a^{n+1}+b^{n+1}}{2}+\frac{ab^n+ba^n}{2}\right).$$  Now I have to prove that $$\frac{1}{2}\left(\frac{a^{n+1}+b^{n+1}}{2}+\frac{ab^n+ba^n}{2}\right)$$ $$\leq\frac{a^{n+1}+b^{n+1}}{2}$$ $$\Leftrightarrow a^{n+1}+b^{n+1}\geq ab^n +ba^n.$$ But I don't know how to prove that one, is it possible to be proved by induction ? Thank you. Cleaning PCBs... using water? I found  where a technician with a lot of experience seems to clean his PCBs using water, and then he just puts it in the sun so it can dry. I was just amazed to see the circuit keeps working as intended, he says he keeps doing this all the time and several times, however I doubt this is normal.  Which leave me some questions:   What are the recommended methods to clean huge PCBs like the one in the video? What happens to charged capacitors when water shorts them out? Solder joints do not get any kind of corrosion? Is the antisolder mask waterproof? Why do I have to enter my password to log in? Using the Stack Exchange login, why do I have to enter my password again and again to log in? Wasn’t the whole purpose of OpenID to do away with this?  This doesn’t even only happen when visiting a new Stack Exchange site – it also happens for sites where I already have an account but which I don’t visit regularly. Here’s what I mean:    Of course I don’t remember my password (it’s a long random string). And so I have to go to my password manager and manually copy the password from openid.stackexchange.com’s account into the login form. – And incidentally, this is what my password manager looks like now:    (And, by the way, those are the only entries in my keychain so far, but by far not the only SE sites where I have an account.)  … I feel that this defeats the purpose of site-wide login and OpenID. In the olden times when MyOpenID was still working properly, I’d simply enter my OpenID on the login form, and the software would take care of the rest – namely, redirect to the OpenID service, (automatically) verify that I was authorised to use it, and redirect back to the original website, having logged me in. It was literally one button click for me. It’s substantially more work now, and I’d like to understand why this regression is necessary.  (Note: of course once a site’s password has landed in my password manager I (probably?) won’t have to enter it again. But still: wasn’t site-wide login and OpenID supposed to solve this already?) Many people (at least in Germany) think that Italians add olive oil to the boiling water for spaghetti. However, Italians tend to tell the opposite (from my experience).  So, can you please state objective reasons for or against adding olive oil to the boiling water for spaghetti? How do I get the path and name of the file that is currently executing? "I go to school tomorrow" to imply that I will attend school tomorrow "I go to school tomorrow" to imply that I will attend school tomorrow. Is the aforementioned expression within inverted commas correct? Reason for the discreteness arising in quantum mechanics? What is the most essential reason that actually leads to the quantization. I am reading the book on quantum mechanics by Griffiths. The quanta in the infinite potential well for e.g. arise due to the boundary conditions, and the quanta in harmonic oscillator arise due to the commutation relations of the ladder operators, which give energy eigenvalues differing by a multiple of $\hbar$. But what actually is the reason for the discreteness in quantum theory? Which postulate is responsible for that. I tried going backwards, but for me it somehow seems to come magically out of the mathematics. How to change the order of the subcategory and category in a post? I guess it is something simple which I'm not able to see. Any help would be much appreciated. Best, M. As part of the definition of a cone spell in the handbook:     A cone's point of origin is not included in the cone's area of effect, unless you decide otherwise.   As an example, what would either case be for a casting of Burning Hands?   Could the caster include themselves? or does this mean that they can exclude the first square of the cone? how is this possible? I know there is some trick, should someone please explain?! Process descendants SQL JOIN - WHERE clause vs. ON clause After reading it, this is not a duplicate of . The answer may be related (or even the same) but the question is different.    What is the difference and what should go in each?  If I understand the theory correctly, the query optimizer should be able to use both interchangeably. So earlier today I came across , and I am having some trouble understanding the solution. The question is as follows:     You throw a fair six-sided die until you get 6. What is the expected   number of throws (including the throw giving 6) conditioned on the event   that all throws gave even numbers?   Quoted from Jimmy Jin in "Elchanan Mossel’s dice problem"  In the paper it goes on to state why a common wrong answer is $3$. Then afterwards explains that this problem has the same answer to,   "What is the expected number of times you can roll only $2$’s or $4$’s until you roll any other number?"  I don't understand why this is the case. If the original problem is asking for specifically a $6$, shouldn't that limit many of the possible sequences?  I also attempted to solve the problem using another method, but got an answer different from both $3$ and the correct answer of $1.5$.  I saw that possible sequences could have been something like:  $$\{6\}$$ $$\{2,6\}, \{4,6\}$$ $$\{2,2,6\}, \{2,4,6\}, \{4,2,6\}, \{4,4,6\}$$ $$\vdots$$  To which I set up the following summation and solved using :  $$\text{Expected Value} =\sum_{n=1}^\infty n\left( {\frac{1}{6}} \right)^n 2^{n-1} = 0.375$$ Obviously this is different and probably incorrect, but I can't figure out where the error in the thought process is.  Any help on understanding this would be greatly appreciated.  A blog post discussing the problem can be found