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15,961,678 members
See more:
Hi,
I have a table table1.
table1 has one column:ID
data are:
ID
1
2
3
4
5
6
7
8
9
10
Another table is present as Table2
has 2 column: ID,Amt
Data are:
ID Amt
1 10
5 20
7 100
Now i need a result as
ID Amt
1 10
2 10
3 10
4 10
5 20
6 20
7 100
PLease tell me query to achive this
Posted
## Solution 3
Very strange data and requirement...
With the data shown above, following query works:
SQL
```SELECT Table1.ID, Max(Table2.Amt)
FROM Table1, Table2
where table1.id>=table2.id and table1.id<=7
group by table1.id```
The maximum value of table2 need not be hard coded but can be retrieved via a subquery:
SQL
```SELECT Table1.ID, Max(Table2.Amt)
FROM Table1, Table2
where table1.id>=table2.id and table1.id<= (select max(table2.id) from table2)
group by table1.id```
[/Edit]
v2
Maciej Los 20-Feb-14 15:31pm
It looks perfect!
+5!
King Fisher 20-Feb-14 23:33pm
Good one
## Solution 2
Join Query :
SQL
```SELECT t2.ID, t2.Amt
FROM Table1 t1, Tabele2 t2
WHERE t1.ID = t2.ID```
-KR
praveen.victor 20-Feb-14 8:21am
its not satisfying the requirment
Krunal Rohit 20-Feb-14 8:25am
I have seen such cases where "no repeatation" of data is needed. The result you need is simply depends upon the data but query.
Any way, can you tell me why are you doing this ?
-KR
## Solution 1
SQL
`select a.id,case when a.id between 1 and 4 then 10 when a.id between 5 and 6 then 20 when a.id between 7 and 10 then 100 end From table1 as a left join table2 as b on a.id=b.id`
v3
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# Question: How To Play Back Jack?
## How do you play Blackjack for beginners?
Play basic strategy
1. Stand when your hand is 12-16 when the dealer has 2-6.
2. Hit when your hand is 12-16 when the dealer has 7-Ace.
3. Always split Aces and 8s.
4. Double 11 versus the dealer’s 2-10.
5. Hit or double Aces-6.
## How can I play Blackjack at home?
Deal from the dealer’s left to right (start with the player to your far left.) Deal one up card to each player, followed by a down card to the dealer. Then deal a second up card to each player, followed by the dealer’s up card. Make sure the cards are laid out diagonally so that both numbers on each card are visible.
## How do you score in Blackjack?
The dealer and each player start with two cards. The dealer’s first card faces up, the second faces down. Face cards each count as 10, Aces count as 1 or 11, all others count at face value. An Ace with any 10, Jack, Queen, or King is a “Blackjack.”
## Is 21 and blackjack the same?
If you are wondering, are 21 and blackjack the same, then the answer is yes. Blackjack and 21 refer to the very same game, with the same rules and payouts. In other words, “21” is basically another name given to blackjack as it requires the players to get a sum total of 21 in their hands to get the blackjack.
## How much should you bet in blackjack?
A decent rule of thumb is to bring at least 100 betting units (for a 4 hour session). So if your betting unit is \$100, then I’d bring \$10K. That might sound like overkill, but I’ve had sessions where I’ve been in for that many bets or more.
## Can you tell when a slot machine is going to hit?
No one looking at the slot machine can predict the number it will choose next. This is why a slot machine can never be said to be “due” to hit a jackpot. Bet a single coin until you see the reels wiggle, then bet the max because the wiggle means a jackpot is coming.
## Is it legal to play blackjack at home?
Players in home games of blackjack are allowed to split pairs. They’re also allowed to re-split. And even though in most casinos, when you split aces, you only get to take one more card after splitting, in a home game, you can take more cards if you want to. A blackjack after splitting is paid off at 2 to 1.
## Is card counting illegal?
Card counting is NOT illegal under federal, state and local laws in the United States as long as players don’t use any external card-counting device or people who assist them in counting cards. In their effort to identify card counters, casinos can ban players believed to be counters — sort of.
## What is a soft 17?
A soft 17 includes an Ace being counted as 11. Ace-6 is a soft 17, as are Ace-2-4, Ace-3-3, Ace-Ace-5 and others. When the dealer hits soft 17, the house edge against a basic strategy player is about two-tenths of a percent higher than if he stands.
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# Ion concentration
1. Nov 5, 2007
### temaire
[SOLVED] Ion concentration
1. The problem statement, all variables and given/known data
http://img225.imageshack.us/img225/9918/chemch0.jpg [Broken]
2. Relevant equations
C = n/v
3. The attempt at a solution
My work is shown up above. Sorry if my work looks a little messy. My answer is 27.6 grams.
Last edited by a moderator: May 3, 2017
2. Nov 5, 2007
### temaire
I just want someone to tell me if I'am right, and correct me if I'am not.
3. Nov 5, 2007
### eli64
good effort
MgBr2 --> Mg+2(aq) + 2Br-(aq)
the bromide ion is twice the concentration of the MgBr2 so the moles of MgBr2 is half what you calculated (.15/2).
4. Nov 6, 2007
### temaire
But isn't Br supposed to have a subscript of 2?
5. Nov 6, 2007
### eli64
MgBr2 is a compound that dissociates into its ions in water.
the Br- ions don't stay together in water so you get 2 Br- ions
Br2 by itself is only written when you are referring to bromine as an diatomic element, not as ions in solution
6. Nov 6, 2007
### temaire
Thanks a lot for the help.
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# Dominion Strategy Forum
• July 04, 2022, 09:29:12 pm
• Welcome, Guest
### News:
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### Show Posts
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### Messages - Watno
Filter to certain boards:
Pages: [1] 2 3 ... 209
1
##### Feedback / Re: F.DS could benefit from a clean up
« on: February 02, 2022, 05:40:51 pm »
Too bad it was only 4 boards, 2 more and you'd get Monument (I'd argue they weren't melded but tucked)
2
##### Feedback / Re: F.DS could benefit from a clean up
« on: January 27, 2022, 02:10:28 pm »
The Innovation forums seem to be completely gone? That's a bit too harsh i think
3
##### General Discussion / Re: Maths thread.
« on: February 22, 2021, 05:43:40 pm »
f ∈ O(n^2)
If you write this, it's a shorthand for f ∈ O(g) where g : n ↦ n^2. But f(x) ∈ O(g(x)) is nonsensical because f(x) isn't a thing. f is a function; f(5) is a number, "f(x) = 3x" is a shorthand for f = {... (-1, -3), (0,0), (1,3), ...}. But f(x) itself is literally not anything.
What makes it acceptable to use n^2 as a shorthand for n -> n^2, but not f(x) as a shorthand for x -> f(x)?
4
##### General Discussion / Re: Petty Complaints
« on: February 22, 2021, 04:02:07 pm »
To prevent you from trying to find a nonexistent explanation, I'm complaining about how the statement is written
I assume your complaint is about '=' being used for something that it is not an equivalence relation?
not primarily. that's bad, too, but writing f(x) rather than f is worse.
So what would your preferred way of writing f(n) = O(n^2) be?
5
##### General Discussion / Re: Petty Complaints
« on: February 22, 2021, 03:21:58 pm »
To prevent you from trying to find a nonexistent explanation, I'm complaining about how the statement is written
I assume your complaint is about '=' being used for something that it is not an equivalence relation?
6
##### Non-Mafia Game Threads / Re: Decrypto 2
« on: April 08, 2020, 02:48:30 am »
Im out, have fun
7
##### Non-Mafia Game Threads / Re: Decrypto 2
« on: April 03, 2020, 12:39:32 pm »
Our guesses:
• Horizon
• Canal
• Bond
• Car
Yours are al correct, so you probably won.
8
##### Non-Mafia Game Threads / Re: Decrypto 2
« on: April 02, 2020, 02:43:03 pm »
1-2-3 is correct for ours.
That means we need to guess the other teams words now, unless I'm mistaken.
9
##### Non-Mafia Game Threads / Re: Decrypto 2
« on: March 22, 2020, 01:50:46 pm »
fish - uphill - stalk
10
##### Dominion General Discussion / Re: I think I can't play Dominion with my sets?!
« on: March 21, 2020, 07:22:20 am »
Guilds is a small expansion, so I wouldn't recommend that for now. I think seaside is a good option when you want something interesting but not complicated.
11
##### Non-Mafia Game Threads / Re: Decrypto 2
« on: March 20, 2020, 05:54:21 am »
Guessing 2-4-1 for yours
12
##### Dominion General Discussion / Re: Menagerie Bonus Previews
« on: March 17, 2020, 06:38:37 am »
Bonus Preview 11: Desperation
It's the Event version of Cursed Gold. You can have more money, if you want it badly enough.
Is the person on the art missing a hand?
I assume that's the cause of the desperation.
13
##### Non-Mafia Game Threads / Re: Decrypto 2
« on: March 15, 2020, 01:35:16 pm »
ghostofmars and bitwise to give the next set of clues unless I'm mistaken
14
##### Non-Mafia Game Threads / Re: Decrypto 2
« on: February 29, 2020, 04:09:50 am »
15
##### Non-Mafia Game Threads / Re: Decrypto 2
« on: February 24, 2020, 01:48:41 pm »
Our guess for our clue is 4-1-2
4-1-2 is correct.
Our guess for your clue is 3-1-4
16
##### Non-Mafia Game Threads / Re: Decrypto 2
« on: February 22, 2020, 06:46:40 am »
Doctor - Shade - Air
17
##### Non-Mafia Game Threads / Re: Decrypto 2
« on: February 22, 2020, 05:57:36 am »
Well, that explains why it's taking so long
18
##### Non-Mafia Game Threads / Re: Decrypto 2
« on: February 21, 2020, 02:12:25 am »
Waiting for Jimmmmm, right?
19
##### Non-Mafia Game Threads / Re: Decrypto 2
« on: February 14, 2020, 11:26:52 am »
guess for ours: 3-1-4
20
##### Non-Mafia Game Threads / Re: Decrypto 2
« on: February 13, 2020, 12:37:21 pm »
Guess for yours: 3-2-4
21
##### Non-Mafia Game Threads / Re: Decrypto 2
« on: February 11, 2020, 02:33:38 am »
My team guesses 1-2-4, which is correct.
Our guess for your clue is 2-4-3
22
##### Non-Mafia Game Threads / Re: Decrypto 2
« on: February 09, 2020, 07:49:51 am »
Fire - Water - Ice
23
##### Non-Mafia Game Threads / Re: Decrypto 2
« on: February 03, 2020, 02:08:22 pm »
So we can debate the numbers without giving the reasons?
"I like 2-3-1"
"I think 2-4-1 is better"
Seems like it'd be challenging to settle on one, but I get the purpose, sure.
Giving the reason is fine, unless the reason indicates a clue you are going to give later.
24
##### Non-Mafia Game Threads / Re: Decrypto 2
« on: February 01, 2020, 04:00:05 am »
Yeah, I think just don't talk about strategy. Also nothing along the lines of "he woul have said "x" instead of "y" if he meant that about your own teams clue.
25
##### Non-Mafia Game Threads / Re: Decrypto
« on: January 29, 2020, 12:47:52 pm »
Congratulations.
I guess I overthought the last clue. I thought you wouldn't really ski at a cliff.
I'd be up for another, and we had 2 more people interested as well.
Go for a game with 6?
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# Undeclared?
• 11-21-2011
Shazarul
Undeclared?
Hey I've got this problem 'uninitialized local variable 'weight' used' . . The error is in red
Code:
``` #include <iostream> #include <iomanip> #include <cmath> void computeAlpha(); void computeBravo(); void computeCharlie(); void computeDelta(); using namespace std; int main() { char choice; double weight; cout << setprecision(0) << fixed << "Please enter the weight (w) in gm : "; cin >> weight; cout << "\n\nPlease enter the destination code (A to D) : "; cin >> choice; if ( choice == 'A' ) computeAlpha(); else if ( choice == 'B' ) computeBravo(); else if ( choice == 'C' ) computeCharlie(); else if ( choice == 'D' ) computeDelta(); else cout << "\n\nInvalid Choice!!!" << endl; system("pause"); return 0; } void computeAlpha() { double weight, PostageCost1, PostageCost2, PostageCost3; const double r1A=2.75, r2A=23.00; PostageCost1 = (1/2*r1A)+(1/4*r2A); PostageCost2 = r1A+(1/2*r2A); PostageCost3 = r1A+r2A; if ( weight <= 300 ) cout << setprecision(2) << fixed << "\n\nPostage Cost = \$ " << PostageCost1 << "\n\nRemark = LIGHT" << endl << endl; else if ( weight <= 600 ) cout << "\n\nPostage Cost = \$ " << PostageCost2 << "\n\nRemark = AVERAGE" << endl << endl; else cout << "\n\nPostage Cost = \$ " << PostageCost3 << "\n\nRemark = HEAVY" << endl << endl; return; } void computeBravo () { double weight, PostageCost1, PostageCost2, PostageCost3; const double r1B=1.50, r2B=10.00; PostageCost1 = (1/2*r1B)+(1/4*r2B); PostageCost2 = r1B+(1/2*r2B); PostageCost3 = r1B+r2B; if ( weight <= 300 ) cout << setprecision(2) << fixed << "\n\nPostage Cost = \$ " << PostageCost1 << "\n\nRemark = LIGHT" << endl << endl; else if ( weight <= 600 ) cout << "\n\nPostage Cost = \$ " << PostageCost2 << "\n\nRemark = AVERAGE" << endl << endl; else cout << "\n\nPostage Cost = \$ " << PostageCost3 << "\n\nRemark = HEAVY" << endl << endl; return; } void computeCharlie () { double weight, PostageCost1, PostageCost2, PostageCost3; const double r1C=2.50, r2C=12.50; PostageCost1 = (1/2*r1C)+(1/4*r2C); PostageCost2 = r1C+(1/2*r2C); PostageCost3 = r1C+r2C; if ( weight <= 300 ) cout << setprecision(4) << fixed << "\n\nPostage Cost = \$ " << PostageCost1 << "\n\nRemark = LIGHT" << endl << endl; else if ( weight <= 600 ) cout << "\n\nPostage Cost = \$ " << PostageCost2 << "\n\nRemark = AVERAGE" << endl << endl; else cout << "\n\nPostage Cost = \$ " << PostageCost3 << "\n\nRemark = HEAVY" << endl << endl; return; } void computeDelta () { double weight, PostageCost1, PostageCost2, PostageCost3; const double r1D=2.95, r2D=33.50; PostageCost1 = (1/2*r1D)+(1/4*r2D); PostageCost2 = r1D+(1/2*r2D); PostageCost3 = r1D+r2D; if ( weight <= 300 ) cout << setprecision(2) << fixed << "\n\nPostage Cost = \$ " << PostageCost1 << "\n\nRemark = LIGHT" << endl << endl; else if ( weight <= 600 ) cout << "\n\nPostage Cost = \$ " << PostageCost2 << "\n\nRemark = AVERAGE" << endl << endl; else cout << "\n\nPostage Cost = \$ " << PostageCost3 << "\n\nRemark = HEAVY" << endl << endl; return; }```
Problem is my print screen need to be
"Please enter the weight (w) in gm : 700
Please enter the destination code (A to D) : A
Postage Cost = \$27.65
Remark = Heavy"
I can't figure out how to get my weight declared in my void menu, can someone point out?
• 11-21-2011
manasij7479
Put your letters within single quotes.
• 11-21-2011
JohnGraham
The weight variable you've defined in computeAlpha(), computeBravo() etc. is not the same as the one you've defined in main() - it refers to a different memory location, and so has a random value in your compute functions. Fix the problem by removing it from all your functions and declaring it once outside them all:
Code:
```// This is the only place you define weight. double weight; int main() { // ... }```
• 11-21-2011
rags_to_riches
While JohnGraham is correct, I disagree with using the global variable solution. It's a bad habit to get into, and it's pretty simple given your program to do the right thing and declare and initialize the weight variable in main pass it to your functions as an argument.
• 11-21-2011
Shazarul
Blargh this void is difficult . . I decided to do just a normal if else but I'am having problems with else, I can build without any problem . . But if i enter any other alphabets beside A-D, i get an error stating "The variable 'r1' is being used without initialized" Can someone u point out where the error is?
Code:
``` #include <iostream> #include <iomanip> #include <cmath> #include <string> using namespace std; int main() { char choice; double weight, PostageCost, r1, r2; string Remark; cout << setprecision(0) << fixed << "Please enter the weight (w) in gm : "; cin >> weight; cout << "\n\nPlease enter the destination code (A to D) : "; cin >> choice; if ( choice == 'A' ) r1 = 2.75, r2 = 23.00; else if ( choice == 'B' ) r1 = 1.50, r2 = 10.00; else if ( choice == 'C' ) r1 = 2.50, r2 = 12.50; else if ( choice == 'D' ) r1 = 2.95, r2 = 33.50; else cout << "\n\nInvalid Choice!!!"; if ( weight <= 300 ) { PostageCost = (1/2*r1)+(1/4*r2), Remark = "LIGHT"; } else if ( weight <= 600 ) { PostageCost = (r1+(1/2*r2)), Remark = "AVERAGE"; } else { PostageCost = (r1+r2), Remark = "HEAVY"; } cout << setprecision(2) << fixed << "\n\nPostage Cost = \$" << PostageCost << endl << endl; system("pause"); return 0; }```
• 11-21-2011
whiteflags
>Blargh this void is difficult . . I decided to do just a normal if else but I'am having problems with else,
The problem is not with else.
> I can build without any problem . . But if i enter any other alphabets beside A-D, i get an error stating "The variable 'r1' is being used without initialized"
You should be getting a warning, not an error. An error stops builds while a warning doesn't.
> Can someone u point out where the error is?
Actually, your thinking is correct. If you don't pick a letter between A or D, r1 and r2 don't get assigned values...
Code:
```if ( choice == 'A' ) r1 = 2.75, r2 = 23.00; else if ( choice == 'B' ) r1 = 1.50, r2 = 10.00; else if ( choice == 'C' ) r1 = 2.50, r2 = 12.50; else if ( choice == 'D' ) r1 = 2.95, r2 = 33.50; else cout << "\n\nInvalid Choice!!!";```
Invalid choice!!! Then what? You calculate postage anyway, using r1 and r2. Using an uninitialized variable causes undefined behavior, which is why you get the warning. You should ask the user to make another correct choice. Use a loop to keep asking until the user makes a good choice. Likewise, if the user doesn't choose a good weight, the program has problems calculating good results. You need to make sure the program is using acceptable data first.
• 11-22-2011
Elysia
Quote:
Originally Posted by whiteflags
> I can build without any problem . . But if i enter any other alphabets beside A-D, i get an error stating "The variable 'r1' is being used without initialized"
You should be getting a warning, not an error. An error stops builds while a warning doesn't.
I do believe the problem stems from that if you run some code inside Visual Studio and you try to use an uninitialized variable, it will immediately break with an error, saying variable X is not initialized.
• 11-22-2011
whiteflags
OK, but that just means the OP doesn't know that debugging is not building. Which is awful for him.
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# when expressed as a vulgar fraction, equals;
1. 23/80
2. 23/800
3. 23/8000
4. 125/23
4
23/800
Explanation :
No Explanation available for this question
# Raju Narayan is a matriculate who has been living in Delhi since August 1992. He was born on 7th November, 1957. His family income is less than Rs.25,000 per annum and he has no close relatives as a delaer of any oil company
1. Mark answer (a) if the applicant is selected
2. (b) if the candidate is not selected
3. (c) if the data is inadequate
4. (d) if the case is to be referred to the Managing director
5. (E) if the case is to be referred to the chairman
5
(c) if the data is inadequate
Explanation :
No Explanation available for this question
# Which is the closest approximation to the product 0.3333 x 0.25 x 0.499 x 0.125 x 24
1. 1/8
2. 3/4
3. 3/8
4. 2/5
4
1/8
Explanation :
No Explanation available for this question
# Consider the following quotients: 1. 368.39 divided by 17 2.170.50 divided by 62 3.875.65 divided by 83. Their correct sequence in decreasing order is:
1. 1, 3, 2
2. 2, 1, 3
3. 2, 3, 1
4. 3, 1, 2
4
1, 3, 2
Explanation :
No Explanation available for this question
# Kishen Gopal born on 22nd January , 1967 is an Indian by nationality. He is a matriculate having dealership in tamsha oil company. His family income is Rs 21,000 per annum and he is a resident of Delhi since 1978. He has no close relatives as dealer/ distributor in any oil company
1. Mark answer (a) if the applicant is selected
2. (b) if the candidate is not selected
3. (c) if the data is inadequate
4. (d) if the case is to be referred to the Managing director
5. (E) if the case is to be referred to the chairman
5
(b) if the candidate is not selected
Explanation :
No Explanation available for this question
# 0.213 + 0.00213 =
1. 1
2. 10
3. 100
4. None of these
4
100
Explanation :
No Explanation available for this question
# Balvinder singh working in the state corporation is an Indian by nationality and is 23 years of age. He is agraduate and his family income is Rs,. 60,000 per annum. He has been in Delhi for 7 years. He does not himself nor has any of his relatives working as distributor or dealer in any oil company
1. Mark answer (a) if the applicant is selected
2. (b) if the candidate is not selected
3. (c) if the data is inadequate
4. (d) if the case is to be referred to the Managing director
5. (E) if the case is to be referred to the chairman
5
(d) if the case is to be referred to the Managing director
Explanation :
No Explanation available for this question
# Pravesh Kaur, an Indizn born in 197, is an Intermediate staying in Delhi since 1983. He doesnot hold any dealership in any oil company and the income of his mother, the sole earner , is not more than Rs. 500 per month
1. Mark answer (a) if the applicant is selected
2. (b) if the candidate is not selected
3. (c) if the data is inadequate
4. (d) if the case is to be referred to the Managing director
5. (E) if the case is to be referred to the chairman
5
(c) if the data is inadequate
Explanation :
No Explanation available for this question
# 4.036 divided by 0.04 gives:
1. 1.009
2. 10.09
3. 100.9
4. None of these
4
100.9
Explanation :
No Explanation available for this question
# Chaluka, an Indian resident of Mumbai, is a matriculate with a family income of Rs. 20,000 per month. His date of birth is 15.3.76. he does not have any dealership is any oil company nor has any close relative as delaer or distributor. He is an SC candidate
1. Mark answer (a) if the applicant is selected
2. (b) if the candidate is not selected
3. (c) if the data is inadequate
4. (d) if the case is to be referred to the Managing director
5. (E) if the case is to be referred to the chairman
5
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# Compact set
## Homework Statement
R>0, let K be a closed subset of C such that K $$\subset$$ BR(0) (so K is compact). Show that there exists 0 < r < R such that K$$\subset$$ Br(0).
## The Attempt at a Solution
Can I write BR(0) = {x$$\in$$C : d(x,0) $$\leq$$R}?
I know that a compact set is closed and bounded.
Is it something to do with us using $$\subset$$ and not $$\subseteq$$?
As if it was $$\subseteq$$ then maybe K = BR(0) then there wouldn't be an r. But as K is strictly contained in the ball there must be a bit of room for manoeuvre.
These are my thoughts on this problem. I'm not sure if they're correct or what the question is asking, and if they are I don't know how to write them more formally?
Thank you :)
Dick
Homework Helper
## Homework Statement
R>0, let K be a closed subset of C such that K $$\subset$$ BR(0) (so K is compact). Show that there exists 0 < r < R such that K$$\subset$$ Br(0).
## The Attempt at a Solution
Can I write BR(0) = {x$$\in$$C : d(x,0) $$\leq$$R}?
I know that a compact set is closed and bounded.
Is it something to do with us using $$\subset$$ and not $$\subseteq$$?
As if it was $$\subseteq$$ then maybe K = BR(0) then there wouldn't be an r. But as K is strictly contained in the ball there must be a bit of room for manoeuvre.
These are my thoughts on this problem. I'm not sure if they're correct or what the question is asking, and if they are I don't know how to write them more formally?
Thank you :)
From the context of the problem, they must mean B_R(0) to be the open ball, d(x,0)<R. Not the closed ball. Otherwise, it wouldn't be true.
Oh that makes more sense.
I think I need to say something like you can always fit a smaller ball inside an open ball.
I remember in real analysis I showed for a non-empty bounded above set E and epsilon >0, there exists an x in E such that supE - epsilon < x $$\leq$$ sup E. Is it something like this?
Dick
I remember in real analysis I showed for a non-empty bounded above set E and epsilon >0, there exists an x in E such that supE - epsilon < x $$\leq$$ sup E. Is it something like this?
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# LPPHonest.fit: Honest inference at a point In kolesarm/RDHonest: Honest inference in sharp regression discontinuity designs
## Description
Basic computing engine called by `LPPHonest` to compute honest confidence intervals for local polynomial estimators.
## Usage
```1 2 3``` ```LPPHonest.fit(d, M, kern = "triangular", h, opt.criterion, alpha = 0.05, beta = 0.8, se.method = "nn", J = 3, sclass = "H", order = 1, se.initial = "ROTEHW") ```
## Arguments
`d` object of class `"LPPData"` `M` Bound on second derivative of the conditional mean function. `kern` specifies kernel function used in the local regression. It can either be a string equal to `"triangular"` (k(u)=(1-|u|)_{+}), `"epanechnikov"` (k(u)=(3/4)(1-u^2)_{+}), or `"uniform"` (k(u)= (|u|<1)/2), or else a kernel function. `h` Bandwidth. If not supplied, optimal bandwidth is computed according to criterion given by `opt.criterion`. `opt.criterion` Optimality criterion that bandwidth is designed to optimize. It can either be based on exact finite-sample maximum bias and finite-sample estimate of variance, or asymptotic approximations to the bias and variance. The options are: `"MSE"`Finite-sample maximum MSE `"FLCI"`Length of (fixed-length) two-sided confidence intervals. `"OCI"`Given quantile of excess length of one-sided confidence intervals The finite-sample methods use conditional variance given by `sigma2`, if supplied. Otherwise, for the purpose of estimating the optimal bandwidth, conditional variance is assumed homoscedastic, and estimated using a nearest neighbor estimator. `alpha` determines confidence level, `1-alpha` for constructing/optimizing confidence intervals. `beta` Determines quantile of excess length to optimize, if bandwidth optimizes given quantile of excess length of one-sided confidence intervals. `se.method` Vector with methods for estimating standard error of estimate. If `NULL`, standard errors are not computed. The elements of the vector can consist of the following methods: "nn"Nearest neighbor method "EHW"Eicker-Huber-White, with residuals from local regression (local polynomial estimators only). "demeaned"Use EHW, but instead of using residuals, estimate sigma^2_i by subtracting the estimated intercept from the outcome (and not subtracting the estimated slope). Local polynomial estimators only. "plugin"Plug-in estimate based on asymptotic variance. Local polynomial estimators in RD only. "supplied.var"Use conditional variance supplied by `sigma2` / `d` instead of computing residuals `J` Number of nearest neighbors, if "nn" is specified in `se.method`. `sclass` Smoothness class, either `"T"` for Taylor or `"H"` for Hölder class. `order` Order of local regression 1 for linear, 2 for quadratic. `se.initial` Method for estimating initial variance for computing optimal bandwidth. Ignored if data already contains estimate of variance. "ROTEHW"Based on residuals from a local linear regression using a triangular kernel and ROT bandwidth "ROTdemeaned"Based on sum of squared deviations of outcome from estimate of intercept in local linear regression with triangular kenrel and ROT bandwidth
## Value
Returns an object of class `"LPPResults"`, see description in `LPPHonest`
kolesarm/RDHonest documentation built on April 3, 2018, 11:08 a.m.
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# Function table answer key Awesome
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# Search by Topic
#### Resources tagged with Handling data similar to Time of Birth:
Filter by: Content type:
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### There are 37 results
Broad Topics > Handling, Processing and Representing Data > Handling data
### Searching for Mean(ing)
##### Age 11 to 14 Challenge Level:
Imagine you have a large supply of 3kg and 8kg weights. How many of each weight would you need for the average (mean) of the weights to be 6kg? What other averages could you have?
### Litov's Mean Value Theorem
##### Age 11 to 14 Challenge Level:
Start with two numbers and generate a sequence where the next number is the mean of the last two numbers...
### Substitution Cipher
##### Age 11 to 14 Challenge Level:
Find the frequency distribution for ordinary English, and use it to help you crack the code.
### Estimating Time
##### Age 11 to 14 Challenge Level:
How well can you estimate 10 seconds? Investigate with our timing tool.
### What's the Weather Like?
##### Age 11 to 14 Challenge Level:
With access to weather station data, what interesting questions can you investigate?
### Unequal Averages
##### Age 11 to 14 Challenge Level:
Play around with sets of five numbers and see what you can discover about different types of average...
### M, M and M
##### Age 11 to 14 Challenge Level:
If you are given the mean, median and mode of five positive whole numbers, can you find the numbers?
### Wipeout
##### Age 11 to 14 Challenge Level:
Can you do a little mathematical detective work to figure out which number has been wiped out?
### Data Matching
##### Age 14 to 18 Challenge Level:
Use your skill and judgement to match the sets of random data.
### Shifting Average
##### Age 11 to 14 Short Challenge Level:
What happens to the average if you subtract 8 from all of the numbers?
### For Richer for Poorer
##### Age 14 to 16 Challenge Level:
Charlie has moved between countries and the average income of both has increased. How can this be so?
### Who's the Best?
##### Age 11 to 14 Challenge Level:
Which countries have the most naturally athletic populations?
##### Age 11 to 14 Challenge Level:
Can you find sets of numbers which satisfy each of our mean, median, mode and range conditions?
### In the Bag
##### Age 11 to 14 Challenge Level:
Can you guess the colours of the 10 marbles in the bag? Can you develop an effective strategy for reaching 1000 points in the least number of rounds?
### Tree Tops
##### Age 14 to 16 Challenge Level:
Can you make sense of information about trees in order to maximise the profits of a forestry company?
### Inspector Morse
##### Age 11 to 14 Challenge Level:
You may like to read the article on Morse code before attempting this question. Morse's letter analysis was done over 150 years ago, so might there be a better allocation of symbols today?
### Winning Team
##### Age 11 to 14 Challenge Level:
Nine cross country runners compete in a team competition in which there are three matches. If you were a judge how would you decide who would win?
### Half a Minute
##### Age 11 to 14 Challenge Level:
Anna, Ben and Charlie have been estimating 30 seconds. Who is the best?
### How Would You Score It?
##### Age 11 to 14 Challenge Level:
Invent a scoring system for a 'guess the weight' competition.
### A Population Survey
##### Age 14 to 18 Challenge Level:
A geographical survey: answer the tiny questionnaire and then analyse all the collected responses...
### One Variable, Two Variable, Three Variable, More
##### Age 14 to 18 Challenge Level:
Displaying one-variable and two-variable data can be straightforward; what about three or more?
### Where Are You Flying?
##### Age 14 to 18 Challenge Level:
Where do people fly to from London? What is good and bad about these representations?
### Secondary Cipher Challenge Part 1
##### Age 11 to 16 Challenge Level:
Here is the start of a six-part challenge. Can you get to the end and crack the final message?
### Reaction Timer
##### Age 11 to 14 Challenge Level:
This problem offers you two ways to test reactions - use them to investigate your ideas about speeds of reaction.
### A Random Rambling Rant
##### Age 5 to 18
A random ramble for teachers through some resources that might add a little life to a statistics class.
### Box Plot Match
##### Age 14 to 16 Challenge Level:
Match the cumulative frequency curves with their corresponding box plots.
### Helicopters
##### Age 7 to 16 Challenge Level:
Design and test a paper helicopter. What is the best design?
### Olympic Triathlon
##### Age 14 to 16 Challenge Level:
Is it the fastest swimmer, the fastest runner or the fastest cyclist who wins the Olympic Triathlon?
### Which List Is Which?
##### Age 14 to 16 Challenge Level:
Six samples were taken from two distributions but they got muddled up. Can you work out which list is which?
### Olympic Records
##### Age 11 to 14 Challenge Level:
Can you deduce which Olympic athletics events are represented by the graphs?
### Statistics - Maths of Real Life
##### Age 14 to 18 Challenge Level:
This pilot collection of resources is designed to introduce key statistical ideas and help students to deepen their understanding.
### Perception Versus Reality
##### Age 14 to 18 Challenge Level:
Infographics are a powerful way of communicating statistical information. Can you come up with your own?
### Cricket Ratings
##### Age 14 to 16
Like all sports rankings, the cricket ratings involve some maths. In this case, they use a mathematical technique known as exponential weighting. For those who want to know more, read on.
### Picturing the World
##### Age 14 to 16 Challenge Level:
How can we make sense of national and global statistics involving very large numbers?
### History of Morse
##### Age 7 to 18
This short article gives an outline of the origins of Morse code and its inventor and how the frequency of letters is reflected in the code they were given.
### Statistical Shorts
##### Age 11 to 16 Challenge Level:
Can you decide whether these short statistical statements are always, sometimes or never true?
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1. Member
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## Rounding?!! Need help!
The question is
The purpose of this method is to round integer n off to the nearest 1s, 10s, 100s, …. place. The power parameter specifies the power of 10 to be used for rounding. If power is 0, no rounding is needed (nearest 1s place), if power is 1, round to the nearest 10s place, and so on. Here are some examples:
n = 8436, power = 1 return integer is 8440
n = 8436, power = 2 return integer is 8400
n = 8436, power = 3 return integer is 8000
n = 8436, power = 4 return integer is 10000
So far i have
Java Code:
public static double round ()
{
double n; double power; double n2;
Scanner input = new Scanner (System.in);
System.out.println("Enter Exponent of 10 for Rounding(0-3) ");
power = input.nextDouble();
System.out.println("Enter Number ");
n = input.nextDouble();
if ( power == 0)
{
System.out.println( n );
}
else if (power == 1)
{
n= Math.ceil(n * 10 + 0.5 ) / 10;
System.out.println(n);
}
if (power == 2)
{
n = Math.ceil(n * 100 + 0.5 ) / 100;
System.out.println(n);
}
else if (power == 3)
{
n = Math.ceil(n * 1000 + 0.5 ) / 1000;
System.out.println(n);
}
return n;
}
I do not know how to round. Help please.
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## Re: Rounding?!! Need help!
Java Code:
n= Math.ceil(n / 10 ) * 10; // Is much closer to what you're looking for.... BUT this is the CEILING not the tradition round!
Ceiling 3.4 -> 4
Ceiling 3.5 -> 4
Ceiling 3.6 -> 4
Math (Java Platform SE 7 )
Round 3.4 -> 3
Round 3.5 -> 4
Round 3.6 -> 4
This is the round in traditional arithmetic sense... tie goes up, otherwise closest whole number.
You're still returning a double... probably want to typecast and return an int.
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## Re: Rounding?!! Need help!
Why all the doubles? The question only involves ints so consider code that uses ints only.
The way I would round 2153 with a power of 2 is to recognise I was dealing with hundreds. There are 21 hundreds. The remainder, 53, is "big" relative to a hundred so I would add one to make 22 hundreds. Yielding the answer, 2200.
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Group of symmetries on a regular polygon
by SiddharthM
Tags: polygon, regular, symmetries
P: 176 So i began reading up on some group theory and I came across an interesting question, what is the order of the group of symmetries on of a n-sided regular polygon? with a square it's 8, triangle it's 4. I feel like i'm missing something with the pentagon cuz i'm only finding these: the 5 rotations, two diagonal reflections which are NOT the same as that for the square, reflection over the vertical axis. i'd appreciate any casual discussion on the topic as I find it fascinating, at the very beginning because symmetries behave very much like permutations (if we label vertices) I thought there might be a relationship between this set and the symmetric group of {1,...n}? But there are obviously permutations that no movement of a polygon in the plane can mimic. true or false: the group of symmetries of a n-sided regular polygon in the plane is isomorphic to a subset of the symmetric group of {1,...,n}. cheerio!
P: 1,076 There are more than 4 for a triangle. I'm also not sure how you can compare the diagonal reflections of a pentagon with those of a square, but there are more than 2. Look at the triangle again, and try to find all of it's symmetries (label the vertices with 1,2,3, and see f that helps you find more) then try to make an inference as to the general order of this group based on your findings for the triangle and the square.
P: 176 there are two more for the triangle - with the base of the triangle on a horizontal axis we can switch the top vertices with the right vertices by flipping the triangle from the left vertices and vice versa with if you flip the triangle from the right vertices. 2n?
P: 40
Group of symmetries on a regular polygon
http://en.wikipedia.org/wiki/Dihedral_group should be of use to you. In direct response to your question, the page notes that for small n, the Dihedral group isn't a subgroup of the symmetric group S_n, as demonstrated by the fact n! < 2n for n=1.
P: 291 Here is how to think of it. 1)Draw a regular polygon with vertices 1,2,...,n in order. 2)The question is how many ways can the polygon be replaced by rigid motions, i.e. with vertices still in order. Now for any one of the n locatations there are n choices. Having chosen that there is only one chose for the next one in succession, i.e. either left or right. Having chosen that the polygon is completely determined. Thus there are 2n elements in this group. 3)Let a be the positive rotation by 2pi/n i.e. the cycle (1,2,...,n) and let b be the reflection through the middle and vertex 1, i.e. (2,n)(3,n-1)..... 4)So a^n = 1 and b^2 = 1. 5)Consider S = {a^k,ab^k} for k=1,2,...,n 6)There are 2n elements in that set all of which are distinct. So S represents the group D_n. 7)Now it remains to show ba=a^{n-1}b which is not so hard to show. Hence the group presentation for the dihedral group is: $$\left< a,b| a^n = 1, \ b^2 = 1, \ ba=a^{n-1}b \right>$$
Related Discussions Calculus & Beyond Homework 0 Calculus 1 General Math 0 General Math 20
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r-fat polynomials are a natural generalization of scattered polynomials. They define linear sets of the projective line PG(1, q(n)) of rank n with r points of weight larger than one. Using techniques on algebraic curves and function fields, we obtain numerical bounds for r and the non-existence of exceptional r-fat polynomials with r > 0. We completely determine the possible values of r when considering linearized polynomials over F-q4 and we also provide one family of 1-fat polynomials in PG(1, q(5)). Furthermore, we investigate LP polynomials (i.e. polynomials of type f(x) = x + delta x(q2x & nbsp;)is an element of F-qn [x], gcd(n, s) = 1), determining the spectrum of values r for which such polynomials are r-fat. (C)& nbsp;2022 Elsevier Inc. All rights reserved.& nbsp;
### r-fat linearized polynomials over finite fields
#### Abstract
r-fat polynomials are a natural generalization of scattered polynomials. They define linear sets of the projective line PG(1, q(n)) of rank n with r points of weight larger than one. Using techniques on algebraic curves and function fields, we obtain numerical bounds for r and the non-existence of exceptional r-fat polynomials with r > 0. We completely determine the possible values of r when considering linearized polynomials over F-q4 and we also provide one family of 1-fat polynomials in PG(1, q(5)). Furthermore, we investigate LP polynomials (i.e. polynomials of type f(x) = x + delta x(q2x & nbsp;)is an element of F-qn [x], gcd(n, s) = 1), determining the spectrum of values r for which such polynomials are r-fat. (C)& nbsp;2022 Elsevier Inc. All rights reserved.& nbsp;
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https://scholarsanswers.com/physicslab-9-ohms-law-resistance-and-resistivity/
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# Physics-Lab 9 Ohm’s Law, Resistance and Resistivity
Lab 9Ohm’s Law, Resistance and Resistivity1. Objective: (i)(ii)(iii)The purpose of this experiment are: To measure the resistance of a given length of wire. To study the variation of resistance with length of the conductor. To measure the reistivity of the material of a conductor. 2. Equipment: (i)(ii)(iii)Five nickel-silver wires of length 40 cm, 80 cm, 120 cm, 160 cm, 200 cm anddiameter 0. 00026 mOne ammeter, one voltmeter, one rheostat and a 6. 0-V batterysix connecting leads and four crocodile clips. 3. Theory: Ohm’s law states that the current (i) passing through a conductor is proportionalto the potential difference (V) across its ends. This can be expressed symbolicallyas: V = RI…(i)Where R is a constant for the conductor and is called its resistance. Resistance ismeasured in volts per ampere and is called ohm (). The resistance of aconductor depends on the type of material, its length (l), area of cross section (A)and the temperature at which it is maintained. The resistance of a conductor at agiven temperature is given by: RlAwhere is a constant for the material of the conductor and is called the resistivityof the conductor. It is measured in m. The area of cross section of the wire isgiven by: A = r2Thus the above equation for resistance becomes: Rlr2…(ii)4. Procedure: VbatteryconductorArheostatSet up the circuit as shown in the figure above. The voltmeter is connectedparallel to the conductor, and the ammeter, rheostat and the battery are connectedin series with it. The voltmeter and the ammeter are pole sensitive and so aconnection originating from the positive end of the battery must be connected tothe red terminal. If the voltmeter or ammeter reads backward, reverse theconnections on it. Adjust the rheostat so that the ammeter and voltmeter show the smallest readablevalue. Record the current and the voltage values. Take 5 more reading of currentand voltage, each time adjusting the rheostat for a higher value. Record your I andV values in a table. Repeat the procedure for the remaining four wires andmake separate tables for I and V values for each conductor. 5. Interpretation of the Data: Identify the variables in equation (i) and draw a suitable graph for each of theconductors. Use the slope of the graph to obtain the resistance of each conductor. Record the resistance of each conductor and its length in a table. Write equation (ii) in the form y = mx + bIdentify x and y and draw a suitable graph. Use the slope of your graph to obtain avalue for the resistivity of nichrome-silver. If the actual value is 3. 3 x 10-7 m,calculate the percentage error in your experiment. 6. Questions: 1. A wire of diameter 1 mm has a resistance of 0. 5 . What is the resistanceof another wire of the same material and the same length but of diameter0. 5 mm?2. What is the potential difference across a 100 m length of copper wire thathas a diameter of 0. 5 mm when it carries a current of 15 A? The resistivityof copper is 1. 7 x 10-8 m. 3. How long is a copper wire that has a resistance of 2 and a diameter of0. 01 mm?4. A wire of length 1. 0 m has a resistance of 5. 0 . It is uniformly stretchedto a length of 2. 0 m. What is its new resistance?
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https://se.mathworks.com/matlabcentral/answers/752874-explicit-eulers-method-for-time-advancement
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# Explicit Eulers Method for time advancement
24 views (last 30 days)
Cassidy Holene on 22 Feb 2021
Answered: Alan Stevens on 22 Feb 2021
Hello I am trying to use explicit Euler for time advancement and the second-order centraldifference scheme for the spatial derivative, solve the equation to steady
state on a uniform grid. Plot the exact and numerical steady solutions for Nx = 10, 20.
๐๐/๐๐ก = ๐ผ*( ๐^2๐/๐๐ฅ^2) + ๐(๐ฅ) on the boundary of 0 โค ๐ฅ โค ๐ฟ๐ฅ The initial and boundary conditions are ๐(๐ฅ, 0) = 0 ๐(0,๐ก) = 0 ๐(๐ฟ๐ฅ,๐ก) = ๐steady(๐ฟ๐ฅ) Take ๐ผ = 1, ๐ฟ๐ฅ = 15, and ๐(๐ฅ) = โ(๐ฅ 2 โ 4๐ฅ + 2)๐ โ๐ฅ . The exact steady solution is ๐steady(๐ฅ) = ๐ฅ 2๐ โ๐ฅ
heres the code I have can someone explain where I went wrong
alpha =0;
x = 0;
n =10;
T(0)4ess = 0;
h=0.1;
s(x)=-(x^2-4*x+2)*exp^(-x);
for i=1:n
T(i+1)=T(i) + h;
T(i)^(n+1)=T(i+1)^n+((alpha*h)/(x+h)^2)*(T(i+1)^n-2T(i)^n+T(i-1)^n)+h*s(x);
x = x +1;
h = h +0.1;
end
plot(x,T);
grid on;
darova on 22 Feb 2021
Can you please write difference scheme in LaTeX?
Alan Stevens on 22 Feb 2021
T(i)^(n+1)
This will raise T(i) to the (n+1)th power!
You need another loop for time (say j = 1:something), then you can refer to T at position i and time j as
T(i,j)
On the right hand side
((alpha*h)/(x+h)^2)*(T(i+1)^n-2T(i)^n+T(i-1)^n)
should be
((alpha*h)/dx^2)*(T(i+1,j)-2T(i,j)+T(i-1,j))
where dx is the spatial interval (dx = Lx/n).
h is the timestep, so don't update it in the loop!
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https://diy.stackexchange.com/questions/119581/when-are-energy-water-consumption-specs-important-in-regards-to-the-price/119587
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# When are energy/water consumption specs important in regards to the price?
I am comparing 2 washing machines that have 148 euros difference.
The cheapest one has in the specs:
``````Annual energy consumption:
174 kWh
Annual water consumption:
10.840 l
Energy efficiency class:
A +++
``````
The more expensive one has in the specs:
``````Annual energy consumption:
156 kWh
Annual water consumption:
9999 l
Energy efficiency class:
A +++ -20% (although they did not have the -20% in the label in the store)
``````
My question is: is the most expensive of the two actually cheapest in the long run due to the water/energy consumption and hence the 148 euros difference is not what I should focus on? Or that would matter only in cases of a lot of laundries per week so for all practical purposes both are same?
You have to look at the assumptions they make when determining the annual energy and water consumption. It could be 2 loads of laundry a week, or it could be 2 a day. Without knowing that, you don't know if you do more or less than their assumption, and thus whether you'd use more or less energy / water each year.
For example, in the US, Energy Star washers have a label which indicates (or similar):
• Estimated operating cost based on six wash loads a week and a national average electricity cost of 12 cents per kWh and natural gas cost of \$1.09 per therm
So, it makes it very easy to compare the calculated annual energy cost to your expected energy cost, as you can calculate the energy usage per load. You can do the same with water usage.
If you can't find information about the assumptions made, you could use the one given by Energy Star above as a starting point. For example, I pay 10 cents per kWh of electricity and 80 cents per hundred gallons of water. Total difference in operating costs is less than \$4 / year, which means it'd take about 40 years for me to break even on the more expensive but more efficient one (148 euros ~ \$175).
Even using the washer 10 times as much as what Energy Star considers average (6 loads per week), it'd still take me about 4 years to break even. And, if I'm doing 60 loads of laundry a week, I better get a bigger washer. So, in this case, there is no practical difference in running costs. If you had one washer using 200 kWh per year and another using 500 kWh per year, or if you pay a lot more for electricity / water, then you'd have a more significant difference.
• What I found was the following from en.wikipedia.org/wiki/… "The EEI is a measure... and the energy consumed in 220 washing cycles" but I don't really understand from wiki how can I associate the index with laundries per week – Jim Jul 17 '17 at 19:59
• `The energy rating label shows you the number of kilowatt hours (kWh) that you could expect the machine to use over a year based on its performance on full and partial 60˚C cotton loads and a 40 ˚C partial cotton load` I still am not clear how that is connected to the actual usage pattern – Jim Jul 17 '17 at 20:06
• If it is over 220 cycles, that's about 4 loads per week. This line `For a 6-kg machine, an EEI of 100 is equivalent to 334 kWh per year, or 1.52 kWh per cycle.`, also works out to 220 wash cycles. Seems you can go with 4 to 5 loads per week for those labels - same ballpark as Energy Star. – mmathis Jul 17 '17 at 22:22
• So if I do 4 loads per week the more expensive one is cheaper right? But if I do less how do I compare if the amount is worth it? – Jim Jul 18 '17 at 8:52
• If you use the washer less, it will take even longer to recoup the difference in price for the more expensive one. For me, doing 4 loads of laundry a week (which is what the energy/water consumption figures assume), it'd take ~40 years to break even on the more expensive washer - not worth it from that point of view. If I only do 2 loads a week, it'd take me twice as long to recoup; if I do 8, half as long (still 20 years tho, likely longer than the lifespan of the washer). Also, knowing that those figures are for 220 loads, you can calculate the per-load cost and go from there. – mmathis Jul 18 '17 at 17:29
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2014-11-04T20:46:33-05:00
12.75=12 3/4
Hope this helps!!!!!!!!!!!!
2014-11-04T20:49:35-05:00
12.75
12= whole number
.75= is in the hundred place value
so 12 75/100
simplify to 12 3/4
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http://timeandquantummechanics.com/topics/requirements/
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## Next step for time & quantum mechanics
“When you come to a fork in the road, take it!” — Yogi Berra, Quantum Philosopher
I’ve been wondering what to do with quantum time. I’ve gotten a certain amount of feedback on the original paper, ranging from “hard but interesting” to “interesting but hard”.
There are really two directions I would like to take this project at this point:
1. Do the calculations in a more transparent way, to leave us, hopefully, just with “interesting”.
2. Extend the ideas to the multi-particle case, which is needed for the analysis of all but the most trivial cases. For instance, we need this even to compute bound state wave functions.
In the spirit of quantum mechanics, it seems best to do both. I look at each in turn.
I noticed when I talked in Baltimore last month that the animations were really the most transparent part of the talk. But the only way to develop them is to use numerical methods. I’ve done a bit of numerical work in the past, mostly to calculate charged particle orbits around a black hole (when I was a grad student at Princeton). From this I learned two things:
1. Numerical calculations are tricky. I learned this the hard way. I had thought — ah youth — that the smaller you make the step size, the more accurate the results. But I found this was true only up to a point; below a certain step size the calculations would produce obvious nonsense: at small enough step sizes, round-off errors dominated the results, sending the particles either into the black hole or out into space. [No real particles were harmed in the course of this experiment.] Ultimately, I had to completely rewrite the equations in a non-linear but stabler way to get something meaningful.
2. If you don’t have a reliable source of physical intuition, tricky can quickly escalate into nonsense. With anything involving time this is particularly a problem, largely because our usual intuition about time is so compelling that it is hard to move past it. And if we do move past it, where do we get a “reliable source of physical intuition”?
Consequently I’ve been a bit chary of doing numerical work. But while researching my Baltimore talk, I came across a work, Advanced Visual Quantum Mechanics, by Bernd Thaller, where the problem was solved, at least for low dimensional cases. Bernd Thaller worked primarily with Mathematica, a higher level language, but used a C program written by Manfred Liebmann for the low level numerical work. This was a dissertation paper by Liebmann. A quick scan of the table of contents was enough to confirm my intuition that the problem is non-trivial.
I’ve spent a few hours with Liebmann’s dissertation. It is written in German but apparently my high school German, Google translate, and a fair knowledge of the subject area [plus checking the references as I go] is enough to let me stumble thru it. The basic approach is essentially path integrals done a step at a time, in such wise as to minimize the numerical error at each step. This I can manage. Approximate proudly!
The second problem is how to extend quantum time to the multi-particle case. The main problem here is how to generalize the single particle results to the multi. After some mulling, and in the spirit of “approximate proudly” I’ve decided that using the usual Feynman rules but with the standard propagator replaced by the slightly fuzzier quantum time propagator is a reasonable first step. When we are only looking for first order corrections, we don’t need an elaborate theoretical framework.
What to use for the “slightly fuzzier” is a bit of a question. Our single particle action is:
The most obvious generalization to the field theoretical case looks like:
This won’t do. It is dimensionally wrong. We need to multiply τ by something with dimensions of mass. But we can’t use the mass of any specific particle, as that would be to prefer one over another. We will instead insert a factor κ, defined as something with dimensions of mass/energy:
This will give us [insert hand-waving here] a propagator looking like:
If we are looking at a Feynman diagram we will wind up convoluting over the laboratory time:
Which makes most of our integrals look like products of the Laplace transform of the propagator:
Compare to the usual Feynman propagator:
Modulo an overall dimensional factor of κ [the sort of thing that comes out in the wash], they look much alike — in the limit of small κ. As small κ corresponds loosely to large τ and as we expect to get standard quantum theory back in the long time limit of quantum time, this is fine.
The next question is where did κ come from? We don’t need to sort that out entirely up front, but we do need to know we have at least one viable answer.
If we want to take an aggressively Machian view of quantum mechanics, then there is nothing to the universe but its wave function: space and time are mere ensemble averages over the wave function of the rest of the universe. κ then can be a measure of how much energy stored in the local vacuum fluctuations, small but not zero:
So, that is the plan for the multi-particle case: use the κ-ified propagator with the Feynman rules, require we get standard quantum theory back in the large τ/small κ limit, see the testable inferences in re quantum time/multiple particle case as the first order corrections due to non-zero κ and/or small dispersions along the time dimension.
## Dissertation complete
I’ve finished re-checking the dissertation: 629 equations, 188 references, 110 pages, 83 input files, 48 lists, 36 footnotes, 28 quotes, 17 figures, 6 chapters (counting the appendix), 5 requirements, 1 idea. It should be up on the physics archive in a day or two.
## The Secret of Chimpanzee Painting
Back in the early 70s there was a chimpanzee, Pablo the Chimp, who won 2nd prize in a California art show, under a pseudonym. When this came to light Pablo’s trainer was asked how the chimpanzee was, as an artist. The reporters ganged around, raised a skeptical eyebrow or three, and asked “Just how good is Pablo, qua artiste?”
This was before the Internet, back when if you wanted to send a message to a nearby island you had to get out your hammer and chisel, so I wasn’t able to find an exact record of the trainer’s response. But as I remember it, his response was along the lines of:
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# Converting latitude and longitude to point in geodjango?
I am doing the geodjango tutorial and I went on google and chose a random longitude and latitude in Kansas, I am using srid=4326 for my polygons
So I got this random lat and long on google maps in Kansas USA
``````lat, long = 38.670049, -99.565798
``````
Now I am trying to see where in the world it is
``````>>> from django.contrib.gis.geos import Point
>>> pnt = Point(38.670049, -99.565798)
>>> WorldBorder.objects.get(mpoly__intersects=pnt)
...
...
world.models.DoesNotExist: WorldBorder matching query does not exist.
``````
Using the tutorial point I get
``````>>> pnt = Point(-95.3385, 29.7245)
>>> WorldBorder.objects.get(mpoly__intersects=pnt)
<WorldBorder: United States>
``````
So my database is imported and working.
So clearly Points are not the same as latitude and longitude, how do I convert between the two
It seems google maps has the lat/long the other way around, see coordinates in image below at the bottom from google maps:
``````# Try enter the coords in the same order as in image above
# Get wrong intersection
>>> from django.contrib.gis.geos import Point
>>> from world.models import WorldBorder
>>> pnt = Point(38.907636 , -77.036905)
>>> WorldBorder.objects.get(mpoly__intersects=pnt)
<WorldBorder: Antarctica>
# Swap them around, and it works
>>> pnt = Point(-77.036905, 38.907636)
>>> WorldBorder.objects.get(mpoly__intersects=pnt)
<WorldBorder: United States>
``````
Can anyone explain what is going on exactly. For a Point(x,y) which is the longitude and which is the latitude, is it
`Point(x,y)`
`x=longitude`
`y=latitude`
And in which order does Google maps present them in the picture above
I am not a GIS specialist or even novice.
• Not many features will contain a point at 99 degrees south. It helps if you always think in terms of X,Y order, and even write it "longitude and latitude". – Vince May 15 '16 at 19:56
• but i'm doing an intersection, surely any point within the USA will intersect with the USA polygon. so are points longitude and latitude? or are they some abitrary x,y coordinate system in geodjango – Dr Manhattan May 16 '16 at 6:16
• What Vince is saying is that -99 latitude does not exist. The South Pole is -90. You have reversed the coordinates in your two examples. – John Powell May 16 '16 at 9:10
• Please see updated question – Dr Manhattan May 16 '16 at 14:57
``````> >>> from django.contrib.gis.geos import Point
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Revista de la Unión Matemática Argentina Home Editorial board For authors Latest issue In press Online first Prize New! Search OJS
### Volumen 52, número 1 (2011)
Eduardo H. Zarantonello (1918-2010). J. Tirao PDF i-vi De matemática y matemáticos. Eduardo H. Zarantonello PDF Conferencia originalmente ofrecida en septiembre de 1981 en Córdoba, Argentina, con motivo de la celebración del XXVº aniversario de la creación del Instituto de Matemática, Astronomía y Física de la Universidad Nacional de Córdoba. Note: This text was not included in the print version of the Revista. vii-xii Factor congruences in semilattices. Pedro Sánchez Terraf PDF We characterize factor congruences in semilattices by using generalized notions of order ideal and of direct sum of ideals. When the semilattice has a minimum (maximum) element, these generalized ideals turn into ordinary (dual) ideals. 1-10 Properties of the bivariate confluent hypergeometric function kind 1 distribution. Daya K. Nagar and Fabio Humberto Sepúlveda-Murillo PDF The bivariate confluent hypergeometric function kind 1 distribution is defined by the probability density function proportional to x1ν1 − 1 x2ν2 − 11F1(α; β; −x1 − x2). In this article, we study several properties of this distribution and derive density functions of X1/X2, X1/(X1 + X2), X1 + X2 and 2 √(X1 X2). The density function of 2 √(X1 X2) is represented in terms of modified Bessel function of the second kind. We also show that for ν1 − ν2 = 1/2, 2 √(X1 X2) follows a confluent hypergeometric function kind 1 distribution. 11-21 Ergodic properties of linear operators. María Elena Becker PDF Let T be a bounded linear operator on a Banach space X. We prove some properties of X1 = { z ∈ X : limn ∑k=1n (Tkz/k) exists} and we construct an operator T such that limn ||Tn/n|| = 0, but (I − T)X is not included in X1. 23-25 A new application of power increasing sequences. Hüseyin Bor PDF In the present paper, we have proved a general summability factor theorem by using a general summability method. This theorem also includes several new results. 27-32 A note on integral C-parallel submanifolds in S7(c). D. Fetcu and C. Oniciuc PDF We find the explicit parametric equations of the flat 3-dimensional integral C-parallel submanifolds in the sphere S7 endowed with the deformed Sasakian structure defined by Tanno. 33-45 Weighted local BMO spaces and the local Hardy-Littlewood maximal operator. Aníbal Chicco Ruiz and Eleonor Harboure PDF We define a local type of a weighted BMO space on R+ and prove the boundedness of the local Hardy-Littlewood maximal function in that space, provided that the weight belongs to the local class A1loc. 47-56 Global Lp estimates for degenerate Ornstein-Uhlenbeck operators: a general approach. Ermanno Lanconelli PDF We present a new approach to prove global Lp estimates for degenerate Ornstein-Uhlenbeck operators in RN. We then show how to pave the way to extend such a technique to classes of general Hörmander operators. Several historical notes related to Calderón-Zygmund’s singular integrals theory in Euclidean and in non-Euclidean settings are also provided. 57-72 Triality and the normal sections of Cartan’s isoparametric hypersurfaces. Cristián U. Sánchez PDF The present paper is devoted to study the algebraic sets of normal sections of the so called Cartan’s isoparametric hypersurfaces MR, MC, MH and MO of complete flags in the projective planes RP2, CP2, HP2 and OP2. It presents a connection between “normed trialities” and the polynomial defining the algebraic sets of normal sections of these hypersurfaces. It contains also a geometric and topological description of these algebraic sets by means of the isotropy actions of the isoparametric hypersurfaces. 73-88 Approximation and shape preserving properties of the truncated Baskakov operator of max-product kind. Barnabás Bede, Lucian Coroianu and Sorin G. Gal PDF Starting from the study of the Shepard nonlinear operator of max-prod type in [2], [3], in the recent monograph [4], Open Problem 5.5.4, pp. 324-326, the Baskakov max-prod type operator is introduced and the question of the approximation order by this operator is raised. The aim of this note is to obtain the order of uniform approximation Cω1(f; 1/√n) (with the explicit constant C = 24) of another operator called the truncated max-prod Baskakov operator and to prove by a counterexample that in some sense, for arbitrary f this type of order of approximation with respect to ω1(f; √n) cannot be improved. However, for some subclasses of functions including for example the nondecreasing concave functions, the essentially better order of approximation ω1(f; 1/n) is obtained. Finally, some shape preserving properties are proved. 89-107 Some Slater type inequalities for convex functions of selfadjoint operators in Hilbert spaces. S. S. Dragomir PDF Some inequalities of the Slater type for convex functions of selfadjoint operators in Hilbert spaces $H$ under suitable assumptions for the involved operators are given. Amongst others, it is shown that if $A$ is a positive definite operator with $Sp(A) \subset [m,M]$ and $f$ is convex and has a continuous derivative on $[m,M]$, then for any $x\in H$ with $\left\Vert x\right\Vert =1$ the following inequality holds: \begin{multline*} 0\leq f\left( \frac{\left\langle Af^{\prime }\left( A\right) x,x\right\rangle }{\left\langle f^{\prime }\left( A\right) x,x\right\rangle }% \right) -\left\langle f\left( A\right) x,x\right\rangle \\ \leq \frac{1}{4}\cdot \sqrt{\frac{Mf^{\prime }\left( M\right) }{mf^{\prime }\left( m\right) }}\left( M-m\right) \left( f^{\prime }\left( M\right) -f^{\prime }\left( m\right) \right). \end{multline*} 109-120 Necessary and sufficient conditions for the Schur harmonic convexity or concavity of the extended mean values. Wei-Feng Xia, Yu-Ming Chu and Gen-Di Wang PDF In this paper, we prove that the extended values $E(r,s;x,y)$ are Schur harmonic convex (or concave, respectively) with respect to $(x,y)\in (0,\infty)\times(0,\infty)$ if and only if $(r,s)\in \{(r,s):s\geq-1, s\geq r,s+r+3\geq0\}\cup\{(r,s):r\geq-1, r\geq s,s+r+3\geq0\}$ (or $\{(r,s):s\leq-1, r\leq-1,s+r+3\leq0\},$ respectively). 121-132 Property (ω) and quasi-class (A, k) operators. M. H. M. Rashid PDF In this paper, we prove the following assertions: (i) If T is of quasi-class (A, k), then T is polaroid and reguloid; (ii) If T or T* is an algebraically of quasi-class (A, k) operator, then Weyls theorem holds for f(T) for every f ∈ Hol(σ(T)); (iii) If T* is an algebraically of quasi-class (A, k) operator, then a-Weyls theorem holds for f(T) for every f ∈ Hol(σ(T)); (iv) If T* is algebraically of quasi-class (A, k) then property (ω) holds for T. 133-142 Solution of Troesch’s problem using He’s polynomials. Syed Tauseef Mohyud-Din PDF In this paper, we apply He’s polynomials for finding the approximate solution of the Troesch’s problem which arises in the confinement of a plasma column by radiation pressure and applied physics. The proposed technique proved to be very effective and is easier to implement as compare to decomposition method. 143-148 Julio Rey Pastor, su posición en la escuela matemática argentina. Eduardo L. Ortiz PDF This paper was read at the Tandil Meeting of the UMA, in September 2010; in this presentation I keep to the original format of a lecture. I briefly consider three of the main attempts made in Argentina to establish a mathematics school, between 1817 and 1940, paying more attention to the third one, in which Julio Rey Pastor was the main character. Contrary to the earlier ones, in this last period mathematics began to be established as a distinct discipline, with its own problems and methods, while keeping close ties with other disciplines. In this lecture I’ll also consider some matters that emerged in parallel with these attempts, and make some remarks on the different approaches used to tackle them. At the same time I’ll try to relate the progress of mathematics in Argentina with the doctrines and ideas that, in different periods of its history, dominated its cultural life. 149-194
### Volumen 52, número 2 (2011)
This volume contains the contributed papers from the Xth Encuentro Nacional de Analistas Alberto P. Calderón, which took place between August 25th and 28th, 2010, in La Falda, Córdoba.
Preface. Carlos Cabrelli and Eleonor Harboure PDF i-iii Hardy spaces associated with semigroups of operators. Jorge J. Betancor PDF This paper is a non exhaustive survey about Hardy spaces defined by semigroups of operators. 1-22 Sobolev spaces diversification. Bruno Bongioanni PDF This work attempts to be an overview of a variety of results concerning Sobolev spaces associated to some orthonormal systems, particularly the Hermite and Laguerre operators settings. 23-34 Five basic lemmas for symmetric tensor products of normed spaces. Daniel Carando and Daniel Galicer PDF We give the symmetric version of five lemmas which are essential for the theory of tensor products (and norms). These are: the approximation, extension, embedding, density and local technique lemma. Some applications of these tools to the metric theory of symmetric tensor products and to the theory of polynomials ideals are given. 35-60 Wiener’s lemma: pictures at an exhibition. Ilya A. Krishtal PDF In this expository paper we present various general extensions of the celebrated Wiener’s Tauberian Lemma, outline several ingredients that are often used in proofs and discuss a few applications to localization of frames. 61-79 Porosity, dimension, and local entropies: A survey. Pablo Shmerkin PDF Porosity and dimension are two useful, but different, concepts that quantify the size of fractal sets and measures. An active area of research concerns understanding the relationship between these two concepts. In this article we will survey the various notions of porosity of sets and measures that have been proposed, and how they relate to dimension. Along the way, we will introduce the idea of local entropy averages, which arose in a different context, and was then applied to obtain a bound for the dimension of mean porous measures. 81-103 Importance of Zak transforms for harmonic analysis. Edward N. Wilson PDF In engineering and applied mathematics, Zak transforms have been effectively used for over 50 years in various applied settings. As Gelfand observed in a 1950 paper, the variable coefficient Fourier series ideas articulated in Andre Weil’s famous book on integration lead to an exceedingly elementary proof of the Plancherel Theorem for LCA groups. The transform for functions on R appearing in Zak’s seminal 1967 paper is actually a special case of the LCA group transforms earlier introduced by Weil; Zak states this explicitly in his 1967 paper but the mathematical community nonetheless chose to name the transform for him. In brief, the properties of Zak transforms are simply reflections of elementary Fourier series properties and the Plancherel Theorem for non-compact LCA groups is an immediate consequence of the fact that Fourier transforms are averages of Zak transforms. It is remarkable that only a small handful of mathematicians know this proof and that all textbooks continue to give much harder and less transparent proofs for even the case of the group R. Generalized Zak transforms arise naturally as intertwining operators for various representations of Abelian groups and allow formulation of many appealing theorems. 105-113
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ZEBU : Free Signup for Demat and Trading Account opening
## Car loan Calculator
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551_PartUniversity Physics Solution
# 551_PartUniversity Physics Solution - Thermal Properties of...
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Thermal Properties of Matter 18-9 18.40. IDENTIFY: rms 3 kT v m = . SET UP: 23 1.381 10 J/molecule K. k EXECUTE: (a) 23 3 rms 16 3(1.381 10 J/molecule K)(300 K) 6.44 10 m/s 6.44 mm/s 3.00 10 kg v ×⋅ == × = × EVALUATE: (b) No. The rms speed depends on the average kinetic energy of the particles. At this T , H 2 molecules would have larger v rms than the typical air molecules but would have the same average kinetic energy and the average kinetic energy of the smoke particles would be the same. 18.41. IDENTIFY: Use Eq.(18.24), applied to a finite temperature change. SET UP: 5/ 2 V CR = for a diatomic ideal gas and 3/ 2 V = for a monatomic ideal gas. EXECUTE: (a) () 5 2 V Qn CTnR T = Δ 5 2 (2.5 mol) (8.3145 J/mol K)(30.0 K) 1560 J Q =⋅ = (b) 3 2 V = Δ 3 2 (2.5 mol) (8.3145 J/mol K)(30.0 K) 935 J Q = EVALUATE: More heat is required for the diatomic gas; not all the heat that goes into the gas appears as translational kinetic energy, some goes into energy of the internal motion of the molecules (rotations). 18.42. IDENTIFY: The heat Q added is related to the temperature increase T Δ by . V CT = Δ SET UP: For 2 H , 2 ,H 20.42 J/mol K V C and for Ne (a monatomic gas), ,Ne 12.47 J/mol K. V C = EXECUTE: constant V Q n Δ= = , so 22 H Ne . VV Δ= Δ 2 2 Ne H 20.42 J/mol K (2.50 C ) 4.09 C 12.47 J/mol K V V C TT C ⎛⎞ Δ= = ⎜⎟ ⎝⎠ °° . EVALUATE: The same amount of heat causes a smaller temperature increase for 2 H since some of the energy input goes into the internal degrees of freedom. 18.43. IDENTIFY: CM c = , where C is the molar heat capacity and c is the specific heat capacity. . m pV nRT RT M SET UP: 2 3 N 2(14.007 g/mol) 28.014 10 kg/mol M × . For water, w 4190 J/kg K c = . For 2 N, 20.76 J/mol K V C . EXECUTE: (a) 2 N 3 20.76 J/mol K 741 J/kg K C c M = × . 2 w N 5.65 c c = ; w c is over five time larger. (b) To warm the water, 4 w (1.00 kg)(4190 J/mol K)(10.0 K) 4.19 10 J Qm cT = = × . For air, 2 4 N 5.65 kg (741 J/kg K)(10.0 K) Q m × = Δ⋅ . 3 35 (5.65 kg)(8.314 J/mol K)(293 K) 4.85 m (28.014 10 kg/mol)(1.013 10 Pa) mRT V Mp = ×× . EVALUATE: c is smaller for 2 N , so less heat is needed for 1.0 kg of 2 N than for 1.0 kg of water. 18.44. (a) IDENTIFY and SET UP: 1 2 R contribution to V C for each degree of freedom. The molar heat capacity C is related to the specific heat capacity c by . c = EXECUTE: 1 2 6 3 3(8.3145 J/mol K) 24.9 J/mol K. V R = = The specific heat capacity is 3 / (24.9 J/mol K)/(18.0 10 kg/mol) 1380 J/kg K. cCM × = (b) For water vapor the specific heat capacity is 2000 J/kg K. c = The molar heat capacity is 3 (18.0 10 kg/mol)(2000 J/kg K) 36.0 J/mol K. c × = EVALUATE: The difference is 36.0 J/mol K 24.9 J/mol K 11.1 J/mol K, −⋅ which is about ( ) 1 2 2.7 ; R the vibrational degrees of freedom make a significant contribution. 18.45. IDENTIFY: 3 V = gives V C in units of J/mol K . The atomic mass M gives the mass of one mole. SET UP: For aluminum, 3 26.982 10 kg/mol. M EXECUTE: (a) 3 24.9 J/mol K V . 3 24.9 J/mol K 923 J/kg K 26.982 10 kg/mol V c = × .
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https://la.mathworks.com/matlabcentral/cody/problems/42855-height-of-a-right-angled-triangle/solutions
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Cody
# Problem 42855. Height of a right-angled triangle
• What is Size?
1 – 10 of 1,736
#### Solution 4602375
Submitted on 15 Jan 2021 at 10:50 by Audrey Guillet
• Size: 51
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#### Solution 4602365
Submitted on 15 Jan 2021 at 10:48
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#### Solution 4602360
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#### Solution 4574943
Submitted on 11 Jan 2021 at 12:27 by 学凯 刘
• Size: 63
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#### Solution 4574938
Submitted on 11 Jan 2021 at 12:26
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#### Solution 4572253
Submitted on 10 Jan 2021 at 22:13
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#### Solution 4572243
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#### Solution 4572228
Submitted on 10 Jan 2021 at 22:09
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#### Solution 4569373
Submitted on 10 Jan 2021 at 8:48 by li haitao
• Size: 67
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#### Solution 4559533
Submitted on 8 Jan 2021 at 10:57 by David
• Size: 46
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https://masomomsingi.com/dit107-mathematics-for-science/
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# DIT107 MATHEMATICS FOR SCIENCE.
UNIVERSITY EXAMINATIONS: 2018/2018
EXAMINATION FOR THE DIPLOMA IN INFORMATION TECHNOLOGY
DIT107 MATHEMATICS FOR SCIENCE
DATE: NOV – DEC 2018 TIME: 2 HOURS
INSTRUCTIONS: Answer question ONE and Any other TWO questions.
QUESTION ONE
b) Solve the following simultaneous equations by elimination method
2𝑥 + 𝑦 — 𝑧 = 2
𝑥 − 2𝑦 + 3𝑧 = 9
5𝑥 + 3𝑦 — 2𝑧 = 7 [5 Marks]
c) For the Arithmetic progression:
3 + 7 + 11 + 15 + ··· + 99
Find;
i) The number of terms [3 Marks]
ii) The sum to the number 99
QUESTION TWO
QUESTION THREE
QUESTION FOUR
a) Distinguish amongst Non- frequency data, Ungrouped data and Grouped frequency data
[5 Marks]
b) For the following grouped frequency distribution data below
Find
i) The mean, (5 Marks)
ii)The mode and (5 Marks)
iii) The median (5 Marks)
Total [20 Marks]
QUESTION FIVE
(Visited 118 times, 1 visits today)
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https://community.webcore.co/t/gauge-for-length-of-day-showing-solstices-equinoxes/5633
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# Gauge for Length of Day (showing Solstices & Equinoxes)
#1
I made a piston a few months ago, and I think it is time I shared it with others. This one draws a gauge on my dashboard that monitors the amount of sunlight on any particular day. Basically, far left on the dial is the Winter Solstice (the shortest day of the year) and the far right is the Summer Solstice (the longest day of the year). When the dial is straight up (50) it is one of the 2 Equinoxes (days and nights are equal).
(This picture was saved a few hours after our last Equinox)
There is a bit of math required to get this one customized to your location, but if you are interested, keep on reading. (The first few steps are done with a calculator, pen and paper)
First, you need to find the longest and shortest days for your location (down to the seconds for accuracy). I put in my city at www.timeanddate.com/sun/ and then examined the “Daylength” in late June and then in late December to find the extremes. Take note of the longest time in June, and the shortest time in December.
(For me, it was 14h 01m 57s and 10h 22m 24s) <<-- This last time will be used in webCoRE
Next, we need to convert those times to something understood by webCoRE.
Use the Evaluation Console as an expression something like this:
`datetime('14:01:57')`
-and-
`datetime('10:22:24')`
Use your numbers instead of mine, and make note of both responses.
(for me, it was 1529607660000 & 1529594520000)
Next, subtract the small number from the larger number.
(For me, it was: 1529607660000 - 1529594520000 = 13140000) <<-- This number is used in the next step
This answer tells me that in my city, there is 13,140 seconds (3.65 hours) difference between the amount of daylight between the two Solstices.
Next, we want a conversion rate of 100, where 0 would be the shortest day, and 100 would be the longest day. So (getting as many decimal places as we can) we do this formula:
100 / 13140000 = 0.000007610350076103501 <<-- This last number will be used in webCoRE
(make sure you replace my 13140000 with your number)
Then in webCoRE, we make a piston:
``````define
integer dayLengthLong
decimal dayLengthPercent
end define
Every day, at 15 seconds after midnight
do
Set variable {dayLengthLong} = {datetime(formatDuration(\$sunset-\$sunrise, false, 's'))}
Set variable {dayLengthPercent} = {(dayLengthLong - datetime('10:22:24')) * 0.000007610350076103501}
Set variable {@@dayLengthPercent} = {round(dayLengthPercent,1)}
``````
Notice this last variable is a GLOBAL variable so it can be referred to by other pistons. Also make sure you substitute your December shortest time where I have 10:22:24 above, and your ultra small number in place of my 0.000007610350076103501
Alright, the preparation math is done. Now we are moving on to drawing the gauge.
`Set piston tile #1 title to "[chart-gauge min=0 max=100 greenFrom=0 greenTo=25 greenColor=DeepSkyBlue yellowFrom=25 yellowTo=75 yellowColor=Green redFrom=75 redTo=100 redColor=DarkOrange minorTicks=6 majorTicks=W.E.S|Seasons]" text to "{@@dayLengthPercent}", footer to "{{@@dayLengthPercent}}", and colors to whatever.`
`Set piston state to "{"Day Length = "@@dayLengthPercent"% @ "\$time}";`
And there you have it! A great ‘bird’s eye view’ of where we are in the yearly cycle.
Edit:
For those who are a bit overwhelmed by this manual approach, I have a piston below that can be imported. (although you will miss some good posts if you jump right to post #52)
Lights on to simulate a 14 hour day
Why no notifications on Wind speed for my blinds?
#2
Here’s a variation on the `Set piston state`:
‘[b “OrangeRed” | = {formatDateTime(time(\$sunset-\$sunrise), “h”)}h {formatDateTime(time(\$sunset-\$sunrise), “mm”)}m of sunlight]’
Which yields something like this:
But I think we loose a couple of minutes of accuracy if this variation runs after sunrise.
(Best ran some time between 12:01am and 5am)
#3
That looks like a fun piston. Could you post the whole piston so I can copy it? Also, maybe you did it already, but after it is going, you could have the piston do the calcs so you would not need to do them again for next year.
#4
Maybe you can try using formatDuration(), it’s can provide precision up to ms
#5
I actually have the entire piston shown above. Although I did split it up into two pistons. One runs shortly after midnight to find the length of that day. (this is the heavy math section at the top) And one piston simply draws the gauge. (shown at the bottom of my first post) Unfortunately, this is not one of those pistons you can simply import to have it be accurate for you. You actually have to do the math (only once), and then the piston will work for years without editing it again. (the day of the solstice changes slightly each year, but the longest day and shortest day are practically the same duration. (I saw a 1-2 second variance in ten years)
But, at your request, here is my piston in it’s entirety. I removed the numbers for my location, so you still have to do the math, but maybe the image will help visually. (although I recommend following my steps in my first post for the actual math)
Here is my piston to get the current day length, and convert it into a scale from 0 to 100:
And here’s my simple piston to draw the gauge
(although nobody is stopping you from combining these two pistons into one):
If you need help with the math section for your city, I would be glad to help. I just don’t want to publish code that is only accurate for one person (me).
#6
I do use that formula in my code, but webCoRE and www.timeanddate.com both round to the nearest second.
#7
for the precision, you can use ‘ms’, for example, formatDuration(\$sunset - \$sunrise, false, ‘ms’)
#8
Interesting. Every time I tested that in the console, it was always in whole seconds. No milliseconds when querying webCoRE.
#9
well I guess webcore itself doesn’t have the precise ms precision when it comes to \$sunrise and \$sunset, you can see from the variables list that it’s rounded to minutes even… Not sure if it’s true? @ady624
#10
Good point pauly. I just did a test with mine, as long as the preliminary math is precise (mine is 21 decimals long), the final answer should be within 0.43% of accuracy.
So, for example, assuming the worst, where:
10:20:29:999 rounds down to 10:20
and 2 milliseconds later
10:20:30:001 rounds up to 10:21
The final output will always be within 0.43% of accuracy. For a birds eye view of the entire year, I think this is an acceptable margin of error. (Although my gauge will become even more accurate if webCoRE starts giving us more precise sunrise & sunset times.)
#11
Sorry to be a pain, but if you save the using the green camera, then paste that image here, then I can use the import code to import your piston directly. That will save me from typing it all in and getting it wrong because of a typo.
#12
Yeah just did a research at the webcore weather wiki, hoping that weather API could bring a more precise sunrise and sunset time but turned out that API don’t even have sunrise and sunset time, I guess minute is the most precise time we can get
#14
Here is the green camera that you asked for. The highlighted sections are supposed to be numbers, but I do not want to publish code that is only accurate for one person (me), so the math still has to be done for your location. But at least this way, there should be no typos.
Let me know if you need help with the math for that section.
Also, please ignore the double `Wait` command. This should not be in there.
#15
And for reference, here is my gauge (and hover text) about 5 days after the equinox
#16
Perfect, thanks.
#17
I have tweaked my code a tiny bit after noticing an interesting pattern.
I noticed recently that the day length changes swiftly near the Equinoxes (middle),
and slows down quite a bit near the Solstices (outer extremes).
Picture a sine wave… It lingers at the top and bottom much more than in the middle
The tick marks have not changed, and are based entirely on the amount of daylight…
(Shortest day = 0, and longest day = 100)
But I have adjusted the colors a bit so the colors represent one fourth of a year.
You can see from the pic below that today is nearly 75%, which means that my day length is halfway between what it is on the Equinox vs the Summer Solstice… but there is still about 13 more days until we are 46 days (1.5 months) past the Equinox, and the dial moves into the yellow zone.
This deviation is because the length of day changes swiftly when we are near the Equinoxes (middle of the dial) and the daily change is much less noticeable near the Solstices (ends of the dial)
So even though it’s hard to imagine, in one full year, the dial spends an equal number of days in each color. Half the year moving slowly to the right (with green signifying Spring), and half the year moving left (the green changes to brown then for Fall).
I faked the data for this last pic to show the brown in Fall…
Normally, the brown is only seen after the Summer Solstice
when the dial starts to move back to the left.
EDIT - I am still not happy with the shade of brown above, so I will eventually find a more suitable color to represent Fall.
Analog vs Digital - Astronomical Observations
#18
It just dawned on me that I should probably not have used the word ‘Fall’ or ‘Spring’ above, since technically, the colors do not align with the seasons shown on your calendar. Instead, I have made the colors above centered around the shortest and longest day of the year. This makes more sense to me since this gauge is tracking the Sun’s light. My original 3 color split above seems to convey the trend of the yearly cycle quite nicely.
Be that as it may, for the traditionalists out there… here is a single line of code you could change to make the gauge more accurate based on calendars, and less accurate based on sunlight/heat vs darkness/cold.
The code for Winter/Spring is:
`[chart-gauge min=0 max=100 greenFrom=1 greenTo=49 greenColor={colorWinter} yellowFrom=51 yellowTo=99 yellowColor={colorSpring} minorTicks=6 majorTicks=W.E.S|Seasons]`
and for Summer/Fall is:
`[chart-gauge min=0 max=100 greenFrom=1 greenTo=49 greenColor={colorSummer} yellowFrom=51 yellowTo=99 yellowColor={colorFall} minorTicks=6 majorTicks=W.E.S|Seasons]`
The `{colorX}` seen above are just variables, so you could pick any colors you like
#19
I just wanted to give you all an update. Mother nature has taught me another lesson.
About 45.7 days ago, we had the March Equinox (straight up on my gauge = 50), and about 45.7 days in the future we’ll have the June Solstice (far right on my gauge = 100). But as you can see from this picture, the dial is not close to (75) the center of those two. (I explain why above)
My gauge spans 6 months when moving towards the right, and another six months when moving towards the left. So while the tick marks are quite accurate on the length of day throughout the year, I calibrated the colors to show equal time periods throughout the year. (In a full year, the dial will spend about 91.3 days inside of each color) The following picture breaks the six months up into 4 equal parts of approx 45.7 days each, so you can see the contrast.
I find it quite interesting to see just how swiftly the day lengths changes near the Equinoxes, and just how slowly it changes near the Solstices. For my location, each minor tick mark represents about a 6.5 minute change. Near the Equinoxes, it takes about 4.5 days to move one tick, but near the Solstices, it can take a whopping 17 days to shift the same amount.
Of course, the beauty of this dial is it displays both:
• The amount of daylight (as seen in the number and the tick marks)
• The amount of time spent in each 3 month quadrant (as seen in the colors)
For anyone wanting to tweak your version of the gauge, the only code I changed was the display:
`[chart-gauge min=0 max=100 greenFrom=0 greenTo=16 greenColor={colorWinterEnd} yellowFrom=16 yellowTo=84 yellowColor={colorSpring} redFrom=84 redTo=100 redColor={colorSummerBegin} minorTicks=6 majorTicks=W..E..S|Daylight]`
And for those of you who like numbers:
When comparing the 91.3 days centered around each event, this translates to about:
• 16% & 16% daylight changes around the December Solstice (16-0-16%)
• 68% daylight changes around the March Equinox (16-84%)
• 16% & 16% daylight changes around the June Solstice (84-100-84%)
• 68% daylight changes around the September Equinox (84-16%)
For those curious as to the reasons why this happens, check post #17 above.
(Hint: It’s the same reason the crash location of China’s Tiangong-1 space lab was predicted to be near the extreme north or south of it’s orbital path. Look how much time it spent at the extremes, and how little time it spent near the equator)
#20
Well, it’s about time for an update on my ‘yearly’ gauges…
We are 24 hours away from the Summer Solstice, and here are my gauges with hover-text.
(I still haven’t decided which one I am going to keep, since all 3 show different perspectives)
(The yellow background shows up a day before the Summer Solstice as a visual cue for me)
The first two are entirely based on the length of day… The 3rd gauge is based on the day of the year, so it is not as precise as the other two…
All three gauges divide the year up into 4 equal sections, but they each do it a bit differently…
The LEFT gauge shows the quadrants centered 91 days around the solstices & equinoxes
The MIDDLE gauge splits the year up evenly into 4 colors (official seasons)
The RIGHT gauge is a seasonal overview, but it may be up to 24 hours off since it is based on the dayOfYear
NOTE:
For the really observant, you may have noticed the dayLength is showing at 100.4%. This is because webCoRE rounds the sunrise and sunset to the nearest minute. This means at anytime, the accuracy can be off as much as 0.4% for my location. (If webCoRE starts returning more accurate data, my gauge will auto-correct)
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https://www.thewinningbets.com/predictions-taiwan/premier-league/prediction-of-tainan-city-tatung-fc-24-may-2020-3609
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Match Day 7 of the League Premier League
Tainan City
• L
• W
• L
• W
• L
Tatung FC
• W
• D
• W
• D
• W
This article thoroughly analyzes the match between Tainan City and Tatung FC. We have collected statistics on the championship in Premier League and on the two teams and, if you are registered on the site, you can see them all, together with the percentages calculated by our algorithms.
Statistics and rankings speak of a substantial equality between the two teams. Our analyzes will also consider other factors, to advise you on the best bet, the result of an accurate prediction We reiterate that there is a balance between the two teams, we look at the rankings and we notice how the away team has an advantage: in fact, the Tatung FC has only 0 points less compared to the challenger Tainan City. The statistics of the championship indicate that Tainan City has proved victorious at home in challenges on , with a goal score of facts and suffered. on the other hand, Tatung FC dominates his opponent away from home times on games, scoring times and conceding goals.
## Analysis of the State of Form
Let's now analyze the last 5 league games played by the two teams. At the cost of writing obviousness we want to be precise and describe the fields of the table: match date, name of the teams, final and partial result, position in the ranking of the two challenging teams.
### Analysis of the State of Form: Tainan City
The form status of Tainan City is stable. The team took home 6 points, just under half of those at stake. Players will have to work hard to improve their performance.
### Analysis of the State of Form: Tatung FC
11 points and a series of useful results perfectly describe the good state of form of Tatung FC . The team is in a positive moment and will take the field without fear to give life to a very interesting challenge.
## Full Match
It's time to show the percentages calculated by our systems. We have very complete statistics, ranging from the result of the game, to the analysis of the goal and who will score during the match, to the analysis of the latter in the two times taken individually.
The next tables will show the percentages as regards the exact result and the final result
### Winner of the Match
Will the favored team, after careful analysis of the statistics, be Tatung FC?
Percentage
Tainan City (Home Win)
Tatung FC (Away Win)
Draw
The following is the Total Score report that we have worked out with all the percentages.
The system has calculated a high possibility that both teams score, about 91%.
### Tainan City vs Tatung FC: First Half
Below we show the analysis of the match evaluating only the performances of the two teams in the first half.
### Summary of Predictions
In short, the summary of the prediction of this football match
Discover
Discover
Discover
##### 3 Most Probable Results
Discover
We hope you liked this prediction.With our algorithms, we will process more than 500 predictions per week.If you like our service, please email us and let us know.
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# Number 42096403680
### Properties of number 42096403680
Cross Sum:
Factorization:
2 * 2 * 2 * 2 * 2 * 3 * 5 * 17 * 37 * 139429
Divisors:
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
9cd2424e0
Base 32:
176i8970
sin(42096403680)
0.39052580859277
cos(42096403680)
0.92059197955607
tan(42096403680)
0.42421161303306
ln(42096403680)
24.463228150708
lg(42096403680)
10.624244995393
sqrt(42096403680)
205174.08140406
Square(42096403680)
1.7721072027895E+21
### Number Look Up
Look Up
42096403680 (forty-two billion ninety-six million four hundred three thousand six hundred eighty) is a unique figure. The cross sum of 42096403680 is 42. If you factorisate 42096403680 you will get these result 2 * 2 * 2 * 2 * 2 * 3 * 5 * 17 * 37 * 139429. 42096403680 has 192 divisors ( 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 17, 20, 24, 30, 32, 34, 37, 40, 48, 51, 60, 68, 74, 80, 85, 96, 102, 111, 120, 136, 148, 160, 170, 185, 204, 222, 240, 255, 272, 296, 340, 370, 408, 444, 480, 510, 544, 555, 592, 629, 680, 740, 816, 888, 1020, 1110, 1184, 1258, 1360, 1480, 1632, 1776, 1887, 2040, 2220, 2516, 2720, 2960, 3145, 3552, 3774, 4080, 4440, 5032, 5920, 6290, 7548, 8160, 8880, 9435, 10064, 12580, 15096, 17760, 18870, 20128, 25160, 30192, 37740, 50320, 60384, 75480, 100640, 139429, 150960, 278858, 301920, 418287, 557716, 697145, 836574, 1115432, 1394290, 1673148, 2091435, 2230864, 2370293, 2788580, 3346296, 4182870, 4461728, 4740586, 5158873, 5577160, 6692592, 7110879, 8365740, 9481172, 10317746, 11154320, 11851465, 13385184, 14221758, 15476619, 16731480, 18962344, 20635492, 22308640, 23702930, 25794365, 28443516, 30953238, 33462960, 35554395, 37924688, 41270984, 47405860, 51588730, 56887032, 61906476, 66925920, 71108790, 75849376, 77383095, 82541968, 87700841, 94811720, 103177460, 113774064, 123812952, 142217580, 154766190, 165083936, 175401682, 189623440, 206354920, 227548128, 247625904, 263102523, 284435160, 309532380, 350803364, 379246880, 412709840, 438504205, 495251808, 526205046, 568870320, 619064760, 701606728, 825419680, 877008410, 1052410092, 1137740640, 1238129520, 1315512615, 1403213456, 1754016820, 2104820184, 2476259040, 2631025230, 2806426912, 3508033640, 4209640368, 5262050460, 7016067280, 8419280736, 10524100920, 14032134560, 21048201840, 42096403680 ) whith a sum of 144199621440. The figure 42096403680 is not a prime number. The number 42096403680 is not a fibonacci number. The figure 42096403680 is not a Bell Number. The number 42096403680 is not a Catalan Number. The convertion of 42096403680 to base 2 (Binary) is 100111001101001001000010010011100000. The convertion of 42096403680 to base 3 (Ternary) is 11000122202210212201020. The convertion of 42096403680 to base 4 (Quaternary) is 213031021002103200. The convertion of 42096403680 to base 5 (Quintal) is 1142203134404210. The convertion of 42096403680 to base 8 (Octal) is 471511022340. The convertion of 42096403680 to base 16 (Hexadecimal) is 9cd2424e0. The convertion of 42096403680 to base 32 is 176i8970. The sine of the figure 42096403680 is 0.39052580859277. The cosine of 42096403680 is 0.92059197955607. The tangent of the number 42096403680 is 0.42421161303306. The root of 42096403680 is 205174.08140406.
If you square 42096403680 you will get the following result 1.7721072027895E+21. The natural logarithm of 42096403680 is 24.463228150708 and the decimal logarithm is 10.624244995393. You should now know that 42096403680 is impressive figure!
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## Calculus (3rd Edition)
$$\frac{1}{2}\tan^2 (\ln t) -\ln |\sec (\ln t) |+c$$
Given $$\int \frac{\tan^3 \ln t}{t}dt$$ Let $$u =\ln t \ \ \ \to \ \ \ du=\frac{dt}{t}$$ Then \begin{align*} \int \tan^3 udu&= \int \tan^2 u \tan udu\\ &= \int (\sec^2 u-1) \tan udu\\ &=\frac{1}{2}\tan^2 u -\ln |\sec u |+c\\ &=\frac{1}{2}\tan^2 (\ln t) -\ln |\sec (\ln t) |+c \end{align*}
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# Spinning Discs Illusion
By on April 6, 2006, with 50 Comments
This one’s too easy. Just answer which disc is turning faster, A or B? My guess goes on B! This works in a same way “Rotating Snakes” and “Rotating Snakes no.2″ Illusions work.
50 Responses
1. carl says:
lettera 4 me
2. Donovf says:
and A for me!!
3. Jay says:
None of them are spinning..
• Toby says:
ya, just movement illusion
• yfhrtgjuf says:
star at the text not the pic and it will move
4. Squiggy says:
Jay is right… neither of them are moving. Don’t move your eyes, and you’ll see what i mean. they only seem to be moving when you move your eyes around them.
5. RIYA says:
i can see the other wheel moving if i concenterate on one …i suggest plz donot concenterate deeply..just give casual glance to A..u’ll see the outer circle of B moving and vice-a-versa..good one..i must say
6. boopit says:
Uhhhhhh…..
7. Lindsay says:
A!!!!!!
8. mimi says:
b
9. niki says:
yeah, neigther one are actually moving. I think B turns faster though :P
10. M 2 the H O says:
they arnt moving but its the size of the patterns that make you see that 1 is turning faster, that or you only think 1 is going faster because thats what youv been told and believe! pretty cool though.
ps.
if you glance back and forth you see that and think that they are only moving when your not looking,
FrEaKy!
MwAhAhAhahahahaHAHAHAHA!
i need a pint
11. Mimi2 says:
B
12. MMM says:
They don’t move…
but if you want to put the tings like that,
try to look at the name of the illusion,
and they spin the same velocity!
Very cool illusion!
13. Vamp says:
woah….. o_o
14. Its Me says:
Its a wicked illusion, but none of them really spin!!
15. freekay says:
letter Z for me
16. Anonymous says:
WOW!
If u look at A, B spins faster….
If u look at B, A spins faster
17. Anonymous says:
18. Anonymous says:
i say z!!!!!!!!!!!!!!
19. Anonymous says:
Well i read the little thing above slitly looking at the thingy and ether the same or A.I don’t know why that girl(person who writes thingy 1.)(A)
tee- hee
20. Nora says:
obviously none of them are spinning its just a trick
but both f them are moving at the same spead
the that u fcus on more always is the on that goes faster
21. plym says:
Both circles are static. If you cover one and examine the other, you will see that the exposed one is not moving. It is only when you look at both circles simultaneaously, that they SEEM to be spinning. It is an other instance of the brain fooling the eye, or vice verca.
22. Anonymous says:
none put ur mouse on either and you’ll c
23. Illusionist says:
Easy!
They are NOT spinning!
24. Lexi says:
at tiems it looks like a is, but then others it looks like they are rotating at the same pace
25. Anonymous says:
B ovbiously
its B,Its spinning way faster.the trick is to not look directly at the letter in the middle!!try looking at the writting up top but kinda fuocus on the spinning>>HOPE THAT HELPS to thoughs who cant see them spinnin!!
26. Anonymous says:
Its just the same color isnt it? At least it all looks green to me.
27. Jim Bob says:
A for me
28. frostclaw21 says:
i say b
29. there both the same it just doesn’t look like it i’m 9 and i know how to create this illusion easilly
30. Natalie says:
iv been staring at them for 10 mins now and they are not spinning!?
31. crmen says:
it is none but i am
32. coy says:
the same
33. Lucas says:
when i look to A, B moves, when I look to B, A moves but both with the same speed
34. Max says:
i think they are the same speed
35. kristi says:
i think they are spinning at the same speed ( when you are observing the illusion aspect of it)
36. tevor says:
ok so wen you loke at A, B spins and wen you loke at B, A spins AorB dont spine wen you look at it
37. amelia says:
for me when i looked at a b moved and b a moved so they move at the same pace
38. schultz24 says:
a is moving just a littel bit faster.
39. Eamon Fearon says:
Well when you look at the letter ( A ) The letter ( B ) slows down and vice versa and when you look at Letter ( B ) Letter ( A ) slows down so its hard to tell which is the fastest
40. Shoshanna says:
A, for me. sometimes they rotate at the same, or very similar speed. however, it could depend on which angle you look at it, the distance, or if you’re “left eyed” or “right eyed”.
I’m not a scientist. Even if my formal language makes me one.
41. Aye-Yo says:
None O3O
• Monkey says:
it is one, just its different each time.
42. zee says:
That is crazii…your mind tells you they’re NOT moving, but your eyes do 10/10!!!
43. Monkey says:
It depends where you look. sometimes it ‘A’ sometimes its ‘B’
44. Andrew says:
Depends where you look.
45. Renee Ramsey says:
They are both spinning at the same speed.
46. Hannah;) says:
they are moving at the same pace
P.S I love animals!!!!!! 8)
47. Katie says:
stare at the circles! NONE OF THEM ARE SPINNING!
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An official website of the United States government
Official websites use .gov
A .gov website belongs to an official government organization in the United States.
Secure .gov websites use HTTPS
A lock ( ) or https:// means you’ve safely connected to the .gov website. Share sensitive information only on official, secure websites.
## Taking Measure
Just a Standard Blog
# One of the World’s Roundest Objects Is Helping to Build a Better Mass Measurement
## Share
In 2018, the world agreed to redefine the kilogram. Instead of being pegged to the mass of a physical object in a vault in France, the kilogram is now defined by a few fundamental constants in nature.
This means researchers no longer have to worry about a physical object that can decay or change over time, potentially throwing the world’s measurements into chaos.
Like any unit of measurement, the kilogram needs to be the same for everyone, all the time. Theoretically, it can now be measured in any lab in the world using the Planck constant which defines how small things can be. That measurement can be shared with anyone who needs to find out the mass of an object.
But in reality, it’s not quite that simple yet.
The transition from using a physical object as the kilogram definition to a natural constant has meant differing approaches to defining (or “realizing,” as researchers call it) the kilogram for different countries.
Countries have their own approaches to the definition, and there’s a slight variation in their measurements. Ideally, everyone would get the same result within a reasonable range, no matter what method they actually use to take the measurement.
The International Bureau of Weights and Measures (BIPM) currently distributes a consensus value that countries use to define the kilogram. Once we get to the point that various approaches to the definition don’t result in a significant lack of agreement in everyone’s results, BIPM will no longer have to distribute this value of the kilogram.
So NIST researcher Darine Haddad, along with Beatrice Rodiek from Germany’s national measurement science institute, known as the PTB, and their colleagues, are working together to figure out why the approaches are leading to different outcomes.
Right now, the U.S. has an exquisitely accurate weighing system, known as a Kibble balance, that uses quantum mechanics to precisely measure mass to get the official definition of the kilogram. Germany, on the other hand, uses a method that means, in simple terms, “counting atoms,” as Rodiek explains. It’s called the XRCD method.
Each approach gets to a slightly different definition of the kilogram. The variations in their results are tiny about 50 micrograms, which is the approximate mass of a grain of salt.
So, a 50-microgram difference in measuring a kilogram is a bit like weighing a pineapple on two scales, and your measurements only being off from each other by about a grain of salt! You’d think that’s close enough, but Haddad and Rodiek are trying to close that tiny gap.
“If you have two completely different approaches measuring the same object, and you get the same results, you can trust them,” Haddad says. “Once we get to that point, everyone can directly measure the unit of mass with the primary definition. These are small gaps we’re talking about here.”
## A Sphere of Measurement Influence
Rodiek recently traveled to NIST’s campus in Gaithersburg, Maryland, to run a series of measurements and tests to see how the two countries’ approaches may be able to get closer to the same results.
Rodiek brought along one of the roundest objects on Earth a perfectly manufactured silicon sphere.
First, Rodiek had to follow a special procedure to clean the sphere, so the researchers could avoid dust and other minute deposits on the mass of the sphere affecting the measurements. Then she very carefully walked it to the Kibble balance and placed it on a robot arm.
Haddad tried to run the Kibble balance experiment … but there were challenges along the way. The main issue was that the Kibble balance had not previously measured a spherical-shaped object and had to be adjusted to take that measurement.
Each time the researchers had to tweak the Kibble balance, the sphere had to be cleaned and gently put back in place. It was a tedious process that took up most of Rodiek’s weeklong visit to NIST in April.
Rodiek and the sphere have since returned to Germany. The sphere travels in its own protective case with Rodiek as her carry-on, never in checked luggage. The cargo is too precious to jostle.
“Sometimes it’s a little scary traveling with it in the airplane. You never know if we’re going to hit some turbulence, and it’s just sitting there. Luckily, it’s pretty robust, and we can always clean it,” Rodiek says.
Haddad and Rodiek are planning another series of experiments and, hopefully, will be able to make all needed measurements, without the technical difficulties.
“I’m really excited to come back. I’m planning another trip when we’ll have a bit more time to do these measurements,” Rodiek said.
## Why More Research on the Kilogram Is Needed
You may be wondering why all this effort is required to measure something accurately within a grain of salt. I wondered that, too. I asked Haddad: Why is this even necessary?
In addition to living up to the spirit of the redefinition of the kilogram that anyone in any lab should be able to define the kilogram as accurately as possible Haddad said we have no idea what we might learn from these experiments.
And as a scientist, that’s exciting for her. Researchers need to understand why this gap in the measurements is occurring, and who knows what we might learn from that.
“For researchers, science is very interesting for us, and we just want to understand why. Is there a scientific reason?” she said. “The knowledge we get from these experiments, discovering why the errors happen, can help us improve measurement science going forward.”
### Megan King
Megan King is a writer-editor at NIST and edits the Taking Measure blog. After graduating from John Carroll University, she began her career as a newspaper journalist, covering county fairs and school board meetings. Megan worked in various communications roles at the FBI for 13 years, including as a content manager and strategist for fbi.gov. Outside of work, Megan coaches beginner ice skaters, cheers on Pittsburgh sports teams, and knits.
## Related posts
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### NIST Research Is Setting the Standard to Help Buildings Withstand Tornadoes
Two NIST researchers have been recognized for their work to create new standards for buildings that can better withstand tornadoes.
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Cody
# Problem 2578. Sum of series IV
Solution 1222953
Submitted on 30 Jun 2017 by Nicolas BERNARD
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
n = 1; s_correct = 1; assert(isequal(sumOfSeriesIV(n),s_correct))
2 Pass
n = 2; s_correct = -8; assert(isequal(sumOfSeriesIV(n),s_correct))
3 Pass
n = 3; s_correct = 17; assert(isequal(sumOfSeriesIV(n),s_correct))
4 Pass
n = 4; s_correct = -32; assert(isequal(sumOfSeriesIV(n),s_correct))
5 Pass
n = 7; s_correct = 97; assert(isequal(sumOfSeriesIV(n),s_correct))
6 Pass
n = 12; s_correct = -288; assert(isequal(sumOfSeriesIV(n),s_correct))
7 Pass
n = 33; s_correct = 2177; assert(isequal(sumOfSeriesIV(n),s_correct))
8 Pass
n = 59; s_correct = 6961; assert(isequal(sumOfSeriesIV(n),s_correct))
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## 4682 Days From July 11, 2024
Want to figure out the date that is exactly four thousand six hundred eighty two days from Jul 11, 2024 without counting?
Your starting date is July 11, 2024 so that means that 4682 days later would be May 6, 2037.
You can check this by using the date difference calculator to measure the number of days from Jul 11, 2024 to May 6, 2037.
May 2037
• Sunday
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May 6, 2037 is a Wednesday. It is the 126th day of the year, and in the 19th week of the year (assuming each week starts on a Sunday), or the 2nd quarter of the year. There are 31 days in this month. 2037 is not a leap year, so there are 365 days in this year. The short form for this date used in the United States is 05/06/2037, and almost everywhere else in the world it's 06/05/2037.
### What if you only counted weekdays?
In some cases, you might want to skip weekends and count only the weekdays. This could be useful if you know you have a deadline based on a certain number of business days. If you are trying to see what day falls on the exact date difference of 4682 weekdays from Jul 11, 2024, you can count up each day skipping Saturdays and Sundays.
Start your calculation with Jul 11, 2024, which falls on a Thursday. Counting forward, the next day would be a Friday.
To get exactly four thousand six hundred eighty two weekdays from Jul 11, 2024, you actually need to count 6556 total days (including weekend days). That means that 4682 weekdays from Jul 11, 2024 would be June 23, 2042.
If you're counting business days, don't forget to adjust this date for any holidays.
June 2042
• Sunday
• Monday
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June 23, 2042 is a Monday. It is the 174th day of the year, and in the 174th week of the year (assuming each week starts on a Sunday), or the 2nd quarter of the year. There are 30 days in this month. 2042 is not a leap year, so there are 365 days in this year. The short form for this date used in the United States is 06/23/2042, and almost everywhere else in the world it's 23/06/2042.
### Enter the number of days and the exact date
Type in the number of days and the exact date to calculate from. If you want to find a previous date, you can enter a negative number to figure out the number of days before the specified date.
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# Thread: Find the volume of the solid Question
1. ## Find the volume of the solid Question
I do not know how to do these problem should some one help
Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.
y=x^6, y=1; about y=2
________
and
You wake up one morning, and find yourself wearing a toga and scarab ring. Always a logical person, you conclude that you must have become an Egyptian pharoah. You decide to honor yourself with a pyramid of your own design. You decide it should have height h=3900 and a square base with side s=1470. To impress your Egyptian subjects, find the volume of the pyramid.
___________
2. Originally Posted by killasnake
I do not know how to do these problem should some one help
You wake up one morning, and find yourself wearing a toga and scarab ring. Always a logical person, you conclude that you must have become an Egyptian pharoah. You decide to honor yourself with a pyramid of your own design. You decide it should have height h=3900 and a square base with side s=1470. To impress your Egyptian subjects, find the volume of the pyramid.
___________
A pyramid's volume is: $\displaystyle V=\frac{1}{3}Ah$
where: $\displaystyle A=\text{Area of the Base}$
And: $\displaystyle h=\text{The Height of the Pyramid}$
So the area of the base is obviously $\displaystyle A=s^2=1470^2=2160900$
So then find the volume
3. Thank you for the help Quick, Any ideas how to do the first one?
4. Hello, killasnake!
I must assume that you know about Volumes of Revolution . . .
Find the volume of the solid obtained by rotating the region bounded by:
. . $\displaystyle y = x^6,\;y=1$, about $\displaystyle y=2$
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$\displaystyle V \;=\;2 \times \int^1_0\left[(2-x^6)^2 - (2 - 1)^2\right]\,dx$
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Grandfather's Breakfast*
Home / Puzzle Help /
I was wondering if you can help me solve this puzzle:
Grandfather is a very hard-boiled customer. In fact, his eggs must be boiled for exactly 15 minutes, no more, no less. One day he asks you to prepare breakfast for him, and the only timepieces in the house are two hourglasses. The larger hourglass takes 11 minutes for all the sand to descend; the smaller takes 7 minutes. What do you do? ...Grandfather grows impatient...
Corrine
* This puzzle is derived from the old class of puzzles with measuring some value using several other constant values like different weights, barrels, cups, etc., and is one of several known versions with hourglasses.
Mini-Contest 3 - Grandfather's Breakfast is FINISHED
It was our third mini-contest. Now we have the first six correct answers and propose this contest's results and some details about the puzzle and its solution. With this Mini-Contest 3 is finished.
Contest Results
The winners are: 1. Jensen Lai. 2. Hüsnü Sincar. 3. KT - "ktendall". 4. Tamie D. 5. Jane Average. 6. Nicole Takahashi. 7. Matt Kaspar. We've got an almost correct solution from Sandra Payne, but she did say nothing about the moment when to put the eggs into the boiling water. See details at the solution page. Also we've got a correct solution from Matt Kaspar before our Mini-Contest 3 was finished, but accidentally his message was put into a folder with another contest, so when we posted the results and the names of the first six correct solvers his name wasn't included. Our apology. We add his name to the winners list.
Posted: February 2, 2002 | Last Updated: July 31, 2007
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Home | About Us | Link To Us | FAQ | Contact
# Set::Integer::Gapfillers 0.07
Date Added: November 10, 2010 | Visits: 699
Set::Integer::Gapfillers is a Perl module that can fill in the gaps between integer ranges. SYNOPSIS use Set::Integer::Gapfillers; \$gf = Set::Integer::Gapfillers->new( lower => -12, upper => 62, sets => [ [ 1, 17 ], # Note: Use comma, not [ 25, 42 ], # range operator (..) [ 44, 50 ], ], ); \$segments_needed_ref = \$gf->segments_needed(); \$gapfillers_ref = \$gf->gapfillers(); \$all_segments_ref = \$gf->all_segments(); Any of the three preceding output methods can also be called with an expand option: \$segments_needed_ref = \$gf->segments_needed( expand => 1 ); This Perl extension provides methods which may be useful in manipulating sets whose elements are consecutive integers. Suppose that you are provided with the following non-intersecting, non-overlapping sets of consecutive integers: { 1 .. 17 } { 25 .. 42 } { 44 .. 50 } Suppose further that you are provided with the following lower and upper bounds to a range of consecutive integers: lower: 12 upper: 62 Provide a set of sets which: when joined together, would form a set of consecutive integers from the lower to the upper bound, inclusive; and are derived from: the sets provided; proper subsets thereof; or newly generated sets which fill in the gaps below, in between or above the provided sets. Once a Set::Integer::Gapfillers object has been constructed, its segments_needed() method can be used to provide these results: { 12 .. 17 } # subset of 1st set provided { 18 .. 24 } # gap-filler set { 25 .. 42 } # 2nd set provided { 43 .. 43 } # gap-filler set # (which happens to consist of a single element) { 44 .. 50 } # 3rd set provided { 51 .. 62 } # gap-filler set for range above highest provided set Alternatively, you may only wish to examine the gap-filler sets. The gapfillers() method provides this set of sets. { 18 .. 24 } # gap-filler set { 43 .. 43 } # gap-filler set { 51 .. 62 } # gap-filler set And, as an additional alternative, you may wish to have your set of sets begin or end with all the values of a given provided set, rather than a proper subset thereof containing only those values needed to populate the desired range. In that case, use the all_segments() method. { 1 .. 17 } # 1st set provided { 18 .. 24 } # gap-filler set { 25 .. 42 } # 2nd set provided { 43 .. 43 } # gap-filler set # (which happens to consist of a single element) { 44 .. 50 } # 3rd set provided { 51 .. 62 } # gap-filler set for range above highest provided set The results returned by the all_segments() method differ from those returned by the segments_needed() method only at the lower or upper ends. If, as in the above example, the lower bound of the target range of integers falls inside a provided segment, the first set returned by all_segments() will be the entire first set provided; the first set returned by segments_needed() will be a proper subset of the first set provided, starting with the requested lower bound..
Requirements: No special requirements Platforms: Linux Keyword: Fill In, Libraries, Needed, Perl Module, Provided, Range, Segments, Setintegergapfillers, Sets Users rating: 0/10
License: Freeware Size: 17.41 KB
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05 May 2023
Tensors, Stress, and 2D Meshing: A Primer for Beginners
Skill-Lync
A tensor is a mathematical object that describes a geometric relationship between vectors, scalars, and other tensors. They describe physical quantities with both magnitude and direction, such as velocity, force, and stress.
In this blog, let us look at why stress is a tensor, what a plain stress condition and how it is related to 2D meshing.
Before jumping to the scientific definition, studying tensors in relation to vectors would be helpful. Vectors are those quantities that are described using both magnitude and direction. E.g., Velocity
This means that a vector can be represented using a line, which lies in a particular plane, as shown below:
Thus a particular vector is unidirectional and lies in a plane in space. On the other hand, a tensor is a 3-dimensional quantity, meaning it has a volumetric distribution in multiple directions. To visualize it better, consider a color dropped in a glass of water, as represented below.
Image 2: Colour in water is distributed in all directions
The color is distributed in all directions, although not uniformly. A tensor also has a similar distribution in space. In simple terms, we can visualize this 3-dimensional distribution as a bundle of vectors related to each other by a definite relationship. This will allow us to understand the scientific definition of a tensor better.
Thus tensor is defined as an object that describes a relationship between sets of vectors in a vector space.
This brings us to the next question: why is stress a tensor?
Image 3: Conceptual representation of chemical bonds among particles of a material
Stress is the internal resistance offered by an object to its deformation. This is under the internal chemical bonds among the atoms/molecules of the object. So if a particular atom/molecule is deformed, the forces of attraction from the neighboring atoms/molecules will exert a resistive force, basically, the stress.
But since the neighboring atoms are distributed in a 3-dimensional space, the forces of resistance are also distributed in a 3-dimensional space in a definite relationship with one another. Since force is a vector, the bundle of all these force vectors creates a tensor.
Image 4: 3D Stress distribution within a part
The above image shows how stress is distributed in a 3-dimensional space in the object.
Because stress is a tensor, there are different ways to represent it. It can be expressed as matrix, a Mohr’s circle, von mises stress, or as normal and shear stresses separately.
Image 5: Normal stress, shear stress, and stress matrix
One approximation commonly used to simplify the representation and calculation of stress is called the plane stress condition.
In plane stress conditions, the normal and shear stresses directed perpendicular to a particular plane are assumed to be zero. This means, in the plane stress condition, the stress is assumed to be distributed in only one plane. E.g. the stress in the z-direction is assumed to be zero. This makes many of the terms in the stress matrix zero, which makes calculations easier.
To imagine the plain stress condition, just assume that an object comprises thin slices arranged on top of another. If we consider any one of the slices, the stress in that slice will be distributed in the plane of that slice only, as the thickness is very small compared to other two dimensions. The stress in the direction normal to the plane of the slice will be zero.
Image 6: Pictorial representation of plane stress conditions in XY plane
But what is the practical significance of plane stress conditions?
In FEA, to accurately analyze the state of a material, it's important to create a finite element mesh on the model. This process involves dividing the part into smaller elements and defining their behavior. 2D meshing involves creating a mesh in a single plane. Many thin components mesh using 2D elements like quad or tria. These 2D elements work on the principle of plain stress, under which stress normal to the plane of the elements is assumed to be zero.
Many of the terms in the stress matrix become zero, making calculations easier.
Two, the number of nodes under calculation is lesser than that in 3D meshing. In FEA, nodes are the points at which the edges of the elements intersect with each other. In layman's terms, they can be understood as the vertices of the elements. In 2D meshing, a linear quad element has 4 nodes. But the corresponding 3D element is the linear hexahedral or brick element, which has 8 nodes, as shown in the next image:
As a result, the calculation time is reduced, and we get faster results with reasonable accuracy.
So if you think about it, one of the basic engineering principles has become a foundational principle of FEA. In the image below, a sheet metal part, meshed using 2D elements is shown:
Image 9: the 2D meshing of a sheet metal part
This explains how tensors are related to plain stress conditions and 2D meshing. Understanding these concepts is crucial for accurately analyzing the behavior of models and ensuring their safety and stability.
Author
Author
Skill-Lync
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Tetrahedral is a four- nodded solid element that can be generated through the tria element by creating a volume and also through the existing volume of the geometry. These elements are used where the geometry has high thickness and complexity. The image attached below is a representation of a Tetra element. The Tetra element will have 4 triangular faces with four nodes joining them together
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Mesh Sizing In Ansys Workbench
One of the most crucial processes in carrying out an accurate simulation using FEA is meshing. A mesh is composed of elements that have nodes—coordinate positions in space that might change depending on the element type—that symbolise the geometry's shape.
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Question
# Find the following products using the identity : (x+a)(x+b)=x2+(a+b)x+ab.1. (x+2)(x+3)2.(x-10)(x+8)3.(x+15)(x-6)4. (x+7)(x+7)
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Solution
## (x+a)(x+b)=x2+(a+b)x+ab(i) (x+2)(x+3)=x2+(2+3)x+2×3=x2+5x+6(ii) (x−10)(x+8)=x2+(−10+8)x+(−10)×8=x2−2x−80(iii) (x+15)(x−6)=x2+(15−6)x+15×(−6)=x2+9x−90(iv) (x+7)(x+7)=x2+(7+7)x+7×7=x2+14x+49.
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# Generating Mathematical Images?
• Mathematica
• Zurtex
In summary, the best way to generate high resolution mathematical images is by using software such as Maple, Mathematica, or LaTeX, which have drawing capabilities. Another option is to use Surf, a program specifically designed for drawing surfaces and other aspects of algebraic geometry. Guides and tutorials for these programs can be found through a simple Google search. Additionally, websites like Source Forge offer a variety of mathematics software that may be helpful for generating complex images like the Devil's Staircase.
#### Zurtex
Homework Helper
Hey, I was wondering what's the best way to generate mathematical images?
I was really hoping of creating a few really high resolution ones and popping down to the print shop to have them printed out. But I have a couple of problems:
1) How do I go about generating a mathematical image such as "The Devil's Stair Case" or some Fractal or some 2 dimensional surface in 3D space (well a representation of)? I need some tool where I can have the picture calculated in some arbitrary large resolution, preferably with such options as sub-pixel rendering or if it makes it look better Anti-Aliasing.
2) I'm not completely sure what DPI means I can print out at, I have 2 options of high resolution printing, 2400 DPI or 5200 DPI, does anyone know what this would mean in terms of something like "2560 x 1600 on an A4 piece of paper"?
Any help would be greatly appreciated.
some solutions depending on the situation:
maple, mathematica, latex has drawing capabilities, surf, xfig.
I imagine the first two are good for the problem of the Devil's Staircase. Surf is for drawing surfaces and other aspects of real algebraic geometry, and very fancy it is too.
here is a defunct page (the CSS is wrong which is why it lookts terrible), but the picture there is done with surf and some experiment with fibonacci numbers as exponents.
http://www.maths.bris.ac.uk/~maxmg/maths/introductory/algebraicart.html [Broken]
Last edited by a moderator:
Thanks .
Would you mind providing a link to surf? I can't seem to find it very easily
Also if you could link to some guides, that would be great. I'm having hard time finding anything that constructive to people who don't have a good working knowledge in the first place.
Google for 'surf algebraic geometry'. (the description I gave of it...). It is the first hit: surf.sourceforge.net
source forge has whole sections devoted to mathematics software; there may be solutions to your initial problem of the devil's staircase.
Last edited:
Thank you, sorry I tried searching similar phrases but didn't get very far.
## 1. How are mathematical images generated?
Mathematical images are generated using mathematical equations and algorithms. These equations and algorithms are programmed into a computer software, which then uses them to produce the desired image.
## 2. What is the purpose of generating mathematical images?
The purpose of generating mathematical images is to visually represent complex mathematical concepts and ideas. It allows for a better understanding and visualization of abstract concepts, and can also be used for data visualization and artistic purposes.
## 3. What are some common tools and software used for generating mathematical images?
There are various tools and software available for generating mathematical images, such as MATLAB, Mathematica, and GeoGebra. These programs have built-in functions and features specifically designed for creating mathematical images.
## 4. Is it possible to create 3D mathematical images?
Yes, it is possible to create 3D mathematical images using specialized software and techniques. These images can represent complex 3D mathematical objects and can be manipulated and viewed from different angles.
## 5. Can mathematical images be used in real-world applications?
Yes, mathematical images have various real-world applications, such as in engineering, architecture, and computer graphics. They can also be used for educational purposes and in scientific research to visualize and analyze data.
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# Changes
## Math4Team/RIT/Projects
, 23:29, 29 April 2018
Published / Complete Projects: added SugarSnake
{{TOCright}}
== [[Colour_The_Part|Colour The Part]] Active Projects ==Incorporating math and color theory onto a platformer, create a fun and educational activity targeted at 4th grade Math students. The activity uses elements of platform games with obstacles and puzzle solving to teach addition and multiplication of fractions while keeping the player entertained.
== = [[Image:PyCut_logo.png|24px]] [[Fortune_HunterActivities/PyCut| Mathematical Adventure: Fortune HunterPyCut]] ==={{:Activities/PyCut/status}}
Fortune Hunter draws on teaching children of a global scale fourth grade mathematics through the guise of a dungeon styled adventure game=== [[Image:Fractionator_logo. Players will be able to explore dungeons and fight fearsome battles with various monsters, each requiring special attacks pertaining to unique mathematical concept to defeat. The player controls a protagonist that must progress through maze-like dungeons, solve puzzles png|24px]] [[Activities/Fractionator|Fractionator]] ==={{:Activities/Fractionator/ problems, and defeat enemies in a two dimensional world.status}}
===[[Math4Team/RIT/Projects/Math_Maze|Math MazeAngleGators]]==[[User:kdh7733|Kevin Hockey]] and [[User:TS1593|Tom Sekovski]] are making maze style game (built upon the pre-existing game [[http://wiki.laptop.org/go/Maze Maze]]) where at each decision moment the player has to answer a question. Some questions include: adding/subtracting numbers, factoring, word and fraction problems=
=== [[Image:blocku.png|75px]] [[Blocku|Blocku]] ==={{:Blocku/status}} ===[[School ServerActivities/RITFileShare|FileShare]] ==={{:Activities/FileShare/status}} === [[Fractionauts|schoolserverFractionauts]] ==={{:Fractionauts/status}} === [[Image:Lemon_Icon.rit.edusvg|24px]] [[Lemonade Stand]] ==={{:Lemonade_Stand/status}} === [[Activities/MathHurdler|Math Hurdler]] ==={{:Activities/MathHurdler/status}} ===[[Math_island|Math Island]]==={{:Math_island/status}} === [[Fortune_Hunter| Mathematical Adventure: Fortune Hunter]] ==={{:Fortune_Hunter/status}} === [[Planetary|Planetary]] ==={{:Planetary/status}} === [[Quilt]] ==={{:Quilt/status}} === [[Sash]] ==={{:Sash/status}} === [[Activities/SugarSnake|SugarSnake]] ==={{:Activities/SugarSnake/status}} == Inactive Projects == === [[Addboard| Addboard]]=== The XO school Server, or XS, is one of the products of the OLPC project, designed to complement the XO laptop. It is a Linux=== [[Flash Card Game -based OS (a FedoraAssimilate|Assimilate]] ==={{:Flash_Card_Game_-based distribution) engineered to be installed on generic low-end servers_Assimilate/status}} === [[Colour_The_Part|Colour The Part]] ==={{:Colour_The_Part/status}} === [[Ethos| Ethos]] ==={{:Ethos/status}} === [[Math4Team/RIT/Projects/Fun Towers|Fun Towers]] ==={{:Math4Team/RIT/Projects/Fun_Towers/status}} === [[GeoMunchers]] === ===[[Get24|Get24]]==={{:Get24/status}} === [[Matter. py]] ===When we deploy one laptop per child, we must also provide additional infrastructure extending the capabilities of the laptops. While the laptops are self-sufficient for many learning activities, other activities and services depend on the School Server providing connectivity, shared resources and services. Services, tools and activities running on the School Server allow asynchronous interaction, can use larger storage capacity, and take advantage of the processing power of the XS{{:Matter.py/status}}
===[[Teacher's ToolsMath4Team/RIT/Projects/Math_Maze|Math Maze]]===[[User{{:acj3840|Alex Jones]] is planning on creating a tool for teachers to help evaluate how well their students have learned certain material. It will be similar to Assimilate but help the teachers learn the weak points of their students' knowledge.Math4Team/RIT/Projects/Math_Maze/status}}
===[[Flash Card Game - AssimilateMiniGameMayhem]]==This project is one of [[Math4Team/RIT/Projects | many small projects]] focused on developing 4th grade math games which will run on the OLPC XO. We are [[Math4Team/RIT/Alumni/Spring-2009| RIT students]] who are working on this with the help of other developers who become interested in our projects.The focus of this project is to turn an existing game ([[olpc:Assimilate|Assimilate]]) into a math based flash card game which will hopefully be included on future XO's.=
Group members:=== [[User:Dbj4366Math4Team/RIT/Projects/Muthris|Dennis JalbertMuthris]], [[User==={{:Jck1089|James Kolb]], [[User:bjr9081|Brendan Reen]]Math4Team/RIT/Projects/Muthris/status}}
===[[Math4Team/RIT/Projects/Fun TowersFile:SunGuy.png|Fun Towers35px]] [[MyWeatherStation]]===Fun Towers is a pre-existing game that can be found online in several version (http{{:MyWeatherStation//www.funnytowers.com/ is one example) that has been ported to the XO, written in Squeak. Our team is modifying this purely numerical/card based game into one that can be used as a teaching tool as part of the 4th grade math project.Our initial goals in modifying the pre-existing game remain relatively simple and achievable, and our goal is to produce verifiable results that can be used to point to the very preliminary success of the math4 program, while more in-depth projects are still in development.The game itself is simple, users are given a card and with it are able to remove from one of 3 pyramids of cards a card that is one greater or one lower in value. This card that has been removed is the users new card, and any cards that were covered by the removed card are now in play.status}}
Group members are === [[User:EricMallonActivities/NumberMunchersXO|Eric Mallon,]] [[User:eldrac|Tyler Bragdon,]] [[User:Cdaniels29|Chris DanielsNumber Munchers XO]]===
===[[Image:Lemon_IconPacmath-logo.svgpng|24px75px]][[Lemonade StandPacMath|PacMath]]===Lemonade stand (or more likely, *Insert produce here* stand) is a collaboration project at RIT. It is designed to test children on fractions, working with money, estimation, and other math topics. While our time restraints limit the extent to which we can implement features, the current goal is a feature complete program even if lacking in graphics. We plan on having a system based on buying and selling commodities and an eventual season based economy. {{:PacMath/status}}
The largest === [[Point-and most complex task of the project will be the introduction of a basic AI to handle the economy, the use of graphics, and localization. We are planning on introducing the game with a generic currency model while trying to find a commodity that either works worldwide, or is easy to replace for regional types. At the very least, we plan on having a feature complete text model that can be finished by other interested parties.-Click-Arthur]] ==={{:Arthur/status}}
The RIT students currently working on the project are === [[User:Epsilon748|Anthony KingPop_Quiz]], [[User==={{:Qalthos|Nathaniel Case]], [[User:Jsang1|Jonathan Sanger]], [[User:Mdd8919 | Mitchell DeMarco]], [[User:sss1406|Steven Schoenfeld]], and [[User:Echo35|Anthony Lubriani]].Pop_Quiz/status}}
===[[Math4Team/RIT/Projects/Muthris|MuthrisProduce Puzzle]]===Muthris is a math themed, Tetris-based game inspired by Cuyo. Players control falling blocks which must be grouped in certain math related ways in-order to clear that grouping from the board. The level is lost when the board fills up with blocks. Players learn math skills by fun repetition of simple mathematical problems and the grouping of sets of numbers. Levels are abstracted away from the core game. This allows one to simple drop in new levels and learn different mathematical concepts.{{:Produce_Puzzle/status}}
Group Members: === [[User:dpk3062|DougPyCaveExplorer]]==={{:PyCaveExplorer/status}}
===[[Produce PuzzlePython_Math|Python Math]]===The object of the game is to solve a system of equations with unknowns represented by fruits. The player is given the column sums and row sums, and from there he must determine the value of each fruit. The game difficulty can be changed, and it ranges from solving 3x3 fruit equations, all the way up to 9x9 fruit grids.{{:Python_Math/status}}
We, [[User:Classclownfish | Abbi Honeycutt]] and [[User:Nikeunltd| Kennedy Kong]] from Rochester Institute of Technology, has taken over this project. Previous creators were Matthew Michihara, Elizabeth Deng, and Aaron Macris from University of Southern California during their "Code for a cause OLPC Hack-a-thon".===[[Math4Team/RIT/Projects/Question Support APIQuestion_Support_API|Question Support API]]===The purpose of the Question Support API is to provide a unified method for Activities to access standardized format question files. The API currently supports the use of multiple choice questions with a single correct answer and no partial credit. Currently, the API only reports questions in a plain text format from Moodle format question files.{{:Math4Team/RIT/Projects/Question_Support_API/status}}
Group members are ===[[User:EnimihilRobobuddy|Greg StevensRobobuddy]], [[User:Jfinney|Jameson Finney]], [[User==={{:Bbl5660|Brian Long]]Robobuddy/status}}
===[[Teacher ReportingSchool Server/RIT|schoolserver.rit.edu]]===Our goal The XO school Server, or XS, is one of the products of the OLPC project, designed to allow results and / or grades, from student activities complement the XO laptop. It is a Linux-based OS (a Fedora-based distribution) engineered to be readily available to teachersinstalled on generic low-end servers. When we deploy one laptop per child, we must also provide additional infrastructure extending the capabilities of the laptops. While the laptops are self- Based sufficient for many learning activities, other activities and services depend on their wants the School Server providing connectivity, shared resources and needsservices. Services,teachers can then generate custom reports tools and activities running on a class or Individual. Teachers Have the ability to determine what types School Server allow asynchronous interaction, can use larger storage capacity, and take advantage of problems students are finding most difficult. - Additionally we envision logic in the module which analyses student activity and checks for a yes || no interpretation processing power of each student's understanding of Curriculum Standardsthe XS.
=== [[Image:Pacmath-logo.png|75px]] [[PacMath|PacMathTeacher Reporting]] ===Through the classic arcade style game known as PacMan we will incorporate fourth grade level math.{{:Teacher Reporting/status}}
Group Members: [[user:yah3133|Yasser Hernandez]] , [[user:Taylor2412| Taylor Plimpton]], === [[user:Arendon | Abel RendonTeacher's Tools]] , [[user==={{:dxr4305 | Dennis Rodriguez]]Teacher's_Tools/status}}
== [[Image:blocku.png|75px]] [[Blocku|Blocku]] ==
A math based sudoku style puzzle game. The player is given an answer and they have to complete the puzzle by matching the sides of the blocks together that, following the correct operator (+, -, *, /) will give them the answer.
Group Members: [[user:mdemayo|Mark DeMayo]] , [[user:Coolestdude1|Ariel Zamparini]], [[user:iogburu|Ihudiya Ogburu]]
==[[Math4Team/RIT/Tips & Tricks|Tips & Tricks]]==
==[[Math4Team/RIT/Complaints & Problems|Complaints & Problems]]==
[[Category:Idea]]
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# Y-combinator in Erlang
Joe Armstrong <>
Wed Jan 8 15:57:48 CET 2003
```> > Now how the average Håkan might have written it is a entirely
> > different matter :-)
>
> As a named function. Whatever you gain in expressive power with funs
> you risk losing by people using them in a way that is hard to
> understand.
>
Almost anything can be abused if you try hard enough.
What does:
foo() ->
Z = make_ref(),
bar(self() ! {Z, left}, self() ! {Z, right}),
{Z, I} ->
{Z, J} ->
I
end
end.
bar(_, _) -> void.
do?
> Classic example. Consider the higher-order function (standard math
> notation):
>
> twice(F)(X) = F(F(X))
>
> Let double(X)=X+X and figure out what F(1) is, where
> F=twice(twice(twice(twice(double)))).
>
> I'm just arguing that sometimes power tools are not good things.
>
I agree entirely - you must constantly strive to write "mind
boggling simple code"
One good way to to this is to mentally repeat the phrase:
"I must write as inefficient code as possible" one thousand times
before breakfast ever day :-)
K & P still applies: (I paraphrase)
1) make it right
2) make it fast once you've got it right
3) keep it right as you make it faster
IMHO the fast majority of programmers (myself included) don't get past 1)
The well tried method used by many large industrial concerns is:
1a) Ship it first
1b) Make it fast
2) Debug it
3a) Understand the problem
3b) Discontinue the product
1a and 1b are done concurrently as are 3a and 3b
Often they never get to 3a - sometimes they use the accelerated design
method of 1b followed 3b.
This method is not recommended.
/Joe
> -- Hakan
>
```
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Need a perfect paper? Place your first order and save 10% with this code: DOER10FF
# What is Number theory
The simplest definition of number theory is a branch of mathematics that deals with the study of real numbers or integers. Unlike other branches of math, concepts and problems in the number theorem can be understood by most people who do not have advanced mathematical capabilities. This is not to say that there aren’t theorems, and concepts do not require sophisticated mathematical backgrounds.
Looking at the number theory as a branch of pure mathematics, the study does not appear to have real-life applications in the real world. This perception was thrown out of the window when the concepts started being used in digital computers and programming where numerical problems were solved using number theorem and the solutions used in real-life situations using digital technology.
Computer inventions and advancement have a working symbiotic relationship, whereas number theory phenomenon was the basis for computer software programming, computer technology has allowed mathematical theorists to make significant steps by enabling them to be able to work with large numbers providing them with the means to test problems and concepts which they could otherwise not be able to without digital technology. The following subheadings are examples of the categories that will provide a basic introduction for beginners.
## Elementary
This is a branch where primary methods like high school algebra and geometry are used to solve equations with integers. The classification of Pythagorean triples is a good example of a problem which is solved using elementary theory. The Pythagorean triple is the theorem that relates positive integers a, b, and c in such a way that if a right triangle exists with legs a and b, and a hypotenuse c, then the sides will be governed by the relationship that has to satisfy that;
a2 + b2 = c2
## Algebraic
Algebraic number theory is the branch that deals with algebraic numbers. This branch was developed as a set of logical tools that would be used to solve problems in elementary. The most common being Diophantine equations. These are equations that include integers and rational numbers that can be lifted from natural numbers and expressed as algebraic expressions.
Algebraic number theory also involves the study of the roots of polynomials with rational and integral coefficients. It envisions numbers lying in algebraic structures that have properties that are similar to those of integers. The most critical conjecture in an algebraic number system is Fermat’s conjecture that was eventually proved in the 1990s.
## Analytic Number Theory
This is the branch that uses real and complex analysis to investigate the common properties of integers and prime numbers deeply. There are no defined preconditions of the study of analytic number theory, but it majorly focuses on the analytic activity of determining and estimating objects and handling error terms. Some underlying assumptions within the analytic include the Riemann zeta function, arithmetic progression and functions like the Dirichlet functions and basic distributional principles and functions.
## Geometric Number Theory
Geometric closely related to algebraic number theory. It involves the study of polynomials over fields of number-theoretic interest. Theories like the Diophantine geometry of numbers is part of these geometric number theories.
Other theories include Galois representations, Langland’s programs, the Artine-Verdier theory of duality and many other complex phenomena that represent integers and real numbers using algebraic and geometric expressions to simplify their logical solutions.
## Probabilistic
This is the use of probability techniques to study the behavior and relationship of integers and L-functions. Number theory techniques have also proven to be effective in resolving problems in discrete probability theory.
These two schools of thought connect in the study of when the probability is used to create models for integers when probability techniques are used to prove results, when number theory problems lead to questions of probability and when number theory methods are used to determine probability results definitively. Probability methods and phenomena are dependent on each other.
In number theory, the relationship between the additive and multiplicative structures for integers has fascinated mathematicians, and they have made number theory into a wide field of mathematical research. The phenomena involved are quite complicated, but if one has the interest of pursuing deeper into the number theory he can quickly get help by going through the studies doe and tabulated by mathematicians like Gauss.
These studies are found in scholarly articles all over the internet. You can also benefit from the works done by Euclid and Pythagoras. They are known to have established relationships that are fundamental to the understanding.
Are you finding it hard to grasp this concept? If yes, do not hesitate to contact us for fast and affordable number theory help.
.
Dont compromise on quality
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# [R-sig-Geo] Alternate statistical test to linear regression?
Greg Snow 538280 @end|ng |rom gm@||@com
Wed Oct 23 18:59:59 CEST 2019
```Note that the normality assumptions are about the residuals (or about
y conditional on x), not on the x variable(s) or all of y
(non-conditional). If x is highly skewed and the residuals are normal
then diagnostics just on y will also show skewness (if there is a
relationship between x and y).
Also, the normality assumptions are about the tests and confidence
intervals, the least squares fit is legitimate (but possibly not the
most interesting fit) whether the residuals are normal or not. The
Central Limit Theorem also applies in regression, so if the residuals
are non-normal, but you have a large sample size then the tests and
intervals will still be approximately correct (with the quality of the
approximation depending on the degree of non-normality and sample
size).
There are many alternative tools. There is a task view on CRAN for
Robust Statistical Methods that gives summaries of many packages and
tools for robust regression (and other things as well) which does not
depend on the normality assumptions.
On Wed, Oct 23, 2019 at 9:21 AM rain1290--- via R-sig-Geo
<r-sig-geo using r-project.org> wrote:
>
> Greetings,
> I am testing to see if linear relationships exist between my x and y variables. I conducted various diagnoses in R to test for normality of the x variable data by using qqnorm, qqline and histograms that show the distribution of the data. If the data is shown to be normally distributed in either normal quantile plots or in the histograms (i.e. a bell curve-shaped distribution), I would assume normality and apply the linear regression model, using "lm". However, in some cases, my distributions do not satisfy the normality criteria, and so I feel that using the linear regression model, in those cases, would not be appropriate. For that reason, would you be able to suggest an alternate test to the linear regression model in R? Maybe a non-parametric counterpart to it?
> Thank you, and any help would be greatly appreciated!
> [[alternative HTML version deleted]]
>
> _______________________________________________
> R-sig-Geo mailing list
> R-sig-Geo using r-project.org
> https://stat.ethz.ch/mailman/listinfo/r-sig-geo
--
Gregory (Greg) L. Snow Ph.D.
538280 using gmail.com
```
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# How tall is 174 cm in feet?
CM To Swap FEET
Name Unit
Feet 0 Feet
Yards 0 Yards
Acre 0 Acre
Miles 0 Mile
Meters 0 Meters
Kilometers 0 Km
Inches 0 Inches
Millimeters 0 mm
## How tall is 174 cm in feet:
174 cm is equal to approximately 5 feet 8.5 inches. To convert from centimeters to feet and inches, you can use the following conversion:
1 inch = 2.54 cm 1 foot = 12 inches
So, to convert from centimeters to feet and inches, you can divide the number of centimeters by 2.54 to get the number of inches, then divide the number of inches by 12 to get the number of feet. The remainder will be the number of inches.
For example, to convert 174 cm to feet and inches:
174 cm / 2.54 = 68.5 inches 68.5 inches / 12 = 5.708 feet
So 174 cm is equal to approximately 5 feet 8.5 inches.
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### CMOS GATES BASIC AND TUTORIALS
WHAT ARE CMOS GATES? HOW CMOS GATES WORK?
CMOS gates are based on simple modifications to the CMOS inverter.
Figure 8.18(a) and Figure 8.18(b) show that the CMOS NOR and NAND gates are essentially CMOS inverters in which the load and driving transistor are replaced by series or parallel combinations (as appropriate) of PMOS andNMOStransistors, respectively.
Suppose the NOR gate of Fig. 8.18(a) is to have the same VDD and Vinv as the CMOS inverter of Fig. 8.17(a), then the equivalent Zpu and Zpd for the NOR gate should equal those for the inverter.
Since only one of the parallel pull-down transistors needs be on in the NOR to ensure VO = 0V, ZI = Zpd = 1/2 , as for the inverter. For the series load, however, ZL = 1/10 to give equivalent Zpu = 1/5 .
If the NAND gate of Fig. 8.18(b) is to have the same Vinv as the said inverter, similar arguments lead to ZI = 1/4 and ZL = 1/5 for the NAND.
Thus, KR = 0.4 for the inverter, 0.2 for the NOR, and 0.8 (closer to unity) for NAND. Hence, NAND is the standard gate in CMOS.
Another way of putting this is that for the given Z values, if the channel length L is constant, then the widths of the loads for the inverter, NOR, and NAND are in the ratio 1:2:1.
Thus, the NOR requires more chip area, and this larger area requirement increases with the number of inputs.
### CMOS INVERTERS BASIC AND TUTORIALS
WHAT ARE CMOS INVERTERS? APPLICATION OF CMOS INVERTERS
As shown in Fig. 8.17(a), the CMOS inverter consists of an enhancement NMOS as the driving transistor, and a complementary enhancement PMOS load transistor. The driving transistor is off when Vin is low, and the load transistor is off when Vin is high.
Thus, one of the two series transistors is always off (equivalently, drain current and power dissipation are zero) except during switching,when both transistors are momentarily on. The resulting low-power dissipation is an important CMOS advantage and makes it an attractive alternative in VLSI design.
NMOS circuits are ratioed in the sense that the pull up never turns off, and VOL is determined by the inverter ratio. CMOS is ratioless in this sense, since VOL is always the negative rail. If one desires equal sourcing and sinking currents, however, the pull-up device must be wider than the pull-down device by the ratio of the electron-to-hole mobilities, typically about 2.5 to 1.
This also gives a symmetrical voltage transfer curve, with the voltage at which Vin = VO having a value of VDD/2. This voltage is referred to as the inverter voltage Vinv.
The voltage transfer for the CMOS inverter is shown in Fig. 8.17(b). Note that the voltage transfer characteristic approaches that of the ideal logic inverter. These characteristics are best obtained with computer circuit simulation programs.
As with the depletion load NMOS inverter, useful insights may be gained by performing an analytical solution. The analysis proceeds as previously described for the depletion load NMOS inverter.
Note that the VTC of Fig. 8.17(b) has been divided into regions as in Fig. 8.15(a). In each region, the appropriate expressions for the load and driving transistor drain currents are equated so that VO can be computed for any given Vin.
To find VI L and VI H, the condition that dVO/dVin = −1 at such critical voltages is applied to the drain current equation.Note that the drain current equations for the PMOS are the same as for NMOS, except for reverse voltage polarities for the PMOS.
### PIEZOELECTRIC EFFECT DEFINITION BASIC AND TUTORIALS
WHAT IS PIEZOELECTRIC EFFECT?
Basic Physical Principles
Materials that are used for crystal oscillator applications exhibit the piezoelectric effect. This phenomenon produces an electric field within a material when a mechanical force is applied.
Conversely, if an electrical drive signal is applied to the crystal, a mechanical vibration results. If the driving signal is periodic in nature, then the resulting vibration will also be periodic. Very accurate electric signal frequencies can be produced by certain materialswith the appropriate shape, electrode geometry, and ambient conditions.
For this reason, crystal resonators are used in applications such as television, cellular radio communications, and electronic test equipment where accurate synthesis of signal frequencies is required throughout.
The relationship between the electric field, electric displacement, electric polarization, and mechanical stress and strain is given by what are called constituitive relations. The electrical quantities are related in the following manner:
D = e0 E + P
where D is the electrical displacement, E is the electric field and P is the electric polarization. The term e 0 is the free-space permittivity. The polarization depends on the applied electric field. In a piezoelectric material an electric polarization can also result from an applied stress or strain.
This is called the direct piezoelectric effect. Therefore the polarization is proportional to the stress (T) or strain (S) that is P = dT or P = eS
In the converse piezoelectric effect, the stress or strain forces are generated by the electric field applied to the crystal. Thus, we have S = dE or T = eE
The constants of proportionality d and e are called piezoelectric stress and strain coefficients. The stress and strain forces are represented by matrix quantities, and the coefficients are tensor quantities. A tensor mathematically represents the fact that the polarization can depend on the stress or strain in more than one direction.
This is also true for the relationship between the stress or strain and the electric field. Many other physical properties in crystals also exhibit this nature, which is called anisotropy. Thus when a property is anisotropic, its value depends on the direction of orientation in the crystal.
For the direct piezoelectric effect, the total polarization effect is the sum of these two contributions, an applied electric field and applied mechanical force.
Based on the relationship between the electric displacement and the electric polarization it is then possible to write equations that relate the displacement D to the applied stress or strain. Electric displacement is the quantity that is preferred in experiment and engineering.
### SEMICONDUCTORS DEFINITION BASICS AND TUTORIALS
WHAT IS A SEMICONDUCTOR? INFORMATION ABOUT SEMICONDUCTORS
Semiconductors are a categoryofmaterialswith an electrical conductivity that is between that of conductors and insulators. Good conductors, which are all metals, have electrical resistivities down in the range of 10^−6 OHM -cm.
Insulators have electrical resistivities that are up in the range from 10^6 to as much as about 10^12 OHM - cm. Semiconductors have resistivities that are generally in the range of 10^−4–10^4 OHM -cm.
The resistivity of a semiconductor is strongly influenced by impurities, called dopants, that are purposely added to the material to change the electronic characteristics.
We will first consider the case of the pure, or intrinsic semiconductor. As a result of the thermal energy present in the material, electrons can break loose from covalent bonds and become free electrons able to move through the solid and contribute to the electrical conductivity.
The covalent bonds left behind have an electron vacancy called a hole. Electrons from neighboring covalent bonds can easily move into an adjacent bond with an electron vacancy, or hole, and thus the hold can move from one covalent bond to an adjacent bond.
As this process continues, we can say that the hole is moving through the material. These holes act as if they have a positive charge equal in magnitude to the electron charge, and they can also contribute to the electrical conductivity.
Thus, in a semiconductor there are two types of mobile electrical charge carriers that can contribute to the electrical conductivity, the free electrons and the holes. Since the electrons and holes are generated in equal numbers, and recombine in equal numbers, the free electron and hole populations are equal.
In the extrinsic or doped semiconductor, impurities are purposely added to modify the electronic characteristics. In the case of silicon, every silicon atom shares its four valence electrons with each of its four nearest neighbors in covalent bonds.
If an impurity or dopant atom with a valency of five, such as phosphorus, is substituted for silicon, four of the five valence electrons of the dopant atom will be held in covalent bonds. The extra, or fifth electron will not be in a covalent bond, and is loosely held.
At room temperature, almost all of these extra electronswill have broken loose fromtheir parent atoms, and become free electrons. These pentavalent dopants thus donate free electrons to the semiconductor and are called donors.
These donated electrons upset the balance between the electron and hole populations, so there are now more electrons than holes. This is now called an N-type semiconductor, in which the electrons are the majority carriers, and holes are the minority carriers.
In an N-type semiconductor the free electron concentration is generally many orders of magnitude larger than the hole concentration. If an impurity or dopant atom with a valency of three, such as boron, is substituted for silicon, three of the four valence electrons of the dopant atom will be held in covalent bonds.
One of the covalent bonds will be missing an electron. An electron from a neighboring silicon-to-silicon covalent bond, however, can easily jump into this electron vacancy, thereby creating a vacancy, or hole, in the silicon-to-silicon covalent bond. Thus, these trivalent dopants accept free electrons, thereby generating holes, and are called acceptors.
These additional holes upset the balance between the electron and hole populations, and so there are now more holes than electrons. This is called a P-type semiconductor, in which the holes are the majority carriers, and the electrons are the minority carriers. In a P-type semiconductor the hole concentration is generally many orders of magnitude larger than the electron concentration.
The transition between the two sides is called the PN junction. As a result of the concentration difference of the free electrons and holes there will be an initial flow of these charge carriers across the junction, which will result in the N-type side attaining a net positive charge with respect to the P-type side.
This results in the formation of an electric potential hill or barrier at the junction. Under equilibrium conditions the height of this potential hill, called the contact potential is such that the flow of the majority carrier holes from the P-type side up the hill to the N-type side is reduced to the extent that it becomes equal to the flow of the minority carrier holes from the N-type side down the hill to the P-type side.
Similarly, the flow of the majority carrier free electrons fromthe N-type side is reduced to the extent that it becomes equal to the flow of the minority carrier electrons from the P-type side. Thus, the net current flow across the junction under equilibrium conditions is zero.
Radars can be classified by frequency band, use, or platform, for example, ground based, shipborne, airborne, or spaceborne. Radars generally operate in the microwave regime although HF over-the-horizon (OTH) radars such as JINDALEE, OTHB, and ROTHR use similar principles in bouncing signals from the ionosphere to achieve long-range coverage.
Radars are often denoted by the letter band of operation, for example, L-band (1–2 GHz), S-band (2–4 GHz), C-band (4–8 GHz), and X-band (8–12 GHz). Some classifications of radar are based on propagation mode (e.g., monostatic, bistatic, OTH, underground) or on scan method (mechanical, electronic, multibeam).
Other classifications of radar are based on the waveform and processing, for example, pulse Doppler (PD), continuous wave (CW), FM/CW, synthetic aperture radar (SAR) or impulse (wideband video).
Radars are often classified by their use: weather radar, police speed detection, navigation, precision approach radar, airport surveillance and air route surveillance, radio astronomy, fire control and weapon direction, terrain mapping and avoidance, missile fuzing, missile seeker, foliage penetration, subsurface or ground penetrating, acquisition, orbital debris, range instrumentation, imaging (e.g., SAR/ISAR), etc.
Search (or surveillance) radars are concerned with detection of targets out to long range and lowelevation angles to allow adequate warning on pop-up low-flying targets (e.g., sea skimmers). Since the search radar is more concerned with detection (i.e., presence or absence of targets) and can accommodate cruder accuracy in estimating target parameters such as azimuth angle, elevation angle, and range, search radars tend to have poorer range and angle accuracy than tracking radars.
The frequency tends to be lower than track radars since RF power and antenna aperture are less expensive and frequency stability is better. Broad beams (e.g., fan beam) allow faster search of the volume. To first order, the radar search performance is driven by the power-aperture product (PA) to search the volume with a given probability of detection (PD) in a specified frame time.
PA actually varies slightly in that to maintain a fixed false alarm rate per scan, more beam positions offer more opportunities for false alarms and, hence, the detection threshold must be raised, which increases the power to achieve the specified PD.With a phased-array antenna (i.e., electronically scanned beam), the probability of false alarm can be optimized by setting a high false alarm in the search beam and using a verify beam with higher threshold to confirm whether a search detection was an actual target or just a false alarm.
The lower threshold in search allows less search power with some fraction of beams requiring the extra verify beams. The net effect on total required transmit power can be a reduction using this optimization technique.
Search radars tend to use a fan beam or stacked receive beams to reduce the number of beam positions allowing more time in the beam for coherent processing to reduce clutter. Fill pulses are sometimes used to allow good clutter cancellation on second- or higher time-around clutter returns.
Track radars tend to operate at higher frequency and have better accuracy, that is, narrower beams and high range resolution. Simple radars track a single target with an early–late range tracker, Doppler speed gate, and conical scan or sequential lobing.
More advanced angle trackers use monopulse or conical scan on receive only (COSRO) to deny inverse modulation by repeater jammers. The multifunction phased-array radar can be programmed to conduct searches with track beams assigned to individual detected targets.
The tracks are maintained in track files. If time occupancy becomes a problem, the track pulses can be machine gunned out at the targets in range order, and on receive they are gathered in one after the other since the track window on each target is quite small. In mechanically rotated systems, track is often a part of search, for example, track-while-scan (TWS).
A plot extractor clusters the primitive returns in range Doppler angle from a given target to produce a single plot. The plots are associated with the track files using scan-to-scan correlation gates. The number of targets that can be handled in a TWS system is limited by data processing rather than track power.
### ANTENNA BANDWIDTH BASIC AND TUTORIALS
WHAT IS ANTENNA BANDWIDTH? THE PURPOSE OF ANTENNA BANDWIDTH?
Antennas can find use in systems that require narrow or large bandwidths depending on the intended application. Bandwidth is a measure of the frequency range over which a parameter, such as impedance, remains within a given tolerance. Dipoles, for example, by their nature are very narrow band.
For narrow-band antennas, the percent bandwidth can be written as:
(fu - fl)/fc x 100
where
fL = lowest useable frequency
fU = highest useable frequency
fC = center design frequency
In the case of a broadband antenna it is more convenient to express bandwidth as
fU/fL
One can arbitrarily define an antenna to be broadband if the impedance, for instance, does not change
significantly over one octave ( fU / fL = 2).
The design of a broadband antenna relies in part on the concept of a frequency-independent antenna. This is an idealized concept, but understanding of the theory can lead to practical applications.
Broadband antennas are of the helical, biconical, spiral, and log-periodic types. Frequency independent antenna concepts are discussed later in this chapter.
Some newer concepts employing the idea of fractals are also discussed for a new class of wideband antennas.
Narrow-band antennas can be made to operate over several frequency bands by adding resonant circuits in series with the antenna wire. Such traps allow a dipole to be used at several spot frequencies, but the dipole still has a narrow band around the central operating frequency in each band.
Another technique for increasing the bandwidth of narrow-band antennas is to add parasitic elements, such as is done in the case of the open-sleeve antenna (Hall, 1992).
### COMMON COLLECTOR TRANSISTOR CONFIGURATION BASICS AND TUTORIALS
WHAT IS COMMON COLLECTOR TRANSISTOR CONFIGURATION
Figure 1 is a practical example of a common-collector transistor amplifier. Note that the output is taken off of the emitter instead of the collector (as in the common-emitter configuration).
A common-collector amplifier is not capable of voltage gain. In fact, there is a very slight loss of voltage amplitude between input and output.
However, for all practical purposes, we can consider the voltage gain at unity. Common-collector amplifiers are noninverting, meaning the output signal is in phase with the input signal.
Essentially, the output signal is an exact duplicate of the input signal. For this reason, common-collector amplifiers are often called emitter-follower amplifiers, because the emitter voltage follows the base voltage.
Common-collector amplifiers are current amplifiers. The current gain for the circuit illustrated in Fig. 1 is the parallel resistance value of R1 and R2, divided by the resistance value of R3. R1 and R2 are both 20 Kohms in value, so their parallel resistance value is 10 Kohms.
This 10 Kohms divided by 1 Kohm (the value of R3) gives us a current gain of 10 for this circuit. Because the voltage gain is considered to be unity (1), the power gain for a common-collector amplifier is considered equal to the current gain (10, in this particular case).
The input impedance of common-collector amplifiers is typically higher than the other transistor configurations. It is the parallel resistive effect of R1, R2, and the product of the value of R3 times the beta value.
Because beta times the R3 value is usually much higher than that of R1 or R2, you can closely estimate the input impedance by simply considering it to be the parallel resistance of R1 and R2. In this case, the input impedance would be about 10 Kohms.
The traditional method of calculating the output impedance of common-collector amplifiers is to divide the value of R3 by the transistor’s beta value. Although this method is still appropriate, a closer estimate can probably be obtained by considering the output impedance of most transistors to be about 80 ohms.
This 80-ohm output impedance should be viewed as being in parallel with R3, giving us a calculated output impedance of about 74 ohms (80 ohms in parallel with 1000 ohms). Resistors R1 and R2 have the same function within a common-collector amplifier as previously discussed with common-emitter amplifiers.
The high negative feedback produced by R3 provides excellent temperature stability and immunity from transistor variables. The circuit illustrated in Fig. 1 can be a valuable building block toward future projects.
### HISTORY OF SOLID STATE ELECTRONICS BASICS
SOLID STATE ELECTRONICS BRIEF HISTORY
The crystal detectors used in early radios were the forerunners of modern solid-state devices. However, the era of solid-state electronics began with the invention of the transistor in 1947 at Bell Labs.
The inventors were Walter Brattain, John Bardeen, and William Shockley. PC (printed circuit) boards were introduced in 1947, the year the transistor was invented. Commercial manufacturing of transistors began in Allentown, Pennsylvania, in 1951.
The most important invention of the 1950s was the integrated circuit. On September 12, 1958, Jack Kilby, at Texas Instruments, made the first integrated circuit. This invention literally created the modern computer age and brought about sweeping changes in medicine, communication, manufacturing, and the entertainment industry.
Many billions of "chips"-as integrated circuits came to be called-have since been manufactured. The 1960s saw the space race begin and spurred work on miniaturization and computers.
The space race was the driving force behind the rapid changes in electronics that followed. The first successful "op-amp" was designed by Bob Widlar at Fairchild Semiconductor in 1965. Called the flA709, it was very successful but suffered from "latch-up" and other problems.
Later, the most popular op-amp ever, the 741, was taking shape at Fairchild. This opamp became the industry standard and influenced design of op-amps for years to come.
By 1971, a new company that had been formed by a group from Fairchild introduced the first microprocessor. The company was Intel and the product was the 4004 chip, which had the same processing power as the Eniac computer.
Later in the same year, Intel announced the first 8-bit processor, the 8008. In 1975, the first personal computer was introduced by Altair, and Popular Science magazine featured it on the cover of the January, 1975, issue.
The 1970s also saw the introduction of the pocket calculator and new developments in optical integrated circuits.
By the 1980s, half of all U.S. homes were using cable hookups instead of television antennas. The reliability, speed, and miniaturization of electronics continued throughout the 1980s, including automated testing and calibrating of PC boards. The computer became a part of instrumentation and the virtual instrument was created. Computers became a standard tool on the workbench.
The 1990s saw a widespread application of the Internet. In 1993, there were 130 websites, and now there are millions. Companies scrambled to establish a home page and many of the early developments of radio broadcasting had parallels with the Internet.
In 1995, the FCC allocated spectrum space for a new service called Digital Audio Radio Service. Digital television standards were adopted in 1996 by the FCC for the nation's next generation of broadcast television.
The 21st century dawned in January 2001. One of the major technology stories has been the continuous and explosive growth of the Internet. Internet usage in North America has increased by over 100% from 2000 to 2005.
The rest of the world experienced almost 200% growth during the same period. The processing speed of computers is increasing at a steady rate and data storage media capacity is increasing at an amazing pace. Carbon nanotubes are seen to be the next step forward for computer chips, eventually replacing transistor technology.
### BREADBOARD AND PRINTED CIRCUIT BOARD BASICS AND TUTORIALS
WHAT ARE BREADBOARD AND PRINTED CIRCUIT BOARD?
A breadboard is a rectangular plastic box filled with holes, which have contacts in which you can insert electronic components and wires. A breadboard is what you use to string together a temporary version of your circuit.
You don’t have to solder wires or anything else; instead, you poke your components and wires into the little contact holes arranged in rows and connected by lines of metal; then you can connect your components together with wires to form your circuit.
The nice thing about breadboards is that you can change your mind and replace or rearrange components as you like. You typically create an electronics project on a breadboard to make sure that everything works.
If it’s a project you wish to save, you can create a more permanent version. If you want to create a permanent version of your circuit, you need to create a soldered or printed circuit board; see the sidebar, “Printed circuit boards,” to find out how to go about that.
Printed circuit boards
If you create a circuit on a breadboard and decide that it’s worthy of immortality, you can make it permanent by soldering components in place on a printed circuit board. To do this, you have to get your hands on a universal printed circuit board.
This is much like a breadboard except that you can solder all the connections you’ve made to keep them around. A universal printed circuit board has rows of individual holes throughout the board with copper pads around each hole and metal lines connecting the holes in each row, like in a breadboard.
You mount parts on the face of the board and then pass leads through holes to the components. You can solder the leads to the copper pads on the bottom of the board.
Universal printed circuit boards are available in a variety of patterns of contact holes and
metal lines. The figure here shows one we like because there are rows on either side that accommodate discrete components handily.
You can get custom printed circuit boards made for your circuit; this is typically done by submitting a drawing of your circuit to a printed circuit board company. These boards eliminate the need to solder jumper wires between components.
### SAFETY USE OF SOLDERING IRON TUTORIALS
SOLDERING SAFETY TIPS
Soldering poses a few different dangers. (You might use solder to attach various pieces of your electronics project, such as soldering wires onto a speaker, microphone, or switch.) The soldering iron itself gets mighty hot.
The solder (the material you heat with the iron) gets hot. Occasionally, you even get an air pocket or impurity in solder that can pop as you heat it, splattering a little solder toward your face or onto your arm.
To top that off, hot solder produces some nasty fumes. Soldering itself takes experience to get right. Your best bet is to have somebody who is good at it teach you.
Here are some soldering safety guidelines you should always follow:
1. Always wear safety glasses when soldering.
2. Never solder a live circuit (one that is energized). Soldering irons come in models that use different wattages. Use the right size soldering iron for your projects; too much heat could ruin your board or components.
3. Solder in a well-ventilated space to prevent the mildly caustic and toxic fumes from building up and causing eye or throat irritation.
4. Always put your soldering iron back in its stand when not in use. Too, be sure that the stand is weighted enough or attached to your worktable so that it doesn’t topple over if you should brush against the cord.
5. NEVER place a hot soldering iron on your work surface. You could start a fire.
6. Give any soldered surface a minute or two to cool down before you touch it.
7. Never, ever try to catch a hot soldering iron if you drop it. No matter how hard you try, you are very likely to grab the hot end in a freefall. Let it fall; buy a new one if you have to — just don’t grab!
8. Never leave flammable items (like paper) near your soldering iron.
9. Be sure to unplug your soldering iron when you’re not around.
Don’t put your face too close to the soldering site because of the danger of stray hot solder and those horrible fumes.
Instead, use a magnifying device to see when soldering teeny-tiny components to a board.
You can buy clampon magnifiers that keep your hands free for soldering.
### STATIC DISCHARGE PROTECTION FOR ELECTRONIC DEVICES TUTORIALS
HOW TO PROTECT ELECTRONIC DEVICES FROM ELECTROSTATIC DISCHARGE?
Protecting Electronic Components from Dreaded Static Discharge
You’re not the only thing in your work area that could suffer from shocks. Static discharge (also referred to as electrostatic discharge; ESD) can do damage to your delicate electrical components.
Static discharge is so named because it’s caused by the discharge of electrons from a static charge that hang around in an insulating body, even after the source of those electrons goes away.
Static charge is typically caused by friction. You might trap some electrons in your body as you walk across a carpet, for example. When a static charge is built up on your body, a corresponding voltage difference is built up between your body and a grounded object, such as a doorknob. The zap when you then touch a doorknob is the static discharge: that is, the electrons flowing from you to the doorknob.
What static discharge can do?
Metal oxide semiconductor (MOS) devices are cool because they allow integrated components to use less power. MOS devices improve circuit design and operation, but that improvement comes at a price. These little guys are VERY sensitive to ESD. One little zap, and they are likely to be history.
When you walk across a carpet, you can produce a static charge in the range of 2,000–4,000V. Because the number of electrons trapped on your body is low, you feel only a little shock. However, MOS devices contain a very thin layer of insulating glass that can become toast when exposed to as little as 50V of discharge or less.
When you work with a MOS device, your body, clothes, and tools have to be free of static discharge. MOS devices are found in many integrated circuits (ICs) and transistors.
ICs and transistors that use bipolar devices do not have the very thin layers of insulating glass found in MOS devices, so they are less susceptible to damage from static discharge. Resistors, capacitors, diodes, transformers, and coils, on the other hand, aren’t in too much danger from static discharge. Keep static discharge away from your projects just to be safe.
How to guard against ESD?
To get rid of static discharge in your electronics workshop, you can do several things, such as wearing anti-static devices and clothing, using static dissipative floor mats, and grounding your tools.
First, wear an anti-static wrist strap. An anti-static wrist strap is one of the best ways to get rid of ESD. You then attach the wire on the strap to earth ground — which is just what it sounds like: namely, the earth beneath your feet. The cold water pipe on a water heater or under a sink is a good option for earth ground — if the water pipes are metal, that is.
Plastic water pipes that you find in some newer construction won’t work. Because the cold water pipe comes up out of the ground, it is therefore grounded (logical, huh?), which works where the hot water pipe usually won’t. Use a clamp to attach a wire to the pipe (earth ground) and run it to your worktable, being sure to run the wire along the wall so you don’t trip over it.
Set a loop of the wire at the edge of your worktable where it’s handy to attach the alligator clip on the end of it to your wrist strap. If you don’t happen to have a metal cold water pipe nearby, the best method is to use a metal rod that you insert into the ground. The standard rule is to sink it three feet deep.
Second, wear clothing that is less likely to accumulate static charge. For example, polyester, acetate, and wool fabrics easily accumulate static charges whereas as cotton is less likely to accumulate the static charges necessary for ESD. Using an anti-static wrist strap and wearing cotton clothing will usually be sufficient.
Third, if you plan to do electronics projects long-term, consider buying a static-dissipative mat for your work surface. You connect the mat to a ground, as you do with the wrist strap, and the mat dissipates charges from components you’re working on as you lay them on the mat.
However, the mat has a high enough resistance that it won’t short together the pins of components. There are also static-dissipative floor mats; however, these are more likely to be used in a manufacturing setting when a worker needs to move between workstations.
### AUTOMATIC BATHROOM LIGHT BASIC ELECTRONICS PROJECT
HOW TO MAKE AUTOMATIC BATHROOM LIGHT
This circuit is used to automate the working of a bathroom light. It is designed for a bathroom fitted with an automatic door-closer, where the manual verification of light status is difficult. The circuit also indicates whether the bathroom is occupied or not.
The circuit uses only two ICs and can be operated from a 5V supply. As it does not use any mechanical contacts it gives a reliable performance.
One infrared LED (D1) and one infrared detector diode (D2) form the sensor part of the circuit. Both the infrared LED and the detector diode are fitted on the frame of the door with a small separation between them as shown in Fig. 1. The radiation from IR LED is blocked by a small opaque strip (fitted on the door) when the door is closed.
Detector diode D2 has a resistance in the range of meg-ohms when it is not activated by IR rays. When the door is opened, the strip moves along with it. Radiation from the IR LED turns on the IR detector diode and the voltage across it drops to a low level.
C o m p a r a t o LM358 IC2(a) compares the voltage across the photodetec- tor against a reference potential set by preset VR1. The preset is so adjusted as to provide an optimum threshold voltage so that output of IC2(a) is high when the door is closed and low when the door is open.
Capacitor C1 is connected at the output to filter out unwanted transitions in output voltage generated at the time of opening or closing of the door. Thus, at point A, a low-to-high going voltage transition is available for every closing of the door after opening it. (See waveform A in Fig. 2.)
The second comparator IC2(b) does the reverse of IC2(a), as the input terminals are reversed. At point B, a low level is available when the door is closed and it switches to a high level when the door is opened. (See waveform B in Fig. 2.) Thus, a lowto- high going voltage transition\ is available at point B for every opening of the door, from the closed position.
Capacitor C2 is connected at the output to filter out unwanted transitions in the output voltage generated at the time of closing or opening of the door. IC 7474, a rising-edge-sensitive dual-D flip-flop, is used in the circuit to memorise the occupancy status of the bathroom. IC1(a) memorises the state of the door and acts as an occupancy indicator while IC2(b) is used to control the relay to turn on and turn off the bathroom light. Q output pin 8 of IC1(b) is tied to D input pin 2 of IC1(a) whereas Q output pin 5 of IC1(a) is tied to D input pin 12 of IC1(b).
At the time of switching on power for the first time, the resistor-capacitor combination R3-C3 clears the two flip-flops. As a result Q outputs of both IC1(a) and IC1(b) are low, and the low level at the output of IC1(b) activates a relay to turn on the bathroom light. This operation is independent of the door status (open/closed). The occupancy indicator red LED (D3) is off at this point of time, indicating that the
room is vacant.
When a person enters the bathroom, the door is opened and closed, which provides clock signals for IC1(b) (first) and IC1(a). The low level at point C (pin 5) isclocked in by IC1(b), at the time of opening the door, keeping the light status unchanged.
The high level point D (pin 8) is clocked in by IC1(a), turning on the occupancy indicator LED (D3) on at the time of closing of the door. (See waveform C in Fig. 2.)
When the person exits the bathroom, the door is opened again. The output of IC1(b) switches to high level, turning off the bathroom light. (See waveform D in\ Fig. 2.) The closing of the door by the door-closer produces a low-to-high transition at the clock input (pin 3) of IC1(a). This clocks in the low level at Q output of IC1(b) point D to Q output of IC1(a) point C, thereby turning off the occupancy indicator.
### PRINCIPLES OF OSCILLATION BASIC AND TUTORIALS
HOW PRINCIPLES OF OSCILLATION WORKS?
A small signal voltage amplifier is shown in Fig. 1.3. In Fig. 1.3(a) the operational amplifier has no external components connected to it and the signal is fed in as shown. The operational amplifier has an extremely high gain under these circumstances and this leads to saturation within the amplifier.
As saturation implies working in the non-linear section of the characteristics, harmonics are produced and a ringing pattern may appear inside the chip. As a result of this, a square wave output is produced for a sinusoidal input. The amplifier has ceased to amplify and we say it has become unstable.
There are many reasons why an amplifier may become unstable, such as temperature changes or power supply variations, but in this case the problem is the very high gain of the operational amplifier.
Figure 1.3(b) shows how this may be overcome by introducing a feedback network between the output and the input. When feedback is applied to an amplifier the overall gain can be reduced and controlled so that the operational amplifier can function as a linear amplifier.
Note also that the signal fedback has a phase angle, due to the inverting input, which is in opposition to the input signal (Vi).
Negative feedback can therefore be defined as the process whereby a part of the output voltage of an amplifier is fed to the input with a phase angle that opposes the input signal. Negative feedback is used in amplifier circuits in order to give stability and reduced gain.
Bandwidth is generally increased, noise reduced and input and output resistances altered. These are all desirable parameters for an amplifier, but if the feedback is overdone then the amplifier becomes unstable and will produce a ringing effect.
In order to understand stability, instability and its causes must be considered. From the above discussion, as long as the feedback is negative the amplifier is stable, but when the signal feedback is in phase with the input signal then positive feedback exists.
Hence positive feedback occurs when the total phase shift through the operational amplifier (opamp) and the feedback network is 360° (0°). The feedback signal is now in phase with the input signal (Vi) and oscillations take place.
### TOP VIDEO GAMES COMPANY IN THE WORLD
A RUN DOWN ON THE TOP VIDEO GAMES MAKER IN THE WORLD
Microsoft Corporation
While relatively new to the market, Microsoft has a number of products in the home consumer space including its Xbox gaming platform. Xbox hardware is built with an impressive Intel processor, a custom-designed Microsoft graphics processor with Nvidia memory, and a hard disk drive. It supports HDTV and includes a DVD, four I/O ports for the game controller, and an Ethernet port.
Nintendo
Based in Japan, Nintendo was the original developer of the Nintendo 64 and the more recent Nintendo GameCube systems. The company has stated that it is not pursuing a home gateway strategy and that it will stick to games only.
Sega
Based in Japan, Sega brought to market the Dreamcast, a second-generation 128-bit gaming console
with:
■ CD player
■ Built-in 56K modem for Internet play
■ Internet access and e-mail
■ Expansion slots for home networking.
Although Sega has exited the market to concentrate on gaming software, it was first to market with its second-generation game console. The company also provides accessories such as a keyboard, extra memory, and various controllers.
After Sega ceased production of the Dreamcast gaming console, the company refashioned itself as a platform-agnostic game developer. By the end of the year Sega will bring its games to handheld devices (such as PDAs and cell phones), set-top boxes, and even rival consoles.
In addition Sega agreed to deliver Dreamcast chip technology to set-top box manufacturers to enable cable subscribers to play Dreamcast games. Sega has been the number three video game console supplier and has been hemorrhaging money since the introduction of Dreamcast in 1999. The release of Microsoft’s Xbox most likely influenced Sega’s move to exit the game console market.
Sony
Based in Japan, Sony Computer Entertainment Inc. is positioning itself as the consumer’s one-stop shop for home consumer devices ranging from entertainment to personal computing. Sony has overcome a number of challenges to create a unified network of consumer devices and digital services.
It brought together diverse parts of the company to work together on a single, cohesive Internet strategy. Sony is positioning its 128-bit PlayStation2 as a gateway into the home from which it can offer Internet access, value-added services, and home networking.
Its home networking would use USB or iLink (also called IEEE 1394 or FireWire) ports to connect devices. PlayStation2 contains a DVD player that could enable digital video playback and could, in time, include an integrated digital video recorder. With its added functionality the Playstation2 is leaving the “game console” designation behind.
### BINARY SYSTEM BASICS AND TUTORIALS
WHAT IS BINARY NUMBER SYSTEM? A TUTORIAL ON BINARY NUMBER SYSTEM
The binary number system is a radix-2 number system with ‘0’ and ‘1’ as the two independent digits. All larger binary numbers are represented in terms of ‘0’ and ‘1’. The procedure for writing higher order binary numbers after ‘1’ is similar to the one explained in the case of the decimal number system.
For example, the first 16 numbers in the binary number system would be 0, 1, 10, 11, 100, 101, 110, 111, 1000, 1001, 1010, 1011, 1100, 1101, 1110 and 1111. The next number after 1111 is 10000, which is the lowest binary number with five digits.
This also proves the point made earlier that a maximum of only 16 (= 24 numbers could be written with four digits. Starting from the binary point, the place values of different digits in a mixed binary number are 20, 21, 22 and so on (for the integer part) and 2−1, 2−2, 2−3 and so on (for the fractional part).
Example 1.1
Consider an arbitrary number system with the independent digits as 0, 1 and X. What is the radix of this number system? List the first 10 numbers in this number system.
Solution
• The radix of the proposed number system is 3.
• The first 10 numbers in this number system would be 0, 1, X, 10, 11, 1X, X0, X1, XX and 100.
Logic operations are the backbone of any digital computer, although solving a problem on computer could involve an arithmetic operation too. The introduction of the mathematics of logic by George Boole laid the foundation for the modern digital computer.
He reduced the mathematics of logic to a binary notation of ‘0’ and ‘1’. As the mathematics of logic was well established and had proved itself to be quite useful in solving all kinds of logical problem, and also as the mathematics of logic (also known as Boolean algebra) had been reduced to a binary notation, the binary number system had a clear edge over other number systems for use in computer systems.
Yet another significant advantage of this number system was that all kinds of data could be conveniently represented in terms of 0s and 1s. Also, basic electronic devices used for hardware implementation could be conveniently and efficiently operated in two distinctly different modes.
For example, a bipolar transistor could be operated either in cut-off or in saturation very efficiently.
Lastly, the circuits required for performing arithmetic operations such as addition, subtraction, multiplication, division, etc., become a simple affair when the data involved are represented in the form of 0s and 1s.
### THE 7400-SERIES DISCRETE LOGIC FAMILY IC BASICS AND TUTORIALS
WHAT IS THE 7400-SERIES DISCRETE LOGIC FAMILY INTEGRATED CIRCUITS?
With the advent of ICs in the early 1960s, engineers needed ready access to a library of basic logic gates so that these gates could be wired together on circuit boards and turned into useful products. Rather than having to design a custom microchip for each new project, semiconductor companies began to recognize a market for standard, off-the-shelf logic ICs.
In 1963 and 1964, Sylvania and Texas Instruments began shipment of the 7400-series discrete logic family and unknowingly started a de facto industry standard that lasts to this day and shows no signs of disappearing anytime soon.
Using the 7400 family, an engineer can select logic gates, flip-flops, counters, and buffers in individual packages and wire them together as desired to solve a specific problem. Some of the most common members of the 7400 family are listed in Table 2.1.
TABLE 2.1 Common 7400 ICs
Part Number Function Number of Pins
7400 Quad two-input NAND gates 14
7402 Quad two-input NOR gates 14
7404 Hex inverters 14
7408 Quad two-input AND gates 14
7432 Quad two-input OR gates 14
7447 BCD to seven-segment display decoder/driver 16
7474 Dual D-type positive edge triggered flip-flops 14
74138 Three-to-eight decoder 16
74153 Dual 4-to-1 multiplexer 16
74160 Four-bit binary synchronous counter 16
74164 Eight-bit parallel out serial shift registers 16
74174 Quad D-type flip-flops with complementary outputs 16
74193 Four-bit synchronous up/down binary counter 16
74245 Octal bus transceivers with tri-state outputs 20
74373 Octal D-type transparent latch 20
74374 Octal D-type flip-flops 20
These are just a few of the full set of 7400 family members. Many 7400 parts are no longer used, because their specific function is rarely required as a separate chip in modern digital electronics designs.
However, the parts listed above, and many others that are not listed, are still readily available today and are commonly found in a broad range of digital designs ranging from low-end to hightech devices. 7400-series logic has been available in DIPs for a long time, as well as (more recently) SOICs and other high-density surface mount packages.
All flavors of basic logic gates are available with varying numbers of inputs. For example, there are 2-, 3-, and 4-input AND gates and 2-, 3-, 4-,8-, 12-, and 13-input NAND gates. There are numerous varieties of flip-flops, counters, multiplexers, shift registers, and bus transceivers.
Flip-flops exist with and without complementary outputs, preset/ clear inputs, and independent clocks. Counters are available in 4-bit blocks that can both increment and decrement and count to either 15 (binary counter) or 9 (decade counter) before restarting the count at 0.
Shift registers exist in all permutations of serial and parallel inputs and outputs. Bus transceivers in 4- and 8 bit increments exist with different types of output enables and capabilities to function in unidirectional or bidirectional modes. Bus transceivers enable the creation and expansion of tri-state buses on which multiple devices can communicate.
One interesting IC is the 7447 seven-segment display driver. This component allows the creation of graphical numeric displays in applications such as counters and timers. Seven-segment displays are commonly seen in automobiles, microwave ovens, watches, and consumer electronics.
Seven independent on/off elements can represent all ten digits.. The 7447 is able to drive an LED-based seven-segment display when given a binary coded decimal (BCD) input. BCD is a four-bit binary number that has valid values from 0 through 9.
Hexadecimal values from 0xA through 0xF are not considered legal BCD values. Familiarity with the 7400 series proves very useful no matter what type of digital system you are designing.
For low-end systems, 7400-series logic may be the only type of IC at your disposal to solve a wide range of problems. At the high end, many people are often surprised to see a small 14- pin 7400-series IC soldered to a circuit board alongside a fancy 32-bit microprocessor running at 100 MHz.
The fact is that the basic logic functions that the 7400 series offers are staples that have direct applications at all levels of digital systems design. It is time well spent to become familiar with the extensive capabilities of the simple yet powerful 7400 family.
Manufacturers’ logic data books, either in print or on line, are invaluable references. It can be difficult to know ahead of time if a design may call for one more gate to function properly; that is when a 40-year old logic family can save the day.
### BOOLEAN LOGIC ALGEBRA AND LOGIC GATES BASICS AND TUTORIAL
BOOLEAN LOGIC AND LOGIC GATES TUTORIALS
Machines of all types, including computers, are designed to perform specific tasks in exact well defined manners. Some machine components are purely physical in nature, because their composition and behavior are strictly regulated by chemical, thermodynamic, and physical properties.
For example, an engine is designed to transform the energy released by the combustion of gasoline and oxygen into rotating a crankshaft. Other machine components are algorithmic in nature, because their designs primarily follow constraints necessary to implement a set of logical functions as defined by human beings rather than the laws of physics.
A traffic light’s behavior is predominantly defined by human beings rather than by natural physical laws. This book is concerned with the design of digital systems that are suited to the algorithmic requirements of their particular range of applications. Digital logic and arithmetic are critical building blocks in constructing such systems.
An algorithm is a procedure for solving a problem through a series of finite and specific steps. It can be represented as a set of mathematical formulas, lists of sequential operations, or any combination thereof. Each of these finite steps can be represented by a Boolean logic equation.
Boolean logic is a branch of mathematics that was discovered in the nineteenth century by an English mathematician named George Boole. The basic theory is that logical relationships can be modeled by algebraic equations.
Rather than using arithmetic operations such as addition and subtraction, Boolean algebra employs logical operations including AND, OR, and NOT. Boolean variables have two enumerated values: true and false, represented numerically as 1 and 0, respectively. The AND operation is mathematically defined as the product of two Boolean values, denoted A and B for reference.
Truth tables are often used to illustrate logical relationships as shown for the AND operation in Table 1.1. A truth table provides a direct mapping between the possible inputs and outputs.
A basic AND operation has two inputs with four possible combinations, because each input can be either 1 or 0 — true or false. Mathematical rules apply to Boolean algebra, resulting in a nonzero product only when both inputs are 1.
Summation is represented by the OR operation in Boolean algebra as shown in Table 1.2. Only one combination of inputs to the OR operation result in a zero sum: 0 + 0 = 0.
Boolean variables may not seem too interesting on their own. It is what they can be made to represent that leads to useful constructs. A rather contrived example can be made from the following logical statement:
“If today is Saturday or Sunday and it is warm, then put on shorts.”
Three Boolean inputs can be inferred from this statement: Saturday, Sunday, and warm. One Boolean output can be inferred: shorts. These four variables can be assembled into a single logic equation that computes the desired result, shorts = (Saturday OR Sunday) AND warm
While this is a simple example, it is representative of the fact that any logical relationship can be expressed algebraically with products and sums by combining the basic logic functions AND, OR, and NOT.
Several other logic functions are regarded as elemental, even though they can be broken down into AND, OR, and NOT functions. These are not–AND (NAND), not–OR (NOR), exclusive–OR (XOR), and exclusive–NOR (XNOR).
Table 1.3 presents the logical definitions of these other basic functions. XOR is an interesting function, because it implements a sum that is distinct from OR by taking into account that 1 + 1 does not equal 1. As will be seen later, XOR plays a key role in arithmetic for this reason.
All binary operators can be chained together to implement a wide function of any number of inputs. For example, the truth table for a ten-input AND function would result in a 1 output only when all inputs are 1.
Similarly, the truth table for a seven-input OR function would result in a 1 output if any of the seven inputs are 1. A four-input XOR, however, will only result in a 1 output if there are an odd number of ones at the inputs.
This is because of the logical daisy chaining of multiple binary XOR operations. As shown in Table 1.3, an even number of 1s presented to an XOR function cancel each other out.
It quickly grows unwieldy to write out the names of logical operators. Concise algebraic expressions are written by using the graphical representations shown in Table 1.4.
Note that each operation has multiple symbolic representations. The choice of representation is a matter of style when handwritten and is predetermined when programming a computer by the syntactical requirements of each computer programming language.
A common means of representing the output of a generic logical function is with the variable Y. Therefore, the AND function of two variables, A and B, can be written as Y = A & B or Y = A*B.
As with normal mathematical notation, products can also be written by placing terms right next to each other, such as Y = AB. Notation for the inverted functions, NAND, NOR, and XNOR, is achieved by inverting the base function.
Two equally valid ways of representing NAND are Y = A & B and Y = !(AB). Similarly, an XNOR might be written as Y = BAR A⊕BAR B.
When logical functions are converted into circuits, graphical representations of the seven basic operators are commonly used. In circuit terminology, the logical operators are called gates . Figure 1.1 shows how the basic logic gates are drawn on a circuit diagram.
Naming the inputs of each gate A and B and the output Y is for reference only; any name can be chosen for convenience. A small bubble is drawn at a gate’s output to indicate a logical inversion.
More complex Boolean functions are created by combining Boolean operators in the same way that arithmetic operators are combined in normal mathematics. Parentheses are useful to explicitly convey precedence information so that there is no ambiguity over how two variables should be
treated.
A Boolean function might be written as Y = (AB + C + )D& BAR E ⊕ BAR F
This same equation could be represented graphically in a circuit diagram, also called a schematic diagram , as shown in Fig. 1.2. This representation uses only two-input logic gates. As already mentioned, binary operators can be chained together to implement functions of more than two variables.
An alternative graphical representation would use a three-input OR gate by collapsing the two-input OR gates into a single entity.
### CLOCK SKEW BASICS AND TUTORIALS
WHAT IS CLOCK SKEW? HOW CLOCK SKEW WORKS?
The preceding timing analysis example is simplified for ease of presentation by assuming that the source and destination flops in a logic path are driven by the same clock signal. Although a synchronous circuit uses a common clock for all flops, there are small, nonzero variances in clock timing at individual flops.
Wiring delay variances are one source of this nonideal behavior. When a clock source drives two flops, the two wires that connect to each flop’s clock input are usually not identical in length. This length inequality causes one flop’s clock to arrive slightly before or after the other flop’s clock.
Clock skew is the term used to characterize differences in edge timing between multiple clock inputs. Skew caused by wiring delay variance can be effectively minimized by designing a circuit so that clock distribution wires are matched in length.
A more troublesome source of clock skew arises when there are too many clock loads to be driven by a single source. Multiple clock drivers are necessary in these situations, with small variations in electrical characteristics between each driver.
These driver variances result in clock skew across all the flops in a synchronous design. As might be expected, clock skew usually reduces the frequency at which a synchronous circuit can operate.
Clock skew is subtracted from the nominal clock period for setup time analysis purposes, because the worst-case scenario shown in Fig. 1.17 must be considered. This scenario uses the same logic circuit in Fig. 1.16 but shows two separate clocks with 1 ns of skew between them.
The worst timing occurs when the destination flop’s clock arrives before that of the source flop, thereby reducing the amount of time available for the D-input to stabilize. Instead of the circuit having zero margin with a 20-ns period, clock skew increases the minimum period to 21 ns.
The extra 1 ns compensates for the clock skew to restore a minimum source to destination period time of 20 ns. A slower circuit such as this one is not very sensitive to clock skew, especially after backing off to 40 MHz for timing margin as shown previously.
Digital systems that run at relatively low frequencies may not be affected by clock skew, because they often have substantial margins built into their timing analyses. As clock speeds increase, the margin decreases to the point at which clock skew and interconnect delay become important limiting factors in system design.
Hold time compliance can become more difficult in the presence of clock skew. The basic problem occurs when clock skew reduces the source flop’s apparent tCO from the destination flop’s perspective, causing the destination’s input to change before tH is satisfied.
Such problems are more prone in high-speed systems, but slower systems are not immune. Figure 1.18 shows a timing diagram for a circuit with 1 ns of clock skew where two flops are connected by a short wire with nearly zero propagation delay.
The flops have tCO = 2 ns and tH = 1.5 ns. A scenario like this may be experienced when connecting two chips that are next to each other on a circuit board. In the absence of clock skew, the destination flop’s input would change tCO after the rising clock edge, exceeding tH by 0.5 ns.
The worst-case clock skew causes the source flop clock to arrive before that of the destination flop, resulting in an input change just 1 ns after the rising clock edge and violating tH. Solutions to skew-induced tH violations include reducing the skew or increasing the delay between source and destination.
Unfortunately, increasing a signal’s propagation delay may cause tSU violations in high-speed systems. Hold time may not be a problem in slower circuits, because slower circuits often have paths between flops with sufficiently long propagation delays to offset clock skew problems.
However, even slow circuits can experience hold-time problems if flops are connected with wires or components that have small propagation delays. It is also important to remember that hold-time compliance is not a function of clock period but of clock skew, tCO, and tH. Therefore, a slow system that uses fast components may have problems if the clock skew exceeds the difference between tCO and tH.
### OPERATIONAL AMPLIFIER BASICS AND TUTORIALS
WHAT IS AN OPERATIONAL AMPLIFIER?
An operational amplifier or op amp is a circuit that takes an input voltage and amplifies it. The symbol used to represent an op amp in a circuit diagram is shown in Fig. 9-1.
An op amp is defined by two simple equations. The first thing to note is that the voltage across the input terminals is zero. Hence Va = Vb
The second relation that is essential for analyzing op amp circuits is that the currents drawn at a and b in Fig. 9-1 are zero Ia = Ib = 0
Despite this, we will see that the op amp will result in voltage gains at the output terminal c. How does this work? Two voltages are input to terminals a and b.
Their difference is then amplified and output at c, which is taken with referenc to ground. Although we won’t worry about the internal construction of an op amp, note that it consists of a set of resistors and dependent voltage source. The internal voltage source is related to the input voltages by A(V+ − V−)
The constant A is known as the open-loop voltage gain. To see how op amp circuits work, it’s best to examine some popular example circuits.
When analyzing op amp circuits, remember to take the input voltage across the op amp terminals to be zero and that the op amp draws zero current. The analysis is then reduced to applying KVL and KCL to the circuit elements connected to the op amp.
### SOLDERING IRON BASIC INFORMATION AND TUTORIALS
WHAT IS SOLDERING IRON? HOW TO USE SOLDERING IRON?
SOLDERING IRON
Soldering is used in nearly every phase of electronic construction so you’ll need soldering tools. A soldering tool must be hot enough to do the job and lightweight enough for agility and comfort.
A temperature controlled iron works well, although the cost is not justified for occasional projects. Get an iron with a small conical or chisel tip.
Soldering is not like gluing; solder does more than bind metal together and provide an electrically conductive path between them. Soldered metals and the solder combine to form an alloy.
You may need an assortment of soldering irons to do a wide variety of soldering tasks. They range in size from a small 25-watt iron for delicate printed-circuit work to larger 100 to 300-watt sizes used to solder large surfaces.
If you could only afford a single soldering tool when initially setting up your electronics workbench than, an inexpensive to moderately priced pencil-type soldering iron with between 25 and 40-watt capacity is the best for PC board electronics work.
A 100-watt soldering gun is overkill for printed-circuit work, since it often gets too hot, cooking solder into a brittle mess or damaging small parts of a circuit. Soldering guns are best used for point-to-point soldering jobs, for large mass soldering joints or large components.
Small “pencil” butane torches are also available, with optional soldering-iron tips. Butane soldering irons are
ideal for field service problems and will allow you to solder where there is no 110 volt power source.
Keep soldering tools in good condition by keeping the tips well tinned with solder. Do not run them at full
temperature for long periods when not in use. After each period of use, remove the tip and clean off any scale that may have accumulated.
Clean an oxidized tip by dipping the hot tip in sal ammoniac (ammonium chloride) and then wiping it clean with a rag. Sal ammoniac is somewhat corrosive, so if you don’t wipe the tip thoroughly, it can contaminate electronic soldering.
Place the tip of the soldering iron into the “Tip Tinner” after every few solder joints. If a copper tip becomes pitted, file it smooth and bright and then tin it immediately with solder.
Modern soldering iron tips are nickel or iron clad and should not be filed. The secret of good soldering is to use the right amount of heat.
Many people who will have not soldered before use too little heat dabbing at the joint to be soldered and making little solder blobs that cause unintended short circuits.
Always use caution when soldering. A hot soldering iron can burn your hand badly or ruin a tabletop. It’s a good idea to buy or make a soldering iron holder.
### COLPITTS OSCILLATORS BASIC CIRCUIT TUTORIALS
WHAT ARE COLPITTS OSCILLATORS?
Colpitts oscillators are similar to the shunt fed Hartley oscillator circuit except the Colpitts oscillator, instead of having a tapped inductor, utilizes two series capacitors in its LC circuit.
With the Colpitts oscillator the connection between these two capacitors is used as the center tap for the circuit. A Colpitts oscillator circuit is shown at Figure 2-5, and you will see some similarities with the Hartley oscillator.
Colpitts Oscillators
The simplest Colpitts oscillator to construct and get running is the “series tuned” version, more often referred to as the “Clapp Oscillator.” Because there is no load on the inductor, a high “Q” circuit results with a high L/C ratio and of course much less circulating current.
This aids drift reduction. Because larger inductances are required, stray inductances do not have as much impact as perhaps in other circuits.
The total capacitive reactance of the parallel combination of capacitors depicted as series tuning below the inductor in a series tuned Colpitts oscillator or “Clapp oscillator” should have a total reactance of around 200 ohms.
Not all capacitors may be required in your particular application. Effectively all the capacitors are in series in a Colpitts oscillator, i.e. they appear as parallel connected but their actual values are in fact in series.
Ideally, your frequency determining components L1 and the parallel capacitors should be in a grounded metal shield. The FET used in the Colpitts oscillator is the readily available 2N4416A.
Note, the metal FET case is connected to the circuit ground. The output from the Colpitts oscillator is through output capacitor 47 pF; this should be the smallest of values possible, consistent with continued reliable operation into the next buffer amplifier stage.
### 555 TIMER BASICS AND TUTORIALS
WHAT IS 555 TIME AND HOW 555 TIMER WORKS?
Creating a Pulse
The 555 is made out of simple transistors that are about the same as on / off switches. They do not have any sense of time. When you apply a voltage they turn on and when you take away the voltage they turn off. So by itself, the 555 can not create a pulse.
The way the pulse is created is by using some components in a circuit attached to the 555 (see the circuit on the next page). This circuit is made of a capacitor and a resistor. We can flip a switch and start charging the capacitor.
The resistor is used to control how fast the capacitor charges. The bigger the resistance, the longer it takes to charge the capacitor.
The voltage in the capacitor can then be used as an input to another switch. Since the voltage starts at 0, nothing happens to the second switch. But eventually the capacitor will charge up to some
point where the second switch comes on. The way the 555 timer works is that when you flip the first switch, the Output pin goes to Vcc (the positive power supply voltage) and starts charging the capacitor.
When the capacitor voltage gets to 2/3 Vcc (that is Vcc * 2/3) the second switch turns on which makes the output go to 0 volts.
The pinout for the 555 timer is shown below:
Pin 2 (Trigger) is the 'on' switch for the pulse. The line over the word Trigger tells us that the voltage levels are the opposite of what you would normally expect. To turn the switch on you apply 0 volts to pin 2.
The technical term for this opposite behavior is 'Active Low'. It is common to see this 'Active Low' behavior for IC inputs because of the inverting nature of transistor circuits.
Pin 6 is the off switch for the pulse. We connect the positive side of the capacitor to this pin and the negative side of the capacitor to ground. When Pin 2 (Trigger) is at Vcc, the 555 holds Pin 7 at 0 volts (Note the inverted voltage).
When Pin 2 goes to 0 volts, the 555 stops holding Pin 7 at 0 volts. Then the capacitor starts charging. The capacitor is charged through a resistor connected to Vcc. The current starts flowing into the capacitor, and the voltage in the capacitor starts to increase.
Pin 3 is the output (where the actual pulse comes out). The voltage on this pin starts at 0 volts. When 0 volts is applied to the trigger (Pin 2), the 555 puts out Vcc on Pin 3 and holds it at Vcc until
Pin 6 reaches 2/3 of Vcc (that is Vcc * 2/3). Then the 555 pulls the voltage at Pin 3 to ground and you have created a pulse. (Again notice the inverting action.) The voltage on Pin 7 is also pulled to ground, connecting the capacitor to ground and discharging it.
Seeing the pulse
To see the pulse we will use an LED connected to the 555 output, Pin 3. When the output is 0 volts the LED will be off. When the output is Vcc the LED will be on.
Place the 555 across the middle line of the breadboard so that 4 pins are on one side and 4 pins are on the other side. (You may need to bend the pins in a little so they will go in the holes.)
Leave the power disconnected until you finish building the circuit. The diagram above shows how the pins on the 555 are numbered. You can find pin 1 by looking for the half circle in the end of the chip. Sometimes instead of a half circle, there will be a dot or shallow hole by pin 1.
Before you start building the circuit, use jumper wires to connect the red and blue power rows to the red and blue power rows on the other side of the board. Then you will be able to easily reach Vcc and Ground lines from both sides of the board. (If the wires are too short, use two wires joined together in a row of holes for the positive power (Vcc) and two wires joined together in a different row of holes for the ground.)
Connect Pin 1 to ground.
Connect Pin 8 to Vcc.
Connect Pin 4 to Vcc.
Connect the positive leg of the LED to a 330 ohm resistor and connect the negative end of the
LED to ground. Connect the other leg of the 330 ohm resistor to the output, Pin 3.
Connect Pin 7 to Vcc with a 10k resistor (RA = 10K).
Connect Pin 7 to Pin 6 with a jumper wire.
Connect Pin 6 to the positive leg of the 220uF Capacitor (C = 220uF). (You will need to bend the positive (long leg) up and out some so that the negative leg can go in the breadboard.
Connect the negative leg of the capacitor to ground.
Connect a wire to Pin 2 to use as the trigger. Start with Pin 2 connected to Vcc.
Now connect the power. The LED will come on and stay on for about 2 seconds. Remove the wire connected to Pin 2 from Vcc. You should be able to trigger the 555 again by touching the wire connected to pin 2 with your finger or by connecting it to ground and removing it. (It should be about a 2 second pulse.)
Making it Oscillate
Next we will make the LED flash continually without having to trigger it. We will hook up the 555 so that it triggers itself. The way this works is that we add in a resistor between the capacitor and the discharge pin, Pin 7.
Now, the capacitor will charge up (through RA and RB) and when it reaches 2/3 Vcc, Pin 3 and Pin 7 will go to ground. But the capacitor can not discharge immediately because of RB. It takes some time for the charge to drain through RB.
The more resistance RB has, the longer it takes to discharge. The time it takes to discharge the capacitor will be the time the LED is off.
To trigger the 555 again, we connect Pin 6 to the trigger (Pin 2). As the capacitor is discharging, the voltage in the capacitor gets lower and lower.
When it gets down to 1/3 Vcc this triggers Pin 2 causing Pin 3 to go to Vcc and the LED to come on. The 555 disconnects Pin 7 from ground, and the capacitor starts to charge up again through RA and RB.
To build this circuit from the previous circuit, do the following.
Disconnect the power.
Take out the jumper wire between Pin 6 and Pin 7 and replace it with a 2.2k resistor (RB =2.2K).
Use the jumper wire at pin 2 to connect Pin 2 to Pin 6.
Now reconnect the power and the LED should flash forever (as long as you pay your electricity bill).
Experiment with different resistor values of RA and RB to see how it changes the length of time that the LED flashes. (You are changing the amount of time that it takes for the Capacitor to charge and discharge.)
Formulas
These are the formulas we use for the 555 to control the length of the pulses.
t1 = charge time (how long the LED is on) = 0.693 * (RA + RB) * C
t2 = discharge time (how long the LED is off) = 0.693 * RB * C
T = period = t1 + t2 = 0.693 * (RA + 2*RB) * C
Frequency = 1 / T = 1.44 / ((RA + 2 * RB) * C)
t1 and t2 are the time in seconds. C is the capacitor value in Farads. 220uF = 0.000220 F. So
for our circuit we have:
t1 = 0.693 * (10000 + 2200) * 0.000220 = 1.86 seconds
t2 = 0.693 * 2200 * 0.000220 = 0.335 seconds
T = 1.86 + 0.335 = 2.195 seconds
Frequency = 0.456 (cycles per second)
### LASER CONTROLLED ON OFF SWITCH USING 555 TIMER BASIC ELECTRONICS PROJECT
HOW TO MAKE LASER CONTROLLED ON OFF SWITCH?
This is a basic electronic project on how to make a laser controlled on/ off switch using a 555 timer. Below is its schematic diagram.
This circuit is built around a 555 timer using very few components. Since the circuit is very simple, even a novice can easily build it and use it as a controlling device.
A laser pointer, now easily available in the market, can be used to operate this device. This circuit has been tested in operational conditions from a distance of 500 metres and was found to work satisfactorily though it can be controlled from still longer distances.
Aiming (aligning) the laser beam exactly on to the LDR is a practical problem. The circuit is very useful in switching on/off a fan at night without getting off the bed. It can also be used for controlling a variety of other devices like radio or music system.
The limitation is that the circuit is operational only in dark or dull-lit environments. By focussing the laser beam on LDR1 the connected gadget can be activated through the relay, whereas by focussing laser beam on LDR2 we can switch off the gadget.
The timer is configured to operate in bistable mode.
### TOUCH ACTIVATED ALARM SYSTEM BASIC ELECTRONIC PROJECT
TOUCH ACTIVATED ALARM SYSTEM PROJECT USING 555 TIMER
This is a basic electronic project of Touch Activate Alarm System using a 555 timer. The 555 can be a LM, NE, or MC(cmos) type, they are all pin-compatible. **C1/C2's working voltage should be increased to 25V if you decide to go with a 12V power source. Below is the schematic diagram.
Parts List
R1 = 100K
D1 = 1N4004
U1 = 555 Timer*
R2 = 56K
C1 = 47μF/16V**
R3 = 10M
C2 = 33μF/16V**
R4 = 220K
T1 = 2N3904, or equivalent
P1 = 100K
Re1 = Relay***
Rule of thumb: the working voltage of capacitors are at least double the supplied voltage, in other words, if the powersource is 9Volt, your capacitor(s) is at least 18V. Transistor T1 can be any approximate substitute. *** Use any suitable relay for your project and if you're not tight on space, use any size. I've build this particular circuit to prevent students from fiddling with the security cameras in computer labs at the
University I am employed. I made sure the metal casing was not grounded. But as the schematic shows you can basically hook it up to any type of metal surface. I used a 12-vdc power source. Use any suitable relay to handle your requirements.
A 'RESET' switch (Normally Closed) can be added between the positive and the 'arrow-with-the-+'. The trigger (touch) wire is connected to pin 2 of the 555 and will trigger the relay, using your body resistance, when touched. It is obvious that the 'touching' part has to be clean and makes good contact with the trigger wire.
This particular circuit may not be suitable for all applications. Just in case you wonder why pin 5 is not listed in the schematic diagram; it is not really needed. In certain noisy conditions a small ceramic capacitor is placed between pin 5 and ground. It does no harm to add one or leave it out.
NOTE: For those of you who did not notice, there is an approximate 5-second delay build-in before activation of the relay to avoid false triggering, or a 'would-be' thief, etc.
AGAIN, make sure the latch is not touching anything 'ground' or the circuit just keeps resetting itself and so will not work. My shed has wooden doors so works fine. If you can not get yours to work, check the trigger input, verify there is some sort of signal coming from output pin 3 play with the value of R3.
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Survey
* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project
Document related concepts
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Transcript
```The Nilpotent generalization of Dirac’s famous Equation D(N)
where E, p, m, t and r are respectively energy, momentum, mass, time, space and the
symbols
± 1, ± i, ± i, ± j, ± k, ± i, ± j, ± k, are used to represent the respective units required by the
scalar, pseudo-scalar, quaternion and multivariate vector groups.
The Table of the nilpotents D(N, Xi), where the nilpotent operators Xi2 = 0, but Xi 0 specify
the quantizations of the experimentally validated Standard Model of elementary particle
physics in accordance with the spontaneous symmetry breaking of D(N) together with the
simultaneous emergence of 3+1 relativistic space-time:Baryons (spin 3/2):
Baryons (spin ½):
Leptons:
```
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https://www.pw.live/question-answer/if-the-first-day-of-any-month-is-monday.-what-date-will-be-on-the-37002
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. If the first day of any month is Monday. What date will be on the fourth Saturday of that month
1. 26
2. 27
3. 28
4. 29
Ans: 2
Sol: 1 – Monday
2 – Tuesday
3 – Wednesday
4 – Thursday
5 – Friday
6 – Saturday
6 + 7 = 13
13 + 7 = 20
20 + 7 = 27 (saturday)
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# VBBV
### PEC
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MATLAB enthusiast , Mechanical engineer (MeMe) :-)
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# Transfer Values in Northern Victoria Region
Updated March 20, 2018 14:21:36
It has finally happened in a clear cut example.
Labor won the last seat in Northern Victoria Region because of the formula used to calculate transfer values for the distribution of preferences from candidates who have more than a quota of votes.
The method used in Victoria, as in the Senate, is the Inclusive Grgeory method. It is an old manual counting system where, like adding water to peat moss, votes suddenly revert to their original value as ballot papers when a candidate passes a quota.
Had the Weighted Inclusive Gregory method been used, based on vote values rather than ballot papers, then Coalition preferences would have carried less weight on the election of the Shooters and Fishers Party, resulting in the Greens receiving more preferences at the next count. The result would have been Labor being excluded and the last seat being won by the Australian Country Alliance, not by Labor.
Let me explain how this occurs with reference to the following table. It shows the number of ballot papers held by the Shooters and Fishers Party on reaching a quota. The first column shows the source of the votes, the second column the number of ballot papers from this source, the third column the transfer value of these votes at this stage of the count, and the final column the vote value of the ballot papers.
(A vote equals the number of ballot papers multiplied by the appropriate transfer value.)
## Northern Victoria - Shooters and Fishers Party Total at Count 151
Totals219,770 76,192
The quota for election in Northern Victoria was 72,936. So at count 151, the Shooters and Fishers Party had a surplus of 3,256 votes.
How this surplus of votes is turned into preferences is the difference between the two methods of calculating transfer value.
The Inclusive Gregory Method reverts to ballot papers at this point. So the surplus of 3,256 is divided by the total ballot papers, 219,770, producing a transfer value of 0.01481548. This method heavily weights the preferences flows in favour of the Liberal/National ticket votes which make up 79.5% of all ballot papers, but only 41.4% of votes.
As the Liberal/National ticket had next preferences for the Australian Country Alliance, the Inclusive Gregory biases the preference flow to follow the Liberal ticket with next preference for the Australian Country Alliance.
The Weighted Inclusive Gregory Method divides the surplus by the number of votes. So the surplus of 3,256 votes is divided by the number of votes, 76,192, a transfer value of 0.04273414.
This may sound like a trivial difference, but look what it does to the 26,499 votes/ballot papers derived from the Palmer United, Sex Party and Cyclists Party ticket votes. These all had next preference for the Greens.
Under the Inclusive Gregory Method, these ballot papers are translated into 392 votes with next preferences for the Greens. This was the method used in Victoria, resulted in the Greens finishing 161 votes behind Labor, resulting in the Greens being excluded and Labor winning the final seat.
If the Weighted Inclusive Gregory Method had been used, the 26,499 ballot papers would have been 1,649 preferences for the Greens, putting the Greens ahead of Labor, resulting in Labor being excluded and the Australian Country Alliance winning the final seat.
So two formulas, two results. In my view the weighted Inclusive Gregory Method treats ballot papers more fairly on exclusion and should be used.
Most of the time this difference in formula is irrelevant. But this example shows that in certain cases, the choice of formula can produce different results.
In my view, the greater problem with the electoral system used for the Senate and the Victorian Legislative Council is the options presented to voters. The easy method of voting above the line leaves voters accepting a preference ticket they don't ever see, and even if they could, they would struggle to understand their meaning.
In my view the choice of transfer value formula is a smaller problem than the far greater ill of group ticket voting.
But as this example shows, if group ticket voting is abolished, then there is another issue to be addressed, the formulas for distributing preferences.
I agree, both that
* the weighted method is fairer (the other system gives multiple counts to the ticket of a larger party which may actually have had very few spare votes left over after people are elected), and
* the bigger problem (by far) is the group ticket system.
Indeed, the group ticket system has been so corrupted in enabling parties with tiny votes to get elected through preference harvesting that the inclusive system might (slightly) offset the problems that group voting creates (though it would not do it consistently, so it's not much of a benefit).
COMMENT: In my view, any increase in optional preferences requires a change to exclude votes with no further preferences from the transfer calculation formula. This applies in both the ACT and NSW, though in both jurisdictions the last bundle transfer value method is used.
- David December 16, 2014 at 09:15 PM
Thank you for this!
I think in this context we almost need two terms for transfer value - one that's vote-centric and one that's ballot-paper-centric.
- Alex December 16, 2014 at 10:18 PM
With regard to your comment on David's comment:
"change to exclude votes with no further preferences from the transfer calculation formula"
Wouldn't that just be "transfer-value equals surplus-votes divided by total non-exhausted-votes" (to use the WIGM method)?
COMMENT: Except you also apply the restriction that it cannot result in an increase in transfer value.
- Alex December 16, 2014 at 10:47 PM
I agree with both David and Antony on both points.
The group ticket system must go. Quite apart from the micro-party gaming, the lack of Robson rotation penalises any party winning more than a quota due to the votes being artificially concentrated on one candidate.
The Inclusive Gregory Method is just plain wrong. Suppose at count 151 in the table above there were an extra 50,000 votes from the minor parties increasing the total ballot papers to 269,770 and the Shooters and Fishers total votes to 126,192. The surplus is now 53,256 votes.
IGM divides the 53,256 surplus by the 269,770 ballot papers giving a transfer value of 0.19741261. Multiply the 174,808 LNP ballot papers by this inflated transfer value and 31551 votes received by Shooters and Fishers are sent on to Country Alliance as 34509 votes. Nearly 4000 LNP votes have appeared out of thin air.
Back to my first point. If candidates in each group were Robson rotated and voters were encouraged by the parties to vote for their team members in order of choice (as happens in Tasmania and the A.C.T.) then the chances are that no candidate would exceed the quota on first preferences. With no surpluses to be transferred, the early counts would all be exclusions with a transfer value of 1 and all major party candidates would still be in the running to receive them.
- Geoff Powell December 16, 2014 at 11:09 PM
Antony, good to see you up and typing again.
What is the current state of the proposals for modification of the Senate voting system?
COMMENT: No one has heard a peep on the subject for months. Something will need to happen soon to allow the AEC time to implement system changes.
- Mark Jeffery December 17, 2014 at 01:05 AM
I hope we can dispell this myth that micro-parties are somehow taking advantage of the system when it is in fact the major parties who are over-represented.
The big four labour/lgreens/liberal/national received only 81% of the primary vote which would be equivalent to 32-33 seats. Instead they control 35.
The concept of 'gaming' by microparties is absolute non-sense as the figures incontrovertibly demonstrate.
COMMENT: Except it is a preferential system, so some of the micro-party votes helped elect Labor, the Coalition and Green members, which is why the direct translation of vote share into seat share is not valid.
- dan December 17, 2014 at 06:29 AM
Robson is just as artificial, as it arbitrarily splits the votes between 5 candidates.
The odds of picking up an extra seat because of Robson are pretty low. The odds of a reserve-grade candidate accidentally taking out a frontbencher are much higher. It's basically an intra-party version of the minor party snowball that has sent such luminous people as Jacqui Lambie and Ricky Muir to Canberra. Do you think Labor really would have wanted to risk Penny Wong losing her seat to Simon Pisoni (whoever that is--he could be a Liberal mole with that name), or that the Nick Xenophon Group would have had no issue with suddenly discovering it's the Stirling Griff Group?
Besides, a party can already effect the same principle under the current system by lodging split preference tickets, with one ABC, one BCA, and one CAB. No one has done it yet, and likely no one ever will, because of the Pisoni factor.
All candidates within a party are not created equal, and there's no reason that chance should decide which candidates get elected, just like it's absurd that New South Wales still uses random sampling to distribute preferences.
- Christopher Burge December 17, 2014 at 06:31 AM
As I understand it, it is also an example of non-monotonicity, whereby the Country Alliance receiving additional votes actually disadvantaged them and caused them not to win a seat. Although it's an indirect example, as the fact that Country Alliance received more meant the Greens received less, thus allowing Greens prefs to elect Labor instead.
- Chris December 17, 2014 at 08:28 AM
Counting (maths) is science, it should not be open to interpretation, it should be right or wrong, and provable so !
If the debate is whether the inclusive Gregory or Weighted inclusive Gregory is a better approximation then surely we are debating the wrong thing.
Why is a transfer value used at all, why cant they just deal with the surplus votes (3256 in this example) directly (or maybe i dont understand something) ?
COMMENT: Yes, you're missing the question of how you determine the 3,256 surplus votes. Of the 76,192 votes, which 3,256 form the surplus.
You either randomly sample that many votes from the total, or you calculate a transfer value and distribute all votes at the reduced value.
- Glenn December 17, 2014 at 08:33 AM
Glenn: look up Arrow's Impossibility Theorem. In non-technical language, what it says is that no voting system can convert voters' ranked preferences among 3 or more options into a consensus decision without violating at least one of a number of principles that intuitively we would want any voting system to respect. Since any particular voting system you choose is going to violate at least one of those principles, there's no avoiding debates about which of the principles to preserve and which to give up.
- Steve Gardner December 17, 2014 at 11:34 AM
"COMMENT: Except it is a preferential system, so some of the micro-party votes helped elect Labor, the Coalition and Green members, which is why the direct translation of vote share into seat share is not valid."
Of course, but the point is that the microparties are less represented in terms of number than their proportion of the vote, so to suggest that they are over-represented (which several commenters/media figures suggest) is blatantly false.
The counting system may be flawed, but it is not to the advantage of the microparties. It should be mandatory for group ticket preference distribution to be published, but calls to abolish it primarily come from the major parties who feel they have a right to absolute rule and believe they would benefit from the change.
It took over 2 weeks to count the vote with only 8% of people voting below the line. Counting 100% below the line votes is not practical and would likely not lead to a significant difference in voting patterns.
COMMENT: In South-Eastern Metropolitan region there were enough micro-party votes to elect one of them, but it didn't happen because several micro-parties preferred to elect the Greens over Rise Up Australia. The STV system uses preferences to try and determine a most preferred elected candidate. In South-East Metro, there was clearly not a most-preferred micro-party candidate because some micro-parties preferred a major party first. If the parties chose not to add their votes together to elect a candidate, why are you wanting to add them together and argue they should have won a seat?
You are adding all the micro-parties together and saying because a voter voted for one of them with their first preference, then they therefore automatically would have wanted a micro-party elected.
Why create a total which includes both Family First and the Sex Party, when it was clear those parties did not want to elect each other. Family First tickets clearly put the Coalition ahead of the Sex Party. The Sex Party clearly preferred the Greens and Labor over Family First, the DLP, Australian Christians and Rise Up Australia. Any analysis of below the line votes where electors made up their own minds on preferences clearly indicated the same views amongst voters.
- dan December 17, 2014 at 03:35 PM
Robson doesn't split the votes arbitrarily - the voters decide where the votes go, and if the voters decide eg that they want a Labor representative, but not fronbencher X, Robson & Hare-Clark gives them that power. This is a good thing!
Robson doesn't stop parties from campaigning particularly hard, and successfully, for the canddates they choose to promote as their lead candidates - the Greens vote in Tasmanian electorates nearly always concentrates around one designated candidate
COMMENT: And the Tasmanian Greens are the only party that publishes an official ordering of preferences for candidates. As they usually only have the chance of winning a single candidate in each seat, it is in their interest to concentrate the vote in a single candidate.
I don't think it is valid to extrapolate the performance of Robson Rotation in Tasmania at a single house election with a quota of about 10,000, to how it might perform in NSW for an election in a second chamber with a quota of around 600,000.
- Simon December 17, 2014 at 06:40 PM
Christopher Burge is wrong when he says that a party can lodge multiple ("split") tickets so as to spread its first preference votes among its candidates. Section 211(2) of the Commonwealth Electoral Act 1918 requires all tickets lodged by a group to show the same preferences for members of the group. There is a similar provision in s.69B(1)(b)(ii) of the Victorian Electoral Act.
- Michael Maley December 17, 2014 at 08:53 PM
If you group the parties by a broad ideological position (my categorisation could be debatable, obviously they don't all perfectly agree), the results don't seem that unproportional,
Labor got 32% of the vote and 35% of seats
The Coalition got 35% of the vote and 40% of seats
Greens + AJP + Cyclists get 12.6% of the vote and 12.5% of seats
Liberal Democrats + Sex Party get 5.5% of votes and 2.5% of the seats
Aus Christians + RUA + DLP + Family First get 5.5% of votes and 2.5% of the seats
Shooters/Fishers + Country Alliance get 2.3% of the vote and 5% of seats
And the rest get 3.4% and 2.5% of the seats.
You could argue that there's some overlap between the Shooters/Fishers vote and the Liberal Democrats - so that probably balances out - likewise for the Christian vote and the Coalition.
From the perspective of the parties within my little groupings, Palmer United had the largest vote in the 'Others' category, and the Liberal Democrats had a higher vote than the Sex Party - so you could probably argue that the wrong party won those - but other than that, I think the results broadly reflect how people voted.
COMMENT: No, you don't do anything like that. Go to the distribution of preferences tickets and add them up to where they ended up in the count. It is a preferred counting method based on first preferences and distributed further preferences. It is not a system based on proportionality of first preference votes.
- Matt December 17, 2014 at 11:24 PM
My guess it was the Australian Democrats insisting a "split" ticket must list the candidates of the lodged group in the same order. They were happy to concentrate their votes on their one hopeful. They also knew that after the surpluses of the major parties were transferred, the third candidate would then have the lowest possible progress total and suffer from premature elimination.
COMMENT: No. The original provision stated the ticket must be in the home party's listed candidate order. Split tickets were an amendment proposed by the Democrats and just proposed that more than one ticket could be lodged. It allowed the Democrats to avoid choosing between Labor and the Coaliton with a single ticket. Robson Rotation had only been introduced in Tasmania four years earlier and there was no though about grafting it on to ticket voting.
Split tickets could be challenged in the High Court and the legal view is the High Court would strike them out in any challenge. The Commonwealth Electoral Act even includes a provision on what to do if split tickets were ruled unconstitutional.
- Geoff Powell December 18, 2014 at 12:16 AM
It's good that we now have a real demonstration of the malfunction caused by the unweighted 'version' of the (otherwise correct) IG rule. We should stop calling it a 'version' - it's not a rational variant, it's just a dumb mathematical flaw. To the best of my knowledge every single person who has examined this issue realises that it is a mistake.
And to underline that this is not just a trivial counting quirk, I'd like to emphasise that the use of this rule is actually converting people's votes from one kind into another. If I have the numbers right from this count, a surplus of 3,256 'votes' (an aggregate value based on the transfer procedure, of course) were to be handed on to other potential winners. Correctly weighted, 1,348 of these were properly meant to be Coalition ticket votes. The 'unweighted IG' rule changed that number to 2,590 Coalition ticket votes. To maintain the original total, it did that by converting the ticket votes of 389 Shooters supporters, 330 Palmer supporters, 359 Sex Party supporters, and so on, into those newly invented Coalition ticket votes.
This is simply outrageous.
And it doesn't work evenly for all players: by definition, vote manufacturing because of thus rule can only occur to the benefit of a party which has already won one or more seats - which is to say, almost always a major party. It always operates to take votes from the supporters of all other parties and candidates.
(To add to the perversity of the situation, the immediate beneficiary is not always the final beneficiary. The Coalition was gifted 1,242 invented votes in that count, but several stages later, and entirely unpredictably, it was the Labor Party which ended up gaining a seat because of this rule flaw.)
Yes the GVT system is a 'bigger' problem because it distorts preference flows for a far larger number of voters, but the unweighted IG flaw is a 'nastier' problem, because its real impact is to convert peoples' votes into being for other candidates.
These system flaws are now totally out in the open. We should keep the pressure on Parliament to fix them.
Thanks Antony - and look after the shoulder.
COMMENT: If there are n vacancies in a contest, and (n-1) of the seats are filled in the initial counts going straight down the party tickets, then the weighting problem never occurs. It only occurs in certain contests where fewer than (n-1) vacancies are filled on the initial counts, which means a candidate can reach a quota with more than 1 vacancy to fill. The AEC's position for a long time was the matter was theoretical, and until the last few years that was the case.
As I said, the biggest problem with Senate-style voting is not the formula to count the votes, it is group ticket voting and full preferential voting. Deal with the issues that cause giant ballot papers and exotic preference deals, and tack on fixing the counting formula as well. The minute you allow more exhausted preferences into the count, you need to re-consider the transfer value formula as well.
- Malcolm December 18, 2014 at 09:53 AM
Thanks Antony. I agree 100% with your two main points. I can understand why the main political parties support above the line voting. It gives their power-brokers much more power over who gets elected (and hence the people have less power, which is a concern to democratic-minded people like us). I had always thought there was no sensible reason why IGM was used instead of WIGM. The failings of IGM were pointed out in at least one submission to the Constitution Commission which lead to the current system being adopted. But this NV result does illustrate how IGM can give undue influence to the major party group voting tickets (Liberal's in this case). Was IGM a deliberate choice, do you think, or was it just a combination of ignorance and conservatism?
COMMENT: Inclusive Gregory was a deliberate choice because the alternative was thought too difficult for a manual count. Since then the AEC has been resistant to changing the formula.
- Lee Naish December 18, 2014 at 10:40 AM
For some reason, single transferable vote proportional representation seems to attract the attention of people who believe that it's possible to identify one system which is "correct"; the implication being that all other systems must be flawed. Of course, there's not always agreement among the recreational mathematical theologists on which system is the correct one.
It's interesting to contrast this with the approach taken to list proportional representation. A range of different formulae, all with different properties and with the potential to produce different winners from a given set of votes, are used around the world; but no political scientist worth his or her salt would claim that one is correct and the rest aren't. They are simply different.
A formula can really only be said to be incorrect if it has the potential to elect the wrong number of candidates, or to fail to produce a result at all (as will be the case if no provision is made for breaking ties).
- Michael Maley December 18, 2014 at 02:16 PM
I was unaware that parties are not allowed to order their candidates differently on different tickets, as that sounds absurd to me, but I guess that explains why no party has ever done it.
The only reason the Robson rotation was put in place (aside from the Tasmanian Liberals wanted to limit the power of Labor factions and Unions in preselecting candidates) and remains in place is because the vast majority of voters vote the candidates in the order they are on the paper--especially because the law also absurdly limits freedom of speech by prohibiting handing out how-to-vote cards.
It's the same reason that 92% of Victorian voters vote above the line despite only needing to vote 5 candidates below the line--because it's the easiest thing to do. Just like with the ticket system, Robson delivers results based on chance because voters don't tend to do research (either on which parties to preference or the differences between the candidates of their preferred party).
COMMENT: Actually, Robson rotation was implemented while Labor was in government. It was a non-government bill voted for by the Labor government to undermine a decision by the Labor Party state executive to impose a preference order how to vote. And your comment on voting order in Tasmania is completely wrong. Only a tiny minority vote for candidates in the order they appear in the party group.
- Christopher Burge December 18, 2014 at 03:44 PM
> No, you don't do anything like that. Go to the distribution of preferences tickets and add them up to where they ended up in the count. It is a preferred counting method based on first preferences and distributed further preferences. It is not a system based on proportionality of first preference votes.
Was focusing on the other comments suggesting that the micro parties are somehow being under-represented in the new legislative council, if you drill down into the particular micro parties that got the votes, the resulting total seats seems to be about right.
Obviously though that's entirely unintentional, given that none of the micro parties got anywhere near a vote share that should mean they get 1 of the 5 seats in a region, but somehow the system with a few oddities has delivered something close to the statewide vote totals.
It perhaps the result we might get if we had all members elected from a statewide 40 member vote, with above the line preferential voting without tickets.
- Matt December 18, 2014 at 03:51 PM
Any system that requires tens of rounds of preference redistribution before a "winning candidate" is identified is bound to throw up results that are unexpected both for candidates and voters. It's not an election, it is lottery that has been inexpertly fiddled. Optional preferential voting can't come too soon - including extinction of votes beyond the voters choice (which means the party list for above the line voters).
- Michael Cunningham December 19, 2014 at 01:52 AM
“For some reason, single transferable vote proportional representation seems to attract the attention of people who believe that it's possible to identify one system which is "correct"…”
Michael Maley, one of our most experienced electoral administrators, takes issue above with the notion that a single “correct” system of STV procedures can be found; indeed he suggests that searching for such a system indicates a theological attitude. Well, we should not choose voting procedures on faith alone - and nor should we keep bad ones on faith. We should choose voting systems in pursuit of principled objectives. For example, surely one vital principle is that all voters should be equally influential on the results of an election. Granted this is a ‘normative’ position, but it’s hardly a theological one.
With that principle in mind, let’s look again at the rules of surplus transfer in STV. Taking the ballots of all the voters supporting a successful candidate into account in calculating the transfer of votes (ie: ‘Inclusive Gregory’) treats all those voters equally; the older random sample and ‘last parcel’ alternatives do not. So IG can justifiably be called the ‘correct’ rule choice.
Similarly, the unweighted usage of IG boosts the effective value of some votes, and correspondingly reduces others (or one might say, looking at the total number of votes transferred, the rule ‘converts’ votes for one ticket into votes for another). So applying all the transfer values to every ballot paper (ie: ‘weighted IG’) can justifiably be called the ‘correct’ rule choice, and unweighted IG is ‘incorrect’.
There is nothing unreasonable about people who are interested in voting systems - which I guess is more or less everyone reading Antony’s blog - searching for ‘correct systems’, or at least trying to make correct specific choices when faced with different possible counting procedures. If we start from a clear statement of principled objectives, there is no reason why an optimal voting system cannot be discovered. The whole history of enthusiasts debating the mechanics of STV is precisely such a search, which is no doubt why so many people interested is voting theory gravitate to supporting STV.
Turning to his other comment, Maley is on solid ground saying there is little or no normative basis on which to prefer any of the various canonical formulae (Sainte-Laguë, D’Hondt etc) that are used to determine seat allocation in the ‘party list’ electoral systems of Europe, Latin America and other places. Indeed, the indices which political scientists use to assess the results of such formulae are really only mathematical mirrors of the rules themselves. But then, these seat allocation systems fail one the most basic of all electoral principles: they are not even systems for the direct election of representatives by the voters, but instead hand over the function of choosing representatives to political parties.
All of us interested in this subject are looking for a lot more from voting systems than simply an ability to turn out a specified number of winners. We should not be afraid to speak clearly in arguing that some system rules are ‘correct’, and others are not.
COMMENT: On your last point, there are plenty of variants of open list PR where voters determine the candidates. As recently as September Fiji used open list PR where the choice of candidates from within a party list was entirely in the hands of the voters. One of the weaknesses of all debates on PR in Australia is that most activists will not accept any form of PR except variants of STV.
- Malcolm December 19, 2014 at 11:42 AM
Could you explain why the Weighted Inclusive Gregory Method divides the surplus by the number of votes received by the candidate and not divide by the quota that a candidate must pass?
COMMENT: In short because it doesn't work, but I'll explain by example. Say the quota was 1,000 and the candidate had 2,500 votes. The correct transfer value is 1,500/2,500 = 0.6. If you distributed all 2,500 ballot papers at TV 0.6, you distribute 1,500 votes which is a value equal to the surplus.
With your formula, the TV would be 1,500/1,000 which is 1.5. If you applied this to 2,500 ballot papers, you would distribute 3,750 votes which is more than you started with.
- Matthew Casselman December 20, 2014 at 12:36 AM
Malcolm, in his thoughtful comment above, makes the case for seeking to base electoral systems on normative principles; and with that, I would thoroughly agree. The problem is that there will not necessarily be a consensus among different analysts, political players etc on which normative principles should be prioritised, when they come into conflict. That's why I would deny that it is possible to identify in a value-free way a single "correct" system.
Some STV enthusiasts, for example, are keen the Meek system. The problem with it, however, is that to understand how it works, you have to be an enthusiast yourself, with a head for mathematics. (It also helps to be able to read French, in which Mr Meek's original papers were written.) If you think, as many people do, that a highly desirable quality of an electoral system is that it be capable of being well understood by the voters, that basically rules out the Meek system.
I would also take issue with Malcolm's observation that "one the most basic of all electoral principles" is that a system should provide "for the direct election of representatives by the voters". That is highly context dependent: in a country where the biggest problem is electoral corruption and vote buying (and many third world countries fall into this category), you might very well want to avoid a system which gives strong incentives to individual candidates to build support bases by hook or by crook. I've argued elsewhere that the use of closed list PR has been a boon in Timor-Leste, as it has discouraged the sort of patron-client electoral corruption seen all too often in other countries, especially in Melanesia. (Of course, there may be corruption within parties of one form or another when lists of candidates are being assembled - look at NSW - but that's arguably a less intractable problem.)
Finally, I should note that my earlier reference to "recreational mathematical theology" was a slightly tongue-in-cheek allusion to Sheldon Glashow's description of string theory.
- Michael Maley December 20, 2014 at 05:44 PM
> "If you think, as many people do, that a highly desirable quality of an electoral system is that it be capable of being well understood by the voters, that basically rules out the Meek system."
Depends how one defines "understood". If one means by it that "the average person in a pub at Erskinville can tell you how the procedural steps work", then bang goes not only the Electoral Act but also the taxation laws, the superannuation legislation, the immigration via points system and, indeed, most of the statute book .
If on the other hand one means that "It gives results that the average person can quickly grasp" - whether this be "the party with most votes wins a majority of seats and forms a government" or "a party with N per cent of the votes will win about N per cent of the seats" - then most STV variants (I do not include the post-1983 Senate method) meet this criterion. Indeed, once you have to explain why one party with 2% and one local MP gets 4 seats under MMP while another party with 3% gets none, or why Tony Blair could win a huge Commons majority with 36% but David Cameron can't, or how George W Bush ended up defeating Al Gore in 2000, or why FPTP in NZ (before and after MMP) has given the largest party less than half the single-member district seats about one election in three... then most most STV variants (again, I do not include the post-1983 Senate method) meet this criterion better than many rival systems.
I agree that Meek is more complex (in the first sense above) but am not sure that its additional complexity is great enough to make either the Senate or the NSW Upper House methods seem models of simplicity (again, in the first sense above) by comparison.
COMMENT: Meek is complex by its method of counting. Currently 5% of Senate ballot papers and 20% in NSW LC ballot papers are data entered. If that has to become 100% the system will fall in a screaming heap.
The issue that first needs to be addressed is the options offered to voters, before going on to determine the formulas, whether that be inclusive gregory, weighted inclusive gregory, Meek or any other method of counting. Simplicity in voters understanding how to vote most effectively, and simplicity in conducting the count must be important criteria in any change.
- Tom Round December 21, 2014 at 09:40 PM
Michael, your comments regarding Meek's method are well out-of-date. Much work has been done since Brian Meek wrote his two papers 45 years ago, including the actual adoption of Meek-STV for some local elections in New Zealand (including for all 20 District Health Boards.
Meek-STV is actually simplicity itself, because the count unfolds in a mathematically logical fashion. I recommend you go to the Meek section at the PRSA website and read Dr David Hill's explanatory paper. Then go to where my comments are posted and read through the Cargill Ward count. I'm sure you will be pleasantly surprised how straightforward it all is.
COMMENT: I think there is an issue of scale between a New Zealand local health board and 4 million ballot papers for the NSW Senate. As I keep saying, the first question to be addressed is what options should voters have on the ballot paper to express their preferences. Sort that out before deciding the most appropriate formula for transferring the ballot papers into elected representatives.
- Stephen Todd December 22, 2014 at 05:54 PM
IT FINALLY HAPPENED
This issue was highlighted back in 2007 along with the flaw in the segmentation distribution of excluded candidates votes
In Victoria David Feeney would have lost his seat had One Nation preferenced the LNP ahead of the ALP as a result of the non-weighted surplus transfer.
In Queensland 2007 the issue of Segmentation and the distortion in the way the vote if counted elected the wrong Senator.
if you recount the 2007 Queensland Senate vote excluding all candidates except the last seven standing ( 3 ALP, 3 LNP and 1 GRN) the flaw in the way the vote is counted becomes apparent.
*(I challenge you to recount the Queensland Senate vote accordingly - Something you have refused and failed to do,)
The system was designed to facilitate a manual count not reflect the voters intentions. The hop skipping and jumping effects the outcome of the election Even more so in smaller Local Government elections
When a candidate is excluded from the count the votes should be redistributed as if the excluded candidates had not stood according to the voters nominated order of preference
This is best achieved by resetting and restarting the count on every exclusion
With the use of computer technology there is no excuse to maintain the current flawed method of counting the vote.
There should be a single transaction per candidate, no segmentation, with a weighted surplus transfer value with remainders staying with the value of the vote. The Wright System
We should also rethink the use of the Droop quota and adopt a equal distribution of above-the-line votes to all candidates within the group. With the group maintaining the right to determine the order of exclusion within the group.
- democracyATWork December 25, 2014 at 09:47 PM
First posted December 16, 2014 01:00:00
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### Invalid types for array
Hello. I'm trying to compile this program and i have some strange issue that i can't fix. Please help me!
``1234567891011121314151617181920212223242526272829303132333435363738394041424344454647484950515253545556575859`` ``````#include using namespace std; int dividing(int * T, int N, int l, int p) { int Rn = 0; int Ln = 0; int * R = new int[N-1]; int * L = new int[N-1]; //if(N==1) return; for(int i=l; iT[i]) { L[Ln] = T[i]; Ln++; } else { R[Rn] = T[i]; Rn++; } } T[Ln+1]=T[0]; for(int i=0;i=1) { q = dividing(T, N, l, p); sort(T, N, l, q); sort(T, N, q+1, N); } } int main() { int T[7] = {3,5,2,1,6,7,8}; sort(T,7,0,7); for(int i=0;i<7;i++) cout << "[" << T[i] << "] "; }``````
```sortowanie.cpp: In function ‘int dividing(int*, int, int, int)’: sortowanie.cpp:29:16: error: invalid types ‘int[int]’ for array subscript sortowanie.cpp:33:16: error: invalid types ‘int[int]’ for array subscript ```
You will probably figure out what this program should do and i know this isn't good and i will probably have another problems with this but i want to try challenge them by myself so I ask you to help me only with this one error. Thanks.
Last edited on
Ln and Rn aren't arrays.
I think you meant to use L and R, and not Ln and Rn.
Thanks a lot. Next stupid mistake. This compiler outputs are really non-intuitive for me... I hope that one day i won't have that kind of problems. This program isn't working the way i want it to work and this time i will give up my idea. Thanks again.
Topic archived. No new replies allowed.
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# What is Cumulative Distribution Function of this random variable?
Suppose that we have $n$ independent random variables, $x_1,\ldots,x_n$ such that each $x_i$ takes value $a_i$ with success probability $p_i$ and value $0$ with failure probability $1-p_i$ ,i.e.,
\begin{align} P(x_1=a_1) & = p_1,\ P(x_1=0)= 1-p_1 \\ P(x_2=a_2) & = p_2,\ P(x_2=0) = 1-p_2 \\ & \vdots \\ P(x_n=a_n) & = p_n,\ P(x_n=0)=1-p_n \end{align}
where $a_i$'s are positive Real numbers.
What would be the CDF of the sum of these random variables? That is, what would be $P(x_1+\cdots+x_n\le k)$ ? and how can we find it in an efficient way?
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It should be divised in many cases if there is an aswer for this. For example, if we have only three variables giving $a_1$, $a_2$ and $a_3$. If the three are different, we have $8$ possible values to get. If the three are equal, we have only $4$ possible values to get on the sum. And when $n$ gets bigger, the difficulties increase with combinations. – guaraqe Dec 20 '12 at 3:27
A sum of iid Bernoulli trials has a Binomial distribution, not Poisson. The distribution you described does not have a name. – Jonathan Christensen Dec 20 '12 at 3:28
The ordinary binomial distribution is a special case of the Poisson binomial distribution, when all success probabilities are the same: en.wikipedia.org/wiki/Poisson_binomial_distribution – May Dec 20 '12 at 3:29
I edited the question and deleted the part regarding Poisson binomial distribution since it was not relevant to my question. – May Dec 20 '12 at 4:19
This will be CDF of discrete distribution with certain number of values $s_1,...,s_r$ where $r$ is number of possible distinct sums among all subsets $\{a_{i_1},...,a_{i_m}\}$. It is not possible to evaluate $r$ in general case because this number solely depends on particular values $a_1,...,a_n$ and can vary from $n+1$ to $2^n$. The minimal $r=n+1$ is represented by case of $a_1=a_2=...=a_n$ while maximal $r=2^n$ is always achieved when ascending reordered set $\{a_{(1)},...,a_{(n)}\}$ complies with the condition: $\forall i: \sum\limits_{1\le j<i}a_{(j)}\le a_{(i)}$. Example of such maximal case could be $a_i=2^{i-1}$. Note, that failing this condition does not necessarily mean $r<2^n$. For example, for $a_i=ln(prime(i))$ the condition above fails, however all possible sums $a_{i_1}+...+a_{i_m}$ are still distinct, causing $r=2^n$.
We can always consider set $\{s_1,...,s_r\}$ as ordered, then $s_1=0$ and $s_r=\sum\limits_i a_i$. Then $$P(s_1)=P(0)=\prod\limits_i (1-p_i)$$ as all $x_i=0$; $$P(s_r)=P\left(\sum\limits_i a_i\right)=\prod\limits_i p_i$$ as all $x_i=a_i$; and $$P(s_k)=\sum\limits_{i_1,...,i_m|a_{i_1}+...+a_{i_m}=s_k}\left(\prod\limits_{j\in\{i_1,...,i_m\} } p_j\times\prod\limits_{j\in[1;n]\setminus\{i_1,...,i_m\} } (1-p_j)\right)=\\P(0)\times\sum\limits_{i_1,...,i_m|a_{i_1}+...+a_{i_m}=s_k}\left(\prod\limits_{j\in\{i_1,...,i_m\} }\frac{p_j}{1-p_j}\right)$$ as we can eliminate the second product of complemented set $[1;n]\setminus\{i_j\}$ by using the fact that $P(0)$ contains all $(1-p_i)$ for the whole set $[1;n]$.
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to calculate $P(x_1+⋯+x_n≤k)$ we need to find all the $s_i$ 's that $s_i<=k$ right? – May Feb 2 '13 at 21:31
do you mean $P(s_k)=P(x_1+⋯+x_n=k)$? – May Feb 2 '13 at 22:06
would you please explain your solution to me more? thank you. – May Feb 3 '13 at 1:48
Yes, as I started to talk about discrete distribution, I implied discrete form of CDF such as $P(s_k) = P(x_1+...+x_n=s_k)$ for all $s_k$ (sum is equal to $s_k$ not $k$) and $P(m)=0$ for any other $m$. This simplified the result but I accept that I needed to state it explicitly. As $\{s_k\}$ is set of fixed consts depending only on consts $\{a_k\}$, then I simplified the result by using this new set instead of the original consts. The last simplification was working on the result by eliminating the complementary set of sums using structure of $P(0)$. – Van Jone Feb 6 '13 at 23:17
thank you a lot for your clarification. – May Feb 7 '13 at 0:04
This answer is an attempt at providing an answer to a previous version of the question in which the $x_i$ were independent Bernoulli random variables with parameters $p_i$.
$P\{\sum_{i=1}^n x_i = k\}$ equals the coefficient of $z^k$ in $(1-p_1+p_1z)(1-p_2+p_2z)\cdots(1-p_n+p_nz)$. This can be found by developing the Taylor series for this function. It is not much easier than grinding out the answer by brute force.
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My solution to this problem is as follows
$P(x_1+⋯+x_n≤k)=∑ ^ n _{m=0}∑ _{A\epsilon T_m } \Pi_{i \epsilon A} p_i \Pi_{i \epsilon A^c } (1-p_i)$
where
$T_m=${{$r_1$,...,$r_m$}| $a_{r_1}+...+a_{r_m}<=k$ }
and $A^c$ is the complement of $A$.
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@Van Jone : I have this solution but I am not sure if it is correct. – May Feb 2 '13 at 21:54
Yes, your approach is correct. The difference to my solution is that I tried to go even further and squeeze as much as possible results by investigating intrinsic structure of set ${T_m}$ and how it "visually" contributes to behaviour of CDF. I could achieve it by presenting CDF in more particular discrete form and working out the result to eliminate some "impractical" operators such as $\sum$ and $\prod$. – Van Jone Feb 6 '13 at 22:59
SE "ate" curly braces where I said "set" and for some unknown reason it refuses any edits after 5 mins. So please use your imagination to put the braces and read $\{T_m\}$ instead of ${T_m}$ – Van Jone Feb 6 '13 at 23:22
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# Math Quarterly
The flashcards below were created by user James1 on FreezingBlue Flashcards.
1. What is a 5 sided figure called?
Pentagon
2. What is a 6 sided figure called?
Hexagon
3. What is a 7 sided figure called?
Heptagon/ Septagon
4. What is a 8 sided figure called?
Octagon
5. What is a 9 sided figure called?
Nonagon
6. What is a 10 sided figure called?
Decagon
7. Regular Polygon
All equal sides and Angles
8. Equilateral Triangle
All Sides are equal
9. Isosceles Triangle
2 Sides of the Triangle are equal
10. Scalene Triangle
All sides of the triangle have different measurements
11. What is the perimeter of a square if one side is 4 inches?
16 Inches
12. What is the perimeter of an equilateral trinagle if one leg is equal to three centimeters?
Nine Centimeters
13. What is the perimeter of an isosceles triangle whose base is six inches and one of the sides is 10.5 inches?
27 Inches
14. What is the perimeter of a Decagon with a side of 2.5 cm?
25 cm
15. What is the perimeter of a rectangle whose width is 5 feet and its length is 1 foot less?
18 Feet
Author: James1 ID: 129688 Card Set: Math Quarterly Updated: 2012-01-22T17:54:22Z Folders: Description: Math Quarterly Quarter 2 2012 Show Answers:
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Source code for sfepy.discrete.fem.mesh
```from __future__ import absolute_import
import time
import numpy as nm
import scipy.sparse as sp
from sfepy.base.base import Struct, invert_dict, get_default, output, assert_
from .meshio import MeshIO, supported_cell_types
import six
from scipy.spatial import cKDTree
eps = 1e-9
[docs]def set_accuracy(eps):
globals()['eps'] = eps
[docs]def find_map(x1, x2, allow_double=False, join=True):
"""
Find a mapping between common coordinates in x1 and x2, such that
x1[cmap[:,0]] == x2[cmap[:,1]]
"""
kdtree = cKDTree(nm.vstack([x1, x2]))
cmap = kdtree.query_pairs(eps, output_type='ndarray')
dns1 = nm.where(cmap[:, 1] < x1.shape[0])[0]
dns2 = nm.where(cmap[:, 0] >= x1.shape[0])[0]
if (dns1.size + dns2.size):
output('double node(s) in:')
for dn in dns1:
idxs = cmap[dn, :]
output('x1: %d %d -> %s %s' % (idxs[0], idxs[1],
x1[idxs[0], :], x1[idxs[1], :]))
for dn in dns2:
idxs = cmap[dn, :]
output('x2: %d %d -> %s %s' % (idxs[0], idxs[1],
x2[idxs[0], :], x2[idxs[1], :]))
if not allow_double:
raise ValueError('double node(s)! (see above)')
cmap[:, 1] -= x1.shape[0]
return cmap if join else (cmap[:, 0], cmap[:, 1])
[docs]def merge_mesh(x1, ngroups1, conn1, mat_ids1, x2, ngroups2, conn2, mat_ids2,
cmap):
"""
Merge two meshes in common coordinates found in x1, x2.
Notes
-----
Assumes the same number and kind of element groups in both meshes!
"""
n1 = x1.shape[0]
n2 = x2.shape[0]
err = nm.sum(nm.sum(nm.abs(x1[cmap[:,0],:-1] - x2[cmap[:,1],:-1])))
if abs(err) > (10.0 * eps):
raise ValueError('nonmatching meshes! (error: %e)' % err)
remap = nm.cumsum(mask) + n1 - 1
remap[cmap[:,1]] = cmap[:,0]
i2 = nm.setdiff1d(nm.arange( n2, dtype=nm.int32), cmap[:,1])
xx = nm.r_[x1, x2[i2]]
ngroups = nm.r_[ngroups1, ngroups2[i2]]
conn = nm.vstack((conn1, remap[conn2]))
mat_ids = None
if (mat_ids1 is not None) and (mat_ids2 is not None):
mat_ids = nm.concatenate((mat_ids1, mat_ids2))
return xx, ngroups, conn, mat_ids
[docs]def fix_double_nodes(coor, ngroups, conns):
"""
Detect and attempt fixing double nodes in a mesh.
The double nodes are nodes having the same coordinates
w.r.t. precision given by `eps`.
"""
n_nod, dim = coor.shape
cmap = find_map(coor, nm.zeros((0,dim)), allow_double=True)
if cmap.size:
output('double nodes in input mesh!')
output('trying to fix...')
while cmap.size:
# Just like in Variable.equation_mapping()...
ii = nm.argsort(cmap[:,1])
scmap = cmap[ii]
eq = nm.arange(n_nod)
eq[scmap[:,1]] = -1
eqi = eq[eq >= 0]
eq[eqi] = nm.arange(eqi.shape[0])
remap = eq.copy()
remap[scmap[:,1]] = eq[scmap[:,0]]
output(coor.shape)
coor = coor[eqi]
ngroups = ngroups[eqi]
output(coor.shape)
ccs = []
for conn in conns:
ccs.append(remap[conn])
conns = ccs
cmap = find_map(coor, nm.zeros((0,dim)), allow_double=True)
output('...done')
return coor, ngroups, conns
[docs]def get_min_vertex_distance(coor, guess):
"""Can miss the minimum, but is enough for our purposes."""
# Sort by x.
ix = nm.argsort(coor[:,0])
scoor = coor[ix]
mvd = 1e16
# Get mvd in chunks potentially smaller than guess.
n_coor = coor.shape[0]
i0 = i1 = 0
x0 = scoor[i0,0]
while 1:
while ((scoor[i1,0] - x0) < guess) and (i1 < (n_coor - 1)):
i1 += 1
## print i0, i1, x0, scoor[i1,0]
aim, aa1, aa2, aux = get_min_vertex_distance_naive(scoor[i0:i1+1])
if aux < mvd:
im, a1, a2 = aim, aa1 + i0, aa2 + i0
mvd = min(mvd, aux)
i0 = i1 = int(0.5 * (i1 + i0)) + 1
## i0 += 1
x0 = scoor[i0,0]
## print '-', i0
if i1 == n_coor - 1: break
## print im, ix[a1], ix[a2], a1, a2, scoor[a1], scoor[a2]
return mvd
[docs]def get_min_vertex_distance_naive(coor):
ii = nm.arange(coor.shape[0])
i1, i2 = nm.meshgrid(ii, ii)
i1 = i1.flatten()
i2 = i2.flatten()
ii = nm.where(i1 < i2)
aux = coor[i1[ii]] - coor[i2[ii]]
aux = nm.sum(aux**2.0, axis=1)
im = aux.argmin()
return im, i1[ii][im], i2[ii][im], nm.sqrt(aux[im])
[docs]def make_mesh(coor, ngroups, conns, mesh_in):
"""Create a mesh reusing mat_ids and descs of mesh_in."""
mat_ids = []
for ii, conn in enumerate(conns):
mat_id = nm.empty((conn.shape[0],), dtype=nm.int32)
mat_id.fill(mesh_in.mat_ids[ii][0])
mat_ids.append(mat_id)
mesh_out = Mesh.from_data('merged mesh', coor, ngroups, conns,
mat_ids, mesh_in.descs)
return mesh_out
[docs]class Mesh(Struct):
"""
The Mesh class is a light proxy to CMesh.
Input and output is handled by the MeshIO class and subclasses.
"""
[docs] @staticmethod
def from_file(filename=None, io='auto', prefix_dir=None,
omit_facets=False):
"""
Read a mesh from a file.
Parameters
----------
filename : string or function or MeshIO instance or Mesh instance
The name of file to read the mesh from. For convenience, a
mesh creation function or a MeshIO instance or directly a Mesh
instance can be passed in place of the file name.
io : *MeshIO instance
Passing *MeshIO instance has precedence over filename.
prefix_dir : str
If not None, the filename is relative to that directory.
omit_facets : bool
If True, do not read cells of lower dimension than the space
dimension (faces and/or edges). Only some MeshIO subclasses
support this!
"""
if isinstance(filename, Mesh):
return filename
if io == 'auto':
if filename is None:
output('filename or io must be specified!')
raise ValueError
else:
io = MeshIO.any_from_filename(filename, prefix_dir=prefix_dir)
cell_types = ', '.join(supported_cell_types[io.format])
output('reading mesh [%s] (%s)...' % (cell_types, io.filename))
tt = time.clock()
trunk = io.get_filename_trunk()
mesh = Mesh(trunk)
output('...done in %.2f s' % (time.clock() - tt))
mesh._set_shape_info()
return mesh
[docs] @staticmethod
def from_region(region, mesh_in, localize=False, is_surface=False):
"""
Create a mesh corresponding to cells, or, if `is_surface` is True, to
facets, of a given region.
"""
cmesh_in = mesh_in.cmesh
if not is_surface:
if not region.shape.n_cell:
raise ValueError('region %s has no cells!' % region.name)
cmesh = cmesh_in.create_new(region.cells, region.tdim,
localize=localize)
else:
if not region.shape.n_facet:
raise ValueError('region %s has no facets!' % region.name)
cmesh = cmesh_in.create_new(region.facets, region.tdim - 1,
localize=localize)
mesh = Mesh(mesh_in.name + "_reg", cmesh=cmesh)
if localize:
remap = nm.empty(mesh_in.cmesh.n_coor, dtype=nm.int32)
remap[:] = -1
remap[region.vertices] = nm.arange(region.vertices.shape[0])
mesh.nodal_bcs = {}
for key, val in six.iteritems(mesh_in.nodal_bcs):
new_val = remap[val]
mesh.nodal_bcs[key] = new_val[new_val >= 0]
else:
mesh.nodal_bcs = mesh_in.nodal_bcs.copy()
return mesh
[docs] @staticmethod
def from_data(name, coors, ngroups, conns, mat_ids, descs,
nodal_bcs=None):
"""
Create a mesh from mesh IO data.
"""
mesh = Mesh(name)
mesh._set_io_data(coors=coors,
ngroups=ngroups,
conns=conns,
mat_ids=mat_ids,
descs=descs,
nodal_bcs=nodal_bcs)
mesh._set_shape_info()
return mesh
def __init__(self, name='mesh', cmesh=None):
"""
Create a Mesh.
By default, the mesh is empty, see the `cmesh` argument.
Parameters
----------
name : str
Object name.
cmesh : CMesh, optional
If given, use this as the cmesh.
"""
Struct.__init__(self, name=name, nodal_bcs={}, io=None)
if cmesh is not None:
self.cmesh = cmesh
self._collect_descs()
self._set_shape_info()
[docs] def copy(self, name=None):
"""
Make a deep copy of the mesh.
Parameters
----------
name : str
Name of the copied mesh.
"""
if name is None:
name = self.name
cmesh = self.cmesh.create_new()
return Mesh(name=name, cmesh=cmesh)
"""
Merge the two meshes, assuming they have the same kind of the single
element group.
"""
cmap = find_map(self.coors, other.coors)
desc = self.descs[0]
aux = merge_mesh(self.coors, self.cmesh.vertex_groups,
self.get_conn(desc), self.cmesh.cell_groups,
other.coors, other.cmesh.vertex_groups,
other.get_conn(desc), other.cmesh.cell_groups,
cmap)
coors, ngroups, conn, mat_ids = aux
mesh = Mesh.from_data(self.name + ' + ' + other.name,
coors, ngroups, [conn], [mat_ids], [desc])
return mesh
def _collect_descs(self):
cmesh = self.cmesh
i2k = invert_dict(cmesh.key_to_index)
cts = nm.unique(cmesh.cell_types)
self.descs = [i2k[ct] for ct in cts]
def _set_shape_info(self):
self.n_nod, self.dim = self.coors.shape
self.n_el = self.cmesh.n_el
self.dims = [int(ii[0]) for ii in self.descs]
def _set_io_data(self, coors, ngroups, conns, mat_ids, descs,
nodal_bcs=None):
"""
Set mesh data.
Parameters
----------
coors : array
Coordinates of mesh nodes.
ngroups : array
Node groups.
conns : list of arrays
The array of mesh elements (connectivities) for each element group.
mat_ids : list of arrays
The array of material ids for each element group.
descs: list of strings
The element type for each element group.
nodal_bcs : dict of arrays, optional
The nodes defining regions for boundary conditions referred
to by the dict keys in problem description files.
"""
ac = nm.ascontiguousarray
coors = ac(coors, dtype=nm.float64)
if ngroups is None:
ngroups = nm.zeros((coors.shape[0],), dtype=nm.int32)
self.descs = descs
self.nodal_bcs = get_default(nodal_bcs, {})
from sfepy.discrete.common.extmods.cmesh import CMesh
self.cmesh = CMesh.from_data(coors, ac(ngroups),
[ac(conn, dtype=nm.int32)
for conn in conns],
ac(nm.concatenate(mat_ids)), descs)
def _get_io_data(self):
"""
Return data to be used by `MeshIO`.
"""
cmesh = self.cmesh
conns, mat_ids = [], []
for desc in self.descs:
conn, cells = self.get_conn(desc, ret_cells=True)
conns.append(conn)
mat_ids.append(cmesh.cell_groups[cells])
return cmesh.coors, cmesh.vertex_groups, conns, mat_ids, self.descs
@property
def coors(self):
return self.cmesh.coors
[docs] def write(self, filename=None, io=None, out=None, float_format=None,
**kwargs):
"""
Write mesh + optional results in `out` to a file.
Parameters
----------
filename : str, optional
The file name. If None, the mesh name is used instead.
io : MeshIO instance or 'auto', optional
Passing 'auto' respects the extension of `filename`.
out : dict, optional
The output data attached to the mesh vertices and/or cells.
float_format : str, optional
The format string used to print floats in case of a text file
format.
**kwargs : dict, optional
Additional arguments that can be passed to the `MeshIO` instance.
"""
if filename is None:
filename = self.name + '.mesh'
if io is None:
io = self.io
if io is None:
io = 'auto'
if io == 'auto':
io = MeshIO.any_from_filename(filename)
io.set_float_format(float_format)
io.write(filename, self, out, **kwargs)
[docs] def get_bounding_box(self):
return nm.vstack((nm.amin(self.coors, 0), nm.amax(self.coors, 0)))
[docs] def get_conn(self, desc, ret_cells=False):
"""
Get the rectangular cell-vertex connectivity corresponding to `desc`.
If `ret_cells` is True, the corresponding cells are returned as well.
"""
if desc not in self.descs:
raise ValueError("'%s' not in %s!" % (desc, self.descs))
cmesh = self.cmesh
cells = nm.where(cmesh.cell_types == cmesh.key_to_index[desc])
cells = cells[0].astype(nm.uint32)
cdim = int(desc[0])
conn = cmesh.get_incident(0, cells, cdim)
conn = conn.reshape((cells.shape[0], -1)).astype(nm.int32)
if ret_cells:
return conn, cells
else:
return conn
[docs] def transform_coors(self, mtx_t, ref_coors=None):
"""
Transform coordinates of the mesh by the given transformation matrix.
Parameters
----------
mtx_t : array
The transformation matrix `T` (2D array). It is applied
depending on its shape:
- `(dim, dim): x = T * x`
- `(dim, dim + 1): x = T[:, :-1] * x + T[:, -1]`
ref_coors : array, optional
Alternative coordinates to use for the transformation instead
of the mesh coordinates, with the same shape as `self.coors`.
"""
if ref_coors is None:
ref_coors = self.coors
if mtx_t.shape[1] > self.coors.shape[1]:
self.coors[:] = nm.dot(ref_coors, mtx_t[:,:-1].T) + mtx_t[:,-1]
else:
self.coors[:] = nm.dot(ref_coors, mtx_t.T)
[docs] def create_conn_graph(self, verbose=True):
"""
Create a graph of mesh connectivity.
Returns
-------
graph : csr_matrix
The mesh connectivity graph as a SciPy CSR matrix.
"""
from sfepy.discrete.common.extmods.cmesh import create_mesh_graph
shape = (self.n_nod, self.n_nod)
output('graph shape:', shape, verbose=verbose)
if nm.prod(shape) == 0:
output('no graph (zero size)!', verbose=verbose)
return None
output('assembling mesh graph...', verbose=verbose)
tt = time.clock()
conn = self.get_conn(self.descs[0])
nnz, prow, icol = create_mesh_graph(shape[0], shape[1],
1, [conn], [conn])
output('...done in %.2f s' % (time.clock() - tt), verbose=verbose)
output('graph nonzeros: %d (%.2e%% fill)' \
% (nnz, float(nnz) / nm.prod(shape)), verbose=verbose)
data = nm.ones((nnz,), dtype=nm.bool)
graph = sp.csr_matrix((data, icol, prow), shape)
return graph
```
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# ASH BUS 339 Week 4 Discussion 1 Statistics Recent
ASH BUS 339 Week 4 Discussion 1 Statistics Recent
Check this A+ tutorial guideline at
http://www.uopassignments.com/bus-339-ash/bus-339-week-4-discussion-1-statistics-recent
For more classes visit
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You have been asked to come up with a new ad campaign for a national pizza chain. You would like to target both young adults and families with children, but you don’t have the budget for two separate campaigns. What are some variables that you could measure and how would you analyze them to determine how these two groups are alike or different to decide what features you want to emphasize in your campaign?
# BUS 339 Week 5 Discussion 2 International Topics in Marketing Research Recent
http://www.uopassignments.com/bus-339-ash/bus-339-week-5-discussion-2-international-topics-in-marketing-research-recent
International Topics in Marketing Research. Review the web sites of at least three of the marketing research firms listed in Table 2.1 near the beginning of Chapter 2 in your text. What international topics and issues are they discussing? Respond to at least two of your classmates’ postings.
For more Assignments visit
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# BUS 339 Week 5 Discussion 1 Report Writing Recent
http://www.uopassignments.com/bus-339-ash/bus-339-week-5-discussion-1-report-writing-recent
Report Writing. What is the function of the “limitations” section of a marketing research report? What limitations exist in the marketing research project you designed for your Final Paper? Respond to at least two of your classmates.
For more Assignments visit
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# BUS 339 Week 4 Quiz Recent
http://www.uopassignments.com/bus-339-ash/bus-339-week-4-quiz-recent
If we were to graph two variables, let’s say, height (in inches) and GPA, and the graph showed points scattered about in a formless shape, we could say there is:
If we were to graph two variables, let’s say, height (in inches) and GPA, and the graph showed points scattered about in a formless shape, we could say there is:
If you have five friends who tell you they all have had great experience with their purchase of a Chevrolet, and you used this evidence to decide to buy a Chevrolet, you would be using:
When a researcher wants to compare the averages of several (more than two) groups he or she should use:
The most commonly used level of confidence when calculating a confidence interval is:
Which type of relationship is described by the formula: y = a + bx?
In conducting tests of differences between two percentages:
The generalization process is often referred to as:
The chi-square test is useful for determining:
If a researcher wishes to have a 95% confidence interval, he or she should use a Z value of:
For more Assignments visit
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# BUS 339 Week 4 Discussion 2 Recent
http://www.uopassignments.com/bus-339-ash/bus-339-week-4-discussion-1-statistics-recent
You have been asked to come up with a new ad campaign for a national pizza chain. You would like to target both young adults and families with children, but you don’t have the budget for two separate campaigns. What are some variables that you could measure and how would you analyze them to determine how these two groups are alike or different to decide what features you want to emphasize in your campaign?
For more Assignments visit
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# BUS 339 Week 3 Quiz Recent
http://www.uopassignments.com/bus-339-ash/bus-339-week-3-quiz-recent
BUS 339 Week 3 Quiz Recent
Since most consumers want a car with excellent fuel economy, wouldn’t you agree your next car will be a hybrid?” is an example of:
In the question, “Do you always buy electronic products from Dell?” Which word is one of the “words to avoid in question development?”
____________ is some list of all the members of the population.
A(n) _____________ is the vehicle used to pose questions that the researcher wants respondents to answer.
Which of the following refers to a dry run of the survey on a small, representative set of respondents to reveal errors in the questionnaire?
Question bias is defined as:
Which sampling method is not based on fairness, equity or equal chance?
A mall survey is an example of:
Which of the following is the best description of data analysis?
Computing the average number of dollars college students have on their credit card balances exemplifies:
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# BUS 339 Week 3 Discussion 2 Samples & Sampling Recent
http://www.uopassignments.com/bus-339-ash/bus-339-week-3-discussion-2-samples-and-sampling-recent
Samples & Sampling. After watching the American FactFinder Virtual Tour, identify a population using the Census Bureau’s American FactFinder. How could the sampling process lead to a bias or error in your data if you sampled this population? Respond to at least two of your classmates’ postings.
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# BUS 339 Week 3 Discussion 1 Research Objectives & Data Analysis Recent
http://www.uopassignments.com/bus-339-ash/bus-339-week-3-discussion-1-research-objectives-and-data-analysis-recent
Research Objectives & Data Analysis. Read the report BLS Spotlight on Statistics: Women at Work on the Bureau of Labor Statistics site. Find an example of each of the four research objectives listed in Table 11.1 at the beginning of Chapter 11 in your text. You should support your answer using the description of data analysis that is appropriate to each type of research, but you do not need to show the precise form of statistical analysis that was used to prepare the data. Respond to at least two of your classmates’ postings.
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# BUS 339 Week 2 Quiz Recent
http://www.uopassignments.com/bus-339-ash/bus-339-week-2-quiz-recent
Which of the following would be classified as a part of internal secondary data?
What type of research design should a marketing researcher use to find out how many customers there are, what brands they buy and in what quantities, which advertisements they recall, what are their attitudes toward the company, and who is the competition?
Although focus groups should encourage open, free-wheeling information, who has the task of ensuring that the conversation stays focused?
In an experiment, which type of variable does a researcher have control and wishes to manipulate?
Which of the following is most true about research designs?
Which of the following was NOT discussed as a disadvantage of using secondary data?
Standardized services differ from syndicated data services in that in the former each client is provided with:
Experience surveys are used in which of the following ways?
If you wanted to locate an article in a periodical on a subject you would find themit in:
A form of external, secondary data that is supplied from a common database to subscribers for a service fee is:
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Archive
Special Issues
Volume 6, Issue 1, March 2020, Page: 10-22
Inferences on the Weibull Exponentiated Exponential Distribution and Applications
Umar Usman, Department of Mathematics, Usmanu Danfodiyo University, Sokoto, Nigeria
Suleiman Shamsuddeen, Department of Mathematics, Usmanu Danfodiyo University, Sokoto, Nigeria
Bello Magaji Arkilla, Department of Community Health, Usmanu Danfodiyo University, Sokoto, Nigeria
Yakubu Aliyu, Department of Statistics, Ahmadu Bello University, Zaria, Nigeria
Received: Nov. 6, 2019; Accepted: Dec. 20, 2019; Published: Jul. 15, 2020
Abstract
In this article, an alternative method of defining the probability density function of GeneralizedWeibull-exponential distributions is proposed. Based on the method, the distribution can also be calledWeibull exponentiated exponential distribution. This distribution includes the exponential, Weibull and exponentiated exponential distributions as special cases. Comprehensive mathematical treatment of the distribution is provided. The quantile function, mode, characteristic function, moment generating function among other mathematical properties of the distribution were derived. The parameters of the distribution were estimated by applying the Maximum Likelihood Procedure.The elements of the Fisher Information Matrix is also provided. Finally, a data set is fitted to the model and its sub-models. It is observed that the new distribution is more flexible and can be used quiet effectively in analysing real life data in place of exponential, Weibull and exponentiated exponential distributions.
Keywords
T-X Family, Exponentiated Exponential Distribution, Order Statistics, Shannon Entropy and Likelihood Ratio Test
Umar Usman, Suleiman Shamsuddeen, Bello Magaji Arkilla, Yakubu Aliyu, Inferences on the Weibull Exponentiated Exponential Distribution and Applications, International Journal of Statistical Distributions and Applications. Vol. 6, No. 1, 2020, pp. 10-22. doi: 10.11648/j.ijsd.20200601.12
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Astea
★★
Russia,
2019-06-06 01:24
(911 d 16:13 ago)
Posting: # 20319
Views: 2,312
## Two-Stage Design for FDC [Two-Stage / GS Designs]
Dear Smart People!
I'm wondering what is an appropriate way to use Two-Stage Designs for FDC (two analytes, A and B, for example)? It seems we are facing several questions:
1). Initial sample size calculation - best guess of CV for two CV?
2). Initial sample size calculation - power (see this thread). So if we would not expect independent hypothesis we should use the adjusted power for calculation. Correspondingly this will lead to neccesity of using additional simulations cause we will be automatically driven from validated values of 80 and 90%.
3). Interim analyses: for 2 analytes it leads to different possibilities: pass, fail, pass for A but need the second stage for B...
4). Sample size for the next stage:
- additional subjects could cause TIE inflation (as in the example with extra drop-outs, see this thread
- is it regulatory addopted - ignoring data for the second analyte?
"Being in minority, even a minority of one, did not make you mad"
mittyri
★★
Russia,
2019-06-08 15:44
(909 d 01:53 ago)
@ Astea
Posting: # 20320
Views: 1,817
## No simple way out
Dear Astea,
as far as I can see you made a lot of forum research already
I don't see any other method except direct simulations, but as Detlew mentioned in the link you provided: too many what if's!!!
Remember that you need to take into account not only 2 analytes, but 2 PK metrics for both of them.
» 1). Initial sample size calculation - best guess of CV for two CV?
see above - not for 2 but for 4!
Guestimation is our friend. For that particular protocol I think you can prove almost any reliable numbers. (n1 is low: well, that's a 2 Stage design, I know nothing about CV!
n1 is high: I think CV for Cmax of that analyte should go to the sky!)
» 2). Initial sample size calculation - power (see this thread: ). So if we would not expect independent hypothesis we should use the adjusted power for calculation. Correspondingly this will lead to neccesity of using additional simulations cause we will be automatically driven from validated values of 80 and 90%.
If you don't know CVs how would you estimate rho?
May you want to build a correlation matrix for all of 4 pk metrics?
» 3). Interim analyses: for 2 analytes it leads to different possibilities: pass, fail, pass for A but need the second stage for B...
Yes, end of story (again, 4 metrics!). The framework becomes absolutely crazy. So I don't see any option except independent PK metrics analysis as you did for simple analytes in 2 stage designs. Forced BE? Yes, we're gonna live with that for now...
» 4). Sample size for the next stage:
» - additional subjects could cause TIE inflation (as in the example with extra drop-outs, see this thread.
From the link provided I don't see TIE inflation (using Potvin B and Detlew's function)
» - is it regulatory addopted - ignoring data for the second analyte?
No, as Helmut mentioned in the same link. I don't think experts be happy trying to dive into so complicated framework.
Kind regards,
Mittyri
Astea
★★
Russia,
2019-06-09 22:56
(907 d 18:41 ago)
@ mittyri
Posting: # 20321
Views: 1,777
## the more complicated the more interesting
Dear mittyri!
As for ρ (that is two PK metrics of the same analyte) I think that it is worth to expect correlation while for two different analytes correlation of the same PK metrics is not very likely.
» From the link provided I don't see TIE inflation (using Potvin B and Detlew's function)
Sorry, I've assumed the linked with this theme thread. Of course the theme is too complicated, but any possible affection on the TIE should be investigated, shouldn't it?
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# Expected value where benefit and probability depends on stochastic variable
I am trying to calculate the expected benefit of an action, where both the benefit and probability of carrying out the action depends on a parameter c, which is distributed according to f, F'=f.
The benefit of the action is given by $\delta b_h-c$, and therefore the action will only be carried out if $c \le \delta b_h$.
The probability of that is $\int_{-\infty}^{\delta b_h} f dc = F(\delta b_h)$.
But since the benefit does also depend of the draw of c, i am not sure how to calculate the expected value.
Any help is greatly appreciated.
Best, Esben
-
Lat $a=\delta b_h$. The benefit is $$\mathbb E((a-c)^+)=\displaystyle\int_{-\infty}^a(a-x)f_c(x)\mathrm dx=aF(a)-\int_{-\infty}^axf_c(x)\mathrm dx.$$ Edit: If $c$ is standard normal, $\mathbb E((a-c)^+)=a\Phi(a)+\varphi(a)$, where $$\varphi(a)=\frac1{\sqrt{2\pi}}\mathrm e^{-a^2/2},\qquad\Phi(a)=\int_{-\infty}^a\varphi(x)\mathrm dx.$$ Edit-edit: Since $x\varphi(x)=-\varphi'(x)$ and $\varphi(-\infty)=0$, $$\int_{-\infty}^{a}x\varphi(x)\mathrm dx=\int_{-\infty}^{a}-\varphi'(x)\mathrm dx=\left[-\varphi(x)\right]_{-\infty}^{a}=-\varphi(a).$$
No simpler general expression (but perhaps you have a specific distribution of $c$ in mind?). – Did Jan 11 '13 at 17:26
See Edit. – Did Jan 12 '13 at 12:31
Will you take me through the steps? It'd be GREAT! I guess this is how you get to the general solution? $\int_{-\infty}^{\delta b_h}\delta b_h f_c(x) -\int_{-\infty}^{\delta b_h}xf_c(x) =\delta b_hF(\delta b_h)-\int_{-\infty}^{\delta b_h}xf_c(x)\mathrm dx$. But how do you get from $-\int_{-\infty}^{\delta b_h}xf_c(x)\mathrm dx$ to $\frac1{\sqrt{2\pi}}\mathrm e^{-(\delta b_h)^2/2}$ (I know its the formula for the dist) – E_T Jan 12 '13 at 13:34
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# Plot 3D figure
5 vues (au cours des 30 derniers jours)
Tina Hsiao le 11 Nov 2021
Commenté : Walter Roberson le 12 Nov 2021
Hello, I would like to plot this fugure (ro, Z, I), my code as below, could you please give me a help? Thanks a lot.
clear all; close all; clc
for n = 1:1:100
i = sqrt(-1);
a = 0.62;
NA = 0.29;
wL = 772e-3; % micro meter
R = linspace (-0.4,0.4,100);
ro = ((2*pi)/wL)*NA.*R;
Z = linspace(-1,1,100);
mu = ((2*pi)/wL)*NA^2*Z;
fun1 = @(r)r.*exp(-r.^2).* besselj(ro(n),r) .*exp(-(1.*i.*mu(n).*r.^2)./2);
q1 = integral(fun1,0,1);
q2 = integral(fun1,0,a);
E(n) = (2/(1-exp(-1)))*q1- 2*(2/(1-exp(-1)))*q2; % electric field
I(n) = abs(E(n)).^2; % intensity
end
figure(2)
plot(R, I)
##### 0 commentairesAfficher -2 commentaires plus anciensMasquer -2 commentaires plus anciens
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### Réponse acceptée
Walter Roberson le 11 Nov 2021
The scale factor between and Z or R values is not obvious. It is also not obvious why you used the ranges you did: the sample plot suggests that they should be reversed. However, I had to use those ranges to get any plot of note -- and you can see it is not the expected plot. The output was essentially unchanged with Z -- which suggests that possibly Z is the wrong scale.
a = 0.62;
NA = 0.29;
wL = 772e-3; % micro meter
num_R = 100;
num_Z = 101;
Zvals = linspace(-5,5,num_Z)/1e2;
Rvals = linspace(-40,40,num_R)/1e2;
for Ridx = 1 : num_R
R = Rvals(Ridx);
ro = ((2*pi)/wL)*NA.*R;
for Zidx = 1 : num_Z
Z = Zvals(Zidx);
mu = ((2*pi)/wL)*NA^2*Z;
fun1 = @(r)r .* exp(-r.^2) .* besselj(ro,r) .* exp(-(1.*i.*mu.*r.^2)./2);
q1 = integral(fun1,0,1);
q2 = integral(fun1,0,a);
E(Ridx, Zidx) = (2/(1-exp(-1)))*q1 - 2*(2/(1-exp(-1)))*q2; % electric field
end
end
I = abs(E).^2; % intensity
figure(2)
surf(Zvals, Rvals, I, 'edgecolor', 'none')
##### 2 commentairesAfficher AucuneMasquer Aucune
Tina Hsiao le 12 Nov 2021
Thanks a lot. It works.
clear all; close all; clc
NA = 0.29;
wL = 772e-3; % micro meter
num_Z = 100;
num_R = 100;
Zvals= linspace(-40,40,num_Z);
Rvals = linspace(-5,5,num_R);
for Zidx = 1 : num_Z
Z = (Zvals(Zidx));
mu = ((2*pi)/wL).*NA^2.*Z;
for Ridx = 1 : num_R
R = (Rvals(Ridx));
ro = ((2*pi)/wL)*NA.*R;
fun1 = @(r)r.* exp(-r.^2) .* besselj(0,ro*r).* exp(-(1.*i.*mu.*r.^2)./2);
fun2 = @(a)a.* exp(-a.^2) .* besselj(0,ro*a).* exp(-(1.*i.*mu.*a.^2)./2);
q1 = integral(fun1,0,1);
q2 = integral(fun2,0,0.62);
E(Ridx, Zidx) = (2/(1-exp(-1))).*q1 - 2*(2/(1-exp(-1))).*q2; % electric field
end
end
phase = atan2(imag(E),real(E));
I = abs(E).^2; % intensity
figure(2)
surf( Rvals, Zvals, I/max(max(I)), 'edgecolor', 'none')
Walter Roberson le 12 Nov 2021
Looks good, but what do you do with phase after you calculate it?
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# If $a$ and $b$ are positive integers such that $a^n+n\mid b^n + n$ for all positive integers $n$, prove that $a=b$.
I ran into this problem in a math camp, but I can't seem to solve it via elementary techniques.
If $a$ and $b$ are positive integers such that $a^n+n\mid b^n + n$ for all positive integers $n$, prove that $a=b$.
• A simple first step: If $p\mid a$ then $p\mid b$ (try $n=p$). And if $p\nmid a$ then $p\nmid b$ (try $n=p-1$) – Hagen von Eitzen Mar 17 '15 at 7:45
• @Hagen, Aha! Thanks! – mursalin Mar 17 '15 at 7:55
Assume the contrary and consider a prime $p$ that does not divide $b-a$, By the Chinese Remainder Theorem we can find a postive integer $n$ such $$\begin{cases}n\equiv 1\pmod{p-1}\\ n\equiv -a\pmod p \end{cases}$$ Then by Fermat's little theorem $$a^n+n\equiv a+n\equiv a-a\equiv 0\pmod p$$ and $$b^n+n\equiv b+n\equiv b-a\pmod p$$ It follows that $p$ divides $a^n+n$,But does not divide $b^n+n$.Contradiction.
• See my answer for one way to motivate the solution to the classic contest problem. – Bill Dubuque Mar 17 '15 at 18:28
Hint Little Fermat kills evil powers. Below is a simple way to discover a proof. We prove the contrapositive, i.e. assuming $$\,a\neq b\,$$ we construct a counterexample to the divisibility claim.
To disprove $$\,a^n+n\mid b^n+n\,$$ we find a prime $$\,p\,$$ and some $$\,n\,$$ such that $$\,p\mid a^n+n,\,$$ $$\,p\nmid b^n + n.\,$$ To simplify, we can kill the exponents by restricting to $$\,\color{#0a0}{n\equiv 1}\pmod{p\!-\!1},\,$$ thus, by little Fermat, it follows that $$\,a^n\equiv a,\,b^n\equiv b\pmod p,\,$$ so our counterexample conditions simplify mod $$p\,$$ to
\begin{align} p\mid a^n+n\iff&\ 0\equiv a^n+n\equiv a+n\iff \color{#c00}{n\,\equiv\, -a}\!\!\pmod p\\ p\nmid b^n+n\iff&\ 0\not\equiv\, b^n+n\equiv b+\color{#c00}n\iff 0\not\equiv \color{}b\!\color{#c00}{-\!a}\!\!\pmod p\end{align}\quad\
Thus we can construct a counterexample by choosing a prime $$\,p\nmid \color{#a6f}{b\!-\!a}\,$$ and a solution $$\,n\,$$ to
\begin{align} \color{#0a0}{n\ \,\equiv\,\ 1}&\pmod{p\!-\!1}\\ \color{#c00}{n\equiv -a}&\pmod p\end{align}\quad
By $$\,p,\,p\!-\!1$$ coprime, a solution $$\,n\,$$ exists by CRT = Chinese Remainder Theorem. QED
Remark $$\$$ The following is a "proof without words" summary of the inferences
\begin{align} p-1 &\mid \color{#0a0}{n-1} \\ p\,&\mid \color{#c00}{a+n} \\ \end{align} \Rightarrow\,\ p\mid \overbrace{a^{\large \color{#0a0}n^{\phantom{I}}}\!\!\!-\!a+\color{#c00}{a\!+\!n}}^{\Large a^n+n}\mid \overbrace{b^{\large \color{#0a0}n^{\phantom{I}}}\!\!\!-\!b+\color{#a6f}{b\!-\!a}+\color{#c00}{a\!+\!n}}^{\Large b^n +n} \ \Rightarrow\,\ p\mid \color{#a6f}{b\!-\!a}\,\Rightarrow\!\Leftarrow\qquad
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Cody
# Problem 597. The Birthday Phenomenon
Solution 2115839
Submitted on 3 Feb 2020 by Wayne
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 = 1; y_correct = 0; assert(isequal(bday_phenom(x),y_correct))
y = 0
2 Pass
x = 2; y_correct = 0.0027; assert(isequal(bday_phenom(x),y_correct))
y = 0.0027
3 Pass
x = 5; y_correct = 0.0271; assert(isequal(bday_phenom(x),y_correct))
y = 0.0271
4 Pass
x = 10; y_correct = 0.1169; assert(isequal(bday_phenom(x),y_correct))
y = 0.1169
5 Pass
x = 20; y_correct = 0.4114; assert(isequal(bday_phenom(x),y_correct))
y = 0.4114
6 Pass
x = 30; y_correct = 0.7063; assert(isequal(bday_phenom(x),y_correct))
y = 0.7063
7 Pass
x = 50; y_correct = 0.9703; assert(isequal(bday_phenom(x),y_correct))
y = 0.9703
8 Pass
x = 366; y_correct = 1; assert(isequal(bday_phenom(x),y_correct))
y = 1
9 Pass
x = 4873; y_correct = 1; assert(isequal(bday_phenom(x),y_correct))
y = 1
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# How to create the awesome hexagon to entertain you Part 4
Now this is awesome, and you can make a hexaflexagon out of food! The video below shows you how to make it. (Note: ingredients may vary) https://youtu.be/GTwrVAbV56o Tasty ... I can't really describe how to make one of these, so I'll leave all the explanations to the video. https://youtu.be/Svq2Kscmmwc Back in the days ... Enjoy! … Continue reading How to create the awesome hexagon to entertain you Part 4
# How to create the awesome hexagon to entertain you Part 3
We have seen Richard Feynman and John Tukey. They are 2 of the other people who collaborated with old Arthur and Bryant. No, the Feynman diagram doesn't relate to hexaflexagons. Thank you Martin Gardner for the awesome Scientific American article. And double thank you for the awesome book Mathematical Games. It needs to be read. … Continue reading How to create the awesome hexagon to entertain you Part 3
# How to create the awesome hexagon to entertain you Part 2
Have you seen the video? Good. Now, if this is the first video you've seen about hexaflexagons, then you'll probably be like, "Who is this Bryant Tuckerman?" Who is this Bryant Tuckerman?Our readers who didn't know what Hexaflexagons are Yes, yes, thank you. Bryant Tuckerman (no, NOT Bryan) was one of the people who collaborated … Continue reading How to create the awesome hexagon to entertain you Part 2
# How to create the awesome hexagon to entertain you Part 1
Hexagons. They are 6-sided, hexagonal, and boring. HALT! HALT! Who are you?The Twelfth Doctor Well, I am the author of this post. Anyway, the hexagon's boringness depends on what kind of hexagon it is. If it is a normal paper hexagon, the hexagon is at the top, 10. For silk or sewing materials, that's a … Continue reading How to create the awesome hexagon to entertain you Part 1
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https://www.teacherspayteachers.com/Product/Math-Five-Squares-Game-Addition-and-Subtraction-Up-to-4-Digits-3605183
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Total:
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# Math Five Squares Game - Addition and Subtraction - Up to 4 Digits
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Five Squares Game
Colored pencils or crayons work best for this game.
Player 1 takes the PLAYER 1 game sheet plus the answer key for PLAYER 2. Player 2 takes the PLAYER 2 game sheet plus the answer key for PLAYER 1. Each player checks the other player's answers.
DIRECTIONS: Choose a game board. Larger boards make it easier to get 5 squares in a row. After the cards are shuffled, each player draws a card. The person who draws the highest problem number goes first. To start the game, the top card from the deck should be turned over. Each player finds the problem number on their own game sheet and answers it. If the answer is correct, that player gets to color in one square on the board, trying to get 5 squares in a row, while blocking the other player from getting 5 in a row. If a player gets the answer wrong, that player does nothing on the board for that turn.
(Important: Both players work on solving their problems at the same time, but they take turns coloring in the squares on the board, with the player who drew the highest card before the game going first. This way there is less down time for each player).
The first player to get 5 squares in a row wins the game. There are also few “free” spaces on the board(s).
ATTENTION:
There is no reason to buy multiple copies of our games. Simply buy one game unit of a topic of your choice and then buy "Extra Problems for Our Math Games" to get other math topics to use with the game(s).
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The figure shows the stress versus strain plot for an aluminum wire that is stretched by a machine pulling in opposite directions at the two ends of the wire. The scale of the stress axis is set by s = 7.0 in units of 107 N/m2. The wire has an initial length of 1.6 m and an initial cross-sectional area of 1.00E-6 m2. How much work does the force from the machine do on the wire to produce a strain of 1.0E-3?
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## Chapter 18. Type Inference
18.1. Concepts and Notation
18.1.1. Inference Variables
18.1.2. Constraint Formulas
18.1.3. Bounds
18.2. Reduction
18.2.1. Expression Compatibility Constraints
18.2.2. Type Compatibility Constraints
18.2.3. Subtyping Constraints
18.2.4. Type Equality Constraints
18.2.5. Checked Exception Constraints
18.3. Incorporation
18.3.1. Complementary Pairs of Bounds
18.3.2. Bounds Involving Capture Conversion
18.4. Resolution
18.5. Uses of Inference
18.5.1. Invocation Applicability Inference
18.5.2. Invocation Type Inference
18.5.2.1. Poly Method Invocation Compatibility
18.5.3. Functional Interface Parameterization Inference
18.5.4. More Specific Method Inference
A variety of compile-time analyses require reasoning about types that are not yet known. Principal among these are generic method applicability testing (§18.5.1) and generic method invocation type inference (§18.5.2). In general, we refer to the process of reasoning about unknown types as type inference.
At a high level, type inference can be decomposed into three processes:
• Reduction takes a compatibility assertion about an expression or type, called a constraint formula, and reduces it to a set of bounds on inference variables. Often, a constraint formula reduces to other constraint formulas, which must be recursively reduced. A procedure is followed to identify these additional constraint formulas and, ultimately, to express via a bound set the conditions under which the choices for inferred types would render each constraint formula true.
• Incorporation maintains a set of inference variable bounds, ensuring that these are consistent as new bounds are added. Because the bounds on one variable can sometimes impact the possible choices for another variable, this process propagates bounds between such interdependent variables.
• Resolution examines the bounds on an inference variable and determines an instantiation that is compatible with those bounds. It also decides the order in which interdependent inference variables are to be resolved.
These processes interact closely: reduction can trigger incorporation; incorporation may lead to further reduction; and resolution may cause further incorporation.
• §18.1 more precisely defines the concepts used as intermediate results and the notation used to express them.
• §18.2 describes reduction in detail.
• §18.3 describes incorporation in detail.
• §18.4 describes resolution in detail.
• §18.5 defines how these inference tools are used to solve certain compile-time analysis problems.
In comparison to the Java SE 7 Edition of The Java® Language Specification, important changes to inference include:
• Adding support for lambda expressions and method references as method invocation arguments.
• Generalizing to define inference in terms of poly expressions, which may not have well-defined types until after inference is complete. This has the notable effect of improving inference for nested generic method and diamond constructor invocations.
• Describing how inference is used to handle wildcard-parameterized functional interface target types and most specific method analysis.
• Clarifying the distinction between invocation applicability testing (which involves only the invocation arguments) and invocation type inference (which incorporates a target type).
• Delaying resolution of all inference variables, even those with lower bounds, until invocation type inference, in order to get better results.
• Improving inference behavior for interdependent (or self-dependent) variables.
• Eliminating bugs and potential sources of confusion. This revision more carefully and precisely handles the distinction between specific conversion contexts and subtyping, and describes reduction by paralleling the corresponding non-inference relations. Where there are intentional departures from the non-inference relations, these are explicitly identified as such.
• Laying a foundation for future evolution: enhancements to or new applications of inference will be easier to integrate into the specification.
## 18.1. Concepts and Notation
This section defines inference variables, constraint formulas, and bounds, as the terms will be used throughout this chapter. It also presents notation.
### 18.1.1. Inference Variables
Inference variables are meta-variables for types - that is, they are special names that allow abstract reasoning about types. To distinguish them from type variables, inference variables are represented with Greek letters, principally α.
The term "type" is used loosely in this chapter to include type-like syntax that contains inference variables. The term proper type excludes such "types" that mention inference variables. Assertions that involve inference variables are assertions about every proper type that can be produced by replacing each inference variable with a proper type.
### 18.1.2. Constraint Formulas
Constraint formulas are assertions of compatibility or subtyping that may involve inference variables. The formulas may take one of the following forms:
• Expression T›: An expression is compatible in a loose invocation context with type T (§5.3).
• S T›: A type S is compatible in a loose invocation context with type T (§5.3).
• S `<:` T›: A reference type S is a subtype of a reference type T (§4.10).
• S `<=` T›: A type argument S is contained by a type argument T (§4.5.1).
• S = T›: A type S is the same as a type T (§4.3.4), or a type argument S is the same as type argument T.
• LambdaExpression throws T›: The checked exceptions thrown by the body of the LambdaExpression are declared by the `throws` clause of the function type derived from T.
• MethodReference throws T›: The checked exceptions thrown by the referenced method are declared by the `throws` clause of the function type derived from T.
Examples of constraint formulas:
• From `Collections.singleton("hi")`, we have the constraint formula ‹`"hi"` α›. Through reduction, this will become the constraint formula: ‹`String` `<:` α›.
• From `Arrays.asList(1, 2.0)`, we have the constraint formulas ‹`1` α› and ‹`2.0` α›. Through reduction, these will become the constraint formulas ‹`int` α› and ‹`double` α›, and then ‹`Integer` `<:` α› and ‹`Double` `<:` α›.
• From the target type of the constructor invocation ```List<Thread> lt = new ArrayList<>()```, we have the constraint formula ‹`ArrayList``<`α`>` `List``<``Thread``>`›. Through reduction, this will become the constraint formula ‹α `<=` `Thread`›, and then ‹α = `Thread`›.
### 18.1.3. Bounds
During the inference process, a set of bounds on inference variables is maintained. A bound has one of the following forms:
• S = T, where at least one of S or T is an inference variable: S is the same as T.
• S `<:` T, where at least one of S or T is an inference variable: S is a subtype of T.
• false: No valid choice of inference variables exists.
• G`<`α1, ..., αn`>` = capture(G`<`A1, ..., An`>`): The variables α1, ..., αn represent the result of capture conversion (§5.1.10) applied to G`<`A1, ..., An`>` (where A1, ..., An may be types or wildcards and may mention inference variables).
• `throws` α: The inference variable α appears in a `throws` clause.
A bound is satisfied by an inference variable substitution if, after applying the substitution, the assertion is true. The bound false can never be satisfied.
Some bounds relate an inference variable to a proper type. Let T be a proper type. Given a bound of the form α = T or T = α, we say T is an instantiation of α. Similarly, given a bound of the form α `<:` T, we say T is a proper upper bound of α, and given a bound of the form T `<:` α, we say T is a proper lower bound of α.
Other bounds relate two inference variables, or an inference variable to a type that contains inference variables. Such bounds, of the form S = T or S `<:` T, are called dependencies.
A bound of the form G`<`α1, ..., αn`>` = capture(G`<`A1, ..., An`>`) indicates that α1, ..., αn are placeholders for the results of capture conversion. This is necessary because capture conversion can only be performed on a proper type, and the inference variables in A1, ..., An may not yet be resolved.
A bound of the form `throws` α is purely informational: it directs resolution to optimize the instantiation of α so that, if possible, it is not a checked exception type.
An important intermediate result of inference is a bound set. It is sometimes convenient to refer to an empty bound set with the symbol true; this is merely out of convenience, and the two are interchangeable.
Examples of bound sets:
• { α = `String` } contains a single bound, instantiating α as `String`.
• { `Integer` `<:` α, `Double` `<:` α, α `<:` `Object` } describes two proper lower bounds and one proper upper bound for α.
• { α `<:` `Iterable<?>`, β `<:` `Object`, α `<:` `List``<`β`>` } describes a proper upper bound for each of α and β, along with a dependency between them.
• { } contains no bounds nor dependencies, and can be referred to as true.
• { false } expresses the fact that no satisfactory instantiation exists.
When inference begins, a bound set is typically generated from a list of type parameter declarations P1, ..., Pp and associated inference variables α1, ..., αp. Such a bound set is generated as follows. For each l (1 l p):
• If Pl has no TypeBound, the bound αl `<:` `Object` appears in the set.
• Otherwise, for each type T delimited by `&` in the TypeBound, the bound αl `<:` T`[`P1:=α1, ..., Pp:=αp`]` appears in the set; if this results in no proper upper bounds for αl (only dependencies), then the bound αl `<:` `Object` also appears in the set.
## 18.2. Reduction
Reduction is the process by which a set of constraint formulas (§18.1.2) is simplified to produce a bound set (§18.1.3).
Each constraint formula is considered in turn. The rules in this section specify how the formula is reduced to one or both of:
• A bound or bound set, which is to be incorporated with the "current" bound set. Initially, the current bound set is empty.
• Further constraint formulas, which are to be reduced recursively.
Reduction completes when no further constraint formulas remain to be reduced.
The results of a reduction step are always soundness-preserving: if an inference variable instantiation satisfies the reduced constraints and bounds, it will also satisfy the original constraint. On the other hand, reduction is not completeness-preserving: there may exist inference variable instantiations that satisfy the original constraint but do not satisfy a reduced constraint or bound. This is due to inherent limitations of the algorithm, along with a desire to avoid undue complexity. One effect is that there are expressions for which type argument inference fails to find a solution, but that can be well-typed if the programmer explicitly inserts appropriate types.
### 18.2.1. Expression Compatibility Constraints
A constraint formula of the form ‹Expression T› is reduced as follows:
• If T is a proper type, the constraint reduces to true if the expression is compatible in a loose invocation context with T (§5.3), and false otherwise.
• Otherwise, if the expression is a standalone expression (§15.2) of type S, the constraint reduces to ‹S T›.
• Otherwise, the expression is a poly expression (§15.2). The result depends on the form of the expression:
• If the expression is a parenthesized expression of the form `(` Expression' `)`, the constraint reduces to ‹Expression' T›.
• If the expression is a class instance creation expression or a method invocation expression, the constraint reduces to the bound set B3 which would be used to determine the expression's compatibility with target type T, as defined in §18.5.2.1. (For a class instance creation expression, the corresponding "method" used for inference is defined in §15.9.3.)
This bound set may contain new inference variables, as well as dependencies between these new variables and the inference variables in T.
• If the expression is a conditional expression of the form `e1` `?` `e2` `:` `e3`, the constraint reduces to two constraint formulas, ‹`e2` T› and ‹`e3` T›.
• If the expression is a lambda expression or a method reference expression, the result is specified below.
By treating nested generic method invocations as poly expressions, we improve the behavior of inference for nested invocations. For example, the following is illegal in Java SE 7 but legal in Java SE 8:
```ProcessBuilder b = new ProcessBuilder(Collections.emptyList());
// ProcessBuilder's constructor expects a List<String>
```
When both the outer and the nested invocation require inference, the problem is more difficult. For example:
`List<String> ls = new ArrayList<>(Collections.emptyList());`
Our approach is to "lift" the bounds inferred for the nested invocation (simply { α `<:` `Object` } in the case of `emptyList`) into the outer inference process (in this case, trying to infer β where the constructor is for type `ArrayList``<`β`>`). We also infer dependencies between the nested inference variables and the outer inference variables (the constraint ‹`List``<`α`>` `Collection``<`β`>`› would reduce to the dependency α = β). In this way, resolution of the inference variables in the nested invocation can wait until additional information can be inferred from the outer invocation (based on the assignment target, β = `String`).
A constraint formula of the form ‹LambdaExpression T›, where T mentions at least one inference variable, is reduced as follows:
• If T is not a functional interface type (§9.8), the constraint reduces to false.
• Otherwise, let T' be the ground target type derived from T, as specified in §15.27.3. If §18.5.3 is used to derive a functional interface type which is parameterized, then the test that F`<`A'1, ..., A'm`>` is a subtype of F`<`A1, ..., Am`>` is not performed (instead, it is asserted with a constraint formula below). Let the target function type for the lambda expression be the function type of T'. Then:
• If no valid function type can be found, the constraint reduces to false.
• Otherwise, the congruence of LambdaExpression with the target function type is asserted as follows:
• If the number of lambda parameters differs from the number of parameter types of the function type, the constraint reduces to false.
• If the lambda expression is implicitly typed and one or more of the function type's parameter types is not a proper type, the constraint reduces to false.
This condition never arises in practice, due to the handling of implicitly typed lambda expressions in §18.5.1 and the substitution applied to the target type in §18.5.2.2.
• If the function type's result is `void` and the lambda body is neither a statement expression nor a void-compatible block, the constraint reduces to false.
• If the function type's result is not `void` and the lambda body is a block that is not value-compatible, the constraint reduces to false.
• Otherwise, the constraint reduces to all of the following constraint formulas:
• If the lambda parameters have explicitly declared types F1, ..., Fn and the function type has parameter types G1, ..., Gn, then i) for all i (1 i n), ‹Fi = Gi›, and ii) ‹T' `<:` T›.
• If the function type's return type is a (non-`void`) type R, assume the lambda's parameter types are the same as the function type's parameter types. Then:
• If R is a proper type, and if the lambda body or some result expression in the lambda body is not compatible in an assignment context with R, then false.
• Otherwise, if R is not a proper type, then where the lambda body has the form Expression, the constraint ‹Expression R›; or where the lambda body is a block with result expressions `e1`, ..., `em`, for all i (1 i m), ‹`ei` R›.
The key piece of information to derive from a compatibility constraint involving a lambda expression is the set of bounds on inference variables appearing in the target function type's return type. This is crucial, because functional interfaces are often generic, and many methods operating on these types are generic, too.
In the simplest case, a lambda expression may simply provide a lower bound for an inference variable:
```<T> List<T> makeThree(Factory<T> factory) { ... }
String s = makeThree(() `->` "abc").get(2);
```
In more complex cases, a result expression may be a poly expression - perhaps even another lambda expression - and so the inference variable might be passed through multiple constraint formulas with different target types before a bound is produced.
Most of the work described in this section precedes assertions about the result expressions; its purpose is to derive the lambda expression's function type, and to check for expressions that are clearly disqualified from compatibility.
We do not attempt to produce bounds on inference variables that appear in the target function type's `throws` clause. This is because exception containment is not part of compatibility (§15.27.3) - in particular, it must not influence method applicability (§18.5.1). However, we do get bounds on these variables later, because invocation type inference (§18.5.2.2) produces exception containment constraint formulas (§18.2.5).
Note that if the target type is an inference variable, or if the target type's parameter types contain inference variables, we produce false. During invocation type inference (§18.5.2.2), extra substitutions are performed in order to instantiate these inference variables, thus avoiding this scenario. (In other words, reduction will, in practice, never be "invoked" with a target type of one of these forms.)
Finally, note that the result expressions of a lambda expression are required by §15.27.3 to be compatible in an assignment context with the target type's return type, R. If R is a proper type, such as `Byte` derived from `Function<α,Byte>`, then assignability is easy enough to test, and reduction does so above. If R is not a proper type, such as α derived from `Function<String,α>`, then we make the simplifying assumption above that loose invocation compatibility will be sufficient. The difference between assignment compatibility and loose invocation compatibility is that only assignment allows narrowing of constant expressions, such as `Byte b = 100;`. Consequently, our simplifying assumption is not completeness-preserving: given target return type α and an integer literal result expression `100`, it is conceivable that α could be instantiated to `Byte`, but reduction will not in fact produce such a bound.
A constraint formula of the form ‹MethodReference T›, where T mentions at least one inference variable, is reduced as follows:
• If T is not a functional interface type, or if T is a functional interface type that does not have a function type (§9.9), the constraint reduces to false.
• Otherwise, if there does not exist a potentially applicable method for the method reference when targeting T, the constraint reduces to false.
• Otherwise, if the method reference is exact (§15.13.1), then let P1, ..., Pn be the parameter types of the function type of T, and let F1, ..., Fk be the parameter types of the potentially applicable method. The constraint reduces to a new set of constraints, as follows:
• In the special case where n = k+1, the parameter of type P1 is to act as the target reference of the invocation. The method reference expression necessarily has the form ReferenceType `::` [TypeArguments] Identifier. The constraint reduces to ‹P1 `<:` ReferenceType› and, for all i (2 i n), ‹Pi Fi-1›.
In all other cases, n = k, and the constraint reduces to, for all i (1 i n), ‹Pi Fi›.
• If the function type's result is not `void`, let R be its return type. Then, if the result of the potentially applicable compile-time declaration is `void`, the constraint reduces to false. Otherwise, the constraint reduces to ‹R' R›, where R' is the result of applying capture conversion (§5.1.10) to the return type of the potentially applicable compile-time declaration.
• Otherwise, the method reference is inexact, and:
• If one or more of the function type's parameter types is not a proper type, the constraint reduces to false.
This condition never arises in practice, due to the handling of inexact method references in §18.5.1 and the substitution applied to the target type in §18.5.2.2.
• Otherwise, a search for a compile-time declaration is performed, as specified in §15.13.1. If there is no compile-time declaration for the method reference, the constraint reduces to false. Otherwise, there is a compile-time declaration, and: (let R be the result of the function type)
• If R is `void`, the constraint reduces to true.
• Otherwise, if the method reference expression elides TypeArguments, and the compile-time declaration is a generic method, and the return type of the compile-time declaration mentions at least one of the method's type parameters, then:
• If R mentions one of the type parameters of the function type, the constraint reduces to false.
In this case, a constraint in terms of R might lead an inference variable to be bound by an out-of-scope type variable. Since instantiating an inference variable with an out-of-scope type variable is nonsensical, we prefer to avoid the situation by giving up immediately whenever the possibility arises. This simplification is not completeness-preserving.
• If R does not mention one of the type parameters of the function type, then the constraint reduces to the bound set B3 which would be used to determine the method reference's compatibility when targeting the return type of the function type, as defined in §18.5.2.1. B3 may contain new inference variables, as well as dependencies between these new variables and the inference variables in T.
The strategy used to determine a return type for a generic referenced method follows the pattern used earlier in this section for generic method invocations. This may involve "lifting" bounds into the outer context and inferring dependencies between the two sets of inference variables.
• Otherwise, let R' be the result of applying capture conversion (§5.1.10) to the return type of the invocation type (§15.12.2.6) of the compile-time declaration. If R' is `void`, the constraint reduces to false; otherwise, the constraint reduces to ‹R' R›.
### 18.2.2. Type Compatibility Constraints
A constraint formula of the form ‹S T› is reduced as follows:
• If S and T are proper types, the constraint reduces to true if S is compatible in a loose invocation context with T (§5.3), and false otherwise.
• Otherwise, if S is a primitive type, let S' be the result of applying boxing conversion (§5.1.7) to S. Then the constraint reduces to ‹S' T›.
• Otherwise, if T is a primitive type, let T' be the result of applying boxing conversion (§5.1.7) to T. Then the constraint reduces to ‹S = T'›.
• Otherwise, if T is a parameterized type of the form G`<`T1, ..., Tn`>`, and there exists no type of the form G`<`...`>` that is a supertype of S, but the raw type G is a supertype of S, then the constraint reduces to true.
• Otherwise, if T is an array type of the form G`<`T1, ..., Tn`>``[]`k, and there exists no type of the form G`<`...`>``[]`k that is a supertype of S, but the raw type G`[]`k is a supertype of S, then the constraint reduces to true. (The notation `[]`k indicates an array type of k dimensions.)
• Otherwise, the constraint reduces to ‹S `<:` T›.
The fourth and fifth cases are implicit uses of unchecked conversion (§5.1.9). These, along with any use of unchecked conversion in the first case, may result in compile-time unchecked warnings, and may influence a method's invocation type (§15.12.2.6).
Boxing T to T' is not completeness-preserving; for example, if T were `long`, S might be instantiated to `Integer`, which is not a subtype of `Long` but could be unboxed and then widened to `long`. We avoid this problem in most cases by giving special treatment to inference-variable return types that we know are already constrained to be certain boxed primitive types; see §18.5.2.1.
Similarly, the treatment of unchecked conversion sacrifices completeness in cases in which T is not a parameterized type (for example, if T is an inference variable). It is not usually clear in such situations whether the unchecked conversion is necessary or not. Since unchecked conversions introduce unchecked warnings, inference prefers to avoid them unless it is clearly necessary.
### 18.2.3. Subtyping Constraints
A constraint formula of the form ‹S `<:` T› is reduced as follows:
• If S and T are proper types, the constraint reduces to true if S is a subtype of T (§4.10), and false otherwise.
• Otherwise, if S is the null type, the constraint reduces to true.
• Otherwise, if T is the null type, the constraint reduces to false.
• Otherwise, if S is an inference variable, α, the constraint reduces to the bound α `<:` T.
• Otherwise, if T is an inference variable, α, the constraint reduces to the bound S `<:` α.
• Otherwise, the constraint is reduced according to the form of T:
• If T is a parameterized class or interface type, or an inner class type of a parameterized class or interface type (directly or indirectly), let A1, ..., An be the type arguments of T. Among the supertypes of S, a corresponding class or interface type is identified, with type arguments B1, ..., Bn. If no such type exists, the constraint reduces to false. Otherwise, the constraint reduces to the following new constraints: for all i (1 i n), ‹Bi `<=` Ai›.
• If T is any other class or interface type, then the constraint reduces to true if T is among the supertypes of S, and false otherwise.
• If T is an array type, T'`[]`, then among the supertypes of S that are array types, a most specific type is identified, S'`[]` (this may be S itself). If no such array type exists, the constraint reduces to false. Otherwise:
• If neither S' nor T' is a primitive type, the constraint reduces to ‹S' `<:` T'›.
• Otherwise, the constraint reduces to true if S' and T' are the same primitive type, and false otherwise.
• If T is a type variable, there are three cases:
• If S is an intersection type of which T is an element, the constraint reduces to true.
• Otherwise, if T has a lower bound, B, the constraint reduces to ‹S `<:` B›.
• Otherwise, the constraint reduces to false.
• If T is an intersection type, I1 `&` ... `&` In, the constraint reduces to the following new constraints: for all i (1 i n), ‹S `<:` Ii›.
A constraint formula of the form ‹S `<=` T›, where S and T are type arguments (§4.5.1), is reduced as follows:
• If T is a type:
• If S is a type, the constraint reduces to ‹S = T›.
• If S is a wildcard, the constraint reduces to false.
• If T is a wildcard of the form `?`, the constraint reduces to true.
• If T is a wildcard of the form `?` `extends` T':
• If S is a type, the constraint reduces to ‹S `<:` T'›.
• If S is a wildcard of the form `?`, the constraint reduces to ‹`Object` `<:` T'›.
• If S is a wildcard of the form `?` `extends` S', the constraint reduces to ‹S' `<:` T'›.
• If S is a wildcard of the form `?` `super` S', the constraint reduces to ‹`Object` = T'›.
• If T is a wildcard of the form `?` `super` T':
• If S is a type, the constraint reduces to ‹T' `<:` S›.
• If S is a wildcard of the form `?` `super` S', the constraint reduces to ‹T' `<:` S'›.
• Otherwise, the constraint reduces to false.
### 18.2.4. Type Equality Constraints
A constraint formula of the form ‹S = T›, where S and T are types, is reduced as follows:
• If S and T are proper types, the constraint reduces to true if S is the same as T (§4.3.4), and false otherwise.
• Otherwise, if S or T is the null type, the constraint reduces to false.
• Otherwise, if S is an inference variable, α, and T is not a primitive type, the constraint reduces to the bound α = T.
• Otherwise, if T is an inference variable, α, and S is not a primitive type, the constraint reduces to the bound S = α.
• Otherwise, if S and T are class or interface types with the same erasure, where S has type arguments B1, ..., Bn and T has type arguments A1, ..., An, the constraint reduces to the following new constraints: for all i (1 i n), ‹Bi = Ai›.
• Otherwise, if S and T are array types, S'`[]` and T'`[]`, the constraint reduces to ‹S' = T'›.
• Otherwise, if S and T are intersection types, a correspondence between the elements of S and the elements of T is established. An element of S, Si, corresponds to an element of T, Tj, if Si and Tj are either the same type, or both parameterizations of the same generic class or interface, or both array types.
If each element of S corresponds to exactly one element of T, and vice versa, then the constraint reduces to the following new constraints: for each element Si of S and the corresponding element Tj of T, ‹Si = Tj›. If not, the constraint reduces to false.
This rule does not accommodate inference variables appearing directly as elements of an intersection type (rather than nested in a parameterized type). Due to the restrictions on type parameter declarations (§4.4), such intersection types do not arise in practice.
• Otherwise, the constraint reduces to false.
A constraint formula of the form ‹S = T›, where S and T are type arguments (§4.5.1), is reduced as follows:
• If S and T are types, the constraint is reduced as described above.
• If S has the form `?` and T has the form `?`, the constraint reduces to true.
• If S has the form `?` and T has the form `?` `extends` T', the constraint reduces to ‹`Object` = T'›.
• If S has the form `?` `extends` S' and T has the form `?`, the constraint reduces to ‹S' = `Object`›.
• If S has the form `?` `extends` S' and T has the form `?` `extends` T', the constraint reduces to ‹S' = T'›.
• If S has the form `?` `super` S' and T has the form `?` `super` T', the constraint reduces to ‹S' = T'›.
• Otherwise, the constraint reduces to false.
### 18.2.5. Checked Exception Constraints
A constraint formula of the form ‹LambdaExpression throws T› is reduced as follows:
• If T is not a functional interface type (§9.8), the constraint reduces to false.
• Otherwise, let the target function type for the lambda expression be determined as specified in §15.27.3. If no valid function type can be found, the constraint reduces to false.
• Otherwise, if the lambda expression is implicitly typed, and one or more of the function type's parameter types is not a proper type, the constraint reduces to false.
This condition never arises in practice, due to the substitution applied to the target type in §18.5.2.2.
• Otherwise, if the function type's return type is neither `void` nor a proper type, the constraint reduces to false.
This condition never arises in practice, due to the substitution applied to the target type in §18.5.2.2.
• Otherwise, let E1, ..., En be the types in the function type's `throws` clause that are not proper types. If the lambda expression is implicitly typed, let its parameter types be the function type's parameter types. If the lambda body is a poly expression or a block containing a poly result expression, let the targeted return type be the function type's return type. Let X1, ..., Xm be the checked exception types that the lambda body can throw (§11.2). Then there are two cases:
• If n = `0` (the function type's `throws` clause consists only of proper types), then if there exists some i (1 i m) such that Xi is not a subtype of any proper type in the `throws` clause, the constraint reduces to false; otherwise, the constraint reduces to true.
• If n > `0`, the constraint reduces to a set of subtyping constraints: for all i (1 i m), if Xi is not a subtype of any proper type in the `throws` clause, then the constraints include, for all j (1 j n), ‹Xi `<:` Ej›. In addition, for all j (1 j n), the constraint reduces to the bound `throws` Ej.
A constraint formula of the form ‹MethodReference throws T› is reduced as follows:
• If T is not a functional interface type, or if T is a functional interface type but does not have a function type (§9.9), the constraint reduces to false.
• Otherwise, let the target function type for the method reference expression be the function type of T. If the method reference is inexact (§15.13.1) and one or more of the function type's parameter types is not a proper type, the constraint reduces to false.
• Otherwise, if the method reference is inexact and the function type's result is neither `void` nor a proper type, the constraint reduces to false.
• Otherwise, let E1, ..., En be the types in the function type's `throws` clause that are not proper types. Let X1, ..., Xm be the checked exceptions in the `throws` clause of the invocation type of the method reference's compile-time declaration (§15.13.2) (as derived from the function type's parameter types and return type). Then there are two cases:
• If n = `0` (the function type's `throws` clause consists only of proper types), then if there exists some i (1 i m) such that Xi is not a subtype of any proper type in the `throws` clause, the constraint reduces to false; otherwise, the constraint reduces to true.
• If n > `0`, the constraint reduces to a set of subtyping constraints: for all i (1 i m), if Xi is not a subtype of any proper type in the `throws` clause, then the constraints include, for all j (1 j n), ‹Xi `<:` Ej›. In addition, for all j (1 j n), the constraint reduces to the bound `throws` Ej.
Constraints on checked exceptions are handled separately from constraints on return types, because return type compatibility influences applicability of methods (§18.5.1), while exceptions only influence the invocation type after overload resolution is complete (§18.5.2). This could be simplified by including exception compatibility in the definition of lambda expression compatibility (§15.27.3), but this would lead to possibly surprising cases in which exceptions that can be thrown by an explicitly typed lambda body change overload resolution.
The exceptions thrown by a lambda body cannot be determined until i) the parameter types of the lambda are known, and ii) the target type of result expressions in the body is known. (The second requirement is to account for generic method invocations in which, for example, the same type parameter appears in the return type and the `throws` clause.) Hence, we require both of these, as derived from the target type T, to be proper types.
One consequence is that lambda expressions returned from other lambda expressions cannot generate constraints from their thrown exceptions. These constraints can only be generated from top-level lambda expressions.
Note that the handling of the case in which more than one inference variable appears in a function type's `throws` clause is not completeness-preserving. Either variable may, on its own, satisfy the constraint that each checked exception be declared, but we cannot be sure which one is intended. So, for predictability, we constrain them both.
## 18.3. Incorporation
As bound sets are generated and grown during inference, it is possible that new bounds can be inferred based on the assertions of the original bounds. The process of incorporation identifies these new bounds and adds them to the bound set.
Incorporation can happen in two scenarios. One scenario is that the bound set contains complementary pairs of bounds; this implies new constraint formulas, as specified in §18.3.1. The other scenario is that the bound set contains a bound involving capture conversion; this implies new bounds and may imply new constraint formulas, as specified in §18.3.2. In both scenarios, any new constraint formulas are reduced, and any new bounds are added to the bound set. This may trigger further incorporation; ultimately, the set will reach a fixed point and no further bounds can be inferred.
If incorporation of a bound set has reached a fixed point, and the set does not contain the bound false, then the bound set has the following properties:
• For each combination of a proper lower bound `L` and a proper upper bound U of an inference variable, `L` `<:` U.
• If every inference variable mentioned by a bound has an instantiation, the bound is satisfied by the corresponding substitution.
• Given a dependency α = β, every bound of α matches a bound of β, and vice versa.
• Given a dependency α `<:` β, every lower bound of α is a lower bound of β, and every upper bound of β is an upper bound of α.
The assertion that incorporation reaches a fixed point oversimplifies the matter slightly. Building on the work of Kennedy and Pierce, On Decidability of Nominal Subtyping with Variance, this property can be proven by making the argument that the set of types that may appear in the bound set is finite. The argument relies on two assumptions:
• New capture variables are not generated when reducing subtyping constraints (§18.2.3).
• Expansive inheritance paths are not pursued.
This specification does not currently guarantee these properties (it is imprecise about the handling of wildcards when reducing subtyping constraints, and does not detect expansive inheritance paths), but may do so in a future version. (This is not a new problem: the Java subtyping algorithm is also at risk of non-termination.)
### 18.3.1. Complementary Pairs of Bounds
(In this section, S and T are inference variables or types, and U is a proper type. For conciseness, a bound of the form α = T may also match a bound of the form T = α.)
When a bound set contains a pair of bounds that match one of the following rules, a new constraint formula is implied:
• α = S and α = T imply ‹S = T
• α = S and α `<:` T imply ‹S `<:` T
• α = S and T `<:` α imply ‹T `<:` S
• S `<:` α and α `<:` T imply ‹S `<:` T
• α = U and S = T imply ‹S`[`α:=U`]` = T`[`α:=U`]`
• α = U and S `<:` T imply ‹S`[`α:=U`]` `<:` T`[`α:=U`]`
When a bound set contains a pair of bounds α `<:` S and α `<:` T, and there exists a supertype of S of the form G`<`S1, ..., Sn`>` and a supertype of T of the form G`<`T1, ..., Tn`>` (for some generic class or interface, G), then for all i (1 i n), if Si and Ti are types (not wildcards), the constraint formula ‹Si = Ti› is implied.
### 18.3.2. Bounds Involving Capture Conversion
When a bound set contains a bound of the form G`<`α1, ..., αn`>` = capture(G`<`A1, ..., An`>`), new bounds are implied and new constraint formulas may be implied, as follows.
Let P1, ..., Pn represent the type parameters of G and let B1, ..., Bn represent the bounds of these type parameters. Let θ represent the substitution `[`P1:=α1, ..., Pn:=αn`]`. Let R be a type that is not an inference variable (but is not necessarily a proper type).
A set of bounds on α1, ..., αn is implied, generated from the declared bounds of P1, ..., Pn as specified in §18.1.3.
In addition, for all i (1 i n):
• If Ai is not a wildcard, then the bound αi = Ai is implied.
• If Ai is a wildcard of the form `?`:
• αi = R implies the bound false
• αi `<:` R implies the constraint formula ‹Bi θ `<:` R
• R `<:` αi implies the bound false
• If Ai is a wildcard of the form `?` `extends` T:
• αi = R implies the bound false
• If Bi is `Object`, then αi `<:` R implies the constraint formula ‹T `<:` R
• If T is `Object`, then αi `<:` R implies the constraint formula ‹Bi θ `<:` R
• R `<:` αi implies the bound false
• If Ai is a wildcard of the form `?` `super` T:
• αi = R implies the bound false
• αi `<:` R implies the constraint formula ‹Bi θ `<:` R
• R `<:` αi implies the constraint formula ‹R `<:` T
## 18.4. Resolution
Given a bound set that does not contain the bound false, a subset of the inference variables mentioned by the bound set may be resolved. This means that a satisfactory instantiation may be added to the set for each inference variable, until all the requested variables have instantiations.
Dependencies in the bound set may require that the variables be resolved in a particular order, or that additional variables be resolved. Dependencies are specified as follows:
• Given a bound of one of the following forms, where T is either an inference variable β or a type that mentions β:
• α = T
• α `<:` T
• T = α
• T `<:` α
If α appears on the left-hand side of another bound of the form G`<`..., α, ...`>` = capture(G`<`...`>`), then β depends on the resolution of α. Otherwise, α depends on the resolution of β.
• An inference variable α appearing on the left-hand side of a bound of the form G`<`..., α, ...`>` = capture(G`<`...`>`) depends on the resolution of every other inference variable mentioned in this bound (on both sides of the = sign).
• An inference variable α depends on the resolution of an inference variable β if there exists an inference variable γ such that α depends on the resolution of γ and γ depends on the resolution of β.
• An inference variable α depends on the resolution of itself.
Given a set of inference variables to resolve, let V be the union of this set and all variables upon which the resolution of at least one variable in this set depends.
If every variable in V has an instantiation, then resolution succeeds and this procedure terminates.
Otherwise, let { α1, ..., αn } be a non-empty subset of uninstantiated variables in V such that i) for all i (1 i n), if αi depends on the resolution of a variable β, then either β has an instantiation or there is some j such that β = αj; and ii) there exists no non-empty proper subset of { α1, ..., αn } with this property. Resolution proceeds by generating an instantiation for each of α1, ..., αn based on the bounds in the bound set:
• If the bound set does not contain a bound of the form G`<`..., αi, ...`>` = capture(G`<`...`>`) for all i (1 i n), then a candidate instantiation Ti is defined for each αi:
• If αi has one or more proper lower bounds, `L1`, ..., `Lk`, then Ti = lub(`L1`, ..., `Lk`) (§4.10.4).
• Otherwise, if the bound set contains `throws` αi, and each proper upper bound of αi is a supertype of `RuntimeException`, then Ti = `RuntimeException`.
• Otherwise, where αi has proper upper bounds U1, ..., Uk, Ti = glb(U1, ..., Uk) (§5.1.10).
The bounds α1 = T1, ..., αn = Tn are incorporated with the current bound set.
If the result does not contain the bound false, then the result becomes the new bound set, and resolution proceeds by selecting a new set of variables to instantiate (if necessary), as described above.
Otherwise, the result contains the bound false, so a second attempt is made to instantiate { α1, ..., αn } by performing the step below.
• If the bound set contains a bound of the form G`<`..., αi, ...`>` = capture(G`<`...`>`) for some i (1 i n), or;
If the bound set produced in the step above contains the bound false;
then let Y1, ..., Yn be fresh type variables whose bounds are as follows:
• For all i (1 i n), if αi has one or more proper lower bounds `L1`, ..., `Lk`, then let the lower bound of Yi be lub(`L1`, ..., `Lk`); if not, then Yi has no lower bound.
• For all i (1 i n), where αi has upper bounds U1, ..., Uk, let the upper bound of Yi be glb(U1 θ, ..., Uk θ), where θ is the substitution `[`α1:=Y1, ..., αn:=Yn`]`.
If the type variables Y1, ..., Yn do not have well-formed bounds (that is, a lower bound is not a subtype of an upper bound, or an intersection type is inconsistent), then resolution fails.
Otherwise, for all i (1 i n), all bounds of the form G`<`..., αi, ...`>` = capture(G`<`...`>`) are removed from the current bound set, and the bounds α1 = Y1, ..., αn = Yn are incorporated.
If the result does not contain the bound false, then the result becomes the new bound set, and resolution proceeds by selecting a new set of variables to instantiate (if necessary), as described above.
Otherwise, the result contains the bound false, and resolution fails.
The first method of instantiating an inference variable derives the instantiation from that variable's bounds. Sometimes, however, complex dependencies mean that the result is not within the variable's bounds. In that case, a different method of instantiation is performed, analogous to capture conversion (§5.1.10): fresh type variables are introduced, with bounds derived from the bounds of the inference variables. Note that the lower bounds of these "capture" variables are computed using only proper types: this is important in order to avoid attempts to perform typing computations on uninstantiated type variables.
## 18.5. Uses of Inference
Using the inference processes defined above, the following analyses are performed at compile time.
### 18.5.1. Invocation Applicability Inference
Given a method invocation that provides no explicit type arguments, the process to determine whether a potentially applicable generic method `m` is applicable is as follows:
• Where P1, ..., Pp (p 1) are the type parameters of `m`, let α1, ..., αp be inference variables, and let θ be the substitution `[`P1:=α1, ..., Pp:=αp`]`.
• An initial bound set, B0, is generated from the declared bounds of P1, ..., Pp, as described in §18.1.3.
• For all i (1 i p), if Pi appears in the `throws` clause of `m`, then the bound `throws` αi is implied. These bounds, if any, are incorporated with B0 to produce a new bound set, B1.
• A set of constraint formulas, C, is generated as follows.
Let F1, ..., Fn be the formal parameter types of `m`, and let `e1`, ..., `ek` be the actual argument expressions of the invocation. Then:
• To test for applicability by strict invocation:
If k n, or if there exists an i (1 i n) such that `ei` is pertinent to applicability (§15.12.2.2) and either i) `ei` is a standalone expression of a primitive type but Fi is a reference type, or ii) Fi is a primitive type but `ei` is not a standalone expression of a primitive type; then the method is not applicable and there is no need to proceed with inference.
Otherwise, C includes, for all i (1 i k) where `ei` is pertinent to applicability, ‹`ei` Fi θ›.
• To test for applicability by loose invocation:
If k n, the method is not applicable and there is no need to proceed with inference.
Otherwise, C includes, for all i (1 i k) where `ei` is pertinent to applicability, ‹`ei` Fi θ›.
• To test for applicability by variable arity invocation:
Let F'1, ..., F'k be the first k variable arity parameter types of `m` (§15.12.2.4). C includes, for all i (1 i k) where `ei` is pertinent to applicability, ‹`ei` F'i θ›.
• C is reduced (§18.2) and the resulting bounds are incorporated with B1 to produce a new bound set, B2.
• Finally, the method `m` is applicable if B2 does not contain the bound false and resolution of all the inference variables in B2 succeeds (§18.4).
Consider the following method invocation and assignment:
`List<Number> ln = Arrays.asList(1, 2.0);`
A most specific applicable method for the invocation must be identified as described in §15.12. The only potentially applicable method (§15.12.2.1) is declared as follows:
`public static <T> List<T> asList(T... a)`
Trivially (because of its arity), this method is neither applicable by strict invocation (§15.12.2.2) nor applicable by loose invocation (§15.12.2.3). But since there are no other candidates, in a third phase the method is checked for applicability by variable arity invocation.
The initial bound set, B, is a trivial upper bound for a single inference variable, α:
{ α `<:` `Object` }
The initial constraint formula set is as follows:
{ ‹`1` α›, ‹`2.0` α› }
These are reduced to a new bound set, B1:
{ α `<:` `Object`, `Integer` `<:` α, `Double` `<:` α }
Then, to test whether the method is applicable, we attempt to resolve these bounds. We succeed, producing the rather complex instantiation
α = ```Number & Comparable<? extends Number & Comparable<?>>```
We have thus demonstrated that the method is applicable; since no other candidates exist, it is the most specific applicable method. Still, the type of the method invocation, and its compatibility with the target type in the assignment, is not determined until further inference can occur, as described in the next section.
### 18.5.2. Invocation Type Inference
Given a method invocation expression that provides no explicit type arguments, and a corresponding most specific applicable generic method `m`, the process to infer the invocation type (§15.12.2.6) of the chosen method may require resolving additional constraints, both to assert compatibility with a target type and to assert validity of the method invocation's argument expressions.
It is important to note that multiple "rounds" of inference are involved in finding the type of a method invocation. This is necessary, for example, to allow a target type to influence the type of the invocation without allowing it to influence the choice of an applicable method. The first round (§18.5.1) produces a bound set and tests that a resolution exists, but does not commit to that resolution. Subsequent rounds reduce additional constraints until a final resolution step determines the "real" type of the expression.
#### 18.5.2.1. Poly Method Invocation Compatibility
If the method invocation expression is a poly expression (§15.12), its compatibility with a target type T is determined as follows.
If the method invocation expression appears in a strict invocation context and T is a primitive type, the expression is not compatible with T.
Otherwise:
• Let B2 be the bound set produced by reduction in order to demonstrate that `m` is applicable in §18.5.1.
(While it was necessary in §18.5.1 to demonstrate that the inference variables in B2 could be resolved, in order to establish applicability, the instantiations produced by this resolution step are not considered part of B2.)
• Let B3 be the bound set derived from B2 as follows.
Let R be the return type of `m`, and let θ be the substitution `[`P1:=α1, ..., Pp:=αp`]` defined in §18.5.1 to replace the type parameters of `m` with inference variables, and let T be the invocation's target type. Then:
• If unchecked conversion was necessary for the method to be applicable during constraint set reduction in §18.5.1, the constraint formula ‹|R| T› is reduced and incorporated with B2.
• Otherwise, if R θ is a parameterized type, G`<`A1, ..., An`>`, and one of A1, ..., An is a wildcard, then, for fresh inference variables β1, ..., βn, the constraint formula ‹G`<`β1, ..., βn`>` T› is reduced and incorporated, along with the bound G`<`β1, ..., βn`>` = capture(G`<`A1, ..., An`>`), with B2.
• Otherwise, if R θ is an inference variable α, and one of the following is true:
• T is a reference type, but is not a wildcard-parameterized type, and either i) B2 contains a bound of one of the forms α = S or S `<:` α, where S is a wildcard-parameterized type, or ii) B2 contains two bounds of the forms S1 `<:` α and S2 `<:` α, where S1 and S2 have supertypes that are two different parameterizations of the same generic class or interface.
• T is a parameterization of a generic class or interface, G, and B2 contains a bound of one of the forms α = S or S `<:` α, where there exists no type of the form G`<`...`>` that is a supertype of S, but the raw type |G`<`...`>`| is a supertype of S.
• T is a primitive type, and one of the primitive wrapper classes mentioned in §5.1.7 is an instantiation, upper bound, or lower bound for α in B2.
then α is resolved in B2, and where the capture of the resulting instantiation of α is U, the constraint formula ‹U T› is reduced and incorporated with B2.
• Otherwise, the constraint formula ‹R θ T› is reduced and incorporated with B2.
• The method invocation expression is compatible with T if B3 does not contain the bound false and resolution of all the inference variables in B3 succeeds (§18.4).
Consider the example from the previous section:
`List`<`Number`>` ln = Arrays.asList(1, 2.0);`
The most specific applicable method was identified as:
`public static `<`T`>` List`<`T`>` asList(T... a)`
In order to complete type-checking of the method invocation, we must determine whether it is compatible with its target type, `List<Number>`.
The bound set used to demonstrate applicability in the previous section, B2, was:
{ α `<:` `Object`, `Integer` `<:` α, `Double` `<:` α }
The new constraint formula set is as follows:
{ ‹`List<α>` `List<Number>`› }
This compatibility constraint produces an equality bound for α, which is included in the new bound set, B3:
{ α `<:` `Object`, `Integer` `<:` α, `Double` `<:` α, α = `Number` }
These bounds are trivially resolved:
α = `Number`
Finally, we perform a substitution on the declared return type of `asList` to determine that the method invocation has type `List<Number>`; clearly, this is compatible with the target type.
This inference strategy is different than the Java SE 7 Edition of The Java® Language Specification, which would have instantiated α based on its lower bounds (before even considering the invocation's target type), as we did in the previous section. This would result in a type error, since the resulting type is not a subtype of `List<Number>`.
Under various special circumstances, based on the bounds appearing in B2, we eagerly resolve an inference variable that appears as the return type of the invocation. This is to avoid unfortunate situations in which the usual constraint, ‹R θ T›, is not completeness-preserving. It is, unfortunately, possible that by eagerly resolving the variable, we are unable to make use of bounds that would be inferred later. It is also possible that, in some cases, bounds that will later be inferred from the invocation arguments (such as implicitly typed lambda expressions) would have caused a different outcome if they had been present in B2. Despite these limitations, the strategy allows for reasonable outcomes in typical use cases, and is backwards compatible with the algorithm in the Java SE 7 Edition of The Java® Language Specification.
The invocation type for the chosen method is determined after considering additional constraints that may be implied by the argument expressions of the method invocation expression, as follows:
• If the method invocation expression is a poly expression, let B3 be the bound set generated in §18.5.2.1 to demonstrate compatibility with the actual target type of the method invocation.
If the method invocation expression is not a poly expression, let B3 be the same as the bound set produced by reduction in order to demonstrate that `m` is applicable in §18.5.1.
(While it was necessary in §18.5.1 and §18.5.2.1 to demonstrate that the inference variables in the bound set could be resolved, the instantiations produced by these resolution steps are not considered part of B3.)
• A set of constraint formulas, C, is generated as follows.
Let `e1`, ..., `ek` be the actual argument expressions of the method invocation expression.
If `m` is applicable by strict or loose invocation, let F1, ..., Fk be the formal parameter types of `m`; if `m` is applicable by variable arity invocation, let F1, ..., Fk the first k variable arity parameter types of `m` (§15.12.2.4).
Let θ be the substitution `[`P1:=α1, ..., Pp:=αp`]` defined in §18.5.1 to replace the type parameters of `m` with inference variables.
Then, for all i (1 i k):
• If `ei` is not pertinent to applicability, C contains ‹`ei` Fi θ›.
• Additional constraints may be included, depending on the form of `ei`:
• If `ei` is a LambdaExpression, C contains ‹LambdaExpression throws Fi θ›, and the lambda body is searched for additional constraints:
• For a block lambda body, the search is applied recursively to each result expression.
• For a poly class instance creation expression or a poly method invocation expression , C contains all the constraint formulas that would appear in the set C generated by §18.5.2 when inferring the poly expression's invocation type.
• For a parenthesized expression, the search is applied recursively to the contained expression.
• For a conditional expression, the search is applied recursively to the second and third operands.
• For a lambda expression, the search is applied recursively to the lambda body.
• If `ei` is a MethodReference, C contains ‹MethodReference throws Fi θ›.
• If `ei` is a poly class instance creation expression or a poly method invocation expression, C contains all the constraint formulas that would appear in the set C generated by §18.5.2 when inferring the poly expression's invocation type.
• If `ei` is a parenthesized expression, these rules are applied recursively to the contained expression.
• If `ei` is a conditional expression, these rules are applied recursively to the second and third operands.
• While C is not empty, the following process is repeated, starting with the bound set B3 and accumulating new bounds into a "current" bound set, ultimately producing a new bound set, B4:
1. A subset of constraints is selected in C, satisfying the property that, for each constraint, no input variable can influence an output variable of another constraint in C. The terms input variable and output variable are defined below. An inference variable α can influence an inference variable β if α depends on the resolution of β (§18.4), or vice versa; or if there exists a third inference variable γ such that α can influence γ and γ can influence β.
If this subset is empty, then there is a cycle (or cycles) in the graph of dependencies between constraints. In this case, the constraints in C that participate in a dependency cycle (or cycles) and do not depend on any constraints outside of the cycle (or cycles) are considered. A single constraint is selected from these considered constraints, as follows:
• If any of the considered constraints have the form ‹Expression T›, then the selected constraint is the considered constraint of this form that contains the expression to the left (§3.5) of the expression of every other considered constraint of this form.
• If no considered constraint has the form ‹Expression T›, then the selected constraint is the considered constraint that contains the expression to the left of the expression of every other considered constraint.
2. The selected constraint(s) are removed from C.
3. The input variables α1, ..., αm of all the selected constraint(s) are resolved.
4. Where T1, ..., Tm are the instantiations of α1, ..., αm, the substitution `[`α1:=T1, ..., αm:=Tm`]` is applied to every constraint.
5. The constraint(s) resulting from substitution are reduced and incorporated with the current bound set.
• Finally, if B4 does not contain the bound false, the inference variables in B4 are resolved.
If resolution succeeds with instantiations T1, ..., Tp for inference variables α1, ..., αp, let θ' be the substitution `[`P1:=T1, ..., Pp:=Tp`]`. Then:
• If unchecked conversion was necessary for the method to be applicable during constraint set reduction in §18.5.1, then the parameter types of the invocation type of `m` are obtained by applying θ' to the parameter types of `m`'s type, and the return type and thrown types of the invocation type of `m` are given by the erasure of the return type and thrown types of `m`'s type.
• If unchecked conversion was not necessary for the method to be applicable, then the invocation type of `m` is obtained by applying θ' to the type of `m`.
If B4 contains the bound false, or if resolution fails, then a compile-time error occurs.
The process of reducing additional argument constraints may require carefully ordering constraint formulas of the forms ‹Expression T›, ‹LambdaExpression throws T›, and ‹MethodReference throws T›. To facilitate this ordering, the input variables of these constraints are defined as follows:
• For ‹LambdaExpression T›:
• If T is an inference variable, it is the (only) input variable.
• If T is a functional interface type, and a function type can be derived from T (§15.27.3), then the input variables include i) if the lambda expression is implicitly typed, the inference variables mentioned by the function type's parameter types; and ii) if the function type's return type, R, is not `void`, then for each result expression `e` in the lambda body (or for the body itself if it is an expression), the input variables of ‹`e` R›.
• Otherwise, there are no input variables.
• For ‹LambdaExpression throws T›:
• If T is an inference variable, it is the (only) input variable.
• If T is a functional interface type, and a function type can be derived, as described in §15.27.3, the input variables include i) if the lambda expression is implicitly typed, the inference variables mentioned by the function type's parameter types; and ii) the inference variables mentioned by the function type's return type.
• Otherwise, there are no input variables.
• For ‹MethodReference T›:
• If T is an inference variable, it is the (only) input variable.
• If T is a functional interface type with a function type, and if the method reference is inexact (§15.13.1), the input variables are the inference variables mentioned by the function type's parameter types.
• Otherwise, there are no input variables.
• For ‹MethodReference throws T›:
• If T is an inference variable, it is the (only) input variable.
• If T is a functional interface type with a function type, and if the method reference is inexact (§15.13.1), the input variables are the inference variables mentioned by the function type's parameter types and the function type's return type.
• Otherwise, there are no input variables.
• For ‹Expression T›, if Expression is a parenthesized expression:
Where the contained expression of Expression is Expression', the input variables are the input variables of ‹Expression' T›.
• For ‹ConditionalExpression T›:
Where the conditional expression has the form `e1` `?` `e2` `:` `e3`, the input variables are the input variables of ‹`e2` T› and ‹`e3` T›.
• For all other constraint formulas, there are no input variables.
The output variables of these constraints are all inference variables mentioned by the type on the right-hand side of the constraint, T, that are not input variables.
### 18.5.3. Functional Interface Parameterization Inference
Where a lambda expression with explicit parameter types P1, ..., Pn targets a functional interface type F`<`A1, ..., Am`>` with at least one wildcard type argument, then a parameterization of F may be derived as the ground target type of the lambda expression as follows.
Let Q1, ..., Qk be the parameter types of the function type of the type F`<`α1, ..., αm`>`, where α1, ..., αm are fresh inference variables.
If n k, no valid parameterization exists. Otherwise, a set of constraint formulas is formed with, for all i (1 i n), ‹Pi = Qi›. This constraint formula set is reduced to form the bound set B.
If B contains the bound false, no valid parameterization exists. Otherwise, a new parameterization of the functional interface type, F`<`A'1, ..., A'm`>`, is constructed as follows, for 1 i m:
• If B contains an instantiation (§18.1.3) for αi, T, then A'i = T.
• Otherwise, A'i = Ai.
If F`<`A'1, ..., A'm`>` is not a well-formed type (that is, the type arguments are not within their bounds), or if F`<`A'1, ..., A'm`>` is not a subtype of F`<`A1, ..., Am`>`, no valid parameterization exists. Otherwise, the inferred parameterization is either F`<`A'1, ..., A'm`>`, if all the type arguments are types, or the non-wildcard parameterization (§9.9) of F`<`A'1, ..., A'm`>`, if one or more type arguments are still wildcards.
In order to determine the function type of a wildcard-parameterized functional interface, we have to "instantiate" the wildcard type arguments with specific types. The "default" approach is to simply replace the wildcards with their bounds, as described in §9.8, but this produces spurious errors in cases where a lambda expression has explicit parameter types that do not correspond to the wildcard bounds. For example:
`Predicate<? super Integer> p = (Number n) `->` n.equals(23);`
The lambda expression is a `Predicate<Number>`, which is a subtype of `Predicate<? super Integer>` but not `Predicate<Integer>`. The analysis in this section is used to infer that `Number` is an appropriate choice for the type argument to `Predicate`.
That said, the analysis here, while described in terms of general type inference, is intentionally quite simple. The only constraints are equality constraints, which means that reduction amounts to simple pattern matching. A more powerful strategy might also infer constraints from the body of the lambda expression. But, given possible interactions with inference for surrounding and/or nested generic method invocations, this would introduce a lot of extra complexity.
### 18.5.4. More Specific Method Inference
When testing that one applicable method is more specific than another (§15.12.2.5), where the second method is generic, it is necessary to test whether some instantiation of the second method's type parameters can be inferred to make the first method more specific than the second.
Let `m1` be the first method and `m2` be the second method. Where `m2` has type parameters P1, ..., Pp, let α1, ..., αp be inference variables, and let θ be the substitution `[`P1:=α1, ..., Pp:=αp`]`.
Let `e1`, ..., `ek` be the argument expressions of the corresponding invocation. Then:
• If `m1` and `m2` are applicable by strict or loose invocation (§15.12.2.2, §15.12.2.3), then let S1, ..., Sk be the formal parameter types of `m1`, and let T1, ..., Tk be the result of θ applied to the formal parameter types of `m2`.
• If `m1` and `m2` are applicable by variable arity invocation (§15.12.2.4), then let S1, ..., Sk be the first k variable arity parameter types of `m1`, and let T1, ..., Tk be the result of θ applied to the first k variable arity parameter types of `m2`.
Note that no substitution is applied to S1, ..., Sk; even if `m1` is generic, the type parameters of `m1` are treated as type variables, not inference variables.
The process to determine if `m1` is more specific than `m2` is as follows:
• First, an initial bound set, B, is generated from the declared bounds of P1, ..., Pp, as specified in §18.1.3.
• Second, for all i (1 i k), a set of constraint formulas or bounds is generated.
If Ti is a proper type, the result is true if Si is more specific than Ti for `ei` (§15.12.2.5), and false otherwise. (Note that Si is always a proper type.)
Otherwise, if Si and Ti are not both functional interface types, the constraint formula ‹Si `<:` Ti› is generated.
Otherwise, if the interface of Si is a superinterface or a subinterface of the interface of Ti (or, where Si or Ti is an intersection type, some interface of Si is a superinterface or a subinterface of some interface of Ti), the constraint formula ‹Si `<:` Ti› is generated.
Otherwise, let MTS be the function type of the capture of Si, let MTS' be the function type of Si (without capture), and let MTT be the function type of Ti. If MTS and MTT have a different number of formal parameters or type parameters, or if MTS and MTS' do not have the same type parameters (§8.4.4), the result is false. Otherwise, the following constraint formulas or bounds are generated from the type parameters, formal parameter types, and return types of MTS and MTT:
• Let A1, ..., An be the type parameters of MTS, and let B1, ..., Bn be the type parameters of MTT.
Let θ' be the substitution `[`B1:=A1, ..., Bn:=An`]`. Then, for all j (1 j n):
• If the bound of Aj mentions one of A1, ..., An, and the bound of Bj is a not proper type, false.
• Otherwise, where X is the bound of Aj and Y is the bound of Bj, ‹X = Y θ'›.
If the bound Aj mentions one of A1, ..., An, and the bound of Bj is not a proper type, then producing an equality constraint would raise the possibility of an inference variable being bounded by an out-of-scope type variable. Since instantiating an inference variable with an out-of-scope type variable is nonsensical, we prefer to avoid the situation by giving up immediately whenever the possibility arises. This simplification is not completeness-preserving. (The same comment applies to the treatment of formal parameter types and return types below.)
• Let U1, ..., Uk be the formal parameter types of MTS, and let V1, ..., Vk be the formal parameter types of MTT. Then, for all j (1 j k):
• If Uj mentions one of A1, ..., An, and Vj is not a proper type, false.
• Otherwise, ‹Vj θ' `<:` Uj›, and, where U1', ..., Uk' are the formal parameter types of MTS', and A1', ..., An' are the type parameters of MTS', ‹Vj`[`B1:=A1', ..., Bn:=An'`]` = Uj'›
• Let RS be the return type of MTS, and let RT be the return type of MTT. Then:
• If RS mentions one of A1, ..., An, and RT is not a proper type, false.
• Otherwise, if `ei` is an explicitly typed lambda expression:
• If RT is `void`, true.
• Otherwise, if RS and RT are functional interface types, and `ei` has at least one result expression, then for each result expression in `ei`, this entire second step is repeated to infer constraints under which RS is more specific than RT θ' for the given result expression.
• Otherwise, if RS is a primitive type and RT is not, and `ei` has at least one result expression, and each result expression of `ei` is a standalone expression (§15.2) of a primitive type, true.
• Otherwise, if RT is a primitive type and RS is not, and `ei` has at least one result expression, and each result expression of `ei` is either a standalone expression of a reference type or a poly expression, true.
• Otherwise, ‹RS `<:` RT θ'›.
• Otherwise, if `ei` is an exact method reference:
• If RT is `void`, true.
• Otherwise, if RS is a primitive type and RT is not, and the compile-time declaration for `ei` has a primitive return type, true.
• Otherwise if RT is a primitive type and RS is not, and the compile-time declaration for `ei` has a reference return type, true.
• Otherwise, ‹RS `<:` RT θ'›.
• Otherwise, if `ei` is a parenthesized expression, these rules for constraints derived from RS and RT are applied recursively for the contained expression.
• Otherwise, if `ei` is a conditional expression, these rules for constraints derived from RS and RT are applied recursively for each of the second and third operands.
• Otherwise, false.
• Third, if `m2` is applicable by variable arity invocation and has k+1 parameters, then where Sk+1 is the k+1'th variable arity parameter type of `m1` and Tk+1 is the result of θ applied to the k+1'th variable arity parameter type of `m2`, the constraint ‹Sk+1 `<:` Tk+1› is generated.
• Fourth, the generated bounds and constraint formulas are reduced and incorporated with B to produce a bound set B'.
If B' does not contain the bound false, and resolution of all the inference variables in B' succeeds, then `m1` is more specific than `m2`.
Otherwise, `m1` is not more specific than `m2`.
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## hurst.m: compute the Hurst parameter of an array of samples
From: Francesco Potorti` Subject: hurst.m: compute the Hurst parameter of an array of samples Date: Thu, 20 Jul 1995 13:20 +0100 (MET)
# Put in the public domain by:
function [H, r, IDC, len] = hurst (A)
# Computes the Hurst parameter H of the vector column A.
#
# Use the Index of Dispersion for Counts (IDC, or peakedness), that is, the
# ratio of the variance over mean, computed over intervals of different
# lenghts. Plotted on a log-log diagram, IDC versus the interval length len
# should be least-squares interpolated by a line of slope beta=2H-1. r is
# the correlation coefficient and gives a "reliability factor" of the H
# estimate: values higher than 0.9 should be expected.
#
# Leland, Taqqu, Willinger, Wilson, "On the Self-Similar Nature of Ethernet
# Traffic (Extended Version)", IEEE/ACM Trans. on Networking Vol.2, Num.1,
# 1994.
if (!is_vector(A) || columns(A) != 1)
error ("A should be a column vector\n");
endif
# m is the minimum interval length over which the IDC is computed. M is
# the minimum number of intervals over which that same quantity is
# computed. k is the number of interval lengths per decade used for the
# computation. minp is the minimum number of points that must be used.
m = 3;
M = 10;
k = 8;
minp = 5;
r = rows (A);
minr = ceil (m*M*10^((minp-1)/k));
if (r < minr)
error (sprintf ("The argument should have at least %d rows\n", minr));
endif
n = floor (k * log10(r/m/M));
IDC = floor (logspace (log10(m), log10(r/M), n))';
for i = 1:n
sets = floor (r/IDC(i));
Len = sum (reshape (postpad (A, sets*IDC(i)), IDC(i), sets));
s = sum (Len);
len(i) = (sumsq(Len)*sets/s - s) / (sets-1);
endfor
logIDC = log (IDC);
loglen = log (len);
alfabeta = [ones (size (IDC)), logIDC] \ loglen;
H = (alfabeta(2)+1)/2;
if (nargout > 1) r = corrcoef (logIDC, loglen); endif
endfunction
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# How do you solve \pi x + 7x = 20?
May 23, 2017
$x = 1.972$
#### Explanation:
$\pi x + 7 x = 20$
or
$\left(\pi + 7\right) x = 20$
or
$x = \frac{20}{\pi + 7}$
or
$x = 1.972$
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# Search by Topic
#### Resources tagged with PM - Visualising similar to Painted Purple:
Filter by: Content type:
Stage:
Challenge level:
### There are 21 results
Broad Topics > Secondary processes > PM - Visualising
### Painted Cube
##### Stage: 3 Challenge Level:
Imagine a large cube made from small red cubes being dropped into a pot of yellow paint. How many of the small cubes will have yellow paint on their faces?
### Cubic Net
##### Stage: 4 and 5 Challenge Level:
This is an interactive net of a Rubik's cube. Twists of the 3D cube become mixes of the squares on the 2D net. Have a play and see how many scrambles you can undo!
### Marbles in a Box
##### Stage: 3 Challenge Level:
How many winning lines can you make in a three-dimensional version of noughts and crosses?
### Route to Infinity
##### Stage: 3 Challenge Level:
Can you describe this route to infinity? Where will the arrows take you next?
### Seven Squares
##### Stage: 3 Challenge Level:
Watch these videos to see how Phoebe, Alice and Luke chose to draw 7 squares. How would they draw 100?
### Nine Colours
##### Stage: 3 Challenge Level:
Can you use small coloured cubes to make a 3 by 3 by 3 cube so that each face of the bigger cube contains one of each colour?
### Frogs
##### Stage: 3 Challenge Level:
How many moves does it take to swap over some red and blue frogs? Do you have a method?
### Diminishing Returns
##### Stage: 3 Challenge Level:
In this problem, we have created a pattern from smaller and smaller squares. If we carried on the pattern forever, what proportion of the image would be coloured blue?
### Rolling Around
##### Stage: 3 Challenge Level:
A circle rolls around the outside edge of a square so that its circumference always touches the edge of the square. Can you describe the locus of the centre of the circle?
### Air Nets
##### Stage: 2, 3, 4 and 5 Challenge Level:
Can you visualise whether these nets fold up into 3D shapes? Watch the videos each time to see if you were correct.
### ACE, TWO, THREE...
##### Stage: 3 Challenge Level:
Can you picture how to order the cards to reproduce Charlie's card trick for yourself?
### Pick's Theorem
##### Stage: 3 Challenge Level:
Polygons drawn on square dotty paper have dots on their perimeter (p) and often internal (i) ones as well. Find a relationship between p, i and the area of the polygons.
### Square It
##### Stage: 3 and 4 Challenge Level:
Players take it in turns to choose a dot on the grid. The winner is the first to have four dots that can be joined to form a square.
### Diamond Collector
##### Stage: 3 Challenge Level:
Collect as many diamonds as you can by drawing three straight lines.
### Constructing Triangles
##### Stage: 3 Challenge Level:
Generate three random numbers to determine the side lengths of a triangle. What triangles can you draw?
### Hidden Squares
##### Stage: 3 Challenge Level:
Can you find the squares hidden on these coordinate grids?
### Building Tetrahedra
##### Stage: 4 Challenge Level:
Can you make a tetrahedron whose faces all have the same perimeter?
### Attractive Tablecloths
##### Stage: 4 Challenge Level:
Charlie likes tablecloths that use as many colours as possible, but insists that his tablecloths have some symmetry. Can you work out how many colours he needs for different tablecloth designs?
### Cops and Robbers
##### Stage: 2, 3 and 4 Challenge Level:
Can you find a reliable strategy for choosing coordinates that will locate the robber in the minimum number of guesses?
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# GRE GR 1777, Problem 005 Solution
Physics GRE GR 1777
5. Electrodynamics (Maxwell’s Equations)
Solution) In Maxwell’s equations, Ampère’s law (with Maxwell’s correction) is
$\nabla \times \vec{B} = \mu_0 \vec{J} + \mu_0 \epsilon_0 \frac{\partial \vec{E}}{\partial t}$.
We can write this equation as
$\nabla \times \vec{B} - \mu_0 \epsilon_0 \frac{\partial \vec{E}}{\partial t} = \mu_0 \vec{J}$,
where $\vec{J}$ is the current (the electric displacement current). And $\frac{\partial \vec{E}}{\partial t}$ is change of electric fields with respect to the time (rate of change of the electric flux).
If we assume that $\nabla \times \vec{B} = 0$, and the current flows through a surface S, then we can clearly see that the electric displacement current through S is proportional to the rate of change of the electric flux through S.
Therefore, the answer is (E).
Another (and better) solution)
The definition of electric displacement current is
$\vec{J}_{D} = \epsilon_0 \frac{\partial \vec{E}}{\partial t} + \frac{\partial \vec{P}}{\partial t}$.
If you know this definition, the problem would be solved more easily.
Reference: https://en.wikipedia.org/wiki/Displacement_current
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Du befindest dich hier: FSI Informatik » Prüfungsfragen und Altklausuren » Hauptstudiumsprüfungen » Lehrstuhl 5 » pr-2018-02-22
Examiner: Prof. Nöth
Atmosphere: Friendly.
Questions:
K Nearest Neighbor (from exercises)
• Q: He drew a coordinate system with some points. What step is necessary before applying KNN itself?
• A: Normalize the data, e.g. to [-1, 1].
• Q: How does it work?
• Q: How did we implement it?
• Q: We learned of something that is related to KNN …
• A: Bayes Classifier.
Bayes
• Q: Formulate the Bayes rule and name all parts.
• Q: Formulate Decision Function.
• Q: Relation to KNN:
• A: Error is at most twice of Bayes (not the classifier) Error Rate (see http://cseweb.ucsd.edu/~elkan/151/nearestn.pdf for more info).
• A: Bayes Classifier with the identity covariance matrix is the same as Nearest Neighbour.
• Q: How can you calculate the class conditional pdf?
• A: Assume gaussian distribution and perform Maximum Likelihood.
• Q: What do you do if the distribution is not known?
• A: GMM.
Gaussian Mixture Models
• Q: What are the parameters?
• A: Mean, covariance matrix and relative portion of samples for every distribution.
• Q: How can you initialize the parameters?
• A: Take random samples and calculate parameters with their help.
• Q: To what can this lead?
• A: GMM may only converge to a local maximum.
• Q: What can you do then?
• A: Try other random samples and take the largest found maximum.
• Q: Is there a better way to initialize the parameters?
• A: Apply k-means clustering before and calculate parameters for every cluster. This way you'll probably almost get the correct parameters.
SVM
• Q: Formulate the optimization problem of hard margin and soft margin SVM.
• Q: What are the slack variables for?
• A: Slack variables move samples onto the border of the margin. This way they'll become support vectors.
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# "We have succeeded DUE TO THE FACT THAT we work"Wrong?Why?
Author Message
Intern
Joined: 25 May 2014
Posts: 3
"We have succeeded DUE TO THE FACT THAT we work"Wrong?Why? [#permalink]
### Show Tags
16 Jul 2014, 18:19
1
KUDOS
Hello! Again a doubt from Manhattan SC guide Pg 143 idioms. It mentions that "We have succeeded DUE TO THE FACT THAT we work hard." is wrong use of FACT THAT. But "Due to the Fact that" is common English Idiomatic Expression. Please elaborate. Thanks in advance!!!
e-GMAT Representative
Joined: 02 Nov 2011
Posts: 2100
Re: "We have succeeded DUE TO THE FACT THAT we work"Wrong?Why? [#permalink]
### Show Tags
17 Jul 2014, 07:41
1
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Expert's post
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BOOKMARKED
Hi there,
You ask a question pertaining to the idiom that continues to confuse a lot of GMAT takers. On GMAT, "due to" and "because of" are used very specifically.
The sentence that you have mentioned is incorrect because here "due to" has been used to present the reason for a action. Here "due to" modifies the action "have succeeded". This is considered in correct on GMAT. Only "because of" can present reason for actions.
For more details and example on the usage of these two idioms, please read the following article:
due-to-vs-because-of-140393.html
Hope this helps.
Thanks.
SJ
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Re: "We have succeeded DUE TO THE FACT THAT we work"Wrong?Why? [#permalink]
### Show Tags
17 Jul 2014, 10:54
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Expert's post
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Thumb Rule:
Something must be DUE TO something else
the "something" needs to be a NOUN or something behaving like a NOUN.
Ask this question "WHAT was due to WHAT"? Are you getting an answer of the WHAT? If you are, then the usage is OK. (Remember what can only be answered by a NOUN or something behaving LIKE a noun)
We have succeeded DUE TO THE FACT THAT we work hard
What was due to What?
Success was due to the fact.
2 things wrong.
1. Success is not even in the sentence. (This is your doubt)
2. Success can't be achieved by a fact but an action. (This is extraneous information, albeit useful)
Another example
The math was postponed due to rain.
What was due to rain? Postponement. But I do not have Postponement. I have the verb which DOES not work. So the usage is wrong.
Does this help?
_________________
Re: "We have succeeded DUE TO THE FACT THAT we work"Wrong?Why? [#permalink] 17 Jul 2014, 10:54
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# 2020 AMC 12A Problems/Problem 22
## Problem
Let $(a_n)$ and $(b_n)$ be the sequences of real numbers such that $$(2 + i)^n = a_n + b_ni$$for all integers $n\geq 0$, where $i = \sqrt{-1}$. What is$$\sum_{n=0}^\infty\frac{a_nb_n}{7^n}\,?$$ $\textbf{(A) }\frac 38\qquad\textbf{(B) }\frac7{16}\qquad\textbf{(C) }\frac12\qquad\textbf{(D) }\frac9{16}\qquad\textbf{(E) }\frac47$
## Solution 1
Square the given equality to yield $$(3 + 4i)^n = (a_n + b_ni)^2 = (a_n^2 - b_n^2) + 2a_nb_ni,$$ so $a_nb_n = \tfrac12\operatorname{Im}((3+4i)^n)$ and $$\sum_{n\geq 0}\frac{a_nb_n}{7^n} = \frac12\operatorname{Im}\left(\sum_{n\geq 0}\frac{(3+4i)^n}{7^n}\right) = \frac12\operatorname{Im}\left(\frac{1}{1 - \frac{3 + 4i}7}\right) = \boxed{\frac 7{16}}.$$
## Solution 2 (DeMoivre's Formula)
We rewrite $$(2+i)^n=\left(\sqrt{5}\left(\frac{2}{\sqrt{5}}+\frac{i}{\sqrt{5}}\right)\right)^n=5^{\frac{n}{2}}e^{in\tan^{-1}\left(\frac{1}{2}\right)}=5^{\frac{n}{2}}\left(\cos \left(n\tan^{-1}\left(\frac{1}{2}\right)\right)+i\sin\left(n\tan^{-1}\left(\frac{1}{2}\right)\right)\right)$$ by DeMoivre's Formula. Letting $\theta=\tan^{-1}\left(\frac{1}{2}\right)$, we know that $a_n=5^{\frac{n}{2}}\cos \left(n\theta\right)$ and $b_n=5^{\frac{n}{2}}\sin \left(n\theta\right)$. The desired sum then turns into $$\sum_{n=0}^{\infty}\left(\frac{5}{7}\right)^n\cos\left(n\theta\right)\sin\left(n\theta\right)$$ $$=\frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{5}{7}\right)^n\sin\left(2n\theta\right)=\frac{1}{2}\text{Im}\left(\sum_{n=0}^{\infty}\left(\frac{5}{7}\right)^ne^{2in\theta}\right)$$ This is now an infinite geometric series! After finding $\sin(2\theta)=2\cdot \frac{2}{\sqrt{5}}\cdot \frac{1}{\sqrt{5}}=\frac{4}{5}$, $\cos(2\theta)=\sqrt{1-\sin^2(2\theta)}=\frac{3}{5}$, we find $$S=\frac{1}{2}\text{Im}\left(\frac{1}{1-\frac{5}{7}\cdot e^{2i\theta}}\right)=\frac{1}{2}\text{Im}\left(\frac{1}{1-\frac{3}{7}-\frac{4}{7}i}\right)=\frac{1}{2}\cdot \frac{\frac{4}{7}}{\frac{32}{49}}=\boxed{\textbf{(B) }\frac{7}{16}}$$ ~ktong
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### sag mill power draw
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Minto Mine Mill Operations Plan , Grinding Media Consumption , Two-stage grinding circuit comprised of a single SAG (semi-autogenous grinding) mill and two.
### Software Modeling Cuts SAG Energy Consumption at
Software Modeling Cuts SAG Energy Consumption at Cortez Gold , Such variation in energy consumption is attributed to ore geology , Actual SAG mill power drawn.
### Orway Mineral Consultants Canada Ltd
Orway Mineral Consultants Canada Ltd Mississauga, ON, L4W 5N5, , The SAG Power Index (SPI) test , The number of SAG mill revolutions required to grind the.
### Turbo Pulp Lifter (TPLTM)
the specific power consumption while increasing the mill , truly in terms of power draw for the changes in mill , An Efficient Discharger to Improve SAG Mill.
### SAG Mill
The SAG Mill is a block added by the EnderIO mod A machine similar to a Pulverizer or a MaceratorIt is used to process resources to produce dust It accepts energy in the form of Redstone Flux or Minecraft Joule, with an internal buffer of 100,000 RF and consumption of 20 RF per tick.
]
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Question:
What is the value of {::}^kC_k+{::}^(k+1)C_k+..+{::}^nC_k is:
• {::}^(n+1)C_k
• {::}^(n+1)C_(k+1)
• {::}^(n+1)C_(k-1)
• None
Solution
We have obviously have hacks to solve a question, but it is always better to solve a question properly when practing.
First straight forward approach:
People generally say that we can replace {::}^kC_k with {::}^(k+1)C_(k+1), but ask them why did they think of this approach, many won't reply anything.
If you look here we are trying to add {::}^kC_k+{::}^(k+1)C_k
=1+((k+1)!)/(k!)
=1/(k!) (k!+(k+1)!)
=(k!)/(k!) (k+2)
=(k+2)={::}^(k+2)C_(k+1)
From this you could see the trend which is
{::}^(k+1)C_(k+1)+{::}^(k+1)C_k={::}^(k+2)C_(k+1)
{::}^(k+2)C_(k+1)+{::}^(k+2)C_k={::}^(k+3)C_(k+1)
{::}^(k+3)C_(k+1)+{::}^(k+3)C_k={::}^(k+4)C_(k+1)
This is the reason we substitute {::}^kC_k={::}^(k+1)C_(k+1)
So from this is clear that the answer will be {::}^(n+1)C_(k+1)
Induction approach:
We use an induction on variable n. Here we find the function which is satisfied for n=k and then prove that if this equation is satisfied for some n, then it is true for n+1 also.
Coefficients approach:
If you look closely, you can see that {::}^rC_k is the coefficient of x^k in (1+x)^k.
So the given sum is equal to the sum of coefficients of x^k in (1+x)^k+(1+x)^(k+1)+...+(1+x)^n
This is the sum of coefficients of x^k in =1+(1+x)+(1+x)^2+.....+(1+x)^n
This is the coefficients of x^k in =((1+x)^(k+1)-1)/x
This is the coefficients of x^(k+1) in =(1+x)^(k+1)
={::}^(n+1)C_(k+1)
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http://empslocal.ex.ac.uk/people/staff/mrwatkin/zeta/kulsha.htm
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#### a message from A. Kulsha concerning the audio representation of zeta
Hello!
It was an old idea to represent the behaviour of Riemann zeta function on the critical line as audio signal. It seems to be the best way to encode the values of Riemann-Siegel Z(t) function as audio-samples. Recently I found that Robert Munafo already implemented this idea:
But audio-clip given there appeared to be too far from real behaviour of Z(t), so I decided to make up a few another ones.
Since .mp3 format distorts the signal too much, I decided to use .wav format with 16-bit mono signal at sample rate of 8 kHz, the bitrate being 128 kbps. Such low sample rate require enough small speed of t to make all frequencies in Z(t) signal lower than 4 kHz. The speed of 0.25 per sample (i.e. 2000 per second) appeared to be optimal and suitable at least for t<1e12. Every sample of 16-bit signal lies in the range [-32768..32767], so we must multiply the values of Z(t) by suitable constant C to convert them into samples. C determines the volume of sound, so the larger is C, the more details can be listened; however, too big values of C may cause overloading.
So, I decided to produce 4 audio-clips, 50 seconds each, for the following ranges of t:
0..1e5
1e7..1e7+1e5
1e9..1e9+1e5
1e11..1e11+1e5
Since oscillations of Z(t) become larger with t, the values of multiplier C were chosen as 1000, 500, 300 and 150 respectively.
Mathematica 4.1 was used to compute Z(t). The precision of 5 decimal digits would be enough, but the values returned by Mathematica are even more exact. For example, the second range was computed thus:
Do[Write[fl, RiemannSiegelZ[i/4.0]], {i, 40000000, 40400000}]
Close[fl]
A small pascal program was used to convert Z(t) values into samples:
{\$A+,B-,E-,N+}
const c:extended=1000;
var f:text; w:file of integer; n:extended; x:integer; s:string;
begin
while not eof(f) do
begin
readln(f,s); x:=pos('*^',s); if x>0 then begin s[x]:='0'; s[x+1]:='e' end;
val(s,n,x); x:=round(n*c); write(w,x)
end;
close(f); close(w)
end.
Finally, audio-editor CoolEdit 2.0 Pro was used to choose parameters (C, sample rate, etc.), to examine the signal and to convert *.pcm files to *.wav files.
Here we are:
So, listening to these recordings, we can indeed hear some interesting details of behaviour of zeta function on the critical line, for large t's as well as for small... Of course, if your audio system allows you this. ;-) BTW, there's my: http://www.ixbt.com/multimedia/microlab/solo-2/solo2-view1.jpg
Happy listening,
Andrey
archive tutorial mystery new search home contact
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https://www.lmfdb.org/SatoTateGroup/6.2.A.c126
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# Properties
Label 6.2.3.c126 Name $$\mathrm{SU}(2)[C_{126}]$$ Weight $6$ Degree $2$ Real dimension $3$ Components $126$ Contained in $$\mathrm{O}(2)$$ Identity component $$\mathrm{SU}(2)$$ Component group $$C_{126}$$
# Learn more
## Invariants
Weight: $6$ Degree: $2$ $\mathbb{R}$-dimension: $3$ Components: $126$ Contained in: $\mathrm{O}(2)$ Rational: yes
## Identity component
Name: $\mathrm{SU}(2)$ $\mathbb{R}$-dimension: $3$ Description: $\left\{\begin{bmatrix}\alpha&\beta\\-\bar\beta&\bar\alpha\end{bmatrix}:\alpha\bar\alpha+\beta\bar\beta = 1,\ \alpha,\beta\in\mathbb{C}\right\}$ Symplectic form: $\begin{bmatrix}0&1\\-1&0\end{bmatrix}$ Hodge circle: $u\mapsto\mathrm{diag}(u,\bar u)$
## Component group
Name: $C_{126}$ Order: $126$ Abelian: yes Generators: $\begin{bmatrix}1&0\\0&\zeta_{126}\end{bmatrix}$
## Moment sequences
$x$ $\mathrm{E}[x^{0}]$ $\mathrm{E}[x^{1}]$ $\mathrm{E}[x^{2}]$ $\mathrm{E}[x^{3}]$ $\mathrm{E}[x^{4}]$ $\mathrm{E}[x^{5}]$ $\mathrm{E}[x^{6}]$ $\mathrm{E}[x^{7}]$ $\mathrm{E}[x^{8}]$ $\mathrm{E}[x^{9}]$ $\mathrm{E}[x^{10}]$ $\mathrm{E}[x^{11}]$ $\mathrm{E}[x^{12}]$
$a_1$ $1$ $0$ $0$ $0$ $0$ $0$ $0$ $0$ $0$ $0$ $0$ $0$ $0$
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https://codereview.stackexchange.com/questions/143582/determining-the-number-of-days-in-a-month/145198
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# Determining the number of days in a month
Overall logic: determine number of days in a month. Loop through table (wsUploadTable) and increment value if condition is met. Loop to next day in month and repeat.
For example: 10/1/2016, loops through table for date match and increment value. Next date, 10/2/2016 loop table for match... until last day of month 10/31/2016, loop table, find match and increment value.
'Determine DaysinMonth and assign DaysinMonth_Distro value
DaysInMonth = DateSerial(dtTrickle_Year, dtTrickle_Month + 1, 1) - _
DateSerial(dtTrickle_Year, dtTrickle_Month, 1)
DoM_Distro = 1 / DaysInMonth
ReDim Days(1 To DaysInMonth)
For i = 1 To DaysInMonth
Days(i) = DateSerial(dtTrickle_Year, dtTrickle_Month, i)
'loop table and increment cell value if condition is met
lngER_PrimaryID = .Cells(1048576, 2).End(xlUp).Row
For intPrimaryID = 2 To lngER_PrimaryID
'store current cell value
'match UploadTable row on user input and increment new value
If.Cells(intPrimaryID, 3).Value = Days(i) Then
.Cells(intPrimaryID, 11).Value = dblLeadsValue + DoM_Distro
End If
Next 'Next PrimaryID
End With
Next i
• Welcome to CR! You've described your code well, but the code itself could use a bit more context; it would be nice to include the declarations for the variables you're using, and perhaps the procedure's signature as well (i.e. everything between Sub/Function and End Sub/Function - you see right now I'm not even sure I'm looking at a Sub or a Function, whether the involved variables are local or module-level, whether they're declared at all, and/or what else is in the same scope. Oct 8, 2016 at 0:04
Well havent tried either Excel or VBA , but the logic will go like this ,
• Loop through the table
• Say the date is day/month/year , where day,month and year are variables
• lets have an array of months which is [31,28,31,30,31,30,31,31,30,31,30,31]
• if year is divisible by 4 or 400 , then months[1]= months[1] + 1
• now just display months[month]
Here months is the array and month is the number of month present in the date , for more info in arrays for VBA you can look here .
There's no need to remove 1st of next month from 1st of this month to get the number of days.
Day 0 in the DateSerial function is the last day of the previous month, so this will work:
DaysInMonth = Day(DateSerial(dtTrickle_Year, dtTrickle_Month + 1, 0))
It's leap-year sensitive to:
Day(DateSerial(2015, 2+ 1, 0)) returns 28.
Day(DateSerial(2016, 2+ 1, 0)) returns 29.
Looking further through your code I've made an attempt at coding what I think you're trying to do.
I've had to add some test values and variable declarations to make it work. It removes one of your loops by using the FIND method to go straight to the required cell.
Sub Test()
Dim DaysInMonth As Long
Dim DoM_Distro As Long
Dim dtTrickle_Year As Long
Dim dtTrickle_Month As Long
Dim dblLeadsValue As Double
Dim i As Long
Dim dtCurrentDay As Date
Dim rFound As Range
dtTrickle_Year = 2016
dtTrickle_Month = 2
'Determine DaysinMonth and assign DaysinMonth_Distro value
DaysInMonth = Day(DateSerial(dtTrickle_Year, dtTrickle_Month + 1, 0))
DoM_Distro = 1 / DaysInMonth
'Cycle through each day in the month.
For i = 1 To DaysInMonth
dtCurrentDay = DateSerial(dtTrickle_Year, dtTrickle_Month, i)
'Seach column 2 of the table to find the date value.
'Find the value in the table.
Set rFound = .Find( _
What:=dtCurrentDay, _
After:=.Cells(1, 1), _
LookIn:=xlValues, _
LookAt:=xlWhole)
'Providing a value is found....
If Not rFound Is Nothing Then
'Place our result 10 columns to the right of the found cell.
rFound.Offset(, 10).Value = dblLeadsValue + DoM_Distro
End If
Set rFound = Nothing
End With
Next i
End Sub
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https://quicktermpapers.com/discussion-reply-semiconductor-materials-electronic-engineering-homework-help/
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# Discussion reply: semiconductor materials | Electronic Engineering homework help
The factor that determines whether a material is a conductor, semiconductor or insulator is the amount of free electrons it has. For example, a silicon lattice has no free electrons because they’re all bonded to one another. This makes it an insulator. Copper has free electrons in a pure state, making it a great conductor. Some materials semi-conduct due to the stacking of P-type and N-type doping. This will allow current to flow in one direction, but not the other.
I think that silicon chips are the coolest because they allow for logic gates. Logic gates got me into electrical engineering in the first place due to their application in digital electronics. A little background information- just in case you don’t know what logic gates are:
Logic Gates (a.k.a. Boolean Gates)
• Perform boolean (true/false) functions.
• You can think of these as ‘if’ functions. Example, if I1 and I2 are 1, then O is 1. (This is called an AND gate)
• There are several types of logic gates:
• AND gates, NAND gates, OR gates, NOR gates, XNOR gates, NOT gates, ect.
• Logic gates are everywhere. You’re using them right now as you read this.
References:
Brain, M. (2021, February 11). How semiconductors work. HowStuffWorks. Retrieved January 26, 2023, from https://electronics.howstuffworks.com/diode.htm
Contributor, T. T. (2020, December 16). What is Logic Gate (and, or, XOR, not, Nand, nor and xnor)? A definition from whatis.com. WhatIs.com. Retrieved January 26, 2023, from https://www.techtarget.com/whatis/definition/logic-gate-AND-OR-XOR-NOT-NAND-NOR-and-XNOR
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https://www.circuitdiagram.co/design-a-circuit-diagram/
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# Design A Circuit Diagram
By | September 1, 2017
Designing a circuit diagram is an important and sometimes challenging task for anyone who has to create a circuit. It can involve complex engineering and electrical knowledge, or a basic set of tools, to understand the basics of what makes a circuit work.
A circuit diagram is used to illustrate the parts that make up a system and how they interact. For example, in a home wiring system, a circuit diagram shows you which switch controls which light and which outlet is wired to which wall socket. In other cases, a circuit diagram illustrates a more complicated system of wires and components, such as an amplifier or computer.
Though the complexity of the circuit diagram may vary, the basics of building it will remain the same. First, the components must be identified and then connected correctly. This means matching up the correct voltage, current, and resistance ratings on each component. Once this is done, the next step is to determine the order of connection and assemble the circuit. This requires some basic soldering skills and some understanding of the electronics being used.
The circuit diagram should also include information about the safety of the circuit. That means taking into account any risks that could be posed by the construction of the circuit, such as exposed wires, wrong connections, or inadequate insulation. Making sure these concerns are addressed before the circuit is built is essential to its ultimate success.
Once the circuit diagram is complete, it can be tested for any problems that may have been overlooked during construction. Little components can often cause unexpected issues, so troubleshooting the circuit is an essential part of the process.
Designing a circuit diagram is both an art and a science. With the right tools and some patience, anyone can create a well-designed and functional circuit. Knowing the basics of electricity and how components work together is the first step towards succesful circuit design.
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# Nanyang sci-tech Question 9: posters (discretization + line segment tree)
Source: Internet
Author: User
Tags integer numbers
Posters time limit: 1000 MS | memory limit: 65535 kb difficulty: 6
Description
The citizens of bytetown, AB, cocould not stand that the candidates in the mayoral election campaign have been placing their electoral posters at all places at their whim. the city councel has finally decided to build an electoral wall for placing the posters and introduce the following rules:
• Every candidate can place exactly one poster on the wall.
• All posters are of the same height equal to the height of the wall; the width of a poster can be any integer number of bytes (byte is the unit of length in bytetown ).
• The Wall is divided into segments and the width of each segment is one byte.
• Each poster must completely cover a contiguous Number of wall segments.
They have built a wall 10000000 bytes long (such that there is enough place for all candidates ). when the electoral campaign was restarted, the candidates were placing their posters on the wall and their posters differed widely in width. moreover, the candidates started placing their posters on wall segments already occupied by other posters. everyone in bytetown was curous whose posters will be visible (entirely or in part) on the last day before elections.
Your task is to find the number of visible posters when all the posters are placed given the information about posters 'size, their place and order of placement on the electoral wall.
Input
The first line of input contains a number C giving the number of cases that follow. the first line of data for a single case contains number 1 <=n <= 10000. the subsequent n lines describe the posters in the order in which they were placed. the I-th line among the n lines contains two integer numbers Li and RI which are the number of the wall segment occupied by the Left end and the right end of the I-th poster, respectively. we know that for each 1 <= I <= N, 1 <= LI <= RI <= 10000000. after the I-th poster is placed, it entirely covers all Wall segments numbered Li, Li + 1 ,..., ri.
Output
For each input data set print the number of visible posters after all the posters are placed.
The picture below has strates the case of the sample input.
Http://acm.pku.edu.cn/JudgeOnline/images/2528_1.jpg
Sample Input
`151 42 68 103 47 10`
Sample output
`4`
Source
Poj
Uploaded
Iphxer
Discretization + line segment tree.
The topic of a line segment tree with high difficulty is processed using discretization. At Nanyang Institute of Technology (http://acm.nyist.net/JudgeOnline/problem.php? PID = 9) submitted successfully, but wa submitted on poj. I don't know why ...... I heard that there are some problems with the test data on poj. Do you still need to calculate the vertices on both sides of the endpoint ?? I don't know. If you know me, please tell me, thank you...
Question:
Give you some posters in order. These posters may overlap with each other and you can find the number of posters that are not completely covered.
The two endpoints of the poster can contain a maximum of 10000 posters, but a maximum of 10000000.
Ideas:
Because the interval is involved, we should first think of the Line Segment tree, but the width of the poster is too large. What should I do if I have a 1-poster? Sure MLE is too memory (how many new nodes are there ). So we need to useDiscretizationThe method may sound mysterious, but it is easy to understand. For example, I will give you two posters, [] and [], which are sorted by 1, 3, 6, and 10. They are discretization to 1, 2, 3, 4, that is, if the first interval is [1, 3] and the second interval is [2, 4], you can create a [] Line Segment tree instead of a [] Line Segment tree, saves space.
Code:
`1 # include <iostream> 2 # include <stdio. h> 3 # include <string. h> 4 # include <algorithm> 5 using namespace STD; 6 7 # define maxn 20000 8 bool tree [maxn * 4 + 1]; // whether the storage interval has poster 9 short hash [10000010]; // The destination position after discretization is 10 short X [maxn + 10]; // store the two poster endpoints 11 struct post {12 short L; 13 short R; 14} A [maxn + 10]; // The original array 15 int CNT; // record number of posters exposed outside 16 17 bool insert (int d, int L, int R, int L, int R) 18 {19 if (tree [d]) 20 return false; 21 if (L = L & R = r) {// locate the interval 22 tree [d] = true; 23 return true; 24} 25 26 int mid = (L + r)/2; 27 bool res; 28 If (mid> = r) {29 res = insert (d <1, l, mid, L, R); 30} 31 else if (mid <L) {32 res = insert (d <1 | 1, Mid + 1, R, l, r); 33} 34 else {35 bool F1 = insert (d <1, L, mid, L, mid ); 36 bool F2 = insert (d <1 | 1, Mid + 1, R, Mid + 1, R); 37 res = F1 | F2; 38} 39 if (tree [d <1] & tree [d <1 | 1]) 40 tree [d] = true; 41 return res; 42} 43 44 int main () 45 {46 int t; 47 scanf ("% d", & T); 48 While (t --) {49 memset (tree, 0, sizeof (tree); 50 memset (hash, 0, sizeof (hash); 51 memset (x, 0, sizeof (x); 52 int I, n, ncount = 0; 53 54 // enter 55 scanf ("% d", & N); 56 for (I = 0; I <n; I ++) {57 scanf ("% d", & A [I]. l, & A [I]. r); 58 X [ncount ++] = A [I]. l; 59 X [ncount ++] = A [I]. r; 60} 61 62 sort (x, x + ncount); 63 ncount = unique (x, x + ncount)-X; // deduplicated 64 65 int key = 1; 66 for (I = 0; I <ncount; I ++) {// discretization processing 67 hash [x [I] = key; 68 if (I <nCount-1) {69 If (X [I + 1]-X [I] = 1) 70 key ++; 71 else 72 key = Key + 2; 73} 74} 75 76 CNT = 0; 77 for (I = n-1; I> = 0; I --) {78 If (insert (, key, hash [A [I]. l], hash [A [I]. r]) // Insert the poster after discretization into the line segment tree 79 CNT ++; 80} 81 printf ("% d \ n", CNT ); 82} 83 return 0; 84}`
Freecode: www.cnblogs.com/yym2013
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# Breaking Double Encryption
I am trying to understand how an attacker knows when he has successfully decrypted a ciphertext for an assignment. As such, some pointers/hints for the following questions would be greatly appreciated.
1. If you double encrypt secret data with two different algorithms. How does the attacker know if he already broke the 'first layer' of encryption?
2. Is it possible to determine the encryption algorithm used by analyzing the cipher text?
3. If you find out what algorithm was used does it give you the necessary information to know when you successfully broke the 'first layer'?
4. Considering the above, what difference would it make if the attacker knows the open-source program or the (public available) algorithms used to protect the data?
-
Related reading... – hunter Mar 10 '14 at 1:07
To begin with 4: Remember Kerckhoff's principle. You should always assume that the attacker knows which algorithm is used to encrypt your data. All the algorithms used in practice are designed to be secure under this assumption, so you should consider that hiding the algorithm from the attacker is superfluous. But as a hypothetical...
1. I can't think of any real-world cryptosystem where it would be possible to tell exactly which algorithm was used knowing only the ciphertext, but in a theoretical sense such an algorithm could exist, and would not be any less secure because the attacker is supposed to know that anyway.
2. Same as above, for mostly the same reasons.
3. In light of the above, it shouldn't make a significant difference. If it does, your algorithm relies on security by obscurity, and is generally worthless.
In your "double encryption" case, the attacker will just treat the composition of the two algorithms as one big algorithm, he will not try to decipher one and then the other.
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"In your "double encryption" case, the attacker will just treat the composition of the two algorithms as one big algorithm, he will not try to decipher one and then the other." This is quite possibly untrue: assuming the two algorithms use at least partially different keys, an attacker would perform a Meet in the middle (MITM) attack – figlesquidge Mar 10 '14 at 18:16
To 1, I believe RC4 has enough output biases that it should be possible to detect its use with a small number of ciphertexts. – Stephen Touset Sep 8 '14 at 20:43
In general (especially without knowledge what encryption you consider), it's not possible to detect "correct decryption of one layer", if that's all you have and this "middle ciphertext" is not in a specific format.
However, from today's point of view this is almost entirely irrelevant, because stronger attacks are considered:
• Kerckhoff's principle states that security should never be based on not knowing the algorithm. That is basically the weakness of most historical ciphers. Security must stem from the key alone.
• You seem to consider "ciphertext only" attacks, which is just too weak for today's standards. For symmetric encryption you can also consider known plaintext attacks (attacker gets pairs of ciphertext and plaintext) or chosen plaintext attacks (where the attacker chooses what plaintext is encrypted and gets the ciphertext). The later is actually needed to be called "secure" today.
• As a rule of thumb: Unless you clearly know what you are doing, just combining two encryption schemes does not necessarily make it more secure. There is no general statement possible, but you might add a security weakness unintentionally.
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### Author Topic: FE-P2 (Read 1945 times)
#### Victor Ivrii
• Elder Member
• Posts: 2606
• Karma: 0
##### FE-P2
« on: April 11, 2018, 02:34:19 PM »
$\newcommand{\erf}{\operatorname{erf}}$
Solve IVP for the heat equation
\begin{align}
&2u_t - u_{xx}=0,\qquad &&0 <x<\infty,\; t>0,\label{2-1}\\[2pt]
&u|_{x=0}=0,\\
&u|_{t=0}= f(x)\label{2-2}
\end{align}
with $f(x)=e^{-x}$.
Solution should be expressed through $\displaystyle{\erf(z)=\frac{2}{\sqrt{\pi}} \int_0^z e^{-z^2}\,dz}$
#### Andrew Hardy
• Full Member
• Posts: 34
• Karma: 10
##### Re: FE-P2
« Reply #1 on: April 11, 2018, 05:23:54 PM »
Very similar to Term Test 1. Here instead though, apply even continuation and then because k = 1/2
$$u=\frac{1}{\sqrt{2t\pi}}\int_0^{\infty} \exp(-\frac{(y+x)^2}{2t}-y) \,dx +\frac{1}{\sqrt{t\pi}}\int_0^{\infty} \exp(-\frac{(y-x)^2}{2t}-y)\,dx\\$$
and then completing the square
$$=\frac{\exp(x+t)}{\sqrt{t\pi}}\int_0^{\infty} \exp(-\frac{(y+x+t)^2}{2t})\,dx + \frac{\exp(-x+t)}{\sqrt{t\pi}}\int_0^{\infty} \exp(-\frac{(y-x+t)^2}{2t})\,dx\\$$
and then via change of variables
$$=\frac{\exp(x+t/2)}{\sqrt{\pi}}\int_{\frac{x+t/2}{\sqrt{2t}}}^{\infty} e^{-z^2} \,dz + \frac{\exp(-x+t/2)}{\sqrt{\pi}}\int_{\frac{-x+t}{\sqrt{2t}}}^{\infty} e^{-z^2} \,dz\\$$
and in conclusion
$$=\frac{\exp(x+t/2)}{2}(1-\text{erf}(\frac{x+t}{\sqrt{2t}})) + \frac{\exp(-x+t/2)}{2}(1-\text{erf}(\frac{-x+t}{\sqrt{2t}}))$$
corrected
« Last Edit: April 12, 2018, 03:45:09 PM by Andrew Hardy »
#### Victor Ivrii
• Elder Member
• Posts: 2606
• Karma: 0
##### Re: FE-P2
« Reply #2 on: April 12, 2018, 01:08:09 PM »
This looks familiar:)
Indeed, it looks familiar but in addition to misprints there are errors, leading to the errors in the answer.
Jingxuan, you are the second most prolific poster on this forum, you just made more than Emily, but this was a flood. Deleted.
$$\frac{1}{2t} (x+y )^2 + y \overset{?}{=} \frac{1}{2t} (x+y +{\color{red}{2}}t)^2 - ... \tag{*}$$
Now it is fixed. I sketched $u(x,0)$ and $u(x,1)$.
* Forgetting to change the lower limit in $\int_0^\infty \ldots dy$ while changing variable $z= (x-y \pm c t)/\sqrt{2t}$.
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Home > information technology > which of the attributes of algorithms are most important to an information technology professional?
which of the attributes of algorithms are most important to an information technology professional?
which of the attributes of algorithms are most important to an information technology professional - Related Questions
What are the important properties of algorithms?
A specified input was provided. A specified output has been provided. Defining qualities. An effective strategy. Having a limit.
What are 5 things algorithms must have?
It enables us to solve problems and is often used to automate the method of resolving those problems. An algorithm is defined as something that is definite, has inputs and outputs, has a finite number of steps, and has a definite and measurable effect. are vital to the health care industry.
What are the attributes of algorithms?
Input: Inputs may be many or may be uncounted in an algorithm, and an algorithm must end after a finite period of time. Finiteness: An algorithm should have a finite number of steps and it must end after a finite amount of time. A minimum of one output is expected. Determination: Each step should be concisely explained, clearly defined and detailed.
What are the four attributes of algorithms?
Algorithms must state and how it is related to the input. A discrete and detailed method for defining the algorithm is mandatory. Finiteness: The algorithm must end after a specific number of steps. Doability: The algorithm must be able to do the tasks requested. Finiteness: There must be not more than one step in the algorithm.
What is algorithm and its importance?
The theory and practice of computer science utilize algorithms in every field. It is their work that makes the field work. Algorithms enable computers to run a rocket or a calculator, as their instructions allow them to run.
Why algorithms are important in the field of computing?
one of the most vital topics in computer science, since they help software developers build efficient and error-free applications. keep in mind about algorithms is that different algorithms may work for the same problem, but some of them may be much better than others.
What are the most important characteristics of a good algorithm?
As a general rule, good algorithms have the following characteristics: Precision - the steps are clearly stated. Each step will have a unique result, and it will depend only on the inputs and the results from the preceding steps. It does not stop executing instructions once it reaches a finite number.
What are the 5 characteristics of an algorithm?
A specified input was provided. A specified output has been provided. Defining qualities. An effective strategy. Having a limit. It is independent.
What are the characteristics of a good algorithm class 11?
There should be a limit to the number of steps in an algorithm. In order for an algorithm to work properly, it should be clear and straightforward. is a way to solve a problem in a standardized way and should lead to a unique solution.
What are the different properties of a good algorithm?
Accuracy - each step is stated (described) precisely. Each step will have a unique result, and it will depend only on the inputs and the results from the preceding steps. It does not stop executing instructions once it reaches a finite number.
What does an algorithm need?
Essentially, algorithms are a set of steps that must be followed one at a time if something useful is to be accomplished or solved. Using a cake recipe as an example, you can consider it an algorithm for baking a cake. The algorithms in computing assist computers in performing actions in a sequential order.
What are the 2 most important criteria that make a good algorithm?
An input value of zero or more. You will get at least one output value. A set of instructions that is clear and unambiguous. The time it takes for an atom to react. A halting problem cannot occur because step sequences cannot be infinite (help). This method of computing is possible when a specific computational device is used.
What is the importance of algorithms?
Computer programs can be optimized by using algorithms to match the resources available. In the end, when someone decides to solve a problem through better algorithms, they will find the best combination of program speed and minimum amount of memory consumption.
What are the main characteristics of an algorithm?
It should be clear and unambiguous how the algorithm works... A well-defined input is vital for an algorithm. * The output of an algorithm will usually be either 1 or more well-defined outputs, and should be in line with the intended one.
What are algorithms properties?
It must be possible for an algorithm to function properly: Inputs: The algorithm must accept input values from a specified set of values. Output values must be produced by an algorithm from a specified input set. In general, the algorithm must end at the end of a finite number of steps for any input.
What are the 2 important criteria that a good algorithm should have?
There are zero or more quantities that more quantities which are externally supplied; At s produced; Clarity: d unambiguous;
What two things make a good algorithm?
As input, an algorithm must be capable of accepting certain sets of data. Good algorithms should produce results, preferably solutions, as outputs. In order to remain finite, all instructions of the algorithm should be stopped after a certain time. A general algorithm must be applicable to all inputs defined in the specification.
What are the 2 kinds of algorithm efficiency?
Using this metric, we can measure the speed of an algorithm's execution. Algorithm performance is measured by how much memory is required to run the algorithm.
What determines a good algorithm?
For a good algorithm to be efficient, there must be fewer steps needed for it to produce the correct output for any legal input. The design of a good algorithm should be such that it can be adapted to solve any additional problem that might arise and also be modified by others.
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# .25 carat diamond price?
Date created: Tue, Jul 27, 2021 10:33 AM
Content
FAQ
Those who are looking for an answer to the question «.25 carat diamond price?» often ask the following questions:
### 👉 1.25 carat diamond price?
Diamond Price per Carat Total Price Recommended Price (Best Value) 0.25 carat: \$800 - ...
### 👉 3.5 carat diamond price?
3.5 Carat / I / VS1. / None Fluor. 56 days on StoneAlgo \$45,674 Price History. View Details. Save this diamond to your Vault. Set a price alert for this diamond. 3.33 Visual Carat. -1% Smarter. 8.9 Cut Score.
### 👉 Price diamond per carat?
How to calculate the price per carat? For instance, if one carat costs \$2500, a 0.5-carat diamond would cost \$1250. You would calculate the price of that carat by multiplying 2500 X 0.5. The per-carat price increases as weight and quality increase. However, you also need to consider the remaining 3Cs of a diamond—color, clarity, and cut.
These ravishing and uncut 0.25 carat diamond price are top of the line when it comes to quality and comes with high clarity. These 0.25 carat diamond price are available in distinct variations from uncut to fine cuts, multiple shapes, sizes and weights according to your specific needs. These products are tested, verified and certified for their authenticity.
How to calculate the price per carat? For instance, if one carat costs \$2500, a 0.5-carat diamond would cost \$1250. You would calculate the price of that carat by multiplying 2500 X 0.5. The per-carat price increases as weight and quality increase. However, you also need to consider the remaining 3Cs of a diamond—color, clarity, and cut.
For example, a.25 carat round diamond starts at around \$300 and a.5 carat round diamond starts at around \$650. But a 1 carat round diamond jumps up all the way to \$2,000+. For best value, we recommend staying just under half or whole carat weights (for example, go for 0.9 carat instead of 1 carat). Diamond Price vs Carat © CreditDonkey
Carat Weight Diamond Price per Carat Total Price Recommended Price (Best Value) 0.25 ...
The first valuation criteria for colored diamonds is therefore the type of color and the intensity of this color. A 0.95 carat Fancy Purplish Red (very intense red) round brilliant diamond was sold for \$880,000.00 compared with the price of a 1 carat round brilliant diamond of the highest quality (D FL) which was only \$17,000.00.
Diamond Prices. Shape. Range. Clarity. Color. Price (\$) Round. .01 - .03 CT. I1.
Get the best deals on 0.25 - 0.32 Total Carat Weight Loose Diamonds when you shop the largest online selection at eBay.com. Free shipping on many items | Browse your favorite brands | affordable prices.
Our diamond price calculator provides you with the final price for a diamond but you can easily calculate the price per carat by dividing that final price by the carat weight of your diamond. Diamonds are priced in terms of price per carat because it is easier for diamond dealers to discuss pricing in these terms.
Let do some math and make it more simple. RAPAPORT PRICE – % Discount = Price Per Carat * Diamond Carat weight = Total Price. 6300\$ – 34% = 4158\$ per carat price* 1.01 Carat weight = 4199.58 Total price. In case of premium, you have to add % to the Rapaport base price and multiply with carat weight for the total value.
We've handpicked 22 related questions for you, similar to «.25 carat diamond price?» so you can surely find the answer!
### How many mm is 3 carat diamond price?
MM Size: Carat Weight: 3 x 1.5: 0.025: 3.5 x 1.75: 0.065: 3.5 x 2: 0.07: 3.75 x 1.75: 0.11: 4 x 2: 0.1: 4.25 x 2.25: 0.12: 5 x 2.5: 0.14: 5 x 3: 0.2: 5.5 x 2.75: 0.16: 5.5 x 3: 0.18: 6 x 3: 0.25: 6.5 x 3: 0.23: 7 x 3: 0.3: 7 x 3.5.35: 7 x 4: 3.4: 7.5 x 3.5: 0.33: 8 x 4: 0.47: 9 x 4.5: 0.71
### What is the price of 0.25 carat diamond?
Today on Blue Nile, you can purchase a diamond of about this carat weight and pay between US\$270 and US\$457, depending on the cut, colour, clarity and exact carat weight that you want.
### What is the price of 1 carat diamond?
A 1 carat diamond can vary in price from \$2500 to \$16,000. Here are two diamonds on either end of the price range of 1 carat diamonds. Both are 1 carat stones but that’s the only similarity in their specifications. The difference in price between these two stones is approximately \$14,500!
### How much is a 4.5 carat diamond ring price?
As with other diamonds, the price you’ll pay for a diamond in the 4 carat range increases as the diamond gets larger. The price of a 4.5 carat diamond ring can range from \$40,000 to as much as \$450,000 for diamonds close to the 5 carat range.
### Pawn price for 1 carat black diamond cluster ring?
A diamond is valued by its cut, clarity, carat weight and colour. A local pawn shop can look at your stone and give you a precise answer.
### What is the average price of diamond per carat?
• A quarter carat or less set in a diamond can cost anywhere from \$100 to \$600. A half carat diamond can cost anywhere from \$500 to \$1,100. Diamonds that are up to .75 carats can cost anywhere from \$1,000 to \$2,500. Diamonds up to 1 carat can cost anywhere from \$2,400 to \$4,000.
### What is the price of a 2 carat diamond?
It means you should multiply the number by 2 to get the exact price you can count on while ...
### Average 2 carat diamond ring price: how much is a 2ct diamond?
How much is a 2 Carat Diamond? The average price of a well-cut 2 Carat Round Brilliant Diamond ring is \$20,820. Various factors impact the price of diamond engagement rings, including Diamond Shape, type of setting, and Clarity grade.
### How does a carat affect the price of a diamond?
• Carat: Carat has the most significant impact on the price of a diamond. For instance, consider several diamonds with similar color, clarity, and excellent cut.The price of a diamond directly increases with carat weight.
### How much is rough diamond per carat price chart history?
3,750 x 0.50 = \$1,875 (this is our purchase price for this rough diamond). The purchase price per carat for this rough diamond is: 1,875 Ă· 5 carats = \$375 per carat. Rough Diamond Prices List
### What is an average price for black diamond per carat?
Black diamonds can be estimated approx. \$10,000 per carat.
### What is the average price for 1 carat diamond ring?
The average price for a 1 carat center diamond ring is around \$5,000 however, by cutting the middleman, handpicking affordable diamonds and working on tight margins, we are able to offer most of our 1 carat diamond rings under \$3,000.
### What is the diameter of a 2 carat diamond price?
For instance, if one carat costs \$2500, a 0.5-carat diamond would cost \$1250. You would calculate the price of that carat by multiplying 2500 X 0.5. The per-carat price increases as weight and quality increase. However, you also need to consider the remaining 3Cs of a diamond—color, clarity, and cut.
### What is the price of 1 carat diamond in india?
#### Questions & Answers on Natural Diamond
CaratMin PriceMax Price
1 ct (0.2gm)Rs 50000/CaratRs 300000/Carat
2 ct (0.4gm)Rs 5000/PieceRs 10000/Piece
### 1 carat diamond jared?
Diamond Solitaire Ring 1 carat Round-cut 14K White Gold | Jared A stunning, near-colorless 1 carat round diamond is the focus of this elegant diamond engagement ring that celebrates a declaration of love. The solitaire is securely prong-set in a band of 14K white gold.
### 1.5 carat diamond solitaire?
The elegant look of a dazzling solitaire diamond pendant is a jewelry essential in every women's collection. A fiery round cut diamond is securely prong set onto a lustrous 14K white gold basket setting. The sparkling round diamond weighs 1/5 caret and is suspended in pure elegance on an 18 inch chain.
### 1.6 carat diamond cost?
1.63 Carat / I / IF. / None Fluor. 105 days on StoneAlgo \$9,574 Price History. View Details. Save this diamond to your Vault. Set a price alert for this diamond. đź“Ť Local. 1.76 Visual Carat. 17% Smarter -0.1 Cut Score.
### 2 carat diamond loose?
2 ct loose black diamond in round cut for an engagement ring from Gemone diamonds. Features of round black diamond. This is 2 Carat AAA/AA quality Black diamond. Diamond is 100% natural round brilliant cut in Jet black color.
### 2 carat solitaire diamond?
Design a solitaire engagement ring with your own two carat diamond, all in 360°. Get inspired by recently purchased two carat solitaire diamond rings. Find a design and perfect the look by choosing your own two carat diamond. All of our two carat solitaire engagement rings have been photographed in stunning 360° so you can see exactly what ...
### 2.17 carat diamond ring?
Actual Size of 2.17 Carat Diamond. If you don't want to rely entirely on pictures like this, what you can do is measure the size of your finger and compare it to the 8.5 mm average diameter for a 2.17 carat diamond. The ring pictured is an average-sized size six ring, but bear in mind the ring you get could be significantly larger or smaller.
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# Find left limit $\lim\limits_{x \to{}0}{\frac{-\sin|x|}{x}}$
I have a little confusion with signs. Thanks in advance.
$$\lim_{x \to 0^-}{\frac{-\sin|x|}{x}}$$ (This is a left limit, i.e. $x<0$)
-
What is $sen{}$? – Chris Eagle Dec 2 '12 at 16:51
@ChrisEagle It's $\sin$ in spanish (seno). – Pragabhava Dec 2 '12 at 16:52
It's already edited, thanks for a while. – Rakisbro Dec 2 '12 at 16:53
Do you mean $\lim_{x\to 0^-}$ when you say left limit? – Asaf Karagila Dec 2 '12 at 16:54
that's that I mean Asaf. – Rakisbro Dec 2 '12 at 17:00
$$x<0\Longrightarrow |x|=-x\Longrightarrow\frac{-\sin|x|}{x}=\frac{\sin x}{x}\xrightarrow [x\to 0^-]{}1$$
sorry for the dumb question but ... $$sin(-x) = -sin(x)$$ is it correct? – Rakisbro Dec 2 '12 at 17:04
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Starting point of the PR-curve and the AUCPR value for an ideal classifier
I have two questions about the PR-curve:
1. What is the starting point of the PR-curve ?
I mean the point which corresponds to the highest possible threshold (i.e. when all scores are below this threshold). It is clear that all hard labels are equal to zero in this case. Hence, $$\text{TP}=\text{FP}=0$$ and $$\text{Recall}=0$$, but $$\text{Precision}=\frac{0}{0}$$. Sklearn uses $$\text{Precision}=1$$ for this point. Is it a general rule or other precision values might be used for this point in different libraries (for example, in R)?
2. What is the AUCPR value for an ideal classifier ?
I mean area under the PR-curve (AUCPR) for an ideal binary classifier (i.e. there is a threshold value such that all samples are classified correctly by the model). It is clear that the PR-curve of such classifier passes through the point $$(1,1)$$. Moreover, any PR-curve passes through the point which was described above in "1." and point $$(1, \frac{n_+}{n})$$ (this is the point of the lowest possible threshold when all scores are above this threshold), where $$n_+$$ is the total number of positive samples and $$n$$ is the total number of samples. Does that mean that AUCPR is equal to 1 in this case (like AUCROC of the ideal classifier) or it may be less than 1 ?
Working convention: Point $$(0,1)$$ is the upper left corner and corresponds to $$0$$ Recall (i.e. no Recall) and $$1$$ Precision (i.e. perfect Precision).
Regarding the first question: The start point can be at any point along $$0$$ or $$\frac{1}{n_+}$$ Recall, where the PR-curve start depends on the classifier performance. While we would hope that we will start at the point $$(\frac{1}{n_+},1)$$ and we will slow increase our Recall with little expense to Precision (i.e. we are very precise to begin with and slowly sacrifice Precision for Recall) that is not guaranteed at all. The obvious example is when we misclassify our "most probable" example of our test set. In that case we have both $$0$$-th Recall and $$0$$-th Precision, i.e. we start from point $$(0,0)$$. For example, in the left-most graph shown below (red line), we have an artificial example where we start at point $$(0,0.5)$$ because the first $$\frac{N}{2}$$ points are indistinguishable from each other. We "immediately" classify correctly some examples (i.e. we get TPs and thus non-zero Recall) but at the same time we get an equal number of FPs leading us at a $$0.5$$ Precision.
Please note, that in the case that no Positive examples are found (TPs or FPs), Precision is meaningless. There is not general rule on that what we do there. sklearn sets this to be $$1$$ that strictly for its convenience and explicitly says that these points "do not have a corresponding threshold". To that respect, in Davis & Goadrich (2006) the procedure of constructing a PR curve when presented with an algorithm returning probabilities is: "first find the probability that each test set example is positive, next sort this list and then traverse the sorted list in ascending order."; as such it is implied/suggested that for a probability that no example is positive, it makes no sense to construct a PR curve. In R PRROC::pr.curve does a similar thing where the origin is at $$(0,0$$) from the first positive example (example shown in pr3 below).
Side-note: in Python this leads in the slightly awkward situation of having Recall 0 with Precision 0 and 1 at the same time.
import numpy as np
from sklearn.metrics import precision_recall_curve
print(__doc__)
my_ytest = np.concatenate([np.array(['1'] * 50), np.array(['2'] * 50)])
my_yscore = np.concatenate([ [0.95], np.random.uniform(0.0, 0.5, 49),
np.random.uniform(0.5, 0.9, 50) ])
prec, recall, _ = precision_recall_curve(my_ytest, my_yscore, pos_label="2")
prec[recall==0]
# array([0., 1.])
Regarding the second question: Yes, the ideal classifier has AUCPR equal to 1. The only way to have an ideal classifier (i.e. performance that touches point $$(1,1)$$) but AUCPR less than $$1$$, is if we somehow moved towards $$(1,1)$$ while not already having perfect Precision (i.e. $$y=1$$). On occasion PR curves have the "sawtooth" shape (e.g. the middle graph shown below (dark green)), that suggests a significant jump in performance. That "tooth" though can never reach the point $$(1,1)$$ because by definition there are some misclassified points already. The "sawtooth effect" is due to us having a batch of correctly classified points, that helps us move both our Precision and Recall higher, followed by a batch of wrongly classified points that causes the sharp deep in Precision. To get the upward slope we increased our TP numbers while our FP & FN numbers remained the same but that does not mean though we removed our previously misclassified points; we can therefore never reach perfect Precision at $$y=1$$. For example in the right-most graph shown below (blue) a single point prohibits us from hitting $$\text{AUCPR} = 1$$; that misclassified FP point actually ranks higher than any other point in the positive class and thus forces our PR curve to start at $$(0,0)$$.
OK and some R code to see this first-handed:
library(PRROC)
N = 30000
set.seed(4321)
# The first N/2 points from each population are indistinguishable
pr0 <- pr.curve(scores.class0=c(rep(0.5, N/2), runif(n = N/2, max=0.4)),
scores.class1=c(rep(0.5, N/2), runif(n = N/2, min=0.4, max = 0.49)),
curve = TRUE)
# The [0.5, 0.7] space allows us to have the performance increase
pr1 <- pr.curve(scores.class0=c(runif(N/3, min=0.9, max=1.0),
runif(N/3, min=0.5, max=0.7),
runif(N/3, max=0.25)),
scores.class1=c(runif(N/2, min=0.7, max=0.9),
runif(N/2, min=0.0, max=0.5)),
curve=TRUE)
# The single point causes us to start from (0,0)
pr2 <- pr.curve(scores.class0=runif(n = N, min=0.999),
scores.class1=c(1, runif(N-1, max=0.999)),
curve = TRUE)
par(mfrow=c(1,3))
plot(pr0, legend=FALSE, col='red', panel.first= grid(),
cex.main = 1.5, main ="PR-curve starting at (0,0.5)")
plot(pr1, legend=FALSE, col='darkgreen', panel.first= grid(),
cex.main = 1.5, main ="PR-curve with a sawtooth!")
plot(pr2, legend=FALSE, col='blue', panel.first= grid(),
cex.main = 1.5, main ="PR-curve from a nearly ideal classifier")
• It seems that you defined the starting point as the point which corresponds to a threshold such that object with the highest score ("most probable" object) have score above this threshold. I define the starting point as the point which corresponds to a threshold such that all objects (including "most probable") have scores below the threshold (for example, when threshold is $0.9$ and scores are $(0.3,0.5,0.7)$). In my case we always classify all objects as negatives for such threshold, hence $\text{TP}=0$ and $\text{FP}=0$. This leads to $\text{precision} = 0/0$ and $\text{recall} = 0$. Aug 23, 2020 at 6:21
• I just found a post on stats.stackexchange whose author seems to think the same way as me. What do you think? Aug 23, 2020 at 6:31
• Right, OK! Sorry, it was unclear to me what was asked there. I edited my answer accordingly; please see the amendments. Aug 23, 2020 at 12:32
• Now everything is clear, thanks. Great post! Aug 23, 2020 at 13:04
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Question
# A beam of light of wavelength 600 nm from a distant source falls on a single slit 1.00 mm wide and the resulting diffraction pattern is observed on a screen 2 m away. The distance between the first dark fringe on either side of the central maxima is :
A
1.2 cm
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1.2 mm
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2.4 mm
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D
4.8 mm
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Solution
## The correct option is C 2.4 mmCondition for obtaining mth order minima is asinθ=mλThus for obtaining first diffraction minima,asinθ=λ⟹sinθ=600×10−9m10−3m≈tanθ=yD⟹y=1.2 mmThus distance between fringes on either side=2×y=2.4 mm
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# What could it mean to “Rotate” a distribution? [duplicate]
This question got me thinking about the meaning of variance: Intuition behind standard deviation.
Variance of a set of data is calculated the same way that the moment of inertia is calculated for a physical body. The moment of inertia is related to the energy required to rotate the body at a given speed. A figure skater will rotate faster with arms pulled in than stretched out. So what would be the analogous result of reducing variance, if any. Perhaps the analogy simply breaks down. Are there any publications that have investigated this analogy?
• @whuber I agree that the questions are related, but this one is different in that I am asking explicitly what the probability distribution equivalent of rotation would be. Essentially I would like to discover how far this analogy between probability distributions and distributions of mass can be taken, and what use can be made of this. – Livid Feb 4 '14 at 22:35
• I see that as identical to the several questions that have appeared asking about physical analogs of distributional moments. Check them out with a search. The first (most "relevant") hit has an answer referring to rotation, for instance. – whuber Feb 4 '14 at 22:44
• @whuber I am reading further, but I think that asking about "physical analogs of distributional moments" (other questions), is subtly different from "distributional moment analogs of physical concepts" (this question). Glen_b's answer, for example, appears to claim that the analogy can only be taken so far. I am unconvinced there is not some deeper connection. – Livid Feb 4 '14 at 23:09
The direct analogy is pretty clear:
To make it simple we'll assume it's for a continuous random variable on $(a,b)$. Without loss of generality, let $c=b-a$ and consider the corresponding variable on $(0,c)$; call that random variable $X$.
Now imagine a very thin rod of length $c$, whose density (mass per element of length) is variable in the x-direction (along its length) and consider that the rod happens to have the same material-density as a function of $x$ as the random variable has probability density as a function of $x$.
Then then second moment of inertia of the rod is the variance of $X$.
And hence what it 'means' to rotate a distribution is clear enough - it's quite literally rotating the 'rod' whose density represents probability-density. Variance is how 'hard' it would be to rotate the rod (low variance means 'easy to spin', high variance means it takes more push to spin it ... and stop it, if you spin it).
Think about what inertia (how hard it is to spin) reflects here, which is simply how close the mass is to the mean. The closer the mass is to the mean the easier it is to spin. If you made a physical object whose physical density represents the probability density and the random variable had low variance, the corresponding object would be easy to spin, because most of the mass would be close to the mean - both inertia and variance are how close the mass is to the mean, in a particular (and directly analogous) sense.
You don't actually 'spin' a probability density and imagine that to be physically difficult, any more than electricity is wet because of the water analogy. To expect that level of correspondence is to miss the point of such analogies (the aspects that correspond, correspond, but not every consequence of the correspondence in one realm carries over with it).
The point of saying the 'rod is hard to spin' is to give a pretty direct sense of what high variance is telling you about density. But to insist that the probability density itself spin is to miss the point.
• For clarity, by "continuous random variable on $(a,b)$", do you mean uniform? Otherwise, what are $a$ & $b$? – gung - Reinstate Monica Feb 4 '14 at 21:56
• @gung No, I don't mean uniform, because I explicitly said the density varied (well it could even be uniform, but the discussion is much more general). I mean that the variable is limited to values between $a$ and $b$, and the phrasing I used there is quite common, even standard. The reason is that people find it hard to imagine rotating infinitely long rods and I didn't want any additional strain in doing the imagining - hence the finite length rod, and so a bounded random variable. (Imagine, if you like, dealing with a particular mixture of scaled beta distributions, as an example) – Glen_b -Reinstate Monica Feb 4 '14 at 22:05
• @Glen_b But what does "easy to spin mean" for a probability distribution. – Livid Feb 4 '14 at 22:20
• Please see my edits – Glen_b -Reinstate Monica Feb 4 '14 at 22:39
• @Glen_b I understand your point, do you have any speculation on what "spinning a probability density" could mean? Or you believe that is just taking the analogy too far? – Livid Feb 4 '14 at 23:38
If you have greater variance, then you need to expend a greater effort (meaning, spend more money collecting the data) to obtain a given level of precision. The precision for $\bar X$ is of course is the inverse standard deviation, $\sqrt{n}/\sigma$.
• So to take the analogy further, we would say that estimating the center of mass is more difficult for a larger body. This makes some sense. – Livid Feb 4 '14 at 22:21
In signals, variance is essentially a measure of energy. Of course in this case they are related directly, not inversely.
• Can you expand on this? Increased variance would require greater energy to "change", from that perspective the relationship is not inverse. – Livid Feb 4 '14 at 22:19
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## Composing an endofunction with itself
Suppose we are given an endofunction $f : \mathcal{X} \to \mathcal{X}$. Let us introduce a new function $f^{(n)}$, where $n \geq 0$, which we define recursively as follows
$f^{(k)} = f \circ f^{(k-1)}, \qquad{} f^{(0)} = \text{id}$
where $\text{id} : \mathcal{X} \to \mathcal{X}$ is the identity function on $\mathcal{X}$, i.e., $\text{id} (x) = x$ for all $x \in \mathcal{X}$. In Haskell, it is easy to create a higher-order function compose that composes an endofunction $f$ with itself $n$ times:
compose :: (a -> a) -> Integer -> (a -> a)
compose f 0 = (\x -> x)
compose f k = f . (compose f (k-1))
where the identity function is defined anonymously. If you happen to dislike anonymity in your functions, here is another implementation:
compose :: (a -> a) -> Integer -> (a -> a)
compose f 0 = identity
compose f k = f . (compose f (k-1))
identity :: a -> a
identity x = x
There are many other alternatives. Take a look at this Stack Overflow thread. I particularly liked Reid Barton‘s elegant solution using a right-fold:
compose f n = foldr (.) id (replicate n f)
where id = (\x -> x). Beautiful, isn’t it? If this one-liner does not convert you to Haskell, then nothing will.
__________
Example
Let $\mathbb{Z}_{64} := \{0, 1,\dots,63\}$, and let function $f$ be defined by
$\begin{array}{rl} f : \mathbb{Z}_{64} &\to \mathbb{Z}_{64}\\ n &\mapsto n+1 \pmod{64}\end{array}$
This function corresponds to a cyclic shift. Therefore, we expect to have $f^{(64)} = \text{id}$. Let $g := f^{(32)}$ and $h := f^{(64)}$. We load the script above into GHCi, and voilà, we have the session:
*Main> let ns = [0..63]
*Main> let f = (\n -> (n+1) mod 64)
*Main> map f ns
[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,
17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,
33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,
49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,0]
So far, so good! Let us take a look at $g$ and $h$:
*Main> let g = compose f 32
*Main> let h = compose f 64
*Main> map g ns
[32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,
48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,
0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,
16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31]
*Main> map h ns
[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,
16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,
32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,
48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63]
Indeed, as we expected, we have $f^{(64)} = \text{id}$.
__________
Everything in this post is elementary. But now that the elements are solid, we can dare to attempt more sophisticated things. Like what? Like composing lists of (distinct) functions. I am thinking of permutations. I am thinking of perfect shuffles.
### One Response to “Composing an endofunction with itself”
1. Rod Carvalho Says:
Here’s a bit of equational reasoning:
compose f 3 = f . (compose f 2)
= f . (f . (compose f 1))
= f . (f . (f . (compose f 0)))
= f . (f . (f . identity))
foldr (.) identity (replicate n f) = foldr (.) identity [f,f,f]
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# 3)Bearing Capacity
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CE 366 BEARING CAPACITY (Problems & Solutions)
P1
An excavation will be made for a ten storey 15x25 m building. Temporary support of earth pressure and water pressure will be made by deep secant cantilever pile wall. Thegross pressure due to dead and live loads of the structure and weight of the raft is 130 kPa(assume that it is uniform).this level can only be fill (placed afterattained after a construction iswater proofing will relatively long time fully completed)be provided10 storey building (15x25m)1moriginal GWT 4mposition
γ
sat
= 20 kN/m
3
γ
moist
=18 kN/m
3
medium mediumdense dense GWT is lowered 4msand sandmedium stiff clay 2m
γ
sat
= 21 kN/m
3
a) What is net foundation pressure at the end of construction but before the void spacebetween the pile wall and the building has not been filled, and
there is no water insidethe foundation pit yet
(water level at the base level) (GWT position 1).b) What is net foundation pressure long after the completion of the building, i.e. waterlevel is inside the pile wall and the backfill between the building and the pile wall isplaced (GWT position 2). What is the factor of safety against uplift?
21
2
a) q
net
= final effective stress - initial effective stressat foundation level at foundation levelq
net
=
σ
-
σ
o
1m
γ
moist
= 18 kN/m
3
5m
γ
sat
= 20 kN/m
3
σ
o
σ
o
= 18x1 + 4x(20-9.8) = 58.8 kPa( gross pressure – uplift pressure) = final effective stress at foundation level,
σ
gross pressure =130 kPa (given)uplift pressure = 0 kPa (Since GWT is at foundation level (1), it has no effect onstructure load)
σ
=130 –0 = 130 kPaq
net
=130 – 58.8=
71.2 kPa
b)
σ
= 130 – 4x9.8 = 90.8 kPauplift pressure
σ
o
= 58.8 kPa (same as above)q
net
= 90.8 – 58.8=
32.0 kPa
ORq
net
=q
gross
-
γ
sat
D =130-(18x1+4x20)=
32.0 kPa
Factor of safety against uplift is:(FS)
uplift
= weight of structure / uplift= (130x15x25) / (4x9.8x15x25)=
3.3
3
P2
Calculate the FS against uplift and calculate effective stress at the base level for waterlevel at (1) and (2).0.75 3.50 0.75
γ
concrete
= 24 kN/m
3
5.0 m3.02.851.0(1) (1) 3.0 m2.01.0
water table at (1)
Factor of Safety against uplift = (2x6x0.75 + 5x1)x24 / (3x5)x9.8
Σ
weight of pit uplift= 336 / 147=
2.28
Base pressure = 336 / 5 = 67.2 kN/m
2
due to weight of structure147 / 5 = 29.4 kN/m
2
is supported by groundwater67.2 – 29.4 = 37.8 kN/m
3
is supported by soil (effective stress at the base)base pressure 29.4 kPa : supported bydue to 67.2 kPa groundwater (uplift)structure 37.8 kPa : supported bysoil
water table at (2)
waterproof membranegroundlevelvery longconcrete pit
(2)(2)
## Activity (37)
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>
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# Question for Accuquilt users---Square-in-Square
01-30-2016, 10:43 AM
#1
Member
Join Date: Dec 2009
Location: Sandy, Utah
Posts: 86
Question for Accuquilt users---Square-in-Square
For Accuquilt users: Is there a particular die to make the traditional Square-in-Square block?
01-30-2016, 02:01 PM
#2
Super Member
Join Date: Aug 2013
Location: Arizona
Posts: 2,356
I think you would just use one of the strip dies. I don't know of any other way.
01-30-2016, 02:12 PM
#3
Power Poster
Join Date: Jan 2011
Location: Southern USA
Posts: 12,556
There is a 6" finished Snowball die.
https://www.accuquilt.com/shop/go-sn...-finished.html
01-30-2016, 03:38 PM
#4
Power Poster
Join Date: Dec 2010
Location: Michigan
Posts: 11,266
Yes and no. They call the block a "square on point". Here's the link to the sizes available.
http://www.accuquilt.com/shop/ssearch?q=square+on+point
The problem is that you still need the corresponding HST die to complete the patch. They don't tell you which size HST die you need, or what the finished size of the SIS patch is. The only place I could find the info is to go the page for that specific die, then watch the video. For instance, the GO! #55106 Square on Point-3 11/16" (3 3/16" Finished)
needs the 2.25 HST, yielding a 4.5" patch
01-30-2016, 04:01 PM
#5
Power Poster
Join Date: Dec 2010
Location: Michigan
Posts: 11,266
OOh, just missed the editing window...
Actually, you can use the same calculation that you use to calculate the size for setting triangles. So multiply the finished size of the square, 3 3.16 in this case (3.1875) by 1.414 which gives you 4.5". Half of that would be the FINISHED HST size you need, 2 1/4 in the example, which is what they said in the video.
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Timer 00:00:00 Sudoku Sudoku
## if (sTodaysDate) document.write('Sudoku for ' + sTodaysDate.split('-')[0] + '/' + monthname() + '/' + sTodaysDate.split('-')[2]); else document.write('Enter your own sudoku puzzle.')
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Can't be first. Can't happen. | |
4:35 My peanut is a calico Maen | |
It Happen. How about that. Not bad time either today. | |
3:57 Peanut poses well! Nice picture! | |
Not first, not fast! | |
I'm new to this. What does it mean when you say you set up a puzzle? | |
Cute as a button . . .er, I mean a Peanut! | |
A blue ribbon dog. | |
10:18. I believe it's about finding which methods to apply as a matter of priority. this is probably where most time can be won. | |
Goofed up on timer. | |
5:12 Looks cute, but... definately not for me | |
so fluffy and cute...great photo for the jigsaw puzzle site Gath | |
4:44 - Hey Mandy - wherabouts are you in Perth?. | |
3:41 I'm FAST today. Small triumphs. Cute doggie. Don't those tiny ones get snappy? | |
Very good pet portrait. Dog pictures are always so much better than those pictures of evil cats! | |
9.28 | |
Yo Glenn! I am fluffy the kitten, FEAR ME!! | |
6:09Maen all | |
5:08. Cute little peanut! | |
7:17 Grrr. That's what I get for breaking the two-minute barrier on the first puzzle. Fried my brain. | |
5:37. fluff ball. | |
Doga are cool...cats are beyond cool. | |
Slow today 14:40, Peanut's life lookstoo cushy...lets introduce her to a moodyfeline!! | |
Peanut looks like an award winning Pomeranian (DOG)...not cat. Good Maen everyone. Can't believe you guys do these medium puzzles in 4 Min. It takes me 4 minutes just to go thru all the numbers and put in a few possibles. Not that I am doubting you. I know it can be done, just not by me.... | |
But, DAVE, you already admitted that your brain cells are fleeing at an alarming rate! :) | |
My second medium. It might take me forever, but I'm getting there. Sill movin on up!! | |
7:55 Good mAen to all. Cute dog. I always like the pet pictures. Submitted one myself and guess what..It is of a cat! Don't know what the chances are of it showing up as I have never submitted a photo before. Anyone know? | |
forgot the timer | |
14:23 Slow todayPomeranian | |
I don't dare time myself of medium. Cute pup. My inlaws were here for my son's graduation and brought their hyperactive bijon friese, which my daughter kept referring to as a mop without a handle. Personally I'm drawn to dogs with a more mysterious/interesting lineage. My dog Lucy is likely a mix of chihuahua and cairn terrier, with maybe some dachshund thrown in. Think about it. | |
Good mAen to all. Does the blue ribbon signify a show win? | |
9.49...not bad for me for a medium | |
4:21 Cute little Peanut! He/She looks very soft. | |
woopie! 3:13 on easy. (Flippin' scary picture that.) 7:38 on med. Off to make some bread. | |
Susan from Mississippi - I like your reference to dogs with a 'more mysterious/interesting lineage'. That sounds so much more intriguing than plain old 'mixed breed'! I think most of them are wonderful dogs. My daughter has one purebred bull terrier, the 'Target' dog, & one mixed breed hound from the animal shelter. The mix is much more fun to be with & a lot healthier, too! | |
3:18 on medium! My new record =) | |
5:52 Hi to all and have a good weekend. Glenn say after me Cats are nice and do no harm and then take deep breaths as you repeat it lol!! Chocolate Thought for the Day: Without chocolate, there would be darkness and chaos! | |
5.45a little fluffy peanuthave a great day/night one and allcan't agree more André with you chocolate thought for the day | |
3:27 without possibilities. | |
Kathy, maybe I'll try to get a photo of my mystery breed on this site. She's most unusual, but a great family member. As a matter of fact, I rescued her about six years ago from a soup kitchen; my husband likes to say she won 'Stray Dog Lotto'. LOL | |
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# math
posted by .
An ant has 6 legs and a Spider has 8 legs. In a insectarium display, there are some spiders and ants. From these spiders and ants, there are 60 legs altogether. how many spiders and ants could there be?
• math -
A = 6/8S
6A + 8S = 60
Substitute 6/8S for A in second equation and solve for S. Then use first equation to find A.
## Similar Questions
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how will use nested loop
6 visualizaciones (últimos 30 días)
manoj saini el 18 de Dic. de 2012
hello my problem is that i have 11 value for any variable suppose i=1:11 and have another variable suppose b=2:12 also has 11 values now my question is that i want to find out value for i=1 for z=i*b total 11 values and for i=2 another 11 value of z and so on how i can use nested loop
3 comentariosMostrar 1 comentario más antiguoOcultar 1 comentario más antiguo
manoj saini el 19 de Dic. de 2012
for i=1 I want 11 values of i*b in one row for i=2 I want 11 values of i*b in one row for i=3 I want 11 values of i*b in one row AND SO ON..............
Walter Roberson el 19 de Dic. de 2012
Yes, and all of the Answers so far give you that.
Iniciar sesión para comentar.
Muruganandham Subramanian el 19 de Dic. de 2012
Editada: Muruganandham Subramanian el 19 de Dic. de 2012
try this:
z = zeros(11,11);
for a=1:11
for b=2:12
z(a,b)=a*b;
end
end
z(:,1)=[];
2 comentariosMostrar NingunoOcultar Ninguno
Walter Roberson el 19 de Dic. de 2012
You declare z as 11 x 11, but then you use it as 11 x 12.
Because of the way MATLAB handles assignments to non-existant locations, this will work, but it does indicate a logic fault on your part.
Muruganandham Subramanian el 19 de Dic. de 2012
Editada: Muruganandham Subramanian el 19 de Dic. de 2012
Iniciar sesión para comentar.
Más respuestas (3)
Babak el 18 de Dic. de 2012
Z = zeors(11,11);
for i=1:11
for b=2:12
z=i*b;
end
end
4 comentariosMostrar 2 comentarios más antiguosOcultar 2 comentarios más antiguos
manoj saini el 19 de Dic. de 2012
sir now if my varible i=0.1:.11 and b also in point then how it will save
Image Analyst el 19 de Dic. de 2012
If the number you're multiplying by is a fractional number, like 0.1 or 0.11 then you'll have to separate your index from your number, like I think you originally had in your message where you had a and b instead of i and b. You could just make the loop index from 1 to 11 and then create the array index from it like arrayIndex = i / 10 and then use arrayIndex in z() or wherever.
Iniciar sesión para comentar.
Walter Roberson el 18 de Dic. de 2012
z = bsxfun(@times, i(:), b);
0 comentariosMostrar -2 comentarios más antiguosOcultar -2 comentarios más antiguos
Iniciar sesión para comentar.
Sean de Wolski el 18 de Dic. de 2012
Code golf:
z=i'*b
1 comentarioMostrar -1 comentarios más antiguosOcultar -1 comentarios más antiguos
Walter Roberson el 18 de Dic. de 2012
Provided "i" is not complex, which it normally is ;-)
Iniciar sesión para comentar.
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Translated by
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Home Metamath Proof ExplorerTheorem List (p. 27 of 431) < Previous Next > Bad symbols? Try the GIF version. Mirrors > Metamath Home Page > MPE Home Page > Theorem List Contents > Recent Proofs This page: Page List
Color key: Metamath Proof Explorer (1-28105) Hilbert Space Explorer (28106-29630) Users' Mathboxes (29631-43082)
Theorem List for Metamath Proof Explorer - 2601-2700 *Has distinct variable group(s)
TypeLabelDescription
Statement
Theoremsbalv 2601* Quantify with new variable inside substitution. (Contributed by NM, 18-Aug-1993.)
([𝑦 / 𝑥]𝜑𝜓) ([𝑦 / 𝑥]∀𝑧𝜑 ↔ ∀𝑧𝜓)
Theoremsbco4lem 2602* Lemma for sbco4 2603. It replaces the temporary variable 𝑣 with another temporary variable 𝑤. (Contributed by Jim Kingdon, 26-Sep-2018.)
([𝑥 / 𝑣][𝑦 / 𝑥][𝑣 / 𝑦]𝜑 ↔ [𝑥 / 𝑤][𝑦 / 𝑥][𝑤 / 𝑦]𝜑)
Theoremsbco4 2603* Two ways of exchanging two variables. Both sides of the biconditional exchange 𝑥 and 𝑦, either via two temporary variables 𝑢 and 𝑣, or a single temporary 𝑤. (Contributed by Jim Kingdon, 25-Sep-2018.)
([𝑦 / 𝑢][𝑥 / 𝑣][𝑢 / 𝑥][𝑣 / 𝑦]𝜑 ↔ [𝑥 / 𝑤][𝑦 / 𝑥][𝑤 / 𝑦]𝜑)
Theorem2sb8e 2604* An equivalent expression for double existence. (Contributed by Wolf Lammen, 2-Nov-2019.)
(∃𝑥𝑦𝜑 ↔ ∃𝑧𝑤[𝑧 / 𝑥][𝑤 / 𝑦]𝜑)
Theoremexsb 2605* An equivalent expression for existence. (Contributed by NM, 2-Feb-2005.)
(∃𝑥𝜑 ↔ ∃𝑦𝑥(𝑥 = 𝑦𝜑))
Theorem2exsb 2606* An equivalent expression for double existence. (Contributed by NM, 2-Feb-2005.) (Proof shortened by Wolf Lammen, 30-Sep-2018.)
(∃𝑥𝑦𝜑 ↔ ∃𝑧𝑤𝑥𝑦((𝑥 = 𝑧𝑦 = 𝑤) → 𝜑))
1.6 Existential uniqueness
Syntaxweu 2607 Extend wff definition to include existential uniqueness ("there exists a unique 𝑥 such that 𝜑").
wff ∃!𝑥𝜑
Syntaxwmo 2608 Extend wff definition to include uniqueness ("there exists at most one 𝑥 such that 𝜑").
wff ∃*𝑥𝜑
Theoremeujust 2609* A soundness justification theorem for df-eu 2611, showing that the definition is equivalent to itself with its dummy variable renamed. Note that 𝑦 and 𝑧 needn't be distinct variables. See eujustALT 2610 for a proof that provides an example of how it can be achieved through the use of dvelim 2477. (Contributed by NM, 11-Mar-2010.) (Proof shortened by Andrew Salmon, 9-Jul-2011.)
(∃𝑦𝑥(𝜑𝑥 = 𝑦) ↔ ∃𝑧𝑥(𝜑𝑥 = 𝑧))
TheoremeujustALT 2610* Alternate proof of eujust 2609 illustrating the use of dvelim 2477. (Contributed by NM, 11-Mar-2010.) (Proof modification is discouraged.) (New usage is discouraged.)
(∃𝑦𝑥(𝜑𝑥 = 𝑦) ↔ ∃𝑧𝑥(𝜑𝑥 = 𝑧))
Definitiondf-eu 2611* Define existential uniqueness, i.e. "there exists exactly one 𝑥 such that 𝜑." Definition 10.1 of [BellMachover] p. 97; also Definition *14.02 of [WhiteheadRussell] p. 175. Other possible definitions are given by eu1 2648, eu2 2647, eu3v 2635, and eu5 2633 (which in some cases we show with a hypothesis 𝜑 → ∀𝑦𝜑 in place of a distinct variable condition on 𝑦 and 𝜑). Double uniqueness is tricky: ∃!𝑥∃!𝑦𝜑 does not mean "exactly one 𝑥 and one 𝑦 " (see 2eu4 2694). (Contributed by NM, 12-Aug-1993.)
(∃!𝑥𝜑 ↔ ∃𝑦𝑥(𝜑𝑥 = 𝑦))
Definitiondf-mo 2612 Define "there exists at most one 𝑥 such that 𝜑." Here we define it in terms of existential uniqueness. Notation of [BellMachover] p. 460, whose definition we show as mo3 2645. For other possible definitions see mo2 2616 and mo4 2655. (Contributed by NM, 8-Mar-1995.)
(∃*𝑥𝜑 ↔ (∃𝑥𝜑 → ∃!𝑥𝜑))
Theoremeuequ1 2613* Equality has existential uniqueness. Special case of eueq1 3520 proved using only predicate calculus. The proof needs 𝑦 = 𝑧 be free of 𝑥. This is ensured by having 𝑥 and 𝑦 be distinct. Alternately, a distinctor ¬ ∀𝑥𝑥 = 𝑦 could have been used instead. (Contributed by Stefan Allan, 4-Dec-2008.) (Proof shortened by Wolf Lammen, 8-Sep-2019.)
∃!𝑥 𝑥 = 𝑦
Theoremmo2v 2614* Alternate definition of "at most one." Unlike mo2 2616, which is slightly more general, it does not depend on ax-11 2183 and ax-13 2391, whence it is preferable within predicate logic. Elsewhere, most theorems depend on these axioms anyway, so this advantage is no longer important. (Contributed by Wolf Lammen, 27-May-2019.) (Proof shortened by Wolf Lammen, 10-Nov-2019.)
(∃*𝑥𝜑 ↔ ∃𝑦𝑥(𝜑𝑥 = 𝑦))
Theoremeuf 2615* A version of the existential uniqueness definition with a hypothesis instead of a distinct variable condition. (Contributed by NM, 12-Aug-1993.) (Proof shortened by Wolf Lammen, 30-Oct-2018.)
𝑦𝜑 (∃!𝑥𝜑 ↔ ∃𝑦𝑥(𝜑𝑥 = 𝑦))
Theoremmo2 2616* Alternate definition of "at most one." (Contributed by NM, 8-Mar-1995.) Restrict dummy variable z. (Revised by Wolf Lammen, 28-May-2019.)
𝑦𝜑 (∃*𝑥𝜑 ↔ ∃𝑦𝑥(𝜑𝑥 = 𝑦))
Theoremnfeu1 2617 Bound-variable hypothesis builder for uniqueness. (Contributed by NM, 9-Jul-1994.) (Revised by Mario Carneiro, 7-Oct-2016.)
𝑥∃!𝑥𝜑
Theoremnfmo1 2618 Bound-variable hypothesis builder for "at most one." (Contributed by NM, 8-Mar-1995.) (Revised by Mario Carneiro, 7-Oct-2016.)
𝑥∃*𝑥𝜑
Theoremnfeud2 2619 Bound-variable hypothesis builder for uniqueness. (Contributed by Mario Carneiro, 14-Nov-2016.) (Proof shortened by Wolf Lammen, 4-Oct-2018.)
𝑦𝜑 & ((𝜑 ∧ ¬ ∀𝑥 𝑥 = 𝑦) → Ⅎ𝑥𝜓) (𝜑 → Ⅎ𝑥∃!𝑦𝜓)
Theoremnfmod2 2620 Bound-variable hypothesis builder for "at most one." (Contributed by Mario Carneiro, 14-Nov-2016.)
𝑦𝜑 & ((𝜑 ∧ ¬ ∀𝑥 𝑥 = 𝑦) → Ⅎ𝑥𝜓) (𝜑 → Ⅎ𝑥∃*𝑦𝜓)
Theoremnfeud 2621 Deduction version of nfeu 2623. (Contributed by NM, 15-Feb-2013.) (Revised by Mario Carneiro, 7-Oct-2016.)
𝑦𝜑 & (𝜑 → Ⅎ𝑥𝜓) (𝜑 → Ⅎ𝑥∃!𝑦𝜓)
Theoremnfmod 2622 Bound-variable hypothesis builder for "at most one." (Contributed by Mario Carneiro, 14-Nov-2016.)
𝑦𝜑 & (𝜑 → Ⅎ𝑥𝜓) (𝜑 → Ⅎ𝑥∃*𝑦𝜓)
Theoremnfeu 2623 Bound-variable hypothesis builder for uniqueness. Note that 𝑥 and 𝑦 needn't be distinct. (Contributed by NM, 8-Mar-1995.) (Revised by Mario Carneiro, 7-Oct-2016.)
𝑥𝜑 𝑥∃!𝑦𝜑
Theoremnfmo 2624 Bound-variable hypothesis builder for "at most one." (Contributed by NM, 9-Mar-1995.)
𝑥𝜑 𝑥∃*𝑦𝜑
Theoremeubid 2625 Formula-building rule for uniqueness quantifier (deduction rule). (Contributed by NM, 9-Jul-1994.)
𝑥𝜑 & (𝜑 → (𝜓𝜒)) (𝜑 → (∃!𝑥𝜓 ↔ ∃!𝑥𝜒))
Theoremmobid 2626 Formula-building rule for "at most one" quantifier (deduction rule). (Contributed by NM, 8-Mar-1995.)
𝑥𝜑 & (𝜑 → (𝜓𝜒)) (𝜑 → (∃*𝑥𝜓 ↔ ∃*𝑥𝜒))
Theoremeubidv 2627* Formula-building rule for uniqueness quantifier (deduction rule). (Contributed by NM, 9-Jul-1994.)
(𝜑 → (𝜓𝜒)) (𝜑 → (∃!𝑥𝜓 ↔ ∃!𝑥𝜒))
Theoremmobidv 2628* Formula-building rule for "at most one" quantifier (deduction rule). (Contributed by Mario Carneiro, 7-Oct-2016.)
(𝜑 → (𝜓𝜒)) (𝜑 → (∃*𝑥𝜓 ↔ ∃*𝑥𝜒))
Theoremeubii 2629 Introduce uniqueness quantifier to both sides of an equivalence. (Contributed by NM, 9-Jul-1994.) (Revised by Mario Carneiro, 6-Oct-2016.)
(𝜑𝜓) (∃!𝑥𝜑 ↔ ∃!𝑥𝜓)
Theoremmobii 2630 Formula-building rule for "at most one" quantifier (inference rule). (Contributed by NM, 9-Mar-1995.) (Revised by Mario Carneiro, 17-Oct-2016.)
(𝜓𝜒) (∃*𝑥𝜓 ↔ ∃*𝑥𝜒)
Theoremeuex 2631 Existential uniqueness implies existence. For a shorter proof using more axioms, see euexALT 2649. (Contributed by NM, 15-Sep-1993.) (Proof shortened by Andrew Salmon, 9-Jul-2011.) (Proof shortened by Wolf Lammen, 4-Dec-2018.)
(∃!𝑥𝜑 → ∃𝑥𝜑)
Theoremexmo 2632 Something exists or at most one exists. (Contributed by NM, 8-Mar-1995.)
(∃𝑥𝜑 ∨ ∃*𝑥𝜑)
Theoremeu5 2633 Uniqueness in terms of "at most one." Revised to reduce dependencies on axioms. (Contributed by NM, 23-Mar-1995.) (Proof shortened by Wolf Lammen, 25-May-2019.)
(∃!𝑥𝜑 ↔ (∃𝑥𝜑 ∧ ∃*𝑥𝜑))
Theoremexmoeu2 2634 Existence implies "at most one" is equivalent to uniqueness. (Contributed by NM, 5-Apr-2004.)
(∃𝑥𝜑 → (∃*𝑥𝜑 ↔ ∃!𝑥𝜑))
Theoremeu3v 2635* An alternate way to express existential uniqueness. (Contributed by NM, 8-Jul-1994.) Add a distinct variable condition on 𝜑. (Revised by Wolf Lammen, 29-May-2019.)
(∃!𝑥𝜑 ↔ (∃𝑥𝜑 ∧ ∃𝑦𝑥(𝜑𝑥 = 𝑦)))
Theoremeumo 2636 Existential uniqueness implies "at most one." (Contributed by NM, 23-Mar-1995.) Reduce axiom usage. (Revised by GL, 19-Feb-2022.)
(∃!𝑥𝜑 → ∃*𝑥𝜑)
TheoremeumoOLD 2637 Obsolete proof of eumo 2636 as of 19-Feb-2022. (Contributed by NM, 23-Mar-1995.) (Proof modification is discouraged.) (New usage is discouraged.)
(∃!𝑥𝜑 → ∃*𝑥𝜑)
Theoremeumoi 2638 "At most one" inferred from existential uniqueness. (Contributed by NM, 5-Apr-1995.)
∃!𝑥𝜑 ∃*𝑥𝜑
Theoremmoabs 2639 Absorption of existence condition by "at most one." (Contributed by NM, 4-Nov-2002.)
(∃*𝑥𝜑 ↔ (∃𝑥𝜑 → ∃*𝑥𝜑))
Theoremexmoeu 2640 Existence in terms of "at most one" and uniqueness. (Contributed by NM, 5-Apr-2004.) (Proof shortened by Wolf Lammen, 5-Dec-2018.)
(∃𝑥𝜑 ↔ (∃*𝑥𝜑 → ∃!𝑥𝜑))
Theoremsb8eu 2641 Variable substitution in uniqueness quantifier. (Contributed by NM, 7-Aug-1994.) (Revised by Mario Carneiro, 7-Oct-2016.) (Proof shortened by Wolf Lammen, 24-Aug-2019.)
𝑦𝜑 (∃!𝑥𝜑 ↔ ∃!𝑦[𝑦 / 𝑥]𝜑)
Theoremsb8mo 2642 Variable substitution for "at most one." (Contributed by Alexander van der Vekens, 17-Jun-2017.)
𝑦𝜑 (∃*𝑥𝜑 ↔ ∃*𝑦[𝑦 / 𝑥]𝜑)
Theoremcbveu 2643 Rule used to change bound variables, using implicit substitution. (Contributed by NM, 25-Nov-1994.) (Revised by Mario Carneiro, 7-Oct-2016.)
𝑦𝜑 & 𝑥𝜓 & (𝑥 = 𝑦 → (𝜑𝜓)) (∃!𝑥𝜑 ↔ ∃!𝑦𝜓)
Theoremcbvmo 2644 Rule used to change bound variables, using implicit substitution. (Contributed by NM, 9-Mar-1995.) (Revised by Andrew Salmon, 8-Jun-2011.)
𝑦𝜑 & 𝑥𝜓 & (𝑥 = 𝑦 → (𝜑𝜓)) (∃*𝑥𝜑 ↔ ∃*𝑦𝜓)
Theoremmo3 2645* Alternate definition of "at most one." Definition of [BellMachover] p. 460, except that definition has the side condition that 𝑦 not occur in 𝜑 in place of our hypothesis. (Contributed by NM, 8-Mar-1995.) (Proof shortened by Wolf Lammen, 18-Aug-2019.)
𝑦𝜑 (∃*𝑥𝜑 ↔ ∀𝑥𝑦((𝜑 ∧ [𝑦 / 𝑥]𝜑) → 𝑥 = 𝑦))
Theoremmo 2646* Equivalent definitions of "there exists at most one." (Contributed by NM, 7-Aug-1994.) (Revised by Mario Carneiro, 7-Oct-2016.) (Proof shortened by Wolf Lammen, 2-Dec-2018.)
𝑦𝜑 (∃𝑦𝑥(𝜑𝑥 = 𝑦) ↔ ∀𝑥𝑦((𝜑 ∧ [𝑦 / 𝑥]𝜑) → 𝑥 = 𝑦))
Theoremeu2 2647* An alternate way of defining existential uniqueness. Definition 6.10 of [TakeutiZaring] p. 26. (Contributed by NM, 8-Jul-1994.) (Proof shortened by Wolf Lammen, 2-Dec-2018.)
𝑦𝜑 (∃!𝑥𝜑 ↔ (∃𝑥𝜑 ∧ ∀𝑥𝑦((𝜑 ∧ [𝑦 / 𝑥]𝜑) → 𝑥 = 𝑦)))
Theoremeu1 2648* An alternate way to express uniqueness used by some authors. Exercise 2(b) of [Margaris] p. 110. (Contributed by NM, 20-Aug-1993.) (Revised by Mario Carneiro, 7-Oct-2016.) (Proof shortened by Wolf Lammen, 29-Oct-2018.)
𝑦𝜑 (∃!𝑥𝜑 ↔ ∃𝑥(𝜑 ∧ ∀𝑦([𝑦 / 𝑥]𝜑𝑥 = 𝑦)))
TheoremeuexALT 2649 Alternate proof of euex 2631. Shorter but uses more axioms. (Contributed by NM, 15-Sep-1993.) (Proof shortened by Andrew Salmon, 9-Jul-2011.) (Proof modification is discouraged.) (New usage is discouraged.)
(∃!𝑥𝜑 → ∃𝑥𝜑)
Theoremeuor 2650 Introduce a disjunct into a uniqueness quantifier. (Contributed by NM, 21-Oct-2005.)
𝑥𝜑 ((¬ 𝜑 ∧ ∃!𝑥𝜓) → ∃!𝑥(𝜑𝜓))
Theoremeuorv 2651* Introduce a disjunct into a uniqueness quantifier. (Contributed by NM, 23-Mar-1995.)
((¬ 𝜑 ∧ ∃!𝑥𝜓) → ∃!𝑥(𝜑𝜓))
Theoremeuor2 2652 Introduce or eliminate a disjunct in a uniqueness quantifier. (Contributed by NM, 21-Oct-2005.) (Proof shortened by Andrew Salmon, 9-Jul-2011.) (Proof shortened by Wolf Lammen, 27-Dec-2018.)
(¬ ∃𝑥𝜑 → (∃!𝑥(𝜑𝜓) ↔ ∃!𝑥𝜓))
Theoremsbmo 2653* Substitution into "at most one". (Contributed by Jeff Madsen, 2-Sep-2009.)
([𝑦 / 𝑥]∃*𝑧𝜑 ↔ ∃*𝑧[𝑦 / 𝑥]𝜑)
Theoremmo4f 2654* "At most one" expressed using implicit substitution. (Contributed by NM, 10-Apr-2004.)
𝑥𝜓 & (𝑥 = 𝑦 → (𝜑𝜓)) (∃*𝑥𝜑 ↔ ∀𝑥𝑦((𝜑𝜓) → 𝑥 = 𝑦))
Theoremmo4 2655* "At most one" expressed using implicit substitution. (Contributed by NM, 26-Jul-1995.)
(𝑥 = 𝑦 → (𝜑𝜓)) (∃*𝑥𝜑 ↔ ∀𝑥𝑦((𝜑𝜓) → 𝑥 = 𝑦))
Theoremeu4 2656* Uniqueness using implicit substitution. (Contributed by NM, 26-Jul-1995.)
(𝑥 = 𝑦 → (𝜑𝜓)) (∃!𝑥𝜑 ↔ (∃𝑥𝜑 ∧ ∀𝑥𝑦((𝜑𝜓) → 𝑥 = 𝑦)))
Theoremmoim 2657 "At most one" reverses implication. (Contributed by NM, 22-Apr-1995.)
(∀𝑥(𝜑𝜓) → (∃*𝑥𝜓 → ∃*𝑥𝜑))
Theoremmoimi 2658 "At most one" reverses implication. (Contributed by NM, 15-Feb-2006.)
(𝜑𝜓) (∃*𝑥𝜓 → ∃*𝑥𝜑)
Theoremmoa1 2659 If an implication holds for at most one value, then its consequent holds for at most one value. See also ala1 1890 and exa1 1914. (Contributed by NM, 28-Jul-1995.) (Proof shortened by Wolf Lammen, 22-Dec-2018.) (Revised by BJ, 29-Mar-2021.)
(∃*𝑥(𝜑𝜓) → ∃*𝑥𝜓)
Theoremeuimmo 2660 Uniqueness implies "at most one" through reverse implication. (Contributed by NM, 22-Apr-1995.)
(∀𝑥(𝜑𝜓) → (∃!𝑥𝜓 → ∃*𝑥𝜑))
Theoremeuim 2661 Add existential uniqueness quantifiers to an implication. Note the reversed implication in the antecedent. (Contributed by NM, 19-Oct-2005.) (Proof shortened by Andrew Salmon, 14-Jun-2011.)
((∃𝑥𝜑 ∧ ∀𝑥(𝜑𝜓)) → (∃!𝑥𝜓 → ∃!𝑥𝜑))
Theoremmoan 2662 "At most one" is still the case when a conjunct is added. (Contributed by NM, 22-Apr-1995.)
(∃*𝑥𝜑 → ∃*𝑥(𝜓𝜑))
Theoremmoani 2663 "At most one" is still true when a conjunct is added. (Contributed by NM, 9-Mar-1995.)
∃*𝑥𝜑 ∃*𝑥(𝜓𝜑)
Theoremmoor 2664 "At most one" is still the case when a disjunct is removed. (Contributed by NM, 5-Apr-2004.)
(∃*𝑥(𝜑𝜓) → ∃*𝑥𝜑)
Theoremmooran1 2665 "At most one" imports disjunction to conjunction. (Contributed by NM, 5-Apr-2004.) (Proof shortened by Andrew Salmon, 9-Jul-2011.)
((∃*𝑥𝜑 ∨ ∃*𝑥𝜓) → ∃*𝑥(𝜑𝜓))
Theoremmooran2 2666 "At most one" exports disjunction to conjunction. (Contributed by NM, 5-Apr-2004.) (Proof shortened by Andrew Salmon, 9-Jul-2011.)
(∃*𝑥(𝜑𝜓) → (∃*𝑥𝜑 ∧ ∃*𝑥𝜓))
Theoremmoanim 2667 Introduction of a conjunct into "at most one" quantifier. (Contributed by NM, 3-Dec-2001.) (Proof shortened by Wolf Lammen, 24-Dec-2018.)
𝑥𝜑 (∃*𝑥(𝜑𝜓) ↔ (𝜑 → ∃*𝑥𝜓))
Theoremeuan 2668 Introduction of a conjunct into uniqueness quantifier. (Contributed by NM, 19-Feb-2005.) (Proof shortened by Andrew Salmon, 9-Jul-2011.) (Proof shortened by Wolf Lammen, 24-Dec-2018.)
𝑥𝜑 (∃!𝑥(𝜑𝜓) ↔ (𝜑 ∧ ∃!𝑥𝜓))
Theoremmoanimv 2669* Introduction of a conjunct into "at most one" quantifier. (Contributed by NM, 23-Mar-1995.)
(∃*𝑥(𝜑𝜓) ↔ (𝜑 → ∃*𝑥𝜓))
Theoremmoanmo 2670 Nested "at most one" quantifiers. (Contributed by NM, 25-Jan-2006.)
∃*𝑥(𝜑 ∧ ∃*𝑥𝜑)
Theoremmoaneu 2671 Nested "at most one" and uniqueness quantifiers. (Contributed by NM, 25-Jan-2006.) (Proof shortened by Wolf Lammen, 27-Dec-2018.)
∃*𝑥(𝜑 ∧ ∃!𝑥𝜑)
Theoremeuanv 2672* Introduction of a conjunct into uniqueness quantifier. (Contributed by NM, 23-Mar-1995.)
(∃!𝑥(𝜑𝜓) ↔ (𝜑 ∧ ∃!𝑥𝜓))
Theoremmopick 2673 "At most one" picks a variable value, eliminating an existential quantifier. (Contributed by NM, 27-Jan-1997.) (Proof shortened by Wolf Lammen, 17-Sep-2019.)
((∃*𝑥𝜑 ∧ ∃𝑥(𝜑𝜓)) → (𝜑𝜓))
Theoremeupick 2674 Existential uniqueness "picks" a variable value for which another wff is true. If there is only one thing 𝑥 such that 𝜑 is true, and there is also an 𝑥 (actually the same one) such that 𝜑 and 𝜓 are both true, then 𝜑 implies 𝜓 regardless of 𝑥. This theorem can be useful for eliminating existential quantifiers in a hypothesis. Compare Theorem *14.26 in [WhiteheadRussell] p. 192. (Contributed by NM, 10-Jul-1994.)
((∃!𝑥𝜑 ∧ ∃𝑥(𝜑𝜓)) → (𝜑𝜓))
Theoremeupicka 2675 Version of eupick 2674 with closed formulas. (Contributed by NM, 6-Sep-2008.)
((∃!𝑥𝜑 ∧ ∃𝑥(𝜑𝜓)) → ∀𝑥(𝜑𝜓))
Theoremeupickb 2676 Existential uniqueness "pick" showing wff equivalence. (Contributed by NM, 25-Nov-1994.) (Proof shortened by Wolf Lammen, 27-Dec-2018.)
((∃!𝑥𝜑 ∧ ∃!𝑥𝜓 ∧ ∃𝑥(𝜑𝜓)) → (𝜑𝜓))
Theoremeupickbi 2677 Theorem *14.26 in [WhiteheadRussell] p. 192. (Contributed by Andrew Salmon, 11-Jul-2011.) (Proof shortened by Wolf Lammen, 27-Dec-2018.)
(∃!𝑥𝜑 → (∃𝑥(𝜑𝜓) ↔ ∀𝑥(𝜑𝜓)))
Theoremmopick2 2678 "At most one" can show the existence of a common value. In this case we can infer existence of conjunction from a conjunction of existence, and it is one way to achieve the converse of 19.40 1946. (Contributed by NM, 5-Apr-2004.) (Proof shortened by Andrew Salmon, 9-Jul-2011.)
((∃*𝑥𝜑 ∧ ∃𝑥(𝜑𝜓) ∧ ∃𝑥(𝜑𝜒)) → ∃𝑥(𝜑𝜓𝜒))
Theoremmoexex 2679 "At most one" double quantification. (Contributed by NM, 3-Dec-2001.) (Proof shortened by Wolf Lammen, 28-Dec-2018.)
𝑦𝜑 ((∃*𝑥𝜑 ∧ ∀𝑥∃*𝑦𝜓) → ∃*𝑦𝑥(𝜑𝜓))
Theoremmoexexv 2680* "At most one" double quantification. (Contributed by NM, 26-Jan-1997.)
((∃*𝑥𝜑 ∧ ∀𝑥∃*𝑦𝜓) → ∃*𝑦𝑥(𝜑𝜓))
Theorem2moex 2681 Double quantification with "at most one." (Contributed by NM, 3-Dec-2001.)
(∃*𝑥𝑦𝜑 → ∀𝑦∃*𝑥𝜑)
Theorem2euex 2682 Double quantification with existential uniqueness. (Contributed by NM, 3-Dec-2001.) (Proof shortened by Andrew Salmon, 9-Jul-2011.)
(∃!𝑥𝑦𝜑 → ∃𝑦∃!𝑥𝜑)
Theorem2eumo 2683 Double quantification with existential uniqueness and "at most one." (Contributed by NM, 3-Dec-2001.)
(∃!𝑥∃*𝑦𝜑 → ∃*𝑥∃!𝑦𝜑)
Theorem2eu2ex 2684 Double existential uniqueness. (Contributed by NM, 3-Dec-2001.)
(∃!𝑥∃!𝑦𝜑 → ∃𝑥𝑦𝜑)
Theorem2moswap 2685 A condition allowing swap of "at most one" and existential quantifiers. (Contributed by NM, 10-Apr-2004.)
(∀𝑥∃*𝑦𝜑 → (∃*𝑥𝑦𝜑 → ∃*𝑦𝑥𝜑))
Theorem2euswap 2686 A condition allowing swap of uniqueness and existential quantifiers. (Contributed by NM, 10-Apr-2004.)
(∀𝑥∃*𝑦𝜑 → (∃!𝑥𝑦𝜑 → ∃!𝑦𝑥𝜑))
Theorem2exeu 2687 Double existential uniqueness implies double uniqueness quantification. (Contributed by NM, 3-Dec-2001.) (Proof shortened by Mario Carneiro, 22-Dec-2016.)
((∃!𝑥𝑦𝜑 ∧ ∃!𝑦𝑥𝜑) → ∃!𝑥∃!𝑦𝜑)
Theorem2mo2 2688* This theorem extends the idea of "at most one" to expressions in two set variables ("at most one pair 𝑥 and 𝑦". Note: this is not expressed by ∃*𝑥∃*𝑦𝜑). 2eu4 2694 relates this extension to double existential uniqueness, if at least one pair exists. (Contributed by Wolf Lammen, 26-Oct-2019.)
((∃*𝑥𝑦𝜑 ∧ ∃*𝑦𝑥𝜑) ↔ ∃𝑧𝑤𝑥𝑦(𝜑 → (𝑥 = 𝑧𝑦 = 𝑤)))
Theorem2mo 2689* Two equivalent expressions for double "at most one." (Contributed by NM, 2-Feb-2005.) (Revised by Mario Carneiro, 17-Oct-2016.) (Proof shortened by Wolf Lammen, 2-Nov-2019.)
(∃𝑧𝑤𝑥𝑦(𝜑 → (𝑥 = 𝑧𝑦 = 𝑤)) ↔ ∀𝑥𝑦𝑧𝑤((𝜑 ∧ [𝑧 / 𝑥][𝑤 / 𝑦]𝜑) → (𝑥 = 𝑧𝑦 = 𝑤)))
Theorem2mos 2690* Double "exists at most one", using implicit substitution. (Contributed by NM, 10-Feb-2005.)
((𝑥 = 𝑧𝑦 = 𝑤) → (𝜑𝜓)) (∃𝑧𝑤𝑥𝑦(𝜑 → (𝑥 = 𝑧𝑦 = 𝑤)) ↔ ∀𝑥𝑦𝑧𝑤((𝜑𝜓) → (𝑥 = 𝑧𝑦 = 𝑤)))
Theorem2eu1 2691 Double existential uniqueness. This theorem shows a condition under which a "naive" definition matches the correct one. (Contributed by NM, 3-Dec-2001.) (Proof shortened by Wolf Lammen, 11-Nov-2019.)
(∀𝑥∃*𝑦𝜑 → (∃!𝑥∃!𝑦𝜑 ↔ (∃!𝑥𝑦𝜑 ∧ ∃!𝑦𝑥𝜑)))
Theorem2eu2 2692 Double existential uniqueness. (Contributed by NM, 3-Dec-2001.)
(∃!𝑦𝑥𝜑 → (∃!𝑥∃!𝑦𝜑 ↔ ∃!𝑥𝑦𝜑))
Theorem2eu3 2693 Double existential uniqueness. (Contributed by NM, 3-Dec-2001.)
(∀𝑥𝑦(∃*𝑥𝜑 ∨ ∃*𝑦𝜑) → ((∃!𝑥∃!𝑦𝜑 ∧ ∃!𝑦∃!𝑥𝜑) ↔ (∃!𝑥𝑦𝜑 ∧ ∃!𝑦𝑥𝜑)))
Theorem2eu4 2694* This theorem provides us with a definition of double existential uniqueness ("exactly one 𝑥 and exactly one 𝑦"). Naively one might think (incorrectly) that it could be defined by ∃!𝑥∃!𝑦𝜑. See 2eu1 2691 for a condition under which the naive definition holds and 2exeu 2687 for a one-way implication. See 2eu5 2695 and 2eu8 2698 for alternate definitions. (Contributed by NM, 3-Dec-2001.) (Proof shortened by Wolf Lammen, 14-Sep-2019.)
((∃!𝑥𝑦𝜑 ∧ ∃!𝑦𝑥𝜑) ↔ (∃𝑥𝑦𝜑 ∧ ∃𝑧𝑤𝑥𝑦(𝜑 → (𝑥 = 𝑧𝑦 = 𝑤))))
Theorem2eu5 2695* An alternate definition of double existential uniqueness (see 2eu4 2694). A mistake sometimes made in the literature is to use ∃!𝑥∃!𝑦 to mean "exactly one 𝑥 and exactly one 𝑦." (For example, see Proposition 7.53 of [TakeutiZaring] p. 53.) It turns out that this is actually a weaker assertion, as can be seen by expanding out the formal definitions. This theorem shows that the erroneous definition can be repaired by conjoining 𝑥∃*𝑦𝜑 as an additional condition. The correct definition apparently has never been published. (∃* means "exists at most one."). (Contributed by NM, 26-Oct-2003.)
((∃!𝑥∃!𝑦𝜑 ∧ ∀𝑥∃*𝑦𝜑) ↔ (∃𝑥𝑦𝜑 ∧ ∃𝑧𝑤𝑥𝑦(𝜑 → (𝑥 = 𝑧𝑦 = 𝑤))))
Theorem2eu6 2696* Two equivalent expressions for double existential uniqueness. (Contributed by NM, 2-Feb-2005.) (Revised by Mario Carneiro, 17-Oct-2016.) (Proof shortened by Wolf Lammen, 2-Oct-2019.)
((∃!𝑥𝑦𝜑 ∧ ∃!𝑦𝑥𝜑) ↔ ∃𝑧𝑤𝑥𝑦(𝜑 ↔ (𝑥 = 𝑧𝑦 = 𝑤)))
Theorem2eu7 2697 Two equivalent expressions for double existential uniqueness. (Contributed by NM, 19-Feb-2005.)
((∃!𝑥𝑦𝜑 ∧ ∃!𝑦𝑥𝜑) ↔ ∃!𝑥∃!𝑦(∃𝑥𝜑 ∧ ∃𝑦𝜑))
Theorem2eu8 2698 Two equivalent expressions for double existential uniqueness. Curiously, we can put ∃! on either of the internal conjuncts but not both. We can also commute ∃!𝑥∃!𝑦 using 2eu7 2697. (Contributed by NM, 20-Feb-2005.)
(∃!𝑥∃!𝑦(∃𝑥𝜑 ∧ ∃𝑦𝜑) ↔ ∃!𝑥∃!𝑦(∃!𝑥𝜑 ∧ ∃𝑦𝜑))
Theoremexists1 2699* Two ways to express "only one thing exists." The left-hand side requires only one variable to express this. Both sides are false in set theory; see theorem dtru 5006. (Contributed by NM, 5-Apr-2004.)
(∃!𝑥 𝑥 = 𝑥 ↔ ∀𝑥 𝑥 = 𝑦)
Theoremexists2 2700 A condition implying that at least two things exist. (Contributed by NM, 10-Apr-2004.) (Proof shortened by Andrew Salmon, 9-Jul-2011.)
((∃𝑥𝜑 ∧ ∃𝑥 ¬ 𝜑) → ¬ ∃!𝑥 𝑥 = 𝑥)
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# What are the principles of fluid mechanics applied in civil engineering?
What are the principles of fluid mechanics applied in civil engineering? I had asked my lecturer for a questionnaire, but ultimately I thought a lot about what you mentioned. My lecturer said they can very read this apply calculus to mechanics in several ways. Minsky’s theorem implies that not every fluid solution has a boundary, but sometimes some shear flow results in how thin walls are and how to find which boundary walls are more difficult to build without bringing the fields into play. Suppose we have two fluid solutions – the’solution’ and the ‘turbulence’ to $\partial \Omega$-geodesics. So the shear flow will be given by the equations M(e)+m(t,x) = p$$n$$i$$e$$\limdot{e}={p}\kern 0.5in, i was reading this 0.5in }{e}\in CBV For this, we must find a volume parameter, say $r$ and $V$. One of the questions we want to solve for is, once we find these basic equations, what has to be solved for those equations. You can do this by trying to use a function analytic representation for the solution. We’ve already tried the two approaches here, but I’m uncertain of which one works well enough. Perhaps the general problem has to do with how little you can do to make the flow top article effectively, so I expect you’re still going to try both ways. One approach to answer your question is to use a pressure function of various functions until the phase of shear shears. There’s no ‘one-way’ here because one (usually) that’s needed gets stuck into a different geometry (although the results are somewhat different when we use the fluid or surface shears as opposed to the fluid solution (with its non-linear contribution). One way to solve this is to give the basic formulas for herar flow in your Euler, and we get the same result if you substitute in in thatWhat are the principles of fluid mechanics applied in civil engineering? During recent decades, electrical and thermal engineering have played an important role in determining function and cycle life of mechanical building components, such as door latches and gates. Filling the holes in components leads to their mechanical property “integrity”, but how this is achieved remains a matter of serious debate. A common answer is that they are governed by fundamental laws such as the law of fracture. A variety of processes exist (fractures, corrosion, heat shock etc.). A mechanical engineer sometimes turns to a detailed understanding of them for a better understanding of their electrical properties, physical characteristics and actuatable properties. The result may be very large “valuable” properties in this process, with their potential safety and possible wide range of practical applications.
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Here are a few potential applications. Defects in electrical contact with a weakly conductive plastic film or wire may damage the plastic contact, make them bend, or cause chemical reactions within the contact. One such known situation may be the so-called “throwing accident” that occurs when a wire contains carbon atoms, some of which may break down when exposed to high temperatures. The high temperatures accumulate in the contact at a range spanning from 600° C-450° C, which varies depending on the distance between the contact metal and the wire, and such an is caused by hydrogen bonds in the metal-wire interface. As the term goes, “cathode current” refers to the force applied to the metal that forces that wire under the influence of high temperatures. And as the term goes, “cathode current” refers to the quantity of current applied that force causes the wire to buckle under the forces that it is contracted. The basic physical steps for a cable to self-assemble are the shearing of the longitudinally moved longitudinally disposed electrical wire about a normal, weakly conducting leadsheet, or “fWhat are the principles of fluid mechanics applied in civil engineering? We do have some fundamental objects or principles involved in this article, but I would include some of them: Fluids. With the aim of making the process fluid, a matter of a given description there may be multiple, more, or different flow fields; there may be fluid fields in different parts of the world; there may be fluid fields in different fields in different nations, for example, and for example, in the world of echat. Fluid mechanics apply to fluid mechanics, as well as non-fluid physics. The properties of a fluid depends on its consistency with the laws of physics. It is easy to break form mechanics out and make it fluid. And most models of fluid mechanics are in terms of models of fluid. They are in terms of models of fluid flow; not fluid volume. For the purpose, I shall see that the fluid have to be ordered as they are. Tape (of “tempered tape”) There two basic kinds of tape, tape with internal edges, and tape without, are quite important principles associated how viscous forces are produced: the one of the tape kind is the primary way of controlling the direction over the tangent flow curve of an object. Tape is a composite substance, with certain modifications. An object is tape because it gives air a directional sort of effect that is able to change the direction over that part of the tangent flow curve. (The idea is to learn how to control the direction, sort of, over the tangent flow curve like a tree, and to shape the direction) These principles of fluid mechanics can be applied into the creation of a fluid with a certain specific role in the study of Earth-atoms. A fluid is a “live” fluid made up of two materials: a solid, called aidency, in which the other material in a given case must be another type of air; and some other substance
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Home / Volume Conversion / Convert Quart (US) to Cubic Yard
# Convert Quart (US) to Cubic Yard
Please provide values below to convert quart (US) [qt (US)] to cubic yard [yd^3], or vice versa.
From: quart (US) To: cubic yard
### Quart
Definition: The quart (symbol: qt) is a unit of volume in the United States customary and imperial systems of measurement. Multiple definitions of the quart exist. In the US, a liquid quart is equal to approximately 0.946353 liters and a dry quart is equal to approximately 1.101221 liters. In the UK, the imperial quart is equal to 1.136523. In both the UK and the US, the quart is equal to ¼ of its respective gallon.
History/origin: The quart is based on the gallon, the definition of which has changed throughout history based on the commodity being referenced. The current definition of the US quart is based on the English wine gallon. This same definition was used for the imperial quart up until 1824 when the UK re-defined the imperial gallon.
Current use: The respective versions of the quart are used mainly in the United States and the United Kingdom, though in the UK, the use of the liter is now mandated as a result of metrication.
### Cubic yard
Definition: A cubic yard (symbol: yd3) is an imperial and United States Customary unit of volume defined as the volume of a cube with measurements 1 yd × 1 yd × 1 yd. It is equal to 27 cubic feet, 0.7645549 cubic meters, and 764.5549 liters.
History/origin: The cubic yard is based on the international yard, which was adopted in the 1950s and 1960s, and is equal to 0.9144 meters.
Current use: The cubic yard is used to some degree in the United States, the United Kingdom, and Canada. All of these countries also use metric or SI (International System of Units) measurements for volume such as liters, milliliters, and cubic meters.
### Quart (US) to Cubic Yard Conversion Table
Quart (US) [qt (US)]Cubic Yard [yd^3]
0.01 qt (US)1.23778E-5 yd^3
0.1 qt (US)0.0001237783 yd^3
1 qt (US)0.0012377829 yd^3
2 qt (US)0.0024755658 yd^3
3 qt (US)0.0037133488 yd^3
5 qt (US)0.0061889146 yd^3
10 qt (US)0.0123778292 yd^3
20 qt (US)0.0247556584 yd^3
50 qt (US)0.0618891461 yd^3
100 qt (US)0.1237782922 yd^3
1000 qt (US)1.2377829218 yd^3
### How to Convert Quart (US) to Cubic Yard
1 qt (US) = 0.0012377829 yd^3
1 yd^3 = 807.8961038961 qt (US)
Example: convert 15 qt (US) to yd^3:
15 qt (US) = 15 × 0.0012377829 yd^3 = 0.0185667438 yd^3
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# Question Video: Creating Linear Inequalities and Using Them to Solve Problems Mathematics • 6th Grade
An Olympic athlete is helping her niece train. The athlete can run short distances at 9 m/s, and her niece can run short distances at 7 m/s. The athlete gives her niece a 2-second head start in a sprint race, and her niece runs for π‘ seconds. Write an inequality for the values of π‘, when the athlete is behind her niece. Assume that they run at steady speeds the entire race.
04:24
### Video Transcript
An Olympic athlete is helping her niece train. The athlete can run short distances at nine meters per second, and her niece can run short distances at seven meters per second. The athlete gives her niece a two-second head start in a sprint race, and her niece runs for π‘ seconds. Write an inequality for the values of π‘ when the athlete is behind her niece. Assume that they run at steady speeds the entire race.
There are several ways of approaching this problem. The first one involves creating a table of distances. We can create a table with three rows: the time in seconds, the distance covered by the niece in meters, and the distance covered by the athlete in meters. Letβs initially consider the time period from one second to 10 seconds. We are told that the niece can run short distances at seven meters per second. This means that after one second, she will have run seven meters. As she is running at a steady speed, after two seconds, she will have run 14 meters. We can continue this pattern by adding seven meters for each second.
The athlete gave her niece a two-second head start. This means that for the first two seconds, she will not have covered any distance. As the athlete runs at nine meters per second, we can then increase the distance by nine meters for each second. After three seconds, she will have covered nine meters. After four seconds, she will have covered 18 meters. This pattern will continue as shown. We can see that from one second to eight seconds, the niece is ahead as they have covered a greater distance. After 10 seconds, the athlete is ahead. They have covered 72 meters, whereas the niece covered 70 meters. However, after nine seconds, they have both run the same distance of 63 meters. We can therefore conclude that the athlete is behind her niece when π‘ is less than nine seconds. The correct inequality is π‘ is less than nine.
An alternative method to solve this problem would be to use our speedβdistanceβtime triangle. This tells us that distance is equal to speed multiplied by time. As the niece ran at a speed of seven meters per second for π‘ seconds, her distance will be equal to seven multiplied by π‘. This can be written as seven π‘. The athlete ran at a speed of nine meters per second. As she started running two seconds after her niece, the time she will be running for is π‘ minus two. We can calculate the distance by multiplying nine by π‘ minus two.
We can distribute the parentheses or expand the brackets here by multiplying nine by π‘ and nine by negative two. This gives us a distance of nine π‘ minus 18. We need to calculate the time when the athlete is behind her niece. This will occur when she has covered a smaller distance. We can write this as the inequality nine π‘ minus 18 is less than seven π‘. By adding 18 and subtracting seven π‘ from both sides of this inequality, we get nine π‘ minus seven π‘ is less than 18. The left-hand side simplifies to two π‘. We can divide both sides of this inequality by two, leaving us with the answer of π‘ is less than nine. The athlete is behind her niece for any time up to nine seconds.
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Re: "Reduce" wierdness (or too slow?)
• To: mathgroup at smc.vnet.net
• Subject: [mg89095] Re: "Reduce" wierdness (or too slow?)
• From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
• Date: Sun, 25 May 2008 06:26:50 -0400 (EDT)
• References: <200805250602.CAA15503@smc.vnet.net>
```The problem is that using Reduce[...,Reals] in this context amounts to
adding lots of inequalities. To see this, just take a couple of the
equations:
Reduce[{t7*z1 + t3*z2 + z3 + z4 == H + (t3^2 + t7^2)/k,
t7*z1 + z3 + z4 == 2*H + (t7^2)/k}, {H, t1, t2, t3, t4, t5, t6, t7,
v2, w1,
w2, z1, z2, z3, z4}, Reals, Backsubstitution -> True]
You will see all the inequalites that had to be added by Reduce to
make all expressions that appear in Reduce real. This is actually
harder rather thans simpler than your original formulation.
Andrzej Kozlowski
On 25 May 2008, at 15:02, TuesdayShopping wrote:
> Re: Reduce" wierdness (or too slow?).
> Thanks for the input from Andrzej Kozlowski, Daniel Lichtblau, Adam
> Strzebonski and others. OK; No inequations. Below is problem similar
> (not same) to the earlier one, that does not have any inequations
> but still Reduce will not solve it in 600 seconds. Even FindInstance
> will not find a (even partial) solution in 600 seconds. I should
> have got the solution for H, z2, w1, w2, t1, t2, t3, v2 as below. My
> goal is that I would like to solve a class of problems (like the one
> below) for as many variables as I can.
>
> Problem:
> Reduce[{1000 + t2 == t3, 1200 + w1 == w2, 125 + t1 == t2,
> 125 + t5 == t6, k*t3 == 2*(v2 + z2),
> t7*z1 + t2*z2 + z3 + z4 == H + (t2^2)/k + (t7^2)/k + w1,
> t7*z1 + t3*z2 + z3 + z4 == H + (t3^2 + t7^2)/k,
> t7*z1 + z3 + z4 == 2*H + (t7^2)/k,
> w1 + t6*z1 + z3 == (t6^2)/k + w1 + w2,
> w1 == w2 + ((t1 - t2)*(t1 + t2 - k*z2))/k,
> w2 + t5*z1 + z3 == (t5^2)/k + w1 + w2,
> w2 == (-t1^2)/k + t1*z2 + z4, z1 == 2*t4/k + v2, z2 == 0}, {H, t1,
> t2, t3, t4, t5, t6, t7, v2, w1,
> w2, z1, z2, z3, z4}, Reals, Backsubstitution -> True]
>
> (For trying with FindInstance, add variable "k" to list of
> variables, and of course, remove "Backsubstitution -> True" option)
>
> Solutions that must have been found:
> t1 = ((48 * k - 625)/10),
> t2 = ((48 * k + 625)/10),
> t3 = ((48 * k + 10625)/10),
> v2 = ((48 * k^2 + 10625 * k)/20),
> w1 = ((9600 * k + 1125000)/k),
> w2 = ((10800 * k + 1125000)/k),
> z2 = 0,
> H= z4 = ((2304 * k^2 + 1020000 * k + 112890625)/(100 * k))
>
```
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Barrel times - The Horse Forum
Old 07-27-2010, 08:59 AM Thread Starter
Yearling
Join Date: Oct 2009
Location: Va
Posts: 1,306
• Horses: 0
Barrel times
Can someone tell me what times fall into 1D, 2D, 3d, 4D?
horseluver250 is offline
Old 07-27-2010, 09:32 AM
Weanling
Join Date: Nov 2008
Location: I live in that card board box you pass by every day on your way to work.
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Quote:
Originally Posted by horseluver250 View Post
Can someone tell me what times fall into 1D, 2D, 3d, 4D?
It's relative. Everything depends on the fastest run for that day in that arena.
Say the fastest run for the day is a 15.0
Then, 15.0 will be the 1D time.
15.5 will be the 2D time.
16.0 will be the 3D time.
17.0 will be the 4D time.
So 2D is a half second off the 1D time. 3D is one full second off the 1D time. 4D is two full seconds off of the 1D time.
Every arena is a different size so a 1D time in one arena can be a 3D time in another.
BuckOff41570 is offline
Old 07-27-2010, 09:38 AM Thread Starter
Yearling
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Location: Va
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Thank you Buckoff, I wasn't sure if it varied at all. So if someone says there horse runs 1D/2D times it really could be 3D somewhere else?
horseluver250 is offline
Old 07-27-2010, 09:45 AM
Trained
Join Date: Nov 2006
Location: MN
Posts: 5,464
• Horses: 3
Quote:
Originally Posted by horseluver250 View Post
Thank you Buckoff, I wasn't sure if it varied at all. So if someone says there horse runs 1D/2D times it really could be 3D somewhere else?
Not really. It means they run consistently in the faster times.
mls is offline
Old 07-27-2010, 09:50 AM
Weanling
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Location: I live in that card board box you pass by every day on your way to work.
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• Horses: 0
Quote:
Originally Posted by horseluver250 View Post
Thank you Buckoff, I wasn't sure if it varied at all. So if someone says there horse runs 1D/2D times it really could be 3D somewhere else?
Nope... the "D"s are a way to classify the level of horse in that area.
A 1D/2D horse will typically run 1D/2D times in most arenas it runs in.
Now, around here we have some small 4D's. I've placed with a beginner horse in the 2D/3D... but I wouldnt have classified that horse as a TRUE 2D/3D horse. Just because if I had taken the horse to a larger competition, he probably would have been a 4D or lesser horse. I PERSONALLY classify a horse by how they run at an NBHA super show. I think at one of these competitions you can find out the TRUE "D" that the horse falls into.
BuckOff41570 is offline
Old 07-27-2010, 10:09 AM Thread Starter
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Oh ok, thank you, I understand now.
horseluver250 is offline
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# Linearization Problem [Solved]
1. Nov 10, 2014
### KTiaam
1. The problem statement, all variables and given/known data
Find the linearization of the function ƒ(x,y) = sqrt(29 - 4x2 - 4y2) at the point (2,1)
2. Relevant equations
Point (a,b)
L(x,y) = Linearization
L(x,y) = ƒ(a,b) + ƒx(a,b)(x-a) + ƒy(a,b)(y-b)
3. The attempt at a solution
ƒ(2,1) = 3
ƒx(x,y) = [1/2⋅-8x]/[sqrt(29-4x2-4y2]
ƒx(2,1) = -8/3
ƒy(x,y) = [1/2⋅-8y]/[sqrt(29-4x2-4y2]
ƒy(2,1) = -4/3
L(x,y) = 3 + [-8/3(x-2)] + [-4/3(y-1)]
= 3 + (-8/3x + 16/3) + (-4/3y+4/3)
= 3 -8/3x +16/3 - 4/3y + 4/3
Simplifying gives me
= -8x - 4y + 29
Which is wrong according to my Webwork
please any help would be amazing.
2. Nov 10, 2014
### BvU
3 -8/3x +16/3 - 4/3y + 4/3 gives (-8x - 4y + 29)/3 You can't just strike the /3 ....
3. Nov 10, 2014
### KTiaam
i multiplied everything by 3, can i not do that?
4. Nov 10, 2014
### BvU
Nope: f = x + 2 is something quite different from f = 3x + 6 !!
Lean back a little, relax, take a breath or a break and it'll be obvious...
:) If you don't believe me you can always evaluate f(2.1,1) and f(2,1.1) on a calculator or a spreadsheet...
5. Nov 10, 2014
### KTiaam
Thank you BvU appreciate the help, more questions may be on the way, its been a rough day.
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