Introducing LaTeX Math Typesetting

[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:

$$\lambda_1=-1$$

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

So:

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

The middle one is easy:

$$0y=0$$

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

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

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

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

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

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

So that's:

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

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:

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

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

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

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

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].<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 [itex](1+3i)[/tex]:<br /> <br /> If:<br /> <br /> $$v_2=\left(\begin{array}{c} 1+0i \\0+0i \\0+i\end{array}\right)$$<br /> <br /> Then:<br /> <br /> $$\overline{v_2}=v_3=\left(\begin{array}{c} 1-0i \\0-i \\0-i\end{array}\right)=\left(\begin{array}{c} 1 \\0 \\-i\end{array}\right)$$<br /> <br /> See how that works?<br /> <br /> Ok that's one.<br /> <br /> 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:<br /> <br /> $$\left(\begin{array}{c} 1 \\0 \\-i\end{array}\right)=\left(\begin{array}{c} 1 \\0 \\0\end{array}\right)+i\left(\begin{array}{c} 0 \\0 \\1\end{array}\right)$$<br /> <br /> Alright, so let's compute the Real part and the Complex part for:<br /> <br /> $$\begin{align*}<br /> e^{(1+3i)t}\left(\begin{array}{c} 1 \\0 \\i\end{array}\right)<br /> &=e^{(1+3i)t}\left[\left(\begin{array}{c} 1 \\0 \\0\end{array}\right)+i\left(\begin{array}{c} 0 \\0 \\1\end{array}\right)\right] \\<br /> &=e^t\left[(Cos(3t)+iSin(3t))\left\{\left(\begin{array}{c} 1 \\0 \\0\end{array}\right)+<br /> i\left(\begin{array}{c} 0 \\0 \\1\end{array}\right)\right\}\right] \\<br /> &=e^t\left[Cos(3t)\left(\begin{array}{c} 1 \\0 \\0\end{array}\right)+iCos(3t)\left(\begin{array}{c} 0 \\0 \\1\end{array}\right)+iSin(3t)\left(\begin{array}{c} 1 \\0 \\0\end{array}\right)-Sin(3t)\left(\begin{array}{c} 0 \\0 \\1\end{array}\right)\right] \\<br /> &=e^t\left[C_1\left\{Cos(3t)\left(\begin{array}{c} 1 \\0 \\0\end{array}\right)-Sin(3t)\left(\begin{array}{c} 0 \\0 \\1\end{array}\right)\right\}+C_2\left\{Cos(3t)\left(\begin{array}{c} 0 \\0 \\1\end{array}\right)+Sin(3t)\left(\begin{array}{c} 1 \\0 \\0\end{array}\right)\right\}\right]<br /> \end{align}$$<br /> <br /> This with the first solution then yields the general solution:<br /> <br /> $$\begin{align*}<br /> \mathbf{X}&=C_1e^{-t}\left(\begin{array}{c} 0 \\1 \\0\end{array}\right) \\<br /> &+e^t\left[C_2\left\{Cos(3t)\left(\begin{array}{c} 1 \\0 \\0\end{array}\right)-Sin(3t)\left(\begin{array}{c} 0 \\0 \\1\end{array}\right)\right\}+C_3\left\{Cos(3t)\left(\begin{array}{c} 0 \\0 \\1\end{array}\right)+Sin(3t)\left(\begin{array}{c} 1 \\0 \\0\end{array}\right)\right\}\right]<br /> \end{align}$$<br /> <br /> That's read as:<br /> <br /> $$x(t)=C_2e^tCos(3t)+C_3e^tSin(3t)$$<br /> <br /> $$y(t)=C_1e^{-t}$$<br /> <br /> $$z(t)=-C_2e^tSin(3t)+C_3e^tCos(3t)$$[/itex][/itex]

 

[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:

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


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:

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

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 $$v^2 = v_0^2 + 2a\delta x$$

 

[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 [tex]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.[/tex]

 

[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:

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

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:

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