Graphing the tangent to a curve in real time with Manipulate
sliding a tangent line along a curve
Why do some actors in credits appear as - Character (As Actor) Limit position rigidbody How can I limit the position of a rigidbody, between two values ​​if I use this code:  public Vector3 dir;  Rigidbody rb;  void Awake() {     rb = GetComponent&lt;Rigidbody&gt;(); }  void FixedUpdate() {     rb.velocity = new Vector3(0f, 0f, 6f);      rb.AddForce(dir * 1f, ForceMode.VelocityChange); }   If I use Vector3.Clamp then the object will shake Why do light disappears the moment we switch off the source (inside the wooden box)? How do I download Minecraft 1.6.4? I tried connecting to a server this morning but it said "outdated client". I understand that that means I have to get Minecraft 1.6.4 but I have no clue how to do it. I tried getting it on Minecraft.net but I couldn't find it. Please tell me how to download it and use it. Show function is continuous at origin $$ \begin{cases}xy^2\sin\left(\frac 1 y\right), \text {if $y$ $\neq$ 0} \\ 0, \text{if $y$ = 0} \end{cases}$$  I'm not sure how to work out this limit as $x$ goes to infinity. If it were just the y part, I know that $y^2$ goes to 0 and $\sin$ is a limited function, so it has to go to 0, but I don't know what to do with the $x$ term. Given a joint density function, what is the conditional expectation $E[Y|x]$? The random variables $X$ and $Y$ have the joint density $$f_{X,Y}(x,y)= \left\{\begin{matrix}e^{-y}, \mbox{ } 0\leq x \leq y \le \infty \\ 0,  \mbox{   otherwise}\end{matrix}\right.$$ Evaluate the conditional expectation $E[Y|x]$.    I'm not sure if I'm doing this correctly. Please let me know if something is off in my solution. Thanks for reading.  My solution: (My expected value is negative, which I'm pretty sure can never be the case. Also in line 4 and 8, are the domains correct?)  $$E[Y|x] = \int_{-\infty}^{\infty}yf_{Y|X}(y|x)dy.$$  $$f_{Y|X}(y|x) = \frac{f_{X,Y}(x,y)}{f_{X}(x)}.$$    $$f_{X}(x) = \int_{-\infty}^{\infty}f_{X,Y}(x,y)dy$$   $$= \int_{0}^{\infty}e^{-y}dy $$  $$=\left [ -e^{-y} \right]\Big|_0^\infty $$  $$ = -e^{-\infty}-(-e^{0})=1. $$ $$f_{X}(x) = 1, x \geq 0.$$ Thus,   $$E[Y|x] = \int_{0}^{\infty}-ye^{-y}dy$$  $$= -\int_{0}^{\infty}ye^{-y}dy$$ After integration by parts,  $$E[Y|x] = \left [ ye^{-y}+e^{-y} \right]\Big|_0^\infty$$ $$=0-[e^{0} + 0]=-1$$ Therefore, $E[Y|x] = -1.$  Thanks in advance for reading this long problem and responding! Show that, for all $n > 1: \frac{1}{n + 1} < \log(1 + \frac1n) < \frac1n.$ I'm learning calculus, specifically derivatives and applications of MVT, and need help with the following exercice:      Show that, for all $n &gt; 1$ $$\frac{1}{n + 1} &lt; \log(1 + \frac1n) &lt; \frac1n.$$    I tried to follow the below steps in order to prove the RHS inequality:   Proving that $f &lt; g$ on $I$ from $a$ to $b$:  Step $1$. Prove that $f' &lt; g'$ on $\operatorname{Int}(I)$.  Step $2$. Show that $f(a) \leq g(a)$ or that $f(a^+) \leq g(a^+)$  Following the above steps, let $f(x) = \log(1 + \frac1x)$ and $g(x) = \frac1x$, for all $x &gt; 1$. One has  $$f'(x) =  -\frac{1}{x^2 + x} \quad \text{and} \quad g'(x) = -\frac{1}{x^2}.$$  We note that, for every $x &gt; 1$, $f'(x) &gt; g'(x)$. Moreover, $f(1^+) = \log(2) &lt; 1 = g(1^+).$   My problem is that I got the wrong inequality sign in Step $1$.    Looking at the solution in my textbook, the author suggests using the MVT but I don't know how to apply it in this case. How do I deal with a runaway bride? I killed my first wife once I found out there is a mercenary you can marry who gives you up to 300 gold a day (in Whiterun). The only requirement, of course, being that I had to hire her first. So I hired her, and then I popped the question, to which she replied the usual: tell the priest to arrange the wedding.   When we reached Riften I told her to wait, just outside the bee and barb, and arranged the wedding. I went through the process, and the second it finished she left, I followed but as I left she was nowhere to be seen. I have tried going back to my house, the pub where I hired her and that same spot in Riften about 6 times each now, some during the day, some during the night and cannot find her at all.   Now this isn't to much of a problem because I can re-marry...oh wait, to remarry, my spouse has to die, so I can't claim gold off her and can't remarry, and the autosaves have been used up fast travelling to the places I thought she might've been. My last save is something like 6 levels ago, and I don't want to have to load it.   What should I do about this situation? Why KL divergence is non-negative? Why is KL divergence non-negative?  From the perspective of information theory, I have such an intuitive understanding:  Say there are two ensembles $A$ and $B$ which are composed of the same set of elements labeled by $x$. $p(x)$ and $q(x)$ are different probability distributions over ensemble $A$ and $B$ respectively.   From the perspective of information theory, $\log_{2}(P(x))$ is the least amount of bits that required for recording an element $x$ for ensemble $A$. So that the expectation $$\sum_{x \in  ensemble}-p(x)\ln(p(x))$$ can be interpreted as at least how many bits that we need for recording an element in $A$ on average.   Since this formula puts a lower bound on the bits that we need on average, so that for a different ensemble $B$ which brings about a different probability distribution $q(x)$, the bound that it gives for each element $x$ will surely not bit that is given by $p(x)$, which means taking the expectation, $$\sum_{x\in ensemble}-p(x)\ln(q(x))$$ this average length will surely be greater than the former one, which leads to $$\sum_{x\in ensemble }p(x)\frac{\ln(p(x))}{\ln(q(x))} &gt; 0$$  I don't put $\ge$ here since $p(x)$ and $q(x)$ are different.  This is my intuitive understanding, is there a purely mathematical way of proving KL divergence is non-negative? The problem can be stated as:  Given $p(x)$ and $q(x)$ are both positive over real line, and $\int_{-\infty}^{+\infty}p(x)dx = 1$, $\int_{-\infty}^{+\infty}q(x)dx = 1$.  Prove $$\int_{-\infty}^{+\infty}p(x)\ln\frac{p(x)}{q(x)}$$ is non-negative.  How can this be proved? Or can this be proved without extra conditions? Unable to mount Windows (NTFS) filesystem due to hibernation How to do a secure erase on a Macbook Pro? I have a Macbook Pro with SSD and I would like to completely erase every single bits of data on it. How would I do that? On a PC, I would just run Darik's Boot And Nuke, but can I run it on a Mac too, or is there a better alternative? How do I remove all loose edges? I have cut a hole in a plane, but I am left with some loose edges.  Is there an easy way to select and delete them all?  See this screenshot:    See this edges siting out with no face? Normal subgroups of infinite symmetric group I recently took a course on group theory, which mentioned that the following proposition is equivalent to the continuum hypothesis: "The infinite symmetric group (i.e. the group of permutations on the set $\mathbb{N}$) has exactly 4 normal subgroups." Does anyone have any references or explanation for this? What are routed events and how it's different from normal events Using my idle computer for something useful (seti@home and the like) Multiple servers acting like a single one with all the hardware? by now I have 10 servers for hpc, power computing oriented. My users need to launch several processes using qmake. The users are used to work with ubuntu 9.10, and the software from the repositories is switable for them.  I've deployed ubuntu 9.10 to all 10 servers (pxe rocks).  By now we work with parallel-ssh and cluster-ssh, which allows as to launch the same process to all servers. With this tools this tools the servers remain as independent but with the same software and the same launched command.  Now we would like to go to next step and see all the servers as a single one with all the resources from the other 9 as if was its resources.   The difference would be substantial in time to process and also time to design the command to launch.  Any advice on wich software to use will be very useful?  Thanks How can I prevent my Wi-Fi connection from glitching? Each time I changed and saved .config/user-dirs.dirs, and after restarted the system, the XDG_DOWNLOAD_DIR are reverted:  diff --git a/.config/user-dirs.dirs b/.config/user-dirs.dirs index 8f6e2db..dcff34e 100644 --- a/.config/user-dirs.dirs +++ b/.config/user-dirs.dirs @@ -7,11 +7,9 @@  #   XDG_PUBLICSHARE_DIR="$HOME/public"  XDG_TEMPLATES_DIR="$HOME/public/template" -  XDG_DESKTOP_DIR="$HOME/private/desktop"  XDG_DOCUMENTS_DIR="$HOME/private/doc"  XDG_PICTURES_DIR="$HOME/private/picture"  XDG_MUSIC_DIR="$HOME/private/music"  XDG_VIDEOS_DIR="$HOME/private/video" - -XDG_DOWNLOAD_DIR="$HOME/var/downloads" +XDG_DOWNLOAD_DIR="$HOME/"   How to make my changes permanent? Do I have to use “do” in any “wh-” question? What did you say?  What you said?   What did you wear? What you wore?    Where did you go? Where you go?    I know that sometimes in English, people do not use “do” in “wh-” questions. I have heard somewhere that people do not use “do” when the answer is a proper noun or a unique thing. Am I right? Trying to understand what is a p value I am trying to understand what a p value is and what is its importance in statistics, from this example.  For $50$ specimens of alloy steel the rock hardness scale measured an average of $62$ with standard deviation $8$ he manufacturer claims that the alloy has average strength of $64$ at significance of $0.01$ can you refute this?  Well I did this problem and saw no at this level you cant refute it , I use the null being that the hardness was $64$ and the alternate that the hardness was less then $64$. From the data we observe a $z=-1.7677$ but the chart is $z=-2.33$ and so we cant refute null at this level.  But now I want to understand what the p value would be ,  I thought it was that the p-value is the probability of getting the observed value of the test or a value with even more evidence against the null if the null is actually true. So using that I just look on the z chart, the probability of seeing Z greater then $1.7677$ is $0.0384$, so is that the p value?