Introducing LaTeX Math Typesetting

calculus1967 · Sep 24, 2005

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 · Sep 26, 2005

[ tex ] a^x_n [ /tex ]

Jeff Ford · Sep 27, 2005

Just a test

a^_x

bomba923 · Sep 29, 2005

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 · Oct 2, 2005

F_{1}+F{2}

just testing

ktoz · Oct 13, 2005

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 · Oct 13, 2005

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 · Oct 14, 2005

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]. 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 (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&#039;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&#039;s the reason we only need calculate the Real and Complex contribution from ONE member of each complex pair. And also, it&#039;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&#039;s compute the Real part and the Complex part for: &lt;br /&gt; \begin{align*}&lt;br /&gt; e^{(1+3i)t}\left(\begin{array}{c} 1 \\0 \\i\end{array}\right)&lt;br /&gt; &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] \\&lt;br /&gt; &amp;amp;=e^t\left[(Cos(3t)+iSin(3t))\left\{\left(\begin{array}{c} 1 \\0 \\0\end{array}\right)+&lt;br /&gt; i\left(\begin{array}{c} 0 \\0 \\1\end{array}\right)\right\}\right] \\&lt;br /&gt; &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] \\&lt;br /&gt; &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]&lt;br /&gt; \end{align}&lt;br /&gt; This with the first solution then yields the general solution: &lt;br /&gt; \begin{align*}&lt;br /&gt; \mathbf{X}&amp;amp;=C_1e^{-t}\left(\begin{array}{c} 0 \\1 \\0\end{array}\right) \\&lt;br /&gt; &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]&lt;br /&gt; \end{align}&lt;br /&gt; That&#039;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 · Oct 15, 2005

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 · Oct 15, 2005

\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 · Oct 15, 2005

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 · Oct 15, 2005

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 · Oct 15, 2005

\heartsuit I LOVE YOU ALEX \heartsuit

dekoi · Oct 15, 2005

\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 · Oct 15, 2005

sorry everyone.. just testing for a lab report

saltydog · Oct 17, 2005

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 · Oct 17, 2005

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

dekoi · Oct 17, 2005

\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 · Oct 17, 2005

\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 · Oct 17, 2005

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

samoth1 · Oct 18, 2005

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 · Oct 21, 2005

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

dekoi · Oct 21, 2005

\bigoplus

asdf

dekoi · Oct 21, 2005

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 · Oct 22, 2005

s

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

abc

lawtonfogle · Nov 1, 2005

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 · Nov 1, 2005

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 · Nov 1, 2005

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 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.

lawtonfogle · Nov 1, 2005

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 · Nov 3, 2005

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:

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

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

Introducing LaTeX Math Typesetting

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