I want to draw a diagram like this. Could you please help me?
It's the first time I'm using Tikz on Latex. I have to draw this block diagram (this is a low resolution snip from a paper)   In Tikz I have done this:   That is very similar but I need some adjustments, like the u_i arrow and the node sum after the H(s) block.  Is there anybody that can help me with this?  my code:      \tikzstyle{block} = [draw, fill=white, rectangle,  minimum height=2em, minimum width=3em] \tikzstyle{sum} = [draw, fill=white, circle, radius=1mm, node distance=1.5cm] \tikzstyle{input} = [coordinate] \tikzstyle{output} = [coordinate] \tikzstyle{pinstyle} = [pin edge={to-,thin,black}]  \begin{tikzpicture}[auto, node distance=2cm,&gt;=latex'] \node [input, name=input] {}; \node [input, above of=input](input1) {}; \node [sum, right of=input] (sum) {}; \node [block, right of=sum] (controller) {$K(s)$}; \node [sum, right of=controller] (sum1) {}; \node [block, right of=sum1] (filterinv) {$H^{-1}(s)$}; \node [block, right of=filterinv] (system) {$G(s)$}; \node [output, right of=system] (output) {}; \node [output, above of=output] (output1) {}; \node [block, below of=filterinv] (filter) {$H(s)$}; \node [block, above of=controller] (delay) {$D(s)$}; \node [sum, below of=sum1] (sum2) {};  \draw [-&gt;] (filterinv) -- node[name=u] {$u_{i}$} (system); \draw [draw,-&gt;] (input) -- node {$s_{i-1}$} (sum); \draw [-&gt;] (sum) -- node {$e_{i}$} (controller); \draw [-&gt;] (system) -- node [name=q] {$q_{i}$}(output); \draw [-&gt;] (q) |- (filter); \draw [-&gt;] (filter) -- node {} (sum2); \draw [-&gt;] (controller) -- node {} (sum1); \draw [-&gt;] (sum1) -- node[name=xi] {$\xi_{i}$} (filterinv); \node [input, below of=sum1](input2) {}; \draw [draw,-&gt;] (input2) -- node {$r_{i}$} (sum2); \draw [-&gt;] (sum2) -| node[pos=0.99] {$-$}      node [near end] {$\tilde{s}_i$} (sum); \draw [-&gt;] (delay) -| node[] {}      node [near end] {} (sum1); \draw [draw,-&gt;] (input1) -- node[pos=0.15] {$u_{i-1}$} (delay); \draw [-&gt;] (u) |- node[pos=0.99] {$u_i$}      node [near end] {} (output1);  \end{tikzpicture}
Ackermann function - how to calculate the number of times it calls itself Purely for my own amusement I've been playing around with the .  The Ackermann function is a non primitive recursive function defined on non-negative integers by:  $A(m,n) = n+1$, if $m=0$  $A(m,n) = A(m-1,1)$, if $m&gt;0$ and $n=0$  $A(m,n) = A(m-1,A(m,n-1))$, if $m&gt;0$ and $n&gt;0$  I'm not interested in values of the function itself, rather I'm interested in the number of recursions the function takes to arrive at a value. For example:  $A(1,2) = A(0,A(1,1)) = A(0,A(0,A(1,0))) = A(0,A(0,A(0,1))) = A(0,A(0,2)) = A(0,3) = 4$  That's 6 steps, so if we define $RA(m,n)$ as the number of recursions to calculate $A(m,n)$, then $RA(1,2) = 6$.  I made a short Excel macro to calculate $RA(m,n)$ for any given $m$ and $n$. The problem I have is that $RA(m,n)$ grows even faster than $A(m,n)$ and I quickly run out of stackspace. The table below shows $RA(m,n)$ for very small values of $m$ and $n$. The numbers shown for $RA(3,10)$ and $RA(4,1)$ are the number of steps before crashing!    A formula for $RA(1,n)$ is trivial, and it is relatively easy to see that $RA(2,n) = 5(n+1) + 4n(n+1)/2$.  So my questions: can anyone see an explicit formula for $RA(3,n)$? More generally, does a formula exist for $RA(m,n)$, in the same way that $A(m,n)$ can be expressed using Knuth arrows?   For the low values shown of $m$ and $n$, $RA(m,n)$ looks about order of magnitude larger than $A(m,n)$. Can we prove that $RA(m,n) &gt; A(m,n)$ for all $m,n$. Intuitively, it should be.  For bonus points, can $RA(m,n)$ be expressed in terms of $A(m,n)$? Do we know of any living forms on Earth that could continue to live on Mars? How often is $1+\prod_{k=1}^n p_k$ not prime? Reference: What is variable scope, which variables are accessible from where and what are "undefined variable" errors? Is there a way to fake a dual (second) monitor Is there a way to fake a dual monitor for testing.    I have only one monitor and I have to make an application that is suitable for two monitors.  The second monitor screen should appear as a window on my primary monitor. Which hydroxyl from either the phosphate or the glycerol is taken during synthesis of a phospholipid head? The Follow Questions and Answers feature is now live across the Network We are happy to announce that the  follow question and answer feature is now live across the Network, including Stack Overflow, all Stack Exchange sites, and all Meta sites. (International Stack Overflow sites will have it turned on in a day or two once we have translations all set.) You can follow any question or answer (that you did not author) by clicking on the [follow] button that is shown in the menu immediately below the post (alongside the [share] button):    After you have followed a post, you will get inbox notifications for all new answers (in the case where you followed a question), comments, edits, and notices. You will not receive notifications for any action that you performed. As was mentioned in the earlier post, we are not making changes at this point to the notifications received by a post owner, or due to @mentions. The initial release had been planned to only include question following. Thanks to many folks who chipped in over the past few weeks, we are happy to release both question and answer following at the same time. These are available in both the regular and mobile views. We are still planning on two more related releases:  Follows profile tab and question listing filter: A tab for follows will be added to your user profile activity page. Each user will be able to view this tab that will provide a listing of followed questions and answers, with standard sorts, and the ability to unfollow from the listing. Users (with the exception of moderators and authorized staff) will not be able to see this information about other users (nor will it be made public in the API, SEDE, or data dumps). We will be renaming the Favorites feature to Bookmarks. As we mentioned in the preview post, the feature will be the same as Favorites, with the name and icon updated to more accurately represent user expectations and usage. And you will be able to both &quot;bookmark&quot; and &quot;follow&quot; a question.  We hope that this new feature will allow you to have better access to the content that you care about and want to keep tabs on. Update (June 11, 2020): The user profile tab and the rename of Favorites to Bookmarks have been released. I am using ArcMap 10.1. The background: I have an .mxd that I intend to roll out across the company as our default template for all future maps/figure production. This .mxd contains two Data Frames; the first (&quot;Main&quot;) is larger (occupying almost the entire page) and is designed to be used to hold the majority of mapped data whilst the second (&quot;Inset&quot;) is much smaller and is to be used as an overview map. The problem: When choosing to import this new .mxd to an existing .mxd (using the Change Layout button on the Layout toolbar) the wizard automatically selects the smaller (&quot;Inset&quot;) Data Frame to occupy the larger part of the page. The screenshot below shows this interface with a red box indicating the two possible locations of each Data Frame:  Whilst it is possible to manually alter the order of the Data Frames using the Move Up or Move Down buttons, how do I change the default option for this setting (i.e. so that the &quot;Main&quot; Data Frame always appears in the larger space - &quot;2&quot;)? The reasoning is that I would like the process of using this new .mxd template to be as simple as possible, this is especially important as we have a wide range of users with varying abilities. Does anyone know how to choose/change this default setting? It does not appear to be in the Data Frame Properties &gt; Data Frame menu where I would expect it to be. Integrating $\int_1^2 \int_0^ \sqrt{2x-x^2} \frac{1}{((x^2+y^2)^2} dydx $ in polar coordinates I'm having a problem converting $\int\limits_1^2  \int\limits_0^ \sqrt{2x-x^2}   \frac{1}{(x^2+y^2)^2}  dy dx $ to polar coordinates.  I drew the graph using my calculator, which looked like half a circle on the x axis.  I know that $\frac{1}{(x^2+y^2)^2}  dydx$ turns to $ \frac{r}{(r^2)^2}drd\theta$, which would be $ \frac{1}{r^3}$  The region of integration in  $\theta $ is I guess $0 \leq  \theta  \leq  \frac{1}{4}  \pi $, but I'm stuck for r. The lower boundary is $x=1=rcos\theta \rightarrow r=1/cos\theta=sec\theta$, but what is the upper boundary?? Is it  $\sqrt{2rcos\theta-r^2cos^2\theta}$? I tried this but I couldn't integrate my equation...  Could someone please help me out?  Also, how can I draw the graph of $y= \sqrt{2x-x^2} $? I could only know how it looked like by using my calculator... How do I get current date/time on the Windows command line in a suitable format for usage in a file/folder name? Update: Now that it's 2016 I'd use PowerShell for this unless there's a really compelling backwards-compatible reason for it, particularly because of the regional settings issue with using date. See @npocmaka's     What's a Windows command line statement(s) I can use to get the current datetime in a format that I can put into a filename?  I want to have a .bat file that zips up a directory into an archive with the current date and time as part of the name, for example, Code_2008-10-14_2257.zip. Is there any easy way I can do this, independent of the regional settings of the machine?  I don't really mind about the date format, ideally it'd be yyyy-mm-dd, but anything simple is fine.  So far I've got this, which on my machine gives me Tue_10_14_2008_230050_91:  rem Get the datetime in a format that can go in a filename. set _my_datetime=%date%_%time% set _my_datetime=%_my_datetime: =_% set _my_datetime=%_my_datetime::=% set _my_datetime=%_my_datetime:/=_% set _my_datetime=%_my_datetime:.=_%  rem Now use the timestamp by in a new ZIP file name. "d:\Program Files\7-Zip\7z.exe" a -r Code_%_my_datetime%.zip Code   I can live with this, but it seems a bit clunky. Ideally it'd be briefer and have the format mentioned earlier.  I'm using Windows Server 2003 and Windows&nbsp;XP Professional. I don't want to install additional utilities to achieve this (although I realise there are some that will do nice date formatting). Are partially charged wands generally purchasable? I have a problem with using "too" in a sentence. In what situation would you use "too" instead of "so" or "very"? Please include an example to illustrate the difference. I am working on vector auto-regression (VARs) and impulse response function (IRFs) estimation based on panel data with 33 individuals over 77 quarters.  How should this type of situation be analyzed?  What algorithm's exist for this purpose?  I would prefer to conduct these analyses in R, so if anyone is familiar with R code or a package designed for this purpose that they could suggest, that would be especially helpful. I am aware that the "Eye of Sauron" in Peter Jackson's movies was not Sauron's actual physical form.  What the answers at '' don't indicate is, despite this, what abilities of vision Sauron may have had.  Did he have the ability to see from afar? If so, what are his limitations or details of this power? Is it some kind of Elvish-sight like Legolas but stronger? Was it a magical power? Or perhaps with the help of the Palantir (I'm not sure of this last one because I dont remember if the Palantir could only see what other Palantíri were seeing or anything he wanted)? Where can I buy an Ubuntu Phone? I need to buy an Ubuntu phone, where can I buy it? It is unavailable everywhere. Is the converse of Cayley-Hamilton Theorem true? The question is motivated from the following problem:     Let $I\neq A\neq -I$, where $I$ is the identity matrix and $A$ is a real $2\times 2$ matrix. If $A=A^{-1}$, then the trace of $A$ is   $$ (A) 2 \quad(B)1 \quad(C)0 \quad (D)-1 \quad (E)-2$$   Since $A=A^{-1}$, $A^2=I$. If the converse of  is true, then $\lambda^2=1$ and thus $\lambda=\pm1$. And then $\rm{trace}(A)=1+(-1)=0$.   Here are my questions:   Is  $C$ the answer to the quoted problem?  Is the converse of Cayley-Hamilton Theorem, i.e.,"for the square real matrix $A$, if $p(A)=0$, then $p(\lambda)$ is the characteristic polynomial of the matrix $A$" true? If it is not, then what's the right method to solve the problem above? Is the matrix diagonalizable for all values of t? Why is this sequence of functions not uniformly convergent? For the following sequence of functions and its limit function, we can see that $f_n(x)$ is clearly pointwise convergent  $$f_n(x) = x^n\text{ }\forall x\in[0,1]\text{ and }\forall n\in\mathbb N^*\\ f(x) = \begin{cases}0&amp;\text{if } x\in[0,1)\\1&amp;\text{if } x=1\end{cases}$$  However, I was wondering why this is not uniformly convergent. The condition for uniform convergence is:  $$|f_n(x) - f(x)| &lt; \epsilon,\ \ \ \forall x \text{ when } n &gt; N$$  Now most sources present an argument along the lines of: assume that $f_n(x)$ is uniformly convergent and that $0 &lt; x &lt; 1$, this means that $x^n&lt;\epsilon$ whenever $n&gt;N$. Specifically, this would mean $x^{N+1}&lt;\epsilon$ for some fixed $N$. But if we now pick $x$ such that $1 &gt; x &gt; ε^{\frac{1}{N+1}}$, then this would lead to a contradiction, therefore $f_n(x)$ is not uniformly convergent.  However, I was wondering why couldn't we take $n$ to infinity. If $0 &lt; x &lt; 1$, then $\lim_{n\rightarrow \infty} |f_n(x) - f(x)| = 0$ (which is less than $\epsilon$). Now since $|f_n(x) - f(x)|$ will always be $0$ if $n &gt; \infty$, then wouldn't this be uniformly convergent? Bound for the correlation of three random variables There are three random variables, $x,y,z$. The three correlations between the three variables are the same. That is,  $$\rho=\textrm{cor}(x,y)=\textrm{cor}(x,z)=\textrm{cor}(y,z)$$  What is the tightest bound you can give for $\rho$? What is the maximum or minimum tire width I can fit on my bicycle