url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3
values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93
values | snapshot_type stringclasses 2
values | language stringclasses 1
value | language_score float64 0.06 1 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
http://www.hammockforums.net/forum/showthread.php/16424-Sag | 1,521,820,518,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257648313.85/warc/CC-MAIN-20180323141827-20180323161827-00773.warc.gz | 376,860,672 | 16,141 | 1. ## Sag
Is there an established and accepted way of expressing the amount of sag in a suspension line or hammock? Expressions like "generous sag" or "more than you would expect" can mean one thing to one person and quite a different thing to another. Yet I haven't come across a post that was more specific.
2. I know with the Warbonnet hammocks its around 30 degrees.
3. Yes, 25 - 30 degrees is the usual starting point as shown in this illustration by dejoha.
Jerry
4. What makes a comfortable sag will be different for everyone...you might start with a 100" ridgeline and a 10' hammock body. That will set the sag of the hammock's body consistently every time, and it's what several of the popular hammocks use.
If you're doing it w/ a DIY hammock, then hanging it so your supports are about 30 degrees will be a good starting point. Check here for a couple of pics of sag to get an idea.
http://www.tothewoods.net/HammockGlossary.html
5. Nice illustration, but doesn't a ridgeline make the hang angle irrelevant? I thought that was the point. It'll make a difference in the amount of force being applied to the trees, but it won't change the way the hammock feels.
Isn't that the whole point of the ridgeline? Or have I completely misunderstood?
6. Ummmmm......sorry, sports fans.........I think I failed to make my question clear. A newcomer doesn't need to read many posts before coming across the suggestion of hanging at about thirty degrees. But to a lot of people, my dearly beloved wife included, the device for measuring degrees is a thermometer, not an inclinometer. Not for me, of course. I'm mathematically inclined. I even know what the word "trigonometry" means. Big deal! I went out and hung my hammock at "thirty degrees." Then I went back to my shop and dug up my inclinometer. To my ego-smashing astonishment, I was off by ten degrees. Do you have an inclinometer?
Is there an established and accepted way of expressing the amount of sag in a suspension line or hammock?
The answer, of course, is Yes: degrees. The question should have been
Is there a more useful way of expressing the amount of sag.....
I realize that a few degrees one way or the other is no big deal but I found that ten degrees is a big deal.
It seems to me that expressing sag in degrees (measured at the tree, not at the hammock, of course) is a disservice to those who are not mathematically inclined or experienced. Is this a non-issue???
7. I think saying 30 degrees from horizontal is the only practical recommendation. If you shoot for 30 and are off by 10 that shouldn't be a problem. But if someone doesn't know to try for 30 and instead thinks they should pull the suspension straight horizontal...well that's a problem.
If someone doesn't know what 30 degrees means they can find out by using the internet.
8. Originally Posted by Boris Losdindawoods
Nice illustration, but doesn't a ridgeline make the hang angle irrelevant? I thought that was the point. It'll make a difference in the amount of force being applied to the trees, but it won't change the way the hammock feels.
Isn't that the whole point of the ridgeline? Or have I completely misunderstood?
No, you've understood correctly. If you use a ridgeline, it makes your hang much more consistent. There are still a few times when it won't be...like when your ridgeline isn't taut, for example...but for the most part that's the function of a structural ridgeline.
It also means that you can get the optimal hang in suboptimal spots...like not having to put your supports 10' off the ground to get the right amount of sag.
But it's possible to hang with the right angle and get the same sag as if a structural ridgeline were used. That's about where the 30 degrees comes in.
9. I'm pretty good at eyeballing things, but how I wrap my brain around it is: Everyone knows 45 degrees, it's halfway between vertical and horizontal. I just picture 45 in my head and go "a smidge" lower, and usually end up close enough to 30 degrees for my satisfaction.
I've considered printing a little card with a 30-degree angle on it, but never cared enough to bother, but since you asked for an easy way to find 30, here ya go! I even had to break out my Soh-Cah-Toa! Hope you find it helpful.
30-degrees.jpg
Acer
10. It's as much art as science, and understanding can only come with experience. zengringo
#### Posting Permissions
• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
• | 1,075 | 4,518 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2018-13 | latest | en | 0.957757 |
http://www.storyboardthat.com/storyboards/maddymonappella/geometry | 1,527,485,615,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794871918.99/warc/CC-MAIN-20180528044215-20180528064215-00404.warc.gz | 473,508,500 | 15,688 | # geometry
More Options: Make a Folding Card
#### Storyboard Description
This storyboard does not have a description.
#### Storyboard Text
• Oh no! I need to figure out this problem and I don't know how to.
• I can help!
• I need to frame my circle mirror with wood and i need to know how much wood to buy.
• So you must find the cicumfrence of the circle which is the length around the mirror.
• I know that the diameter of the mirror is 14 inches so how would i get the circumference?
• Use the formula which is C=2πr
• So, to get the radius you take half of the diameter which would be 7 and plug that into r
• C=2π(7) which equals
Explore Our Articles and Examples
Try Our Other Websites!
Photos for Class – Search for School-Safe, Creative Commons Photos (It Even Cites for You!)
Quick Rubric – Easily Make and Share Great-Looking Rubrics
abcBABYart – Create Custom Nursery Art | 228 | 898 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2018-22 | latest | en | 0.901857 |
http://mathhelpforum.com/algebra/172010-how-should-i-solve-these-equations.html | 1,524,810,102,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125949489.63/warc/CC-MAIN-20180427060505-20180427080505-00157.warc.gz | 207,779,523 | 10,261 | # Thread: How should I solve these equations?
1. ## How should I solve these equations?
Hey all,
I am trying to solve these 2 equations but I can't.
$\displaystyle 27^x + 12^x = 2 * 8^3$
and,
$\displaystyle 3 * 16 ^x + 37 * 36^x = 26 * 81^x$
2. Originally Posted by mehdi
Hey all,
I am trying to solve these 2 equations but I can't.
$\displaystyle 27^x + 12^x = 2 * 8^3$
and,
$\displaystyle 3 * 16 ^x + 37 * 36^x = 26 * 81^x$
I can show you how to do the 2nd one - but it is kind of tricky:
$\displaystyle 3 * 16 ^x + 37 * 36^x = 26 * 81^x$
$\displaystyle 37 \cdot 6^{2x}+3 \cdot 2^{4x} = 26 \cdot 3^{4x}$
$\displaystyle 37 \cdot 2^{2x} \cdot 3^{2x}+3 \cdot 2^{4x} = 26 \cdot 3^{4x}$ divide by $\displaystyle 2^{2x}$
$\displaystyle 37 \cdot 3^{2x}+3 \cdot 2^{2x} = 26 \cdot 3^{4x} \cdot \dfrac1{2^{2x}}$ divide by $\displaystyle 3^{2x}$
$\displaystyle 37 +3 \cdot \left(\dfrac23\right)^{2x} = 26 \cdot 3^{2x} \cdot \dfrac1{2^{2x}} = 26 \cdot \left(\dfrac32\right)^{2x}$
Now substitute $\displaystyle y = \left(\dfrac32\right)^{2x}$. The equation becomes:
$\displaystyle 37+3 \cdot \dfrac1y=26 \cdot y$ multiply by y:
$\displaystyle 26y^2-37y-3=0$
yields $\displaystyle y = \frac32~\vee~y=-\frac1{13}$
Since y is a power with a positive base the 2nd solution can't be correct.
Now re-substitute:
$\displaystyle \left(\dfrac32\right)^{2x} = \dfrac32$
That means $\displaystyle 2x = 1~\implies~\boxed{x = \dfrac12}$
3. ...
4. Hello, mehdi!
Another approach to the second problem . . .
$\displaystyle 3\cdot16^x + 37\cdot36^x \:=\: 26\cdot81^x$
We have: .$\displaystyle 3(2^4)^x + 37(2^2\!\cdot\!3^3)^x - 26(3^4)^x \;=\;0$
. . . . . . .$\displaystyle 3(2^{4x}) + 37(2^{2x})(3^{2x}) - 26(3^{4x}) \;=\;0$
Factor: . $\displaystyle \bigg[2^{2x} + 13\!\cdot\!3^{2x}\bigg]\,\bigg[3\!\cdot\!2^{2x} - 2\!\cdot\!3^{2x}\bigg] \;=\;0$
We have two equations to solve:
$\displaystyle 2^{2x} + 13\!\cdot\!3^{2x} \:=\:0 \quad\Rightarrow\quad 2^{2x} \:=\:\text{-}13\!\cdot\!3^{2x}$
. . $\displaystyle \dfrac{2^{2x}}{3^{2x}} \:=\:\text{-}13 \quad\Rightarrow\quad \left(\dfrac{2}{3}\right)^{2x} \:=\:\text{-}13\;\hdots\;\text{ no real roots}$
$\displaystyle 3\!\cdot\!2^{2x} - 2\!\cdot\!3^{2x} \:=\:0 \quad\Rightarrow\quad 3\!\cdot\!2^{2x} \:=\:2\!\cdot\!3^{2x}$
. . $\displaystyle \dfrac{2^{2x}}{3^{2x}} \:=\:\dfrac{2}{3} \quad\Rightarrow\quad \left(\dfrac{2}{3}\right)^{2x} =\:\left(\dfrac{2}{3}\right)^1 \quad\Rightarrow\quad 2x \:=\:1$
. . $\displaystyle \text{Therefore: }\:\boxed{x \:=\:\frac{1}{2}}$ | 1,102 | 2,513 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2018-17 | latest | en | 0.526904 |
https://discuss.codecademy.com/t/isnt-there-a-problem-in-this-exercise/67715 | 1,537,645,606,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267158633.40/warc/CC-MAIN-20180922182020-20180922202420-00121.warc.gz | 474,279,026 | 4,312 | # Isn't there a problem in this exercise?
#1
See the code below, I don't understand that middle_third = [8,9,10,11,12,13,14]
Previously, we've been told that in slicing: [start:end:stride]
start = where it starts inclusive
end = where it ends exclusive
stride = the space between items in the sliced list
So why do I get [8,9,10,11,12,13,14] instead of [7,8,9,10,11,12,13] ? Something I'm missing?
``````to_21 = range(1,22)
odds = to_21[::2]
middle_third = to_21[7:14:1]``````
#2
What is the index of the first item in a `list`?
#3
You start at 7, what's at index 7?
#4
Of course we put the indexes and not the items in that slice!
Thanks @ionatan and @appylpye!
#5
This topic was automatically closed 7 days after the last reply. New replies are no longer allowed. | 250 | 780 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2018-39 | latest | en | 0.929854 |
https://gmatclub.com/forum/how-many-3-digit-positive-integers-can-be-represented-as-the-sum-of-ex-257077.html | 1,547,715,801,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583658901.41/warc/CC-MAIN-20190117082302-20190117104302-00239.warc.gz | 544,842,351 | 51,193 | GMAT Question of the Day - Daily to your Mailbox; hard ones only
It is currently 17 Jan 2019, 01:03
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
## Events & Promotions
###### Events & Promotions in January
PrevNext
SuMoTuWeThFrSa
303112345
6789101112
13141516171819
20212223242526
272829303112
Open Detailed Calendar
• ### The winning strategy for a high GRE score
January 17, 2019
January 17, 2019
08:00 AM PST
09:00 AM PST
Learn the winning strategy for a high GRE score — what do people who reach a high score do differently? We're going to share insights, tips and strategies from data we've collected from over 50,000 students who used examPAL.
• ### Free GMAT Strategy Webinar
January 19, 2019
January 19, 2019
07:00 AM PST
09:00 AM PST
Aiming to score 760+? Attend this FREE session to learn how to Define your GMAT Strategy, Create your Study Plan and Master the Core Skills to excel on the GMAT.
# How many 3-digit positive integers can be represented as the sum of ex
Author Message
Intern
Joined: 19 Jun 2017
Posts: 33
Location: Armenia
Concentration: Entrepreneurship, Economics
GMAT 1: 760 Q51 V46
GPA: 4
How many 3-digit positive integers can be represented as the sum of ex [#permalink]
### Show Tags
08 Jan 2018, 08:57
1
00:00
Difficulty:
(N/A)
Question Stats:
67% (01:29) correct 33% (02:47) wrong based on 3 sessions
### HideShow timer Statistics
How many 3-digit positive integers can be represented as the sum of exactly nine different powers of 2?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
please, give kudos, if you liked the question
--== Message from the GMAT Club Team ==--
THERE IS LIKELY A BETTER DISCUSSION OF THIS EXACT QUESTION.
This discussion does not meet community quality standards. It has been retired.
If you would like to discuss this question please re-post it in the respective forum. Thank you!
To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.
Manager
Joined: 17 Mar 2015
Posts: 86
Re: How many 3-digit positive integers can be represented as the sum of ex [#permalink]
### Show Tags
08 Jan 2018, 09:08
AniK97 wrote:
How many 3-digit positive integers can be represented as the sum of exactly nine different powers of 2?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
please, give kudos, if you liked the question
OA: E.
511=2^8+2^7+2^6+2^5+2^4+2^3+2^2+2^1+2^0
767=2^9+2^7+2^6+2^5+2^4+2^3+2^2+2^1+2^0
895=2^9+2^8+2^6+2^5+2^4+2^3+2^2+2^1+2^0
959=2^9+2^8+2^7+2^5+2^4+2^3+2^2+2^1+2^0
991=2^9+2^8+2^7+2^6+2^4+2^3+2^2+2^1+2^0
--== Message from the GMAT Club Team ==--
THERE IS LIKELY A BETTER DISCUSSION OF THIS EXACT QUESTION.
This discussion does not meet community quality standards. It has been retired.
If you would like to discuss this question please re-post it in the respective forum. Thank you!
To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.
Re: How many 3-digit positive integers can be represented as the sum of ex &nbs [#permalink] 08 Jan 2018, 09:08
Display posts from previous: Sort by | 1,108 | 3,676 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2019-04 | latest | en | 0.889466 |
https://aptitude.gateoverflow.in/6831/nielit-2019-feb-scientist-d-section-d-22 | 1,675,326,132,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499967.46/warc/CC-MAIN-20230202070522-20230202100522-00590.warc.gz | 115,526,140 | 20,544 | 306 views
A train travelling from Delhi to Ambala meets with an accident after $1$ hr. It is stopped for $\frac{1}{2}$ hr, after which it proceeds at fourth-fifth of its usual rate, arriving at Ambala at $2$ hr late. If the train has covered $80$ km more before the accident, it would have been just $1$ hr late. The usual speed of the train is :
1. $20$ km/hr
2. $30$ km/hr
3. $40$ km/hr
4. $50$ km/hr
Option (A)
Solution to this problem is here.
272 points
1 | 140 | 466 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2023-06 | latest | en | 0.943711 |
https://www.geeksforgeeks.org/some-tricks-to-solve-problems-on-impartial-games/?ref=rp | 1,674,865,453,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499468.22/warc/CC-MAIN-20230127231443-20230128021443-00627.warc.gz | 800,472,536 | 24,872 | # Some Tricks to solve problems on Impartial games
• Difficulty Level : Easy
• Last Updated : 05 Apr, 2019
How to solve problems that fall in the Finders keepers category in ‘Game Theory’?
Note- Finders keepers game fall into the category of ‘Impartial Games’ in ‘Game Theory’.
What are ‘Impartial Games’?
Let a game is being played between two players ‘A’ and ‘B’. A game between them is said to be ‘Impartial’ if both the players have same set of moves. Which move to choose at a particular moment of the game depends on the state of the game.
FINDERS KEEPERS GAME
Let two players A and B playing a game on a pile of coins. Each player has to remove a minimum of ‘a’ coins or maximum of ‘b’ coins from the pile in his turn until there are less than ‘a’ coins left.
Types:
1. Finder-Winner -> In this format the player to play the last move wins.
2. Keeper-Loser -> In this format the player to play the last move loses.
STRATEGIES
Note- If there are less than ‘a’ coins available then last player to move must pick all coins.
1. FINDER-WINNER: Let ‘A’ be the player to start. For a game of FINDER-WINNER the main strategy is to reduce the present number coins to a multiple (a+b) for ‘A’ to win the game. Else he loses and ‘B’ wins.
2. KEEPER-LOSER: Let ‘A’ starts the game. For ‘A’ to win the game the main condition for winning it is to reduce the pile of coins to – (a+b)k + [1, a], i.e. the number of coins can be reduced to more then a multiple of (a+b) by [1 upto a]. If he can do so he wins else ‘B’ wins.
Examples:
1. Let A and B play a game of FINDER-WINNER on a set of 20 coins. A player in his move can remove a minimum of 5 coins and a maximum of 7 coins. predict the Winner of the Game If A starts first and if both players play optimally.
Given number of coins = 20.
Minimum coins that can be removed = 5 – Let it be ‘a’.
Maximum coins that can be removed = 7 – Let it be ‘b’.
According to given strategy for FINDER-WINNER:
For A to win: A must remove the number of coins to a multiple of (a + b). Nearest multiple of (a + b) for 20 is 12. So A can deduct either 5 or 6 or 7 coins from the set so he cant reach 12 anyhow. So for B to make a move, he will be left with 15 or 14 or 13 coins. Now for A’s next move corresponding states are (10, 9, 8), (9, 8, 7) and (8, 7, 6). Since B is also playing optimally he will keep the number of coins for A’s move greater than 7 so that A cant win. Hence, he will get at least 1 coin to be removed in the last move and WIN.
2. Let A and B play a game of Keeper-Loser on a set of 20 coins. A player in his move can remove a minimum of 2 coins and a maximum of 5 coins. Predict the Winner of the Game If A starts first and if both players play optimally.
Given number of coins = 20.
Minimum coins that can be removed = 2 – Let it be ‘a’.
Maximum coins that can be removed = 5 – Let it be ‘b’.
According to given strategy for KEEPER-LOSER :
For A to win: A must remove the number of coins to a multiple of (a + b)k + [1, a]. Here a + b = 7.
Now, for playing optimally lets see what A will do in further rounds. If A reduces the set of coins to lets say 15 which is of the form (2 + 5) * 2 + 1 as it is reachable by removing 5 coins from a set of 20 coins. From here B can take any number of coins from [2, 5] lets name it as c. The main strategy of A now will be to remove 7 – c in his moves.
Why is it so?
Lets say after 15, B reduces the set to 10 by taking away 5 coins. Now optimal strategy for A is to remove 2 coins. If he does so, the set reduces to 8 coins. Now from here B can reduce the set to (6, 5, 4, 3) by removing (2, 3, 4, 5) coins respectively. From here again, the main idea for A is to take away 7 – c coins which will keep 1 coin for B to remove. Hence A wins.
My Personal Notes arrow_drop_up | 1,070 | 3,789 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2023-06 | latest | en | 0.949015 |
https://brainanswerph.com/math/question2383626 | 1,627,353,711,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046152168.38/warc/CC-MAIN-20210727010203-20210727040203-00513.warc.gz | 171,116,551 | 14,990 | , 28.10.2019 15:29, calmaaprilgrace
# 7.what fraction of the figure is shaded? what percent?
### Iba pang mga katanungan: Math
Math, 28.10.2019 16:29, axelamat70
Mr. winston wants to start a lending corporation. he wants to determine what interest rate should he offer to earn 140 in just two months if the customer borrows 8.000 mathematics business
Kabuuang mga Sagot: 1
Math, 28.10.2019 19:29, snow01
1if a varies directly as b and a= 12 when b=15, find a when b=35 2 if y is directly proportional to x and y=24 when x and y=24 when x=6, find y when x= 15. 3 if (h) is proportional to (g) and h=27 when g=9, find h when g=24 4 if (r) varies inverserly as (s) and r =8 when s=12, find r when s=4. 5 if z varies jointly as x and y, and z=60 when x=12 and y=10, find the z when x=14 and y=14. a. constant of variation b. equation of variation
Kabuuang mga Sagot: 1
Math, 28.10.2019 21:29, homersoncanceranguiu
Anong ibig sabihin ng 200 off
Kabuuang mga Sagot: 2
Math, 28.10.2019 21:29, kimashleybartolome
What is thew word form of 73.0873
Kabuuang mga Sagot: 1
Do you know the correct answer?
7.
what fraction of the figure is shaded? what percent? ...
### Questions in other subjects:
Science, 26.10.2020 10:55
Science, 26.10.2020 10:55
Questions on the website: 14763448 | 462 | 1,281 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2021-31 | latest | en | 0.761229 |
http://www.numbersaplenty.com/1023041 | 1,601,480,747,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402127075.68/warc/CC-MAIN-20200930141310-20200930171310-00218.warc.gz | 193,340,819 | 3,389 | Search a number
1023041 is a prime number
BaseRepresentation
bin11111001110001000001
31220222100102
43321301001
5230214131
633532145
711460425
oct3716101
91828312
101023041
11639698
12414055
1329a866
141c8b85
151531cb
hexf9c41
1023041 has 2 divisors, whose sum is σ = 1023042. Its totient is φ = 1023040.
The previous prime is 1023037. The next prime is 1023047. The reversal of 1023041 is 1403201.
Adding to 1023041 its reverse (1403201), we get a palindrome (2426242).
It is a happy number.
1023041 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 863041 + 160000 = 929^2 + 400^2 .
It is a cyclic number.
It is not a de Polignac number, because 1023041 - 22 = 1023037 is a prime.
It is a super-2 number, since 2×10230412 = 2093225775362, which contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (1023047) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 511520 + 511521.
It is an arithmetic number, because the mean of its divisors is an integer number (511521).
21023041 is an apocalyptic number.
It is an amenable number.
1023041 is a deficient number, since it is larger than the sum of its proper divisors (1).
1023041 is an equidigital number, since it uses as much as digits as its factorization.
1023041 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 24, while the sum is 11.
The square root of 1023041 is about 1011.4548927164. The cubic root of 1023041 is about 100.7622089480.
The spelling of 1023041 in words is "one million, twenty-three thousand, forty-one". | 545 | 1,816 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2020-40 | latest | en | 0.88396 |
https://testbook.com/question-answer/the-code-used-to-reduce-the-error-due-to-ambiguity--5f98759f45abb7f45f3ea3ba | 1,638,296,976,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964359065.88/warc/CC-MAIN-20211130171559-20211130201559-00325.warc.gz | 635,214,191 | 32,131 | # The code used to reduce the error due to ambiguity in reading of a binary optical encoder is
This question was previously asked in
TANGEDCO AE EE 2015 Official Paper
View all TANGEDCO Assistant Engineer Papers >
1. Octal code
2. Excess-3 code
3. Gray code
4. BCD code
Option 3 : Gray code
Free
CT 1: विलोम शब्द
6832
10 Questions 30 Marks 10 Mins
## Detailed Solution
Concept:
Gray codes are less error-prone because they only change in one-bit position at a time. Hence, they are considered as the minimum error code.
Explanation:
Decimal Binary Gray code Octal Excess – 3 code 0 0000 0000 000 0011 1 0001 0001 001 0100 2 0010 0011 010 0101 3 0011 0010 011 0110 4 0100 0110 100 0111 5 0101 0111 101 1000 6 0110 0101 110 1001 7 0111 0100 111 1010 8 1000 1100 001 000 1011
From Table:
Distance between two successive values In Octal, Binary and Excess-3 code ≥ 1
Distance between two successive values in gray code is 1
Hence minimum error code is gray code. | 319 | 970 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2021-49 | latest | en | 0.715303 |
http://math.stackexchange.com/questions/tagged/asymptotics?sort=active&pagesize=30 | 1,467,225,702,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783397795.31/warc/CC-MAIN-20160624154957-00193-ip-10-164-35-72.ec2.internal.warc.gz | 218,725,860 | 28,379 | # Tagged Questions
Questions involving asymptotic analysis, including growth of functions, Big-O notation, Big-Omega and Big-Theta notations.
0answers
30 views
0answers
26 views
### The asymptotic behaviour of $\sum_{1\leq k\leq N-1}\int_{p_k}^{p_{k+1}}\log x d[x]$, where $p_n$ is the nth prime number
Let $p_k$ is the kth prime number and consider for $N\geq 2$ the arithmetic function $$f(N)=\sum_{k=1}^{N-1}\int_{p_k}^{p_{k+1}}\log(x) d[x]$$ where $[x]$ is the integer part function (provide us in ...
2answers
41 views
2answers
42 views
0answers
20 views
### Determining bounds for a sum with nested infinite series
I am computing the inner product of the characters of the trivial and the $k$-th irreducible two dimensional representations of the dihedral group $D_n$ of order $2 n$ when $n$ is even. The ...
1answer
20 views
1answer
17 views
### multiple of an integer and asymptotics
Let us suppose that we have a positive integer $N$. We take the integer $\lceil \log_2 N \rceil$. Does there always exist an integer $X \geq N$ such that the following both conditions are satisfied: ...
2answers
343 views
0answers
50 views
### How quickly can we find a value that has large multiplicative order modulo $n$?
If we're trying to find an element modulo $n$ that has multiplicative order at least $\sqrt{n}$, how quickly can we do this? We don't know if $n$ is prime or composite, only that $n$ definitely has a ...
2answers
37 views
### $(x + y + xy)/2 = f( f^{[-1]}(x) + f^{[-1]}(y) )$
Consider $(x + y + xy)/2 = f( f^{[-1]}(x) + f^{[-1]}(y) )$ Where $f^{[-1}]$ denotes the functional inverse of $f$. How to find $f$ ? How about the more General idea of finding $f$ for a given $g$? ...
0answers
89 views
### Find a non-constant real-analytic function $f(x)$ such that for $x\in\Bbb R,\;f(2^x) = f(4^x + 2^{x+1} + 2) - f(4^x + 1)$
Let $f(x)$ be a non-constant real-analytic function and for real $x$ it satisfies : $f(2^x) = f(4^x + 2^{x+1} + 2) - f(4^x + 1)$ Before you ask if this simplifies by writing $2^x = y$ note that \$2^... | 665 | 2,057 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2016-26 | longest | en | 0.797911 |
https://www.saonastudios.com/answers/17742675-if-jim-could-drive-a-jetson-s-flying-car-at-a-constant-speed-of-490-km-hr-across-oceans-and | 1,709,162,789,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474746.1/warc/CC-MAIN-20240228211701-20240229001701-00047.warc.gz | 968,715,730 | 8,190 | If Jim could drive a Jetson's flying car at a constant speed of 490 km/hr across oceans and space, approximately how long (in millions of years, in 106 years) would he take to drive to a nearby star that is 4.5 light-years away? Use 9.461 × 1012 km/light-year and 8766 hours per year (365.25 days). unanswered
109.5 million years
Explanation:
The question asked us to find the time.
Remember that
Rate of velocity = distance / time, and this,
time taken = distance/rate
Due to the confusing nature of the units, we would have to be converting them to a more uniform one.
1 km is equal to 9.461*10^12 km/light-year, that's if we try to convert km to light year.
Since the speed is in km, the distance has to be in km also, and therefore, we convert ly to km:
4.5 light-years = 9.461*10^12 km/light-year) = 42.57*10^13 km
We that this value as our distance, in km.
Also,
Time = distance/speed
Time = 45.57*10^13 km / 490 km/hr = 9.3*10^11 hr
Now the next step is to convert hours to years, using the conversion factor 8766 hr/yr.
time (in years) = 9.6*10^11 hr / 8766 hr/yr) = 10.95*10^7 years
the final step is to divide the time in years by 10^6 years/million years, which gives the final answer as the trip takes 109.5 million years.
Related Questions
A 0.060 ???????? tennis ball, moving with a speed of 5.28 m/???? , has a head-on collision with a 0.080 ???????? ball initially moving in the same direction at a speed of 3.00 m/ ???? . Assume that the collision is perfectly elastic. Determine the velocity (speed and direction) of both the balls after the collision.
Explanation:
It is given that,
Mass of the tennis ball,
Initial speed of tennis ball,
Mass of ball,
Initial speed of ball,
In case of elastic collision, the momentum remains conserved. The momentum equation is given by :
are final speed of tennis ball and the ball respectively.
..............(1)
We know that the coefficient of restitution is equal to 1. It is given by :
.................(2)
On solving equation (1) and (2) to find the values of velocities after collision.
So, the speed of both balls are 5.28 m/s and 3 m/s respectively. Hence, this is the required solution.
Two identical small charged spheres are a certain distance apart, and each one initially experiences an electrostatic force of magnitude F due to the other. With time, charge gradually leaks off of both spheres. Calculate the magnitude of the electrostatic force, when each of the spheres has lost half of its initial charge. (Your answer will be a function of F, since no values are giving)
1/4F
Explanation:
We already know thatThe electrostatic force is directly proportional to the product of the charge, from Coulomb's law.
So F α Qq
But if it is now half the initial charges, then
F α (1/2)Q *(1/2)q
F α (1/4)Qq
Thus the resultant charges are each halved is (1/4) and the first initial force experienced at full charge.
Thus the answer will be 1/4F
1. A bicyclist starts at 2.5 m/s and accelerates along a straight path to a speed of 12.5 m/s ina time of 4.5 seconds. What is the bicyclist's acceleration to the nearest tenth of a m/s??
2.2m/s
Explanation:
a=v-u/t
12.5-2.5/4.5=2.222
~2.2m/s
The Golden Gate Bridge in San Francisco has a main span of length 1.28 km, one of the longest in the world. Imagine that a steel wire with this length and a cross-sectional area of 3.10 ✕ 10^−6 m^2 is laid on the bridge deck with its ends attached to the towers of the bridge, on a summer day when the temperature of the wire is 43.0°C. When winter arrives, the towers stay the same distance apart and the bridge deck keeps the same shape as its expansion joints open. When the temperature drops to −10.0°C, what is the tension in the wire? Take Young's modulus for steel to be 20.0 ✕ 10^10 N/m^2. (Assume the coefficient of thermal expansion of steel is 11 ✕ 10−6 (°C)−1.)
361.46 N
Explanation:
= Coefficient of thermal expansion =
Y = Young's modulus for steel =
A = Area =
= Original length = 1.28 km
= Change in temperature =
Length contraction is given by
Also,
The tension in the wire is 361.46 N
For a very rough pipe wall the friction factor is constant at high Reynolds numbers. For a length L1 the pressure drop over the length is p1. If the length of the pipe is then doubled, what is the relation of the new pressure drop p2 to the original pressure drop p1 at the original mass flow rate?
Explanation:
Given that all other factors remain constant. The pressure drop across the pipeline is directly proportional to the length.
i.e ∆p ~ L
Therefore,
∆p2/L2 = ∆p1/L1
Since L2 = 2 * L1
∆p2/2*L1 = ∆p1/L1
Eliminating L1 we have,
∆p2/2 = ∆p1
Multiplying both sides by 2
∆p2 = 2 * ∆p1 | 1,282 | 4,709 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2024-10 | latest | en | 0.935715 |
https://reddingvwclub.org/and-pdf/1434-geometric-symmetry-in-patterns-and-tilings-pdf-476-840.php | 1,642,466,773,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300658.84/warc/CC-MAIN-20220118002226-20220118032226-00554.warc.gz | 567,639,628 | 7,067 | and pdfSunday, December 20, 2020 5:38:20 PM1
# Geometric Symmetry In Patterns And Tilings Pdf
File Name: geometric symmetry in patterns and tilings .zip
Size: 1702Kb
Published: 20.12.2020
About Blog Contact Connect. Ask right here Symmetry and Pattern Design Resources. Teaching or studying symmetry? Learning pattern design?
## Geometric symmetry in patterns and tilings
A pattern is a regularity in the world, in human-made design, or in abstract ideas. As such, the elements of a pattern repeat in a predictable manner. A geometric pattern is a kind of pattern formed of geometric shapes and typically repeated like a wallpaper design. Any of the senses may directly observe patterns. Conversely, abstract patterns in science , mathematics , or language may be observable only by analysis. Direct observation in practice means seeing visual patterns, which are widespread in nature and in art.
## Penrose tiling
A Penrose tiling is an example of an aperiodic tiling. Here, a tiling is a covering of the plane by non-overlapping polygons or other shapes, and aperiodic means that shifting any tiling with these shapes by any finite distance, without rotation, cannot produce the same tiling. However, despite their lack of translational symmetry , Penrose tilings may have both reflection symmetry and fivefold rotational symmetry. Penrose tilings are named after mathematician and physicist Roger Penrose , who investigated them in the s. There are several different variations of Penrose tilings with different tile shapes.
Classification of designs by symmetry group; Classification of designs by symmetry group and design unit; Classification of discrete patterns; Classification of isohedral tilings. This book encompasses a wide range of mathematical concepts relating to regularly repeating surface decoration from basic principles of symmetry to more complex issues of graph theory, group theory and topology. It presents a comprehensive means of classifying and constructing patterns and tilings. The classification of designs is investigated and discussed forming a broad basis upon which designers may build their own ideas. A wide range of original illustrative material is included. In a complex area previously best understood by mathematicians and crystallographers, the author develops and applies mathematical thinking to the context of regularly repeating surface-pattern design in a manner accessible to artists and designers. Design construction is covered from first principles through to methods appropriate for adaptation to large-scale screen-printing production.
## Symmetry and Tilings: An Exploration
Students are directed to read through a Web-based tutorial on Symmetry and Tilings in the form of an short and colorful article entitled Tilings and Tesselations; afterwards, they answer several questions on tilings tessellations , tiling terminology, types of symmetry isometries , periodic tilings and Penrose tilings. In addition, they are given opportunity to use an interactive Java applet in which various types of symmetry can be sketched and explored in the form of wallpaper groups, frieze groups and rosette groups. In a subsequent think-pair-share activity or write-pair-share activity, they analyze some tilings and apply their newly obtained knowledge. Your Account.
Меган! - завопил он, грохнувшись на пол. Острые раскаленные иглы впились в глазницы. Он уже ничего не видел и только чувствовал, как тошнотворный комок подкатил к горлу. Его крик эхом отозвался в черноте, застилавшей. Беккер не знал, сколько времени пролежал, пока над ним вновь не возникли лампы дневного света.
Единственным кандидатом в подозреваемые был Грег Хейл, но Сьюзан могла поклясться, что никогда не давала ему свой персональный код. Следуя классической криптографической процедуре, она выбрала пароль произвольно и не стала его записывать. То, что Хейл мог его угадать, было исключено: число комбинаций составляло тридцать шесть в пятой степени, или свыше шестидесяти миллионов.
Он закрыл глаза, и воспоминания хлынули бурным потоком. | 897 | 4,069 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2022-05 | latest | en | 0.90431 |
http://www.popflock.com/learn?s=Explicit_symmetry_breaking | 1,582,503,007,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145859.65/warc/CC-MAIN-20200223215635-20200224005635-00516.warc.gz | 206,901,188 | 10,880 | Explicit Symmetry Breaking
Get Explicit Symmetry Breaking essential facts below. View Videos or join the Explicit Symmetry Breaking discussion. Add Explicit Symmetry Breaking to your PopFlock.com topic list for future reference or share this resource on social media.
Explicit Symmetry Breaking
In theoretical physics, explicit symmetry breaking is the breaking of a symmetry of a theory by terms in its defining equations of motion (most typically, to the Lagrangian or the Hamiltonian) that do not respect the symmetry. Usually this term is used in situations where these symmetry-breaking terms are small, so that the symmetry is approximately respected by the theory. An example is the spectral line splitting in the Zeeman effect, due to a magnetic interaction perturbation in the Hamiltonian of the atoms involved.
Explicit symmetry breaking differs from spontaneous symmetry breaking. In the latter, the defining equations respect the symmetry but the ground state (vacuum) of the theory breaks it.[1]
Explicit symmetry breaking is also associated with electromagnetic radiation. A system of accelerated charges results in electromagnetic radiation when the geometric symmetry of the electric field in free space is explicitly broken by the associated electrodynamic structure under time varying excitation of the given system. This is quite evident in an antenna where the electric lines of field curl around or have rotational geometry around the radiating terminals in contrast to linear geometric orientation within a pair of transmission lines which does not radiate even under time varying excitation.[2]
## References
1. ^ Castellani, E. (2003) "On the meaning of Symmetry Breaking" in Brading, K. and Castellani, E. (eds) Symmetries in Physics: New Reflections, Cambridge: Cambridge University Press
2. ^ Sinha & Amaratunga (2016) "Explicit Symmetry Breaking in Electrodynamic Systems and Electromagnetic Radiation" Morgan Claypool, Institute of Physics, UK | 383 | 1,977 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2020-10 | latest | en | 0.869919 |
http://forums.wolfram.com/mathgroup/archive/2006/Sep/msg00148.html | 1,717,076,850,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971667627.93/warc/CC-MAIN-20240530114606-20240530144606-00213.warc.gz | 10,417,720 | 8,772 | RE: Re: Dot Product in Cylindrical Coordinates
• To: mathgroup at smc.vnet.net
• Subject: [mg69302] RE: [mg69276] Re: Dot Product in Cylindrical Coordinates
• From: "David Park" <djmp at earthlink.net>
• Date: Wed, 6 Sep 2006 04:28:12 -0400 (EDT)
• Sender: owner-wri-mathgroup at wolfram.com
```And actually in the second example I should have used CrossProduct, in which
case it even more doesn't work.
David Park
Paul,
I want to expand a little more on my previous posting.
The point is that the VectorAnalysis does expect that the components of a
vector are given in the orthornormal frame in Curl and Divergence. So it is
a little quixotic to switch the context when using DotProduct and
CrossProduct.
Needs["Calculus`VectorAnalysis`"]
SetCoordinates[Cylindrical[\[Rho], \[Phi], z]];
Here is a theorem from vector calculus:
div[g x f] == curl[g].f - curl[f].g
Which dot product is meant? If we use the ordinary Dot product the theorem
is true.
Div[Cross[{g\[Rho][\[Rho], \[Phi], \[Theta]],
g\[Phi][\[Rho], \[Phi], \[Theta]],
gz[\[Rho], \[Phi], \[Theta]]}, {f\[Rho][\[Rho], \[Phi], \[Theta]],
f\[Phi][\[Rho], \[Phi], \[Theta]], fz[\[Rho], \[Phi], \[Theta]]}]]
==
DotProduct[
Curl[{g\[Rho][\[Rho], \[Phi], \[Theta]],
g\[Phi][\[Rho], \[Phi], \[Theta]],
gz[\[Rho], \[Phi], \[Theta]]}], {f\[Rho][\[Rho], \[Phi],
\[Theta]],
f\[Phi][\[Rho], \[Phi], \[Theta]], fz[\[Rho], \[Phi], \[Theta]]}] -
DotProduct[
Curl[{f\[Rho][\[Rho], \[Phi], \[Theta]],
f\[Phi][\[Rho], \[Phi], \[Theta]],
fz[\[Rho], \[Phi], \[Theta]]}], {g\[Rho][\[Rho], \[Phi],
\[Theta]],
g\[Phi][\[Rho], \[Phi], \[Theta]], gz[\[Rho], \[Phi], \[Theta]]}];
% // Simplify
True
If instead we use the VectorAnalysis DotProduct the theorem is not true.
Div[Cross[{g\[Rho][\[Rho], \[Phi], \[Theta]],
g\[Phi][\[Rho], \[Phi], \[Theta]],
gz[\[Rho], \[Phi], \[Theta]]}, {f\[Rho][\[Rho], \[Phi], \[Theta]],
f\[Phi][\[Rho], \[Phi], \[Theta]],
fz[\[Rho], \[Phi], \[Theta]]}]] ==
DotProduct[
Curl[{g\[Rho][\[Rho], \[Phi], \[Theta]],
g\[Phi][\[Rho], \[Phi], \[Theta]],
gz[\[Rho], \[Phi], \[Theta]]}], {f\[Rho][\[Rho], \[Phi], \
\[Theta]], f\[Phi][\[Rho], \[Phi], \[Theta]], fz[\[Rho], \[Phi],
\[Theta]]}] -
DotProduct[
Curl[{f\[Rho][\[Rho], \[Phi], \[Theta]],
f\[Phi][\[Rho], \[Phi], \[Theta]],
fz[\[Rho], \[Phi], \[Theta]]}], {g\[Rho][\[Rho], \[Phi], \
\[Theta]], g\[Phi][\[Rho], \[Phi], \[Theta]], gz[\[Rho], \[Phi],
\[Theta]]}];
% // Simplify
(output omitted)
This is certainly a 'feature' and I think it is an error in design. The user
should at least be warned about the change in context and meaning. The
documentation uses the phrase 'in the default coordinate system' in both
DotProduct and in Curl so one could become easily confused.
David Park
From: Paul Abbott [mailto:paul at physics.uwa.edu.au]
To: mathgroup at smc.vnet.net
Sergio Miguel Terrazas Porras <sterraza at uacj.mx> wrote:
> When I calculate the dot product of vectors {1,Pi/4,0} and {2,0,1} in
> Cylindrical Coordinates Mathematica 5.1 returns the result Sqrt[2], when
the
> result should be 2.
Notwithstanding several of the other responses, the result _is_ Sqrt[2].
When you write {1,Pi/4,0}, surely you mean
{rho, phi, z} == {1, Pi/4, 0}
and _not_ that
{x, y, z} == {1, Pi/4, 0} ?
Needs["Calculus`VectorAnalysis`"]
and selecting Cylindrical coordinates,
SetCoordinates[Cylindrical];
then in cartesian coordinates, this point is
p1 = CoordinatesToCartesian[{1, Pi/4, 0}]
{1/Sqrt[2], 1/Sqrt[2], 0}
Similarly,
p2 = CoordinatesToCartesian[{2, 0, 1}]
{2, 0, 1}
Hence the dot product of the coordinate vectors (relative to the origin
{0,0,0}), computed in cartesian coordinates, is
p1 . p2
Sqrt[2]
This is the same result that you got, presumably using,
DotProduct[ {1, Pi/4, 0}, {2, 0, 1} ]
Sqrt[2]
Of course, if you really mean
{x, y, z} == {1, Pi/4, 0}
then there is no need to load Calculus`VectorAnalysis`: the dot product
is just
{1, Pi/4, 0} . {2, 0, 1}
2
Note that Dot is _not_ modified when this package is loaded so
Jean-Marc's response,
Needs["Calculus`VectorAnalysis`"]
SetCoordinates[Cylindrical];
{1, Pi/4, 0} . {2, 0, 1}
is bogus -- the first two lines have no effect on the third.
Modifying Andrzej's code, we have
Simplify[(JacobianMatrix[] . p1) . (JacobianMatrix[] . p2)]
Sqrt[2]
Cheers,
Paul
_______________________________________________________________________
Paul Abbott Phone: 61 8 6488 2734
School of Physics, M013 Fax: +61 8 6488 1014
The University of Western Australia (CRICOS Provider No 00126G)
AUSTRALIA http://physics.uwa.edu.au/~paul
```
• Prev by Date: Re: Re: generalized foldlist problem - part 2
• Next by Date: Fit rectangle to set of points
• Previous by thread: Re: RE: Re: Dot Product in Cylindrical Coordinates
• Next by thread: Re: Dot Product in Cylindrical Coordinates | 1,648 | 4,885 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2024-22 | latest | en | 0.543728 |
https://quantumcomputing.stackexchange.com/questions/11435/h-ei-pi-4-sqrtinot | 1,713,868,607,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818474.95/warc/CC-MAIN-20240423095619-20240423125619-00500.warc.gz | 426,362,606 | 40,528 | # $H = e^{i\pi/4} \sqrt{iNOT}$?
In the paper Valley qubit in Gated MoS$$_2$$ monolayer quantum dot, a description of how a $$NOT$$ gate would be performed on a qubit in the described device is given.
The authors say that in the described implementation the operation performed is an $$iNOT$$ gate, and that the Hadamard operation can be implemented by performing half of the $$iNOT$$ Rabi transition. Particularly, the authors say that $$H = e^{i\pi/4}\sqrt{iNOT}$$, (actually they say $$H = e^{i\pi/4}\sqrt{NOT}$$, but I assume this is a typo).
Most generally, my question is: Does $$H = e^{i\pi/4}\sqrt{iNOT}$$? I cannot work it out.
My confusion may stem from my lack of understanding concerning why the implemented operation corresponds to $$iNOT$$ instead of simply $$NOT$$. My understanding is that $$iNOT = i\sigma_x$$. I'd appreciate any insight you have on this as well. Thank you.
It cannot be the case that $$H=e^{i\pi/4}\sqrt{iNOT}$$. Whatever your interpretation of $$iNOT$$ (I'd agree with your definition), just square the thing. $$H^2=I$$, the identity, and so it is certainly not the case that $$I=e^{i\pi/2}iNOT.$$
It is true, however, that if you perform the sequence that would give you an operation such as $$NOT$$ or $$iNOT$$, and you only evolve for half the time, that gives you something that achieves an equivalent result to the Hadamard. It's usually referred to as a beam-splitter. For example, if a theorist writes about a Mach-Zehnder interferometer, they usually write down a sequence of Hadamard - phase - Hadamard, whereas an experimentalist will use beam splitter - phase - beam splitter. It achieves the same practical task (e.g. identify if the phase is 0 or $$\pi$$) although the interpretation of the results is different as which output is which gets switched. | 471 | 1,804 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 17, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2024-18 | latest | en | 0.928054 |
http://nrich.maths.org/public/leg.php?code=32&cl=2&cldcmpid=8586 | 1,503,080,359,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886105086.81/warc/CC-MAIN-20170818175604-20170818195604-00651.warc.gz | 283,029,582 | 9,956 | # Search by Topic
#### Resources tagged with Multiplication & division similar to Dice in a Corner:
Filter by: Content type:
Stage:
Challenge level:
### There are 165 results
Broad Topics > Calculations and Numerical Methods > Multiplication & division
### Dice and Spinner Numbers
##### Stage: 2 Challenge Level:
If you had any number of ordinary dice, what are the possible ways of making their totals 6? What would the product of the dice be each time?
### Arranging the Tables
##### Stage: 2 Challenge Level:
There are 44 people coming to a dinner party. There are 15 square tables that seat 4 people. Find a way to seat the 44 people using all 15 tables, with no empty places.
### How Much Did it Cost?
##### Stage: 2 Challenge Level:
Use your logical-thinking skills to deduce how much Dan's crisps and ice-cream cost altogether.
### Zargon Glasses
##### Stage: 2 Challenge Level:
Zumf makes spectacles for the residents of the planet Zargon, who have either 3 eyes or 4 eyes. How many lenses will Zumf need to make all the different orders for 9 families?
### Shapes in a Grid
##### Stage: 2 Challenge Level:
Can you find which shapes you need to put into the grid to make the totals at the end of each row and the bottom of each column?
### Journeys in Numberland
##### Stage: 2 Challenge Level:
Tom and Ben visited Numberland. Use the maps to work out the number of points each of their routes scores.
### Asteroid Blast
##### Stage: 2 Challenge Level:
A game for 2 people. Use your skills of addition, subtraction, multiplication and division to blast the asteroids.
### Code Breaker
##### Stage: 2 Challenge Level:
This problem is based on a code using two different prime numbers less than 10. You'll need to multiply them together and shift the alphabet forwards by the result. Can you decipher the code?
### Twenty Divided Into Six
##### Stage: 2 Challenge Level:
Katie had a pack of 20 cards numbered from 1 to 20. She arranged the cards into 6 unequal piles where each pile added to the same total. What was the total and how could this be done?
### The Pied Piper of Hamelin
##### Stage: 2 Challenge Level:
This problem is based on the story of the Pied Piper of Hamelin. Investigate the different numbers of people and rats there could have been if you know how many legs there are altogether!
### Rabbits in the Pen
##### Stage: 2 Challenge Level:
Using the statements, can you work out how many of each type of rabbit there are in these pens?
### A Square of Numbers
##### Stage: 2 Challenge Level:
Can you put the numbers 1 to 8 into the circles so that the four calculations are correct?
### Being Thoughtful - Primary Number
##### Stage: 1 and 2 Challenge Level:
Number problems at primary level that require careful consideration.
### X Is 5 Squares
##### Stage: 2 Challenge Level:
Can you arrange 5 different digits (from 0 - 9) in the cross in the way described?
### Calendar Calculations
##### Stage: 2 Challenge Level:
Try adding together the dates of all the days in one week. Now multiply the first date by 7 and add 21. Can you explain what happens?
### Today's Date - 01/06/2009
##### Stage: 1 and 2 Challenge Level:
What do you notice about the date 03.06.09? Or 08.01.09? This challenge invites you to investigate some interesting dates yourself.
### Oh! Harry!
##### Stage: 2 Challenge Level:
A group of children are using measuring cylinders but they lose the labels. Can you help relabel them?
### It Figures
##### Stage: 2 Challenge Level:
Suppose we allow ourselves to use three numbers less than 10 and multiply them together. How many different products can you find? How do you know you've got them all?
### Magic Constants
##### Stage: 2 Challenge Level:
In a Magic Square all the rows, columns and diagonals add to the 'Magic Constant'. How would you change the magic constant of this square?
### A-magical Number Maze
##### Stage: 2 Challenge Level:
This magic square has operations written in it, to make it into a maze. Start wherever you like, go through every cell and go out a total of 15!
### Doplication
##### Stage: 2 Challenge Level:
We can arrange dots in a similar way to the 5 on a dice and they usually sit quite well into a rectangular shape. How many altogether in this 3 by 5? What happens for other sizes?
### Route Product
##### Stage: 2 Challenge Level:
Find the product of the numbers on the routes from A to B. Which route has the smallest product? Which the largest?
### Six Ten Total
##### Stage: 2 Challenge Level:
This challenge combines addition, multiplication, perseverance and even proof.
### Multiply Multiples 3
##### Stage: 2 Challenge Level:
Have a go at balancing this equation. Can you find different ways of doing it?
### Multiply Multiples 2
##### Stage: 2 Challenge Level:
Can you work out some different ways to balance this equation?
### The Puzzling Sweet Shop
##### Stage: 2 Challenge Level:
There were chews for 2p, mini eggs for 3p, Chocko bars for 5p and lollypops for 7p in the sweet shop. What could each of the children buy with their money?
### Exploring Number Patterns You Make
##### Stage: 2 Challenge Level:
Explore Alex's number plumber. What questions would you like to ask? What do you think is happening to the numbers?
### A Mixed-up Clock
##### Stage: 2 Challenge Level:
There is a clock-face where the numbers have become all mixed up. Can you find out where all the numbers have got to from these ten statements?
### How Old?
##### Stage: 2 Challenge Level:
Cherri, Saxon, Mel and Paul are friends. They are all different ages. Can you find out the age of each friend using the information?
### Multiply Multiples 1
##### Stage: 2 Challenge Level:
Can you complete this calculation by filling in the missing numbers? In how many different ways can you do it?
### Mystery Matrix
##### Stage: 2 Challenge Level:
Can you fill in this table square? The numbers 2 -12 were used to generate it with just one number used twice.
### Highest and Lowest
##### Stage: 2 Challenge Level:
Put operations signs between the numbers 3 4 5 6 to make the highest possible number and lowest possible number.
### Machines
##### Stage: 2 Challenge Level:
What is happening at each box in these machines?
### Sending Cards
##### Stage: 2 Challenge Level:
This challenge asks you to investigate the total number of cards that would be sent if four children send one to all three others. How many would be sent if there were five children? Six?
### Clock Face
##### Stage: 2 Challenge Level:
Where can you draw a line on a clock face so that the numbers on both sides have the same total?
### All the Digits
##### Stage: 2 Challenge Level:
This multiplication uses each of the digits 0 - 9 once and once only. Using the information given, can you replace the stars in the calculation with figures?
##### Stage: 2 Challenge Level:
Use the information to work out how many gifts there are in each pile.
### Zios and Zepts
##### Stage: 2 Challenge Level:
On the planet Vuv there are two sorts of creatures. The Zios have 3 legs and the Zepts have 7 legs. The great planetary explorer Nico counted 52 legs. How many Zios and how many Zepts were there?
### Forgot the Numbers
##### Stage: 2 Challenge Level:
On my calculator I divided one whole number by another whole number and got the answer 3.125. If the numbers are both under 50, what are they?
### Make 100
##### Stage: 2 Challenge Level:
Find at least one way to put in some operation signs (+ - x ÷) to make these digits come to 100.
### Book Codes
##### Stage: 2 Challenge Level:
Look on the back of any modern book and you will find an ISBN code. Take this code and calculate this sum in the way shown. Can you see what the answers always have in common?
### Cows and Sheep
##### Stage: 2 Challenge Level:
Use your logical reasoning to work out how many cows and how many sheep there are in each field.
### Become Maths Detectives
##### Stage: 2 Challenge Level:
Explore Alex's number plumber. What questions would you like to ask? Don't forget to keep visiting NRICH projects site for the latest developments and questions.
### How Do You Do It?
##### Stage: 2 Challenge Level:
This group activity will encourage you to share calculation strategies and to think about which strategy might be the most efficient.
### Fingers and Hands
##### Stage: 2 Challenge Level:
How would you count the number of fingers in these pictures?
### It Was 2010!
##### Stage: 1 and 2 Challenge Level:
If the answer's 2010, what could the question be?
##### Stage: 2 Challenge Level:
What happens when you add the digits of a number then multiply the result by 2 and you keep doing this? You could try for different numbers and different rules.
### Escape from the Castle
##### Stage: 2 Challenge Level:
Skippy and Anna are locked in a room in a large castle. The key to that room, and all the other rooms, is a number. The numbers are locked away in a problem. Can you help them to get out? | 2,085 | 9,056 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2017-34 | latest | en | 0.885671 |
https://www.coursehero.com/file/6913/problem05-23/ | 1,493,035,120,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917119356.19/warc/CC-MAIN-20170423031159-00259-ip-10-145-167-34.ec2.internal.warc.gz | 891,526,204 | 41,970 | problem05_23
University Physics with Modern Physics with Mastering Physics (11th Edition)
This preview shows page 1. Sign up to view the full content.
5.23: a) For the net force to be zero, the applied force is N. 0 . 22 ) s m (9.80 kg) 2 . 11 ( ) 20 . 0 ( 2 k k k = = = = = mg μ n μ f F b) The acceleration is
This is the end of the preview. Sign up to access the rest of the document.
Unformatted text preview: , k g μ and , 2 2 v ax = so , 2 k 2 g v x = or m. 13 . 3 = x...
View Full Document
This document was uploaded on 02/04/2008.
Ask a homework question - tutors are online | 187 | 587 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2017-17 | longest | en | 0.852756 |
https://en.wikipedia.org/wiki/Truncated_power_function | 1,477,289,687,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719468.5/warc/CC-MAIN-20161020183839-00043-ip-10-171-6-4.ec2.internal.warc.gz | 841,105,632 | 9,675 | # Truncated power function
Jump to: navigation, search
In mathematics, the truncated power function[1] with exponent ${\displaystyle n}$ is defined as
${\displaystyle x_{+}^{n}={\begin{cases}x^{n}&:\ x>0\\0&:\ x\leq 0.\end{cases}}}$
In particular,
${\displaystyle x_{+}={\begin{cases}x&:\ x>0\\0&:\ x\leq 0.\end{cases}}}$
and interpret the exponent as conventional power.
## Relations
• Truncated power functions can be used for construction of B-splines.
• ${\displaystyle x\mapsto x_{+}^{0}}$ is the Heaviside function.
• ${\displaystyle \chi _{[a,b)}(x)=(b-x)_{+}^{0}-(a-x)_{+}^{0}}$ where ${\displaystyle \chi }$ is the indicator function.
• Truncated power functions are refinable. | 227 | 694 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 6, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2016-44 | latest | en | 0.673702 |
https://duetosymmetry.com/notes/note-on-a-dimension-dependent-Weyl-identity/ | 1,722,735,404,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640388159.9/warc/CC-MAIN-20240804010149-20240804040149-00384.warc.gz | 169,824,485 | 6,840 | # Note on a (dimension-dependent) Weyl identity
There is a nice 4-dimensional Weyl identity that I can never seem to remember off the top of my head; so I decided I need to write this note so I don’t have to remember it in the future. The identity is:
\begin{align} \label{eq:theIdentity} C_{a}{}^{cde}C_{bcde} = \frac{1}{4}g_{ab} C^{cdef}C_{cdef} \,. \end{align}
This identity comes in handy for simplifying the equations of motion of Einstein-dilaton-Gauss-Bonnet gravity.
So, how do we work out this identity so we don’t have to remember it? I gave a hint when I said this is a dimension-dependent identity: it’s specific to dimension 4. Most1 dimension-dependent identities come from antisymmetrizing some expression over more than $d$ indices, where $d$ is the dimension. Because there are only $d$ coordinates, antisymmetrizing over more than $d$ slots will automatically vanish (i.e. more than $d$ vectors must be linearly-dependent in a $d$-dimensional vector space).
This particular identity comes from antisymmetrizing over $[abcdf]$ in the expression:
\begin{align} \label{eq:antisymmetrize} 0 = C^{ab}{}_{[ab}C^{cd}{}_{cd}\delta^e{}_{f]} \,. \end{align}
Obviously this will produce $5!=120$ terms in a sum, but the vast majority of them vanish because Weyl is tracefree on every pair of indices.
In fact, up to index order, there are only 5 combinations that are not obviously vanishing. The first Weyl tensor vanishes unless the lower indices are two of $\{c,d,f\}$; similarly, the second Weyl tensor vanishes unless the lower indices are two of $\{a,b,f\}$. This allows us to enumerate all of the possibilities in short order. After applying the antisymmetry of Weyl on the latter two indices, there are exactly 5 index permutations of $\{abcdf\}$ that are allowed, and they are (lexicographically):
\begin{align*} C^{ab}{}_{cd}C^{cd}{}_{ab}\delta^e{}_{f} \,, C^{ab}{}_{cd}C^{cd}{}_{af}\delta^e{}_{b} \,, C^{ab}{}_{cd}C^{cd}{}_{bf}\delta^e{}_{a} \,,\\ C^{ab}{}_{cf}C^{cd}{}_{ab}\delta^e{}_{d} \,, C^{ab}{}_{df}C^{cd}{}_{ab}\delta^e{}_{c} \,. \end{align*}
All that remains to do is to contract the $\delta$ indices and figure out the signature of each permutation to get the sign correct. Doing so gives the identity \eqref{eq:theIdentity} (with an index raised and indices renamed).
Of course, all of this is much easier with the xAct/xTensor package. I highly recommend this to find e.g. the Riemann form of the above identity. It comes from the same expression as \eqref{eq:antisymmetrize} but replacing $C$ with $R$, and now there are many more combinations that do not vanish, but instead produce Ricci terms. Explicitly, we get the unwieldy
\begin{align} R_{ebcd} R_{f}{}^{bcd} = \frac{1}{4} g_{ef} R_{abcd}R^{abcd} - g_{ef} R_{ab}R^{ab} + 2 R^{bc} R_{ebfc} + 2 R_{eb} R_f{}^b + \frac{1}{4} g_{ef} R^2 - R R_{ef} \,. \end{align}
While I was hunting for my lost identity, I came across a nice paper2 on the more general topic of dimension-dependent identities. This included, for example, the fact that the Cayley-Hamilton theorem can be derived from a dimension-dependent identity. Specifically, consider an $n$-dimensional vector space $V$ and the matrix $T^a{}_b$ with indices in $V$ and $V^*$. Then the Cayley-Hamilton theorem for $T$ can be written as
\begin{align} T^{i_1}{}_{[i_1} T^{i_2}{}_{i_2} \cdots T^{i_n}{}_{i_n} \delta^{a}{}_{b]} = 0 \,, \end{align}
where there are $n$ copies of $T$, and an antisymmetrization over $n+1$ indices.
This doesn’t obviously look like the Cayley-Hamilton theorem, but: it is a linear combination of various matrix powers of $(T^k)^a{}_b$, with coefficients determined by various traces of other powers of $T$ in a very specific way (they turn out to be elementary symmetric polynomials of the eigenvalues of $T$).
An example makes this a bit easier to see. Let’s expand this for $n=4$ dimensions. Again, I recommend using xTensor. For shorthand, let me write the matrix power
\begin{align} (T^k)^a{}_b \equiv T^a{}_{i_1} T^{i_1}{}_{i_2} \cdots T^{i_{k-1}}{}_b \end{align}
and denote the trace with $[T] \equiv T^a{}_a$. Then the four-dimensional Cayley-Hamilton theorem says
\begin{align} 0 ={}& 5 T^{i_1}{}_{[i_1} T^{i_2}{}_{i_2} T^{i_3}{}_{i_3} T^{i_4}{}_{i_4} \delta^{a}{}_{b]} \\ 0={}& (T^4)^a{}_b - (T^3)^a{}_b [T] + (T^2)^a{}_b \tfrac{1}{2} ([T]^2 - [T^2]) \\ &{}+ T^a{}_b \tfrac{1}{6} (-[T]^3 + 3 [T] [T^2] - 2 [T^3]) \nonumber\\ &{}+ \delta^a{}_b \tfrac{1}{24} ([T]^4 - 6 [T]^2 [T^2] + 3 [T^2]^2 + 8 [T] [T^3] - 6 [T^4]) \,. \nonumber \end{align}
What are all those strange combinations of traces? Well, use the identity that $[T^k] = \sum_{i=1}^n \lambda_i^k$ (easiest proved for diagonalizable matrices by going into the diagonal basis) where $\lambda_i$ is the i’th eigenvalue. Then you will readily verify that these are in fact the elementary symmetric polynomials of the eigenvalues of $T$, explicitly:
\begin{align*} \tfrac{1}{24} ([T]^4 - 6 [T]^2 [T^2] + 3 [T^2]^2 + 8 [T] [T^3] - 6 [T^4]) &= \lambda_1 \lambda_2 \lambda_3 \lambda_4 \\ \tfrac{1}{6} ([T]^3 - 3 [T] [T^2] + 2 [T^3])&= \lambda_1 \lambda_2 \lambda_3 + \lambda_1 \lambda_2 \lambda_4 + \lambda_1 \lambda_3 \lambda_4 + \lambda_2 \lambda_3 \lambda_4 \\ \tfrac{1}{2} ([T]^2 - [T^2]) &= \lambda_1 \lambda_2 + \lambda_1 \lambda_3 + \lambda_1 \lambda_4 + \lambda_2 \lambda_3 + \lambda_2 \lambda_4 + \lambda_3 \lambda_4 \\ [T] &= \lambda_1 + \lambda_2 + \lambda_3 + \lambda_4 \end{align*}
So this is indeed the characteristic polynomial of $T$! (Aside: equations such as these are examples of Newton-Girard identities).
# Footnotes
1. I think there may be more complicated ones which come from Garnir relations, but I know nothing about these.
2. Edgar and Höglund, J.Math.Phys. 43 (2002) 659-677 arXiv:gr-qc/0105066
Tags:
Categories:
Updated: | 1,953 | 5,804 | {"found_math": true, "script_math_tex": 32, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 6, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2024-33 | latest | en | 0.836844 |
https://somme2016.org/trendy/what-is-figural-analogy/ | 1,680,254,473,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949598.87/warc/CC-MAIN-20230331082653-20230331112653-00631.warc.gz | 605,794,528 | 10,174 | What is figural analogy?
What is figural analogy?
In figural analogies, a set of geometric shapes is presented. The first figure is related to the second figure, in the same way the third figure will be related to one of the answers. The objective is to solve the analogy problem.
How do you pass an analogy test?
Tips for solving Analogies
1. The only way to become better at verbal analogies is through practice.
2. Try to determine the relationship between the first pair of words.
3. Turn the analogies into sentences.
4. Go through tough problems systematically.
5. Read all of the answer choices first before making a decision.
What is figural analysis?
As Bill Routt reminds us in his admirable article on the figural in film, figural analysis is a form of hermeneutics involving the historical relation between signs and events, between the text’s present condition of meaning and its capacity to draw on and summon forth the past through the power of signs.
What is figural classification?
Figural Classification Questions and Answers Candidates are required to choose that one figure, which does not belong to the common group. This type of approach of finding the odd figure known as figural classification.
What is figural series?
Figure series problems are generally asked in competitive exams to test reasoning ability of candidates. It tests ability of candidate to imagine quick and proper solution. The figure on question may contain various geometrical figures, numbers, symbols or other pictures.
What are the 7 types of relationship in analogy?
Association — object/characteristic, cause and effect, function or purpose, sequential order, etc. Mathematical — equality, inequality, proportion (ratio, fraction, percent), etc. Logical (non-semantic) — letter patterns, phonetics.
What are the six types of analogy?
Analogies 1-six-types-of-analogies
• • SYNONYMS • ANTONYMS • OBJECT/ACTION • SOURCE/PRODUCT • PART/WHOLE • ANIMAL/HABITAT Analogies 1.
• Analogies An analogy compares two pairs of words that are related in the same way.
What is figural classification in reasoning?
What is Geeta’s rank in the class?
20th
So, Geeta’s rank in the class is 20th. Thus, both the statements are needed to answer the question.
What are Figural Analogies?
Figural Analogies – These questions are basically what they sound like – they are analogies, but instead of comparing words, they compare shapes. They will test your ability to understand and compare shapes and figures to each other.
What is the importance of analog topics in verbal reasoning?
For SSC, UPSC, RRB and Banking competitive examinations, the figure Analog Topics covered in Verbal Reasoning are very important. In questions based on analogy, a specific relation is given and another similar relation is identified from the given alternatives.
What are figural series questions?
Figural Series – In Figural Series questions, you will be presented with a sequence of figures and shapes, with one space left blank. You will have to choose which shape or figure fits into that blank space.
What kind of questions are on the shapes and figures test?
These multiple choice questions will test your abstract cognitive abilities. This means you will have to be able to understand and manipulate shapes and have a strong understanding of spatial orientations. You will be presented with abstract patterns of shapes and figures, and will be asked to figure out what comes next. | 714 | 3,474 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2023-14 | latest | en | 0.941421 |
https://www.abstract-polytopes.com/atlas/1728/30413/d12.html | 1,702,322,369,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679516047.98/warc/CC-MAIN-20231211174901-20231211204901-00088.warc.gz | 698,732,819 | 2,789 | Questions?
See the FAQ
or other info.
# Polytope of Type {2,12,12}
This page is part of the Atlas of Small Regular Polytopes
Atlas Canonical Name : {2,12,12}*1728e
if this polytope has a name.
Group : SmallGroup(1728,30413)
Rank : 4
Schlafli Type : {2,12,12}
Number of vertices, edges, etc : 2, 36, 216, 36
Order of s0s1s2s3 : 4
Order of s0s1s2s3s2s1 : 2
Special Properties :
Degenerate
Universal
Orientable
Flat
Related Polytopes :
Facet
Vertex Figure
Dual
Facet Of :
None in this Atlas
Vertex Figure Of :
None in this Atlas
Quotients (Maximal Quotients in Boldface) :
2-fold quotients : {2,12,6}*864f
3-fold quotients : {2,4,12}*576
4-fold quotients : {2,12,6}*432a
6-fold quotients : {2,4,6}*288
12-fold quotients : {2,4,6}*144
27-fold quotients : {2,4,4}*64
54-fold quotients : {2,2,4}*32, {2,4,2}*32
108-fold quotients : {2,2,2}*16
Covers (Minimal Covers in Boldface) :
None in this atlas.
Permutation Representation (GAP) :
```s0 := (1,2);;
s1 := ( 4, 5)( 6, 7)( 9, 11)( 12, 21)( 13, 23)( 14, 22)( 15, 25)( 16, 24)
( 17, 26)( 18, 29)( 19, 28)( 20, 27)( 31, 32)( 33, 34)( 36, 38)( 39, 48)
( 40, 50)( 41, 49)( 42, 52)( 43, 51)( 44, 53)( 45, 56)( 46, 55)( 47, 54)
( 58, 59)( 60, 61)( 63, 65)( 66, 75)( 67, 77)( 68, 76)( 69, 79)( 70, 78)
( 71, 80)( 72, 83)( 73, 82)( 74, 81)( 85, 86)( 87, 88)( 90, 92)( 93,102)
( 94,104)( 95,103)( 96,106)( 97,105)( 98,107)( 99,110)(100,109)(101,108);;
s2 := ( 3, 4)( 6, 21)( 7, 23)( 8, 22)( 9, 14)( 10, 13)( 11, 12)( 15, 29)
( 16, 28)( 17, 27)( 18, 20)( 24, 25)( 30, 31)( 33, 48)( 34, 50)( 35, 49)
( 36, 41)( 37, 40)( 38, 39)( 42, 56)( 43, 55)( 44, 54)( 45, 47)( 51, 52)
( 57, 85)( 58, 84)( 59, 86)( 60,102)( 61,104)( 62,103)( 63, 95)( 64, 94)
( 65, 93)( 66, 92)( 67, 91)( 68, 90)( 69,110)( 70,109)( 71,108)( 72,101)
( 73,100)( 74, 99)( 75, 87)( 76, 89)( 77, 88)( 78,106)( 79,105)( 80,107)
( 81, 98)( 82, 97)( 83, 96);;
s3 := ( 3, 64)( 4, 65)( 5, 63)( 6, 60)( 7, 61)( 8, 62)( 9, 59)( 10, 57)
( 11, 58)( 12, 82)( 13, 83)( 14, 81)( 15, 78)( 16, 79)( 17, 80)( 18, 77)
( 19, 75)( 20, 76)( 21, 73)( 22, 74)( 23, 72)( 24, 69)( 25, 70)( 26, 71)
( 27, 68)( 28, 66)( 29, 67)( 30, 91)( 31, 92)( 32, 90)( 33, 87)( 34, 88)
( 35, 89)( 36, 86)( 37, 84)( 38, 85)( 39,109)( 40,110)( 41,108)( 42,105)
( 43,106)( 44,107)( 45,104)( 46,102)( 47,103)( 48,100)( 49,101)( 50, 99)
( 51, 96)( 52, 97)( 53, 98)( 54, 95)( 55, 93)( 56, 94);;
poly := Group([s0,s1,s2,s3]);;
```
Finitely Presented Group Representation (GAP) :
```F := FreeGroup("s0","s1","s2","s3");;
s0 := F.1;; s1 := F.2;; s2 := F.3;; s3 := F.4;;
rels := [ s0*s0, s1*s1, s2*s2, s3*s3, s0*s1*s0*s1,
s0*s2*s0*s2, s0*s3*s0*s3, s1*s3*s1*s3,
s3*s1*s2*s3*s1*s2*s3*s1*s2*s3*s1*s2,
s1*s2*s3*s2*s1*s2*s1*s2*s1*s2*s3*s2*s1*s2*s1*s2,
s1*s2*s3*s2*s3*s2*s1*s2*s1*s2*s1*s3*s2*s1*s2*s1*s2*s3*s1*s2*s1*s2,
s1*s2*s3*s2*s1*s2*s3*s2*s1*s2*s3*s2*s1*s2*s3*s2*s1*s2*s3*s2*s1*s2*s3*s2 ];;
poly := F / rels;;
```
Permutation Representation (Magma) :
```s0 := Sym(110)!(1,2);
s1 := Sym(110)!( 4, 5)( 6, 7)( 9, 11)( 12, 21)( 13, 23)( 14, 22)( 15, 25)
( 16, 24)( 17, 26)( 18, 29)( 19, 28)( 20, 27)( 31, 32)( 33, 34)( 36, 38)
( 39, 48)( 40, 50)( 41, 49)( 42, 52)( 43, 51)( 44, 53)( 45, 56)( 46, 55)
( 47, 54)( 58, 59)( 60, 61)( 63, 65)( 66, 75)( 67, 77)( 68, 76)( 69, 79)
( 70, 78)( 71, 80)( 72, 83)( 73, 82)( 74, 81)( 85, 86)( 87, 88)( 90, 92)
( 93,102)( 94,104)( 95,103)( 96,106)( 97,105)( 98,107)( 99,110)(100,109)
(101,108);
s2 := Sym(110)!( 3, 4)( 6, 21)( 7, 23)( 8, 22)( 9, 14)( 10, 13)( 11, 12)
( 15, 29)( 16, 28)( 17, 27)( 18, 20)( 24, 25)( 30, 31)( 33, 48)( 34, 50)
( 35, 49)( 36, 41)( 37, 40)( 38, 39)( 42, 56)( 43, 55)( 44, 54)( 45, 47)
( 51, 52)( 57, 85)( 58, 84)( 59, 86)( 60,102)( 61,104)( 62,103)( 63, 95)
( 64, 94)( 65, 93)( 66, 92)( 67, 91)( 68, 90)( 69,110)( 70,109)( 71,108)
( 72,101)( 73,100)( 74, 99)( 75, 87)( 76, 89)( 77, 88)( 78,106)( 79,105)
( 80,107)( 81, 98)( 82, 97)( 83, 96);
s3 := Sym(110)!( 3, 64)( 4, 65)( 5, 63)( 6, 60)( 7, 61)( 8, 62)( 9, 59)
( 10, 57)( 11, 58)( 12, 82)( 13, 83)( 14, 81)( 15, 78)( 16, 79)( 17, 80)
( 18, 77)( 19, 75)( 20, 76)( 21, 73)( 22, 74)( 23, 72)( 24, 69)( 25, 70)
( 26, 71)( 27, 68)( 28, 66)( 29, 67)( 30, 91)( 31, 92)( 32, 90)( 33, 87)
( 34, 88)( 35, 89)( 36, 86)( 37, 84)( 38, 85)( 39,109)( 40,110)( 41,108)
( 42,105)( 43,106)( 44,107)( 45,104)( 46,102)( 47,103)( 48,100)( 49,101)
( 50, 99)( 51, 96)( 52, 97)( 53, 98)( 54, 95)( 55, 93)( 56, 94);
poly := sub<Sym(110)|s0,s1,s2,s3>;
```
Finitely Presented Group Representation (Magma) :
```poly<s0,s1,s2,s3> := Group< s0,s1,s2,s3 | s0*s0, s1*s1, s2*s2,
s3*s3, s0*s1*s0*s1, s0*s2*s0*s2, s0*s3*s0*s3,
s1*s3*s1*s3, s3*s1*s2*s3*s1*s2*s3*s1*s2*s3*s1*s2,
s1*s2*s3*s2*s1*s2*s1*s2*s1*s2*s3*s2*s1*s2*s1*s2,
s1*s2*s3*s2*s3*s2*s1*s2*s1*s2*s1*s3*s2*s1*s2*s1*s2*s3*s1*s2*s1*s2,
s1*s2*s3*s2*s1*s2*s3*s2*s1*s2*s3*s2*s1*s2*s3*s2*s1*s2*s3*s2*s1*s2*s3*s2 >;
```
to this polytope | 2,820 | 4,862 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2023-50 | longest | en | 0.270214 |
https://www.h2g2.com/edited_entry/A476606 | 1,708,864,804,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474595.59/warc/CC-MAIN-20240225103506-20240225133506-00239.warc.gz | 815,049,837 | 9,125 | # Trigonometry
Trigonometry is the branch of mathematics that concerns itself with the measurement of triangles. Mathematicians have been enamoured of triangles for thousands of years to the point where entire textbooks and classes are devoted to their measurement.
Despite the mathematical tomes dedicated to triangle measurement, there are only a few key things you really need to know in order to have a functional grasp of trigonometry.
### Some Basics
• The three angles in a triangle add up to 180°.
• One of the angles in a right-angled triangle1 is equal to 90°.
• The hypotenuse of a right-angled triangle is the side opposite the 90° angle. It is also the longest side of a right-angled triangle.
### Pythagoras' Theorem
Pythagoras was an ancient Greek philosopher-mathematician who lived around 500 BC. Among his claims to fame is the abstraction of mathematics2. His best known achievement, though, was to borrow a theorem that had been well-known among the Babylonians for 1,000 years, and name it after himself.
This theorem states the relationship in a right-angled triangle between the hypotenuse and the other two sides.
Pythagoras' Theorem states:
c2 = a2 + b2
Where c is the length of the hypotenuse, and a and b are the lengths of the other two sides.
You can see this in Diagram One, where Pythagoras' Theorem shows that:
52 = 42 + 32
Thanks to some Babylonian ingenuity and Pythagoras' promotion, you can calculate the length of any side of a right-angled triangle, provided you already know the length of the other two sides. For example, if we only knew the lengths of the two shortest sides in the triangle in Diagram One, then we could use Pythagoras' Theorem to come up with:
c2 = 42 + 32
c2 = 16 + 9 = 25
To give c = 5.
### SOHCAHTOA
About the same time Pythagoras was busy finding out the length of the sides of a right-angled triangle, the Mayans were upgrading the Mayan calendar and needed tools to describe the angles of triangles. From their work we derive trigonometric functions, the most important of which are sine, cosine, and tangent.
The Mayans realised that various properties of the angles of right triangles could be expressed as ratios between the sides of the triangles.
The Mayans also determined that for one of the non-right angles in a right-angled triangle, which we'll call θ3, these properties4 hold:
• The sine of θ is equal to the length of the side opposite the angle, divided by the length of the hypotenuse.
• The cosine of θ is equal to the length of the side adjacent to the angle, divided by the length of the hypotenuse.
• The tangent of θ is equal to the length of the side opposite the angle, divided by the length of the side adjacent to the angle.
Or, to state things more succinctly, referring to Diagram Two:
• Sin θ = Opposite/Hypotenuse
A perfect mnemonic device for remembering these relationships is SOHCAHTOA.
SOHCAHTOA
SOHCAHTOA gives us the ability to determine the value of any angle in a right-angled triangle, given the lengths of any two of the sides.
For example, let's consider the angle created where the hypotenuse c meets side b in Diagram One above, and let's pretend we only know the lengths of a and b (though a similar process could be done with any two sides and a different trigonometric function). We can find the value of this angle using any pocket calculator with trigonometric functions, as follows.
First, we only have the lengths of the opposite (a) and adjacent (b) sides, so let's look at SOHCAHTOA to see which of the three equations we can use. It's the TOA part, and we get:
Tan θ = Opp/Adj = 3/4 = 0.75
To convert from the tangent of the angle to the angle itself, you need the 'inverse' tangent, which is normally shown on calculators as tan-1. Try it out and you'll find that the value of the angle in Diagram One is 36.87° (or 0.644 radians, if you're into that sort of thing).
### The Unit Circle
The unit circle is simply a circle whose centre is on the Cartesian origin (0,0), and whose radius is 1.
The most interesting thing about the unit circle, as far as trigonometry is concerned, is that it gives us the values of sine and cosine5 for any angle.
In Diagram Three we're measuring the angle θ between the x-axis of the Cartesian plane and a line that extends from the origin. Now, here's the really interesting thing; the sine of the angle is equal to the y-coordinate of the point on the unit circle where the line crosses, and the cosine of the angle is equal to the x-coordinate. This is true for any line extending from the origin.
Why is this? Well, the line segment from the origin to the point where it crosses the unit circle forms the hypotenuse of a right-angled triangle. Because the radius of the circle is 1, the length of the hypotenuse is likewise 1. SOHCAHTOA's rules then boil down to:
Sin θ = Opposite
In other words:
Sin θ = y
Cos θ = x
Tan θ = y/x
1Also sometimes known as a 'right triangle'.2Pythagoras realised, for instance, that numbers are actually abstract objects. He would assert: 16 is just as real a concept as 16 pomegranates.3The Greek letter theta. On some browsers, notably Netscape, this might not display properly - sorry if this is the case, but that's browsers for you. Perhaps the next version will work better... but in the meantime, Internet Explorer 5 displays this perfectly.4If you're wondering how you calculate the sine, cosine and tangent of angles, then the easy answer is 'use a calculator'. The actual methods used to calculate the values of trigonometric functions are beyond the scope of this entry, but the humble calculator is proof that these methods do exist, and that they work.5And tangent, since the tangent of an angle is equal to the sine divided by the cosine. | 1,357 | 5,781 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2024-10 | latest | en | 0.948591 |
http://docplayer.net/13340833-Business-interruption-factsheet.html | 1,539,660,517,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583509996.54/warc/CC-MAIN-20181016030831-20181016052331-00171.warc.gz | 94,439,622 | 27,028 | # Business Interruption Factsheet
Save this PDF as:
Size: px
Start display at page:
## Transcription
6 Gross profit sum insured calculation - an example From Financial year end Gross Profit = Turnover 1,000,000 Closing stock and work in progress 10,000 Total 1,010,000 Less Uninsured Working Expenses (generally) Purchases 350,000 Packing and Freight 50,000 Bad Debts 5,000 Opening Stock and Work in Progress 5,000 Total 410,000 Gross Profit 600,000 Projecting Forward* 78,038 Sum insured for Insurance Year /16 678,038 *If you are working from figures from your last year end accounts you need to make an adjustment to bring the figures up to date, including an allowance for the insurance year ahead and then allow for growth/trends in the indemnity period. For convenience, the example assumes 5% annual growth, so there is 15,000 (2.5%) catch up to the start of the insurance year, 30,750 (5%) for the insurance year and a further 32,288 (5%) for a 12 month indemnity period. For a 24 months indemnity period you would need to add another 5% and then double the sum insured.
B u s i n e s s I n t erruption Calculation Guide 1. G R O S S P R O F I T - D I F F E R E N C E M E T H O D Gross Profit is defined as the amount by which the sum of - a) the turnover and the amount of
### Introduction to Business Interruption Insurance. Module 5: Increased Costs and Savings
Introduction to Module 5: Increased Costs and Savings Increase in Cost of Working There is an automatic provision for increased costs in the wording of the Insured Gross Profit item, which is highlighted
### Underinsurance in the commercial market
Retirement Investments Insurance Health Underinsurance in the commercial market What does it mean for businesses? Enter Contents What is underinsurance? What does it mean if I m underinsured? How big is
### Introduction to Business Interruption Insurance. Module 4: Gross Profit as the Sum Insured
Introduction to Module 4: The previous module described how a claim in respect of reduction in turnover is adjusted (reduced) for savings in expenses by multiplying the lost sales by the Rate of Gross
### Frequently Asked Questions (Including Definitions)
My Business Income Consultation from Chubb Frequently Asked Questions (Including Definitions) Frequently Asked Questions How To Use The Tool Click on the question that you would like to have answered.
### Underinsurance in the commercial market
Insurance Fully Managed Claims Service Financial Advice Underinsurance in the commercial market What does it mean for businesses? Enter Created by Contents What is underinsurance? What does it mean if
### Business Interruption Insurance
So! You have suffered a financial loss, have you..? And you think it is covered by your Business Interruption insurance policy? This article introduces you to Business Interruption (BI) insurance and the
### Airmic review of the supply chain insurance market Review of recent developments in the supply chain insurance market
REPORT Airmic review of the supply chain insurance market Review of recent developments in the supply chain insurance market 1. Executive summary Increasingly complex supply chains, together with greater
### PROFITS INSURANCE PROPOSAL FORM
SACOS INSURANCE COMPANY LTD P.O Box 636, Republic of Seychelles Telephone:295000, Telefax:224495 E-Mail:sacl@seychelles.net NOTES PROFITS INSURANCE PROPOSAL FORM (Perils (It is necessary that there is
### BUSINESS PACKAGE PROPOSAL
BUSINESS PACKAGE PROPOSAL RM Insurance Company (PRIVATE) LIMITED RM 317A INDEX SECTION S. PAGES 1 Fire 1 2 Business Interruption 2 3 3 All Risks 3 4 Theft 4 5 Money 4 6 Glass 5 7 Goods in Transit 5 8 Liability
### Your business plan. helping you with your business planning and forecasting. Name of business. Date when completed
Your business plan helping you with your business planning and forecasting Name of business Date when completed Please read the following notes before completing the form. (You may wish to talk to an accountant
### Renewable Energy Insurance When You Need Specialist Advice
Renewable Energy Insurance When You Need Specialist Advice The Renewable Energy Industry is diverse and fast evolving, however, with advances in technology comes new risk. Insurance for the renewables
### An Insurance package suitable for businesses in the professional services industry for example solicitors, accountants etc.
OFFICE INSURANCE Summary of Cover June 2005 Edition An Insurance package suitable for businesses in the professional services industry for example solicitors, accountants etc. Why choose AXA s Offices
### 6.3 PROFIT AND LOSS AND BALANCE SHEETS. Simple Financial Calculations. Analysing Performance - The Balance Sheet. Analysing Performance
63 COSTS AND COSTING 6 PROFIT AND LOSS AND BALANCE SHEETS Simple Financial Calculations Analysing Performance - The Balance Sheet Analysing Performance Analysing Financial Performance Profit And Loss Forecast
### Business Interruption (BI) Insurance Time to Reassess
Business Interruption (BI) Insurance Time to Reassess Over 90% of businesses that did not have business interruption insurance failed within two years Barry Woods, Commercial Underwriting Manager, QBE
### FLOODING INCIDENTS INSURANCE CLAIMS GUIDANCE
FLOODING INCIDENTS INSURANCE CLAIMS GUIDANCE INTRODUCTION Flooding caused by storms and heavy seas continues to affect many communities and businesses around the UK. Marsh is committed to helping communities
### Builder s Risk Structuring the Right Coverage, Avoiding and Resolving Disputes
Builder s Risk Structuring the Right Coverage, Avoiding and Resolving Disputes American Bar Association (ABA) Section of Litigation Insurance Coverage Tucson, Arizona March 6, 2015 Builder s Risk Speakers
### Indemnity Basis of Settlement. Practical Problems in Adjusting Losses. By the CILA Property Special Interest Group
Indemnity Basis of Settlement. Practical Problems in Adjusting Losses. By the CILA Property Special Interest Group INTRODUCTION All insurance policies are based on providing the policyholder with an indemnity,
### Resource 2.7 Introduction to Insurance Cover for Business What insurance is compulsory for businesses? Employers' liability insurance
Page 1 of 5 Introduction to Insurance Cover for Business All businesses need to be insured against risks, such as the theft of equipment or workrelated injury to staff. If you run a small business you
Insurance for your small business @BritishInsurers As an employer, you are legally required to have Employers Liability Insurance. If you use motor vehicles for your business, you are legally required
Business-critical Insurance Identifying those insurances that support the business and its strategy Guide 2015 Contents Introduction... 4 Categories of insurance... 5 Determining which insurance covers
### Business Interruption Insurance. Module 1: The Basics of BI
Module 1: Introduction Business Interruption ( BI ) insurance and claims are built on a few basic principles and if these are well understood the mystery of the subject is resolved. This article is the
### Contract Works and What you need to Know
ACN 002 566 743 Brookvale Insurance Brokers Pty Ltd Contract Works and What you need to Know 1. The Need for Contract Works Insurance The growth of large-scale construction projects has been accompanied
### Schedule 12 - Insurances
RFP Version There are two parts to this Schedule 12: Part A Insurances applicable to the D&C Phase; and Part B Insurances during the O&M Phase. [State Note: The State in its discretion may procure any
### An Insurance package suitable for businesses in the professional services industry for example solicitors, accountants etc.
Office Insurance Summary of Cover September 201 Edition An Insurance package suitable for businesses in the professional services industry for example solicitors, accountants etc. Why choose AXA s Offices
### Business Income/Extra Expense Calculator
Business Income and Extra Expense Calculator Navigate throughout this workbook calculator tool by clicking on desired button below. Each module you enter has a button to bring you back to this main Menu
### Business Interruption Insurance what your should know right now
Business Interruption Insurance what your should know right now If a natural or man-made disaster interrupted your company's or client's business would it survive? What may have seemed like protection
### BUSINESS INCOME (AND EXTRA EXPENSE) COVERAGE FORM
COMMERCIAL PROPERTY CP 00 30 10 00 BUSINESS INCOME (AND EXTRA EXPENSE) COVERAGE FORM Various provisions in this policy restrict coverage. Read the entire policy carefully to determine rights, duties and
### KEY GUIDE. Setting up a new business
KEY GUIDE Setting up a new business The business idea You have a business idea, but can you turn it into a viable enterprise? Is there a market for the goods you will produce or the services you will supply?
### Step 1. Step 2 REVIEW YOUR POLICY PROMPT NOTICE TO INSURANCE CARRIER
business income insurance checklist Each year, many organizations suffer a temporary interruption of their business operations due to the damage and destruction caused by hurricanes and other causes of
### Risk in Construction
Risk in Construction Market Research prepared for June 2014 In association with the Chartered Institution of Civil Engineering Surveyors, The Institution of Civil Engineers and the UK s leading Design
### Ensuring Proper Protection of Your Company s Assets and Exposures
Ensuring Proper Protection of Your Company s Assets and Exposures Know the Worth of Your Assets The development of accurate insurable values is an excellent way to verify the physical worth of your assets
### A business insurance package available for self employed persons and small businesses (including limited companies) with up to 5 employees.
6758_WLI354NZ_POL.qxd 17/8/04 8:49 Page 1 VENTURE PLUS Summary of Cover A business insurance package available for self employed persons and small businesses (including limited companies) with up to 5
### An introduction to insurance cover for businesses
An introduction to insurance cover for businesses All businesses need to be insured against potential risks such as contractual disputes, accidental damage, fire, flood or theft and claims for negligence.
### WORKING CAPITAL MANAGEMENT
CHAPTER 9 WORKING CAPITAL MANAGEMENT Working capital is the long term fund required to run the day to day operations of the business. The company starts with cash. It buys raw materials, employs staff
### Cash Flow Forecasting & Break-Even Analysis
Cash Flow Forecasting & Break-Even Analysis 1. Cash Flow Cash Flow Projections What is cash flow? Cash flow is an estimate of the timing of when the cash associated with sales will be received and when
### PROPERTY OWNERS INSURANCE PROPOSAL
PROPERTY OWNERS INSURANCE PROPOSAL Please use BLOCK CAPITALS and tick boxes where applicable. Where requested enter further details in the space provided. If there is insufficient space please continue
### My Business Income Consultation from Chubb. Help Documents. User Guide. Frequently Asked Questions. Script
My Business Income Consultation from Chubb Help Documents Select one of the links below for additonal information: User Guide Frequently Asked Questions Script My Business Income Consultation from Chubb
### Fire? Loss of Profits? 3. What company covers the machinery to be insured under a Machinery Breakdown Policy?
Questionnaire and Proposal for Loss of Profits following Machinery Breakdown Insurance Name Postal address Nature of business: 1. What works of your business are to be insured against Machinery Loss of
### Shops Insurance. Summary of Cover September 2013 Edition. An Insurance Package for Retailers and Franchises. Why choose AXA s Shops Insurance Package?
Shops Insurance Summary of Cover September 201 Edition An Insurance Package for Retailers and Franchises. Why choose AXA s Shops Insurance Package? AXA s Shops Insurance Package offers you generous levels
### BUSINESS INSURANCE PROPOSAL FORM
Head Office, Insurance Centre, 30 Samora Machel Avenue, P.O. Box 1256, Harare, Zimbabwe Tel: 263-4-701133, 704911-4, 700346, Fax: 263-4-704134, 700083 BUSINESS INSURANCE PROPOSAL FORM Name of Proposer
### CPD Spotlight Quiz September 2012. Working Capital
CPD Spotlight Quiz September 2012 Working Capital 1 What is working capital? This is a topic that has been the subject of debate for many years and will, no doubt, continue to be so. One response to the
### Chapter. Working capital
Chapter 10 Working capital 1 10.1 Working capital Working capital is the capital available for conducting the day-to-day operations of the business and consists of current assets and current liabilities.
### Summary of Cover September 2013 Edition
Surgeries Insurance Policy Summary Surgeries Insurance Summary of Cover September 201 Edition An insurance package for doctors, dentists and vets. Why choose AXA s Surgeries Insurance Package? AXA s Surgeries
### An introduction to our General Insurance Broking Service. We ll develop a customised insurance and risk management solution to protect your business
An introduction to our General Insurance Broking Service We ll develop a customised insurance and risk management solution to protect your business We ve been helping Australians create secure financial
### Business Income Insurance Policy: End the Confusion
1740H Dell Range Blvd., #300 I Cheyenne, Wyoming 82009 PHONE: 307.433.8180 I FAX: 307.634.9497 EMAIL: info@bisimplified.com www.bisimplified.com ARTICLES: THE 411 Business Income Insurance Policy: End
Business Insurance business, assets, liability I need to feel sure my insurance company measures up to the job! your partner for business insurance Are we the right business for you? Are we the right business
### LIFETIME MORTGAGE LUMP SUM
LIFETIME MORTGAGE LUMP SUM Terms and Conditions (version 5) This is an important document. Please keep it in a safe place. LV= Lifetime Mortgage lump sum Terms and Conditions Welcome to LV=, and thank
### LIABILITY INSURANCE SUMMARY OF COVER
LIABILITY INSURANCE SUMMARY OF COVER This gives only a summary of the cover provided and it does not give details of all the terms, conditions and exclusions. A full policy wording is available on request.
### A BRIEFING NOTE FROM MICHAEL WHITTON OF EDWIN COE LLP DEPRECIATION A MISSED OPPORTUNITY
A BRIEFING NOTE FROM MICHAEL WHITTON OF EDWIN COE LLP DEPRECIATION A MISSED OPPORTUNITY 1. The appeal in Synergy Health (UK) Limited v- CGU Insurance Plc (t/a Norwich Union) and Others arose in relation
### Steps to recovery. Queensland flood crisis. 19 January 2011
Steps to recovery Queensland flood crisis 19 January 2011 Immediate considerations Business interruption You may have experienced operational outages, IT interruptions as well as disruption to your manufacturing
### FINAL ACCOUNTS FINAL ACCOUNTS AND THE TRIAL BALANCE
12 FINAL ACCOUNTS For most businesses, the final accounts, which are produced at the end of each financial year, comprise: trading account profit and loss account balance sheet Final accounts can be presented
### Business Insurance Cover. Information for business owners
Business Insurance Cover Information for business owners IMPORTANT NOTE This brochure summarises some of the cover provided by AMP s Business Insurance policy. It is important to note that limits, excesses,
### KEY GUIDE. Setting up a new business
KEY GUIDE Setting up a new business The business idea Some of the decisions and actions that you take when starting a business can have significant effects for some time. The foundations you put in place
### Incisive Business Guide to Business Insurance
Incisive Business Guide to Business Insurance Business Insurance Guide Summary This guide from Incisive Business outlines the features and benefits for your business from taking out business insurance.
### COMMERCIAL INSURANCE POLICY
COMMERCIAL INSURANCE POLICY Today s business owner is faced with a broad array and multitude of risks on a day to day basis, both insurable and uninsurable. The control and management of these risks is
### THE INSURANCE POLICY: SIMPLIFIED!
THE INSURANCE POLICY: SIMPLIFIED! INSURANCE CONTRACT: COMMERCIAL PROPERTY AND BUSINESS INCOME The insurance policy is a contract and as such, is governed by its conditions, provisions, and exclusions.
### CHANGES IN THE RISK. If yes, what is the new name? If yes, what has changed? If yes, what has changed? If yes, how has it changed?
Loss management, risk assessment and technical advisers to the general insurance and business communities MEMBER COMPANY OF THE LMI GROUP Risk Management International Pty Ltd Melbourne, Sydney, Brisbane,
### A L L B R I G H T B I S H O P R O W L E Y L I M I T E D T E R M S OF B U S I N E S S
GENERAL POINTS A L L B R I G H T B I S H O P R O W L E Y L I M I T E D T E R M S OF B U S I N E S S 1. Allbright Bishop Rowley Limited (or ABR) is a General insurance broker based at 2-3 The Courtyard,
### Engage Education Foundation
2015 End of Year Seminar Exam Engage Education Foundation Units 3 and 4 Accounting Practice Exam Solutions Stop! Don t look at these solutions until you have attempted the exam. Any questions? Check the
### An insurance package suitable for small to medium sized businesses in the construction industry with a:-
Contractors Choice Summary of Cover An insurance package suitable for small to medium sized businesses in the construction industry with a:- Turnover up to 1.5m at new business and 2m at renewal Maximum
### Putting a business idea into practice
Putting a business idea into practice Objectives when starting up What is an objective? An objective is what the business is trying to achieve eg make profit Why do entrepreneurs start a business? There
### Chubb Group of Insurance Companies
Chubb Group of Insurance Companies Business Income, Extra Expense and Research and Development Income Worksheet For Ongoing-Business with/without R&D Activities This worksheet will help you determine the
### BUSINESS CASE STUDIES
BUSINESS CASE STUDIES Below you will find 2 Case Studies of 2 very different organisations. An small Advertising Firm; & a well established Construction Equipment Hire Business. Most business owners have
### A Sole Trader. Accounts. 31 December 2007
Accounts 31 December 2007 Approval statement I approve these accounts which comprise the Profit and Loss Account, Balance Sheet and related notes. I acknowledge my responsibility for the accounts, including
### 5.2 BUDGETING; CASH FLOW FORECASTS. Introduction To Budgets And Cash Flow Forecasts. Cash Flow Forecasts. Budget And Cash Flow Exercises
52 FUNDING 5 BUDGETING; CASH FLOW FORECASTS Introduction To Budgets And Cash Flow Forecasts Cash Flow Forecasts Budget And Cash Flow Exercises Cash Flow Exercises P 168 INTRODUCTION TO BUDGETS AND CASH
### Insurance Specialists to the building and allied trades
www.tradedirectinsurance.co.uk Insurance Specialists to the building and allied trades Welcome Welcome to Trade Direct Insurance, we hope that you find our brochure informative and that our range of policies
### Trustee indemnity insurance
This information sheet provides guidance and advice for village halls and similar community buildings and how to manage their employees and volunteers Last updated September 2013 Contents Contents Page
### WORKING FROM HOME. Summary of Cover
WORKING FROM HOME Summary of Cover An insurance package for your business available for many occupations where the business is home based. Why choose AXA s Working From Home Insurance Package? Tailor-made
### BUSINESS PLAN TEMPLATE
PARRY SOUND AREA COMMUNITY BUSINESS & DEVELOPMENT CENTRE INC. A Community Futures Development Corporation 1A Church Street, Parry Sound, ON P2A 1Y2 Telephone: (705)746-4455 Toll Free: 1-888-746-4455 Fax:
### Accounts Information for Sole Traders. Turnover less than 320,000 ( 250,000) Revenue On-Line Service
Accounts Information for Sole Traders Turnover less than 320,000 ( 250,000) File Returns and Accounts Information online at www.revenue.ie Revenue On-Line Service Paper Filers (Accounts Information - Income
### NERA Analysis of Energy Supplier Margins
7 December 2009 NERA Analysis of Energy Supplier Margins By Graham Shuttleworth Even though wholesale energy prices have fallen recently, gas and electricity suppliers are earning very little margin on
### Key Person Insurance. A succinct guide for business owners. Author - Paul Benson, B. Bus, CFP, SSA
Key Person Insurance A succinct guide for business owners Author - Paul Benson, B. Bus, CFP, SSA Guidance Financial Services provides specialist financial planning advice for business owners and the self-employed.
### NonProfit 101. Notes: Session 1B: Insurance, What you do not know can hurt you! From Survivability to Sustainability. June 10, 2015 Session 1B page 1
Session 1B: Insurance, What you do not know can hurt you! Notes: June 10, 2015 Session 1B page 1 Session 1B: Insurance, What you do not know can hurt you! Notes: June 10, 2015 Session 1B page 2 June 10,
### Summary of Cover Commercial Retail Policy
Summary of Cover Commercial Retail Policy The information provided in this policy summary is key information you should read. This policy summary does not contain the full terms and conditions of the cover,
### Please provide details of your claims within the last 5 years Date Incident details Amount paid Amount reserved
Combined Liability Insurance Client name: Operating address: Company reg. number: Business description: Year business established: Current Insurer: Current Broker: Renewal date: Premium: Please provide
### Cashflow Management. What is cashflow
Cashflow Management This Fact File Information Sheet looks at the key elements of cashflow, and how effective cashflow management will help protect the financial security of a business. It outlines the
### General BI Subjects. The Insurance of Wages
General BI Subjects No other epense receives as much attention in Business Interruption ( BI ) insurance and claims, as does Wages. Following a damage event, when turnover is down, there are many competing
### SOLE TRADER FINAL ACCOUNTS
6 SOLE TRADER FINAL ACCOUNTS CASE STUDY Starting out in business Olivia Boulton used to work as a buyer of kitchen and cookware goods for a large department store in central London. She was good at her
### SECTION 4: CLAIMS MANAGEMENT
SECTION 4 INDEX SECTION 4: CLAIMS MANAGEMENT PART 1 Adjusters Reports: (all perils) B. All reports; front page format; B. Full reports: i. Main body format; ii. Explanatory notes. C. Abbreviated reports
Perrys Chartered Accountants Buy-to-let guide about tax Introduction As a buy-to-let landlord it is important you know about tax and how it affects you and your investment. This is why Perrys Chartered
### OVERVIEW OF BUSINESS INTERRUPTION INSURANCE
OVERVIEW OF BUSINESS INTERRUPTION INSURANCE ANDREW C.M. BOURNE, CPA, CMA, DIFA, CFF Partner, BDO Canada LLP - FINANCIAL ADVISORY SERVICES 2 March 2015 AGENDA Property Insurance vs. Business Interruption
### A guide to business cash flow management
A guide to business cash flow management Contents 01. Cash flow management 01 02. Practical steps to managing cash flow 04 03. Improving everyday cash flow 06 04. How to manage cash flow surpluses and
### A simple guide to Life Insurance
A simple guide to Life Insurance Your guide to Life, Income Protection, Trauma, and Total & Permanent Disability Insurance. Lifebroker Pty Ltd ABN 92 083 177 011 Introduction Why do I need personal insurance?...
### Insurance for small businesses
Simply Business THE SIMPLY BUSINESS GUIDE TO Insurance for small businesses Guide #2 The UK s biggest business insurance provider 2 3 Contents 01 Finding a provider The role of insurance brokers and aggregators,
### STARTER GUIDE TO CONTRACTING/FREELANCING
STARTER GUIDE TO CONTRACTING/FREELANCING This guide has been developed by the Partners of Forbes Young and is meant to be used as an informative and useful resource for people new to contracting or freelancing
SME BUSINESS INSURANCE MADE SIMPLE Working with the profession to simplify the language of insurance PUTTING INSURANCE TO WORK Business insurance is about protecting your livelihood. For SMEs in particular,
### Stream 4. Business interruption: traditional risks, new complexities
Stream 4 Business interruption: traditional risks, new complexities NEW COMPLEXITIES IN BUSINESS INTERRUPTION (BI) LONDON MARKET CLAIMS CONFERENCE 15 OCTOBER 2015 New Complexities in Business Interruption
HOW TO BUY A BUSINESS Buying a business is possibly the most important financial decision you will make. It may consume all your spare cash, cause you to borrow a significant sum of money, determine your
### Leasing Business Premises: Occupier Guide
Leasing Business Premises: Occupier Guide This document is one of three component parts of the Code for Leasing Business Premises Copyright The Joint Working Group on Commercial Leases, 2007. Any of the
### Q&A. Questions and Answers on Insurance for Housing Co-operatives. What insurance does our housing co-operative need?
Q&A Questions and Answers on Insurance for Housing Co-operatives What insurance does our housing co-operative need? You need at least five kinds of insurance perhaps more, depending on your circumstances.
### The real costs of accidents and ill health at work
The real costs of accidents and ill health at work This leaflet: explains how expensive work accidents and ill health can be; provides a real-life example to show this; suggests simple methods to work
### Sports Liability Insurance. for Clubs & Associations, Sports Coaching Businesses, Sports Coaches & Trainers and Event/Camp Organisers
Sports Liability Insurance for Clubs & Associations, Sports Coaching Businesses, Sports Coaches & Trainers and Event/Camp Organisers Contents Choose sportsinsurance4u from General & Medical Healthcare
### MOTOR TRADE INSURANCE GUIDE
MOTOR TRADE INSURANCE GUIDE 1 Personalised Policy Each policy is different because each business is different. We tailor your policy to your business to ensure you only pay for what you need to 2 Flexible
### THIS APPLICATION IS FOR A CLAIMS MADE POLICY
PEMBROKE SYNDICATE 4000 MISCELLANEOUS ERRORS & OMISSIONS INSURANCE APPLICATION THIS APPLICATION IS FOR A CLAIMS MADE POLICY ALL QUESTIONS MUST BE ANSWERED COMPLETELY. DO NOT LEAVE ANY SPACE BLANK. INDICATE
### IndustryEdge for hotels and restaurants OUR KNOWLEDGE IS YOUR EDGE
IndustryEdge for hotels and restaurants OUR KNOWLEDGE IS YOUR EDGE OUR KNOWLEDGE IS YOUR EDGE IndustryEdge At Travelers, we recognise that no two industries are the same and that dealing with the complexities
### You should read this Summary of Cover in conjunction with the Policy Schedule and Policy Wording.
Summary of Cover iprism Office Policy About Your Policy This document provides a summary of the cover provided by the iprism Office policy. This summary is provided to you for information purposes only
### RESTAURANTS, WINE BARS AND PUBLIC HOUSES INSURANCE
RESTAURANTS, WINE BARS AND PUBLIC HOUSES INSURANCE Summary of Cover June 2005 Edition An insurance package for Restaurants, Wine Bars and Public Houses. Why choose AXA s Restaurants, Wine Bars and Public | 5,887 | 27,440 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2018-43 | latest | en | 0.876797 |
https://aiuta.org/en/how-many-prime-numbers-are-there-between-45-and-726-or-7-some-saying-6-and-some-saying-is-7-i.41958.html | 1,555,727,201,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578528481.47/warc/CC-MAIN-20190420020937-20190420042937-00419.warc.gz | 333,708,211 | 6,599 | Mathematics
How many prime numbers are there between 45 and 72? 6 or 7 ?? Some saying 6 and some saying is 7. I am confused. I calculate it and it shows 7. Help me with the answer PLZ ....
yassautumn
4 years ago
47,53,59,61,67,71
6 numbers
ghhhh
4 years ago
Prime numbers mean that other numbers don't go in the actual number such as 17. No numbers go into 17.
The answer to your question is 47, 53, 59, 61, 67, 71 so 6 numbers :) | 139 | 439 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2019-18 | latest | en | 0.915771 |
https://www.agisoft.com/forum/index.php?topic=11164.0;prev_next=next | 1,726,474,375,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651682.69/warc/CC-MAIN-20240916080220-20240916110220-00000.warc.gz | 587,655,193 | 7,465 | # Forum
### Author Topic: script inserting points with Z label (Read 1688 times)
#### geodezja3D
• Newbie
• Posts: 7
##### script inserting points with Z label
« on: June 17, 2024, 01:25:51 PM »
Hi,
New to scripting so I need some help.
I would like the script to insert grid of points with Z coordinate as labels. I generated code as below but it does not work.
import Metashape
import math
doc = Metashape.app.document
chunk = doc.chunk
if not chunk:
grid_step = 1.0
region = chunk.region
bbox = region.bounding_box
min_point = bbox.min
max_point = bbox.max
x_steps = math.ceil((max_point.x - min_point.x) / grid_step)
y_steps = math.ceil((max_point.y - min_point.y) / grid_step)
points = []
for i in range(x_steps + 1):
for j in range(y_steps + 1):
x = min_point.x + i * grid_step
y = min_point.y + j * grid_step
# Szukamy wysokości dla danej współrzędnej (x, y) na modelu 3D
ray_origin = Metashape.Vector([x, y, max_point.z])
ray_direction = Metashape.Vector([0, 0, -1])
point = chunk.model.pickPoint(ray_origin, ray_direction)
if point is not None:
point_coord = point.coord
height = point_coord.z
points.append((Metashape.Vector([x, y, height]), str(height)))
for point, name in points:
marker.label = name
Could you correct this?
#### Alexey Pasumansky
• Agisoft Technical Support
• Hero Member
• Posts: 15029
##### Re: script inserting points with Z label
« Reply #1 on: June 20, 2024, 05:30:42 PM »
Hello geodezja3D,
Do you want to create markers or point shapes?
Best regards,
Alexey Pasumansky,
Agisoft LLC
#### geodezja3D
• Newbie
• Posts: 7
##### Re: script inserting points with Z label
« Reply #2 on: June 21, 2024, 03:23:21 PM »
Hello! I would like to create point shapes.
Hello geodezja3D,
Do you want to create markers or point shapes?
#### Alexey Pasumansky
• Agisoft Technical Support
• Hero Member
• Posts: 15029
##### Re: script inserting points with Z label
« Reply #3 on: June 24, 2024, 01:05:31 PM »
Hello geodezja3D,
Please see the example below that creates a grid of point shapes with elevation calculated based on DEM altitude in corresponding XY location:
Code: [Select]
`import Metashapedef convert_meters_to_deg(value, chunk): crs = chunk.crs if chunk.crs.authority != 'EPSG::4326': return Metashape.Vector([value, value]) T = chunk.transform.matrix origin = chunk.region.center origin_geoc = T.mulp(origin) origin_geog = crs.project(origin_geoc) #longitude v_x = (crs.unproject(origin_geog + Metashape.Vector([1E-5,0,0])) - origin_geoc).norm() * 1E5 #latitude v_y = (crs.unproject(origin_geog + Metashape.Vector([0,1E-5,0])) - origin_geoc).norm() * 1E5 return Metashape.Vector([value / v_x, value / v_y]) def create_grid(): doc = Metashape.app.document chunk = doc.chunk if not chunk: print("Empty project, script aborted") return 0 if not chunk.elevation: print("Missing DEM in the active chunk, script aborted") return 0 STEP = Metashape.app.getFloat("Please input the grid step (in meters) for point shapes:", 10) if STEP <= 0: print("Wrong number! Step value should be positive. Script aborted.") return 0 DEM = chunk.elevation if not chunk.shapes: chunk.shapes = Metashape.Shapes() chunk.shapes.crs = elevation.crs shapes = chunk.shapes limits = [[DEM.left, DEM.right], [DEM.bottom, DEM.top]] points = chunk.shapes.addGroup() points.label = "Automatic point shapes (step {:.2f} meters)".format(STEP) step = convert_meters_to_deg(STEP, chunk) x = limits[0][0] y = limits[1][0] while x < limits[0][1]: y = limits[1][0] while y < limits[1][1]: altitude = DEM.altitude(Metashape.Vector([x, y])) if altitude == -32767: y += step.y continue #skipping point point = chunk.shapes.addShape() point.group = points point.geometry = Metashape.Geometry.Point(Metashape.Vector([x,y,altitude])) point.label = "{:.3f}".format(altitude) y += step.y x += step.x return 1label = "Custom Menu/Create grid of points from DEM"print("To execute this script press {}".format(label))Metashape.app.addMenuItem(label, create_grid)`
Best regards,
Alexey Pasumansky,
Agisoft LLC
#### geodezja3D
• Newbie
• Posts: 7
##### Re: script inserting points with Z label
« Reply #4 on: June 26, 2024, 04:14:52 PM »
Works great. Thank you! | 1,211 | 4,164 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-38 | latest | en | 0.655474 |
http://www.cs.unca.edu/~brock/classes/Spring2001/255/homework/sol01.html | 1,544,607,382,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823817.62/warc/CC-MAIN-20181212091014-20181212112514-00284.warc.gz | 347,157,320 | 1,413 | # Spring 2001 CSCI 255 Homework 1 Solution
## Problem 1
Convert the following numbers from decimal into ten-bit twos-complement notation.
431 0110101111 -400 1001110000 1 0000000001 -11 1111110101
## Problem 2
Convert the following ten-bit twos-complement numbers into decimal numbers.
0011001100 204 1110001110 -114 0000000000 0 1111111100 -4
## Problem 3
What are the largest and smallest numbers that can be expressed as a six-bit twos-complement number?
The smallest is -32 (10000) and the largest is 31 (011111). In general, if n is the width of the twos complement number, the smallest possible number is -2n-1 and the largest possible number is 2n-1-1,
## Problem 4
Express the following decimal numbers in base 6:
13 21 20 32 | 219 | 748 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2018-51 | latest | en | 0.735817 |
https://puzzling.stackexchange.com/users/49874/george-menoutis?tab=answers | 1,642,472,611,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300658.84/warc/CC-MAIN-20220118002226-20220118032226-00404.warc.gz | 541,903,541 | 24,169 | George Menoutis
• Member for 3 years, 7 months
• Last seen this week
There are
This is a proof that Omega Krypton's answer of 38 is optimal, as long as these words are nonexistent (single implication): EEVEN,EINE,ENINE,FIUR,FIVR,FOVE,NIGE,NINH,OWO,,SEGEN,SEVET,SEVHN,SIVEN,TNO,...
I think it's I can dance and swirl in romantic ballrooms But I'm just as good a companion in the gloom. Dizzying, flying far above the ground, But I'll always land, safe and sound. My best friend ...
I guess it's Beauty I can bring, as I make a sound like thunder, Over the air I sing, as I am torn asunder. Of the many wondrous things I can do, My favorite is bring light to your view. And of ...
Let's start low: 40. Let's name the players A,B,C,D,E,F,G,H,I,J and define "n-next" of a player to be the player who is positioned n places afterwards, with round wrapping. Eg, the 2-next of A is C, ...
I say which is Generally, For example: For any algorithm,
This is an extension of Dr Xorile's observations, which fully answers the question that "there is no optimal strategy as long as knowdledge(and/or guessing) of your opponent distrubution is missing". ...
Just a visual cue of Gareth's answer
I guess this is more probable to add a clarification in the riddle's body rather than be accepted as an answer.
I vote for Reasoning: Additionally: I cannot for the love of all create spoiler quote newlines!
I think it's the Let me first clarify a possible loophole in the typical answer: If this is a case, Additionally, On the other hand,
OK, I'll go high-risk here: Because: Alternatively:
I think I've this somewhere around here. You are positioned at Correction:
Another simple solution:
This is easy, and all answers are already fine. I'll try to gather all the conditions that needed to be true in order for the perpetual motion machine to have been true. Obviously, the "catch" (what ...
Solved: Maybe a bit too solved, there...
Disclaimer: This is a proof of p_sutherland's answer. If you consider this correct, please accept their answer. Spoiler-splitter Spoiler-splitter Spoiler-splitter
Hmmm. Could it be because and of course
Here goes then.... This is a reference to Hasbro's board game "Hotels" (actually "Hotel" in some regions). Each picture refers to one of the game's hotel chains. The first picture is a photo of ...
Explanation: .
Maybe it is I come in every shape and every form In society I'm nothing special I'm the norm Without me there would be no control They would die, the things filled with petrol If a being placed ...
I already agree with some answers presented. Let me provide a graphical one. This is the 3-set venn diagram of all posible sets an element can be a member of: I assume "some" means "1 or more". Thus, ...
This is a partial answer. The trick is to After we establish this, it's just a matter of which will take a maximum of
Alright, wrong and I love it: You're dead, don't you see it? Just stop hanging around! You tell all my secrets, yet make not a sound. Gotten under my skin—you think that you're cute! Unfeeling, ...
I think it's Reasoning: So, we have: Which also leads to Amy's tactic:
The minimum amount requested is Because This, then, means Which leaves | 787 | 3,219 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2022-05 | latest | en | 0.951943 |
https://docslib.org/doc/4925911/5-phase-transitions | 1,701,789,416,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100551.2/warc/CC-MAIN-20231205140836-20231205170836-00716.warc.gz | 262,521,827 | 44,030 | <<
5. Transitions
A is an abrupt, discontinuous change in the properties of a system. We’ve already seen one example of a phase transition in our discussion of Bose-Einstein . In that case, we had to look fairly closely to see the discontinuity: it was lurking in the derivative of the capacity. In other phase transitions — many of them already familiar — the discontinuity is more manifest. Examples include condensing to and water to .
In this section we’ll explore a couple of phase transitions in some detail and extract some lessons that are common to all transitions.
5.1 - Transition Recall that we derived the van der Waals for a gas (2.31)inSection 2.5.WecanwritethevanderWaalsequationas
k T a p = B (5.1) v b v2 where v = V/N is the per . In the literature, you will also see this equation written in terms of the particle ⇢ =1/v.
On the right we fix T at di↵erent values p and sketch the graph of p vs. V determined by the . These curves are isotherms — line of constant . T>Tc As we can see from the diagram, the isotherms T=Tc take three di↵erent shapes depending on the value of T .Thetopcurveshowstheisotherm T
–135– At some intermediate temperature, the wiggle must flatten out so that the bottom curve looks like the top one. This happens when the maximum and minimum meet to form an inflection point. Mathematically, we are looking for a to dp/dv = d2p/dv2 =0.itissimpletocheckthatthesetwoequationsonlyhaveasolutionatthe critical temperature T = Tc given by 8a k T = (5.2) B c 27b
Let’s look in more detail at the T
First look at the middle solution. This has p some fairly weird properties. We can see from the T 0. |T This means that if we apply a to the con- tainer to squeeze the gas, the decreases. The gas doesn’t push back; it just relents. But if v we expand the gas, the pressure increases and the b gas pushes harder. Both of these properties are Figure 35: telling us that the gas in that state is unstable. If we were able to create such a state, it wouldn’t hand around for long because any tiny perturbation would to a rapid, change in its density. If we want to find states which we are likely to observe in then we should look at the other two .
The solution to the left on the graph has v slightly bigger than b.But,recallfrom our discussion of Section 2.5 that b is the closest that the atoms can get. If we have v b,thentheatomsareverydenselypacked.Moreover,wecanalsoseefromthe ⇠ graph that dp/dv is very large for this solution which means that the state is very | | dicult to compress: we need to add a great deal of pressure to change the volume only slightly. We have a name for this state: it is a liquid.
You may recall that our original derivation of the van der Waals equation was valid only for densities much lower than the liquid state. This means that we don’t really trust (5.1) on this solution. Nonetheless, it is interesting that the equation predicts the existence of and our plan is to gratefully accept this gift and push ahead to explore what the van der Waals tells us about the liquid-gas transition. We will see that it captures many of the qualitative features of the phase transition.
–136– The last of the three solutions is the one on the right in the figure. This solution has v b and small dp/dv .Itisthegas state.Ourgoalistounderstandwhathappens | | in between the liquid and gas state. We know that the naive, middle, solution given to us by the van der Waals equation is unstable. What replaces it?
5.1.1 Phase Equilibrium Throughout our derivation of the van der Waals equation in Section 2.5,weassumed that the system was at a fixed density. But the presence of two solutions — the liquid and gas state — allows us to consider more general configurations: part of the system could be a liquid and part could be a gas.
How do we figure out if this indeed happens? Just because both liquid and gas states can exist, doesn’t mean that they can cohabit. It might be that one is preferred over the other. We already saw some conditions that must be satisfied in order for two systems to sit in equilibrium back in Section 1.Mechanicalandthermalequilibrium are guaranteed if two systems have the same pressure and temperature respectively. But both of these are already guaranteed by construction for our two liquid and gas solutions: the two solutions sit on the same isotherm and at the same value of p.We’re left with only one further requirement that we must satisfy which arises because the two systems can exchange . This is the requirement of chemical equilibrium,
µliquid = µgas (5.3)
Because of the relationship (4.22)betweenthechemicalpotentialandtheGibbsfree , this is often expressed as
gliquid = ggas (5.4) where g = G/N is the per particle.
Notice that all the equilibrium conditions involve only intensive quantities: p, T and µ. This means that if we have a situation where liquid and gas are in equilibrium, then we can have any number Nliquid of atoms in the liquid state and any number Ngas in the gas state. But how can we make sure that chemical equilibrium (5.3)issatisfied?
Maxwell Construction
We want to solve µliquid = µgas.Wewillthinkofthechemicalpotentialasafunctionof p and T : µ = µ(p, T ). Importantly, we won’t assume that µ(p, T )issinglevaluedsince that would be assuming the result we’re trying to prove! Instead we will show that if we fix T ,thecondition(5.3)canonlybesolvedforaveryparticularvalueofpressure
–137– p.Toseethis,startintheliquidstateatsomefixedvalueofp and T and travel along the isotherm. The infinitesimal change in the is @µ dµ = dp @p T However, we can get an expression for @µ/@ p by recalling that arguments involving extensive and intensive variables tell us that the chemical potential is proportional to the Gibbs free energy: G(p, T, N)=µ(p, T )N (4.22). Looking back at the variation of the Gibbs free energy (4.21)thentellsusthat @G @µ = N = V (5.5) @p @p N,T T Integrating along the isotherm then tells us the chem- p ical potential of any point on the curve, T
Ishouldconfessthatthere’ssomethingslightlydodgyabouttheMaxwellconstruc- tion. We already argued that the part of the isotherm with dp/dv > 0su↵ersan and is unphysical. But we needed to trek along that part of the curve to derive our result. There are more rigorous arguments that give the same answer.
For each isotherm, we can determine the pressure at which the liquid and gas states are in equilibrium. The gives us the co-existence curve,shownbythedottedlinein Figure 37.Insidethisregion,liquidandgascanbothexistatthesametemperature and pressure. But there is nothing that tells us how much gas there should be and how much liquid: atoms can happily move from the liquid state to the gas state. This means that while the density of gas and liquid is fixed, the average density of the system is not. It can vary between the gas density and the liquid density simply by changing the amount of liquid. The upshot of this argument is that inside the co-existence curves, the isotherms simply become flat lines, reflecting the fact that the density can take any value. This is shown in graph on the right of Figure 37.
–138– p p
T=Tc T=Tc
v v
Figure 37: The co-existence curve in red, resulting in constant pressure regions consisting of a harmonious of vapour and liquid.
To illustrate the of this situation, suppose that we sit at some fixed density ⇢ =1/v and cool the system down from a high temperature to T
Meta-Stable States We’ve understood what replaces the unstable region of the van der Waals . But we seem to have p removed more states than anticipated: parts of the Van der Waals isotherm that had dp/dv < 0arecontainedin the co-existence region and replaced by the flat pressure lines. This is the region of the p-V phase diagram that is contained between the two dotted lines in the figure to v the right. The outer dotted line is the co-existence curve. The inner dotted curve is constructed to pass through the Figure 39: stationary points of the van der Waals isotherms. It is called the spinodial curve.
–139– The van der Waals states which lie between the spinodial curve and the co-existence curve are good states. But they are meta-stable. One can show that their Gibbs free energy is higher than that of the liquid-gas equilibrium at the same p and T . However, if we compress the gas very slowly we can coax the system into this state. It is known as a supercooled vapour. It is delicate. Any small disturbance will cause some amount of the gas to condense into the liquid. Similarly, expanding a liquid beyond the co- existence curve results in an meta-stable, superheated liquid.
5.1.2 The Clausius-Clapeyron Equation
We can also choose to plot the liquid-gas phase dia- p critical point gram on the p T plane. Here the co-existence region is squeezed into a line: if we’re sitting in the gas phase and liquid increase the pressure just a little bit at at fixed T
Either side of the line, all particles are either in the gas or liquid phase. We know from (5.4)thattheGibbsfreeenergies(perparticle)ofthesetwostatesareequal,
gliquid = ggas
So G is continuous as we move across the line of phase transitions. Suppose that we sit on the line itself and move along it. How does g change? We can easily compute this from (4.21),
dG = S dT + V dp dg = s dT + v dp liquid liquid liquid ) liquid liquid liquid where, just as g = G/N an v = V/N,theentropydensityiss = S/N.Equatingthis with the free energy in the gaseous phase gives
dg = s dT + v dp = dg = s dT + v dp liquid liquid liquid gas gas gas This can be rearranged to gives us a nice expression for the slope of the line of phase transitions in the p T plane. It is dp s s = gas liquid dT v v gas liquid
–140– We usually define the specific
L = T (s s ) gas liquid This is the energy released per particle as we pass through the phase transition. We see that the slope of the line in the p T plane is determined by the ratio of latent heat released in the phase transition and the discontinuity in volume. The result is known as the Clausius-Clapeyron equation, dp L = (5.6) dT T (v v ) gas liquid There is a classification of phase transitions, due originally to Ehrenfest. When the nth derivative of a thermodynamic potential (either F or G usually) is discontinuous, we say we have an nth order phase transition. In practice, we nearly always deal with first, second and (very rarely) third order transitions. The liquid-gas transition releases latent heat, which means that S = @F/@T is discontinuous. Alternatively, we can say that V = @G/@p is discontinuous. Either way, it is a first order phase transition. The Clausius-Clapeyron equation (5.6)appliestoanyfirstordertransition.
As we approach T Tc, the discontinuity dimin- p ! ishes and S S .AtthecriticalpointT = T we liquid ! gas c have a second order phase transition. Above the critical solid point, there is no sharp distinction between the gas phase liquid and liquid phase.
For most simple materials, the phase diagram above is gas T part of a larger phase diagram which includes at Tc smaller or higher . A generic ver- sion of such a phase diagram is shown to the right. The Figure 41: van der Waals equation is missing the physics of solidifica- tion and includes only the liquid-gas line.
An Approximate Solution to the Clausius-Clapeyron Equation We can solve the Clausius-Clapeyron solution if we make the following assumptions:
The latent heat L is constant. • v v ,sov v v .Forwater,thisisanerroroflessthan0.1% • gas liquid gas liquid ⇡ gas Although we derived the phase transition using the van der Waals equation, now • we’ve got equation (5.6)we’llpretendthegasobeystheidealgaslawpv = kBT .
–141– With these assumptions, it is simple to solve (5.6). It reduces to
dp Lp L/k T B = 2 p = p0e dT kBT ) 5.1.3 The Critical Point Let’s now return to discuss some aspects of life at the critical point. We previously worked out the critical temperature (5.2) by looking for solutions to simultaneous equa- tions @p/@v = @2p/@v2 =0.There’saslightlymoreelegantwaytofindthecritical point which also quickly gives us pc and vc as well. We rearrange the van der Waals equation (5.1)togetacubic,
pv3 (pb + k T )v2 + av ab =0 B
For TTc there is just one. Precisely at T = Tc, the three roots must therefore coincide (before two move o↵onto the complex plane). At the critical point, this curve can be written as
p (v v )3 =0 c c Comparing the coecients tells us the values at the critical point, 8a a k T = ,v=3b, p= (5.7) B c 27b c c 27b2 The Law of Corresponding States We can invert the relations (5.7)toexpresstheparametersa and b in terms of the critical values, which we then substitute back into the van der Waals equation. To this end, we define the reduced variables, T v p T¯ = , v¯ = p¯ = Tc vc pc
The advantage of working with T¯,¯v andp ¯ is that it allows us to write the van der Waals equation (5.1) in a form that is universal to all , usually referred to as the law of corresponding states
8 T¯ 3 p¯ = 3 v¯ 1/3 v¯2
Moreover, because the three variables Tc, pc and vc at the critical point are expressed in terms of just two variables, a and b (5.7), we can construct a combination of them
–142– Figure 42: The co-existence curve for gases. Data is plotted for Ne, Ar,Kr,Xe,N2, O2, CO and CH4. which is independent of a and b and therefore supposedly the same for all gases. This is the universal ratio, p v 3 c c = =0.375 (5.8) kBTc 8
Comparing to real gases, this number is a little high. Values range from around 0.28 to 0.3. We shouldn’t be too discouraged by this; after all, we knew from the beginning that the van der Waals equation is unlikely to be accurate in the liquid regime. Moreover, the fact that gases have a critical point (defined by three variables
Tc, pc and vc)guaranteesthatasimilarrelationshipwouldholdforanyequationof state which includes just two parameters (such as a and b)butwouldmostlikelyfail to hold for equations of state that included more than two parameters.
Dubious as its theoretical foundation is, the law of corresponding states is the first suggestion that something remarkable happens if we describe a gas in terms of its reduced variables. More importantly, there is striking experimental evidence to back this up! Figure 42 shows the Guggenheim plot, constructed in 1945. The co-existence curve for 8 di↵erent gases in plotted in reduced variables: T¯ along the vertical axis; ⇢¯ =1/v¯ along the horizontal. The gases vary in complexity from the simple Ne to the CH4. As you can see, the co-existence curve for all gases is essentially the same, with the chemical make-up largely forgotten. There is clearly something interesting going on. How to understand it?
Critical Exponents We will focus attention on physics close to the critical point. It is not immediately
–143– obvious what are the right questions to ask. It turns out that the questions which have the most interesting answer are concerned with how various quantities change as we approach the critical point. There are lots of ways to ask questions of this type since there are many quantities of interest and, for each of them, we could approach the critical point from di↵erent directions. Here we’ll look at the behaviour of three quantities to get a feel for what happens.
First, we can ask what happens to the di↵erence in (inverse) densities v v as gas liquid we approach the critical point along the co-existence curve. For T
¯ (3¯vliquid 1)(3¯vgas 1)(¯vliquid +¯vgas) T = 2 2 8¯vgasv¯liquid Notice that as we approach the critical point,v ¯ , v¯ 1andtheequationabove gas liquid ! tells us that T¯ 1asexpected.WecanseeexactlyhowweapproachT¯ =1by ! expanding the right right-hand side for small ✏ v¯ v¯ .Todothisquickly,it’s ⌘ gas liquid best to notice that the equation is symmetric inv ¯gas andv ¯liquid,soclosetothecritical point we can writev ¯ =1+✏/2and¯v =1 ✏/2. Substituting this into the gas liquid equation above and keeping just the leading order term, we find 1 T¯ 1 (¯v v¯ )2 ⇡ 16 gas liquid
Or, re-arranging, as we approach Tc along the co-existence curve,
v v (T T )1/2 (5.9) gas liquid ⇠ c This is the answer to our first question.
Our second variant of the question is: how does the volume change with pressure as we move along the critical isotherm. It turns out that we can answer this question without doing any . Notice that at T = Tc,thereisauniquepressureforagiven 2 2 volume p(v, Tc). But we know that @p/@v = @ p/@v =0atthecriticalpoint.Soa Taylor expansion around the critical point must start with the cubic term,
p p (v v )3 (5.10) c ⇠ c This is the answer to our second question.
–144– Our third and final variant of the question concerns the compressibility, defined as 1 @v = (5.11) v @p T We want to understand how changes as we approach T T from above. In fact, we ! c met the compressibility before: it was the feature that first made us nervous about the van der Waals equation since is negative in the unstable region. We already know that at the critical point @p/@v =0.SoexpandingfortemperaturesclosetoTc,we |Tc expect @p = a(T T )+... @v c T ;v=vc This tells us that the compressibility should diverge at the critical point, scaling as 1 (T T ) (5.12) ⇠ c We now have three answers to three questions: (5.9), (5.10)and(5.12). Are they right?! By which I mean: do they agree with experiment? Remember that we’re not sure that we can trust the van der Waals equation at the critical point so we should be nervous. However, there is also reason for some confidence. Notice, in particular, that in order to compute (5.10)and(5.12), we didn’t actually need any details of the van der Waals equation. We simply needed to assume the existence of the critical point and an analytic Taylor expansion of various quantities in the neighbourhood. Given that the answers follow from such general grounds, one may hope that they provide the correct answers for a gas in the neighbourhood of the critical point even though we know that the approximations that went into the van der Waals equation aren’t valid there. Fortunately, that isn’t the case: physics is much more interesting than that!
The experimental results for a gas in the neighbourhood of the critical point do share one feature in common with the discussion above: they are completely independent of the atomic make-up of the gas. However, the scaling that we computed using the van der Waals equation is not fully accurate. The correct results are as follows. As we approach the critical point along the co-existence curve, the densities scale as
v v (T T ) with 0.32 gas liquid ⇠ c ⇡ (Note that the exponent has nothing to do with inverse temperature. We’re just near the end of the course and running out of letters and is the canonical name for this exponent). As we approach along an isotherm,
p p (v v ) with 4.8 c ⇠ c ⇡
–145– Finally, as we approach Tc from above, the compressibility scales as
(T T ) with 1.2 ⇠ c ⇡ The quantities , and are examples of critical exponents.Wewillseemoreofthem shortly. The van der Waals equation provides only a crude first approximation to the critical exponents.
Fluctuations We see that the van der Waals equation didn’t do too badly in capturing the dynamics of an interacting gas. It gets the qualitative behaviour right, but fails on precise quantitative tests. So what went wrong? We mentioned during the derivation of the van der Waals equation that we made certain approximations that are valid only at low density. So perhaps it is not surprising that it fails to get the numbers right near the critical point v =3b.Butthere’sactuallyadeeperreasonthatthevanderWaals equation fails: fluctuations.
This is simplest to see in the grand . Recall that back in Section 1 that we argued that N/N 1/pN,whichallowedustohappilyworkinthe ⇠ even when we actually had fixed . In the context of the liquid-gas transition, fluctuating particle number is the same thing as fluctuating density ⇢ = N/V .Let’srevisitthecalculationofN near the critical point. Using (1.45)and(1.48), the grand canonical partition function can be written as log = V p(T,µ), so the average particle number (1.42)is Z @p N = V h i @µ T,V We already have an expression for the variance in the particle number in (1.43), 1 @ N N 2 = h i @µ T,V Dividing these two expressions, we have N 2 1 @ N @µ 1 @ N = h i = h i N V @µ @p V @p T,V T,V T,V But we can re-write this expression using the general relationship between partial derivatives @x/@y @y/@z @z/@x = 1. We then have |z |x |y N 2 1 @ N 1 @V = h i N @V V @p p,T N,T
–146– This final expression relates the fluctuations in the particle number to the compress- ibility (5.11). But the compressibility is diverging at the critical point and this means that there are large fluctuations in the density of the fluid at this point. The result is that any simple equation of state, like the van der Waals equation, which works only with the average volume, pressure and density will miss this key aspect of the physics.
Understanding how to correctly account for these fluctuations is the subject of .Ithascloselinkswiththerenormalization and conformal field which also arise in particle physics and . You will meet some of these ideas in next year’s course. Here we will turn to a di↵erent phase transition which will allow us to highlight some of the key ideas.
5.2 The The Ising model is one of the touchstones of ; a simple system that exhibits non-trivial and interesting behaviour.
The Ising model consists of N sites in a d-dimensional lattice. On each lattice site lives a quantum that can sit in one of two states: spin up or spin down. We’ll call th the eigenvalue of the spin on the i lattice site si.Ifthespinisup,si =+1;ifthespin is down, s = 1. i The spins sit in a magnetic field that endows an energy advantage to those which point up,
N E = B s B i i=1 X (A comment on notation: B should be properly denoted H.We’restickingwithB to avoid confusion with the Hamiltonian. There is also a factor of the magnetic moment which has been absorbed into the definition of B). The lattice system with energy EB is equivalent to the two-state system that we first met when learning the techniques of statistical back in Section 1.2.3. However, the Ising model contains an addi- tional complication that makes the sysem much more interesting: this is an interaction between neighbouring spins. The full energy of the system is therefore,
E = J s s B s (5.13) i j i ij i Xh i X The notation ij means that we sum over all “nearest neighbour” pairs in the lattice. h i The number of such pairs depends both on the d and the type of lattice.
–147– We’ll denote the number of nearest neighbours as q.Forexample,ind =1alattice has q =2;ind =2,asquarelatticehasq =4.Asquarelatticeind has q =2d.
If J>0, neighbouring spins prefer to be aligned ( or ). In the context of "" ## magnetism, such a system is called a ferromagnet.IfJ<0, the spins want to anti- align ( ). This is an anti-ferromagnet. In the following, we’ll choose J>0although "# for the level of discussion needed for this course, the di↵erences are minor.
We work in the canonical ensemble and introduce the partition function
E[s ] Z = e i (5.14) s X{ i} While the e↵ect of both J>0andB =0istomakeitenergeticallypreferableforthe 6 spins to align, the e↵ect of temperature will be to randomize the spins, with winnning out over energy. Our interest is in the average spin, or average 1 1 @log Z m = s = (5.15) N h ii N @B i X The Ising Model as a Lattice Gas Before we develop techniques to compute the partition function (5.14), it’s worth point- ing out that we can drape slightly di↵erent words around the mathematics of the Ising model. It need not be interpreted as a system of spins; it can also be thought of as a lattice description of a gas.
To see this, consider the same d-dimensional lattice as before, but now with particles hopping between lattice sites. These particles have hard cores, so no more than one can sit on a single lattice site. We introduce the variable n 0, 1 to specify whether i 2{ } agivenlatticesite,labelledbyi, is empty (ni =0)orfilled(ni =1).Wecanalso introduce an attractive force between atoms by o↵ering them an energetic reward if they sit on neighbouring sites. The Hamiltonian of such a lattice gas is given by E = 4J n n µ n i j i ij i Xh i X where µ is the chemical potential which determines the overall particle number. But this Hamiltonian is trivially the same as the Ising model (5.13)ifwemaketheidentification s =2n 1 1, 1 i i 2{ } The chemical potenial µ in the lattice gas plays the role of magnetic field in the spin system while the magnetization of the system (5.15)measurestheaveragedensityof particles away from half-filling.
–148– 5.2.1 Mean Field Theory For general lattices, in arbitrary dimension d,thesum(5.14) cannot be performed. An exact solution exists in d =1and,whenB =0,ind =2.(Thed =2solutionis originally due to Onsager and is famously complicated! Simpler solutions using more modern techniques have since been discovered).
Here we’ll develop an approximate method to evaluate Z known as mean field theory. We write the interactions between neighbouring spins in term of their deviation from the average spin m,
s s =[(s m)+m][(s m)+m] i j i j =(s m)(s m)+m(s m)+m(s m)+m2 i j j i The mean field approximation means that we assume that the fluctuations of spins away from the average are small which allows us to neglect the first term above. Notice that this isn’t the statement that the variance of an individual spin is small; that can never be true because s takes values +1 or 1so s2 =1andthevariance (s m)2 is i h i i h i i always large. Instead, the mean field approximation is a statement about fluctuations between spins on neighbouring sites, so the first term above can be neglected when summing over ij .Wecanthenwritetheenergy(5.13)as h i P E = J [m(s + s ) m2] B s mf i j i ij i Xh i X 1 = JNqm2 (Jqm+ B) s (5.16) 2 i i X where the factor of Nq/2 in the first term is simply the number of nearest neighbour pairs ij .Thefactoror1/2istherebecause ij is a sum over pairs rather than a sum of individualh i sites. (If you’re worried about thish i formula, you should check it for P P a simple square lattice in d =1andd =2dimensions).Asimilarfactorinthesecond term cancelled the factor of 2 due to (si + sj).
We see that the mean field approximation has removed the interactions. The Ising model reduces to the two state system that we saw way back in Section 1. The result of the interactions is that a given spin feels the average e↵ect of its neighbour’s spins through a contribution to the e↵ective magnetic field,
Be↵ = B + Jqm
Once we’ve taken into account this extra contribution to Be↵ ,eachspinactsindepen-
–149– m m tanh tanh m0
−m0
Figure 43: tanh(Jqm) for Jq< 1 Figure 44: tanh(Jqm) for Jq> 1 dently and it is easy to write the partition function. It is
1 JNqm2 B B N Z = e 2 e e↵ + e e↵ 1 JNqm2 N N = e 2 2 cosh Be↵ (5.17)
However, we’re not quite done. Our result for the partition function Z depends on Be↵ which depends on m which we don’t yet know. However, we can use our expression for Z to self-consistently determine the magnetization (5.15). We find,
m =tanh(B + Jqm)(5.18)
We can now solve this equation to find the magnetization for various values of T and B: m = m(T,B). It is simple to see the nature of the solutions using graphical methods.
B=0 Let’s first consider the situation with vanishing magnetic field, B =0.Thefigures above show the graph linear in m compared with the tanh function. Since tanh x ⇡ x 1 x3+...,theslopeofthegraphneartheoriginisgivenbyJq.Thisthendetermines 3 the nature of the solution.
The first graph depicts the situation for Jq < 1. The only solution is m =0. • This means that at high temperatures kBT>Jq,thereisnoaveragemagne- tization of the system. The entropy associated to the random temperature flu- cutations wins over the energetically preferred ordered state in which the spins align.
The second graph depicts the situation for Jq > 1. Now there are three solu- • tions: m = m and m =0.Itwillturnoutthatthemiddlesolution,m =0,is ± 0 unstable. (This solution is entirely analogous to the unstable solution of the van der Waals equation. We will see this below when we compute the free energy).
–150– For the other two possible solutions, m = m ,themagnetizationisnon-zero. ± 0 Here we see the e↵ects of the interactions begin to win over temperature. Notice that in the limit of vanishing temperature, , m 1. This means that all !1 0 ! the spins are pointing in the same direction (either up or down) as expected.
The critical temperature separating these two cases is •
kBTc = Jq (5.19)
The results described above are perhaps rather surprising. Based on the intuition that things in physics always happen smoothly, one might have thought that the magneti- zation would slowly to zero as T . But that doesn’t happen. Instead the !1 magnetization turns o↵abruptly at some finite value of the temperature T = Tc,with no magnetization at all for higher temperatures. This is the characteristic behaviour of a phase transition.
B = 0 6 For B =0,wecansolvetheconsistencyequation(5.18)inasimilarfashion.Thereare 6 acoupleofkeydi↵erencestotheB = 0 case. Firstly, there is now no phase transition 6 at fixed B as we vary temperature T .Instead,forverylargetemperaturesk T Jq, B the magnetization goes smoothly to zero as
B m as T ! kBT !1
At low temperatures, the magnetization again asymptotes to the state m 1which !± minimizes the energy. Except this time, there is no ambiguity as to whether the system chooses m =+1orm = 1. This is entirely determined by the sign of the magnetic field B.
In fact the low temperature behaviour requires slightly more explanation. For small values of B and T ,thereareagainthreesolutionsto(5.18). This follows simply from continuity: there are three solutions for T
–151– m m +1 +1 B>0 T T
−1 −1 B<0
Figure 45: Magnetization with B =0 Figure 46: Magnetization at B = 0. 6 and the phase transtion
The net result of our discussion is depicted in the figures above. When B =0 there is a phase transition at T = Tc.ForT
5.2.2 Critical Exponents It is interesting to compare the phase transition of the Ising model with that of the liquid-gas phase transition. The two are sketched in the Figure 47 above. In both cases, we have a first order phase transition and a quantity jumps discontinuously at T
T = Tc.
We can calculate critical exponents for the Ising model. To compare with our discus- sion for the liquid-gas critical point, we will compute three quantities. First, consider the magnetization at B =0.Wecanaskhowthismagnetizationdecreasesaswetend towards the critical point. Just below T = Tc, m is small and we can Taylor expand (5.18)toget
1 m Jqm (Jqm)3 + ... ⇡ 3
–152– critical point p B liquid
critical point gas T T
Tc Tc
Figure 47: A comparison of the phase diagram for the liquid-gas system and the Ising model.
The magnetization therefore scales as
m (T T )1/2 (5.20) 0 ⇠± c This is to be compared with the analogous result (5.9)fromthevanderWaalsequation. We see that the values of the exponents are the same in both cases. Notice that the derivative dm/dT becomes infinite as we approach the critical point. In fact, we had already anticipated this when we drew the plot of the magnetization in Figure 45.
Secondly, we can sit at T = Tc and ask how the magnetization changes as we approach B =0.Wecanreadthiso↵from(5.18). At T = Tc we have Jq =1andthe consistency condition becomes m =tanh(B/Jq + m). Expanding for small B gives B 1 B 3 B 1 m + m + m + ... + m m3 + (B2) ⇡ Jq 3 Jq ⇡ Jq 3 O ✓ ◆ So we find that the magnetization scales as
m B1/3 (5.21) ⇠ Notice that this power of 1/3isagainfamiliarfromtheliquid-gastransition(5.10) where the van der Waals equation gave v v (p p )1/3. gas liquid ⇠ c Finally, we can look at the ,definedas @m = N @B T This is analogous to the compressibility of the gas. We will ask how changes as we approach T T from above at B =0.Wedi↵erentiate(5.18)withrespecttoB to ! c get N Jq = 1+ cosh2 Jqm N ✓ ◆
–153– We now evaluate this at B = 0. Since we want to approach T T from above, we can ! c also set m =0intheaboveexpression.EvaluatingthisatB =0givesusthescaling
N 1 = (T T ) (5.22) 1 Jq ⇠ c Once again, we see that same that the van der Waals equation gave us for the gas (5.12).
5.2.3 Validity of Mean Field Theory The phase diagram and critical exponents above were all derived using the mean field approximation. But this was an unjustified approximation. Just as for the van der Waals equation, we can ask the all-important question: are our results right? There is actually a version of the Ising model for which the mean field theory is exact: it is the d = dimensional lattice. This is unphysical (even for a string 1 theorist). Roughly speaking, mean field theory works for large d because each spin has alargenumberofneighboursandsoindeedseessomethingclosetotheaveragespin. But what about dimensions of interest? Mean field theory gets things most dramati- cally wrong in d = 1. In that case, no phase transition occurs. We will derive this result below where we briefly describe the exact solution to the d =1Isingmodel.Thereis a general lesson here: in low dimensions, both thermal and quantum fluctuations are more important and invariably stop systems forming ordered phases. In higher dimensions, d 2, the crude features of the phase diagram, including the existence of a phase transition, given by mean field theory are essentially correct. In fact, the very existence of a phase transition is already worthy of comment. The defining feature of a phase transition is behaviour that jumps discontinuously as we vary or B.Mathematically,thefunctionsmustbenon-analytic.Yetallpropertiesof the theory can be extracted from the partition function Z which is a sum of smooth, analytic functions (5.14). How can we get a phase transition? The loophole is that Z is only necessarily analytic if the sum is finite. But there is no such guarantee when the number of lattice sites N .WereachasimilarconclusiontothatofBose- !1 Einstein condensation: phase transitions only strictly happen in the thermodynamic limit. There are no phase transitions in finite systems. What about the critical exponents that we computed in (5.20), (5.21)and(5.22)? It turns out that these are correct for the Ising model defined in d 4. (We will briefly sketch why this is true at the end of this Chapter). But for d =2andd =3,the critical exponents predicted by mean field theory are only first approximations to the true answers.
–154– For d =2,theexactsolution(whichgoesquitesubstantiallypastthiscourse)gives the critical exponents to be,
1 m (T T ) with = 0 ⇠ c 8 m B1/ with =15 ⇠ 7 (T T ) with = ⇠ c 4 The biggest surprise is in d = 3 dimensions. Here the critical exponents are not known exactly. However, there has been a great deal of numerical work to determine them. They are given by
0.32 , 4.8 , 1.2 ⇡ ⇡ ⇡ But these are exactly the same critical exponents that are seen in the liquid-gas phase transition. That’s remarkable! We saw above that the mean field approach to the Ising model gave the same critical exponents as the van der Waals equation. But they are both wrong. And they are both wrong in the same, complicated, way! Why on would a system of spins on a lattice have anything to do with the phase transition between a liquid and gas? It is as if all memory of the microscopic physics — the type of particles, the nature of the interactions — has been lost at the critical point. And that’s exactly what happens.
What we’re seeing here is evidence for universality.Thereisasingletheorywhich describes the physics at the critical point of the liquid gas transition, the 3d Ising model and many other systems. This is a theoretical physicist’s dream! We spend a great deal of time trying to throw away the messy details of a system to focus on the elegant essentials. But, at a critical point, Nature does this for us! Although critical points in two dimensions are well understood, there is much that we don’t know about critical points in three dimensions. This, however, is a story that will have to wait for another day.
5.3 Some Exact Results for the Ising Model
This subsection is something of a diversion from our main interest. In later subsec- tions, we will develop the idea of mean field theory. But first we pause to describe some exact results for the Ising model using techniques that do not rely on the mean field approximation. Many of the results that we derive have broader implications for systems beyond the Ising model.
–155– As we mentioned above, there is an exact solution for the Ising model in d =1 dimension and, when B =0,ind = 2 dimensions. Here we will describe the d =1 solution but not the full d =2solution.Wewill,however,deriveanumberofresults for the d =2Isingmodelwhich,whilefallingshortofthefullsolution,nonetheless provide important insights into the physics.
5.3.1 The Ising Model in d =1Dimensions We start with the Ising chain, the Ising model on a one dimensional line. Here we will see that the mean field approximation fails miserably, giving qualitatively incorrect results: the exact results shows that there are no phase transitions in the Ising chain.
The energy of the system (5.13)canbetriviallyrewrittenas
N B N E = J s s (s + s ) i i+1 2 i i+1 i=1 i=1 X X We will impose periodic boundary conditions, so the spins live on a circular lattice with s s .Thepartitionfunctionisthen N+1 ⌘ 1 N B Z = ... exp Js s + (s + s ) (5.23) i i+1 2 i i+1 s1= 1 sN = 1 i=1 ✓ ◆ X± X± Y The crucial observation that allows us to solve the problem is that this partition function can be written as a product of matrices. We adopt notation from quantum mechanics and define the 2 2matrix, ⇥ B s T s exp Js s + (s + s ) (5.24) h i| | i+1i⌘ i i+1 2 i i+1 ✓ ◆ The row of the matrix is specified by the value of s = 1andthecolumnbys = 1. i ± i+1 ± T is known as the transfer matrix and, in more conventional notation, is given by
J+B J e e T = J J B e e !
The sums over the spins and product over lattice sites in (5.23)simplytellus si i to multiply the matrices defined in (5.24) and the partition function becomes P Q Z =Tr(s T s s T s ... s T s )=TrT N (5.25) h 1| | 2ih 2| | 3i h N | | 1i
–156– where the trace arises because we have imposed periodic boundary conditions. To com- plete the story, we need only compute the eigenvalues of T to determine the partition function. A quick calculation shows that the two eigenvalues of T are
= eJ cosh B e2J cosh2 B 2sinh2J (5.26) ± ± q where, clearly, <+. The partition function is then N N N N N Z = + + = + 1+ N + (5.27) ⇡ ✓ + ◆ where, in the last step, we’ve used the simple fact that if + is the largest eigenvalue N N then /+ 0forverylargeN. ⇡ The partition function Z contains many quantities of interest. In particular, we can use it to compute the magnetisation as a function of temperature when B =0.This, recall, is the quantity which is predicted to undergo a phase transition in the mean field approximation, going abruptly to zero at some critical temperature. In the d =1 Ising model, the magnetisation is given by
1 @ log Z 1 @ m = = + =0 N @B @B B=0 + B=0 We see that the true physics for d =1isverydi ↵erentthanthatsuggestedbythe mean field approximation. When B = 0, there is no magnetisation! While the J term in the energy encourages the spins to align, this is completely overwhelmed by thermal fluctuations for any value of the temperature.
There is a general lesson in this calculation: thermal fluctuations always win in one dimensional systems. They never exhibit ordered phases and, for this reason, never exhibit phase transitions. The mean field approximation is bad in one dimension.
5.3.2 2d Ising Model: Low Temperatures and Peierls Droplets Let’s now turn to the Ising model in d =2dimensions.We’llworkonasquarelattice and set B =0.Ratherthantryingtosolvethemodelexactly,we’llhavemoremodest goals. We will compute the partition function in two di↵erent limits: high temperature and low temperature. We start here with the low temperature expansion.
E The partition function is given by the sum over all states, weighted by e .Atlow temperatures, this is always dominated by the lowest lying states. For the Ising model,
–157– we have
Z = exp J s s 0 i j1 s ij X{ i} Xh i @ A The low temperature limit is J , where the partition function can be approxi- !1 mated by the sum over the first few lowest energy states. All we need to do is list these states.
The ground states are easy. There are two of them: spins all up or spins all down. For example, the ground state with spins all up looks like
Each of these ground states has energy E = E = 2NJ. 0 The first excited states arise by flipping a single spin. Each spin has q =4nearest neighbours – denoted by red lines in the example below – each of which to an energy cost of 2J.TheenergyofeachfirstexcitedstateisthereforeE1 = E0 +8J.
There are, of course, N di↵erent spins that we we can flip and, correspondingly, the first has a degeneracy of N.
To proceed, we introduce a diagrammatic method to list the di↵erent states. We draw only the “broken” bonds which connect two spins with opposite orientation and, as in the diagram above, denote these by red lines. We further draw the flipped spins as red dots, the unflipped spins as blue dots. The energy of the state is determined simply by the number of red lines in the diagram. Pictorially, we write the first excited state as
E1 = E0 +8J Degeneracy = N
–158– The next lowest state has six broken bonds. It takes the form
E2 = E0 +12J Degeneracy = 2N where the extra factor of 2 in the degeneracy comes from the two possible orientations (vertical and horizontal) of the graph.
Things are more interesting for the states which sit at the third excited level. These have 8 broken bonds. The simplest configuration consists of two, disconnected, flipped spins
E = E +16J 3 0 (5.28) Degeneracy = 1 N(N 5) 2
The factor of N in the degeneracy comes from placing the first graph; the factor of N 5arisesbecausetheflippedspininthesecondgraphcansitanywhereapartfrom on the five vertices used in the first graph. Finally, the factor of 1/2arisesfromthe interchange of the two graphs.
There are also three further graphs with the same energy E3.Theseare
E3 = E0 +16J Degeneracy = N
and
E3 = E0 +16J Degeneracy = 2N
–159– where the degeneracy comes from the two orientations (vertical and horizontal). And, finally,
E3 = E0 +16J Degeneracy = 4N
where the degeneracy comes from the four orientations (rotating the graph by 90).
Adding all the graphs above together gives us an expansion of the partition function J in power of e 1. This is ⌧
2NJ 8J 12J 1 2 16J Z =2e 1+Ne +2Ne + (N +9N)e + ... (5.29) 2 ✓ ◆ where the overall factor of 2 originates from the two ground states of the system. We’ll make use of the specific coecients in this expansion in Section 5.3.4.Before we focus on the physics hiding in the low temperature expansion, it’s worth making a quick comment that something quite nice happens if we take the log of the partition function,
8J 12J 9 16J log Z =log2+2NJ + Ne +2Ne + Ne + ... 2 The thing to notice is that the N 2 term in the partition function (5.29)hascancelled out and log Z is proportional to N,whichistobeexpectedsincethefreeenergyof the system is extensive. Looking back, we see that the N 2 term was associated to the disconnected diagrams in (5.28). There is actually a general lesson hiding here: the partition function can be written as the exponential of the sum of connected diagrams. We saw exactly the same issue arise in the expansion in (2.37).
Peierls Droplets Continuing the low temperature expansion provides a heuristic, but physically intuitive, explanation for why phase transitions happen in d 2dimensionsbutnotind =1. As we flip more and more spins, the low energy states become droplets,consistingofa region of space in which all the spins are flipped, surrounded by a larger in which the spins have their original alignment. The energy cost of such a droplet is roughly
E 2JL ⇠
–160– where L is the perimeter of the droplet. Notice that the energy does not scale as the area of the droplet since all spins inside are aligned with their neighbours. It is only those on the edge which are misaligned and this is the reason for the perimeter scaling. To understand how these droplets contribute to the partition function, we also need to know their degeneracy. We will now argue that the degeneracy of droplets scales as
Degeneracy e↵L ⇠ for some value of ↵.Toseethis,considerfirstlytheproblemofarandomwalkona2d square lattice. At each step, we can move in one of four directions. So the number of paths of length L is
#paths 4L = eL log 4 ⇠ Of course, the perimeter of a droplet is more constrained that a random walk. Firstly, the perimeter can’t go back on itself, so it really only has three directions that it can move in at each step. Secondly, the perimeter must return to its starting point after L steps. And, finally, the perimeter cannot self-intersect. One can show that the number of paths that obey these conditions is
#paths e↵L ⇠ where log 2 <↵
The fact that both energy and entropy scale with L means that there is an interesting competition between them. At temperatures where the droplets are important, the partition function is schematically of the form
↵L 2JL Z e e ⇠ L X For large (i.e. low temperature) the partition function converges. However, as the temperature increases, one reaches the critical temperature 2J k T (5.30) B c ⇡ ↵ where the partition function no longer converges. At this point, the entropy wins over the energy cost and it is favourable to populate the system with droplets of arbitrary sizes. This is the how one sees the phase transition in the partition function. For temperature above Tc,thelow-temperatureexpansionbreaksdownandtheordered magnetic phase is destroyed.
–161– We can also use the droplet argument to see why phase transitions don’t occur in d =1dimension.Onaline,theboundaryofanydropletalwaysconsistsofjust two points. This means that the energy cost to forming a droplet is always E =2J, regardless of the size of the droplet. But, since the droplet can exist anywhere along the line, its degeneracy is N.Thenetresultisthatthefreeenergyassociatedtocreating adropletscalesas
F 2J k T log N ⇠ B and, as N ,thefreeenergyisnegativeforanyT>0. This means that the system !1 will prefer to create droplets of arbitrary length, randomizing the spins. This is the intuitive reason why there is no magnetic ordered phase in the d =1Isingmodel.
5.3.3 2d Ising Model: High Temperatures We now turn to the 2d Ising model in the opposite limit of high temperature. Here we expect the partition function to be dominated by the completely random, disordered configurations of maximum entropy. Our goal is to find a way to expand the partition function in J 1. ⌧ We again work with zero magnetic field, B =0andwritethepartitionfunctionas
Z = exp J s s = eJsisj 0 i j1 s ij s ij X{ i} Xh i X{ i} Yh i @ A Jsisj There is a useful way to rewrite e which relies on the fact that the product sisj only takes 1. It doesn’t take long to check the following identity: ±
Jsisj e =coshJ + sisj sinh J
=coshJ (1 + sisj tanh J)
Using this, the partition function becomes
Z = cosh J (1 + sisj tanh J) s ij X{ i} Yh i qN/2 =(coshJ) (1 + sisj tanh J)(5.31) s ij X{ i} Yh i where the number of nearest neighbours is q =4forthe2dsquarelattice.
–162– With the partition function in this form, there is a natural expansion which suggests itself. At high temperatures J 1 which, of course, means that tanh J 1. ⌧ ⌧ But the partition function is now naturally a product of powers of tanh J. This is somewhat analogous to the cluster expansion for the interacting gas that we met in Section 2.5.3. As in the cluster expansion, we will represent the expansion graphically.
We need no graphics for the leading order term. It has no factors of tanh J and is simply
Z (cosh J)2N 1=2N (cosh J)2N ⇡ s X{ i} That’s simple.
Let’s now turn to the leading correction. Expanding the partition function (5.31), each power of tanh J is associated to a nearest neighbour pair ij .We’llrepresent h i this by drawing a line on the lattice:
i j = sisj tanh J
But there’s a problem: each factor of tanh J in (5.31)alsocomeswithasumoverall spins s and s . And these are +1 and 1 which means that they simply sum to zero, i j s s =+1 1 1+1=0 i j s ,s Xi j How can we avoid this? The only way is to make sure that we’re summing over an even 2 number of spins on each site, since then we get factors of si =1andnocancellations. Graphically, this means that every site must have an even number of lines attached to it. The first correction is then of the form
1 2 4 4 4 =(tanhJ) s1s2 s2s3 s3s4 s4s1 =2(tanh J) 3 4 s X{ i} There are N such terms since the upper left corner of the square can be on any one of the N lattice sites. (Assuming periodic boundary conditions for the lattice). So including the leading term and first correction, we have
Z =2N (cosh J)2N 1+N(tanh J)4 + ...
We can go further. The next terms arise from graphs of length 6 and the only possibil- ities are rectangles, oriented as either landscape or portrait. Each of them can sit on
–163– one of N sites, giving a contribution
+ =2N(tanh J)4
Things get more interesting when we look at graphs of length 8. We have four di↵erent types of graphs. Firstly, there are the trivial, disconnected pair of squares 1 = N(N 5)(tanh J)8 2 Here the first factor of N is the possible positions of the first square; the factor of N 5 arises because the possible location of the upper corner of the second square can’t be on any of the vertices of the first, but nor can it be on the square one to the left of the upper corner of the first since that would give a graph that looks like which has three lines coming o↵the middle site and therefore vanishes when we sum over spins. Finally, the factor of 1/2comesbecausethetwosquaresareidentical.
The other graphs of length 8 are a large square, a rectangle and a corner. The large square gives a contribution
= N(tanh J)8
There are two orientations for the rectangle. Including these gives a factor of 2,
=2N(tanh J)8
Finally, the corner graph has four orientations, giving
=4N(tanh J)8
Adding all contributions together gives us the first few terms in high temperature expansion of the partition function
Z =2N (cosh J)2N 1+N(tanh J)4 +2N(tanh J)6 ⇣ 1 + (N 2 +9N)(tanh J)8 + ... (5.32) 2 ⌘
–164– There’s some magic hiding in this expansion which we’ll turn to in Section 5.3.4. First, let’s just see how the high energy expansion plays out in the d =1dimensionalIsing model.
The Ising Chain Revisited Let’s do the high temperature expansion for the d =1Ising chain with periodic boundary conditions and B =0.Wehavethe same partition function (5.31)andthesameissuethatonlygraphs with an even number of lines attached to each vertex contribute. But, for the Ising chain, there is only one such term: it is the closed loop. This means that the partition function is
Z =2N (cosh J)N 1+(tanhJ)N Figure 48: In the limit N ,(tanhJ)N 0athightemperaturesandeventhecontribution !1 ! from the closed loop vanishes. We’re left with
Z =(2coshJ)N
This agrees with our exact result for the Ising chain given in (5.27), which can be seen by setting B =0in(5.26)sothat+ =2coshJ.
5.3.4 Kramers-Wannier Duality In the previous sections we computed the partition function perturbatively in two extreme regimes of low temperature and high temperature. The physics in the two cases is, of course, very di↵erent. At low temperatures, the partition function is dominated by the lowest energy states; at high temperatures it is dominated by maximally disordered states. Yet comparing the partition functions at low temperature (5.29)andhigh temperature (5.32)revealsanextraordinaryfact:theexpansionsarethesame!More concretely, the two series agree if we exchange
2J e tanh J (5.33) ! Of course, we’ve only checked the agreement to the first few orders in perturbation theory. Below we shall prove that this miracle continues to all orders in perturbation theory. The symmetry of the partition function under the interchange (5.33) is known as Kramers-Wannier duality. Before we prove this duality, we will first just assume that it is true and extract some consequences.
–165– We can express the statement of the duality more clearly. The Ising model at tem- perature is related to the same model at temperature ˜,definedas
2˜J e =tanhJ (5.34) This way of writing things hides the symmetry of the transformation. A little algebra shows that this is equivalent to 1 sinh 2˜J = sinh 2J Notice that this is a hot/cold duality. When J is large, ˜J is small. Kramers-Wannier duality is the statement that, when B =0,thepartitionfunctionsoftheIsingmodel at two temperatures are related by 2N (cosh J)2N Z[]= Z[˜] 2e2N˜J N 1 N =2 (cosh J sinh J) Z[˜](5.35) This means that if you know the of the Ising model at one temperature, then you also know the thermodynamics at the other temperature. Notice however, that it does not say that all the physics of the two models is equivalent. In particular, when one system is in the ordered phase, the other typically lies in the disordered phase.
One immediate consequence of the duality is that we can use it to compute the exact critical temperature Tc.Thisisthetemperatureatwhichthepartitionfunction in singular in the N limit. (We’ll discuss a more refined criterion in Section !1 5.4.3). If we further assume that there is just a single phase transition as we vary the temperature, then it must happen at the special self-dual point = ˜.Thisis 2J kBT = 2.269 J log(p2+1) ⇡ The exact solution of Onsager confirms that this is indeed the transition temperature. It’s also worth noting that it’s fully consistent with the more heuristic Peierls droplet argument (5.30)sincelog2< log(p2+1)< log 3.
Proving the Duality So far our evidence for the duality (5.35) lies in the agreement of the first few terms in the low and high temperature expansions (5.29)and(5.32). Of course, we could keep computing further and further terms and checking that they agree, but it would be nicer to simply prove the equality between the partition functions. We shall do so here.
–166– The key idea that we need can actually be found by staring hard at the various graphs that arise in the two expansions. Eventually, you will realise that they are the same, albeit drawn di↵erently. For example, consider the two “corner” diagrams
vs
The two graphs are dual.Theredlinesinthefirstgraphintersecttheblacklinesin the second as can be seen by placing them on top of each other:
The same pattern occurs more generally: the graphs appearing in the low temperature expansion are in one-to-one correspondence with the dual graphs of the high tempera- ture expansion. Here we will show how this occurs and how one can map the partition functions onto each other.
Let’s start by writing the partition function in the form (5.31)thatwemetinthe high temperature expansion and presenting it in a slightly di↵erent way,
Z[]= (cosh J + sisj sinh J) s ij X{ i} Yh i kij = Ckij [J](sisj)
s ij kij =0,1 X{ i} Yh i X where we have introduced the rather strange variable kij associated to each nearest neighbour pair that takes values 0 and 1, together with the functions.
C0[J]=coshJ and C1[J]=sinhJ
The variables in the original Ising model were spins on the lattice sites. The observation that the graphs which appear in the two expansions are dual suggests that it might be
–167– profitable to focus attention on the links between lattice sites. Clearly, we have one link for every nearest neighbour pair. If we label these links by l,wecantriviallyrewrite the partition function as
kl Z = Ckl [J](sisj)
kl=0,1 l s X Y X{ i} Notice that the strange label kij has now become a variable that lives on the links l rather than the original lattice sites i.
At this stage, we do the sum over the spins si. We’ve already seen that if a given spin, say si, appears in a term an odd number of times, then that term will vanish when we sum over the spin. Alternatively, if the spin si appears an even number of times, then the sum will give 2. We’ll say that a given link l is turned on in configurations with kl =1andturnedo↵whenkl =0.Inthislanguage,aterminthesumoverspin si contributes only if an even number of links attached to site i are turned on. The partition function then becomes
N Z =2 Ckl [J] (5.36) kl l Constrained X Y Now we have something interesting. Rather than summing over spins on lattice sites, we’re now summing over the new variables kl living on links. This looks like the partition function of a totally di↵erent physical system, where the degrees of freedom live on the links of the original lattice. But there’s a catch – that big “Constrained” label on the sum. This is there to remind us that we don’t sum over all kl configurations; only those for which an even number of links are turned on for every lattice site. And that’s annoying. It’s telling us that the kl aren’t really independent variables. There are some constraints that must be imposed.
Fortunately, for the 2d square lattice, there is a simple ~ s5 ~ s1 s way to solve the constraint. We introduce yet more variables,s ˜i 4 which, like the original spin variables, take values 1. However, ± s s s thes ˜i do not live on the original lattice sites. Instead, they live 2 k12 1 4 on the vertices of the dual lattice. For the 2d square lattice, the ~ ~ s2 s3 dual vertices are drawn in the figure. The original lattice sites s are in white; the dual lattice sites in black. 3 Figure 49: The link variables kl are related to the two nearest spin vari- abless ˜i as follows: 1 k = (1 s˜ s˜ ) 12 2 1 2
–168– 1 k = (1 s˜ s˜ ) 13 2 2 3 1 k = (1 s˜ s˜ ) 14 2 3 4 1 k = (1 s˜ s˜ ) 15 2 1 4
Notice that we’ve replaced four variables kl taking values 0, 1withfourvariables˜si taking values 1. Each set of variables gives 24 possibilities. However, the map is not ± one-to-one. It is not possible to construct for all values of kl using the parameterization in terms ofs ˜i.Toseethis,weneedonlylookat 1 k + k + k + k =2 (˜s s˜ +˜s s˜ +˜s s˜ +˜s s˜ ) 12 13 14 15 2 1 2 2 3 3 4 1 4 1 =2 (˜s +˜s )(˜s +˜s ) 2 1 3 2 4 =0, 2, or 4
In other words, the number of links that are turned on must be even. But that’s exactly what we want! Writing the kl in terms of the auxiliary spinss ˜i automatically solves the constraint that is imposed on the sum in (5.36). Moreover, it is simple to check that for every configuration k obeying the constraint, there are two configurations of s˜ . { l} { i} This means that we can replace the constrained sum over k with an unconstrained { l} sum over s˜ .Theonlypricewepayisanadditionalfactorof1/2. { i}
1 N Z[]= 2 C 1 [j] (1 s˜ s˜ ) 2 2 i j s˜ ij X{ i} Yh i
Finally, we’d like to find a simple expression for C0 and C1 in terms ofs ˜i.That’seasy enough. We can write
Ck[J]=coshJ exp (k log tanh J) 1 =(sinhJ cosh J)1/2 exp s˜ s˜ log tanh J 2 i j ✓ ◆ Substituting this into our newly re-written partition function gives
N 1 1/2 1 Z[]=2 (sinh J cosh J) exp s˜ s˜ log tanh J 2 i j s˜i ij ✓ ◆ X{ } Yh i
N 1 N 1 =2 (sinh J cosh J) exp log tanh J s˜ s˜ 02 i j1 s˜ ij X{ i} Xh i @ A
–169– But this final form of the partition function in terms of the dual spinss ˜i has exactly the same functional form as the original partition function in terms of the spins si.More precisely, we can write
N 1 N Z[]=2 (sinh 2J) Z[˜] where
2˜J e =tanhJ as advertised previously in (5.34). This completes the proof of Kramers-Wannier duality in the 2d Ising model on a square lattice.
The concept of duality of this kind is a major feature in much of modern theoretical physics. The key idea is that when the temperature gets large there may be a di↵erent set of variables in which a theory can be written where it appears to live at low tem- perature. The same idea often holds in quantum , where duality maps strong coupling problems to weak coupling problems.
The duality in the Ising model is special for two reasons: firstly, the new variables s˜i are governed by the same Hamiltonian as the original variables si.Wesaythatthe Ising model is self-dual. In general, this need not be the case — the high temperature limit of one system could look like the low-temperature limit of a very di↵erent system. Secondly, the duality in the Ising model can be proven explicitly. For most systems, we have no such luck. Nonetheless, the idea that there may be dual variables in other, more dicult theories, is compelling. Commonly studied examples include the exchange particles and vortices in two dimensions, and electrons and magnetic monopoles in three dimensions.
5.4 Landau Theory
We saw in Sections 5.1 and 5.2 that the van der Waals equation and mean field Ising model gave the same (sometimes wrong!) answers for the critical exponents. This suggests that there should be a unified way to look at phase transitions. Such a method was developed by Landau. It is worth stressing that, as we saw above, the Landau approach to phase transitions often only gives qualitatively correct results. However, its advantage is that it is extremely straightforward and easy. (Certainly much easier than the more elaborate methods needed to compute critical exponents more accurately).
–170– The Landau theory of phase transitions is based around the free energy. We will illustrate the theory using the Ising model and then explain how to extend it to di↵erent systems. The free energy of the Ising model in the mean field approximation is readily attainable from the partition function (5.17), 1 1 N F = log Z = JNqm2 log (2 cosh B )(5.37) 2 e↵ So far in this course, we’ve considered only systems in equilibrium. The free energy, like all other thermodynamic potentials, has only been defined on equilibrium states. Yet the equation above can be thought of as an expression for F as a function of m. Of course, we could substitute in the equilibrium value of m given by solving (5.18), but it seems a shame to throw out F (m)whenitissuchanicefunction.Surelywecan put it to some use!
The key step in Landau theory is to treat the function F = F (T,V ; m)seriously. This means that we are extending our viewpoint away from equilibrium states to a whole class of states which have a constant average value of m.Ifyouwantsomewords to drape around this, you could imagine some external magical power that holds m fixed. The free energy F (T,V ; m) is then telling us the equilibrium properties in the presence of this magical power. Perhaps more convincing is what we do with the free energy in the absence of any magical constraint. We saw in Section 4 that equilibrium is guaranteed if we sit at the minimum of F .LookingatextremaofF ,wehavethe condition @F =0 m =tanhB @m ) e↵ But that’s precisely the condition (5.18)thatwesawpreviously.Isn’tthatnice!
In the context of Landau theory, m is called an order parameter. When it takes non- zero values, the system has some degree of order (the spins have a preferred direction in which they point) while when m =0thespinsarerandomisedandhappilypointin any direction.
For any system of interest, Landau theory starts by identifying a suitable order parameter. This should be taken to be a quantity which vanishes above the critical temperature at which the phase transition occurs, but is non-zero below the critical temperature. Sometimes it is obvious what to take as the order parameter; other times less so. For the liquid-gas transition, the relevant order parameter is the di↵erence in densities between the two phases, v v .Formagneticorelectricsystems,the gas liquid order parameter is typically some form of magnetization (as for the Ising model) or
–171– the polarization. For the Bose-Einstein condensate, superfluids and superconductors, the order parameter is more subtle and is related to o↵-diagonal long-range order in the one-particle density matrix11,althoughthisisusuallyratherlazilysimplifiedtosay that the order parameter can be thought of as the macroscopic wavefunction 2. | | Starting from the existence of a suitable order parameter, the next step in the Landau programme is to write down the free energy. But that looks tricky. The free energy for the Ising model (5.37) is a rather complicated function and clearly contains some detailed information about the physics of the spins. How do we just write down the free energy in the general case? The trick is to assume that we can expand the free energy in an analytic power series in the order parameter. For this to be true, the order parameter must be small which is guaranteed if we are close to a critical point (since m =0forT>Tc). The nature of the phase transition is determined by the kind of terms that appear in the expansion of the free energy. Let’s look at a couple of simple examples.
5.4.1 Second Order Phase Transitions We’ll consider a general system (Ising model; liquid-gas; BEC; whatever) and denote the order parameter as m. Suppose that the expansion of the free energy takes the general form
2 4 F (T ; m)=F0(T )+a(T )m + b(T )m + ... (5.38)
One common reason why the free energy has this form is because the theory has a symmetry under m m,forbiddingtermswithoddpowersofm in the expansion. ! For example, this is the situation in the Ising model when B =0.Indeed,ifwe expand out the free energy (5.37)fortheIsingmodelforsmallm using cosh x ⇡ 1+ 1 x2 + 1 x4 + ... and log(1 + y) y 1 y2 + ... we get the general form above with 2 4! ⇡ 2 explicit expressions for F0(T ), a(T )andb(T ), NJq N3J 4q4 F (T ; m)= Nk T log 2 + (1 Jq) m2 + m4 + ... Ising B 2 24 ✓ ◆ ✓ ◆
The leading term F0(T )isunimportantforourstory.Weareinterestedinhowthefree energy changes with m. The condition for equilibrium is given by @F =0 (5.39) @m But the solutions to this equation depend on the sign of the coecients a(T )and
11See, for example, the book “Quantum Liquids” by Anthony Leggett
–172– F F
m m
Figure 50: Free energy when a(T ) > 0 Figure 51: Free energy when a(T ) < 0 b(T ). Moreover, this sign can change with temperature. This is the essence of the phase transitions. In the following discussion, we will assume that b(T ) > 0forallT . (If we relax this condition, we have to also consider the m6 term in the free energy which leads to interesting results concerning so-called tri-critical points).
The two figures above show sketches of the free energy in the case where a(T ) > 0 and a(T ) < 0. Comparing to the explicit free energy of the Ising model, a(T ) < 0 when T>Tc = Jq/kB and a(T ) < 0whenT 0, we have just asingleequilibriumsolutionto(5.39)atm = 0. This is typically the situation at high temperatures. In contrast, at a(T ) < 0, there are three solutions to (5.39). The solution m =0clearlyhashigherfreeenergy:thisisnowtheunstablesolution.The two stable solutions sit at m = m .Forexample,ifwechoosetotruncatethefree ± 0 energy (5.38)atquarticorder,wehave
a m0 = T 0tom =0ata(T ) < 0. This describes a second order phase 6 transition occurring at Tc,definedbya(Tc)=0.
Once we know the equilibrium value of m, we can then substitute this back into the free energy F (T ; m)in(5.38). This gives the thermodynamic free energy F (T )of the system in equilibrium that we have been studying throughout this course. For the quartic free energy, we have
F (T ) T>T F (T )= 0 c (5.40) 2 ( F0(T ) a /4bT
Because a(Tc) = 0, the equilibrium free energy F (T )iscontinuousatT = Tc.Moreover, the entropy S = @F/@T is also continuous at T = T . However, if you di↵erentiate the c
–173– 2 equilibrium free energy twice, you will get a term a0 /b which is generically not vanishing at T = Tc. This means that the C = T@S/@T changes discontinuously at T = Tc,asbefitsasecondorderphasetransition.Awordofwarning:ifyouwantto compute equilibrium quantities such as the heat capacity, it’s important that you first substitution in the equilibrium value of m and work with (5.40)ratherthani(5.38). If you don’t, you miss the fact that the magnetization also changes with T .
We can easily compute critical exponents within the context of Landau theory. We need only make further assumptions about the behaviour of a(T )andb(T )inthe vicinity of Tc.IfweassumethatnearT = Tc,wecanwrite b(T ) b ,a(T ) a (T T )(5.41) ⇡ 0 ⇡ 0 c then we have a m 0 (T T )1/2 T
Notice that we didn’t put any discontinuities into the free energy. Everything in F (T ; m) was nice and smooth. When Taylor expanded, it has only integer powers of m and T as shown in (5.38)and(5.41). But the minima of F behave in a non-analytic fashion as seen in the expression for m0 above. Landau’s theory of phase transitions predicts this same critical exponent for all values of the dimension d of the system. But we’ve already mentioned in previous contexts that the critical exponent is in fact only correct for d 4. We will understand how to derive this criterion from Landau theory in the next section.
Spontaneous Symmetry Breaking As we approach the end of the course, we’re touching upon a number of ideas that become increasingly important in subsequent developments in physics. We already briefly met the idea of universality and critical phenomena. Here I would like to point out another very important idea: spontaneous symmetry breaking.
The free energy (5.38)isinvariantundertheZ symmetry m m.Indeed,we 2 ! said that one common reason that we can expand the free energy only in even powers of m is that the underlying theory also enjoys this symmetry. But below Tc, the system must pick one of the two ground states m =+m or m = m . Whichever choice it 0 0 makes breaks the Z2 symmetry. We say that the symmetry is spontaneously broken by the choice of ground state of the theory.
–174– Spontaneous symmetry breaking has particularly dramatic consequences when the symmetry in question is continuous rather than discrete. For example, consider a situation where the order parameter is a complex number and the free energy is given by (5.38)withm = 2.(Thisise↵ectivelywhathappensforBECs,superfluidsand | | superconductors). Then we should only look at the m>0solutionssothattheground state has 2 =+m .Butthisleavesthephaseof completely undetermined. So | | 0 there is now a continuous choice of ground states: we get to sit anywhere on the circle parameterised by the phase of . Any choice that the system makes spontaneously breaks the U(1) rotational symmetry which acts on the phase of .Somebeautiful results due to Nambu and Goldstone show that the much of the physics of these systems can be understood simply as a consequence of this symmetry breaking. The ideas of spontaneous symmetry breaking are crucial in both condensed physics and particle physics. In the latter context, it is intimately tied with the Higgs mechanism.
5.4.2 First Order Phase Transitions Let us now consider a situation where the expansion of the free energy also includes odd powers of the order parameter
2 3 4 F (T ; m)=F0(T )+↵(T )m + a(T )m + (T )m + b(T )m + ...
For example, this is the kind of expansion that we get for the Ising model free energy (5.37)whenB =0,whichreads 6
JNq 2 N 2 N 4 FIsing(T ; m)= NkBT log 2 + m (B + Jqm) + 3 (B + Jqm) + ... 2 2kBT 24(kBT ) Notice that there is no longer a symmetry relating m m:theB field has a ! preference for one sign over the other.
If we again assume that b(T ) > 0foralltemperatures,thecrudeshapeofthefree energy graph again has two choices: there is a single minimum, or two minima and a local maximum.
Let’s start at suitably low temperatures for which the situation is depicted in Figure 52. The free energy once again has a double well, except now slightly skewed. The local maximum is still an unstable point. But this time around, the minima with the lower free energy is preferred over the other one. This is the true ground state of the system. In contrast, the point which is locally, but not globally, a minimum corresponds to a meta-stable state of the system. In order for the system to leave this state, it must first fluctuate up and over the energy barrier separating the two.
–175– F F F B<0 B=0 B>0
m
m m
Figure 52: The free energy of the Ising model for B<0, B = 0 and B>0.
In this set-up, we can initiate a first order phase transition. This occurs when the coecient of the odd terms, ↵(T )and(T )changesignandthetruegroundstate changes discontinuously from m<0tom>0. In some systems this behaviour occurs when changing temperature; in others it could occur by changing some external parameter. For example, in the Ising model the first order phase transition is induced by changing B.
At very high temperature, the double well poten- F tial is lost in favour of a single minimum as depicted in the figure to the right. There is a unique ground state, al- beit shifted from m =0bythepresenceofthe↵(T )term m above (which translates into the magnetic field B in the Ising model). The temperature at which the meta-stable ground state of the system is lost corresponds to the spin- odial point in our discussion of the liquid-gas transition. Figure 53:
One can play further games in Landau theory, looking at how the shape of the free energy can change as we vary temperature or other parameters. One can also use this framework to give a simple explanation of the concept of . You can learn more about these from the links on the course webpage.
5.4.3 Lee-Yang Zeros You may have noticed that the flavour of our discussion of phase transitions is a little di↵erent from the rest of the course. Until now, our philosophy was to derive everything from the partition function. But in this section, we dumped the partition function as soon as we could, preferring instead to work with the macroscopic variables such as the free energy. Why didn’t we just stick with the partition function and examine phase transitions directly?
–176– The reason, of course, is that the approach using the partition function is hard! In this short section, which is somewhat tangential to our main discussion, we will describe how phase transitions manifest themselves in the partition function.
For concreteness, let’s go back to the classical interacting gas of Section 2.5, although the results we derive will be more general. We’ll work in the grand canonical ensemble, with the partition function N N z 3 U(r ) (z,V,T)= z Z(N,V,T)= d r e j
U(rjk)=0 for rjk N , and the grand partition function is therefore V Z afinitepolynomialinthefugacityz,oforderNV .Butifthepartitionfunctionisa finite polynomial, there can’t be any discontinuous behaviour associated with a phase transition. In particular, we can calculate pV = k T log (5.43) B Z which gives us pV as a smooth function of z.Wecanalsocalculate @ N = z log (5.44) @z Z which gives us N as a function of z.Eliminatingz between these two functions (as we did for both and fermions in Section 3) tells us that pressure p is a smooth function of density N/V .We’renevergoingtogetthebehaviourthatwederivedfrom the Maxwell construction in which the plot of pressure vs density shown in Figure 37 exhibits a discontinous derivative.
The discussion above is just re-iterating a statement that we’ve alluded to several times already: there are no phase transitions in a finite system. To see the discontinuous behaviour, we need to take the limit V .AtheoremduetoLeeandYang12 gives !1 us a handle on the analytic properties of the partition function in this limit. 12This theorem was first proven for the Ising model in 1952. Soon afterwards, the same Lee and Yang proposed a model of parity violation in the weak interaction for which they won the 1957 .
–177– The surprising insight of Lee and Yang is that if you’re interested in phase transitions, you should look at the zeros of in the complex z-plane. Let’s firstly look at these Z when V is finite. Importantly, at finite V there can be no zeros on the positive real axis, z>0. This follows follows from the defintion of given in (5.42)whereitisasum Z of positive quantities. Moreover, from (5.44), we can see that is a monotonically Z increasing function of z because we necessarily have N>0. Nonetheless, is a Z polynomial in z of order NV so it certainly has NV zeros somewhere in the complex z-plane. Since ?(z)= (z?), these zeros must either sit on the real negative axis or Z Z come in complex pairs.
However, the statements above rely on the fact that is a finite polynomial. As we Z take the limit V ,themaximumnumberofparticlesthatwecanfitinthesystem !1 diverges, N ,and is now defined as an infinite series. But infinite series can do V !1 Z things that finite ones can’t. The Lee-Yang theorem says that as long as the zeros of Z continue to stay away from the positive real axis as V ,thennophasetransitions !1 can happen. But if one or more zeros happen to touch the positive real axis, life gets more interesting.
More concretely, the Lee-Yang theorem states:
Lee-Yang Theorem: The quantity • 1 ⇥= lim log (z,V,T) V V Z !1 ✓ ◆ exists for all z>0. The result is a continuous, non-decreasing function of z which is independent of the shape of the box (up to some sensible assumptions such as 1/3 Surface Area/V V which ensures that the box isn’t some stupid fractal ⇠ shape). Moreover, let R be a fixed, volume independent, region in the complex z plane which contains part of the real, positive axis. If R contains no zero of (z,V,T) Z for all z R then ⇥is a an analytic function of z for all z R.Inparticular,all 2 2 derivatives of ⇥are continuous.
In other words, there can be no phase transitions in the region R even in the V !1 limit. The last result means that, as long as we are safely in a region R,taking derivatives of with respect to z commutes with the limit V .Inotherwords,we !1 are allowed to use (5.44)towritetheparticledensityn = N/V as
@ p @⇥ lim n =limz = z V V @z k T @z !1 !1 ✓ B ◆
–178– However, if we look at points z where zeros appear on the positive real axis, then ⇥will generally not be analytic. If d⇥/dz is discontinuous, then the system is said to undergo afirstorderphasetransition.Moregenerally,ifdm⇥/dzm is discontinuous for m = n, but continuous for all m
A Made-Up Example Ideally, we would like to start with a Hamiltonian which exhibits a first order phase transition, compute the associated grand partition function and then follow its zeros Z as V . However, as we mentioned above, that’s hard! Instead we will simply !1 make up a partition function which has the appropriate properties. Our choice is Z somewhat artificial,
(z,V )=(1+z)[↵V ](1 + z[↵V ]) Z Here ↵ is a constant which will typically depend on temperature, although we’ll suppress this dependence in what follows. Also,
[x]=Integerpartofx
Although we just made up the form of ,itdoeshavethebehaviourthatonewould Z expect of a partition function. In particular, for finite V ,thezerossitat
z = 1andz = e⇡i(2n+1)/[↵V ] n =0, 1,...,[↵V ] 1 As promised, none of the zeros sit on the positive real axis.However, as we increase V , the zeros become denser and denser on the unit circle. From the Lee-Yang theorem, we expect that no phase transition will occur for z =1butthatsomethinginteresting 6 could happen at z =1.
Let’s look at what happens as we send V .Wehave !1 1 ⇥= lim log (z,V ) V !1 V Z 1 =lim [↵V ]log(1+z)+log(1+z[↵V ]) V !1 V ↵ log(1 + z) z < 1 = | | ( ↵ log(1 + z)+↵ log z z > 1 | | We see that ⇥is continuous for all z as promised. But it is only analytic for z =1. | |6
–179– We can extract the physics by using (5.43)and(5.44)toeliminatethedependence on z.Thisgivesustheequationofstate,withpressurep as a function of n = V/N. For z < 1, we have | | ↵ p = ↵k T log n [0,↵/2) ,p 1, we have | | 2↵n p = ↵k T log n (3↵/2, 2↵) ,p>kT log 2 B (2↵ n)2 2 B ✓ ◆
They key point is that there is a jump in particle density of n = ↵ at p = ↵kBT log 2. Plotting this as a function of p vs v =1/n,wefindthatwehaveacurvethatisqualita- tively identical to the pressure-volume plot of the liquid-gas phase diagram under the co-existence curve. (See, for example, figure 37). This is a first order phase transition.
5.5 Landau-Ginzburg Theory
Landau’s theory of phase transition focusses only on the average quantity, the order parameter. It ignores the fluctuations of the system, assuming that they are negligible. Here we sketch a generalisation which attempts to account for these fluctuations. It is known as Landau-Ginzburg theory.
The idea is to stick with the concept of the order parameter, m.Butnowweallow the order parameter to vary in space so it becomes a function m(~r ). Let’s restrict ourselves to the situation where there is a symmetry of the theory m m so we ! need only consider even powers in the expansion of the free energy. We add to these agradienttermwhoseroleistocapturesthefactthatthereissomesti↵nessinthe system, so it costs energy to vary the order parameter from one point to another. (For the example of the Ising model, this is simply the statement that nearby spins want to be aligned). The free energy is then given by
F [m(~r )] = ddr a(T )m2 + b(T )m4 + c(T )( m)2 (5.45) r Z ⇥ ⇤ where we have dropped the constant F0(T )piecewhichdoesn’tdependontheorder parameter and hence plays no role in the story. Notice that we start with terms quadratic in the gradient: a term linear in the gradient would violate the rotational symmetry of the system.
–180– We again require that the free energy is minimised. But now F is a functional – it is afunctionofthefunctionm(~r ). To find the stationary points of such objects we need to use the same kind of variational methods that we use in Lagrangian mechanics. We write the variation of the free energy as
F = ddr 2am m +4bm3 m +2c m m r ·r Z ⇥ ⇤ = ddr 2am +4bm3 2c 2m m r Z ⇥ ⇤ where to go from the first line to the second we have integrated by parts. (We need to remember that c(T )isafunctionoftemperaturebutdoesnotvaryinspaceso that doesn’t act on it). The minimum of the free energy is then determined by r setting F =0whichmeansthatwehavetosolvetheEuler-Lagrangeequationsfor the function m(~r ),
c 2m = am +2bm3 (5.46) r The simplest solutions to this equation have m constant, reducing us back to Landau theory. We’ll assume once again that a(T ) > 0forT>Tc and a(T ) < 0forTT and m = m = a/2b for c ± 0 ± T
Domain Walls Suppose that we have T0livesinm =+m .Thisisexactly 0 0 the situation that we already met in the liquid-gas transition and is depicted in Figure 38. It is also easy to cook up the analogous configuration in the Ising model. The two regions in which the spins point up or down are called domains. The place where these regions meet is called the domain wall.
We would like to understand the structure of the domain wall. How does the system interpolate between these two states? The transition can’t happen instantaneously because that would result in the gradient term ( m)2 giving an infinite contribution r to the free energy. But neither can the transition linger too much because any point at 2 which m(~r )di↵erssignificantlyfromthevaluem0 costs free energy from the m and m4 terms. There must be a happy medium between these two.
–181– To describe the system with two domains, m(~r )mustvarybutitneedonlychange in one direction: m = m(x). Equation (5.46)thenbecomesanordinarydi↵erential equation, d2m am 2bm3 = + dx2 c c This equation is easily solved. We should remember that in order to have two vacua,
T
a m = m0 tanh x r 2c ! where m = a/2b is the constant ground state solution for the spin. As x , 0 !±1 the tanh function tends towards 1whichmeansthatm m .Sothissolution p ± !± 0 indeed interpolates between the two domains as required. We learn that the width of the domain wall is given by 2c/a.Outsideofthisregion,themagnetisationrelaxes exponentially quickly back to the ground state values. p We can also compute the cost in free energy due to the presence of the domain wall. To do this, we substitute the solution back into the expression for the free energy (5.45). The cost is not proportional to the volume of the system, but instead proportional to the area of the domain wall. This means that if the system has linear size L then the free energy of the ground state scales as Ld while the free energy required by the wall d 1 scales only as L .Itissimpletofindtheparametricdependenceofthisdomainwall energy without doing any integrals; the energy per unit area scales as ca3/b. Notice that as we approach the critical point, and a 0, the two vacua are closer, the width ! p of the domain wall increases and its energy decreases.
5.5.1 Correlations One of the most important applications of Landau-Ginzburg theory is to understand the correlations between fluctuations of the system at di↵erent points in space. Suppose that we know that the system has an unusually high fluctuation away from the average at some point in space, let’s say the origin ~r =0. What is the e↵ect of this on nearby points?
There is a simple way to answer this question that requires us only to solve the di↵erential equation (5.46). However, there is also a more complicated way to derive the same result which has the advantage of stressing the underlying physics and the role played by fluctuations. Below we’ll start by deriving the correlations in the simple manner. We’ll then see how it can also be derived using more technical machinery.
–182– We assume that the system sits in a given ground state, say m =+m0,andimagine small perturbations around this. We write the magnetisation as
m(~r )=m0 + m(~r )(5.47)
If we substitute this into equation (5.46)andkeeponlytermslinearinm,wefind 2a c 2m + m =0 r c where we have substituted m2 = a/2b to get this result. (Recall that a<0in 0 the ordered phase). We now perturb the system. This can be modelled by putting a delta-function source at the origin, so that the above equation becomes 2a 1 c 2m + m = d(0) r c 2c where the strength of the delta function has been chosen merely to make the equation somewhat nicer. It is straightforward to solve the asymptotic behaviour of this equa- tion. Indeed, it is the same kind of equation that we already solved when discussing the Debye-H¨uckel model of screening. Neglecting constant factors, it is
r/⇠ e m(~r ) (d 1)/2 (5.48) ⇠ r This tells us how the perturbation decays as we move away from the origin. This equation has several names, reflecting the fact that it arises in many contexts. In liquids, it is usually called the Ornstein-Zernicke correlation. It also arises in particle physics as the Yukawa potential. The length scale ⇠ is called the correlation length
c ⇠ = (5.49) r 2a The correlation length provides a measure of the distance it takes correlations to decay. Notice that as we approach a critical point, a 0andthecorrelationlengthdiverges. ! This provides yet another hint that we need more powerful tools to understand the physics at the critical point. We will now take the first baby step towards developing these tools.
5.5.2 Fluctuations The main motivation to allow the order parameter to depend on space is to take into the account the e↵ect of fluctuations. To see how we can do this, we first need to think alittlemoreaboutthemeaningofthequantityF [m(~r )] and what we can use it for.
–183– To understand this point, it’s best if we go back to basics. We know that the true free energy of the system can be equated with the log of the partition function (1.36). We’d like to call the true free energy of the system F because that’s the notation that we’ve been using throughout the course. But we’ve now called the Landau-Ginzburg functional F [m(~r )] and, while it’s closely related to the true free energy, it’s not quite the same thing as we shall shortly see. So to save some confusion, we’re going to change notation at this late stage and call the true free energy A.Equation(1.36)thenreads A = k T log Z,whichwewritethisas B
A En e = Z = e n X We would like to understand the right way to view the functional F [m(~r )] in this frame- work. Here we give a heuristic and fairly handwaving argument. A fuller treatment involves the ideas of the renormalisation group.
The idea is that each microstate n of the system can be associated to some specific | i function of the spatially varying order parameter m(~r ). To illustrate this, we’ll talk in the language of the Ising model although the discussion generalises to any system. There we could consider associate a magnetisation m(~r )toeachlatticesitebysimply averaging over all the spins within some distance of that point. Clearly, this will only lead to functions that take values on lattice sites rather than in the continuum. But if the functions are suitably well behaved it should be possible to smooth them out into continuous functions m(~r )whichareessentiallyconstantondistancescalessmaller than the lattice spacing. In this way, we get a map from the space of microstates to the magnetisation, n m(~r ). But this map is not one-to-one. For example, if the | i 7! averaging procedure is performed over enough sites, flipping the spin on just a single site is unlikely to have much e↵ect on the average. In this way, many microstates map onto the same average magnetisation. Summing over just these microstates provides a first principles construction of the F [m(~r )],
F[m(~r )] En e = e (5.50) n m(~r ) X| Of course, we didn’t actually perform this procedure to get to (5.45): we simply wrote it down the most general form in the vicinity of a critical point with a bunch of unknown coecients a(T ), b(T )andc(T ). But if we were up for a challenge, the above procedure tells us how we could go about figuring out those functions from first principles. More importantly, it also tells us what we should do with the Landau-Ginzburg free energy. Because in (5.50) we have only summed over those states that correspond to a particular
–184– value of m(~r ). To compute the full partition function, we need to sum over all states. But we can do that by summing over all possible values of m(~r ). In other words,
F[m(~r )] Z = Dm(~r ) e (5.51) Z This is a tricky beast: it is a functional integral.Weareintegratingoverallpossible function m(~r ), which is the same thing as performing an infinite number of integrations. (Actually, because the order parameters m(~r ) arose from an underlying lattice and are suitably smooth on short distance scales, the problem is somewhat mitigated).
The result (5.51) is physically very nice, albeit mathematically somewhat daunting. It means that we should view the Landau-Ginzburg free energy as a new e↵ective Hamiltonian for a continuous variable m(~r ). It arises from performing the partition function sum over much of the microscopic information, but still leaves us with a final sum, or integral, over fluctuations in an averaged quantity, namely the order parameter.
To complete the problem, we need to perform the function integral (5.51). This is hard. Here “hard” means that the majority of unsolved problems in theoretical physics can be boiled down to performing integrals of this type. Yet the fact it’s hard shouldn’t dissuade us, since there is a wealth of rich and beautiful physics hiding in the path integral, including the deep reason behind the magic of universality. We will start to explore some of these ideas in next year’s course on Statistical Field Theory.
–185– | 24,175 | 91,318 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2023-50 | longest | en | 0.920282 |
http://hvac-talk.com/vbb/showthread.php?91387-i-m-98&mode=hybrid | 1,495,544,321,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607636.66/warc/CC-MAIN-20170523122457-20170523142457-00267.warc.gz | 164,825,370 | 30,333 | # Thread: i'm 98%
#### Hybrid View
1. NOTE: This is freaky.....you have got to try it!
At the end of this message, you are asked a question.
Answer it immediately. Don't stop and think about it.
Just say the first thing that pops into your mind.
This is a fun "test"... AND kind of spooky at the same time! Give it a try, then e-mail it around (including back to me) and you'll see how many people you know fall into the same percentage as you. Be sure to put in the subject line if you are among the 98% or the 2%. You'll understand what that means after you finish taking the test".
Now... just follow the instructions as quickly as possible.
Do not go to the next calculation before you have finished the previous one..
You do not ever need to write or remember the answers, just do it using your mind.
You'll be surprised.
Start:
How much is:
15 + 6
3 + 56
89 + 2
12 + 53
75 + 26
25 + 52
63 + 32
I know! Calculations are hard work, but it's nearly over..
Come on, one more!
123 + 5
QUICK! THINK ABOUT A COLOR AND A TOOL!
Scroll further to the bottom....
A bit more...
You just thought about a red hammer ! , didn't you?
If this is not your answer, you are among 2% of people who have a different, if not abnormal, mind.
98% of the folks would answer a red hammer while doing this exercise.
If you do not believe this, pass it around and you'll see.
Be sure to put in the subject line if you are among the 98% or the 2% and send to everyone, including the person that sent it to
2. Professional Member
Join Date
Jul 2004
Location
midwest
Posts
2,868
Post Likes
98% here. Pretty cool
3. Professional Member*
Join Date
Feb 2006
Location
Float'N Vally, MS
Posts
2,498
Post Likes
2%
Yellow Screwdriver...
Guess I've been working with the 5/16 too much.
4. 2%here
5. Regular Guest
Join Date
Jul 2005
Posts
150
Post Likes
## 98 %
I thought red hammer, that freaked me out,
is there a scientific explanation for that???????
please reply if you know.
6. Professional Member
Join Date
Oct 2004
Posts
100
Post Likes
I guess I am odd, 2% here
7. 2%, i thought blue hammer
8. Professional Member
Join Date
Jul 2013
Location
Ohio
Posts
57
Post Likes
Originally Posted by t527ed
2%, i thought blue hammer
x2
9. Professional Member
Join Date
Dec 2004
Location
Newnan, GA
Posts
100
Post Likes
2% Red Screwdriver
10. Regular Guest
Join Date
May 2005
Posts
996
Post Likes
green needlenose pliers
11. Professional Member
Join Date
Jul 2004
Location
midwest
Posts
2,868
Post Likes
Originally posted by outside rep
2% orange pipe wrench
Your favorite color
12. Professional Member
Join Date
Jun 2003
Location
Chicago, IL
Posts
6,067
Post Likes
blue wrench
13. Professional Member
Join Date
Dec 2002
Location
davenport, iowa
Posts
778
Post Likes
blue wrench
Page 1 of 13 1234567811 ... Last
#### Posting Permissions
• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•
## Related Forums
The place where Electrical professionals meet. | 840 | 3,033 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2017-22 | longest | en | 0.956736 |
https://howtodiscuss.com/t/what-does-tfti-mean/56524 | 1,627,240,067,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151760.94/warc/CC-MAIN-20210725174608-20210725204608-00277.warc.gz | 314,528,570 | 6,448 | # What Does Tfti Mean
## What Does Tfti Mean
### What is TFTI?
TFTI is a text and social media acronym meaning thank you for inviting and is usually used sarcastically when someone has not been invited. It can also be used to give thanks for the information.
### And what are TF and TI in chemistry?
Tf: final temperature. Ten: initial temperature. Tf Ti = concept of temperature variation (Delta T). Heat problems with more than one substance.
### Second, what is TF physics?
TF stands for TimeFrequency (Physics) Science, Medicine, Engineering, etc.
Q = mc∆T.
### Q = thermal energy (Joule, J) m = mass of a substance (kg) c = specific heat (unit J / kg ∙ K) Δ is a symbol means changeWhat does qh2o mean?
By definition, the change in the internal energy of the system, E, is the final energy of the system minus the original energy of the system: E = Final - Final.
### IS?
And for the environment it is the same, but with the opposite sign!
### How do I find TF?
Add the temperature change to the original temperature of the substance to find the final heat. For example, if your water was originally 24 degrees Celsius, the final temperature would be: 24 + 6 or 30 degrees Celsius.
### In the end, is Delta less initial?
This says: Daylight saving time equals the end time minus the start time, or delta t equals the last t minus the start time.
### What is the equation for calorimetry?
Calculate the heat obtained by the calorimeter, Q, according to the equation Q = m * c * Delta (T), where m is the mass of water calculated in step 2, c is the water capacity of the water, or 4,184 Joules per. Grams per degree Celsius, J / gC and delta (T) represent the temperature change calculated in step 1.
### What is the conductivity of the water?
then. Eh = 8.7 x 105 J. This is the value of Lf for. Water, the value changes. for different materials.
### What is the heat capacity of the water?
The specific heat of water is 1 calorie / gram ° C = 4.166 joule / gram ° C, which is higher than any other common substance. Therefore, water plays a very important role in temperature regulation.
Element titanium
### Q is delta H?
Q (heat) can be said to be energy in transport. Enthalpy (Delta H), on the other hand, is the state of the system, the total heat content. Both can deal with heat (qp) (Q at constant pressure) = (Delta H), but heat and enthalpy are always related to energy, not specifically to heat.
### What is Q in Q MCT Chemistry?
Yes, your answer is correct ... for the most part. In the equation Q = mcΔt: Q = thermal energy (joule), water, at about 4.166 joule / gram × ° C) and t = the temperature change in ° C (9 ° C in this negative problem because exothermic, orothermic same energy from).
### Q Is A a joule or a kilojoule?
To calculate the amount of heat released in a chemical reaction, use the equation Q = mc ΔT, where Q is the heat energy transferred (in joules), m is the mass of the heated liquid (in kg), c is the specific value of the heat capacity of the liquid (Joules per kilograde Celsius), and ΔT is the variation of
### What does Delta H mean?
In chemistry, the letter H stands for the enthalpy of the system. Enthalpy describes the sum of the internal energy of a system plus the product of the pressure and volume of the system. The delta symbol is used to represent changes. Hence, Delta H represents the enthalpy change of a system in a reaction.
### What is C Mcdeltat Q?
The specific heat capacity of a material (c) is the amount of heat energy that causes a temperature change of 1 K or 1 ° C per kg of this material.
### What is the formula for specific heat?
Specific heat is the amount of heat needed to increase one gram of a substance by one degree Celsius or Kelvin. The formula for specific heat is the amount of heat absorbed or emitted = mass x specific heat x temperature change. specific heat calories.
### What does Q ml mean in physics?
The latent specific heat (L) of a material ... is a measure of the thermal energy (Q) per mass (m) that is released or absorbed during a phase change. is defined by the formula Q = ml. it is often referred to simply as the latent heat of the material. uses the SI unit joule per kilo [J / kg]. | 1,031 | 4,227 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2021-31 | latest | en | 0.887352 |
http://mathhelpforum.com/calculus/207341-how-find-absolute-extrema.html | 1,529,737,165,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864943.28/warc/CC-MAIN-20180623054721-20180623074721-00054.warc.gz | 206,044,500 | 11,325 | # Thread: How to find absolute extrema.
1. ## How to find absolute extrema.
Well I have a problem like:
f(x) = 3x2/3 - 2x, Interval: [-1, 1]
So this is what I did first:
Found the derivative: 2x-1/3-2
0 = 2x-1/3-2
x = 1
OR
f'(x) = 2/x1/3 - 2
x does not equal 0
Critical numbers: x = 1, x = 0
Used a chart:
x [-1, 0] 0 (0, 1) 1
f(x) negative undefined positive 0
Minimum occurs at x=0 since it goes from negative to positive
But I'm a bit confused on the maximum
I plugged in the points f(-1) = 3(-1)2/3 - 2(-3)
=5, so the point is (-1, 5)
Then I plugged in 1
f(1) = 3(1)2/3 - 2(1)
=1, so the point is (1, 1)
So I'm a bit confused, since at the back of my book,
The minimum is: (0, 0)
While the only maximum is (-1, 5)
So how come (1, 1) didn't get included as well
2. ## Re: How to find absolute extrema.
Originally Posted by Chaim
Well I have a problem like:
f(x) = 3x2/3 - 2x, Interval: [-1, 1]
So I'm a bit confused, since at the back of my book,
The minimum is: (0, 0)
While the only maximum is (-1, 5)
So how come (1, 1) didn't get included as well
The title of the thread says it all: absolute extrema.
Is $\displaystyle (1,1)$ an absolute extreme point?
3. ## Re: How to find absolute extrema.
Originally Posted by Chaim
Well I have a problem like:
f(x) = 3x2/3 - 2x, Interval: [-1, 1]
So this is what I did first:
Found the derivative: 2x-1/3-2
0 = 2x-1/3-2
x = 1
OR
f'(x) = 2/x1/3 - 2
x does not equal 0
Critical numbers: x = 1, x = 0
Used a chart:
x [-1, 0] 0 (0, 1) 1
f(x) negative undefined positive 0
Minimum occurs at x=0 since it goes from negative to positive
But I'm a bit confused on the maximum
I plugged in the points f(-1) = 3(-1)2/3 - 2(-3)
=5, so the point is (-1, 5)
Then I plugged in 1
f(1) = 3(1)2/3 - 2(1)
=1, so the point is (1, 1)
So I'm a bit confused, since at the back of my book,
The minimum is: (0, 0)
While the only maximum is (-1, 5)
So how come (1, 1) didn't get included as well
Always keep the original function in mind as you choose critical points (or anything else for that matter.)
So. Critical points.
Solve $\displaystyle 2x^{2/3} - 2x = 0$ for critical points.
Factor the x:
$\displaystyle x \left ( 2x^{-1/3} - 2 \right ) = 0$
The solutions are $\displaystyle x = 0$ or $\displaystyle \left ( 2x^{-1/3} - 2 \right ) = 0$
You can solve this in any way you like, and I also get $\displaystyle x = \pm 1$.
So we have 3 critical points $\displaystyle x = -1,~0~, 1$
We also need to check two more things. We need to see if the derivative has any critical points. In this case that would be anything were the derivative does not exist or is 0. You already have critical points at $\displaystyle x = \pm 1$ so these don't give anything new. The other thing to check is the value of the function (and its derivative) are at the ends of your domain. Again, this doesn't give anything new.
You did the rest of it correctly.
-Dan
4. ## Re: How to find absolute extrema.
The point (1,1) is a local maximum, but it's not the absolurte maximum within teh range [-1,1]. Look at the attached graph - clearly the value of the function at x=-1 is greater than at x=1:
,
,
,
### Locate the absolute extrema of the function on the closed interval. y = 3x2/3 − 2x, [−1, 1]
Click on a term to search for related topics. | 1,124 | 3,269 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2018-26 | latest | en | 0.904793 |
http://www.infomutt.com/e/es/escape_velocity.html | 1,611,795,501,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704835583.91/warc/CC-MAIN-20210128005448-20210128035448-00279.warc.gz | 138,492,104 | 3,348 | Main Page | See live article | Alphabetical index
Escape velocity can also mean Escape Velocity (computer game).
An escape velocity is the minimum speed at which an object without propulsion can move away from a source of a gravitational field indefinitely, barring other factors such as friction. (This definition may need modification for the practical problem of two or more sources in some cases. In any case, the object is assumed to be a point with a mass that is negligible compared with that of the source of the field, usually an excellent approximation.) It is commonly described as the speed needed to break free from a gravitational field, but this is inaccurate because gravitational fields are infinite in extent.
One somewhat counterintuitive feature of escape velocity is that it is independent of direction, so that "velocity" is a misnomer; it is a scalar quantity and would more accurately be called "escape speed".
The simplest way of deriving the formula for escape velocity is to use conservation of energy.
Defined a bit more formally "escape velocity" is the initial speed required to go from an initial point in a gravitational potential field to infinity with a residual velocity of zero, relative to the field. In common usage, the initial point is a point on the surface of a planet or moon. It is a theoretical quantity, because it assumes that an object is fired into space like a bullet. Instead propulsion is almost always used to get into "space". It is usually in "space" that the idea gets a more concrete meaning. On the surface of Earth the escape velocity is about 11 kilometres per second. However, at 9000 km from the surface in "space," it is slightly less than 7.1 km/s. Continual acceleration from the surface to attain that speed at that height is possible. At no time would the "escape velocity" of 11 km/s be attained; yet at that height, even with zero propulsion now, the object can move away from Earth indefinitely.
For a simple case of escape velocity from a single body, escape velocity can be calculated as follows:
where is the escape velocity, G is the gravitational constant, M is the mass of the body being escaped from, "m" is the mass of the escaping body (cancels out), and r is the distance between the center of the body and the point for which escape velocity is being calculated. | 477 | 2,349 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2021-04 | latest | en | 0.940446 |
https://www.physicsforums.com/threads/first-peerson-to-define-work-as-w-f-d.357363/ | 1,722,997,635,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640667712.32/warc/CC-MAIN-20240807015121-20240807045121-00144.warc.gz | 731,987,254 | 17,257 | # First peerson to define work as W = F d.
• carvajal57
In summary, Gustave-Gaspard Coriolis (1792-1843) was the first person to define work as W = Fd, and his contributions to the concept were examined in detail by I Grattan-Guinness in 1984.
carvajal57
Please, can someone tell me who was the first person to define work as we currently use, that is, W = F d? Thank you.
mc
carvajal57 said:
Please, can someone tell me who was the first person to define work as we currently use, that is, W = F d? Thank you.
mc
According the article below, it was Gustave-Gaspard Coriolis (1792-1843) who:
"The modern statement: "work is force times path" is due to French physicist Gustave-Gaspard Coriolis,[1] who gave the correct formula for change in kinetic energy associated with work[/quote]
http://en.citizendium.org/wiki/Work_(Physics)
The acclaimed website, MacTutor History of Mathematics Archive,
http://www-history.mcs.st-and.ac.uk/
has the following biography on him:
http://www-history.mcs.st-and.ac.uk/Biographies/Coriolis.html
Last edited by a moderator:
From the MacTutor biography, we learn of an article from 1984 that examines in detail the relative contributions of Poncelet, Navier and Coriolis to the concept of "work":
"I Grattan-Guinness, Work for the workers : advances in engineering mechanics and instruction in France, 1800-1830, Ann. of Sci. 41 (1) (1984), 1-33."
## What is the significance of defining work as W = F d?
The equation W = F d is significant because it provides a quantifiable way to measure work, which was previously thought of as a vague concept. It allows for a clear understanding and comparison of the amount of work being done in different situations.
## Who was the first person to define work as W = F d?
The first person to define work as W = F d was the French mathematician and physicist, Gaspard-Gustave Coriolis, in 1829.
## Is the equation W = F d still valid in modern science?
Yes, the equation W = F d is still considered valid in modern science and is used extensively in physics, engineering, and other scientific fields.
## What are the units of measurement for each variable in the equation W = F d?
The unit for work (W) is joules (J), the unit for force (F) is newtons (N), and the unit for distance (d) is meters (m).
## Can the equation W = F d be applied to all types of work?
No, the equation W = F d is specifically applicable to work done by a constant force in a straight line. For more complex situations, other equations and principles must be used.
• Mechanics
Replies
13
Views
1K
• Mechanics
Replies
3
Views
802
• Mechanics
Replies
3
Views
327
• Mechanics
Replies
9
Views
967
• Mechanics
Replies
24
Views
2K
• Mechanics
Replies
16
Views
1K
• Mechanics
Replies
9
Views
836
• Mechanics
Replies
9
Views
10K
• Mechanics
Replies
3
Views
915
• Mechanics
Replies
12
Views
2K | 765 | 2,854 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2024-33 | latest | en | 0.961433 |
http://www.pixelstech.net/article/1332511256-Decision-Trees-in-C%23 | 1,503,469,354,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886117874.26/warc/CC-MAIN-20170823055231-20170823075231-00578.warc.gz | 659,272,189 | 17,617 | Today's Question: What are you most afraid of as a programmer?
# Decision Trees in C#
César Souza 2012-03-23 10:00:56 10,568 0 8
Decision trees are simple predictive models which map input attributes to a target value using simple conditional rules. Trees are commonly used in problems whose solutions must be readily understandable or explainable by humans, such as in computer-aided diagnostics and credit analysis.
## Introduction
Decision Trees give a direct and intuitive way for obtaining the classification of a new instance from a set of simple rules. Because of this characteristic, decision trees find wide use in situations in which the interpretation of the obtained results and of the reasoning process is crucial, such as in computer-aided diagnostics (CAD) and in financial credit analysis. Consumer credit analysis is an interesting example because, in many countries, one can not simply reject credit without giving a justification, justification of which is trivial to extract from a decision tree.
The tree on the right has been generated using the Context Free software based on the grammar shared by davdrn.
### Learning decision trees
Decision trees can be simply drawn by hand based on any prior knowledge the author may have. However, their real power becomes apparent when trees are learned automatically, through some learning algorithm.
The most notable and classics examples to decision tree learning are the algorithms ID3 (Quinlan, 1986) and the C4.5 (Quinlan, 1993). Both are examples of greedy algorithms, performing local optimum decisions in the hope of producing a most general tree. Such algorithms are based on the principle of the Occam's razor, favoring simpler or smaller trees in the hope that such smaller trees could retain more generalization power. This preference is formalized through the specification of an hypothesis ordering criteria such as the information gain. The information gain measures the, as the name implies, gain of information in using each of the attributes as a next factor to consider during decision. The information gain can be defined as:
However, the information gain has a bias towards attributes with a high number of possible values (Mitchell, 1997). A way to overcome this bias is to select new selection attributes based on alternative criteria, such as the gain ratio (Quinlan, 1986), defined as:
In the GainRatio, the SplitInformation term attenuates the importance given to attributes with many, uniformly distributed, possible values.
### Iterative Dichotomizer 3 (ID3)
The algorithm presented below is a slightly different version of the original ID3 algorithm as presented by Quinlan. The modifications are to support multiple output labels. In each recursion of the algorithm, the attribute which bests classifiers the set of instances (or examples, or input-output pairs, or data) is selected according to some selection criteria, such as the InfoGain or the GainRatio.
• ID3(instances, target_attribute, attributes)
• Create a new root node to the tree.
• If all instances have the target_attribute belonging to the same class c,
• Return the tree with single root node with label c.
• If attributes is empty, then
• Return the tree with single root node with the most common label of the target_attribute in instances.
• Else
• A ← the attribute in attributes which best classifies instances
• root decision attribute ← A
• Foreach possible value vi of A,
• Add a new ramification below root, corresponding to the test A = vi
• Let instancesvi be the subset of instances with the value vi for A
• If instancesvi is empty then
• Below this ramification, add a new leaf node with the most common value of target_attribute in instances.
• Else below this ramification, add the subtree given by the recursion:
ID3(instancesvi, target_attribute, attributes – { A })
• End
• Return root
### Difficulties and disadvantages of decision tree learning
Despite relying on the Occam’s Razor to guide the learning, neither ID3 or C4.5 algorithms are not guaranteed to produce the smaller or more general tree possible. This happens because their learning is based solely on heuristics and not in truly optimality criteria. The following example, from (Esmeir & Markovitch, 2007) illustrates the learning of the concept xor (exclusive or) by the ID3 algorithm. In this example, A3 and A4 are irrelevant attributes, having absolutely no influence on the target answer. However, yet being irrelevant, ID3 will select one of them to belong to the tree. In fact, ID3 will select the irrelevant attribute A4 to be the root of the learned tree.
A1 A2 A3 A4 label 1 0 0 1 + 0 1 0 0 + 0 0 0 0 - 1 1 0 0 - 0 1 1 1 + 0 0 1 1 - 1 0 1 1 +
One complication of decision tree learning is that the problem to find the smaller consistent tree (in the sense of classifying correctly all training samples) is known to be NP-complete (Hyafil & Rivest, 1976). Moreover, the separating decision surface generated by such trees are always formed by parallel cuts alongside the attribute space axis, which could be a severely suboptimal solution (Bishop, 2007, p. 666). The example given by Bishop illustrates well this problem: for example, to separate classes whose optimum decision margin are on 45 degrees from one of the axis, it will be needed a high number of parallel cuts in comparison with a single cut on the diagonal such as could be given by any linear decision machine. Another disadvantage of traditional decision tree learning algorithms is that most methods require only a constant learning time, and, as such, do not allow for trading extra training time for a better solutions. The work of (Esmeir & Markovitch, 2007) is dedicated to overcome such problem.
The following picture shows an example on how learning by decision trees is limited to cuts parallel to the axis, as described by Bishop. The Ying-Yang classification problem is a classical example of a non-linearly separable decision problem. Decision trees, albeit not being linear classifiers, have difficulty classifying this set with simple thresholds.
Top-Left: Ying-Yang non-linear classification problem. Top-Right: Decision surface extracted by a decision tree. Bottom: Decision threshold rules extracted from the tree.
However, despite all of those shortcomings, decision trees plays major roles as base of many successful algorithms. One interesting application and of notorious success is given in the field of object detection in images. The first real-time face detecting algorithm (Viola & Jones, 2001) is based on a degenerated decision tree (also known as a cascade). The body recognition and segmentation algorithm used by the Xbox 360 gaming platform used to process depth images generated by its companion sensor Kinect is equally based on the use of decision trees (Shotton, et al., 2011). As well is the case of the FAST algorithm for interest point detection in images (Rosten & Drummond, 2006).
I should also make the note that both the Viola-Jones and the FAST algorithms are available in the Accord.NET Framework for immediate use (the Viola-Jones (HaarObjectDetector) has also been recently updated to support multiple threads, so feel free to take a look and experiment with it!).
In sum, its possible to say that great part of the applicability of decision trees came from the simple fact that they are extremely fast to evaluate. One of the reasons for this feature is being easily translated to sets of conditional instructions in a conventional computer program. The decision trees now available in the Accord.NET Framework make full use of this fact and enables the user to actually compile the decision trees to native code on-the-fly, augmenting even more its performance during classification.
## Source code
The code presented in this section is actually part of the Accord.NET Framework. The Accord.NET Framework is a framework extension to the already very popular AForge.NET Framework, adding and providing new algorithms and techniques for machine learning, computer vision and even more.
The Accord.MachineLearning.DecisionTree namespace is comprised of the following classes:
• DecisionTree, the main class representing a decision tree, with methods such as Compute to compute the tree classification given an input vector;
• DecisionNode, the node class for the decision tree, which may or may not have children nodes contained under a collection of children represented by a DecisionBranchNodeCollection;
• DecisionVariable, a class to specify the nature of each variable processable by the tree, such as if the variable is continuous, discrete, which are their expected or valid ranges;
• DecisionBranchNodeCollection, a class to contain children nodes together with information about which attribute of the data should be compared with the child nodes during reasoning.
Whose relationships can be seen in the following class diagram:
The learning algorithms available are either the ID3 algorithm discussed above, or its improved version C4.5 (which can handle continuous variables, but at this time does not yet support pruning), both proposed and published by Ross Quinlan.
Despite the bit complicated look, usage is rather simple as it will be shown in the next section.
## Using the code
Consider, for example, the famous Play Tennis example by Tom Mitchell (1998):
` DataTable data = new DataTable("Mitchell's Tennis Example"); data.Columns.Add("Day", "Outlook", "Temperature", "Humidity", "Wind", "PlayTennis"); data.Rows.Add( "D1", "Sunny", "Hot", "High", "Weak", "No" ); data.Rows.Add( "D2", "Sunny", "Hot", "High", "Strong", "No" ); data.Rows.Add( "D3", "Overcast", "Hot", "High", "Weak", "Yes" ); data.Rows.Add( "D4", "Rain", "Mild", "High", "Weak", "Yes" ); data.Rows.Add( "D5", "Rain", "Cool", "Normal", "Weak", "Yes" ); data.Rows.Add( "D6", "Rain", "Cool", "Normal", "Strong", "No" ); data.Rows.Add( "D7", "Overcast", "Cool", "Normal", "Strong", "Yes" ); data.Rows.Add( "D8", "Sunny", "Mild", "High", "Weak", "No" ); data.Rows.Add( "D9", "Sunny", "Cool", "Normal", "Weak", "Yes" ); data.Rows.Add( "D10", "Rain", "Mild", "Normal", "Weak", "Yes" ); data.Rows.Add( "D11", "Sunny", "Mild", "Normal", "Strong", "Yes" ); data.Rows.Add( "D12", "Overcast", "Mild", "High", "Strong", "Yes" ); data.Rows.Add( "D13", "Overcast", "Hot", "Normal", "Weak", "Yes" ); data.Rows.Add( "D14", "Rain", "Mild", "High", "Strong", "No" );`
In the aforementioned example, one would like to infer if a person would play tennis or not based solely on four input variables. Those variables are all categorical, meaning that there is no order between the possible values for the variable (i.e. there is no order relationship between Sunny and Rain, one is not bigger nor smaller than the other, but are just distinct). Moreover, the rows, or instances presented above represent days on which the behavior of the person has been registered and annotated, pretty much building our set of observation instances for learning.
In order to try to learn a decision tree, we will first convert this problem to a more simpler representation. Since all variables are categories, it does not matter if they are represented as strings, or numbers, since both are just symbols for the event they represent. Since numbers are more easily representable than text string, we will convert the problem to use a discrete alphabet through the use of a codebook.
A codebook effectively transforms any distinct possible value for a variable into an integer symbol. For example, “Sunny” could as well be represented by the integer label 0, “Overcast” by “1”, Rain by “2”, and the same goes by for the other variables. So:
``` // Create a new codification codebook to
// convert strings into integer symbols Codification codebook = new Codification(data);```
Now we should specify our decision tree. We will be trying to build a tree to predict the last column, entitled “PlayTennis”. For this, we will be using the “Outlook”, “Temperature”, “Humidity” and “Wind” as predictors (variables which will we will use for our decision). Since those are categorical, we must specify, at the moment of creation of our tree, the characteristics of each of those variables. So:
` DecisionVariable[] attributes = { new DecisionVariable("Outlook", 3), // 3 possible values (Sunny, overcast, rain) new DecisionVariable("Temperature", 3), // 3 possible values (Hot, mild, cool) new DecisionVariable("Humidity", 2), // 2 possible values (High, normal) new DecisionVariable("Wind", 2) // 2 possible values (Weak, strong) };`
` int classCount = 2; // 2 possible output values for playing tennis: yes or no`
Let's now proceed and create our DecisionTree:
` DecisionTree tree = new DecisionTree(attributes, classCount); `
Now we have created our decision tree. Unfortunately, it is not really very useful, since we haven't taught it the problem we are trying to predict. So now we must instantiate a learning algorithm to make it useful. For this task, in which we have only categorical variables, the simplest choice is to use the ID3 algorithm by Quinlan. Let’s do it:
``` // Create a new instance of the ID3 algorithm ID3Learning id3learning = new ID3Learning(tree); // Translate our training data into integer symbols using our codebook: DataTable symbols = codebook.Apply(data);
int[][] inputs = symbols.ToIntArray("Outlook", "Temperature", "Humidity", "Wind");
int[] outputs = symbols.ToIntArray("PlayTennis").GetColumn(0); // Learn the training instances!
id3learning.Run(inputs, outputs); ```
At this point, the tree has been created. A suitable representation of the learned tree could be given by the following diagram:
However, more than just a diagram, we can also go and generate a .NET Expression Tree describing our decision tree. Expression trees represent code in the form of a tree of expressions, which can then be read, modified or compiled. By compiling the DecisionTree's expression, we are able to generate code on-the-fly and let the JIT compile it down to native code at runtime, greatly improving the performance of our decision tree:
``` // Convert to an expression tree
var expression = tree.ToExpression(); // Compiles the expression to IL
var func = expression.Compile();```
And here is the resulting C# code obtained by compiling the expression into a lambda function, dumping the function into an dynamic assembly, opening and inspecting this assembly using ILSpy:
` public static int Compute(double[] input) { if (input[0] == 0.0) { if (input[2] == 0.0) { return 0; } if (input[2] == 1.0) { return 1; } throw new ArgumentException( "Input contains a value outside of expected ranges.", "input"); } else { if (input[0] == 1.0) { return 1; } if (input[0] != 2.0) { throw new ArgumentException( "Input contains a value outside of expected ranges.", "input"); } if (input[3] == 0.0) { return 1; } if (input[3] == 1.0) { return 0; } throw new ArgumentException( "Input contains a value outside of expected ranges.", "input"); } }`
I suspect I should have included more input validation on the generated code. However, the release is now closed and will be made available for download soon. If there is need, more robust code generation may added as a feature in later versions.
## Conclusion
Decision trees are useful tools when the problem to be solved needs to be quickly interpreted and understood by humans. Another suitable use is when the decision needs to be fast. However, decision trees, at least those trained by simple training algorithms such as ID3 and C4.5 can perform quite poorly depending on the problem. As it happens with all machine learning techniques, it is futile to believe there is a one true classifier which would act perfectly on all possible imaginable scenarios. As always, it is important to know our tools and know in which situation each technique would work better.
## References
• Bishop, C. M., 2007. Pattern Recognition and Machine Learning (Information Science and Statistics). 1st ed. 2006. Corr. 2nd printing ed. s.l.:Springer
• Fayyad, U. M. & Irani, K. B., 1992. On the Handling of Continuous-Valued Attributes in Decision Tree Generation. Machine Learning, January, 8(1), pp. 87-102.
• Quinlan, J. R., 1986. Induction of decision trees. Machine Learning, 1(1), pp. 81-106.
• Quinlan, J. R., 1993. C4.5: Programs for Machine Learning (Morgan Kaufmann Series in Machine Learning). 1 ed. s.l.:Morgan Kaufmann.
• Shotton, J. et al., 2011. Real-Time Human Pose Recognition in Parts from Single Depth Images. s.l., s.n.
• Viola, P. & Jones, M., 2001. Robust Real-time Object Detection. s.l., s.n.
• Mitchell, T. M., 1997. Decision Tree Learning. In:: Machine Learning (McGraw-Hill Series in Computer Science). s.l.:McGraw Hill.
• Mitchell, T. M., 1997. Machine Learning (McGraw-Hill Series in Computer Science). Boston(MA): WCB/McGraw-Hill.
• Esmeir, S. & Markovitch, S., 2007. Anytime Learning of Decision Trees. J. Mach. Learn. Res., May, Volume 8, pp. 891-933.
• Hyafil, L. & Rivest, R. L., 1976. Constructing Optimal Binary Decision Trees is NP-complete. Information Processing Letters, 5(1), pp. 15-17. | 4,136 | 17,673 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2017-34 | latest | en | 0.923793 |
https://it.mathworks.com/matlabcentral/answers/260958-how-to-save-image-of-the-result-imfindcircle | 1,621,345,375,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989637.86/warc/CC-MAIN-20210518125638-20210518155638-00351.warc.gz | 324,489,937 | 27,965 | MATLAB Answers
# how to save image of the result imfindcircle
36 views (last 30 days)
raja gopal on 19 Dec 2015
Answered: Michal Heker on 24 Sep 2020
i have used imfindcircle to find circle object in image. do you know how to save the image result of the circle object detected?
##### 1 CommentShowHide None
Image Analyst on 19 Dec 2015
What kind of result do you want? The function imwrite() saves images - why can't you use that? Please explain why that won't work but only after you read this.
Sign in to comment.
### Accepted Answer
harjeet singh on 20 Dec 2015
dear raja, most probable you are using viscircles to highlighter the circles, this commands uses superimposing of graph over the image not actually changes the pixel values.
use this program to save circles with markings
clear all
close all
clc
A = imread('coins.png');
figure(1)
imshow(A)
drawnow
[centers, radii, metric] = imfindcircles(A,[15 30]);
centersStrong5 = centers(1:5,:);
radiiStrong5 = radii(1:5);
metricStrong5 = metric(1:5);
%/////// this command makes plot superimpose on the image not the pixels change permanently
viscircles(centersStrong5, radiiStrong5,'EdgeColor','b');
imwrite(A,'image_out_1.jpg');
%/////// for changing the values of the pixels in the image for making circle //////
B(:,:,1)=A;
B(:,:,2)=A;
B(:,:,3)=A;
for i=1:length(radiiStrong5)
theta=0:1:360;
r=round(centersStrong5(i,1) + radiiStrong5(i)*sin(theta));
c=round(centersStrong5(i,2) + radiiStrong5(i)*cos(theta));
for j=1:length(r)
B(c(j),r(j),:)=[0 0 255];
end
end
figure(2)
imshow(B)
drawnow
imwrite(B,'image_out_2.jpg');
##### 4 CommentsShowHide 3 older comments
raja gopal on 3 Feb 2016
hello how to make marking circle become thicker? please help :(
Sign in to comment.
### More Answers (2)
Alejandro Navarro on 20 Mar 2019
Hello.
instead of saving the output of a viscircles + image to a file, I would like to assign it as a variable. I have tried "imfuse(Image,Circles)" but this does not work because the circles are not an image, but a "matlab.graphics.primitive.Group". How would I do that related task?
Thanks
##### 0 CommentsShowHide -1 older comments
Sign in to comment.
Michal Heker on 24 Sep 2020
How about making a circles mask with imdilate and a disk element structure?
Something like this:
BW = zeros(size(I,1),size(I,2));
for i = 1:length(points)
landmarksMask(points(i,2),(points(i,1)) = 1;
end
BW = imdilate(BW, strel('disk',3));
And duplicate your image with BW.
##### 0 CommentsShowHide -1 older comments
Sign in to comment.
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
Translated by | 753 | 2,657 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2021-21 | latest | en | 0.752448 |
https://www.ciese.org/curriculum/tideproj/teacherstranded | 1,718,830,807,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861832.94/warc/CC-MAIN-20240619185738-20240619215738-00808.warc.gz | 616,800,290 | 2,760 | A CIESE Realtime Data Project
# Stranded - Teacher Guide: Lesson Plans
## Stranded
Objectives
Students will be able to:
• predict the occurrence of low tides
• determine how much water will be needed to sustain a crew of eight stranded people
Materials
computers with Internet access
calculators
Background
The novel by Nathaniel Philbrick, In the Heart of the Sea: The Tragedy of the Whaleship Essex, chronicles the tragedy of the crew of the Essex, specifically when the crew was stranded on a island in the Pacific where the crew had to calculate when the fresh water would be exposed enough to gather water to supply the crew.
In this lesson, students will be charged to determine similar calculations using real time data.
Procedure
Problem Statement
You and the 7 other crew members of your boat are stranded on an island. The only source of fresh water is an aquifer whose outlet is on the rocky coast and is only accessible during an extremely low tide, lower than -2.0 feet to MLLW.
Your challenge is to determine when the low tide will occur to collect fresh water. Calculate how much water people will need per day to survive (see: daily water requirements). Then, determine when the next low tide will occur so that you can calculate the total amount of fresh water needed to sustain the crew.
1. You and your 7 crew members are stranded on Kodiak Island for the next 3 months. Will there be a tide low enough to collect fresh water? How much will you need to collect to sustain the entire crew until the end of the three month period, or until the next low tide, which ever event arrives first.
Assessment
The prediction results. | 365 | 1,660 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2024-26 | latest | en | 0.908617 |
https://serc.carleton.edu/sp/library/games/examples/48565.html | 1,720,848,540,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514490.70/warc/CC-MAIN-20240713051758-20240713081758-00225.warc.gz | 418,066,041 | 21,847 | # Transaction Analysis Tic-Tac-Toe
Initial Publication Date: August 17, 2010
## Summary
Transaction analysis is the process of identifying the specific effects of economic events on the accounting equation. Each transaction at a minimum has a dual effect on the accounting equation and it must always remain in balance. Students enrolled in principles of financial accounting courses need practice applying these concepts and playing tic-tac-toe helps to accomplish this task in a novel way. The game utilizes PowerPoint hyperlinked slides to emulate the game of tic-tac-toe. Instructors can also modify the tic-tac-toe questions to serve their individual needs.
Used this activity? Share your experiences and modifications
## Learning Goals
As a result of completing this worksheet, students should be able to:
• Analyze the effect of business transactions on the expanded accounting equation:
Assets = Liabilities + [Stockholders Equity + Revenues – Expenses]
## Context for Use
This worksheet is designed to be used after students studying principles of financial accounting have been introduced to the following concepts:
• Mathematical formulas that form the income statement, statement of retained earnings, and balance sheet.
• Accounting equation and the accounts comprising each category
• A basic understanding of transaction analysis prior to the recognition of adjusting entries. No adjusting entry transaction analysis items are included in this game.
Implementation options.
Transaction Analysis Tic-Tac-Toe can be used:
1. As an in-class game for drill and practice or review with students divided into two teams. As an alternative, if students have laptop computers they can bring to class, then multiple teams can be formed with a pair of students playing the game.
2. As a supplementary activity completed by students for drill and practice outside of class.
## Description and Teaching Materials
Transaction Analysis Tic-Tac-Toe PowerPoint slides. (Rules and instruction for using the presentation to play tic-tac-toe are included on slides 2-3.) Transaction Analysis Tic-Tac-Toe (revised 6/21/11) (PowerPoint 2007 (.pptx) 259kB Jun22 11)
Use this Tic-Tac-Toe Template to enter your own questions and responses.Tic-Tac-Toe Template (PowerPoint 2007 (.pptx) 259kB Jun22 11)
## Teaching Notes and Tips
• Instructors need to practice running the tic-tac-toe game to become adept at where to click on each screen so that the game executes properly. Once the game is launched, clicking on the white areas of any screen will cause the game sequencing to malfunction. Similarly, displaying the wrong winner of a square cannot be corrected.
### How the game is intended to be played once launched.
• Each tic-tac-toe square links to a 3-response multiple choice question. Selecting a response hyperlinks to positive or negative feedback. Clicking the return button takes players back to either the question screen or tic-tac-toe game board depending on its use.
• The calling team selects any square. If the item is answered correctly, the calling team wins the square. Selection passes to the competitive team.
• Should the calling team miss the question, the competing team is allowed to steal the square by answering correctly. If the competing team also selects the incorrect response, the square goes to the calling team. Under both circumstances, the next square selected passes to the competitive team.
• Process is repeated until a team achieves a tic-tac-toe.
## Assessment
Playing the tic-tac-toe provides for formative assessment as students can monitor their understanding of specific transaction analysis as the game progresses in class.
## References and Resources
Kimmel, P. D., Weygandt, J. J. & Kieso, D. E. (2009). Accounting Tools for Business Decision Making, 3rd edition, John Wiley & Sons, Inc. | 789 | 3,847 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2024-30 | latest | en | 0.925536 |
http://www.physicsforums.com/showthread.php?t=593525 | 1,386,494,224,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163059081/warc/CC-MAIN-20131204131739-00061-ip-10-33-133-15.ec2.internal.warc.gz | 500,358,684 | 7,404 | WHY do vernier calliper works???
by babita
Tags: calliper, vernier
P: 61 1. The problem statement, all variables and given/known data This is not exactly a problem but i think this is the most appropriate place to post this. I know how to read a vernier calliper. But i want to know WHY it works. I mean how it gives us the true length. 2. Relevant equations 3. The attempt at a solution I have been thinking on it for a long time and i'm not able to conclude why we see coinciding lines. Also if the 0's on vernier and main scale are marked exactly at the edges wouldnt it be easier to use? please explain...please dont write just a line or two...i'm going mad Attached Thumbnails
P: 953 When you align example 3mm, the lower part align to 0. You can see that unit 1 of lower scale is 1/10 nearer to main scale This continue 2->2/10......9->9/10 If you move the vernier scale, the distance moved can be measured from initial position to final position. The one that align must be the easiest to calculate the distance moved.
P: 61 well i didnt really understand that.....but anyways i've got the logic We're comparing two lengths, one is having one more no of divisions than the other. and so on. (since u already know i'm skipping explanation)
Related Discussions General Physics 4 General Physics 8 Mechanical Engineering 2 Engineering, Comp Sci, & Technology Homework 0 Introductory Physics Homework 5 | 339 | 1,411 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2013-48 | longest | en | 0.906637 |
https://crbcentral.com/new-south-wales/simple-random-sampling-examples-pdf.php | 1,586,262,263,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371799447.70/warc/CC-MAIN-20200407121105-20200407151605-00553.warc.gz | 421,626,225 | 7,035 | Simple random sampling examples pdf
New South Wales - 2019-11-20
# Sampling pdf examples random simple
## Application of Simple Random Sampling (SRS) in eDiscovery. Scalable Simple Random Sampling and Strati ed Sampling.
1 Simple Random Sampling The goal is to estimate the mean and the variance of a variable of interest in a nite population by collecting a random sample from it.. random-number tables, to obtain simple random samples. The built-in programs for doing so are called random-number generators. When using randomnumber generators, be aware of whether they provide samples with replacement or samples without replacement. 4. Other Sampling Designs Simple random sampling is the most natural and easily understood method of probability sampling — it ….
Stratified Random Sampling •Sometimes in survey sampling certain amount of information is known about the elements of the popu- lation to be studied. •For instance, information may be available on the geographical location of the area, e.g. if it is an inner city, a suburban or a rural area. •Census information will provide a wealth of other information about the area, for instance, its Scalable Simple Random Sampling and Strati ed Sampling both kand nare given and hence the sampling prob-ability p= k=n. Then, in Section 3.4, we consider the
Scalable Simple Random Sampling and Strati ed Sampling both kand nare given and hence the sampling prob-ability p= k=n. Then, in Section 3.4, we consider the Stratified Random Sampling •Sometimes in survey sampling certain amount of information is known about the elements of the popu- lation to be studied. •For instance, information may be available on the geographical location of the area, e.g. if it is an inner city, a suburban or a rural area. •Census information will provide a wealth of other information about the area, for instance, its
1 Simple Random Sampling The goal is to estimate the mean and the variance of a variable of interest in a nite population by collecting a random sample from it.. Rice-15149 book March 16, 2006 12:53 7.3 Simple Random Sampling 203 7.3.1 The Expectation and Variance of the Sample Mean We will denote the sample size by n (n is less than N) and the values of the sample.
“2.3 Simple Random Sampling Montana State University”.
Simple random sampling Stratified sampling Cluster sampling Systematic sampling RSMichael 2-8 Simple Random Sampling The preferred method – probability is highest that sample is representative of population than for any other sampling method. Every member of a population has an equal chance of being selected. Least chance of sample bias. 5 RSMichael 2-9 Proportional Stratified Sampling.
Stratified Random Sampling... A stratified random sample is obtained by separating the population into mutually exclusive sets, or strata, and then drawing simple random samples from each stra-. With systematic sampling, the target population is partitioned into H > 1 non- overlapping subpopulations of strata. If the population size consists of N discrete elements, then under stratified. Simple random sampling is a sample of individuals that exist in a population; the individuals are randomly selected from the population and placed into a sample..
With systematic sampling, the target population is partitioned into H > 1 non- overlapping subpopulations of strata. If the population size consists of N discrete elements, then under stratified (simple random sampling) or a known probability of being selected (stratified random sampling). The The sample is referred to as representative because the characteristics of a properly drawn sample …
## How to find free pdf books online
How to Find Free Books Online YouTube. 18/11/2018 · Children's Books Online is a repository of public domain children's books, most with illustrations. The books are organized by reading level (there's even a section for adult readers!), and some even have audio files of the book being read aloud. Be aware that the image quality of some of the books is grainy. …
## Nist Handbook 105-1 Pdf
Procedures and General Requirements NIST Page. ASM Handbook, Volume 2: Properties and Selection: Nonferrous Alloys and Special-Purpose Materials Properties of Wrought Aluminum and Aluminum Alloys / 63 Table 3 Typical mechanical properties of 1060 aluminum Tensile Yield Shear strength strength Elongation(a), Hardness, strength Fatigue limit(c) … | 972 | 4,441 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2020-16 | latest | en | 0.851928 |
http://mail.scipy.org/pipermail/numpy-discussion/2009-May/042801.html | 1,406,124,162,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1405997878518.58/warc/CC-MAIN-20140722025758-00018-ip-10-33-131-23.ec2.internal.warc.gz | 241,080,538 | 2,351 | # [Numpy-discussion] asarray() and PIL
Christopher Barker Chris.Barker@noaa....
Wed May 27 11:43:55 CDT 2009
```cp wrote:
>>> arr=asarray(img)
>>> arr.shape
>>> (1600,1900,3)
>
>> No, it means that you have 1600 rows, 1900 columns and 3 colour channels.
>
> According to scipy documentation at
> http://pages.physics.cornell.edu/~myers/teaching/ComputationalMethods/python/arrays.html
> you are right.
>
> In this case I import numpy where according to
> http://www.scipy.org/Tentative_NumPy_Tutorial and the Printing Arrays paragraph
> (also in http://www.scipy.org/Numpy_Example_List#reshape reshape example) the
> first number is the layer, the second the rows and the last the columns.
>
> Are all the above valid or am I missing something?
I'm not sure what part of those docs you're referring to -- but they are
probably both right. What you are missing is that numpy doesn't define
for you what the axis mean, they just are:
the zeroth axis is of length 1600 elements
the first axis is of length 1900 elements
the second axis is of length 3 elements
what they represent is up to your application. In the case of importing
from PIL, it is (height, width, rgb) (I think height and width get
swapped due to how memory is laid out in PIL vs numpy)
by default, numpy arrays are stored and worked with in C-array order, so
that array layout has the pixels together in memory as rgb triples.
Depending on what you are doing you may not want that. You can work with
them any way you want:
sub_region = arr[r1:r2, c1:c2, :]
all_red = arr[:,:,0]
but if you tend to work with all the red, or all the blue, etc, then it
might be easier to re-arrange it to:
(rgb, height, width)
If you google a bit, you'll find various notes about working with image
data in numpy.
HTH,
-Chris
--
Christopher Barker, Ph.D.
Oceanographer
Emergency Response Division
NOAA/NOS/OR&R (206) 526-6959 voice
7600 Sand Point Way NE (206) 526-6329 fax
Seattle, WA 98115 (206) 526-6317 main reception
Chris.Barker@noaa.gov
``` | 556 | 2,039 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2014-23 | longest | en | 0.890675 |
http://www.earthsciweek.org/forteachers/2012/ModelingEarthquakeWaves_Apr_cont.html | 1,394,231,987,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1393999651577/warc/CC-MAIN-20140305060731-00083-ip-10-183-142-35.ec2.internal.warc.gz | 309,670,320 | 1,994 | Modeling Earthquake Waves
Source: American Geophysical Union. Adapted with permission from Investigating Earth Systems, American Geosciences Institute.
An earthquake occurs when massive rock layers slide past each other. This motion makes enormous vibrations, which travel from the site of the earthquake in waves.
The waves (seismic waves) travel all the way through the Earth. Seismologists can record these waves when they reach Earth’s surface using seismographs.
Earthquakes generate three kinds of waves: Compressional waves (P waves) travel the fastest. Shear (S) waves travel more slowly than P waves. Surface waves are the slowest of the three.
Materials
• Flat, smooth surface
• Notebook and pen
• Safety goggles
Procedure
1. Work with a partner. Put on safety goggles before starting this activity. With a partner, stretch out a Slinky™ on the floor about as far as it can go without making a permanent bend in the metal.
2. Have one partner make waves by gathering several coils at the end of the Slinky™ and then releasing the coils, while still keeping hold of the end of the Slinky™. Observe the direction of wave movement relative to the Slinky™. Does it move in the same direction (parallel to the Slinky™) or in the opposite direction (perpendicular to the Slinky™)? Record your observations.
3. This kind of wave is called a P (primary) or compressional wave. (To compress means to squeeze together.) From your observations, explain why this is a good name for this wave. You may wish to use diagrams to illustrate your answer.
4. Stretch out the Slinky™ again. This time, have one partner make waves by moving the Slinky™ from side to side (left to right or right to left). Again observe the direction of wave movement, relative to the Slinky™. Does the wave move in the same direction (parallel to the Slinky™) or in the opposite direction (perpendicular to the Slinky™)? Record your observations.
5. This kind of wave is called a secondary or shear (S) wave. (To shear means to slide one thing sideways past another thing.) From your observations, explain why this is a good name for this wave.
6. Stretch out the Slinky™ a third time. This time, move one end of the Slinky™ up and down to generate a wave. This shows how the surface waves from earthquakes behave. What effect could this type of wave have on houses anchored to the Earth’s surface? Why is that? | 530 | 2,397 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2014-10 | longest | en | 0.89033 |
http://www.blogjava.net/ytl-zlq/archive/2011/05/06/349702.html | 1,660,405,242,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571959.66/warc/CC-MAIN-20220813142020-20220813172020-00405.warc.gz | 64,742,936 | 8,117 | # Ytl's Java Blog
BlogJava :: 首页 :: 新随笔 :: 联系 :: 聚合 :: 管理
# Binary search algorithm
Generally, to find a value in unsorted array, we should look through elements of an array one by one, until searched value is found. In case of searched value is absent from array, we go through all elements. In average, complexity of such an algorithm is proportional to the length of the array.
Situation changes significantly, when array is sorted. If we know it, random access capability can be utilized very efficientlyto find searched value quick. Cost of searching algorithm reduces to binary logarithm of the array length. For reference, log2(1 000 000) ≈ 20. It means, that in worst case, algorithm makes 20 steps to find a value in sorted array of a million elements or to say, that it doesn't present it the array.
## Algorithm
Algorithm is quite simple. It can be done either recursively or iteratively:
1. get the middle element;
2. if the middle element equals to the searched value, the algorithm stops;
3. otherwise, two cases are possible:
• searched value is less, than the middle element. In this case, go to the step 1 for the part of the array, before middle element.
• searched value is greater, than the middle element. In this case, go to the step 1 for the part of the array, after middle element.
Now we should define, when iterations should stop. First case is when searched element is found. Second one is when subarray has no elements. In this case, we can conclude, that searched value doesn't present in the array.
## Examples
Example 1. Find 6 in {-1, 5, 6, 18, 19, 25, 46, 78, 102, 114}.
Step 1 (middle element is 19 > 6): -1 5 6 18 19 25 46 78 102 114
Step 2 (middle element is 5 < 6): -1 5 6 18 19 25 46 78 102 114
Step 3 (middle element is 6 == 6): -1 5 6 18 19 25 46 78 102 114
Example 2. Find 103 in {-1, 5, 6, 18, 19, 25, 46, 78, 102, 114}.
Step 1 (middle element is 19 < 103): -1 5 6 18 19 25 46 78 102 114
Step 2 (middle element is 78 < 103): -1 5 6 18 19 25 46 78 102 114
Step 3 (middle element is 102 < 103): -1 5 6 18 19 25 46 78 102 114
Step 4 (middle element is 114 > 103): -1 5 6 18 19 25 46 78 102 114
Step 5 (searched value is absent): -1 5 6 18 19 25 46 78 102 114
## Complexity analysis
Huge advantage of this algorithm is that it's complexity depends on the array size logarithmically in worst case. In practice it means, that algorithm will do at most log2(n) iterations, which is a very small number even for big arrays. It can be proved very easily. Indeed, on every step the size of the searched part is reduced by half. Algorithm stops, when there are no elements to search in. Therefore, solving following inequality in whole numbers:
n / 2iterations > 0
resulting in
iterations <= log2(n).
It means, that binary search algorithm time complexity is O(log2(n)).
## Code snippets.
You can see recursive solution for Java and iterative for python below.
### Java
int binarySearch(int[] array, int value, int left, int right) {
if (left > right)
return -1;
int middle = left + (right-left) / 2;
if (array[middle] == value)
return middle;
if (array[middle] > value)
return binarySearch(array, value, left, middle - 1);
else
return binarySearch(array, value, middle + 1, right);
}
### Python
def biSearch(L,e,first,last):
if last - first < 2: return L[first] == e or L[last] == e
mid = first + (last-first)/2
if L[mid] ==e: return True
if L[mid]> e :
return biSearch(L,e,first,mid-1)
return biSearch(L,e,mid+1,last) | 1,142 | 3,593 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2022-33 | latest | en | 0.71355 |
http://exampleproblems.com/wiki/index.php?title=Main_Page&diff=44755&oldid=44727 | 1,660,768,679,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573104.24/warc/CC-MAIN-20220817183340-20220817213340-00131.warc.gz | 17,926,281 | 8,502 | # Difference between revisions of "Main Page"
Jump to navigation Jump to search
Welcome to ExampleProblems.com!
ExampleProblems.com is the first user-supported free content online database of higher mathematics example problems.
<a href="https://www.wirelessamberalerts.org"><img src="https://www.wirelessamberalerts.org/images/head_logo.gif" alt="Wireless Amber Alerts" border=0></a>
Please choose your subject to browse our database of examples:
MATH BOOKS New alternative section- Dream art Please add one problem! Contributing - each visitor's pleasure. Register, log in, and see the editing info below. Please leave me feedback and suggestions here: Feedback Must see - ASCII cows
Register - User number 500 and the winner of the 10,912 digit prime number is Jumbya!! The number is 3507!-1. Congratulations! -Todd
Register - User number 1000 and the winner of the 895,932 digit prime number is Fisc_Gen!! The number is $\displaystyle 2^{2976221}-1$ . Congratulations! -Todd
We are almost 5000 users strong and person number 2000 will win a free 1,521,561 digit prime number!!!! So Register now.
It's like asking why Beethoven's Ninth Symphony is beautiful. If you don't see why, someone can't tell you. I know numbers are beautiful. If they aren't beautiful, nothing is. - Paul Erdős
#### To Edit
Register. Log in. Click the 'edit' button at the top of any screen. Follow the example of the problems that are already up.
The math code works just like TeX. I suggest looking up symbols and functions in other problems and seeing how it's done there.
Here's a good list of symbols: http://meta.wikimedia.org/wiki/Help:Formula
Here is a bunch of TeX documentation: http://www.google.com/search?q=TeX+documentation
Thank You for contributing!
-Todd | 418 | 1,766 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 1} | 2.765625 | 3 | CC-MAIN-2022-33 | latest | en | 0.809808 |
http://www.helpwithnumbers.net/newstats1tips/boxplots | 1,631,870,407,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780055632.65/warc/CC-MAIN-20210917090202-20210917120202-00481.warc.gz | 90,098,643 | 8,925 | New Stats 1 Tips >
### Creating Box Plots
posted Mar 5, 2010, 2:43 PM by Prof Kiernan [ updated Feb 6, 2014, 5:59 AM ]
How to make a Box plot by hand: 1. Find the 5-number summary (smallest value, Q1, Median, Q3, & the largest value) 2. Construct a scale with values that include the minimum and maximum values 3. Construct a box extending from the first quartile to the third quartile and draw a line in the box at the median value 4. Draw lines extending outward from the box to the minimum and the maximum data values and you're done! How to make a Box Plot using the calculator:Put data into list #1: Press STAT Press ENTER (1:Edit) Put in variable under list 1 L1 Press ENTER (continue the last two steps until you have added all of the variables in your list)Get the 5 number summary: Press STAT Press CALC Press 1 (1: 1-Var Stat) Press 2nd Press L1 (or the name of the list that you have put your information in)Press ENTERPress Down arrow (the last 5 things in the list are the 5 number summary)To see the Box Plot:Press 2nd Press Y=Press ENTERPress ENTERPress DOWN ARROW Press RIGHT ARROW (till you highlight the box plot with the dots)Press ENTERPress ZOOMPRESS 9 (9: zoom stat)Voila! You have your box plot!To find the Q1, Median, and Q3 by hand:Put your data in a list from small to largeFind the median (the value in the middle of the dataset) if there are 2 data values average them together to get the medianFind the first and second Quartiles (Q1 & Q3) the same way:make 2 separate data sets and find the middle value of each new datasets.Find the Inter Quartile Range: Q3-Q1Apply the Interquartile range to the formula below to ensure there are no outliersIf there are outliers use the IQR (inter quartile range) to get the min and max values accepted as normal values Put any outliers on the plot as dots. | 491 | 1,869 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2021-39 | longest | en | 0.76803 |
https://www.physicsforums.com/threads/finding-the-electric-potential-of-point-charge.236383/ | 1,579,645,656,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250606226.29/warc/CC-MAIN-20200121222429-20200122011429-00480.warc.gz | 1,031,097,674 | 21,106 | Finding the electric potential of point charge
[SOLVED] Finding the electric potential of point charge
Homework Statement
I want to find the electric potential at a distance r from a point charge. Here's what I'm doing:
We have
$$V = - \int {E \cdot {\rm{d}}{\bf{l}}}$$
I find that
$$V = - \int_\infty ^r {k_e \frac{q}{{r'^2 }}{\bf{r}} \cdot {\rm{d}}{\bf{l}}} = - \int_\infty ^r {k_e \frac{q}{{r'^2 }}\cos \left( \phi \right){\rm{dl}}} = \int_\infty ^r {k_e \frac{q}{{r'^2 }}{\rm{dl}}} = - k_e \frac{q}{{r }}$$
I used that I integrate from infinity to r, so dl is pointing opposite to E. I get a negative potential, which is of course wrong for a point charge. Where's my error?
Related Advanced Physics Homework Help News on Phys.org
Doc Al
Mentor
E and dl (which is dr) both point outward, so $\cos\phi = 1$.
Why does dl point outward when I am integrating from infinity to r?
Doc Al
Mentor
Why does dl point outward when I am integrating from infinity to r?
Because r increases outward, regardless of the direction of integration. Note that:
$$\int_a^b f(x) dx = - \int_b^a f(x) dx$$
(I'm sure you're well aware of that!)
I don't get it, and it's elementary calculus.
So dl always points in the direction of increasing r?
Actually, if I use
$$\int_a^b f(x) dx = - \int_b^a f(x) dx$$
it works out perfectly, because I integrate from r to infinity, so the angle between E and dl is zero. But again, I cannot see why dl is also in the direction of E when we go from infinity to r.
Electric potential is the work it takes per unit charge to bring that charge from zero potential (usually infinity) to the point in question.
In a line integral, dl is never negative. The limits (the beginning and end of the path) define whether the work done is negative.
$$V=\frac{W_{_{by applied}}}{q_{_{test}}}$$
$$V=\frac{\int_a^b \vec{F}_{_{app}}\cdot\vec{dl}}{q_{_{test}}}$$
Since the applied force is opposite to coulumb repulsion, we can change that to:
$$V=\frac{\int_a^b -\vec{F}_{_{e}}\cdot\vec{dl}}{q_{_{test}}}$$
or
$$V=-q_{_{test}}\frac{\int_a^b \vec{E}\cdot\vec{dl}}{q_{_{test}}}$$
$$V=-\int_a^b \vec{E}\cdot\vec{dl}$$
In this case, we start at infinity and end at r, so the limits become
$$V=-\int_\infty^r \vec{E}\cdot\vec{dl}$$
And then, solving it:
$$V=-\int_\infty^r \left(E\hat{r}\right)\cdot \left(dr\hat{r}\right)$$
$$V=-\int_\infty^r E dr\left(\hat{r}\cdot\hat{r}\right)$$
$$V=-\int_\infty^r E dr$$
$$V=-\int_\infty^r k\frac{q}{r^2} dr$$
$$V=-\left[-k\frac{q}{r}\right]_\infty^r$$
$$V=\left[k\frac{q}{r}\right]-\left[k\frac{q}{\infty}\right]$$
Or simply
$$V=k\frac{q}{r}$$
-Ataman
So I should never care which way E and dl are pointing to eachother? The only thing that can make the minus-sign go away is if I reverse the limits?
Doc Al
Mentor
Let me see if I can clarify things a bit.
First, let's understand the meaning of the minus sign in the line integral definition of potential:
$$V = - \int {E \cdot {\rm{d}}{\bf{l}}}$$
If I move along the curve of the line integral in such a way that I oppose the electric field (and thus $\vec{E}\cdot\vec{d\ell}$ is negative), that represents a positive increase in potential: the external minus sign cancels the minus sign of the dot product to give a positive potential.
The tricky part is translating the line integral (and dot product) into an ordinary definite integral. In so doing you parameterize the curve in terms of the variable r. Note that dr is a positive change in the direction of increasing r. Now imagine integrating from infinity to some point r = a. In this case E points outward, in the direction of increasing r. As I move inward along r, I'm actually moving in the negative r direction thus my differential element of distance is -dr. So if I integrate in that direction I get a positive value for potential:
$$V = \int_{\infty}^{a}E(r) (-dr) = - \int_{\infty}^{a}E(r) dr$$
$$V = - \int_{\infty}^{a}\frac{kq}{r^2} dr = \frac{kq}{a}$$
Of course, if I reverse direction I get a negative potential difference.
(I hope this helps a bit rather than adds to the confusion.)
It actually helped.
In my book it says
$$V_a - V_b = \int_{a}^{b}Er \cdot dr$$
Now I want to test it all. Using your reasoning, I want to find the potential between two plates a and b separated by a distance d. Lets just say that E goes from b -> a so the zero-potential point is at a.
$$V_b - V_a = \int_{d}^{0}E(r) dr = -Ed$$
Now where is my error?
Doc Al
Mentor
Since E points from b to a, opposite to the direction of your integration, the field should actually be -E. (Assuming the field is a constant of magnitude E, the component in the direction of r is -E.)
So:
$$V_a - V_b = \int_{a}^{b}E(r) \cdot dr = \int_{a}^{b}(-E) dr = -Ed$$
Thus, taking V_a as 0:
$$V_a - V_b = - V_b = -Ed$$
$$V_b = Ed$$
Ok, now I am only confused about one thing.
Why do you not include a minus in
$$V = \int_{\infty}^{a}E(r) (-dr) = - \int_{\infty}^{a}E(r) dr$$?
I mean, the minus in $$V = - \int {E \cdot {\rm{d}}{\bf{l}}}$$ is not included in your above expression?
EDIT: I actually asked my professor about this question today, but he was kinda busy, but from what I could tell, I am not to worry about the direction of dl when using the formula
$$V = - \int {E \cdot {\rm{d}}{\bf{l}}}$$
but only worry about the direction of E with respect to the coordinate-system. Am I correct?
Last edited:
Doc Al
Mentor
I don't include the outside minus sign because the direction of integration is opposite to the direction of the field.
The only thing that I worry about is whether my end point is at a higher or lower potential than the starting point. You can tell that by the direction of the field.
Last edited:
Ok, now I am thoroughly confused.
You first remove the minus sign because we integrate in the opposite direction of the electric field, and then you dot then and get a minus. Isn't that doing the same thing twice? In the example with the parallel plates you just dotted and got the minus.
I really appreciate you guys helping me, and being so patient. But sometimes I wish consistency in physics-litterature could be explained a little better.
Doc Al
Mentor
let's get serious ;-)
To be honest with you, I never attempt to do the line integral to ordinary integral step "formally", I just know the appropriate sign that the answer must have. But I'm going to give it one more shot, as I think I might be confusing you (and myself) and would like to straighten it out.
EDIT: I actually asked my professor about this question today, but he was kinda busy, but from what I could tell, I am not to worry about the direction of dl when using the formula
$$V = - \int {E \cdot {\rm{d}}{\bf{l}}}$$
but only worry about the direction of E with respect to the coordinate-system. Am I correct?
I think I know what he means and that's a good way to approach it. Let's take a simple example and do it every which way to be sure we have it straight. Let's assume $\vec{E}$ has constant magnitude E (a positive number). From here on, we will let the coordinate system do the work.
Options:
(A) Electric field' points right
(B) Electric field' points left
(1) Coordinate system (variable r) goes left to right
(2) Coordinate system (variable r) goes right to left
Let's say point a is to the left of point b.
Case A-1:
Let's find Vb - Va.
$$V_b - V_a = -\int_{a}^{b}\vec{E}\cdot d\ell = -\int_{a}^{b}E dr = - E (b-a) = -Ed$$
Since the field is in the direction of the coordinate system (r), the dot product is positive. This answer makes sense.
If we integrated from b to a, we'd get -E(a-b) = +Ed. This answer makes sense, since that's calculating Va - Vb.
Case B-1:
$$V_b - V_a = -\int_{a}^{b}\vec{E}\cdot d\ell = -\int_{a}^{b}-E dr = E (b-a) = Ed$$
This makes sense, since now the field opposes the direction of r.
Case A-2:
$$V_b - V_a = -\int_{a}^{b}\vec{E}\cdot d\ell = -\int_{a}^{b}-E dr = E (b-a) = -Ed$$
The only difference here is that we changed the coordinates system to point the other way. The potential difference doesn't care about our coordinate choices, so the answer is the same as A-1. Note that the dot product is negative, since the field opposes the direction of r. But the integral of the distance is also negative, since now a > b.
Case B-2:
$$V_b - V_a = -\int_{a}^{b}\vec{E}\cdot d\ell = -\int_{a}^{b}E dr = - E (b-a) = Ed$$
Again, this should make sense: The field and coordinate system are aligned, but the distance integral is negative.
Hopefully this helps a bit (and I didn't make typos). Using this systematic approach (defining things with respect to the coordinate system) should work like a machine for all cases.
If you still have questions, ask away. Let's get this one done.
I like the approach in post #15 - very systematic, which is what I really need it this case. I agree and understand 100% what you have written. I will now apply it to the point-charge example, where we set the direction that points away from the positive point charge to be positive:
$$V_r - V_\infty = - \int_{\infty}^{r}\vec{E}\cdot d\vec{r} = \int_{\infty}^{r}Edr$$
Since E points in the positive direction and dr points in the negative, we get the above signs, and now I have my pickle. What part of my reasoning is wrong?
Last edited:
Doc Al
Mentor
Since E points in the positive direction and dr points in the negative, we get the above signs, and now I have my pickle. What part of my reasoning is wrong?
No, dr does not point in the negative direction. With this systematic approach, all that matters is whether E points in the same direction as the coordinate system (variable r). And it does, so:
$$V_r - V_\infty = - \int_{\infty}^{r}\vec{E}\cdot d\vec{r} = - \int_{\infty}^{r}Edr = - \int_{\infty}^{r}\frac{kq}{r^2}dr = \frac{kq}{r}$$
Ok, now I think I have it, but there is one final thing.
When finding the potential difference, should I do it with the limits:
$$V_{highest} - V_{lowest}$$?
Last edited:
Doc Al
Mentor
When finding the potential difference, should I do it with the limits:
$$V_{highest} - V_{lowest}$$?
It doesn't matter. Vb-Va means the potential of point b with respect to point a. If Vb > Va, you'll get a positive number; if Vb < Va, it will be negative. Obviously, Vb-Va = - (Va-Vb).
Often, when finding the potential difference all you care about is the magnitude.
Redbelly98
Staff Emeritus
Homework Helper
Pardon me for jumping in.
In a path integral, the vector dl points in the direction one is integrating. I.e. from the lower integration limit towards the upper limit.
dr will always point away from the origin, regardless of what the integration limits are.
So if the lower integration limit is $$\infty$$, and the integral's path is toward the origin, then dl = -dr.
Ok, I got it now. Thanks all for helping me.
So the bottom line is: Doing it our systematic way, all I have to care about is which way the electric field E is point with respect to the coordinate system, and nothing else?
Doc Al
Mentor
So the bottom line is: Doing it our systematic way, all I have to care about is which way the electric field E is point with respect to the coordinate system, and nothing else?
Right. Realize that this is entirely consistent with what Redbelly98 just said, which is what I attempted to explain earlier. (I did a rather poor job of it, I'm afraid.) Stick to this method and you'll be fine.
Very nice, thanks for taking the time to clear things up. And the examples in post #15 are just excellent. | 3,271 | 11,484 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2020-05 | longest | en | 0.861952 |
https://www.gkrecall.com/2020/06/data-interpretation-radar-di-fir-sbi-po.html | 1,611,183,331,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703522133.33/warc/CC-MAIN-20210120213234-20210121003234-00128.warc.gz | 806,584,681 | 41,291 | # gkrecall
We brings you daily Current affairs, daily Current Affairs Quiz, weekly Current Affairs, weekly Current Affairs Quiz to enhance your preparation for upcoming exams. We also provide articles related to quant, reasoning, English, ssc etc .
## Breaking
SBI Clerk and SBI PO Prelims exam are going to be held in the upcoming months. We have already provided you with the PDFs of many topics of Quantitative Aptitude like Simplification/Approximation, Number Series,datainterpretation.important Arithmetic questions to prepare for SBI Clerk and PO Pre exam. Practicing these questions will help you to know about the level of the questions. To increase your speed and accuracy, enhance your calculations.
START TEST
(1 - 5)The pie-chart shows sources of income for an NGO. The total income is Rs.40 crore. The bar chart gives the expenditure incurred on various items A – Food for poor, B – Education to illiterate, C – Mid-day deal programme, D – General Expenses, E – Eye Camp expenses, F – Integrated Street Children Programme.(in Crores)
Total Expenditure = Rs.39 crore
1.What Percentage of money is saved by the NGO?
A. 1.5%
B. 2.5%
C. 1.8%
D. 3.5%
2.If the industrialist stops donation and the expenditure pattern remains the same, then what will be the decrease in money spent for mid-day meal programme?
A. 0.77 crore
B. 0.87 crore
C. 0.93 crore
D. 0.94 crore
3.What is the ratio of expenditure on food for poor and mid-day meal programmes together to that of grant from central government?
A. 1:3
B. 2:3
C. 4:5
D. 5:4
4.The “General expenses” is how many times “income from investment”?
A. 0.75
B. 0.25
C. 1.25
D. 2
5.Suppose in the next year, grant from central government increase by 10%, foreign contribution decreases by 10% and other income amounts remain same. If the expense pattern remains same, what is the percent increase in “Food for Poor”?
A. 2%
B. 3%
C. 1%
D. 4%
(6 - 10)Study the following Radar Graph and answer the given questions
Number of new countries registered (in thousand) in four countries in various years.
6.In which year is the average of the number of companies registered the maximum?
A. 2008
B. 2010
C. 2009
D. 2011
7.The number of companies registered in 2015 is what percentage more or less than the number of companies registered in the year 2008?
A. 44.44%
B. 43.75%
C. 22.22%
D. 11.11%
8.what is the ratio of the number of companies registered in Japan to that in China during 2008 to 2015?
A. 39:40
B. 23:24
C. 40:41
D. 24:23
9.The total number of companies registered in France is how much more or less than the total number of companies registered in China?
A. 34.14%
B. 33.13%
C. 29.16%
D. 11.11%
10.In which country is the average number of companies registered the maximum?
A. Germany
B. Japan
C. China
D. France
SOLUTIONS:
Explanation :
Total income = Rs.40 crore
Total Expenditure = Rs.39 crore
% of money is saved by the NGO =
[(40-39)/40]*100 = 2.5%
Explanation :
Decrease in Income = 15% of 40 crore
New Income = 85% of 40 crore = 34 crore
percentage of expenditure on midday meal = (5/39)*100 = 12.8%
New Expenditure after gradual decrease is expenditure with respect to income = (39/40)*34 = 33.15crore
Expenditure on midday meal = (12.8/100)*33.15 = 4.24 crore
Decrease in Expenditure on midday meal = 5 – 4.24 = 0.77 crore
Explanation :
Expenditure on food for poor = 10 crore
Expenditure on food for mid-day meal programme = 5 crore
Total Expenditure = 15 crore
Grant from central Government = 30% of 40 crore = 12 crore
Ratio = 15:12 => 5:4
Explanation :
General Expenses = 3 crore
Income from Investment = 10% of 40 crore = 4 crore
“General expenses” is x times “income from investment” then
3 = 4x
x = 3/4 = 0.75
Explanation :
Grant from central government = 30% of 40 crore = 12 crore
After increase 10% = 110% of 12crore = 13.2 crore
foreign contribution = 20% of 40 crore = 8 crore
After 10% decrease in foreign contribution = 90% of 8 crore = 7.2 crore
Total increase in donation = [13.2 + 7.2] – [12 + 8] = 0.4 crore
Gradual increase in expenditure = (39/40)*40.4 = 39.39 crore
Gradual increase in A =(10/39)*39.4 = 10.10 crore
% increase = (0.10/10)*100 = 1%
Explanation :
Average of the number of companies registered in 2008 = [5 + 20 + 15 + 40]*1000/4 = 20000
Average of the number of companies registered in 2009 = [10 + 30 + 40 + 35]*1000/4 = 28750
Average of the number of companies registered in 2010 = [5 + 25 + 15 + 20]*1000/4 = 16250
Average of the number of companies registered in 2011 = [25 + 20 + 20 + 35]*1000/4 = 25000
Explanation :
Number of companies registered in 2015 = [20 + 20 + 40 + 35]*1000 = 115000
Number of companies registered in 2008 = [5 + 20 + 15 + 40]*1000 = 80000
% = [(115000 – 80000)/ 80000]*100 = 43.75% more
Explanation :
Number of companies registered in Japan during 2008 to 2015 = 230*1000 = 230000
Number of companies registered in China during 2008 to 2015 = 240*1000 = 240000
Ratio = 23:24 | 1,510 | 4,935 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2021-04 | latest | en | 0.887229 |
https://e-eduanswers.com/mathematics/question2794634 | 1,628,182,505,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046156141.29/warc/CC-MAIN-20210805161906-20210805191906-00643.warc.gz | 229,306,778 | 18,008 | Ataxi charges a flat rate of \$3.00, plus an additional \$0.50 per mile. carl will only take the taxi home if the cost is under \$10, otherwise
, 18.10.2019 06:30, vasquez8518
# Ataxi charges a flat rate of \$3.00, plus an additional \$0.50 per mile. carl will only take the taxi home if the cost is under \$10, otherwise he will take a bus. carl is 15 miles from home. explain how to write and solve an inequality to determine if carl will take the taxi or a bus.
### Other questions on the subject: Mathematics
Mathematics, 21.06.2019 15:00, donmak3833
Which statement it’s true ? 9.07> 9.008 3.37< 3.368 50.1< 5.01 123.466> 132.465
Mathematics, 21.06.2019 17:40, Arealbot
Find the value of ax 4 ; a = 2, x = 1. select one: a. 2 b. 4 c. 1 d. 8 | 266 | 750 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2021-31 | latest | en | 0.716954 |
https://www.gcsescience.com/pfm29.htm | 1,713,008,591,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816734.69/warc/CC-MAIN-20240413114018-20240413144018-00100.warc.gz | 767,250,410 | 4,122 | gcsescience.com 29 gcsescience.com
Forces and Motion
The Stopping Distance of a Car - Velocity.
The total stopping distance = thinking distance + braking distance.
Both the thinking distance and the braking distance
are changed as the velocity of a car changes.
See also the calculation of the force need to stop a moving car
using kinetic energy or momentum.
How does Velocity affect the Braking Distance of a Car?
The braking distance of a car increases as the velocity increases.
The two calculations below show how
doubling the velocity changes the braking distance of a car.
Q1
. A car is moving with a velocity (speed) of 10 m/s.
When the brakes are applied the car slows down
and has a constant negative acceleration of -2 m/s2.
What is its braking distance?
A1. Find how much time the car took to stop.
Then find the average velocity of the car.
Then calculate the braking distance.
Use a = (v-u) ÷ t
to find the time (
t) for how long it takes the car to stop.
a = - 2
v =
u = 10
t = (v-u) ÷ a
t = (0 - 10) ÷ -2
t = 10 ÷ 2
t = 5 seconds.
For an object that has a constant acceleration,
the average velocity = (initial velocity + final velocity) ÷ 2
= (10 + 0) ÷ 2
= 5 m/s.
As velocity = distance ÷ time
then distance = velocity x time
The braking distance of the car
= 5 x 5
= 25 m.
Q2
. The same car is now moving with twice the velocity at 20 m/s.
When the brakes are applied,
the car has the same constant negative acceleration of -2 m/s2.
What is its braking distance?
A2. Use the same method as above, a = (v-u) ÷ t
a = -2
v =
u = 20
t = (v-u) ÷ a
t = (0 - 20) ÷ -2
t = 20 ÷ 2
t = 10 seconds.
The average velocity = (initial velocity + final velocity) ÷ 2
= (20 + 0) ÷
= 10 m/s.
Since velocity = distance ÷ time
distance = velocity x time
The braking distance of the car
= 10 x 10
= 100 m.
Notice that doubling the velocity of the car from 10 to 20 m/s
has more than doubled the braking distance.
In fact the braking distance goes up x4 when the velocity goes up x2.
This is because of
the effect of velocity on the kinetic energy of the car
.
gcsescience.com Physics Quiz Index Force Quiz gcsescience.com
Copyright © 2015 gcsescience.com. All Rights Reserved. | 659 | 2,321 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2024-18 | latest | en | 0.891344 |
tondofulfillment.com | 1,726,130,514,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651440.11/warc/CC-MAIN-20240912074814-20240912104814-00070.warc.gz | 550,528,676 | 24,026 | # What is Dimensional Weight?
Dimensional weight, also known as "dim weight," is a pricing technique used by shipping carriers to account for the size of a package, rather than just its weight. This can be especially important when shipping large, lightweight items that take up a lot of space in the carrier's vehicle or cargo hold.
Here's how it works: shipping carriers have a set "dimensional factor" that they use to calculate the dimensional weight of a package. This factor is typically based on the size of the package and the type of shipping service being used.
To calculate dimensional weight, the carrier takes the package's length, width, and height, and multiplies these measurements by the dimensional factor. The resulting number is the package's dimensional weight, and it is used to determine the shipping cost for the package.
For example, let's say you have a package that is 18 inches long, 12 inches wide, and 8 inches tall, and you are using a shipping service with a dimensional factor of 166. The package's dimensional weight would be calculated as follows:
Dimensional weight = (length x width x height) / dimensional factor
Dimensional weight = (18 x 12 x 8) / 166
Dimensional weight = 2,208 / 166
Dimensional weight = 13.3 pounds
In this example, the package's actual weight is 10 pounds, but because its dimensional weight is higher, the carrier would charge you for shipping based on the dimensional weight of 13.3 pounds.
It's important to note that carriers have different dimensional factors, so the dimensional weight of a package can vary depending on the carrier and the shipping service being used. Additionally, some carriers have minimum weight requirements for packages, so if the actual weight of a package is below the minimum, the carrier will charge for the minimum weight instead of the actual weight.
Understanding dimensional weight is important for businesses that ship packages, as it can affect the cost of shipping. By being aware of dimensional weight, businesses can take steps to optimize the size and weight of their packages to reduce shipping costs. This can include using smaller boxes or packaging materials, or consolidating multiple items into a single package.
In summary, dimensional weight is a pricing technique used by shipping carriers to account for the size of a package, rather than just its weight. It is calculated by multiplying the package's length, width, and height by the carrier's dimensional factor, and is used to determine the shipping cost for the package. Understanding and optimizing dimensional weight can help businesses reduce their shipping costs and improve their bottom line. | 531 | 2,677 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2024-38 | latest | en | 0.94386 |
https://orion.readthedocs.io/en/v0.1.16/_modules/orion/algo/tpe.html | 1,656,398,339,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103355949.26/warc/CC-MAIN-20220628050721-20220628080721-00002.warc.gz | 502,930,767 | 19,864 | # Source code for orion.algo.tpe
```# -*- coding: utf-8 -*-
"""
Tree-structured Parzen Estimator Approach
=========================================
"""
import logging
import numpy
from scipy.stats import norm
from orion.algo.base import BaseAlgorithm
from orion.core.utils.points import flatten_dims, regroup_dims
logger = logging.getLogger(__name__)
[docs]def compute_max_ei_point(points, below_likelis, above_likelis):
"""Compute ei among points based on their log likelihood and return the point with max ei.
:param points: list of point with real values.
:param below_likelis: list of log likelihood for each point in the good GMM.
:param above_likelis: list of log likelihood for each point in the bad GMM.
"""
max_ei = -numpy.inf
point_index = 0
for i, (lik_b, lik_a) in enumerate(zip(below_likelis, above_likelis)):
ei = lik_b - lik_a
if ei > max_ei:
max_ei = ei
point_index = i
return points[point_index]
[docs]def ramp_up_weights(total_num, flat_num, equal_weight):
:param total_num: total number of observed trials.
:param flat_num: the number of the most recent trials which
get the full weight where the others will be applied with a linear ramp
from 0 to 1.0. It will only take effect if equal_weight is False.
:param equal_weight: whether all the observed trails share the same weights.
"""
if total_num < flat_num or equal_weight:
return numpy.ones(total_num)
ramp_weights = numpy.linspace(1.0 / total_num, 1.0, num=total_num - flat_num)
flat_weights = numpy.ones(flat_num)
return numpy.concatenate([ramp_weights, flat_weights])
# pylint:disable=assignment-from-no-return
mus, low, high, prior_weight=1.0, equal_weight=False, flat_num=25
):
"""Return the sorted mus, the corresponding sigmas and weights with adaptive kernel estimator.
This adaptive parzen window estimator is based on the original papers and also refer the use of
prior mean in `this implementation
<https://github.com/hyperopt/hyperopt/blob/40bd617a0ca47e09368fb65919464ba0bfa85962/hyperopt/tpe.py#L400-L468>`_.
:param mus: list of real values for observed mus.
:param low: real value for lower bound of points.
:param high: real value for upper bound of points.
:param prior_weight: real value for the weight of the prior mean.
:param equal_weight: bool value indicating if all points with equal weights.
:param flat_num: int value indicating the number of the most recent trials which
get the full weight where the others will be applied with a linear ramp
from 0 to 1.0. It will only take effect if equal_weight is False.
"""
mus = numpy.asarray(mus)
prior_mu = (low + high) * 0.5
prior_sigma = (high - low) * 1.0
size = len(mus)
if size > 1:
order = numpy.argsort(mus)
sorted_mus = mus[order]
prior_mu_pos = numpy.searchsorted(sorted_mus, prior_mu)
weights = ramp_up_weights(size, flat_num, equal_weight)
mixture_mus = numpy.zeros(size + 1)
mixture_mus[:prior_mu_pos] = sorted_mus[:prior_mu_pos]
mixture_mus[prior_mu_pos] = prior_mu
mixture_mus[prior_mu_pos + 1 :] = sorted_mus[prior_mu_pos:]
mixture_weights = numpy.ones(size + 1)
mixture_weights[:prior_mu_pos] = weights[:prior_mu_pos]
mixture_weights[prior_mu_pos] = prior_weight
mixture_weights[prior_mu_pos + 1 :] = weights[prior_mu_pos:]
sigmas = numpy.ones(size + 1)
sigmas[0] = mixture_mus[1] - mixture_mus[0]
sigmas[-1] = mixture_mus[-1] - mixture_mus[-2]
sigmas[1:-1] = numpy.maximum(
(mixture_mus[1:-1] - mixture_mus[0:-2]),
(mixture_mus[2:] - mixture_mus[1:-1]),
)
sigmas = numpy.clip(
sigmas, prior_sigma / max(10, numpy.sqrt(size)), prior_sigma
)
else:
if prior_mu < mus[0]:
mixture_mus = numpy.array([prior_mu, mus[0]])
sigmas = numpy.array([prior_sigma, prior_sigma * 0.5])
mixture_weights = numpy.array([prior_weight, 1.0])
else:
mixture_mus = numpy.array([mus[0], prior_mu])
sigmas = numpy.array([prior_sigma * 0.5, prior_sigma])
mixture_weights = numpy.array([1.0, prior_weight])
weights = mixture_weights / mixture_weights.sum()
return mixture_mus, sigmas, weights
[docs]class TPE(BaseAlgorithm):
"""Tree-structured Parzen Estimator (TPE) algorithm is one of Sequential Model-Based
Global Optimization (SMBO) algorithms, which will build models to propose new points based
on the historical observed trials.
Instead of modeling p(y|x) like other SMBO algorithms, TPE models p(x|y) and p(y),
and p(x|y) is modeled by transforming that generative process, replacing the distributions of
the configuration prior with non-parametric densities.
The TPE defines p(x|y) using two such densities l(x) and g(x) while l(x) is distribution of
good points and g(x) is the distribution of bad points. New point candidates will be sampled
with l(x) and Expected Improvement (EI) optimization scheme will be used to find the most
promising point among the candidates.
- `Algorithms for Hyper-Parameter Optimization
<https://papers.nips.cc/paper/4443-algorithms-for-hyper-parameter-optimization.pdf>`_
- `Making a Science of Model Search: Hyperparameter Optimizationin Hundreds of Dimensions
for Vision Architectures <http://proceedings.mlr.press/v28/bergstra13.pdf>`_
Parameters
----------
space: `orion.algo.space.Space`
Optimisation space with priors for each dimension.
seed: None, int or sequence of int
Seed to sample initial points and candidates points.
Default: ``None``
n_initial_points: int
Number of initial points randomly sampled. If new points
are requested and less than `n_initial_points` are observed,
the next points will also be sampled randomly instead of being
sampled from the parzen estimators.
Default: ``20``
n_ei_candidates: int
Number of candidates points sampled for ei compute.
Default: ``24``
gamma: real
Ratio to split the observed trials into good and bad distributions.
Default: ``0.25``
equal_weight: bool
True to set equal weights for observed points.
Default: ``False``
prior_weight: int
The weight given to the prior point of the input space.
Default: ``1.0``
full_weight_num: int
The number of the most recent trials which get the full weight where the others will be
applied with a linear ramp from 0 to 1.0. It will only take effect if equal_weight
is False.
"""
requires_type = None
requires_dist = "linear"
requires_shape = "flattened"
# pylint:disable=too-many-arguments
def __init__(
self,
space,
seed=None,
n_initial_points=20,
n_ei_candidates=24,
gamma=0.25,
equal_weight=False,
prior_weight=1.0,
full_weight_num=25,
):
if n_initial_points < 2:
n_initial_points = 2
logger.warning(
"n_initial_points %s is not valid, set n_initial_points = 2",
str(n_initial_points),
)
if n_ei_candidates < 1:
n_ei_candidates = 1
logger.warning(
"n_ei_candidates %s is not valid, set n_ei_candidates = 1",
str(n_ei_candidates),
)
super(TPE, self).__init__(
space,
seed=seed,
n_initial_points=n_initial_points,
n_ei_candidates=n_ei_candidates,
gamma=gamma,
equal_weight=equal_weight,
prior_weight=prior_weight,
full_weight_num=full_weight_num,
)
@property
def space(self):
"""Return transformed space of TPE"""
return self._space
@space.setter
def space(self, space):
"""Set the space of TPE and initialize it"""
self._space = space
self._initialize()
def _initialize(self):
"""Initialize TPE once the space is transformed"""
for dimension in self.space.values():
if dimension.type != "fidelity" and dimension.prior_name not in [
"uniform",
"reciprocal",
"int_uniform",
"int_reciprocal",
"choices",
]:
raise ValueError(
"TPE now only supports uniform, loguniform, uniform discrete "
f"and choices as prior: {dimension.prior_name}"
)
shape = dimension.shape
if shape and len(shape) != 1:
raise ValueError("TPE now only supports 1D shape.")
[docs] def seed_rng(self, seed):
"""Seed the state of the random number generator.
:param seed: Integer seed for the random number generator.
"""
self.rng = numpy.random.RandomState(seed)
@property
def state_dict(self):
"""Return a state dict that can be used to reset the state of the algorithm."""
_state_dict = super(TPE, self).state_dict
_state_dict["rng_state"] = self.rng.get_state()
_state_dict["seed"] = self.seed
return _state_dict
[docs] def set_state(self, state_dict):
"""Reset the state of the algorithm based on the given state_dict
:param state_dict: Dictionary representing state of an algorithm
"""
super(TPE, self).set_state(state_dict)
self.seed_rng(state_dict["seed"])
self.rng.set_state(state_dict["rng_state"])
[docs] def suggest(self, num=None):
"""Suggest a `num` of new sets of parameters. Randomly draw samples
from the import space and return them.
Parameters
----------
num: int, optional
Number of points to sample. If None, TPE will sample all random points at once, or a
single point if it is at the Bayesian Optimization stage.
:param num: how many sets to be suggested.
.. note:: New parameters must be compliant with the problem's domain
`orion.algo.space.Space`.
"""
# Only sample up to `n_initial_points` and after that only sample one at a time.
num = min(num, max(self.n_initial_points - self.n_suggested, 1))
samples = []
candidates = []
while len(samples) < num and self.n_suggested < self.space.cardinality:
if candidates:
candidate = candidates.pop(0)
if candidate:
self.register(candidate)
samples.append(candidate)
elif self.n_observed < self.n_initial_points:
candidates = self._suggest_random(num)
else:
candidates = self._suggest_bo(max(num - len(samples), 0))
if not candidates:
break
if samples:
return samples
return None
def _suggest(self, num, function):
points = []
ids = set(self._trials_info.keys())
while len(points) < num:
for candidate in function(num - len(points)):
candidate_id = self.get_id(candidate)
if candidate_id not in ids:
points.append(candidate)
if len(ids) >= self.space.cardinality:
return points
return points
def _suggest_random(self, num):
def sample(num):
return self.space.sample(
num, seed=tuple(self.rng.randint(0, 1000000, size=3))
)
return self._suggest(num, sample)
def _suggest_bo(self, num):
def suggest_bo(num):
return [self._suggest_one_bo() for _ in range(num)]
return self._suggest(num, suggest_bo)
def _suggest_one_bo(self):
point = []
below_points, above_points = self.split_trials()
below_points = list(map(list, zip(*below_points)))
above_points = list(map(list, zip(*above_points)))
idx = 0
for i, dimension in enumerate(self.space.values()):
shape = dimension.shape
if not shape:
shape = (1,)
if dimension.type == "real":
dim_samples = self._sample_real_dimension(
dimension,
shape[0],
below_points[idx : idx + shape[0]],
above_points[idx : idx + shape[0]],
)
elif dimension.type == "integer" and dimension.prior_name in [
"int_uniform",
"int_reciprocal",
]:
dim_samples = self.sample_one_dimension(
dimension,
shape[0],
below_points[idx : idx + shape[0]],
above_points[idx : idx + shape[0]],
self._sample_int_point,
)
elif dimension.type == "categorical" and dimension.prior_name == "choices":
dim_samples = self.sample_one_dimension(
dimension,
shape[0],
below_points[idx : idx + shape[0]],
above_points[idx : idx + shape[0]],
self._sample_categorical_point,
)
elif dimension.type == "fidelity":
# fidelity dimension
dim_samples = [point[i] for point in self.space.sample(1)]
else:
raise NotImplementedError()
idx += shape[0]
point += dim_samples
return self.format_point(point)
# pylint:disable=no-self-use
[docs] def sample_one_dimension(
self, dimension, shape_size, below_points, above_points, sampler
):
"""Sample values for a dimension
:param dimension: Dimension.
:param shape_size: 1D Shape Size of the Real Dimension.
:param below_points: good points with shape (m, n), m=shape_size.
:param above_points: bad points with shape (m, n), m=shape_size.
:param sampler: method to sample one value for upon the dimension.
"""
points = []
for j in range(shape_size):
new_point = sampler(dimension, below_points[j], above_points[j])
points.append(new_point)
return points
def _sample_real_dimension(self, dimension, shape_size, below_points, above_points):
"""Sample values for real dimension"""
if dimension.prior_name in ["uniform", "reciprocal"]:
return self.sample_one_dimension(
dimension,
shape_size,
below_points,
above_points,
self._sample_real_point,
)
else:
raise NotImplementedError()
def _sample_loguniform_real_point(self, dimension, below_points, above_points):
"""Sample one value for real dimension in a loguniform way"""
return self._sample_real_point(
dimension, below_points, above_points, is_log=True
)
def _sample_real_point(self, dimension, below_points, above_points, is_log=False):
"""Sample one value for real dimension based on the observed good and bad points"""
low, high = dimension.interval()
if is_log:
low = numpy.log(low)
high = numpy.log(high)
below_points = numpy.log(below_points)
above_points = numpy.log(above_points)
below_points,
low,
high,
self.prior_weight,
self.equal_weight,
flat_num=self.full_weight_num,
)
above_points,
low,
high,
self.prior_weight,
self.equal_weight,
flat_num=self.full_weight_num,
)
gmm_sampler_below = GMMSampler(
self, below_mus, below_sigmas, low, high, below_weights
)
gmm_sampler_above = GMMSampler(
self, above_mus, above_sigmas, low, high, above_weights
)
candidate_points = gmm_sampler_below.sample(self.n_ei_candidates)
lik_blow = gmm_sampler_below.get_loglikelis(candidate_points)
lik_above = gmm_sampler_above.get_loglikelis(candidate_points)
new_point = compute_max_ei_point(candidate_points, lik_blow, lik_above)
if is_log:
new_point = numpy.exp(new_point)
return new_point
def _sample_int_point(self, dimension, below_points, above_points):
"""Sample one value for integer dimension based on the observed good and bad points"""
low, high = dimension.interval()
choices = range(low, high + 1)
below_points = numpy.array(below_points).astype(int) - low
above_points = numpy.array(above_points).astype(int) - low
sampler_below = CategoricalSampler(self, below_points, choices)
candidate_points = sampler_below.sample(self.n_ei_candidates)
sampler_above = CategoricalSampler(self, above_points, choices)
lik_below = sampler_below.get_loglikelis(candidate_points)
lik_above = sampler_above.get_loglikelis(candidate_points)
new_point = compute_max_ei_point(candidate_points, lik_below, lik_above)
new_point = new_point + low
return new_point
def _sample_categorical_point(self, dimension, below_points, above_points):
"""Sample one value for categorical dimension based on the observed good and bad points"""
choices = dimension.interval()
below_points = [choices.index(point) for point in below_points]
above_points = [choices.index(point) for point in above_points]
sampler_below = CategoricalSampler(self, below_points, choices)
candidate_points = sampler_below.sample(self.n_ei_candidates)
sampler_above = CategoricalSampler(self, above_points, choices)
lik_below = sampler_below.get_loglikelis(candidate_points)
lik_above = sampler_above.get_loglikelis(candidate_points)
new_point_index = compute_max_ei_point(candidate_points, lik_below, lik_above)
new_point = choices[new_point_index]
return new_point
[docs] def split_trials(self):
"""Split the observed trials into good and bad ones based on the ratio `gamma``"""
sorted_trials = sorted(
(point for point in self._trials_info.values() if point[1] is not None),
key=lambda x: x[1]["objective"],
)
sorted_points = [list(points) for points, results in sorted_trials]
split_index = int(numpy.ceil(self.gamma * len(sorted_points)))
below = sorted_points[:split_index]
above = sorted_points[split_index:]
return below, above
[docs] def observe(self, points, results):
"""Observe evaluation `results` corresponding to list of `points` in
space.
A simple random sampler though does not take anything into account.
"""
super(TPE, self).observe(points, results)
[docs]class GMMSampler:
"""Gaussian Mixture Model Sampler for TPE algorithm
Parameters
----------
tpe: `TPE` algorithm
The tpe algorithm object which this sampler will be part of.
mus: list
mus for each Gaussian components in the GMM.
Default: ``None``
sigmas: list
sigmas for each Gaussian components in the GMM.
low: real
Lower bound of the sampled points.
high: real
Upper bound of the sampled points.
weights: list
Weights for each Gaussian components in the GMM
Default: ``None``
"""
def __init__(self, tpe, mus, sigmas, low, high, weights=None):
self.tpe = tpe
self.mus = mus
self.sigmas = sigmas
self.low = low
self.high = high
self.weights = weights if weights is not None else len(mus) * [1.0 / len(mus)]
self.pdfs = []
self._build_mixture()
def _build_mixture(self):
"""Build the Gaussian components in the GMM"""
for mu, sigma in zip(self.mus, self.sigmas):
self.pdfs.append(norm(mu, sigma))
[docs] def sample(self, num=1, attempts=10):
"""Sample required number of points"""
point = []
for _ in range(num):
pdf = numpy.argmax(self.tpe.rng.multinomial(1, self.weights))
new_points = list(
self.pdfs[pdf].rvs(size=attempts, random_state=self.tpe.rng)
)
while True:
if not new_points:
raise RuntimeError(
f"Failed to sample in interval ({self.low}, {self.high})"
)
pt = new_points.pop(0)
if self.low <= pt <= self.high:
point.append(pt)
break
return point
[docs] def get_loglikelis(self, points):
"""Return the log likelihood for the points"""
points = numpy.array(points)
weight_likelis = [
numpy.log(self.weights[i] * pdf.pdf(points))
for i, pdf in enumerate(self.pdfs)
]
weight_likelis = numpy.array(weight_likelis)
# (num_weights, num_points) => (num_points, num_weights)
weight_likelis = weight_likelis.transpose()
# log-sum-exp trick
max_likeli = numpy.nanmax(weight_likelis, axis=1)
point_likeli = max_likeli + numpy.log(
numpy.nansum(numpy.exp(weight_likelis - max_likeli[:, None]), axis=1)
)
return point_likeli
[docs]class CategoricalSampler:
"""Categorical Sampler for discrete integer and categorical choices
Parameters
----------
tpe: `TPE` algorithm
The tpe algorithm object which this sampler will be part of.
observations: list
Observed values in the dimension
choices: list
Candidate values for the dimension
"""
def __init__(self, tpe, observations, choices):
self.tpe = tpe
self.obs = observations
self.choices = choices
self._build_multinomial_weights()
def _build_multinomial_weights(self):
"""Build weights for categorical distribution based on observations"""
weights_obs = ramp_up_weights(
len(self.obs), self.tpe.full_weight_num, self.tpe.equal_weight
)
counts_obs = numpy.bincount(
self.obs, minlength=len(self.choices), weights=weights_obs
)
counts_obs = counts_obs + self.tpe.prior_weight
self.weights = counts_obs / counts_obs.sum()
[docs] def sample(self, num=1):
"""Sample required number of points"""
samples = self.tpe.rng.multinomial(n=1, pvals=self.weights, size=num)
assert samples.shape == (num,) + (len(self.weights),)
samples_index = samples.argmax(-1)
assert samples_index.shape == (num,)
return samples_index
[docs] def get_loglikelis(self, points):
"""Return the log likelihood for the points"""
return numpy.log(numpy.asarray(self.weights)[points])
``` | 4,697 | 18,849 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2022-27 | latest | en | 0.622933 |
http://gmatclub.com/forum/manhattan-gmat-cats-6-including-free-sample-or-6-on-top-of-106268.html#p833746 | 1,477,370,016,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719877.27/warc/CC-MAIN-20161020183839-00239-ip-10-171-6-4.ec2.internal.warc.gz | 109,071,636 | 47,950 | Find all School-related info fast with the new School-Specific MBA Forum
It is currently 24 Oct 2016, 21:33
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
Events & Promotions
Events & Promotions in June
Open Detailed Calendar
Manhattan GMAT CATs- 6 including free sample? or 6 on top of
Author Message
Intern
Joined: 13 Nov 2010
Posts: 23
Followers: 0
Kudos [?]: 1 [0], given: 1
Manhattan GMAT CATs- 6 including free sample? or 6 on top of [#permalink]
Show Tags
14 Dec 2010, 12:20
Took free sample a bit ago and now have the prep books. It states I get 6 CAT exams and I haven't used any of these yet. Do I have 5 tests or 6 remaining? In order words, did the free sample count as one of the 6 offered by MGMAT?
GMAT Forum Moderator
Status: Accepting donations for the mohater MBA debt repayment fund
Joined: 05 Feb 2008
Posts: 1866
Location: United States
Concentration: Operations, Finance
Schools: Ross '14 (M)
GMAT 1: 610 Q0 V0
GMAT 2: 710 Q48 V38
GPA: 3.54
WE: Accounting (Manufacturing)
Followers: 58
Kudos [?]: 751 [0], given: 234
Re: Manhattan GMAT CATs- 6 including free sample? or 6 on top of [#permalink]
Show Tags
14 Dec 2010, 12:51
Six including the free one/sample.
That being said, the MGMAT folks WILL reset the tests for you after you finish the six. There is a chance of repeat questions after resetting (MGMAT guarantees no repeats in the tests as you progress 1-6).
_________________
Strategy Discussion Thread | Strategy Master | GMAT Debrief| Please discuss strategies in discussion thread. Master thread will be updated accordingly. | GC Member Write Ups
GMAT Club Premium Membership - big benefits and savings
Re: Manhattan GMAT CATs- 6 including free sample? or 6 on top of [#permalink] 14 Dec 2010, 12:51
Similar topics Replies Last post
Similar
Topics:
2 The lowest cost route to 6 Manhattan GMAT CATs 0 10 Feb 2016, 09:29
1 Manhattan GMAT 6 exams for 10\$ 2 16 Apr 2013, 18:50
3 MAnhattan 6 CATs gmat practice tests 3 18 Jul 2011, 11:56
3 How to get the 6 free Manhattan GMAT CATs? 6 19 Oct 2009, 23:58
Has anyone done the 6 adaptive CATs in Manhattan GMAT? 1 21 Nov 2006, 23:30
Display posts from previous: Sort by
Manhattan GMAT CATs- 6 including free sample? or 6 on top of
Moderators: HiLine, WaterFlowsUp
Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 820 | 2,903 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2016-44 | latest | en | 0.892857 |
https://calculator.academy/thermal-expansion-calculator/ | 1,686,446,156,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224646652.16/warc/CC-MAIN-20230610233020-20230611023020-00490.warc.gz | 185,773,036 | 55,138 | Enter the thermal expansion coefficient, length, and change in temperature into the calculator to determine the change in length from thermal expansion.
## Thermal Expansion Formula
The following formula describes the relationship of thermal expansion.
dL = a * L * dT
• Where dL is the change in length
• a is the coefficient of thermal expansion
• L is the original length
• dT is the change in temperature.
## Thermal Expansion Definition
Thermal expansion is defined as the total increase in length or a dimension due to an increase in the temperature of the material.
## Thermal Expansion Formula
How to calculate thermal expansion.
1. First, determine the coefficient of thermal expansion.
This can be looked up on a table based on the material or calculated through an experiment.
2. Next, measure the original length.
Determine the original length or dimension of the object.
3. Next, measure the change in temperature.
Calculate or measure the change in temperature.
4. Finally, calculate the change in length.
Using the formula for thermal expansion, calculate the change in length.
## FAQ
What is thermal expansion?
Thermal expansion is the change in length or another dimension of an object when it undergoes a change in temperature.
What causes thermal expansion?
Thermal expansion is caused by the increased outward force of particles when their temperature is increased. The extra movement of particles causes objects to change in size. | 283 | 1,472 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2023-23 | latest | en | 0.863618 |
https://www.gamedev.net/forums/topic/422907-bijective-parameterized-surfaces/ | 1,503,297,952,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886107720.63/warc/CC-MAIN-20170821060924-20170821080924-00485.warc.gz | 917,979,386 | 23,055 | # Bijective parameterized surfaces
## Recommended Posts
Bourbaki 122
Greetings, Does anyone happen to know if there exists something similar to NURBS that features an inverse function? sufrace: R^2 -> R^3 forall x \in R^2, surface^-1|surface(x): R^3 -> R^2 the surface would be non intersecting in R^3 and it is continous (should be at least C^2+) Maybe someone knows if theres something like a picking routine for NURBS that does something similar numerically. Thanks N.
##### Share on other sites
Wasting Time 411
You have a parameterized surface (x(s,t),y(s,t),z(s,t)) and you want to know what (s,t) generates a surface point (x0,y0,z0). This gives you three equations in two unknowns: x(s,t) = x0, y(s,t) = y0, z(s,t) = z0. Generally, this is over constrained but knowing the point is on the surface, one of the three equations is redundant (the three equations are "functionally dependent").
For the sake of argument, suppose that x(s,t) = x0 and y(s,t) = y0 are the functionally independent equations.
For a NURBS surface, the x(s,t) = u(s,t)/w(s,t) and y(s,t) = v(s,t)/w(s,t), where u, v, and w are polynomials in s and t. You then have u(s,t) - x0*w(s,t) = 0 and v(s,t) - y0*w(s,t) = 0, which are two polynomial equations in two unknowns. One of the variables may be eliminated, say t, to obtain p(s) = 0, a higher-degree polynomial equation of a single variable. (Search on "resultant of polynomials" and/or "elimination theory").
The mathematics is clear but numerical construction of polynomial roots is not robust when using a fixed-size floating-point arithmetic system. Arbitrary precision systems are better to use for finding the roots. Once you (numerically) construct a root s of p(s) = 0, you can replace this in u(s,t) - x0*w(s,t) = 0 and solve for t. Verify that this (s,t) pair satisfies v(s,t) = y0*w(s,t) = 0 (obtaining p(s) can introduce extraneous solutions).
For picking, you can set up a similar system of polynomial equations. You have a line (or ray or segment) r*(x1,y1,z1)+(x0,y0,z0). To intersect the surface, you need x(s,t) = r*x1+x0, y(s,t) = r*y1+y0, and z(s,t) = r*z1+z0. For NURBS, introduce again u(s,t), v(s,t), and w(s,t). You get three polynomial equations in three variables (r, s, and t). Elimination theory again allows you to obtain a single polynomial (of higher-degree). Etc. Etc. | 666 | 2,342 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2017-34 | longest | en | 0.881837 |
https://www.numere-romane.ro/cum_se_scrie_numarul_arab_cu_numerale_romane.php?nr_arab=1151987&nr_roman=(M)(C)(L)MCMLXXXVII&lang=en | 1,638,055,119,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358323.91/warc/CC-MAIN-20211127223710-20211128013710-00303.warc.gz | 1,043,643,854 | 9,914 | Convert number: 1,151,987 in Roman numerals, how to write?
Latest conversions of Arabic numbers to Roman numerals
1,151,987 = (M)(C)(L)MCMLXXXVII Nov 27 23:18 UTC (GMT) 1,014,431 = (M)(X)M(V)CDXXXI Nov 27 23:18 UTC (GMT) 100,989 = (C)CMLXXXIX Nov 27 23:18 UTC (GMT) 1,010 = MX Nov 27 23:18 UTC (GMT) 48,609 = (X)(L)(V)MMMDCIX Nov 27 23:18 UTC (GMT) 1,262,022 = (M)(C)(C)(L)(X)MMXXII Nov 27 23:18 UTC (GMT) 166,675 = (C)(L)(X)(V)MDCLXXV Nov 27 23:18 UTC (GMT) 1,541,959 = (M)(D)(X)(L)MCMLIX Nov 27 23:18 UTC (GMT) 3,600,154 = (M)(M)(M)(D)(C)CLIV Nov 27 23:18 UTC (GMT) 154,929 = (C)(L)M(V)CMXXIX Nov 27 23:17 UTC (GMT) 654,548 = (D)(C)(L)M(V)DXLVIII Nov 27 23:17 UTC (GMT) 22,307 = (X)(X)MMCCCVII Nov 27 23:17 UTC (GMT) 59,118 = (L)M(X)CXVIII Nov 27 23:17 UTC (GMT) converted numbers, see more...
The set of basic symbols of the Roman system of writing numerals
• (*) M = 1,000,000 or |M| = 1,000,000 (one million); see below why we prefer this notation: (M) = 1,000,000.
(*) These numbers were written with an overline (a bar above) or between two vertical lines. Instead, we prefer to write these larger numerals between brackets, ie: "(" and ")", because:
• 1) when compared to the overline - it is easier for the computer users to add brackets around a letter than to add the overline to it and
• 2) when compared to the vertical lines - it avoids any possible confusion between the vertical line "|" and the Roman numeral "I" (1).
(*) An overline (a bar over the symbol), two vertical lines or two brackets around the symbol indicate "1,000 times". See below...
Logic of the numerals written between brackets, ie: (L) = 50,000; the rule is that the initial numeral, in our case, L, was multiplied by 1,000: L = 50 => (L) = 50 × 1,000 = 50,000. Simple.
(*) At the beginning Romans did not use numbers larger than 3,999; as a result they had no symbols in their system for these larger numbers, they were added on later and for them various different notations were used, not necessarily the ones we've just seen above.
Thus, initially, the largest number that could be written using Roman numerals was:
• MMMCMXCIX = 3,999. | 732 | 2,138 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2021-49 | latest | en | 0.907676 |
https://www.physicsforums.com/threads/amperes-law-vs-biot-savart-law.236980/ | 1,670,578,227,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711394.73/warc/CC-MAIN-20221209080025-20221209110025-00143.warc.gz | 973,152,429 | 13,922 | Ampere's law vs. Biot-Savart law
kasse
What's the difference? Is Ampere's law a special case (when the conductor carrying the current is a straight wire?)
Mephisto
Biot-Savart law is the more important and fundamental one. Ampere's law can be derived from it but can only be used in some very specific situations, but not only in straight wire cases. It can be applied to sheets of current, solenoids, etc.
kasse
Biot-Savart law is the more important and fundamental one. Ampere's law can be derived from it but can only be used in some very specific situations, but not only in straight wire cases. It can be applied to sheets of current, solenoids, etc.
OK, thanks. Btw, isn't the last calculation wrong here: http://planetphysics.org/encyclopedia/QuarterLoopExampleOfBiotSavartLaw.html
There should be no "pi" in the answer if I'm right.
Nick89
Ampere's law is usually only applied to problems that have certain symmetry.
For example, a straight wire is symmetric about it's center axis, a sheet is symmetric, a solenoid is symmetric, etc...
Some arbitrary wire is not symmetric however and Ampere's law can generally not be used there.
Also if you only have a short solenoid or wire for example (a short length in relation to it's radius) then Ampere's law does not give a very accurate answer either, because it ignores the fact that the magnetic field lines curve away from the wire at the ends; it assumes the magnetic field is nearly constant.
About the link, if the calculation up until the integral is correct then yes, the pi should cancel indeed. I haven't checked what they did before that though, but I think it's correct (so the pi is wrong). | 396 | 1,667 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2022-49 | latest | en | 0.938702 |
https://www.jiskha.com/display.cgi?id=1365786989 | 1,503,346,431,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886109525.95/warc/CC-MAIN-20170821191703-20170821211703-00237.warc.gz | 926,872,460 | 3,558 | # maths- urgent
posted by .
show that √Π+1 is irrational
• maths- urgent -
If it were rational, then it follows that pi must be rational. However, we know that pi is irrational, therefore sqrt(pi) + 1 must be irrational.
• maths- urgent -
is there any proof
• maths- urgent -
Show that the reciprocal of 3+2√2 is an irrational number
## Similar Questions
1. ### irrational numbers?!
what are rational and irrational numbers?
2. ### algebra/trig
1. which number below is irrational? a)√4/9 b) √20 c)√121 Why is the number you chose irrational?
3. ### maths- urgent
show that √Π+1 is irrational if there is any proof
4. ### Math
1a) Prove that there exist irrational numbers a and b so that a^b is rational. look at the numbers √2^√2 and (√2^√2)^√2 1b) use similar idea to prove that there exists a rational number a and an irrational …
There exist irrational numbers a and b so that a^b is rational. look at the numbers √2^√2 and (√2^√2)^√2 a) use similar idea to prove that there exists a rational number a and an irrational number b so …
6. ### THE HOMEWORK IS DUE TODAY- HELP! HELP! PLEASE
There exist irrational numbers a and b so that a^b is rational. look at the numbers √2^√2 and (√2^√2)^√2 a) use similar idea to prove that there exists a rational number a and an irrational number b so …
7. ### math
There exist irrational numbers a and b so that a^b is rational. look at the numbers √2^√2 and (√2^√2)^√2 a) use similar idea to prove that there exists a rational number a and an irrational number b so …
8. ### Maths
Show that the reciprocal of 3+2√2 is an irrational number
9. ### Maths
Between which two integers does each irrational number lie?
10. ### Maths
Between which two integers does each irrational number lie?
More Similar Questions | 469 | 1,773 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2017-34 | latest | en | 0.874045 |
http://asesoriacastellon.com/lib/differentialgleichungen-losungsmethoden-und-losungen-band-ii-partielle | 1,606,756,731,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141216897.58/warc/CC-MAIN-20201130161537-20201130191537-00439.warc.gz | 7,318,340 | 9,500 | Differentialgleichungen, Lösungsmethoden und Lösungen Band by E. Kamke
By E. Kamke
Read or Download Differentialgleichungen, Lösungsmethoden und Lösungen Band II Partielle Differentialgleichunge erster Ordnung für eine gesuchte Funktion mit 16 Figuren, 3 verbesserte Auflage (Mathematik und ihre Anwendungen in Physik und Technik, Reihe A, Band 18) PDF
Best mathematics_1 books
Mathematics, Affect and Learning: Middle School Students' Beliefs and Attitudes About Mathematics Education
This booklet examines the ideals, attitudes, values and feelings of scholars in Years five to eight (aged 10 to fourteen years) approximately arithmetic and arithmetic schooling. essentially, this ebook makes a speciality of the improvement of affective perspectives and responses in the direction of arithmetic and arithmetic studying. additionally, it appears scholars boost their extra detrimental perspectives of arithmetic throughout the heart institution years (Years five to 8), and so right here we be aware of scholars during this severe interval.
Additional resources for Differentialgleichungen, Lösungsmethoden und Lösungen Band II Partielle Differentialgleichunge erster Ordnung für eine gesuchte Funktion mit 16 Figuren, 3 verbesserte Auflage (Mathematik und ihre Anwendungen in Physik und Technik, Reihe A, Band 18)
Example text
For k = 0 define Pk to be the (n + 1) × (n + 1) matrix 1 −iα/k whose upper left 2 × 2 block is , where 0 < α < k is a parameter iα/k 1 to be chosen below, and with the remaining diagonal entries being 1, and all other entries being 0. Then the eigenvalues of Pk are (k + α)/k, 1 and (k − α)/k, so that Pk is positive definite, and close to the identity for large k. We take −Ck := −ikA + B as above. Then C∗k Pk + Pk Ck = −ik(APk − Pk A) − (BPk + Pk B) , and its upper left 3 × 3 block reads ⎛ ⎞ 2a1 α −iα/k a2 α ⎝ iα/k 2 − 2a1 α 0 ⎠ .
Second, the signs appearing in (2) generate cancellations and a very peculiar behavior2 of the QRW, as one can see in the result below extracted from [6]: Theorem 1 (Grimmett/Janson/Scudo’03) For any ψ ∈ Ω which is a finite sum of localized states, ψ X n n→∞ −→ Y, in distribution, n where Y is a real random variable of density f (y) = 2 In ⎧ ⎨ ⎩ 1 π(1 − y 2 ) 1 − 2y 2 0, comparison with the classical random walk. , if y ∈ [− otherwise. √ √ 2 2 , ], 2 2 (4) Hydrodynamic Limit of Quantum Random Walks 45 4 Hydrodynamic Limit for a System of Independent Quantum Random Walks We turn now our attention to a system of independent QRW’s.
N − 1 . On Linear Hypocoercive BGK Models 25 −h 4 λ 1 5 h0 λ λ − − 58 −h 2 λ Fig. 1 Functions appearing in the eigenvalue equation of −ikL1 + L2 ; solid blue curve: h 0 (λ); red dash-dotted curve: −h 2 (λ); purple dashed line: −h 4 (λ) For example, with n = 4, ⎛ 0 1 0 0 ⎜ 1 0 √3/2 0 ⎜ √ √ 3/2 L1 = ⎜ ⎜ 0 3/2 √ 0 ⎝0 0 3/2 0 0 0 0 1 ⎞ 0 0⎟ ⎟ 0⎟ ⎟. 13) towards f ∞ = m. We shall focus on the example with order n = 4, but the other cases behave similarly. 526948302245121... which has the least negative real part, and hence determines the exponential decay rate of f±1 (t). | 965 | 3,085 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2020-50 | longest | en | 0.51103 |
http://tectonicpuzzel.eu/kenken-puzzle-multiplication-online.html | 1,600,622,804,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400198287.23/warc/CC-MAIN-20200920161009-20200920191009-00106.warc.gz | 120,008,643 | 9,003 | Kenken puzzles online to learn multiplication
WWW.TECTONICPUZZEL.EU
# LOGIC PUZZLES FOR KIDS
## KENKEN DIVISION TABLES (8X8)
1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 LEV 1 (10) 2 (10) PUZZLE
With our special KenKen 8x8 puzzles you can fully practice the division tables. With the exception of table 10. These puzzles will drastically improve your knowledge of division and multiplication.
We have made a small adjustment to the classic kenken rules to focus specific mathematics topics
There are dividends in the first cell of each cage. Use your knowledge of the division tables to find quotient and divisor of each dividend and place them in the right cell.
In our version cages always consist of 2 cells. These cells can be connected to each other horizontal, vertical or diagonal.
There are 2 levels to choose from:
• Level 1: division tables 1 to 8.
• Level 2: division tables 2 to 9.
We have also developed a smaller version, namely a kenken 6x6. Don't forget to check our kenken with fractions.
# KENKEN ONLINE
TIP: use the arrow keys to move the cursor in the grid.
Follow the steps below to solve a kenken puzzle from our database.
• LEV: choose the starting level of your kenken puzzle.
• PUZZLE: fill in the number of the puzzle you want to solve. | 338 | 1,271 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2020-40 | latest | en | 0.814991 |
https://sciencedocbox.com/Physics/72566380-Edexcel-gce-mechanics-m3-advanced-advanced-subsidiary.html | 1,627,427,014,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153491.18/warc/CC-MAIN-20210727202227-20210727232227-00283.warc.gz | 523,964,015 | 15,397 | Size: px
Start display at page:
Transcription
2 1. A particle of mass 0.8 kg is attached to one end of a light elastic string of natural length 0.6 m. The other end of the string is attached to a fixed point A. The particle is released from rest at A and comes to instantaneous rest 1.1 m below A. Find the modulus of elasticity of the string. (4) 2 *P40094A0228*
3 Question 1 continued Q1 (Total 4 marks) *P40094A0328* 3 Turn over
4 2. A particle P is moving in a straight line with simple harmonic motion. The centre of the oscillation is the fixed point C, the amplitude of the oscillation is 0.5 m and the time to complete one oscillation is 2 π seconds. The point A is on the path of P and 0.2 m from C. 3 Find (a) the magnitude and direction of the acceleration of P when it passes through A, (b) the speed of P when it passes through A, (3) (2) (c) the time P takes to move directly from C to A. (3) 4 *P40094A0428*
5 Question 2 continued Q2 (Total 8 marks) *P40094A0528* 5 Turn over
6 3. A particle P is moving in a straight line. At time t seconds, P is at a distance x metres from a fixed point O on the line and is moving away from O with speed 10 1 ms. x + 6 (a) Find the acceleration of P when x = 14 (4) Given that x = 2 when t = 1, (b) find the value of t when x = 14 (6) 6 *P40094A0628*
7 Question 3 continued *P40094A0728* 7 Turn over
8 Question 3 continued 8 *P40094A0828*
9 Question 3 continued Q3 (Total 10 marks) *P40094A0928* 9 Turn over
10 4. A light elastic string AB has natural length 0.8 m and modulus of elasticity 19.6 N. The end A is attached to a fixed point. A particle of mass 0.5 kg is attached to the end B. The particle is moving with constant angular speed rad s 1 in a horizontal circle whose centre is vertically below A. The string is inclined at 60 to the vertical. (a) Show that the extension of the string is 0.4 m. (5) (b) Find the value of. (5) 10 *P40094A01028*
11 Question 4 continued *P40094A01128* 11 Turn over
12 Question 4 continued 12 *P40094A01228*
13 Question 4 continued Q4 (Total 10 marks) *P40094A01328* 13 Turn over
14 5. Above the Earth s surface, the magnitude of the gravitational force on a particle due to the Earth is inversely proportional to the square of the distance of the particle from the centre of the Earth. The Earth is modelled as a sphere of radius R and the acceleration due to gravity at the Earth s surface is g. A particle P of mass m is at a height x above the surface of the Earth. (a) Show that the magnitude of the gravitational force acting on P is 2 mgr 2 ( R+ x) (3) A rocket is fired vertically upwards from the surface of the Earth. When the rocket is at gr height 2R above the surface of the Earth its speed is. You may assume that air 2 resistance can be ignored and that the engine of the rocket is switched off before the rocket reaches height R. Modelling the rocket as a particle, (b) find the speed of the rocket when it was at height R above the surface of the Earth. (9) 14 *P40094A01428*
15 Question 5 continued *P40094A01528* 15 Turn over
16 Question 5 continued 16 *P40094A01628*
17 Question 5 continued Q5 (Total 12 marks) *P40094A01728* 17 Turn over
18 6. A particle P of mass m is attached to one end of a light inextensible string of length l. The other end of the string is attached to a fixed point O. The particle is hanging in equilibrium at the point A, vertically below O, when it is set in motion with a horizontal speed 1 ( 11gl ). When the string has turned through an angle and the string is still taut, 2 the tension in the string is T. ( ) (a) Show that T = 3mg cosθ (8) At the instant when P reaches the point B, the string becomes slack. Find (b) the speed of P at B, (3) (c) the maximum height above B reached by P before it starts to fall. (4) 18 *P40094A01828*
19 Question 6 continued *P40094A01928* 19 Turn over
20 Question 6 continued 20 *P40094A02028*
21 Question 6 continued Q6 (Total 15 marks) *P40094A02128* 21 Turn over
22 7. y Diagram NOT accurately drawn R O 2 6 x Figure 1 The shaded region R is bounded by the curve with equation y = x( 6 x), the x-axis and 2 the line x = 2, as shown in Figure 1. The unit of length on both axes is 1 cm. A uniform solid P is formed by rotating R through 360 about the x-axis. (a) Show that the centre of mass of P is, to 3 significant figures, 1.42 cm from its plane face. (9) The uniform solid P is placed with its plane face on an inclined plane which makes an angle with the horizontal. Given that the plane is sufficiently rough to prevent P from sliding and that P is on the point of toppling when = 1 (b) find the angle. (4) Given instead that P is on the point of sliding down the plane when = and that the coefficient of friction between P and the plane is 0.3, (c) find the angle (3) 22 *P40094A02228*
23 Question 7 continued *P40094A02328* 23 Turn over
24 Question 7 continued 24 *P40094A02428*
25 Question 7 continued *P40094A02528* 25 Turn over
26 Question 7 continued 26 *P40094A02628*
27 Question 7 continued Q7 (Total 16 marks) TOTAL FOR PAPER: 75 MARKS END *P40094A02728* 27
28 BLANK PAGE 28 *P40094A02828*
### Paper Reference. Mechanics M3 Advanced/Advanced Subsidiary. Monday 10 June 2013 Morning Time: 1 hour 30 minutes
Centre No. Candidate No. Paper Reference 6 6 7 9 0 1 Surname Signature Paper Reference(s) 6679/01 Edexcel GCE Mechanics M3 Advanced/Advanced Subsidiary Monday 10 June 2013 Morning Time: 1 hour 30 minutes
### Paper Reference. Mechanics M1 Advanced/Advanced Subsidiary. Friday 6 June 2014 Afternoon Time: 1 hour 30 minutes
Centre No. Candidate No. Paper Reference(s) 6677/01 Edexcel GCE Mechanics M1 Advanced/Advanced Subsidiary Friday 6 June 2014 Afternoon Time: 1 hour 30 minutes Materials required for examination Mathematical
Centre No. Candidate No. Paper Reference 6 6 7 9 0 1 Paper Reference(s) 6679/01 Edexcel GCE Mechanics M3 Advanced/Advanced Subsidiary Wednesday 13 May 2015 Morning Time: 1 hour 30 minutes Surname Signature
### Paper Reference R. Mechanics M1 Advanced/Advanced Subsidiary. Friday 6 June 2014 Afternoon Time: 1 hour 30 minutes
Centre No. Candidate No. Paper Reference(s) 6677/01R Edexcel GCE Mechanics M1 Advanced/Advanced Subsidiary Friday 6 June 2014 Afternoon Time: 1 hour 30 minutes Materials required for examination Mathematical
Centre No. Candidate No. Paper Reference(s) 6679/01 Edexcel GCE Mechanics M3 Advanced/Advanced Subsidiary Friday 28 January 2011 Morning Time: 1 hour 30 minutes Materials required for examination Mathematical
### Paper Reference. Paper Reference(s) 6679/01 Edexcel GCE Mechanics M3 Advanced/Advanced Subsidiary
Centre No. Candidate No. Paper Reference 6 6 7 9 0 1 Paper Reference(s) 6679/01 Edexcel GCE Mechanics M3 Advanced/Advanced Subsidiary Friday 6 June 2008 Afternoon Time: 1 hour 30 minutes Surname Signature
### Paper Reference. Mechanics M3 Advanced/Advanced Subsidiary. Friday 29 January 2010 Morning Time: 1 hour 30 minutes
Centre No. Candidate No. Paper Reference 6 6 7 9 0 1 Paper Reference(s) 6679/01 Edexcel GCE Mechanics M3 Advanced/Advanced Subsidiary Friday 29 January 2010 Morning Time: 1 hour 30 minutes Surname Signature
### Paper Reference. Mechanics M1 Advanced/Advanced Subsidiary. Wednesday 3 June 2015 Morning Time: 1 hour 30 minutes
Centre No. Candidate No. Paper Reference(s) 6677/01 Edexcel GCE Mechanics M1 Advanced/Advanced Subsidiary Wednesday 3 June 2015 Morning Time: 1 hour 30 minutes Materials required for examination Mathematical
Centre No. Candidate No. Paper Reference 6 6 7 8 0 1 Paper Reference(s) 6678/01 Edexcel GCE Mechanics M2 Advanced/Advanced Subsidiary Thursday 31 May 2012 Morning Time: 1 hour 30 minutes Materials required
Centre No. Candidate No. Paper Reference(s) 6677/01 Edexcel GCE Mechanics M1 Advanced/Advanced Subsidiary Wednesday 3 June 2015 Morning Time: 1 hour 30 minutes Materials required for examination Mathematical
Centre No. Candidate No. Paper Reference(s) 6678/01R Edexcel GCE Mechanics M2 Advanced/Advanced Subsidiary Thursday 6 June 2013 Morning Time: 1 hour 30 minutes Materials required for examination Mathematical
Centre No. Candidate No. Paper Reference(s) 6677/01 Edexcel GCE Mechanics M1 Advanced/Advanced Subsidiary Wednesday 16 May 2012 Morning Time: 1 hour 30 minutes Materials required for examination Mathematical
Centre No. Candidate No. Paper Reference 6 6 7 7 0 1 Paper Reference(s) 6677/01 Edexcel GCE Mechanics M1 Advanced/Advanced Subsidiary Friday 20 January 2012 Afternoon Time: 1 hour 30 minutes Surname Signature
Paper Reference(s) 6679 Edexcel GCE Mechanics M3 Advanced/Advanced Subsidiary Monday 12 January 2004 Afternoon Time: 1 hour 30 minutes Materials required for examination Answer Book (AB16) Mathematical
### Paper Reference. Mechanics M1 Advanced/Advanced Subsidiary. Friday 15 January 2010 Afternoon Time: 1 hour 30 minutes
Centre No. Candidate No. Paper Reference 6 6 7 7 0 1 Paper Reference(s) 6677/01 Edexcel GCE Mechanics M1 Advanced/Advanced Subsidiary Friday 15 January 2010 Afternoon Time: 1 hour 30 minutes Surname Signature
### PhysicsAndMathsTutor.com
PhysicsAndMathsTutor.com January 2007 4. Figure 2 O θ T v P A (3ag) A particle P of mass m is attached to one end of a light inextensible string of length a. The other end of the string is attached to
Centre No. Candidate No. Surname Signature Paper Reference(s) 6677/01 Edexcel GCE Mechanics M1 Advanced/Advanced Subsidiary Wednesday 23 January 2013 Morning Time: 1 hour 30 minutes Materials required
Centre No. Candidate No. Paper Reference(s) 6677/01 Edexcel GCE Mechanics M1 Advanced/Advanced Subsidiary Monday 13 May 2013 Afternoon Time: 1 hour 30 minutes Materials required for examination Mathematical
### Paper Reference. Advanced/Advanced Subsidiary. Thursday 7 June 2007 Morning Time: 1 hour 30 minutes. Mathematical Formulae (Green)
Centre No. Candidate No. Paper Reference(s) 6677/01 Edexcel GCE Mechanics M1 Advanced/Advanced Subsidiary Thursday 7 June 2007 Morning Time: 1 hour 30 minutes Materials required for examination Mathematical
### Paper Reference. Paper Reference(s) 6677/01 Edexcel GCE Mechanics M1 Advanced/Advanced Subsidiary
Centre No. Candidate No. Paper Reference 6 6 7 7 0 1 Paper Reference(s) 6677/01 Edexcel GCE Mechanics M1 Advanced/Advanced Subsidiary Friday 11 January 2008 Morning Time: 1 hour 30 minutes Materials required
### physicsandmathstutor.com Paper Reference Mechanics M1 Advanced/Advanced Subsidiary Friday 11 January 2008 Morning Time: 1 hour 30 minutes
Centre No. Candidate No. Paper Reference 6 6 7 7 0 1 Paper Reference(s) 6677/01 Edexcel GCE Mechanics M1 Advanced/Advanced Subsidiary Friday 11 January 2008 Morning Time: 1 hour 30 minutes Materials required
### Paper Reference. Further Pure Mathematics FP2 Advanced/Advanced Subsidiary. Friday 6 June 2014 Afternoon Time: 1 hour 30 minutes
Centre No. Candidate No. Surname Signature Paper Reference(s) 6668/01 Edexcel GCE Further Pure Mathematics FP2 Advanced/Advanced Subsidiary Friday 6 June 2014 Afternoon Time: 1 hour 30 minutes Materials
Centre No. Candidate No. Surname Signature Paper Reference(s) 6668/01 Edexcel GCE Further Pure Mathematics FP2 Advanced/Advanced Subsidiary Thursday 23 June 2011 Morning Time: 1 hour 30 minutes Materials
### Paper Reference. Paper Reference(s) 6677/01 Edexcel GCE Mechanics M1 Advanced/Advanced Subsidiary
Centre No. Candidate No. Paper Reference(s) 6677/01 Edexcel GCE Mechanics M1 Advanced/Advanced Subsidiary Tuesday 13 January 2009 Morning Time: 1 hour 30 minutes Materials required for examination Mathematical
### Paper Reference. Paper Reference(s) 6678/01 Edexcel GCE Mechanics M2 Advanced/Advanced Subsidiary
Centre No. Candidate No. Paper Reference 6 6 7 8 0 1 Surname Signature Paper Reference(s) 6678/01 Edexcel GCE Mechanics M2 Advanced/Advanced Subsidiary Thursday 7 June 2007 Morning Time: 1 hour 30 minutes
### Edexcel GCE Core Mathematics C2 Advanced Subsidiary
Centre No. Candidate No. Paper Reference 6 6 6 4 0 1 Paper Reference(s) 6664/01 Edexcel GCE Core Mathematics C2 Advanced Subsidiary Friday 13 January 2012 Morning Time: 1 hour 30 minutes Materials required
### Edexcel GCE Core Mathematics C2 Advanced Subsidiary
Centre No. Candidate No. Paper Reference(s) 6664/01R Edexcel GCE Core Mathematics C2 Advanced Subsidiary Friday 24 May 2013 Morning Time: 1 hour 30 minutes Materials required for examination Mathematical
### Paper Reference R. Mechanics M5 Advanced/Advanced Subsidiary. Wednesday 18 June 2014 Afternoon Time: 1 hour 30 minutes
Centre No. Candidate No. Surname Paper Reference(s) 6681/01R Edexcel GCE Mechanics M5 Advanced/Advanced Subsidiary Wednesday 18 June 2014 Afternoon Time: 1 hour 30 minutes Materials required for examination
### Edexcel GCE Core Mathematics C2 Advanced Subsidiary
Centre No. Candidate No. Paper Reference(s) 6664/01 Edexcel GCE Core Mathematics C2 Advanced Subsidiary Friday 24 May 2013 Morning Time: 1 hour 30 minutes Materials required for examination Mathematical
### Edexcel GCE Core Mathematics C2 Advanced Subsidiary
Centre No. Candidate No. Paper Reference(s) 6664/01 Edexcel GCE Core Mathematics C2 Advanced Subsidiary Friday 24 May 2013 Morning Time: 1 hour 30 minutes Materials required for examination Mathematical
### Paper Reference. Paper Reference(s) 6677/01 Edexcel GCE Mechanics M1 Advanced/Advanced Subsidiary
Centre No. Candidate No. Paper Reference(s) 6677/01 Edexcel GCE Mechanics M1 Advanced/Advanced Subsidiary Friday 12 January 2007 Morning Time: 1 hour 30 minutes Materials required for examination Mathematical
### Paper Reference. Paper Reference(s) 6677/01 Edexcel GCE Mechanics M1 Advanced/Advanced Subsidiary. Friday 22 May 2009 Morning Time: 1 hour 30 minutes
Centre No. Candidate No. Paper Reference(s) 6677/01 Edexcel GCE Mechanics M1 Advanced/Advanced Subsidiary Friday 22 May 2009 Morning Time: 1 hour 30 minutes Materials required for examination Mathematical
### Edexcel GCE Core Mathematics C4 Advanced
Centre No. Candidate No. Paper Reference 6 6 6 6 0 1 Paper Reference(s) 6666/01 Edexcel GCE Core Mathematics C4 Advanced Monday 28 January 2013 Morning Time: 1 hour 30 minutes Materials required for examination
### Paper Reference. Mechanics M2 Advanced/Advanced Subsidiary. Friday 29 January 2010 Morning Time: 1 hour 30 minutes
Centre No. Candidate No. Paper Reference(s) 6678/01 Edexcel GCE Mechanics M2 Advanced/Advanced Subsidiary Friday 29 January 2010 Morning Time: 1 hour 30 minutes Materials required for examination Mathematical
Centre No. Candidate No. Surname Signature Paper Reference(s) 6669/01R Edexcel GCE Further Pure Mathematics FP3 Advanced/Advanced Subsidiary Monday 24 June 2013 Afternoon Time: 1 hour 30 minutes Materials
### Edexcel GCE Core Mathematics C4 Advanced
physicsandmathstutor.com Centre No. Candidate No. Paper Reference 6 6 6 6 0 1 Paper Reference(s) 6666/01 Edexcel GCE Core Mathematics C4 Advanced Monday 28 January 2013 Morning Time: 1 hour 30 minutes
### PhysicsAndMathsTutor.com. Advanced/Advanced Subsidiary. You must have: Mathematical Formulae and Statistical Tables (Blue)
Write your name here Surname Other names Pearson Edexcel International Advanced Level Centre Number Mechanics M3 Advanced/Advanced Subsidiary Candidate Number Monday 27 January 2014 Morning Time: 1 hour
### Edexcel GCE Core Mathematics C2 Advanced Subsidiary
Centre No. Candidate No. Paper Reference(s) 6664/01 Edexcel GCE Core Mathematics C2 Advanced Subsidiary Monday 10 January 2011 Morning Time: 1 hour 30 minutes Materials required for examination Mathematical
Centre No. Candidate No. Surname Signature Paper Reference(s) 6667/01 Edexcel GCE Further Pure Mathematics FP1 Advanced/Advanced Subsidiary Wednesday 22 June 2011 Morning Time: 1 hour 30 minutes Materials
### Edexcel GCE Core Mathematics C2 Advanced Subsidiary
Centre No. Candidate No. Paper Reference 6 6 6 4 0 1 Paper Reference(s) 6664/01 Edexcel GCE Core Mathematics C2 Advanced Subsidiary Thursday 24 May 2012 Morning Time: 1 hour 30 minutes Materials required
### Edexcel GCE Core Mathematics C4 Advanced
physicsandmathstutor.com Centre No. Candidate No. Paper Reference(s) 6666/01 Edexcel GCE Core Mathematics C4 Advanced Thursday 21 June 2012 Afternoon Time: 1 hour 30 minutes Materials required for examination
### Edexcel GCE Core Mathematics C2 Advanced Subsidiary
physicsandmathstutor.com Centre No. Candidate No. Paper Reference 6 6 6 4 0 1 Paper Reference(s) 6664/01 Edecel GCE Core Mathematics C2 Advanced Subsidiary Friday 13 January 2012 Morning Time: 1 hour 30
### Paper Reference. Paper Reference(s) 6678/01 Edexcel GCE Mechanics M2 Advanced/Advanced Subsidiary
Centre No. Candidate No. Surname Paper Reference(s) 6678/01 Edexcel GCE Mechanics M2 Advanced/Advanced Subsidiary Wednesday 21 May 2008 Afternoon Time: 1 hour 30 minutes Materials required for examination
Centre No. Candidate No. Paper Reference(s) 6665/01 Edexcel GCE Core Mathematics C3 Advanced Friday 12 June 2015 Morning Time: 1 hour 30 minutes Materials required for examination Mathematical Formulae
### Edexcel GCE Core Mathematics C3 Advanced
Centre No. Candidate No. Paper Reference 6 6 6 5 0 1 Paper Reference(s) 6665/01 Edexcel GCE Core Mathematics C3 Advanced Thursday 14 June 01 Morning Time: 1 hour 30 minutes Materials required for examination
### Paper Reference. Paper Reference(s) 6668/01 Edexcel GCE Further Pure Mathematics FP2 Advanced/Advanced Subsidiary
Centre No. Candidate No. Paper Reference 6 6 6 8 0 1 Surname Signature Paper Reference(s) 6668/01 Edexcel GCE Further Pure Mathematics FP2 Advanced/Advanced Subsidiary Thursday 24 June 2010 Morning Time:
### Paper Reference. Core Mathematics C2 Advanced Subsidiary. Thursday 22 May 2014 Morning Time: 1 hour 30 minutes
Centre No. Candidate No. Paper Reference 6 6 6 4 0 1 Paper Reference(s) 6664/01 Edexcel GCE Core Mathematics C2 Advanced Subsidiary Thursday 22 May 2014 Morning Time: 1 hour 30 minutes Materials required
### Paper Reference. Core Mathematics C3 Advanced. Monday 16 June 2014 Morning Time: 1 hour 30 minutes
Centre No. Candidate No. Paper Reference(s) 6665/01 Edexcel GCE Core Mathematics C3 Advanced Monday 16 June 2014 Morning Time: 1 hour 30 minutes Materials required for examination Mathematical Formulae
### Edexcel GCE Core Mathematics C4 Advanced
Centre No. Candidate No. Paper Reference 6 6 6 6 0 1 Surname Signature Paper Reference(s) 6666/01 Edexcel GCE Core Mathematics C4 Advanced Wednesday 25 January 2012 Afternoon Time: 1 hour 30 minutes Materials
### Edexcel GCE Core Mathematics C2 Advanced Subsidiary
Centre No. Candidate No. Paper Reference 6 6 6 4 0 1 Paper Reference(s) 6664/01 Edexcel GCE Core Mathematics C2 Advanced Subsidiary Monday 14 January 2013 Morning Time: 1 hour 30 minutes Materials required
Centre No. Candidate No. Paper Reference(s) 6666/01 Edexcel GCE Core Mathematics C4 Advanced Tuesday 16 June 015 Afternoon Time: 1 hour 30 minutes Materials required for examination Mathematical Formulae
Centre No. Candidate No. Paper Reference(s) 6666/01 Edexcel GCE Core Mathematics C4 Advanced Tuesday 16 June 2015 Afternoon Time: 1 hour 30 minutes Materials required for examination Mathematical Formulae
### Paper Reference R. Core Mathematics C4 Advanced. Wednesday 18 June 2014 Afternoon Time: 1 hour 30 minutes
Centre No. Candidate No. Surname Paper Reference(s) 6666/01R Edexcel GCE Core Mathematics C4 Advanced Wednesday 18 June 2014 Afternoon Time: 1 hour 30 minutes Materials required for examination Mathematical
### Paper Reference R. Core Mathematics C4 Advanced. Wednesday 18 June 2014 Afternoon Time: 1 hour 30 minutes
Centre No. Candidate No. Surname Paper Reference(s) 6666/01R Edexcel GCE Core Mathematics C4 Advanced Wednesday 18 June 2014 Afternoon Time: 1 hour 30 minutes Materials required for examination Mathematical
### Paper Reference. Paper Reference(s) 6664/01 Edexcel GCE Core Mathematics C2 Advanced Subsidiary. Monday 2 June 2008 Morning Time: 1 hour 30 minutes
Centre No. Candidate No. Paper Reference 6 6 6 4 0 1 Paper Reference(s) 6664/01 Edexcel GCE Core Mathematics C2 Advanced Subsidiary Monday 2 June 2008 Morning Time: 1 hour 30 minutes Materials required
### Paper Reference. Further Pure Mathematics FP3 Advanced/Advanced Subsidiary. Monday 24 June 2013 Afternoon Time: 1 hour 30 minutes
Centre No. Candidate No. Surname Signature Paper Reference(s) 6669/01 Edexcel GCE Further Pure Mathematics FP3 Advanced/Advanced Subsidiary Monday 24 June 2013 Afternoon Time: 1 hour 30 minutes This is
### Paper Reference. Core Mathematics C4 Advanced. Tuesday 18 June 2013 Morning Time: 1 hour 30 minutes
Centre No. Candidate No. Paper Reference(s) 6666/01 Edexcel GCE Core Mathematics C4 Advanced Tuesday 18 June 2013 Morning Time: 1 hour 30 minutes Materials required for examination Mathematical Formulae
### Edexcel GCE Core Mathematics C2 Advanced Subsidiary
Centre No. Candidate No. Paper Reference 6 6 6 4 0 1 Paper Reference(s) 6664/01 Edexcel GCE Core Mathematics C2 Advanced Subsidiary Thursday 26 May 2011 Morning Time: 1 hour 30 minutes Materials required
### physicsandmathstutor.com Paper Reference Core Mathematics C4 Advanced Monday 25 January 2010 Morning Time: 1 hour 30 minutes
Centre No. Candidate No. Paper Reference(s) 6666/0 Edexcel GCE Core Mathematics C4 Advanced Monday 25 January 200 Morning Time: hour 30 minutes Materials required for examination Mathematical Formulae
### Paper Reference. Paper Reference(s) 6665/01 Edexcel GCE Core Mathematics C3 Advanced. Tuesday 15 June 2010 Morning Time: 1 hour 30 minutes
Centre No. Candidate No. Paper Reference(s) 6665/01 Edexcel GCE Core Mathematics C3 Advanced Tuesday 15 June 2010 Morning Time: 1 hour 30 minutes Materials required for examination Mathematical Formulae
### Paper Reference. Core Mathematics C2 Advanced Subsidiary. Wednesday 20 May 2015 Morning Time: 1 hour 30 minutes
Centre No. Candidate No. Paper Reference(s) 6664/01 Edexcel GCE Core Mathematics C2 Advanced Subsidiary Wednesday 20 May 2015 Morning Time: 1 hour 30 minutes Materials required for examination Mathematical
### PhysicsAndMathsTutor.com. Paper Reference. Core Mathematics C2 Advanced Subsidiary. Wednesday 20 May 2015 Morning Time: 1 hour 30 minutes
Centre No. Candidate No. Paper Reference(s) 6664/01 Edexcel GCE Core Mathematics C2 Advanced Subsidiary Wednesday 20 May 2015 Morning Time: 1 hour 30 minutes Materials required for examination Mathematical
### Edexcel GCE Core Mathematics C2 Advanced Subsidiary
Centre No. Candidate No. Paper Reference 6 6 6 4 0 1 Paper Reference(s) 6664/01 Edexcel GCE Core Mathematics C2 Advanced Subsidiary Thursday 26 May 2011 Morning Time: 1 hour 30 minutes Materials required
### physicsandmathstutor.com Paper Reference Core Mathematics C3 Advanced Thursday 15 January 2009 Morning Time: 1 hour 30 minutes
Centre No. Candidate No. Paper Reference(s) 6665/01 Edexcel GCE Core Mathematics C3 Advanced Thursday 15 January 2009 Morning Time: 1 hour 30 minutes Materials required for examination Mathematical Formulae
### You must show sufficient working to make your methods clear to the Examiner. Answers without working may gain no credit.
Paper Reference(s) 6679 Edexcel GCE Mechanics M3 dvanced/dvanced Subsidiary Monday 19 May 2003 Morning Time: 1 hour 30 minutes Materials required for examination nswer ook (16) Mathematical Formulae (Lilac)
### Paper Reference. Paper Reference(s) 6665/01 Edexcel GCE Core Mathematics C3 Advanced. Friday 6 June 2008 Afternoon Time: 1 hour 30 minutes
Centre No. Candidate No. Paper Reference(s) 6665/01 Edexcel GCE Core Mathematics C3 Advanced Friday 6 June 2008 Afternoon Time: 1 hour 30 minutes Materials required for examination Mathematical Formulae
Centre No. Candidate No. Surname Signature Paper Reference(s) 6669/01 Edexcel GCE Further Pure Mathematics FP3 Advanced/Advanced Subsidiary Friday 24 June 2011 Afternoon Time: 1 hour 30 minutes Materials
### Paper Reference. Further Pure Mathematics FP1 Advanced/Advanced Subsidiary. Friday 30 January 2009 Afternoon Time: 1 hour 30 minutes
Centre No. Candidate No. Surname Signature Paper Reference(s) 6667/01 Edexcel GCE Further Pure Mathematics FP1 Advanced/Advanced Subsidiary Friday 30 January 2009 Afternoon Time: 1 hour 30 minutes Materials
### Paper Reference. Paper Reference(s) 6667/01 Edexcel GCE Further Pure Mathematics FP1 Advanced/Advanced Subsidiary
Centre No. Candidate No. Surname Signature Paper Reference(s) 6667/01 Edexcel GCE Further Pure Mathematics FP1 Advanced/Advanced Subsidiary Friday 30 January 2009 Afternoon Time: 1 hour 30 minutes Materials
### Paper Reference. Further Pure Mathematics FP3 Advanced/Advanced Subsidiary. Tuesday 26 June 2007 Afternoon Time: 1 hour 30 minutes
Centre No. Candidate No. Paper Reference 6 6 7 6 0 1 Surname Signature Paper Reference(s) 6676/01 Edexcel GCE Further Pure Mathematics FP3 Advanced/Advanced Subsidiary Tuesday 26 June 2007 Afternoon Time:
### Advanced/Advanced Subsidiary. You must have: Mathematical Formulae and Statistical Tables (Pink)
Write your name here Surname Other names Pearson Edexcel GCE Centre Number Mechanics M1 Advanced/Advanced Subsidiary Candidate Number Wednesday 8 June 2016 Morning Time: 1 hour 30 minutes You must have:
### Paper Reference. Further Pure Mathematics FP1 Advanced/Advanced Subsidiary. Wednesday 17 June 2009 Morning Time: 1 hour 30 minutes
Centre No. Candidate No. Surname Signature Paper Reference(s) 6667/01 Edexcel GCE Further Pure Mathematics FP1 Advanced/Advanced Subsidiary Wednesday 17 June 2009 Morning Time: 1 hour 30 minutes Materials
### Advanced/Advanced Subsidiary. You must have: Mathematical Formulae and Statistical Tables (Blue)
Write your name here Surname Other names Pearson Edexcel International Advanced Level Centre Number Mechanics M1 Advanced/Advanced Subsidiary Candidate Number Monday 25 January 2016 Afternoon Time: 1 hour
### physicsandmathstutor.com Paper Reference Mechanics M1 Advanced/Advanced Subsidiary Thursday 12 January 2006 Afternoon Time: 1 hour 30 minutes
Centre No. Candidate No. physicsandmathstutor.com Paper Reference 6 6 7 7 0 1 Surname Signature Paper Reference(s) 6677/01 Edexcel GCE Mechanics M1 Advanced/Advanced Subsidiary Thursday 12 January 2006
### Advanced/Advanced Subsidiary. You must have: Mathematical Formulae and Statistical Tables (Blue)
Write your name here Surname Other names Pearson Edexcel International Advanced Level Centre Number Mechanics M1 Advanced/Advanced Subsidiary Candidate Number Monday 25 January 2016 Afternoon Time: 1 hour
### Paper Reference. Paper Reference(s) 6664/01 Edexcel GCE Core Mathematics C2 Advanced Subsidiary
Centre No. Candidate No. Paper Reference(s) 6664/01 Edexcel GCE Core Mathematics C2 Advanced Subsidiary Wednesday 9 January 2008 Afternoon Time: 1 hour 30 minutes Materials required for examination Mathematical
PhysicsAndMathsTutor.com Centre No. Candidate No. Surname Signature Paper Reference(s) 6669/01 Edexcel GCE Further Pure Mathematics FP3 Advanced/Advanced Subsidiary Monday 4 June 013 Afternoon Time: 1
### Paper Reference. Paper Reference(s) 6664/01 Edexcel GCE Core Mathematics C2 Advanced Subsidiary
Centre No. Candidate No. Paper Reference 6 6 6 4 0 1 Paper Reference(s) 6664/01 Edexcel GCE Core Mathematics C2 Advanced Subsidiary Monday 11 January 2010 Morning Time: 1 hour 30 minutes Materials required
### Paper Reference. Advanced/Advanced Subsidiary. Friday 24 June 2005 Morning Time: 1 hour 30 minutes
Centre No. Candidate No. Paper Reference 6 6 7 8 0 1 Paper Reference(s) 6678/01 Edexcel GCE Mechanics M2 Advanced/Advanced Subsidiary Friday 24 June 2005 Morning Time: 1 hour 30 minutes Surname Signature
### Paper Reference. Further Pure Mathematics FP3 Advanced/Advanced Subsidiary. Monday 22 June 2015 Morning Time: 1 hour 30 minutes
Centre No. Candidate No. Surname Signature Paper Reference(s) 6669/01 Edexcel GCE Further Pure Mathematics FP3 Advanced/Advanced Subsidiary Monday 22 June 2015 Morning Time: 1 hour 30 minutes Materials
Centre No. Candidate No. Surname Signature Paper Reference(s) 6669/01 Edexcel GCE Further Pure Mathematics FP3 Advanced/Advanced Subsidiary Monday 22 June 2015 Morning Time: 1 hour 30 minutes Materials
### Paper Reference R. Statistics S2 Advanced/Advanced Subsidiary. Tuesday 24 June 2014 Morning Time: 1 hour 30 minutes
Centre No. Candidate No. Paper Reference(s) 6684/01R Edexcel GCE Statistics S2 Advanced/Advanced Subsidiary Tuesday 24 June 2014 Morning Time: 1 hour 30 minutes Materials required for examination Mathematical
### Paper Reference R. Statistics S2 Advanced/Advanced Subsidiary. Tuesday 24 June 2014 Morning Time: 1 hour 30 minutes
Centre No. Candidate No. Paper Reference(s) 6684/01R Edexcel GCE Statistics S2 Advanced/Advanced Subsidiary Tuesday 24 June 2014 Morning Time: 1 hour 30 minutes Materials required for examination Mathematical
### Paper Reference. Paper Reference(s) 6676/01 Edexcel GCE Further Pure Mathematics FP3 Advanced/Advanced Subsidiary
Centre No. Candidate No. Surname Signature Paper Reference(s) 6676/01 Edexcel GCE Further Pure Mathematics FP3 Advanced/Advanced Subsidiary Wednesday 3 February 2010 Morning Time: 1 hour 30 minutes Materials
### physicsandmathstutor.com Paper Reference Core Mathematics C2 Advanced Subsidiary Wednesday 9 January 2008 Afternoon Time: 1 hour 30 minutes
physicsandmathstutor.com Centre No. Candidate No. Paper Reference(s) 6664/01 Edexcel GCE Core Mathematics C2 Advanced Subsidiary Wednesday 9 January 2008 Afternoon Time: 1 hour 30 minutes Materials required
### physicsandmathstutor.com Paper Reference Core Mathematics C2 Advanced Subsidiary Monday 11 January 2010 Morning Time: 1 hour 30 minutes
Centre No. Candidate No. Paper Reference 6 6 6 4 0 1 Paper Reference(s) 6664/01 Edexcel GCE Core Mathematics C2 Advanced Subsidiary Monday 11 January 2010 Morning Time: 1 hour 30 minutes Materials required
### Edexcel GCE Core Mathematics C3 Advanced
Centre No. Candidate No. Paper Reference 6 6 6 5 0 1 Paper Reference(s) 6665/01 Edexcel GCE Core Mathematics C3 Advanced Thursday 16 June 2011 Afternoon Time: 1 hour 30 minutes Materials required for examination
Centre No. Candidate No. Paper Reference(s) 6683/01R Edexcel GCE Statistics S1 Advanced/Advanced Subsidiary Friday 17 May 2013 Morning Time: 1 hour 30 minutes Materials required for examination Mathematical
### Paper Reference. Core Mathematics C4 Advanced. Wednesday 18 June 2014 Afternoon Time: 1 hour 30 minutes
Centre No. Candidate No. Surname Paper Reference(s) 6666/01 Edexcel GCE Core Mathematics C4 Advanced Wednesday 18 June 2014 Afternoon Time: 1 hour 30 minutes Materials required for examination Mathematical
### Advanced/Advanced Subsidiary. You must have: Mathematical Formulae and Statistical Tables (Pink)
Write your name here Surname Other names Pearson Edexcel GCE Centre Number Mechanics M4 Advanced/Advanced Subsidiary Candidate Number Wednesday 15 June 2016 Morning Paper Reference Time: 1 hour 30 minutes
### Paper Reference. Core Mathematics C3 Advanced. Wednesday 20 January 2010 Afternoon Time: 1 hour 30 minutes. Mathematical Formulae (Pink or Green)
Centre No. Candidate No. Surname Signature Paper Reference(s) 6665/01 Edexcel GCE Core Mathematics C3 Advanced Wednesday 20 January 2010 Afternoon Time: 1 hour 30 minutes Materials required for examination
### Paper Reference. Statistics S1 Advanced/Advanced Subsidiary. Friday 5 June 2015 Morning Time: 1 hour 30 minutes
Centre No. Candidate No. Paper Reference(s) 6683/01 Edexcel GCE Statistics S1 Advanced/Advanced Subsidiary Friday 5 June 2015 Morning Time: 1 hour 30 minutes Materials required for examination Mathematical
Centre No. Candidate No. Paper Reference(s) 6677/01 Edexcel GCE Mechanics M1 Advanced/Advanced Subsidiary Wednesday 16 May 2012 Morning Time: 1 hour 30 minutes Materials required for examination Mathematical
### Edexcel GCE Core Mathematics C3 Advanced
Centre No. Candidate No. Paper Reference(s) 6665/01 Edexcel GCE Core Mathematics C3 Advanced Monday 24 January 2011 Morning Time: 1 hour 30 minutes Materials required for examination Mathematical Formulae
### Advanced/Advanced Subsidiary. You must have: Mathematical Formulae and Statistical Tables (Blue)
Write your name here Surname Other names Pearson Edexcel International Advanced Level Centre Number Mechanics M1 Advanced/Advanced Subsidiary Candidate Number Tuesday 20 January 2015 Morning Time: 1 hour
### Edexcel GCE Core Mathematics C1 Advanced Subsidiary
Centre No. Candidate No. Paper Reference(s) 6663/01 Edexcel GCE Core Mathematics C1 Advanced Subsidiary Monday 13 May 2013 Afternoon Time: 1 hour 30 minutes Materials required for examination Mathematical
### PhysicsAndMathsTutor.com. Advanced/Advanced Subsidiary. You must have: Mathematical Formulae and Statistical Tables (Blue)
Write your name here Surname Other names Pearson Edexcel International Advanced Level Centre Number Mechanics M1 Advanced/Advanced Subsidiary Candidate Number Wednesday 8 June 2016 Morning Time: 1 hour
### Paper Reference. Paper Reference(s) 6666/01 Edexcel GCE Core Mathematics C4 Advanced. Monday 18 June 2007 Morning Time: 1 hour 30 minutes
Centre No. Candidate No. Paper Reference 6 6 6 6 0 1 Paper Reference(s) 6666/01 Edexcel GCE Core Mathematics C4 Advanced Monday 18 June 2007 Morning Time: 1 hour 30 minutes Materials required for examination | 8,847 | 33,424 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2021-31 | latest | en | 0.869481 |
http://brainmass.com/economics/supply-and-demand/102805 | 1,386,639,692,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164004057/warc/CC-MAIN-20131204133324-00010-ip-10-33-133-15.ec2.internal.warc.gz | 24,379,814 | 7,634 | # Calculating price elasticity of demand, profit-maximizing, sketching marginal cost and average cost curves
1. Huntsville, Alabama is engaging in a bumper sticker advertising campaign. Monthly sales data suggest that the following demand relationship exists:
Q = 6,000 - 2,000P
Where: Q = bumper-sticker sales
P = price
A. How many bumper stickers could be sold at \$2 each?
B. What would the price have to be in order to sell 5,000 bumper stickers?
C. At what price would bumper sticker sales equal zero?
D. How many bumper stickers could be given away?
E. Calculate the point price elasticity of demand at a price of \$1.
2. Pile Buffers, Inc., has enjoyed substantial economic profits derived from patents covering the manufacturing of a special shock absorber used in driving piles at construction sites. Market demand relations for the shock absorber are:
P = \$5,000 - \$0.05Q
Fixed costs are zero, because research and development expenses have
been fully amortized during previous periods. Average variable costs are
constant at \$4,000 per unit.
A. Calculate the profit-maximizing price/output combination and economic profits if Pile Buggers enjoys an effective monopoly on shock absorber due to patent protection
B. Calculate the price/output combination and total economic profits that would result if competitors offer clones that make the pile driver shock absorber market perfectly competitive.
3. Increased productivity is the main way in which individuals, firms, and entire economies can improve their real economic situation. Please address the following questions regarding productivity.
A. What is implied by "increased productivity?"
B. How is productivity measured and what problems are generally encountered when attempting to quantify changes
C. Discuss "productivity in the 1990s,"?reasons for dramatic increases and the effect of this on keeping inflation very low with low unemployment rates
4. Theoretically a firm in a perfectly competitive market cannot make above "normal profits" in the long run.
A. Discuss reasons why this is the case and how the long run adjustment process takes place
B. Given a cost function of: TC = Q2 address the following
1) Sketch the marginal cost and average cost curves
2) In perfect competition, what it the theoretical final outcome of such a cost function
5. Can monopolies lose money?
A. Graph this possibility and explain what action a monopoly can take to remedy the problem
B. Discuss two reasons why a monopoly, acting in their self-interest, might elect to charge a lower price and produce more output than the theoretical model to maximize profits would suggest.
See attached file for full problem description.
This question has the following supporting file(s):
###### File Viewer (Click To Zoom)
Solution Summary
Monopoly case is given.
\$3.99
• Plain text
• Cited sources when necessary
• Attached file(s)
• Monopoly - produce more than required.doc
Rohtas Kumar, MBA
Rating 4.9/5
Active since 2004
BEng, Birla Institute of Technology and Science, India
MBA, Delhi University
CFA (Indian accrediation not US) , Institute of Chartered Financial Analysts of India
Ph.D, Indian Institute of Management Bangalore, India
Responses 3669 | 688 | 3,232 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2013-48 | latest | en | 0.9385 |
https://www.whizwriters.com/finance-20/ | 1,590,793,506,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347406785.66/warc/CC-MAIN-20200529214634-20200530004634-00545.warc.gz | 958,397,830 | 8,922 | # FINANCE
72. One-year Treasury bill rates in 20XX averaged 5.15% and inflation for the year was 7.3%. If investors had expected the same inflation rate as that realized, calculate the real interest rate for 20XX according to the Fisher effect.
A. 0.00%
B. -2.15%
C. 2.15%
D. 3.95%
73. Assume that you observe the following rates on long-term bonds:
U.S. Treasury bonds = 4.15%
AAA Corporate bonds = 6.2%
BBB Corporate bonds = 7.15%
The main reason for the differences in the interest rates is:
74. Which of the following statements is correct?
A. According to the unbiased expectations theory, the return for holding a 2-year bond to maturity is equal to the nominal rate divided by the real interest rate.
B. The rate on a 10-year Corporate can never be less than the rate on a 10-year Treasury.
C. We usually observe the inverted yield curve.
D. The rate on a 3-year Treasury can never be less than the rate on a 15-year Treasury.
75. One-year interest rates are 3%. The market expects one-year rates to be 5% one year from now. The market also expects one-year rates to be 7% two years from now. Assume that the unbiased expectations theory holds. Which of the following is correct?
A. The yield curve is downward sloping.
B. The yield curve is flat.
C. The yield curve is upward sloping.
D. We need the maturity risk premiums to be able to answer this question.
76. Which of the following statements is correct?
A. If the unbiased expectations theory is correct, we could see an inverted yield curve.
B. If a yield curve is inverted, long-term bonds have higher yields than short-term bonds.
C. If the maturity risk premium is zero, the yield curve would be flat.
D. If the unbiased expectations theory is correct, the maturity risk premium is zero.
77. The Wall Street Journal states that the yield curve for Treasuries is downward sloping and there is no liquidity premium or maturity risk premium. Given this information, which of the following statements is correct?
A. A 30-year corporate bond must have a higher yield than a 5-year corporate bond.
B. A 5-year corporate bond must have a higher yield than a 30-year Treasury bond.
C. A 5-year Treasury bond must have a higher yield than a 5-year corporate bond.
D. All of these statements are correct.
78. Which of the following statements is correct?
A. An IPO is an example of a primary market transaction.
B. Money markets are subject to wider price fluctuations and are therefore more risky than capital market instruments.
C. A direct transfer of funds is more efficient than utilizing financial institutions.
D. The market segmentation theory argues that the different investors have different risk preferences which determine the shape of the yield curve.
79. In 20XX, the 10-year Treasury rate was 4.5% while the average 10-year Aaa corporate bond debt carried an interest rate of 6.0%. What is the average default risk premium on Aaa corporate bonds?
A. 0.75%
B. 1.5%
C. 1.95%
D. 2.25%
80. Which of the following statements is correct?
A. The default risk premium of Baa 20-year corporate bonds over Aaa 20-year corporate bonds does not vary.
B. The market segmentation theory assumes that borrowers and investors do not want to shift from one maturity sector to another without an interest rate premium.
C. Real interest rates are the rates that are quoted in the news.
D. All of these statements are correct.
81. All of the following are types of financial institutions except _______.
A. Insurance companies
B. Pension funds
C. Thrifts
D. Federal Reserve Bank
82. All of the following are benefits that financial institutions provide to our economy except _________.
A. Increased liquidity
B. Increased monitoring
C. Increased dollar amount of funds flowing from suppliers to fund users
D. Increased price risk
83. All of the following are factors that affect nominal interest rates except ___________.
A. Time to maturity
B. Real interest rate
C. Convertibility features
D. Foreign exchange
84. Which of the following statements is correct?
A. A flat yield curve occurs when the yield-to-maturity is virtually unaffected by the term-to-maturity.
B. Real interest rates are generally lower than nominal interest rates.
C. Liquidity risk is the risk that a security may be difficult to sell on short notice for its true value.
D. All of these statements are correct.
85. Which of the following statements is incorrect?
A. Governments affect foreign exchange rates indirectly by altering prevailing interest rates within their own countries.
B. Foreign currency exchange rates vary with the day-to-day demand and supply of the two foreign currencies.
C. Central governments can intervene in foreign exchange markets directly and value their currency at high rates relative to another currency.
D. All of these statements are correct.
86. The theory that argues that individual investors and financial institutions have specific maturity preferences is called the ______________.
A. Market segmentation theory
B. Unbiased expectations theory
C. Liquidity preference theory
D. Inverted forward theory
Order now and get 10% discount on all orders above \$50 now!!The professional are ready and willing handle your assignment.
ORDER NOW »» | 1,221 | 5,224 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2020-24 | latest | en | 0.934929 |
https://ask.libreoffice.org/en/question/32227/problem-using-offset-with-concat-to-current-cell/?sort=votes | 1,624,132,064,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487649688.44/warc/CC-MAIN-20210619172612-20210619202612-00273.warc.gz | 121,926,663 | 15,206 | We will be migrating from Ask to Discourse on the first week of August, read the details here
# Problem using offset with concat to current cell. [closed]
I want to concatenate the contents of two cells immediately to the left of my current cell, and place the result in my current cell, by using the offset functions. Is this possible? I cannot figure it out. I tried using address(row(),column()) to reference the current cell and offsetting from that value, but just get error 504.
Formula goes into cell L10. concatenate(offset(L10,0,-2),offset(L10,0,-1)) works fine, but I need to remove direct reference to current cell, i.e. L10.
tried replacing L10 with address(row(),column()) .. does not work. also tried cell("address") .. does not work either. Probably to do with result format but I do not know solution. 20 years ago I would have pursued like a dog with a bone, but age has slowed me down and I cannot endure long nights anymore. Any help joyfully received.
edit retag reopen merge delete
### Closed for the following reason the question is answered, right answer was accepted by Alex Kemp close date 2016-02-20 16:01:32.271773
Hi @stevenolan, I think there is no way, it's possible use offset e.g. to change a range to sum, but not to join strings. Maybe explaning what you want achieve.
( 2014-04-01 00:36:21 +0200 )edit
Sort by » oldest newest most voted
Thanks guys. Turns out to be a pretty stupid question. It was late at night. I just wanted to concatenate the adjacent two cells to the left of my current cell to my current cell. Example:
J10 = "abc" K10 = "def" J20 = "ghi" K20 = "jkl"
originally I had L10 = concatenate(J10,K10) : Result = "abcdef" If I cut and paste L10 to L20 then L20 became concatenate(J10,K10) : Result = "abcdef" I did not want this I wanted L20 to become concatenate(J20,K20) : Result = "ghijkl"
Solution was L10 = concatenate(offset(L10,0,-2),offset(L10,0,-1))
If I now cut and paste L10 to L20 then L20 becomes concatenate(offset(L20,0,-2),offset(L20,0,-1)) : Result = "ghijkl"
So, problem solved.
Again, thank you all.
more
1. Select cell C3
2. Press Ctrl+F3 or choose Insert - Names - Define
3. Press button Add, set Name (for example MyConcat), Range =A1&B1, press Add and OK
4. Use formula =MyConcat anywhere
more | 605 | 2,284 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2021-25 | latest | en | 0.944138 |
http://www.chegg.com/homework-help/questions-and-answers/uniform-16-kg-rod-074-m-long-suspended-rest-ceiling-springs-end-rod-springs-hang-straight--q1273303 | 1,394,454,642,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394010779425/warc/CC-MAIN-20140305091259-00080-ip-10-183-142-35.ec2.internal.warc.gz | 275,195,420 | 7,038 | # A uniform rod.
0 pts ended
A uniform 1.6-kg rod that is 0.74 m long is suspended at rest from the ceiling by two springs, one at each end of the rod. Both springs hang straight down from the ceiling. The springs have identical lengths when they are unstretched. Their spring constants are 54 N/m and 36 N/m. Find the angle that the rod makes with the horizontal. | 91 | 366 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2014-10 | latest | en | 0.934968 |
https://goodcalculators.com/acid-test-quick-ratio-calculator/ | 1,722,666,808,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640361431.2/warc/CC-MAIN-20240803060126-20240803090126-00892.warc.gz | 230,142,047 | 24,274 | # Acid Test Ratio (Quick Ratio) Calculator
You can use this acid test ratio calculator to compute a company's acid-test ratio. The acid test ratio, which is also referred to as the quick ratio or liquid ratio, provides an indication of an organization's immediate short-term liquidity.
You can determine the acid test ratio in three simple steps:
• Select the currency in which you want to perform the calculation (optional)
• Input a value for the company's liquid assets and liquid liabilities
• Click the "Calculate Acid Test Ratio" button to generate the ratio.
Acid Test Ratio | Quick Ratio Calculator
## Calculating the Acid Test Ratio
The acid test ratio (ATR), which is also commonly referred to as the liquid ratio or quick ratio, is a measure of the ratio of an organization's liquid assets to its liquid liabilities and provides an indication of how quickly a business could repay debt. It differs from the current ratio on the basis that it doesn't take into account inventories and prepaid expenses because these do not represent liquid assets. For instance, if an organization has an ATR of 0.65:1, it can pay only \$0.65 for every \$1 that it owes to its creditors and other agents. If its ratio is 1.16:1, it means that it has \$1.16 available for every \$1 that it owes.
Companies should seek to have a ratio higher than 1:1. If the ATR falls below this ratio, an organization does not have sufficient funds available to cover.
A ratio of 1:1 is considered acceptable. Higher ratios indicate that a firm is financially stable. Ratios lower than 1:1 mean that a business does not have enough liquid assets to cover its liabilities.
The ATR is determined by dividing the total value of liquid assets – which include cash, accounts receivable, short-term investments, and other assets that can readily be converted into cash form – by the total short-term liabilities. Inventory is not included as a liquid asset because it cannot be quickly and easily converted into cash form without incurring some form of loss. Pre-paid assets are also not considered to be a liquid asset.
A firm's short-term liabilities include accounts payable, short-term loans, income tax due, and accrued expenses that the organization has yet to pay off. Accrued expenses can include any fraction of a long-term loan that is due for repayment within the next 12 months.
## Calculating the Acid Test Ratio: An Example
Let's say an organization has the following liquid assets: \$10,000 cash, \$3,000 in receivables, \$1,000 in short-term investments. As such, its total liquid assets are \$14,000.
Let's say the firm also has the following short-term liabilities: \$5,000 in due income tax, \$4,000 in accounts payable, and \$1,000 in other expenses. Its total short-term liabilities are \$10,000.
Acid test ratio = liquid assets / short-term liabilities
14000 / 10000 = 1.4
As such, the acid test ratio for this organization is 1.4:1.
This ratio indicates that the company is in a good financial position because it has enough liquid assets available to service its short-term liabilities.
You may also be interested in our CAC (Customer Acquisition Cost) Calculator | 690 | 3,174 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2024-33 | latest | en | 0.949819 |
https://socratic.org/questions/how-do-you-solve-the-following-linear-system-x-3y-4x-13-y-1-2x-3-2 | 1,716,223,095,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058291.13/warc/CC-MAIN-20240520142329-20240520172329-00281.warc.gz | 478,877,276 | 6,319 | # How do you solve the following linear system: x - 3y = 4x - 13 , y = 1/2x - 3/2 ?
Feb 26, 2017
$x = 3.89 \text{ , } y = 0.45$
#### Explanation:
$x - 3 y = 4 x - 13 \text{ (1)}$
$\textcolor{red}{y} = \frac{1}{2} x - \frac{3}{2} \text{ (2)}$
$\text{rearrange the equation (1) above}$
$x - 4 x = 3 y - 13$
$- 3 x = 3 y - 13$
$\text{divide both sides by 3}$
$\frac{- \cancel{3} x}{\cancel{3}} = \frac{3 y - 13}{3}$
$- x = \frac{\cancel{3} y}{\cancel{3}} - \frac{13}{3}$
$- x = y - \frac{13}{3}$
$\textcolor{red}{y} = - x + \frac{13}{3} \text{ (3)}$
$\text{the equations (3) and (2) are equal}$
$\text{we can write as ;}$
$\frac{1}{2} x - \frac{3}{2} = - x + \frac{13}{3}$
$\frac{1}{2} x + x = \frac{13}{3} + \frac{3}{2}$
$\frac{x}{2} + \textcolor{g r e e n}{\frac{2}{2}} x = \textcolor{g r e e n}{\frac{2}{2}} \cdot \frac{13}{3} + \textcolor{g r e e n}{\frac{3}{3}} \frac{3}{2}$
$\frac{3 x}{2} = \frac{26}{6} + \frac{9}{6}$
$\frac{3 x}{2} = \frac{35}{6}$
$\text{if "a/b=c/d" then } a \cdot d = b \cdot c$
$3 x \cdot 6 = 2 \cdot 35$
$18 x = 70$
$x = \frac{70}{18}$
$x = 3.89$
$\text{use (2)}$
$y = \frac{1}{2} x - \frac{3}{2}$
$y = \frac{1}{2} \cdot 3.89 - \frac{3}{2}$
$y = \frac{3.89 - 3}{2}$
$y = \frac{0.89}{2}$
$y = 0.45$ | 615 | 1,254 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 29, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.75 | 5 | CC-MAIN-2024-22 | latest | en | 0.332174 |
https://www.mathworks.com/matlabcentral/cody/problems/6-select-every-other-element-of-a-vector/solutions/190122 | 1,508,212,086,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187820700.4/warc/CC-MAIN-20171017033641-20171017053641-00437.warc.gz | 1,198,478,442 | 11,495 | Cody
# Problem 6. Select every other element of a vector
Solution 190122
Submitted on 12 Jan 2013 by Miha
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% x = rand(1,10); actual = everyOther(x); expected = x(1:2:length(x)); assert(isequal(actual, expected))
2 Pass
%% x = rand(1,100); actual = everyOther(x); expected = x(1:2:length(x)); assert(isequal(actual, expected))
3 Pass
%% x = ['A' 'long' 'time' 'ago' 'in' 'a' 'galaxy' 'far' 'far' 'away']; actual = everyOther(x); expected = x(1:2:length(x)); assert(isequal(actual, expected)) | 204 | 667 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2017-43 | latest | en | 0.629208 |
http://ecoursesonline.iasri.res.in/mod/page/view.php?id=1750 | 1,725,870,550,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651092.31/warc/CC-MAIN-20240909071529-20240909101529-00589.warc.gz | 10,998,361 | 7,766 | ## LESSON 6. Theory of Wave and particle
De-Broglie wavelength
The expression of the De-Broglie wave associated with a material particle can be derived on comparison of radiation.
Suppose, according to Plank theory of radiation the energy of photon (quantum) is
E = hv
But ν = ${c \over \lambda }$ c - velocity of light
E = ${{hc} \over \lambda }$…… (1) $\lambda$ - wavelength of light
But according to energy and mass relation E = mc2…… (2)
Equating (1) and (2)
mc2 = ${{hc} \over \lambda }$
Hence $\lambda$ = ${h \over {mc}}$……. (3)
We know that c- velocity of light (photon) so
mc = P ……. (4) Associated with photon (radiation)
Now suppose a material particle of mass –m and material particle is moving with velocity ν
Hence, the momentum of particle P = mν
The wavelength ($\lambda$) is associated with particle (material particle)
$\lambda$${h \over {mc}}$ = ${h \over P}$…….(5)
Let the kinetic energy of moving material particle is
E = ${{m{v^2}} \over 2}$
E = ${{m{v^2}} \over 2}$ × ${m \over m}$
E = ${{{m^2}{v^2}} \over {2m}}$
E = ${{{P^2}} \over {2m}}$ where, P = mν
E = ${{m{v^2}} \over 2}$ = ${{{m^2}{v^2}} \over {2m}}$ = ${{{P^2}} \over {2m}}$
P = $\sqrt {2mE}$…….(6)
De-Broglie wavelength
$\lambda$ = ${h \over {\sqrt {2mE} }}$……(7)
De-Broglie wavelength associated with electron
Suppose
Electron is at rest
It has having mass- m0
Charge of electron- e
Now electron associated by potential voltage and obtained the velocity ν from its rest position.
KE = ${{{m_0}{v^2}} \over 2}$ = $eV$
ν2 = ${{2eV} \over {{m_0}}}$
ν = $\sqrt {{{2eV} \over {{m_0}}}$
De-Broglie wavelength is associated with electron
$\lambda$ = ${h \over {{m_0}v}}$
$\lambda$ = ${h \over {\sqrt {2{m_0}eV} }}$
h - Plank’s constant- 6.625 × 10-34 Joule.sec
m0 - Rest mass of electron- 9.11 × 10-31 kg
E - Charge of electron- 1.62 × 10-19C
V - 100 Volts
$\lambda$ = 1.226 Å ……(8)
The wavelength is associated with an electron accelerated to 100 volts is
1.226 Å
Properties of matter wave
• Lighter in the particle, greater is the wavelength associated with it
• $\lambda$ = ${h \over {mv}}$ = ${h \over P}$
• Smaller is the velocity of the particle, greater is the wavelength associated with it.
• Suppose,ν = 0, then $\lambda$ = ∞ ; wave becomes indeterminate
• Suppose ν = ∞ then $\lambda$ = 0; wave becomes a zero. These conditions suggest, the matter waves are generated by the motion of particles.
• These waves ( $\lambda$ = ∞ and $\lambda$ = 0) are produced whether the particles are charged particles or the particles are un-changed.
• These waves are not electromagnetic waves but they are a new kind of waves (electromagnetic waves are produced only by motion of charged particle). The velocity of material particle is not a constant while the velocity of electromagnetic wave is constant.
• The velocity of matter waves is greater than the velocity of light. This can be provide as under
The wave velocity ω is ν × $\lambda$
So, ω = ${{{c^2}} \over v}$
As a particle velocity ν can’t exceed c hence ω is greater than velocity of light.
• The wave and particle aspect of moving bodies can never appear together in the same experiment. What we can say is that waves have particle like properties and particle have wave like properties and the concepts are inseparably linked. Matter wave representation is only a symbolic representation.
• The wave nature of matter introduced an uncertainty in the location of the position of the particle because a wave can’t be said exactly at this point or exactly at that point. However, where the wave is large(strong) there is a good chance of finding the particle while, where the wave is small there is very small(weak) chance of finding the particle | 1,136 | 3,774 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2024-38 | latest | en | 0.77681 |
https://sites.google.com/site/learningtodrawbuildings/creatinga3dsketchfrom2dplanspart2 | 1,532,057,596,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676591481.75/warc/CC-MAIN-20180720022026-20180720042026-00178.warc.gz | 765,090,166 | 7,648 | Creating a 3D Sketch From 2D Plans: Part 2
Adding A Roof To The Guest House
I knew I'd find you here! You just can't get enough of this stuff.
Your guest house needs a pitched roof because: A. City zoning requires it. - B. You are tired of shoveling snow off it and you can't get your lazy guests to do it. - C. The neighbors say it's ugly and reminds them of a shoebox. Ha Ha
This is what a roof looks like in elevation. It's called a Gable Roof because it has triangle shaped ends (more Architectural terms to impress your friends).
Here's a cool shortcut! On the short side draw two diagonal lines, corner to corner, as in the example. Where they intersect is the exact middle of that wall.
If you did that on the 2D elevation it works the same way. The top of the roof is 5 high or half the wall height of 10. On your perspective, draw a vertical line up from the center point (Centerline or C/L). The top of the roof will be half the height of the wall as shown on the elevation. For now make a good guess where that top point is on your perspective and complete the triangle gable end as shown below.
The section called Shortcuts will help you with an exact way to find the height of the roof.
Remember how we drew the hidden opposite wall in Freehand Sketching? Do the same on your guest house.
Draw converging dashed lines, where they intersect is the hidden inside back corner of the house. I removed the door and windows to see this step better.
Draw the top of the roof.
Using the super cool shortcut you learned, find the center of the wall and complete the roof.
This drawing is also known as a Wire Frame. It is similar to what is produced with more advanced Perspective Projection, which will be covered in a future advanced lesson.
Also Computer Aided Design (CAD) programs can produce wire frames from 2D plans. Now you have learned the basics which take the mystery away. If your plans are to go into Design, Architecture or Engineering, this knowledge will be extremely beneficial.
Want more? Part three coming up.
Back Home | 478 | 2,102 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2018-30 | latest | en | 0.931645 |
https://www.audiosciencereview.com/forum/index.php?threads/simulation-overview-of-the-double-bass-array-configuration.37943/ | 1,716,430,094,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058588.75/warc/CC-MAIN-20240523015422-20240523045422-00027.warc.gz | 562,184,917 | 27,924 | # Simulation Overview of the 'Double Bass Array' Configuration.
#### René - Acculution.com
##### Senior Member
Technical Expert
I will give a short introduction via simulations in COMSOL Multiphysics to the ‘Double Array Bass’ subwoofer setup and how the response will differ between an idealized situation of two walls moving compared to having a finite number of subwoofers.
Imagine first a 3x4x5 meters room. We can calculate the associated room modes as shown below.
These modes are an inherent feature of the room geometry and air being the medium inside it. They exist independent of any sources and are calculated without any regard to sourcing. Whether a mode is a problem or not will be evaluated when a source(s) is activated. Let us now put a subwoofer in the corner and evaluate the sound pressure level in 6 preselected places at a height of 1.1 meters.
The responses in the points vary quite a bit.
Some distinct issues can be seen at frequencies where there is a mode, but it would be wrong to say that a mode will always cause a resonance; that will depend on the source type and characteristics, and placement of both source(s) and your ears.
Many different multi-sub approaches could now be tried out but let us focus on the so-called Double Bass Array method. Here, the strategy basically is to turn the room into a tube. Tube acoustics is a little more involved than room acoustics (and a good topic if you want to really understand room acoustics better), but there is one major thing to understand: “The plane wave will always propagate”. Tube modes differ from room modes in that the former type really exist at all frequencies at once and will either propagate if excited above an associated modal frequency or die out as an evanescent wave below it. The only wave that will always propagate is the plane wave, and one way to achieve a propagating plane wave is to excite both the front wall and the rear wall in a pistonic-like fashion as shown in the animation below.
The front wall has its particular displacement, and the rear wall has a modified displacement to account for the phase shift coming from the wave travelling the room distance. You now effectively sit in an infinite tube with only a travelling wave and no reflection. Let us look at the response for this case.
We have a one-dimensional sound field, with a travelling wave and no reflection. If the excitation is the entirety of the walls in a pistonic matter, this will be the case at all frequencies(!), no matter how high.
Now comes the practical implementation. With 4 subwoofers at the front wall, and 4 subwoofers at the rear wall with a phase modification to their response, a placement can be found that tries to emulate the full wall moving. Each driver sits 1/4 of the width/hight of the room away from the wall and are spaced 1/2 of that apart to have the mirror sources be spaced equally. We assume that the subwoofers are very flat or placed in the wall somehow as shown below.
The corresponding responses actually look very good still at the low frequencies we are considering here.
With the simulation setup in place, you can now experiment away and for example investigate what happens if the four upper subwoofers are not connected.
You could argue that they should be placed halfway up now, but I am going with this setup, since you probably have to place them near the floor in most situations anyway. In general, as you stray more and more away from the ‘walls moving’ situation, the response curves will naturally take a hit.
One can consider dropping the subwoofers at the rear wall and replace them with damping material. To investigate via simulations you will have to have some idea of that setup with damping material parameter and such.
You will need more than a phase knob
on your subwoofers to do this, as there is a time delay involved and that will not give you one phase for all frequencies, so some kind of DSP is needed.
Hopefully you have gotten the idea behind the DBA principle, and of course ask questions if something is not clear.
René Christensen, Denmark, BSEE, MSc (Physics), PhD (Microacoustics), FEM and BEM simulations specialist in/for loudspeaker, hearing aid, and consultancy companies. Own company Acculution, blog at acculution.com/blog
Last edited:
#### Cbdb2
##### Major Contributor
The subs on the second wall are supposed to "remove that wall" so theres no reflection? So there active absorbers similar to an out of phase woofer right beside an in phase one? Why are there no modes in the other 2 directions? What happens in a real room with walls that have a good amount of frequency dependent low end absorption? And I guess this removes all the room gain.
Last edited:
OP
#### René - Acculution.com
##### Senior Member
Technical Expert
The subs on the second wall are supposed to "remove that wall" so theres no reflection? So there active absorbers similar to an out of phase woofer right beside an in phase one? Why are there no modes in the other 2 directions? What happens in a real room with walls that have frequency dependent low end absorption?
Yes they effectively works as an absorbing layer. The rear wall has a specialised displacement that follows the displacement that the travelling wave has and would have were the room infinitely long. But an out of phase sub next to one of the subs will be a very different situation where you create a dipole. The point of the DBA is to create the travelling wave in a confined space.
The wave only travels in one direction and so no modes are excited in the idealised tube situation. Looking at higher frequencies you will see that a finite number of subs will not be enough and modal behaviour will become more important, but you will be crossing over to your mains anyway.
I would need to know what the absorption is. I can try and put a thick layer of foam but then more and more thinks could be tried out.
#### JPA
##### Active Member
Forum Donor
Good post. I'm very interested in DBA because I currently have a house under construction that will have a dedicated home theater room. I'm planning to experiment with both Welti multi-sub configurations and DBA. I'll report on my observations and measurements when the room is ready, but that's several months down the road.
The main problem with DBA aside from the cost of the multiple subs is that efficiency is low. You pay the full cost of the rear subs and amplification, but they don't contribute to the bass SPL at the listening positions. I was wondering the other day how a "hybrid" or "part-time" DBA implementation would sound. In this scheme, you would use the rear subs only to cancel the room modes, and let the other frequencies go uncancelled. You'd set it up by measuring the frequency response at the rear wall, identify the modes, then use a parametric equalizer to create a peak for each mode at the rear subs. That would cause them activate and cancel only the modes, and leave the other frequencies alone.
Can that be simulated in COMSOL Multiphysics?
#### voodooless
##### Grand Contributor
Forum Donor
Cool sims. I’ve looked up some previous threads on this, notably one from @ppataki. Interesting was post 2:
Specifically this image:
Now in that case the rear sub was delayed about 16ms. So here it looks like the rear delayed wave still reaches the listener. It is about 10 dB less loud though. So what gives?
So here is my gut reaction: by the time the front wave gets to the rear, it’s attenuated (inverse square law). Now if the rear sub plays at the same volume as the front, it will not just cancel out the front, but will still send energy back. So to really cancel out, you’ll need to attenuate the rear sub as much as the front sub looses over the distance. Am I right in this?
Sadly, I have an L-shaped room . It would require a shitload of drivers, and multiple delays to get that working.
Last edited:
OP
#### René - Acculution.com
##### Senior Member
Technical Expert
Good post. I'm very interested in DBA because I currently have a house under construction that will have a dedicated home theater room. I'm planning to experiment with both Welti multi-sub configurations and DBA. I'll report on my observations and measurements when the room is ready, but that's several months down the road.
The main problem with DBA aside from the cost of the multiple subs is that efficiency is low. You pay the full cost of the rear subs and amplification, but they don't contribute to the bass SPL at the listening positions. I was wondering the other day how a "hybrid" or "part-time" DBA implementation would sound. In this scheme, you would use the rear subs only to cancel the room modes, and let the other frequencies go uncancelled. You'd set it up by measuring the frequency response at the rear wall, identify the modes, then use a parametric equalizer to create a peak for each mode at the rear subs. That would cause them activate and cancel only the modes, and leave the other frequencies alone.
Can that be simulated in COMSOL Multiphysics?
It is a crazy concept when it comes to implementation, so it would be very interesting to follow your work.
Well you can simulate most things in Comsol Multiphysics. One could consider letting Comsol do the optimisation of placement, level and phase based on certain objectives.
OP
#### René - Acculution.com
##### Senior Member
Technical Expert
Cool sims. I’ve looked up some previous threads on this, notably one from @ppataki. Interesting was post me 2:
Specifically this image:
View attachment 234727
Now in that case the rear sub was delayed about 16ms. So here it looks like the rear delayed wave still reaches the listener. It is about 10 dB less loud though. So what gives?
So here is my gut reaction: by the time the front wave gets to the rear, it’s attenuated (inverse square law). Now if the rear sub plays at the same volume as the front, it will not just cancel out the front, but will still send energy back. So to really cancel out, you’ll need to attenuate the rear sub as much as the front sub looses over the distance. Am I right in this?
Sadly, I have an L-shaped room . It would require a shitload of drivers, and multiple delays to get that working.
There are no losses in the simulated room here, but a real room would have losses that take out energy of the wave. Perfect transmission can probably not be expected in real life. Also, the subs need to sit flush on the wall for this idealised case.
#### ppataki
##### Major Contributor
Cool sims. I’ve looked up some previous threads on this, notably one from @ppataki. Interesting was post 2:
Specifically this image:
View attachment 234727
Now in that case the rear sub was delayed about 16ms. So here it looks like the rear delayed wave still reaches the listener. It is about 10 dB less loud though. So what gives?
So here is my gut reaction: by the time the front wave gets to the rear, it’s attenuated (inverse square law). Now if the rear sub plays at the same volume as the front, it will not just cancel out the front, but will still send energy back. So to really cancel out, you’ll need to attenuate the rear sub as much as the front sub looses over the distance. Am I right in this?
Sadly, I have an L-shaped room . It would require a shitload of drivers, and multiple delays to get that working.
@voodooless In my case this experiment was kind of a failure since I have an odd shaped room (6 corners instead of 4....) and I tried it with 2 subs only
I read somewhere that the rear subs shall have the same volume as the fronts
But I guess this is all down to experimentation at the end of the day....
In my current room I would not try this again but in a regular shaped room definitely! (with at least 4 subs)
#### Flaesh
##### Senior Member
I just saw this thread .
BTW, I registered here mainly in the hope of talking about DBA, but here, as well as on local forums and diyaudiocom, almost no one builds and only doubt the usefulness of the idea.
Last edited:
#### Flaesh
##### Senior Member
I was wondering the other day how a "hybrid" or "part-time" DBA implementation would sound. In this scheme, you would use the rear subs only to cancel the room modes, and let the other frequencies go uncancelled.
It seems wrong.
The goal is a travelling plane wave and no reflection.
Last edited:
#### Flaesh
##### Senior Member
I read somewhere that the rear subs shall have the same volume as the fronts
In real room rear subs shall have slightly less volume and same [as possible] FR and PhR as the fronts. Simple IIR eq may help if different subs used.
OP
#### René - Acculution.com
##### Senior Member
Technical Expert
It seems wrong.
The goal is a travelling plane wave and no reflection.
And the room modes can never be 'cancelled'. They always exist, regardless of source setup. The question is to what degree the modes are excited by the sources.
#### JPA
##### Active Member
Forum Donor
It seems wrong.
The goal is a travelling plane wave and no reflection.
Well, I have two goals. The first is to eliminate or greatly reduce room modes. The second is to achieve a high ratio of dinosaur stomp volume to subwoofer count.
Besides, I'm concerned about how effective DBA is in a real HT with tiered seating and human bodies to disrupt that nice laminar flow of bass waves.
#### Flaesh
##### Senior Member
how effective DBA is
Very effective in cuboidal room . Have you room plan\drawing\pics? Asymmetric openings in fron and rear walls are the worst. And what upper frequency for subs do you need?
SBA can be an alternative. Nils Öllerer aka FoLlGoTt used both in his HT.
"Source to sink" (in Fazenda's terms) systems with subs on floor aren't DBA, while subjectively may be rated good; I haven't opportunity to listen SS.
OP
#### René - Acculution.com
##### Senior Member
Technical Expert
Very effective in cuboidal room . Have you room plan\drawing\pics? Asymmetric openings in fron and rear walls are the worst. And what upper frequency for subs do you need?
SBA can be an alternative. Nils Öllerer aka FoLlGoTt used both in his HT.
"Source to sink" (in Fazenda's terms) systems with subs on floor aren't DBA, while subjectively may be rated good; I haven't opportunity to listen SS.
I can set up an optimisation routine in Consol that will find the best subwoofer settings for a quasi plane wave for a given room.
#### andyc56
##### Active Member
I will give a short introduction via simulations in COMSOL Multiphysics to the ‘Double Array Bass’ subwoofer setup and how the response will differ between an idealized situation of two walls moving compared to having a finite number of subwoofers.
[...]
Hopefully you have gotten the idea behind the DBA principle, and of course ask questions if something is not clear.
I have a question. In Beranek's original book from the '50s, there is an equation (4.17) for the acoustic pressure at a point at a given distance from a rigid piston in an infinite baffle under anechoic conditions. That pressure is proportional to j * omega * u, where u is the velocity amplitude of the piston's (assumed sinusoidal) motion. In the time domain, this of course translates to the time derivative of the piston's velocity (that is, its acceleration). So for a constant acoustic pressure vs. frequency, the acceleration amplitude of the piston must be constant with frequency as well. Of course, this is the same as saying the displacement amplitude must be inversely proportional to the square of the frequency for a constant acoustic pressure vs. frequency at a fixed distance and location (since acceleration must be integrated with respect to time twice to get displacement). So in the case of radiation from a piston in an infinite baffle, for the same acoustic pressure at half of some reference frequency, the piston's displacement amplitude must be 4 times its value at the reference frequency.
Your plot of the SPL response of the ideal DBA shows a curious downward slope vs. frequency, maybe -6 dB/octave or so? So my question is this: what is the assumed displacement amplitude vs. frequency of the "pulsating wall" in this simulation? Is it chosen to give a constant acceleration amplitude vs. frequency (that which would give a flat anechoic response in the "radiation from a piston" case)?
Thanks.
OP
#### René - Acculution.com
##### Senior Member
Technical Expert
I have a question. In Beranek's original book from the '50s, there is an equation (4.17) for the acoustic pressure at a point at a given distance from a rigid piston in an infinite baffle under anechoic conditions. That pressure is proportional to j * omega * u, where u is the velocity amplitude of the piston's (assumed sinusoidal) motion. In the time domain, this of course translates to the time derivative of the piston's velocity (that is, its acceleration). So for a constant acoustic pressure vs. frequency, the acceleration amplitude of the piston must be constant with frequency as well. Of course, this is the same as saying the displacement amplitude must be inversely proportional to the square of the frequency for a constant acoustic pressure vs. frequency at a fixed distance and location (since acceleration must be integrated with respect to time twice to get displacement). So in the case of radiation from a piston in an infinite baffle, for the same acoustic pressure at half of some reference frequency, the piston's displacement amplitude must be 4 times its value at the reference frequency.
Your plot of the SPL response of the ideal DBA shows a curious downward slope vs. frequency, maybe -6 dB/octave or so? So my question is this: what is the assumed displacement amplitude vs. frequency of the "pulsating wall" in this simulation? Is it chosen to give a constant acceleration amplitude vs. frequency (that which would give a flat anechoic response in the "radiation from a piston" case)?
Thanks.
I am in the process of moving so I will get back to this later. The sound field to look up is the that of a tube, not an infinite baffle.
#### Flaesh
##### Senior Member
under anechoic conditions.
*BA (DBA or SBA) radiate in (approx.) PWT, plane wave tube. So anechoic spherical radiation equations are not veri useful for this. The simplest and accessible for all Hornresp can give you an idea if you consider infinite horn. Subs may be as low Q as available.
Funny intermediate case is linear BA, but it isn't very practical. For Single Linear BA on floor at front wall not only the back, but also the top surface (the latter is usually called the ceiling)) in the room must be LF absorber. DLBA is not realistic at all, except that the room has such a shape:
#### Flaesh
##### Senior Member
So for a constant acoustic pressure vs. frequency, the acceleration amplitude of the piston must be constant with frequency as well.
For plane wave the velocity amplitude of the piston must be constant with frequency, I think so.
p. s.:
Last edited:
#### andyc56
##### Active Member
I am in the process of moving so I will get back to this later. The sound field to look up is the that of a tube, not an infinite baffle.
I am not trying to argue that this response shape can be predicted from the radiation of a piston in an infinite baffle into half-space. My question was about the nature of the stimulus used in the simulation itself.
So why should I care about "radiation of a piston in an infinite baffle into half-space" then? Because that's how the on-axis pressure at a point at distance d from the piston is calculated using the Thiele/Small theory. For an ideal sealed or infinite-baffle subwoofer with zero voice-coil inductance, in the T/S theory the passband is the frequency region for which the acceleration amplitude is constant with frequency.
For such a sub, with an applied voltage having a constant amplitude vs. frequency:
• The displacement amplitude vs. frequency has a low-pass characteristic
• The velocity amplitude vs. frequency has a bandpass characteristic
• The acceleration amplitude vs. frequency has a high-pass characteristic (same as the pressure response of the sub itself)
This nature then becomes relevant when such a sub is placed in multiples in a DBA. One might rightly wonder how the subs' response according to the T/S theory is modified by the DBA. I would argue that, to determine such an effect, such a simulation would best be done with a stimulus having a constant acceleration amplitude with frequency.
Replies
69
Views
6K
Replies
19
Views
1K
Replies
4
Views
582
Replies
25
Views
1K
Replies
4
Views
667 | 4,550 | 20,611 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-22 | longest | en | 0.945306 |
https://eikaiwa.dmm.com/uknow/questions/35977/ | 1,721,902,685,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763857355.79/warc/CC-MAIN-20240725084035-20240725114035-00890.warc.gz | 188,357,479 | 15,768 | AI講師ならいつでも相談可能です！
AI講師に聞く
5人に1人が75歳以上って英語でなんて言うの？
( NO NAME )
2018/01/21 12:43
43
16087
• Every 1 in ５ people are 75 or older.
• Every 1 out of ５ people are 75 or older.
• One in five people are 75 or older.
5人に1人が75歳か75歳以上。 Every 1 in 5 people are 75 or older. Every 1 out of 5 people are 75 or older. 5人に1人 1 in 5 one in five every 1 in 5 ご参考になれば幸いです。
• One is five people are 75 or older.
• 1 in 5 people are over 75.
"One is five people are 75 or older." and "1 in 5 people are over 75." are both short ways of expressing that in every five people one is 75 or older.
どちらの回答例も、5人に1人が75歳以上という意味の短文です。
• In every five people, one is 75 years old or above
• The ratio of people who are 75 years old and above is 1:5
*In every five people, one is 75 years old or above. - This means that it is possible that in a group of 5 people amongst them one is 75 years. *The ratio of people who are 75 years old and above is 1:5- Ratio means proportion. It is usually used to refer to probability. 1:5 read as one is to five.
*In every five people, one is 75 years old or above. ５人に１人は７５歳以上です。 ５人中の１人は７５歳だという可能性を表しています。 *The ratio of people who are 75 years old and above is 1:5 ７５歳以上の人口割合は1:5です。 ratioとは割合という意味です。通常、確率を表します。 1:5はone is to fiveと読みます。
• Twenty percent of people are over 75 years old.
• Eighty percent of people are under 75 years of age.
You may either describe the number of older people whose age is past the threshold of 75 as being 'twenty percent,' or, as a statistic of age distribution for a country, you could refer to the 'eighty percent' of people who are under 75. "My grandfather is one of the twenty percent of pensioners aged over 75."
75歳を過ぎた高齢者の数が "twenty percent"(20%)であるまたは、その国の年齢分布を考えて、75歳以下の人が"eighty percent"(80%)であるということもできます。 "My grandfather is one of the twenty percent of pensioners aged over 75." （私のおじいちゃんは20%いる75歳以上の年金受給者のひとりです。）
• 1 in 5 people are over 75.
• One in every five people is 75 years or older.
• 25 percent of people are 75 years or over.
These sentences give different examples of how you can say '1 in 5'. It is common in English to just say an age as '75' or whatever the number is, without adding 'years' or 'years old'. The terms 'over', 'or older' and 'or over' all describe that the age includes 75 years and then goes above this as well. Therefore, they are not just 75 and they are not younger than 75.
これらの文は「5人中1人」をいろいろな言い方で表しています。 英語では年齢を、'years' や 'years old' を加えず単に '75' などと表すのが一般的です。 'over' と 'or older' 'or over' これらは全て、75歳を含みその上の範囲をいいます。ですから、単に75歳でもないですし、74歳以下でもありません。
• Every 1 in 5 people are 75 or older
• Every 1 out of 5 people are 75 or older
If you are talking about statistics then you can say 'every 1 in 5' means one person out of 5 people or a fifth (1/5)
• One in five people are 75 years old or older in Japan.
One in five' or 'one-fifth' is the correct way to say this amount. It is also important to clarify which population you are referring to unless it is already very obvious from the previous conversation.
One in five'（5分の1）または 'One-fifth'（同）がこの数量の正しい言い方です。 また、それまでの会話で明白になっている場合を除いて、どこの人口について言っているのか明確にすることも重要です。
43
16087
• 役に立った：43
• PV：16087
AI講師なら
いつでも相談可能です！
AI講師に聞く
アンカーランキング
• 週間
• 月間
• 総合
メニュー | 1,135 | 3,222 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2024-30 | latest | en | 0.809527 |
https://solvedlib.com/n/question-131-pts425-kg-ball-is-dropped-from-restat-the-top,13115303 | 1,670,154,194,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710972.37/warc/CC-MAIN-20221204104311-20221204134311-00223.warc.gz | 556,053,976 | 20,160 | # Question 131 pts425 kg ball Is dropped from restat the top ofa 125 m tall building How much kinetic energy
###### Question:
Question 13 1 pts 425 kg ball Is dropped from restat the top ofa 125 m tall building How much kinetic energy does It have when it is 5.12 mabove the ground? 3142 J 2937 J 1869 3324 Nung ot the bhott
#### Similar Solved Questions
##### A health journal found that 56.2% of college students who livedin coed dormitories consumed alcohol weekly (comparedwith 26.5% who lived in single-sex dormitories). Arandom sample of 150 students who live in coed dormitories wasselected. Complete parts a through d below.a. What is the probability that more than 50% of the studentsin the sample consume alcohol weekly?b. What is the probability that less than 65% of the studentsin the sample consume alcohol weekly?c. What impact would changing the
A health journal found that 56.2% of college students who lived in coed dormitories consumed alcohol weekly (compared with 26.5% who lived in single-sex dormitories). A random sample of 150 students who live in coed dormitories was selected. Complete parts a through d below. a. What is the probabili...
##### PLEASE SOLVE USING EXCEL. PLEASE INCLUDE STEP BY STEP ON HOW TO SOLVE THIS QUESTION ON...
PLEASE SOLVE USING EXCEL. PLEASE INCLUDE STEP BY STEP ON HOW TO SOLVE THIS QUESTION ON EXCEL.. THANK YOU!!!! 1. DEVELOP SPREADSHEET MODELS FOR FORECASTING ATTENDANCE USING SINGLE MOVING AVERAGE AND SINGLE EXPONENTIAL SMOOTHING. (completed) 2. Using MAD, MSE, AND MAPE as guidance, find the best numbe...
##### If C1and C3 = 1then C + C2+ C2+C3
If C1 and C3 = 1then C + C2 + C2 +C3...
##### 28. Show that the systemX1 + X2 + X3 = YI, 2x1 + 3x2 +X3 =Y2, 3x1 + 5x2 + X3 =Y3,has an infinite number of solutions provided that (y1, Y2, Y3) lies on the plane with equation Y1 L 2y2 + Y3 = 0.
28. Show that the system X1 + X2 + X3 = YI, 2x1 + 3x2 +X3 =Y2, 3x1 + 5x2 + X3 =Y3, has an infinite number of solutions provided that (y1, Y2, Y3) lies on the plane with equation Y1 L 2y2 + Y3 = 0....
##### 4. (a) (6 marks) Let A be a square matrix with eigenvector v, and corresponding eigenvalue...
4. (a) (6 marks) Let A be a square matrix with eigenvector v, and corresponding eigenvalue 1. Let c be a scalar. Show that A-ch has eigenvector v, and corresponding eigenvalue X-c. (b) (8 marks) Let A = (33) i. Find the eigenvalues of A. ii. For one of the eigenvalues you have found, calculate the c...
##### Mike also needs to understand the biological effects of insulin when it is released in response to high blood glucose: To help him with this, use your biochemical knowledge of metabolism to complete the following rable Table 1).Table 1. The biological effects of insulin. Target How is this enzymes_ activity regulated (eg What is the effect on both enzymeltransport- phosphorylation, dephosphorylation, etc)? activity and levels of pathway products:Glucose transportersFatty acid synthaseGlycogen sy
Mike also needs to understand the biological effects of insulin when it is released in response to high blood glucose: To help him with this, use your biochemical knowledge of metabolism to complete the following rable Table 1). Table 1. The biological effects of insulin. Target How is this enzymes_...
##### Using Zu Odservalions tne toliowing regression oulpuliS ODlainea Trom eslimaling >PoPixPzaP3xaCoefficients 15 _ 58Standard Error 28Stat P-value 4 .75 000 000 -2.20 043 4.50 000Interceptxa3 . 80Compute Y for x= 11and d=1; then compute for x = 11and d= 0. (Round intermediate calculations to at least 4 decimal places and final answers to 2 decimal places:)11 and d =11 and d = 0
Using Zu Odservalions tne toliowing regression oulpuliS ODlainea Trom eslimaling > Po Pix Pza P3xa Coefficients 15 _ 58 Standard Error 28 Stat P-value 4 .75 000 000 -2.20 043 4.50 000 Intercept xa 3 . 80 Compute Y for x= 11and d=1; then compute for x = 11and d= 0. (Round intermediate calculations...
##### Question 84 ptsTwo hundred undergraduate students were randomly selected from a university that has 47,000 students in total Systolic blood pressure was tested on the 200 students. The sample standard deviation is 8.6 mmHg What is the estimated standard error of the sample mean?0.970.790 1.790.61
Question 8 4 pts Two hundred undergraduate students were randomly selected from a university that has 47,000 students in total Systolic blood pressure was tested on the 200 students. The sample standard deviation is 8.6 mmHg What is the estimated standard error of the sample mean? 0.97 0.79 0 1.79 0...
##### Use residues and the method of indented paths to evaluate the following in- proper integral: Sin / ds 7(+4) Justify your answer completely: Hint: Use the function F(:) = 7+48' and the contourYRYps12
Use residues and the method of indented paths to evaluate the following in- proper integral: Sin / ds 7(+4) Justify your answer completely: Hint: Use the function F(:) = 7+48' and the contour YR Yps 12...
##### Help ercises i Brokia Electronics manufactures three cell phone models, which differ only in the components...
Help ercises i Brokia Electronics manufactures three cell phone models, which differ only in the components included: Basic, Photo, and UrLife. Production takes place in two departments, Assembly and Special Packaging. The Basic and Photo models are complete after Assembly. The UrLife model goes fro...
##### Old Country Links, Inc.. produces sausages in three production departments-Mixing, Casing and Curing, and Packaging. In...
Old Country Links, Inc.. produces sausages in three production departments-Mixing, Casing and Curing, and Packaging. In the Mixing Department, meats are prepared and ground and then mixed with spices. The spiced meat mixture is then transferred to the Casing and Curing Department, where the mixture ...
##### 8.4.5-TQuestion HelpA simple random sample of 30 men frorn minute_ normally The normal disuibuted population range esults in a standard deviation of 8 pulse rales of adulls beats per thumb Is typically given as 60 Io 100 beats per minute. If the range applied to that normal range result is rule of with standard 0.05 significance level l0 deviation of 10 beals per minute Use the sample results test the claim that pulse rales of men nave minute: Complete parts (a) through (d) below , standard devi
8.4.5-T Question Help A simple random sample of 30 men frorn minute_ normally The normal disuibuted population range esults in a standard deviation of 8 pulse rales of adulls beats per thumb Is typically given as 60 Io 100 beats per minute. If the range applied to that normal range result is rule of...
##### 7:25 Question 17 of 40 Which of the following molecules is unsaturated? A) A B)B C)...
7:25 Question 17 of 40 Which of the following molecules is unsaturated? A) A B)B C) C D) D V Learn More in OpenStax Periodic Table...
##### What is your understanding of tissue classification
What is your understanding of tissue classification...
##### (25 pts Disgonclize the following matrix A jE possible; Zbat j8, fod an invertible Noatrix and diagonal matrix D such that A = PDP-1 To A =
(25 pts Disgonclize the following matrix A jE possible; Zbat j8, fod an invertible Noatrix and diagonal matrix D such that A = PDP-1 To A =...
##### What is the pressure (in atm and mmHg) of 5.00 g of oxygen gas at 35oC...
What is the pressure (in atm and mmHg) of 5.00 g of oxygen gas at 35oC that is contained in a 255 mL container? Can u please show dymentional analysis?...
##### ACE-I are known to be renoprotective (protect the kidney) Is BB renoprotective ??? YES or NO...
ACE-I are known to be renoprotective (protect the kidney) Is BB renoprotective ??? YES or NO and WHY (ONLY 3 lines) Is CCB renoprotective ?? YES or NO and WHY (only 3 lines)...
##### Find an equation of the line tangent to the following curve at the point (a,fla))f(x) =4e* + 13x; a =0
Find an equation of the line tangent to the following curve at the point (a,fla)) f(x) =4e* + 13x; a =0...
##### Please ignore the work it is incorrect 4. The Peacock Electric Company thinks that 40% of...
please ignore the work it is incorrect 4. The Peacock Electric Company thinks that 40% of their customers would be interested in bundling solar power purchasing with their electric bill. A random sample of 400 households reveals that 110 of the households are interested in this. What is the critica...
##### Zkeme Iloq 01 Jaiem a41 PJ/y} auo JO} axe} ! Illm Buo| Moy 'sunjjo uoneinjji? ou !i Pue 'Mw Osejo 31e1 a41 Je ABjaua Bupjnpoud SI 3403 asoym Jonjeaj auijooj Jujuoinjunyiew O1ui padwnp aje JoSI Je Jalem Bui/oo) Ajuaajaua jo 8XsOT Xvz JI "9 peappnu
zkeme Iloq 01 Jaiem a41 PJ/y} auo JO} axe} ! Illm Buo| Moy 'sunjjo uoneinjji? ou !i Pue 'Mw Osejo 31e1 a41 Je ABjaua Bupjnpoud SI 3403 asoym Jonjeaj auijooj Jujuoinjunyiew O1ui padwnp aje JoSI Je Jalem Bui/oo) Ajuaajaua jo 8XsOT Xvz JI "9 peappnu...
##### An article will be written on a 24 cm" part of a paper. Since [.5 cm from the top and bottom and cm from the right and left will be left blank; how many em" should the area of this paper be at least? ( 15 p.)
An article will be written on a 24 cm" part of a paper. Since [.5 cm from the top and bottom and cm from the right and left will be left blank; how many em" should the area of this paper be at least? ( 15 p.)...
##### Can 50mm, 13mm and 12mm be a right triangle?
Can 50mm, 13mm and 12mm be a right triangle?...
##### Find the number of units _ x, that produces the maximum prolit P If C(x) =20 - 48x and p-84 - 2x.36 units9 units48 units192 units
Find the number of units _ x, that produces the maximum prolit P If C(x) =20 - 48x and p-84 - 2x. 36 units 9 units 48 units 192 units...
##### 3= 7Fad Ret k hiargl A8c , If 4-32" use +e lcud Aines b tind b , a = 3.8 inABC , A-30 B-76 and a-8cm Zn hargl Rnd tbe lergth 4 sice € using te law { Sius -Slale Csmpllke Sck Y= 241ex-5) _ Gnph oe Pevica and Phase ckfh: He ampltxd_
3= 7Fad Ret k hiargl A8c , If 4-32" use +e lcud Aines b tind b , a = 3.8 in ABC , A-30 B-76 and a-8cm Zn hargl Rnd tbe lergth 4 sice € using te law { Sius - Slale Csmpllke Sck Y= 241ex-5) _ Gnph oe Pevica and Phase ckfh: He ampltxd_... | 2,967 | 10,175 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2022-49 | latest | en | 0.85727 |
http://www.cs.rug.nl/svcg/People/JasperVanDeGronde | 1,550,455,372,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247484020.33/warc/CC-MAIN-20190218013525-20190218035525-00357.warc.gz | 351,221,905 | 9,425 | Scientific Visualization and Computer Graphics > CS > BI > FSE > RUG
# Jasper van de Gronde
• room number: 491 (Bernoulliborg, building 5161)
• e-mail: j.j.van.de.gronde rug.nl
## Research interests
I am interested in all sorts of image processing topics, including compressed sensing, linear and morphological filters, and deep learning, with a special interest in applications on large/high-dimensional and non-scalar data. Making proper use of vector- or tensor-valued data, possibly as an intermediate step, also interests me greatly, as does the interplay between (mostly) linear approaches and morphological approaches. My PhD thesis was on morphological operators for tensor images, exploring shape and structure in movement and direction dependence. I defended my thesis on 30 June 2015, and am currently a post-doc.
Tensor images (or volumes) are just like ordinary images or volumes, except that at each point we have a vector or matrix (tensors are essentially a generalization of vectors and matrices). To visualize a tensor image with matrices at each position, we typically use "glyphs": for each position a small glyph is drawn whose shape represents the matrix (S). The glyph is often a unit sphere deformed in such a way that each vector v on the unit sphere is given the magnitude v.(S.v). Below you can interact with a live demonstration of this technique (requires a WebGL capable browser, I recommend Firefox, and possibly some patience):
WebGL based demonstration of tensor field glyphs Rotating Click and drag. Zooming Press +/- or use mousewheel. Glyph resizing Hold Ctrl and zoom. Press 'u' to make all glyphs approximately the same size. Cutting plane Press 'c'. The plane will be parallel to the screen and through the origin. In this case, the matrices describe stresses at atomic positions in a nanowire (courtesy of S.S.R. Saane, Micromechanics group, RuG). Blue shows positive values(/radii), while orange shows negative values.
## Demos
Apart from the rendering above, here is some more code you can try out both in the browser and using Node.js:
There are also some demos that can only be run off-line:
Various projects not (necessarily) associated with a particular publication can be found on Github (also see the SciJS repository). | 487 | 2,278 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2019-09 | longest | en | 0.890597 |
https://thestudyofthehousehold.com/qa/question-what-are-the-types-of-probability-distribution.html | 1,627,710,403,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154053.17/warc/CC-MAIN-20210731043043-20210731073043-00242.warc.gz | 563,076,870 | 18,382 | # Question: What Are The Types Of Probability Distribution?
## What are four common types of continuous distribution?
Types of Continuous Probability DistributionBeta distribution,Cauchy distribution,Exponential distribution,Gamma distribution,Logistic distribution,Weibull distribution.Jan 25, 2021.
## What is standard probability distribution?
The Standard Normal Distribution Table. The standard normal distribution table provides the probability that a normally distributed random variable Z, with mean equal to 0 and variance equal to 1, is less than or equal to z. It does this for positive values of z only (i.e., z-values on the right-hand side of the mean).
## What are the three major distribution strategies?
Types of Distribution Strategies with ExamplesDirect Distribution Strategy.Indirect Distribution Strategy.Intensive Distribution Strategy.Exclusive Distribution Strategy.Aug 9, 2019
## What are different types of distribution strategies?
1) Indirect distribution.2) Direct distribution.3) Intensive distribution.4) Selective distribution.5) Exclusive distribution.May 25, 2018
## What is the first property of probability distribution?
General Properties of Probability Distributions The sum of all probabilities for all possible values must equal 1. Furthermore, the probability for a particular value or range of values must be between 0 and 1. Probability distributions describe the dispersion of the values of a random variable.
## What is the best distribution channel for a new product?
These are the 8 most important distribution channels to know:Direct sales. … Retailer. … Intensive distribution. … Selective distribution. … Exclusive distribution. … Dual distribution. … Wholesaler. … Channel partners or value-added resellers.May 10, 2021
## What are the 3 levels of distribution?
The Three Types of DistributionIntensive Distribution: As many outlets as possible. The goal of intensive distribution is to penetrate as much of the market as possible.Selective Distribution: Select outlets in specific locations. … Exclusive Distribution: Limited outlets.
## What are the 4 types of distribution?
There are four types of distribution channels that exist: direct selling, selling through intermediaries, dual distribution, and reverse logistics channels.
## What is an example of a continuous probability distribution?
Continuous probability distribution: A probability distribution in which the random variable X can take on any value (is continuous). Because there are infinite values that X could assume, the probability of X taking on any one specific value is zero. … The normal distribution is one example of a continuous distribution.
## What is a probability distribution table?
A probability distribution table links every outcome of a statistical experiment with the probability of the event occurring. The outcome of an experiment is listed as a random variable, usually written as a capital letter (for example, X or Y).
## What is the difference between discrete and continuous probability distribution?
A discrete distribution means that X can assume one of a countable (usually finite) number of values, while a continuous distribution means that X can assume one of an infinite (uncountable) number of different values.
## How many types of distribution are there in statistics?
List of Statistical DistributionsArcsine Distribution.Bates Distribution.Bernoulli Distribution.Beta Binomial Distribution.Beta Distribution.Beta Geometric Distribution (Type I Geometric)Binomial Distribution.Bimodal Distribution.More items…•Oct 12, 2013
## What are the types of discrete probability distribution?
Discrete Probability DistributionsBernoulli Distribution. … Binomial Distribution. … Hypergeometric Distribution. … Negative Binomial Distribution. … Geometric Distribution. … Poisson Distribution. … Multinomial Distribution.Jan 25, 2021
## What is discrete probability distribution example?
A discrete probability distribution counts occurrences that have countable or finite outcomes. This is in contrast to a continuous distribution, where outcomes can fall anywhere on a continuum. Common examples of discrete distribution include the binomial, Poisson, and Bernoulli distributions.
## What is a discrete probability distribution What are the two conditions?
In the development of the probability function for a discrete random variable, two conditions must be satisfied: (1) f(x) must be nonnegative for each value of the random variable, and (2) the sum of the probabilities for each value of the random variable must equal one.
## What is the first property of discrete probability distribution?
A discrete probability distribution function has two characteristics: Each probability is between zero and one, inclusive. The sum of the probabilities is one.
## What is the physical distribution?
Physical distribution is the set of activities concerned with efficient movement of finished goods from the end of the production operation to the consumer. … Physical distribution is part of a larger process called “distribution,” which includes wholesale and retail marketing, as well the physical movement of products.
## What are examples of distribution channels?
Distribution channels include wholesalers, retailers, distributors, and the Internet. In a direct distribution channel, the manufacturer sells directly to the consumer. Indirect channels involve multiple intermediaries before the product ends up in the hands of the consumer. | 1,031 | 5,521 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2021-31 | latest | en | 0.820611 |
https://www.teachtasticiep.com/iep-goal/iep-goal-to-add-three-numbers-using-doubles | 1,719,248,988,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865401.1/warc/CC-MAIN-20240624151022-20240624181022-00389.warc.gz | 867,809,747 | 159,546 | top of page
I'm a paragraph. Click here to add your own text and edit me. It's easy.
Learning Standard
### 1.NBT.C.4
Add within 100, including adding a two-digit number and a one-digit number, and adding a two-digit number and a multiple of 10, using concrete models or drawings and strategies based on place value, properties of operations, and/or the relationship between addition and subtraction; relate the strategy to a written method and explain the reasoning used. Understand that in adding two-digit numbers, one adds tens and tens, ones and ones; and sometimes it is necessary to compose a ten.
Target IEP Goal
By (date), when given exercises with addition strategies up to 20, the student will add three numbers using doubles, improving number and operations in base ten skills from 0/10 work samples out of ten consecutive trials to 8/10 work samples in ten consecutive trials.
Teach Tastic IEP goals written to be SMART: Specific, Measurable, Attainable, Results-Oriented, and Time-Bound.
IEP Goal Objectives
1
Add Doubles
By (date), when given exercises with addition strategies up to 20, the student will add doubles, improving operations and algebraic thinking skills from 0/10 work samples out of ten consecutive trials to 8/10 work samples in ten consecutive trials.
1
2
Complete Doubles Addition with Models
By (date), when given exercises with addition strategies up to 20, the student will complete doubles addition with models, improving operations and algebraic thinking skills from 0/10 work samples out of ten consecutive trials to 8/10 work samples in ten consecutive trials.
1
3
Complete Addition Facts (sums up to 10)
By (date), when given exercises with addition up to 10, the student will complete addition facts (sums up to 10), improving operations and algebraic thinking skills from 0/10 work samples out of ten consecutive trials to 8/10 work samples in ten consecutive trials.
TeachTastic Center: Complete Addition Facts (Sums up to 10) 3714 Write the Room: Complete Addition Facts (Sums up to 10) 3714 Addition Sums up to 10 Worksheet Pack 3714
4
Count On to Add (sums up to 10)
By (date), when given exercises with addition strategies up to 10, the student will count on to add (sums up to 10), improving operations and algebraic thinking skills from 0/10 work samples out of ten consecutive trials to 8/10 work samples in ten consecutive trials.
1
bottom of page | 559 | 2,425 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2024-26 | latest | en | 0.90811 |
http://technet.microsoft.com/en-us/library/cc723005.aspx | 1,408,512,029,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1408500800168.29/warc/CC-MAIN-20140820021320-00016-ip-10-180-136-8.ec2.internal.warc.gz | 196,776,666 | 22,041 | # STDEVA
Archived content. No warranty is made as to technical accuracy. Content may contain URLs that were valid when originally published, but now link to sites or pages that no longer exist.
Estimates standard deviation based on a sample. The standard deviation is a measure of how widely values are dispersed from the average value (the mean). Text and logical values such as TRUE and FALSE are included in the calculation.
Syntax
STDEVA ( value1 ,value2, ...)
Value1,value2, ... are 1 to 30 values corresponding to a sample of a population.
Remarks
• STDEVA assumes that its arguments are a sample of the population. If your data represents the entire population, you must compute the standard deviation using STDEVPA.
• Arguments that contain TRUE evaluate as 1; arguments that contain text or FALSE evaluate as 0 (zero). If the calculation must not include text or logical values, use the STDEV spreadsheet function instead.
• The standard deviation is calculated using the "nonbiased" or "n-1" method.
• STDEVA uses the following formula:
Example
Suppose 10 tools stamped from the same machine during a production run are collected as a random sample and measured for breaking strength.
St1
St2
St3
St4
St5
St6
St7
St8
St9
St10
Formula
Description (Result)
1345
1301
1368
1322
1310
1370
1318
1350
1303
1299
=STDEVA([St1], [St2], [St3], [St4], [St5], [St6], [St7], [St8], [St9], [St10])
Standard deviation of breaking strength for all the tools (27.46391572)
(1500 characters remaining) | 387 | 1,531 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2014-35 | longest | en | 0.873837 |
https://www.lumoslearning.com/llwp/resources/pinterest_listings.html?id=1754945 | 1,600,765,157,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400204410.37/warc/CC-MAIN-20200922063158-20200922093158-00432.warc.gz | 937,616,934 | 41,528 | Pinned By -
#### OVERVIEW:
is a educational Infographics - By .It helps students in grades HS practice the following standards HSN.RN.A.1.
This page not only allows students and teachers to get information about the book but also find engaging Sample Questions, Videos, Pins, Worksheets, Apps related to the following topics.
1. HSN.RN.A.1 : Explain how the definition of the meaning of rational exponents follows from extending the properties of integer exponents to those values, allowing for a notation for radicals in terms of rational exponents. For example, we define 51/3 to be the cube root of 5 because we want (51/3)3 = 5(1/3)3 to hold, so (51/3)3 must equal 5.
### RELATED PINS
Rate this Pin?
0
0 Ratings & 0 Reviews
5
0
0
4
0
0
3
0
0
2
0
0
1
0
0 | 213 | 765 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2020-40 | latest | en | 0.873746 |
https://pdfslide.us/documents/stochastic-calculus-for-finance-i-blue-ridge-community-seminarpdffilesshrevesteven.html | 1,686,131,740,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224653631.71/warc/CC-MAIN-20230607074914-20230607104914-00231.warc.gz | 494,420,579 | 74,689 | of 60 /60
Steven E. Shreve Stochastic Calculus for Finance I Student’s Manual: Solutions to Selected Exercises December 14, 2004 Springer Berlin Heidelberg NewYork Hong Kong London Milan Paris Tokyo
# Stochastic Calculus for Finance I - Blue Ridge Community ...academic.brcc.edu/crow/Projects/Quant Seminar/PdfFiles/Shreve... · Steven E. Shreve Stochastic Calculus for Finance I
Embed Size (px)
### Text of Stochastic Calculus for Finance I - Blue Ridge Community ...academic.brcc.edu/crow/Projects/Quant...
Steven E. Shreve
Stochastic Calculus for Finance I
Student’s Manual: Solutions to Selected
Exercises
December 14, 2004
Springer
Berlin Heidelberg NewYork
Hong Kong London
Milan Paris Tokyo
Contents
1 The Binomial No-Arbitrage Pricing Model . . . . . . . . . . . . . . . . 1
1.7 Solutions to Selected Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
2 Probability Theory on Coin Toss Space . . . . . . . . . . . . . . . . . . . . 7
2.9 Solutions to Selected Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
3 State Prices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
3.7 Solutions to Selected Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
4 American Derivative Securities . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
4.9 Solutions to Selected Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
5 Random Walk . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
5.8 Solutions to Selected Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
6 Interest-Rate-Dependent Assets . . . . . . . . . . . . . . . . . . . . . . . . . . . 53
6.9 Solutions to Selected Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53
1
The Binomial No-Arbitrage Pricing Model
1.7 Solutions to Selected Exercises
Exercise 1.2. Suppose in the situation of Example 1.1.1 that the option sellsfor 1.20 at time zero. Consider an agent who begins with wealth X0 = 0and at time zero buys ∆0 shares of stock and Γ0 options. The numbers ∆0
and Γ0 can be either positive or negative or zero. This leaves the agent witha cash position of −4∆0 − 1.20Γ0. If this is positive, it is invested in themoney market; if it is negative, it represents money borrowed from the moneymarket. At time one, the value of the agent’s portfolio of stock, option andmoney market is
X1 = ∆0S1 + Γ0(S1 − 5)+ − 5
4(4∆0 + 1.20Γ0) .
Assume that both H and T have positive probability of occurring. Show thatif there is a positive probability that X1 is positive, then there is a positiveprobability that X1 is negative. In other words, one cannot find an arbitragewhen the time-zero price of the option is 1.20.
Solution. Considering the cases of a head and of a tail on the first toss, andutilizing the numbers given in Example 1.1.1, we can write:
X1(H) = 8∆0 + 3Γ0 −5
4(4∆0 + 1.20Γ0),
X1(T ) = 2∆0 + 0 · Γ0 −5
4(4∆0 + 1.20Γ0)
Adding these, we get
X1(H) + X1(T ) = 10∆0 + 3Γ0 − 10∆0 − 3Γ0 = 0,
or, equivalently,
2 1 The Binomial No-Arbitrage Pricing Model
X1(H) = −X1(T ).
In other words, either X1(H) and X1(T ) are both zero, or they have oppositesigns. Taking into account that both p > 0 and q > 0, we conclude that ifthere is a positive probability that X1 is positive, then there is a positiveprobability that X1 is negative.
Exercise 1.6 (Hedging a long position - one period.). Consider a bankthat has a long position in the European call written on the stock price inFigure 1.1.2. The call expires at time one and has strike price K = 5. InSection 1.1, we determined the time-zero price of this call to be V0 = 1.20.At time zero, the bank owns this option, which ties up capital V0 = 1.20.The bank wants to earn the interest rate 25% on this capital until time one,i.e., without investing any more money, and regardless of how the coin tossingturns out, the bank wants to have
5
4· 1.20 = 1.50
at time one, after collecting the payoff from the option (if any) at time one.Specify how the bank’s trader should invest in the stock and money marketto accomplish this.
Solution. The trader should use the opposite of the replicating portfoliostrategy worked out in Example 1.1.1. In particular, she should short 1
2 shareof stock, which generates \$2 income. She should invest this in the moneymarket. At time one, if the stock goes up in value, the bank has an optionworth \$3, has \$
(54 ·2)
= \$2.50 in the money market, and must pay \$4 to coverthe short position in the stock. This leaves the bank with \$1.50, as desired.On the other hand, if the stock goes down in value, then at time one the bankhas an option worth \$0, still has \$2.50 in the money market, and must pay\$1 to cover the short position in stock. Again, the bank has \$1.50, as desired.
Exercise 1.8 (Asian option). Consider the three-period model of Example1.2.1, with S0 = 4, u = 2, d = 1
2 , and take the interest rate r = 14 , so that
p = q = 12 . For n = 0, 1, 2, 3, define Yn =
∑n
k=0 Sk to be the sum of thestock prices between times zero and n. Consider an Asian call option thatexpires at time three and has strike K = 4 (i.e., whose payoff at time three is(
14Y3 − 4
)+). This is like a European call, except the payoff of the option is
based on the average stock price rather than the final stock price. Let vn(s, y)denote the price of this option at time n if Sn = s and Yn = y. In particular,
v3(s, y) =(
14y − 4
)+.
(i) Develop an algorithm for computing vn recursively. In particular, write aformula for vn in terms of vn+1.
(ii) Apply the algorithm developed in (i) to compute v0(4, 4), the price of theAsian option at time zero.
1.7 Solutions to Selected Exercises 3
(iii) Provide a formula for δn(s, y), the number of shares of stock which shouldbe held by the replicating portfolio at time n if Sn = s and Yn = y.
Solution.
(i), (iii) Assume that at time n, Sn = s and Yn = y. Then if the (n + 1)-sttoss results in H, we have
Sn+1 = us, Yn+1 = Yn + Sn+1 = y + us.
If the (n + 1)-st toss results in T , we have instead
Sn+1 = ds, Yn+1 = Yn + Sn+1 = y + ds.
Therefore, formulas (1.2.16) and (1.2.17) take the form
vn(s, y) =1
1 + r[pvn+1(us, y + us) + qvn+1(ds, y + ds)],
δn(s, y) =vn+1(us, y + us) − vn+1(ds, y + ds)
us − ds.
(ii) We first list the relevant values of v3, which are
v3(32, 60) = (60/4 − 4)+ = 11,
v3(8, 36) = (36/4 − 4)+ = 5,
v3(8, 24) = (24/4 − 4)+ = 2,
v3(8, 18) = (18/4 − 4)+ = 0.50,
v3(2, 18) = (18/4 − 4)+ = 0.50,
v3(2, 12) = (12/4 − 4)+ = 0,
v3(2, 9) = (9/4 − 4)+ = 0,
v3(.50, 7.50) = (7.50/4 − 4)+ = 0.
We next use the algorithm from (i) to compute the relevant values of v2:
v2(16, 28) =4
5
[1
2v3(32, 60) +
1
2v3(8, 36)
]= 6.40,
v2(4, 16) =4
5
[1
2v3(8, 24) +
1
2v3(2, 18)
]= 1,
v2(4, 10) =4
5
[1
2v3(8, 18) +
1
2v3(2, 12)
]= 0.20,
v2(1, 7) =4
5
[1
2v3(2, 9) +
1
2v3(.50, 7.50)
]= 0.
4 1 The Binomial No-Arbitrage Pricing Model
We use the algorithm again to compute the relevant values of v1:
v1(8, 12) = 45
[12v2(16, 28) + 1
2v2(4, 16)]
= 2.96,
v1(2, 6) = 45
[12v2(4, 10) + 1
2v2(1, 7)]
= 0.08.
Finally, we may now compute
v0(4, 4) =4
5
[1
2v1(8, 12) +
1
2v1(2, 6)
]= 1.216.
Exercise 1.9 (Stochastic volatility, random interest rate). Consider abinomial pricing model, but at each time n ≥ 1, the “up factor” un(ω1ω2 . . . ωn),the “down factor” dn(ω1ω2 . . . ωn), and the interest rate rn(ω1ω2 . . . ωn) areallowed to depend on n and on the first n coin tosses ω1ω2 . . . ωn. The initialup factor u0, the initial down factor d0, and the initial interest rate r0 are notrandom. More specifically, the stock price at time one is given by
S1(ω1) =
u0S0 if ω1 = H,d0S0 if ω1 = T,
and, for n ≥ 1, the stock price at time n + 1 is given by
Sn+1(ω1ω2 . . . ωnωn+1) =
un(ω1ω2 . . . ωn)Sn(ω1ω2 . . . ωn) if ωn+1 = H,dn(ω1ω2 . . . ωn)Sn(ω1ω2 . . . ωn) if ωn+1 = T.
One dollar invested in or borrowed from the money market at time zero growsto an investment or debt of 1 + r0 at time one, and, for n ≥ 1, one dollar in-vested in or borrowed from the money market at time n grows to an investmentor debt of 1 + rn(ω1ω2 . . . ωn) at time n + 1. We assume that for each n andfor all ω1ω2 . . . ωn, the no-arbitrage condition
0 < dn(ω1ω2 . . . ωn) < 1 + rn(ω1ω2 . . . ωn) < un(ω1ω2 . . . ωn)
holds. We also assume that 0 < d0 < 1 + r0 < u0.
(i) Let N be a positive integer. In the model just described, provide analgorithm for determining the price at time zero for a derivative securitythat at time N pays off a random amount VN depending on the result ofthe first N coin tosses.
(ii) Provide a formula for the number of shares of stock that should be held ateach time n (0 ≤ n ≤ N − 1) by a portfolio that replicates the derivativesecurity VN .
(iii) Suppose the initial stock price is S0 = 80, with each head the stock priceincreases by 10, and with each tail the stock price decreases by 10. Inother words, S1(H) = 90, S1(T ) = 70, S2(HH) = 100, etc. Assume theinterest rate is always zero. Consider a European call with strike price 80,expiring at time five. What is the price of this call at time zero?
1.7 Solutions to Selected Exercises 5
Solution.
(i) We adapt Theorem 1.2.2 to this case by defining
p0 =1 + r0 − d0
u0 − d0, q0 =
u0 − 1 − r0
u0 − d0,
and for each and and for all ω1ω2 . . . ωn,
pn(ω1ω2 . . . ωn) =1 + rn(ω1ω2 . . . ωn) − dn(ω1ω2 . . . ωn)
un(ω1ω2 . . . ωn) − dn(ω1ω2 . . . ωn),
qn(ω1ω2 . . . ωn) =un(ω1ω2 . . . ωn) − 1 − rn(ω1ω2 . . . ωn)
un(ω1ω2 . . . ωn) − dn(ω1ω2 . . . ωn).
In place of (1.2.16), we define for n = N − 1, N − 2, . . . , 1,
Vn(ω1ω2 . . . ωn) =1
1 + r
[pn(ω1ω2 . . . ωn)Vn+1(ω1ω2 . . . ωnH)
+qn(ω1ω2 . . . ωn)Vn+1(ω1ω2 . . . ωnT )],
and for the the case n = 0 we adopt the definition
V0 =1
1 + r
[p0V1(H) + q0V1(T )
].
(ii) The number of shares of stock that should be held at time n is still givenby (1.2.17):
∆n(ω1 . . . ωn) =Vn+1(ω1 . . . ωnH) − Vn+1(ω1 . . . ωnT )
Sn+1(ω1 . . . ωnH) − Sn+1(ω1 . . . ωnT ).
The proof that this hedge works, i.e., that taking the position ∆n in thestock at time n and holding it until time n+1 results in a portfolio whosevalue at time n + 1 is Vn+1, is the same as the proof given for Theorem1.2.2.
(iii) If the stock price at a particular time n is x, then the stock price at thenext time is either x + 10 or x − 10. That means that the up factor isun = x+10
xand the down factor is dn = x−10
x. The corresponding risk-
neutral probabilities are
pn =1 − dn
un − dn
=1 − x−10
xx+10
x− x−10
x
=1
2,
qn =un − 1
un = dn
=x+10
x− 1
x+10x
− x−10x
=1
2.
Because these risk-neutral probabilities do not depend on the time n noron the coin tosses ω1 . . . ωn, we can easily compute the risk-neutral prob-
ability of an arbitrary sequence ω1ω2ω3ω4ω5 to be(
12
)5= 1
32 .
6 1 The Binomial No-Arbitrage Pricing Model
There are three ways for the call with strike 80 to expire in the moneyat time 5: either the five tosses result in five heads (S5 = 130), result infour heads and one tail (S5 = 110), or result in three heads and two tails(S5 = 90). The risk-neutral probability of five heads is 1
32 . If a tail occurs,it can occur on any toss, and so there are five sequences that have fourheads and one tail. Therefore, the risk-neutral probability of four headsand one tail is 5
32 . Finally, if there are two tails in a sequence of five tosses,there 10 ways to choose the two tosses that are tails. Therefore, the risk-neutral probability of three heads and two tails is 10
32 . The time-zero priceof the call is
V0 =1
32· (130 − 80) +
5
32· (110 − 80) +
10
32· (90 − 80) = 9.375.
2
Probability Theory on Coin Toss Space
2.9 Solutions to Selected Exercises
Exercise 2.2. Consider the stock price S3 in Figure 2.3.1.
(i) What is the distribution of S3 under the risk-neutral probabilities p = 12 ,
q = 12 .
(ii) Compute ES1, ES2, and ES3. What is the average rate of growth of the
stock price under P?
(iii) Answer (i) and (ii) again under the actual probabilities p = 23 , q = 1
3 .
Solution.
(i) The distribution of S3 under the risk-neutral probabilities p and q is
32 8 2 .50p3 3p2q 3pq2 q3
With p = 12 , q = 1
2 , this becomes
32 8 2 .50.125 .375 .375 .125
(ii) By Theorem 2.4.4,
ES3
(1 + r)3= E
S2
(1 + r)2= E
S1
(1 + r)= ES0 = S0 = 4.
Therefore,
8 2 Probability Theory on Coin Toss Space
ES1 = (1 + r)S0 = (1.25)(4) = 5,
ES2 = (1 + r)2S0 = (1.25)2(4) = 6.25,
ES3 = (1 + r)3S0 = (1.25)3(4) = 7.8125.
In particular, we see that
ES3 = 1.25 · ES2, ES2 = 1.25 · ES1, ES1 = 1.25 · S0.
Thus, the average rate of growth of the stock price under P is the sameas the interest rate of the money market.
(iii) The distribution of S3 under the probabilities p and q is
32 8 2 .50p3 3p2q 3pq2 q3
With p = 23 , q = 1
3 , this becomes
32 8 2 .50.2963 .4444 .2222 .0371
To compute the average rate of growth, we reason as follows:
EnSn+1 = En
(Sn
Sn+1
Sn
)= SnEn
(Sn+1
Sn
)= (pu + qd)Sn.
In our case,
pu + qd =2
3· 2 +
1
3· 12 = 1.5.
In other words, the average rate of growth of the stock price under theactual probabilities is 50%. Finally, taking expectations, we have
ESn+1 = E(EnSn+1
)= 1.5 · ESn,
so that
ES1 = 1.5 · ES0 = 6,
ES2 = 1.5 · ES1 = 9,
ES3 = 1.5 · ES2 = 13.50.
Exercise 2.3. Show that a convex function of a martingale is a submartin-gale. In other words, let M0,M1, . . . ,MN be a martingale and let ϕ be aconvex function. Show that ϕ(M0), ϕ(M1), . . . , ϕ(MN ) is a submartingale.
Solution Let an arbitrary n with 0 ≤ n ≤ N − 1 be given. By the martingaleproperty, we have
EnMn+1 = Mn,
2.9 Solutions to Selected Exercises 9
and henceϕ(EnMn+1) = ϕ(Mn).
On the other hand, by the conditional Jensen’s inequality, we have
Enϕ(Mn+1) ≥ ϕ(EnMn+1).
Combining these two, we get
Enϕ(Mn+1) ≥ ϕ(Mn),
and since n is arbitrary, this implies that the sequence of random variablesϕ(M0), ϕ(M1), . . . , ϕ(MN ) is a submartingale.
Exercise 2.6 (Discrete-time stochastic integral). Suppose M0,M1, . . . ,MN is a martingale, and let ∆0,∆1, . . . ,∆N−1 be an adapted process. Definethe discrete-time stochastic integral (sometimes called a martingale transform)I0, I1, . . . , IN by setting I0 = 0 and
In =n−1∑
j=0
∆j(Mj+1 − Mj), n = 1, . . . , N.
Show that I0, I1, . . . , IN is a martingale.
Solution. Because In+1 = In +∆n(Mn+1 −Mn) and In, ∆n and Mn dependon only the first n coin tosses, we may “take out what is known” to write
En[In+1] = En
[In + ∆n(Mn+1 − Mn)
]= In + ∆n
(En[Mn+1] − Mn
).
However, En[Mn+1] = Mn, and we conclude that En[In+1] = In, which is themartingale property.
Exercise 2.8. Consider an N -period binomial model.
(i) Let M0,M1, . . . ,MN and M ′0,M
′1, . . . ,M
′N be martingales under the risk-
neutral measure P. Show that if MN = M ′N (for every possible outcome
of the sequence of coin tosses), then, for each n between 0 and N , we haveMn = M ′
n (for every possible outcome of the sequence of coin tosses).
(ii) Let VN be the payoff at time N of some derivative security. This is arandom variable that can depend on all N coin tosses. Define recursivelyVN−1, VN−2, . . . , V0 by the algorithm (1.2.16) of Chapter 1. Show that
V0,V1
1 + r, . . . ,
VN−1
(1 + r)N−1,
VN
(1 + r)N
is a martingale under P.
10 2 Probability Theory on Coin Toss Space
(iii) Using the risk-neutral pricing formula (2.4.11) of this chapter, define
V ′n = En
[VN
(1 + r)N−n
], n = 0, 1, . . . , N − 1.
Show that
V ′0 ,
V ′1
1 + r, . . . ,
V ′N−1
(1 + r)N−1,
VN
(1 + r)N
is a martingale.
(iv) Conclude that Vn = V ′n for every n (i.e., the algorithm (1.2.16) of Theorem
1.2.2 of Chapter 1 gives the same derivative security prices as the risk-neutral pricing formula (2.4.11) of Chapter 2).
Solution.
(i) We are given that Mn = M ′N . For n between 0 and N − 1, this equality
and the martingale property imply
Mn = En[MN ] = En[M ′N ] = Mn.
(ii) For n between 0 and N − 1, we compute the following conditional expec-tation:
En
[Vn+1
(1 + r)n+1
](ω1ω2 . . . ωn)
= pVn+1(ω1ω2 . . . ωnH)
(1 + r)n+1+ q
Vn+1(ω1ω2 . . . ωnT )
(1 + r)n+1
=Vn(ω1ω2 . . . ωn)
(1 + r)n,
where the second equality follows from (1.2.16). This is the martingalproperty for Vn
(1+r)n .
(iii) The martingale property forV ′
n
(1+r)n follows from the iterated conditioning
property (iii) of Theorem 2.3.2. According to this property, for n between0 and n − 1,
En
[V ′
n+1
(1 + r)n+1
]= En
[1
(1 + r)n+1En+1
[VN
(1 + r)N−(n+1)
]]
=1
(1 + r)nEn
[En+1
[VN
(1 + r)N−n
]]
=1
(1 + r)nEn
[VN
(1 + r)N−n
]
=V ′
n
(1 + r)n.
2.9 Solutions to Selected Exercises 11
(iv) Since the processes in (ii) and (iii) are martingales under the risk-neutralprobability measure and they agree at the final time N , they must agreeat all earlier times because of (i).
S0 = 4r0 = 1
4
S1(H) = 8
r1(H) = 1
4
S1(T ) = 2
r1(T ) = 1
2
S2(HH) = 12
S2(HT ) = 8
S2(TH) = 8
S2(TT ) = 2
Fig. 2.8.1. A stochastic volatility, random interest rate model.
Exercise 2.9 (Stochastic volatility, random interest rate). Considera two-period stochastic volatility, random interest rate model of the typedescribed in Exercise 1.9 of Chapter 1. The stock prices and interest rates areshown in Figure 2.8.1.
(i) Determine risk-neutral probabilities
P(HH), P(HT ), P(TH), P(TT ),
such that the time-zero value of an option that pays off V2 at time two isgiven by the risk-neutral pricing formula
V0 = E
[V2
(1 + r0)(1 + r1)
].
(ii) Let V2 = (S2 − 7)+. Compute V0, V1(H), and V1(T ).
(iii) Suppose an agent sells the option in (ii) for V0 at time zero. Compute theposition ∆0 she should take in the stock at time zero so that at time one,regardless of whether the first coin toss results in head or tail, the valueof her portfolio is V1.
(iv) Suppose in (iii) that the first coin toss results in head. What position∆1(H) should the agent now take in the stock to be sure that, regardless
12 2 Probability Theory on Coin Toss Space
of whether the second coin toss results in head or tail, the value of herportfolio at time two will be (S2 − 7)+?
Solution.
(i) For the first toss, the up factor is u0 = 2 and the down factor is d0 = 12 .
Therefore, the risk-neutral probability of a H on the first toss is
p0 =1 + r0 − d0
u0 − d0=
1 + 14 − 1
2
2 − 12
=1
2,
and the risk-neutral probability of T on the first toss is
q0 =u0 − 1 − r0
u0 − d0=
2 − 1 − 14
2 − 12
=1
2.
If the first toss results in H, then the up factor for the second toss is
u1(H) =S2(HH)
S1(H)=
12
8=
3
2,
and the down factor for the second toss is
d1(H) =S2(HT )
S1(H)=
8
8= 1.
It follows that the risk-neutral probability of getting a H on the secondtoss, given that the first toss is a H, is
p1(H) =1 + r1(H) − d1(H)
u1(H) − d1(H)=
1 + 14 − 1
32 − 1
=1
2,
and the risk-neutral probability of T on the second toss, given that thefirst toss is a H, is
q1(H) =u1(H) − 1 − r1(H)
u1(H) − d1(H)=
32 − 1 − 1
432 − 1
=1
2,
If the first toss results in T , then the up factor for the second toss is
u1(T ) =S2(TH)
S1(T )=
8
2= 4,
and the down factor for the second toss is
d1(H) =S2(TT )
S1(T )=
2
2= 1.
It follows that the risk-neutral probability of getting a H on the secondtoss, given that the first toss is a T , is
2.9 Solutions to Selected Exercises 13
p1(T ) =1 + r1(T ) − d1(T )
u1(T ) − d1(T )=
1 + 12 − 1
4 − 1=
1
6,
and the risk-neutral probability of T on the second toss, given that thefirst toss is a T , is
q1(T ) =u1(T ) − 1 − r1(T )
u1(T ) − d1(T )=
4 − 1 − 12
4 − 1=
5
6.
The risk-neutral probabilities are
P(HH) = p0p1(H) = 12 · 1
2 = 14 ,
P(HT ) = p0q1(H) = 12 · 1
2 = 14 ,
P(TH) = q0p1(T ) = 12 · 1
6 = 112 ,
P(TT ) = q0q1(T ) = 12 · 5
6 = 512 .
(ii) We compute
V1(H) =1
1 + r1(H)
[p1(H)V2(HH) + q1(H)V2(HT )
]
=4
5
[1
2· (12 − 7)+ +
1
2· (8 − 7)+
]
= 2.40,
V1(T ) =1
1 + r1(T )
[p1(T )V2(TH) + q1(T )V2(TT )
]
=2
3
[1
6· (8 − 7)+ +
5
6· (2 − 7)+
]
= 0.111111,
V0 =1
1 + r0[p0V1(H) + q0V1(T )]
=4
5
[1
2· 2.40 +
1
2· 0.1111
]
= 1.00444.
We can confirm this price by computing according to the risk-neutralpricing formula in part (i) of the exercise:
14 2 Probability Theory on Coin Toss Space
V0 = E
[V2
(1 + r0)(1 + r1)
]
=V2(HH)
(1 + r0)(1 + r1(H))· P(HH) +
V2(HT )
(1 + r0)(1 + r1(H))· P(HT )
+V2(TH)
(1 + r0)(1 + r1(T ))· P(TH) +
V2(TT )
(1 + r0)(1 + r1(T ))· P(TT )
=(12 − 7)+
(1 + 14 )(1 + 1
4 )· 1
4+
(8 − 7)+
(1 + 14 )(1 + 1
4 )· 1
4
+(8 − 7)+
(1 + 14 )(1 + 1
2 )· 1
12+
(2 − 7)+
(1 + 14 )(1 + 1
2 )· 5
12
= 0.80 + 0.16 + 0.04444 + 0
= 1.00444.
(iii) Formula (1.2.17) still applies and yields
∆0 =V1(H) − V1(T )
S1(H) − S1(T )=
2.40 − 0.111111
8 − 2= 0.381481.
(iv) Again we use formula (1.2.17), this time obtaining
∆1(H) =V2(HH) − V2(HT )
S2(HH) − S2(HT )=
(12 − 7)+ − (8 − 7)+
12 − 8= 1.
Exercise 2.11 (Put–call parity). Consider a stock that pays no dividendin an N -period binomial model. A European call has payoff CN = (SN −K)+
at time N . The price Cn of this call at earlier times is given by the risk-neutralpricing formula (2.4.11):
Cn = En
[CN
(1 + r)N−n
], n = 0, 1, . . . , N − 1.
Consider also a put with payoff PN = (K − SN )+ at time N , whose price atearlier times is
Pn = En
[PN
(1 + r)N−n
], n = 0, 1, . . . , N − 1.
Finally, consider a forward contract to buy one share of stock at time N forK dollars. The price of this contract at time N is FN = SN −K, and its priceat earlier times is
Fn = En
[FN
(1 + r)N−n
], n = 0, 1, . . . , N − 1.
(Note that, unlike the call, the forward contract requires that the stock bepurchased at time N for K dollars and has a negative payoff if SN < K.)
2.9 Solutions to Selected Exercises 15
(i) If at time zero you buy a forward contract and a put, and hold them untilexpiration, explain why the payoff you receive is the same as the payoffof a call; i.e., explain why CN = FN + PN .
(ii) Using the risk-neutral pricing formulas given above for Cn, Pn, and Fn
and the linearity of conditional expectations, show that Cn = Fn +Pn forevery n.
(iii) Using the fact that the discounted stock price is a martingale under therisk-neutral measure, show that F0 = S0 − K
(1+r)N .
(iv) Suppose you begin at time zero with F0, buy one share of stock, borrowingmoney as necessary to do that, and make no further trades. Show thatat time N you have a portfolio valued at FN . (This is called a static
replication of the forward contract. If you sell the forward contract forF0 at time zero, you can use this static replication to hedge your shortposition in the forward contract.)
(v) The forward price of the stock at time zero is defined to be that value ofK that causes the forward contract to have price zero at time zero. Theforward price in this model is (1 + r)NS0. Show that, at time zero, theprice of a call struck at the forward price is the same as the price of a putstruck at the forward price. This fact is called put–call parity.
(vi) If we choose K = (1 + r)NS0, we just saw in (v) that C0 = P0. Do wehave Cn = Pn for every n?
Solution
(i) Consider three cases:
Case I: SN = K. Then CN = PN = FN = 0;
Case II: SN > K. Then PN = 0 and CN = SN − K = FN ;
Case III: SN < K. Then CN = 0 and PN = K − SN = −FN .
In all three cases, we see that CN = FN + PN .
(ii)
Cn = En
[CN
(1 + r)N−n
]= En
[FN + PN
(1 + r)N−n
]
= En
[FN
(1 + r)N−n
]+ En
[PN
(1 + r)N−n
]= Fn + Pn.
(iii)
F0 = E
[FN
(1 + r)N
]= E
[SN − K
(1 + r)N
]
= E
[SN
(1 + r)N
]− E
[K
(1 + r)N
]= S0 −
K
(1 + r)N.
16 2 Probability Theory on Coin Toss Space
(iv) At time zero, your portfolio value is
F0 = S0 + (F0 − S0).
At time N , the value of the portfolio is
SN + (1 + r)N (F0 − S0) = SN + (1 + r)N
(− K
(1 + r)N
)
= SN − K = FN .
(v) First of all, if K = (1 + r)NS0, then, by (iii), F0 = 0. Further, if F0 = 0,then, by (ii), C0 = F0 + P0 = P0.
(vi) No. This would mean, in particular, that CN = PN , and hence (SN −K)+ = (K −SN )+, which in turn would imply that SN (ω) = K for all ω,which is not the case for most values of ω.
Exercise 2.13 (Asian option). Consider an N -period binomial model. AnAsian option has a payoff based on the average stock price, i.e.,
VN = f
(1
N + 1
N∑
n=0
Sn
),
where the function f is determined by the contractual details of the option.
(i) Define Yn =∑n
k=0 Sk and use the Independence Lemma 2.5.3 to showthat the two-dimensional process (Sn, Yn), n = 0, 1, . . . , N is Markov.
(ii) According to Theorem 2.5.8, the price Vn of the Asian option at time n issome function vn of Sn and Yn; i.e.,
Vn = vn(Sn, Yn), n = 0, 1, . . . , N.
Give a formula for vN (s, y), and provide an algorithm for computingvn(s, y) in terms of vn+1.
Solution
(i) Note first that
Sn+1 = Sn · Sn+1
Sn
, Yn+1 = Yn + Sn · Sn+1
Sn
,
and whereas Sn and Yn depend only on the first n tosses, Sn+1
Sndepends
only on toss n + 1. According to the Independence Lemma 2.5.3, for anyfunction hn+1(s, y) of dummy variables s and y, we have
2.9 Solutions to Selected Exercises 17
En [hn+1(Sn+1, Yn+1)] = En
[hn+1
(Sn · Sn+1
Sn
, Yn + Sn · Sn+1
Sn
)]
= hn(Sn, Yn),
where
hn(s, y) = Ehn+1
(s · Sn+1
Sn
, y + s · Sn+1
Sn
)
= phn+1(su, y + su) + qhn+1(sd, y + sd).
Because En [hn+1(Sn+1, Yn+1)] can be written as a function of (Sn, Yn),the two-dimensional process (Sn, Yn), n = 0, 1, . . . , N , is a Markov process.
(ii) We have the final condition VN (s, y) = f(
yN+1
). For n = N −1, . . . , 1, 0,
we have from the risk-neutral pricing formula (2.4.12) and (i) above that
Vn =1
1 + rEnVn+1 =
1
1 + rEnvn+1(Sn+1, Yn+1) = vn(Sn, Yn),
where
vn(s, y) =1
1 + r
[pvn+1(su, y + su) + qvn+1(sd, y + sd)
].
3
State Prices
3.7 Solutions to Selected Exercises
Exercise 3.1. Under the conditions of Theorem 3.1.1, show the followinganalogues of properties (i)–(iii) of that theorem:
(i′) P(
1Z
> 0)
= 1;
(ii′) E1Z
= 1;
(iii′) for any random variable Y ,
EY = E
[1
Z· Y]
.
In other words, 1Z
facilitates the switch from E to E in the same way Z
facilitates the switch from E to E.
Solution
(i′) Because P(ω) > 0 and P(ω) > 0 for every ω ∈ Ω, the ratio
1
Z(ω)=
P(ω)
P(ω)
is defined and positive for every ω ∈ Ω.
(ii′) We compute
E1
Z=∑
ω∈Ω
1
Z(ω)P(ω) =
ω∈Ω
P(ω)
P(ω)P(ω) =
ω∈Ω
P(ω) = 1.
20 3 State Prices
(iii′) We compute
E
[1
Z· Y]
=∑
ω∈Ω
P(ω)
P(ω)Y (ω)P(ω) =
ω∈Ω
Y (ω)P(ω) = EY.
Exercise 3.3. Using the stock price model of Figure 3.1.1 and the actualprobabilities p = 2
3 , q = 13 , define the estimates of S3 at various times by
Mn = En[S3], n = 0, 1, 2, 3.
Fill in the values of Mn in a tree like that of Figure 3.1.1. Verify that Mn,n = 0, 1, 2, 3, is a martingale.
Solution We note that M3 = S3. We compute M2 from the formula M2 =E2[S3]:
M2(HH) = 23S3(HHH) + 1
2S3(HHT ) = 23 · 32 + 1
3 · 8 = 24,
M2(HT ) = 23S3(HTH) + 1
3S3(HTT ) = 23 · 8 + 1
3 · 2 = 6,
M2(TH) = 23S3(THH) + 1
3S3(THT ) = 23 · 8 + 1
3 · 2 = 6,
M2(TT ) = 23S3(THH) + 1
2S3(THT ) = 23 · 2 + 1
3 · 0.50 = 1.50.
We next compute M1 from the formula M1 = E1[S3]:
M1(H) =4
9S3(HHH) +
2
9S3(HHT ) +
2
9S3(HTH) +
1
9S3(HTT )
=4
9· 32 +
2
9· 8 +
2
9· 8 +
1
9· 2
= 18,
M1(T ) =4
9S3(THH) +
2
9S3(THT ) +
2
9S3(TTH) +
1
9S3(TTT )
=4
9· 8 +
2
9· 2 +
2
9· 2 +
1
9· 0.50
= 4.50.
Finally, we compute
M0 = E[S3]
=8
27S3(HHH) +
4
27S3(HHT ) +
4
27S3(HTH) +
4
27S3(THH)
+2
27S3(HTT ) +
2
27S3(THT ) +
2
27S3(TTH) +
1
27S3(TTT )
=8
27· 32 +
4
27· 8 +
4
27· 8 +
4
27· 8 +
2
27· 2 +
2
27· 2 +
2
27· 2 +
1
27· 0.50
= 13.50.
3.7 Solutions to Selected Exercises 21
ZZ
ZZ
ZZ
!!
!!
!!
aa
aa
aa
ZZ
!!
!!
!!
aa
aa
aa
M0 = 13.50
M1(T ) = 4.50
M1(H) = 18
M2(TT ) = 1.50
M2(HT ) = M2(TH) = 6
M2(HH) = 24
S3(HHH) = 32
S3(HHT ) = S3(HTH)
= S3(THH) = 8
S3(HTT ) = S3(THT )
= S3(TTH) = 2
S3(TTT ) = .50
Fig. 3.7.1. An estimation martingale.
We verify the martingale property. We have M2 = E2[M3] because M3 =S3 and we used the formula M2 = E2[S3] to compute M2. We must check thatM1 = E1[M2] and M0 = E0[M1] = E[M1], which we do below:
E1[M2](H) = 23M2(HH) + 1
3M2(HT ) = 23 · 24 + 1
3 · 6 = 18 = M1(H),
E1[M2](T ) = 12M2(TH) + 1
2M2(TT ) = 23 · 6 + 1
3 · 1.50 = 4.50 = M1(T ),
M0 = 23M1(H) + 1
3M1(T ) = 23 · 18 + 1
3 · 4.50 = 13.50 = M0.
Exercise 3.5 (Stochastic volatility, random interest rate). Considerthe model of Exercise 2.9 of Chapter 2. Assume that the actual probabilitymeasure is
P(HH) =4
9, P(HT ) =
2
9, P(TH) =
2
9, P(TT ) =
1
9.
The risk-neutral measure was computed in Exercise 2.9 of Chapter 2.
(i) Compute the Radon-Nikodym derivative Z(HH), Z(HT ), Z(TH) and
Z(TT ) of P with respect to P
(ii) The Radon-Nikodym derivative process Z0, Z1, Z2 satisfies Z2 = Z. Com-pute Z1(H), Z1(T ) and Z0. Note that Z0 = EZ = 1.
(iii) The version of the risk-neutral pricing formula (3.2.6) appropriate for thismodel, which does not use the risk-neutral measure, is
22 3 State Prices
V1(H) =1 + r0
Z1(H)E1
[Z2
(1 + r0)(1 + r1)V2
](H)
=1
Z1(H)(1 + r1(H))E1[Z2V2](H),
V1(T ) =1 + r0
Z1(T )E1
[Z2
(1 + r0)(1 + r1)V2
](T )
=1
Z1(T )(1 + r1(T ))E1[Z2V2](T ),
V0 = E
[Z2
(1 + r0)(1 + r1)V2
].
Use this formula to compute V1(H), V1(T ) and V0 when V2 = (S2 − 7)+.Compare to your answers in Exercise 2.6(ii) of Chapter 2.
Solution
(i) In Exercise 2.9 of Chapter 2, the risk-neutral probabilities are
P(HH) =1
4, P(HT ) =
1
4, P(TH) =
1
12, P(TT ) =
5
12.
Therefore, the Radon-Nikodym derivative is
Z(HH) =P(HH)
P(HH)=
1
4· 9
4=
9
16, Z(HT ) =
P(HT )
P(HT )=
1
4· 9
2=
9
8,
Z(TH) =P(TH)
P(TH)=
1
12· 9
2=
3
8, Z(TT ) =
P(TT )
P(TT )=
5
12· 9
1=
15
4,
(ii)
Z1(H) = E1[Z2](H)
= Z2(HH)Pω2 = H given that ω1 = H+Z2(HT )Pω2 = T given that ω1 = H
= Z2(HH)P(HH)
P(HH) + P(HT )+ Z2(HT )
P(HT )
P(HH) + P(HT )
=9
16·
49
49 + 2
9
+9
29
49 + 2
9
=3
4,
3.7 Solutions to Selected Exercises 23
Z1(T ) = E1[Z2](T )
= Z2(TH)Pω2 = H given that ω1 = T+Z2(TT )Pω2 = T given that ω1 = T
= Z2(TH)P(TH)
P(TH) + P(TT )+ Z2(TT )
P(TT )
P(TH) + P(TT )
=3
29
29 + 1
9
+15
19
29 + 1
9
=3
2,
Z0 = E0[Z1]
= E[Z1]
= Z1(H)(P(HH) + P(HT )
)+ Z1(T )
(P(TH) + P(TT )
)
=3
4·(
4
9+
2
9
)+
3
2·(
2
9+
1
9
)
= 1.
We may also check directly that EZ = 1, as follows:
EZ = Z(HH)P(HH) + Z(HT )P(HT ) + Z(TH)P(TH) + Z(TT )P(TT )
=9
16· 4
9+
9
8· 2
9+
3
8· 2
9+
15
4· 1
9
=1
4+
1
4+
1
12+
5
12= 1.
(iii) We recall that
V2(HH) = 5, V2(HT ) = 1, V2(TH) = 1, V2(TT ) = 0.
We computed in part (ii) that
Pω2 = H given that ω1 = H =49
49+ 2
9
=2
3,
Pω2 = T given that ω1 = H =29
49+ 2
9
=1
3,
Pω2 = H given that ω1 = T =29
29+ 1
9
=2
3,
Pω2 = T given that ω1 = T =19
29+ 1
9
=1
3.
Therefore,
24 3 State Prices
V1(H) =1
Z1(H)(1 + r1(H)
)E1[Z2V2](H)
=
(3
4· 5
4
)−1 [Z2(HH)V2(HH)Pω2 = H given that ω1 = H
+Z2(HT )V2(HT )Pω2 = T given that ω1 = H]
=16
15
[9
16· 5 · 2
3+
9
8· 1 · 1
3
]
= 2.40,
V1(T ) =1
Z1(T )(1 + r1(T )
)E1[Z2V2](T )
=
(3
2· 3
2
)−1 [Z2(TH)V2(TH)Pω2 = H given that ω1 = T
+Z2(TT )V2(TT )Pω2 = T given that ω1 = T]
=4
9
[3
8· 1 · 2
3+
15
4· 0 · 1
3
]
= 0.111111,
and
V0 = E
[Z2V2
(1 + r0)(1 + r1)
]
=Z2(HH)V2(HH)
(1 + r0)(1 + r1(H))P(HH) +
Z2(HT )V2(HT )
(1 + r0)(1 + r1(H))P(HT )
+Z2(TH)V2(TH)
(1 + r0)(1 + r1(T ))P(TH) +
Z2(TT )V2(TT )
(1 + r0)(1 + r1(T ))P(TT )
=
(5
4· 5
4
)−19
16· 5 · 4
9+
(5
4· 5
4
)−19
8· 1 · 2
9+
(5
4· 3
2
)−13
8· 1 · 2
9
+
(5
4· 3
2
)−115
4· 0 · 1
9
=16
25· 5
4+
16
25· 1
4+
8
15· 1
12= 1.00444.
Exercise 3.6. Consider Problem 3.3.1 in an N -period binomial model withthe utility function U(x) = ln x. Show that the optimal wealth process cor-responding to the optimal portfolio process is given by Xn = X0
ζn, n =
0, 1, . . . , N , where ζn is the state price density process defined in (3.2.7).
Solution From (3.3.25) we have
XN = I
(λZ
(1 + r)N
)= I(λζN ).
3.7 Solutions to Selected Exercises 25
When U(x) = ln x, U ′(x) = 1x
and the inverse function of U ′ is I(y) = 1y.
Therefore,
XN =1
λζN
We must choose λ to satisfy (3.3.26), which in this case takes the form
X0 = E[ζNI(λζN )
]=
1
λ.
Substituting this into the previous equation, we obtain
XN =X0
ζN
.
Because Xn
(1+r)n is a martingale under the risk-neutral measure P, we have
Xn
(1 + r)n= En
[XN
(1 + r)N
]=
1
Zn
En
[ZNXN
(1 + r)N
]=
1
Zn
En [ζNXN ] =X0
Zn
.
Therefore,
Xn =(1 + r)nX0
Zn
=X0
ζn
.
Exercise 3.8. The Lagrange Multiplier Theorem used in the solution of Prob-lem 3.3.5 has hypotheses that we did not verify in the solution of that problem.In particular, the theorem states that if the gradient of the constraint function,which in this case is the vector (p1ζ1, . . . , pmζm), is not the zero vector, thenthe optimal solution must satisfy the Lagrange multiplier equations (3.3.22).This gradient is not the zero vector, so this hypothesis is satisfied. However,even when this hypothesis is satisfied, the theorem does not guarantee thatthere is an optimal solution; the solution to the Lagrange multiplier equa-tions may in fact minimize the expected utility. The solution could also beneither a maximizer nor a minimizer. Therefore, in this exercise, we outline adifferent method for verifying that the random variable XN given by (3.3.25)maximizes the expected utility.
We begin by changing the notation, calling the random variable givenby (3.3.25) X∗
N rather than XN . In other words,
X∗N = I
(1 + r)NZ
), (3.6.1)
where λ is the solution of equation (3.3.26). This permits us to use the nota-tion XN for an arbitrary (not necessarily optimal) random variable satisfying(3.3.19). We must show that
EU(XN ) ≤ EU(X∗N ). (3.6.2)
26 3 State Prices
(i) Fix y > 0, and show that the function of x given by U(x)−yx is maximizedby y = I(x). Conclude that
U(x) − yx ≤ U(I(y)) − yI(y) for every x. (3.6.3)
(ii) In (3.6.3), replace the dummy variable x by the random variable XN
and replace the dummy variable y by the random variable λZ(1+r)N . Take
expectations of both sides and use (3.3.19) and (3.3.26) to conclude that(3.6.2) holds.
Solution
(i) Because U(x) is concave, and for each fixed y > 0, yx is a linear functionof x, the difference U(x) − yx is a concave function of x. The derivativeof this function is U ′(x) − y, and this is zero if and only if U ′(x) = y,which is equivalent to x = I(y). A concave function has its maximum atthe point where its derivative is zero. The inequality (3.6.2) is just thisstatement.
(ii) Making the suggested replacements, we obtain
U(XN ) − λZXN
(1 + r)N≤ U
(I
(λZ
(1 + r)N
))− λZ
(1 + r)NI
(λZ
(1 + r)N
).
Taking expectations under P and using the fact that Z is the Radon-Nikodym derivative of P with respect to P, we obtain
EU(XN ) − λEXN
(1 + r)N
≤ EU
(I
(λZ
(1 + r)N
))− λE
[Z
(1 + r)NI
(λZ
(1 + r)N
)].
From (3.3.19) and (3.3.26), we have
EXN
(1 + r)N= X0 = E
[Z
(1 + r)NI
(λZ
(1 + r)N
)].
Cancelling these terms on the left- and right-hand sides of the above equation,we obtain (3.6.2).
4
American Derivative Securities
4.9 Solutions to Selected Exercises
Exercise 4.1. In the three-period model of Figure 1.2.2 of Chapter 1, let theinterest rate be r = 1
4 so the risk-neutral probabilities are p = q = 12 .
(i) Determine the price at time zero, denoted V P0 , of the American put that
expires at time three and has intrinsic value gP (s) = (4 − s)+.
(ii) Determine the price at time zero, denoted V C0 , of the American call that
expires at time three and has intrinsic value gC(s) = (s − 4)+.
(iii) Determine the price at time zero, denoted V S0 , of the American straddle
that expires at time three and has intrinsic value gS(s) = gP (s) + gC(s).
(iv) Explain why V S0 < V P
0 + V C0 .
Solution
(i) The payoff of the put at expiration time three is
V P3 (HHH) = (4 − 32)+ = 0,
V P3 (HHT ) = V P
3 (HTH) = V P3 (THH) = (4 − 8)+ = 0,
V P3 (HTT ) = V P
3 (THT ) = V P3 (TTH) = (4 − 2)+ = 2,
V P3 (TTT ) = (4 − 0.50)+ = 3.50.
Because 11+r
p = 11+r
q = 25 , the value of the put at time two is
28 4 American Derivative Securities
V P2 (HH) = max
(4 − S2(HH)
)+,2
5V P
3 (HHH) +2
5V P
3 (HHT )
= max
(4 − 16)+,
2
5· 0 +
2
5· 0
= max0, 0= 0,
V P2 (HT ) = max
(4 − S2(HT )
)+,2
5V P
3 (HTH) +2
5V P
3 (HTT )
= max
(4 − 4)+,
2
5· 0 +
2
5· 2
= max0, 0.80= 0.80,
V P2 (TH) = max
(4 − S2(TH)
)+,2
5V P
3 (THH) +2
5V P
3 (THT )
= max
(4 − 4)+,
2
5· 0 +
2
5· 2
= max0, 0.80= 0.80,
V P2 (TT ) = max
(4 − S2(TT )
)+,2
5V P
3 (TTH) +2
5V P
3 (TTT )
= max
(4 − 1)+,
2
5· 2 +
2
5· 3.50
= max3, 2.20= 3.
At time one the value of the put is
V P1 (H) = max
(4 − S1(H)
)+,2
5V P
2 (HH) +2
5V P
2 (HT )
= max
(4 − 8)+,
2
5· 0 +
2
5· 0.80
= max0, 0.32= 0.32,
V P1 (T ) = max
(4 − S1(T )
)+,2
5V P
2 (TH) +2
5V P
2 (TT )
= max
(4 − 2)+,
2
5· 0.80 +
2
5· 3
= max2, 1.52= 2.
The value of the put at time zero is
4.9 Solutions to Selected Exercises 29
V P0 = max
(4 − S0)
+,2
5V P
1 (H) +2
5V P
1 (T )
= max
(4 − 4)+,
2
5· 0.32 +
2
5· 2
= max0, 0.928= 0.928.
(ii) The payoff of the call at expiration time three is
V C3 (HHH) = (32 − 4)+ = 28,
V C3 (HHT ) = V C
3 (HTH) = V C3 (THH) = (8 − 4)+ = 4,
V C3 (HTT ) = V C
3 (THT ) = V C3 (TTH) = (2 − 4)+ = 0,
V C3 (TTT ) = (0.50 − 4)+ = 0.
Because 11+r
p = 11+r
q = 25 , the value of the call at time two is
V C2 (HH) = max
(S2(HH) − 4
)+,2
5V C
3 (HHH) +2
5V C
3 (HHT )
= max
(16 − 4)+,
2
5· 28 +
2
5· 4
= max12, 12.8= 12.8,
V C2 (HT ) = max
(S2(HT ) − 4
)+,2
5V C
3 (HTH) +2
5V C
3 (HTT )
= max
(4 − 4)+,
2
5· 4 +
2
5· 0
= max0, 1.60= 1.60,
V C2 (TH) = max
(S2(TH) − 4
)+,2
5V C
3 (THH) +2
5V C
3 (THT )
= max
(4 − 4)+,
2
5· 4 +
2
5· 0
= max0, 1.60= 1.60,
V C2 (TT ) = max
(S2(TT ) − 4
)+,2
5V C
3 (TTH) +2
5V C
3 (TTT )
= max
(1 − 4)+,
2
5· 0 +
2
5· 0
= max0, 0= 0.
At time one the value of the call is
30 4 American Derivative Securities
V C1 (H) = max
(S1(H) − 4
)+,2
5V C
2 (HH) +2
5V C
2 (HT )
= max
(8 − 4)+,
2
5· 12.8 +
2
5· 1.60
= max4, 5.76 = 5.76,
V C1 (T ) = max
(S1(T ) − 4
)+,2
5V C
2 (TH) +2
5V C
2 (TT )
= max
(2 − 4)+,
2
5· 1.60 +
2
5· 0
= max0, 0.64 = 0.64.
The value of the call at time zero is
V C0 = max
(S0 − 4)+,
2
5V C
1 (H) +2
5V C
1 (T )
= max
(4 − 4)+,
2
5· 5.76 +
2
5· 0.64
= max0, 2.56= 2.56.
(iii) Note that gS(s) = |s − 4|. The payoff of the straddle at expiration timethree is
V S3 (HHH) = |32 − 4| = 28,
V S3 (HHT ) = V S
3 (HTH) = V S3 (THH) = |8 − 4| = 4,
V S3 (HTT ) = V S
3 (THT ) = V S3 (TTH) = |2 − 4| = 2,
V S3 (TTT ) = |0.50 − 4| = 3.50.
We see that the payoff of the straddle is the payoff of the put given inthe solution to (i) plus the payoff of the call given in the solution to (ii).Because 1
1+rp = 1
1+rq = 2
5 , the value of the straddle at time two is
V S2 (HH) = max
∣∣S2(HH) − 4∣∣, 2
5V S
3 (HHH) +2
5V S
3 (HHT )
= max
|16 − 4|, 2
5· 28 +
2
5· 4
= max12, 12.8= 12.8,
V S2 (HT ) = max
∣∣S2(HT ) − 4∣∣+,
2
5V S
3 (HTH) +2
5V S
3 (HTT )
= max
(4 − 4)+,
2
5· 4 +
2
5· 2
= max0, 2.40= 2.40,
4.9 Solutions to Selected Exercises 31
V S2 (TH) = max
(S2(TH) − 4
)+,2
5V S
3 (THH) +2
5V S
3 (THT )
= max
(4 − 4)+,
2
5· 4 +
2
5· 2
= max0, 2.40= 2.40,
V S2 (TT ) = max
∣∣S2(TT ) − 4∣∣, 2
5V S
3 (TTH) +2
5V S
3 (TTT )
= max
|1 − 4|, 2
5· 2 +
2
5· 3.50
= max3, 2.20= 3.
One can verify in every case that V S2 = V P
2 + V C2 . At time one the value
of the straddle is
V S1 (H) = max
∣∣S1(H) − 4∣∣, 2
5V S
2 (HH) +2
5V S
2 (HT )
= max
|8 − 4|, 2
5· 12.8 +
2
5· 2.40
= max4, 6.08= 6.08,
V S1 (T ) = max
∣∣S1(T ) − 4∣∣, 2
5V S
2 (TH) +2
5V S
2 (TT )
= max
|2 − 4|, 2
5· 2.40 +
2
5· 3
= max2, 2.16= 2.16.
We have V S1 (H) = 6.08 = 0.32 + 5.76 = V P
1 (H) + V C1 (H), but V S
1 (T ) =2.16 < 2 + 0.64 = V P
1 (T ) + V C1 (T ).
The value of the straddle at time zero is
V S0 = max
|S0 − 4|, 2
5V S
1 (H) +2
5V S
1 (T )
= max
|4 − 4|, 2
5· 6.08 +
2
5· 2.16
= max0, 3.296= 3.296.
We have V S0 = 3.296 < 0.928 + 2.56 = V P
0 + V C0 .
32 4 American Derivative Securities
(iv) For the put, if there is a tail on the first toss, it is optimal to exercise attime one. This can be seen from the equation
V P1 (T ) = max
(4 − S1(T )
)+,2
5V P
2 (TH) +2
5V P
2 (TT )
= max
(4 − 2)+,
2
5· 0.80 +
2
5· 3
= max2, 1.52= 2,
which shows that the intrinsic value at time one if the first toss resultsin T is greater than the value of continuing. On the other hand, for thecall the intrinsic value at time one if there is a tail on the first toss is(S1(T ) − 4)+ = (2 − 4)+ = 0, whereas the value of continuing is 0.64.Therefore, the call should not be exercised at time one if there is a tail onthe first toss.
The straddle has the intrinsic value of a put plus a call. When it is ex-ercised, both parts of the payoff are received. In other words, it is notan American put plus an American call, because these can be exercisedat different times whereas the exercise of a straddle requires both theput payoff and the call payoff to be received. In the computation of thestraddle price
V S1 (T ) = max
∣∣S1(T ) − 4∣∣, 2
5V S
2 (TH) +2
5V S
2 (TT )
= max
|2 − 4|, 2
5· 2.40 +
2
5· 3
= max2, 2.16= 2.16,
we see that it is not optimal to exercise the straddle at time one if the firsttoss results in T . It would be optimal to exercise the put part, but not thecall part, and the straddle cannot exercise one part without exercising theother. Greater value is achieved by not exercising both parts than wouldbe achieved by exercising both. However, this value is less than would beachieved if one could exercise the put part and let the call part continue,and thus V S
1 (T ) < V P1 (T ) + V C
1 (T ). This loss of value at time one resultsin a similar loss of value at the earlier time zero: V S
0 < V P0 + V C
0 .
Exercise 4.3. In the three-period model of Figure 1.2.2 of Chapter 1, let theinterest rate be r = 1
4 so the risk-neutral probabilities are p = q = 12 . Find
the time-zero price and optimal exercise policy (optimal stopping time) forthe path-dependent American derivative security whose intrinsic value at each
time n, n = 0, 1, 2, 3, is(4 − 1
n+1
∑n
j=0 Sj
)+
. This intrinsic value is a put on
the average stock price between time zero and time n.
4.9 Solutions to Selected Exercises 33
Solution The intrinsic value process for this option is
G0 = (4 − S0)+
= (4 − 4)+ = 0,
G1(H) =(4 − S0+S1(H)
2
)+
=(4 − 4+8
2
)+= 0,
G1(T ) =(4 − S0+S1(T )
2
)+
=(4 − 4+2
2
)+= 1,
G2(HH) =(4 − S0+S1(H)+S2(HH)
3
)+
=(4 − 4+8+16
3
)+= 0,
G2(HT ) =(4 − S0+S1(H)+S2(HT )
3
)+
=(4 − 4+8+4
3
)= 0,
G2(TH) =(4 − S0+S1(T )+S2(TH)
3
)+
=(4 − 4+2+4
3
)+= 0.6667,
G2(TT ) =(4 − S0+S1(T )+S2(TT )
3
)+
=(r − 4+2+1
3
)= 1.6667.
At time three, the intrinsic value G3 agrees with the option value V3. In otherwords,
V3(HHH) = G3(HHH)
=
(4 − S0 + S1(H) + S2(HH) + S3(HHH)
4
)+
=
(4 − 4 + 8 + 16 + 32
4
)+
= 0,
V3(HHT ) = G3(HHT )
=
(4 − S0 + S1(H) + S2(HH) + S3(HHT )
4
)+
=
(4 − 4 + 8 + 16 + 8
4
)+
= 0,
V3(HTH) = G3(HTH)
=
(4 − S0 + S1(H) + S2(HT ) + S3(HTH)
4
)+
=
(4 − 4 + 8 + 4 + 8
4
)+
= 0,
V3(HTT ) = G3(HTT )
=
(4 − S0 + S1(H) + S2(HT ) + S3(HTT )
4
)+
=
(4 − 4 + 8 + 4 + 2
4
)+
= 0,
34 4 American Derivative Securities
V3(THH) = G3(THH)
=
(4 − S0 + S1(T ) + S2(TH) + S3(THH)
4
)+
=
(4 − 4 + 2 + 4 + 8
4
)+
= 0,
V3(THT ) = G3(THT )
=
(4 − S0 + S1(T ) + S2(TH) + S3(THT )
4
)+
=
(4 − 4 + 2 + 4 + 2
4
)+
= 1,
V3(TTH) = G3(TTH)
=
(4 − S0 + S1(T ) + S2(TT ) + S3(TTH)
4
)+
=
(4 − 4 + 2 + 1 + 2
4
)+
= 1.75,
V3(TTT ) = G3(TTT )
=
(4 − S0 + S1(T ) + S2(TT ) + S3(TTT )
4
)+
=
(4 − 4 + 2 + 1 + 0.50
4
)+
= 2.125.
We use the algorithm of Theorem 4.4.3, noting that p1+r
= q1+r
= 25 , to obtain
V2(HH) = max
G2(HH),
2
5V3(HHH) +
2
5V3(HHT )
= max
0,
2
5· 0 +
2
5· 0
= 0,
V2(HT ) = max
G2(HT ),
2
5V3(HTH) +
2
5V3(HTT )
= max
0,
2
5· 0 +
2
5· 0
= 0,
4.9 Solutions to Selected Exercises 35
V2(TH) = max
G2(TH),
2
5V3(THH) +
2
5V3(THT )
= max
0.6667,
2
5· 0 +
2
5· 1
= max0.6667, 0.40= 0.6667,
V2(TT ) = max
G2(TT ),
2
5V3(TTH) +
2
5V3(TTT )
= max
1.6667,
2
5· 1.75 +
2
5· 2.125
= max1.6667, 1.55= 1.6667.
Continuing, we have
V1(H) = max
G1(H),
2
5V2(HH) +
2
5V2(HT )
= max
0,
2
5· 0 +
2
5· 0
= 0,
V1(T ) = max
G1(T ),
2
5V2(TH) +
2
5V2(TT )
= max
1,
2
5· 0.6667 +
2
5· 1.6667
= max1, 0.9334= 1,
V0 = max
G0,
2
5V1(H) +
2
5V1(T )
= max
0,
2
5· 0 +
2
5· 1
= max0, 0.40= 0.40.
To find the optimal exercise time, we work forward. Since V0 > G0, oneshould not exercise at time zero. However, V1(T ) = G1(T ), so it is optimal toexercise at time one if there is a T on the first toss. If the first toss results inH, the option is destined always be out of the money. With the intrinsic value
Gn =(4 − 1
n+1
∑n
j=1 Sj
)+
defined in the exercise, it does not matter what
exercise rule we choose in this case. If the payoff were(4 − 1
n+1
∑n
j=1 Sj
), so
that exercising out of the money is costly (as one would expect in practice),then one should allow the option to expire unexercised.
36 4 American Derivative Securities
Exercise 4.5. In equation (4.4.5), the maximum is computed over all stop-ping times in S0. List all the stopping times in S0 (there are 26), and fromamong those, list the stopping times that never exercise when the option isout of the money (there are 11). For each stopping time τ in the latter set,compute E
[Iτ≤2
(45
)τGτ
]. Verify that the largest value for this quantity is
given by the stopping time of (4.4.6), the one which makes this quantity equalto the 1.36 computed in (4.4.7).
Solution A stopping time is a random variable, and we can specify a stoppingtime by listing its values τ(HH), τ(HT ), τ(TH), and τ(TT ). The stoppingtime property requires that τ(HH) = 0 if and only if τ(HT ) = τ(TH) =τ(TT ) = 0. Similarly, τ(HH) = 1 if and only if τ(HT ) = 1 and τ(TH) = 1if and only if τ(TT ) = 1. The 26 stopping times in the two-period binomialmodel are tabulated below.
StoppingTime HH HT TH TT
τ1 0 0 0 0τ2 1 1 1 1τ3 1 1 2 2τ4 1 1 2 ∞τ5 1 1 ∞ 2τ6 1 1 ∞ ∞τ7 2 2 1 1τ8 2 2 2 2τ9 2 2 2 ∞τ10 2 2 ∞ 2τ11 2 2 ∞ ∞τ12 2 ∞ 1 1τ13 2 ∞ 2 2τ14 2 ∞ 2 ∞τ15 2 ∞ ∞ 2τ16 2 ∞ ∞ ∞τ17 ∞ 2 1 1τ18 ∞ 2 2 2τ19 ∞ 2 2 ∞τ20 ∞ 2 ∞ 2τ21 ∞ 2 ∞ ∞τ22 ∞ ∞ 1 1τ23 ∞ ∞ 2 2τ24 ∞ ∞ 2 ∞τ25 ∞ ∞ ∞ 2τ26 ∞ ∞ ∞ ∞
The intrinsic value process for this option is given by
4.9 Solutions to Selected Exercises 37
G0 = 1, G1(H) = −3, G1(T ) = 3,
G2(HH) = −11, G2(HT ) = G2(TH) = 1, G2(TT ) = 4.
The stopping times that take the value 1 when there is an H on the first tossare mandating an exercise out of the money (G1(H) = −3). This rules out τ2–τ6. Also, the stopping times that take the value 2 when there is an HH on thefirst two tosses are mandating an exercise out of the money (G2(HH) = −11).This rules out τ7–τ16. For all other exercise situations, G is positive, so theoption is in the money. This leaves us with τ1 and the ten stopping timesτ17–τ26. We evaluate the risk-neutral expected payoff of these eleven stoppingtimes.
E
[Iτ1≤2
(4
5
Gτ1
]= G0 = 1,
E
[Iτ17≤2
(4
5
Gτ17
]=
1
4· 16
26G2(HT ) +
1
2· 4
5G1(T )
=4
25· 1 +
2
5· 3 = 1.36,
E
[Iτ18≤2
(4
5
Gτ18
]=
1
4· 16
26G2(HT ) +
1
4· 16
25G2(TH) +
1
4· 16
25G2(TT )
=4
25· 1 +
4
25· 1 +
4
25· 4 = 0.96,
E
[Iτ19≤2
(4
5
Gτ19
]=
1
4· 16
26G2(HT ) +
1
4· 16
25G2(TH)
=4
25· 1 +
4
25· 1 = 0.32,
E
[Iτ20≤2
(4
5
Gτ20
]=
1
4· 16
26G2(HT ) +
1
4· 16
25G2(TT )
=4
25· 1 +
4
25· 4 = 0.80,
E
[Iτ21≤2
(4
5
Gτ21
]=
1
4· 16
26G2(HT )
=4
25· 1 = 0.16,
E
[Iτ22≤2
(4
5
Gτ22
]=
1
2· 4
5G1(T )
=2
5· 3 = 1.20,
E
[Iτ23≤2
(4
5
Gτ23
]= +
1
4· 16
25G2(TH) +
1
4· 16
25G2(TT )
=4
25· 1 +
4
25· 4 = 0.80,
38 4 American Derivative Securities
E
[Iτ24≤2
(4
5
Gτ24
]= +
1
4· 16
25G2(TH)
=4
25· 1 = 0.16,
E
[Iτ25≤2
(4
5
Gτ25
]= +
1
4· 16
25G2(TT )
= +4
25· 4 = 0.64,
E
[Iτ26≤2
(4
5
Gτ26
]= 0.
The largest value, 1.36, is obtained by the stopping time τ17.
Exercise 4.7. For the class of derivative securities described in Exercise 4.6whose time-zero price is given by (4.8.3), let Gn = Sn − K. This derivativesecurity permits its owner to buy one share of stock in exchange for a paymentof K at any time up to the expiration time N . If the purchase has not beenmade at time N , it must be made then. Determine the time-zero value andoptimal exercise policy for this derivative security. (Assume r ≥ 0.)
Solution Set Yn = 1(1+r)n (Sn − K), n = 0, 1, . . . , N . We assume r ≥ 0.
Because the discounted stock price is a martingale under the risk-neutralmeasure and K
(1+r)n+1 is not random, we have
En[Yn+1] = En
[Sn+1
(1 + r)n+1
]− En
[K
(1 + r)n+1
]
=Sn
(1 + r)n− K
(1 + r)n+1
≥ Sn
(1 + r)n− K
(1 + r)n.
This shows that Yn, n = 0, 1, . . . , N , is a submartingale. According to Theorem4.3.3 (Optional Sampling—Part II), EYN∧τ ≤ EYN whenever τ is a stoppingtime. If τ is a stopping time satisfying τ(ω) ≤ N for every sequence of coin
tosses ω, this becomes EYτ ≤ EYN , or equivalently,
E
[1
(1 + r)τGτ
]≤ E
[1
(1 + r)NGN
].
Therefore,
V0 = maxτ∈S0,τ≤N
E
[1
(1 + r)τGτ
]≤ E
[1
(1 + r)NGN
].
On the other hand, because the stopping time that is equal to N regardlessof the outcome of the coin tossing is in the set of stopping times over whichthe above maximum is taken, we must in fact have equality:
4.9 Solutions to Selected Exercises 39
V0 = maxτ∈S0,τ≤N
E
[1
(1 + r)τGτ
]= E
[1
(1 + r)NGN
].
Hence, it is optimal to exercise at the final time N regardless of the outcomeof the coin tossing.
5
Random Walk
5.8 Solutions to Selected Exercises
Exercise 5.2. (First passage time for random walk with upwarddrift) Consider the asymmetric random walk with probability p for an upstep and probability q = 1 − p for a down step, where 1
2 < p < 1 so that0 < q < 1
2 . In the notation of (5.2.1), let τ1 be the first time the random walkstarting from level 0 reaches the level 1. If the random walk never reaches thislevel, then τ1 = ∞.
(i) Define f(σ) = peσ + qe−σ. Show that f(σ) > 1 for all σ > 0.
(ii) Show that when σ > 0, the process
Sn = eσMn
(1
f(σ)
)n
is a martingale.
(iii) Show that for σ > 0,
e−σ = E
[Iτ1<∞
(1
f(σ)
)τ1]
.
Conclude that Pτ1 < ∞ = 1.
(iv) Compute Eατ1 for α ∈ (0, 1).
(v) Compute Eτ1.
Solution.
(i) The function f(σ) satisfies f(0) = 1 and f ′(σ) = peσ−qe−σ, f ′′(σ) = f(σ).Since f is always positive, f ′′ is always positive and f is convex. Underthe assumption p > 1/2 > q, we have f ′(0) = p−q > 0, and the convexityof f implies that f(σ) > 1 for all σ > 0.
42 5 Random Walk
(ii) We compute
En[Sn+1] =
(1
f(σ)
)n+1
En
[eσ(Mn+Xn+1)
]
=
(1
f(σ)
)n+1
eσMnE[eσXn+1
]
=
(1
f(σ)
)n+1
eσMn(peσ + qe−σ
)
= eσMn
(1
f(σ)
)n
= Sn.
(iii) Because Sn∧τ1is a martingale starting at 1, we have
1 = ESn∧τ1= EeσMn∧τ1
(1
f(σ)
)n∧τ1
. (5.8.1)
For σ > 0, the positive random variable eσMn∧τ1 is bounded above by eσ,and 0 < 1/f(σ) < 1. Therefore,
limn→∞
eσMn∧τ1
(1
f(σ)
)n∧τ1
= Iτ1<∞eσ
(1
f(σ)
.
Letting n → ∞ in (5.8.1), we obtain
1 = E
[Iτ1<∞e
σ
(1
f(σ)
)τ1]
,
or equivalently,
e−σ = E
[Iτ1<∞
(1
f(σ)
)τ1]
. (5.8.2)
This equation holds for all positive σ. We let σ ↓ 0 to obtain the formula
1 = EIτ1<∞ = Pτ1 < ∞.
(iv) We now introduce α ∈ (0, 1) and solve the equation α = 1f(σ) for e−σ.
This equation can be written as
αpeσ + αqe−σ = 1,
which may be rewritten as
αq(e−σ
)2 − e−σ + αp = 0,
and the quadratic formula gives
5.8 Solutions to Selected Exercises 43
e−σ =1 ±
√1 − 4α2pq
2αq. (5.8.3)
We want σ to be positive, so we need e−σ to be less than 1. Hence wetake the negative sign, obtaining
e−σ =1 −
√1 − 4α2pq
2αq.
Substituting this into (5.8.2), we obtain the formula
Eατ1 =1 −
√1 − 4α2pq
2αq. (5.8.4)
(v) Differentiating (5.8.4) with respect to α leads to
Eτατ−1 =1 −
√1 − 4α2pq
2α2q√
1 − 4α2pq,
and letting α ↑ 1, we obtain
Eτ1 =1 −√
1 − 4pq
2q√
1 − 4pq=
1 −√
(1 − 2q)2
2q√
(1 − 2q)2=
1 − (1 − 2q)
2q(1 − 2q)=
1
1 − 2q=
1
p − q.
Exercise 5.4 (Distribution of τ2). Consider the symmetric random walk,and let τ2 be the first time the random walk, starting from level 0, reachesthe level 2. According to Theorem 5.2.3,
Eατ2 =
(1 −
√1 − α2
α
)2
for all α ∈ (0, 1).
Using the power series (5.2.21), we may write the right-hand side as
(1 −
√1 − α2
α
)2
=2
α· 1 −
√1 − α2
α− 1
= −1 +
∞∑
j=1
2
)2j−2 (2j − 2)!
j!(j − 1)!
=
∞∑
j=2
2
)2j−2 (2j − 2)!
j!(j − 1)!
=
∞∑
k=1
2
)2k (2k)!
(k + 1)!k!.
(i) Use the above power series to determine Pτ2 = 2k, k = 1, 2, . . . .
44 5 Random Walk
(ii) Use the reflection principle to determine Pτ2 = 2k, k = 1, 2, . . . .
Solution
(i) Since the random walk can reach the level 2 only on even-numbered steps,Pτ2 = 2k − 1 = 0 for k = 1, 2, . . . . Therefore,
Eατ2 =∞∑
k=1
α2kPτ2 = 2k =
∞∑
k=1
2
)2k (2k)!
(k + 1)!k!.
Equating coefficients, we conclude that
Pτ = 2k =(2k)!
(k + 1)!k!
(1
2
)2k
, k = 1, 2, . . . .
(ii) For k = 1, 2, . . . , the number of paths that reach or exceed level 2 by time2k is equal to the number of paths that are at level 2 at time 2k plus thenumber of paths that exceed level 2 at time 2k plus the number of pathsthat reach the level 2 before time 2k but are below level 2 at time 2k. Apath is of the last type if and only if its reflected path exceeds level 2 attime 2k. Thus, the number of paths that reach or exeed level 2 by time 2kis equal to the number of paths that are at level 2 at time 2k plus twicethe number of paths that exeed level 2 at time 2k. For the symmetricrandom walk, every path has the same probability, and hence
Pτ2 ≤ 2k = PM2k = 2 + 2PM2k ≥ 4= PM2k = 2 + PM2k ≥ 4 + PM2k ≤ −4= 1 − PM2k = 0 − PM2k = −2
= 1 − (2k)!
k!k!
(1
2
)2k
− (2k)!
(k + 1)!(k − 1)!
(1
2
)2k
.
Consequently,
Pτ2 = 2k= Pτ2 ≤ 2k − Pτ2 ≤ 2k − 2
=
(1
2
)2k−2 [(2k − 2)!
(k − 1)!(k − 1)!+
(2k − 2)!
k!(k − 2)!
]
−(
1
2
)2k [(2k)!
k!k!+
(2k)!
(k + 1)!(k − 1)!
]
=
(1
2
)2k [4(2k − 2)!
(k − 1)!(k − 1)!− (2k)!
k!k!
]
+
(1
2
)2k [4(2k − 2)!
k!(k − 2)!− (2k)!
(k + 1)!(k − 1)!
]
5.8 Solutions to Selected Exercises 45
=
(1
2
)2k [2k · 2k(2k − 2)! − 2k(2k − 1)(2k − 2)!
k!k!
]
+
(1
2
)2k [(2k + 2)(2k − 2)(2k − 2)! − 2k(2k − 1)(2k − 2)!
(k + 1)!(k − 1)!
]
=
(1
2
)2k [4k2(2k − 2)! − (4k2 − 2k)(2k − 2)!
k!k!
]
+
(1
2
)2k [(4k2 − 4)(2k − 2)! − (4k2 − 2k))(2k − 2)!
(k + 1)!(k − 1)!
]
=
(1
2
)2k [2k(2k − 2)!
k!k!+
2(k − 2)(2k − 2)!
(k + 1)!(k − 1)!
]
=
(1
2
)2k(2k − 2)!
(k + 1)!k!
[(2k(k + 1) + 2(k − 2)k
]
=
(1
2
)2k(2k − 2)!
(k + 1)!k!2k(2k − 1)
=
(1
2
)2k(2k)!
(k + 1)!k!.
Exercise 5.7. (Hedging a short position in the perpetual Americanput). Suppose you have sold the perpetual American put of Section 5.4 andare hedging the short position in this put. Suppose at the current time thestock price is s and the value of your hedging portfolio is v(s). Your hedge isto first consume the amount
c(s) = v(s) − 4
5
[1
2v(2s) +
1
2v(s
2
)](5.7.3)
and then take a position
δ(s) =v(2s) − v
(s2
)
2s − s2
(5.7.4)
in the stock. (See Theorem 4.2.2 of Chapter 4. The processes Cn and ∆n inthat theorem are obtained by replacing the dummy variable s by the stockprice Sn in (5.7.3) and (5.7.4); i.e., Cn = c(Sn) and ∆n = δ(Sn).) If you hedgethis way, then regardless of whether the stock goes up or down on the nextstep, the value of your hedging portfolio should agree with the value of theperpetual American put.
(i) Compute c(s) when s = 2j for the three cases j ≤ 0, j = 1 and j ≥ 2.
(ii) Compute δ(s) when s = 2j for the three cases j ≤ 0, j = 1 and j ≥ 2.
46 5 Random Walk
(iii) Verify in each of the three cases s = 2j for j ≤ 0, j = 1 and j ≥ 2 thatthe hedge works (i.e., regardless of whether the stock goes up or down,the value of your hedging portfolio at the next time is equal to the valueof the perpetual American put at that time).
Solution
(i) Throughout the following computations, we use (5.4.6):
v(2j) =
4 − 2j , if j ≤ 1,4
2j, if j ≥ 1.
For j ≤ 0, we have
c(2j) = v(2j) − 4
5
[1
2v(2j+1) +
1
2v(2j−1)
]
= 4 − 2j − 2
5
[4 − 2j+1 + 4 − 2j−1
]
= 4 − 2j − 2
5
[8 − 5 · 2j−1
]
=4
5.
For j = 1, we have
c(2) = v(2) − 4
5
[1
2v(4) +
1
2v(1)
]
= 2 − 2
5[1 + 3]
=2
5.
Finally, for j ≥ 2,
c(2j) = v(2j) − 4
5
[1
2v(2j+1) +
1
2v(2j−1)
]
=4
2j− 2
5
[4
2j+1+
4
2j−1
]
=4
2j− 2
5
[4
2j+1+
16
2j+1
]
= 0.
(ii) For j ≤ 0, we have
δ(2j) =v(2j+1) − v(2j−1)
2j+1 − 2j−1=
(4 − 2j+1
)−(4 − 2j−1
)
2j+1 − 2j−1= −1.
5.8 Solutions to Selected Exercises 47
For j = 1, we have
δ(2) =v(4) − v(1)
4 − 1=
1 − 3
4 − 1= −2
3.
Finally, for j ≥ 2,
δ(2j) =v(2j+1) − v(2j−1)
2j+1 − 2j−1
=
4
2j+1− 4
2j−1
2j+1 − 2j−1
=4 − 16
2j+1· 1
(4 − 1)2j−1
= − 4
22j.
(iii) If the stock price is 2j , where j ≤ 0, then we have a portfolio valuedat v(2j) = 4 − 2j . We consume c(2j) = 4
5 and short one share of stock(δ(2j) = −1). This creates a cash position of
4 − 2j − 4
5+ 2j =
16
5,
which is invested in the money market. When we go to the next period,the money market investment grows to 5
4 · 165 = 4. If the stock goes up to
2j+1, our short stock position has value −2j+1, so our portfolio is valuedat 4− 2j+1, which is the option value in this case. If the stock goes downto 2j−1, our portfolio is valued at 4− 2j−1, which is the option value alsoin this case.
When the stock price is 2j with j ≤ 0, the owner of the option shouldexercise, and we are prepared for that by being short one share of stock.When she exercises, she will deliver the share of stock, which will coverour short position. But when she exercises, we must pay her 4, so we aremaintaining a cash position of 4. At the beginning of any period in whichshe fails to exercise, we get to consume the present value of the interestthat will accrue to this cash position over the next period.
If the stock price is 2, then we have a portfolio valued at v(2) = 2. Weconsume c(2) = 2
5 and short 23 of a share of stock (δ(2) = − 2
3 ). Thiscreates a cash position of
2 − 2
5+
2
3· 2 =
44
15,
which is invested in the money market. When we go to the next period,the money market investment grows to 5
4 · 4415 = 11
3 . If the stock goes up
48 5 Random Walk
to 4, the short stock position has value − 23 · 4 = − 8
3 , so the portfolio isvalued at 11
3 − 83 = 1, which is the option value in this case. If the stock
goes down to 1, the short stock position has value − 23 · 1 = − 2
3 , so theportfolio is valued at 11
3 − 23 = 3, which is the option value also in this
case.
Finally, if the stock price is 2j , where j ≥ 2, then we have a portfolio valuedat 4
2j . We consume c(2j) = 0 and short 422j shares of stock (δ(2j) = − 4
22j ).This creates a cash position of
4
2j+
4
22j· 2j =
8
2j,
which is invested in the money market. When we go to the next period,the money market investment grows to
5
4· 8
2j=
10
2j.
If the stock goes up to 2j+1, the short stock position has value
− 4
22j· 2j+1 = − 8
2j,
so the portfolio is valued at
10
2j− 8
2j=
2
2j=
4
2j+1,
which is the option value in this case. If the stock goes down to 2j−1, theshort stock position has value
− 4
22j· 2j−1 = − 2
2j,
so the portfolio is valued at
10
2j− 2
2j=
8
2j=
4
2j−1,
which is the option value also in this case.
Exercise 5.9. (Provided by Irene Villegas.) Here is a method for solvingequation (5.4.13) for the value of the perpetual American put in Section 5.4.
(i) We first determine v(s) for large values of s. When s is large, it is notoptimal to exercise the put, so the maximum in (5.4.13) will be given bythe second term,
4
5
[1
2v(2s) +
1
2v(s
2
)]=
2
5v(2s) +
2
5v(s
2
).
5.8 Solutions to Selected Exercises 49
We thus seek solutions to the equation
v(s) =2
5v(2s) +
2
5v(s
2
). (5.7.5)
All such solutions are of the form sp for some constant p, or linear com-binations of functions of this form. Substitute sp into (5.7.5), obtain aquadratic equation for 2p, and solve to obtain 2p = 2 or 2p = 1
2 . Thisleads to the values p = 1 and p = −1, i.e., v1(s) = s and v2(s) = 1
sare
solutions to (5.7.5).
(ii) The general solution to (5.7.5) is a linear combination of v1(s) and v2(s),i.e.,
v(s) = As +B
s. (5.7.6)
For large values of s, the value of the perpetual American put must begiven by (5.7.6). It remains to evaluate A and B. Using the second bound-ary condition in (5.4.15), show that A must be zero.
(iii) We have thus established that for large values of s, v(s) = Bs
for someconstant B still to be determined. For small values of s, the value of theput is its intrinsic value 4 − s. We must choose B so these two functionscoincide at some point, i.e., we must find a value for B so that, for somes > 0,
fB(s) =B
s− (4 − s)
equals zero. Show that, when B > 4, this function does not take the value0 for any s > 0, but, when B ≤ 4, the equation fB(s) = 0 has a solution.
(iv) Let B be less than or equal to 4 and let sB be a solution of the equationfB(s) = 0. Suppose sB is a stock price which can be attained in the model(i.e., sB = 2j for some integer j). Suppose further that the owner of theperpetual American put exercises the first time the stock price is sB orsmaller. Then the discounted risk-neutral expected payoff of the put isvB(S0), where vB(s) is given by the formula
vB(s) =
4 − s, if s ≤ sB ,Bs, if s ≥ sB .
(5.7.7)
Which values of B and sB give the owner the largest option value?
(v) For s < sB , the derivative of vB(s) is v′B(s) = −1. For s > sB , this
derivative is v′B(s) = − B
s2 . Show that the best value of B for the optionowner makes the derivative of vB(s) continuous at s = sB (i.e., the twoformulas for v′
B(s) give the same answer at s = sB).
50 5 Random Walk
Solution
(i) With v(s) = sp, (5.7.5) becomes
sp =2
5· 2psp +
2
5· 1
2psp,
Mulitplication by 2p
sp leads to
2p =2
5(2p)
2+
2
5,
and we may rewrite this as
(2p)2 − 5
2· 2p + 1 = 0,
a quadratic equation in 2p. The solution to this equation is
2p =1
2
(5
2±√
25
4− 4
)=
1
2
(5
2± 3
2
).
We thus have either 2p = 2 or 2p = 12 , and hence either p = 1 or p = −1.
(ii) With v(s) = As + Bs, we have lims→∞ v(s) = ±∞ unless A = 0. The
boundary condition lims→∞ v(s) = 0 implies therefore that A is zero.
(iii) Since v(s) is positive, we must have B > 0. We note that lims↓0 fB(s) =lims→∞ fB(s) = ∞. Therefore, fB(s) takes the value zero for some s ∈(0,∞) if and only if its minimium over (0,∞) is less than or equal to zero.To find the minimizing value of s, we set the derivative of fB(s) equal tozero:
−B
s2+ 1 = 0.
This results in the critical point sc =√
B. We note that the second deriva-tive, 2B
s3 , is positive on (0,∞), so the function is convex, and hence fB
attains a minimum at sc. The minimal value of fB on (0,∞) is
fB(sc) = 2√
B − 4.
This is positive if B > 4, in which case fB(s) = 0 has no solution in(0,∞). If B = 4, then fB(sc) = 0 and sc is the only solution to theequation fB(s) = 0 in (0,∞). If 0 < B < 4, then fB(sc) < 0 and theequation fB(s) = 0 has two solutions in (0,∞).
(iv) Since vB(s) = Bs
for all large values of s, we maximize this by choosing
B as large as possible, i.e., B = 4. For values of B < 4, the curve Bs
liesbelow the curve 4
s(see Figure 5.8.1), and values of B > 4 are not possible
because of part (iii).
5.8 Solutions to Selected Exercises 51
(iv) We see from the tangency of the curve y = 4s
with the intrinsic valuey = 4− s at the point (2,2) in Figure 5.8.1 that y = 4
sand y = 4− s have
the same derivative at s = 2. Indeed,
d
ds
4
s
∣∣∣∣s=2
= − 4
s2
∣∣∣∣s=2
= −1,
and as noted in the statement of the exercise,
d
ds(4 − s) = −1.
1 2 3
y
s
4
4
y = 4
s
y = 3
sy = 4− s
(2, 2)
Fig. 5.8.1. The curve y = B
sfor B = 3 and B = 4.
6
Interest-Rate-Dependent Assets
6.9 Solutions to Selected Exercises
Exercise 6.1. Prove parts (i), (ii), (iii) and (v) of Theorem 2.3.2 when con-ditional expectation is defined by Definition 6.2.2. (Part (iv) is not true in theform stated in Theorem 2.3.2 when the coin tosses are not independent.)
Solution We take each of parts (i), (ii), (iii) and (v) of Theorem 2.3.2 in turn.
(i) Linearity of conditional expectations. According to Definition 6.2.2,
En[c1X + c2Y ](ω1 . . . ωn)
=∑
ωn+1,...,ωN
(c1X(ω1 . . . ωnωn+1 . . . ωN ) + c2Y (ω1 . . . ωnωn+1 . . . ωN )
)
×Pωn+1 = ωn+1, . . . , ωN = ωN |ω1 = ω1, . . . , ωn = ωn= c1
ωn+1,...,ωN
X(ω1 . . . ωnωn+1 . . . ωN )
×Pωn+1 = ωn+1, . . . , ωN = ωN |ω1 = ω1, . . . , ωn = ωn+c2
ωn+1,...,ωN
Y (ω1 . . . ωnωn+1 . . . ωN )
×Pωn+1 = ωn+1, . . . , ωN = ωN |ω1 = ω1, . . . , ωn = ωn= c1En[X](ω1 . . . ωn) + c2En[Y ](ω1 . . . ωn).
(ii) Taking out what is known. If X depends only on the first n cointosses, then
54 6 Interest-Rate-Dependent Assets
En[XY ](ω1 . . . ωn)
=∑
ωn+1,...,ωN
X(ω1 . . . ωn)Y (ω1 . . . ωnωn+1 . . . ωN )
×Pωn+1 = ωn+1, . . . , ωN = ωN |ω1 = ω1, . . . , ωn = ωn= X(ω1 . . . ωn)
ωn+1,...,ωN
Y (ω1 . . . ωnωn+1 . . . ωN )
×Pωn+1 = ωn+1, . . . , ωN = ωN |ω1 = ω1, . . . , ωn = ωn= X(ω1 . . . ωn)En[Y ](ω1 . . . ωn)
(iii) Iterated conditioning. If 0 ≤ n ≤ m ≤ N , then because Em[X] de-pends only on the first m coin tosses and
ωm+1,...,ωN
Pωn+1 = ωn+1, . . . , ωm = ωm, ωm+1 = ωm+1, . . . , ωN = ωN
|ω1 = ω1, . . . , ωn = ωn= Pωn+1 = ωn+1, . . . , ωm = ωm|ω1 = ω1, . . . , ωn = ωn,
we have
En
[Em[X]
](ω1 . . . ωn)
=∑
ωn+1,...,ωN
Em[X](ω1 . . . ωm)
×Pωn+1 = ωn+1, . . . , ωm = ωm, ωm+1 = ωm+1, . . . , ωN = ωN
|ω1 = ω1, . . . , ωn = ωn=
ωn+1,...,ωm
Em[X](ω1 . . . ωm)∑
ωm+1,...,ωN
×Pωn+1 = ωn+1, . . . , ωm = ωm, ωm+1 = ωm+1, . . . , ωN = ωN
|ω1 = ω1, . . . , ωn = ωn=
ωn+1,...ωm
Em[X](ω1 . . . ωm)
×Pωn+1 = ωn+1, . . . , ωm = ωm|ω1 = ω1, . . . , ωn = ωn=
ωn+1,...,ωm
ωm+1,...,ωN
X(ω1 . . . ωmωm+1 . . . ωN )
×Pωm+1 = ωm+1, . . . , ωN = ωN |ω1 = ω1, . . . , ωm = ωm×Pωn+1 = ωn+1, . . . , ωm = ωm|ω1 = ω1, . . . , ωn = ωn,
where we have used the definition
6.9 Solutions to Selected Exercises 55
Em[X](ω1 . . . ωm)
=∑
ωm+1,...,ωN
X(ω1 . . . ωmωm+1 . . . ωN )
×Pωm+1 = ωm+1, . . . , ωN = ωN |ω1 = ω1, . . . , ωm = ωm
in the last step. Using the fact that
Pωm+1 = ωm+1, . . . , ωN = ωN |ω1 = ω1, . . . , ωm = ωm×Pωn+1 = ωn+1, . . . , ωm = ωm|ω1 = ω1, . . . , ωn = ωn
= Pωn+1 = ωn+1, . . . , ωm = ωm, ωm+1 = ωm+1, . . . , ωN = ωN
|ω1 = ω1, . . . , ωn = ωn= Pωn+1 = ωn+1, . . . , ωN = ωN |ω1 = ω1, . . . , ωn = ωn,
we may write the last term in the above formula for En
[Em[X]
](ω1 . . . ωn)
as
En
[Em[X]
](ω1 . . . ωn)
=∑
ωn+1,...,ωm
ωm+1,...,ωN
X(ω1 . . . ωmωm+1 . . . ωN )
×Pωn+1 = ωn+1, . . . , ωN = ωN |ω1 = ω1, . . . , ωn = ωn=
ωn+1,...,ωN
X(ω1 . . . ωN )
×Pωn+1 = ωn+1, . . . , ωN = ωN |ω1 = ω1, . . . , ωn = ωn= En[X](ω1 . . . ωn).
(iv) Conditional Jensen’s inequality. This follows from part (i) just likethe proof of part (v) in Appendix A.
Exercise 6.2. Verify that the discounted value of the static hedging portfolioconstructed in the proof of Theorem 6.3.2 is a martingale under P.
Solution The static hedging portfolio in Theorem 6.3.2 is, at time n, to shortSn
Bn,mzero coupon bonds maturing at time m and to hold one share of the
asset with price Sn. The value of this portfolio at time k, where n ≤ k ≤ m,is
Xk = Sk − Sn
Bn,m
Bk,m, k = n, n + 1, . . . ,m.
For n ≤ k ≤ m − 1, we have
Ek [Dk+1Xk+1] = Ek[Dk+1Sk+1] −Sn
Bn,m
Ek [Dk+1Bk+1,m] .
56 6 Interest-Rate-Dependent Assets
Using the fact that the discounted asset price is a martingale under therisk-neutral measure and also using (6.2.5) first in the form Dk+1Bk+1,m =
Ek+1[Dm] and then in the form DkBk,m = Ek[Dm], we may rewrite this as
Ek[Dk+1Xk+1] = DkSk − Sn
Bn,m
Ek
[Ek+1[Dm]
]
= DkSk − Sn
Bn,m
Ek[Dm]
= DkSk − Sn
Bn,m
DkBk,m
= DkXk.
This is the martingale property.
Exercise 6.4. Using the data in Example 6.3.9, this exercise constructs ahedge for a short position in the caplet paying (R2 − 1
3 )+ at time three. Weobserve from the second table in Example 6.3.9 that the payoff at time threeof this caplet is
V3(HH) =2
3, V3(HT ) = V3(TH) = V3(TT ) = 0.
Since this payoff depends on only the first two coin tosses, the price of thecaplet at time two can be determined by discounting:
V2(HH) =1
1 + R2(HH)V3(HH) =
1
3, V2(HT ) = V2(TH) = V2(TT ) = 0.
Indeed, if one is hedging a short position in the caplet and has a portfoliovalued at 1
3 at time two in the event ω1 = H,ω2 = H, then one can simplyinvest this 1
3 in the money market in order to have the 23 required to pay off
the caplet at time three.
In Example 6.3.9, the time-zero price of the caplet is determined to be221 (see (6.3.10)).
(i) Determine V1(H) and V1(T ), the price at time one of the caplet in theevents ω1 = H and ω1 = T , respectively.
(ii) Show how to begin with 221 at time zero and invest in the money market
and the maturity two bond in order to have a portfolio value X1 at timeone that agrees with V1, regardless of the outcome of the first coin toss.Why do we invest in the maturity two bond rather than the maturitythree bond to do this?
(iii) Show how to take the portfolio value X1 at time one and invest in themoney market and the maturity three bond in order to have a portfoliovalue X2 at time two that agrees with V2, regardless of the outcome of thefirst two coin tosses. Why do we invest in the maturity three bond ratherthan the maturity two bond to do this?
6.9 Solutions to Selected Exercises 57
Solution.
(i) We determine V1 by the risk-neutral pricing formula. In particular,
V1(H) =1
D1(H)E1[D2V2](H)
= Pω2 = H|ω1 = HD2(HH)V2(HH)
+Pω2 = T |ω1 = HD2(HT )V2(HT )
=2
3· 6
7· 1
3+
1
3· 6
7· 0 =
4
21,
V1(T ) =1
D1(T )E1[D2V2](T ) = 0.
(ii) We compute the number of shares of the time-two maturity bond by theusual formula:
∆0 =V1(H) − V1(T )
B1,2(H) − B1,2(T )=
421 − 067 − 5
7
=4
3.
It is straight-forward to verify that this works. Set X0 = V0 = 221 and compute
X1(H) = ∆0B1,2(H) + (1 + R0)(X0 − ∆0B0,2)
=4
3· 6
7+
2
21− 4
3· 11
14=
4
21= V1(H),
X1(T ) = ∆0B1,2(T ) + (1 + R0)(X0 − ∆0B0,2)
=4
3· 5
7+
2
21− 4
3· 11
14= 0 = V1(T ).
We do not use the maturity three bond because B1,3(H) = B1,3(T ), and thisbond therefore provides no hedge against the first coin toss.
(iii) In the event of a T on the first coin toss, the caplet price is zero, thehedging portfolio has zero value, and no further hedging is required. In theevent of a H on the first toss, we hedge by taking in the maturity three bondthe position
∆1(H) =V2(HH) − V2(HT )
B2,3(HH) − B2,3(HT )=
13 − 012 − 1
= −2
3.
It is straight-forward to verify that this works. We compute
X2(HH) = ∆1(H)B2,3(HH) + (1 + R1(H))(X1(H) − ∆1(H)B1,3(H))
= −2
3· 1
2+
7
6
(4
21+
2
3· 4
7
)=
1
3= V2(HH),
X2(HT ) = ∆1(H)B2,3(HT ) + (1 + R1(H))(X1(H) − ∆1(H)B1,3(H))
= −2
3· 1 +
7
6
(4
21+
2
3· 4
7
)= 0 = V2(HT ).
58 6 Interest-Rate-Dependent Assets
We do not use the maturity two bond because B2,2(HH) = B2,2(HT ), andthis bond therefore provides no hedge against the second coin toss.
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
##### Shreve, Steven - Stochastic Calculus for Finance I: The Binomial Asset Pricing Model
Economy & Finance
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents | 29,463 | 72,140 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2023-23 | latest | en | 0.314268 |
https://headshotsmarathon.org/blog/what-is-the-relation-between-time-domain-and-frequency-domain/ | 1,719,255,503,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865482.23/warc/CC-MAIN-20240624182503-20240624212503-00458.warc.gz | 256,857,734 | 14,741 | # What is the relation between time domain and frequency domain?
## What is the relation between time domain and frequency domain?
There is the difference between time domain and frequency domain are given below,
Time domain Frequency domain
A time domain graph shows how a signal changes over time. The frequency domain graph shows how much of the signal lies within each given frequency band over a range of frequencies.
What is a Fourier transform frequency domain?
The term Fourier transform refers to both the frequency domain representation and the mathematical operation that associates the frequency domain representation to a function of space or time.
### How do you find the frequency of a Fourier transform?
Let X = fft(x) . Both x and X have length N . Suppose X has two peaks at n0 and N-n0 . Then the sinusoid frequency is f0 = fs*n0/N Hertz….
1. Replace all coefficients of the FFT with their square value (real^2+imag^2).
2. Take the iFFT.
3. Find the largest peak in the iFFT.
What is relation between time and frequency response?
The period T is the reciprocal of a frequency, T = 1 / f.
#### How do you convert a time domain to a frequency domain in Matlab?
1. nfft = length(y);
2. f = (0:1/nfft:1-1/nfft)*fs; % define frequency-domain.
3. figure; % figure should be written before subplot to open new figure.
4. subplot(2,1,1); % subplot(2,1,4) will give error beacause for a 2×1 vector valid indeces are 1&2, 4 is wrong.
Is frequency inversely proportional to time?
The very definition of a frequency is the rate at which something occurs over time. Therefore by shortening the time interval in which something occurs you are increasing its frequency, and by increasing the time interval you decrease the frequency. Hence, Frequency is inversely proportional to time.
## How are t w and f related to each other?
ω = angular frequency of the wave. T = time period of the wave. f = ordinary frequency of the wave….What is Angular Frequency?
Formula ω = 2 π T = 2 π f
What is time and frequency domain analysis?
Frequency-domain analysis is widely used in such areas as communications, geology, remote sensing, and image processing. While time-domain analysis shows how a signal changes over time, frequency-domain analysis shows how the signal’s energy is distributed over a range of frequencies.
### How do you find the frequency in a frequency distribution?
To do this, divide the frequency by the total number of results and multiply by 100. In this case, the frequency of the first row is 1 and the total number of results is 10. The percentage would then be 10.0. The final column is Cumulative percentage.
What is relationship between time and frequency?
The number of times a cycle is completed in a second is the frequency. The time taken to complete one vibration is called time period. Frequency and time period is inversely proportional, the number of vibrations per second is frequency. f = 1 t f=\frac{1}{t} f=t1 or t = 1 f t=\frac{1}{f} t=f1.
#### What is relation between time period and frequency?
RELATIONSHIP BETWEEN THE TIME PERIOD (T) and FREQUENCY (ƒ): By the initial time period is the. time taken to complete 1 vibration. Or. In time T, number of vibration= 1. ∴ In 1 second, the frequency or number of vibrations = 1 / T.
How do you transform frequency domain data to time domain?
Transform frequency-domain data to the time domain. Frequencies are linear and equally spaced. To transform frequency-domain iddata object f_data to time-domain iddata object t_data , use: t_data = ifft (f_data) Converting iddata data into the form of an idfrd frequency response is a type of estimation.
## What is the Fourier transform of a signal?
The Fourier transform decomposes a signal into the weights of orthonormal basis functions: sines and cosines of different frequencies. An analogy would be decomposing a vector in the x-y plane into its x and y components, which are the weights applied to the unit vectors along the x axis and y axis (the orthonormal basis vectors of the x-y plane).
What is time-domain and frequency-domain data?
Frequency-domain data consists of either transformed input and output time-domain signals or system frequency response sampled as a function of the independent variable frequency. For detailed information about representing time-domain and frequency-domain data in MATLAB ®, see Representing Data in MATLAB Workspace.
### What is frequency domain analysis in Electrical Engineering?
Frequency domain analysis and Fourier transforms are a cornerstone of signal and system analysis. These ideas are also one of the conceptual pillars within electrical engineering. Among all of the mathematical tools utilized in electrical engineering, frequency domain analysis is arguably the most far-reaching.
### How do you make Boh?How do you make Boh?
How do you make Boh? HOW TO MAKE BASIC CHILI BOH OR CABE GILING REMOVE THE SEEDS FROM DRIED CHILIES. Rinse the chili with some water to get rid of
### Whats the difference between flan and quiche?Whats the difference between flan and quiche?
Whats the difference between flan and quiche? What’s the Difference Between Flan and Quiche? Flan, in the British sense, is an open-faced pie or tart with a savory or sweet
### Who developed Kanata?Who developed Kanata?
Who developed Kanata? Bill Teron Modern Kanata is largely the creation of Bill Teron, a developer and urban planner who purchased over 1,200 hectares (3,000 acres) of rural land and | 1,209 | 5,496 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2024-26 | latest | en | 0.923478 |
https://www.lexic.us/definition-of/octal_notation | 1,539,832,321,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583511642.47/warc/CC-MAIN-20181018022028-20181018043528-00025.warc.gz | 1,000,287,242 | 8,112 | ### Definition of Octal notation
1. Noun. Any mathematical notation that uses 8 different characters (usually the digits 0 to 7).
Generic synonyms: Mathematical Notation
### Octal Notation Pictures
Click the following link to bring up a new window with an automated collection of images related to the term: Octal Notation Images
### Lexicographical Neighbors of Octal Notation
octagonsoctagynousoctahedraoctahedraloctahedralityoctahedrallyoctahedriteoctahedronoctahedronsoctahydrate octahydratesoctahydroxyanthraquinoneoctakis-octaloctal digitoctal notation (current term)octal number systemoctal numeration systemoctalogyoctaloop octaloopsoctalsoctameroctamer binding proteinoctamer motifoctamericoctamerousoctamersoctameteroctameters
### Literary usage of Octal notation
Below you will find example usage of this term as found in modern and/or classical literature:
1. Recreations in Agriculture, Natural-history, Arts, and Miscellaneous Literature by James Anderson (1803)
"•octal notation. The figures at the side are supernumerary. In the first example, for the sake of illustration, I have placed a dot at the right hand of' ..."
2. Who Goes There (2004)
"Thus the decimal number 4012, which is 7654 in octal notation, would be represented by A=7=A4+A2+A1, B=6-B4+B2, C=5=C4+C1, and D=4=D4. ..." | 356 | 1,313 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2018-43 | longest | en | 0.788725 |
https://www.studypool.com/discuss/514986/finding-the-increase?free | 1,508,445,007,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823462.26/warc/CC-MAIN-20171019194011-20171019214011-00454.warc.gz | 982,364,108 | 13,866 | ##### Finding the % increase
label Mathematics
account_circle Unassigned
schedule 1 Day
account_balance_wallet \$5
12 Teachers to 48 teachers
May 5th, 2015
12 teachers to 48 teachers.
1. If you mean that number of teachers increased from 48 to 60 (48+12) then -
% increase = ((60-48)/48)*100 %
= (12/48)*100 %
= 25% increase
2. If you mean that number of teachers increased from 12 to 48 then -
% increase = ((48-12)/12)*100 %
= 300% increase
Let me know if you need further help.
May 5th, 2015
...
May 5th, 2015
...
May 5th, 2015
Oct 19th, 2017
check_circle | 183 | 572 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2017-43 | latest | en | 0.895732 |
https://www.doubtnut.com/question-answer/in-a-quadrilateral-a-b-c-d-given-that-a-d90o-prove-that-a-c2-b-d2a-d2-b-c2--1410332 | 1,627,930,311,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154356.39/warc/CC-MAIN-20210802172339-20210802202339-00292.warc.gz | 755,815,515 | 70,781 | Home
>
English
>
Class 10
>
Maths
>
Chapter
>
Triangles
>
In a quadrilateral A B C D ...
# In a quadrilateral A B C D , given that /_A+/_D=90o . Prove that A C^2+B D^2=A D^2+B C^2 .
Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.
Updated On: 28-9-2020
Apne doubts clear karein ab Whatsapp par bhi. Try it now.
Watch 1000+ concepts & tricky questions explained!
26.2 K+
1.3 K+
Image Solution
3719
108.7 K+
238.0 K+
2:22
1415094
7.1 K+
68.8 K+
1:42
88357424
5.6 K+
40.5 K+
3:29
1414752
4.1 K+
82.6 K+
2:22
1410335
4.0 K+
19.3 K+
5:20
31344010
24.8 K+
51.5 K+
5:01
51238242
2.5 K+
50.7 K+
2:06
88357108
1.4 K+
27.1 K+
4:13
34139770
1.2 K+
23.4 K+
2:07
76129594
1.5 K+
29.6 K+
4:58
39181313
1.3 K+
25.7 K+
51238417
4.5 K+
32.7 K+
1:33
88357422
1.4 K+
29.0 K+
7:34
1410328
2.6 K+
18.8 K+
1:50
3221
47.8 K+
93.4 K+
6:18
Class 10th
Triangles
Class 10th
Triangles
Class 10th
Triangles
Class 10th
Triangles | 435 | 1,023 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2021-31 | latest | en | 0.53548 |
https://isabelle.in.tum.de/repos/isabelle/diff/0bc590051d95/src/HOL/Presburger.thy | 1,582,265,796,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145443.63/warc/CC-MAIN-20200221045555-20200221075555-00263.warc.gz | 410,258,843 | 2,343 | src/HOL/Presburger.thy
changeset 23146 0bc590051d95 parent 22801 caffcb450ef4 child 23148 ef3fa1386102
``` 1.1 --- a/src/HOL/Presburger.thy Thu May 31 11:00:06 2007 +0200
1.2 +++ b/src/HOL/Presburger.thy Thu May 31 12:06:31 2007 +0200
1.3 @@ -9,10 +9,14 @@
1.4 header {* Presburger Arithmetic: Cooper's Algorithm *}
1.5
1.6 theory Presburger
1.7 -imports NatSimprocs "../SetInterval"
1.8 +imports "Integ/NatSimprocs" SetInterval
1.9 uses
1.10 - ("cooper_dec.ML") ("cooper_proof.ML") ("qelim.ML")
1.11 - ("reflected_presburger.ML") ("reflected_cooper.ML") ("presburger.ML")
1.12 + ("Tools/Presburger/cooper_dec.ML")
1.13 + ("Tools/Presburger/cooper_proof.ML")
1.14 + ("Tools/Presburger/qelim.ML")
1.15 + ("Tools/Presburger/reflected_presburger.ML")
1.16 + ("Tools/Presburger/reflected_cooper.ML")
1.17 + ("Tools/Presburger/presburger.ML")
1.18 begin
1.19
1.20 text {* Theorem for unitifying the coeffitients of @{text x} in an existential formula*}
1.21 @@ -1047,15 +1051,15 @@
1.22 show ?thesis by (simp add: 1)
1.23 qed
1.24
1.25 -use "cooper_dec.ML"
1.26 -use "reflected_presburger.ML"
1.27 -use "reflected_cooper.ML"
1.28 +use "Tools/Presburger/cooper_dec.ML"
1.29 +use "Tools/Presburger/reflected_presburger.ML"
1.30 +use "Tools/Presburger/reflected_cooper.ML"
1.31 oracle
1.32 presburger_oracle ("term") = ReflectedCooper.presburger_oracle
1.33
1.34 -use "cooper_proof.ML"
1.35 -use "qelim.ML"
1.36 -use "presburger.ML"
1.37 +use "Tools/Presburger/cooper_proof.ML"
1.38 +use "Tools/Presburger/qelim.ML"
1.39 +use "Tools/Presburger/presburger.ML"
1.40
1.41 setup "Presburger.setup"
1.42
``` | 626 | 1,621 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2020-10 | latest | en | 0.406862 |
https://zefdamen.nl/CropCircles/Reconstructions/2014/NettleHill14/nettlehill2014en.htm | 1,721,712,440,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518014.29/warc/CC-MAIN-20240723041947-20240723071947-00216.warc.gz | 926,028,754 | 3,239 | Reconstruction of the
Nettle Hill 16-08-2014 formation
1. Draw a circle. Draw and extend the horizontal and vertical centerlines.
2. Construct the inscribed dodecagon (regular 12-sided polygon) of circle 1, pointing up.
3. Construct the inscribed circle of dodecagon 2.
4. Construct the inscribed regular 24-sided polygon of circle 3, pointing up. Number the angular points 1 - 24, counterclockwise, starting at the top.
5. Draw the diagonal of polygon 4, between angular points nrs. 1 and 10 (diagonal 1-10).
6. Construct a circle concentric to circle 1, tangent to line 5.
7. Construct the inscribed regular 24-sided polygon of circle 6, pointing up.
8. Construct the inscribed circle of polygon 7.
9. Repeat step 5 twenty-three times, relative to all other angular points of polygon 4, as shown.
10. Construct a circle concentric to circle 1, passing through the intersection of line 5 and the horizontal centerline.
11. Point A is the intersection of line 5 and diagonal 9 6-21. Construct a circle concentric to circle 1, passing through A. See detail.
12. Point B is the intersection of diagonals 9 3-18 and 7-22. Construct a circle centered at A, passing through B. See detail.
13. Copy circle 12 to the center of circle 1.
14. Point C is the intersection of diagonals 9 5-20 and 6-21, and point D is the intersection of diagonals 9 2-17 and 3-18. Construct a circle centered at C, passing through D. See detail.
15. Copy circle 14 to the center of circle 1.
16. Construct two circles, centered at the upper and lower intersections of circle 15 and the vertical centerline respectively, both passing through the righthand intersection of circle 15 and the horizontal centerline.
17. For clarity reasons, some of the results of previous steps are removed temporarily.
Draw the connecting line between the intersections of circle 8 and upper circle 16.
18. Construct a circle concentric to circle 1, tangent to line 17.
19. Construct the inscribed regular 24-sided polygon of circle 15, pointing up. Number the angular points similar to polygon 4.
20. Draw the twenty-four rays of polygon 19.
21. Draw seven lines, between angular points 19 nrs. 1 and 7 (1-7), 2-8, 3-9, 4-10, 5-11, 6-12, and 7-13.
22. Copy lines 21, while mirroring with respect to the vertical centerline.
23. Construct a "two-points" circle (defined by the two end-points of a centerline) between the lefthand intersections of circles 8 and 10 with the horizontal centerline.
24. Copy circle 23 five times, to the center of circle 1, and to the four intersections of circle 11 with the horizontal and vertical centerlines.
25. Construct two pairs of parallel lines, one horizontal, one vertical, tangent to circles 24 at both sides.
26. Copy circle 23 four times, to the four intersections of central circle 24 with the horizontal and vertical centerlines, and move each of these circles (copy and delete original) to their own corresponding intersection, away from the center of circle 1, as shown. See detail.
27. Construct two pairs of parallel lines, one horizontal, one vertical, tangent to circles 26 at the outer sides, extending upto circle 11 in both directions, as shown.
28. Construct a circle concentric to circle 1, passing through the intersection of lower circle 16 and ray 20 (to angular point 19) nr. 22.
29. Construct a "two-points" circle between the righthand intersections of the horizontal centerline and the lower horizontal line 25 with lower circle 16.
30. Copy circle 29 to the righthand intersection of circle 28 and the horizontal centerline, and move this circle to its own lower intersection with circle 28.
31. Extend ray 20 nr. 2 upto circle 10.
32. Construct a "two-points" circle between the intersections of circles 8 and 10 with line 31.
33. Point E is the intersection of diagonals 9 6-21 and 7-22. Construct a circle concentric to circle 32, passing through E. See detail.
34. Copy circle 33 three times, to the corresponding positions relative to rays 20 nrs. 8, 14 and 20.
35. Construct a circle centered at the intersection of diagonal 9 5-14 and the vertical centerline, tangent to circle 8 at the lower side. See detail.
36. Copy circle 35 to the lower intersection of circle 1 and the vertical centerline.
37. Construct a circle concentric to circle 1, tangent to circle 36 at the lower side.
38. Construct a circle centered at the lower intersection of circle 10 and the vertical centerline, tangent to circle 8 at the lower side.
39. Copy circle 38 to the lower intersection of circle 37 and the vertical centerline, and move this circle to its own lefthand intersection with circle 37.
40. Copy circle 33 to the center of circle 39. See detail.
41. Circles 1, 3, 8, 10, 11, 13, 15, 16, 18, 30, 33, 34, 39 and 40, and lines 5, 9, 20, 21, 22, 25 and 27, are used for the final reconstruction.
42. Remove all parts not visible within the formation itself.
pdf-file
43. Colour all areas corresponding to standing...
pdf-file
44. ...or to flattened crop, and finish the reconstruction of the Nettle Hill formation of 16-08-2014.
pdf-file
45.
Courtesy the Crop Circle Connector
The final result, matched with two aerial images.
Copyright © 2014, Zef Damen, The Netherlands
Personal use only, commercial use prohibited. | 1,350 | 5,284 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2024-30 | latest | en | 0.830056 |
https://puzzling.stackexchange.com/questions/47102/malware-in-the-computer | 1,568,625,670,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514572516.46/warc/CC-MAIN-20190916080044-20190916102044-00498.warc.gz | 639,524,426 | 27,710 | # Malware in the computer [closed]
You are IT staff working in a company and there are a few amount of computers which have some malwares in them.
• If you clean 3 computers every day starting today, you will be finished at the end of the day on Sunday.
• If you clean 5 computers every day starting today, you will be finished at the end of the day on Friday.
What day is it today?
## closed as off-topic by Peregrine Rook, Mithrandir, Alconja, Beastly Gerbil, IAmInPLSDec 25 '16 at 21:46
This question appears to be off-topic. The users who voted to close gave this specific reason:
• "This question is off-topic as it appears to be a mathematics problem, as opposed to a mathematical puzzle. For more info, see "Are math-textbook-style problems on topic?" on meta." – Peregrine Rook, Mithrandir, Alconja, Beastly Gerbil, IAmInPLS
If this question can be reworded to fit the rules in the help center, please edit the question.
## 2 Answers
Since we are finished at the end of the day,
it follows that the number of computers is divisible by 3 and 5. However, we do not know which multiple of 15 the number of computers is.
However,
if we let $a$ be the number of days cleaning 3 computers at a time, and $b$ be the number of days cleaning 5 computers at a time, we see $a\equiv b+2 \pmod 7$, due to the number of days between Friday and Sunday.
Also,
$5b\equiv 3a\equiv 3(b+2) \pmod 7$, so rearranging gives $2b \equiv 6 \pmod 7$, or $b \equiv 3 \pmod 7$. That means 3 days and some weeks will pass since starting to clean computers 5 at a time. Hence, 3 days and some weeks before the end of Friday would be the start of Wednesday.
In conclusion:
Without knowing the total number of computers, today is Wednesday.
• Yeah I got the same answer. – Beastly Gerbil Dec 24 '16 at 15:27
• @BeastlyGerbil Well, but your working assumes that the number of computers is the smallest possible. – Element118 Dec 24 '16 at 22:32
• anything bigger would result in you not finishing this week – Beastly Gerbil Dec 25 '16 at 9:59
Today is
Wednesday
We take the
Lowest Common Multiple (LCM) of 5 and 3 which is 15. We divide that by the computers we clean that day, and then take it away from that day.
So
15/3 = 5. Sunday - 5 days = Wednesday
15/5 = 3. Friday - 3 days = Wednesday
I have included Sunday/Friday in the sums above, as it says 'you will be finished at the end of the day' meaning you worked on that day too.
• @Oray am I correct? – Beastly Gerbil Dec 24 '16 at 13:51 | 697 | 2,492 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2019-39 | longest | en | 0.941735 |
https://halshs.archives-ouvertes.fr/halshs-01954353 | 1,579,635,443,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250605075.24/warc/CC-MAIN-20200121192553-20200121221553-00313.warc.gz | 469,723,809 | 10,044 | # Working on and with division in early China, 3 rd century BCE-7 th century CE
Abstract : When compared with the mathematical manuscripts from the earliest decades of the Chinese empire recently excavated, the oldest mathematical book in Chinese handed down through the written tradition, The Nine Chapters (1st century C.E.), presents interesting differences. The texts of its procedures regularly make use of theoretical terms that are not found in the manuscripts, or at least not with the same theoretical meaning. Some of these terms are read by the commentators as related to the reasons why the procedures are correct: “make communicate 通 tong," “equalize tong,” lü 率—a concept of number as defined relatively to other numbers. Others point out that a certain structure has been identified in a set of operations. This is the case primarily for one of the terms by means of which division is prescribed 除 chu, which is the basis for a whole terminological complex absent from the earliest manuscripts. I do not mean to say that theoretical terms are completely absent from the manuscripts. Writings on Mathematical Procedures, excavated from a tomb sealed ca 186 B.C.E., contains some such terms. Moreover, where they occur is interesting. However, the extent and the nature of the phenomenon cannot be compared to what we have in The Nine Chapters. Interestingly enough, a great deal of these terms occur in relation to division or related procedures, predominantly the rule of three. These clues strongly suggest that division has been an important topic of reflection in some Chinese milieus between the 3rd century B.C.E. and the first century C.E. The hypothesis is supported by many other facts. Division, for instance, with closely related procedures like the rule of three, is a major topic in the manuscripts, in which many procedures are devoted to its various types of execution. One has to bear in mind that the operands to which division is applied were usually complex measured quantities. Division was also the only operation for which operands are designated by technical terms. Further, elementary operations, such as changes of units, essential to the practice of division can be correlated to theoretical discussions developed in the commentaries on The Nine Chapters and to the emergence of mathematical concepts. The article suggests that the execution of division was carried out through mainly four basic phases. On one of them, the transformation of the operands into the smallest possible integers with respect to a single measuring unit, there seems to me a great continuity of practice between the manuscripts and The Nine Chapters, the main difference being the introduction of technical terms to designate procedures. On the next two elementary operations, through which the result is produced bit by bit, we can perceive a break between the time when the known manuscripts were written and the time of the compilation of The Nine Chapters. I argue that this break might have been correlated to a change in the number system with which integers were written down with counting rods. The Nine Chapters bears witness to the emergence of a system of practice with the instrument of computation, centered on division and the then opposed operation of multiplication, a system that remained in use for centuries. Finally, the fourth phase related to the statement of the result, using fractions. It evidences transformations and provides evidence on how the concept of fraction to which ancient Chinese sources testify might have emerged in the context of the ancient way of computing divisions.
Keywords :
Document type :
Book sections
Domain :
Cited literature [5 references]
https://halshs.archives-ouvertes.fr/halshs-01954353
Contributor : Karine Chemla <>
Submitted on : Thursday, December 13, 2018 - 3:52:55 PM
Last modification on : Thursday, February 21, 2019 - 12:22:03 PM
Long-term archiving on: Thursday, March 14, 2019 - 3:12:15 PM
### File
Chemla_DivisionSAW Workshop_20...
Files produced by the author(s)
### Identifiers
• HAL Id : halshs-01954353, version 1
### Citation
Karine Chemla. Working on and with division in early China, 3 rd century BCE-7 th century CE. Karine Chemla; Agathe Keller; Christine Proust. Cultures of computation and quantification in the ancient world, In press. ⟨halshs-01954353⟩
Record views | 897 | 4,375 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2020-05 | latest | en | 0.973422 |
https://www.formulaconversion.com/formulaconversioncalculator.php?convert=gigameterspersecond_to_decimetersperhour | 1,721,503,156,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517515.18/warc/CC-MAIN-20240720174732-20240720204732-00896.warc.gz | 683,567,981 | 14,436 | # Gigameters/second to decimeters/hour (Gm/s to dm/hr) Metric conversion calculator
Welcome to our gigameters/second to decimeters/hour (Gm/s to dm/hr) conversion calculator. You can enter a value in either the gigameters/second or decimeters/hour input fields. For an understanding of the conversion process, we include step by step and direct conversion formulas. If you'd like to perform a different conversion, just select between the listed Speed units in the 'Select between other Speed units' tab below or use the search bar above. Tip: Use the swap button to switch from converting gigameters/second to decimeters/hour to decimeters/hour to gigameters/second.
## decimeters/hour (dm/hr)
(not bookmarks)
Swap
< == >
1 Gm/s = 35999999997120 dm/hr 1 dm/hr = 0 Gm/s
Algebraic Steps / Dimensional Analysis Formula
Gm/s * 3600 Gm/hr 1 Gm/s * 10000000000 dm/hr1 Gm/hr = dm/hr
Direct Conversion Formula
Gm/s * 35999999997120 dm/hr1 Gm/s = dm/hr
feet/hour 0 feet/second 0 kilometers/hour 0 kilometers/second 0 knots 0 meters/hour 0 meters/second 0 miles/hour 0 miles/second 0
If you would like to switch between Speed units, select from the tables below
centimeters/hour centimeters/second decimeters/hour decimeters/second dekameters/hour dekameters/second feet/hour feet/second gigameters/hour gigameters/second hectometers/hour hectometers/second kilometers/hour kilometers/second knots megameters/hour megameters/second meters/hour meters/second micrometers/hour micrometers/second miles/hour miles/second millimeters/hour millimeters/second yards/hour yards/second < == > centimeters/hour centimeters/second decimeters/hour decimeters/second dekameters/hour dekameters/second feet/hour feet/second gigameters/hour gigameters/second hectometers/hour hectometers/second kilometers/hour kilometers/second knots megameters/hour megameters/second meters/hour meters/second micrometers/hour micrometers/second miles/hour miles/second millimeters/hour millimeters/second yards/hour yards/second | 529 | 2,002 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2024-30 | latest | en | 0.70219 |
https://www.physicsforums.com/threads/projectile-motion-water-hose-question.835403/ | 1,531,975,987,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590493.28/warc/CC-MAIN-20180719031742-20180719051742-00485.warc.gz | 959,129,174 | 17,465 | # Homework Help: Projectile Motion Water hose Question
Tags:
1. Oct 1, 2015
### jasonchiang97
1. The problem statement, all variables and given/known data
A water hose is used to fill a large cylindrical storage tank of diameter D and height 2D . The hose shoots the water at 45 ∘ above the horizontal from the same level as the base of the tank and is a distance 6D away
2. Relevant equations
Vx=Vt
D=vit+1/2at^2
3. The attempt at a solution
I tried plugging in the values into the equation and I end up with√6dG<V<√√7dG
2. Oct 1, 2015
### haruspex
You should not apply equations without understanding the context in which they are valid. The equation you tried does not allow you to use the height of the tank, so it cannot be appropriate. When is it appropriate?
3. Oct 1, 2015
### jasonchiang97
Right, okay I think I'm only allowed to use the equation when the object is landing at the same level ground as it started off on.
4. Oct 1, 2015
### haruspex
Right. Or, more precisely, when d is the distance to a point where it is at the same level it started at.
5. Oct 1, 2015
### Staff: Mentor
What is the question? What are you trying to find?
6. Oct 1, 2015
7. Oct 1, 2015
### jasonchiang97
oops. Sorry I thought I posted that.
8. Oct 1, 2015
### Staff: Mentor
Okay, so what are the scenarios for the water stream that you need to investigate? That is, what are the significant characteristics of the trajectory in each case?
9. Oct 1, 2015
### @navin
OK. From where is the hose 6D away? From the centre of the tank? Or from where the tank begins?
10. Oct 1, 2015
### Staff: Mentor
Clever student's strategy: Unless the problem states a condition explicitly, choose the simplest situation to work with and declare it as an assumption. Just make sure that it's really undefined and not subtly implied somehow in the problem statement.
11. Oct 1, 2015
### @navin
You can't use dg = v^2(sin∆) here as this is equation gives us the maximum range of the projectile.
12. Oct 1, 2015
### haruspex
See post #3.
13. Oct 1, 2015
### haruspex
So what equation should you use here?
14. Oct 1, 2015
### jasonchiang97
I used the 2D equations and plugged in t=Vx/Vcos45 into d=vtsin45+1/2at^2 and solved.
Thanks
15. Oct 1, 2015
### haruspex
16. Oct 1, 2015
yep | 672 | 2,298 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2018-30 | latest | en | 0.910173 |
https://www.physicsforums.com/threads/have-a-proof-re-cyclic-groups-need-a-little-explaining.383244/ | 1,713,723,991,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817790.98/warc/CC-MAIN-20240421163736-20240421193736-00365.warc.gz | 853,044,162 | 16,606 | Have a proof re. cyclic groups, need a little explaining
• bennyska
In summary, the author argues that if a group has finite order, then its subgroups have finite order as well. He provides a proof.
bennyska
Homework Statement
Let a,b be elements of a group G. show that if ab has finite order, then ba has finite order.
The Attempt at a Solution
provided proof:
Let n be the order of ab so that (ab)n = e. Multiplying this equation on the left by b and on the right by a, we find that (ba)n+1 = bea = (ba)e. Cancellation of the first factor ba from both sides shows that (ba)n = e, so the order of ba is < n. If the order of ba were less than n, a symmetric argument would show that the order of ab is less than n, contrary to our choice of n. Thus ba has order n also.
okay. so, this is from a chapter on cyclic groups, so I'm assuming that it has to do with, duh, cyclic groups. i know cyclic groups are abelian, and if that were to be assumed from the beginning, i believe this problem would be relatively easy. although, actually, it probably wouldn't even be a problem, since it would be true from the beginning (since we could just say ab=ba, and abn=ban
so:
Let n be the order of ab so that (ab)n = e.
got that, that makes sense, i believe that's the definition of order of an element.
WAIT. here's the crux, something i (maybe) just realized as i typed this sentence. since we define the order of ab such that abn=e, does that imply that G is cyclic, and thus abelian? that would make things easier, if I'm right. so far, a quick internet search has failed me on an answer.
If the group is abelian then ab=ba and there is really nothing to prove. The proof you gave holds regardless of whether the group is abelian or not.
okay, i was just confused as how b(ab)na could turn into (ba)n+1 without it being abelian. i see how it's from letting (ab)n = e. thanks.
bennyska said:
okay, i was just confused as how b(ab)na could turn into (ba)n+1 without it being abelian. i see how it's from letting (ab)n = e. thanks.
It's actually just from regrouping the terms.
yeah, actually, i still don't get it. how can you regroup the terms if it's not commutative? or at least it's not assumed to be commutative.
we have (ab)(ab)...(ab) n times. so b(ab)na = b(ab)(ab)...(ab)a...
Oh! got it! thanks!
1. What is a cyclic group?
A cyclic group is a group that is generated by a single element. This means that all the elements in the group can be obtained by repeatedly applying the group operation to the generator element. In simpler terms, a cyclic group is a group that forms a cycle when its elements are multiplied together.
2. How do you prove that a group is cyclic?
To prove that a group is cyclic, you need to show that there exists an element in the group that can generate all other elements through repeated application of the group operation. This can be done by finding an element that, when raised to different powers, produces all the elements in the group.
3. What is the order of a cyclic group?
The order of a cyclic group is the number of elements in the group. This is equal to the number of times the generator element needs to be multiplied with itself to produce all the elements in the group. The order of a cyclic group is always finite.
4. Can a cyclic group be infinite?
Yes, a cyclic group can be infinite. This happens when the generator element has infinite order, meaning that it can be multiplied with itself an infinite number of times to produce all the elements in the group. An example of an infinite cyclic group is the group of all integers under addition.
5. How are cyclic groups used in mathematics?
Cyclic groups are used in various areas of mathematics, including algebra and number theory. They are particularly useful in studying symmetry and patterns, and in solving equations in abstract algebra. They also have applications in cryptography, as they can be used to generate large, random numbers.
• Calculus and Beyond Homework Help
Replies
2
Views
2K
• Calculus and Beyond Homework Help
Replies
1
Views
1K
• Calculus and Beyond Homework Help
Replies
9
Views
1K
• Calculus and Beyond Homework Help
Replies
5
Views
1K
• Calculus and Beyond Homework Help
Replies
2
Views
959
• Calculus and Beyond Homework Help
Replies
5
Views
1K
• Calculus and Beyond Homework Help
Replies
1
Views
2K
• Calculus and Beyond Homework Help
Replies
11
Views
2K
• Calculus and Beyond Homework Help
Replies
3
Views
1K
• Calculus and Beyond Homework Help
Replies
1
Views
496 | 1,119 | 4,516 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2024-18 | latest | en | 0.970485 |
http://www.wolfram.com/customer-stories/analysts-better-represent-financial-structures-with-mathematica-symbolics.html | 1,534,406,834,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221210559.6/warc/CC-MAIN-20180816074040-20180816094040-00264.warc.gz | 611,243,876 | 9,884 | # Analysts Better Represent Financial Structures with Mathematica Symbolics
"Mathematica makes it possible to manipulate the Black-Scholes pricing formula directly as a symbolic entity. This greatly simplifies computing the various risk characteristics of an option—a task that is both tedious and error-prone when performed in a more traditional manner—and facilitates extensions of the model that are made 'on the fly.'"
Now, more than ever, financial computation involves more than pure number crunching, and the new financial leaders are looking to Mathematica for help.
Investment houses regularly "slice and dice" securities such as mortgages, government bonds, and even junk bonds to meet specific risk/return objectives. While spreadsheet programs and C/C++ remain useful, symbolic programming languages--i.e., languages that manipulate both the numbers and the symbols that represent the financial structures--are increasingly needed to deal with the increasing complexity of the financial world.
Dr. Ross Miller,1 one of the first to take a symbolic approach to financial analysis, used Mathematica to create a relatively simple model that is traditionally numerical in nature: the Black-Scholes option pricing model.
"Mathematica makes it possible to manipulate the Black-Scholes pricing formula directly as a symbolic entity. This greatly simplifies computing the various risk characteristics of an option--a task that is both tedious and error-prone when performed in a more traditional manner--and facilitates extensions of the model that are made 'on the fly.'"
Mathematica then helped pick up where the Black-Scholes option pricing model left off. Although the Black-Scholes model gives a closed-form solution for pricing European options, many other options cannot be priced by this model. Such options include American options and options whose terminal payoffs are dependent on the path followed by the stock price. Using the binary and binomial option pricing models and the Monte Carlo method, three financial engineers2 used Mathematica to price such options.
"Without Mathematica, others have had to use approximations to pricing formulas for options that are difficult to price. Mathematica enabled us to price many of these options directly."
1. Dr. Ross Miller is the author of the book Computer-Aided Financial Analysis and has published several journal articles on information transfer in financial markets and on the design of advanced electronic market systems. He was the head of research for the National Westminster Bank and served as a member of the technical staff of the General Electric Corporate Research and Development Center before starting his own consulting company specializing in investment and risk management.
2. Simon Benninga, Raz Steinmetz, and John Stroughair. Dr. Benninga is the author of Numerical Methods in Finance, and his research interests include options and futures markets. Raz Steinmetz and John Stroughair were graduate students at the Wharton School of the University of Pennsylvania when this research was conducted.
Share Email Embed
### Related Stories
Get started with Wolfram technologies, or work with us to apply computational expertise to your projects.
Questions? Comments? Get in touch: 1-800-WOLFRAM, or email us » | 625 | 3,304 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2018-34 | longest | en | 0.902157 |
https://mathoverflow.net/questions/385045/unique-bipartite-perfect-matchings-and-cycles | 1,624,508,084,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488550571.96/warc/CC-MAIN-20210624015641-20210624045641-00477.warc.gz | 343,682,440 | 30,198 | Unique bipartite perfect matchings and cycles?
Given a graph $$G$$ which is bipartite and balanced and has unique perfect matching let $$G^{e}$$ be $$G$$ without edge $$e$$. Let $$G\cup G_{\pi,\pi'}$$ be union of $$G$$ and $$G_{\pi,\pi'}$$ where $$G_{\pi,\pi'}$$ is $$G$$ but having vertices of permuted by permutation $$\pi,\pi'$$ and an edge is in the union iff it is in either $$G$$ or its permutation.
NOTATION $$\pi\in S_n$$ permutes color $$1$$ vertices and $$\pi′\in S_n$$ color $$2$$ vertices. The graph $$H=G_{\pi,\pi'}$$ has new edge $$(i,j)$$ if $$(\pi^{-1}(i),\pi'^{-1}(j))$$ is an edge in $$G$$. Remember we are specifying union and so the new permuted graph is technically considered 'different' and so we can specify union.
Eg: Consider graph having two vertices of color $$1$$ and $$2$$ and edge is $$(1,2)$$ and $$(1,1)$$. The permutation $$\pi$$ flips $$1$$ and $$2$$ of color $$1$$. $$G_{\pi,id}$$ has edges $$(2,2)$$ and $$(2,1)$$.
Is there a statement similar to "If $$e$$ belongs to the unique perfect matching then it is true at every $$\pi,\pi'$$ satisfying $$\pi\pi'^{-1}\neq id$$ and $$\pi^{-1}\pi'\neq id$$ the graph $$G^e\cup G^e_{\pi,\pi'}$$ has a vertex which is not part of a cycle"?
Since union of perfect matchings in bipartite graphs is disjoint union of cycles the converse is correct. If the graph is not of unique perfect matching I think we can produce a counterexample.
• It looks interesting but it is hard to follow what you mean by $G_{\pi, \pi'}$ – Mike Feb 27 at 0:15
• $\pi$ permutes color $1$ vertices and $\pi'$ color $2$. Remember we are specifying union and so the new permuted graph is technically considered 'different' and so we can specify union. – 1.. Feb 27 at 2:54
• Thanks for the edits, it does seem clearer now... – Mike Feb 27 at 3:33
Counterexample: Let $$G$$ be a path on 10 vertices $$y_1,y_2, \ldots y_{10}$$. This has a unique matching and this matching includes $$e=y_5y_6$$. Then $$G\setminus \{e\}$$ is 2 paths w $$5$$ vertices each; $$y_1y_2y_3y_4y_5$$ and $$y_6y_7y_8y_9y_{10}$$. So let $$\pi_1$$ be the permutation on $$\{y_1,y_3, y_5,y_7,y_9\}$$ that transposes $$y_1$$ and $$y_5$$ and leaves each of $$y_3$$, $$y_7$$, $$y_9$$ fixed. Let $$\pi_2$$ be the permutation on $$\{y_2,y_4,y_6,y_8, y_{10}\}$$ that transposes $$y_6$$ and $$y_{10}$$ and leaves each of $$y_2,y_4,y_8$$ fixed. Then every edge in $$H \doteq G^e \cup G^e_{\pi_1,\pi_2}$$ is in a cycle. Indeed, the first component of $$H$$ is the path $$y_1y_2y_3y_4y_5$$ plus the edges $$y_1y_4$$ and $$y_2y_5$$. The second component of $$H$$ is isomorphic to the first. | 894 | 2,604 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 60, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2021-25 | longest | en | 0.804205 |
https://www.crazyengineers.com/threads/crompton-placement-papers-and-campus-placement-process.46401 | 1,716,835,581,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059045.25/warc/CC-MAIN-20240527175049-20240527205049-00621.warc.gz | 634,660,385 | 8,202 | # Crompton Placement Papers And Campus Placement Process
I heard mostly Crompton visits campus for Electrical Engineers (B.E. or B.Tech).
The placement process is same everywhere:
1. Written Test
2. Group Discussion
3. Interview
Here are a few of the questions appearing in their aptitude test:
## Replies
• Ankita Katdare
Some of it's technical questions are given below. @Electrical Engineers: Please try to answer them.
1) Magnetic compass indicates
a) true north pole b) alionic earth pole c)geographic north pole d)magnetic north pole.
2) For a cantilever beam with uniformly varying load, shape of bending moment curve is
a)parabolic b) hyperbolic c) straight line d) cubic.
3) For perfectly elastic bodies, coefficient of restitution is
a) 0 b) 0.5 c) 1 d) infinity.
4) For 1st order lever, mechanical advantage is
a)<1 b)>1 c) =1 d)none.
5) 18:4:1 in H.S.S represent
a) 18tu : 4vn : 1 cr. Other choices r just interchanged.
6) Due to intense braking of #-Link-Snipped-#
#-Link-Snipped-#an automobile reaction of wheels on road
Will be equally distributed to all 4 wheels.
Will be more at rear wheels
Inertia will be on the front wheels.
Not remember.
If you appl#-Link-Snipped-#y brakes on the wheels of a bicycle then
a. normal reaction on the wheel increases
b. normal reaction on the wheel decreases
c. normal reaction remains same as there is no radial component of frictional force
7) Thermal efficiency of I.C engines will be a)20-25% b)30-35% c)60-75% d) 45- 55%.
8) When a mass is supported by spring of spring constant k, which is cut in 4 equal pieces and the connected in parallel way the equivalent spring constant is
a)k/4 b) 16k c) 64k d) k.
9) When a mass is supported on roof of a elevator and elevator moves with an acceleration of 4.9 m/s2 shows a spring reading of 15kg, then the weight of the body is
a) 10 b) 15 c) 30 d)20.
10.Mass production of bolts and rivets is by
a. hot extrusion b. forging c. cold heading d. cold peening
11.why doesn't mercury stick to glass tube
a .cohesive force>adhesive force
b .cohesive force
c .cohesive force= adhesive force
d .none of the above
12.How are bolts and nuts manufactured?
a.head stamping b .cold extrusion c .cold peening d .forging
13.Eulers formula is used for column of length
a. >80 b. <80 c. >90 d. >110
14.If a body weighing 40kg floats in water with 40% of volume immersed, the sp. gravity of the body is
1 b. 0.16 c. 0.25 d.
15.Axial claw clutches a.Disengages b. engages c. engages only with load d. operates only with load
16.If center distance of a involute mating gear changes then the pressure angle
Increases b. decreases c. remains unaltered
17.Normalising is
a.heated above critical temp and cooled in air .
b. heated above critical temp and quenched in oil
c. heated below critical temp and cooled in air
18.The best suited material for permanent magnet
a. Alnico b. silicon
19.Hookes joint is used for
a. parallel and intersecting shafts b. non parallel non intersecting
c. nonparallel and intersecting d. parallel and non intersecting
20.To have a const.velocity ratio the two gears should have (or) angular velocity of the gear will remain the same if the .......... cuts the line joining ........ at a fixed point ans law of gearing
[/COLOR]
You are reading an archived discussion.
## Goldman Sachs Campus Placement Process
Goldman Sachs Campus Placement Process is as follows: • Goldman Sachs occasionally gives campus presentations prior to the scheduled recruitment day. These are good opportunities to learn about the firm...
## I-Flex Campus Placement Papers & Recruitment Process
I-Flex is the Oracle Financial Services Company. The placement process consists of aptitude test & interview. Here are a few sample papers for placement:
## Eligibilty For AIEEE Spot Round
All candidates eligible for central counseling, except those who have been admitted to an institute till 5th round of allotment, are eligible for spot round. All eligible candidates who did...
## Masters in engineering management
HEY PEOPLE .... WATS THE SCOPE OF THIS BRANCH IN THE FUTURE .... ANY SUGGESTIONS ??? AS I AM PLANNING TO PURSUE THIS NOW ... FROM CANADA CIAOXXX 😁
## Enable google.com/related - Browsing Assistant - Try it out!
Google Related is a browsing assistant tool that gives additional information about the topics you are browsing on the web. For example, if you’re browsing a page about a restaurant... | 1,121 | 4,446 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2024-22 | latest | en | 0.821535 |
https://math.answers.com/other-math/What_is_one-half_to_the_sixth_power | 1,726,238,503,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651523.40/warc/CC-MAIN-20240913133933-20240913163933-00356.warc.gz | 356,124,272 | 47,433 | 0
# What is one-half to the sixth power?
Updated: 4/28/2022
Wiki User
13y ago
1/64
Joshua Knowlton
Lvl 2
2y ago
Wiki User
13y ago
1/64
Earn +20 pts
Q: What is one-half to the sixth power?
Submit
Still have questions?
Related questions
### What is onehalf of a quart and what is the answer from onehalf to pints?
One half of a quart is one pint.
### Which is the area of this triangle?
The formula for the area of a triangle is: onehalf(b)(h) b:base and h:height For example: A triangle is 4 inches tall and 3 inches wide. What is the area of the triangle? Answer: onehalf(b)(h) onehalf(3)(4) onehalf(12) 6 So, the area of this particular triangle is 6 inches squared.
350
### What is nine to the sixth power?
nine to the sixth power is 531441.
### What is 99 to the sixth power?
99 to the sixth power is 941,480,149,401
### What 0.3 to the sixth power?
0.3 to the sixth power equals 0.000729
### What is 1.028 to the sixth power?
1.028 to the sixth power is 1.18020836358(rounded).
### What is 25 to the sixth power?
25 to the sixth power equates to 244,140,625
### What is 27 to the sixth power?
27 to the sixth power equals 387,420,489
### What does 10 to the sixth power equal?
10 to the sixth power equals 1,000,000
### What is -1 to the sixth power?
-1 to the sixth power equals one.
91,000,000 | 403 | 1,332 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2024-38 | latest | en | 0.924928 |
https://proofwiki.org/wiki/Pointwise_Multiplication_on_Real-Valued_Functions_is_Commutative | 1,596,457,462,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735810.18/warc/CC-MAIN-20200803111838-20200803141838-00538.warc.gz | 457,058,801 | 9,317 | Pointwise Multiplication on Real-Valued Functions is Commutative
Definition
Let $S$ be a non-empty set.
Let $f, g: S \to \R$ be real-valued functions.
Let $f \times g: S \to \R$ denote the pointwise product of $f$ and $g$.
Then:
$f \times g = g \times f$
Proof
$\, \displaystyle \forall x \in S: \,$ $\displaystyle \map {\paren {f \times g} } x$ $=$ $\displaystyle \map f x \times \map g x$ Definition of Pointwise Multiplication of Real-Valued Functions $\displaystyle$ $=$ $\displaystyle \map g x \times \map f x$ Real Multiplication is Commutative $\displaystyle$ $=$ $\displaystyle \map {\paren {g \times f} } x$ Definition of Pointwise Multiplication of Real-Valued Functions
$\blacksquare$ | 212 | 705 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2020-34 | latest | en | 0.572466 |
https://mathvoices.ams.org/mathmedia/tonys-take-january-2022/ | 1,722,782,876,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640404969.12/warc/CC-MAIN-20240804133418-20240804163418-00510.warc.gz | 311,731,634 | 19,100 | # Tony’s Take January 2022
## This month’s topics:
### Gödel’s incompleteness theorem in The Guardian
Alex Bellos’s Monday puzzle in The Guardian for January 10, 2022 was derived from explanations of Gödel’s incompleteness theorem due to the logician Raymond Smullyan. (Smullyan published Forever Undecided: A Puzzle Guide to Gödel in 1987). The setting for the puzzle, as Bellos presents it, is a hypothetical island he calls “If.” Natives of If are either “Alethians” or “Pseudians;” they are indistinguishable except that Alethians always tell the truth, while Pseudians always lie. This sounds like the traditional Liar Problem, but there is a wrinkle: on the island is a ledger where every native is listed, along with his or her tribe. Anyone can consult this ledger. You arive on If and a person, Kurt, comes up to you stating: “You will never have concrete evidence that confirms I am an Alethian.” The puzzle: is Kurt an Alethian, a Pseudian or neither? Think about it before checking Bellos’s solution and before reading on.
Now comes the connection with Gödel’s incompleteness theorem, which states, as Bellos puts it, “that there are mathematical statements that are true but not formally provable.” Suppose you are the first non-native ever to visit If, so you know that everyone you meet is Alethian or Pseudian. Kurt pops up and says, just as before, “You will never have concrete evidence that confirms I am an Alethian.” But now just as he speaks the Ledger burns to ashes.
Where are we? Kurt cannot be a Pseudian, because with no Ledger that statement has to be true. So you know Kurt is an Alethian. But you can never have concrete evidence of that fact because if you did, his statement would be false, and it can’t be false since he is Alethian. Think about it. [Thanks to Jonathan Farley for bringing this item to my attention. -TP]
### “Fun with Math” in The New Yorker
Dan Rockmore’s contribution to “Talk of the Town” in the January 17, 2022 issue of the The New Yorker was an item titled “Fun with Math.” He recounts attending “a recent evening of math dinner theater” organized by Cindy Lawrence, director of New York’s National Museum of Mathematics. The event featured Peter Winkler, a mathematics professor at Dartmouth and expert on math puzzles. Among the puzzles and phenomena that Winkler served up for discussion during dinner:
• “On average, how many cards does it take to get to a jack in a shuffled deck of fifty-two cards?”
• “What’s the best way to use two coin tosses to determine which of two coins, one fair and one ‘biased,’ is fair?”
• Simpson’s paradox in statistics, best described by an example: for Berkeley’s graduate programs in 1973, overall “men were admitted at a higher rate than women, but, program by program, women were admitted at a higher rate.” (See MinutePhysics for a more detailed explanation.)
Apropos of Simpson’s paradox Marilyn Simons, a guest with a PhD in economics, remarked, “I think that, to a lot of us who even think we know statistics, the way we process statistics is not deeply informed.” Elsewhere Rockmore quotes her as saying that her husband Jim (identified as “a financier and a former mathematician”) doesn’t like puzzles: “He says that if he works that hard he wants to get a theorem out of it.”
### Rock-paper-scissors and evolutionary game theory
“Non-Hermitian topology in rock-paper-scissors games” by the three Tsukuba physicists Tsuneya Yoshida, Tomonari Mizoguchi and Yasuhiro Hatsugai was published January 12, 2022 in Scientific Reports. This is a physics article, but it applies a nice piece of mathematics, evolutionary game theory, to the familiar rock-paper-scissors game.
The game consists of two players; at a signal each shows a clenched fist (“rock”), a flat hand (“paper”) or a vertical hand with the first two fingers displayed (“scissors”). The winner (rock smashes scissors, scissors cut paper, paper covers rock) gets one point, the loser loses one. If both players show the same symbol, each gets zero.
The article contains this image:
Here R, P and S have to stand for rock, paper and scissors, but how is this diagram related to the game? We need to make a detour into evolutionary game theory. This is a method for simulating the process of evolution in populations. Here the population is split among three subspecies; let’s call them Ravens, a fraction $s_1$ of the population, Penguins with $s_2$, and Swifts with $s_3$, where the fractions $s_1, s_2, s_3$ add up to 1. These correspond to the three “pure strategies” in the game: at every encounter, a Raven will play “rock,” a Penguin will play “paper” and a Swift, “scissors.” The state vector ${\bf s}=(s_1, s_2, s_3)$ encapsulates the current mix in the population.
Evolution occurs in time. Suppose the population is in state ${\bf s}$ at some moment. Where will it be just a litle later? For example, suppose ${\bf s}=(\frac{1}{2},\frac{1}{2},0)$. That means half the population are Ravens and half are Penguins. So a Raven will meet a Penguin with probability $\frac{1}{2}$, and can expect to lose half a point. Likewise a Penguin will meet a Raven with probability $\frac{1}{2}$ and so can expect to gain half a point. The Penguins have an advantage. If the object is to model evolution, then the Penguin’s advantage in that state should translate into their population increasing at the expense of the Ravens. That gives a clue to the meaning of the red arrow at the point $(\frac{1}{2},\frac{1}{2},0)$: the population mix at that point is shifting to the right. More Penguins, fewer Ravens.
To make this more precise, keeping the language of evolution, we measure the fitness of one of the groups at some state ${\bf s}=(s_1,s_2,s_3)$ of the population by the expected gain or loss in points at the next encounter. So the fitness of the Ravens at state ${\bf s}$ will be the probability of meeting a Penguin times $-1$ plus the probability of meeting a Swift times 1. We write this as
$F(\mbox{Ravens}|{\bf s})= -s_2 + s_3.$ Similarly $F(\mbox{Penguins}|{\bf s})= s_1 – s_3$ and $F(\mbox{Swifts}|{\bf s})= -s_1 + s_2.$
Finally we set up a dynamical system by stating that the proportion of the population in any group will increase or decrease exponentially with growth coefficient equal to the fitness of that group (which can be positive or negative) at that instant in time. Writing that statement as a differential equation gives the replicator equation for rock-paper-scissors:
$$\frac{ds_1}{dt}= s_1(-s_2 + s_3), ~~\frac{ds_2}{dt}= s_2(s_1 – s_3), ~~\frac{ds_3}{dt}= s_3(-s_1 + s_2).$$
In vector form, the equivalent equation is
$$\frac{d{\bf s}}{dt}= (s_1(-s_2 + s_3), s_2(s_1 – s_3), s_3(-s_1 + s_2)).$$
Now we can interpret the first image in this item, which shows the state space for rock-paper-scissors as an evolutionary game. The arrows represent the direction of evolution, with the magnitude encoded by color saturation. The central cross marks the equilibrium $(\frac{1}{3},\frac{1}{3},\frac{1}{3})$; the blue loop is the solution curve obtained by numerically integrating the replicator equation starting at the point ${\bf s} = \frac{1}{3}(1-\delta, 1+\delta/2, 1+\delta/2)$, with $\delta=0.1$.
For this game, all the solution curves are closed loops. In fact, they are level curves of the function $f(s_1,s_2,s_3)=s_1s_2s_3$. This follows from the equality $\displaystyle{\frac{d{\bf s}}{dt}\cdot \nabla f= 0}$, which is actually fun to check. Try it.
The evolutionary game described here is just the starting point for the article by Yoshida et al. They consider perturbations of this system that break its symmetry—surprisingly, the resulting phenomena have parallels in condensed-matter physics.
### “What physics owes to math”
On January 12, 2022 Le Monde ran a guest article with the title “What physics owes to math” (text in French) written by two physicists, Jean Farago and Wiebke Drenckhan, from the Institut Charles-Sadron in Strasbourg. The authors begin with the famous quote from Galileo about the Book of Nature being written in the language of mathematics, and go on to observe: “The millenia elapsed between the birth of mathematics and its use in physics demonstrate that this contiguity between natural phenomena and the mathematical laws of our human rationality was far from being obvious.”
Farago and Drenckhan mention that one of the most antonishing examples of the “intimate” relationship between mathematics and physics comes from complex numbers. Starting in the 16th century, mathematicians found that calculating solutions to polynomial equations with whole-number coefficients required the use of an ‘imaginary’ number $i$ with $i^2=-1.$ “How could anything be more abstract than this fictitious number, given that ordinary numbers always have a positive square ($2^2=(-2)^2=4$)!” But fast-forward to 1929 and Schrödinger’s equation $i\hbar\partial_t\psi=H\psi$, which doesn’t work without it. We read that no one was more surprised by “this irruption of $i$ in the corpus of physical laws” than Schrödinger himself, and that he described his reaction, in a footnote, by quoting an unnamed Viennese physicist, “known for his ability to always find the mot juste, the cruder the juster,” and who compared the appearance of $i$ in that equation to one’s involuntary (but welcome) emission of a burp. Our authors add: “This shows us that a contiguity can sometimes also exist between humor and physics.” [My translations. -TP]
“What physics owes to math” could have mentioned an article from Nature last month: “Quantum theory based on real numbers can be experimentally falsified,” written by an international team with corresponding author Miguel Navascués (IQOQI, Vienna). Physical experiments are expressed in terms of probabilities, which are real numbers. So why can’t there be a “real” quantum theory? The authors show that complex numbers are actually needed, by devising “a Bell-like experiment, the successful realization of which would disprove real quantum theory, in the same way as standard Bell experiments disproved local physics.” | 2,489 | 10,093 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2024-33 | latest | en | 0.946353 |
https://stats.stackexchange.com/questions/100613/poisson-regression-in-r-glm-concept-and-practice | 1,563,331,746,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525009.36/warc/CC-MAIN-20190717021428-20190717043428-00316.warc.gz | 563,880,088 | 31,299 | # Poisson regression in R glm(): concept and practice [closed]
I have a dataset data about a group of persons developing events with Poisson distribution. These events are uniform and can be recurrent in some of the subjects. Their observation time is different. I tried to use R to calculate the incidence rate (in 1000 person years) of the event in the whole group and specific subgroups (e.g. male and female).
no.of.event is the total number of observed events and time.in.years is the total observation time in years.
As I do not understand Poisson regression well, I would like to ask:
1. If no.of.event equal to the total cumulative number of all events observed?
2. And time.in.years be the time from the start of observation till study end/ default or death? Or be the time from the start of observation till the development of event of interest?
3. In those persons with recurrent events what would be time.in.years? Say if someone has developed 3 events and did not default, is the time counted from the start of observation till the development of third events? Or from the start till the end of observation?
4. Is the following R script right?
model <- glm(no.of.event ~1, offset(time.in.years), family="poisson", data=data)
5. Should incidence (cases per 1000 person years) be calculated by this?
exp(model$coef[1])*1000 6. And the 95% confidence interval be like this? exp(model$coef[1] + 1.96*sqrt(diag(vcov(model)))[1])
exp(model$coef[1] - 1.96*sqrt(diag(vcov(model)))[1]) 7. If I want to calculate specific incidence, say among males and females and I have a column in the data frame called sex with 1 coded for males, should I write something like this? model <- glm(no.of.event~1, offset(time.in.years), family="poisson", data=data[data$sex==1, ])
## closed as too broad by kjetil b halvorsen, whuber♦Mar 31 at 21:21
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question. | 523 | 2,166 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2019-30 | longest | en | 0.892099 |
http://metamath.tirix.org/mpests/riotaex | 1,718,461,220,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861594.22/warc/CC-MAIN-20240615124455-20240615154455-00882.warc.gz | 21,114,081 | 1,854 | # Metamath Proof Explorer
## Theorem riotaex
Description: Restricted iota is a set. (Contributed by NM, 15-Sep-2011)
Ref Expression
Assertion riotaex ${⊢}\left(\iota {x}\in {A}|{\psi }\right)\in \mathrm{V}$
### Proof
Step Hyp Ref Expression
1 df-riota ${⊢}\left(\iota {x}\in {A}|{\psi }\right)=\left(\iota {x}|\left({x}\in {A}\wedge {\psi }\right)\right)$
2 iotaex ${⊢}\left(\iota {x}|\left({x}\in {A}\wedge {\psi }\right)\right)\in \mathrm{V}$
3 1 2 eqeltri ${⊢}\left(\iota {x}\in {A}|{\psi }\right)\in \mathrm{V}$ | 203 | 520 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 4, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2024-26 | latest | en | 0.482431 |
https://www.solutioninn.com/study-help/physics-11th-edition/a-person-looking-out-the-window-of-a-stationary-train-844226 | 1,717,027,282,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059412.27/warc/CC-MAIN-20240529230852-20240530020852-00059.warc.gz | 858,490,847 | 16,701 | # A person looking out the window of a stationary train notices that raindrops are falling vertically down
## Question:
A person looking out the window of a stationary train notices that raindrops are falling vertically down at a speed of 5.0 m/s relative to the ground. When the train moves at a constant velocity, the raindrops make an angle of 25° when they move past the window, as the drawing shows. How fast is the train moving?
Fantastic news! We've Found the answer you've been seeking! | 108 | 497 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-22 | latest | en | 0.928817 |
http://www.perlmonks.org/?node_id=938310 | 1,469,551,354,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824995.51/warc/CC-MAIN-20160723071024-00081-ip-10-185-27-174.ec2.internal.warc.gz | 633,494,517 | 6,717 | The stupid question is the question not asked PerlMonks
### Re^2: Challenge: Generate fixed size combination across groups without duplicates
by Limbic~Region (Chancellor)
on Nov 16, 2011 at 04:30 UTC ( #938310=note: print w/ replies, xml ) Need Help??
GrandFather,
Ah yes, recursion - that thing my brain finds so confusing. Thanks - this appears to be perfect.
Cheers - L~R
• Comment on Re^2: Challenge: Generate fixed size combination across groups without duplicates
Replies are listed 'Best First'.
Re^3: Challenge: Generate fixed size combination across groups without duplicates
by GrandFather (Sage) on Nov 16, 2011 at 05:17 UTC
The real trick is adding the "no-op" entry to each group.
True laziness is hard work
GrandFather,
No, I assure you that my brain does not think recursively. I have never encountered any other programmer or a math aficionado with this affliction. That makes it amazingly frustrating when people say - it is simple, just think of it as repeating a simple process until a termination condition is met. Of course I understand that and I don't usually having a problem understanding someone else's recursive solution. My problem is that my brain just doesn't think that way - I don't know how else to explain.
On the bright side, as soon as I stopped thinking about the problem thanks to your solution, I discovered how to do it iteratively using a variation on Arbitrarily Nested Loops (odometer model). If I get a chance to code it up I will but the key was to have a set of empty slots representing the combination that you copied values from as you rotate wheels on the dial. It becomes a very simple process at that point.
Cheers - L~R
a la 'Enigma' but going it one better?
Create A New User
Node Status?
node history
Node Type: note [id://938310]
help
Chatterbox?
and the web crawler heard nothing...
How do I use this? | Other CB clients
Other Users?
Others examining the Monastery: (14)
As of 2016-07-26 16:42 GMT
Sections?
Information?
Find Nodes?
Leftovers?
Voting Booth?
What is your favorite alternate name for a (specific) keyboard key?
Results (237 votes). Check out past polls. | 514 | 2,140 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2016-30 | longest | en | 0.918123 |
http://jwilson.coe.uga.edu/EMT668/EMAT6680.2003.fall/Brock/Assignmnet%2012/Assign12.htm | 1,544,811,069,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376826145.69/warc/CC-MAIN-20181214162826-20181214184826-00318.warc.gz | 137,492,833 | 3,521 | # Lisa Brock
Assignment 12
Using Excel to Explore the Fibonacci Sequence
Enter a 1 in cells A1 and A2. In cell A3 type =A1+A2. Click on cell A3. The outline of the cell should now be highlighted. Place the pointer on the bottom right corner of the cell. The pointer should turn into a square with arrows on two corners. Click and hold down the button on the mouse. Drag the box down about 30 rows. Release the mouse button. The cells A1 through A30 should now represent the Fibonacci sequence.
Now let’s look at the ratio of each pair of adjacent terms. In cell B2 type =A2/A1. Just like before, highlight the cell and click and hold on the bottom right corner. Drag the formula down.
As you can see, the ratio of adjacent terms quickly converges to the golden ratio.
Now let’s investigate the ratio of every second term. In cell B3 type =A3/A1. Just like before, highlight the cell and click and hold on the bottom right corner. Drag the formula down.
As you can see, the ratio of every second term quickly converges to the square of the golden ratio.
Let’s see what happens for the ratio of every third term. What do you think it will be?
It should be no surprise that the ratio of every third term is the golden ratio cubed.
It appears that the ratio of every nth term of the Fibonacci sequence is the golden ration raised to the n power. | 328 | 1,366 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2018-51 | latest | en | 0.836976 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.