Testing LaTeX

**Isomorphism** · May 2nd 2008, 10:44 AM

Originally Posted by Moo

And what does \right. stand for ?

It adjusts its format according to the brackets,paranthesis, etc according to the size of the expression contained within it.

For example:
Without \left and \right
If you type [\frac{\frac12}{\frac12}] = 1 you get $[\frac{\frac12}{\frac12}] = 1$

The square brackets are small compared to the expression inside and they look ugly

However with \right and \left
If you type \left[\frac{\frac12}{\frac12}\right] = 1 you get $\left[\frac{\frac12}{\frac12}\right] = 1$

The square brackets are now properly reformatted and looks nice.

**Moo** · May 2nd 2008, 12:10 PM

Yeah Iso, I was talking about the dot after \right, actually

Thanks for explanations guys !

**flyingsquirrel** · May 3rd 2008, 02:23 PM

Is it possible to have equations aligned like this :

$\begin{array}{lll} \exp x & = & 1+x+\frac{x^2}{2}+\ldots+\frac{x^k}{k!}+\ldots\\ \exp x &=&\sum_{n=0}^{\infty}\frac{x^n}{n!}\\ & = & 42 \end{array}$

... but with a nice symbol for the sum and a bigger size ? (\displaystyle doesn't seem to work :/)

That is to say something which looks like that :

$\begin{aligned} \exp x & = & 1+x+\frac{x^2}{2}+\ldots+\frac{x^k}{k!}+\ldots\\ \exp x &=&\sum_{n=0}^{\infty}\frac{x^n}{n!}\\ & = & 42 \end{aligned}$

... but aligned on the left ?

**Krizalid** · May 3rd 2008, 02:39 PM

Originally Posted by Isomorphism

However with \right and \left
If you type \left[\frac{\frac12}{\frac12}\right] = 1 you get $\left[\frac{\frac12}{\frac12}\right] = 1$

Improve this by addin' \dfrac, this yields

$\left[\frac{\dfrac12}{\dfrac12}\right] = 1$

Originally Posted by flyingsquirrel

Is it possible to have equations aligned like this :

$\begin{array}{lll} \exp x & = & 1+x+\frac{x^2}{2}+\ldots+\frac{x^k}{k!}+\ldots\\ \exp x &=&\sum_{n=0}^{\infty}\frac{x^n}{n!}\\ & = & 42 \end{array}$

... but with a nice symbol for the sum and a bigger size ? (\displaystyle doesn't seem to work :/)

That is to say something which looks like that :

$\begin{aligned} \exp x & = & 1+x+\frac{x^2}{2}+\ldots+\frac{x^k}{k!}+\ldots\\ \exp x &=&\sum_{n=0}^{\infty}\frac{x^n}{n!}\\ & = & 42 \end{aligned}$

... but aligned on the left ?

aligned environment will do it:

$\begin{aligned} \exp x&=1+x+\frac{x^{2}}{2}+\cdots +\frac{x^{k}}{k!}+\cdots \\ & =\sum\limits_{n\,=\,0}^{\infty }{\frac{x^{n}}{n!}} \\ & =42. \end{aligned}$

**Chris L T521** · May 15th 2008, 04:47 PM

I myself am still trying to figure this language out, and I came across something that seems pretty useful to me.

Click here for LaTex Mathematical Symbols.

**Moo** · June 22nd 2008, 04:40 AM

$\mathcal{Aa \quad Bb \quad Cc \quad Dd \quad Ee \quad Ff \quad Gg \quad Hh}$

$\mathcal{Ii} \qquad \text{shortcuts ?}$

$\mathcal{Jj \quad Kk \quad Ll \quad Mm}$

$\mathcal{Nn} \qquad \iiiint \rightarrow \left\{\begin{array}{ll} \text{This is the maximum :} \\ \text{Quadruple integral} \end{array} \right.$

$\mathcal{Oo \quad Pp \quad Qq \quad Rr \quad Ss \quad Tt \quad Uu \quad Vv \quad Ww \quad Xx \quad Yy \quad Zz}$

$\mathcal{0 \quad 1 \quad 2 \quad 3 \quad 4 \quad 5 \quad 6 \quad 7 \quad 8 \quad 9}$

**masters** · June 22nd 2008, 02:19 PM

Originally Posted by Isomorphism

It adjusts its format according to the brackets,paranthesis, etc according to the size of the expression contained within it.

For example:
Without \left and \right
If you type [\frac{\frac12}{\frac12}] = 1 you get $[\frac{\frac12}{\frac12}] = 1$

The square brackets are small compared to the expression inside and they look ugly

However with \right and \left
If you type \left[\frac{\frac12}{\frac12}\right] = 1 you get $\left[\frac{\frac12}{\frac12}\right] = 1$

The square brackets are now properly reformatted and looks nice.

So what happened to this? I only got reformatted parentheses once instead of twice:

$\left(-\frac{1}{4}, 0\right) \cup \left(0, \infty\right)$

**Moo** · June 22nd 2008, 02:27 PM

Originally Posted by masters

So what happened to this? I only got reformatted parentheses once instead of twice:

$\left(-\frac{1}{4}, 0\right) \cup \left(0, \infty\right)$

Because it's formatted with respect to the size of what is in the brackets.

You would have to use \big, \bigg, \Big, \Bigg (I think the last 2 are giving "bold" brackets).

**masters** · June 22nd 2008, 02:38 PM

Originally Posted by masters

So what happened to this? I only got reformatted parentheses once instead of twice:

$\left(-\frac{1}{4}, 0\right) \cup \left(0, \infty\right)$

[tex]\left(-\frac{1}{4}, 0\right) \cup \left(0, \infty\right)[/tex]

causes this to appear.

$\left(-\frac{1}{4}, 0\right) \cup \left(0, \infty\right)$

So if I use:

[tex]\left(-\frac{1}{4}, 0\right) \cup \bigg(0, \infty\bigg)[/tex]

I get:

$\left(-\frac{1}{4}, 0\right) \cup \bigg(0, \infty\bigg)$

Thanks Moo. I'm still learning.

**Moo** · June 22nd 2008, 02:41 PM

Originally Posted by masters

[tex]\left(-\frac{1}{4}, 0\right) \cup \left(0, \infty\right)[/tex]

causes this to appear.

$\left(-\frac{1}{4}, 0\right) \cup \left(0, \infty\right)$

So if I use:

[tex]\left(-\frac{1}{4}, 0\right) \cup \bigg(0, \infty\bigg)[/tex]

I get:

$\left(-\frac{1}{4}, 0\right) \cup \bigg(0, \infty\bigg)$

Thanks Moo. I'm still learning.

Nice smilie !

If you want it to be the same size :

[tex]\big(-\frac 14,0\big) \cup \big(0, \infty\big)[/tex] (note that you can write \frac ab, when there is only one term in the fraction ^^).

$\big(-\frac 14,0\big) \cup \big(0, \infty\big)$

[tex]\bigg(-\frac 14,0\bigg) \cup \bigg(0, \infty\bigg)[/tex]

$\bigg(-\frac 14,0\bigg) \cup \bigg(0, \infty\bigg)$

I would say this is useful when one has some difficulties to remember that the left comes before the right

**nikhil** · June 28th 2008, 07:02 PM

$ \sin^{-1}x $

**Mathstud28** · June 30th 2008, 08:50 PM

If $f=f\left(x,y,z,\cdots\right)$

Then $\nabla{f}=\left\langle\frac{\partial{f}}{\partial{ x}},\frac{\partial{f}}{\partial{y}},\frac{\partial {f}}{\partial{z}}\cdots\right\rangle$

and $\mathbf{D}_{\mathbf{u}}=\nabla{f}\cdot\mathbf{\hat {u}}$

How would you do qunitic integration?

$\iiiint\int{dxdydzdtdu}$

That doesn't look very nice

.

I think I would actually prefer

$\int\int\int\int\int{dxdydzdtdu}$

Ahh..ohh...symmetry

------------------------------------------------------------------------
Which do you think looks better?

$\int_a^{b}f(x)dx$

or

$\int\limits_a^b{f(x)}dx$ ?

**nikhil** · July 1st 2008, 12:57 AM

$ \sin^{-1}x+\sin^{-1}y=\sin^{-1}(x\sqrt{1-y^2}+y\sqrt{1-x^2}) $
or $ \sqrt{1-x^2}\sqrt{1-y^2}>=xy $
and $ |xy|<=\sqrt{1-x^2}\sqrt{1-y^2} $

**nikhil** · July 1st 2008, 02:26 AM

$ \sin^{-1}x=A \sin^{-1}y=B $
$ \sin^{-1}x+\sin^{-1}y=\sin^{-1}(x\sqrt{1-y^2}+y\sqrt{1-x^2}) $
but range of LHS is -pi to pi and range of RHS is -pi/2 to pi/2.so this equation will be true if $ \sin^{-1}x+\sin^{-1}y $ lies between -pi/2 to pi/2. This happens if cos(A+B)>=0 as its >=0 between -pi/2 to pi/2
so $ \sqrt{1-x^2}\sqrt{1-y^2}-xy>=0 $ or $ \sqrt{1-x^2}\sqrt{1-y^2}>=xy $
If x,y>=0 or x,y <= on squaring there will be no change is sign therfor on squaring we get
$ x^{2}+y^{2}<=1 $
also if xy<0 but $ |xy|<=\sqrt{1-x^2}\sqrt{1-y^2} $ on squaring there will be again no change of equality sign and we will get
$ x^{2}+y^{2}<=1 $
but if xy<0 and $ |xy|>\sqrt{1-x^2}\sqrt{1-y^2} $ on squaring there will be change of equality sign and we will get
$ x^{2}+y^{2}>1 $
so finally we get
$ \sin^{-1}x+\sin^{-1}y=\sin^{-1}(x\sqrt{1-y^2}+y\sqrt{1-x^2}) $
if
1) $ x^{2}+y^{2}<=1 $ and
a)x,y>=0
b)x,y<=0
c)xy<0 and $ |xy|<=\sqrt{1-x^2}\sqrt{1-y^2} $
2) $ x^{2}+y^{2}>1 $ if
xy<0 and $ |xy|>\sqrt{1-x^2}\sqrt{1-y^2} $
but all the books that I have read say nothing about $ |xy|>\sqrt{1-x^2}\sqrt{1-y^2} $ and $ |xy|<=\sqrt{1-x^2}\sqrt{1-y^2} $ . Are they not necesrary???

**Mathstud28** · July 1st 2008, 08:45 AM

Originally Posted by nikhil

$ \sin^{-1}x=A \sin^{-1}y=B $
$ \sin^{-1}x+\sin^{-1}y=\sin^{-1}(x\sqrt{1-y^2}+y\sqrt{1-x^2}) $
but range of LHS is -pi to pi and range of RHS is -pi/2 to pi/2.so this equation will be true if $ \sin^{-1}x+\sin^{-1}y $ lies between -pi/2 to pi/2. This happens if cos(A+B)>=0 as its >=0 between -pi/2 to pi/2
so $ \sqrt{1-x^2}\sqrt{1-y^2}-xy>=0 $ or $ \sqrt{1-x^2}\sqrt{1-y^2}>=xy $
If x,y>=0 or x,y <= on squaring there will be no change is sign therfor on squaring we get
$ x^{2}+y^{2}<=1 $
also if xy<0 but $ |xy|<=\sqrt{1-x^2}\sqrt{1-y^2} $ on squaring there will be again no change of equality sign and we will get
$ x^{2}+y^{2}<=1 $
but if xy<0 and $ |xy|>\sqrt{1-x^2}\sqrt{1-y^2} $ on squaring there will be change of equality sign and we will get
$ x^{2}+y^{2}>1 $
so finally we get
$ \sin^{-1}x+\sin^{-1}y=\sin^{-1}(x\sqrt{1-y^2}+y\sqrt{1-x^2}) $
if
1) $ x^{2}+y^{2}<=1 $ and
a)x,y>=0
b)x,y<=0
c)xy<0 and $ |xy|<=\sqrt{1-x^2}\sqrt{1-y^2} $
2) $ x^{2}+y^{2}>1 $ if
xy<0 and $ |xy|>\sqrt{1-x^2}\sqrt{1-y^2} $
but all the books that I have read say nothing about $ |xy|>\sqrt{1-x^2}\sqrt{1-y^2} $ and $ |xy|<=\sqrt{1-x^2}\sqrt{1-y^2} $ . Are they not necesrary???

Hmm, this is interesting. I have never seen this identity before. I assume it is derived by converting the inverse sine function to the complex logarithmic form. But anyways, very interesting.

Also note that

$\leq$ is gotten by \leq

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