New answers tagged

2

Take a look here: MathJax basic tutorial and quick reference MathJax is how it's all done. If you already know Latex you're $75\%$ of the way there.


4

You received two or more answers, in fact one positively scored answer would suffice, so you can't just delete your question. It now involves effort from other people, and that would be unfair to them. That downgrades your delete vote to the same delete vote of every other user (moderators and SE employees excluded, of course). However you lack the privilege ...


3

As in Xander's answer, / can be used as a delimiter like ()[]|, by which I mean its adjusted by the commands \left, \right, \middle, and also \big, \Big, \bigg, and \Bigg. For instance, $A/B\big/C\Big/D\bigg/E\Bigg/$ gives $A/B\big/C\Big/D\bigg/E\Bigg/$. PS the slash slanting the other way can be achieved with $\Bigg\backslash$$\Bigg\backslash$. Another way ...


5

Using the pmatrix environment (or bmatrix for square braces, but I find the round braces more aesthetically pleasing), the expression with a vinculum may be typeset with the code $$ \frac{% \begin{matrix}% A & B \\% C & D% \end{pmatrix} + 1% }{% \begin{pmatrix} D & E \\ F & G \end{pmatrix}% }. $$ This is rendered as $$ \frac{%...


5

\frac{\left[\begin{array}{cc}A&B\\C&D\end{array}\right]+1}{\left[\begin{array}{cc}E&F\\G&H\end{array}\right]} $$\frac{\left[\begin{array}{cc}A&B\\C&D\end{array}\right]+1}{\left[\begin{array}{cc}E&F\\G&H\end{array}\right]}$$ Anything you can create in MathJax (as far as I know) you can put into a fraction using \frac{Top}{...


4

Use Approach0 Use Approach0 to search for any possible duplicates of your question before asking. We get hundreds of questions every day. If your question is about arithmetic, geometry, algebra-precalculus, combinatorics, probability / statistics (first-year university or below), calculus (first-year university or below, such as Calculus I and II), or from a ...


-2

Roots: For the $n$th root of $m$, usse \sqrt[n]{m}, to get $\sqrt[n]{m}$. E.g., $\sqrt[3]{64} = 4$.


3

Use \sqrt[n]{m} to get the $n$th root of m: $\sqrt[n]{m}.$


3

$$\sqrt[3]{64}$$ produces $$\sqrt[3]{64}$$ This is standard LaTeX syntax, see e.g. this page. It might be confusing since 'sqrt' is an abbreviation of 'square root', but it works nevertheless.


2

Here is what to do. It is the same as many other sites. First, before you log out, make sure the email associated to your account is the one you want. (Otherwise, change it as Glorfindel explained.) Next, log out. Then choose "log in". There is an option "forgot my password" to click. That will email a link to you using the email ...


4

Not your password, no. Stack Exchange hashes your password before it's stored, and there's no way for them to retrieve it, all they can do it compare the result when they apply the same procedure to the password you enter when you log in. (If you didn't know this, I encourage you to read more about the topic; it's a practical application of theoretical ...


4

Some explanation when something like this happens is given in this post on Mathematics Stack Exchange: Why is my total reputation less than my monthly reputation? (You might be also interested in other posts linked there, for example: What are private reputation events?) I will quote the answer posted there: The discrepancy is by design: Total reputation ...


4

A couple of comments: If you suspect that a problem has been plagiarized, please flag it for moderator attention. Please include a link to the original source in the flag, as this will make it easier to track down. In general, it is not appropriate to post copyrighted material onto Math SE. Please also note that flags can sometimes take time to be dealt ...


-1

I keep two text files at home, here are selections. I group them by what I see as the topic. ANYONE can do this. Sometimes an OP wants lots of examples in a topic I have frequently answered. I just copy a relevant section and paste into an answer window, say it is too long for a comment and make it CW. Vieta Jumping/ Hurwitz/Markov Grundlö...


5

Yes, there's an advanced search option for this: just add inbookmarks:mine to your search query. You can even search within another user's bookmarks; I don't have any bookmarks on the main site, so when I search for inbookmarks:mine integration I get no results, but this must be the results you get: The ability to group them has been requested (a couple of ...


5

https://math.stackexchange.com/users/229831/?tab=votes&sort=upvote Only you have access to this page, though, not even the moderators can see into its content. Anyone else can swap the number for their user number.


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