content stringlengths 86 994k | meta stringlengths 288 619 |
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Please help me with a component problem.
You are given vectors A= 4.7i - 7.0j and B= - 3.2i+ 6.9j . A third vector C lies in the xy-plane. Vector C is perpendicular to vector A and the scalar product of vector C with vector B is 16.0.
What is the x-component of vector C?
What is the y-component of vector C?
I'm not sure wither to use a dot product to solve this or use the cross product. I haven't yet learned how to use the cross product of two vectors so I don't think its that. How would i go about
solving this problem? I thought about setting vector B equal to 16 but that didn't work out to well. Can someone please help me? Thank you! | {"url":"http://www.physicsforums.com/showthread.php?t=188085","timestamp":"2014-04-21T07:16:08Z","content_type":null,"content_length":"22269","record_id":"<urn:uuid:c72b904c-efc8-4937-a582-71fdb4fa72cf>","cc-path":"CC-MAIN-2014-15/segments/1397609539665.16/warc/CC-MAIN-20140416005219-00184-ip-10-147-4-33.ec2.internal.warc.gz"} |
UA:Armor as Damage Reduction
From D&D Wiki
This material is published under the OGL
Armor as Damage Reduction
In the abstract combat system of the D&D game, a character's armor defends him by reducing the chance that an attack will deal damage. That system simplifies the realities of battle in order to
streamline combat resolution. An attack that fails due to a character's armor or natural armor doesn't really fail to connect, but rather fails to connect with enough force to deal any damage.
(That's why touch attacks ignore a character's armor and natural armor - the touch attack only needs to connect to deliver its effect, and need not actually breach the target's armor.)
If you're willing to add a layer of complexity to your combats, consider this variant. In this system, armor reduces the amount of damage dealt by an attack instead of merely turning would-be hits
into misses. Armor still prevents some hits outright, but also reduces the deadliness of attacks that do connect. In essence, the system "gives up" some of armor's ability to turn hits into misses in
exchange for a small reduction in the damage dealt by any given attack.
Armor Damage Reduction Values
In this system, armor offers two benefits against attacks: an armor bonus to Armor Class, which functions just like the armor bonus in the standard D&D rules but is usually lower in value; and damage
reduction. See Table: Armor and Damage Reduction for the armor bonus and DR values for common armor types. (All other armor statistics, such as maximum Dexterity bonus, armor check penalty, and
arcane spell failure chance, are unchanged.)
For armors not covered on Table: Armor and Damage Reduction, you can determine the new armor values and damage reduction based on the standard D&D armor bonus. To determine the armor's damage
reduction, divide the armor's normal armor bonus by 2 (rounding down). To determine the armor's new armor bonus, subtract the damage reduction from the normal armor bonus. For example, studded
leather has a normal armor bonus of +3. That gives it a damage reduction of 1/- (half of 3, rounded down) and a new armor bonus of +2 (3 minus 1).
1 Add any enhancement bonus to this value
Magic Armor
An armor's enhancement bonus (if any) increases its armor bonus to Armor Class, but has no effect on the armor's damage reduction. A +3 chain shirt, for example, adds +5 to Armor Class and grants
damage reduction 2/-.
Stacking Damage Reduction
The damage reduction granted by armor stacks with other damage reduction of the same type (that is, damage reduction that has a dash after the number). A 7th level barbarian wearing a breastplate has
DR 3/- (1/- from his class level and 2/- from his armor). A fighter wearing full plate armor who is the target of a stoneskin spell, however, has DR 4/- from the armor and 10/adamantine from the
spell (see page 292 of the Dungeon Master's Guide, under Damage Reduction, for rules on characters with multiple types of damage reduction).
Shields function normally in this variant, granting their full shield bonus to Armor Class. Unlike with armor, a shield's effectiveness is measured wholly by its ability to keep an attack from
connecting with your body.
Natural Armor
A creature's natural armor also provides a modicum of damage reduction. Divide the monster's natural armor bonus (not including any enhancement bonus) by 5 to determine the monster's damage reduction
. The same value is subtracted from the monster's natural armor bonus to Armor Class to find the monster's new Armor Class. These calculations are summarized in Table: Natural Armor and Damage
If the creature already has damage reduction, either add the value granted from natural armor (if the existing damage reduction is of the same type) or treat it as a separate DR value (if it is of a
different type). See page 292 of the Dungeon Master's Guide for rules on creatures with multiple types of damage reduction.
For example, a mummy normally has a natural armor bonus of +10. This gives it DR 2/-, and its natural armor bonus is reduced by 2 points to +8 (making its Armor Class 18). Since the mummy already has
DR 5/- as a special quality, its total damage reduction becomes DR 7/-.
A mature adult red dragon has a natural armor bonus of +24. This gives it DR 4/-, and its natural armor bonus is reduced by 4 points to +20 (making its Armor Class 28). The dragon's existing damage
reduction is 10/magic, so the two damage reduction values remain separate.
Finally, a frost giant has a +9 natural armor bonus, so it gains DR 1/- from natural armor. The chain shirt it wears gives it an additional DR 2/-. If that frost giant were a 7th-level barbarian, the
barbarian class levels would give it DR 1/-. These three values add up to DR 4/-. The frost giant's Armor Class would be 20 (10, +8 natural armor bonus, +2 chain shirt).
Table: Natural Armor and Damage Reduction
Natural Armor Bonus Damage Reduction Subtract from Natural Armor
0-4 none 0
5-9 1/- 1
10-14 2/- 2
15-19 3/- 3
20-24 4/- 4
25-29 5/- 5
30-34 6/- 6
35-39 7/- 7
40-44 8/- 8
Back to Main Page → Variant Rules → Adventuring
Open Game Content ( place problems on the discussion page).
This is Open Game Content from Unearthed Arcana. It is covered by the Open Game License v1.0a, rather than the GNU Free Documentation License 1.3. To distinguish it, these items will have this
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Homework Help
Posted by Laura on Monday, May 21, 2012 at 6:03pm.
I need to state the period and 2 consecutive asymptotes on the graph for the following questions.
1: y = -3 tan pi*x
period: pi (?)
asymptotes: ?
2: y = 2 sec 4x
period: ?
asymptotes: ?
3: y = csc (x/3)
period: ?
asymptotes: ?
4: y = 3 cot (pi*x/2)
period: ?
asymptotes: ?
• TRIG - Anonymous, Tuesday, July 3, 2012 at 1:38pm
I ca help you with the first one as this is the only one I understand so far. Since you have pi*x in the tan, you have to take the pi and divide by the k of the function, which in this case is
pi. This gets you a perio of one, meaning that there is a point at every single interval. The aymptotes are in between each interval. Since you have to get half way between each quarter of pi,
there is an asymptote at pi/8 and then add pi/4 to find the other asymptotes there of. A few I found are pi/8, which is the first one, and (3*pi)/8. Hope it helps. If I find out how to do the
other three problems, I'll help! :D
• TRIG - Hack, Tuesday, January 14, 2014 at 6:10pm
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Lesson Tutor Lesson Plan: Solving for x from our basic algebra series.
Objective(s): By the end of this lesson the student will be able to:
Pre Class Assignment: Completion of Basic Algebra - Lesson 2 Resources/Equipment/Time Required: Outline:
A quick review:
1. When you write algebraic expressions, use +, -, and = signs. For division, use / , the same way you know that when you see a fraction, it means to divide the top number by the bottom number.
2. For multiplication, write the expression with no symbol or sign between them as the X (multiplication) symbol can be confused with the variable x. For example 3 times the variable y should be
written 3y. You can also use parentheses to indicate multiplication. This is especially useful in longer problems such as (3y)(4-2x).
3. When you want to multiply something AFTER another expression has been done first, use parenthesis. For example, if you want to add x and y and THEN multiply the result by 7, write it this way: 7(x
+ y).
4. To translate from language to a math expression, read the sentence carefully. Then decide what operations it will take to reach a solution. Write this into an algebraic expression.
Something new:
1. To take something OUT of parenthesis, do the operation one number at a time. For example, 7(x+y). First, multiply 7 times x. Then multiply 7 times y. The result is 7x + 7y.
2. When you solve for x, you want to "isolate" the x on one side of the equal sign. To do this, use the opposite sign of the number you want to move and do the same thing to BOTH sides of the
For example: 8x + 2 = 50
8x + 2 - 2 = 50 - 2 (Subtract 2 from both sides of the = sign)
8x = 48 (Divide by 8 to solve because that is opposite of multiplication)
x = 6
And now a fun problem to make you really think. Solve using algebra. Translate into an expression and solve.
You can do it!
Assignment(s) including Answer key:
George is 4 years older than Jon, who is 4 years older than Jim, who is 4 years older than Sam, who is 1/2 the age of George. How old is each boy? Hint: let x represent George's age.
Answer Key : Click Here
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Equation of locus of points representing z
September 7th 2009, 07:22 PM #1
Apr 2009
Equation of locus of points representing z
Hi, i got this question and dont fully understand how they solved it:
You have to find the locus of points representing z if:
my first thought would be to replace z with (x+iy) but what then?
Then take the absolute value of x + iy - 3i = x + i(y - 3) and x + iy + 3i = x + i(y + 3).
So you have to simplify something of the form $\sqrt{A} - \sqrt{B} = 2 \Rightarrow \sqrt{A} = 2 + \sqrt{B}$.
Square both sides, simplify, square both sides again and simplify. You will get a single branch of a hyperbola (there is an implicit restriction on z you will need to be careful of that. This
restriction means that both branches do no get included).
Of course, if you're familiar with the locus definition of a hyperbola then there's a much simpler geometric approach that can be taken.
Okay, seems quite simple now. Thanks. This is what i have done:
And in this case it will only be the bottom branch of the graph.
i have also looked at the locus definition of the hyperbola, didnt think of defining it that way. makes it much simpler just by substituting in:
you get the final equation.
September 7th 2009, 08:05 PM #2
September 8th 2009, 02:58 PM #3
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Draw the graph of y=-2f(-x+3)+1
Number of results: 15,692
This problem can be done almost by inspection if you just draw a graph of the function and think about what each question means in terms of the graph. For (a), g(x) is the area of a curve that starts
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The cost of producing a number of items x is given by C = mx + b, in which b is the fixed cost and m is the variable cost (the cost of producing one more item). (a) If the fixed cost is $40 and the
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Monday, July 23, 2007 at 1:13am by jen
-(8) < n <= -(-5) Removing the parenthesis from terms preceeded by a negative sign is equivalent to multiplying each term by negative 1. In other words, you change the sign of each term. -8 < n <= 5
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Saturday, August 7, 2010 at 1:49pm by Henry
Draw this on a graph. Put point A on the positive y-axis. From A, draw a line that looks to represent 80 degrees, to the x-axis at point B. The distance from the origin to B is what you want to find.
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Saturday, April 6, 2013 at 10:23pm by rbowh
Math (functions)
Oh, sorry I wasn't clear. I didn't mention that it's supposed to be graphed from y = f(x). Also, we just write out how to obtain the graph, not the exact coordinates. So I think from y=-2f(4(x-2)) +
8 -Vertically stretched by a factor of 2, and horizontally stretched by a ...
Sunday, March 30, 2008 at 7:38pm by Lucy
Can you please tell me how to draw a scientific line graph about Conposition Of Air ? ? ?
Wednesday, November 18, 2009 at 2:00pm by Lilii
You could draw a graph and count squares, or compute the integral of f(x) - g(x) between the intersection points
Tuesday, July 20, 2010 at 12:26pm by drwls
i need to draw a graph using this equation 8x x 4= they dont say what x equals
Thursday, January 20, 2011 at 7:46pm by john
1. Using graph paper, draw a 4-inch line from the origin parallel to the positive x-axis. 2. Draw a 3-inch vertical line from the origin parallel to the positive y-axis. 3. Complete the rectangle
with the hor. sides 4 inches long and the ver sides 3 inches long. 4.Draw a ...
Monday, July 11, 2011 at 10:51am by Henry
algebra 1
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graph that looks something like this...
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Monday, January 16, 2012 at 11:47am by drwls
We use / to indicate division, which of course is the fraction bar, and placing the expression to be divided in brackets so yours would be [4 - (-3) + 1]/[2 - 2÷(-2)] = 8/3 I put (4 - (-3) + 1)/((2 -
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Calculus: drwls, please
This web site does not have a way of showing graphs or images, so I am not able to draw the graph for you. If the function really is (1/x) -4 and not 1/(x-4), then it will have a value of y = -4 at
very large and very small values of x. As you approach x=0 from the left, the ...
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i'm not sure how to do this. can someone help, please? thanks! consider the function f(x0 = x^2 + 2x on the interval [-2, 2] a. draw a sketch of the graph of f(x). find the average rate of change on
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Sunday, November 27, 2011 at 8:36pm by zoe
I disagree with your math. Isn't -5-1=-6? m=-3/-6= 1/2 Now, graphing the line. mark this point on a graph paper (1,0). Now mark this point (-5,-3). Put ruler on the graph paper such that the edge
touches each point. Draw the connecting line.
Saturday, September 12, 2009 at 7:33am by bobpursley
math (graphing)
y ≥ (2/3)x-4 graph y = (2/3)x - 4, then shade in the region ABOVE that line had it been y > (2/3)x-4 the boundary is the line y = (2/3)x - 4, so draw a dotted line instead of a solid line in general
if y > mx + b draw a dotted line for y = mx + b and shade in the ...
Sunday, May 3, 2009 at 5:45am by Reiny
4th grade
Draw this out on a piece of graph paper. Then you'll see which answer is correct.
Monday, April 19, 2010 at 4:16pm by Ms. Sue
programming concepts
here's a little perl program which should do the job. You can probably adapt it to the language of your choice. my @rates = [ {amount=>150,us=>0,canada=>0}, {amount=>100,us=>12,canada=>15}, {amount=>
50,us=>9,canada=>12}, {amount=>0,us=>6,canada...
Monday, February 10, 2014 at 1:15pm by Steve
working with data
Making a scatter plot is about the same as a line graph. Except you plot your points and then draw a straight line. Your line may not hit any points. The year would go on the bottom. I would go by
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Probability and Statistics (Continuation)
Katie, you were right, there should have been 3 successes and 2 failures, which would be 6( .9^3)(.1^2) Don't know why I miscounted. in my "prob( a specific 2F,2S, with S at end) = (.9^2)(.1^2)(.1) =
.00081 " I had the first part correct, with 2F's and 3S's, but somehow put in...
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Tuesday, June 28, 2011 at 11:44am by Anonymous
Δx/3*[f(x0)+4f(x1)+2f(x2)+4f(x3)+........+2f(xn−2)+4f(xn−1)+f(xn)] = 30/3 * (76 + 4*118 + 2*130 + 4*143 + 2*139 + 4*136 + 2*137 + 4*139 + 2*130 + 4*122 + 60)
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I have a strong feeling you meant (x-4)(x+2)/( (x-3)(x+5) ) ≥ 0 I good and fast way is to look at the graph of y = (x-4)( (x+2)/(x-3)(x+5) ) http://www.wolframalpha.com/input/?i=
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3 beakers are being heated (small, medium and large). consider the first two. both beakers are filled with water and heated for 10 minutes. after about 3 minutes the water in the small beaker the
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We do not draw graphs here. You will have to do that yourself, or use a graph plotting calculator. There are some online.
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Diploma Math
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P(x) =(2x+1)(x+3)(x-3)/((x-3)(x+2) = (2x+1)(x+3)/(x+2) When x = 3 , in new P(3) = 7(6)/5 = 42/5 so there is a hole at (3, 42/5) using P(x) = (2x+1)(x+3)/(x+2) = (2x^2 + 7x + 3)/(x+2) = 2x+3 - 3/(x+2)
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Below is the graph of a polynomial function f with real coefficients. Use the graph to answer the following questions about f. All local extrema of f are shown in the graph. I really need help with
this one but I can't post the graph.
Friday, February 4, 2011 at 5:58pm by Rachal
college alg
critical values: x=0 , x=-8, x=5, x=-5 and x=2 So want to see where the graph of y = x^2(8+x)(x-5)/ ((x+5)(x-2)) lies above or on the x-axis investigate the following domains 1. x < -8 2. x between
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Wednesday, September 5, 2012 at 8:31pm by Reiny
Use the Fundamental Theorem of Calculus to find the area of the region bounded by the x-axis and the graph of y = 4 x3 − 4 x. Answer: (1) Use the Fundamental Theorem of Calculus to find the average
value of f(x) = e0.9 x between x = 0 and x = 2. Answer: (2) Draw the ...
Monday, December 6, 2010 at 11:42pm by Erika
Draw the various choices and see. For example draw a trapezoid. Can you do it so it has no right angles?(of course) Can it have congruent diagonals (go ahead draw it) Could it be a rhombus?
Saturday, January 30, 2010 at 2:36pm by Damon
-4 < X < 4. The inequality states that X IS greater than -4 but less than 4. In other words, X includes all real numbers between -4 and 4, but -4 and 4 are not included. To graph this inequality on a
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A student travels 6.0 m [E] in 3.0 s and then 10.0 m [N] in 4.0 s. Draw a vector diagram to determine the resultant displacement. Q: How do you know where the starting point is on the graph?
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Draw the demand curve for the A-Phone. Explain how the graph, price, and quantity demanded will change if the following occurs:
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Draw a graph and you will see that it is true. Each drop in price (P) of one unit results in an increase in quantity (Q) of 100.
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given f(x)=x^3-x^2-8x+12,draw a graph of f,showing all intercept with axes and turning point
Sunday, May 19, 2013 at 12:32pm by Kay-Cee
Original quanity | new quantity . |price. |quanity supplied demanded demanded 40 12$ 80 45 11$ 75 50 10$ 70 55 9$ 65 60 8$ 60 65 7$ 55 70 6$ 50 75 5$ 45 1. what is the equilibrium price and quanity?
explain 2.what will occur if the price is instially set a $12 3.what will ...
Monday, February 2, 2009 at 8:58pm by sean
it looks like you starting statement was (x-4)(x+2) ≤ 0 there are two critical values on the x-axis. At x=4 and at x=-2 Mark those with a solid dot. Now you have 3 sections of that graph 1. the part
to the left of -2 2. between -2 and 4 3. to the right of 4 So simply ...
Thursday, April 3, 2008 at 5:02pm by Reiny
How do you make a coordinate plane graph? I can't go and get my text book because its winter break! Could someone help me here? I need to know how to make one using an equation. specifically these
equations y = 6x - 2 y = -3x + 5 y = ½x + 3 please I need help now!!! do you ...
Thursday, December 21, 2006 at 1:26pm by Austin
your equation "falls apart" when p = 100 because you would be dividing by zero So mathematically, the cost would be infinitely large (that is, it is not possible) look at this graph and see what
happens at p = 100 http://www.wolframalpha.com/input/?i=plot+C%3D80%2C000p%2F%...
Friday, February 21, 2014 at 11:39pm by Reiny
The graph of the equation y=4-x consists of all the points in the coordinate plane that satisfy the equation. List 5 points that satisfy y= 4-x. Also, what do you think is the minimum number of
points you need to plot in order to draw the graph of y=4-x. Please explain Set y=0...
Tuesday, November 21, 2006 at 6:40pm by victoria
Slope-Intercept Form: 5 x + 2 y = 8 Subtract 5 x to both sides 5 x + 2 y - 5 x = 8 - 5 x 2 y = - 5 x + 8 Divide both sides by 2 y = - 5 x / 2 + 4 For graph : In google type: functions graphs online
When you see list of results click on: rechneronline.de/function-graphs/ When ...
Tuesday, March 26, 2013 at 12:48am by Bosnian
3. write the function whose graph is the graph of y=sqrt of x, but is shifted to the left 3 units y=____ 4. write the function whose graph is the graph of y=|x|, but is shifted down 8 units y=___ 5.
write the function whose graph is the graph is the graph of y=(x+4)^2, but is ...
Friday, May 24, 2013 at 10:07am by sharday
I will assume that the axis of the wedge is vertically upward. The force on the ball at each contact point is directed toward the center of the sphere at each contact point, because there is no
friction. The contact force angle is inclined (180-110)/2 = 35 degrees to the ...
Monday, January 9, 2012 at 10:57pm by drwls
"The graph of y = g(t) is provided below. Based on the graph, where is ln(g(x)) continuous?" I did not include the graph but I would like to know in what ways does ln effect the continuity of a
graph. Thanks.
Monday, October 14, 2013 at 4:25pm by Josh
If you wat to see graph of your function in google type: "function graphs online" When you see list of results click on: rechneronline.de/function-graphs When page be open in blue rectacangle type: x
^3-1.5x^2 Then click option Draw You will see graph of your function
Wednesday, February 16, 2011 at 10:48pm by Bosnian
Adv. Math.
Describe how the graph of y= abs(x-2) is related to the parent graph? My answer: The graph is shifted 2 units to the right (or in the positive direction) in comparison to the parent graph.
Tuesday, March 31, 2009 at 2:18pm by Maria
Algebra II
Graph 16x^2 + 9y^2 = 144 I know that you cannot graph this but the graph of this equation should be a circle I think. I have points (3,0), (0,4), (-3, 0), and (0,-4) graphed. Is there any more points
that I could use to graph? Thanks.
Sunday, March 9, 2008 at 2:07pm by Lucy
( (-1)^5/(-2)^-3 )^2 = (-1/(1/-8))^2 = 8^2 = 64 http://www.google.ca/#hl=en&source=hp&q=%28+%28-1%29^5%2F%28-2%29^-3+%29^2+&btnG=Google+Search&meta=&aq=f&oq=%28+%28-1%29^5%2F%28-2%29^-3+%29^2+&fp=
Thursday, March 4, 2010 at 7:28pm by Reiny
I assume you know how to plot a graph. Choose 10 or more x values between -pi and pi, compute f(x) for each, and graph them with f(x) on the y axis. Draw a smooth line between the points. The x
intercepts are where y = 0. That would be whenever cos x = 0 or 1. You should know ...
Friday, February 27, 2009 at 1:22am by drwls
intermediate algebra
Don't know what you mean by "solve" here. They have defined a function in two pieces. What do you need to do with it? I also suspect a typo, since "x<0" and "x less than or equal to zero" are
basically the same. If you can graph y=2x then just draw the line, omitting ...
Wednesday, June 19, 2013 at 8:51pm by Steve
College Physics
Again, we do not see the graph. However, if the graph is force versus distance, the area of the graph from distance =0 to 3 will give the work done. Be careful with the units, which you can find from
the graph's axes.
Tuesday, October 13, 2009 at 6:34pm by MathMate
It's hard to show here the actual graph so I'll just tell you what to do. There are plenty of ways to graph them, but this is, for me, the simplest: 1. First, find the x- and y-intercepts. To the
x-intercept, let y = 0 and solve for x. To find the y-intercept, let x = 0 and ...
Sunday, October 6, 2013 at 9:52pm by Jai
first, you have to get the value of x where |x^2 - 1| becomes zero, thus x = -1 and 1 since the boundaries given are only from 0 to 4, we consider only what happens to the graph at x = 1 (because 1
is within the boundaries) first, draw or imagine the graph. the graph of x^2-1 ...
Monday, November 8, 2010 at 9:09pm by jai
when a mass attached to a spring is released from rest 3.0 cm from its equilibrium position, it oscillates with a frequency f. If this mass were instead released from rest 6.0 cm from its equilibrium
position, it would oscillate with frequency.. A.2f B.sqrt 2f C.f D.f/2 I can ...
Monday, November 16, 2009 at 6:07pm by collegekid
pre algebra
remember, you need only two points to draw a straight line. A quick way to find any two points is to use the intercepts. for 3x+y=9 let x=0 ---> y = 9 let y=0 ---> x = 3 , so we have two points (0,9)
and (3,0) for -2y = x-8 let x=0 ---> y = 4 let y=0 ---> x = 8 so ...
Tuesday, June 9, 2009 at 4:53pm by Reiny
Choose the three true statements about the graph of the quadratic function y = x2 − 3x − 4. Options A) The graph is a parabola with a minimum point. B) The graph is a parabola with a maximum point.
C) The point (2, 2) lies on the graph. D) The point (1, 6) lies on ...
Saturday, June 5, 2010 at 2:28pm by S Rey
how can i calculate the slope of the best-fit line in a length v.s. resistance(x=length;y=resistance) graph where the plotted points are: (5.1,1.6)(11.0,3.8)(16.0,4.5)(18.0,5.9)(23.0,7.5)?
Personally, I would graph the points, and draw a best fit line, then measure the slope ...
Sunday, May 20, 2007 at 6:13pm by koeit
Managerial Economics
I am trying to understand how to variate between MC and MB using the theory of optimization the circumstance which a waste site could be made too clean. Note: Good answers are dispassionate and
employ economic analysis. Draw a graph, put cost/benefits on the y-axis, % of waste...
Friday, May 25, 2007 at 7:01pm by Sharon Williams
I assume you want to find a point (x, y) that fits the conditions. Find the distance between (-2,0) and (3,1) but keep it in x,y form. (3-(-2),1-0) = (5,1) It may be helpful now for you to draw a
graph to see whether to add (5,1) to (3,1) or subtract (5,1) from (-2,0). However...
Tuesday, June 30, 2009 at 8:47am by Marth
college algebra, Please help!!
Answer the following function. f(x)=2x^2-x-1 A. Is the point (-2,9) on the graph of f? B. If x equals 2, what is fx? What point(S) are on the graph of f? c. if f(x)= -1, what is x? what point(s) are
on the graph of f? d. what is the domain of f? e. List the x-intercpts, if any...
Friday, November 9, 2012 at 1:11pm by ladybug
college algebra, Please help!!
Answer the following function. f(x)=2x^2-x-1 A. Is the point (-2,9) on the graph of f? B. If x equals 2, what is fx? What point(S) are on the graph of f? c. if f(x)= -1, what is x? what point(s) are
on the graph of f? d. what is the domain of f? e. List the x-intercpts, if any...
Friday, November 9, 2012 at 1:11pm by ladybug
answer the questions about the following function f(x)= 10x^2/x^4+25 a. is the point (-sqrt 5,1) on the graph b. if x=3, what is f(x)? what point is on the graph of f? c. if f(x)=1, what is x? what
points are on the graph? d. what is the domain of f? e. list the x-intersepts? ...
Sunday, December 16, 2012 at 9:48am by Help--Please help!!!!
College Algebra
1.The graph of y = - 1/2 |x - 5| - 3 can be obtained from the graph of y = |x| by which transformations? What is the horizontal shift? By what factor is the graph stretched or shrunk vertically and
how is it reflected? What is the vertical shift? 2. Find the following for the ...
Saturday, August 4, 2012 at 3:38pm by Kameesha
Gemoetry question
draw a pair of ax-axes on graph paper,each scaled from -8 to 8.plot the points r(-5,2),c(-4,3),h(-1,4)and s(-3,1)and connect them to form a quadrilateral
Saturday, August 4, 2007 at 12:24pm by shantia robinson
Your question is incomplete. Are you supposed to draw a graph of tha available supply of lobster (y) vs price(x)? If so, use the data you were provided.
Tuesday, January 1, 2008 at 11:33am by drwls
Analytical Chemistry
Slow and very time consuming. Does not require an indicator. Must draw a graph and determine the equivalence point graphically.
Thursday, June 10, 2010 at 8:55am by DrBob222
You do the graphs. I assume the x axis will be time. The acceleration is zero and the velocity is constant, 0.75 m/s. Size and mass are not needed to draw the graph
Wednesday, December 5, 2012 at 11:52pm by drwls
pre calculus
2. Graph the following function using transformations. Be sure to graph all of the stages on one graph. State the domain and range. y= -2 lxl +2 For example, if you were asked to graph y= x^2 + 1
using transformations, you would show the graph of y= x^2 and the graph shifted ...
Tuesday, April 2, 2013 at 11:00am by michelle
Which type of graph would be most appropriate to show the students in the debate club, the computer club and the ecology club and the students who are in more than one of these clubs? A bar graph,
line graph, Bennett diagram, or a circle graph? Is it a circle graph?
Monday, April 18, 2011 at 8:46pm by Nick
math please help
Take the real-life situation and create an equation or inequality that could be used for analysis, prediction, or decision making. Then, draw a graph to depict the variables in your situation (refer
to problem 40 on p. 649). Use your graph and what you know about linear ...
Tuesday, January 15, 2008 at 7:26pm by Anonymous
dy = (1/9) e^(x/9) dx at x=0, dx=-.03, we have dy = (1/9)(1)(-.03) = -1/300 What his means is that if you draw the graph of y=e^(x/9) and then draw the tangent line at x=0, if you move along that
line horizontally to the left by .03 (dx), you will move vertically down by 1/300...
Wednesday, June 26, 2013 at 10:54pm by Steve
algebra 2
solve: 3+2(1-x)>6 i got -11/2 no Here is what I got: 3+2(1-x)>6 3 + 2 - 2x > 6 -2x > 1 x < -1/2 what about the 3+2? i have to graph the solution set on a number line. What about the 3+2 ???? What a
strange question re this problem What level of math are you ...
Friday, July 27, 2007 at 7:25pm by anonymous
You want to make a graph to show how you spend your time each day. What is an advantage of choosing a circle graph for this data? A.A circle graph shows how each category of time relates to the total
amount of time. B.A circle graph is easier to make. C.It is easy to calculate...
Monday, March 4, 2013 at 7:47pm by Cassie
This is fairly easy on a polar graph. Draw vectors1 and 2, and a line representing the direction of movement. graphically add 1 and 2, getting a resultant r. Now r added to some third vector must
equal someting along the line for the movement at 15N of E. You will need a polar...
Wednesday, January 16, 2008 at 3:23pm by bobpursley
first graph y = 2x + 8,, now, using the boundaries given, find the value of y at x = -4, and find the value of y at x = 1: at x=-4, y = 2(-4) + 8 = 0 at x=1, y = 2(1) + 8 = 10 now, look for these
points on your graph and draw a vertical line (this is only a guide) from this ...
Monday, November 8, 2010 at 9:11pm by jai
In general, you can translate the graph of a function to the right (+x direction) by h if you transform the function from f(x) to f(x-h). A reflection about the x-axis is done by transforming from f
(x) to -f(x). A stretching by a factor of k in the y-direction requires a ...
Tuesday, September 15, 2009 at 2:08pm by MathMate
algebra 2
As I metioned before, you will have to do the graphing yourself. Draw the line y = -x + 4 The solution region will be above that line. Also draw the line y = x - 6. The solution region is also above
that line on the graph. The solution region is above two lines that intersect ...
Sunday, November 8, 2009 at 1:48am by drwls
College Algebra
Graph the following function using transformations. Be sure to graph all of the stages on one graph. State the domain and range. y = -SQRT x - 6. My instructor is saying that I have to do more than
one graph. Can anybody help me I am lost!
Friday, July 27, 2012 at 1:53pm by Kameesha
y = 1/2x + 3 Slope: m = 1/2 y-intercept is when x = 0 y = mx + b y = 1/2(0)+3 y = 0 + 3 y = 3 y-intercept is 3 x-intercept is when y = 0 0 = 1/2x + 3 > move 1/2x to the left and change sign. 0 -1/2x
= 3 -1/2x = 3 > divide on both sides by -1/2 x = -6 x-intercept is -6 ...
Saturday, January 11, 2014 at 1:01am by Chelle
5th grade math
If you have 78.1% nitrogen,20.9% oxygen,.9% Argon and small amounts of other gases Which graph type would best display the data? Line graph bar graph pie graph stem and leaf plot
Thursday, December 2, 2010 at 10:59pm by jill
You are driving down the road at a constant speed. another car going a bit faster catches up with you and passes you. draw a position graph for both vehicles on the same set of axes and note the
point on the graph where the other vehicles passes you. if you have been passed by...
Tuesday, September 4, 2012 at 8:38am by Emily
Use the even odd properties to find the exact value of the expression. It may be helpful to draw the unit circle graph sin(-pi)
Sunday, November 18, 2007 at 6:26pm by kelly
Calculus- Help please
Does that mean 1/(x-4) or (1/x) -4 ? They are not the same. Get or draw yourself some graph paper, pick different values of x, and use the value of f(x) that you compute for y. For example, when x =
5, f(x) = 1/(5-4) = 1
Friday, April 4, 2008 at 1:18am by drwls
11th grade
Draw a straight horizonal line that passes through y = -2 on the (vertical) y axis. It should extend over the full width of the graph.
Thursday, February 11, 2010 at 2:02pm by drwls
Math/ Physiology
Why do you draw a linear graph of absorbance versus concentration even though your experimental values deviated slightly from a straight line?
Monday, May 16, 2011 at 3:27pm by Jerome
A car weighing 1200kg was moving at a speed of 15m/s.breaks were applied and it stopped 6s latter.draw a graph showing the results
Tuesday, April 17, 2012 at 1:29am by Oscar
The graph of cos2x looks just like the graph of cos x, but oscillates twice as fast. So, since cos x crosses at -90 and 90, cos 2x crosses at -135, -45, 45, 135 deg. The sin graph looks just like sin
x, but is shifted to the left 30 deg. (That is, x+30 is zero when x=-30. In ...
Monday, September 19, 2011 at 12:02pm by Steve
Take the real-life situation and create an equation or inequality that could be used for analysis, prediction, or decision making. Then, draw a graph to depict the variables in your situation. Use
your graph and what you know about linear inequalities to discuss the ...
Monday, July 2, 2007 at 11:03pm by Evevlyn
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monoids-0.1.17: Monoids, specialized containers and a general map/reduce framework Source code Contents Index
Portability portable
Data.Group Stability experimental
Maintainer Edward Kmett <ekmett@gmail.com>
Extends Monoid to support Group operations
module Data.Monoid.Additive
class Monoid a => Group a where Source
Minimal complete definition: gnegate or minus
minus :: a -> a -> a Source
gsubtract :: a -> a -> a Source
gnegate :: Group a => a -> a Source
gsubtract :: Group a => a -> a -> a Source
minus :: Group a => a -> a -> a Source
Produced by Haddock version 2.4.1 | {"url":"http://hackage.haskell.org/package/monoids-0.1.17/docs/Data-Group.html","timestamp":"2014-04-21T04:46:31Z","content_type":null,"content_length":"10197","record_id":"<urn:uuid:ee323c12-ec5a-4e23-addb-94bb7534d358>","cc-path":"CC-MAIN-2014-15/segments/1397609539493.17/warc/CC-MAIN-20140416005219-00536-ip-10-147-4-33.ec2.internal.warc.gz"} |
Kids.Net.Au - Encyclopedia > Girth
graph theory
, the
of a graph is the length of the shortest cycle contained in the graph. If the graph doesn't contain any cycles, its girth is defined to be
For example, a 4-cycle (square) has girth 4. A grid has girth 4 as well, and a triangular mesh has girth 3.
The Heawood graph in the figure on the left is the smallest trivalent graph with girth 6; the Petersen graph on the right is the smallest trivalent graph with girth 5.
In common usage, girth refers to the
of a cylindrical object such as a tree trunk
All Wikipedia text is available under the terms of the GNU Free Documentation License | {"url":"http://encyclopedia.kids.net.au/page/gi/Girth","timestamp":"2014-04-18T23:28:19Z","content_type":null,"content_length":"12588","record_id":"<urn:uuid:389d5230-4bf5-4f55-9b1e-a2a81c4a929c>","cc-path":"CC-MAIN-2014-15/segments/1398223202457.0/warc/CC-MAIN-20140423032002-00481-ip-10-147-4-33.ec2.internal.warc.gz"} |
When is the realization of a simplicial space compact ?
up vote 1 down vote favorite
Suppose $X$ is a simplicial space of dimension $M$ (i.e. all simplices above dimension $M$ are degenerate). The claim is:
$|X|$ is compact. iff $X_n$ is compact for each $n$.
Suppose each $X_n$ is compact. Then $|X|$ is by definition a quotient of a compact space (you don't have to include the simplices above dimension $M$ in the realization). I wonder, whether the
converse is true.
Here is one motivating example. Equip the unit interval with the structure of a simplicial space in the following way:
Let $X_0$ be the Cantor-set and let the nondegenerate simplices in $X_1$ are just all the intervals, that get removed in the construction of the cantor set. One can regard $X_1$ as a subspace of $[0;
1]$ using the map (of sets) $X_1\rightarrow [0;1]$ , that sends every point in the Cantor set to the corresponding point in the unit interval and that sends each of the small intervals to its
barycenter. Equip $X_1$ with the subspace topology using this map.
The geometric realization of this space is the unit interval, which is compact and $X_0,X_1$ are also compact ($X_1$ is a closed subset of the unit interval).
This question arose in the context of this question. I realized, that I don't have a good criterium to say, when a subspace of the geometric realization of a simplicial space is compact.
gn.general-topology simplicial-stuff ct.category-theory
Doesn't $\Delta^n \times X_n$ sit inside the realisation as a closed subset? – Oscar Randal-Williams Mar 26 '10 at 18:41
Why is the realization in your example an interval and not a space with uncountably many components (most of which are points)? Do you construct a simplicial set and talk about its usual
realization as a CW-complex? If not, please define your terms. – Sergei Ivanov Mar 26 '10 at 21:51
More specifically, your simplicial space is a simplicial object in what category? – Sergei Ivanov Mar 26 '10 at 23:07
It is a simplicial object in the category of spaces. You can view simplicial sets as simplicial spaces by the use of the discrete topology. The geometric realization of a simplicial space is
defined analogously to the one of a simplicial set as a quotient of the union of $X_n\times \Delta^n$ (using the same equivalence relation). If each $X_n$ is Hausdorff, the realization is equipped
with a filtration of closed subspaces (the skeleta). In general this filtration does not give the structure of a CW-complex: You can just take any non-CW-complex Y and define X to be the constant-Y
functor. – HenrikRüping Mar 27 '10 at 9:14
@Oscar: The definition of the geometric realisation is just $|X|:=\amalg_{n=1}^M X_n\times \Delta^n/\sim$, where the equivalence relation is just the same as for simplicial sets. So there is a map
$X_n\times \Delta^n\rightarrow |X|$, which is not injective. But if i precompose with the map $X_n\rightarrow X_n\times \Dalta^n$, that picks the midpoint in the second coordinate, I get another
map, which is hopefully injective. If it is, one has to check, that its image is a closed subset of |X| and that the topology on $X_n$ is the subspace topology. I will have a try. – HenrikRüping
Mar 27 '10 at 9:43
add comment
1 Answer
active oldest votes
Ok I checked the idea mentioned in the comments above.
As $X_n$ is a closed subspace of $X_M$ (use any degeneracy; it has a left inverse; composing both the other way round yields a projection and projections have closed images in the
Hausdorff setting), it is enough to show, that $X_M$ is compact.
So consider the "generalized midpoint map"
$i: X_M\rightarrow \coprod_{n\le M} X_n\times \Delta^n\rightarrow |X|\qquad x\mapsto [x,c]$, where $c\in\Delta^M$ denotes the barycenter.
For example, if $M=2$ and $x$ is a nondegenerate one simplex in $X$, there are two degeneracies $s_0,s_1:X_1\rightarrow X_2$. Then one gets in |X|:
So in the realisation one really picks several midpoints of a simplex, that doesn't have maximal dimension.
up vote 0
down vote But the map $i$ is still injective. Any simplex $x\in X_M$ might be written in a unique way as $x=s(y)$, where $y\in X_n$ is nondegenerate and $s$ is a degeneracy map (it corresponds to
accepted a surjective map $[M]\rightarrow [n]$ in $\Delta^{Op}$, $\Delta^{Op}$ can be identified with the category whose objects are the sets $\{0,\ldots,n\}$ and whose morphisms are
nondecreasing maps).
Then $i(x)=[y,s(c)]=[y,\frac{s^{-1}(1)}{3},\ldots,\frac{s^{-1}(1)}{3}]$. Now the right side has normal form (There is a normal form for points in $|X|$. This can be found in every book
about simplicial sets and it works the same way for simplicial spaces). Given any other $x'=s'(y')$ with $i(x)=i(x')$, we get
$[y,\frac{|s^{-1}(1)|}{M+1},\ldots,|\frac{s^{-1}(1)|}{M+1}]=[y',\frac{|s'^{-1}(1)|}{M+1},\ldots,\frac{|s'^{-1}(n)|}{M+1}]$. As this is in normal form, we get $y=y'$ and $|s'^{-1}(i)|=|s
^{-1}(i)|$. But such nondecreasing maps are characterized by the size of the preimages, so $s=s'$ and hence $x=x'$. So the map $i$ is injective.
We still have to show, that $i$ is a homeo onto its image and that the image of $i$ is closed. Then $X_M$ is a closed subspace of a compact space and hence compact. Both will follow
from the next claim:
For any closed subset $A\subset X_M$ the saturation $A'$ of $A\times \{c\}$ is still a closed subset of $\coprod_{n\le M} X_n\times \Delta^n$. This is an even worse calculation than the
last one. One has to consider for each degeneracy map $s:X_N\rightarrow X_M$ the preimage $s^{-1}(A)\times \{s(c)\}$ and the set $B_s$ of all points that can be simplified to one of
those points. Then one can show, that $B_s$ is closed and $A'=\bigcup_s B_s$. As there are only finitely many degeneracy maps (below dim $M$) this is a finite union and hence closed.
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MathGroup Archive: March 2013 [00076]
[Date Index] [Thread Index] [Author Index]
Re: Mathematica and Lisp
• To: mathgroup at smc.vnet.net
• Subject: [mg130057] Re: Mathematica and Lisp
• From: Richard Fateman <fateman at cs.berkeley.edu>
• Date: Thu, 7 Mar 2013 04:00:17 -0500 (EST)
• Delivered-to: l-mathgroup@mail-archive0.wolfram.com
• Delivered-to: l-mathgroup@wolfram.com
• Delivered-to: mathgroup-newout@smc.vnet.net
• Delivered-to: mathgroup-newsend@smc.vnet.net
• References: <kgse4s$jam$1@smc.vnet.net> <20130303072215.D75466867@smc.vnet.net> <CAEtRDSdZ37Q7yCggT8AGWkKncfF57V57Et5JxjQ6i1yBnSVePQ@mail.gmail.com> <kh163q$sa3$1@smc.vnet.net>
<kh4d4r$70k$1@smc.vnet.net> <kh6c93$b26$1@smc.vnet.net> <kh77sk$q9c$1@smc.vnet.net>
On 3/6/2013 3:04 AM, David Bailey wrote:
<snip ... Fateman advocates preference for FullForm for some uses>
> So perhaps we should extend your principle to maths itself? Why risk
> students getting confused about the meaning of a + b c + d or f(a+b)-
> better to teach students to use a notation equivalent to FullForm!
This happens, to some extent, already. Students are advised to use
parentheses to try to limit ambiguity. I've seen power(x,y) instead
of x^y if x and especially y are long strings. Most mathematical
functions are probably parenthesized prefix, eg sin cos tan atan sinh
cosh.... There are a few where delimiters are kind of built in because
they are underlines or overlines like complex conjugate or
fractions. It is only
in the case of + and * that infix becomes so prevalent. Mathematica
of course makes a hash of this because sin(x) is a multiplication.
But in any case it is a false impression that mathematics beyond
(say) high-school level is presented in unambiguous formats.
> principle would be even more useful when they got to calculus, where
> notations like dy/dx and integrals are hopelessly ambiguous in that the
> terminating dx looks superficially as if it could commute with the
> integrand!
Actually this is not the problem, because context is a very big
help, and we know that dy/dx is not the same as (d*y)/(d*x). After
all, the d's would cancel producing y/x.
The real problem in integral tables and such is the REST of the
integral representation. You will see sin n x or x/ 2 pi .
These notations probably often encourage students to perform
> invalid manipulations - but even so, most people value them!
Not the students. It is just another burden on them. Like the
people who try to enrich the calculus courses with (say) Mathematica.
The majority of students, according to surveys, do not view it
as enrichment but merely another thing to learn. But that is
another story.
> I guess Mathematicians themselves realised why operator notation is so
> useful a long time back. It reduces the clutter and helps people to
> concentrate on what matters. Ultimately the choice between FullForm and
> operator form is a psychological question - not a math or computer
> science one. Those of us who do a lot of programming, also value
> operators that assist with that task too.
The nice thing about programming is that it gives you the opportunity
to name and re-use things like functions and subroutines. It allows you
to do this to essentially arbitrary depth. Hence "Solve" does whatever
it does, but you only have to remember one name. Mathematics allows
you to do this, but you must keep the notation either on paper or in
your head, and it is not usually "in the computer". In either case
the use of appropriate notation helps, and some would say is essential
in progress.
That does not mean that all notation is good. Some notation
increases the mental burden and may even increase the clutter. I have
read papers that advance the art and science not one millimeter, but
do nothing but lay claim to definitions and notations. Some editors
mistakenly view this as progress.
> The Mathematica language offers users a lot of choice - which you seem
> to abhor because some people don't choose to use it your way!
No, I have no problem in people making choices. I complain that the
design of the programming language EF is poor, at least in places. It
encourages incorrect choices (bugs) and the production of
unnecessarily difficult-to-read programs. Again I remind you that
I am talking about EF, not the library of
tools (like Solve).
> David Bailey
> http://www.dbaileyconsultancy.co.uk
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Xiaodong Wang, New York US
Xiaodong Wang, New York, NY US
application Description Published
Method for Scheduling Heterogeneous Traffic in B3G/4G Cellular Networks with Multiple Channels - A method includes tracking average user throughput, packet delay and jitter for
every user that is serviced in an OFDM cellular system; using feedback to determine a potential schedule set responsive to user requirements for data and voice traffic
20080219145 responsive to minimum rate guarantee for data flows, and maximum tolerable delay and jitter guarantees for voice flows; with multiple flows for each user, first determining 09-11-2008
contending flow for each user responsive to aggregate rate feedback; obtaining search space for the user requirements of rate, delay and jitter responsive to corresponding
optimizing strategies; determining individual rate, delay and jitter related schedules providing maximum incremental or marginal utility; and calculating final schedule of
users on channels providing maximum incremental or marginal utility among parameter specific schedules form the determining step.
Group MMSE-DFD with Rate (SINR) Feedback and Without Pre-Determined Decoding Order for Reception on a Cellular Downlink - In accordance with the invention, a method includes
the steps of: i) initializing with channel matrix estimates and inner codes of all co-channel transmitter sources in a wireless network, modulation and coding schemes of all
sources not of interest; ii) converting each channel matrix estimate into an effective channel matrix responsive to the inner code of the corresponding transmitter source; iii)
20080225763 selecting iteratively from a first set of transmitter sources transmitting at fixed rates, a transmitter source which maximizes a first metric; iv) computing iteratively a 09-18-2008
filter for the transmitter source which maximizes the first metric; v) selecting iteratively from a second set of transmitter sources of interest, a transmitter source which
maximizes a second metric; vi) computing iteratively a rate and a filter for the transmitter source which maximizes the second metric; and vii) obtaining an ordered set of
indices of all transmitter sources that will be decoded along with their corresponding filters, and feedback rates for all transmitter sources of interest.
OPTIMAL RESOURCE ALLOCATION IN A MULTI-HOP OFDMA WIRELESS NETWORK WITH COOPERATIVE RELAYING - An optimal resource allocation strategy for OFDMA multi-hop wireless networks is
20080225774 disclosed. The system allocates one or more resources in a multi-hop network by solving one or more higher-layer sub-problem; solving one or more physical layer and media 09-18-2008
access control (PHY/MAC) layer sub-problems per tone per time slot with one of cooperative relaying of radio signals or spatial reusing of radio spectrum; updating prices; and
allocating radio resources based on the PHY/MAC layer sub-problems.
Group MMSE-DFD with Order and Filter Computation for Reception on a Cellular Downlink - A method for decoding in a wireless downlink channel, where all dominant transmitting
sources use inner codes from a particular set, including the steps of: estimating a channel matrix seen from each dominant transmitter source in response to a pilot or preamble
signal transmitted by each such source; converting each estimated channel matrix into an effective channel matrix responsive to the inner code of the corresponding transmitting
source; obtaining the received observations in a linear equivalent form whose output is an equivalent of the received observations and in which the effective channel matrix
20080225781 corresponding to each dominant transmitting source inherits the structure of its inner code; i) determining an order for processing each of the transmitting sources; ii) 09-18-2008
computing a filter for each transmitting source that will be decoded; iii) demodulating and decoding each transmitting source responsive to the determined order from step i)
assuming perfect cancellation of signals of preceding or previously decoded transmitting sources; and iv) re-encoding the decoded message of each transmitting source, except
the source decoded last, responsive to the modulation and coding scheme employed by the source and the corresponding effective channel matrix and subtracting it from the
received observations in the equivalent linear form.
Max-Log Receiver for Multiple-Input Multiple-Output (MIMO) Systems - A method includes the steps of i) listing out all possibilities for a first symbol of a two stream signal;
ii) determining a second symbol of the two stream signal for each of the first symbol listed out, iii) evaluating a metric for each of the first symbol and second symbol pair,
20080225974 iv) listing out all possibilities for second symbol, v) determining a first symbol for each choice of the second symbol listed out, vi) evaluating a metric for each of the 09-18-2008
second symbol and first symbol pair, vii) determining an exact maximum log likelihood ratio for all bits using the metrics, and viii) decoding codeword(s) in the two stream
signal using the determined exact maximum log likelihood ratio for all bits.
Group MMSE-DFD with Rate (SINR) Feedback and Pre-Determined Decoding Order for Reception on a Cellular Downlink - A method for decoding and rate assignment in a wireless
channel, where all dominant transmitter sources use inner codes from a particular set, comprising the steps of: i) estimating channel matrices seen from all dominant
transmitter sources in response to a pilot or preamble signal transmitted by each such source; ii) converting each estimated channel matrix into an effective channel matrix
20080225979 responsive to the inner code of the corresponding transmitter source; iii) obtaining the received observations in a linear equivalent form (linear model) whose output is an 09-18-2008
equivalent of the received observations and in which the effective channel matrix corresponding to each dominant transmitter source inherits the structure of its inner code;
iv) processing the transmitter sources according to the specified (or pre-determined) order of decoding; v) for each transmitter source, assuming perfect cancellation of
signals of preceding transmitter sources; vi) computing a signal-to-interference-noise-ratio SINR responsive to the effective channel matrix of the transmitter source and the
covariance matrix of the noise plus signals from remaining transmitter sources; and vii) feeding back all computed SINRs to respective transmitter sources.
Group LMMSE Demodulation Using Noise and Interference Covariance Matrix for Reception on a Cellular Downlink - A method for filtering in a wireless downlink channel, where all
dominant transmitting sources use inner codes from a particular set, includes the steps of estimating a channel matrix seen from a desired transmitter source in response to a
20080227397 pilot or preamble signal; converting the estimated channel matrix into an effective channel matrix responsive to the inner code of the desired transmitting source; estimating a 09-18-2008
covariance matrix of noise plus interference in a linear model whose output is an equivalent of the received observations and in which the effective channel matrix
corresponding to each dominant transmitting source inherits the structure of its inner code; computing a signal-to-noise-interference-ratio SINR responsive to the covariance
matrix and the effective channel matrix corresponding to the desired source; and feeding back the computed SINR to the transmitter source.
STATIC AND DIFFERENTIAL PRECODING CODEBOOK FOR MIMO SYSTEMS - Systems and methods are disclosed to generate a codebook for channel state information by generating a random
20080232501 codebook; partitioning channel state information into a set of nearest neighbors for each codebook entry based on a distance metric; and updating the codebook by finding a 09-25-2008
centroid for each partition.
Methods and Systems for Providing Feedback for Beamforming - Methods and systems for providing feedback for beamforming are provided. In some embodiments, methods for providing
feedback for beamforming are provided, the methods comprising: initializing a weighting vector; determining a perturbation vector; determining a plurality of weighting vectors
20080285665 to be applied to data based on the weighting vector and the perturbation vector; applying the plurality of weighting vectors to the data to provide weighted data; transmitting 11-20-2008
the weighted data to a receiver; receiving from the receiver a feedback signal based on at least one of the plurality of weighting vectors; and updating the plurality of
weighting vectors based on the feedback signal.
20080285666 Methods and Systems for Digital Wireless Communication - Methods and systems for digital wireless communication are provided. 11-20-2008
MULTI-CELL INTERFERENCE MITIGATION VIA COORDINATED SCHEDULING AND POWER ALLOCATION IN DOWNLINK ODMA NETWORKS - A multi-cell Orthogonal Frequency-Division Multiple Access
(OFDMA) based wireless system and method with full spectral reuse co-channel interference mitigation via base station coordination in a downlink channel includes a plurality of
20080298486 base stations configured to handle communications with mobile units. A central controller is configured to mitigate interference between base stations via jointly optimizing 12-04-2008
coordinated scheduling and power allocation in accordance with a sub-optimal iterative solution. Five methods provide the solution, which include: 1) Improved Iterative
Water-Filling (I-IWF); 2) Iterative Spectrum Balancing (ISB); 3) Successive Convex Approximation for Low-complexity (SCALE); 4) Opportunistic Base Station Selection (OBSS) and
5) Per-tone binary power control (PT-BPC).
SYSTEM AND METHOD FOR SCHEDULING IN RELAY-ASSISTED WIRELESS NETWORKS - A scheduling system and method for use with relay-assisted wireless networks includes accessing feedback
from mobile stations in a network and arranging users associated with a relay station in a list in accordance with marginal utilities. A determination of whether the users in
20090003259 the list can be eliminated from feedback overhead by testing conditions for feedback reduction is made. A diversity schedule is generated by employing a weighted bipartite 01-01-2009
graph with relay channels and access channels and performing a matching method. A transmission schedule is generated for channel usage in accordance with multi-user and channel
diversity for mobile users and spatial reuse of channels across relay and access hops by incorporating rate feedback and interference for the mobile stations and the relay
stations based upon the matching method applied to a new weighted graph which accounts for traffic loads and fairness as well.
EFFICIENT LOW COMPLEXITY HIGH THROUGHPUT LDPC DECODING METHOD AND OPTIMIZATION - A decoder and method for iteratively decoding of low-density parity check codes (LDPC)
20090083609 includes, in a code graph, performing check node decoding by determining messages from check nodes to variable nodes. In the code graph, variable node decoding is performed by 03-26-2009
determining messages from the variable nodes to the check nodes. The variable node decoding is independent from degree information regarding the variable nodes. Decoded results
are output.
Wireless Communication Rate Allocation on a Gaussian Interference Channel - There is provided a method for allocating transmission rates in a wireless network, includes the
steps of associating transmitters with corresponding receivers for communicating on an interference channel in the wireless network, and allocating a transmission rate to each
transmitter for decoding by its corresponding receiver, the allocated transmission rate being equal to a desired rate of a fixed user rate and being no less than a minimum rate
20090129328 of a variable user rate. The step of allocating can include a sequential allocation that meets the minimum rate requirement of all users and assigns excess rates to variable 05-21-2009
rate users in a sequential fashion according to specified priorities. The step of allocating can include a parallel symmetric rate allocation when all variable rate users have
the same priority and that meets minimum rate requirements of all users and is symmetric fair. The step of allocating can include a parallel iterative rate allocation with the
sequence of rate allocations obtained for each user being non-decreasing and a rate allocation vector obtained after each iteration meets minimum rate requirements for all
users and is max-min fair when all variable rate users have the same priority.
Anti-Jamming Piecewise Coding Method for Parallel Inference Channels - A method for encoding includes encoding K blocks of information for transmission on N subchannels
responsive to a number of redundant blocks M according to one of i) employing a single parity check code when the number of redundant blocks M is about 1; ii) employing a code
exhibited by a code graph having one third of variable nodes are connected to one of the check nodes, another one third of variable nodes is connected to the other check node
20090210757 and the remaining one third of variable nodes is connected to both check nodes, when the number of redundant blocks M is 2; iii) employing a first process for determining a 08-20-2009
code for the K blocks of information, when the number of redundant blocks M is about 3 together with K blocks of information less than about 150 or the number of redundant
blocks M is about 4 together with K blocks of information less than about 20; and iv) employing a second process for determining a code for the K blocks of information with
redundant block M values other than for steps i), ii) and iii).
TWO-STAGE LOW-COMPLEXITY MAX-LOG BIT-LEVEL LLR CALCULATOR AND METHOD - A demodulator and demodulation method includes a bit/symbol hard demodulator configured to obtain hard
20090231028 bit or symbol information from a received signal. At least one lookup table is configured to reference coefficients for computation of log-likelihood ratios (LLRs) from the 09-17-2009
hard bit or symbol information. A log-likelihood ratio calculation module is configured to compute bit-level LLRs from the coefficients and the received signal.
AUCTION BASED RESOURCE ALLOCATION IN WIRELESS SYSTEMS - Systems and methods to assign one or more resources in a multi-user cellular Orthogonal Frequency-Division Multiple
20090232074 Access (OFDMA) uplink includes specifying a resource allocation problem for one or more resources; converting the resource allocation problem into an assignment problem; 09-17-2009
solving the assignment problem through an auction; and allocating one or more resources to cellular users to maximize a system utility.
RECEIVER WITH PREFILTERING FOR DISCRETE FOURIER TRANSFORM-SPREAD-ORTHOGONAL FREQUENCY DIVISION MULTIPLEXING (DFT-S-OFDM) BASED SYSTEMS - A receiver for discrete Fourier
transform-spread-orthogonal frequency division multiplexing (DFT-S-OFDM) based systems, including a prefilter for received signal codeword(s); and a log-likelihood ratio LLR
20090262872 module responsive to the prefilter; wherein the prefilter includes a pairing and whitening module that based on channel estimates and data rate enables the LLR module to 10-22-2009
perform either a Serial-In-Serial-Out (SISO) based log likelihood ratio processing of an output from the paring and whitening module or a two-symbol max-log soft output
demodulator (MLSD) based log likelihood ratio processing of an output from the pairing and whitening module.
Rateless Coding for Multiuser Interference Relay Channel - An interference relay channel (IRC) utilizes a single relay to assist communications between multiple
20090270028 source-destination links under the half-duplex (HD) constraint. It is assumed that each source has an independent message and its transmitted signal may cause interference at 10-29-2009
the other destinations. It is also assumed that each node only estimates its backward channels and has no knowledge of its forward channels as well as the other links. The role
of the relay is to generate signals to cooperate with the intended signal and mitigate the interference at all destinations.
MULTI-RESOLUTION PRECODING CODEBOOK - Systems and methods are for generating a codebook by: generating a multi-resolution codebook by selecting a common precoder index from a
20090274225 low resolution codebook for a group of adjacent resource blocks (RB)s and for each RB within the group, selecting a high-resolution codebook to fine-tune each RB precoder; and 11-05-2009
generating feedback for the multi-resolution codebook by quantizing channel state variations.
Hybrid ARQ Transmission Method With Channel State Information - A method for decoding of multiple wireless signals by a chase combining hybrid-automatic-repeat-request CC-HARQ
receiver includes demodulating wireless signals received from respective mobile sources using an effective channel matrix and decision statistics; updating
20090276679 log-likelihood-ratios LLRs and decoding the received codewords using the corresponding updated LLRs; determining set of correctly decoded codewords using a cyclic redundancy 11-05-2009
check; updating the effective channel matrix and decision statistics responsive to the step of determining; and resetting the effective channel matrix and decision statistics
in the event that the number of decoding errors for a codeword exceeds its maximum limit after storing the updated LLRs of all remaining erroneously decoded codewords for which
the number of decoding errors is below the respective maximum limit.
COGNITIVE RADIO, ANTI-JAMMING CODING RETRANSMISSION METHODS AND SYSTEMS - Implementations of the present principles include methods and systems for retransmitting un-recovered
20090282309 information within a cognitive radio, anti-jamming system. In accordance with aspects of the present principles, encoding schemes may be optimized for retransmission by 11-12-2009
utilizing a jamming rate and a number of un-recovered packets to minimize packet loss and thereby enhance throughput. In addition, rateless encoding features may be employed to
re-encode un-recovered portions of an information sequence for efficient retransmission.
COORDINATED LINEAR BEAMFORMING IN DOWNLINK MULTI-CELL WIRELESS NETWORKS - System and methods are disclosed for optimizing wireless communication for a plurality of mobile
20090296650 wireless devices. The system uses beamforming vectors or precoders having a structure optimal with respect to the weighted sum rate in a multi-cell orthogonal frequency 12-03-2009
division multiple access (OFDMA) downlink. A plurality of base stations communicate with the mobile devices and all base stations perform a distributed non-convex optimization
exploiting the determined structure.
Systems and Methods for Adaptive Hybrid Automatic Retransmission Requests - A turbo coded hybrid ARQ with both IR and repetition existing in one or more transmission blocks is
20090319855 designed. The mixed combining is performed at the receiver—Chase combining is performed for the repetitions first, followed by code combining for all code bits based on Chase 12-24-2009
combining outputs. The packet error probability is obtained by applying the union-Bhattacharyya (UB) bound for parallel channels. The throughput optimization of hybrid ARQ is
determined as well as the optimal assignment rates by solving the optimization using a genetic algorithm.
DISTRIBUTED BEAMFORMING AND RATE ALLOCATION IN MULTI-ANTENNA COGNITIVE RADIO NETWORKS - Systems and methods are disclosed for designing beamforming vectors for and allocating
20090323619 transmission rates to secondary users in a wireless cognitive network with secondary (cognitive) users and primary (license-holding) users by performing distributed beamforming 12-31-2009
design and rate allocation for the secondary users to maximize a minimum weighted secondary rate; and granting simultaneous spectrum access to the primary and secondary users
subject to one or more co-existence constraints.
System and Method for Wireless Transmission Using Hybrid ARQ Based on Average Mutual Information Per Bit - A method and system for packet transmission in a hybrid automatic
repeat request (ARQ) system. A modulation and a block length for a transmission are determined based on the average mutual information per bit. The average mutual information
per bit is computed based on a current channel signal-to-noise ratio and a plurality of previous transmissions, each being transmitted with a respective coded block length,
20100042876 modulation form, and signal-to-noise ratio. A block error rate is computed for the potential block lengths and modulations based on the average mutual information per bit, and 02-18-2010
a throughput of the current transmission is determined based on the block error rate. The modulation form and the block length of the transmission are determined based on an
analysis of the throughput. If the receiver cannot decode the current transmission, the transmitter repeats the computation to determine the modulation and the block length for
DISTRIBUTED MESSAGE-PASSING BASED RESOURCE ALLOCATION IN WIRELESS SYSTEMS - Systems and methods are disclosed to allocate resources in discrete Fourier transform spread
20100091729 orthogonal frequency division multiple access (DFT-S-OFDMA) networks, which involve determining a reward for each user when assigned a frequency chunk (FC) of subcarriers, 04-15-2010
where each FC is a set of contiguous subcarriers; splitting each user into one or more sub-users, with each sub-user having identical rewards; and assigning resources with a
message-passing based FC allocation.
MU-MIMO-OFDMA SYSTEMS AND METHODS FOR SERVICING OVERLAPPING CO-SCHEDULED USERS - Methods and systems for conveying or transmitting to any given user in an OFDMA-MU-MIMO system
scheduling information of other co-scheduled users to permit the user to perform error-correction on received data and/or interference reduction on its received signals. The
20100157924 scheduling information can include resource block assignment, modulation constellations employed, coding rates employed, power levels utilized and precoder matrix indices used. 06-24-2010
Further, the scheduling information can be conveyed in part through dedicated reference symbol layers or pilot streams. Moreover, a base station may transmit a preliminary
estimate of the total number of users the base station expects to schedule, or an upper-bound on the total number of users, to the MU-MIMO users to permit the MU-MIMO users to
determine preferred precoder matrix indices and indications of channel quality indices.
MAX-LOG STACK DECODER - A method for demodulating signals in a multi-input multi-output (MIMO) receiver includes obtaining a transformed vector by a coordinate transformation
20100158150 of a received observation vector using a unitary matrix determined through QR decomposition of an estimated channel matrix; maintaining a list containing nodes along with a 06-24-2010
cost metric for each node; using the list to generate soft-outputs in the form of log-likelihood ratios (LLRs) for selected symbols of interest, based on the transformed vector
and a lower triangular matrix determined through QR decomposition of the estimated channel matrix and the constellations to which the input symbols belong.
Apparatus and Method for Multilayer Space-Time-Frequency Precoding for a MIMO-OFDM Wireless Transmission System - In a wireless wideband MIMO-OFDM transmission system, a method
includes converting a coded bit sequence to parallel data layers, responsive to channel encoding and interleaving of an information sequence to provide the coded bit sequence;
20100232535 passing each data layer through a respective repetition encoder, independently interleaving respective spread data sequences from the respective repetition encoder, and 09-16-2010
amplifying the respective interleaved outputs responsive to power allocation of a respective layer of multiple layers for both I and Q channels for being combined to form
complex symbols for transmission through respective multiple antennas.
CONSTELLATION RE-ARRANGEMENT AND BIT GROUPING - Methods and systems for subpacket generation using a convolutional turbo code in hybrid automatic repeat request
20100271929 re-transmissions that includes separating a codeword into subblocks of bits, interleaving the subblocks, and performing a permutation to group the bit streams and rearrange a 10-28-2010
symbol constellation such that bits are assigned to bit positions based on a number of re-transmissions.
LDPC Hard Decision Decoder for High-Speed Wireless Data Communications - A method for low-density parity-check hard decision decoding includes computing, for every decoding
20110010600 iteration, a discrepancy of extrinsic messages responsive to channel inputs of a receiver, performing a flipping of the channel inputs responsive to a comparison of the 01-13-2011
discrepancy of extrinsic messages to a flipping threshold, the flipping threshold for each decoding iteration being determined based on a threshold computation responsive to a
channel error probability estimation in a first iteration of a decoding of the channel inputs, and check node decoding responsive to the flipping of channel inputs
ROBUST LINEAR PRECODER DESIGNS FOR MULTI-CELL DOWNLINK TRANSMISSION - Methods and systems for optimizing the utilities of receiver devices in a wireless communication network
20110059705 are disclosed. Precoder design formulations that maximize a minimum worst-case rate or a worst-case sum rate are described for both full base station cooperation and limited 03-10-2011
base station cooperation scenarios. In addition, optimal equalizers are also selected to optimize the worst-case sum rate.
Methods and Systems for Providing Feedback for Beamforming and Power Control - Methods and systems for providing feedback for beamforming and power control are provided. In
some embodiments, the methods comprise: calculating a threshold associated with a subcarrier; receiving the subcarrier containing an information symbol at the receiver;
20110069774 determining a channel estimate of the subcarrier and a weighting vector for the information symbol; based at least on the channel estimate and the weighting vector, determining 03-24-2011
a power level of the subcarrier; comparing the power level to the threshold; generating a feedback signal indicating a first energy level to be used for the subcarrier based on
the comparison; subsequent to generating the feedback signal, determining a second power level of the subcarrier; comparing the second power level to the threshold; and
generating a second feedback signal to the transmitter indicating a second energy level to be used for the subcarrier.
Message-Wise Unequal Error Protection - Message-wise unequal error protection is provided using codeword flipping to separate special and ordinary codewords without discarding
any codewords. Special messages are encoded to ensure the codeword weight is less than a certain threshold weight. Ordinary messages are encoded to ensure the codeword weight
20110093760 is greater than the threshold weight. The bits of the codeword are flipped to enforce the weight criterion. Ordinary and special messages are encoded using different encodings 04-21-2011
to provide different levels of error protection. Upon receipt, codewords are separated into special and ordinary codewords for appropriate decoding. If a codeword is of
indeterminate type, it is iteratively processed as both a special codeword and an ordinary codeword. The decoding result of each process is periodically checked to determine
which decoding result satisfies decoding criteria.
Transmission for Half-Duplex Relay in Fading Channel and Rateless Code Configuration - In one aspect of the invention, a method for transmission in a wireless communication
20110170457 system includes selecting by a signal destination one of a source-destination direct transmission, a decode-forward relay transmission, and a compress-forward transmission, 07-14-2011
responsive to channel gains between a signal source and a relay, between the signal source and the signal destination, and between the relay and the signal destination;
informing a selected transmission mode from the selecting step to the signal source and the relay; and operating in the selected transmission mode by the relay.
LT DECODING AND RETRANSMISSION FOR WIRELESS BROADCAST - Methods and systems for doped rateless retransmission include receiving ratelessly coded symbols. An attempt is made to
20110246848 decode the coded symbols using a processor by creating an associated code graph that represents the structure of the rateless code used by the symbols. If the decoding attempt 10-06-2011
fails, an input node is selected from the code graph using a metric that gauges the number and degree of connections to the input node based on the code graph structure. The
selected input node is then requested for retransmission of the selected input node by a feedback channel.
Method and Systems for Conveying Scheduling Information of Overlapping Co-Scheduled Users in an OFDMA-MU-MIMO System - Methods and systems for conveying or transmitting to any
given user in an OFDMA-MU-MIMO system scheduling information of other co-scheduled users to permit the user to perform error-correction on received data and/or interference
20120057557 reduction on its received signals. The scheduling information can include resource block assignment, modulation constellations employed, coding rates employed, power levels 03-08-2012
utilized and precoder matrix indices used. Further, the scheduling information can be conveyed in part through dedicated reference symbol layers or pilot streams. Moreover, a
base station may transmit a preliminary estimate of the total number of users the base station expects to schedule, or an upper-bound on the total number of users, to the
MU-MIMO users to permit the MU-MIMO users to determine preferred precoder matrix indices and indications of channel quality indices.
Method and Systems for Conveying Scheduling Information of Overlapping Co-Scheduled Users in an OFDMA-MU-MIMO System - Methods and systems for conveying or transmitting to any
given user in an OFDMA-MU-MIMO system scheduling information of other co-scheduled users to permit the user to perform error-correction on received data and/or interference
20120057558 reduction on its received signals. The scheduling information can include resource block assignment, modulation constellations employed, coding rates employed, power levels 03-08-2012
utilized and precoder matrix indices used. Further, the scheduling information can be conveyed in part through dedicated reference symbol layers or pilot streams. Moreover, a
base station may transmit a preliminary estimate of the total number of users the base station expects to schedule, or an upper-bound on the total number of users, to the
MU-MIMO users to permit the MU-MIMO users to determine preferred precoder matrix indices and indications of channel quality indices.
Codebook Method for a Multiple Input Multiple Output Wireless System - A method for wireless encoding includes encoding wireless multiple input and multiple output signals in
20120140850 accordance with a codebook being one of a discrete codebook restricting elements of codebook entries to be within a predetermined finite set of complex numbers and a constant 06-07-2012
amplitude codebook including each entry in its codebook having equal column norm and equal row norm. In a preferred embodiment the digital codebook further includes restricting
elements of a finite set in the discrete codebook to be in the form of k
COORDINATED LINEAR BEAMFORMING IN DOWNLINK MULTI-CELL WIRELESS NETWORKS - System and methods are disclosed for optimizing wireless communication for a plurality of mobile
20120163332 wireless devices. The system uses beamforming vectors or precoders having a structure optimal with respect to the weighted sum rate in a multi-cell orthogonal frequency 06-28-2012
division multiple access (OFDMA) downlink. A plurality of base stations communicate with the mobile devices and all base stations perform a distributed non-convex optimization
exploiting the determined structure.
Robust Linear Precoder Designs for Multi-Cell Downlink Transmission - Methods and systems for optimizing the utilities of receiver devices in a wireless communication network
20120170676 are disclosed. Precoder design formulations that maximize a minimum worst-case rate or a worst-case sum rate are described for both full base station, cooperation and limited 07-05-2012
base station cooperation scenarios. In addition, optimal equalizers are also selected to optimize the worst-case sum rate.
COORDINATED LINEAR BEAMFORMING IN DOWNLINK MULTI-CELL WIRELESS NETWORKS - System and methods are disclosed for optimizing wireless communication for a plurality of mobile
20120170677 wireless devices. The system uses beamforming vectors or precoders having a structure optimal with respect to the weighted sum rate in a multi-cell orthogonal frequency 07-05-2012
division multiple access (OFDMA) downlink. A plurality of base stations communicate with the mobile devices and all base stations perform a distributed non-convex optimization
exploiting the determined structure.
MULTI-RESOLUTION PRECODING CODEBOOK - Systems and methods are for generating a codebook by: generating a multi-resolution codebook by selecting a common precoder index from a
20120224649 low resolution codebook for a group of adjacent resource blocks (RB)s and for each RB within the group, selecting a high-resolution codebook to fine-tune each RB precoder; and 09-06-2012
generating feedback for the multi-resolution codebook by quantizing channel state variations.
Codebook Method for a Multiple Input Multiple Output Wireless System - A method for wireless encoding includes encoding wireless multiple input and multiple output signals in
20120250791 accordance with a codebook being one of a discrete codebook restricting elements of codebook entries to be within a predetermined finite set of complex numbers and a constant 10-04-2012
amplitude codebook including each entry in its codebook having equal column norm and equal row norm. In a preferred embodiment the digital codebook further includes restricting
elements of a finite set in the discrete codebook to be in the form of k
Codebook Method for a Multiple Input Multiple Output Wireless System - A method for wireless encoding includes encoding wireless multiple input and multiple output signals in
20120275535 accordance with a codebook being one of a discrete codebook restricting elements of codebook entries to be within a predetermined finite set of complex numbers and a constant 11-01-2012
amplitude codebook including each entry in its codebook having equal column norm and equal row norm. In a preferred embodiment the digital codebook further includes restricting
elements of a finite set in the discrete codebook to be in the form of k
Scheduling Information of Overlapping Co-Scheduled Users in an OFDMA-MU-MIMO System - A user equipment (UE) used in a multi-user (MU)-multiple input multiple output (MIMO)
orthogonal frequency division multiple access (OFDMA) system is disclosed. The UE includes a receiving unit to receive from a base station an indication of an estimate of or an
20120281662 upper-bound on the total number of MU-MIMO user equipments (|S|) that are scheduled on a sub-band by the base station, wherein the sub-band includes one or more resource units, 11-08-2012
a calculation unit to calculate channel quality based on the indication of the estimate of or the upper-bound on the total number of MU-MIMO user equipments, and a transmission
unit to transmit to the base station an indication of the channel quality. Other methods and apparatuses also are disclosed.
MULTI-RESOLUTION PRECODING CODEBOOK - Systems and methods are for generating a codebook by: generating a multi-resolution codebook by selecting a common precoder index from a
20130094453 low resolution codebook for a group of adjacent resource blocks (RB)s and for each RB within the group, selecting a high-resolution codebook to fine-tune each RB precoder; and 04-18-2013
generating feedback for the multi-resolution codebook by quantizing channel state variations.
Multi-user downlink linear MIMO precoding system - A method implemented in a base station used for a downlink multi-user (MU) multi-input multi-output (MIMO) system is
20130170445 disclosed. The method includes receiving an indication of a quantized matrix from each of a plurality of scheduled user equipments, precoding data streams for the plurality of 07-04-2013
scheduled user equipments, transmitting the precoded data to the plurality of scheduled user equipments. Other methods and some apparatuses for wireless communications also are
MULTI-RESOLUTION PRECODING CODEBOOK - Systems and methods are for generating a codebook by: generating a multi-resolution codebook by selecting a common precoder index from a
20130230116 low resolution codebook for a group of adjacent resource blocks (RB)s and for each RB within the group, selecting a high-resolution codebook to fine-tune each RB precoder; and 09-05-2013
generating feedback for the multi-resolution codebook by quantizing channel state variations.
Methods, Systems, and Media for Detecting Usage of a Radio Channel - Methods, systems, and media for detecting usage of a radio channel are provided. In some embodiments,
methods for detecting usage of a radio channel are provided, the methods comprising: collecting noise samples on the radio channel from a radio receiver; determining a noise
20140024316 empirical cumulative distribution function using a hardware processor; collecting signal samples on the radio channel from the radio receiver; determining a signal empirical 01-23-2014
cumulative distribution function using a hardware processor; calculating a largest absolute difference between the noise empirical cumulative distribution function and the
signal empirical cumulative distribution function using a hardware processor; and determining that the radio channel is being used when the largest absolute difference is
greater than a threshold using a hardware processor.
Patent applications by Xiaodong Wang, New York, NY US | {"url":"http://www.faqs.org/patents/inventor/xiaodong-wang-new-york-us-1/","timestamp":"2014-04-17T00:09:54Z","content_type":null,"content_length":"48921","record_id":"<urn:uuid:9df6c3f4-6736-4c04-9a13-fb984d4a0809>","cc-path":"CC-MAIN-2014-15/segments/1397609525991.2/warc/CC-MAIN-20140416005205-00294-ip-10-147-4-33.ec2.internal.warc.gz"} |
metalogic :: Elementary logic
Article Free Pass
Elementary logic
An area that is perhaps of more philosophical interest is that of the nature of elementary logic itself. On the one hand, the completeness discoveries seem to show in some sense that elementary logic
is what the logician naturally wishes to have. On the other hand, he is still inclined to ask whether there might be some principle of uniqueness according to which elementary logic is the only
solution that satisfies certain natural requirements on what a logic should be. The development of model theory has led to a more general outlook that enabled the Swedish logician Per Lindström to
prove in 1969 a general theorem to the effect that, roughly speaking, within a broad class of possible logics, elementary logic is the only one that satisfies the requirements of axiomatizability and
of the Löwenheim-Skolem theorem. Although Lindström’s theorem does not settle satisfactorily whether or not elementary logic is the right logic, it does seem to suggest that mathematical findings can
help the logician to clarify his concepts of logic and of logical truth.
A particularly useful tool for obtaining new models from the given models of a theory is the construction of a special combination called the “ultraproduct” of a family of structures (see below
Ultrafilters, ultraproducts, and ultrapowers)—in particular, the ultrapower when the structures are all copies of the same structure (just as the product of a[1], . . . , a[n] is the same as the
power a^n, if a[i] = a for each i). The intuitive idea in this method is to establish that a sentence is true in the ultraproduct if and only if it is true in “almost all” of the given structures
(i.e., “almost everywhere”—an idea that was present in a different form in Skolem’s construction of a nonstandard model of arithmetic in 1933). It follows that, if the given structures are models of
a theory, then their ultraproduct is such a model also, because every sentence in the theory is true everywhere (which is a special case of “almost everywhere” in the technical sense employed).
Ultraproducts have been applied, for example, to provide a foundation for what is known as “nonstandard analysis” that yields an unambiguous interpretation of the classical concept of infinitesimals
—the division into units as small as one pleases. They have also been applied by two mathematicians, James Ax and Simon B. Kochen, to problems in the field of algebra (on p-adic fields).
Nonelementary logic and future developments
There are also studies, such as second-order logic and infinitary logics, that develop the model theory of nonelementary logic. Second-order logic contains, in addition to variables that range over
individual objects, a second kind of variable ranging over sets of objects so that the model of a second-order sentence or theory also involves, beyond the basic domain, a larger set (called its “
power set”) that encompasses all the subsets of the domain. Infinitary logics may include functions or relations with infinitely many arguments, infinitely long conjunctions and disjunctions, or
infinite strings of quantifiers. From studies on infinitary logics, William Hanf, an American logician, was able to define certain cardinals, some of which have been studied in connection with the
large cardinals in set theory. In yet another direction, logicians are developing model theories for modal logics—those dealing with such modalities as necessity and possibility—and for the
intuitionistic logic.
There is a large gap between the general theory of models and the construction of interesting particular models such as those employed in the proofs of the independence (and consistency) of special
axioms and hypotheses in set theory. It is natural to look for further developments of model theory that will yield more systematic methods for constructing models of axioms with interesting
particular properties, especially in deciding whether certain given sentences are derivable from the axioms. Relative to the present state of knowledge, such goals appear fairly remote. The gap is
not unlike that between the abstract theory of computers and the basic properties of actual computers.
Do you know anything more about this topic that you’d like to share? | {"url":"http://www.britannica.com/EBchecked/topic/377696/metalogic/65881/Elementary-logic","timestamp":"2014-04-19T06:36:40Z","content_type":null,"content_length":"92398","record_id":"<urn:uuid:88965ad6-3eb8-4e5a-8168-be5963aee77f>","cc-path":"CC-MAIN-2014-15/segments/1397609535775.35/warc/CC-MAIN-20140416005215-00631-ip-10-147-4-33.ec2.internal.warc.gz"} |
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Multidimensional Fixed-Point Theorems in Partially Ordered Complete Partial Metric Spaces under (
Abstract and Applied Analysis
Volume 2013 (2013), Article ID 634371, 12 pages
Research Article
Multidimensional Fixed-Point Theorems in Partially Ordered Complete Partial Metric Spaces under ()-Contractivity Conditions
University of Jaén, Campus Las Lagunillas s/n, 23071 Jaén, Spain
Received 17 March 2013; Accepted 17 June 2013
Academic Editor: Abdelouahed Hamdi
Copyright © 2013 A. Roldán et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any
medium, provided the original work is properly cited.
We study the existence and uniqueness of coincidence point for nonlinear mappings of any number of arguments under a weak ()-contractivity condition in partial metric spaces. The results we obtain
generalize, extend, and unify several classical and very recent related results in the literature in metric spaces (see Aydi et al. (2011), Berinde and Borcut (2011), Gnana Bhaskar and Lakshmikantham
(2006), Berzig and Samet (2012), Borcut and Berinde (2012), Choudhury et al. (2011), Karapınar and Luong (2012), Lakshmikantham and Ćirić (2009), Luong and Thuan (2011), and Roldán et al. (2012)) and
in partial metric spaces (see Shatanawi et al. (2012)).
1. Introduction
The notion of coupled fixed point was introduced by Guo and Lakshmikantham [1] in 1987. In a recent paper, Gnana Bhaskar and Lakshmikantham [2] introduced the concept mixed monotone property for
contractive operators of the form , where is a partially ordered metric space, and then established some coupled fixed-point theorems. After that, many results appeared on coupled fixed-point theory
in different contexts (see, e.g., [3–6]). Later, Berinde and Borcut [7] introduced the concept of tripled fixed point and proved tripled fixed-point theorems using mixed monotone mappings (see also [
Very recently, Roldán et al. [11] proposed the notion of coincidence point between mappings in any number of variables and showed some existence and uniqueness theorems that extended the mentioned
previous results for this kind of nonlinear mappings, not necessarily permuted or ordered, in the framework of partially ordered complete metric spaces, using a weaker contraction condition, that
also generalized other works by Berzig and Samet [12], Karapınar and Berinde [13].
Partial metric spaces were firstly introduced by Matthews in [14] as an attempt to generalize the metric spaces by establishing the condition that the distance between a point to itself (which is not
necessarily zero) is less or equal than the distance between that point and another point of the space. In the mentioned papers, Matthews studied topological properties of partial metric spaces and
stated a modified version of a Banach contraction mapping principle on this kind of spaces. After Matthews' pioneering work, the theory of partial metric spaces and particularly the field of
fixed-point theorems have expansively been developed due to the increasing interest in this area and motivated by its possible applications (see [15, 16] and references therein).
In this paper, our main aim is to study a weaker contractivity condition for nonlinear mappings of any number of arguments. This condition can be particularized in a variety of forms that let us
extend the previously mentioned results and other recent ones in this field (see [2, 5, 7, 9, 11, 12, 16–20]). We also notice that our results cannot be obtained by the very recent paper of Haghi et
al. [21] (for more details see Remark 26).
2. Preliminaries
Preliminaries and notation about coincidence points can also be found in [11]. Let be a positive integer. Henceforth, will denote a nonempty set, and will denote the product space . Throughout this
paper, and will denote nonnegative integers and . Unless otherwise stated, “for all ” will mean “for all ”, and “for all ” will mean “for all ”. Let .
A metric on is a mapping satisfying, for all : (i) if, and only if, ;(ii).
From these properties, we can easily deduce that and for all . The last requirement is called the triangle inequality. If is a metric on , we say that is a metric space (for short, an MS).
Definition 1 (see [22]). A triple is called a partially ordered metric space if is a MS and is a partial order on .
Definition 2 (see [2]). An ordered MS is said to have the sequential -monotone property if it verifies(i)if is a nondecreasing sequence and , then for all ;(ii)if is a nonincreasing sequence and ,
then for all .
If is the identity mapping, then is said to have the sequential monotone property.
Henceforth, fix a partition of two non-empty subsets of ; that is, and We will denote If is a partially ordered space, , and , we will use the following notation:
Let and be two mappings.
Definition 3 (see [11]). One says that and are commuting if for all .
Definition 4 (see [11]). Let be a partially ordered space. One says that has the mixed -monotone property (with respect to ) if is -monotone nondecreasing in arguments of and -monotone nonincreasing
in arguments of ; that is, for all and all ,
Henceforth, let be mappings from into itself, and let be the -tuple .
Definition 5 (see [11]). A point is called a -coincidence point of the mappings and if If is the identity mapping on , then is called a -fixed point of the mapping .
Remark 6. If and are commuting and is a -coincidence point of and , then also is a -coincidence point of and .
Definition 7 (see [14]). A partial metric on is a mapping verifying, for all :;;;.
In this case, is a partial metric space (for short, a PMS).
Example 8 (see, e.g., [14]). Let , and define on by for all . Then, is a partial metric space.
Example 9 (see [14]). Let , and define = . Then, is a partial metric space.
Example 10 (see [14]). Let , and define by Then, is a partial metric space.
Example 11 (see, e.g., [23, 24]). Let and be a metric space and a partial metric space, respectively. Functions given by define partial metrics on , where is an arbitrary function and .
Obviously, if is a MS and we define , then is a PMS. Indeed, a partial metric on verifies(i); (ii); (iii),
but the condition does not necessarily hold. For a partial metric on , the mappings given by for all , are (usual) metrics on . On a PMS, the concepts of convergence, Cauchy sequences, completeness,
and continuity are defined as follows.
Definition 12 (see [14, 25, 26]). Let be a sequence on a PMS .(i)-converges to (and one will write ) if .(ii) is called -Cauchy if exists (and it is finite).(iii) is said to be -complete if every
-Cauchy sequence in -converges to a point such that .(iv)A mapping is said to be -continuous at if, for every , there exists such that .
We have used the previous notation because we need to distinguish between -convergence and -convergence on and usual convergence for real sequences.
Lemma 13 (see [14, 25, 26]). Let be a sequence on a PMS .(1) is -Cauchy if, and only if, it is -Cauchy.(2) if, and only if, and ; that is, (3) is complete if, and only if, the MS is complete.(4)If
and , then for all .
3. Auxiliary Results
We will use the following results about real sequences in the proof of our main theorems.
Lemma 14. Let be real lower bounded sequences such that . Then, there exists and a subsequence such that .
Proof. Let for all . As is convergent, it is bounded. As for all and , then every is bounded. As is a real bounded sequence, it has a convergent subsequence . Consider the subsequences ; that are
real bounded sequences and the sequence that also converges to . As is a real bounded sequence, it has a convergent subsequence . Then, the sequences , , also are real bounded sequences, , and .
Repeating this process times, we can find subsequences , , (where ) such that for all . And . But so , and there exists such that . Therefore, there exists and a subsequence such that .
Lemma 15. Let be a sequence of nonnegative real numbers which has not any subsequence converging to zero. Then, for all , there exist and such that for all .
Proof. Suppose that the conclusion is not true. Then, there exists such that, for all , there exists verifying . Let be such that . For all , take . Then, there exists verifying . Taking limit when ,
we deduce that . Then, has a subsequence converging to zero (maybe, reordering ), but this is a contradiction.
Lemma 16. If is a sequence in a MS that is not Cauchy, then there exist and two subsequences and such that, for all ,
Proof. We know that If this condition is not true, then Let . Then, there exists such that and . Let , and consider the numbers Since , between the previous numbers there exists a first nonnegative
integer such that but for all . In particular, .
Now, let . Then, there exists such that and . Let , and consider the numbers Since , between the previous numbers there exists a first nonnegative integer such that but for all . In particular, .
Repeating this process, we can find two subsequences and such that, for all :
Definition 17. Let be the family of all continuous, nondecreasing mappings such that if, and only if, .
These mappings are known as altering distance functions (see [27]). Note that every selected commutes with ; that is, for all .
Lemma 18. If and , then .
Proof. As there exists , then . If the conclusion is not true, there exists such that, for all , there exists verifying . This means that has a subsequence such that . As is nondecreasing, for all .
Therefore, has a subsequence lower bounded by , but this is impossible since .
With regards to coincidence points, it is possible to consider the following simplification. If is a permutation of , and we reorder (4), then we deduce that every coincidence point may be seen as a
coincidence point associated to the identity mapping on (see, for instance, [28]).
Lemma 19. Let be a permutation of , and let and = . Then, a point is a -coincidence point of the mappings and if, and only if, is a -coincidence point of the mappings and .
Therefore, in the sequel, without loss of generality, we will only consider -coincidence points where , that is, that verify for all . We also show some preliminary results on PMS.
Lemma 20. Let be a sequence on a PMS , and let .(1)If and , then , and for all .(2)If and , then .
Proof. Since and , then = . Therefore, = = , so . Since is continuous, then for all , and item 4 of Lemma 13 implies that .
Item 2 of Lemma 13 shows that .
Remark 21. Although the limit in a MS is unique, the -limit in a PMS is not necessarily unique. For instance, let as in Example 10. Then, is a complete PMS (see [14]). Consider for all . Then, but
whenever .
Definition 22. Let , let be a PMS, let be a mapping, and let . We will say that is -continuous at if, for all sequences on such that for all , for all and , we have that and . One will say that is -
continuous if it is continuous at every point .
Lemma 23. If is a PMS, and is -continuous at , then is -continuous at .
Proof. Let sequences on such that for all , for all , and . Item 1 of Lemma 20 implies that for all . Since is -continuous at , then . Item 2 of Lemma 13 assures us that and Then, is -continuous at .
4. Main Results
In the following result, we show sufficient conditions to ensure the existence of -coincidence points, where .
Theorem 24. Let be a complete PMS, and let a partial order on . Let be an -tuple of mappings from into itself verifying if and if . Let and be two mappings such that has the mixed -monotone property
on , and is -continuous and commuting with . Assume that there exist such that for which for all . Suppose either is -continuous or has the sequential -monotone property. If there exist verifying for
all , then and have, at least, one -coincidence point.
Proof. The proof is divided into seven steps. The first two steps are the same as in the proof of Theorem 9 in [11], since the contractivity condition does not play any role in these parts of the
Step1. There exist sequences such that = for all and all .
Step2. for all and all .
Step3. We claim that for all (i.e., ).
Indeed, define for all . As for all and all , then condition (17) implies that, for all and all :Therefore, for all , . This means that the sequence is nonincreasing and lower bounded. Hence, it is
convergent; that is, there exists such that . We are going to show that . Since Lemma 14 assures that there exist and a subsequence such that . Repeating (18), for all , Consider the sequence Suppose
that this sequence has no subsequence converging to zero. Using , Lemma 15 assures us that there exists and such that for all . It follows that Then, (20) says to us Taking limit in , we deduce that
, which is impossible. Therefore, the sequence in (21) must have a subsequence converging to zero. Since and are continuous, taking limit when in (20) using this subsequence, we deduce that , so .
Then, we have just proved that . Therefore, , and Lemma 18 assures that , which means that for all since for all and all .
Step4. for all (i.e., ). It is the same proof of Step3.
Since for all and , joining Steps 3 and 4, it follows that Step5. Every sequence is -Cauchy. We reason by contradiction. Suppose that are not -Cauchy () and are -Cauchy, being . By Lemma 16, for
all , there exists and subsequences and such that Now, let and . Since are -Cauchy, for all , there exists such that if , then . Since by Step4, there exists such that if , then . Define . If ,
then Therefore, we have proved that there exists such that if , then
Next, let such that . Let such that , and define . Consider the numbers until finding the first positive integer verifying Now let such that , and define . Consider the numbers until finding the
first positive integer verifying Repeating this process, we can find sequences such that, for all , Note that by (27), for all , so for all and all . Furthermore, for all , Therefore, for all and all
, Next, for all , let be an index such that Then, for all , Applying the contractivity condition (17), it follows, for all , Consider the sequence: If this sequence has a subsequence that converges
to zero, then we can take limit when in (36) using this subsequence, so that we would have , which is impossible since . Therefore, the sequence (37) has no subsequence converging to zero. In this
case, taking in Lemma 15, there exist and such that , for all . It follows that, for all , . Thus, by (36), Fix any and we are going to prove that . Indeed, by Step3 and (24), since are sequences
converging to zero, we can find such that if , then Therefore, (33) implies that, for all and for all such that , Then, (38) guarantees that . This means that for all . If we take (where ), we deduce
that for all . Since is continuous, we have that , which is impossible since . This contradiction finally proves that every sequence is -Cauchy.
Since is -complete, then is -complete (item 3 of Lemma 13). Then, there exist such that for all . Furthermore, = = = for all . Since is -continuous, then and for all . Item 1 of Lemma 20 shows that
for all . Therefore, for all , . Moreover, for all and all , .
Step6. Suppose that is -continuous. In this case, we know that and for all and which implies that and , for all . Item 1 of Lemma 20 assures us that, for all , Since the limit in a MS is unique, we
deduce that for all , so is a -coincidence point of and .
Step7. Suppose that has the sequential -monotone property. In this case, by Step2, we know that for all and all . This means that the sequence is monotone. As , we deduce that for all and all.
This condition implies that, for all and all , (the first case occurs when and the second one when ). Then, by (17), for all , | {"url":"http://www.hindawi.com/journals/aaa/2013/634371/","timestamp":"2014-04-16T13:46:14Z","content_type":null,"content_length":"1047843","record_id":"<urn:uuid:ec085dd5-1744-4f66-90e1-a766c043ce87>","cc-path":"CC-MAIN-2014-15/segments/1397609537308.32/warc/CC-MAIN-20140416005217-00371-ip-10-147-4-33.ec2.internal.warc.gz"} |
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[SOLVED] Diagonalizable
May 1st 2010, 09:59 AM #1
MHF Contributor
Mar 2010
[SOLVED] Diagonalizable
If A is nxn diagonalizable matrix, then each vector in $\mathbb{R}^n$ can be written as a linear comination of eigenvectors of A.
This is true but how do I prove this?
Since A is diagonalizable, there exist a diagonal matrix D and an invertible matrix P such that $A= PDP^{-1}$, $D= P^{-1}AP$
Suppose $\lambda$ is an eigenvalue of A. That means that there exist a non-zero vector v such that $Av= \lambda v$. Let $u= P^{-1}v$. Since P is invertible, u is non-zero also and $v= Pu$. But
now, $Du= P^{-1}APu= P^{-1}Av= \lambda P^{-1}v= \lambda u$.
That is, $\lambda$ is also an eigenvalue of the diagonal matrix D and, since a diagonal matix has its eigenvalues on its diagonal, is one of the numbers on the diagonal of D. Further the
eigenvectors of a diagonal matrix are simply the standard basis vectors <1, 0, 0...>, etc.
That means that set of all eigenvectors u, corresponding to the eigenvalues of D are independent and their number is equal to the dimension of the space.. Now, v= Pu and since P is invertible, it
maps independent sets of vectors into independent sets of vectors. Since their number is equal to the dimension of the space, it follow that they are a basis for the space.
May 2nd 2010, 04:51 AM #2
MHF Contributor
Apr 2005
May 2nd 2010, 08:44 AM #3
MHF Contributor
Mar 2010 | {"url":"http://mathhelpforum.com/advanced-algebra/142484-solved-diagonalizable.html","timestamp":"2014-04-20T12:33:50Z","content_type":null,"content_length":"38608","record_id":"<urn:uuid:1c1c9acf-95ea-4047-b143-e65adb6ac06c>","cc-path":"CC-MAIN-2014-15/segments/1397609538423.10/warc/CC-MAIN-20140416005218-00248-ip-10-147-4-33.ec2.internal.warc.gz"} |
Math Forum Discussions - Re: Teleprompter Code
Date: Mar 23, 2013 3:24 AM
Author: Christoph Lhotka
Subject: Re: Teleprompter Code
Dear Bruce,
I changed the end of your code into
fY[k_, dt_] := Module[{}, Pause[dt]; Y[[Mod[k, Length[Y]] + 1]]];
Animate[fY[k, dt], {k, 1, \[Infinity], 1}, AnimationRate -> 100,
ContentSize -> {700, 50}],
{dt, {0.05, 0.25, 0.5}}]
fY does now return the content with index k after dt seconds.
With this we can set AnimationRate to a high value, to
run the animation smoothly, and control the speed of the
'string stream' using dt.
Remark: Neither ListAnimate nor Animate do produce
smooth animations on my PC (Mathematica 9, Ubuntu 12.04),
with the available options. Without the workaround they
are not really useful for me.
PS: Here is the complete code:
X = X = StringJoin@ConstantArray[" ", 25] <>
"Let's define calculus terms using this sketch in which you're \
hiking over hills and valleys. You may want to pause this clip now \
in order to study the sketch and when you're ready, hit play. \t\t \
If the shape of the land is regarded as a function f(x), the sketch \
says that you start at a point whose x-coordinate is a and that you \
end at a point whose x-coordinate is b. \tWhile hiking, you'll \
always track the altitude--sometimes you're above sea level and \
sometimes you're not. (open new layer)\t\t\t At a blue point, you're \
at the hi-point for that part of the hike. (close/open layer)\t\t\
\t At a red point, you're at the lo-point for that part of the hike. \
(show both layers) \t\t\t The y-value of each blue point is called \
a maximum (WRITE THIS OUT in blue) and the y-value of each red point \
is called a minimum (WRITE THIS OUT in red). The blue points' \
y-values are collectively called the maxima (in BLUE) and likewise \
(in RED) minima for
the red points. If you're wondering why the green peak isn't \
blue, it's because the hike ends before you reach that peak,\t ";
L = StringLength@X;
subLength = 80;
Y = (y = StringTake[X, {#, Min[# + subLength, L]}];
Style[y, Blue, Bold, 12, FontFamily -> "Courier"]) & /@ Range@L;
fY[k_, dt_] := Module[{}, Pause[dt]; Y[[Mod[k, Length[Y]] + 1]]];
Animate[fY[k, dt], {k, 1, \[Infinity], 1}, AnimationRate -> 100,
ContentSize -> {700, 50}],
{dt, {0.05, 0.25, 0.5}}]
On 03/22/2013 09:18 AM, bruce.colletti@gmail.com wrote:
> The Washington DC-Area Mathematica Special Interest Group has been collaborating to build teleprompter code. The code below is the brainchild of Dan Martinez and we've been tinkering with it. We've also looked at ot her code, notably from the Great Guru himself, Harry Calkins.
> Here's my question about the code below: why does ListAnimate inexplicably pause sometimes? The text should run smoothly until the end.
> Tech Support says there's an issue with ListAnimate, one tied to AnimationRate: if it's "good", the passage will scroll smoothly. Otherwise not. I can get it to run smoothly when the rate is high--but then I can't follow the words.
> Can this code be fixed so that the text always runs smoothly for any rate? Thanks.
> Bruce
> X=StringJoin@ConstantArray[" ",25] <>"Let's define calculus terms using this sketch in which you're hiking over hills and valleys. You may want to pause this clip now in order to study the sketch and when you're ready, hit play. \t\t If the shape of the land is regarded as a function f(x), the sketch says that you start at a point whose x-coordinate is a and that you end at a point whose x-coordinate is b. \tWhile hiking, you'll always track the altitude--sometimes you're above sea level and sometimes you're not. (open new layer)\t\t\t At a blue point, you're at the hi-point for that part of the hike. (close/open layer)\t\t\t At a red point, you're at the lo-point for that part of the hike. (show both layers) \t\t\t The y-value of each blue point is called a maximum (WRITE THIS OUT in blue) and the y-value of each red point is called a minimum (WRITE THIS OUT in red). The blue points' y-values are collectively called the maxima (in BLUE) and likewise (in RED) minima for
> the red points. If you're wondering why the green peak isn't blue, it's because the hike ends before you reach that peak,\t ";
> L=StringLength@X;
> subLength=80;
> Clear@Y;
> Y=(y=StringTake[X,{#,Min[#+subLength,L]}];
> Style[y,Blue,Bold,12,FontFamily->"Courier"])&/@Range@L;
> ListAnimate[Y,AnimationRate->20,ContentSize->{700,50},AnimationRepetitions->\[Infinity]] | {"url":"http://mathforum.org/kb/plaintext.jspa?messageID=8730119","timestamp":"2014-04-19T00:12:56Z","content_type":null,"content_length":"5809","record_id":"<urn:uuid:549ed007-84f4-4b63-af88-290562f1d4c6>","cc-path":"CC-MAIN-2014-15/segments/1398223202774.3/warc/CC-MAIN-20140423032002-00427-ip-10-147-4-33.ec2.internal.warc.gz"} |
ML Aggarwal Solutions for ICSE students.
// July 11th, 2010 // M.L. Aggarwal
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Linear Transformations
Definition. Let V and W be vector spaces over a field F. A linear transformation
Note that u and v are vectors, whereas k is a scalar (number).
You can break the definition down into two pieces:
Conversely, it is clear that if these two equations are satisfied then f is a linear transformation.
The notation
Example. The function
is a function
Sometimes, the outputs are given variable names, e.g.
This is the same as saying
You plug numbers into f in the obvious way:
Since you can think of f as taking a 2-dimensional vector as its input and producing a 3-dimensional vector as its output, you could write
But I'll supress some of the angle brackets when there's no danger of confusion.
Example. Define
I'll show that f is a linear transformation the hard way. First, I need two 2-dimensional vectors:
I also need a real number k.
Notice that I don't pick specific numerical vectors like general vectors and numbers.
I must show that
I'll compute the left side and the right side, then show that they're equal. Here's the left side:
And here's the right side:
This was a pretty disgusting computation, and it would be a shame to have to go through this every time. I'll come up with a better way of recognizing linear transformations shortly.
Example. The function
is not a linear transformation from
To prove that a function is not a linear transformation --- unlike proving that it is --- you must come up with specific, numerical vectors u and v and a number k for which the defining equation is
false. There is often no systematic way to come up with such a counterexample; you simply try some numbers till you get what you want.
I'll take
Since not a linear transformation.
Example. Let differentiable at a point
In this definition, if
Since f produces outputs in
It turns out that the matrix of
This matrix is called the Jacobian matrix of f at a.
For example, suppose
The next lemma gives an easy way of constructing --- or recognizing --- linear transformations.
Theorem. Let F be a field, and let A be an
is a linear transformation.
Proof. This is pretty easy given the rules for matrix arithmetic. Let
Therefore, f is a linear transformation.
This result says that any function which is defined by matrix multiplication is a linear transformation. Later on, I'll show that for finite-dimensional vector spaces, any linear transformation can
be thought of as multiplication by a matrix.
Example. Define
I'll show that f is a linear transformation the easy way. Observe that
f is given by multiplication by a matrix of numbers, exactly as in the lemma. (
Lemma. Let F be a field, and let V and W be vector spaces over F. Suppose that
Proof. (a) Put
(b) I know that
The lemma gives a quick way of showing a function is not a linear transformation.
Example. Define
Since g does not take the zero vector to the zero vector, it is not a linear transformation.
Be careful! If can't conclude that f is a linear transformation. For example, I showed that the function does take the zero vector to the zero vector.
Next, I want to prove the result I mentioned earlier: Every linear transformation on a finite-dimensional vector space can be represented by matrix multiplication. I'll begin by reviewing some
Definition. The standard basis vectors for
I showed earlier that
Lemma. Every vector
Example. In
Now here is the converse to the Theorem I proved earlier.
Theorem. If
Proof. Regard the standard basis vectors
I claim that A is the matrix I want. To see this, take a vector
By the standard basis vector lemma above,
Then I can use the fact that f is a linear transformation --- so f of a sum is the sum of the f's, and constants (like the
On the other hand,
To get the last equality, think about how matrix multiplication works.
Linear transformations and matrices are not quite identical. If a linear transformation is like a person, then a matrix for the transformation is like a picture of the person --- the point being that
there can be many different pictures of the same person. You get different "pictures" of a linear transformation by changing coordinates --- something I'll discuss later.
Example. Define
I already know that f is a linear transformation, and I found its matrix by inspection. Here's how it would work using the theorem. I feed the standard basis vectors into f:
I construct a matrix with these vectors as the columns:
It's the same matrix as I found by inspection.
You can combine linear transformations to obtain other linear transformations. First, I'll consider sums and scalar multiplication.
Definition. Let
a. The sum
b. The product
Lemma. Let
Proof. I'll prove the first part by way of example and leave the proof of the second part to you.
If composite of f and g is
Note that things go from left to right in the picture, but that they go right to left in "
Lemma. Let
Proof. Let
Now suppose
f and g can be represented by matrices; I'll use
What's the matrix for the composite
The matrix for
Example. Suppose
are linear transformations. Thus,
Write them in matrix form:
The matrix for the composite transformation
To write it out in equation form, multiply:
That is,
Example. The idea of composing transformations can be extended to affine transformations. For the sake of this example, you can think of an affine transformation as a linear transformation plus a
translation (a constant). This provides a powerful way of doing various geometric constructions.
For example, I wanted to write a program to generate self-similar fractals. There are many self-similar fractals, but I was interested in those that can be constructed in the following way. Start
with an initiator, which is a collection of segments in the plane. For example, this initiator is a square:
Next, I need a generator. It's a collection of segments which start at the point
The construction proceeds in stages. Start with the initiator and replace each segment with a scaled copy of the generator. (There is an issue here of which way you "flip" the generator when you copy
it, but I'll ignore this for simplicity.) Here's what you get by replacing the segments of the square with copies of the square hump:
Now keep going. Take the current figure and replace all of its segments with copies of the generator. And so on. Here's what you get after around 4 steps:
Roughly, self-similarity means that if you enlarge a piece of the figure, the enlarged piece looks like the original. If you imagine carrying out infinitely many steps of the construction above,
you'd get a figure which would look "the same" no matter how much you enlarged it --- which is a very crude definition of a fractal. If you're interested in this stuff, you should look at Benoit
Mandelbrot's classic book ([1]) --- it has great pictures!
What does this have to do with transformations? The idea is that to replace a segment of the current figure with a scaled copy of the generator, you need to stretch the generator, rotate it, then
translate it. Here's a picture with a different initiator and generator:
Stretching by a factor of k amounts to multiplying by the matrix
This is a linear transformation.
Rotation through an angle
This is also a linear transformation.
Finally, translation by a vector
Thus, to stretch by k, rotate through
By thinking of the operations as linear or affine transformations, it is very easy to write down the formula.
If inverse of f is a linear transformation
That is,
Lemma. If A and B are
then A and B are inverses --- that is,
Proof. Since
Put the vectors in the equations above into the columns of a matrix. They form the identity:
The same reasoning shows
Proposition. If
Remark. If I use
Proof. Let A be the matrix of f and let B be the matrix of
Of course, a linear transformation may not be invertible, and the last result gives an easy test --- a linear transformation is invertible if and only if its matrix is invertible. You should know
almost half a dozen conditions which are equivalent to the invertibility of a matrix.
Example. Suppose
The matrix of f is
Therefore, the matrix of
Hence, the inverse transformation is
So, for example,
You can check that
Example. Let
Here are some elements of
You can represent polynomials in
I'm writing the coefficients with the powers increasing to make it easy to extend this to higher powers. For example,
Returning to
Do you see what it is?
If I switch back to polynomial notation, it is
Of course, D is just differentiation.
Now you know from calculus that:
• The derivative of a sum is the sum of the derivatives.
• Constants can be pulled out of derivatives.
This means that D is a linear transformation
To find the matrix of a linear transformation (relative to the standard basis), apply the transformation to the standard basis vectors. Use the results as the columns of your matrix.
In vector form,
As I apply D, I'll translate to polynomial notation to make it easier for you to follow:
Therefore, the matrix of D is
Example. Construct a linear transformation which maps the unit square
The idea is to send vectors for the square's sides --- namely,
The following matrix does it:
Therefore, the transformation is
1. Benoit Mandelbrot, The fractal geometry of nature. New York: W. H. Freeman and Company, 1983. [ISBN 0-7167-1186-9]
Send comments about this page to: Bruce.Ikenaga@millersville.edu.
Copyright 2008 by Bruce Ikenaga | {"url":"http://www.millersville.edu/~bikenaga/linear-algebra/lintrans/lintrans.html","timestamp":"2014-04-21T09:43:28Z","content_type":null,"content_length":"48200","record_id":"<urn:uuid:690d3123-946d-4dad-b983-d784948c603c>","cc-path":"CC-MAIN-2014-15/segments/1397609539705.42/warc/CC-MAIN-20140416005219-00535-ip-10-147-4-33.ec2.internal.warc.gz"} |
Symmetric products of smooth non-proper curves over generalized Jacobians
up vote 7 down vote favorite
Does anyone know a written reference for the following fact?
For large n, $\operatorname{Sym}^n X \to \operatorname{Jac}^nX$ is a vector bundle, where $X$ is a smooth, non-proper curve, and $\operatorname{Jac}X$ is its generalized Jacobian, so $\operatorname
{Jac}^nX = \operatorname{Pic}^n X^+$ where $X^+$ is the one-point compactification given by the quotient of the smooth compactification $Xc$ by $Xc -X$.
(I know how to prove it-- I would like to be able to cite a reference for it.)
add comment
1 Answer
active oldest votes
I think I should have said "affine bundle" instead of "vector bundle." I still haven't found a reference, but I wrote a proof in the appendix of:
up vote 4 down vote http://www.math.harvard.edu/~kwickelg/papers/delta2real.pdf
add comment
Not the answer you're looking for? Browse other questions tagged ag.algebraic-geometry or ask your own question. | {"url":"http://mathoverflow.net/questions/116499/symmetric-products-of-smooth-non-proper-curves-over-generalized-jacobians/117153","timestamp":"2014-04-21T09:51:37Z","content_type":null,"content_length":"49393","record_id":"<urn:uuid:5e2be5f5-b2c7-4082-99bb-398e42ae7705>","cc-path":"CC-MAIN-2014-15/segments/1397609539705.42/warc/CC-MAIN-20140416005219-00123-ip-10-147-4-33.ec2.internal.warc.gz"} |
How to align equation to the last line
On 3-Feb-2006, "Jay Freedman" wrote:
I don't think there's anything you can do in a style, but you can ask
Design Science (
), the
company that supplies the Equation Editor to Microsoft. You should
also look at their commercial product MathType, which (among
many other enhancements) can make its own equation numbers.
Niwrad, Jay's correct. His solution of using the table is the only way I
know of to do it in Word. ("It" being aligning the number and the equation
as you described.) Now if you're making web pages out of the Word document,
that's another story all together, but I don't think that's the case.
Jay mentioned MathType. He's right -- MathType will do the numbering &
referencing automatically. Even using MathType though, the only way I know
of to align the bottom of the equation number with the bottom of the
equation is to use a table.
Bob Mathews bobm at dessci.com
Director of Training
FREE fully-functional 30-day evaluation of MathType 5
Design Science, Inc. -- "How Science Communicates"
MathType, WebEQ, MathPlayer, MathFlow, Equation Editor, TeXaide
niwrad wrote:
"Jay Freedman" wrote:
On Fri, 3 Feb 2006 03:26:19 -0800, niwrad
I have numbered the equations on its right in my Word document and
they are aligned to the center when the equations have more than
one line. I would like to align the equation numbers to the last
line of the equation (or the bottom of the object) and wonder how I
can do so.
The most reliable method is to use a two-column borderless table,
with the equations in the left (wide) column and the equation
numbers in the right (narrow) column. Select the right column, right
click it, select Cell Alignment, and choose the bottom right
Jay Freedman
Microsoft Word MVP FAQ:
Email cannot be acknowledged; please post all follow-ups to the
newsgroup so all may benefit.
Thanks for the response.
I have created a particular formatting style to automatically number
the equations and align them horizontally and it's been very useful
other than for this stated issue. So I'm wondering if it's possible
to maintain this and just alter some formatting setting such that the
equation numbers on the right can be align to the bottom of the
equation object.
I don't think there's anything you can do in a style, but you can ask
Science (
), the company that supplies
Equation Editor to Microsoft. You should also look at their commercial
product MathType, which (among many other enhancements) can make its own
equation numbers.
Jay Freedman
Microsoft Word MVP FAQ:
Email cannot be acknowledged; please post all follow-ups to the newsgroup
so all may benefit. | {"url":"http://www.wordbanter.com/showthread.php?t=43151","timestamp":"2014-04-20T13:19:37Z","content_type":null,"content_length":"64488","record_id":"<urn:uuid:6e3c9c7a-9105-44e8-8657-38b55e4af0dd>","cc-path":"CC-MAIN-2014-15/segments/1398223203422.8/warc/CC-MAIN-20140423032003-00199-ip-10-147-4-33.ec2.internal.warc.gz"} |
NotInInterval Procedure
8.4.9 NotInInterval Procedure
The NotInInterval procedure provides a way to check whether or not a value lies outside of a specified interval. The interval is considered to be a closed interval that includes the end-points. For
an array-valued X, all the values must satisfy the condition for the result to be true.
Note that I would rather call this routine ``Not_In_Interval'', but the F90 standard does not allow underscores in defined operator names (i.e. ``.Not_In_Interval.'' is not allowed).
Calling syntax:
Logical = X .NotInInterval. (/Int1, Int2/) ,
Logical = X .NotInInterval. Interval or
Logical = NotInInterval(X, Interval)
Input variables:
X Input integer or real, scalar or array variable.
Interval A vector of length 2 that specifies the extents of the interval to be checked.
(/Int1, Int2/) A means of expressing an interval without declaring a vector.
Output variable:
NotInInterval Logical which is true if X is in the closed interval, which includes the end-points.
The NotInInterval code listing contains additional documentation.
Michael L. Hall | {"url":"http://www.lanl.gov/Caesar/node72.html","timestamp":"2014-04-17T12:29:31Z","content_type":null,"content_length":"6138","record_id":"<urn:uuid:7d365c44-0c84-4cd3-8bfb-9351600954d3>","cc-path":"CC-MAIN-2014-15/segments/1398223201753.19/warc/CC-MAIN-20140423032001-00238-ip-10-147-4-33.ec2.internal.warc.gz"} |
Triple integration: f(x,t,z) = (2x - 1)yz^2, S bounded by...
I am having difficulty setting up triple integrals. Perhaps an illustration would help.
Suppose f(x,t,z) = (2x - 1)yz^2, S is the solid in the first octant bounded by x + y = 1 and z^2 = x^2 = y^2.
I really don't understand how to set this up. Could anyone explain it to me? Thanks for any help you can be.
Re: Triple integration: f(x,t,z) = (2x - 1)yz^2, S bounded by...
redrick wrote:I am having difficulty setting up triple integrals. Perhaps an illustration would help.
Suppose f(x,t,z) = (2x - 1)yz^2, S is the solid in the first octant bounded by x + y = 1 and z^2 = x^2 = y^2.
I really don't understand how to set this up. Could anyone explain it to me? Thanks for any help you can be.
you obviously have at least one typo in the above post...I would guess 3 typos.
Re: Triple integration: f(x,t,z) = (2x - 1)yz^2, S bounded by...
You are right Martingale, there were 2 mistakes.
I am having difficulty setting up triple integrals. Perhaps an illustration would help.
Suppose f(x,y,z) = (2x - 1)yz^2, S is the solid in the first octant bounded by x + y = 1 and z^2 = x^2 + y^2.
I think when I am freaked out a bit,
Re: Triple integration: f(x,t,z) = (2x - 1)yz^2, S bounded by...
redrick wrote:You are right Martingale, there were 2 mistakes.
I am having difficulty setting up triple integrals. Perhaps an illustration would help.
Suppose f(x,y,z) = (2x - 1)yz^2, S is the solid in the first octant bounded by x + y = 1 and z^2 = x^2 + y^2.
I think when I am freaked out a bit,
looks like...
Re: Triple integration: f(x,t,z) = (2x - 1)yz^2, S bounded by...
Martingale wrote:
looks like...
I am not sure what you are saying at all. x + y = 1 means that x = 1 - y and y = 1-x
and z^2 = x^2 + y^2 means that z = (x^2 + y^2)^1/2
I don't understand where you are getting the less than symbols and what that has to do with the integration process?
Re: Triple integration: f(x,t,z) = (2x - 1)yz^2, S bounded by...
redrick wrote:
Martingale wrote:
looks like...
I am not sure what you are saying at all. x + y = 1 means that x = 1 - y and y = 1-x
and z^2 = x^2 + y^2 means that z = (x^2 + y^2)^1/2
I don't understand where you are getting the less than symbols and what that has to do with the integration process?
you are not integrating on the boundary. so if you are integrating over the region that is in the 1st octant and is bounded by the curves
then $z=\sqrt{x^2+y^2}$
is the largest $z$ can be and 0 is the smallest...therefore $0\leq z\leq \sqrt{x^2+y^2}$ ...and so on... | {"url":"http://www.purplemath.com/learning/viewtopic.php?p=1080","timestamp":"2014-04-18T21:57:53Z","content_type":null,"content_length":"28683","record_id":"<urn:uuid:e5752980-2481-47e3-8403-394bfe787ba1>","cc-path":"CC-MAIN-2014-15/segments/1398223205375.6/warc/CC-MAIN-20140423032005-00608-ip-10-147-4-33.ec2.internal.warc.gz"} |
Finding minima of a complex function
July 20th 2010, 04:29 AM #1
Jan 2010
Finding minima of a complex function
Hello all,
I have a complex function
How can I prove that it has a minimum at certain point.
Can anyone give me some suggestions for finding the minimum.
I have a number of questions:
1. Have you tried the usual Calc I method?
2. What is D(s)?
3. What is d(s)?
4. What are any constraints on any of the parameters, especially the domain for x, and the allowed values of m? (We can see at a glance that 1 is not in the domain.)
5. What does it mean to sum over D(s)? Do you mean there's a dummy variable running over the values in D(s)?
1.I have differentiated the equation.
But i need not get the extreme point.I need to prove that there exists a global minimum for the function.
2.D(s) is just a set representation.there are no variables in D(s)
3.d(s) for particular set of node.
4.m > 0 and $P_w > 0$
5.the value $\frac{1}{1-P_w}$ sums for all the nodes in D(s)
Your answers are a bit mystifying.
1. In the OP, you asked how you can prove the function has a minimum at a certain point (this means you're talking about a local minimum). In your latest post in this thread, you're asking how
you can show that the function has a global minimum. Which is it?
2. Are you saying that the summation variable for $D(s)$ does not show up in your expression anywhere? If it doesn't, then you could simply remove the summation symbol and replace it with
multiplication by the cardinality of $D(s)$. The same thing goes for the product over $d(s)$: if nothing in the expression depends on $r$ explicitly, then you can replace the product with
exponentiation to the power of the cardinality of $d(s)$ thus:
Is it your opinion that this expression is equivalent to your original expression? Here, I've used $|D(s)|$ for the cardinality of $D(s)$, and $|d(s)|$ for the cardinality of $d(s)$.
3. What are "nodes" in this context?
4. What is $P_{w}$? There is no appearance of that symbol in the original expression. How does it relate to your original expression?
5. What's the difference, if any, between $D(s)$ and $d(s)$?
6. Is this function given to you, or have you derived it as part of a bigger problem? If the latter, could you please type up that problem, so I can have more context? Thanks!
7. A comment: the $1/(1-x)$ factors are very troubling for proving there is a global minimum. If $m$ is allowed to be odd, then it seems to me that
$\displaystyle{\lim_{x\to 1^{+}}F(x)=-\infty},$
which would prove that there is no global minimum, unless $-\infty$ is an acceptable answer. On the other hand, if $m$ is only allowed to be even, then you still have a chance of proving that
there is a finite global minimum.
8. What is $a$?
July 20th 2010, 04:38 AM #2
July 21st 2010, 08:38 PM #3
Jan 2010
July 22nd 2010, 02:00 AM #4 | {"url":"http://mathhelpforum.com/advanced-applied-math/151455-finding-minima-complex-function.html","timestamp":"2014-04-16T16:10:28Z","content_type":null,"content_length":"45309","record_id":"<urn:uuid:03db123d-cbef-4fed-8383-e48cd910cc47>","cc-path":"CC-MAIN-2014-15/segments/1397609535745.0/warc/CC-MAIN-20140416005215-00443-ip-10-147-4-33.ec2.internal.warc.gz"} |
Edge Detection
Sobel Filtered Common Edges: Jim
Sobel Filtered Common Edges: Roger
We see that although it does do better for some features (ie. the nose), it still suffers from mismapping some of the lines. A morph constructed using individually selected points would still work
better. It should also be noted that this method suffers the same drawbacks as the previous page; difficulties due to large contrast between images and the inability to handle large translations of
features. Another method of detecting edges is using wavelets. Specifically a two-dimensional Haar wavelet transform of the image produces essentially edge maps of the vertical, horizontal, and
diagonal edges in an image. This can be seen in the figure of the transform below, and the following figure where we have combined them to see the edges of the entire face.
Haar Wavelet Transformed Image
Edge Images Generated from the Haar Wavelet Transform
And here are the maps of common control points generated by the feature extraction algorithm for the Jim-Roger morph.
Haar Filtered Common Edges: Jim
Haar Filtered Common Edges: Roger
Although the Haar filter is nearly equivalent to the gradient and Laplacian edge detection methods, it does offer the ability to easily extend our edge detection to multiscales as demonstrated in
this figure. | {"url":"http://www.owlnet.rice.edu/~elec539/Projects97/morphjrks/moredge.html","timestamp":"2014-04-18T00:13:08Z","content_type":null,"content_length":"5033","record_id":"<urn:uuid:526e62e7-fd46-4d34-8a4b-a0487a073622>","cc-path":"CC-MAIN-2014-15/segments/1398223206118.10/warc/CC-MAIN-20140423032006-00364-ip-10-147-4-33.ec2.internal.warc.gz"} |
Polylogarithmic Approximation for Edit Distance and the Asymmetric Query Complexity
The 51st Annual IEEE Symposium on Foundations of Computer Science (FOCS 2010).
We present a near-linear time algorithm that approximates the edit distance between two strings within a polylogarithmic factor; specifically, for strings of length $n$ and every fixed $\epsilon>0$,
it can compute a $(\log n)^{O(1/\epsilon)}$-approximation in $n^{1+\epsilon}$ time. This is an exponential improvement over the previously known factor, $2^{\tilde O(\sqrt{\log n})}$, with a
comparable running time [Ostrovsky, Rabani J.ACM 2007; Andoni, Onak STOC 2009]. Previously, no efficient polylogarithmic approximation algorithm was known for any computational task involving edit
distance (e.g., nearest neighbor search or sketching).
This result arises naturally in the study of a new asymmetric query model. In this model, the input consists of two strings $x$ and $y$, and an algorithm can access $y$ in an unrestricted manner,
while being charged for querying every symbol of $x$. Indeed, we obtain our main result by designing an algorithm that makes a small number of queries in this model. We then provide also a
nearly-matching lower bound on the number of queries.
Our lower bound is the first to expose hardness of edit distance stemming from the input strings being “repetitive”, which means that many of their substrings are approximately identical.
Consequently, our lower bound provides the first rigorous separation between edit distance and Ulam distance, which is edit distance on non-repetitive strings, i.e., permutations. | {"url":"http://onak.pl/papers/focs_2010-approximating_edit_distance.html","timestamp":"2014-04-18T13:15:30Z","content_type":null,"content_length":"4633","record_id":"<urn:uuid:206826fb-4d6a-4c81-b4d5-c54554652b1a>","cc-path":"CC-MAIN-2014-15/segments/1398223211700.16/warc/CC-MAIN-20140423032011-00136-ip-10-147-4-33.ec2.internal.warc.gz"} |
Types Of Differential Equations And Definitions
Odinary Differential Equations
An Ordinary Differential Equation is a differential equation that depends on only one independent varialble.
For example
Partial Differential Equations
A Partial Differential Equation is differential equation in which the dependent varialble depends on two or more independent variables.
For example
The Laplace's equation
Order of a Differential Equation
The order of a differential is the order of the highest derivative entering the equation.
For example
The equation
Linear Differential Equation
A first-order differential equation is linear if it can be written in the form
For example
Nonlinear Differential Equation
It is a differential equation whose right hand side is not a linear function of the dependent variable.
For example
Homogeneous Differential Equation
A linear first-order differential equation is homogeneous if its right hand side is zero , that is
For example
Nonhomogeneous Differential Equation
A linear first-order differential equation is nonhomogeneous if its right-hand side is non-zero that is
For example | {"url":"http://people.cs.uct.ac.za/~kmaposa/DE/types.html","timestamp":"2014-04-17T01:51:12Z","content_type":null,"content_length":"3719","record_id":"<urn:uuid:50c3b163-6fc3-418f-b8e6-bb68bf3342e6>","cc-path":"CC-MAIN-2014-15/segments/1398223202548.14/warc/CC-MAIN-20140423032002-00497-ip-10-147-4-33.ec2.internal.warc.gz"} |
Posts about surprising on The Math Less Traveled
Tag Archives: surprising
Patrick Vennebush of Math Jokes 4 Mathy Folks recently wrote about the following procedure that yields surprising results. Choose some positive integer . Now, starting with consecutive integers,
raise each integer to the th power. Then take pairwise differences by … Continue reading | {"url":"http://mathlesstraveled.com/tag/surprising/","timestamp":"2014-04-18T23:20:09Z","content_type":null,"content_length":"47886","record_id":"<urn:uuid:782060e0-5d8e-4254-866a-b848d35dc8d9>","cc-path":"CC-MAIN-2014-15/segments/1397609535535.6/warc/CC-MAIN-20140416005215-00059-ip-10-147-4-33.ec2.internal.warc.gz"} |
mathematics: slope poem
Author Message
nljuyth Posted: Friday 29th of Dec 08:58
Hi , I have been trying to solve equations related to mathematics: slope poem but I don’t seem to be getting any success. Does any one know about pointers that might aid me?
AllejHat Posted: Friday 29th of Dec 16:31
I don’t think I know of any website where you can get your answers of mathematics: slope poem checked within hours. There however are a couple of websites which do offer help , but one
has to wait for at least 24 hours before expecting any reply .What I know for sure is that, this program called Algebrator, that I used during my college career was really good and I was
quite satisfied with it. It almost gives the type of solutions you need.
cufBlui Posted: Saturday 30th of Dec 09:44
I would just add a note to what has been posted above. Algebrator no doubt is the most useful tool one could have. Always use it as a guide and a means to learn and never to copy .
janceld Posted: Sunday 31st of Dec 09:00
Amazing! This sounds extremely useful to me. I was looking for such tool only. Please let me know where I can purchase this software from?
From: KÖLN,
sxAoc Posted: Sunday 31st of Dec 20:26
Accessing the program is effortless. All you want to know about it is accessible at http://www.algebra-help.org/solving-equations-with-log-terms-and-other-terms.html. You are assured
satisfaction. And besides , there is a money-back guarantee. Hope this is the end of your hunt.
Matdhejs Posted: Monday 01st of Jan 11:07
I remember having often faced difficulties with quadratic formula, angle complements and function domain. A truly great piece of math program is Algebrator software. By simply typing in a
problem from workbook a step by step solution would appear by a click on Solve. I have used it through many algebra classes – Basic Math, College Algebra and College Algebra. I greatly
recommend the program.
From: The | {"url":"http://www.algebra-help.org/basic-algebra-help/distance-of-points/mathematics-slope-poem.html","timestamp":"2014-04-18T18:11:26Z","content_type":null,"content_length":"50791","record_id":"<urn:uuid:acc1e7d0-4e34-479a-868a-9996f00396a3>","cc-path":"CC-MAIN-2014-15/segments/1397609535095.7/warc/CC-MAIN-20140416005215-00170-ip-10-147-4-33.ec2.internal.warc.gz"} |
[Tutor] math question
Alan Gauld alan.gauld at blueyonder.co.uk
Fri Apr 23 16:41:34 EDT 2004
> tutorial. I understand that the sqrt(2) can't be written as an exact
> value. But the sqrt(2)^2 certainly can be written in an exact value.
No it can't. The mathematically correct result can be but not
the evaluation of the expression. Thats because to evaluate
the expression Python must work out each factor.
In other words, Python knows how to do arithmetic it doesn't
know how to do math.
> seems strange to me (I'm no mathematician and I suspect
> this is a philosophy question rather than a Python question)
> that the language can't figure this out
Python is just a program like any other. Sit down and try to
figure out how you would write that kind of knowledge into
a python program. Then try to figure out how to write it
such that it didn't slow down arith,metic operations to a crawl.
> should see the code and know a difference between adding
> .33 + .33 + .333 and 1/3 + 1/3 + 1/3
The only way the computer can do that is to parse the values,
apply some expert system rules:
If the inverse of a number, x, is added to itself x times
then the result is x.
Now try writing a program to implement that one rule.
Now extend it to cover:
A) If a number, x, has its square root multiplied by its square
root, anywhere within an expression, substitute x for the result.
B) If the square root of a number, x, is raised to the power 2 return
Check your solution caters for
sqrt(2) * sqrt(2) => 2
sqrt(2) * 3 * sqrt(2) => 2*3 = 6
Now extend it for the zillions of other "obvious" mathematical
rules, (and don't forget to keep the performance real snappy
for number crunching apps)
You'll find it is much easier (and faster) to just evaluate
each term as you come to it (taking account of precedence
and parentheses etc) and produce a final result that way.
> hard time explaining to my daughter when I was helping her
> with a math/geometry problem and showed her how to use the
> Python shell as a kind of calculator
This is not a Python issue its a binary storage issue...
But if you *print* the results (which uses the str() function)
instead of relying on the Python repr() function you should
find it all looks ok:
>>> from math import sqrt
>>> print sqrt(2) * sqrt(2)
>>> sqrt(2) * sqrt(2)
See the difference?
(This is one reason I use print in all the examples in my online
tutor despite the extra typing - it mostly avoids revealing these
nasty issues early on!)
Alan G
Author of the Learn to Program web tutor
More information about the Tutor mailing list | {"url":"https://mail.python.org/pipermail/tutor/2004-April/029373.html","timestamp":"2014-04-19T06:19:34Z","content_type":null,"content_length":"4996","record_id":"<urn:uuid:e0c4299e-b9ae-407b-8e27-02678b5129f0>","cc-path":"CC-MAIN-2014-15/segments/1398223206120.9/warc/CC-MAIN-20140423032006-00472-ip-10-147-4-33.ec2.internal.warc.gz"} |
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Biot–Savart law
From Citizendium, the Citizens' Compendium
In physics, more particularly in electrodynamics, the law first formulated by Jean-Baptiste Biot and Félix Savart describes the magnetic induction B (proportional to the magnetic field H) caused by a
direct electric current in a wire. Biot and Savart interpreted their measurements by an integral relation. Laplace gave a differential form of their result, which now often is also referred to as the
Biot-Savart law, or sometimes as the Biot-Savart-Laplace law. By integrating Laplace's equation over an infinitely long wire, the original integral form of Biot and Savart is obtained.
Ørsted's discovery
The Danish physicist Ørsted noticed in April 1820, while experimenting with the Voltaic pile, an effect of an electric current on a permanent magnet. He wrote a Latin publication which he sent round
Europe. François Arago demonstrated the discovery at the French Académie des Sciences (11 September 1820), which inspired Jean-Baptiste Biot and André-Marie Ampère to investigate the effect further.
At a meeting of the Académie des Sciences on 30 October 1820, Biot and Savart announced that the magnetic force exerted by a long conductor on a magnetic pole falls off with the reciprocal of the
distance and is orientated perpendicular to the wire (la force qui sollicite la molécule est perpendiculaire à cette ligne et à l'axe de fil. Son intensité est réciproque à la simple distance).
Simultaneously they published this in a short note^[1]. Later Laplace gave an improved and generalized mathematical formulation of their result.
Laplace's formula
The infinitesimal magnetic induction $\scriptstyle d\vec{\mathbf{B}}$ at point $\scriptstyle \vec{\mathbf{r}}$ (see figure on the right) is given by the following formula due to Laplace,
$d\vec{\mathbf{B}} = k \frac{i d\vec{\mathbf{s}} \times \vec{\mathbf{r}}} {|\vec{\mathbf{r}}|^3},$
where the magnetic induction is given as a vector product, i.e., is perpendicular to the plane spanned by $\scriptstyle d\vec{\mathbf{s}}$ and $\scriptstyle \vec{\mathbf{r}}$. The electric current i
is constant in time. The piece of the wire contributing to the magnetic induction can be seen as a vector of infinitesimal length ds and with direction tangent to the wire. The constant k depends on
the units chosen. In SI units k is the magnetic constant (vacuum permeability) divided by 4π. The magnetic constant μ[0] = 4π ×10^−7 N/A^2 (newton divided by ampere squared). In Gaussian units k = 1/
c (one over the speed of light).
If we remember the fact that the vector $\scriptstyle \vec{\mathbf{r}}$ has dimension length, we see that this equation is an Inverse-square law.
Infinite straight conductor
Take a straight, infinitely long, conducting wire transporting the current i. Historically, this is the case considered by Biot and Savart. The magnetic field H at a distance R from the wire is (in
SI units)
$\qquad H = \frac{i}{ 2\pi R},$
where in vacuum the magnetic field and the magnetic induction are related by B = μ[0]H.
In order to prove the expression for H, we write, using R = rsinα (see the figure),
$d\vec{\mathbf{s}} \times \vec{\mathbf{r}} = \hat{\mathbf{e}} \,r\sin\alpha\, ds = \hat{\mathbf{e}}\, R\,ds,$
where $\scriptstyle \hat{\mathbf{e}}$ is a unit vector along $\scriptstyle \mathbf{B}$, which is perpendicular to the plane spanned by the wire and the vector $\scriptstyle \vec{\mathbf{R}}$
perpendicular to the wire. Note that if $\scriptstyle d\vec{\mathbf{s}}$ is moved along the wire, all contributions from the segments to the magnetic induction are along this unit vector. Hence, if
we integrate over the wire we add up all these contributions, so that
$\vec{\mathbf{B}} = \hat{\mathbf{e}} \,i R k \int_{-\infty}^{\infty} \frac{ds}{(s^2+R^2)^{3/2}}$
where, by the Pythagorean theorem,
$|\vec{\mathbf{r}}|^2 = s^2 + R^2.$
Substitution of y = s / R and y = cotφ = cosφ / sinφ, successively, gives finally,
$|\vec{\mathbf{B}}| = \frac{ik}{R} \int_{-\infty}^{\infty} \frac{dy}{(y^2+1)^{3/2}} = \frac{ik}{R} \int_{0}^{\pi} \sin\phi \, d\phi = \frac{2 ik}{R},$
where i is the current and R is the distance of the point of observation to the wire (length of vector $\scriptstyle \vec{\mathbf{R}}$).
The constant k depends on the choice of electromagnetic units and is μ[0]/4π = 10^−7 henry/m in rationalized SI units. In Gaussian units k = 1/c. This equation gives the original formulation of Biot
and Savart.
The SI dimension of B is T (tesla). A field of 1 T corresponds to 10000 G (gauss).
Magnetic field on axis of circular current
The Biot-Savart expression can be integrated in
cylinder coordinates
for a point
on the
-axis. This gives (see figure on the right) a vector parallel to the
-axis of magnitude
$B = 2\pi k \frac{i r^2}{(R^2+r^2)^{3/2}}.$
Note that the direction of the electric current i and the magnetic induction B are related according to the right-hand screw rule: the screw is driven into the direction of B and rotated along i. The
factor k is μ[0]/4π for SI units and 1/c for Gaussian units.
Generalized Biot–Savart–Laplace law
From hereon vectors are indicated by bold letters, arrows on top are omitted.
In the above we wrote i for the current, which is equal to the current density J times the cross section A of the wire. If the current density J is not constant over the cross section, i.e., J = J(r'
), we must use a surface integral over the cross section A. Rather than introducing a surface element, we multiply immediately by ds and obtain an infinitesimal volume element $\scriptstyle d\mathbf
{r}' \equiv dV$,
$i d\mathbf{s} \rightarrow \iint_{A} \mathbf{J}(\mathbf{r}') d\mathbf{r}',$
where we defined the current density J as a vector parallel to the line segment ds and of magnitude J(r'). The volume element has height |ds| and base an infinitesimal surface element of the cross
section A. The B-field at point r, due to a volume V = As of the current becomes,
$\mathbf{B}(\mathbf{r}) = k \iiint_{V} \frac{\mathbf{J}(\mathbf{r}') \times (\mathbf{r}-\mathbf{r'})} {|\mathbf{r}-\mathbf{r'}|^3}d\mathbf{r}' .$
Note that:
$\frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3} = -\boldsymbol{abla} \left( \frac{1}{|\mathbf{r}-\mathbf{r'}|} \right),$
where we choose the nabla operator, the gradient, to act on the unprimed coordinates and hence it may be moved outside the integral, giving the following generalized form of the Biot-Savart-Laplace
law for the magnetic induction at point r:
$\mathbf{B}(\mathbf{r}) = k \boldsymbol{abla} \times \iiint_{V} \frac{\mathbf{J}(\mathbf{r}')} {|\mathbf{r}-\mathbf{r'}|}d\mathbf{r}', \qquad\qquad\qquad\qquad\qquad\qquad\qquad(1).$
Here $\scriptstyle \boldsymbol{abla} \times$ is the curl and V is a finite volume of the current generating the B-field. The total B-field is obtained by having V encompass all current (integrating
over all current).
Consistency with Maxwell equations
The expression for B given in Eq. (1) is a solution of the Maxwell equations with zero displacement current. This is of interest as it confirms the generally accepted notion that the Maxwell
equations form a set of postulates for classical electrodynamics. All electrodynamic results, including Biot-Savart's law, must be derivable from them. We will show that the Biot-Savart-Laplace law
can indeed be seen as a consequence of the Maxwell postulates, although Biot and Savart made their discovery some forty-five years before Maxwell.
The Helmholtz decomposition of B(r) is,
$\mathbf{B}(\mathbf{r}) = \mathbf{B}_\perp(\mathbf{r}) + \mathbf{B}_\parallel(\mathbf{r})$
\begin{align} \mathbf{B}_\perp(\mathbf{r}) &= \frac{1}{4\pi}\boldsymbol{abla} \times \iiint \frac{\boldsymbol{abla}'\times \mathbf{B}(\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|} d\mathbf{r}' \\ \
mathbf{B}_\parallel(\mathbf{r}) &= -\frac{1}{4\pi}\boldsymbol{abla} \iiint \frac{\boldsymbol{abla}'\cdot \mathbf{B}(\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|} d\mathbf{r}' \\ \end{align}
are, respectively, the perpendicular (transverse, divergence-free) and parallel (longitudinal, curl-free) components. The operator ∇ acts on unprimed coordinates and ∇' acts on primed coordinates.
The pertinent Maxwell's equations are in SI units
\begin{align} \boldsymbol{abla}'\cdot \mathbf{B}(\mathbf{r}') &= 0 \\ \boldsymbol{abla}'\times \mathbf{B}(\mathbf{r}') & = \mu_0 \mathbf{J}(\mathbf{r}') \end{align}
$\mathbf{B}(\mathbf{r}) =k \boldsymbol{abla} \times \iiint \frac{\mathbf{J}(\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|} d\mathbf{r}', \qquad\hbox{with}\qquad k= \frac{\mu_0 }{4\pi}.$
1. ↑ J.-B. Biot and F. Savart, Note sur le Magnétisme de la pile de Volta, Annales Chim. Phys. vol. 15, pp. 222-223 (1820), free online version | {"url":"http://en.citizendium.org/wiki/Biot%E2%80%93Savart_law","timestamp":"2014-04-17T01:04:47Z","content_type":null,"content_length":"44895","record_id":"<urn:uuid:3a642e55-ce9b-4d2d-8198-e7292209d8aa>","cc-path":"CC-MAIN-2014-15/segments/1398223211700.16/warc/CC-MAIN-20140423032011-00375-ip-10-147-4-33.ec2.internal.warc.gz"} |
Type Cell array of strings characterizing the model structure.
For a SISO model sys, the property sys.Type contains a single string specifying the structure of the system.
For a MIMO model with Ny outputs and Nu inputs, sys.Type is an Ny-by-Nu cell array of strings specifying the structure of each input/output pair in the model. For
example, type{i,j} specifies the structure of the subsystem sys(i,j) from the jth input to the ith output.
The strings are made up of a series of characters that specify aspects of the model structure, as follows.
┃ Characters │ Meaning ┃
┃ Pk │ A process model with k poles (not including an integrator). k is 0, 1, 2, or 3. ┃
┃ Z │ The process model includes a zero (T[z] ≠ 0). ┃
┃ D │ The process model includes a time delay (deadtime) (T[d] ≠ 0). ┃
┃ I │ The process model includes an integrator (1/s). ┃
┃ U │ The process model is underdamped. In this case, the process model includes a complex pair of poles ┃
If you create an idproc model sys using the idproc command, sys.Type contains the strings that you specify with the type input argument.
If you obtain an idproc model by identification using procest, then sys.Type contains the strings describing the model structures that you specified for that
In general, you cannot change the type string of an existing model. However, you can change whether the model contains an integrator using the property sys.Integration.
Kp,Tp1,Tp2,Tp3,Tz,Tw,Zeta,Td Values of process model parameters.
If you create an idproc model using the idproc command, the values of all parameters present in the model structure initialize by default to NaN. The values of
parameters not present in the model structure are fixed to 0. For example, if you create a model, sys, of type 'P1D', then Kp, Tp1, and Td are initialized to NaN and are
identifiable (free) parameters. All remaining parameters, such as Tp2 and Tz, are inactive in the model. The values of inactive parameters are fixed to zero and cannot
be changed.
For a MIMO model with Ny outputs and Nu inputs, each parameter value is an Ny-by-Nu cell array of strings specifying the corresponding parameter value for each input/
output pair in the model. For example, sys.Kp(i,j) specifies the Kp value of the subsystem sys(i,j) from the jth input to the ith output.
For an idproc model sys, each parameter value property such as sys.Kp, sys.Tp1, sys.Tz, and the others is an alias to the corresponding Value entry in the Structure
property of sys. For example, sys.Tp3 is an alias to the value of the property sys.Structure.Tp3.Value.
Default: For each parameter value, NaN if the process model structure includes the particular parameter; 0 if the structure does not include the parameter.
Integration Logical value or matrix denoting the presence or absence of an integrator in the transfer function of the process model.
For a SISO model sys, sys.Integration = true if the model contains an integrator.
For a MIMO model, sys.Integration(i,j) = true if the transfer function from the jth input to the ith output contains an integrator.
When you create a process model using the idproc command, the value of sys.Integration is determined by whether the corresponding type string contains I.
NoiseTF Coefficients of the noise transfer function.
sys.NoiseTF stores the coefficients of the numerator and the denominator polynomials for the noise transfer function H(s) = N(s)/D(s).
sys.NoiseTF is a structure with fields num and den. Each field is a cell array of N[y] row vectors, where N[y] is the number of outputs of sys. These row vectors specify
the coefficients of the noise transfer function numerator and denominator in order of decreasing powers of s.
Typically, the noise transfer function is automatically computed by the estimation function procest. You can specify a noise transfer function that procest uses as an
initial value. For example:
NoiseNum = {[1 2.2]; [1 0.54]};
NoiseDen = {[1 1.3]; [1 2]};
NoiseTF = struct('num', {NoiseNum}, 'den', {NoiseDen});
sys = idproc({'p2'; 'p1di'}); % 2-output, 1-input process model
sys.NoiseTF = NoiseTF;
Each vector in sys.NoiseTF.num and sys.NoiseTF.den must be of length 3 or less (second-order in s or less). Each vector must start with 1. The length of a numerator
vector must be equal to that of the corresponding denominator vector, so that H(s) is always biproper.
Default: struct('num',{num2cell(ones(Ny,1))},'den',{num2cell(ones(Ny,1))})
Structure Information about the estimable parameters of the idproc model.
sys.Structure includes one entry for each parameter in the model structure of sys. For example, if sys is of type 'P1D', then sys includes identifiable parameters Kp,
Tp1, and Td. Correspondingly, sys.Structure.Kp, sys.Structure.Tp1, and sys.Structure.Td contain information about each of these parameters, respectively.
Each of these parameter entries in sys.Structure contains the following fields:
● Value — Parameter values. For example, sys.Structure.Kp.Value contains the initial or estimated values of the K[p] parameter.
NaN represents unknown parameter values.
For SISO models, each parameter value property such as sys.Kp, sys.Tp1, sys.Tz, and the others is an alias to the corresponding Value entry in the Structure property
of sys. For example, sys.Tp3 is an alias to the value of the property sys.Structure.Tp3.Value.
For MIMO models, sys.Kp{i,j} is an alias to sys.Structure(i,j).Kp.Value, and similarly for the other identifiable coefficient values.
● Minimum — Minimum value that the parameter can assume during estimation. For example, sys.Structure.Kp.Minimum = 1 constrains the proportional gain to values greater
than or equal to 1.
● Maximum — Maximum value that the parameter can assume during estimation.
● Free — Logical value specifying whether the parameter is a free estimation variable. If you want to fix the value of a parameter during estimation, set the
corresponding Free = false. For example, to fix the dead time to 5:
sys.Td = 5;
sys.Structure.Td.Free = false;
● Scale — Scale of the parameter's value. Scale is not used in estimation.
● Info — Structure array for storing parameter units and labels. The structure has Label and Unit fields.
Use these fields for your convenience, to store strings that describe parameter units and labels.
Structure also includes a field Integration that stores a logical array indicating whether each corresponding process model has an integrator. sys.Structure.Integration
is an alias to sys.Integration.
For a MIMO model with Ny outputs and Nu input, Structure is an Ny-by-Nu array. The element Structure(i,j) contains information corresponding to the process model for the
(i,j) input-output pair.
NoiseVariance The variance (covariance matrix) of the model innovations e.
An identified model includes a white, Gaussian noise component e(t). NoiseVariance is the variance of this noise component. Typically, the model estimation function
(such as procest) determines this variance.
For SISO models, NoiseVariance is a scalar. For MIMO models, NoiseVariance is a N[y]-by-N[y] matrix, where N[y] is the number of outputs in the system.
Report Information about the estimation process.
Report contains the following fields:
● InitialCondition — Whether estimation estimated initial conditions or fixed them at zero.
● Fit — Quantitative quality assessment of estimation, including percent fit to data and final prediction error.
● Parameters — Estimated values of model parameters and input offset, and their covariances.
● OptionsUsed — Options used during estimation (see procestOptions).
● RandState — Random number stream state at start of estimation.
● Status — Whether model was obtained by construction, estimated, or modified after estimation.
● Method — Name of estimation method used.
● DataUsed — Attributes of data used for estimation, such as name and sampling time.
● Termination — Termination conditions for the iterative search scheme used for prediction error minimization, such as final cost value or stopping criterion.
InputDelay Input delays. InputDelay is a numeric vector specifying a time delay for each input channel. Specify input delays in the time unit stored in the TimeUnit property.
For a system with Nu inputs, set InputDelay to an Nu-by-1 vector, where each entry is a numerical value representing the input delay for the corresponding input channel.
You can also set InputDelay to a scalar value to apply the same delay to all channels.
Default: 0 for all input channels
OutputDelay Output delays.
For identified systems, like idproc, OutputDelay is fixed to zero.
Ts Sampling time. For idproc, Ts is fixed to zero because all idproc models are continuous time.
TimeUnit String representing the unit of the time variable. For continuous-time models, this property represents any time delays in the model. For discrete-time models, it
represents the sampling time Ts. Use any of the following values:
● 'nanoseconds'
● 'microseconds'
● 'milliseconds'
● 'seconds'
● 'minutes'
● 'hours'
● 'days'
● 'weeks'
● 'months'
● 'years'
Changing this property changes the overall system behavior. Use chgTimeUnit to convert between time units without modifying system behavior.
Default: 'seconds'
InputName Input channel names. Set InputName to a string for single-input model. For a multi-input model, set InputName to a cell array of strings.
Alternatively, use automatic vector expansion to assign input names for multi-input models. For example, if sys is a two-input model, enter:
sys.InputName = 'controls';
The input names automatically expand to {'controls(1)';'controls(2)'}.
When you estimate a model using an iddata object, data, the software automatically sets InputName to data.InputName.
You can use the shorthand notation u to refer to the InputName property. For example, sys.u is equivalent to sys.InputName.
Input channel names have several uses, including:
● Identifying channels on model display and plots
● Extracting subsystems of MIMO systems
● Specifying connection points when interconnecting models
Default: Empty string '' for all input channels
InputUnit Input channel units. Use InputUnit to keep track of input signal units. For a single-input model, set InputUnit to a string. For a multi-input model, set InputUnit to a
cell array of strings. InputUnit has no effect on system behavior.
Default: Empty string '' for all input channels
InputGroup Input channel groups. The InputGroup property lets you assign the input channels of MIMO systems into groups and refer to each group by name. Specify input groups as a
structure. In this structure, field names are the group names, and field values are the input channels belonging to each group. For example:
sys.InputGroup.controls = [1 2];
sys.InputGroup.noise = [3 5];
creates input groups named controls and noise that include input channels 1, 2 and 3, 5, respectively. You can then extract the subsystem from the controls inputs to all
outputs using:
Default: Struct with no fields
OutputName Output channel names. Set OutputName to a string for single-output model. For a multi-output model, set OutputName to a cell array of strings.
Alternatively, use automatic vector expansion to assign output names for multi-output models. For example, if sys is a two-output model, enter:
sys.OutputName = 'measurements';
The output names to automatically expand to {'measurements(1)';'measurements(2)'}.
When you estimate a model using an iddata object, data, the software automatically sets OutputName to data.OutputName.
You can use the shorthand notation y to refer to the OutputName property. For example, sys.y is equivalent to sys.OutputName.
Output channel names have several uses, including:
● Identifying channels on model display and plots
● Extracting subsystems of MIMO systems
● Specifying connection points when interconnecting models
Default: Empty string '' for all input channels
OutputUnit Output channel units. Use OutputUnit to keep track of output signal units. For a single-output model, set OutputUnit to a string. For a multi-output model, set
OutputUnit to a cell array of strings. OutputUnit has no effect on system behavior.
Default: Empty string '' for all input channels
OutputGroup Output channel groups. The OutputGroup property lets you assign the output channels of MIMO systems into groups and refer to each group by name. Specify output groups as
a structure. In this structure, field names are the group names, and field values are the output channels belonging to each group. For example:
sys.OutputGroup.temperature = [1];
sys.InputGroup.measurement = [3 5];
creates output groups named temperature and measurement that include output channels 1, and 3, 5, respectively. You can then extract the subsystem from all inputs to the
measurement outputs using:
Default: Struct with no fields
Name System name. Set Name to a string to label the system.
Default: ''
Notes Any text that you want to associate with the system. Set Notes to a string or a cell array of strings.
Default: {}
UserData Any type of data you wish to associate with system. Set UserData to any MATLAB^® data type.
Default: []
SamplingGrid Sampling grid for model arrays, specified as a data structure.
For arrays of identified linear (IDLTI) models that are derived by sampling one or more independent variables, this property tracks the variable values associated with
each model. This information appears when you display or plot the model array. Use this information to trace results back to the independent variables.
Set the field names of the data structure to the names of the sampling variables. Set the field values to the sampled variable values associated with each model in the
array. All sampling variables should be numeric and scalar valued, and all arrays of sampled values should match the dimensions of the model array.
For example, if you collect data at various operating points of a system, you can identify a model for each operating point separately and then stack the results
together into a single system array. You can tag the individual models in the array with information regarding the operating point:
nominal_engine_rpm = [1000 5000 10000];
sys.SamplingGrid = struct('rpm', nominal_engine_rpm)
where sys is an array containing three identified models obtained at rpms 1000, 5000 and 10000, respectively.
Default: [] | {"url":"http://www.mathworks.com/help/ident/ref/idproc.html?nocookie=true","timestamp":"2014-04-20T11:14:36Z","content_type":null,"content_length":"78658","record_id":"<urn:uuid:16b228c8-ceef-4661-90fa-8985269be282>","cc-path":"CC-MAIN-2014-15/segments/1397609538423.10/warc/CC-MAIN-20140416005218-00200-ip-10-147-4-33.ec2.internal.warc.gz"} |
discrete question
November 28th 2005, 01:12 PM #1
Nov 2005
help, i'd be impressed
Assume that F: A to B is a function and S, a subset of B, is a subset. Prove that f(f-1(S))= S intersected with f(A).
[ f(f-1)) is f and inverse of f]
Last edited by ummmm; November 28th 2005 at 01:28 PM.
Assume that F: A to B is a function and S, a subset of B, is a subset. Prove that f(f-1(S))= S intersected with f(A).
[ f(f-1)) is f and inverse of f]
If $f(A)\ eq \ B$, $\ f^{-1}(S)$ may not even be defined! (As we may choose $S$ to be $B-f(A)$)
Or have I missed something.
Of course I have missed something:
choose $S$ to be $B-f(A)$,
$f^{-1}(S)\ =\ f^{-1}(B-f(A))\ =\ \emptyset$ -the empty set.
Back to the problem:
(we need only consider the case where $f^{-1}(S)\ eq\ \emptyset$ as the result holds trivally otherwise)
$f^{-1}(S)\ =\ \{x;\ x \epsilon A \mbox{ and } f(x) \epsilon S \}$,
so if $a\ \epsilon \ f^{-1}(S)$ then $f(a)\ \epsilon \ S$ and $f(a)\ \epsilon \ f(A)$
$\Rightarrow\ f(f^{-1}(S))\ \subseteq\ S$ and $f(f^{-1}(S))\ \subseteq\ f(A)$,
i.e. $\ f(f^{-1}(S))\ \subseteq\ S\cap f(A)$
Now suppose $s \epsilon\ (S\cap f(A))$ then clearly
$s\ =\ f(f^{-1}(s))$ and so $\ s \epsilon f(f^{-1}(S))\$
$\Rightarrow \ S\cap f(A)\ \subseteq\ f(f^{-1}(S))$
$S\cap f(A)\ =\ f(f^{-1}(S))$
Last edited by CaptainBlack; November 29th 2005 at 12:05 AM.
i think f(a)=b
i think f(a) has to equal b because there if a function from a to b.
B is merely the codomain for f
f(A) may be a proper subset of B
All that is necessary is that if x and y are in A and f(x) and f(y) are distinct, then x and y are distinct.
The proof on its own:
(we need only consider the case where $f^{-1}(S)\ eq\ \emptyset$ as the result holds trivally otherwise)
$f^{-1}(S)\ =\ \{x;\ x \epsilon A \mbox{ and } f(x) \epsilon S \}$,
so if $a\ \epsilon \ f^{-1}(S)$ then $f(a)\ \epsilon \ S$ and $f(a)\ \epsilon \ f(A)$
$\Rightarrow\ f(f^{-1}(S))\ \subseteq\ S$ and $f(f^{-1}(S))\ \subseteq\ f(A)$,
i.e. $\ f(f^{-1}(S))\ \subseteq\ S\cap f(A)$
Now suppose $s \epsilon\ (S\cap f(A))$ then clearly
$s\ =\ f(f^{-1}(s))$ and so $\ s \epsilon f(f^{-1}(S))\$
$\Rightarrow \ S\cap f(A)\ \subseteq\ f(f^{-1}(S))$
$S\cap f(A)\ =\ f(f^{-1}(S))$
November 28th 2005, 01:42 PM #2
Grand Panjandrum
Nov 2005
November 28th 2005, 02:48 PM #3
Nov 2005
November 28th 2005, 05:00 PM #4
Nov 2005
November 29th 2005, 12:40 AM #5
Nov 2005
December 1st 2005, 09:07 PM #6
Grand Panjandrum
Nov 2005 | {"url":"http://mathhelpforum.com/discrete-math/1369-discrete-question.html","timestamp":"2014-04-20T09:57:56Z","content_type":null,"content_length":"49852","record_id":"<urn:uuid:b4eb6ef5-9517-4a69-a08c-ab8a68d190b9>","cc-path":"CC-MAIN-2014-15/segments/1397609538110.1/warc/CC-MAIN-20140416005218-00183-ip-10-147-4-33.ec2.internal.warc.gz"} |
Integration Problem
Is there any way to integrate the following?: $\int \frac{dx}{\sqrt{a^{-1}-x^{-1}}}$ (Where "a" is a constant.) Any help would be appreciated.
Fist we would need to simplfy to get $\frac{1}{\sqrt{\frac{1}{a}-\frac{1}{x}}}=\frac{1}{\sqrt{\frac{x-a}{ax}}}=\frac{\sqrt{ax}}{\sqrt{x-a}}$ So we have the integral $\sqrt{a}\int \frac{\sqrt{x}}{\
sqrt{x-a}}dx$ Now we let $x=a\sec^2(t) \implies dx=2a\sec(t)[\sec(t)\tan(t)]dt$ $\sqrt{a}\int \frac{\sqrt{x}}{x-a}dx=\sqrt{a}\int \frac{\sqrt{a}\sec(t)}{\sqrt{a\sec^2(t)-a}}(2a\sec^2(t)\tan(t)dt)=$
$2a^{\frac{3}{2}}\int \sec^{3}(t)dt$ Here is how you integrate $\sec^{3}(t)$ Integral of secant cubed - Wikipedia, the free encyclopedia So we get $2a^{3/2}\left[\frac{1}{2}\sec(t)\tan(t)+\frac{1}{2}
\ln|\sec(t)+\ tan(t)| \right]$ From here we just need to back substitue $x=a\sec^2(t) \iff \frac{x}{a}=\sec^2(t) \iff \frac{x}{a}=\tan^2(t)+1$ $\tan^2(t)=\frac{x-a}{a} \iff \tan(t)=\sqrt{\frac{x-a}
{a}}$ and of course $\sec(t) =\sqrt{\frac{x}{a}}$ I will leave the rest to you | {"url":"http://mathhelpforum.com/calculus/90617-integration-problem.html","timestamp":"2014-04-20T17:16:13Z","content_type":null,"content_length":"39706","record_id":"<urn:uuid:664840af-fa21-49b9-8f58-562d329bef3e>","cc-path":"CC-MAIN-2014-15/segments/1397609538824.34/warc/CC-MAIN-20140416005218-00619-ip-10-147-4-33.ec2.internal.warc.gz"} |
difference between foldLeft and reduceLeft in Scala
up vote 46 down vote favorite
i have learned the basic difference between foldLeft and reduceLeft
• initial value has to be passed
• takes first element of the collection as initial value
• throws exception if collection is empty
is there any other difference ?
any specific reason to have two methods with similar functionality ?
scala functional-programming fold higher-order-functions
add comment
5 Answers
active oldest votes
Few things to mention here, before giving the actual answer:
• Your question doesn't have anything to do with left, it's rather about the difference between reducing and folding
• The difference is not the implementation at all, just look at the signatures.
• The question doesn't have anything to do with Scala in particular, it's rather about the two concepts of functional programming.
Back to your question:
Here is the signature of foldLeft (could also have been foldRight for the point I'm going to make):
def foldLeft [B] (z: B)(f: (B, A) => B): B
And here is the signature of reduceLeft (again the direction doesn't matter here)
def reduceLeft [B >: A] (f: (B, A) => B): B
These two look very similar and thus caused the confusion. reduceLeft is a special case of foldLeft (which by the way means that you sometimes can express the same thing by using
up vote 86 down vote either of them).
When you call reduceLeft say on a List[Int] it will literally reduce the whole list of integers into a single value, which is going to be of type Int (or a supertype of Int,
hence [B >: A]).
When you call foldLeft say on a List[Int] it will fold the whole list (imaging rolling a piece of paper) into a single value, but this value doesn't have to be even related to
Int (hence [B]).
Here is an example:
def listWithSum(numbers: List[Int]) = numbers.foldLeft((List[Int](), 0)) {
(resultingTuple, currentInteger) =>
(currentInteger :: resultingTuple._1, currentInteger + resultingTuple._2)
This method takes a List[Int] and returns a Tuple2[List[Int], Int] or (List[Int] -> Int). It calculates the sum and returns a tuple with a list of integers and it's sum. By the
way the list is returned backwards, because we used foldLeft instead of foldRight.
add comment
reduceLeft is just a convenience method. It is equivalent to
up vote 43 down vote list.tail.foldLeft(list.head)(_)
Good, concise answer :) Might want to correct the spelling of reducelft though – hanxue Nov 15 '13 at 4:40
@hanxue: thx, done – Kim Stebel Nov 15 '13 at 8:16
add comment
foldLeft is more generic, you can use it to produce something completely different than what you originally put in. Whereas reduceLeft can only produce an end result of the same type
or super type of the collection type. For example:
List(1,3,5).foldLeft(0) { _ + _ }
List(1,3,5).foldLeft(List[String]()) { (a, b) => b.toString :: a }
The foldLeft will apply the closure with the last folded result (first time using initial value) and the next value.
up vote 21 down
vote reduceLeft on the other hand will first combine two values from the list and apply those to the closure. Next it will combine the rest of the values with the cumulative result. See:
List(1,3,5).reduceLeft { (a, b) => println("a " + a + ", b " + b); a + b }
If the list is empty foldLeft can present the initial value as a legal result. reduceLeft on the other hand does not have a legal value if it can't find at least one value in the
add comment
The basic reason they are both in Scala standard library is probably because they are both in Haskell standard library (called foldl and foldl1). If reduceLeft wasn't, it would quite
up vote 2 down often be defined as a convenience method in different projects.
add comment
For reference, reduceLeft will error if applied to an empty container with the following error.
java.lang.UnsupportedOperationException: empty.reduceLeft
Reworking the code to use
myList foldLeft(List[String]()) {(a,b) => a+b}
up vote 1 down vote is one potential option. Another is to use the reduceLeftOption variant which returns an Option wrapped result.
myList reduceLeftOption {(a,b) => a+b} match {
case None => // handle no result as necessary
case Some(v) => println(v)
add comment
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Backward lasing yields a perfect absorber
It is well known that an amplifying medium embedded in an optical cavity, i.e., a laser oscillator, can emit coherent electromagnetic radiation when the gain coefficient of photons in the medium
reaches a threshold value that balances the light leakage out of the cavity [1]. The emitted radiation escapes from the cavity in the form of outgoing monochromatic waves, usually along preferred
spatial directions [see Fig. 1(a)]. The frequency $ω0$ and spatial pattern of the emitted waves are those of the electromagnetic mode sustained by the leaky cavity, which experiences the largest
amplification in the gain medium.
A less familiar process is the time-reversed counterpart of a laser: by replacing the gain with absorption, the same optical system supports a purely incoming radiation pattern with complete
absorption and zero reflection [see Fig. 1(b)]. Such an optical system, referred to as a coherent perfect absorber (CPA), has been theoretically proposed in an article appearing in Physical Review
Letters by Yidong Chong, Li Ge, Hui Cao, and Douglas Stone at Yale University in the US [2]. In this device, even small single-pass absorption can lead to perfect overall absorption, and optical
media that normally do not absorb radiation well at certain frequencies, can be made to do so.
When light hits a material, three things can happen: light can be reflected, as by a mirror; it can be transmitted, as with window glass; or it can be absorbed and turned, for example, into heat. The
time-reverse of the lasing process ensures that a coherent light shining on a dissipative medium at a specific wavelength and with appropriate spatial pattern is neither reflected nor transmitted,
rather it is fully absorbed. This kind of coherent perfect absorption extends and explains in an elegant way previous known methods of absorption enhancement in integrated optical devices [3, 4]. As
compared to other perfect optical absorbers, such as those based on engineered metamaterials [5], the CPA principle proposed by Chong et al. [2] works for any kind of absorbing medium, is not based
on resonant material absorption, and does not require material nanostructuring.
Perfect absorption of photons by a dissipative optical medium as proposed by the authors is reminiscent of the problem of perfect absorbing potentials for matter waves investigated in computational
quantum physics, for example, in reactive scattering calculations or other molecular collision studies [6, 7, 8, 9]. Perfect absorption of optical and matter waves share a similar mathematical
description, involving the search for poles and zeros of the underlying scattering matrix, which relates the incoming and outgoing electromagnetic wave modes. Just as in a laser at the onset of
oscillation, coherent perfect absorption provides a clear physical signature of certain spectral singularities that arise in the underlying non-Hermitian Hamiltonian [10, 11]. However, the picture of
coherent perfect absorption as time-reversed laser oscillation provides a simple and elegant explanation of the underlying physics that does not require any advanced knowledge of scattering or
spectral theory.
In a dielectric optical cavity filled by a gain medium, the spatial profile of the lasing mode is found as an eigenfunction of the Helmholtz equation for the complex electric field amplitude $E$,
which in the simplest one-dimensional spatial case reads
is the complex refractive index of the dielectric medium, and
is the speed of light in a vacuum. The electric field inside and outside the cavity is simply given by
. The imaginary part
of the refractive index accounts for light amplification
or absorption
in the medium. For a laser, the outgoing radiation pattern
satisfies Eq. (1) for a real frequency
when the (positive) imaginary part
of the refractive index reaches a threshold value. Outside the resonator, the field has the form of outgoing waves with appropriate amplitudes
and phases
, i.e.,
If we take the complex conjugate of Eq. (1), i.e., if we consider the time-reversed process of lasing at threshold, it follows that $E*$ again satisfies the Helmholtz equation but in a medium with a
refractive index $n*(x)=n′-in′′$, i.e., in a lossy medium with the amount of loss exactly equal to the amount of gain in the original lasing medium. Since complex conjugation of the field pattern
$E*$ replaces outgoing waves with incoming waves, the dielectric lossy medium inside the cavity fully absorbs the incoming radiation pattern at frequency $ω0$, and thus acts as a perfect absorber. In
other words, the combination of cavity effects and material absorption enables the incident radiation to be fully turned into heat, or re-radiated, without being reflected. Just like laser radiation,
however, perfect absorption only occurs within narrow frequency bands, requires spatial mode matching and phase control of the incoming waves, and is observed for special values of the dissipation.
Nevertheless, Chong et al. [2] have shown that such conditions can be realized using, for example, an indirect band-gap semiconductor (like silicon) in a waveguide configuration.
The simplest CPA they propose is a two-port device consisting of a silicon slab of thickness $a$ and (imaginary) refractive index $n=n′-in′′$, embedded in a waveguide or fiber with (real) refractive
index $n0$. The waveguide is excited from the two input sides by two counterpropagating coherent fields with the same intensity, same frequency $ω$, and phase difference $Δθ=θ+-θ-$ [see the inset of
Fig. 1(c)]. The spectral behavior of the light intensity reflected from the two ports of the device, normalized to the total intensity of incident waves, is shown in Fig. 1(c) for coherent input
excitation with $Δθ=0$ (blue curve), $Δθ=π$ (red curve), and for incoherent beam excitation (dashed black curve). For coherent excitation, resonant peaks, leading to strong absorption in the
material, are clearly visible, which correspond to the various resonant modes supported by the silicon slab with either odd or even spatial symmetry. The exact condition of time reversal of lasing,
and thus perfect absorption, is attained at a special frequency $ω0$ solely, which is the wavelength of the lasing mode that one would observe when the loss is replaced by gain. Conversely, for
incoherent light excitation the resonant structure of the absorption greatly disappears. For the silicon slab design considered by the authors [2], the spectral peaks of perfect (or enhanced)
absorption occur at around $940$-$nm$ wavelength. It is remarkable that CPA enables not only a resonant enhancement of absorption, but also a reduction of absorption, as compared to incoherent
excitation, for appropriate choice of the phase difference $Δθ$ [2] [see Fig. 1(c)].
The main limitations of CPA are the narrow-band operation and the need for coherent (phase and spatially controlled) excitation fields. Such severe conditions make CPA unlikely in applications where
absorption of broadband and incoherent light is needed (such as in solar photovoltaic or stealth applications). However, owing to the strong phase-dependence of the amount of absorbed radiation in
the medium, CPA can be ingeniously exploited to build a compact absorptive multichannel interferometer, as discussed by Chong et al. [2]. Unlike an ordinary interferometer, a CPA-based interferometer
does not shift the input beams between possible output channels, but causes them to be absorbed within the material. CPA-based interferometers could thus have potential applications in the
realization of transducers, modulators, or optical switches, for example, in on-chip integrated optical circuits based on $Si$ waveguide/resonator technology.
1. A. E. Siegman, Lasers (University Science Books, Mill Valley, CA, 1986)[Amazon][WorldCat].
2. Y. D. Chong, L. Ge, H. Cao, and A. D. Stone, Phys. Rev. Lett. 105, 053901 (2010).
3. M. Cai, O. Painter, and K. J. Vahala, Phys. Rev. Lett. 85, 74 (2000).
4. J. R. Tischler, M. S. Bradley, and V. Bulovic, Opt. Lett. 31, 2045 (2006).
5. N. I. Landy, S. Sajuyigbe, J. J. Mock, D. R. Smith, and W. J. Padilla, Phys. Rev. Lett. 100, 207402 (2008).
6. J. P. Palao, J. G. Muga, and R. Sala, Phys. Rev. Lett. 80, 5469 (1998).
7. J. G. Muga, J. P. Palao, B. Navarro, and I. L. Egusquiza, Phys. Rep. 395, 357 (2004).
8. O. Shemer, D. Brisker, and N. Moiseyev, Phys. Rev. A 71, 032716 (2004).
9. X. Antoine, A. Arnold, C. Besse, M. Ehrhardt, and A. Schädle, Comm. Comp. Phys. 4, 729 (2008).
10. A. Mostafazadeh, Phys. Rev. Lett. 102, 220402 (2009).
11. S. Longhi, Phys. Rev. Lett. 105, 013903 (2010). | {"url":"http://physics.aps.org/articles/print/v3/61","timestamp":"2014-04-18T08:10:10Z","content_type":null,"content_length":"23963","record_id":"<urn:uuid:aa339428-ee83-4545-ab0c-bdadeba6080e>","cc-path":"CC-MAIN-2014-15/segments/1398223206770.7/warc/CC-MAIN-20140423032006-00214-ip-10-147-4-33.ec2.internal.warc.gz"} |
MATH 099. Developmental Mathematics, 1-2 credits
Online developmental course to prepare students for college-level mathematics courses. Topics include algebraic operations, expressions, and equations; factoring; graphing and solving linear
equations. Students will develop mastery of the course outcomes by solving exercises on an adaptive, online learning system. MATH 099 is a Pass/Fail course and the credit hours do not count towards
the 120-hour requirement.
Placement: ACT Math score of 21 or less
MATH 131. Thinking Mathematically, 3 credits
This course is a survey of topics in applied mathematics stressing the connections between contemporary mathematics and modern society. Topics include critical thinking, financial management,
statistical reasoning, probability, math in politics, and math in art. This course meets the general education requirement in mathematics.
Prerequisite: MATH 099 or ACT Math score of at least 22
MATH 171. Elementary Functions, 3 credits
Study of algebraic, exponential, logarithmic, and trigonometric functions; their graphs, properties, and applications. Graphing calculator strongly recommended.
Prerequisite: MATH 099 or ACT Math score of at least 22
MATH 191. Calculus & Analytic Geometry I, 4 credits
Review of algebra and functions of a single variable; limits, continuity, differentiability, and integrability. Applications of limits, derivatives, differentials and integrals to solutions of
physical and social problems.
Prerequisite: MATH 171 or ACT Math score of at least 28
MATH 192. Calculus & Analytic Geometry II, 4 credits
Techniques of integration; polar coordinates; sequences and series. Modeling with differential equations. Introduction to partial differentiation and multiple integration.
Prerequisite: MATH 192 or score of 3 or higher on the AP Calculus AB Exam
MATH 210. Theory of Arithmetic, 3 credits
Limited to candidates for elementary teaching licensure or certificate. Topics include problem solving strategies, sets and elementary number theory and number systems, probability and statistics,
informal geometry and measurement.
Prerequisite: MATH 099 or ACT Math score of at least 22
MATH 211. Mathematics Concepts for Teachers, 3 credits
The course gives prospective elementary school teachers insights into the application of mathematical reasoning, critical thinking skills, and topics related to mathematical content standards --
algebra, geometry, measurement, and data analysis and probability. This includes the process standards of problem solving, reasoning and proof, connections, communication and representation within
each content standard identified in "Principles and Standards for School Mathematics" (NCTM, 2000).
Prerequisite: MATH 210; EDUC 284; admission to Teacher Education Program
MATH 220. Introduction to Logic & Proof, 3 credits, Writing Intensive
Introduction to Logic and Proof is designed to help students develop skills in reading and understanding elementary mathematical proofs, and in expressing their own mathematical ideas through formal
writing. Emphasis will be on precision and style. Math topics include: Logical connectives and quantifiers; types of proof; elementary set theory; functions; integers and induction; equivalence
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Prerequisite: MATH 171 or ACT Math score of at least 28
MATH/CSCI 281. Discrete Structures, 3 credits
Introduction to discrete mathematics as it is used in computer science. Topics include propositional and the predicate logic, simple circuit logic, elementary number theory, sequences and summations,
methods of proof (direct, by contradiction, by contraposition, by induction), set theory, graph theory, combinatorics, and discrete probability.
MATH 290. Elementary Linear Algebra, 3 credits
Matrices and systems of linear equations. Determinants. Lines and planes in three-space. Vector spaces and linear transformations. Characteristic equations, eigenvalues and eigenvectors.
Prerequisite: MATH 191 or a score of 3 or higher on the AP Calculus AB Exam
MATH 291. Calculus & Analytic Geometry III, 4 credits
Limits and continuity of functions of several variables, partial derivatives, directional derivatives, multiple integration, vectors, planes and and vector fields. Green's Theorem
Prerequisite: MATH 192 or a score of 3 or higher on the AP Calculus BC Exam
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A Calculus-based introduction to probability and the application of mathematical principles to the collection, analysis, and presentation of data. Modern probability concepts, discrete/continuous
models, and applications; estimation and statistical inference through modern parametric, nonparametric, and simulation/randomization methods; maximum likelihood; Bayesian methods. This course
prepares students for the preliminary P/1 exam of the Society of Actuaries (SOA) and Casualty Actuarial Society (CAS).
Prerequisite: MATH 191
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The development, application, and evaluation of statistical models to analyze data for decision-making. Univariate and multivariate general linear models (ANOVA, ANCOVA, MANOVA, linear regression),
generalized linear models (logistic and Poisson regression), and nonlinear models. The course focuses on experimental design and model estimation (including robust and randomization-based methods),
fit, and interpretation. Students are also introduced to multivariate techniques, including multidimensional scaling, principal components analysis, cluster analysis, and structural equation
Prerequisite: A previous statistics course, such as MATH 300 or STAT 213, that introduces statistical inference.
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Case study approach to topics in the statistical analysis of data. Collecting, coding, validating data; exploratory data analysis; effective quantitative displays; survey/experimental design and
sampling; power and error rates; measurement theory; multivariate statistical methods; data mining techniques.
Prerequisite: MATH 300 or STAT 213 or another statistics course
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First-order ordinary differential equations, linear second order and higher differential equations, series solutions, systems of differential equations and their applications, matrix methods for
linear systems, existence and uniqueness theorems.
Prerequisite: MATH 290
MATH 340. Secondary Math Methods, 3 credits
Limited to secondary teaching certificate candidates. Current issues, approaches, and materials in school mathematics teaching, including philosophy and objectives, curricula, local/state/national
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experience and before student teaching.
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Euclidean and non-Euclidean systems. Axiomatic approach.
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Basic elements of real analysis for students of mathematics. Topics include limits of functions, continuity, and metric spaces.
Prerequisite: MATH 192, 220, 290
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Definitions and basic properties of groups. Homomorphisms, normal subgroups, quotient groups and direct products. Rings, integral domains and fields. Ideals, quotient rings and polynomials
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The use of mathematical applications and markup languages
Prerequisite: Departmental approval
MATH 400. Topics in Mathematics, 1-3 credits
Prerequisite: Departmental Approval | {"url":"http://www.sau.edu/Academic_Programs/Mathematics/Courses.html","timestamp":"2014-04-20T14:02:04Z","content_type":null,"content_length":"16501","record_id":"<urn:uuid:b859efb4-4dc5-4acf-93ed-8d5d08244377>","cc-path":"CC-MAIN-2014-15/segments/1397609538787.31/warc/CC-MAIN-20140416005218-00570-ip-10-147-4-33.ec2.internal.warc.gz"} |
A New Equal Area Method to Calculate and Represent Physiolog... : Anesthesiology
QUANTIFICATION of physiologic dead space (Vd
) provides important insight regarding the efficiency of ventilation and its relation to pulmonary perfusion.
Respiratory dead space measurement has found wide applications in respiratory physiology, clinical anesthesia, and critical care medicine. It has been used in the diagnosis of pulmonary embolism
and as a predictor of lung volume during controlled ventilation.
A physiologic dead space to tidal volume ratio higher than 0.6 was associated with a 1.5-fold increase in mortality rate in infants with congenital diaphragmatic hernia.
In a prospective study of adults with acute respiratory distress syndrome, patients who died showed a significantly higher mean dead space fraction compared with survivors (0.63
0.54, respectively;
< 0.05).
et al.^6
found similar results in critically ill children with lung injury. Routine monitoring of dead space to tidal volume ratio in pediatric patients has been demonstrated to permit earlier extubation and
to reduce unexpected extubation failures.
The lack of a simple, practical method for the calculation of dead space components simultaneously has hindered the clinical application of dead space measurement.
The total respiratory dead space can be portioned into two parts: the anatomical dead space (airway, serial, or Fowler dead space, Vd
), and alveolar or parallel dead space (Vd
Because alveolar dead space cannot be measured directly, it is commonly estimated by subtracting the anatomical dead space from the physiologic dead space.
Anatomical dead space is usually calculated by a simple equal area graphical method developed by Fowler.
A graphical method for calculating and representing anatomical, physiologic, and alveolar dead space was reported by Fletcher
et al.
in which the dead spaces are represented by areas of trapezoids.
The calculation of alveolar and physiologic dead spaces by a graphical method similar to the Fowler equal area method and the representation of dead spaces on a linear scale would provide better
physiologic insight and clinical representation than the Bohr-Enghoff or Fletcher methods, which are apparently unrelated to the Fowler method.
In this article, we propose a new method, similar to the Fowler equal area method, for calculating and representing physiologic space, and demonstrate its relation to the classic Bohr-Enghoff
equation. Our method facilitates simple visual comparison of anatomical, alveolar, and physiologic dead spaces on a linear scale.
Materials and Methods
Physiologic Dead Space
Physiologic dead space is usually calculated by the Bohr equation, modified by Enghoff
where Vd[phys]be is physiologic dead space calculated by the Bohr-Enghoff equation, Fēco[2] is mixed expired concentration of carbon dioxide, Faco[2] is the carbon dioxide fraction of a gas in
equilibrium with arterial blood, and Vt is tidal volume. Arterial and mixed expired carbon dioxide partial pressures can also be used in this equation instead of Faco[2] and Fēco[2].
In our new equal area method, which is similar to that of Fowler, a vertical line
is drawn intersecting the expirogram at
such that areas A
and A
are equal (
fig. 1
). Then,
Vd[phys]ea = ok,
where Vd[phys]ea is physiologic dead space calculated by the new equal area method and ok is the distance between the origin and the point k. Vd[phys]ea may be shown to equal Vd[phys]be.
Equation 1
can be interpreted graphically as follows (
fig. 1
). The mean expired carbon dioxide fraction is given by
where A
is the area under the expirogram, which is the total expired carbon dioxide (Vco
). Substituting
equation 2
Equation 3 Equation 4
Image Tools Image Tools
equation 4
Area A[icdk] can be expressed as
Equation 5 Equation 6 Equation (Uncited)
Image Tools Image Tools Image Tools
equation 6
equation 5
and simplifying yields
It is thus demonstrated that the vertical line that makes area A[ojk] equal to area A[ichj] intersects the x-axis at a point (k) that represents the physiologic dead space.
Anatomical Dead Space
In the Fowler equal area method (
fig. 1
), line
is fitted to phase III of the expirogram and extrapolated to
. The vertical line
intersects the expirogram and the x-axis at points
respectively, making areas A
and A
equal. The point
on the expired volume axis is the anatomical dead space.
Alveolar Dead Space
figure 1
, the difference between physiologic dead space (
) and anatomical dead space (
) is
, which therefore represents alveolar dead space, Vd
. Hence anatomical, physiologic, and alveolar dead spaces are calculated by similar equal areas principles and displayed graphically on the same x-axis.
Fletcher Graphical Method and Representation
Equation (Uncited)Image Tools
In the Fletcher method,
physiologic, anatomical, and alveolar dead spaces are calculated as follows:
Equation 8 Equation 10
Image Tools Image Tools
where X is the area of trapezoid
, Y is the area of trapezoid
, Z is the area of rectangle
fig. 1
), and f indicates the Fletcher method.
Equation 8
is analytically identical to the Bohr-Enghoff equation (
equation 1
), and
equation 10
is analytically identical to the Fowler equal area method.
Clinical Study
After approval by the ethics committee of the Royal Prince Alfred Hospital (Sydney, NSW, Australia) and written informed consent by the patients, 10 patients (6 male and 4 female) with American
Society of Anesthesiologists physical status II or III who were undergoing lower limb vascular surgery were enrolled in this study. None of the patients had clinically significant lung disease. All
patients received a radial artery cannula for clinical hemodynamic monitoring and blood gas sampling. Other routine monitoring included pulse oximetry, electrocardiography, pharyngeal temperature,
and capnometry. Anesthesia was induced with 2 mg/kg propofol, 2 μg/kg fentanyl, and 0.8 mg/kg rocuronium intravenously, and a cuffed endotracheal tube was inserted. Anesthesia was maintained with
inhalational isoflurane, rocuronium, and fentanyl, and the patients were mechanically ventilated (Cato anesthetic machine; Dräger, Lübeck, Germany). Blood pressure was maintained within 20% of
baseline with a low-dose infusion of metaraminol (0–0.05 mg/min). The baseline respiratory parameters for ventilation were as follows: tidal volume, 10 ml/kg; respiratory frequency, 10 breaths/min;
inspiratory-to-expiratory ratio, 1:1.7; end-inspiratory hold, 10%; end-expiratory pressure, 0 cm H[2]O; and inspired oxygen concentration, 35%.
The airway gas flow, carbon dioxide fraction, and airway pressure were measured by a NICO monitor (Novametrix Medical Systems Inc., Wallingford, CT). The airway configuration from patient to machine
was as follows: endotracheal tube, flexible endotracheal connector, airway filter, mainstream infrared carbon dioxide analyzer, pneumotachograph, pressure monitor, sidestream oxygen analyzer, and Y
piece of the anesthetic circuit. Partial pressure of carbon dioxide, oxygen, gas flow rate, and airway pressure signals in the airway were logged at a frequency of 300 Hz through a 12-bit
analog-to-digital converter (DAQPad-1200; National Instruments Corporation, Austin, TX) and recorded by a computer running MATLAB (Mathworks, Natick, MA). Barometric pressure was measured using an
electronic barometer (Vaisala PTB100A; Helsinki, Finland), which was calibrated by an ISO 17025–accredited laboratory. The pneumotachograph was calibrated by a second-order polynomial method.
The carbon dioxide analyzer was calibrated with known concentrations of carbon dioxide and was cross-calibrated with the blood gas analyzer (ABL 700; Radiometer, Copenhagen, Denmark). The
synchronization of flow and carbon dioxide signals of the NICO monitor was verified by rapid injection of carbon dioxide at 1 l/min into the airway immediately upstream of the carbon dioxide analyzer
while 5 l/min of oxygen was flowing.
In each patient, the ventilation parameters, which were known to affect physiologic dead space,
were adjusted one at a time from the baseline in random order to one of the following settings: tidal volume 80, 100, and 120% of baseline tidal volume; end-expiratory pressure 0, 5, and 10 cm H
O; inspiratory-to-expiratory ratio 1:1.7, 1:1, and 2:1; inspiratory hold 10, 30, and 50% of the inspiratory time. After 15 min at each setting, 10 carbon dioxide expirograms were recorded for
analysis, and at the same time, arterial blood was drawn into a heparinized syringe (PICO70; Radiometer) and stored in ice slush for blood gas analysis.
Data Analysis
was calculated according to Faco
= Paco
− P
), where P
is barometric pressure, P
is the saturated water vapor pressure at body temperature,
and Paco
is arterial partial pressure of carbon dioxide corrected to the patient's body temperature.
Total expired carbon dioxide volume (Vco
= A
) was calculated by integrating the expired carbon dioxide concentration with respect to expired volume. For each carbon dioxide expirogram, physiologic dead space was calculated by using the
Bohr-Enghoff equation (
equation 1
), the new equal area method, and the Fletcher area method (
equation 8
). All methods were implemented without interpolating between data points. The points
were assigned by selecting the sampled data points that minimized the difference between the areas. The average and SD of each set of 10 dead spaces were calculated.
The bias and limits of agreement of the new equal area method compared with the Bohr-Enghoff equation and Fletcher methods were assessed by Bland and Altman analysis.
The limits of agreement were defined as the mean difference ± 2 SD and describe the range that includes 95% of the differences between the two methods compared. All other results are reported as mean
± SD.
< 0.05 was considered to be statistically significant. All calculations were performed by software written in MATLAB.
The ages of the patients were 67 ± 12 yr (range, 44–79 yr), the weights were 76.1 ± 9.2 kg (range, 51–90 kg), and the body mass indices were 26.6 ± 2.9 kg/m^2 (range, 23.6–30.4 kg/m^2). A total of
120 sets of 10 expirograms were obtained from the 10 patients. Calculated Vd[anat], Vd[phys]be, Vd[phys]f, and Vd[phys]ea were 197.7 ± 33.2, 313.6 ± 80.1, 313.6 ± 80.1, and 313.5 ± 80.1 ml,
respectively. Intraindividual dead space measurements varied by −15.9 ± 5.1 to 18.4 ± 7.3% due to changes in tidal volume, by 0 to 6.7 ± 1.7% due to changes in end-expiratory pressure, by 0 to −9.7 ±
2.9% due to changes in inspiratory-to-expiratory ratio, and by 0 to −18.4 ± 6.9% due to changes in inspiratory hold. The mean intraindividual coefficient of variation of the 120 sets of 10 dead space
measurements at each ventilation setting was 2.0% (range, 0.7–5.0%).
Bland-Altman analysis shows that the differences between Vd
ea and Vd
be ranged from −1.65 to 0.79 ml (mean, −0.07 ml; limits of agreement, −1.27 to 1.13 ml;
fig. 2A
). Differences between Vd
ea and Vd
f ranged from −2.08 to 1.19 ml (mean, −0.09 ml; limits of agreement, −1.52 to 1.34 ml;
fig. 2B
). The differences were all randomly distributed over the range of dead spaces, and the mean differences were not statistically significantly different from zero in either comparison. Pearson
correlation analysis showed correlation coefficients of 0.999 (
< 0.05) between Vd
be and Vd
ea and between Vd
f and Vd
ea. Analysis of variance showed that (Vd
be − Vd
ea) and (Vd
f − Vd
ea) were not statistically different from zero (
> 0.05) over all ventilator settings (means, −0.07 and −0.09 ml; ranges, −0.25 to 0.11 and −0.30 to 0.15 ml, respectively).
Fig. 3 Table 1
Image Tools Image Tools
Figure 3
shows typical carbon dioxide expirograms from four different patients with anatomical and physiologic dead spaces calculated by the Fowler equal area method and our new equal area method,
respectively. The patients' data and dead spaces are shown in
table 1
. The expirograms for patients A and D are at the extremes of the 10 patients studied.
This study introduces a new equal area method for the calculation, representation, and visualization of physiologic dead space based on a principle similar to the Fowler equal area method. The new
method is analytically identical to the Bohr-Enghoff equation and yields numerical results that do not differ significantly from those calculated directly by the Bohr-Enghoff equation. Advantages of
this new method include the following: (1) the graphical representation on a linear scale of the relations between all of the dead space volumes and fractions, including Vd
, Vd
, Vd
, Vt, Vd
/Vt, Vd
/Vt, Vd
/Vt, and Vd
, which is especially helpful for the visualization of dead spaces during bedside monitoring of patients; (2) the use of a principle similar to that used in the Fowler equal area method, which makes
the calculation of anatomical and physiologic dead spaces on the carbon dioxide expirogram consistent; (3) the use of a more straightforward method than the partitioning of areas on a carbon dioxide
expirogram proposed by Fletcher
et al.^11
; (4) the equal area method is simpler than the classic Douglas bag method, an advantage it shares with other open-circuit methods, although all three methods require arterial blood sampling. In
addition, this new equal area method emphasizes the relations that exist between the various respiratory dead spaces and thus will greatly assist in understanding and teaching of their concepts. Once
synchronized flow and carbon dioxide signals are digitized, the new method requires a few lines of code more than the open-circuit Bohr-Enghoff method, but it is not difficult to implement. With the
advance of computing technology, the calculation and visualization of the dead spaces can be easily implemented for clinical application.
Theoretically, the physiologic dead space obtained by the three methods, our new equal area method, the classic Bohr-Enghoff equation, and the Fletcher method, should be identical. The small
differences are due to quantization of the data in time by the analog-to-digital converter and the fact that our software did not interpolate between data points when locating the division between
the equal areas. The differences between the results due to quantization would be bigger if the analog-to-digital sample rate were lower and smaller if the sample rate were higher, but could be
eliminated almost completely if the software interpolated between data points when determining the equal areas.
This new equal method for calculating physiologic dead space has been evaluated in patients without severe lung diseases. It should be noted that the difference between physiologic dead space
calculated by our new equal area method and the Bohr-Enghoff equation method is not affected by the alteration of phase II of a carbon dioxide expirogram and is thus independent of changes in the
carbon dioxide expirogram caused by lung diseases. Compared with the Bohr-Enghoff equation method, this new method has similar sensitivity to measurement signal noise and the synchronization of
carbon dioxide and flow data, which could affect the accuracy of dead space measurement.
The delay between carbon dioxide analyzer and flow signals and the rise time of the carbon dioxide analyzer, especially in sidestream carbon oxide analyzers, should be corrected to reduce error.
In conclusion, this new equal area method for calculating, displaying, and visualizing physiologic dead space is easy to understand and yields the same results as the classic Bohr-Enghoff equation.
All the three dead spaces—physiologic, anatomical, and alveolar—together with their relations to expired volume, can be displayed conveniently on the x-axis of a carbon dioxide expirogram to
demonstrate their values and relations. | {"url":"http://journals.lww.com/anesthesiology/Fulltext/2006/04000/A_New_Equal_Area_Method_to_Calculate_and_Represent.13.aspx","timestamp":"2014-04-21T07:54:24Z","content_type":null,"content_length":"240039","record_id":"<urn:uuid:9812deb3-99ec-49a5-9915-3dc70a80c048>","cc-path":"CC-MAIN-2014-15/segments/1398223207985.17/warc/CC-MAIN-20140423032007-00400-ip-10-147-4-33.ec2.internal.warc.gz"} |
Reply to comment
May 2002
When visitors to Paris admire the elegant tapering curves of the Eiffel Tower, they probably aren't aware that they are observing a triumph of mathematics over the forces of nature. The distinctive
shape of the Tower, although beautiful, was actually designed mathematically by Eiffel to withstand the forces of the Parisian winds. Eiffel said of the tower, which stood as the world's tallest
building for 41 years:
"What phenomenon did I have to give primary concern in designing the Tower? It was wind resistance. Well, I hold that the curvature of the monument's four outer edges, which is as the
mathematical calculations have dictated it should be, will give a great impression of strength and beauty."
The mathematical calculations dictating the shape of the Tower have been examined in a recent article in the American Journal of Physics by Joseph Gallant.
The force of the wind (dF) produces a torque around the bottom left corner of the tower which is countered by the force of the Tower's weight (dW)
By balancing the maximum torque generated by the wind with the torque generated by the Tower's own weight, Gallant has derived an equation which describes the shape of the Tower:
Eiffel, experienced in designing open lattice structures, allowed for a large safety margin by designing the Tower to withstand wind pressures of 4 kN/m^2. The fastest winds recorded at the Tower
reached a speed of 214 km/h in 1999 and would have produced pressures of just 2.28 kN/m^2.
The Eiffel Tower was built in 1889 for the Universal Exposition as a monument to the scientific achievements of the 18th century. And, thanks to the mathematical and engineering prowess of its
designer Eiffel, it still stands as a witness to these scientific achievements over a century later.
But is it stronger than me?! | {"url":"http://plus.maths.org/content/comment/reply/2750","timestamp":"2014-04-20T13:52:40Z","content_type":null,"content_length":"25327","record_id":"<urn:uuid:718a81ff-674b-41bf-b2f2-e9b0b279a858>","cc-path":"CC-MAIN-2014-15/segments/1398223207046.13/warc/CC-MAIN-20140423032007-00164-ip-10-147-4-33.ec2.internal.warc.gz"} |
Using Two Equidistant Points to Determine a Perpendicular Bisector
You can use two equidistant points to determine the perpendicular bisector of a segment. (To “determine” something means to fix or lock in its position, basically to show you where something is.)
Here’s the theorem.
Two equidistant points determine the perpendicular bisector: If two points are each (one at a time) equidistant from the endpoints of a segment, then those points determine the perpendicular bisector
of the segment. (Here’s an easy way to think about it: If you have two pairs of congruent segments, then there’s a perpendicular bisector.)
This theorem is a royal mouthful, so the best way to understand it is visually. Consider the kite-shaped diagram in the above figure.
The theorem works like this: If you have one point (like X) that’s equally distant from the endpoints of a segment (W and Y) and another point (like Z) that’s also equally distant from the endpoints,
then the two points (X and Z) determine the perpendicular bisector (line XZ) of that segment (segment WY).
Here’s a “SHORT” proof that shows how to use this equidistance theorem as a shortcut so you can skip showing that triangles are congruent.
You can do this proof using congruent triangles, but it’d take you about nine steps and you’d have to use two different pairs of congruent triangles.
Statement 1:
Reason for statement 1: Given.
Statement 2:
Reason for statement 2: If angles, then sides.
Statement 3:
Reason for statement 3: Given.
Statement 4:
Reason for statement 4: If two points (S and O) are each equidistant from the endpoints of a segment (segment RH), then they determine the perpendicular bisector of that segment.
Statement 5:
Reason for statement 5: Definition of bisect.
Statement 6:
Reason for statement 6: Definition of midpoint. | {"url":"http://www.dummies.com/how-to/content/using-two-equidistant-points-to-determine-a-perpen.html","timestamp":"2014-04-16T05:08:11Z","content_type":null,"content_length":"54567","record_id":"<urn:uuid:169b3954-7404-42ad-a6dc-d730f4e68cc5>","cc-path":"CC-MAIN-2014-15/segments/1397609521512.15/warc/CC-MAIN-20140416005201-00154-ip-10-147-4-33.ec2.internal.warc.gz"} |
Discharge of a capacitor
I suppose that dVc/dt doesn't direction into account. Why doesn't it though? At first I considered this, but then I dismissed it, knowing that it's perfectly fine to have a rate of change w.r.t
another variable negative. The equation should be:
[tex]|I| = C |\frac{dV_{c}}{dt}|[/tex]
Anyway, thanks very much for your help.
The defining equation for the capacitor does take direction into account. In the definition, the current direction is defined to be INTO the capacitor, and resulting voltage change is positive. Thus,
when the voltage change is NEGATIVE, the current will come OUT of the capacitor. | {"url":"http://www.physicsforums.com/showthread.php?p=3627294","timestamp":"2014-04-18T18:15:10Z","content_type":null,"content_length":"32336","record_id":"<urn:uuid:30e5e619-a3b1-4e0d-8a4c-c5d175d10d9e>","cc-path":"CC-MAIN-2014-15/segments/1397609535095.7/warc/CC-MAIN-20140416005215-00245-ip-10-147-4-33.ec2.internal.warc.gz"} |
The Cosmological Constant
Annu. Rev. Astron. Astrophys. 1992. 30: 499-542
Copyright © 1992 by . All rights reserved
3.5 Growth of Linear Perturbations
In all the homogeneous and isotropic cosmologies, linear cold matter perturbations Peebles 1980). An explicit expression for the amplitude of a growing perturbation (Heath 1977) is
where a' is the dummy integration variable, and da / da or a', in our case given explicitly by Equation 9. Equation 28 is normalized so that the fiducial case of [M] = 1, [] = 0 gives the familiar
scaling a) = a, with coefficient unity.
Different values of [M], [] lead to different linear growth factors from early times (a a = 1, da / d[0]([M], []) we have
(The remarkable approximation formula - good to a few percent in regions of plausible [M], [] - follows from Lahav et al 1991 and Lightman & Schechter 1990.) Figure 7 shows numerical values for [0]
([M], []) for the region in the ([], [tot]) plane previously seen in Figures 1 and 4. One sees that as [M] is reduced from unity, both along the line [] = 0 and along the line [tot] = 1, the growth
of perturbations is suppressed, but somewhat less suppressed in the [tot] = 1 case. The reason is that, for fixed [M], linear growth effectively stopped at a redshift (1 + z) = [M]^-1 in the open
case (when the universe became curvature dominated), but, more recently, at (1 + z) = [M]^-1/3 in the flat case (when the universe became
Figure 7. Growth factor for linear perturbations, as contours in the ([M], [tot]) plane, normalized to unity for the case [M] = 1, []= 0. There is relatively less suppression of growth as [M] is
decreased along the line [tot] = 1 than along the line []= 0; but for credible values of [M] the difference is not a large factor. Perturbation growth approaches [M] it occurs at too small a redshift
to explain quasars (see Figure 1).
To the right of the line [tot] = 1 in Figure 7, one sees values of [0] ([M], []) that are greater than 1, in fact approaching infinity at the loitering cosmology line (cf Figure 1 and discussion
above). Loitering cosmologies allow the arbitrarily large growth of linear perturbations, since the perturbations continue to grow during the (arbitrarily long) loiter time.
Related to the growth of linear perturbations is the relation between peculiar velocity v and peculiar acceleration g, or (as a special case) the radial infall velocity v[rad] around a spherical
perturbation of radius a,
the relation being
where <Peebles 1980, Section 14). One can calculate accurately by taking the derivative of Equation 28, using Equation 9, and solving the resulting integral numerically. Lahav et al (1991), however,
give an approximation formula valid for all redshifts z,
Figure 8 plots f(z) for our standard models A-E. One sees that, at small redshifts, peculiar velocities depend almost entirely on [M] and are quite insensitive to []. This is because they are driven
by the matter perturbations in primarily the most recent Hubble time. Looking back to redshifts z [], allowing in principle for observational tests (but see Lahav et al 1991 for caveats).
Figure 8. Peculiar velocities around fixed-density condensations as a function of redshift. for models B-E, relative to model A. For z []. | {"url":"http://ned.ipac.caltech.edu/level5/Carroll/Carroll3_5.html","timestamp":"2014-04-18T10:43:22Z","content_type":null,"content_length":"10650","record_id":"<urn:uuid:21c31e1e-5c0b-49b4-b7b2-aad3b2af188d>","cc-path":"CC-MAIN-2014-15/segments/1397609533308.11/warc/CC-MAIN-20140416005213-00587-ip-10-147-4-33.ec2.internal.warc.gz"} |
characterization of trees in terms of products of transpositions
up vote 4 down vote favorite
Suppose a simple graph has $n$ vertices and $m$ edges. If the vertices are labelled, then each edge then corresponds to a transposition in a natural way. A theorem in Godsil and Royle's Algebraic
Graph Theory, section 3.10, asserts the following:
If the graph is not a tree, then some product of the $m$ distinct transpositions is not an $n$-cycle.
For example, consider $K_4-e$, i.e. the graph with vertex set $\[4\]$ and edges $\lbrace 1, 2\rbrace, \lbrace 1, 3\rbrace, \lbrace1, 4\rbrace, \lbrace 2, 3\rbrace$, and $\lbrace2, 4\rbrace$. Then $(1
2)(1 4)(1 3)(2 4)(2 3) = (1 4 2 3)$, a 4-cycle. However, $(1 2)(1 4)(1 3)(2 3)(2 4) = (1 3).$
Unfortunately for me, the proof is left as an exercise, which I cannot solve. Can anyone help?
add comment
1 Answer
active oldest votes
Notation: αβ means apply α then β.
If the graph is not a tree, then either it contains a cycle or it contains less than n−1 edges. In the latter case, we get a contradiction since less than n−1 transpositions cannot
multiply to a big cycle.
So suppose that the graph contains a cycle (12...k), and assume that for each order of the edges, you get a big cycle. In particular, this holds for orderings where you take the cycle
last. Fix some ordering of the other edges, and denote the interim product (without the edges of the cycle) by π. So for each product σ of the edges of the cycle, πσ is a long cycle.
up vote 4 down
vote accepted For each i<j, there is some ordering of the cycle which produces a permutation containing the transposition (ij). Indeed, take (i i+1)(i+1 i+2)...(j-2 j-1) (j j+1) ... (k 1) (1 2) ...
(i-1 i) (j-1 j).
Thus for each i<j, π(ij)τ is a big cycle, where τ does not involve i or j. In particular, π(i)≠j. Since this is true for all i≠j, it follows that π must be the identity. We get a
contradiction since σ is not a big cycle.
You should add the easy case of a non-connected graph. – Benoît Kloeckner Jul 20 '10 at 7:39
3 If the graph is not connected, then either there are less than n-1 edges or there's a cycle. Less than n-1 transpositions cannot multiply to a big cycle. – Yuval Filmus Jul 20 '10
at 15:54
I do not follow your fourth paragraph. Consider the graph with edges $(51)(53)(54)(52)(12)(23)(41)(34)$. It contains a cycle on $\lbrace 1, 2, 3, 4 \rbrace$. In your notation, let $
\pi = (51)(53)(54)(52) = (51342)$. Let $i=1$ and $j=4$. Then $\sigma = (12)(23)(41)(34)$ which does indeed equal $(14)(23)$. Thus $\tau = (2 3)$, which does not contain 1 or 4.
However $\pi (i j) \tau = (1 2 5 4 3)$, a big cycle. Yet 1 and 4 are in the same cycle of $\pi$. – user7760 Jul 20 '10 at 21:30
I modified the proof, hopefully now it does work. – Yuval Filmus Jul 20 '10 at 22:16
add comment
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Carmichael Numbers
Recall Carmichael numbers are composite numbers that almost always fool the Fermat primality test. We can show that Carmichael numbers must have certain properties.
First we show they cannot be of the form $n=pq$ where $p,q$ are distinct primes with $p>q$. By the Chinese Remainder Theorem we have ${\mathrm{ℤ}}_{n}={\mathrm{ℤ}}_{p}×{\mathrm{ℤ}}_{q}
$. Then
Suppose $a$ is not a multiple of $p$. Then ${a}^{n-1}=\left({a}^{p-1}{\right)}^{q}{a}^{q-1}={a}^{q-1}\left(modp\right)$ (by Fermat we have ${a}^{p-1}=1\left(modp\right)$). Then if $a$ passes the
Fermat test, we must have ${a}^{q-1}=1\left(modp\right)$ and hence ${a}^{d}=1$, where $d=\mathrm{gcd}\left(p-1,q-1\right)$. Since ${\mathrm{ℤ}}_{p}^{*}$ is cyclic, there are exactly $d$ choices
for $a$ that satisfy ${a}^{d}=1$.
Since $q$ is strictly smaller than $p$, the greatest common divisor of $p-1$ and $q-1$ is at most $\left(p-1\right)/2$. Thus at most half the choices for $a$ can fool the Fermat test. For primes
$p,q$ used in RSA, $d$ is small with high probability, so the product $pq$ will almost never fool the Fermat test, as mentioned earlier.
Next suppose $n$ is not squarefree, that is $n={p}^{k}r$ for some prime $p$ and $k\ge 2$. Then by the Chinese Remainder Theorem, ${\mathrm{ℤ}}_{n}={\mathrm{ℤ}}_{{p}^{k}}×{\mathrm{ℤ}}_
{r}$. Any $a∈{\mathrm{ℤ}}_{{p}^{k}}$ satisfies ${a}^{\varphi \left({p}^{k}\right)}=1$ by Euler’s Theorem, so if ${a}^{n-1}=1$ as well, then we must have ${a}^{d}=1$ where
$d=\mathrm{gcd}\left(n-1,\varphi \left({p}^{k}\right)\right)=\mathrm{gcd}\left({p}^{k}r-1,{p}^{k-1}\left(p-1\right)\right).$
Since ${\mathrm{ℤ}}_{{p}^{k}}$, is cyclic, exactly $d$ elements $a∈{\mathrm{ℤ}}_{{p}^{k}}$ satisfy ${a}^{d}=1$. As $p$ cannot divide ${p}^{k}r-1$, the largest possible value for $d$
is $p-1$, giving an upper bound of $\left(p-1\right)/\left({p}^{k}-1\right)\le 1/4$ probability that $n$ will pass the Fermat test.
Hence if $n$ is a Carmichael number, then $n$ is squarefree and is the product of at least three distinct primes.
Solovay-Strassen Test
We can improve this by checking instead that
This is known as the Solovay-Strassen test, which we examine here for historical interest. Recall that the Jacobi symbol can be evaluated quickly using quadratic reciprocity.
Why does this help? From above we know that if $n$ is not squarefree then $n$ fails the Fermat test with probability at least $3/4$. So we need only consider the case when $n$ is squarefree but
composite, say $n=pr$ where $p$ is prime.
By the Chinese Remainder Theorem any $x∈{\mathrm{ℤ}}_{n}$ can be written as $\left(a,b\right)∈{\mathrm{ℤ}}_{p}×{\mathrm{ℤ}}_{r}$. For any nonzero $a∈{\mathrm{&
Zopf;}}_{p}$ we have ${a}^{p-1}=1$ by Fermat, so if ${a}^{n-1}=1$ then we have ${a}^{d}=1$ where $d=\mathrm{gcd}\left(p-1,n-1\right)$.
If $d<p-1$ then $d$ is at most $\left(p-1\right)/2$. Since ${\mathrm{ℤ}}_{p}^{*}$ is cyclic, this means at most $\left(p-1\right)/2$ elements of $a∈{\mathrm{ℤ}}_{p}$ satisfy ${a}^
{d}=1$. That is, the probability $a$ will pass the Fermat test is at most $1/2$.
On the other hand, if $d=p-1$, then this implies $p-1∣N-1$. Since $N-1=pr-1=\left(p-1\right)r+r-1$ we have $p-1∣r-1$, so write $r-1=s\left(p-1\right)$. Then
${a}^{pr-1}={a}^{\left(p-1\right)r}{a}^{\left(p-1\right)s}=1.$ $\left(\frac{x}{n-1}\right)=\left(\frac{a}{p}\right)\left(\frac{b}{r}\right)$
and since there is a $1/2$ chance that $a$ is a quadratic residue, there is a $1/2$ chance that $\left(x∣n-1\right)=-1$ and fail the test.
In other words, for any composite $n$ (even a Carmichael number) the probability $n$ passes the Solovay-Strassen test is at most $1/2$. This was the first algorithm discovered for finding large
The Miller-Rabin test surpasses the Solovay-Strassen test in every way: the probability a composite number $n$ passes is only $1/4$, and no Jacobi symbol computations are required. Moreover, any $a$
that exposes the compositeness of $n$ in the Solovay-Strassen test also causes the Miller-Rabin test to fail. | {"url":"http://crypto.stanford.edu/pbc/notes/numbertheory/carmichael.xhtml","timestamp":"2014-04-17T07:35:32Z","content_type":null,"content_length":"21741","record_id":"<urn:uuid:d3974bcd-5dd3-4881-a740-4eeb8f617089>","cc-path":"CC-MAIN-2014-15/segments/1397609526311.33/warc/CC-MAIN-20140416005206-00395-ip-10-147-4-33.ec2.internal.warc.gz"} |
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Math Tools Discussion: Roundtable, hands-on interactive software
Discussion: Roundtable
Topic: hands-on interactive software
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Subject: Geometry Projects
Author: Cathi
Date: Mar 19 2003
I hadn't seen the Four points on a Circle article in NCTM; having just read it
this morning, I found it interesting. I like to carry the exploration further,
though, and Sketchpad is very helpful in doing that hands-on.
R&D worldwide sounds interesting (I never heard of it ? Research and
Development?). What is the Question/Problem in this activity?
Re. Pythagorean Theorem; yes, that?s a good one. There are so many on the web!
What do they do with these proofs, after finding them?
Reply to this message Quote this message when replying?
yes no
Post a new topic to the Roundtable Discussion discussion
Discussion Help | {"url":"http://mathforum.org/mathtools/discuss.html?context=dtype&do=r&msg=10149","timestamp":"2014-04-19T15:18:23Z","content_type":null,"content_length":"15528","record_id":"<urn:uuid:590e23d7-6d18-409a-b605-2780583166db>","cc-path":"CC-MAIN-2014-15/segments/1397609537271.8/warc/CC-MAIN-20140416005217-00228-ip-10-147-4-33.ec2.internal.warc.gz"} |
Statistics 102
Spring Semester, 2000
• I will have a review session for questions related to the final exam next Monday evening (May 8), from 8:00 pm -- 9:30 pm in SH-DH 351 (not room 215 as previously announced).
• I will have office hours Tuesday, May 9, from 3-5 and on Wednesday from 2-3:30.
• I am missing scores for some of you for one or two assignments. To make sure there is no error in the recording of grades, please let me know if you have the missing assignment. If you didn't
seem to do any assignments, I omitted your name from the list. I have updated the list to include PS #5 , which quite a few of you evidently did not do.
│Name1 │ps1, ps3 │
│Name2 │ps2, ps5 │
If there's an error in this list, simply bring me the marked assignment, and I will fix the grades.
• Some review notes on regression are here . Its 10 pages long, so I do not have copies to distribute. Have a look, and print it if you want a paper copy. Evidently, the printed version is a lot
more easy to read than when you view the document with Acrobat.
• The final exam is on Wednesday, May 10
The make-up final exam is scheduled for Friday, September 8, 2000, from 4-6 pm. If you do not take the final exam on May 10, you will be required to take the make-up final next fall.
• The average for the second exam was 85 with an SD of 14. The median was 91 and mode was 92.
• The mean for the first midterm was about 80 with SD = 13. You can figure out your relative standing from these summaries since the grades are close to normal - albeit a bit skewed.
• I have the impression that some of you are not aware of the grading distribution for the material in this class. I have given them when asked, but not published it here. It was in the class
syllabus. Here it is:
□ Homework 10%
□ Project 10%
□ First test 25%
□ Second test 25%
□ Final 30%
• For questions related to the project please look at the list of "frequently asked questions" here .
• For JMP questions, you should always see the TAs at the Stat Lab for help. They are there most of the day M-F and can show you how to use JMP for doing the calculations. Their schedule is on the
class web page.
Other materials (including the syllabus and assignments) are available from the class web page.
For the datasets from the regression casebook, go to this link.
Supplemental Handouts
Extra handouts, including some JMP files.
1. Paired testing handout (computer retail advertising) Software Ads jmp file
2. Claimed color percentages for M&Ms (Does not load on some systems!)
3. Example of chi-square data file Illustrates the use of JMP-IN to compute chi-square values from tables.
4. Mon (March 6): One-way analysis of variance
5. Wed (March 8): Using one-way anova (Mileage comparison JMP data )
6. Mon (March 20): Two-way anova with some review of one-way (Web page experiment JMP data )
7. Wed (March 22): Using two-way anova with a sample of test flight data
8. Mon (March 27): Introduction to regression
9. Wed (March 29): Introduction to inference in regression
10. Mon (April 3): Review for second midterm
11. Wed (April 5): Beta and regression
This example introduces the notion of 'beta' as used in finance. Interpret the results of the calculations with a bit of common sense, as we have not taken into account the cost of borrowing or
12. Mon (April 10): Prediction and outliers in regression
13. Wed (April 12): Introduction to multiple regression
A reading for this class covers an automobile design case .
14. Mon (April 17): Multiple regression and collinearity.
15. Wed (April 19): Categorical variables in regression.
16. Mon (April 24): Categorical variables, continued
17. Wed (April 26): Summary regression modeling
18. Mon (May 8): Review
□ Review of methods
☆ Confidence intervals
☆ Comparisons of two groups (t-test)
☆ Comparisons of several groups (anova)
○ Anova summary and F-test
○ Multiple comparisons (Tukey, Hsu)
☆ Regression
○ R2, anova table, and F-ratio
○ Regression coefficients, SE, and t-ratio
■ Marginal vs partial coefficients
■ Transformations
■ Collinearity (VIF)
○ Categorical predictors, interaction
○ Prediction, RMSE, and prediction intervals
○ Assumptions and residual diagnostics
○ Leverage points and outliers
□ Sample data analyses
□ Questions on prior exams | {"url":"http://www-stat.wharton.upenn.edu/~stine/stat102/index.html","timestamp":"2014-04-19T19:53:52Z","content_type":null,"content_length":"9627","record_id":"<urn:uuid:bb384e03-ea36-47db-a608-83f20ed33d3f>","cc-path":"CC-MAIN-2014-15/segments/1398223210034.18/warc/CC-MAIN-20140423032010-00536-ip-10-147-4-33.ec2.internal.warc.gz"} |
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find the equation of the line connecting the two points (4,9) and (3,2). write in answer in slope-intercept form - WyzAnt Answers
find the equation of the line connecting the two points (4,9) and (3,2). write in answer in slope-intercept form
I dont have any additional details.
Tutors, please
to answer this question.
First, we want to calculate the slope.
m=(9-2)/(4-3) = 7
We know the form of a line in slope-intercept form is y=mx+b, and we have found m to be 7.
Therefore, y=7x+b
Let us use the point (3,2) to find b.
2=7(3)+b, or 2=21+b
Solving for b, we get b=-19.
Therefore, the equation of the line in slope-intercept form is y=7x-19. | {"url":"http://www.wyzant.com/resources/answers/2635/find_the_equation_of_the_line_connecting_the_two_points_4_9_and_3_2_write_in_answer_in_slope_intercept_form","timestamp":"2014-04-16T05:42:43Z","content_type":null,"content_length":"44426","record_id":"<urn:uuid:b96ff800-77a2-4ae3-b23a-dc365673073b>","cc-path":"CC-MAIN-2014-15/segments/1397609521512.15/warc/CC-MAIN-20140416005201-00390-ip-10-147-4-33.ec2.internal.warc.gz"} |
The Switch Has Been In Position A For A Long, Long ... | Chegg.com
Image text transcribed for accessibility: The switch has been in position a for a long, long time. At t = 0s, Sw moves to position b. What is the initial value of v(t)? v(0) = What is the value of
the time constant? tau = What is the initial value of i(t)? io(t) = What is the algebraic expression for v(t)? What is the algebraic expression for io(t)?
Electrical Engineering | {"url":"http://www.chegg.com/homework-help/questions-and-answers/switch-position-long-long-time-t-0s-sw-moves-position-b-initial-value-v-t-v-0-value-time-c-q3162060","timestamp":"2014-04-19T23:52:12Z","content_type":null,"content_length":"20570","record_id":"<urn:uuid:5f7d91e6-c476-4dcf-a46f-38dac0123daa>","cc-path":"CC-MAIN-2014-15/segments/1397609537754.12/warc/CC-MAIN-20140416005217-00152-ip-10-147-4-33.ec2.internal.warc.gz"} |
A question with panle model
April 19th 2011, 04:32 PM #1
Nov 2009
In my field, a number of studies aggragated amount panel data (means data with station and time) and put them into a panel model to estimate regression coefficients. And then the model will apply
to a large region for projecting future.
However, panel model actually restricted all stations to obey the same functional relationship. But to me, will this give some risks to the stations that did not obey the estimated relationships?
I am thinking if panel model method may reject heterogeneity and exaggerate homogeneity.
If it is true, is it OK if the fitted model was used to apply to a large region?
Any idea is welcomed. THANKS. | {"url":"http://mathhelpforum.com/advanced-statistics/178113-question-panle-model.html","timestamp":"2014-04-20T16:43:01Z","content_type":null,"content_length":"29640","record_id":"<urn:uuid:710340be-ac73-466c-a9da-8e7e06bb9af2>","cc-path":"CC-MAIN-2014-15/segments/1397609538824.34/warc/CC-MAIN-20140416005218-00656-ip-10-147-4-33.ec2.internal.warc.gz"} |
An Algorithm for Three-Dimensional Orthogonal Graph Drawing
Wood, David R. (1998) An Algorithm for Three-Dimensional Orthogonal Graph Drawing. In: Graph Drawing 6th International Symposium, GD’ 98, August 13-15, 1998, Montréal, Canada , pp. 332-346 (Official
URL: http://dx.doi.org/10.1007/3-540-37623-2_25).
Full text not available from this repository.
In this paper we present an algorithm for 3-dimensional orthogonal graph drawing based on the movement of vertices from an initial layout along the main diagonal of a cube. For an n-vertex m-edge
graph with maximum degree six, the algorithm produces drawings with bounding box volume at most 2.37n^3 and with a total of 7m/3 bends, using no more than 4 bends per edge route. For maximum degree
five graphs the bounding box has volume n^3 and each edge route has two bends. These results establish new bounds for 3-dimensional orthogonal graph drawing algorithms and improve on some existing
Repository Staff Only: item control page | {"url":"http://gdea.informatik.uni-koeln.de/337/","timestamp":"2014-04-20T18:23:31Z","content_type":null,"content_length":"37428","record_id":"<urn:uuid:7c7a4cc4-a232-4216-938a-375469f2c983>","cc-path":"CC-MAIN-2014-15/segments/1397609539066.13/warc/CC-MAIN-20140416005219-00090-ip-10-147-4-33.ec2.internal.warc.gz"} |
$(0,1)$-Category theory
Monoidal categories
With symmetry
With duals for objects
With duals for morphisms
With traces
Closed structure
Special sorts of products
In higher category theory
A quantale is a closed monoidal suplattice. Equivalently, it is a monoid object in the closed symmetric monoidal category of suplattices where the morphisms are the set maps that preserve arbitrary
joins. This means it is a poset having all joins and an associative, unital tensor product $\otimes$ which distributes over joins (the internal-homs then come automatically by the adjoint functor
theorem).The internal-homs in a quantale are sometimes called residuations and written $x\backslash y$ and $y/x$. Unitality is skipped by some authors; in that case we can talk about subclass of
unital quantales.
As a semigroup (monoid if unital) in suplattices, a quantale is essentially the same thing as a 1-object quantaloid, i.e., a 1-object category enriched in suplattices.
Quantales and Frames
Additional conditions often imposed on a quantale include:
• Commutativity: $x\otimes y = y\otimes x$
• Idempotence: $x\otimes x = x$
• Affineness: the unit for $\otimes$ is the top element: $1=\top$.
If all three of commutativity, idempotence, and affineness are assumed, they force $\otimes$ to be the meet and therefore the quantale to be a frame. General quantales are sometimes considered to be
a “noncommutative” version of a frame, whose opposite category would be a category of “noncommutative locales.”
(This is the origin of the name “quantale,” a portmanteau of “quantum” and “locale”. Note, though, that quantales seem to be generally treated in the literature more as “quantum frames” than “quantum
locales,” and in particular their morphisms usually go in the “frame direction.” Possibly this can be explained by the fact that in the past, it was common to use the word “locale” for what we now
call a “frame” and simply distinguish between “locale homomorphisms” (now called “frame homomorphisms”) and “continuous maps.” The name “quantale” was introduced by C.J. Mulvey.)
The following construction gives a simple means for passing from commutative affine quantales to frames:
Let $(Q, \cdot, 1)$ be a commutative affine quantale, and let $Idem(Q)$ be the subposet of elements $x \cdot x = x$. Then $Idem(Q)$ is a frame, where the meet operation is given by multiplication in
$Q$. The functor $Idem$ is right adjoint to the forgetful functor from commutative affine quantales to frames.
Notice that $x \cdot x \leq x \cdot 1 = x$ for any $x \in Q$, so the interest is in the other condition $x \leq x x$. If $x, y$ are idempotent, we easily have $x y$ idempotent using commutativity,
and $x y \leq x 1 = x$ and $x y \leq 1 y = y$ by affineness. Thus $z \leq x y$ implies $z \leq x$ and $z \leq y$. Conversely, if $z$ is idempotent and $z \leq x$ and $z \leq y$, we have
$z \leq z z \leq x y$
and we now conclude that $\cdot$ is the meet operation on $Idem(Q)$. Next, we show that $Idem(Q)$ is closed under taking joins in $Q$: if $x_i$ is a collection of idempotents, we have
$x_i \leq x_i x_i \leq (\bigvee_i x_i) (\bigvee_i x_i)$
for all $i$, whence
$\bigvee_i x_i \leq (\bigvee_i x_i) (\bigvee_i x_i),$
which is all we need. Since joins in $Idem(Q)$ are calculated just as they are in $Q$, and since multiplication in $Q$ distributes over arbitrary joins, we have that binary meets distribute over
arbitrary joins in $Idem(Q)$.
Finally, if $A$ is a frame and $Q$ is a commutative affine quantale, it is clear that a quantale map $f \colon A \to Q$ takes elements in $A$ (which are idempotent under meet) to idempotents in $Q$.
Hence $f$ factors uniquely through $Idem(Q) \hookrightarrow Q$, and the map $A \to Idem(Q)$ is a frame map. This shows that $Idem$ is the right adjoint as claimed.
In fact, we may also observe that the forgetful functor from commutative affine quantales to commutative quantales also has a right adjoint, just be passing from a commutative quantale to the
principal downset given by the quantale unit. (However, the forgetful functor from commutative quantales to quantales does not have a right adjoint.)
Enrichment over quantales
A different way of thinking about quantales views them as a (0,1)-categorical analogue of a cosmos (in the sense of Benabou). In particular, one can then study enriched categories over a quantale. A
classic example is Lawvere metric spaces, seen as categories enriched in the quantale $([0, \infty], \geq)$ with $+$ taken as tensor product.
Enrichment is often particularly interesting for $*$-quantales (see below), where one can study $*$-enriched categories.
Quantales are a surprisingly commonplace structure in computer science. A very simple example is the powerset of strings (i.e., the powerset of the free monoid over some set of characters $\Sigma$).
The order is the inclusion order on sets, and meet and join are just intersection and union, respectively. Taking $\epsilon$ to be empty string, and $a \cdot b$ to the join of two string, the
quantalic operations are then:
• $1 = \{\epsilon\}$
• $L \otimes M = \{ l\cdot m \;|\; l \in L, m \in M \}$
This example generalizes as follows: given any monoidal preorder $M$ (for instance, a monoid equipped with the discrete order, as in the previous example), the collection of down-closed subsets of
$M$ carries a quantale structure given by Day convolution with respect to categories enriched in $\mathbf{2} = TV$, the Heyting algebra of truth values. Explicitly, if $e$ denotes the unit of $M$ and
$\cdot$ the multiplication, then
• $1 = \{x \in M: x \leq e\}$
• $L \otimes M = \{x \in M: \exists_{l \in L} \exists_{m \in M} x \leq l \cdot m\}$
Another class of examples: internal homs $\hom_{sLat}(X, X)$ in the closed monoidal category of suplattices. For example, when the suplattice $X$ is a power set $P(S)$, one may identify $\hom_{sLat}
(P(S), P(S))$ with the poset of binary relations $P(S \times S)$, ordered by inclusion and where the quantalic multiplication is relational composition.
Quantales, as monoids in the symmetric monoidal category $sLat$, can be tensored to produce new quantales.
A $*$-quantale is a quantale $Q$ equipped with an additional structure of an involution
$* : Q \to Q$
for which $(x \otimes y)^* = y^* \otimes x^*$ and $1^* = 1$, where $1$ denotes the monoidal unit. (The operator is assumed to be covariant with respect to the poset structure.)
An example of a $*$-quantale is the quantale of binary relations on a set $S$, where the $*$-operation is relational opposite:
• $R^* = \{(y, x): (x, y) \in R\}$
Another example is obtained by taking the quantale of down-closed subsets of a $*$-monoidal poset $M$ (which is the same thing as a $*$-monoid? in the cartesian monoidal category of posets), with the
quantale structure given by Day convolution as described above, and the $*$-operator obtained by cocontinuously extending the $*$-operator on $M$. Explicitly,
A $*$-enriched category over a $*$-quantale $Q$ is a category $(X, d: X \times X \to Q)$ enriched in the underlying quantale, such that
$d(y, x) = d(x, y)^*$
This notion can also be expressed in terms of lax morphisms of $*$-quantales; see below.
Relation to linear logic
A commutative quantale is in particular a symmetric monoidal category (a symmetric monoidal (0,1)-category). As such it may be thought of as a model for linear logic in the general sense. Precisely
if it has a dualizing object then it is a star-autonomous category and hence a model for linear logic in the original sense. (see e.g. Yetter 90, page 43). Indeed, quantales have been argued to
provide models for quantum logic, see there for more.
Morphisms of quantales
There is a variety of notions of morphism of quantale, just as there is a variety of notions of morphism between closed monoidal categories. All the notions considered here are morphisms between the
underlying sup-lattices, in other words preserve arbitrary joins, hence are left adjoints as functors between the underlying categories.
• At the weak end of the scale, one may consider lax morphisms of quantales, i.e., (lax) monoidal functors of quantales seen as monoidal categories.
□ An important example of this is that categories enriched in a monoidal poset $M$, such as Lawvere metric spaces, amount to the same thing as lax quantale morphisms of the form $2^d: 2^{M} \to
2^{X \times X}$ where the domain is the quantale of upward-closed subsets of $M$ with the Day convolution structure, and the codomain is the quantale of binary relations on $X$, with
multiplication being relational composition.
• A stronger notion is of strong morphisms of quantales seen as monoidal categories. As noted above, all quantale morphisms considered here are already left adjoints in $Cat$, and if the adjunction
lifts to $MonCat$ (the 2-category of monoidal categories, lax monoidal functors, and monoidal transformations), then the left adjoint is strong monoidal. This often occurs in practice.
• An even stronger notion is where the morphisms also strongly preserve the closed structure, i.e., the internal homs or residuations. (An example is to be developed for buildings.)
• There are corresponding notions of morphisms of $*$-quantales, where in each case morphisms strongly respect the $*$ operations. For instance, the notion of $*$-enriched category over a $*$
-monoidal poset $M$ can be equivalently recast as a lax morphism between $*$-quantales, $2^d: 2^M \to 2^{X \times X}.$
The initial paper to use the term `quantale’ was
Discussion of how quantales serve as a model for linear logic and quantum logic is in
• David Yetter, Quantales and (noncommutative) linear logic, Journal of Symbolic Logic 55 (1990), 41-64.
A monograph on quantales:
• Kimmo I. Rosenthal, Quantales and their applications, Pitman Res. Notes in Math. Series 234, Longman 1990
Connections to operator algebras and etale groupoids is discussed in
• Pedro Resende, Étale groupoids and their quantales, Adv. Math. 208 (2007) 147-209; also published electronically: doi; math/0412478
• M.C. Protin, P. Resende, Quantales of open groupoids, J. Noncommut. Geom. 6 (2012) 199–247.
• P. Resende, Lectures on ́tale groupoids, inverse semigroups and quantales, Lecture Notes for the GAMAP IP Meeting, Antwerp, 4–18 September, 2006, 115 pp.; pdf
• P. Resende, Groupoid sheaves as quantale sheaves, J. Pure Appl. Algebra 216 (2012), 41–70; arxiv/0807.4848 doi
• D. Kruml, J.W. Pelletier, P. Resende, J. Rosický, On quantales and spectra of $C^\ast$-algebras, Appl. Categ. Structures 11 (2003) 543–560.
• D. Kruml, P. Resende, On quantales that classify $C^\ast$-algebras, Cah. Topol. Geom. Differ. Categ. 45 (2004) 287–296.
• F. Borceux, J. Rosický, G. Van den Bossche, Quantales and $C^\ast$-algebras, J. London Math. Soc. 40 (1989) 398–404 doi
Sheaves on a quantale
• Francis Borceux, Rosanna Cruciani, Sheaves on a quantale, Cahiers de Topologie et Géométrie Différentielle Catégoriques (1993) 34:3, page 209-228 pdf | {"url":"http://www.ncatlab.org/nlab/show/quantale","timestamp":"2014-04-16T04:21:37Z","content_type":null,"content_length":"67292","record_id":"<urn:uuid:ff52028a-738f-415d-a786-e595d9aaf682>","cc-path":"CC-MAIN-2014-15/segments/1397609521512.15/warc/CC-MAIN-20140416005201-00605-ip-10-147-4-33.ec2.internal.warc.gz"} |
2008 apm formula sheet [Archive] - Actuarial Outpost
I couldn't find the formula sheet for the 2008 apm exam on the soa website. Can anyone please be kind enough to post me the link? Thanks!
I'll bet $5 there's not a question on valuing a CDS Hull style. It was asked on FET and the same folks are on the EC for APMV. I bet they mix it up...that said, I still know how to work the problem.
I am actually betting GMAB option values are not going to be tested again this year... i don't think I can do those without forumlas in the formula "sheet". | {"url":"http://www.actuarialoutpost.com/actuarial_discussion_forum/archive/index.php/t-135281.html","timestamp":"2014-04-19T19:35:51Z","content_type":null,"content_length":"33777","record_id":"<urn:uuid:e277ca37-f560-4a53-aa43-396a995f3870>","cc-path":"CC-MAIN-2014-15/segments/1397609537376.43/warc/CC-MAIN-20140416005217-00288-ip-10-147-4-33.ec2.internal.warc.gz"} |
Supply Chains and Fuzzy Demand
Submitted by Daniel Dumke on Wed, 2011-05-04 17:33
Risk in supply chains can be included in several different ways into the decision making process.
No Risk
A statement in many supply chain models is that some/most/all parameters of the model are fixed (e.g. fixed demand, zero probability of a hurricane).
The result is, if the real value of this parameter diverges from the assumptions, the results of the model will be flawed to a certain degree (up to completely unusable).
Another allegedly more complete way of including risks into modeling is to assign specific probabilities or even stochastic distributions the model’s parameters. In this way using a normal
distribution, would represent the uncertainty about the future values of the parameter. Nonetheless, this approach has theoretical and practical flaws as well. From a practical standpoint, the time
needed for solving the resulting model (especially with realistic model sizes) increases dramatically and may even be insolvable. And from a theoretical point of view, the assigned distributions may
also prove to be wrong in reality.
Fuzzy Sets
Fuzzy logic is another way of introducing risk considerations into a model. This article is about the theoretical application of this logic to a supply chain problem.
Fuzzy sets are a construct which can be used to a limited amount of uncertainty into a system’s model.
A fuzzy demand variable, as an example, could have four possible manifestations of (14, 16, 18, 19). This can be interpreted as the demand can have one of those values without further specification
of the individual probabilities.
Model and Results
Wen and Iwamura (2008) developed a facility location allocation model where the demand parameter has been defined as a fuzzy variable.
Since the demand in this case is not deterministic anymore, decisions have been based on an expected value.
The authors used the Hurwicz criterion, a decision function based on the weighted average of the worst and best outcome of a given decision, as the objective function to evaluate the “best”
They also suggest and test an solution heuristic. A possible solution of this location problem is shown in figure 1.
Figure 1: Exemplary Results: Location of Customers (dot) and Facilities (diamond) (Wen and Iwamura, 2008)
One advantage of the concept of fuzzy sets is that it may be easier for an analyst or expert to specify a limited number of possible demand values than fill the parameters an appropriate distribution
The disadvantages lie in the further assumptions of the fuzzy sets.
• The fuzzy variables are defined to be trapezoidal, ie. probabilities are assigned to the possible manifestations eg. (14, 16, 18, 19) in such way that the probability for the extreme values are
lower than the center values. And this is basically the representation of a discrete distribution function.
• There is no evidence for fuzzy logic as better representing the uncertainty in a real supply chain system
• The data shows that the computational efficiency does not seem to be very good
Furthermore, fuzzy logic does not allow to represent extreme and unlikely events in a adequate manner. In the case of demand, there may be the possibility of extreme high or low demand which could
lead to a change in the optimal locations.
Wen, M., & Iwamura, K. (2008). Fuzzy facility location-allocation problem under the Hurwicz criterion European Journal of Operational Research, 184 (2), 627-635 DOI: 10.1016/j.ejor.2006.11.029 | {"url":"http://scrmblog.com/review/supply-chains-and-fuzzy-demand","timestamp":"2014-04-17T12:29:21Z","content_type":null,"content_length":"37333","record_id":"<urn:uuid:aa668462-78a2-4860-a007-4f263140e06a>","cc-path":"CC-MAIN-2014-15/segments/1397609530131.27/warc/CC-MAIN-20140416005210-00570-ip-10-147-4-33.ec2.internal.warc.gz"} |
Statistical Inference
August 12th 2010, 11:06 AM #1
Aug 2010
Along a road 1 km long are 3 people distributed at random. Find the probability that no 2 people are less than a distance of d km apart, when
d ≤ ½.
(I think this might have something to do with uniform distribution but I have no idea how to solve this problem, please help......Thanks.) | {"url":"http://mathhelpforum.com/advanced-statistics/153528-statistical-inference.html","timestamp":"2014-04-18T10:45:47Z","content_type":null,"content_length":"28917","record_id":"<urn:uuid:3614b3ce-6051-4b84-a682-b0da8168abda>","cc-path":"CC-MAIN-2014-15/segments/1397609533308.11/warc/CC-MAIN-20140416005213-00517-ip-10-147-4-33.ec2.internal.warc.gz"} |
Post a reply
hi bobbym here Agnishom solve all the problems and he also given hints. i think still you have some confusion.I will explain clearly along with formulas
solution: [(x^2)^2+2(x^2)(12)+(12)^2]-16x^2 Formula:(a+b)^2=a^2+2ab+b^2
here (x^2+12)^2 = (x^2)^2+2(x^2)(12)+12^2
If u solve this, the result be x^4+24x^2+144
But in the given question we have 8x^2, we got 24x^2 for that we are subtracting 16x^2 from our result
so the equation be [(x^2)^2+2(x^2)(12)+12^2]-16x^2
(x^2+12)^2-(4x)^2 Formula(a+b)(a-b)=a^2-b^2
here a=x^2+12 b=4x, [x^2+12+4x][x^2+12-4x] | {"url":"http://www.mathisfunforum.com/post.php?tid=19652&qid=276084","timestamp":"2014-04-20T08:42:25Z","content_type":null,"content_length":"22030","record_id":"<urn:uuid:2eae4385-6281-4de1-9cfa-ed4b6e78ce5c>","cc-path":"CC-MAIN-2014-15/segments/1398223203841.5/warc/CC-MAIN-20140423032003-00426-ip-10-147-4-33.ec2.internal.warc.gz"} |
Crystola, CO Math Tutor
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Fachbereich Mathematik
27 search hits
A: New Wavelet Methods for Approximating Harmonic Functions; B: Satellite Gradiometry - from Mathematical and Numerical Point of View (1995)
Willi Freeden Michael Schreiner
Some new approximation methods are described for harmonic functions corresponding to boundary values on the (unit) sphere. Starting from the usual Fourier (orthogonal) series approach, we propose
here nonorthogonal expansions, i.e. series expansions in terms of overcomplete systems consisting of localizing functions. In detail, we are concerned with the so-called Gabor, Toeplitz, and
wavelet expansions. Essential tools are modulations, rotations, and dilations of a mother wavelet. The Abel-Poisson kernel turns out to be the appropriate mother wavelet in approximation of
harmonic functions from potential values on a spherical boundary.
Direct Coupling of Fluid and Kinetic Equations: I (1995)
J. Schneider
Second Order Scheme for the Spatially Homogeneous Boltzmann Equation with Maxwellian Molecules (1995)
Jens Struckmeier Konrad Steiner
In the standard approach, particle methods for the Boltzmann equation are obtained using an explicit time discretization of the spatially homogeneous Boltzmann equation. This kind of
discretization leads to a restriction of the discretization parameter as well as on the differential cross section in the case of the general Boltzmann equation. Recently, it was shown, how to
construct an implicit particle scheme for the Boltzmann equation with Maxwellian molecules. The present paper combines both approaches using a linear combination of explicit and implicit
discretizations. It is shown that the new method leads to a second order particle method, when using an equiweighting of explicit and implicit discretization.
Numerical Simulation of the Stationary One-Dimensional Boltzmann Equation by Particle Methods (1995)
Jens Struckmeier A.V. Bobylev
The paper presents a numerical simulation technique - based on the well-known particle methods - for the stationary, one-dimensional Boltzmann equation for Maxwellian molecules. In contrast to
the standard splitting methods, where one works with the instationary equation, the current approach simulates the direct solution of the stationary problem. The model problem investigated is the
heat transfer between two parallel plates in the rarefied gas regime. An iteration process is introduced which leads to the stationary solution of the exact - space discretized - Boltzmann
equation, in the sense of weak convergence.
Normalized Coprime Factorizations in Continuous and Discrete Time - A Joint State-Space Approach (1995)
Jörg Hoffmann
Based on state-space formulas for coprime factorizations over ... and an algebraic characterization of J-inner functions, normalized doubly-coprime factorizations for different classes of
continuous- and discrete-time transfer functions are derived by using a single general construction method. The parametrization of the factors is in terms of the stabilizing solutions of general
degenerate continuous- respectively discrete-time Riccati equations, which are obtained by examining state-space representations of J-normalized factor matrices.
Asymptotic-Induced Domain Decomposition Methods for Kinetic and Drift Diffusion Semiconductor Equations (1995)
Axel Klar
This paper deals with domain decomposition methods for kinetic and drift diffusion semiconductor equations. In particular accurate coupling conditions at the interface between the kinetic and
drift diffusion domain are given. The cases of slight and strong nonequilibrium situations at the interface are considered and some numerical examples are shown.
A Kinetic Model for Vehicular Traffic Derived from a Stochastic Microscopic Model (1995)
R. Wegener Axel Klar
A way to derive consistently kinetic models for vehicular traffic from microscopic follow the leader models is presented. The obtained class of kinetic equations is investigated. Explicit
examples for kinetic models are developed with a particular emphasis on obtaining models, that give realistic results. For space homogeneous traffic flow situations numerical examples are given
including stationary distributions and fundamental diagrams.
Locally Supported Kernels for Spherical Spline Interpolation (1995)
Michael Schreiner
By the use of locally supported basis functions for spherical spline interpolation the applicability of this approximation method is spread out since the resulting interpolation matrix is sparse
and thus efficient solvers can be used. In this paper we study locally supported kernels in detail. Investigations on the Legendre coefficients allow a characterization of the underlying Hilbert
space structure. We show now spherical spline interpolation with polynomial precision can be managed with locally supported kernels, thus giving the possibility to combine approximation
techniques based on spherical harmonic expansions with those based on locally supported kernels.
On a New Condition for Strictly Positive Definite Functions on Spheres (1995)
Michael Schreiner
Recently, Xu and Cheney (1992) have proved that if all the Legendre coefficients of a zonal function defined on a sphere are positive then the function is strictly positive definite. It will be
shown in this paper, that even if finitely many of the Legendre coefficients are zero, the strict positive definiteness can be assured. The results are based on approximation properties of
singular integrals, and provide also a completely different proof of the results ofXu and Cheney.
Equidistribution on the Sphere (1995)
Willi Freeden J. Cui
A concept of generalized discrepancy, which involves pseudodifferential operators to give a criterion of equidistributed pointsets, is developed on the sphere. A simply structured formula in
terms of elementary functions is established for the computation of the generalized discrepancy. With the help of this formula five kinds of point systems on the sphere, namely lattices in polar
coordinates, transformed 2-dimensional sequences, rotations on the sphere, triangulation, and sum of three squares sequence, are investigated. Quantitative tests are done, and the results are
compared with each other. Our calculations exhibit different orders of convergence of the generalized discrepancy for different types of point systems. | {"url":"https://kluedo.ub.uni-kl.de/solrsearch/index/search/searchtype/collection/id/15997/start/0/rows/10/yearfq/1995/doctypefq/preprint","timestamp":"2014-04-20T00:54:02Z","content_type":null,"content_length":"42477","record_id":"<urn:uuid:0ba07124-f42c-4b91-a960-bde886cb584c>","cc-path":"CC-MAIN-2014-15/segments/1398223202548.14/warc/CC-MAIN-20140423032002-00320-ip-10-147-4-33.ec2.internal.warc.gz"} |
Finding shortest / cheapest paths in a dense graph
npm install graph-paths
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26 downloads in the last month
graph-paths is a small module for finding cheapest (shortest) paths in a dense graph (network).
$ npm install graph-paths
var cheapest_paths = require('graph-paths').cheapest_paths;
* The network is defined by a matrix describing the cost of getting from node i to node j.
* If there is no way from node i to node j, then the cost is infinite.
var costs = [
var cheapest_paths_from_0 = cheapest_paths(costs, 0);
console.log("cheapest paths from node #0 to all other nodes:");
console.log("cheapest path from node #0 to node #3:"); | {"url":"https://www.npmjs.org/package/graph-paths","timestamp":"2014-04-18T21:42:44Z","content_type":null,"content_length":"7394","record_id":"<urn:uuid:f6f5f9fc-be0d-42c6-90e2-a1ff1fd71424>","cc-path":"CC-MAIN-2014-15/segments/1397609535095.9/warc/CC-MAIN-20140416005215-00442-ip-10-147-4-33.ec2.internal.warc.gz"} |
Laveen Trigonometry Tutor
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...Kelly, and I am well versed in mathematics, physics, and the medical sciences. I got my undergraduate degrees in physics and economics and my doctor of naturopathic medicine degree from SCNM. I
believe that a good tutor gets to know their student and understands what THEIR problems are and attends to them instead of just trying to hammer all the concepts into their head.
14 Subjects: including trigonometry, chemistry, physics, calculus
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12 Subjects: including trigonometry, chemistry, geometry, accounting | {"url":"http://www.purplemath.com/laveen_trigonometry_tutors.php","timestamp":"2014-04-17T15:45:59Z","content_type":null,"content_length":"24192","record_id":"<urn:uuid:0c45c4ee-c32e-4116-8b1d-9e63ae64f558>","cc-path":"CC-MAIN-2014-15/segments/1397609530136.5/warc/CC-MAIN-20140416005210-00173-ip-10-147-4-33.ec2.internal.warc.gz"} |
Geometry With Pizzazz Worksheets
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MIDDLE SCHOOL MATH WITH PIZZAZZ! BOOK A 63 Creative Publications ANSWERS How Do Clocks Communicate? Do each exercise below. Find your answer in the answer column and notice the letter next to it.
Geometry Name _____ Worksheet: Congruent Triangles SSS & SAS Date _____ Period ___ Write a congruence statement between triangles and state the postulate implied.
Create your own worksheets like this one with Infinite Geometry. Free trial available at KutaSoftware.com. Title: 3-Parallel Lines and Transversals Author: Mike Created Date:
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Create your own worksheets like this one with Infinite Geometry. Free trial available at KutaSoftware.com. Title: 7-Similar Polygons Author: Mike Created Date:
ID: A 1 G.G.69: Quadrilaterals in the Coordinate Plane 1: Investigate the properties of quadrilaterals in the coordinate plane, using the distance, midpoint, and slope formulas
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equipartition (plural equipartitions)
1. the division of something into equal parts
2. (mathematics, of a graph) the partition of its vertex set into sets whose sizes differ from each other by no more than 1
Derived termsEdit
equipartition (third-person singular simple present equipartitions, present participle equipartitioning, simple past and past participle equipartitioned)
1. (transitive) To divide into equal parts.
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Finding the rate at which a project has maximum NPV
July 9th 2012, 05:29 AM
Finding the rate at which a project has maximum NPV
How do you find the interest rate at which an investment project has a maximum NPV?
And how would you use such a rate to make a decision about the investment?
I assume it would be acceptable if the rate, at which project has maximum NPV, is higher than the company's cost of capital (WACC) (is this so or am I wrong on this one)
Would you also compare such a rate with IRR?
July 14th 2012, 01:00 AM
Re: Finding the rate at which a project has maximum NPV
Can't teach here...no blackboard and the likes!
Ever heard of Googling?
NPV and IRR -- Measures for Evaluating Investments
July 30th 2012, 05:31 AM
Re: Finding the rate at which a project has maximum NPV
Better put some sample problems so that you can fully understand the things to be discuss, you can actually understand that easily if you will put some samples.
this website | {"url":"http://mathhelpforum.com/business-math/200790-finding-rate-project-has-maximum-npv-print.html","timestamp":"2014-04-16T21:08:38Z","content_type":null,"content_length":"4640","record_id":"<urn:uuid:3c40a4cb-65d6-44d5-84fd-da176e75a93e>","cc-path":"CC-MAIN-2014-15/segments/1397609524644.38/warc/CC-MAIN-20140416005204-00040-ip-10-147-4-33.ec2.internal.warc.gz"} |
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book recommendation on data analysis and statistics
up vote 4 down vote favorite
I am looking for a book on data analysis and statistics.
My objective is to better analyse and understand data over time (like trends or events) and extract useful information from raw statistics. I am not especially interested in forecasting.
Any recommendations?
st.statistics books reference-request
1 Community-wiki'd, as per usual for book recommendation requests. – Scott Morrison♦ Apr 26 '10 at 0:30
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4 Answers
active oldest votes
Again, recommendations. depends on the preknowledge of the questiones, but for the stated purpose, time series and trends analysis, it will be difficult to find an better introduction than
""The Analysis of Time Series: An Introduction, Sixth Edition (Chapman & Hall/CRC Texts in Statistical Science) " by Chris Chatfield. That is a very simple read with a lot of good
up vote 2 practical . There is no lack of more advanced books which can be attacked when Chatfield is mastered.
down vote
add comment
I have used Bluman's book on Elementary Statisics at the last two colleges where I taught. The book is designed for students with little more than a 9th grade mathematics background and
is very effective.
A number of years ago, David Moore published a book on elementary statistics for the American Statistical Association. The book has accompanying tapes that are to be used to introduce the
up vote 2 major topics. This book is also excellent. I believe that if you contact the ASA and tell them that you are an instructor, you can get a FREE copy of the tapes and book for viewing.
down vote
There are alot of dogs out there too.
add comment
An oldie but goody is John Tukey's book Exploratory Data Analysis. It won't tell you everything you need to know, but it has some good basic content.
up vote 2 down vote
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"Statistical Analysis with R" & "R Graphs Cookbook" serve as an excellent resource for tackling problems in Data Analysis and Statistics using R.
If you are a beginner to R then the first book would be just perfect for you to learn all you need to, about programming in R.
up vote 2 down vote
The R Graphs Cookbook contains more detailed recepies for creating the most useful graphs using R.
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Rates Of Change Question
May 17th 2007, 12:37 PM #1
May 2007
Rates Of Change Question
"Water is being poured into a conical vessel of semi vertical angle of 30 degrees at the rate of 100Pi cm cubed per minute. find the rate at which the depth pf water is increasing when the water
is 5cm deep."
I can't really seem to get my head around this question.
As far as I can make out it involves a little bit of trigonometery which I'm terrible at. The only thing I can make out so far is that I'm looking for dV/dT.
All help really appreciated!
The question is worth quiet a bit of my exam.
"Water is being poured into a conical vessel of semi vertical angle of 30 degrees at the rate of 100Pi cm cubed per minute. find the rate at which the depth pf water is increasing when the water
is 5cm deep."
I can't really seem to get my head around this question.
As far as I can make out it involves a little bit of trigonometery which I'm terrible at. The only thing I can make out so far is that I'm looking for dV/dT.
All help really appreciated!
The question is worth quiet a bit of my exam.
If the depth of water is d, then the radius of th surface is r=d/sqrt(3) since
the semi vertical angle is 30 degrees. So the volume of water is:
V = (1/3) pi r^2 d = (1/3) pi d^3/ sqrt(3)
(volume of cone formula used)
Now we are told that dV/dt = 100 pi cc/min. so we differentiate the formula
for V wrt t:
dV/dt = [1/(3 sqrt(3)] pi 3 d^2 dd/dt
so when d=5 cm we have:
dd/dt = 4 sqrt(3) cm/min
which is the rate of change of depth.
If the depth of water is d, then the radius of th surface is r=d/sqrt(3) since
the semi vertical angle is 30 degrees. So the volume of water is:
V = (1/3) pi r^2 d = (1/3) pi d^3/ sqrt(3)
(volume of cone formula used)
Now we are told that dV/dt = 100 pi cc/min. so we differentiate the formula
for V wrt t:
dV/dt = [1/(3 sqrt(3)] pi 3 d^2 dd/dt
so when d=5 cm we have:
dd/dt = 4 sqrt(3) cm/min
which is the rate of change of depth.
Amazing! There's just one simple thing I don't understand if you could explain it to me it'd be brilliant. When you say "3 sqrt(3)" is that 3 (square root sign)3
And when you say "(1/3) pi d^3/ sqrt(3)" is that 1/3 (multiplied by) pi d(to the power of 3) ALL OVER the square root of 3?
Sorry, I know I sound like an idiot but I'm not used to Maths Speak on the internet..
Thanks a million for the help.
Last edited by Fionnan; May 17th 2007 at 02:57 PM.
The above is because the sides of a 30,60,90 triangle are in the ratio
So the volume of water is:
V = (1/3) pi r^2 d = (1/3) pi d^3/ sqrt(3)
(volume of cone formula used)
volume of a cone is 1/3 of the area of the base times the height, and
r=d/sqrt(3), so:
V = (1/3) pi d^3/3 = pi d^3/9
So there is a mistake in the formula for V
Now we are told that dV/dt = 100 pi cc/min. so we differentiate the formula
for V wrt t:
dV/dt = [1/(3 sqrt(3)] pi 3 d^2 dd/dt
So this last should be:
dV/dt = [1/9] pi 3 d^2 dd/dt
100 pi = (1/3) pi d^2 dd/dt
300/d^2 = dd/dt
so when d=5 cm we have:
dd/dt = 4 sqrt(3) cm/min
which is the rate of change of depth.
So the final answer is:
dd/dt = 12 cm/min
May 17th 2007, 01:57 PM #2
Grand Panjandrum
Nov 2005
May 17th 2007, 02:46 PM #3
May 2007
May 17th 2007, 08:20 PM #4
Grand Panjandrum
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Expected value of probability distribution
June 4th 2010, 10:34 PM
Expected value of probability distribution
The number $X$ of particles emitted as the result of an experiment is a random variable with probability distribution:
$P(X=k)=\left(\frac{1}{2}\right)^{k+1}\ for\ k \geq 0$
What is the expected number of particles emitted during one experiment
This is how I've gone so far:
$=\left(\frac{1}{2}\right)^2+2\left(\frac{1}{2}\rig ht)^3+ 3\left(\frac{1}{2}\right)^4+...$
I'm confused as to how to evaluate this sum
June 5th 2010, 02:49 AM
mr fantastic
The number $X$ of particles emitted as the result of an experiment is a random variable with probability distribution:
$P(X=k)=\left(\frac{1}{2}\right)^{k+1}\ for\ k \geq 0$
What is the expected number of particles emitted during one experiment
This is how I've gone so far:
$=\left(\frac{1}{2}\right)^2+2\left(\frac{1}{2}\rig ht)^3+ 3\left(\frac{1}{2}\right)^4+...$
I'm confused as to how to evaluate this sum
It is well known that the sum of an infinite geometric series is $\sum_{k=0}^{+\infty} r^k = \frac{1}{1-r}$ for |r| < 1. Therefore:
$\frac{d}{dr} \left(\sum_{k=0}^{+\infty} r^k\right) = ....$
$\Rightarrow \sum_{k=0}^{+\infty}\frac{d}{dr} \left( r^k \right) = ....$
$\Rightarrow \sum_{k=0}^{+\infty}k r^{k-1} = ....$
$\Rightarrow \sum_{k=0}^{+\infty}k r^{k+1} = r^2(....)$
June 5th 2010, 02:55 AM
im pretty rusty on infinite sums and it would be interesting to know if there is a closed form summation of that.
Nevertheless, you dont need to do anything difficult to solve this problem. You can rewrite as follows:
$P(x=k) = 0.5^k \times 0.5$
This is the PF of a geometric distribution with p=0.5. The mean of a geometric distribution is a standard result: $\frac{1}{p}$
edit too slow :D
June 5th 2010, 03:05 AM
oops i had the wrong form of the geometric distribution.
its $\frac{1-p}{p}$
June 5th 2010, 03:06 AM
It is well known that the sum of an infinite geometric series is $\sum_{k=0}^{+\infty} r^k = \frac{1}{1-r}$ for |r| < 1. Therefore:
$\frac{d}{dr} \left(\sum_{k=0}^{+\infty} r^k\right) = ....$
$\Rightarrow \sum_{k=0}^{+\infty}\frac{d}{dr} \left( r^k \right) = ....$
$\Rightarrow \sum_{k=0}^{+\infty}k r^{k-1} = ....$
$\Rightarrow \sum_{k=0}^{+\infty}k r^{k+1} = r^2(....)$
Yes, but for my case,
$\left(\sum_{k=0}^{+\infty} r^k\right) = 2$
So if you differentiate that, it just becomes 0
June 5th 2010, 03:11 AM
mr fantastic
No, the starting point is the first line of my previous reply. Yuo are trying to find a general formula. The, after you find it, you substitute r = 1/2.
June 5th 2010, 03:34 AM
Ok, so is this correct:
$\frac{d}{dr} \left(\sum_{k=0}^{+\infty} r^k\right) = \frac{d}{dr} \left(\frac{1}{1-r}\right)$ for $|r|<1$.
$\Rightarrow \sum_{k=0}^{+\infty}\frac{d}{dr} \left( r^k \right) = <br /> \sum_{k=0}^{+\infty}k r^{k-1}$
$\Rightarrow \sum_{k=0}^{+\infty}k r^{k+1} = \frac{r^2}{(1-r)^2}$ where $r=\frac{1}{2}$
$=\frac{\left(\frac{1}{2}\right)^2}{\left(\frac{1}{ 2}\right)^2}$
June 5th 2010, 03:42 AM
mr fantastic
Ok, so is this correct:
$\frac{d}{dr} \left(\sum_{k=0}^{+\infty} r^k\right) = \frac{d}{dr} \left(\frac{1}{1-r}\right)$ for $|r|<1$.
$\Rightarrow \sum_{k=0}^{+\infty}\frac{d}{dr} \left( r^k \right) = <br /> \sum_{k=0}^{+\infty}k r^{k-1}$
$\Rightarrow \sum_{k=0}^{+\infty}k r^{k+1} = \frac{r^2}{(1-r)^2}$ where $r=\frac{1}{2}$
$=\frac{\left(\frac{1}{2}\right)^2}{\left(\frac{1}{ 2}\right)^2}$
Yes. (Verification is simple using the well known formula given here: Geometric distribution - Wikipedia, the free encyclopedia).
June 5th 2010, 03:50 AM
Yes. (Verification is simple using the well known formula given here: Geometric distribution - Wikipedia, the free encyclopedia).
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Q-factor of an inductor
Hi all,
For a resonant system, there is usually a transfer of energy into another kind of energy back an forth (kinetic to potential; electric to magnetic, etc). for an LC tank or an RLC circuit, we know
that the energy is transfered from as an electric field between the capacitor's plate to a magnetic field around an inductor coil.
But, how come we can define a q-factor for an inductor alone?
Thank you very much for your help :) | {"url":"http://www.physicsforums.com/showthread.php?t=288391","timestamp":"2014-04-16T19:13:55Z","content_type":null,"content_length":"27978","record_id":"<urn:uuid:ec96cfbb-70fd-4884-b952-aed6d796649c>","cc-path":"CC-MAIN-2014-15/segments/1398223202774.3/warc/CC-MAIN-20140423032002-00565-ip-10-147-4-33.ec2.internal.warc.gz"} |
Does the Border (Boundary) Points of a convex body make a concave function?
up vote 2 down vote favorite
Let $\mathbb{S}$ be a closed and bounded convex body in 2-D with some non-empty intersection with positive quadrant and let it also contain origin. Let $c>0$ be the right-most point on the x-axis
such that $(c,y)\in \mathbb{S}$ for some $y$. Define the function f(x)=\max_{(x,y)~\in~\mathbb{S}}y ~~,x\in[0,c] Clearly,for a given $x'$, $f(x')$ is the northernmost point in the vertical strip $x=
x'$, is $f(x)$ a concave function? (or does it have some nice properties.).
If you look at $f(x)$, it is the pointwise supremum of an affine function. And also, all the examples I can imagine is concave.
real-analysis convex-polytopes linear-algebra
3 Isn't it an instant corollary to the very definition of a convex set? (also, it can be the whole $[b;c]$ interval, and never mind $0$). – Wlodzimierz Holsztynski May 3 '13 at 2:11
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1 Answer
active oldest votes
As Wlodzimierz points out, the answer is trivially yes. Maybe it would help to recall that $S$ is compact, so that for each $x$ there exists $y$ such that $(x,y) \in S$ and $f(x) = y$.
So if $f(x_1) = y_1$ and $f(x_2) = y_2$ then the point $\frac{1}{2}((x_1, y_1) + (x_2, y_2)) = (\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2})$ belongs to $S$, and hence $f(\frac{x_1 + x_2}
{2}) \geq \frac{y_1 + y_2}{2}$. Also it's easy to see that $f$ is continuous.
up vote 2 down But it's interesting to note that continuity fails in ${\bf R}^3$. Let $S$ be the convex hull of the circle in the $xy$-plane with center $(1,0,0)$ and radius $1$, and the point $
vote accepted (0,0,1)$. So $S$ is a cone. Now if we set $f(x,y) = {\rm max}_{(x,y,z) \in S} z$ for $(x,y)$ belonging to the disc with center $(1,0)$ and radius $1$, we find that $f(x,y) = 0$ on the
boundary circle except at the point $(0,0)$, where it takes the value $1$.
@Nik -- you mean, of course, that the continuity fails (while concavity still holds, and in any dimension). (Just a clarification). – Wlodzimierz Holsztynski May 3 '13 at 5:46
@Wlodzimierz: yes. I've edited my answer to clarify this. – Nik Weaver May 3 '13 at 7:51
+1, nice. I got the proof. I am not a mathematician and it is not obvious to me as it is to you people. Can you give an intuitive explanation for this? – dineshdileep May 4 '13 at
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Not the answer you're looking for? Browse other questions tagged real-analysis convex-polytopes linear-algebra or ask your own question. | {"url":"http://mathoverflow.net/questions/129478/does-the-border-boundary-points-of-a-convex-body-make-a-concave-function","timestamp":"2014-04-18T13:43:13Z","content_type":null,"content_length":"57048","record_id":"<urn:uuid:cd2dea04-45b4-4fca-a43f-e4d900dc0863>","cc-path":"CC-MAIN-2014-15/segments/1397609533689.29/warc/CC-MAIN-20140416005213-00261-ip-10-147-4-33.ec2.internal.warc.gz"} |
Ecology and Society: Adaptive Harvesting in a Multiple Species Coral Reef Food Web
Fig. 11. Alternate measures of resilience: the proportion of the plane space of the coral/algae phase in the basin of attraction of the coral-dominated state (11A) and the minimum Euclidean distance
between the coral-dominated equilibrium state and the boundary of its basin of attraction (11B). At a population of 500, there is no coral-dominated state. Instead, there is an algae-dominated state
and a state characterized by stable limit cycles. In this case, the Euclidean distance is a measure from the stationary state of the stable limit cycle. | {"url":"http://www.ecologyandsociety.org/vol13/iss1/art17/figure11.html","timestamp":"2014-04-16T08:21:20Z","content_type":null,"content_length":"1609","record_id":"<urn:uuid:034d449d-c67b-47f9-b9b1-17983228d807>","cc-path":"CC-MAIN-2014-15/segments/1398223206770.7/warc/CC-MAIN-20140423032006-00340-ip-10-147-4-33.ec2.internal.warc.gz"} |
Multiplying and Adding Fractions
Date: 04/01/97 at 23:31:39
From: Eric Pelfrey
Subject: Multiplying and Adding Fractions
(1/3 + 2/5) * 3/4. I can't find the answer!
Please help me.
Date: 04/02/97 at 06:26:04
From: Doctor Mitteldorf
Subject: Re: Multiplying and Adding Fractions
Dear Eric,
People spend a lot of time in school studying fractions before they
take on a problem as tough as this one.
Here's a way you might think about it. First, you might break up
the multiplication: the 3/4 multiplies the sum of the other two
fractions, but you could just as well make it multiply each one
1/3 * 3/4 + 2/5 * 3/4
The first part is 1/3 of 3/4. Well, this is easy, since 1/3 of 3
anythings is 1 anything. If you have 3 fourths, you have 1/4, 1/4
and 1/4. So a third of that is just 1/4.
The next part is harder. 2/5 of 3/4. I'd think of my 2/5 as 4/10 to
start with, if I were you. Then it's just 3/4 of 4/10 that you want.
Well, 1/4 of 4 tenths is just 1 tenth - same way we did the other one.
If 1/4 of 4 tenths is 1 tenth, then 2/4 of it makes 2/10 and 3/4 of it
makes 3 tenths. So the answer is 3/10.
Here's what we've got so far:
(1/3 + 2/5) * 3/4
1/3 * 3/4 + 2/5 * 3/4
1/4 + 3/10
Now what's left is to add 1/4 and 3/10. I don't know if you've
studied this yet - there's a trick that they teach you in 6th grade,
that goes like this:
Think of the 1/4 as so many twentieths. It's the same as 5/20.
Think of the 3/10 as so many twentieths. It's the same as 6/20.
Now you can just add them up.
1/4 + 3/10 is the same as 5/20 + 6/20, which is 11/20. That's the
Whew - that had a lot of steps. Did you get them all? Anything we
should go over again?
The one thing I'd be asking if I were you: How did I know to change
the 1/4 and 3/10 to twentieths? Where did I get the number 20? Are
there any other numbers I could have used instead of 20?
What do you think?
-Doctor Mitteldorf, The Math Forum
Check out our web site! http://mathforum.org/dr.math/ | {"url":"http://mathforum.org/library/drmath/view/58933.html","timestamp":"2014-04-17T08:31:45Z","content_type":null,"content_length":"6836","record_id":"<urn:uuid:b74d2ac9-9b6d-40a4-a724-69e6c81c9920>","cc-path":"CC-MAIN-2014-15/segments/1397609535095.7/warc/CC-MAIN-20140416005215-00057-ip-10-147-4-33.ec2.internal.warc.gz"} |
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tructures & Algorithms - (
1. The space factor when determining the efficiency of algorithm is measured by
2. The complexity of the average case of an algorithm is
3. The indirect change of the values of a variable in one module by another module is called
4. Which of the following data structure is linear data structure?
5. Each array declaration need not give, implicitly or explicitly, the information about
6. Which of the following case does not exist in complexity theory
8. Two main measures for the efficiency of an algorithm are
10. The time factor when determining the efficiency of algorithm is measured by
11. The Worst case occur in linear search algorithm when
13. The elements of an array are stored successively in memory cells because
16. Which of the following data structure is not linear data structure?
19. Finding the location of the element with a given value is:
20. The operation of processing each element in the list is known as | {"url":"http://www.proprofs.com/quiz-school/story.php?title=data-structures-algorithms-mcqs-objective-set-1","timestamp":"2014-04-16T13:05:00Z","content_type":null,"content_length":"185385","record_id":"<urn:uuid:2ba3905e-5589-463f-8fbb-9dab0d54d19d>","cc-path":"CC-MAIN-2014-15/segments/1398223206647.11/warc/CC-MAIN-20140423032006-00485-ip-10-147-4-33.ec2.internal.warc.gz"} |
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I need help with one more trig prob, on how to set it up
April 26th 2009, 11:59 AM #1
Apr 2009
I need help with one more trig prob, on how to set it up
Question: The world's tallest fountain is in fountain hills, Arizona. If the angle of elevation to the top of the fountain from a point 755 ft from its base is 36.5 degrees, find the height of
the fountain.
I don't know how to put this in a picture.
Draw a horizontal line for the ground.
Draw a vertical line for the height of the water in the fountain's spray.
Draw a slanty line from the top of the vertical line, down to somewhere along the horizontal line, indicating the line of sight.
Find the triangle in your picture.
Label the intersection of the horizontal line and the vertical line as a right angle.
Label the other base angle with the given angle measure.
Label the height as "h".
Label the base with the given length.
Now that you have your picture, solve for the height.
April 26th 2009, 12:15 PM #2
MHF Contributor
Mar 2007 | {"url":"http://mathhelpforum.com/trigonometry/85772-i-need-help-one-more-trig-prob-how-set-up.html","timestamp":"2014-04-18T08:48:53Z","content_type":null,"content_length":"32852","record_id":"<urn:uuid:4d41ec83-ad4e-4ddc-b433-f75269cf22d0>","cc-path":"CC-MAIN-2014-15/segments/1398223206647.11/warc/CC-MAIN-20140423032006-00130-ip-10-147-4-33.ec2.internal.warc.gz"} |
Beta: The Alpha and Omega to Risk Analysis?
When people use this word, they might mean "second," as opposed to "first." Or they could be referring to a particle in physics, a protein conformation in biology, or a test in computer programming.
But when investors use it, they mean "risk."
Technically, according to sources such as the Fool glossary or Investor Words, beta is the measure of a security's volatility compared with the volatility of the stock market as a whole, often
measured by the S&P 500 index. It comes from the capital asset pricing model. In a sense, it indicates how much broad market moves affect a stock's price.
A beta greater than 1 means the security is more volatile than the overall market, while a value less than 1 means the security is less volatile. Some stocks, such as Boeing (NYSE: BA ) , have a
beta near 1, which means they track the movement of the overall stock market fairly closely. Boeing's current beta is 0.91. Other stocks, such as Altria Group (NYSE: MO ) , have a beta less than 1
-- 0.50 in this case -- and thus have less volatility than the overall stock market. Still others, such as JDS Uniphase (Nasdaq: JDSU ) , have a large beta -- 3.22 for this company -- and have
wilder price swings than the overall stock market. Cash has a beta of 0, since it's always worth the same (excluding inflation effects), regardless of how the stock market moves.
How is beta calculated?
Well, for those who really want to get into it, there are several descriptions available, including this one. Suffice it to say that it's a statistical calculation of how a company's stock price
returns vary when compared with an index's returns.
For us more visual learners, a simpler way to calculate beta is the following.
• First, determine the price fluctuations for the index over some time period -- for instance, the end of each week for the past five years.
• Second, determine the price fluctuations for the stock over the same period.
• Third, make a graph with the index's fluctuations on the X-axis and the stock's fluctuations on the Y-axis. For instance, on Oct. 28, 2005, the S&P 500 had moved up by 1.6% from Oct. 21 -- from
1179.59 to 1198.41. Boeing's price moved down by 0.58% between those two weeks -- from $66.02 to $65.64. This would correspond to a point of (+1.60, - 0.58) on the graph.
• Repeat this procedure for the entire range of dates. (Hint: Use a spreadsheet to do all this.)
• Finally, fit a straight line to the plot and look at the slope. That is the beta. For an example of this using five years for Boeing, look here, where beta was calculated to be +1.20.
The easiest way to "calculate" it, though, is to let someone else do it. Many summary financial websites list beta as one of the company's key statistics. The values given earlier are from MSN
Money's website.
What does it all mean?
As mentioned above, beta is a measure of how volatile a stock's price is compared with the overall market. If the market rises 10%, then one could expect JDS Uniphase's price to rise by 32% (3.22
times 10%). On the other hand, Altria could be expected to rise only 5%.
This is also the case for downward movements. If the market drops by 5%, then these two companies could be expected to drop by 16% and 2.5%, respectively. In the capital asset pricing model, beta is
a measure of the company's market risk. That is, it measures how the company's price will probably react if the market goes up or down by some percentage.
What are some shortcomings to the calculation?
First, the value one gets from any of the above calculation methods is heavily dependent upon several considerations. These include how long a timeframe is used (one year, three years, etc.), how
often the prices are compared (weekly, monthly, etc.), and even when within the period the calculation begins (e.g., end of the month versus middle of the month). For example, if I calculate JDS
Uniphase's beta using closing weekly prices vs. the S&P 500 index over the past five years, I get 2.51. If instead I calculate it over the past three years, I get 2.18.
This also explains the difference astute readers may have noticed between my beta calculation for Boeing of 1.20 and the 0.91 beta number for Boeing I mention at the beginning of the article. MSN
Money's data provider for the 0.91 figure uses end-of-month data for the past five years, whereas my example used end-of-week data. Timeframe really matters.
Second, companies change over time, and, therefore, the beta does as well. Take Boeing, for example. Is it the same company it was five years ago? If not, then one shouldn't use a beta based on five
years of prices. Further, as companies become larger and more diversified, their beta tends to get closer to 1.
When Boeing first started out in Seattle, it built only airplanes (although for a time, it also built furniture). Over time, it expanded and started to manufacture joint direct attack munitions and
harpoon missiles, sold computer services, built interplanetary probes, and even managed housing projects and built desalinization plants. Obviously, it has become more diversified. In 1971, its beta
was 1.81. Today, its beta is 0.91.
Third, choice of index for comparison is important. Should one use the S&P 500, the Russell 2000, the Wilshire 5000, or the NYSE Composite? What if the company is foreign? Ideally, one should use as
broad an index as possible -- one that is reflective of the investor making the analysis about the company. Most information sources calculate a stock's beta based on the S&P 500, which is a
large-cap index. If an investor's portfolio is small cap in nature, however, a beta based on the Russell 2000 index would be more useful.
Finally, there's the R-squared, which ties into index choice. R-squared is a measure of how close all the data points are to the line itself. One can always fit a straight line to scattered data
points. But the line is more meaningful -- and thus, the beta is, too -- if the data points tend to cluster around the line rather than being all over the place. An R-squared of 1 (or 100%) means all
the points are exactly on the line. An R-squared of 0 means that the points are totally random and there is no correlation. One will get a very poor R-squared if one uses a non-relevant index. For
instance, the price of gold has a very poor correlation to the S&P 500 index and, thus, a very low R-squared -- the average R-squared for gold index funds versus the S&P 500 is about 0.03. In such a
circumstance of extremely low R-squared, beta loses all usefulness in predicting price movement.
People familiar with modern portfolio theory (MPT) will point out here that a high R-squared is important. For MPT, that is true. A fund with an R-squared of 0.90, for instance, is said to have 90%
of its price volatility occur due to the market and, therefore, is a well-designed fund for tracking that particular index.
We aren't discussing MPT, though. For individual companies, the R-squared value can vary wildly and does not have to be greater than the accepted 0.75 that MPT defines as being the threshold number
for a useful beta number. Individual companies do not, generally, track very closely with the market as a whole, unless they are very big and very diversified. Thus, an R-squared value of 0.34, as in
the above calculation of Boeing's beta, is perfectly valid. Nevertheless, it's true that an extremely low R-squared value should indicate a lack of faith in the calculated beta, maybe because of poor
index choice.
Are there other types of risk besides beta?
For many people, a wildly swinging stock price (e.g., JDS Uniphase) is perceived to be more risky than one that plods along (Altria). This perception has led many investors to equate beta with
overall risk. So people who don't want a lot of exposure to risk choose lower-beta stocks as their investments of choice.
In reality, however, there are other kinds of risk -- not just those associated with the overall market. Think of company-specific risk. Take the popular restaurant chain Cheesecake Factory (Nasdaq:
CAKE ) , with its beta of 0.70. Less variable than the market seems all right, doesn't it? But if that is the only consideration of risk you make, then you're ignoring the possibilities that the
popularity of this chain will wane or that the company will saturate its locations and lose profitability. Is The Cheesecake Factory still a popular restaurant? Can the company open multiple
locations in the same city without cannibalizing other locations? The answers are not found by looking at the beta.
How do you take into account these other risks? As a previous article pointed out, the answer is "information." By digging into a company's annual and quarterly reports, by looking at other companies
in the same industry, and by reading press releases and news articles about the company, you can amass a wealth of knowledge that will give you a sound basis for assigning different levels of
probability to different scenarios.
Note that in a diversified portfolio, company-specific risk is muted since factors that affect one company specifically would not likely affect another one. Thus, the best way to guard against
company-specific risk is to diversify one's portfolio across many stocks in different industries.
In addition, a stock's correlation with the overall stock market is not the only measure of correlation an investor needs to know about. One can also mute overall wild swings in a portfolio by
choosing equities that are not highly correlated with each other.
Most investors should probably have a portion of their portfolio in conservative companies with lower betas -- that is, stocks with a low-price volatility. But beta shouldn't be your only look into
the risk of investing in each company. Take advantage of the information out there, whether it be here at The Motley Fool or at other sites such as Yahoo! Finance, MSN Money, or the company's own
website. Armed with this knowledge and a healthy appreciation of the other risks present at a company, one can make intelligent investment choices and move even further away from the image that
investing is no more than gambling.
Related statistical Foolishness:
Fool contributor Jim Mueller doesn't really use beta in his own investing decisions, but he does look at company- and industry-specific risks. Besides, he likes to learn about different aspects of
investing. He does not own shares in any company mentioned here. Check out the Fool's fascinating disclosure policy!
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The Matthew Maths Show
Another hiatus in blog posts: I've been attending Prof Shahn Majid's course in Noncommutative Geometry at the
London Taught Course Center
Additionally, this whole month I've been busy writing my MSci project and preparing for my presentation. I still have so much to do, and little more than a week left! Exam revision comes afterwards,
so I don't expect to have any more exciting maths post during that period either. Though I am hoping to write a post or two about 1) Ergodic theory 2) Cornelissen and Marcolli's paper on Bost-Connes
systems and isomorphisms of number fields, 3) Bora Yaklinoglu's proof of a full arithemtic subalgebra for BC-systems (assuming I get a chance to see it!) and/or 4) Things I've learned from the LTCC
In the meantime, I thought I'd post a link to draft of my
slides for my MSci presentation
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Why is the transfer map Tate-dual to restriction ?
up vote 4 down vote favorite
In one of their papers (before Theorem 7.2), Benson and Carlson state that the transfer map is Tate-dual to the restriction homomorphisms (also see Remark 1.3 of this recent paper).
More precisely: If $H \le G$ are finite groups and $k$ a field those characteristic divides the order of $H$, then the there should be a commutative diagramm $$\begin{array}{ccc} \hat{H}^{-s-1}(G,k)
& \cong & \text{Hom}_k(\hat{H}^sS(G,k),k) \newline res^G_H \downarrow & & \downarrow (tr^G_H)^\ast \newline \hat{H}^{-s-1}(H,k) & \cong & \text{Hom}_k(\hat{H}^sS(H,k),k) \end{array}$$ where the
horizontal isomorphisms are Tate duality.
Does anyone know a reference with a proof or can provide a proof of this statement ? Thanks in advance.
group-cohomology reference-request homological-algebra
If you pass to a broader context then you can use degree-shifting. Namely, consider the category of all finite-dimensional $k$-linear representations $V$ of $G$, so the cup product pairing in Tate
cohomology ${\rm{H}}^s(G,V) \times {\rm{H}}^{-s-1}(G,V^{\ast}) \rightarrow {\rm{H}}^{-1}(G,k) = k$ is a perfect duality (if I remember correctly), and likewise for cohomology of $H$. Then by the
two-sided "erasability" of Tate cohomology and variation in $V$ you should be able to degree-shift the problem to the case $s = 0$, which can be analyzed by inspection. – user29720 Dec 10 '12 at
To clarify, at the end of my preceding comment I meant "the case $s = 0$ with coefficients in $V$ and $V^{\ast}$" (not just $s = 0$ with coefficients in $k$, which is too restrictive). – user29720
Dec 10 '12 at 5:43
Do you mean $\hat{H}$ instead of $H$ in your diagramm ? – Ralph Dec 11 '12 at 15:31
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1 Answer
active oldest votes
I don't know of a reference, but the duality in question can be proved by results from Brown's book on group cohomology. I'll show the case $s\ge 0$. First note that for each integer $j$
there is an isomorphism
$$\psi: \hat{H}^j(G,k) \xrightarrow{\sim} \text{Hom}_k(\hat{H}_j(G,k),k)$$
(Brown, VI.7.2) and for $s\ge 0$ there is an isomorphism $$\varphi: \hat{H}_{-s-1}(G,k) \xrightarrow{\sim} H^s(G,k)$$
(Brown, VI.4). Denote the $k$-dual of a vector space or of a homomorphism by $(-)^\ast$. Tate duality is then the composition $$t = (\varphi^{-1})^\ast\circ \psi: \hat{H}^{-s-1}(G,k) \
xrightarrow{\sim} \hat{H}_{-s-1}(G,k)^\ast \xrightarrow{\sim} H^s(G,k)^\ast.$$ Hence we have to show the commutativity of the diagramm
$$\begin{array}{ccccc} \hat{H}^{-s-1}(G,k) & \xrightarrow{\psi} & \hat{H}_{-s-1}(G,k)^\ast & \xleftarrow{\varphi^\ast} & H^s(G,k)^\ast \newline {\scriptstyle \widehat{res}} \downarrow & &
\downarrow \scriptstyle res^\ast & & \downarrow \scriptstyle tr^\ast\newline \hat{H}^{-s-1}(H,k) & \xrightarrow{\psi} & \hat{H}_t(H,k)^\ast & \xleftarrow{\varphi^\ast} & H^s(H,k)^\ast \end
up vote 3 ($t$ stands for $-s-1$ which the editor doesn't accept!?) The commutativity of the left hand square follows right from the definition of the maps and the right hand square commutes if we
down vote can show the commutativity of the following square: $$\begin{array}{ccc} \hat{H}_{-s-1}(G,k) & \xrightarrow{\varphi} & H^s(G,k)\newline {\scriptstyle \widehat{res}} \uparrow & & \uparrow \
accepted scriptstyle tr^G_H \newline \hat{H}_{-s-1}(H,k) & \xrightarrow{\varphi} & H^s(H,k) \end{array}\tag{1}$$ In order to describe $\varphi$ on chain level, let $P \to k$ be a projective
resolution over $kG$ and let $F$ be a complete resolution such that $F_i=P_i$ and $F_{-i-1} = P_i^\ast$ for $i \ge 0$. Then $\varphi$ is induced by the composition $$\varphi: F_{-i-1}\
otimes_{kG}k=P_i^\ast \otimes_{kG}k \xrightarrow{\alpha\otimes id} \text{Hom}_{kG}(P_i,kG) \otimes_{kG} k\xrightarrow{\beta}\text{Hom}_{kG}(P_i,k)$$ where $\alpha(f)(x)=\sum_{g \in G}f(g^
{-1}x)g$ (Brown, VI.3.4) and $\beta(f \otimes a)(x)=f(x)a$ (Brown, I.8.3). Hence $$\varphi(f \otimes a)(x)=\sum_{g \in G}f(g^{-1}x)(ga)=\sum_{g \in G}f(g^{-1}x)a=tr^G_E(f)(x)a\tag{2}$$
where $f \in P_i^\ast, a \in k, x \in P_i$ and $E=\{1\}$.
On chain level $(1)$ is given by the diagramm $$\begin{array}{ccc} P_i^\ast \otimes_{kG} k & \xrightarrow{\varphi_G} & \text{Hom}_{kG}(P_i,k) \newline {\scriptstyle \kappa} \uparrow & & \
uparrow \scriptstyle tr^G_H \newline P_i^\ast \otimes_{kH} k & \xrightarrow[\varphi_H]{} & \text{Hom}_{kH}(P_i,k) \newline \end{array}\tag{3}$$
where $\kappa(f \otimes_H a)=f \otimes_G a$. With $f,a,x$ as above, we obtain $$(tr^G_H \circ \varphi_H)(f \otimes_H a)(x)=\sum_{g \in G/H}\varphi_H(f\otimes_H a)(g^{-1}x) \overset{(2)}{=}
\sum_{g \in G/H}tr^H_E(f)(g^{-1}x)a$$ $$\qquad=tr^G_H(tr^H_E(f))(x)a=tr^G_E(f)(x)a$$ $$\qquad\qquad=\varphi_G(f \otimes_G a)(x)=(\varphi_G \circ res)(f \otimes_H a)(x)$$ Thus the
commutativity of $(3)$ is shown. QED
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Blaise Pascal (1623 - 1662)
From `A Short Account of the History of Mathematics' (4th edition, 1908) by W. W. Rouse Ball.
Among the contemporaries of Descartes none displayed greater natural genius than Pascal, but his mathematical reputation rests more on what he might have done than on what he actually effected, as
during a considerable part of his life he deemed it his duty to devote his whole time to religious exercises.
Blaise Pascal was born at Clermont on June 19, 1623, and died at Paris on Aug. 19, 1662. His father, a local judge at Clermont, and himself of some scientific reputation, moved to Paris in 1631,
partly to prosecute his own scientific studies, partly to carry on the education of his only son, who had already displayed exceptional ability. Pascal was kept at home in order to ensure his not
being overworked, and with the same object it was directed that his education should be at first confined to the study of languages, and should not include any mathematics. This naturally excited the
boy's curiosity, and one day, being then twelve years old, he asked in what geometry consisted. His tutor replied that it was the science of constructing exact figures and of determining the
proportions between their different parts. Pascal, stimulated no doubt by the injunction against reading it, gave up his play-time to this new study, and in a few weeks had discovered for himself
many properties of figures, and in particular the proposition that the sum of the angles of a triangle is equal to two right angles. I have read somewhere, but I cannot lay my hand on the authority,
that his proof merely consisted in turning the angular points of a triangular piece of paper over so as to meet in the centre of the inscribed circle: a similar demonstration can be got by turning
the angular points over so as to meet at the foot of the perpendicular drawn from the biggest angle to the opposite side. His father, struck by this display of ability, gave him a copy of Euclid's
Elements, a book which Pascal read with avidity and soon mastered.
At the age of fourteen he was admitted to the weekly meetings of Roberval, Mersenne, Mydorge, and other French geometricians; from which, ultimately, the French Academy sprung. At sixteen Pascal
wrote an essay on conic sections; and in 1641, at the age of eighteen, he constructed the first arithmetical machine, an instrument which, eight years later, he further improved. His correspondence
with Fermat about this time shews that he was then turning his attention to analytical geometry and physics. He repeated Torricelli's experiments, by which the pressure of the atmosphere could be
estimated as a weight, and he confirmed his theory of the cause of barometrical variations by obtaining at the same instant readings at different altitudes on the hill of Puy-de-Dôme.
In 1650, when in the midst of these researches, Pascal suddenly abandoned his favourite pursuits to study religion, or, as he says in his Pensées, ``contemplate the greatness and the misery of man'';
and about the same time he persuaded the younger of his two sisters to enter the Port Royal society.
In 1653 he had to administer his father's estate. He now took up his old life again, and made several experiments on the pressure exerted by gases and liquids; it was also about this period that he
invented the arithmetical triangle, and together with Fermat created the calculus of probabilities. He was meditating marriage when an accident again turned the current of his thoughts to a religious
life. He was driving a four-in-hand on November 23, 1654, when the horses ran away; the two leaders dashed over the parapet of the bridge at Neuilly, and Pascal was saved only by the traces breaking.
Always somewhat of a mystic, he considered this a special summons to abandon the world. He wrote an account of the accident on a small piece of parchment, which for the rest of his life he wore next
to his heart, to perpetually remind him of his covenant; and shortly moved to Port Royal, where he continued to live until his death in 1662. Constitutionally delicate, he had injured his health by
his incessant study; from the age of seventeen or eighteen he suffered from insomnia and acute dyspepsia, and at the time of his death was physically worn out.
His famous Provincial Letters directed against the Jesuits, and his Pensées, were written towards the close of his life, and are the first example of that finished form which is characteristic of the
best French literature. The only mathematical work that he produced after retiring to Port Royal was the essay on the cycloid in 1658. He was suffering from sleeplessness and toothache when the idea
occurred to him, and to his surprise his teeth immediately ceased to ache. Regarding this as a divine intimation to proceed with the problem, he worked incessantly for eight days at it, and completed
a tolerably full account of the geometry of the cycloid.
I now proceed to consider his mathematical works in rather greater detail.
His early essay on the geometry of conics, written in 1639, but not published till 1779, seems to have been founded on the teaching of Desargues. Two of the results are important as well as
interesting. The first of these is the theorem known now as ``Pascal's Theorem,'' namely, that if a hexagon be inscribed in a conic, the points of intersection of the opposite sides will lie in a
straight line. The second, which is really due to Desargues, is that if a quadrilateral be inscribed in a conic, and a straight line be drawn cutting the sides taken in order in the points A, B, C,
and D, and the conic in P and Q, then
PA.PC : PB.PD = QA.QC : QB.QD.
Pascal employed his arithmetical triangle in 1653, but no account of his method was printed till 1665. The triangle is constructed as in the figure below, each horizontal line being formed form the
one above it by making every number in it equal to the sum of those above and to the left of it in the row immediately above it; ex. gr. the fourth number in the fourth line, namely, 20, is equal to
1 + 3 + 6 + 10.
The numbers in each line are what are now called figurate numbers. Those in the first line are called numbers of the first order; those in the second line, natural numbers or numbers of the second
order; those in the third line, numbers of the third order, and so on. It is easily shewn that the mth number in the nth row is (m+n-2)! / (m-1)!(n-1)!
Pascal's arithmetical triangle, to any required order, is got by drawing a diagonal downwards from right to left as in the figure. The numbers in any diagonal give the coefficients of the expansion
of a binomial; for example, the figures in the fifth diagonal, namely 1, 4, 6, 4, 1, are the coefficients of the expansion m things taken n at a time, which he stated, correctly, to be (n+1)(n+2)
(n+3) ... m / (m-n)!
Perhaps as a mathematician Pascal is best known in connection with his correspondence with Fermat in 1654 in which he laid down the principles of the theory of probabilities. This correspondence
arose from a problem proposed by a gamester, the Chevalier de Méré, to Pascal, who communicated it to Fermat. The problem was this. Two players of equal skill want to leave the table before finishing
their game. Their scores and the number of points which constitute the game being given, it is desired to find in what proportion they should divide the stakes. Pascal and Fermat agreed on the
answer, but gave different proofs. The following is a translation of Pascal's solution. That of Fermat is given later.
The following is my method for determining the share of each player when, for example, two players play a game of three points and each player has staked 32 pistoles.
Suppose that the first player has gained two points, and the second player one point; they have now to play for a point on this condition, that, if the first player gain, he takes all the money
which is at stake, namely, 64 pistoles; while, if the second player gain, each player has two points, so that there are on terms of equality, and, if they leave off playing, each ought to take 32
pistoles. Thus if the first player gain, then 64 pistoles belong to him, and if he lose, then 32 pistoles belong to him. If therefore the players do not wish to play this game but to separate
without playing it, the first player would say to the second, ``I am certain of 32 pistoles even if I lose this game, and as for the other 32 pistoles perhaps I will have them and perhaps you
will have them; the chances are equal. Let us then divide these 32 pistoles equally, and give me also the 32 pistoles of which I am certain.'' Thus the first player will have 48 pistoles and the
second 16 pistoles.
Next, suppose that the first player has gained two points and the second player none, and that they are about to play for a point; the condition then is that, if the first player gain this point,
he secures the game and takes the 64 pistoles, and, if the second player gain this point, then the players will be in the situation already examined, in which the first player is entitled to 48
pistoles and the second to 16 pistoles. Thus if they do not wish to play, the first player would say to the second, ``If I gain the point I gain 64 pistoles; if I lose it, I am entitled to 48
pistoles. Give me then the 48 pistoles of which I am certain, and divide the other 16 equally, since our chances of gaining the point are equal.'' Thus the first player will have 56 pistoles and
the second player 8 pistoles.
Finally, suppose that the first player has gained one point and the second player none. If they proceed to play for a point, the condition is that, if the first player gain it, the players will
be in the situation first examined, in which the first player is entitled to 56 pistoles; if the first player lose the point, each player has then a point, and each is entitled to 32 pistoles.
Thus, if they do not wish to play, the first player would say to the second, ``Give me the 32 pistoles of which I am certain, and divide the remainder of the 56 pistoles equally, that is divide
24 pistoles equally.'' Thus the first player will have the sum of 32 and 12 pistoles, that is, 44 pistoles, and consequently the second will have 20 pistoles.
Pascal proceeds next to consider the similar problems when the game is won by whoever first obtains m + n points, and one player has m while the other has n points. The answer is obtained using the
arithmetical triangle. The general solution (in which the skill of the players is unequal) is given in many modern text-books on algebra, and agrees with Pascal's result, though of course the
notation of the latter is different and less convenient.
Pascal made an illegitimate use of the new theory in the seventh chapter of his Pensées. In effect, he puts his argument that, as the value of eternal happiness must be infinite, then, even if the
probability of a religious life ensuring eternal happiness be very small, still the expectation (which is measured by the product of the two) must be of sufficient magnitude to make it worth while to
be religious. The argument, if worth anything, would apply equally to any religion which promised eternal happiness to those who accepted its doctrines. If any conclusion may be drawn from the
statement, it is the undersirability of applying mathematics to questions of morality of which some of the data are necessarily outside the range of an exact science. It is only fair to add that no
one had more contempt than Pascal for those who changes their opinions according to the prospect of material benefit, and this isolated passage is at variance with the spirit of his writings.
The last mathematical work of Pascal was that on the cycloid in 1658. The cycloid is the curve traced out by a point on the circumference of a circular hoop which rolls along a straight line.
Galileo, in 1630, had called attention to this curve, the shape of which is particularly graceful, and had suggested that the arches of bridges should be built in this form. Four years later, in
1634, Roberval found the area of the cycloid; Descartes thought little of this solution and defied him to find its tangents, the same challenge being also sent to Fermat who at once solved the
problem. Several questions connected with the curve, and with the surface and volume generated by its revolution about its axis, base, or the tangent at its vertex, were then proposed by various
mathematicians. These and some analogous question, as well as the positions of the centres of the mass of the solids formed, were solved by Pascal in 1658, and the results were issued as a challenge
to the world, Wallis succeeded in solving all the questions except those connected with the centre of mass. Pascal's own solutions were effected by the method of indivisibles, and are similar to
those which a modern mathematician would give by the aid of the integral calculus. He obtained by summation what are equivalent to the integrals of
This page is included in a collection of mathematical biographies taken from A Short Account of the History of Mathematics by W. W. Rouse Ball (4th Edition, 1908).
Transcribed by
D.R. Wilkins
School of Mathematics
Trinity College, Dublin | {"url":"http://www.maths.tcd.ie/pub/HistMath/People/Pascal/RouseBall/RB_Pascal.html","timestamp":"2014-04-19T03:23:10Z","content_type":null,"content_length":"15542","record_id":"<urn:uuid:410b9204-1e8b-4221-b752-c092435c889b>","cc-path":"CC-MAIN-2014-15/segments/1397609539493.17/warc/CC-MAIN-20140416005219-00507-ip-10-147-4-33.ec2.internal.warc.gz"} |
- -
Please install Math Player to see the Math Symbols properly
Click on a 'View Solution' below for other questions:
e Does the figure possess any line of symmetry? e View Solution
e The word DICE has _________line of symmetry. e View Solution
e Is AB a line of symmetry for the figure? e View Solution
e Count the number of lines of symmetry for the figure. View Solution
e How many lines of symmetry can be drawn to the figure? e View Solution
e Which of the figures can have more number of lines of symmetry? e View Solution
e How many lines of symmetry can be drawn through the figure? e View Solution
e How many lines of symmetry does an equilateral triangle have? e View Solution
e Is the vertical line drawn through the center of the figure a line of symmetry? View Solution
e Does the figure possess any line of symmetry? View Solution
e How many lines of symmetry does the figure has? View Solution
e How many lines of symmetry does a square have? View Solution
e How many vowels in the English alphabet have at least one line of symmetry? e View Solution
e Which of the following words have a vertical line of symmetry, when the letters of these words are arranged one below the other? e View Solution
e Does the figure have any lines of symmetry? View Solution
e How many lines of symmetry does the figure have? View Solution
e How many lines of symmetry does the figure shown have? e View Solution
e Which of the figures do not have a line of symmetry? e View Solution
e Which of the following line segments is the line of symmetry to the figure shown? e View Solution
e How many lines of symmetry can be drawn to the figure? e View Solution
e Does the figure possess any line of symmetry? View Solution
e How many lines of symmetry does a circle have? View Solution | {"url":"http://www.icoachmath.com/solvedexample/sampleworksheet.aspx?process=/__cstlqvxbefxaxbgdgjxkjhgh&.html","timestamp":"2014-04-18T20:47:32Z","content_type":null,"content_length":"60791","record_id":"<urn:uuid:e890844d-16b2-43dd-8798-341c54ced1c7>","cc-path":"CC-MAIN-2014-15/segments/1397609535095.9/warc/CC-MAIN-20140416005215-00571-ip-10-147-4-33.ec2.internal.warc.gz"} |
William W. Cohen's Papers: Formal Results
1. William Cohen (2009): Graph Walks and Graphical Models in SCS Technical Report Collection.
2. William W. Cohen, Matthew Hurst & Lee S. Jensen (2003): A Flexible Learning System for Wrapping Tables and Lists in HTML Documents in Web Document Analysis: Challenges and Opportunities, ed.
Antonacopoulos & Hu, Word Scientific Publishing. (Originally published as: William W. Cohen, Matthew Hurst & Lee S. Jensen (2002): A Flexible Learning System for Wrapping Tables and Lists in HTML
Documents in WWW 2002: 232-241; Lee S. Jensen & William W. Cohen (2001): A Structured Wrapper Induction System for Extracting Information from Semi-Structured Documents in Proc. of the IJCAI-2001
Workshop on Adaptive Text Extraction and Mining).
3. William W. Cohen, David McAllester, and Henry Kautz (2000): Hardening Soft Information Sources in KDD 2000: 255-259.
4. William W. Cohen (1998): Hardness Results for Learning First-Order Representations and Programming by Demonstration in Machine Learning 30(1): 57-87 (1998). (Originally published as: William W.
Cohen (1996): The Dual DFA Learning Problem: Hardness Results for Programming by Demonstration and Learning First-Order Representations (Extended Abstract) in COLT 1996: 29-40).
5. William W. Cohen (1995): Pac-learning non-recursive prolog clauses in Artif. Intell. 79(1): 1-38 (1995).
6. William W. Cohen and C. David Page Jr (1995): Polynomial learnability and inductive logic programming: Methods and results in New Generation Comput. 13(3&4): 369-409 (1995).
7. William W. Cohen (1995): Pac-learning recursive logic programs: Efficient algorithms in J. Artif. Intell. Res. (JAIR) 2: 501-539 (1995).
8. William W. Cohen (1995): Pac-learning recursive logic programs: Negative results in J. Artif. Intell. Res. (JAIR) 2: 541-573 (1995).
9. William W. Cohen (1994): Pac-learning nondeterminate Clauses in AAAI 1994: 676-681.
10. L. Thorn McCarty and William W. Cohen (1994): The case for explicit exceptions in Methods of Logic in Computer Science, 1(1).
11. William W. Cohen and Haym Hirsh (1994): Learning the CLASSIC description logic: Theoretical and experimental results in KR 1994: 121-133.
12. William W. Cohen and Haym Hirsh (1994): Learnability of description logics with equality constraints in Machine Learning 17(2-3): 169-199 (1994). (Originally published as: William W. Cohen and
Haym Hirsh (1992): Learnability of Description Logics in COLT 1992: 116-127).
13. William W. Cohen, Russell Greiner, and Dale Schuurmans (1994): Probabilistic hill-climbing in Computational learning theory and natural learning systems (Volume II), MIT Press..
14. William W. Cohen (1994): Incremental abductive EBL in Machine Learning 15(1): 5-24 (1994).
15. William W. Cohen (1993): Cryptographic limitations on learning one-clause logic programs in AAAI 1993: 80-85.
16. William W. Cohen (1993): Pac-learning a restricted class of recursive logic programs in AAAI 1993: 86-92.
17. William W. Cohen, Alex Borgida, and Haym Hirsh (1992): Computing least common subsumers in description logics in AAAI 1992: 754-760.
18. William W. Cohen (1992): Using distribution-free learning theory to analyze solution path caching mechanisms in Computational Intelligence 8: 336-375 (1992).
19. William W. Cohen (1993): Learnability of Restricted Logic Programs in Proc. of the Third International Workshop on Inductive Logic Programming (ILP-93).
[Selected papers| By topic: Matching/Data Integration| Text Categorization| Topic Modeling| Rule Learning| Explanation-Based Learning| Formal Results| Inductive Logic Programming| Information
Extraction| Collaborative Filtering| Applications| Intelligent Tutoring| Learning in Graphs| By year: All papers] | {"url":"http://www.cs.cmu.edu/~wcohen/pubs-f.html","timestamp":"2014-04-18T05:57:43Z","content_type":null,"content_length":"7123","record_id":"<urn:uuid:003cff0c-3369-4f57-aeb5-2a4d83640396>","cc-path":"CC-MAIN-2014-15/segments/1397609532573.41/warc/CC-MAIN-20140416005212-00629-ip-10-147-4-33.ec2.internal.warc.gz"} |
Interactive Statistics Tutorials in Stata
Christopher Ferrall
Queen's University
Journal of Statistics Education v.3, n.3 (1995)
Copyright (c) 1995 by Christopher Ferrall, all rights reserved. This text may be freely shared among individuals, but it may not be republished in any medium without express written consent from the
author and advance notification of the editor.
Key Words: Teaching aids; Econometrics; Monte Carlo experiments.
This paper discusses a set of programs written in the statistical package Stata that is designed to support interactive student tutorials. The tutorial package has several desirable features,
including customized tutorials, full student interaction, checking of student answers, repetition of practice problems using randomly chosen values, and a simple way to gauge student comprehension
even when students run the tutorials at home. As an example, a tutorial used in an undergraduate econometrics class is discussed. The example illustrates Monte Carlo experiments on the linear
regression model that allow students to demonstrate the validity of various formulas for the sampling distribution of ordinary least squares estimators.
1. Introduction
1 There is strong evidence that computer-assisted tutorials can help create an effective environment for learning statistics. In her review of the literature, Garfield (1995) indicates that
self-directed learning, feedback on student performance, and reduced computational burden are important elements of statistical tutorials. When applying these lessons to statistics, an effective
computer-assisted tutorial should:
1. Be interactive AND statistical. A rich interface must be combined with the capability to perform serious statistical calculations in the background. Since doing statistics necessarily means
using a statistical package, the tutorials should teach the material and how to use an underlying statistics package simultaneously. Relying on pedagogical statistical packages means that
students must learn (and purchase) other packages in more advanced classes or to complete research projects.
2. Lessen the drudgery of learning statistics. Allow students to enter expressions as answers rather than having to do side calculations. In numerical exercises, provide practice problems that
can be repeated with different values that are randomly chosen.
3. Give students feedback on their performance during the tutorial and assess student comprehension after the tutorial.
2 Identifying the features that make computer-assisted tutorials effective learning tools is akin to understanding the demand side of a market. Computer-assisted tutorials must also be effective from
the instructional, or supply, side. For instance, tutorial software that requires specialized computer hardware may not be feasible to implement even if it is an excellent teaching tool.
3 Two of the most costly aspects of teaching are coordinated class time and instructor development time. Computer-assisted tutorials can either reduce or increase reliance on these resources. To
reduce the demands on computer laboratories and instructor development time, computer-assisted tutorials should have two additional properties:
4. Be tutor-less and independent of specific hardware. If tutorial instructions can be provided electronically, a teaching assistant need not be present, and students can run tutorials in their
own time on their own machines.
5. Be programmable. Instructors must be able to customize the material covered in the tutorials as well as share tutorials with other instructors.
4 Roughly one-third to one-half of the students taking my undergraduate econometrics class own computers that can run statistical software. The tutor-less aspect of a tutorial system substantially
reduces the burden placed on campus computer capacity. A programmable tutorial system initially entails more development time than a completely "canned" package. However, if the choice is between
programming a tutorial and no computer-assisted tutorial that fits the course, then programming may be more attractive. In the long run, programmed tutorials can be shared and built upon so that
overall development time falls substantially.
5 This paper describes a system for creating and running statistical tutorials with the imaginative name "Tutor." Tutor is an attempt to implement the desirable features suggested above. An example
of a tutorial used in my undergraduate econometrics course, which is one step in a longer tutorial that introduces Monte Carlo experiments, demonstrates features and capabilities of Tutor.
2. How Difficult and Costly is Tutor to Implement?
6 Tutor is a set of programs written in the statistical package Stata (compatible with versions 3.0 and higher). Stata has several built-in features that make it possible to program interactive
sessions. The programs in Tutor simply make it easier to write interactive tutorials. Tutorials themselves are Stata programs. Some programs may simply display instructions and explanations on the
screen. Other programs that make up a tutorial may go through practice problems or allow students to fit data. To write customized tutorials an instructor needs to know enough Stata to replicate the
structure of sample tutorials already written.
7 The only requirement for using Tutor is a computer running Stata. Stata runs on most computer platforms, including DOS, Windows, OS/2, Macintosh, and most Unix platforms. An affordable version of
DOS and Macintosh Stata called Student Stata maintains all but a few features of the professional version, enabling students to purchase their own copy at a reasonable price. Stata also has
structured programming features that make it possible to write interactive programs. Hamilton (1993) is a student manual that accompanies Student Stata. Several other guides and textbooks that use
Stata are available.
8 Tutor does not use a graphical interface. From the student point of view, graphical interfaces are usually preferred, but relying on them would make Tutor less portable. As written, Tutor runs on
anything from a DOS 286 machine to a Unix workstation.
9 Affordability, portability, and programmability make Stata a good, but certainly not unique, platform for developing tutorial software for a single course. Since the student and professional
versions of Stata are fully compatible, students can make a seamless transition from learning statistics with Stata to doing statistics with Stata. From this broader perspective, tutorials written in
Stata or other professional packages are preferable to specialized tutorial packages.
3. How Students Use Tutor
10 To a student the command "tutor" looks like any other Stata command. Tutor defines several commands that help students go through the tutorial. Students download the Tutor commands once at the
beginning of the term and then download the tutorials as they become available. After entering Stata, a student starts a tutorial (say week5) by simply typing
. tutor week5
11 This loads the programs associated with Tutor and then loads the commands specific to week5. Then an introductory screen is displayed:
Screen 1. Start Up Screen
Welcome to ......
--------- . . --------- .-------. .-----.
| | | | | | | |
| | | | | | |-----\
| |_______| | |_______| | \
A set of programs and interactive tutorials
in econometrics written and developed by
Chris Ferrall
Department of Economics
Queen's University
Press any key to continue.
Do you want to see a summary of available tutor commands (y/n)?. y
Do you want to record what comes to screen in a file (y/n)?. y
12 The second question allows students to store output that comes to the screen in a file so that they do not have to keep notes while going through the tutorial. For instance, one tutorial goes
through the elements required to fully describe a hypothesis test. The elements are displayed on a single screen, and examples are given. Students can cut and paste the screen into their homework
assignments as a template for reporting their own tests.
13 The summary of Tutor commands look just like a Stata help screen (which it is):
Screen 2: The Tutor Help Screen
Here are the commands tutor understands:
. tutor <tutorial name>
. next
. goto <number>
. intro
. thelp
. quiz
tutor loads the named tutorial, displays its intro screen, goes to step 1
next takes you to the next step in the tutorial
goto takes you to the step in the tutorial you specify
intro re-displays the intro screen for the current tutorial
thelp displays this list of commands
quiz display the quiz question for this week if there is one.
14 Each of these commands is a simple Stata program which allows a student to navigate through steps in the tutorial. Next, the introduction to the tutorial is displayed (using the Tutor command
Screen 3: Introduction to a Tutorial
Tutorial Number 5
Steps in this Tutorial:
1. Introduction to Monte Carlo Experiments
2. Are OLS Estimates Unbiased Under A1-A5?
3. Specify and Run Your Own OLS Monte Carlo Experiment
4. List of Questions to Answer Using Monte Carlo
5. Displays the Currently Loaded OLS Monte Carlo Experiment
There is a quiz defined for this tutorial.
15 This screen summarizes for the student what the tutorial is going to cover. The tutorial is organized into steps. Each step is a Stata program. Tutorials can be a mix of simple steps that rely on
the textbook and Stata manual to teach the mechanics of Stata and complicated steps that demonstrate statistical results interactively or that give the students practice problems.
4. Commands to Support Instructor-Written Tutorials
16 To design a tutorial using Tutor an instructor must
• Choose what each step in the tutorial will do.
• Define a program for each step (called step1, step2, etc.) and give each step a title.
• Create each step using built-in Tutor commands and original code.
17 Besides the navigation commands that students see, Tutor includes several commands to support the writing of tutorials. Some of the important primitive commands are:
Command Purpose
------- -------
inprompt <ans> <prompt> Asks for input with <prompt> and checks against
the correct answer <ans>. (Rounds to two
decimal places to allow for approximations and
round off.) Lets the student guess until he/she
gets the correct answer or gives up, and pauses
to perform Stata commands. Confirms the answer
or provides the correct answer.
quiz Resets the seed of the random number generator
using the student's identification number.
Calls the program doquiz to display the quiz
for the tutorial.
quizans Runs doquiz for each student number, computes
the correct answers and stores a table of
identification numbers and answers. (This
command is given only to the teaching
assistant to check answers submitted by
18 The inprompt command relies on Stata's display command and its request ( ) option. These two elements allow Stata to ask for and display information interactively. Keyboard input is stored as a
string in the argument passed to request. The Tutor command inprompt evaluates the input. For example, a step in a tutorial may include the command
inprompt ans "What is the value of the test statistic?"
which displays the prompt and waits for the answer. Students can pause the tutorial to run Stata commands before entering their guesses. Any keyboard input equal to the correct answer will be
accepted. For instance, if ans=5, then correct responses include "5" and "10/2" and "mnx/se" if mnx and se are variables with current values 10 and 2. The value in ans may be randomly determined as
the student re-runs the tutorial. Inprompt lets the student try to get the correct answer as many times as the student wishes.
19 The quiz command allows the instructor to ask questions whose answers are randomly determined for each student. The correct answer depends upon a number entered by the student; the seed of a
pseudo-random number generator is set equal to a function of this input. (I assign a number to each student at the start of the course.) Students are instructed to e-mail their answers to the
teaching assistant. The quizans function runs the quiz question for each student number and creates a table of correct values for comparison to the student answers. This system provides a simple
means of gauging student comprehension even though students may be running the tutorial completely on their own.
20 These commands help make it easier to write steps of a tutorial. Because a step is itself a program, some steps of specific tutorials have become inherent commands in Tutor. Some of these include:
xdist Creates a univariate discrete distribution table
and asks students to compute selected moments of
the distribution.
xydist Does the same thing as xdist but with a bivariate
joint distribution. Independent and non-independent
distributions are randomly created.
ttable Displays statistical tables (standard normal and t)
to the screen.
5. Example: Designing and Running Monte Carlo Experiments
21 An example of how a tutorial works is Step 3 of the tutorial above. It allows the student to specify and run Monte Carlo experiments on the simple linear regression model:
Y = b1 + b2 * X + u
The student chooses values for b1 and b2, the distribution of u, the sample values of X, the sample size, and the number of Monte Carlo replications. For simplicity, X takes on equally spaced
values between LowX and UpX in each sample. Varying LowX and UpX alters the mean and variance of X, and the uniform spacing of X makes it possible to calculate important values such as \sum (X-\
barX)*(X-\barX) as functions of LowX, UpX, and N. This in turn makes it possible to inform students of the theoretical distribution of estimators before setting the sample. Students choose the
degree of normality in the error term by setting a parameter t where
u ~ N(0,sigma^2) with probability 1 - t, and
u ~ U(-sqrt(3)sigma,sqrt(3)sigma) with probability t.
Here U(a,b) denotes the uniform distribution on the interval (a,b). When t = 0, u is normal. Screen 4 shows how to run the default experiment. (Assumption A.6 is the assumption that the error
terms are normally distributed.)
Screen 4: Entering Parameters of the Experiment
STEP 3 Specify and Run an OLS Monte Carlo Experiment
See Steps 1 and 2 for explanation and example.
First you need to set up the parameters of the experiment:
Enter TRUE b1 to use (press ENTER to leave it equal to 1.5).
Enter TRUE b2 term to use (press ENTER to leave equal to -.6).
sigma is the square root of the variance of u.
Enter the TRUE sigma to use (press ENTER to leave equal to 3).
Enter the sample size (press ENTER to leave equal to 15).
Enter number of artificial samples (press ENTER to leave equal to 80).
Enter LOWER bound of X values (press ENTER to leave it equal to -2).
Enter UPPER bound of X values (press ENTER to leave equal to 2).
To satisfy Assumption A.6 enter 0 next or otherwise a number between 0 & 1.
Enter % of u's to be NOT normal (press ENTER to leave equal to 0).
Enter y next if this is the second of two related experiments.
Append results to a previous experiment for comparison (y/n)?.
22 After specifying the experiment the student reviews it before running the simulations.
Screen 5: Reviewing the Experiment to be Run
You have specified the following Monte Carlo Experiment
The Currently Loaded OLS Monte Carlo Experiment
Population Regression Function:
Y = 1.5 + -.6 * X + u
Summary of the PRF:
True beta1 = 1.5 True beta2 = -.6
Var(u) = 9 % of u's Uniformly Distr. = 0%
% of u's Normally Distr. = 100%
Summary of the Sample and the Experimental Design:
sample size (N) = 15 replications (# of samples) = 80
Smallest X = -2 Largest X = 2
So in each artificial sample
X1 = -2.0000 X2 = -1.7140 ... X15 = 2.0000
Mean of X = 0 Sum of X*X = 22.857
Sum of (X-meanX)*(X-meanX) = 22.857
Do you want to run it or not (y/n)?. y
23 The student can see the results of each iteration of the experiment to get a feel for how sampling variation moves the estimated parameters around as the population parameters remain the same.
After each iteration the student can choose to stop viewing the results and let the experiment run to completion automatically.
Screen 6: Output for One Monte Carlo Experiment
Population Regression Function:
Y = 1.5 + -.6 * X + u
Sample Regression Function in Sample #1
Y = 2.07 + -.11 * X + e
Source | SS df MS Number of obs = 15
---------+------------------------------ F( 1, 13) = 0.02
Model | .256067518 1 .256067518 Prob > F = 0.8869
Residual | 158.147646 13 12.1652035 R-square = 0.0016
---------+------------------------------ Adj R-square = -0.0752
Total | 158.403713 14 11.314551 Root MSE = 3.4879
y | Coef. Std. Err. t P>|t| [95% Conf. Interval]
x | -.105844 .7295393 -0.145 0.887 -1.681918 1.47023
_cons | 2.071906 .9005629 2.301 0.039 .1263584 4.017454
Notice the difference between the OLS ESTIMATES and the population PARAMETERS.
Press any key to see graph.
24 The graph in Figure 1 shows the the population and sample regression lines and the sample data. Once students are satisfied with what is happening, they can stop looking at each sample and let the
experiment run automatically. The results are then loaded and stored as a Stata data set. The student might be asked to demonstrate that the formulas for the variance of the estimated regression
coefficients are correct by looking at the sample variation across experiments. The Stata describe and summarize commands display the main features of the experimental data.
Figure 1. Monte Carlo Graph.
Screen 7: Summary of Monte Carlo Results
. describe
Contains data from mmm1.dta
Obs: 80 (max= 646) Monte Carlo Results, N=15
Vars: 7 (max= 99)
Width: 28 (max= 200)
1. expno float %9.0g Experiment # (either 1 or 2)
2. sample float %9.0g sample number
3. sighat float %9.0g estimate of sigma
4. b1hat float %9.0g estimate of true_b1
5. se1 float %9.0g estimated standard error of b1h
6. b2hat float %9.0g estimate of true_b2
7. se2 float %9.0g estimated standard error of b2h
Sorted by:
. summarize
Variable | Obs Mean Std. Dev. Min Max
expno | 80 1 0 1 1
sample | 80 40.5 23.2379 1 80
sighat | 80 2.851439 .5865814 1.277367 4.358493
b1hat | 80 1.628286 .7532563 -.1775271 3.810717
se1 | 80 .7362385 .1514547 .3298146 1.125358
b2hat | 80 -.5371064 .6468333 -2.19868 .7345619
se2 | 80 .5964213 .1226923 .2671804 .9116443
25 The true variance of the ordinary least squares estimator of b2 is
Var(u)/\sum(X-\barX)^2 = 9/22.857 = .394.
From the summary table, we can see that the sample standard deviation of b2hat across the 80 samples is 0.647 whose square (0.418) is close to the theoretical variance. Furthermore, the mean
value of the estimated standard error of b2hat (0.596) is close to both the actual and theoretical values. Seeing the connection between sampling variation and theoretical formulas is one of the
most difficult concepts to understand in statistics. This engine for Monte Carlo experiments and the integrated tutorial program give the students a much better chance of understanding the
meaning of the formulas derived in lecture.
26 The quiz for the tutorial containing this step might use the Monte Carlo engine described above to generate experimental results. For example, the quiz could ask the student to determine the
proportion of samples in which a 95% confidence interval contains the true value of b1, or it could ask how many times a hypothesis test about b1 is rejected by the data. Students are then led to see
directly the correct interpretation of statistical inference.
27 Besides the Monte Carlo exercise, tutorials have been written to demonstrate the central limit theorem, to practice calculating conditional means and variances, to let students visually minimize
the sum of squared residuals in a regression, to practice performing hypotheses tests, and to learn how to use the Stata commands required to complete homework assignments such as computing standard
errors of forecasts.
6. Conclusion
28 This paper has discussed Tutor, a program for designing interactive statistical tutorials using Stata. Tutor uses features of Stata to create an interactive environment that can be customized by
the instructor. An advantage of Tutor over free-standing educational software is that it is written within the confines of a professional statistical package that students can use beyond their
introductory class. Tutor therefore provides continuity between learning and doing statistics.
29 This paper has also argued that instructional software should be evaluated in broad terms. Tutorials and other learning aids should be evaluated for both their learning effectiveness and their
cost effectiveness. The high cost of developing and implementing computer-aided tutorials perhaps explains their slow adoption in the light of the evidence that their learning effectiveness can be
great. Ultimately, what should emerge is a system for designing tutorials that can be shared with and modified by other teachers so as to lower costs and to pool the talent of many instructors. The
tutorial system introduced in this paper is a step in this direction.
I would like to thank Joanne Roberts for her teaching assistance and the students in Economics 351 at Queen's University for their patience, enthusiasm, and pure ability. Tutor and the tutorials used
in my class are available by anonymous ftp at highway61.econ.queensu.ca in the directory pub/tutor or through the World Wide Web at http://highway61.econ.queensu.ca/pub/tutor.
Garfield, J. (1995), "How Students Learn Statistics," International Statistical Review, 63(1), 35-48. Also available at http://www.geom.umn.edu/docs/snell/chance/teaching_aids/isi/isi.html
Hamilton, L. C. (1993), Statistics With Stata 3.0, Belmont, CA: Duxbury.
Christopher Ferrall
Department of Economics
Queen's University
Kingston, Ontario K7K 3N6
The following Tutor files are needed to run the week5 tutorial.
The file tutor.readme gives instructions for running Tutor.
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A Small Example of Applicative Functors with Scalaz
I recently blogged about Functors and mentioned a mysterious beast called Applicative Functor. Here is a simple, complete example that shows how you can use Applicative Functors with the Scalaz
library, with references for further reading.
What’s the Problem?
Given a function that takes multiple arguments A,B,… that returns a Result, how can that function by applied to arguments M[A], M[B],… and get M[Result]?
where M is an Option, List, etc.
Applicative Functors
To understand the mechanics of how applicative functors solve our problem, please refer to the references below, in particular Heiko Seeberger’s Applicatives are generalized functors.
What I present below is an example of use. I hope you can take it away and benefit from it without necessarily understanding the theory. However, I thoroughly recommend further reading and study, its
fun and you will gain much deeper insights.
So without further ado, here’s the code:
import scalaz._
import Scalaz._
object Applicatives extends App {
* Lets assume some Options which must be applied to a function.
val x:Option[Int] = 2.some // scalaz enrichment for options
val y:Option[Int] = 3.some
val z:Option[Int] = 5.some
// Lets add them without applicative functors
val usingFor = for (theX <- x; theY <- y; theZ <- z)
yield theX + theY + theZ
val usingMaps = x flatMap
(theX => y flatMap
(theY => z map
(theZ => theZ + theY + theX)))
/* With scalaz we can do the following instead of for or maps
* First we need to put the function in the right form, curried.
* To understand why please read the references I've given below.
val addInts = ( (a:Int, b:Int, c:Int) => a + b + c ).curried
// apply the function to x, y and z
val sum = x <*> (y <*> (z map addInts)) // Some(10)
// Scalaz offers an alternative syntax that is easier to use
(x |@| y |@| z) {_ + _ + _} // Some(10)
* If one of the options is a none
* then the result of the whole expression will be none.
(x |@| none[Int]) {_ + _} // None
* The function can be any method, including 'apply'
case class Person(age: Int, height:Double, name: String)
* Person.apply method is a function (Int, Double, String) => Person
(some(2) |@| some(1.1) |@| none[String]) {Person.apply _}
// none
(some(4) |@| some(1.1) |@| "Angelica".some) {Person.apply _}
// Some(Person(4, 1.1, Angelica))
* The beauty of this is that it works with ANY higher kind, eg. List
* or your own types!
val l1 = 1 :: 2 :: Nil
val l2 = 3 :: 4 :: Nil
(l1 |@| l2) {_ + _} // List(4, 5, 5, 6)
(<xml/> |@| <xml2/>){_ ++ _} // List(<xml></xml><xml2></xml2>)
Further Reading
Recent Comments
• bhericher on A Small Example of Kleisli Arrows
• channing on A Small Example of Kleisli Arrows
• channing on A Small Example of Kleisli Arrows
• bhericher on A Small Example of Kleisli Arrows
• channing on A Small Example of Kleisli Arrows | {"url":"http://www.casualmiracles.com/2012/01/16/a-small-example-of-applicative-functors-with-scalaz/","timestamp":"2014-04-21T09:35:31Z","content_type":null,"content_length":"26465","record_id":"<urn:uuid:7c9faf44-c2c3-4217-9ffa-3c2331882550>","cc-path":"CC-MAIN-2014-15/segments/1397609539705.42/warc/CC-MAIN-20140416005219-00432-ip-10-147-4-33.ec2.internal.warc.gz"} |
The Kunen inconsistency and definable classes
up vote 5 down vote favorite
There is a tension between (1) interpreting proper class talk in set theory as talk about first-order formulas and satisfaction; and (2) taking it to be an interesting and non-trivial result that
there is no (non-trivial) elementary embedding from V into V and/or taking it to be an open question whether there can be such an e.e. in the absence of choice. Basically, there is a very simple
proof that there can be no definable e.e. from V into V (see Suzuki (1999)).
This tension was recently highlighted by Hamkins, Kirmayer, and Perlmutter (2012) (and pointed out here and here). There, the resolution was to give up on (1), since accepting it "does not convey the
full power of the [Kunen's] theorem" (p. 1873). But this is perhaps the only place I've seen this issue addressed. For instance, Kanamori seems to hold both (1) and (2) in The Higher Infinite: "By
“class” in the ZFC context is meant definable class,... $x \in M$ is merely [a] facon de parler" (p. 33); and "[t]he following unresolved question [i.e. whether there could be an e.e. from V into V
in the absence of choice] is therefore of foundational interest" (p. 324).
My question is: how do other set theorists prefer to resolve this tension?
Hamkins, J., Kirmayer, G., Perlmutter, N. (2012) ``Generalizations of the Kunen inconsistency". Annals of Pure and Applied Logic, 163, 1872–1890.
Suzuki, A. (1999) No elementary embedding from V into V is definable from parameters. Journal of Symbolic Logic 64, 1591-1594.
set-theory large-cardinals
Sam, your question is that you want to get a sense of the range of positions on the matter? – Joel David Hamkins May 3 '13 at 11:40
Hi Joel, I just wanted to know where set theorists stand on the question. In particular, I'm interested if any set theorists confronted with the result simply take the open question to now be
settled. – Sam Roberts May 3 '13 at 11:47
add comment
2 Answers
active oldest votes
My perspective on this issue is that there are a variety of ways to take the claim of the Kunen inconsistency, and we needn't pick a particular one as the only right one. Rather, we
gain a fuller perspective of the result by understanding the full robust context including all of the interpretations.
• Kunen proved his result in Kelly-Morse set theory, in large part in order that he could formalize what it means for a class function $j:V\to V$ to be (fully) elementary. In KM, we
can prove that that there is a satisfaction class, a truth predicate for first-order truth, and with this class (which is definable) one can express the elementarity of $j$ as a
single second-order assertion.
• Meanwhile, using the observation (Gaifman) that any cofinal $\Sigma_1$-elementary embedding is $\Sigma_n$-elementary for any meta-theoretic natural number $n$, we can formalize the
result in GBC as the claim that no class $j$ is a nontrivial cofinal $\Sigma_1$-elementary embedding. Thus, this kind of elementarity of $j$ becomes expressible as a first-order
assertion about $j$.
• We don't actually need full GBC, since for example global choice is not used, but only the usual AC for sets, and so this argument can be formalized in GB+AC.
• But actually, we don't need the full second-order part of GB, but only the ability to refer to the class $j$. So we can formalize the argument in $\text{ZFC}(j)$, the theory using
ZFC where the axioms of replacement is allowed to use formulas in which the class $j$ appears. (But we only insist on elementarity of $j$ in the language without $j$.) This theory
is used and suffices to show, for example, that the supremum of the critical sequence $\lambda=\sup_n\kappa_n$ exists.
• If one intends to rule out only definable class embeddings $j$, that is, ones which are classes in the ZFC sense of being first-order definable from set parameters, then as you
mentioned, there is an easy argument ruling them out, and this argument does not use AC. I do not know any set theorist, however, who takes this result as an answer to the question
up vote 8 of whether one can prove the Kunen inconsistency in ZF. Rather, this example reveals the issues of formalization, and shows us that it may be important to take more care in our
down vote formal treatment of the result.
• Meanwhile, a purely first-order version of the Kunen inconsistency is formalizable in ZFC, with no talk of classes of any kind, as the claim that there is no nontrivial $j:V_{\
lambda+2}\to V_{\lambda+2}$ for any $\lambda$. This version still uses AC, and it is open in ZF. It avoids the set/class issues underlying your question by noting that the Kunen
inconsistency proof establishes more by restricting to $V_{\lambda+2}$. This set version of the result implies the full result in any set theory capable of showing that a purported
class $j$ must have a closure point $\lambda$.
• The wholeness axiom gets around the issue of the previous point by stating the theory ZFC + "$j:V\to V$ is nontrivial and elementary in the language with a function symbol for $j$.
Elementarity is expressed by the scheme $\forall x[\varphi(x)\iff \varphi(j(x))]$.
• Various weakenings and strengthenings of the wholeness axiom are realized by making further claims about $j$, such as whether it has a critical point, whether it moves an ordinal,
etc. Also, one can make claims about the extent to which $j$ may appear in the ZFC axioms. Officially, $j$ is allowed in the separation axiom but not in replacement, and so models
of WA are not able to prove the supremum of the critical sequence exists.
• If one uses merely the model-theoretic concept of embedding, one would be considering $j:V\to V$ for which $x\in y\iff j(x)\in j(y)$. But now the point is that ZFC proves that there
are nontrivial embeddings. For example, we can inductively define $j(y)=\{j(x)\mid x\in y\}\cup\{\{\emptyset,y\}\}$, and prove that this is a nontrivial embedding $j:V\to V$. (See
my paper Every countable model of set theory embeds into its own constructible universe, to appear in the JML, for more information.)
I prefer to understand the Kunen inconsistency in the rich context of all these results, rather than pick just one perspective and say that that perspective is the right one.
Thanks, Joel! Clearly there are many hypotheses in the vicinity. For my purposes, though, it suffices that Kunen's result is considered non-trivial and/or that it's taken to be an
open question whether there could be an embedding in the absence of choice when we formulate things in terms of classes. My impression is now that set theorists are divided on this.
Does that sound right? – Sam Roberts May 10 '13 at 9:51
One more question (if that's ok!): although we can't prove the existence of a satisfaction class in NBG, don't we have the resources to define satisfaction with a $\Delta^1_1$
formula? – Sam Roberts May 10 '13 at 10:12
I don't think set theorists are divided on those points. Most set theorists agree with your characterization. The situation, however, is that set theorists have regrettably often been
somewhat sloppy with the formalization of the Kunen result, and one most often hears it asserted as "There is no nontrivial embedding from $V$ to $V$", which is ambiguous for the
reasons I've said. – Joel David Hamkins May 10 '13 at 10:25
About satisfaction classes, it is consistent with NGB that there is none, since if one takes only the first-order definable classes, then one has NGB but there is no satisfaction
class there by Tarski's theorem. But meanwhile, in any $\omega$-standard model, the satisfaction class is implicitly definable by a first-order formula (it is the unique class
satisfying the Tarskian truth conditions). – Joel David Hamkins May 10 '13 at 10:28
Ah, I see. So you're in the minority on that question! Re. satisfaction: sure, there is no satisfaction class, but I thought it was entirely straightforward to show that there is a ($
\Delta^1_1$) formula $\Phi(x, y)$ such that NBG proves $\Phi(\ulcorner\phi\urcorner, a) \leftrightarrow \phi(a(0),...,a(n))$, where $\phi$'s free variables are among $x_0,..,x_n$.
Indeed, I thought it was provable that $\Phi$ satisfies the general Tarski clauses. – Sam Roberts May 10 '13 at 11:25
add comment
I won't try to say what set theorists generally do, but I usually handle the problem as follows. Most of the time, I work in ZFC and I use "class" to mean a class definable with set
parameters. This is adequate most of the time --- for example, when I want to talk about $V$, $L$, $L[U]$, $L(\mathbb R)$, the elementary embeddings arising from measures, extenders, etc. In
situations where it isn't adequate, for example in saying how Kunen's theorem goes beyond Suzuki's, I would work in ZFC with the assumption that there is an inaccessible cardinal $\kappa$,
up vote and I would (temporarily) use "set" to mean an element of $V_\kappa$ and use "class" to mean a subset of $V_\kappa$. (As long as I don't need anything of even higher rank than classes, this
7 down is pretty much equivalent to working in Morse-Kelley set-class theory. But, once I'm working in a world that goes beyond what I'm calling sets, I figure I might as well continue the
vote cumulative hierarchy naturally rather than stopping after just one layer of non-sets.)
Thanks, Andreas! If it's not too much trouble, it would help me to get clearer on what considerations are at play here to know what you think about plural quantification plato.stanford.edu/
entries/plural-quant in this context. For instance, suppose that plural quantification allowed one to unproblematically simulate MK over V (not merely over some $V_\alpha$). Would you then
rather work in that setting when considering Reinhardt, supercompact etc cardinals or the question how Kunen's result goes beyond Suzuki's? – Sam Roberts May 10 '13 at 10:01
@Sam: I know very little about plural quantification, but my immediate reaction is that it's likely to introduce some awkwardness into these issues. Like MK, its formalization would
presumably rely on the pairing function in $V$ in order to code elementary embeddings (class-sized functions) as sets. In the MK context, this is quite familiar --- some people will even
tell you (whether in MK or ZF) that functions are sets of ordered pairs (as opposed to being coded as sets of ordered pairs), but in plural quantification, the same idea sounds a bit
strange. (continued in next comment) – Andreas Blass May 10 '13 at 14:37
To assert the existence of an elementary embedding we'd say "There are some ordered pairs such that ..." followed by a description of what it means for those pairs to constitute an
elementary embedding. Part of my difficulty here may be simply that this stuff is familiar in the set-based context but not in the plural quantification context. But in addition to that, it
seems to me that such use of plural quantification loses a lot of the naturality that makes plural quantification attractive in the first place. – Andreas Blass May 10 '13 at 14:41
Thanks, Andreas! I suppose it's a matter of familiarity - although one could simply work in MK but rely ultimately on a plural interpretation, it's also pretty straightforward to work
directly in plural logic. There a (proper class sized) function will be some ordered pairs. In any case, the reason I brought up plural quantification was because I was wondering if the
reluctance to employ MK class quantification was to do with ontological commitment to set-like entities which aren't actually sets. Whether or not plural quantification is natural, it at
least doesn't suffer from that problem. – Sam Roberts May 10 '13 at 19:36
add comment
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SPRING 2014 BMED 2400 A
Contents INTRODUCTION TO BIOENGINEERING STATISTICS
Syllabus Statistical Methods in Biomedical Engineering
Calendar Distributions and Models, Inference, Fundamentals of Regression and Experimental Design.
Calendar Philosophy
Homework This course is concerned with the use of statistical inference for the modeling and analysis of data from a variety of sources under the umbrella of biomedical engineering research.
Additional The orientation is applied rather than theoretical, but such theory as is necessary for a proper understanding of the methods will be covered. Data from engineering, biology, and
Material medical practice will be analyzed during the course. Coverage will include: design of biomedical studies, sample size problem, prediction via a statistical model, testing hypotheses,
Exams Bayesian methods, etc.
Text Software Support
The course will be supported by MATLAB and WinBUGS
1. Data and Descriptive Statistics
2. Distributions as Models for Observations
3. Normal Distribution and Relatives
4. Estimation and Testing Statistical Hypothesis
5. Bayesian Approaches
6. Two Sample Problems
7. One- and Two-Way ANOVA. Block Design. Examples of more complex experimental designs.
8. Some Non-parametric Procedures
9. Tables and Chi-Square Theory
10. Regression Linear, Logistic and Poisson
11. Case studies
Copyright © 2014 Brani Vidakovic. The course pages may not be reposted without written permission and may not be reprinted for profit. | {"url":"http://zoe.bme.gatech.edu/~bv20/bmed2400/","timestamp":"2014-04-17T06:46:20Z","content_type":null,"content_length":"4302","record_id":"<urn:uuid:ea8e6e2b-24e9-46f1-a459-1890cf603ab2>","cc-path":"CC-MAIN-2014-15/segments/1398223203841.5/warc/CC-MAIN-20140423032003-00064-ip-10-147-4-33.ec2.internal.warc.gz"} |
Linear Interpolation
From GPWiki
Linear interpolation is a method for constructing new data points between two existing data points in a linear fashion (analytically speaking using linear polynomials, geometrically speaking on a
stright line between two points).
The formula for such an interpolation is
$\mathbf{p}(u) = \mathbf{a} + u \times (\mathbf{b} - \mathbf{a}) \;\;\;\; 0 \leq u \leq 1$
where a and b are the two points to interpolate between and u indicates the ratio along the line from a to b of the desired point. If u lies outside the range [0, 1] this is known as extrapolation.
It may be helpful to note the following:
$\mathbf{p}(0) = \mathbf{a} \;\;\;\; \mathbf{p}(1) = \mathbf{b}$
This formula applies to points of any dimensionality.
For example, in R^3:
$\mathbf{a} = (x_0, y_0, z_0) \;\;\;\; \mathbf{b} = (x_1, y_1, z_1)$
$x(u) = x_0 + u \times (x_1 - x_0)$
$y(u) = y_0 + u \times (y_1 - y_0)$
$z(u) = z_0 + u \times (z_1 - z_0)$
A new point p at distance u would then be
$\mathbf{p}(u) = (x(u), y(u), z(u))$
Sometimes it is necessary to find both coordinates of a point on a line when only one coordinate is known (in addition to two points on the line). This can be accomplished by expanding the
two-dimensional case. Given the desired x coordinate, we can find the corresponding y value:
$\mathbf{a} = (x_0, y_0) \;\;\;\; \mathbf{b} = (x_1, y_1)$
$x(u) = x_0 + u \times (x_1 - x_0)$
$y(u) = y_0 + u \times (y_1 - y_0)$
Solving the first equation for u in terms of x:
$u(x) = \frac{x - x_0}{x_1 - x_0}$
and substituting into the second equation:
$y(x) = y_0 + (x - x_0) \frac{y_1 - y_0}{x_1 - x_0}$
TODO: Insert image to demonstrate usage.
This article or section is incomplete and may require expansion and/or cleanup.
Please improve this article or discuss the issue on the talk page.
Linear interpolation has a few common applications. It is frequently used as a way to interpolate between values in a table. It can also be used as a way to animate an object moving between two
points by using time as the ratio u.
TODO: Discuss bilinear interpolation.
See also | {"url":"http://content.gpwiki.org/index.php/Linear_Interpolation","timestamp":"2014-04-19T08:13:20Z","content_type":null,"content_length":"21006","record_id":"<urn:uuid:e267ec62-a7f4-4e0f-af4f-2d90710b5224>","cc-path":"CC-MAIN-2014-15/segments/1398223202548.14/warc/CC-MAIN-20140423032002-00015-ip-10-147-4-33.ec2.internal.warc.gz"} |
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Math Forum - Ask Dr. Math Archives: High School Discrete Math
This page:
discrete math
Dr. Math
See also the
Dr. Math FAQ:
permutations and
Internet Library:
discrete math
About Math
basic algebra
linear algebra
linear equations
Complex Numbers
Discrete Math
Fibonacci Sequence/
Golden Ratio
conic sections/
coordinate plane
practical geometry
Negative Numbers
Number Theory
Square/Cube Roots
Browse High School Discrete Math
Stars indicate particularly interesting answers or good places to begin browsing.
Selected answers to common questions:
Four-color map theorem.
How many handshakes?
Squares in a checkerboard.
Tournament scheduling.
Venn diagrams.
A student sees a palindrome in the date 01 02 2010, and wonders how to generate all such palindromic dates. Building on another math doctor's work with date arithmetic, Doctor Carter shares a
program written in C, then goes on to explain the purpose of each line of code.
All Derfs are Enajs. One-third of all Enajs are Derfs. Half of all Sivads are Enajs. One Sivad is a Derf. Eight Sivads are Enajs. The number of Enajs is 90. How many Enajs are neither Derfs nor
My daughter is doing a tree diagram using terms related to math "numbers." Could you please explain in lay terms what surds are?
A woman has 15 friends. For 35 days she wants to have dinner with 3 friends a day, arranging it so that each pair of friends will come only once. Is this possible?
What is discrete math? How do you use it?
Show that every odd square leaves a remainder 1 when divided by 8... Prove that n^5-n is divisible by 30... Suppose m is a positive integer divisible by 11...
A unique and interesting use for base 32 or "duotrigesimal" numbers.
Any suggestions for units on electoral math, for use with students in grades 7-9, with plenty of entry points for both beginning and advanced students?
Can you help me make a schedule to staff an ice cream parlor?
Is there an equivalence class containing exactly 271 elements?
Let X = {people in the world} and R be a relation on the set X... find the equivalence classes.
I am looking for a few simple applications of Eulerian and Hamiltonian tours - real-world applications for 8th grade students.
How can I prove that the number of persons who have shaken an odd number of hands is even?
There are three ways to go from Town A to Town B, and four ways to go from Town B to Town C. How many different ways are there to go from Town A to Town C, passing through Town B?
We are a small discrete math class of eight students studying logical arguments. Two arguments we have examined are "modus tollens" and "modus ponens." We understand the arguments but would like
to know what the terms mean in English.
What is the probability of my being related to a famous person of the past?
Can you prove that the factorial of a number (greater than 1) can never be a perfect square?
Find the smallest number (integer) that has 30 factors.
We have to make a sequence of numbers, all different, each of which is a factor or a multiple of the one preceding it.
I'd like to know if Fermat's problem is solved, and when chess is likely to be solved.
At some stops, the SLU Express bus picks up 5 people. At other stops, it picks up 2 and lets off 5...
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My question is about trying to find a formula between numbers.
Howlers are fractions like 16/64; when you cross out the 6 on the top and the bottom, you are left with 1/4, which is the simplified fraction. How can I find all 2-digit, 3-digit and 4-digit
Given that twelve is the least positive integer with six different positive factors (1,2,3,4,6,12), what is the least positive integer with exactly twenty-four positive factors?
How many ways are there to get from top left to bottom right on a square when there are three lines going across each way?
My formula works with the exception of the first term.
What does a five-set Venn diagram look like?
What do 'floor' and 'ceiling' mean in mathematics?
Find a formula connecting any (k+1) coefficients in the nth row of the Pascal Triangle with a single coefficient in the (n+k)th row.
How many triangles can you draw on a square grid of dots of size x*x?
Other than trial and error is there any scientific or mathematical way to solve the Four Color Problem? How about even explaining it in layman's terms?
Divide a circle in eighths. Use 4 colors to color the segments. Colors may be repeated as long as you use all 4 colors at least once. What are the total combinations possible?
If you wish to color in each "country" or "space" on a map in such a way that no two contiguous countries or spaces have the same color, what is the minimum number of colors you can use?
Do we need more than four colors to color a two-dimensional map?
The game of NIM is played with a bunch of beans....
Given some restrictions, calculate the number of possible 8-character passwords.
What math rule would I need to follow if I wanted to generate all possible combinations in a 50 number draw 6 lottery?
In a graph with infinite "points," if we colour the lines with two colors we'll have either a red or a blue infinite chain of lines, an infinite number of points, all of them joined to each other
with the same colour...
What are graphs with three vertices? Could you give me some examples?
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Plato's Ghost is the first book to examine the development of mathematics from 1880 to 1920 as a modernist transformation similar to those in art, literature, and music. Jeremy Gray traces the growth
of mathematical modernism from its roots in problem solving and theory to its interactions with physics, philosophy, theology, psychology, and ideas about real and artificial languages. He shows how
mathematics was popularized, and explains how mathematical modernism not only gave expression to the work of mathematicians and the professional image they sought to create for themselves, but how
modernism also introduced deeper and ultimately unanswerable questions.
Plato's Ghost evokes Yeats's lament that any claim to worldly perfection inevitably is proven wrong by the philosopher's ghost; Gray demonstrates how modernist mathematicians believed they had
advanced further than anyone before them, only to make more profound mistakes. He tells for the first time the story of these ambitious and brilliant mathematicians, including Richard Dedekind, Henri
Lebesgue, Henri Poincaré, and many others. He describes the lively debates surrounding novel objects, definitions, and proofs in mathematics arising from the use of naïve set theory and the revived
axiomatic method--debates that spilled over into contemporary arguments in philosophy and the sciences and drove an upsurge of popular writing on mathematics. And he looks at mathematics after World
War I, including the foundational crisis and mathematical Platonism.
Plato's Ghost is essential reading for mathematicians and historians, and will appeal to anyone interested in the development of modern mathematics.
"In Plato's Ghost, he has . . . present[ed] us with an ambitious and in many respects remarkable synthesis of the modern transformation of mathematics via structural and set-theoretic notions,
together not only with its logic and philosophy but also with related developments in artificial languages and psychology. . . . I can certainly recommend Plato's Ghost highly as a rich resource and
point of departure for readers who want to learn more about this exciting period in the development of modern mathematics."--Solomon Feferman, American Scientist
"This accessible, rigorous volume belongs in every serious library."--J. McCleary, Choice
"In a book aimed at the educated public, the author presents an impressive amount of data--both of the kind mathematicians with some awareness of the history of their subject may be aware of, and of
an entirely different kind, coming from the outskirts of mathematics, from philosophy, from physics, or from the popularization of mathematics, which will likely be new even to historians of
mathematics."--Victor V Pambuccian, Mathematical Reviews
"It is . . . no small assertion to say that the book under review, Plato's Ghost, is [Gray's] most far-reaching and ambitious work to date. . . . [T]here is a wealth of valuable data here which, if
not fully processed and pigeonholed, is at least tagged and cataloged in a helpful way. Plato's Ghost provides an insightful and informative resource for anyone doing mathematics today who has
wondered how (and perhaps why) the subject has come to possess the features it has today. The book gives us a lot to think about, which is exactly what a good history should do."--Jeremy Avigad,
Mathematical Intelligencer
More reviews
Table of Contents
Another Princeton book authored or coauthored by Jeremy Gray:
Subject Areas: | {"url":"http://press.princeton.edu/titles/8833.html","timestamp":"2014-04-19T06:55:20Z","content_type":null,"content_length":"18530","record_id":"<urn:uuid:a0c5707c-f34a-490f-a00d-d1894aeb59dc>","cc-path":"CC-MAIN-2014-15/segments/1397609536300.49/warc/CC-MAIN-20140416005216-00518-ip-10-147-4-33.ec2.internal.warc.gz"} |
Another DC Circuit (But more complicated with a SWITCH and Capacitor!)
1. The problem statement, all variables and given/known data
For the circuit shown below, the capacitors are initially uncharged. At t=0, the switch S is closed.
a) Determine the current in each resistor immediately after the switch is closed.
b) Determine the current in each resistor a very long time after the switch is closed.
c) Determine the voltage across the capacitor a very long time after the switch is closed.
After the switch has been closed for a very long time, it is reopened.
d) Determine the initial current I[0] through the 100 ohm resistor, as a function of time after the switch is reopened.
e) Determine I(t), the current through the 100 ohm resistor, as a function of time after the switch is reopened.
f) How long after the switch is reopened does the charge on the capacitor fall to 10% of its fully charged state?
2. Relevant equations
I through a capacitor is I = Cdv/dt = Q/C
Kirchoff's Loops Rules
3. The attempt at a solution
a) Immediately after the switch is closed, the capacitor behaves ls a short circuit and hence no current flows through the 100 ohm resistor.
I in the 200 ohm resistor = V/R=15V/200ohms=.075A
b) long after the switch is closed, the capacitor is fully charged and behanves as an open circuit. All of the current passes through both resistors.
Left Loop Clockwise:
Right Loop Counter clockwise: Q/C-100ΩI[1]-0
I[1] is the only current passing through both the resistors
I= 20A
as for the rest of the parts, I want to make sure I get A and B right before I move on. | {"url":"http://www.physicsforums.com/showthread.php?t=394138","timestamp":"2014-04-19T02:22:39Z","content_type":null,"content_length":"54008","record_id":"<urn:uuid:18357c02-b7fd-4fa8-8ca0-36e9ff4c4ab8>","cc-path":"CC-MAIN-2014-15/segments/1397609535745.0/warc/CC-MAIN-20140416005215-00172-ip-10-147-4-33.ec2.internal.warc.gz"} |
Number of results: 36
math pre-agebra
Monday, October 7, 2013 at 7:24pm by cayla
Pre- Agebra
what does the prefix centi- mean?????
Thursday, December 18, 2008 at 7:35pm by Taylor
simplify -14xy / 42x^5y^8
Tuesday, March 1, 2011 at 9:13pm by Jackster
(square root of x-10) - 3x = 1
Wednesday, March 27, 2013 at 9:32pm by Cedric
college agebra
Friday, May 3, 2013 at 9:08pm by jessica
math pre-agebra
Monday, October 7, 2013 at 7:24pm by Ms. Sue
Agebra 1
Is it possible for the additive inverse of a number to be its reciprocal?
Tuesday, October 6, 2009 at 3:46am by Shavaleir
Agebra 1
25d + .30(7+7)d = Cost Insert 4 for d and calculate.
Monday, November 26, 2012 at 9:38pm by PsyDAG
math pre-agebra
1356 - 284 = ?
Monday, October 7, 2013 at 7:24pm by Ms. Sue
agebra 2 honors
subsitution with 3 equations (solve for x,y,z) x+y-2z=5 -x-y+z=2 -x+y+32=4
Wednesday, December 7, 2011 at 8:01pm by nitra
12th grade agebra 2
find all real roots: x^4-6x^2+8=0 and the other polynomial is: 8x^3+1=0
Tuesday, April 20, 2010 at 12:24am by anna
Hey, im a little confused with this question: 2^n x 4^(n+1) / 8^(n-2) I have to change all bases to 2 then simplify fully. Thanks everoyne.
Tuesday, March 11, 2008 at 10:45pm by jack
Agebra 1
Look at the time at which you posted this question. Can you guess why no one has answered your question?
Tuesday, October 6, 2009 at 3:46am by Writeacher
Agebra 1
Its not like i have all night i do have a baby to get up with in 4 hours. Can sombody please help me??????????
Tuesday, October 6, 2009 at 3:46am by Shavaleir
If you are comparing similar geometrical figures, the answer is b). Your question seems to be about geometry, not agebra.
Wednesday, December 15, 2010 at 10:53pm by drwls
Pre- Agebra
I walk 3.5 kilometers every day. How far do I walk in days? Write this answer in meters!
Thursday, December 18, 2008 at 7:35pm by Taylor
Pre- Agebra
a kilometer is 1000 meters. 3.5km is 3500 meters. A centi is 1/100. 345cm=3.45m
Thursday, December 18, 2008 at 7:35pm by bobpursley
Pre- Agebra
1 km = 1,000 meters Multiply: 3.5 * 1000 = ? meters walked per day
Thursday, December 18, 2008 at 7:35pm by Ms. Sue
12th grade agebra 2
for the first one, all real roots are 2, -2, and the square root of 2. i'm not sure of the second one. hope this helps.
Tuesday, April 20, 2010 at 12:24am by Megan
oh no Vitaliy, this person is in 7th grade pre-agebra or 8th grade pre-algebra.
Thursday, February 5, 2009 at 6:58pm by haha
4^(n+1) = 2^[2*(n+1)]= 2^(2n+2) 8^(n-2)= (2^3)^(n-2) = 2^(3n-6) If the 8^(n-2) is supposed to be in a denominator, 2^n * 4^(n+1) ____________ 8^(n-2) = 2^(n + 2n + 2 -3n +6) = 2^8 = 256 no matter
what n is. Pick a value of n and prove it for yourself.
Tuesday, March 11, 2008 at 10:45pm by drwls
x feet = short piece. 12-x = long piece. 3x+8 = 12-x Solve for x.
Sunday, September 27, 2009 at 6:22pm by DrBob222
Can someone please help me.. find an equation that models the path of a satelite if its path is a hyperbola, a=55,000km and c=81,000km assume that the center of the hyperbola is the origin and the
tranverse axis is horizontal
Wednesday, February 27, 2013 at 10:07am by lee
A 12ft. board is cut into two pieces so that one piece is 8 ft. longer than 3 times the shorter piece. If the shorter piece is x feet long, find the lengths of both pieces
Sunday, September 27, 2009 at 6:22pm by Ursula
Hmmm. Have you any clues? what is the correct answer? see any mistakes in my agebra? If you go to wolframalpha.com and type in derivative y/(x+4y) = x^4 – 4 you will see the value of y', which is
what I got above. So, if my answer is wrong, I must have erred in my evaluation ...
Thursday, March 14, 2013 at 1:57pm by Steve
math pre-agebra
how do i write this equation cars the marton family took a vacation that covered a total distance of 1356 miles. the return of the trip was 284 miles shorter than the first part of the trip. how long
was the return trip?
Monday, October 7, 2013 at 7:24pm by cayla
Factor out 14xy(-)/14xy(3x^4y^7) now cancel out 14xy your left with -1/3x^4y^7
Tuesday, March 1, 2011 at 9:13pm by zane
12th grade agebra 2
For the first one : You put X = x˛ Then you have (variable change) : X˛ - 6X + 8 = 0 And you use common tools to solve that (use delta). You find X = 2 or X = 4. You have x˛ = X, so x = sqrt(X) or x
= -sqrt(X). Here, the solutions are sqrt(2), -sqrt(2), 2, -2. For the second ...
Tuesday, April 20, 2010 at 12:24am by Nolan
Agebra 1
by the same token, do you expect these volunteer tutors to be up all night ? Students posting here come from the west to the east coast of the US and Canada, and I have even answered questions from
students from Britain, Australia and New Zealand. As to your question, there is...
Tuesday, October 6, 2009 at 3:46am by Reiny
x + (8 + 3x) = 12 4x + 8 = 12 4x + 8 - 8 = 12 - 8 4x = 4 x = 1 One piece is 1 foot long The other piece is 11 feet long Does that make sense? (3 * 1) + 8 = 11
Sunday, September 27, 2009 at 6:22pm by Ms. Sue
L=laser tag, V=video games 3L+5V=17 (1) 4L+7V=23 (2) Since the question asks only about the video games, we want to eliminate the L's. We can do this by multiplying the first equation by 4 and the
second equation by -3. 12L+20V=68 (3) -12L-21V=-69 (4) Add 3 and 4 together. -V...
Friday, November 5, 2010 at 9:14pm by Jen
agebra 2 honors
substitution is a poor choice for your equations, elimination is the obvious way to go add 1st and 2nd --> -z = 7 z = -7 sub that into 2nd --> -x - y = 9 sub z=-7 into 3rd --> -x + y = 25 (I will
assume the 32 is a typo, and you meant 3z ) now add these last two ...
Wednesday, December 7, 2011 at 8:01pm by Reiny
Agebra 1
A rental car charges $25.00 per day plus $.30 for every mile the car is driven. Dale rents a car while his own car is being repaired, and he only drives it to and from work each day. Dale drives 7
miles each way to and from work. Write an expression to represents Dale's cost ...
Monday, November 26, 2012 at 9:38pm by israel oyedapo
Agebra Word Problem
What a steal of a Deal! Hunter went to TV City to buy a new TV. When he got there, he found that the manager was offering a 25% discount on every TV in the store. The one Hunter wanted was last
year's model so he got an additional 10% discount on his purchase. When he got to ...
Tuesday, December 6, 2011 at 2:53pm by Linda
Sam and Chris went to “Lots O Fun” to play laser tag and video games for Chris’s birthday. Sam played 3 games of laser tag, 5 video games, and spent $17 total. Chris played 4 games of laser tag, 7
video games, and spent $23 total. How much does one video game cost to play?
Friday, November 5, 2010 at 9:14pm by Mariel
college agebra
(3x^2 - 2x - 1)(x^2 + x + 5) = 3x^4 + 3x^3 + 15x^2 + -2x^3 - 2x^2 ..... etc (you should have 9 terms in this line) can you see what I am doing? it is just an expanded version of the old FOIL concept
I am sure you can finish it There will be all kinds of like terms, make sure ...
Friday, May 3, 2013 at 9:08pm by Reiny | {"url":"http://www.jiskha.com/search/index.cgi?query=Agebra","timestamp":"2014-04-19T13:36:50Z","content_type":null,"content_length":"15386","record_id":"<urn:uuid:f388933b-95f5-48f4-a13a-2a0dbd52d209>","cc-path":"CC-MAIN-2014-15/segments/1398223206147.1/warc/CC-MAIN-20140423032006-00184-ip-10-147-4-33.ec2.internal.warc.gz"} |
Homework Help
Posted by Anon on Wednesday, September 18, 2013 at 7:19pm.
Calculate the wavelength of a proton with energy 2.45 eV.
Please some help!
• Chemistry - DrBob222, Wednesday, September 18, 2013 at 9:09pm
2.45 eV = 3.92E-19 joules.
Then E = hc/wavelength.
E = joules
h = Planck's constant
solve for wavelength in meters.
• Chemistry - Anon, Thursday, September 19, 2013 at 8:51am
Can you write the result because i found 5.067e-7 and it's wrong.I try it many time in diffrent ways baut I finished with the some result!
• Chemistry - Anon, Thursday, September 19, 2013 at 2:11pm
Finally I found it.We have to do this :
E = joule
h = Planck's constant
m = kg
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❝I forgot what I sent you❞
Anonymous:What's your major in college?
… Y’see, now, y’see, I’m looking at this, thinking, squares fit together better than circles, so, say, if you wanted a box of donuts, a full box, you could probably fit more square donuts
in than circle donuts if the circumference of the circle touched the each of the corners of the square donut.
So you might end up with more donuts.
But then I also think… Does the square or round donut have a greater donut volume? Is the number of donuts better than the entire donut mass as a whole?
A round donut with radius R[1] occupies the same space as a square donut with side 2R[1]. If the center circle of a round donut has a radius R[2] and the hole of a square donut has a side 2R
[2], then the area of a round donut is πR[1]^2 - πr[2]^2. The area of a square donut would be then 4R[1]^2 - 4R[2]^2. This doesn’t say much, but in general and throwing numbers, a full box
of square donuts has more donut per donut than a full box of round donuts.
The interesting thing is knowing exactly how much more donut per donut we have. Assuming first a small center hole (R[2] = R[1]/4) and replacing in the proper expressions, we have a 27,6%
more donut in the square one (Round: 15πR[1]^2/16 ≃ 2,94R[1]^2, square: 15R[1]^2/4 = 3,75R[1]^2). Now, assuming a large center hole (R[2] = 3R[1]/4) we have a 27,7% more donut in the square
one (Round: 7πR[1]^2/16 ≃ 1,37R[1]^2, square: 7R[1]^2/4 = 1,75R[1]^2). This tells us that, approximately, we’ll have a 27% bigger donut if it’s square than if it’s round.
tl;dr: Square donuts have a 27% more donut per donut in the same space as a round one.
Thank you donut side of Tumblr.
(Source: nimstrz) | {"url":"http://itsbooey.tumblr.com/","timestamp":"2014-04-18T02:58:27Z","content_type":null,"content_length":"36591","record_id":"<urn:uuid:11b29e86-4d2e-47e0-b6df-4de11671ac4a>","cc-path":"CC-MAIN-2014-15/segments/1397609532480.36/warc/CC-MAIN-20140416005212-00313-ip-10-147-4-33.ec2.internal.warc.gz"} |
gcd(a,b)=pi^min{si,ti} and lcm(a,b)=pi^max{si,ti} proof?
October 8th 2008, 11:43 PM
gcd(a,b)=pi^min{si,ti} and lcm(a,b)=pi^max{si,ti} proof?
a>1 , b>1 are two integers(whole numbers)
s>=0 , t>=0
for each i=1,2,3,....,r is mi=min{si,ti} and ni=max{si,ti}.
show that : gcd(a,b)=(P1^m1)*...*(Pr^mr)
I've tried to ask the question as the way it is
Hope I'm clear enogh.
I would be glad if you explain the answers step by step.
Thanks in advance.
October 9th 2008, 05:25 AM
Use definition!
For example by definition $g=gcd(a,b)$
1) if g is a factor of a and b and i
2) if d is another factor of a and b then g divides d.
since $m_i\leq s_i and m_i\leq t_i$ then $p_i^{m_i}$divides both $p_i^{s_i}$ and $p_i^{t_i}$. Hence $p_1^{m_1}\cdots p_k^{m_k}$ divides both a and b.
Now supposed d divides $a=p_1^{s_1}\cdots p_k^{s_k}$ and let q be any of prime factor of d. Since d divides a then q must be one of $\{p_1,\cdots, p_k\}$. Then the prime factors of d are also $\
{p_1,\cdots, p_k\}$. So we can write $d=p_1^{u_1}\cdots p_k^{u_k}$ where $u_i\leq s_i$. By doing the same thing with respect to b, we also have $u_i\leq t_i$. Hence $u_i\leq min\{s_i,t_i\}$. Then
it follows that g divides d.
The same (similar) argument also works for lcm. | {"url":"http://mathhelpforum.com/number-theory/52776-gcd-b-pi-min-si-ti-lcm-b-pi-max-si-ti-proof-print.html","timestamp":"2014-04-20T09:29:13Z","content_type":null,"content_length":"8065","record_id":"<urn:uuid:7532c280-d118-4ef7-b0ac-80e76478bbc7>","cc-path":"CC-MAIN-2014-15/segments/1397609538110.1/warc/CC-MAIN-20140416005218-00160-ip-10-147-4-33.ec2.internal.warc.gz"} |
Limit Evaluation
Date: 7/7/96 at 16:21:41
From: Anonymous
Subject: Limit Evaluation
I'm in the first year of electronics in Buenos Aires and I have a
problem with a limit. I have no idea how to begin:
Lim Root x of (x)
X -> oo
Limit of root X of x, with X -> infinity.
Date: 7/7/96 at 19:3:28
From: Doctor Anthony
Subject: Re: Limit Evaluation
I believe you are wanting the xth root of x, which is better shown as
Let y = x^(1/x)
Now, with a variable as a power, it is usual to take logs.
Take logs (base e) of both sides
ln(y) = (1/x)ln(x) = ln(x)/x
Now as x -> infinity this becomes inf/inf and we can use l'Hopital's
rule to see what the ratio is becoming as x approaches its limit.
LT(as x->infin.)ln(y) = (diff. of top line)/(diff. of bottom line)
= (1/x)/1
= 1/x
->0 as x -> infin.
If ln(y) -> 0 then y -> 1 and so Lt(as x-> infin.) x^(1/x) = 1
-Doctor Anthony, The Math Forum
Check out our web site! http://mathforum.org/dr.math/
Date: 7/7/96 at 20:58:21
From: Anonymous
Subject: Re: Limit Evaluation
Thank you Mr. Anthony, this is the best WWW of math!!
Hernan Gabriel Zapata. (archie@giga.com.ar) | {"url":"http://mathforum.org/library/drmath/view/53348.html","timestamp":"2014-04-19T12:46:35Z","content_type":null,"content_length":"6116","record_id":"<urn:uuid:553c3fb3-477f-4feb-8984-331085abc660>","cc-path":"CC-MAIN-2014-15/segments/1397609537186.46/warc/CC-MAIN-20140416005217-00454-ip-10-147-4-33.ec2.internal.warc.gz"} |