Is $f'(x)$ Riemann integrable if it is bounded
Sufficient condition(s) for $f'$ to be Riemann integrable
I installed the ubuntu 12.04 server edition for my server pc . i had installed lamp server. i need to change the var/www location to my secondary hard disk location. i was configured so many time to at gedit /etc/apache2/sites-available/default here is my code  &lt;VirtualHost *:80&gt;     ServerAdmin webmaster@localhost      #DocumentRoot /var/www     DocumentRoot /media/myserver/     &lt;Directory /&gt;         Options FollowSymLinks         AllowOverride None     &lt;/Directory&gt;     #&lt;Directory /var/www/&gt;         &lt;Directory /media/myserver/&gt;         Options Indexes FollowSymLinks MultiViews         AllowOverride None         Order allow,deny         allow from all     &lt;/Directory&gt;      ScriptAlias /cgi-bin/ /usr/lib/cgi-bin/     &lt;Directory "/usr/lib/cgi-bin"&gt;         AllowOverride None         Options +ExecCGI -MultiViews +SymLinksIfOwnerMatch         Order allow,deny         Allow from all     &lt;/Directory&gt;      ErrorLog ${APACHE_LOG_DIR}/error.log      # Possible values include: debug, info, notice, warn, error, crit,     # alert, emerg.     LogLevel warn      CustomLog ${APACHE_LOG_DIR}/access.log combined      Alias /doc/ "/usr/share/doc/"     &lt;Directory "/usr/share/doc/"&gt;         Options Indexes MultiViews FollowSymLinks         AllowOverride None         Order deny,allow         Deny from all         Allow from 127.0.0.0/255.0.0.0 ::1/128     &lt;/Directory&gt;  &lt;/VirtualHost&gt;   and also used  sudo chown -R var/www /media/myserver/   and  chmod -R 755 /media/myserver/   still i was not able to connect my /media/myserver and my browser show the following message  Forbidden  You don't have permission to access / on this server.   Please tell anyone how to mount myserver at my var/www, thanks advance How do I read an entire file into a std::string in C++? Connecting GSM SIM900A module with Ardunio I have bought an SIM900A Mini V4 GSM module. Here is the module,   I want to configure with arduino to send and receive SMS.   Do I have to connect TXD of this module--Tx of Arduino &amp; RXD of this module--Rx of Arduino or other way around?  There is no datasheet or documents on this new module.  When I search about RXD &amp; TXD, I found this in      TxD means "Data to be transmitted over the comms link" - which will clearly be an input to the modem;    Similarly, RxD means "Data that has been received from the comms link" - which will clearly be an output from the modem.    But I couldn't send or receive SMS using this configurations.  As I am new to this, It would be very helpful if anyone help on this.  Thank you in advance. I suspected that installing xfce4 cause some issues in Ubuntu so I removed them using apt-get remove xfce4 xfce4-goodeis.  But when I ran 'dpkg -l' I still can see some libraries and packages for xfce4:  $ dpkg -l | grep -i xfce rc  libexo-1-0:amd64                              0.10.2-2                                 amd64        Library with extensions for Xfce rc  libgarcon-1-0                                 0.2.1-1                                  amd64        freedesktop.org compliant menu implementation for Xfce rc  libxfce4ui-1-0                                4.11.0-0ubuntu1~ppa0.13.10.1             amd64        widget library for Xfce - Gtk+2 variant rc  libxfce4ui-common                             4.11.0-0ubuntu1~ppa0.13.10.1             all          common files for libxfce4ui rc  libxfce4util6                                 4.10.1-1                                 amd64        Utility functions library for Xfce4 rc  libxfcegui4-4                                 4.10.0-2                                 amd64        Basic GUI C functions for Xfce4 rc  libxfconf-0-2                                 4.10.0-2                                 amd64        Client library for Xfce4 configure interface rc  mousepad                                      0.3.0-2                                  amd64        simple Xfce oriented text editor rc  thunar                                        1.6.3-1ubuntu1                           amd64        File Manager for Xfce rc  xfce4-appfinder                               4.10.1-1                                 amd64        Application finder for the Xfce4 Desktop Environment rc  xfce4-clipman                                 2:1.2.3-2ubuntu1                         amd64        clipboard history utility rc  xfce4-mixer                                   1:4.10.0-1ubuntu2                        amd64        Xfce mixer application rc  xfce4-notes                                   1.7.7-3ubuntu2                           amd64        Notes application for the Xfce4 desktop rc  xfce4-panel                                   4.10.1-1ubuntu1                          amd64        panel for Xfce4 desktop environment rc  xfce4-power-manager                           1.2.0-2ubuntu1                           amd64        power manager for Xfce desktop rc  xfce4-session                                 4.10.1-1ubuntu1                          amd64        Xfce4 Session Manager rc  xfce4-settings                                4.11.1-0ubuntu1~ppa0.13.10.1             amd64        graphical application for managing Xfce settings rc  xfce4-terminal                                0.6.2-3ubuntu1.1                         amd64        Xfce terminal emulator rc  xfce4-volumed                                 0.2.0-0ubuntu1                           amd64        volume keys daemon rc  xfdesktop4                                    4.11.2-0ubuntu1~ppa0.13.10.1             amd64        xfce desktop background, icons and root menu manager rc  xfwm4                                         4.11.1-0ubuntu1~ppa0.13.10.1             amd64        window manager of the Xfce project   I copy-pasted all these packages\libraries to form an apt-get remove command to remove all of these dependencies, and the command ran normally, but running dpkg -l | grep -i xfce showed these dependencies again !  Why are these dependencies still installed ? With powershell get exactly the same application list as in Add/Remove programms Two homologous but not homotopic loops on a closed surface of genus greater than one Suppose $S$ is a closed surface with genus greater than $1$. Give an example of two loops on $S$ which are not homotopic, but are homologous to each other.   I need help solving this question. I have read the Wikipedia page on it and I realize that it does not contain the necessary knowledge to visualize this problem.  And I was also wondering the relation between the fundamental group $\pi_1(S)$ and the first homology group $H_1(S)$? Some restrictions is that $S$ is a closed surface and the genus is greater than $1$. when I look at my Nexus 5 with android 5.1. battery screen, I can see something like this:    Screenshots (click images for larger variants)  Sum of all per-app percentages in list is 10. However as you can see, there is 60% battery consumed (40% remaining). So, what is the deal here? Who consumed 50% battery? Or what do the percentages next to apps mean?  As you can see, screen was on just for few short times, the same for awake, and yet battery life about 18 hours? I dont get it.  Also, I found this issue reported on AOSP, but noone from google seems to care: Do separate autofocus motors exist for entry level Nikon DSLRs? I mistakenly typed cd // instead of cd /. To my surprise, current directory showed up as //.   What is that directory? Why does it exist?  apple@snipped $ pwd /home/apple apple@snipped $ cd / apple@snipped $ pwd / apple@snipped $ cd // apple@snipped $ pwd // apple@snipped $ cd /// apple@snipped $ pwd / Err 80 appears when I turn on my Canon EOS 5D Mark II Proving that $\sum\limits_{n = 0}^{2013} a_n z^n \neq 0$ if $a_0 > a_1 > \dots > a_{2013} > 0$ and $|z| \leq 1$ I'm going to teach a preparation course for the complex analysis qualifying exam from my university (which basically consists of me doing some problems from past exams) and I'm trying to solve some questions from previous exams. One of the questions is the following:           Prove that $\displaystyle{\sum_{n = 0}^{2013} a_n z^n \neq 0}$ if the coefficients  satisfy $a_0 &gt; a_1 &gt; \dots &gt; a_{2013} &gt; 0$ and $|z| \leq 1$.      I tried to approach the problem by using the reverse triangle inequality to try to isolate the leading term, which I thought should be the constant coefficient, as follows:  \begin{array} . \left|  \sum_{n = 0}^{2013} a_n z^n \right| &amp;= \left| a_0 +  \sum_{n = 1}^{2013} a_n z^n \right| \\ &amp;\geq |a_0| - \left|  \sum_{n = 1}^{2013} a_n z^n \right|\\ &amp;\geq |a_0| -  \sum_{n = 1}^{2013} |a_n z^n|\\ &amp;\geq a_0 -  \sum_{n = 1}^{2013} a_n |z^n|\\ &amp;\geq a_0 -  \sum_{n = 1}^{2013} a_0 |z^n|\\ &amp;= a_0\left( 1  -  \sum_{n = 1}^{2013} |z^n| \right) \end{array}  Then for $|z| &lt; 1$, we have that this last expression is equal to  $$ a_0\left( 1  -  |z| \frac{|z|^{2013} - 1}{|z| - 1} \right) $$  and my hope was to prove that this was always positive, nevertheless, after failing to prove it, I plotted the expression inside the parenthesis as a function of $|z|$, and found out that it changes sign in the interval $(0, 1)$.  Question  Can my approach somehow be made to work? And if not, how can I prove that the sum $\displaystyle{\sum_{n = 0}^{2013} a_n z^n \neq 0}$ under the given conditions?  Thank you very much for any help. How to convert a currency string to a double with jQuery or Javascript? Please give me feedback on my answer to this question. Question: For all $ n\geq1:\binom{2n}{0}+\binom{2n}{2}+\binom{2n}{4}+\cdots+\binom{2n}{2k}+\cdots+\binom{2n}{2n} $   is equal to $ \binom{2n}{1}+\binom{2n}{3}+\binom{2n}{5}+\cdots+\binom{2n}{2k+1}+\cdots+\binom{2n}{2n-1} $. Answer: False, since;  Let $a=1,  b=-1$  and $n=1$  ,$k=0$   , using Binomial expression.  $$(a+b)^n=  \sum   \binom{n}{k}  a^{k}b^{n-k}$$  Then $(1+-1)^{2(1)}=\binom{2}{0}1^{0}(-1)^{2-0}+\binom{2}{1}1^{1}(-1)^{2-1}+\binom{2}{2}1^{2}(-1)^{2-2}=\binom{2}{0}-\binom{2}{1}+\binom{2}{2}$  Thus :For all $n\geq1:  \binom{2n}{0}+\binom{2n}{2}+\binom{2n}{4}+\cdots+\binom{2n}{2k}+\cdots+\binom{2n}{2n}   \ne \binom{2n}{1}+\binom{2n}{3}+\binom{2n}{5}+\cdots+\binom{2n}{2k+1}+\cdots+\binom{2n}{2n-1}$  . If I leave fully-cooked food (particularly meat) out at warm temperature - say on the counter or in a crock pot that's been turned off -  how long will it stay safe to eat?  Does it make any difference if I re-cook the food afterward? How do I convert my PhD dissertation so that it can be published as a book? I have just completed a PhD by research, which I think has some merits in being available as a book (This is, of course, my personal opinion). The research cuts across several fields (e.g. sociology, public policy).  I am wondering what I need to do to take the next step.  In particular:   How do you turn an academic piece like the dissertation (which is written for the examiners) into a book (which is for the general public)? How do I find publishers who specialise in my field? Should I use Named Pipes, or .NET Remoting to communicate with a running process on my machine? Hotspot and Internet access together — why not? If I try to set up a hotspot in GNOME, I get the following message:  If you have a connection to the Internet other than wireless, you can use it to share your internet connection with others. Switching on the wireless hotspot will disconnect you from &lt;network name&gt;. It is not possible to access the internet through your wireless while the hotspot is active.  What are the technical reasons why I can't set up an Hotspot and be connected to the Internet at the same time? For a given inner product space $V$, I would like to prove the inequality $$\|\mathbf{x}+\mathbf{y}\|^2 + \|\mathbf{x}-\mathbf{y}\|^2 = 2(\|\mathbf{x}\|^2 + \|\mathbf{y}\|^2).$$  We see that $$\|\mathbf{x}+\mathbf{y}\|^2 + \|\mathbf{x}-\mathbf{y}\|^2 = \langle\mathbf{x}+\mathbf{y},\mathbf{x}+\mathbf{y}\rangle + \langle\mathbf{x}-\mathbf{y},\mathbf{x}-\mathbf{y}\rangle,$$  which by linearity $$=\langle\mathbf{x},\mathbf{x}+\mathbf{y}\rangle + \langle\mathbf{y},\mathbf{x}+\mathbf{y}\rangle + \langle\mathbf{x},\mathbf{x}-\mathbf{y}\rangle - \langle\mathbf{y},\mathbf{x}-\mathbf{y}\rangle.$$  I've been trying to figure out how to go about this proof using linearity in the second argument of an inner product, but my textbook does not say that linearity necessarily holds in the second component. If it does, how would I go about the rest of this proof? If not, should I be potentially using some aspect of conjugate symmetry to prove this statement? Excel function that evaluates a string as if it were a formula? KernelEvents::REQUEST is not fired on cached pages