Tom McCurdy
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[tex]ax^n[/tex] hello just testing
Introducing LaTeX Math Typesetting
- Context: LaTeX
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Vance said:
In LaTeX, you can use \includegraphics
see http://www-group.slac.stanford.edu/techpubs/help/figures/latex2efigs.html
and http://www.linmpi.mpg.de/~daly/latex/grf.htm
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h8ter
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test again...
[tex]\sqrt[3]{27}=3[/tex]
[tex]\sqrt[3]{27}=3[/tex]
Vance
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robphy said:
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james11
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[ tex ] a^x_n [ /tex ]
poolwin2001
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[tex]x_1 *<br />
x_2<br />
x_3[/tex]
Emc2brain
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[ tex ] a^x_n [ /tex ]
Emc2brain
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[tex]a^x_n[/tex ]<br />
<br />
trial 2 (w/o spaces)[/tex]
Emc2brain
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Eke! Oh Well..i'll Try This Later
humanino
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That would be great if we could draw Feynman graphs ! Is it too difficult to handle ?
cyber ghega
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[tex]A=x^_n[/tex]
but
[tex]f(t)=alpha_n[/tex]
but
[tex]f(t)=alpha_n[/tex]
james11
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[ tex ]a^x_n[ /tex ]
james11
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[tex]a^x_n[/tex]
Trying without the spaces!
Trying without the spaces!
james11
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test
[tex]P=2pieG/3*P^2*R^2[/tex]
[tex]P=2pieG/3*P^2*R^2[/tex]
james11
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test!
P= \frac{2 \p G}{3}p^2r^2
P= \frac{2 \p G}{3}p^2r^2
james11
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Lets try again!
[tex]P= \frac{2 \p G}{3}p^2r^2[\tex][/tex]
[tex]P= \frac{2 \p G}{3}p^2r^2[\tex][/tex]
james11
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Lets try again!
[tex]P= \frac{2 \p G}{3}p^2r^2[/tex]
[tex]P= \frac{2 \p G}{3}p^2r^2[/tex]
james11
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just testing!
[tex] <br /> P= \frac{2 {\p} G}{3}p^2r^2[/tex]
[tex] <br /> P= \frac{2 {\p} G}{3}p^2r^2[/tex]
james11
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[tex]P= \frac{2\pG}{3}p^2r^2[/tex]
james11
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[tex]P= \frac{2{\pG}}{3}p^2r^2[/tex]
james11
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[tex]P= \frac{2{\p[G]}}{3}p^2r^2[/tex]
[tex]P= \frac{2{\p[G]}}{3}p^2r^2[/tex]
[tex]P= \frac{2{\p[G]}}{3}p^2r^2[/tex]
james11
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[tex]P= \frac{2{\p}{G}{3}p^2r^2[/tex]
james11
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[tex]P= \frac{2\pieG}{3}p^2r^2[/tex]
james11
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[tex]P= \frac{\2pieG}{3}p^2r^2[/tex]
[tex]2\pieG}{[/tex]
[tex]\2pieG[/tex]
[tex]2\pieG}{[/tex]
[tex]\2pieG[/tex]
brianparks
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Testing...
[tex]\Vec{v}(x,y,z)=\frac{1}{x^2+y^2+z^2}(x\Vec{i}+y\Vec{j}+z\Vec{k})[/tex]
[tex]\frac{\partial v_x}{\partial x} =-\frac{2x^2}{(x^2+y^2+z^2)^2}+\frac{1}{x^2+y^2+z^2}= \frac{-2x^2}{r^4} + \frac{1}{r^2}[/tex]
In the same way,
[tex]\frac{\partial v_y}{\partial y} =-\frac{2y^2}{r^4}+\frac{1}{r^2}[/tex]
and
[tex]\frac{\partial v_z}{\partial z} =-\frac{2z^2}{r^4}+\frac{1}{r^2}[/tex]
The divergence is
[tex]\mbox{Div}(\Vec{v})=\frac{\partial v_x}{\partial x} +\frac{\partial v_y}{\partial y} +\frac{\partial v_z}{\partial z}=[/tex]
[tex]=\frac{-2x^2-2y^2-2z^2}{r^4}+\frac{3}{r^2}=\frac{-2r^2}{r^4}+\frac{3}{r^2}=\frac{1}{r^2}[/tex]
It is the function
[tex]\Vec{v}(\Vec{r}) = \frac{\Vec{r}}{r^3}[/tex]
whose divergence is zero everywhere except at the origin.
[tex]\Vec{v}(x,y,z)=\frac{1}{x^2+y^2+z^2}(x\Vec{i}+y\Vec{j}+z\Vec{k})[/tex]
[tex]\frac{\partial v_x}{\partial x} =-\frac{2x^2}{(x^2+y^2+z^2)^2}+\frac{1}{x^2+y^2+z^2}= \frac{-2x^2}{r^4} + \frac{1}{r^2}[/tex]
In the same way,
[tex]\frac{\partial v_y}{\partial y} =-\frac{2y^2}{r^4}+\frac{1}{r^2}[/tex]
and
[tex]\frac{\partial v_z}{\partial z} =-\frac{2z^2}{r^4}+\frac{1}{r^2}[/tex]
The divergence is
[tex]\mbox{Div}(\Vec{v})=\frac{\partial v_x}{\partial x} +\frac{\partial v_y}{\partial y} +\frac{\partial v_z}{\partial z}=[/tex]
[tex]=\frac{-2x^2-2y^2-2z^2}{r^4}+\frac{3}{r^2}=\frac{-2r^2}{r^4}+\frac{3}{r^2}=\frac{1}{r^2}[/tex]
It is the function
[tex]\Vec{v}(\Vec{r}) = \frac{\Vec{r}}{r^3}[/tex]
whose divergence is zero everywhere except at the origin.