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Asymptotes of Rational Functions
Sub Ratio of two polynomial functions is defined as a rational function where a polynomial is a finite length expression made up of basic mathematical operations (addition, subtraction,
Topics multiplication and non-negative exponents) which contains variables and constants. A function which evaluates a polynomial is known as polynomial function. One argument function named as g, is
said to be a polynomial if it satisfies,
g(a) = $y_{n}a^{n}+ y_{n-1}a^{n}+ .......+ y_{2}a^{2}+ y_{1}a + y_{0}$
Example: g(a) = a$^{3}$ – a, the function 'a' is said to be rational function only if it can be written as, g(a) = $\frac{R(a)}{S(a)}$ where R and S are polynomial functions in a and S is not
a zero polynomial. The set of all points ‘a’ for which the denominator S(a) is not zero comes under the domain of g. One can assume that this rational function is written in its lowest degree
i.e. R and S have many positive degree. Polynomial functions with S(a) = 1, is said to be rational functions. Functions that can be written in this form are not rational functions.
Asymptote is a straight line that the curve approaches but does not cross. Rational functions exhibit three types of asymptotes.
Vertical asymptotes:
In vertical asymptote we will have vertical lines near which the function f(x) becomes infinite.
If the denominator of a rational function has more factors of (x - a) than the numerator, then the rational function will have a vertical asymptote at x = a. Only the denominator matters in vertical
Horizontal asymptote:
Horizontal line is an asymptote only to the far left and the far right of the graph. There won't be asymptotes in the middle.
f(x) =
$\frac{ax^{n}+....}{bx^{m} + .......}$
Numerator will have nth degree polynomial and denominator will have mth degree polynomial.
│When n < m│x axis is the horizontal asymptote. │
│When n = m│Horizontal asymptote is the line Y = $\frac{a}{b}$ │
│When n > m│No horizontal asymptote │
Oblique asymptote:
A rational function will have an oblique or slant asymptote when the degree of the numerator is greater than that of the numerator. The quotient obtained by division will yield the oblique asymptote.
A function that contains two polynomials in fraction form or written in ratios is known as rational functions. Let's see mathematical representation of rational function. In case of polynomial with
one variable 'a' is said to be rational function if and only if it is written in the form:
=> f (a) =
, here both 'R' and 'S' are polynomial functions in 'a' and value of 'S(a)' is not zero. Here Domain of function (f) can be defined as set of all points of 'a' for which value of denominator is not
zero. Each polynomial function can follow property of a rational function with S(a) = 1. If a function is written in form of f(a) = sin (a) then this is not a rational function.
Example of rational function is
$\frac{x + 1}{ x^{2} – 1}$
. Here one is divided by the other like a ratio.
To find the derivative of a rational function, we can use the reciprocal rule of the derivative which is a combination of quotient rule and product rule.
We do differentiation in order to reduce the equation with respect to a certain variable. Differentiating functions with real numbers except for those involve finding derivatives of rational
functions or expressions including fractions are comparatively easy.
For example,
suppose the function is given as:
f(x) =
Where, H(x) and G(x) are two mathematical expressions and are present in form of fraction. Now we will discuss how to find derivatives of rational functions? To resolve this we have to follow a
formula given as:
$\frac{d (f(x))}{dx}$
= [(G(x) (d (
) – H(x) (d (
) / G
To understand it better let’s consider an example of a function: f(x) =
$\frac{(x + 1)}{ (x + 2)}$
Here, let us say: H(x) = (x + 1) and G(x) = (x + 2).
Applying the above formula directly to differentiate the function f(x) we can write the derivative as:
$\frac{d (f(x))}{dx}$
= [((x + 2) (
$\frac{d (x + 1)}{dx}$
) – (x + 1) (
$\frac{d (x + 2)}{dx}$
)) / (x + 2)
$\frac{d (f(x))}{dx}$
= [
$\frac{(x + 2) (1) – (x + 1) (1)}{(x^2 + 4 + 4x)}$
$\frac{d (f(x))}{dx}$
= [
$\frac{(x + 2 - x - 1)}{(x^2 + 4 + 4x)}$
$\frac{d (f(x))}{dx}$
$\frac{1}{x^2 + 4 + 4x}$
In a proper rational function the degree of numerator will be less than the degree of denominator.
Examples:$\frac{x}{x^{2} - 2}$
$\frac{1}{x + 8}$
$\frac{x^{5}}{3x^{4} -2}$
In all the examples above, the degree of numerator is less than the degree of denominator.
In improper rational function the degree of numerator is greater than or equal to the denominator. Improper rational function can be simplified by polynomial long division.
Examples: $\frac{x^{5} - 4}{x^{2} + 5}$
$\frac{5x^{3} - 7}{8x^{3} - 4}$
$\frac{x^{4} - 5x} {23x + 27}$
To find the inverse of a rational function, we have to follow some steps:
If we have f(x) = G(x)/ H(x), replace f(x) by y.
Solve the equation for x in terms of y.
When we can find the value or expression for x in terms of y replace x by f$^{-1}$(y), which is the inverse of the given rational function.
Find the inverse of the rational function f(x) =
$\frac{x-7}{7x - 5}$Solution:
f(x) = y =
$\frac{x - 7}{7x - 5}$
y (7x - 5) = x - 7
7xy - 5y = x - 7
7xy - x = 5y - 7
x(7y - 1) = 5y - 7
x =
$\frac{5y - 7}{7y -1}$
x = f$^{-1}$(y) =
$\frac{5y -7}{7y -1}$
Both rational functions and rational expressions are undefined for numbers that make the denominator zero. These numbers should be excluded from being possible values for the variable. Hence, for all
rational expressions or functions, the domain consists of all real numbers except those numbers that make the denominator equal to zero.
While graphing a rational function we need to remember the following steps:
Step 1: Look for the common factor of the numerator. This will give the holes of the curve.
Step 2: Find the vertical asymptotes by finding the zeroes of the denominator.
Step 3: Look for the degree of the numerator and the denominator. If degree of numerator is greater than or equal to that of the denominator, there will be horizontal asymptotes.
Step 4: Find the x - intercepts by equating the rational expression to 0.
Step 5: Find the y - intercepts by plugging in x = 0.
Step 6: Prepare the table with the values to the right and left of the restricted values obtained from the step 1.
Step 7: Graph the function by drawing the asymptotes, plot the x and y intercepts and then, plot the points from the table.Given below is an example:
Graph the following function y =
$\frac{3x + 5}{x - 2}$Solution:
Find the vertical asymptotes by finding the zeroes of the denominator.
$\Rightarrow$ x - 2 = 0
$\Rightarrow$ x = 2
From the given equation we see that the numerator and the denominator have the same degree so the asymptotes will be horizontal. Horizontal asymptote is found by dividing the leading coefficients.
y =
y = 3
Find any x or y intercepts
Put x = 0
y =
$\frac{3(0) + 5}{(0) - 2}$
= -2.5
When y = 0,
We have 3x + 5 =0
$\Rightarrow$ x =
= - 1.67
Pick few more x values to find the corresponding y values.
│x │y = $\frac{3x + 5}{x - 2}$ │
│-2│0.25 │
│-1│-0.67 │
│1 │-8 │
│5 │6.67 │
Solved Examples
Question 1:
Find the derivative of
$\frac{x^{2} + 2x - 5}{x^{3} - 3x + 2}$ Solution:
$\frac{(x^{3} - 3x + 2)(2x+2)-(x^{2}+2x - 5)(3x^{2}-3)}{(x^{3}-3x + 2)^2}$
= $\frac{2x^{4}-6x^{2}+4x + 2x^{3}-6x + 4-[3x^{4}+6x^{3}-15x^{2}-3x^{2}-6x + 15]}{(x^{3}-3x + 2)^2}$
= $\frac{2x^{4}+2x^{3}-6x^{2}-2x + 4-[3x^{4}+6x^{3}-18x^{2}-6x + 15]}{(x^{3}-3x + 2)^2}$
= $\frac{-x^{4}-4x^{3}+12x^{2}+4x - 11}{(x^{3}-3x + 2)^2}$
Question 2:
Find domain and vertical asymptote of f(x) =
$\frac{1}{x^{2}+12x +32}$ Solution:
Equate the denominator to zero to find the domain of the given function
x$^{2}$ + 12x + 32 = 0
(x + 4) (x + 8) = 0
x = -4 and x = - 8
Therefore, domain is all real except - 4 and - 8.
Vertical asymptotes are x = - 4 and x = -8 | {"url":"http://math.tutorcircle.com/precalculus/rational-function.html","timestamp":"2014-04-19T22:05:23Z","content_type":null,"content_length":"38022","record_id":"<urn:uuid:648eea8d-12bc-4c10-a642-dc262aff97ab>","cc-path":"CC-MAIN-2014-15/segments/1397609537754.12/warc/CC-MAIN-20140416005217-00247-ip-10-147-4-33.ec2.internal.warc.gz"} |
Analog and Digital Time
Date: 11/15/2004 at 01:02:00
From: natalie
Subject: what is analog time
What is 4:05 in analog time? I don't get what analog time is. Is it
something like 24 hours? Or is it something else?
Date: 11/15/2004 at 14:41:08
From: Doctor Peterson
Subject: Re: what is analog time
Hi, Natalie.
Numbers can be represented in either analog or digital form. "Analog"
means that we are making an analogy between the number and some
quantity we can sense, such as a position on a graph or the angle of a
needle on a gauge. "Digital" means that we are showing a numerical
form (digits), that you have to read.
For example, on older cars, the speedometer is analog (a needle whose
position represents the speed), while the odometer is digital (numbers
click into place to show how far you have driven). Some new cars have
digital speedometer and odometer, both shown electronically.
Each type is useful in a different way. Analog representation makes
it easy to see how large a number is at a glance; digital
representation tells you the exact number, as closely as the
instrument can determine, so you can more easily write down the value.
Old clocks are analog: they show the time by the positions of the
hands, so that you can quickly see when it is getting close to a
certain time. Many new clocks are digital, showing "4:05" (or
whatever time it is) directly so you can say it or write it down, but
may have to think a bit to decide how close it is to 3:55. (Until
digital clocks were invented, we never called analog clocks analog; we
just called them clocks!)
So you are being asked to show what position the hands of an old-style
clock would be in at that time.
If you have any further questions, feel free to write back.
- Doctor Peterson, The Math Forum | {"url":"http://mathforum.org/library/drmath/view/66828.html","timestamp":"2014-04-18T23:44:17Z","content_type":null,"content_length":"6654","record_id":"<urn:uuid:73ebc4f2-68f0-4f5f-ab05-bf6fb1afb2f2>","cc-path":"CC-MAIN-2014-15/segments/1397609535535.6/warc/CC-MAIN-20140416005215-00417-ip-10-147-4-33.ec2.internal.warc.gz"} |
Finding distance between two points using two angles
January 18th 2011, 07:04 AM #1
Jan 2011
Finding distance between two points using two angles
Hello, I'm looking for help in learning to solve this kind of problem. I want to know the distance between point A and point B, the length of X, and the length of Y (assuming that X and Y
intersect at a right angle), using the information I have to start with (everything in blue).
If anybody can make sense of this, please help! I'm stumped, I'm an absolute idiot in mathematics (especially trigonometry) but I'm pretty sure I can follow along if the steps are basic enough.
Thanks in advance!
Is there no more information given, such as a right angle, parallel lines?
Also, from what I see, 10 and 5 are lengths, right?
All of the information given is above: the numbers in black are lengths (you are correct), both of the length 5 lines are at 90 degrees with line X, I'm sorry I forgot to mention that.
Last edited by Choscura; January 18th 2011 at 08:59 AM.
I tried to get AB but I didn't find much, however, I found another method, involving putting this on a cartesian plane, and solving simultaneously to get the coordinates of B. Knowing the
coordinates of A, AB can then easily be obtained. How about it? Do you think you can use that?
yes, that would work, I think. Looking at it it looks like the 5" line across X from Y is directly proportional to Y when the two lines from that 5" line intersect the ends of Y forming a
proportional triangle- the tricky bit is finding the size of the proportions (if this makes sense....)
I'm still absolutely stumped by this, I can look at it and feel the answer staring me back in the face but I can't figure out how to get something that works at all, much less every single time.
Please, post the cartesian coordinate solution and maybe this can finally make sense.
Okay, put the whole thing on cartesian plane.
You'll notice that the 10, 10, 10 triangle at the lower left is an equilateral.
Put the origin in the middle of the base of this triangle. A then becomes point $(2.5, 2.5\sqrt3)$ (see if you can get that)
Then, at coordinates (5, 0), there is this line going up to B. We can get the gradient of this line, knowing that:
$\tan70 = m_1$
Hence, we get the equation of this line, let's name it line 1.
Now, there is another line from the vertex of the equilateral to B. That vertex has coordinates $(0, 5\sqrt3)$
The gradient of line 2 is $\tan(90 - (50+30)) = m_2$ (can you see how?)
You have a gradient and a point, you get the equation of line 2.
Equate both lines, to get the point B.
You know point B and point A, you can find AB.
Hello, I'm looking for help in learning to solve this kind of problem. I want to know the distance between point A and point B, the length of X, and the length of Y (assuming that X and Y
intersect at a right angle), using the information I have to start with (everything in blue).
If anybody can make sense of this, please help! I'm stumped, I'm an absolute idiot in mathematics (especially trigonometry) but I'm pretty sure I can follow along if the steps are basic enough.
Thanks in advance!
1. I've modified you drawing a little bit. (see attachment)
2. The grey triangle EFG is an equilateral triangle with the side length 10. Thus the 3 interior angles are 60°.
3. From this you can determine the interior angles of the triangle GFB (values in green). Use the Sine rule to determine $|\overline{GB}|$ and $|\overline{FB}|$.
4. AB is a median of the triangle FGB. Use Cosine rule to determine it's length.
January 18th 2011, 08:31 AM #2
January 18th 2011, 08:44 AM #3
Jan 2011
January 18th 2011, 09:13 AM #4
January 18th 2011, 09:21 AM #5
Jan 2011
January 18th 2011, 09:29 AM #6
January 18th 2011, 11:32 AM #7 | {"url":"http://mathhelpforum.com/trigonometry/168687-finding-distance-between-two-points-using-two-angles.html","timestamp":"2014-04-21T06:13:51Z","content_type":null,"content_length":"52023","record_id":"<urn:uuid:e912147b-a30e-4f36-abbb-087a3c89e965>","cc-path":"CC-MAIN-2014-15/segments/1397609539493.17/warc/CC-MAIN-20140416005219-00559-ip-10-147-4-33.ec2.internal.warc.gz"} |
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[R] R Error: System is computationally singular
I am trying to fit a zero-inflated Poisson model using zeroinfl() from the
pscl library. I have 5 covariates (4 continuous, 1 categorical); the
categorical variable has 7 levels. I have had success fitting models that
contain only the continuous covariates; however, when I add the categorical
variable to any of the models (or if I run it by itself) I get the following
Error in solve.default(as.matrix(fit$hessian)) :
system is computationally singular: reciprocal condition number =
The code I am using is:
f1 <- formula(LOCS ~ as.factor(LCOVER) + D_ROADS + D_WATER + D_EDGE +
ZIP1 <- zeroinfl(f1, dist="poisson", link = "logit", data = FAWNS)
There is no correlation between my covariates. Also, I tried reducing my
categorical covariate to 3 levels and still receive the same error. Can
anyone suggest why I may be getting this error when I add the categorical
I appreciate your time and input. Thank you,
Nathan Svoboda
Graduate Research Assistant
Carnivore Ecology Lab
Mississippi State University | {"url":"http://grokbase.com/t/r/r-help/125rrjvex4/r-r-error-system-is-computationally-singular","timestamp":"2014-04-18T21:02:28Z","content_type":null,"content_length":"66599","record_id":"<urn:uuid:2f6d7927-572b-4085-a5c9-77937cb40463>","cc-path":"CC-MAIN-2014-15/segments/1398223202457.0/warc/CC-MAIN-20140423032002-00091-ip-10-147-4-33.ec2.internal.warc.gz"} |
Fractions of Money b
Number of results: 11,736
Harry spent one third of his money on a hat, 1/6 of his money on socks, and $12.00 on a tie. How much did Harry spend?
Saturday, March 16, 2013 at 5:24pm by Christine
math: percent and fractions
What are "nice" fractions? The problems? If a shirt was originally marked $30 and the store is offering 20% off, how much money will you save if you buy it? If you invest $1,000 and earn %50 on it in
one year, what is the percent interest you've earned?
Sunday, April 6, 2008 at 10:33pm by Ms. Sue
math (fractions)
Three people share a sum of money. The first receives 2/7 and the second receives a 2/5 share of the money. What fracion does the third receive?
Monday, January 20, 2014 at 1:03pm by Mack
fractions, decimals,and percents
JJ -- I gave you two uses of fractions and two uses of percents, plus money -- $1.25, $0.99, and so on.
Monday, November 17, 2008 at 9:15pm by Ms. Sue
Fractions of Money b.
What is two thirds of £33?
Saturday, March 2, 2013 at 12:25pm by hind
Maybe it will help if we see it with spaces and get rid of what seem to be typos.. 1. 16 nickels a.circle 3 of the nickels b.how much money is that 3 nickels = 15 cents 2. b. says circle 5/6 of the
dimes then it says how much money is that ? I assume there are 6 dimes. You ...
Wednesday, February 18, 2009 at 9:18pm by Ms. Sue
We use decimals when we use money. We use fractions when we follow a recipe; i.e. 1/2 cup sugar, 1/4 teaspoon salt, etc. Then we have to know how to multiply and divide fractions when we increase or
reduce the size of the recipe. How about decimals for sports statistics? How ...
Tuesday, October 12, 2010 at 4:14pm by Ms. Sue
By the way, fractions are also a great part of standard music notation. You need to know fractions to read musical notes.
Monday, July 27, 2009 at 1:14am by mathland
Like fractions are fractions with the same denominator. You can add and subtract like fractions easily - simply add or subtract the numerators and write the sum over the common denominator. Before
you can add or subtract fractions with different denominators, you must first ...
Tuesday, January 17, 2012 at 9:07pm by Laruen
math -fractions
ordering fractions 1/2,7/8,9/10,1/3,3/5,1/4 write the above fractions in order.
Monday, January 10, 2011 at 4:20pm by arjel
I know but wouldn't you have to go through confusing fractions and multiply by fractions to get a whole number of x? I am trying to follow the teachers directions which do not include
Tuesday, December 14, 2010 at 8:08pm by David
what is the difference and similarities if solving an equation with varibles in the equation if it had 1 or 2 or even 3 variables? Here is one: Money left in Wallet= money earned + money found -
money lost - money spent. I don't understand? can you please explain...
Saturday, January 13, 2007 at 1:33am by jasmine20
math fractions
ADAM WANTS TO COMPARE THE FRACTIONS 3/12,1/6,AND1/3.HE WANTS TO ORDER THEM FROM LEAST TO GREATEST AND REWRITE THEM SO THEY ALL HAVE THE SAME DENOMINATOR?EXPLAIN HOW ADAM CAN REWRITE THE FRACTIONS?If
anyone could helpme explain this to my 4 th grader.i am really bad with fractions
Tuesday, February 26, 2013 at 4:49pm by ttuuffyy
8th grade
youtube(dot)com/watch?v=BeCQWUl1p00&feature=related be sure to change your mix fractions into improper fractions when you divide fractions, you multiply the reciprocal
Sunday, September 12, 2010 at 9:14pm by Anonymous
Algebra 1-Fractions
Or, eliminating fractions, I should say. So, I need some help. See, I am really not a big fan of fractions. But I need to eliminate fractions to do a math problem. First one is 1/2-x=3/8. I know how
to find LCD, then multiply both sides, distributive property, etc etc. But ...
Monday, September 16, 2013 at 8:16pm by Breighton
6 = 5 4/4 5 4/4 = 3/4 = 5 3/4 You could find the common denominator for those fractions and convert them to equivalent fractions. But the easier way is to convert these fractions to decimals. 1/3 =
0.33 4/9 = 0.44 and so on
Monday, April 18, 2011 at 7:51pm by Ms. Sue
( egyptian fractions are fractions where the numerator can only be one) find two egyptian fractions where when added together it equalls 11/32
Tuesday, May 8, 2012 at 8:02pm by livy
3rd grade math
Change all fractions to equivalent fractions with a denominator of 60. Either that or change these fractions to decimals.
Wednesday, March 20, 2013 at 7:02pm by Ms. Sue
3rd grade math
Change all fractions to equivalent fractions with a denominator of 60. Either that or change these fractions to decimals.
Wednesday, March 20, 2013 at 7:03pm by Ms. Sue
Algebra 1
c. multiplying x/6-5/8=4 by 6 did not eliminate all the fractions/ What could you have multiplied to get rid of all the fractions? Explain how you got your answer and write the equivelent equation
that has no fractions. HELP ME PLEASE!!!!!!! I don't understand this.
Monday, May 11, 2009 at 8:06pm by Kelsie
Fractions of Money b.
1) 1/3 * 18 = 18/3 = 6 6) 3 is a third of 9 I'll be glad to check your answers for the other problems.
Saturday, March 2, 2013 at 12:25pm by Ms. Sue
Applied Business Math/Colleg student
2/3+1/6+11/12= First you need to change these fractions to equivalent fractions with a common denominator. If you post the equivalent fractions, we'll be glad to help you solve this problem.
Saturday, January 29, 2011 at 8:19pm by Ms. Sue
Shelly and Marcom are selling popcorn for their music club. Each of them received a case of popcorn to sell. Shelly has sold 7/8 of her case and Marcon has sold 5/6 of his case. Which of the
following explains how to find the portion of popcorn they have sold together? A.Add ...
Thursday, November 8, 2012 at 7:45pm by Jerald
The simplest way is to change the fractions to decimals and multiply. 13.25 * 11.5 = ? If your teacher wants you to use the fractions, then change the two numbers to mixed fractions. 53/4 * 23/2 =
1219/8 = 152 3/8
Saturday, July 25, 2009 at 4:04pm by Ms. Sue
Convert all of these fractions to equivalent fractions with a common denominator. An easier way is to use a calculator to convert each of these fractions to decimals. Then you can compare them
easily. 7/10 = 0.7 5/12 = 0.42 1/2 = 0.5 5/16 = 0.31
Tuesday, February 15, 2011 at 10:25pm by Ms. Sue
When we add or subtract fractions, we must have a common denonimator. 2 1/4 = 2 2/8 When I added the two fractions, I added the whole numbers and the numerators. That is 5 9/8. But since 9/8 is
larger than 1, I simplified the fraction to 6 1/8. Which part doesn't the child ...
Thursday, March 14, 2013 at 5:09pm by Ms. Sue
simple equations ( math)
You want to solve for the amount of money that Peter has. So you let your variable equal that amount of money. Now, you know that Peter's money + Joseph's money = $100. Can you express the amount of
money Joseph has in terms of x, the amount of money Peter has?
Sunday, January 10, 2010 at 2:07pm by Marth
b. What do think the fractions that are expressed as terminating decimals have in common? Think about equivalent fractions and common multiples. c. Do these fractions follow the same pattern as what
you decided about the first set of fractions? d. Why or why not? Note: The ...
Monday, August 2, 2010 at 11:22am by Ms. Sue
Change the fractions to equivalent fractions with the same denominator. 2/3 = 16/24 3/8 = 9/24 5/12 = 10/24 Add the numerators. Simplify the answer. What do you get?
Tuesday, May 28, 2013 at 7:17pm by Ms. Sue
It is possible to divide fractions by fractions. You can write a unit rate dealing with fractions by using fraction division. For example: If ½ of the apples are rotten in every ¾ of the boxes then
the unit rate is: ⅔ rotten apples per box. The only difference is that ...
Thursday, September 5, 2013 at 8:24pm by Graham
Actually, that is the best reason. I use the following criterium. If I see one of the variables having a coefficient of 1 OR -1, I solve for that variable and use substitution, resulting in no
fractions, unless the equation contains fractions to begin with. As a matter of fact...
Monday, February 18, 2008 at 9:52pm by Reiny
1. Why is representative money more useful than commodity money? B. Representative money has value because the government says it does. C. Representative money exists in unlimited supply, so more
people use it. D. Representative money is portable, durable, divisible, and ...
Monday, June 4, 2012 at 10:03pm by Anonymous
Susan had some money. she spent one-sixth of her money on Saturday.on Sunday,she spent one-half of the remaining money and gave $20 to her niece. she spent the rest of her money at an average of $15
for the next 5days.how much money did Susan have at first?
Tuesday, December 27, 2011 at 12:03am by Da S
Separate the fractions 2/6,2/5,6/13,1/25,7/8and 9/29into two categories: those that can be written as a terminating decimal and those that cannot. Write an explanation of how you made your decisions.
b. Form a conjecture about which fractions can be expressed as terminating ...
Sunday, September 27, 2009 at 7:22pm by Anonymous
Every month, a girl gets allowance. Assume last year she had no money, and kept it up to now. Then she spends 1/2 of her money on clothes, then 1/3 of the remaining money on games, and then 1/4 of
the remaining money on toys. After she bought all of that, she had $7777 left. ...
Thursday, April 30, 2009 at 9:41am by xxx
Fractions don't format well here. Try using a/b for fractions. I'll do #1 and yo can post your own answers for the others, which we will be happy to check. #1. 7 2/3 + 8 5/6 One way is to add the
whole numbers, then add the fractions: 7+8 + 2/3 + 5/6 15 + 2/3 + 5/6 Now, set ...
Tuesday, December 20, 2011 at 2:14pm by Steve
during column chromatography, if four fractions are done, how do we know what is in each beaker and why. I know that beakers 1 and 3 will contain the majority of the fractions being separated but
what about beakers 2 and 4. How do we know that they are the residues of the ...
Monday, October 22, 2007 at 9:16pm by Del
: Fractions are an important part of your daily lives. Describe some practical applications for fractions in your daily life and some challenges that you have experienced regarding the use of
Tuesday, May 18, 2010 at 10:30am by ree
fractions from least to greatest
Yes, I can. You'll be able to do it too after you change these fractions to equivalent fractions with the same denominator. Hint: The common denominator is 60. 3/4 = 45/60
Wednesday, November 7, 2012 at 5:07pm by Ms. Sue
money spent in meat shop=4xRs money spent in drugstore= xRs money spent in book store=x-15Rs money remaining=5Rs sum=4x+x+x-15+5 =6x-10
Tuesday, February 9, 2010 at 3:32pm by jagadheeswar
Help me on this one :( Express y= (7-3x-x^2)/[((1-x)^2)(2+x)] in partial fractions. Hence, prove that if x^3 and higher powers of x may be neglected, then y=(1/8)(28+30x+41x^2) I did the first part
of expressing it in partial fractions. (Since it's very difficult to type out ...
Wednesday, March 3, 2010 at 10:24am by Keira
1. Will you lend me some money? 2. He lent me some money. 3. Can I borrow you some money? 4. Can I borrow some money from you? 5. I borrowed some money from him. 6. He borrowed some money from me.
(Are the sentences grammatical? Would you like to check them?)
Friday, November 20, 2009 at 2:38pm by rfvv
math -fractions
Change the fractions to equivalent fractions with the same denominator. 7/8 = 21/24 5/6 = 20/24 Follow the same directions I posted before -- except you could draw rectangles, rather than circles.
Tuesday, January 4, 2011 at 6:25pm by Ms. Sue
PUC raised money for the recent disaster victims.They decided to allocate the money in manner: 1/3 of the money goes for medical supplies, 1/4 of what was left brought tents,2/3 of the remaining went
to water purification systems, the rest of the money was spent on shipping ...
Wednesday, September 21, 2011 at 6:14pm by Michelle
PUC raised money for the recent disaster victims.They decided to allocate the money in manner: 1/3 of the money goes for medical supplies, 1/4 of what was left brought tents,2/3 of the remaining went
to water purification systems, the rest of the money was spent on shipping ...
Wednesday, September 21, 2011 at 7:01pm by Michelle
You need to change the fractions to have a common denominator. Check this site to see how to change fractions so they have a common denominator. http://www.themathpage.com/ARITH/
Wednesday, February 6, 2008 at 3:26pm by Ms. Sue
Sarah was given a sum of money. She spent the same amount of money each day. She spent 2/7 of her money in 6 days. After another 5 days, she had $20 left. How much money did she have at first.
Sunday, May 19, 2013 at 5:51am by Sarah
hi to all of you teachers i hope you will help me to get the answers for this question below.. and thanks god bless you. Perk spent 1/4 of his money on a food and another 3/10 of his money on a
drink. a. What fraction of his money did he spend in total? b. What fraction of ...
Sunday, February 13, 2011 at 4:07pm by oscar
maths urgent
1/5 of J's money is equal to 1/3 of S's money. The difference in their amount is 1/2 of A's money. If Adam gives $120 to S, S will have the same amount of money as J. How much do the 3 people have
altogether?(please show me a model method, not algebra)
Friday, December 30, 2011 at 2:00pm by Mayyday
Change these fractions to equivalent fractions with a common denominator. 3/7 = 27/63 1/9 = 7/63 2/3 = 42/63
Thursday, February 16, 2012 at 8:12pm by Ms. Sue
I got part a. I do not understand the rest. b. form a conjecture about which fractions can be expressed as terminating decimals. c. test your conjecture on the following fractions; 6/12, 7/15, 28/
140, and 0/7. d. use the idea of equivalent fractions and common multiples to ...
Monday, August 2, 2010 at 11:22am by Betty
If Will gives Molly $9 dollars he will have the same amount of money as her. If Molly give Will $9, the ratio of money she has to the money will has will be 1:2. How much money does will have in the
Wednesday, February 5, 2014 at 8:18pm by Dylan
math -fractions
Change these fractions to equivalent fractions with the same denominator. 1/2 = 60/120 7/8 = 105/120 9/10 = 108/120 1/3 = 40/120 3/5 = 72/120 1/4 = 30/120 Now you can arrange them in order.
Monday, January 10, 2011 at 4:20pm by Ms. Sue
Snow spent 5/8 of her money on books and another 1/6 of her money on stationeries. What fraction of Kathy’s money was left?
Sunday, February 13, 2011 at 3:35am by rowena
Your pay stub deducts money for FICA. What does this mean? A. Money is being withheld for personal exemptions and deductions. B. Money is being withheld for excise and estate taxes. C. Money is being
withheld to fund Social Security and Medicare. D. Money is being withheld for...
Friday, November 30, 2012 at 10:43am by bob
A gift of money
Yes this is the question and if you take the offer you get the money. and the money has to be on your stomach for 10 min.
Wednesday, September 15, 2010 at 6:29pm by Anonymous
social studies
Economics has to do with money -- the way people earn money and how they spend money.
Wednesday, October 27, 2010 at 7:13pm by Ms. Sue
the sum of two fractions is 7/12. therre difference is 1/12. they have the same denominators. what are the two fractions?
Thursday, January 20, 2011 at 3:03pm by jayjay
4th grade
Equivalent fractions are fractions that may look different, but are equal to each other. Two equivalent fractions may have a different numerator and a different denominator. a/b=c/d
Thursday, May 6, 2010 at 5:19pm by Luke
social studies
Money! Money! Money! China wants the huge amounts of money it can earn with trade and tourism. It learned that cooperation with other countries is far more profitable than isolation. And now that
China is making so much more money, many of its people are able to live much more...
Tuesday, March 11, 2008 at 9:14pm by Ms. Sue
math 4th grade
You can find these decimals by long division. The other way to solve this is to find a common denominator. Then convert all of the fractions to fractions with a common denominator. That's complicated
with these five fractions.
Tuesday, April 13, 2010 at 9:50pm by Ms. Sue
A merchant visited 3 fairs. At the first, he doubled his money and spent $30; at the second he tripled his money and spent $54; at the third, he quadrupled his money and spent $72, and then had $48
left. How much money had he at the start?
Wednesday, April 3, 2013 at 11:50am by Barb
I need to complete a project based on fiat money. I need to design my own money and decide what amount of bills and/or money I want to produce for my economy. I understand that producing more money
causing inflation, but I need to know: How much money I should intially produce...
Friday, September 17, 2010 at 6:28pm by Tom
he spend (1/4+1/2)=3/4 of his total money. so has remaining money=(1-3/4)=1/4 of total money. he had total money=$58*4=$432
Sunday, March 10, 2013 at 3:03am by saiko
She spends 3/5 of her money, so she had 2/5 of it left Then she spends (3/5) of her remaining money, which would leave her with (2/5)(2/5) = 4/25 of her money so 4/25 of her money equals 8 1/25 of
her money equals 8/4 then 25/25 of the money equals 25(8/4) = 50
Wednesday, May 12, 2010 at 6:44am by Reiny
A man has $32 and decides to plant a garden. He cannot stand the thought of spending all of his money on any single day, so each day he goes to the store and spends half of his money on tomato
plants. When he no longer has enough money to buy additional plants, how much money ...
Thursday, August 2, 2012 at 1:55pm by Anonymous
contemporary math
Julius deposited his money in a money market account paying 1.05% compounded monthly. How much total money will Juliushave after 5 years?
Thursday, July 4, 2013 at 9:23am by Anonymous
0k i dont know the answer for these percents, decimals, and fractions. You have to change decimals to percents, fractions to decimals, and percents to fractions. 0.23 3/100 32 1/2% 0.25 3/5 75% 1/8
0.835 10% 95% 4% 120% 0.3333.... 1.05 1/6 If you can please help me anyone =\
Friday, March 14, 2008 at 1:23am by Kenya
A woman goes to a store that sells used books and spends half of her money on books. Each day, she returns to the store and again spends half of her remaining money on books because she does not want
to spend all of her money in one day. When she no longer has enough money to ...
Thursday, August 2, 2012 at 5:06pm by Anonymous
How do I do this problem: Chelsea sold baked goods to raise money for various charitable organizations. She gave a third of the money raised to the American Red Cross. Then gave a fourth of that
money to United Way. THEN gave half of the money to Twin city Missions. If Twin ...
Wednesday, September 30, 2009 at 1:10pm by McKenzie
Math repost for Grace
Check this site. http://themathpage.com/arith/add-fractions-subtract-fractions-1.htm
Friday, September 28, 2007 at 6:50pm by Ms. Sue
Hypothetical Economy: -Money Supply= $200 billion -Quantity of money demanded for transactions=$150 Billion -Quantity of money demanded as an asset=$10 billion at 12% interest -increaseing by $10
billion for each 2 percentage point fall in the interest rate. A. What is the ...
Monday, January 29, 2007 at 8:26pm by Frank
fractions, decimals,and percents
Recipes call for 1/2 cup, 1/4 teaspoon, etc. Decimals? Think about money. Percents? Computer downloads use percetages to show you much of the download is completed.
Monday, November 17, 2008 at 9:15pm by Ms. Sue
english essay, please help me to correct.
can you help to correct my answer and english grammar. thank you so much. question below: 1. How does money is importance in our society right now? answer: The importance of money in our society
right is more than anything else because everything is needed of money. money is ...
Monday, October 3, 2011 at 5:34pm by jessie
division of fractions
Division of fractions is sometimes said to be multiplication of inverses. Elaborate with an explanation and example. If you have a/b divided by c/d this is the same as a/b * d/c If e.g., we have 2/3
divided by 7/9 then we have 2/3 * 9/7 = 6/7 When dividing fractions the rule ...
Saturday, October 7, 2006 at 9:12pm by Brenda
In A Year, Seema Earns RS 1,50,000. Find The Ratio Of Money That Seema Earns To The Money She Saves And Money That She Saves To The Money She Spends?
Saturday, December 19, 2009 at 12:56pm by naisha
kailash gave one-third of his money to keshav.keshav gave three-fourth of tha money he received from kailash to saket.if sanket got rs. 900 less than the money kailsh had ,how much money did kesav
get from kailash.
Tuesday, May 31, 2011 at 12:48am by shweta
Math Fractions
What is the first step to adding multiple fractions (3) with differenct denominators?
Wednesday, July 8, 2009 at 1:40pm by Angie
ordering numbers fractions
The common denominator is 42. What are the equivalent fractions for 5/6 and 7/21?
Sunday, January 23, 2011 at 10:28pm by Ms. Sue
write the following fractions in increasing order 45/44 5/4/and 8/13
Monday, September 12, 2011 at 12:53pm by michael
rounding fractions to 0, 0.5, or 1. where would 0.599 lay? Can you round fractions down?
Thursday, January 26, 2012 at 6:16pm by lexi
Decimals to Fractions(simplify the fractions)
0.2=1/5? 0.9=9/10 0.80=4/5? 0.55=11/20
Tuesday, October 9, 2012 at 5:46pm by Jerald
Decimals to Fractions(simplify the fractions)
1.5=1/2 0.5=1/2 0.06=3/50 0.75=3/4 2.25=1/4 0.60=3/5
Tuesday, October 9, 2012 at 5:46pm by Jerald
The least common denominator is 238. Change these fractions to equivalent fractions.
Thursday, November 29, 2012 at 7:09pm by Ms. Sue
If Im naming fractions are these fractions in the right order from least to greatest? 1/4 1/3 1/2?
Thursday, January 31, 2013 at 5:52pm by Jerald
4th grade math fractions
three different improper fractions that equal 4 1/2
Tuesday, March 23, 2010 at 8:24pm by rita
I don't see where you need to use fractions to find the total. 9 + 6 = 15
Tuesday, February 11, 2014 at 9:41am by PsyDAG
4th grade math
Do you know how to change these fractions to equivalent fractions with the same denominator?
Monday, March 24, 2014 at 6:14pm by Ms. Sue
Adding Fractions
Before you can add fractions, you have to have common denominators. That is, both fractions need to have the same number in the denominator (the bottom number). The easiest way to find a common
denominator is to multiply the two denominators. In this case 3 * 8 = 24. http://...
Monday, March 14, 2011 at 10:43pm by Writeacher
Determine whether each of the following would lead to an increase, a decrease, or no change in the quantity of money people wish to hold. Also determine whether there is a shift in the money demand
curve or a movement along a given money demand curve a. A decrease in the price...
Wednesday, October 11, 2006 at 6:49pm by jada
Math- Fractions
I know this seems easy, but i stink at fractions. What is .105 as a fraction?
Monday, November 19, 2007 at 7:59pm by Ariana
Dividing fractions
and how do you reduce fractions isnt it dive the bottom number by the top
Saturday, October 25, 2008 at 3:22pm by kate
Joanne, do you want me to help Change the fractions to equivalent fractions with the same denominator??
Sunday, November 7, 2010 at 3:47pm by Erin
x - 1/3 = 4/5 x = 4/5 + 1/3 Convert the fractions to equivalent fractions with a common denominator. Add.
Sunday, November 7, 2010 at 8:36pm by Ms. Sue
I would change the mixed fractions to improper fractions first.
Monday, June 27, 2011 at 4:11pm by bobpursley
Change the fractions to equivalent fractions with a common denominator or to decimals.
Tuesday, January 31, 2012 at 6:04pm by Ms. Sue
how to you compare 8 fractions with different denominators including improper fractions?
Tuesday, February 14, 2012 at 4:47pm by john
If both fractions are portions of the 27, you have none left.
Wednesday, March 21, 2012 at 9:42pm by PsyDAG
Equivalent Fractions
Jerald -- do the same thing to number 5 that you did to the other fractions.
Wednesday, October 3, 2012 at 5:52pm by Ms. Sue
Change all of these fractions to equivalent fractions with a denominator of 12.
Wednesday, February 6, 2013 at 9:02pm by Ms. Sue
Pages: 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | Next>> | {"url":"http://www.jiskha.com/search/index.cgi?query=Fractions+of+Money+b","timestamp":"2014-04-20T04:54:25Z","content_type":null,"content_length":"38196","record_id":"<urn:uuid:8741d8a9-6bb7-4a13-a405-267057578b61>","cc-path":"CC-MAIN-2014-15/segments/1397609537864.21/warc/CC-MAIN-20140416005217-00427-ip-10-147-4-33.ec2.internal.warc.gz"} |
nonuniformly charged spherical surface
August 27th 2012, 12:07 PM
nonuniformly charged spherical surface
A sphere of radius a in free space is nonuniformly charged over its surface such that the charge density is given by ρ[s](θ) = ρ[s0] sin 2θ, where ρ[s0] is a constant and 0≤θ≤∏. Compute the total
charge of the sphere.
So I know
ρ[s] = dQ/dS
Integrating the surface charge density function will give me the charge Q. My question is how would you set up this integral?
∫ρ[s0] sin 2θ dS
integrating 0 to ∏
Or would this involve much more than that such as a triple integral?
Any help getting this set up would be great! Thanks! (Rofl)
August 27th 2012, 04:00 PM
Re: nonuniformly charged spherical surface
Since you are integrating over the surface, you should have a double integral. If you use spherical coordinates, one angle goes from 0 to $2\pi$, the other goes from 0 to $\pi$, all the while the
radius is constant.
The surface element for a constant radius is
$dS = r^2 \sin \theta d\theta d\phi$
Can you set up the integral now?
August 27th 2012, 05:06 PM
Re: nonuniformly charged spherical surface
Yes after integration I got the total charge Q=0 C.
Thanks for the help. | {"url":"http://mathhelpforum.com/advanced-math-topics/202611-nonuniformly-charged-spherical-surface-print.html","timestamp":"2014-04-18T23:44:55Z","content_type":null,"content_length":"5475","record_id":"<urn:uuid:b0b70c74-324e-431c-a6d9-e822868e063d>","cc-path":"CC-MAIN-2014-15/segments/1397609535535.6/warc/CC-MAIN-20140416005215-00419-ip-10-147-4-33.ec2.internal.warc.gz"} |
Android Forums - View Single Post - 6÷2(1+2) = ?
Originally Posted by
From a previous wiki reference -
Similarly, care must be exercised when using the slash ('/') symbol. The string of characters "1/2x" is interpreted by the above conventions as (1/2)x. The contrary interpretation should be written
explicitly as 1/(2x).
Substitute (1+2) for x and it's case closed.
That one I can follow.
6/2(x+y) being equal to 3(x+y) I have trouble following.
substituting (1+2) for x in that example is still not the same, as you only have the multiplier.
So you can follow my logic, I was taught 6/2(x+y)
which is same as
2x +2y
Sounds like something to send to school with my son, and see what answer his algebra IV professor sends back
EDIT: never mind - I see your reference . almost 5am here.
1/2x = 1/2(x), 6/2x = 6/2(x).
Still curious as to what his prof will say | {"url":"http://androidforums.com/2638458-post153.html","timestamp":"2014-04-17T08:47:46Z","content_type":null,"content_length":"14279","record_id":"<urn:uuid:12cd1136-4433-44ff-891a-f4c886ee209d>","cc-path":"CC-MAIN-2014-15/segments/1397609526311.33/warc/CC-MAIN-20140416005206-00221-ip-10-147-4-33.ec2.internal.warc.gz"} |
Scalar Triple Product
Next: Vector Triple Product Up: Vector Algebra and Vector Previous: Rotation
Scalar Triple Product
Consider three vectors scalar triple product is defined volume of the parallelepiped defined by vectors A.106. This volume is independent of how the triple product is formed from
So, the ``volume'' is positive if i.e., if
The triple product is also invariant under any cyclic permutation of
but any anti-cyclic permutation causes it to change sign,
The scalar triple product is zero if any two of
If i.e.,
Forming the dot product of this equation with
Analogous expressions can be written for i.e., provided that the three vectors are non-coplanar.
Next: Vector Triple Product Up: Vector Algebra and Vector Previous: Rotation Richard Fitzpatrick 2011-03-31 | {"url":"http://farside.ph.utexas.edu/teaching/336k/Newton/node157.html","timestamp":"2014-04-18T23:23:40Z","content_type":null,"content_length":"12581","record_id":"<urn:uuid:48567c52-15cc-4a42-a01d-f8083fc6937d>","cc-path":"CC-MAIN-2014-15/segments/1398223206647.11/warc/CC-MAIN-20140423032006-00382-ip-10-147-4-33.ec2.internal.warc.gz"} |
Table of Contents
Study Skills: Your Brain Can Learn Mathematics
Chapter R: Prealgebra Review
R.1 Fractions
R.2 Decimals and Percents
1. The Real Number System
1.1 Exponents, Order of Operations, and Inequality
Study Skills: Homework: How, Why, and When
1.2 Variables, Expressions, and Equations
Study Skills: Taking Lecture Notes
1.3 Real Numbers and the Number Line
Study Skills: Using Study Cards
1.4 Adding Real Numbers
1.5 Subtracting Real Numbers
1.6 Multiplying and Dividing Real Numbers
Summary Exercises on Operations with Real Numbers
1.7 Properties of Real Numbers
1.8 Simplifying Expressions
Study Skills: Reviewing a Chapter
Study Skills: Preparing for Tests
2. Equations, Inequalities, and Applications
2.1 The Addition Property of Equality
2.2 The Multiplication Property of Equality
2.3 More on Solving Linear Equations
Study Skills: Using Study Cards Revisited
2.4 An Introduction to Applications of Linear Equations
2.5 Formulas and Additional Applications from Geometry
2.6 Ratio, Proportion, and Percent
Summary Exercises on Solving Applied Problems
2.7 Solving Linear Inequalities
Study Skills: Managing Your Time
Study Skills: Tips for Taking Math Tests
3. Graphs of Linear Equations and Inequalities; Functions
3.1 Reading Graphs; Linear Equations in Two Variables
3.2 Graphing Linear Equations in Two Variables
3.3 Slope of a Line
3.4 Equations of Lines
Summary Exercises on Linear Equations and Graphs
3.5 Graphing Linear Inequalities in Two Variables
3.6 Introduction to Functions
Study Skills: Analyzing Your Test Results
4. Systems of Linear Equations and Inequalities
4.1 Solving Systems of Linear Equations by Graphing
4.2 Solving Systems of Linear Equations by Substitution
4.3 Solving Systems of Linear Equations by Elimination
Summary Exercises of Solving Systems of Linear Equations
4.4 Applications of Linear Systems
4.5 Solving Systems of Linear Inequalities
5. Exponents and Polynomials
5.1 Adding and Subtracting Polynomials
5.2 The Product Rule and Power Rules for Exponents
5.3 Multiplying Polynomials
5.4 Special Products
5.5 Integer Exponents and the Quotient Rule
Summary Exercises on the Rules for Exponents
5.6 Dividing a Polynomial by a Monomial
5.7 Dividing a Polynomial by a Polynomial
5.8 An Application of Exponents: Scientific Notation
6. Factoring and Applications
6.1 Factors; The Greatest Common Factor
6.2 Factoring Trinomials
6.3 Factoring Trinomials by Grouping
6.4 Factoring Trinomials Using FOIL
6.5 Special Factoring Techniques
6.6 A General Approach to Factoring
6.7 Solving Quadratic Equations by Factoring
6.8 Applications of Quadratic Equations
Study Skills: Preparing for Your Math Final Exam
7. Rational Expressions and Functions
7.1 Rational Expressions and Functions; Multiplying and Dividing
Study Skills: Making a Mind Map
7.2 Adding and Subtracting Rational Expressions
7.3 Complex Functions
7.4 Equations with Rational Expressions and Graphs
Summary Exercises on Rational Expressions and Equations
7.5 Applications of Rational Expressions
7.6 Variation
8. Equations, Inequalities, and Systems Revisited
8.1 Review of Solving Linear Equations and Inequalities
8.2 Set Operations and Compound Inequalities
8.3 Absolute Value Equations and Inequalities
Summary Exercises on Solving Linear and Absolute Value Equations and Inequalities
8.4 Review of Systems of Linear Equations in Two Variables
8.5 Systems of Linear Equations in Three Variables; Applications
8.6 Solving Systems of Linear Equations by Matrix Methods
9. Roots, Radicals, and Root Functions
9.1 Radical Expressions and Graphs
9.2 Rational Exponents
9.3 Simplifying Radical Expressions
9.4 Adding and Subtracting Radical Expressions
9.5 Multiplying and Dividing Radical Expressions
Summary Exercises on Operations with Radicals and Rational Exponents
9.6 Solving Equations with Radicals
9.7 Complex Numbers
10. Quadratic Equations, Inequalities, and Functions
10.1 Solving Quadratic Equations by the Square Root Property
10.2 Solving Quadratic Equations by Completing the Square
10.3 Solving Quadratic Equations by the Quadratic Formula
10.4 Equations Quadratic in Form
Summary Exercises on Solving Quadratic Equations
10.5 Formulas and Further Applications
10.6 Graphs of Quadratic Functions
10.7 More about Parabolas and Their Applications
10.8 Polynomial and Rational Inequalities
11. Inverse, Exponential, and Logarithmic Functions
11.1 Inverse Functions
11.2 Exponential Functions
11.3 Logarithmic Functions
11.4 Properties of Logarithms
11.5 Common and Natural Logarithms
11.6 Exponential and Logarithmic Equations; Further Applications
12. Nonlinear Functions, Conic Sections, and Nonlinear Systems
12.1 Additional Graphs of Functions; Operations and Composition
12.2 The Circle and the Ellipse
12.3 The Hyperbola and Other Functions Defined by Radicals
12.4 Nonlinear Systems of Equations
12.5 Second-Degree Inequalities and Systems of Inequalities
Appendix A. Review of Exponents, Polynomials, and Factoring
Appendix B. Strategies for Problem Solving
Appendix C. Synthetic Division
Additional Course Materials
Purchase Info
ISBN-10: 0-321-57569-5
ISBN-13: 978-0-321-57569-2
Format: Book
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Integral Curves Equations
February 14th 2010, 12:41 AM
Awsom Guy
Integral Curves Equations
For a curve y=h(x), the second derivative h''(x)=4, find the equation of the curve given that when x=1, y=-10 and when x=2, y=-9.
I know that y'= 4x + C
But I don't what to do next and how to place these points to find the equation.
Any help io good.
February 14th 2010, 12:44 AM
find y.
then, use the given points to set up two simultaneous equations to find the arbitrary constants of integration.
February 14th 2010, 12:47 AM
Awsom Guy
Don't I already have y, I can't integrate this equation as I don't know enough to get to y, I know how to get to y'.
February 14th 2010, 12:50 AM
no, you found y', you need to integrate that to get to y. you will get another arbitrary constant, call it D. then solve for C and D using the points you were given.
February 14th 2010, 12:52 AM
Awsom Guy
Do I sub in the x values aswell.
February 14th 2010, 12:56 AM
February 14th 2010, 12:58 AM
Awsom Guy
but I will have town unknown values C and D how am I meant to figure out the two values:
The answers:
-12=C+D and
How does this help.
February 14th 2010, 01:07 AM
your second equation is not right. and you've solved simultaneous equations before, yeah? That's definitely something you'd have to have done before calculus.
February 14th 2010, 01:23 AM
Awsom Guy
What is the second equation then?
February 14th 2010, 01:25 AM
February 14th 2010, 01:27 AM
Awsom Guy
I got:
y=2x^2 + C + D.
February 14th 2010, 01:31 AM
February 14th 2010, 01:32 AM
Awsom Guy
How does that help though
February 14th 2010, 01:41 AM
February 14th 2010, 01:45 AM
Awsom Guy
yes I got it thanks. | {"url":"http://mathhelpforum.com/calculus/128739-integral-curves-equations-print.html","timestamp":"2014-04-18T18:03:08Z","content_type":null,"content_length":"12008","record_id":"<urn:uuid:c2e8c9bf-45a7-413c-8eda-a74aed301931>","cc-path":"CC-MAIN-2014-15/segments/1397609533957.14/warc/CC-MAIN-20140416005213-00105-ip-10-147-4-33.ec2.internal.warc.gz"} |
An indeterminate limit question
March 22nd 2013, 08:33 AM #1
Junior Member
Mar 2013
An indeterminate limit question
Suppose lim x→1 (sqrt(ax+b)-2)/(x-1)=1, then a-b = ?
The answer key solved this by seeing the limit as 0/0, and therefore, sqrt(ax+b)-2 as 0.
What's the assumption, sqrt(ax+b)-2=0, based on?
Thanks in advance!
Re: An indeterminate limit question
If you substitute the value $x=1$ in the limit then the denominator is equal to $0$. This means the numerator can't be equal to some $L \in \mathbb{R}\setminus \{0\}$ as the limitvalue would be
infinite then (i.e it would not exist). Hence the only possibility is that the numerator is equal to $0$ as $\frac{0}{0}$ is undefined.
Last edited by Siron; March 22nd 2013 at 09:26 AM.
March 22nd 2013, 09:24 AM #2 | {"url":"http://mathhelpforum.com/calculus/215284-indeterminate-limit-question.html","timestamp":"2014-04-20T12:26:25Z","content_type":null,"content_length":"32979","record_id":"<urn:uuid:57d5d8e4-8073-4c1b-bd28-698f146f4c7a>","cc-path":"CC-MAIN-2014-15/segments/1397609538423.10/warc/CC-MAIN-20140416005218-00562-ip-10-147-4-33.ec2.internal.warc.gz"} |
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On The Measurement Of The Hydraulic Properties Of The Environmental Medium
STP1415: On The Measurement Of The Hydraulic Properties Of The Environmental Medium
Gordji, SS
Statistical Consultant & Assistant Professor, University of Mississippi, University, MS
Pirouzian, L
Graduate Student, University of Mississippi, University, MS
Pages: 7 Published: Jan 2002
Contaminates behave differently in the vadose zone than they do in the saturated zone and the governing equations for them are more involved. The hydraulic conductivity and porosity are the most
important parameters affecting water and contaminates movements in the vadose and the saturated zones. To estimate these parameters certain methods are examined for measuring the hydraulic properties
where the soil permeability is low and, where possible, the results are compared with measured data. Unsaturated hydraulic conductivity (UHC), unlike saturated hydraulic conductivity (SHC), is
difficult to obtain especially for clayey soils and fractured rocks. Most available studies either overestimate or underestimate the UHC, especially for values close to the surface where the UHC
values are several orders less than their corresponding values for SHC. The goal of this paper is to adjust and modify some empirical equations for obtaining UHC and compare the results with those
already available in the literature. This is achieved through the use of non-linear least squares. Values that represent parameter b, in the Campbell's power function is allowed to vary in a given
range instead of having a single value. For every value in this range the non-linear least squares equations is calculated and the b value with the smallest error term is picked to represent the
empirical parameter for the Campbell's power function. Results indicate adjustment to the reported empirical parameter b is needed to minimize the sums of squares and to obtain a closer match between
the measured and the experimental data. That is the sum of errors for the non-linear least squares between measured and empirical data is minimum somewhere near the values that they suggested.
Therefore, for the clayey soil and fractured rocks the empirical parameters for the power function are modified to reflect the minimum sums of squares for non-linear least squares equation. To fine
tune b further, non-linear least squares are used to find the best fitted value of b using the empirical equation that relates depth (h) to water content. We are in the process of classifying
fractured rocks and are testing them to see how their hydraulic conductivity compares with clayey, sandy and silty soils. This is done through the application of fractal geometry and use of
Sierpinski carpet. The empirical and the theoretical relationships used here are simple and practical and may directly be applied to obtain corresponding UHC. The equations that are utilized here are
based on the theory or have their roots in statistical modeling. Results obtained from this investigation should be compared with the field data before they are applied toward solving Richards'
vadose zone, saturated zone, permeability, hydraulic conductivity, least squares
Paper ID: STP10619S
Committee/Subcommittee: D18.04
DOI: 10.1520/STP10619S | {"url":"http://www.astm.org/DIGITAL_LIBRARY/STP/PAGES/STP10619S.htm","timestamp":"2014-04-21T02:50:09Z","content_type":null,"content_length":"15741","record_id":"<urn:uuid:363900d4-bace-4f53-bb87-81a48dfcdbf6>","cc-path":"CC-MAIN-2014-15/segments/1397609539447.23/warc/CC-MAIN-20140416005219-00402-ip-10-147-4-33.ec2.internal.warc.gz"} |
Homework Help
Posted by tanisha on Wednesday, March 26, 2008 at 10:08pm.
*Fe2O3(s)+3CO(s)-->2Fe(s)+3CO2(g)(delta h is -23kJ)
*3Fe2O3(s)+CO(g)-->2Fe3O4(s)+CO2(g)(delta h is -39kJ)
*Fe3O4(s)+CO(g)-->3FeO(s)+CO2 (delta h is 18kJ)
find delta h for the desired reaction.
which reactions do i reverse and then how do i cancel the out the ones i dont need?
• chemistry - DrBob222, Wednesday, March 26, 2008 at 10:35pm
Start with the end equation.
FeO(s) + CO(g) ==> Fe(s) + CO2(g)
I want an Fe(s) on the right and CO2(g) on the right. The first equation has that so write that one as is (at least for the time being).
I would reverse equations 2 and 3
and multiply equation 1 by 3, equation 2 by 1, and equation 3 by 2. Then add everything on the left side and everything on the right side, then cancel those molecules/atoms that are on both
sides. If I didn't goof I obtained the equation you want by doing that.
• chemistry - tanisha, Wednesday, March 26, 2008 at 11:02pm
ok this is what i have gotten:
but the desired reaction is in a one to one ratio, and also the CO(g) is on the left side in the desired reaction. does that mean i have to switch over again?
• chemistry - DrBob222, Wednesday, March 26, 2008 at 11:34pm
I don't think so. Let's see how it plays out. I'll go through some of it and let you finish.
3Fe2O3 at top left cancels with 3Fe2O3 on right in equation 2.
2Fe3O4 on left in equation 2 cancels with 2Fe3O4 on right of equation 3. Then 9CO in equation 1(left side) cancels with 1 CO on right of equation 2 and 1 CO on right of equation 3 to leave 6 CO
on left side. For CO2, we have 2CO2 on left of equation 3 and 1 CO2 on left of equation 2 which subtracts from 9 CO2 on right of equation 1 leaving 6 CO2 on right side. If I've done all that
right (you should confirm all of this), that leaves 6FeO + 6CO ==> 6Fe + 6 CO2. So whatever delta H adds up, just divide by 6 to get delta H for the equation you want (1:1:1:1).
• chemistry - tanisha, Wednesday, March 26, 2008 at 11:52pm
i got -1/3 for delta H. thank you very much, you have been a big help.
• chemistry - DrBob222, Thursday, March 27, 2008 at 11:43am
You might want to check your work. Since you multiplied equation 1 by 3, did you multiply delta H by 3? And you multiplied equation 3 by 2, did you multiply delta H by 2? If those are delta Hs
for the reaction as shown, then the multiplications are in order and that will change the final answer.
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PACK Function
The PACK function packs an array into a vector under the control of a mask.
PACK (array, mask [, vector] )
Required Arguments
array is an INTENT(IN) array can be of any type.
mask is INTENT(IN) and must be of type LOGICAL. mask must be conformable with array.
Optional Arguments
vector is an INTENT(IN) array of rank one, and must be the same type and kind as array. It must have at least as many elements as there are true elements in array. If mask is scalar with value true,
vector must have at least as many elements as array.
The result is an array of rank one with the same type and kind as array.
If vector is present, the result size is the size of vector.
If vector is absent, the result size is the number of true elements in mask unless mask is scalar with the value true, in which case the size is the size of array.
The value of element i of the result is the ith true element of mask, in array-element order. If vector is present and is larger than the number of true elements in mask, the elements of the result
beyond the number of true elements in mask are filled with values from the corresponding elements of vector.
Example integer :: c(3,3)=reshape((/0,3,2,4,3,2,5,1,2/),shape(c))
integer :: cc(9)=-1
write(*,'(3i3)') c ! writes 0 3 2
! 4 3 2
! 5 1 2
write(*,*) pack(c,mask=(c > 2))
! writes 3 4 3 5
write(*,*) pack(c,mask=(c > 2),vector=cc)
! writes 3 4 3 5 -1 -1 -1 -1 -1
write(*,*) pack(c,.true.) ! writes 0 3 2 4 3 2 5 1 2
See Also | {"url":"http://www.lahey.com/docs/lfpro73help/f95arpackfn.htm","timestamp":"2014-04-18T13:08:15Z","content_type":null,"content_length":"6064","record_id":"<urn:uuid:cdc22684-62e5-4936-ad53-05570968baaa>","cc-path":"CC-MAIN-2014-15/segments/1398223203422.8/warc/CC-MAIN-20140423032003-00225-ip-10-147-4-33.ec2.internal.warc.gz"} |
How I design a top-down raglan pullover
There's one special thing about raglan sweaters that I see overlooked in so many designs.
This one special thing is the depth of the back neck. This is determined by how many stitches are cast on for the sleeves at the neck edge. Half of the sleeve cast-on measurement determines the back
neck depth (and the front neck depth if it is worked the same as the back). This measurement, when added to the length of the raglan shaping determines the total length from the shoulder to the
underarm, and is important to include for the desired armhole length to be accurate.
For example, if you have 4" worth of stitches cast-on at the top of the sleeve, that would result in a 2" back neck depth. And, if the raglan shaping is worked over 5.5", the total armhole depth is
7.5". This is an important thing to keep in mind when writing the raglan shaping, because a 5.5" armhole is a lot smaller than a 7.5" armhole.
That said, I'm inspired to share how I write a top-down raglan pullover pattern from A to Z. At the end of all this there is a link to download the free pattern, and the excel spreadsheet I used to
create this tutorial. If you happen to knit this sweater, I'd love if you'd post some photos on
! There is also a
Knit-A-Long group
for you to join in the social-knitting fun!
First, I'll begin with rudimentary measurements. I say rudimentary because if stitch patterns are involved, often the measurements will change slightly in order to account for matching the stitch
counts of the stitch patterns. For this sample pattern, it's all knit in Stockinette Stitch—making things easy-peasy.
The most important measurements to know when beginning a raglan pattern are the finished bust circumference and the circumference of the upper arm.
Bust Measurements: 32 (36, 40, 44, 48, 52, 56)"
Upper Arm Circumference: 13 (14, 14.75, 15.75, 16.5, 17.5, 18.25)"
And the gauge. We MUST know the gauge… of course.
Since I'm making this up, I'll use a gauge that I see commonly used for worsted weight yarns:
5 stitches and 7 rounds per inch.
Next we'll figure out how many stitches we'll need at the bust, and at the upper sleeve.
And… this is where I get Excel involved. If you are unfamiliar with how to use excel spreadsheets for sizing your patterns. Check out
this post
I made once upon a time.
The first thing I'll do is set up my excel spreadsheet with all the descriptions of what I think I might need to know to determine the raglan shaping.
Basically, I'll need to know the number of stitches to cast on and how many of those stitches are used for each back, front and sleeves and what each of the measurements are. For the raglan shaping,
I'll often have 2 or more sets of shaping—increasing at each raglan every row on the body and sleeves, and every other row on the body and sleeves. We may need to add shaping that happens just on the
body or just on the sleeves—but I won't know if we'll need that until later, so I'll leave it out for now.
Then I fill in the most important information (bust and underarm measurements).
And, determine how many stitches are needed to get to those measurements.
The body stitch counts all turned out to be a whole number. But, you can see that some sizes of the sleeves have decimals. I'll round those numbers to the nearest whole number and adjust the
measurement accordingly. As you can see, it didn't change much. When working with a stitch pattern, be sure that these numbers match whatever the necessary multiple of stitches for that pattern.
I like to have a few cast-on stitches worked at the underarm. I find that it is more flattering, and forms the natural curve of the body more than a sweater without it. I'll usually cast-on about 1/
2" – 3" of stitches at each underarm. If I plan on having waist shaping on the body I'll cast-on an even number of stitches so it's easy to place a marker at the center on each side. This information
gets filled in next.
When I subtract the number of cast-on stitches from the body and the sleeves, I learn how many stitches I'll end up with at the base of the raglan shaping. That number is shown highlighted in purple,
I'll walk you through the chart below. Read it from the bottom up, for the first size, in the left-hand column.
For the sleeve. There are 65 stitches total, minus 4 cast-on stitches results in 61 stitches for each sleeve before the cast-on stitches.
There are 160 stitches on the body and 4 stitches cast-on at each underarm, so, the number of stitches on each back and front before the cast-on would be determined by subtracting each set of 4
cast-on stitches from 160, which gives us 152. Then divide that in half to get the front and the back = 76 stitches each front and back.
When we add 76 (front) + 61 (one sleeve) + 76 (back) + 61 (second sleeve), we get 274.
Next, I'll figure out how much shaping needs to happen between the neck and the bust in order for the neck width and depth to fit comfortably.
This is one of the reasons I love Excel. What I do here is set up the formula, then try out different numbers for each size until I get to the measurement that I want.
In the chart below, I fill in a formula at the neck edge, where each front, back and sleeves are determined for the cast-on. For each the body and sleeve I'll take the number of stitches at the end
of the shaping and subtract 2 stitches for each repeat of the increase row. This tells me how many stitches I'll need before increasing that number of stitches.
I'll also add the formula (highlighted in yellow, at the top) to tell me what these stitch counts will measure. This is what I use as an indicator to determine the perfect number of stitches.
Once the formula is filled in and I drag it across for all the sizes, I begin entering numbers in the lower light-blue highlighted line, and the number of cast-on stitches for the front/back and
sleeves will change accordingly.
My goal is usually to make the neck no wider than 9.5".
For the neck width to have a comfortable fit, it looks like we'll need to work 19 (23, 26, 30, 34, 37, 40) increase rounds on the front and the back, increasing 2 stitches each front and back on each
of those rounds.
But… look at the stitch count for the cast-on edge of the sleeve. I don't like the look of that. It seems that 19 (23, 26, 30, 34, 37, 40) increases won't work out so well for the sleeves. I usually
like this measurement to stay relatively the same across all the sizes, or increase in size from the smallest to largest size. This does the opposite.
To fix this, we'll need to work more sleeve increases for the smaller sizes, and fewer sleeve increases for the larger size. So, I've added 2 more shaping sections to my spreadsheet—one section for
the smaller sizes that will work some rounds with shaping only on the sleeves, and another section for the larger sizes that will work some rounds with shaping only on the body. I add these extra
rounds at the neck edge, incorporating them between rounds that have shaping on both the body and the sleeves.
So, the basic structure of the raglan shaping begins at the neck with increases on the body and sleeves on every round, then it changes to alternating between 1 round of body and sleeve increases and
1 round of shaping on either body OR sleeves (depending on the size), then it changes to working increases on the body and sleeves every other round, with a round worked even in between.
The reason I place the increases every round at the neck edge, is because of the shape of the body. When there are more increases at a faster rate at the neck edge, the shaping of the sweater expands
quickly from the neck to the shoulders, then gradually from the shoulders to the underarm. If it is worked the other way around, the shaping doesn't flatter the body as well. See the diagram below
for a visual.
Then I adjust the formula for the cast-on stitches to include the shaping that may happen during these sections.
Because we know how many times to increase on the body, and we don't want to mess that up while playing with the number of individual body and sleeve increases, I turn the number of increases on the
body into a formula to include the body increases worked during the individual body/sleeve shaping.
Rather than doing this for one size and dragging it across, this needs to be done for each size individually as follows: For the first size, there are a total of 19 increases needed for the body. So,
this formula is created by subtracting the number of body increases from each of the individual sections from 19. Then for the following sizes, from 23, 26, 30, and so on…
Now I can play with adding numbers into the light blue rows to see how the number of stitches at the top of the sleeve adjusts, and tweak them until I have a measurement that I like.
You can see that with the 7 (5, 3, 1, 0, 0, 0) extra sleeve increase rounds and 0 (0, 0, 0, 2, 3, 6) rounds with only body increases that the number of stitches for the body cast-on remains the same,
and the sleeve cast-on stitches are now pretty similar to each other. You'll also see that the number of body/sleeve increase rounds that are worked every other round have been adjusted accordingly
because of the formula we entered to maintain the correct number of body increases.
And, now we know how many stitches we'll need to cast on to begin knitting the yoke. I've also double checked the measurement of the total number of cast-on stitches to be sure it's large enough to
fit over the head—which it is. Yay!
The next thing we need to do is determine how many times to work the body and sleeve increases every row at the neck edge. This will determine the total number of rounds used for the raglan shaping
and thus give us the raglan and armhole depth.
Let's fill in some formulas to make this easy to play with.
Beginning at the top of my file, I begin by filling in the places where we will count rounds. Line 17 shows the round where we place the markers for the raglan, then line 20 will be filled in with
how many rows are worked with increases on the body and sleeve every round.
So, on line 21 we add them together to get the total number of rounds worked to here.
On line 22, we divide the total number of rounds used by the row gauge to get that measurement.
For each of the increase sections, we'll do the same thing. Line 31 shows 1 round worked with increases on the body and sleeves, then line 32 shows 1 round worked with increases on the sleeves only.
The light blue (line 33) shows how many times we will repeat those 2 rounds. So, on line 34 we add lines 31 and 32 together and multiply them by line 33 for the total number of rounds worked in this
On line 35, we divide the rounds for this section by the row gauge, and add them to line 22 to get the total measurement to here from the cast-on.
Repeat entering row and measurement formulas for each shaping section, being sure to add the measurement for each section to the previous measurement so you'll have the total measurement from the
cast-on at the end of each shaping section. When you get to the end of the last shaping section you'll have the total measurement of the raglan shaping from the neck to the underarm (line 62 shown
below). Add this measurement to the neck depth measurement (half of the sleeve cast-on measurement) to determine the total armhole depth.
Ideally, I like the armhole depths to range from 7–10", so it looks like we'll need to reduce them a bit. We do this by working more body/sleeve increases every round at the neck edge, and fewer
every other round at the underarm edge.
We don't want to mess up the number of total body/sleeve increases, so we'll need to update the formula on line 60 (the every other round body/sleeve increases) to exclude the number of times we work
the body/sleeve increases on every round.
Similar to when we adjusted this line previously, we'll want to adjust each size individually so the total number of increases (19 [23, 26, 30, 34, 37, 40] stitches) remains in tact.
And, we'll need to adjust the formula for the Front/back cast-on stitches and the Sleeve stitches so they don't change from the numbers we have them set at. This is done by subtracting 2 x the number
of body/sleeve every-row increases from the formula that is already there, because there are 2 stitches increased on each back, front and sleeve per increase round.
Now we can begin trying numbers in line 20 (body/sleeve increases every row), while watching the measurement of the armhole depth, adjusting the line 20 numbers to give us an ideal armhole depth
We're ALMOST there. The very last thing we need to do is fill in the stitch counts for each of the shaping sections, and make sure that they work out correctly from the neck cast-on to the underarms.
Cross your fingers!
Beginning at the neck cast-on edge, I fill in a formula on line 24, which adds 8 stitches for each of the body/sleeve increase rounds worked on every round.
To double check that, I add 2 stitches to each Front/Back cast-on stitches and 2 stitches to each Sleeve cast-on stitch count.
For line 28, add the Front/Back stitches to the Sleeve stitches twice (because there are 2 of each).
Line 24 and line 28 should match. That's why they're both pink.
This step gets repeated for each shaping section, being sure to add the accurate number of stitches per repeat. For example, for the second section, there is 1 round with body and sleeve shaping
(that's 8 stitches increased), and 1 round with only sleeve shaping (that's 4 more stitches), so the total stitches will add 12 stitches per increase round. And, for the body, there are 2 stitches
each repeat for the front/back, and on the sleeves there are 4 stitches each repeat. When you add together the Front/Back and Sleeve stitches, they
match the total number of stitches figured from the total stitches of the last section. If they don't—there may be something up with your formula in regards to the number of stitches increased for
each sleeve and body per repeat.
From this point on, for a really easy pattern, you could knit the body and sleeves completely straight for whatever length you want, trying the sweater on as you go… and you're done.
Here's a pattern
written out in detail for all the stuff we just did.
And, if you want to download the excel spreadsheet, to see the whole thing in detail.
Here's that too
Please consider donating any amount (even just $1) to support the free information Kristen offers.
13 comments:
1. Such a huge mystery for me...SOLVED! If you were here I would give you a huge hug! Thank you so much. I know how much time and detail you have put into this and I sincerely appreciate it!
1. I'm glad it has helped you out!
2. Wauw - I'm really impressed and glad to find another excel-fan. I will definitely take a closer look at your spreadsheets. It seems very useful - can't hardly wait to "play" with it.
Thanks for sharing!
1. Great! I hope you have fun with it! :)
3. Dear Kristen,
I have always loved working raglan sweaters from the neck down, but your instructions are so thorough and easy to follow. Yours is a major contribution and I would like to applaud you. Thank you
so much for this information. I have downloaded it into my computer and will have it available any time I need it.
Thanks for sharing your knowledge and experience.
Jean Ritchey
1. Jean, I'm so happy to share what I know with others, and I'm glad you think this is useful!
4. This is awesome! I'm a bit of a nerd (ok, a really big one) and I've often wondered how I could create something like this. Knitting is an art, but the engineering-part of my brain is constantly
questioning why so many designs can't just be explained as formulas. Great job and thank you for sharing!
1. I think many simple designs can be explained as formulas—though it's not the norm, so people don't tend to do it very often. The more complicated a design gets, the more challenging it is to
make into a formula, and so many of the designs coming out these days are pretty complicated. IMO.
5. A knitter who does excel! I am in a happy place.
1. Me too! XOXO Happy Excelling.
6. I have been using Excel for years to ensure I understand the pattern and to create a clear, clean pattern customized just for me! Thanks so much for sharing you work, I really appreciate it!!!
1. You're very welcome. Knitting is about 90% (don't quote me on that %, it's an estimate) math, and that's what Excel is great at, so it's pretty natural, I think, to use them together.
7. Thank you so much for these instructions. I just ordered Your first book. With adding these instructions I will be able to knit easy Things that fits well. | {"url":"http://kristentendyke.blogspot.com/2013/06/how-i-design-top-down-raglan-pullover.html","timestamp":"2014-04-17T03:51:37Z","content_type":null,"content_length":"162587","record_id":"<urn:uuid:6ec0e0af-7dd3-4f9c-a2b5-8f1927e1ed30>","cc-path":"CC-MAIN-2014-15/segments/1397609526252.40/warc/CC-MAIN-20140416005206-00238-ip-10-147-4-33.ec2.internal.warc.gz"} |
15 projects tagged "OS Independent"
NOVAS (Naval Observatory Vector Astrometry Subroutines) is an integrated package of subroutines for the computation of a wide variety of common astrometric quantities and transformations. It can
provide the instantaneous coordinates (apparent, topocentric, or astrometric place) of any star or planet, and also provides general astrometric utility transformations, such as those for precession,
nutation, aberration, parallax, etc. It is useful for data reduction programs, telescope control systems, and simulations. | {"url":"http://freecode.com/tags/os-independent?page=1&with=2906&without=","timestamp":"2014-04-19T12:58:19Z","content_type":null,"content_length":"64665","record_id":"<urn:uuid:9e3a2cfc-fbf4-4371-bf3a-28fd391a7736>","cc-path":"CC-MAIN-2014-15/segments/1397609537186.46/warc/CC-MAIN-20140416005217-00616-ip-10-147-4-33.ec2.internal.warc.gz"} |
Rainbow Squares
Rainbow Squares are another neat Math Model designed for elementary, secondary, or adult learners as an individual or group learning puzzle. Six squares are each made up of three different
rainbow colored pieces that can be put together in a variety of ways. Each of these pieces can also be used to form other squares using two, four, five, or six pieces that are easily constructed
by younger learners and recorded as colored shapes. Older learners can describe the area of each piece as a fraction or percentage of one square. They can also make configurations of several
pieces and copy the outline for other learners to replicate like super Tangrams.
Rainbow Squares come with a guide to the activities described above and are packaged in a handmade wooden box, so just putting this Math Model away requires some skill! | {"url":"http://www.prainbow.com/Rainbow_Squares.html","timestamp":"2014-04-18T18:19:05Z","content_type":null,"content_length":"2799","record_id":"<urn:uuid:67e7c9aa-d1a3-4326-aa5c-802bb148ab2c>","cc-path":"CC-MAIN-2014-15/segments/1397609535095.7/warc/CC-MAIN-20140416005215-00074-ip-10-147-4-33.ec2.internal.warc.gz"} |
Combinatorics: A Guided Tour
Combinatorics is mathematics of enumeration, existence, construction, and optimization questions concerning finite sets. This text focuses on the first three types of questions and covers basic
counting and existence principles, distributions, generating functions, recurrence relations, Pòlya theory, combinatorial designs, error correcting codes, partially ordered sets, and selected
applications to graph theory including the enumeration of trees, the chromatic polynomial, and introductory Ramsey theory. The only prerequisites are single-variable calculus and familiarity with
sets and basic proof techniques.
The text emphasizes the brands of thinking that are characteristic of combinatorics: bijective and combinatorial proofs, recursive analysis, and counting problem classification. It is flexible enough
to be used for undergraduate courses in combinatorics, second courses in discrete mathematics, introductory graduate courses in applied mathematics programs, as well as for independent study or
reading courses.
Table of Contents
Before You Go
1. Principles of Combinatorics
2. Distributions and Combinatorial Proofs
3. Algebraic Tools
4. Famous Number Families
5. Counting Under Equivalence
6. Combinatorics on Graphs
7. Designs and Codes
8. Partially Ordered Sets
Hints and Answers to Selected Exercises
List of Notation
Excerpt: (p.150)
A university has 120 incoming freshman that still have to be assigned to on-campus housing. The only remaining dorm holds 105 students and contains 42 doubles (rooms housing two students) and seven
triples (three students). In how many ways can the university select 105 students to house in this dorm and then arrange those students into roommate pairs and triples, without yet assigning them to
In the previous exercise, suppose the university gets approval to house temporarily the remaining 15 students among the dorm's three lounges. Each lounge will house five students. How many ways are
there for the university to assign all 120 students to rooms?
About the Author
David R. Mazur (Western New England College, in Springfield, Mass.), who received his Ph.D. in mathematics from Johns Hopkins University in 1999, was a 2000-2001 Project NExT fellow. Mazur (
dmazur@wnec.edu) now serves the MAA as a consultant to Project NExT and as a member of its Membership Committee. This is Mazur's first book. He will be doing his first book signing at the Joint
Mathematics Meetings in San Francisco in January 2010.
MAA Review
Combinatorics: A Guided Tour is an undergraduate textbook designed for a first course or reading course in combinatorics. A student learning from this book should have taken one year of calculus and
an introduction to proofs. The book takes a more theoretical approach in presenting combinatorics, in contrast to some of the more application-based combinatorics books on the market. | {"url":"http://www.maa.org/publications/books/combinatorics-a-guided-tour?device=mobile","timestamp":"2014-04-19T15:43:44Z","content_type":null,"content_length":"23618","record_id":"<urn:uuid:a27447b5-9ad3-4be2-a2dd-f2a230b2b4c3>","cc-path":"CC-MAIN-2014-15/segments/1397609537271.8/warc/CC-MAIN-20140416005217-00330-ip-10-147-4-33.ec2.internal.warc.gz"} |
the centers of
Exploring the relationship of the centers of
triangles with the centers of other
external geometric figures
Angie Head
After doing many explorations with the centers of a triangle in GSP, we will now look at the relationship between these centers and the centers of other externally constructed geometric figures. We
will compare the centers of a triangle with the centers of externally constructed equilateral triangles, externally constructed squares, and externally constructed isosceles triangles.
First, lets construct an equilateral triangle ABC. Then lets construct equilateral triangles on each side of our triangle. We get the following construction.
Now, we need to locate the centroid of triangle ABC. From prior constructions, we know that in order to locate the centroid of a triangle we first need to find the midpoint of each side of the
triangle. Next, we need to join the midpoints of each side to the opposite vertex. We also know that these medians are concurrent. Now, we need to find the centroids of the external triangles, using
the same method. We get the following construction.
By observing the above construction, one can see the relationship of the centroids of the external triangles with the centroid of the original triangle. In this construction, the points A', B', and
C' are the centroids of the external equilateral triangles. Also by observing this, you can see that the lines AA', BB', and CC' all pass through the centroid of the original triangle, point G. Since
G is the point of intersection of these median segments, they are concurrent at this point.
Now, let's see if this holds true for any triangle ABC. Let's construct a scalene triangle and its external equilateral triangles on each side. Now we need to locate the centroid of each of these
triangles. G is the centroid for triangle ABC and A', B', and C' are the centroids of the external triangles. If the lines AA', BB', and CC' intersect at G, then G is the point of concurrency.
By observing the above construction, one notices that G is not the point of concurrency in this case. Let's construct the rest of the rest of our centers (i.e. the incenter (I), the orthocenter (H),
and the circumcenter (C)) to see if one of these points is the point of concurrency for these lines. It looks like the incenter is the point of currency, but through further investigation we notice
that none of these points are the point of concurrency. Hence, the point of concurrency does not lie on Euler's Line.
We can further our investigation of the centers of triangles by constructing a square externally to each side of the triangle ABC. Next, we need to find the centers A', B', and C' of each square and
construct the lines AA', BB', and CC'.
By observation, one notices that these lines do not intersect at the centroid G of the triangle. Hence G is not the point of concurrency. One can also observe that the point of concurrency is not any
of the points on Euler's line (i.e. it is not the incenter, the orthocenter, the cicumcenter, the centroid).
We started our investigation of the centers of triangles by observing equilateral triangles that were constructed off of each side of the equilateral triangle ABC, where A', B', and C' were the
centroids of the external triangles. Now, we are going to explore these same triangles but now A', B', and C' are the external vertices of the external equilateral triangles.
As in the previous investigation, the lines AA', BB', and CC' are concurrent and the point of concurrency is the centroid G of triangle ABC.
What happens if we begin with a scalene triangle ABC instead of an equilateral triangle ABC?
From observing the above construction, you can see that the lines AA', BB', and CC' are concurrent, but the point of concurrency is not any of the centers of the triangle ABC.
Now, we are going to see what happens to the point of concurrency when we construct isosceles triangles having a height equal to the side that it is constructed on.
As you can observe, the lines AA', BB', and CC' are concurrent, but their point of concurrency does not lie on Euler's line.
Now, lets investigate what happens to the point of concurrency when we construct equilateral triangles toward the center of the original triangle ABC. In this investigation, A', B', and C' are once
again the centroids of the equilateral triangles.
By observation, it is obvious that these triangles are not concurrent through the centroids of each of these triangles. They are also not concurrent to any of the other centers of the original
triangle. There is one exception to this. When the original triangle is an equilateral triangle, then they are all concurrent through the centroids of each of the triangles.
In the above construction, G is the centroid for all of the triangles.
One can continue further investigations of these figures. One can see if any of the triangles, angles, or sides are similar or congruent. For example, consider the following construction.
This construction consists of isosceles triangles constructed on each side of the triangle ABC. Since the external triangles are isosceles, the triangles AB'L and CB'L are congruent, the triangle
AC'N is congruent to BC'N, and triangle BA'M is congruent to CA'M. These are only a few of the extensions that one could investigate.
From the above investigations, we saw that the centers of triangles are related to the centers of other figures that are constructed externally to the original triangle. Sometimes the centers of the
original triangle are the point of concurrency for the lines that are are constructed from the center of the external figure to the opposite vertex of the original triangle and sometimes they are
Click here to go to Angela Head's EMT 668 Page | {"url":"http://jwilson.coe.uga.edu/EMT668/EMT668.Student.Folders/HeadAngela/writeup3/writeup3.html","timestamp":"2014-04-17T01:12:21Z","content_type":null,"content_length":"7624","record_id":"<urn:uuid:38530b43-de1a-47f4-81f1-81b6f6bb3d6e>","cc-path":"CC-MAIN-2014-15/segments/1397609526102.3/warc/CC-MAIN-20140416005206-00403-ip-10-147-4-33.ec2.internal.warc.gz"} |
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Here's the question you clicked on:
In the figure below, the following is true: ABD congruent to CDB and DBC congruent to BDA. How can you justify that ABD congruent to CDB?
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Did you mean to ask why one of the givens is true?
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no exactly what the question says
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hint : BD = BD by reflexive property
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these are the options SAS SSS ASA CPCTC
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you need to figure out based on congruent sides and angles
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ABD congruent to CDB do u understand this ?
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Your only options are SAS SSS ASA CPCTC. I would say CPCTC. We are talking about congruent triangles the whole time. You could prove SSS, but since it is only talking about the triangles being
congruent, the best answer would be CPCTC. Do you need more of an explanation because I'm not sure I can provide a good one, but I can try.
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is replying to Can someone tell me what button the professor is hitting...
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Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.
This is the testimonial you wrote.
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Noncommutative geometry
As I explained in a previous post, it is only because one drops commutativity that, in the calculus, variables with continuous range can coexist with variables with countable range. In the classical
formulation of variables, as maps from a set X to the real numbers, we saw above that discrete variables cannot coexist with continuous variables.
The uniqueness of the separable infinite dimensional Hilbert space cures that problem, and variables with continuous range coexist happily with variables with countable range, such as the
infinitesimal ones. The only new fact is that they do not commute.
One way to understand the transition from the commutative to the noncommutative is that in the latter case one needs to care about the ordering of the letters when one is writing.
As an example, use the "commutative rule" to simplify the following cryptic message I received from a friend :"Je suis alençonnais, et non alsacien. Si t'as besoin d'un conseil nana, je t'attends au
coin annales. Qui suis-je?"
It is Heisenberg who discovered that such care was needed when dealing with the coordinates on the phase space of microscopic systems.
At the philosophical level there is something quite satisfactory in the variability of the quantum mechanical observables. Usually when pressed to explain what is the cause of the variability in the
external world, the answer that comes naturally to the mind is just: the passing of time. But precisely the quantum world provides a more subtle answer since the reduction of the wave packet which
happens in any quantum measurement is nothing else but the replacement of a "q-number" by an actual number which is chosen among the elements in its spectrum. Thus there is an intrinsic variability
in the quantum world which is so far not reducible to anything classical. The results of observations are intrinsically variable quantities, and this to the point that their values cannot be
reproduced from one experiment to the next, but which, when taken altogether, form a q-number.
Heisenberg's discovery shows that the phase-space of microscopic systems is noncommutative inasmuch as the coordinates on that space no longer satisfy the commutative rule of ordinary algebra. This
example of the phase space can be regarded as the historic origin of noncommutative geometry. But what about spacetime itself ? We now show why it is a natural step to pass from a commutative
spacetime to a noncommutative one.
The full action of gravity coupled with matter admits a huge natural group of symmetries. The group of invariance for the Einstein-Hilbert action is the group of diffeomorphisms of the manifold and
the invariance of the action is simply the manifestation of its geometric nature. A diffeomorphism acts by permutations of the points so that points have no absolute meaning.
The full group of invariance of the action of gravity coupled with matter is however richer than the group of diffeomorphisms of the manifold since one needs to include something called ``the group
of gauge transformations" which physicists have identified as the symmetry of the matter part.
This is defined as the group of maps from the manifold to some fixed other group, G, called the `gauge group', which as far as we known is: G=U(1).SU(2).SU(3). The group of diffeomorphisms acts on
the group of gauge transformations by permutations of the points of the manifold and the full group of symmetries of the action is the semi-direct product of the two groups (in the same way, the
Poincaré group which is the invariance group of special relativity, is the semi-direct product of the group of translations by the group of Lorentz transformations). In particular it is not a simple
group (a simple group is one which cannot be decomposed into smaller pieces, a bit like a prime number cannot be factorized into a product of smaller numbers) but is a ``composite" and contains a
huge normal subgroup.
Now that we know the invariance group of the action, it is natural to try and find a space X whose group of diffeomorphisms is simply that group, so that we could hope to interpret the full action as
pure gravity on X. This is the old Kaluza-Klein idea. Unfortunately this search is bound to fail if one looks for an ordinary manifold since by a mathematical result, the connected component of the
identity in the group of diffeomorphisms is always a simple group, excluding a semi-direct product structure as that of the above invariance group of the full action of gravity coupled with matter.
But noncommutative spaces of the simplest kind readily give the answer, modulo a few subtle points. To understand what happens note that for ordinary manifolds the algebraic object corresponding to a
diffeomorphism is just an automorphism of the algebra of coordinates i.e. a transformation of the coordinates that does not destroy their algebraic relations. When an involutive algebra A is not
commutative there is an easy way to construct automorphisms.
One takes a unitary element u of the algebra i.e. such that u u*=u*u=1. Using u one obtains an automorphism called inner, by the formula x -> uxu*.
Note that in the commutative case this formula just gives the identity automorphism (since one could then permute x and u*). Thus this construction is interesting only in the noncommutative case.
Moreover the inner automorphisms form a subgroup denoted Int(A) which is always a normal subgroup of the group of automorphisms of A.
In the simplest example, where we take for A the algebra of smooth maps from a manifold M to the algebra of matrices of complex numbers, one shows that the group Int(A) in that case is (locally)
isomorphic to the group of gauge transformations i.e. of smooth maps from M to the gauge group G= PSU(n) (quotient of SU(n) by its center). Moreover the relation between inner automorphisms and all
automorphisms becomes identical to the exact sequence governing the structure of the above invariance group of the full action of gravity coupled with matter.
It is quite striking that the terminology coming from physics: internal symmetries agrees so well with the mathematical one of inner automorphisms. In the general case only automorphisms that are
unitarily implemented in Hilbert space will be relevant but modulo this subtlety one can see at once from the above example the advantage of treating noncommutative spaces on the same footing as the
ordinary ones. The next step is to properly define the notion of metric for such spaces and we shall indulge, in the next post, in a short historical description of the evolution of the definition of
the ``unit of length" in physics. This will prepare the ground for the introduction to the spectral paradigm of noncommutative geometry.
14 comments:
Well... I want to know what the riddle means! Any hints as to where commutativity should be applied?
Dear Alain, thank you for this post.
I'm going to ask a question which is probably terribly naive and possibly a bit crazy. As I understand, classical Kaluza-Klein theory suffers form the drawback of instability of the extra compact
dimensions which would tend to shrink down to singularities. Now, correct me if i'm wrong, but the finite NC part A_F of the algebra C(M)\tensor A_F can be seen as the algebra of functions on a
virtual discrete space. Could this virtual space be interpreted as the remnant of a shrunk down ancestral continuous space now in the quantum regime ? If so, could the particular structure of A_F
relfect the topology of this continuous space ?
Guy on the street, just try to permute some letters and get 4 times a name which is not so difficult to guess.......what can you come up with starting with "non alsacien" for instance?
Dear Fabien
Your question is pertinent. The role of the finite space is now much better understood from the very recent papers with A. Chamseddine: "why the Standard Model" and "A dress for SM the beggar"
which are on the hep-th arXiv. My intention is to use this blog, this summer holidays, to explain their content in details, but one step at a time. So far I just wanted to explain why it is
natural to consider NC spacetimes and not be so dependent on the "point-set" commutative view of spaces. So even if one cannot exclude that the finite space F is, as you suggest, a "remnant of a
shrunk down ancestral continuous space" it will give us a lot more freedom to drop the dependence on the commutative view.
Dear Alain,
thank you for your inspiring post.
Let me briefly outline a recent understanding about gravity based on noncommutative geometry.
Recent developments from string theory imply that gravity may be emergent from gauge theories in noncommutative spacetime or large N gauge theories like as the AdS/CFT duality.
As your motto, geometry is emergent from (noncommutative) algebra. In deep spaces, algebra seems to be more fundamental than geometry. Geometry is simply a consequence of coarse-graining
approximations of an underlying algebra.
An important point is that noncommutative spacetime is also a basically (noncommutative) phase space.
So a noncommutative space endows the group of canonical transformations -symplectomorphism- which preserves the algebra of noncommutative spacetime. This symmetry is infinite dimensional,
therefore there is a radical enhancement of spacetime symmetry compared to commutative one. More interestingly, it turns out that this symplectomorphism can be identified with a gauge symmetry in
NC spacetime. This fact already implies that the transition from the commutative to the noncommutative spacetime goes with a radical change of physics since there is an infinite dimensional
Since the inner automorphism (its infinitesimal version) in noncommutative spacetime as you mentioned in your post defines a (generalized) derivation, as a result, gauge fields in noncommutative
spacetime actually define vector fields on some manifold M (as an emergent geometry) and these vector fields form an orthonomal basis of M and so define a metric on M.
A whole point for the emergent gravity is due to the Darboux theorem (as noticed above, noncommutative spacetime is a basically phase space). The curvature F=dA of U(1) bundle - electromagnetic
fields - in noncommutative spacetime appears as a deformation of the original symplectic structure of background noncommutative spacetime. The Darboux theorem says that this dynamical deformation
of symplectic structure in terms of gauge fields can always be (locally) translated into coordinate transformations, i.e., diffeomorphisms. In a sense, the Darboux theorem in noncommutative
spacetime precisely plays a role of equivalence principle in Einstein relativity.
One can make this argument be more precise in the context of deformation quantization a la Kontsevich. The Darboux theorem appears as an automorphism of (noncommutative) C*-algebra which can be
identified with the diffeomorphism between gauge equivalent star products.
Further musing about noncommutative field theory may lead to beautiful and surprising picutres about Einstein gravity as an emergent phenomenon from noncommuative spacetime.
Hyun Seok
Your comment on the equivalence principle is very interesting. Could you possibly provide us with references?
This is a question for Hyun Seok. In your comment you mention NC `phase space. ' Do you know any precise definition of a NC phase space, e.g. a NC symplectic manifold? Can anyone comment on this?
Dear anonymous,
A famous example of NC phase space is quantum mechanics. Quantum mechanics is by definition the formulation of mechanics in "NC phase space". So a paricle phase space in quantum mechanics is an
example of NC symplectic manifold.
Dear kea,
unfortunately, this picture has not been condensed yet into a compact form although it is ubiquitous in recent string theory papers. But you may consult the following papers: arXiv:0704.0929
[hep-th]; hep-th/0612231; hep-th/0611174. Sorry, they are mine -.-.
Dear anonymous and Hyun Seok. Here is what I think about your question and answer. In Connes' notion of `spectral triple' we have a remarkable extension of the idea of spin Riemannian manifolds
to a noncommutative setting. It is based on a non-trivial spectral realization of the distance in Riemannian geometry using the Dirac operator. The definition of a NC symplectic manifold,
whatever it is, would similarly be based on a spectral realization of some key features of symplectic manifolds. I am not sure what it is, but a simple minded approach, e.g. by trying to mimic
the definition of an alternating 2-form on the tangent bundle I don't think would take us anywhere here.
I believe that the extension of the "symplectic" framework to the NC world is simply the notion of the first order term in a deformation of the NC-algebra. This is quite clear in the commutative
case where a symplectic structure (or more generally Poisson structure) is just the first term in the expansion of the deformed product. Thus it is a semi-classical form of the deformation. In
the NC case there are many examples where it is natural to use a similar starting point for deformations (for instance in Rankin-Cohen brackets generalized to deformations of NC projective
About the finite space of the Standard Model, and while we wait to hear more from it in this blog, I have been reading a bit more on the history of extra dimensional spaces of string theory.
Gliozzi-Scherk-Olive, in 1977 "Supersymmetry, Supergravity theories and the dual spinor model" stress their need of Majorana-Weil spinors and then "The requeriment ... gives thus the following
condition on D: D=2[mod8]. The first non trivial dimension (2 is somewhat trivial) is thus D=10, which is a nice way to recover the critical dimension of the NSR model"
This is to be compared with Green-Schwarz 1984, "Covariant description of superstrings", which focuses in triality and then gives a non periodic analisis: "This is precisely the identity that
arises in the proof of supersymmetry for super Yang-Mills theories. Therefore, at the classical level, the possibilities for superstring actions are in one to one correspondence with super
Yang-Mill theories: they are D=10 with Majorana-Weyl spinors, D=6 with Weyl..., D=4 ... Majorana, and D=3 Majorana, as well as their dimensionally reduced forms. However, quantum mechanically
only the D=10 theory is consistent, unless a Polyakov-type interpretation is possible in the other cases". This view of super, but non-quantum, strings was reinforced with the advent of branes
(d-dimensional objects in D-dimensional space), where codimension (D-d) works more or less in the same way, generating four "ladders" of admissible branes. The equivalent result for super
yang-mils appears in 1977 (again!) by Brink-Schwarz-Scherk.
It is not straightforward (to me) how the usual critical dimension, due to quantization of the [super]string, relates to the above considerations. It is interesting to note that some researchers
developed the concept of "W3-gravity" and "W3-strings", where the corresponding dimensions were not k+2, but 3k+2 (for k=1,2,4,8): 5,8,14,26 and then it marks the bosonic critical dimension. Also
G. Sierra, in 1987, tried to see this jump as a move from "Jordan algebras" to "Freudenthal triple systems". Perhaps "stringers" moved away from mod 8 periodicity because there is not a clear
meaning available for the intermediate D=18. Moreover, the mod 8 game depends on fermions, and the other critical dimension, D=26, is supposedly a bosonic object.
Dear Alain,
you alluded in your text to Thurston's theorem about the simplicity of the identity component of diffeo groups. As far as I know, this works for closed manifolds. With a noncompact Lorentzian
manifold M as a base space, I guess one could still imagine to cook up a bundle X over M such that diff(X) is the semi-direct product of diff(M) with the SM gauge group G. Or is there a way to
prevent this also ?
I see that in the fourth comment above Alain Connes said that
"My intention is to use this blog, this summer holidays, to explain [the] content [of articles on the NCG standard model] in details".
Maybe these discussions haven't appeared on the blog so far, or did I miss them? (just asking)
Over at the n-category Café I am having a discussion with Jacques Distler on the noncommutative standard model. He raised a couple of questions that would be good to hear the answer to from an | {"url":"http://noncommutativegeometry.blogspot.com/2007/07/noncommutative-spacetime.html","timestamp":"2014-04-17T15:26:50Z","content_type":null,"content_length":"123458","record_id":"<urn:uuid:7a45b476-0791-4f38-969d-6cfa698d4d7b>","cc-path":"CC-MAIN-2014-15/segments/1397609530136.5/warc/CC-MAIN-20140416005210-00622-ip-10-147-4-33.ec2.internal.warc.gz"} |
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Chapter 5
Chapter 5: Hypothesis Testing
Related links:
This page contains links to online calculators that compute either critical values or p-values for commonly used distribution functions.
This site provides a very nice discussion of the central limit theorem. A quicktime movie clip illustrates how quickly a binomial distribution tends toward a bell-shaped normal distribution as
the number of drawings increase.
This java applet provides a graph of a normal distribution function. The user is able to change the mean and the standard deviation and see how the distribution shifts.
This page contains a description of the normal distribution function. Quicktime movies illustrate the effect of changing the mean and the variance of the distribution.
Galton's board is a simple mechanical device that illustrates how the sum of outcomes from a binomial distribution can be approximated by a normal density function. The Java applet on this page
provides a graphical depiction of the functioning of this device.
This page contains a description of the normal density function and a Java Applet that allows the user to view the effect of changes in the mean and variance of a normal distribution on the
position and shape of the density function. Two other applets illustrate the use of the normal density function to construct confidence intervals.
The first applet on this page allows the user to find areas within any region under the normal density curve. The second applet can be used to find the z-score corresponding to any given
probability value for a normal cumulative density function.
This page provides a Java applet that illustrates the relationship between a t-distribution and a standard normal distribution for alternative degrees of freedom. You can use this applet to see
how quickly the t-distribution converges to a standard normal distribution as the size of a sample rises. On this page you can also learn how a brewing company was responsible for providing us
with the use of t-statistics.
This page contains a collection of online java applets written by John Marden (supported by grants from the National Science Foundation and the Sloan Center for Asynchronous Learnings
Environments). The "Box Models" applet illustrates the concept of confidence intervals and the central limit theorem.
This online statistical text contains a solid discussion of statistics. While it becomes rather advanced fairly quickly, the introductory section provides a good discussion of many of the basic
statistical concepts discussed in this chapter.
This online regression package, created by Elmer G. Wiens, allows the user to estimate multiple regression models online (including models with parameter restrictions). T-statistics are provided
for parameter estimates.
This is an online hyperlinked statistics text that discusses most of the statistical concepts that you will need for this course.
John Kane - kane@oswego.edu
Department of Economics, SUNY-Oswego, Oswego, NY 13126 | {"url":"http://www.oswego.edu/~kane/econometrics/chap5.htm","timestamp":"2014-04-18T16:17:44Z","content_type":null,"content_length":"5451","record_id":"<urn:uuid:8aee11c4-6d56-4f8b-a05b-f89e0321747e>","cc-path":"CC-MAIN-2014-15/segments/1397609533957.14/warc/CC-MAIN-20140416005213-00215-ip-10-147-4-33.ec2.internal.warc.gz"} |
Feudal Warlord IV-[Winner:Shaneback]
Hello everybody
Congratulations to
for his great victory.
This is a Five round tournament on
Feudal War
Once games are sent you have 72 hours to join.After that your kicked and opponents advanced.(depending on reserves.)
Round 1:5 games with 5 people.First eliminated out.
Round 2:5 games with 4 people.First eliminated out.
Round 3:5 games with 3 people.First eliminated out.
Semi-Finals:5 games with 2 people.First eliminated out
Finals:1 game with five people.Winner champion.
*If you dont understand above its basically you playing one game a round
*Groups stick together till the finals
No question marks
Last edited by naxus on Tue Mar 15, 2011 7:31 pm, edited 6 times in total.
Haggis_McMutton wrote:2. Anyone else find it kind of funny that naxus is NK'd right after insisting that we're all paranoid?
Re: Feudal Warlord IV-[0/25]
i'll join but i have one question - how are you determining the groups ?
random, score, score on that map, signup order, ?
Re: Feudal Warlord IV-[1/25]
Its random
Haggis_McMutton wrote:2. Anyone else find it kind of funny that naxus is NK'd right after insisting that we're all paranoid?
Re: Feudal Warlord IV-[1/25]
We are what we repeatedly do. Excellence, therefore, is not an act but a habit.
Re: Feudal Warlord IV-[1/25]
In please
Re: Feudal Warlord IV-[1/25]
In please
Re: Feudal Warlord IV-[1/25]
added till here
Haggis_McMutton wrote:2. Anyone else find it kind of funny that naxus is NK'd right after insisting that we're all paranoid?
Re: Feudal Warlord IV-[10/25]
In please.
Re: Feudal Warlord IV-[10/25]
in please | {"url":"http://www.conquerclub.com/forum/viewtopic.php?p=2908442","timestamp":"2014-04-19T23:01:32Z","content_type":null,"content_length":"157134","record_id":"<urn:uuid:e3fd22cf-c957-4144-815c-015dff3eb40e>","cc-path":"CC-MAIN-2014-15/segments/1398223206147.1/warc/CC-MAIN-20140423032006-00610-ip-10-147-4-33.ec2.internal.warc.gz"} |
The Open Door Web Site : IB Physics : Measurements : The VERNIER Scale
Measurements : Appendix I
The VERNIER Scale
A Vernier scale is a small, moveable scale placed next to the main scale of a measuring instrument. It is named after its inventor, Pierre Vernier (1580 - 1637). It allows us to make measurements to
a precision of a small fraction of the smallest division on the main scale of the instrument. (In the first example below the "small fraction" is one tenth.) Vernier scales are found on many
instruments, for example, spectroscopes, supports for astronomical telescopes etc. One specific example, the Vernier calliper, is considered below.
Using a Vernier Scale
Figure 1 shows a Vernier scale reading zero. Notice that 10 divisions of the Vernier scale have the same length as 9 divisions of the main scale.
In the following examples we will assume that the smallest division on the main scale is 1mm so the divisions on the Vernier scale are 0·9mm each. The position of the zero of the Vernier scale tells
us the number of cm and mm in our measurement. For example, in figure 2, the reading is a little over 1·2cm.
To find a more precise reading, consider figure 3 (which is a magnified view of part of figure 2).
We are, in effect, trying to find the distance, x.
To find x, find the mark on the Vernier scale which most nearly coincides with a mark on the main scale. In figure 3 it is obviously the third mark.
Now, it is clear that ............x = d - d’
Remembering that each division on the main scale is 1mm and that each division on the Vernier scale is 0·9mm, we have:
x = 3mm - 3(0·9)mm = 3(0·1)mm
Therefore, the reading in the example is: 1·23cm
Similarly, if it had been, for example, the seventh mark on the Vernier scale which had been exactly opposite a mark on the main scale, the reading would be: 1·27cm
Hence, the level of precision of an instrument which has a Vernier scale depends on the difference between the size of the smallest division on the main scale and the size of the smallest division on
the Vernier scale.
In the example above, this difference is 0·1mm so measurements made using this instrument should be stated as: reading ±0·1mm.
Another instrument might have a scale like the one shown in figure 4.
Therefore, the precision is: 1mm - (49/50)mm = (1/50)mm = 0·02mm.
Results of measurements made using this instrument should therefore be stated as: reading ±0·02mm.
This principle is used in the Vernier calliper shown below.
The diagrams below illustrate how to use a Vernier calliper to measure:
A. the internal diameter of a hollow cylinder
B. the external dimensions of an object
C. the depth of a hole in a piece of metal.
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Donna Danna's Eclipse Earthquakes
Kevin Heckle (3 March 2011)
"Donna Danna's Eclipse Earthquakes"
I would agree that there certainly seems to be a correlation with lunar eclipses and earthquakes. If you envision our earth's crust as a thin liquid layer, it only stands to reason that the tidal
effects exerted on our oceans by the moon would also affect the earth's crust. From the data I sorted through involving the conjuction of Earth, Venus and the Sun, there definitely was a
correlation between the conjuction of EVS and the severity of earthquakes. An eclipse of the moon is really the same thing as earth and an inner planet conjunction (especially transits). Its just
that the moon is much closer and obvious. I would guess the reason there are not always earthquakes during an eclipse is that the moon makes the tidal cycle every day due to the rotation of the
earth and once a month due to its orbit. One month or 29.5305888 days is really the CONJUCTION period of the Earth, Moon and the Sun whereas the specific conjuction period of the Earth, Venus and
the Sun happens every 583.92 days. I would suppose that the more regular period of the moon would cause fault line disturbances to be more settled as those faults get shaken out by the tidal
effect every day or every month. The conjucntion of earth and an inner planet like Venus doesn't happen as often and it seems that whatever fault lines that tidal effect hits on isn't as settled
as the Eclipse earthquake, and consequently is either more violent or more coincidental.
Now factor in the morning star (Venus) and moon conjunction with the earth and the sun every 2923 days ( 8 years, 99 months, 13 orbits of Venus, 5 EVS conjuctions). What tidal effect does that
have on the earth's crust?
I've spoke of this before. What do the numbers in Daniel have in common? The answer is the earth year, lunar conjuction period and the Venus conjuction period.
1260+1290+1335 = 3885 1335-1290-1260= (-1215)
By the tropical earth year, every fifth 2919.601 day conjuction moves clockwise (OPPOSITE or negative the orbital direction of all the planets) through the Mazzaroth about 2.3 degrees. In
1215.0276 years, or (152) 2919.601 day Earth-Venus-Sun conjucntions, the event returns to nearly the same place in the Mazzaroth by tropical reckoning. At this point, the EVS conjucntion is one
lunar year ahead of the full 1216 years if Venus were slower. This averaged is 1215.5138 years. There are also (5 )- 243 year transit periods of Venus in 1215 years
Take the sum of Daniel's numbers 3885 divided by the difference 1215 results in the result is 3.1975. The vision in Daniel 12 of the 'time, times and a half' where he raises one arm to the sky
north and the other to the sky south is clearly a reference to the half year (solstice to solstice or equinox to equinox) of 182.62 days. 182.62 x 3.1975 = 583.92 days, just like the conjuction
period of Earth Venus and the sun.
Take the average of the amount of time the EVS conjunction is ahead at 1215.0276 years vs. 1216 years and it is 1215.5138.
Divide 3885 by this number 1215.5138 and the result is 3.1961792. Now add the tropical year of 365.2421896 days plus the lunar year (29.5305888 x 12) of 354.36705 days which is a total of
719.60923 days. Multiply 719.60923 x 3.1961792. What is the result? It is exactly 2300!!!! The other Daniel mystery number in chapter 8. " After 2300 mornings and evenings the temple is
cleansed". According to the Law, what day was the Temple to be cleansed? NISAN 1, which is the SPRING EQUINOX.
What does 2300 have to do with reconciling the equinox? 2300 mornings and evenings (the full average cycle of Morning Star rising to Morning Star rising is that 583.92 days). Ideally, the average
would be 584.38748 days. 5 conjunctions would equal exactly 8 tropical years. 2300 x 584.38748 ideal days equals exactly 3680 years. The equinox moves due to an effect of the sun and moon's
gravity and it is called precession of the equinox. IF 3680 years is one day of the 'great' week that it takes precession of the equinox to come full circle, then 3680 years x 7 (great days) is
25,760 years. This value is within a few years of NASA's best guess!!! A sabbath or heptad (7) of 2300 mornings and evenings will reconcile the equinox, returning it to its exact place as it had
been 25,760 years earlier!
Other interesting facts: there are (1290 + 1335) 2625 lunar years from the date of Daniel's Vision until the summer solstice in 2012; the average of 1260, 1290 and 1335 is 1295. The average of
1295 and 1215 (reverse difference) is 1255. A conjuction of Venus upon the Equinox will recur exactly on the equinox 1255 years later, repeatedly, even though the equinox precedes about 17.8
degrees each 1255 years!!!; the average between the tropical and sidereal year is about 365.25. Add it to the calendar year of 365 days which is 730.25. Multiply this by the ecclesiastical year
(by sabbaths or sevens) of 364 days and the result is 265,811 days, which is also a season (91 days) times the Octaeteris average (2921 days). Multiplied by a time, times and a half (3.5) it is
930,338 days or 2,547 solar years.
Signs in the sun, moon and stars causing Earthquakes and Floods in diverse places!!!
Kevin Heckle | {"url":"http://www.fivedoves.com/letters/march2011/kevinh33.htm","timestamp":"2014-04-18T16:37:18Z","content_type":null,"content_length":"6569","record_id":"<urn:uuid:92d08b63-adc6-4a3a-891b-177848f9f575>","cc-path":"CC-MAIN-2014-15/segments/1398223210034.18/warc/CC-MAIN-20140423032010-00335-ip-10-147-4-33.ec2.internal.warc.gz"} |
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Re: kinetic problem
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Re: kinetic problem
Long time no talk. Hope everything is going right for you.
> My student Hui Tan and I have encountered the problem when using kinetics
with phreeqc. After reaching eqbm after about 0.1 year, it oscillates
between zero and negative rates --- meaning feldspar is precipitating. I
suspect that this is a numerical problem. Do you have any ideas how we can
fix or trick it?
Specify -tol 1e-12 in the KINETICS definition. This tolerance reduces the
oscillations to the 1e-15 range. By default PHREEQC integrates the
equations to a precision of 1e-8 moles. In this case, you need a smaller
> We have math profs here who are expert. If you describe
clear enough what is the problem, they may be able to help devise some
numerical solutions.
It would be advantageous to use an implicit method for integrating the rate
equations. I worked on this once with an implicit Runge-Kutta method, but
did not finish. This approach would help with stiff equations where the
rates differ by orders of magnitude, like when you want one reaction to be
nearly equilibrium and others slow. You may run into problems with stiff
equations with your set of rates, I'm not sure.
David Parkhurst (dlpark@xxxxxxxx)
U.S. Geological Survey
Box 25046, MS 413
Denver Federal Center
Denver, CO 80225
Project web page: http://wwwbrr.cr.usgs.gov/projects/GWC_coupled
Project Home Page
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Please note that some U.S. Geological Survey (USGS) information accessed through this page may be preliminary in nature and presented prior to final review and approval by the Director of the USGS.
This information is provided with the understanding that it is not guaranteed to be correct or complete and conclusions drawn from such information are the sole responsibility of the user.
Any use of trade, product, or firm names in this publication is for descriptive purposes only and does not imply endorsement by the U.S. Government.
The URL of this page is: http://wwwbrr.cr.usgs.gov/projects/GWC_coupled/phreeqc/mail/msg00646.html
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c# station
When a computer boots, it “loads” the operating system. If you want to use a program, you must find it either on the Start menu or from its directory and take the necessary action to open it. Such a
program uses numbers, characters, meaningful words, pictures, graphics, etc, that are part of the program. As these things are numerous, so is the size of the program, and so is the length of time
needed to come up. Your job as a programmer is to create such programs and make them available to the computer, then to people who want to interact with the machine.
To write your programs, you will be using alphabetic letters that are a, b, c, d, e, f, g, h, I, j, k, l, m, n, o, p, q, r, s, t, v, w, x, y, z, A, B, C, D, E, F, G, H, I, J, K, L, M, N, O, P, Q, R,
S, T, U, V, W, X, Y, Z. You will also use numeric symbols 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. Additionally, you will use characters that are not easily readable but are part of the common language; they
are ` ~ ! @ # $ % ^ & * ( ) _ + - = : “ < > ; ‘ , . /. Some of these symbols are used in the C# language while some others are not. When creating your programs, you will be combining letters and/or
symbols to create English words or language instructions.
Some of the instructions you will give to the computer could consist of counting the number of oranges, converting water to soup, or making sure that a date occurs after January 15. After typing an
instruction, the compiler would translate it to machine language. The computer represents any of your instructions as a group of numbers. Even if you ask the computer to use an orange, it would
translate it into a set of numbers. As you give more instructions or create more words, the computer stores them in its memory using a certain amount of space for each instruction or each item you
There are three numeric systems that will be involved in your programs, with or without your intervention.
The Binary System
When dealing with assignments, the computer considers a piece of information to be true or to be false. To evaluate such a piece, it uses two symbols: 0 and 1. When a piece of information is true,
the computer gives it a value of 1; otherwise, its value is 0. Therefore, the system that the computer recognizes and uses is made of two symbols: 0 and 1. As the information in your computer is
greater than a simple piece, the computer combines 0s and 1s to produce all sorts of numbers. Examples of such numbers are 1, 100, 1011, or 1101111011. Therefore, because this technique uses only two
symbols, it is called the binary system.
When reading a binary number such as 1101, you should not pronounce "One Thousand One Hundred And 1", because such a reading is not accurate. Instead, you should pronounce 1 as One and 0 as zero or
o. 1101 should be pronounced One One Zero One, or One One o One.
The sequence of the symbols of the binary system depends on the number that needs to be represented.
The Decimal System
The numeric system that we are familiar with uses ten symbols that are 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. Each of these symbols is called a digit. Using a combination of these digits, you can display
numeric values of any kind, such as 240, 3826 or 234523. This system of representing numeric values is called the decimal system because it is based on 10 digits.
When a number starts with 0, a calculator or a computer ignores the 0. Consequently, 0248 is the same as 248; 030426 is the same as 30426. From now on, we will represent a numeric value in the
decimal system without starting with 0: this will reduce, if not eliminate, any confusion.
Decimal Values: 3849, 279, 917293, 39473
Non- Decimal Values: 0237, 0276382, k2783, R3273
The decimal system is said to use a base 10. This allows you to recognize and be able to read any number. The system works in increments of 0, 10, 100, 1000, 10000, and up. In the decimal system, 0
is 0*100 (= 0*1, which is 0); 1 is 1*100 (=1*1, which is 1); 2 is 2*100 (=2*1, which is 2), and 9 is 9*100 (= 9*1, which is 9). Between 10 and 99, a number is represented by left-digit * 101 +
right-digit * 100. For example, 32 = 3*101 + 2*100 = 3*10 + 2*1 = 30 + 2 = 32. In the same way, 85 = 8*101 + 5*100 = 8*10 + 5*1 = 80 + 5 = 85. Using the same logic, you can get any number in the
decimal system. Examples are:
2751 = 2*103 + 7*102 + 5*101 + 1*100 = 2*1000 + 7*100 + 5*10 + 1 = 2000 + 700 + 50 + 1 = 2751
67048 = 6*104 + 7*103 + 0*102 + 4*101 + 8*100 = 6*10000 + 7*1000+0*100+4*10+8*1 = 67048
Another way you can represent this is by using the following table:
etc Add 0 to the preceding value 1000000 100000 10000 1000 100 10 0
When these numbers get large, they become difficult to read; an example is 279174394327. To make this easier to read, you can separate each thousand fraction with a comma. Our number would become
279,174,394,327. You can do this only on paper, never in a program: the compiler would not understand the comma(s).
The Hexadecimal System
While the decimal system uses 10 digits (they are all numeric), the hexadecimal system uses sixteen (16) symbols to represent a number. Since the family of Latin languages consists of only 10 digits,
we cannot make up new ones. To compensate for this, the hexadecimal system uses alphabetic characters. After counting from 0 to 9, the system uses letters until it gets 16 different values. The
letters used are a, b, c, d, e, and f, or their uppercase equivalents A, B, C, D, E, and F. The hexadecimal system counts as follows: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, a, b, c, d, e, and f; or 0, 1, 2,
3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F. To produce a hexadecimal number, you use a combination of these sixteen symbols.
Examples of hexadecimal numbers are 293, 0, df, a37, c23b34, or ffed54. At first glance, the decimal representation of 8024 and the hexadecimal representation of 8024 are the same. Also, when you see
fed, is it a name of a federal agency or a hexadecimal number? Does CAB represent a taxi, a social organization, or a hexadecimal number?
From now on, to express the difference between a decimal number and a hexadecimal one, each hexadecimal number will start with 0x or 0X. The number will be followed by a valid hexadecimal
combination. The letter can be in uppercase or lowercase.
Legal Hexadecimals: 0x273, 0xfeaa, 0Xfe3, 0x35FD, 0x32F4e
Non-Hex Numbers: 0686, ffekj, 87fe6y, 312
There is also the octal system but we will not use it anywhere in our applications. | {"url":"http://csharp-station.blogspot.com/","timestamp":"2014-04-19T07:13:00Z","content_type":null,"content_length":"47926","record_id":"<urn:uuid:762e85aa-2710-4091-b946-20a733631ee7>","cc-path":"CC-MAIN-2014-15/segments/1398223202457.0/warc/CC-MAIN-20140423032002-00617-ip-10-147-4-33.ec2.internal.warc.gz"} |
2014 49ers Salaries, As of 3/24/14, The 49ers are $4,057,335 under the '14 Cap
Oct 28, 2009 at 5:43 PM
Originally posted by Jakemall:
This needs an update
i know i know i know.
working on it.
Not sure on Crabs exact Terms, like Bonus and how it was pro rated.
Oct 28, 2009 at 5:49 PM
Nov 1, 2009 at 3:29 PM
updated? i didnt know Crabtree played for free
Nov 5, 2009 at 1:49 PM
Originally posted by AB83Rules:
So we have 55 mil in cap space?
Nov 5, 2009 at 5:48 PM
Originally posted by BirdmanJr:
Originally posted by AB83Rules:
So we have 55 mil in cap space?
Yeah that def doesn't seem right to me. How could we have that much? Clements salary jumps up next year and we have given out quite a few extensions in the last year or so.
If that is true and I hope it is we need to extend Willis/Gore/V.Davis/A.Franklin like yesterday.
Dec 28, 2009 at 3:03 PM
Originally posted by Jakemall:
This needs an update
Dec 28, 2009 at 9:29 PM
Updated, made it official 20010 salary thread now.
Also has 2011 numbers. But no cap totals as there is no CBA, but if there is a cba for 2011, the Cap would likely be in the 148M range.
Jan 1, 2010 at 12:20 AM
so we seriously have 55M in cap space? After we get some extensions done with Willis/Gore/Davis/Franklin/Brooks/Goldston we really need to go out and get a couple top tier FA's on the OL/DB/OLB
Jan 7, 2010 at 7:48 PM
updated after recent signees of Zeiger, Boone, Burnett and Mitchell to future contracts.
Mar 5, 2010 at 11:15 PM
do you know what happens to Patrick Willis' contract if there is a lockout in 2011? And does that mean he has 1 or 2 more years left, I always get confused
Mar 7, 2010 at 4:46 PM
Originally posted by teeohh:
do you know what happens to Patrick Willis' contract if there is a lockout in 2011? And does that mean he has 1 or 2 more years left, I always get confused
If im correct the 2011 yr if a lockout become his 2012 salary, and is signed thru then.
Mar 29, 2010 at 9:06 AM
Originally posted by AB83Rules:
Originally posted by teeohh:
do you know what happens to Patrick Willis' contract if there is a lockout in 2011? And does that mean he has 1 or 2 more years left, I always get confused
If im correct the 2011 yr if a lockout become his 2012 salary, and is signed thru then.
As always great job ab83 your always on your game to keep us 100% informed!
Thanks bro!
Mar 29, 2010 at 9:59 AM
Originally posted by frankie:
Originally posted by AB83Rules:
Originally posted by teeohh:
do you know what happens to Patrick Willis' contract if there is a lockout in 2011? And does that mean he has 1 or 2 more years left, I always get confused
If im correct the 2011 yr if a lockout become his 2012 salary, and is signed thru then.
As always great job ab83 your always on your game to keep us 100% informed!
Thanks bro!
you got it bud, trying my best to keep everyone informed.
Mar 29, 2010 at 10:08 AM
Originally posted by AB83Rules:
Originally posted by frankie:
Originally posted by AB83Rules:
Originally posted by teeohh:
do you know what happens to Patrick Willis' contract if there is a lockout in 2011? And does that mean he has 1 or 2 more years left, I always get confused
If im correct the 2011 yr if a lockout become his 2012 salary, and is signed thru then.
As always great job ab83 your always on your game to keep us 100% informed!
Thanks bro!
you got it bud, trying my best to keep everyone informed.
And ya do a great job at it!
Apr 25, 2010 at 2:57 PM | {"url":"http://www.49erswebzone.com/forum/niners/98000-2014-49ers-salaries-4057335-under-cap/page10/","timestamp":"2014-04-18T11:09:16Z","content_type":null,"content_length":"47702","record_id":"<urn:uuid:3c53c42f-e58d-472f-b232-50e0e20d0956>","cc-path":"CC-MAIN-2014-15/segments/1397609533308.11/warc/CC-MAIN-20140416005213-00311-ip-10-147-4-33.ec2.internal.warc.gz"} |
Recent publications
• O. DeWolfe, S. Gubser, and C. Rosen, "Minding the gap in ${\cal N}=4$ super-Yang-Mills," arXiv:1312.7347 [hep-th] (2013).
• A. Ficnar, S. Gubser, and M. Gyulassy, "Shooting String Holography of Jet Quenching at RHIC and LHC," arXiv:1311.6160 [hep-ph] (2013).
• A. Ficnar and S. Gubser, "Finite momentum at string endpoints," arXiv:1306.6648 [hep-th] (2013).
• O. DeWolfe, S. Gubser, C. Rosen, and D. Teaney, "Heavy ions and string theory," arXiv:1304.7794 [hep-th] (2013).
• A. Chandran, A. Nanduri, S. Gubser, and S. Sondhi, "On equilibration and coarsening in the quantum $O(N)$ model at infinite $N$," Phys. Rev. B88, 024306, arXiv:1304.2402 [cond-mat.stat-mech]
• R. Caldwell and S. Gubser, "A Brief History of Curvature," Phys. Rev. D87, 063523, arXiv:1302.1201 [astro-ph.CO] (2013).
• S. Gubser, "Complex deformations of Bjorken flow," Phys. Rev. C87, 014909, arXiv:1210.4181 [hep-th] (2013).
• S. Gubser, "The gauge-string duality and heavy ion collisions," Foundations of Physics 43, 140-155, arXiv:1103.3636 [hep-th] (2013).
• O. DeWolfe, S. Gubser, and C. Rosen, "Fermi surfaces in ${\cal N}=4$ Super-Yang-Mills theory," Phys. Rev. D86, 106002, arXiv:1207.3352 [hep-th] (2012).
• S. Gubser and J. Ren, "Analytic fermionic Green's functions from holography," Phys. Rev. D86, 046004, arXiv:1204.6315 [hep-th] (2012).
• A. Chandran, A. Erez, S. Gubser, and S. Sondhi, "The Kibble-Zurek Problem: Universality and the Scaling Limit," Phys. Rev. B86, 064304, arXiv:1202.5277 [cond-mat.stat-mech] (2012).
• H. Bantilan, F. Pretorius, and S. Gubser, "Simulation of Asymptotically $AdS_5$ Spacetimes with a Generalized Harmonic Evolution Scheme," Phys. Rev. D85, 084038, arXiv:1201.2132 [hep-th] (2012).
• O. DeWolfe, S. Gubser, and C. Rosen, "Fermi Surfaces in Maximal Gauged Supergravity," Phys. Rev. Lett. 108, 251601, arXiv:1112.3036 [hep-th] (2012). | {"url":"http://www.princeton.edu/physics/research/high-energy-theory/gubser-group/publications/","timestamp":"2014-04-16T09:51:56Z","content_type":null,"content_length":"11047","record_id":"<urn:uuid:b301ce3d-21c1-447f-9b3f-d0a5376a0c42>","cc-path":"CC-MAIN-2014-15/segments/1397609521558.37/warc/CC-MAIN-20140416005201-00464-ip-10-147-4-33.ec2.internal.warc.gz"} |
Thermal Physics
Science Reasoning Center - Thermal Physics
There are two Thermal Physics passages that target students' science reasoning abilities. They are ...
This passage describes the phase changes that occur in a sample of matter as it is heated from a temperature below its melting point to a temperature above its boiling point. In addition to the two
paragraphs describing the state changes, a heating curve graph is included. Questions target a student's ability to use the model presented in the body of text to interpret the graph, to connect
information in the body of text to the graph, and to compare various points on the graph to one another in terms of the state of matter that is present and the process (state change or temperature
increase) that is occurring.
This passage describes in quantitative terms the linear expansion that materials undergo when heated. The passage includes an equation, a table of coefficients of expansion, and a graph. Questions
target a student's ability to use an equation to make predictions, to draw conclusions that are consistent with a model, to select points on a graph, to combine data from a table and a graph in order
to compare the expansion of a material under various circumstances, to extrapolate outside the range of values provided by a graph, and to combine data from a table and a graph to make predictions
about the amount of expansion a given length of material would undergo.
On the Resource CD:
A Resource CD is being prepared to assist classroom teachers in using this section of the website more effectively within their classrooms. The Physics Classroom hopes to release the CD during the
Fall semester of of 2014.
The Thermal Physics section of the CD will include four different passages, including these two passages. The passages are accompanied by questions, answers with thorough explanations, and a table
that associates each question with a specific science reasoning skill (or college readiness standard). Each passage includes considerably more questions than those presented here in the Science
Reasoning Center. In addition to PDF files, the information on the CD is also available as Microsoft Word files, allowing teachers the ability to pick and choose questions from the large bank of
questions associated with each passage.
More information about the CD can be found at the
Resource CD page | {"url":"http://www.physicsclassroom.com/reasoning/thermalphysics","timestamp":"2014-04-18T05:53:32Z","content_type":null,"content_length":"49799","record_id":"<urn:uuid:cc16e00e-6657-47ee-a590-0c60fb9d1615>","cc-path":"CC-MAIN-2014-15/segments/1398223207985.17/warc/CC-MAIN-20140423032007-00601-ip-10-147-4-33.ec2.internal.warc.gz"} |
t believe someone makes
Good ideas with bad execution, or good execution of what should be bad ideas - an analysis of inferior, off-beat or malfunctioning products, and how other people's failures can help us design better
I can’t believe someone makes… Numerist clocks
Enough with the faceless watches, you say (passim here, here, and here)? Well, check out this wall clock instead. One for very special mathematicians.
No more 9 o’clock, but 21 [(4)] o’clock. No more 6 o’clock, but 3! o’clock… See what we are doing here? Base 4 and factorials were never so much fun!
The crème de la calculation is ol’ midnight (or midday, as some know it): 3?1728.
Okay, whats quarter-past seven in numerist time? (see below for a full ‘cheat list’ for the hours).
Unusable, surely, but as with most items in this series, I’m glad someone does make this beauty! The Geek Clock will set you back $25 (there’s also a wrist watch version available – “It’s Half-Past a
Binomial Coefficient”). Check out the video:
“Cheat Sheet”
12 – A radical
1 – Legendre’s constant is a mathematical constant occurring in a formula conjectured by Adrien-Marie Legendre to capture the asymptotic behavior of the prime-counting function. Its value is now
known to be exactly 1.
2 – A joke in the math world: An infinite number of mathematicians walk into a bar. The first one orders a beer. The second orders half a beer. The third, a quarter of a beer. The bartender says,
“You’re all idiots,” and pours two beers.
3 – A unicode character XML “numeric character reference.”
4 – Modular arithmetic, also known as clock arithmetic, is a system of arithmetic for integers, where numbers “wrap around” after they reach a certain value. The modular multiplicative inverse of 2
(mod 7) is the integer /a/ such that 2*/a/ is congruent to 1 modulo 7.
5 – The Golden Mean…reworked a little.
6 – Three factorial (3*2*1=6)
7 – A repeating decimal that is proven to be exactly equal to 7 with Cauchy’s Convergence Test.
8 – Graphical representation of binary code.
9 – An example of a base-4 number, which uses the digits 0, 1, 2 and 3 to represent any real number.
10 – A Binomial Coefficient, also known as the choose function. 5 choose 2 is equal to 5! divided by (2!*(5-2)!)
11 – A hexadecimal, or base-16, number.
Previous I can’t believe posts:
* I can’t believe someone makes… Yet More USB nonsense
* I can’t believe someone makes… Coca Cola powered cell phones
* I can’t believe someone makes… Cassette Tape Ties
Related posts | {"url":"http://www.electronicsweekly.com/made-by-monkeys/i-cant-believe-someone-makes/i-cant-believe-someone-makes-n-2010-05/","timestamp":"2014-04-17T15:38:22Z","content_type":null,"content_length":"69719","record_id":"<urn:uuid:d0329733-ea65-4901-bd62-a170dcd44fde>","cc-path":"CC-MAIN-2014-15/segments/1398223206672.15/warc/CC-MAIN-20140423032006-00136-ip-10-147-4-33.ec2.internal.warc.gz"} |
Patent US6141669 - Pseudorandom binary sequence block shifter
This invention relates to linear feedback shift registers (LFSR's), and more particularly to methods and apparatus for rapidly determining the state of an LFSR at any point in the past relative to a
current state, and to synchronizing the LFSR's in base and mobile stations in a CDMA communication system.
LFSR's are used in many applications for generating pseudorandom numbers, which in turn may be used for such purposes as encryption, or synchronization of data transmissions.
A well known such use is in the CDMA (Code Division Multiple Access) scheme employed in the IS-95 standard for cellular telephone transmission. Both the base station and the mobile stations employ a
42-bit LFSR to generate a periodic code with period 2.sup.42 -1 bits (known to those in the cellular telephone art as "the Long Code"). For a base station to communicate with a mobile station, their
Long Codes must be synchronized. Due to range uncertainty (which affects transmission time) and randomized delays introduced to avoid multiple user traffic collisions, at a given time a mobile
station's Long Code and a base station's Long Code may be different, and thus require synchronization in order to communicate. (Typically, the base station's code is ahead of the mobile station's
Synchronization requires a "searcher" in the base station to determine the number of shifts (bit offsets) by which the base station's LFSR leads the mobile station's LFSR. The two LFSRs must then be
syunchronized. The base station's LFSR cannot simply be backward-shifted to accomplish this, because the feedback mechanisms are not susceptible of reversal.
One conventional method for accomplishing synchronization in the case where the mobile station's LFSR is ahead of the base station's is to run the base station's LFSR forward at a clock rate many
times (typically, 48 times) higher than normal until it has advanced to the desired state.
In the case where the base station's LFSR leads the mobile station's LFSR, the base station's LFSR is simply stopped until the mobile LFSR "catches up" to it. These conventional methods have the
disadvantage of requiring extra hardware for the high-speed forward shift, and the disadvantage of wasting time while the LFSR is stopped.
Another conventional method is to store the base station LFSR output over a sufficient number of bits (typically, 1400) to enable matching with the mobile station's current bit pattern. The obvious
drawback is that 1400 flip-flops are required, with their resultant increase in cost and power consumption.
Yet another conventional method is to use a mirror-image LFSR in addition to the main LFSR with feedback mechanisms such that it shifts backward through the reverse of the sequence through which the
main LFSR shifts forward. This has the obvious drawback of necessitating a great deal of additional circuitry.
While most conventional solutions are hardware-based, some recent algorithms have been posited for determining future states of LFSR's by mathematical methods susceptible of implementation in
software or firmware. See the work of Arthur H. M. Ross, Ph.D., at Internet web page http://www.cdj.org/a.sub.-- ross/LFSR.html (date unknown), or the work of M. Serra at http:/www.csr.uvic.ca/home/
mserra/CApaper/node4.html (Jun. 24, 1996). These algorithms, however, are not able to determine past states of LFSRs.
Accordingly, there exists a need for a method of determining a past state of a LFSR that does not require significant additional hardware and that does not cause the waste of significant amounts of
It is thus an object of the present invention to provide a method of resetting a LFSR that eliminates the drawbacks of conventional CDMA systems.
It is a further object of the present invention to provide a method for resetting a LFSR that has fixed overhead for shifting backward or forward independent of shift length.
It is a further object of the present invention to provide a practical method of performing very large shifts, which are impractical in conventional systems.
It is a further object of the present invention to provide a method of resetting a LFSR that may be easily embedded in firmware in an application-specific integrated circuit (ASIC) and ideally in
parallel operation where multiple shift positions are needed, as in multi-user detection.
These and other objects of the invention will become apparent to those skilled in the art from the following description thereof.
In accordance with the teachings of the present invention, these and other objects may be accomplished by the present systems and methods of generating spreading codes which are resistant to the
effects of time delays in CDMA systems. An embodiment of the present invention includes a method of determining an inverse transition matrix which when used to multiply in modulo-2 arithmetic the
current state of an LFSR yields the previous state of the LFSR, and multiplying the state of the LFSR by the inverse transition matrix in modulo-2 arithmetic once for each desired backward shift of
the LFSR, and loading the state thus determined back into the LFSR. Another embodiment of the invention uses apparatus to determine the inverse transition matrix, to multiply the LFSR contents by the
inverse transition matrix one for each backward shift required, and to load the result back into the LFSR.
The invention will next be described in connection with certain exemplary embodiments; however, it should be clear to those skilled in the art that various modifications, additions and subtractions
can be made without departing from the spirit or scope of the claims.
The invention will be more clearly understood by reference to the following detailed description of an exemplary embodiment in conjunction with the accompanying drawings, in which:
FIG. 1 depicts a generic 5-bit LFSR.
FIG. 2 depicts a Type I LFSR of arbitrary length.
FIG. 3 depicts a Type II LFSR of arbitrary length.
FIG. 4 illustrates the transition matrix for the LFSR of FIG. 2.
FIG. 5 illustrates the transition matrix for the LFSR of FIG. 3.
FIG. 6 illustrates the inverse transition matrix for the LFSR of FIG. 2.
FIG. 7 illustrates the inverse transition matrix for the LFSR of FIG. 3.
FIG. 8 illustrates the transition matrix for a 42-bit CDMA LFSR.
FIG. 9 illustrates the inverse transition matrix for a 42-bit CDMA LFSR.
FIG. 10 depicts a basic operation of modulo-2 matrix multiplication.
A preferred embodiment of the present invention is associated with the 42-bit LFSR employed for producing the Long Code used for synchronizing data transmission between a base station and a mobile
station in an IS-95 CDMA cellular communications system.
To service a mobile user, the base station's LFSR must be set to the same value as the mobile user's. A searcher algorithm in the base station determines the amount of time difference between the two
LFSR's. Time alignment of the base station Long Code generator LFSR includes resetting the binary value stored in the shift register to a value of a previous or future time, according to the searcher
An example of an embodiment of the present invention will now be considered using a five-bit LFSR, an example of which is shown in FIG. 1. Those in the art will recognize this as a Type I, or Galois,
LFSR since the exclusive OR is located in a series path (as opposed to a Type II, or Fibonacci, LFSR which would have the exclusive OR in the feedback paths). Although the ensuing discussion applies
to the Type I LFSR, the principles apply equally to the Type II LFSR.
The state transition equations for the LFSR of FIG. 1 are
z.sub.1 '=z.sub.5
z.sub.2 '=z.sub.1
z.sub.3 '=(z.sub.2 XOR z.sub.5)
z.sub.4 '=z.sub.3
z.sub.5 '=z.sub.4
where z.sub.i is the present state of cell i and z.sub.i ' is the state after clocking the shift register one time. In matrix form, this can be expressed as:
[z.sub.1 ', z.sub.2 ', z.sub.3 ', z.sub.4 ', z.sub.5 '].sup.T =A X [z.sub.1, z.sub.2, z.sub.3, z.sub.4, z.sub.5 ].sup.T
where all operations are carried out in modulo 2 arithmetic, T indicates transpose and A is the transition matrix: ##EQU1## Because of the modulo-2 arithmetic that must be employed, the inverse
transition matrix can not be determined from the transition matrix by the conventional means used in decimal arithmetic for determining an inverse matrix. This becomes evident by taking a
conventional inverse transition matrix, multiplying it by the transition matrix, and finding that the result is not the identity matrix.
The inverse transition matrix can be determined empirically by determining what is necessary to transit the LFSR back by one shift. For the LFSR of FIG. 1, the inverse transition matrix is: ##EQU2##
Obtaining the next state of the LFSR using the transition matrix could be written as: ##EQU3## Similarly, obtaining the previous state (one clock pulse ago) can now be written as: ##EQU4## Block
jumps (jumps of more than one clock pulse) may easily be obtained by raising the transition matrix to a power equal to the desired number of clock pulses prior to performing the multiplication. That
is: ##EQU5## yields the state of the LFSR N clock pulses in the future, while ##EQU6## yields the state of the LFSR N clock pulses in the past.
It will now be shown how to find the inverse transition matrix for a Type I or Type II LFSR of arbitrary length >2.
A Type I (Galois) LFSR of arbitrary length is shown in FIG. 2. A Type II (Fibonacci) LFSR of arbitrary length is shown in FIG. 3. Generally, the number of taps and tap positions are not the same for
Galois and Fibonacci versions. The original state of each of the LFSR is the binary sequence {S.sub.k }.sub.k for k=1, . . . , N. The set of state transition equation for Type 1 (Galois) LFSR is in
general, as follows: ##EQU7## where k, . . . , m are positive integers less than N, corresponding to tap positions in the Type 1 Galois LFSR. Note that "+" is the Exclusive--OR or modulo two
operation The transition matrix representing these state transition equations is given in FIG. 4.
The set of state transition equations for Type 2 (Fibonacci) LFSR is in general, as follows: ##EQU8## where h, . . . , j are positive integers less than N.
The state transition matrix, G, for one step forward in the Galois LFSR is shown in FIG. 4. The N LFSR) matrix has a subdiagonal of ones, and ones in the Nth column corresponding to tap positions.
All other elements are zero.
The state transition matrix, F, for one step forward in the Fibonacci LFSR is shown in FIG. 5. The N ones in the first row corresponding to tap positions. All other elements are zero.
Determination of inverse transition matrix for generalized Type I:
The problem of determining G.sup.-1, the inverse of the Galois transition matrix is equivalent to finding the permutations needed to reverse the corresponding state equations.
The matrix G maps the sequence {S.sub.k }.sub.k into the next step {S.sub.k.sup.+ }.sub.k.
Substituting the S.sub.k.sup.+ with the equivalent S.sub.k terms, the one-step inverse transition matrix operations can be shown as follows: ##EQU9##
The first row of G.sup.-1 is all zeros except for a one in column 2 to map S.sub.1 to the first position in the present state vector, on the right hand side. The second row of G.sup.-1 is all zeros
except for a one in column 3 to map S.sub.2 to the second position in the present state vector, on the right-hand side:
Row 1 of G.sup.-1 =[0 1.sub.1,2 0 0 . . . 0]
Row 2 of G.sup.-1 =[0 0 1.sub.2,3 0 . . . 0]
Row m of G.sup.-1 =[0 . . . 1.sub.m,m+ . . . ], for 1≦m ≦N
Row k of G.sup.-1 =[1.sub.k,1 0 . . . 1.sub.k,k+1 . . . 0], since S.sub.1.sup.+ =S.sub.N and S.sub.N +(S.sub.K +S.sub.N)=S.sub.K
Row k+1 of G.sup.-1 =[0 . . . 1.sub.k+1,k+2 . . . 0]
This is true for all k, for 1<k<N, where each k corresponds to a tap in the LFSR.
The resulting structure of the inverse transition matrix G.sup.-1. for a Type I (Galois) LFSR is given in FIG. 6. Then, G.sup.-1 *G=N is the identity matrix of rank N. The product is obtained using
modulo 2 arithmetic.
Determination of inverse transition matrix for generalized Type II:
The problem of determining F.sup.-1, the inverse of the Fibonacci transition matrix is equivalent to finding the permutations needed to reverse the corresponding state equations.
The matrix F maps the sequence {S.sub.k }.sub.k into the next step {S.sub.k.sup.+ }.sub.k.
Substituting the S.sub.k.sup.+ with the equivalent S.sub.k terms, the one-step inverse transition matrix operations can be shown as follows: ##EQU10##
The first row of F.sup.-1 is all zeros except for a one in column 2 to map S.sub.1 to the first position in the present state vector, on the right hand side. The second row of F.sup.-1 is all zeros
except for a one in column 3 to map S.sub.2 to the second position in the present state vector, on the right-hand side.
Row 1 of F.sup.-1 =[0 1.sub.1,2 0 0 . . . 0]
Row 2 of F.sup.-1 =[0 0 1.sub.2,3 0 . . . 0]
In general the pattern is
Row w of F.sup.-1 =[0 . . . 1.sub.w,w+1 . . . ], for 1≦w≦N, and
Row N of F.sup.-1 =[1.sub.N,1 0 . . . 1.sub.N,h . . . 1.sub.N,j . . . 0]
Row k+1 of G.sup.-1 =[0 . . . 1.sub.k+1,k+2 . . . 0] for 1<h<j<N
The resulting structure of the inverse transition matrix F.sup.-1 for a Type II (Fibonacci) LFSR is given in FIG. 7. Then, F.sup.-1 *F=I.sub.N obtained using modulo 2 arithmetic.
Application to the IS-95 CDMA Long Code LFSR:
The forward transition matrix A for the IS-95 Long Code is given in FIG. 8. A.sup.-1, the inverse of A for shifting backwards in time is shown in FIG. 9. By direct multiplication using modulo 2
arithmetic, A*A.sup.-1 =I.sub.42
Considering the one-step forward case, the shifting and XOR operation can be emulated as a matrix operation:
S.sub.1 =A
S.sub.1 =42-bit contents of LFSR after one clock period,
S.sub.0 =initial contents of LFSR
A is the transition matrix of FIG. 8.
Similarly, the one-step backward case can be emulated by:
S.sub.-1 =A.sup.-1
The 42-bit contents of the LFSR can be shifted up to 2.sup.42 -1 clock pulses in a single matrix operation. For example, had the value of
A.sup.7 =A*A*A*A*A*A*A
been precalculated and prestored, emulation of shifting 7 clock pulses into the future could be performed as:
S.sub.7 =A.sup.7
It may be beneficial to precalculate and prestore such matrices, such as power-of-two numbers of shifts:
A.sup.2, A.sup.4, A.sup.8, A.sup.16, . . . A.sup.-2, A.sup.-4, A.sup.-8, A.sup.-16
A shift emulation of some arbitrary number of clock pulses can then be performed by a few well-chosen matrix operations in succession. While an example involving powers of two has been considered, it
will be apparent to those skilled in the art that any power can be used and that matrices of arbitrary powers can be stored.
In a similar vein, with some hardware implementations of an LFSR, for shifts of fewer than 42 clock pulses it may be faster to actually shift the LFSR than to perform the matrix multiplies. In such
cases, shifts by large numbers of clock pulses may be speeded up by calculating and prestoring values for multiples of 42 shifts:
A.sup.42, A.sup.84, A.sup.126, A.sup.168, . . . A.sup.-42, A.sup.-84, A.sup.-126, A.sup.-168
Shifting can then be emulated by matrix mathematics unless or until there are fewer than 42 forward shifts remaining to be performed, and they can then be performed by actual shifting. The number 42
has been chosen here since that is the number of bits in the LFSR under the IS-95 standard, but other numbers may be chosen and still fall within the scope of the invention.
The invention may be practiced in host hardware equipped with computational capability, such as a minicomputer or microcomputer, in which a program may be employed to determine the inverse transition
matrix and to perform matrix multiplications with it so as to obtain past states of an LFSR. Firmware may also be employed, enabling practicing the invention in, for example, an ASIC
(application-specific integrated circuit).
Modulo-2 matrix multiplication can be speeded up using parity checker hardware instead of actually performing a row-by-column multiplication. FIG. 10 depicts the multiplication of the PN vector by a
row of a transition matrix. This process must be executed for each row of the transition matrix. In modulo-2 arithmetic, the process is equivalent to:
overlaying the row and PN vector;
ANDing the corresponding bits; and
summing the resultant 1's in modulo-2 addition.
In modulo-2 arithmetic, the latter step can yield a result of either zero or one, and zero if the number of 1's is even, and one if the number of 1's is odd. This determination of whether the number
of 1's is even or odd is the function performed by a parity checker. If host hardware includes a parity checker it can be invoked to assist in the matrix multiply, saving 42 row-by-column
multiplications each time.
It will thus be seen that the invention efficiently attains the objects set forth above, among those made apparent from the preceding description. In particular, the invention provides rapid
determination of a past state of a linear feedback shift register without significant additional hardware.
It will be understood that changes may be made in the above construction and in the foregoing sequences of operation without departing from the scope of the invention. It is accordingly intended that
all matter contained in the above description or shown in the accompanying drawings be interpreted as illustrative rather than in a limiting sense.
It is also to be understood that the following claims are intended to cover all of the generic and specific features of the invention as described herein, and all statements of the scope of the
invention which, as a matter of language, might be said to fall therebetween. | {"url":"http://www.google.co.uk/patents/US6141669","timestamp":"2014-04-20T13:32:39Z","content_type":null,"content_length":"98256","record_id":"<urn:uuid:4eadcac2-cc58-46e2-88f4-0242d51dbd5f>","cc-path":"CC-MAIN-2014-15/segments/1398223202774.3/warc/CC-MAIN-20140423032002-00451-ip-10-147-4-33.ec2.internal.warc.gz"} |
Full Load Amps - OnlineConversion Forums
Originally Posted by twinspa
I'm designing the electrical feed to a well. I need to size the transformer & cable feeding the main switchboard based on the KVA. I also need to run my voltage drop calcs based on the KW.
I have a single line consisting of multiple pumps, a 100A panel for misc power needs (lighting, recaptacles, etc...) & a unit heater.
I have the HP of the pumps & a panel schedule for the 100A panel. But the only info I have for the heater is 19 FLA. I don't know the BTU's or even the model number to look up more info.
Is there any way to determing the KVA or KW based solely on FLA? Or will I need to keep digging for more info?
The only other thing you need to know is the rated voltage; I don't know if you are using 115 V, 230 V or some other voltage rated components. Power = Voltage * Amperage, or P = V*I.
However, for ac circuits, there is an additional complication. The voltage and the current may not be exactly in phase; this is called "power factor." If the electrical phase angle between current
and voltage is theta (I can't make the Greek letter here), then P = V*I*cos (theta). But for many purposes, the product of volts and amps is used, and called VA. If divided by 1000, then kVA. NOTE:
On a heater, current and voltage should be in phase, Power factor mostly applies to motor loads.
Actually just the current, and the resistance of the wires suffices to calculate the voltage drop, but what is allowed may depend on the nominal voltage. | {"url":"http://forum.onlineconversion.com/showthread.php?t=6284","timestamp":"2014-04-18T08:03:36Z","content_type":null,"content_length":"46588","record_id":"<urn:uuid:6235bc24-7cad-416d-92a3-22b58f320df6>","cc-path":"CC-MAIN-2014-15/segments/1398223206647.11/warc/CC-MAIN-20140423032006-00154-ip-10-147-4-33.ec2.internal.warc.gz"} |
Mean Value Theorem Question
1. The problem statement, all variables and given/known data
Let f(x)=log(x)+sin(x) on the positive real line. Use the mean value theorem to assure that for all M>0, there exists positive numbers a and b such that f(b)-f(a)/b-a=M
2. Relevant equations
3. The attempt at a solution
I know that as x→0, f'(x) gets arbitrarily large and as x→∞, f'(x) gets arbitrarily small, so for every M>0 there exists a and b such that f'(a)-M<0 and f'(b)-M>0. f'(x)-M would then have a root by
the intermediate value theorem. From here, I don't know where to go. I know that as h→0, f(r+h)-f(r)/h→f'(r), where r is the root of f'(x)-M, so as h→0, the intervals (r,r+h) are such that f(b)-f(a)/
b-a gets arbitrarily close to M by the Mean Value Theorem, but I don't know where to go. Any ideas?
I don't know if MVT is necessarily the best way to prove this. In fact, I think you need a converse MVT to prove it, and I don't think the converse of the MVT holds in general.
This is far more simply proved using IVT (Intermediate Value Theorem). First observe that [itex]\lim_{x \rightarrow 0} f(x) = -\infty[/itex]. Also observe that there is a local maximum value for f(x)
around x = 2. Proving this is easy by observing the behaviour of f'(x) in the neighborhood of x = 2. Calculate f'(2) and f'(2.1) and apply IVT to f'(x) to prove that there has to be a stationary
point between these two x-values, which is immediately seen to be a local maximum. Call the x-value at that local maximum [itex]x_0[/itex].
Now, consider secant lines originating from x in the open interval (0,x
). Clearly, the secant lines have gradients in the open interval (0,∞). Since the curve y = f(x) is continous, the IVT guarantees that you can always choose a positive x value less than x
to give you any positive gradient for the secant terminating at x
. Hence, simply letting b = x
and letting a be a particular positive x smaller than x
will suffice to complete the proof. | {"url":"http://www.physicsforums.com/showthread.php?t=617843","timestamp":"2014-04-18T03:15:46Z","content_type":null,"content_length":"51200","record_id":"<urn:uuid:76906d2d-e2b4-4e16-9bf8-a82121a375f0>","cc-path":"CC-MAIN-2014-15/segments/1397609532480.36/warc/CC-MAIN-20140416005212-00559-ip-10-147-4-33.ec2.internal.warc.gz"} |
MathGroup Archive: November 2007 [00252]
[Date Index] [Thread Index] [Author Index]
Re: Re: Can you get a package back to a notebook easily?
• To: mathgroup at smc.vnet.net
• Subject: [mg83074] Re: [mg83032] Re: Can you get a package back to a notebook easily?
• From: Murray Eisenberg <murray at math.umass.edu>
• Date: Fri, 9 Nov 2007 05:20:04 -0500 (EST)
• Organization: Mathematics & Statistics, Univ. of Mass./Amherst
• References: <fgen3e$flh$1@smc.vnet.net> <fgs7k8$3gb$1@smc.vnet.net> <200711081108.GAA26985@smc.vnet.net>
• Reply-to: murray at math.umass.edu
1. Converting notebook to a package.
This assumes you've written code in the notebook that includes the
appropriate extra stuff for good package functionality...
ThisFunc::usage="describe thisFunc";
ThatFunc::usage="How to use thatFunc";
(* definitions of objects here *)
All these cells (except possibly comments), but not text cells, should
be made, by the Cell > Properties menu item, into Initialization Cells
that are Active.
Save the notebook with name pkgName.nb. When asked if you want to
automatically save this as a package, too, answer yes. Automatically,
wherever the notebook MyNB.nb is saved, you'll get the corresponding
pkgName.m package file in the same directory.
2. Move (or copy) both the .nb file and the .m file into a subdirectory,
named topName, of InstallationDir\AddOns\Applications\ (where
InstallationDir refers to the top-level directory of the Mathematica
tree -- the value of the Mathematica system variable $InstallationDirectory.
That's it.
When you want to revise the package, revise the "auto-save" notebook you
thereby created in step 1, and when you save the notebook, it will
automatically update the package accordingly.
3. To use the package,
Re Step 2, of course you can keep a copy of the original saved notebook
and auto-saved package .m file elsewhere and work on revisions there
before moving the new versions as explained in Step 2.
This is just a brief (and, I hope, entirely correct) outline of a
process that has many variants, e.g., with respect to where you may put
packages, or how to make packages load automatically whenever you open
AES wrote:
> In article <fgs7k8$3gb$1 at smc.vnet.net>,
> Szabolcs Horv=E1t <szhorvat at gmail.com> wrote:
>> David Reiss wrote:
>>> Actually, in Version 6+ the answer is no.
>>> The initialization cells are saved, and some other classes of cells
>>> are contained as comments with special information that allows them to
>>> still be viewed when the .m file opened in Mathematica.
>>> Try it out and then open the .m file in a text editor to see this...
>> Can you give an example? Before I replied, I tried it with a notebook
>> that only had input cells and text cells. Only the initialization input
>> cells were saved.
>> Szabolcs
> Or better, after developing and testing in notebook form a lengthy
> module that I want to convert to package form, I'd like to have a
> specific set of instructions on how to:
> * Convert the notebook to a package
> * Save/install the package [To where? And how?]
> * Save/preserve the notebook also [Where? And how?]
> so that I can when needed, edit or modify the original notebook and
> repeat the process.
> The problem is not so much "how to write a package"; it's finding, all
> in one place, the above specific instructions on how to manage it
Murray Eisenberg murray at math.umass.edu
Mathematics & Statistics Dept.
Lederle Graduate Research Tower phone 413 549-1020 (H)
University of Massachusetts 413 545-2859 (W)
710 North Pleasant Street fax 413 545-1801
Amherst, MA 01003-9305
• References: | {"url":"http://forums.wolfram.com/mathgroup/archive/2007/Nov/msg00252.html","timestamp":"2014-04-18T21:15:29Z","content_type":null,"content_length":"29321","record_id":"<urn:uuid:d7502f16-d12d-43fa-a41e-5a0f1f9379e2>","cc-path":"CC-MAIN-2014-15/segments/1398223203422.8/warc/CC-MAIN-20140423032003-00151-ip-10-147-4-33.ec2.internal.warc.gz"} |
math - Can someone please help to solve
Number of results: 314,528
someone please help please help don't understand solve each literal equation for the indicated variable. (solve for s) d=c-s ______ n
Saturday, December 8, 2007 at 12:07am by allen
someone please help i hav posted my questions like 5 times and no one is answering, can someone please solve these inequalitys 6y+1<19 2r-8>6 b-5>-2 2y+1<-5 4x-6>-10
Tuesday, March 18, 2008 at 7:48pm by Julie
Algebra......Help Quick!!
12a^3bc-18a^2bc^2+6a^2c, x^2+20x+36, 12x^4-8x^3-4x^2, x^2+36x, 9x^2+6x+1, 6x^3+11x-10, x^2+3x-6..............can someone please help me solve some of these problems if not all. I have forgotten how
to solve them can someone please help..Thanks
Tuesday, September 2, 2008 at 8:22pm by Carrie
Will someone please help me solve this problem. Solve and Check 3x^2=2x=5
Sunday, April 21, 2013 at 3:19pm by timothy
actually, in a addition to the question i posted a few minutes ago, can someone please help me with a few actual specific questions? thanks so much i'm very confused and i've been working on these
all day. *Literal Equations* ~solve for x ax + bx = 8ab ~solve for y my - ny = m...
Friday, October 26, 2007 at 2:58pm by kim
Solve for x: x^4-15X^2+14=0 How would I solve this problem? Could someone please show me the steps?
Tuesday, January 10, 2012 at 9:32pm by Emma
8th grade science
Can someone please help me solve this question please because I'm having trouble with this question and can someone please explain to me how they got this answer? Thank you :D
Wednesday, November 13, 2013 at 9:59pm by gina
Algebra 1
Can someone help me solve this problems please? Thank you very much 7x+9>1x+3= Can someone help me undertand the steps to solving this problem, please.
Monday, January 3, 2011 at 4:12pm by Erudita
6th grade math
can someone please help me with this question?And can someone also teach me step by step how to solve this question???Please???
Monday, February 13, 2012 at 5:19pm by Kate
solve for indicated letter. f=2g, for g what is the solution for g? can someone please tell me how to start this? or how to solve this question?? I would appreciate it much
Wednesday, November 11, 2009 at 1:37pm by Kylie
I am having trouble figuring out how to solve these logarithms. Could someone please help! log2(log4x)=1 and solve for x and y: (1/2)^x+y= 16 logx-y8=-3
Sunday, January 23, 2011 at 10:58pm by Chloe
I am having trouble figuring out how to solve these logarithms. Could someone please help! log2(log4x)=1 and solve for x and y: (1/2)^x+y= 16 logx-y8=-3
Sunday, January 23, 2011 at 11:20pm by Chloe
Physics- can someone Please help?
I'm asking nicely. =D (I can almost solve the problem if only someone could tell me what do I put into the force equation for q2 -pretty frustrating that I can "almost" solve it but I need just one
Wednesday, April 2, 2008 at 12:33am by ~christina~
Can someone please help with this problem? Solve by the elimination method. 7r-9s= -58 9r+7s=74 I know that I need to multiply the first equation by 9 and the second equation by 7. The equations look
like this: 63r-81s= -522 63r+49s=518 After this I get confused. Could someone...
Saturday, April 17, 2010 at 12:19pm by B.B.
grade 9 math
Can someone please help me solve this equation. And please explain how to break this down. 4b-6+b-9= 0
Tuesday, April 12, 2011 at 7:45pm by brooke
solve and graph the following inequality. -8p-5<-13 Can someone please help me solve this because i tried it but im not quite sure if i solved it right.
Wednesday, June 10, 2009 at 3:34pm by Kimberly
Math-Advanced Functions
I am having trouble figuring out how to solve this logarithms. Could someone please help! log2(log4x)=1 and solve for x and y: (1/2)^x+y= 16 logx-y8=-3
Sunday, January 23, 2011 at 7:59pm by Chloe
solve equation then check for extraneous solution please someone explain this to me. 6/6-4x/=8+4 then one that is solve the inequality s/x+3/4<2 I don't understand these problems and then it says to
Sunday, August 26, 2012 at 5:38pm by kevin
please someone tell me how to solve this. im very confused and im homeschooled so my teachers are no help and my mom cant do math. please please help!!!
Thursday, January 17, 2013 at 9:46am by Corie
college algebra
Solve the following inequality. x^3+9x^2-108 less or equal to 0. Then write your solution in interval notation. I am sooo confused and do not have any idea how to do this. Can someone please show me
step by step please??? Thanks in advance. Someone who is confused.
Tuesday, December 25, 2012 at 6:27am by allison-please help!!!
Q: Corbin is playing a board game that requires rolling two number cubes to move a game piece. He needs to roll a sum of six on his next term and then a sum of ten to land on to the next two bonus
spaces. what is the probability that Corbin will roll a sum of six and then a ...
Saturday, March 9, 2013 at 10:07pm by dingbat
Can someone please show me step by step how to solve this, I've been stuck. Find the common ration r of the geometric series an that satisfies the following conditions: c=4, S=16 I would reallyy
appreciate it if someone could help me
Wednesday, June 12, 2013 at 4:59pm by mysterychicken
Can someone please help me with this word problem!?! I know the answer which is 784 feet. But I have no clue how to solve it. Can someone please show how to solve this problem? The word problem is:
The melt in Your Mouth Chocolate Factory is a rectangular building. The ...
Tuesday, September 4, 2012 at 9:18pm by aaron
My teacher said we are supossed to solve this using systems of equations: Let a, b, and c be real numbers such that a-7b+8c=4 and 8a+4b-c=7. Then a^2-b^2+c^2= ? can someone please help?! i don't know
what method to use to solve this or where to start
Sunday, September 13, 2009 at 6:45pm by chrissy
would someone please solve this
Wednesday, June 2, 2010 at 5:14pm by Hardy
Math (Algebra)
Could I have someone check my work please? If I did something wrong please explain to me where I went wrong. Thank You. Example of the product rule to solve would be, (5^3)(6^3) Example of the
quotient rule to solve would be, (4^2) / (8^2) Example of the power rule to solve ...
Sunday, July 18, 2010 at 12:09pm by Lynn
Can someone help me solve this please A*.3=B A-B=150,000
Wednesday, December 17, 2008 at 1:34pm by Matthew
The height is x someone please solve the problem
Wednesday, October 9, 2013 at 8:36pm by Angel
Can someone please explain how to solve 3x^2+7x-15=0 by using the quadratic equation? I know that the equation is: ax^2+bx+c=0 but I get confused about how to set the problem up and solve. Thanks.
Saturday, September 26, 2009 at 10:08am by B.B.
Please Solve: 4-x=-5 I thought the answer was -9 but that doesn't seem to be it. Can someone please help? Thanks! This is what I have: 6. 4-x=-5 The problem to solve is: 4-x=-5 Subtract 4 from both
sides x = -9 Then divie by -1 Would the answer be 9? To Check: 4- - 9 = -5
Saturday, June 27, 2009 at 2:22am by Angie
Can someone please help me with math problem and explain to me step by step how to solve it so I will know how solve other problems like this Thank you in advance Multiply and then, if possible,
simplify by factoring /7x /21y
Tuesday, February 28, 2012 at 7:16pm by Anonymous
Math SOmeone please please help
Find the derivative of the function. g(u) = (5+u^2)^5(3-9u^2)^8 Could someone please explain the steps that would lead me to the answer? I'm completely stuck.
Monday, February 21, 2011 at 11:15pm by Anonymous
Algebra II- please help!
I just cannot seem to be able to grasp the concept of solving polynomial inequalities. Can someone please explain, step by step, how to solve them? Here's a problem I can't solve. Please use this as
an example: (x-2)(x-5)<0 I cannot thank you enough for helping me with this...
Wednesday, December 10, 2008 at 4:13pm by Erin
is (-1, 5) a soultion for each system? x + y = 4 x=-1 can someone please explain how to solve this? thanks.
Sunday, September 16, 2007 at 7:34pm by Amber
Math 9
5 x 10^n = 5 Can someone please tell me the steps to solve for n? thank you!
Friday, February 5, 2010 at 10:39pm by Lisa
Math- Precalc
could someone please help me solve this problem... e^x - 12e^(-x) - 1
Monday, November 15, 2010 at 7:16pm by Alex
Can someone solve this and explain it please 4^log2 (2log2 5)
Monday, May 23, 2011 at 10:43pm by Chuck
can someone solve this please!!! 100/x = 120/176
Thursday, January 23, 2014 at 2:36pm by Kitty
Whats a consisnent. like suppose 2,3,6,9,8 Can someone help please? Can someone help please: What are conssitent numbers in math? i need help on my math its algebra
Tuesday, December 5, 2006 at 6:29pm by Paris
can someone please help me solve this solve by completeting the square 1/2 ”¼X ”½^2 -7x=16
Wednesday, October 2, 2013 at 6:36pm by Rick
Greg, what is that?? That has nothing to do with the problem I am trying to solve. Someone please help!
Wednesday, December 2, 2009 at 11:02pm by Anonymous
Could someone please show me how to solve this Prove tan(-č) = -tanč
Saturday, August 7, 2010 at 12:42pm by Seven
The square root of 27b^11 Can someone explain to me how to solve this?
Wednesday, May 4, 2011 at 9:26pm by Please Help
Solve for B: A=1/2(b+B) h show work. I am struggling with the above equation can someone please help? Thanks
Sunday, November 10, 2013 at 5:14pm by A
3/x + 5x/12 = 9/4.... i need to solve for x but i dont know what im doing.. can someone help me please
Wednesday, June 4, 2008 at 3:03pm by Alicia
Can someone please help me solve this? What are the odds in favor of drawing a spade and a heart?
Friday, October 9, 2009 at 10:01pm by Aleah
Can someone explain to me how to do this please!!! Solve the system of equations graphically: x+y>3 x-y<6
Thursday, May 15, 2008 at 8:47pm by Anonymous
Managerial Economics
Can someone please help me solve this equation: Qs=1,050 and Qd=2000-2.5P. Solve for the equilibrium price 'P'
Friday, August 23, 2013 at 11:28am by Kenesha
Can someone please help me with this? I'm not looking for the answer, just a detailed explanation on how to solve this. Thank you in advance. Solve: 16(t - 1) + 10 > 8(t +2) + 4(t - 1) + 4t
Sunday, April 3, 2011 at 2:35pm by Lace
MATH *
0=-16t^2+80t+96 Solve for t. Can someone please show me the steps on how to do this I can not seem to get it right. Thanks!!
Monday, December 2, 2013 at 7:22pm by Steph
math someone please help
m = y/x = 7-4/5-1 = 3/4 y = mx + b Plug in 1 for x and 4 for y and solve for b. ============= another way. Substitute x and y values for the equation y = mx + b. 4=m(1) + b 7=m(5) + b You have two
equations and two unknowns. Solve for m and b, then put in y = mx + b form.
Monday, November 5, 2007 at 11:46pm by DrBob222
Solve for S: r^2s-5s=7. Solve for b: L= 4a^2b. I'm confused about how to solve these problems. Could someone show me the steps?
Monday, September 12, 2011 at 3:23am by Sue
Physics Someone help please!
An example of the neutron absorption reaction is . 1 0 n+10 5 B-->7 3 Li+ 4 2 He The rest masses of each particle in atomic mass units (u) are 1 0 n= 1 u 10 5 B=10 u 7 3 Li =7 u 4 2 He= 4 u where 1u=
1.66x10^-27 kg. calculate Q Q= in MeV I have the same question please ...
Wednesday, May 8, 2013 at 3:53pm by Amy
Could you please help me solve this question: Solve this system of equations using elementary operations. 1. x/3 + y/4 + z/5 =14 2. x/4 + y/5 + z/3 =-21 3. x/5 + y/3 + z/4 =7 So far, I have
multiplied all three equations by 60, to turn them into whole numbers. Then everytime I...
Tuesday, May 25, 2010 at 6:24pm by Darren
ALGEBRA 1
Can someone please check on these for me please .Thanks Use the Quadratic Formula to solve 1.
Thursday, April 19, 2007 at 12:53pm by DANIELLE
Will someone please help me solve this problem. Simplify by factoring. ć250 this is a radical sign in front of 250.
Sunday, April 14, 2013 at 7:42pm by Michael
solve by the quadriatic formula, Im still trying to figure how to do this can someone please help me. (5x+1)^2 + 13=0
Thursday, October 3, 2013 at 11:14pm by jerri
91. Find the domain of: y=-3x+1 a. all negative numbers b. x>0 c. all real numbers d. x=3 Can someone please show the work and how to solve this problem. There are many of these in my lessons and I
don't understand how to solve them. Do you solve for x?
Friday, May 7, 2010 at 11:37am by y912f
Will someone please help me slove this problem? Use the elimination method to solve the system of equations. 5u+2v=-18 3u+v=-5 I have tried to solve this problem several times and have not gotten the
correct answer please help. Thank you.
Thursday, February 21, 2013 at 10:19am by Sade C
Someone please help me with this question. What would happen to an equation if you solve from right to left instead of left to right? And please provide an example.
Thursday, May 5, 2011 at 7:50pm by sammyg
Could someone please help me with this math problem. I am lost on trying to solve this math problem. Add. Simplify if possible. s+r/sr^2 + 2s+r/s^2r Thanks.
Sunday, May 16, 2010 at 8:13pm by B.B.
could someone please help me solve these two math problems? Which of the following numbers are integers? 2/13,570,0.096,-670,33.3? write in the lowest terms 2 1/4+4 1/7?
Thursday, September 23, 2010 at 4:28am by kiki83
adult education
I need help with someone to go over my eassy with me can someone please help me.This is my first time using this site so I don't know how it really works SOMEONE PLEASE HELP ME!!
Wednesday, October 27, 2010 at 5:02pm by jennifer
Please help me with the following by checking it. Solve for x: √(2x^2+5)=2 A. ±1 B. √2 - (This is my answer.) C. ±2 D. None of the above. I am not sure about my answer, can someone check it please.
Any help will be greatly appreciated.:)
Monday, April 7, 2014 at 10:55pm by Brady
Solve by elimination method. 2r-5s=-33 5r+2s=48 Solve by substitution method. 5x+6y=5 x-9y=40 I always get these two confused... can someone please help me with these two problems?? I would really
appreciate it..
Wednesday, November 11, 2009 at 3:00pm by Amanda
Solve by elimination method. 2r-5s=-33 5r+2s=48 Solve by substitution method. 5x+6y=5 x-9y=40 I always get these two confused... can someone please help me with these two problems?? I would really
appreciate it..
Wednesday, November 11, 2009 at 3:00pm by Amanda
can someone please help me solve this problem......-3x=3/2 3/2 is a fraction. i tink the answer is -2 while my friend thinks it is -.5 who is rite?
Sunday, October 14, 2007 at 9:25pm by Ann
Solve by using the multiplication principle -2x> 1 over 9 I have tried this problem and can get no where with it can someone please help me????
Wednesday, January 19, 2011 at 7:59pm by Brad
Algebra 2
Completing the square method allows you to solve any quadratic equation. For each of the following determine what number completes the square. I cannot find my notes on completing the square, can
someone please help with these two problems? 1. x^2+8x 2. y^2+10y i dont know ha...
Monday, May 21, 2007 at 11:32am by Christian
Math Analysis
6. Prove that f(x) = (3/4)x^4 + 2 and g(x) = (4sqrt(108x-216))/3 are inverses. I know you solve it with composite functions f(g(x)) and g(f(x)), and I solved f(g(x)), but I can't seem to solve g(f
(x)) to equal x, like it should. Can someone show me how to solve this? Thanks :)
Saturday, October 24, 2009 at 8:42pm by Emily
solve using the addition principle. -6+x>5 I have not done one of these problems with the less and greater sign... can someone please help me? thanks
Thursday, September 24, 2009 at 8:12pm by Anonymous
Hard math
Solve by using the multiplication principle -2x> 1 over 9 I have tried this problem and can get no where with it can someone please help me????
Wednesday, January 19, 2011 at 8:22pm by Sally
Johns football game last for 50 minutes. What reduced fraction of an hour is that? Can someone please explain to me how to solve this question?
Tuesday, November 1, 2011 at 1:46pm by Mindy
Algebra 1
Hi! I'm doing my math homework, but I got confused. Can someone please help me how to solve simplest radical forms with an example? With square roots?
Thursday, March 20, 2014 at 8:45pm by Bob
Could someone please show me how to solve this Prove tan(-theta) = -tan(theta)
Saturday, August 7, 2010 at 5:30pm by Please Answer
Could someone please help me solve this If f(2)=f'(2)=g'(2)=g(2)=2, find (fg)'(2). How would I solve this, I am completely lost as to what I would do.
Monday, February 22, 2010 at 10:18pm by Taryn
Algebra 1
f(x)=x^2+3 Solve for f(2)and f(-5) I need help, can someone please help me by explaining step by step how to solve this problem, I am most grateful to you. Thank you before hand.
Wednesday, January 19, 2011 at 9:59pm by Esther
Algebra 1
f(x)=x^2+3 Solve for f(2)and f(-5) I need help, can someone please help me by explaining step by step how to solve this problem, I am most grateful to you. Thank you before hand.
Wednesday, January 19, 2011 at 9:59pm by Esther
math/ algebra 1
can someone please help me explain how they got this answer and can someone please show me the steps into solving this question? Q: Four more than the quotient of a number and three is at least nine.
Sunday, November 10, 2013 at 4:34pm by ellie
solve the following pair of simultaneous equation: (x/a)+(y/b)=1 (x/b)+(y/a)=1 can someone help please?i have no idea how to solve it =( x/(ab) + y/b^2 = 1/b x/(ab) + y/a^2 = 1/a Subtract one from
the other y(1/b^2 - 1/a^2) = 1/b - 1/a y(1/b + 1/a)(1/b - 1/a) = (1/b - 1/a) ...
Wednesday, July 11, 2007 at 4:18am by Emma
I don't want to get this wrong for this homework that i have to turn in can someone please relook at it and tell me if it is correct plz.... solve. x^2+3x-18 my answer is x=3 and x=-6 yes.
Thursday, March 1, 2007 at 1:09pm by jas20
Could someone please tell me how to solve this problem: A pressure of 25.7 inHg would be how many kilopascals? Could you please list the steps on solving it. Thanks
Thursday, November 24, 2011 at 11:23am by Sandy
math urgent please...please...
I couldn't do g). someone help i need it in the morning. help me please. do it for me please. Show me the steps too please...please...
Sunday, December 15, 2013 at 12:56pm by kavi
can someone please help me solve the following problems Solve the following: 1. 2/7 - 1/3 = ? 2. 4/5 x 15/2 = ? 3. 2 1/2 ÷ 3/4 = ?
Wednesday, January 7, 2009 at 6:33pm by sandra
My teacher said we are supossed to solve this using systems of equations: Let a, b, and c be real numbers such that a-7b+8c=4 and 8a+4b-c=7. Then a^2-b^2+c^2= ? can someone please help?! i don't know
what method to use to solve this or where to start
Sunday, September 13, 2009 at 7:46pm by chrissy
Solve. x-6/x-9=3/x-9. I came up with a answer of -2/3 but I don't think that is the right answer. Could someone please help? Thanks.
Sunday, May 16, 2010 at 12:12pm by B.B.
-8<3x-5<-4 unsure how to solve can someone please give me the steps?
Tuesday, March 23, 2010 at 2:57pm by KiKi
I've done a few google searches, but how can I find someone who's influencing poverty in the state of Georgia? i need to find anyone who's doing anything to solve the problem of poverty, in any way.
Can someone please help me..?
Friday, September 30, 2011 at 3:46pm by y912f
Intermediate Algebra
solve equation (x+1)(x-2)=54 Eric stated that the solution would be (x+1)=54==> x=53 or (x-2)=54==> x=56 However at least one of these solutions fails to work when substituted back in the orginial
equation. Why is that? Please help Eric to understand better, solve the ...
Monday, June 18, 2012 at 8:45pm by ladybug Please Help!!
Algebra(please help)
Will someone please help me withthis question I couldn't solve it. What is the length of the segment whose endpoints are A(2, 3) and B(10, 7)? a)2 Sqrroot29 b)2 Sqrroot41 c)4 Sqrroot2 d)2 Sqrroot5
Tuesday, June 19, 2012 at 9:39am by Liz
6th grade math
can someone please explain to me what the Answer is and also can someone please explain to me how to slove this question step by step??? Please??!!
Monday, February 13, 2012 at 5:33pm by Kate
Math help please
1. Solve for x. 4(x - b) = x b = 4/3x b = 3/4x x = 4/3b x = 3/4b 2. Solve for y, then find the value of y when given x = 2. 6x = 7 - 4y -12 7/4 19/4 24 I need help solving these. Can you please
explain how to solve these problems?
Wednesday, October 2, 2013 at 6:52pm by Charlotte
What are the odds in favor of getting at least one head in three successive flips of a coin? Can someone help me figure out how to solve this, please?
Friday, November 13, 2009 at 11:34am by Anonymous
0.3x-0.2y=4 0.5x+0.3y=-7/17 Can anyone Please help me solve using the elimination system..... Someone tried to explain it but I still do not get it at all
Saturday, March 13, 2010 at 7:19pm by Holly
can someone correct this for me please... Solve for x: 6x minus 1 divided by 5 minus 2x divided by 3 equals 3 my answer: x = 6 Solve for x: 3/7x = 15 My answer: x = 35 correct, as best I can tell
from the description. Thank you.
Monday, January 1, 2007 at 4:50am by jasmine20
Can someone please solve this? x+5=12 x7=35 3=5x(8-3) 12-4+2(3x3)
Monday, May 2, 2011 at 7:40pm by Robert Wayne
Can someone help me solve the following....using elimination methods 1) 5x+7y=43 -4x+y=25 2)9x-9y=36 7y-3x=-14 I think the answers are would the answer to number 1 be (-4.26,1.52)? would the answer
to number 2 be (-0.5,-6.5)? Can someone please verify or tell me what I did ...
Saturday, March 13, 2010 at 7:19pm by Holly
Can someone help me
multipy both sides by 44x Then take the square root of each side. Math question here. Can someone solve this for me?1/x=x/44
Monday, May 21, 2007 at 7:03pm by bobpursley
Pre-Algebra [urgent!]
I can't find the solution to this problem; can someone please show me how to solve it step-by-step? Solve the equation for y in terms of x: 1. -8x + 5y = 10 Thanks! -MC
Monday, May 4, 2009 at 4:58pm by mysterychicken
Math algebra honours
Can someone please answer my question it is the other one named algbra honurs i am kinda new to this website so i dont know if you postt it twice or not can someone please help thankyou
Sunday, September 26, 2010 at 5:59pm by Jacytiopa (i know its weird but my mom picked it
Pages: 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | Next>> | {"url":"http://www.jiskha.com/search/index.cgi?query=math+-+Can+someone+please+help+to+solve","timestamp":"2014-04-19T23:06:27Z","content_type":null,"content_length":"36958","record_id":"<urn:uuid:43c5ce50-cf13-4a5b-9535-61653753c52d>","cc-path":"CC-MAIN-2014-15/segments/1397609537754.12/warc/CC-MAIN-20140416005217-00111-ip-10-147-4-33.ec2.internal.warc.gz"} |
Payoff Matrix Another Example
A selection of articles related to payoff matrix another example.
Original articles from our library related to the Payoff Matrix Another Example. See Table of Contents for further available material (downloadable resources) on Payoff Matrix Another Example.
Body Mysteries >> Sexuality
Payoff Matrix Another Example is described in multiple online sources, as addition to our editors' articles, see section below for printable documents, Payoff Matrix Another Example books and
related discussion.
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matrices? So let us implement it to other matrices using dominance and study the importance follow are: Rule 1: If all the elements in a row ( say ith row ) of a pay off matrix are less than
or Now, consider some simple examples.
and the payoff matrix A = (aij) demonstrates the amount Player II gives to Player I under each combination of strategies. For example: II plays j. 1.
The effects of payoff-matrix multiplication, payoff-matrix addition, the presence of long-run gains versus long-run For example, no matter how advanced an animal's visual or exemplars that
overlap with exemplars from other distributions.
and payoffs in a matrix and can then calculate the best single strategy or combination of another illustration of the power of matrix algebra and linear programming. Paper, Scissors, Rock is
an example of a two-person zero sum game.
sets of strategies available to the two players and the payoff matrix. The set ..
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Great care has been taken to prepare the information on this page. Elements of the content come from factual and lexical knowledge databases, realmagick.com library and third-party sources. We
appreciate your suggestions and comments on further improvements of the site. | {"url":"http://www.realmagick.com/payoff-matrix-another-example/","timestamp":"2014-04-17T21:46:35Z","content_type":null,"content_length":"29585","record_id":"<urn:uuid:e28ddf24-4050-472d-87a6-76d3441b761e>","cc-path":"CC-MAIN-2014-15/segments/1397609532128.44/warc/CC-MAIN-20140416005212-00581-ip-10-147-4-33.ec2.internal.warc.gz"} |
All solutions for trig equation
September 13th 2011, 05:18 AM
All solutions for trig equation
Find all solutions for each of the following in ther given intervals:
$(a) tan^2(x) = 3,~ -2\pi<x\le\pi$
$(b) sin^2(x) = \frac{1}{2},~ \frac{\pi}{2}\le x\le2\pi$
$tan(x) = \sqrt3$
Do I find only the values that are positive $\sqrt{3}$ in the given interval? i.e., $\frac{\pi}{3}$, $-\frac{2\pi}{3}$, $-\frac{5\pi}{3}$ or do I just find every value where $tan(x) = \sqrt3$?,
i.e, 60, 180-60, -60, -180-(-60), -180+(-60), -360-(-60)
same deal accept with $\frac{\pi}{4}$
September 13th 2011, 05:45 AM
Prove It
Re: All solutions for trig equation
Find all solutions for each of the following in ther given intervals:
$(a) tan^2(x) = 3,~ -2\pi<x\le\pi$
$(b) sin^2(x) = \frac{1}{2},~ \frac{\pi}{2}\le x\le2\pi$
$tan(x) = \sqrt3$
Do I find only the values that are positive $\sqrt{3}$ in the given interval? i.e., $\frac{\pi}{3}$, $-\frac{2\pi}{3}$, $-\frac{5\pi}{3}$ or do I just find every value where $tan(x) = \sqrt3$?,
i.e, 60, 180-60, -60, -180-(-60), -180+(-60), -360-(-60)
same deal accept with $\frac{\pi}{4}$
Well first of all $\displaystyle \tan^2{x} = 3 \implies \tan{x} = \pm \sqrt{3}$, not $\displaystyle \sqrt{3}$...
September 13th 2011, 05:52 AM
Re: All solutions for trig equation
(Doh) ah ok, thanks. So it is everything. | {"url":"http://mathhelpforum.com/trigonometry/187891-all-solutions-trig-equation-print.html","timestamp":"2014-04-19T12:31:35Z","content_type":null,"content_length":"9033","record_id":"<urn:uuid:7e5f9977-5043-43fd-bb3e-6f85e784c990>","cc-path":"CC-MAIN-2014-15/segments/1397609537186.46/warc/CC-MAIN-20140416005217-00629-ip-10-147-4-33.ec2.internal.warc.gz"} |
Math Mayhem
September 1st, 2010 by Steven Pomeroy |
No Comments
| Filed in
Math Games
Math Software
Math Games: Professor Pi’s Math Mayhem Beta 2.0
I have uploaded version Beta 2.0 of Math Mayhem to our math games section. Please let me know what you like about it/don’t like about it if you try it out. Also, please report any bugs if you find
I have introduced some power ups in this version, and a “How to Play” section (yea – instructions are a good thing). Also, I added two more levels for a total of three. I don’t want to add too many
levels in case there are bugs to hunt down. If all goes well with this release, I will work on the first full-scale roll-out (i.e., I’ll add a bunch of levels).
Anyway, please give it a try, and I hope you like it!
Professor Pi’s Math Mayhem Beta 1.0
File Name:
math mayhem beta 2 install.exe
1.31 MB
Installed File Name:
math mayhem beta 2.exe
Installed File Size:
1.21 MB
Operating Systems:Windows XP, Vista, and Windows 7
Please note that this application was tested under these operating systems, but further testing is required. If this application fails to run on your computer, please let me know which operating
system you have.
Tags: educational games, Math Games, math mayhem, math software, math software game, professor pi, professor pi's math mayhem | {"url":"http://mathtricks.org/tag/math-mayhem/","timestamp":"2014-04-19T02:50:44Z","content_type":null,"content_length":"24473","record_id":"<urn:uuid:20f886f8-311e-4f64-8bfa-da3daa0becaf>","cc-path":"CC-MAIN-2014-15/segments/1397609535745.0/warc/CC-MAIN-20140416005215-00220-ip-10-147-4-33.ec2.internal.warc.gz"} |
Linear Algebra: Linear Transformations(1)
December 24th 2008, 10:25 PM
Linear Algebra: Linear Transformations(1)
I am teaching myself Linear Algebra following P.R.Halmos' "Finite Dimensional Vector Spaces".
I am stuck at a couple of problems:
Problem (1)
Prove that corresponding to every linear transformation A on a finite dimensional vector space V, there exists an invertible linear transformation P such that APA = A.(Or equivalently prove that
PA is a projection)
Problem (2)
If A,B,C are linear transformations on a finite dimensional vector space, then does (AB - BA)^2 commutes with C always?
What happens when the dimension of the vector space is 2?
Problem (3)
From the basic definition of a determinant of a linear transformation(see below), prove that $\text{det }(A \otimes B) = \text{det }(A) ^m \text{det }(B) ^n$, where A and B are linear
transformations on vector spaces of dimensions n and m respectively
(I know the proof of a particular case when A and B are matrices, using eigenvalues. But I would love to see a proof using the below definition).
Definition of Determinant of a linear transformation:
If W is the space of all alternating n-linear forms on an n-dimensional vector space V,then we know that it is one-dimensional. Thus given a linear transformation A, we have a scalar $\delta$
such that for every w in W, $w(Ax_1,Ax_2,.....,Ax_n) = \delta w(x_1,x_2,....,x_n)$ $\forall x_1,x_2,...x_n$ vectors in V. Then the det(A) is defined to be the scalar $\delta$.
Thank you,
December 25th 2008, 12:05 AM
I am teaching myself Linear Algebra following P.R.Halmos' "Finite Dimensional Vector Spaces".
I am stuck at a couple of problems:
Problem (1)
Prove that corresponding to every linear transformation A on a finite dimensional vector space V, there exists an invertible linear transformation P such that APA = A.(Or equivalently prove that
PA is a projection)
the proof of this is in the book and it's quite easy! see Theorem 3, page 94.
Problem (2)
If A,B,C are linear transformations on a finite dimensional vector space, then does (AB - BA)^2 commutes with C always?
no. it's easier to work with matrices rather than transformations here. they're the same anyway: $A=e_{12}, \ B=e_{21}, \ C=e_{13}.$ (here $e_{ij}$ has 1 as its (i,j)-entry and 0 everywhere
What happens when the dimension of the vector space is 2?
in this case, the claim is true: let $AB-BA=X.$ clearly $\text{tr}(X)=0.$ thus by Cayley Hamilton: $X^2 = \alpha I,$ for some scalar $\alpha.$ now it's obvious that $X^2C=CX^2,$ for all $C.$
December 25th 2008, 12:47 AM
in this case, the claim is true: let http://www.mathhelpforum.com/math-he...02eee9a9-1.gif clearly http://www.mathhelpforum.com/math-he...bf2b19c6-1.gif thus by Cayley Hamilton: http://
www.mathhelpforum.com/math-he...3dd0e211-1.gif for some scalar http://www.mathhelpforum.com/math-he...96a7ebaf-1.gif now it's obvious that http://www.mathhelpforum.com/math-he...d22b4911-1.gif
for all C.
Wow! This proof is fantastic (Cool).
As of now, AB and BA are linear transformations and he has not defined trace or proved Cayley-Hamilton theorem. So probably there is a elementary proof of this result.But I know both the results
for matrices.
Nevertheless its a wonderful proof. Thank you NonCommAlg, you (Rock)
If you have any idea on how to do the third one, can you tell me?
Thanks again,
December 25th 2008, 01:21 AM
here's a direct way for Problem 2, part 2: let $\{e_1,e_2 \}$ be a basis for our vector space. let $A(e_1)=ae_1 + be_2, \ A(e_2)=ce_1 + de_2, \ \ B(e_1)=a'e_1+b'e_2, \ B(e_2)=c'e_1 + d'e_2.$ see
$(AB-BA)(e_1)=re_2, \ \ (AB-BA)(e_2)=se_1,$ for some scalars $r,s.$ thus: $(AB-BA)^2(e_1)=rse_1, \ (AB-BA)^2(e_2)=rse_2.$ hence: $(AB-BA)^2=rsI$ and the result follows. | {"url":"http://mathhelpforum.com/advanced-algebra/66004-linear-algebra-linear-transformations-1-a-print.html","timestamp":"2014-04-18T09:18:24Z","content_type":null,"content_length":"14016","record_id":"<urn:uuid:0696224f-2fe7-4d9b-82ba-f8532a1f3982>","cc-path":"CC-MAIN-2014-15/segments/1397609533121.28/warc/CC-MAIN-20140416005213-00048-ip-10-147-4-33.ec2.internal.warc.gz"} |
[SciPy-Dev] Generalized eigenproblem with rank deficient matrices
[SciPy-Dev] Generalized eigenproblem with rank deficient matrices
Charles R Harris charlesr.harris@gmail....
Sun Sep 4 10:29:19 CDT 2011
On Sun, Sep 4, 2011 at 7:53 AM, Nils Wagner <nwagner@iam.uni-stuttgart.de>wrote:
> Hi all,
> how can I solve the eigenproblem
> A x = \lambda B x
> where both matrices are rank deficient ?
I'd do eigh and transform the problem to something like:
U * A * U^t * x= \lambda D * x
where D is diagonal. Note that the solutions may not be unique and \lambda
can be arbitrary, as you can see by studying
A = B = array([[1, 0], [0, 0]])
Where there are solutions for arbitrary \lambda. Likewise, there may be no
solutions under the requirement that x is non-zero:
A = array([[1, 1], [1, 0]]),
B = array([[1, 0], [0, 0]])
The usual case where B is positive definite corresponds to finding extrema
on a compact surface x^t * B *x = 1, but the surface is no longer compact
when B isn't positive definite. Note that these cases are all sensitive to
roundoff error.
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Re: st: Beta coefficients are not equal to coefficients on standardized
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Re: st: Beta coefficients are not equal to coefficients on standardized variables?
From Maarten Buis <maartenlbuis@gmail.com>
To statalist@hsphsun2.harvard.edu
Subject Re: st: Beta coefficients are not equal to coefficients on standardized variables?
Date Sat, 16 Jun 2012 10:14:46 +0200
On Fri, Jun 15, 2012 at 5:22 PM, Roberto Liebscher wrote:
> There is one thing that makes me puzzling about the - beta - option in
> regression commands. In a simple example using the lifeexp dataset I first
> used the built-in function - beta - :
> Then I standardized the variables by hand and re-ran the regression with the
> new variables:
> Now the coefficients are slightly different. For example the coefficient on
> gnppc_std is 0.6608475 whereas it has been 0.6506803 in the first
> calculation.
> Is this caused by rounding errors? Or is there any other explanation for
> this?
The main reason is that you standardized within the wrong sample. The
-beta- option will standardize within the sample used by -regress-,
i.e. ignore all observations where at least one variable contained a
missing value. You standardized within the entire dataset. After that
there is still a little difference, which you can take away by
creating the standardized variables as -double-.
*---------------- begin example ------------------
sysuse lifeexp, clear
regress lexp gnppc popgrowth, beta
// standardize in the false sample
egen popgrowth_std_fs = std(popgrowth)
egen lexp_std_fs = std(lexp)
egen gnppc_std_fs = std(gnppc)
regress lexp_std_fs gnppc_std_fs popgrowth_std_fs
// get the right sample
gen byte touse = !missing(popgrowth,lexp,gnppc)
egen popgrowth_std = std(popgrowth) if touse
egen lexp_std = std(lexp) if touse
egen gnppc_std = std(gnppc) if touse
regress lexp_std gnppc_std popgrowth_std
// extra precision
egen double popgrowth_std_d = std(popgrowth) if touse
egen double lexp_std_d = std(lexp) if touse
egen double gnppc_std_d = std(gnppc) if touse
regress lexp_std_d gnppc_std_d popgrowth_std_d
*----------------------- end example --------------------
(For more on examples I sent to the Statalist see:
http://www.maartenbuis.nl/example_faq )
Hope this helps,
Maarten L. Buis
Institut fuer Soziologie
Universitaet Tuebingen
Wilhelmstrasse 36
72074 Tuebingen
* For searches and help try:
* http://www.stata.com/help.cgi?search
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/ | {"url":"http://www.stata.com/statalist/archive/2012-06/msg00786.html","timestamp":"2014-04-20T19:56:38Z","content_type":null,"content_length":"9752","record_id":"<urn:uuid:1ae35f16-02fa-4ae6-b7f0-4cd03602f222>","cc-path":"CC-MAIN-2014-15/segments/1397609539066.13/warc/CC-MAIN-20140416005219-00034-ip-10-147-4-33.ec2.internal.warc.gz"} |
What is the Schouten bracket for the Chevalley-Eilenberg complex with coefficients in a nontrivial module?
up vote 5 down vote favorite
Let $\mathfrak g$ be a Lie algebra. The Chevalley-Eilenberg complex is defined to be $\wedge^* \mathfrak g$ with differential $d\colon \wedge^* \mathfrak g\to \wedge^{*-1}\mathfrak g$ defined by $$d
(a_1\wedge\cdots \wedge a_k)=\sum_{i,j}(-1)^{i+j-1}[a_i,a_j] a_1\wedge \cdots\wedge\hat{a_i}\wedge\cdots\wedge\hat{a}_j\wedge\cdots\wedge a_k.$$ The differential $d$ is not a derivation with respect
to the exterior product $\wedge$, but the deviation from being a derivation is a binary operation which defines a graded Lie algebra structure on $\wedge^* \mathfrak g$: If $\underline{a},\underline
{b}\in\wedge^*\mathfrak g$, let $$[\underline{a},\underline{b}]_{s}=d(\underline{a}\wedge\underline{b})-d\underline{a}\wedge b+\underline{a}\wedge d\underline{b}$$ (I'm omitting some signs.) This
bracket operation vanishes once you take homology, since if $d\underline{a}=d\underline{b}=0$ then it is obvious that $d(\underline{a}\wedge\underline{b})=[\underline{a},\underline{b}]_s$. However, I
was talking to Jim Stasheff several years ago, and he mentioned that the Schouten bracket doesn't necessarily vanish on Lie algebra homology if there are coefficients in a nontrivial $\mathfrak g$
module, $M$. However, I don't know what the definition of the Schouten bracket is in this case. The Chevalley-Eilenberg complex is easy enough to understand: $\wedge^*\mathfrak g\otimes M$, where the
differential includes terms where the $a_i$ act on $M$, but the obvious generalization of the above construction fails since two elements of $M$ somehow need to get combined into one element. So my
basic question is how you define a Schouten bracket on the Chevalley-Eilenberg complex with coefficients in a nontrivial $\mathfrak g$-module?
lie-groups abstract-algebra lie-algebra-cohomology
The analog thing for associative algebras, which is the Gerstenhaber bracket, is not defined for not-regular values (which is the analogue of trivial values) – Mariano Suárez-Alvarez♦ Oct 4 '10 at
By "values" I mean "coefficients"... Sometime ago I decided that homology takes coefficients in the module, and cohomology takes values in the module :) – Mariano Suárez-Alvarez♦ Oct 4 '10 at
A truly terrible way to get at this bracket is as follows. If $\mathfrak g$ acts on $M$, then it also acts on the dual space $M^*$, which you should think of as a geometric space, and so there is
1 a map $\mathfrak g \to \Gamma(\rm TM^*)$ (sections of tangent bundle). The Schouten bracket on $\wedge^\bullet\mathfrak g\otimes M$ is the pullback of said bracket on $\wedge^\bullet\Gamma(\rm TM^
*)$ to $\mathfrak g$, and restricted to those sections that are linear in the base $M^*$. As I said, this is a terrible way to get at this bracket. – Theo Johnson-Freyd Oct 5 '10 at 2:46
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1 Answer
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Theo Johnson-Freyd:
A truly terrible way to get at this bracket is as follows. If $g$ acts on $M$, then it also acts on the dual space $M^*$, which you should think of as a geometric space, and so there
is a map $g\to\Gamma(TM^*)$ (sections of tangent bundle). The Schouten bracket on $\wedge^* g\otimes M$ is the pullback of said bracket on $\wedge^*\Gamma(TM^*)$ to $g$ , and
up vote 0 down restricted to those sections that are linear in the base $M^*$ . As I said, this is a terrible way to get at this bracket.
vote accepted
(JC: I'm trying to clear the unanswered question backlog.)
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Math and Text
Mathematics as science
Carl Friedrich Gauss referred to mathematics as "the Queen of the Sciences". In the original Latin Regina Scientiarum, as well as in German Königin der Wissenschaften, the word corresponding to
science means (field of) knowledge. Indeed, this is also the original meaning in English, and there is no doubt that mathematics is in this sense a science. The specialization restricting the meaning
to natural science is of later date. If one considers science to be strictly about the physical world, then mathematics, or at least pure mathematics, is not a science. Albert Einstein has stated
that "as far as the laws of mathematics refer to reality, they are not certain; and as far as they are certain, they do not refer to reality."
Many philosophers believe that mathematics is not experimentally falsifiable,[citation needed] and thus not a science according to the definition of Karl Popper. However, in the 1930s important work
in mathematical logic showed that mathematics cannot be reduced to logic, and Karl Popper concluded that "most mathematical theories are, like those of physics and biology, hypothetico-deductive:
pure mathematics therefore turns out to be much closer to the natural sciences whose hypotheses are conjectures, than it seemed even recently."[11] Other thinkers, notably Imre Lakatos, have applied
a version of falsificationism to mathematics itself.
An alternative view is that certain scientific fields (such as theoretical physics) are mathematics with axioms that are intended to correspond to reality. In fact, the theoretical physicist, J. M.
Ziman, proposed that science is public knowledge and thus includes mathematics. In any case, mathematics shares much in common with many fields in the physical sciences, notably the exploration of
the logical consequences of assumptions. Intuition and experimentation also play a role in the formulation of conjectures in both mathematics and the (other) sciences. Experimental mathematics
continues to grow in importance within mathematics, and computation and simulation are playing an increasing role in both the sciences and mathematics, weakening the objection that mathematics does
not use the scientific method. In his 2002 book A New Kind of Science, Stephen Wolfram argues that computational mathematics deserves to be explored empirically as a scientific field in its own
The opinions of mathematicians on this matter are varied. While some in applied mathematics feel that they are scientists, those in pure mathematics often feel that they are working in an area more
akin to logic and that they are, hence, fundamentally philosophers. Many mathematicians feel that to call their area a science is to downplay the importance of its aesthetic side, and its history in
the traditional seven liberal arts; others feel that to ignore its connection to the sciences is to turn a blind eye to the fact that the interface between mathematics and its applications in science
and engineering has driven much development in mathematics. One way this difference of viewpoint plays out is in the philosophical debate as to whether mathematics is created (as in art) or
discovered (as in science). It is common to see universities divided into sections that include a division of Science and Mathematics, indicating that the fields are seen as being allied but that
they do not coincide. In practice, mathematicians are typically grouped with scientists at the gross level but separated at finer levels. This is one of many issues considered in the philosophy of
Mathematical awards are generally kept separate from their equivalents in science. The most prestigious award in mathematics is the Fields Medal, established in 1936 and now awarded every 4 years. It
is usually considered the equivalent of science's Nobel prize. Another major international award, the Abel Prize, was introduced in 2003. Both of these are awarded for a particular body of work,
either innovation in a new area of mathematics or resolution of an outstanding problem in an established field. A famous list of 23 such open problems, called "Hilbert's problems", was compiled in
1900 by German mathematician David Hilbert. This list achieved great celebrity among mathematicians, and at least nine of the problems have now been solved. A new list of seven important problems,
titled the "Millennium Prize Problems", was published in 2000. Solution of each of these problems carries a $1 million reward, and only one (the Riemann hypothesis) is duplicated in Hilbert's
Mathematics (colloquially, maths, or math in American English) is the body of knowledge centered on concepts such as quantity, structure, space, and change, and the academic discipline which studies
them; Benjamin Peirce called it "the science that draws necessary conclusions".[1] It evolved, through the use of abstraction and logical reasoning, from counting, calculation, measurement, and the
study of the shapes and motions of physical objects. Mathematicians explore such concepts, aiming to formulate new conjectures and establish their truth by rigorous deduction from appropriately
chosen axioms and definitions.
Knowledge and use of basic mathematics have always been an inherent and integral part of individual and group life. Refinements of the basic ideas are visible in ancient mathematical texts
originating in ancient Egypt, Mesopotamia, Ancient India, and Ancient China with increased rigour later introduced by the ancient Greeks. From this point on, the development continued in short bursts
until the Renaissance period of the 16th century where mathematical innovations interacted with new scientific discoveries leading to an acceleration in understanding that continues to the present
Today, mathematics is used throughout the world in many fields, including science, engineering, medicine and economics. The application of mathematics to such fields, often dubbed applied
mathematics, inspires and makes use of new mathematical discoveries and has sometimes led to the development of entirely new disciplines. Mathematicians also engage in pure mathematics for its own
sake without having any practical application in mind, although applications for what begins as pure mathematics are often discovered later | {"url":"http://www.mathandtext.blogspot.com/","timestamp":"2014-04-16T22:38:38Z","content_type":null,"content_length":"23025","record_id":"<urn:uuid:2e193fc0-7378-47d0-bcf6-b989be860021>","cc-path":"CC-MAIN-2014-15/segments/1397609525991.2/warc/CC-MAIN-20140416005205-00294-ip-10-147-4-33.ec2.internal.warc.gz"} |
Universitatis Iagellonicae Acta Mathematica Issue 44
1. Józef Siciak
A note on pluripolar extensions of univalent functions
Issue 44 (2006), pp. 7-14
Article (PDF 242K)
2. Sabrina Brusadin, Gianluca Gorni
The degree of the inverse of a polynomial automorphism
Issue 44 (2006), pp. 15-19
Article (PDF 190K)
3. V.M. Fedorchuk, V.I. Fedorchuk
First-order differential invariants of the splitting subgroups of the Poincare group P91,4)
Issue 44 (2006), pp. 21-30
Article (PDF 187K)
4. Sławomir Bakalarski
Jacobian problem for factorial varieties
Issue 44 (2006), pp. 31-34
Article (PDF 157K)
5. Tomasz Bielaczyc
Generic properties of iterated function systems with place dependent probabilities
Issue 44 (2006), pp. 35-46
Article (PDF 234K)
6. Agnieszka Lipieta
The strong unicity constant for projections
Issue 44 (2006), pp. 47-68
Article (PDF 311K)
7. Dominik Mielczarek
Minimal projections onto spaces of symmetric matrices
Issue 44 (2006), pp. 69-82
Article (PDF 255K)
8. Lucjan Sapa
Existence and uniqueness of a classical solution of Fourier's first problem for nonlinear parabolic-elliptic systems
Issue 44 (2006), pp. 83-95
Article (PDF 270K)
9. Lech Zaręba
Existence of a solution for the initial boundary value problem for a high order parabolic equation in an unbounded domain
Issue 44 (2006), pp. 97-114
Article (PDF 287K)
10. Marcin Ziomek
A geometrical version of the Moore theorem in the case of infinite dimensional Banach spaces
Issue 44 (2006), pp. 115-122
Article (PDF 195K)
11. Marcin Ziomek
A remark on the Moore theorem
Issue 44 (2006), pp. 123-126
Article (PDF 176K) | {"url":"http://www.emis.de/journals/UIAM/toc-44.html","timestamp":"2014-04-21T07:48:51Z","content_type":null,"content_length":"6689","record_id":"<urn:uuid:e88871a8-5759-41d8-9297-bf0ae9162d5d>","cc-path":"CC-MAIN-2014-15/segments/1397609539665.16/warc/CC-MAIN-20140416005219-00582-ip-10-147-4-33.ec2.internal.warc.gz"} |
Re: [TenTec] NEC, ground, grounds, and radials
On 1/8/2011 9:09 AM, Jack Mandelman wrote:
> Forget about the formulas! None of the discussed formulas is
> sufficiently accurate of the wide ranges of S/d discussed. All are
> approximations that are reasonably valid only over their limited
> domains.
The log formula gives industry accepted precision for impedances above
about 250 or 275 ohms. Spacing about 4 times diameter out to arbitrarily
large spacing. The inverse hyperbolic cosh formula matches the impedance
for all that range yet also matches the near zero impedance with the
conductor spacing is infinitesimal at the low Z domain boundary.
> A much more practical engineering approach would be to apply a
> finite-element analysis of LaPlace's equation using the boundary
> conditions appropriate to the geometry of interest. For these types
> of problems a 2-D quasi-static analysis provides much better accuracy
> than any formulas presented.
Are you sure that the finite element stepped analysis is better than
Harold Wheeler's 2d analysis of 1938? That gave a new formula based on
the inv hyp cos for capacitance and the effects of inductance have been
neglected taking Z0 = sqrt (L/C). Are you sure that he didn't do exactly
a 2 D static analysis to come up with that formula? I think he did
without the discreteness of finite element approximations. Harold
Wheeler was a transmission line guru of the highest class with many
papers to his credit, but when interviewed later in life he was most
proud of the inverse hyperbolic cosh formula for capacitance of closely
spaced parallel wires. His career ran another 30 or 40 years after that
paper and he had plenty of time to revisit it but didn't change the
results while investigating many other transmission line shapes and
publishing curves and formulas for their parameters.
Its conceptually easy to check the available formulae, simply construct
samples of transmission lines and measure their characteristic
impedance, probably as quarter wave transformers with a known load. Or
by varying the line or the load to present the same impedance when a
quarter and a half wave long. Then you might have to work out the
complications of the balun you measure the balanced line through or
create a balanced bridge and the effects of spacers and conductor
mountings at the ends and any intermediate points. 1/2" copper tubing
would be stiff enough to allow measuring at 6 meters without needing
intermediate supports. Perhaps the most precise measuring instrument
would use the line with a traveling differential voltage probe loosely
coupled like a slotted line without shielding. Perhaps that voltage
probe could be made arbitrarily small to include a microwave transmitter
to send a measure of the detected voltage to the user without having to
use wires to mess up the balance of the line under test.
73, Jerry, K0CQ
> Jack K1VT
>> Correct!
>> I believe it was Jerry who pointed me to the accurate formula a
>> couple of years ago on this very List. I was considering how to
>> build low impedance balanced line with a Zo in the range 50 Ohms to
>> 70 Ohms for a Hexbeam application. One look at the published curves
>> told me it couldn't be done; but Jerry put me right!
>> Steve G3TXQ
>> On 08/01/2011 02:05, Dr. Gerald N. Johnson wrote:
>>> / Like my curves its center to center. The upper trace is using
>>> the 276/ / log formula and the lower one is the 120 inv
>>> hyperbolic cosine function./ / Mine has the spacings on a log
>>> scale his is a liner scale. The upper/ / trace formula is
>>> increasingly inaccurate below 300 ohms impedance with/ / 87,000
>>> percent error at a spacing just a hair wider than the
>>> conductors/ / touching. The wrong formula says you can't get a
>>> Z0 less than 87 ohms/ / (as published in QST last summer again),
>>> but the correct formula gets/ / down to practically zero Z0 just
>>> before the conductors make contact./ /
>>> http://www.geraldj.networkiowa.com/papers/CSVHF2010/lztl1.JPG// /
>>> http://www.geraldj.networkiowa.com/papers/CSVHF2010/lztl2.JPG//
>>> / 73, Jerry, K0CQ/
>>> / On 1/7/2011 7:46 PM, Jack Mandelman wrote:/ /> Steve,/ />
>>> In your plot how are you defining S, the spacing between
>>> conductors? Is it/ /> center to center? Is it consistent for
>>> both the approximated and actual/ /> curves? It appears that
>>> only the actual curve uses the center to center/ /> definition;
>>> in the limit as S-->d the inner edge to inner edge spacing
>>> goes/ /> to zero, and Zo also goes to zero./ />/ /> Jack
>>> K1VT/
TenTec mailing list | {"url":"http://lists.contesting.com/_tentec/2011-01/msg00207.html?contestingsid=6aip81fkhok4tsaluuml9nptc6","timestamp":"2014-04-21T02:52:48Z","content_type":null,"content_length":"13983","record_id":"<urn:uuid:0e6d6fd1-9e6e-4618-bd20-6cf4594a4af1>","cc-path":"CC-MAIN-2014-15/segments/1397609539447.23/warc/CC-MAIN-20140416005219-00118-ip-10-147-4-33.ec2.internal.warc.gz"} |
[Haskell-cafe] type classes
Dan Doel dan.doel at gmail.com
Wed Jul 2 21:57:22 EDT 2008
On Wednesday 02 July 2008, Cotton Seed wrote:
> Hi everyone,
> I'm working on a computational algebra program and I've run into a problem.
> In my program, I have types for instances of algebraic objects, e.g. ZModN
> for modular integers, and types for the objects themselves, e.g. ZModNTy
> for the ring of modular integers.
> Now, I want to subclass ZModNTy from something like
> class RingTy a b where
> order :: a -> Integer
> units :: a -> [b]
> where `a' represents algebraic object, and `b' the type of instances of
> that object. I want an instance
> instance RingTy ZModNTy ZModN where ...
> but then code that only uses order fails with errors like
> No instance for (RingTy ZModNTy b)
> arising from a use of `order' at Test2.hs:16:8-15
> since there is no constraint on the second type variable.
> I think what I really want is
> class RingTy a where
> order :: a b -> Integer
> units :: a b -> [b]
> but this doesn't work either since ZModNTy is not parametric in its type
> like, say, `Polynomial a' is.
> Is this a common problem? Is there a standard way to handle it?
Correct me if I'm wrong, but wouldn't the a uniquely determine the b? In that
case, you'd probably want a functional dependency:
class RingTy a b | a -> b where
order :: a -> Integer
units :: a -> [b]
This solves the problem with order, because with multi-parameter type classes,
all the variables should be determined for a use of a method. Since b is not
involved with order, it could be anything, so it's rather ambiguous. The
functional dependency solves this by uniquely determined b from a, so order
is no longer ambiguous.
Alternately, with the new type families, this can become:
class RingTy a where
type RingElem a :: *
order :: a -> Integer
units :: a -> [RingElem a]
Or something along those lines.
Hope that helps.
-- Dan
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Poway Trigonometry Tutor
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Two sample test of Hypothesis
June 24th 2009, 02:38 AM
Two sample test of Hypothesis
The Tampa Bay Area Chamber of Commerce wanted to know whether the mean weekly salary of nurses was larger than that of school teachers. To investigate, they collected the following information on
the amounts earned last week by a sample of school teachers and nurses.
Teachers: 845, 826, 827, 875, 784, 809, 802, 820, 829, 830, 842, 832.
Nurses: 841, 890, 821, 771, 850, 859, 825, 829.
Is it reasonable to conclude that the mean weekly salary of nurses is higher? Use the .01 significance level. What is the p-value.
June 24th 2009, 06:38 AM
You preform a t test on the sample means.
You need to assume normality of both populations.
BUT we (you and me) need to know if you're assuming equal populatoin variances or not.
If they are equal you pool the two sample variances.
June 24th 2009, 06:50 AM
I would say that we are assuming equal population variances.
June 24th 2009, 06:55 AM
you can test to see if the variances are equal via an F test.
If the pop st deviations are equal then use ...
Unequal sample sizes, equal variance at http://en.wikipedia.org/wiki/T-test
June 27th 2009, 09:46 AM
N: salary of nurses
T: salary of teachers
You are testing
H0 : E(T) = E(N)
H1 : E(T) > E(N), E(T) - E(N) > 0
with 1% significance (or 99% confidence).
Some parts are in portuguese. But i think you cant get it anyway. (GL is Degrees of Freedom).
You assume variances are different, because the populations are of small dimension, and you'd have to estimate the common variance.
There's another way which is to setup an interval of confidence of 99%.
You do it for the difference of the expected values.
GL is degrees of freedom
NA is size of population A
SA is the estimator of the variance of population A (since it is of small dimension we have to use the estimator, which you calculate the same way as the variance)
alpha on this case will be 0.01
If the interval goes from positive values to positives values [+,+] you know the expected value of the 1st is bigger than the second, and vice versa. Otherwise you can't conclude anything.
For the p value, you have to go with the first method, so this second one would be just to answer the first part of the question. | {"url":"http://mathhelpforum.com/advanced-statistics/93611-two-sample-test-hypothesis-print.html","timestamp":"2014-04-21T00:10:20Z","content_type":null,"content_length":"6671","record_id":"<urn:uuid:cce340fd-0600-434e-9edc-dee46aab32c5>","cc-path":"CC-MAIN-2014-15/segments/1397609539337.22/warc/CC-MAIN-20140416005219-00389-ip-10-147-4-33.ec2.internal.warc.gz"} |
[Haskell-cafe] Tutorial: Curry-Howard Correspondence
Dan Weston westondan at imageworks.com
Thu Oct 18 16:59:49 EDT 2007
> The function needs to be total. You seem to be using Haskell to execute
> a function to see if it terminates as a proof of totality. Is that
> fair? This approach might work for some simple examples, but if the
> function doesn't terminate immediately then what? I would assume that
> proof of
> totality would be an obligation of the person who constructed the
> function.
>> thm2 :: (Prop a -> Prop a) -> Prop a
>> thm2 f = Prop undefined
Prop is not total, but thm2 is (if I am not mistaken) total. Every
non-_|_ f results in a non-bottom result (namely, Prop _|_).
My (still hesitant) assertion is that the totality must be transitive to
all sub-expressions for the function to be considered a valid proof. The
easiest way to get Haskell to do this work for you is to make everything
strict and just evaluate the function.
Tim Newsham wrote:
> I'm fairly new to this and struggling to follow along so I might
> be a little off base here...
>> I assume you mean then that it is a valid proof because it halts for
>> *some* argument? Suppose I have:
>> thm1 :: (a -> a) -> a
>> thm1 f = let x = f x in x
>> There is no f for which (thm1 f) halts (for the simple reason that _|_
>> is the only value in every type), so thm1 is not a valid theorem.
>> Now we reify our propositions (as the tutorial does) in a constructor:
>> data Prop a = Prop a
>> thm2 :: (Prop a -> Prop a) -> Prop a
>> thm2 f = Prop undefined
>> fix :: (p -> p) -> p
>> fix f = let x = f x in x
>> instance Show (Prop a) where
>> show f = "(Prop <something>)"
>> *Prop> :t thm2
>> thm2 :: (Prop a -> Prop a) -> Prop a
>> *Prop> thm2 (fix id)
>> (Prop <something>)
>> Wow! thm2 halts. Valid proof. We have a "proof" (thm2 (fix id)) of a
>> "theorem" (((Prop a) -> (Prop a)) -> (Prop a)), assuming that can
>> somehow be mapped isomorphically to ((a -> a) -> a), thence to
>> intuitionist logic as ((a => a) => a).
> The function needs to be total. You seem to be using Haskell to execute
> a function to see if it terminates as a proof of totality. Is that
> fair? This approach might work for some simple examples, but if the
> function doesn't terminate immediately then what? I would assume that
> proof of
> totality would be an obligation of the person who constructed the
> function. Using Haskell to prove termination for some simple cases might
> be fair (in which case I see your point about avoiding a non-lazy
> constructor), but it doesn't seem to be very general anyway. If you're
> relying on the programmer to provide proof of totality, then I don't
> see the harm in using "Prop a" instead of "a".
>> That "somehow" in the tutorial seems to be an implied isomorphism from
>> Prop a to a, so that proofs about (Prop a) can be interpreted as
>> proofs about a. I hope to have shown that unless without constructors
>> strict in their arguments that this is not valid.
>> My hypothesis was that
>> data Prop a = Prop !a
>> justified this isomorphism. Or am I still just not getting it?
> That would allow you to use the dynamic property (I observed termination)
> to verify the static property (the function is total) in some situations,
> right? But the static property of totality shouldn't rely on evaluation
> strategy, right?
>> Dan
> Tim Newsham
> http://www.thenewsh.com/~newsham/
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Topological properties of sets definable in weakly o-minimal structures
A first order structure M equipped with a dense linear ordering is called weakly o-minimal iff all definable subsets of M are finite unions of convex sets. In the first part of the talk we will
discuss some properties of the topological dimension of sets definable in weakly o-minimal structures. This will constitute a basis for the second part which will be focused on the problem of
topologisation of groups, group actions and fields definable in weakly o-minimal structures. | {"url":"http://www.newton.ac.uk/programmes/MAA/wencel.html","timestamp":"2014-04-19T22:47:06Z","content_type":null,"content_length":"2441","record_id":"<urn:uuid:435b2b41-97c0-48ea-9692-394111d856e2>","cc-path":"CC-MAIN-2014-15/segments/1397609537754.12/warc/CC-MAIN-20140416005217-00237-ip-10-147-4-33.ec2.internal.warc.gz"} |
Is there a good metric under which a sequence of compact sets can converge to an infinite dimensional set?
up vote 3 down vote favorite
I have a sequence of finite-dimensional, compact sets in $L^2(\Omega)$, where $\Omega\subset \mathbf{R}^2$ is closed and bounded. The dimension grows monotonically with the sequence, and there is no
assumption that the sets are nested or anything. I would like to say something about the "limiting set", but sequence doesn't converge under standard "set metrics" like the Hausdorff metric (for
example, in the Hausdorff psuedometric, a limit of compact sets, if it exists, is compact, so long as the underlying metric space is complete).
Does anyone know a way to talk about convergence of finite-dimensional compact sets to an infinite-dimensional set?
P.S. This is my first post. I read the rules, but my apologies in advance for any mistakes.
infinite-dim-manifolds gn.general-topology ap.analysis-of-pdes
1 There are stronger norms defined on subsets of $L^2$ whose unit balls are compact in $L^2$. If in your problem the sets are uniformly bounded in such a norm, maybe you could say something? –
Anthony Quas Nov 15 '12 at 20:35
2 "Infinite-dimensional" does not imply "non-compact". Some sequences with dimension going to infinity actually converge in the Hausdorff metric. – Sergei Ivanov Nov 15 '12 at 22:19
@Sergei: Hilbert cube! – Nik Weaver Nov 16 '12 at 1:09
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3 Answers
active oldest votes
Ok. Suppose $\mathcal{F}_n \subset L^2(\Omega)$ such that $\mathcal{F}_n$ is spanned by $n$ linearly independent "vectors" with bounded norm, which means $L^2(\Omega)$ functions such
that $||f||\leq C$ for all $f \in \mathcal{F}_n$. Then being finite-dimensional subspaces these sets will be compact in the strong topology.
up vote 1 However, in general, the limit will almost never be compact, since in this case it will recover a ball in $L^2(\Omega)$ which is not compact in the strong topology. You are going to need
down vote to make some more assumptions on the relationship between the sets to recover some information on the limit.
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As a general idea, without any other information on a sequence, it seems rather unlikely to find a natural metric in which it converges. Yet a metric in which it has a convergent
subsequence is a more reasonable task; then you may prove that your sequence actually converges there, by means of additional informations on the particular sequence.
up vote 1 Here, if these compact sets $C_n$ are included in a closed ball $B$ of $L^2(\Omega)$, you may use the Hausdorff distance induced from the weak topology, which makes $B$ a metric compact
down vote space. So $\mathcal{H}(B)$ is a compact metric space, and your sequence has a convergent subsequence there, that converges to a weakly compact set. Otherwise you may just consider the trace
on each ball, $C_n\cap \bar B( 0,r)$, and get by a standard diagonal argument a subsequence $C_ {n _ k}$ that converges on the Hausdorff distance of the weak topology on bounded sets.
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If you do not have a limit, but want to talk about it you say "ultralimit".
Let $(K_n)$ be your sequence of compact sets (in any complete metric space not nesessury $L^2(\Omega)$). Consider sequence of functions $$f_n=\mathop{\rm dist}\nolimits_{K_n}.$$ It is a a
up vote 0 sequence of 1-Lipschtz functions, so you can pass to its ultralimit $f_\omega$ for a fixed in advance ultrafilter $\omega$. The zero set $K_\omega$ of $f_\omega$ can be considered as the
down vote ultralimit of $K_n$.
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Richard Hoagland - Not Credible
Did Aliens use the same units of measure we did? Surely and logicaly the answer is a giant NO! I'll do some of my math and see is he's on to something or not.
In this picture he uses to discribe his point that the numbers are too perfect for it to be a natural occurance. Except the numbers he uses are going to produce the answer he wanted. Iapetus has an
orbital tilt of not ~16° but 7.52°. This inacuaracy in his data may have been due to an unreliable source of information (himself) or he used a protractor against the comupter screen. As for it's
diameter, which he should have used SI units, kilometers (to make his calculations with formulas correct but also make him sound more like a scientist). In miles 892.29 miles is the accurate number,
it should be in kilometers which is 1436 kilometers. He uses Saturn's radiis did he tell you what the radii is? No. He just says that is 60 of them far. It's radii is 60268 kilometers.
60 * 60268 = 3,616,080 KM < This is the orbital radii
Iapetus' actual orbital radii 3,561,300 it has a good half a million kms over his answer.
Accuracy = (3616080 - 3561300) / 3561300 = %accuacy
Accuracy = 1.5%
I'm not going to lie to you people, but he is close, very close, only 1.5% off what is the accepted value I have assigned, this shocked me, big time. Run that again substituting miles.
Accuracy = (2246927.940826 - 2212889.226915) / 2212889.226915 = %accuacy
Accuracy = 1.5%
If his calculation was off by 15% then we have a BS. 1.5% is amazingly accurate.
I guess math works both ways, right? 1 point for Hoagie
What about that inaccuracy in in the orbital angle? The use of a "~" means in other words anywhere on the number line resonable close to 16, or in his math from 0.00 - what ever I feel like. How do
you get ~16° from 7.52° which is the know fact? Did aliens use degrees? How about Radians? or Gradians? Did they even use the same system of measument of angles we do? Or the same way of counting
altogether? We are using high school math, aliens are bound to have a more sophisticated form of it. Okay enough of that. Lets see what we learned can change his answer in the picture.
~15 * ~60 = ~900 < scary alien math
7.52 * 60 = 451.2 < sounds natural to me
Hoagland: 1 - GoldEagle: 2
I'll be back with more so stay tuned. | {"url":"http://www.abovetopsecret.com/forum/thread140532/pg2","timestamp":"2014-04-18T03:01:21Z","content_type":null,"content_length":"66672","record_id":"<urn:uuid:ad895d31-f746-4cea-994a-9bba9cf19003>","cc-path":"CC-MAIN-2014-15/segments/1397609532480.36/warc/CC-MAIN-20140416005212-00271-ip-10-147-4-33.ec2.internal.warc.gz"} |
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Guess a number with at most one wrong answer
up vote 11 down vote favorite
Consider a game where one player picks an integer number between 1 and 1000 and other has to guess it asking yes/no questions.
If the second player always gives correct answers than it's clear that in worst case it's enought to ask 10 questions. And 10 is the smallest such number.
What if the second player is allowed to give wrong answers? I'm interested in a case when the second player is allowed to give at most one wrong answer.
I know the strategy with 15 guesses in worst case. Consider a number in range [1..1000] as 10 bits. At first you ask the values of all 10 bits ("Is it true that $i$-th bit is zero?"). After that you
get some number. Ask if this number is the number first player guessed. And if not you have to find where he gave wrong answer. There are 11 positions. Using the similar argument you can do it in 4
Is it possible to ask less then 15 questions in worst case?
4 What if the person lies when you go for the number on Question 11? – Bruce Westbury Jul 17 '10 at 8:02
This is a standard ECC (en.wikipedia.org/wiki/Error-correcting_code) problem. – BlueRaja Jul 18 '10 at 7:22
BlueRaja -- no, it's not: see Peter Shor's comment to falagar's answer. – JBL Jul 18 '10 at 14:37
You can see the difference between adaptive and non-adaptive (i.e. ECC's) in shreevatsa's answer below. – Peter Shor Jul 18 '10 at 18:43
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5 Answers
active oldest votes
Yes, there is a way to guess a number asking 14 questions in worst case. To do it you need a linear code with length 14, dimension 10 and distance at least 3. One such code can be
built based on Hamming code (see http://en.wikipedia.org/wiki/Hamming_code).
Here is the strategy.
Let us denote bits of first player's number as $a_i$, $i \in [1..10]$. We start with asking values of all those bits. That is we ask the following questions: "is it true that i-th
bit of your number is zero?" Let us denote answers on those questions as $b_i$, $i \in [1..10]$.
Now we ask 4 additional questions:
Is it true that $a_{1} \otimes a_{2} \otimes a_{4} \otimes a_{5} \otimes a_{7} \otimes a_{9}$ is equal to zero? ($\otimes$ is sumation modulo $2$).
Is it true that $a_{1} \otimes a_{3} \otimes a_{4} \otimes a_{6} \otimes a_{7} \otimes a_{10}$ is equal to zero?
Is it true that $a_{2} \otimes a_{3} \otimes a_{4} \otimes a_{8} \otimes a_{9} \otimes a_{10}$ is equal to zero?
Is it true that $a_{5} \otimes a_{6} \otimes a_{7} \otimes a_{8} \otimes a_{9} \otimes a_{10}$ is equal to zero?
Let $q_1$, $q_2$, $q_3$ and $q_4$ be answers on those additional questions. Now second player calculates $t_{i}$ ($i \in [1..4]$) --- answers on those questions based on bits $b_j$
which he previously got from first player.
Now there are 16 ways how bits $q_i$ can differ from $t_i$. Let $d_i = q_i \otimes t_i$ (hence $d_i = 1$ iff $q_i \ne t_i$).
up vote 24 down
vote accepted Let us make table of all possible errors and corresponding values of $d_i$:
position of error -> $(d_1, d_2, d_3, d_4)$
no error -> (0, 0, 0, 0)
error in $b_1$ -> (1, 1, 0, 0)
error in $b_2$ -> (1, 0, 1, 0)
error in $b_3$ -> (0, 1, 1, 0)
error in $b_4$ -> (1, 1, 1, 0)
error in $b_5$ -> (1, 0, 0, 1)
error in $b_6$ -> (0, 1, 0, 1)
error in $b_7$ -> (1, 1, 0, 1)
error in $b_8$ -> (0, 0, 1, 1)
error in $b_9$ -> (1, 0, 1, 1)
error in $b_{10}$ -> (0, 1, 1, 1)
error in $q_1$ -> (1, 0, 0, 0)
error in $q_2$ -> (0, 1, 0, 0)
error in $q_3$ -> (0, 0, 1, 0)
error in $q_4$ -> (0, 0, 0, 1)
All the values of $(d_1, d_2, d_3, d_4)$ are different. Hence we can find where were an error and hence find all $a_i$.
2 +1- This is really clever! – Dylan Wilson Jul 17 '10 at 8:32
5 This is a non-adaptive strategy (meaning the questions don't depend on previous answers. For one lie, I think non-adaptive and adaptive strategies give the same answer, but this
is no longer the case for more than one lie. See the survey article by Pelc referenced in another answer. – Peter Shor Jul 17 '10 at 14:47
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BTW, this problem is known as the Ulam(-Renyi) liar problem or Ulam's searching game (or just "playing Twenty Questions with a liar"), and has an extensive literature. The following is a
survey as of 2002:
• Andrzej Pelc, Searching games with errors--fifty years of coping with liars, Theoretical Computer Science, Volume 270 (2002), pp. 71-109
In particular, with 1 lie allowed, to guess a number in {1…n} where n is even, the number of queries needed is the smallest integer q which satisfies n ≤ 2^q/(q+1), which for n=1000 is
indeed 14. There are alternative solutions to the one-lie game in more recent papers like this and this. As observed by Peter Shor in a comment above, the general adaptive strategy when
multiple lies are allowed does not look like Hamming codes.
Edit: Since this has been bumped up, I may as well mention a nice result in the more general setting, proved by Joel Spencer and Peter Winkler in their paper Three Thresholds for a Liar.
up vote It is traditional to name the two players Paul and Carole, where Paul (named after Paul Erdős) is the one who asks the questions, and Carole (an anagram of oracle) is the one who answers
17 down them. Paul asks $q$ questions in all, and Carole is allowed to lie a fraction $r$ of the time. We will consider three progressively harder (for Paul) versions of what this means. In Version
vote A, Carole is allowed to lie at most $\lfloor ri\rfloor$ times to the first $i$ questions, for all $i$. In Version B, Carole is only required to lie at most $\lfloor rq \rfloor$ times in
total — she can choose to exhaust all her lies at the beginning, for instance. In Version C (nonadaptive), Paul must ask all his questions in one batch, and Carole can choose up to $\lfloor
rq \rfloor$ ones to lie to.
Note that if no lies are allowed ($r = 0$), the number of questions needed is exactly $\lceil \log_2 n\rceil$, and that, intuitively, if $r$ is too large, Paul cannot guess correctly at all.
Specifically, they show that:
• In version A, Paul wins with $\Theta(\log n)$ questions if $r < 1/2$, but Carole wins if $r \ge 1/2$.
• In version B, Paul wins with $\Theta(\log n)$ questions if $r < 1/3$, but Carole wins if $r \ge 1/3$.
• In version C, Paul wins with $\Theta(\log n)$ questions if $r < 1/4$, Carole wins if $r > 1/4$, and if $r = 1/4$, Paul wins but needs $\Theta(n)$ questions.
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I wrote an expository paper on this kind of problem, http://www.austms.org.au/Publ/Gazette/2009/May09/TechPaperMeyerson.pdf
up vote 7 down vote
Gerry, this discloses the info about yourself! I would be happy to add +1 more for your openness. – Wadim Zudilin Jul 17 '10 at 14:32
7 Wadim, considering the subject of this thread, it is of course possible that in alleging that I wrote the paper, I have given my one wrong answer! – Gerry Myerson Jul 18 '10 at
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The other answers are truly excellent and have settled the intended question. For a bit of fun, however, allow me to mention the following paradoxical solution.
Namely, with a certain precise and reasonable understanding of the rules of your game, which I shall presently give, I claim that no additional questions are required for the lie-telling game
over the truth-telling game. In particular, in your case 10 questions still suffice!
Specifically, to be a bit more definite about what it means to give a wrong answer, I propose that the rules should allow that the second player, at most once during the game, decides that a
given round will be a lie-telling round, for which he will privately ponder the correct truthful answer, but then give as his answer precisely the opposite of the correct answer. So if a
truthful answer would have been Yes, then on this lie-telling round he says No, and conversely. (In particular, in this version of the game, the wrong answer is not a random answer in any
sense, although it could be that the choice of which round is to be a lie-telling round is determined randomly.) On the other rounds, he tells the truth. Secondly, I note that you didn't
insist that the questions of player 1 must have a particular form.
With these rules for the liar game, I claim that no additional questions over the fully truthful case are required to determine the secret number.
The reason is a simple logic trick: if in the fully truthful game, one would want on a round to ask a question $Q$, then in this liar game, one should instead ask the question $P$:
up vote • If I were to have asked $Q$ on this round, would you have said Yes?
6 down
vote Consider how the second player will react. First, if he is on a truth-telling round, then he will give the same answer to this question that he would have given to $Q$. If in contrast he is
on a lie-telling round, then he ponders question $P$, and considers that if the first player had asked $Q$ on this round and a truthful answer had been Yes, then he would have said No, since
this is a lie-telling round, and so a truthful answer to $P$ is No, but since it is a lie-telling round, he answers Yes to $P$. Similarly, if a truthful answer to $Q$ would have been No, then
a lie-telling answer to $Q$ would be Yes, and so a truthful answer to $P$ would be Yes, but since it is a lie-telling round, he answers No. Thus, because of the double-negation effect, the
lie-telling answer to $P$ is the same as the truth-telling answer to $Q$.
Therefore, the first player can in effect gain exactly the same information from the second player in the liar game that he can in the fully truth-telling game.
The same argument shows that, in fact, it doesn't matter how often the second player decides to lie, as long as he lies by stating each time the exact opposite of a truthful answer. Indeed,
the second player could randomly decide for each round whether he will lie or tell the truth on that round, but the double-negation trick of question $P$ allows the first player nevertheless
to gain exactly the same information, and so no additional questions beyond the truth-telling case are required, even if the second player decides randomly at the beginning of every round
whether to lie or tell the truth on that round.
One could alternatively use the question: Does $Q$ hold if and only if this is a truth-telling round? – Joel David Hamkins Jul 18 '10 at 3:01
Nice trick. :-) The standard formulation of the game avoids this possibility (so that lies do matter) by requiring that questions must be of the form "Does the number lie in set A?" for
some $A \subseteq [n]$. – shreevatsa Jul 18 '10 at 4:08
Shreevatsa, in that case, I would make `A_P=\{\,n\,|\,n\in A_Q\iff\text{this is truth-telling round }\}$'. – Joel David Hamkins Jul 18 '10 at 11:35
Yeah, you're right; ignore my previous comment. It has nothing to do with the form of the question; it's rather that the notion of a "lie-telling round" which forces Carole to commit to
being a "liar" even internally is too restrictive and self-referential (like the "one who always lies" logic puzzles). To lie here is to give an answer other than the truth, and your
restriction on the round effectively takes away that option... I guess the original questioner's statement of "at most one wrong answer" is better after all. :-) – shreevatsa Jul 18 '10 at
1 Shreevatsa, you could say that player 1 must list the elements of $A$ explicitly, since my description of $A$ is a set that only player 2 can compute, and then your remark would regain its
effect. – Joel David Hamkins Jul 19 '10 at 1:21
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Some more references.
Joel Spencer's web page has several downloadable papers on searching with lies:
up vote 1 down
vote Ivan Niven, "Coding Theory applied to a Problem of Ulam", http://www.jstor.org/pss/2689543 , gives the Hamming code approach.
Andrszej Pelc, who solved the original Ulam liar problem with $n=10^6$ and one optional lie, also has a number of papers on extensions of the problem to more lies and to other models
of searching with noisy queries: http://w3.uqo.ca/pelc/search.html
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Fat-Tailed Butterflies Invade the Finance Faculty Lounge
Book Review
Mandelbrot, Benoit and Hudson, Richard L. "The (mis)Behavior of Markets: A Fractal View of Risk, Ruin, and Reward," New York: Perseus Books Group, 2004, pp. 328, Cloth. US$27.50.
Most literate humans have by now heard of "chaos theory." There has even been a Hollywood movie, "the Butterfly Effect," intended to illustrate the great clich? of the field, the notion that the flap
of a butterfly's wings in Brazil can stimulate a tornado in Texas.
We owe that illustration to chaos pioneer Edward Lorenz, a meteorologist at the Massachusetts Institute of Technology. He concluded in a 1962 paper that there are limits to weather prediction--since
it is impossible to know how every "seagull" everywhere will flap its wings. (In later speeches and essays he substituted the more picturesque butterfly.)
In its study of such limits of the knowable, chaos theory has come to incorporate fractal geometry, the study of figures with a kind of complexity that proves independent of scale. In a fractal
figure, an attempt to break down a whole into simpler parts is foiled when the parts turn out to be just as complex as the whole and to be themselves composed of parts equally complex--and so on
indefinitely. Fractal geometry turns its practitioners into Seussian elephants, looking for tiny but complete worlds on each speck of dust, in the expectation that those worlds will contain their own
specks of dusts, with their own still smaller worlds.
The best-known fractal figure is the "Mandelbrot set," which takes its name from the author of this book, Polish (French-educated) mathematician Benoit Mandelbrot. It's called a "set" because it's
defined by the points on a plane that satisfy an equation. The equation itself, although relatively simple, contains a feedback loop that results in the intricacies of this shape--with its arms made
out of swirling galaxies that themselves prove to contain arms made out of swirling galaxies.
In this book, Mr. Mandelbrot addresses the "modern house of finance." He believes that the building is cracked, and the problems go to the foundations. Fractal geometry, he writes (with the
assistance of Richard Hudson, former managing editor of the Wall Street Journal's European edition) will repour those foundations.
It is a common observation by now that bell curves in finance always have "fat tails," i.e. the odds of an extraordinary event are far greater than one would expect were every tick in asset price a
step in a random walk, on the model proposed by Louis Bachelier more than a century ago. Finance theorists haven't done anything fundamental about these fat tails. They have conceded their existence
and sought to tack them on as amendments to the underlying random-walk model, the very model that makes them an anomaly.
By way of repouring the foundation, though, Messrs. Mandelbrot and Hudson propose, first, that we should regard the volatility of any market as an instance of a "power law," i.e. a correlation in
which the size of a price change varies with a power of the frequency of the change. The bell curve would be a special case, a power law in which the power equals -2. Far greater volatility exists in
assets that show what is known as a Cauchy distribution, in which the power equals -1. The absence of any correlation between size and frequency of change (an asset manager's nightmare in which a
crash is just as likely at any moment as a one-penny tick) would be a power of zero. Empirical research shows that the actual power (which these authors call alpha and usually state without the
negative sign, which can be taken as implied) varies between 2 and 1, between "mild" and "wild." In the case of cotton prices, for example, alpha is 1.7.
Furthermore, asset prices exhibit a constant volatility across time scales. "A month looks like a day, one set of days like another," just as a fractal geometer would expect.
The authors also propose that the assumption of the independence of jumps be abandoned. In this case, the commonsense of traders has always conflicted with finance theory. Traders routinely speak of
"hot streaks," or "momentum." "All illusion!" the theorists have replied, in the spirit of Bachelier and the random walk theory. Each step a drunk takes as he staggers about a lamppost is independent
of the step he took before, or the one he will take next.
Here, Messrs Mandelbrot and Hudson side with the traders, and they use illustrations that remind me a bit of that famous butterfly. "Think of a small country, like Sweden, where every big company
does business, directly or indirectly, with every other one. Volvo does something that affects Saab--say, launches a new car model that steals market share. Saab comes back with a fancier car, making
satellite-location services standard ... so Ericsson starts selling more Global Positioning System receivers. And so it spins on, throughout the Swedish economy--and spilling gradually into
neighboring Finland ... and as far around the world in ever-diminishing ripples as we can measure it." This mutual dependence of the commercial world will naturally be echoed in, and then feed back
into that world from, financial markets, the authors suggest. Furthermore, the dependence of future upon past events does much work in creating those apparently unlikely events, cascading bad fortune
of the sort that destroyed Long-Term Capital Management. The butterflies are fat tailed, one might say, in defiance of entomology.
But then, what is to be done? What does all this mean for, say, the chief financial officer of a corporation, trying to hedge his company's currency position? What does it mean for the hedge funds or
speculators who might be tempted to strike a deal with that CFO? What does it mean for the ordinary working stiff trying to invest some of his savings wisely in anticipation of retirement? It means
very little, as yet. The authors acknowledge that their "multifractal model of asset returns" is still in its infancy. Indeed, although they acknowledge that fractals are now in vogue in some
financial circles, they are wary of this, "as with any fashion there is often more show than substance." They believe that much fundamental research is necessary before a new model can replace the
Since I am almost comically underqualified to pass judgments on the merits of this book, I convened (through cyberspace) a panel of experts, consisting of authors who have written well-received books
on related subjects, and I asked my panels' opinion. Perhaps the consensus opinion was best expressed by Salih Neftci, author of "An Introduction to the Mathematics of Financial Derivatives," (1996).
Mr. Neftci agreed that financial data do exhibit more that just random walk behavior. He agreed, too, power laws "do not turn out to be integers, but fractions. This effect is fairly significant."
The fractal model, then, "is a useful avenue to pursue but it is also one of several possible approaches."
Mr. Neftci's views might be especially significant in that he, a professor at the Graduate School of the City University of New York, is a pillar of the establishment that Messrs. Mandelbrot and
Hudson propose to overthrow. We can perhaps take it as a given, then, that this particular establishment, the masters of the house of modern finance, are well aware of the cracks to which their
critics point and don't even quarrel with the proposition that a new foundation might be necessary. They are, though, sensibly curious about what exactly the new contractor proposes to do and want to
see at least a full blueprint before their existing abode is razed.
Contact Bob Keane with questions or comments at: bkeane@investmentadvisor.com</a.
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Complex Problem
February 5th 2009, 07:26 AM #1
Junior Member
Mar 2008
Complex Problem
Let X be the path with parametrisation
f(t)=2-i-3e^(it) (t E [1/2pi, 2pi]
Sketch X
I think I know how to sketch the graph once I get the parametric equations, but I cant seem to get my head around them.
I know that the parametric equation for x will have 2 in it and the one for y will have the -i which will become -1, but its the 3e^(it) value that is confusing me. Do I need to do something to
this value to make it more simplified? Any help would be great?
$f(t)=2-i-3 (\cos t + i \sin t)$
$x(t)=2 - 3 \cos t$
$y(t)=-1-3 \sin t$
Thanks for your help. I am finally able to sketch the graph
February 5th 2009, 08:26 AM #2
MHF Contributor
Nov 2008
February 6th 2009, 04:39 AM #3
Junior Member
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Mathematics Placement Information
Interpreting test results:
Basic Algebra
Score Placement
9 or below places into MATH 100
10 or above places out of MATH 100
Functions and Graphs
Score Placement
412 or above, with a trig subscore of 9 or above places into MATH 128
Trig subscore of 11 or above places into MATH 128
411 or below, with a trig subscore of 10 or below places into MATH 127
Options for retaking placement exams:
For students who place into Math 100:
A student who has not yet matriculated may attempt to place out of MATH 100 by scheduling with the Registrar’s office to take the placement test given the day before the first day of classes. After
matriculating, a student may attempt to exempt MATH 100 by arranging with the Freshmen Dean to take the competency exam on the second day of classes in the student's second or third semester.
Potential majors in Astronomy, Chemistry, Computer Science, Cooperative Engineering, Mathematics, Medical Technology, or Physics should schedule MATH 100 in the first semester, if they do not
successfully place out of, or exempt, MATH 100.
Potential majors in Accounting, Biology, Business, Psychology, or Sociology should schedule MATH 100 as soon as possible, if they do not successfully place out of, or exempt, MATH 100.
For students who place into Math 127:
A student may retake the Functions and Graphs placement exam once prior to matriculation. Contact the Registrar’s Office for details. After matriculating, this student may take the three-hour Math
127 exemption exam, one time. Contact the Math Department Chairperson for details.
Potential majors in Astronomy, Chemistry, Computer Science, Cooperative Engineering, Mathematics, or Physics should schedule MATH 127 in the first semester, if they do not successfully place out of,
or exempt, MATH 127.
Mathematics Distribution Requirement:
For the official description of the Mathematics distribution requirement, see College Catalog, Section "The Academic Program", The Distribution Program, Section E. | {"url":"http://www.lycoming.edu/mathematics/placement.aspx","timestamp":"2014-04-16T07:23:53Z","content_type":null,"content_length":"14965","record_id":"<urn:uuid:39f1a829-a638-41bf-8808-ac41e9476c84>","cc-path":"CC-MAIN-2014-15/segments/1398223201753.19/warc/CC-MAIN-20140423032001-00325-ip-10-147-4-33.ec2.internal.warc.gz"} |
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Chapter 12 - What is Malted Grain?
12.3 Extraction and Maximum Yield
All of these grains can be used to produce the fermentable sugars that make up the wort. But to brew the same beer recipe consistently, we need to be able to quantify how much yield we can expect
from each type of grain. Under laboratory conditions, each grain will yield a typical amount of fermentable and non-fermentable sugars that is referred to as its percent extraction or maximum yield.
This number ranges from 50 - 80% by weight, with some wheat malts hitting as high as 85%. This means that 80% (for example) of the malt's weight is soluble in the laboratory mash. (The other 20%
represents the husk and insoluble starches.) In the real world, we brewers will never hit this target, but it is useful for comparison.
The reference for comparison is pure sugar (sucrose) because it yields 100% of its weight as soluble extract when dissolved in water. (One pound of sugar will yield a specific gravity of 1.046 when
dissolved in 1 gallon of water.) To calculate the maximum yield for the malts and other adjuncts, the percent extraction for each is multiplied by the reference number for sucrose-46 points/pound/
gallon (ppg).
For example, let's look at a typical pilsner base malt. Most light base malts have a maximum yield of 80% by weight of soluble materials. So, if we know that sugar will yield 100% of its weight as
soluble sugar and that it raises the gravity of the wort by 46 ppg, then the maximum increase in gravity we can expect from pilsner base malt, at 80% solubility, is 80% of 46 or 37 ppg.
The typical maximum yields for the malts are listed in Table 9. You may be wondering how useful the maximum yield number of a malt can be if you can never expect to hit it. The answer is to apply a
scaling factor to the maximum yield and derive a number we will usually achieve - a typical yield. | {"url":"http://howtobrew.com/section2/chapter12-3.html","timestamp":"2014-04-17T13:08:41Z","content_type":null,"content_length":"17596","record_id":"<urn:uuid:032cc52a-28b6-47d2-99de-457af1da4d0b>","cc-path":"CC-MAIN-2014-15/segments/1397609530131.27/warc/CC-MAIN-20140416005210-00367-ip-10-147-4-33.ec2.internal.warc.gz"} |
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Find all points of inflection: f(x)=(1/12)x^4-2x^2+15. I know that f"(x)=x^2-4 but what to do after that? Is (2, 0), (-2, 0) the answer?
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to get the x-coordinate of the point of inflection, set the 2nd derivative to zero, and solve x. whatever x you get, sub it in f(x) to give you the y-coordinate.
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A clustering technique for the identification of piecewise affine systems
Results 1 - 10 of 34
- In ECCV , 2006
"... Abstract. We cast the problem of motion segmentation of feature trajectories as linear manifold finding problems and propose a general framework for motion segmentation under affine projections
which utilizes two properties of trajectory data: geometric constraint and locality. The geometric constra ..."
Cited by 69 (0 self)
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Abstract. We cast the problem of motion segmentation of feature trajectories as linear manifold finding problems and propose a general framework for motion segmentation under affine projections which
utilizes two properties of trajectory data: geometric constraint and locality. The geometric constraint states that the trajectories of the same motion lie in a low dimensional linear manifold and
different motions result in different linear manifolds; locality, by which we mean in a transformed space a data and its neighbors tend to lie in the same linear manifold, provides a cue for
efficient estimation of these manifolds. Our algorithm estimates a number of linear manifolds, whose dimensions are unknown beforehand, and segment the trajectories accordingly. It first transforms
and normalizes the trajectories; secondly, for each trajectory it estimates a local linear manifold through local sampling; then it derives the affinity matrix based on principal subspace angles
between these estimated linear manifolds; at last, spectral clustering is applied to the matrix and gives the segmentation result. Our algorithm is general without restriction on the number of linear
manifolds and without prior knowledge of the dimensions of the linear manifolds. We demonstrate in our experiments that it can segment a wide range of motions including independent, articulated,
rigid, non-rigid, degenerate, non-degenerate or any combination of them. In some highly challenging cases where other state-of-the-art motion segmentation algorithms may fail, our algorithm gives
expected results. 2 1
- In Proc. of IEEE Conference on Decision and Control , 2003
"... We propose an algebraic geometric solution to the identification of a class of linear hybrid systems. We show that the identification of the model parameters can be decoupled from the inference
of the hybrid state and the switching mechanism generating the transitions, hence we do not constraint the ..."
Cited by 37 (11 self)
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We propose an algebraic geometric solution to the identification of a class of linear hybrid systems. We show that the identification of the model parameters can be decoupled from the inference of
the hybrid state and the switching mechanism generating the transitions, hence we do not constraint the switches to be separated by a minimum dwell time. The decoupling is obtained from the so-called
hybrid decoupling constraint, which establishes a connection between linear hybrid system identification, polynomial factorization and hyperplane clustering. In essence, we represent the number of
discrete states n as the degree of a homogeneous polynomial p and the model parameters as factors of p. We then show that one can estimate n from a rank constraint on the data, the coe#cients of p
from a linear system, and the model parameters from the derivatives of p. The solution is closed form if and only if n 4. Once the model parameters have been identified, the estimation of the hybrid
state becomes a simpler problem. Although our algorithm is designed for noiseless data, we also present simulation results with noisy data. 1
- Automatica , 2004
"... This paper addresses the problem of identification of hybrid dynamical systems, by focusing the attention on hinging hyperplanes (HHARX) and Wiener piecewise affine (W-PWARX) autoregressive
exogenous models. In particular, we provide algorithms based on mixed-integer linear or quadratic programming ..."
Cited by 22 (4 self)
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This paper addresses the problem of identification of hybrid dynamical systems, by focusing the attention on hinging hyperplanes (HHARX) and Wiener piecewise affine (W-PWARX) autoregressive exogenous
models. In particular, we provide algorithms based on mixed-integer linear or quadratic programming which are guaranteed to converge to a global optimum. For the special case where switches occur
only seldom in the estimation data, we also suggest a way of trading off between optimality and complexity by using a change detection approach. 1
- ERCIM News , 2003
"... The functioning and development of living organisms is controlled by large and complex networks of genes, proteins, small molecules, and their interactions, so-called genetic regulatory
networks. The concerted efforts of genetics, molecular biology, biochemistry, and physiology have led to the accum ..."
Cited by 19 (1 self)
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The functioning and development of living organisms is controlled by large and complex networks of genes, proteins, small molecules, and their interactions, so-called genetic regulatory networks. The
concerted efforts of genetics, molecular biology, biochemistry, and physiology have led to the accumulation of enormous amounts of data on the molecular components of genetic regulatory networks and
their interactions. Notwithstanding the advances in the mapping of the network structure, surprisingly little is understood about how the dynamic behavior of the system emerges from the interactions
between the network components. This has incited an increasingly large group of researchers to turn from the structure to the behavior of genetic regulatory networks, against the background of a
broader movement nowadays often referred to as systems biology
, 2005
"... Abstract. Recent advances of experimental techniques in biology have led to the production of enormous amounts of data on the dynamics of genetic regulatory networks. In this paper, we present
an approach for the identification of PieceWise-Affine (PWA) models of genetic regulatory networks from exp ..."
Cited by 17 (2 self)
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Abstract. Recent advances of experimental techniques in biology have led to the production of enormous amounts of data on the dynamics of genetic regulatory networks. In this paper, we present an
approach for the identification of PieceWise-Affine (PWA) models of genetic regulatory networks from experimental data, focusing on the reconstruction of switching thresholds associated with
regulatory interactions. In particular, our method takes into account geometric constraints specific to models of genetic regulatory networks. We show the feasibility of our approach by the
reconstruction of switching thresholds in a PWA model of the carbon starvation response in the bacterium Escherichia coli. 1
"... This tutorial paper is concerned with the identification of hybrid models, i.e. dynamical models whose behavior is determined by interacting continuous and discrete dynamics. Methods
specifically aimed at the identification of models with a hybrid structure are of very recent date. After discussing ..."
Cited by 10 (0 self)
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This tutorial paper is concerned with the identification of hybrid models, i.e. dynamical models whose behavior is determined by interacting continuous and discrete dynamics. Methods specifically
aimed at the identification of models with a hybrid structure are of very recent date. After discussing the main issues and difficulties connected with hybrid system identification, and giving an
overview of the related literature, this paper focuses on four different approaches for the identification of switched affine and piecewise affine models, namely an algebraic procedure, a Bayesian
procedure, a clustering-based procedure, and a bounded-error procedure. The main features of the selected procedures are presented, and possible interactions to still enhance their effectiveness are
- In Proceedings of IEEE American Control Conference , 2004
"... We consider the problem of identifying the orders and the model parameters of PWARX hybrid models from noiseless input/output data. We cast the identification problem in an algebraic geometric
framework in which the number of discrete states corresponds to the degree of a multivariate polynomial p a ..."
Cited by 8 (4 self)
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We consider the problem of identifying the orders and the model parameters of PWARX hybrid models from noiseless input/output data. We cast the identification problem in an algebraic geometric
framework in which the number of discrete states corresponds to the degree of a multivariate polynomial p and the orders and the model parameters are encoded on the factors of p. We derive a rank
constraint on the input/output data from which one can estimate the coefficients of p. Given p, we show that one can estimate the orders and the parameters of each ARX model from the derivatives of p
at a collection of regressors that minimize a certain objective function. Our solution does not require previous knowledge about the orders of the ARX models (only an upper bound is needed), nor does
it constraint the orders to be equal. Also the switching mechanism can be arbitrary, hence the switches need not be separated by a minimum dwell time. We illustrate our approach with an algebraic
example of a switching circuit and with simulation results in the presence of noisy data.
, 2002
"... Abstract. A new connectionist model for the solution of piecewise linear regression problems is introduced; it is able to reconstruct both continuous and non continuous real valued mappings
starting from a finite set of possibly noisy samples. The approximating function can assume a different linear ..."
Cited by 7 (0 self)
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Abstract. A new connectionist model for the solution of piecewise linear regression problems is introduced; it is able to reconstruct both continuous and non continuous real valued mappings starting
from a finite set of possibly noisy samples. The approximating function can assume a different linear behavior in each region of an unknown polyhedral partition of the input domain. The proposed
learning technique combines local estimation, clustering in weight space, multicategory classification and linear regression in order to achieve the desired result. Through this approach piecewise
affine solutions for general nonlinear regression problems can also be found. 1
"... Abstract — We consider the problem of recursively identifying the parameters of a switched ARX (SARX) model from input/output data under the assumption that the number of models, the model
orders and the switching sequence are unknown. Our approach exploits the fact that applying a polynomial embedd ..."
Cited by 4 (0 self)
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Abstract — We consider the problem of recursively identifying the parameters of a switched ARX (SARX) model from input/output data under the assumption that the number of models, the model orders and
the switching sequence are unknown. Our approach exploits the fact that applying a polynomial embedding to the input/output data leads to a lifted ARX model whose dynamics are linear on the so-called
hybrid model parameters and independent of the switching sequence. In principle, one can use a standard recursive algorithm to identify such hybrid parameters. However, when the number of models and
the model orders are unknown the embedded regressors may not be persistently exciting, hence the estimates of the hybrid parameters may not converge exponentially to a constant vector. Nevertheless,
we show that these estimates still converge to a vector that depends continuously on the initial condition. By identifying the hybrid model parameters starting from two different initial conditions,
we show that one can build two homogeneous polynomials whose derivatives at a regressor give an estimate of the parameters of the ARX model generating that regressor. After properly enforcing some of
the entries of the hybrid model parameters to be zero, such estimates are shown to converge exponentially to the true ARX model parameters under suitable persistence of excitation conditions on the
input/output data. Although our algorithm is designed for the case of perfect input/output data, our experiments also show its performance with noisy data. I.
- In Marinaro,M. and Tagliaferri,R. (eds.) Neural Nets: WIRN Vietri-01 , 2001
"... A new learning algorithm for solving piecewise linear regression problems is proposed. It is able to train a proper multilayer feedforward neural network so as to reconstruct a target function
assuming a different linear behavior on each set of a polyhedral partition of the input domain. The propose ..."
Cited by 3 (0 self)
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A new learning algorithm for solving piecewise linear regression problems is proposed. It is able to train a proper multilayer feedforward neural network so as to reconstruct a target function
assuming a different linear behavior on each set of a polyhedral partition of the input domain. The proposed method combine local estimation, clustering in weight space, classification and regression
in order to achieve the desired result. A simulation on a benchmark problem shows the good properties of this new learning algorithm. 1 | {"url":"http://citeseerx.ist.psu.edu/showciting?cid=3143","timestamp":"2014-04-16T07:06:21Z","content_type":null,"content_length":"40623","record_id":"<urn:uuid:94706aa2-0661-4354-978d-6ec2837a6f9d>","cc-path":"CC-MAIN-2014-15/segments/1397609521512.15/warc/CC-MAIN-20140416005201-00540-ip-10-147-4-33.ec2.internal.warc.gz"} |
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Ivan's article - III
After discussing heat losses in combined cycles and the three basic principles for GT CC cycles to achieve maximum efficiency, Ivan Rice talks about a specific strategy to inch aeroderivative
combined cycle efficiencies up to 60%.
Fuel gas heating is independent of any particular CC arrangement and is being used by some of the latest combined cycles. The cycle efficiency of the Combo 5 LM 6000s is increased right at one
percentage point. An average specific heat value of .640 BTU/Lb/oF obtained from the methane (natural gas) Mollier diagram between 100 and 600 o F at 500 psia has been used in the calculations.
(An LMS100 aeroderivative gas turbine)
The boiler supplementary firing temperatures have been calculated to be 1290 o F for 120 MW of steam turbine power and 1347 o F for 125 MW, both for 1050 o F steam. A rather high specific heat value
of .275 was used for the exhaust flow due to the extra water from Sprint and the water of combustion of the fuel.
The decrease in steam flow can be readily calculated by applying the various ratios of the TSRs involved. The effect of topping power at 3413 BTU/KWH can likewise be easily calculated which equates
to the fuel added. These calculations are based on the efficiency equation of:
CC Efficiency = Work Output/Heat Input
with the heat (fuel) input known of the 5 LM 6000s and the output work (GT plus ST) known of the referenced cycle being used as a starting point. Incremental additions of input and output have been
introduced to obtain individual results.
As of now, computer runs have not been done on this. Computer programs for combined cycles are most useful and helpful to obtain accurate and exact data but they cannot think through a process and
come up with new combined cycle approaches. The thermodynamic reasoning presented above regarding steam turbine topping power approaching 3413 BTU/KWH and the lower TSRs for the Combo 5 LM 6000 power
plant with supplementary fired steam super heating only in the central unit should be verified by computer runs for the two cycles. The ThermoFlow and the Gate computer programs have to be
manipulated for the split steam cycle as they were designed primarily for the conventional combined cycle.
Another example of steam splitting is the application of 2 uprated LMS 100 125 MW units as anticipated by using the 80E1 gas generator and modifying the first compressor and power turbine and using 1
uprated LM 6000 unit whereby only the HRSG of the LM 6000 is supplementary fired for heating all the initial superheat and reheat steam for 1250 psia 1100/1100 o F steam conditions. This Combo 3
system, using the current 80C2 100 MW LMS 100 and applying 1250 psia and 1050/1050 o F steam conditions, yields a CC efficiency of about 56 % with about 85 MW of steam turbine power produced for a
total of around 335 MW.
A low pressure inter cooler boiler with 2 to 4 psig steam pressure and 300 o F output can be used to produce admission steam to the steam turbine. Considering 40,000 LB/Hr per LMS 100, the steam
turbine will generate about 4 MW where the TSR is 18.5 Lb/KWH for 2" Hg condenser pressure for the two LMS 100s. The CC cycle efficiency is raised about one percentage point.
The CC efficiency increases to about 59 % with added steam turbine output and decreased fuel input when
(1) steam conditions are raised to 1100/1100 o F,
(2) fuel gas heating is incorporated and
(3) the IC low pressure admission steam is included.
The intercooler heat loss, if not incorporated in the cycle some way, will degrade the overall combined cycle efficiency by about 2 percentage points. Therefore, a workable solution has to be
incorporated. Perhaps the ORC propane cycle or a low steam pressure admission to the LPT section of the steam turbine can be applied to use up an optimum portion of the intercooler heat rejection and
inch up closer to a 60 % cycle efficiency level. The projected new 80E1 LMS 100 Combo 3 should top 60 % efficiency at an increase of about 70 MW to exceed 400 MW total output. Additional refinements
in each example can be made but are not discussed herein.
In the concluding part of this series, the author talks about the significance of flexible features of a proposed multi-cycle unit.
(This article is the third part of a series by the author)
Ivan G. Rice was past chairman of the South Texas Section of ASME (1974 - 75), past chairman of the ASME Gas Turbine Division (now IGTI) (1975 - 76). A Life Fellow Member of ASME and Life Member of
NSPE/TSPE, he has authored many articles and ASME papers on gas turbines, inter-cooling, reheat, HRSGs, steam cooling and steam injection. | {"url":"http://www.turbomachinerymag.com/blog/content/ivans-article-iii","timestamp":"2014-04-18T13:07:17Z","content_type":null,"content_length":"23718","record_id":"<urn:uuid:6699b1cb-4cb5-499c-86dd-c60c0e320bbf>","cc-path":"CC-MAIN-2014-15/segments/1398223204388.12/warc/CC-MAIN-20140423032004-00220-ip-10-147-4-33.ec2.internal.warc.gz"} |
piecewise problem
February 10th 2009, 12:00 PM #1
Junior Member
Feb 2009
A man walks for 45 minutes at a rate of 3 mph, then jogs for 75 minutes at a rate of 5 mph, then sits and rests for 30 minutes, and finally walks for 1 hour and half.Find the rule of the function
that expresses this distance traveled as a function of time [ Caution: Don't mix up the units of time ; use either minutes or hours , not both]
could someone help me with this problem? I sketched the graph .the ranges I got are from 0 to .75 , .75 to 2, 2 to 2.5 and 2.5 to 4 .However, I don't know how to write the rules.
Thanks in advance.
Last edited by vance; February 10th 2009 at 02:13 PM.
Here are some hints for you graph.
We're graphing distance as a function of time. So the x axis should represent time, and the y axis should represent distance.
You know the "rate" at which distance is changing. This rate should be the _____ of your function for a certain time (do you know which word goes in the blank?).
Bonus hint in case you don't know the word:
d = 3t (this is a formula for a man who is walking at 3 miles per hour).
well, I alredy know the 1st and 3rd rule, but I'm still triying to get the other two ones.
February 10th 2009, 03:30 PM #2
Jan 2009
February 10th 2009, 03:41 PM #3
Junior Member
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Search Results
Below results based on the criteria 'forecasting'
Total number of records returned: 11
Automated Production of High-Volume, Near-Real-Time Political Event Data
Schrodt, Philip
Uploaded 08-30-2010
event data
Keywords natural language processing
1 Paper open source
This paper summarizes the current state-of-the-art for generating high-volume, near-real-time event data using automated coding methods, based on recent efforts for the DARPA
Integrated Crisis Early Warning System (ICEWS) and NSF-funded research. The ICEWS work expanded by more than two orders of magnitude previous automated coding efforts, coding of
Abstract about 26-million sentences generated from 8-million stories condensed from around 30 gigabytes of text. The actual coding took six minutes. The paper is largely a general ``how-to''
guide to the pragmatic challenges and solutions to various elements of the process of generating event data using automated techniques. It also discusses a number of ways that this
could be augmented with existing open-source natural language processing software to generate a third-generation event data coding system.
Racing Horses: Constructing and Evaluating Forecasts in Political Science
Brandt, Patrick
Freeman, John R.
Schrodt, Philip
Uploaded 07-27-2011
2 political conflict
Paper scoring rules
Keywords model training
forecast density
verification rank histogram
probability integral transform
We review methods for forecast evaluations and how they can be used in political sciences. We examine how forecast densities are more useful summaries of forecasted variables than
Abstract point metrics. We also cover how continuous rank probability scores, probability integral transforms, and verification rank histograms can be used to calibrate and evaluate forecast
performance. Finally, we present two illustrations, one a simulation and the other a comparison of forecasting models for the China-Taiwan (cross-straits) conflict.
Dynamic Bayesian Forecasting of Presidential Elections in the States
Linzer, Drew
Uploaded 07-16-2012
Keywords Forecasting
Public Opinion
3 Elections
Paper I present a dynamic Bayesian forecasting model that enables early and accurate prediction of U.S. presidential election outcomes at the state level. The method systematically
combines information from historical forecasting models in real time with results from the large number of state-level opinion surveys that are released publicly during the campaign.
The result is a set of forecasts that are initially as good as the historical model, then gradually increase in accuracy as Election Day nears. I employ a hierarchical specification
Abstract to overcome the limitation that not every state is polled on every day, allowing the model to borrow strength both across states and, through the use of random-walk priors, across
time. The model also filters away day-to-day variation in the polls due to sampling error and national campaign eects, which enables daily tracking of voter preferences towards the
presidential candidates at the state and national levels. Simulation techniques are used to estimate the candidates' probability of winning each state and, consequently, a majority
of votes in the Electoral College. I apply the model to pre-election polls from the 2008 presidential campaign and demonstrate that the victory of Barack Obama was never
realistically in doubt. The model is currently ready to be deployed for forecasting the outcome of the 2012 presidential election. Project website: votamatic.org
Moving Mountains: Bayesian Forecasting As Policy Evaluation
Brandt, Patrick T.
Freeman, John R.
Uploaded 04-24-2002
Bayesian vector autoregression
4 Paper Keywords VAR
policy evaluation
conditional forecasting
Many policy analysts fail to appreciate the dynamic, complex causal nature of political processes. We advocate a vector autoregression (VAR) based approach to policy analysis that
Abstract accounts for various multivariate and dynamic elements in policy formulation and for both dynamic and specification uncertainty of parameters. The model we present is based on recent
developments in Bayesian VAR modeling and forecasting. We present an example based on work in Goldstein et al. (2001) that illustrates how a full accounting of the dynamics and
uncertainty in multivariate data can lead to more precise and instructive results about international mediation in Middle Eastern conflict.
How Factual is your Counterfactual?
King, Gary
Zeng, Langche
Uploaded 07-12-2001
Keywords causality
5 forecasting
Paper democracy
Inferences about counterfactuals are essential for prediction, answering ``what if'' questions, and estimating causal effects. However, when the counterfactuals posed are too far
from the data at hand, conclusions drawn from well-specified statistical analyses become based on speculation and convenient but indefensible model assumptions rather than empirical
evidence. Yet, standard model outputs do not reveal the degree of model-dependence, and so this problem can be hard to detect, regardless of its severity. We develop easy-to-apply
Abstract methods to evaluate counterfactuals that do not require sensitivity testing over specified classes of models. One analysis with these methods applies to the class of all models, for
any smooth conditional expectation function, and to the set of all possible dependent variables, given only the choice of a set of explanatory variables. We illustrate by studying
the scholarly literatures that try to assess the effects of changes in the degree of democracy in a country (on any dependent variable); we find widespread evidence that scholars are
inadvertently drawing conclusions based more on their hypotheses than on their empirical evidence.
Forecasting Conflict in the Balkans using Hidden Markov Models
Schrodt, Philip A.
Uploaded 08-24-2000
event data
Keywords hidden Markov models
This study uses hidden Markov models (HMM) to forecast conflict in the former Yugoslavia for the period January 1991 through January 1999. The political and military events reported
in the lead sentences of Reuters news service stories were coded into the World Events Interaction Survey (WEIS) event data scheme. The forecasting scheme involved randomly selecting
eight 100-event "templates" taken at a 1-, 3- or 6-month forecasting lag for high-conflict and low-conflict weeks. A separate HMM is developed for the high-conflict-week sequences
6 Paper and the low-conflict-week sequences. Forecasting is done by determining whether a sequence of observed events fit the high-conflict or low-conflict model with higher probability.
Models were selected to maximize the difference between correct and incorrect predictions, evaluated by week. Three weighting schemes were used: unweighted (U), penalize false
positives (P) and penalize false negatives (N). There is a relatively high level of convergence in the estimates‹the best and worst models of a given type vary in accuracy by only
about 15% to 20%. In full-sample tests, the U and P models produce at overall accuracy of around 80%. However, these models correctly forecast only about 25% of the high-conflict
weeks, although about 60% of the cases where a high-conflict week has been forecast turn out to have high conflict. In contrast, the N model has an overall accuracy of only about 50%
Abstract in full-sample tests, but it correctly forecasts high-conflict weeks with 85% accuracy in the 3- and 6-month horizon and 92% accuracy in the 1-month horizon. However, this is
achieved by excessive predictions of high-conflict weeks: only about 30% of the cases where a high-conflict week has been forecast are high-conflict. Models that use templates from
only the previous year usually do about as well as models based on the entire sample. The models are remarkably insensitive to the length of the forecasting horizon‹the drop-off in
accuracy at longer forecasting horizons is very small, typically around 2%-4%. There is also no clear difference in the estimated coefficients for the 1-month and 6-month models. An
extensive analysis was done of the coefficient estimates in the full-sample model to determine what the model was "looking at" in order to make predictions. While a number of
statistically significant differences exist between the high and low conflict models, these do not fall into any neat patterns. This is probably due to a combination of the large
number of parameters being estimated, the multiple local maxima in the estimation surface, and the complications introduced by the presence of a number of very low probability event
categories. Some experiments with simplified models indicate that it is possible to use models with substantially fewer parameters without markedly decreasing the accuracy of the
predictions; in fact predictions of the high conflict periods actually increase in accuracy quite substantially.
The Problem with Quantitative Studies of International Conflict
Beck, Nathaniel
King, Gary
Zeng, Langche
Uploaded 07-15-1998
7 logit
Paper Keywords neural networks
Bayesian analysis
Despite immense data collections, prestigious journals, and sophisticated analyses, empirical findings in the literature on international conflict are frequently unsatisfying.
Statistical results appear to change from article to article and specification to specification. Very few relationships hold up to replication with even minor respecification.
Abstract Accurate forecasts are nonexistent. We provide a simple conjecture about what accounts for this problem, and offer a statistical framework that better matches the substantive issues
and types of data in this field. Our model, a version of a ``neural network'' model, forecasts substantially better than any previous effort, and appears to uncover some structural
features of international conflict.
Forecasting Parliamentary Outcomes in Multiparty Elections: Hungary 1998
Benoit, Kenneth
Uploaded 08-16-1998
computer simulation
Keywords election forecasting
8 electoral systems
Paper Hungary
Forecasting seat outcomes in legislative elections in countries with stable, two-party systems is sufficiently challenging as to have proven elusive for much of democratic
experience. Forecasting an election in a relatively new democracy with a fluid multi-party system, therefore, would seem on its face to be a hopeless objective. In this paper I
attempt to demonstrate that election forecasting in such an environment is in fact quite feasible, using data from previous elections, opinion poll research, and computer simulation
Abstract models to predict the outcome of the Hungarian parliamentary elections which took place in May 1998. First, I discuss the general problems with election forecasting, and then outline
a strategy for dealing with each. I outline a forecasting method in detail, which I apply to Hungary's case to generate a prediction published in December 1997. The remainder of the
paper compares the actual results of the election to the author's forecasts published before the election, identifying areas for improvement in the basic forecasting model but also
proving that accurate forecasting of final outcomes in multiparty elections is possible in practice.
Estimating the Probability of Events That have Never Occurred: When Does Your Vote Matter?
Gelman, Andrew
King, Gary
Boscardin, John
Uploaded 10-27-1997
conditional probability
decision analysis
electoral campaigning
Keywords political science
presidential elections
9 Paper rare events
rational choice
subjective probability
voting power
Researchers sometimes argue that statisticians have little to contribute when few realizations of the process being estimated are observed. We show that this argument is incorrect
even in the extreme situation of estimating the probabilities of events so rare that they have never occurred. We show how statistical forecasting models allow us to use empirical
data to improve inferences about the probabilities of these events. Our application is estimating the probability that your vote will be decisive in a U.S. presidential election, a
problem that has been studied by political scientists for more than two decades. The exact value of this probability is of only minor interest, but the number has important
Abstract implications for understanding the optimal allocation of campaign resources, whether states and voter groups receive their fair share of attention from prospective presidents, and
how formal ``rational choice'' models of voter behavior might be able to explain why people vote at all. We show how the probability of a decisive vote can be estimated empirically
from state-level forecasts of the presidential election and illustrate with the example of 1992. Based on generalizations of standard political science forecasting models, we
estimate the (prospective) probability of a single vote being decisive as about 1 in 10 million for close national elections such as 1992, varying by about a factor of 10 among
states. Our results support the argument that subjective probabilities of many types are best obtained via empirically-based statistical prediction models rather than solely
mathematical reasoning. We discuss the implications of our findings for the types of decision analyses that are used in public choice studies.
Estimating the Probability of Events That have Never Occurred: When Does Your Vote Matter?
Gelman, Andrew
King, Gary
Boscardin, John
Uploaded 02-14-1997
conditional probability
decision analysis
electoral campaigning
Keywords political science
presidential elections
10 Paper rare events
rational choice
subjective probability
voting power
Researchers sometimes argue that statisticians have little to contribute when few realizations of the process being estimated are observed. We show that this argument is incorrect
even in the extreme situation of estimating the probabilities of events so rare that they have never occurred. We show how statistical forecasting models allow us to use empirical
data to improve inferences about the probabilities of these events. Our application is estimating the probability that your vote will be decisive in a U.S. presidential election, a
problem that has been studied by researchers in political science for more than two decades. The exact value of this probability is of only minor interest, but the number has
Abstract important implications for understanding the optimal allocation of campaign resources, whether states and voter groups receive their fair share of attention from prospective
presidents, and how formal ``rational choice'' models of voter behavior might be able to explain why people vote at all. We show how the probability of a decisive vote can be
estimated empirically from state-level forecasts of the presidential election and illustrate with the example of 1992. Based on generalizations of standard political science
forecasting models, we estimate the (prospective) probability of a single vote being decisive as about 1 in 10 million for close national elections such as 1992, varying by about a
factor of 10 among states. Our results support the argument that subjective probabilities of many types are best obtained via empirically-based statistical prediction models rather
than solely mathematical reasoning. We discuss the implications of our findings for the types of decision analyses that are used in public choice studies.
Demographic Forecasting
Girosi, Federico
King, Gary
Uploaded 07-10-2003
Keywords forecasting
We introduce a new framework for forecasting age-sex-country-cause-specific mortality rates that incorporates considerably more information, and thus has the potential to forecast
11 much better, than any existing approach. Mortality forecasts are used in a wide variety of academic fields, and for global and national health policy making, medical and
Paper pharmaceutical research, and social security and retirement planning. As it turns out, the tools we developed in pursuit of this goal also have broader statistical implications, in
addition to their use for forecasting mortality or other variables with similar statistical properties. First, our methods make it possible to include different explanatory
Abstract variables in a time series regression for each cross-section, while still borrowing strength from one regression to improve the estimation of all. Second, we show that many existing
Bayesian (hierarchical and spatial) models with explanatory variables use prior densities that incorrectly formalize prior knowledge. Many demographers and public health researchers
have fortuitously avoided this problem so prevalent in other fields by using prior knowledge only as an ex post check on empirical results, but this approach excludes considerable
information from their models. We show how to incorporate this demographic knowledge into a model in a statistically appropriate way. Finally, we develop a set of tools useful for
developing models with Bayesian priors in the presence of partial prior ignorance. This approach also provides many of the attractive features claimed by the empirical Bayes
approach, but fully within the standard Bayesian theory of inference. The latest version of this manuscript is available at http://gking.harvard.edu. | {"url":"http://polmeth.wustl.edu/mediaResults.php?txtSearch=forecasting&urlKeyword=1","timestamp":"2014-04-19T12:05:57Z","content_type":null,"content_length":"56640","record_id":"<urn:uuid:12ced6d7-d409-49da-a038-5ffd8c9b78a9>","cc-path":"CC-MAIN-2014-15/segments/1398223210034.18/warc/CC-MAIN-20140423032010-00181-ip-10-147-4-33.ec2.internal.warc.gz"} |
Curious about what's out there? Students who are comfortable with mathematics and physics and who want to understand the nature of the Solar System and other planetary systems, stars, galaxies and/or
the universe are encouraged to consider majoring in astronomy.
The knowledge acquired and the analytical skills developed provide excellent broad-based training for careers in industry, education and government as well as preparation for graduate study in
astronomy and astrophysics, science education, engineering, law and medicine.
The Department of Astronomy offers two undergraduate degree options. The Bachelor of Science (B.S.) in astronomy is designed for students who intend to pursue careers in a scientific or technical
field by continuing to study astronomy, astrophysics or physics at the graduate level or to commence study in some related field such as planetary science.
The Bachelor of Arts (B.A.) in astronomy is broader and less specialized than the B.S., with the aim of developing and sharpening analytical and quantitative reasoning while at the same time
cultivating broader knowledge that can be applied to a variety of careers, including business, law, the health professions and teaching.
Coursework for the Major
Students are required to take 10 core courses in mathematics, physics and astronomy. Students pursuing the B.A. have fewer required astronomy and physics courses, which offers greater flexibility for
taking courses in other disciplines.
All required courses must be completed with minimum grades of C.
Required Courses for Both Degrees
• Analytic Geometry and Calculus 1, 2, 3 (MAC 2311, 2312 and 2313)
• Physics with Calculus 1 and 2 with labs (PHY 2048/2048L and PHY 2049/2049L)
• Astronomy and Astrophysics 1 and 2 (AST 3018 and 3019)
• Techniques of Observational Astronomy (AST 3722C)
Additional required coursework for the B.A.:
• Two astronomy courses at the 3000/4000 level
• Two additional courses (6 credits minimum) from:
□ AST or PHY courses at the 3000/4000 level
□ AST 2003 Introduction to the Solar System, AST 2037 Life in the Universe,
GLY 2010C Physical Geology, GLY 2042 Planetary Geology, GLY 3105C Evolution of Earth and Life, MCB 3703 Astrobiology, PHY 2464 Physical Basis of Music, PHY 3101 Introduction to Modern
Physics, PHZ 4710 Introduction to Biophysics or SWS 2007 World of Water
Additional required coursework for the B.S.:
• Four 3-credit AST courses at the 4000 level (prerequisites: AST 3018 and AST 3019)
• Differential Equations (MAP 2302)
• Modern Physics (PHY 3101)
• Mechanics 1 and 2 (PHY 3221 and 4222)
• Electromagnetism 1 and 2 (PHY 3323 and 4324)
• Choose one course from PHY 3513 Thermal Physics, PHY 4424 Optics 1, PHY 4523 Statistical Physics or PHY 4604 Introductory Quantum Mechanics 1
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Recommended Coursework for Graduate Study
Students should talk with the undergraduate coordinator and plan to take:
• PHY 4604 and
• Additional courses from COP 2271 Computer Programming for Engineers, MAA 4402 Functions of a Complex Variable, MAS 3114 Computational Linear Algebra, PHY 3513 Thermal Physics, PHY 4424 Optics 1,
PHY 4523 Statistical Physics and STA 3032 Engineering Statistics
Relevant Minors and/or Certificates
UFTeach Program: There is a severe shortage of qualified secondary science teachers in Florida and nationwide. Students interested in becoming part of this high-demand profession should see the
undergraduate coordinator about the UFTeach program. UFTeach students can complete the UFTeach minor in science teaching along with their B.A. in astronomy and have the coursework and preparation for
professional teacher certification in Florida when they graduate.
Students pursuing the B.S. are encouraged to engage in research with astronomy faculty by signing up for at least three credits of AST 4905; no more than six credits can be counted toward the
required 4000-level courses.
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Critical Tracking
To graduate with this major, students must complete all university, college and major requirements. For degree requirements outside of the major, refer to CLAS Degree Requirements — Structure of a
CLAS Degree.
Equivalent critical-tracking courses as determined by the State of Florida Common Course Prerequisites may be used for transfer students.
Critical-tracking requirements are for both the B.S. and B.A. degrees. All students must meet these criteria to remain on track for the major.
Semester 1
• 2.0 UF GPA for semesters 1-5
• Complete MAC 1147 or MAC 2311
Semester 2
Semester 3
• Complete MAC 2312, PHY 2048 and PHY 2048L with a 2.5 critical-tracking GPA
Semester 4
• Complete MAC 2313, PHY 2049 and PHY 2049L with 2.5 critical-tracking GPA
Semester 5
• Complete AST 3018 with a 2.5 critical-tracking GPA. UF freshmen and sophomores should take AST 3018 by semester 4.
Bachelor of Arts
Bachelor of Science
Bachelor of Arts
Recommended Semester Plan
Students are expected to complete the writing and math requirement while in the process of taking the courses below. Students are required to complete HUM 2305 The Good Life (GE-H) in semester 1 or
2. Students are also expected to complete the general education international (GE-N) and diversity (GE-D) requirements concurrently with another general education requirement (typically, GE-C, H or
Semester 1 Credits
HUM 2305 What is the Good Life (GE-H) 3
MAC 2311 Analytical Geometry and Calculus 1 (GE-M) 4
Biological Sciences (GE-B) 3
Composition (GE-C, WR) 3
Social and Behavioral Sciences (GE-S) 3
Total 16
Semester 2 Credits
MAC 2312 Analytical Geometry and Calculus 2 4
PHY 2048 Physics with Calculus 1 (GE-P) 3
PHY 2048L Physics with Calculus 1 Laboratory (GE-P) 1
Biological Science (GE-B) 3
Humanities (GE-H) 3
Total 14
Semester 3 Credits
AST 3018 Astronomy and Astrophysics 1 3
MAC 2313 Analytical Geometry and Calculus 3 4
PHY 2049 Physics with Calculus 2 (GE-P) 3
PHY 2049L Physics with Calculus 2 Laboratory (GE-P) 1
Social and Behavioral Sciences (GE-S) 3
Total 14
Semester 4 Credits
AST 3019 Astronomy and Astrophysics 2 3
AST 3722C Observational Techniques 1 3
Elective 3
Humanities (GE-H) 3
Social and Behavioral Sciences (GE-S) 3
Total 15
Semester 5 Credits
AST course (3000/4000 level) 3
Elective (3000/4000 level, not in major) 3
Electives 6
Foreign language 3-5
Total 15-17
Semester 6 Credits
AST course (3000/4000 level) 3
Elective 3
Electives (3000/4000 level, not in major) 6
Foreign language 3-5
Total 15-17
Semester 7 Credits
AST or PHY or approved major course 3
Composition (GE-C, WR) 3
Elective (or complete foreign language if 4-3-3 option) 3
Electives 6
Total 15
Semester 8 Credits
AST or PHY or approved major course 3
Electives 13
Total 16
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Bachelor of Science
Recommended Semester Plan
Students are expected to complete the writing and math requirement while in the process of taking the courses below. Students are required to complete HUM 2305 The Good Life (GE-H) in semester 1 or
2. Students are also expected to complete the general education international (GE-N) and diversity (GE-D) requirements concurrently with another general education requirement (typically, GE-C, H or
Semester 1 Credits
HUM 2305 What is the Good Life (GE-H) 3
MAC 2311 Analytic Geometry and Calculus 1 (GE-M) 4
Biological Science (GE-B) 3
Composition (GE-C, WR) 3
Social and Behavioral Sciences (GE-S) 3
Total 16
Semester 2 Credits
MAC 2312 Analytic Geometry and Calculus 2 (GE-M) 4
PHY 2048 Physics with Calculus 1 (GE-P) 3
PHY 2048L Physics with Calculus 1 Laboratory (GE-P) 1
Biological Science (GE-B) 3
Humanities (GE-H) 3
Total 14
Semester 3 Credits
AST 3018 Astronomy and Astrophysics 1 (GE-P) 3
MAC 2313 Analytic Geometry and Calculus 3 4
PHY 2049 Physics with Calculus 2 (GE-P) 3
PHY 2049L Physics with Calculus 2 Laboratory (GE-P) 1
Social and Behavioral Sciences (GE-S) 3
Total 14
Semester 4 Credits
AST 3019 Astronomy and Astrophysics 2 (GE-P) 3
AST 3722C Techniques of Observational Astronomy 1 3
MAP 2302 Elementary Differential Equations 3
Humanities (GE-H) 3
Social and Behavioral Sciences (GE-S) 3
Total 15
Semester 5 Credits
AST course (4000 level) 3
PHY 3101 Introduction to Modern Physics 3
PHY 3221 Mechanics 1 3
Elective 3
Foreign language 3-5
Total 15-17
Semester 6 Credits
AST course (4000 level) 3
PHY 3323 Electromagnetism 1 3
PHY 4222 Mechanics 2 3
Elective 3
Foreign language 3-5
Total 15-17
Semester 7 Credits
AST course (4000 level) 3
PHY 4324 Electromagnetism 2 3
PHY 3513 Thermal Physics 1 or
PHY 4424 Optics 1 or 3
PHY 4523 Statistical Physics or
PHY 4604 Introductory Quantum Mechanics 1
Composition (GE-C, WR) 3
Elective (or complete foreign language if 4-3-3 option) 3
Total 15
Semester 8 Credits
AST course (4000 level) 3
Electives 6
Electives (3000 level, non-major if needed) 7-3
Total 16-12
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Therefore (thing)
@@@ @@@
@@@@@ @@@@@
@@@@@ @@@@@
@@@ @@@
In mathematics, "therefore" is represented as a triangular set of three dots. This may have developed from alchemy: some alchemical formulas involved burning a substance down into a fine white
powdery ash, which resembled salt. The "therefore" symbol may have stemmed from this process: something is burned, therefore we have a "salt" (as far as alchemy is concerned; not the NaCl kind).
Further, the alchemical symbol for salt is in fact a set of three dots. This roughly corresponds to the symbol's significance in mathematical proofs. Usage of the symbol in a mathematical context
dates back to at least 1659, from Teusche Algebra by Johann Rahn (source).
How to insert the symbol:
A handful of other logic symbols originated with this one: for example, the because symbol is an upside-down version of it, and the ratio and proportion symbols are also sets of dots (resembling a
colon and double-colon, respectively).
Swap says re Therefore: It's a funny thing that in modern mathematics the symbol is practically no longer used, where by "modern" I mean "university". It belongs to a logical style of argument that
dates back to Aristotle, and ever since the so-called crisis in foundations of mathematical logic, other symbols became more fashionable, such as the "implies" symbol, ⇒ | {"url":"http://everything2.com/user/chisel/writeups/Therefore","timestamp":"2014-04-20T11:17:13Z","content_type":null,"content_length":"23464","record_id":"<urn:uuid:f8649552-6cbe-404d-84eb-c66877f24753>","cc-path":"CC-MAIN-2014-15/segments/1397609538423.10/warc/CC-MAIN-20140416005218-00629-ip-10-147-4-33.ec2.internal.warc.gz"} |
Prove Functional Completeness
January 31st 2010, 11:38 AM #1
Junior Member
Aug 2009
Prove Functional Completeness
Hello, I believe I just need a nudge in the right direction with the following problem:
A set of logical operators is functionally complete if every compound proposition is logically equivalent to one using only the operators in the set. Prove that NOT and OR form a functionally
complete set of operators.
I am not quite sure how to approach the problem. It seems to be the case that two statements can be equivalent by using only the two operators. In this case OR and NOT.
Thanks for any and all help.
I believe a set of operators is functionally complete if every boolean operation (AND, OR, and NOT) can be implemented using them. You already have OR and NOT, so all you need to do is prove that
you can implement AND using OR and NOT.
You can use deMorgan's Law: NOT((NOT a) OR (NOT b)) = a AND b
Thank you for your help. That makes complete sense.
February 2nd 2010, 05:44 PM #2
Feb 2010
February 3rd 2010, 09:10 PM #3
Junior Member
Aug 2009 | {"url":"http://mathhelpforum.com/discrete-math/126461-prove-functional-completeness.html","timestamp":"2014-04-18T12:58:33Z","content_type":null,"content_length":"33884","record_id":"<urn:uuid:3046b20e-5caf-4bf4-9055-c4830cc86909>","cc-path":"CC-MAIN-2014-15/segments/1397609533308.11/warc/CC-MAIN-20140416005213-00051-ip-10-147-4-33.ec2.internal.warc.gz"} |
Re: st: interpreting marginal effects of poisson
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Re: st: interpreting marginal effects of poisson
From Scott Cunningham <scunning@gmail.com>
To statalist@hsphsun2.harvard.edu
Subject Re: st: interpreting marginal effects of poisson
Date Wed, 19 Apr 2006 09:21:05 -0400
My last email was not written in a way that was easy to read, so I'm reposting just the questions I had. The set up, though, is that I have a Poisson model with both discrete and continuous
covariates, but I'm having trouble interpreting the continuous coefficients. First I'll post the questions, then I'll post the data and output itself.
On Apr 18, 2006, at 10:10 PM, Scott Cunningham wrote:
Question: The coefficient on sr is -.6870002, and it is the variable of interest. Am I correct in the following interpretations:
Q1. The constant (_cons) represents the log of the mean number of sex partners for the reference cell, which in this case is Black men living in homes where parents are not married (hhd1=1 if
biological parents are married). Since exp{.5387156)=1.7138, we see that on the average these men have 1.7 sex partners at this point in their lives. Is this the correct interpretation?
Q2. The hhd1 variable reflects the state of the family in which the Black male lives. As we move from hhd1=0 to hhd1=1, the log of the mean decreases by .7, which means that the number of sex
partners gets multiplied by exp{-0.300639)=0.740345. This means that Black males with married biological parents have 25% fewer sex partners than their counterparts whose parents are not married.
Is this the correct interpretation?
Q3. The sr variable is continuous. It is the ratio of eligible Black males (of a certain age range) to eligible Black females (of like age range) at the state level, and will take on a value from
as low as 0.3 to 2.5. I am unsure of how to interpret the marginal effect of a change in the sex ratio ("sr") on sex partners. Am I correct that a one unit increase in the sex ratio causes recent
sex partners for Black males to fall by 50%? If so, what is a "one unit" when we are talking about a continuous random variable? Is it a one unit increase in the standard deviation? I've been
unable to find this information from my reference books, and do not currently own the Stata book on categorical variables. But if anyone can provide basic help here, I'd appreciate it.
Q4: I am able to estimate a model with state, year and individual fixed effects using OLS with FE (-xtreg-), but not Poisson (- xtpoisson- nor -poisson-). Specifically, I can estimate the Poisson
model with year and individual effects, but not with year, state and individual fixed effects. When I include the state effects, the likelihood iteratations hang up, even after letting it go for
thousands of iterations. It has hung up on a single likelihood, for that matter, and does not appear to be moving closer towards convergence. What could be causing this?
Description of Data:
Individual-level survey data from waves 1998, 2000 and 2002. It is a balanced panel dataset, and I am focusing currently just on Black American males aged 12-17 in 1998 (and thus age over the course
of the survey). The relevant variables are:
rp: "recent sex partners"
sr: "sex ratio"
age2: age-squared
hgc: "highest grade completed"
hhd1: "household dummy variable for bio. parents still married"
Description of Model:
I am estimating a model in which the number of recent sex partners men have is a function of their age, age-squared, the relative availability of men and women in the mating market, their education
attainment, whether their biological parents are married, and controls for individual, year and state fixed effects. I'm modeling this as a Poisson distribution. I'm having trouble taking the
coefficients and creating marginal effects. (FYI, I have downloaded - sposta- utilities, and have used -prchange- but at this point, am trying to manually create the marginal effects and decipher
. xi:poisson rp sr age2 hgc hhd1 i.year, robust
i.year _Iyear_1998-2002 (naturally coded; _Iyear_1998 omitted)
Iteration 0: log pseudolikelihood = -10504.806
Iteration 1: log pseudolikelihood = -10504.805
Poisson regression Number of obs = 2277
Wald chi2(6) = 60.92
Prob > chi2 = 0.0000
Log pseudolikelihood = -10504.805 Pseudo R2 = 0.0347
------------------------------------------------------------------------ ------
| Robust
rp | Coef. Std. Err. z P>|z| [95% Conf. Interval]
------------- +----------------------------------------------------------------
sr | -.6870002 .3183816 -2.16 0.031 -1.311017 -.0629838
age2 | .0045666 .0012833 3.56 0.000 . 0020515 .0070818
hgc | .000394 .0365008 0.01 0.991 -. 0711463 .0719343
hhd1 | -.300639 .117171 -2.57 0.010 -.5302899 -.0709881
_Iyear_2000 | .065989 .1566268 0.42 0.674 -. 240994 .372972
_Iyear_2002 | -.3705945 .1926353 -1.92 0.054 -.7481528 . 0069637
_cons | .5387156 .4899692 1.10 0.272 -. 4216065 1.499038
------------------------------------------------------------------------ ------
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Set Products Question
August 23rd 2013, 02:07 PM #1
Set Products Question
I'm beginning a section on Product Topologies and I have to do a few problems, but I'm not seeing the difference between these two.
Let $\mathcal{A}$ and $\mathcal{B}$ be two nonempty families of sets. Prove that:
$\bigcap \mathcal{A} \times \bigcap \mathcal{B} = \bigcap\{A \times B : A \in \mathcal{A}, \ B \in \mathcal{B}\}$
Suppose that $A_1,\cdots ,A_n$ and $B_1,\cdots ,B_n$ are sets. Prove that:
$(A_1\times B_1)\cap\cdots\cap (A_n\times B_n) = (A_1\cap\cdots\cap A_n) \times (B_1\cap\cdots\cap B_n)$
Aren't these asking for the same thing??
Re: Set Products Question
Hey Aryth.
Does your combination contain all permutations Ai X Bj for all possible i and j or is i = j? (The way you have written it is unclear).
Re: Set Products Question
To be honest I can't answer that question. The problems are exactly as I have typed them. In the proofs... I imagine that the equations would hold either way (Sets under intersections commute, so
I can order them however I like on the right hand side and arrive at any necessary permutation as long as it's $A_i \times B_j$).
Re: Set Products Question
I'm beginning a section on Product Topologies and I have to do a few problems, but I'm not seeing the difference between these two.
Let $\mathcal{A}$ and $\mathcal{B}$ be two nonempty families of sets. Prove that:
$\bigcap \mathcal{A} \times \bigcap \mathcal{B} = \bigcap\{A \times B : A \in \mathcal{A}, \ B \in \mathcal{B}\}$
Suppose that $A_1,\cdots ,A_n$ and $B_1,\cdots ,B_n$ are sets. Prove that:
$(A_1\times B_1)\cap\cdots\cap (A_n\times B_n) = (A_1\cap\cdots\cap A_n) \times (B_1\cap\cdots\cap B_n)$
Aren't these asking for the same thing??
No, they are not. The second assumes a finite number of sets (n) while the first does not.
August 23rd 2013, 06:25 PM #2
MHF Contributor
Sep 2012
August 23rd 2013, 06:35 PM #3
August 27th 2013, 02:39 PM #4
MHF Contributor
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"sum over labelings" representations of graph polynomials
up vote 5 down vote favorite
It seems that there's a general way to go from "recursive" definition of a graph polynomials to "subset expansion" formulas.
Furthermore, polynomials with subset expansion formulas often have a representation as a sum over all possible vertex labelings of some "local interactions" model.
For instance, generating function for Eulerian subgraphs becomes Ising model partition function, generating function of independent sets becomes partition function of the hard-core model and Potts
model has both "sum over labelings" and "sum over subgraphs" representation.
1. Are there other interesting examples of graph polynomials with "sum over labelings" representation?
2. When is it possible to get this representation of a graph polynomial? More specifically, to represent it as a sum over labelings of some quantity that is a product of functions each depending
only on the variables corresponding to some edge of the graph. IE, if a graph has edges (1,2),(2,3) the term being summed over has to factor into f(x1,x2)*g(x2,x3)
Motivation: there's a general algorithm to efficiently compute "sum over labelings" for functions decomposing over graph or hyper-graph of bounded tree-width, and it's interesting to see which graph
polynomials I can use it for
algebraic-graph-theory graph-theory
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Coend computation
up vote 9 down vote favorite
$F:A^{\mbox{op}} \to \mbox{Set}$
and define
$G_a:A\times A^{\mbox{op}} \to \mbox{Set}$
$G_a(b,c) = \mbox{hom}(a,b) \times F(c)$.
I think the coend of $G_a$,
ought to be $F(a)$--it's certainly true when A is discrete, since then hom is a delta function. But my colimit-fu isn't good enough to actually compute the thing and verify it's true. Can someone
walk me through the computation, please?
colimits ct.category-theory
is this a category-theoretic analogue of the main theorem of the fundamental theorem of calculus? – Martin Brandenburg Apr 6 '10 at 0:13
2 @Martin: an FToC analogue would require a 'boundary' computation of some sort. This seems more like a distribution computation, where the hom works like $\delta(a-b)$, so the integral over the
whole space is just evaluation. – Jacques Carette Apr 6 '10 at 3:20
add comment
1 Answer
active oldest votes
Hi Mike. This is what's often called the Density Formula, or (at the n-Lab) the coYoneda Lemma (I think), or (by Australian ninja category theorists) simply the Yoneda Lemma. (But
Australian ninja category theorists call everything the Yoneda Lemma.) In any case, it's a kind of dual to the ordinary Yoneda Lemma.
But you asked to be walked through it. First: yes, it is $F(a)$. Another way of writing your coend $$ \int^A G_a $$ is as $$ \int^{b \in A} G_a(b, b) = \int^b \mathrm{hom}(a,b) \times F
(b). $$ I claim this is canonically isomorphic to $F(a)$. I'll prove this by showing that for an arbitrary set $S$, the homset $\mathrm{hom}(\mathrm{this}, S)$ is canonically isomorphic
to $\mathrm{hom}(F(a), S)$. The claim will then follow from the ordinary Yoneda Lemma.
up vote 17
down vote So, let $S$ be a set. Then $$ \mathrm{Set}(\int^b \mathrm{hom}(a, b) \times F(b), S) & \cong \int_b \mathrm{Set}(\mathrm{hom}(a, b) \times F(b), S) \\ &\cong \int_b \mathrm{Set}(\mathrm
accepted {hom}(a, b), \mathrm{Set}(F(b), S)) \\ &\cong \mathrm{Nat}(\hom(a, -), \mathrm{Set}(F(-), S)) \\ &\cong \mathrm{Set}(F(a), S) $$ I don't know how much of this you'll want explaining, so
I'll just say it briefly for now. If you want further explanation, just ask. The first isomorphism is kinda the definition of colimit. The second is the usual exponential transpose/
currying operation. The third is maybe the most important: it's a fundamental fact about ends that if $F, G: C \to D$ are functors then $$ \mathrm{Nat}(F, G) = \int_c D(F(c), G(c)). $$
The fourth and final isomorphism is the ordinary Yoneda Lemma applied to the functor $\mathrm{Set}(F(-), S)$.
I'm a little confused by your notation: specifically, what is the category $\mathrm{Nat}$? From the types its objects are functors $A^{op} \to \mathrm{Set}$, so it's the functor
category whose objects are the natural transformations between those functors? – Neel Krishnaswami Apr 6 '10 at 14:14
FWIW You can formalise the steps in the construction of this isomorphism as a Haskell program. Here's one direction, with f being the relevant morphism: hpaste.org/fastcgi/hpaste.fcgi/
view?id=24722 I suspect that by interpreting Haskell code as the internal language of some family of categories then this becomes a perfectly good definition of the isomorphisms for
mathematical purposes too, and not just a statement about some Haskell functions. – Dan Piponi Apr 6 '10 at 16:53
Thanks! The third step is really the one I need to understand, so I'll think about that for a while and come back if I need more help. – Mike Stay Apr 6 '10 at 18:30
Neel, I guess Nat is a slightly unspecific notation, like hom. The first occurrence of Nat meant "hom in $[A^{op}, \mathrm{Set}]$". Here $[A^{op}, \mathrm{Set}]$ is the category whose
1 objects are functors $A^{op} \to \mathrm{Set}$. So: Nat is not a category; it means "hom" in that functor category. I might equally well have written "$[A^{op}, \mathrm{Set}]$" in
place of "Nat". (In fact I prefer to; I was making a perhaps misguided effort to be more widely comprehensible.) Similarly, the second occurrence of "Nat" could be replaced by "$[C, D]
$". – Tom Leinster Apr 6 '10 at 20:08
1 Ah, thanks -- I've just never seen Nat used like that before. Amusingly, the $[A^{op}, \mathrm{Set}]$ notation is the one I've seen before. :) – Neel Krishnaswami Apr 6 '10 at 22:43
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GMAT Math: Calculating Combinations
First, a few practice questions. Remember — no calculator!
1) A radio station has to choose three days of the seven in a week to broadcast a certain program, and that set will repeat each week. The program can be broadcast equally on any of the seven
weekdays —- weekdays vs. weekends don’t matter at all —- nor does it matter whether the days the program airs are adjacent or not. Absolutely any three of the seven weekdays can be chosen. How
many different three-day combinations of the seven weekdays can be constructed?
1. 9
2. 15
3. 21
4. 35
5. 56
2) Claudia can choose any two of four different candles and any 8 of 9 different flowers for a centerpiece arrangement. Given these choices, how many candle + flower groupings can she select?
1. 54
2. 72
3. 96
4. 144
5. 432
3) A newly-wed couple is using a website to design an eBook Wedding Album to distribute to their friends and families. The template they have chosen has places for 3 large photos and 19 smaller
photos. The couple has 6 large photos they could use for those three slots, and 21 smaller photos they could use for those 19 slots. Given these choices, how many different possible albums could
they create?
1. 3,150
2. 4,200
3. 5,040
4. 20,520
5. 84,000
In this post, we’ll discuss how to handle questions like this — without a calculator.
Mathematically, a combination is a group of things, irrespective of order. For example, {A, B, D} and {D, A, B} and {B, A, D} are all the same combination — order doesn’t matter at all. The
expression nCr (read “n choose r”) is the expression for the number of combinations of r things, r choices, you can make from a pool of n unique items. For example,
6C3 = the number of combinations of three one can choose from a pool of six unique items.
In a previous post about combinations, I give the following formula for nCr
where the exclamation point (“!”) is the factorial symbol — n! means the product of all the positive integers from n down to 1. Using this formula, we could compute the value of 6C3
So, it turns out, there are twenty ways to pick a set of three items from a pool of six unique items. That’s one way to calculate nCr, but it’s not the only way.
Pascal’s Triangle
The mathematician and philosopher Blaise Pascal (1623 – 1662) created a magical triangular array of numbers known now as Pascal’s Triangle:
How does this pattern work? Well, of course, the edges are diagonals of 1′s. Every inside number is the sum of the two numbers above it in the previous row, diagonally to the left and diagonally to
the right. For example, the 2 is the sum 1+1; both 3′s are the sum 1+2; both 4′s are the sums 1+3; the 6 is the sum 3+3, etc. They often show Pascal’s Triangle to grade school students to give them
practice with addition.
Despite its relatively easy origins, Pascal’s Triangle is a treasure trove of miraculous mathematical properties. Most relevant for us right now is: Pascal’s Triangle is, among other things, an
array of all possible nCr’s.
nCr = the rth entry of the nth row of Pascal’s Triangle
In that definition, we have to be careful — we have to start counting at zero instead of one. The top 1 on Pascal’s Triangle is the zeroth row, zeroth entry, 0C0 = 1 (a relatively meaningless number
in terms of combinations!) The next row (1, 1) is the first row, and the next row is the second row (1, 2, 1), etc. Notice that the second number in any row (as well as the penultimate number in
any row) equals the row number. The first number (always 1) is actually the zeroth entry, so that second number would actually be the first entry — the first entry of the nth row always equals n. In
other words
nC1 = n
That makes sense: if we have n different items, we have exactly n ways of selecting any one item. Those entries, the first entries of each row, line along a diagonal down the left side of the
triangle. Because of the complete symmetry of the triangle, this always equals the numbers on the corresponding diagonals on the right side, which would be the (n-1) entries of each row. Thus:
nC1 = nC(n-1) = n
When you have to figure out nCr when n & r are both relatively small numbers, it may be easier simply to jot down the first few rows of Pascal’s Triangle. For example, with what we have showing, we
can see that 6C3, the 3rd entry of the 6th row, is 20 — the same as the answer we found via the factorials formula.
Things get even more interesting when we move to the next diagonal in, shown in green here:
These numbers, the set of the second entries in each row, are the triangular numbers. Among other things, the second entry in the n row is the sum of the first n-1 positive integers. For example
3 = 2 + 1
6 = 3 + 2 + 1
10 = 4 + 3 + 2 + 1 etc.
The formula for this is:
Because we have a formula, we can calculate this for much higher numbers. For 21C2, we would have to write out everything to the twenty-first row of Pascal’s Triangle, an arduous undertaking.
Rather, we could simply use the formula
Notice that the symmetry of Pascal’s Triangle also provides tremendous insight into the nature of the nCr numbers. First of all, in any row, the second entry, the triangular number in that row, must
be equal to the third-to-last entry of the row, that is, the (n-2) entry of the row. Thus
Thus, via the triangular numbers, we have a formula, not only for the second entry of each row, but also for the third-to-last entry of every row. Thus, it’s very easy to figure out the first three
or last three numbers in any row. More generally, symmetry guarantees that:
nCr = nC(n-r)
If you think about combinations this makes sense: if we have a pool of n unique items, then every time we choose a unique set of r items, we necessarily exclude a corresponding unique set of (n-r)
items. In other words, there is necessarily a 1-to-1 correspondence between unique sets of r elements and unique sets of the other (n-r) elements —- because there’s a 1-to-1 correspondence, the
number of each must always be the same. This is precisely what that equation says.
That discussion was liberally peppered with hints about how to do the above three questions. If you had trouble with them on the first pass, you may want to give them a second look before proceeding
to the explanations below. Here’s an additional question from inside Magoosh:
4) http://gmat.magoosh.com/questions/847
Practice question explanations
1) Behind the story, we are really being asked to evaluate 7C3. We could use the factorial formula, but above we conveniently happen to have Pascal’s Triangle written out to the seventh row. We see
that 7C3, the third entry of the seventh row, is 35. Answer = D.
2) For this one, we have to use the Fundamental Counting Principle (FCP) as well as information about combinations. For the flowers, we want 9C8, which by the symmetry of Pascal’s Triangle, has to
equal 9C1, the first entry in the row, which of course equals the row number.
9C8 = 9C1 = 9
That’s the number of flower combinations. For the candles, 4C2, we read the second entry of the fourth row of Pascal’s Triangle.
4C2 = 6
Now, by the FCP, we multiply these for the total number of centerpiece arrangements: 6*9 = 54. Answer = A
3) For the large photos, we need 6C3, which we calculated in the article:
6C3 = 20
For the smaller photos, we need 21C19, which by symmetry must equal 21C2, and we have a formula for that. In fact, in the article above, we already calculated that 21C2 = 210.
Now, by the FCP, we just multiply these: total number of possible albums = 20*210 = 4200. Answer = B
4 Responses to GMAT Math: Calculating Combinations
1. Vijay Rajan August 22, 2013 at 4:51 pm #
Thank you for the very informative posts. I have a question about question #2 at the beginning of this blog. Please tell me what am I missing here. Here it goes:
Should we use permutation instead of combination for the centerpiece arrangement question? She has to use 2 candles from 4 choices. I don’t know much about the centerpiece arrangements but, hear
me out please! There are 10 slots to be filled with 2 different types of candles and 8 different types of flowers. And there are 10 empty slots – A, B, C, D, E, F, G, H, I, and J, okay? Wouldn’t
the order in which flowers and candles fill these slots matter? Wouldn’t a centerpiece in which a rose scented candle is put in slot A be different from a centerpiece different from a centerpiece
in which the same rose scented candle is put in slot B? So, that way there will be 4 X 3 X 9! arrangements. What am I missing here?
□ Mike August 23, 2013 at 1:21 pm #
Dear Vijay,
You know, for that question, I didn’t want people considering how the centerpiece might be arranged. FWIW, centerpieces are usually somewhat circular, and definitely are not candles & flowers
all in a row. Given two candles and eight flowers, how many different centerpieces could one design with different spacial configurations? That question is truly mind-boggling, and has no
straightforward answer.
I changed the text of the question, to make more clear what is actually being asked. Does it make sense now?
☆ Vijay Rajan August 23, 2013 at 3:25 pm #
Yes, sir! Thank you!
○ Mike August 23, 2013 at 4:02 pm #
you are more than welcome. Best of luck to you.
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instantaneous acceleration calculator
Best Results From Wikipedia Yahoo Answers Youtube
From Wikipedia
In physics, acceleration is the rate of change of velocity over time. In one dimension, acceleration is the rate at which something speeds up or slows down. However, since velocity is a vector,
acceleration describes the rate of change of both the magnitude and the direction of velocity. Acceleration has the dimensionsL T^−2. In SI units, acceleration is measured in meters per second
per second (m/s^2).
Proper acceleration, the acceleration of a body relative to a free-fall condition, is measured by an instrument called an accelerometer.
In common speech, the term acceleration is used for an increase in speed (the magnitude of velocity); a decrease in speed is called deceleration. In physics, a change in the direction of velocity
also is an acceleration: for rotary motion, the change in direction of velocity results in centripetal (toward the center) acceleration; where as the rate of change of speed is a tangential
In classical mechanics, for a body with constant mass, the acceleration of the body is proportional to the net force acting on it (Newton's second law):
\mathbf{F} = m\mathbf{a} \quad \to \quad \mathbf{a} = \mathbf{F}/m
where F is the resultant force acting on the body, m is the mass of the body, and a is its acceleration.
Average and instantaneous acceleration
Average acceleration is the change in velocity (Δ'v) divided by the change in time (Δt). Instantaneous acceleration is the acceleration at a specific point in time which is for a very short
interval of time as Δt approaches zero.
The velocity of a particle moving on a curved path as a function of time can be written as:
\mathbf{v} (t) =v(t) \frac {\mathbf{v}(t)}{v(t)} = v(t) \mathbf{u}_\mathrm{t}(t) ,
with v(t) equal to the speed of travel along the path, and
\mathbf{u}_\mathrm{t} = \frac {\mathbf{v}(t)}{v(t)} \ ,
a unit vector tangent to the path pointing in the direction of motion at the chosen moment in time. Taking into account both the changing speed v(t) and the changing direction of u[t], the
acceleration of a particle moving on a curved path on a planar surface can be written using thechain rule of differentiation and the derivative of the product of two functions of time as:
\mathbf{a} & = \frac{d \mathbf{v}}{dt} \\ & = \frac{\mathrm{d}v }{\mathrm{d}t} \mathbf{u}_\mathrm{t} +v(t)\frac{d \mathbf{u}_\mathrm{t}}{dt} \\ & = \frac{\mathrm{d}v }{\mathrm{d}t} \mathbf{u}_\mathrm
{t}+ \frac{v^2}{R}\mathbf{u}_\mathrm{n}\ , \\ \end{alignat}
where u[n] is the unit (inward) normal vector to the particle's trajectory, and R is its instantaneous radius of curvature based upon the osculating circle at time t. These components are called the
tangential accelerationand the radial acceleration or centripetal acceleration (see alsocircular motion and centripetal force).
Extension of this approach to three-dimensional space curves that cannot be contained on a planar surface leads to the Frenet-Serret formulas.
Special cases
Uniform acceleration
Uniform or constant acceleration is a type of motion in which the velocity of an object changes by an equal amount in every equal time period.
A frequently cited example of uniform acceleration is that of an object in free fall in a uniform gravitational field. The acceleration of a falling body in the absence of resistances to motion is
dependent only on the gravitational field strength g (also called acceleration due to gravity). By Newton's Second Law the force, F, acting on a body is given by:
\mathbf {F} = m \mathbf {g}
Due to the simple algebraic properties of constant acceleration in the one-dimensional case (that is, the case of acceleration aligned with the initial velocity), there are simple formulae that
relate the following quantities: displacement, initial velocity, final velocity, acceleration, and time:
\mathbf {v}= \mathbf {u} + \mathbf {a} t
\mathbf {s}= \mathbf {u} t+ \over {2}} \mathbf {a}t^2 = \over {2}}
\mathbf{s} = displacement
\mathbf{u} = initial velocity
\mathbf{v} = final velocity
\mathbf{a} = uniform acceleration
t = time.
In the case of uniform acceleration of an object that is initially moving in a direction not aligned with the acceleration, the motion can be resolved into two orthogonal parts, one of constant
velocity and the other according to the above equations. As Galileo showed, the net result is parabolic motion, as in the trajectory of a cannonball, neglecting air resistance.
Circular motion
An example of a body experiencing acceleration of a uniform magnitude but changing direction is uniform <
From Yahoo Answers
Question:A train started from rest and moved with constant acceleration, At one time it was travelling at 30m/s and 160metres further on it was travelling at 50m/s, how do I find: 1. Acceleration
2.time required to travel the 160m mentioned
Answers:2. time required to travel the 160 m x = .5(vi + vf) * t x = 160 m vi = 30 m/s vf = 50 m/s t = ? 160 = .5(30 + 50) * t 160 = 40 * t t = 160 / 40 = 4 s 1. acceleration over the 160 m vf = vi +
a * t 50 = 30 + a * 4 a = (50 - 30) / 4 = 5 m/s/s
Question:I'm trying to simulate a car accelerating but my results are way off. I'm calculating the accelerating force as (power in watts) / ( (velocity in m/s) * (mass in kg) ) is that right? Si
units please.
Answers:use a Cray supercomputer
Question:another question,..lol's......anybody can help me for this question?this is my assignment in physics.
Answers:Instantaneous acceleration is acceleration (change of velocity/time) at any given time. In layman's terms, it is the acceleration of a certain body (or particle) at any particular given or
chosen instant. Hope this helps.
Question:I am confused! I know instantaneous acceleration is tangent to acceleration graph. But then does that mean instantaneous velocity equal to acceleration since acceleration is tangent of
Answers:Instantaneous acceleration is the tangent of the **velocity** graph, not the tangent of the acceleration graph. That's probably why you're confused. If you have a graph of acceleration versus
time, and you want to know the instantaneous acceleration at any particular time, just read the graph at that time - you don't need to measure the tangent or calculate the slope. You only need to use
the tangent or slope if you need to figure out the acceleration from the graph of velocity versus time.
From Youtube
AP Physics B: Instantaneous Acceleration :A free lecture from our AP Physics B series at www.educator.com Other subjects include Algebra, Geometry, Pre-Calculus, Pre-Algebra, Trigonometry, Calculus,
Biology, Chemistry, Statistics, and Computer Science. -All lectures are broken down by individual topics -No more wasted time -Just search and jump directly to the answer
Instantaneous Rates of Change: Algebraic Model :Introduction to the algebraic model for calculating instantaneous rates of change. Introduction to the limit concept. | {"url":"http://www.edurite.com/kbase/instantaneous-acceleration-calculator","timestamp":"2014-04-16T13:31:28Z","content_type":null,"content_length":"76690","record_id":"<urn:uuid:ce0672ff-eb65-44ff-9726-17af894f940a>","cc-path":"CC-MAIN-2014-15/segments/1398223206147.1/warc/CC-MAIN-20140423032006-00047-ip-10-147-4-33.ec2.internal.warc.gz"} |
What is ABSOLUTE VALUE IN EXCEL 2010?
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Find Missing Number
This is a discussion on Find Missing Number within the Brainteasers forums, part of the Brain Quiz category; Fill in the missing number in the series. 3,10,7,8,--,12,9,16....
hi, the missing number is 7. Sol 1. 3 10 7 8 -- 12 9 16 so, lets name them as a=3, b=10, c=7, d=8, e=?, f=12, g=9, h= 16 a+9 gives you f(12), and b-1 gives you g(9), adn c+9 gives you h(16), so in
that fashion d-1, should give you e, which will be 7 as the value of d=8. Thanks Sushma
3, 10, 7, 8 3*2, 10+2, 7+2, 8*2 6, 12, 9, 16 So the missing number is 6
The missing number is 5 Here is how i found out There are two sequences interwined one is a sequence of even numbers and another one is a sequence of odd numbers Consider the odd numbers and take
three numbers at a time the lowest number is 3 and the first set of 3 odd numbers would be 3,5 and 7. Write them in the order of 1st, third and second ie (o1,o3,o2) this will give 3,7 and 5. Next set
would be 9,13 and 11 and so on and so forth Now consider the even number. each set has three numbers in the increasing order The lowest numbers so far here is 8. Suppose the first set consists of
(a1,a2,a3) then rewrite them by (a2,a1,a3). So a set (8,10,12) will become (10,8,12) followed by (16,14,18) and so on and so forth. Now combine these two sequences odd one first and in between any
two odd numbers insert the even numbers and the series would unravel 3,10,7,8,5,12,9,16,13,14,11,18,15,22,19,20,17,24.... | {"url":"http://www.geekinterview.com/talk/11943-find-missing-number.html","timestamp":"2014-04-21T01:01:02Z","content_type":null,"content_length":"43171","record_id":"<urn:uuid:d3e4373d-e219-46f3-b726-84fa95710fa9>","cc-path":"CC-MAIN-2014-15/segments/1398223204388.12/warc/CC-MAIN-20140423032004-00118-ip-10-147-4-33.ec2.internal.warc.gz"} |
optimize with respect to domain shape
up vote 0 down vote favorite
Let $\Gamma$ be the set of all closed $C^2$ curves in the plane which enclose unit area and let $\Omega$ be the set of all subsets of $\mathbb{R}^2$ that are enclosed by some curve in $\Gamma$. Now
let $f: \mathbb{R}^2\rightarrow\mathbb{R}$ be a real-valued function on the plane. How can we find the set $\omega\in\Omega$ with boundary $\gamma\in\Gamma$ such that $ \int_\omega\ f\ $ is
A related question (with a more physics-style interpretation): Let $\Omega$ and $\Gamma$ be as before. Let $u(\vec{x})$ be the real-valued function on the plane that solves the PDE $$ \Delta u(\vec
{x}) = 0\ \text{for } \vec{x}\in\omega $$
$$ u(\vec{x}) = 1\ \text{for } \vec{x}\in \partial\omega, $$
where $\omega$ is some set in $\Omega$. How can we find the $\omega$ that minimizes the quantity $\int_{\gamma}|du/dn|^2$ where $\gamma$ is the boundary of $\omega$ with unit outward normal $n$ ?
I'm more interested in finding out which branch of mathematics studies questions like this and what concepts/tools are important to approach questions like this. Any references or suggestions to
similar problems are greatly appreciated. (A friend suggested I tag this as geometric measure theory, but I don't know how appropriate that is)
oc.optimization-control calculus-of-variations geometric-measure-theory
add comment
1 Answer
active oldest votes
I'd say these belongs to the family of shape optimization problems in the Calculus of Variations (of whom the most famous example is the isoperimetric problem). As to the first one, you
should make some assumption on $f$ to ensure existence (for e.g. $f(x):=\frac{1}{1+|x|^2}$ gives a non attained infimum, equal to 0). Natural assumptions are e.g.: $f$ is smooth, coercive,
up vote and its only critical point is its global minimum. This should give as minimizer a level set of $f$ -the last assumption is just to get a regular simple curve. (The latter problem seems less
2 down easy and should require at least some small computation. I guess the solution is a circle). PS: Say, what about any $\omega$ as minimizer, as $u=1$ constant in $\omega$ ? :-) The latter
vote problem was a joke, wasn't it?
add comment
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Electrical computers: arithmetic processing and calculating
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2010 > patent applications in patent subcategories. category listing
20100191786 - Digital signal processing block with preadder stage: A digital signal processing block with a preadder stage for an integrated circuit is described. The digital signal processing block
includes a preadder stage and a control bus. The control bus is coupled to the preadder stage for dynamically controlling operation of the preadder stage. The preadder stage includes: a... Agent:
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20100191788 - Multiplier with shifter: A digital system has a memory configured to hold operands and a multiply-shift unit coupled to the memory and configured to receive a first operand and a second
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20100191787 - Sequential multiplier: A sequential multiplier for multiplying a binary multiplier and a binary multiplicand to produce a final product. A first logic circuit generates a control signal
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20100191789 - Method for regulating an actual value of a variable characterizing a position of an actuator, computer program product, computer program, and recording medium: A method is provided for
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20100191790 - System and method for correlation scoring of signals: Systems, methods and computer readable storage media are provided for identifying, in a signal of interest, signal segments
matching a reference signal segment. A processor coupled to memory is adapted to perform operations including: converting the reference signal segment to a first vector characterized by n pairs of
data points,... Agent: Agilent Technologies Inc.
20100191791 - Method and apparatus for evaluation of multi-dimensional discrete fourier transforms: A device and method for evaluating multidimensional discrete Fourier transforms (DFT) by
eliminating transpose operations by transforming every dimension concurrently. At least one computing node is enabled to evaluate a DFT of one of a multidimensional input data set and a subgroup of
the input data set, wherein the subgroup... Agent: Perry + Currier Inc.
20100191792 - Signal processing with fast s-transforms: The ability to examine the frequency content of a signal is critical in a variety of fields, and many techniques have been proposed to fill
this need, including the Fourier and wavelet family of transforms. One of these, the S-transform, is a Fourier based transform that provides simultaneous time and... Agent: Fulbright & Jaworski
20100191793 - Symbolic computation using tree-structured mathematical expressions: A method for performing symbolic computations on a mathematical expression. The mathematical expression may be
converted to a tree structure having one or more parent nodes and one or more child nodes. Each parent node may be a mathematical operation. Each child node may be a mathematical expression on
which... Agent: Microsoft Corporation
07/22/2010 > patent applications in patent subcategories. category listing
20100185715 - Method and device for transform computation: A method of operating a data-processing unit to produce a transform comprises calculating first and second output data values based at least
on first and second input data values. The method comprises reading the first and second input data values from locations of a first buffer, the locations being determined... Agent: Potomac Patent
Group PLLC
20100185716 - Eigenvalue decomposition apparatus and eigenvalue decomposition method: The present invention provides an eigenvalue decomposition apparatus that can perform processing in parallel at
high speed and high accuracy. The eigenvalue decomposition apparatus comprises a matrix dividing portion 14 that repeatedly divides a symmetric tridiagonal matrix T into two symmetric tridiagonal
matrices, an eigenvalue decomposition portion 15 that performs... Agent: Kenealy Vaidya LLP
07/15/2010 > patent applications in patent subcategories. category listing
20100179974 - Signal processing method for hierarchical empirical mode decomposition and apparatus therefor: A signal processing method for performing hierarchical empirical mode decomposition
(H-EMD) and an apparatus therefor are provided. In an embodiment, when empirical mode decomposition is performed on an input signal, an artificial assisting signal and the input signal are combined
to assist the search for extrema and frequency reduction is... Agent: Thomas, Kayden, Horstemeyer & Risley, LLP
20100179975 - Method for decomposing barrel shifter, decomposed circuit and control method thereof: A method for decomposing a barrel shifter decomposes N, the number of digits of input word, into N1
to Nm, and utilizes m layers of shifter circuit layer, which are composed of a plurality of barrel shifters, such that each barrel shifter performs a shifting procedure to obtain the desired...
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20100179976 - Semiconductor device performing operational processing: A semiconductor device includes a decoder receiving first multiplier data of 3 bits indicating a multiplier to output a shift
flag, an inversion flag, and an operation flag in accordance with Booth's algorithm, and a first partial product calculation unit receiving first multiplicand data of 2 bits indicating a
multiplicand,... Agent: Mcdermott Will & Emery LLP
20100179977 - Sampled filter with finite impulse response: According to the invention, there is proposed an FIR filter comprising a transconductance amplifier with controllable gain (AGM), at least
one sampling capacitor (CE) intended to receive an output current (di) from the amplifier and to periodically accumulate the charges produced by N successive samples of this current, and means...
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20100179978 - Fft-based parallel system with memory reuse scheme: A method may include storing N number of Fast Fourier Transform (FFT) data points into x-memories, N and x being integers greater
than one, and the x-memories having a total memory capacity equivalent to store the N number of FFT data points, and reading K FFT data points of the... Agent: Harrity & Harrity, LLP
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20100174764 - Reuse of rounder for fixed conversion of log instructions: A method for converting a signed fixed point number into a floating point number that includes reading an input number
corresponding to a signed fixed point number to be converted, determining whether the input number is less than zero, setting a sign bit based upon whether the input number is... Agent: Cantor
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20100174765 - Performing variable and/or bitwise shift operation for a shift instruction that does not provide a variable or bitwise shift option: Some embodiments present a method of performing a
variable shift operation. This method can be used by a microprocessor that does not allow variable shift operation for certain operand sizes. The method simulates a shift instruction that shifts an
operand by a shift count. The method identifies a first shift... Agent: Adeli & Tollen, LLP
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20100174767 - Efficient filtering with a complex modulated filterbank: A filter apparatus for filtering a time domain input signal to obtain a time domain output signal, which is a representation of
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signals from... Agent: Nxp, B.v. Nxp Intellectual Property & Licensing
20100174769 - In-place fast fourier transform processor: An N-point Fast Fourier Transform (FFT) using mixed radix stages with in-place data sample storage may be performed by decomposing N into a
product of R sequential mixed radix stages of radix-r(i). N data samples are partitioned into at least B memory banks, where B is equal to a largest... Agent: Texas Instruments Incorporated
07/01/2010 > patent applications in patent subcategories. category listing
20100169396 - Efficient computation for eigenvalue decomposition and singular value decomposition of matrices: For eigenvalue decomposition, a first set of at least one variable is derived based on a
first matrix being decomposed and using Coordinate Rotational Digital Computer (CORDIC) computation. A second set of at least one variable is derived based on the first matrix and using a look-up
table. A second... Agent: Qualcomm Incorporated
20100169397 - Mobile terminal and unit converting method thereof: A mobile terminal and its unit conversion method are disclosed. When a unit conversion function is selected through a menu
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20100169398 - Method and apparatus having a measured value input for applying a measured value: The invention pertains to a device such as a sensor, operator device, communication device, or a liquid
level metering device, with a measured value input to apply a measured value. The device includes at least a first memory region to provide for an adjustment factor, and a computer, which is...
Agent: Lackenbach Siegel, LLP
20100169400 - Partially random permutation sequence generator: Embodiments of the present disclosure provide methods, systems, and apparatuses related to a partially random permutation sequence
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20100169399 - Personal identification number (pin) generation between two devices in a network: A method of generating a Personal Identification Number (PIN) between a first device and a second
device in a network is provided. The method includes securely receiving information of input choices of the second device and random numbers assigned to the input choices at the first device. At the
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20100169401 - Filter for network intrusion and virus detection: Methods and apparatus to perform string matching for network packet inspection are disclosed. In some embodiments there is a set of
string matching slice circuits, each slice circuit of the set being configured to perform string matching steps in parallel with other slice circuits. Each slice circuit may include an... Agent:
Larry Mennemeier C/o Intellevate, LLC
20100169402 - Fast fourier transform processor: An FFT processor is disclosed, which includes a first multi-pipelined MDC unit, a second multi-pipelined MDC unit and a switching network. The first
multi-pipelined MDC unit and the second multi-pipelined MDC unit respectively employ a plurality of MDC circuits to change the positions of the delayers thereof in parallel way.... Agent: Jianq Chyun
Intellectual Property Office
20100169403 - System for matrix partitioning in large-scale sparse matrix linear solvers: A system for solving large-scale matrix equations comprises a plurality of field programmable gate arrays
(FPGAs), a plurality of memory elements, a plurality of memory element controllers, and a plurality of processing elements. The FPGAs may include a plurality of configurable logic elements and a
plurality of configurable storage elements.... Agent: Hovey Williams LLP
20100169404 - Flexible accumulator in digital signal processing circuitry: A multiplier-accumulator (MAC) block can be programmed to operate in one or more modes. When the MAC block implements at
least one multiply-and-accumulate operation, the accumulator value can be zeroed without introducing clock latency or initialized in one clock cycle. To zero the accumulator value, the most
significant bits (MSBs)... Agent: Ropes & Gray LLP
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Posts by
Total # Posts: 521
Science ASAP
2. Ferns have roots, stems, and leaves just like seed plants. Why are they classified separately? (1 point) Their leaves are compound and feathery. They produce cones to reproduce. Their roots are
only shallow rhizomes. They produce spores instead of seeds. 3. The first plants...
Math Ms. Sue please
thank you!
Math Ms. Sue please
4. Which of the following is true about a trend line for data? (1 point) The minimum data point always lies on the trend line. Every data point must lie on the trend line. The trend line describes
the pattern in the data if one exists. The trend line includes the effect of all...
Math for Ms. Sue please! Last math questions!
Thank you Ms. Sue!
Math for Ms. Sue please! Last math questions!
1. What is the term for data that are grouped closely together? (1 point) outlier linear positive clustering 2. What association would you expect if graphing height and weight? (1 point) positive
nonlinear negative none of these 4. What association would you expect if graphing...
math Ms. Sue please
thaaaaank you!!!!
math Ms. Sue please
oh outside temperature and heating bill!
math Ms. Sue please
3. Which of the following examples would show a negative trend? (1 point) height and weight of students test scores and height of students outside temperature and heating bill none of these Is the
answer " test scores and height of students"?
health Ms. Sue please
Oh and yes, I agree with your answer, bobpursley
health Ms. Sue please
Thank you. But bobpursley, is Ms. Sue's answer right for #2?
health Ms. Sue please
1. The body's functions are slowed when alcohol reaches where? (1 point) The stomach The heart The liver The brain 2. Which statement is NOT true? (1 point) Regular use of alcohol can lead to
addiction. Alcoholism can be cured. Someone cannot cause someone else to drink. A...
Ms. Sue please science
14. The ______ system uses all of your body s senses to help maintain homeostasis is the answer control?
health science please ASAP
thank you
health science please ASAP
2. One of the purposes of sweat glands is to release waste products, such as water and _____. (1 point) blood dirt sugar salt 11. A human baby develops inside the mother s _________ for ______
months. (1 point) vagina; six pelvis; one uterus; nine ovary; eight My answers:...
science help ASAP please
The ______ system uses all of your body s senses to help maintain homeostasis. Is the answer "control"? Or is it "endocrine"?
math help please ASAP
ohmygosh i am so sorry
math help please ASAP
The ______ system uses all of your body s senses to help maintain homeostasis. Is the answer "control"? Or is it "endocrine"?
Math Ms. Sue please
Thank you!
Math Ms. Sue please
6. Which is the best measure of central tendency for the type of data below the mean, the median, or the mode? Explain. Hours of sleep each night Median; there will be outliers Range; there are no
outliers Mode; the data are non-numeric Mean; the outliers are limited My a...
Health Ms. Sue please
That is ok! Thank you for checking them!
Health Ms. Sue please
oh actually #2 was "pituitary gland; procreation".
Health Ms. Sue please
Health Ms. Sue please
1. What occurs in a woman's body during a menstrual cycle? (1 point) The egg matures in the ovary. The uterine lining builds up. The egg releases from the ovary. all of the above 2. The _______
produces the hormone that develops the reproductive system, which is responsibl...
Math for Ms. Sue please last questions
actually Ms. Sue the answer for 2 was 191...why?
Math for Ms. Sue please last questions
Oh I see! I was doing 620 - 420 = 210! Thank you!
Math for Ms. Sue please last questions
I am confused! Why is the answer wrong? Can you please explain why!
Math for Ms. Sue please last questions
okay thank you. Hold on while I fix it.
Math for Ms. Sue please last questions
2. In a school of 464 students, 89 students are in the band, 215 students are on sports teams, and 31 students participate in both activities. How many students are involved in neither band nor
sports? (1 point) 160 students 191 students 249 students 433 students 3. In a marke...
Math Ms. Sue please
Thank you!
Math Ms. Sue please
2. Find the outlier in the data set and tell how it affects the mean. 4, 4, 6, 2, 14, 1, 1 (1 point) 6; it raises the mean by about 1. 6; it lowers the mean by about 1. 14; it raises the mean by
about 1.9. 14; it lowers the mean by about 1.9. 5. 22.6 is...
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I am confused! Is 17 240 or 160??
Math for Ms. Sue please!
I am not sure about 17 either. Do you know how to solve it?
Math for Ms. Sue please!
1. Name the solid with the description as a figure that has one base that is a rectangle and four lateral surfaces that are triangles. (1 point) triangular pyramid cone rectangular prism rectangular
pyramid 2. A solid with two parallel and congruent bases cannot be which of th...
gas, liquid, and solid.
math Ms. Sue please ASAP
Please help!
math Ms. Sue please ASAP
3. Find the surface area for a cylinder with a height of 24 and a radius of 6. Use 3.14 for pi and round to the nearest whole number. 904 yd^2 1,017 yd^2 1,130 yd^2 2,713 yd^2 4. Find the volume for
a cylinder which has a height of 24 and a radius of 6. Use 3.14 for pi and rou...
Um, I did, actually the right answer was 180 in.^2
A pyramid has a height of 5 in. and a surface area of 90 in^2. Find the surface area of a similar pyramid with a height of 10 in. Round to the nearest tenth, if necessary. 360 in^2 180 in^2 22.5 in^2
3.6 in^2
Math for Ms. Sue please! Last math questions!
Okay, the answer was 180 in.^2
Math for Ms. Sue please! Last math questions!
Okay thank you anyway! The answer is "2 cm" for #4.
Math for Ms. Sue please! Last math questions!
For numbers 1 3, find the indicated measurement of the figure described. Use 3.14 for pi and round to the nearest tenth. Find the surface area of a sphere with a radius of 8 cm. 267.9 cm^2 803.8 cm^2
2143.6 cm^2 201.0 cm^2 Find the surface area of a sphere with a diameter...
Math Ms. Sue please
Whoops! I am so sorry! Didn't see them! Yeah I thought it was 8 I solved the problem backwards and that is what I got. Thank you ever so much!
Math Ms. Sue please
**I meant can you delete it once you are done checking it. Thank you.
Math Ms. Sue please
Can you delete this post please! I need to be careful because students in K-12 have been cheating off of my posts. Sorry for the inconvenience!
Math Ms. Sue please
Is answer #5 "8 cm"?
Math Ms. Sue please
108! Thank you!
Math Ms. Sue please
Please help ASAP
Math Ms. Sue please
Find the volume of a square pyramid with a base length of 9 cm and a height of 4 cm. 324 cm^3 108 cm^3 36 cm^3 152 cm^3 Is the answer 152?
Ms. Sue please.....
Thank you.
Ms. Sue please.....
Oh okay.....oh well. Do you know who I could contact to make this happen?
Ms. Sue please.....
Thank you! Wait what about those math posts I posted yesterday...can you delete those? How do you delete posts?
Ms. Sue please.....
Please delete these if you can: 1. "7th grade math Ms. Sue please last questions! Posted by Delilah on Wednesday, February 13, 2013 at 9:40pm." 2. "7th grade math please help Ms. Sue ASAP Posted by
Delilah on Wednesday, February 13, 2013 at 9:19pm." 3. &quo...
Ms. Sue please.....
Okay thank you. Hold on a few seconds....
Ms. Sue please.....
If you are a member, can it be deleted?
Ms. Sue please.....
I found out that my posts were helping other students cheat.
Ms. Sue please.....
Huh....that's weird. I am a member, though.
7th grade math Ms. Sue please last questions!
15. Find the area of a triangle with a base of 9.3 ft. and the height is 3.5 ft. 32.55 ft² 24.81 ft² 16.275 ft² 11.0 ft² 16. Find the area of a parallelogram with a base of 15 in. and the height is 7
in. 105 in² 110 in² 52.5 in² 210 in² ...
7th grade math please help Ms. Sue ASAP
Okay thank you!
7th grade math please help Ms. Sue ASAP
Okay thank you. But how did you get 720?
7th grade math please help Ms. Sue ASAP
14. Find the value of the missing angle. (It is a hexagon. In the picture, going counter-clockwise, the angles are: 152, 85, 125, 135, 85, and "x" degree angles) 720º 120º 128º 138º I need help
finding "x"! So far, the angles they have p...
Yes, the answer is "their". So your answer is correct.
7th grade math please help Ms. Sue
Thank you, Ms. Sue and "drwls"!
7th grade math please help Ms. Sue
Okay, I think that is why I was confused. But #9 is still right, no?
7th grade math please help Ms. Sue
9. Name the triangles that are classified by angles. (1 point) right, scalene, isosceles scalene, isosceles, equilateral acute, right, obtuse obtuse, isosceles, acute 11. Name the quadrilaterals that
have four equal sides. (1 point) rhombus, square square, rectangle parallelog...
7th grade social studies
Okay thank you!
7th grade social studies
Which consumers use this natural resource? Which products are made using this natural resource? How is the environment impacted by the use of the natural resource? How important is the natural
resource to the country? Why? **I really have researched, I have spent 2 hours alrea...
7th grade social studies
Okay will do! Is it okay if you help me on a few others? Please!
7th grade social studies
Okay thanks! So 1. Jericho, Ekati, Diavik, Snap Lake, and Victor 2. Yes it has to be processed (I will explain why) 3. Diamonds are imported; the Kimberly Process Certification Scheme is also used.
**I will write everything in greater detail, but those are my basic ideas.
7th grade social studies
I am doing a research on diamonds and need help. I am having trouble finding a source that answers my questions besides Wikipedia. My teachers will not let us use Wikipedia. Can you please provide
links to these questions? 1. Where is the natural resource (diamonds) located? I...
7th grade language arts Ms. Sue please
I didn't pick that one because I didn't think that evolution would be related to dog behavior.
7th grade language arts Ms. Sue please
I suppose not. Thank you!
7th grade language arts Ms. Sue please
oh no! Actually it was "a webpage on the evolution of dogs written by a college professor"
7th grade language arts Ms. Sue please
1. Which of the following is not a part of evaluating sources? (1 point) checking the author's credentials questioning possible bias making sure the information is current agreeing with the author's
opinions 2. Which of the following websites would you most likely use ...
7th grade science help ASAP 2 questions!
4. Most digestion takes place in the _____. (1 point) large intestine mouth small intestine stomach 5. The large intestine maintains ___________ by absorbing water from the _______, or undigested
mass. (1 point) homeostasis; chyme digestion; saliva mechanical digestion; enzym...
7th grade math please help Ms. Sue
Thank you ever so much! :-)
7th grade math please help Ms. Sue
4. Find the area for a triangle: base 14, height 5.5. (1 point) 19.5 cm² 77 cm² 38.5 cm² 770 cm² 5. Find the area of a parallelogram: base 5.2, height 2.3. (1 point) 10.6 cm² 12.96 cm² 5.98 cm² 11.96
cm² My answers: 4. 38.5 cm² 5. 1...
7th grade health please help Ms. Sue!
Thank you so much!
7th grade health please help Ms. Sue!
Okay thanks. Is #4 "Tobacco affects the whole body"?
7th grade health please help Ms. Sue!
1. Which substance found in tobacco is a drug that speeds up the heartbeat? (1 point) Tar Carbon monoxide Nicotine Emphysema 2. Which substance found in tobacco causes emphysema, a disease of the
lungs? (1 point) Tar Carbon monoxide Nicotine Secondhand smoke 3. Which substance...
7th grade math please help Ms. Sue ASAP
12. Write a sequence of transformations that maps triangle ABC onto triangle A''B''C''. It is a graph, so here are the coordinates: A is 1,8 B is 3,12 C is 4,4 A'' is 3, -3 B'' is 5, -6 C'' is 6, 2
Please help! I have no idea how...
7th grade math Ms. Sue please
2. Find the sum of the interior angles of a nonagon. (1 point) 140° 1,620° 1,260° 1,450° 4. Find the measure of each interior angle of a polygon with 12 sides. (1 point) 1,800° 150° 180° 145° 5. Four
of the angles of a pentagon measure 85°, ...
7th grade math Ms. Sue please
4. List all of the quadrilaterals with exactly one pair of parallel sides. (1 point) parallelogram; trapezoid rhombus; parallelogram; rectangle trapezoid rhombus;, square; rectangle 5. List all of
the quadrilaterals with four right angles. (1 point) trapezoid; rectangle; rhomb...
7th grade language arts Ms. Sue pleaselast queston
Thanks ever so much, Ms. Sue!
7th grade language arts Ms. Sue pleaselast queston
3. Which of the following words shares the base word of "destination"? (1 point) destroy destiny destabilize destruction Is the answer "destiny"?
Sorry I do not know this question, but if you want Ms. Sue to help you, type her name in the box where you put "Math". So your title would be "Math for Ms. Sue please" Just a suggestion! That way she
can help you faster!
7th grade language arts Ms. Sue please
oops sorry! 5. Which of the following describes the underlined part of this sentence from "LAFFF"? IN SPITE OF ALL I COULD DO, my grades were nothing compared to Peter's. (1 point) simple sentence
compound sentence independent clause dependent clause 6. Which of ...
7th grade language arts Ms. Sue please
5. Which of the following describes the underlined part of this sentence from "LAFFF"? In spite of all I could do, my grades were nothing compared to Peter's. (1 point) simple sentence compound
sentence independent clause dependent clause 6. Which of the followin...
7th grade science bobpursley
Okay I will! Thanks!
7th grade science bobpursley
5. Change in the hereditary features of a type of organism over time is ______. (1 point) growth biogenesis spontaneous generation evolution 10. Gregor Mendel discovered some alleles can cover up the
presence of others; these alleles are called __________. (1 point) My answers...
7th grade science ASAP
1. How many chromosomes do human sex cells have? (1 point) 22 43 46 23 2. The actual genetic makeup of an organism for a particular trait is called the ______. The physical characteristic that trait
gives the organisms is called the ______. (1 point) genotype; phenotype genome...
7th grade science
1. How many chromosomes do human sex cells have? (1 point) 22 43 46 23 2. The actual genetic makeup of an organism for a particular trait is called the ______. The physical characteristic that trait
gives the organisms is called the ______. (1 point) genotype; phenotype genome...
7th grade science ms sue please
okay thanks anyway!
7th grade science ms sue please
1. How many chromosomes do human sex cells have? (1 point) 22 43 46 23 2. The actual genetic makeup of an organism for a particular trait is called the ______. The physical characteristic that trait
gives the organisms is called the ______. (1 point) genotype; phenotype genome...
7th grade art ASAP please
1. This occurs when an animal, object, or idea is given human form or characteristics. (1 point) literalism symbolism personification allusion 2. This occurs when you make reference to a well-known
character, person, place, or event from life or literature. (1 point) literalis...
7th grade health true or false
11. Disagreements are a normal part of life. (1 point) True False 12. Talking to a trustworthy adult is the best way to stop sexual abuse and get help. (1 point) True False 13. Some conflicts are
made worse by peer negotiation. (1 point) True False 14. Feeling negatively about...
7th grade health please
1. The process of talking directly to the other person to resolve a conflict (1 point) conflict negotiation assault compromise 2. Forced sexual intercourse (1 point) gang assault rape neglect 3. A
disagreement between people with opposing viewpoints, interests, or needs (1 poi...
7th grade social studies ms sue last question
7th grade social studies ms sue last question
9. Many people in developing nations get their food by practicing (2 points) subsistence farming. commercial farming. mechanized farming. developing farming. is the answer "subsistence farming"?
7th grade Social Studies Ms. Sue please!
This homework question was removed due to a copyright claim submitted by Connections Education. My answers: 1. organize information. 2. enough detail. 3. Earth's tilt and orbit 4. new landforms. 5.
latitude. 6. polar. 7. birth rates and death rates. 8. They cannot find wor...
7th grade language arts Ms. Sue please
1. Trying to spread ideas that help one cause while hurting another is called (1 point) marketing. semantic slanting. propaganda. 2. How would you revise the title of "Take the Junk out of Marketing
Food to Kids" to avoid semantic slanting? (1 point) Marketing Health...
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[SciPy-Dev] import scipy convention in docstring Examples
David Goldsmith d.l.goldsmith@gmail....
Sun Jun 13 14:54:33 CDT 2010
On Sun, Jun 13, 2010 at 12:43 PM, David Goldsmith
> On Sun, Jun 13, 2010 at 12:41 PM, Warren Weckesser <
> warren.weckesser@enthought.com> wrote:
>> David Goldsmith wrote:
>> > Hi! For the Examples in the numpy docstrings, we have the convention
>> > that we do not need to include "import numpy" (because that is done by
>> > our automated example tester) but when we're using a numpy namespace
>> > object, we need to prefix it with "np." (because that's how our
>> > automated tester is importing it). Is there a similar convention for
>> > docstring Examples in scipy?
>> >
>> According to http://projects.scipy.org/numpy/wiki/CodingStyleGuidelines,
>> "The examples may assume that import numpy as np is executed before the
>> example code in numpy, and import scipy as sp in scipy. " But sp does
>> not appear to be widely used; I can only find it used in
>> ndimage.geometric_transform and ndimage.map_coordinates.
Sorry, hit send too soon. :( As I was saying: thanks Warren - obviously, I
need to do a little better research before I post. As far as the low
occurrence, however, this actually does not surprise me much: aside from the
simple dearth of scipy docstrings from which such a search would produce
results, unlike the numpy examples, where it is common to want/need
something in the numpy-level namespace, scipy is structured such that it
will probably be rare that things in the scipy-level namespace are
needed/wanted - I mainly asked to just get it announced: in the Examples in
the scipy docstrings, explicitly import the namespaces you need, EXCEPT for
scipy itself - if you need something from that namespace, it will already be
there for you, abbreviated as "sp."
Thanks again,
> Thanks, Warren, I d
>> Warren
>> > DG
>> > ------------------------------------------------------------------------
>> >
>> > _______________________________________________
>> > SciPy-Dev mailing list
>> > SciPy-Dev@scipy.org
>> > http://mail.scipy.org/mailman/listinfo/scipy-dev
>> >
>> _______________________________________________
>> SciPy-Dev mailing list
>> SciPy-Dev@scipy.org
>> http://mail.scipy.org/mailman/listinfo/scipy-dev
> --
> Mathematician: noun, someone who disavows certainty when their uncertainty
> set is non-empty, even if that set has measure zero.
> Hope: noun, that delusive spirit which escaped Pandora's jar and, with her
> lies, prevents mankind from committing a general suicide. (As interpreted
> by Robert Graves)
Mathematician: noun, someone who disavows certainty when their uncertainty
set is non-empty, even if that set has measure zero.
Hope: noun, that delusive spirit which escaped Pandora's jar and, with her
lies, prevents mankind from committing a general suicide. (As interpreted
by Robert Graves)
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Logic translation
October 30th 2007, 10:32 AM #1
Logic translation
I got this question wrong on my test, and I just can't understand why. I was supposed to translate this sentence into sentential logic. (I posted it in discrete math b/c there is no section for
logic, but logic is covered in discrete math)
If my plans work out, then even though I got a D on the first test I will go into the final with a good average. (use P,D,G as propositions)
My translation:
$P\to G$
Expected translation:
$P\to (D$& $G)$
On the answer key, my professor added a note which said "even though" = "and"
I guess when I read it, "even though I got a D" means that I already got a D, so getting a D must be true regardless of what my plans are. So in my interpretation, D is always true (because it
already happened). In his interpretation, D is true if P is true, meaning both P and D could be false.
I want to go argue the point with him, but I figured I'd see what you guys think first, b/c you guys are considerably more educated than I am. Also, if you disagree with me, can you post an easy
to understand reason why, and if you do agree with me, can you suggest a method of validating my interpretation.
I agree with your instructor.
I read it as: "If my plans work out, then even with a D I will go into the final with a good average."
I got this question wrong on my test, and I just can't understand why. I was supposed to translate this sentence into sentential logic. (I posted it in discrete math b/c there is no section for
logic, but logic is covered in discrete math)
If my plans work out, then even though I got a D on the first test I will go into the final with a good average. (use P,D,G as propositions)
My translation:
$P\to G$
Expected translation:
$P\to (D$& $G)$
On the answer key, my professor added a note which said "even though" = "and"
I guess when I read it, "even though I got a D" means that I already got a D, so getting a D must be true regardless of what my plans are. So in my interpretation, D is always true (because it
already happened). In his interpretation, D is true if P is true, meaning both P and D could be false.
I want to go argue the point with him, but I figured I'd see what you guys think first, b/c you guys are considerably more educated than I am. Also, if you disagree with me, can you post an easy
to understand reason why, and if you do agree with me, can you suggest a method of validating my interpretation.
You have a case. "Even though" does mean and, but I think your professor is attaching it to the wrong clause. The sentence should read better:
Even though I got a D, if my plans work out, then I will go into the final with a good average. It's D & (P -> G).
The statement as read should obviously be false if you didn't get a D on first test, but if your professor's result is correct, then the statement would hold true in a world where your plans
didn't work out, you passed the test, and went into the final with a poor average (because False->(False & False) is true).. But in our better answer, False & (False->False) = False & True =
False, which is how it should be under those circumstances.
Consider a statement with the same structure whose meanings are even more obvious: If I can find a partner, even though it's raining, I am going to play tennis. It is absurd to interpret this as
anything other than It's raining AND if I can find a parter, I'm going to play tennis., not If I can find a partner, then it is raining and I'll play tennis. In the latter case, the statement
would be true in some conditions even if it wasn't raining.
Go get your points.
October 30th 2007, 12:23 PM #2
October 30th 2007, 12:41 PM #3
Jun 2006
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Free Quarks ? Don't Be Fooled!
These days I am preparing a three-hour course of statistics for particle physicists which I will give at a winter school in a couple of months. This stimulating task forces me to find nice and simple
examples of good and bad applications of basic statistics. Stuff with high didactical value, and hopefully also entertaining.
For today, I am happy with a simple illustration of why to be a physicist you need to know basic Statistics. The example is of course based on a real analysis in particle physics. It is based on a
claim made in 1969 by McCusker and Cairns that they had observed the track of a special charged particle in a bubble chamber exposed to energetic air showers. The track appeared to ionize the gas of
the chamber less than half as much as what it should have: 110 droplets along a unit-length path instead of 229.
The figure on the right, taken from PRL 23 (1969), page 658, shows a bunch of parallel tracks caused by a shower of charged particles from a energetic cosmic ray. Among the tracks there is one which
is much fainter than the others (I leave you to guess which one it is): that is the one which McCuster and Cairns claimed to be due to a fractionally charged particle.
The question one must answer, in order to start fiddling with the idea that the track in question is due to a free quark or some other exotic thing produced in the very high-energy interaction in the
atmosphere, is how likely it is that one observed, among 55,000 tracks, one track with 110 or fewer droplets per unit path length, when the average expected number is 229 (the latter determined by
studying the total sample of tracks).
I can hear some of you radiating confidence as you think: "Ah! A simple problem in Statistics... The number of droplets is a Poisson variable, since the Poisson distribution describes the probability
of finding n events in a given time, if these occur independently from a constant rate process". Very good then: what you have in mind is the formula
P(n) = [mu^n e^(-mu)]/n!
where mu is the average number in the given time (or given path length, in our case). If for a track we observe n=110 when mu=229, we may ask "What is the probability that a track produces less than
111 droplets, if the expected number of droplets is 229?". The required probability is the sum of P(n) with n running from 0 to 110, and the result is P(n<=110) = 1.6 x 10^(-18).
Okay, then if we have 55,000 of these, the total probability to observe one or more of these is P(>=1 weird track in 55000) = 1-[1-P(n<=110)]^(55000), which is in the whereabouts of 10^-13. One in
ten thousand billions! That must surely be a fractional-charge particle ! We are surpassing the five-sigma "observation-level" significance head, shoulders, and tail here.
Hmmm, not so fast.
If you are a good physicist, you know that a single scattering of a charged particle off a nucleus in the vapour of the chamber produces more than a single droplet. In fact, this number is four, on
average. The droplet production is itself a Poisson process. Fine, so we have two separate Poisson processes - particle scattering, and droplet formation. Does it change the picture ? No, if you are
a good physicist who does not know squat about Statistics.
If you have at least a good hunch of basic Statistics, however, you know that what you are describing is not a simple Poisson process, but a compound Poisson process. And the two are rather different
things. The fact that scatterings yield an average of four droplets in fact dramatically increases the likelihood of tracks with small droplet multiplicity: by using the correct distribution function
of the compound Poisson (where now mu=4, lambda*mu=229, and N is the number of scatterings):
If we then ask what is the chance to have seen at least one such low-ionization track in the 55,000 sample, the probability is now 1-(1-P')^(55000), which is a striking 92.5% ! Ooooops! We should
have rather been surprised to NOT have observed such a low-ionization track!
The summary of this example is simple to spell: You may know your detector and the underlying physics as well as you know your ***, but if you do not know basic Statistics you are going to be fooled
The McCusker and Cairns PRL article soon received its due rebuttal [R.Adair and H.Kasha, PRL 23, 1355 (1969)]. For an evidence of quarks, one would have to wait five more years...
Vladimir Kalitvia... | 11/07/11 | 11:25 AM
Hi Vladimir,
all charged particles except electrons above some energy reach a regime called "MIP", for minimum-ionizing particle. In these conditions their energy loss is quite similar. The particles recorded
were in the core of high-energy showers, and all in the MIP regime.
As for "stability" of the tracks I do not know what you refer to. A track is a trajectory, and as such it has a direction and a curvature, but no dynamical properties. If you are talking about the
collection of droplets, if you need a description of the way droplets condense along charged particle paths I suggest a book on tracking detectors... There are no stability concerns that I know of.
Tommaso Dorigo | 11/07/11 | 14:00 PM
Thanks, Tommaso, for your answer. Yes, I meant droplet formation, growth, and disappearing dynamics.
As to the charge "measuring", it is better done in magnetic fields.
So your explanation of a thinner track is a rare statistical fluctuation. Then, more frequent fluctuations must be present amongst those 55000 tracks too.
Vladimir Kalitvia... | 11/07/11 | 14:28 PM
So your explanation of a thinner track is a rare statistical fluctuation. Then, more frequent fluctuations must be present amongst those 55000 tracks too.
That is exactly what crossed my mind. In other words, the argument of
the chance to have seen at least one such low-ionization track in the 55,000 sample, the probability is now 1-(1-P')^(55000), which is a striking 92.5% !
is unconvincing if there is only one such track and not at least many more in a continuum of ionization track densities. As it is presented now, it seems somebody stopped arguing once he had the
result he needed. But the argument is not finished. If this is the only track of low density, the probability is certainly not 92.5%!
Sascha Vongehr | 11/07/11 | 22:30 PM
Indeed, there must have been exactly 27500 tracks with lower-than-average ionization. But the data we have, from the short article I quoted above, is that they found one low-ionization track, what
was its specific droplet density, and what was the expectation value for that. I do not venture to speculate what the authors have done with the rest of their data.
In general, however, please note that the fact that one expects four droplet per interaction is just a model. As such, it approximates things. Not all tracks have the same ionization, as mentioned
above, and not all scatterings are equal.
What do we get from those observations ? We get that the fixing of mu=4 and using a double Poisson is in itself a poor description of the physics. But it is a much better description than the
original one. Since the original claim of free quarks, and the calculated exceedingly low probability, was done with a poor statistical model, while a distinctly better model (although arguably still
not perfect) finds no statistical significant effect, the argument is over: if one wanted to resurrect the observation one would have to do a better homework.
Tommaso Dorigo | 11/08/11 | 04:33 AM
they found one low-ionization track ... and what was the expectation value for that.
So you are charging scientific misconduct rather than bad statistics? Surely if they found a whole continuum of lower density tracks, they should not have portrayed this one as special, even without
doing any statistics. Assuming they were not cheating, the ones presented were the only ones found (sure they would have been proud to present more if they have had them). The expectation value for
that is one in gazillions.
Sascha Vongehr | 11/08/11 | 05:22 AM
Tommaso Dorigo | 11/08/11 | 05:56 AM
And another insult. "Pathetic". Thank you TD. I could have called your 92.5% pathetic, but hey, that would have been an insult, right. What is it with you two southern Europe guys here thinking you
can insult other people and at the same time be all pissed off claiming others like me or Lubos are insulting if they plainly point out your mistakes? Oh right, that is your version of the scientific
Sascha Vongehr | 11/08/11 | 06:23 AM
Doug Sweetser | 11/09/11 | 00:05 AM
Paolo Ciafaloni | 11/07/11 | 11:38 AM
Tommaso Dorigo | 11/07/11 | 14:28 PM
"it's the first time I learn something from a Science 2.0 blog"
Wow - you surely like to insult people a lot in your comments and articles. Some people who cannot bear hardness of hard science arguments claim I do, but you are just here a short while and have
insulted so many people already with under the belt throws, I am a saint compared to that. Maybe you like to tone it down a little?
Sascha Vongehr | 11/07/11 | 22:41 PM
Ionization is (well, was) measured in bubble chambers by counting with a microscope the number of droplets along the particle path.
Thanks for this great educational article Tommaso. I read recently here and here that bubble chambers are no longer just history, that they are being rebuilt and resurrected to aid in the search for
dark matter :-
The Tevatron collider shut down at the end of September, but Fermilab physicists are still active in the ongoing search to directly detect dark matter. To aid in the research they're resurrecting
bubble chambers and fixed target experiments dating back to the 1970s. ANALYSIS: Where is Dark Matter Hiding? Bubble chambers are basically vessels filled with superheated liquid to detect
particles moving through it. A new experiment underway aims to achieve better calibration for the bubble chambers used in the Chicago land Observatory for Underground Particle Physics (COUPP)
experiment, located 350 feet underground in a Chicago tunnel. It's called the COUPP Iodine Recoil Threshold Experiment (CIRTE), and it's designed to improve the sensitivity of the COUPP detector.
There are lots of different experiments, using a variety of approaches, designed to search for dark matter particles, with some promising -- if hotly debated -- preliminary results. But bubble
chambers were nearly extinct in the field before COUPP leader Juan Collar hit upon the notion of using them to search for dark matter. They're great as neutrino detectors, too. In 2007, COUPP
installed a new germanium-based neutrino detector 330 feet below ground in the sewers of Chicago, renting this unusual lab space from the city. The design was modified to detect WIMPs instead of
Bubble chambers sound like fun things to experiment with, though maybe a bit slow, and I love the pictures, however I couldn't help wondering if fire extinguishers all over the world containing
iodotrifluoromethane are also all inadvertently acting as unrecognised bubble chambers, without the digital cameras in situ, after reading this part?
COUPP's "detector" is a glass jar filled with a liter or so of a fire-extinguishing liquid (iodotrifluoromethane). When a WIMP hits a nucleus of one of those atoms, it triggers an evaporation of
a small amount of that liquid, producing a tiny bubble.It's initially too tiny to see, but it grows, and that growth can be recorded with digital cameras."The bubbles in the fluid are slow enough
that high-speed cameras will capture the changes through continuous still shots. We're making the world's most boring movie," Peter Cooper, the Fermilab physicist heading up CITRE, told Symmetry
My latest forum article 'Australian Researchers Discover Potential Blue Green Algae Cause & Treatment of Motor Neuron Disease (MND)&(ALS)' Parkinsons's and Alzheimer's can be found at http://
Helen Barratt | 11/07/11 | 14:36 PM
Hi Helen,
yes, bubble chambers are not dead - I was thinking at HEP when I wrote that. They are very beautiful detectors, the only ones that allow a visualization of the physical reactions without the need of
computer reconstruction (emulsions also might be argued to provide the same feature, but that's another story).
Tommaso Dorigo | 11/07/11 | 14:33 PM
They are very beautiful detectors
What about the fire-extinguishers Tommaso? Are they also acting as bubble chambers?
My latest forum article 'Australian Researchers Discover Potential Blue Green Algae Cause & Treatment of Motor Neuron Disease (MND)&(ALS)' Parkinsons's and Alzheimer's can be found at http://
Helen Barratt | 11/07/11 | 14:58 PM
Tommaso Dorigo | 11/07/11 | 15:05 PM
Vladimir Kalitvia... | 11/07/11 | 15:23 PM
I just wondered because this article at a physics lectures site says :-
'you can have a superheated liquid at room temperature (20oC) like the CO2 in a fire extinguisher. Liquid CO2 at 20oC has a vapor pressure of 100 atmospheres (100 kg/cm2! To contain that we need
a strong steel container. If the valve is broken off, the awesome flashing of liquid to gas makes the cylinder into a rocket that can penetrate a wall! Video:'
My latest forum article 'Australian Researchers Discover Potential Blue Green Algae Cause & Treatment of Motor Neuron Disease (MND)&(ALS)' Parkinsons's and Alzheimer's can be found at http://
Helen Barratt | 11/07/11 | 16:19 PM
Technically, the liquid CO[2] in a fire extinguisher is not a superheated liquid, until "the valve is broken off". The distinction is that not only is temperature involved (20°C), but pressure as
Superheated liquid means that you have a liquid under temperature and pressure conditions where the phase diagram for the material indicates you should have a gas. So, for the given temperature and
pressure, one should have a gas, but, instead, one has arranged things (very carefully) so as to have a liquid instead.
(Of course, there are also cases where one may wish to go the other direction—with a supercooled gas, say. One may also have supercooled liquids, and superheated solids.)
David Halliday | 11/07/11 | 16:50 PM
Thanks David, so what about a pressurised fire extinguisher filled with the same fire-extinguishing liquid, iodotrifluoromethane as used in the COUPP detector? Will it occasionally be behaving like a
bubble chamber as electrically charged particles pass through it or only when and if the fire extinguisher valve is broken off?
My latest forum article 'Australian Researchers Discover Potential Blue Green Algae Cause & Treatment of Motor Neuron Disease (MND)&(ALS)' Parkinsons's and Alzheimer's can be found at http://
Helen Barratt | 11/07/11 | 17:08 PM
From the description you quoted in your previous post, the material, in the bubble chamber, must be in a superheated state (otherwise, the bubbles will collapse, rather than grow). So, if this is
correct, then no: Such fire extinguishers will not be acting as "bubble chambers". Even when one uses the fire extinguisher, thus lowering its internal pressure, one is highly unlikely to reach a
superheated state (too much perturbation occurring, and too many nucleating contaminants, typically).
Technically, even "if the fire extinguisher valve is broken off" one still doesn't have a superheated liquid state, but an almost explosive vaporization (boiling) of the liquid into a gas, all due to
the decompression one has instigated.
David Halliday | 11/07/11 | 17:38 PM
This stimulating task forces me to find nice and simple examples of good and bad applications of basic statistics.
Well, since your example above is not convincing, what about this one: Corrupting Bayesian updating and entering a prior of P = 0 (for example: Orthodoxy demands FTL neutrinos are impossible,
period!) in order to kill all experiments' significance to squat? That is a very simple example of "bad applications of basic statistics".
Sascha Vongehr | 11/07/11 | 22:36 PM
Tommaso Dorigo | 11/08/11 | 04:24 AM
You are off-topic Sascha.
Did you or did you not ask for examples of bad fundamental statistics? You know that my official work is mostly on statistics, right? Well, probably not, since you do not care about critical people.
So let me tell you, even my PhD was on nontrivial cluster collision statistics and since then I have published on several different issues where people in Nanotechnology use bad statistics.
Can you discuss anything else than neutrinos these days ?
These are my four recent articles (not counting the blog entry): On AI personal assistants, on a book about the IPCC and climategate, on ecological problems due to cats, and on the physical basis of
influence from the future. You are welcome to read my articles. Why don't you inform yourself first before being dismissive?
Sascha Vongehr | 11/08/11 | 05:15 AM
WHO KNOWS THAT THE MUON AND TAUON ARE RESP. DOWN- AND CHARM QUARKS.
Leo Vuyk (not verified) | 11/08/11 | 03:31 AM
From my point of view, there is another reason why this article is important: it underlines the fact that data, by themselves (or a figure like the one commented here), do not "speak" to us. In order
to interpret experimental data we always need tools (like statistics for instance) and models (like a model for droplet formation for instance). In other words, I think that sometimes it is very hard
to distinguish "theory" from "experiment": they both conspire to give the final interpretation.
I have one question: does your analysys imply that also the probability for higher inonization tracks is increased if one takes into account the underlying droplet formation process?
Cheers Paolo
PS I have been unable to spot the fainter track in the image :(
Paolo Ciafaloni | 11/08/11 | 07:01 AM
Tommaso Dorigo | 11/08/11 | 08:11 AM
How they did things in the old days: here are some pensioned off apparatus at DESY – “C” is their original bubble chamber.
Robert H. Olley / Quondam Physics Department / University of Reading / England
Robert H Olley | 11/08/11 | 11:45 AM
Tommaso Dorigo | 11/08/11 | 11:50 AM
In what sense are top quarks not "free quarks" if they hadronize? Is the point that we do not observe them direclty, and instead only observe their decay products?
ohwilleke (not verified) | 11/08/11 | 13:00 PM
typo correction "if they don't hadronize"
ohwilleke (not verified) | 11/08/11 | 13:01 PM
Hi Ohwilleke,
indeed, the top quark is the only quark that lives its life free of QCD infrared slavery. The price to pay for this luxury is that it lives a very short life!
It is the width of the top quark (the inverse of its lifetime) what guarantees that the top quark does not hadronize before decaying: G_t is 1.5 GeV, while Lambda_QCD is 0.2 GeV. So the scale of its
lifetime is one order of magnitude shorter than the scale of time needed for QCD interactions.
But saying that the top quark is free does not lead us very far. QCD still applies. Yes, we cannot observe them directly, because we do not have cameras with a 1/10000000000000000000000000 sec
Tommaso Dorigo | 11/08/11 | 16:30 PM
Vladimir Kalitvia... | 11/09/11 | 07:44 AM
Tommaso Dorigo | 11/09/11 | 07:56 AM
I am a bit puzzled by your formula for P(n). Did you mean:
P(n) = [mu^n e^(-mu)]/n!
Chris Austin (not verified) | 11/08/11 | 15:05 PM
Tommaso Dorigo | 11/08/11 | 16:32 PM
Nice catch... I believe the same happened in OPERA, where the proton extraction functions were averaged, and the information on its -horizontal- spread was washed away.
freemeson (not verified) | 11/09/11 | 07:00 AM
A nit: it is not clear whether it is a cloud chamber or a bubble chamber picture. A cloud chamber uses supersaturated gas a bubble chamber uses supercooled liquid, no?
Your statistics discussion is interesting, but I would bet on an expansion cycle effect, i.e. a left over from the previous cycle. bubble chambers are certainly cycled and I expect cloud chambers
would also have been, a clean film so to speak. It could also be an effect of the time distribution of the shower, mgcht have been a late track or the first one, when the expansion was starting .Back
in the 70's I worked a lot with bubble chamber pictures and do vaguely remember ghost tracks
anna v (not verified) | 11/10/11 | 00:26 AM
Tommaso Dorigo | 11/10/11 | 05:27 AM
Awesome article T. Keep the coming...
Αλέξανδρος (not verified) | 11/11/11 | 08:39 AM
Tommaso Dorigo | 11/11/11 | 08:42 AM | {"url":"http://www.science20.com/quantum_diaries_survivor/free_quarks_dont_be_fooled-84364","timestamp":"2014-04-17T07:36:22Z","content_type":null,"content_length":"91224","record_id":"<urn:uuid:a674349c-6f0b-43d6-9bd7-b81a59bba02f>","cc-path":"CC-MAIN-2014-15/segments/1398223206120.9/warc/CC-MAIN-20140423032006-00319-ip-10-147-4-33.ec2.internal.warc.gz"} |
NPV Function
Returns a Double specifying the net present value of an investment based on a series of periodic cash flows (payments and receipts) and a discount rate.
Required. Double specifying discount rate over the length of the period, expressed as a decimal.
Required. Array of Double specifying cash flow values. The array must contain at least one negative value (a payment) and one positive value (a receipt).
Exception type Error number Condition
ArgumentException 5 ValueArray is Nothing, rank of ValueArray <> 1, or Rate = -1
See the "Error number" column if you are upgrading Visual Basic 6.0 applications that use unstructured error handling. (You can compare the error number against the Number Property (Err Object).)
However, when possible, you should consider replacing such error control with Structured Exception Handling Overview for Visual Basic.
The net present value of an investment is the current value of a future series of payments and receipts.
The NPV function uses the order of values within the array to interpret the order of payments and receipts. Be sure to enter your payment and receipt values in the correct sequence.
The NPV investment begins one period before the date of the first cash flow value and ends with the last cash flow value in the array.
The net present value calculation is based on future cash flows. If your first cash flow occurs at the beginning of the first period, the first value must be added to the value returned by NPV and
must not be included in the cash flow values of ValueArray.
The NPV function is similar to the PV function (present value) except that the PV function allows cash flows to begin either at the end or the beginning of a period. Unlike the variable NPV cash flow
values, PV cash flows must be fixed throughout the investment.
This example uses the NPV function to return the net present value for a series of cash flows contained in the array values(). The return value, stored in FixedRetRate, represents the fixed internal
rate of return.
' Define money format.
Dim MoneyFmt As String = "###,##0.00"
' Define percentage format.
Dim PercentFmt As String = "#0.00"
Dim values(4) As Double
' Business start-up costs.
values(0) = -70000
' Positive cash flows reflecting income for four successive years.
values(1) = 22000
values(2) = 25000
values(3) = 28000
values(4) = 31000
' Use the NPV function to calculate the net present value.
' Set fixed internal rate.
Dim FixedRetRate As Double = 0.0625
' Calculate net present value.
Dim NetPVal As Double = NPV(FixedRetRate, values)
' Display net present value.
MsgBox("The net present value of these cash flows is " & _
Format(NetPVal, MoneyFmt) & ".")
Namespace: Microsoft.VisualBasic
Module: Financial
Assembly: Visual Basic Runtime Library (in Microsoft.VisualBasic.dll) | {"url":"http://msdn.microsoft.com/en-US/library/4k3y7xeh(d=printer,v=vs.80).aspx","timestamp":"2014-04-19T18:26:58Z","content_type":null,"content_length":"55160","record_id":"<urn:uuid:2a69e529-f999-44a8-844f-6bd9c4ffc602>","cc-path":"CC-MAIN-2014-15/segments/1397609537308.32/warc/CC-MAIN-20140416005217-00243-ip-10-147-4-33.ec2.internal.warc.gz"} |
Another full day spent working with Jean-Michel Marin on the new edition of Bayesian Core (soon to be Bayesian Essentials with R!) and the remaining hierarchical Bayes chapter… I have reread and
completed the regression and GLM chapters, sent to very friendly colleagues for a last round of comments. Now, I am essentially idle, waiting
Core minus one!
Jean-Michel Marin visited me in Paris last week and, besides taking part in Pierre’s PhD defence, we made enough progress to close two more chapters of the new edition of Bayesian Core (soon to be
Bayesian Essentials with R!) This follows the good work session we had in Carnon where we also completed two chapters
Carnon [and Core, end]
Yet another full day working on Bayesian Core with Jean-Michel in Carnon… This morning, I ran along the canal for about an hour and at last saw some pink flamingos close enough to take pictures (if
only to convince my daughter that there were flamingos in the area!). Then I worked full-time on the spatial
non-stationary AR(10)
In the revision of Bayesian Core on which Jean-Michel Marin and I worked together most of last week, having missed our CIRM break last summer (!), we have now included an illustration of what happens
to an AR(p) time series when the customary stationarity+causality condition on the roots of the associated polynomial is not satisfied.
quantum forest
Thanks to a link on R-bloggers, I was introduced to Luis Apiolaza’s blog, Quantum Forest, which covers data analyses and R comments he encounters in his research as a quantitative forester/
geneticist. And he works at the University of Canterbury, Christchurch, where I first taught from Bayesian Core in 2006. Which may be why he chose
understanding computational Bayesian statistics: a reply from Bill Bolstad
Bill Bolstad wrote a reply to my review of his book Understanding computational Bayesian statistics last week and here it is, unedited except for the first paragraph where he thanks me for the
opportunity to respond, “so readers will see that the book has some good features beyond having a “nice cover”.” (!) I simply processed
understanding computational Bayesian statistics
I have just finished reading this book by Bill Bolstad (University of Waikato, New Zealand) which a previous ‘Og post pointed out when it appeared, shortly after our Introducing Monte Carlo Methods
with R. My family commented that the cover was nicer than those of my own books, which is true. Before I launch into
Number of components in a mixture
I got a paper (unavailable online) to referee about testing for the order (i.e. the number of components) of a normal mixture. Although this is an easily spelled problem, namely estimate k in I came
to the conclusion that it is a kind of ill-posed problem. Without a clear definition of what a component is,
JSM 2011 [3]
Monday August 01 was the first full day of JSM 2011 and full is the appropriate word to describe the day! It started for me at 7am with a round table run by Marc Suchard on parallel computing (or at
3am if I am considering the time I woke up!). I was rather out of
Core not in CiRM
Despite not enjoying this year the optimal environment of CiRM, we are still making good progress on the revision (or the R vision) of Bayesian Core. In the past two days, we went over Chapters 1
(Introduction), 2 (Normal Models), 5 (Capture-Recapture Experiments), and 6 (Mixture Models), with Chapters 3 (Regression), 4 (Generalised Linear Models) | {"url":"http://www.r-bloggers.com/tag/bayesian-core/","timestamp":"2014-04-21T09:51:57Z","content_type":null,"content_length":"40199","record_id":"<urn:uuid:dfcbe2f0-c6af-4448-982b-88b08f5421af>","cc-path":"CC-MAIN-2014-15/segments/1397609539705.42/warc/CC-MAIN-20140416005219-00558-ip-10-147-4-33.ec2.internal.warc.gz"} |
Vacaville Statistics Tutor
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Krein condition
From Encyclopedia of Mathematics
A condition in terms of the logarithmic normalized integral
used to derive non-uniqueness or uniqueness of the moment problem for absolutely continuous probability distributions (cf. also Absolute continuity; Probability distribution). In (a1), distribution
The general question of interest is: Does the moment sequence
It is essential to note that the quantity
Hamburger moment problem.
In this problem, the support of
For this problem the following Krein conditions are used:
The following is true:
if (a2) holds, then
if, in addition to (a3), the Lin condition below is satisfied, then
Here, the following Lin condition is used:
Stieltjes moment problem.
In this problem, the support of
In this case one uses the following Krein conditions
The following is true:
if (a5) holds, then
if, in addition to (a6), the Lin condition below is satisfied, then
From these four assertions one can derive several interesting results. In particular, one can easily show that the log-normal distribution is M-indeterminate. This fact was discovered by Th.J.
Stieltjes in 1894 (in other terms; see [a1], [a3]), and was later given in a probabilistic setting by others, see e.g. [a4].
Examples in probability theory.
Suppose random variable with a normal distribution. Then:
the distribution of
the distribution of
the distribution of [a2]. A proof of this result based on the Krein or Krein–Lin technique is given in [a12].
Suppose the random variable normal distribution; an exponential distribution; a gamma-distribution; a logistic distribution; or an inverse Gaussian distribution (cf. also Gauss law). Then in each of
these cases the distribution of [a12].
A more general problem is to describe classes of functions of random variables (not just powers) preserving or destroying the determinacy of the probability distributions of the given variables.
There is a more general form of the Krein condition, which requires instead of (a2) that
where [a8].
The Krein condition, in conjunction with the Lin condition, is used for absolutely continuous distributions whose densities are positive in both Hamburger and Stieltjes problems. [a7] contains an
extension of the Krein condition for indeterminacy as well as a discrete analogue applicable to distributions concentrated on the integers.
The Krein condition can also be used for other purely analytic problems, see [a3] and [a9].
The book [a1] is the basic source describing the progress in the moment problem, providing also an intensive discussion on the Krein condition. For distributions on the real line, this condition was
introduced by M.G. Krein in 1944, see [a5]. For recent (1998) developments involving the Krein condition see [a3], [a6], [a7], [a9], [a10]. Several applications of the Krein condition are given in
[a11] and [a12].
[a1] N.I. Akhiezer, "The classical moment problem" , Hafner (1965) (In Russian)
[a2] C. Berg, "The cube of a normal distribution is indeterminate" Ann. of Probab. , 16 (1988) pp. 910–913
[a3] C. Berg, "Indeterminate moment problems and the theory of entire functions" J. Comput. Appl. Math. , 65 (1995) pp. 27–55
[a4] C.C. Heyde, "On a property of the lognormal distribution" J. R. Statist. Soc. Ser. B , 29 (1963) pp. 392–393
[a5] M.G. Krein, "On one extrapolation problem of A.N. Kolmogorov" Dokl. Akad. Nauk SSSR , 46 : 8 (1944) pp. 339–342 (In Russian)
[a6] G.D. Lin, "On the moment problem" Statist. Probab. Lett. , 35 (1997) pp. 85–90
[a7] H.L. Pedersen, "On Krein's theorem for indeterminacy of the classical moment problem" J. Approx. Th. , 95 (1998) pp. 90–100
[a8] Yu.V. Prohorov, Yu.A. Rozanov, "Probability theory" , Springer (1969) (In Russian)
[a9] B. Simon, "The classical moment problem as a self-adjoint finite difference operator" Adv. Math. , 137 (1998) pp. 82–203
[a10] E.V. Slud, "The moment problem for polynomial forms of normal random variables" Ann. of Probab. , 21 (1993) pp. 2200–2214
[a11] J. Stoyanov, "Counterexamples in probability" , Wiley (1997) (Edition: Second)
[a12] J. Stoyanov, "Krein condition in probabilistic moment problems" Bernoulli , to appear (1999/2000)
How to Cite This Entry:
Krein condition. J. Stoyanov (originator), Encyclopedia of Mathematics. URL: http://www.encyclopediaofmath.org/index.php?title=Krein_condition&oldid=16214
This text originally appeared in Encyclopedia of Mathematics - ISBN 1402006098 | {"url":"http://www.encyclopediaofmath.org/index.php?title=Krein_condition","timestamp":"2014-04-18T13:08:28Z","content_type":null,"content_length":"31401","record_id":"<urn:uuid:49c131b8-cde9-4a26-8533-782499aca711>","cc-path":"CC-MAIN-2014-15/segments/1398223202548.14/warc/CC-MAIN-20140423032002-00229-ip-10-147-4-33.ec2.internal.warc.gz"} |
Questions about amping :D
post #1 of 12
4/20/13 at 9:35pm
Thread Starter
Hello all!
So instead of turning around, going up and down I decided to straight off ask the question although this is completely, not my style XD
So my question is, let's say a "X" headphone maximum power capacity is 100mw and 32 Ohm. But the amp output is 300mw at 32Ohm. Will the headphone take in only 100mw or 300mw? Will it damage anything
Power delivered depends on the source level, amp gain, any volume controls anywhere including the amp, any amp output impedance, headphone impedance, etc.
The more power the headphones receive, the louder the sound they produce. Max power capacity for pretty much anything except maybe some really cheap sets, is generally way too loud for actual
listening. As in, you'd get hearing damage really quickly if you put it on your head. Furthermore, distortion levels would be way up there. The max power rating is I think a continuous rating and
not a peak rating though, usually.
If you're really maxing out the amp and putting 300 mW into the headphones, then yeah, the headphones would receive 300 mW. The coils may overheat and otherwise be damaged (if it can only handle 100
mW). Driver could run out of excursion. I think. I'm not entirely sure on the actual power ratings set and their typical levels relative to the physical limitations.
Edited by mikeaj - 4/20/13 at 10:19pm
Seems legit~ thanks! But what if let's say headphone like HD600 requires at least 100mw (example only, not according to specs) and I drive it with 10mw, will the driver weaken? I know it will result
in lower volume but other than that, anything else will happen? XD
"require at least 100 mW" for what? Let's assume to reach 100 dB SPL.
If your amp only supplies 10 mW then you will only reach 90 dB SPL. Nothing else will happen. Well, assuming all else being equal, the headphone drivers will produce a bit less distortion at lower
I think your idea from "underpowering" headphones stems from clueless people that will tell you that you need lots of headroom (excess gain) for an amp to drive a headphone properly. That's absolute
Just like anything else in audio, excess gain is also a tradeoff. Too much and you reduce the usable volume control range (look at the mag ni thread where some people are stuck at 8:00 to 9:00
o'clock) which goes hand in hand with channel balance problems. High gain also produces more noise, distortion ..
Unless you listen at high SPLs, most of the time most headphones only need a fraction of a milliwatt of power. And ideally, your volume control should allow you to go up to 14:00 o'clock.
Edited by xnor - 4/21/13 at 4:09am
Originally Posted by xnor
"require at least 100 mW" for what? Let's assume to reach 100 dB SPL.
If your amp only supplies 10 mW then you will only reach 90 dB SPL. Nothing else will happen. Well, assuming all else being equal, the headphone drivers will produce a bit less distortion at lower
I think your idea from "underpowering" headphones stems from clueless people that will tell you that you need lots of headroom (excess gain) for an amp to drive a headphone properly. That's absolute
Just like anything else in audio, excess gain is also a tradeoff. Too much and you reduce the usable volume control range (look at the mag ni thread where some people are stuck at 8:00 to 9:00
o'clock) which goes hand in hand with channel balance problems. High gain also produces more noise, distortion ..
Unless you listen at high SPLs, most of the time most headphones only need a fraction of a milliwatt of power. And ideally, your volume control should allow you to go up to 14:00 o'clock.
So you mean "underpowering" is not true? I trust you! Your answer is always very logical XD
How do you calculate how much mW to get how many dB? Curiosity comes**
Billson ^_^
Check the sensitivity of your headphones, usually specified as X dB SPL @ 1 mW or @ 1 V.
To get X+10 dB you need 10 times the power (10 mW) or about 3.2 times the voltage (3.2 V) and 1/10 or 1/3.2 for -10 dB.
HD600 is about 102 dB @ 1V or 97 dB @ 1 mW.
87 dB = 0.1 mW
77 dB = 0.01 mW
edit: corrected sensitivity, see below
Edited by xnor - 4/21/13 at 10:47am
wow. so really you barely need ANY power if your a quite listener?
and you say "headroom" is a lie? i was under the impression that you need some extra, unused, power to allow for voltage swings and big, or sudden climaxes in the music?
You, with most headphones you don't need much power at all. Most are around 100 dB @ 1 mW but some IEMs are crazy efficient reaching almost 120 dB @ 1mW.
I don't say headroom is a lie. What I say is that you do not need lots of headroom and it also doesn't guarantee that the amp is any good else we'd all be using monoblocks with our headphones...
As said before, a rule of thumb is that the volume control should be between 10:00 and 14:00 (2 pm) most of the time. That way you still have some excess gain for quiet songs. If you normalize tracks
using something like ReplayGain you need even less excess gain because loud and quiet songs will be approximately equally loud.
I'm pretty sure they aren't that sensitive, and you probably looked up the wrong phones or typoed a 0 into a 1.
I think listed is 102 dB @ 1V or 97 dB @ 1 mW. These lower numbers are also very close to what InnerFidelity got.
97 dB = 1 mW
87 dB = 0.1 mW
77 dB = 0.01 mW
Overall points still stand of course.
Edited by mikeaj - 4/21/13 at 8:55am
Let's take a simple example:
The DAC outputs 2 V for full-scale signals. The amps gain is 5x (about 14 dB). The headphone sensitivity is 95 dB @ 1 mW and nominal impedance 300 ohms. We want to listen at 75 dB SPL.
So the volume control needs to be set so that the amp outputs about 0.01 mW. To keep things simple let's take a full-scale pure tone (sine wave) as signal.
P = V*V/R
--> V = sqrt(P*R) = sqrt(0.00001 * 300) = 0.055 V
full-scale output: 2 V * 5 = 10 V
20*log10(0.055 / 10) = -45 dB ... or a voltage gain of 0.0055x and on a normal volume control that is somewhere around 8 o'clock so very close to the off position. That sucks.
Now let's assume you're listening to real music and the average* RMS amplitude of a track is -20 dB relative to a full-scale sine wave. Then we're at -25 dB (0.056x) which leaves plenty headroom for
all the peaks that come close to full-scale. The volume control is still somewhere below 10 o'clock!
*) this needs to be a pretty dynamic song and the RMS amplitude can shortly hit -5 dB from time to time, which would be about 90 dB SPL. True peaks of course could reach 0 dB or even surpass that.
Compressed metal songs can reach an average of only -7 dB or even louder..
edit: lowered headphone sensitivity
Edited by xnor - 4/21/13 at 10:58am
thank you very much!
You are of course right, it's 102 not 112 dB/V. Thanks.
Made some corrections and changes above.
Edited by xnor - 4/21/13 at 10:52am
post #2 of 12
4/20/13 at 10:18pm
post #3 of 12
4/21/13 at 2:54am
Thread Starter
post #4 of 12
4/21/13 at 3:57am
post #5 of 12
4/21/13 at 5:34am
Thread Starter
post #6 of 12
4/21/13 at 5:45am
post #7 of 12
4/21/13 at 5:55am
post #8 of 12
4/21/13 at 8:42am
post #9 of 12
4/21/13 at 8:53am
post #10 of 12
4/21/13 at 9:02am
post #11 of 12
4/21/13 at 9:42am
post #12 of 12
4/21/13 at 10:47am | {"url":"http://www.head-fi.org/t/660713/questions-about-amping-d","timestamp":"2014-04-23T13:35:38Z","content_type":null,"content_length":"155162","record_id":"<urn:uuid:5ef8930b-19fd-4c10-a984-94e3b244cbdd>","cc-path":"CC-MAIN-2014-15/segments/1398223202548.14/warc/CC-MAIN-20140423032002-00127-ip-10-147-4-33.ec2.internal.warc.gz"} |
Group of local complementation as a coxeter group
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Can the group generated by local complementations, ${lc_i|i=1,\cdots,n}$ on simple graphs on $n$ vertices, be categorized as a coxeter group? After all these obey: $$\langle lc_i| (lc_i lc_j)^{m_
{ij}}=1\rangle \quad where \quad m_{ij}=\{ 2,3,6\}$$
If so, which classification do they belong to?
A few definitions:
Local complementation, $lc_k$, of vertex $k$ replaces the subgraph induced on the neighborhood, $N_k$ of $k$, by its complement. The transformation on the adjacency matrix, $A_G$ of a graph, $G$ is:
$lc_k: (A_G)_{ij} \rightarrow (A_G)_{ij}+(A_G)_{ik}(A_G)_{jk}+Diagonal[(A_G)_{ij}+(A_G)_{ik}(A_G)_{jk}]$.
Where the addition is $mod(2)$.
We did some numerical work in this regard, see my question "the group of local complementation on simple graphs". We find an interesting pattern which leads us to believe there must be an (if not
trivial) interesting underlying group structure.
gr.group-theory graph-theory algebraic-graph-theory
How do you know that those are the only relations among the $lc_i$'s? – Gjergji Zaimi Apr 3 '11 at 3:08
To be a (quotient of a) Coxeter group you would also need the relations $lc_i^2=1$ i.e. $m_{ii}=1$. – Derek Holt Apr 3 '11 at 9:42
@Derek: If I understand it correctly a local complementation is the operation of selecting a vertex and changing the subgraph of this vertex and all its neighbors into its complement. Therefore
local complementations are involutions by definition. – Johannes Hahn Apr 3 '11 at 13:28
1 Suggestions from a non-specialist: 1) Define "local complementation" for people closer to groups than to graphs, making the involutive property explicit. 2) Indicate what range of examples you've
looked at so far, since at first sight there's no reason to expect that you always get Coxeter groups. 3) If there are Coxeter groups in the picture, would that have any immediate implications
from the viewpoint of graph theory? – Jim Humphreys Apr 3 '11 at 15:46
@Zaimi they can easily be worked out. The $m_{ij}=2$ is for when $i=j$ or $i$ and $j$ are disconnected, and $m_{ij}=3$ when $i$ and $j$ are disconnected. So for a given $(lc_i lc_j)^{m_{ij}}=
1$,$m_{ij}=2,3 or 6$. – P.H. Apr 3 '11 at 20:42
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[Python-ideas] matrix operations on dict :)
Sturla Molden sturla at molden.no
Wed Feb 8 17:21:12 CET 2012
On 08.02.2012 13:12, julien tayon wrote:
> * Because it naturally provides matrix. And matrix are an easy way to
> formalize and standardize tree manipulations which are a growing
> concern in real life computer craft.
No, it naturally provides a hash-table, which is a simple in-memory
database, not a matrix.
> at my very personnal opinion, mathematical signs should have been
> reserved in all langages to operation analog to mathematics. And
> linear algebrae is one of the most accpeted behaviour for these
> symbols.
There is a world beyond linear algebra. Sometimes we need to do things
that cannot easily be fit into the semantics of matrix operations. And
for those that only can think in terms of matrices there are languages
called Matlab, Scilab, and Octave.
> It is stupid to code matrix with an hash, I just say as there is a
> strong analogy between dict and vectors,
No there is not. A vector is ordered, a hash-table (dict) is unordered.
- In a vectorlike structure, e.g. a Python list, element i+1 is stored
subsequently to element i.
- In a hash-table, e.g. a Python dict, element hash(i+1) is not stored
subsequently to element hash(i).
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Spectral decomposition of R matrix -> Wenzl projectors?
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Just curious: if you take a R matrix from knot theory and apply a spectral decomposition (see. e.g. my following post Matrix decomposition the other way) you'll get projectors: $T_i*T_j=T_i*\delta_
Are they related to the Wenzl projectors?
knot-theory linear-algebra
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