1. ## nb

$\displaystyle z=(4+\sqrt15)^1/3$

2. ## ?

$\displaystyle z=\sqr[3]{4+\sqrt{15}}$

3. ## v

$\displaystyle z=\sqrt[3]{(4+\sqrt15)}$

4. Originally Posted by i_zz_y_ill
$\displaystyle z=(4+\sqrt15)^1/3$
Always enclose your latex code with and when you want to write 1+ characters as a power, enclose them with brackets {}

5. ## g

$\displaystyle \frac{(2n)!}{n!}$

6. ## v

$\displaystyle 2^n.1.2.3.4.....(2n-1)$

7. ## g

$\displaystyle (1+x)^n$

$\displaystyle (1+x)^m(1+x)^n=(1+x)^(n+m)$

$\displaystyle r_{x{_m$

8. $\displaystyle \left{ \frac{1}{2} \right}$

anybody know how to get the curly brackets {} to show, within:

\left{ \frac{1}{2} \right} for example

9. Originally Posted by lllll
$\displaystyle \left{ \frac{1}{2} \right}$

anybody know how to get the curly brackets {} to show, within:

\left{ \frac{1}{2} \right} for example
Do either:

$$\left\{ \frac{1}{2} \right\}$$, which gives us $\displaystyle \left\{ \frac{1}{2} \right\}$

OR

$$\{ \frac{1}{2} \}$$, which gives us $\displaystyle \{ \frac{1}{2} \}$

Note the difference between these. The first one looks better...the second one seems odd....

--Chris

10. Originally Posted by Chris L T521
Do either:

$$\left\{ \frac{1}{2} \right\}$$, which gives us $\displaystyle \left\{ \frac{1}{2} \right\}$

OR

$$\{ \frac{1}{2} \}$$, which gives us $\displaystyle \{ \frac{1}{2} \}$

Note the difference between these. The first one looks better...the second one seems odd....

--Chris
Better still (in my opinion), do

$$\left\{ \tfrac{1}{2} \right\}$$, which gives us $\displaystyle \left\{ \tfrac{1}{2} \right\}$

The textstyle fractions usually (not always) look better than displaystyle when dealing with single-digit numbers in the fraction.

11. Still better :
$$\left\{\tfrac12\right\}$$ $\displaystyle \left\{\tfrac12\right\}$ (that was the lazy way)

12. Originally Posted by Moo
Still better :
$$\left\{\tfrac12\right\}$$ $\displaystyle \left\{\tfrac12\right\}$ (that was the lazy way)
I think, $$\{1/2\}$$, which gives $\displaystyle \{1/2\}$, looks better when in line.

13. $\displaystyle \begin{array}{rcll} x & \equiv & 6 & \text{mod } 9 \\ x & \equiv & 3 & \text{mod } 13 \\ x & \equiv & 5 & \text{mod } 16 \end{array}$

14. Originally Posted by Jones
$\displaystyle \begin{array}{rcll} x & \equiv & 6 & \text{mod } 9 \\ x & \equiv & 3 & \text{mod } 13 \\ x & \equiv & 5 & \text{mod } 16 \end{array}$
... or you can use \pmod, as in $$x \equiv 6 \pmod{9}$$, to get $\displaystyle x \equiv 6 \pmod{9}$.

Personally, I think that \pmod leaves too much space before the first parenthesis, so I put in a couple of small backspaces (\!\!), to get $\displaystyle x \equiv 6\!\! \pmod{9}$.

15. Originally Posted by Opalg
... or you can use \pmod, as in $$x \equiv 6 \pmod{9}$$, to get $\displaystyle x \equiv 6 \pmod{9}$.

Personally, I think that \pmod leaves too much space before the first parenthesis, so I put in a couple of small backspaces (\!\!), to get $\displaystyle x \equiv 6\!\! \pmod{9}$.

Hehe, how would i get a left bracket first to indicate an equation system, (or congruence system if you like)

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