Time Complexity of Fibonacci Algorithm
Computational complexity of Fibonacci Sequence
How to navigate to C drive in bash on WSL-Ubuntu? Let $A$ be a diagonalizable matrix and $\lambda$ be an eigenvalue of A. Prove that rank($\lambda I - A$) = rank(($\lambda I - A)^2$). Let $A$ be a diagonalizable matrix and $\lambda$ be an eigenvalue of A. Prove that rank($\lambda I - A$) = rank(($\lambda I - A)^2$).  The hint they provided is that for any invertible matrix B, rank(AB) = rank (A) and rank(BA) = rank(A). Why does Ridders' method work as well as it does? I've just read section 9.2.1 in Numerical Recipes Ed. 3 (Press et al. 2007), which describes Ridders' method of root finding. I understand that allowing for some curvature of the function by linearizing three points with an exponential will probably get you a better approximation and more rapid convergence, but I have no idea how they get that it converges quadratically, and the motivation behind choosing an exponential (rather than e.g. a polynomial) is not clear to me.  So, specifically, I'm wondering (1) why linearize with an exponential, and (2) why the method converges quadratically.  I tried looking for Ridders' original paper "A new algorithm for computing a single root of a real continuous function", but the only copy I could find was in a journal that my school's library does not give me free access to.  UPDATE:  I've realized that the fact that an exponential is strictly positive means that weighting the three points with an exponential won't change their signs. Hence, the root of the linearization will have to land inside the original bracket.  Polynomials, rationals, sinusoids, and logarithms all have positive and negative bits, so using them probably runs the risk of getting bad interpolated roots. One of these squared might be usable, but maybe non-monoticity gives some sort of undesirable effect. Since this is a complete list of all simple functions that I am aware of, exponentials seem to be the only easily available weighting function with this property.  A constant function is also technically single-signed and monotone, but it can't be used to linearize stuff. Install ffmpeg in Ubuntu 12.04 LTS Background :  I used ffmpeg to merge two videos side by side. First I install ffmpeg by following command as it suggest by Ubuntu itself.  sudo apt-get install libav-tools   or   sudo apt-get install ffmpeg   But then I realize it is not the correct version since it doesn't support for -filter_complex and few other functions.   After a few hours researcher I found  and I succeed with video conversion as well. As per that article, it is suggesting we need to build it our own.        Question :  What is easiest way to install that ffmpeg version (which is support for -filter_complex ) without following all those steps ? appears to have been question banned because of two unpopular Meta suggestions. (He says so in )  Look at the two contributions that appear to have gotten him banned. Like those of many others, they may not be of perfect quality, but they are not actively harmful to the site, either. As it stands, you can get question blocked on Meta simply for asking questions that aren't popular with the Meta crowd.  Why does the team insist on keeping the question ban active on Meta?  Why can't real problem users be dealt with through manual moderator action? Don't tell me it's a problem of scale; as far as I can see, actually toxic users are very rare on Meta.  People will frequently complain that Meta actively discourages the expression of unpopular viewpoints and suggestions. With the threat of question bans in place, I find those complaints difficult to refute. I hate that. Why does Windows only show about 3.5 GB of my 4 GB of RAM? Hungerford's book of algebra has exercise $6$ chapter $3$ section $6$ [Probably impossible with the tools at hand.]:     Let $p \in \mathbb{Z}$ be a prime; let $F$ be a field  and let $c \in  F$. Then $x^p - c$ is irreducible in $F[x]$ if and only if $x^p - c$   has no root  in $F$. [Hint: consider two cases: char$(F) = p$ and   char$(F)$ different of $p$.]   I have attempted this a lot. Anyone has an answer? $ d_1,d_2\mid n\iff {\rm lcm}(d_1,d_2)\mid n\ $ [LCM Universal Property] What does the dot mean If $E/F$ is algebraic and every $f\in F[X]$ has a root in $E$, why is $E$ algebraically closed? How to calculate $\lim_{n\to\infty}(1+1/n^2)(1+2/n^2)\cdots(1+n/n^2)$? How to Join Games with Minecraft Pocket Edition 0.7 To show that ${x_n}$ is convergent where $|x_{n+1} - x_n|< \frac{1}{n^2}$ Let ${x_n}$ be a sequence in $R$ such that $|x_{n+1} - x_n|&lt; \frac{1}{n^2}$ for all $n \in N$.  Show that the sequence is convergent.  If it were $|x_{n+1} - x_n|= \frac{1}{n^2}$, could take help of the fact $\sum \frac{1}{n^2}$ is convergent and of triangular inequality.  But what to do here? Please help. Thanks in advance. How does editing work? It is possible for users to edit an existing post.   How do edits work? Who may edit a question or answer? How can you tell what has been changed between edits?      For more information, see "" in the .     Related: Failed to Login : Invalid session (Try restarting your game) Minecraft Error When I try to connect to servers or join up on lan worlds in Minecraft it says "Failed to Login : Invalid session (Try restarting your game)". Multiplayer used to work fine but now keeps coming up with this error. Can anybody help? Exercise on local and global symmetry for a given Lagrangian density Proving that free modules are flat (without appealing projective modules) Suppose $R\neq 0$ is a commutative ring with $1$. Let $M$ be a free $R$-module. I would like to prove that $M$ is a flat $R$-module. Everywhere I have looked (mostly online) this is proved by first proving that every free module is projective, and then proving that every projective module is flat. Unfortunately, Atiyah &amp; Macdonald's "Introduction to Commutative Algebra" (Chapter 2) does not discuss projective modules. But the result that every free module is flat comes very handy in the exercises.   So my question is,     Is it possible to prove that every free module is flat just by   definitions and without appealing to projective modules?   Thanks! Eclipse is giving me a warning of the following form:     Type safety: Unchecked cast from Object to HashMap   This is from a call to an API that I have no control over which returns Object:  HashMap&lt;String, String&gt; getItems(javax.servlet.http.HttpSession session) {   HashMap&lt;String, String&gt; theHash = (HashMap&lt;String, String&gt;)session.getAttribute("attributeKey");   return theHash; }   I'd like to avoid Eclipse warnings, if possible, since theoretically they indicate at least a potential code problem.  I haven't found a good way to eliminate this one yet, though.  I can extract the single line involved out to a method by itself and add @SuppressWarnings("unchecked") to that method, thus limiting the impact of having a block of code where I ignore warnings.  Any better options?  I don't want to turn these warnings off in Eclipse.  Before I came to the code, it was simpler, but still provoked warnings:  HashMap getItems(javax.servlet.http.HttpSession session) {   HashMap theHash = (HashMap)session.getAttribute("attributeKey");   return theHash; }   Problem was elsewhere when you tried to use the hash you'd get warnings:  HashMap items = getItems(session); items.put("this", "that");  Type safety: The method put(Object, Object) belongs to the raw type HashMap.  References to generic type HashMap&lt;K,V&gt; should be parameterized. Prove divisibility law $\,b\mid a,c\,\Rightarrow\, b\mid ka + lc$ for all $k,l\in\Bbb Z$ What is the recommended CIDR when creating VPC on AWS? I have been creating AWS VPCs and I am wondering if there is a recommended CIDR value when creating VPCs. What are the factors that I must consider when choosing a CIDR and does the CIDR value affect the performance of the network?