forum.texasmath.org

by **SumitG** » Fri Sep 05, 2008 6:01 pm

I can't seem to figure out how to write piecewise functions. I thought that there was a piecewise environment, but every time I try to use it I get a bunch of errors. Help please?

by **stupidityismygam** » Sun Sep 07, 2008 8:45 pm

Code: Select all: f(x)=\begin{cases}x+1&\text{if } x\geq 0\\-x&\text{if } x<0\end{cases}

Should print $f(x)=\begin{cases}x+1&\text{if } x\geq 0\\-x&\text{if } x<0\end{cases}$

by **SumitG** » Mon Sep 08, 2008 5:48 am

hmm...earlier yesterday I found a different way to do it,
something like f(x)=\left\{\begin{array}{c l} 2x+1 & x<{0} \\ 2x+1 & x\geq{0} \end{array}\right.
But this works well too. Thanks.

by **SumitG** » Wed Sep 10, 2008 7:26 pm

Has anyone gotten eps graphics to show up on a pdf? I can get them to show up on a dvi, but whenever I try to make one show up on a pdf it just displays the file name.

by **peterliang** » Sat Dec 27, 2008 10:21 pm

If is a real and , then thet complete set of values of for which is real, is:

$\text{(A)} \ x\le -2~\text{or}~x\ge3 \qquad \text{(B)}$ $\ x\le 2~\text{or}~x\ge3 \qquad \text{(C)}$ $\ x\le -3 ~\text{or}~x\ge 2 \qquad \text{(D)} \ -3\le x \le 2\qquad \text{(E)} \ \-2\le x \le 3$

by **B3ndythumbs** » Sun Dec 28, 2008 12:39 pm

does anyone have an ownage thing?

by **ɐʇʇodɐʇsıɯ** » Tue Dec 30, 2008 4:18 pm

by **mttl** » Sun Mar 01, 2009 3:29 pm

by **MistaPotta** » Mon Apr 13, 2009 10:46 am

peterliang wrote: $2^{69}\ mod\ 7 = 8^{23}\ mod\ 7 = (8\ mod\ 7)^{23}\ mod\ 7 = 1^{23}\ mod\ 7 = 1$

$2^{68}\ mod\ 7 = 8^{23}\ mod\ 7 = (8\ mod\ 7)^{23}\ mod\ 7 = 1^{23}\ mod\ 7 = 1$

mttl wrote: $\displaystyle{-\!\!\!\!\!\!\!\int_E \equiv \frac{1}{\mathrm{vol}\,E} \int_E}$

$Mf(x) := \sup_{B \ni x}\,-\!\!\!\!\!\!\int_B |f|\,d\mu$

$\cfrac{1}{1+\cfrac{e^{-2\pi \sqrt5}}{1+\cfrac{e^{-4\pi \sqrt5}}{1+\cfrac{e^{-6\pi \sqrt5}}{\ddots}}}} = \left(\dfrac{\sqrt5}{1+\sqrt[5]{5^{3/4}(\frac{\sqrt5-1}2)^{5/2}-1}}-\dfrac{\sqrt5-1}{2}\right)e^{2\pi/\sqrt5}$

$\displaystyle{-\!\!\!\!\!\!\int_E \equiv \frac{1}{\mathrm{vol}\,E} \int_E}$

$Mf(x) := \sup_{B \ni x}\,-\!\!\!\!\!\int_B |f|\,d\mu$

$\cfrac{1}{1+\cfrac{e^{-2\pi \sqrt7}}{1+\cfrac{e^{-4\pi \sqrt5}}{1+\cfrac{e^{-6\pi \sqrt5}}{\ddots}}}} = \left(\dfrac{\sqrt5}{1+\sqrt[5]{5^{3/4}(\frac{\sqrt5-1}2)^{5/2}-1}}-\dfrac{\sqrt5-1}{2}\right)e^{2\pi/\sqrt5}$

by **MistaPotta** » Mon Apr 13, 2009 10:47 am

Is it a problem of background?

MistaPotta wrote:
peterliang wrote: $2^{69}\ mod\ 7 = 8^{23}\ mod\ 7 = (8\ mod\ 7)^{23}\ mod\ 7 = 1^{23}\ mod\ 7 = 1$

$2^{68}\ mod\ 7 = 8^{23}\ mod\ 7 = (8\ mod\ 7)^{23}\ mod\ 7 = 1^{23}\ mod\ 7 = 1$

mttl wrote: $\displaystyle{-\!\!\!\!\!\!\!\int_E \equiv \frac{1}{\mathrm{vol}\,E} \int_E}$

$Mf(x) := \sup_{B \ni x}\,-\!\!\!\!\!\!\int_B |f|\,d\mu$

$\cfrac{1}{1+\cfrac{e^{-2\pi \sqrt5}}{1+\cfrac{e^{-4\pi \sqrt5}}{1+\cfrac{e^{-6\pi \sqrt5}}{\ddots}}}} = \left(\dfrac{\sqrt5}{1+\sqrt[5]{5^{3/4}(\frac{\sqrt5-1}2)^{5/2}-1}}-\dfrac{\sqrt5-1}{2}\right)e^{2\pi/\sqrt5}$

$\displaystyle{-\!\!\!\!\!\!\int_E \equiv \frac{1}{\mathrm{vol}\,E} \int_E}$

$Mf(x) := \sup_{B \ni x}\,-\!\!\!\!\!\int_B |f|\,d\mu$

$\cfrac{1}{1+\cfrac{e^{-2\pi \sqrt7}}{1+\cfrac{e^{-4\pi \sqrt5}}{1+\cfrac{e^{-6\pi \sqrt5}}{\ddots}}}} = \left(\dfrac{\sqrt5}{1+\sqrt[5]{5^{3/4}(\frac{\sqrt5-1}2)^{5/2}-1}}-\dfrac{\sqrt5-1}{2}\right)e^{2\pi/\sqrt5}$

by **MistaPotta** » Mon Apr 13, 2009 10:50 am

A change to PNG...

$2^{67}\ mod\ 7 = 8^{23}\ mod\ 7 = (8\ mod\ 7)^{23}\ mod\ 7 = 1^{23}\ mod\ 7 = 1$

$\displaystyle{-\!\!\!\!\!\!\!\!\!\int_E \equiv \frac{1}{\mathrm{vol}\,E} \int_E} [tex]Mf(x) := \sup_{B \ni x}\,-\!\!\!\!\!\!\!\!\!\int_B |f|\,d\mu$

$\cfrac{1}{1+\cfrac{e^{-2\pi \sqrt3}}{1+\cfrac{e^{-4\pi \sqrt5}}{1+\cfrac{e^{-6\pi \sqrt5}}{\ddots}}}} = \left(\dfrac{\sqrt5}{1+\sqrt[5]{5^{3/4}(\frac{\sqrt5-1}2)^{5/2}-1}}-\dfrac{\sqrt5-1}{2}\right)e^{2\pi/\sqrt5}$

MistaPotta wrote:Is it a problem of background?

MistaPotta wrote:
peterliang wrote: $2^{69}\ mod\ 7 = 8^{23}\ mod\ 7 = (8\ mod\ 7)^{23}\ mod\ 7 = 1^{23}\ mod\ 7 = 1$

$2^{68}\ mod\ 7 = 8^{23}\ mod\ 7 = (8\ mod\ 7)^{23}\ mod\ 7 = 1^{23}\ mod\ 7 = 1$

mttl wrote: $\displaystyle{-\!\!\!\!\!\!\!\int_E \equiv \frac{1}{\mathrm{vol}\,E} \int_E}$

$Mf(x) := \sup_{B \ni x}\,-\!\!\!\!\!\!\int_B |f|\,d\mu$

$\cfrac{1}{1+\cfrac{e^{-2\pi \sqrt5}}{1+\cfrac{e^{-4\pi \sqrt5}}{1+\cfrac{e^{-6\pi \sqrt5}}{\ddots}}}} = \left(\dfrac{\sqrt5}{1+\sqrt[5]{5^{3/4}(\frac{\sqrt5-1}2)^{5/2}-1}}-\dfrac{\sqrt5-1}{2}\right)e^{2\pi/\sqrt5}$

$\displaystyle{-\!\!\!\!\!\!\int_E \equiv \frac{1}{\mathrm{vol}\,E} \int_E}$

$Mf(x) := \sup_{B \ni x}\,-\!\!\!\!\!\int_B |f|\,d\mu$

$\cfrac{1}{1+\cfrac{e^{-2\pi \sqrt7}}{1+\cfrac{e^{-4\pi \sqrt5}}{1+\cfrac{e^{-6\pi \sqrt5}}{\ddots}}}} = \left(\dfrac{\sqrt5}{1+\sqrt[5]{5^{3/4}(\frac{\sqrt5-1}2)^{5/2}-1}}-\dfrac{\sqrt5-1}{2}\right)e^{2\pi/\sqrt5}$