Eigenvalues and Eigenvectors Order
Can Eigenvalues[] and Eigenvectors[] be assumed to return the same ordering?
Install java Runtime Environment How do I base64 encode (decode) in C? I have binary data in an unsigned char variable. I need to convert them to PEM base64 in c. I looked in openssl library but i could not find any function. Does any body have any idea? Interaction term positive while components are negative? I am confused about the interpretation of my interaction term , I have been through similar questions but I am still not sure about the interpretation that I should give to my output .  So this is my model , Y  stands for earnings management , X1 = CSR , X2=CEO duality , as you can see both of the coefficients are negative CSR reduces earnings management manipulation and Duality in my sample reduces em as well  which is in line with the literature and confirm my hypothesis , but to my surprise my interaction term is positive and significant does that mean that CSR combined with duality increase EM ? Or that CEO duality moderates the effect of CSR on EM knowing that in papers that I've read, they only use&quot; moderate &quot;when the interaction term and the components are all negative . Could you please help me ? I am supposed to comment my results but I don&quot;t understand this . Thank you Proving of Integral $\int_{0}^{\infty}\frac{e^{-bx}-e^{-ax}}{x}dx = \ln\left(\frac{a}{b}\right)$ Prove that      $$ \int_{0}^{\infty}\frac{e^{-bx}-e^{-ax}}{x}\,dx = \ln\left(\frac{a}{b}\right) $$   My Attempt:  Define the function $I(a,b)$ as  $$ I(a,b) = \int_{0}^{\infty}\frac{e^{-bx}-e^{-ax}}{x}\,dx $$  Differentiate both side with respect to $a$ to get  $$ \begin{align} \frac{dI(a,b)}{da} &amp;= \int_{0}^{\infty}\frac{0-e^{-ax}(-x)}{x}\,dx\\ &amp;= \int_{0}^{\infty}e^{-ax}\,dx\\ &amp;= -\frac{1}{a}(0-1)\\ &amp;= \frac{1}{a} \end{align} $$  How can I complete the proof from here? Feature request: Add pronouns to usercards? One of the questions people have raised about the new rules about pronouns is how users will know what pronouns are preferred.   We've had a trolling question which got  when the user included their preferred pronouns at the bottom of the question, and I agree that particularly on Stackoverflow (rather than sites like Interpersonal Skills or Workplace), this should be considered as 'noise' since it isn't the strict Q&amp;A this site is about. (The codez should work whatever your gender).  Currently, the suggested way of displaying your pronouns is to put it in your profile's 'About Me' section. However, this is suboptimal, since it requires opening a new page.  I suggest adding a 'preferred pronouns' field in the user profile, and displaying this on a user's usercard. There's currently a 'new contributor' label, so it could be similar to that, with she/her pronouns or whatever they choose.  In comments, it could be appended to the reputation score that currently gets shown when you hover over a name:   4,305 reputation, she/her pronouns  The default would be blank. No flair updates, no hover-in-comments update. If a user fills in this field in their profile, then it will show. What's the preferred method of writing AJAX-enabled plugins? Conjugacy Class in a matrix group with columns of standard basis I want to find the conjugacy classes of this group: $$G=\{(a_1,a_2,a_3):a_i=\pm e_j\}\subset GL_3(\mathbb C)$$ I know that $S_3$ is one of the subgroups of this group, so I can check which elements of $S_3$ continues to be conjugate in our group, I know that in $Z(G)$ there is two elements $I,-I$. I know that I have to check the diagonal matrices on which ones are conjugate, but I don't know how to finish and find the others conjugacy class of this group, and I don't Know how to find the number of the elements of the conjugacy class. I'll be thankful for help Is it possible to connect two monitors on a laptop, one through HDMI and the other through DisplayPort? For example the Alienware m17x offers 3 connections, HDMI, DisplayPort, and VGA. Is it possible to connect two independent external monitors, one through the DisplayPort and one through the HDMI port? K&R Exercise 1.16 - Limitation on line length Windows 8 disk usage is 100% How to urlencode data for curl command? ... namely that if $n &gt; 1$ and $(n − 1)!\equiv−1\pmod{n}$, then $n$ is prime.  This is for a number theory class I'm in at Penn State. My idea is to follow accordingly, but I can't get it concretely..  Reason by contradiction, suppose that $n$ is not prime.   Then $n = ab$, where $1 &lt; a, b &lt; n$  In particular $(n-1)! = ai$ for some $i\in\Bbb{Z}$. $\\$ If $(n-1)! = nj - 1$ for some $j\in\Bbb{Z}$ then it follows that a probably doesn't divide $-1$? Unique pair of positive integers $(p,n)$ satisfying $p^3-p=n^7-n^3$ where $p$ is prime Q. Find all pairs $(p,n)$ of positive integers where $p$ is prime and $p^3-p=n^7-n^3$.   Rewriting the given equation as $p(p+1)(p-1)=n^3(n^2+1)(n+1)(n-1)$, we see that $p$ must divide one of the factors $n,n+1,n-1,n^2+1$ on the $\text{r.h.s}$.   Now, the $\text{l.h.s}$ is an increasing function of $p$ for $p\ge1$. This implies that for any given $n\ge1$, there is exactly one real $p$ for which $\text{l.h.s}=\text{r.h.s}$. For $p=n^2$, we get $\text{l.h.s}=n^6-n^2&lt;n^7-n^3=\text{r.h.s}.$ This means that either $p&gt;n^2$ or $p&lt;n^2$ must hold.  Assuming $p&gt;n^2$, it follows that the prime $p$ cannot divide any of $n,n+1,n-1$. So $p$ must divide $n^2+1$ and hence $p=n^2+1\quad (\because p&gt;n^2)$.  Substituting the value of $p$ in the given equation we get, $n^2+2=n^3-n\implies n^3-n^2-n=2$. As the factor $n$ on the $\text{l.h.s}$ must divide $2$, the above equation has a unique integer solution $n=2$.  Finally, we get $(5,2)$ as the solution to the given equation.      But how do I conclude this is the only solution possible? Also, why does'nt $p&lt;n^2$ (the case which I ignored) hold? As a bonus question, I would like to ask for any alternative/elegant solution (possibly using congruence relations) to the problem. Why is my SSH login slow? I'm seeing delays in SSH Logins. Specifically, there are 2 spots where I see a range from instantaneous to multi-second delays.   Between issuing the ssh command and getting a login prompt and between entering the passphrase and having the shell load   Now, specifically I'm looking at ssh details only here. Obviously network latency, speed of the hardware and OSes involved, complex login scripts, etc can cause delays.  For context I ssh to a vast multitude of linux distributions and some Solaris hosts using mostly Ubuntu, CentOS, and MacOS X as my client systems.  Almost all of the time, the ssh server configuration is unchanged from the OS's default settings.  What ssh server configurations should I be interested in?  Are there OS/kernel parameters that can be tuned? Login shell tricks?  Etc? Upgrading from Yosemite to El Capitan, but not Sierra We have an older iMac (Model A1311, Mid 2011). It has been running Yosemite. We have run into some of the WIFI issues associated with Yosemite.   Is it possible to upgrade to El Capitan but not Sierra?   El Cap has been out longer. Its my wife's computer, its older, and I really don't want to be debugging Sierra until a few  more "dot" releases come out. QGIS apply style to shapefile I have a script in file bla.sh and it is executable. When I click on it, the script is executed and the window is closed. I'd like the window to stay open.  Something like command cmd /k** command in Windows.  P.S. I don't want to use pause, but I want to able to write more commands after the script was executed. Here there is my proof (quite short and easy) of a rather straightforward result. The text of this question comes from a previous question of mine, where I ended up working on a wrong inequality. Here things are fixed. Still, I would like to know:  if it is sound, because absolute value always creates me some problem, and if there is a shorter (neater) way to get the result (maybe without using contradiction).    Proposition: $|d(x,y)-d(y,z)| \leq d(x,z)$. Proof: We proceed by cases.  Case 1: $d(x,y)-d(y,z) \geq 0$. Assume by contradiction that $d(x,y)-d(y,z) &gt; d(x,z)$. Hence, $d(x,y)&gt; d(y,z) + d(x,z)$, contradicting the triangle inequality.  Case 2: $d(x,y)-d(y,z) &lt; 0$. Assume by contradiction that $ - d(x,y) + d(y,z) &gt; d(x,z)$. Hence, $d(x,y) &lt; d(y,z) - d(x,z)$. By applying the triangle inequality, we have that $d(y,x) \geq d(y,z) + d(x,z)$, which is by symmetry equivalent to  $d(x,y) \geq d(y,z) + d(x,z)$, obtaining the desired contradiction. $\square$     As always, any feedback is welcome. Thank you for your time. Dynamic Object Intellisense Example where $f\circ g$ is bijective, but neither $f$ nor $g$ is bijective Can anyone come up with an explicit example of two functions $f$ and $g$ such that: $f\circ g$ is bijective, but neither $f$ nor $g$ is bijective?  I tried the following: $$f:\mathbb{R}\rightarrow \mathbb{R^{+}} $$ $$f(x)=x^{2}$$  and  $$g:\mathbb{R^{+}}\rightarrow \mathbb{R}$$ $$g(x)=\sqrt{x}$$  $f$ is not injective, and $g$ is not surjective, but $f\circ g$ is bijective  Any other examples?