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Math Help - Testing LaTeX

  1. #31
    Bar0n janvdl's Avatar
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    By the way:

    \ creates a space in Latex when something doesn't immediately follow it.

    For example

    James Jarvis (with the [tex] wrapped around of course)

    James Jarvis

    But James \ Jarvis

    James \ Jarvis
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  2. #32
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    Sorry! Im trying to write v= K x ln(1/x) K being constant and x being algibraic x.

    Thanks.
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  3. #33
    Bar0n janvdl's Avatar
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    Quote Originally Posted by james jarvis View Post
    Sorry! Im trying to write v= K x ln(1/x) K being constant and x being algibraic x.

    Thanks.
    V = K \times ln( \frac{1}{x} )

     V = K \times ln( \frac{1}{x} )
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  4. #34
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    V = K x ln( \frac{1}{x} )
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  5. #35
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    Krizalid's Avatar
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    I'd write it as

    V=K\cdot\ln\frac1x. Or V=K\times\ln\frac1x.

    Click on the LaTeX images to see their codes.
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  6. #36
    Bar0n janvdl's Avatar
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    Quote Originally Posted by Krizalid View Post
    I'd write it as

    V=K\cdot\ln\frac1x. Or V=K\times\ln\frac1x.

    Click on the LaTeX images to see their codes.
    It seems both x's are letters, not a multiplication sign
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  7. #37
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by janvdl View Post
    V = K \times ln( \frac{1}{x} )

     V = K \times ln( \frac{1}{x} )
    use \ln as opposed to ln

    and use \left( \frac 1x \right) as opposed to ( \frac 1x )

    you will get \ln \left( \frac 1x \right) which looks nicer than ln( \frac 1x)

    personally, i like using ~ for space, so i'd type James~Jarvis
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  8. #38
    Super Member PaulRS's Avatar
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    f(x)|^b_a=f(b)-f(a)

    I'd like to know the correct code for the long bar

    Thank you
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  9. #39
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    Hello, Paul!

    f(x)|^b_a=f(b)-f(a)

    I'd like to know the correct code for the long bar

    I use: .f(x) \bigg | ^b _a

    . . and get: .  f(x) \bigg|^b_a

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  10. #40
    Math Engineering Student
    Krizalid's Avatar
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    Quote Originally Posted by PaulRS View Post
    f(x)|^b_a=f(b)-f(a)

    I'd like to know the correct code for the long bar
    Ya te dieron una posibilidad, la otra es que uses

    \Big| ; \big| o como es común \left| para abrir y luego \right| para cerrar. En realidad esto último se adecúa a la expresión (tamaño de ésta), y se extiende.

    Eventualmente, usaría f(x)\Big|^b_a

    Donde aplicamos \Big|.

    Saludos
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  11. #41
    Newbie JoFaSs's Avatar
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    First time using LaTex and first post!
    Testing...

    \frac {x^{n+1}}{x^{n-1}}

    log_{7}7^k=k

    x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

    \int_{0}^{50}(x)=1250

    f(x)=3+\frac{x-2}{x+3}

    \frac{d}{dx}3x^k=3kx^{k-1}

    This is
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  12. #42
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    <br />
\frac{\sqrt{3}}{2}<br />
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  13. #43
    Senior Member ecMathGeek's Avatar
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    IFF~ \frac{d}{dx}F(x)=f(x) for a\leq x\leq b

    \lim_{n\rightarrow\infty}\sum^n_{i=1}{f\left(x+\le  ft(\frac{b-a}{n}\right)i\right)}\left(\frac{b-a}{n}\right)~=~\int^b_a{f(x)}dx~=~F(x)\bigg |^b_a~=~F(b)-F(a)
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  14. #44
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    log_2\;x\;+\;log_2\;y\;=\;3\quad\Rightarrow x\;+\;y\;=\;8

    6\cdot6=36

    <br />
\sum ^{\infty}_{i=1}<br />
    Last edited by OzzMan; February 9th 2008 at 12:45 AM.
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  15. #45
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    Testing
    mmm I get an error, back to the books
    <br />
\<br />
\int_0^a {\sqrt {(1 + t} ^2 )dt} <br />
\<br />
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