- #301
Tom McCurdy
- 1,017
- 1
ax^n hello just testing
LaTeX Introducing LaTeX Math Typesetting
- Thread starter chroot
- Start date
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- Tags
- Introducing Latex Typesetting
Click For Summary
Physics Forums has integrated LaTeX mathematical typesetting into its platform, allowing users to create visually appealing mathematical expressions using markup similar to HTML. Users can include LaTeX graphics in posts by wrapping their code in [tex] or [itex] tags for display or inline formatting, respectively. A PDF guide with essential LaTeX commands and symbols is available, along with links to additional resources. The community is encouraged to experiment with the system and share examples, while also being mindful of server load when generating graphics. This addition aims to enhance the clarity and professionalism of mathematical discussions on the forum.
- #302
- 7,300
- 2,813
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
Last edited by a moderator:
- #303
h8ter
- 32
- 0
test again...
\sqrt[3]{27}=3
\sqrt[3]{27}=3
- #304
Vance
- 180
- 0
robphy said:
Last edited by a moderator:
- #305
james11
- 22
- 0
[ tex ] a^x_n [ /tex ]
- #306
- 10,266
- 45
james11,
Try it without the spaces.
- Warren
Try it without the spaces.
- Warren
- #307
poolwin2001
- 179
- 0
x_1 *<br />
x_2<br />
x_3<br />
- #308
Emc2brain
- 22
- 0
[ tex ] a^x_n [ /tex ]
- #309
Emc2brain
- 22
- 0
a^x_n[/tex ]<br />
<br />
trial 2 (w/o spaces)
- #310
Emc2brain
- 22
- 0
Eke! Oh Well..i'll Try This Later
- #311
- 10,266
- 45
Emc2brain,
You still have a space in your trial 2.
- Warren
You still have a space in your trial 2.
- Warren
- #312
humanino
- 2,523
- 8
That would be great if we could draw Feynman graphs ! Is it too difficult to handle ?
- #313
cyber ghega
- 1
- 0
A=x^_n
but
f(t)=alpha_n
but
f(t)=alpha_n
- #314
james11
- 22
- 0
[ tex ]a^x_n[ /tex ]
- #315
james11
- 22
- 0
a^x_n
Trying without the spaces!
Trying without the spaces!
- #316
james11
- 22
- 0
test
P=2pieG/3*P^2*R^2
P=2pieG/3*P^2*R^2
- #317
james11
- 22
- 0
test!
P= \frac{2 \p G}{3}p^2r^2
P= \frac{2 \p G}{3}p^2r^2
- #318
james11
- 22
- 0
Lets try again!
P= \frac{2 \p G}{3}p^2r^2[\tex]
P= \frac{2 \p G}{3}p^2r^2[\tex]
- #319
james11
- 22
- 0
Lets try again!
P= \frac{2 \p G}{3}p^2r^2
P= \frac{2 \p G}{3}p^2r^2
- #320
james11
- 22
- 0
just testing!
<br /> <br /> P= \frac{2 {\p} G}{3}p^2r^2
<br /> <br /> P= \frac{2 {\p} G}{3}p^2r^2
- #321
james11
- 22
- 0
P= \frac{2\pG}{3}p^2r^2
- #322
james11
- 22
- 0
P= \frac{2{\pG}}{3}p^2r^2
- #323
james11
- 22
- 0
P= \frac{2{\p[G]}}{3}p^2r^2
P= \frac{2{\p[G]}}{3}p^2r^2
P= \frac{2{\p[G]}}{3}p^2r^2
- #324
james11
- 22
- 0
P= \frac{2{\p}{G}{3}p^2r^2
- #325
james11
- 22
- 0
P= \frac{2\pieG}{3}p^2r^2
- #326
james11
- 22
- 0
P= \frac{\2pieG}{3}p^2r^2
2\pieG}{
\2pieG
2\pieG}{
\2pieG
- #327
- 10,266
- 45
james11,
Use \pi to represent \pi, rather than using "pie."
- Warren
Use \pi to represent \pi, rather than using "pie."
- Warren
- #328
Janitor
Science Advisor
- 1,107
- 1
- #329
brianparks
- 24
- 0
Testing...
\Vec{v}(x,y,z)=\frac{1}{x^2+y^2+z^2}(x\Vec{i}+y\Vec{j}+z\Vec{k})
\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}
In the same way,
\frac{\partial v_y}{\partial y} =-\frac{2y^2}{r^4}+\frac{1}{r^2}
and
\frac{\partial v_z}{\partial z} =-\frac{2z^2}{r^4}+\frac{1}{r^2}
The divergence is
\mbox{Div}(\Vec{v})=\frac{\partial v_x}{\partial x} +\frac{\partial v_y}{\partial y} +\frac{\partial v_z}{\partial z}=
=\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}
It is the function
\Vec{v}(\Vec{r}) = \frac{\Vec{r}}{r^3}
whose divergence is zero everywhere except at the origin.
\Vec{v}(x,y,z)=\frac{1}{x^2+y^2+z^2}(x\Vec{i}+y\Vec{j}+z\Vec{k})
\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}
In the same way,
\frac{\partial v_y}{\partial y} =-\frac{2y^2}{r^4}+\frac{1}{r^2}
and
\frac{\partial v_z}{\partial z} =-\frac{2z^2}{r^4}+\frac{1}{r^2}
The divergence is
\mbox{Div}(\Vec{v})=\frac{\partial v_x}{\partial x} +\frac{\partial v_y}{\partial y} +\frac{\partial v_z}{\partial z}=
=\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}
It is the function
\Vec{v}(\Vec{r}) = \frac{\Vec{r}}{r^3}
whose divergence is zero everywhere except at the origin.
- #330
- 10,266
- 45
Test
\vec F = m \vec a
\vec F = m \vec a
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