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Wednesday March 1, 2006

stock market trademarks

Index	Value	Change Net / %	Market Status
NASDAQ Composite Index	2281.39	25.79 -1.12%	Closed
NASDAQ-100 Index	1670.57	25.09 -1.48%	Closed
American Stock Exchange Composite Index	1847.22	7.69 0.42%	Closed
Dow Jones Industrial Average Index	10993.41	104.14 -0.94%	Closed
Standard and Poor's 500 Index	1280.66	13.46 -1.04%	Closed
New York Stock Exchange Composite Index	8060.61	79.06 -0.97%	Closed
30 year Treasury Bond Index	45.03	0.40 -0.88%	Closed
13 Week T-Bill Index	45.07	0.15 0.33%	Closed
CBOE Gold Index	128.25	1.91 -1.47%	Closed

NASDAQ-100 1670.57 25.09 -1.48%

TBill 45.07 0.15 0.33%

Posted by jesse edwards at 12:48 AM - No Comments Add a Comment

Monday February 27, 2006

RETIREMENT MATH

All logarithm functions have some useful properties: AND ALL BLOGS DO ALSO . IN THIS CASEBLOGSTREAM WILL ACCEPT THIS RETIREMENT FORMULA IN ADVANCED HTML . IN THE REGULAR FORMAT ONLY THE WORDS APPEAR . IN BLOLINES THE MATH SHOWS UP WITH NO PROBLEM .

All logarithm functions have some useful properties:

Properties of logarithms
(M and N are positive.)

$log(MN) = log(M) + log(N)$		$ln(MN) = ln(M) + ln(N)$	logarithm of a product
$log(M/N) = log(M) - log(N)$		$ln(M/N) = ln(M) - ln(N)$	logarithm of a quotient
$log(M^p) = p log(M)$		$ln(M^p) = p ln(M)$	logarithm of a power
$log(1) = 0$		$ln(1) = 0$	logarithm of +1
$log(10) = 1$		$ln(e) = 1$	logarithm of the base

Be careful not to make the mistake of trying to use these properties when they do not apply. None of the following forms can be rewritten using the properties above.

$log(M+N)$	$ln(M+N)$	no rules for the logarithm of a sum
$log(M)/log(N)$	$ln(M)/ln(N)$	no rules for the quotient of logarithms

Relationship between logarithms and exponentials

These follow from the definition of the logarithms as the inverse functions for the corresponding exponential functions.

Logarithms and exponentials

$log(N) = x <=> 10^x = N$		$ln(N) = x <=> e^x = N$
$log(10^x) = x$		$ln(e^x) = x$
$10^{log(x)} = x$		$e^{ln(x)} = x$

Examples
For each of these, note what properties of logarithms are being used.

$log(10,000) = log(10^4) = 4$ (the power of 10 needed to get 10,000 is 4).
$ln(1)=0$ (the power of e needed to get 1 is 0).
$log(x) = -3$ means $10^(-3) = x$
$ln(x) = sqrt(2)$ means $e^(sqrt(2)) = x$
$10^3 = 1000$ means $log(1000) = 3$
$10^(-2) = 0.01$ means $log(0.01) = -2$
$log(10x) = log(10) + log(x) = 1 + log(x)$
$ln(e^2/sqrt(x)) = ln(e^2) - ln(sqrt(x)) = 2ln(e) - ln(x^(1/2)) = 2*1 - (1/2) ln(x) = 2 - (1/2) ln(x)$
$3( 2log(M) + (4/3)log(N) ) = 6 log(M) + 4 log(N) = log(M^6) + log(N^4) = log(M^6 N^4)$
$e^(3ln(x)) + 2ln(e^(-x)) = e^(ln(x^3)) + 2(-x) = x^3 - 2x$
$log(10^(2x)) + log(sqrt(10)) = 2x log(10) + log(10^(1/2)) = 2x + (1/2)log(10) = 2x + 1/2$

Practice THE ART OF RETIREMENT . INDIA IS AGOOD PLACE TO START .

Posted by jesse edwards at 12:42 AM - No Comments Add a Comment

Tuesday February 21, 2006

from greatest journal

jesarchives

$=== mass slider #1 | |--- __ | \ | \ field | \ . (*) | \ . source | | . | . | |__________________=== mass slider #2$

Posted on: Mon, Feb 20 2006 7:57 PM

from blogtastic MATH SERIES

By jesarchives

XY PLANE AND ORBIT SPACE

February 20, 2006 @ 8:15 am · Filed under Uncategorized ·

The orbit space

We now have a variety,[ OF TRAVEL OPTIONS . FIRST OF ALL CHECK ON THE POSTAL RATES , THEM AFTER THAT CHECK WITH THE MONEY CHANGERS .] the xy-plane, which is divided into orbits such as {(a,b),(b,a)}. We shall, by example, show that there exists another plane, the st-plane, which has points that are in a one-to-one correspondence with the orbits in the xy-plane. This means that every point of the st-plane corresponds to an orbit in the xy-plane. This st-plane is called an orbit space. It can be shown that the relationship between the xy-coordinates and st-coordinates is s = x + y and t = xy.

FIND THE TRAIN STATION IN THIS IMAGE . . . . . . . . . . . . . X MARKS THE SPOT . . . . .

Figure 1: Each point in the st-plane corresponds to an orbit in the xy-plane.

We first pick a point in the st-plane and then calculate the corresponding points in the xy-plane. We than prove that the two points are in the same orbit.

Remember that s = x + y and t = xy. Substituting y = t/x into s = x + y gives

s = x + t/x

which is the same as

x² - xs + t = 0.

17.   Simplify:

6

192
+

5

12



2967

7125





204



58

6



1682

6



841

12



58

3
convert the currency into the coin of your realm .

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[Monday, February 20, 2006 at 7:54 pm]

Subject: math series

Monday, February 20, 2006

Sighting Gravity
Following is a description of the "gravity sextant", which is a
simple (though not very practical) device for determining the
exponent of a spherically symmetrical force law of the form
F = 1/r^c.

The sextant (or perhaps "quadrant" would be more correct) consists
of a right angled frame subtended by a protractor, similar to a
normal sextant. At the end of each arm of the right angled frame
is a mass held by a spring in a slider. Both of the sliders are
parallel to one of the arms, as illustrated below:

Oriented so that the plane of the sextant contains the center of the
field source, rotate the sextant until the displacements of the masses
in the two sliders are equal, and then sight the field source on a
line through the corner of the frame, and mark the angle on the
protractor.

The angle "theta" between the line of sight and the slider direction
is related to the exponent c in the force law according to the equation

tan^2(theta) + (c+1)tan(theta) - c = 0

For example, if the angle reads 29.3165 degrees, then we have c=2, so
we know the field obeys an inverse square force law. On the other
hand, if the angle reads 22.5 degrees, then c=1, and the field obeys
a simple inverse law. If the angle reads 32.8524 degrees, then the
field obeys an inverse cube law. now this is all beyond me . but what do i know

i just copied it from somewhere and pasted it somewhere else . i like the different

look in the different formats . process , progress , project . the doing is as fun as the thinking .

leave a comment

will it light up and link up ?

Posted by jesse edwards at 12:45 AM - No Comments Add a Comment

Monday February 20, 2006

from the math pages

$=== mass slider #1 | |--- __ | \ | \ field | \ . (*) | \ . source | | . | . | |__________________=== mass slider #2$
it worked in bloglines . and the text box showed up . so lets see what happens here .

Posted by jesse edwards at 10:59 PM - No Comments Add a Comment

mailer demon math test

From: VISUAL ARTS CLUB BAND <jesarchives@HOTMAIL.COM> To: jesse@thedancingbazaar.com Subject: [TRAVEL BAZAAR] 2/20/2006 06:52:04 PM Sent: Tuesday, February 21, 2006 2:57 AM ï»¿ Trigonometry ï¿½ If sin = and 0 = = , then tan = a) b) c) d) e) Logarithmic and Exponential Functions ï¿½ Log = a) 81 b) 9 c) 3 d) e) Word Problems ï¿½ If is of of a certain number, than that number is: a) b) c) d) e) Level 4 Answers 1-d, 2-a, 3-b, 4-c, 5-b, 6-e, 7-c, 8-b, Trigonometry ï¿½ If sin = and 0 = = , then tan = a) b) c) d) e) Logarithmic and Exponential Functions ï¿½ Log = a) 81 b) 9 c) 3 d) e) Word Problems ï¿½ If is of of a certain number, than that number is: a) b) c) d) e) Level 4 Answers 1-d, 2-a, 3-b, 4-c, 5-b, 6-e, 7-c, 8-b, Â -- Posted by VISUAL ARTS CLUB BAND to TRAVEL BAZAAR at 2/20/2006 06:52:04 PM

Considering that it's impossible to "square the circle" by Euclidean
methods, it's interesting that some regions whose boundaries are
circular arcs CAN be "squared" by Euclidean methods - meaning that we
can construct a square with the same area using just straightedge and
compass - whereas most such regions cannot. Hippocrates of Chios was
the first to demonstrate such "quadratures" (around 440 BC) for lunes.
It turns out that only five particular lunes can be "squared". Three
of these were described by Hippocrates himself, and two more were
discovered in the mid 1700's. These last two are often credited to
Euler in 1771, but according to Heath all five squarable lunes were
given in a dissertation by Martin Johan Wallenius in 1766. It's now
known (Tschebatorew and Dorodnow) that these five cases are the ONLY
lunes that are squarable by Euclidean methods. from math pages dot com

so let the guessing begin .

Posted by jesse edwards at 10:04 PM - No Comments Add a Comment

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Author: jesse edwards

From venice beach , USA

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