LaTeX Introducing LaTeX Math Typesetting

[JamesU]

\sqrt{this is Very cool} \Upsilon \varrho \varsigma \overrightarrow{it} \overleftarrow{is}

 

[Castilla]

f(x,y) = \int_y^{x}t^{xy}dt

Nice.

 

[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

<br /> \mathop {\lim }\limits_{x \to \infty } f(x) = 0<br />

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

 

[tomfitzyuk]

<br /> T = \frac{1}{f} = 2 \pi \sqrt \frac{l}{g}<br />

 

[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| &lt; R}

 

[wr1015]

F_{1}+F{2}

just testing

 

[ktoz]

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

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

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

level: 4
<br /> \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}\\<br />

level: n
<br /> = \frac{(m + n)!(a(n + 1) + cm)}{m!(n + 1)!}<br />

As product
<br /> \frac{(m + n)!(a(n + 1) + cm)}{m!(n + 1)!} = (a(n + 1) + cm) \prod_{j=1}^n \frac{m + j}{j + 1}<br />

 

[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:

<br /> \mathbf{M}v=\lambda v<br />

So for:

<br /> \lambda_1=-1<br />

<br /> \left(<br /> \begin{array}{ccc} 1 &amp; 0 &amp; 3 \\<br /> 0 &amp; -1 &amp; 0 \\<br /> -3 &amp; 0 &amp; 1<br /> \end{array}<br /> \right)<br /> \left(<br /> \begin{array}{c} x \\<br /> y \\<br /> z<br /> \end{array}<br /> \right)=-1<br /> \left(<br /> \begin{array}{c} x \\<br /> y \\<br /> z<br /> \end{array}<br /> \right)<br />

So:

<br /> x+3z=-x\\<br /> -y=-y\\<br /> -3x+z=-z<br />

The middle one is easy:

<br /> 0y=0<br />

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

<br /> 2x+3z=0<br /> -3x+2z=0<br />

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

<br /> v_1=<br /> \left(<br /> \begin{array}{c} 0 \\<br /> 1 \\<br /> 0<br /> \end{array}<br /> \right)<br />

For:
<br /> \lambda_2=(1+3i)<br />

<br /> \left(<br /> \begin{array}{ccc} 1 &amp; 0 &amp; 3 \\<br /> 0 &amp; -1 &amp; 0 \\<br /> -3 &amp; 0 &amp; 1<br /> \end{array}<br /> \right)<br /> \left(<br /> \begin{array}{c} x \\<br /> y \\<br /> z<br /> \end{array}<br /> \right)=(1+3i)<br /> \left(<br /> \begin{array}{c} x \\<br /> y \\<br /> z<br /> \end{array}<br /> \right)<br />

So that's:

<br /> x+3z=(1+3i)x<br /> -y=(1+3i)y<br /> -3x+z=(1+3i)z<br />

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

or:

z=ix

so let x=1 and z=i.

Thus:

<br /> v_2=<br /> \left(<br /> \begin{array}{c} 0 \\<br /> 1 \\<br /> 0<br /> \end{array}<br /> \right)<br />

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

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

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

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 \lambda_3[/tex].<br /> <br /> So above I calculated the eigenvector for (1+3i) to be:<br /> <br /> v_2=\left(\begin{array}{c} 1 \\0 \\i\end{array}\right)<br /> <br /> Therefore to calculate the eigenvector for (1-3i), conjugate the eigenvector for (1+3i)[/tex]:&lt;br /&gt; &lt;br /&gt; If:&lt;br /&gt; &lt;br /&gt; v_2=\left(\begin{array}{c} 1+0i \\0+0i \\0+i\end{array}\right)&lt;br /&gt; &lt;br /&gt; Then:&lt;br /&gt; &lt;br /&gt; \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)&lt;br /&gt; &lt;br /&gt; See how that works?&lt;br /&gt; &lt;br /&gt; Ok that&amp;#039;s one.&lt;br /&gt; &lt;br /&gt; 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&amp;#039;s the reason we only need calculate the Real and Complex contribution from ONE member of each complex pair. And also, it&amp;#039;s VERY convenient to write the eigenvectors as:&lt;br /&gt; &lt;br /&gt; \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)&lt;br /&gt; &lt;br /&gt; Alright, so let&amp;#039;s compute the Real part and the Complex part for:&lt;br /&gt; &lt;br /&gt; &amp;lt;br /&amp;gt; \begin{align*}&amp;lt;br /&amp;gt; e^{(1+3i)t}\left(\begin{array}{c} 1 \\0 \\i\end{array}\right)&amp;lt;br /&amp;gt; &amp;amp;amp;=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] \\&amp;lt;br /&amp;gt; &amp;amp;amp;=e^t\left[(Cos(3t)+iSin(3t))\left\{\left(\begin{array}{c} 1 \\0 \\0\end{array}\right)+&amp;lt;br /&amp;gt; i\left(\begin{array}{c} 0 \\0 \\1\end{array}\right)\right\}\right] \\&amp;lt;br /&amp;gt; &amp;amp;amp;=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] \\&amp;lt;br /&amp;gt; &amp;amp;amp;=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]&amp;lt;br /&amp;gt; \end{align}&amp;lt;br /&amp;gt;&lt;br /&gt; &lt;br /&gt; This with the first solution then yields the general solution:&lt;br /&gt; &lt;br /&gt; &amp;lt;br /&amp;gt; \begin{align*}&amp;lt;br /&amp;gt; \mathbf{X}&amp;amp;amp;=C_1e^{-t}\left(\begin{array}{c} 0 \\1 \\0\end{array}\right) \\&amp;lt;br /&amp;gt; &amp;amp;amp;+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]&amp;lt;br /&amp;gt; \end{align}&amp;lt;br /&amp;gt;&lt;br /&gt; &lt;br /&gt; That&amp;#039;s read as:&lt;br /&gt; &lt;br /&gt; x(t)=C_2e^tCos(3t)+C_3e^tSin(3t)&lt;br /&gt; &lt;br /&gt; y(t)=C_1e^{-t}&lt;br /&gt; &lt;br /&gt; 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}<br /> +\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^{&#039;})^2}dx


So:

F(x,y,y^{&#039;})=e^{y/h}\sqrt{1+(y^{&#039;})^2}

and therefore:

\frac{\partial F}{\partial y}=\frac{e^{y/h}\sqrt{1+(y^{&#039;})^2}}{h}

and:

\frac{\partial F}{\partial y^{&#039;}}=\frac{e^{y/h}y^{&#039;}}{\sqrt{1+(y^{&#039;})^2}}

and so:

<br /> \begin{align*}<br /> \frac{d}{dx}\left(\frac{\partial F}{\partial y^{&#039;}}\right)&amp;=\frac{d}{dx}\left[\frac{e^{y/h}y^{&#039;}}{\sqrt{1+(y^{&#039;})^2}}\right] \\<br /> <br /> &amp;=\frac{d}{dx}\left[e^{y/h}y^{&#039;} \cdot \frac{1}{\sqrt{1+(y^{&#039;})^2}}\right] \\<br /> <br /> &amp;=\left[e^{y/h}y^{&#039;}\cdot\frac{-1/2}{(1+(y^{&#039;})^2)^{3/2}}\cdot 2 y^{&#039;}y^{&#039;&#039;} \\<br /> <br /> &amp;+\frac{1}{\sqrt{1+(y^{&#039;})^2}}\left(e^{y/h}y^{&#039;&#039;}+y^{&#039;}\frac{1}{h}y^{&#039;}e^{y/h}\right) \\<br /> <br /> &amp;=\frac{e^{y/h}y^{&#039;&#039;}}{\sqrt{1+(y^{&#039;})^2}}-\frac{e^{y/h}(y^{&#039;})^2y^{&#039;&#039;}}{(1+(y^{&#039;})^2)^{3/2}}+<br /> \frac{e^{y/h}(y^{&#039;})^2}{h\sqrt{1+(y^{&#039;})^2}}<br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> \end{align}<br />

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^{&#039;}}\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

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

 

[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

 

[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
Solving for a we get a = \frac {v^2 - V_0^2} {2\Delta x}
Taking that \Delta x = .5 and v_0 = 0
We have a = \frac {v^2} {meters*}
Now a_exp stands for experimental value of a
and a_t [/tex[ stands for theoretical value of a<br /> so a_exp = \frac {v^2} {meters*}<br /> and a_t = \frac {m_2 \cdot g} {m_1 + m_2}<br /> * meters is added so units sovle correctly.

 

[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
Solving for a we get a = \frac {v^2 - V_0^2} {2\Delta x}
Taking that \Delta x = .5 and v_0 = 0
We have a = \frac {v^2} {meters*}
Now a_exp stands for experimental value of a
and a_t stands for theoretical value of a
so a_exp = \frac {v^2} {meters*}
and a_t = \frac {m_2 \cdot g} {m_1 + m_2}
* meters is added so units sovle correctly.

 

[saltydog]

I've achieved oneness with the integral Tide . . . Here it is with your scalling factor in case other people are following this:

Above, after letting:

\sigma^2=s

and completing the square, we obtain:

2e^{-\pi/t}\int e^{t(\sigma-\sqrt{\pi}/t)^2}\sigma d\sigma

Now, in order to remove the t in the exponent, we let:

v=\sqrt{t}\sigma

so that:

dv=\sqrt{t}d\sigma,\quad \sigma=\frac{v}{\sqrt{t}},\quad d\sigma=\frac{dv}{\sqrt{t}}

Substituting this scalling factor into the exponent:

<br /> \begin{align*}<br /> t\left[\frac{v^2}{t}-\frac{2v\sqrt{\pi}}{t\sqrt{t}}+\frac{\pi}{t^2}\right]&amp;=<br /> v^2-2v\sqrt{\pi/t}+\frac{\pi}{t} \\<br /> &amp;=(v-\sqrt{\pi/t})^2<br /> \end{align}<br />

substituting this into the integral:

2e^{-\pi/t}\int e^{(v-\sqrt{\pi/t})^2}\left(\frac{v}{\sqrt{t}}\right)\frac{dv}{\sqrt{t}}

Simplifying:

2\frac{e^{-\pi/t}}{t}\int e^{(v-\sqrt{\pi/t})^2}dv;\quad v=\sqrt{t}\sigma;\quad \sigma^2=s

 

[saltydog]

I wish to finally evaluate:

<br /> y(t)=\frac{e^{-\pi/t}}{2 i t^{3/2}}\mathop\lim\limits_{R\to\infty}\left\{\text{Erfi}\left(v-\sqrt{\pi/t}\right)_{\sqrt{Rt}e^{-\pi i/4}}^{\sqrt{Rt}e^{\pi i/4}}\right\}\tag{1}<br />

First note:

<br /> \text{Erfi}(z)=\frac{\text{Erf(iz)}}{i}=\frac{1}{i}\frac{2}{\sqrt{\pi}}\int_0^{iz} e^{-w^2}dw<br />

Using this relation, I can re-write the above limit as:

<br /> \frac{2}{i\sqrt{\pi}}\left\{\mathop\lim\limits_{R\to\infty}\int_0^{i(\sqrt{Rt}e^{\pi i/4})} e^{-w^2}dw-\mathop\lim\limits_{R\to\infty}\int_0^{i(\sqrt{Rt}e^{-\pi i/4})} e^{-w^2}dw\right\}<br />

Using Euler's relation, I convert the upper limits to real and imaginary parts and multiply by i:

<br /> \begin{align*}<br /> i\sqrt{Rt}e^{\pi i/4}&amp;=i\left[\left(\sqrt{Rt/2}-\sqrt{\pi/2}\right)+i\sqrt{Rt/2}\right] \\<br /> &amp;=-\sqrt{Rt/2}+i\left(\sqrt{Rt/2}-\sqrt{\pi/t}\right)<br /> \end{align}<br />

Now, as R goes to infinity, this quantity goes to -\infty+i\infty.

And the lower limit:

<br /> \begin{align*}<br /> i\sqrt{Rt}e^{-\pi i/4}&amp;=i\left[\sqrt{Rt/2}+\left(\sqrt{Rt/2}-\sqrt{\pi/2}\right)\right] \\<br /> &amp;=\sqrt{Rt/2}+i\left(\sqrt{Rt/2}-\sqrt{\pi/2}\right)<br /> \end{align}<br />

as R goes to infinity, this limit goes to \infty+i\infty.

Thus the integrals can be rewritten as:

<br /> \frac{1}{i}\frac{2}{\sqrt{\pi}}\int_0^{-\infty+i\infty} e^{-w^2}dw-<br /> \frac{1}{i}\frac{2}{\sqrt{\pi}}\int_0^{\infty+i\infty} e^{-w^2}dw<br />

I choose to evaluate these line integrals along a parametric path using the following relation from Complex Analysis:

<br /> \int_C f(z)dz=\int_c udx-vdy+i\int_c vdx+udy<br />

Thus the first step is to express the integrand in terms of a real and imaginary part. Relying on Euler's formula:

<br /> \begin{align*}<br /> e^{-w^2}&amp;=e^{-(x+yi)^2} \\<br /> &amp;=e^{-(x^2-y^2+2xyi)} \\<br /> &amp;=e^{-(x^2-y^2)}Cos(2xy)+ie^{-(x^2-y^2)}(-Sin(2xy))\\<br /> &amp;=u+iv<br /> \end{align}<br />

For the first integral, the path goes from the origin to the point -\infty+i\infty

Thus if I let:

x=-t,\quad dx=-dt

y=t,\quad dy=dt

Making this substitution, note that the exponent term is 1 and we have for the first integral (with t going from 0 to infinity):

<br /> \begin{align*}<br /> \int_0^{-\infty+i\infty} e^{-w^2}dw&amp;=\left(-\int_0^\infty Cos(-2t^2)dt+\int_0^\infty Sin(-2t^2)dt\right) \\<br /> &amp;+i\left(\int_0^\infty Cos(-2t^2)dt+\int_0^\infty Sin(-2t^2)dt\right) \\<br /> &amp;=\left(-\frac{\sqrt{\pi}}{4}-\frac{\sqrt{\pi}}{4}\right)+i\left(\frac{\sqrt{\pi}}{4}-\frac{\sqrt{\pi}}{4}\right) \\<br /> &amp;=-\frac{\sqrt{\pi}}{2}<br /> \end{align}<br />


The same analysis can be made (with the path going from 0 to \infty+i\infty for the second integral with the results that the line integral in that case is:

<br /> \frac{\sqrt{\pi}}{2}

Substituting these values into (1), I obtain:

<br /> \frac{1}{i}\frac{2}{\sqrt{\pi}}\int_0^{-\infty+i\infty} e^{-w^2}dw=\frac{1}{i}\frac{2}{\sqrt{\pi}}(-\frac{\sqrt{\pi}}{2})=-\frac{1}{i}\frac{i}{i}=i<br />

and:

<br /> \frac{1}{i}\frac{2}{\sqrt{\pi}}\int_0^{\infty+i\infty} e^{-w^2}dw=\frac{1}{i}\frac{2}{\sqrt{\pi}}(\frac{\sqrt{\pi}}{2})=\frac{1}{i}\frac{i}{i}=-i<br />

So that the limit is 2i.


Substituting this into the last expression of the inverse transform, we obtain:

<br /> y(x)=\frac{e^{-\pi/t}}{2i t^{3/2}}(2i)=\frac{e^{-\pi/t}}{t^{3/2}}

 

[opticaltempest]

Testing TeX output from Maple 10

\lim _{n\rightarrow \infty } \left(\frac {27}{n^2}\ \sum _{i=1}^{n} i \right )<br /> <br />

 

[Arust]

Thank you, it's very useful

 

[TinTinn Tabulation]

I'm just TeX'ing away on the thin ice of a new Day !
<br /> \int _{0}^{1}\!{\frac {1}{\sqrt {{x}^{2}+{y}^{2}+{z}^{2}}}}{dx}=-1/2\,\ln \left( {y}^{2}+{z}^{2} \right) +\ln \left( 1+\sqrt {1+{y}^{2}+{z<br /> }^{2}} \right)<br />
Oo-er, me ickle preview, it preview-th not !

 

[Ratzinger]

a^2 I tried to preview my post but the preview only says LaTex graphic is being generated. Reload this page in a moment. But when I reload nothing happens. What can be done?

\phi (x) = \frac{1}{{(2\pi )^{3/2} }}\int {d^4 p\delta (p^2 } - m^2 )A(p)e^{ - ip \cdot x}

 

[ranger]

Is it possible to show when you cancel out terms i.e show a line through the terms being canceled?

 

[krab]

R = \frac{V}{I} = \frac{V}{small no.}

I notice a lot of this sort of error. LaTeX math typesetting cannot "know" that "small no." is not the symbols s times m times a times ... If Text appears within math symbols, it should be formatted as such. Then it is in regular type and interword spaces are observed. This is done with mbox, as for the above example:
R = \frac{V}{I} = \frac{V}{\mbox{small no.}}
And of course similarly for units: it's not
64 ft^2,
rahter, it is
64\mbox{ ft}^2

 

[twoflower]

Why can't I insert newline in tex code? I tried \\ but no success...

 

[robphy]

Is it possible to show when you cancel out terms i.e show a line through the terms being canceled?

Here's a crude way
a\hspace{-1ex}\slash +b=a\hspace{-1ex}\slash +c

I have a fancier way with a macro that requires a little trial and error.

 

[xouper]

For a function f : \mathbb{N} \mapsto \{0,1\}f(n)=\left\{\begin{array}{cc}0,&amp;\mbox{ if }f(n)=1\\1,&amp;\mbox{ if } f(n)=0\end{array}\right.