I have a problem with derivation of some function. This is related to computing the Poincare Map of Logistic Population model with Periodic harvesting. This function is:  $\phi(t,x_0) = x_0 + \int_{0}^{t} f(s,\phi(s,x_0)) ds \ (1)$  where $\phi(t,x_0)$ is a function: $\mathbb{R}  \times \mathbb{R} \rightarrow \mathbb{R}$.  I'm trying to differentiate  equation of this function both sides  with respect to $x_{0}$. In book from this equation comes from the result of this differentiation is:  $\frac{\partial \phi}{\partial x_0} (t,x_0) = 1 + \int_0^{t} \frac{\partial f}{\partial x_0} (s, \phi(s,x_0)) \cdot \frac{\partial \phi}{\partial x_0} (s, x_0)ds$.  I can't understand expression under the integral. In this book it is explained that it is clear from chain rule. When I'm trying to apply the chain rule I have expression under integral such as:  $\frac{\partial }{\partial x_0} f(s, \phi(s,x_0)) = \frac{\partial }{\partial \phi(s,x_0) } f(s, \phi(s,x_0)) \cdot \frac{\partial }{\partial x_0} \phi(s,x_0) $  Could someone explain to me why in my expression first I have derivative respect by $\partial \phi(s,x_0)$ and then by $\partial x_0?$ In solution there are in both expressions $\partial x_0$ Why?  Maybe important is here additional information that differentiable function $(1)$ is solution of differential equation:  $\frac{\partial \phi}{\partial t} (t,x_0) = f(t,\phi,x_0))$, where $\phi(0,x_0)=x(0).$   I will be grateful for your help Best regards
I have a problem with derivation of some function. This is related to computing the Poincare Map of Logistic Population model with Periodic harvesting. This function is:  $\phi(t,x_0) = x_0 + \int_{0}^{t} f(s,\phi(s,x_0)) ds $,  where $\phi(t,x_0)$ is a function: $\mathbb{R}  \times \mathbb{R} \rightarrow \mathbb{R}$.  I'm trying to differentiate  equation of this function both sides  with respect to $x_{0}$. In book from this equation comes from the result of this differentiation is:  $\frac{\partial \phi}{\partial x_0} (t,x_0) = 1 + \int_0^{t} \frac{\partial f}{\partial x_0} (s, \phi(s,x_0)) \cdot \frac{\partial \phi}{\partial x_0} (s, x_0)ds$.  I can't understand expression under the integral. In this book it is explained that it is clear from chain rule.Could someone explain to me this transformation in easy way? I will be grateful for your help Best regards
Solve for $x$ in $6x^2-25x+12+\frac{25}{x}+\frac{6}{x^2}=0$ Highlight \verb in inline text Is it possible to add a list of hosts that are only specific to a certain user? Perhaps a user-specific hosts file?  This mechanism should also complement the entries in the /etc/hosts file. First player rolls a 6-sided die 100 times and the second player 101 times. What are the odds of the second player getting more odd numbers than the first one?  I tried solving this via listing the options which led to a nasty sum that I wasn't able to solve neither by hand nor by using wolfram alpha. Clearly there has to be a more clever solution using the properties of a binomial distribution, but so far I haven't come across anything useful. How can I find the points at which two circles intersect? Given the radius and $x,y$ coordinates of the center point of two circles how can I calculate their points of intersection if they have any? Make square image Is there such a thing as a switch that can be actuated automatically? I'm interested in a toggle switch that can be toggled without user input. That is, the physical state can be toggled electrically. It would need to incorporate some sort of motor or magnetic actuator. My google-fu is weak in this area, and I haven't been able to come up with the correct combination of search terms.  Note that I'm not talking about a relay; I'm talking about a normal toggle switch:    Except that it can be toggled without user input. As in, the lever physically moves. Like a  except that the mechanism is internal to the switch. Elementary properties of closure Hi everyone I'd like to know if the following is correct. I really appreciate any suggestion. (Honestly the only one that matters me is the second property the others are easy, I think) Thanks.  Definition: Let $X\subset \mathbb{R}$ and let $x'\in \mathbb{R}$, we say that $x'$ is an adherent point of $X$ iff $\,\forall\varepsilon&gt;0\,\exists x\in X \text{ s.t.} \; d(x',x)\le \varepsilon$. We say that $\overline{X}$ is the closure of $X$ iff contain all the adherent points of $X$.  Lemma (Elementary properties of Closures): Let $X$ and $Y$ be arbitrary subset of $\mathbb{R}$. Then $X\subset \overline{X}, \; \overline{X\cup Y}=\overline{X}\cup\overline{Y},\; \overline{X\cap Y}\subset\overline{X}\cap\overline{Y}$. If $X\subset Y$ then $\overline{X}\subset \overline{Y}$.  Proof:  Clearly if $x\in X$,  $x$ is $\varepsilon$-adherent to $x$ (indeed is $0$-close to itself) for every $\varepsilon$, hence $x\in \overline X$.   Let $z'$ be an adherent point of $X\cup Y$, then there exists some $z\in X\cup Y$ such that $d(z',z)$ for a given $\varepsilon&gt;0$. Suppose  $z\in X$, hence $z'$ is an adherent point of $X$, i.e., $z'\in \overline{X}\subset \overline{X}\cup \overline{Y}$. Similarly when  $z\in Y$, $z'\in \overline{Y}\subset \overline{X}\cup \overline{Y}$. Thus $ \overline{X\cup Y}\subset\overline{X}\cup \overline{Y}$. We  know that $X\cup Y \subset\overline{X\cup Y}\subset\overline{X}\cup \overline{Y}$. Then $\overline{\overline{X\cup Y}}=\overline{X}\cup \overline{Y}$. Since $\overline{\overline{X\cup Y}} = \overline{X\cup Y}$ we're done.  Let $z'\in \overline{X\cap Y}$, i.e., $z'$ is an adherent point of $X\cap Y$. So, there is some $z\in X\cap Y \subset X$ such that $d(z',z)\le \varepsilon$ for any given $\varepsilon&gt;0$. It follows that $z'$ is an adherent point of $X$, i.e., $z'\in \overline{X}$. Similarly since $z\in X\cap Y \subset Y$ we have $z'\in \overline{Y}$. Hence $z'\in \overline{X}\cap\overline{Y}$.   Let $x'\in \overline{X}$ we'd like to show that $x'\in \overline{Y}$. Let $\varepsilon&gt;0$ be arbitrary then $\exists x\in X\subset Y$ such that $d(x',x)\le \varepsilon$, so $x'$ is an adherent point of $Y$, i.e., $x'\in \overline{Y}$ as desired. $\Box$     Claim 1: If $X\subset Y \subset \overline{X}$. Then $\overline{Y}=\overline{X}$.    Proof claim: Let $y'$ be an adherent point of $Y$, i.e., $y'\in \overline{Y}$ and let $\varepsilon&gt;0$ be given. Thus, $\exists y \in Y\subset \overline{X}$ s.t. $d(y,y')\le \varepsilon$. So, $y$ is an adherent point of $X$. Thus $\exists x\in X$ s.t. $d(x,y)\le \varepsilon$ and therefore  we must have $d(x,y')\le 2\varepsilon$ which shows that $y'$ is an adherent point of $X$, i.e., $\overline{Y}\subset \overline{X}$. Conversely if $x'$ is an adherent point of $X$. Then $\exists x\in X\subset Y$ s.t. $d(x,x')\le \varepsilon$, and so is adherent to $Y$. $\Box$     Claim 2: $\overline{\overline{X}} = \overline{X}$    Proof: It will suffice to show that if $x''\in \overline{\overline{X}}$, then $x''$ is an adherent point of $X$. Let $\varepsilon &gt;0$ be arbitrary and let $x''$ be an adherent point of $\overline{X}$. Then there is some $x' \in \overline{X}$ such that $d(x'',x')\le \varepsilon$. Since $x' \in \overline{X}$ by definition is an adherent point of $X$, so there is some $x\in X$ for which $d(x,x')\le \varepsilon$. Thus $d(x,x'')\le d(x,x')+d(x',x'')\le 2\varepsilon$ it follows that $x''$ is an adherent point of $X$ as desired. $\Box$ Lower bounds for ${{2n} \choose {n}}$ What are common lower bounds for ${{2n} \choose {n}}$?    Edit:  I made a mistake in my original question.  It doesn't change my question but there is no reason for me to include the mistake. Mouse cursor flickering and disappearing I can't use the terminal while gedit command is running I recently installed 12.04. When I try to edit a file with gedit, I can't use the terminal until I close the editing file or I have to open a new terminal. But I think I didn't have this problem with 11.04, however I'm not sure. Is there anyway to avoid this and to use same terminal while editing files. Bootstrap CSS Active Navigation Proving $\lim_{n\to\infty} a^{\frac{1}{n}}=1$ by definition of limit given a sequence $a_n=a^{\frac{1}{n}}$ for $n\in\mathbb{N}^*$, $a\in\mathbb{R},a&gt;1$ then proof that $\lim\limits_{n\to+\infty}a_n=1$ by definition.   proof:  given $a_n=a^{\frac{1}{n}}$ for $a\in\mathbb{R},a&gt;1$.  for $n\in\mathbb{N}^*,n+1&gt;n\Rightarrow \frac{1}{n+1}&lt;\frac{1}{n}$ and because $a&gt;1$ we gets $a^{\frac{1}{n+1}}&lt;a^{\frac{1}{n}}$, since $\frac{1}{n+1}&gt;0$, then $1=a^{0}&lt;a^{\frac{1}{n+1}}&lt;a^{\frac{1}{n}}\Rightarrow 1&lt;a_{n+1}&lt;a_{n}$.  then we proof that $\lim\limits_{n\to+\infty}a_n=1$ we need to show that $\forall\epsilon&gt;0,\exists N,\forall n&gt;N,|a_n-1|&lt;\epsilon$  then for $\epsilon&gt;0$, choose $N=\frac{1}{\log_a(\epsilon+1)}$, since $a&gt;1$ we gets that $$\forall n&gt;N,1&lt;a^{\frac{1}{n}}&lt;a^\frac{1}{N}\Rightarrow0&lt;a^{\frac{1}{n}}-1&lt;a^{\frac{1}{N}}-1\Rightarrow |a^{\frac{1}{n}}-1|&lt;|a^{\frac{1}{N}}-1|=|a^{\log_a(\epsilon+1)}-1|=|\epsilon+1-1|=\epsilon\Rightarrow |a^{\frac{1}{n}}-1|&lt;\epsilon$$  wich implies that $\lim\limits_{n\to+\infty}a_n=1\square$  my proof to the limit is correct? I'm trying to install wifi drivers for a particular mini wifi card i bought. The output of lsusb is:  0bda:f179 Realtek Semiconductor Corp   I've searched everywhere for drivers, but could not find anything. Any help?  Thanks How to install Google Chrome I was working on a Beamer presentation, and noticed that, if I open a group scope and use a \tiny, the font size will bet set, but there will be a large space between lines. However, if I issue a \par at the end of the paragraph, it works as intended:  \documentclass[t]{beamer} \newcommand{\shortlipsum} {Lorem ipsum dolor sit amet, consectetuer adipiscing elit. Ut purus elit, vestibulum ut, placerat ac, adipiscing vitae, felis. Curabitur dictum gravida mauris. Nam arcu libero, nonummy eget, consectetuer id, vulputate a, magna. Donec vehicula augue euneque. Pellentesque habitant morbi tristique senectus et netus et malesuada fames ac turpis egestas. Mauris ut leo. Cras viverra}  \begin{document} \begin{frame} \shortlipsum  {\tiny \shortlipsum\par   % &lt;--- removing this \par will break line spacing } \end{frame} \end{document}   Here's a comparison between not using and using \par:    Am I doing it the right way? Why is \par the right way? Is there a way to head/tail a document and get the reverse output; because you don't know how many lines there are in a document?  I.e. I just want to get everything but the first 2 lines of foo.txt to append to another document. Is there a reason I would not add /usr/local/sbin, /usr/sbin, /sbin to my path on Debian? how does capacitor work ? how do they conduct electricity ? Sorry, i am novice in electronics, trying to understand basic fundamental of electronics.  1. How does capacitor conduct current when two plates aren't physically connected. Suppose $n$ is an even positive integer and $H$ is a subgroup of $\mathbb Z/n \mathbb Z$. Prove that either every element of $H$ is... Suppose $n$ is an even positive integer and $H$ is a subgroup of $\mathbb Z/n\mathbb Z$. Prove that either every element of $H$ is even or exactly half of its elements are even. (Gallian, Contemporary Abstract Algebra, Exercise 24, Chapter 3.)