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Math Help - Testing LaTeX

  1. #76
    Lord of certain Rings
    Isomorphism's Avatar
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    Quote Originally Posted by Moo View Post
    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.
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  2. #77
    Moo
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    Yeah Iso, I was talking about the dot after \right, actually

    Thanks for explanations guys !
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  3. #78
    Super Member flyingsquirrel's Avatar
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    Is it possible to have equations aligned like this :

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

    ... but aligned on the left ?
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  4. #79
    Math Engineering Student
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    Quote Originally Posted by Isomorphism View Post

    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

    Quote Originally Posted by flyingsquirrel View Post
    Is it possible to have equations aligned like this :

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

    ... but aligned on the left ?
    aligned environment will do it:

    \begin{aligned}<br />
   \exp x&=1+x+\frac{x^{2}}{2}+\cdots +\frac{x^{k}}{k!}+\cdots  \\ <br />
 & =\sum\limits_{n\,=\,0}^{\infty }{\frac{x^{n}}{n!}} \\ <br />
 & =42. <br />
\end{aligned}
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  5. #80
    Rhymes with Orange Chris L T521's Avatar
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    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.
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  6. #81
    Moo
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    \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}
    Last edited by Moo; June 22nd 2008 at 11:25 AM.
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  7. #82
    A riddle wrapped in an enigma
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    Quote Originally Posted by Isomorphism View Post
    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)
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  8. #83
    Moo
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    Quote Originally Posted by masters View Post
    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).
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  9. #84
    A riddle wrapped in an enigma
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    Quote Originally Posted by masters View Post
    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.
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  10. #85
    Moo
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    Quote Originally Posted by masters View Post
    [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
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  11. #86
    Senior Member nikhil's Avatar
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    Testing

    <br />
\sin^{-1}x<br />
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  12. #87
    MHF Contributor Mathstud28's Avatar
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    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?
    Last edited by Mathstud28; June 30th 2008 at 09:45 PM.
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  13. #88
    Senior Member nikhil's Avatar
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    <br />
\sin^{-1}x+\sin^{-1}y=\sin^{-1}(x\sqrt{1-y^2}+y\sqrt{1-x^2})<br />
    or <br />
\sqrt{1-x^2}\sqrt{1-y^2}>=xy<br />
    and <br />
|xy|<=\sqrt{1-x^2}\sqrt{1-y^2}<br />
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  14. #89
    Senior Member nikhil's Avatar
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    <br />
\sin^{-1}x=A <br />
\sin^{-1}y=B<br />
    <br />
\sin^{-1}x+\sin^{-1}y=\sin^{-1}(x\sqrt{1-y^2}+y\sqrt{1-x^2})<br />
    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 <br />
\sin^{-1}x+\sin^{-1}y<br />
lies between -pi/2 to pi/2. This happens if cos(A+B)>=0 as its >=0 between -pi/2 to pi/2
    so <br />
\sqrt{1-x^2}\sqrt{1-y^2}-xy>=0<br />
or <br />
\sqrt{1-x^2}\sqrt{1-y^2}>=xy<br />
    If x,y>=0 or x,y <= on squaring there will be no change is sign therfor on squaring we get
    <br />
x^{2}+y^{2}<=1<br />
    also if xy<0 but <br />
|xy|<=\sqrt{1-x^2}\sqrt{1-y^2}<br />
on squaring there will be again no change of equality sign and we will get
    <br />
x^{2}+y^{2}<=1<br />
    but if xy<0 and <br />
|xy|>\sqrt{1-x^2}\sqrt{1-y^2}<br />
on squaring there will be change of equality sign and we will get
    <br />
x^{2}+y^{2}>1<br />
    so finally we get
    <br />
\sin^{-1}x+\sin^{-1}y=\sin^{-1}(x\sqrt{1-y^2}+y\sqrt{1-x^2})<br />
    if
    1) <br />
x^{2}+y^{2}<=1<br />
and
    a)x,y>=0
    b)x,y<=0
    c)xy<0 and <br />
|xy|<=\sqrt{1-x^2}\sqrt{1-y^2}<br />
    2) <br />
x^{2}+y^{2}>1<br />
if
    xy<0 and <br />
|xy|>\sqrt{1-x^2}\sqrt{1-y^2}<br />
    but all the books that I have read say nothing about <br />
|xy|>\sqrt{1-x^2}\sqrt{1-y^2}<br />
and <br />
|xy|<=\sqrt{1-x^2}\sqrt{1-y^2}<br />
. Are they not necesrary???
    Last edited by nikhil; July 1st 2008 at 02:41 AM.
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  15. #90
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by nikhil View Post
    <br />
\sin^{-1}x=A <br />
\sin^{-1}y=B<br />
    <br />
\sin^{-1}x+\sin^{-1}y=\sin^{-1}(x\sqrt{1-y^2}+y\sqrt{1-x^2})<br />
    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 <br />
\sin^{-1}x+\sin^{-1}y<br />
lies between -pi/2 to pi/2. This happens if cos(A+B)>=0 as its >=0 between -pi/2 to pi/2
    so <br />
\sqrt{1-x^2}\sqrt{1-y^2}-xy>=0<br />
or <br />
\sqrt{1-x^2}\sqrt{1-y^2}>=xy<br />
    If x,y>=0 or x,y <= on squaring there will be no change is sign therfor on squaring we get
    <br />
x^{2}+y^{2}<=1<br />
    also if xy<0 but <br />
|xy|<=\sqrt{1-x^2}\sqrt{1-y^2}<br />
on squaring there will be again no change of equality sign and we will get
    <br />
x^{2}+y^{2}<=1<br />
    but if xy<0 and <br />
|xy|>\sqrt{1-x^2}\sqrt{1-y^2}<br />
on squaring there will be change of equality sign and we will get
    <br />
x^{2}+y^{2}>1<br />
    so finally we get
    <br />
\sin^{-1}x+\sin^{-1}y=\sin^{-1}(x\sqrt{1-y^2}+y\sqrt{1-x^2})<br />
    if
    1) <br />
x^{2}+y^{2}<=1<br />
and
    a)x,y>=0
    b)x,y<=0
    c)xy<0 and <br />
|xy|<=\sqrt{1-x^2}\sqrt{1-y^2}<br />
    2) <br />
x^{2}+y^{2}>1<br />
if
    xy<0 and <br />
|xy|>\sqrt{1-x^2}\sqrt{1-y^2}<br />
    but all the books that I have read say nothing about <br />
|xy|>\sqrt{1-x^2}\sqrt{1-y^2}<br />
and <br />
|xy|<=\sqrt{1-x^2}\sqrt{1-y^2}<br />
. 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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