Note: I think my question has a poor title but I am not sure how to fix it, so please correct if you think of something better.  In my office our phones are equipped with a "daisy-chain" port so that users may plug in both their phone and PC into a single network port.  One of our users messed up and connected the daisy-chain port back into the network (as there were two data ports). This created a ton of problems very quickly and it was only due to dumb luck that I discovered the misconfiguration.  Even while the phone was doubly-connected I was able to connect to other network machines and even browse the web, albeit extremely slowly. There was packet loss and high latency while pinging other machines on the network.  I am wondering:   What was happening while the phone/switch was doubly-connected? How would you diagnose something like this without unplugging stuff from the main switch or patch panel?
What is likely to happen when you plug two ends of a network cable to a single switch/router? Will this create problems on the network, or just be ignored?
Can Tamron 16-300mm Macro Lens fit to Canon EOS 600D? I use Canon EOS 600D and I want a lens with a larger range. I found out Tamron 16-300 f/3.5-6.3 Di II VC PZD Macro Lens, I added it into my wishlist. I did some research on it and I'm scared that it couldn't fit to my Canon EOS 600D.  I'm still a beginner... I'm confused with EF, EF-S and APS-C/DX. Can someone please explain it to me? For some reason all of a sudden when I update cells in Excel, my equations are not being updated.  For example, I have a simple equation =SUM(C4:C16).  If I change any of the cells in C4-C16, the answer doesn't get updated...  I don't know what happened I've never seen this fail before.  Any ideas? If $\sum_{n_0}^{\infty} a_n$ diverges prove that $\sum_{n_0}^{\infty} \frac{a_n}{a_1+a_2+...+a_n} = +\infty $ I've tried to use the limit test: $\lim_{n \rightarrow \infty} \frac{a_n}{b_n} = C$ in the series of $a_n$ and $\frac{a_n}{a_1+a_2+...+a_n}$ but it was inconclusive.  The question does not especify the sign of $a_n$ but proving it for positives would already help.  I think the sign can change as long as the partial sums never get to be zero. Like if the series has only negative numbers the result seems to stil hold. Filling a polygon with a pattern of insets Sources missing in /etc/apt/sources.list, how do I get them back? Understanding a probability paradox Three prisoners are informed by their jailer that one of them has been chosen at random to be executed and the other two are to be freed. Prisoner A asks the jailer to tell him privately which of his fellow prisoners will be set free, claiming that there would be no harm in divulging this information because he already knows that at least one of the two will go free. The jailer refuses to answer the question, pointing out that if A knew which of his fellow prisoners were to be set free, then his own probability of being executed would rise from 1 3 to 1 2 because he would then be one of two prisoners. What do you think of the jailer’s reasoning?  If the jailer refuses to say anything, then the probability that prisoner $A$ is excecuted is $\frac{1}{3}$. If the jailer says to prisoner $A$ that prisoner $B$ will walk free, then $2$ prisoners remain to be considered, $A$ and $C$. One dies, one does not. Heads or tails essentially. $\frac{1}{2}$ ought to be the conditional probability that $A$ dies given that $B$ walks free no? Apparently not though, allegedly the correct answer is still $\frac{1}{3}$. Even my attempt to calculate the correct answer yielded the result $\frac{1}{2}$.   Let $A_D$ and $C_D$ respectively denote the event of $A$ and $C$ dying. Let $B_F$ denote the event that $B$ walks free. Assume that the jailer tells prisoner $A$ that prisoner $B$ will walk free.   Here's my attempt.  $$P(A_D\mid B_F)=\frac{P(A_D\cap B_F)}{P(B_F)}=\frac{P(A_D\cap B_F)}{P((B_F\cap A_D)\cup (B_F\cap C_D))}=\frac{P(A_D)P(B_F\mid A_D)}{P(A_D)P(B_F\mid A_D)+P(C_D)P(B_F\mid C_D)}=\frac{\frac{1}{3}\times 1}{\frac{1}{3}\times 1+\frac{1}{3}\times 1}=\frac{1}{2}$$  What am I doing wrong?  Edit 1: Intuitively I am still troubled but I understand now that $B_F$ may occur even though the jailer does not necessarally say $B$.   Edit 2: I suppose that it makes some sense if the faiths of the prisoners had already been decided before prisoner $A$ asked the jailer the question. If the jailer decides to reveal one of the others who will walk free then that must've been the case. I am moving from one job to another (postdoc to tenure track) this summer and I have papers in several stages of being written, these go from handwritten notes with all the necessary arguments to under revision at a journal. All the papers will have my new contact information put on them but where is the line for changing affiliation? It seems obvious to me that the one that is still in the handwritten notes stage will get the new affiliation and that the ones under revision won't. So somewhere in between is a line.   Where is it?  There are some similar questions so far (see  and  but they do not seem to be quite the same situation since I am not changing fields and the appointments are not concurrent. How to make a file executable? How can I make my file so that I can double click on it, and it runs. It is a .sh script, but I also have files that say:  executable (application/x-executable)  in the description of what they are.  I can't run any of these from terminal, or by double clicking.  If possible, I would like a way using either the GUI or a Terminal, but not a combination of the two.  Here is a screenshot of what I see when I right click then go on properties. The file first:    And the shell script here:    NB: I accept that this is a duplicate (I was looking for it, and couldn't find it, so asked + answered it, hoping that I would find it) however, I don't think the question about .desktop files is a duplicate. I'm so puzzled about this:   $$a^{b^c} = a^{(b^c)} \neq (a^b)^c = a^{(bc)}.$$  Why isn't $a^{b^c}$ equal to $a^{(bc)}$? Why is $a^{b^c}$ instead equal to $a^{(b^c)}$? And how is it possible that $(a^b)^c = a^{(bc)}$?  My mind is pretty much exploding from trying to understand this. If $\gcd(a,b)=1$, then $\gcd(a+b,a^2 -ab+b^2)=1$ or $3$. Hint:  $a^2 -ab +b^2 = (a+b)^2 -3ab.$  I know we can say that there exists an $x,y$ such that $ax + by = 1$.  So in this case, $(a+b)x + ((a+b)^2 -3ab)y =1.$  I thought setting $x = (a+b)$ and $y = -1$ would help but that gives me $3ab =1.$  Any suggestions? How do I read / convert an InputStream into a String in Java? Why don't the Ubuntu repositories have the latest versions of software? How is the value of $\pi$ ( Pi ) actually calculated? When I was a child I was taught $\pi$ (Circumference/Diameter) is an irrational number  and can be approximated to $22/7$ but $= 3.(142857)(\ldots)$.      But where does this value comes from?  In short How do I derive $\pi$? What is the probability of a biased coin coming up heads given that a liar is claiming that the coin came up heads? A biased coin is tossed.   Probability of Head - $\frac{1}{8}$  Probability of Tail - $\frac{7}{8}$  A liar watches the coin toss. Probability of his lying is $\frac{3}{4}$ and telling the truth is $\frac{1}{4}$. He says that that the outcome is Head. What is the probability that the coin has truly turned Head?  My Attempt:  I used the formula: $$P(A \mid B) = \frac{P(A\cap B)}{P(B)}$$  $\ \ \ \ \ \ $P(it is head GIVEN liar said it's head) = P(it's head AND liar said it's head) / P(liar said it's head)  or, P(it is head GIVEN liar said it's head) = $\frac{ \frac{1}{8} \frac{1}{4} }{ \frac{7}{8}\frac{3}{4} + \frac{1}{8}\frac{1}{4} }$ [using a probability tree will be helpful here]  or, P(it is head GIVEN liar said it's head) = $\frac{1}{22}$  The Question: Is the method I used wrong in any way? Some others I have talked to are saying the answer will be $\frac{1}{4}$. Their reasoning is this: since the liar lies 3 times out of 4 and he said it is head, then the probability of it being head is 1/4. So who is right? What will be the answer? Why is setting a variable before a command legal in bash? I've just encountered several answers such as to  that use the construct:  while IFS=, read xx yy zz;do     echo $xx $yy $zz done &lt; input_file   where the IFS variable is set before the read command.  I've been reading through the  but can't figure out why this is legal.  I tried  $ x="once upon" y="a time" echo $x $y   from the bash command prompt but got nothing echoed. Can someone point me to where that syntax is defined in the reference that allows the IFS variable to be set in that way? Is it a special case or can I do something similar w/ other variables? How can I center a too wide table? I have a document containing a table which is slightly too wide for the page. But instead of growing to the right side only, I would like to have it centered on the page.  I have tried to use the center environment but this doesn't seem to help. How do I embed a gather environment within a custom environment? I've tried defining the following custom environment, but I get errors:     \begin{gather} ended by \end{solution}   etc...   Here's a MWE showing my problem:  \documentclass{article}  \usepackage[fleqn]{mathtools}  \newenvironment{solution}{%  &lt;-Solution   \par\addvspace{1em}   \itshape{Solution:}\par\begin{gather}}   {\end{gather}\par\normalfont\hfill\qed\par\addvspace{\medskipamount}}%  \begin{document}   \begin{gather}     testing   \end{gather}    %% This does not work:   \begin{solution}     testing   \end{solution} \end{document} Calculate the last digit of $3^{347}$ Suppose $\int_{[a,b]}f=0,\text{ then }f(x)=0 \forall x\in[a,b]$ Let $a&lt;b$ be real numbers. Let $f:[a,b]\to\mathbb{R}$ be a continuous non-negative function. Suppose $\int_{[a,b]}f=0,\text{ then }f(x)=0 \forall x\in[a,b]$    Proof: Suppose for the sake of contradiction $\exists x_0\in[a,b] \text{ such that } f(x_0)= D&gt; 0$  Since the function is continuous at $x_0$, we have $$\forall \epsilon&gt;0, \exists \delta&gt;0 \text{ such that } |f(x)-f(x_0)|\leq\epsilon\text{ whenever } x\in[a,b]\cap(x_0-\delta,x_0+\delta)$$ choose $\delta=\delta_1$ such that $\epsilon=D/2$, then we can say: $$-D/2 \leq f(x)-f(x_0)\leq D/2$$ $$-D/2 \leq f(x) - D$$ $$f(x)\geq D/2&gt;0$$ This implies that $\int_{[a,b]\cap(x_0-\delta_1,x_0+\delta_1)}f &gt;0$  Now since $\int_{[a,b]} f = 0$. we have  $$\int_{[a,b]}f=\int_{[a,x_0-\delta_1]}f+ \int_{(x_0-\delta_1, x_0+\delta_1)} f + \int_{[x_0+\delta_1,b]}f=0$$ $$\int_{[a,x_0-\delta_1]}f+\int_{[x_0+\delta_1,b]}f&lt;-D/2,$$ but this is a contradiction since $f(x)\geq 0$ and hence $\int_{[a,x_0-\delta_1]}f+\int_{[x_0+\delta_1,b]}f\geq 0$  Is my proof correct? Removing all fraction symbols like “¼” and “½” from a string