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 I'm the 1st 初来乍到
02-26-2010, 01:46 PM
Post: #1
 tester Posts: 6 Joined: Feb 2010 Reputation: 0
I'm the 1st 初来乍到
tester/tester is a testing account for anyone to use without registration.

$$x = a_0 + \frac{1}{\displaystyle a_1 + \frac{1}{\displaystyle a_2 + \frac{1}{\displaystyle a_3 + a_4}}}$$
$$\begin{pmatrix} a_{11} & \cdots & a_{1n}\\ \vdots & \ddots & \vdots\\ a_{m1} & \cdots & a_{mn} \end{pmatrix}$$
Interesting! 有点意思

$$a x^2 + b x + c = 0$$ has roots
$$x = \displaystyle {\frac {-b \pm \sqrt{b^2-4ac}}{2a}}$$
Let $$\delta = b^2-4ac$$, then when $$\delta = 0$$, the equation gets a two-fold root $$-b/(2a)$$
while $$\delta > 0$$ we get two real roots. Finally when $$\delta < 0$$ we get two complex roots.
Of course this is true when $$a,b$$ and $$c$$ are reals.

What we can say if $$a,b$$ and $$c$$ are not all reals?

We can still say that if and only if $$\delta = 0$$ the equation get a two-fold root.

Here we may test a problem with Firefox: Whenever you insert something here by javascript, it will scroll the text to the top in this textarea. To see this, we can add some text here to make the post longer
03-18-2010, 03:09 AM (This post was last modified: 03-18-2010 03:11 AM by tester.)
Post: #2
 tester Posts: 6 Joined: Feb 2010 Reputation: 0
RE: I'm the 1st 初来乍到
$x^2+2x+1=0$
$\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$$x = a_0 + \frac{1}{\displaystyle a_1 + \frac{1}{\displaystyle a_2 + \frac{1}{\displaystyle a_3 + a_4}}}$$
$$\begin{pmatrix} a_{11} & \cdots & a_{1n}\\ \vdots & \ddots & \vdots\\ a_{m1} & \cdots & a_{mn} \end{pmatrix}$$
08-18-2010, 01:50 AM
Post: #3
 tester Posts: 6 Joined: Feb 2010 Reputation: 0
RE: I'm the 1st 初来乍到
Code:
\sum_{a}^{b}
will not trigger MathJax to render but
Code:
$\sum_{a}^{b}$
will be displayed as
$\sum_{a}^{b}$
05-30-2012, 02:59 AM
Post: #4
 admin Administrator Posts: 26 Joined: Feb 2010 Reputation: 0
Test 1.6.8 (upgraded from 1.4.7!)
This is just to test new version 1.6.8...

$$\sum_1^\infty i = \infty,\quad \sqrt[3]{8} = 2$$
08-04-2012, 11:00 AM
Post: #5
 elim Moderator Posts: 578 Joined: Feb 2010 Reputation: 0
RE: I'm the 1st 初来乍到
Is this thread can be replied by guests?
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