Introducing LaTeX Math Typesetting

[Castilla]

Testing...

$${\int_1^\sqrt{x}} t^{x}dt$$

 

[Castilla]

Testing...

If the functions $$f(x,t)$$ and
$$\frac{\partial}{\partial t}f(x,t)$$ are continuous in
$$[a,b] \times [c,d]$$, then $$\frac{d}{dt}\int_a^{b}f(x,t)dx = \int_a^{b}\frac{\partial}{\partial t}f(x,t)dx.$$

 

[JamesU]

$$\sqrt{Opposite^2 + Adjacent^2} = Hypotenuse$$

 

[beautiful1]

I too am just testing
$$\int dx \int dy \exp (-a (x+y)^2 +ib(x-y)) sinc(cx+dy) sinc(dx+cy)$$

 

[JamesU]

$$\sqrt{cool} = \sqrt{me} . \ me^2=me. \ cool^2=awesome. \ so \ me = awesome$$

 

[fliptomato]

QUESTION...

In the archives for this forum, someone suggested the following site for a script on how to put in wick contractions into latex:

http://www.fzu.cz/~kolorenc/tex.php

The post suggested using simple_wick.tex, saying it worked very well.

I agree that it works... but I can't figure out how to use it. That is, the example works well, and I can toy around and get random results, but I can't figure out what the logical rules are to use this script. Any ideas?


Flip

 

[Neohaven]

As x tends to infinity, the limit of 1/x is such that

$$\mathop {\lim }\limits_{x \to \infty } f(x) = 0$$

Works kinda well, doesn't it? Too bad it doesn't work on my computer...

 

[tomfitzyuk]

$$T = \frac{1}{f} = 2 \pi \sqrt \frac{l}{g}$$

 

[calculus1967]

Let the function $f$ be continuos on the closed interval $[a, b]$, and assume that $f(x) \geq 0$ for all $x$ in $[a, b]$. If $S$ is the solid of revolution obtained by revolving about the $x$ axis the region bounded by the curve $y = f(x)$, the $x$ axis, and the lines $x = a$ and $x = b$, and if $V$ is the number of cubic units in the volume of $S$, then

$$V = \pi \int^b_{a}[f(x)]^2 dx$$

 

[blimkie]

[ tex ] a^x_n [ /tex ]

 

[Jeff Ford]

Just a test

$$a^_x$$

 

[bomba923]

I think we all know Taylor expansion:

$$\boxed{f\left( x \right) = \sum\limits_{n = 0}^\infty {\frac{{f^{\left( n \right)} \left( a \right)}}{{n!}}\left( {x - a} \right)^n } ,\left| {x - a} \right| < R}$$

 

[wr1015]

$$F_{1}+F{2}$$

just testing

 

[ktoz]

level: 1
$$\sum_{j=0}^m a + cj = \frac{(m + 1)(2a + cm)}{2}$$

level: 2
$$\sum_{j=0}^m \frac{(j + 1)(2a + cj)}{2} = \frac{(m + 1)(m + 2)(3a + cm)}{6}\\$$

level: 3
$$\sum_{j=0}^m \frac{(j + 1)(j + 2)(3a + cj)}{6} = \frac{(m + 1)(m + 2)(m + 3)(4a + cm)}{24}\\$$

level: 4
$$\sum_{j=0}^m \frac{(j + 1)(j + 2)(j + 3)(4a + cj)}{24} = \frac{(m + 1)(m + 2)(m + 3)(m + 4)(5a + cm)}{120}\\$$

level: n
$$= \frac{(m + n)!(a(n + 1) + cm)}{m!(n + 1)!}$$

As product
$$\frac{(m + n)!(a(n + 1) + cm)}{m!(n + 1)!} = (a(n + 1) + cm) \prod_{j=1}^n \frac{m + j}{j + 1}$$

 

[saltydog]

Did I mention Mathematica has an Eigenvector[Matrix] command? However, I'm not good at calculating eigenvectors so I really should do a few by hand:

The eigenvector equation is simple:

$$\mathbf{M}v=\lambda v$$

So for:

$$\lambda_1=-1$$

$$\left( \begin{array}{ccc} 1 & 0 & 3 \\ 0 & -1 & 0 \\ -3 & 0 & 1 \end{array} \right) \left( \begin{array}{c} x \\ y \\ z \end{array} \right)=-1 \left( \begin{array}{c} x \\ y \\ z \end{array} \right)$$

So:

$$x+3z=-x\\ -y=-y\\ -3x+z=-z$$

The middle one is easy:

$$0y=0$$

That means y can be anything so let y=1.
The other two:

$$2x+3z=0 -3x+2z=0$$

The simple thing here, since we're looking for ANY eigenvector, is to just pick the zero solution and thus:

$$v_1= \left( \begin{array}{c} 0 \\ 1 \\ 0 \end{array} \right)$$

For:
$$\lambda_2=(1+3i)$$

$$\left( \begin{array}{ccc} 1 & 0 & 3 \\ 0 & -1 & 0 \\ -3 & 0 & 1 \end{array} \right) \left( \begin{array}{c} x \\ y \\ z \end{array} \right)=(1+3i) \left( \begin{array}{c} x \\ y \\ z \end{array} \right)$$

So that's:

$$x+3z=(1+3i)x -y=(1+3i)y -3x+z=(1+3i)z$$

For the middle one, the only way ay=y is if y=0. The other two yield:
$$3z=3ix$$

or:

z=ix

so let x=1 and z=i.

Thus:

$$v_2= \left( \begin{array}{c} 0 \\ 1 \\ 0 \end{array} \right)$$

Same dif for the other eigenvalue which yields:
$$v_3= \left( \begin{array}{c} 1 \\ 0 \\ -i \end{array} \right)$$

Mathematica returns equivalent eigenvectors.
Thus we are led to the solution in matrix form:

$$\mathbf{x}=c_1e^{-t} \left( \begin{array}{c} 0 \\ 1 \\ 0 \end{array} \right)+ c_2e^{(1+3i)t} \left( \begin{array}{c} 1 \\ 0 \\ i \end{array} \right)+ c_3e^{(1-3i)t} \left( \begin{array}{c} 1 \\ 0 \\ -i \end{array} \right)$$

You ever work a problem and the answer is just as difficult as the question?

 

[saltydog]

After reviewing, I wish to clear up two points in my efforts to solve this equation:

1. There is no need to directly calculate the eigenvector of $\lambda_3$:

The complex conjugate of an eigenvector for $\lambda_2$ is an eigenvector for [itex]\lambda_3[/tex].

So above I calculated the eigenvector for $(1+3i)$ to be:

$$v_2=\left(\begin{array}{c} 1 \\0 \\i\end{array}\right)$$

Therefore to calculate the eigenvector for $(1-3i)$, conjugate the eigenvector for [itex](1+3i)[/tex]:

If:

$$v_2=\left(\begin{array}{c} 1+0i \\0+0i \\0+i\end{array}\right)$$

Then:

$$\overline{v_2}=v_3=\left(\begin{array}{c} 1-0i \\0-i \\0-i\end{array}\right)=\left(\begin{array}{c} 1 \\0 \\-i\end{array}\right)$$

See how that works?

Ok that's one.

2. The complex eigenvalues both yield the same solution! That is, the solution from (1+3i) is the same solution as that from (1-3i). Go figure. I did. So that's the reason we only need calculate the Real and Complex contribution from ONE member of each complex pair. And also, it's VERY convenient to write the eigenvectors as:

$$\left(\begin{array}{c} 1 \\0 \\-i\end{array}\right)=\left(\begin{array}{c} 1 \\0 \\0\end{array}\right)+i\left(\begin{array}{c} 0 \\0 \\1\end{array}\right)$$

Alright, so let's compute the Real part and the Complex part for:

$$\begin{align*} e^{(1+3i)t}\left(\begin{array}{c} 1 \\0 \\i\end{array}\right) &=e^{(1+3i)t}\left[\left(\begin{array}{c} 1 \\0 \\0\end{array}\right)+i\left(\begin{array}{c} 0 \\0 \\1\end{array}\right)\right] \\ &=e^t\left[(Cos(3t)+iSin(3t))\left\{\left(\begin{array}{c} 1 \\0 \\0\end{array}\right)+ i\left(\begin{array}{c} 0 \\0 \\1\end{array}\right)\right\}\right] \\ &=e^t\left[Cos(3t)\left(\begin{array}{c} 1 \\0 \\0\end{array}\right)+iCos(3t)\left(\begin{array}{c} 0 \\0 \\1\end{array}\right)+iSin(3t)\left(\begin{array}{c} 1 \\0 \\0\end{array}\right)-Sin(3t)\left(\begin{array}{c} 0 \\0 \\1\end{array}\right)\right] \\ &=e^t\left[C_1\left\{Cos(3t)\left(\begin{array}{c} 1 \\0 \\0\end{array}\right)-Sin(3t)\left(\begin{array}{c} 0 \\0 \\1\end{array}\right)\right\}+C_2\left\{Cos(3t)\left(\begin{array}{c} 0 \\0 \\1\end{array}\right)+Sin(3t)\left(\begin{array}{c} 1 \\0 \\0\end{array}\right)\right\}\right] \end{align}$$

This with the first solution then yields the general solution:

$$\begin{align*} \mathbf{X}&=C_1e^{-t}\left(\begin{array}{c} 0 \\1 \\0\end{array}\right) \\ &+e^t\left[C_2\left\{Cos(3t)\left(\begin{array}{c} 1 \\0 \\0\end{array}\right)-Sin(3t)\left(\begin{array}{c} 0 \\0 \\1\end{array}\right)\right\}+C_3\left\{Cos(3t)\left(\begin{array}{c} 0 \\0 \\1\end{array}\right)+Sin(3t)\left(\begin{array}{c} 1 \\0 \\0\end{array}\right)\right\}\right] \end{align}$$

That's read as:

$$x(t)=C_2e^tCos(3t)+C_3e^tSin(3t)$$

$$y(t)=C_1e^{-t}$$

$$z(t)=-C_2e^tSin(3t)+C_3e^tCos(3t)$$

 

[dekoi]

$$avg=\frac{n_1+n_2+n_3+n_4+n_5}{n}$$$$dv_n=average-reading_n$$
$$\delta=\sqrt{\frac{dv_1^2+dv_2^2+dv_3^2+dv_4^2+dv_5^2}{n-1}}$$
$$\delta_{avg}=\frac{\delta}{\sqrt{n}}$$$$v_n=\frac{d_n}{t_n}$$$$\Delta v_n=v_n\sqrt{(\frac{\Delta L}{l})^2+(\frac{\Delta t_n}{t_n})^2}$$$$\rho_n=m_n(\frac{3}{4\pi})(\frac{d_n}{2})^{-3}$$$$\Delta\rho_n=6\sqrt{\frac{9\Delta d_n^2m_n^2}{\pi^2d_n^8}+\frac{\Delta m_n^2}{\pi^2d_n^6}}$$

$$\eta_n=(\frac{2g}{9v_n})(\frac{d_n}{2})^2(\rho_n-\rho_l)$$

 

[dekoi]

$$\Delta\eta_n=\frac{2}{9}\sqrt{\frac{4\Delta r_n^2g^2r_n^2(\rho_n-\rho_l)^2}{v_n^2}+\frac{\Delta\rho_n^2g^2r_n^4}{v_n^2}+\frac{\Delta g^2r_n^4(\rho_n-\rho_l)^2}{v_n^2}+\frac{\Delta\rho_l^2g^2r_n^4}{v_n^2}+\frac{\Delta v_n^2g^2r_n^4(\rho_n-\rho_l)^2}{v_n^4}}$$

 

[dekoi]

$$m_{avg}=\frac{19.837+19.839+19.840+19.841+19.840}{5}=19.84 g$$

$$v_G=\frac{0.50}{24.25}= 0.0206 m/s$$

$$\rho_G=0.01984(\frac{3}{4\pi})(\frac{0.02451}{2})^{-3}= 2573 kg/m^3$$

$$\eta_G=(\frac{2g}{9(0.0206)})(\frac{0.02451}{2})^2(2573-1013)= 24.77 kg/ms$$

 

[dekoi]

$$dv_1=24.51-24.70= -0.19 mm$$
$$dv_2=24.51-24.40= 0.11 mm$$
$$dv_3=24.51-24.44= 0.07 mm$$
$$dv_4=24.51-24.32= 0.19 mm$$
$$dv_5=24.51-24.68= -0.17mm$$
$$\delta_G=\sqrt{\frac{-0.19^2+0.11^2+0.07^2+0.19^2+-0.17^2}{5-1}}= 0.1718 mm$$
$$\delta_{avg of G}=\frac{0.1718}{\sqrt{5}}= 0.07683 mm$$

$$\Delta v_G=0.0206\sqrt{(\frac{0.003}{0.50})^2+(\frac{\0.1566}{24.25})^2}= 0.00018 m/s$$

$$\Delta\rho_G=6\sqrt{\frac{9(0.00007683)^2(0.01984)^2}{\pi^20.02451^8}+\frac{0.00005^2}{\pi^20.02451^6}}= 25.05 kg/m^3$$

 

[dekoi]

$$\heartsuit I LOVE YOU ALEX \heartsuit$$

 

[dekoi]

$$\Delta\eta_G=\frac{2}{9}\sqrt{\frac{4(0.000038415)^2(9.8)^2(0.012255)^2(2573-1013)^2}{0.0206^2}+\frac{(25.05)^2(9.8)^2(0.012255)^4}{0.0206^2}$$

$$\sqrt{adfsadfasf+\frac{(0.005)^2(0.012255)^4(2573-1013)^2}{0.0206^2}+\frac{(5)^2(9.8)^2(0.012255)^4}{0.0206^2} +\frac{(0.00018)^2(9.8)^2(0.012255)^4(2573-1013)^2}{0.0206^4}}= 0.4854 kg/ms}$$

 

[dekoi]

sorry everyone.. just testing for a lab report

 

[saltydog]

First, let's clean up the functional: Really, if we want to minimize that integral, we can just move U across the integral sign right, and let's put the exponential in the numerator:

$$\mathbf{T}[y(x)]=\frac{1}{U}\int_{p_1}^{p_2} e^{y/h}\sqrt{1+(y^{'})^2}dx$$


So:

$$F(x,y,y^{'})=e^{y/h}\sqrt{1+(y^{'})^2}$$

and therefore:

$$\frac{\partial F}{\partial y}=\frac{e^{y/h}\sqrt{1+(y^{'})^2}}{h}$$

and:

$$\frac{\partial F}{\partial y^{'}}=\frac{e^{y/h}y^{'}}{\sqrt{1+(y^{'})^2}}$$

and so:

$$\begin{align*} \frac{d}{dx}\left(\frac{\partial F}{\partial y^{'}}\right)&=\frac{d}{dx}\left[\frac{e^{y/h}y^{'}}{\sqrt{1+(y^{'})^2}}\right] \\ &=\frac{d}{dx}\left[e^{y/h}y^{'} \cdot \frac{1}{\sqrt{1+(y^{'})^2}}\right] \\ &=\left[e^{y/h}y^{'}\cdot\frac{-1/2}{(1+(y^{'})^2)^{3/2}}\cdot 2 y^{'}y^{''} \\ &+\frac{1}{\sqrt{1+(y^{'})^2}}\left(e^{y/h}y^{''}+y^{'}\frac{1}{h}y^{'}e^{y/h}\right) \\ &=\frac{e^{y/h}y^{''}}{\sqrt{1+(y^{'})^2}}-\frac{e^{y/h}(y^{'})^2y^{''}}{(1+(y^{'})^2)^{3/2}}+ \frac{e^{y/h}(y^{'})^2}{h\sqrt{1+(y^{'})^2}} \end{align}$$

So, once we obtain the partials, then we substitute them into the Euler equation and equate the expression to zero. Now, can you please substitute these expressions into:

$$\frac{\partial F}{\partial y}-\frac{d}{dx}\left(\frac{\partial F}{\partial y^{'}}\right)=0$$

and post the results?

Also, with regards to h=1: I just worked the problem with that value and obtained the results you indicated. Perhaps it works for any value of h. Not sure.

 

[dekoi]

$$\heartsuit I\; Love\; You\; Alex! \heartsuit$$

 

[dekoi]

$$\Delta\eta_G=\frac{2}{9}\sqrt{\frac{4(0.000038415) ^2(9.8)^2(0.012255)^2(2573-1013)^2}{0.0206^2}+\frac{(25.05)^2(9.8)^2(0.012255 )^4}{0.0206^2}+\frac{(0.005)^2(0.012255)^4(2573-1013)^2}{0.0206^2}+\frac{(5)^2(9.8)^2(0.012255)^4} {0.0206^2}+\frac{(0.00018)^2(9.8)^2(0.012255)^4(2573-1013)^2}{0.0206^4}}= 0.4854 kg/ms}$$

 

[dekoi]

$$\overrightarrow{F_f}=-6\pi\eta(\frac{d}{2})\overrightarrow{v}$$

$$\overrightarrow{mg}=\frac{4}{3}\pi(\frac{d}{2})^3\rho_s\overrightarrow g$$

$$\overrightarrow{F_b}=-\frac{4}{3}\pi(\frac{d}{2})^3\rho_l\overrightarrow g$$

$$\overrightarrow{F_b}+\overrightarrow{F_f}+\overrightarrow{mg} = 0$$

 

[dekoi]

$$avg=\frac{x_1+x_2+x_3+...+x_n}{n}$$

 

[samoth1]

test

$$\int_{0}^{1} x dx = \left[ \frac{1}{2}x^2 \right]_{0}^{1} = \frac{1}{2}$$
Let $$\theta$$ be a (p+1)-form defined on the ranges of all the cubes of a p-chain, where p>0. Then $$\int_{C} d \theta = \int_{\partial C} \theta$$
$$Let\; \theta\; be\; a (p+1)-form\; defined\; on\; the\; ranges\; of\; all\; the\; cubes\; of\; a\; p-chain,\; where\; p>0.\; Then \int_{C} d \theta = \int_{\partial C} \theta.$$

 

[dekoi]

$$e=\frac{f}{a}$$
$$e=\sqrt{1-\frac{b^2}{a^2}}$$

 

[dekoi]

$$\bigoplus$$



asdf

 

[dekoi]

$$F_b=m_ba_b$$
$$F_e=m_ea_e$$
$$F_b=F_e$$
$$m_ba_b=m_ea_e$$
$$\frac{m_b}{m_e}=\frac{a_e}{a_b}$$

 

[spacetime]

s

$$\nabla \cross E=0$$
$$\nabla \cdot E =\frac {\rho}{\epsilon_0}$$

abc

 

[lawtonfogle]

Taken $$\sum F = m_1 \cdot a$$
We have $$\sum F_x = m_1 \cdot a_x$$ and $$\sum F_y = m_1 \cdot a_y$$
With $$a_y = 0$$
We We have $$\sum F_x = m_1 \cdot a_x$$ and $$\sum F_y = 0$$
So $$T = m_1 \cdot a_x$$
Taken $$\sum F = m_2 \cdot a$$
We have $$\sum F_y = m_2 \cdot a_y$$
Which is $$m_2 \cdot g - T = m_2 \cdot a$$
Inserting $$T = m_1 \cdot a_x$$ into $$m_2 \cdot g - T = m_2 \cdot a$$ and solving for $$a$$,
We get $$a = \frac {m_2 \cdot g} {m_1 + m_2}$$
Inserting this into $$T = m_1 \cdot a$$,
we get $$T = \frac {m_1 \cdot m_2 \cdot g} {m_1 = m_2}$$Now, taken $$v^2 = v_0^2 + 2a\deltax$$

 

[lawtonfogle]

Taken $$\sum F = m_1 \cdot a$$
We have $$\sum F_x = m_1 \cdot a_x$$ and $$\sum F_y = m_1 \cdot a_y$$
With $$a_y = 0$$
We We have $$\sum F_x = m_1 \cdot a_x$$ and $$\sum F_y = 0$$
So $$T = m_1 \cdot a_x$$
Taken $$\sum F = m_2 \cdot a$$
We have $$\sum F_y = m_2 \cdot a_y$$
Which is $$m_2 \cdot g - T = m_2 \cdot a$$
Inserting $$T = m_1 \cdot a_x$$ into $$m_2 \cdot g - T = m_2 \cdot a$$ and solving for $$a$$,
We get $$a = \frac {m_2 \cdot g} {m_1 + m_2}$$
Inserting this into $$T = m_1 \cdot a$$,
we get $$T = \frac {m_1 \cdot m_2 \cdot g} {m_1 = m_2}$$


Now, taken $$v^2 = v_0^2 + 2a\delta x$$