I want to know how to add an image logo as website title. for example if we open our gmail, it shows gmail logo in the title bar how to achieve this..  &lt;title&gt; &lt;img src="as.jpg"&gt;&lt;/img&gt; Hello user  &lt;/title&gt;
What is currently the best way to get a favicon to display in all browsers that currently support it?  Please include:   Which image formats are supported by which browsers. Which lines are needed in what places for the various browsers.
Defining the Product of Ideals So I am using cryptopp to encrypt a file in a project and I am using AES-GCM. I have a 256bit key and a 128bit ivec and a 128bit tag that I use to encrypt files.  However, I want to know where the authentication tag is stored? I notice if I specify a tag size of 0 then the encrypted file size is the same number of bytes as the original file.  Is the authentication tag simply appended to the end of the file? How do I install Java? How to convert a std::string to const char* or char*? Find remainder of division of $x^3$ by $x^2-x+1$ I am stuck at my exam practice here.      The remainder of the division of $x^3$ by $x^2-x+1$ is ..... and that of $x^{2007}$ by $x^2-x+1$ is .....   I tried the polynomial remainder theorem but I am not sure if I did it correctly.  By factor theorem definition, provided by ,     the remainder of the division of a polynomial $f(x)$ by a linear polynomial  $x-r$ is equal to $f(r)$.   So I attempted to find $r$ by factorizing $x^2-x+1$ first but I got the complex form $x=\frac{1\pm\sqrt{3}i}{2}=r$.   $f(r)$ is then $(\frac{1+\sqrt{3}i}{2})^3$ or $(\frac{1-\sqrt{3}i}{2})^3$ which do not sound right.  However, the answer key provided is $-1$ for the first question and also $-1$ for the second one. Please help. Eclipse doesn't start on Ubuntu 18.04 I installed eclipse via Ubuntu Software. When I start it, it take some seconds and then comes an error window showing the path of the error.  The following is the .log file:  !SESSION Wed May 02 12:53:28 BRT 2018 ------------------------------------------ !ENTRY org.eclipse.equinox.launcher 4 0 2018-05-02 12:53:28.587 !MESSAGE Exception launching the Eclipse Platform: !STACK java.lang.ClassNotFoundException: org.eclipse.core.runtime.adaptor.EclipseStarter     at java.base/java.net.URLClassLoader.findClass(URLClassLoader.java:466)     at java.base/java.lang.ClassLoader.loadClass(ClassLoader.java:566)     at java.base/java.lang.ClassLoader.loadClass(ClassLoader.java:499)     at org.eclipse.equinox.launcher.Main.invokeFramework(Main.java:626)     at org.eclipse.equinox.launcher.Main.basicRun(Main.java:584)     at org.eclipse.equinox.launcher.Main.run(Main.java:1438)     at org.eclipse.equinox.launcher.Main.main(Main.java:1414)   What is the problem and how can I fix it? How could Argus Filch clean up the whole castle by himself without magic? Calculate the limit of two interrelated sequences? I'm given two sequences:  $$a_{n+1}=\frac{1+a_n+a_nb_n}{b_n},b_{n+1}=\frac{1+b_n+a_nb_n}{a_n}$$  as well as an initial condition $a_1=1$, $b_1=2$, and am told to find: $\displaystyle \lim_{n\to\infty}{a_n}$.  Given that I'm not even sure how to approach this problem, I tried anyway. I substituted $b_{n-1}$ for $b_n$ to begin the search for a pattern. This eventually reduced to:  $$a_{n+1}=\frac{a_{n-1}(a_n+1)+a_n(1+b_{n-1}+a_{n-1}b_{n-1})}{1+b_{n-1}+a_{n-1}b_{n-1}}$$  Seeing no pattern, I did the same once more:  $$a_{n+1}=\frac{a_{n-2}a_{n-1}(a_n+1)+a_n\left(a_{n-2}+(a_{n-1}+1)(1+b_{n-2}+a_{n-2}b_{n-2})\right)}{a_{n-2}+(a_{n-1}+1)(1+b_{n-2}+a_{n-2}b_{n-2})}$$  While this equation is atrocious, it actually reveals somewhat of a pattern. I can sort of see one emerging - though I'm unsure how I would actually express that. My goal here is generally to find a closed form for the $a_n$ equation, then take the limit of it.  How should I approach this problem? I'm totally lost as is. Any pointers would be very much appreciated!  Edit:  While there is a way to prove that $\displaystyle\lim_{n\to\infty}{a_n}=5$ using $\displaystyle f(x)=\frac{1}{x-1}$, I'm still looking for a way to find the absolute form of the limit, $\displaystyle\frac{1+2a+ab}{b-a}$. No sound in Ubuntu 13.04, only Dummy output device listed My sounds suddenly stopped playing on Ubuntu 13.04 64bit. I can only see Dummy output device in the Sound Settings and no sound is played during testing.  I can get the sound temporarily working by killing one of the strange processes, but I have to repeat the same after reboot  jan@jan-HP-Desktop:~$ ps -ef|grep pulseaudio 112       1405     1  0 09:38 ?        00:00:00 /usr/bin/pulseaudio --start --log-target=syslog 112       2132  1405  0 09:38 ?        00:00:00 /usr/lib/pulseaudio/pulse/gconf-helper jan       2575     1  0 09:39 ?        00:00:00 /usr/bin/pulseaudio --start --log-target=syslog jan       2597  2575  0 09:39 ?        00:00:00 /usr/lib/pulseaudio/pulse/gconf-helper jan       3160  2975  0 09:40 pts/2    00:00:00 grep --color=auto pulseaudio  jan@jan-HP-Desktop:~$ sudo kill 1405   Is there a permanent solution for this problem? TikZ is undergoing lots of development with new features being added, but the versions distributed with standard TeX releases are often a little old. How do I get the latest version? If I want to be sure that I get the latest stable version, how do I do that?    Note: I'm not asking this question for myself, but because a question like this seems to be useful for pointing to it when a new feature of TikZ is mentioned in an answer to some other question. Feel free to post answers for any operating system/TeX distribution combination. Can "but also" be split in "not only ... but also"? I think I generally agree with the statement  When using not only . . . but also in a sentence, parallelism should be the goal. It means that the words following both parts of this correlative conjunction (i.e., not only and but also) should belong to the same parts of speech.   However, the Cambridge dictionary has an example that is not 100% parallel in my eyes, as &quot;but also&quot; is split:  Not only did she forget my birthday, but she also didn’t even apologise for forgetting it.   What I would expect, with maximum parallelism, is  Not only did she forget my birthday, but also she didn’t even apologise for forgetting it.  Is my expectation wrong? Is my formulation acceptable? Disable Tabbing in Firefox 3, Chrome 4 and IE8? Tabbed Browsing may be nice for reading news sites in one sitting, but I still think they are a usability nightmare if you need to keep the page open for a long time, e.g., for testing it while developing or as a reference for something.  The nice thing about the Windows Taskbar (before Windows 7 ) is that it's easy and natural to remember where things are. I can have 30 tabs in it, but because I know what each one is, I know where I have to click to open something - it's the most efficient way of working I've encountered so far.  Sadly, Tabbed Browsing completely breaks this efficiency because it is not possible to specifically click on a tab in the taskbar, I have to click on the browser, search the tab and then click it, which means at least 2 clicks.  Long story, short question: Is there an extension or setting to completely disable tabbing on Firefox 3, Chrome 4 and Internet Explorer 8 and have one Window/Taskbar entry for each open page? I know that in Chrome I can just drag a tab out of a Window to create a new one, but if there is a general way, that would be easier.  Edit: IE8 conveniently offers an option for that: Tools - Options - Tabs - Enable Tabbed Browsing. How to typeset a file path? When a 0-1-matrix with exactly two 1’s on each column and on each row is non-degenerated? [1] Let $A$ be an $n\times n$ matrix with entries in the set $\{0,1\}$ which has exactly two ones in each column and two ones in each row.  Give necessary and sufficient conditions for the rank of $A$ to be $n$. Let $F$ be a Galois extension over $\mathbb{Q}$ with $[F:\mathbb{Q}]=2^n$, then all elements in $F$ are constructible Let $F\subseteq\mathbb{C}$ be a Galois extension of $\mathbb{Q}$ such that $[F:\mathbb{Q}]=2^n$; then all elements in $F$ are constructible.    Added.  Here is what I have so far.  Since there exist the finite normal extension $F\subseteq\mathbb{C}$ over $\mathbb{Q}$ which contains an element say, s, then s has a degree of power of 2. From group theory we have 2 propositons we could use. 'A group $G$ is solvable iff it has a normal series whose factors are abelian', and also 'If $G$ is solvable and $N\triangleleft G$, then $G/N$ is solvable'. Then we can say that the galois group F has a normal series $\{e\}=G_0\triangleleft G_1\triangleleft\cdots\triangleleft G_n=\mathrm{Gal}(F/\mathbb{Q})$, whose factors have order 2. Thus there must exist intermediate fields $\mathbb{Q}=F_0 \subseteq F_1 \subseteq F_2 \subseteq\cdots\subseteq F_n=F$, such that $[F_{j+1}:F_j]=2$ for all $j$. Now this is where I got stuck. Next I need to somehow show that $F_{j+1}=F_j(s_j)$, where $s^2_j$ is an element in $F_j$. Then I think I can conclude that if $s_j$ is constructible than every element in $F_{j+1}$ is constructible... then with $F_n$ is constructible... Much as I loathe Microsquish's Word, it does allow you to insert into a header or footer the automated text "Page X of Y."   When using InDesign, it's easy to set up the "Page X" part on a master page. But is there any placeholder character for the Y? That is, it will count the number of pages in the document, either in an absolute or a relative-to-the-section fashion? Every now and again I narrowly miss sending out something with a footer which reads "Page 4 of 3," and I would love to know if there's some way around that. Exactly half of the elements of $\mathcal{P}(A)$ are odd-sized Let $A$ be a non-empty set and $n$ be the number of elements in $A$, i.e. $n:=|A|$.  I know that the number of elements of the power set of $A$ is $2^n$, i.e.  $|\mathcal{P}(A)|=2^n$.  I came across the fact that exactly half of the elements of $\mathcal{P}(A)$ contain an odd number of elements, and half of them an even number of elements.  Can someone prove this? Or hint at a proof? Proof by induction of summation inequality: $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots+\frac1{2^n}\ge 1+\frac{n}2$ Prove by induction the summation of $\frac1{2^n}$ is greater than or equal to $1+\frac{n}2$.  We start with $$1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots+\frac1{2^n}\ge 1+\frac{n}2$$ for all positive integers.   I have resolved that the following attempt to prove this inequality is false, but I will leave it here to show you my progress. In my proof, I need to define P(n), work out the base case for n=1, and then follow through with the induction step. Strong mathematical induction may be used.  This is equivalent to $$\sum_{k=0}^n\frac1{2^k}\ge 1+\frac{n}2\;.$$  Let $P(n)$ be summation shown above.   Base case for $n=1$, the first positive integer,   $$\sum_{k=0}^1\frac1{2^k}=\frac1{2^0}+\frac1{2^1}=1+\frac12=\frac32\ge 1+\frac12=\frac32\;,$$  so base case is true.  Induction step: Assume $P(n)$ is true and implies $P(n+1)$. Thus   $$\sum_{k=0}^{n+1}\frac1{2^k}\ge\frac1{2^{n+1}}+\sum_{k=0}^n\frac1{2^k}\ge 1+\frac{n+1}2\;.$$  This can be written as   $$\sum_{k=0}^{n+1}\frac1{2^k}\ge \frac1{2^{n+1}}+1+\frac{n}2\ge 1+\frac{n+1}2\;.$$  I work the math out but I get stuck contradicting my statement. Please show your steps hereafter so I can correct my mistakes. Profile Gravatar get change Do not know whether someone else observe it or not but today I found my profile Gravatar changed. It is now different than earlier I have. Is it because today I have updated my profile information?        My previous gravatar looks like: If two coins are flipped and one gets head, what is the probability that both get head?