I want to solve part b) of the problem below.  Another user has provided the picture below the problem which describes the map transformation.  I am still uncertain about how to produce this image.  I understand the picture depicting the domain, since $A$ is restricted by $x&gt;0$, but how do we then get the image $f(A)$ to wrap around the way it does.  Since the image consists of those points $(x^2-y^2,2xy)$, and in complex form, $(x^2-y^2+i2xy)$, why do we get that $y\neq 0$ for $x\leq 0$ in the image?  it's only $x$ that's restricted in the image, not the real part $(x^2-y^2)$, right?  Thanks!
I'm trying to prove the problem below - which comes from Munkres' "Analysis on Manifolds" book in the section on the Inverse function theorem.  Since its in the chapter on the Inverse Function Theorem I figured I'd start by showing that $f$ satisfies the conditions of the theorem.  Writing the Jacobian shows that it's both $C^r$ and we get $\det f'(x,y)=4x^2+4y^2\neq0$ when $x&gt;0$.  So we can apply the theorem but I'm unsure of how to proceed to show that $f$ is $1$-$1$.  And I didn't see how to use the hint they provided.  Thanks!      Let $f\colon \mathbf R^2\to \mathbf R^2$ be defined by the equation   $$f(x,y)=(x^2-y^2,2xy).$$   (a) Show that $f$ is one-to-one on the set of all $(x,y)$ with $x&gt;0$. [Hint: If $f(x,y)=f(a,b)$, then $\|f(x,y)\|=\|f(a,b)\|$.]
How can I prove these are the adjoint operators? What is the definition of a timelike and spacelike singularity? I'm trying modeling things like this using oracle:  An account can have 1~5 members  So I'll have a account table and a member table with a FK of account. But is that possible to check if the account have less than 5 members?  I've tried constraint and trigger but oracle doesn't allow sub queries in constraint or condition of trigger like  CHECK( SELECT ...) or IF count(*) &gt;= 5  which make things getting hard. I am currently studying  answer on the Moore-Penrose pseudoinverse and the Euclidean norm by the user &quot;Etienne dM&quot;. The first point of their answer proceeds as follows:  Let $x$ be $A^+y$.  Let me begin by the second point. For all $z$, we have: \begin{align} \lVert Az-b \rVert_2^2 &amp;= \lVert Ax-b \rVert_2^2 + \lVert A(z-x) \rVert_2^2 + 2 (z-x)^TA^T(Ax-b)\\ &amp; \geq  \lVert Ax-b \rVert_2^2 + 2 (z-x)^TA^T(Ax-b) \end{align} Moreover, because $(AA^+)^T = AA^+$, $$ A^T(Ax-b) = ((AA^+)A)^Tb - A^Tb = 0$$ Thus, we prove that for all $z$, $\rVert Az-b \lVert_2^2 \geq\rVert Ax-b \lVert_2^2$, that is to say $A^+b$ is as close as possible to $y$ in term of the Euclidian norm $\lVert Ax-b\rVert_2$.   I realise that $x = A^+ y$, but I don't understand how any of this implies that &quot;$A^+ b$ is as close as possible to $y$ in terms of the Euclidean norm $\lVert Ax-b\rVert_2$&quot;. I would greatly appreciate it if people would please take the time to explain this to me. Is there a characterization of the nonconstant entire functions $f$ that satisfy $|f(z)|=1$ for all $|z|=1$?   Clearly, $f(z)=z^n$ works for all $n$. Also, it's not difficult to show that if $f$ is such an entire function, then $f$ must vanish somewhere inside the unit disk. What else can be said about those functions?  Thank you. How do you run as a different user from the Start Menu in Windows 10? Running an application as a different user (e.g. domain admin account) from the start menu (by shift + right-clicking the application) used to be an option in Windows 7 &amp; XP.  However, I can't find that option in Windows 10. The workaround seems to be either 1) to find the application in Windows Explorer (shift + right-click) or 2) use runas.exe from the command line.  However, in order to use those workarounds, I have look up the executable name first. It's a bit difficult because I don't have the name of every RSAT tool or executable name memorized.  (e.g. "Active Directory Users and Computers" is dsa.msc, "Routing and Remote Acces" is rrasmgmt.msc)  Is there a simpler way to do this? How to start applications such as xcompmgr at start up? I've been trying to make xcompmgrto autostart in Lubuntu 11.10 but it just doesn't, I have to start it manually.  I have added the next lines to the autostart script in ~/.config/openbox/  #Run a composite manager xcompmgr &amp;   But whenever I log in it doesn't do anything, I don't really need any special parameters for it I just want it to start at log in.  How can I make it run automatically at login? Sometimes it takes too long for the network manager to update to see new wireless AP. Is it possible to change the time interval between checks? Why are polymorphic values not inferred in Haskell? Using launchd to update MacPorts I want to use launchd to update my MacPorts tree periodically (i.e. port selfupdate.  I have created /Library/LaunchDaemons/org.macports.selfupdate.plist:   $ ls -l /Library/LaunchDaemons/org.macports.selfupdate.plist -rw-r--r--  1 root  wheel  669  8 Nov 12:49 /Library/LaunchDaemons/org.macports.selfupdate.plist   It has the following contents:   &lt;?xml version="1.0" encoding="UTF-8"?&gt; &lt;!DOCTYPE plist PUBLIC "-//Apple//DTD PLIST 1.0//EN" "http://www.apple.com/DTDs/PropertyList-1.0.dtd"&gt; &lt;plist version="1.0"&gt; &lt;dict&gt; &lt;key&gt;Label&lt;/key&gt; &lt;string&gt;org.macports.selfupdate&lt;/string&gt; &lt;key&gt;ProgramArguments&lt;/key&gt; &lt;array&gt;     &lt;string&gt;/opt/local/bin/port&lt;/string&gt;     &lt;string&gt;selfupdate&lt;/string&gt; &lt;/array&gt; &lt;key&gt;StandardOutPath&lt;/key&gt; &lt;string&gt;/var/log/org.macports.selfupdate.log&lt;/string&gt; &lt;key&gt;StandardErrorPath&lt;/key&gt; &lt;string&gt;/var/log/org.macports.selfupdate.err&lt;/string&gt; &lt;key&gt;RunAtLoad&lt;/key&gt; &lt;true/&gt; &lt;key&gt;StartInterval&lt;/key&gt; &lt;integer&gt;604800&lt;/integer&gt;&lt;!-- 1 week --&gt; &lt;/dict&gt; &lt;/plist&gt;   I've loaded it and verified that it's loaded:  $ launchctl list | grep macports -   78  org.macports.selfupdate   But it doesn't look like it actually run.    $ port info wireshark Warning: port definitions are more than two weeks old, consider updating them by running 'port selfupdate'. wireshark @1.12.8_3 (net) ...   What am I missing? "Can’t help but" vs. "can help but" Draw power from a portable mobile charger to power circuits I usually power my beginner prototype circuits using mb102 power supply board, but it needs an external elimeter to draw power and then regulate it.  I was looking for a portable solution for my 5 volt and 3.3 volt circuit needs and came up with using a portable battery charger for mobile phones.  I tried powering up a simple transistor and LED circuit with it but it wouldn't power-up.   I feel that there is some load sensing circuit in the portable charger which doesn't detect these low power circuits.  I also tried connecting different resistors to the charger, the charger outputs 5 Volts for a few seconds and then again shuts down and doesn't stay ON continuously.  It would be really comfortable to use a portable charger since it could be charged using a phone charger and would be very handy.  What can I do to make the charger output power on low loads? How do I stop program from skipping over getline? How to copy the filename quickly in windows Interior of cartesian product is cartesian product of interiors I have to prove that:  $$Int(A\times B) = Int(A)\times Int(B)$$  Where $A\subset M$ and $B\subset N$, both $M$ and $N$ metric spaces.  The problem is that the exercise does not specify the metric, so I need to try to prove it using a generic metric.  If $x\in Int(A\times B)$, then an open ball can be centered at $(x,y)$ such that it is contained in $A\times B$. Therefore, this open ball, $B((x,y),r)\subset A\times B$. It means that $d((x,y),a)&lt;r$ for any $a\in A\times B$ for some $r&gt;0$. I guess that if I can take from here that $d(x,a)&lt;r$ and $d(y,a)&lt;r$ I can prove thhat $Int(A\times B)\subset Int(A)\times Int(B)$, but I still have to prove the other way around.   Any ideas? Which are safe and unsafe regions in Iran? Now there's a lot of propaganda in the west that leads us to believe Iran is dangerous and travellers shouldn't go there.  Obviously there are major political disagreements between several governments and the Iranian government and this should be taken into account.  But at a human level I have met many Iranians travelling elsewhere in the world and have found them to be friendly, charming, and hospitable. I have also met several western travellers who spent time in Iran, including hitchhiking around the country, who only have wonderful things to say about the country and its people.  As a traveller I've learned to trust the most what I hear from other travellers about a destination, especially ones which prefer similar ways of travel to myself. I've also learned to trust the opinions of inhabitants of countries bordering on the destination country the least.  Yet some people still insist that Iran is not safe for a traveller, or that .  I can look up for myself the general travel advisories issued by governments and disclaimers in WikiVoyage. So please don't regurgitate those an answers.  I want to know from travellers who've been there, from Iranians, or from references to specific incidents targeting foreigners/westers about which parts of Iran are no problem, and which parts should be avoided, even by experienced intrepid travellers.  (I realize any advice won't apply to Israeli citizens unfortunately. I'm Australian.) My wife wants a 4th baby, but I don't What's the best way of accessing field in the enclosing class from the nested class? I have two expressions of type Expression&lt;Func&lt;T, bool&gt;&gt; and I want to take to OR, AND or NOT of these and get a new expression of the same type  Expression&lt;Func&lt;T, bool&gt;&gt; expr1; Expression&lt;Func&lt;T, bool&gt;&gt; expr2;  ...  //how to do this (the code below will obviously not work) Expression&lt;Func&lt;T, bool&gt;&gt; andExpression = expr AND expr2 I am doing exercise 3.24 of Marcus which is the following.           Let $L,K$ be number fields with $L/K$ a finite extension (of degree $[L:K] = n$) with $R = \mathcal{O}_K$ and $S = \mathcal{O}_L$. A  prime $P$ of $R$ is said to be totally ramified in $L$ iff $PS = Q^n$ for some prime $Q$ of $S$.           (a) Show that if $P$ is totally ramified in $L$ then it is totally ramified in $M$ for any intermediate field $M$, $K \subseteq M \subseteq L$.          (b) If $P$ is totally ramified in $L$ and unramified in another extension $L'$ of $K$ then $L \cap L'= K$.           (c) Give a new proof that $\Bbb{Q}(w)$ where $w = e^{2\pi i/m}$ has degree $\varphi(m)$ over $\Bbb{Q}$ (First do it for $m = p^r$ by using the fact that $(p) = (1- w)^{\varphi(m)}$ and then build up to any $m$ by using (b) above.      Now I have done (a) and (b) but when I come to (c) I fail to see how the problem is essentially circular. Even in the case $m = p^r$, I would have to use the fact that the ring of integers of $\Bbb{Q}(w)$ is $\Bbb{Z}[w]$. The proof of this fact already uses knowledge of $[\Bbb{Q}(w) : \Bbb{Q}]$ being equal to $\varphi(m)$. Am I missing something here?  Thanks.