The basic vector command in Latex is

\( \vec{v}    :=   \)      \vec{v} 

Saturday, 05 July 2014 00:00

Long Division with Polynomials


How do we divide polynomials?

Suppose, for example that you are given the problems:

Express \(\frac{x^{2} - 1 } {x - 1 } \) in terms of \(x\).

Find \(\frac{x^{3} + 2x^{2} + x}{x + 1}\) in terms of \(x.\)

What do we do?

When we view stars from the surface of the Earth, we a looking up through a ''sea'' of air, the atmosphere, about three hundred miles thick.

Stars are suns, like our own sun, and we see them because they emit particles of light called photons.

The closest star to our solar system, Proxima Centauri, is more than a parsec from our Sun

(1 parsec \(\approx 10 \times 10^{13}\) miles \( \approx 3 \times 10^{18}\) cm),

and most stars are much farther away (hundreds or thousands of parsecs--the Milky Way Galaxy has a radius of about eight thousand parsecs).

[To learn more about parsecs and stellar parallax, please click here.]

[To learn more about cosmic distance scales, please click here.]

Because stars are so far away their apparent angular diameters--that is, the number of degrees they cover in an arc across the celestial sphere as seen from Earth--are tiny. And so they are essentially point sources of light.

When a stream of photons from a star is scattered by the molecules in the atmosphere the photons that are scattered do not reach our eyes, and so the star appears to twinkle.

From space, above the atmosphere; or on the surface of the Moon, which has no atmosphere, stars do not appear to twinkle.

Planets are part of our solar system and so are much closer than the closest star. The Kuiper belt, of which Pluto is a member, is about \( 2 \times 10^{-4}\) parsecs away from the sun (that is, about 2 thousandths of a parsec).

Planets have angular diameters that are much larger than those of stars, and so even if some of the photons they reflect towards Earth are scattered by our atmosphere, there are plenty of other photons that still get through the atmosphere, and so planets do not appear to twinkle.


Consider a particle that moves in the plane in a circle of radius \(a\) described by the equation:

\[ g(x, y) = x^{2} + y^{2} - a^{2} = 0.\]

A possible parametrization of this equation is:

\[ x(t) = a \cos{(t)}, \]

\[ y(t) = a \sin{(t)}\]

where \(t\) is time.

Then the particle revolves counterclockwise about the origin on a circle of radius \(a, \) with a velocity vector:

\[ v = ( \frac{dx}{dt} , \frac{dy}{dt} ) = ( - a \sin{(t)}, a \cos{(t)} ) = ( - y, x).\]

Note that the particle's speed is constant.

\[ \text{speed } := |v| := \sqrt{ v_{x}^{2} + v_{y}^{2}} = \sqrt{ a^{2} \sin^{2}{(t)} + a^{2} \cos^{2}{(t)} } = \sqrt{ a^{2} } = a. \]

Now suppose that there is a temperature gradient in the plane described by the equation:

\[ T( x, y) = c_{0} + cy.\]

Where \(c_{0}\) and \(c\) are constants.

We want to know where on the circle the particle will experience the most extreme (greatest, least) temperatures.

We cannot use \(dT=0,\) because

\[ dT = \frac{ \partial T}{\partial x} dx + \frac{\partial T}{\partial y}dy = \frac{\partial T}{\partial y}dy = c dy \]

and \(\frac{\partial T}{\partial y} = c\) is a constant.


The technique that allows us to find the extremal values of \(T\) subject to the constraint \(g = 0,\) is the use of Lagrange multipliers.

Saturday, 17 May 2014 00:00

The Maxwell Distribution





Suppose we are given a certain quantity of an ideal gas at some fixed temperature, and we want to know what sort of distribution of velocities to associate with this gas.
That is, given a range of velocities, \[\Delta v = v_\beta - v_\alpha, \]
what is the number of molecules, \[\Delta n,  \]with velocities in the region of phase space \[\Delta v= \Delta v_x \Delta v_y \Delta v_z? \]

Consider two stars that appear close together in the night sky. Suppose that one star is relatively close to our solar system (what this means exactly we will come to shortly) while the second star is extremely distant.

As the Earth revolves around the sun, the apparent positions of the two stars will shift. The far distant star will not suffer any noticeable change in position, but the nearer star will be seen to move around the distant star in an ellipse.

Let \(\alpha\) be the maximum angular separation between the two stars expressed in radians.

If \(d_{o}\) is one A. U., i.e. the distance from the Earth to the Sun, and \(d_{\star},\) is the distance from the Sun to the star, then \( d_{\star} \alpha \approx d_{0}.\)

Because in practice the distances to stars are so great, and the angles so small, this approximation is an excellent one and we can write:

\[ d_{\star} = \frac{ d_{0}}{\alpha} = \frac{1 (A.U.)}{\alpha} \]

In this way, if we can measure the angle of parallax, \(\alpha\) we can find the distance to a star.


Here is a table with truth values for various binary relations on Boolean algebras. 


Changes in percent can be counter-intuitive.

For example, if the value of a stock falls by fifty percent on one day, and then rises by fifty percent on the next, it will not have recovered all of its value.

It will in fact be seventy-five percent of its original value.

It would need to increase by one hundred percent to recover all its value.

What is Percent?

A percent is a special ratio or fraction that always has denominator equal to one hundred.

For example,

\[ \frac{1}{100},\]

\[ \frac{10}{100},\]

\[ \frac{50}{100},\]

are all percents, called one percent, ten percent, and fifty percent, respectively.

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